4

Disclaimer: I'm not a mathematician, but here's my attempt at a mathy version of my question

Start with a noiseless, discretely sampled expected signal $I(x_n)$. Construct a Poisson-noisy measurement of this signal $P(I(x_n))$, by drawing samples from the Poisson distribution with expectation value $I(x_n)$ at each pixel. If I repeat my measurement many times at pixel $x_n$, by definition I expect the histogram of my measurement to approach the Poisson distribution with expectation value $I(x_n)$.

Now consider the discrete Fourier transform of the noisy signal, $DFT(P(I(x_n)))$. If I measure $P(I(x_n))$ many times, and look at the histogram of values of one particular pixel of the DFT, what distribution do I expect this histogram to approach? How does this distribution depend on which frequency of the DFT I look at?

Backstory (why am I asking this question?):

I do a lot of image processing, and my images are typically corrupted by Poisson noise. When we want to make claims about the resolution of an image, the signal-to-noise versus frequency of the image is important. I know how to characterize signal versus frequency for my images, but I have been assuming that noise versus frequency is roughly constant. Is this actually true for Poisson-noisy images?

EDIT: The paper "Fourier-space properties of photon-limited noise in focal plane array data, calculated with the discrete Fourier transform" https://doi.org/10.1364/JOSAA.18.000777 seems fairly relevant to this question.

Andrew
  • 307

2 Answers2

2

Denote the sampled random values as $x_n$ which are assumed to be uncorrelated and each distributed according to a Poissonian distribution with mean value $o_n$.

The discrete Fourier transform $h_k$ of $x_n$ is nothing but a weighted sum of these random values:

$$h_k = \sum_{n=0}^{N-1} x_n e^{-i 2 \pi n k /N}$$

Especially the real and imaginary part of $h_k$ will tend to be Gaussian-like distributed (central limit theorem) with mean as the weighted mean of the samples and the variance being the weighted sum of the variances of the sample values (samples values being independent):

$$E(\Re h_k) = \sum_{n=0}^{N-1} o_n \cos(2 \pi n k /N)$$ $$V(\Re h_k) = \sum_{n=0}^{N-1} o_n \cos^2(2 \pi n k /N)$$

where we used that the variance of a Poisson distribution equals the mean. The imaginary part can be treated equivalently.

Some key observations:

  • Real and imaginary parts of $h_k$ are correlated (cosine and sine weighting).
  • The $h_k$ are not independent (since the fourier transform mingles different locations $x_n$)
  • The relationship between variance and the mean of $h_k$ depends on $k$, there signal to noise is not constant. For example in a region where no signal frequency is transmitted you still get noise (independent random values give white noise which extends to all frequencies).
  • If the overall brightness of the signal increases by a factor $M$, not only the relative standard deviation of the $x_n$ goes down with $\sqrt M$ but also the relative standard deviation of $h_k$
NoDataDumpNoContribution
  • 172
1

Since Poisson noise is added pixel to pixel independently this noise can be assumed as a two dimensional random process therefore direct Fourier transform can not be used for finding the spectral content .IMHO i feel the spectral content is best estimated by finding the Fourier transform of the auto correlation function for two dimensional Poisson random process. please ref this IEEE paper "Bispectral analysis of two-dimensional random processes"

raghu
  • 171
  • 2
  • 6