Solutions Manual to accompany
Communication Systems An Introduction to Signals and Noise in Electrical Communication Fourth Edition
A. Bruce Carlson Rensselaer Rensselaer Polytechni Polytechnicc Institute 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
cn
=
Ae jφ
T 0 / 2
∫
− T 0 / 2
T 0
e
j 2 π ( m−n )f 0t
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
2 0
T 0 / 2
∫
T 0 / 4
3 2 A / 3π
(− A)cos
4 0
2π nt T0
2A πn dt = sin 2 πn
5 2 A / 5π
±180°
0
arg cn
dt +
6 0
7 2 A / 7π
±180°
0
2.1-3
= A/ 2 2 T /2 2 At 2π nt A A cn = A− cos dt = sin π n − (cos π n − 1) ∫ πn T0 0 T0 T0 (π n) 2 c0
=
v (t)
0
n
0 1 2 3 4 5 0.5 A 0.2 A 0 0.02 A 0 0.01 A
cn
0
arg cn
0
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
dt =
2 A sin (π − π n ) 2t / T0 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 ) 2
=0 T 0 / 2
∫
A sin
cn
1 2 A / π
2 0
arg cn
−90°
n
T0
0
2π nt T0
dt = − j
3 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
2π t
2 A sin (π
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 m jA / 2 n = ±1 = − j A [sinc(1 − n ) − sinc(1 + n ) ] = 2 otherwise 0 2
2π nt
2.1-7 cn
=
1
T / 2 v ( t) e− jnω t dt + T v(t )e − jnω t dt ] ∫ T / 2 T 0 ∫0 0
0
0
0
0
where
∫
T0
T 0 / 2
v(t )e − jnω0 t dt =
T 0 / 2
∫ 0
v (λ + T0 /2) e− jnω 0λ e− jnω 0T 0 / 2 dλ T 0 / 2
since e jnπ
= −e jnπ ∫ 0 = 1 for even n, cn = 0
v (t )e − jnω0 t dt
for even n
2-2
2.1-8 P = c0
2
∞
+ 2 ∑ cn = Af0τ + 2 2
2
Af0τ sinc f 0τ
2
+2
Af0τ sinc2 f 0τ
2
+2
Af0τ sinc3 f 0τ
2
+L
n=1
1
where f
>
f
>
f
>
τ
= 4 f 0 A2
1 + 2sinc2 1 + 2sinc2 1 + 2sinc2 3 = 0.23A2 16 4 2 4 A2 2 1 2 1 2 3 2 5 2 3 2 7 P = 1 2sinc 2sinc 2sinc 2sinc 2sinc 2sinc + + + + + + = 0.24 A2 16 4 2 4 4 2 4 A2 2 1 2 1 P= 1 2sinc 2sinc + + = 0.21A2 16 4 2
1
P=
τ 2
τ 1 2τ
2.1-9
0 cn = 2 2 π n
n even n odd 2
2
4 t 1 2 T / 2 4t a) P = 1 dt 1 dt − = − = T0 ∫−T / 2 T0 T0 ∫ 0 T0 3 2 2 2 4 4 4 P′ = 2 2 + 2 2 + 2 = 0.332 so P′ / P = 99.6% 2 π 9π 25π 1
T0 / 2
0
0
b) v′(t) =
8
π
cos ω 0 t + 2
8 9π 2
cos3ω 0t +
8 2 cos 5ω 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 π π 3
so P′ / P = 93.3%
(cont.) 2-3
4 4 4 cos ( ω0 t − 90° ) + cos ( 3ω0 t − 90° ) + cos ( 5ω 0t − 90° ) π 3π 5π 4 4 4 = sin (ω 0t ) + sin ( 3ω 0t ) + sin ( 5ω0t ) π 3π 5π
b) v′(t ) =
2.1-11 2
n=0 t 1 / 2 1 P = ∫ dt = cn = T0 0 T 0 3 1 / 2π n n ≠ 0 4 4 ∞ 2 2 1 1 1 = 1 P = 2 ∑ 2 = + + + L π 14 34 54 3 n odd π n 1
T 0
1
Thus,
12
1
+
22
+
1 32
+L=
4π 2 1
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 A π + = [ sinc( f τ − 1 /2) + sinc( f τ + 1/2) ] π + 2π f 2 − π 2 2 f 2 ) (τ ) (τ
V( f ) = 2
τ/2
∫
A cos
(cont.)
2-4
2.2-2 V( f ) = −j2
τ /2
∫ 0
A sin
2π t cos2π ftdt τ
sin 2π −2π f τ sin 2π + 2π f τ )2 ( τ ) 2 Aτ (τ = − j2 A − = − j [ sinc( f τ − 1) − sinc( f τ + 1) ] 2π −2π f 2π + 2π f 2 2 2 (τ ) ( ) τ
2.2-3 V( f ) = 2
A − A t cos ω tdt = 2 Aτ 2sin 2 ωτ − 1+ 1 = 2 2 ∫ 0 τ (ωτ ) τ
Aτ sinc 2 f τ
2.2-4
= − j
τ
t
2 Aτ
τ
(ωτ ) 2
∫
A sin ω tdt = − j
A
(sinc2 f τ − cos2π f τ )
V ( f ) = − j2
0
π f
(sin ωτ − ωτ cos ωτ )
2.2-5 f Π 2W 2W 2 ∞ ∞ 1 ∞ 2 f 1 1 sinc2 Wt dt df df = Π = = ∫−∞ ∫−∞ 2W 2W ∫ −∞ 4W 2 2W
v (t) = sinc2Wt ↔
1
2-5
2.2-6
E = E ′ E
∫
∞
0
2
=
π
2
− bt
( Ae ) dt = arctan
2π W b
A2
A2
W
E ′ = 2 ∫ 0
b 2 + (2π f )2 50% W = b / 2π
2b
= 84%
W
df =
A2
πb
arctan
2πW
b
= 2b / π
2.2-7
∫
∞
−∞
v (t )w( t) dt =
∫
∞
−∞
v( t)
∞
df dt
∫ W ( f ) e
jω t
−∞
∞ ∞ ∞ = ∫−∞ W ( f ) ∫−∞ v (t) e− j (− ω ) t dt df = ∫ −∞ W ( f )V (− f ) df ∞ ∞ V ( − f ) = V * ( f ) when v (t) is real, so ∫ v 2 ( t ) dt =∫ V ( f −∞ −∞
)V * ( f )df
∞
= ∫ −∞ V ( f )
2
df
2.2-8 ∗
∗
∞ w( t)e j 2π ft dt = ∞ w( t) e− j 2π ( − f )t dt = W ∗ ( f ) = w ( t ) e dt ∫−∞ ∫−∞ ∫ −∞ Let z ( t ) = w∗ ( t) so Z ( f ) = W ∗ ( − f ) and W ∗ ( f ) = Z (− f ) ∞
∗
Hence
− j 2π ft
∫
∞
−∞
v( t )z( t ) dt
∞
= ∫ −∞ V ( f ) Z (− f ) df
2.2-9 t 1 f Π ↔ A sinc Af so sinc At ↔ Π A A A v (t) = sinc
2t τ
τ ↔ V ( f ) = Π 2
f τ 2
for A =
2 τ
2.2-10 πt t B cos Π τ τ
↔ Bτ [ sinc( f τ − 1/2) + sinc( f τ + 1/2) ] 2 π (− f ) − f πf Bτ f so Π = B cos Π [sinc(tτ − 1/2) + sinc(tτ + 1 /2 )] ↔ B cos τ τ 2 τ τ Let B = A and τ = 2W ⇒ z (t ) = AW [sinc(2Wt − 1/2) + sinc(2Wt + 1/2) ]
2.2-11 2π t t B sin Π τ τ
↔ − j Bτ sinc( f τ − 1) + sinc( f τ + 1) [ ] 2 Bτ 2π (− f ) − f 2π f f Π = − B sin Π so − j [ sinc(tτ − 1) + sinc( tτ + 1)] ↔ B sin τ τ 2 τ τ Let B = − jA and τ = 2W ⇒ z (t ) = AW [ sinc(2Wt − 1) + sinc(2Wt + 1)]
2-6
2.2-12 e
−b t
↔
∞
∫−∞ ( e− Thus,
2b
+ (2π f )
2
b
2 π a t
2
)
dt =
2
⇒ e−2π a t ↔
1 2π a
a 2 + f 2
1 π
2
= 2 ∫ 0 (a 2 + x2 ) 2 a ∞
dx
1 2π a
df
=
=
a /π
+ (2π f ) a + f 2 2 a ∞ df = 2 ∫ 0 2 2 2 π ( a + f ) 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− jω T
+ V ( f )e jωT = 2 Aτ sinc f τ cos2π fT
2.3-2 z (t ) = v (t − 2T ) + 2v( t ) + v( t + 2T ) where v(t ) = aΠ ( t /τ ) ↔ Aτ sinc f τ Z ( f ) = V ( f )e − j 2ωT
+ V ( f ) + V ( f ) e j2ωT = 2 Aτ (sinc
f τ )(1 + cos4π fT )
2.3-3 z (t ) = v (t − 2T ) − 2v( t) + v( t + 2 T ) where v(t ) = aΠ ( t / τ ) ↔ Aτ sinc f τ Z ( f ) = V ( f )e − j 2ωT
− 2V ( f ) + V ( f ) e j2ωT = 2 Aτ (sinc
2.3-4
t − T + ( B− A)Π t − T / 2 T 2T V ( f ) = 2 AT sinc2 fTe − j ωT + ( B − A)T sinc fT e− jω T / 2 v (t) = AΠ
2-7
f τ )(cos4π fT − 1)
2.3-5
t − 2T + ( B− A)Π t − 2T 2T 4T − j 2 ωT V ( f ) = 4 AT sinc4 fTe + 2( B − A)T sinc2 fTe− j 2ωT v (t) = AΠ
2.3-6 Let w (t ) = v( at ) ↔ W ( f ) =
1 a
V ( f / a)
Then z (t ) = v[a (t − td / a)] = w(t − td / a) so Z( f ) = W ( f ) e− jω td / a
=
1 a
V ( f / a) e− jω td / a
2.3-7 F v (t) e
jω ct
∞
∞
= ∫ v(t )e jω t e− jω t dt =∫ v (t )e − j 2π ( f − f ) t dt =V ( f − f c ) −∞ −∞ c
2.3-8 v (t) = AΠ(t /τ )cos ωc t with ω c V( f ) =
Aτ 2
−
sinc( f
f c )τ
+
Aτ 2
c
= 2 π fc = π / τ +
sinc( f
2.3-9 v( t) = AΠ (t /τ )cos(ωc t − π /2) with ω c V( f ) =
e− jπ / 2
=
f c )τ
Aτ 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 τ = − j A [sinc( f τ − 1) − sinc( f τ + 1) ] 2
+
fc )τ
2.3-10 − t
z (t ) = v (t) c o s ω ct Z ( f ) =
1 V( f 2
v(t ) = Ae
− fc ) +
1 V (f 2
+
↔
f c ) =
2 A 1 + (2π f ) 2
1+
A 4π ( f 2
−
fc )
2
+
1+
A 4π ( f 2
+
f c )2
2.3-11 v (t) = Ae −t for t ≥ 0
z (t ) = v ()c t os(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
+
f c ) =
− jA / 2 1 + j 2π ( f −
A/2 j − 2π ( f
↔
+ f c )
2-8
A 1 + j 2π f fc )
+
jA / 2 1 + j 2π ( f
+
f c )
2.3-12 t Π ↔ 2 A sinc2 f τ τ τ d d sin2π f τ 2A (2πτ )2 f cos2π f τ − 2πτ si n 2π f τ Z ( f ) = 2 A = 2 df df 2π f τ (2π f τ ) 1 d − jA V( f ) = Z( f ) = ( sinc2 f τ − cos2π f τ ) − j 2π df π f v (t) = t z(t )
z (t ) =
A
2.3-13 − b t
z (t ) = tv(t ) Z ( f ) =
v( t) = Ae
1
d
− j 2π
df
↔
2 Ab
+ (2π f ) 2 j 2 Abf 2 Ab = 2 b 2 + (2π f ) 2 b2 + (2π f ) 2 b
2
2.3-14 v(t ) = Ae −t for t ≥ 0
z (t ) = t 2v( t) Z ( f ) =
1
↔
A b + j2π f
A 2A = b + j 2π f 3 [ b + j2π f ]
d
2 ( − j 2π f ) df
2.3-15 v (t) = e−π (bt) ( a)
d dt
2
↔ V( f ) =
1 b
v (t) = −2π b 2 te −π ( bt )
2
(b ) t e− π (bt )
2
↔
1
d
− j2π
df
e−π ( f / b)
↔
2
j 2π f
V( f ) =
b f jb
Both results are equivalent to bte 2.4-1 y (t) = 0
e−π ( f / b)
e−π ( f / b )
−π ( bt )2
2
2
↔ − jf e−π ( f / b)
2
t < 0 t
= ∫ 0 Aλ d λ = 2
At 2 2
= ∫ 0 Aλ d λ = 2A
0 < t < 2 t > 2
2-9
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