数字通信第14---15章答案new

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1、CHAPTER14Problem14.1:BasedontheinfoaboutthescatteringfunctionweknowthatthemultipathspreadisTm=1ms,andtheDopplerspreadisBd=0.2Hz.(a)(i)T=10−3secm(ii)Bd=0.2Hz1(iii)(∆t)≈=5seccBd(iv)(∆f)≈1=1000HzcTm(v)TB=2·10−4md(b)(i)Frequencynon-selectivechannel:Thismeansth

2、atthesignaltransmittedoverthechannelhasabandwidthlessthat1000Hz.(ii)Slowlyfadingchannel:thesignalingintervalTisT<<(∆t).c(iii)Thechannelisfrequencyselective:thesignaltransmittedoverthechannelhasabandwidthgreaterthan1000Hz.(c)Thesignaldesignproblemdoesnothav

3、eauniquesolution.WeshoulduseorthogonalM=4FSKwithasymbolrateof50symbols/sec.HenceT=1/50sec.Forsignalorthogonality,weselectthefrequencieswithrelativeseparation∆f=1/T=50Hz.Withthisseparationweobtain10000/50=200frequencies.Sincefourfrequenciesarerequirestotran

4、smit2bits,wehaveupto50th−orderdiversityavailable.Wemayusesimplerepetition-typediversityoramoreefficientblockorconvolutionalcodeofrate≥1/50.Thedemodulatormayusesquare-lawcombining.Problem14.2:(a)32P2h=p+3p(1−p)wherep=1,and¯γisthereceivedSNR/cell.2+¯γcc(b)For¯

5、γ=100,P≈10−6+3·10−4≈3·10−4c2hFor¯γ=1000,P≈10−9+3·10−6≈3·10−6c2h

L2L−11(c)Since¯γc>>1,wemayusetheapproximation:P2s≈Lγ¯c,whereListheorderof283diversity.ForL=3,wehave:10P≈10−5,γ¯=1002scP2s≈3⇒P≈10−8,γ¯=1000γ¯c2sc(d)Forhard-decisiondecoding:LLkL−kL/2P2h

6、=p(1−p)≤[4p(1−p)]kk=L+12wherethelatteristheChernoffbound,Lisodd,andp=1.Forsoft-decisiondecoding:2+¯γcL2L−11P2s≈Lγ¯cProblem14.3:
a22E(a)Forafixedchannel,theprobabilityoferroris:Pe(a)=QN0.Wenowaveragethisconditionalerrorprobabilityoverthepossiblevaluesof

7、α,whicharea=0,withprobability0.1,anda=2withprobability0.9.Thus:8E8EPe=0.1Q(0)+0.9Q=0.05+0.9QN0N0E(b)AsN0→∞,Pe→0.05(c)Whenthechannelgainsa1,a2arefixed,theprobabilityoferroris:(a2+a2)2EP(a,a)=Q12e12N0Averagingovertheprobabilitydensityfunctionp(a1,a2)=

8、p(a1)·p(a2),weobtaintheaverageprobabilityoferror:

P=(0.1)2Q(0)+2·0.9·0.1·Q8E+(0.9)2Q16Ee
N0
N0=0.005+0.18Q8E+0.81Q16EN0N0(d)AsE→∞,P→0.005N0e284Problem14.4:(a)T=1sec⇒(∆f)≈1=1HzmcTmB=0.01Hz⇒(∆t)≈1=100s