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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 reproduced 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

Ae jφ T0

cn =

 Ae jφ n = m j 2 π ( m−n )f 0t jφ e dt = Ae sinc( m − n ) =  ∫− T0 / 2 0 otherwise T0 / 2

2.1-2

c0 v (t ) = 0 cn =

2 T0

n cn

∫

T0 / 4

0

0 0

A cos

T0 / 2 2π nt 2π nt 2A πn dt + ∫ ( − A)cos dt = sin T / 4 0 T0 T0 πn 2

1 2A/π

2 0

4 0

5 2 A / 5π

±180°

0

arg cn

3 2 A / 3π

6 0

7 2 A / 7π

±180°

0

2.1-3

c0 = v (t ) = A / 2 cn =

2 T0

n

∫

T0 /2

0

 2 At  2π nt A A dt = sin π n − (cos π n − 1) A−  cos T0  T0 πn (π n) 2 

0 1 2 3 4 5 6 0.5A 0.2A 0 0.02A 0 0.01A 0

cn

0

arg cn

0

0

0

2.1-4

c0 =

2 T0

∫

T0 / 2

0

A cos

2π t =0 T0

(cont.)

2-1

2 cn = T0

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

T0 / 2

∫

T0 / 2

0

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

= 2.1-5

c0 = v (t ) = 0 cn = − j

2 T0

∫

T0 / 2

0

n cn

1 2A/π

arg cn

−90°

A sin 2 0

2π nt A dt = − j (1 − cos π n ) T0 πn 3 2 A / 3π

4

5 2 A / 5π

−90°

−90°

2.1-6

c0 = v(t ) = 0 2 cn = − j T0 = −j

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

T0 / 2

∫

T0 / 2

0

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

2.1-7 cn =

T0 1  T0 / 2 v ( t) e− jnω0 t dt + ∫ v(t )e − jnω 0t dt ] ∫ T0 / 2 T0  0

where

∫

T0

T0 / 2

v(t )e − jnω0 t dt = ∫

T0 / 2

0

v (λ + T0 /2) e− jnω 0λ e− jnω 0T0 / 2 d λ

= −e jnπ ∫

T0 / 2

0

v (t )e − jnω0 t dt

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 = 4 f0 τ 1 A2 f > P= τ 16

where

1 + 2sinc2 1 + 2sinc2 1 + 2sinc2 3  = 0.23 A2  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 2 1 A  1 1 f > P= 1 + 2sinc2 + 2sinc 2  = 0.21A2  2τ 16  4 2 2.1-9  0  cn =  2 2  π n    1 a) P = T0

n even n odd 2

 4t  2 ∫−T0 / 2  1 − T0  dt = T0 T0 / 2

2

2

2

∫

T0 / 2

0

 4t  1  1 −  dt = 3  T0 

2

 4   4   4  P′ = 2  2  + 2  2  + 2  = 0.332 so P′ / P = 99.6% 2  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  cn =  − j 2  π n 1 a) P = T0

n even n odd  2 2  2  2  2 2  ′ ∫−T0 / 2 (1) dt = 1 P = 2  π  +  3π  +  5π   = 0.933 so P′ / P = 93.3%   T0 / 2

2

(cont.) 2-3

4 4 4 cos ( ω0t − 90° ) + cos ( 3ω0t − 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 1 P= T0

2

∫

T0

0

n=0 1 / 2 cn =  1 / 2π n n ≠ 0

 t  1   dt = 3  T0 

∞  2  2  1 1 1  1 P =2∑   = 2    4 + 4 + 4 +L = π  1 3 5  3 n odd  π n  4

4

1 1 1 4π 2 + + + L = 12 22 32 2

Thus,

1 1  π2 3− 4= 6  

2.1-12 2 P= T0

2

∫

T0 / 2

0

 4t  1  1 −  dt = 3  T0 

n even 0 cn =  2  (2/ π n) n odd

∞ 1 2 1  1  P =   + 2∑   = + 2 4 4π 2 n =1  2π n  2

2

1 1 1  1  2 + 2 + 2 +L = 2 3 1  3

1 1 1 π4 1 π 4 Thus, 4 + 4 + 4 + L = = 1 3 5 2 ⋅ 24 3 96 2.2-1 πt cos2π ftdt τ τ  π  sin τ −2π f 2 sin = 2A  + π −2π f 2 2  τ 

V ( f ) = 2∫

τ /2

0

A cos

( (

) )

( πτ 2π f ) τ2  Aτ [ sinc( f τ −1/2) + sinc( f τ + 1/2) ] = (πτ 2π f )  2 + +



(cont.)

2-4

2.2-2 2π t cos2π ftdt 0 τ  2π −2π f τ sin sin  τ 2− = − j2 A  2 π 2  2 τ −2π f 

V ( f ) = − j 2∫

τ /2

A sin

( (

) )

( 2τπ 2π f ) τ2  = − j Aτ sinc( f τ −1) − sinc( f τ + 1) [ ] ( 2τπ 2π f )  2 +

+

2.2-3 τ t 2 Aτ   ωτ V ( f ) = 2∫  A − A  cos ω tdt = 2sin 2  2  0 τ (ωτ )    2

  2  − 1 + 1 = Aτ sinc f τ  

2.2-4 τ t 2 Aτ V ( f ) = − j 2∫ A sin ω tdt = − j (sin ωτ − ωτ cos ωτ ) 0 τ (ωτ ) 2 A = −j (sinc2 f τ − cos2π f τ ) πf

2.2-5

v (t ) = sinc2Wt ↔

∫

∞

−∞

1  f  Π  2W  2W 

sinc2Wt dt = ∫ 2

∞

−∞

2

∞ 1 1 1  f  Π df = ∫ df =  2 −∞ 4W 2W  2W  2W

2-5

2.2-6 W A2 A2 A2 2πW E′ = 2∫ 2 df = arctan 2 0 0 b + (2π f ) 2b πb b E′ 2 2πW 50% W = b / 2π = arctan = E π b 84% W = 2b / π

E=∫

∞

( Ae − bt ) dt = 2

2.2-7 ∞  ∞ W ( f ) e jω t df dt v ( t ) w ( t ) dt = v ( 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 ) df 2

2.2-8 ∗

∗

∞ ∞ ∫−∞ w∗ (t)e− j 2π ft dt =  ∫−∞ w(t )e j 2π ft dt  =  ∫−∞ w( t)e− j 2π ( − f )t dt  = W ∗ ( f ) Let z ( t ) = w∗ ( t ) so Z ( f ) = W ∗ ( − f ) and W ∗ ( f ) = Z ( − f ) ∞

Hence

∫

∞

−∞

∞

v( t )z( t) dt = ∫ V ( f ) Z ( − f ) df −∞

2.2-9 1  f  t  Π   ↔ A sinc Af so sinc At ↔ Π   A  A  A 2t τ  fτ  2 v (t ) = sinc ↔ V ( f ) = Π   for A = τ 2  2  τ 2.2-10 πt  t  Bτ B cos Π   ↔ [ sinc( f τ − 1/2) + sinc( f τ + 1/2) ] τ τ  2 Bτ π (− f )  − f  πf  f  so Π Π  [sinc(tτ − 1/2) + sinc(tτ + 1/2) ] ↔ B cos  = B cos 2 τ τ  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  Bτ B sin Π  ↔− j [sinc( f τ − 1) + sinc( f τ + 1) ] τ 2 τ  Bτ 2π (− f )  − f  2π f  f  so − j Π Π  [ sinc(tτ − 1) + sinc( tτ + 1) ] ↔ B sin  = − B sin 2 τ τ  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 ↔

∫

∞

−∞

2b 4π a a /π ⇒ e−2π a t ↔ = 2 2 2 2 b + (2π f ) (2π a ) + (2π f ) a + f2 2

(e

−2 π a t

Thus,

∫

)

2

∞

0

2

∞ 1 a /π a dt = =∫ df = 2   2 2 2π a −∞ a + f π 

(a

∞

0

(a

df

2

+ f2)

2

2

dx 2

∫

+ x2 )

2

1 π  1 π =   = 3 2  a  2π a 4 a

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 + 2T ) 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 f τ )(cos4π fT − 1)

2.3-4  t −T   t −T / 2  v (t ) = AΠ   + ( B − A)Π    2T   T  − j ωT V ( f ) = 2 AT sinc2 fTe + ( B − A)T sinc fTe− jω T / 2

2-7

2.3-5  t − 2T   t − 2T  v (t ) = AΠ   + ( B − A)Π    4T   2T  V ( f ) = 4 AT sinc4 fTe − j 2ωT + 2( B − A)T sinc2 fTe− j 2ωT 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 td / a =

1 V ( f / a) e− jω td / a a

2.3-7 ∞

∞

−∞

−∞

F  v (t) e jω ct  = ∫ v(t )e jω ct e− jω t dt =∫ v (t) e− j 2π ( f − fc ) t dt =V ( f − f c ) 2.3-8 v (t ) = AΠ (t /τ )cos ωc t with ω c = 2π f c = π / τ

V( f ) =

Aτ Aτ Aτ sinc( f − f c )τ + sinc( f + f c )τ = [ sinc( fτ − 1/2) + sinc( f τ + 1/2) ] 2 2 2

2.3-9 v( t ) = AΠ (t /τ )cos(ωc t − π /2) with ω c = 2π f c = 2π / τ

e− jπ / 2 e jπ / 2 Aτ sinc( f − f c )τ + Aτ sinc( f + f c )τ 2 2 Aτ = −j [sinc( f τ − 1) − sinc( f τ + 1) ] 2

V( f ) =

2.3-10

2A 1 + (2π f ) 2 1 1 A A Z ( f ) = V ( f − fc ) + V ( f + fc ) = + 2 2 2 2 2 1 + 4π ( f − f c ) 1 + 4π ( f + f c )2 z (t ) = v (t) c o s ω ct

v(t ) = Ae − t ↔

2.3-11 z (t ) = v ()cos( t ωc t − π /2)

v (t ) = Ae −t for t ≥ 0 ↔

A 1 + j 2π f

e − jπ / 2 e jπ / 2 − jA / 2 jA / 2 V ( f − fc) + V ( f + fc ) = + 2 2 1 + j 2π ( f − f c ) 1 + j 2π ( f + f c ) A/2 A/2 = − j − 2π ( f − f c ) j − 2π ( f + f c )

Z( f ) =

2-8

2.3-12

  ↔ 2 A sinc2 f τ  d d  sin2π f τ  2A  (2πτ )2 f cos2π f τ − 2πτ sin2π f τ  Z ( f ) = 2A  =  2  df df  2π f τ  (2π f τ )

v (t ) = t z (t ) z (t ) =

A t Π τ  τ

1 d − jA Z( f ) = ( sinc2 f τ − cos2π f τ ) − j 2π df πf

V( f ) = 2.3-13

z (t ) = tv (t ) v( t ) = Ae

−b t

↔

2 Ab 2 b + (2π f ) 2

1 d  2 Ab  j 2 Abf =  2 2 2 2 − j 2π df  b + (2π f )   b + (2π f ) 2   

Z( f ) =

2.3-14

z (t ) = t 2v( t ) v (t ) = Ae −t for t ≥ 0 ↔ 1

Z( f ) =

( − j 2π f )

2

d df

A b + j2π f

 A  2A  b + j 2π f  = 3   [ b + j 2π f ]

2.3-15 2 2 1 v (t ) = e−π (bt) ↔ V ( f ) = e −π ( f / b) b 2 d j 2π f −π ( f / b)2 ( a) v (t ) = −2π b 2te −π ( bt ) ↔ e dt b 2 1 d f −π ( f / b )2 (b ) te− π (bt) ↔ V( f ) = e − j 2π df jb

Both results are equivalent to bte −π ( bt ) ↔ − jf e−π ( f / b) 2

2.4-1 y (t ) = 0

2

t