Revision as of 14:01, 31 July 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} $

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

File:.jpg

$ 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} $

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

File:.jpg

$ \widetilde{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:} $

sol2 here

$ \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(\sqrt{(r cos\theta - z sin\theta - 1)^2, (r sin\theta + z cos \theta - 1)^2) 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{blue}\text{Solution 2:} $

sol2 here

$ \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) = f_c[rect(\frac{f}{2f_c}) - \Lambda(\frac{f}{f_c})]. $

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

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

sol2 here

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

Go to

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

Back to ECE Qualifying Exams (QE) page