Sistemas de comunicatcon carlson

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 ELECTRICALCOMMUNICATION, 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 classroomuse 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 fordistance learning. www.mhhe.com

Chapter 2 2.1-1

cn =
2.1-2

Ae jφ T0

 Ae jφ n = m e j2π ( m−n )f 0t dt = Ae jφ sinc( m − n ) =  ∫− T0 / 2 0 otherwise
T0 / 2

c0 v (t ) = 0 cn = 2 T0

∫

T0 / 4

0

A cos 1 2A/π 0

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

n cn
arg cn

0 0

2 0

3 2 A / 3π

4 0

5 2 A / 5π 0

6 0

7 2 A / 7π±180°

±180°

2.1-3

c0 = v (t ) = A / 2 cn = 2 T0

∫

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 

n
cn
arg cn

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

2.1-4

c0 =

2 T0

∫

T0 / 2

0

A cos

2π t =0 T0

(cont.)

2-1

2 cn = T0 =
2.1-5

∫

T0 / 2

0

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

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

c0 = v (t ) = 0 cn = − j 2 T0

∫

T0 / 2

0

A sin 2 0

2π nt A dt = − j (1 − cos π n ) T0 πn 3 2 A / 3π 4 5 2 A / 5π

n cn
arg cn 2.1-6

1 2A/π

−90°

−90°

−90°

c0 = v(t ) = 0
2 cn = − j T0 = −j
2.1-7cn =
T0 1  T0 / 2 − jnω t − jnω t ∫0 v ( t) e 0 dt + ∫T0 / 2 v(t )e 0 dt ]   T0

∫

T0 / 2

0

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

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

where

∫

T0

T0 / 2

v(t )e − jnω0 t dt = ∫

T0 / 2

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

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

= −e jnπ ∫

v (t )e − jnω0 t dt

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 ∞

where

1 = 4 f0 τ 1 A2 f > P= τ 16

1 + 2sinc2 1 + 2sinc2 1 + 2sinc2 3  = 0.23 A2  4 2 4   2 2 A  2 1 2 1 2 3 2 5 23 2 7 2 f > P= 1 + 2sinc 4 + 2sinc 2 + 2sinc 4 + 2sinc 4 + 2sinc 2 + 2sinc 4 = 0.24 A τ 16   2 1 A  2 1 2 1 2 f > P= 1 + 2sinc 4 + 2sinc 2  = 0.21A 2τ 16   2.1-9  0  cn =  2 2  π n    1 a) P = T0
T0 / 2

n even n odd
2

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

∫

T0 / 2

0 2

 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  πn 
1 a) P = T0

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

(cont.) 2-3

b) v′(t ) =

4 4 4 cos ( ω0t − 90° ) + cos ( 3ω0t − 90° )...