This may be an elementary question, but I'm not sure how DFT/FFT is used to approximate regular Fourier transforms.

Consider the radial distribution function $g(r)$. The structure factor is defined as $$ S(q) = 1 + \rho \int_V d\vec{r} e^{-i\vec{q}\vec{r}} g(r) $$ and consider only the integral part of the above. The integral part is the Fourier transform of $g(r)$.

Now I don't know if there is any straightforward way to calculate that part using a DFT/FFT, what I would do is the following.

First convert to spherical coordinates, $$ \int_V d\vec{r} e^{-i\vec{q}\vec{r}} g(r) = 2\pi \int_{-1}^{1}d\cos{\theta} \int_{0}^{r_{max}}dr r^2 e^{-iqr \cos\theta}g(r) = \frac{4 \pi}{q} \int_{0}^{r_{max}}dr r \sin{qr} g(r) $$ We can write this as, $$ Im\left(\frac{4 \pi}{q} \int_{0}^{r_{max}}dr g(r)re^{iqr}\right) $$ Then, discretization gives, $$ Im\left(\frac{4 \pi}{q} \sum_{n=0}^{k}\frac{r_{max}}{k} g(n\frac{r_{max}}{k})n\frac{r_{max}}{k}e^{iqn\frac{r_{max}}{k}}\right) = Im\left(\frac{4 \pi r_{max}^2}{k^2q} \sum_{n=0}^{k} n g_n e^{-i 2\pi \frac{q'}{k} n}\right) $$ Now if I haven't done any mistakes, the sum is the discrete Fourier transform definition, with $x_n = ng_n$ and I set $q' = \frac{qr_{max}}{2\pi}$. Now I can use DFT/FFT to calculate it.

As you see, it took quite a few calculations to arrive at the correspondence. The question thus is if there is any direct correspondence between regular Fourier transform and DFT/FFT so that the above intermediate steps can be avoided.


1 Answer 1


In two dimensions, you would have ended with a Hankel transform. There exists some techniques for fast Hankel transforms, which could also be applicable to your problem. One trick is to turn the problem into a convolution, for example by transforming to logarithmic coordinates. Then the FFT can be used for the convolution.

It seems that you arrived at a formulation of your problem which is even more directly accessible to an FFT approach.

Regarding your question about the direct correspondence between Fourier transform and DFT/FFT, you are mixing dimensions here. Why should a three dimensional radial symmetric Fourier transform have a direct correspondence with a one dimensional DFT/FFT? There are correspondences and the FFT can be used for fast evaluations, but of course these correspondences are not direct.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .