数字信号处理-基于计算机的方法第四版答案8-11章.pdf

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 Sanjit 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 1 and 1.5 1+K +0.5. From the first inequality we have 1.5 K 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+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 with 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 d1+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 variables 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) where 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) = 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 X1(z) X2(z) = 1 H21(z)G12(z)H12(z) G12(z) H21(z) G21(z)1 H12(z)G21(z) X1(z) X2(z) . The cross-talk is eliminated if V1(z) is a function of either X1(z) or X2(z), and similarly, if V2(z) is a function of either X1(z) or X2(z). From the above equation it follows that if H12(z) = G12(z), and H21(z) = G21(z), then V1(z) = 1 H21(z)G12(z)()X1(z),and V2(z) = 1 H12(z)G21(z)()X2(z). Alternately, if G12(z) = H21 1(z), and G21(z) = H12 1(z), then V1(z) = H12(z)H21(z) 1 H21(z) X 2(z), and V2(z) = H12(z)H21(z) 1 H12(z) X 1(z). 8.11 Analyzing Figure P8.8, we get wn = A Xn+CDun() and yn = C ABxn+ un(). A direct implementation of these two equations leads to the structure shown below which has no delay-free loop. 8.12 (a) Figure P8.9(a) with internal adder output variables is shown below: 6 The outputs of the 4 adders are given by Y(z) = z1W3(z)+1W1(z), W1(z) = X(z)+2W3(z), W2(z) =W1(z)+ k2W2(z), and W3(z) = k1W2(z)+ z1W2(z). From the 3rd equation given above we obtain W2(z) = W1(z) 1 k2 and from the 4th equation given above we obtain W3(z) = (z1+ k1)W2(z) = (z1+ k1) 1 k2 W1(z). Substituting the expression for W3(z) in the first equation given above we arrive at Y(z) = z1(z1+ k1) 1 k2 W1(z)+1W1(z) = z1(z1+ k1)+(1 k2)1 1 k2 W1(z). Substituting the expression for W3(z) given above in the second equation we get W1(z) = X(z)+2 (z1+ k1) 1 k2 W1(z) which leads to W1(z) = (1 k2)X(z) 1 k22z12k1 . Substituting this expression for W1(z) in the last expression for Y(z) we finally arrive at the expression for input-output relation of the structure of Figure P8.9(a) Y(z) = z1(z1+ k1)+(1 k2)1 1 k22z12k1 X(z). It can be seen that there are two delay-free loops in the structure: one going through the multipliers k1 and 2, and the other one going through the multiplier k2. To develop the equivalent realization without any delay-free loop we first remove the two unit delays as shown below: X(z) Y(z) k1 k2 z 1 _ 2 1 z 1 _ W (z) 1 W (z) 2 W (z) 3 X(z) Y(z) k1 k22 1 1 X (z) 1 Y (z)2 X (z) 2 Y (z) 7 We then remove the output variable Y(z) and the two inputs to its adder as shown below: As can be seen from the above structure, there are two delay-free loops with loop gains 2k1 and k2. The corresponding expression for the determinant is =12k1 k2. The above structure is next redrawn as indicated below: The transfer matrix of the above three-pair is given by Y(z) Y1(z) Y2(z) = t11t12t13 t21t22t23 t31t32t33 = 1 1(1 k2)12(1 k2) k10(1 k2) 102 X(z) X1(z) X2(z) . A realization of the above structure without any delay-free loop is thus as shown below: (b) Figure P8.9(b) is redrawn below with the dashed block containing the two delays. X(z) k1 k22 1 Y (z)2 X (z) 2 Y (z) X(z) k2 2 1 Y (z)2 X (z) 2 Y (z) 1 k /k2 Y(z) X(z) Y (z) 1 X (z) 1 X (z) 2 Y (z) 2 t1121 t 23 t 33 t 31 t 13 t 8 Figure P8.9(b) with delays removed is as shown below The only delay-free loop goes through the multipliers and . The corresponding loop-gain is and the determinant is =1loopgain =1. The transfer matrix of the above two-pair is given by Y(z) Y1(z) = t11t12 t21t22 X(z) X1(z) = 1 1 1 X(z) X1(z) . A realization of the structure of Figure P8.9(b) without any delay-free loop is thus as shown below: 8.13 (a) H(z) = (1+0.4z1)4(10.2z1)2 =1+1.2z1+0.36z20.64z30.0384z4+0.001z6. A direct form realization of H(z) is shown below: Y(z) z1z1 X(z) Y (z) 1 + X (z) 1 Y(z) X(z) Y (z) 1 + X (z) 1 Y(z) z1z1 X(z) Y (z) 1 + X (z) 1 1/ 1/ 9 The transposed form of the above structure yields another direct form realization as indicated below: (b) A realization in the form of cascade of six first-order sections is shown below: (c) A realization in the form of cascade of three second-order sections is shown below: (d) A realization in the form of cascade of two third-order sections is shown below: (e) A realization in the form of cascade of two first-order sections and two second- order sections is shown below: z 1 _ z 1 _ z 1 _ + z 1 _ + xn yn z 1 _ + z 1 _ 1.20.360.001 0.0384 _ 0.064 _ xn yn z1z1z1z1z1z1 1.2 0.36 0.001 0.0384 _ 0.064 _ xn yn z 1 _ + z 1 _ + z 1 _ + z 1 _ + z 1 _ + z 1 _ + 0.4 0.4 0.4 0.4 _0.2 _0.2 xn yn z 1 _ + z 1 _ + z 1 _ + z 1 _ + 0.8 0.8 0.16 0.16 z 1 _ + z 1 _ + 0.12 _ 0.12 _ z 1 _ z 1 _ + z 1 _ + xn yn z 1 _ z 1 _ + z 1 _ + 1.20.480.064 0.016 0.12 _ xn yn z 1 _ + z 1 _ + z 1 _ + z 1 _ + 0.8 0.8 0.16 0.16 z 1 _ + z 1 _ + 0.12 _ 0.12 _ 10 8.14 H(z) = E0(z4)+ z1E1(z4)+ z2E2(z4)+ z3E3(z4) where, E0(z) = a+ez1, E1(z) = b+ fz1, E2(z) = c + gz1, E3(z) = d + hz1. A direct 4-band polyphase realization of H(z) is shown below which requires 18 unit delays and hence is a non-canonic structure. A minimum-delay realization of the above structure is shown below which requires 7 unit delays and hence it is a canonic structure. 8.15 H(z) = E0(z3)+ z1E1(z3)+ z2E2(z3) where, E0(z) = a+ dz1+ gz2, E1(z) = b+ez1+ hz2, E2(z) = c + fz1. A direct 3- band polyphase realization of H(z) is shown below which requires 17 unit delays and hence is a non-canonic structure. xnyn z 1 _ z 1 _ 4 _ a + ez 4 _ c + gz 4 _ b + fz 4 _ d + hz + + z 1 _ + xn yn z 4 _z 1 _ + + z 1 _ + z 1 _ + + + + a b c d e f g h 11 A minimum-delay realization of the above structure is shown below which requires 8 unit delays and hence it is a non-canonic structure. 8.16 H(z) = E0(z2)+ z1E1(z2) where, E0(z) = a+cz1+ez2+ gz3, E1(z) = b+ dz1+ fz2+ hz3. A direct 2-band polyphase realization of H(z) is shown below which requires 20 unit delays and hence is a non-canonic structure. xn yn z 1 _ + z 1 _ + 3 _ a + dz + gz 6 _ 3 _ b + ez + hz 6 _ 3 _ c + fz xn yn z 3 _ z 1 _ + + z 1 _ + + + a b c d e f g h z 3 _ xn yn z 1 _ + 2 _ a + cz + ez + gz 4 _ 6 _ 2 _ b + dz + fz + hz 4 _ 6 _ 12 A minimum-delay realization of the above structure is shown below which requires 7 unit delays and hence it is a canonic structure. 8.17 8.18 8.19 G(z) = zN /2 H(z). A canonic realization of both G(z) and H(z) is shown below for N =6. xn yn z 2 _ + z 1 _ + a b c d e f g h z 2 _ z 2 _ + + + z 1 _ + h0h1 z 1 _ z 1 _ z 1 _ + + xn h3 z 1 _ z 1 _ + + yn h2 z 1 _ z 1 _ + + _1 _1 _1_1 z 1 _ h0h1h2 z 1 _ z 1 _ z 1 _ z 1 _ + h3 + + z 1 _ z 1 _ + z 1 _ z 1 _ + h4 + _1 _1_1_1 _1 xn yn 13 8.20 Without any loss of generality, assume M =5 which means N =11. In this case, the transfer function is given by = z5h5+ h4(z+ z1)+ h3(z2+ z2)+ h2(z3+ z3)+ h1(z4+ z4)+ h0(z5+ z5) . Now, the recursion relation for the Chebyshev polynomial is given by Tr(x) = 2xTr1(x) Tr2(x), r 2 with T0(x) =1 and T1(x) = x. Hence, T2(x) = 2xT1(x) T0(x) = 2x21, T3(x) = 2xT2(x) T1(x) = 2x(2x21) x = 4x3 3x, T4(x) = 2xT3(x) T2(x) = 2x(4x3 3x) (2x21) = 8x48x2+1, T5(x) = 2xT4(x) T3(x) = 2x(8x48x2+1) (4x3 3x) =16x520x3+5x. We can thus rewrite the expression inside the square brackets given above as h5+2h4T1 z+z1 2 +2h3T2 z+z1 2 +2h2T3 z+z1 2 +2h1T4 z+z1 2 +2h0T5 z+z1 2 = h5+2h4 z+z1 2 +2h3 2 z+z1 2 2 1 +2h2 4 z+z1 2 3 3 z+z1 2 +2h1 8 z+z1 2 4 8 z+z1 2 2 +1 +2h0 16 z+z1 2 5 20 z+z1 2 3 +5 z+z1 2 =an z+z1 2 n n=0 5 , where a0 = h52h3+2h1, a1 = 2h46h2+10h0, a2 = 4h316h1, a3 = 8h2 40h1, a4 =16h1,and a5 = 32h0. A realization of H(z) = z5an z+z1 2 n n=0 5 = a0z5+ a1 1+z2 2 z4 +a2 1+z2 2 2 z3 +a3 1+z2 2 3 z2+ a4 1+z2 2 4 z1+ a5 1+z2 2 5 is shown below: 14 8.21 Consider H(z) = P(z) D(z) = P1(z) D1(z) P2(z) D2(z) P3(z) D3(z) . Assume all zeros of P(z) and D(z) are complex. Note that the numerator of the first stage can be one of the 3 factors, P1(z), P2(z), and P3(z). Likewise, the numerator of the second stage can be one of the remaining 2 factors, and the numerator of the third stage is the remaining factor. Similarly, that the denominator of the first stage can be one of the 3 factors, D1(z), D2(z), and D3(z). Likewise, the denominator of the second stage can be one of the remaining 2 factors, and the denominator of the third stage is the remaining factor. Hence, there are (3!)2= 36 different types of cascade realizations. If the zeros of P(z) and D(z) are all real, then P(z) has 6 real zeros and D(z) has 6 real zeros. In this case, then there are (6!)2= 518400 different types of cascade realizations. 8.22 H(z) = Pi(z) Di(z) i=1 K . Here, the numerator of the first stage can be chosen in K 1 ways, the numerator of the second stage can be chosen in K1 1 ways, and so on, until there is only one possible choice for the numerator of the Kth stage. Likewise, the denominator of the first stage can be chosen in K 1 ways, the denominator of the second stage can be chosen in K1 1 ways, and so on, until there is only one possible choice for the denominator of the Kth stage. Hence, the total number of possible cascade realizations are equal to K 1 2 K1 1 2 K2 1 2 2 1 2 1 1 2 = (K!)2. 8.23 A realization of the ringing delay is shown below: The second-order direct form II structure with each delay replaced by the above ringing delay is shown below: + z D _ 15 8.24 (a) A direct form II realization of H1(z) is shown below: (b) A direct form II realization of H2(z) is shown below: 8.25 A direct form II realization of H(z) is shown below: + z D _ + z D _ + + + + 1 p 1 _d 2 p 0 p 2 d _ z 1 _ z 1 _ + + + xn 2 0.6 0.18 _ 0.9 _ yn z 1 _ z 1 _ + + + xn 3 0.15 _ yn 0.6 _ 0.8 + z 1 _ z 1 _ z 1 _ + + xnyn 3 4.5 _2.9 _2.2 0.81 _5.1 16 A direct form II t realization of H(z) is shown below: 8.26 (a) A direct form II realization of H1(z) is shown below: A direct form II t realization of H1(z) is shown below: (b) A direct form II realization of H2(z) is shown below: A direct form II t realization of H2(z) is shown below: + + + z 1 _ + z 1 _ z 1 _ xn yn 3 4.5 _2.9 _2.2 _5.1 0.81 z1z1 z1 3 0.6 _0.5 _4 1.5 0.3 _1 xn yn + z 1 _ + z 1 _ 0.3 _1 _4 1.5 + + z 1 _ 3 _ 0.5 1.8 z1 z1 z1 z1 3 1.5 _ 4.9 _0.8 4.2 0.6 17 (c) A direct form II realization of H3(z) is shown below: A direct form II t realization of H3(z) is shown below: 8.27 The transfer function of the structure of Figure P8.11 is H(z) = (1 a0zM) p0+ p1z1+ p2z2 1+ d1z1+ d2z2 . (a) M-point moving average filter: HMA(z) = 1 M 1 zM 1 z1 . Choose a0=1, p0=1/M, p1= 0, p2= 0, d1= 1, d2= 0. (b) First difference differentiator: HFD(z) =1+ z1. Choose a0= 0, p0=1, p1=1, p2= 0, d1= 0, d2= 0. (c) Central difference differentiator: HCD(z) = 0.5(1 z2). + z 1 _ z 1 _ + + + z 1 _ z 1 _ + 3 1.5 _ 4.9 4.2 0.6 _ 0.8 z1z1 z1 4 0.475 0.75 4/3 _ 0.25 1/12 2.917 _ xn yn + z 1 _ + z 1 _ + + z 1 _ 4 _1.9 0.75 _2.917 _0.25 + 4/3 1/12 18 Choose a0= 0, p0= 0.5, p1= 0.5, p2= 0, d1= 0, d2= 0. (d) Running sum integrator: HRS(z) = 1 1 z1 . Choose a0= 0, p0=1, p1= 0, p2= 0, d1= 1, d2= 0. (e) Leaky integrator: HLI(z) = 1(1)z1 . Choose a0= 0, p0=, p1= 0, p2= 0, d1= (1), d2= 0. (f) DC blocker: HDC(z) = 1 z1 1z1 . Choose a0= 0, p0=1, p1= 1, p2= 0, d1= , d2= 0. (g) Trapezoidal integrator: HTI(z) = 0.5+0.5z1 1 z1 . Choose a0= 0, p0= 0.5, p1= 0.5, p2= 0, d1= 1, d2= 0. (h) First-order allpass delay network: A1(z) = + z1 1+z1 . Choose a0= 0, p0=, p1= 0.5, p2= 0, d1= , d2= 0. (i) Second-order allpass delay network: A2(z) = +z1+ z2 1+z1+z2 . Choose a0= 0, p0=, p1=, p2=1, d1= , d2=. (j) Simpson integrator: HSI(z) = 1 3 + 4 3 z1+ 1 3 z2 1 z2 . Choose a0= 0, p0=1/3, p1= 4/3, p2=1/3, d1= 0, d2= 1. 8.28 (a) H(z) = 10.6z1 1+0.25z1 0.2+ z1 1+0.3z1 2 1+0.25z1 = 0.4 +1.76z11.2z2 1+0.8z1+0.2125z2+0.0187z3 . (b) yn = 0.4xn+1.76xn 11.2xn 20.8yn 10.21