capacity_4.m

本代码给出基于MIMO系统下的系统容量分析

clear
%%%%%%%%%%%%%%%%%%%%%%%%%%% 4x4
Nt=4;
Nr=4;
max_snr=30;
c44=zeros(1,max_snr+1);
H=RayleighCH(Nr,Nt);
w=H*H';
u=eye(Nr,Nr);
for snr_in_db=0:max_snr
    SNR=10^(snr_in_db/10);
v=u+w/Nt*SNR;
delt=det(v);
c44(snr_in_db+1)=log2(delt);
end
%%%%%%%%%%%%%%%%%%%%%%%%%% 4x1
Nt=4;
Nr=1;
max_snr=30;
c41=zeros(1,max_snr+1);
h=randn(Nr*Nt,1)/sqrt(2)+j*randn(Nr*Nt,1)/sqrt(2);
w1=(abs(h(1))^2+abs(h(2))^2+abs(h(3))^2+abs(h(4))^2);
for snr_in_db=0:max_snr
    SNR=10^(snr_in_db/10);
c41(snr_in_db+1)=log2(1+SNR*w1/4);
end

%%%%%%%%%%%%%%%%%%%%%%%% 1x4
Nt=1;
 Nr=4;
max_snr=30;
c14=zeros(1,max_snr+1);
h=randn(Nr*Nt,1)/sqrt(2)+j*randn(Nr*Nt,1)/sqrt(2);
w=(abs(h(1))^2+abs(h(2))^2+abs(h(3))^2+abs(h(4))^2);
for snr_in_db=0:max_snr
    SNR=10^(snr_in_db/10);
c14(snr_in_db+1)=log2(1+SNR*w);
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%siso
Nt=1;
Nr=1;
max_snr=30;
c11=zeros(1,max_snr+1);
h=randn(Nr*Nt,1)/sqrt(2)+j*randn(Nr*Nt,1)/sqrt(2);
w=(abs(h)^2);
for snr_in_db=0:max_snr
    SNR=10^(snr_in_db/10);
c11(snr_in_db+1)=log2(1+SNR*w);
end
plot(0:max_snr,c44,'-*',0:max_snr,c41,'b',0:max_snr,c14,'-+',0:max_snr,c11,'r--')
legend('4Tx4Rx','4Tx1Rx','1Tx4Rx','1Tx1Rx')
xlabel('SNR (dB)')
ylabel('信道容量 (b/s/Hz)')