TEST BANK FOR Digital Communications 5th Edition By Proakis Salehi
Exam (elaborations)
TEST BANK FOR Digital Communications 5th Edition By Proakis Salehi (Instructor Solution Manual)
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TEST BANK FOR Digital Communications 5th Edition By Proakis Salehi
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Harvard University
Exam (elaborations) TEST BANK FOR Digital Communications 5th Edition By Proakis Salehi (Instructor Solution Manual)
Solutions Manual
for
Digital Communications, 5th Edition
(Chapter 2) 1
Prepared by
Kostas Stamatiou
January 11, 2008
1PROPRIETARY MATERIAL.
c The McGraw-Hill Companies, Inc...
TEST BANK FOR Digital Communications 5th Edition By Proakis Salehi
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, Solutions Manual
for
Digital Communications, 5th Edition
(Chapter 2) 1
Prepared by
Kostas Stamatiou
January 11, 2008
1
PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this
Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written
permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by
McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using
it without permission.
, 2
Problem 2.1
a. Z ∞
1 x(a)
x̂(t) = da
π −∞ t−a
Hence : R∞
x(a)
−x̂(−t) = − π1 −∞ −t−a da
R −∞
= − π1 ∞ x(−b)
−t+b (−db)
R
1 ∞ x(b)
= − π −∞ −t+b db
R
1 ∞ x(b)
= π −∞ t−b db = x̂(t)
where we have made the change of variables : b = −a and used the relationship : x(b) = x(−b).
b. In exactly the same way as in part (a) we prove :
x̂(t) = x̂(−t)
c. x(t) = cos ω0 t, so its Fourier transform is : X(f ) = 21 [δ(f − f0 ) + δ(f + f0 )] , f0 = 2πω0 .
Exploiting the phase-shifting property (2-1-4) of the Hilbert transform :
e. The positive frequency content of the new signal will be : (−j)(−j)X(f ) = −X(f ), f > 0, while
ˆ
the negative frequency content will be : j · jX(f ) = −X(f ), f < 0. Hence, since X̂(f ) = −X(f ),
ˆ = −x(t).
we have : x̂(t)
f. Since the magnitude response of the Hilbert transformer is characterized by : |H(f )| = 1, we
have that : X̂(f ) = |H(f )| |X(f )| = |X(f )| . Hence :
Z ∞ 2
Z ∞
X̂(f ) df = |X(f )|2 df
−∞ −∞
PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the
limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a
student using this Manual, you are using it without permission.
, 3
and using Parseval’s relationship :
Z ∞ Z ∞
2
x̂ (t)dt = x2 (t)dt
−∞ −∞
g. From parts (a) and (b) above, weR note that if x(t) is even, x̂(t) is odd and vice-versa. Therefore,
∞
x(t)x̂(t) is always odd and hence : −∞ x(t)x̂(t)dt = 0.
where we have used the fact that since Xl (f − f0 ) and Yl (−f − f0 ) do not overlap, Xl (f −
f0 )Yl (−f − f0 ) = 0 and similarly Xl (−f − f0 )Yl (f − f0 ) = 0.
2. Putting y(t) = x(t) we get the desired result from the result of part 1.
PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the
limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a
student using this Manual, you are using it without permission.
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