What is the $\mathcal Z$-transform of Bessel function $J_0(\alpha n)$ sequence

Question

What is the $\mathcal Z$-transform of the sequence $J_0(\alpha n)$ for $n \in \mathbb{Z}$?

The Fourier transform of zero$^{\rm th}$ order Bessel function $J_0(\alpha x)$ is known to be $\frac{2}{\sqrt{\alpha^2 - \omega^2}}$ for $|\omega| < \alpha$. This has a pole at $\omega = \alpha$. Does this imply that the $\mathcal Z$-transform will also have a pole on the unit circle?

EDIT:

The problem I'm looking at involves discrete samples of Bessel function i.e. $J_0(n)$. How should I proceed to determine its $\mathcal Z$-transform?

Answer 1

The Taylor expansion for the Bessel function of the first kind and 0th order is $$ J_{0}\left ( x \right )= \sum_{m=0}^{\infty }\frac{(-1)^{m}}{(m!)^{^{2}}}\left ( \frac{1}{2}x \right )^{2m} $$

(see http://en.wikipedia.org/wiki/Bessel_function)

So you can basically approximate this as the Z-transform of a polynomial.

Answer 2

You can apply the definition of the $\mathcal Z$-transform to an equivalent expression of the Bessel function, or to an approximation.

The equivalent function can be: \begin{align} J_0(x) &= \frac 1\pi\cos\left(x\cos\phi\right)d\phi\\ &=\frac 1\pi\int_0^\pi\left(1-\frac{x^2\cos^2\phi}{2!}+\frac{x^4\cos^4\phi}{4!}-\frac{x^6\cos^6\phi}{6!}+\cdots\right)d\phi \end{align}

Update:

More information about equivalent expressions is here.