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Solution Manual For Digital Signal Processing: Principles, Algorithms and Applications (4th Edition) by John G. Proakis | Latest Version

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Solution Manual For Digital Signal Processing: Principles, Algorithms and Applications (4th Edition) by John G. Proakis | Latest Edition

Institution
Digital Signal Processing
Course
Digital Signal Processing

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SOLUTION MANUAL FOR
Digital Signal Processing: Principles, Algorithms and
Applications
By John G. Proakis
5th Edition

,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

1
= 200 ⇒ Periodic With Np = 200.
(b) F = 105 ( 2π) = 17⇒ Periodic With Np = 7.
30π 1

(c) F = 3π

= 3 2⇒ Periodic With Np = 2.
(d) F = 2π ⇒ Non-Periodic.
3

10 ( 2π) = 10⇒ Periodic With Np = 10.
F = 62π 1 31
(e)


1.3

(a) Periodic With Period Tp = .
5
(b) F = 2π ⇒ Non-Periodic.
5
1
(c) F = 12π ⇒ Non-Periodic.
(d) Cos( N ) Is Non-Periodic; Cos( Πn ) Is Periodic; Their Product Is Non-Periodic.
8 8
(e) Cos( Πn2 ) Is Periodic With Period Np=4
Sin( Πn8 ) Is Periodic With Period Np=16
Cos( Πn + Π ) Is Periodic With Period Np=8
4 3
Therefore, X(N) Is Periodic With Period Np=16. (16 Is The Least Common Multiple Of 4,8,16).


1.4
2πk K
(a) W = Implies That F = . Let
N N

Α = Gcd Of (K, N ), I.E.,

K = K′Α, N = N ′Α.
Then,

F = K , Which Implies That
N′
N
N′ = .
Α
3

,(b)

N = 7
K = 01234567
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)

3



2




1
−−−> xa(t)




0



−1




−2



−3
0 5 10 15 20 25 30
−−−> T (Ms)


Figure 1.5-1:


X(N) = Xa(Nt )
= Xa(N/Fs)
= 3sin(Πn/3) ⇒
1 Π
F = ( )
2π 3
1
= , Np = 6
6
4

, T (Ms)



-3




Figure 1.5-2:

(c) Refer n
To Fig 1.5-2 ,
x(n) = 0, √3 ,2√3 ,20, — √3 2, — √32 , Np = 6.
(d) Yes.
100π
X(1) = 3 = 3sin( ) ⇒ Fs = 200 Samples/Sec.
Fs

1.6
(a)

X(N) = Acos(2πf0n/Fs + Θ)
= Acos(2π(T/Tp)N + Θ)
But T/Tp = F ⇒ X(N) Is Periodic If F Is Rational.
(b) If X(N) Is Periodic, Then F=K/N Where N Is The Period. Then,
K Tp
Td = ( T ) = K( )T = Ktp.
F T
Thus, It Takes K Periods (Ktp) Of The Analog Signal To Make 1 Period (Td) Of The Discrete Signal.
(c) Td = Ktp ⇒ Nt = Ktp ⇒ F = K/N = T/Tp ⇒ F Is Rational ⇒ X(N) Is Periodic.


1.7
(a) Fmax = 10khz ⇒ Fs ≥ 2fmax = 20khz.
(b) For Fs = 8khz, Ffold = Fs/2 = 4khz ⇒ 5khz Will Alias To 3khz.
(c) F=9khz Will Alias To 1khz.


1.8
(a) Fmax = 100khz, Fs ≥ 2fmax = 200hz.
(b) Ffold = F2s = 125hz.

5

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Course
Digital Signal Processing

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