Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I stuck with a rather simple question, but i can't fix it. Hope you can help me.

Hankel transform for a cylindrically symmetric problem is given by the following formula $g(\rho) = 2\pi \int \limits_{0}^{\infty} r f(r) J_l(2 \pi r \rho) dr$

Inverse transform looks the same. It is obvious that in a real system we deal with a bounded region $[0,r_{max}]$. Reading an article concerning numerical implementation of HT I met the change of variables: $x=r/b$ and $y=\rho/\beta$

where $b$ and $\beta$ are maximum values of spatial ($r$) and frequency ($\rho$) domain, respectively.

So, authors obtained the following expression

$g(y) = 2\pi \gamma \frac{b}{\beta} \int \limits_0^1 x f(x) J_l (2 \pi \gamma x y)dx$

where $\gamma = b \beta$

It is not obvious for me how to move from $g(\rho) = g(\beta y)$ to $g(y)$. And the same with $f(bx) \to f(x)$ under the integal.

share|improve this question
    
I found an answer. This change of variables is described in Goudman's Introduction to Fourier Optics. –  jacksonslsmg4 Sep 28 '12 at 8:30

1 Answer 1

SO, $r=bx$, where $b$ is some constant, then $dr=b\cdot dx$ (meaning exactly that the differentiate of $r$ by $x$, $\displaystyle\frac{dr}{dx} = b$). So, $$g(y)=2\pi \underset{bx=0}{\overset{bx=b}\int} bx f(bx)J_l(2\pi bx y)\cdot b dx $$ Moving $g(\rho)$ to $g(y)$ was just substituting $y$ in the place of the variable. But, for moving $f(bx)$ to $f(x)$, I guess, needs some assumption on $f$ (for example, being linear), else it's not changing..

share|improve this answer
    
No, it's not linear. In general $f(r)$ describes electric field distribution in a laser pulse. i don't understand how they move from $r$ to $x$ in $f$'s argument. –  jacksonslsmg4 Sep 26 '12 at 11:36

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.