Verifying Channel Response Identity

Question

If $x[n]$ is input signal, $y[m]$ is output signal, $h[m]$ is impulse response of the channel, then the following relation holds

$\lvert Y(\omega)\rvert = \lvert H(\omega)\rvert^2 . \lvert X(\omega)\rvert$

where $F(\omega)$ is the DFT of $f[n]$.

To verify this, I wrote following MATLAB code

x = randn(1,1000);
h = [1 2 3 4 5];
y = conv(x,h);
plot((abs(fft(h,1024)).^2).*(abs(fft(x,1024))));
hold on
plot(abs(fft(y,1024)),'r')

Result shows that second plot is at a much lower level than first plot. When I multiply second plot by $1024$ i.e. NFFT then both the plots overlap.

What is wrong with the original equation? I remember learning it that way in class.

Answer 1

Just change the code to the following:

x = randn(1,1000);
h = [1 2 3 4 5];
y = conv(x,h);
plot((abs(fft(h,1024))).*(abs(fft(x,1024)))); % It's |H(w)||X(w)|
hold on
plot(abs(fft(y,1024)),'--r')

By mistake you raised the DFT of an impulse response to the second power. You could see that magnitudes are bit off, but peaks and valleys are more-less at the same point.

Additionally I've noticed in my MATLAB 2014a, that you get an error after running your original code. That's because vector x was a row vector, and y a column one.

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This part is to answer Drazick, claiming that this approach will produce some errors (already present in my solution).

In @jojek code there is no equality since he neglected the phase by applying the abs operator on both signals before the multiplication (As opposed to the Convolution Theorem).

If the filter had a more harsh phase his result would have much bigger error.

First of all please read my MATLAB code, maybe you will notice --r switch in second plot function. I intentionally used a dashed line, otherwise we would have just red plot and no blue at all.

But let's cut to the chase. Code is now slightly modified, as random signal is being filtered by long filter with random coefficients. This is what we get for comparison of $|H(\omega)X(\omega)|$ and $|H(\omega)||X(\omega)|$ with convolution in time domain:

Let's zoom in:

So all three are equal. Also script provides maximum difference between two methods to be around $10^{-13}$. Below script to reproduce the results:

clc, close all
x = randn(1,1000);
h = randn(1,200); % some crazy filter
y = conv(x,h);
N = length(y);

Y =  abs(fft(h,N)).*abs(fft(x,N)); % |H(w)||X(w)|
YY = abs(fft(h,N).*fft(x,N));      % |H(w)X(w)|

plot(YY,'go'); 
hold on
plot(Y,'b')    
plot(abs(fft(y,N)),'--r')
grid on
legend({'|H(w)X(w)|','|H(w)||X(w)|','conv'})

display(sprintf('Max error: %e', max(abs(YY-Y))))

Answer 2

I think you made a mistake between the power spectrum and the Fourier Transform. This is the right form of the Convolution Theory (Nothing is squared):

$$ \lvert Y(\omega)\rvert = \lvert H(\omega) X(\omega)\rvert $$

Try this and it will work for you on MATLAB.

MATLAB Code

% Calculating the Magnitude of the Filtered Signal

vFilterCoeff = [1; 2; 3; 4; 5];
vInputSignal = randn(1000, 1);

vOutputSginal = conv(vInputSignal, vFilterCoeff);

fftNumPoints = 1024;

vOutputSignalDftMagnitude = abs(fft(vOutputSginal, fftNumPoints));
vFilteredSignalDftMagnitude = abs(fft(vFilterCoeff, fftNumPoints) .* fft(vInputSignal, fftNumPoints));

hFigure = figure();
hAxes   = axes();
plot([vOutputSignalDftMagnitude, vFilteredSignalDftMagnitude]);

The result is:

As can be seen there's a full equality.