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1、信号与系统奥本海姆英文课后答案chapteChapter 4 Answers 4.1 (a)Let x(t)=l-2(t-1)u(t-1).then the Fourier transform x(jw)ofx(t) is : X(jw)=l-2(t-1)u(t-1)l-jwtdt-1X(jw) is as shown in figure s4.1. (b) Let =l-2(t-1)-jwtldt=l-jw/(2+jw)x(t)=lX(jw)=l-jw-2t-1.then the Fourier transform X(jw)ofx(t) is : 11-jw-2t-1l-jwtdt=l-2
2、(t-1)l-jwtdt+l2(t-1)l-jwtdt -jw=l/(2+jw)+l/(2-jw)=4lX(jw) is as shown in figure s4.1 /4+w2(-)X(jw) X(jw) 2 1/2 0 0 w (a) (b) figure s4.1 4.2 (a) Let x1(t)=d(t+1)+d(t-1).then the Fourier transform X1(jw)ofx(t) is : X1(jw)=d(t+1)+d(t-1)l-jwtdt=ljw+l-jw=2cosw -w X1(jw) is as sketched in figure s4.2. (b
3、)the signal x2(t)=u(-2-t)+u(t-2) is as shown in the figure below .Clearly, du(-2-t)+u(t-2)=d(t-2)-d(t+2) dtTherefore =e-2jw-e2jw=-2jsin(2w)x1(jw) is as sketched in figure s4.2. x2(jw)=d(t-2)-d(t+2)e-jwtdt-X2(jw) X 1(jw)2 2 0 -p p w -3p/2 -p/2 p/2 3p/2 w figure s4.2 4.3 (a) the signal x1(t)=sin(2pt+p
4、/4) is periodic with a fundamental periodic of T=1. This translations to a fundamental frequency of w0=2p .the nonzero Fourier series coefficients of this signals of this signal may be found by writing it in the form 1j(2pt+p/4)(e-e-j(2pt+p/4) 2j1jp/4)j2pt1-jp/4)-j2pt=ee-ee2j2jx1(t)= therefore, the
5、nonzero Fourier series coefficients of x1(t)are a1=1ejp/4)ej2pt,a2=1e-jp/4)e-j2pt 2j2jForm Section 4.2 we know that for periodic signals, the Fourier transforms consists of train of impulse occurring at kw0.Furthermore, the area under each impulse is2p times the Fourier series coefficients 80 x1(jw)
6、=2pa1d(w-w0)+2pa-1d(w+w0) =(p/j)ejp/4d(w-2p)-(p/j)e-jp/4d(w+2p) ak.Therefore, for x1(t) the corresponding Fourier transforms x1(jw) is given by (b) The signal x2(t)=1+cos(6pt+p/8)is periodic with a fundamental period of T=1/3 .This translates to a fundamental frequency of w0=6p .The nonzero Fourier
7、series coefficients of this signal may be found by writing it in the form 1x2(t)=1+(ej(6pt+p/8)+e-j(6pt+p/8) 2 =1+1ejp/8ej2p6t1/-8j+e-jpe2pt6Therefore ,The nonzero Fourier series coefficients of x2(t) are a=1,a=1ejp/8ej6pt,a=1e-jp/8e-j6pt. 012-12Form Section 4.2 we know that for periodic signals, th
8、e Fourier transforms consists of train of impulse occurring at kw0 .Furthermore, the area under each impulse is 2p times the Fourier series coefficients ak .Therefore, for x2(t) the corresponding Fourier transforms x2(jw) is given by x2(jw)=2pa0d(w)+2pa1d(w-w0)+2pa-1d(w+w0) =2pd(w)+pejp/8d(w-6p)+pe-
9、jp/8d(w+6p) 4.4 (a) The inverse Fourier transforms is x(t)=(1/2p)2pd(w)+pd(w-4p)+pd(w+4p)ejwtdw 1- =(1/2p)2pejqt+pej4pt+pe-j4pt j4pt-j4pt=1+(1/2)e+(1/2)e =1+cos(4pt) (b) The inverse Fourier transforms is x2(t)=(1/2p)X2(jw)ejwtdw - =(1/2p)2ejwtdw+-2ejwtdw 0-220 =(ej2t-1)/(pjt)-(1-e-j2t)/(pjt) =-(4jsi
10、2ntp)/t( )4.5 Form the given information x(t)=(1/2p)X(jw)ejwtdw -=(1/2p)X(jw)e-3jX(ejw)wtejwdw =(1/2p)2e-2/3w+pejwdw -3=The signal x(t) is zero when3(t-2/3) is a nonzero integer multiple of p this gives t=kp+2/3,forkIandk0 24.6. Throughout this problem ,we assume that FTx(t) X1(jw) (a) Using the tim
11、e reversal property (Sec. 4.3.5), we have FTx(-t) X1(-jw) Using the time shifting property (Sec. 4.3.2) on this, we have FTFT-jwtjwtx(-t+1) eX1(-jw) and x(-t-1) eX1(-jw) therefore 81 2sin3(t-2/3) p(t-2/3)x1 (t)= x(-t+1)+ x(-t-1) eX1(-jw)+e(b) Using the time scaling property (Sec. 4.3.5), we have FTx
12、(3t) 1/3X(jw/3) FT-jwtjwtFTX1(-jw)2 X(-jw)cosw FTeUsing the time shifting property on this, we have x2 (3(t -2) 1/3X(jw/3) FT(c) Using the differentiation in time property (Sec. 4.3.4), we have dx(t)jwX(jw) -2jwdt4.7 (a)Since X1(-jw) is not conjugate symmetric ,the corresponding signal x1 (t) is not
13、 real Since X1(-jw) is neither even nor odd ,the corresponding signal x1 (t) is neither even nor odd (b) the FT of a real and odd signal is purely imaginary and odd. therefore ,we may conclude that the FT of a purely imaginary and odd signal is real and odd . since X2(jw)is real and odd we may there
14、fore conclude that the corresponding signal x2(t) is purely imaginary and odd. (c) Consider a signal y3 (t) whose magnitude of the FT is | Y3(jw)|=A(w),and whose phase of the FT is | and Y3(jw)=-Y3(-jw),we may conclude that the signal y3 (t) is real (d) Since X4(jw) is both real and even , correspon
15、ding signal x4 (t) is real and even 4.8 (a)The signal x(t) is as shown in Figure S4.8. x(t) y (t) 1 1 We may express this signal as tFigure S4.8 x(t) =y(t)dt -1/2 1/2 t -1/2 1/2 t -dt22Using the time shifting property, we have x3=dx(t-1)dt2FTApplying this property again, we have dx(t)-w2X(jw) FT-jwt
16、 -w2X(jw)e2Y3(jw)=2w. Since |Y3(jw)Where y (t) is the rectangular pulse shown in S4.8 Using the integration property of FT we have FTx(t) X(jw) =1Y(jw)+pY(j0)s(w) jwy(t) 1/2 we know from 4.2 that x(t) 2sin(w/2) 1 Y(jw)=wTherefore X(jw)= 2sin(w/2)+ps(w) jw2(b) if g(t)=x(t)-(1/2) ps(w)= 2sin(w/2) jw2
17、-1 1 t -1 1 t 4.9 (a) the signal x(t) is plotted in figure S4.9 x(t)= -y(t)dt-u(t-1/2) jwtusing the result obtain ed in part (a) of the previous problem ,the FT X(jw)of x(t) is -jwX(jw)=2sin(w/2)+ps(w)-FTu(t-1/2)= sinw-e 22jwjw(b) the even part of x(t) is given by evx(t)=(x(t)+x(-t)/2 This is as sho
18、wn in the 4.9 Therefore sinw FTevx(t)=wNow the real part of answer to part (a) is ejw1sinw Re-=Rej(cosw-jsinjw)=jwww(c) the FT of the odd part of x(t) is same as j times imaginary part of the answer to part (a),we have sinwe-jwsinwcosw Im2-=-2+jwjwwwTherefore ,the desired result is 82 FTOdd part of
19、x(t)= sinw=cosw 2jwjw4.10 we know from table 4.2 that FT sintRectangular function y(jw) see figure s4.10 ptTherefore FT (sint)2(1/2p) Rectangular function y(jw)*Rectangular function y(jw) ptThis is a triangular function y1(jw) as show in the figure s4.10 X(jw) Y1(jw) Y(jw) 1 j/2p 1/p w0 -2 w -2 2 0
20、0 -1 1 w -j/2p Using table 4.1 we may write Figure S4.10 dy(jw)sintFT t ( )x(jw)=j 1ptdw this is as show in the figure above .x(jw) may be expressed mathematically as j/2p,-2w0 x(jw)=-j/2p,0w0. we may conclude that x(t)=2|t|e for t0 -ttherefore x(t)=2teu(t) 4.16 (a) we may write x(t) = -|t|- sin(kp/
21、4)d(t-kp/4)kp/4sin(t)=pd(t-kp/4)pt- 84 Therefore，g(t) = pd(t-kp/4) - (b) Since g(t) is an impulse train , its Fourier transform G(jw) is also an impulse train From Table 4.2, G(jw)=p2pd(w-2pk) p/4-p/4=8pk=-d(w-8k)We see that G(jw) is periodic with a period of 8.Using the multiplication property, We
22、know that X(jw)=1FTsint*G(jw) 2pptIf we denote FTsint by A(jw),then ptX(jw)=1/2pA(jw)*8p=4A(jw-8k)k=-k=-d(w-8k) X(jw) may thus be viewed as a replication of 4A(jw) every 8 rad/sec. this is obviously Periodic. Using Table 4.2 , we obtain 1, |w|1 0, otherwise A(jw)=Therefore, we may specify X(jw) over
23、 one period as 4 , |w|1 X(jw)= 0, 1|w|4 4.17. (a) From Table 4.1 ,we know that a real and odd signal signal x(t) has a purely imaginary and odd Fourier transform X(jw). Let us now consider the purely imaginary and odd signal jx(t) ,using linearity ,we obtain the Fourier transform of this signal to b
24、e jX(jw) . The function jX(jw) will clearly be real and odd. Therefore, the given statement is false. (b) An odd Fourier transform corresponds to an odd signal, while an even Fourier transform Corresponds to an even signal. .The convolution of an even Fourier transform with an odd Fourier may be vie
25、wed in the time domain as a multiplication of an even and odd signal Such a multiplication will always result in a an odd time signal .The Fourier transform of this odd signal will always be odd ,Therefore ,the given statement is true. 4.18. Using Table 4.2, we see that the rectangular pulse x1(t) s
26、hown in Figure S4.18 has a Fourier transform X1(jw) = sin(3w)/w. Using the convolution property of the Fourier transform, We may write sin(3w)FT x2(t)=x1(t)*x1(t)x2(jw)=x1(jw)x1(jw)=(w)2 The signal x2(t) is shown in Figure S4.18. Using the shifting property ,we also note that FT1jwsin(3w)2 1 x(t+1)2
27、22e(w)FT21-jwAnd 1 x(t-1)e222wAdding the two above equation ,we obtain w)2FT1 h(t)=1cos(w)(sin(32x2(t+1)+2x2(t-1)w) The signal h(t) is as shown in Figure S4.18 .we note that h(t) has the given Fourier transform H(jw) h(t) 5/4 x(2x1(t) s/2 1/2 1/4 t -3 0 3 -7 -5 0 5 7 t -6 0 6 t Figure S4.18 Mathemat
28、ically h(t) may be expressed as 5 , |t|1 4|t| -4+3, 1|t|5 2h(t)= |t|7 -8+8 5|t|7 sin(3w) 85 0 otherwise 4.19 we know that (jw) H(jw)=Y X(jw). Since it is given that y(t)=e-3tu(t)-e-4tu(t), we can compute Y(jw) to be 1 Y(jw)=3+1jw-4+1jw=(3+jw)(4 +jw) Since H(jw) = 1/(3+jw), we have Y(jw) X(jw)=X(jw)=
29、1/(4+jw) Taking the inverse Fourier transform of X(jw), we have x(t)=e-4tu(t) 4.20 From the answer to Problem 3.20 we know that the frequency response of the circuit is H(jw)=21 -w+jw+1Breaking this up into partial fractions, we may write Using the Fourier transform pairs provided in Table 4.2, we o
30、btain the Fourier transform of H(jw) to be 22H(jw)=-1j31-32-1j+jw+1+1j32-1j+jw h(t)=-Simplifying , h(t) = -e3e-1t2(-1+232j)t+e(-1-232j)tu(t). 23sin(32t)u(t). 4.21. (a) The given signal is e-atcos(w0t)u(t)=ee12-atjw0tu(t)+ee12-at-jw0tu(t) Therefore . X(jw) = 2(a-jw0+jw)(b). the given signal is 1+2(a+
31、jw0+jw)=1a+jw(a+jw)2+w02x(t) we have e-3tsin(2t)u(t)+esin(2t)u(-t).(3+jw)2+4 23t2j1/2jFTx1(t)=e-3tsin(2t)u(t)X1(jw)=3-1/-j2+jw3+j2+jw= Also j1/2jFTx2(t)=e3tsin(2t)u(-t)=-x1(-t)X2(jw)=-X1(-jw)=3-1/2-j2-jw3+j2-jw Therefore X(jw)=X1(jw)+X2(jw)=3j9+(w+2)2-9+(w-2)23j. (c) Using the Fourier transform anal
32、ysis equation (4.9) we have 86 (d) Using the Fourier transform analysis equation (4.9) we have X(jw)=1-jwT 1-aesin(w)sin(w)w2psinwX(jw)=2sin+-wp-wp+w=w(p2-w2)2(e) we have x(t)=(1/2j)te-2tej4tu(t)-(1/2j)te-2te-j4tu(t) Therefore 2j2j X(jw)=(2-1/ -(2+1/j4+jw)2j4-jw)2FT1pt(f). We have x1(t)=sinX1(jw)= p
33、tAlso x2(t)=|w|p 0otherwisesin2p(t-1)FTe-2w|w|2p X2(jw)=p(t-1)otherwise0FT x(t)=x1(t)x2(t) X(jw)=21pX1(jw)*X2(jw)Therefore , e-jw wp (1/2p)(3p+w)e-jw,-3pw-p X(jw)= (1/p2)p(-3we-jw),pw3p 0 otherwise (g) Using the Fourier transform analysis eq. (4.9) we obtain 2jsinw X(jw)=wcos2w-w if x(t)=1k=-d(t-2k), Then X(t)=2x1(t)+x1(t-1). Therefore X(jw)=X1(jw)2+e-w=pd(w-kp)2+(-1)k. -Using the Fourier tr
