fsk modulator and demodulator
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i have this code and although i added noise to it the bit error rate is still zero.. any clue?!!
%%%%%%%%%%%%%fsk mod and demod%%%%%%%%%%%%%%%%%%%%
M = 2;
k = log2(M);
EbNo = 5;
Fs = 16;
nsamp = 17;
freqsep = 8;
n=100;
msg = randint(n,1,M); % Random signal
txsig = fskmod(msg,M,freqsep,nsamp,Fs); % Modulate.
ab=abs(txsig);
ps=(sum(ab.^2))/n;
snr=30;
pn=10.^(-0.1.*snr).*ps;
noise= sqrt(pn)*randn(1,n);
G1=randn(1,n); %generation of Gaussian noise
G2=randn(1,n);
v= sqrt(power(G1,2)+ power(G2,2));
A=v(2);
theta=2*pi*rand;
msg_rx = A*exp(j*theta)*txsig + noise(3); %flat fading
msg_rrx = fskdemod(msg_rx,M,freqsep,nsamp,Fs); % Demodulate
[num,BER] = biterr(msg,msg_rrx) % Bit error rate
BER_theory = berawgn(EbNo,'fsk',M,'noncoherent') % Theoretical BER
Accepted Answer
More Answers (1)
Walter Roberson
on 21 Apr 2011
You calculate v=sqrt(G1.^2+G2.^2) where G1 and G2 are 1 x n. You then take v(2) and throw away the rest of v. What is the point of doing all of that, when you could just do
A = sqrt(randn^2+randn^2);
This hints that you are doing something wrong. As does the fact that you use only noise(3) when noise is 1 x n .
You need to check out the magnitude of noise(3) and compare it to the magnitude of A*exp(j*theta)*txsig -- if noise(3) is very small then it would be as if you had not added the noise, merely phase-shifted the signal.
4 Comments
Salma
on 21 Apr 2011
Walter Roberson
on 21 Apr 2011
It doesn't matter what noise(17) is, as you only use noise(3)
Salma
on 21 Apr 2011
Salma
on 29 Apr 2011
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