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: PreT E X, 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. 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Pearson Education, Upper Saddle River, New Jersey Contents Chapter 2 1 Chapter 3 42 Chapter 4 71 Chapter 5 114 Chapter 6 128 Chapter 7 161 Chapter 8 213 Chapter 9 250 Chapter 10 283 iii Chapter 2 Problem 2.1 1)  2 =  ∞ −∞      x(t) − N  i=1 α i φ i (t)      2 dt =  ∞ −∞  x(t) − N  i=1 α i φ i (t)    x ∗ (t) − N  j=1 α ∗ j φ ∗ j (t)   dt =  ∞ −∞ |x(t)| 2 dt − N  i=1 α i  ∞ −∞ φ i (t)x ∗ (t)dt − N  j=1 α ∗ j  ∞ −∞ φ ∗ j (t)x(t)dt + N  i=1 N  j=1 α i α ∗ j  ∞ −∞ φ i (t)φ ∗ j dt =  ∞ −∞ |x(t)| 2 dt + N  i=1 |α i | 2 − N  i=1 α i  ∞ −∞ φ i (t)x ∗ (t)dt − N  j=1 α ∗ j  ∞ −∞ φ ∗ j (t)x(t)dt Completing the square in terms of α i we obtain  2 =  ∞ −∞ |x(t)| 2 dt − N  i=1      ∞ −∞ φ ∗ i (t)x(t)dt     2 + N  i=1     α i −  ∞ −∞ φ ∗ i (t)x(t)dt     2 The first two terms are independent of α’s and the last term is always positive. Therefore the minimum is achieved for α i =  ∞ −∞ φ ∗ i (t)x(t)dt which causes the last term to vanish. 2) With this choice of α i ’s  2 =  ∞ −∞ |x(t)| 2 dt − N  i=1      ∞ −∞ φ ∗ i (t)x(t)dt     2 =  ∞ −∞ |x(t)| 2 dt − N  i=1 |α i | 2 Problem 2.2 1) The signal x 1 (t) is periodic with period T 0 =2. Thus x 1,n = 1 2  1 −1 Λ(t)e −j2π n 2 t dt = 1 2  1 −1 Λ(t)e −jπnt dt = 1 2  0 −1 (t +1)e −jπnt dt + 1 2  1 0 (−t +1)e −jπnt dt = 1 2  j πn te −jπnt + 1 π 2 n 2 e −jπnt      0 −1 + j 2πn e −jπnt     0 −1 − 1 2  j πn te −jπnt + 1 π 2 n 2 e −jπnt      1 0 + j 2πn e −jπnt     1 0 1 π 2 n 2 − 1 2π 2 n 2 (e jπn + e −jπn )= 1 π 2 n 2 (1 − cos(πn)) 1 When n = 0 then x 1,0 = 1 2  1 −1 Λ(t)dt = 1 2 Thus x 1 (t)= 1 2 +2 ∞  n=1 1 π 2 n 2 (1 − cos(πn)) cos(πnt) 2) x 2 (t) = 1. It follows then that x 2,0 = 1 and x 2,n =0, ∀n =0. 3) The signal is periodic with period T 0 =1. Thus x 3,n = 1 T 0  T 0 0 e t e −j2πnt dt =  1 0 e (−j2πn+1)t dt = 1 −j2πn +1 e (−j2πn+1)t     1 0 = e (−j2πn+1) − 1 −j2πn +1 = e − 1 1 − j2πn = e − 1 √ 1+4π 2 n 2 (1 + j2πn) 4) The signal cos(t) is periodic with period T 1 =2π whereas cos(2.5t) is periodic with period T 2 =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)= ∞  n=1 α n cos(2π n T 0 t) = ∞  n=1 α n cos( n 2 t) By equating the coefficients of cos( n 2 t) of both sides we observe that a n = 0 for all n unless n =2, 5 in which case a 2 = a 5 = 1. Hence x 4,2 = x 4,5 = 1 2 and x 4,n = 0 for all other values of n. 5) The signal x 5 (t) is periodic with period T 0 =1. Forn =0 x 5,0 =  1 0 (−t +1)dt =(− 1 2 t 2 + t)     1 0 = 1 2 For n =0 x 5,n =  1 0 (−t +1)e −j2πnt dt = −  j 2πn te −j2πnt + 1 4π 2 n 2 e −j2πnt      1 0 + j 2πn e −j2πnt     1 0 = − j 2πn Thus, x 5 (t)= 1 2 + ∞  n=1 1 πn sin 2πnt 6) The signal x 6 (t) is periodic with period T 0 =2T . We can write x 6 (t)as x 6 (t)= ∞  n=−∞ δ(t − n2T ) − ∞  n=−∞ δ(t − T −n2T ) 2 = 1 2T ∞  n=−∞ e jπ n T t − 1 2T ∞  n=−∞ e jπ n T (t−T ) = ∞  n=−∞ 1 2T (1 − e −jπn )e j2π n 2T t However, this is the Fourier series expansion of x 6 (t) and we identify x 6,n as x 6,n = 1 2T (1 − e −jπn )= 1 2T (1 − (−1) n )=  0 n even 1 T n odd 7) The signal is periodic with period T . Thus, x 7,n = 1 T  T 2 − T 2 δ  (t)e −j2π n T t dt = 1 T (−1) d dt e −j2π n T t     t=0 = j2πn T 2 8) The signal x 8 (t) is real even and periodic with period T 0 = 1 2f 0 . Hence, x 8,n = a 8,n /2or x 8,n =2f 0  1 4f 0 − 1 4f 0 cos(2πf 0 t) cos(2πn2f 0 t)dt = f 0  1 4f 0 − 1 4f 0 cos(2πf 0 (1+2n)t)dt + f 0  1 4f 0 − 1 4f 0 cos(2πf 0 (1 − 2n)t)dt = 1 2π(1+2n) sin(2πf 0 (1+2n)t)| 1 4f 0 1 4f 0 + 1 2π(1 − 2n) sin(2πf 0 (1 − 2n)t)| 1 4f 0 1 4f 0 = (−1) n π  1 (1+2n) + 1 (1 − 2n)  9) The signal x 9 (t) = cos(2πf 0 t)+|cos(2πf 0 t)| is even and periodic with period T 0 =1/f 0 .Itis equal to 2 cos(2πf 0 t) in the interval [− 1 4f 0 , 1 4f 0 ] and zero in the interval [ 1 4f 0 , 3 4f 0 ]. Thus x 9,n =2f 0  1 4f 0 − 1 4f 0 cos(2πf 0 t) cos(2πnf 0 t)dt = f 0  1 4f 0 − 1 4f 0 cos(2πf 0 (1 + n)t)dt + f 0  1 4f 0 − 1 4f 0 cos(2πf 0 (1 − n)t)dt = 1 2π(1 + n) sin(2πf 0 (1 + n)t)| 1 4f 0 1 4f 0 + 1 2π(1 − n) sin(2πf 0 (1 − n)t)| 1 4f 0 1 4f 0 = 1 π(1 + n) sin( π 2 (1 + n)) + 1 π(1 − n) sin( π 2 (1 − n)) Thus x 9,n is zero for odd values of n unless n = ±1 in which case x 9,±1 = 1 2 . When n is even (n =2l) then x 9,2l = (−1) l π  1 1+2l + 1 1 − 2l  3 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)= a 0 2 + ∞  n=1  a n cos(2π n T 0 t)+b n sin(2π n T 0 t)  The even part of x(t)is x e (t)= x(t)+x(−t) 2 = 1 2  a 0 + ∞  n=1 a n (cos(2π n T 0 t) + cos(−2π n T 0 t)) +b n (sin(2π n T 0 t) + sin(−2π n T 0 t))  = a 0 2 + ∞  n=1 a n cos(2π n T 0 t) The last is true since cos(θ) is even so that cos(θ) + cos(−θ)=2cosθ whereas the oddness of sin(θ) provides sin(θ) + sin(−θ) = sin(θ) − sin(θ)=0. The odd part of x(t)is x o (t)= x(t) − x(−t) 2 − ∞  n=1 b n sin(2π n T 0 t) Problem 2.4 a) The signal is periodic with period T .Thus x n = 1 T  T 0 e −t e −j2π n T t dt = 1 T  T 0 e −(j2π n T +1)t dt = − 1 T  j2π n T +1  e −(j2π n T +1)t     T 0 = − 1 j2πn + T  e −(j2πn+T) − 1  = 1 j2πn + T [1 − e −T ]= T −j2πn T 2 +4π 2 n 2 [1 − e −T ] If we write x n = a n −jb n 2 we obtain the trigonometric Fourier series expansion coefficients as a n = 2T T 2 +4π 2 n 2 [1 − e −T ],b n = 4πn T 2 +4π 2 n 2 [1 − e −T ] b) The signal is periodic with period 2T . Since the signal is odd we obtain x 0 =0. Forn =0 x n = 1 2T  T −T x(t)e −j2π n 2T t dt = 1 2T  T −T t T e −j2π n 2T t dt = 1 2T 2  T −T te −jπ n T t dt = 1 2T 2  jT πn te −jπ n T t + T 2 π 2 n 2 e −jπ n T t      T −T = 1 2T 2  jT 2 πn e −jπn + T 2 π 2 n 2 e −jπn + jT 2 πn e jπn − T 2 π 2 n 2 e jπn  = j πn (−1) n 4 The trigonometric Fourier series expansion coefficients are: a n =0,b n =(−1) n+1 2 πn c) The signal is periodic with period T .Forn =0 x 0 = 1 T  T 2 − T 2 x(t)dt = 3 2 If n = 0 then x n = 1 T  T 2 − T 2 x(t)e −j2π n T t dt = 1 T  T 2 − T 2 e −j2π n T t dt + 1 T  T 4 − T 4 e −j2π n T t dt = j 2πn e −j2π n T t     T 2 − T 2 + j 2πn e −j2π n T t     T 4 − T 4 = j 2πn  e −jπn − e jπn + e −jπ n 2 − e −jπ n 2  = 1 πn sin(π n 2 )= 1 2 sinc( n 2 ) Note that x n = 0 for n even and x 2l+1 = 1 π(2l+1) (−1) l . The trigonometric Fourier series expansion coefficients are: a 0 =3,,a 2l =0,,a 2l+1 = 2 π(2l +1) (−1) l ,,b n =0, ∀n d) The signal is periodic with period T .Forn =0 x 0 = 1 T  T 0 x(t)dt = 2 3 If n = 0 then x n = 1 T  T 0 x(t)e −j2π n T t dt = 1 T  T 3 0 3 T te −j2π n T t dt + 1 T  2T 3 T 3 e −j2π n T t dt + 1 T  T 2T 3 (− 3 T t +3)e −j2π n T t dt = 3 T 2  jT 2πn te −j2π n T t + T 2 4π 2 n 2 e −j2π n T t      T 3 0 − 3 T 2  jT 2πn te −j2π n T t + T 2 4π 2 n 2 e −j2π n T t      T 2T 3 + j 2πn e −j2π n T t     2T 3 T 3 + 3 T jT 2πn e −j2π n T t     T 2T 3 = 3 2π 2 n 2 [cos( 2πn 3 ) − 1] The trigonometric Fourier series expansion coefficients are: a 0 = 4 3 ,a n = 3 π 2 n 2 [cos( 2πn 3 ) − 1],b n =0, ∀n 5 e) The signal is periodic with period T . Since the signal is odd x 0 = a 0 =0. Forn =0 x n = 1 T  T 2 − T 2 x(t)dt = 1 T  T 4 − T 2 −e −j2π n T t dt + 1 T  T 4 − T 4 4 T te −j2π n T t dt + 1 T  T 2 T 4 e −j2π n T t dt = 4 T 2  jT 2πn te −j2π n T t + T 2 4π 2 n 2 e −j2π n T t      T 4 − T 4 − 1 T  jT 2πn e −j2π n T t      − T 4 − T 2 + 1 T  jT 2πn e −j2π n T t      T 2 T 4 = j πn  (−1) n − 2 sin( πn 2 ) πn  = j πn  (−1) n − sinc( n 2 )  For n even, sinc( n 2 )=0andx n = j πn . The trigonometric Fourier series expansion coefficients are: a n =0, ∀n, b n =  − 1 πl n =2l 2 π(2l+1) [1 + 2(−1) l π(2l+1) ] n =2l +1 f) The signal is periodic with period T .Forn =0 x 0 = 1 T  T 3 − T 3 x(t)dt =1 For n =0 x n = 1 T  0 − T 3 ( 3 T t +2)e −j2π n T t dt + 1 T  T 3 0 (− 3 T t +2)e −j2π n T t dt = 3 T 2  jT 2πn te −j2π n T t + T 2 4π 2 n 2 e −j2π n T t      0 − T 3 − 3 T 2  jT 2πn te −j2π n T t + T 2 4π 2 n 2 e −j2π n T t      T 3 0 + 2 T jT 2πn e −j2π n T t     0 − T 3 + 2 T jT 2πn e −j2π n T t     T 3 0 = 3 π 2 n 2  1 2 − cos( 2πn 3 )  + 1 πn sin( 2πn 3 ) The trigonometric Fourier series expansion coefficients are: a 0 =2,a n =2  3 π 2 n 2  1 2 − cos( 2πn 3 )  + 1 πn sin( 2πn 3 )  ,b n =0, ∀n Problem 2.5 1) The signal y(t)=x(t − t 0 ) is periodic with period T = T 0 . y n = 1 T 0  α+T 0 α x(t − t 0 )e −j2π n T 0 t dt = 1 T 0  α−t 0 +T 0 α−t 0 x(v)e −j2π n T 0 (v + t 0 )dv = e −j2π n T 0 t 0 1 T 0  α−t 0 +T 0 α−t 0 x(v)e −j2π n T 0 v dv = x n e −j2π n T 0 t 0 6 where we used the change of variables v = t − t 0 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 )e j2πf 0 t e j2πf 0 T so that y(t) is periodic if T = T 0 (the period of x(t)) and f 0 T = k for some k in Z. In this case y n = 1 T 0  α+T 0 α x(t)e −j2π n T 0 t e j2πf 0 t dt = 1 T 0  α+T 0 α x(t)e −j2π (n−k) T 0 t dt = x n−k 3) The signal y(t) is periodic with period T = T 0 /α. y n = 1 T  β+T β y(t)e −j2π n T t dt = α T 0  β+ T 0 α β x(αt)e −j2π nα T 0 t dt = 1 T 0  βα+T 0 βα x(v)e −j2π n T 0 v dv = x n where we used the change of variables v = αt. 4) y n = 1 T 0  α+T 0 α x  (t)e −j2π n T 0 t dt = 1 T 0 x(t)e −j2π n T 0 t     α+T 0 α − 1 T 0  α+T 0 α (−j2π n T 0 )e −j2π n T 0 t dt = j2π n T 0 1 T 0  α+T 0 α x(t)e −j2π n T 0 t dt = j2π n T 0 x n Problem 2.6 1 T 0  α+T 0 α x(t)y ∗ (t)dt = 1 T 0  α+T 0 α ∞  n=−∞ x n e j2πn T 0 t ∞  m=−∞ y ∗ m e − j2πm T 0 t dt = ∞  n=−∞ ∞  m=−∞ x n y ∗ m 1 T 0  α+T 0 α e j2π(n−m) T 0 t dt = ∞  n=−∞ ∞  m=−∞ x n y ∗ m δ mn = ∞  n=−∞ x n y ∗ n Problem 2.7 Using the results of Problem 2.6 we obtain 1 T 0  α+T 0 α x(t)x ∗ (t)dt = ∞  n=−∞ |x n | 2 Since the signal has finite power 1 T 0  α+T 0 α |x(t)| 2 dt = K<∞ Thus,  ∞ n=−∞ |x n | 2 = K<∞. The last implies that |x n |→0asn →∞. To see this write ∞  n=−∞ |x n | 2 = −M  n=−∞ |x n | 2 + M  n=−M |x n | 2 + ∞  n=M |x n | 2 7 [...]... 2 βi B2 i=1 1 + 2B 2 2 αi i=1 n 2 βi = i=1 1 2 1 A + B2 = 1 2A2 2B 2 Thus, 1 AB n n αi βi ≤ 1 ⇒ i=1 1 2 n αi βi ≤ 2 αi i=1 i=1 1 2 n 2 βi i=1 Equality holds if αi = kβi , for i = 1, , n ∗ ∗ 2) The second equation is trivial since |xi yi | = |xi ||yi | To see this write xi and yi in polar jθxi jθyi ∗ coordinates as xi = ρxi e and yi = ρyi e Then, |xi yi | = |ρxi ρyi ej(θxi −θyi ) | = ρxi ρyi = |xi . SOLUTIONS MANUAL Communication Systems Engineering Second Edition John G. Proakis Masoud Salehi Prepared by Evangelos Zervas Upper Saddle. Managing Editor: Vince O’Brien Managing Editor: David A. George Production Editor: Barbara A. Till Composition: PreT E X, Inc. Supplement Cover Manager: