Solution Manual for Communication Systems 5th Edition by A. Bruce Carlson & Paul B. Crilly , ISBN: 9780073380407 |All Chapters Verified| Guide A+

,Chapter 2

2.1-1
Ae jφ  Ae jφ n = m
∫− T
T
j 2 π ( m−n )f 0t jφ
cn = e dt = Ae sinc( m − n ) = 
T0 0 otherwise

2.1-2




c0 v (t ) = 0
2 T 2π nt T 2π nt 2A πn
cn =
T0 ∫
0
A cos
T0
dt + ∫ ( − A)cos
T T0
dt =
πn
sin
2

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




c0 = v (t ) = A / 2
2 T0 /2  2 At  2π nt A A
cn =
T0 ∫ 0
A−
 T0 
 cos
T0
dt =
πn
sin π n −
(π n) 2
(cos π n − 1)


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 T 2π t
c0 =
T0 ∫
0
A cos
T0
=0 (cont.)



2-1

, 2 A  sin (π − π n ) 2t / T0 sin (π + π n ) 2t / T0 
T
2 T 2π t 2π nt
cn =
T0 ∫0
A cos
T0
cos
T0
dt = 
T0  4(π − π n) / T0
+
4(π + π n ) / T0  0


A/2 n = ±1
[ sinc(1 − n) + sinc(1 + n )] = 
A
=
2  0 otherwise

2.1-5




c0 = v (t ) = 0
2 T 2π nt A
cn = − j
T0 ∫0
A sin
T0
dt = − j
πn
(1 − cos π n )


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

2.1-6




c0 = v(t ) = 0
2 A  sin (π − π n ) 2t / T0 sin ( π + π n ) 2t / T0 
T
2 T 2π t 2π nt
cn = − j
T0 ∫0
A sin
T0
sin
T0
dt = − j 
T0  4(π − π n ) / T0
−
4(π + π n )/ T0  0


m jA / 2 n = ±1
[sinc(1 − n ) − sinc(1 + n ) ] = 
A
= −j
2  0 otherwise

2.1-7
1  T
v ( t) e− jnω0 t dt + ∫ v(t )e − jnω 0t dt ]
T0
cn = ∫
T0  0 T

T0 T
where ∫T
v(t )e − jnω0 t dt = ∫
0
v (λ + T0 /2) e− jnω 0λ e− jnω 0T d λ
T
= −e jnπ ∫ v (t )e − jnω0 t dt
0

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




2-2

, 2.1-8

∞
P = c0 + 2 ∑ cn = Af 0τ + 2 Af 0τ sinc f 0τ + 2 Af 0τ sinc2 f 0τ + 2 Af 0τ sinc3 f 0τ + L
2 2 2 2 2 2

n =1

1
where = 4 f0
τ
1 1 + 2sinc2 1 + 2sinc2 1 + 2sinc2 3  = 0.23 A2
A2
f > P= 
τ 16 4 2 4 
2 A2  1 1 3 5 3 7
f > P= 1 + 2sinc2 + 2sinc2 + 2sinc2 + 2sinc2 + 2sinc 2 + 2sinc 2  = 0.24 A2
τ 16  4 2 4 4 2 4
1 A 
2
1 1
f > P= 1 + 2sinc2 + 2sinc 2  = 0.21A2
2τ 16  4 2

2.1-9
 0 n even

cn =  2 2
 π n  n odd
 
2 2
1 T  4t  2 T  4t  1
a) P =
T0 ∫−T  1 − T0  dt = T0 ∫
0
 1 −  dt =
 T0  3
2 2 2
 4   4   4 
P′ = 2  2  + 2  2  + 2  2 
= 0.332 so P′ / P = 99.6%
π   9π   25π 
8 8 8
b) v′(t ) = 2 cos ω 0t + 2 cos3ω 0t + cos5ω 0t
π 9π 25π 2




2.1-10
 0 n even

cn =  − j 2
 π n n odd

1  2 2  2  2  2 2 
∫−T (1) dt = 1 P = 2  π  +  3π  +  5π   = 0.933 so P′ / P = 93.3%
T
a) P = ′
2

T0  


(cont.)


2-3