1. Proakis And Manolakis Edition 2 Solution Manual Pdf

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 1 (a) f = 0.01π 2π = 200 ⇒ periodic with Np = 200. 30π 1 (b) f = 105 ( 2π ) = 17 ⇒ periodic with Np = 7. 3π (c) f = 2π = 32 ⇒ periodic with Np = 2. 3 (d) f = 2π ⇒ non-periodic. 1 31 (e) f = 62π 10 ( 2π ) = 10 ⇒ periodic with Np = 10. 1.3 (a) Periodic with period Tp = 2π 5.

5 ⇒ non-periodic. (b) f = 2π 1 (c) f = 12π ⇒ non-periodic. N (d) cos( 8 ) is non-periodic; cos( πn 8 ) is periodic; Their product is non-periodic. (e) cos( πn ) is periodic with period Np =4 2 sin( πn ) is periodic with period N p =16 8 π cos( πn + ) 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 (a) w = 2πk N implies that f = k N.

Let 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.

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Proakis and Dimitris G. ISBN 0-13-187374-1.

3 10 0 t (ms) 20 -3 Figure 1.5-2: (c)Refer nto fig 1.5-2 o x(n) = 0, √32, √32, 0, − √32, − √32, Np = 6. 100π x(1) = 3 = 3sin( ) ⇒ Fs = 200 samples/sec. Fs 1.6 (a) x(n) = Acos(2πF0 n/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, Tp k 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 ⇒ N T = 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.

Proakis And Manolakis Edition 2 Solution Manual Pdf

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. 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(Fs t) = −2sin(480πt). 1.10 (a) Number of bits/sample Fs Ffold = log2 1024 = 10. 10, 000 bits/sec = 10 bits/sample = 1000 samples/sec. (b) Fmax = FN = = 1800π 2π 900Hz 2Fmax = 1800Hz. (c) f1 = = (d) △ = xmax −x f2 = But f2 = = Hence, x(n) = min = m−1 5−(−5) 1023 = 600π 1 ( ) 2π Fs 0.3; 1800π 1 ( ) 2π Fs 0.9; 0.9 0.5 ⇒ f2 = 0.1.

3cos(2π)(0.3)n + 2cos(2π)(0.1)n 10 1023. 1.11 x(n) = xa (nT )     250πn 100πn + 2sin = 3cos 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.

ISBN 0-13-187374-1. Fs = 5KHz, F0=500Hz Fs = 5KHz, F0=2000Hz 1 1 0.5 0.5 0 0 −0.5 −0.5 −1 0 50 −1 0 100 Fs = 5KHz, F0=3000Hz 1 0.5 0.5 0 0 −0.5 −0.5 50 100 Fs = 5KHz, F0=4500Hz 1 −1 0 50 −1 0 100 50 100 Figure 1.15-1: (b) Refer to fig 1.15-2. Y(n) is a sinusoidal signal. By taking the even numbered samples, the sampling frequency is reduced to half i.e., 25kHz which is still greater than the nyquist rate.

The frequency of the downsampled signal is 2kHz. 1.16 (a) for levels = 64, using truncation refer to fig 1.16-1. For levels = 128, using truncation refer to fig.