Revision as of 16:32, 30 May 2013

sLecture

↳ Topic 2: Tomographic Reconstruction

↳ Intro

↳ CT

↳ PET

↳ Co-ordinate Rotation

↳ Radon Transform

↳ Fourier Slice Theorem

↳ Convolution Back Projection

Excerpt from Prof. Bouman's Lecture

Accompanying Lecture Notes

introduction Convolution back projection (CBP) or forward back projection. do not need any calculation in frequency domain so it is faster

definition?

Summary of CBP algorithm

1. Measure projections $ p_{\theta}(r) $.
2. Filter the projections to obtain $ g_{\theta}(r) = h(r)*p_{\theta}(r) $.
3. Back project filtered projections
   

$ f(x,y) = \int_0^{\pi}g_{\theta}(x\cos(\theta)+x\sin(\theta))d\theta $

find images like the copyrighted one using impulse?

In order to compute the inverse CSFT of $ F(u,v) $ in polar coordinates, we must use the Jacobian of the polar coordinate transformation.
$ du dv = |\rho|d\theta d\rho $ where $ u = \rho\cos(\theta) $ and $ v = \rho\sin(\theta) $

This results in the expression
$ f(x,y) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}F(u,v)e^{2\pi j(xu+yv)}dudv $

Using the Fourier slice theorem, we can replace F(x,y) with its representation in polar coordinates, $ P_{\theta}(\rho) $
$ \begin{align} \Rightarrow f(x,y) &= \int_{-\infty}^{\infty}\int_0^{\pi}P_{\theta}(r)e^{2\pi j(x\rho\cos(\theta) +y\rho\sin(\theta))}dudv \\ &= \int_0^{\pi}\underbrace{[\int_{-\infty}^{\infty}|\rho|P_{\theta}e^{2\pi j\rho(x\cos(\theta) +y\sin(\theta))}d\rho]}_{g_{\theta}(x\cos(\theta) + y\sin(\theta))}d\theta \end{align} $

Then $ g_{\theta}(t) $ is given by
$ \begin{align} g_{\theta}(t) &= \int_{\infty}^{\infty}|\rho|P_{\theta}(\rho)e^{2\pi j\rho t}d\rho \\ &= CTFT^{-1}\{|\rho|P_{\theta}(\rho)\} \\ &= h(t) * p_{\theta}(r) \end{align} $

where $ h(t) = CTFT^{-1}\{|\rho|\} $, and
$ f(x,y) = \int_0^{\pi}g_{\theta}(x\cos(\theta) + y\sin(\theta))d\theta $

A closer look at Projection Filter

1. At each angle, projections are filtered.
$ g_{\theta}(r) = h(r)*p_{\theta}(r) \ $. 2. the frequency response of the filter is given by
$ H(\rho) = |\rho| $</br> it's like a high pass filter. 3. But the real filters must be bandlimited to $ |\rho| $ ≤ $ f_c $ for some cutoff frequency $ f_c $.

Fig 1: Frequency response of $ H(\rho) = |\rho| $

So
$ \begin{align} H(\rho) &= f_c[rect(f/(2f_c))-\wedge(f/f_c)] \\ h(r) &= f_c^2[2sinc(t2f_c)-sinc^2(tf_c)] \end{align} $

Back projection function is
$ f(x,y) = \int_0^{\pi}b_{\theta}(x,y)d\theta $

where
$ b_{\theta}(x,y) = g_{\theta}(x\cos(\theta)+y\sin(\theta)) $

Consider the set of points $ (x,y) $ such that
$ r = x\cos(\theta)+y\sin(\theta) $

This set looks like

Fig 2: Graphical representation of the set $ \{(x,y):r=x\cos(\theta)+y\sin(\theta)\} $

Along this line, $ b_{\theta}(x,y) = g_{\theta}(r) $.

For each angle $ \theta $ back projection is constant along the lines of angle $ \theta $ and take on value $ g_{\theta}(r) $.

Fig 3: Geometric interpretation

Complete back projection is formed by integrating (summing) back projections for angles ranging from $ 0 $ to $ \pi $.
$ \begin{align} f(x,y) &= \int_0^{\pi}b_{\theta}(x,y)d\theta \\ &\approx \frac{\pi}{M} \sum_{m=0}^{M-1} b_{\frac{m\pi}{M}}(x,y) \end{align} $

Back projection "smears" values of $ g(r) $ back over the image, and then adds smeared images for each angle.

Fig 1: $ P_{\theta}(\rho) $ is $ F(u,v) $ in polar coordinates

References

• C. A. Bouman. ECE 637. Class Lecture. Digital Image Processing I. Faculty of Electrical Engineering, Purdue University. Spring 2013.

Questions and comments

If you have any questions, comments, etc. please post them on this page

Back to the "Bouman Lectures on Image Processing" by Maliha Hossain