0
$\begingroup$

I have two signals Sa and Sb, both affected by the same noise. I don't know how much noise is mixed into each one, and the signal+noise is transformed using some (unknown) nonlinear function before I get it. If the original signals are Sa and Sb and the noise is N, what I actually have Ta = f1(Sa + k1*N) and Ta = f2(Sb + k2*N). f1 and f2 are different. I would like to get Sa and Sb, or possibly f1(Sa) and f2(Sb): ie the transforms are not especially of interest, but removing the shared noise is.

The transform functions f1 and f2 are monotonic and reasonably well behaved. I think they are closely approximated by f(x) = c1*(x^c2 + c3) where c2 is typically 0.9-1.1. In some cases f1 is approximately the same as f2, so I'm interested in a solution for f1 same as f2 also.

The signals are actually images; so there is quite a lot of data to work with, typically 1-4 million data points each in a pair of signals.

This seems like a source separation or possibly deconvolution/guided filtering problem, but I can't really think of a good approach. I can't even come up in with a good solution in the limited case when there are no transform functions... it is an under-determined problem, because given Ta = Sa+k1*N and Tb = Sb+k2*N, any arbitrary choice of N and k1 and k2 works for Sa = Ta-k1*N and Sb = Tb-k2*N. I think there have to be some regularizing assumptions, like minimizing total variation of Sa, Sb and N, or minimizing cross-correlation, or some kind of other entropy based metric.

EDIT As requested, here is an image pair. The highlighted black dot is not really in the images, it is due to debris in the optics, and present in both images in the exact same position (although: it has somewhat lower intensity in the second image).

enter image description here enter image description here enter image description here

$\endgroup$
  • $\begingroup$ Can you provide an example of such a pair of images? $\endgroup$ – Jazzmaniac Feb 26 '15 at 10:08
  • $\begingroup$ noise is not stochastic? the functions f1 and f2 are spatially invariant? that should be pretty easy, right? you have a bunch of data points... $\endgroup$ – thang Feb 27 '15 at 8:20
  • $\begingroup$ @thang: Yes and yes. In a nutshell, it is plausible to say that any given point could have a noise spike if both real images contain the inverse of the noise; something has to rule that out (as very unlikely). Conversely, if there is overlap in both images which is not noise, something has to tell us it isn't noise. I guess I'm not seeing what you're thinking of, could you explain? $\endgroup$ – Alex I Feb 27 '15 at 10:39
  • $\begingroup$ I take it back. It's not easy. You have to make assumptions about the structure of N. The problem is that at each pixel, you have two data points (Ta, Tb), but there are many unknowns (k1,k2,ac1,ac2,ac3,bc1,bc2,bc3). Maybe try to expand the formulas to see if there's a clever trick you can do, but as it is, it's a severely under determined system. You'll need to regularize with some prior information... $\endgroup$ – thang Feb 27 '15 at 10:52
  • $\begingroup$ @thang: exactly, assumptions abou N and the signals both. I think that is how source separation always works. $\endgroup$ – Alex I Feb 27 '15 at 13:35
2
$\begingroup$

For me, if all you want to deal with is the possible 'debris in the optics,' why not get a reference background image? This will help you to easily identify which are 'noise' caused by your optics. Assume you can get a perfect background, namely everything remains the same except for your cells ( maybe you can get this by taking another image after removing all cells ), then simply subtract the background image from the original one will work.

$\endgroup$
  • $\begingroup$ Great idea. Unfortunately, the reference image is not available - all I have is matched pairs (or often larger sets), all with real images. $\endgroup$ – Alex I Feb 27 '15 at 10:40

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.