通信原理习题课2005

(25/10)=316.2,thebandwidthisB=3200-300=Example1-1Atelephonetouch-tonekeypadhasthedigits0to9withprobabilityofsendingbeing0.099each,plus“*”and“#”withprobabilityofsendingbeing0.005each.Ifthekeysarepressedatarateof2keys/putethedatarateforthissource.mm(1)bitsSolution：HPjIjPjlog2j1j1Pj111log10210(0.099)log100.0992(0.005)log100.0053.38bits/keyUsingdefineabove,whereT=1/2=0.5s/key,then:HR3.38/0.56.76bits/sTExample1-2Acomputeruserplanstobuyahigher-speedmodemforsendingdataoverhisorheranalogtelephoneline.Thetelephonelinehasasignal-to-noiseratio(SNR)of25dBandpassesaudiofrequenciesovertherange,from300to3200Hz,Calculatethemaximumdataratethatcouldbesentoverthetelephonelinewhenthereareonerrorsatthereceivingend.Solution：Inthetermofapowerratio,theSNR=102900Hz，weget:CBlog21S(1316.2)24097bits/s2900log2NProblem:Cana28.8kbit/smodemworkonthistelephoneline?Anda14.4kbits/smodem?Example1-3某信息原的符号集由A、B、C、D和E组成，设每一符号独立出现，其出现的概率分别为1/4、1/8、1/8、3/16和5/16。试求该信息源符号的平均信息量。mmlog2(1)bits，Solution知识点：HPjIjPjj1j1Pj将p(A)=1/4,p(B)=1/8,p(c)=1/8,p(D)=3/16,和p(E)=5/16分别代入即可求解。mmHPjIjPjlog2(1)j1j1Pj1log211log211log213log235log25448888161616162.23bit/symbolExample1-4已知某四进制数字信号传输系统的信息速率为接收到216个错误码元，试计算该系统的误符号率

2400bit/s，接收端在Ps的值。

30分钟内共Solution

由于

Rb

2400bit/s，

M=4，

因此

Rs

Rb

/(log2

M)

2400/2

1200

baud。30分钟共传输的的码元数为：Rs*30*602,160,0001因此系统的误码率为Psreceivingerrorsymbolsduringunittime216104transmissiontotalsymbolsduringunittime2160000Example1-5已知二进制信号在3分钟内共传达72000码元，试求其码元速率和信息速率各是多少？若是码元宽度保持不变，但该为八进制数字信号，则其码元速率和信息速率又各是多少？Solution(1)Rs72000/(3*60)400baud由于M=2则RbRs(log2M)400bit/s若是采用八进制，即M=8，由于码元宽度没有变化，故Rs400baudRbRs(log2M)1200bit/sExample1-6已知X、Y两个八进制数字传输系统，码元速率同样，在接收端4分钟内，X收到m个错误码元，Y共收到m+3个错误比特，是比较两个系统哪个性能较好？为什么？Solution比较两个通信系统的性能利害，主要看它们的有效性和可靠性。在这两个通信系统中，它们的传输码率同样，即有效性是同样的。下面来看它们的可靠性。由题意可知，M=8。依照误码率定义可知：Rxsmm240RsRs*4*60(m3)/(log28)m3RysRs*4*60720Rs讨论，(1)当m=1时，Rxs141Y系统；，Rys，即X系统的性能优于240Rs720Rs180Rs(2)当m2时，能够得出RysRxs，即Y系统的性能优于X系统。2Example2-1thePSD(PowerSpectralDensity)ofabasebandsignals(t)isshowninthefollowingfigure,theabsolutebandwidth、3dBbandwidthandfirstzero-crossingbandwidthis___________respectly(A)f3f2,(f1+f3)/2Ps(f)|(B)f3,f2,(f1+f2)/245°(C)f2,f3,(f1+f2)/2(D)f3,(f1+f2)/2,f20f1f2f3fSolution(D)Example2-2TheampilitudespectraofaRaisedcosine-rolloffsignalisshowninthefollowingFigure,its6-dBbandwidthis______(A)(BT+f1)/4(B)(BT+f1)/2|S(f)|(B)(BT-f1)(D)f145°Solution(B)0f1BTf议论：dB和信号带宽是本章的重点。Example2-3nn已知一个低通信号m(t)的频谱M(f)为1|f|,|f|200HzM(f)200otherwise若是以fs300Hz的速率对m(t)进行理想采样，试画出采样后信号的频谱图；(2)若是用fs400Hz的速率采样，其频谱又怎样？并重做上题。知识点：察看理想采样定理Solutionfs300Hz采样后信号的频谱会发生混叠(2)fs400Hz恰好知足采样定理，采样后信号的频谱不会发生混叠。(见通5p255)Example2-4已知一基带信号m(t)cos2t2cos4t，试确定其最小的采样频次拟使采样后信号不失真。Solution基带信号可看作低通信号，其最高频次为fm2Hz，依照采样定理可知，采样频次fs2fm4Hz3Example3-1TheinformationinananalogvoltagewaveformistobetransmittedoveraPCMsystemwitha0.1%accuracy(fullscale).Theanalogwaveformhasanabsolutebandwidthof100Hzandanamplituderangeof–10to+10V.Determinetheminimumsamplingrateneeded.DeterminethenumberofbitsneededineachPCMword.DeterminetheminimumbitraterequiredinthePCMsignal.DeterminetheminimumabsolutechannelbandwidthrequiredfortransmissionofthisPCMsignal.Solution:(a)Determinetheminimumsamplingrateneeded.fs2B2(100)200samples/secDeterminethenumberofbitsneededineachPCMword.Usingtheresultsgiveninprob.3-7.n3.32log10503.32log10508.96P0.19bits/wordDeterminetheminimumbitraterequiredinthePCMsignal.nbitsfswordsKbitsRsec200(9)1.8wordsecDeterminetheminimumabsolutechannelbandwidthrequiredfortransmissionofthisPCMsignal.ForbinaryPCMD=Req.(3-74)D2B,forBmin,r=01rBD900Hz2Example3-2An850-MbyteharddiskisusedtostorePCMdata,Supposethatavoice-frequencysignalissampledat8ksamples/sandtheencodedPCMistohaveanaverageSNRofatleast30dB.HowmanyminutesofVFconversationcanbestoredontheharddisk?知识点：S/N）dB6.02nwhere（α=4.77-20log(V/xrms)(foruniformquantizing)α≈4.77-20log[ln(1+)]μ(for-μlawcompanding)α≈4.77-20log[1+lnA](forA-lawcompanding)4（S/N）M2，where＝out14(M2-1)PePe0SolutionbecauseV＝850Mbytefs＝8kHz。outM2M230dB（S/N）14(M2-1)PethusM21031000M31.6usingM＝32＝25,thatistosay,n=5bit/sampleRnfs8*540kpbs5kbyte/sV850*103T170000sR5Example3-3设信号m(t)9Acost，其中，A10V。若m(t)被平均量化为40个电平，试确定所需的二进制码组的位数N和量化间隔。知识点：察看平均量化Solution由题意可知，M=40表示40个电平需要的二进制码组的位数为：Nlog2M1log24016量化间隔为2A2*10/400.5VMExample3-4设二进制符号序列为，试以矩形脉冲为例，分别画出相应的单极性码、双极性码，单极性归零码、双极性归零码、二进制差分码和八电平码的波形。知识点：察看基带信号波形的种类及其特点。需要注意的是二进制信号和基带信号的电位之间的对应关系及归零波形和不归零波形之间的差别，特别需要掌握多元码波形的画法和特点。5单极性波形E0

110010001110双极性波形E-E单极性归零E波形-0双极性归零E波形-EExample3-5设某基带传输系统拥有