Filtered backprojection reconstruction of the standard radon transform uses a ramp-like filter (Ram-Lak) with frequency dependence of |f|, before back-projecting. This can be understood from the determinant of the Jacobian for transforming from Cartesian to circular coordinates.
The exponential radon transform (Tretiak and Metz: Article The Exponential Radon Transform
) can be inverted using a similar filter, except all frequencies below the scale of 1/mu, where mu is the attenuation length are zeroed. Is there an intuitive explanation for this similar to that of the radon transform?
Methods for inverting the Radon transform
https://statweb.stanford.edu/~candes/math262/Hw/hw2.pdf
http://orbit.dtu.dk/files/5529668/Bin
Mostafa M. Hamed thanks but your answer is for the standard radon transform and not the exponential radon. Do you have any resources for the exponential that a cursory google search will not bring up? Thanks :)
Article Fast Laplace Transforms for the Exponential Radon Transform
Let T be any imaging operator and T*, it's dual. All FBP methods roughly take the following form: f=T*(w#Tf),where I am using # to indicate convolution and w to indicate a kernel corresponding to some filter. One can usually show that:
T*(w#Tf) = (T*w)#f. Now the idea is that if you can show T*w is an approximate identity (read : almost like a Dirac-delta distribution), then convolving T*w with f will give you (almost) f. For the exponential Radon transform, if you use the Ramp filter along with the "modified' Ram-Lak filter (frequencies cut below 1/mu), then it leads to showing that T*w is indeed an approximate identity. For more on this refer to page 49 of Natterer's book: The mathematics of computerized tomography. I hope this helps.