In my understanding, the deconvolution process with FFT is to use divide operator instead of the deconvolution operator, that is ---aries44's post----------- In order to convolve two functions 'a' and 'b', we can take their Fourier Transform(FT) and multiply them in Fourier domain i.e. C= FT(a) * FT(b) c = IFT(C) and then Inverse Ft(IFT) of 'C' gives us the convolution of 'a' and 'b'. Now if we want to deconvolve 'a' from 'c' to get 'b' we can do B = FT(c)/FT(a) b = IFT(B) -------------------------- when i read the paper by Wiess, he used a normalization function g, and a delta function delta, to implement the deconvolution, (in equation (6) and (7)) what do the equations mean??? thank you! in this paper Deriving intrinsic images from image sequences Weiss Y. proceedings ICCV 2001 download paper here: http://www.ai.mit.edu/courses/6.899/papers/13_02.PDF

# Deconvolution procedure by Weiss method?

Started by ●March 31, 2006

Reply by ●April 3, 20062006-04-03

anthony wrote:> In my understanding, the deconvolution process with FFT is to use divide > operator instead of the deconvolution operator, that is > ---aries44's post----------- > In order to convolve two functions 'a' and 'b', we can take their Fourier > Transform(FT) and multiply them in Fourier domain i.e. > C= FT(a) * FT(b) > c = IFT(C) > and then Inverse Ft(IFT) of 'C' gives us the convolution of 'a' and 'b'. > Now if we want to deconvolve 'a' from 'c' to get 'b' we can do > B = FT(c)/FT(a) > b = IFT(B) > -------------------------- > > when i read the paper by Wiess, he used a normalization function g, > and a delta function delta, to implement the deconvolution, > (in equation (6) and (7)) > what do the equations mean???(6) tells you how to calculate r-hat from the specified summation using an unspecified g. (7) tells you what g is without having to use the division sign. He doesn't seem to have a symbol for deconvolution/division/inversion. g is specified as the convolutional inverse of the summation it is shown convolved with. i.e. it is the sequence which when convolved with the summation yields the delta function. So your B, above, is the FT of the summation in (6) (what you call c) deconvolved by the summation in (7) (which you call a). Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein

Reply by ●April 3, 20062006-04-03

anthony skrev:> In my understanding, the deconvolution process with FFT is to use divide > operator instead of the deconvolution operator, that is > ---aries44's post----------- > In order to convolve two functions 'a' and 'b', we can take their Fourier > Transform(FT) and multiply them in Fourier domain i.e. > C= FT(a) * FT(b) > c = IFT(C) > and then Inverse Ft(IFT) of 'C' gives us the convolution of 'a' and 'b'. > Now if we want to deconvolve 'a' from 'c' to get 'b' we can do > B = FT(c)/FT(a) > b = IFT(B) > -------------------------- > > when i read the paper by Wiess, he used a normalization function g, > and a delta function delta, to implement the deconvolution, > (in equation (6) and (7)) > what do the equations mean???It means that the spectrum division B=FT(c)/FT(a) is ill-posed. If one of the coefficients in the sequence or image FT(a) vanishes, the corresponding coefficient in B goes to infinity, and the corresponding wave pattern will completely dominate the image b. So instead of using the naive approach above, one tries to find some filter that collapses the impulse response you want to remove, back into a delta function. If you apply the same filter on the corrupted image, you (hopefully) get an improvement while avoiding the bad effects of the spectrum division. Rune