Line 118: Line 118:
  
  
−
<math>
+
<font face="serif"><span style="font-size: 19px;"><math>
 
\text{ Similar to the solution 1 to part a) we define u: } u = r cos\theta - z sin\theta - 1  
 
\text{ Similar to the solution 1 to part a) we define u: } u = r cos\theta - z sin\theta - 1  
−
</math>
+
</math></span></font>
  
  
−
<math>
+
<font face="serif"><span style="font-size: 19px;"><math>
 
\text{ Following the same logic as in part a) we obtain the final answer:}
 
\text{ Following the same logic as in part a) we obtain the final answer:}
−
</math>
+
</math></span></font>
  
  
Line 161: Line 161:
  
 
<math>
 
<math>
−
P_{\theta}(\rho)  = \int_{-\infty}^{+\infty}{rect(\sqrt{(r cos\theta - z sin\theta)^2, (r sin\theta + z cos \theta)^2)} dz}
+
P_{\theta}(\rho)  = \int_{-\infty}^{+\infty}{rect(\sqrt{(r cos\theta - z sin\theta)^2 + (r sin\theta + z cos \theta)^2)} dz}
 
</math>
 
</math>
  
Line 168: Line 168:
 
= \int_{-\sqrt{\frac{1}{4} - r^2}}^{\sqrt{\frac{1}{4} - r^2}}{1 dz}
 
= \int_{-\sqrt{\frac{1}{4} - r^2}}^{\sqrt{\frac{1}{4} - r^2}}{1 dz}
 
</math>
 
</math>
 +
  
 
<math>
 
<math>
 
= \left\{\begin{matrix}
 
= \left\{\begin{matrix}
−
&\sqrt{1 - 4r^2}, &\text{ if }|r| \leq \frac{1}{2}
+
\sqrt{1 - 4r^2}, &\text{ if }|r| \leq \frac{1}{2}
 
\\
 
\\
−
&0, &\text{ otherwise}
+
0, &\text{ otherwise}
 
\end{matrix}\right.
 
\end{matrix}\right.
 
</math>
 
</math>
−
 
  
 
----
 
----
Line 192: Line 192:
 
= \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}
 
= \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}
 
</math>
 
</math>
 +
  
 
<font face="serif"><span style="font-size: 19px;"><math>
 
<font face="serif"><span style="font-size: 19px;"><math>
Line 213: Line 214:
 
= \int_{-\sqrt{\frac{1}{4} - (r - (cos\theta + sin\theta))^2}}^{\sqrt{\frac{1}{4} - (r - (cos\theta + sin\theta))^2}}{1 dz}
 
= \int_{-\sqrt{\frac{1}{4} - (r - (cos\theta + sin\theta))^2}}^{\sqrt{\frac{1}{4} - (r - (cos\theta + sin\theta))^2}}{1 dz}
 
</math>
 
</math>
 +
  
 
<math>
 
<math>
 
= \left\{\begin{matrix}
 
= \left\{\begin{matrix}
−
&\sqrt{1 - 4(r - (cos\theta + sin\theta))^2}, &\text{ if }|r| \leq \frac{1}{2}
+
\sqrt{1 - 4(r - (cos\theta + sin\theta))^2}, &\text{ if }|r| \leq \frac{1}{2}
 
\\
 
\\
−
&0, &\text{ otherwise}
+
0, &\text{ otherwise}
 +
\end{matrix}\right.
 +
</math>
 +
 
 +
 
 +
<math>
 +
= \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.
 
\end{matrix}\right.
 
</math>
 
</math>
Line 229: Line 240:
 
\tilde{p}_\theta(r) = p_{\theta}(r - \sqrt{1+1} cos(\theta - tan^{-1}(\frac{1}{1})))
 
\tilde{p}_\theta(r) = p_{\theta}(r - \sqrt{1+1} cos(\theta - tan^{-1}(\frac{1}{1})))
 
</math>
 
</math>
 +
  
 
<math>
 
<math>
 
= p_\theta(r - \sqrt{2} cos(\theta - \frac{\pi}{4}))
 
= p_\theta(r - \sqrt{2} cos(\theta - \frac{\pi}{4}))
 
</math>
 
</math>
 +
  
 
<math>
 
<math>
 
\text{ where } P_\theta(r) = \left\{\begin{matrix}
 
\text{ where } P_\theta(r) = \left\{\begin{matrix}
−
&\sqrt{1 - 4r^2}, &\text{ if }|r| \leq \frac{1}{2}
+
\sqrt{1 - 4r^2}, &\text{ if }|r| \leq \frac{1}{2}
 
\\
 
\\
−
&0, &\text{ else}
+
0, &\text{ else}
 
\end{matrix}\right.
 
\end{matrix}\right.
 
</math>
 
</math>

Revision as of 18:12, 2 August 2012

ECE Ph.D. Qualifying Exam in "Communication, Networks, Signal, and Image Processing" (CS)

Question 1, August 2011, Part 1

Part 1,2]

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


$ {\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) } $


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


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


$ \text{ Now } $


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


$ P_{\theta}(\rho) = \int_{-\infty}^{+\infty}{\delta(g(u)) \delta(u) \frac{du}{|sin\theta|}} = \frac{\delta(u=0)}{|sin\theta|} $


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

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

.QE 11 CS5 2 a sol2.PNG

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

$ = \delta(r) $

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


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


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


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

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

QE 11 CS5 2 b sol2.PNG

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


$ = \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:} $

QE 11 CS5 2 c sol2.PNG

$ P_{\theta}(\rho) = \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}(\rho) = 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}(\rho) = \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. $


$ = \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{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 ], \text{ for some cut-off, } f_c $

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

$ \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} $

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

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