Simply I am curious does convolution theorem still hold for in NUFFT case?
The convolution theorem says
$F \{f* g\}=F \{f\} \cdot F \{g\} $
where $F\{\}$ is the Fourier operator performing Fourier transform. It simplys says the product of two signal in one domain is equivalent of the convolution of the two in the other (transformed) domain.
I am just wondering, does this hold for non-uniform Fourier transform (NUFFT)?
Ex.
I have a 2D Fourier spectrum of an image, and I sample it in following fashion (blue stars are samplings, along radial direction):
Obviously the sampling is non-Cartesian. It is equivalent to sample the orginal continuous Fourier spectrum with a bianry mask who non-zero elements are essentially radial lines.
So the signal I have in Fourier domain is the product of the original signal and the radial mask. So in this case, is it equivalent to convolve the image with the Fourier transform of the radial mask?
Is following correct?
$M\cdot F \{img\}=F \{M\} * img $
$M: radial mask$
$F\{\}$: Fourier operator
$img$: Image
$*$: Convolution
$\cdot$: Multiplication
Thanks a lot.
1) If you are talking about sampling a CTFT (or interpolating from a DFT) in a way you choose (as in your example) :
Using a very crude approach, it is possible to say as long as your samples are dense enough to interpolate the signal in a lossless manner (which is not possible in real life usually, e.g. time limited signals and finite bandwidth) and your samples coincide in the frequency domain why not(!).
Just consider underlying continuous FTs. You would multiply each point with another corresponding point which belongs to a continuum. What you are only doing is picking some points from that continuum. In fact you may choose whatever basis (orthonormal) as you please for your sampling.
2) If you are talking about polar FFT it is a different thing. (Check Natalie Baddour's https://issuu.com/nbaddour/docs/2d-fourier-polar-chapter page 42-43)
3) Regarding "Is the following correct?" part, at that point what you mean by $F$ operator is ambiguous, however, even if a consistent definition is our premise RHS must be inverse transform of $M$.
If your $F$ operation takes a FT then samples a signal and if your $F^{-1}$ reconstructs FT signal with "appropriate"(may not be easy to find for an arbitrary signal space) reconstruction filter (kernel) and takes a IFT you are good to go.
So if we approach your problem in a robot like cold logical manner:
i)No it is not true because to start with RHS must be the inverse transform.
ii)If we ignore that and if you are applying forward and backwards transforms on continuous FTs (CTFT or DTFT) after your samplings/reconstructions then answer is yes
iii) If you want to use a discrete transformation $(x,y)\rightarrow(w_r,w_\theta)$ whose basis functions are $w_rexp(jw_\theta)$ then this may not even be a unitary transform so we cannot call this Fourier transform.