Time-domain convolution doesn't match Frequency-domain multiplication in MATLAB

I wonder why these two lines of codes generate very different results:

subplot(1,2,1);
semilogx(abs(X.*HPF));
subplot(1,2,2);
semilogx(abs(fft(conv(ifft(X),ifft(HPF),'same'))));

I mean, regardless of the actual values of X and HPF, spectrum derived from multiplying X and HPF should have the same results as the spectrum derived from convoluting their time-domain impulse responses (which is derived by ifft). But my results show that they are quite different. In my case x is the input signal that have a few sine wave components at different frequencies and HPF is a high-pass filter, but the two expressions should always yield the same results right? I'm confused.

Answer 1

[Matt J]

Note that conv() implements linear convolution, not circular convolution.

[Tianjia]

Thank you very much! I think I can understand your point. Dividing length(x) part is easy. But how to conduct circular convolution in MATLAB?

Update:

I tried cconv() function in Signal Processing Toolbox, which according to MathWorks is capable of doing cyclic convolution. But the convolution result is still incorrect.

[Matt J]

Dividing length(x) part is easy

I made a mistake. You would need to do that if you were convolving in the frequency domain, which isn't the case in what you've shown.

But how to conduct circular convolution in MATLAB?

Here's what I use,

function z=cyconv(x,y)
%Non-Fourier domain cyclic convolution
%
%  z=cyconv(x,y)

 siz=num2cell(size(x));

 subs=cellfun(@(n)[2:n,1:n],siz,'uni',0);

 x=x(subs{:});

 z=convn(x,y,'valid');
 
end

[Matt J]

I tried cconv() function in Signal Processing Toolbox, which according to MathWorks is capable of doing cyclic convolution. But the convolution result is still incorrect.

With my implementation, you can see that the results are consistent.

X=rand(1,8);

HPF=rand(1,8);

abs(X.*HPF)

ans = 1×8

0.3504 0.0729 0.4264 0.0397 0.1455 0.1869 0.4023 0.1709

abs( fft(   cyconv(  ifft(X),ifft(HPF)   )  ) )

ans = 1×8

0.3504 0.0729 0.4264 0.0397 0.1455 0.1869 0.4023 0.1709

[Tianjia]

I just found out another, easier way. Use cconv(x, hpf, N) and the result is correct! Thank you for reminding me of the circular and linear thing! I've been thinking about this for almost a week.

[Askic V]

Thank you very much for your answer @Matt J. I also started to investigate this.