How to do BPSK modulation using Rayleigh fadding channel?

Given a signal y= h(x) + n where n is AWGN noise and x is binary signal, h is complex channel constant. We need to determine y and then obtain x from it using Cholaski Inverse Method.

clear
N = 10^6 % number of bits or symbols

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

Eb_N0_dB = [-3:35]; % multiple Eb/N0 values

for ii = 1:length(Eb_N0_dB)

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

     % Channel and noise Noise addition
     y = h.*s + 10^(-Eb_N0_dB(ii)/20)*n; 

     % equalization
     yHat = y./h;

     % receiver - hard decision decoding
     ipHat = real(yHat)>0;

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

end

In the line

   yHat = y./h;

it is a equalization process done by simple method. I need to do the same using Cholaski Decomposition. Any help will be highly apprecialbe.

Answer 1

[possibility]

I am not sure how this is related to Cholesky decomposition. But let me describe a solution for it.

Assume, y=hs+n where h is the 1-by-N channel vector. If you know the channel at the receiver, you can do the following:

h'y=h'(hs+n)

h'y=h'hs + h'n % Since h'h is a N-by-N square matrix, inverse can be taken now. Multiply both sides with inv(h'h)

inv(h'h) h' y = inv(h'h) h'hs + inv(h'h)h'n

inv(h'h) h' y = s + inv(h'h)h'n = yhat

Hence, multiplying the received signal, y, with inv(h'h) h' results in the noisy version of s.

You can do it as follows:

yhat = inv(h'h)*h'*y

I couldn't see now whether it has to be "transpose" or "conjugate transpose". But this method is well-known as least squares.

[sangeet sagar]

The noisy signal obtained at the receiving end( as named yHat) is nxn matrix whereas the sinal sent by us over the channel is a 1x n matrix so how can we convert the obtained yHat to 1xn