Let's say you have an original image and a version of the same image that may have been convoluted with a Gaussian blur. How could you demonstrate that the Gaussian blur has been applied and calculate the blur radius?
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ I'm assuming you mean "convoluted" instead of "deconvoluted"? $\endgroup$– Roger RowlandJun 25, 2013 at 6:19
-
$\begingroup$ I also have the same question about it. According to Roger Rowland: "the blur kernel can be recovered by calculating H(u,v)=G(u,v)/F(u,v) and then taking inverse the Fourier transform of H to obtain the PSF as an intensity." my question: how big the size of inverse of H(u,v)? and how to measure the radius? $\endgroup$– user25838Jan 16, 2017 at 14:51
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
If you have the blurred and unblurred images, what I think you are asking is how you can recover the point spread function (PSF).
In the absence of noise, this is theoretically possible by considering the operation in the frequency domain. The blurred was introduced by a convolution of the unblurred image with a Gaussian kernel. In Fourier space, convolution is applied by multiplication of the image with the kernel, so if you have two of the operands (the source and the result), it should be possible to recover the blur kernel by division of the respective Fourier transforms.
This is effectively what some deconvolution operations attempt to do when the PSF is unknown or is a mixture of defocus and motion blur (e.g. blind deconvolution, Wiener filter, etc.).
So, in the frequency domain, the blur can be expressed as
G(u,v) = H(u,v)F(u,v) + N(u,v)
Where G is the Fourier transform of the blurred image, H is the blurring function, F is the source image and N is additive noise. Assuming the noise is negligible (unrealistic in most real world situations), the blur kernel can be recovered by calculating
H(u,v) = G(u,v) / F(u,v)
and then taking the inverse Fourier transform of H to obtain the PSF as an intensity image. From this you could measure the radius.
