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I've been tasked with creating a 32 x 32 half-band low-pass image filter in MATLAB. My thinking is to generate the ideal filter mask in the frequency domain and compute the corresponding convolution mask using the inverse FFT. I first generate the filter in the frequency domain.

filter = zeros(32);
filter (1:8, 1:8) = 1;
filter (1:8, 24:32) = 1;
filter (24:32, 1:8) = 1;
filter (24:32, 24:32) = 1;

This filter turns out to be the following in the frequency domain. I've confirmed my MATLAB code produces this pattern which is symmetric. enter image description here

Note, I'm assuming I need to define the mask with the frequency ranges from 0 -> 2pi rad/sec, hence putting the "ones" in the corners.

I then generate the convolution mask using the "iift2" MATLAB function. However, this mask turns out to be complex which raises some red flags. I would expect it to be a real filter.

mask = ifft2(filter)

When I convolve this filter mask with my image however, I get some weird artifacts.

image_filtered = imfilter(image_orig, mask);

enter image description here

I'm curious if there is some flaw in my thinking here.

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    $\begingroup$ Does this answer your question? Why is it a bad idea to filter by zeroing out FFT bins? $\endgroup$ Feb 7, 2021 at 20:50
  • $\begingroup$ all the things said for 1D filters there 100% directly translate to 2D filters, plus you forgot the circular nature of DFT convolution. $\endgroup$ Feb 7, 2021 at 20:51
  • $\begingroup$ also "ideal low-pass filter" has a specific meaning, and that is a low-pass filter with infinitely steep transition, which would be infinite in size and hence can not be implemented. $\endgroup$ Feb 7, 2021 at 20:52
  • $\begingroup$ @MarcusMüller I'm more interested in whether or not the fourier mask I've defined is correct and why I would be getting a complex convolution mask when performing the IFFT. Ideally, I would like the IFFT to create a 2D signal that entirely real. $\endgroup$
    – Izzo
    Feb 8, 2021 at 18:06
  • $\begingroup$ again, no, you didn't define that mask correctly for any practical filtering problem. $\endgroup$ Feb 8, 2021 at 18:24

1 Answer 1

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First, I will replace some of the coefficients with 0.5 and 0.25.
In order to understand this, please have a look at:

In order to keep the symmetry in place, what you need to have (I will show in 16x16 filter, but it is the same):

enter image description here

The reason to the 0.5 value is above. The reason for the 0.25 values is the separability property of the DFT, basically multiplying 0.5 * 0.5.

In order to verify it, the trick is to apply fftshift(). We expect to see a symmetric centered filter.

enter image description here

The center is at 7 and indeed around it we see a symmetric filter as required.

The inverse DFT, using ifft2() will generate the time filter, yet in coordinate system which starts at (0, 0):

enter image description here

Applying the fftshift() will show use the filter as we want.

enter image description here

Indeed looks like a 2D Sinc / Dirichlet Kernel.

If you calculate the maximum imaginary element of the inverse DFT you will see it is within the numerical error of the calculation, so basically zero as expected.

Remark: This is not a good way to generate filters. You better build a 1D filters according to the needs of each axis and then use outer product to generate the 2D filter.

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