数字信号处理-基于计算机的方法(第四版)答案 8-11章

1、 1 SOLUTIONS MANUAL to accompany Digital Signal Processing: A Computer-Based Approach Fourth Edition Sanjit K. Mitra Prepared by Chowdary Adsumilli, John Berger, Marco Carli, Hsin-Han Ho, Rajeev Gandhi, Martin Gawecki, Chin Kaye Koh, Luca Lucchese, Mylene Queiroz de Farias, and Travis Smith Copyright 2011 by Sanjit K. Mitra. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of S

2、anjit K. Mitra, including, but not limited to, in any network or other electronic Storage or transmission, or broadcast for distance learning. 2 Chapter 8 Part 1 8.1 Analysis yields Y(z) = G(z) X(z) C(z)Y(z)(). Hence, H(z) = Y(z) X(z) = G(z) 1+G(z)C(z) = 2 (1+2K)+ 3z1 . The overall transfer function H(z) is given by H(z) = z2 1+1.5z1+(K +0.5)z2 . The transfer function is stable if K +0.5 0. Hence H(z) is stable if 0 K 0.5. 8.2 The overall transfer function H(z) is given by H(z) = z1 1+(1.5+K)z1+

3、0.5z2 . Thus The transfer function is stable if 1.5+K 1.5 which is satisfied if 3 K 0. 8.3 From the results of Problem 8.3, we have H(z) = G(z) 1+G(z)C(z) which can be solved yielding C(z) = G(z) H(z) G(z)H(z) . Substituting the expressions for G(z) and H(z) in this expression we get C(z) = 1.2+0.4667z11.8133z2 4.2867z3 3.735z41.9275z50.9z6 1+2.3667z1+ 3.65z2+ 3.7617z3+2.9217z4+1.49z5+0.56z6 . Pole-zero plots of G(z),C(z),and H(z)obtained using zplane can be easily obtained. 8.4 The structure wi

4、th internal variables is shown below. Analysis of this structure yields U(z) = KX(z)+ d2z1V(z), V(z) =U(z) d1z1V(z), Y(z) = d1z1V(z) z2V(z) d1V(z). Eliminating the internal variables we arrive at H(z) = Y(z) X(z) = K(d2+ d1z1+ z2) 1+d1z1 d2z2 . d1 2 d _1 +z 1 _ z 1 _ + z 1 _ z 1 _ _1 K X(z) Y(z) U(z) V(z) 3 (a) Since the transfer function is second-order, the structure is non-canonic. (b) H(e j0) =K(d2+ d1+1) 1+ d1 d2 = K. Hence the structure has a unity gain at = 0 if K =1. (c) H(e j ) = K(d2 d

5、1+1) 1 d1 d2 = K. Hence the structure has a unity gain at = if K =1. (d) If we let D(z) =1+ d1z1 d2z2, then H(z) = Kz2D(z1) D(z) . Now, H(z)H(z1) = Kz2D(z1) D(z) Kz2D(z) D(z1) = K2. This implies H(e j)2 = H(z)H(z1) z=e j = K2, or in other words the transfer function has a constant magnitude for all values of . 8.5 The structure with internal variables is shown below. Analysis of this structure yields V(z) = 2X(z)+U(z), U(z) = k1Y(z)+1X(z), Y(z) = z1V(z)+ k2z1U(z). Eliminating the internal variab

6、les we arrive at H(z) = Y(z) X(z) = (1+ k2)1+2z1 1+ k1(1+ k2)z1 . For stability we require k1(1+ k2) 1. 8.6 An equivalent representation of the structure of Figure P8.4 with internal variables is shown below. Analysis of this structure yields U(z) = X(z)+0(z11)W(z), W(z) =1(z11)U(z) Y(z), Y(z) =U(z) 2(z11)Y(z). z 1 _ z 1 _ + k1 k2 1 2 X(z) Y(z)+ U(z) V(z) X(z)Y(z) 1 _ U(z) W(z) 0 (z 1) _ _1 1 (z 1) _ _1 2 (z 1) _ _1 4 Eliminating the internal variables we arrive at H(z) = Y(z) X(z) = 1 D(z) wher

7、e D(z) =1(0+2) 01+012+(0+2)+201 3012z1 + 01+ 3012z2012z3. 8.7 An equivalent representation of the structure of Figure P8.4 with internal variables is shown below. Let Ti(z) = iz1 1iz1 , i =1, 2, 3. Then analysis of the structure yields V(z) = X(z)+T2(z)U(z), W(z) = T1(z)V(z), U(z) =W (z)+T3(z)V(z), Y(z) =0X(z)+W(z). Eliminating the internal variables we arrive at H(z) = Y(z) X(z) = 1 D(z) where 8.8 The structure with internal variables is shown below. Analysis of this structure yields (1): W(z)

8、= X(z)+ k1Y(z),(2): U(z) = 1 1 z1 W(z)+k2Y(z), and (3): Y(z) = k1 1 z1 U(z). Substituting Eq. (2) in Eq. (3) we get (4): Y(z) = k1 1 z1 1 1 z1 W (z)+ k2Y(z) = k1 (1 z1)2 W (z) k1k2 1 z1 Y(z). Substituting Eq. (1) in Eq. (4) we then get Y(z) = k1 (1 z1)2 X(z)+ k1Y(z) k1k2 1 z1 Y(z) = k1 (1 z1)2 X(z) k1 1 z1 k1+ k2 k2z1 1 z1 Y(z), or, X(z) Y(z) 0 1 _1 z _1 z 1 _ 1 2 _1 z _1 z 2 _ 1 V(z) W(z) 3 _1 z _1 z 3 _ 1 U(z) 5 1+ k1(k1+ k2 k2z1) (1 z1)2 Y(z) = k1 (1 z1)2 X(z). Hence, H(z) = Y(z) X(z) = k1 1+ k1(k1+ k2)(2+ k1k2)z1+ z2 . 8.10 From Figure P8.7(a), the input-output relation of the channel is given by Y1(z) Y2(z) = 1H12(z) H21(z)1 X1(z) X2(z) . Likewise, the input-output relation of the channel separation circuit of Figure P8.7(b) is given by V1(z) V2(z) = 1G12(z) G21(z)1 Y1(z) Y2(z) . Hence, the overall system is characterized by V1(z) V2(z) = 1G12(z) G21(z)1 1H12(z) H21(z)1

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