I am convolving the same image with a 2D Gabor over different gaussian noise masks that are generated in every trial. The convolution naturally takes time, is there any way to speed up the process by decomposing the image and the noise, and proving that I only need to compute the convolution (for the image and gabor, once)?

For example, given the image $I$, and the noise $n$, and the gabor filter $g$:

$g*(I+n) = g*I + g*n$.

That being said, I only would need to compute $g*I$ once. However since noise changes consistenttly, I'd have to compute $g*n$ many times for different iterations. Can I make a computational shortcut in which I can do $g*n = n'$, where $n'$ is a gaussian distribution as well, so that I only need to generate different guassian masks, without the need to convolve the gabor filter over gaussian noise. Is actually valid, and if so, co can I prove it? I've tried doing it by 2D Fourier domain analysis, but I don't really know how to pose the problem since I would be convolving a function by a random variable, not sure what to do here! I do know that a gabor itself is the convolution of a sine wave and a gaussian. How should I go about this problem so I can show the equality, or at least show that they are approximate?:

$g*(I+n) \approx g*I + m$

where $m$ is another gaussian noise template (just like $n$). Maybe I can show that if the expected values are the same $E(m) = E(g*n)$, that would work? (although there might be some loop holes there) I've also hear about stuff called 'gabor noise', not sure if that's what $g*n$ reduces to...

  • $\begingroup$ Consider more applied-oriented sites: Signal Processing, Computational Science $\endgroup$ – user147263 Aug 1 '15 at 0:50
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    $\begingroup$ Caclulate $g \ast I$ once and store it in memory. Take the FFT of $g$ and store it in memory, call it $G$. Generate the noise in the frequency domain, call it $N$, then $g \ast (I + n) = ( g \ast I ) + \operatorname{FFT}^{-1}( G \circ N)$. I can't say if this will be "fast enough," but it will be much quicker than your current approach. $\endgroup$ – AnonSubmitter85 Aug 1 '15 at 19:12

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