This document is a solutions manual for a textbook on communication systems. It provides step-by-step solutions to problems from each chapter of the textbook. The problems cover topics such as signal representations using Fourier series and integrals, power calculations for periodic signals, and bandpass signal representations. The solutions demonstrate techniques for analyzing and working with signals commonly encountered in electrical communication systems.
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Solutions Manual for Communication Systems textbook
1. 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
3. 2-1
Chapter 2
2.1-1
0
0
0
/2
2 ( )
/2
0
sinc( )
0 otherwise
jj
T
j m n f t j
n T
Ae n mAe
c e dt Ae m n
T
φφ
π φ−
−
=
= = − =
∫
2.1-2
0 0
0
0
/ 4 / 2
0 / 4
0 0 0
( ) 0
2 2 2 2
cos ( )cos sin
2
T T
n
T
c v t
nt nt A n
c A dt A dt
T T T n
π π π
π
=
= + − =∫ ∫
n 0 1 2 3 4 5 6 7
nc 0 2 /A π 0 2 / 3A π 0 2 / 5A π 0 2 / 7A π
arg nc 0 180± ° 0 180± °
2.1-3
0
0
/2
20
0 0 0
( ) / 2
2 2 2
cos sin (cos 1)
( )
T
n
c v t A
At nt A A
c A dt n n
T T T n n
π
π π
π π
= =
= − = − −
∫
n 0 1 2 3 4 5 6
nc 0.5A 0.2A 0 0.02A 0 0.01A 0
arg nc 0 0 0 0
2.1-4
0 / 2
0
0
0 0
2 2
cos 0
T t
c A
T T
π
= =∫ (cont.)
4. 2-2
( ) ( )
[ ]
0
0
/2
/2
0 0
0
0 0 0 0 0 0 0
sin 2 / sin 2 /2 2 2 2
cos cos
4( ) / 4( ) /
/ 2 1
sinc(1 ) sinc(1 )
0 otherwise2
T
T
n
n t T n t Tt nt A
c A dt
T T T T n T n T
A nA
n n
π π π ππ π
π π π π
− +
= = +
− +
= ±
= − + + =
∫
2.1-5
0
0
/ 2
0
0 0
( ) 0
2 2
sin (1 cos )
T
n
c v t
nt A
c j A dt j n
T T n
π
π
π
= =
= − = − −∫
n 1 2 3 4 5
nc 2 /A π 0 2 / 3A π 2 / 5A π
arg nc 90− ° 90− ° 90− °
2.1-6
0 ( ) 0c v t= =
( ) ( )
[ ]
0
0
/2
/2
0 0
0
0 0 0 0 0 0 0
sin 2 / sin 2 /2 2 2 2
sin sin
4( )/ 4( )/
/ 2 1
sinc(1 ) sinc(1 )
0 otherwise2
T
T
n
n t T n t Tt nt A
c j A dt j
T T T T n T n T
jA nA
j n n
π π π ππ π
π π π π
− +
= − = − −
− +
= ±
= − − − + =
∫
m
2.1-7
]
0 0
0 0
0
/ 2
0 / 2
0
1
( ) ( )
T T
jn t jn t
n
T
c v t e dt v t e dt
T
ω ω− −= +
∫ ∫
0 0
0 0 0 0
0
0
0
/ 2
/ 2
0
/ 2 0
/ 2
0
where ( ) ( /2)
( )
T T
jn t jn jn T
T
T
jn tjn
v t e dt v T e e d
e v t e dt
ω ω λ ω
ωπ
λ λ− − −
−
= +
= −
∫ ∫
∫
since 1 for even , 0 for evenjn
ne n c nπ
= =
5. 2-3
2.1-8
2 2 2 2 2 2
0 0 0 0 0 0 0 0
1
0
2
2 2 2 2
2
2 2 2 2 2 2
2 2 sinc 2 sinc2 2 sinc3
1
where 4
1 1 1 3
1 2sinc 2sinc 2sinc 0.23
16 4 2 4
2 1 1 3 5 3 7
1 2sinc 2sinc 2sinc 2sinc 2sinc 2sinc
16 4 2 4 4 2 4
n
n
P c c Af Af f Af f Af f
f
A
f P A
A
f P
τ τ τ τ τ τ τ
τ
τ
τ
∞
=
= + = + + + +
=
> = + + + =
> = + + + + + +
∑ L
2
2
2 2 2
0.24
1 1 1
1 2sinc 2sinc 0.21
2 16 4 2
A
A
f P A
τ
=
> = + + =
2.1-9
0 0
0
2
2 2
/ 2 / 2
/ 2 0
0 0 0 0
2 2 2
2 2 2
0 02 2 2
0 even
2
odd
n
41 2 4 1
a) 1 1
3
4 4 4
2 2 2 0.332 so / 99.6%
9 25
8 8 8
b) ( ) cos cos3 cos5
9 25
n
T T
T
n
c
n
t t
P dt dt
T T T T
P P P
v t t t
π
π π π
ω ω ω
π π π
−
=
= − = − =
′ ′= + + = =
′ = + +
∫ ∫
0t
2.1-10
( )
0
0
2 2 2
/ 2 2
/ 2
0
0 even
2
odd
1 2 2 2
a) 1 1 2 0.933 so / 93.3%
3 5
n
T
T
n
c j
n
n
P dt P P P
T
π
π π π−
= −
′ ′= = = + + = =
∫
(cont.)
6. 2-4
( ) ( ) ( )
( ) ( ) ( )
0 0 0
0 0 0
4 4 4
b) ( ) cos 90 cos 3 90 cos 5 90
3 5
4 4 4
sin sin 3 sin 5
3 5
v t t t t
t t t
ω ω ω
π π π
ω ω ω
π π π
′ = − ° + − ° + − °
= + +
2.1-11
0
2
0
0 0
1/2 01 1
1/2 03
T
n
nt
P dt c
n nT T π
=
= = =
≠
∫
4 4
4 4 4
odd
2 2 1 1 1 1
2 2
1 3 5 3n
P
nπ π
∞
= = + + + =
∑ L
2 2
2 2 2
1 1 1 4 1 1
Thus,
1 2 3 2 3 4 6
π π
+ + + = − =
L
2.1-12
0
2
/ 2
20
0 0
0 even2 4 1
1
(2/ ) odd3
T
n
nt
P dt c
n nT T π
= − = =
∫
2 2
2 2 2 2
1
1 1 1 2 1 1 1 1
2
2 2 4 4 1 2 3 3n
P
nπ π
∞
=
= + = + + + + =
∑ L
4 4
4 4 4 4
1 1 1 1
Thus,
1 3 5 2 2 3 96
π π
+ + + = =
⋅
L
2.2-1
( )
( )
( )
( )
[ ]
/ 2
0
( ) 2 cos cos2
sin 2 sin 2
2 22 sinc( 1/2) sinc( 1/2)
22 2 2 2
t
V f A ftdt
f f
A
A f f
f f
τ
π π
τ τ
π π
τ τ
π
π
τ
τ τ
π π
τ
τ τ
π π
+
+
=
−
= + = − + +
−
∫
(cont.)
7. 2-5
2.2-2
( )
( )
( )
( )
[ ]
/ 2
0
2 2
2 2
2
( ) 2 sin cos2
sin 2 sin 2
2 22 sinc( 1) sinc( 1)
22 2 2 2
t
V f j A ftdt
f f
A
j A j f f
f f
τ
π π
τ τ
π π
τ τ
π
π
τ
τ τ
π π
τ
τ τ
π π
+
+
= −
−
= − − = − − − +
−
∫
2.2-3
2 2
20
2
( ) 2 cos 2sin 1 1 sinc
( ) 2
t A
V f A A tdt A f
τ τ ωτ
ω τ τ
τ ωτ
= − = − + =
∫
2.2-4
20
2
( ) 2 sin (sin cos )
( )
(sinc2 cos2 )
t A
V f j A tdt j
A
j f f
f
τ τ
ω ωτ ωτ ωτ
τ ωτ
τ π τ
π
= − = − −
= − −
∫
2.2-5
2
2
2
1
( ) sinc2
2 2
1 1 1
sinc2
2 2 4 2
f
v t Wt
W W
f
Wt dt df df
W W W W
∞ ∞ ∞
−∞ −∞ −∞
= ↔ Π
= Π = =
∫ ∫ ∫
8. 2-6
2.2-6
( )
2 2 2
2
2 20 0
2
2 arctan
2 (2 )
W
bt A A A W
E Ae dt E df
b b f b b
π
π π
∞
−
′= = = =
+∫ ∫
50% / 22 2
arctan
84% 2 /
W bE W
W bE b
ππ
ππ
=′
= =
=
2.2-7
( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
j t
j t
v t w t dt v t W f e df dt
W f v t e dt df W f V f df
ω
ω
∞ ∞ ∞
−∞ −∞ −∞
∞ ∞ ∞
− −
−∞ −∞ −∞
=
= = −
∫ ∫ ∫
∫ ∫ ∫
22
( ) *( ) when ( ) is real, so ( ) ( ) *( ) ( )V f V f v t v t dt V f V f df V f df
∞ ∞ ∞
−∞ −∞ −∞
− = = =∫ ∫ ∫
2.2-8
2 2 2 ( )
( ) ( ) ( ) ( )
Let ( ) ( ) so ( ) ( ) and ( ) ( )
Hence ( ) ( ) ( ) ( )
j ft j ft j f t
w t e dt w t e dt w t e dt W f
z t w t Z f W f W f Z f
v t z t dt V f Z f df
π π π
∗ ∗
∞ ∞ ∞
∗ − − − ∗
−∞ −∞ −∞
∗ ∗ ∗
∞ ∞
−∞ −∞
= = =
= = − = −
= −
∫ ∫ ∫
∫ ∫
2.2-9
1
sinc so sinc
2 2
( ) sinc ( ) for
2 2
t f
A Af At
A A A
t f
v t V f A
τ τ
τ τ
Π ↔ ↔ Π
= ↔ = Π =
2.2-10
[ ]
[ ]
[ ]
cos sinc( 1/2) sinc( 1/2)
2
( )
so sinc( 1/2) sinc( 1/2) cos cos
2
Let and 2 ( ) sinc(2 1/2) sinc(2 1/2)
t t B
B f f
B f f f f
t t B B
B A W z t AW Wt Wt
π τ
τ τ
τ τ
τ π π
τ τ
τ τ τ τ
τ
Π ↔ − + +
− −
− + + ↔ Π = Π
= = ⇒ = − + +
2.2-11
[ ]
[ ]
[ ]
2
sin sinc( 1) sinc( 1)
2
2 ( ) 2
so sinc( 1) sinc( 1) sin sin
2
Let and 2 ( ) sinc(2 1) sinc(2 1)
t t B
B j f f
B f f f f
j t t B B
B jA W z t AW Wt Wt
π τ
τ τ
τ τ
τ π π
τ τ
τ τ τ τ
τ
Π ↔ − − + +
− −
− − + + ↔ Π = − Π
= − = ⇒ = − + +
9. 2-7
2.2-12
( )
( )
( )
2
2 2 2 2 2 2
2
2
2
22 2 0 2 2
2
2 30 2 2
2 4 /
(2 ) (2 ) (2 )
1 /
2
2
1 1
Thus,
2 2 4
b t a t
a t
b a a
e e
b f a f a f
a a df
e dt df
a a f a f
dx
a a aa x
π
π
π π
π π π
π
π π
π π
π
− −
∞ ∞ ∞−
−∞ −∞
∞
↔ ⇒ ↔ =
+ + +
= = =
+ +
= =
+
∫ ∫ ∫
∫
2.3-1
( ) ( ) ( ) where v( ) ( / ) sinc
so Z( ) ( ) ( ) 2 sinc cos2j T j T
z t v t T v t T t A t A f
f V f e V f e A f fTω ω
τ τ τ
τ τ π−
= − + + = Π ↔
= + =
2.3-2
2 2
( ) ( 2 ) 2 ( ) ( 2 ) where v( ) ( / ) sinc
( ) ( ) ( ) ( ) 2 (sinc )(1 cos4 )j T j T
z t v t T v t v t T t a t A f
Z f V f e V f V f e A f fTω ω
τ τ τ
τ τ π−
= − + + + = Π ↔
= + + = +
2.3-3
2 2
( ) ( 2 ) 2 ( ) ( 2 ) where ( ) ( / ) sinc
( ) ( ) 2 ( ) ( ) 2 (sinc )(cos4 1)j T j T
z t v t T v t v t T v t a t A f
Z f V f e V f V f e A f fTω ω
τ τ τ
τ τ π−
= − − + + = Π ↔
= − + = −
2.3-4
/ 2
/ 2
( ) ( )
2
( ) 2 sinc2 ( ) sincj T j T
t T t T
v t A B A
T T
V f AT fTe B A T fTeω ω− −
− −
= Π + − Π
= + −
10. 2-8
2.3-5
2 2
2 2
( ) ( )
4 2
( ) 4 sinc4 2( ) sinc2j T j T
t T t T
v t A B A
T T
V f AT fTe B A T fTeω ω− −
− −
= Π + − Π
= + −
2.3-6
/ /
1
Let ( ) ( ) ( ) ( / )
1
Then ( ) [ ( / )] ( / ) so ( ) ( ) ( / )d dj t a j t a
d d
w t v at W f V f a
a
z t v a t t a w t t a Z f W f e V f a e
a
ω ω− −
= ↔ =
= − = − = =
2.3-7
2 ( )
( ) ( ) ( ) ( )c c cj t j t j f f tj t
cv t e v t e e dt v t e dt V f fω ω πω
∞ ∞ − −−
−∞ −∞
= = = − ∫ ∫F
2.3-8
[ ]
( ) ( / )cos with 2 /
( ) sinc( ) sinc( ) sinc( 1/2) sinc( 1/2)
2 2 2
c c c
c c
v t A t t f
A A A
V f f f f f f f
τ ω ω π π τ
τ τ τ
τ τ τ τ
= Π = =
= − + + = − + +
2.3-9
[ ]
/ 2 /2
( ) ( / )cos( /2) with 2 2 /
( ) sinc( ) sinc( )
2 2
sinc( 1) sinc( 1)
2
c c c
j j
c c
v t A t t f
e e
V f A f f A f f
A
j f f
π π
τ ω π ω π π τ
τ τ τ τ
τ
τ τ
−
= Π − = =
= − + +
= − − − +
2.3-10
2
2 2 2 2
2
( ) ()cos ( )
1 (2 )
1 1
( ) ( ) ( )
2 2 1 4 ( ) 1 4 ( )
t
c
c c
c c
A
z t v t t v t Ae
f
A A
Z f V f f V f f
f f f f
ω
π
π π
−
= = ↔
+
= − + + = +
+ − + +
2.3-11
/ 2 / 2
( ) ()cos( /2) ( ) for 0
1 2
/ 2 / 2
( ) ( ) ( )
2 2 1 2 ( ) 1 2 ( )
/ 2 / 2
2 ( ) 2 ( )
t
c
j j
c c
c c
c c
A
z t v t t v t Ae t
j f
e e jA jA
Z f V f f V f f
j f f j f f
A A
j f f j f f
π π
ω π
π
π π
π π
−
−
= − = ≥ ↔
+
−
= − + + = +
+ − + +
= −
− − − +
11. 2-9
2.3-12
( )
2
2
( ) ( ) ( ) 2 sinc2
sin2 2
( ) 2 (2 ) cos2 2 sin2
2 (2 )
1
( ) ( ) sinc2 cos2
2
A t
v t t z t z t A f
d d f A
Z f A f f f
df df f f
d jA
V f Z f f f
j df f
τ
τ τ
π τ
πτ π τ πτ π τ
π τ π τ
τ π τ
π π
= = Π ↔
= = −
−
= = −
−
2.3-13
2 2
22 2 2 2
2
( ) ( ) ( )
(2 )
1 2 2
( )
2 (2 ) (2 )
b t Ab
z t tv t v t Ae
b f
d Ab j Abf
Z f
j df b f b f
π
π π π
−
= = ↔
+
= =
− + +
2.3-14
( ) [ ]
2
2 3
( ) ( ) ( ) for 0
2
1 2
( )
22 2
t A
z t t v t v t Ae t
b j f
d A A
Z f
df b j fj f b j f
π
ππ π
−
= = ≥ ↔
+
= = +− +
2.3-15
2 2
2 2
2 2
2 2
( ) ( / )
2 ( ) ( / )
( ) ( / )
( ) ( / )
1
( ) ( )
2
( ) ( ) 2
1
( ) ( )
2
Both results are equivalent to
bt f b
bt f b
bt f b
bt f b
v t e V f e
b
d j f
a v t b te e
dt b
d f
b te V f e
j df jb
bte jf e
π π
π π
π π
π π
π
π
π
− −
− −
− −
− −
= ↔ =
= − ↔
↔ =
−
↔ −
2.4-1
2
0
2
0
( ) 0 0
0 2
2
2 2
t
y t t
At
A d t
A d A t
λ λ
λ λ
= <
= = < <
= = >
∫
∫
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