1、是协方差矩阵为的零均值循环对称复高斯变换噪声向量。向量必须满足已限制总的发送能量。可以看出,i=1,2,rMIMO信道的容量是单个平行SISO信道容量之和,由下式给出(i=1,2,r)反映了第i个子信道的发送能量,且满足可以在子信道中分配可变的能量来最大化互信息。现在互信息最大化问题就变成了:最大化目标在变量中是凹的,用拉格朗日法最大化。最佳能量分配政策注水算法:Step1:迭代计数p=1,计算Step2:用计算,i=1,2,r-p+1Step3:若分配到最小增益的信道能量为负值,即设,p=p+1,转至Step1.若任意非负,即得到最佳注水功率分配策略。1.2 发送端知道信道时的信道容量% i
2、n this programe a highly scattered enviroment is considered. The% Capacity of a MIMO channel with nt transmit antenna and nr recieve% antenna is analyzed. The power in parallel channel (after% decomposition) is distributed as water-filling algorithmclear allclose allclcnt_V = 1 2 3 2 4;nr_V = 1 2 2
3、3 4;N0 = 1e-4;B = 1;Iteration = 1e2; % must be grater than 1e2SNR_V_db = -10:3:20;SNR_V = 10.(SNR_V_db/10);color = b;rgkm;notation = -o-sfor(k = 1 : 5) nt = nt_V(k); nr = nr_V(k); for(i = 1 : length(SNR_V) Pt = N0 * SNR_V(i); for(j = 1 : Iteration) H = random(rayleigh,1,nr,nt); S V D = svd(H); landa
4、s(:,j) = diag(V); Capacity(i,j) PowerAllo = WaterFilling_alg(Pt,landas(:,j),B,N0); end f1 = figure(1); hold on plot(SNR_V_db,mean(Capacity),notation(k,:),color,color(k,:) clear landasendf1 = figure(1)legend_str = ;for( i = 1 : length(nt_V) legend_str = legend_str ;. nt = ,num2str(nt_V(i), , nr = ,nu
5、m2str(nr_V(i);legend(legend_str)grid onset(f1,1 1 1)xlabel(SNR in dB)ylabel(Capacity bits/s/Hz注水算法子函数function Capacity PowerAllo = WaterFilling_alg(PtotA,ChA,B,N0);% WaterFilling in Optimising the Capacity%=% InitializationChA = ChA + eps;NA = length(ChA); % the number of subchannels allocated toH =
6、 ChA.2/(B*N0); % the parameter relate to SNR in subchannels% assign the power to subchannelPowerAllo = (PtotA + sum(1./H)/NA - 1./H;while(length(find(PowerAllo 0) IndexN = find(PowerAllo 0); MP = length(IndexP); PowerAllo(IndexN) = 0; ChAT = ChA(IndexP); HT = ChAT.2/(B*N0); PowerAlloT = (PtotA + sum
7、(1./HT)/MP - 1./HT; PowerAllo(IndexP) = PowerAlloT;PowerAllo = PowerAllo. Capacity = sum(log2(1+ PowerAllo. .* H);注意:的奇异值,所以对H奇异值分解后要平方ChA.21.3 发送端不知道信道时的信道容量功率均等发送,信道容量的表达式为:osCapacity(i,j)=log2(det(eye(nr)+Pt/(nt*B*N0)* H*H);f2= figure(2);f2= figure(2)set(f2,1.4 已知信道和未知信道容量比较notation_uninf= Capaci
8、ty_uninf(i,j)=log2(det(eye(nr)+Pt/(nt*B*N0)* H*Hplot(SNR_V_db,mean(Capacityhold onplot(SNR_V_db,mean(Capacity_uninf),notation_uninf (k,:clear landas由图形中可以看出:1. 在小信噪比时,相同信噪比下利用CSI的功率注水算法获得容量优于未知CSI的平均功率分配算法;相同容量下已知CSI信噪比比未知CSI时的信噪比小3dB.2. 当信噪比增大到一定程度时,功率注水算法所获得的信道容量将收敛到平均功率分配的信道容量。Welcome ToDownload !欢迎您的下载,资料仅供参考!