Chapter 4 Solution

Chapter 4 Solution - Chapter 4 Answers 4.1. (a) Let $(t) =...

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Unformatted text preview: Chapter 4 Answers 4.1. (a) Let $(t) = e‘2(“1)u(t— 1). Then the Fourier transform X(jw) of x(t) is: m ‘ X(jw) = / e'2(t'1)u(t - l)e"“"dt -W I! m . / e—2(t—-l)e-_1wtdt 1 = e-J‘w/(z + jw) |X(jw)| is as shown in Figure 84.1. (b) Let $(t) = e'm'”. Then the Fourier transform X(jw) of a:(t) is: m . X(jw) = / e"2|"1le”“’tdt -oo 00 . 1 . / e—2(t-1)e—Jwtdt + / 62(t—1)e—Jwtdt l --00 = e‘jw/(z +21») + e‘jw/(z - M = 4e'jw/(4+w2) ll |X(jw)| is as shown in Figure 84.1. lfla‘ ml luau“ 2. Va. 0 N O (a) Figure 54.1 (b) 4.2. (a) Let 2:1(t) = 6(t + 1) + 6(t — 1). Then the Fourier transform X1(jw) of z(t) is: m . X1(jw) = / [6(t + 1) + 6(t—1)]e'3“"dt —oo = el‘“ +e‘j‘” = 2cosw [Xx(jw)| is as sketched in Figure 84.2. (b) The signal 2:2(t) = u(—2 — t) + u(t — 2) is as shown in the figure below. Clearly, $544.» — t) + u(t — 2)} = m —'2) — at + 2) 129 4.3. Therefore, [00 [6(t — 2) - 6(t + 2)]e‘jw‘dt X2010) = e41“ — 6”“ = —2j sin(2w) IX1(jw)| is as sketched in Figure 84.2. —IV,_ 0 Iv,L as]; Figure 84.2 - 5 "/1 (a) The signal 2:1(t) = sin(21rt + 1r/4) is periodic with a fundamental period of T = 1. This translates to a fundamental frequency of we = 21r. The nonzero Fourier series coefficients of this signal may be found by writing it in the form 1 . . = _ (21rt+1r/4) _ —J(21rt+1r/4) 21(t) 2], (e’ e ) ___ Lew/46in: _ _1_e—j1r/4e—j21rt 23' 2j Therefore, the nonzero Fourier series coefficients of z‘(t) are 1 ~ . 1 a = fern/46121” a_ ____. __:e 1 i l From Section 4.2, we know that for periodic signals, the Fourier transform consists of a train of impulses occurring at kwo. Furthermore, the area under each impulse is 27r times the Fourier series coefficient ak. Therefore, for 11(t), the corresponding Fourier transform X1(jw) is given by X1(jw) = 21ra16(w — mg) + 21ra_16(w + (do) = (7r/j)ei"/‘6(w — 2n) — (1r/j)e‘j”/46(w + 21) (b) The signal 12(t) = 1+cos(61rt+1r/ 8) is periodic with a fundamental period of T = 1/3. This translates to a fundamental frequency of wo = 61r. The nonzero Fourier series coefficients of this signal may be found by writing it in the form —j1r/4e—j21rt 1 . . _ (61t+1r/8) _ —J(61rt+1r/8) l + 2 (e’ e ) l + gal/881.611+ ée-jr/Be—jfi‘nt $20) = 130 Therefore, the desired result. is sin in cos w .7-‘T{Odd part ofz(t) = — jw 4.10. (a) We know from Table 4.2 that ' t if? (E) Rectangular function Y(jw) [See Figure 84.10] Therefore . 2 t (5:: ) FT (1/2”) [Rectangular function Y(jw) 1‘: Rectangular function YUM] This is a triangular function Y1(jw) as shown in the Figure 84.10. 16“) Figure $4.10 Using Table 4.1, we may write —j/21r, 0 S :02 j/21r, ——2 S w < 0 0. otherwise 00 V- 4 co 2 _.1_ - 2 _; /_ t(m) dt—21r/:°o|X(]w)|dw—21r3 135 Since, H(jw) = 1/(3 + jw), we have You) z HUD) Taking the inverse Fourier transform of X (jw), we have X00) = 1/(4 +210) 22(t) = e’“u(t). 4.20. From the answer to Problem 3.20, we know that the frequency response of the circuit is , 1 Hum) _ —w2 +jw + 1' Breaking this up into partial fractions, we may write 1 —1 —-l H (in!) = -—-. —-————— + ————— J~/§ %—32@j+jw %+32@j+jw Using the Fourier transform pairs provided in Table 4.2, we obtain the FOIU‘K’!‘ *rilnSfOI‘m of H ( jw) to be 1 1 13- -1-£3' ht=_——_(-§+21)t++e(2 nglut' ( ) j\/§ [ e l ( ) Simplifying, h(t) = %§e'%‘sin(—?t)u(t). 4.21. (a) The given signal is 8—01 . 1 . cos(wot)u(t) = e'°'e""°tu(t) + Ee‘“'e"“’°‘u(t). 1 2 Therefore, 1 1 X0“) = ' ' (b) The given signal is a:(t) = 6—31 sin(2t)u(t) + €3t$in(2t)u("t)‘ We have 1/23' 1/21' _ -32 - F‘T - = _______ _ _________ $1(t)——e srn(2t)u(t) (——-)X1(_7w) 3_j2+jw 3+j2+jw Also, -32- _ __ __ FT - =__ _- =_______‘__________;____. “(n—e sm(2t)u( t)— m t)<—+X2(Jw) 2m 1w) 3_ 1.2 w 3+ 1.2 _, jw Therefore, X('w)-—X('w+X('w)-—-——————3j ______3j J — l] 2.7 9+(w_2)2' 141 (c) Using the Fourier transform analysis equation (4.9) we have (d) Using the Fourier transform analysis equation (4.9) we have 1 (e) We have . I $(t) = (1/2j)te"2‘e’“u(t) — (1/2j)te'2‘e"“u(t). Therefore, / . __ l/2j 1 2;“ X0“) _ (2—j4+jw)2 (2+j4—jw)2' (f) We have _ sm1rt FT . _ 1, [cu] < 7r x1“) — 1rt H X1 (3w) _ { 0, otherwise ' Also _ 2 (t 1) 2“ I I 2 $111 7I’ - FT . _ e" , w < 7r $20) — 1r(t — 1) H quw) _ { 0, otherwise ' FT . 1 . . 1(1) = z1(t)$2(t) H X001) = ‘2;{X1(Jw) * X2(Jw)}- Therefore, _ e‘J‘“, lwl < 1r . __ (1/21r)(37r + w)e'j“’, ~31r < w < —7r X0“) _ (1/27r)(31r — w)e‘j“’, 1r < w < 31r ' 0 otherwise i (g) Using the Fourier transform analysis eq. (4.9) we obtain X(jw) = 3]— [cosZw — Egg]. 1.: w (b) If 00 mt) = 2 6a — 2k). k=-oo then :1:(t) = 211(t) + z1(t — 1). Therefore, qu) = X1(jw)[2 + e'“’] = 1r 2 6(a) — k1r)[2 + (-1)'=]. k=—oo 142 (i) Using the Fourier transform analysis eq. (4.9) we obtain 1 26"” 2e’j‘” — 2 X ' = __ + _____I (1w) jw -W2 .1102 (j) :z:(t) is periodic with period 2. Therefore, X(jw) = 1: Z X(jk1r)6(w — k1r), Ic=-oo where X (jw) is the Fourier transform of one period of z:(t). That is, X_ . 1 1 _ e—2(l+ju)) 8-2“ _ e—2(l+jw)] (Fl-:3 ‘73:?" l—jw - ejz’”, t < 3 432' (a) x“) = { 0 Lt'herwise (b) a:(t) = ée’jflaflt — 4) + -;—eJ"'/36(t + 4). (c) The Fourier transform synthesis eq. (4.8) may be written as w) = 51;] |X(jw)|ej<x(j“')ej“"dw. From the given figure we have ___1_ sin(t-3) cos(t—3)-—1 $(t)_7r[ t—3 + (t-—3)2 (d) $(t) = sint + ,3—rcos(21rt) (e) Using the Fourier transform synthesis equation (4.8), x“) __ coth + sint — sin2t _ j1rt jrrt2 ' 4.23. For the given signal so“), we use the Fourier transform analysis eq. (4.8) to evaluate the corresponding Fourier transform 1 _.. e—(l+jw) X0001): 1+jw (i) We know that $1“) = zo(t) + 10(4)- Using the linearity and time reversal properties of the Fourier transform we have 2 -— 2e‘loosw — 2mg“l sinw 1+u)2 X1(jw) = XOUW) + Xo(—.'iw) = 143 (ii) We know that z2(t) = In“) - 10(4). Using the linearity and time reversal properties of the Fourier transform we have —2w + 2e"l sinw + 2m:—1 l+m2 X2010) = Xo(jw) - Xo(—jw) =j[ 0°30] _ (iii) We know that $3(t) = 1:0(t) + 1:0(t + 1). Using the linearity and time shifting properties of the Fourier transform we have 1+ ej‘” -— (‘0 + 6‘") X3(jw) = XoUuJ) + ej”Xo(—jw) = 1 +11“ (iv) We know that 174(t) = t$o(t). Using the differentiation in frequency property X ('w) — 1X (2») 4 .7 - .7 dw o J - Therefore, _ l — 26“e'3“’ —jwe’1e”“’ 4.24. (a) (i) For 'Re{X(jw)} to be 0, the signal a:(t) must be real and odd. Therefore, signals in figures (a) and (c) have this property. (ii) For Im{X(jw)} to be 0, the signal z(t) must be real and even. Therefore, signals in figures (e) and (f) have this property. (iii) For there to exist a real 0: such that ejWXUw) is real, we require that z(t + a) be a real and even signal. Therefore, signals in figures (a), (b), (e), and (1") have this property. (iv) For this condition to be true, x(0) = 0. Therefore, signals in figures (a), (b), (c), (d), and (f) have this property. (v) For this condition to be true the derivative of :r:(t) has to be zero at t = 0. Therefore, signals in figures (b), (c), (e), and (1') have this property. (vi) For this to be true, the signal a:(t) has to be periodic. Only the signal in figure (a) has this property. (b) For a signal to satisfy only properties (i), (iv), and (v), it must be real and odd, and z(t) = 0, :c'(0) = 0. The signal shown below is an example of that. 144 Figure $4.24 4.25. (a) Note that y(t) = z(t + 1) is a real and even signal. Therefore. Y(jw) is also real and even. This implies that <IY(jw) = 0. Also, since Y(jw) = eJ“X(jw), we know that <IX(jw) = —w. (b) We have X(j0) =/ z(t)dt = 7. (c) We have [00X(jw)dw = 27m:(0) :3 47!. (d) Let Y(jw) == bridge”? The corresponding signal y(t) is (t)_ 1, —3<t<—1 y _ 0, otherwise ' Then the given integral is co / X(jw)Y(jw)dw = 21r{:c(t) at y(t)}¢=0 = 77r. —oo (e) We have N (X) / |X(jw)|2dw = 21r/ |a:(t)|2dt = 267r. -oo —00 (f) The inverse Fourier transform of Re{X(jw)} is the £v{z(t)} which is [2:(t) + x(—t)]/2. This is as shown in the figure below. Ewing ‘3 *2 l g. 3 Figure $4.25 4.26. (a) (i) We have Y(jw) = X(J'w)H(J'w)= [4:110] (1/4 (1/4) + (1/2) 4+jw"2+jw (2+jw)2 II 145 4.27. Taking the inverse Fourier transform we obtain 1 I 1 y“) = 4-e'“u(t) " 33-251“) + ate—2‘11“) (ii) We have YUw) X(jw)H(jw) = _ (1/4) (1/4) (1/4) (1/4) — . + ——-.-—2 - ——. + ——.—2 2+Jw (2+Jw) 4+Jw (4+Jw) Taking the inverse Fourier transform we obtain _ _1_ —2t l -22 _ _1_ -4t 1 —42 y(t) — 4e u(t) + 4te u(t) 4e u(t) + 4te u(t). (iii) We have Y(:iw) X(J'w)H(J'w) 1 1 [1m] [l—jw] 1/2 1/2 1+jw l—jw ll Taking the inverse Fourier transform, we obtain y(t) = gee-"'- (b) By direct convolution of :r(t) with h(t) we obtain 0, t < 1 y(t)= 1—e-(H), 1<t55 . e-(‘-5) - «fit-1), t> 5 Taking the Fourier transform of y(t), 2653‘” sin(2w) w(l + jw) 8-1.2” e‘j“’2 sin(2w) [l + jw] w X(J'w)H(jw) WW) II (a) The Fourier transform X (jw) is co ‘ 2 . 3 i / $(t)e'1“"dt=/e’JW‘dt—fe‘wtdt —oo 1 2 25in(::/2) {1 _ e—ju}e—j3w/2 X010) II 146 (c) We have 1 1 X‘J“)=4+jw‘m' Therefore, 1 (4 + J'w)(2 +110) ‘ Finding the partial fraction expansion of Y(jw) and taking the inverse Fourier trans- form, YUM) = X (J'w)H (1w) = y(t) = ée‘muU) — ée'“u(t). 4.35. (a) From the given information, mm,» = _.____V“+“’ =1, W Also, LL} <IH(jw) = —-tan’1 2 — tan‘1 2 = —2 tan’1 —. a a 0 Also, H(jw) = —1 + a :“jw => h(t) = -6(t) + 2ae—atu(t). (b) If a = 1, we have lHlel = 1, <H(jw) = --2tan'l w. Therefore, 2 t 1r 1r 1r y(t) _ cos(—\/r3. — 3) — cos(t — E) + cosh/5t — Y). 4.36. (a) The frequency response is , Y 'w 3 3 + in How) = (J ) , ( J ) X00) _ (4 + jw)(2 +10). (b) Finding the partial fraction expansion of answer in part (a) and taking its inverse Fourier transform, we obtain h(t) = [e"“ + e‘z‘] u(t). MIC»: (c) We have Y(jw) __ (9+ 3jw) X(jw) ‘ 8 +6jw ~w2' Cross-multiplying and taking the inverse Fourier transform, we obtain d’yU) dy(t) drit) dtz +6—dt +8y(t)=3 dt + 9:1:(t). 153 ...
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This note was uploaded on 02/06/2012 for the course EE 3610 taught by Professor Liang during the Fall '10 term at LSU.

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Chapter 4 Solution - Chapter 4 Answers 4.1. (a) Let $(t) =...

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