1
$\begingroup$

In theory of noise,Typically, the standard deviation of noise has 1 and mean has 0. I think that the reason of noise mean has 0 that we can assume that all noise signal go to zero when we sum it all. But I can't understand standard deviation of noise has 1 in image noise. They are always 0~255. There is no negative value.

What does zero-mean noise mean?

Especially, but I have got an idea as following. I can't catch quite a bit. So would you please let me know how to illustrate following answer?

"for an image intensity which is always non-negative, you can simply modify your added N(0,1) standard gaussian noise in 3 steps: 1-scale it so that your noise has whatever std you wish. 2-add whatever proper mean you need such as 128. 3-clip the noise for below zero and above 255 samples. This is simple bt It will help you get what you need at least initially."

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Your question is too basic, and it gives the impression that you haven't done enough study of the subject on your own. I'll provide a few hints:

  • To understand noise, you first need to specify its probability distribution. I assume you're refering to Gaussian noise in your question, but not all noise is Gaussian. In particular, in digital capture of an image, there are several kinds of noise present.

  • Gaussian noise is specified by its mean and its variance ($\sigma_n^2$) or its standard deviation ($\sigma_n$). The physical interpretation of the variance is that it is related to the noise power. There is nothing special about $\sigma_n=1$.

  • Most computer random generators (for instance, Matlab's randn) generate samples of a Gaussian random variable with zero mean and variance 1. It is easy to convert this random variable to another Gaussian random variable with any specified mean and variance.

  • You need to understand the effects of clipping Gaussian noise. Think about a sensor photosite being pulled or pushed beyond its dynamic range by the noise. However, as I said in another answer, the resulting noise is no longer Gaussian.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ @gmotree I don't fully understand that darktable page. It assumes familiarity with their approach and terminology and does not explain very well -- for example, they don't label the axes in their plots. I believe they want a constant $\sigma$ across the image, not necessarily a value of 1. Clipping happens because a photosite has a certain dynamic range, and it can't output values outside of it. Similarly, the ADC will clip values below and above its own dynamic range. $\endgroup$
    – MBaz
    Commented Jun 1, 2015 at 15:39
  • $\begingroup$ I think if you do explain more about" clipping of gaussian noise " and "that is no longger gaussian distribute" then it will help to them who don't know this . $\endgroup$
    – gmotree
    Commented Jun 2, 2015 at 0:15
  • $\begingroup$ Sir, is this "the meaning of clipping the gaussian distribution is that the reduction dynamic range" correct? What does it have meaning? $\endgroup$
    – gmotree
    Commented Jun 2, 2015 at 0:25
  • 1
    $\begingroup$ @gmotree Regarding clipping's effect on the noise distribution, see this question: dsp.stackexchange.com/questions/23718/… $\endgroup$
    – MBaz
    Commented Jun 2, 2015 at 1:02
  • 1
    $\begingroup$ @gmotree When exposed to light, a detector's photosites "count" the photons that arrive. If less than a low threshold of photons arrive, the count is 0. If more than a high threshold arrive, the count is the maximum. The difference between the thresholds is the detector's dynamic range. So, the detector clips if the number of photons is too low or too high. There is noise present, too, which can cause the detector to clip when the number of photons is close to the threshold, but still within the dynamic range. $\endgroup$
    – MBaz
    Commented Jun 2, 2015 at 1:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.