Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have employed the fourier(projection) slice theorem in matlab. I have a 3D image, P(x,y,z) defines their pixel intensities at a given location int he image volume, it is discrete and uniform. I take the FFT of this image and get a 3D volume in the frequency domain. I then take a 2d slice from this 3D volume at an arbitrary angle making sure that the centre of the slice and the centre of the 3D FFT image volume pass through the same point. I then inverse FFT this 2d extracted plane to get a projection of my 3d volume.

I have noticed that I get an overlapping of artifacts but they are shifted by bit, also their intensity is reduced. If I sample at a higher rate the shift becomes greater to a point where it doesn't overlap anymore. Why does sampling at a higher rate increase the shift of the overlapped image? What can I do to stop the artifacts from being produced?

share|cite|improve this question
I've removed the filters tag, since it is intended for filters in set-theoretical and order-theoretical sense; see the tag description. – Martin Sleziak Sep 17 '12 at 12:48
Did you ever find out why? I'm wondering if it has to do with fft padding somehow? – Brian Jul 20 '14 at 19:05
@B-Brock I assumed it had something do with the fact that it was a discrete function being sampled and thus in the frequency domain its power spectrum is repeated periodically, if you don't sample with enough resolution the spectrum overlaps with the adjacent ones(Nyquist sampling theorem I believe). If you were to sample at a higher rate then the you avoid this overlapping and when doing the inverse you avoid the energy from the adjacent spectrums. This however is only my guess at why it worked, I am not sure if this is correct. – user1084113 Jul 21 '14 at 14:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.