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Radon Inverse Transform: Filtered Back Projection (FBP) 📂Tomography

Radon Inverse Transform: Filtered Back Projection (FBP)

Theorem

There is a formula that holds for f:R2→Rf : \mathbb{R}^{2} \to \mathbb{R}.

Description

Also known as The filtered back projection formula.

Given the Radon transform Rf\mathcal{R}f of ff, it is said that ff can be obtained using the Fourier transform and back projection. This means, applying a Fourier transform to the Radon transform, multiplying by ∣S∣|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)=F2−1F2f(x,y) f(x,y)={\mathcal{F}_2}^{-1}\mathcal{F}_2 f(x,y)

Here, F2\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.

  • 14π2∫−∞∞∫−∞∞F2f(X,Y)ei(xX+yY)dXdY \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)(X,Y) in polar coordinates (S,θ)(S,\theta). Then X=Scos⁡θX=S\cos\theta, Y=Ssin⁡θY=S\sin\theta. And the following holds.

  • ∣∂X∂S∂X∂θ∂Y∂S∂Y∂θ∣=∣S∣ \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∣dSdθdXdY=|S|dSd\theta, and if we represent the above integration in polar coordinates, it looks like this.

  • 14π2∫0π∫−∞∞F2f(Scos⁡θ,Ssin⁡θ)eiS(xcos⁡θ+ysin⁡θ)∣S∣dSdθ \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

Fourier Slice Theorem

F2f(Scos⁡θ, Ssin⁡θ)=F(Rf)(S, θ) \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.

  • 12π12π∫0π∫−∞∞F(Rf)(S, θ)eiS(xcos⁡θ+ysin⁡θ)∣S∣dSdθ \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.

  • 12π∫0πF−1[∣S∣F(Rf)(S, θ)](xcos⁡θ+ysin⁡θ, θ)dθ \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

Back Projection

Bf(x,y):=∫0πf(xcos⁡θ+ysin⁡θ, θ)dθ \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.

  • 12B{F−1[∣S∣F(Rf)(S, θ)]}(x,y) \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)=12B{F−1[∣S∣F(Rf)(S, θ)]}(x,y) 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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