solutions manual communication systems engineering proakis j (2002)

www.elsolucionario.net SOLUTIONS MANUAL Second Edition John G Proakis Masoud Salehi Prepared by Evangelos Zervas Upper Saddle River, New Jersey 07458 www.elsolucionario.net Communication Systems Engineering 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 ISBN Pearson Pearson Pearson Pearson Pearson Pearson Pearson Pearson Pearson 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 Educac`ıon de Mexico, S.A de C.V Education—Japan, Tokyo Education Malaysia, Pte Ltd Education, Upper Saddle River, New Jersey www.elsolucionario.net www.elsolucionario.net www.elsolucionario.net Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter .1 42 71 114 128 161 213 250 10 283 iii www.elsolucionario.net Contents www.elsolucionario.net Chapter Problem 2.1 1) ∞ = −∞ x(t) − ∞ αi φi (t) x∗ (t) − x(t) − i=1 ∞ |x(t)|2 dt − N N αi αj∗ i=1 j=1 ∞ αi i=1 + ∞ −∞ αj∗ φ∗j (t) dt N φi (t)x∗ (t)dt − αj∗ j=1 ∞ −∞ φ∗j (t)x(t)dt φi (t)φ∗j dt −∞ N −∞  N j=1 N −∞ =  N −∞ = αi φi (t) dt i=1 ∞ = N ∞ N |x(t)|2 dt + |αi |2 − i=1 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 ∞ = −∞ ∞ N |x(t)|2 dt − i=1 −∞ φ∗i (t)x(t)dt + N αi − i=1 ∞ −∞ φ∗i (t)x(t)dt 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 ∞ = −∞ ∞ = −∞ N |x(t)|2 dt − i=1 N |x(t)|2 dt − ∞ −∞ φ∗i (t)x(t)dt |αi |2 i=1 Problem 2.2 1) The signal x1 (t) is periodic with period T0 = Thus 1 Λ(t)e−jπnt dt −1 −1 1 = (t + 1)e−jπnt dt + (−t + 1)e−jπnt dt −1 −jπnt j −jπnt j −jπnt te e = + 2e + πn π n 2πn −1 −1 1 j −jπnt j −jπnt te e − + 2 e−jπnt + πn π n 2πn 0 1 − (ejπn + e−jπn ) = 2 (1 − cos(πn)) π n2 2π n2 π n x1,n = 2 Λ(t)e−j2π t dt = n www.elsolucionario.net www.elsolucionario.net When n = then x1,0 = Thus x1 (t) = −1 Λ(t)dt = ∞ 1 +2 (1 − cos(πn)) cos(πnt) 2 π n2 n=1 2) x2 (t) = It follows then that x2,0 = and x2,n = 0, ∀n = 3) The signal is periodic with period T0 = Thus = = T0 T0 et e−j2πnt dt = e(−j2πn+1)t dt e(−j2πn+1) − 1 e(−j2πn+1)t = −j2πn + −j2πn + e−1 e−1 =√ (1 + j2πn) − j2πn + 4π 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) n=1 By equating the coefficients of cos( n2 t) of both sides we observe that an = for all n unless n = 2, in which case a2 = a5 = Hence x4,2 = x4,5 = 12 and x4,n = for all other values of n 5) The signal x5 (t) is periodic with period T0 = For n = x5,0 = (−t + 1)dt = (− t2 + t) = For n = x5,n = (−t + 1)e−j2πnt dt j te−j2πnt + 2 e−j2πnt 2πn 4π n j = − 2πn = − Thus, x5 (t) = + j −j2πnt e 2πn ∞ 1 + sin 2πnt 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=−∞ δ(t − T − n2T ) www.elsolucionario.net x3,n = www.elsolucionario.net = 2T ∞ n ejπ T t − n=−∞ ∞ = 2T ∞ n ejπ T (t−T ) n=−∞ n (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 − e−jπn ) = (1 − (−1)n ) = 2T 2T x6,n = n even n odd T 7) The signal is periodic with period T Thus, = n δ (t)e−j2π T t dt T T −2 n j2πn d = (−1) e−j2π T t T dt T2 t=0 8) The signal x8 (t) is real even and periodic with period T0 = 4f0 x8,n = 2f0 2f0 Hence, x8,n = a8,n /2 or cos(2πf0 t) cos(2πn2f0 t)dt − 4f1 = f0 4f0 − 4f1 cos(2πf0 (1 + 2n)t)dt + f0 4f0 cos(2πf0 (1 − 2n)t)dt − 4f1 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) + 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 cos(2πf0 t) in the interval [− 4f10 , 4f10 ] and zero in the interval [ 4f10 , 4f30 ] Thus x9,n = 2f0 4f0 − 4f1 cos(2πf0 t) cos(2πnf0 t)dt = f0 4f0 − 4f1 cos(2πf0 (1 + n)t)dt + f0 = = 4f0 − 4f1 cos(2πf0 (1 − n)t)dt 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 π sin( (1 + n)) + sin( (1 − n)) π(1 + n) π(1 − n) Thus x9,n is zero for odd values of n unless n = ±1 in which case x9,±1 = (n = 2l) then 1 (−1)l + x9,2l = π + 2l − 2l When n is even www.elsolucionario.net T x7,n = www.elsolucionario.net 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) T0 T0 n=1 The even part of x(t) is xe (t) = = x(t) + x(−t) ∞ n n a0 + an (cos(2π t) + cos(−2π t)) T0 T0 n=1 n n t) + sin(−2π t)) T0 T0 ∞ a0 n + an cos(2π t) T n=1 = The last is true since cos(θ) is even so that cos(θ) + cos(−θ) = cos θ whereas the oddness of sin(θ) provides sin(θ) + sin(−θ) = sin(θ) − sin(θ) = The odd part of x(t) is x(t) − x(−t) ∞ xo (t) = − bn sin(2π n=1 n t) T0 Problem 2.4 a) The signal is periodic with period T Thus xn = T T e−t e−j2π T t dt = n T T e−(j2π T +1)t dt n T n 1 e−(j2π T +1)t = − e−(j2πn+T ) − n T j2π T + j2πn + T T − j2πn [1 − e−T ] = [1 − e−T ] j2πn + T T + 4π n2 = − = If we write xn = an −jbn we obtain the trigonometric Fourier series expansion coefficients as an = 2T [1 − e−T ], T + 4π n2 bn = 4πn [1 − e−T ] T + 4π n2 b) The signal is periodic with period 2T Since the signal is odd we obtain x0 = For n = xn = = = = = 2T 2T 2T T −T T x(t)e−j2π 2T t dt = −T n n −jπ T t te 2T T −T dt jT −jπ n t T −jπ n t T + T te e πn π n2 jT t −j2π n t 2T dt e T T2 T −T jT jπn T jπn −jπn −jπn e e + e + − e 2T πn π n2 πn π n2 j (−1)n πn www.elsolucionario.net +bn (sin(2π www.elsolucionario.net The trigonometric Fourier series expansion coefficients are: bn = (−1)n+1 an = 0, πn c) The signal is periodic with period T For n = T x0 = T x(t)dt = − T2 If n = then xn = = T = j −j2π n t T e 2πn − T2 T − T2 n e−j2π T t dt + n T T T − T4 e−j2π T t dt n j −j2π n t T e + 2πn − T2 T − T4 n n j e−jπn − ejπn + e−jπ − e−jπ 2πn n n sin(π ) = sinc( ) πn 2 = = Note that xn = for n even and x2l+1 = coefficients are: a0 = 3, x(t)e−j2π T t dt , a2l = 0, l π(2l+1) (−1) , a2l+1 = The trigonometric Fourier series expansion (−1)l , π(2l + 1) , bn = 0, ∀n d) The signal is periodic with period T For n = x0 = T T x(t)dt = If n = then xn = T T = T2 − + = n + T T2 T n e−j2π T t dt + T x(t)e−j2π T t dt = 2T T T 3 −j2π n t T dt te T T n (− t + 3)e−j2π T t dt 2T T T jT −j2π n t T −j2π n t T + T te e 2πn 4π n2 T2 n jT −j2π n t T + te e−j2π T t 2 2πn 4π n j −j2π n t T e 2πn 2T T + jT −j2π n t T e T 2πn T 2T T 2T 3 2πn ) − 1] [cos( 2 2π n The trigonometric Fourier series expansion coefficients are: a0 = , an = π n2 [cos( 2πn ) − 1], bn = 0, ∀n www.elsolucionario.net T T www.elsolucionario.net e) The signal is periodic with period T Since the signal is odd x0 = a0 = For n = T T xn = − T2 T T + T x(t)dt = T − T4 − T2 −e−j2π T t dt n T −j2π n t T dt + te T T T e−j2π T t dt n T2 jT −j2π n t T −j2π n t T + T te e 2πn 4π n2 − T jT −j2π n t T e 2πn = − T4 − T2 T − T4 jT −j2π n t T e 2πn + T T T sin( πn j j n ) (−1)n − = (−1)n − sinc( ) πn πn πn For n even, sinc( n2 ) = and xn = j πn an = 0, ∀n, The trigonometric Fourier series expansion coefficients are: − πl bn = π(2l+1) [1 + n = 2l n = 2l + 2(−1)l π(2l+1) ] f ) The signal is periodic with period T For n = T x0 = T − T3 x(t)dt = For n = xn = = n ( t + 2)e−j2π T t dt + T − T3 T − T3 T jT −j2π n t T −j2π n t T + T te e 2πn 4π n2 − T = n (− t + 2)e−j2π T t dt T jT −j2π n t T −j2π n t T + T te e 2πn 4π n2 T2 + T T jT −j2π n t T e T 2πn − T3 + jT −j2π n t T e T 2πn T 3 1 2πn 2πn ) + ) − cos( sin( 2 π n πn 3 The trigonometric Fourier series expansion coefficients are: a0 = 2, an = 2πn 2πn 1 − cos( ) + sin( ) , πn 3 π n2 Problem 2.5 1) The signal y(t) = x(t − t0 ) is periodic with period T = T0 yn = = α+T0 T0 T0 −j2π Tn t x(t − t0 )e α α−t0 +T0 −j2π Tn x(v)e dt (v + t0 )dv α−t0 −j2π Tn t0 = e T0 α−t0 +T0 α−t0 −j2π Tn t0 = xn e −j2π Tn v x(v)e dv bn = 0, ∀n www.elsolucionario.net = www.elsolucionario.net ✉ ✉ ✉ ✉ ✉ ✉ ✉ ✉ -7 -5 -3 -1 ✦€€ ✦✦ ✦ €€ €€ ✦✦ ✉ ✉ -7 -3 € € ✉ ✉ ✉ ✉ -5 -1 ✉ ◗ 0   ◗◗   ◗ 0★★❝❝1 ★ ✉ ❝ ✉ ✉ ✉ ✉ -7 -3 ✉ -1 ✉ ✉ ✉ -5 -7,1 ✲✈ ✈ ❍ -5,3 ❍❍   ✒   ❍ -5,3  ❍ ❥ ❍ ✈ ❍  -7,1 ✟ ✯✈ ✟ ❅❍❍✟   ❅✟❍-3,5 ✟✟❅-1,7 ❍   ❥ ❍ ✈ ✯✈ ✟ ✟ ✟ -1,7 ✟❅ ✟✟-3,5 ❅ ❘✈ ❅ ✲ ✈ The operation of the Viterbi algorithm for the decoding of the sequence {−.2, 1.1, 6, 4, −3, −4.8, 3.3} is shown schematically in the next figure It has been assumed that we start at the all zero state and that a sequence of zeros terminates the input bit stream in order to clear the encoder The numbers at the nodes indicate the minimum Euclidean distance, and the branches have been marked with the decoded transmitted symbol The paths that have been purged are marked with an X -.2 1.1 1.44 1.45 26.45 -3 7.05 -4.8 15.45 3.3 11.09 14.38 ✈ ✈ ✈ ✈ X ✈ X ✈ X ✈ X ◗ ◗ ◗ ✓ X◗◗ 3✓✓◗◗ -5 ◗3 ◗ ✓✓ ◗ ✓✓ ✓✓ -5✓ ◗ 10.24 ◗ 5.05 ◗ ◗ 10.45 15.05 ◗ 11.05 ✓ ✈ ✓X ✓ ✈ ✓-5 ◗✈ ◗✈ ◗✈ ◗ ✈ ✓◗ ✈ ✈ ◗ ◗ X◗ ✓ 1✑ ◗ ✓ ✓ ✑ ✓ X✑◗ ✓ ❙ ❙ ◗5 ❙◗ ❙ ◗ -3 ◗✑ ◗✑ ✑ ✓ ❙✑◗11.45 ✓ ❙ ◗25.45 ✓ ❙✑◗ 6.05 ✓ ❙✑◗11.05 ✓ ◗ ◗✓ ✑ ✈ ✈ ❙ ◗✑ ✈ ❙7 ◗✓ ✈ ❙5 ◗ ✈ ✈ ✈14.29 ✈ ✓ ✓ ❙ ◗✓ ✑ X X -1 -1 ✑ ✑ ✑ ✑ X ❙ ❙ 14.65 ✑ ❙ ❙ ✑ ✑ ✑ ✑ ❙6.05 ✑ X❙7.05 ✑ ❙7.05 ✑ ❙❙ ✑ ❙ ✑ ✑ ✈ ✈ ✈ ✈ ✑ ✈ -3 ✈ ✈ ✈ ❙ ❙ X Transmitted sequence: 282 -5 -3 www.elsolucionario.net 2) The next figure shows one frame of the trellis used to decode the received sequence Each branch consists of two transitions which correspond to elements in the same coset in the final partition level www.elsolucionario.net Chapter 10 Problem 10.1 1) The wavelength λ is λ= 3 × 108 m m = 10 10 Hence, the Doppler frequency shift is fD = ± 100 Km/hr u 100 × 103 × 10 =± Hz = ±92.5926 Hz = ± λ × 3600 10 m 2) The maximum difference in the Doppler frequency shift, when the vehicle travels at speed 100 km/hr and f = GHz, is ∆fD max = 2fD = 185.1852 Hz This should be the bandwidth of the Doppler frequency tracking loop 3) The maximum Doppler frequency shift is obtain when f = GHz + MHz and the vehicle moves towards the transmitter In this case λmin = × 108 m = 0.2997 m 109 + 106 and therefore 100 × 103 = 92.6853 Hz 0.2997 × 3600 Thus, the Doppler frequency spread is Bd = 2fD max = 185.3706 Hz fD max = Problem 10.2 1) Since Tm = second, the coherence bandwidth Bcb = = 0.5 2Tm Hz and with Bd = 0.01 Hz, the coherence time is Tct = (2) Since the channel bandwidth W (3) Since the signal duration T = 100/2 = 50 2Bd seconds bcb , the channel is frequency selective Tct , the channel is slowly fading 283 www.elsolucionario.net The plus sign holds when the vehicle travels towards the transmitter whereas the minus sign holds when the vehicle moves away from the transmitter www.elsolucionario.net (4) The ratio W/Bcb = 10 Hence, in principle up to tenth order diversity is available by subdividing the channel bandwidth into 10 subchannels, each of width 0.5 Hz If we employ binary PSK with symbol duration T = 10 seconds, then the channel bandwidth can be subdivided into 25 subchannels, each of bandwidth T2 = 0.2 Hz We may choose to have 5th order frequency diversity and for each transmission, thus, have parallel transmissions Thus, we would have a data rate of bits per signal interval, i.e., a bit rate of 1/2 bps By reducing the order of diversity, we may increase the data rate, for example, with no diversity, the data rate becomes 2.5 bps (5) To answer the question we may use the approximate relation for the error probability given by (10.1.37), or we may use the results in the graph shown in Figure 10.1.10 For example, for binary PSK with D = 4, the SNR per bit required to achieve an error probability of 10−6 is 18 dB This the total SNR per bit for the four channels (with maximal ration combining) Hence, the SNR per bit per channel is reduced to 12 dB (a factor of four smaller) α −α2 /2σα , e σα p(α) = α>0 otherwise 0, Hence, the probability of error for the binary FSK and DPSK with noncoherent detection averaged over all possible values of α is ∞ P2 = = But, ∞ −c αNEb α −α2 /2σα2 e e dα σα2 2σα2 ∞ −α2 αe cEb + 12 N0 2σα dα x2n+1 e−ax dx = n! , 2an+1 (a > 0) so that with n = we obtain P2 = 2σα2 2 Eb 2σα N0 −α2 αe cEb + 12 N0 2σα dα = = where ρ¯b = ∞ c EbN2σ0 α = +1 [c¯ ρb + 1] With c = (DPSK) and c = P2 = 2σα2 2 (FSK) we have 2(1+¯ ρb ) , 2+¯ ρb , 284 DPSK FSK cEb N0 + 2σα www.elsolucionario.net Problem 10.3 The Rayleigh distribution is www.elsolucionario.net Problem 10.4 (a) cos 2πf1 t ❄ ✲ ×♥✲ Matched Filter     ( )2 ❄ +✐ ✲ ×♥✲ Matched Filter ✻ (     ( )2 )2 ✲ sin 2πf1 t cos 2πf2 t ❄ ✲ ×♥✲ Matched Filter ❄ ❄ ✲ ×♥✲ Matched Filter ✻ sin 2πf2 t     ( )2 ( )2 +✐ ✲ +♥ ✻ ✻ ✻ sample at t = kT cos 2πf1 t ❄ ✲ ×♥✲ Matched Filter ❄     Detector select the larger ❄ output ✲ +✐ ✲ ×♥✲ Matched Filter ✻ r2 (t) ✻     ( )2 ❄ ✲ sin 2πf1 t +♥ cos 2πf2 t ❄ ✲ ×♥✲ Matched Filter ✲ ✻     ( )2 ❄ +✐ ✲ ×♥✲ Matched Filter ✻ ✻     ( )2 sin 2πf2 t (b) The probability of error for binary FSK with square-law combining for D = is given in Figure 10.1.10 The probability of error for D = is also given in Figure 10.1.10 Note that an increase in SNR by a factor of 10 reduces the error probability by a factor of 10 when D = and by a factor of 100 when D = Problem 10.5 √ (a) r is a Gaussian random variable If Eb is the transmitted signal point, then E(r) = E(r1 ) + E(r2 ) = (1 + k) Eb ≡ mr and the variance is σr2 = σ12 + k σ22 285 www.elsolucionario.net r1 (t) ✻     www.elsolucionario.net The probability density function of r is f (r) = √ (r−mr ) − 2σr e 2πσr and the probability of error is P2 = = −∞ f (r) dr r −m σ √ 2π r −∞ x2 dx m2r σr2 = Q where e− The value of k that maximizes this ratio is obtained by differentiating this expression and solving for the value of k that forces the derivative to zero Thus, we obtain k= σ12 σ22 Note that if σ1 > σ2 , then k > and r2 is given greater weight than r1 On the other hand, if σ2 > σ1 , then k < and r1 is given greater weight than r2 When σ1 = σ2 , k = In this case 2Eb m2r = σr2 σ1 (b) When σ22 = 3σ12 , k = 13 , and (1 + 13 )2 Eb m2r = = 2 σr2 σ1 + (3σ1 ) Eb σ12 On the other hand, if k is set to unity we have 4Eb Eb m2r = = 2 σr σ1 + 3σ1 σ1 Therefore, the optimum weighting provides a gain of 10 log = 1.25 dB Problem 10.6 1) The probability of error for a fixed value of a is  Pe (a) = Q  2a2 E N0   since the given a takes two possible values, namely a = and a = with probabilities 0.1 and 0.9, respectively, the average probability of error is Pe = 0.1 +Q 8E N0 = 0.05 + Q 286 8E N0 www.elsolucionario.net m2r (1 + k)2 Eb = σr2 σ12 + k σ22 www.elsolucionario.net (2) As NE0 → ∞, Pe → 0.05 (3) The probability of error for fixed values of a1 and a2 is   Pe (a1 , a2 ) = Q  2(a21 + a22 )E  N0 In this case we have four possible values for the pair (a1 , a2 ), namely, (0, 0), (0, 2), (2, 0), and (2, 2), with corresponding probabilities ).01, 0.09, 0.09 and 0.81 Hence, the average probability of error is 8E 16E 0.01 + 0.18Q Pe = + 0.81Q N0 N0 E N0 → ∞, Pe → 0.005, which is a factor of 10 smaller than in (2) Problem 10.7 We assume that the input bits 0, are mapped to the symbols -1 and respectively The terminal phase of an MSK signal at time instant n is given by θ(n; a) = π k ak + θ0 k=0 where θ0 is the initial phase and ak is ±1 depending on the input bit at the time instant k The following table shows θ(n; a) for two different values of θ0 (0, π), and the four input pairs of data: {00, 01, 10, 11} θ0 0 0 π π π π b0 0 1 0 1 b1 1 1 a0 -1 -1 1 -1 -1 1 a1 -1 -1 -1 -1 θ(n; a) −π 0 π π π 2π Problem 10.8 1) The envelope of the signal is |s(t)| = |sc (t)|2 + |ss (t)|2 = 2Eb πt cos2 Tb 2Tb = 2Eb Tb Thus, the signal has constant amplitude 287 + πt 2Eb sin2 Tb 2Tb www.elsolucionario.net (4) As www.elsolucionario.net 2) The signal s(t) has the form of the four-phase PSK signal with πt , 2Tb gT (t) = cos ≤ t ≤ 2Tb Hence, it is an MSK signal A block diagram of the modulator for synthesizing the signal is given in the next figure ✲× ❧ ✻ ✛ Serial / ✲ Parallel data an Demux Serial ✲× ❧ ✻ ✛ ❦ ❦ ❄ s(t) ❧ ✲ ❄ cos(2πfc t) + π ✻ − πt ❄ cos( 2Tb ) − π2 ❄ ❧ ✲× ❄ ❧ ✲× a2n+1 3) A sketch of the demodulator is shown in the next figure t = 2Tb ✲× ❧ r(t) ✲ ✻ ✛ ✲× ❧ ❦ ✻ ✛ ✲ 2Tb (·)dt ❄ ❧ ✲× Threshold ❄ ❦ πt cos(2πfc t)) cos( 2T ) b ❄ ❄ − π2 − π2 ❄ ❧ ✲× ❅ ❅ Parallel to✲ Serial t = 2Tb ❅ ✲ 2Tb (·)dt ❅ ✻ Threshold Problem 10.9 Since p = 2, m is odd (m = 1) and M = 2, there are Ns = 2pM = phase states, which we denote as Sn = (θn , an−1 ) The 2p = phase states corresponding to θn are 3π π Θs = 0, , π, 2 and therefore, the states Sn are (0, 1), (0, −1), π ,1 , π , −1 , (π, 1), (π, −1), 3π ,1 , 3π , −1 Having at our disposal the state (θn , an−1 ) and the transmitted symbol an , we can find the new phase state as π an (θn , an−1 ) −→ (θn + an−1 , an ) = (θn+1 , an ) The following figure shows one frame of the phase-trellis of the partial response CPM signal 288 www.elsolucionario.net a2n www.elsolucionario.net (θn , an−1 ) (θn+1 , an ) ✉ ✉ ◗ ◗     ✁✁ ◗  ✁ ✉ (0, −1) ✉ ◗   ✁ ◗ ❚   .◗ ( π2 , 1) ✉ ◗✉ ❚   .✁.✁ .◗ ❚   ◗ ◗ ✁   ❚   ✉ ( π2 , −1)   ◗ ✁   ✉ ❚ ◗ ✁❚   ◗ ◗   (π, 1) ✉ ◗✉ ◗  ✁ ❚ ◗ ✁ ◗ ❚     ❚  .✁ ◗   ✉ (π, −1)   ✁ .◗ ❚ ✉   ◗ ✁ ❚ ◗   ( 3π ◗ ✉   .◗ ❚✉ , 1) ✁ .  ( 3π ✉ ✉ , −1)   (0, 1) (0, 1) (0, −1) ( π2 , 1) ( π2 , −1) (π, 1) (π, −1) ( 3π , 1) ( 3π , −1) (π,1) ✓✏ €€1 ✶ ✒✑ ( π2 ,1) 1✏✏✏ ( 3π ,1) €€ ✓✏ ✓✏ q € ❆❑❆ ✏✏ ❆ ❆ ❆ ✏✏✏ ✒✑ ✒✑ ✏ ❆ ✣ ❆ ✡ ❆❑ ✏✏❆ ❆❆ 1 ✡ ❆❑❆ ❆ ❆ ✏✏ ❆ ✏ ✡ -1❆ ❆ ❆1 ✏✏ -1❆ ❆ ✡ ✏✏✏❆ ❆ ❆ ❆ ✓✏ ✓✏ ❆ ❆❯ ✮ ✏ ❆ ❆ ❆ (0,1) (0,−1) ❆ ❆❆ ✶ ✏ ✒✑ ❆ ❆-1 ❆✏✏✏ ✒✑ ❆ ❆ ✏ ❑ ❆ ✏ ❆❆ ✁ ❆❆ ✏✏ ❆❆ -1 ✏ ❆ ✁ -1 ❆ ❆ ✏ ❆❆ -1 ❆✏✏ ❆ ✁ ❆ ✓✏ ✓✏ ❆ ✏ ❆ ❆ ❆❯ ✁☛ ❆❯ ✏ ❆❆ ✒✑ -1 ✏✏✒✑ ✐ € €€ ❆ ❆ ✏ ❆❯ ✏ €€❆✓✏ ( 3π ,−1) ✮✏ ( π ,−1) 2 ✒✑ (π,−1) Problem 10.10 1) For a full response CPFSK signal, L is equal to If h = 23 , then since m is even, there are p terminal phase states If h = 34 , the number of states is Ns = 2p 2) With L = and h = 23 , the number of states is Ns = p22 = 12 When L = and h = 34 , the number of states is Ns = 2p22 = 32 Problem 10.11 (a) The coding gain is Rc d H = × 10 = (7dB) (b) The processing gain is W/R, where W = 107 Hz and R = 2000bps Hence, 107 W = = × 103 (37dB) R × 103 289 www.elsolucionario.net The following is a sketch of the state diagram of the partial response CPM signal www.elsolucionario.net (c) The jamming margin given by (10.3.43) is PJ Ps W R = dB dB + (CG)dB − Eb J0 dB = 37 + − 10 = 34dB Problem 10.12 The probability of error for DS spread spectrum with binary PSK may be expressed as 2W/Rb PJ /PS P2 = Q Therefore, 2Eb J0 P2 = Q which is identical to the performance obtained with a non-spread signal Problem 10.13 We assume that the interference is characterized as a zero-mean AWGN process with power spectral density J0 To achieve an error probability of 10−5 , the required Eb /J0 = 10 Then, by using the relation in (10.3.40) and (10.3.44), we have W/R PN /PS = W/R Nu −1 W/R = Eb J0 W = R = Eb J0 (Nu − 1) Eb J0 (Nu − 1) where R = 104 bps, Nu = 30 and Eb /J0 = 10 Therefore, W = 2.9 × 106 Hz The minimum chip rate is 1/Tc = W = 2.9 × 106 chips/sec Problem 10.14 To achieve an error probability of 10−6 , we require Eb J0 = 10.5dB dB Then, the number of users of the CDMA system is Nu = = W/Rb Eb /J0 1000 11.3 +1 + = 89 users If the processing gain is reduced to W/Rb = 500, then Nu = 500 + = 45users 11.3 290 www.elsolucionario.net where W/R is the processing gain and PJ /PS is the jamming margin If the jammer is a broadband, WGN jammer, then P J = W J0 PS = Eb /Tb = Eb Rb www.elsolucionario.net Problem 10.15 (a) We are given a system where (PJ /PS )dB = 20 dB, R = 1000 bps and (Eb /J0 )dB = 10 dB Hence, using the relation in (10.3.40) we obtain W R dB PJ PS dB = + Eb J0 dB W R = 1000 W = 1000R = 106 Hz = 30 dB α∗ = 0.7 0.7 = = 0.07 Eb /J0 10 The corresponding probability of error for this worst-case jamming is P2 = 0.082 0.082 = 8.2 × 10−3 = Eb /J0 10 Problem 10.16 The radio signal propagates at the speed of light, c = × 108 m/ sec The difference in propagation delay for a distance of 300 meters is Td = 300 = 1à sec ì 108 The minimum bandwidth of a DS spread spectrum signal required to resolve the propagation paths is W = M Hz Hence, the minimum chip rate is 106 chips per second Problem 10.17 (a) We have Nu = 15 users transmitting at a rate of 10, 000 bps each, in a bandwidth of W = M Hz The b /J0 is E J0 = W/R Nu −1 = 106 /104 14 = 100 14 = 7.14 (8.54 dB) (b) The processing gain is 100 (c) With Nu = 30 and Eb /J0 = 7.14, the processing gain should be increased to W/R = (7.14) (29) = 207 Hence, the bandwidth must be increased to W = 2.07M Hz 291 www.elsolucionario.net (b) The duty cycle of a pulse jammer for worst-case jamming is www.elsolucionario.net Problem 10.18 (a) The length of the shift-register sequence is L = 2m − = 215 − = 32767 bits For binary FSK modulation, the minimum frequency separation is 2/T, where 1/T is the symbol (bit) rate The hop rate is 100 hops/ sec Since the shift register has N = 32767 states and each state utilizes a bandwidth of 2/T = 200 Hz, then the total bandwidth for the FH signal is 6.5534 M Hz (b) The processing gain is W/R We have, (c) If the noise is AWG with power spectral density N0 , the probability of error expression is Eb N0 P2 = Q =Q W/R PN /PS Problem 10.19 (a) If the hopping rate is hops/bit and the bit rate is 100 bits/sec, then, the hop rate is 200 hops/sec The minimum frequency separation for orthogonality 2/T = 400Hz Since there are N = 32767 states of the shift register and for each state we select one of two frequencies separated by 400 Hz, the hopping bandwidth is 13.1068 M Hz (b) The processing gain is W/R, where W = 13.1068 M Hz and R = 100bps Hence W = 0.131068 M Hz R (c) The probability of error in the presence of AWGN is given by (10.3.61) with N = chips per hop Problem 10.20 a) The total SNR for three hops is 20 ∼ 13 dB.Therefore the SNR per hop is 20/3 The probability of a chip error with noncoherent detection is − Ec p = e 2N0 where Ec /N0 = 20/3 The probability of a bit error is Pb = − (1 − p)2 = − (1 − 2p + p2 ) = 2p − p2 Ec − 2N = e = 0.0013 292 − Ec − e N0 www.elsolucionario.net 6.5534 × 106 W = = 6.5534 × 104 bps R 100 www.elsolucionario.net b) In the case of one hop per bit, the SNR per bit is 20, Hence, Ec − 2N e −10 e = = 2.27 × 10−5 Pb = Therefore there is a loss in performance of a factor 57 AWGN due to splitting the total signal energy into three chips and, then, using hard decision decoding Problem 10.21 (a) We are given a hopping bandwidth of GHz and a bit rate of 10 kbs Hence, (b) The bandwidth of the worst partial-band jammer is α∗ W, where α∗ = 2/ (Eb /J0 ) = 0.2 Hence α∗ W = 0.4GHz (c) The probability of error with worst-case partial-band jamming is P2 = e−1 (Eb /J0 ) = e−1 10 = 3.68 × 10−2 Problem 10.22 The processing gain is given as W = 500 (27 dB) Rb The (Eb /J0 ) required to obtain an error probability of 10−5 for binary PSK is 9.5 dB Hence, the jamming margin is PJ W = − JE0b PS Rb dB dB dB = 27 − 9.5 = 17.5 dB Problem 10.23 If the jammer is a pulse jammer with a duty cycle α = 0.01, the probability of error for binary PSK is given as 2W/Rb P2 = αQ PJ /PS For P2 = 10−5 , and α = 0.01, we have Q 2W/Rb PJ /PS 293 = 10−3 www.elsolucionario.net × 109 W = = × 105 (53dB) R 104 www.elsolucionario.net Then, 500 W/Rb = =5 PJ /PS PJ /PS and PJ = 100 (20 dB) PS Problem 10.24 ∞ cn p (t − nTc ) c (t) = n=−∞ The power spectral density of c (t) is given by Sc (f ) = Sc (f ) |P (f )|2 Tc |P (f )|2 = (ATc )2 sin c2 (f Tc ) , Tc = 1µ sec and Sc (f ) is the power spectral density of the sequence {cn } Since the autocorrelation of the sequence {cn } is periodic with period L and is given as    L, Rc (m) = m = 0, ±L, ±2L,   −1, otherwise then, Rc (m) can be represented in a discrete Fourier series as Rc (m) = L−1 rC (k) ej2πmk/L , m = 0, 1, , L − L k=0 where {rc (k)} are the Fourier series coefficients, which are given as L−1 rc (k) = Rc (m) e−j2πkm/L , k = 0, 1, , L − m=0 and rc (k + nL) = rc (k) for n = 0, ±1, ±2, The latter can be evaluated to yield rc (k) = L + − = L−1 −j2πkm/L m=0 e 1, k = 0, ±L, ±2L, L + 1, otherwise The power spectral density of the sequence {cn } may be expressed in terms of {rc (k)} These coefficients represent the power in the spectral components at the frequencies f = k/L Therefore, we have ∞ k Sc (f ) = rc (k) δ f − L k=−∞ LTc Finally, we have Sc (f ) = LTc ∞ rc (k) P k=−∞ 294 k LTc δ f− k LTc www.elsolucionario.net where www.elsolucionario.net Problem 10.25 Without loss of generality, let us assume that L1 < L2 Then, the period of the sequence obtained by forming the modulo-2 sum of the two periodic sequences is L3 = kL2 where k is the smallest integer multiple of L2 such that kL2 /L1 is an integer For example, suppose that L1 = 15 and L2 = 63 Then, we find the smallest multiple of 63 which is divisible by L1 = 15, without a remainder Clearly, if we take k = periods of L2 , which yields a sequence of L3 = 315, and divide L3 by L1 , the result is 21 Hence, if we take 21L1 and 5L2 , and modulo-2 add the resulting sequences, we obtain a single period of length L3 = 21L, = 5L2 of the new sequence Problem 10.26 L = 210 − = 1023 Since Tb = LTc , then the processing gain is L= Tb = 1023 (30dB) Tc (b) The jamming margin is PJ PS dB W Rb dB = − Eb J0 dB = 30 − 10 = 20dB where Jav = J0 W ≈ J0 /Tc = J0 × 106 Problem 10.27 At the bit rate of 270.8 Kbps, the bit interval is Tb = 10−6 = 3.69µsec 2708 a) For the suburban channel model, the delay spread is µsec Therefore, the number of bits affected by intersymbol interference is at least The number may be greater than if the signal pulse extends over more than one bit interval, as in the case of partial response signals, such as CPM b) For the hilly terrain channel model, the delay spread is approximately 20 µ sec Therefore, the number of bits affected by ISI is at least The number may be greater than if the signal pulse extends over more than one bit interval 295 www.elsolucionario.net (a) The period of the maximum length shift register sequence is www.elsolucionario.net Problem 10.28 In the case of the urban channel model, the number of RAKE receiver taps will be at least If the signal pulse extends over more than one bit interval, the number of RAKE taps must be further increased to account for the ISI over the time span of the signal pulse For the hilly terrain channel model, the minimum number of RAKE taps is at least but only three will be active, one for the first arriving signal and for the delayed arrivals If the signal pulse extends over more than one bit interval, the number of RAKE taps must be further increased to account for the ISI over the same span of the signal pulse For this channel, in which the multipath delay characteristic is zero in the range of µsec to 15 µsec, as many as RAKE taps between the first signal arrival and the delayed signal arrivals will contain no signal components fm = 105 × 108 vf0 = × = 83.3Hz c 3600 38 For a train travelling at a speed of 200 Km/hr, fm = 166.6Hz The corresponding spread factors are Tm Bd = Tm fm = 5.83 × 10−4 , automobile 1.166 × 10−3 , train The plots of the power spectral density for the automobile and the train are shown below 0.08 0.07 0.06 0.05 0.04 0.03 0.02 train automobile 0.01 -200 -150 -100 -50 50 100 150 200 fm (Hz) 296 www.elsolucionario.net Problem 10.29 For an automobile travelling at a speed of 100 Km/hr, ... [Yl (f )] = = 2j − 12 (f + 1)ej2πf t df + 2j (−f + 1)ej2πf t df 1 f ej2πf t + 2 ej2πf t 2j j2πt 4π t − 12 + 1 j2 πf t e 2j j2πt 1 1 j2 πf t f ej2πf t + 2 ej2πf t e + 2j j2πt 4π t 2j j2πt 1 sin πt... 1 Λ(t)e? ?j? ?nt dt −1 −1 1 = (t + 1)e? ?j? ?nt dt + (−t + 1)e? ?j? ?nt dt −1 ? ?j? ?nt j ? ?j? ?nt j ? ?j? ?nt te e = + 2e + πn π n 2πn −1 −1 1 j ? ?j? ?nt j ? ?j? ?nt te e − + 2 e? ?j? ?nt + πn π n 2πn 0 1 − (ejπn + e? ?j? ?n ) =... (f )] = − 12 Y (f )ej2πf t df (f + 1)ej2πf t df + (−f + 1)ej2πf t df 1 f ej2πf t + 2 ej2πf t j2 πt 4π t = + − 12 j2 πf t e j2 πt 1 j2 πf t f ej2πf t + 2 ej2πf t e + j2 πt 4π t j2 πt 1 [1 − cos(πt)]