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Radon Transform--Cylinder

\begin{figure}\begin{center}\BoxedEPSF{radon_cylinder.epsf scaled 890}\end{center}\end{figure}

Let the 2-D cylinder function be defined by

\begin{displaymath}
f(x,y)\equiv\cases{
1 & for $r<R$\cr
0 & for $r>R$.\cr}
\end{displaymath} (1)

Then the Radon transform is given by
\begin{displaymath}
R(p, \tau)=\int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y) \delta[y-(\tau+px)]\,dy\,dx,
\end{displaymath} (2)

where
\begin{displaymath}
\delta(x)={1\over 2\pi} \int_{-\infty}^\infty e^{-ikx}
\end{displaymath} (3)

is the Delta Function.
$R(p, \tau) = {1\over 2\pi} \int_0^{2\pi}\int_0^R \int_{-\infty}^\infty e^{-ik(r\sin\theta-pr\cos\theta)}r\,dr\,d\theta\,dk$
$ = {1\over 2\pi} \int_{-\infty}^\infty e^{ik\tau}\int_0^{2\pi}\int_0^R e^{-ikr(\sin\theta-p\cos\theta)}r\,dr\,d\theta\,dk.$

(4)
Now write
\begin{displaymath}
\sin\theta-p\cos\theta=\sqrt{1+p^2}\,\cos(\theta+\phi)\equiv \sqrt{1+p^2}\,\cos\theta',
\end{displaymath} (5)

with $\phi$ a phase shift. Then


$\displaystyle R(p, \tau)$ $\textstyle =$ $\displaystyle {1\over 2\pi}\int_{-\infty}^\infty e^{ik\tau}\int_0^R\left({\int_0^{2\pi} e^{-ik\sqrt{1+p^2}\,r\cos\theta'}\,d\theta'}\right)\,r\,dr\,dk$  
  $\textstyle =$ $\displaystyle {1\over 2\pi} \int_{-\infty}^\infty e^{ik\tau} \int_0^R 2\pi J_0(k\sqrt{1+p^2}\, r)r\,dr\,dk$  
  $\textstyle =$ $\displaystyle \int_{-\infty}^\infty e^{ik\tau} \int_0^R J_0(k\sqrt{1+p^2}\,r)r\,dr\,dk.$ (6)

Then use
\begin{displaymath}
\int_0^z t^{n+1}J_n(t)\,dt=z^{n+1}J_{n+1}(z),
\end{displaymath} (7)

which, with $n=0$, becomes
\begin{displaymath}
\int_0^z tJ_0(t)\,dt=zJ_1(z).
\end{displaymath} (8)

Define
$\displaystyle t$ $\textstyle \equiv$ $\displaystyle k\sqrt{1+p^2}\,r$ (9)
$\displaystyle dt$ $\textstyle =$ $\displaystyle k\sqrt{1+p^2}\,dr$ (10)
$\displaystyle r\,dr$ $\textstyle =$ $\displaystyle {t\,dt\over k^2(1+p^2)},$ (11)

so the inner integral is


$\displaystyle \int_0^{R\sqrt{1+p^2}} J_0(t) {t\,dt\over k^2(1+p^2)}$ $\textstyle =$ $\displaystyle {1\over k^2(1+p^2)} kR\sqrt{1+p^2} J_1(kR\sqrt{1+p^2}\,)$ (12)
  $\textstyle =$ $\displaystyle {J_1(kR\sqrt{1+p^2}\,)\over k\sqrt{1+p^2}} R,$ (13)

and the Radon transform becomes


$\displaystyle R(p, \tau)$ $\textstyle =$ $\displaystyle {R\over \sqrt{1+p^2}} \int_{-\infty}^\infty {e^{ik\tau}J_1(kR\sqrt{1+p^2}\,)\over k}\,dk$  
  $\textstyle =$ $\displaystyle {2R\over \sqrt{1+p^2}} \int_0^\infty {\cos(k\tau)J_1(kR\sqrt{1+p^2}\,)\over k}\,dk$  
  $\textstyle =$ $\displaystyle \left\{\begin{array}{ll} {2\over 1+p^2} \sqrt{R^2(1+p^2)-\tau^2} ...
...tau^2<R^2(1+p^2)$}\\  0 & \mbox{for $\tau^2\geq R^2(1+p^2)$.}\end{array}\right.$ (14)

Converting to $R'$ using $p=\cot\alpha$,
$\displaystyle R'(r, \alpha)$ $\textstyle =$ $\displaystyle {2\over\sqrt{1+\cot^2\alpha}} \sqrt{(1+\cot^2\alpha)R^2-r^2\csc^2\alpha}$  
  $\textstyle =$ $\displaystyle {2\over \csc\alpha}\sqrt{\csc^2\alpha R^2-r^2\csc^2\alpha}$  
  $\textstyle =$ $\displaystyle 2\sqrt{R^2-r^2},$ (15)

which could have been derived more simply by
\begin{displaymath}
R'(r, \alpha) = \int_{-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}} dy.
\end{displaymath} (16)



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© 1996-9 Eric W. Weisstein
1999-05-25