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I'm looking into equalisation for FSK over HF channels for Digital Voice applications. As a starting point, I am trying to understand the CMA algorithm.

I found a nice CMA Matlab example for BPSK in a previous post here on this site.

I have modified the simulation to use 2FSK but it doesn't converge to the same result. The first plot is for BPSK modulation, the 2nd plot for FSK. The combination of the channel and equaliser impulse response is converging to 1 for BPSK.

CMA with BPSK

CMA with FSK

Any suggestions of how I can get this simulation working for FSK would be much appreciated!

Here is the source code. the variable "tx_type" switches the tx modulation source.

N = 20000;          % # symbols
h = [1 0.45 -0.2];  % channel impulse response
h = h/norm(h);
Le = 20;            % equalizer length
mu = 1E-3;          % step size
snr = 30;           % snr in dB
M = 10;             % oversample rate

tx_type = "bpsk";   % select modulation type here "bpsk" or "fsk"

if strcmp(tx_type, "bpsk")
  s0 = round( rand(N,1) )*2 - 1;     % BPSK signal
  s0M = zeros(N*M,1);                % oversampled BPSK signal
  k = 1;
  for i=1:M:N*M
   s0M(i:i+M-1) = s0(k);
    k ++;
  end
end

if strcmp(tx_type, "fsk")
  tx_bits = round(rand(1,N));

  % continuous phase FSK modulator

  w1 = pi/4;
  w2 = pi/2;
  tx_phase = 0;
  tx = zeros(M*N,1);

  for i=1:N
    for k=1:M
      if tx_bits(i)
        tx_phase += w2;
      else
        tx_phase += w1;
      end
      tx((i-1)*M+k) = exp(j*tx_phase);
    end
  end

  s0M = real(tx);
end

s = filter(h,1,s0M);                % filtered signal

% add Gaussian noise at desired snr

n = randn(N*M,1);
vs = var(s);
vn = vs*10^(-snr/10);
n = sqrt(vn)*n;
r = s + n;          % received signal

e = zeros(N*M,1);   % error
w = zeros(Le,1);    % equalizer coefficients
w(Le)=1;            % actual filter taps are flipud(w)!

yd = zeros(N*M,1);

for i = 1:N*M-Le,
    x = r(i:Le+i-1);
    y = w'*x;
    yd(i)=y;
    e(i) = y^2 - 1;
    w = w - mu * e(i) * y * x;
end

np = 100;           % # sybmols to plot (last np will be plotted); np < N!

figure(1); clf;
subplot(311), plot(e.*e), title('error')
subplot(312), stem(conv(flipud(w),h)), title('equalized channel impulse response')
subplot(313);
plot(1:np,s0M(N-np+1:N),1:np,yd(N-np+1-Le+1:N-Le+1))
title('transmitted and equalized signal'), legend('transmitted','equalized'), axis([0,np,-1.5,2])

Thanks,

David

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The confusion comes from the fact that what is tagged as "transmitted" isn't the real transmitted waveform but its baseband representation, which are $\left\{+1,-1\right\}$ symbols in 1 dimension. For 2-FSK, that representation would be $\left\{\left(+1,0\right), \left(0,+1\right)\right\}$ symbols in 2 dimensions.

A common convenient way to handle such 2 dimensional symbols is to represent them with complex number, mapping the coordinate $\left(+1,0\right)$ to the purely real value $+1$ and the coordinate $\left(0,+1\right)$ to the purely imaginary value $j=\sqrt{-1}$. Using that mapping, the symbol generation code for 2-FSK becomes:

if strcmp(tx_type, "fsk")
  tx_bits = round(rand(1,N));
  s0 = zeros(N,1);
  s0M = zeros(N*M,1);

  for i=1:N
    if tx_bits(i)
      s0(i) = 1i;
    else
      s0(i) = 1;
    end
  end

  k = 1;
  for i=1:M:N*M
   s0M(i:i+M-1) = s0(k);
    k ++;
  end
end

The problem then is that the CMA weight update rule from the original post is given for a purely real signal. The adjusted weight update rule for complex signals is given by:

$$ \begin{align} y &= \vec w_{n}^H x\\ e_n &= \left|y\right|^2 - 1 \\ \vec w_{n+1} &= \vec w_{n} - \mu \vec \nabla e_n \\ &= \vec w_{n} - \mu' e_n \Re\left\{y^\star x\right\} \end{align} $$

This can be implemented with:

for i = 1:N*M-Le,
    x = r(i:Le+i-1);
    y = w'*x;
    yd(i)=y;
    e(i) = abs(y)^2 - 1;
    w = w - mu * e(i) * real(conj(y) * x);
end

Plotting the results (showing the real part and imaginary parts in separate graphs) should give you something like the following:

CMA equalization simulation for 2-FSK

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  • $\begingroup$ Thank you, that fill sin some pieces of the puzzle. I recall that mapping from BER derivations for FSK. $\endgroup$ – David Rowe Jan 23 '17 at 19:05
  • $\begingroup$ OK, so I need to operate the equaliser on the real (or possibly complex) passband signals from a channel, i.e two tones f1 and f2 in noise. Could you please tell me (or point me at a link) that explains how take the real passband signal and convert it to the mapping above? Is it perhaps the output from the FSK rx filters? $\endgroup$ – David Rowe Jan 23 '17 at 19:13
  • $\begingroup$ Or looking at it another way - can the CMA simulation operate on bandpass BPSK or FSK signals where multipath occurs on the bandpass signal? $\endgroup$ – David Rowe Jan 23 '17 at 23:12
  • $\begingroup$ Yes, the output from the FSK frontend should provide you with two signals (one for $f_1$ and another $f_2$). Provided some care is taken to avoid cross-contamination (so that those two received signals remain orthogonal), those can be mapped to the real & imaginary parts respectively. $\endgroup$ – SleuthEye Jan 24 '17 at 0:26
  • $\begingroup$ I don't have an authoritative detailed reference link handy at the moment, but a quick search turned these slides from University of Maryland and these lecture notes from University of Cape Town, providing an outline of some common FSK receiver structures. $\endgroup$ – SleuthEye Jan 24 '17 at 0:42
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OK when the complex version of the weight update is used, it works for bandpass FSK signals. I guess that makes sense, a complex bandpass FSK signal has constant modulus. Figures below show it is equalising a simulated multipath channel.

Problem for my application is it only works for high SNRs, I want to run 800 bit/s 4FSK down to Eb/No=6 dB, SNR = 6+10*log10(800/3000) = 0.26dB in a 3000 Hz noise bandwidth. Suspect I'll need pilot symbols for that.

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