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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?

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  • $\begingroup$ I'm curious, what is the application for this? $\endgroup$ – nibot Mar 23 '12 at 10:17
  • $\begingroup$ @nibot I am working with isotropic noise model and for 2D case, the noise covariance matrix elements are zeroth order Bessel functions of first kind. An the eigenvalues of the cov. matrix happens to be related to the Z-transform of the Bessel function sequence. $\endgroup$ – sauravrt Mar 24 '12 at 12:27
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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.

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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.

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    $\begingroup$ The approximation for $J_0(x)$ is missing the integral sign in the first step. I am unable to see you to got the approximate Z-transform. I had another idea, using the approximation $J_0(x) = \sqrt(\frac{2}{x \pi} cos(x - \pi/4)$. I tried this approach and ended up with Z-transform involving PolyLogarithmic function. (Used Mathematica). $\endgroup$ – sauravrt Mar 21 '12 at 1:02
  • $\begingroup$ I believe the approximation that he's talking about is an approximation for the modified Bessel function of the first kind $I_0(z)$ (if memory serves me). The $z$ is the argument to the function, not $z$ as in $z$-transform. He's pointing out that instead of evaluating the $z$-transform sum directly, you could use some other form that is either equivalent or approximately equivalent to the function of interest that might be easier to transform. $\endgroup$ – Jason R Mar 21 '12 at 1:42
  • $\begingroup$ Your appreciation about the approximation was true. I have edited my answer. $\endgroup$ – Luis Andrés García Mar 21 '12 at 7:25

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