Hello, I am working with images that have the quality and resolution of ultrasound images. I am not sure what types of noise are in the images, other than that there's surely Gaussian noise, there's some Salt and Pepper noise that I removed with medfilt2, and likely noise caused by the CCD. I am trying to further filter the image to suppress the noise for segmentation purposes using wiener2. However, I'd like to estimate the noise variance in a homogeneous region of the image rather than only using the default estimate of Gaussian noise. Can anyone tell me how to do this? I know I can use the var function to calculate variance in one direction, but what about getting total variance for my M x N region selected? Should I just choose the maximum after the var function is used, or is there some other way to do this? Thank you for any help you can provide!
MATLAB: Estimating Noise Variance in a Homogeneous Area of an Image
image analysisimage processingnoisenoise removalvariance
Related Solutions
The question is, I know you are singular, but just HOW singular are you? ;-) It seems to me there are several questions of interest.
- Is A numerically singular at all? Cond gives us a measure of that.
- What is the rank of the null space? Null gives us an opinion about that, but not a metric.
A measure of how close to entrance into the null space is to consider the tolerance that null would use. That depends on the largest singular value. Any singular value that lies below some fraction of the largest singular value, results in the corresponding singular vector getting dumped into the null space.
tol = max(m,n) * eps(max(S));
So I would look at the smallest singular value that lies above that tolerance. Then compute something like this:
min(S(S > tol))/tol
So how close is that value to the tolerance for entrance into the null space?
Note that this applies to a matrix that is numerically full rank too. For example...
A = vander(1:10);[m,n] = size(A);tol = max(m,n) * eps(max(S))tol = 2.38418579101562e-06min(S(S > tol))/tolans = 216.509958943425
So A is close to singular, but not there yet.
But if I push things a bit harder...
A = vander(1:15);[m,n] = size(A);S = svd(A)S = 3.15839869049552e+16 66664300967337.7 322942655858.374 2723747542.51185 36045801.6655343 713911.28106567 20777.0142721889 891.531934617208 58.0952625517871 6.2867119286002 1.42076230762098 0.36061079623119 0.0322030740265822 0.00106225764978576 1.22304251430976e-05tol = max(m,n) * eps(max(S))tol = 60
So it looks like 7 singular values fall below tol. Easy to compute the rank of the null space.
sum(S<tol)ans = 7
And the next one is not that far behind.
min(S(S > tol))/tolans = 14.8588655769535
Or, perhaps report it in terms of the log10 of that, which gives the answer in terms of how many powers of 10 does it lie above the tolerance.
log10(min(S(S > tol))/tol)ans = 1.17198565380708
Another one. I recall that magic squares (as produced by magic, for order > 2) are singular for even order.
A = magic(4);[m,n] = size(A);S = svd(A)S = 34 17.8885438199983 4.47213595499958 4.17280694870505e-16tol = max(m,n) * eps(max(S))tol = 2.8421709430404e-14min(S(S > tol))/tolans = 1.5735e+14
So the rank of the null space was 1, but the next singular value is nowhere close to that tolerance.
So I'd think you need to report two measures.
1. How many singular values fall below tol?
2. How close is the next one to tolerance failure?
You need to give more context than that. Increasing the variance does not necessarily add noise. For example, merely stretching the contrast will increase the variance. Of course if you add something not originally in the scene, there is a possibility that doing that will increase the variance, though adding a negative image will, or could, decrease variance.
Best Answer