I have an image with size $256*256$ then i get the Fourier transform of the image, If i only take the central region of the Fourier transform whose size is $64*64$ and get the inverse FT ,What is the effect on the output image$(64*64)$?

Down-sampling, Blurring, or both.


Have you fftshifted your results or not? The answers will be different:

If you fftshift your $256$ x $256$ 2-dimensional Fourier transform result, retain only the $64$ x $64 $ central co-efficients, inverse fftshift, and then inverse fourier transform the result, the effect will be a blurring of your original image.

This is because you have culled our your images' high frequency components, hence, the effect will be a blurring.

If however you have not fftshifted, then you are doing the opposite, and will actually be doing a high pass filtering. (Your image will certainly not be blurred).

By the way, hard thresholding fourier co-efficients like this is certainly used, but will give you ringing effects. However, those effects might be negligible and/or removable based on your application.

  • $\begingroup$ so i won't get aliased image, only blurring. do you have any idea how to remove this blurring by wiener filtering? $\endgroup$
    – HforHesham
    Dec 13 '13 at 19:03
  • $\begingroup$ @HforHesham If you only remove fourier co-efficients, you will not be causing aliasing, just a filtering operation. If you have a question about Weiner filtering the best thing to do would be to make a new question about it. $\endgroup$ Dec 13 '13 at 19:15

By taking an FFT, retaining a sub-spectrum, then taking the inverse FFT you do actually perform two operations: low-pass filtering then downsampling. Reducing the size of tje spectrum can be seen as first setting to 0 everything outside the 64-by-64 area, then cropping out this 64-by-64 piece.

Let's detail a bit what happens:

  1. First FFT: you compute $\hat{I} = \mathcal{F}(I)$, where $I$ is the original image (256-by-256).
  2. Set to 0 the coefficients outside the 64-by-64 area: $LPF(\hat{I})(f) = \hat{I}(f)$ if $f$ is in the 64-by-64 region, 0 otherwise. This is a low-pass filter (LPF) since you are actually removing high frequencies.
  3. Downsampling: crop out the frequency region of interest: $\hat{X} = LPF(\hat{I})_{64 \times 64}$.
  4. Invert FFT: you obtain a smaller image $X = \mathcal{F}^{-1}(\hat{X})$.

This downsampling procedure is however not common because low-pass filters with smoother decays are preferred.


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