Image Processing Noise differences

6 vues (au cours des 30 derniers jours)
UJJWAL
UJJWAL le 19 Déc 2011
Déplacé(e) : DGM le 20 Fév 2023
Hi,
Suppose I want to add white gaussian noise to an image. I propose to do by following means :-
a) imnoise(I,'gaussian',0,0.25);
b) I = awgn(I,var(I(:))/0.25);
c) I = I + 0.25*randn(size(I));
Here I is a certain image.
What is difference between using the above statements ??

Réponse acceptée

Wayne King
Wayne King le 19 Déc 2011
J = imnoise(I,'gaussian',0,0.25);
J = I+0.5*randn(size(I));
For awgn(), your function syntax assumes the power of the input is 0 dBW, so you would need to do.
denom = -(var(I(:))/(10*log10(0.25)));
I = awgn(I,var(I(:))/denom);

Plus de réponses (2)

Wayne King
Wayne King le 19 Déc 2011
Hi, in
I = I +0.25*randn(size(t));
you get noise with a standard deviation of 0.25, not variance. If you want noise with a variance of 0.25, then you must do
I = I +0.5*randn(size(t));
that would be equivalent to:
imnoise(I,'gaussian',0,0.25);
The variance of a constant times a random variable is the constant squared times the variance of the random variable.
Finally, the actual variance of the additive Gaussian noise in:
I = awgn(I,var(I(:))/0.25);
depends on I, so it's not clear that you are really getting a variance of 0.25. For example:
I = randn(256,256);
Because var(I(:)) = 1.0551 (in this particular example)
Your call of
I = awgn(I,var(I(:))/0.25);
results in an additive WGN process with variance:
10^(-4.2203/10) = 0.3784
which is greater than you think.
  1 commentaire
UJJWAL
UJJWAL le 19 Déc 2011
Déplacé(e) : DGM le 20 Fév 2023
Ok . So suppose the problem is to add a noise with a variance of 0.25 and mean of 0 and the noise is gaussian and additive.
What are the equivalent statements using imnoise, awgn and the first one to introduce such a noise ??

Connectez-vous pour commenter.


Shaveta Arora
Shaveta Arora le 24 Fév 2016
How to add gaussian noise of variance 10 by both methods?
  1 commentaire
Image Analyst
Image Analyst le 24 Fév 2016
Hint from the help:
Create a vector of 1000 random values drawn from a normal distribution with a mean of 500 and a standard deviation of 5.
a = 5;
b = 500;
y = a.*randn(1000,1) + b;
For you, a would be sqrt(10) and b would be 0, so
[rows, columns] = size(grayImage);
noisyArray = sqrt(10)*randn(rows, columns);
output = double(grayImage) + noisyArray;

Connectez-vous pour commenter.

Catégories

En savoir plus sur Image Processing Toolbox dans Help Center et File Exchange

Produits

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by