Mapping the value of a sample in a 2-D DFT to cycles/pixel

If I have an image and its 2-D DFT of that image, what is the mapping between the value of the DFT at (u,v), and the frequency in the spatial domain in the x and y components, in cycles/pixel? I want to be able to say "There is a peak in the DFT at (u,v), which corresponds to this frequency". I don't think the answer is u cycles/pixel, since the values shift with different image sizes, even if the image content remains the same.

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1 Answer

Well the Nyquist frequency is 0.5 cycles / pixel and the range of frequencies can be considered to range from -0.5 to +0.5 cycles / pixel. So if the dimension in a given axis is N pixels then each pixel is 1 / N cycles per pixel.

Depending on your particular DFT/FFT implementation you may have zero frequency in the middle of the image (M/2, N/2) or in the corner (0, 0). For the former case the spatial frequency at u, v will be:

(M/2 - u) / M, (N/2 - v) / N

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So, am I correct in assuming that the sundamental frequency is the image itself? And one copy of the image is equivalent to 1 wave cycle? If this is correct, how do I compare two distinct images? can I compare the 1/N_1 of one image with 1/N_2 if a second image? - Does this warrant being a new question? –  Lord Loh. Feb 2 at 22:31
No - the larger features of the image will tend to account for the lower frequency components, while edges, fine details and patterns (textures) etc will account for the higher frequency components. There is generally no "fundamental" frequency unless it's just an image containing a regular pattern, e.g. a checkerboard. Try looking at some DFT's of a few simple images and comparing them to get more of a feel for how spatial domain features map to the frequency domain. –  Paul R Feb 3 at 7:06