数字信号处理-基于计算机的方法(第四版)答案 8-11章
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€ y[0] + y[1]z−1 + y[2]z
€ or € − 2
1 , 2
computation of the coefficients of
require
+ y[3]z−3 + y[4]z−4 = h[0] + h[1]z−1 + h[2]z−2 x[0] + x[1]z−1 + x[2]z−2 .
D = 2〈2−1 〉 5 = 6.
€
€
⎧0 ≤ n1 ≤ 4, Thus, n = 〈5 n1 + 2 n 2 〉10 , ⎨ ⎩0 ≤ n 2 ≤ 2, ⎧0 ≤ k ≤ 4, k = 〈5 k1 + € 6 k 2 〉10 , ⎨ €1 ⎩0 ≤ k 2 ≤ 2. The corresponding€ index mappings are indicated below:
I3 ( z) (1 − z0 z−1 )(1 − z1z−1 )(1 − z2 z−1 )(1 − z4 z−1 ) 2 −1 1 = = z (1 + z−1 )(1 − z−2 ), − 1 − 1 − 1 − 1 4 I3 ( z3 ) (1 − z0 z3 )(1 − z1z3 )(1 − z2 z3 )(1 − z4 z3 ) 3
I1 ( z) (1 − z0 z−1 )(1 − z2 z−1 )(1 − z3z−1 )(1 − z4 z−1 ) 2 1 = = − z−1 (1 − z−1 )(1 − z−2 ), − 1 − 1 − 1 − 1 3 4 I1 ( z1 ) (1 − z0 z1 )(1 − z2 z1 )(1 − z3z1 )(1 − z4 z1 )
€
I1( z) = (1 − z0 z−1 )(1 − z2 z−1 )(1 − z3z−1 )(1 − z4 z−1 ), I 2 ( z) = (1 − z0 z−1 )(1 − z1z−1 )(1 − z3z−1 )(1 − z4 z−1 ),
€ € € € €
€ €
I 3( z) = (1 − z0 z−1 )(1 − z1z−1 )(1 − z2 z−1 )(1 − z4 z−1 ), I 4 ( z) = (1 − z0 z−1 )(1 − z1z−1 )(1 − z2 z−1 )(1 − z3z−1 ). Therefore, I0 ( z) (1 − z1z−1 )(1 − z2 z−1 )(1 − z3z−1 )(1 − z4 z−1 ) 1 −1 1 = = z (1 − z−1 )(1 − z−2 ), − 1 − 1 − 1 − 1 12 2 I0 ( z0 ) (1 − z1z0 )(1 − z2 z0 )(1 − z3z0 )(1 − z4 z0 )
1
Chapter 11 – Part 2
11.40 (a) N = 6. Choose N1 = 2 and N 2 = 3.
D = 2〈2−1 〉 3 = 4.
Thus,
€
€
€
The corresponding index mappings are indicated below:
(b) N = 10. Choose N1 = 2 and N 2 = 5.
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
5 4
1 4
2 3
1 4
1 4
€
2 5 2 1 ⎛ 1 ⎞ +⎜ − Y ( z0 ) + Y ( z1 ) − Y ( z2 ) + Y ( z3 ) − Y ( z4 )⎟z−2 ⎝ 24 ⎠ 3 4 3 24 1 1 1 ⎛ 1 ⎞ +⎜ − Y ( z0 ) + Y ( z1 ) − Y ( z3 ) + Y ( z4 )⎟ z−3 ⎝ 12 ⎠ 6 6 12 1 1 1 1 1 ⎛ ⎞ +⎜ Y ( z0 ) − Y ( z1 ) + Y ( z2 ) − Y ( z3 ) + Y ( z4 )⎟z−4 . € ⎝ 24 ⎠ 6 4 6 24 Substituting the expressions for and Y ( z 4 ), in the above € equation, we then arrive at the expressions for the coefficients { y[ n ]} in terms of the coefficients {h[ n ]} and { x[ n ]}. Thus, €
1 1 ⎛ 1 ⎞ ⎛ 1 ⎞ = Y ( z1 ) + ⎜ − Y ( z0 ) + Y ( z2 )⎟ z−1 + ⎜ Y ( z0 ) − Y ( z1 ) + Y ( z2 )⎟z−2 ⎝ 2 ⎠ ⎝ 2 ⎠ 2 2 1 1 ⎛ ⎞ = h[0] x[0] + ⎜ − ( h[0] − h[1])( x[0] − x[1]) + ( h[0] + h[1])( x[0] + x[1])⎟z−1 ⎝ 2 ⎠ 2
1 −1 (z 12
€ €
€
Hence, Y ( z) =
− z−2 − z−3 + z−4 )Y ( z0 ) − (z−1 − z−2 − z−3 + z−4 )Y ( z1 )
1 2
1 4
2 3
1 4
1 4
6
+(1 − z−2 + z−4 )Y ( z2 ) + (z −1 + z−2 − z−3 − z−4 )Y ( z3 )
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.
Y ( z1) = Y ( ∞ ) = y[0] = H ( ∞ ) X ( ∞ ) = h[0] x[0],
€
€
From Eqs. (6.114) and (6.115), we arrive at where
€
Therefore,
I0 ( z) (1 − z1z−1 )(1 − z2 z−1 ) 1 = = − z−1 (1 − z−1 ), − 1 − 1 2 I0 ( z0 ) (1 − z1z0 )(1 − z2 z0 ) I1 ( z) (1 − z0 z−1 )(1 − z2 z−1 ) = = (1 − z−2 ), and −1 −1 I1 ( z1 ) (1 − z0 z1 )(1 − z2 z1 )
(
)(
)
Now,
€
Y ( z3 ) = Y (1) = ( h[0] + h[1] + h[2])( x[0] + x[1] + x[2]),
€来自百度文库
From Eqs. (6.114) and (6.115), we arrive at I (z) I (z) I (z) I (z) I (z) Y ( z) = 0 Y ( z0 ) + 1 Y ( z1 ) + 2 Y ( z2 ) + 3 Y ( z3) + 4 Y ( z4 ), I0 ( z 0 ) I1 ( z1 ) I2 ( z 2 ) I3 ( z 3 ) I4 ( z 4 ) where I 0 ( z) = (1 − z1z−1 )(1 − z2 z−1 )(1 − z3z−1 )(1 − z4 z−1 ),
€
3
(d) N = 15. Choose
€
€
A = 3, B = 5, C = 3〈 3−1 〉 5 = 6, ⎧0 ≤ n1 ≤ 4, D = 5〈5−1 〉 3 = 10. Thus, n = 〈 3n1 + 5 n 2 〉15, ⎨ ⎩0 ≤ n 2 ≤ 2, ⎧0 ≤ k1 ≤ 4, € k = 〈6 k1 + 10 k 2 〉15, ⎨ ⎩0 ≤ k 2 ≤ 2. The corresponding index mappings are indicated below: €
y[1] =
1 2 2 1 Y ( z0 ) − Y ( z1 ) + Y ( z3 ) − Y ( z4 ) 12 3 3 12
€ = h[0] x[1] + h[1] x[0], €
€
€
€
€
1 1 1 1 y[3] = − Y ( z0 ) + Y ( z1 ) − Y ( z3 ) + Y ( z4 ) = h[1] x[2] + h[2] x[1], 12 6 6 12 1 1 1 1 1 y[4] = Y ( z0 ) − Y ( z1 ) + Y ( z2 ) − Y ( z3 ) + Y ( z4 ) = h[2] x[2]. 24 6 4 6 24 1 2 5 1 1 Hence, ignoring the multiplications by , , , , , and computation 12 3 4 4 6
I4 ( z) (1 − z0 z−1 )(1 − z1z−1 )(1 − z2 z−1 )(1 − z3z−1 ) 1 1 = = − z−1 (1 + z−1 )(1 − z−2 ). 1 −1 −1 −1 12 2 I4 ( z4 ) (1 − z0 z− 4 )(1 − z1z 4 )(1 − z 2 z 4 )(1 − z 3z 4 )
2
(c) N = 12. Choose N1 = 3 and N 2 = 4.
D = 3〈 3−1 〉 4 = 3.
€
⎧ 0 ≤ n1 ≤ 2, Thus, n = 〈 4 n1 + 3n 2 〉12 , ⎨ ⎩0 ≤ n 2 ≤ 3, ⎧ 0 ≤ k ≤ 2, k = 〈 4 k1 +€ 3k 2 〉12 , ⎨ € 1 ⎩0 ≤ k 2 ≤ 3. The corresponding€ index mappings are indicated below:
€ € €
€
I2 ( z) (1 − z0 z−1 )(1 − z1z−1 ) 1 −1 = = z (1 + z−1 ). Hence, 1 −1 2 I2 ( z0 ) (1 − z0 z− 2 )(1 − z1z 2 )
1 2 1 2
Y ( z) = − z−1 (1 − z−1 )Y ( z0 ) + (1 − z−2 )Y ( z1 ) + z−1 (1 + z−1 )Y ( z2 )
€ €
5
1 ⎛ 1 ⎞ +⎜ ( h[0] − h[1])( x[0] − x[1]) − h[0] x[0] + ( h[0] + h[1])( x[0] + x[1])⎟ z−2 ⎝ 2 ⎠ 2
Ignoring the multiplications by
€
the values of Y ( z0 ), Y ( z1 ), and Y ( z2 ) , which can be evaluated using only 3 multiplications. (b)
and
4
− jθ o − jφ o 11.41 z = α . Hence, AoVo e e = α . Since Ao = 1, Vo = 1/ α ,
is real, we have
z) = H ( z) X ( z) or y[0] + y[1]z−1 + y[2]z−2 = ( h[0] + h[1]z−1 )( x[0] + x[1]z−1 ). € 11.42 (a) Y (€ Now, €
€ or € − 2
1 , 2
computation of the coefficients of
require
+ y[3]z−3 + y[4]z−4 = h[0] + h[1]z−1 + h[2]z−2 x[0] + x[1]z−1 + x[2]z−2 .
D = 2〈2−1 〉 5 = 6.
€
€
⎧0 ≤ n1 ≤ 4, Thus, n = 〈5 n1 + 2 n 2 〉10 , ⎨ ⎩0 ≤ n 2 ≤ 2, ⎧0 ≤ k ≤ 4, k = 〈5 k1 + € 6 k 2 〉10 , ⎨ €1 ⎩0 ≤ k 2 ≤ 2. The corresponding€ index mappings are indicated below:
I3 ( z) (1 − z0 z−1 )(1 − z1z−1 )(1 − z2 z−1 )(1 − z4 z−1 ) 2 −1 1 = = z (1 + z−1 )(1 − z−2 ), − 1 − 1 − 1 − 1 4 I3 ( z3 ) (1 − z0 z3 )(1 − z1z3 )(1 − z2 z3 )(1 − z4 z3 ) 3
I1 ( z) (1 − z0 z−1 )(1 − z2 z−1 )(1 − z3z−1 )(1 − z4 z−1 ) 2 1 = = − z−1 (1 − z−1 )(1 − z−2 ), − 1 − 1 − 1 − 1 3 4 I1 ( z1 ) (1 − z0 z1 )(1 − z2 z1 )(1 − z3z1 )(1 − z4 z1 )
€
I1( z) = (1 − z0 z−1 )(1 − z2 z−1 )(1 − z3z−1 )(1 − z4 z−1 ), I 2 ( z) = (1 − z0 z−1 )(1 − z1z−1 )(1 − z3z−1 )(1 − z4 z−1 ),
€ € € € €
€ €
I 3( z) = (1 − z0 z−1 )(1 − z1z−1 )(1 − z2 z−1 )(1 − z4 z−1 ), I 4 ( z) = (1 − z0 z−1 )(1 − z1z−1 )(1 − z2 z−1 )(1 − z3z−1 ). Therefore, I0 ( z) (1 − z1z−1 )(1 − z2 z−1 )(1 − z3z−1 )(1 − z4 z−1 ) 1 −1 1 = = z (1 − z−1 )(1 − z−2 ), − 1 − 1 − 1 − 1 12 2 I0 ( z0 ) (1 − z1z0 )(1 − z2 z0 )(1 − z3z0 )(1 − z4 z0 )
1
Chapter 11 – Part 2
11.40 (a) N = 6. Choose N1 = 2 and N 2 = 3.
D = 2〈2−1 〉 3 = 4.
Thus,
€
€
€
The corresponding index mappings are indicated below:
(b) N = 10. Choose N1 = 2 and N 2 = 5.
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
5 4
1 4
2 3
1 4
1 4
€
2 5 2 1 ⎛ 1 ⎞ +⎜ − Y ( z0 ) + Y ( z1 ) − Y ( z2 ) + Y ( z3 ) − Y ( z4 )⎟z−2 ⎝ 24 ⎠ 3 4 3 24 1 1 1 ⎛ 1 ⎞ +⎜ − Y ( z0 ) + Y ( z1 ) − Y ( z3 ) + Y ( z4 )⎟ z−3 ⎝ 12 ⎠ 6 6 12 1 1 1 1 1 ⎛ ⎞ +⎜ Y ( z0 ) − Y ( z1 ) + Y ( z2 ) − Y ( z3 ) + Y ( z4 )⎟z−4 . € ⎝ 24 ⎠ 6 4 6 24 Substituting the expressions for and Y ( z 4 ), in the above € equation, we then arrive at the expressions for the coefficients { y[ n ]} in terms of the coefficients {h[ n ]} and { x[ n ]}. Thus, €
1 1 ⎛ 1 ⎞ ⎛ 1 ⎞ = Y ( z1 ) + ⎜ − Y ( z0 ) + Y ( z2 )⎟ z−1 + ⎜ Y ( z0 ) − Y ( z1 ) + Y ( z2 )⎟z−2 ⎝ 2 ⎠ ⎝ 2 ⎠ 2 2 1 1 ⎛ ⎞ = h[0] x[0] + ⎜ − ( h[0] − h[1])( x[0] − x[1]) + ( h[0] + h[1])( x[0] + x[1])⎟z−1 ⎝ 2 ⎠ 2
1 −1 (z 12
€ €
€
Hence, Y ( z) =
− z−2 − z−3 + z−4 )Y ( z0 ) − (z−1 − z−2 − z−3 + z−4 )Y ( z1 )
1 2
1 4
2 3
1 4
1 4
6
+(1 − z−2 + z−4 )Y ( z2 ) + (z −1 + z−2 − z−3 − z−4 )Y ( z3 )
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.
Y ( z1) = Y ( ∞ ) = y[0] = H ( ∞ ) X ( ∞ ) = h[0] x[0],
€
€
From Eqs. (6.114) and (6.115), we arrive at where
€
Therefore,
I0 ( z) (1 − z1z−1 )(1 − z2 z−1 ) 1 = = − z−1 (1 − z−1 ), − 1 − 1 2 I0 ( z0 ) (1 − z1z0 )(1 − z2 z0 ) I1 ( z) (1 − z0 z−1 )(1 − z2 z−1 ) = = (1 − z−2 ), and −1 −1 I1 ( z1 ) (1 − z0 z1 )(1 − z2 z1 )
(
)(
)
Now,
€
Y ( z3 ) = Y (1) = ( h[0] + h[1] + h[2])( x[0] + x[1] + x[2]),
€来自百度文库
From Eqs. (6.114) and (6.115), we arrive at I (z) I (z) I (z) I (z) I (z) Y ( z) = 0 Y ( z0 ) + 1 Y ( z1 ) + 2 Y ( z2 ) + 3 Y ( z3) + 4 Y ( z4 ), I0 ( z 0 ) I1 ( z1 ) I2 ( z 2 ) I3 ( z 3 ) I4 ( z 4 ) where I 0 ( z) = (1 − z1z−1 )(1 − z2 z−1 )(1 − z3z−1 )(1 − z4 z−1 ),
€
3
(d) N = 15. Choose
€
€
A = 3, B = 5, C = 3〈 3−1 〉 5 = 6, ⎧0 ≤ n1 ≤ 4, D = 5〈5−1 〉 3 = 10. Thus, n = 〈 3n1 + 5 n 2 〉15, ⎨ ⎩0 ≤ n 2 ≤ 2, ⎧0 ≤ k1 ≤ 4, € k = 〈6 k1 + 10 k 2 〉15, ⎨ ⎩0 ≤ k 2 ≤ 2. The corresponding index mappings are indicated below: €
y[1] =
1 2 2 1 Y ( z0 ) − Y ( z1 ) + Y ( z3 ) − Y ( z4 ) 12 3 3 12
€ = h[0] x[1] + h[1] x[0], €
€
€
€
€
1 1 1 1 y[3] = − Y ( z0 ) + Y ( z1 ) − Y ( z3 ) + Y ( z4 ) = h[1] x[2] + h[2] x[1], 12 6 6 12 1 1 1 1 1 y[4] = Y ( z0 ) − Y ( z1 ) + Y ( z2 ) − Y ( z3 ) + Y ( z4 ) = h[2] x[2]. 24 6 4 6 24 1 2 5 1 1 Hence, ignoring the multiplications by , , , , , and computation 12 3 4 4 6
I4 ( z) (1 − z0 z−1 )(1 − z1z−1 )(1 − z2 z−1 )(1 − z3z−1 ) 1 1 = = − z−1 (1 + z−1 )(1 − z−2 ). 1 −1 −1 −1 12 2 I4 ( z4 ) (1 − z0 z− 4 )(1 − z1z 4 )(1 − z 2 z 4 )(1 − z 3z 4 )
2
(c) N = 12. Choose N1 = 3 and N 2 = 4.
D = 3〈 3−1 〉 4 = 3.
€
⎧ 0 ≤ n1 ≤ 2, Thus, n = 〈 4 n1 + 3n 2 〉12 , ⎨ ⎩0 ≤ n 2 ≤ 3, ⎧ 0 ≤ k ≤ 2, k = 〈 4 k1 +€ 3k 2 〉12 , ⎨ € 1 ⎩0 ≤ k 2 ≤ 3. The corresponding€ index mappings are indicated below:
€ € €
€
I2 ( z) (1 − z0 z−1 )(1 − z1z−1 ) 1 −1 = = z (1 + z−1 ). Hence, 1 −1 2 I2 ( z0 ) (1 − z0 z− 2 )(1 − z1z 2 )
1 2 1 2
Y ( z) = − z−1 (1 − z−1 )Y ( z0 ) + (1 − z−2 )Y ( z1 ) + z−1 (1 + z−1 )Y ( z2 )
€ €
5
1 ⎛ 1 ⎞ +⎜ ( h[0] − h[1])( x[0] − x[1]) − h[0] x[0] + ( h[0] + h[1])( x[0] + x[1])⎟ z−2 ⎝ 2 ⎠ 2
Ignoring the multiplications by
€
the values of Y ( z0 ), Y ( z1 ), and Y ( z2 ) , which can be evaluated using only 3 multiplications. (b)
and
4
− jθ o − jφ o 11.41 z = α . Hence, AoVo e e = α . Since Ao = 1, Vo = 1/ α ,
is real, we have
z) = H ( z) X ( z) or y[0] + y[1]z−1 + y[2]z−2 = ( h[0] + h[1]z−1 )( x[0] + x[1]z−1 ). € 11.42 (a) Y (€ Now, €