when i tried this program getting the error as Subscripted assignment dimension mismatch Error in ==> mrczf at 79 EbN0EffSim(ii,jj) = (mean(abs(yHat))+simBer) ;.

clear N = 10^6; % number of bits or symbols Eb_N0_dB = [0:25]; % multiple Eb/N0 values nTx = 2; %number of transmitters nRx = 2; %number of receivers for ii = 1:length(Eb_N0_dB)

    % Transmitter
    ip = rand(1,N)>0.5; % generating 0,1 with equal probability
    s = 2*ip-1; % BPSK modulation 0 -> -1; 1 -> 0

    sMod1 = kron(s,ones(nRx,1)); % 
    sMod1 = reshape(sMod1,[nRx,nTx,N/nTx]); % grouping in [nRx,nTx,N/NTx ] matrix
    %conj vi du 1 ham 1+j se tra lai 1-j va nguoc lai
    %reshape: sap xep lai cac phan tu cua 1 ma tran 3x4 thang 1 ma tran 2x6
    %voi thu tu uu tien giu nguyen cot.

    h1 = 1/sqrt(2)*[randn(nRx,nTx,N/nTx) + j*randn(nRx,nTx,N/nTx)]; % Rayleigh channel
    n1 = 1/sqrt(2)*[randn(nRx,N/nTx) + j*randn(nRx,N/nTx)]; % white gaussian noise, 0dB variance

    % Channel and noise Noise addition
    y1 = squeeze(sum(h1.*sMod1,2)) + 10^(-Eb_N0_dB(ii)/20)*n1;
    %B = squeeze(A) returns an array B with the same elements as A, but
    %with all singleton dimensions removed.
    % Receiver

    % Forming the Zero Forcing equalization matrix W = inv(H^H*H)*H^H
    % H^H*H is of dimension [nTx x nTx]. In this case [2 x 2] 
    % Inverse of a [2x2] matrix [a b; c d] = 1/(ad-bc)[d -b;-c a]

    h1Cof = zeros(2,2,N/nTx)  ; 
    h1Cof(1,1,:) = sum(h1(:,2,:).*conj(h1(:,2,:)),1);  % d term
    h1Cof(2,2,:) = sum(h1(:,1,:).*conj(h1(:,1,:)),1);  % a term
    h1Cof(2,1,:) = -sum(h1(:,2,:).*conj(h1(:,1,:)),1); % c term
    h1Cof(1,2,:) = -sum(h1(:,1,:).*conj(h1(:,2,:)),1); % b term
    h1Den = ((h1Cof(1,1,:).*h1Cof(2,2,:)) - (h1Cof(1,2,:).*h1Cof(2,1,:))); % ad-bc term
    h1Den = reshape(kron(reshape(h1Den,1,N/nTx),ones(2,2)),2,2,N/nTx);  % formatting for division
    h1Inv = h1Cof./h1Den; % inv(H^H*H)

    h1Mod =  reshape(conj(h1),nRx,N); % H^H operation

    y1Mod1 = kron(y1,ones(1,2)); % formatting the received symbol for equalization
    y1Mod1 = sum(h1Mod.*y1Mod1,1); % H^H * y 
    y1Mod1 =  kron(reshape(y1Mod1,2,N/nTx),ones(1,2)); % formatting
    y1Hat1 = sum(reshape(h1Inv,2,N).*y1Mod1,1); % inv(H^H*H)*H^H*y

    % receiver - hard decision decoding
    ipHat1 = real(y1Hat1)>0;

    % counting the errors
    nErr(ii) = size(find([ip- ipHat1]),2);

end

simBer = nErr/N; % simulated ber

for jj = 1:length(nRx)

    for ii = 1:length(Eb_N0_dB)

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

        % Channel and noise Noise addition
        sD = kron(ones(nRx(jj),1),s);%sD = 1; 
        y = h.*sD + 10^(-Eb_N0_dB(ii)/20)*n;

        % maximal ratio combining
        yHat =  sum(conj(h).*y,1); 

        % effective SNR
        EbN0EffSim(ii,jj) = (mean(abs(yHat))+simBer) ;

    end
end

%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(nRx, EbN0EffSim,'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, ZF)'); xlabel('Average Eb/No,dB'); ylabel('Bit Error Rate'); title('BER for BPSK modulation with 2x2 MIMO and ZF equalizer (Rayleigh channel)');

Answer 1

[Walter Roberson]

simerr is a vector. I did not examine whether yhat is a vector. You add something to simerr so the results have to be a vector. You try to store that complete vector in the single numeric array position you designated on the left hand side of the assignment