Proakis Fundamentals of Communication Systems 2nd Edition Solution

SOLUTIONS MANUAL Communication Systems EngineeringSecond EditionJohn G. Proakis Masoud

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SOLUTIONS MANUAL Communication Systems Engineering Second Edition John G. Proakis Masoud Salehi Prepared by Evangelos Zervas Upper Saddle River, New Jersey 07458 Publisher: Tom Robbins Editorial Assistant: Jody McDonnell Executive Managing Editor: Vince O'Brien Managing Editor: David A. George Production Editor: Barbara A. Till Composition: PreTEX, Inc. Supplement Cover Manager: Paul Gourhan Supplement Cover Design: PM Workshop Inc. Manufacturing Buyer: Ilene Kahn c 2002 Prentice Hall
by Prentice-Hall, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. No part of this book may be reproduced in any form or by any means, without permission in writing from the publisher. The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Printed in the United States of America 10 9 ISBN Pearson Pearson Pearson Pearson Pearson Pearson Pearson Pearson Pearson 8 7 6 5 4 3 2 1 0-13-061974-6 Education Ltd., London Education Australia Pty. Ltd., Sydney Education Singapore, Pte. Ltd. Education North Asia Ltd., Hong Kong Education Canada, Inc., Toronto Educacı̀on de Mexico, S.A. de C.V. Education—Japan, Tokyo Education Malaysia, Pte. Ltd. Education, Upper Saddle River, New Jersey Contents Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 iii Chapter 2 Problem 2.1 1) 2
2
∞  N     = αi φi (t) dt x(t) −  −∞  i=1   
∞ N N   x(t) − αi φi (t) x∗ (t) − α∗ φ∗ (t) dt = j j −∞
∞ = −∞ + i=1 |x(t)|2 dt − αi αj∗ i=1 j=1 = −∞ N 
∞ αi i=1 N  N 
∞ j=1
∞ −∞ |x(t)|2 dt + N  −∞ φi (t)x∗ (t)dt − N  αj∗ j=1
∞ −∞ φ∗j (t)x(t)dt φi (t)φ∗j dt |αi |2 − i=1 N 
∞ αi −∞ i=1 φi (t)x∗ (t)dt − N  αj∗ j=1
∞ −∞ φ∗j (t)x(t)dt Completing the square in terms of αi we obtain
2 =
∞ −∞ |x(t)|2 dt − 2 2
∞ N 
∞ N      ∗ ∗   + αi −  φ (t)x(t)dt φ (t)x(t)dt i i     i=1 −∞ i=1 −∞ The first two terms are independent of α's and the last term is always positive. Therefore the minimum is achieved for
∞ αi = φ∗i (t)x(t)dt −∞ which causes the last term to vanish. 2) With this choice of αi 's 2

∞ = −∞
∞ = −∞ 2 N 
∞   ∗  |x(t)| dt − φi (t)x(t)dt  2 |x(t)|2 dt − i=1 N  −∞ |αi |2 i=1 Problem 2.2 1) The signal x1 (t) is periodic with period T0 = 2. Thus

n 1 1 1 1 Λ(t)e−j2π 2 t dt = Λ(t)e−jπnt dt 2 −1 2 −1

1 0 1 1 = (t + 1)e−jπnt dt + (−t + 1)e−jπnt dt 2 −1 2 0   1 j −jπnt 1 −jπnt 0 j −jπnt 0 te = + 2 2e  + 2πn e  2 πn π n −1 −1 1 1  1 j −jπnt 1 j −jπnt  te e − + 2 2 e−jπnt  +  2 πn π n 2πn 0 0 1 1 1 − (ejπn + e−jπn ) = 2 2 (1 − cos(πn)) π 2 n2 2π 2 n2 π n x1,n = 1 When n = 0 then x1,0 Thus x1 (t) = 1 = 2
1 −1 Λ(t)dt = 1 2 ∞  1 1 +2 (1 − cos(πn)) cos(πnt) 2 2 π n2 n=1 2) x2 (t) = 1. It follows then that x2,0 = 1 and x2,n = 0, ∀n = 0. 3) The signal is periodic with period T0 = 1. Thus x3,n = = = 1 T0
T0
1 et e−j2πnt dt = 0 e(−j2πn+1)t dt 0 1  e(−j2πn+1) − 1 1 e(−j2πn+1)t  = −j2πn + 1 −j2πn + 1 0 e−1 e−1 =√ (1 + j2πn) 1 − j2πn 1 + 4π 2 n2 4) The signal cos(t) is periodic with period T1 = 2π whereas cos(2.5t) is periodic with period T2 = 0.8π. It follows then that cos(t) + cos(2.5t) is periodic with period T = 4π. The trigonometric Fourier series of the even signal cos(t) + cos(2.5t) is cos(t) + cos(2.5t) = = ∞  αn cos(2π n=1 ∞  n t) T0 n αn cos( t) 2 n=1 By equating the coefficients of cos( n2 t) of both sides we observe that an = 0 for all n unless n = 2, 5 in which case a2 = a5 = 1. Hence x4,2 = x4,5 = 12 and x4,n = 0 for all other values of n. 5) The signal x5 (t) is periodic with period T0 = 1. For n = 0
1 x5,0 = 0  1 1 1 (−t + 1)dt = (− t2 + t) = 2 2 0 For n = 0
1 x5,n = (−t + 1)e−j2πnt dt 0   1 1 j −j2πnt 1 j te−j2πnt + 2 2 e−j2πnt  + e  2πn 4π n 2πn 0 0 j = − 2πn = − Thus, x5 (t) = ∞ 1 1  + sin 2πnt 2 n=1 πn 6) The signal x6 (t) is periodic with period T0 = 2T . We can write x6 (t) as x6 (t) = ∞  δ(t − n2T ) − n=−∞ ∞  n=−∞ 2 δ(t − T − n2T ) = = ∞ ∞ n n 1  1  ejπ T t − ejπ T (t−T ) 2T n=−∞ 2T n=−∞ ∞  n 1 (1 − e−jπn )ej2π 2T t 2T n=−∞ However, this is the Fourier series expansion of x6 (t) and we identify x6,n as 1 1 (1 − e−jπn ) = (1 − (−1)n ) = 2T 2T x6,n = 0 n even n odd 1 T 7) The signal is periodic with period T . Thus,
x7,n = = T n 1 2  δ (t)e−j2π T t dt T T −2  1 j2πn d −j2π n t  T = (−1) e  T dt T2 t=0 8) The signal x8 (t) is real even and periodic with period T0 =
x8,n = 2f0
= f0 1 4f0 0 − 4f1 Hence, x8,n = a8,n /2 or cos(2πf0 t) cos(2πn2f0 t)dt − 4f1 1 4f0 1 2f0 .
cos(2πf0 (1 + 2n)t)dt + f0 0 1 4f0 cos(2πf0 (1 − 2n)t)dt − 4f1 0 1 1 1 1 4f 4f sin(2πf0 (1 − 2n)t)| 10 sin(2πf0 (1 + 2n)t)| 10 + 2π(1 + 2n) 2π(1 − 2n) 4f0 4f0  n 1 1 (−1) + π (1 + 2n) (1 − 2n) = = 9) The signal x9 (t) = cos(2πf0 t) + | cos(2πf0 t)| is even and periodic with period T0 = 1/f0 . It is equal to 2 cos(2πf0 t) in the interval [− 4f10 , 4f10 ] and zero in the interval [ 4f10 , 4f30 ]. Thus
x9,n = 2f0
= f0 1 4f0 − 4f1 cos(2πf0 t) cos(2πnf0 t)dt 0 1 4f0 − 4f1
cos(2πf0 (1 + n)t)dt + f0 0 = = 1 4f0 − 4f1 cos(2πf0 (1 − n)t)dt 0 1 1 1 1 4f 4f sin(2πf0 (1 + n)t)| 10 + sin(2πf0 (1 − n)t)| 10 2π(1 + n) 2π(1 − n) 4f0 4f0 π 1 π 1 sin( (1 + n)) + sin( (1 − n)) π(1 + n) 2 π(1 − n) 2 Thus x9,n is zero for odd values of n unless n = ±1 in which case x9,±1 = (n = 2l) then  1 1 (−1)l + x9,2l = π 1 + 2l 1 − 2l 3 1 2. When n is even Problem 2.3 It follows directly from the uniqueness of the decomposition of a real signal in an even and odd part. Nevertheless for a real periodic signal  x(t) = ∞ a0  n n + an cos(2π t) + bn sin(2π t) 2 T0 T0 n=1 The even part of x(t) is x(t) + x(−t)  2 ∞  n n 1 a0 + an (cos(2π t) + cos(−2π t)) 2 T0 T0 n=1 xe (t) = = n n t) + sin(−2π t)) T0 T0 ∞  a0 n + an cos(2π t) 2 T 0 n=1 +bn (sin(2π = The last is true since cos(θ) is even so that cos(θ) + cos(−θ) = 2 cos θ whereas the oddness of sin(θ) provides sin(θ) + sin(−θ) = sin(θ) − sin(θ) = 0. The odd part of x(t) is x(t) − x(−t) 2 ∞ xo (t) =  − bn sin(2π n=1 n t) T0 Problem 2.4 a) The signal is periodic with period T . Thus xn = 1 T
T e−t e−j2π T t dt = n 0 1 T
T e−(j2π T +1)t dt n 0 T    n 1 1 −(j2π T +1)t   = − e =− e−(j2πn+T ) − 1 n  T j2π T + 1 j2πn + T 0 1 T − j2πn [1 − e−T ] = 2 = [1 − e−T ] j2πn + T T + 4π 2 n2  If we write xn = an −jbn 2 we obtain the trigonometric Fourier series expansion coefficients as an = 2T [1 − e−T ], T 2 + 4π 2 n2 bn = 4πn [1 − e−T ] T 2 + 4π 2 n2 b) The signal is periodic with period 2T . Since the signal is odd we obtain x0 = 0. For n = 0
xn = = = = = T n 1 1 x(t)e−j2π 2T t dt = 2T −T 2T
T n 1 te−jπ T t dt 2 2T −T 1 2T 2  
T t −j2π n t 2T dt e −T T  jT −jπ n t T 2 −jπ n t T T + T te e  πn π 2 n2 −T 1 jT 2 −jπn T 2 −jπn jT 2 jπn T 2 jπn e e + e + − e 2T 2 πn π 2 n2 πn π 2 n2 j (−1)n πn 4  The trigonometric Fourier series expansion coefficients are: bn = (−1)n+1 an = 0, 2 πn c) The signal is periodic with period T . For n = 0 1 x0 = T
T 2 3 2 x(t)dt = − T2 If n = 0 then
T 2 xn = 1 T = 1 T = j −j2π n t  2 j −j2π n t  4 T T e e +   T 2πn 2πn − T2 −4 − T2 T 2
− T2 n e−j2π T t dt + n 1 T
T 4 − T4 T e−j2π T t dt n T  n n j  −jπn e − ejπn + e−jπ 2 − e−jπ 2 2πn n 1 n 1 sin(π ) = sinc( ) πn 2 2 2 = = Note that xn = 0 for n even and x2l+1 = coefficients are: a0 = 3, x(t)e−j2π T t dt , a2l = 0, 1 l π(2l+1) (−1) . , a2l+1 = The trigonometric Fourier series expansion 2 (−1)l , π(2l + 1) , bn = 0, ∀n d) The signal is periodic with period T . For n = 0
T 1 x0 = T x(t)dt = 0 2 3 If n = 0 then xn = 1 T
T 1 + T = 3 T2 n 0
 3 − 2 T 2T 3 T 3
T 1 3 3 −j2π n t T dt te T 0 T
n n 1 T 3 −j2π T t e dt + (− t + 3)e−j2π T t dt 2T T 3 T x(t)e−j2π T t dt =  T jT −j2π n t T 2 −j2π n t  3 T + T te e  2πn 4π 2 n2 0   jT −j2π n t T 2 −j2π n t T T + T te e  2T 2πn 4π 2 n2 3  2T  j −j2π n t  3 3 jT −j2π n t T T T e + + e T  2T 2πn T 2πn 3 3 = 3 2πn ) − 1] [cos( 2 2 2π n 3 The trigonometric Fourier series expansion coefficients are: 4 a0 = , 3 an = 3 π 2 n2 [cos( 5 2πn ) − 1], 3 bn = 0, ∀n e) The signal is periodic with period T . Since the signal is odd x0 = a0 = 0. For n = 0
1 T xn = − T2 1 T + 4 T2 = T 2
 1 − T
1 x(t)dt = T T 4 − T4 T 4 n 4 −j2π n t 1 T dt + te T T
T 2 T 4 e−j2π T t dt n  T jT −j2π n t T 2 −j2π n t  4 T + T te e  T 2πn 4π 2 n2 −4   −e−j2π T t dt − T2 jT −j2π n t − 4 1 T + e  T 2πn T −2 T T jT −j2π n t  2 T e T 2πn 4   2 sin( πn j j n 2 ) (−1)n − = (−1)n − sinc( ) πn πn πn 2 = For n even, sinc( n2 ) = 0 and xn = an = 0, ∀n, j πn . The trigonometric Fourier series expansion coefficients are: 1 − πl bn = 2 π(2l+1) [1 + n = 2l n = 2l + 1 2(−1)l π(2l+1) ] f ) The signal is periodic with period T . For n = 0 x0 = 1 T
T 3 − T3 x(t)dt = 1 For n = 0 xn = = 1 T
0 n 3 1 ( t + 2)e−j2π T t dt + T − T3 T 3 T2  3 − 2 T
0 T 3 n 3 (− t + 2)e−j2π T t dt T  jT −j2π n t T 2 −j2π n t 0 T + T te e  T 2πn 4π 2 n2 −3   T jT −j2π n t T 2 −j2π n t  3 T + T te e  2πn 4π 2 n2 0  + T 2 jT −j2π n t 0 2 jT −j2π n t  3 T T + e e   T 2πn T 2πn − T3 0  = 3 1 1 2πn 2πn ) + ) − cos( sin( 2 2 π n 2 πn 3 3 The trigonometric Fourier series expansion coefficients are: a0 = 2,  2πn 2πn 1 1 − cos( ) + sin( ) , 2 3 πn 3 3 an = 2 2 2 π n Problem 2.5 1) The signal y(t) = x(t − t0 ) is periodic with period T = T0 .
1 α+T0 −j2π Tn t 0 dt x(t − t0 )e T0 α
1 α−t0 +T0 −j2π Tn 0 (v + t0 )dv = x(v)e T0 α−t0
α−t0 +T0 −j2π Tn t0 1 −j2π Tn v 0 0 dv = e x(v)e T0 α−t0 yn = −j2π Tn t0 = xn e 0 6 bn = 0, ∀n where we used the change of variables v = t − t0 2) For y(t) to be periodic there must exist T such that y(t + mT ) = y(t). But y(t + T ) = x(t + T )ej2πf0 t ej2πf0 T so that y(t) is periodic if T = T0 (the period of x(t)) and f0 T = k for some k in Z. In this case
yn = = 1 α+T0 −j2π Tn t j2πf0 t 0 e x(t)e dt T0 α
(n−k) 1 α+T0 −j2π T t 0 x(t)e dt = xn−k T0 α 3) The signal y(t) is periodic with period T = T0 /α.
yn = =
T0 n 1 β+T α β+ α −j2π nα t T0 dt y(t)e−j2π T t dt = x(αt)e T β T0 β
1 βα+T0 −j2π Tn v 0 dv = xn x(v)e T0 βα where we used the change of variables v = αt. 4)
yn 1 α+T0  −j2π Tn t 0 dt = x (t)e T0 α α+T0
1 α+T0 n −j2π Tn t 1 −j2π Tn t  0 0 dt x(t)e − (−j2π )e =  T0 T0 α T0 α
n 1 α+T0 n −j2π Tn t 0 dt = j2π = j2π x(t)e xn T0 T0 α T0 Problem 2.6 1 T0
α+T0 1 T0 x(t)y ∗ (t)dt = α
α+T0  ∞ α ∞  = ∞  ∞  j2πn t T0 n=−∞ − j2πm t T ∗ ym e 0 dt m=−∞ ∗ xn ym n=−∞ m=−∞ ∞ ∞   = xn e 1 T0
α+T0 e j2π(n−m) t T0 dt α ∞  ∗ xn ym δmn = n=−∞ m=−∞ xn yn∗ n=−∞ Problem 2.7 Using the results of Problem 2.6 we obtain 1 T0
α+T0 ∞  x(t)x∗ (t)dt = α |xn |2 n=−∞ Since the signal has finite power Thus, ∞ n=−∞ |xn 1 T0 |2
α+T0 α |x(t)|2 dt = K < ∞ = K < ∞. The last implies that |xn | → 0 as n → ∞. To see this write ∞  n=−∞ |xn |2 = −M  |xn |2 + n=−∞ M  n=−M 7 |xn |2 + ∞  n=M |xn |2

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Proakis Fundamentals of Communication Systems 2nd Edition Solution

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