数字信号处理-基于计算机的方法(第四版)答案 8-11章

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
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D = 2〈2−1 〉 5 = 6.
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⎧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:
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 )
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, €
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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
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the values of Y ( z0 ), Y ( z1 ), and Y ( z2 ) , which can be evaluated using only 3 multiplications. (b)
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], €
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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
(
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Now,
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Y ( z3 ) = Y (1) = ( h[0] + h[1] + h[2])( x[0] + x[1] + x[2]),
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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 ),
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 )
1 −1 (z 12
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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 )
€ 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 .
Y ( z1) = Y ( ∞ ) = y[0] = H ( ∞ ) X ( ∞ ) = h[0] x[0],
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From Eqs. (6.114) and (6.115), we arrive at wh ( 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 )
5 4
1 4
2 3
1 4
1 4
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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, €
2
(c) N = 12. Choose N1 = 3 and N 2 = 4.
D = 3〈 3−1 〉 4 = 3.
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⎧ 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:
1
Chapter 11 – Part 2
11.40 (a) N = 6. Choose N1 = 2 and N 2 = 3.
D = 2〈2−1 〉 3 = 4.
Thus,
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The corresponding index mappings are indicated below:
(b) N = 10. Choose N1 = 2 and N 2 = 5.
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3
(d) N = 15. Choose
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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: €