Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Have the stochastic following process f(t) been studied in mathematics ? It is stationary, Gaussian, f(t) - complex independent Gaussians N(0,1). The autocorrelation is given by the zeroth-order Bessel function of the first kind: $J_{0} (\tau)$.

In radio wave propagation it is called Rayleigh fading or sometimes Jakes fading model. And it is often used in signal processing. So I wonder that it might be some studies of this process in mathematics, which might give me some new point of view on it.

In particular I hope for the following. There should be some natural and mathematically clearly formulated reason (model) which will lead to Bessel function auto-correlation. In signal processing this is said as "radio wave amplitudes" autocorrelate with Bessel function. But can we avoid "radio waves" ? Can we just formulate some simple mathematical model from which we can derive this autocorrelation from something like a central limit theorem or some other mathematically clear reason. I think this should be known, but I am not expert in the field.

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

Jakes is a bit beyond me, but I'm fairly sure that the Bessel function $J_0$ emerges, because of the integral presentation $$ J_0(x)=\frac1{2\pi}\int_{-\pi}^{\pi}e^{ix\sin t}dt. $$ If you have two plane waves of the same frequency, the other reflected so that the two copies have angular separation $\theta$. Then their correlation would vary like $e^{ix\cos\theta}$, because the the projection of the wavelength of the other wave along the direction of propagation of the other gets multiplied by $\cos\theta$. Now treat $\theta$ as a random variable uniformly distributed over the circle and compute the average.

IOW, I think that the Bessel function emerges as a consequence of the underlying geometry as opposed to being a design parameter. Hopefully a more knowledgeable person can answer. This was just a bit too long to fit into a comment.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.