樊昌信-通信原理(第六版)同步辅导及习题全解
HP(x )log P(xi)(bit/HH=log2 n(biH(r)f(x)Inf(x)dxRB2= RBN log2n (B)(bit/s)。R, (bit/s)尺BN(Baud),Rb=RBN×log2N(bit/2·32400 BaudP(x4)16(2)1hTT×RP(x5)=P(x6)=P(x2)=P(x8)P(xi)loge p(ci)〔2×,log21og28+log2+4×log282.875bit/R=RB×H=2400×2.875=6900bit/(2)Ⅰ=T×R=3600×6900=2.484×10bit(3)ogImx=T×RB×Hmx=3600×2400×3=2.592×10bit2】A、B、C、DT=216161616(2)A→00,B01,C→11,D→10(1)T=2 msR500BT、2×10H= log2 M= log2 4= 2bit/Rb=RBH=500×2=1000bi②TR500BP(a )log2 p(xi)16X log21616X log1616X log×log0.25+0.453+0.524+0.522=1.75bit/Rb=RBH=500×1.75=875bit/(2)msTs=2T6=22005)A、B、C、D00A,01B,10C,11B=100B2×5×10Rb= RBlog2 4= 200 bit/sH(X3log2 5+=l0g24+i1010(0.2×2.32+0.5×2+0.3×1.74)bit/1.985bit/Rb=RB·H(X)=(100×1.985)bit/s=198.5bit/2005A、B、C、DE1/4,1/8,1/8,3/165/16;1000B(1):P(A)=P(B)=P(C)=P(D)=P(E)=16516gg2.23bit/Rb=Rg·H(X)=(1000×2.23)bi2.23×103bit/Ⅰ=(2.23×103×3600)bit028m bit(2)max(log,5)bit2.322bitImax=(2.322×1000×3600)bit=8.359Mbit1-521-12○1-1E0.105,x0.002。E6gg:3.25 bit(E)0.105x)=0.002gg8.97bi0○1-2A、B、C、DE1/4、1/8、1/8、3/165/16。p(ti)log2 p354-810g38-80g28-160g216160g216g24og2816g 2g223 bit/A、B、C、D1/4、1/8、1/81/2H(x)∑p(xi log2 p(xi)114082481.75bit/◎1A、B、C、D00A,01B,102)∑P(x,)log2H(x)=log,n(bPa PRPc PH(x)H(x)RRH(x)=log, 4=2 bitms×5R100 BaudRb=RB×H(x)=200bit/(2)P( log2 p(xi)g 2g244gg1.985bit/Rb=RB×H(x)=100×1.985=198.5bi1/3(2)1/3log2 p(x)bit∑P(x)log21/3”,P +PPP=1/4,P=3/4。log2(4/3)=0.415bitlog, 4= 2bitgH(x)=(3/4)I+(1/4)I=0.81bit128161/321121/2241000∑P(x,)log2P(x)biH(x)H(xRP(zilog p(xi)16×og112×1326.405bit/R1000 BaudRb=RBN×H(x)=1000×6.405=6405bit/s1-7300RB01RH(x)=log2 nbiR=RBN×H(x)bit/sR300Baud。H(x)=log2 2=1 bit/Rb=RB×H(x)=300×1=300bit/s1000 Baud∑p(x)og2p(x,)bitHCx)=log nbit/p(A),力(B),p(C),力(D),p(ER=RBN×H(x)bit/s,Ⅰ=RbT(bit)H(x)=2.23biRb=RB×H(x)=1000×2.23=2230biⅠ=R6×T=2230×3600=8.028×10°bitR,=RB×H(x)=1000×2.322=2322bit/sⅠ=Rb×T=2322×3600359×106bitRB RmsRBORBR2= 1bit/RB2000 Baud.5×10R=RN×H(x)=2000×1=2000bt/sH(x)=log2 4= 2bit/RB=1/(0.5×10-3)=2000BaudRb=RBN×H(x)=2000×2=4000bit/s10
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