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Download Digital Communication: Solutions Manual by Edward A. Lee, David G. Messerschmitt PDF

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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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.

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