# Fourier transform artifacts

My starting point in what follows is a radially symmetric random field. Taking the Fourier transform of this (and plotting it in logarithm to highlight the patterns), I obtain the following image in Fourier space:

As you can see, there is a radially symmetric part of concentric circles, superimposed with a cross-pattern. Now, I do not understand this last part, but I highly suspect this to be an artifact that is not supposed to be there...

It would not surprise me if this is a problem that more people have run into this problem, but I have not been able to find an answer yet.

So, bottom line: Why is there a cross-pattern in my image?

• Please include a picture of the random field too. Sep 19 '15 at 21:24
• not saying that this was done with MATLAB, but if the fftshift() has already been done with this (so DC is in the middle and not at the corners), i would say that you get what you have because of DC and low frequency components that exist purely in the left-right and top-down directions and not diagonally. maybe there is a checkerboard or similar pattern that is square with the two spacial axes.  just a guess. Sep 20 '15 at 0:56

## 2 Answers

The cross pattern is typically a border effect, due to the periodicity induced by the standard implementation and hypotheses behind the Fast Fourier transform, when the image lacks periodicity from the right to the left, and the bottom to the top. In other words: if two opposite borders lacks continuity in values (when glued together), artifacts show.

• As an aside: if I use an apodizing mask, I seem to be losing power on all scales. I tried to compensate for this by multiplying the power spectrum by the ratio of the original image size to the size if we ignore the tapered edges. However, I'm still losing (some) power. Would you know how to solve this? (just to be clear: If the original image is 1000 pixels, and we use a taper width of 100 pix on each side, I multiply the final power spectrum by (1000/800)**2. to try and recover the final image. Perhaps I should make a new question about this altogether. Sep 29 '15 at 13:36
• Could you share the picture, to help people find more suitable windows? Sep 29 '15 at 14:05
• I will make a new question about this, probably cleaner. Thanks in advance. Sep 29 '15 at 16:15
• dsp.stackexchange.com/questions/26122/… Sep 29 '15 at 16:51

The 2D square rectangular window is not radially symmetric (because it has non-zero corners that stick out farther on the diagonals). Try a round window of some sort before the 2D FFT.

• You seemed to double post your answer, so I moved one to a comment.
– Peter K.
Sep 20 '15 at 22:54