Direct Inverse Filtering an image, spatially convolved with a blurring function without any noise, doees not give my original image back?

Question

I am trying to learn about inverse filtering of images and am trying it out on a test image in MATLAB.

The test image is an 8x8 checkerboard, with 64x64 pixels, generated by the MATLAB code:

g = checkerboard(8);

I am using a 3x3 averaging filter, h, and convolving it with the test image.

I am not adding noise and I want to see the results of direct inverse filtering the distorted image.

I am applying the deconvolution function in the frequency domain by:

1) Padding the 3x3 averaging mask h to generate a 64x64 mask hp.

2) Centering the coefficients (0.111...) of hp to maintain symmetry.

3) Multiplying hp by (-1)^x+y (since I want the fft of the mask to give me a centered Fourier transform, hpc.

4) Taking the FFt of the padded, centered mask and multiplying (-1)^(u+v) (since I centered the mask in the spatial domain in step 2) to give me H.

5) Taking the distorted, and frequency shifted test image and dividing it by H, (ie. Y = G./H). (Note: I am not padding the image since the mask was padded, and wraparound is not a huge concern here)

6) Obtaining the inverse filtered image inv = (-1)^x+y * real(ifft2(Y))

My question: Why can't I obtain the original (or something very close to it)?

Answer 1

I think I found out the answer to my own question.

Since I'm using an averaging filter mask (some constant multiplied by a matrix of ones), the Fourier Transform of such a filter is an impulse at zero frequency.

When I pad the mask filter in step 1 to make the dimensions of the mask equal to the image, I am 'truncating' the filter, ie. windowing the mask, which is akin to multiplying the Fourier Transform of the mask with a sinc function.

As a result, the Frequency Domain filter I use in step 5, ie. H, is not an impulse but a sinc function. That's the reason why my results are not ideal.

On the other hand, if I were to use an impulse in step 5, only the zero-frequency value would remain, and my result would be a constant image.

In short, it's a lose-lose situation.