research article conflict set and waveform modelling...

Research ArticleConflict Set and Waveform Modelling forPower Amplifier Design

Anamarija Juhas and Ladislav A Novak

Department of Power Electronics and Communication Engineering Faculty of Technical Sciences University of Novi Sad21000 Novi Sad Serbia

Correspondence should be addressed to Ladislav A Novak ladislavunsacrs

Received 19 July 2014 Accepted 7 February 2015

Academic Editor Ben T Nohara

Copyright copy 2015 A Juhas and L A Novak This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

Various classes of nonnegative waveforms containing dc component fundamental and 119896th harmonic (119896 ge 2) which proved to beof interest in waveform modelling for power amplifier (PA) design are considered in this paper In optimization of PA efficiencynonnegative waveforms with maximal amplitude of fundamental harmonic and those with maximal coefficient of cosine term offundamental harmonic (optimal waveforms) play an important role Optimal waveforms have multiple global minima and this factclosely relates the problem of optimization of PA efficiency to the concept of conflict set There is also keen interest in findingdescriptions for various classes of suboptimal waveforms such as nonnegative waveforms with at least one zero nonnegativewaveforms with maximal amplitude of fundamental harmonic for prescribed amplitude of 119896th harmonic nonnegative waveformswith maximal coefficient of cosine part of fundamental harmonic for prescribed coefficients of 119896th harmonic and nonnegativecosine waveforms with at least one zero Closed form descriptions for all these suboptimal types of waveforms are provided in thispaper Suboptimal waveforms may also have multiple global minima and therefore be related to the concept of conflict set Fourcase studies of usage of closed form descriptions of nonnegative waveforms in PA modelling are also provided

1 Introduction

Theorigin of the concept of conflict set goes back to J CMax-well (Maxwell 1831ndash1879) who informally introduced mostof features of what today is called conflict set [1] Fromthis reason Maxwell set or Maxwell stratum is also used assynonyms for conflict set Roughly speaking conflict setassociated with a smooth function 119891 with 119898 parameters isthe set of 119898-tuples in parameter space for which 119891 has mul-tiple global minima Conflict set is also intimately related tosingularity theory and catastrophe theory [1]

Although without explicit reference many max-minmin-max engineering design problems related to nonsmoothoptimizations in parameter spaces (eg see [2]) includingproblems related to the optimization of efficiency of poweramplifiers (PAs) (eg see [3ndash12]) are connected to the con-cept of conflict set The concept of conflict set has been alsoused in mathematics (eg see [13ndash16]) and physics (eg see[17 18]) including subjects like black holes [19]

Nonnegative waveforms with maximal amplitude of fun-damental harmonic and those with maximal coefficient ofcosine term of fundamental harmonic (optimal waveforms)have multiple global minima and therefore are closely relatedto the concept of conflict setThe suboptimal waveforms suchas

(i) nonnegative waveforms with at least one zero(ii) nonnegative waveforms with maximal amplitude of

fundamental harmonic for prescribed amplitude of119896th harmonic

(iii) nonnegative waveforms with maximal coefficient ofcosine part of fundamental harmonic for prescribedcoefficients of 119896th harmonic

(iv) nonnegative cosine waveforms with at least one zeromay also havemultiple global minima [9 11 12] and thereforebe related to the concept of conflict set as well These subop-timal waveforms are clearly of interest in shapingmodelling

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015 Article ID 585962 29 pageshttpdxdoiorg1011552015585962

2 Mathematical Problems in Engineering

drain (collectorplate) waveforms in PA design (eg see [3ndash12 20 21])

Fejer in his seminal paper [22] provided general descrip-tion of all nonnegative trigonometric polynomials with 119899

consecutive harmonics in terms of 2119899 + 2 parameters satis-fying one nonlinear constraint He also derived closed formsolution to the problem of finding maximum possible ampli-tude of the first harmonic of nonnegative cosine polynomialswith consecutive harmonics

Fuzik [3] (see also [10]) considered cosine polynomialswith dc fundamental and 119896th harmonic for arbitrary 119896 ge 2

and provided closed form solution for coefficients of optimalwaveform Rhodes in [7] provided closed form expressionfor maximum possible amplitude of fundamental harmonicof nonnegative waveforms containing consecutive odd har-monics A subclass of nonnegative cosine waveforms withdc fundamental and third harmonic having factorized formdescription has been considered in [23]

High efficiency PA with arbitrary output harmonic ter-minations has been analysed in [9] along with maximal effi-ciency fundamental output power and load impedance

Factorized form of nonnegative waveforms up to secondharmonic with at least one zero has been suggested in [11] inthe context of continuous class BJ mode of PA operation

General description of all nonnegative waveforms up tosecond harmonic in terms of four independent parametershas been provided in [12] This includes nonnegative wave-forms with at least one zero as a special case

End point of conflict set normally corresponds to so-called maximally flat waveform which also belongs to classof suboptimal waveforms First comprehensive usage of max-imally flat waveforms in the context of analysis of PA goesto Raab [20] General description of maximally flat wave-forms with arbitrary number of harmonics has been pre-sented in [21] along with closed form expressions for effi-ciency of class-F and inverse class-F PA with maximally flatwaveforms Description of maximally flat cosine waveformswith consecutive harmonics has been presented in [8] in thecontext of finite harmonic class-C PA

In this paper we provide general descriptions of a num-ber of optimal and suboptimal nonnegative waveforms con-taining dc component fundamental and an arbitrary 119896thharmonic 119896 ge 2 and show how they are related to theconcept of conflict set According to our best knowledge thispaper provides the very first usage of conflict set in the courseof solving problems related to optimization of PA efficiencyMain results are stated in six propositions (Propositions 1 69 18 22 and 26) four corollaries (Corollaries 2ndash5) twentyremarks and three algorithms Four case studies of usage ofclosed form descriptions of nonnegative waveforms in PAefficiency analysis are considered in detail in Section 7

This paper is organized in the following way In Section 2we introduce concepts of minimum function and gain func-tion (Section 21) conflict set (Section 22) and parameterspace (Section 23) In Sections 3ndash6 we provide generaldescriptions of various classes of nonnegative waveformscontaining dc component fundamental and 119896th harmonicwith at least one zero General case of nonnegative waveformswith at least one zero is presented in Section 31The case with

exactly two zeros is considered in Section 32 An algorithmfor calculation of coefficients of fundamental harmonic ofnonnegative waveforms with two zeros for prescribed coeffi-cients of 119896th harmonic is presented in Section 33 Descrip-tion of nonnegative waveforms with maximal amplitude offundamental harmonic for prescribed amplitude of 119896th har-monic is provided in Section 4 Nonnegative waveforms withmaximal coefficient of cosine part of fundamental harmonicfor prescribed coefficients of 119896th harmonic are consideredin Section 51 An illustration of results of Section 51 forparticular case 119896 = 3 is given in Section 52 Section 61 isdevoted to nonnegative cosine waveforms with at least onezero and arbitrary 119896 ge 2 whereas Section 62 considerscosine waveforms with at least one zero for 119896 = 3 InSection 7 four case studies of application of descriptions ofnonnegative waveforms with fundamental and 119896th harmonicin PA modelling are presented In the Appendices list ofsome finite sums of trigonometric functions widely usedthroughout the paper and brief account of the Chebyshevpolynomials are provided

2 Minimum Function Gain Function andConflict Set

In this sectionwe considerminimum function and gain func-tion (Section 21) conflict set (Section 22) and parameterspace (Section 23) in the context of nonnegative waveformswith fundamental and 119896th harmonic

We start with provision of a brief account of the factsrelated to the concepts of minimum function and conflict setFor this purpose let us denote by 119891(119909 119906) a family of smoothfunctions of 119899 variables depending on 119898 parameters where119909 isin 119877

119899 is 119899-tuple of variables and 119906 isin 119877119898 is 119898-tuple of

parameters The minimum function 119865 119877119898

rarr 119877 associatedwith the function 119891 is defined as 119865(119906) = min

119909119891(119909 119906)

Therefore the domain of theminimum function is parameterspace of the function 119891 The minimum function 119865(119906) is con-tinuous but not necessarily smooth function of parameters[13 24] It is a smooth function if 119891(119909 119906) possesses uniqueglobal minimum at nondegenerate critical point [13] (criticalpoint is degenerate if at least first two consecutive derivativesare equal to zero) In this context the conflict set can bedefined as the set of the parameters for which function 119891

has global minimum at a degenerate critical point orandmultiple global minima [13]

For a wide class ofminimum functions when the numberof parameters is not greater than four the behaviour ofminimum function in a neighbourhood of any point can bedescribed by one of ldquonormal formsrdquo from a finite list as statedin [24] For example for smooth function 119891 119877 times 119877

2

rarr 119877the minimum function 119865(119906

1 1199062) = min

119909119891(119909 119906

1 1199062) near the

origin can be locally reduced to one of the following threenormal forms [25] minus|119906

1|min(119906

1 1199062 1199061+ 1199062) or min

119909(1199094

+

11990611199092

+ 1199062119909) In this example the conflict set is the set of all

points (1199061 1199062) for which minimum function 119865(119906

1 1199062) is not

differentiable because function 119891(119909 1199061 1199062) possesses at least

two global minima [25]

Mathematical Problems in Engineering 3

21 Minimum Function and Gain Function In what followswe consider family of waveforms of type

119908 (120591 120574 119860 120572) = 1 minus 120574 (cos 120591 + 119860 cos (119896120591 + 120572)) (1)

where 120591 stands for 120596119905 120574 gt 0 119896 ge 2 119860 gt 0 and 120572 isin [0 2120587)Waveforms of type (1) include all possible shapes which canoccur but not all possiblewaveforms containing fundamentaland 119896th harmonic However shifting of waveforms of type(1) along 120591-axis could recover all possible waveforms withfundamental and 119896th harmonic

The problem of finding nonnegative waveform of type(1) having maximum amplitude of fundamental harmonicplays an important role in optimization of PA efficiency Thisextremal problem can be reformulated as problem of findingnonnegative waveform from family (1) having maximumpossible value of coefficient 120574 Nonnegative waveform offamily (1) with maximum possible value of coefficient 120574 iscalled ldquooptimalrdquo or ldquoextremalrdquo waveform

Furthermore let us introduce an auxiliary waveform

119891 (120591 119860 120572) = minus cos 120591 minus 119860 cos (119896120591 + 120572) (2)

which is smooth function of one variable 120591 and two parame-ters119860 and 120572 In terms of 119891(120591 119860 120572) the above extremal prob-lem reduces to the problem of finding maximum possiblevalue of coefficient 120574 that satisfies

1 + 120574119891 (120591 119860 120572) ge 0 (3)

Clearly for any prescribed pair (119860 120572) there is a uniquemaximal value of coefficient 120574 for which inequality (3) holdsfor all 120591 This maximal value of 120574 associated with the pair(119860 120572) we denote it by 119866(119860 120572) and call it ldquogain functionrdquo

Let

119865min (119860 120572) = min120591

119891 (120591 119860 120572) (4)

be theminimum function associatedwith119891(120591 119860 120572) Accord-ing to (3) 119866(119860 120572) and 119865min(119860 120572) satisfy the following rela-tion 1 + 119866(119860 120572)119865min(119860 120572) = 0 Since 119865min(119860 120572) is obviouslynonzero it follows immediately that

119866 (119860 120572) = minus1

119865min (119860 120572) (5)

A relation analogue to (5) for 119896 = 2 (fundamental andsecond harmonic) has been derived in [4] According toour best knowledge it was the first appearance of gainfunction expressed via associated minimum function Theconsideration presented in [4] has been restricted to theparticular case when 120572 = 120587 The same problem for 120572 = 120587 andarbitrary 119896 ge 2 has been investigated in [3] (see also [10])

According to above consideration the problem of finding3-tuple (120574 119860 120572) with maximum possible value of 120574 for which(3) holds is equivalent to the problem of finding maximumvalue of gain function

120574max = max119860120572

119866 (119860 120572) = minus1

max119860120572

119865min (119860 120572)(6)

14

12

1

08

06

04

G1 G2

001

02

0304

0506 0

1205872

12058731205872

2120587

Parameter 120572

Parameter A

Gai

n fu

nctio

n

k = 2

Figure 1 Graph of 119866(119860 120572) for 119896 = 2 Points 1198661and 119866

2denote

beginning of the ridge and maximum of gain function respectively

and corresponding pair (119860lowast 120572lowast) that satisfies

120574max = max119860120572

119866 (119860 120572) = 119866 (119860lowast

120572lowast

)

max119860120572

119865min (119860 120572) = 119865min (119860lowast

120572lowast

)

(7)

Thus the optimal waveform 119908lowast

(120591) is determined by parame-ters 120574max 119860

lowast and 120572lowast that is

119908lowast

(120591) = 119908 (120591 120574max 119860lowast

120572lowast

) (8)

Optimal waveform has two global minima (this claim willbe justified in Section 4 Remark 21) Consequently the pair(119860lowast

120572lowast

) which corresponds to maximum of gain function119866(119860 120572) belongs to conflict set in (119860 120572) parameter space

Figure 1 shows graph of gain function 119866(119860 120572) for 119896 = 2Notice that it has sharp ridge and that maximum of gainfunction (point 119866

2) lies on the ridge This maximum corre-

sponds to the optimal waveform (solution of the consideredextremal problem) The beginning of the ridge (point 119866

1)

corresponds to the waveform which possesses global mini-mum at degenerate critical point that is corresponds tomax-imally flat waveform (eg see [21]) Gain function 119866(119860 120572)

is not differentiable on the ridge and consequently is notdifferentiable at the point where it has global maximumThisexplains why the approach based on critical points does notwork and why conflict set is so important in the consideredproblem

Positions of global minima of 119891(120591 119860 120572) for 119896 = 2 arepresented in Figure 2 According to Proposition 1 conflict setis the ray defined by119860 gt 14 and120572 = 120587Waveforms119891(120591 119860 120572)with parameters that belong to the conflict set have two globalminimaThewaveform corresponding to the end point of theray (119860 = 14 and 120572 = 120587) has global minimum at degeneratecritical point (so-called maximally flat waveform [21])

Nonnegative waveforms of type (1) with 120574 = 119866(119860 120572) =

minus1119865min(119860 120572) have at least one zero To show that it issufficient to see that 119908(120591 120574 119860 120572) = 0 for 120591 satisfying 119891(120591119860 120572) = 119865min(119860 120572)


001

02

0304

05

06 01205872

12058731205872

2120587

Parameter 120572

Parameter A

1205872

1205874

0

minus1205874

minus1205872Posit

ions

of g

loba

l min

k = 2

Figure 2 Positions of global minima of 119891(120591 119860 120572) for 119896 = 2

The problem of finding maximum value of fundamentalharmonic cosine part of nonnegative waveform of the form

119908119886(120591 120574119886 119887 119860 120572) = 1 minus 120574

119886(cos 120591 + 119887 sin 120591 + 119860 cos (119896120591 + 120572))

(9)

where 120574119886gt 0 is also related to the problem of finding max-

imumof theminimum functionOptimalwaveformof family(9) has two global minima (this claim will be justified inSection 5 Remark 25) and therefore corresponding 3-tupleof parameters belongs to the conflict set in parameter spaceof family (9)

Let us introduce an auxiliary waveform

119891119886(120591 119887 119860 120572) = minus cos 120591 minus 119887 sin 120591 minus 119860 cos (119896120591 + 120572) (10)

and corresponding minimum function 119865119886min(119887 119860 120572) =

min120591119891119886(120591 119887 119860 120572) Inequality 119908

119886(120591 120574119886 119887 119860 120572) ge 0 can be

rewritten as 1 + 120574119886119865119886min(119887 119860 120572) ge 0 and therefore the

highest value of 120574119886is attained for 120574

119886= minus1119865

119886min(119887 119860 120572) Itimmediately follows that nonnegative waveform of type (9)with 120574

119886= minus1119865

119886min(119887 119860 120572) has zero for 120591 satisfying 119891119886(120591

119887 119860 120572) = 119865119886min(119887 119860 120572)

22 Conflict Set Historically conflict set came into beingfrom the problems in which families of smooth functions(such as potentials distances and waveforms) with two com-peting minima occurThe situation when competing minimabecome equal refers to the presence of conflict set (Maxwellset Maxwell strata) in the associated parameter space

There are many facets of conflict set For example in theproblem involving distances between two sets of points theconflict set is the intersections between iso-distance lines[14] Conflict set also arises in the situation when two wavefronts coming from different objects meet [15 25] In thestudy of black holes conflict set is the line of crossover of thehorizon formed by the merger of two black holes [19] In theclassical Euler problem conflict set is a set of points wheredistinct extremal trajectories with the same value of the costfunctional meet one another [18]

Conflict set is very difficult to calculate both analyticallyand numerically (eg see [15]) because of apparent nondif-ferentiability in some directions In optimization of PA effi-ciency some authors already reported difficulties in findingoptimum via standard analytical tools [4 5]

In this section we consider conflict set in the context offamily of waveforms of type (2) for arbitrary 119896 ge 2 In thiscontext for prescribed integer 119896 ge 2 conflict set is said to bea set of all pairs (119860 120572) for which 119891(120591 119860 120572) possesses multipleglobal minima

Suppose that 1205911015840 and 12059110158401015840 are the positions of global min-

ima of 119891(120591 119860 120572) Then the conflict set is specified by the fol-lowing set of relations

119891 (1205911015840

119860 120572) = 119891 (12059110158401015840

119860 120572) (11)

1198911015840

(1205911015840

119860 120572) = 0 1198911015840

(12059110158401015840

119860 120572) = 0 (12)

11989110158401015840

(1205911015840

119860 120572) gt 0 11989110158401015840

(12059110158401015840

119860 120572) gt 0 (13)

(forall120591) 119891 (120591 119860 120572) ge 119891 (1205911015840

119860 120572) (14)

Relations (12) and (13) say that 119891(120591 119860 120572) has minima at 1205911015840and 12059110158401015840 while relations (11) and (14) imply that these minimaare equal and global

The following proposition describes the conflict set offamily of waveforms of type (2)

Proposition 1 Conflict set of family of waveforms of type (2)is the set of all pairs (119860 120572) such that 119860 gt 1119896

2 and 120572 = 120587

The proof of Proposition 1 which is provided at the endof this section also implies that the following four corollarieshold

Corollary 2 The conflict set has end point at (119860 120572) =

(11198962

120587) This end point corresponds to the maximally flatwaveform [21]

Corollary 3 Waveforms of type (2) with parameters thatbelong to conflict set have two global minima at plusmn120591

Δ where

0 lt 120591Δle 120587119896

Corollary 4 Every waveform with fundamental and 119896th har-monic has either one or two global minima

Corollary 5 Conflict set can be parameterised in terms of 120591Δ

as follows

120572 = 120587 119860 (120591Δ) =

sin 120591Δ

119896 sin 119896120591Δ

0 lt 120591Δle120587

119896 (15)

Notice that119860(120591Δ) ismonotonically increasing function on inter-

val 0 lt 120591Δle 120587119896

Proof of Proposition 1 Without loss of generality we canrestrict our consideration to the interval minus120587 lt 120591 le 120587 This isan immediate consequence of the fact that 119891(120591 119860 120572) is aperiodic function

Suppose that 1205911015840 and 12059110158401015840 where 1205911015840 lt 12059110158401015840 are points at which

119891(120591 119860 120572) has two equal global minima Then conflict set is


specified by relations (11)ndash(14) From (11)ndash(13) it follows thatrelations

119891 (1205911015840

119860 120572) minus 119891 (12059110158401015840

119860 120572) = 0

1198911015840

(1205911015840

119860 120572) + 1198911015840

(12059110158401015840

119860 120572) = 0

1198911015840

(1205911015840

119860 120572) minus 1198911015840

(12059110158401015840

119860 120572) = 0

11989110158401015840

(1205911015840

119860 120572) + 11989110158401015840

(12059110158401015840

119860 120572) gt 0

(16)

also hold Let

120591119904119903=(1205911015840

+ 12059110158401015840

)

2 120591

Δ=(12059110158401015840

minus 1205911015840

)

2

(17)

be a pair of points associated with (1205911015840 12059110158401015840) Clearly

minus120587 lt 120591119904119903lt 120587 (18)

0 lt 120591Δlt 120587 (19)

12059110158401015840

= 120591119904119903+ 120591Δ 120591

1015840

= 120591119904119903minus 120591Δ (20)

The first and second derivatives of 119891(120591 119860 120572) are equal to

1198911015840

(120591 119860 120572) = sin 120591 + 119896119860 sin (119896120591 + 120572)

11989110158401015840

(120591 119860 120572) = cos 120591 + 1198962119860 cos (119896120591 + 120572) (21)

By using (20)-(21) system (16) can be rewritten as

sin 120591119904119903sin 120591Δ+ 119860 sin (119896120591

119904119903+ 120572) sin 119896120591

Δ= 0 (22)

sin 120591119904119903cos 120591Δ+ 119896119860 sin (119896120591

119904119903+ 120572) cos 119896120591

Δ= 0 (23)

cos 120591119904119903sin 120591Δ+ 119896119860 cos (119896120591

119904119903+ 120572) sin 119896120591

Δ= 0 (24)

cos 120591119904119903cos 120591Δ+ 1198962

119860 cos (119896120591119904119903+ 120572) cos 119896120591

Δgt 0 (25)

From (19) it follows that sin 120591Δgt 0 Multiplying (24) and

(25) with minus cos 120591Δand sin 120591

Δgt 0 respectively and sum-

ming the resulting relations we obtain 119896119860 cos(119896120591119904119903

+

120572)[119896 sin 120591Δcos 119896120591

Δminus sin 119896120591

Δcos 120591Δ] gt 0 The latest relation

immediately implies that

119896 sin 120591Δcos 119896120591

Δminus sin 119896120591

Δcos 120591Δ

= 0 (26)

Equations (22) and (23) can be considered as a system oftwo linear equations in terms of sin 120591

119904119903and 119860 sin(119896120591

119904119903+ 120572)

According to (26) the determinant of this system is nonzeroand therefore it has only trivial solution

sin 120591119904119903= 0 sin (119896120591

119904119903+ 120572) = 0 (27)

According to (18) sin 120591119904119903= 0 implies

120591119904119903= 0 (28)

According to (20) 120591119904119903= 0 implies

12059110158401015840

= 120591Δ 120591

1015840

= minus120591Δ (29)

Furthermore 120591119904119903= 0 and sin(119896120591

119904119903+ 120572) = 0 imply that sin120572 =

0 From (29) it follows that 120591Δis position of global minimum

of 119891(120591 119860 120572) Clearly 119891(120591Δ 119860 120572) le 119891(0 119860 120572) which together

with sin120572 = 0 leads to

1 minus cos 120591Δ+ 119860 cos120572 (1 minus cos 119896120591

Δ) le 0 (30)

From 120591Δ

= 0 (see (19)) 119860 gt 0 and (30) it follows that cos 120572 lt

0 which together with sin120572 = 0 yields

120572 = 120587 (31)

Since 120591Δis position of global minimum it follows that

119891(120591Δ 119860 120587) le 119891(120587119896 119860 120587) Accordingly 119860(1 + cos 119896120591

Δ) le

cos 120591Δminus cos120587119896 which together with 119860 gt 0 implies that

cos 120591Δminus cos120587119896 ge 0 This relation along with (19) yields

0 lt 120591Δle120587

119896 (32)

Substitution of (31) and (28) in (24) leads to

119860 =sin 120591Δ

119896 sin 119896120591Δ

(33)

Notice that sin 119896120591Δ sin 120591

Δis monotonically decreasing func-

tion on interval (32)Therefore parameter119860 is monotonicallyincreasing function on the same interval with lim

120591Δrarr0+119860 =

11198962 Consequently 119860 gt 1119896

2 which completes theproof

23 Parameter Space In parameter space of family of wave-forms (2) there are two subsets playing important role in theclassification of the family instancesThese are conflict set andcatastrophe set

Catastrophe set is subset of parameter space of waveform119891(120591 119860 120572) It consists of those pairs (119860 120572) for which thecorresponding waveforms 119891(120591 119860 120572) have degenerate criticalpoints at which first and second derivatives are equal to zeroThus for finding catastrophe set we have to consider thefollowing system of equations

1198911015840

(120591119889 119860 120572) = 0

11989110158401015840

(120591119889 119860 120572) = 0

(34)

where 120591119889is a degenerate critical point of waveform119891(120591 119860 120572)

Conflict set in parameter space of waveform119891(120591 119860 120572) asshown in Proposition 1 is the ray described by 119860 gt 1119896

2 and120572 = 120587 It is intimately connected to catastrophe set

In what follows in this subsection we use polar coordinatesystem (119860 cos120572 119860 sin120572) instead of Cartesian coordinatesystem (119860 120572) Examples of catastrophe set and conflict setfor 119896 le 5 plotted in parameter space (119860 cos120572 119860 sin120572) arepresented in Figure 3 Solid line represents the catastropheset while dotted line describes conflict set The isolated pickpoints (usually called cusp) which appear in catastrophecurves correspond to maximally flat waveforms with max-imally flat minimum andor maximally flat maximumThereare two such picks in the catastrophe curves for 119896 = 2 and


00

00

k = 2 k = 3

k = 4 k = 5

Acos 120572

Acos 120572

Acos 120572

Acos 120572

A sin 120572

A sin 120572A sin 120572

A sin 120572

Figure 3 Catastrophe set (solid line) and corresponding conflict set(dotted line) for 119896 le 5 In each plot white triangle dot correspondsto optimal waveform and white circle dot corresponds to maximallyflat waveform

119896 = 4 and one in the catastrophe curves for 119896 = 3 and 119896 = 5Notice that the end point of conflict set is the cusp point

Catastrophe set divides the parameter space (119860 cos120572119860 sin120572) into disjoint subsets In the cases 119896 = 2 and 119896 =

3 catastrophe curve defines inner and outer part For 119896 gt

3 catastrophe curve makes partition of parameter space inseveral inner subsets and one outer subset (see Figure 3)

Notice also that multiplying 119891(120591 119860 120572) with a positiveconstant and adding in turn another constant which leads towaveform of type 119908(120591 120574 119860 120572) (see (1) and (2)) do not makeimpact on the character of catastrophe and conflict sets Thisis because in the course of finding catastrophe set first andsecond derivatives of 119891(120591 119860 120572) are set to zero Clearly (34) interms of 119891(120591 119860 120572) are equivalent to the analogous equationsin terms of119908(120591 120574 119860 120572) Analogously in the course of findingconflict set we consider only the positions of global minima(these positions forwaveforms119891(120591 119860 120572) and119908(120591 120574 119860 120572) arethe same)

3 Nonnegative Waveforms with atLeast One Zero

In what follows let us consider a waveform containing dccomponent fundamental and 119896th (119896 ge 2) harmonic of theform

119879119896(120591) = 1 + 119886

1cos 120591 + 119887

1sin 120591 + 119886

119896cos 119896120591 + 119887

119896sin 119896120591 (35)

The amplitudes of fundamental and 119896th harmonic of wave-form of type (35) respectively are

1205821= radic11988621+ 11988721 (36)

120582119896= radic1198862119896+ 1198872119896 (37)

As it is shown in Section 21 nonnegative waveforms withmaximal amplitude of fundamental harmonic or maximalcoefficient of fundamental harmonic cosine part have atleast one zero It is also shown in Section 22 (Corollary 4)that waveforms of type (35) with nonzero amplitude offundamental harmonic have either one or two globalminimaConsequently if nonnegative waveform of type (35) withnonzero amplitude of fundamental harmonic has at least onezero then it has at most two zeros

In Section 31 we provide general description of nonnega-tive waveforms of type (35) with at least one zero In Sections32 and 33 we consider nonnegative waveforms of type (35)with two zeros

31 General Description of Nonnegative Waveforms with atLeast One Zero The main result of this section is presentedin the following proposition

Proposition 6 Every nonnegative waveform of type (35) withat least one zero can be expressed in the following form

119879119896(120591) = [1 minus cos (120591 minus 120591

0)] [1 minus 120582

119896119903119896(120591)] (38)

where119903119896(120591) = (119896 minus 1) cos 120585

+ 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(39)

providing that

120582119896le [(119896 minus 1) cos 120585 + 119896 sin(120585 minus 120585119896)

sin(120585119896)]

minus1

(40)

10038161003816100381610038161205851003816100381610038161003816 le 120587 (41)

Remark 7 Function on the right hand side of (40) is mono-tonically increasing function of |120585| on interval |120585| le 120587 (formore details about this function see Remark 15) From (57)and (65) it follows that relation

0 le 120582119896le 1 (42)

holds for every nonnegative waveform of type (35) Noticethat according to (40) 120582

119896= 1 implies |120585| = 120587 Substitution

of 120582119896

= 1 and |120585| = 120587 into (55) yields 119879119896(120591) = 1 minus

cos 119896(120591 minus 1205910) Consequently 120582

119896= 1 implies that amplitude

1205821of fundamental harmonic is equal to zero

Remark 8 Conversion of (38) into additive form leads to thefollowing expressions for coefficients of nonnegative wave-forms of type (35) with at least one zero

1198861= minus (1 + 120582

119896cos 120585) cos 120591

0minus 119896120582119896sin 120585 sin 120591

0 (43)

1198871= minus (1 + 120582

119896cos 120585) sin 120591

0+ 119896120582119896sin 120585 cos 120591

0 (44)

119886119896= 120582119896cos (119896120591

0minus 120585) (45)

119887119896= 120582119896sin (119896120591

0minus 120585) (46)

providing that 120582119896satisfy (40) and |120585| le 120587


2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

1205823 = radic2201205823 = radic281205823 = radic24

Figure 4 Nonnegative waveforms with at least one zero for 119896 = 31205910= 1205876 and 120585 = 31205874

Three examples of nonnegative waveforms with at leastone zero for 119896 = 3 are presented in Figure 4 (examples ofnonnegative waveformswith at least one zero for 119896 = 2 can befound in [12]) For all three waveforms presented in Figure 4we assume that 120591

0= 1205876 and 120585 = 31205874 From (40) it follows

that 1205823le radic24 Coefficients of waveform with 120582

3= radic220

(dotted line) are 1198861= minus08977 119887

1= minus03451 119886

3= 005

and 1198873= minus005 Coefficients of waveform with 120582

3= radic28

(dashed line) are 1198861= minus09453 119887

1= minus01127 119886

3= 0125


3= radic24

(solid line) are 1198861= minus10245 119887

1= 02745 119886

3= 025 and

1198873= minus025 First two waveforms have one zero while third

waveform (presented with solid line) has two zeros

Proof of Proposition 6 Waveform of type (35) containingdc component fundamental and 119896th harmonic can be alsoexpressed in the form

119879119896(120591) = 1 + 120582

1cos (120591 + 120593

1) + 120582119896cos (119896120591 + 120593

119896) (47)

where1205821ge 0120582

119896ge 0120593

1isin (minus120587 120587] and120593

119896isin (minus120587 120587] It is easy

to see that relations between coefficient of (35) and param-eters of (47) read as follows

1198861= 1205821cos1205931 119887

1= minus1205821sin1205931 (48)

119886119896= 120582119896cos120593119896 119887

119896= minus120582119896sin120593119896 (49)

Let us introduce 120585 such that10038161003816100381610038161205851003816100381610038161003816 le 120587 120585 = (119896120591

0+ 120593119896) mod2120587 (50)

Using (50) coefficients (49) can be expressed as (45)-(46)Let us assume that 119879

119896(120591) is nonnegative waveform of type

(35) with at least one zero that is 119879119896(120591) ge 0 and 119879

119896(1205910) = 0

for some 1205910 Notice that conditions 119879

119896(120591) ge 0 and 119879

119896(1205910) = 0

imply that 1198791015840119896(1205910) = 0 From 119879

119896(1205910) = 0 and 1198791015840

119896(1205910) = 0 by

using (50) it follows that

1205821cos (120591

0+ 1205931) = minus (1 + 120582

119896cos 120585)

1205821sin (1205910+ 1205931) = minus119896120582

119896sin 120585

(51)

respectively On the other hand 1205821cos(120591 + 120593

1) can be rewrit-

ten as

1205821cos (120591 + 120593

1) = 1205821cos (120591

0+ 1205931) cos (120591 minus 120591

0)

minus 1205821sin (1205910+ 1205931) sin (120591 minus 120591

0)

(52)

Substitution of (51) into (52) yields

1205821cos (120591 + 120593

1) = minus (1 + 120582

119896cos 120585) cos (120591 minus 120591

0)

+ 119896120582119896sin 120585 sin (120591 minus 120591

0)

(53)

According to (50) it follows that cos(119896120591 + 120593119896) = cos(119896(120591 minus

1205910) + 120585) that is

cos (119896120591 + 120593119896) = cos 120585 cos 119896 (120591 minus 120591

0) minus sin 120585 sin 119896 (120591 minus 120591

0)

(54)

Furthermore substitution of (54) and (53) into (47) leads to

119879119896(120591) = [1 minus cos (120591 minus 120591

0)] [1 + 120582

119896cos 120585]

minus 120582119896[1 minus cos 119896 (120591 minus 120591

0)] cos 120585

+ 120582119896[119896 sin (120591 minus 120591

0) minus sin 119896 (120591 minus 120591

0)] sin 120585

(55)

According to (A2) and (A4) (see Appendices) there is com-mon factor [1minuscos(120591minus120591

0)] for all terms in (55) Consequently

(55) can be written in the form (38) where

119903119896(120591) = minus cos 120585 + [

1 minus cos 119896 (120591 minus 1205910)

1 minus cos (120591 minus 1205910)] cos 120585

minus [119896 sin (120591 minus 120591

0) minus sin 119896 (120591 minus 120591

0)

1 minus cos (120591 minus 1205910)

] sin 120585

(56)

From (56) by using (A2) (A4) and cos 120585 cos 119899(120591 minus 1205910) minus

sin 120585 sin 119899(120591 minus 1205910) = cos(119899(120591 minus 120591

0) + 120585) we obtain (39)

In what follows we are going to prove that (40) also holdsAccording to (38) 119879

119896(120591) is nonnegative if and only if

120582119896max120591

119903119896(120591) le 1 (57)

Let us first show that position of global maximumof 119903119896(120591)

belongs to the interval |120591 minus 1205910| le 2120587119896 Relation (56) can be

rewritten as

119903119896(120591) = 119903

119896(1205910minus2120585

119896) + 119902119896(120591) (58)

where

119903119896(1205910minus2120585

119896) = (119896 minus 1) cos 120585 + 119896 sin (120585 minus 120585119896)

sin (120585119896) (59)

119902119896(120591) =

1

1 minus cos (120591 minus 1205910)

sdot [cos 120585 minus cos(119896(120591 minus 1205910+120585

119896))

+119896 sin 120585sin (120585119896)

(cos(120591 minus 1205910+120585

119896) minus cos(120585

119896))]

(60)


For |120585| lt 120587 relation sin 120585 sin(120585119896) gt 0 obviously holds Fromcos 119905 gt cos 1199051015840 for |119905| le 120587119896 lt |119905

1015840

| le 120587 it follows that positionof global maximum of the function of type [119888 cos 119905 minus cos(119896119905)]for 119888 gt 0 belongs to interval |119905| le 120587119896 Therefore position ofglobal maximum of the expression in the square brackets in(60) for |120585| lt 120587 belongs to interval |120591 minus 120591

0+ 120585119896| le 120587119896 This

inequality together with |120585| lt 120587 leads to |120591minus1205910| lt 2120587119896 Since

[1 minus cos(120591 minus 1205910)]minus1 decreases with increasing |120591 minus 120591

0| le 120587 it

follows that 119902119896(120591) for |120585| lt 120587 has global maximum on interval

|120591minus1205910| lt 2120587119896 For |120585| = 120587 it is easy to show thatmax

120591119902119896(120591) =

119902119896(1205910plusmn 2120587119896) = 0 Since 119903

119896(120591) minus 119902

119896(120591) is constant (see (58))

it follows from previous considerations that 119903119896(120591) has global

maximum on interval |120591 minus 1205910| le 2120587119896

To find max120591119903119896(120591) let us consider first derivative of 119903

119896(120591)

with respect to 120591 Starting from (56) first derivative of 119903119896(120591)

can be expressed in the following form

119889119903119896(120591)

119889120591= minus119904 (120591) sdot sin(

119896 (120591 minus 1205910)

2+ 120585) (61)

where

119904 (120591) = [sin(119896 (120591 minus 120591

0)

2) cos(

120591 minus 1205910

2)

minus 119896 cos(119896 (120591 minus 120591

0)

2) sin(

120591 minus 1205910

2)]

sdot sinminus3 (120591 minus 1205910

2)

(62)

Using (A6) (see Appendices) (62) can be rewritten as

119904 (120591) = 2

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos((119896 minus 2119899) (120591 minus 120591

0)

2) (63)

From 119899(119896 minus 119899) gt 0 and |119896 minus 2119899| lt 119896 119899 = 1 (119896 minus 1) itfollows that all summands in (63) decrease with increasing|120591 minus 1205910| providing that |120591 minus 120591

0| le 2120587119896 Therefore 119904(120591) ge 119904(120591

0plusmn

2120587119896) = 119896sin2(120587119896) gt 0 for |120591 minus 1205910| le 2120587119896 Consequently

119889119903119896(120591)119889120591 = 0 and |120591minus120591

0| le 2120587119896 imply that sin(119896(120591minus120591

0)2+

120585) = 0From |120585| le 120587 |120591minus120591

0| le 2120587119896 and sin(119896(120591minus120591

0)2+120585) = 0

it follows that 120591minus1205910+120585119896 = minus120585119896 or |120591minus120591

0+120585119896| = (2120587minus|120585|)119896

and therefore cos(119896(120591 minus 1205910+ 120585119896)) = cos 120585 Since cos(120585119896) ge

cos(2120587 minus |120585|)119896 it follows that max120591119902119896(120591) is attained for 120591 =

1205910minus2120585119896 Furthermore from (60) it follows that max

120591119902119896(120591) =

119902119896(1205910minus 2120585119896) = 0 which together with (58)-(59) leads to

max120591

119903119896(120591) = 119903

119896(1205910minus2120585

119896)

= (119896 minus 1) cos 120585 + 119896 sin (120585 minus 120585119896)sin (120585119896)

(64)

Both terms on the right hand side of (64) are even functionsof 120585 and decrease with increase of |120585| |120585| le 120587 Thereforemax120591119903119896(120591) attains its lowest value for |120585| = 120587 It is easy to

show that right hand side of (64) for |120585| = 120587 is equal to 1which further implies that

max120591

119903119896(120591) ge 1 (65)

From (65) it follows that (57) can be rewritten as 120582119896

le

[max120591119903119896(120591)]minus1 Finally substitution of (64) into 120582

119896le

[max120591119903119896(120591)]minus1 leads to (40) which completes the proof

32 Nonnegative Waveforms with Two Zeros Nonnegativewaveforms of type (35) with two zeros always possess twoglobal minima Such nonnegative waveforms are thereforerelated to the conflict set

In this subsection we provide general description of non-negative waveforms of type (35) for 119896 ge 2 and exactly twozeros According to Remark 7 120582

119896= 1 implies |120585| = 120587 and

119879119896(120591) = 1 minus cos 119896(120591 minus 120591

0) Number of zeros of 119879

119896(120591) = 1 minus

cos 119896(120591minus1205910) on fundamental period equals 119896 which is greater

than two for 119896 gt 2 and equal to two for 119896 = 2 In the followingproposition we exclude all waveforms with 120582

119896= 1 (the case

when 119896 = 2 and 1205822= 1 is going to be discussed in Remark 10)

Proposition 9 Every nonnegative waveform of type (35) withexactly two zeros can be expressed in the following form

119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(66)

where

119888119899= [sin(120585 minus 119899120585

119896) cos(120585

119896)

minus (119896 minus 119899) cos(120585 minus 119899120585

119896) sin(120585

119896)]

sdot sinminus3 (120585119896)

(67)

120582119896= [(119896 minus 1) cos 120585 + 119896 sin(120585 minus 120585119896)

sin(120585119896)]

minus1

(68)

0 lt10038161003816100381610038161205851003816100381610038161003816 lt 120587 (69)

Remark 10 For 119896 = 2 waveforms with 1205822= 1 also have

exactly two zeros These waveforms can be included in aboveproposition by substituting (69) with 0 lt |120585| le 120587

Remark 11 Apart from nonnegative waveforms of type (35)with two zeros there are another two types of nonnegativewaveforms which can be obtained from (66)ndash(68) These are

(i) nonnegative waveforms with 119896 zeros (correspondingto |120585| = 120587) and

(ii) maximally flat nonnegative waveforms (correspond-ing to 120585 = 0)

Notice that nonnegative waveforms of type (35) with120582119896= 1 can be obtained from (66)ndash(68) by setting |120585| =

120587 Substitution of 120582119896

= 1 and |120585| = 120587 into (66) alongwith execution of all multiplications and usage of (A2) (seeAppendices) leads to 119879

119896(120591) = 1 minus cos 119896(120591 minus 120591

0)


Also maximally flat nonnegative waveforms (they haveonly one zero [21]) can be obtained from (66)ndash(68) by setting120585 = 0 Thus substitution of 120585 = 0 into (66)ndash(68) leads tothe following form of maximally flat nonnegative waveformof type (35)

119879119896(120591) =

[1 minus cos (120591 minus 1205910)]2

3 (1198962 minus 1)

sdot [119896 (1198962

minus 1)

+ 2

119896minus2

sum

119899=1

(119896 minus 119899) ((119896 minus 119899)2

minus 1) cos 119899 (120591 minus 1205910)]

(70)

Maximally flat nonnegative waveforms of type (35) for 119896 le 4

can be expressed as

1198792(120591) =

2

3[1 minus cos(120591 minus 120591

0)]2

1198793(120591) =

1

2[1 minus cos(120591 minus 120591

0)]2

[2 + cos (120591 minus 1205910)]

1198794(120591) =

4

15[1 minus cos (120591 minus 120591

0)]2

sdot [5 + 4 cos (120591 minus 1205910) + cos 2 (120591 minus 120591

0)]

(71)

Remark 12 Every nonnegative waveform of type (35) withexactly one zero at nondegenerate critical point can bedescribed as in Proposition 6 providing that symbol ldquolerdquoin relation (40) is replaced with ldquoltrdquo This is an immediateconsequence of Propositions 6 and 9 and Remark 11

Remark 13 Identity [1minus cos(120591minus1205910)][1minus cos(120591minus120591

0+2120585119896)] =

[cos 120585119896 minus cos(120591 minus 1205910+ 120585119896)]

2 implies that (66) can be alsorewritten as

119879119896(120591) = 120582

119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]

2

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(72)

Furthermore substitution of (67) into (72) leads to

119879119896(120591)

= 120582119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]

sdot [(119896 minus 1) sin 120585sin (120585119896)

minus 2

119896minus1

sum

119899=1

sin (120585 minus 119899120585119896)sin (120585119896)

cos 119899(120591 minus 1205910+120585

119896)]

(73)

Remark 14 According to (A6) (see Appendices) it followsthat coefficients (67) can be expressed as

119888119899= 2

119896minus119899minus1

sum

119898=1

119898(119896 minus 119899 minus 119898) cos((119896 minus 119899 minus 2119898) 120585119896

) (74)

Furthermore from (74) it follows that coefficients 119888119896minus2

119888119896minus3

119888119896minus4

and 119888119896minus5

are equal to119888119896minus2

= 2 (75)

119888119896minus3

= 8 cos(120585119896) (76)

119888119896minus4

= 8 + 12 cos(2120585119896) (77)

119888119896minus5

= 24 cos(120585119896) + 16 cos(3120585

119896) (78)

For example for 119896 = 2 (75) and (68) lead to 1198880= 2 and

1205822= 1(2+cos 120585) respectively which from (72) further imply

that

1198792(120591) =

2 [cos(1205852) minus cos(120591 minus 1205910+ 1205852)]

2

[2 + cos 120585] (79)

Also for 119896 = 3 (75) (76) and (68) lead to 1198881= 2 119888

0=

8 cos(1205853) and 1205823= [2(3 cos(1205853) + cos 120585)]minus1 respectively

which from (72) further imply that

1198793(120591) =

2 [cos (1205853) minus cos (120591 minus 1205910+ 1205853)]

2

[3 cos (1205853) + cos 120585]

sdot [2 cos(1205853) + cos(120591 minus 120591

0+120585

3)]

(80)

Remark 15 According to (A5) (see Appendices) relation(68) can be rewritten as

120582119896= [(119896 minus 1) cos 120585 + 119896

119896minus1

sum

119899=1

cos((119896 minus 2119899)120585119896

)]

minus1

(81)

Clearly amplitude 120582119896of 119896th harmonic of nonnegative wave-

form of type (35) with exactly two zeros is even functionof 120585 Since cos((119896 minus 2119899)120585119896) 119899 = 0 (119896 minus 1) decreaseswith increase of |120585| on interval 0 le |120585| le 120587 it follows that120582119896monotonically increases with increase of |120585| Right hand

side of (68) is equal to 1(1198962

minus 1) for 120585 = 0 and to one for|120585| = 120587 Therefore for nonnegative waveforms of type (35)with exactly two zeros the following relation holds

1

1198962 minus 1lt 120582119896lt 1 (82)

The left boundary in (82) corresponds to maximally flatnonnegative waveforms (see Remark 11) The right boundaryin (82) corresponds to nonnegative waveforms with 119896 zeros(also see Remark 11)

Amplitude of 119896th harmonic of nonnegative waveform oftype (35) with two zeros as a function of parameter 120585 for 119896 le5 is presented in Figure 5

Remark 16 Nonnegative waveform of type (35) with twozeros can be also expressed in the following form

119879119896(120591) = 1 minus 120582

119896

119896 sin 120585sin (120585119896)

cos(120591 minus 1205910+120585

119896)

+ 120582119896cos (119896 (120591 minus 120591

0) + 120585)

(83)


1

08

06

04

02

0minus1 minus05 0 05

Am

plitu

de120582k

1

Parameter 120585120587

k = 2k = 3

k = 4k = 5

Figure 5 Amplitude of 119896th harmonic of nonnegative waveformwith two zeros as a function of parameter 120585

where 120582119896is given by (68) and 0 lt |120585| lt 120587 From (83) it follows

that coefficients of fundamental harmonic of nonnegativewaveform of type (35) with two zeros are

1198861= minus1205821cos(120591

0minus120585

119896) 119887

1= minus1205821sin(120591

0minus120585

119896) (84)

where 1205821is amplitude of fundamental harmonic

1205821=

119896 sin 120585sin (120585119896)

120582119896 (85)

Coefficients of 119896th harmonic are given by (45)-(46)Notice that (68) can be rewritten as

120582119896= [cos(120585

119896)

119896 sin 120585sin(120585119896)

minus cos 120585]minus1

(86)

By introducing new variable

119909 = cos(120585119896) (87)

and using the Chebyshev polynomials (eg see Appendices)relations (85) and (86) can be rewritten as

1205821= 119896120582119896119880119896minus1

(119909) (88)

120582119896=

1

119896119909119880119896minus1

(119909) minus 119881119896(119909)

(89)

where119881119896(119909) and119880

119896(119909) denote the Chebyshev polynomials of

the first and second kind respectively From (89) it followsthat

120582119896[119896119909119880119896minus1

(119909) minus 119881119896(119909)] minus 1 = 0 (90)

which is polynomial equation of 119896th degree in terms of var-iable 119909 From 0 lt |120585| lt 120587 and (87) it follows that

cos(120587119896) lt 119909 lt 1 (91)

Since 120582119896is monotonically increasing function of |120585| 0 lt |120585| lt

120587 it follows that 120582119896is monotonically decreasing function of

119909 This further implies that (90) has only one solution thatsatisfies (91) (For 119896 = 2 expression (91) reads cos(1205872) le

119909 lt 1) This solution for 119909 (which can be obtained at leastnumerically) according to (88) leads to amplitude 120582

1of

fundamental harmonicFor 119896 le 4 solutions of (90) and (91) are

119909 = radic1 minus 1205822

21205822

1

3lt 1205822le 1

119909 =1

23radic1205823

1

8lt 1205823lt 1

119909 = radic1

6(1 + radic

51205824+ 3

21205824

)1

15lt 1205824lt 1

(92)

Insertion of (92) into (88) leads to the following relationsbetween amplitude 120582

1of fundamental and amplitude 120582

119896of

119896th harmonic 119896 le 4

1205821= radic8120582

2(1 minus 120582

2)

1

3lt 1205822le 1 (93)

1205821= 3 (

3radic1205823minus 1205823)

1

8lt 1205823lt 1 (94)

1205821= radic

32

27(radic2120582

4(3 + 5120582

4)3

minus 21205824(9 + 7120582

4))

1

15lt 1205824lt 1

(95)

Proof of Proposition 9 As it has been shown earlier (seeProposition 6) nonnegative waveform of type (35) with atleast one zero can be represented in form (38) Since weexclude nonnegative waveforms with 120582

119896= 1 according to

Remark 7 it follows that we exclude case |120585| = 120587Therefore inthe quest for nonnegative waveforms of type (35) having twozeros we will start with waveforms of type (38) for |120585| lt 120587It is clear that nonnegative waveforms of type (38) have twozeros if and only if

120582119896= [max120591

119903119896(120591)]minus1

(96)

and max120591119903119896(120591) = 119903

119896(1205910) According to (64) max

120591119903119896(120591) =

119903119896(1205910) implies |120585| = 0 Therefore it is sufficient to consider

only the interval (69)Substituting (96) into (38) we obtain

119879119896(120591) =

[1 minus cos (120591 minus 1205910)] [max

120591119903119896(120591) minus 119903

119896(120591)]

max120591119903119896(120591)

(97)


Expression max120591119903119896(120591) minus 119903

119896(120591) according to (64) and (39)

equals

max120591

119903119896(120591) minus 119903

119896(120591) = 119896

sin ((119896 minus 1) 120585119896)sin (120585119896)

minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(98)

Comparison of (97) with (66) yields

max120591

119903119896(120591) minus 119903

119896(120591) = [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(99)

where coefficients 119888119899 119899 = 0 119896 minus 2 are given by (67) In

what follows we are going to show that right hand sides of(98) and (99) are equal

From (67) it follows that

1198880minus 1198881cos(120585

119896) = 119896

sin (120585 minus 120585119896)sin (120585119896)

(100)

Also from (67) for 119899 = 1 119896minus3 it follows that the followingrelations hold

(119888119899minus1

+ 119888119899+1

) cos(120585119896) minus 2119888

119899= 2 (119896 minus 119899) cos(120585 minus 119899120585

119896)

(119888119899minus1

minus 119888119899+1

) sin(120585119896) = 2 (119896 minus 119899) sin(120585 minus 119899120585

119896)

(101)

From (99) by using (75) (76) (100)-(101) and trigonometricidentities

cos(120591 minus 1205910+2120585

119896) = cos(120585

119896) cos(120591 minus 120591

0+120585

119896)

minus sin(120585119896) sin(120591 minus 120591

0+120585

119896)

cos(120585 minus 119899120585

119896) cos(119899(120591 minus 120591

0+120585

119896))

minus sin(120585 minus 119899120585

119896) sin(119899(120591 minus 120591

0+120585

119896))

= cos (119899 (120591 minus 1205910) + 120585)

(102)

we obtain (98) Consequently (98) and (99) are equal whichcompletes the proof

33 Nonnegative Waveforms with Two Zeros and PrescribedCoefficients of 119896thHarmonic In this subsectionwe show thatfor prescribed coefficients 119886

119896and 119887119896 there are 119896 nonnegative

waveforms of type (35) with exactly two zeros According to

(37) and (82) coefficients 119886119896and 119887119896of nonnegative waveforms

of type (35) with exactly two zeros satisfy the followingrelation

1

1198962 minus 1lt radic1198862119896+ 1198872119896lt 1 (103)

According to Remark 16 the value of 119909 (see (87)) that cor-responds to 120582

119896= radic1198862

119896+ 1198872119896can be determined from (90)-

(91) As we mentioned earlier (90) has only one solutionthat satisfies (91) This value of 119909 according to (88) leadsto the amplitude 120582

1of fundamental harmonic (closed form

expressions for 1205821in terms of 120582

119896and 119896 le 4 are given by (93)ndash

(95))On the other hand from (45)-(46) it follows that

1198961205910minus 120585 = atan 2 (119887

119896 119886119896) + 2119902120587 119902 = 1 (119896 minus 1) (104)

where function atan 2(119910 119909) is defined as

atan 2 (119910 119909) =

arctan(119910

119909) if 119909 ge 0

arctan(119910

119909) + 120587 if 119909 lt 0 119910 ge 0

arctan(119910

119909) minus 120587 if 119909 lt 0 119910 lt 0

(105)

with the codomain (minus120587 120587] Furthermore according to (84)and (104) the coefficients of fundamental harmonic of non-negative waveforms with two zeros and prescribed coeffi-cients of 119896th harmonic are equal to

1198861= minus1205821cos[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

1198871= minus1205821sin[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

(106)

where 119902 = 0 (119896minus 1) For chosen 119902 according to (104) and(66) positions of zeros are

1205910=1

119896[120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

1205910minus2120585

119896=1

119896[minus120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

(107)

From (106) and 119902 = 0 (119896minus1) it follows that for prescribedcoefficients 119886

119896and 119887119896 there are 119896 nonnegative waveforms of

type (35) with exactly two zerosWe provide here an algorithm to facilitate calculation

of coefficients 1198861and 1198871of nonnegative waveforms of type

(35) with two zeros and prescribed coefficients 119886119896and 119887

119896

providing that 119886119896and 119887119896satisfy (103)


2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0

q = 1

q = 2

Figure 6 Nonnegative waveforms with two zeros for 119896 = 3 1198863=

minus015 and 1198873= minus02

Algorithm 17 (i) Calculate 120582119896= radic1198862119896+ 1198872119896

(ii) identify 119909 that satisfies both relations (90) and (91)(iii) calculate 120582

1according to (88)

(iv) choose integer 119902 such that 0 le 119902 le 119896 minus 1(v) calculate 119886

1and 1198871according to (106)

For 119896 le 4 by using (93) for 119896 = 2 (94) for 119896 = 3 and (95)for 119896 = 4 it is possible to calculate directly 120582

1from 120582

119896and

proceed to step (iv)For 119896 = 2 and prescribed coefficients 119886

2and 1198872 there are

two waveforms with two zeros one corresponding to 1198861lt 0

and the other corresponding to 1198861gt 0 (see also [12])

Let us take as an input 119896 = 3 1198863= minus015 and 119887

3= minus02

Execution of Algorithm 17 on this input yields 1205823= 025 and

1205821= 11399 (according to (94)) For 119902 = 0 we calculate

1198861= minus08432 and 119887

1= 07670 (corresponding waveform is

presented by solid line in Figure 6) for 119902 = 1 we calculate1198861= minus02426 and 119887

1= minus11138 (corresponding waveform is

presented by dashed line) for 119902 = 2 we calculate 1198861= 10859

and 1198871= 03468 (corresponding waveform is presented by

dotted line)As another example of the usage of Algorithm 17 let us

consider case 119896 = 4 and assume that1198864= minus015 and 119887

4= minus02

Consequently 1205824= 025 and 120582

1= 09861 (according to (95))

For 119902 = 0 3we calculate the following four pairs (1198861 1198871) of

coefficients of fundamental harmonic (minus08388 05184) for119902 = 0 (minus05184 minus08388) for 119902 = 1 (08388 minus05184) for 119902 =2 and (05184 08388) for 119902 = 3 Corresponding waveformsare presented in Figure 7

4 Nonnegative Waveforms with MaximalAmplitude of Fundamental Harmonic

In this section we provide general description of nonnegativewaveforms containing fundamental and 119896th harmonic withmaximal amplitude of fundamental harmonic for prescribedamplitude of 119896th harmonic

The main result of this section is presented in the fol-lowing proposition

3

2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0q = 1

q = 2q = 3



Proposition 18 Every nonnegativewaveformof type (35)withmaximal amplitude 120582

1of fundamental harmonic and pre-

scribed amplitude 120582119896of 119896th harmonic can be expressed in the

following form

119879119896(120591) = [1 minus cos (120591 minus 120591

0)]

sdot [1 minus (119896 minus 1) 120582119896minus 2120582119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(108)

if 0 le 120582119896le 1(119896

2

minus 1) or

119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(109)

if 1(1198962 minus 1) le 120582119896le 1 providing that 119888

119899 119899 = 0 119896 minus 2 and

120582119896are related to 120585 via relations (67) and (68) respectively and

|120585| le 120587

Remark 19 Expression (108) can be obtained from (38) bysetting 120585 = 0 Furthermore insertion of 120585 = 0 into (43)ndash(46)leads to the following expressions for coefficients ofwaveformof type (108)

1198861= minus (1 + 120582

119896) cos 120591

0 119887

1= minus (1 + 120582

119896) sin 120591

0

119886119896= 120582119896cos (119896120591

0) 119887

119896= 120582119896sin (119896120591

0)

(110)

On the other hand (109) coincides with (66) Thereforethe expressions for coefficients of (109) and (66) also coincideThus expressions for coefficients of fundamental harmonic ofwaveform (109) are given by (84) where 120582

1is given by (85)

while expressions for coefficients of 119896th harmonic are givenby (45)-(46)

Waveforms described by (108) have exactly one zerowhile waveforms described by (109) for 1(1198962 minus 1) lt 120582

119896lt 1


14

12

1

08

06

04

02

00 05 1

Amplitude 120582k

Am

plitu

de1205821

k = 2

k = 3

k = 4

Figure 8 Maximal amplitude of fundamental harmonic as a func-tion of amplitude of 119896th harmonic

have exactly two zeros As we mentioned earlier waveforms(109) for 120582

119896= 1 have 119896 zeros

Remark 20 Maximal amplitude of fundamental harmonic ofnonnegative waveforms of type (35) for prescribed amplitudeof 119896th harmonic can be expressed as

1205821= 1 + 120582

119896 (111)

if 0 le 120582119896le 1(119896

2

minus 1) or

1205821=

119896 sin 120585119896 sin 120585 cos (120585119896) minus cos 120585 sin (120585119896)

(112)

if 1(1198962 minus 1) le 120582119896le 1 where 120585 is related to 120582

119896via (68) (or

(86)) and |120585| le 120587From (110) it follows that (111) holds Substitution of (86)

into (85) leads to (112)Notice that 120582

119896= 1(119896

2

minus 1) is the only common point ofthe intervals 0 le 120582

119896le 1(119896

2

minus 1) and 1(1198962

minus 1) le 120582119896le

1 According to (111) 120582119896= 1(119896

2

minus 1) corresponds to 1205821=

1198962

(1198962

minus1) It can be also obtained from (112) by setting 120585 = 0The waveforms corresponding to this pair of amplitudes aremaximally flat nonnegative waveforms

Maximal amplitude of fundamental harmonic of non-negative waveform of type (35) for 119896 le 4 as a function ofamplitude of 119896th harmonic is presented in Figure 8

Remark 21 Maximum value of amplitude of fundamentalharmonic of nonnegative waveform of type (35) is

1205821max =

1

cos (120587 (2119896)) (113)

This maximum value is attained for |120585| = 1205872 (see (112)) Thecorresponding value of amplitude of 119896th harmonic is 120582

119896=

(1119896) tan(120587(2119896)) Nonnegative waveforms of type (35) with1205821= 1205821max have two zeros at 1205910 and 1205910 minus 120587119896 for 120585 = 1205872 or

at 1205910and 1205910+ 120587119896 for 120585 = minus1205872

14

12

1

08

06

04

02

0minus1 minus05 0 05 1

Am

plitu

de1205821


k = 2k = 3k = 4

Figure 9 Maximal amplitude of fundamental harmonic as a func-tion of parameter 120585

To prove that (113) holds let us first show that the fol-lowing relation holds for 119896 ge 2

cos( 120587

2119896) lt 1 minus

1

1198962 (114)

From 119896 ge 2 it follows that sinc(120587(4119896)) gt sinc(1205874) wheresinc 119909 = (sin119909)119909 and therefore sin(120587(4119896)) gt 1(radic2119896)By using trigonometric identity cos 2119909 = 1 minus 2sin2119909 weimmediately obtain (114)

According to (111) and (112) it is clear that 1205821attains its

maximum value on the interval 1(1198962 minus 1) le 120582119896le 1 Since

120582119896is monotonic function of |120585| on interval |120585| le 120587 (see

Remark 15) it follows that 119889120582119896119889120585 = 0 for 0 lt |120585| lt 120587

Therefore to find critical points of 1205821as a function of 120582

119896

it is sufficient to find critical points of 1205821as a function of

|120585| 0 lt |120585| lt 120587 and consider its values at the end points120585 = 0 and |120585| = 120587 Plot of 120582

1as a function of parameter 120585

for 119896 le 4 is presented in Figure 9 According to (112) firstderivative of 120582

1with respect to 120585 is equal to zero if and only

if (119896 cos 120585 sin(120585119896) minus sin 120585 cos(120585119896)) cos 120585 = 0 On interval0 lt |120585| lt 120587 this is true if and only if |120585| = 1205872 Accordingto (112) 120582

1is equal to 119896

2

(1198962

minus 1) for 120585 = 0 equal to zerofor |120585| = 120587 and equal to 1 cos(120587(2119896)) for |120585| = 1205872 From(114) it follows that 1198962(1198962minus1) lt 1 cos(120587(2119896)) and thereforemaximum value of 120582

1is given by (113) Moreover maximum

value of 1205821is attained for |120585| = 1205872

According to above consideration all nonnegative wave-forms of type (35) having maximum value of amplitude offundamental harmonic can be obtained from (109) by setting|120585| = 1205872 Three of them corresponding to 119896 = 3 120585 = 1205872and three different values of 120591

0(01205876 and1205873) are presented

in Figure 10 Dotted line corresponds to 1205910= 0 (coefficients

of corresponding waveform are 1198861= minus1 119887

1= 1radic3 119886

3= 0

and 1198873= minusradic39) solid line to 120591

0= 1205876 (119886

1= minus2radic3 119887

1= 0

1198863= radic39 and 119887

3= 0) and dashed line to 120591

0= 1205873 (119886

1= minus1

1198871= minus1radic3 119886

3= 0 and 119887

3= radic39)

Proof of Proposition 18 As it has been shown earlier (Propo-sition 6) nonnegative waveform of type (35) with at least


2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

1205910 = 01205910 = 12058761205910 = 1205873

Figure 10 Nonnegative waveforms with maximum amplitude offundamental harmonic for 119896 = 3 and 120585 = 1205872

one zero can be represented in form (38) According to (43)(44) and (36) for amplitude 120582

1of fundamental harmonic of

waveforms of type (38) the following relation holds

1205821= radic(1 + 120582

119896cos 120585)2 + 11989621205822

119896sin2120585 (115)

where 120582119896satisfy (40) and |120585| le 120587

Because of (40) in the quest of finding maximal 1205821for

prescribed 120582119896 we have to consider the following two cases

(Case i)120582119896lt [(119896minus1) cos 120585 + 119896 sin(120585minus120585119896) sin(120585119896)]minus1

(Case ii)120582119896= [(119896minus1) cos 120585+119896 sin(120585minus120585119896) sin(120585119896)]minus1

Case i Since 120582119896lt [(119896 minus 1) cos 120585 + 119896 sin(120585 minus 120585119896) sin(120585119896)]minus1

implies 120582119896

= 1 according to (115) it follows that 1205821

= 0Hence 119889120582

1119889120585 = 0 implies

2120582119896sin 120585 [1 minus (1198962 minus 1) 120582

119896cos 120585] = 0 (116)

Therefore 1198891205821119889120585 = 0 if 120582

119896= 0 (Option 1) or sin 120585 = 0

(Option 2) or (1198962 minus 1)120582119896cos 120585 = 1 (Option 3)

Option 1 According to (115) 120582119896= 0 implies 120582

1= 1 (notice

that this implication shows that 1205821does not depend on 120585 and

therefore we can set 120585 to zero value)

Option 2 According to (115) sin 120585 = 0 implies 1205821= 1 +

120582119896cos 120585 which further leads to the conclusion that 120582

1is

maximal for 120585 = 0 For 120585 = 0 120582119896lt [(119896 minus 1) cos 120585 + 119896 sin(120585 minus

120585119896) sin(120585119896)]minus1 becomes 120582119896lt 1(119896

2

minus 1)

Option 3 This option leads to contradiction To show thatnotice that (119896

2

minus 1)120582119896cos 120585 = 1 and 120582

119896lt [(119896 minus

1) cos 120585 + 119896 sin(120585 minus 120585119896) sin(120585119896)]minus1 imply that (119896 minus 1) cos 120585 gtsin(120585minus120585119896) sin(120585119896) Using (A5) (see Appendices) the latestinequality can be rewritten assum119896minus1

119899=1[cos 120585minuscos((119896minus2119899)120585119896)] gt

0 But from |119896 minus 2119899| lt 119896 119899 = 1 (119896 minus 1) and |120585| le 120587

it follows that all summands are not positive and therefore(119896minus1) cos 120585 gt sin(120585minus120585119896) sin(120585119896) does not hold for |120585| le 120587

Consequently Case i implies 120585 = 0 and 120582119896lt 1(119896

2

minus 1)Finally substitution of 120585 = 0 into (38) leads to (108) whichproves that (108) holds for 120582

119896lt 1(119896

2

minus 1)

Case ii Relation120582119896= [(119896minus1) cos 120585+119896 sin(120585minus120585119896) sin(120585119896)]minus1

according to Proposition 9 and Remark 11 implies that cor-responding waveforms can be expressed via (66)ndash(68) for|120585| le 120587 Furthermore 120582

119896= [(119896 minus 1) cos 120585 + 119896 sin(120585 minus

120585119896) sin(120585119896)]minus1 and |120585| le 120587 imply 1(1198962 minus 1) le 120582119896le 1

This proves that (109) holds for 1(1198962 minus 1) le 120582119896le 1

Finally let us prove that (108) holds for 120582119896= 1(119896

2

minus

1) According to (68) (see also Remark 11) this value of 120582119896

corresponds to 120585 = 0 Furthermore substitution of 120582119896=

1(1198962

minus 1) and 120585 = 0 into (109) leads to (70) which can berewritten as

119879119896(120591) =

[1 minus cos (120591 minus 1205910)]

(1 minus 1198962)

sdot [119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(117)

Waveform (117) coincides with waveform (108) for 120582119896

=

1(1 minus 1198962

) Consequently (108) holds for 120582119896= 1(1 minus 119896

2

)which completes the proof

5 Nonnegative Waveforms with MaximalAbsolute Value of the Coefficient of CosineTerm of Fundamental Harmonic

In this sectionwe consider general description of nonnegativewaveforms of type (35) with maximal absolute value ofcoefficient 119886

1for prescribed coefficients of 119896th harmonicThis

type of waveform is of particular interest in PA efficiencyanalysis In a number of cases of practical interest eithercurrent or voltage waveform is prescribed In such casesthe problem of finding maximal efficiency of PA can bereduced to the problem of finding nonnegative waveformwith maximal coefficient 119886

1for prescribed coefficients of 119896th

harmonic (see also Section 7)In Section 51 we provide general description of nonneg-

ative waveforms of type (35) with maximal absolute value ofcoefficient 119886

1for prescribed coefficients of 119896th harmonic In

Section 52 we illustrate results of Section 51 for particularcase 119896 = 3

51 Nonnegative Waveforms with Maximal Absolute Value ofCoefficient 119886

1for 119896 ge 2 Waveforms 119879

119896(120591) of type (35) with

1198861ge 0 can be derived from those with 119886

1le 0 by shifting

by 120587 and therefore we can assume without loss of generalitythat 119886

1le 0 Notice that if 119896 is even then shifting 119879

119896(120591) by

120587 produces the same result as replacement of 1198861with minus119886

1

(119886119896remains the same) On the other hand if 119896 is odd then

shifting 119879119896(120591) by 120587 produces the same result as replacement

of 1198861with minus119886

1and 119886119896with minus119886

119896

According to (37) coefficients of 119896th harmonic can beexpressed as

119886119896= 120582119896cos 120575 119887

119896= 120582119896sin 120575 (118)


where

|120575| le 120587 (119)

Conversely for prescribed coefficients 119886119896and 119887

119896 120575 can be

determined as

120575 = atan 2 (119887119896 119886119896) (120)

where definition of function atan 2(119910 119909) is given by (105)The main result of this section is stated in the following

proposition

Proposition 22 Every nonnegative waveform of type (35)withmaximal absolute value of coefficient 119886

1le 0 for prescribed

coefficients 119886119896and 119887119896of 119896th harmonic can be represented as

119879119896(120591)

= [1 minus cos 120591]


119896minus1

sum

119899=1

(119896 minus 119899) (119886119896cos 119899120591 + 119887

119896sin 119899120591)]

(121)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1+119886

119896 where 120575 = atan 2(bk

119886119896) or

119879119896(120591) = 120582

119896[1 minus cos(120591 minus (120575 + 120585)

119896)]

sdot [1 minus cos(120591 minus (120575 minus 120585)

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120575

119896)]

(122)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 where 119888

119899 119899 = 0

119896minus2 and 120582119896= radic1198862119896+ 1198872119896are related to 120585 via relations (67) and

(68) respectively and |120585| le 120587

Remark 23 Expression (121) can be obtained from (38) bysetting 120591

0= 0 and 120585 = minus120575 and then replacing 120582

119896cos 120575 with

119886119896(see (118)) and 120582

119896cos(119899120591 minus 120575) with 119886

119896cos 119899120591 + 119887

119896sin 119899120591

(see also (118)) Furthermore insertion of 1205910= 0 and 120585 =

minus120575 into (43)ndash(46) leads to the following relations betweenfundamental and 119896th harmonic coefficients of waveform(121)

1198861= minus (1 + 119886

119896) 119887

1= minus119896119887

119896 (123)

On the other hand expression (122) can be obtained from(66) by replacing 120591

0minus120585119896with 120575119896 Therefore substitution of

1205910minus 120585119896 = 120575119896 in (84) leads to

1198861= minus1205821cos(120575

119896) 119887

1= minus1205821sin(120575

119896) (124)

where 1205821is given by (85)

The fundamental harmonic coefficients 1198861and 1198871of wave-

form of type (35) with maximal absolute value of coefficient1198861le 0 satisfy both relations (123) and (124) if 119886

119896and 119887119896satisfy

1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) For such waveforms

relations 1205910= 0 and 120585 = minus120575 also hold

Remark 24 Amplitude of 119896th harmonic of nonnegativewaveform of type (35) with maximal absolute value of coeffi-cient 119886

1le 0 and coefficients 119886

119896 119887119896satisfying 1 + 119886

119896=

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) is

120582119896=

sin (120575119896)119896 sin 120575 cos (120575119896) minus cos 120575 sin (120575119896)

(125)

To show that it is sufficient to substitute 119886119896= 120582119896cos 120575 (see

(118)) into 1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896)

Introducing new variable

119910 = cos(120575119896) (126)

and using the Chebyshev polynomials (eg see Appendices)relations 119886

119896= 120582119896cos 120575 and (125) can be rewritten as

119886119896= 120582119896119881119896(119910) (127)

120582119896=

1

119896119910119880119896minus1

(119910) minus 119881119896(119910)

(128)

where119881119896(119910) and119880


the first and second kind respectively Substitution of (128)into (127) leads to

119886119896119896119910119880119896minus1

(119910) minus (1 + 119886119896) 119881119896(119910) = 0 (129)

which is polynomial equation of 119896th degree in terms of var-iable 119910 From |120575| le 120587 and (126) it follows that

cos(120587119896) le 119910 le 1 (130)

In what follows we show that 119886119896is monotonically increas-

ing function of 119910 on the interval (130) From 120585 = minus120575 (seeRemark 23) and (81) it follows that 120582minus1

119896= (119896 minus 1) cos 120575 +

119896sum119896minus1

119899=1cos((119896 minus 2119899)120575119896) ge 1 and therefore 119886

119896= 120582119896cos 120575 can

be rewritten as

119886119896=

cos 120575(119896 minus 1) cos 120575 + 119896sum119896minus1

119899=1cos ((119896 minus 2119899) 120575119896)

(131)

Obviously 119886119896is even function of 120575 and all cosines in (131)

are monotonically decreasing functions of |120575| on the interval|120575| le 120587 It is easy to show that cos((119896 minus 2119899)120575119896) 119899 =

1 (119896 minus 1) decreases slower than cos 120575 when |120575| increasesThis implies that denominator of the right hand side of(131) decreases slower than numerator Since denominator ispositive for |120575| le 120587 it further implies that 119886

119896is decreasing

function of |120575| on interval |120575| le 120587 Consequently 119886119896is

monotonically increasing function of 119910 on the interval (130)Thus we have shown that 119886

119896is monotonically increasing

function of 119910 on the interval (130) and therefore (129) hasonly one solution that satisfies (130) According to (128) thevalue of 119910 obtained from (129) and (130) either analyticallyor numerically leads to amplitude 120582

119896of 119896th harmonic


1

05

0

minus05

minus1

minus1 minus05 0 05 1

Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4

Figure 11 Plot of (119886119896 119887119896) satisfying 1 + 119886

119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4

By solving (129) and (130) for 119896 le 4 we obtain

119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8

119910 =radicradic2 minus 4119886

4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)

Insertion of (132) into (128) leads to the following explicitexpressions for the amplitude 120582

119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3

minus1 le 1198863le1

8 (134)

1205824=1

4(minus1 minus 119886

4+ radic2 minus 4119886

4+ 1011988624) minus1 le 119886

4le

1

15

(135)

Relations (133)ndash(135) define closed lines (see Figure 11) whichseparate points representing waveforms of type (121) frompoints representing waveforms of type (122) For given 119896points inside the corresponding curve refer to nonnegativewaveforms of type (121) whereas points outside curve (andradic1198862119896+ 1198872119896le 1) correspond to nonnegative waveforms of type

(122) Points on the respective curve correspond to the wave-forms which can be expressed in both forms (121) and (122)

Remark 25 Themaximum absolute value of coefficient 1198861of

nonnegative waveform of type (35) is

100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)

This maximum value is attained for |120585| = 1205872 and 120575 = 0

(see (124)) Notice that |1198861|max is equal to the maximum value

1205821max of amplitude of fundamental harmonic (see (113))

Coefficients of waveform with maximum absolute value ofcoefficient 119886

1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)

Waveformdescribed by (137) is cosinewaveformhaving zerosat 120587(2119896) and minus120587(2119896)

In the course of proving (136) notice first that |1198861|max le

1205821max holds According to (123) and (124) maximum of |119886

1|

occurs for 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 From (124)

it immediately follows that maximum value of |1198861| is attained

if and only if 1205821= 1205821max and 120575 = 0 which because of

120575119896 = 1205910minus120585119896 further implies 120591

0= 120585119896 Sincemaximumvalue

of 1205821is attained for |120585| = 1205872 it follows that corresponding

waveform has zeros at 120587(2119896) and minus120587(2119896)

Proof of Proposition 22 As it was mentioned earlier in thissection we can assume without loss of generality that 119886

1le 0

We consider waveforms119879119896(120591) of type (35) such that119879

119896(120591) ge 0

and119879119896(120591) = 0 for some 120591

0 Fromassumption that nonnegative

waveform 119879119896(120591) of type (35) has at least one zero it follows

that it can be expressed in form (38)Let us also assume that 120591

0is position of nondegenerate

critical point Therefore 119879119896(1205910) = 0 implies 1198791015840

119896(1205910) = 0 and

11987910158401015840

119896(1205910) gt 0 According to (55) second derivative of 119879

119896(120591) at

1205910can be expressed as 11987910158401015840

119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840

119896(1205910) gt 0 it follows immediately that

1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)

Let us further assume that 119879119896(120591) has exactly one zeroThe

problem of finding maximum absolute value of 1198861is con-

nected to the problem of finding maximum of the minimumfunction (see Section 21) If waveforms possess unique globalminimum at nondegenerate critical point then correspond-ing minimum function is a smooth function of parameters[13] Consequently assumption that 119879

119896(120591) has exactly one

zero at nondegenerate critical point leads to the conclusionthat coefficient 119886

1is differentiable function of 120591

0 First

derivative of 1198861(see (43)) with respect to 120591

0 taking into

account that 1205971205851205971205910= 119896 (see (50)) can be expressed in the

following factorized form

1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)


From (138) and (139) it is clear that 12059711988611205971205910= 0 if and only if

sin 1205910= 0 According toRemark 12 assumption that119879

119896(120591)has

exactly one zero implies 120582119896lt 1 From (51) (48) and 120582

119896lt 1

it follows that 1198861cos 1205910+ 1198871sin 1205910lt 0 which together with

sin 1205910= 0 implies that 119886

1cos 1205910lt 0 Assumption 119886

1le 0

together with relations 1198861cos 1205910lt 0 and sin 120591

0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)

Insertion of 1205910= 0 into (38) leads to

119879119896(120591)

= [1 minus cos 120591]

sdot [1 minus (119896 minus 1) 120582119896cos 120585 minus 2120582

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)

Substitution of 1205910= 0 into (45) and (46) yields 119886

119896= 120582119896cos 120585

and 119887119896

= minus120582119896sin 120585 respectively Replacing 120582

119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)

immediately leads to (121)Furthermore 119886

119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)

According to (38)ndash(40) and (142) it follows that (141) is non-negative if and only if

120582119896[(119896 minus 1) cos 120575 + 119896 sin (120575 minus 120575119896)

sin (120575119896)] lt 1 (143)

Notice that 119886119896= 120582119896cos 120575 implies that the following relation

holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)

Finally substitution of (144) into (143) leads to 119896120582119896[sin 120575

sin(120575119896)] cos(120575119896) lt 1 + 119886119896 which proves that (121) holds

when 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) lt 1 + 119886

119896

Apart from nonnegative waveforms with exactly one zeroat nondegenerate critical point in what follows we will alsoconsider other types of nonnegative waveforms with at leastone zero According to Proposition 9 and Remark 11 thesewaveforms can be described by (66)ndash(68) providing that 0 le|120585| le 120587

According to (35) 119879119896(0) ge 0 implies 1 + 119886

1+ 119886119896ge 0

Consequently 1198861le 0 implies that |119886

1| le 1 + 119886

119896 On the other

hand according to (123) |1198861| = 1 + 119886

119896holds for waveforms

of type (121) The converse is also true 1198861le 0 and |119886

1| =

1 + 119886119896imply 119886

1= minus1 minus 119886

119896 which further from (35) implies

119879119896(0) = 0 Therefore in what follows it is enough to consider

only nonnegativewaveformswhich can be described by (66)ndash(68) and 0 le |120585| le 120587 with coefficients 119886

119896and 119887119896satisfying

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896

For prescribed coefficients 119886119896and 119887119896 the amplitude 120582

119896=

radic1198862119896+ 1198872119896of 119896th harmonic is also prescribed According to

Remark 15 (see also Remark 16) 120582119896is monotonically

decreasing function of 119909 = cos(120585119896) The value of 119909 can beobtained by solving (90) subject to the constraint cos(120587119896) le119909 le 1 Then 120582

1can be determined from (88) From (106) it

immediately follows that maximal absolute value of 1198861le 0

corresponds to 119902 = 0 which from (104) and (120) furtherimplies that

120575 = 1198961205910minus 120585 (145)

Furthermore 119902 = 0 according to (107) implies that waveformzeros are

1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)

Substitution of 1205910= (120575 + 120585)119896 into (66) yields (122) which

proves that (122) holds when 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge

1 + 119886119896

In what follows we prove that (121) also holds when119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886


119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)

As we mentioned earlier relation (142) holds for all wave-forms of type (121) Substituting (142) into (147) we obtain


sin (120585119896)] = 1 (148)

This expression can be rearranged as

120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)

On the other hand for waveforms of type (122) according to(68) relations (148) and (149) also hold Substitution of 120591

0=

(120575 + 120585)119896 (see (145)) and (67) into (122) leads to

119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)

Furthermore substitution of (142) into (145) implies that1205910

= 0 Finally substitution of 1205910

= 0 and (149) into(150) leads to (141) Therefore (141) holds when 119896120582

119896[sin 120575

sin(120575119896)] cos(120575119896) = 1 + 119886119896 which in turn shows that (121)

holds when 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896 This

completes the proof


52 Nonnegative Waveforms with Maximal Absolute Valueof Coefficient 119886

1for 119896 = 3 Nonnegative waveform of type

(35) for 119896 = 3 is widely used in PA design (eg see [10])In this subsection we illustrate results of Section 51 for thisparticular case The case 119896 = 2 is presented in detail in [12]

Coefficients of fundamental harmonic of nonnegativewaveform of type (35) with 119896 = 3 and maximal absolutevalue of coefficient 119886

1le 0 for prescribed coefficients 119886

3and

1198873(1205823= radic11988623+ 11988723) according to (123) (124) (134) (94) and

(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(

3radic1205823minus 1205823) and 120575 = atan 2(119887

3 1198863) if [(1 minus

21198863)3]3

le 1205822

3le 1The line 1205822

3= [(1minus2119886

3)3]3 (see case 119896 = 3

in Figure 11) separates points representing waveforms withcoefficients satisfying (151) from points representing wave-forms with coefficients satisfying (152) Waveforms describedby (151) for 1205822

3lt [(1 minus 2119886

3)3]3 have exactly one zero at

1205910= 0 Waveforms described by (151) and (152) for 1205822

3= [(1 minus

21198863)3]3 also have zero at 120591

0= 0 These waveforms as a rule

have exactly two zeros However there are two exceptionsone related to the maximally flat nonnegative waveform withcoefficients 119886

1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which

has only one zero and the other related to the waveform withcoefficients 119886

1= 0 119886

3= minus1 and 119887

1= 1198873= 0 which has three

zerosWaveforms described by (152) for [(1minus21198863)3]3

lt 1205822

3lt

1 have two zeros Waveforms with 1205823= 1 have only third

harmonic (fundamental harmonic is zero)Plot of contours of maximal absolute value of coefficient

1198861 1198861le 0 for prescribed coefficients 119886

3and 1198873is presented

in Figure 12 According to Remark 25 the waveform withmaximum absolute value of 119886

1le 0 is fully described with

the following coefficients 1198861

= minus2radic3 1198863

= radic39 and1198871= 1198873= 0 This waveform has two zeros at plusmn1205876

Two examples of nonnegative waveforms for 119896 = 3

and maximal absolute value of coefficient 1198861 1198861le 0 with

prescribed coefficients 1198863and 1198873are presented in Figure 13

One waveform corresponds to the case 12058223lt [(1 minus 2119886

3)3]3

(solid line) and the other to the case 12058223gt [(1 minus 2119886

3)3]3

(dashed line)Thewaveform represented by solid line has onezero and its coefficients are 119886

3= minus01 119887

3= 01 119886

1= minus09

and 1198871= minus03 Dashed line corresponds to the waveform

having two zeros with coefficients 1198863= minus01 119887

3= 03 119886

1=

minus08844 and 1198871= minus06460 (case 1205822

3gt [(1 minus 2119886

3)3]3)

6 Nonnegative Cosine Waveforms withat Least One Zero

Nonnegative cosine waveforms have proved to be of impor-tance for waveform modelling in PA design (eg see [10])In this section we consider nonnegative cosine waveforms

1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11

Figure 12 Contours ofmaximal absolute value of coefficient 1198861 1198861le

0 as a function of 1198863and 1198873

2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03

Figure 13 Nonnegative waveforms for 119896 = 3 and maximal absolutevalue of 119886

1 1198861le 0 with prescribed coefficients 119886

3and 1198873

containing fundamental and 119896th harmonic with at least onezero

Cosine waveform with dc component fundamental and119896th harmonic can be obtained from (35) by setting 119887

1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)

In Section 61 we provide general description of non-negative cosine waveforms of type (153) with at least onezero We show that nonnegative cosine waveforms with atleast one zero coincide with nonnegative cosine waveformswith maximal absolute value of coefficient 119886

1for prescribed

coefficient 119886119896 In Section 62 we illustrate results of Section 61

for particular case 119896 = 3

61 Nonnegative Cosine Waveforms with at Least One Zerofor 119896 ge 2 Amplitudes of fundamental and 119896th harmonic


of cosine waveform of type (153) are 1205821= |1198861| and 120582

119896=

|119886119896| respectively According to (42) for nonnegative cosine


minus1 le 119886119896le 1 (154)

This explains why 119896th harmonic coefficient 119886119896in Proposi-

tion 26 goes through interval [minus1 1]Waveforms (153) with 119886

1ge 0 can be obtained from

waveforms with 1198861le 0 by shifting by 120587 and therefore with-

out loss of generality we can assume that 1198861le 0

Proposition 26 Each nonnegative cosine waveform of type(153) with 119886

1le 0 and at least one zero can be represented as

119879119896(120591) = [1 minus cos 120591] [1 minus (119896 minus 1) 119886

119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591

0minus (119896 minus 119899) cos ((119896 minus 119899) 120591

0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1

Remark 27 Identity [1minuscos(120591minus1205910)][1minuscos(120591+120591

0)] = [cos 120591

0minus

cos 120591]2 implies that (156) can be rewritten as

119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)

Remark 28 All nonnegative cosine waveforms of type (153)with at least one zero and 119886

1le 0 except one of them can be

represented either in form (155) or form (156)This exceptionis maximally flat cosine waveform with 119886

1lt 0 which can be

obtained from (155) for 119886119896= 1(119896

2

minus 1) or from (156) for 1205910=

0 Maximally flat cosine waveform with 1198861lt 0 can also be

obtained from (70) by setting 1205910= 0 Furthermore setting

1205910= 0 in (71) leads to maximally flat cosine waveforms for

119896 le 4 and 1198861lt 0

Remark 29 Nonnegative cosine waveform of type (155) with1198861lt 0 and minus1 lt 119886

119896le 1(119896

2

minus 1) has exactly one zero at120591 = 0 Nonnegative cosine waveform described by (156) with1198861lt 0 and 1(1198962 minus 1) lt 119886

119896lt 1 has two zeros at plusmn120591

0 where

0 lt |1205910| lt 120587119896 For 119886

119896= ∓1 nonnegative cosine waveform

of type (153) reduces to 119879119896(120591) = 1 ∓ cos 119896120591 (clearly these two

waveforms both have 119896 zeros)

Remark 30 Transformation of (155) into an additive formleads to the following relation

1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2

minus1) Similarly transformation of (156)leads to the following relation

1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)

where 119886119896is given by (158) 1(1198962minus1) le 119886

119896le 1 and |120591

0| le 120587119896

Notice that coefficients of maximally flat cosine waveformnamely 119886

119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962

minus1) satisfy relation(162) They also satisfy relation (163) for 120591

0= 0

Remark 31 Nonnegative cosine waveforms of type (153) withat least one zero coincide with nonnegative cosine waveformswith maximal absolute value of coefficient 119886

1for prescribed

coefficient 119886119896

In proving that Remark 31 holds notice that expression(155) can be obtained from (121) by setting 119887

119896= 0 Fur-

thermore if 119886119896

ge 0 then 120582119896

= 119886119896 which together

with 119887119896

= 0 and (118) implies 120575 = 0 In this case119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 1198962119886

119896le 1 +

119886119896 On the other hand if 119886

119896lt 0 then 120582

119896= minus119886

119896 which

together with 119887119896= 0 and (118) implies |120575| = 120587 In this case

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896

Therefore every nonnegative cosine waveform of type (155)has maximal absolute value of coefficient 119886

1for prescribed

coefficient 119886119896 when minus1 le 119886

119896le 1(119896

2

minus 1)Let us now show that expression (156) can be obtained

from (122) by setting 119887119896= 0 and 119886

119896gt 0 For waveforms

of type (122) according to (118) 119887119896= 0 and 119886

119896gt 0 imply

120575 = 0 and 120582119896= 119886119896 Substitution of 120582

119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2

minus1)Furthermore substitution of 120575 = 0 into (145) yields 120591

0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads

to (156) Therefore every nonnegative cosine waveform oftype (156) has maximal absolute value of coefficient 119886

1for

prescribed coefficient 119886119896 when 1(1198962 minus 1) le 119886

119896le 1

Proof of Proposition 26 Let us start with nonnegative cosinewaveform of type (153) with 120582

119896= |119886119896| = 1 According to

Remark 7 120582119896= |119886119896| = 1 implies that 120582

1= |1198861| = 0

Substitution of 119886119896

= minus1 into (155) and using (A2) (seeAppendices) lead to 119879

119896(120591) = 1 minus cos 119896120591 Consequently (155)

holds for 119886119896= minus1 On the other hand substitution of 119886

119896= 1

into (158) yields |1205910| = 120587119896 Furthermore substitution of

119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along

with performing all multiplications and using (A2) leads to


119879119896(120591) = 1 + cos 119896120591 Consequently (156)ndash(158) hold for 119886

119896= 1

and |1205910| = 120587119896

It is easy to see that 120582119896= |119886119896| lt 1 and 119879

119896(1205910) = 0 for some

1205910imply 120582

1= |1198861| = 0 Therefore in what follows we assume

that |119886119896| = 1 and 119886

1lt 0

Cosine waveforms are even functions of 120591 Therefore ifnonnegative cosine waveform has exactly one zero it has to beeither at 0 or at 120587 On the other hand if nonnegative cosinewaveform with 119886

1= 0 has exactly two zeros then these zeros

are placed at plusmn1205910 such that 120591

0is neither 0 nor 120587

In order to prove that (155) holds for minus1 lt 119886119896le 1(1minus119896

2

)let us start by referring to the description (38) of nonnegativewaveformswith at least one zero As wementioned earlier fornonnegative cosine waveformwith exactly one zero (denotedby 1205910) it is either 120591

0= 0 or 120591

0= 120587 Therefore in both cases

sin 1205910= 0 Substitution of sin 120591

0= 0 into (43) together with

1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0

also holds which further implies 120582119896= 119886119896= 0 or sin 120585 =

0 In the case when 120582119896= 119886119896= 0 from (164) and (43) we

obtain 1198861= minus1 which further implies that 119879

119896(120591) = 1 minus cos 120591

Consequently (155) holds for 119886119896= 0 In the case when sin 120585 =

0 from (164) and (45) we obtain 119886119896= 120582119896if 120585 = 0 or 119886

119896= minus120582119896

if 120585 = 120587 Relations 119886119896= 120582119896and 120585 = 0 according to (40) imply

that 0 le 119886119896le 1(1 minus 119896

2

) Substitution of 120585 = 0 120582119896= 119886119896 and

(164) into (38) leads to (155) which proves that (155) holdsfor 0 le 119886

119896le 1(1 minus 119896

2

) On the other hand relations 119886119896=

minus120582119896and 120585 = 120587 according to (40) imply that minus1 lt 119886

119896le 0

Substitution of 120585 = 120587 120582119896= minus119886119896 and (164) into (38) also leads

to (155) which proves that (155) also holds for minus1 lt 119886119896le 0

Consequently (155) holds for minus1 lt 119886119896le 1(1 minus 119896

2

)In what follows we first prove that (156)-(157) hold for

1(1 minus 1198962

) lt 119886119896lt 1 For this purpose let us start with non-

negative waveforms with two zeros described by (66) As wementioned before nonnegative cosine waveforms with twozeros have zeros at 120591

0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871

= 0 according to (84) implycos(1205910minus 120585119896) = 1 and therefore

120585

119896= 1205910 (165)

From 120585119896 = 1205910and 0 lt |120585| lt 120587 it follows that 0 lt |120591

0| lt

120587119896 Insertion of 120585119896 = 1205910into (45) yields 119886

119896= 120582119896 Relations

119886119896= 120582119896and (82) imply that 1(1 minus 1198962) lt 119886

119896lt 1 Substitution

of 120582119896= 119886119896and 120585119896 = 120591

0into (66)ndash(68) leads to (156)ndash(158)

which proves that (156)ndash(158) hold for 1(1 minus 1198962

) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896

Finally substitution of 119886119896= 1(1 minus 119896

2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962

) which in turn proves that (156) holds for 119886119896=

1(1 minus 1198962

) and 1205910= 0 This completes the proof

62 Nonnegative CosineWaveforms with at Least One Zero for119896 = 3 In this subsection we consider nonnegative cosinewaveforms with at least one zero for 119896 = 3 (for case 119896 = 2

see [12])Cosine waveform with fundamental and third harmonic

reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)

For 1198861le 0 and minus1 le 119886

3le 18 according to (155) non-

negative cosine waveform of type (167) with at least one zerocan be expressed as

1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879

3(120591) it immediately follows that for

1198861ge 0 and minus18 le 119886

3le 1 119879

3(120591) can be expressed as

1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886

3le 1 from (158) it follows that 119886

3=

[8cos31205910]minus1 This relation along with (160) and (157) further

implies that 1198793(120591) can be expressed as

1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)

providing that |1205910| le 1205873 From 119879

3(120591 + 120587) = 2 minus 119879

3(120591) it

follows that (170) also holds for 1198861ge 0 and minus1 le 119886

3le minus18

providing that 1205910isin [21205873 41205873]

Maximally flat nonnegative cosinewaveformof type (167)with 119886

1lt 0 (minimum at 120591

0= 0) reads 119879

3(120591) = [1 minus

cos 120591]2[1 + (12) cos 120591] Dually maximally flat nonnegativecosine waveform with 119886

1gt 0 (minimum at 120591

0= 120587) reads

1198793(120591) = [1 + cos 120591]2[1 minus (12) cos 120591]In what follows we provide relations between coefficients

1198861and 1198863of nonnegative cosine waveforms of type (167) with

at least one zeroFor 1198861le 0 conversion of (168) into an additive form

immediately leads to the following relation

1198861= minus1 minus 119886

3for minus 1 le 119886

3le1

8 (171)

Conversion of (170) into an additive form leads to 1198861

=

minus31198863(2 cos 2120591

0+ 1) which can be also expressed as 119886

1=

minus31198863(4cos2120591

0minus 1) For 119886

1le 0 relations |120591

0| le 1205873 119886

1=

minus31198863(4cos2120591

0minus 1) and 119886

3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)

Similarly for 1198861ge 0 conversion of (169) into an additive form

leads to the following relation

1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)

Figure 14 Parameter space of cosine waveforms for 119896 = 3

For waveform of type (170) with 1198861ge 0 relations 120591

0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863

1003816100381610038161003816 + 1198863] for minus 1 le 1198863le minus

1

8 (174)

Every cosine waveform of type (167) corresponds to apair of real numbers (119886

1 1198863) and vice versa Points (119886

1 1198863)

in grey area in Figure 14 correspond to nonnegative cosinewaveforms for 119896 = 3 The points at the boundary of grey areacorrespond to nonnegative cosinewaveformswith at least onezero A number of shapes of nonnegative cosine waveformswith 119896 = 3 and at least one zero plotted on interval [minus120587 120587]are also presented in Figure 14 The boundary of grey area inFigure 14 consists of four line segments described by relations(171)ndash(174) The common point of line segments (172) and(173) is cusp point with coordinates 119886

1= 0 and 119886

3= 1

Another cusp point with coordinates 1198861= 0 and 119886

3= minus1

is the common point of line segments (171) and (174) Thecommon point of line segments (171)-(172) has coordinates(minus98 18) and common point of line segments (173)-(174)has coordinates (98 minus18) These points are representedby white circle dots and they correspond to maximallyflat cosine waveforms (eg see [21]) White triangle dotswith coordinates (2radic3 minusradic39) and (minus2radic3radic39) refer tothe nonnegative cosine waveforms with maximum value ofamplitude of fundamental harmonic

7 Four Case Studies of Usage of NonnegativeWaveforms in PA Efficiency Analysis

In this section we provide four case studies of usage ofdescription of nonnegative waveforms with fundamental and119896th harmonic in PA efficiency analysis In first two casestudies to be presented in Section 71 voltage is nonnegativewaveform with fundamental and second harmonic with atleast one zero In remaining two case studies to be consideredin Section 72 voltage waveform contains fundamental andthird harmonic

i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in

Figure 15 Generic PA circuit diagram

Let us consider generic PA circuit diagram as shown inFigure 15 We assume here that voltage and current wave-forms at the transistor output are

V (120579) = 1 + 1198861V cos 120579 + 1198871V sin 120579 + 119886119896V cos 119896120579 + 119887119896V sin 119896120579

119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)

where 120579 stands for 120596119905 Both waveforms are normalized in thesense that dc components of voltage and current are 119881dc =

1 and 119868dc = 1 respectively Under assumption that blockingcapacitor 119862

119887behaves as short-circuit at the fundamental and

higher harmonics current and voltage waveforms at the loadare

V119871(120579) = 119886

1V cos 120579 + 1198871V sin 120579 + 119886119896V cos 119896120579 + 119887119896V sin 119896120579

119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)

In terms of coefficients of voltage and current waveforms theload impedance at fundamental harmonic is 119911

1= minus(119886

1V minus

1198951198871V)1198861119894 whereas load impedance at 119896th harmonic is 119911

119896=

minus(119886119896Vminus119895119887119896V)119886119896119894 All other harmonics are short-circuited (119911

119899=

0 for 119899 = 1 and 119899 = 119896) Time average output power of PA (egsee [10]) with waveform pair (175) at fundamental frequencycan be expressed as

1198751= minus

11988611198941198861V

2 (177)

For normalized waveforms (175) with 119881dc = 1 and 119868dc = 1dc power is 119875dc = 1 Consequently PA efficiency 120578 = 119875

1119875dc

(eg see [10 26]) is equal to

120578 = minus11988611198941198861V

2 (178)

Thus time average output power 1198751of PA with pair of nor-

malized waveform (175) is equal to efficiency (178)Power utilization factor (PUF) is defined [26] as ldquothe

ratio of power delivered in a given situation to the power


delivered by the same device with the same supply voltagein Class A moderdquo Since the output power in class-A modeis 1198751class-A = max[V(120579)] sdotmax[119894(120579)]8 (eg see [9]) it follows

that power utilization factor PUF = 11987511198751class-A for PA with

pair of normalized waveforms (175) can be expressed as

PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)

71 NonnegativeWaveforms for 119896 = 2 in PAEfficiencyAnalysisIn this subsection we provide two case studies of usage ofdescription of nonnegative waveforms with fundamental andsecond harmonic (119896 = 2) in PA efficiency analysis For moreexamples of usage of descriptions of nonnegative waveformswith fundamental and second harmonic in PA efficiencyanalysis see [12]

Case Study 71 In this case study we consider efficiency of PAfor given second harmonic impedance providing that voltageis nonnegative waveform with fundamental and second har-monic and current is ldquohalf-sinerdquo waveform frequently used inefficiency analysis of classical PA operation (eg see [10])

Standard model of current waveform for classical PAoperation has the form (eg see [10 26])

119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)

where 120572 is conduction angle and 119868119863

gt 0 Since 119894119863(120579) is

even function it immediately follows that its Fourier seriescontains only dc component and cosine terms

119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)

The dc component of the waveform (180) is

119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)

where sinc119909 = (sin119909)119909 The coefficient of the fundamentalharmonic component reads

1198681=119868119863120572

2120587(1 minus sinc120572) (183)

and the coefficient of 119899th harmonic component can bewrittenin the form

119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)

For ldquohalf-sinerdquo current waveform conduction angle is equalto 120587 (class-B conduction angle) According to (182) thisfurther implies that 119868dc = 119868

119863120587 To obtain normalized form of

waveform (180) we set 119868dc = 1which implies that 119868119863= 120587 Fur-

thermore substitution of 120572 = 120587 and 119868119863= 120587 in (180) leads to

119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)

Similarly substitution of 119868119863= 120587 and 120572 = 120587 into (183) and

(184) leads to the coefficients of waveform (185) Coefficientsof fundamental and second harmonic respectively are

1198861119894=120587

2 119886

2119894=2

3 (186)

On the other hand voltage waveform of type (35) for 119896 =2 reads


(187)

This waveform contains only fundamental and second har-monic and therefore all harmonics of order higher thantwo are short-circuited (119911

119899= 0 for 119899 gt 2) For current

voltage pair (185) and (187) load impedance at fundamentalharmonic is 119911

1= minus(119886

1V minus 1198951198871V)1198861119894 whereas load impedance

at second harmonic is 1199112= minus(119886

2Vminus1198951198872V)1198862119894 According to ourassumption the load is passive and therefore Re119911

1 gt 0 and

Re1199112 ge 0 which further imply 119886

11198941198861V lt 0 and 119886

21198941198862V le 0

respectivelyIt is easy to see that problem of findingmaximal efficiency

of PA with current-voltage pair (185) and (187) for prescribedsecond harmonic impedance can be reduced to the problemof finding voltage waveform of type (187) with maximal coef-ficient |119886

1V| for prescribed coefficients of second harmonic(see Section 5)

The following algorithm (analogous to Algorithm 22presented in [12]) provides the procedure for calculation ofmaximal efficiency with current-voltage pair (185) and (187)for prescribed second harmonic impedance The definitionof function atan 2(119910 119909) which appears in the step (iii) of thefollowing algorithm is given by (105)

Algorithm 32 (i) Choose 1199112= 1199032+1198951199092such that |119911

2| le 1|119886

2119894|

(ii) calculate 1198862V minus 1198951198872V = minus119911

21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582

2V le 1 minus 1198862V then calculate 119886

1V = minus1 minus 1198862V and

1198871V = minus2119887

2V else calculate 1205821V = radic81205822V(1 minus 1205822V) 1205790V minus 120585V2 =

(12)atan2(1198872V 1198862V) 1198861V = minus120582

1V cos(1205790V minus 120585V2) and 1198871V =

minus1205821V sin(1205790V minus 120585V2)(iv) calculate efficiency 120578 = minus119886

11198941198861V2

(v) calculate 1199111= minus(119886

1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1

In this case study coefficients of fundamental and secondharmonic of current waveform are given by (186) Maximalefficiency of PA associated with the waveform pair (185)and (187) as a function of normalized second harmonicimpedance 119911

2119899= 1199112Re119911

1 is presented in Figure 16(a)

As can be seen from Figure 16(a) efficiency of 078 isachieved at the edge of Smith chart where second harmonicimpedance has small resistive part Corresponding PUFcalculated according to (179) is presented in Figure 16(b)Peak efficiency 120578 = 1205874 = 07854 and peak value of PUF = 1

are attained when second harmonic is short-circuited (whichcorresponds to ideal class-B operation [10 26])

For example for second harmonic impedance 1199112= 01 minus

11989505 and currentwaveform (185) fromAlgorithm32 it followsthat 2120582

2V le 1 minus 1198862V Furthermore according to step (iii)

of above algorithm maximal efficiency of PA is attained


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)

Figure 16 (a) Contours of maximal efficiency of PA and (b) contours of corresponding PUF associated with the waveform pair (185) and(187) as functions of normalized second harmonic impedance 119911

2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)

Figure 17 Waveform pair (185) and (187) that provides maximalefficiency for 119911

2= 01 minus 11989505

with voltage waveform of type (187) with coefficients 1198862V =

minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see

Figure 17) Corresponding efficiency PUF and normalizedsecond harmonic impedance are 120578 = 07330 PUF = 07572and 1199112119899= 01683 minus 11989508415 respectively

On the other hand for second harmonic impedance 1199112=

01 minus 11989508 and current waveform (185) from Algorithm 32 itfollows that 2120582

2V gt 1 minus 1198862V Then according to step (iii) of

above algorithm maximal efficiency is attained with voltagewaveform of type (187) with coefficients 119886

2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887

1V = 10572 (see Figure 18)Efficiency PUF andnormalized secondharmonic impedanceare 120578 = 07330 PUF = 06332 and 119911

2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508

Case Study 72 As another case study let us consider the effi-ciency of PA providing that current waveform is nonnegativecosine waveform up to third harmonic with maximum valueof amplitude of fundamental harmonic [22] (see also [8])

119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5

5cos 2120579 + 5 minus radic5

10cos 3120579

(188)

and voltage waveform is nonnegative waveform of type(187) Load impedances at fundamental second and thirdharmonic are 119911

1= minus(119886

1Vminus1198951198871V)1198861119894 1199112 = minus(1198862Vminus1198951198872V)1198862119894 and

1199113= 0 respectively According to our assumption the load

is passive and therefore Re1199111 gt 0 and Re119911

2 ge 0 which

further imply 11988611198941198861V lt 0 and 119886

21198941198862V le 0 respectively

Because current waveform (188) contains only cosineterms and voltage waveform is the same as in previous case


08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)

Figure 19 (a) Contours of maximal efficiency of PA and (b) contours of corresponding PUF associated with the waveform pair (187)-(188)as functions of normalized second harmonic impedance 119911

2119899= 1199112Re119911

1

study the procedure for calculation of maximal efficiency ofPA with waveform pair (187)-(188) is the same as presentedin Algorithm 32 In this case study the coefficients of funda-mental and second harmonic of current waveform are 119886

1119894=

(1 + radic5)2 and 1198862119894= 2radic55 respectively

Maximal efficiency of PA associated with the waveformpair (187)-(188) as a function of normalized secondharmonicimpedance 119911

2119899= 1199112Re119911


Efficiency of 08 is achieved at the edge of Smith chart wheresecond harmonic impedance has small resistive part Thetheoretical upper bound 120578 = (1 + radic5)4 asymp 08090 isattained when second harmonic is short-circuitedWhen thisupper bound is reached both second and third harmonic areshort-circuited which implies that we are dealing with finiteharmonic class-C [6 8] or dually when current and voltageinterchange their roles with finite harmonic inverse class-C[6 9] Corresponding PUF calculated according to (179) ispresented in Figure 19(b) Peak value of PUF asymp 08541 isattained when second harmonic is short-circuited

For example for second harmonic impedance 1199112

=


2V le 1 minus 1198862V Furthermore according to step

(iii) of Algorithm 32 maximal efficiency of PA is attainedwith voltage waveform of type (187) with coefficients 119886

2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see


On the other hand for 1199112= 005 minus 11989507 and current

waveform (187) it follows that 21205822V gt 1minus119886

2VThen accordingto step (iii) of Algorithm 32 themaximal efficiency is attainedwith voltage waveform of type (187) with coefficients 119886

2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see

Figure 21) Efficiency PUF and normalized second harmonic

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)

Figure 20 Waveform pair (187)-(188) that provides maximal effi-ciency for 119911

2= 007 minus 11989504

impedance are 120578 = 07538 PUF = 05314 and 1199112119899= 00868minus

11989512156 respectively

72 Nonnegative Waveforms for 119896 = 3 in PA EfficiencyAnalysis In this subsection we provide another two casestudies of usage of description of nonnegative waveforms inPA efficiency analysis this time with fundamental and thirdharmonic (119896 = 3)

Case Study 73 Let us consider current-voltage pair such thatvoltage is nonnegative waveform with fundamental and thirdharmonic


(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507

and current is nonnegative cosine waveform given by (188)Load impedances at fundamental second and third har-monic are 119911

1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus

1198951198873V)1198863119894 respectively According to our assumption the load


3 ge 0 which


31198941198863V le 0

In this subsection we consider the problem of findingmaximal efficiency of PA with waveform pair (188)-(189)for given third harmonic impedance As we mentionedearlier problem of finding maximal efficiency of PA withcurrent-voltage pair (188)-(189) for prescribed third har-monic impedance can be reduced to the problem of findingvoltage waveform of type (189) withmaximal coefficient |119886

1V|for prescribed coefficients of third harmonic (see Section 52)

The following algorithm provides the procedure forcalculation of maximal efficiency with current-voltage pair(188)-(189) The definition of function atan 2(119910 119909) whichappears in step (iii) of the following algorithm is given by(105)


3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822

3V le (1 minus 21198863V)3 then calculate 119886

1V = minus1 minus 1198863V

and 1198871V = minus3119887

3V else calculate 1205821V = 3(3radic1205823V minus 120582

3V) 1205790V minus120585V3 = (13)atan 2(119887

3V 1198863V) 1198861V = minus1205821V cos(1205790V minus 120585V3) and

1198871V = minus120582

1V sin(1205790V minus 120585V3)(iv) calculate efficiency 120578 = minus119886

11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1

In this case study coefficients of fundamental and thirdharmonic of current waveform are 119886

1119894= (1 + radic5)2 and

1198863119894= (5 minus radic5)10 respectively For the waveform pair (188)-

(189) maximal efficiency of PA as a function of normalizedthird harmonic impedance 119911

3119899= 1199113Re119911

1 is presented in

Figure 22 Efficiency of 08 is reached when third harmonicimpedance has small resistive part Peak efficiency 120578 = (1 +

radic5)4 asymp 08090 is achieved when third harmonic is short-circuited

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2

Figure 22 Contours of maximal efficiency of PA associated withthe waveform pair (188)-(189) as a function of normalized thirdharmonic impedance 119911

3119899= 1199113Re119911

1

For the present case study in what follows we show thatpower utilization factor is proportional to efficiency For volt-age waveform of type (189) it is easy to see that V(120579 + 120587) =

2minusV(120579) holdsThis relation along with the fact that waveformV(120579) that provides maximal efficiency has at least one zeroimplies that max[V(120579)] = 2 On the other hand currentwaveform (188) is cosine waveform with positive coefficientsand therefore max[119894(120579)] = 119894(0) = 2 + 4radic5 Consequentlyaccording to (179) the following relation holds

PUFCase study 73 = 2 (5 minus 2radic5) 120578 = 10557120578 (190)

Clearly the ratio PUF120578 is constant and therefore in this casestudy PUF can be easily calculated from the correspondingefficiency Accordingly peak efficiency and peak value ofPUFCase study 73 = 3radic52 minus 52 = 08541 are attained forthe same voltage waveform (when third harmonic is short-circuited)

In the first example current waveform (188) and 1199113=

02 minus 11989505 imply that 2712058223V le (1 minus 2119886

3V)3 Then according

to Algorithm 33 the voltage waveform of type (189) thatprovides maximal efficiency has the following coefficients1198863V = minus00553 119887

3V = minus01382 1198861V = minus09447 and 119887

1V =

04146 (see Figure 23) Efficiency PUF and normalized thirdharmonic impedance are 120578 = 07643 PUF = 08069 and1199113119899= 03425 minus 11989508564 respectivelyIn the second example current waveform (188) and 119911

3=

01 minus 11989511 imply that 2712058223V gt (1 minus 2119886

3V)3 Then according



1V =

05807 (see Figure 24) Efficiency PUF and normalized thirdharmonic impedance are 120578 = 07598 PUF = 08021 and1199113119899= 01723 minus 11989518952 respectively


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511

Case Study 74 In this case study let us consider current-voltage pair where current is normalized waveform of type(180) with conduction angle 120572 = 115120587 (207∘) and voltageis nonnegative waveform of type (189) Substitution of 120572 =

115120587 and 119868dc = 1 into (182) leads to 119868119863

= 22535Furthermore substitution of 120572 = 115120587 and 119868

119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)

Similarly substitution of 120572 = 115120587 and 119868119863

= 22535 into(183) and (184) for 119899 = 3 yields coefficients of fundamentaland third harmonic of waveform (191)

1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074

Figure 25 Contours of maximal efficiency of PA associated withthe waveform pair (189) and (191) as a function of normalized thirdharmonic impedance 119911

3119899= 1199113Re119911

1

study the procedure for calculation of maximal efficiencyof PA with waveform pair (189)ndash(191) is the same as thatpresented in Algorithm 33 In this case study the coefficientsof fundamental and third harmonic of current waveform aregiven by (192)

For the waveform pair (189) and (191) maximal efficiencyof PA as a function of normalized third harmonic impedance1199113119899= 1199113Re119911

1 is presented in Figure 25 Efficiency of 084

is obtained in vicinity of 1199113119899= 23685 (corresponding to 119911

3=

18750) Peak efficiency 120578 asymp 08421 is achieved for voltagewaveform of type (189) with coefficients 119886

1V = minus2radic3 1198863V =

radic39 and 1198871V = 1198873V = 0

In the course of finding power utilization factor noticethat currentwaveformof type (191) attains itsmaximumvaluefor 120579 = 0 Insertion of max[119894(120579)] = 119894(0) = 278 andmax[V(120579)] = 2 for voltage waveform of type (189) into (179)leads to

PUFCase study 74 = 1439120578 (193)

Again the ratio PUF120578 is constant and PUF can be easilycalculated from the corresponding efficiency Accordinglypeak value of PUFCase study 74 asymp 12118 and peak efficiencyare attained for the same voltage waveform



3V)3 Then according to

Algorithm 33 voltage waveform of type (189) which providesmaximal efficiency has coefficients 119886

3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887

1V = minus00616 (see Figure 26)Efficiency PUF and normalized third harmonic impedanceare 120578 = 08042 PUF = 11572 and 119911

3119899= 13228 minus 11989502646

respectivelyIn second example current waveform (191) and 119911

3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion

In this paper we consider a problem of finding general de-scriptions of various classes of nonnegative waveforms withfundamental and 119896th harmonic These classes include non-negative waveforms with at least one zero nonnegative wave-forms with maximal amplitude of fundamental harmonic forprescribed amplitude of 119896th harmonic nonnegative wave-forms withmaximal coefficient of cosine part of fundamentalharmonic for prescribed coefficients of 119896th harmonic andnonnegative cosine waveforms with at least one zero Mainresults are stated in six propositions (Propositions 1 6 918 22 and 26) four corollaries (Corollaries 2ndash5) twentyremarks and three algorithms Four case studies of usageof closed form descriptions of nonnegative waveforms in PAefficiency analysis are considered in detail in Section 7

Appendices

Here we provide a list of finite sums of trigonometric func-tions used in this paper (Appendix A) and brief account ofthe Chebyshev polynomials (Appendix B)

A List of Some Finite Sums ofTrigonometric Functions

Dirichlet kernel (eg see [27]) is as follows

119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)

Fejer kernel (eg see [27]) can be expressed in the fol-lowing equivalent forms

119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591

=(1 minus cos 119896120591)119896 (1 minus cos 120591)

(A2)

Lagrangersquos trigonometric identity (eg see [28]) is as fol-lows

1198781(120591) =

119896minus1

sum

119899=1

sin 119899120591 = sin (1198961205912) sin ((119896 minus 1) 1205912)sin (1205912)

(A3)

In what follows we show that the following three trigono-metric identities also hold

2

119896minus1

sum

119899=1

(119896 minus 119899) sin 119899120591 = 119896 sin 120591 minus sin 1198961205911 minus cos 120591

(A4)

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591 = sin (119896 minus 1) 120591sin 120591

(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591

=sin (119896120591) cos 120591 minus 119896 cos (119896120591) sin 120591

2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which

immediately leads to (A4)


Identity (A5) can be obtained as follows

sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591

minus 119890minus119895(119896minus1)120591

119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962

minus (119896 minus 2119899)2 it follows that 4119878

4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591

2 which leads to (A6)

B The Chebyshev Polynomials

The Chebyshev polynomials of the first kind 119881119899(119909) can be

defined by the following relation (eg see [29])

119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)

The Chebyshev polynomials of the second kind 119880119899(119909) can be


119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)

The Chebyshev polynomials satisfy the following recur-rence relations (eg see [29])

1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)

The first few Chebyshev polynomials of the first and secondkind are 119881

2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This work is supported by the SerbianMinistry of EducationScience and Technology Development as a part of ProjectTP32016

References

[1] V I Arnolrsquod V S Afrajmovich Y S Ilrsquoyashenko and L PShilrsquonikov Dynamical Systems V Bifurcation Theory and Catas-trophe Theory Springer Berlin Germany 1994

[2] E Polak ldquoOn the mathematical foundations of nondifferen-tiable optimization in engineering designrdquo SIAM Review vol29 no 1 pp 21ndash89 1987

[3] N S Fuzik ldquoBiharmonic modes of a tuned RF power amplifierrdquoRadiotehnika vol 25 no 7 pp 62ndash71 1970 (Russian)

[4] P Colantonio F Giannini G Leuzzi and E Limiti ldquoClass Gapproach for low-voltage high-efficiency PA designrdquo Interna-tional Journal of RF and Microwave Computer-Aided Engineer-ing vol 10 no 6 pp 366ndash378 2000

[5] F H Raab ldquoMaximum efficiency and output of class-F poweramplifiersrdquo IEEE Transactions on Microwave Theory and Tech-niques vol 49 no 6 pp 1162ndash1166 2001

[6] FH Raab ldquoClass-E class-C and class-F power amplifiers basedupon a finite number of harmonicsrdquo IEEE Transactions onMicrowaveTheory and Techniques vol 49 no 8 pp 1462ndash14682001

[7] J D Rhodes ldquoOutput universality inmaximum efficiency linearpower amplifiersrdquo International Journal of Circuit Theory andApplications vol 31 no 4 pp 385ndash405 2003

[8] A Juhas and L A Novak ldquoComments on lsquoClass-E class-Cand classF power amplifier based upon a finite number ofharmonicsrsquordquo IEEE Transactions on Microwave Theory and Tech-niques vol 57 no 6 pp 1623ndash1625 2009

[9] M Roberg and Z Popovic ldquoAnalysis of high-efficiency poweramplifiers with arbitrary output harmonic terminationsrdquo IEEETransactions on Microwave Theory and Techniques vol 59 no8 pp 2037ndash2048 2011

[10] A Grebennikov N O Sokal and M J Franco Switchmode RFPower Amplifiers ElsevierAcademic Press San Diego CalifUSA 2nd edition 2012

[11] T Canning P J Tasker and S C Cripps ldquoContinuous modepower amplifier design using harmonic clipping contourstheory and practicerdquo IEEE Transactions on Microwave Theoryand Techniques vol 62 no 1 pp 100ndash110 2014

[12] A Juhas and L A Novak ldquoGeneral description of nonnegativewaveforms up to second harmonic for power amplifier mod-ellingrdquoMathematical Problems in Engineering vol 2014 ArticleID 709762 18 pages 2014

[13] V I Arnolrsquod V V Goryunov O V Lyashko and V A VasilrsquoevDynamical Systems VIIImdashSingularity Theory II ApplicationsSpringer Berlin Germany 1993

[14] D Siersma ldquoProperties of conflict sets in the planerdquo BanachCenter Publications Polish Academy of Sciences vol 50 no 1 pp267ndash276 1999 Proceedings of the Banach Center Symposiumon Geometry and Topology of Caustics (Caustics rsquo98) WarsawPoland

[15] M van Manen The geometry of conflict sets [Dissertation]Universiteit Utrecht UtrechtThe Netherlands 2003 httpigi-tur-archivelibraryuunldissertations2003-0912-123058c4pdf

[16] Y L Sachkov ldquoMaxwell strata and symmetries in the problemofoptimal rolling of a sphere over a planerdquo Sbornik Mathematicsvol 201 no 7-8 pp 1029ndash1051 2010

[17] I A Bogaevsky ldquoPerestroikas of shocks and singularities ofminimum functionsrdquoPhysicaDNonlinear Phenomena vol 173no 1-2 pp 1ndash28 2002

[18] Y L Sachkov ldquoMaxwell strata in the Euler elastic problemrdquoJournal of Dynamical andControl Systems vol 14 no 2 pp 169ndash234 2008

[19] M Siino and T Koike ldquoTopological classification of black holesgeneric Maxwell set and crease set of a horizonrdquo InternationalJournal ofModern Physics D Gravitation Astrophysics Cosmol-ogy vol 20 no 6 pp 1095ndash1122 2011


[20] F H Raab ldquoClass-F power amplifiers with maximally flat wave-formsrdquo IEEETransactions onMicrowaveTheory and Techniquesvol 45 no 11 pp 2007ndash2012 1997

[21] A Juhas and L A Novak ldquoMaximally flat waveforms with finitenumber of harmonics in class-F power amplifiersrdquo Mathemat-ical Problems in Engineering vol 2013 Article ID 169590 9pages 2013

[22] L Fejer ldquoUber trigonometrische polynomerdquo Journal fur dieReine und Angewandte Mathematik vol 1916 no 146 pp 53ndash82 1916 (German)

[23] S C Cripps ldquoBessel Waives [microwave bytes]rdquo IEEE Micro-wave Magazine vol 10 no 7 pp 30ndash36 117 2009

[24] L N Bryzgalova ldquoSingularities of the maximum of para-metrically dependent functionrdquo Functional Analysis and ItsApplications vol 11 no 1 pp 49ndash51 1977

[25] V I Arnold A A Davydov V A Vassiliev and V MZakalyukin Mathematical Models of Catastrophes Control ofCatastrophic Process Encyclopedia of Life Support Systems(EOLSS) EOLSS Publishers Oxford UK 2006

[26] S C Cripps RF Power Amplifiers for Wireless CommunicationsArtech House Norwood Mass USA 2nd edition 2006

[27] A Zygmund Trigonometric Series vol 1 Cambridge UniversityPress Cambridge UK 2nd edition 1959

[28] A Jeffrey and H DaiHandbook of Mathematical Formulas andIntegrals ElsevierAcademic Press San Diego Calif USA 4thedition 2008

[29] J C Mason and D C Handscomb Chebyshev PolynomialsChapman amp Hall CRC Press Boca Raton Fla USA 2003

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


drain (collectorplate) waveforms in PA design (eg see [3ndash12 20 21])

Fejer in his seminal paper [22] provided general descrip-tion of all nonnegative trigonometric polynomials with 119899

consecutive harmonics in terms of 2119899 + 2 parameters satis-fying one nonlinear constraint He also derived closed formsolution to the problem of finding maximum possible ampli-tude of the first harmonic of nonnegative cosine polynomialswith consecutive harmonics

Fuzik [3] (see also [10]) considered cosine polynomialswith dc fundamental and 119896th harmonic for arbitrary 119896 ge 2

and provided closed form solution for coefficients of optimalwaveform Rhodes in [7] provided closed form expressionfor maximum possible amplitude of fundamental harmonicof nonnegative waveforms containing consecutive odd har-monics A subclass of nonnegative cosine waveforms withdc fundamental and third harmonic having factorized formdescription has been considered in [23]

High efficiency PA with arbitrary output harmonic ter-minations has been analysed in [9] along with maximal effi-ciency fundamental output power and load impedance

Factorized form of nonnegative waveforms up to secondharmonic with at least one zero has been suggested in [11] inthe context of continuous class BJ mode of PA operation

General description of all nonnegative waveforms up tosecond harmonic in terms of four independent parametershas been provided in [12] This includes nonnegative wave-forms with at least one zero as a special case

End point of conflict set normally corresponds to so-called maximally flat waveform which also belongs to classof suboptimal waveforms First comprehensive usage of max-imally flat waveforms in the context of analysis of PA goesto Raab [20] General description of maximally flat wave-forms with arbitrary number of harmonics has been pre-sented in [21] along with closed form expressions for effi-ciency of class-F and inverse class-F PA with maximally flatwaveforms Description of maximally flat cosine waveformswith consecutive harmonics has been presented in [8] in thecontext of finite harmonic class-C PA

In this paper we provide general descriptions of a num-ber of optimal and suboptimal nonnegative waveforms con-taining dc component fundamental and an arbitrary 119896thharmonic 119896 ge 2 and show how they are related to theconcept of conflict set According to our best knowledge thispaper provides the very first usage of conflict set in the courseof solving problems related to optimization of PA efficiencyMain results are stated in six propositions (Propositions 1 69 18 22 and 26) four corollaries (Corollaries 2ndash5) twentyremarks and three algorithms Four case studies of usage ofclosed form descriptions of nonnegative waveforms in PAefficiency analysis are considered in detail in Section 7

This paper is organized in the following way In Section 2we introduce concepts of minimum function and gain func-tion (Section 21) conflict set (Section 22) and parameterspace (Section 23) In Sections 3ndash6 we provide generaldescriptions of various classes of nonnegative waveformscontaining dc component fundamental and 119896th harmonicwith at least one zero General case of nonnegative waveformswith at least one zero is presented in Section 31The case with

exactly two zeros is considered in Section 32 An algorithmfor calculation of coefficients of fundamental harmonic ofnonnegative waveforms with two zeros for prescribed coeffi-cients of 119896th harmonic is presented in Section 33 Descrip-tion of nonnegative waveforms with maximal amplitude offundamental harmonic for prescribed amplitude of 119896th har-monic is provided in Section 4 Nonnegative waveforms withmaximal coefficient of cosine part of fundamental harmonicfor prescribed coefficients of 119896th harmonic are consideredin Section 51 An illustration of results of Section 51 forparticular case 119896 = 3 is given in Section 52 Section 61 isdevoted to nonnegative cosine waveforms with at least onezero and arbitrary 119896 ge 2 whereas Section 62 considerscosine waveforms with at least one zero for 119896 = 3 InSection 7 four case studies of application of descriptions ofnonnegative waveforms with fundamental and 119896th harmonicin PA modelling are presented In the Appendices list ofsome finite sums of trigonometric functions widely usedthroughout the paper and brief account of the Chebyshevpolynomials are provided

2 Minimum Function Gain Function andConflict Set

In this sectionwe considerminimum function and gain func-tion (Section 21) conflict set (Section 22) and parameterspace (Section 23) in the context of nonnegative waveformswith fundamental and 119896th harmonic

We start with provision of a brief account of the factsrelated to the concepts of minimum function and conflict setFor this purpose let us denote by 119891(119909 119906) a family of smoothfunctions of 119899 variables depending on 119898 parameters where119909 isin 119877

119899 is 119899-tuple of variables and 119906 isin 119877119898 is 119898-tuple of

parameters The minimum function 119865 119877119898

rarr 119877 associatedwith the function 119891 is defined as 119865(119906) = min

119909119891(119909 119906)

Therefore the domain of theminimum function is parameterspace of the function 119891 The minimum function 119865(119906) is con-tinuous but not necessarily smooth function of parameters[13 24] It is a smooth function if 119891(119909 119906) possesses uniqueglobal minimum at nondegenerate critical point [13] (criticalpoint is degenerate if at least first two consecutive derivativesare equal to zero) In this context the conflict set can bedefined as the set of the parameters for which function 119891

has global minimum at a degenerate critical point orandmultiple global minima [13]

For a wide class ofminimum functions when the numberof parameters is not greater than four the behaviour ofminimum function in a neighbourhood of any point can bedescribed by one of ldquonormal formsrdquo from a finite list as statedin [24] For example for smooth function 119891 119877 times 119877

2

rarr 119877the minimum function 119865(119906

1 1199062) = min

119909119891(119909 119906

1 1199062) near the

origin can be locally reduced to one of the following threenormal forms [25] minus|119906

1|min(119906

1 1199062 1199061+ 1199062) or min

119909(1199094

+

11990611199092

+ 1199062119909) In this example the conflict set is the set of all

points (1199061 1199062) for which minimum function 119865(119906

1 1199062) is not

differentiable because function 119891(119909 1199061 1199062) possesses at least

two global minima [25]



119908 (120591 120574 119860 120572) = 1 minus 120574 (cos 120591 + 119860 cos (119896120591 + 120572)) (1)






1 + 120574119891 (120591 119860 120572) ge 0 (3)


Let

119865min (119860 120572) = min120591

119891 (120591 119860 120572) (4)


119866 (119860 120572) = minus1

119865min (119860 120572) (5)



120574max = max119860120572

119866 (119860 120572) = minus1

max119860120572

119865min (119860 120572)(6)

14

12

1

08

06

04

G1 G2

001

02

0304

0506 0

1205872

12058731205872

2120587

Parameter 120572

Parameter A

Gai

n fu

nctio

n

k = 2


2denote



120574max = max119860120572

119866 (119860 120572) = 119866 (119860lowast

120572lowast

)

max119860120572

119865min (119860 120572) = 119865min (119860lowast

120572lowast

)

(7)




119908lowast

(120591) = 119908 (120591 120574max 119860lowast

120572lowast

) (8)


120572lowast





1)







001

02

0304

05

06 01205872

12058731205872

2120587

Parameter 120572

Parameter A

1205872

1205874

0

minus1205874

minus1205872Posit

ions

of g

loba

l min

k = 2



119908119886(120591 120574119886 119887 119860 120572) = 1 minus 120574

119886(cos 120591 + 119887 sin 120591 + 119860 cos (119896120591 + 120572))

(9)






min120591119891119886(120591 119887 119860 120572) Inequality 119908

119886(120591 120574119886 119887 119860 120572) ge 0 can be



119886= minus1119865


119886= minus1119865


119887 119860 120572) = 119865119886min(119887 119860 120572)







119891 (1205911015840

119860 120572) = 119891 (12059110158401015840

119860 120572) (11)

1198911015840

(1205911015840

119860 120572) = 0 1198911015840

(12059110158401015840

119860 120572) = 0 (12)

11989110158401015840

(1205911015840

119860 120572) gt 0 11989110158401015840

(12059110158401015840

119860 120572) gt 0 (13)

(forall120591) 119891 (120591 119860 120572) ge 119891 (1205911015840

119860 120572) (14)




2 and 120572 = 120587



(11198962



Δ where

0 lt 120591Δle 120587119896



as follows

120572 = 120587 119860 (120591Δ) =

sin 120591Δ

119896 sin 119896120591Δ

0 lt 120591Δle120587

119896 (15)


val 0 lt 120591Δle 120587119896






119891 (1205911015840

119860 120572) minus 119891 (12059110158401015840

119860 120572) = 0

1198911015840

(1205911015840

119860 120572) + 1198911015840

(12059110158401015840

119860 120572) = 0

1198911015840

(1205911015840

119860 120572) minus 1198911015840

(12059110158401015840

119860 120572) = 0

11989110158401015840

(1205911015840

119860 120572) + 11989110158401015840

(12059110158401015840

119860 120572) gt 0

(16)

also hold Let

120591119904119903=(1205911015840

+ 12059110158401015840

)

2 120591

Δ=(12059110158401015840

minus 1205911015840

)

2

(17)


minus120587 lt 120591119904119903lt 120587 (18)

0 lt 120591Δlt 120587 (19)

12059110158401015840

= 120591119904119903+ 120591Δ 120591

1015840

= 120591119904119903minus 120591Δ (20)


1198911015840

(120591 119860 120572) = sin 120591 + 119896119860 sin (119896120591 + 120572)

11989110158401015840

(120591 119860 120572) = cos 120591 + 1198962119860 cos (119896120591 + 120572) (21)


sin 120591119904119903sin 120591Δ+ 119860 sin (119896120591

119904119903+ 120572) sin 119896120591

Δ= 0 (22)

sin 120591119904119903cos 120591Δ+ 119896119860 sin (119896120591

119904119903+ 120572) cos 119896120591

Δ= 0 (23)

cos 120591119904119903sin 120591Δ+ 119896119860 cos (119896120591

119904119903+ 120572) sin 119896120591

Δ= 0 (24)

cos 120591119904119903cos 120591Δ+ 1198962

119860 cos (119896120591119904119903+ 120572) cos 119896120591

Δgt 0 (25)





+

120572)[119896 sin 120591Δcos 119896120591

Δminus sin 119896120591



119896 sin 120591Δcos 119896120591

Δminus sin 119896120591

Δcos 120591Δ

= 0 (26)


119904119903and 119860 sin(119896120591

119904119903+ 120572)


sin 120591119904119903= 0 sin (119896120591

119904119903+ 120572) = 0 (27)


120591119904119903= 0 (28)


12059110158401015840

= 120591Δ 120591

1015840

= minus120591Δ (29)


119904119903+ 120572) = 0 imply that sin120572 =





Δ) le 0 (30)

From 120591Δ



120572 = 120587 (31)



Δ) le



0 lt 120591Δle120587

119896 (32)


119860 =sin 120591Δ

119896 sin 119896120591Δ

(33)




120591Δrarr0+119860 =

11198962 Consequently 119860 gt 1119896




1198911015840

(120591119889 119860 120572) = 0

11989110158401015840

(120591119889 119860 120572) = 0

(34)






00

00

k = 2 k = 3

k = 4 k = 5

Acos 120572

Acos 120572

Acos 120572

Acos 120572

A sin 120572

A sin 120572A sin 120572

A sin 120572









119879119896(120591) = 1 + 119886

1cos 120591 + 119887

1sin 120591 + 119886

119896cos 119896120591 + 119887

119896sin 119896120591 (35)


1205821= radic11988621+ 11988721 (36)

120582119896= radic1198862119896+ 1198872119896 (37)





119879119896(120591) = [1 minus cos (120591 minus 120591

0)] [1 minus 120582

119896119903119896(120591)] (38)

where119903119896(120591) = (119896 minus 1) cos 120585

+ 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(39)

providing that


sin(120585119896)]

minus1

(40)

10038161003816100381610038161205851003816100381610038161003816 le 120587 (41)


0 le 120582119896le 1 (42)



of 120582119896






1198861= minus (1 + 120582

119896cos 120585) cos 120591

0minus 119896120582119896sin 120585 sin 120591

0 (43)

1198871= minus (1 + 120582

119896cos 120585) sin 120591

0+ 119896120582119896sin 120585 cos 120591

0 (44)

119886119896= 120582119896cos (119896120591

0minus 120585) (45)

119887119896= 120582119896sin (119896120591

0minus 120585) (46)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms






3= radic220


1= minus03451 119886

3= 005


3= radic28


1= minus01127 119886

3= 0125


3= radic24


1= 02745 119886

3= 025 and




119879119896(120591) = 1 + 120582

1cos (120591 + 120593

1) + 120582119896cos (119896120591 + 120593

119896) (47)

where1205821ge 0120582

119896ge 0120593

1isin (minus120587 120587] and120593



1198861= 1205821cos1205931 119887

1= minus1205821sin1205931 (48)

119886119896= 120582119896cos120593119896 119887

119896= minus120582119896sin120593119896 (49)


0+ 120593119896) mod2120587 (50)




119896(1205910) = 0


119896(120591) ge 0 and 119879

119896(1205910) = 0

imply that 1198791015840119896(1205910) = 0 From 119879

119896(1205910) = 0 and 1198791015840

119896(1205910) = 0 by


1205821cos (120591

0+ 1205931) = minus (1 + 120582

119896cos 120585)

1205821sin (1205910+ 1205931) = minus119896120582

119896sin 120585

(51)


1) can be rewrit-

ten as

1205821cos (120591 + 120593

1) = 1205821cos (120591

0+ 1205931) cos (120591 minus 120591

0)


0)

(52)


1205821cos (120591 + 120593

1) = minus (1 + 120582

119896cos 120585) cos (120591 minus 120591

0)

+ 119896120582119896sin 120585 sin (120591 minus 120591

0)

(53)


1205910) + 120585) that is

cos (119896120591 + 120593119896) = cos 120585 cos 119896 (120591 minus 120591


0)

(54)


119879119896(120591) = [1 minus cos (120591 minus 120591

0)] [1 + 120582

119896cos 120585]


0)] cos 120585

+ 120582119896[119896 sin (120591 minus 120591

0) minus sin 119896 (120591 minus 120591

0)] sin 120585

(55)




119903119896(120591) = minus cos 120585 + [

1 minus cos 119896 (120591 minus 1205910)


minus [119896 sin (120591 minus 120591

0) minus sin 119896 (120591 minus 120591

0)

1 minus cos (120591 minus 1205910)

] sin 120585

(56)



0) + 120585) we obtain (39)



120582119896max120591

119903119896(120591) le 1 (57)



rewritten as

119903119896(120591) = 119903

119896(1205910minus2120585

119896) + 119902119896(120591) (58)

where

119903119896(1205910minus2120585


sin (120585119896) (59)

119902119896(120591) =

1

1 minus cos (120591 minus 1205910)


119896))

+119896 sin 120585sin (120585119896)

(cos(120591 minus 1205910+120585

119896) minus cos(120585

119896))]

(60)



1015840


0+ 120585119896| le 120587119896 This



0| le 120587 it



120591119902119896(120591) =

119902119896(1205910plusmn 2120587119896) = 0 Since 119903

119896(120591) minus 119902

119896(120591) is constant (see (58))




119896(120591)



119889119903119896(120591)

119889120591= minus119904 (120591) sdot sin(

119896 (120591 minus 1205910)

2+ 120585) (61)

where

119904 (120591) = [sin(119896 (120591 minus 120591

0)

2) cos(

120591 minus 1205910

2)

minus 119896 cos(119896 (120591 minus 120591

0)

2) sin(

120591 minus 1205910

2)]


2)

(62)


119904 (120591) = 2

119896minus1

sum

119899=1


0)

2) (63)


0| le 2120587119896 Therefore 119904(120591) ge 119904(120591

0plusmn


119889119903119896(120591)119889120591 = 0 and |120591minus120591


0)2+

120585) = 0From |120585| le 120587 |120591minus120591

0| le 2120587119896 and sin(119896(120591minus120591

0)2+120585) = 0


0+120585119896| = (2120587minus|120585|)119896




120591119902119896(120591) =


max120591

119903119896(120591) = 119903

119896(1205910minus2120585

119896)


(64)



max120591

119903119896(120591) ge 1 (65)


le


119896le




119896= 1 implies |120585| = 120587 and

119879119896(120591) = 1 minus cos 119896(120591 minus 120591


119896(120591) = 1 minus



119896= 1 (the case



119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(66)

where

119888119899= [sin(120585 minus 119899120585

119896) cos(120585

119896)


119896) sin(120585

119896)]


(67)


sin(120585119896)]

minus1

(68)

0 lt10038161003816100381610038161205851003816100381610038161003816 lt 120587 (69)









119896(120591) = 1 minus cos 119896(120591 minus 120591

0)



119879119896(120591) =

[1 minus cos (120591 minus 1205910)]2

3 (1198962 minus 1)

sdot [119896 (1198962

minus 1)

+ 2

119896minus2

sum

119899=1

(119896 minus 119899) ((119896 minus 119899)2

minus 1) cos 119899 (120591 minus 1205910)]

(70)


can be expressed as

1198792(120591) =

2

3[1 minus cos(120591 minus 120591

0)]2

1198793(120591) =

1

2[1 minus cos(120591 minus 120591

0)]2

[2 + cos (120591 minus 1205910)]

1198794(120591) =

4

15[1 minus cos (120591 minus 120591

0)]2


0)]

(71)



0+2120585119896)] =



119879119896(120591) = 120582

119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]

2

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(72)


119879119896(120591)

= 120582119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]


minus 2

119896minus1

sum

119899=1

sin (120585 minus 119899120585119896)sin (120585119896)

cos 119899(120591 minus 1205910+120585

119896)]

(73)


119888119899= 2


sum

119898=1


) (74)


119888119896minus3

119888119896minus4

and 119888119896minus5


= 2 (75)

119888119896minus3

= 8 cos(120585119896) (76)

119888119896minus4

= 8 + 12 cos(2120585119896) (77)

119888119896minus5

= 24 cos(120585119896) + 16 cos(3120585

119896) (78)



that

1198792(120591) =


2

[2 + cos 120585] (79)


0=



1198793(120591) =


2

[3 cos (1205853) + cos 120585]


0+120585

3)]

(80)


120582119896= [(119896 minus 1) cos 120585 + 119896

119896minus1

sum

119899=1

cos((119896 minus 2119899)120585119896

)]

minus1

(81)





1

1198962 minus 1lt 120582119896lt 1 (82)




119879119896(120591) = 1 minus 120582

119896

119896 sin 120585sin (120585119896)

cos(120591 minus 1205910+120585

119896)

+ 120582119896cos (119896 (120591 minus 120591

0) + 120585)

(83)


1

08

06

04

02


Am

plitu

de120582k

1


k = 2k = 3

k = 4k = 5




1198861= minus1205821cos(120591

0minus120585

119896) 119887

1= minus1205821sin(120591

0minus120585

119896) (84)


1205821=

119896 sin 120585sin (120585119896)

120582119896 (85)


120582119896= [cos(120585

119896)

119896 sin 120585sin(120585119896)


(86)


119909 = cos(120585119896) (87)


1205821= 119896120582119896119880119896minus1

(119909) (88)

120582119896=

1

119896119909119880119896minus1

(119909) minus 119881119896(119909)

(89)

where119881119896(119909) and119880



120582119896[119896119909119880119896minus1

(119909) minus 119881119896(119909)] minus 1 = 0 (90)


cos(120587119896) lt 119909 lt 1 (91)





1of


119909 = radic1 minus 1205822

21205822

1

3lt 1205822le 1

119909 =1

23radic1205823

1

8lt 1205823lt 1

119909 = radic1

6(1 + radic

51205824+ 3

21205824

)1

15lt 1205824lt 1

(92)



119896of


1205821= radic8120582

2(1 minus 120582

2)

1

3lt 1205822le 1 (93)

1205821= 3 (

3radic1205823minus 1205823)

1

8lt 1205823lt 1 (94)

1205821= radic

32

27(radic2120582

4(3 + 5120582

4)3

minus 21205824(9 + 7120582

4))

1

15lt 1205824lt 1

(95)




120582119896= [max120591

119903119896(120591)]minus1

(96)

and max120591119903119896(120591) = 119903

119896(1205910) According to (64) max

120591119903119896(120591) =



119879119896(120591) =


120591119903119896(120591) minus 119903

119896(120591)]

max120591119903119896(120591)

(97)



119896(120591) according to (64) and (39)

equals

max120591

119903119896(120591) minus 119903

119896(120591) = 119896

sin ((119896 minus 1) 120585119896)sin (120585119896)

minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(98)


max120591

119903119896(120591) minus 119903

119896(120591) = [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(99)




1198880minus 1198881cos(120585

119896) = 119896

sin (120585 minus 120585119896)sin (120585119896)

(100)


(119888119899minus1

+ 119888119899+1

) cos(120585119896) minus 2119888

119899= 2 (119896 minus 119899) cos(120585 minus 119899120585

119896)

(119888119899minus1

minus 119888119899+1


119896)

(101)


cos(120591 minus 1205910+2120585

119896) = cos(120585

119896) cos(120591 minus 120591

0+120585

119896)


0+120585

119896)

cos(120585 minus 119899120585

119896) cos(119899(120591 minus 120591

0+120585

119896))

minus sin(120585 minus 119899120585

119896) sin(119899(120591 minus 120591

0+120585

119896))

= cos (119899 (120591 minus 1205910) + 120585)

(102)







1

1198962 minus 1lt radic1198862119896+ 1198872119896lt 1 (103)


119896= radic1198862







1198961205910minus 120585 = atan 2 (119887

119896 119886119896) + 2119902120587 119902 = 1 (119896 minus 1) (104)


atan 2 (119910 119909) =

arctan(119910

119909) if 119909 ge 0

arctan(119910

119909) + 120587 if 119909 lt 0 119910 ge 0

arctan(119910

119909) minus 120587 if 119909 lt 0 119910 lt 0

(105)


1198861= minus1205821cos[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

1198871= minus1205821sin[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

(106)


1205910=1

119896[120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

1205910minus2120585

119896=1

119896[minus120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

(107)






119896



2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0

q = 1

q = 2





1according to (88)




1from 120582

119896and






3= minus02



1198861= minus08432 and 119887








4= minus02


1= 09861 (according to (95))






3

2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0q = 1

q = 2q = 3






following form

119879119896(120591) = [1 minus cos (120591 minus 120591

0)]


119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(108)

if 0 le 120582119896le 1(119896

2

minus 1) or

119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(109)


119899 119899 = 0 119896 minus 2 and


|120585| le 120587


1198861= minus (1 + 120582

119896) cos 120591

0 119887

1= minus (1 + 120582

119896) sin 120591

0

119886119896= 120582119896cos (119896120591

0) 119887

119896= 120582119896sin (119896120591

0)

(110)


1is given by (85)



119896lt 1


14

12

1

08

06

04

02

00 05 1

Amplitude 120582k

Am

plitu

de1205821

k = 2

k = 3

k = 4



119896= 1 have 119896 zeros


1205821= 1 + 120582

119896 (111)

if 0 le 120582119896le 1(119896

2

minus 1) or

1205821=


(112)


119896via (68) (or



119896= 1(119896

2


119896le 1(119896

2

minus 1) and 1(1198962

minus 1) le 120582119896le

1 According to (111) 120582119896= 1(119896

2


1198962

(1198962




1205821max =

1

cos (120587 (2119896)) (113)


119896=


at 1205910and 1205910+ 120587119896 for 120585 = minus1205872

14

12

1

08

06

04

02


Am

plitu

de1205821


k = 2k = 3k = 4



cos( 120587

2119896) lt 1 minus

1

1198962 (114)







119896







1is equal to 119896

2

(1198962








1= 1radic3 119886

3= 0


0= 1205876 (119886


1= 0

1198863= radic39 and 119887


0= 1205873 (119886

1= minus1

1198871= minus1radic3 119886

3= 0 and 119887

3= radic39)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

1205910 = 01205910 = 12058761205910 = 1205873





1205821= radic(1 + 120582

119896cos 120585)2 + 11989621205822

119896sin2120585 (115)







implies 120582119896


= 0Hence 119889120582

1119889120585 = 0 implies

2120582119896sin 120585 [1 minus (1198962 minus 1) 120582

119896cos 120585] = 0 (116)

Therefore 1198891205821119889120585 = 0 if 120582

119896= 0 (Option 1) or sin 120585 = 0



1= 1 (notice





1is



2

minus 1)


2

minus 1)120582119896cos 120585 = 1 and 120582

119896lt [(119896 minus






2


119896lt 1(119896

2

minus 1)







2

minus



1(1198962


119879119896(120591) =

[1 minus cos (120591 minus 1205910)]

(1 minus 1198962)

sdot [119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(117)


=

1(1 minus 1198962


2













119896(120591) of type (35) with


1le 0 by shifting



119896(120591) by


1




1and 119886119896with minus119886

119896


119886119896= 120582119896cos 120575 119887

119896= 120582119896sin 120575 (118)


where

|120575| le 120587 (119)


119896 120575 can be

determined as

120575 = atan 2 (119887119896 119886119896) (120)


proposition




119879119896(120591)

= [1 minus cos 120591]


119896minus1

sum

119899=1

(119896 minus 119899) (119886119896cos 119899120591 + 119887

119896sin 119899120591)]

(121)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1+119886

119896 where 120575 = atan 2(bk

119886119896) or

119879119896(120591) = 120582

119896[1 minus cos(120591 minus (120575 + 120585)

119896)]


119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120575

119896)]

(122)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 where 119888

119899 119899 = 0





119896cos 120575 with

119886119896(see (118)) and 120582

119896cos(119899120591 minus 120575) with 119886

119896cos 119899120591 + 119887

119896sin 119899120591



1198861= minus (1 + 119886

119896) 119887

1= minus119896119887

119896 (123)



1205910minus 120585119896 = 120575119896 in (84) leads to

1198861= minus1205821cos(120575

119896) 119887

1= minus1205821sin(120575

119896) (124)




119896and 119887119896satisfy





119896 119887119896satisfying 1 + 119886

119896=

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) is

120582119896=


(125)


(118)) into 1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896)


119910 = cos(120575119896) (126)



119886119896= 120582119896119881119896(119910) (127)

120582119896=

1

119896119910119880119896minus1

(119910) minus 119881119896(119910)

(128)

where119881119896(119910) and119880



119886119896119896119910119880119896minus1

(119910) minus (1 + 119886119896) 119881119896(119910) = 0 (129)


cos(120587119896) le 119910 le 1 (130)



119896= (119896 minus 1) cos 120575 +

119896sum119896minus1


119896= 120582119896cos 120575 can

be rewritten as

119886119896=


119899=1cos ((119896 minus 2119899) 120575119896)

(131)




119896is decreasing







1

05

0

minus05

minus1


Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4


119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4


119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8


4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)


119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3


8 (134)

1205824=1



4+ 1011988624) minus1 le 119886

4le

1

15

(135)





100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)





1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)




1|


119896 From (124)








1le 0


119896(120591) ge 0

and119879119896(120591) = 0 for some 120591






119896(1205910) = 0 and

11987910158401015840


119896(120591) at


119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840


1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)







0 First


0 taking into



1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)




119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119908 (120591 120574 119860 120572) = 1 minus 120574 (cos 120591 + 119860 cos (119896120591 + 120572)) (1)






1 + 120574119891 (120591 119860 120572) ge 0 (3)


Let

119865min (119860 120572) = min120591

119891 (120591 119860 120572) (4)


119866 (119860 120572) = minus1

119865min (119860 120572) (5)



120574max = max119860120572

119866 (119860 120572) = minus1

max119860120572

119865min (119860 120572)(6)

14

12

1

08

06

04

G1 G2

001

02

0304

0506 0

1205872

12058731205872

2120587

Parameter 120572

Parameter A

Gai

n fu

nctio

n

k = 2


2denote



120574max = max119860120572

119866 (119860 120572) = 119866 (119860lowast

120572lowast

)

max119860120572

119865min (119860 120572) = 119865min (119860lowast

120572lowast

)

(7)




119908lowast

(120591) = 119908 (120591 120574max 119860lowast

120572lowast

) (8)


120572lowast





1)







001

02

0304

05

06 01205872

12058731205872

2120587

Parameter 120572

Parameter A

1205872

1205874

0

minus1205874

minus1205872Posit

ions

of g

loba

l min

k = 2



119908119886(120591 120574119886 119887 119860 120572) = 1 minus 120574

119886(cos 120591 + 119887 sin 120591 + 119860 cos (119896120591 + 120572))

(9)






min120591119891119886(120591 119887 119860 120572) Inequality 119908

119886(120591 120574119886 119887 119860 120572) ge 0 can be



119886= minus1119865


119886= minus1119865


119887 119860 120572) = 119865119886min(119887 119860 120572)







119891 (1205911015840

119860 120572) = 119891 (12059110158401015840

119860 120572) (11)

1198911015840

(1205911015840

119860 120572) = 0 1198911015840

(12059110158401015840

119860 120572) = 0 (12)

11989110158401015840

(1205911015840

119860 120572) gt 0 11989110158401015840

(12059110158401015840

119860 120572) gt 0 (13)

(forall120591) 119891 (120591 119860 120572) ge 119891 (1205911015840

119860 120572) (14)




2 and 120572 = 120587



(11198962



Δ where

0 lt 120591Δle 120587119896



as follows

120572 = 120587 119860 (120591Δ) =

sin 120591Δ

119896 sin 119896120591Δ

0 lt 120591Δle120587

119896 (15)


val 0 lt 120591Δle 120587119896






119891 (1205911015840

119860 120572) minus 119891 (12059110158401015840

119860 120572) = 0

1198911015840

(1205911015840

119860 120572) + 1198911015840

(12059110158401015840

119860 120572) = 0

1198911015840

(1205911015840

119860 120572) minus 1198911015840

(12059110158401015840

119860 120572) = 0

11989110158401015840

(1205911015840

119860 120572) + 11989110158401015840

(12059110158401015840

119860 120572) gt 0

(16)

also hold Let

120591119904119903=(1205911015840

+ 12059110158401015840

)

2 120591

Δ=(12059110158401015840

minus 1205911015840

)

2

(17)


minus120587 lt 120591119904119903lt 120587 (18)

0 lt 120591Δlt 120587 (19)

12059110158401015840

= 120591119904119903+ 120591Δ 120591

1015840

= 120591119904119903minus 120591Δ (20)


1198911015840

(120591 119860 120572) = sin 120591 + 119896119860 sin (119896120591 + 120572)

11989110158401015840

(120591 119860 120572) = cos 120591 + 1198962119860 cos (119896120591 + 120572) (21)


sin 120591119904119903sin 120591Δ+ 119860 sin (119896120591

119904119903+ 120572) sin 119896120591

Δ= 0 (22)

sin 120591119904119903cos 120591Δ+ 119896119860 sin (119896120591

119904119903+ 120572) cos 119896120591

Δ= 0 (23)

cos 120591119904119903sin 120591Δ+ 119896119860 cos (119896120591

119904119903+ 120572) sin 119896120591

Δ= 0 (24)

cos 120591119904119903cos 120591Δ+ 1198962

119860 cos (119896120591119904119903+ 120572) cos 119896120591

Δgt 0 (25)





+

120572)[119896 sin 120591Δcos 119896120591

Δminus sin 119896120591



119896 sin 120591Δcos 119896120591

Δminus sin 119896120591

Δcos 120591Δ

= 0 (26)


119904119903and 119860 sin(119896120591

119904119903+ 120572)


sin 120591119904119903= 0 sin (119896120591

119904119903+ 120572) = 0 (27)


120591119904119903= 0 (28)


12059110158401015840

= 120591Δ 120591

1015840

= minus120591Δ (29)


119904119903+ 120572) = 0 imply that sin120572 =





Δ) le 0 (30)

From 120591Δ



120572 = 120587 (31)



Δ) le



0 lt 120591Δle120587

119896 (32)


119860 =sin 120591Δ

119896 sin 119896120591Δ

(33)




120591Δrarr0+119860 =

11198962 Consequently 119860 gt 1119896




1198911015840

(120591119889 119860 120572) = 0

11989110158401015840

(120591119889 119860 120572) = 0

(34)






00

00

k = 2 k = 3

k = 4 k = 5

Acos 120572

Acos 120572

Acos 120572

Acos 120572

A sin 120572

A sin 120572A sin 120572

A sin 120572









119879119896(120591) = 1 + 119886

1cos 120591 + 119887

1sin 120591 + 119886

119896cos 119896120591 + 119887

119896sin 119896120591 (35)


1205821= radic11988621+ 11988721 (36)

120582119896= radic1198862119896+ 1198872119896 (37)





119879119896(120591) = [1 minus cos (120591 minus 120591

0)] [1 minus 120582

119896119903119896(120591)] (38)

where119903119896(120591) = (119896 minus 1) cos 120585

+ 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(39)

providing that


sin(120585119896)]

minus1

(40)

10038161003816100381610038161205851003816100381610038161003816 le 120587 (41)


0 le 120582119896le 1 (42)



of 120582119896






1198861= minus (1 + 120582

119896cos 120585) cos 120591

0minus 119896120582119896sin 120585 sin 120591

0 (43)

1198871= minus (1 + 120582

119896cos 120585) sin 120591

0+ 119896120582119896sin 120585 cos 120591

0 (44)

119886119896= 120582119896cos (119896120591

0minus 120585) (45)

119887119896= 120582119896sin (119896120591

0minus 120585) (46)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms






3= radic220


1= minus03451 119886

3= 005


3= radic28


1= minus01127 119886

3= 0125


3= radic24


1= 02745 119886

3= 025 and




119879119896(120591) = 1 + 120582

1cos (120591 + 120593

1) + 120582119896cos (119896120591 + 120593

119896) (47)

where1205821ge 0120582

119896ge 0120593

1isin (minus120587 120587] and120593



1198861= 1205821cos1205931 119887

1= minus1205821sin1205931 (48)

119886119896= 120582119896cos120593119896 119887

119896= minus120582119896sin120593119896 (49)


0+ 120593119896) mod2120587 (50)




119896(1205910) = 0


119896(120591) ge 0 and 119879

119896(1205910) = 0

imply that 1198791015840119896(1205910) = 0 From 119879

119896(1205910) = 0 and 1198791015840

119896(1205910) = 0 by


1205821cos (120591

0+ 1205931) = minus (1 + 120582

119896cos 120585)

1205821sin (1205910+ 1205931) = minus119896120582

119896sin 120585

(51)


1) can be rewrit-

ten as

1205821cos (120591 + 120593

1) = 1205821cos (120591

0+ 1205931) cos (120591 minus 120591

0)


0)

(52)


1205821cos (120591 + 120593

1) = minus (1 + 120582

119896cos 120585) cos (120591 minus 120591

0)

+ 119896120582119896sin 120585 sin (120591 minus 120591

0)

(53)


1205910) + 120585) that is

cos (119896120591 + 120593119896) = cos 120585 cos 119896 (120591 minus 120591


0)

(54)


119879119896(120591) = [1 minus cos (120591 minus 120591

0)] [1 + 120582

119896cos 120585]


0)] cos 120585

+ 120582119896[119896 sin (120591 minus 120591

0) minus sin 119896 (120591 minus 120591

0)] sin 120585

(55)




119903119896(120591) = minus cos 120585 + [

1 minus cos 119896 (120591 minus 1205910)


minus [119896 sin (120591 minus 120591

0) minus sin 119896 (120591 minus 120591

0)

1 minus cos (120591 minus 1205910)

] sin 120585

(56)



0) + 120585) we obtain (39)



120582119896max120591

119903119896(120591) le 1 (57)



rewritten as

119903119896(120591) = 119903

119896(1205910minus2120585

119896) + 119902119896(120591) (58)

where

119903119896(1205910minus2120585


sin (120585119896) (59)

119902119896(120591) =

1

1 minus cos (120591 minus 1205910)


119896))

+119896 sin 120585sin (120585119896)

(cos(120591 minus 1205910+120585

119896) minus cos(120585

119896))]

(60)



1015840


0+ 120585119896| le 120587119896 This



0| le 120587 it



120591119902119896(120591) =

119902119896(1205910plusmn 2120587119896) = 0 Since 119903

119896(120591) minus 119902

119896(120591) is constant (see (58))




119896(120591)



119889119903119896(120591)

119889120591= minus119904 (120591) sdot sin(

119896 (120591 minus 1205910)

2+ 120585) (61)

where

119904 (120591) = [sin(119896 (120591 minus 120591

0)

2) cos(

120591 minus 1205910

2)

minus 119896 cos(119896 (120591 minus 120591

0)

2) sin(

120591 minus 1205910

2)]


2)

(62)


119904 (120591) = 2

119896minus1

sum

119899=1


0)

2) (63)


0| le 2120587119896 Therefore 119904(120591) ge 119904(120591

0plusmn


119889119903119896(120591)119889120591 = 0 and |120591minus120591


0)2+

120585) = 0From |120585| le 120587 |120591minus120591

0| le 2120587119896 and sin(119896(120591minus120591

0)2+120585) = 0


0+120585119896| = (2120587minus|120585|)119896




120591119902119896(120591) =


max120591

119903119896(120591) = 119903

119896(1205910minus2120585

119896)


(64)



max120591

119903119896(120591) ge 1 (65)


le


119896le




119896= 1 implies |120585| = 120587 and

119879119896(120591) = 1 minus cos 119896(120591 minus 120591


119896(120591) = 1 minus



119896= 1 (the case



119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(66)

where

119888119899= [sin(120585 minus 119899120585

119896) cos(120585

119896)


119896) sin(120585

119896)]


(67)


sin(120585119896)]

minus1

(68)

0 lt10038161003816100381610038161205851003816100381610038161003816 lt 120587 (69)









119896(120591) = 1 minus cos 119896(120591 minus 120591

0)



119879119896(120591) =

[1 minus cos (120591 minus 1205910)]2

3 (1198962 minus 1)

sdot [119896 (1198962

minus 1)

+ 2

119896minus2

sum

119899=1

(119896 minus 119899) ((119896 minus 119899)2

minus 1) cos 119899 (120591 minus 1205910)]

(70)


can be expressed as

1198792(120591) =

2

3[1 minus cos(120591 minus 120591

0)]2

1198793(120591) =

1

2[1 minus cos(120591 minus 120591

0)]2

[2 + cos (120591 minus 1205910)]

1198794(120591) =

4

15[1 minus cos (120591 minus 120591

0)]2


0)]

(71)



0+2120585119896)] =



119879119896(120591) = 120582

119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]

2

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(72)


119879119896(120591)

= 120582119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]


minus 2

119896minus1

sum

119899=1

sin (120585 minus 119899120585119896)sin (120585119896)

cos 119899(120591 minus 1205910+120585

119896)]

(73)


119888119899= 2


sum

119898=1


) (74)


119888119896minus3

119888119896minus4

and 119888119896minus5


= 2 (75)

119888119896minus3

= 8 cos(120585119896) (76)

119888119896minus4

= 8 + 12 cos(2120585119896) (77)

119888119896minus5

= 24 cos(120585119896) + 16 cos(3120585

119896) (78)



that

1198792(120591) =


2

[2 + cos 120585] (79)


0=



1198793(120591) =


2

[3 cos (1205853) + cos 120585]


0+120585

3)]

(80)


120582119896= [(119896 minus 1) cos 120585 + 119896

119896minus1

sum

119899=1

cos((119896 minus 2119899)120585119896

)]

minus1

(81)





1

1198962 minus 1lt 120582119896lt 1 (82)




119879119896(120591) = 1 minus 120582

119896

119896 sin 120585sin (120585119896)

cos(120591 minus 1205910+120585

119896)

+ 120582119896cos (119896 (120591 minus 120591

0) + 120585)

(83)


1

08

06

04

02


Am

plitu

de120582k

1


k = 2k = 3

k = 4k = 5




1198861= minus1205821cos(120591

0minus120585

119896) 119887

1= minus1205821sin(120591

0minus120585

119896) (84)


1205821=

119896 sin 120585sin (120585119896)

120582119896 (85)


120582119896= [cos(120585

119896)

119896 sin 120585sin(120585119896)


(86)


119909 = cos(120585119896) (87)


1205821= 119896120582119896119880119896minus1

(119909) (88)

120582119896=

1

119896119909119880119896minus1

(119909) minus 119881119896(119909)

(89)

where119881119896(119909) and119880



120582119896[119896119909119880119896minus1

(119909) minus 119881119896(119909)] minus 1 = 0 (90)


cos(120587119896) lt 119909 lt 1 (91)





1of


119909 = radic1 minus 1205822

21205822

1

3lt 1205822le 1

119909 =1

23radic1205823

1

8lt 1205823lt 1

119909 = radic1

6(1 + radic

51205824+ 3

21205824

)1

15lt 1205824lt 1

(92)



119896of


1205821= radic8120582

2(1 minus 120582

2)

1

3lt 1205822le 1 (93)

1205821= 3 (

3radic1205823minus 1205823)

1

8lt 1205823lt 1 (94)

1205821= radic

32

27(radic2120582

4(3 + 5120582

4)3

minus 21205824(9 + 7120582

4))

1

15lt 1205824lt 1

(95)




120582119896= [max120591

119903119896(120591)]minus1

(96)

and max120591119903119896(120591) = 119903

119896(1205910) According to (64) max

120591119903119896(120591) =



119879119896(120591) =


120591119903119896(120591) minus 119903

119896(120591)]

max120591119903119896(120591)

(97)



119896(120591) according to (64) and (39)

equals

max120591

119903119896(120591) minus 119903

119896(120591) = 119896

sin ((119896 minus 1) 120585119896)sin (120585119896)

minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(98)


max120591

119903119896(120591) minus 119903

119896(120591) = [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(99)




1198880minus 1198881cos(120585

119896) = 119896

sin (120585 minus 120585119896)sin (120585119896)

(100)


(119888119899minus1

+ 119888119899+1

) cos(120585119896) minus 2119888

119899= 2 (119896 minus 119899) cos(120585 minus 119899120585

119896)

(119888119899minus1

minus 119888119899+1


119896)

(101)


cos(120591 minus 1205910+2120585

119896) = cos(120585

119896) cos(120591 minus 120591

0+120585

119896)


0+120585

119896)

cos(120585 minus 119899120585

119896) cos(119899(120591 minus 120591

0+120585

119896))

minus sin(120585 minus 119899120585

119896) sin(119899(120591 minus 120591

0+120585

119896))

= cos (119899 (120591 minus 1205910) + 120585)

(102)







1

1198962 minus 1lt radic1198862119896+ 1198872119896lt 1 (103)


119896= radic1198862







1198961205910minus 120585 = atan 2 (119887

119896 119886119896) + 2119902120587 119902 = 1 (119896 minus 1) (104)


atan 2 (119910 119909) =

arctan(119910

119909) if 119909 ge 0

arctan(119910

119909) + 120587 if 119909 lt 0 119910 ge 0

arctan(119910

119909) minus 120587 if 119909 lt 0 119910 lt 0

(105)


1198861= minus1205821cos[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

1198871= minus1205821sin[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

(106)


1205910=1

119896[120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

1205910minus2120585

119896=1

119896[minus120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

(107)






119896



2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0

q = 1

q = 2





1according to (88)




1from 120582

119896and






3= minus02



1198861= minus08432 and 119887








4= minus02


1= 09861 (according to (95))






3

2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0q = 1

q = 2q = 3






following form

119879119896(120591) = [1 minus cos (120591 minus 120591

0)]


119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(108)

if 0 le 120582119896le 1(119896

2

minus 1) or

119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(109)


119899 119899 = 0 119896 minus 2 and


|120585| le 120587


1198861= minus (1 + 120582

119896) cos 120591

0 119887

1= minus (1 + 120582

119896) sin 120591

0

119886119896= 120582119896cos (119896120591

0) 119887

119896= 120582119896sin (119896120591

0)

(110)


1is given by (85)



119896lt 1


14

12

1

08

06

04

02

00 05 1

Amplitude 120582k

Am

plitu

de1205821

k = 2

k = 3

k = 4



119896= 1 have 119896 zeros


1205821= 1 + 120582

119896 (111)

if 0 le 120582119896le 1(119896

2

minus 1) or

1205821=


(112)


119896via (68) (or



119896= 1(119896

2


119896le 1(119896

2

minus 1) and 1(1198962

minus 1) le 120582119896le

1 According to (111) 120582119896= 1(119896

2


1198962

(1198962




1205821max =

1

cos (120587 (2119896)) (113)


119896=


at 1205910and 1205910+ 120587119896 for 120585 = minus1205872

14

12

1

08

06

04

02


Am

plitu

de1205821


k = 2k = 3k = 4



cos( 120587

2119896) lt 1 minus

1

1198962 (114)







119896







1is equal to 119896

2

(1198962








1= 1radic3 119886

3= 0


0= 1205876 (119886


1= 0

1198863= radic39 and 119887


0= 1205873 (119886

1= minus1

1198871= minus1radic3 119886

3= 0 and 119887

3= radic39)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

1205910 = 01205910 = 12058761205910 = 1205873





1205821= radic(1 + 120582

119896cos 120585)2 + 11989621205822

119896sin2120585 (115)







implies 120582119896


= 0Hence 119889120582

1119889120585 = 0 implies

2120582119896sin 120585 [1 minus (1198962 minus 1) 120582

119896cos 120585] = 0 (116)

Therefore 1198891205821119889120585 = 0 if 120582

119896= 0 (Option 1) or sin 120585 = 0



1= 1 (notice





1is



2

minus 1)


2

minus 1)120582119896cos 120585 = 1 and 120582

119896lt [(119896 minus






2


119896lt 1(119896

2

minus 1)







2

minus



1(1198962


119879119896(120591) =

[1 minus cos (120591 minus 1205910)]

(1 minus 1198962)

sdot [119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(117)


=

1(1 minus 1198962


2













119896(120591) of type (35) with


1le 0 by shifting



119896(120591) by


1




1and 119886119896with minus119886

119896


119886119896= 120582119896cos 120575 119887

119896= 120582119896sin 120575 (118)


where

|120575| le 120587 (119)


119896 120575 can be

determined as

120575 = atan 2 (119887119896 119886119896) (120)


proposition




119879119896(120591)

= [1 minus cos 120591]


119896minus1

sum

119899=1

(119896 minus 119899) (119886119896cos 119899120591 + 119887

119896sin 119899120591)]

(121)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1+119886

119896 where 120575 = atan 2(bk

119886119896) or

119879119896(120591) = 120582

119896[1 minus cos(120591 minus (120575 + 120585)

119896)]


119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120575

119896)]

(122)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 where 119888

119899 119899 = 0





119896cos 120575 with

119886119896(see (118)) and 120582

119896cos(119899120591 minus 120575) with 119886

119896cos 119899120591 + 119887

119896sin 119899120591



1198861= minus (1 + 119886

119896) 119887

1= minus119896119887

119896 (123)



1205910minus 120585119896 = 120575119896 in (84) leads to

1198861= minus1205821cos(120575

119896) 119887

1= minus1205821sin(120575

119896) (124)




119896and 119887119896satisfy





119896 119887119896satisfying 1 + 119886

119896=

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) is

120582119896=


(125)


(118)) into 1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896)


119910 = cos(120575119896) (126)



119886119896= 120582119896119881119896(119910) (127)

120582119896=

1

119896119910119880119896minus1

(119910) minus 119881119896(119910)

(128)

where119881119896(119910) and119880



119886119896119896119910119880119896minus1

(119910) minus (1 + 119886119896) 119881119896(119910) = 0 (129)


cos(120587119896) le 119910 le 1 (130)



119896= (119896 minus 1) cos 120575 +

119896sum119896minus1


119896= 120582119896cos 120575 can

be rewritten as

119886119896=


119899=1cos ((119896 minus 2119899) 120575119896)

(131)




119896is decreasing







1

05

0

minus05

minus1


Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4


119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4


119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8


4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)


119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3


8 (134)

1205824=1



4+ 1011988624) minus1 le 119886

4le

1

15

(135)





100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)





1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)




1|


119896 From (124)








1le 0


119896(120591) ge 0

and119879119896(120591) = 0 for some 120591






119896(1205910) = 0 and

11987910158401015840


119896(120591) at


119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840


1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)







0 First


0 taking into



1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)




119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










001

02

0304

05

06 01205872

12058731205872

2120587

Parameter 120572

Parameter A

1205872

1205874

0

minus1205874

minus1205872Posit

ions

of g

loba

l min

k = 2



119908119886(120591 120574119886 119887 119860 120572) = 1 minus 120574

119886(cos 120591 + 119887 sin 120591 + 119860 cos (119896120591 + 120572))

(9)






min120591119891119886(120591 119887 119860 120572) Inequality 119908

119886(120591 120574119886 119887 119860 120572) ge 0 can be



119886= minus1119865


119886= minus1119865


119887 119860 120572) = 119865119886min(119887 119860 120572)







119891 (1205911015840

119860 120572) = 119891 (12059110158401015840

119860 120572) (11)

1198911015840

(1205911015840

119860 120572) = 0 1198911015840

(12059110158401015840

119860 120572) = 0 (12)

11989110158401015840

(1205911015840

119860 120572) gt 0 11989110158401015840

(12059110158401015840

119860 120572) gt 0 (13)

(forall120591) 119891 (120591 119860 120572) ge 119891 (1205911015840

119860 120572) (14)




2 and 120572 = 120587



(11198962



Δ where

0 lt 120591Δle 120587119896



as follows

120572 = 120587 119860 (120591Δ) =

sin 120591Δ

119896 sin 119896120591Δ

0 lt 120591Δle120587

119896 (15)


val 0 lt 120591Δle 120587119896






119891 (1205911015840

119860 120572) minus 119891 (12059110158401015840

119860 120572) = 0

1198911015840

(1205911015840

119860 120572) + 1198911015840

(12059110158401015840

119860 120572) = 0

1198911015840

(1205911015840

119860 120572) minus 1198911015840

(12059110158401015840

119860 120572) = 0

11989110158401015840

(1205911015840

119860 120572) + 11989110158401015840

(12059110158401015840

119860 120572) gt 0

(16)

also hold Let

120591119904119903=(1205911015840

+ 12059110158401015840

)

2 120591

Δ=(12059110158401015840

minus 1205911015840

)

2

(17)


minus120587 lt 120591119904119903lt 120587 (18)

0 lt 120591Δlt 120587 (19)

12059110158401015840

= 120591119904119903+ 120591Δ 120591

1015840

= 120591119904119903minus 120591Δ (20)


1198911015840

(120591 119860 120572) = sin 120591 + 119896119860 sin (119896120591 + 120572)

11989110158401015840

(120591 119860 120572) = cos 120591 + 1198962119860 cos (119896120591 + 120572) (21)


sin 120591119904119903sin 120591Δ+ 119860 sin (119896120591

119904119903+ 120572) sin 119896120591

Δ= 0 (22)

sin 120591119904119903cos 120591Δ+ 119896119860 sin (119896120591

119904119903+ 120572) cos 119896120591

Δ= 0 (23)

cos 120591119904119903sin 120591Δ+ 119896119860 cos (119896120591

119904119903+ 120572) sin 119896120591

Δ= 0 (24)

cos 120591119904119903cos 120591Δ+ 1198962

119860 cos (119896120591119904119903+ 120572) cos 119896120591

Δgt 0 (25)





+

120572)[119896 sin 120591Δcos 119896120591

Δminus sin 119896120591



119896 sin 120591Δcos 119896120591

Δminus sin 119896120591

Δcos 120591Δ

= 0 (26)


119904119903and 119860 sin(119896120591

119904119903+ 120572)


sin 120591119904119903= 0 sin (119896120591

119904119903+ 120572) = 0 (27)


120591119904119903= 0 (28)


12059110158401015840

= 120591Δ 120591

1015840

= minus120591Δ (29)


119904119903+ 120572) = 0 imply that sin120572 =





Δ) le 0 (30)

From 120591Δ



120572 = 120587 (31)



Δ) le



0 lt 120591Δle120587

119896 (32)


119860 =sin 120591Δ

119896 sin 119896120591Δ

(33)




120591Δrarr0+119860 =

11198962 Consequently 119860 gt 1119896




1198911015840

(120591119889 119860 120572) = 0

11989110158401015840

(120591119889 119860 120572) = 0

(34)






00

00

k = 2 k = 3

k = 4 k = 5

Acos 120572

Acos 120572

Acos 120572

Acos 120572

A sin 120572

A sin 120572A sin 120572

A sin 120572









119879119896(120591) = 1 + 119886

1cos 120591 + 119887

1sin 120591 + 119886

119896cos 119896120591 + 119887

119896sin 119896120591 (35)


1205821= radic11988621+ 11988721 (36)

120582119896= radic1198862119896+ 1198872119896 (37)





119879119896(120591) = [1 minus cos (120591 minus 120591

0)] [1 minus 120582

119896119903119896(120591)] (38)

where119903119896(120591) = (119896 minus 1) cos 120585

+ 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(39)

providing that


sin(120585119896)]

minus1

(40)

10038161003816100381610038161205851003816100381610038161003816 le 120587 (41)


0 le 120582119896le 1 (42)



of 120582119896






1198861= minus (1 + 120582

119896cos 120585) cos 120591

0minus 119896120582119896sin 120585 sin 120591

0 (43)

1198871= minus (1 + 120582

119896cos 120585) sin 120591

0+ 119896120582119896sin 120585 cos 120591

0 (44)

119886119896= 120582119896cos (119896120591

0minus 120585) (45)

119887119896= 120582119896sin (119896120591

0minus 120585) (46)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms






3= radic220


1= minus03451 119886

3= 005


3= radic28


1= minus01127 119886

3= 0125


3= radic24


1= 02745 119886

3= 025 and




119879119896(120591) = 1 + 120582

1cos (120591 + 120593

1) + 120582119896cos (119896120591 + 120593

119896) (47)

where1205821ge 0120582

119896ge 0120593

1isin (minus120587 120587] and120593



1198861= 1205821cos1205931 119887

1= minus1205821sin1205931 (48)

119886119896= 120582119896cos120593119896 119887

119896= minus120582119896sin120593119896 (49)


0+ 120593119896) mod2120587 (50)




119896(1205910) = 0


119896(120591) ge 0 and 119879

119896(1205910) = 0

imply that 1198791015840119896(1205910) = 0 From 119879

119896(1205910) = 0 and 1198791015840

119896(1205910) = 0 by


1205821cos (120591

0+ 1205931) = minus (1 + 120582

119896cos 120585)

1205821sin (1205910+ 1205931) = minus119896120582

119896sin 120585

(51)


1) can be rewrit-

ten as

1205821cos (120591 + 120593

1) = 1205821cos (120591

0+ 1205931) cos (120591 minus 120591

0)


0)

(52)


1205821cos (120591 + 120593

1) = minus (1 + 120582

119896cos 120585) cos (120591 minus 120591

0)

+ 119896120582119896sin 120585 sin (120591 minus 120591

0)

(53)


1205910) + 120585) that is

cos (119896120591 + 120593119896) = cos 120585 cos 119896 (120591 minus 120591


0)

(54)


119879119896(120591) = [1 minus cos (120591 minus 120591

0)] [1 + 120582

119896cos 120585]


0)] cos 120585

+ 120582119896[119896 sin (120591 minus 120591

0) minus sin 119896 (120591 minus 120591

0)] sin 120585

(55)




119903119896(120591) = minus cos 120585 + [

1 minus cos 119896 (120591 minus 1205910)


minus [119896 sin (120591 minus 120591

0) minus sin 119896 (120591 minus 120591

0)

1 minus cos (120591 minus 1205910)

] sin 120585

(56)



0) + 120585) we obtain (39)



120582119896max120591

119903119896(120591) le 1 (57)



rewritten as

119903119896(120591) = 119903

119896(1205910minus2120585

119896) + 119902119896(120591) (58)

where

119903119896(1205910minus2120585


sin (120585119896) (59)

119902119896(120591) =

1

1 minus cos (120591 minus 1205910)


119896))

+119896 sin 120585sin (120585119896)

(cos(120591 minus 1205910+120585

119896) minus cos(120585

119896))]

(60)



1015840


0+ 120585119896| le 120587119896 This



0| le 120587 it



120591119902119896(120591) =

119902119896(1205910plusmn 2120587119896) = 0 Since 119903

119896(120591) minus 119902

119896(120591) is constant (see (58))




119896(120591)



119889119903119896(120591)

119889120591= minus119904 (120591) sdot sin(

119896 (120591 minus 1205910)

2+ 120585) (61)

where

119904 (120591) = [sin(119896 (120591 minus 120591

0)

2) cos(

120591 minus 1205910

2)

minus 119896 cos(119896 (120591 minus 120591

0)

2) sin(

120591 minus 1205910

2)]


2)

(62)


119904 (120591) = 2

119896minus1

sum

119899=1


0)

2) (63)


0| le 2120587119896 Therefore 119904(120591) ge 119904(120591

0plusmn


119889119903119896(120591)119889120591 = 0 and |120591minus120591


0)2+

120585) = 0From |120585| le 120587 |120591minus120591

0| le 2120587119896 and sin(119896(120591minus120591

0)2+120585) = 0


0+120585119896| = (2120587minus|120585|)119896




120591119902119896(120591) =


max120591

119903119896(120591) = 119903

119896(1205910minus2120585

119896)


(64)



max120591

119903119896(120591) ge 1 (65)


le


119896le




119896= 1 implies |120585| = 120587 and

119879119896(120591) = 1 minus cos 119896(120591 minus 120591


119896(120591) = 1 minus



119896= 1 (the case



119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(66)

where

119888119899= [sin(120585 minus 119899120585

119896) cos(120585

119896)


119896) sin(120585

119896)]


(67)


sin(120585119896)]

minus1

(68)

0 lt10038161003816100381610038161205851003816100381610038161003816 lt 120587 (69)









119896(120591) = 1 minus cos 119896(120591 minus 120591

0)



119879119896(120591) =

[1 minus cos (120591 minus 1205910)]2

3 (1198962 minus 1)

sdot [119896 (1198962

minus 1)

+ 2

119896minus2

sum

119899=1

(119896 minus 119899) ((119896 minus 119899)2

minus 1) cos 119899 (120591 minus 1205910)]

(70)


can be expressed as

1198792(120591) =

2

3[1 minus cos(120591 minus 120591

0)]2

1198793(120591) =

1

2[1 minus cos(120591 minus 120591

0)]2

[2 + cos (120591 minus 1205910)]

1198794(120591) =

4

15[1 minus cos (120591 minus 120591

0)]2


0)]

(71)



0+2120585119896)] =



119879119896(120591) = 120582

119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]

2

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(72)


119879119896(120591)

= 120582119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]


minus 2

119896minus1

sum

119899=1

sin (120585 minus 119899120585119896)sin (120585119896)

cos 119899(120591 minus 1205910+120585

119896)]

(73)


119888119899= 2


sum

119898=1


) (74)


119888119896minus3

119888119896minus4

and 119888119896minus5


= 2 (75)

119888119896minus3

= 8 cos(120585119896) (76)

119888119896minus4

= 8 + 12 cos(2120585119896) (77)

119888119896minus5

= 24 cos(120585119896) + 16 cos(3120585

119896) (78)



that

1198792(120591) =


2

[2 + cos 120585] (79)


0=



1198793(120591) =


2

[3 cos (1205853) + cos 120585]


0+120585

3)]

(80)


120582119896= [(119896 minus 1) cos 120585 + 119896

119896minus1

sum

119899=1

cos((119896 minus 2119899)120585119896

)]

minus1

(81)





1

1198962 minus 1lt 120582119896lt 1 (82)




119879119896(120591) = 1 minus 120582

119896

119896 sin 120585sin (120585119896)

cos(120591 minus 1205910+120585

119896)

+ 120582119896cos (119896 (120591 minus 120591

0) + 120585)

(83)


1

08

06

04

02


Am

plitu

de120582k

1


k = 2k = 3

k = 4k = 5




1198861= minus1205821cos(120591

0minus120585

119896) 119887

1= minus1205821sin(120591

0minus120585

119896) (84)


1205821=

119896 sin 120585sin (120585119896)

120582119896 (85)


120582119896= [cos(120585

119896)

119896 sin 120585sin(120585119896)


(86)


119909 = cos(120585119896) (87)


1205821= 119896120582119896119880119896minus1

(119909) (88)

120582119896=

1

119896119909119880119896minus1

(119909) minus 119881119896(119909)

(89)

where119881119896(119909) and119880



120582119896[119896119909119880119896minus1

(119909) minus 119881119896(119909)] minus 1 = 0 (90)


cos(120587119896) lt 119909 lt 1 (91)





1of


119909 = radic1 minus 1205822

21205822

1

3lt 1205822le 1

119909 =1

23radic1205823

1

8lt 1205823lt 1

119909 = radic1

6(1 + radic

51205824+ 3

21205824

)1

15lt 1205824lt 1

(92)



119896of


1205821= radic8120582

2(1 minus 120582

2)

1

3lt 1205822le 1 (93)

1205821= 3 (

3radic1205823minus 1205823)

1

8lt 1205823lt 1 (94)

1205821= radic

32

27(radic2120582

4(3 + 5120582

4)3

minus 21205824(9 + 7120582

4))

1

15lt 1205824lt 1

(95)




120582119896= [max120591

119903119896(120591)]minus1

(96)

and max120591119903119896(120591) = 119903

119896(1205910) According to (64) max

120591119903119896(120591) =



119879119896(120591) =


120591119903119896(120591) minus 119903

119896(120591)]

max120591119903119896(120591)

(97)



119896(120591) according to (64) and (39)

equals

max120591

119903119896(120591) minus 119903

119896(120591) = 119896

sin ((119896 minus 1) 120585119896)sin (120585119896)

minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(98)


max120591

119903119896(120591) minus 119903

119896(120591) = [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(99)




1198880minus 1198881cos(120585

119896) = 119896

sin (120585 minus 120585119896)sin (120585119896)

(100)


(119888119899minus1

+ 119888119899+1

) cos(120585119896) minus 2119888

119899= 2 (119896 minus 119899) cos(120585 minus 119899120585

119896)

(119888119899minus1

minus 119888119899+1


119896)

(101)


cos(120591 minus 1205910+2120585

119896) = cos(120585

119896) cos(120591 minus 120591

0+120585

119896)


0+120585

119896)

cos(120585 minus 119899120585

119896) cos(119899(120591 minus 120591

0+120585

119896))

minus sin(120585 minus 119899120585

119896) sin(119899(120591 minus 120591

0+120585

119896))

= cos (119899 (120591 minus 1205910) + 120585)

(102)







1

1198962 minus 1lt radic1198862119896+ 1198872119896lt 1 (103)


119896= radic1198862







1198961205910minus 120585 = atan 2 (119887

119896 119886119896) + 2119902120587 119902 = 1 (119896 minus 1) (104)


atan 2 (119910 119909) =

arctan(119910

119909) if 119909 ge 0

arctan(119910

119909) + 120587 if 119909 lt 0 119910 ge 0

arctan(119910

119909) minus 120587 if 119909 lt 0 119910 lt 0

(105)


1198861= minus1205821cos[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

1198871= minus1205821sin[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

(106)


1205910=1

119896[120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

1205910minus2120585

119896=1

119896[minus120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

(107)






119896



2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0

q = 1

q = 2





1according to (88)




1from 120582

119896and






3= minus02



1198861= minus08432 and 119887








4= minus02


1= 09861 (according to (95))






3

2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0q = 1

q = 2q = 3






following form

119879119896(120591) = [1 minus cos (120591 minus 120591

0)]


119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(108)

if 0 le 120582119896le 1(119896

2

minus 1) or

119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(109)


119899 119899 = 0 119896 minus 2 and


|120585| le 120587


1198861= minus (1 + 120582

119896) cos 120591

0 119887

1= minus (1 + 120582

119896) sin 120591

0

119886119896= 120582119896cos (119896120591

0) 119887

119896= 120582119896sin (119896120591

0)

(110)


1is given by (85)



119896lt 1


14

12

1

08

06

04

02

00 05 1

Amplitude 120582k

Am

plitu

de1205821

k = 2

k = 3

k = 4



119896= 1 have 119896 zeros


1205821= 1 + 120582

119896 (111)

if 0 le 120582119896le 1(119896

2

minus 1) or

1205821=


(112)


119896via (68) (or



119896= 1(119896

2


119896le 1(119896

2

minus 1) and 1(1198962

minus 1) le 120582119896le

1 According to (111) 120582119896= 1(119896

2


1198962

(1198962




1205821max =

1

cos (120587 (2119896)) (113)


119896=


at 1205910and 1205910+ 120587119896 for 120585 = minus1205872

14

12

1

08

06

04

02


Am

plitu

de1205821


k = 2k = 3k = 4



cos( 120587

2119896) lt 1 minus

1

1198962 (114)







119896







1is equal to 119896

2

(1198962








1= 1radic3 119886

3= 0


0= 1205876 (119886


1= 0

1198863= radic39 and 119887


0= 1205873 (119886

1= minus1

1198871= minus1radic3 119886

3= 0 and 119887

3= radic39)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

1205910 = 01205910 = 12058761205910 = 1205873





1205821= radic(1 + 120582

119896cos 120585)2 + 11989621205822

119896sin2120585 (115)







implies 120582119896


= 0Hence 119889120582

1119889120585 = 0 implies

2120582119896sin 120585 [1 minus (1198962 minus 1) 120582

119896cos 120585] = 0 (116)

Therefore 1198891205821119889120585 = 0 if 120582

119896= 0 (Option 1) or sin 120585 = 0



1= 1 (notice





1is



2

minus 1)


2

minus 1)120582119896cos 120585 = 1 and 120582

119896lt [(119896 minus






2


119896lt 1(119896

2

minus 1)







2

minus



1(1198962


119879119896(120591) =

[1 minus cos (120591 minus 1205910)]

(1 minus 1198962)

sdot [119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(117)


=

1(1 minus 1198962


2













119896(120591) of type (35) with


1le 0 by shifting



119896(120591) by


1




1and 119886119896with minus119886

119896


119886119896= 120582119896cos 120575 119887

119896= 120582119896sin 120575 (118)


where

|120575| le 120587 (119)


119896 120575 can be

determined as

120575 = atan 2 (119887119896 119886119896) (120)


proposition




119879119896(120591)

= [1 minus cos 120591]


119896minus1

sum

119899=1

(119896 minus 119899) (119886119896cos 119899120591 + 119887

119896sin 119899120591)]

(121)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1+119886

119896 where 120575 = atan 2(bk

119886119896) or

119879119896(120591) = 120582

119896[1 minus cos(120591 minus (120575 + 120585)

119896)]


119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120575

119896)]

(122)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 where 119888

119899 119899 = 0





119896cos 120575 with

119886119896(see (118)) and 120582

119896cos(119899120591 minus 120575) with 119886

119896cos 119899120591 + 119887

119896sin 119899120591



1198861= minus (1 + 119886

119896) 119887

1= minus119896119887

119896 (123)



1205910minus 120585119896 = 120575119896 in (84) leads to

1198861= minus1205821cos(120575

119896) 119887

1= minus1205821sin(120575

119896) (124)




119896and 119887119896satisfy





119896 119887119896satisfying 1 + 119886

119896=

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) is

120582119896=


(125)


(118)) into 1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896)


119910 = cos(120575119896) (126)



119886119896= 120582119896119881119896(119910) (127)

120582119896=

1

119896119910119880119896minus1

(119910) minus 119881119896(119910)

(128)

where119881119896(119910) and119880



119886119896119896119910119880119896minus1

(119910) minus (1 + 119886119896) 119881119896(119910) = 0 (129)


cos(120587119896) le 119910 le 1 (130)



119896= (119896 minus 1) cos 120575 +

119896sum119896minus1


119896= 120582119896cos 120575 can

be rewritten as

119886119896=


119899=1cos ((119896 minus 2119899) 120575119896)

(131)




119896is decreasing







1

05

0

minus05

minus1


Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4


119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4


119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8


4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)


119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3


8 (134)

1205824=1



4+ 1011988624) minus1 le 119886

4le

1

15

(135)





100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)





1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)




1|


119896 From (124)








1le 0


119896(120591) ge 0

and119879119896(120591) = 0 for some 120591






119896(1205910) = 0 and

11987910158401015840


119896(120591) at


119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840


1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)







0 First


0 taking into



1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)




119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119891 (1205911015840

119860 120572) minus 119891 (12059110158401015840

119860 120572) = 0

1198911015840

(1205911015840

119860 120572) + 1198911015840

(12059110158401015840

119860 120572) = 0

1198911015840

(1205911015840

119860 120572) minus 1198911015840

(12059110158401015840

119860 120572) = 0

11989110158401015840

(1205911015840

119860 120572) + 11989110158401015840

(12059110158401015840

119860 120572) gt 0

(16)

also hold Let

120591119904119903=(1205911015840

+ 12059110158401015840

)

2 120591

Δ=(12059110158401015840

minus 1205911015840

)

2

(17)


minus120587 lt 120591119904119903lt 120587 (18)

0 lt 120591Δlt 120587 (19)

12059110158401015840

= 120591119904119903+ 120591Δ 120591

1015840

= 120591119904119903minus 120591Δ (20)


1198911015840

(120591 119860 120572) = sin 120591 + 119896119860 sin (119896120591 + 120572)

11989110158401015840

(120591 119860 120572) = cos 120591 + 1198962119860 cos (119896120591 + 120572) (21)


sin 120591119904119903sin 120591Δ+ 119860 sin (119896120591

119904119903+ 120572) sin 119896120591

Δ= 0 (22)

sin 120591119904119903cos 120591Δ+ 119896119860 sin (119896120591

119904119903+ 120572) cos 119896120591

Δ= 0 (23)

cos 120591119904119903sin 120591Δ+ 119896119860 cos (119896120591

119904119903+ 120572) sin 119896120591

Δ= 0 (24)

cos 120591119904119903cos 120591Δ+ 1198962

119860 cos (119896120591119904119903+ 120572) cos 119896120591

Δgt 0 (25)





+

120572)[119896 sin 120591Δcos 119896120591

Δminus sin 119896120591



119896 sin 120591Δcos 119896120591

Δminus sin 119896120591

Δcos 120591Δ

= 0 (26)


119904119903and 119860 sin(119896120591

119904119903+ 120572)


sin 120591119904119903= 0 sin (119896120591

119904119903+ 120572) = 0 (27)


120591119904119903= 0 (28)


12059110158401015840

= 120591Δ 120591

1015840

= minus120591Δ (29)


119904119903+ 120572) = 0 imply that sin120572 =





Δ) le 0 (30)

From 120591Δ



120572 = 120587 (31)



Δ) le



0 lt 120591Δle120587

119896 (32)


119860 =sin 120591Δ

119896 sin 119896120591Δ

(33)




120591Δrarr0+119860 =

11198962 Consequently 119860 gt 1119896




1198911015840

(120591119889 119860 120572) = 0

11989110158401015840

(120591119889 119860 120572) = 0

(34)






00

00

k = 2 k = 3

k = 4 k = 5

Acos 120572

Acos 120572

Acos 120572

Acos 120572

A sin 120572

A sin 120572A sin 120572

A sin 120572









119879119896(120591) = 1 + 119886

1cos 120591 + 119887

1sin 120591 + 119886

119896cos 119896120591 + 119887

119896sin 119896120591 (35)


1205821= radic11988621+ 11988721 (36)

120582119896= radic1198862119896+ 1198872119896 (37)





119879119896(120591) = [1 minus cos (120591 minus 120591

0)] [1 minus 120582

119896119903119896(120591)] (38)

where119903119896(120591) = (119896 minus 1) cos 120585

+ 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(39)

providing that


sin(120585119896)]

minus1

(40)

10038161003816100381610038161205851003816100381610038161003816 le 120587 (41)


0 le 120582119896le 1 (42)



of 120582119896






1198861= minus (1 + 120582

119896cos 120585) cos 120591

0minus 119896120582119896sin 120585 sin 120591

0 (43)

1198871= minus (1 + 120582

119896cos 120585) sin 120591

0+ 119896120582119896sin 120585 cos 120591

0 (44)

119886119896= 120582119896cos (119896120591

0minus 120585) (45)

119887119896= 120582119896sin (119896120591

0minus 120585) (46)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms






3= radic220


1= minus03451 119886

3= 005


3= radic28


1= minus01127 119886

3= 0125


3= radic24


1= 02745 119886

3= 025 and




119879119896(120591) = 1 + 120582

1cos (120591 + 120593

1) + 120582119896cos (119896120591 + 120593

119896) (47)

where1205821ge 0120582

119896ge 0120593

1isin (minus120587 120587] and120593



1198861= 1205821cos1205931 119887

1= minus1205821sin1205931 (48)

119886119896= 120582119896cos120593119896 119887

119896= minus120582119896sin120593119896 (49)


0+ 120593119896) mod2120587 (50)




119896(1205910) = 0


119896(120591) ge 0 and 119879

119896(1205910) = 0

imply that 1198791015840119896(1205910) = 0 From 119879

119896(1205910) = 0 and 1198791015840

119896(1205910) = 0 by


1205821cos (120591

0+ 1205931) = minus (1 + 120582

119896cos 120585)

1205821sin (1205910+ 1205931) = minus119896120582

119896sin 120585

(51)


1) can be rewrit-

ten as

1205821cos (120591 + 120593

1) = 1205821cos (120591

0+ 1205931) cos (120591 minus 120591

0)


0)

(52)


1205821cos (120591 + 120593

1) = minus (1 + 120582

119896cos 120585) cos (120591 minus 120591

0)

+ 119896120582119896sin 120585 sin (120591 minus 120591

0)

(53)


1205910) + 120585) that is

cos (119896120591 + 120593119896) = cos 120585 cos 119896 (120591 minus 120591


0)

(54)


119879119896(120591) = [1 minus cos (120591 minus 120591

0)] [1 + 120582

119896cos 120585]


0)] cos 120585

+ 120582119896[119896 sin (120591 minus 120591

0) minus sin 119896 (120591 minus 120591

0)] sin 120585

(55)




119903119896(120591) = minus cos 120585 + [

1 minus cos 119896 (120591 minus 1205910)


minus [119896 sin (120591 minus 120591

0) minus sin 119896 (120591 minus 120591

0)

1 minus cos (120591 minus 1205910)

] sin 120585

(56)



0) + 120585) we obtain (39)



120582119896max120591

119903119896(120591) le 1 (57)



rewritten as

119903119896(120591) = 119903

119896(1205910minus2120585

119896) + 119902119896(120591) (58)

where

119903119896(1205910minus2120585


sin (120585119896) (59)

119902119896(120591) =

1

1 minus cos (120591 minus 1205910)


119896))

+119896 sin 120585sin (120585119896)

(cos(120591 minus 1205910+120585

119896) minus cos(120585

119896))]

(60)



1015840


0+ 120585119896| le 120587119896 This



0| le 120587 it



120591119902119896(120591) =

119902119896(1205910plusmn 2120587119896) = 0 Since 119903

119896(120591) minus 119902

119896(120591) is constant (see (58))




119896(120591)



119889119903119896(120591)

119889120591= minus119904 (120591) sdot sin(

119896 (120591 minus 1205910)

2+ 120585) (61)

where

119904 (120591) = [sin(119896 (120591 minus 120591

0)

2) cos(

120591 minus 1205910

2)

minus 119896 cos(119896 (120591 minus 120591

0)

2) sin(

120591 minus 1205910

2)]


2)

(62)


119904 (120591) = 2

119896minus1

sum

119899=1


0)

2) (63)


0| le 2120587119896 Therefore 119904(120591) ge 119904(120591

0plusmn


119889119903119896(120591)119889120591 = 0 and |120591minus120591


0)2+

120585) = 0From |120585| le 120587 |120591minus120591

0| le 2120587119896 and sin(119896(120591minus120591

0)2+120585) = 0


0+120585119896| = (2120587minus|120585|)119896




120591119902119896(120591) =


max120591

119903119896(120591) = 119903

119896(1205910minus2120585

119896)


(64)



max120591

119903119896(120591) ge 1 (65)


le


119896le




119896= 1 implies |120585| = 120587 and

119879119896(120591) = 1 minus cos 119896(120591 minus 120591


119896(120591) = 1 minus



119896= 1 (the case



119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(66)

where

119888119899= [sin(120585 minus 119899120585

119896) cos(120585

119896)


119896) sin(120585

119896)]


(67)


sin(120585119896)]

minus1

(68)

0 lt10038161003816100381610038161205851003816100381610038161003816 lt 120587 (69)









119896(120591) = 1 minus cos 119896(120591 minus 120591

0)



119879119896(120591) =

[1 minus cos (120591 minus 1205910)]2

3 (1198962 minus 1)

sdot [119896 (1198962

minus 1)

+ 2

119896minus2

sum

119899=1

(119896 minus 119899) ((119896 minus 119899)2

minus 1) cos 119899 (120591 minus 1205910)]

(70)


can be expressed as

1198792(120591) =

2

3[1 minus cos(120591 minus 120591

0)]2

1198793(120591) =

1

2[1 minus cos(120591 minus 120591

0)]2

[2 + cos (120591 minus 1205910)]

1198794(120591) =

4

15[1 minus cos (120591 minus 120591

0)]2


0)]

(71)



0+2120585119896)] =



119879119896(120591) = 120582

119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]

2

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(72)


119879119896(120591)

= 120582119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]


minus 2

119896minus1

sum

119899=1

sin (120585 minus 119899120585119896)sin (120585119896)

cos 119899(120591 minus 1205910+120585

119896)]

(73)


119888119899= 2


sum

119898=1


) (74)


119888119896minus3

119888119896minus4

and 119888119896minus5


= 2 (75)

119888119896minus3

= 8 cos(120585119896) (76)

119888119896minus4

= 8 + 12 cos(2120585119896) (77)

119888119896minus5

= 24 cos(120585119896) + 16 cos(3120585

119896) (78)



that

1198792(120591) =


2

[2 + cos 120585] (79)


0=



1198793(120591) =


2

[3 cos (1205853) + cos 120585]


0+120585

3)]

(80)


120582119896= [(119896 minus 1) cos 120585 + 119896

119896minus1

sum

119899=1

cos((119896 minus 2119899)120585119896

)]

minus1

(81)





1

1198962 minus 1lt 120582119896lt 1 (82)




119879119896(120591) = 1 minus 120582

119896

119896 sin 120585sin (120585119896)

cos(120591 minus 1205910+120585

119896)

+ 120582119896cos (119896 (120591 minus 120591

0) + 120585)

(83)


1

08

06

04

02


Am

plitu

de120582k

1


k = 2k = 3

k = 4k = 5




1198861= minus1205821cos(120591

0minus120585

119896) 119887

1= minus1205821sin(120591

0minus120585

119896) (84)


1205821=

119896 sin 120585sin (120585119896)

120582119896 (85)


120582119896= [cos(120585

119896)

119896 sin 120585sin(120585119896)


(86)


119909 = cos(120585119896) (87)


1205821= 119896120582119896119880119896minus1

(119909) (88)

120582119896=

1

119896119909119880119896minus1

(119909) minus 119881119896(119909)

(89)

where119881119896(119909) and119880



120582119896[119896119909119880119896minus1

(119909) minus 119881119896(119909)] minus 1 = 0 (90)


cos(120587119896) lt 119909 lt 1 (91)





1of


119909 = radic1 minus 1205822

21205822

1

3lt 1205822le 1

119909 =1

23radic1205823

1

8lt 1205823lt 1

119909 = radic1

6(1 + radic

51205824+ 3

21205824

)1

15lt 1205824lt 1

(92)



119896of


1205821= radic8120582

2(1 minus 120582

2)

1

3lt 1205822le 1 (93)

1205821= 3 (

3radic1205823minus 1205823)

1

8lt 1205823lt 1 (94)

1205821= radic

32

27(radic2120582

4(3 + 5120582

4)3

minus 21205824(9 + 7120582

4))

1

15lt 1205824lt 1

(95)




120582119896= [max120591

119903119896(120591)]minus1

(96)

and max120591119903119896(120591) = 119903

119896(1205910) According to (64) max

120591119903119896(120591) =



119879119896(120591) =


120591119903119896(120591) minus 119903

119896(120591)]

max120591119903119896(120591)

(97)



119896(120591) according to (64) and (39)

equals

max120591

119903119896(120591) minus 119903

119896(120591) = 119896

sin ((119896 minus 1) 120585119896)sin (120585119896)

minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(98)


max120591

119903119896(120591) minus 119903

119896(120591) = [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(99)




1198880minus 1198881cos(120585

119896) = 119896

sin (120585 minus 120585119896)sin (120585119896)

(100)


(119888119899minus1

+ 119888119899+1

) cos(120585119896) minus 2119888

119899= 2 (119896 minus 119899) cos(120585 minus 119899120585

119896)

(119888119899minus1

minus 119888119899+1


119896)

(101)


cos(120591 minus 1205910+2120585

119896) = cos(120585

119896) cos(120591 minus 120591

0+120585

119896)


0+120585

119896)

cos(120585 minus 119899120585

119896) cos(119899(120591 minus 120591

0+120585

119896))

minus sin(120585 minus 119899120585

119896) sin(119899(120591 minus 120591

0+120585

119896))

= cos (119899 (120591 minus 1205910) + 120585)

(102)







1

1198962 minus 1lt radic1198862119896+ 1198872119896lt 1 (103)


119896= radic1198862







1198961205910minus 120585 = atan 2 (119887

119896 119886119896) + 2119902120587 119902 = 1 (119896 minus 1) (104)


atan 2 (119910 119909) =

arctan(119910

119909) if 119909 ge 0

arctan(119910

119909) + 120587 if 119909 lt 0 119910 ge 0

arctan(119910

119909) minus 120587 if 119909 lt 0 119910 lt 0

(105)


1198861= minus1205821cos[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

1198871= minus1205821sin[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

(106)


1205910=1

119896[120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

1205910minus2120585

119896=1

119896[minus120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

(107)






119896



2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0

q = 1

q = 2





1according to (88)




1from 120582

119896and






3= minus02



1198861= minus08432 and 119887








4= minus02


1= 09861 (according to (95))






3

2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0q = 1

q = 2q = 3






following form

119879119896(120591) = [1 minus cos (120591 minus 120591

0)]


119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(108)

if 0 le 120582119896le 1(119896

2

minus 1) or

119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(109)


119899 119899 = 0 119896 minus 2 and


|120585| le 120587


1198861= minus (1 + 120582

119896) cos 120591

0 119887

1= minus (1 + 120582

119896) sin 120591

0

119886119896= 120582119896cos (119896120591

0) 119887

119896= 120582119896sin (119896120591

0)

(110)


1is given by (85)



119896lt 1


14

12

1

08

06

04

02

00 05 1

Amplitude 120582k

Am

plitu

de1205821

k = 2

k = 3

k = 4



119896= 1 have 119896 zeros


1205821= 1 + 120582

119896 (111)

if 0 le 120582119896le 1(119896

2

minus 1) or

1205821=


(112)


119896via (68) (or



119896= 1(119896

2


119896le 1(119896

2

minus 1) and 1(1198962

minus 1) le 120582119896le

1 According to (111) 120582119896= 1(119896

2


1198962

(1198962




1205821max =

1

cos (120587 (2119896)) (113)


119896=


at 1205910and 1205910+ 120587119896 for 120585 = minus1205872

14

12

1

08

06

04

02


Am

plitu

de1205821


k = 2k = 3k = 4



cos( 120587

2119896) lt 1 minus

1

1198962 (114)







119896







1is equal to 119896

2

(1198962








1= 1radic3 119886

3= 0


0= 1205876 (119886


1= 0

1198863= radic39 and 119887


0= 1205873 (119886

1= minus1

1198871= minus1radic3 119886

3= 0 and 119887

3= radic39)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

1205910 = 01205910 = 12058761205910 = 1205873





1205821= radic(1 + 120582

119896cos 120585)2 + 11989621205822

119896sin2120585 (115)







implies 120582119896


= 0Hence 119889120582

1119889120585 = 0 implies

2120582119896sin 120585 [1 minus (1198962 minus 1) 120582

119896cos 120585] = 0 (116)

Therefore 1198891205821119889120585 = 0 if 120582

119896= 0 (Option 1) or sin 120585 = 0



1= 1 (notice





1is



2

minus 1)


2

minus 1)120582119896cos 120585 = 1 and 120582

119896lt [(119896 minus






2


119896lt 1(119896

2

minus 1)







2

minus



1(1198962


119879119896(120591) =

[1 minus cos (120591 minus 1205910)]

(1 minus 1198962)

sdot [119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(117)


=

1(1 minus 1198962


2













119896(120591) of type (35) with


1le 0 by shifting



119896(120591) by


1




1and 119886119896with minus119886

119896


119886119896= 120582119896cos 120575 119887

119896= 120582119896sin 120575 (118)


where

|120575| le 120587 (119)


119896 120575 can be

determined as

120575 = atan 2 (119887119896 119886119896) (120)


proposition




119879119896(120591)

= [1 minus cos 120591]


119896minus1

sum

119899=1

(119896 minus 119899) (119886119896cos 119899120591 + 119887

119896sin 119899120591)]

(121)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1+119886

119896 where 120575 = atan 2(bk

119886119896) or

119879119896(120591) = 120582

119896[1 minus cos(120591 minus (120575 + 120585)

119896)]


119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120575

119896)]

(122)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 where 119888

119899 119899 = 0





119896cos 120575 with

119886119896(see (118)) and 120582

119896cos(119899120591 minus 120575) with 119886

119896cos 119899120591 + 119887

119896sin 119899120591



1198861= minus (1 + 119886

119896) 119887

1= minus119896119887

119896 (123)



1205910minus 120585119896 = 120575119896 in (84) leads to

1198861= minus1205821cos(120575

119896) 119887

1= minus1205821sin(120575

119896) (124)




119896and 119887119896satisfy





119896 119887119896satisfying 1 + 119886

119896=

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) is

120582119896=


(125)


(118)) into 1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896)


119910 = cos(120575119896) (126)



119886119896= 120582119896119881119896(119910) (127)

120582119896=

1

119896119910119880119896minus1

(119910) minus 119881119896(119910)

(128)

where119881119896(119910) and119880



119886119896119896119910119880119896minus1

(119910) minus (1 + 119886119896) 119881119896(119910) = 0 (129)


cos(120587119896) le 119910 le 1 (130)



119896= (119896 minus 1) cos 120575 +

119896sum119896minus1


119896= 120582119896cos 120575 can

be rewritten as

119886119896=


119899=1cos ((119896 minus 2119899) 120575119896)

(131)




119896is decreasing







1

05

0

minus05

minus1


Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4


119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4


119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8


4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)


119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3


8 (134)

1205824=1



4+ 1011988624) minus1 le 119886

4le

1

15

(135)





100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)





1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)




1|


119896 From (124)








1le 0


119896(120591) ge 0

and119879119896(120591) = 0 for some 120591






119896(1205910) = 0 and

11987910158401015840


119896(120591) at


119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840


1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)







0 First


0 taking into



1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)




119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










00

00

k = 2 k = 3

k = 4 k = 5

Acos 120572

Acos 120572

Acos 120572

Acos 120572

A sin 120572

A sin 120572A sin 120572

A sin 120572









119879119896(120591) = 1 + 119886

1cos 120591 + 119887

1sin 120591 + 119886

119896cos 119896120591 + 119887

119896sin 119896120591 (35)


1205821= radic11988621+ 11988721 (36)

120582119896= radic1198862119896+ 1198872119896 (37)





119879119896(120591) = [1 minus cos (120591 minus 120591

0)] [1 minus 120582

119896119903119896(120591)] (38)

where119903119896(120591) = (119896 minus 1) cos 120585

+ 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(39)

providing that


sin(120585119896)]

minus1

(40)

10038161003816100381610038161205851003816100381610038161003816 le 120587 (41)


0 le 120582119896le 1 (42)



of 120582119896






1198861= minus (1 + 120582

119896cos 120585) cos 120591

0minus 119896120582119896sin 120585 sin 120591

0 (43)

1198871= minus (1 + 120582

119896cos 120585) sin 120591

0+ 119896120582119896sin 120585 cos 120591

0 (44)

119886119896= 120582119896cos (119896120591

0minus 120585) (45)

119887119896= 120582119896sin (119896120591

0minus 120585) (46)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms






3= radic220


1= minus03451 119886

3= 005


3= radic28


1= minus01127 119886

3= 0125


3= radic24


1= 02745 119886

3= 025 and




119879119896(120591) = 1 + 120582

1cos (120591 + 120593

1) + 120582119896cos (119896120591 + 120593

119896) (47)

where1205821ge 0120582

119896ge 0120593

1isin (minus120587 120587] and120593



1198861= 1205821cos1205931 119887

1= minus1205821sin1205931 (48)

119886119896= 120582119896cos120593119896 119887

119896= minus120582119896sin120593119896 (49)


0+ 120593119896) mod2120587 (50)




119896(1205910) = 0


119896(120591) ge 0 and 119879

119896(1205910) = 0

imply that 1198791015840119896(1205910) = 0 From 119879

119896(1205910) = 0 and 1198791015840

119896(1205910) = 0 by


1205821cos (120591

0+ 1205931) = minus (1 + 120582

119896cos 120585)

1205821sin (1205910+ 1205931) = minus119896120582

119896sin 120585

(51)


1) can be rewrit-

ten as

1205821cos (120591 + 120593

1) = 1205821cos (120591

0+ 1205931) cos (120591 minus 120591

0)


0)

(52)


1205821cos (120591 + 120593

1) = minus (1 + 120582

119896cos 120585) cos (120591 minus 120591

0)

+ 119896120582119896sin 120585 sin (120591 minus 120591

0)

(53)


1205910) + 120585) that is

cos (119896120591 + 120593119896) = cos 120585 cos 119896 (120591 minus 120591


0)

(54)


119879119896(120591) = [1 minus cos (120591 minus 120591

0)] [1 + 120582

119896cos 120585]


0)] cos 120585

+ 120582119896[119896 sin (120591 minus 120591

0) minus sin 119896 (120591 minus 120591

0)] sin 120585

(55)




119903119896(120591) = minus cos 120585 + [

1 minus cos 119896 (120591 minus 1205910)


minus [119896 sin (120591 minus 120591

0) minus sin 119896 (120591 minus 120591

0)

1 minus cos (120591 minus 1205910)

] sin 120585

(56)



0) + 120585) we obtain (39)



120582119896max120591

119903119896(120591) le 1 (57)



rewritten as

119903119896(120591) = 119903

119896(1205910minus2120585

119896) + 119902119896(120591) (58)

where

119903119896(1205910minus2120585


sin (120585119896) (59)

119902119896(120591) =

1

1 minus cos (120591 minus 1205910)


119896))

+119896 sin 120585sin (120585119896)

(cos(120591 minus 1205910+120585

119896) minus cos(120585

119896))]

(60)



1015840


0+ 120585119896| le 120587119896 This



0| le 120587 it



120591119902119896(120591) =

119902119896(1205910plusmn 2120587119896) = 0 Since 119903

119896(120591) minus 119902

119896(120591) is constant (see (58))




119896(120591)



119889119903119896(120591)

119889120591= minus119904 (120591) sdot sin(

119896 (120591 minus 1205910)

2+ 120585) (61)

where

119904 (120591) = [sin(119896 (120591 minus 120591

0)

2) cos(

120591 minus 1205910

2)

minus 119896 cos(119896 (120591 minus 120591

0)

2) sin(

120591 minus 1205910

2)]


2)

(62)


119904 (120591) = 2

119896minus1

sum

119899=1


0)

2) (63)


0| le 2120587119896 Therefore 119904(120591) ge 119904(120591

0plusmn


119889119903119896(120591)119889120591 = 0 and |120591minus120591


0)2+

120585) = 0From |120585| le 120587 |120591minus120591

0| le 2120587119896 and sin(119896(120591minus120591

0)2+120585) = 0


0+120585119896| = (2120587minus|120585|)119896




120591119902119896(120591) =


max120591

119903119896(120591) = 119903

119896(1205910minus2120585

119896)


(64)



max120591

119903119896(120591) ge 1 (65)


le


119896le




119896= 1 implies |120585| = 120587 and

119879119896(120591) = 1 minus cos 119896(120591 minus 120591


119896(120591) = 1 minus



119896= 1 (the case



119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(66)

where

119888119899= [sin(120585 minus 119899120585

119896) cos(120585

119896)


119896) sin(120585

119896)]


(67)


sin(120585119896)]

minus1

(68)

0 lt10038161003816100381610038161205851003816100381610038161003816 lt 120587 (69)









119896(120591) = 1 minus cos 119896(120591 minus 120591

0)



119879119896(120591) =

[1 minus cos (120591 minus 1205910)]2

3 (1198962 minus 1)

sdot [119896 (1198962

minus 1)

+ 2

119896minus2

sum

119899=1

(119896 minus 119899) ((119896 minus 119899)2

minus 1) cos 119899 (120591 minus 1205910)]

(70)


can be expressed as

1198792(120591) =

2

3[1 minus cos(120591 minus 120591

0)]2

1198793(120591) =

1

2[1 minus cos(120591 minus 120591

0)]2

[2 + cos (120591 minus 1205910)]

1198794(120591) =

4

15[1 minus cos (120591 minus 120591

0)]2


0)]

(71)



0+2120585119896)] =



119879119896(120591) = 120582

119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]

2

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(72)


119879119896(120591)

= 120582119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]


minus 2

119896minus1

sum

119899=1

sin (120585 minus 119899120585119896)sin (120585119896)

cos 119899(120591 minus 1205910+120585

119896)]

(73)


119888119899= 2


sum

119898=1


) (74)


119888119896minus3

119888119896minus4

and 119888119896minus5


= 2 (75)

119888119896minus3

= 8 cos(120585119896) (76)

119888119896minus4

= 8 + 12 cos(2120585119896) (77)

119888119896minus5

= 24 cos(120585119896) + 16 cos(3120585

119896) (78)



that

1198792(120591) =


2

[2 + cos 120585] (79)


0=



1198793(120591) =


2

[3 cos (1205853) + cos 120585]


0+120585

3)]

(80)


120582119896= [(119896 minus 1) cos 120585 + 119896

119896minus1

sum

119899=1

cos((119896 minus 2119899)120585119896

)]

minus1

(81)





1

1198962 minus 1lt 120582119896lt 1 (82)




119879119896(120591) = 1 minus 120582

119896

119896 sin 120585sin (120585119896)

cos(120591 minus 1205910+120585

119896)

+ 120582119896cos (119896 (120591 minus 120591

0) + 120585)

(83)


1

08

06

04

02


Am

plitu

de120582k

1


k = 2k = 3

k = 4k = 5




1198861= minus1205821cos(120591

0minus120585

119896) 119887

1= minus1205821sin(120591

0minus120585

119896) (84)


1205821=

119896 sin 120585sin (120585119896)

120582119896 (85)


120582119896= [cos(120585

119896)

119896 sin 120585sin(120585119896)


(86)


119909 = cos(120585119896) (87)


1205821= 119896120582119896119880119896minus1

(119909) (88)

120582119896=

1

119896119909119880119896minus1

(119909) minus 119881119896(119909)

(89)

where119881119896(119909) and119880



120582119896[119896119909119880119896minus1

(119909) minus 119881119896(119909)] minus 1 = 0 (90)


cos(120587119896) lt 119909 lt 1 (91)





1of


119909 = radic1 minus 1205822

21205822

1

3lt 1205822le 1

119909 =1

23radic1205823

1

8lt 1205823lt 1

119909 = radic1

6(1 + radic

51205824+ 3

21205824

)1

15lt 1205824lt 1

(92)



119896of


1205821= radic8120582

2(1 minus 120582

2)

1

3lt 1205822le 1 (93)

1205821= 3 (

3radic1205823minus 1205823)

1

8lt 1205823lt 1 (94)

1205821= radic

32

27(radic2120582

4(3 + 5120582

4)3

minus 21205824(9 + 7120582

4))

1

15lt 1205824lt 1

(95)




120582119896= [max120591

119903119896(120591)]minus1

(96)

and max120591119903119896(120591) = 119903

119896(1205910) According to (64) max

120591119903119896(120591) =



119879119896(120591) =


120591119903119896(120591) minus 119903

119896(120591)]

max120591119903119896(120591)

(97)



119896(120591) according to (64) and (39)

equals

max120591

119903119896(120591) minus 119903

119896(120591) = 119896

sin ((119896 minus 1) 120585119896)sin (120585119896)

minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(98)


max120591

119903119896(120591) minus 119903

119896(120591) = [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(99)




1198880minus 1198881cos(120585

119896) = 119896

sin (120585 minus 120585119896)sin (120585119896)

(100)


(119888119899minus1

+ 119888119899+1

) cos(120585119896) minus 2119888

119899= 2 (119896 minus 119899) cos(120585 minus 119899120585

119896)

(119888119899minus1

minus 119888119899+1


119896)

(101)


cos(120591 minus 1205910+2120585

119896) = cos(120585

119896) cos(120591 minus 120591

0+120585

119896)


0+120585

119896)

cos(120585 minus 119899120585

119896) cos(119899(120591 minus 120591

0+120585

119896))

minus sin(120585 minus 119899120585

119896) sin(119899(120591 minus 120591

0+120585

119896))

= cos (119899 (120591 minus 1205910) + 120585)

(102)







1

1198962 minus 1lt radic1198862119896+ 1198872119896lt 1 (103)


119896= radic1198862







1198961205910minus 120585 = atan 2 (119887

119896 119886119896) + 2119902120587 119902 = 1 (119896 minus 1) (104)


atan 2 (119910 119909) =

arctan(119910

119909) if 119909 ge 0

arctan(119910

119909) + 120587 if 119909 lt 0 119910 ge 0

arctan(119910

119909) minus 120587 if 119909 lt 0 119910 lt 0

(105)


1198861= minus1205821cos[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

1198871= minus1205821sin[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

(106)


1205910=1

119896[120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

1205910minus2120585

119896=1

119896[minus120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

(107)






119896



2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0

q = 1

q = 2





1according to (88)




1from 120582

119896and






3= minus02



1198861= minus08432 and 119887








4= minus02


1= 09861 (according to (95))






3

2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0q = 1

q = 2q = 3






following form

119879119896(120591) = [1 minus cos (120591 minus 120591

0)]


119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(108)

if 0 le 120582119896le 1(119896

2

minus 1) or

119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(109)


119899 119899 = 0 119896 minus 2 and


|120585| le 120587


1198861= minus (1 + 120582

119896) cos 120591

0 119887

1= minus (1 + 120582

119896) sin 120591

0

119886119896= 120582119896cos (119896120591

0) 119887

119896= 120582119896sin (119896120591

0)

(110)


1is given by (85)



119896lt 1


14

12

1

08

06

04

02

00 05 1

Amplitude 120582k

Am

plitu

de1205821

k = 2

k = 3

k = 4



119896= 1 have 119896 zeros


1205821= 1 + 120582

119896 (111)

if 0 le 120582119896le 1(119896

2

minus 1) or

1205821=


(112)


119896via (68) (or



119896= 1(119896

2


119896le 1(119896

2

minus 1) and 1(1198962

minus 1) le 120582119896le

1 According to (111) 120582119896= 1(119896

2


1198962

(1198962




1205821max =

1

cos (120587 (2119896)) (113)


119896=


at 1205910and 1205910+ 120587119896 for 120585 = minus1205872

14

12

1

08

06

04

02


Am

plitu

de1205821


k = 2k = 3k = 4



cos( 120587

2119896) lt 1 minus

1

1198962 (114)







119896







1is equal to 119896

2

(1198962








1= 1radic3 119886

3= 0


0= 1205876 (119886


1= 0

1198863= radic39 and 119887


0= 1205873 (119886

1= minus1

1198871= minus1radic3 119886

3= 0 and 119887

3= radic39)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

1205910 = 01205910 = 12058761205910 = 1205873





1205821= radic(1 + 120582

119896cos 120585)2 + 11989621205822

119896sin2120585 (115)







implies 120582119896


= 0Hence 119889120582

1119889120585 = 0 implies

2120582119896sin 120585 [1 minus (1198962 minus 1) 120582

119896cos 120585] = 0 (116)

Therefore 1198891205821119889120585 = 0 if 120582

119896= 0 (Option 1) or sin 120585 = 0



1= 1 (notice





1is



2

minus 1)


2

minus 1)120582119896cos 120585 = 1 and 120582

119896lt [(119896 minus






2


119896lt 1(119896

2

minus 1)







2

minus



1(1198962


119879119896(120591) =

[1 minus cos (120591 minus 1205910)]

(1 minus 1198962)

sdot [119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(117)


=

1(1 minus 1198962


2













119896(120591) of type (35) with


1le 0 by shifting



119896(120591) by


1




1and 119886119896with minus119886

119896


119886119896= 120582119896cos 120575 119887

119896= 120582119896sin 120575 (118)


where

|120575| le 120587 (119)


119896 120575 can be

determined as

120575 = atan 2 (119887119896 119886119896) (120)


proposition




119879119896(120591)

= [1 minus cos 120591]


119896minus1

sum

119899=1

(119896 minus 119899) (119886119896cos 119899120591 + 119887

119896sin 119899120591)]

(121)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1+119886

119896 where 120575 = atan 2(bk

119886119896) or

119879119896(120591) = 120582

119896[1 minus cos(120591 minus (120575 + 120585)

119896)]


119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120575

119896)]

(122)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 where 119888

119899 119899 = 0





119896cos 120575 with

119886119896(see (118)) and 120582

119896cos(119899120591 minus 120575) with 119886

119896cos 119899120591 + 119887

119896sin 119899120591



1198861= minus (1 + 119886

119896) 119887

1= minus119896119887

119896 (123)



1205910minus 120585119896 = 120575119896 in (84) leads to

1198861= minus1205821cos(120575

119896) 119887

1= minus1205821sin(120575

119896) (124)




119896and 119887119896satisfy





119896 119887119896satisfying 1 + 119886

119896=

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) is

120582119896=


(125)


(118)) into 1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896)


119910 = cos(120575119896) (126)



119886119896= 120582119896119881119896(119910) (127)

120582119896=

1

119896119910119880119896minus1

(119910) minus 119881119896(119910)

(128)

where119881119896(119910) and119880



119886119896119896119910119880119896minus1

(119910) minus (1 + 119886119896) 119881119896(119910) = 0 (129)


cos(120587119896) le 119910 le 1 (130)



119896= (119896 minus 1) cos 120575 +

119896sum119896minus1


119896= 120582119896cos 120575 can

be rewritten as

119886119896=


119899=1cos ((119896 minus 2119899) 120575119896)

(131)




119896is decreasing







1

05

0

minus05

minus1


Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4


119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4


119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8


4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)


119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3


8 (134)

1205824=1



4+ 1011988624) minus1 le 119886

4le

1

15

(135)





100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)





1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)




1|


119896 From (124)








1le 0


119896(120591) ge 0

and119879119896(120591) = 0 for some 120591






119896(1205910) = 0 and

11987910158401015840


119896(120591) at


119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840


1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)







0 First


0 taking into



1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)




119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms






3= radic220


1= minus03451 119886

3= 005


3= radic28


1= minus01127 119886

3= 0125


3= radic24


1= 02745 119886

3= 025 and




119879119896(120591) = 1 + 120582

1cos (120591 + 120593

1) + 120582119896cos (119896120591 + 120593

119896) (47)

where1205821ge 0120582

119896ge 0120593

1isin (minus120587 120587] and120593



1198861= 1205821cos1205931 119887

1= minus1205821sin1205931 (48)

119886119896= 120582119896cos120593119896 119887

119896= minus120582119896sin120593119896 (49)


0+ 120593119896) mod2120587 (50)




119896(1205910) = 0


119896(120591) ge 0 and 119879

119896(1205910) = 0

imply that 1198791015840119896(1205910) = 0 From 119879

119896(1205910) = 0 and 1198791015840

119896(1205910) = 0 by


1205821cos (120591

0+ 1205931) = minus (1 + 120582

119896cos 120585)

1205821sin (1205910+ 1205931) = minus119896120582

119896sin 120585

(51)


1) can be rewrit-

ten as

1205821cos (120591 + 120593

1) = 1205821cos (120591

0+ 1205931) cos (120591 minus 120591

0)


0)

(52)


1205821cos (120591 + 120593

1) = minus (1 + 120582

119896cos 120585) cos (120591 minus 120591

0)

+ 119896120582119896sin 120585 sin (120591 minus 120591

0)

(53)


1205910) + 120585) that is

cos (119896120591 + 120593119896) = cos 120585 cos 119896 (120591 minus 120591


0)

(54)


119879119896(120591) = [1 minus cos (120591 minus 120591

0)] [1 + 120582

119896cos 120585]


0)] cos 120585

+ 120582119896[119896 sin (120591 minus 120591

0) minus sin 119896 (120591 minus 120591

0)] sin 120585

(55)




119903119896(120591) = minus cos 120585 + [

1 minus cos 119896 (120591 minus 1205910)


minus [119896 sin (120591 minus 120591

0) minus sin 119896 (120591 minus 120591

0)

1 minus cos (120591 minus 1205910)

] sin 120585

(56)



0) + 120585) we obtain (39)



120582119896max120591

119903119896(120591) le 1 (57)



rewritten as

119903119896(120591) = 119903

119896(1205910minus2120585

119896) + 119902119896(120591) (58)

where

119903119896(1205910minus2120585


sin (120585119896) (59)

119902119896(120591) =

1

1 minus cos (120591 minus 1205910)


119896))

+119896 sin 120585sin (120585119896)

(cos(120591 minus 1205910+120585

119896) minus cos(120585

119896))]

(60)



1015840


0+ 120585119896| le 120587119896 This



0| le 120587 it



120591119902119896(120591) =

119902119896(1205910plusmn 2120587119896) = 0 Since 119903

119896(120591) minus 119902

119896(120591) is constant (see (58))




119896(120591)



119889119903119896(120591)

119889120591= minus119904 (120591) sdot sin(

119896 (120591 minus 1205910)

2+ 120585) (61)

where

119904 (120591) = [sin(119896 (120591 minus 120591

0)

2) cos(

120591 minus 1205910

2)

minus 119896 cos(119896 (120591 minus 120591

0)

2) sin(

120591 minus 1205910

2)]


2)

(62)


119904 (120591) = 2

119896minus1

sum

119899=1


0)

2) (63)


0| le 2120587119896 Therefore 119904(120591) ge 119904(120591

0plusmn


119889119903119896(120591)119889120591 = 0 and |120591minus120591


0)2+

120585) = 0From |120585| le 120587 |120591minus120591

0| le 2120587119896 and sin(119896(120591minus120591

0)2+120585) = 0


0+120585119896| = (2120587minus|120585|)119896




120591119902119896(120591) =


max120591

119903119896(120591) = 119903

119896(1205910minus2120585

119896)


(64)



max120591

119903119896(120591) ge 1 (65)


le


119896le




119896= 1 implies |120585| = 120587 and

119879119896(120591) = 1 minus cos 119896(120591 minus 120591


119896(120591) = 1 minus



119896= 1 (the case



119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(66)

where

119888119899= [sin(120585 minus 119899120585

119896) cos(120585

119896)


119896) sin(120585

119896)]


(67)


sin(120585119896)]

minus1

(68)

0 lt10038161003816100381610038161205851003816100381610038161003816 lt 120587 (69)









119896(120591) = 1 minus cos 119896(120591 minus 120591

0)



119879119896(120591) =

[1 minus cos (120591 minus 1205910)]2

3 (1198962 minus 1)

sdot [119896 (1198962

minus 1)

+ 2

119896minus2

sum

119899=1

(119896 minus 119899) ((119896 minus 119899)2

minus 1) cos 119899 (120591 minus 1205910)]

(70)


can be expressed as

1198792(120591) =

2

3[1 minus cos(120591 minus 120591

0)]2

1198793(120591) =

1

2[1 minus cos(120591 minus 120591

0)]2

[2 + cos (120591 minus 1205910)]

1198794(120591) =

4

15[1 minus cos (120591 minus 120591

0)]2


0)]

(71)



0+2120585119896)] =



119879119896(120591) = 120582

119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]

2

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(72)


119879119896(120591)

= 120582119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]


minus 2

119896minus1

sum

119899=1

sin (120585 minus 119899120585119896)sin (120585119896)

cos 119899(120591 minus 1205910+120585

119896)]

(73)


119888119899= 2


sum

119898=1


) (74)


119888119896minus3

119888119896minus4

and 119888119896minus5


= 2 (75)

119888119896minus3

= 8 cos(120585119896) (76)

119888119896minus4

= 8 + 12 cos(2120585119896) (77)

119888119896minus5

= 24 cos(120585119896) + 16 cos(3120585

119896) (78)



that

1198792(120591) =


2

[2 + cos 120585] (79)


0=



1198793(120591) =


2

[3 cos (1205853) + cos 120585]


0+120585

3)]

(80)


120582119896= [(119896 minus 1) cos 120585 + 119896

119896minus1

sum

119899=1

cos((119896 minus 2119899)120585119896

)]

minus1

(81)





1

1198962 minus 1lt 120582119896lt 1 (82)




119879119896(120591) = 1 minus 120582

119896

119896 sin 120585sin (120585119896)

cos(120591 minus 1205910+120585

119896)

+ 120582119896cos (119896 (120591 minus 120591

0) + 120585)

(83)


1

08

06

04

02


Am

plitu

de120582k

1


k = 2k = 3

k = 4k = 5




1198861= minus1205821cos(120591

0minus120585

119896) 119887

1= minus1205821sin(120591

0minus120585

119896) (84)


1205821=

119896 sin 120585sin (120585119896)

120582119896 (85)


120582119896= [cos(120585

119896)

119896 sin 120585sin(120585119896)


(86)


119909 = cos(120585119896) (87)


1205821= 119896120582119896119880119896minus1

(119909) (88)

120582119896=

1

119896119909119880119896minus1

(119909) minus 119881119896(119909)

(89)

where119881119896(119909) and119880



120582119896[119896119909119880119896minus1

(119909) minus 119881119896(119909)] minus 1 = 0 (90)


cos(120587119896) lt 119909 lt 1 (91)





1of


119909 = radic1 minus 1205822

21205822

1

3lt 1205822le 1

119909 =1

23radic1205823

1

8lt 1205823lt 1

119909 = radic1

6(1 + radic

51205824+ 3

21205824

)1

15lt 1205824lt 1

(92)



119896of


1205821= radic8120582

2(1 minus 120582

2)

1

3lt 1205822le 1 (93)

1205821= 3 (

3radic1205823minus 1205823)

1

8lt 1205823lt 1 (94)

1205821= radic

32

27(radic2120582

4(3 + 5120582

4)3

minus 21205824(9 + 7120582

4))

1

15lt 1205824lt 1

(95)




120582119896= [max120591

119903119896(120591)]minus1

(96)

and max120591119903119896(120591) = 119903

119896(1205910) According to (64) max

120591119903119896(120591) =



119879119896(120591) =


120591119903119896(120591) minus 119903

119896(120591)]

max120591119903119896(120591)

(97)



119896(120591) according to (64) and (39)

equals

max120591

119903119896(120591) minus 119903

119896(120591) = 119896

sin ((119896 minus 1) 120585119896)sin (120585119896)

minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(98)


max120591

119903119896(120591) minus 119903

119896(120591) = [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(99)




1198880minus 1198881cos(120585

119896) = 119896

sin (120585 minus 120585119896)sin (120585119896)

(100)


(119888119899minus1

+ 119888119899+1

) cos(120585119896) minus 2119888

119899= 2 (119896 minus 119899) cos(120585 minus 119899120585

119896)

(119888119899minus1

minus 119888119899+1


119896)

(101)


cos(120591 minus 1205910+2120585

119896) = cos(120585

119896) cos(120591 minus 120591

0+120585

119896)


0+120585

119896)

cos(120585 minus 119899120585

119896) cos(119899(120591 minus 120591

0+120585

119896))

minus sin(120585 minus 119899120585

119896) sin(119899(120591 minus 120591

0+120585

119896))

= cos (119899 (120591 minus 1205910) + 120585)

(102)







1

1198962 minus 1lt radic1198862119896+ 1198872119896lt 1 (103)


119896= radic1198862







1198961205910minus 120585 = atan 2 (119887

119896 119886119896) + 2119902120587 119902 = 1 (119896 minus 1) (104)


atan 2 (119910 119909) =

arctan(119910

119909) if 119909 ge 0

arctan(119910

119909) + 120587 if 119909 lt 0 119910 ge 0

arctan(119910

119909) minus 120587 if 119909 lt 0 119910 lt 0

(105)


1198861= minus1205821cos[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

1198871= minus1205821sin[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

(106)


1205910=1

119896[120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

1205910minus2120585

119896=1

119896[minus120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

(107)






119896



2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0

q = 1

q = 2





1according to (88)




1from 120582

119896and






3= minus02



1198861= minus08432 and 119887








4= minus02


1= 09861 (according to (95))






3

2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0q = 1

q = 2q = 3






following form

119879119896(120591) = [1 minus cos (120591 minus 120591

0)]


119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(108)

if 0 le 120582119896le 1(119896

2

minus 1) or

119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(109)


119899 119899 = 0 119896 minus 2 and


|120585| le 120587


1198861= minus (1 + 120582

119896) cos 120591

0 119887

1= minus (1 + 120582

119896) sin 120591

0

119886119896= 120582119896cos (119896120591

0) 119887

119896= 120582119896sin (119896120591

0)

(110)


1is given by (85)



119896lt 1


14

12

1

08

06

04

02

00 05 1

Amplitude 120582k

Am

plitu

de1205821

k = 2

k = 3

k = 4



119896= 1 have 119896 zeros


1205821= 1 + 120582

119896 (111)

if 0 le 120582119896le 1(119896

2

minus 1) or

1205821=


(112)


119896via (68) (or



119896= 1(119896

2


119896le 1(119896

2

minus 1) and 1(1198962

minus 1) le 120582119896le

1 According to (111) 120582119896= 1(119896

2


1198962

(1198962




1205821max =

1

cos (120587 (2119896)) (113)


119896=


at 1205910and 1205910+ 120587119896 for 120585 = minus1205872

14

12

1

08

06

04

02


Am

plitu

de1205821


k = 2k = 3k = 4



cos( 120587

2119896) lt 1 minus

1

1198962 (114)







119896







1is equal to 119896

2

(1198962








1= 1radic3 119886

3= 0


0= 1205876 (119886


1= 0

1198863= radic39 and 119887


0= 1205873 (119886

1= minus1

1198871= minus1radic3 119886

3= 0 and 119887

3= radic39)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

1205910 = 01205910 = 12058761205910 = 1205873





1205821= radic(1 + 120582

119896cos 120585)2 + 11989621205822

119896sin2120585 (115)







implies 120582119896


= 0Hence 119889120582

1119889120585 = 0 implies

2120582119896sin 120585 [1 minus (1198962 minus 1) 120582

119896cos 120585] = 0 (116)

Therefore 1198891205821119889120585 = 0 if 120582

119896= 0 (Option 1) or sin 120585 = 0



1= 1 (notice





1is



2

minus 1)


2

minus 1)120582119896cos 120585 = 1 and 120582

119896lt [(119896 minus






2


119896lt 1(119896

2

minus 1)







2

minus



1(1198962


119879119896(120591) =

[1 minus cos (120591 minus 1205910)]

(1 minus 1198962)

sdot [119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(117)


=

1(1 minus 1198962


2













119896(120591) of type (35) with


1le 0 by shifting



119896(120591) by


1




1and 119886119896with minus119886

119896


119886119896= 120582119896cos 120575 119887

119896= 120582119896sin 120575 (118)


where

|120575| le 120587 (119)


119896 120575 can be

determined as

120575 = atan 2 (119887119896 119886119896) (120)


proposition




119879119896(120591)

= [1 minus cos 120591]


119896minus1

sum

119899=1

(119896 minus 119899) (119886119896cos 119899120591 + 119887

119896sin 119899120591)]

(121)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1+119886

119896 where 120575 = atan 2(bk

119886119896) or

119879119896(120591) = 120582

119896[1 minus cos(120591 minus (120575 + 120585)

119896)]


119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120575

119896)]

(122)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 where 119888

119899 119899 = 0





119896cos 120575 with

119886119896(see (118)) and 120582

119896cos(119899120591 minus 120575) with 119886

119896cos 119899120591 + 119887

119896sin 119899120591



1198861= minus (1 + 119886

119896) 119887

1= minus119896119887

119896 (123)



1205910minus 120585119896 = 120575119896 in (84) leads to

1198861= minus1205821cos(120575

119896) 119887

1= minus1205821sin(120575

119896) (124)




119896and 119887119896satisfy





119896 119887119896satisfying 1 + 119886

119896=

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) is

120582119896=


(125)


(118)) into 1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896)


119910 = cos(120575119896) (126)



119886119896= 120582119896119881119896(119910) (127)

120582119896=

1

119896119910119880119896minus1

(119910) minus 119881119896(119910)

(128)

where119881119896(119910) and119880



119886119896119896119910119880119896minus1

(119910) minus (1 + 119886119896) 119881119896(119910) = 0 (129)


cos(120587119896) le 119910 le 1 (130)



119896= (119896 minus 1) cos 120575 +

119896sum119896minus1


119896= 120582119896cos 120575 can

be rewritten as

119886119896=


119899=1cos ((119896 minus 2119899) 120575119896)

(131)




119896is decreasing







1

05

0

minus05

minus1


Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4


119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4


119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8


4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)


119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3


8 (134)

1205824=1



4+ 1011988624) minus1 le 119886

4le

1

15

(135)





100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)





1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)




1|


119896 From (124)








1le 0


119896(120591) ge 0

and119879119896(120591) = 0 for some 120591






119896(1205910) = 0 and

11987910158401015840


119896(120591) at


119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840


1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)







0 First


0 taking into



1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)




119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











1015840


0+ 120585119896| le 120587119896 This



0| le 120587 it



120591119902119896(120591) =

119902119896(1205910plusmn 2120587119896) = 0 Since 119903

119896(120591) minus 119902

119896(120591) is constant (see (58))




119896(120591)



119889119903119896(120591)

119889120591= minus119904 (120591) sdot sin(

119896 (120591 minus 1205910)

2+ 120585) (61)

where

119904 (120591) = [sin(119896 (120591 minus 120591

0)

2) cos(

120591 minus 1205910

2)

minus 119896 cos(119896 (120591 minus 120591

0)

2) sin(

120591 minus 1205910

2)]


2)

(62)


119904 (120591) = 2

119896minus1

sum

119899=1


0)

2) (63)


0| le 2120587119896 Therefore 119904(120591) ge 119904(120591

0plusmn


119889119903119896(120591)119889120591 = 0 and |120591minus120591


0)2+

120585) = 0From |120585| le 120587 |120591minus120591

0| le 2120587119896 and sin(119896(120591minus120591

0)2+120585) = 0


0+120585119896| = (2120587minus|120585|)119896




120591119902119896(120591) =


max120591

119903119896(120591) = 119903

119896(1205910minus2120585

119896)


(64)



max120591

119903119896(120591) ge 1 (65)


le


119896le




119896= 1 implies |120585| = 120587 and

119879119896(120591) = 1 minus cos 119896(120591 minus 120591


119896(120591) = 1 minus



119896= 1 (the case



119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(66)

where

119888119899= [sin(120585 minus 119899120585

119896) cos(120585

119896)


119896) sin(120585

119896)]


(67)


sin(120585119896)]

minus1

(68)

0 lt10038161003816100381610038161205851003816100381610038161003816 lt 120587 (69)









119896(120591) = 1 minus cos 119896(120591 minus 120591

0)



119879119896(120591) =

[1 minus cos (120591 minus 1205910)]2

3 (1198962 minus 1)

sdot [119896 (1198962

minus 1)

+ 2

119896minus2

sum

119899=1

(119896 minus 119899) ((119896 minus 119899)2

minus 1) cos 119899 (120591 minus 1205910)]

(70)


can be expressed as

1198792(120591) =

2

3[1 minus cos(120591 minus 120591

0)]2

1198793(120591) =

1

2[1 minus cos(120591 minus 120591

0)]2

[2 + cos (120591 minus 1205910)]

1198794(120591) =

4

15[1 minus cos (120591 minus 120591

0)]2


0)]

(71)



0+2120585119896)] =



119879119896(120591) = 120582

119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]

2

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(72)


119879119896(120591)

= 120582119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]


minus 2

119896minus1

sum

119899=1

sin (120585 minus 119899120585119896)sin (120585119896)

cos 119899(120591 minus 1205910+120585

119896)]

(73)


119888119899= 2


sum

119898=1


) (74)


119888119896minus3

119888119896minus4

and 119888119896minus5


= 2 (75)

119888119896minus3

= 8 cos(120585119896) (76)

119888119896minus4

= 8 + 12 cos(2120585119896) (77)

119888119896minus5

= 24 cos(120585119896) + 16 cos(3120585

119896) (78)



that

1198792(120591) =


2

[2 + cos 120585] (79)


0=



1198793(120591) =


2

[3 cos (1205853) + cos 120585]


0+120585

3)]

(80)


120582119896= [(119896 minus 1) cos 120585 + 119896

119896minus1

sum

119899=1

cos((119896 minus 2119899)120585119896

)]

minus1

(81)





1

1198962 minus 1lt 120582119896lt 1 (82)




119879119896(120591) = 1 minus 120582

119896

119896 sin 120585sin (120585119896)

cos(120591 minus 1205910+120585

119896)

+ 120582119896cos (119896 (120591 minus 120591

0) + 120585)

(83)


1

08

06

04

02


Am

plitu

de120582k

1


k = 2k = 3

k = 4k = 5




1198861= minus1205821cos(120591

0minus120585

119896) 119887

1= minus1205821sin(120591

0minus120585

119896) (84)


1205821=

119896 sin 120585sin (120585119896)

120582119896 (85)


120582119896= [cos(120585

119896)

119896 sin 120585sin(120585119896)


(86)


119909 = cos(120585119896) (87)


1205821= 119896120582119896119880119896minus1

(119909) (88)

120582119896=

1

119896119909119880119896minus1

(119909) minus 119881119896(119909)

(89)

where119881119896(119909) and119880



120582119896[119896119909119880119896minus1

(119909) minus 119881119896(119909)] minus 1 = 0 (90)


cos(120587119896) lt 119909 lt 1 (91)





1of


119909 = radic1 minus 1205822

21205822

1

3lt 1205822le 1

119909 =1

23radic1205823

1

8lt 1205823lt 1

119909 = radic1

6(1 + radic

51205824+ 3

21205824

)1

15lt 1205824lt 1

(92)



119896of


1205821= radic8120582

2(1 minus 120582

2)

1

3lt 1205822le 1 (93)

1205821= 3 (

3radic1205823minus 1205823)

1

8lt 1205823lt 1 (94)

1205821= radic

32

27(radic2120582

4(3 + 5120582

4)3

minus 21205824(9 + 7120582

4))

1

15lt 1205824lt 1

(95)




120582119896= [max120591

119903119896(120591)]minus1

(96)

and max120591119903119896(120591) = 119903

119896(1205910) According to (64) max

120591119903119896(120591) =



119879119896(120591) =


120591119903119896(120591) minus 119903

119896(120591)]

max120591119903119896(120591)

(97)



119896(120591) according to (64) and (39)

equals

max120591

119903119896(120591) minus 119903

119896(120591) = 119896

sin ((119896 minus 1) 120585119896)sin (120585119896)

minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(98)


max120591

119903119896(120591) minus 119903

119896(120591) = [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(99)




1198880minus 1198881cos(120585

119896) = 119896

sin (120585 minus 120585119896)sin (120585119896)

(100)


(119888119899minus1

+ 119888119899+1

) cos(120585119896) minus 2119888

119899= 2 (119896 minus 119899) cos(120585 minus 119899120585

119896)

(119888119899minus1

minus 119888119899+1


119896)

(101)


cos(120591 minus 1205910+2120585

119896) = cos(120585

119896) cos(120591 minus 120591

0+120585

119896)


0+120585

119896)

cos(120585 minus 119899120585

119896) cos(119899(120591 minus 120591

0+120585

119896))

minus sin(120585 minus 119899120585

119896) sin(119899(120591 minus 120591

0+120585

119896))

= cos (119899 (120591 minus 1205910) + 120585)

(102)







1

1198962 minus 1lt radic1198862119896+ 1198872119896lt 1 (103)


119896= radic1198862







1198961205910minus 120585 = atan 2 (119887

119896 119886119896) + 2119902120587 119902 = 1 (119896 minus 1) (104)


atan 2 (119910 119909) =

arctan(119910

119909) if 119909 ge 0

arctan(119910

119909) + 120587 if 119909 lt 0 119910 ge 0

arctan(119910

119909) minus 120587 if 119909 lt 0 119910 lt 0

(105)


1198861= minus1205821cos[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

1198871= minus1205821sin[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

(106)


1205910=1

119896[120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

1205910minus2120585

119896=1

119896[minus120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

(107)






119896



2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0

q = 1

q = 2





1according to (88)




1from 120582

119896and






3= minus02



1198861= minus08432 and 119887








4= minus02


1= 09861 (according to (95))






3

2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0q = 1

q = 2q = 3






following form

119879119896(120591) = [1 minus cos (120591 minus 120591

0)]


119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(108)

if 0 le 120582119896le 1(119896

2

minus 1) or

119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(109)


119899 119899 = 0 119896 minus 2 and


|120585| le 120587


1198861= minus (1 + 120582

119896) cos 120591

0 119887

1= minus (1 + 120582

119896) sin 120591

0

119886119896= 120582119896cos (119896120591

0) 119887

119896= 120582119896sin (119896120591

0)

(110)


1is given by (85)



119896lt 1


14

12

1

08

06

04

02

00 05 1

Amplitude 120582k

Am

plitu

de1205821

k = 2

k = 3

k = 4



119896= 1 have 119896 zeros


1205821= 1 + 120582

119896 (111)

if 0 le 120582119896le 1(119896

2

minus 1) or

1205821=


(112)


119896via (68) (or



119896= 1(119896

2


119896le 1(119896

2

minus 1) and 1(1198962

minus 1) le 120582119896le

1 According to (111) 120582119896= 1(119896

2


1198962

(1198962




1205821max =

1

cos (120587 (2119896)) (113)


119896=


at 1205910and 1205910+ 120587119896 for 120585 = minus1205872

14

12

1

08

06

04

02


Am

plitu

de1205821


k = 2k = 3k = 4



cos( 120587

2119896) lt 1 minus

1

1198962 (114)







119896







1is equal to 119896

2

(1198962








1= 1radic3 119886

3= 0


0= 1205876 (119886


1= 0

1198863= radic39 and 119887


0= 1205873 (119886

1= minus1

1198871= minus1radic3 119886

3= 0 and 119887

3= radic39)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

1205910 = 01205910 = 12058761205910 = 1205873





1205821= radic(1 + 120582

119896cos 120585)2 + 11989621205822

119896sin2120585 (115)







implies 120582119896


= 0Hence 119889120582

1119889120585 = 0 implies

2120582119896sin 120585 [1 minus (1198962 minus 1) 120582

119896cos 120585] = 0 (116)

Therefore 1198891205821119889120585 = 0 if 120582

119896= 0 (Option 1) or sin 120585 = 0



1= 1 (notice





1is



2

minus 1)


2

minus 1)120582119896cos 120585 = 1 and 120582

119896lt [(119896 minus






2


119896lt 1(119896

2

minus 1)







2

minus



1(1198962


119879119896(120591) =

[1 minus cos (120591 minus 1205910)]

(1 minus 1198962)

sdot [119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(117)


=

1(1 minus 1198962


2













119896(120591) of type (35) with


1le 0 by shifting



119896(120591) by


1




1and 119886119896with minus119886

119896


119886119896= 120582119896cos 120575 119887

119896= 120582119896sin 120575 (118)


where

|120575| le 120587 (119)


119896 120575 can be

determined as

120575 = atan 2 (119887119896 119886119896) (120)


proposition




119879119896(120591)

= [1 minus cos 120591]


119896minus1

sum

119899=1

(119896 minus 119899) (119886119896cos 119899120591 + 119887

119896sin 119899120591)]

(121)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1+119886

119896 where 120575 = atan 2(bk

119886119896) or

119879119896(120591) = 120582

119896[1 minus cos(120591 minus (120575 + 120585)

119896)]


119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120575

119896)]

(122)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 where 119888

119899 119899 = 0





119896cos 120575 with

119886119896(see (118)) and 120582

119896cos(119899120591 minus 120575) with 119886

119896cos 119899120591 + 119887

119896sin 119899120591



1198861= minus (1 + 119886

119896) 119887

1= minus119896119887

119896 (123)



1205910minus 120585119896 = 120575119896 in (84) leads to

1198861= minus1205821cos(120575

119896) 119887

1= minus1205821sin(120575

119896) (124)




119896and 119887119896satisfy





119896 119887119896satisfying 1 + 119886

119896=

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) is

120582119896=


(125)


(118)) into 1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896)


119910 = cos(120575119896) (126)



119886119896= 120582119896119881119896(119910) (127)

120582119896=

1

119896119910119880119896minus1

(119910) minus 119881119896(119910)

(128)

where119881119896(119910) and119880



119886119896119896119910119880119896minus1

(119910) minus (1 + 119886119896) 119881119896(119910) = 0 (129)


cos(120587119896) le 119910 le 1 (130)



119896= (119896 minus 1) cos 120575 +

119896sum119896minus1


119896= 120582119896cos 120575 can

be rewritten as

119886119896=


119899=1cos ((119896 minus 2119899) 120575119896)

(131)




119896is decreasing







1

05

0

minus05

minus1


Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4


119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4


119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8


4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)


119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3


8 (134)

1205824=1



4+ 1011988624) minus1 le 119886

4le

1

15

(135)





100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)





1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)




1|


119896 From (124)








1le 0


119896(120591) ge 0

and119879119896(120591) = 0 for some 120591






119896(1205910) = 0 and

11987910158401015840


119896(120591) at


119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840


1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)







0 First


0 taking into



1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)




119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119879119896(120591) =

[1 minus cos (120591 minus 1205910)]2

3 (1198962 minus 1)

sdot [119896 (1198962

minus 1)

+ 2

119896minus2

sum

119899=1

(119896 minus 119899) ((119896 minus 119899)2

minus 1) cos 119899 (120591 minus 1205910)]

(70)


can be expressed as

1198792(120591) =

2

3[1 minus cos(120591 minus 120591

0)]2

1198793(120591) =

1

2[1 minus cos(120591 minus 120591

0)]2

[2 + cos (120591 minus 1205910)]

1198794(120591) =

4

15[1 minus cos (120591 minus 120591

0)]2


0)]

(71)



0+2120585119896)] =



119879119896(120591) = 120582

119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]

2

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(72)


119879119896(120591)

= 120582119896[cos 120585119896

minus cos(120591 minus 1205910+120585

119896)]


minus 2

119896minus1

sum

119899=1

sin (120585 minus 119899120585119896)sin (120585119896)

cos 119899(120591 minus 1205910+120585

119896)]

(73)


119888119899= 2


sum

119898=1


) (74)


119888119896minus3

119888119896minus4

and 119888119896minus5


= 2 (75)

119888119896minus3

= 8 cos(120585119896) (76)

119888119896minus4

= 8 + 12 cos(2120585119896) (77)

119888119896minus5

= 24 cos(120585119896) + 16 cos(3120585

119896) (78)



that

1198792(120591) =


2

[2 + cos 120585] (79)


0=



1198793(120591) =


2

[3 cos (1205853) + cos 120585]


0+120585

3)]

(80)


120582119896= [(119896 minus 1) cos 120585 + 119896

119896minus1

sum

119899=1

cos((119896 minus 2119899)120585119896

)]

minus1

(81)





1

1198962 minus 1lt 120582119896lt 1 (82)




119879119896(120591) = 1 minus 120582

119896

119896 sin 120585sin (120585119896)

cos(120591 minus 1205910+120585

119896)

+ 120582119896cos (119896 (120591 minus 120591

0) + 120585)

(83)


1

08

06

04

02


Am

plitu

de120582k

1


k = 2k = 3

k = 4k = 5




1198861= minus1205821cos(120591

0minus120585

119896) 119887

1= minus1205821sin(120591

0minus120585

119896) (84)


1205821=

119896 sin 120585sin (120585119896)

120582119896 (85)


120582119896= [cos(120585

119896)

119896 sin 120585sin(120585119896)


(86)


119909 = cos(120585119896) (87)


1205821= 119896120582119896119880119896minus1

(119909) (88)

120582119896=

1

119896119909119880119896minus1

(119909) minus 119881119896(119909)

(89)

where119881119896(119909) and119880



120582119896[119896119909119880119896minus1

(119909) minus 119881119896(119909)] minus 1 = 0 (90)


cos(120587119896) lt 119909 lt 1 (91)





1of


119909 = radic1 minus 1205822

21205822

1

3lt 1205822le 1

119909 =1

23radic1205823

1

8lt 1205823lt 1

119909 = radic1

6(1 + radic

51205824+ 3

21205824

)1

15lt 1205824lt 1

(92)



119896of


1205821= radic8120582

2(1 minus 120582

2)

1

3lt 1205822le 1 (93)

1205821= 3 (

3radic1205823minus 1205823)

1

8lt 1205823lt 1 (94)

1205821= radic

32

27(radic2120582

4(3 + 5120582

4)3

minus 21205824(9 + 7120582

4))

1

15lt 1205824lt 1

(95)




120582119896= [max120591

119903119896(120591)]minus1

(96)

and max120591119903119896(120591) = 119903

119896(1205910) According to (64) max

120591119903119896(120591) =



119879119896(120591) =


120591119903119896(120591) minus 119903

119896(120591)]

max120591119903119896(120591)

(97)



119896(120591) according to (64) and (39)

equals

max120591

119903119896(120591) minus 119903

119896(120591) = 119896

sin ((119896 minus 1) 120585119896)sin (120585119896)

minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(98)


max120591

119903119896(120591) minus 119903

119896(120591) = [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(99)




1198880minus 1198881cos(120585

119896) = 119896

sin (120585 minus 120585119896)sin (120585119896)

(100)


(119888119899minus1

+ 119888119899+1

) cos(120585119896) minus 2119888

119899= 2 (119896 minus 119899) cos(120585 minus 119899120585

119896)

(119888119899minus1

minus 119888119899+1


119896)

(101)


cos(120591 minus 1205910+2120585

119896) = cos(120585

119896) cos(120591 minus 120591

0+120585

119896)


0+120585

119896)

cos(120585 minus 119899120585

119896) cos(119899(120591 minus 120591

0+120585

119896))

minus sin(120585 minus 119899120585

119896) sin(119899(120591 minus 120591

0+120585

119896))

= cos (119899 (120591 minus 1205910) + 120585)

(102)







1

1198962 minus 1lt radic1198862119896+ 1198872119896lt 1 (103)


119896= radic1198862







1198961205910minus 120585 = atan 2 (119887

119896 119886119896) + 2119902120587 119902 = 1 (119896 minus 1) (104)


atan 2 (119910 119909) =

arctan(119910

119909) if 119909 ge 0

arctan(119910

119909) + 120587 if 119909 lt 0 119910 ge 0

arctan(119910

119909) minus 120587 if 119909 lt 0 119910 lt 0

(105)


1198861= minus1205821cos[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

1198871= minus1205821sin[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

(106)


1205910=1

119896[120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

1205910minus2120585

119896=1

119896[minus120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

(107)






119896



2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0

q = 1

q = 2





1according to (88)




1from 120582

119896and






3= minus02



1198861= minus08432 and 119887








4= minus02


1= 09861 (according to (95))






3

2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0q = 1

q = 2q = 3






following form

119879119896(120591) = [1 minus cos (120591 minus 120591

0)]


119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(108)

if 0 le 120582119896le 1(119896

2

minus 1) or

119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(109)


119899 119899 = 0 119896 minus 2 and


|120585| le 120587


1198861= minus (1 + 120582

119896) cos 120591

0 119887

1= minus (1 + 120582

119896) sin 120591

0

119886119896= 120582119896cos (119896120591

0) 119887

119896= 120582119896sin (119896120591

0)

(110)


1is given by (85)



119896lt 1


14

12

1

08

06

04

02

00 05 1

Amplitude 120582k

Am

plitu

de1205821

k = 2

k = 3

k = 4



119896= 1 have 119896 zeros


1205821= 1 + 120582

119896 (111)

if 0 le 120582119896le 1(119896

2

minus 1) or

1205821=


(112)


119896via (68) (or



119896= 1(119896

2


119896le 1(119896

2

minus 1) and 1(1198962

minus 1) le 120582119896le

1 According to (111) 120582119896= 1(119896

2


1198962

(1198962




1205821max =

1

cos (120587 (2119896)) (113)


119896=


at 1205910and 1205910+ 120587119896 for 120585 = minus1205872

14

12

1

08

06

04

02


Am

plitu

de1205821


k = 2k = 3k = 4



cos( 120587

2119896) lt 1 minus

1

1198962 (114)







119896







1is equal to 119896

2

(1198962








1= 1radic3 119886

3= 0


0= 1205876 (119886


1= 0

1198863= radic39 and 119887


0= 1205873 (119886

1= minus1

1198871= minus1radic3 119886

3= 0 and 119887

3= radic39)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

1205910 = 01205910 = 12058761205910 = 1205873





1205821= radic(1 + 120582

119896cos 120585)2 + 11989621205822

119896sin2120585 (115)







implies 120582119896


= 0Hence 119889120582

1119889120585 = 0 implies

2120582119896sin 120585 [1 minus (1198962 minus 1) 120582

119896cos 120585] = 0 (116)

Therefore 1198891205821119889120585 = 0 if 120582

119896= 0 (Option 1) or sin 120585 = 0



1= 1 (notice





1is



2

minus 1)


2

minus 1)120582119896cos 120585 = 1 and 120582

119896lt [(119896 minus






2


119896lt 1(119896

2

minus 1)







2

minus



1(1198962


119879119896(120591) =

[1 minus cos (120591 minus 1205910)]

(1 minus 1198962)

sdot [119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(117)


=

1(1 minus 1198962


2













119896(120591) of type (35) with


1le 0 by shifting



119896(120591) by


1




1and 119886119896with minus119886

119896


119886119896= 120582119896cos 120575 119887

119896= 120582119896sin 120575 (118)


where

|120575| le 120587 (119)


119896 120575 can be

determined as

120575 = atan 2 (119887119896 119886119896) (120)


proposition




119879119896(120591)

= [1 minus cos 120591]


119896minus1

sum

119899=1

(119896 minus 119899) (119886119896cos 119899120591 + 119887

119896sin 119899120591)]

(121)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1+119886

119896 where 120575 = atan 2(bk

119886119896) or

119879119896(120591) = 120582

119896[1 minus cos(120591 minus (120575 + 120585)

119896)]


119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120575

119896)]

(122)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 where 119888

119899 119899 = 0





119896cos 120575 with

119886119896(see (118)) and 120582

119896cos(119899120591 minus 120575) with 119886

119896cos 119899120591 + 119887

119896sin 119899120591



1198861= minus (1 + 119886

119896) 119887

1= minus119896119887

119896 (123)



1205910minus 120585119896 = 120575119896 in (84) leads to

1198861= minus1205821cos(120575

119896) 119887

1= minus1205821sin(120575

119896) (124)




119896and 119887119896satisfy





119896 119887119896satisfying 1 + 119886

119896=

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) is

120582119896=


(125)


(118)) into 1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896)


119910 = cos(120575119896) (126)



119886119896= 120582119896119881119896(119910) (127)

120582119896=

1

119896119910119880119896minus1

(119910) minus 119881119896(119910)

(128)

where119881119896(119910) and119880



119886119896119896119910119880119896minus1

(119910) minus (1 + 119886119896) 119881119896(119910) = 0 (129)


cos(120587119896) le 119910 le 1 (130)



119896= (119896 minus 1) cos 120575 +

119896sum119896minus1


119896= 120582119896cos 120575 can

be rewritten as

119886119896=


119899=1cos ((119896 minus 2119899) 120575119896)

(131)




119896is decreasing







1

05

0

minus05

minus1


Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4


119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4


119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8


4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)


119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3


8 (134)

1205824=1



4+ 1011988624) minus1 le 119886

4le

1

15

(135)





100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)





1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)




1|


119896 From (124)








1le 0


119896(120591) ge 0

and119879119896(120591) = 0 for some 120591






119896(1205910) = 0 and

11987910158401015840


119896(120591) at


119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840


1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)







0 First


0 taking into



1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)




119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










1

08

06

04

02


Am

plitu

de120582k

1


k = 2k = 3

k = 4k = 5




1198861= minus1205821cos(120591

0minus120585

119896) 119887

1= minus1205821sin(120591

0minus120585

119896) (84)


1205821=

119896 sin 120585sin (120585119896)

120582119896 (85)


120582119896= [cos(120585

119896)

119896 sin 120585sin(120585119896)


(86)


119909 = cos(120585119896) (87)


1205821= 119896120582119896119880119896minus1

(119909) (88)

120582119896=

1

119896119909119880119896minus1

(119909) minus 119881119896(119909)

(89)

where119881119896(119909) and119880



120582119896[119896119909119880119896minus1

(119909) minus 119881119896(119909)] minus 1 = 0 (90)


cos(120587119896) lt 119909 lt 1 (91)





1of


119909 = radic1 minus 1205822

21205822

1

3lt 1205822le 1

119909 =1

23radic1205823

1

8lt 1205823lt 1

119909 = radic1

6(1 + radic

51205824+ 3

21205824

)1

15lt 1205824lt 1

(92)



119896of


1205821= radic8120582

2(1 minus 120582

2)

1

3lt 1205822le 1 (93)

1205821= 3 (

3radic1205823minus 1205823)

1

8lt 1205823lt 1 (94)

1205821= radic

32

27(radic2120582

4(3 + 5120582

4)3

minus 21205824(9 + 7120582

4))

1

15lt 1205824lt 1

(95)




120582119896= [max120591

119903119896(120591)]minus1

(96)

and max120591119903119896(120591) = 119903

119896(1205910) According to (64) max

120591119903119896(120591) =



119879119896(120591) =


120591119903119896(120591) minus 119903

119896(120591)]

max120591119903119896(120591)

(97)



119896(120591) according to (64) and (39)

equals

max120591

119903119896(120591) minus 119903

119896(120591) = 119896

sin ((119896 minus 1) 120585119896)sin (120585119896)

minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(98)


max120591

119903119896(120591) minus 119903

119896(120591) = [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(99)




1198880minus 1198881cos(120585

119896) = 119896

sin (120585 minus 120585119896)sin (120585119896)

(100)


(119888119899minus1

+ 119888119899+1

) cos(120585119896) minus 2119888

119899= 2 (119896 minus 119899) cos(120585 minus 119899120585

119896)

(119888119899minus1

minus 119888119899+1


119896)

(101)


cos(120591 minus 1205910+2120585

119896) = cos(120585

119896) cos(120591 minus 120591

0+120585

119896)


0+120585

119896)

cos(120585 minus 119899120585

119896) cos(119899(120591 minus 120591

0+120585

119896))

minus sin(120585 minus 119899120585

119896) sin(119899(120591 minus 120591

0+120585

119896))

= cos (119899 (120591 minus 1205910) + 120585)

(102)







1

1198962 minus 1lt radic1198862119896+ 1198872119896lt 1 (103)


119896= radic1198862







1198961205910minus 120585 = atan 2 (119887

119896 119886119896) + 2119902120587 119902 = 1 (119896 minus 1) (104)


atan 2 (119910 119909) =

arctan(119910

119909) if 119909 ge 0

arctan(119910

119909) + 120587 if 119909 lt 0 119910 ge 0

arctan(119910

119909) minus 120587 if 119909 lt 0 119910 lt 0

(105)


1198861= minus1205821cos[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

1198871= minus1205821sin[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

(106)


1205910=1

119896[120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

1205910minus2120585

119896=1

119896[minus120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

(107)






119896



2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0

q = 1

q = 2





1according to (88)




1from 120582

119896and






3= minus02



1198861= minus08432 and 119887








4= minus02


1= 09861 (according to (95))






3

2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0q = 1

q = 2q = 3






following form

119879119896(120591) = [1 minus cos (120591 minus 120591

0)]


119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(108)

if 0 le 120582119896le 1(119896

2

minus 1) or

119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(109)


119899 119899 = 0 119896 minus 2 and


|120585| le 120587


1198861= minus (1 + 120582

119896) cos 120591

0 119887

1= minus (1 + 120582

119896) sin 120591

0

119886119896= 120582119896cos (119896120591

0) 119887

119896= 120582119896sin (119896120591

0)

(110)


1is given by (85)



119896lt 1


14

12

1

08

06

04

02

00 05 1

Amplitude 120582k

Am

plitu

de1205821

k = 2

k = 3

k = 4



119896= 1 have 119896 zeros


1205821= 1 + 120582

119896 (111)

if 0 le 120582119896le 1(119896

2

minus 1) or

1205821=


(112)


119896via (68) (or



119896= 1(119896

2


119896le 1(119896

2

minus 1) and 1(1198962

minus 1) le 120582119896le

1 According to (111) 120582119896= 1(119896

2


1198962

(1198962




1205821max =

1

cos (120587 (2119896)) (113)


119896=


at 1205910and 1205910+ 120587119896 for 120585 = minus1205872

14

12

1

08

06

04

02


Am

plitu

de1205821


k = 2k = 3k = 4



cos( 120587

2119896) lt 1 minus

1

1198962 (114)







119896







1is equal to 119896

2

(1198962








1= 1radic3 119886

3= 0


0= 1205876 (119886


1= 0

1198863= radic39 and 119887


0= 1205873 (119886

1= minus1

1198871= minus1radic3 119886

3= 0 and 119887

3= radic39)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

1205910 = 01205910 = 12058761205910 = 1205873





1205821= radic(1 + 120582

119896cos 120585)2 + 11989621205822

119896sin2120585 (115)







implies 120582119896


= 0Hence 119889120582

1119889120585 = 0 implies

2120582119896sin 120585 [1 minus (1198962 minus 1) 120582

119896cos 120585] = 0 (116)

Therefore 1198891205821119889120585 = 0 if 120582

119896= 0 (Option 1) or sin 120585 = 0



1= 1 (notice





1is



2

minus 1)


2

minus 1)120582119896cos 120585 = 1 and 120582

119896lt [(119896 minus






2


119896lt 1(119896

2

minus 1)







2

minus



1(1198962


119879119896(120591) =

[1 minus cos (120591 minus 1205910)]

(1 minus 1198962)

sdot [119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(117)


=

1(1 minus 1198962


2













119896(120591) of type (35) with


1le 0 by shifting



119896(120591) by


1




1and 119886119896with minus119886

119896


119886119896= 120582119896cos 120575 119887

119896= 120582119896sin 120575 (118)


where

|120575| le 120587 (119)


119896 120575 can be

determined as

120575 = atan 2 (119887119896 119886119896) (120)


proposition




119879119896(120591)

= [1 minus cos 120591]


119896minus1

sum

119899=1

(119896 minus 119899) (119886119896cos 119899120591 + 119887

119896sin 119899120591)]

(121)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1+119886

119896 where 120575 = atan 2(bk

119886119896) or

119879119896(120591) = 120582

119896[1 minus cos(120591 minus (120575 + 120585)

119896)]


119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120575

119896)]

(122)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 where 119888

119899 119899 = 0





119896cos 120575 with

119886119896(see (118)) and 120582

119896cos(119899120591 minus 120575) with 119886

119896cos 119899120591 + 119887

119896sin 119899120591



1198861= minus (1 + 119886

119896) 119887

1= minus119896119887

119896 (123)



1205910minus 120585119896 = 120575119896 in (84) leads to

1198861= minus1205821cos(120575

119896) 119887

1= minus1205821sin(120575

119896) (124)




119896and 119887119896satisfy





119896 119887119896satisfying 1 + 119886

119896=

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) is

120582119896=


(125)


(118)) into 1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896)


119910 = cos(120575119896) (126)



119886119896= 120582119896119881119896(119910) (127)

120582119896=

1

119896119910119880119896minus1

(119910) minus 119881119896(119910)

(128)

where119881119896(119910) and119880



119886119896119896119910119880119896minus1

(119910) minus (1 + 119886119896) 119881119896(119910) = 0 (129)


cos(120587119896) le 119910 le 1 (130)



119896= (119896 minus 1) cos 120575 +

119896sum119896minus1


119896= 120582119896cos 120575 can

be rewritten as

119886119896=


119899=1cos ((119896 minus 2119899) 120575119896)

(131)




119896is decreasing







1

05

0

minus05

minus1


Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4


119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4


119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8


4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)


119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3


8 (134)

1205824=1



4+ 1011988624) minus1 le 119886

4le

1

15

(135)





100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)





1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)




1|


119896 From (124)








1le 0


119896(120591) ge 0

and119879119896(120591) = 0 for some 120591






119896(1205910) = 0 and

11987910158401015840


119896(120591) at


119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840


1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)







0 First


0 taking into



1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)




119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119896(120591) according to (64) and (39)

equals

max120591

119903119896(120591) minus 119903

119896(120591) = 119896

sin ((119896 minus 1) 120585119896)sin (120585119896)

minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)

(98)


max120591

119903119896(120591) minus 119903

119896(120591) = [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(99)




1198880minus 1198881cos(120585

119896) = 119896

sin (120585 minus 120585119896)sin (120585119896)

(100)


(119888119899minus1

+ 119888119899+1

) cos(120585119896) minus 2119888

119899= 2 (119896 minus 119899) cos(120585 minus 119899120585

119896)

(119888119899minus1

minus 119888119899+1


119896)

(101)


cos(120591 minus 1205910+2120585

119896) = cos(120585

119896) cos(120591 minus 120591

0+120585

119896)


0+120585

119896)

cos(120585 minus 119899120585

119896) cos(119899(120591 minus 120591

0+120585

119896))

minus sin(120585 minus 119899120585

119896) sin(119899(120591 minus 120591

0+120585

119896))

= cos (119899 (120591 minus 1205910) + 120585)

(102)







1

1198962 minus 1lt radic1198862119896+ 1198872119896lt 1 (103)


119896= radic1198862







1198961205910minus 120585 = atan 2 (119887

119896 119886119896) + 2119902120587 119902 = 1 (119896 minus 1) (104)


atan 2 (119910 119909) =

arctan(119910

119909) if 119909 ge 0

arctan(119910

119909) + 120587 if 119909 lt 0 119910 ge 0

arctan(119910

119909) minus 120587 if 119909 lt 0 119910 lt 0

(105)


1198861= minus1205821cos[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

1198871= minus1205821sin[

atan 2 (119887119896 119886119896) + 2119902120587

119896]

(106)


1205910=1

119896[120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

1205910minus2120585

119896=1

119896[minus120585 + atan 2 (119887

119896 119886119896) + 2119902120587]

(107)






119896



2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0

q = 1

q = 2





1according to (88)




1from 120582

119896and






3= minus02



1198861= minus08432 and 119887








4= minus02


1= 09861 (according to (95))






3

2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0q = 1

q = 2q = 3






following form

119879119896(120591) = [1 minus cos (120591 minus 120591

0)]


119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(108)

if 0 le 120582119896le 1(119896

2

minus 1) or

119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(109)


119899 119899 = 0 119896 minus 2 and


|120585| le 120587


1198861= minus (1 + 120582

119896) cos 120591

0 119887

1= minus (1 + 120582

119896) sin 120591

0

119886119896= 120582119896cos (119896120591

0) 119887

119896= 120582119896sin (119896120591

0)

(110)


1is given by (85)



119896lt 1


14

12

1

08

06

04

02

00 05 1

Amplitude 120582k

Am

plitu

de1205821

k = 2

k = 3

k = 4



119896= 1 have 119896 zeros


1205821= 1 + 120582

119896 (111)

if 0 le 120582119896le 1(119896

2

minus 1) or

1205821=


(112)


119896via (68) (or



119896= 1(119896

2


119896le 1(119896

2

minus 1) and 1(1198962

minus 1) le 120582119896le

1 According to (111) 120582119896= 1(119896

2


1198962

(1198962




1205821max =

1

cos (120587 (2119896)) (113)


119896=


at 1205910and 1205910+ 120587119896 for 120585 = minus1205872

14

12

1

08

06

04

02


Am

plitu

de1205821


k = 2k = 3k = 4



cos( 120587

2119896) lt 1 minus

1

1198962 (114)







119896







1is equal to 119896

2

(1198962








1= 1radic3 119886

3= 0


0= 1205876 (119886


1= 0

1198863= radic39 and 119887


0= 1205873 (119886

1= minus1

1198871= minus1radic3 119886

3= 0 and 119887

3= radic39)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

1205910 = 01205910 = 12058761205910 = 1205873





1205821= radic(1 + 120582

119896cos 120585)2 + 11989621205822

119896sin2120585 (115)







implies 120582119896


= 0Hence 119889120582

1119889120585 = 0 implies

2120582119896sin 120585 [1 minus (1198962 minus 1) 120582

119896cos 120585] = 0 (116)

Therefore 1198891205821119889120585 = 0 if 120582

119896= 0 (Option 1) or sin 120585 = 0



1= 1 (notice





1is



2

minus 1)


2

minus 1)120582119896cos 120585 = 1 and 120582

119896lt [(119896 minus






2


119896lt 1(119896

2

minus 1)







2

minus



1(1198962


119879119896(120591) =

[1 minus cos (120591 minus 1205910)]

(1 minus 1198962)

sdot [119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(117)


=

1(1 minus 1198962


2













119896(120591) of type (35) with


1le 0 by shifting



119896(120591) by


1




1and 119886119896with minus119886

119896


119886119896= 120582119896cos 120575 119887

119896= 120582119896sin 120575 (118)


where

|120575| le 120587 (119)


119896 120575 can be

determined as

120575 = atan 2 (119887119896 119886119896) (120)


proposition




119879119896(120591)

= [1 minus cos 120591]


119896minus1

sum

119899=1

(119896 minus 119899) (119886119896cos 119899120591 + 119887

119896sin 119899120591)]

(121)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1+119886

119896 where 120575 = atan 2(bk

119886119896) or

119879119896(120591) = 120582

119896[1 minus cos(120591 minus (120575 + 120585)

119896)]


119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120575

119896)]

(122)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 where 119888

119899 119899 = 0





119896cos 120575 with

119886119896(see (118)) and 120582

119896cos(119899120591 minus 120575) with 119886

119896cos 119899120591 + 119887

119896sin 119899120591



1198861= minus (1 + 119886

119896) 119887

1= minus119896119887

119896 (123)



1205910minus 120585119896 = 120575119896 in (84) leads to

1198861= minus1205821cos(120575

119896) 119887

1= minus1205821sin(120575

119896) (124)




119896and 119887119896satisfy





119896 119887119896satisfying 1 + 119886

119896=

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) is

120582119896=


(125)


(118)) into 1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896)


119910 = cos(120575119896) (126)



119886119896= 120582119896119881119896(119910) (127)

120582119896=

1

119896119910119880119896minus1

(119910) minus 119881119896(119910)

(128)

where119881119896(119910) and119880



119886119896119896119910119880119896minus1

(119910) minus (1 + 119886119896) 119881119896(119910) = 0 (129)


cos(120587119896) le 119910 le 1 (130)



119896= (119896 minus 1) cos 120575 +

119896sum119896minus1


119896= 120582119896cos 120575 can

be rewritten as

119886119896=


119899=1cos ((119896 minus 2119899) 120575119896)

(131)




119896is decreasing







1

05

0

minus05

minus1


Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4


119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4


119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8


4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)


119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3


8 (134)

1205824=1



4+ 1011988624) minus1 le 119886

4le

1

15

(135)





100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)





1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)




1|


119896 From (124)








1le 0


119896(120591) ge 0

and119879119896(120591) = 0 for some 120591






119896(1205910) = 0 and

11987910158401015840


119896(120591) at


119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840


1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)







0 First


0 taking into



1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)




119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0

q = 1

q = 2





1according to (88)




1from 120582

119896and






3= minus02



1198861= minus08432 and 119887








4= minus02


1= 09861 (according to (95))






3

2

1

0

Wav

efor

ms

0 1 2 3 4

Angle 120591120587

q = 0q = 1

q = 2q = 3






following form

119879119896(120591) = [1 minus cos (120591 minus 120591

0)]


119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(108)

if 0 le 120582119896le 1(119896

2

minus 1) or

119879119896(120591) = 120582

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos(120591 minus 120591

0+2120585

119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120591

0+120585

119896)]

(109)


119899 119899 = 0 119896 minus 2 and


|120585| le 120587


1198861= minus (1 + 120582

119896) cos 120591

0 119887

1= minus (1 + 120582

119896) sin 120591

0

119886119896= 120582119896cos (119896120591

0) 119887

119896= 120582119896sin (119896120591

0)

(110)


1is given by (85)



119896lt 1


14

12

1

08

06

04

02

00 05 1

Amplitude 120582k

Am

plitu

de1205821

k = 2

k = 3

k = 4



119896= 1 have 119896 zeros


1205821= 1 + 120582

119896 (111)

if 0 le 120582119896le 1(119896

2

minus 1) or

1205821=


(112)


119896via (68) (or



119896= 1(119896

2


119896le 1(119896

2

minus 1) and 1(1198962

minus 1) le 120582119896le

1 According to (111) 120582119896= 1(119896

2


1198962

(1198962




1205821max =

1

cos (120587 (2119896)) (113)


119896=


at 1205910and 1205910+ 120587119896 for 120585 = minus1205872

14

12

1

08

06

04

02


Am

plitu

de1205821


k = 2k = 3k = 4



cos( 120587

2119896) lt 1 minus

1

1198962 (114)







119896







1is equal to 119896

2

(1198962








1= 1radic3 119886

3= 0


0= 1205876 (119886


1= 0

1198863= radic39 and 119887


0= 1205873 (119886

1= minus1

1198871= minus1radic3 119886

3= 0 and 119887

3= radic39)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

1205910 = 01205910 = 12058761205910 = 1205873





1205821= radic(1 + 120582

119896cos 120585)2 + 11989621205822

119896sin2120585 (115)







implies 120582119896


= 0Hence 119889120582

1119889120585 = 0 implies

2120582119896sin 120585 [1 minus (1198962 minus 1) 120582

119896cos 120585] = 0 (116)

Therefore 1198891205821119889120585 = 0 if 120582

119896= 0 (Option 1) or sin 120585 = 0



1= 1 (notice





1is



2

minus 1)


2

minus 1)120582119896cos 120585 = 1 and 120582

119896lt [(119896 minus






2


119896lt 1(119896

2

minus 1)







2

minus



1(1198962


119879119896(120591) =

[1 minus cos (120591 minus 1205910)]

(1 minus 1198962)

sdot [119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(117)


=

1(1 minus 1198962


2













119896(120591) of type (35) with


1le 0 by shifting



119896(120591) by


1




1and 119886119896with minus119886

119896


119886119896= 120582119896cos 120575 119887

119896= 120582119896sin 120575 (118)


where

|120575| le 120587 (119)


119896 120575 can be

determined as

120575 = atan 2 (119887119896 119886119896) (120)


proposition




119879119896(120591)

= [1 minus cos 120591]


119896minus1

sum

119899=1

(119896 minus 119899) (119886119896cos 119899120591 + 119887

119896sin 119899120591)]

(121)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1+119886

119896 where 120575 = atan 2(bk

119886119896) or

119879119896(120591) = 120582

119896[1 minus cos(120591 minus (120575 + 120585)

119896)]


119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120575

119896)]

(122)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 where 119888

119899 119899 = 0





119896cos 120575 with

119886119896(see (118)) and 120582

119896cos(119899120591 minus 120575) with 119886

119896cos 119899120591 + 119887

119896sin 119899120591



1198861= minus (1 + 119886

119896) 119887

1= minus119896119887

119896 (123)



1205910minus 120585119896 = 120575119896 in (84) leads to

1198861= minus1205821cos(120575

119896) 119887

1= minus1205821sin(120575

119896) (124)




119896and 119887119896satisfy





119896 119887119896satisfying 1 + 119886

119896=

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) is

120582119896=


(125)


(118)) into 1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896)


119910 = cos(120575119896) (126)



119886119896= 120582119896119881119896(119910) (127)

120582119896=

1

119896119910119880119896minus1

(119910) minus 119881119896(119910)

(128)

where119881119896(119910) and119880



119886119896119896119910119880119896minus1

(119910) minus (1 + 119886119896) 119881119896(119910) = 0 (129)


cos(120587119896) le 119910 le 1 (130)



119896= (119896 minus 1) cos 120575 +

119896sum119896minus1


119896= 120582119896cos 120575 can

be rewritten as

119886119896=


119899=1cos ((119896 minus 2119899) 120575119896)

(131)




119896is decreasing







1

05

0

minus05

minus1


Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4


119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4


119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8


4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)


119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3


8 (134)

1205824=1



4+ 1011988624) minus1 le 119886

4le

1

15

(135)





100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)





1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)




1|


119896 From (124)








1le 0


119896(120591) ge 0

and119879119896(120591) = 0 for some 120591






119896(1205910) = 0 and

11987910158401015840


119896(120591) at


119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840


1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)







0 First


0 taking into



1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)




119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










14

12

1

08

06

04

02

00 05 1

Amplitude 120582k

Am

plitu

de1205821

k = 2

k = 3

k = 4



119896= 1 have 119896 zeros


1205821= 1 + 120582

119896 (111)

if 0 le 120582119896le 1(119896

2

minus 1) or

1205821=


(112)


119896via (68) (or



119896= 1(119896

2


119896le 1(119896

2

minus 1) and 1(1198962

minus 1) le 120582119896le

1 According to (111) 120582119896= 1(119896

2


1198962

(1198962




1205821max =

1

cos (120587 (2119896)) (113)


119896=


at 1205910and 1205910+ 120587119896 for 120585 = minus1205872

14

12

1

08

06

04

02


Am

plitu

de1205821


k = 2k = 3k = 4



cos( 120587

2119896) lt 1 minus

1

1198962 (114)







119896







1is equal to 119896

2

(1198962








1= 1radic3 119886

3= 0


0= 1205876 (119886


1= 0

1198863= radic39 and 119887


0= 1205873 (119886

1= minus1

1198871= minus1radic3 119886

3= 0 and 119887

3= radic39)



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

1205910 = 01205910 = 12058761205910 = 1205873





1205821= radic(1 + 120582

119896cos 120585)2 + 11989621205822

119896sin2120585 (115)







implies 120582119896


= 0Hence 119889120582

1119889120585 = 0 implies

2120582119896sin 120585 [1 minus (1198962 minus 1) 120582

119896cos 120585] = 0 (116)

Therefore 1198891205821119889120585 = 0 if 120582

119896= 0 (Option 1) or sin 120585 = 0



1= 1 (notice





1is



2

minus 1)


2

minus 1)120582119896cos 120585 = 1 and 120582

119896lt [(119896 minus






2


119896lt 1(119896

2

minus 1)







2

minus



1(1198962


119879119896(120591) =

[1 minus cos (120591 minus 1205910)]

(1 minus 1198962)

sdot [119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(117)


=

1(1 minus 1198962


2













119896(120591) of type (35) with


1le 0 by shifting



119896(120591) by


1




1and 119886119896with minus119886

119896


119886119896= 120582119896cos 120575 119887

119896= 120582119896sin 120575 (118)


where

|120575| le 120587 (119)


119896 120575 can be

determined as

120575 = atan 2 (119887119896 119886119896) (120)


proposition




119879119896(120591)

= [1 minus cos 120591]


119896minus1

sum

119899=1

(119896 minus 119899) (119886119896cos 119899120591 + 119887

119896sin 119899120591)]

(121)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1+119886

119896 where 120575 = atan 2(bk

119886119896) or

119879119896(120591) = 120582

119896[1 minus cos(120591 minus (120575 + 120585)

119896)]


119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120575

119896)]

(122)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 where 119888

119899 119899 = 0





119896cos 120575 with

119886119896(see (118)) and 120582

119896cos(119899120591 minus 120575) with 119886

119896cos 119899120591 + 119887

119896sin 119899120591



1198861= minus (1 + 119886

119896) 119887

1= minus119896119887

119896 (123)



1205910minus 120585119896 = 120575119896 in (84) leads to

1198861= minus1205821cos(120575

119896) 119887

1= minus1205821sin(120575

119896) (124)




119896and 119887119896satisfy





119896 119887119896satisfying 1 + 119886

119896=

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) is

120582119896=


(125)


(118)) into 1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896)


119910 = cos(120575119896) (126)



119886119896= 120582119896119881119896(119910) (127)

120582119896=

1

119896119910119880119896minus1

(119910) minus 119881119896(119910)

(128)

where119881119896(119910) and119880



119886119896119896119910119880119896minus1

(119910) minus (1 + 119886119896) 119881119896(119910) = 0 (129)


cos(120587119896) le 119910 le 1 (130)



119896= (119896 minus 1) cos 120575 +

119896sum119896minus1


119896= 120582119896cos 120575 can

be rewritten as

119886119896=


119899=1cos ((119896 minus 2119899) 120575119896)

(131)




119896is decreasing







1

05

0

minus05

minus1


Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4


119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4


119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8


4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)


119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3


8 (134)

1205824=1



4+ 1011988624) minus1 le 119886

4le

1

15

(135)





100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)





1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)




1|


119896 From (124)








1le 0


119896(120591) ge 0

and119879119896(120591) = 0 for some 120591






119896(1205910) = 0 and

11987910158401015840


119896(120591) at


119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840


1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)







0 First


0 taking into



1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)




119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

1205910 = 01205910 = 12058761205910 = 1205873





1205821= radic(1 + 120582

119896cos 120585)2 + 11989621205822

119896sin2120585 (115)







implies 120582119896


= 0Hence 119889120582

1119889120585 = 0 implies

2120582119896sin 120585 [1 minus (1198962 minus 1) 120582

119896cos 120585] = 0 (116)

Therefore 1198891205821119889120585 = 0 if 120582

119896= 0 (Option 1) or sin 120585 = 0



1= 1 (notice





1is



2

minus 1)


2

minus 1)120582119896cos 120585 = 1 and 120582

119896lt [(119896 minus






2


119896lt 1(119896

2

minus 1)







2

minus



1(1198962


119879119896(120591) =

[1 minus cos (120591 minus 1205910)]

(1 minus 1198962)

sdot [119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910))]

(117)


=

1(1 minus 1198962


2













119896(120591) of type (35) with


1le 0 by shifting



119896(120591) by


1




1and 119886119896with minus119886

119896


119886119896= 120582119896cos 120575 119887

119896= 120582119896sin 120575 (118)


where

|120575| le 120587 (119)


119896 120575 can be

determined as

120575 = atan 2 (119887119896 119886119896) (120)


proposition




119879119896(120591)

= [1 minus cos 120591]


119896minus1

sum

119899=1

(119896 minus 119899) (119886119896cos 119899120591 + 119887

119896sin 119899120591)]

(121)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1+119886

119896 where 120575 = atan 2(bk

119886119896) or

119879119896(120591) = 120582

119896[1 minus cos(120591 minus (120575 + 120585)

119896)]


119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120575

119896)]

(122)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 where 119888

119899 119899 = 0





119896cos 120575 with

119886119896(see (118)) and 120582

119896cos(119899120591 minus 120575) with 119886

119896cos 119899120591 + 119887

119896sin 119899120591



1198861= minus (1 + 119886

119896) 119887

1= minus119896119887

119896 (123)



1205910minus 120585119896 = 120575119896 in (84) leads to

1198861= minus1205821cos(120575

119896) 119887

1= minus1205821sin(120575

119896) (124)




119896and 119887119896satisfy





119896 119887119896satisfying 1 + 119886

119896=

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) is

120582119896=


(125)


(118)) into 1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896)


119910 = cos(120575119896) (126)



119886119896= 120582119896119881119896(119910) (127)

120582119896=

1

119896119910119880119896minus1

(119910) minus 119881119896(119910)

(128)

where119881119896(119910) and119880



119886119896119896119910119880119896minus1

(119910) minus (1 + 119886119896) 119881119896(119910) = 0 (129)


cos(120587119896) le 119910 le 1 (130)



119896= (119896 minus 1) cos 120575 +

119896sum119896minus1


119896= 120582119896cos 120575 can

be rewritten as

119886119896=


119899=1cos ((119896 minus 2119899) 120575119896)

(131)




119896is decreasing







1

05

0

minus05

minus1


Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4


119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4


119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8


4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)


119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3


8 (134)

1205824=1



4+ 1011988624) minus1 le 119886

4le

1

15

(135)





100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)





1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)




1|


119896 From (124)








1le 0


119896(120591) ge 0

and119879119896(120591) = 0 for some 120591






119896(1205910) = 0 and

11987910158401015840


119896(120591) at


119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840


1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)







0 First


0 taking into



1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)




119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










where

|120575| le 120587 (119)


119896 120575 can be

determined as

120575 = atan 2 (119887119896 119886119896) (120)


proposition




119879119896(120591)

= [1 minus cos 120591]


119896minus1

sum

119899=1

(119896 minus 119899) (119886119896cos 119899120591 + 119887

119896sin 119899120591)]

(121)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1+119886

119896 where 120575 = atan 2(bk

119886119896) or

119879119896(120591) = 120582

119896[1 minus cos(120591 minus (120575 + 120585)

119896)]


119896)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899(120591 minus 120575

119896)]

(122)

if 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896 where 119888

119899 119899 = 0





119896cos 120575 with

119886119896(see (118)) and 120582

119896cos(119899120591 minus 120575) with 119886

119896cos 119899120591 + 119887

119896sin 119899120591



1198861= minus (1 + 119886

119896) 119887

1= minus119896119887

119896 (123)



1205910minus 120585119896 = 120575119896 in (84) leads to

1198861= minus1205821cos(120575

119896) 119887

1= minus1205821sin(120575

119896) (124)




119896and 119887119896satisfy





119896 119887119896satisfying 1 + 119886

119896=

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) is

120582119896=


(125)


(118)) into 1 + 119886119896= 119896120582119896[sin 120575 sin(120575119896)] cos(120575119896)


119910 = cos(120575119896) (126)



119886119896= 120582119896119881119896(119910) (127)

120582119896=

1

119896119910119880119896minus1

(119910) minus 119881119896(119910)

(128)

where119881119896(119910) and119880



119886119896119896119910119880119896minus1

(119910) minus (1 + 119886119896) 119881119896(119910) = 0 (129)


cos(120587119896) le 119910 le 1 (130)



119896= (119896 minus 1) cos 120575 +

119896sum119896minus1


119896= 120582119896cos 120575 can

be rewritten as

119886119896=


119899=1cos ((119896 minus 2119899) 120575119896)

(131)




119896is decreasing







1

05

0

minus05

minus1


Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4


119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4


119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8


4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)


119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3


8 (134)

1205824=1



4+ 1011988624) minus1 le 119886

4le

1

15

(135)





100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)





1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)




1|


119896 From (124)








1le 0


119896(120591) ge 0

and119879119896(120591) = 0 for some 120591






119896(1205910) = 0 and

11987910158401015840


119896(120591) at


119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840


1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)







0 First


0 taking into



1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)




119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










1

05

0

minus05

minus1


Coefficient ak

Coe

ffici

entb

k

radica2k+ b2

kle 1

k = 2k = 3k = 4


119896= 119896120582

119896[sin 120575 sin(120575

119896)] cos(120575119896) for 119896 le 4


119910 = radic1 + 1198862

2 (1 minus 1198862) minus1 le 119886

2le1

3

119910 = radic3

4 (1 minus 21198863) minus1 le 119886

3le1

8


4+ 1011988624minus 2 (1 minus 119886

4)

4 (1 minus 31198864)

minus1 le 1198864le

1

15

(132)


119896 119896 le 4

1205822=1

2(1 minus 119886

2) minus1 le 119886

2le1

3 (133)

1205822

3= [

1

3(1 minus 2119886

3)]

3


8 (134)

1205824=1



4+ 1011988624) minus1 le 119886

4le

1

15

(135)





100381610038161003816100381611988611003816100381610038161003816max =

1

cos (120587 (2119896)) (136)





1 1198861lt 0 are

1198861= minus

1

cos (120587 (2119896)) 119886

119896=1

119896tan( 120587

(2119896))

1198871= 119887119896= 0

(137)




1|


119896 From (124)








1le 0


119896(120591) ge 0

and119879119896(120591) = 0 for some 120591






119896(1205910) = 0 and

11987910158401015840


119896(120591) at


119896(1205910) = 1 minus 120582

119896(1198962

minus 1) cos 120585 Since11987910158401015840


1 minus 120582119896(1198962

minus 1) cos 120585 gt 0 (138)







0 First


0 taking into



1205971198861

1205971205910

= sin 1205910[1 minus 120582

119896(1198962

minus 1) cos 120585] (139)




119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra












119896(120591)has


119896lt 1




1le 0


0= 0 further

implies 1198861

= 0 and

1205910= 0 (140)


119879119896(120591)

= [1 minus cos 120591]


119896

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899120591 + 120585)]

(141)


119896= 120582119896cos 120585

and 119887119896


119896cos 120585 with

119886119896and 120582

119896cos(119899120591 + 120585) with (119886

119896cos 119899120591 + 119887

119896sin 119899120591) in (141)


119896= 120582119896cos 120585 119887

119896= minus120582

119896sin 120585 and (118)

imply that

120575 = minus120585 (142)



sin (120575119896)] lt 1 (143)


holds


sin (120575119896)]

= minus119886119896+ 119896120582119896

sin 120575sin (120575119896)

cos(120575119896)

(144)




119896



1+ 119886119896ge 0


1| le 1 + 119886

119896 On the other




1| =

1 + 119886119896imply 119886






119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1 + 119886

119896


119896=







120575 = 1198961205910minus 120585 (145)


1205910=(120575 + 120585)

119896 120591

1015840

0= 1205910minus2120585

119896=(120575 minus 120585)

119896 (146)



1 + 119886119896



119896=

120582119896cos 120575 into 119896120582

119896[sin 120575 sin(120575119896)] cos(120575119896) = 1 + 119886

119896leads to


sin (120575119896)] = 1 (147)



sin (120585119896)] = 1 (148)


120582119896

119896 sin ((119896 minus 1) 120585119896)sin 120585119896

= 1 minus (119896 minus 1) 120582119896cos 120585 (149)


0=


119879119896(120591)

= 120582119896[1 minus cos (120591 minus 120591

0)]

sdot [119896 sin ((119896 minus 1) 120585119896)

sin 120585119896minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos (119899 (120591 minus 1205910) + 120585)]

(150)




119896[sin 120575



119896 This

completes the proof







3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra















3and


(120) are equal to

1198861= minus1 minus 119886

3 119887

1= minus3119887

3 (151)

if 12058223le [(1 minus 2119886

3)3]3

1198861= minus1205821cos(120575

3) 119887

1= minus1205821sin(120575

3) (152)

where 1205821= 3(


3 1198863) if [(1 minus

21198863)3]3

le 1205822


3= [(1minus2119886

3)3]3 (see case 119896 = 3


3lt [(1 minus 2119886



3= [(1 minus

21198863)3]3 also have zero at 120591



1= minus98 119886

3= 18 and 119887

1= 1198873= 0 which


1= 0 119886




lt 1205822

3lt














3)3]3


3)3]3


3= minus01 119887

3= 01 119886

1= minus09



3= 03 119886

1=


3gt [(1 minus 2119886

3)3]3)



1

05

0

minus05

minus1


Coefficient a3

Coe

ffici

entb

3

02

04

06

08

10

11



2

1

0

0 1 2 3 4

Angle 120591120587

Wav

efor

ms

a3 = minus01 b3 = 01

a3 = minus01 b3 = 03



3and 1198873



1= 119887119896=

0 that is

119879119896(120591) = 1 + 119886

1cos 120591 + 119886

119896cos 119896120591 (153)


1for prescribed






119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119896=



minus1 le 119886119896le 1 (154)









119896minus 2119886119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(155)

if minus1 le 119886119896le 1(119896

2

minus 1) or

119879119896(120591) = 119886

119896[1 minus cos (120591 minus 120591

0)] [1 minus cos (120591 + 120591

0)]

sdot [1198880+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591]

(156)

where

119888119899=sin ((119896 minus 119899) 120591

0) cos 120591


0) sin 120591

0

sin31205910

(157)

119886119896=

sin 1205910

119896 sin (1198961205910) cos 120591

0minus cos (119896120591

0) sin 120591

0

(158)

100381610038161003816100381612059101003816100381610038161003816 le

120587

119896 (159)

if 1(1198962 minus 1) le 119886119896le 1


0)] = [cos 120591

0minus


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]2 [119888

0+ 2

119896minus2

sum

119899=1

119888119899cos 119899120591] (160)


119879119896(120591) = 119886

119896[cos 120591

0minus cos 120591]

sdot [(119896 minus 1) sin 119896120591

0

sin 1205910

minus 2

119896minus1

sum

119899=1

sin ((119896 minus 119899) 1205910)

sin 1205910

cos 119899120591]

(161)




1lt 0 which can be


2





119896 le 4 and 1198861lt 0


119896le 1(119896

2



0 where

0 lt |1205910| lt 120587119896 For 119886





1198861= minus1 minus 119886

119896 (162)

where minus1 le 119886119896le 1(119896

2


1198861= minus119886119896

119896 sin 1198961205910

sin 1205910

(163)


119896le 1 and |120591

0| le 120587119896


119896= 1(119896

2

minus1) and 1198861= minus1198962

(1198962


0= 0


1for prescribed



119896= 0 Fur-


ge 0 then 120582119896


with 119887119896


119896becomes 1198962119886

119896le 1 +


119896lt 0 then 120582

119896= minus119886

119896 which


119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) le 1 + 119886

119896becomes 0 le 1 + 119886

119896


1for prescribed


119896le 1(119896

2





119896gt 0 imply


119896= 119886119896and 120575 = 0 into

119896120582119896[sin 120575 sin(120575119896)] cos(120575119896) ge 1+119886

119896leads to 119886

119896ge 1(119896

2


0= 120585119896

Insertion of 120582119896= 119886119896 120575 = 0 and 120591

0= 120585119896 into (122) leads


1for


119896le 1


119896= |119886119896| = 1 According to


1= |1198861| = 0





119896= 1


119886119896= 1 and 120591

0= 120587119896 (or 120591

0= minus120587119896) into (156) along




119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119896= 1

and |1205910| = 120587119896


119896(1205910) = 0 for some

1205910imply 120582


that |119886119896| = 1 and 119886

1lt 0






2


0= 0 or 120591




1198861

= 0 and 120582119896= |119886119896| lt 1 leads to

1205910= 0 (164)

Clearly 1205910= 0 119887

1= 0 and 119887

119896= 0 according to (44) and (46)

imply 120582119896sin 120585 = 0 Since 120582

119896= |119886119896| it follows that |119886

119896| sin 120585 = 0




119896(120591) = 1 minus cos 120591



119896= minus120582119896


that 0 le 119886119896le 1(1 minus 119896

2



119896le 1(1 minus 119896

2



119896le 0




2


1(1 minus 1198962



0and minus120591

0 such that 120591

0= 0 and 120591

0= 120587

Relations 1198861

lt 0 and 1198871


120585

119896= 1205910 (165)


0| lt


119896= 120582119896 Relations



of 120582119896= 119886119896and 120585119896 = 120591



) lt 119886119896lt 1

and 0 lt |1205910| lt 120587119896


2

) and 1205910= 0 into

(161) leads to

119879119896(120591) =

[1 minus cos 120591](1 minus 1198962)

[119896 (119896 minus 1) minus 2

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591]

(166)


=

1(1 minus 1198962


1(1 minus 1198962




reads

1198793(120591) = 1 + 119886

1cos 120591 + 119886

3cos 3120591 (167)




1198793(120591) = (1 minus cos 120591) [1 minus 2119886

3(1 + 2 cos 120591 + cos 2120591)] (168)

From 1198793(120591 + 120587) = 2 minus 119879


1198861ge 0 and minus18 le 119886

3le 1 119879


1198793(120591) = (1 + cos 120591) [1 + 2119886

3(1 minus 2 cos 120591 + cos 2120591)] (169)

For 1198861le 0 and 18 le 119886


3=



1198793(120591) =

[cos 1205910minus cos 120591]2 [2 cos 120591

0+ cos 120591]

2cos31205910

(170)


3(120591 + 120587) = 2 minus 119879

3(120591) it


3le minus18




0= 0) reads 119879

3(120591) = [1 minus



0= 120587) reads





1198861= minus1 minus 119886


3le1

8 (171)


=

minus31198863(2 cos 2120591


1=

minus31198863(4cos2120591



0| le 1205873 119886

1=

minus31198863(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= minus3 [ 3radic119886

3minus 1198863] for 1

8le 1198863le 1 (172)



1198861= 1 minus 119886

3for minus

1

8le 1198863le 1 (173)


1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










1

15

15

05

0

minus05

minus15minus15

minus1


Coefficient a1

Coe

ffici

enta

3 a1= minus

3(3radica3minus a3)

a1 =

minus1 minus a3

a1 =

1 minusa3

a1 =3(3radic|a3

| +a3)



0isin [21205873

41205873] 1198861= minus3119886

3(4cos2120591


3= [8cos3120591

0]minus1 lead to

1198861= 3 [

3radic10038161003816100381610038161198863


1

8 (174)



1 1198863)


1= 0 and 119886

3= 1


3= minus1




i

v Load

Vdc

Idc

Lch

vL

iL

Cb

+ +in




119894 (120579) = 1 + 1198861119894cos 120579 +

infin

sum

119899=2

119886119899119894cos 119899120579

(175)





V119871(120579) = 119886


119894119871(120579) = minus119886

1119894cos 120579 minus

infin

sum

119899=2

119886119899119894cos 119899120579

(176)


1= minus(119886

1V minus


119896=


119899=


1198751= minus

11988611198941198861V

2 (177)


1119875dc


120578 = minus11988611198941198861V

2 (178)








PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra













PUF =8120578

max [V (120579)] sdotmax [119894 (120579)] (179)




119894119863(120579) =

119868119863[cos 120579 minus cos(120572

2)] |120579| le

120572

2

0120572

2le |120579| le 120587

(180)


gt 0 Since 119894119863(120579) is


119894119863(120579) = 119868dc +

infin

sum

119899=1

119868119899cos 119899120579 (181)


119868dc =119868119863120572

2120587[sinc(120572

2) minus cos(120572

2)] (182)


1198681=119868119863120572

2120587(1 minus sinc120572) (183)


119868119899=119868119863

119899120587[sin ((119899 minus 1) 1205722)

(119899 minus 1)minussin ((119899 + 1) 1205722)

(119899 + 1)] 119899 ge 2

(184)





119894 (120579) =

120587 cos 120579 |120579| lt120587

2

0120587

2lt |120579| le 120587

(185)



1198861119894=120587

2 119886

2119894=2

3 (186)



(187)




1= minus(119886




1 gt 0 and


11198941198861V lt 0 and 119886

21198941198862V le 0






2| le 1|119886

2119894|


21198862119894and 120582

2V = radic11988622V + 1198872

2V(iii) if 2120582



1198871V = minus2119887


(12)atan2(1198872V 1198862V) 1198861V = minus120582



11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199112119899 = 1199112Re119911

1


2119899= 1199112Re119911









+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

078

02

05

1 2 5

075 07 065

06 05

075

07

065

06

05

120578 lt 05

(a)

099

095

091

083 075

067

0 59 051

051

099

095

091

083

075 0

67 059

+j5

+j1

+j2

+j5

minusj5

minusj1

minusj2

minusj5

infin

+j2

minusj2

0 02

05

1 2 5

120578 lt 05

(b)


2119899= 1199112Re119911

1

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989505


minus00667 1198872V = minus03333 119886

1V = minus09333 and 1198871V = 06667 (see






2V = minus00667 1198872V =

minus05333 1198861V = minus09333 and 119887


2119899= 01683 minus 11989513465

respectively

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 01 minus 11989508


119894 (120579) = 1 +1 + radic5

2cos 120579 + 2radic5


10cos 3120579

(188)


1= minus(119886




2 ge 0 which





08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










08

075

07

06

06

065

065

05

05

075 0

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

02

05

1 2 5

(a)

085

08 0

75 07 0

6

05 04

08

075

0405

060

7

120578 lt 05

+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2 5

(b)


2119899= 1199112Re119911

1


1119894=



2119899= 1199112Re119911




=




2V =

minus00626 1198872V = minus03578 119886

1V = minus09374 and 1198871V = 07155 (see





2V =

minus00447 1198872V = minus06261 119886

1V = minus09318 and 1198871V = 10007 (see


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 007 minus 11989504






(189)


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


2= 005 minus 11989507


1= minus(119886

1V minus 1198951198871V)1198861119894 1199112 = 0 and 119911

3= minus(119886

3V minus



3 ge 0 which


31198941198863V le 0





3| le 1|119886

3119894|


31198863119894and 120582

3V = radic11988623V + 1198872

3V(iii) if 271205822



and 1198871V = minus3119887


3V) 1205790V minus120585V3 = (13)atan 2(119887


1198871V = minus120582


11198941198861V2


1V minus 1198951198871V)1198861119894 and 1199113119899 = 1199113Re119911

1


1119894= (1 + radic5)2 and



3119899= 1199113Re119911

1 is presented in



+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0

08

08

075

075

07

07

06

05

02

05

1 2


3119899= 1199113Re119911

1







3V)3 Then according



1V =


3=


3V)3 Then according



1V =



2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 02 minus 11989505

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 01 minus 11989511




119863= 22535 into

(180) leads to

119894 (120579)

=

22535 [cos 120579 minus cos(1151205872

)] |120579| le115120587

2

0115120587

2le |120579| le 120587

(191)



1198861119894= 14586 119886

3119894= minus01026 (192)


+j2

+j5

+j1

+j2

+j5

minusj2

minusj5

minusj1

minusj2

minusj5

infin0 02

05

1 2

074

076

08

084

082

078

076

074


3119899= 1199113Re119911

1





3=



radic39 and 1198871V = 1198873V = 0


PUFCase study 74 = 1439120578 (193)






3V = 01026 1198873V =

00205 1198861V = minus11026 and 119887


3119899= 13228 minus 11989502646


3=




3V = 01540 1198873V =


2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra










2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 1 minus 11989502

2

3

4

1

0

2 3 410

Wav

efor

ms

Angle 120579120587

i(120579)

(120579)


3= 15 minus 11989512

01232 1198861V = minus11255 and 119887


3119899= 19439 minus 11989515552

respectively

8 Conclusion


Appendices




119863119896minus1

(120591) = 1 + 2

119896minus1

sum

119899=1

cos 119899120591 = sin ((2119896 minus 1) 1205912)sin (1205912)

(A1)


119865119896minus1

(120591) =1

119896

119896minus1

sum

119899=0

119863119899(120591) = 1 +

2

119896

119896minus1

sum

119899=1

(119896 minus 119899) cos 119899120591


(A2)


1198781(120591) =

119896minus1

sum

119899=1


(A3)


2

119896minus1

sum

119899=1


(A4)

119896minus1

sum

119899=1


(A5)

119896minus1

sum

119899=1

119899 (119896 minus 119899) cos (119896 minus 2119899) 120591


2sin3120591

(A6)

Denote 1198782(120591) = 2sum

119896minus1

119899=1(119896 minus 119899) sin 119899120591 119878

3(120591) = sum

119896minus1

119899=1cos(119896 minus

2119899)120591 and 1198784(120591) = sum

119896minus1

119899=1119899(119896 minus 119899) cos(119896 minus 2119899)120591

Notice that 1198782(120591) = 2119896119878

1(120591) + 119889119863

119896minus1(120591)119889120591 which




sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra











sin (119896 minus 1) 120591sin 120591

=119890119895(119896minus1)120591


119890119895120591 minus 119890minus119895120591

= 119890119895119896120591

119890minus2119895120591


1 minus 119890minus2119895120591

= 119890119895119896120591

119896minus1

sum

119899=1

119890minus2119895119899120591

=

119896minus1

sum

119899=1

119890119895(119896minus2119899)120591

=

119896minus1

sum

119899=1

cos (119896 minus 2119899) 120591

(A7)

From 4119899(119896 minus 119899) = 1198962


4(120591) =

1198962

1198783(120591) + 119889

2

1198783(120591)119889120591





119881119899(119909) = cos 119899120591 when 119909 = cos 120591 (B1)



119880119899(119909) =

sin (119899 + 1) 120591sin 120591

when 119909 = cos 120591 (B2)


1198810(119909) = 1 119881

1(119909) = 119909

119881119899+1

(119909) = 2119909119881119899(119909) minus 119881

119899minus1(119909)

1198800(119909) = 1 119880

1(119909) = 2119909

119880119899+1

(119909) = 2119909119880119899(119909) minus 119880

119899minus1(119909)

(B3)


2(119909) = 2119909

2

minus 1 1198813(119909) = 4119909

3

minus 3119909 1198814(119909) = 8119909

4

minus

81199092

+ 1 1198802(119909) = 4119909

2

minus 1 1198803(119909) = 8119909

3

minus 4119909 and 1198804(119909) =

161199094

minus 121199092

+ 1



Acknowledgment


References






































Volume 2014




Journal of











Journal of


Function Spaces






Algebra



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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