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Introduction

The Fourier Slice Theorem elucidates how the projections measured by a medical imaging device can be used to reconstruct the object being scanned. From those projections a Continuous Time Fourier Transform (CTFT) is taken. Then according to the theorem, an inverse Continuous Space Fourier Transform (CSFT) can be used to form the original object,$ f(x,y) $. There are two proofs that will be demonstrated.

Fourier Slice Theorem

Given:

$ (x,y): $= the coordinates of the system the original object resides in (as seen in Figure 1)

$ (r,z): $= the coordinates of the system the projection resides in rotated at an angle $ \theta $ relative to the object's coordinate system (as seen in Figure 2)

$ \rho: $= the frequency variable corresponding to $ r $

$ u: $= the frequency variable corresponding to $ x $

$ v: $= the frequency variable corresponding to $ y $

The Fourier Slice Theorem (FST) states that if
$ \begin{align} P_{\theta}({\rho}) &= CTFT \{p_\theta(r)\} \\ F(u,v) &= CSFT\{f(x,y)\} \end{align} $

Then
$ P_{\theta}({\rho}) = F(\rho\cos(\theta),\rho\sin(\theta)) \ $

This means that $ P_{\theta}({\rho}) $ is $ F(u,v) $ in polar coordinates. Basically

Proof

Method 1

From the derivation of the Radon transform, we have
$ p_{\theta}(r) = \int_{-\infty}^{\infty}f(\mathbf{A_{\theta}} \begin{bmatrix} r \\ z \end{bmatrix}) dz $

$ \begin{align} \Rightarrow P_{\theta}(\rho) &= CTFT \{p_\theta(r)\}\\ &= \int_{-\infty}^{\infty} p_{\theta}(r)e^{-j2\pi r\rho}dr \\ &= \int_{-\infty}^{\infty}[\int_{-\infty}^{\infty} f(\mathbf{A_{\theta}} \begin{bmatrix}r \\ z \end{bmatrix})dz]e^{-j2\pi r\rho}dr \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(\mathbf{A_{\theta}} \begin{bmatrix}r \\ z \end{bmatrix})e^{-j2\pi r\rho}dzdr \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(\mathbf{A_{\theta}}*\mathbf{A_{-\theta}} \begin{bmatrix}x \\ y \end{bmatrix})e^{-j2\pi r\rho}dzdr \ \ (*)\\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y)e^{-j2\pi r\rho} \left | \frac{\partial (r,z)}{\partial (x,y)} \right \vert dxdy \ \ (**) \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y)e^{-j2\pi (x\rho\cos(\theta) + y\rho\sin(\theta))}dxdy \\ &= F(\rho\cos(0),\rho\sin(0)) \end{align} $

$ * $ Start a change of variable where
$ \begin{bmatrix}r \\ z\end{bmatrix} = \mathbf{A_{-\theta}} \begin{bmatrix}x \\ y\end{bmatrix} $
Note that from this relationship, $ r = x\cos(\theta) + y\sin(\theta) $ and $ z = -x\sin(\theta) + y\cos(\theta) $

$ ** $ Continue the changing of variables with the Jacobian. Note:

$ \begin{align} \left | \frac{\partial (r,z)}{\partial (x,y)} \right \vert &= d \end{end} $

Method 2

Let $ \theta = 0 $. Then notice that when $ \theta = 0 $, the $ x $ and $ r $ axes line up, as do the $ y $ and $ z $ axes. From the derivation of the Radon transform, we get the following definition equation.

$ p_{\theta}(r) = \int_{-\infty}^{\infty} f(r\cos(\theta)-z\sin(\theta),r\sin(\theta)+z\cos(\theta))dz $ $ \begin{align} p_{0}(r) &= \int_{-\infty}^{\infty} f(r\cos(0)-z\sin(0),r\sin(0)+z\cos(0))dz \\ &= \int_{-\infty}^{\infty} f(r,z)dz \\ \end{align} $
Now let's take the CTFT of both sides.
$ \begin{align} P_0(\rho) &= \int_{-\infty}^{\infty} p_0(r)e^{-j2\pi r\rho}dr \\ &= \int_{-\infty}^{\infty}[\int_{-\infty}^{\infty} f(r,z)dz]e^{-j2\pi r\rho}dr \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(r,z)e^{-j2\pi r\rho}e^{-j2\pi z*0}drdz \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y)e^{-j2\pi(x\rho + y0)}dxdy \\ &= F(\rho,0) \\ &= F(\rho\cos(0),\rho\sin(0)) \end{align} $

Since the FST holds true under $ \theta = 0 $, by the rotation property of the CSFT, the FST must hold true for any $ \theta $.

References:
[1] C. A. Bouman. ECE 637. Class Lecture. Digital Image Processing I. Faculty of Electrical Engineering, Purdue University. Spring 2013.

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