Image Processing Noise differences
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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 ??
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채택된 답변
Wayne King
2011년 12월 19일
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);
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추가 답변 (2개)
Wayne King
2011년 12월 19일
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.
Shaveta Arora
2016년 2월 24일
How to add gaussian noise of variance 10 by both methods?
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Image Analyst
2016년 2월 24일
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;
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