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SOLUTIONS MANUAL to accompany Digital Signal Processing: A Computer-Based Approach Third Edition
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SOLUTIONS MANUAL to accompany Digital Signal Processing: A Computer-Based Approach Third Edition
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, SOLUTIONS MANUALto accompany
Digital Signal Processing:
A Computer-Based Approach
Third Edition
Sanjit K. Mitra
Prepared by
Chowdary Adsumilli, John Berger, Marco Carli,
Hsin-Han Ho, Rajeev Gandhi, Chin Kaye Koh,
Luca Lucchese, and Mylene Queiroz de Farias
Not for sale. 1
, Chapter 2
2.1 (a) x1 1 = 22.85, x1 2 = 9.1396, x1 ∞ = 4.81,
(b) x 2 1 = 18.68, x 2 2 = 7.1944, x 2 ∞ = 3.48.
1, n ≥ 0, 1, n < 0,
2.2 µ[ n] = ⎧⎨ Hence, µ[ −n − 1] = ⎧⎨ Thus, x[ n] = µ[ n] + µ[ − n − 1].
⎩0, n < 0. ⎩0, n ≥ 0.
n
2.3 (a) Consider the sequence defined by x[ n] = ∑ δ[ k ]. If n < 0, then k = 0 is not included
k = −∞
in the sum and hence, x[n] = 0 for n < 0. On the other hand, for n ≥ 0, k = 0 is included
in the sum, and as a result, x[n] =1 for n ≥ 0. Therefore,
1, n ≥ 0,
x[ n] = ∑ δ[ k ] = ⎧⎨
n
= µ[ n].
k = −∞ ⎩0, n < 0,
1, n ≥ 0, 1, n ≥ 1,
(b) Since µ[ n] = ⎧⎨ it follows that µ[ n − 1] = ⎧⎨ Hence,
⎩0 , n < 0, ⎩ , n < 1.
0
1, n = 0,
µ[ n] − µ[ n − 1] = ⎧⎨ = δ[ n].
⎩0, n ≠ 0,
2.4 Recall µ[ n] − µ[ n − 1] = δ[ n]. Hence,
x[ n] = δ[ n] + 3δ[ n − 1] − 2δ[ n − 2] + 4δ[ n − 3]
= (µ[ n] − µ[ n − 1]) + 3(µ[ n − 1] − µ[ n − 2]) − 2(µ[ n − 2] − µ[ n − 3]) + 4(µ[ n − 3] − µ[ n − 4])
= µ[ n] + 2µ[ n − 1] − 5µ[ n − 2] + 6µ[ n − 3] − 4µ[ n − 4].
2.5 (a) c[ n] = x[ − n + 2] = {2 0 − 3 − 2 1 5 − 4},
↑
(b) d[ n] = y[ − n − 3] = {− 2 7 8 0 − 1 − 3 6 0 0},
↑
(c) e[ n] = w[ − n] = {5 − 2 0 − 1 2 2 3 0 0},
↑
(d) u[ n] = x[ n] + y[ n − 2] = {− 4 5 1 − 2 3 − 3 1 0 8 7 − 2},
↑
(e) v[ n] = x[ n] ⋅ w[ n + 4] = {0 15 2 − 4 3 0 − 4 0},
↑
(f) s[ n] = y[ n] − w[ n + 4] = {− 3 4 − 5 0 0 10 2 − 2},
↑
(g) r[ n] = 3.5 y[ n] = {21 − 10.5 − 3.5 0 2.8 24.5 − 7}.
↑
2.6 (a) x[ n] = −4δ[ n + 3] + 5δ[ n + 2] + δ[ n + 1] − 2δ[ n] − 3δ[ n − 1] + 2δ[ n − 3],
y[ n] = 6δ[ n + 1] − 3δ[ n] − δ[ n − 1] + 8δ[ n − 3] + 7δ[ n − 4] − 2δ[ n − 5],
w[ n] = 3δ[ n − 2] + 2δ[ n − 3] + 2δ[ n − 4] − δ[ n − 5] − 2δ[ n − 7] + 5δ[ n − 8],
(b) Recall δ[ n] = µ[ n] − µ[ n − 1]. Hence,
x[ n] = −4(µ[ n + 3] − µ[ n + 2]) + 5(µ[ n + 2] − µ[ n + 1]) + (µ[ n + 1] − µ[ n])
− 2(µ[ n] − µ[ n − 1]) − 3(µ[ n − 1] − µ[ n − 2]) + 2(µ[ n − 3] − µ[ n − 4])
Not for sale. 2
, = −4µ[ n + 3] + 9µ[ n + 2] − 4µ[ n + 1] − 3µ[ n] − µ[ n − 1] + 3µ[ n − 2] + 2µ[ n − 3] − 2µ[ n − 4],
2.7 (a)
_ x[n-1] _ x[n-2]
x[n] z 1 z 1
h[0] h[1] h[2]
+ + y[n]
From the above figure it follows that y[ n] = h[ 0] x[ n] + h[1] x[ n − 1] + h[2] x[ n − 2].
(b)
h[0] w[n]
x[n] + + y[n]
_1 _1
z z
β11 β12
x[n _ 1] + w[n _ 1] +
_ _
z 1 z 1
β 21 β 22
x[n _ 2] w[n _ 2]
From the above figure we get w[ n] = h[0]( x[ n] + β11 x[ n − 1] + β 21 x[ n − 2]) and
y[ n] = w[ n] + β12 w[ n − 1] + β 22 w[ n − 2]. Making use of the first equation in the second
we arrive at
y[ n] = h[0]( x[ n] + β11 x[ n − 1] + β 21 x[ n − 2])
+ β12 h[0]( x[ n − 1] + β11 x[ n − 2] + β 21 x[ n − 3])
+ β 22 h[0]( x[ n − 2] + β11 x[ n − 3] + β 21 x[ n − 4])
= h[0]( x[ n] + (β11 + β12 ) x[ n − 1] + (β 21 + β12 β11 + β 22 ) x[ n − 2]
+ (β12 β 21 + β 22 β11 ) x[ n − 3] + β 22 β 21 x[ n − 4]).
(c) Figure P2.1(c) is a cascade of a first-order section and a second-order section. The
input-output relation remains unchanged if the ordering of the two sections is
interchanged as shown below.
0.6 u[n] y[n+1]
w[n]
x[n] + + +
_ _
z 1 z 1
_ 0.3
0.8 0.4
+ + y[n]
w[n _ 1]
_
z 1
_ 0.5
0.2
w[n _ 2]
Not for sale. 3
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