2x2,3x3,4x4 MIMO ML or sphere decoder performance ???

Salam alikom

H r u all?? I hope u r all fine. Well, I want to know which is better 2x2 or 3x3 or 4x4 MIMO receiver whether we used ML or sphere decoder detector??

I mean, is the performance improves with number of antennas increase or decrease (nxn MIMO systems)

please help me with performance BER curves and reason behind ur answers (why??)

thanks,

Salam alikom

MIMO implement to increase the diversity. By increasing the anttennas u can get a lower BER performance

generating bit error curve is not a big deal, I can implement it. but i am new to mimo ofdm.

Salam alikom

I wanted to the answer for Spatial multiplexing techniques not Diversity ones as ML,SD,ZF receivers

thanks,

Salam alikom

For spatial multiplexing MIMO,
4x4 is better than 3x3
3x3 is better than 2x2

Hi
i have Q ??

why we use 1/sqrt(2) when we find rayleigh channel???

if possible send me the code of this , mrtn023@gmail.com , i will try to ans if i can

% Transmitter
ip = rand(1,N)>0.5; % generating 0,1 with equal probability
s = 2*ip-1; % BPSK modulation 0 -> -1; 1 -> 0
sMod = kron(s,ones(nRx,1)); %
sMod = reshape(sMod,[nRx,nTx,N/nTx]); % grouping in [nRx,nTx,N/NTx ] matrix

h = 1/sqrt(2)*[randn(nRx,nTx,N/nTx) + j*randn(nRx,nTx,N/nTx)]; % Rayleigh channel
n = 1/sqrt(2)*[randn(nRx,N/nTx) + j*randn(nRx,N/nTx)]; % white gaussian noise, 0dB variance

here man why use 1/sqrt(2) for n and h ???

% Channel and noise Noise addition
y = squeeze(sum(h.*sMod,2)) + 10^(-Eb_N0_dB(ii)/20)*n;

also here why divide by20?? and for what the minus (-) ??


thanks man

% Maximum Likelihood Receiver
% ----------------------------
% if [s1 s2 ] = [+1,+1 ]
sHat1 = [1 1];
sHat1 = repmat(sHat1,[1 ,N/2]);
sHat1Mod = kron(sHat1,ones(nRx,1));
sHat1Mod = reshape(sHat1Mod,[nRx,nTx,N/nTx]);
zHat1 = squeeze(sum(h.*sHat1Mod,2)) ;
J11 = sum(abs(y - zHat1),1);

% if [s1 s2 ] = [+1,-1 ]
sHat2 = [1 -1];
sHat2 = repmat(sHat2,[1 ,N/2]);
sHat2Mod = kron(sHat2,ones(nRx,1));
sHat2Mod = reshape(sHat2Mod,[nRx,nTx,N/nTx]);
zHat2 = squeeze(sum(h.*sHat2Mod,2)) ;
J10 = sum(abs(y - zHat2),1);

% if [s1 s2 ] = [-1,+1 ]
sHat3 = [-1 1];
sHat3 = repmat(sHat3,[1 ,N/2]);
sHat3Mod = kron(sHat3,ones(nRx,1));
sHat3Mod = reshape(sHat3Mod,[nRx,nTx,N/nTx]);
zHat3 = squeeze(sum(h.*sHat3Mod,2)) ;
J01 = sum(abs(y - zHat3),1);

% if [s1 s2 ] = [-1,-1 ]
sHat4 = [-1 -1];
sHat4 = repmat(sHat4,[1 ,N/2]);
sHat4Mod = kron(sHat4,ones(nRx,1));
sHat4Mod = reshape(sHat4Mod,[nRx,nTx,N/nTx]);
zHat4 = squeeze(sum(h.*sHat4Mod,2)) ;
J00 = sum(abs(y - zHat4),1);

% finding the minimum from the four alphabet combinations
rVec = [J11;J10;J01;J00];
[jj dd] = min(rVec,[],1);

% mapping the minima to bits
ref = [1 1; 1 0; 0 1; 0 0 ];
ipHat = zeros(1,N);
ipHat(1:2:end) = ref(dd,1);
ipHat(2:2:end) = ref(dd,2);

% counting the errors
nErr(ii) = size(find([ip- ipHat]),2);

end

simBer = nErr/N; % simulated ber
EbN0Lin = 10.^(Eb_N0_dB/10);
theoryBer_nRx1 = 0.5.*(1-1*(1+1./EbN0Lin).^(-0.5));
p = 1/2 - 1/2*(1+1./EbN0Lin).^(-1/2);
theoryBerMRC_nRx2 = p.^2.*(1+2*(1-p));

close all
figure
semilogy(Eb_N0_dB,theoryBer_nRx1,'bp-','LineWidth',2);
hold on
semilogy(Eb_N0_dB,theoryBerMRC_nRx2,'kd-','LineWidth',2);
semilogy(Eb_N0_dB,simBer,'mo-','LineWidth',2);
axis([0 25 10^-5 0.5])
grid on
legend('theory (nTx=1,nRx=1)', 'theory (nTx=1,nRx=2, MRC)', 'sim (nTx=2, nRx=2, ML)');
xlabel('Average Eb/No,dB');
ylabel('Bit Error Rate');
title('BER for BPSK modulation with 2x2 MIMO and ML equalizer (Rayleigh channel)');

normalize channel power to unity

abdulmonem89 said:

Hi
i have Q ??

why we use 1/sqrt(2) when we find rayleigh channel???

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