Solutions Manual for Communication Systems 5th Edition by A. Bruce Carlson & Paul B. Crilly – Complete Exam & Study Guide

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SOLUTION MANUAL
Communication systems: An introduction to signals and noise in
electrical communication 5th Edition by A. Bruce Carlson & Paul B. Crilly

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Chapter 2

2.1-1
jϕ Ae jϕ n =m
cn = Ae dt =Ae sinc(m −n) =

T
0/ 2 j2π ( m−n )f 0t jϕ
e
−T0 /2
T0 0 otherwise

2.1-2




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


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


2.1-3




2
c =  2 At  2πnt A A

T0 /2
n
T 0  A− cos dt = sinπn − 2 (cosπ n −1)
T T πn
0  0  0 (π n)

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 2π t
c = Acos =0 (cont.)
0
T 
0 T
0 0

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2 T /2 2πt 2πnt 2A sin (π−π n)2t /T sin (π+π n)2t / T T /2
0


cn =  Acos 0
cos dt =  0
+
0

T0 0 T0 T0 T  4(π −π
n =1 n) / T 4(π +πn)/ T 
A A / 2 0 0 0
= sinc(1 −n) +sinc(1 +n)
0


=
2 0 otherwise

2.1-5




2 T0 /2 2πnt A
c =−j Asin dt =−j (1−cosπ n)
n
T 
0 T πn
0 0


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

2.1-6




2 A sin (π−π n)2t / T0 −sin (π+π n)2t / T 
T0/2
2 2πt 2πnt
T0 /2
cn =−j dt =−j  
T 
0
Asin
T
sin
T T 4(π −π n)/ T 4(π +π n)/ T 0
A0 0 0
mjA / 02  0 0 0
=−j sinc(1−n ) −sinc(1+n) n =1
=
2  0 otherwise

2.1-71
c = T0 /2 −jnω0 t
T0
−jnω0t

n 0 v(t) e dt +T0/ 2 v(t)e dt 
T 
 0
T0
v(t)e−jnω0t dt = T0 /2
v(λ +T /2) e−jnω0λ e−jnω0T0/ 2dλ

where
T 
0
0
T
=−ejnπ −jnω0 t
v(t )e
0
dt
since e jnπ =1 for even n, c =0 for even n
n

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2.1-8

+2c
 2 2 2 2 2
P =c 2
=Af τ +2 Af τ sinc f τ +2 Af τ sinc2 f τ +2 Af τ sinc3 f τ +⋯
0 n 0 0 0 0 0 0 0
n=1
1
where =4 f
0
τ
1 A2  2 1 21 2 3 2
f  P = 1+2sinc +2sinc +2sinc =0.23A
τ 16  4 2 4
2 A 
2
21 21 23 25 23 2 7
2
f  P = 1+2sinc +2sinc +2sinc +2sinc +2sinc +2sinc =0.24A
τ 16  4 2 4 4 2 4
1 A2  2 1 2 1 2
f  P = 1+2sinc +2sinc =0.21A
2τ 16  4 2

2.1-9
 0 n even

cn =2  n odd
2


 
πn 
a)  1
P= T /2  4t 2 T /2  4t 
2 2

T −T/ 2
1
 1−
dt = 
0 0


 1− dt =
0
0
0
 T0  T0  T0  3
 4    
4 2
4  2

2

P =2 2 +2 2
+2  2
=0.332 so P / P =99.6%
π  9π  25π 
b) 8 8 8
v(t) = cosω t + cos3ω t + cos5ω t
0
π 2
9π 2
0
25π 2
0




2.1-10
0 n even

c =
n −j2
πn n odd
a) P = 1 T /2 (1) dt =1 P=2 2 2 2 2 2 2 =0.933 so P/ P =93.3%
0 2


T −T/ 2   + +  
0
0

π 3π  5π 


(cont.)