hankel transform (fourier bessel of zero order) question

[ElDonAntonio]

Is there anyone who would know by any chance the steps to find the hankel transform (also called fourier-bessel of zero order) of a function? I know the equation of the Hankel transform, but I just don't understand how to simplify it.
I'm actually trying to find the hankel transform of a dirac's delta. I found the answer on the web, but I'd like to know how to get to it myself.

 

[RaynorWolfcastle]

Mathworld is your friend

I hope that helps. I have no idea how that transform works (I'm only familiar with Fourier and Laplace) but Mathworld is usually pretty good at explaining this kind of stuff.

 

[ElDonAntonio]

thanks RaynorWolfcastle, I appreciate it.
Actually I saw that page before, and that's where I found the answer for the transform of a dirac's delta (pi*a*J0(2pi*a*q)). My problem is i can't seem to get to this answer by starting with the formula of the Hankel transform. How to integrate the damn Bessel series is over me.

Any clues?

 

[RaynorWolfcastle]

Well here's some insight (that you probably already know, but here goes. The integral of Dirac's delta over 0 to infinity is 1 by definition since delta is 0 everywhere but at 0. Now if you imagine that J0 is actually just a weight given to the delta function, then you find that int( delta*J0,0,infinity) = J0 at 0. And so you have that the hankel transform of the delta function is 2*Pi*J0(0)

This explanation isn't very rigorous but I'm pretty sure it's right, because it's very similar to how you find the Fourier transform of the delta function

 

[ElDonAntonio]

right! I see what you mean! not sure I can really demonstrate it in a rigorous way, but I'll give it a shot.

thanks again RaynorWolfcastle!

 

[RaynorWolfcastle]

Originally posted by: ElDonAntonio
right! I see what you mean! not sure I can really demonstrate it in a rigorous way, but I'll give it a shot.

thanks again RaynorWolfcastle!

Actually, if you want to show it rigorously just say that delta(x) is identically 0 at every point where x != 0. Therefore J0*delta(x) is therefore found to be identically 0 at every point where x !=0. Then show that J0(0) = K , where K is some expression that is independent of the variable of integration x. You can then move K outside yout integral and integrate delta(x), which is 1 by definition.

Just put that in nice mathematical expressions and you're done 🙂. Out of curosity, what is the Hankel Transformation used for?

 

[ElDonAntonio]

that's pretty much what I just wrote down on my paper, we thought about it at the same time 🙂 It's actually simpler than that, no need to mention the delta's value is 0, simply that it's integral is 1.

As I'm calculating the transform for the delta(r-r0), I get 2*pi * int(r0*J0(2pi*r0*q)*delta(r-r0) dr ), and everything becomes a constant (and can be moved out of the integral) except for the delta function. The integral of that is easy, as it results in 1. You're a smart fellow!

I'm taking a medical imaging class (biomedical engineering), and I believe the Hankel transform is used to do sampling with a radially symmetric kernel (eg a circle). I'm not really sure how that's applied to real-world problems though. We just had one class and I think the professor is trying to scare off people (too many people in the class), so we just saw a bunch of barely-comprehensible math, and nothing really about medical imaging yet. That was one question in the homework we have.

Now I have to do the same thing (Hankel transform) for a rectangle...stick around, I might be back 😀

 

[ElDonAntonio]

I think I just understood how to do it for the rectangle. The rectangle function will simply change the bounds of integration. And the integral of J0 ends up being a term in J1 (can easily be shown if you integrate the Bessel series of 0 order).