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It's pretty clear that zero-padding an image before performing Fourier transform simply enlarges the magnitude image (stretching it to the new, padded size).
What I can't understand is how it affects the phase... At least to me it does not look like simple enlargement of the phase image. Could somebody clarify?
Symmetric zero-padding (in the center of an image around the N/2,N/2 sample) does not affect the FFT phase result. Or after an 2d fftshift before the 2DFFT, symmetric zero-padding around the edges (circularly around the 0,0 sample) does not add phase shift.
An FFT phase measure the even to odd ratio around 0,0. This won't change with any padding that maintains the identical even and odd decomposition of the input. An fftshift moves the phase reference to the middle of a 2D image, which may make more sense with reference to an odd and even function decomposition of the original input.
Figured this one out - I'll post an answer in case someone else has the same question.
There is an additional phase shift that occurs when an image is zero-padded. That is, a constant factor multiplied by the frequency needs to be added to the phase image. This factor depends on the padding width and the image size.
