ME37500FINALEXAMThursday,May5,2016

8:00–10:00amEE129

Section(circleone):Chiu(11:30am)Deng(1:30pm)Zhang(3:30m)

HWID:____________

Name:___________________________

Instructions(1) Thisisaclosedbookexamination,butyouareallowedthreesingle-sidedhand-written

8.5×11cribsheets.

(2) YoucanonlyusetheapprovedMEcalculatorfortheexam.(3) WriteyournameandHWIDonthetopofeachpage.

(4) Youmustclearlyidentifyyouranswerstoreceivecredit.InPartA,youmustshowallofyourworktoreceiveanycredit.Leavingmultiplesolutionsandderivationswithoutidentifyingtheonethatistobeconsideredwillnotreceiveanycredit.

(5) Youmustwriteneatlyandshouldusealogicalformattosolvetheproblems.Youareencouragedtoreally“think”abouttheproblemsbeforeyoustarttosolvethem.

(6) AtableofLaplacetransformpairsandpropertiesofLaplacetransformsisonthelastpageoftheexam.

PARTA PARTBProblem Score Problem Score

1.(20) 1.(10) 2.(30) 2.(10)

3.(30) 3.(15)

4.(20) 4.(10) 5.(15)

6.(10)

7.(10) 8.(10)

9.(10)

Total(A) /100 Total(B) /100

TOTAL(A+B) /200

PART A Name: HWID:

Page 2 of 18

Problem1(20points)Considertheclosed-loopsysteminthefiguretotherightwithalooptransferfunctionL(s)oftheform

L(s)=K ⋅G(s) ,whereKisaconstantgain.TheBodediagramforG(s)(NOTincludingK)isgivenbelow

(A) (10points)UsingtheaboveinformationfindtheminimalvalueofgainKtoensurethe

magnitudeofthesteady-stateerror(e=r–y)foraunitstepdisturbance,d=1andzeroinput,r=0,islessthan0.001.

e = r − y = −y ⇒ E(s) = −Y(s) = − 11+ L(s)

⋅R(s)

eSS = lims→0

sE(s) = lims→0− s 1

1+ L(s)⋅R(s) = − lim

s→0s 1

1+ L(s)⋅ 1s= − lim

s→0

11+ L(s)

= − lims→0

11+ K ⋅ G(s)

= − 11+ K ⋅ G(0)

eSS = −1

1+ K ⋅ G(0)< 0.001 = 1

1000

From Bode plot: G(0) = 20dB = 10 ⇒ − 11+ K ⋅ G(0)

= 11+ 10K

< 0.001 = 11000

⇒ 1+ 10K > 1000 ⇒ 10> 999 ⇒ K > 99.9

PART A Name: HWID:

Page 3 of 18

Problem1(continued)

(B) (10points)Ignoredisturbance,i.e.letd=0andletK＝1.Whatisthemagnitudeofthesteady-stateerrorforinputr=sin(9t),i.e.asinusoidalinputwithfrequency9[rad/sec]andmagnitude1?(Hint:youcanapproximate1+L( jω) ≈ L( jω) .)

e = r − y ⇒ E(s) = 11+ L(s)

⋅R(s)

For r = sin(9t) ⇒ eSS(t) =1

1+ L( jω)ω=9⋅ sin(9t +φ)

eSS(t) =1

1+ L( jω)ω=9= 1

1+ L( jω)ω=9

≈ 1L( jω)

ω=9

= 1L( j 9)

From Bode plot: L( j 9) = 32dB = 39.8 or 40

⇒ eSS(t) =1

39.8≈ 0.025

PART A Name: HWID:

Page 4 of 18

Problem2(30Points)

Giventheunityfeedbacksystemshowntotheright,theplanttransferfunctionG(s)isgivenby

G(s)= 50

s (s /5)+1( ) andthecontrollerC(s)=1.

(A) (10points)Thegaincross-overfrequencyis15.4[rad/sec].Calculatethephasemarginoftheclosed-loopsystem?

Gain cross-over frquency ω gc = 15.4 [rad/sec]

At ω gc, !G( jω gc) = 0− 90° −!( jω gc / 5+ 1) = 0− 90° − tan−115.4 5

1

⎛

⎝⎜

⎞

⎠⎟

= 0− 90° − 72° = −162°

Phase Margin PM = 180° +!G( jω gc) = 180° − 162° = 18° = 0.314 [rad]

PART A Name: HWID:

Page 5 of 18

Problem2(Continue)

(B) (10points)Designaleadcompensatoroftheform

CL(s)=sω1

+1sω2

+1whereα =ω2

ω1>1

thatwillcreateaphaselead(φm)of55°atthegaincross-overfrequencyof15.4[rad/sec].Youwillneedtofindω1,ω2,andα.

For a lead compensator, amount of maximum phase lead φm is

φm = sin−1α − 1α + 1⎛⎝⎜

⎞⎠⎟ ⇒ α =

1+ sinφm1− sinφm

For φm = 55° ⇒ α =1+ sin55°1− sin55°

= 1+ 0.8191− 0.819

= 1.8190.181

= 10.06 ≈ 10

ωm =ω gc = 15.4=ω1 α ⇒ ω1 =ωmα= 15.4

10= 4.87

ω2 =α ⋅ω1 = 10 ⋅ω1 = 48.7

⇒α = 10,ω1 = 4.87 [rad/sec], and ω2 = 48.7 [rad/sec]

PART A Name: HWID:

Page 6 of 18

Problem2(Continue)

(C) (10points)Withthecontrollerdesignedinpart(B),thenewclosed-loopsystem’sgaincross-overfrequencyisat40[rad/sec]withaphasemarginof50°.Iftheactualplanttransferfunctionincludedatime-delayofTD[sec],i.e.

G(s)=G(s) ⋅e−TDs = 50

s (s /5)+1( )⋅e−TDs

DeterminethelargesttimedelayTDtheclosed-loopsystemwouldremainstable.(Hint:considertheadditionalphaselagintroducedbythetimedelayandbesuretouseconsistentunitsforthephaseangles.)

Gain cross-over frequency ω gc = 40 [rad/sec]

Phase margin PM = 50°

To ensure stability, additional phase lab due to delay =TDω

need to be less than the phase margin at ω gc, i.e.

TDω gc < PM ⇒ 40 ⋅TD < 50° =50180

⋅ π = 0.873

⇒ TD <50π

40 ⋅ 180= 0.022 ⇒ TD < 0.022 [sec]

PART A Name: HWID:

Page 7 of 18

Problem3(30Points)

Givenastate-spacerepresentationofasingleinputsingleoutputsystem:

!x = 0 1−6 −5

⎡

⎣⎢

⎤

⎦⎥x+ 01

⎡

⎣⎢

⎤

⎦⎥u

y = 1 2⎡⎣ ⎤⎦x

(A) (10points)Thedesiredclosed-loopsystemshouldhavea2%settlingtimelessthan1secondandapercentovershootlessthan4.32%.Onthecomplexplanebelow,clearlyidentify(intersectionwithreal-axisandtheappropriateangles)theregionforacceptableclosed-looppoles.

2% settling time: 4ζωn

< 1 ⇒ ζωn > 4

⇒ Real part of the roots < -4

%OS = 100e−ζ π

1−ζ 2 = 4.32%

⇒ζπ

1−ζ 2= − ln 4.32

100⎛⎝⎜

⎞⎠⎟

⇒ζ

1−ζ 2= 1 ⇒ ζ 2 = 1

2

⇒ ζ = 12= 0.707

⇒ sin−1ζ = 45°

45° -4

PART A Name: HWID:

Page 8 of 18

Problem3(Continue)

(B) (10points)Forthesystem,

ifthedesiredclosed-looppolesistobe−5± j5 3 ,findthestatefeedbackgainvectorK,whereu=−K⋅x+r,andristhereferenceinput

A = 0 1−6 −5

⎡

⎣⎢

⎤

⎦⎥ , B = 01

⎡

⎣⎢

⎤

⎦⎥ , K = K1 K2⎡⎣

⎤⎦

A− BK = 0 1−6 −5

⎡

⎣⎢

⎤

⎦⎥−

01

⎡

⎣⎢

⎤

⎦⎥ K1 K2⎡⎣

⎤⎦ =

0 1−6− K1 −5− K2

⎡

⎣⎢

⎤

⎦⎥

det[sI−(A− BK)]= s2 +(5+ K2)s +(6+ K1) = 0

Desired CL Characteristic Eq: DCL(s) = (s + 5)2 +(5 3)2 = s2 + 10s + 100= 0

DCL(s) = det[sI−(A− BK)]= s2 + 10s + 100= s2 +(5+ K2)s +(6+ K1)

⇒5+ K2 = 106+ K1 = 100

⇒K2 = 5K1 = 94

⇒ K = 94 5⎡⎣⎤⎦

!x = 0 1−6 −5

⎡

⎣⎢

⎤

⎦⎥x+ 01

⎡

⎣⎢

⎤

⎦⎥u

y = 1 2⎡⎣ ⎤⎦x

PART A Name: HWID:

Page 9 of 18

Problem3(Continue)

(C) (10points)Findthesteady-stategainoftheclosed-looptransferfunctionfromreferenceinputrtotheoutputyforthestate-feedbackcontrolyoufindinpart(B).(Hint:calculatesteady-stategainoftheclosed-looptransferfunction,youmaybeabletofinditinthestandardcontrollablecanonicalform)

Mathod 1:Closed-loop system:

!x = 0 1−6− K1 −5− K2

⎡

⎣⎢

⎤

⎦⎥x+ 0

1⎡

⎣⎢

⎤

⎦⎥r

= 0 1−100 −10

⎡

⎣⎢

⎤

⎦⎥x+ 01

⎡

⎣⎢

⎤

⎦⎥r

y = 1 2⎡⎣⎤⎦x

⇒ GCL(s) =2s + 1

s2 + 10s + 100⇒ Steady-state gain = GCL(0) =

1100

= 0.01

Mathod 2:Closed-loop system:

GCL(s) = C sI− A+ BK⎡⎣ ⎤⎦−1

B = 1 2⎡⎣⎤⎦Adj[sI− A+ BK]det[sI− A+ BK]

01

⎡

⎣⎢

⎤

⎦⎥

= 1 2⎡⎣⎤⎦

x 1x s

⎡

⎣⎢

⎤

⎦⎥

det[sI− A+ BK]01

⎡

⎣⎢

⎤

⎦⎥ =

2s + 1s2 + 10s + 100

⇒ GCL(s) =2s + 1

s2 + 10s + 100⇒ Steady-state gain = GCL(0) =

1100

= 0.01

PART A Name: HWID:

Page 10 of 18

Problem4(20points)

Considerthefollowingfirstordersystemanditsstatespacerepresentation:

G(s)= Y(s)

U(s) =4s+2 ⇔

!x =−2 ⋅ x+4 ⋅uy = x

(A) (10points)Designastate-feedback+integralcontrol,i.e.

u(t)=−k ⋅ x+kI ⋅ (r(τ )− y(τ ))dτ0t∫ ,

sothattheresultingclosed-looppolesareat−5± j5 3 .FindkandkI.

!x = −2x + 4u

y = x

Let u = −Kx + KI xI and xI = (r − y)dτ∫⇒ !xI = r − y = r − x

Augmented syste:

!x!xI

⎡

⎣⎢

⎤

⎦⎥ = −2 0

−1 0⎡

⎣⎢

⎤

⎦⎥

xxI

⎡

⎣⎢

⎤

⎦⎥+ 4

0⎡

⎣⎢

⎤

⎦⎥ −K KI⎡⎣

⎤⎦

xxI

⎡

⎣⎢

⎤

⎦⎥+ 0

1⎡

⎣⎢

⎤

⎦⎥r

= −2 0−1 0

⎡

⎣⎢

⎤

⎦⎥+

−4K 4KI0 0

⎡

⎣⎢

⎤

⎦⎥

⎡

⎣⎢

⎤

⎦⎥

xxI

⎡

⎣⎢

⎤

⎦⎥+ 0

1⎡

⎣⎢

⎤

⎦⎥r

= −2− 4K 4KI−1 0

⎡

⎣⎢

⎤

⎦⎥

xxI

⎡

⎣⎢

⎤

⎦⎥+ 0

1⎡

⎣⎢

⎤

⎦⎥r

Closed-loop Characteristic Eq:

det(sI− AE) =s + 2+ 4K −4KI

1 s= s2 +(2+ 4K)s + 4KI

Desired CL Char. Eq. DCL(s) = (s + 5)2 +(5 3)2 = s2 + 10s + 100= s2 +(2+ 4K)s + 4KI

⇒ 2+ 4K = 104KI = 100

⇒ K = 2KI = 25

PART A Name: HWID:

Page 11 of 18

Problem4(continued)

(B) (10points)Assumethatinpart(B)theclosed-looptransferfunctionbetweenthereferenceinputrandtheoutputyassociatedwiththestate-space+integralfeedbackcontrolis

GCL(s)=

Y(s)R(s) =

64s2+8s+64 .

IftheyouaretoimplementacontrollerC(s)intheblockdiagramtotherighttoachievethesameclosed-looptransferfunctionGCL(s),findthecontrollertransferfunctionC(s).

64s2 + 8s + 64

=C(s) 4

s + 21+ C(s) 4

s + 2

=4C(s)

(s + 2)+ 4C(s)

Solve for C(s)

⇒ 4(s2 + 8s + 64)C(s) = 64(s + 2)+ 256C(s)

⇒ 4s(s + 8)C(s)+ 256C(s) = 64(s + 2)+ 256C(s)

⇒ 4s(s + 8)C(s) = 64(s + 2)

⇒ C(s) =64(s + 2)4s(s + 8)

=16(s + 2)s(s + 8)

PART B Name: HWID:

Page 12 of 18

Problem1(10points)–ZeroandstepresponseGiventhefollowingtwoclosed-looptransferfunctions

G1(s)=

(s+1)(s2+ s+1) and

G2(s)=(0.1s+1)(s2+ s+1) ,

whichonewillhavealargerovershootforastepinput?Brieflyexplainwhy?

G1(s) has larger overshoot

Both transfer functions have the same poles at - 12±- 3

2G1(s) have zero at -1 and G2(s) has zero at -10zero at -1 is closer to the poles, hence G1(s) will have larger overshoot.

Problem2(10points)–SensitivityGivenatransferfunction

T(s)= A

τ s2+ s+kA,

determinethesensitivityofthetransferfunctionT(s)tochangesintheparameterA,i.e.SAT(s)

SAT (s ) = AT (s )⋅ ∂T (s )∂A

= AT (s )

⋅ (τ s2 + s + AK) − AK

(τ s 2 + s + AK)2= A

A(τ s 2 + s + AK)

⋅ τ s2 + s

(τ s 2 + s + AK)2

= τ s2 + s

(τ s 2 + s + AK)=

s (τ s + 1)(τ s 2 + s + AK)

PART B Name: HWID:

Page 13 of 18

Problem3(15points)–RouthstabilitytestConsiderthefollowingclosed-loopcharacteristicequation

1+K 1

s(s2+ s+1) =0

(A) (10points)UsetheRouthtesttodeterminetherangeofKforwhichtheclosed-loopsystemisstable.

Closed-loop Characteristic Eq.

DCL(s) = 1+ K1

s(s2 + s + 1)= 0 ⇒ s(s2 + s + 1)+ K = 0

⇒ s(s2 + s + 1)+ K = s3 + s2 + s + K = 0

Routh test:

s3 : 1 1s2 : 1 Ks : 1− K1: K

⇒ Stability ⇒ 1− K > 0K > 0

⇒ 0< K < 1

(B) (5points)Findthejω-axiscrossingpoleswhenKisatthestabilitylimit.

At statbility limit, substitute s = jω into CLCE

⇒ ( jω)3 +( jω)2 +( jω)+ K = 0

⇒ − jω 3 −ω 2 +( jω)+ K = 0

⇒Real part: K−ω 2 = 0

Imaginary part: ω −ω 3 = 0⇒ K =ω

2

ω(1−ω 2) = 0

⇒ K = 1ω = 1

⇒ jω-axis cross poles are ± j

PART B Name: HWID:

Page 14 of 18

Problem4(10points)–RootlocusandPIDcontrolA positioning system with the following transfer function

1( )( 1)

G ss s

=+

is to be put under PID control. There can be 4 different PID controllers, namely, P control, PI control, PD control, and PID control. The following set of four Root Loci are the open-loop transfer function of G(s) with the 4 (P, PI, PD, PID) types of controllers.

You are asked to match the 4 different Root Loci below with their corresponding controllers by circle the appropriate controller: Root Locus Controller

(A) ⇔ P PI PD PID

(B) ⇔ P PI PD PID

(C) ⇔ P PI PD PID

(D) ⇔ P PI PD PID

PART B Name: HWID:

Page 15 of 18

Problem5(10points)–DirectpoleplacementGiventhefollowingfeedbackcontrolsystem

UsedirectpoleplacementdeterminethepolynomialsD(s)andN(s)sothattheclosed-loopcharacteristicequationisDCL(s)=s3+4s2+5s+8=0

CLCE: s D(s)(s + 1)+ 2N(s) = s3 + 4s2 + 5s + 8

⇒D(s) = (s + a) 1st orderN(s) = (bs + c) 1st order

⇒ s(s + a)(s + 1)+ 2(bs + c) = s3 + 4s2 + 5s + 8

⇒ s3 +(1+ a)s2 +(a + 2b)s + 2c = s3 + 4s2 + 5s + 8

⇒1+ a = 4

a + 2b = 52c = 8

⇒a = 3b = 1c = 4

⇒ C(s) = s + 4s(s + 3)

PART B Name: HWID:

Page 16 of 18

Problem6(15points)–State-spacerepresentationThefollowingthirdordersystemdifferentialequation:

!!!y(t)+4 !!y(t)+5 !y(t)+6 y(t)=3!!u(t)+2!u(t)+u(t) canberepresentedasthefollowingblockdiagramandstate-spacerepresentation.Fillintheemptyblocksandmatrixelements.Thestatevariablesarenotedintheblockdiagram.

and

!x1

!x2

!x3

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

=

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

x1

x2

x3

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

+

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

u

y = ⎡⎣⎢

⎤

⎦⎥

x1

x2

x3

⎡

⎣

⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

Y(s)U(s)

= 3s2 + 2s + 1

s3 + 4s2 + 5s + 6

Let: X(s) = 1s3 + 4s2 + 5s + 6

U(s) ⇒ Y(s) = (3s2 + 2s + 1)X(s)

4

3

2

1

5

6

0 0 0 0

0 0 1

11

-4 -5 -6

1 2 3

PART B Name: HWID:

Page 17 of 18

Problem7(10points)–ModulationanddemodulationAconstantvaluedsignalXDCundergoesamplitudemodulationusinga4kHzcarriersignalwithamplitude1.TheresultingAMsignaliseM(t)=XDC·sin(8000πt).Todemodulate,theAMsignalisfirstmultipliedbyitscarriersignal,resultingine’M(t)=XDC·sin2(8000πt).Whatarethefrequencycomponentsandtheassociatedmagnitudesofthesignale’M(t)?

e'M(t) = XDC ⋅ sin2(8000πt) = XDC ⋅12− 1

2cos(2⋅ 8000πt)⎡

⎣⎢⎤⎦⎥=

XDC2−

XDC2

cos(2⋅ 8000πt)

⇒DC component: 0 [rad/sec] magnitude: XDC

2

16000π [rad/sec] or 8000 Hz: magnitude: XDC2

Problem8(10points)–KarnaughmapThetruthtableofathreeinputoneoutputcombinationallogicisgivebythetruthtabletotherightwithA,B,andCastheinputsandYistheoutput.CompletetheKarnaughmapbelowtogenerateaminimalrepresenationofYasabinaryfunctionoftheinputsA,B,andC.

Y =B + AC + AC

00 01 11 10

0

1

ABC

A B C Y

0 0 0 1

0 0 1 1

0 1 0 1

0 1 1 0

1 0 0 1

1 0 1 1

1 1 0 0

1 1 1 10

0 1 1

1 1 1

1

PART B Name: HWID:

Page 18 of 18

Problem9(10points)–GainmarginandphasemarginTheBodediagramofthelooptransferfunctionL(s)ofthefeedbacksystemtotherightisshownbelow.Identifythegainandphasecross-overfrequenciesandthecorrespondinggainmargin(GM)andphasemargin(PM).(YouneedtomarkontheBodediagramthetwocross-overfrequenciesandthecorrespondingmagnitudeandphaseinformationyouusedtocalculatetheGMandPM.)

Gain cross-over frequency ω gc = 0.4 [rad/sec]: PM = 180° +(−80°) = 100°

Phase cross-over frequency ωpc = 4 [rad/sec]: GM = 0−(−10dB) = 10dB= 3.16

~-80°

ωgc = 0.4 rad/s

~-10 dB

ωpc = 4 rad/s