SOLUTIONS MANUAL
Communication Systems Engineering
Second Edition
John G. Proakis
Masoud Salehi
Prepared by Evangelos Zervas
Upper Saddle River, New Jersey 07458
Publisher: Tom Robbins
Editorial Assistant: Jody McDonnell
Executive Managing Editor: Vince O’Brien
Managing Editor: David A. George
Production Editor: Barbara A. Till
Composition: PreT
E
X, Inc.
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Supplement Cover Design: PM Workshop Inc.
Manufacturing Buyer: Ilene Kahn
c
2002 Prentice Hall
by Prentice-Hall, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved. No part of this book may be reproduced in any form or by any means, without permission in
writing from the publisher.
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the
development, research, and testing of the theories and programs to determine their effectiveness. The author and
publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation
contained in this book. The author and publisher shall not be liable in any event for incidental or consequential
damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Printed in the United States of America
10987654321
ISBN 0-13-061974-6
Pearson Education Ltd., London
Pearson Education Australia Pty. Ltd., Sydney
Pearson Education Singapore, Pte. Ltd.
Pearson Education North Asia Ltd., Hong Kong
Pearson Education Canada, Inc., Toronto
Pearson Educac`ıon de Mexico, S.A. de C.V.
Pearson Education—Japan, Tokyo
Pearson Education Malaysia, Pte. Ltd.
Pearson Education, Upper Saddle River, New Jersey
Contents
Chapter 2.........................................1
Chapter 3 ....................................... 42
Chapter 4 ....................................... 71
Chapter 5 ......................................114
Chapter 6 ......................................128
Chapter 7 ......................................161
Chapter 8 ......................................213
Chapter 9 ......................................250
Chapter 10 .....................................283
iii
Chapter 2
Problem 2.1
1)
2
=
∞
−∞
x(t) −
N
i=1
α
i
φ
i
(t)
2
dt
=
∞
−∞
x(t) −
N
i=1
α
i
φ
i
(t)
x
∗
(t) −
N
j=1
α
∗
j
φ
∗
j
(t)
dt
=
∞
−∞
|x(t)|
2
dt −
N
i=1
α
i
∞
−∞
φ
i
(t)x
∗
(t)dt −
N
j=1
α
∗
j
∞
−∞
φ
∗
j
(t)x(t)dt
+
N
i=1
N
j=1
α
i
α
∗
j
∞
−∞
φ
i
(t)φ
∗
j
dt
=
∞
−∞
|x(t)|
2
dt +
N
i=1
|α
i
|
2
−
N
i=1
α
i
∞
−∞
φ
i
(t)x
∗
(t)dt −
N
j=1
α
∗
j
∞
−∞
φ
∗
j
(t)x(t)dt
Completing the square in terms of α
i
we obtain
2
=
∞
−∞
|x(t)|
2
dt −
N
i=1
∞
−∞
φ
∗
i
(t)x(t)dt
2
+
N
i=1
α
i
−
∞
−∞
φ
∗
i
(t)x(t)dt
2
The first two terms are independent of α’s and the last term is always positive. Therefore the
minimum is achieved for
α
i
=
∞
−∞
φ
∗
i
(t)x(t)dt
which causes the last term to vanish.
2) With this choice of α
i
’s
2
=
∞
−∞
|x(t)|
2
dt −
N
i=1
∞
−∞
φ
∗
i
(t)x(t)dt
2
=
∞
−∞
|x(t)|
2
dt −
N
i=1
|α
i
|
2
Problem 2.2
1) The signal x
1
(t) is periodic with period T
0
=2. Thus
x
1,n
=
1
2
1
−1
Λ(t)e
−j2π
n
2
t
dt =
1
2
1
−1
Λ(t)e
−jπnt
dt
=
1
2
0
−1
(t +1)e
−jπnt
dt +
1
2
1
0
(−t +1)e
−jπnt
dt
=
1
2
j
πn
te
−jπnt
+
1
π
2
n
2
e
−jπnt
0
−1
+
j
2πn
e
−jπnt
0
−1
−
1
2
j
πn
te
−jπnt
+
1
π
2
n
2
e
−jπnt
1
0
+
j
2πn
e
−jπnt
1
0
1
π
2
n
2
−
1
2π
2
n
2
(e
jπn
+ e
−jπn
)=
1
π
2
n
2
(1 − cos(πn))
1
When n = 0 then
x
1,0
=
1
2
1
−1
Λ(t)dt =
1
2
Thus
x
1
(t)=
1
2
+2
∞
n=1
1
π
2
n
2
(1 − cos(πn)) cos(πnt)
2) x
2
(t) = 1. It follows then that x
2,0
= 1 and x
2,n
=0, ∀n =0.
3) The signal is periodic with period T
0
=1. Thus
x
3,n
=
1
T
0
T
0
0
e
t
e
−j2πnt
dt =
1
0
e
(−j2πn+1)t
dt
=
1
−j2πn +1
e
(−j2πn+1)t
1
0
=
e
(−j2πn+1)
− 1
−j2πn +1
=
e − 1
1 − j2πn
=
e − 1
√
1+4π
2
n
2
(1 + j2πn)
4) The signal cos(t) is periodic with period T
1
=2π whereas cos(2.5t) is periodic with period
T
2
=0.8π. It follows then that cos(t) + cos(2.5t) is periodic with period T =4π. The trigonometric
Fourier series of the even signal cos(t) + cos(2.5t)is
cos(t) + cos(2.5t)=
∞
n=1
α
n
cos(2π
n
T
0
t)
=
∞
n=1
α
n
cos(
n
2
t)
By equating the coefficients of cos(
n
2
t) of both sides we observe that a
n
= 0 for all n unless n =2, 5
in which case a
2
= a
5
= 1. Hence x
4,2
= x
4,5
=
1
2
and x
4,n
= 0 for all other values of n.
5) The signal x
5
(t) is periodic with period T
0
=1. Forn =0
x
5,0
=
1
0
(−t +1)dt =(−
1
2
t
2
+ t)
1
0
=
1
2
For n =0
x
5,n
=
1
0
(−t +1)e
−j2πnt
dt
= −
j
2πn
te
−j2πnt
+
1
4π
2
n
2
e
−j2πnt
1
0
+
j
2πn
e
−j2πnt
1
0
= −
j
2πn
Thus,
x
5
(t)=
1
2
+
∞
n=1
1
πn
sin 2πnt
6) The signal x
6
(t) is periodic with period T
0
=2T . We can write x
6
(t)as
x
6
(t)=
∞
n=−∞
δ(t − n2T ) −
∞
n=−∞
δ(t − T −n2T )
2
