digital-signal-processing-fourth-edition-by-j-proakis-and-d-manolakis-instructor-solution-manual-

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
= 1200⇒ periodic with Np = 200.
(b) f = 105( 2π) = 1 7⇒ periodic with Np = 7.
30π 1

(c) f = 3π2π= 3 2⇒ periodic with Np = 2.
(d) f = 32π ⇒ non-periodic.
f = 62π10 ( 2π) = 10⇒ periodic with Np = 10.
1 31
(e)


1.3
2π
(a) Periodic with period Tp = .
5
(b) f = 2π ⇒ non-periodic.
5
1
(c) f = n12π ⇒ non-periodic.
(d) cos( ) 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




© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws
as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in
writing from the publisher. For the exclusive use of adopters of the book Digital Signal Processing, Fourth Edition, by John G.
Proakis and Dimitris G. Manolakis. ISBN 0-13-187374-1.

, (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




© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws
as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in
writing from the publisher. For the exclusive use of adopters of the book Digital Signal Processing, Fourth Edition, by John G.
Proakis and Dimitris G. Manolakis. ISBN 0-13-187374-1.

, t (ms)



-3




Figure 1.5-2:

(c) Refer n
to fig 1.5-2 ,
x(n) = 0, √3 2, √3 2, 0, − √32 , − √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




© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws
as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in
writing from the publisher. For the exclusive use of adopters of the book Digital Signal Processing, Fourth Edition, by John G.
Proakis and Dimitris G. Manolakis. ISBN 0-13-187374-1.

, 1.9
(a) Fmax = 360Hz, FN = 2Fmax = 720Hz.
(b) Ffold = F2s = 300Hz.
(c)

x(n) = xa(nT )
= xa(n/Fs)
= sin(480πn/600) + 3sin(720πn/600)
x(n) = sin(4πn/5) − 3sin(4πn/5)
= −2sin(4πn/5).

Therefore, w = 4π/5.
(d) ya(t) = x(Fst) = −2sin(480πt).


1.10
(a)

Number of bits/sample = log21024 = 10.
[10, 000 bits/sec]
Fs =
[10 bits/sample]
= 1000 samples/sec.
Ffold = 500Hz.

(b)
1800π
Fmax =
2π
= 900Hz
FN = 2Fmax = 1800Hz.

(c)
600π 1
f1 = ( )
2π Fs
= 0.3;
1800π 1
f2 = ( )
2π Fs
= 0.9;
But f2 = 0.9 > 0.5 ⇒ f2 = 0.1.
Hence, x(n) = 3cos[(2π)(0.3)n] + 2cos[(2π)(0.1)n]
xmax−xmin 5−(−5)
(d) △ = m−1
= 1023
= 10
1023
.


1.11


x(n) = xa(nT )
100πn 250πn
= 3cos + 2sin
200 200

6




© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws
as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in
writing from the publisher. For the exclusive use of adopters of the book Digital Signal Processing, Fourth Edition, by John G.
Proakis and Dimitris G. Manolakis. ISBN 0-13-187374-1.