# What is the advantage of using Bessel function as the basis in a Hankel Transform?

I came across literature which says that the Fourier - Bessel (Hankel) transformation is better than a STFT or a Wigner - Ville distribution as the Bessel function forms an orthonormal basis and they occupy finite bandwidth. This is supposed to translate into better time and frequency resolutions. Isn't it the same with the regular fourier transform ? so why not use an STFT itself to transform the signal into time frequency domain ?

• Could you provide information on where you came across such claims? – Laurent Duval Jan 25 '18 at 7:31
• Hey Laurent, I may be wrong as well. I am just starting off, and want to understand better: pdfs.semanticscholar.org/468a/… – Asp Jan 25 '18 at 13:28

The Hankel Transform includes the Bessel function as part of its definition. I wouldn’t say the Bessel has an advantage. I would say that the Hankel Transform has an advantage where a two dimensional signal has radial symmetry, or saying it another way, the signal in $x$ and $y$ can be expressed as a product of two functions in $r$ and $\theta$ through a change of variables. $x=r\cos \theta$ $y=r\sin \theta$
$$f(x,y)dx dy \Rightarrow g(r) h( \theta) r dr d \theta$$
The Hankel Transform is equivalent to a Fourier Transform on the radius $r$ variable.