数字信号处理第二版或第一版蔡宝坤课后答案110章.pdf

1 Appendix FAnswers To Partial Problems Answers to Partial Problems in Chapter 2 2.1 01 2 1 1 2 3 12 )( 1 n,n n, n, ny n n . 2.2 1 0 1 2 1 2 )( 1 n, n, ny n . 2.3. 2 2 1 21 20 )( 2 n, n, n, nx n ; 1 2 1 11 10 )( 1 n, n, n, nx n . 2.4 (1) Periodic.200N; (2) Periodic.2N; (3) Periodic.14N; (4) Nonperiodic. 2.8 (1) For 0 nn,0)(ny; (2) For1 00 Nnnn, ) 1( )( 0 11 0 00 ,nn , ny nn nnnn ; (3) For1 0 Nnn, )( 0 0 1 ,N , ny nn NN Nnn . 2.9 0n )1 ( 1 0 1 1 )( , n, ny n . 2.10 (1) Unstable, causal, linear, time- invariant and dynamic; (2) Stable, causal, nonlinear, time- invariant and memoryless; (3) Stable, causal, linear, time- varying and memoryless; (4) Unstable, causal, linear, time- varying and dynamic; (5) Stable for bounded)(tg, causal, linear, time- varying and memoryless; (6) Stable. If0 0 n, then the system is causal. It is linear and time- invariant. If 0 0 n, the system is dynamic; (7) Stable, causal, nonlinear, time- invariant and memoryless. 0 0 0) )y( ( nnnn 0 0 0 0 2 1 n,n, . 1N, ,(y y 1N N 2 2.11) 5 2 5 4 () 1()(jny n ss . 2.12)( 4 1 8 2 1 8)(nuny nn . 2.14 (1)( 3 1 2)()(nunnh n ; (2)( 311 311 21)( 0 0 0 1 nu e/ e/ eny j n j nj . Answers to Partial Problems in Chapter 3 3.1 (1)21 )21 ( )(/|z| , /z z zX; (2)21 )21 ( )(/|z| , /z z zX; (3)2 1/2 32 2 1 3 1 5 627 5 627 6 25 )(|z|, zzzz zz zX. 3.2 (1)1 1cos2 )sin( )( 2 0 2 0 3 |z| , zz zz zX; (2)r|z| , rzrz zrz AzX cos2 )cos(cos )( 2 0 2 0 2 . 3.3 (1)43 :ROC1/|z|;43 1/2 :ROC2/|z|;21 :ROC3/|z|; (2) Three possible pole- zero plots with corresponding ROC are, respectively, shown as follows: Figure F3.3 (3) For43 /|z|,)( 4 3 13 12 )( 2 1 13 4 26 1 Re2)(nunujjnx nn . 4 4/ /; ; (2) Three possible pole- zero plots with corresponding ROC are, respectively, shown(2) Three possible pole- zero plots with corresponding ROC are, respectively, shown 2 |z| | 2 r|z| ,|z r |z| ,|z ) 2 1/2 :2 1/2| (2) Three possible pole- zero plots with corresponding ROC are, respectively, shown(2) Three possible pole- zero plots with corresponding ROC are, respectively, shown 3 For43 21/|z|/,) 1( 4 3 13 12 )( 2 1 13 4 26 1 Re2)(nunujjnx nn . For21 /|z|,) 1( 4 3 13 12 ) 1( 2 1 13 4 26 1 Re2)(nunujjnx nn 3.4 (1) 1( 4 1 7)(8)(nunnx n ; (2) 1( 11 )( 1 )(nu aa an a nx n ; (3) 1(2)( 2 1 )(nununx n . 3.6 (1) 1( 2 1 )()(nunnx n ; (2) 1(2)( 1 nunx n ; (3)( 2 3 3 6 1 2)(nunx nn ; (4)( 2 1 3 1 ) 1( 2 1 3 1 )( 2)2( nununx nn ; (5) 1( 2 1 ) 1()( 1 nunnx n . 3.7 (1)( 4 3 3 7 )( 3 4 )(nunnh n ; (2)( 4 3 13 8 3 1 13 8 )(nuny nn . 3.9 (1)()( jj eXeY; (2)()( jj eXeY; (3)()()( jjj eHeXeY; (4)deHeXeY jjj )()( 2 1 )( )( ; (5) d edX jeY j j )( )(; (6)( 2 1 )( 2 1 )( ) 2 ( 2 j /jj eXeXeY; (7)()( 2jj eXeY; (8)deXeXeY jjj )()( 2 1 )( )( . 3.10 (1)6)( 0j eX; (2)2)( j eX; 3 3 3 3 1 1 1313 8 8 n () )() ); )( )u(; ; 3 1313 8 n ) ); ; )( ( j j ( ( ( j j 4 (3)4)(deX j ; (4)28| )( 2 deX| j ; (5)316| )( 2 d d edX | j . 3.111 4 sin22)(nnyss. 3.12 (1) 2 51 | 2 51 2 51 1 )( 2 |z, zz z zz z zH; (2) The pole- zero plot of)(zHwith its ROC is shown as follows: Figure F3.10 (3)( 2 51 5 1 )( 2 51 5 1 )(nununh nn ; (4) The system is unstable, because the ROC does not include the unit circle in the z- plane. If we want to obtain a stable system, the ROC should be selected as 2 51 | 2 51 |z. The corresponding impulse response is )( 2 51 5 1 ) 1( 2 51 5 1 )(nununh nn . 3.13 (1)s.T 050 0 ,rad 40 0 ; (2) n )020() 2 80cos()(n.tn.txs; (3) The sampled sequence is) 2 80cos()(n.nxand its period5N. 3.14 (1) ROC:2 | 21|z/. The corresponding impulse response)(nhis a two- sided sequence; (2) There are 2 possible two- sided impulse responses whose ROCs are, respectively, given by2 | 21|z/and3 | 2|z; 2 2 5 5 5 (4) The system is unstable, because the ROC does not include the unit circle in the(4) The system is unstable, because the ROC does not include the unit circle in the z- plane. If we want to obtain a stable system, the ROC should be selected asz- plane. If we want to obtain a stable system, the ROC should be selected as Figure F3.10Figure F3.10 11 1 )( )( ( (4) The system is unstable, because the ROC does not include the unit circle in the(4) The system is unstable, because the ROC does not include the unit circle in the z- plane. If we want to obtain a stable system, the ROC should be selected asz- plane. If we want to obtain a stable system, the ROC should be selected as 1 | | 5 (3) No, it is not possible. 3.15 (1)1 | 1 1 )(|z, z z zH; (2) ROCy:1 | |z; (3)( 2 1 3 1 ) 1( 3 1 )(nuny n n . Answers to Partial Problems in Chapter 4 4.1 mNk,j mNk, mNk,j mNk, kX 322 22 122 10 )(,minteger any for . 4.2 (1) The periodic sequence is plotted as follows: Figure F4.2(a) (2) The principal- value sequence of the DFS)(kX is 525552 2363023610102361023635)( /j/j/j/j e.e.e.e.kX The magnitude and phase sequences are, respectively, plotted as follows: Figure F4.2(b) Figure F4.2(c) The magnitude and phase sequences are, respectively, plotted as follows:The magnitude and phase sequences are, respectively, plotted as follows: (2) The principal- value sequence of the DFS(2) The principal- value sequence of the DFSX X( The magnitude and phase sequences are, respectively, plotted as follows:The magnitude and phase sequences are, respectively, plotted as follows: 6 4.3 (1)()(kRkX N ; (2)()()(kRkNkX N ; (3)()( )2( 0 kRekX N N/jkn ; (4)( 1 1 )( 2 kR e kX N N/kj N ; (5) 2 2 )1( 0 0 2 2 )1( 0 0 00 )/2 2 sin( )2sin( )/2 2 sin( )2sin( )( / N k Nj/ N k Nj e N k /N e N k /N kX; ) 10(Nk (6)( )/2 2 sin( )2sin( )( 2 2 )1( 0 0 0 kRe N k /N kX N / N k Nj ; (7) 11 1 0 2 ) 1( )( Nk, W N k, NN kX k N . 4.4)() 2 cos()( 0 nRnk N nx N . 4.5 (1) These sequences are, respectively, plotted as follows: Figure F4.5(a) and (b) (2)()( 1 nxnx NN )()2()( 12 nRnxnx NNN Figure F4.5(c) 4.6 (1) Omitted; 4.5 (1) These sequences are, respectively, plotted as follows:4.5 (1) These sequences are, respectively, plotted as follows: 7 (2) 130 13914 85213 401 00, )()()( 21 n n,n n, n n,n n nxnxny; (3)()( 1 nxny NN 5311353113)( 2 nx N ; (4) Because the length of the circular convolution sum)(nyNis less than the length of the linear convolution sum)(ny,)(nyNis not equal to)(ny. However, if we select the length of the circular convolution sum as141 21 NNN, then )(nyNis exactly equal to)(ny. 4.7 (1) 21 1 )( /e eX j j ; (2)( 2 1 1023 1024 )( 101 nRnx n . 4.8 (1) Omitted; (2) Omitted. 4.9 (1)( 1 1 )(DFT)(kR aW a nxkX N k N N rr ;)()(nRanx N n r ; )( 1 1 )(DFT)(kR bW b nxkX N k N N ii ;)()(nRbnx N n i . (2)()(DFT)(kRnxkX Nrr ;)()()(nRnnx Nr ; )()(DFT)(kNRnxkX Nii ;)()()(nRnNnx Ni . 4.10 (1)s.Tr 10; (2)Hz 5kfh; (3)(samples) 1024N. Answers to Partial Problems in Chapter 5 5.1 (1) 5.766656s; (2) 0.03072s. 5.2 ),()(IFFT)( 21 kjXkXnx)(Re)( 1 nxnxand)(Im)( 2 nxnx. 5.3 (1) Let)()()()( 21 nRnjxnxny N , where)( 1n xand)( 2n xare the even- numberedandodd- numberedsequencescontainedinthepoint-2N sequence)(nx, respectively. Then,)()()( 2 1 )( 1 kRkYkYkX NNN and )()()( 2 1 )( 2 kRkYkYkX NNN . Thus, thepoint-2NDFT is obtained by 10 )()()( )()()( 2 2 1 2 2 1 Nk kXWkXNkX kXWkXkX k N k N ; (2) From the above result, we have ; (samples 4 (samples Answers to Partial Problems in Chapter 5Answers to Partial Problems in Chapter 5 ()() );(x x)( ( i i )( )(x x)( )( r r );)( (x)( ( i ). Answers to Partial Problems in Chapter 5Answers to Partial Problems in Chapter 5 ; ; 8 10 )()( 2 1 )( )()( 2 1 )( 2 2 1 Nk WNkXkXkX NkXkXkX k N Let)()()( 21 kjXkXkY. Then,)(IFFTRe)( 1 kYnx and)(IFFTIm)( 1 kYnx. Thus, 120 odd ), 2 1 ( even ), 2 ( )( 2 1 Nn n n x n n x nx. Answers to Partial Problems in Chapter 6 6.1 The direct- form II structure of the filter is as follows: Figure F6.1 6.2 (1) The direct- form II: Figure F6.2(a) (2) The cascade form: Figure F6.2(b) 9 (3) The parallel form: Figure F6.2(c) 6.3 The parallel structure: Figure F6.3 6.4 The direct- form I and II structures: Figure F6.4(a) Direct- form I structure Figure F6.4(b) Direct- form II structure 6.5 The linear- phase structure: 6.4 The direct- form I and II structures:6.4 The direct- form I and II structures: Figure F6.3Figure F6.3 6.4 The direct- form I and II structures:6.4 The direct- form I and II structures: 10 Figure F6.5 6.6 The linear- phase structure: Figure F6.6 6.7 (1) Two different direct forms are as follows: Figure F6.7(a) Direct- form I structure. Figure F6.7(b) Direct- form II structure. (2) Cascade of five first- order sections: Figure F6.7(c) Desired cascade structure. (3) Cascade of one first- order section and two second- order sections: 6.7 (1) Two different direct forms are as follows:6.7 (1) Two different direct forms are as follows: Figure F6.7(a) Direct- form I structure.Figure F6.7(a) Direct- form I structure. 11 Figure F6.7(d) Desired cascade structure. All of the above realizations have the same computational complexity. Every realization uses only5Nmultipliers and5Ntwo- input adders. 6.8 (1)|.|z|, z. z.z. zH40 401 060501 )( 1 21 ; (2)2(060) 1(50)() 1(40)(nx.nx.nxny.ny; (3) The canonic direct- form realization: Figure F6.8(a) The canonic realization. (4) The parallel form realization: Figure F6.8(b) The parallel realization. (5)()40125(0) 1(510)(8750)(nun.n.nh n ; (6) 1(0.4)10)()( 1 nu.nny n . 6.9 (1) The flow graph of a direct- form nonrecursive implementation of the system: Figure F6.9(a) Figure F6.8(a) The canonic realization.Figure F6.8(a) The canonic realization. 12 (2) 7766554433221 1)(zazazazazazaazzH 0 1 1 1 88 |z|, az za ; (3) The desired cascade structure: Figure F6.9(b) (4) The implementation in part (3) is recursive. However, the overall system is FIR.; (5) The implementation in part (3) requires the most storage and arithmetic. 6.10 The desired transposed structure: Figure F6.10 Answers to Partial Problems in Chapter 7 7.1. 7.2 2 21 2 210 1 )( jj jj j eaea ebebb eH, 211 b/baand 202 b/ba. 7.3 dd d aT d aT d aT d ezbTez bTezT zH 221 1 )cos(21 )cos(1 )(. 7.4 21 1 60650559511 0897010 )( z.z. z zH. 7.5 d Ts N N d ez z dz d z N AT zH 0 1 )!1( )(. 7.6 1 21 201 05010150 )( z. z.z zH. 7.7 21 21 40980136211 068401368006840 )( z.z. z.z zH. Answers to Partial Problems in Chapter 8 8.1 Omitted. 8.2 (1)/21( ),( 2 2sin 2 1 )(NnR /n /n nh N . (2) Omitted. aTaT d d 2 2 j ea ea2 2e j b b2 j e j ezez aTaT zTz d d 1 1 2 2 1 1 e e 10 0 1 5955 1 1 5595595 1 1 Answers to Partial Problems in Chapter 7Answers to Partial Problems in Chapter 7 2 , 1 ba a1 1 aTd daTd cos(os( d bTbTd cos(c s bT 08970 0 9708970 0897 13 8.3 (1) The preliminarily designed impulse response of the filter is 10 ),(350sin550sin 1 )(Nnnwn.n. n nh where66N,32.5and)( 1 2 0.46cos-0.54)(nR N n nw N ; (2) Omitted. 8.4 (1) The preliminarily designed impulse response of the filter is 10 ),(50sinsin 1 )(Nnnwn.n n nh where37N,18and)65326537Kaiser()Kaiser()(.,Nnw; (2) Omitted. 8.5 (1)The preliminarily designed impulse response of the filter is 10 ),(550sin350sinsin 1 )(Nnnwn.n.n n nhwhere 67N,33and)( 1 2 0.46cos-0.54)(nR N n nw N (2) Omitted. 8.6 (1)10 |,)(| )(| 12 NnkHkHand 10 ,)(arg)(arg 12 NnkkHkH; (2) These two filters are both linear phase. Two filter outputs have a common time delay with 3.5 samples; (3) Omitted. 8.7 (1) Omitted; (2) odd ),( even ),( )/21( /2) 1sin() 1( )( 1 Nnh Nnh Nn N nh bp bp n bs . Answers to Partial Problems in Chapter 9 9.1 (1) 3 2 )(limny n ; (2) 2 1 )0(y, 8 5 ) 1 (y, 32 21 )2(y, 128 85 )3(y, 512 341 )4(y, 2048 1365 )5(y; 3 2 )(y; 21(0.1000)0( 2 / y ,85(0.1010) 1 ( 2 / y ,85(0.1010)2( 2 / y , 85(0.1010)3( 2 /) y ,85(0.1010)4( 2 / y ,85(0.1010)5( 2 / y ;85)(/ y ; (3)0)(limny n ;0)( y . 9.2 )1 (3 2 4 2 2 a b f . 9.39b. 9.4 (1)1615(0.1111)0( 2 / y ,168(0.1000) 1 ( 2 / y ,164(0.1000)2( 2 / y , 162(0.0110)3( 2 / y ,161(0.0001)4( 2 / y ,161(0.0001)5( 2 / y ; ),( ),( ( h h( ( ( bpbp Answers to Partial Problems in Chapter 9Answers to Partial Problems in Chapter 9 ) (2) These two filters are both linear phase. Two filter outputs have a common time(2) These two filters are both linear phase. Two filter outputs have a common time )/21 2 /2 h h Answers to Partial Problems in Chapter 9Answers to Partial Problems in Chapter 9 3 3 2 ; ; 14 For4n, the filter runs into the zero- input limit cycle with a constant output 161/; (2)1615(0.1111)0( 2 / y ,1611(0.1011) 1 ( 2 / y ,168(0.1000)2( 2 / y , 166(0.0110)3( 2 / y ,165(0.0101)4( 2 / y ,164(0.0100)5( 2 / y , 163(0.0011)6( 2 / y ,162(0.0010)7( 2 / y ,162(0.0010)8( 2 / y ; For7n, the filter runs into the zero- input limit cycle with a constant output162 /. 9.5 (1) Stable; (2) Unstable. 9.6 (1) The direct form I and II realization structures of the system are shown in Figure F9.6(a) and (b), respectively; Figure p9.6 (2)2(4572