| Line 67: | Line 67: | ||
<math>\color{blue}\text{Solution 1:}</math> | <math>\color{blue}\text{Solution 1:}</math> | ||
| + | <math> | ||
| + | P_{\theta}(\rho) = \int_{-\infty}^{+\infty}{\delta(r cos\theta - z sin\theta - 1, r sin\theta + z cos \theta - 1) dz} | ||
| + | </math> | ||
| + | |||
| + | |||
| + | <math> | ||
| + | = z \text{ when } \left\{\begin{matrix} | ||
| + | r cos\theta - z sin\theta = 1 | ||
| + | \\ | ||
| + | r sin \theta + z cos \theta = 1 | ||
| + | \end{matrix}\right. | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | = \frac{r cos\theta - 1}{sin \theta}, \theta > 0 | ||
| + | </math> | ||
| Line 83: | Line 99: | ||
<math>\color{blue}\text{Solution 1:}</math> | <math>\color{blue}\text{Solution 1:}</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} | ||
| + | </math> | ||
| + | |||
| + | |||
| + | <math> | ||
| + | = \int_{-\sqrt{\frac{1}{4} - r^2}}^{\sqrt{\frac{1}{4} - r^2}}{1 dz} | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | = \left\{\begin{matrix} | ||
| + | &\sqrt{1 - 4r^2}, &\text{ if }|r| \leq \frac{1}{2} | ||
| + | \\ | ||
| + | &0, &\text{ otherwise} | ||
| + | \end{matrix}\right. | ||
| + | </math> | ||
| + | |||
---- | ---- | ||
Revision as of 11:42, 31 July 2012
ECE Ph.D. Qualifying Exam in "Communication, Networks, Signal, and Image Processing" (CS)
Question 1, August 2011, Part 1
$ \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:} $
here put sol.2
$ \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:} $
sol2 here
$ \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:} $
$ \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:} $
$ \color{blue}\text{Solution 2:} $
sol2 here
"Communication, Networks, Signal, and Image Processing" (CS)- Question 1, August 2011
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- Part 1: solutions and discussions
- Part 2: solutions and discussions
