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- TL;DR Summary
- Computing Fourier transforms is simple in Cartesian co-ordinates but how do you do it for polar co-ordinates?

The 2D Fourier transform is given by: [tex]\hat{f}(k,l)=\int_{\mathbb{R}^{2}}f(x,y)e^{-ikx-ily}dxdy[/tex]

In terms of polar co-ordinates: [tex]\hat{f}(\rho,\phi)=\int_{0}^{\infty}\int_{-\pi}^{\pi}rf(r,\theta)e^{-ir\rho\cos(\theta-\phi)}drd\theta[/tex]

For Fourier transforms in cartesian co-ordinates, relating the Fourier transform of a derivative of a function to the Fourier transform of the function. However, what happens with the polar Fourier transform? I've done a simple calculation for the [itex]r-[/itex]derivative and I get:

[tex]\widehat{\frac{\partial f}{\partial r}}=-\rho\sum_{n\in\mathbb{Z}}\int_{0}^{\infty}rf_{n}(r)J_{n-1}(r\rho)dr[/tex]

This is after I expanded [itex]f(r,\theta)[/itex] as a Fourier series:[tex]f(r,\theta)=\sum_{n\in\mathbb{Z}}f_{n}(r)e^{in\theta}[/tex]

I feel a little lost at this point. Can anyone suggest anything?

In terms of polar co-ordinates: [tex]\hat{f}(\rho,\phi)=\int_{0}^{\infty}\int_{-\pi}^{\pi}rf(r,\theta)e^{-ir\rho\cos(\theta-\phi)}drd\theta[/tex]

For Fourier transforms in cartesian co-ordinates, relating the Fourier transform of a derivative of a function to the Fourier transform of the function. However, what happens with the polar Fourier transform? I've done a simple calculation for the [itex]r-[/itex]derivative and I get:

[tex]\widehat{\frac{\partial f}{\partial r}}=-\rho\sum_{n\in\mathbb{Z}}\int_{0}^{\infty}rf_{n}(r)J_{n-1}(r\rho)dr[/tex]

This is after I expanded [itex]f(r,\theta)[/itex] as a Fourier series:[tex]f(r,\theta)=\sum_{n\in\mathbb{Z}}f_{n}(r)e^{in\theta}[/tex]

I feel a little lost at this point. Can anyone suggest anything?