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I am trying to deblur an image using MatLab. I have the point spread function (PSF) that the images was blurred with. Furthermore, I know there is noise that is Gaussian distributed, and the signal to noise ratio (SNR) is very high (>20).

Matlab has a couple of deconvolution functions that use direct filtering (regularized filter and Weiner filter), which do not yield satisfactory results.

MatLab has also the Lucy-Richardson (LR) iterative algorithm that, in my case, does a good job in deblurring the image (judged visually).

My question is: is it theoretically sound to use the LR method when the noise in the image has a Gaussian distribution ?

The LR assumes Poisson noise in the blurred image - but does this mean that it performs best with poisson noise, but may also be adequate to use for other types of noise - so I can expect similar results if I want to deblur similar images in the future?

Or does it mean that the LR may randomly yield non-sense results if used for images that have other types of noise than Poisson ?

Image Analyst
on 14 May 2016

Well no and yes. Maybe not theoretically the same, however you know, or you should know, that Poisson noise for anything over an expected value of about 8-10 looks virtually identical to Gaussian noise. Just prove it to yourself by doing a Poisson curve for, say 100 and then check the differences between that and a Gaussian fitted to it. The differences will be very small.

And you pretty much never have just 10 photons per pixel unless you are doing very low exposure photon noise limited experiments like in astronomy or radiography. So for most intents and purposes, Poisson noise manifests itself as Gaussian noise, and so the Gaussian noise you have can be considered Poisson noise. The bottom line is how the image looks and if it improves it enough for you to get the required measurements out of it. And it sounds like it's working for you, so go with it.

Image Analyst
on 17 May 2016

Rozh, noise and signal to noise ratio are different. In general a higher signal also means higher noise. Let's say I have a photon noise limited signal with an exposure of 1 ms. It will look very "noisy" even though the noise is very low, say around 5. Now let's say we increase the exposure to 100 ms. We're collecting 100 times as many photon so the signal will be 100 times more. Depending on what kind of noise we have, it could be 100 times greater also. Or it could be some lesser number, but it will most likely be more than the noise of 5 we had with the exposure of 1 ms. So more signal and more noise go hand-in-hand. But the signal to noise ratio may actually increase even though the noise is more. Usually it goes up because usually with more exposure you get a better signal and it "appears" less noisy.

If you had a true Poisson noise process the mean would equal the standard deviation even if the actual values are multiplied by some scaling factor such as in an electronic amplifier and a-to-d converter. However you said that your signal > 10 * SD, so we have like a narrow hump way out there from the origin, not like the humps you see for Poisson distributions where the hump is very broad and goes about way back to the origin. As an extreme to help you visualize it, what if your signal and noise were 1000 + or - 5. Well that's certainly not Poisson - a spread of +/- 5 way out at 1000 certainly does not look like a Poisson distribution. Let's say your signal is s and let's say it's Poisson so the variance is also s. So the SD is sqrt(s) and the SNR is s / sqrt(s) which equals sqrt(s). But your SNR is 10. So is your signal 100? So I say that no, your noise does not seem to be Poisson.

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