By Edward A. Lee, David G. Messerschmitt
This complement includes labored out strategies to the bankruptcy finish challenge units present in Digital verbal exchange, moment Edition, ISBN 0-7923-9391-0.
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Extra resources for Digital Communication: Solutions Manual
8 MHz. 5-20. (a) The spectrum will become asymmetric about the carrier, with more power concentrated at frequencies near (OOc + kv) (b) The spectrum will fill out in the middle and get smaller near (OOc ± kv), but still be close to symmetric about the carrier frequency. 5-21. 1 < 3 Hz. 5 parts per million accuracy. 5-22. 148) CHAPTER 6: SOLUTIONS TO PROBLEMS (t-1. 4), we can write H (j(0) =G (joo)B (j ro). 280) Using the result of appendix 3-A, we can write the power spectrum of the received signal as 1 SR(jro)= T IH(jro) I 2SA (e j 6)T) = a2 ; IG(jro)B(jro)1 2 = a2 ; IG (joo)1 28 (joo).
The othec two can be found by finding the observation at which the receiver is indifferent between a decision +3 and a decision +1. 1) = Iy _11 2 _ 2021n(0A). 179) Solving this yields y =2 + ~a2In(4). 180) The thresholds are therefore at ± (2 + 'ha2In(4». 1 yields a decision £ = 1. 9-4. ;. 181) 9-S. (a) 51 =[1,O,OJ, 52 =[0,1 ,OJ, 53 =[0,0,1] . 184) transmitted] S 3Q (2,p) . 185) Pr[ error 15i Note that in both cases Pr[ error 15i transmitted] is independent of i , so 9-6. 2Jt e ···e , [_1_] from independence.
26/2. 5, where q is the sampled output of the matched filter. 75. 75, and selects =0 otherwise. 75 as expected. a a 9-11. (a) Wehave $1(t)=O $2(t)=h(t) 115 1 11 2 =0 1152 11 2 =p,,(0). 197) Taking the logarithm of both sides will not change the inequality, so we get equivalently lnIO<;)< p~~) . 198) In view Of the monotonicity of lnl0('), the ML detector compares K 2 to some threshold v, where that threshold depends on the SNR. The receiver is the same as a passband PAM coherent receiver, consisting of a demodulator, matched filter, and sampler.