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Part 2

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 $ \color{blue}\text{Consider an image } f(x,y) \text{ with a forward projection} $

                $ \color{blue} p_{\theta}(r) = \mathcal{FP}\left \{ f(x,y) \right \} $

                             $ \color{blue} = \int_{-\infty}^{\infty}{f \left ( r cos(\theta) - z sin(\theta),r sin(\theta) + z cos(\theta) \right )dz}. $

$ \color{blue} \text{Let } F(\mu,\nu) \text{ be the continuous-time Fourier transform of } f(x,y) \text{ given by} $
              $ \color{blue} F(u,v) = \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{f(x,y)e^{-j2\pi(ux,vy)}dx}dy} $

$ \color{blue} \text{and let } P_{\theta}(\rho) \text{ be the continuous-time Fourier transform of } p_{\theta}(r) \text{ given by} $
              $ \color{blue} P_{\theta}(\rho) = \int_{-\infty}^{\infty}{p_{\theta}(r)e^{-j2\pi(\rho r)}dr}. $

$ \color{blue}\text{a) Calculate the forward projection }p_{\theta}(r) \text{, for } f(x,y) = \delta(x,y). $

$ \color{blue}\text{Solution 1:} $

$ p_{\theta}(r) = \int_{-\infty}^{+\infty}{\delta(r cos\theta - z sin\theta, r sin\theta + z cos \theta) dz} $

$ = z \text{ when } \left\{\begin{matrix} r cos \theta - z sin \theta = 0 \\ r sin \theta + z cos \theta = 0 \end{matrix}\right. $

$ = \frac{r cos\theta}{sin \theta}, \theta > 0 $

$ {\color{red} \text{This answer is incorrect. The correct answer is as following:}} $

$ {\color{green} \text{Recall:}} $

$ {\color{green} \text{i) } \int_{-\infty}^{+\infty}{f(g(t)) \delta (t) dt} = f(g(t=0)) \int_{-\infty}^{+\infty}{\delta (t) dt} } $

$ {\color{green} \text{ii) } \int_{-\infty}^{+\infty}{\delta (\alpha t) dt} = \int_{-\infty}^{+\infty}{\delta (u) \frac{du}{|\alpha|}} = \frac{1}{|\alpha|} } $

$ {\color{green} \text{iii) } \delta() \text{ function is separable: } \delta(x,y) = \delta(x) \cdot \delta(y) } $

$ \color{green} \text{ Define } u = r cos\theta - z sin\theta $

$ \color{green} \Rightarrow dz = \frac{du}{|sin\theta|} $

$ \color{green} \text{ Now } $

$ \color{green} p_{\theta}(r) = \int_{-\infty}^{+\infty}{\delta(r cos\theta - z sin\theta, r sin\theta + z cos \theta) dz} $

$ \color{green} p_{\theta}(r) = \int_{-\infty}^{+\infty}{\delta(g(u)) \delta(u) \frac{du}{|sin\theta|}} = \frac{\delta(u=0)}{|sin\theta|} $

$ \color{green} = \frac{\delta(\frac{r}{sin\theta})}{|sin\theta|} = \frac{|sin\theta|}{|sin\theta|} \delta(r) = \delta(r) $

$ \color{blue}\text{Solution 2:} $

$ p_{\theta}(r) = \int_{-\infty}^{+\infty}{\delta(r cos\theta - z sin\theta, r sin\theta + z cos \theta) dz} $

$ = \delta(r) $

$ {\color{green} \text{Here, the student uses the intuitive solution: in this case the answer does not depend on } \theta \text{, since the image just contains a peak at origin. } } $

$ \color{blue}\text{b) Calculate the forward projection }p_{\theta}(r) \text{, for } f(x,y) = \delta(x-1,y-1). $

$ \color{blue}\text{Solution 1:} $

$ p_{\theta}(r) = \int_{-\infty}^{+\infty}{\delta(r cos\theta - z sin\theta - 1, r sin\theta + z cos \theta - 1) dz} $

$ = z \text{ when } \left\{\begin{matrix} r cos \theta - z sin \theta = 1 \\ r sin \theta + z cos \theta = 1 \end{matrix}\right. $

$ = \frac{r cos\theta - 1}{sin \theta}, \theta > 0 $

$ {\color{red} \text{This answer is incorrect. The correct answer is as following:}} $

$ \color{green} \text{ Similar to the solution 1 to part a) we define u: } u = r cos\theta - z sin\theta - 1 $

$ \color{green} \text{ Following the same logic as in part a) we obtain the final answer:} $

$ \color{green} p_{\theta}(r) = \delta(r - (cos\theta + sin \theta)) = \delta(r - \sqrt{2} cos (\theta - \frac{\pi}{4})) $

$ \color{blue}\text{Solution 2:} $

$ \tilde{p}_\theta(r) = p_{\theta}(r - \sqrt{1+1} cos(\theta - tan^{-1}(\frac{1}{1}))) $

$ = p_\theta(r - \sqrt{2} cos(\theta - \frac{\pi}{4})) $

$ = \delta(r - \sqrt{2} cos(\theta - \frac{\pi}{4})) $

$ {\color{green} \text{Again, the student uses the intuitive solution: in this case the answer does depend on } \theta \text{, since the peak is shifted from the origin to the point } (1,1). } $

$ \color{blue}\text{c) Calculate the forward projection }p_{\theta}(r) \text{, for } f(x,y) = rect \left(\sqrt[]{x^2+y^2} \right). $

$ \color{blue}\text{Solution 1:} $

$ p_{\theta}(r) = \int_{-\infty}^{+\infty}{rect(\sqrt{(r cos\theta - z sin\theta)^2 + (r sin\theta + z cos \theta)^2)} dz} $

$ \color{green} \text{Recall should be added:} $

$ \color{green} rect(t) = \left\{\begin{matrix} 1, for |t|\leq \frac{1}{2} \\ 0, otherwise \end{matrix}\right. $

$ \color{green} \text{therefore: } $

$ p_{\theta}(r) = \int_{-\sqrt{\frac{1}{4} - r^2}}^{\sqrt{\frac{1}{4} - r^2}}{1 dz} $

$ = \left\{\begin{matrix} \sqrt{1 - 4r^2}, &\text{ if }|r| \leq \frac{1}{2} \\ 0, &\text{ otherwise} \end{matrix}\right. $

$ \color{blue}\text{Solution 2:} $

$ p_{\theta}(r) = \int_{-\infty}^{+\infty}{f(r cos\theta - z sin\theta, r sin\theta + z cos \theta) dz} $

$ = \int_{-\sqrt{\frac{1}{4} - r^2}}^{\sqrt{\frac{1}{4} - r^2}}{1 dz} = \sqrt{1 - 4r^2}, \text{ if }|r| \leq \frac{1}{2} $

$ \text{ else } p_{\theta}(r) = 0 $

$ \color{blue}\text{d) Calculate the forward projection }p_{\theta}(r) \text{, for } f(x,y) = rect \left(\sqrt[]{(x-1)^2+(y-1)^2} \right). $

$ \color{blue}\text{Solution 1:} $

$ p_{\theta}(r) = \int_{-\infty}^{+\infty}{rect \left( \sqrt{(r cos\theta - z sin\theta - 1)^2 + (r sin\theta + z cos \theta - 1)^2} \right) dz} $

$ = \int_{-\sqrt{\frac{1}{4} - (r - (cos\theta + sin\theta))^2}}^{\sqrt{\frac{1}{4} - (r - (cos\theta + sin\theta))^2}}{1 dz} $

$ = \left\{\begin{matrix} \sqrt{1 - 4(r - (cos\theta + sin\theta))^2}, &\text{ if }|r| \leq \frac{1}{2} \\ 0, &\text{ otherwise} \end{matrix}\right. $

$ \color{green} \text{ To make it more clear, the following form could be obtained:} $

$ \color{green} = \left\{\begin{matrix} \sqrt{1 - 4(r - \sqrt{2} cos (\theta - \frac{\pi}{4}))^2}, &\text{ if }|r| \leq \frac{1}{2} \\ 0, &\text{ otherwise} \end{matrix}\right. $

$ \color{blue}\text{Solution 2:} $

$ \tilde{p}_\theta(r) = p_{\theta}(r - \sqrt{1+1} cos(\theta - tan^{-1}(\frac{1}{1}))) $

$ = p_\theta(r - \sqrt{2} cos(\theta - \frac{\pi}{4})) $

$ \text{ where } p_\theta(r) = \left\{\begin{matrix} \sqrt{1 - 4r^2}, &\text{ if }|r| \leq \frac{1}{2} \\ 0, &\text{ else} \end{matrix}\right. $

$ {\color{green} \text{Here, the student uses the results from solutions to part b and c.} } $

$ \color{blue}\text{e) Describe in precise detail, the steps required to perform filtered back projection (FBP) reconstruction of } f(x,y). $

$ \color{blue}\text{Solution 1:} $

$ 1. \text{ Compute } \rho_{\theta}(r) $

$ 2. \text{ Compute FT of step 1.} $

$ 3. \text{ Multiply step 2 by the filter } H(\rho) = |\rho| = f_c \left [ rect(\frac{f}{2f_c}) - \Lambda(\frac{f}{f_c}) \right ] $

$ 4. \text{ Compute inverseFT of step 3.} $

$ \color{red} \text{ This answer is not complete. We need the final step:} $

$ \color{red} 5. \text{ Back project } g_{\theta}(r) \text{ (obtained from step 4) and get: } $

$ \color{red} f(x,y) = \int_{0}^{\pi}{g_\theta(xcos\theta + ysin\theta)d\theta} $

$ \color{blue}\text{Solution 2:} $

$ 1. \text{ Measure the projections } \rho_{\theta}(r) \text{ at various angles} $

$ 2. \text{ Filter the projections } \rho_{\theta}(r) \text{ with } h(r) \text{, where } H(\rho) = |\rho| \text{ and get } g_{\theta}(r) $

$ 3. \text{ Back project } g_{\theta}(r) \text{ along } r = xcos\theta + ysin\theta \text{ and get } $

$ f(x,y) = \int_{0}^{\pi}{g_\theta(xcos\theta + ysin\theta)d\theta} $

Solution 3:

a)

$ p_\theta(r)=\begin{cases}1, & \text{if } r=0\\0, & \text{otherwisr}\end{cases}=\delta(r) $.

b)

• If $ 0\leq\theta<\frac{\pi}{4} $, $ d=\sqrt{2}\text{cos}(\frac{\pi}{4}-\theta)=\sqrt{2}[\text{cos}\frac{\pi}{4}\text{cos}\theta+\text{sin}\frac{\pi}{4}\text{sin}\theta]=\text{cos}\theta+\text{sin}\theta $.

• If $ \frac{\pi}{4}\leq \theta < \frac{\pi}{2} $, $ d=\sqrt{2}\text{cos}(\theta-\frac{\pi}{4})=\sqrt{2}\text{cos}(\frac{\pi}{4}-\theta)=\text{cos}\theta+\text{sin}\theta $.

• If $ \frac{\pi}{2}\leq \theta < \frac{3\pi}{4} $, $ d=\sqrt{2}\text{sin}(\frac{\pi}{4}-(\theta-\frac{\pi}{2}))=\sqrt{2}\text{sin}(\frac{3\pi}{4}-\theta)=\sqrt{2}[\text{sin}\frac{3\pi}{4}\text{cos}\theta-\text{cos}\frac{3\pi}{4}\text{sin}\theta]=\text{cos}\theta+\text{sin}\theta $.

• If $ \frac{3\pi}{4}\leq \theta < 2\pi $, $ d=-\sqrt{2}\text{sin}(\theta-\frac{\pi}{2}-\frac{\pi}{4})=-\sqrt{2}\text{sin}(\theta-\frac{3\pi}{4})=-\sqrt{2}[\text{sin}\theta\text{cos}\frac{3\pi}{4}-\text{cos}\theta\text{sin}\frac{3\pi}{4}]=\text{cos}\theta+\text{sin}\theta $.

$ \Rightarrow p_\theta(r)=\delta(r-(\text{cos}\theta+\text{sin}\theta)) $ for all $ \theta $.

c)

Since $ f(x,y) $ is symmetric to $ (0,0) $

$ \Rightarrow p_\theta(r)=p_0(r)=\int_{-\sqrt{\frac{1}{4}-r^2}}^{\sqrt{\frac{1}{4}-r^2}}1\cdot dz=2\int_0^{\sqrt{\frac{1}{4}-r^2}}1\cdot dz\text{, for }\vert r\vert<\frac{1}{2} $

$ \Rightarrow p_\theta(r)=2\sqrt{\frac{1}{4}-r^2}\text{rect}(r) $

d)

From part (b), we know that, if $ f(x,y) $ has $ (1,1) $ shift in $ x-y $ plane, the $ p_\theta(r) $ would have $ (\text{cos}\theta+\text{sin}\theta) $ shift in $ r $-direction.

$ \Rightarrow p_\theta(r)=2\sqrt{\frac{1}{4}-(r-(\text{cos}\theta+\text{sin}\theta))^2}\text{rect}(r-(\text{cos}\theta+\text{sin}\theta)) $

e)

• First, measure the forward projection $ p_\theta(r) $ for sufficient angles $ \theta s $.
• Second, filter the forward projection with $ h(r): g_\theta(r)=h(r)\ast p_\theta(r) $, where $ h(r)=\text{CTFT}^{-1}\left\{\vert\rho\vert\right\} $.
• Finally, back project filtered projection to reconstruct $ f(x,y) $. $ f(x,y)=\int_0^\pi g_\theta(x\text{cos}\theta+y\text{sin}\theta)d\theta $.

Solution 4:

a)

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

When $ f(x,y)=\delta(x,y) $, $ F(\mu,\nu)=1 $.

So $ p_\theta(r)=\delta_\theta(r) $.

Comments:

b)

When $ f(x,y)=\delta(x-1,y-1) $, $ F(\mu,\nu)=e^{-j\mu}\cdot e^{-j\nu} $.

So $ P_\theta(\rho)=e^{-j\rho\text{cos}\theta}\cdot e^{-j\rho\text{sin}\theta} $, $ p_\theta(r)=\delta(r-\text{cos}\theta-\text{sin}\theta) $.

c)

$ p_\theta(r)=\int_{-\sqrt{\frac{1}{4}-r^2}}^{\sqrt{\frac{1}{4}-r^2}}1dz=\sqrt{1-4r^2} $

So $ p_\theta(r)=\begin{cases} \sqrt{1-4r^2}, & |r|<\frac{1}{2}\\0, & |r|>\frac{1}{2}\end{cases} $.

d)

$ r_0=\sqrt{2}\cdot\left(\frac{\pi}{4}-\theta\right) $

$ p_\theta(r)=\int_{-\infty}^{\infty}f(r\text{cos}\theta-z\text{sin}\theta, r\text{sin}\theta+z\text{cos}\theta)dz=\int_{-\sqrt{\frac{1}{4}-(r-r_0)^2}}^{\sqrt{\frac{1}{4}-(r-r_0)^2}}adz $

So $ p_\theta(r)=\begin{cases}\sqrt{1-4\left(r-\sqrt{2}\left(\frac{\pi}{4}-\theta\right)\right)}, & |r-r_0|\leq\frac{1}{2}\\0, & \text{otherwise}\end{cases} $.

Comments:

e)

$ \begin{align} f(x,y) & =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}F(\mu,\nu)e^{2\pi j(x\mu+y\nu)}d\mu d\nu\\ & = \int_{-\infty}^{\infty}\int_{0}^{\pi}p_\theta(\rho)e^{2\pi j(x\rho\text{cos}\theta+y\rho\text{sin}\theta)}|\rho|d\theta d\rho\\ & = \int_{0}^{\pi}\underbrace{\int_{-\infty}^{\infty}|\rho|p_\theta(\rho)e^{2\pi j(x\rho\text{cos}\theta+y\rho\text{sin}\theta)}d\rho}_{g_\theta(x\text{cos}\theta+y\text{sin}\theta)}d\theta \end{align} $

So $ g_\theta(t) $ is given by:

$ g_\theta(t)=\int_{-\infty}^{\infty}|\rho|p_\theta(\rho)e^{2\pi j\rho t}d\rho=\text{CTFT}^{-1}\{|\rho|p_\theta(\rho)\}=h(t)\ast p_\theta(r) $

So what we do is:

1. measure projection $ p_\theta(r) $;
2. filter the projection $ g_\theta(r)=h(r)\ast p_\theta(r) $;
3. back project filtered projection: $ f(x,y)=\int_0^{\pi}g_\theta(x\text{cos}\theta)+y\text{sin}\theta)d\theta $.

Related Problems

a) Calculate the forward projection $ p_\theta(r) $, for $ f(x,y)=\text{rect}(x,y) $, as $ \theta=0 $ and $ \theta=\pi/4 $.

b) Calculate the forward projection $ p_\theta(r) $, for $ f(x,y)=\delta(x-a,y-b) $.

"Communication, Networks, Signal, and Image Processing" (CS)- Question 5, August 2011

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• Part 1: solutions and discussions
• Part 2: solutions and discussions

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