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 re produced in print form solely for classroom use with COMMUNICATION SYSTEMS: AN INTRODUCTION TO SIGNALS AND NOISE IN ELECTRICAL 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 cn
=
Ae jφ
T 0 / 2
∫
− T 0 / 2
T 0
e
j2 π ( m− n ) f 0 t
dt
= Ae
jφ
Ae jφ n= m sinc( m − n ) = 0 otherwise
2.1-2
c0 v (t )
=0
2
T0 / 4
cn
=
T0
n
∫
0
0 0
cn
A cos
2π nt T0
1 2 A / π
dt +
∫
T 0 / 4
2 3 0 2 A / 3π
( − A)cos
2π nt T0
2A πn dt = sin 2 πn
4 5 0 2 A / 5π
±180°
0
arg cn
T 0 / 2
6 0
7 2 A / 7π
±180°
0
2.1-3
= A/2 2π 2 T /2 2 At cn = A− cos T ∫ T0 0 T0 0 c0
=
v (t )
0
n
0 1 2 0.5 A 0.2 A 0
cn
0
arg cn
0
nt dt =
A A sin π n − (cos π n − 1) (π n) 2 πn
3 4 5 0.02 A 0 0.01 A 0
6 0
0
2.1-4
c0
=
2 T0
T 0 / 2
∫ 0
A cos
2π t T 0
=0
(cont.)
2-1
cn
=
2 T0
T 0 / 2
∫
A cos
0
2π t T0
cos
2π nt T0
2 A sin (π − π n ) 2t / T0 dt = T0 4(π − π n) / T0
T 0 / 2
+ π n ) 2t / T0 + 4(π + π n) / T0 0 sin (π
n = ±1
A / 2 = A [sinc(1 − n) + sinc(1 + n)] = 2 0
otherwise
2.1-5
c0
=
cn
=−j
v (t )
n
2 T0
=0 T 0 / 2
∫ 0
cn
1 2 A / π
arg cn
−90°
A sin
2π nt T0
dt = − j
2 3 0 2 A / 3π
4
A
πn
(1 − cos π n)
5 2 A / 5π
−90°
−90°
2.1-6
c0
=
v( t )
=0 T 0 / 2
− π n ) 2t / T0 sin ( π + π n ) 2t / T0 cn = − j ∫ A sin sin dt = − j − T0 0 T0 T0 T0 4(π − π n )/ T0 4(π + π n) / T0 0 =n ±1 /2 m jA = − Aj [sinc(1 − )n− sinc(1 + )n] = 2 otherwise 0 2
2π t
T 0 / 2
2 A sin (π
2π nt
2.1-7 cn
=
1
T / 2 v ( t) e− T 0 ∫0
where
0
∫
T0
T 0 / 2
v(t )e −
ωjn 0
T 0
∫ v(t )e dt ] dt = ∫ v (λ + T /2) e e = −e ∫ v (t )e dt
ω jn0
t
t + dt
−
ωjn0 t
T 0 / 2
T 0 / 2
−
ωjn 0λ
0
0
jnπ
T 0 / 2
−
j nω0 t
0
jn since e π
=1
for even n , cn
=0
for even n
2-2
−
2 ωjn 0 0 /T
d λ
2.1-8 P
=
c0
2
f
>
+ 2 ∑ cn = Af0τ + 2 2
2
n=1
1
where
>
∞
τ
2
+2
Af0τ sinc2 f 0τ
2
+2
2
Af0τ sinc3 f0τ
+L
= 4 f 0
1 f
=
2
P=
τ τ
Af0τ sinc f 0τ
>f 1
2τ
A2 1 1 3 P 1 + 2sinc2 + 2sinc2 + 2sinc 2 = 0.23 16 4 2 4
2
A
1 + 2sinc2 1 + 2sinc2 1 + 2sinc 2 3 + 2sinc 2 5 + 2sinc 2 3 + 2sinc 2 7 = 0.24 A2 16 4 2 4 4 2 4 A2 = P 1+ 2sinc2 1 + 2sinc 2 1 = 0.21 2 A 16 4 2 A2
2.1-9
0 cn = 2 2 π n
n even n odd 2
2
4 t 1 2 T / 2 4t dt dt − = − = a) P = 1 1 T0 ∫−T / 2 T0 T0 ∫ 0 T0 3 2 2 2 4 4 4 P′ = 2 2 + 2 + 2 = 0.332 so P′ / P = 99.6% 2 2 π 9π 25π 1
T0 / 2
0
0
b) v′(t) =
8
8
π
9π
cos ω 0t + 2
cos3ω 0t + 2
8 2 cos5ω 0t 25π
2.1-10
0 cn = − j 2 π n a) P =
1 T 0
n even n odd T 0 / 2
∫
−T 0 / 2
2
(1)
dt
=1
2 2 2 2 2 2 P′ = 2 + + 5π = 0.933 3 π π
so P′ / P = 93.3%
(cont.) 2-3
b) v′(t ) =
=
4
cos ( ω0t − 90° ) +
π 4
sin (ω 0t ) +
π
4
4 3π
cos ( 3ω0t − 90° ) +
sin ( 3ω 0t ) +
3π
4 5π
4 5π
cos ( 5ω 0t − 90° )
sin ( 5ω 0t )
2.1-11 2
n=0 t 1 / 2 1 P = ∫ dt = cn = 3 T0 0 T 0 1 / 2π n n ≠ 0 4 4 ∞ 2 2 1 1 1 = 1 = + + + L 2 P =2∑ π 14 34 54 3 n odd π n
1
T 0
1
Thus,
12
1
+
22
+
1 32
4π 2 1
+L=
1 π 2
3 − 4 =
2
6
2.1-12 P
=
2 T0
2
T 0 / 2
∫ 0
4t 1 1 − T dt = 3 0
2
cn
0 = 2 ( 2/ π n)
n even n odd
2
∞ 1 1 = 1 + 2 1 + 1 + 1 + L = 1 P = + 2∑ 4 4π 2 1 2 2 2 32 3 2 n =1 2π n
Thus,
1 4
1
+
1 4
3
+
1 4
5
+ L=
π4 1 2⋅ 2 3 4
=
π4 96
2.2-1
π t cos2π ftdt 0 τ sin π −2π f τ sin π +2π f τ )2 (τ ) 2 Aτ (τ =2 Aπ + = [ sinc( τ − 1f/2) + sinc( τ + 1/2) f ] π + 2π f 2 2 2 2 − π f ) (τ ) (τ τ / 2
∫
V( f ) = 2
A cos
(cont.)
2-4
2.2-2 V( f ) = −j2
=−
τ / 2
∫ 0
A sin
2π t
τ
cos2π ftdt
sin 2π −2π f τ sin 2π + 2π f τ )2 ( τ )2 (τ A − =− 2π −2π f 2π + 2π f 2 2 (τ ) (τ )
2j
Aτ j [ sinc( τf − 1) − sinc( τf + 1) ] 2
2.2-3
A − A t cos ω tdt = 2 Aτ 2sin 2 ωτ − 1 + 1 = Aτ sinc 2 2 2 ∫ 0 τ (ωτ )
V( f ) = 2
τ
2.2-4 τ
∫
V ( f ) = − j2
=−
0
t
2 Aτ
τ
(ωτ ) 2
A sin ω tdt = − j
A j (sinc2 τf − cos2π τf ) π f
(sin ωτ − ωτ cos ωτ )
2.2-5 f Π 2W 2W 2 ∞ ∞ 1 ∞ 2 1 1 f sinc2 = Π = = Wt dt df df ∫−∞ ∫−∞ 2W 2W ∫ −∞ 4W 2 2W
v (t)
= sinc2Wt ↔
1
2-5
fτ
2.2-6 ∞
=E ∫0 ( E ′ E
2
=
π
− bt
Ae )
2
dt =
2πW
arctan
A2
b
A2
W
′ E= 2 ∫ 0
2b
50% = 84%
b2
A2 πW df = arctan 2 πb b
+ (2π f )2
= b / 2π = 2b / π
W W
2.2-7 ∞
∞ ∞ W ( f ) e jω t df dt v t = ( ) ∫−∞ ∫−∞ ∫ −∞ ∞ ∞ ∞ = ∫−∞ W ( f ) ∫−∞ v (t) e− j (− ω ) t dt df = ∫ −∞ W ( f )V (− f ) df ∞ ∞ ∞ 2 V ( − f ) = V * ( f ) when v(t) is real, so ∫ v2 ( t) dt =∫ V ( f )V * ( f ) df = ∫ V ( f ) df −∞ −∞ −∞
v (t )w( t) dt
2.2-8 ∗
∗
ft ∞ w(t )e 2jπ dt ∞ w( t) e− 2jπ ( − )f dtt = W ∗ ( f ) = = w ( t ) e dt ∫−∞ ∫−∞ ∫ −∞ Let z( t) = w∗ ( t) so Z ( f ) = W∗ ( − f ) and W ∗ ( f ) = Z( − f )
∞
∗
−
∫
Hence
2jπ
∞
2.2-9 t
Π ↔ A
sinc A 2t
τ
2.2-10 πt t cos Π B τ τ so
τ ↔ V ( f ) = Π 2
↔ Bτ [ sinc( 2
f τ
2
1
f Π A A
for A =
2
B= A and τ
−
j
=2
↔ −
W
τ
τ −f 1/2) + sinc( τ +f 1/2) ]
⇒
(z )t = AW[sinc(2 Wt− 1/2) + sinc(2 Wt+ 1/2) ]
[ sinc( τt − 1) + sinc( τt + 1)] ↔
B= − jAand τ
π ( − )f − f π f f Π = B cos Π τ τ τ τ
Bτ j [sinc( τf − 1) + sinc( τf + 1) ] 2
τB 2
2
[sinc(tτ − 1/2) + sinc(tτ + 1 /2 )] ↔ B cos
2.2-11 2π t t sin B Π τ τ
Let
At ↔
Afso sinc
τB
Let
so
∞
= ∫ −∞ V ( f ) Z ( − f ) df
v( t) z( t) dt
−∞
v (t) = sinc
ft
=2
W
⇒
Bsin
2π (− )f
τ
(z )t = AW[ sinc(2
2-6
2π f f − f Π = − Bsin Π τ τ τ Wt− 1) + sinc(2 Wt+ 1) ]
2.2-12 e
−b t
↔
∞
∫−∞ ( e− Thus,
2b
+ (2π f )
2
2
b
2 π a t
2
)
=
dt
⇒ e−2π a t ↔
1 2π a
a 2 + f 2
1 π
2
= 2 ∫ 0 (a 2 + x 2 ) 2 a dx
∞
1 2π a
df
=
a / π
=
+ (2π f ) a + f 2 2 a ∞ df = 2 ∫ 0 2 π ( a 2 + f 2 ) 2
(2π a )
a / π
∞
=∫−∞
4π a 2
2
π 4 a3
2.3-1 z(t ) = v( t − T ) + v( t + T ) where v( t) = AΠ ( t /τ ) ↔ Aτ sinc f τ so Z( f ) = V( f ) e−
j T ω
+ V( f ) e ωj T = 2 Aτ sinc
fτ cos2π fT
2.3-2 z( t ) = v( t − 2 T ) + 2 v( t) + v( t + 2T) where v( t) = aΠ ( t /τ ) ↔ Aτ sinc f τ −
Z( f ) = V ( f ) e
2j ω T
+ V( f ) + V( f ) e 2jω T = 2 Aτ (sinc
f τ )(1 + cos4π fT)
2.3-3 z(t ) = v ( t − 2T ) − 2 v( t) + v( t + 2 T ) where v( t) = aΠ ( t / τ ) ↔ Aτ sinc f τ Z( f ) = V ( f ) e−
2j ω T
− 2 V ( f ) + V( f ) e 2jω T = 2 Aτ (sinc
fτ )(cos4π fT − 1)
2.3-4
t − T + ( B − A)Π t − T / 2 T 2T V ( f ) = 2 AT sinc2 fTe − ωj T+ ( B − A)T sinc f T e− ωj /T2 v (t) = AΠ
2-7
2.3-5
t − 2T + ( B− A)Π t − 2T 2T 4T − 2jω T V ( f ) = 4 AT sinc4 fTe + 2( B − A)T sinc2 fTe− v (t) = AΠ
2jω T
2.3-6 Let w(t ) = v( at ) ↔ W ( f ) =
1
V ( f / a)
a
Then z(t ) = v[ a(t − td / a)] = w( t − td / a) so Z( f ) = W ( f ) e−
jω dt / a
=
1 a
V ( f / a) e−
j d t / a ω
2.3-7 F v (t) e
jω c t
∞
∞
= ∫ v( t)e ωj t e− jω t dt =∫ v( t) e− j2π ( f− f ) tdt =V ( f − fc ) −∞ −∞ c
2.3-8 v (t) = AΠ (t /τ )cos ωc t with ω c V( f ) =
τA 2
sinc( f
−
fc )τ
+
c
= 2 π fc = π / τ
τA 2
+
sinc( f
2.3-9 v( t ) = AΠ (t /τ )cos(ωc t − π /2) with ω c V( f ) =
e− jπ / 2
=−
=
fc )τ
τA 2
] [sinc( fτ −1/2) + sinc( f τ + 1/2)
= 2π fc = 2 π / τ
e jπ / 2
Aτ sinc( f − fc )τ + Aτ sinc( f 2 2 Aτ j [sinc( τf − 1) − sinc( τf + 1) ] 2
+
f c )τ
2.3-10 − t
z( t) = v( )t c o s ω c t Z( f ) =
1 2
V( f
−
v( t) = Ae
fc ) +
1 2
V( f
+
↔
fc ) =
2 A 1 + (2π f ) 2 A
1 + 4π ( f− fc ) 2
2
+
A
1 + 4π ( f + cf )2 2
2.3-11 v( t) = Ae−t for t ≥ 0
z( t) = v()cos( t ωc t − π /2) Z( f ) =
=
e − jπ / 2
2
V( f
−
fc ) +
A / 2 j− 2π ( f − fc )
−
e jπ / 2
2
V ( f + fc ) =
↔
− jA / 2 1 + j2π ( f −
A/2 j − 2π ( f + fc )
2-8
A
1 + j 2π f fc )
+
jA / 2
1 + j2π ( f
+
fc )
2.3-12 v (t) = t z (t ) d df
Z( f) = 2
V( f ) =
1 2jπ
−
t Π ↔ 2 A sinc2 f τ τ τ d sin2π f τ 2A (2πτ )2 fcos2π A = 2 df 2π f τ (2π f τ ) d − jA Z( f ) = ( sinc2 f τ − cos2π f τ ) z (t ) =
A
π
df
τf − 2πτ si n2π τf
f
2.3-13 − b t
z( t) = tv( t)
v( t) = Ae
df
1
Z ( f ) =
−
2 Ab
d
j2π
↔
b2 + (2π
2 Ab
+ (2π f ) 2 j 2 Abf = 2 f) 2 b2 + (2π f ) 2 b
2
2.3-14 v( t) = Ae−t for t ≥ 0
z( t) = t2 v( t)
1
Z ( f ) =
d
2 ( − j2π f) df
b+
↔
A b + j2π f
2A = 3 j 2π f [ b+ j2π f] A
2.3-15 −π (bt ) 2
v (t) = e
( a)
d dt
↔ V ( f ) = 1 e−π ( f / b)
2
b
v (t) = −2π b 2 te −π ( bt )
2
(b ) t e− π (bt )
2
↔
1
−
d
2jπ
↔
j 2π f
V( f ) = df
b f
Both results are equivalent to bte 2.4-1 (y) t= 0
t
= ∫ 0
2
= ∫ 0
At 2
A λ dλ
=
Aλ dλ
=2A
2
e−π ( f / b)
e−π ( f / b ) jb
−π ( bt )2
2
2
↔ − jf e−π ( f / b)
2
0 < t< 2 t > 2
2-9 please click here. To get the fully access of the document
