I am learning Wiener filtering for 2D images by myself. From one book, it reads

It should be noted that the study of the more general problem of signal denoising dates back to at least Norbert Wiener in the 1940s. The celebrated Wiener filter provides the optimal solution to the recovery of Gaussian signals contaminated by AWGN. The derivation of Wiener filtering, based on the so-called orthogonality principle, represents an elegant solution and the only known situation where constraining to linear solutions does not render any sacrifice on the performance.

It mentions Gaussian signals. I know Gaussian noise. What is Gaussian signal? I cannot find the definition of Gaussian signal via google. Does it mean one signal is contaminated by Gaussian noise?


The stochastic description of signals is realized by random processes. The book you cite actually speaks of a Gaussian random process. By definition, every random variable drawn from that process has a Gaussian probabilty density function. Mathematically speaking: let $\mathbf X(n)$ be the random process representing the signal, then for every $n_1$ the random variable $$ X=\mathbf X(n_1) $$ is Gaussian distributed. This in turn means, that $X$ has the probability density function $$ p_\mathrm X (x) = \frac{1}{\sqrt{2\pi\sigma^2}}\mathrm e^{-\frac{(x-\mu)^2}{2\sigma^2}}, $$ where $\mu$ and $\sigma$ are the mean and standard deviation of $X$, respectively.

  • $\begingroup$ Every random variable drawn from that process has a Gaussian probability density function. Do they have the mean and standard deviation. In another book, it reads "Often, an image is considered to be the realization of a spatial stochastic process [704]. It is common to assume that this stochastic process is stationary, which means that the statistics are the same everywhere in the image." $\endgroup$ Dec 25 '15 at 3:02
  • $\begingroup$ From the link, azimadli.com/vibman/stationarysignals.htm, "The first natural division of all signals is into either stationary or non-stationary categories. Stationary signals are constant in their statistical parameters over time.". Does it mean that each pixel obey probability distribution with same mean and standard deviation? $\endgroup$ Dec 25 '15 at 5:22
  • $\begingroup$ This does not follow from stationarity which tells us something about the temporal behaviour of the process rather than its spatial properties $\endgroup$
    – Deve
    Dec 25 '15 at 8:39

The goal of the Wiener filter is to compute a statistical estimation of an unknown signal.

The basic theoretical model is that at every time point a Gaussian random number is produced.

My answer was copied from: https://en.m.wikipedia.org/wiki/Wiener_filter http://sepwww.stanford.edu/sep/prof/pvi/tsa/paper_html/node6.html

  • $\begingroup$ In my view, an signal is always deterministic. Can statistical estimation be applied to an signal? I think statistical estimation can only be applied to random variables. Maybe I am wrong. $\endgroup$ Dec 24 '15 at 11:46
  • $\begingroup$ Practically, most measured physical phenomena are not 100% deterministic. As for statistical signal estimations, that's an entire subject. Check out this link: en.m.wikipedia.org/wiki/Statistical_signal_processing $\endgroup$
    – Michael
    Dec 24 '15 at 12:32
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    $\begingroup$ In general, a signal is not deterministic. In image processing, if the signal was deterministic, it would mean, that you study only one specific image. Usually, you are interested in a more general approach. $\endgroup$
    – Deve
    Dec 24 '15 at 12:35
  • $\begingroup$ @Deve Do you mean that each observed pixel value from one image is considered as sum of two random variables? One is image signal random variable, and the other is noise random variable. The parameters of image signal random variables differs from pixel to pixel. Random variables from different pixels obey identical probability distribution. $\endgroup$ Dec 25 '15 at 2:12
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    $\begingroup$ These are all valid but separate questions that could be addressed better in a separate question than in the comments $\endgroup$
    – Deve
    Dec 25 '15 at 8:37

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