无线通信Andrea-goldsmith-第10章答案

Chapter101.(a)(AAH)T=(AH)T:AT=(AT)TAT=AAH)(AAH)H=AAHForAAH;¸=¸,i.e.eigen-valuesarerealAAH=Q¤QH(b)XHAAHX=(XHA)(XHA)H=kXHAk¸0)AAHispositivesemide¯nite.(c)IM+AAH=IM+Q¤QH=Q(I+¤)QHAHpositivesemide¯nite)¸i¸08i)1+¸i08i)IM+AAHpositivede¯nite(d)det[IM+AAH]=det[IM+Q¤QH]=det[Q(IM+¤M)QH]=det[IM+¤M]=¦Rank(A)i=1(1+¸i)det[IN+AHA]=det[IN+eQ¤eQH]=det[eQ(IN+¤N)eQH]=det[IN+¤N]=¦Rank(A)i=1(1+¸i)*AAHandAHAhavethesameeigen-value)det[IM+AAH]=det[IN+AHA]2.H=U§VTU=24¡0:47930:8685¡0:1298¡0:5896¡0:4272¡0:6855¡0:6508¡0:25130:716435§=241:70340000:71520000:130235V=2664¡0:34580:68490:4263¡0:57080:21910:0708¡0:7116¡0:61090:0145¡0:21980:3311¡0:901737753.H=U§VTLetU=2410010035V=2410010035§=·1002¸)H=24100020000354.Checktherankofeachmatrixrank(HI)=3)multiplexinggain=3rank(H2)=4)multiplexinggain=45.C=RHXi=1log2µ1+¸i½Mt¶ConstraintPVi=½P¸i=constant)@C@¸i=½Mtln21(1+¸i½Mt)¡½Mtln21(1+¸i½Mt)=0)¸i=¸j)whenallRHsingularvaluesareequal,thiscapacityismaximized.6.(a)AnymethodtoshowH¼U¤Visacceptable.Forexample:D=24:13:08:11:05:09:14:23:13:1035where:dij=¯¯¯Hij¡HHij¯¯¯£100(b)precoding¯lterM=V¡1shaping¯lterF=U¡1F=24¡:5195¡:3460¡:7813¡:0251¡:9078:4188¡:8540:2373:462935M=24¡:2407¡:8894:3887¡:4727¡:2423¡:8473¡:8478:3876:362235ThusY=F(H)MX+FN=U¤U¤VV¤X+U¤N=¤X+U¤N(c)PiP=1°o¡1°ifor1°i1°o,0else°i=¸i2PNoB=94.5fori=1,6.86fori=2,.68fori=3Assume°2°0°3since°3=.68isclearlytoosmallfordatatransmissionPPiP=1)2°0¡1°1¡1°2=1)°0=1:73P1P=:5676P2P=:4324C=B£log2¡1+°1P1P¢+log2¡1+°2P2P¢¤=775.9kbps(d)Withequalweightbeamforming,thebeamformingvectorisgivenbyc=1p(3)[111].TheSNRisthengivenby:SNR=cHHHHcN0B=(:78)(100)=78:(1)Thisgivesacapacityof630.35kbps.TheSNRachievedwithbeamformingissmallerthanthebestchannelinpart(c).Ifwehadchosenctoequaltheeigenvectorcorrespondingtothebesteigenvalue,thentheSNRwithbeamformingwouldbeequaltothelargestSNRinpart(c).ThebeamformingSNRforthegivencisgreaterthanthetwosmallesteigenvaluesinpart(c)becausethechannelmatrixhasonelargeeigenvalueandtwoverysmalleigenvalues.7.C=maxBlog2det[IM°+HRXHH]RX:T°(RX)=½Ifthechannelisknowntothetransmitter,itwillperformanSVDdecompositionofHasH=U§VHRXHH=(U§V)RX(U§V)HByHadamard'sinequalitywehavethatforA2n£ndet(A)·¦ni=1Aiiwithequalityi®Aisdiagonal.WechooseRXtobediagonal,say=­thendet(IMR+HRXHH)=det(I+­§2))C=maxPi½i·½B¾ilog2(1+¸i½i)wherep¸iarethesingularvalues.8.ThecapacityofthechannelisfoundbythedecompositionofthechannelintoRHparallelchannels,whereRHistherankofthechannelmatricH.C=max½i:Pi½i·½XiBlog2(1+¸i½i)wherep¸iaretheRHnon-zerosingularvaluesofthechannelmatrixHand½istheSNRconstraint.°i=¸i½ThentheoptimalpowerallocationisgivenasPiP=½1°0¡1°i°i¸°00°i°0(2)forsomecut-o®value°0.TheresultingcapacityisgivenasC=Xi:°i¸°0Blog2(°i=°0)ForH=266411¡1111¡1¡11111111¡13775RH=3,°1=80,°2=40,°3=40.We¯rstassumethat°0islessthantheminimum°iwhichis40.°0=31+P3i=11°iwhichgives°0=2.8236mini°i,hencetheassumptionwascorrect.CB=12:4732bits/sec/HzForH=2664111¡111¡111¡1111¡1¡1¡13775RH=4,°1=40,°2=40,°3=40,°4=40.We¯rstassumethat°0islessthantheminimum°iwhichis40.°0=41+P4i=11°iwhichgives°0=3.6780mini°i,hencetheassumptionwascorrect.CB=13:7720bits/sec/Hz9.H=266664h11:::h1Mt:::::::::::::::hMr1:::hMrMt377775Mr£MtDenoteG=HHTlimMt!11MtGii=limMt!11Mt[hi1:::hiMt]266664hi1:::hiMt377775=limMt!11MtMtXj=1khijk2=Ejkhijk2=¾2=18ilimMt!1;i6=j1MtGij=limMt!11Mt[hi1:::hiMt]266664hj1:::hjMt377775=limMt!11MtMtXk=1hikhjk=Ekhikhjk=Ek(hik)Ek(hjk)=08i;j;i6=j)limM!11MHHT=IM)limM!1Blog2dethIM+½MHHTi=Blog2det[IM+½IM]=Blog2[1+½]detIM=MBlog2[1+½]10.We¯ndthecapacitybyrandomlygenerating103channelinstantiationsandthenaveragingoverit.Weassumethatdistributionisuniformovertheinstantiations.MATLABCODEclear;clc;Mt=1;Mr=1;rho_dB=[0:25];rho=10.^(rho_dB/10);fork=1:length(rho)fori=1:100H=wgn(Mr,Mt,0,'dBW','complex');[F,L,M]=svd(H);forj=1:min(Mt,Mr)sigma(j)=L(j,j);endsigma_used=sigma(1:rank(H));gamma=rho(k)*sigma_used;%%Nowwedowaterfilling\gammatemp=gamma;gammatemp1=gammatemp;gamma0=1e3;whilegamma0gammatemp1(length(gammatemp1));gammatemp1=gammatemp;gamma0=length(gammatemp1)/(1+sum(1./gammatemp1));gammatemp=gammatemp(1:length(gammatemp)-1);endC(i)=sum(log2(gammatemp1./gamma0));endCergodic(k)=mean(C);end05101520250510152025ρ(dB)CergodicMt=Mr=3Mt=2Mr=3Mt=Mr=2Mt=2Mr=1Mt=Mr=1Figure1:Problem1011.We¯ndthecapacitybyrandomlygenerating104channelinstantiationsandthenaveragingoverit.Weassumethatdistributionisuniformovertheinstantiations.MATLABCODEclear;clc;Mt=1;Mr=1;rho_dB=[0:30];rho=10.^(rho_dB/10);fork=1:length(rho)fori=1:1000H=wgn(Mr,Mt,0,'dBW','complex');[F,L,M]=svd(H);forj=1:min(Mt,Mr)sigma(j)=L(j,j);endsigma_used=sigma(1:rank(H));gamma=rho(k)*sigma_used;C(i)=sum(log2(1+gamma/Mt));endCout(k)=mean(C);pout=sum(CCout(k))/length(C);whilepout.01Cout(k)=Cout(k)-.1;pout=sum(CCout(k))/length(C);endifCout(k)0;Cout(k)=0;endend0510152025300510152025ρ(dB)CoutageMt=Mr=3Mt=2Mr=3Mt=Mr=2Mt=2Mr=1Mt=Mr=1Figure2:Problem1112.P(u?nX)=PÃMrXi=1uiniX!=MrXi=1uiP(niX)=P(niX))thestatisticsofu?narethesameasthestatisticsofeachoftheseelements13.§x=ku?Hvxk=ku?Hvk2kxk2=vHHHu?Hu?Hvkxk2=vHHHHvkxk2=vHQHQvkxk2·¸maxkxk2withequalitywhenu,varethepr