Solutions Manual – Digital Communications, 5th Edition (John G. Proakis & Masoud Salehi) | Prepared by Kostas Stamatiou | Complete Step-by-Step Solutions

Solutions Manual
For
Digital Communications, 5th Edition


Prepared by
Kostas Stamatiou

, 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 :

1 1
X̂(f ) = [−jδ(f − f0 ) + jδ(f + f0 )] = [δ(f − f0 ) − δ(f + f0 )] = F −1 {sin 2πf0 t}
2 2j

Hence, x̂(t) = sin ω0 t.

d. In a similar way to part (c) :

1 1
x(t) = sin ω0 t ⇒ X(f ) = [δ(f − f0 ) − δ(f + f0 )] ⇒ X̂(f ) = [−δ(f − f0 ) − δ(f + f0 )]
2j 2
1
⇒ X̂(f ) = − [δ(f − f0 ) + δ(f + f0 )] = −F −1 {cos 2πω0 t} ⇒ x̂(t) = − cos ω0 t
2


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.




Problem 2.2




1. Using relations

1 1
X(f ) = Xl (f − f0 ) + Xl (−f − f0 )
2 2
1 1
Y (f ) = Yl (f − f0 ) + Yl (−f − f0 )
2 2
and Parseval’s relation, we have
Z ∞ Z ∞
x(t)y(t) dt = X(f )Y ∗ (f ) dt
−∞ −∞
Z ∞  ∗
1 1 1 1
= Xl (f − f0 ) + Xl (−f − f0 ) Yl (f − f0 ) + Yl (−f − f0 ) df
−∞ 2 2 2 2
Z ∞ Z ∞
1 1
= Xl (f − f0 )Yl∗ (f − f0 ) df + Xl (−f − f0 )Yl (−f − f0 ) df
4 −∞ 4 −∞
Z
1 ∞ 1
= Xl (u)Yl∗ (u) du + Xl∗ (v)Y (v) dv
4 −∞ 4
Z ∞ 
1 ∗
= Re Xl (f )Yl (f ) df
2 −∞
Z ∞ 
1 ∗
= Re xl (t)yl (t) dt
2 −∞

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.