Talk:Phase correlation

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I have trouble understanding this page. There's a description of a method, and a proof, but it isn't very clear what the method is supposed to achieve. So there's no "theorem" or "claim" to prove. Certainly the claim needs to be stated more clearly, preferably both in an informal (for understanding) and a formal way. Without this the "proof" should be downgraded to a "motivation" or "why it works" kind of section. Thanks & regards. akay 09:34, 19 July 2006 (UTC)

The introduction clearly states that this is a method to determine the relative translative movement between two images, which is exactly what it does. The proof section provides a derivation for the algorithm, and proves the exactness for continous & noise-free images. Feel free to make the text more verbose it you think that will make the article more clear. You should also check out the refered paper if you want to dig more deeply into the subject. --Fredrik Orderud 23:40, 23 August 2006 (UTC)


I'm not sure what the example is supposed to show. Is the vector from <0 0> to <30 33> (where the prominent white dot appears) supposed to represent the displacement of image 2 relative to image 1? 23, August 2006

Yes, the coordinates of the white peak reveals how much the two images are translated relatively to each other. This is because δ(x − Δx,y − Δy) is nonzero only at the coordinate (\Delta x,\ \Delta y). --Fredrik Orderud 23:40, 23 August 2006 (UTC)

Contents

[edit] Proof

The proof presented is flawed. The second line only applies if the image is circularly shifted, i.e. only if:

i_b(x,y) \ \stackrel{\mathrm{def}}{=}\ i_a((x - \Delta x) \bmod M, (y - \Delta y) \bmod N)

Either this needs to be stated, or the proof needs to be rewritten in terms of the continuous Fourier transform or the discrete-time Fourier transform.

Oli Filth 22:51, 1 May 2007 (UTC)

What's more, the expression that's referred to as a "normalised cross correlation" isn't a correlation at all, it's simply a multiplication. It's the spatial-domain result PC = δ(x + Δx,y + Δy) which is the cross-correlation. Oli Filth 08:54, 2 May 2007 (UTC)

Answer : anyway, in practice, images will be noisy and will never be the circular translated of one another, so the proof will never apply to a practical case, and the only interesting thing is the idea behind the algorithm, that's why it's useless to clutter the proof with a circular shift of the pictures. Say we suppose that both images are on a black background and translated by a small shift, so that they are both in the frame, then the theorem applies, and it gives us some kind of justification so as to why the algorithm is supposed to work. That's all we are asking for (since, again, in practice the images will never be actual translations of each other). —The preceding unsigned comment was added by 129.199.224.240 (talk • contribs) 11:12, 24 June 2007.

Yes, obviously the images are unlikely to be circularly-shifted translations of each other in a real application. However, we can't go around presenting unsound (i.e. incorrect) mathematics like this, especially with something as frequently-misunderstood and frequently-misapplied such as the Fourier transform. Oli Filth 11:18, 24 June 2007 (UTC)
I've now updated the article, hopefully addressing all of my concerns! Oli Filth 13:23, 24 June 2007 (UTC)

[edit] Please generalize

This article is too narrow. Phase correlation is also used in other fields besides image processing. Please generalize it and make the image processing application a subsection. —Keenan Pepper 03:34, 19 July 2007 (UTC)

[edit] Include a description of sub-pixel methods?

I found this article very informative. However, you mention sub-pixel methods, but don't specify how this might be done. Also, I think it would be helpful to comment on how this method relates to motion estimation by the correlation method - increased accuracy and reduced robustness? 196.2.111.133 15:55, 20 July 2007 (UTC)

[edit] Motivation

First, I would like to mention that the phase correlation is a private case of the normalized correlation, where we assume that two images have approximately the same Fourier magnitude. Second, one major benefit of this technique is the fact that you don't actually need to multiply the Fourier transforms of the two images and then divide by the magnitudes, but instead you can DISCARD the two magnitudes and just subtract the phases - which is much faster. —Preceding unsigned comment added by 212.25.107.145 (talk) 09:50, 7 January 2008 (UTC)