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1、信号与系统奥本海姆英文课后答案chapterChapter 10 Answers 10.1 (a)The given summation may be written as 11-1-jwn (r)nen=-122jw,by replacing z with re. If r1 22 And the function within the summation grows towards infinity with increasing n. Also , the summation dose not converge. But if r1,then the summation converge
2、s. 2 (b) The given summation may be written as jw 1(1r-1)ne-jwn , by replacing z with re. If r1, then 2r1 n=-1222 And the function within the summation grows towards infinity with increasing n ,. Also , the summation dose not converge. But if r1, then the function within the summation grows towards
3、infinity with increasing n ,. Also , the summation dose not converge. But if r1. The second summation 2jwconverge for r2. Therefore, the sum of these two summations converges for 1/2r1 51251-1z-1510.18. (a) using the analysis of example10.18,we may show that 1-6Z-1+8Z-2 H(Z)=211-Z-2+-239ZSince h(z)=
4、y(z)/x(z), we may write 21Y(Z)1-Z-1+Z-2=X(Z)1-6Z-18Z-2 39Taking the inverse z-transform we obtain yn-21yn-1+n-2=xn-6xn-1+8xn-2 39 (b) H (Z) has only two poles, these are both at z=1/3. Since the system is causal, the ROC of H (Z) will be the form z1/3. Since the ROC includes the unit circle, the sys
5、tem is Stable. 10.19. (a) The unilateral z-transform is 1x(Z)=nun+5z-n N=041 =nz-nn=0411 ,z141-z-14(b) The unilateral z-transform is 178 x(z)=(sn+3+sn+2n-n)z-nn=0=(0+sn+snzn=0-n=2,All z The unilateral z-transform is 1nx(z)=z-n 21=nz-nn=0211=,z211-z-1210.20.Applying the unilateral z-transform to give
6、n difference equation, we have z-1y(z)+y-1+2y(z)=y(z). (a) For the zero-input response, assume that xn=0. Since we are given that y-1=2, -1 -1zy(z)+y(-1)+2y(z)=0y(z)=11+z-12.Taking the inverse unilateral z-transform, Yn= -(-1)nmn. 2(b) For the zero-state response set y-1=0. Also, we have 1n11 x(z)=m
7、zmn=,z.1-1221-z2Therefore, 12y(z)=.1-12+z-11-z4We use partial fraction expansion followed by the inverse unilateral z-transform to obtain yn=1(-1)nmn+1(1)nmn. 3264( c) The total response is the sum of the zero-input response. This is 2111yn=-(-)nmn+nmn. 326410.21.the pole zero plots are all shown in
8、 figure S10.21. 5 (a) For xn=sn+5,x(z)=z, all z. The Fourier transform exists because the ROC includes the unit circle. (b) For x n =sn-5, X (z)=z, all z expect 0. The Fourier transform exists because the ROC includes the unit circle. n(c) For x n=(-1)mn, X(Z)=xn-nz n=-5=(-1)nz-n =1/(1+z-1),z1 n=0 T
9、he Fourier transform does not exist because the ROC does not include the unit circle (d) For x n=(1)n+1mn+3, 2X(z)=n=-xnz-n 179 =n=-3(2)1n+1z-n1=n-2z-n+3n=024z31=,z12(1-z-1)2(e) For x n=(-1)nm-n-2, 3x(z)=n=-xnz -nn=-=(-1/3)-2nz-n-1-nn)zn=23-1=-n-2zn+2n=03=(=9z2/(1+3z),z1/3=3z,z1/3(1+(1/3)-1)The Four
10、ier transform does not exist because the ROC does not include the unit circle. (f) For xn=(1/4)nm-n+3, x(z)=n=-3xnzn-n-nn=-(1/4)z-nznn=-3(1/4)=(1/4)-n+3zn-3n=0=(1/64)z-3/(1-4z),z1/4=(1/16)z-4/(1-(1/4)z-1),z1/4 The Fourier transform does not exist because the ROC includes the unit circle. (g) Conside
11、r x1(z)=2nm-n. x1(z)=n=-xnz -n1=n=-(2)z0n-n=(2)-nznn=0=1/(1-(1/2)z),z2=-2z-1/(1-2z-1),z1/4The z-transform of the overall sequence x n=x1n+x2n is x(z)=-2z-1z-1/4+,(1/4)z1/3 The Fourier transform exists because the ROC includes the unit circle. IM IM IM n4 (d)(c) n Sn5ordu3m0 Re(b) ReRe (a)I IM IM (e)
12、 ndIM(g)(h) (f)2ordu3u0 3rdorduku0 Re 1 Re R e Figure S10.21 10.22. ( a ) Using the z-transform analysis equation, -44-33-22-11001-1 X(z)=(1/2)z+(1/2)z+(1/2)z+(1/2)z+(1/2)z+(1/2)z IM Re IM Re +(1/2)2z-2+(1/2)3z-3+(1/2)4z-4This may be express as 9-9 X(z)=(1/2)-4z41-(1/2)z. 1-(1/2)z-1This has four zer
13、o at z=0 and 8 more zero distributed on a circle of radius 1/2. The ROC is the entire z plane. (Although form an inspection of expression for X(z) it seems like these is a pole at 1/2 which cancels with this pole.) Since the ROC includes the unit circle ,the Fourier transform exists. (b) Consider th
14、e sequence x1n=(1/2)nun+2nu-n-1. Now, (1/2)nun And z 1,z(1/2)-11-(1/2)z-1,z2. -11-2z(2)nu-n-1 Therefore, x1(z)=z11-,(1/2)z2. 1-(1/2)z-11-2z-1Note that x n=nx1n. Therefore, d(1/2)Z-12Z-1X(z)=-zX1(Z)=-+. dz(1-(1/2)Z-1)2(1-2Z-1)2The ROC is (1/2)z1/ 2-11-(1/2z)And x2n=(2)nu-n-1 X(Z)=-zdX1(z)+zdX2(z)=-dz
15、dzz X(Z)=-21 ,Z21-2Z-1Using the differentiation property, we get (1/2)z-12z-1-. -12-12(1-(1/2)z)(1-2z)The ROC is (1/2)z2. Therefore, the Fourier transform exists. (d) The sequence may be written as ej(2pn/6)+(p/4)xn=4u-n-1. 2ne-jp/41Now ,z4 21-4e-j2p/6z-14nej(2pn/6)+(p/4)u-n-1 And n-j(2pn/6)+(p/4)zj
16、p/4e1 ,z4 21-4ej2p/6z-14eTherefore, u-n-1 ze-jp/41 ,z4 21-4e-j2p/6z-1ejp/41X (z)= + ,z4 j2p/6-121-4ezThe ROC is z1. 2113-1xn=-un+un. 2222nn Performing long-division in order to get a right-sided sequence. we obtain X(z)=1-z-1+1-21-31-41-5z-z+z-z+. 441616 This may be rewritten as 31111111X(z)=1-z-1+z
17、-2-z-3+.-1+z-1+z-2+z-3+. 22482248 Therefore. . 1- xn=-11un+3. un22-nn22 (ii) The partial fraction of the given X(z) is 13 2+2X(z)=111-z-11+z-122z1. 2n31xn=-2dn+un 22 Performing long-division in order to get a right-sided sequence. we obtain 1313 X(z)=-+z-1+z-2+z-3+. 2488 This may be rewritten as 311
18、 X(z)=-2+1+z-1+z-2. 224 Therefore. 3-1xn=-2dn+un22n (iv) The partial fraction of the given x(z) is 3X(z)=-2+21-11-z2Since the ROC is z1.It follow that 21. xn=2un2(b) Carrying out long division on X(z),we get 11 X(z)=1-z-1+z-2-z-2+. 24Using the analysis equation (10.3),we get 1 xn=dn-(-)n-1un-1. 2(c)
19、 We may write X(z) as Xz=3z-1111-z-1-z-248=(1-3z-11-11z)(1+z-1)24. The partial fraction of X(z)is . 44Xz=-111-z-11+z-1242Since xn is absolute summable, the ROC must be z1 in order to include the unit circle. It follow that 11 Xz=42un-4(-)2un 2410.25. (a) The partial fraction equation of 1Xz=X(z)-1-1
20、1-z-11-z2 is 2. Since xn is absolute summable, the ROC must be z1 in order to include the unit circle. It follow that 1 xn=-2un+2un. 2(b) X(z) may be rewritten as Z2Xz=1(Z-)(Z-1)21. Using partialfraction expansion, we may rewrite this as . Xz=2z2z-12-2zz=2z2-+1z-1z-1z-2 If xn is right-side, then the
21、 ROC for this signal is z1.Using this fact, we may Find the inverse z-transform of the term within square bracket above to be yn=-(1)nun+un. 2Note that Xz=2zXz.Therefore , xn=2yn-1.This gives 1 xn=-2n+1un+1+2un+1 2Noting that x-1=0.we may rewrite this as 1xn=-nun+2un 2 This is the answer that we obt
22、ainly in part (a). 10.26 Form part (b) of the previous problem , 184 Xz=Z21(Z-)(Z-1)2. (b) Form part (b) of the previous problem , zzXz=2z-z-12+z-1. (c) If xn is left-side, then the ROC for this signal is z1.Using this fact, we may 2.Find the inverse z-transform of the term within square bracket abo
23、ve to be 1yn=nu-n-1-u-n-1. Note that Xz=2zXz.Therefore , xn=2yn-1.This gives 21xn=2n+1u-n-2-2u-n-2. 210.27 We perform long-division on X(z) so as to obtain a right-seder sequence. This gives us X(z)=z3+4z2+5z+. Therefore ,comparing this with eq.(10.3) we get x(-3)=1 x(-2)=4 x(-1)=5 And x(n)=0for n1
24、20 Figure S10.29 w And zx2nx2(z)=z1 3 Using the time shifting property, we get 185 zx1n+3z3x1(z) |z|1 2Using the time reversal and properties, we get zx1-n+1z-1x1(z-1) z3 Now, using the convolution property, we get zyn=x1n+3*x2-n+1Y(z)=z2X1(z)X2(z-1) 1 |z|1. 3jpjpSince xn is right-sided, the ROC mus
25、t be n10.32. (a) We are given that hn=aun and xn=un-un-Ntherefore yn=xn*hn=k=-hn-kxk=ak=0N-1n-kun-k Now. yn may be evaluated to be 0, nN-1k=0Simplifying 0, nN-1()()(b)Using Table 10.2 we get H(z)= And 11-az-1, |z|a| Therefore X(z)= 1-z, All z, -11-zY(z)=X(z)H(z)= -1-N1(1-z)(1-az)-1-The ROC is |z|a|,
26、 Consider P(z)= -1(1-z)(1-az)-1z-N-1,1-1With ROC |z|a|, The partial fraction expansion of P(z) is P(z)= 1(1-a)1(1-a-1) Therefore, 1-z-1(1-z)(1-az)+1-az-1,186 Pn= 1un+1-a11-a-1aun,nNow, note that Therefore, Y(z)=P(z)1- z-N, 11un-un-N+-11-a1-ayn=pn-pn-N=aun-ann-Nun-N, This may be written as 0, nN-1()(
27、)This is the same as the result of part(a). 10.33. (a) Taking the z-transform of both sides of the give difference equation and simplifying, We get Y(z)1H(z)=X(z)=11-2z-11+4z-2.The poles of H(z) are at (1/4) j(34).Since hn is causal, the ROC has to be |z|(14)+j(34)|=. (b) We have X(z)=11, |z|.1-121-
28、z2Therefore, 1Y(z)=H(z)X(z)=1-11(1-z)(1-22-11-2.z+4z)The ROC of Y(z) will be the intersection of the ROCs of X(z) and H(z).This implies that the ROC of Y(z) is |z|1/2.The partial fraction expansion of Y(z) is Y(z)=1+1-111-z1-22nzz-1-121+4z-2.Using Table 10.2 we get 1un+yn=22312npnsinun.210.34.(a) Ta
29、king the z-transform lf both sides of the give difference equation and simplifying, we get -1Y(z) zH(z)=.X(z)The poles of H(z) are at z=(1/2)(11-z-z-1-252.H(z) has a zero at z=0.The pole-zero plot for H(z) is as )shown in Figure S10.34.since hn is causal, ROC for H(z) has to be |z|(1/2)+(b) The partial fraction expansion of H(z) is H(z)=-+.1+5-11-5-11-1-zz22n(52. )515Therefore, hn=-11+525un+n11-525un.n1-5 2Figure S10.34 1+52 (c). Now assuming that the ROC is (5 hn=-11+522-(1/2)|z|(1/2)+)(52, )we get 5nu-n-1+11-525un.10.35. Taking the z-tran
