信号与系统奥本海姆英文版课后答案chapter10.pdf

Chapter 10 Answers 10.1 (a)The given summation may be written as 1 1 1 1 () 2 2 jwn n rne = ,by replacing z with . If r And the function within the summation grows towards infinity with increasing n Also , the summation dose not converge. But if r 1 2 ,then the summation converges. (b) The given summation may be written as 1 1 1 1 () 2 2 jwn n rne = , by replacing z with . If r jw re 1 2 , then 21r And the function within the summation grows towards infinity with increasing n ,. Also , the summation dose not converge. But if r1, then the jw re function within the summation grows towards infinity with increasing n ,. Also , the summation dose not converge. But if r jw re 1 2 . The second summation converge for r2. Therefore, the sum of these two summations converges for 1/2 10.18. (a) using the analysis of example10.18,we may show that 12 2 2 1 68 ( ) 21 1 39 ZZ H Z Z Z + = + Since h(z)=y(z)/x(z), we may write 121 21 ( )1( )1 68 39 Y ZZZX ZZZ += 2 Taking the inverse z-transform we obtain 21 12 6 18 2 39 y ny nnx nx nx n+=+ (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 system is Stable. 10.19. (a) The unilateral z-transform is 0 1 ( )( )5 4 nn N x Zu n = =+ z 0 1 ( ) 4 nn n z = = 1 11 , 1 4 1 ( ) 4 z z (b) The unilateral z-transform is 178 0 0 ( )( 3 2 ) (0 2, nn n n n x znnn nn z = = =+ =+ = z All z The unilateral z-transform is 0 1 1 ( )( ) 2 1 ( ) 2 11 , 12 1 2 nn nn n x zz z z z = = = = 10.20.Applying the unilateral z-transform to given difference equation, we have 1 ( ) 12 ( )( ).z y zyy zy z += (a) For the zero-input response, assume that xn=0. Since we are given that y-1=2, 1 1 1 ( )( 1)2 ( )0( ). 1 1( ) 2 z y zyy zy z z += + Taking the inverse unilateral z-transform, Yn= 1 () 2 n n . (b) For the zero-state response set y-1=0. Also, we have 1 11 ( )( ) ,. 1 22 1 2 n x zznz z = 1 Therefore, 1 1 12 ( )()(). 1 2 1 4 y z z z = + We use partial fraction expansion followed by the inverse unilateral z-transform to obtain 111 1 () ( ) . 326 4 nn y nnn=+ ( c) The total response is the sum of the zero-input response. This is 211 1 () ( ) . 326 4 nn y nnn= + 10.21.the pole zero plots are all shown in figure S10.21. (a) For 5,x nn=+ 5 ( )x zz=, all z. The Fourier transform exists because the ROC includes the unit circle. (b) For x n =5,n X (z)=, all z expect 0. 5 z The Fourier transform exists because the ROC includes the unit circle. (c) For x n= ( 1) , n n ( ) n n X Zx n z = = 0 ( 1)n n n z = = 1 1/(1),1zz =+ The Fourier transform does not exist because the ROC does not include the unit circle (d) For x n=1 1 ( )3, 2 n n + + ( ) n n X zx n z = = 179 1 3 23 0 3 1 1 ( ) 2 1 ( ) 2 41 , 1 2 (1 ( ) 2 nn n nn n z z z z z + = + = = = = (e) For x n= 1 ()2, 3 n n ( ) n n x zx n z = = 2 2 22 0 2 1 ( 1/3) 1 () 3 1 () 3 9/(1 3 ),1/3 3 ,1/3 (1 (1/3) ) nn n nn n nn n z z z zzz z z = = + = = = = =+ The z-transform of the overall sequence x n= 12 x nx n+ is 11 11 2/4 ( ),(1/4)2 (1 2)1 (1/4) zz x zz zz = + The Fourier transform exists because the ROC includes the unit circle. 0 53 nordu I e R M I M I e R n S e R ( ) c 4n M I e R M I Figure S10.21 10.22. ( a ) Using the z-transform analysis equation, 443322110011 223344 ( )(1/2)(1/2)(1/2)(1/2)(1/2)(1/2) (1/2)(1/2)(1/2) X zzzzzzz zzz =+ + This may be express as 99 44 1 1 (1/2) ( )(1/2). 1 (1/2) z X zz z = This has four zero 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 the sequence 1 (1/2) 21. nn x nu nun=+ Now, (1/2) nu n z 1 1 ,(1/2 1 (1/2) z z ) And (2)1 nu n z 1 1 ,2 12 z. z And 2 (2)1 n x nun= z 2 1 1 ( ),2 12 XZZ Z = . 1 131 2 222 nn x nu n = + u n . Performing long-division in order to get a right-sided sequence. we obtain ( ) 5 1234 1111 1. 441616 X zzzzzz = +. This may be rewritten as ( ) 123123 31111111 1.1. 22482248 X zzzzzzz =+ Therefore. 1 131 2 222 nn x nu nu n = + . (ii) The partial fraction of the given ( )X z is 11 13 22 ( ) 11 11 22 X z zz =+ + Since the ROC is 1 z . 3 1 2 2 2 n x nnu = + n Performing long-division in order to get a right-sided sequence. we obtain ( ) 123 1313 2488 X zzzz = + This may be rewritten as ( ) 12 311 21 224 X zzz = + . Therefore. 31 2 22 n x nnu = + n (iv) The partial fractioen of the giv n ( )x z is ( ) 1 3 2 2 1 1 2 X z z = + Since the ROC is 1 2 z .It follow that 2 1 ( ) 2 x nu=n . (b) Carrying out long division on ( )X z,we get ( ) 122 11 1 24 X zzzz = + . Using the analysis equation (10.3),we get 1 1 ()1. 2 n x nnu n = (c) We may write ( )X z as 11 121 33 1111 1(1)(1 4824 zz X z zzzz = + 1) . The partial fraction of is ( )X z 11 44 11 11 24 X z zz = + . Since x n is absolute summable, the ROC must be 1 2 z in order to include the unit circle. It follow that 22 11 4( ) 4() 24 X zu nu n= 10.25. (a) The partial fraction equation of ( )X z is 1 1 12 1 1 1 2 X z z z = . Since x n is absolute summable, the ROC must be 1z in order to include the unit circle. It follow that 2 1 ( ) 2 2 x nu nu= +n . (b) ( )X z may be rewritten as 2 1 ()(1 2 Z X z ZZ = ) . Using partialfraction expansion, we may rewrite this as 22 12 22 11 11 zz X zzz zz zz =+ 22 . If x n is right-side, then the ROC for this signal is 1z .Using this fact, we may Find the inverse z-transform of the term within square bracket above to be 1 ( ) 2 n y nu nu n= + . Note that .Therefore , 2 X zzX z= 2 1.x ny n=This gives 1 1 2( )12 1 2 n x nu nu n + = + Noting that .we may rewrite this as 10x= 1 ( ) 2 2 n x nu nu= +n This is the answer that we obtainly in part (a). 10.26 （a）Form part (b) of the previous problem , 184 2 1 ()(1 2 Z X z ZZ = ) . (b) Form part (b) of the previous problem , 2 1 1 2 zz X zz z z =+ . (c) If x n is left-side, then the ROC for this signal is 1 2 z And 22 1 1 ( ) 1 1 3 z x nxz z = 1 3 z Using the time shifting property, we get 185 3 1 3( z 1 )x nz+ x z 1 | 2 z Using the time reversal and properties, we get 11 11 1( z )xnz x + z 3z . 10.32. (a) We are given that and n h na u n= x nu nu nN=therefore 1 0 * N n k kk y nx nh nh nk x kau nk = = Now. yn may be evaluated to be 0 1 k=0 0, n Simplifying () () 11 1 0, n (b)Using Table 10.2 we get H(z)= 1 1 , 1az |z|a| And X(z)= 1 1 , 1 N z z All z, Therefore Y(z)=X(z)H(z)= 1111 1 , (1)(1)(1)(1) N aa z zzzz The ROC is |z|a|, Consider P(z)= 11 1 (1)(1)a zz With ROC |z|a|, The partial fraction expansion of P(z) is P(z)= 1 11 1 (1)1 (1) , 11 a aa zz + Therefore, 186 Pn= 1 11 , 1 1 n u nu n a a a + Now, note that Y(z)=P(z) 1- N z , Therefore, 1 11 , 1 1 nn N y np np nNu nu nNu nu nN a aa a =+ This may be written as () () 11 1 0, n 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 12 ( )1 ( ). 11 ( ) 1 24 Y z Hz Xz zz = + The poles of H(z) are at (1/4) ( 3 4).jSince hn is causal, the ROC has to be | | (1 4)( 3 4) |1 2zj+= （）. (b) We have 1 11 ( ),|. 1 2 1 2 Xzz z = Therefore, 11 1 ( )( )( ). 111 (1)(1) 224 Y zHz Xz zzz = + 2 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 1 11 21 ( ). 111 11 224 Yz z zzz 2 =+ + Using Table 10.2 we get 2 sin . 23 11 22 nn n y nu nu n =+ 10.34.(a) Taking the z-transform lf both sides of the give difference equation and simplifying, we get 1 12 ( ) ( ). ( ) 1 Y z H z X z z zz = The poles of H(z) are at z= () (1H(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| /2)5 2 . () /2)5 2 .+(1 (b) The partial fraction expansion of H(z) is 11 1515 ( ). 1515 11 22 H z zz = + + Therefore, 11 . 55 1515 22 nn h nu nu n= + + (c). Now assuming that the ROC is ()() 5 2(1/2) | (1/2)5 2 , we get z2,then ( ) 1 22 . 33 1 2 2 n n nu n h = + u n If the ROC is 1/20. It follows that x n= n.it can similarly be 2 n h and 3 10.36. Taking the z-transform of both side of the given difference equation and simplifying, we get n h satisfy the difference equation. 1 11 ( )1 ( ). 1010 ( ) 1 2 Y z H z X z z z zzz = + 33 The partial fraction expansion of H(z) is 1 1 3 83 8 ( ). 1 1 3 1 3 H z z z =+ Since H(z) corresponds to a stable system, the ROC has ro be (1/3)2/3. The ROC includes the unit circle and hence the system is stable. 10.38.(a) 1 1 ,nn f e = (b) 2 2 ,nn f e = (c) Using the results of part (a) and (d), we may redraw the block-diagram as shown in Figure S10.38. yn xn 1 z -1/21/4 1 z -21/4 xn yn -2/3 -z/4 1/8 -1/2 1 z 1 z xn yn 5/3 1 z -14/3 4 -1/2 1/4 1 z Figure S10.38 (d) Using the approach shown in the examples in the textbook we may draw the block-diagram of 11 1( ) 1(1 4) 1(1 2)z Hz =+ z and 1 2( ) 12 1 (1 4)z 1 Hzz = as shown in the dotted boxes in the figure below. H(z) is the cascade of these two systems. (e) Using the approach show in the examples shown in the textbook, we may draw the block-diagram of 1( ) 4z H = , 1 2( ) 5 3 1(1/2)z Hz =+ and 1 3( ) 14 3 1 (1/4)z Hz = as shown in the dotted boxes in the figure below. H(z) is the parallel combination of 1( ) zH , 2( ) zH ,. 3( ) zH 10.39. (a) The direct form block diagram may be drawn as shown in part(a-i) of Figure S10.39 by noting that 1 123 1 ( ). 51651 1 3361836 z H zzzz = + 4 The cascade block-diagram is as shown in part (a-ii) of Figure S10.39. Figure S10.39 A B C D 1 z 1 z 1 z 1 z xn yn r 1 z xn yn 1 z 1 z 1 z 189 Part(a): A=5/3,B=11/36.C=15/54,D=1/36 Part(b): A=3/2,B=-1.C=5/4,D=-1/2 Part(b): A=2,B=-z/4.C=3/4,D=-1/8 Part(a): =1, =-1/4,r=2/3,=-1/9 Part(a): =1, =-1/2,r=1/2=-1 Part(a): =1, =-1/2,r=1,=-1/4 Note that 1 111 1111 ( ). 1122 1111 2233 z H zzzz = 1 Therefore, may be drawn as a cascade of four systems for which the coefficient multipliers are all real. 1( ) zH (b)The direct form block diagram may be draw as shown in part (b-i) of Figure S10.39 by noting that n 2 123 1 ( ). 351 12 242 z H zzzz = + 4 The cascade block-diagram is as shown in part(b-ii) of Figure S10.39. Now that 2 11 11 1111 ( ). (1)1 115115 11 11 22 44 z jj jj H zz zz = + + Therefore, cannot be drawn as cascade of four systems for which the coefficient multipliers are all real. 1( ) zH (c) The direct form block diagram may be drawn as shown in part(c-i) of the Figure S10.39 by noting that 3 123 1 ( ). 731 12 448 z H zzzz = + 4 The cascade block-diagram is as shown in part(c-ii) of the Figure S10.39. Note that ( ) () 3 111 1 1111 1111 111 1 222 2 z ij zzz z h = + Therefore,cannot be drawn as a cascade of four systems for which the coefficient ( ) 1 Hz Multipliers are all real. 10.40.The definition of the unilateral z-transform is ( ) 0 n n xzxnz = = (a) since xn=n-5 is zero in the range 0 n ,x(z)=0. (b) The unilateral of Laplace transform of xn= n-5 Is ( ) 5 0 n-5 n n xzze = = (c) The unilateral of Laplace transform of xn=( ) 1 n u n is () 1 0 1 ( )1 1 n n n x zu n z z = = + ,z1 (d)The unilateral of Laplace transform of (1/2)3 n x nu n=+ is 0 0 1 ()(1 / 2 )3 (1 / 2 ) 1 1(1 / 2 ) nn n nn n xzunz z z = = =+ = = ,z1/2. 190 (e)Since xn=(-1/3)u-n-2 is zero in the range 0n , x(z)=0. (f) The unilateral of Laplace transform of (1/4)3 n x nun= + is 0 0 12 ()(1 / 4)3 (1 / 4) 111 1 41664 nn n nn n 3 x zun z zzz z = = =+ = =+ ,All z. (g) The unilateral of Laplace transform of 2(1/4)1 nn x nunu n=+ is 0 0 1 ()2(1 / 4 )1 (1 / 4 ) 1 1 1 4 nn n nn n n xzunu n z z z = = =+ = = ,All z. (h) The unilateral of Laplace transform of 2 (1/3)2 n x nu n = is 2 0 2 0 2 1 ()(1 / 3 )2 (1 / 3 ) 1 1 2 n n nn n n xzun zz z z = = = = = z z1/2. 10.41.from the given information, 1 1 0 0 1 ( )(1/ 2)1 (1/ 2)(1/ 2) 1/ 2 1(1/ 2) nn n nn n xzu nz z z + = = =+ = + z1/2. And () 2 0 0 1 (1 / 4 ) ( 1 / 4 ) 1 1( 1 / 4 ) nn n nn n xzunz z z = = = = = z 1/4. Using Table 10.2 and the time shift property we get 1 1 () 1 1 z Xz z = 2 ,z1/2. And 2 1 1 ( ), 1 1 4 X zz z = 1/4. (a) We have 12 11 ( )( )( ) 11 (1)(1) 24 z G zXz Xz zz = . The ROC is |z|(1/2).The partial fraction expansion of G(z) is 11 21 ( ) 11 11 24 G zz zz = . Using Table 10.2 and the time shift property ,we get 191 11 11 2( )1( )1 24 nn g nu nu n + =+ . (b) We have 12 11 1/2 ( )( )( ). 11 (1)(1) 24 Q zXz Xz zz = The ROC of Q(z) is |z|(1/2).The partial fraction expansion of y(z) is 11 121 ( ). 11 2 11 24 Q z zz = Therefore, 11 1 ( ) ( ) . 22 4 nn q nu nu n= Clearly, qn gn for n0. 10.42. (a)Taking the unilateral z-transform of both sides of the given difference equation .we get 1 ( )3( )3 1 .Y zz Y zYX z += Setting x(z) =0,we get 1 3 ( ). 13 Y z z = + The inverse unilateral z-transform gives the zero-input response 1 3( 3) ( 3) . nn zi ynu nu n + = = Now, since it is given that (1/2) , n x nu=nwe have 1 1 ( ), 1 1 X z z = 2 |z|1/2. Setting y-1 to be zero ,we get 1 1 1 ( )3( ). 1 1 2 Y zz Y z z += Therefore, 11 1 ( ). 1 (1)(13) 2 Y z zz = + The partial fraction expansion o f is ( )Y z 1 1 1/76/7 ( ). 1 1 3 1 2 The inverse unilateral z-transform of both sides of the given difference equation ,we get Y z z z =+ + 11 111 ( )( ) 1( )( ).Y z 222 z Y zyX zz X z = = Setting , we get ( )0X z= Y(z)=0. The inverse unilateral z-transform gives the zero-input response 0. zi yn= Now , since it is given that ,x nu n=we have 1 1 ( ), 1 X z z = |z|1. Setting y-1 to be zero, we get 1 1 11 11(1/2 ( )( ) ) 211 z z Y z zz = .Y z Figure S10.46 /4 IM Re Therefore, 1 1 ( ). 1z The inverse unilateral z-transform gives the zero-state response Y z = . zs ynu n= (c) Taking the unilateral z-transform of both sides of the given difference equation ,we get 11 111 ( )( ) 1( )( ).Y z 222 z Y zyX zz X z = 192 Setting , we get ( )0,X z= 1 1/2 ( ). 1 1 2 Y z z = 10.46.Taking the z-transform of both sides of the difference equation relating xn and sn and simplifying, we get ( ) ( ) 88 88 1 8 ( )1 X zze H zz e s zz = = The system has an 8th order pole at z=0 and 8 zeros distributed around a circle of radius e . This is show in Figure S10.46. The ROC is everywhere on the z-plane except at z=0. (b)We have ( ) ( ) ( ) ( ) ( )( ) 2 1 1Y zS z Hz X zX zHz = Therefore, ( ) 8 2 8888 1 1 z Hz z eze = There are two possible ROCs for :( ) 2 Hzze p orze f. If the ROC is ze p,then the ROC dose not include the unit circle. This in turn implies that the system would be unstable and anti-causal. If the ROC is ze f,then the ROC includes the unit circle. This in turn implies that the system would be stable and causal. (d) We have ( ) 2 88 1 1 Hz z e = . We need to choose the ROC to be ze f in order to get s stable system. Now consider ( ) 18 1 1 P z z e = With ROC ze f.Taking the inverse z-transform, we get 8 n p neu n =. Now, note that ( ) () 8 2 HzP z=. From Table 10.1 we know that 10.47. (a) From Clue 1,we have H(-2)=0.From Clue 2, we know that when /8,0, 8, 10, n p nen =K 0, otherwise 2 h n = ( ) 1 11 , 1 2 1 2 X zz z = f We have ( ) 1 1 1, 1 4 1 4 Y zz z = + f . Therefore, ( ) ( ) ( ) 11 1 11 11 142 ,. 1 4 1 4 zz Y z H zz X z z + = f Substituting z=-2 in the above equation and nothing that N(-2)=0,we get 9 8 = . (b)The response to the signal xn=1=1. Therefore , n 1 (1) 4 y nH= . 193 10.48. from the pole-zero diagram, we many write ( ) /4/4 2 3 /43 /4 11 22 11 22 jj jj zeze HzB zeze = And ( ) 3 /43 /4 2 /4/4 11 22 11 22 jj jj zeze HzB zeze = Where A and B are constants. now note that ( ) 211 33 22 j BB HzHzeHz AA = Using the property 10.53 of the z-transform(see Table 10.1),we get 21 2 . 3 n B hnh n A = We may rewrite this as 21 ,h ng n h n= Where .Note that since both and are causal. We may assume that gn=0 for n 01 r 01/r n (r) -0 . Therefore , (r) 01/r n 01/r n (r) for n.Substituting this in 01/r 1 N 1 N eq. , we get ) 149.10(s r(= = n