1
$\begingroup$

Lets say I have a system that is trying to find a small image (assume all images are grayscale) within in an image by using correlation. So this system has the baseline image, and I input 5 different images (the five input images are bigger or have more data points than the baseline image) and do correlation of my baseline images against the five input images. From this I would get 5 "tables" of correlation values, and from each tables I select the highest value as the metric to how similar the input image is to the baseline image.

Now lets say I want to upgrade my system to do the correlation in the fourier domain. I pad the smaller image with zeros, do the fft, multiply the data by the conjugate of the baseline to do correlation, and I get the power spectral density at each point through some PSD estimator method, like Bartlett, welch, blackman-tukey, etc.

From my understanding, the more spread out, or the less dense the power spectrum is over a set number of frequency bins, the less correlated the two signals are, and vice versa.

But my question is, in the frequency domain, particularly for a two-dimensional signal, would the metric for measuring how spread the power is simply be the maximum value within the table I would attain after the correlation and PSD estimate, as in the spatial domain case?

Also if you could cite your answers whether it be a book or article, that would be much appreciated

Improve this question
$\endgroup$
10
  • $\begingroup$ i ain't an image-processing person, but my understanding of correlating one piece of data to another piece of data (of the same size) is that the result is number, not a table of numbers. if you have a reference image (that's what i would call it) and 5 other images that you will correlate the reference to, the result is 5 (likely different) numbers. $\endgroup$
    – robert bristow-johnson
    Sep 12, 2015 at 22:07
  • $\begingroup$ See my edit on what type of correlation I am speaking of $\endgroup$
    – IkeJoka
    Sep 12, 2015 at 23:20
  • $\begingroup$ okay, so are the 5 test images scaled up from where you might expect to see the same image in the reference? is it like you have a good, high-resolution image of a gun (or a person's face) and you want to find that object, scaled arbitrarily, in some larger image with many other objects? then you will have a table indeed and it would have 3 dimensions: x position, y position, and scale factor. and you would want to pick the maximum correlation value out of that. $\endgroup$
    – robert bristow-johnson
    Sep 12, 2015 at 23:43
  • $\begingroup$ Sorry I should have clarified that these are grayscale images $\endgroup$
    – IkeJoka
    Sep 13, 2015 at 0:10
  • $\begingroup$ @robert bristow-johnson thanks for your answer $\endgroup$
    – IkeJoka
    Sep 13, 2015 at 23:43

1 Answer 1

Reset to default
1
$\begingroup$

From my understanding, the more spread out, or the less dense the power spectrum is over a set number of frequency bins, the less correlated the two signals are, and vice versa.

I'm afraid you have this backward. Narrow peaks in the original (spatial) domain correspond to strong cross-correlations. Narrow peaks in the spatial domain will be relatively spread out in the Fourier domain. The results of the cross-correlation are usually interpreted in the spatial domain (rather than the Fourier domain).

Transform the image and the (zero-padded) template, multiply in the Fourier domain, and then transform back to the spatial domain. Peaks in the spatial domain correspond to (the possibility of) matches between the template and the image.

Improve this answer
$\endgroup$
5
  • $\begingroup$ but then how do you explain fig9.1 from this book? dsp-book.narod.ru/302.pdf $\endgroup$
    – IkeJoka
    Sep 13, 2015 at 1:12
  • $\begingroup$ Sorry, didn't realize that "enter" posted the comment. Ah, I think I see the issue: from page 264, just before Figure 9.1, "In general, the more correlated or predictable a signal, the more concentrated its power spectrum, and conversely the more random or unpredictable a signal, the more spread its power spectrum." This quote is talking about the periodicity of a single function and it is true. A periodic signal has a sharp power spectrum. Your original question however talks about cross-correlation of two different functions. Two different meanings of "correlation". $\endgroup$
    – Austin A.
    Sep 13, 2015 at 2:10
  • $\begingroup$ Take a look at this article on the cross-correlation theorem on Mathworld. To take a cross-correlation, take the Fourier transform, multiply, and then inverse transform. The result is interpreted in the original domain. $\endgroup$
    – Austin A.
    Sep 13, 2015 at 2:12
  • $\begingroup$ Yes, but I want to interpret the results in the fourier domain now $\endgroup$
    – IkeJoka
    Sep 13, 2015 at 23:39
  • $\begingroup$ I can't help but notice that the comment thread on your original post is telling you to do exactly this. You want to know what the cross correlation is. The cross correlation equals the inverse transform of the frequency domain product. Period. If you really want to analyze this in the frequency domain, then you are stuck with the complicated question of analyzing the Fourier transform of the cross correlation rather than the easy question analyzing the cross correlation. Why make your life complicated?! $\endgroup$
    – Austin A.
    Sep 14, 2015 at 14:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.