Asked by UJJWAL
on 19 Dec 2011

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 ??

Answer by Wayne King
on 19 Dec 2011

Accepted answer

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);

Log in to comment.

Answer by Wayne King
on 19 Dec 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.

Log in to comment.

Answer by Shaveta Arora
on 24 Feb 2016

How to add gaussian noise of variance 10 by both methods?

Image Analyst
on 24 Feb 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;

Log in to comment.

Related Content

MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi

Learn moreOpportunities for recent engineering grads.

Apply Today
## 0 Comments

Log in to comment.