Solution Manual For Digital Signal Processing, 4th Edition

Solution Manual For Digital Signal Processing, 4th Edition makes solving textbook exercises easier with step-by-step solutions and helpful tips.

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Chapter 1
1.1
(a) One dimensional, multichannel, discrete time, and digital.
(b) Multi dimensional, single channel, continuous-time, analog.
(c) One dimensional, single channel, continuous-time, analog.
(d) One dimensional, single channel, continuous-time, analog.
(e) One dimensional, multichannel, discrete-time, digital.
1.2
(a) f = 0.01π
2π = 1
200 periodic with Np = 200.
(b) f = 30π
105 ( 1
2π ) = 1
7 periodic with Np = 7.
(c) f = 3π
2π = 3
2 periodic with Np = 2.
(d) f = 3
2π non-periodic.
(e) f = 62π
10 ( 1
2π ) = 31
10 periodic with Np = 10.
1.3
(a) Periodic with period Tp = 2π
5 .
(b) f = 5
2π non-periodic.
(c) f = 1
12π non-periodic.
(d) cos( n
8 ) is non-periodic; cos( πn
8 ) is periodic; Their product is non-periodic.
(e) cos( πn
2 ) is periodic with period Np=4
sin( πn
8 ) is periodic with period Np=16
cos( πn
4 + π
3 ) is periodic with period Np=8
Therefore, x(n) is periodic with period Np=16. (16 is the least common multiple of 4,8,16).
1.4
(a) w = 2πk
N implies that f = k
N . Let
α = GCD of (k, N ), i.e.,
k = kα, N = N α.
Then,
f = k
N , which implies that
N = N
α .
3
(b)
N = 7
k = 0 1 2 3 4 5 6 7
GCD(k, N ) = 7 1 1 1 1 1 1 7
Np = 1 7 7 7 7 7 7 1
(c)
N = 16
k = 0 1 2 3 4 5 6 7 8 9 10 11 12 . . . 16
GCD(k, N ) = 16 1 2 1 4 1 2 1 8 1 2 1 4 . . . 16
Np = 1 6 8 16 4 16 8 16 2 16 8 16 4 . . . 1
1.5
(a) Refer to fig 1.5-1
(b)
0 5 10 15 20 25 30
−3
−2
−1
0
1
2
3
−−−> t (ms)
−−−> xa(t)
Figure 1.5-1:
x(n) = xa(nT )
= xa(n/Fs)
= 3sin(πn/3)
f = 1
2π ( π
3 )
= 1
6 , Np = 6
4

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