# MIMO Rayleigh fading Channel Capacity

### Hamid Ramezani (view profile)

02 Oct 2006 (Updated )

maximum capacity of a mimo channel is considered here.

MIMO_System.m
```% in 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 algorithm

% the pdf of the matrix lanada elements is depicted too.

clear all
close all
clc

nt_V = [1 2 3 2 4];
nr_V = [1 2 2 3 4];

N0 = 1e-4;
B  = 1;
Iteration = 1e4; % must be grater than 1e2

SNR_V_db = [-10:3:20];
SNR_V    = 10.^(SNR_V_db/10);

color = ['b';'r';'g';'k';'c'];
notation = ['-o';'->';'<-';'-^';'-s'];

for(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);
landas(:,j)  = diag(V);
[Capacity(i,j) PowerAllo] = WaterFilling_alg(Pt,landas(:,j),B,N0);
end
end

f1 = figure(1);
hold on
plot(SNR_V_db,mean(Capacity'),notation(k,:),'color',color(k,:))

f2 = figure(2);
hold on
[y,x] = hist(reshape(landas,[1,min(nt,nr)*Iteration]),100);
plot(x,y/Iteration,'color',color(k,:));
clear landas
end

f1 = figure(1)

legend_str = [];
for( i = 1 : length(nt_V))
legend_str =[ legend_str ;...
{['nt = ',num2str(nt_V(i)),' , nr = ',num2str(nr_V(i))]}];
end
legend(legend_str)
grid on
set(f1,'color',[1 1 1])
xlabel('SNR in dB')
ylabel('Capacity bits/s/Hz')

f2 = figure(2)
legend(legend_str)
grid on
set(f2,'color',[1 1 1])
ylabel('pdf of elements in matrix landa in svd decomposition of marix H')
```