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1、CHAPTER13Problem13.1:16Ec2πTcg(t)=cost−,0≤t≤Tc3TcTc2∞−j2πftG(f)=−∞g(t)edt=16EcTccos2πt−Tce−j2πftdt3Tc0Tc2Butcos2πt−Tc=11+cos2πt−Tc.ThenTc22Tc2116Ec4EcTc24EcTcG(0)=Tc=⇒G(0)=23Tc33and2216σm=4wmJavG(0)=EcTcwmJav,Ec=RcEb3Hence,MRcEbPM≤Q3wmm=2JavTcThisisanimprovementof1.76dBovertherectang
2、ularpulse.Problem13.2:ThePNspreadspectrumsignalhasabandwidthWandtheinterferencehasabandwidthW1,whereW>>W1.Uponmultiplicationofthereceivedsignalr(t)withthePNreferenceatthereceiver,wehavethefollowing(approximate)spectralcharacteristics266✻WS0=WST1/Tb0bSpectrumofInterferenceJ0W1/W0✛✲W✛✲1/TbæAftermultip
3、licationwiththePNreference,theinterferencepowerinthebandwidth1/Tboccu-piedbythesignalisJ0W11J0W1=WTbWTbPriortomultiplication,thenoisepowerisJ0W.Therefore,inthebandwidthoftheinformation-bearingsignal,thereisareductionintheinterferencepowerbyafactorWT=Tb=L,whichbTccisjusttheprocessinggainofthePNsp
4、readspectrumsignal.Problem13.3:TheconcatenationofaReed-Solomon(31,3)codewiththeHadamard(16,5)coderesultsinanequivalentbinarycodeofblacklengthn=n1n2=31×16=496bits.Thereare15informationbitsconveyedbyeachcodeword,i.e.k=k1k2=15.Hence,theoverallcoderateisRc=15/496,whichistheproductofthetwocoderates.Themi
5、nimumdistancesareReed−Solomoncode:Dmin=31−3+1=29Hadamardcode:d=n2=8min2Hence,theminimumdistanceoftheoverallcodeisdmin=28×8=232.AunionboundontheprobabilityoferrorbasedontheminimumdistanceofthecodeisEbPM≤(M−1)Q2RcdminJavTcwhereM=215=32768.Also,E=ST.Thus,bavb15SavTbkPM≤2Q2dminJavTcn267ButkTb=nTca
6、nddmn=232.Hence,15464PM≤2QJav/SavDuetothelargenumberofcodewords,thisunionboundisveryloose.AmuchtighterboundisM2wmPM≤Qm=2Jav/Savbuttheevaluationoftheboundrequirestheuseoftheweightdistributionoftheconcatenatedcode.Problem13.4:Forhard-decisiondecodingwehavedmin/2m+12m−2PM≤(M−1)[4p(1−p)]=2[4p(1−p)]
7、wherep=Q2W/RR=Q2Sav.Notethatinthepresenceofastrongjammer,theJav/SavcJavprobabilitypislarge,i.ecloseto1/2.Forsoft-decisiondecoding,theerrorprobabilityboundis2W/RPM≤(M−1)QRcdminJav/SavWn1Wese
