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Let <math>\theta = 0</math>. Then notice that when <math>\theta = 0</math>, the <math>x</math> and <math>r</math> axes line up, as do the <math>y</math> and <math>z</math> axes. From the [[HonorsContractECE438CoordinateAndRadon|derivation of the Radon transform]], we get the following definition equation. <br /> | Let <math>\theta = 0</math>. Then notice that when <math>\theta = 0</math>, the <math>x</math> and <math>r</math> axes line up, as do the <math>y</math> and <math>z</math> axes. From the [[HonorsContractECE438CoordinateAndRadon|derivation of the Radon transform]], we get the following definition equation. <br /> | ||
| − | <math>p_{\theta}(r) = \int_{-\infty}^{\infty} f(r\cos(\theta)-z\sin(\theta),r\sin(\theta)+z\cos(\theta))dz | + | <math>p_{\theta}(r) = \int_{-\infty}^{\infty} f(r\cos(\theta)-z\sin(\theta),r\sin(\theta)+z\cos(\theta))dz |
</math> | </math> | ||
<math> | <math> | ||
| Line 61: | Line 61: | ||
\end{align} | \end{align} | ||
</math><br /> | </math><br /> | ||
| + | Now let's take the CTFT of both sides.<br /> | ||
| + | <math> | ||
| + | \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) | ||
| + | \end{align} | ||
| + | </math> | ||
---- | ---- | ||
Revision as of 17:38, 19 December 2014
Fourier Slice Theorem (FST)
A slecture by ECE student Sahil Sanghani
Partly based on the ECE 637 material of Professor Bouman.
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...
Proof
Method 1
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) \end{align} $
References:
[1] C. A. Bouman. ECE 637. Class Lecture. Digital Image Processing I. Faculty of Electrical Engineering, Purdue University. Spring 2013.
