مساعدة في رسالة ماجستير OFDM PAM CODED AND UNCODED

نا طالب ماجستير في هندسة الاتصالات موضوع بحثي عن ofdm pam coded and uncoded in awgn and raleigh channel انا خلصت الجزء uncoded بينما تبقي لي coded with convolution codeعشان كده عايز اي حد في المنتدي يساعدني في الجزء ده ويضيفلي Convolution code علي الكود ده
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clear all
nFFT=64
nDSC = 52; % number of data subcarriers
nBitPerSym = 52; % number of bits per OFDM symbol (same as the number of subcarriers for BPSK)
nSym = 10^4; % number of symbols

EbN0dB = [0:60]; % bit to noise ratio
EsN0dB = EbN0dB + 10*log10(nDSC/nFFT) + 10*log10(64/80); % converting to symbol to noise ratio
% ht=modem.pammod(4);
% hr=modem.pamdemod(4);
N = 10^6 % number of bits or symbols
M=4; % Constellation order
ht=modem.pammod('M', M);
hr=modem.pamdemod('M', M);
for ii = 1:length(EbN0dB)

% Transmitter
ipBit = randi([0 M-1],1,nBitPerSym*nSym) ; % random 1's and 0's
% % % ipMod = 2*ipBit-1; % BPSK modulation 0 --> -1, 1 --> +1
ipMod = pammod(ipBit,M);
ipMod = reshape(ipMod,nBitPerSym,nSym).'; % grouping into multiple symbolsa

% Assigning modulated symbols to subcarriers from [-26 to -1, +1 to +26]
xF = [zeros(nSym,6) ipMod(:,[1:nBitPerSym/2]) zeros(nSym,1) ipMod(:,[nBitPerSym/2+1:nBitPerSym]) zeros(nSym,5)] ;

% Taking FFT, the term (nFFT/sqrt(nDSC)) is for normalizing the power of transmit symbol to 1
xt = (nFFT/sqrt(nDSC))*ifft(fftshift(xF.')).';

% Appending cylic prefix
xt = [xt(:,[49:64]) xt];

% multipath channel
nTap = 10;
ht = 1/sqrt(2)*1/sqrt(nTap)*(randn(nSym,nTap) + j*randn(nSym,nTap));

% computing and storing the frequency response of the channel, for use at recevier
hF = fftshift(fft(ht,64,2));

% convolution of each symbol with the random channel
for jj = 1:nSym
xht(jj, = conv(ht(jj,,xt(jj,);
end
xt = xht;

% Concatenating multiple symbols to form a long vector
xt = reshape(xt.',1,nSym*(80+nTap-1));

% Gaussian noise of unit variance, 0 mean
nt = 1/sqrt(2)*[randn(1,nSym*(80+nTap-1)) + j*randn(1,nSym*(80+nTap-1))];

% Adding noise, the term sqrt(80/64) is to account for the wasted energy due to cyclic prefix
yt = sqrt(80/64)*xt + 10^(-EsN0dB(ii)/20)*nt;

% Receiver
yt = reshape(yt.',80+nTap-1,nSym).'; % formatting the received vector into symbols
yt = yt(:,[17:80]); % removing cyclic prefix

% converting to frequency domain
yF = (sqrt(nDSC)/nFFT)*fftshift(fft(yt.')).';

% equalization by the known channel frequency response
yF = yF./hF;

% extracting the required data subcarriers
yMod = yF(:,[6+[1:nBitPerSym/2] 7+[nBitPerSym/2+1:nBitPerSym] ]);

% PAM demodulation
% +ve value --> 1, -ve value --> -1

ipBitHat=pamdemod(yMod,M);
% converting modulated values into bits
% ipBitHat = (ipModHat+1)/2;
ipBitHat = reshape(ipBitHat.',nBitPerSym*nSym,1).';

nErr(ii) = size(find(ipBitHat - ipBit),2); % couting the number of errors

end

simBer = nErr/(nSym*nBitPerSym);
EbN0Lin = 10.^(EbN0dB/10);
theoryBer = 0.5.*(1-sqrt(EbN0Lin./(EbN0Lin+1)));

close all; figure
semilogy(EbN0dB,theoryBer,'bs-','LineWidth',2);
hold on
semilogy(EbN0dB,simBer,'mx-','LineWidth',2);
axis([0 35 10^-5 1])
grid on
legend('Rayleigh-Theory', 'Rayleigh-Simulation');
xlabel('Eb/No, dB')
ylabel('Bit Error Rate')
title('BER for PAM using OFDM in a 10-tap Rayleigh channel')[/right]
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