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当前位置:首页 > 高等教育 > 大学课件 > 通信原理教程(第三版)樊昌信部分课后习题答案
2.1X(t)()2cos(2),XtttP(=0)=0.5P(=/2)=0.5E[X(t)]XR(0,1)E[X(t)]=P(=0)2cos(2)t+P(=/2)2cos(2)=cos(2)sin22tttcost2.2X(t)()2cos(2),Xttt/2/2/2/21()lim()()1lim2cos(2)*2cos2()TXTTTTTRXtXtdtTttdtT222cos(2)jtjtee2222()()()(1)(1)jfjtjtjfXPfRedeeedff2.6X(t)=AcostR(tt+)=E[X(t)X(t+)]=cos*cos()EAtAt221coscos(2)cos()22AAEtRP=R(0)=22A2.10tnk-e2kRnk(1)fPnP(2)nRfPn(1)222()()2(2)kjjnnkkPfRedeedkf20kRPn(2)()nRfPn2-22-2nR2k0fPn10f2.16LC2-4002n(1)(2)(1)LCH(f)=2221221422jfCfLCjfLjfC20021()()()21inPPHLC00()exp()4CnCRLL(2)20000(0)()(0)4CnRRRL3.1()5cos1000cttm(t)=1+cos200ttttctmts1000cos5200cos1tttttt800cos1200cos251000cos51000cos200cos51000cos5400400456006004550050025fffffffS3-1LC2-4LC3.310kHZ2kHZ5kHZmf=2kHZf=5kHZ52.52fmfmf2()2(52)14kHZmBff3.863()10cos(2*1010cos2*10)sttt1231S(f)254560050040004005006006()2*102000sin2000tt200010*10kHZ2f2331010*1010fmfmf10rad3FMPM2(1)FMfmBmfB=2(10+1)*31022kHZ5.11101001000001AMI3HDBAMI3HDB5.510)(tg[5-2T12Tf110100010010111000001001011OTTtA)(tg5-25.5115-2102,21)(TttTAtg)(tg42)(2wTSaATwG2110/PPP)(1tg=)(tg,)(2tg=0)()(1fGtG0)(,2fGTmfmSaAwTSaTATmfTmGTwTSaTATTmfTmGPTfGPPTTmfTmGPTmPGTfGfGPPTfPs216416214441)1(1)()1(1)1(1)()()1(1)(42422422222212215-35.35.522TmfmSaAwPv216)(42m=1f=1/TTfSaATfSaAwPv12161216)(4242fT1T2T3T4T5O162vA162TA)(fPsf=1/Tf=1/T42424242422216216AAASaASaAS5.6)(tg5-413=T/T434112Tf15-45.61)(tg02/t1)(tg)(tg33)()(TfSaTfSafGTmfmSafGTTmfTmGPTmPGTfGfGPPTfPmm3361)(43)1(1)()()1(1)(222212215-5)(tg12/02/2/Tt2/TT/1T/2T/3T/4T/5T/6T/8T/9T/7f012/T36/1)(fP5-55.62TmfmSafPmv3361)(21m,Tf1TfSaTfSafPv1336113361)(22/Tf1222833/3/sin3613/3/sin361vP5.8)(fH5-7120B2fR5-75.815-25)(fH=0f/100fff0t,/1)(TTttg)()(2fTTSafG,,),()(),j()(0fTtftgfGtgfG)()(020tfSafth1O0f0ff)(fH20B2fR0B2fRf)(fH][0,0ff][0,0ff)(fH5.9,02/1),2cos1()(000fffHBRT)(fH5-8)(fH001412121W01212WRB02/1BSRT5.23310BRBaud5-14(a)(b)(c))(fH02/102/104/10020(1)310BRBaud(a)310aBHz/1000/10001/aBbRBBaudHz(c)310cBHz/1000/10001/cBcRBBaudHzabc(b)(c)(2)(a)(b)(c)32333323()2*10(2*10)()2*10(2*10)()10(10)abchtSathtSathtSat(a)(c)21/t(b)1/t(a)(c)(3)(b)(a)(c)(c)6.64DPSK2400b/s1800Hz4DPSKBd1200224004log2bBRR5.1120018006.82FSK10MHz10.4MHz6102BdV40A180106nW/Hz1212FSK221reePHz104.41022104.0266601BRffB3.3104.410621016002104026181202622BnArn72e1031.121reP22FSK721019.02112erfc21reerrrP6.10-2PSK2PSK-5
本文标题:通信原理教程(第三版)樊昌信部分课后习题答案
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