Solutions Manual to accompany Communication Systems An Introduction to Signals and Noise in Electrical Communication Fourth Edition

Solutions Manual
to accompany


Communication
Systems
An Introduction to Signals and Noise in
Electrical Communication
Fourth Edition


A. Bruce Carlson
Rensselaer Polytechnic Institute

Paul B. Crilly
University of Tennessee

Janet C. Rutledge
University of Maryland at Baltimore

,Solutions Manual to accompany
COMMUNICATION SYSTEMS: AN INTRODUCTION TO SIGNALS AND NOISE IN ELECTRICAL COMMUNICATION,
FOURTH EDITION
A. BRUCE CARLSON, PAUL B. CRILLY, AND JANET C. RUTLEDGE

Published by McGraw-Hill Higher Education, an imprint of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas,
New York, NY 10020. Copyright © The McGraw-Hill Companies, Inc., 2002, 1986, 1975, 1968. All rights reserved.

The contents, or parts thereof, may be reproduced in print form solely for classroom use with COMMUNICATION SYSTEMS: AN
INTRODUCTION TO SIGNALS AND NOISE IN ELE CTRICAL COMMUNICATION, provided such reproductions bear copyright
notice, but may not be reproduced in any other form or for any other purpose without the prior written consent of The McGraw-Hill
Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance
learning.

www.mhhe.com

,Chapter 2

2.1-1
jf Ae jf n =m
cn = Ae dt =Ae sinc(m − n) =
+e
T0/2
j2p ( m−n )f 0t jf
− T0 /2
T0 0 otherwise

2.1-2




2 =0
c0 v(t) T 2p nt T0 /2 2p nt 2A pn
c = Acos dt + (− A)cos dt = sin

T 
n
0 T T /4
0
T pn 2
0 0 0


n 0 1 2 3 4 5 6 7
cn 0 2A/p 0 2 A / 3p 0 2 A / 5p 0 2 A / 7p
arg cn 0 180 0 180


2.1-3




c0 = 2v(t) =A /2
c = T /2  0 2 At  2p nt A A
n   T0 
T0 0 
A − cos
dt = sinp n − (cosp n −1)
 T0 pn (p n)2
n 0 1 2 3 4 5 6
cn 0.5A 0.2A 0 0.02A 0 0.01A 0
arg cn 0 0 0 0


2.1-4




2 T0 /2 2p t
c = Acos =0 (cont.)
0
T 
0 T
0 0




2-1

, 2 2pt 2 A sin (p −p n ) 2t / T sin (p +p n) 2t / T T / 2
2p nt 
T /2 0

cn =  Acos dt = 
0
cos 0
+ 0

T0 0 T0 T0 T0  4(p −p n) / T0 4(p + p n) / T0 0
A
= sinc(1 − n) + sinc(1 + n) A / 2 n =1
=
2 0 otherwise

2.1-5




c0 = v(t)
2 =0T0 /2 2p nt A
c =− j Asin dt =− j (1− cosp n)
T 
n
0 T pn
0 0


n 1 2 3 4 5
cn 2A/p 0 2 A / 3p 2 A / 5p
arg cn −90 −90 −90

2.1-6




c0 = v(t) =0
2 A sin (p −p n ) 2t / T0 sin (p +p n ) 2t / T 
2 T
c =− j T /2 2p t 2p nt
dt =− j −
0
Asin sin  0 
n
T 
0
T T T 4(p −p n)/ T 4(p +p n)/ T
0  0
m jA / 2
0 0 0 0 0
A
=− j sinc(1−n ) − sinc(1+ n) = n =1

2  0 otherwise

2.1-71
c =  T0 /2 − jnw0 t
T0
− jnw 0t

n

T0  0 v(t) e dt + 
T
v(t)e dt 


T0 T0 /2
where T /2
v(t)e − jnw0 t dt =  v(l + T0 /2) e− jnw l e− jnw 0 0T0 /2
dl
0 0
T0 /2
=−e v(t )e− jnw0 t dt

jnp

0

since e jnp =1 for even n, cn =0 for even n




2-2