Complete Solution Manual for Digital Signal Processing: A Computer-Based Approach, 4th Edition by Sanjit K. Mitra.

ALL 14 CHAPTERS COVERED




SOLUTION MANUAL

, Chapter 2
2.1 (a) x1 1 = 22.85, x1 2 = 9.1396, x1  = 4.81,


(b) x2 1 = 18.68, x2 2 = 7.1944, x2  = 3.48.


2.2 [n] = 1,
0,
n  0, Hence, [−n − 1] = 1, n  0,
0, Thus, x[n] = [n] + [−n − 1].
 n  0.  n  0.
n


2.3 (a) Consider the sequence defined bỵ x[n] =  [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,
n
[k] = 1, n  0, = [n].
x[n] =
 0, n  0,
k =− 

(b) Since [n] = 1,
0,
n  0, it follows that [n − 1] = 1, n  1,
0, Hence,
 n  0,  n  1.

n = 0, = [n].
[n] − [n − 1] = 1,
0,
 n  0,

2.4 Recall [n] − [n − 1] = [n]. Hence,
x[n] = [n] + 3[n − 1] − 2[n − 2] + 4[n − 3]
= ([n] − [n − 1]) + 3([n − 1] − [n − 2]) − 2([n − 2] − [n − 3]) + 4([n − 3] − [n − 4])
= [n] + 2[n − 1] − 5[n − 2] + 6[n − 3] − 4[n − 4].

2.5 (a) c[n] = x[−n + 2] = {2 0 − 3 − 2 1 5 − 4},

(b) d[n] = ỵ[−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] + ỵ[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},



Not for sale. 2

, (f) s[n] = ỵ[n] − w[n + 4] = {− 3 4 − 5 0 0 10 2 − 2},


(g) r[n] = 3.5 ỵ[n] = {21 − 10.5 − 3.5 0 2.8 24.5 − 7}.




2.6 (a) x[n] = −4[n + 3] + 5[n + 2] + [n + 1] − 2[n] − 3[n − 1] + 2[n − 3],

ỵ[n] = 6[n + 1] − 3[n] − [n − 1] + 8[n − 3] + 7[n − 4] − 2[n − 5],
w[n] = 3[n − 2] + 2[n − 3] + 2[n − 4] − [n − 5] − 2[n − 7] + 5[n − 8],

(b) Recall [n] = [n] − [n − 1]. Hence,
x[n] = −4([n + 3] − [n + 2]) + 5([n + 2] − [n + 1]) + ([n + 1] − [n])
− 2([n] − [n − 1]) − 3([n − 1] − [n − 2]) + 2([n − 3] − [n − 4])




Not for sale. 2

, = −4[n + 3] + 9[n + 2] − 4[n + 1] − 3[n] − [n − 1] + 3[n − 2] + 2[n − 3] − 2[n − 4],

2.7 (a)
_1 1] _1 2]
x[n]


h[0] h[1] h[2]
ỵ[n]

From the above figure it follows that ỵ[n] = h[0]x[n] + h[1]x[n − 1] + h[2]x[n − 2].


(b)
h[0] w[n]
x[n] ỵ[n]
_1 _1
11 12
x[n _ 1] + w[n _ 1]
_1 _1
21 22
x[n _ 2] w[n 2]

From the above figure we get w[n] = h[0](x[n] + 11x[n − 1] + 21x[n − 2]) and

ỵ[n] = w[n] + 12 w[n − 1] + 22w[n − 2]. Making use of the first equation in the second
we arrive at

ỵ[n] = h[0](x[n] + 11x[n − 1] + 21x[n − 2])
+ 12h[0](x[n − 1] + 11x[n − 2] + 21x[n − 3])
+ 22h[0](x[n − 2] + 11x[n − 3] + 21x[n − 4])
= h[0](x[n] + (11 + 12 )x[n − 1] + (21 + 1211 + 22 )x[n − 2]
+ (1221 + 2211 )x[n − 3] + 2221x[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 ỵ[n+1]
w[n] u[n]
x[n] + +
1
_ 0.3
0.8 0.4
ỵ[n]
w[n _ 1]


0.2
_
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