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I've just come across a paper that interrelates the covariance matrix of time discrete signals to their autocorrelation function (or a time-delay, respectively), i.e. $$\mathbf{C} = E\{\mathbf{x}\left(t\right)\mathbf{x}^H\left(t-\tau\right)\}.$$

Coming from the image processing field, I have usually just worked for the $\tau = 0$ case, whereas $t$ was just the pixel index instead of a time stamp.

Now, I am wondering how I can transfer the concept of this time delay $\tau$ to image processing.

Side-note: Since I don't use box-car based image processing and determine the pixels for this covariance matrix estimation adaptively, what I have basically is a $\left(K\times N\right)$ data matrix containing $N$ pixels with $K$ "measurements" each.

How can I calculate the time-delayed covariance matrix described in the formula above in my case?

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    $\begingroup$ You've already figured out that $t$ maps to the pixel index, so analogously, $\tau$ (a time delay) would map to a pixel offset. Is that what you're looking for? $\endgroup$ – Jason R Jun 5 '13 at 12:33
  • $\begingroup$ Yes, I guess so. However, I don't know how I can apply this pixel offset to the data I have. Do you think it'd be possible to use a subset of the N available pixels as $\mathbf{y}\left(t\right)$ and a shifted and overlapping subset for $\mathbf{y}\left(t-\tau\right)$? This would imply I keep the order of the N pixels and see them as realizations of a time series, i.e. all in all I have N pixels available, but both $\mathbf{y}\left(t\right)$ and $\mathbf{y}\left(t-\tau\right)$ consist of less than N pixels? $\endgroup$ – Michael Jun 5 '13 at 14:26
  • $\begingroup$ That's probably how I would first approach the problem. $\endgroup$ – Jason R Jun 5 '13 at 14:49
  • $\begingroup$ Last question to fully get it: Considering you have, e.g. $N=100$ pixels and $\tau = 10$. How would you solve the "border handling"? If all pixels were numbered from $1$ to $100$, would $t$ start from $11$ and go to $100$, so that $t-\tau$ could start from $1$ and go to $90$? $\endgroup$ – Michael Jun 6 '13 at 16:13

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