通信原理概论课后习题答案 马殿光

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1、1.1Use1-6d1=2h1d2=2h2h1istheheightoftransmitterantennameasuredinfeeth2istheheightofreceiverantennameasuredinfeetd1andd2aremeasuredinmilesd=d1+d2Sinceh1=1200ftd=60milesthenh2=60.6ft2221.2AccordingtoFigure1-3d+re=(re+h)WherereistheradiusoftheEarth,re=5280miles22222d+re=re+rhhe+22

2、22Sinceh
re,thend+re=re+2rhe=>d=2rhe=>d=2rheSince1mile=5280feet,hismeasuredinfeet,then1miled=2´hfeet´5280mile´=>d=2hmile5280feet1.3Use1-6d1=2h1d2=2h2d=d1+d2d1=d2Sinced=25miles,thenh1=h2=78.125ft1.4Use1-6d1=2h1d2=2h2d=d1+d2Sinceh1=60ft,h2=4ft,thend=13.8miles1.5P1=P2=0.2P3=P4=0.3

3、Use1-8,theaverageinformationofsourceis41H=∑Pjlog()1.97=bitsj=1Pj1.9P0=0.25P1=P2=0.15P3=P4=P5=P6=P7=P8=0.07P10=0.03Use1-8,theaverageinformationofsourceis101H=∑Pjlog()=3.08bitsj=0Pj1.15Use1-10SSR=Blog(12+)SNRdB()10log(=)NNa)SinceSNR=45dB,B=1200300-=900Hz,then45S10R=Blog(12+)=90

4、0log´110+=13.45Kbps2Nb)SinceSNR=45dB,B=320015001700-=Hz,then45S10R=Blog(12+)1700log=´110+=25.41Kbps2Nc)SinceSNR=45dB,B=3200300-=2900Hz,then45S10R=Blog(12+)=2900log´110+=43.35Kbps2N2.2a)Use2-5forperiodicwaveformwithperiodT0T01+a22Vdc=∫T0vtdtT()(0=4)T0-+a21321Leta

5、=1thenVdc=∫vtdt()=-volts4-14b)Use2-1121T/221T0/22Vrms==lim∫vtdt()=∫vtdt()=0.816voltsT(0=4)T->¥-T/2-T0/2TT022Vrms0.816c)P===0.667mwR10002.6Use2-132-141T/2T/2222P==lim∫wtdt()E=PT=lim∫wtdt()T->¥T-T/2T->¥-T/2a)Use2-521T/221T/21T0/22T0P=lim∫wtdt()=lim∫P(tT/0)dt=lim∫1

6、dt=lim=0T->¥T-T/2T->¥T-T/2T->¥T-T0/2T->¥TE=PT=T0SinceEisnonzeroandfinite,thenitisenergysignalb)1T/221T0/22P=lim∫wtdt()=lim∫coswotdtT->¥T-T/2T->¥T-T0/21T0/211T0=lim∫+cos2wotdt=lim=0T->¥T-T0/222T->¥2TE=PT=T0SinceEisnonzeroandfinite,thenitisenergysignalc)21T/221T0/2221T0/21P=lim

7、∫T/2wtdt()=∫T0/2coswotdt=∫T0/2(1cos2+wot)dtT->¥T-T0-T0-21T0/2121T0/2111=∫(12cos2+wot+cos2wotdt)=∫(12cos2+wot++cos4wotdt)T0-T0/24T0-T0/24221T0/23113=∫+cos2wot+cos4wotdt)=T0-T0/282883E=PT=limT(infinite)T®¥8SincePisnonzeroandfinite,thenitispowersignal2.43a)Use2-681T/2Rw()t

8、==lim∫wtwt()(+t)dt-T/2T®¥T1T/2=lim∫(512cos