數(shù)字信號處理-基于計算機的方法第3章答案

1、n兀ds3-2(a)SketchthenaturallysampledPAMwaveformthatresultsfromsamplinga1-kHzsinewaveata4-kHzrate.Repeatpart(a)forthecaseofaflat-toppedPAMwaveform.Solution:3-4(a)Showthatananalogoutputwaveform(whichisproportionaltotheoriginalinputanalogwaveform)mayberecoveredfromanaturallysampledPAMwaveformbyusingthed

2、emodulationtechniqueshowedinFig.3-4.FindtheconstantofproportionalityC,thatisobtainedwiththisdemodulationtechnique,wherew(t)istheoriqinalwaveformandCw(t)istherecoveredwaveform.NotethatCisafunctionofn,wheretheoscillatorfrequencyisnfs.t-kTSolution:s(t)=為口k=gZsink兀de-jkok兀dk=s乙cej/kk=sw(t)=w(t)|d+2dSsZs

3、ink兀dcoskotk兀dsk=1sink兀dcoskotk兀dsk=1v(t)=w(t)cosnot1ss(賈sink兀dt)dcosnot+2dcoskotcosnotsk兀dssk=1n豐ksinn兀d+2dcos2notcos2nt=s11+_cos2ntafterLPF:w(t)=dosinn兀dw(t)=cw(t)22ssinn兀d/.c=dn兀d3-7InabinaryPCMsystem,ifthequantizingnoiseisnottoexceedpercent土pofthepeak-to-peakanaloglevel,showthatthenumberofbitsin

4、eachPCMwordneedstobe50、50、logm=3.32log1010、P丿10、P丿nlog210(Hint:LookatFig.3-8c.)Solution:BinaryPCMM=levels2nforP1nlVq100PPWeneedV2Pstepsize=5=ppVM100PP1PPnloga3-8log(a)abbTheinformationinananalogvoltagelog(10)log250()10plog(x)log(x)=b=log(b)log(x)waveformistobetransmittedoveraPCMsystemwitha0.1%accura

5、cy(fullscale).土Theanalogwaveformhasanabsolutebandwidthof100Hzandanamplituderangeof-0to+10V.Determinetheminimumsamplingrateneeded.DeterminethenumberofbitsneededineachPCMword.DeterminetheminimumbitraterequiredinthePCMsignal.DeterminetheminimumabsolutechannelbandwidthrequiredfortransmissionofthisPCMsig

6、nal.Solution:(a)Determinetheminimumsamplingrateneeded.fs=2B=2(100)=200samples/sec(b)DeterminethenumberofbitsneededineachPCMword.Usingtheresultsgiveninprob.3-7.55T=0.1%4V土丄=土0.1%2V2VM=500and29=5125005n=9bitsTaPCMwordDeterminetheminimumbitraterequiredinthePCMsignal.R=(nbitsA(word丿fwordsA-s*sec丿=(9)200

7、1.8KbitssecDeterminetheminimumabsolutechannelbandwidthrequiredfortransmissionofthis60sec/min60sec/minPCMsignal.ForbinaryPCMD=RnB-900Hz23-9An850-MbyteharddiskisusedtostorePCMdata.Supposethatavoice-frequency(VF)signalissampledat8ksamples/sandtheencodedPCMistohaveanaverageSNRofatleast30dB.Howmanyminute

8、sofVFconversation(i.e.,PCMdata)canbestoredontheharddisk?Solution:(S)一=M210lgM230dBM=2n(kbits)(byte)40Isec丿(8bits丿R二850Mbytes850 x106IN丿(8ksamples、(5bits.sec丿(sample丿6.02n=30Tn=5kbits40sec二5kbytessec=sec=170 x103sec5kbytes5x103sec=47hrs,13minnT=170 x103sec=170X103S=2,833min3-10Ananalogsignalwithaband

9、widthof4.2MHzistobeconvertedintobinaryPCMandtransmittedoverachannel,Thepeak-signalquantizingnoiseratioatthereceiveroutputmustbeatleast55dB.(a)IfweassumethatandthattherePe=0isnoISI,whatwillbethewordlengthandthenumberofquantizingstepsneeded?(b)Whatwillbetheequivalentbitrate?Whatwillbethechannelnullban

10、dwidthrequiredifrectangularpulseshapesareused?Solution:IfweassumethatandthatthereisnoP=0eISI,whatwillbethewordlengthandthenum-berofquantizingstepsneeded?Using(3-18),|dB=6.02n+4.7755nn8.34nusen=9bits(N丿peakv7/wordlengthM=2n=29=512quantizingsteps(b)f=2f=2(4.2MHz)=8.4Msamples/secsanalog8.4Msamples9bits

11、八Sample丿ForrectangularpulseshapesecMbits=75.6-secB=R=75.6MHznull3-12Givenanaudiosignalwithspectralcomponentsinthefrequencyband300to3000Hz,assumethatasamplingrateof7KHzwillbeusedtogenerateaPCMsignal.DesignanappropriatePCMsystemasfollows:DrawablockdiagramofthePCMsystem,includingthetransmitter,channel,

12、receiver.Specifythenumberofuniformquantizationstepsneededandthechannelnullbandwidthrequired,assumethatthepeaksignal-to-noiseratioatthereceiveroutputneedstobeatleast30dBandthatpolarNRZsignalingisused.DiscusshownonuniformquantizationcanM19.6useM=20M二2nn=5beusedtoimprovetheperformanceofthesystem.Soluti

13、on:(a)略(b)dBN丿Peak-6.02n+4.7730nn4.10nusen=5bitsV-wordlengthM-2n-25-32quantizingstepsf-7Ksamples/secs7Ksamples5bits、sec丿、Sample丿35竺secB-R-35(KHz)null(c)uniformquantizing:forallsamples,thequantizingnoisepoweristhesame52N-12SsmallsignalTdNbigsignalS個TINuniformquantizingisnotgoodforsmallsignal.Nonunifo

14、rmquantizing:samplesarenonlinearprocessed,SmallsignalisamplifiedSTTN(orsmallsignal-usingsmallstepsizeS)TTN3-14InaPCMsystem,thebitserrorrateduetochannelnoiseis10-4.Assumethatpeaksignal-to-noiseratioonthereceivedanalogsignalneededtobeatleast30dB.Findtheminimumnumberofquantizingstepsthatcanbeusedtoenco

15、detheanalogsignalintoaPCMsignal.Iftheoriginalanalogsignalhadanabsolutebandwidthof2.7kHz,whatisthenullbandwidthofPCMsignalforthepolarNRZsignalingcase.Solution:(a)Pe二10-4(s)一30dB(N丿PKout(S)3M2一=fx1000(N丿1+4IM2-1丿PePKoutf=2x2.7=5.4KzsR二nf=5x5.4=27Kb/sB二R=27KHzsnull3-17Fora4bitPCMsystem,calculateandsket

16、chaplotoftheoutputSNR(indecibels)asafunctionoftherelativeinputlevel,2?！?,v)rmsforAPCMsystemthatuseslaw10compandingAPCMsystemthatusesuniformquantizationWhichofthesesystemisbettertouseinpractice?Why?Solution:n=4bitsaPCMword(a)(S/N)=6.02n+4.77dB-20lgin(1+卩)=6.02x4+4.77-20lgln(1+10)=21.25dB(b)(S/N)dB=6.

17、02n+4.77+20lg(x/V)=6.02x4+4.77+20lg(x/V)rms$rms=28.85+20lg(x201gXrms/V-30-20-15-10-5-3.01-1.158.8513.8518.8523.8525.843-19Amultileveldigitalcommunicationsystemsendsoneof16possiblelevelsoverthechannelevery0.8ms.(a)Whatisthenumberofbitscorrespondingtoeachlevel?(b)Whatisthebaudrate?Whatisthebitrate?Sol

18、ution:(a)Whatisthenumberofbitscorrespondingtoeachlevel?L=21=16nl=4bits/level(b)Whatisthebaudrate?N1symbolD=1,250baudT0.8x10-3sec(c)Whatisthebitrate?R=ID=4(1,250)=5kbits/sec3-20Amultileveldigitalcommunicationsystemistooperateatadatarateof9,600bits/s.If4-bitwordsareencodedintoeachlevelfortransmissiono

19、verthechannel,whatistheminimumrequiredbandwidthforthechannel?(b)Repeatpart(a)forthecaseof8-bitencodingintoeachlevel.Solution:(a)If4-bitwordsareencodedintoeachlevelfortransmissionoverthechannel.Whatisthemin-imumrequiredbandwidthforthechannel?Repeatpart(a)forthecaseof8-bitencodingintoeachlevel.9600D=1

20、200baud8B=600Hzmin11BD=(1200)=600Hz222b2b3-24Considerarandomdatapatternconsistingofbinary1sand0s,wheretheprobabilityofobtainingeitherabinary1orabinary0is1.CalculatethePSDforthefollowingtypesofsignalingformatsasafunctionofT,thetimeneededtosend1bitofbdata:(a)PolarRZsignalingwherethepulsewidthis(b)Manc

21、hesterRZsignalingwherethepulsewidthis1-T4b.Whatisthefirstnullbandwidthofthesesignals?Whatisthespectralefficiencyforeachofthesesignalingcases?Solution:(a)PolarRZsignalingwherethepulsewidth1isF(f)=Ff(t)=Tsin(兀fT/2)bb2兀fT/2b1b:1Ta:+AV,依概率一nn2andthedataareindependentfrombittobit10TAV,依概率For:aann+kandI=2

22、(-A2R(0)=工(aa)P二A2xl+(A)2x=A222For:疋and7=400011011(_A)AA(A)AA概率1/41/41/41/442=1=A2x1/4+(-A)Axl/4+xl/4+(-Afxl/4=0ThusR(k)=A2,k二00,k豐0(3-40)/、F(f)2乎p(f)=-Lr(k2沢kfTsSTsk=g(336a)(T、bTI2丿b2sin2(兀fT/2)bA2(兀fT/2)2bA2Tsin2(兀fT/2)bb4(兀fT/2)2b2R4R2Thefirst-nullbandwidthisB=2RandnullTbthebandwidthefficiencyis(

23、b)ManchesterRZsignalingwherethe1Pulsewidthis一-Tb.WhatisthefirstnuHband-widthofthesesignals?Whatisthespectralefficiencyforeachofthesesignalingcases?22Equation(3-36)canalsobeusedtoevaluatethePSDforRZManchestersignalingwherethepulseshapeisshowninthefigure.F(f)=T廠sin(Kfr)、Tje2Tj-e2(j2tsin(兀fT).(sinUsing

24、(3-40)and(3-36),ManchestersignalingisthePSDfor4A2T2P(f)=TbsinGift)GfT)2sin(兀fT)2f1t，thisbecomes4b(兀)sinfTI4b丿(兀)fTI4b丿(兀sinfTI4b丿Thefirst-nullbandwidthisandthenullTbspectralefficiencyis=1(bits/sec)/Hz.43-29Thedatastream01101000101appearsattheinputofadifferentialencoder.Dependingontheinitialstart-upc

25、onditionoftheencoder,findouttwopossibledifferentialencodeddatastreamsthatcanappearattheoutput.Solution:n-10123456789100110100010101001111001乞110I100001103-30Createapracticalblockdiagramforadifferentialencodingsystem.Explainhowthesystemworkbyshowingtheencodinganddecodingforthesequence001111010001.Ass

26、umethatthereferencedigitisabinary1.Showthaterrorpropagationcannotoccur.Solution:d必務一1紅DifferentialencoderDifferentialdecodern-101*34567891011001111010000陷11101011000011110101100001*0011110100003-34TheinformationinananalogwaveformisfirstencodedintobinaryPCMandthenconvertedtoamultilevelsignalfortransm

27、issionoverthechannel.Thenumberofmultilevelsiseight.Assumethattheanalogsignahasabandwidthof2700Hzandistobereproducedatthereceiveroutputwithanaccuracyof+1%(fullscall).DeterminetheminimumbitrateofthePCMsignal.Determinetheminimumbaudrateofthemultilevelsignal.Determinetheminimumabsolutechannelbandwidthre

28、quiredfortransmissionofthisPCMsignal.Solution:6士7612V=1%t=TM=50Tn=62V4V1006(a)R=nf=6x2x2700=32.4kb/ss(b)L=21L=8l=3R32.4D=10.8kBdl3D(c)B=5.4kHzmin23-35Abinarywaveformof9600bits/sisconvertedintoanoctal(Multilevel)waveformthatispassthroughachannelwitharaisedcosine-rolloffNyquistfiltercharacteristic.The

29、channelhasaconditioned(equalized)phaseresponseoutto2.4kHz.(a)Whatisthebaudrateofthemultilevelsignal?(b)Whatistherollofffactorofthefiltercharacteristic?=B=fQ(l+r)Solution:R9600L=8Tl=3D=3200Bdl3DB=f(1+r)=(1+r)=2.4kHztr=0.5023-37Abinarycommunicationsystemusespolarsignal.Theoverallimpulseresponseissinxd

30、esignedtobeofthetype,asgivenbyxEq(3-67),sothattherewillbenoISI.Thebitrateisr=f=300bit/ssWhatisthebaudrateofthepolarsignal?Plotthewaveformofpolarsignalatthesystemoutputwhentheinputbinarydatais01100101Canyoudiscernthedatabylookingatthispolarwaveform?Solution:兀1H(f)=nefSh(t)=sin兀fstsfB二厶=150Hzt21=DTS(b

31、)3-43Usingtheresultsofprob.3-42,demonstratethatthefollowingfiltercharacteristicsdoordonotsatisfyNyquists2=2f=).0T0(a)H(f)=Llne2(1、fTI2(b)H(f)=Llne2Solution:(1)-f=化n（f）I20丿2/丿(2fTI30丿C)H(f)=初criterionforeliminatingISI(f(b)H(f)=eTkn2(2)-fTI30丿Tkn3-45AnanalogsignalistobeconvertedintoaPCMsignalthatisabi

32、narypolarNRZlinecode.Thesignalistransmittedoverachannelthatisabsolutelybandlimitedto4kHz.AssumethatthePCMquantizerhas16stepsandthattheoverallequivalentsystemtransferfunctionisoftheraisedcosine-rollofftypewithr=0.5.FindthemaximumPCMbitratethatcanbesupportedbythissystemwithoutintroducingISI.Findthemax

33、imumbandwidththatcanbepermittedfortheanalogsignal.Solution:量化器:M=16n=4TaPCMwordB=4kHzr=0.5TB(a)R=D=2f=2x匚=5.33kb/s01+rH0.50-5fofol-5foFf(b)R=nfn-2BsanalogR5.33x1000B=667Hzanalog2n2x43-47Multileveldatawithanequivalentbitrateof2,400bits/sissentoverachannelusingafour-levellinecodethathasarectangularpul

34、seshapeattheoutputofthetransmitter.Theoveralltransmissionsystem(i.e.,thetransmitter,channel,andreceiver)hasanr=0.5raisedcosine-rolloffNyquistfiltercharacteristic.(a)Findthebaudrateofthereceivedsignal.(b)Findthe6-dBbandwidthforthistransmissionsystem.Findtheabsolutebandwidthforthesystem.Solution:(a)Fi

35、ndthebaudrateofthereceivedsignal.L=21=4n/=22400D=R/1=1200Baud2(b)Findthe6-dBbandwidthforthistransmissionsystem.B6dB11=D=(1200)22=600HzFindtheabsolutebandwidthforthesystem.113B=B=(1+r)D=_(1+0.5)(1200)=(1200)=900HzabsoluteT2243-54Oneanalogwaveformw1(t)isbandlimitedto3kHz,andanother,w2(t),isbandlimited

36、to9kHz.ThesetwosignalsaretobesentbyTDMoveraPAM-typesystem.(a)Determinetheminimumsamplingfrequencyforeachsignal,anddesignaTDMcommutatoranddecommutatortoaccommodatethesignals.(b)Drawsometypicalwaveformsforw1(t)andw2(t),andsketchthecorrespondingTDMPAMwaveform.Solution:(a)Determinetheminimumsamplingfreq

37、uencyforeachsignal,anddesignaTDMcommutatoranddecommutatortoaccommodatethesignals.TDM(t):B=3kHz1nf=6ksamples/secs1nf=18ksamples/secs2(t):B=9kHz2(b)Drawsometypicalwaveformsforw1(t)andw2(t),andsketchthecorrespondingTDMPAMwaveform.3-56Twenty-threeanalogsignals,eachwithabandwidthof3.4kHz,aresampledatan8-kHzrateandmultiplexedtogetherwithasynchronizationchannel(8kHz)intoaTDMPAMsignal.ThisTDMsignalispassedthroughachannelwithanoverallraisedcosine-rollof