Radon Inverse Transform: Filtered Back Projection (FBP)
Theorem
There is a formula that holds for $f : \mathbb{R}^{2} \to \mathbb{R}$.
Description
Also known as The filtered back projection formula.
Given the Radon transform $\mathcal{R}f$ of $f$, it is said that $f$ can be obtained using the Fourier transform and back projection. This means, applying a Fourier transform to the Radon transform, multiplying by $|S|$, then applying the inverse Fourier transform, and finally, back projection, is known as the inverse Radon transform.
Proof
By the theorem of Inverse Fourier Transform, the following holds.
- $$ f(x,y)={\mathcal{F}_2}^{-1}\mathcal{F}_2 f(x,y) $$
Here, $\mathcal{F}_2$ is a 2-dimensional Fourier transform. According to the definition of inverse Fourier transform, the right-hand side of the above equation is as follows.
- $$ \dfrac{1}{4\pi^2} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \mathcal{F}_2f(X,Y)e^{i(xX+yY)}dXdY $$
Let’s represent the Cartesian coordinates $(X,Y)$ in polar coordinates $(S,\theta)$. Then $X=S\cos\theta$, $Y=S\sin\theta$. And the following holds.
- $$ \begin{vmatrix} \frac{\partial X}{\partial S} & \frac{\partial X}{\partial \theta} \\ \frac{\partial Y}{\partial S} & \frac{\partial Y}{\partial \theta} \end{vmatrix} =|S| $$
Therefore, $dXdY=|S|dSd\theta$, and if we represent the above integration in polar coordinates, it looks like this.
- $$ \dfrac{1}{4\pi^2} \int_{0}^{\pi} \int_{-\infty}^{\infty} \mathcal{F}_2f(S\cos\theta,S\sin\theta)e^{iS(x\cos\theta+y\sin\theta)}|S|dSd\theta $$
$$ \mathcal{F}_2 f(S \cos\theta,\ S \sin\theta)=\mathcal{F}(\mathcal{R}f)(S ,\ \theta) $$
Then, by the Fourier Slice Theorem, the above equation is as follows.
- $$ \dfrac{1}{2\pi}{\color{blue}\dfrac{1}{2\pi}} \int_{0}^{\pi} {\color{blue}\int_{-\infty}^{\infty} \mathcal{F}(\mathcal{R}f)(S, \ \theta)e^{iS(x\cos\theta+y\sin\theta)}|S|dS}d\theta $$
The part painted blue is as follows, by the definition of inverse Fourier transform.
- $$ \dfrac{1}{2\pi} \int_{0}^{\pi} {\color{blue} \mathcal{F}^{-1} \Big[ |S| \mathcal{F}(\mathcal{R}f)(S,\ \theta) \Big] (x\cos\theta+y\sin\theta,\ \theta) } d\theta $$
$$ \mathcal{B}f(x,y) := \int_{0}^\pi f(x\cos\theta+y\sin\theta,\ \theta) d\theta $$
And the above equation is as follows, by the definition of back projection.
- $$ \dfrac{1}{2} \mathcal{B} \left\{ \mathcal{F}^{-1} \Big[ |S|\mathcal{F}(\mathcal{R}f)(S,\ \theta) \Big]\right\} (x,y) $$
Thus, the following is obtained.
- $$ f(x,y)= \dfrac{1}{2} \mathcal{B} \left\{ \mathcal{F}^{-1} \Big[ |S|\mathcal{F}(\mathcal{R}f)(S,\ \theta) \Big]\right\} (x,y) $$
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