395 The Discrete-TIme Fourier Transform Chap 5

ion to its consequences for the properties of discrete-time Fourier series . .j 1 be useful in reducing the complexity of the calculations involved in det u. !r series representations. This is illustrated in the following example. er.

pie 5.16 Isider the following periodic signal with a period of N = 9:

! sin(57TnI9) x[n) = ~ sin(7TnI9) ' n # multiple of9

(5.72){ 9' n = multiple of 9

:hapter 3, we found that a rectangular square wave has Fourier coefficients in a fonn ;h as in eq. (5.72). Duality, then, suggests that the coefficients for x[n) must be in the n of a rectangular square wave. To see this more precisely, let g[n) be a rectangular are wave with period N = 9 &uch that

I, Inl:S 2 g[n) = { 0, 2 < Inl :S 4.

~ Fourier series coefficients bk for g[n) can be determined from Example 3.12 as

! sin(57TkI9) k # multiple of 9

b = 9 sin(7TkI9) , k

5{ k ,= multiple of 99'

The Fourier series analysis equation (3.95) for g[n) can now be written as

I 2 = 9.2:(l)e-J27TnkJ9

n= -2

bk

'rchanging the names of the variables k and n and noting that x[n1 = bn , we find that

I 2 x[n) = 9.2:(l)e-J27Tnk!9.

k= -2

ing k' = - k in the sum on the right side, we obtain

I 2 x[n) - '" e+ J27Tnk'/99 L .

k' =-2

Illy, moving the factor 1/9 inside the summation, ' we see that the right side of thiS tion has the form of the synthesis equation (3.94) for x[n). We thus conclude that

Fourier coefficients of x[n) are given by

a = {1/9, /k/:S 2 k 0, 2 < Ikl :S 4

of course, are periodic with period N = 9

Sec. 5.7 Duality

5.7.2 Duality between the Discrete-Time Fourier Transform and the Continuous-Time Fourier Series

In addition to the duality for the discrete Fourier series, there is a duality between the discrete-time Fourier trans/onn and the continuous-time Fourier series. Specifically, let us compare the continuous-time Fourier series equations (3.38) and (3.39) with the discrete­time Fourier transform equations (5.8) and (5.9). We repeat these equations here for con­venience:

[eq. (5.8)] x[n] = - X(eJW)eJwndw,If ' . 27T 211"

(5.73)

[eq. (5.9)] X(e jW ) = +00L x[n]e- jwn, (5.74)

n= -00

[eq. (3.38)] x(t) = +00L akejkwol, (5.75)

k= -00

[eq. (3.39)] ak = ~ f x(t)e- jkwoldt.T T

(5.76)

Note that eqs. (5.73) and (5.76) are very similar, as are eqs. (5.74) and (5.75), and in fact, we can interpret eqs. (5.73) and (5.74) as a Fourier series representation of th!! periodic frequency response X(e jUJ ). In particular, since X(e jUJ ) is a periodic function of w

with period 27T, it has a Fourier series representation as a weighted sum of harmonically related periodic exponential functions of w. all of which have the common period of 27T. That is, X(e jW ) can be represented in a Fourier series as a weighted sum of the signals

jwne , n = 0, ± 1, ±2, ... '. From eq. (5.74), we see It,at the nth Fourier coefficient in this expansion-i.e., the coefficient multiplying ejwn-is x[ - n]. Furthermore, since the period of X(e jW ) is 27T, eq. (5.73) can be interpreted as the Fo~er series analysis equation for the Fourier series coefficient x[n]-i.e., for the coefficient multiplying e- jwn in the expression for X(e jW ) in eq. (5.74). The use of this duality relationship is best illustrated with an example.

Example 5.1 7

The duality between the discrete-time Fourier transform synthesis equation and the continuous-time Fourier series analysis equation may be exploited to determine the discrete-time Fourier transform of the sequence

x[n] = sine7TnI2) 7Tn

.. To use duality, we first must identify a continuous-time signal get) with period T = 27T and Fourier coefficients ak = x[k]. From Example 3.5, we know that if g(t) is a periodic square wave with period 27T (or, equivalently, with fundamental frequency Wo = 1) and with

I, ItI :S TI , get) = { 0, TI < ItI :S 7T

then the Fourier series coefficients of g(t) are

----

396 The Discrete-Time Fourier Transform Chap. 5

ak sin(kT.)

k1r

Consequently, if we take T. equation for get) is

= 'TT/2, we will have ak = x[kJ. In this case the anal ' YS.s

rr rr12 sin('TTkl2) = ~ I g(t)e-jk1dt = ~I (1)e- jk1dt.

'TTk 2'TT - rr 2'TT -rrl2

Renaming k as nand t as w, we hav!':

sin ('TTnl2) 1 Irrl22'TT (l)e-jlUUdw (5.77)

Replacing n by -n on both sides of eq. (5.77) and no.ting that the sinc function is even we obtain •

sin ('TTnl2) 1 Irrl2 'd

'TTn - rrl2 •

(1 )eJnw--'------'- = - w. 'TTn 2'TT -rrl2

The right-hand side of this equation has the form of the Fourier transform synthesis equation for x[n], where

X(e jW ) = {I Iwl:$ 'TT12 o 'TT12 < Iwl :$ 'TT'

In Table 5.3, we present a compact summary of the Fourier series and Fourier trans­form expressions tor both continuous-time and discrete-time signals, and we also indicate the duality relationships that apply in each case.

TABLE 5.3 SUMMARY OF FOURIER SERIES AND TRANSFORM EXPRESSIONS

Continuous time Discrete time

Time domain Frequency domain Time domain Frequency domain I

x(l) = I G, = I

x[nj = I ax = L'_(N) G,elk(Z"iN)n kL,_(N) x[nje-I,(Z"'N)'IL;-:_7J akeJlcwot I

I To Iro x(l)e - I"'O' I IIFourier

Series continuous time I

I discrete frequency ~ discrete time ~ discrete frequency

periodic in time I aperiodic in frequency ~".I.' periodic in time periodic in frequency I ~~ I ­

x(l) = I

I X(jw) = x[n] = I

I X(e j ",) = II L+oo~ x(l)e -IW'dl L;: - 00 x[nje- I""i; L+oooo X(jw) eIW'dw i; Iz" X(e IW)e lw, II

Fourier II

discrete time I continuous frequencyTransform continuous time ~ continuous frequency I periodic in frequencyaperiodic in time Iaperiodic in time , aperiodic in frequency

5.8 SYSTEMS CHARACTERIZED BY LINEAR CONSTANT-COEFFICIENT DIFFERENCE EOUATIONS

A general linear constant-coefficient difference equation for an LTI system with input .t[II] and output y[n] is of the form

Sec. 5.B Systems Characterized by Linear Constant-Coefficient Difference EI

N M L aky[n - k] = L bkx[n - kl. k=O k=O

The class of systems described by -such difference equations is qu useful one. In this section, we take advantage of several of the prop time Fourier transform to determine the frequency response H(e j6

described by such an equation. The approach we follow closely pal in Section 4.7 for continuous-time LTI systems described by lineru differential equations.

There are two related ways in which to determine H(e jW ). The we illustrated in Section 3.11 fot several simple difference equation fact that complex exponentials are eigenfunctions of LTI systems. SI e

jwn is the input to an LTI system, then the output must be of the form tuting these expressions into eq. (5.78) and performing some algebra, H(e

jW ). In this section, we follow a second approach making use of the

ity, and time-shifting properties of the discrete-time Fourier transform and H(eF") denote the Fourier transforms of the input x[n], output) sponse h[n], respectively. The convolution property, eq. (5.48), ofthe ( transform then implies that ft' .' ~,

jW H(e jW ) = Y(e ). ___ X(e JW )

Applying the Fourier transform to both sides of eq. <5.78) and m time-shifting properties, we obtain the expression ,

N M

Lake- jkwY(ejw) = Lbke-jWX(ejW),

k =O k=O

or equivalently,

H(e jW ) = Y(e jW

) ",M b e- jkwL,k =O k

X(e jW ) ",N 'w'L,k=Oake ­J

\

Comparing eq. (5.80) with eq. (4.76), we see that, as in the case of conI is a ratio of polynomials, but in discrete time the polynomials are iJ The coefficients of the numerator polynomial are the same coefficieJ right side of eq. (5.78), and the coefficients of the denominator poly! as appear on the left side of that equation. Therefore, the frequency system specified by eq. (5.78) can be written down by inspection.

The difference equation (5.78) is generally referred to as an 1 equation, as it involves delays in the output y[n] of up to N time step nator of H(e jW

) in eq. (5 .80) is an Nth-order polynomial in e- jw

Example 5.18

Consider the causal LTI system that is characterized by the differ

y[nJ - ay[n - 1] = x[n],

The Discrete-Time Fourier Transform Chap 5

sin(kT1) ak = ---'k:-7T­

equently, if we take Tl = 7T/2, we will have ak = x[kJ. In this case the anal . IOn or g t. fi ( ) . IS YSIS

sin (7Tkl2) 1 J1T () -jk'd _ 1 J1Tf2 (1) -jk'd-.- = - g t e t - - e t. 27T -1T 27T -1Tf2

ning k as n and t as w, we haVf~

sin(7Tnl2) = ~ J1Tf2 (l)e-j'""dw. 7Tn 27T -1Tf2 (5.77)

cing n by -n on both sides of eq. (5.77) and no.ting that the sinc fucction is even tain ,

sin (7Tnl2) 1 J 1Tf2 .--'--- = - (l)e1""'dw.

7Tn 27T -1T12

ght-hand side of this equation has the fonn of the Fourier transfonn synthesis on for x[nJ, where

JW. {I Iwl:s 7T12 X(e ) = 0 7T/2 < Iwl :s 7T

.3, we present a compact summary of the Fourier series and Fourier trans ­s for both continuous-time and discrete-time signals, and we also indicate !onships that apply in each case.

ER SERIES AND TRANSFORM EXPRESSIONS

time Discrete time

Frequency domain Time domain Frequency domain at - x[nj = ak = *ITo x(t)e- Itwo' Lk _(N) ake}k(hIN)n ~ Lk _(N) x[nje - It(I,,'N)n

discrete frequency ~" discrete time ~ discre. te frequency apenodic in frequency ~<7/-' periodic in time ~ periodic in frequency

"'~ X(jw) = x[nj = X(elw ) = ~ II" X(e}W)e}wn " L+~oc xU)e - IWldt L;= _ ~ x[nje - I",n

continuous frequency discrete time continuous frequencyaperiodic in frequency aperiodic in time periodic in frequency

~ED BY LINEAR CONSTANT-COEFFICIENT S

ffi ". ·th· put x[nJnstant-coe clent dIfference equatIOn for an LTI system WI In of the form

Sec. 5.8 Systems Characterized by Linear Constant-Coefficient Difference Equations 397

N M

2: aky[n - k] = 2: bkx[n - kl. (S.78) k=O k=O

The class of systems described by such difference equations is quite an importat·t and useful one. In this section, we take advantage of several of the properties of the discrete­time Fourier transform to determine the frequency response H(e jW ) for an LTI system described by such an equation. The approach we follow closely parallels the discussion in Section 4.7 for continuous-time LTI systems described by linear constant-coefficient differential equations.

There are two related ways in which to determine H(e jW ). The first of these, which we illustrated in Section 3.11 for several simple difference equations, explicitly uses the fact that complex exponentials are eigenfunctions of LTI systems. Specifically, if x[n] = ejwn is the input to an LTI system, then the output must be of the form H(ejW)ejwn. Substi­tuting these expressions into eq. (S.78) and performing some algebra allows us to solve for H(e jW ). In this section, we follow a second approach making use of the convolution, linear­ity, and time-shifting properties of the discrete-time Fourier transform. Let X (e jW ), Y (e jW ), and H(e jW ) denote the Fourier transforms of the input x[n], output y[n], and impulse re­sponse h[n], respectively. The convolution property, eq. (S.48), of the discrete-time Fourier transform then implies that r' . " . ~

. Y(e jW ) HJ::-W) = X(e jW )' (S.79)

Applying the Fourier transform to both sides of eq. (5.78) and using the linearity and time-shifting properties, we obtain the expression,

N ML ake- jkw Y(e jW ) = L bke- jkw X(e jW ), k=O k=O

or equivalently,

Y( jW) ,M b - jkwkeH(ejW) = _e_._ = L...k=O . (S.80)

X(e)W) ,N ake- jkwL...k=O

\

Comparing eq. (S.80) with eq. (4.76), we see that, as in the case of continuous time, H(e jW ) is a ratio of polynomials, but in discrete time the polynomials are in the variable e- jw. The coefficients of the numerator polynomial are the same coefficients as appear on the right side of eq. (S.78), and the coefficients of the denominator polynomial are the same as appear on the left side of that equation. Therefore, the frequency response of the LTI system specified by eq. (S.78) can be written down by inspection.

The difference equation (S.78) is generally referred to as an Nth-order difference equation, as it involves delays in the output y[n] of up to N time steps. Also, the denomi­nator of H(e jW ) in eq. (S.80) is an Nth-order polynomial in e- jw

Example 5.18

Consider the causal LTI system that is characterized by the difference equation

y[nJ - ay[n - 1] = x[n], (5.81)

.

The Discrete-TIme Fourier Transform398 Chap. S

with lal < 1. From eq. (S.80). the frequency response of this system is

jWH(e ) = 1 _ ar jw (5.82)

Comparing this with Example S .1. we recognize it as the Fourier transform of the quence a" u[ n]. Thus. the impulse response of the system is se..

h[n] = a"u[n]. (5.83)

Example 5. 19

Consider a causal LTI system that is characterized by the difference equation

y[n] - 43 y[n - 1] + 8"

1 y [n - 2] = 2x[n]. (5 .84)

From eq. (S.80). the frequency response is

2 H(e jW

) = 3· I _ j2w (S.85)1 - -rJW + ge4

As a tirst step in obtaining the impulse response. we factor the de~ominator ofeq. (S.85):

H(e Jw. ) == 2

. (5.86)(1 - ~r jW)(1 - ~r jW)

H(e jW ) can be expanded by the method of partial fractions. as in Example AJ in the appendix. The result of this expansion is

4 2jW (5.87)H(e ) = I -jw 1 _ !e-jw1 - ze 4

The inverse transform of each term can be recognized by inspection. with the result thai

(5.88)h[n] = 4 G)" urn] - 2 (~r urn].

The procedure followed in Example 5.19 is identical in style to that used in contin­uous time. Specifically, after expanding H (e jW ) by the method of partial fractions, we ~an find the inverse transform of each term by inspection. The same approach can be appb~ to the frequency response of any LTI system described by a linear constant-coefficient~­ference equation in order to determine the system impulse response. Also, as illustrat~ ~ the next example, if the Fourier transform X(e jW ) of the input to such a system is a ra~o 0

pplynomials in e- jw. then Y(e jW ) is as well. In this case, we can use the same techmque to find the response y[n] to the input x[n].

Example 5.20

Consider the LTI system of Example 5.19, and let the input to this system be

x[n] = (~r urn].

Sec. 5.9 Summary

Then. using eq. (S.80) and Example 5.1 or S.18. we 0

Y(eiw) = H(eJW)X(eiw ) = [ . 2 (1 - ~rJW)(1 ­

2 (1 - ! r iW)(1 - !

2 4

As described in the appendix. the form of the partial-

Y(e iw ) = BII + ~ 1 - !e - iw (1 - !riw

4 4

where the constants B 11 • B12•and B21 can be detennir in the appendix. This particular expansion is worked the values obtained are

Bll = -4. BI2 = -2.

so that

4 2iw Y(e ) = - I -J'w - (1 - !riI - :Ie 4

The first and third terms are of the same type as th, while the second term is of the same form as one SI

these examples or from Table S.2. we can invert each the inverse transform .

{ (1)" (1\"y[n] = -4 4 - 2(n + 1) 4)

5.9 SUMMARY

In this chapter, we have paralleled Chapter 4 as we deve discrete-time signals and examined many of its important r ter, we have seen a great many similarities between con Fourier analysis, and we have also seen some important d lationship between Fourier series and Fourier transforms ogous to that in continuous time. In particular, our derivat transform for aperiodic signals from the discrete-time F very much the same as the corresponding continuous-time of the properties of continuous-time transforms have exac the other hand, in contrast to the continuous-time case, the of an aperiodic signal is always periodic with period 21T differences such as these, we have desclibed the duality. representations of continuous-time and discrete-time sign

The most important similiarities between continuous ysis are in their uses in analyzing and representing signal the convolution property provides us witll the basis for th LTI systems. We have already seen some of the utility oftJ

!

The Discrete-TIme Fourier Transform Chap. 5

with lal < 1. From eq. (5.80), the frequency response of this system is

H(eJW ) = 1

.1-ae- Jw (5.82)

Comparing this with Example 5.1, we recognize it as the Fourier transform of the quence a"u[n]. Thus, the impulse response of the system is se.

h[n] = a"u[n]. (5.83)

lCample 5. 1 9

Consider a causal LTI system that is characterized by the difference equation

y[n] - 43 y[n - I] + 81

y[n - 2] = 2x[n]. (5.84)

From eq. (5.80), the frequency response is

H(eJW. ) = 2

. (5.85)1 - 'ie-jw + !e- j2uJ4 8

As a tirst step in obtaining the impulse response, we factor the denominator ofeq. (5.85):

H(e Jw. ) :, 2

. (5.86)(1 - !e- jW)(l - !e- jW)

2 4

H(e jW ) can be expanded by the method of partial fractions, as in Example A.3 in the appendix. The result of this expansion is

4 2jW (5.87)H(e ) = ,- jw 1 _ ! e- jw1 - 'i e 4

The inverse transform of each term can be recognized by inspection, with the result that

(5.88)h[n] = 4 G)" urn] - 2 (~Jurn].

he procedure followed in Example 5.19 is identical in style to that used in contin­ne. Specifically, after expanding H(e jW ) by the method of partial fractions, we can inverse transform of each term by inspection. The same approach can be appli~

'equency response of any LTI system described by a linear constant-coefficient eli!­equation in order to determine the system impulse response. Also, as illustrated In

t example, if the Fourier transform X (e jW ) of the input to such a system is a ratio of nials in e- j w, then Y{e jW ) is as well. In this case, we can use the same technique the response y[n) to the input x[n).

xample 5.20

Consider the LTI system of Example 5.19, and let the input to this system be

x[n] = (~)" urn].

Sec. 5.9 Summary

Then, using eq. (5.80) and Example 5.1 or 5.18, we obtain

jW JW Y(e ) = H(eJW)X(e ) = [(1 -~e - jw~l- ~e - Jw)][I- le -J;;;]

(5.89) 2

( I - ! e-Jw)(1 - ! e- tw )2 . 2 4

As described in the appendix, the form of the partial-fraction expansion in this case is

Y(e JW ) = B" + B12 + ~ (5 .90) 1 - ! e- jw (1 - !e- jw )2 1- !e- JW

4 4 2

where the constants B 11, B 12, and B21 can be detennined using the techniques described in the appendix. This particular expansion is worked out in detail in Example AA, and the values obtained are

Bll = -4, B 12 = -2, B21 = 8,

so that

JW. 4 2 + 8 . (5.91)Y(e ) = - 1- !e- jw - (1- ie - jwj2 1 - ~e - JW 4 .

The first and third terms are of the same type as those encountered in Example 5.19, while the second tern! is of the same form as one seen in Example 5.13. Either from these examples or from Table 5.2, we can invert each of the terms in eq. (5.91) to obtain the inverse transform .

y[n] = {- 4 (4I)" - 2(n + 1) (14)\" + 8 (I2)"} urn]. (5.92)

5.9 SUMMARY

In this chapter, we have paralleled Chapter 4 as we developed the Fourier transform for discrete-time signals and examined many of its important properties. Throughout the chap­ter, we have seen a great many similarities between continuous-time and discrete-time Fourier analysis, and we have also seen some important differences. For example, the re­lationship between Fourier series and Fourier transfonns in discrete time is exactly anal­ogous to that in continuous time. In particular, our derivation of the discrete-time Fourier transfonn for aperiodic signals from the discrete-time Fourier series representations is very much the same as the corresponding continuous-time derivation. Furthermore, many of the properties of continuous-time transforms have exact discrete-time counterparts. On the other hand, in contrast to the continuous-time case, the discrete-time Fourier transfonn of an aperiodic signal is always periodic with period 2n. In addition to similarities and differences such as these, we have desclibed the duality relationships among the Fourier representations of continuous-time and discrete-time signals.

The most important similiarities between continuous- and discrete-time Fourier anal­ysis are in their uses in analyzing and representing signals and LTI systems. Specifically, the convolution property provides us with the basis for the frequency-domain analysis of LTI systems. We have already seen some of the utility of this approach in our discussion of

_ __ _

The Discrete-Time Fourier Transform400 Chap. 5

filtering in Chapters 3-5 and in nul' examination of systems described by linear cOnst coefficient differential or differenl'e equations . and we will gain a further appreciation~t' its utility in Chapter 6, in which we examine tiltering and time-versus-frequency issue Or

more detail. In addition, the multiplication properties in cuntinuous and discrete time sIn

essential to our development of sampling in Chapter 7 and communications in Chapte~e

The first section of problems belongs to the basic category and the answers are provided in the back of the book. The remaining three sections contain problems belonging to the basic, advanced, and extension categories, respectively.

BASIC PROBLEMS WITH ANSWERS

5.1. Use the Fourier transform analysis equation (5 .9) to calculate the Fourier transforms of: (a) (~)II 'uln - I] (b) (~);II 1

Sketch and label one period of the magnitude of each Fourier transform.

5.2. Use the Fourier transform analysis equation (5.9) to calculate the Fourier transforms of: (a) lJ[n - I] + o[n + I] (b) o[n + 2] - o[n - 2]

Sketch and label one period of the magnitude of each Fourier rransfoffil .

5.3. Determine the Fourier transform for - 7T ~ W < 7T in the case of each of the fol­lowing peliodic signals: (a) sin(1n +~) (b) 2 + cos(~n + i)

5.4. Use the Fourier transform synthesis equation (5.8) to determine the inverse Fourier transforms of: (a) X, (e jW ) = "'>Z= _x{27TO(W - 27Tk) + 7TO(W - ~ - 27Tk) + 7TO(W + ~ - 27Tk

(b) X~(ejW) = { 2},. 0 < w ~ 7T -2}, - 7T < w ~ 0 .

er 5.5. Use the Fourier transform synthesis equation (5.8) to determine the inverse FoUO

transform of X(e lW ) == IX(ejW)lejH(eIW), where

3w0 Iwl <. {1. ~ ~ and -r.X(e jW ) = IX(elW)1 = 0, ~ ~ Iwl ~ 7T 2

Use your answer to determine the values of n for which x[nl = 0

5.6. Given that x[n] has Fourier transform X(e jW ). express the Fourier transforms of following signals in terms of X(e lW ). You may use the Fourier transform proper1

listed in Table 5 .1. (a) XI [11] = x[1 - II] + x[-1 - 11] (b) X2[11] ="1 "!~ ,1 11 1

(C)tl[lI] = (11 _-1)2 x [lI]

Chap. 5 Problems

5:7. For each of the following Fourier transforms, use Fouri 5.1) to deterntine whether the corresponding time-do nary, or neither and (ii) even, odd, or neither. Do this, of any of the given transforms.

. . ~'O(a) X,(e lW ) = e-lw.L.k=,(sinkw) (b) X2(e jW ) = } sin(w)cos(Sw) (c) X3(e jW ) = A(w) + ejB(w) where

I, o ~ Iwl ~ i A(w) = 0, and B{ :!!. < Iwl ~ 7T8

5.8. Use Tables 5.1 and 5.2 to help determine x[n] when if

. 1 (Sin ~w)X(e lW ) = . . -._2_ + 57TlJ(W),

- e- lW sm ~ 2 I

5.9. The following four facts are given about a particular si form X(e jW ):

1. x[n] = 0 for n > O. 2. x[O] > O. 3. ~m{X(ejW)} = sinw -sin2w.

4. 2~ t'1T IX(e jw )i2 dw = 3. Fr0 Deterntine x[n].

5.10. Use Tables 5.1 and 5.2 in conjunction with the fact th2

X(e jo ) = Lex>

x[n] II::;:-:X;

to deterntine the numerical value of

'" (l)nA = Ln 2 . n=O

5.11. Consider a signal g[n] with Fourier transform G(e jW ).

g[n] = x(2)[n],

where the signal x[n] has a Fourier transform X(e jW ). such that 0 < a < 27T and G(elW) = G(ej(w-cr».

5.12. Let

, V[lI] sinin)- * (Sinwcn - ( 7T1I 7Tn

where * denotes convolution and lWei :s 7T