(New page: = ECE Ph.D. Qualifying Exam in "Communication, Networks, Signal, and Image Processing" (CS) = = Question 1, August 2011, Part 1 = :[[ECE...)
 
Line 6: Line 6:
  
 
----
 
----
 +
&nbsp;<font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue}\text{Consider an image } f(x,y) \text{ with a forward projection}
 +
</math></span></font>
  
−
&nbsp;<font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue}\text{1. } \left( \text{25 pts} \right) \text{ Let X, Y, and Z be three jointly distributed random variables with joint pdf} f_{XYZ}\left ( x,y,z \right )= \frac{3z^{2}}{7\sqrt[]{2\pi}}e^{-zy} exp \left [ -\frac{1}{2}\left ( \frac{x-y}{z}\right )^{2} \right ] \cdot 1_{\left[0,\infty \right )}\left(y \right )\cdot1_{\left[1,2 \right]} \left ( z \right) </math></span></font>  
+
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>\color{blue}
 +
p_{\theta}(r) = \mathcal{FP}\left \{ f(x,y) \right \}
 +
</math><br>  
  
−
'''<math>\color{blue}\left( \text{a} \right) \text{ Find the joint probability density function } f_{YZ}(y,z).</math>'''<br>  
+
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>\color{blue}  
 +
= \int_{-\infty}^{\infty}{f \left ( r cos(\theta) - z sin(\theta),r sin(\theta) + z cos(\theta) \right )dz}.
 +
</math>
  
−
===== <math>\color{blue}\text{Solution 1:}</math> =====
+
<math>\color{blue}
 +
\text{Let } F(\mu,\nu) \text{ be the continuous-time Fourier transform of } f(x,y) \text{ given by}
 +
</math><br>
 +
&nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; &nbsp;<math>\color{blue}
 +
F(u,v) = \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{f(x,y)e^{-j2\pi(ux,vy)}dx}dy}
 +
</math><br>
  
−
<math> f_{YZ}\left (y,z \right )=\int_{-\infty}^{+\infty}f_{XYZ}\left(x,y,z \right )dx </math>&nbsp;
+
<math>\color{blue}
 +
\text{and let } P_{\theta}(\rho) \text{ be the continuous-time Fourier transform of } p_{\theta}(r)  \text{ given by}
 +
</math><br>
 +
&nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; &nbsp;<math>\color{blue}
 +
P_{\theta}(\rho)  = \int_{-\infty}^{\infty}{p_{\theta}(r)e^{-j2\pi(\rho r)}dr}.
 +
</math><br>
  
−
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math> =\frac{3z^{2}}{7\sqrt[]{2\pi}}e^{-zy}\int_{-\infty}^{+\infty}exp\left[-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2} \right ]dx\cdot 1_{[0,\infty)}
 
−
\left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right )</math><br>
 
  
−
<math>\text{But}\int_{-\infty}^{+\infty}exp\left[-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2} \right ]dx \text{looks like the Gaussian pdf, so} </math>
 
  
−
<math> =\frac{3z^{2}}{7\sqrt[]{2\pi}}e^{-zy}
+
<math>\color{blue}\text{a) Calculate the forward projection }p_{\theta}(r) \text{, for } f(x,y) = \delta(x,y).
−
\underset{\sqrt[]{2\pi}z}{\underbrace{\frac{7\sqrt[]{2\pi}z}{7\sqrt[]{2\pi}z}  \int_{-\infty}^{+\infty}exp\left[-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2} \right ]dx}}\cdot 1_{[0,\infty)}
+
</math><br>
−
\left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right )
+
 
−
</math>
+
===== <math>\color{blue}\text{Solution 1:}</math> =====
 +
 
  
−
<math>
 
−
=\frac{3z^{2}}{7}e^{-zy}\cdot 1_{[0,\infty)}
 
−
\left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right )
 
−
</math>
 
  
 
----
 
----
Line 37: Line 47:
 
----
 
----
  
−
<math>\color{blue}\left( \text{b} \right) \text{Find}  
+
<math>\color{blue}\text{b) Calculate the forward projection }p_{\theta}(r) \text{, for } f(x,y) = \delta(x-1,y-1).
−
f_{x}\left( x|y,z\right )
+
 
</math><br>  
 
</math><br>  
  
 
<math>\color{blue}\text{Solution 1:}</math>  
 
<math>\color{blue}\text{Solution 1:}</math>  
  
−
<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math>
 
−
= \frac{f_{XYZ}\left( x,y,z\right )}{f_{YZ}\left(y,z \right )}
 
−
</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;">
 
−
</span></font>
 
  
−
'''<font face="serif"><math>
 
−
= \frac{e^{-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2}}}{\sqrt[]{2\pi}z}
 
−
</math>&nbsp;&nbsp;</font>'''
 
  
 
----
 
----
Line 59: Line 61:
 
----
 
----
  
−
<math>\color{blue}\left( \text{c} \right) \text{Find}
 
−
f_{Z}\left( z\right )
 
−
</math><br>
 
  
−
<math>\color{blue}\text{Solution 1:}</math>  
+
<math>\color{blue}\text{c) Calculate the forward projection }p_{\theta}(r) \text{, for } f(x,y) = rect \left(\sqrt[]{x^2+y^2} \right).
 +
</math><br>
  
−
<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math>
 
−
=\int_{0}^{+\infty}{f_{YZ}\left(y,z \right )dy}
 
−
</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;">
 
−
</span></font>
 
  
−
'''<font face="serif"><math>
+
<math>\color{blue}\text{Solution 1:}</math>  
−
=\frac{3z^{2}}{7}\cdot1_{\left[1,2 \right ]}(z)
+
 
−
</math>&nbsp;&nbsp;</font>'''
+
  
 
----
 
----
Line 81: Line 77:
 
----
 
----
  
−
<math>\color{blue}\left( \text{d} \right) \text{Find}
+
<math>\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).
−
f_{Y}\left(y|z \right )
+
 
</math><br>  
 
</math><br>  
  
 
<math>\color{blue}\text{Solution 1:}</math>  
 
<math>\color{blue}\text{Solution 1:}</math>  
  
−
<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math>
+
<math>
−
=\frac{f_{YZ}\left(y,z \right )}{f_{Z}(z)}</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;">
+
−
</span></font>  
+
  
−
'''<font face="serif"><math>
+
</math>
−
=e^{-zy}z\cdot1_{\left[0,\infty \right )}(y)
+
−
</math>&nbsp;&nbsp;</font>'''
+
  
 
----
 
----
Line 101: Line 92:
 
sol2 here
 
sol2 here
 
----
 
----
−
<math>\color{blue}\left( \text{e} \right) \text{Find}
+
<math>\color{blue}\text{e) Describe in precise detail, the steps required to perform filtered back projection (FBP) reconstruction of } f(x,y).
−
f_{XY}\left(x,y|z \right )
+
 
</math><br>  
 
</math><br>  
 +
  
 
<math>\color{blue}\text{Solution 1:}</math>  
 
<math>\color{blue}\text{Solution 1:}</math>  
  
−
<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math>
+
<math>
−
=\frac{f_{XYZ}\left(x,y,z \right )}{f_{Z}(z)}
+
−
</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;">
+
−
</span></font>  
+
  
−
'''<font face="serif"><math>
+
</math>  
−
=\frac{e^{-zy}}{\sqrt[]{2\pi}}e^{-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2}}\cdot1_{\left[0,\infty \right )}(y)
+
−
</math>&nbsp;&nbsp;</font>'''
+
  
 
----
 
----
Line 127: Line 113:
 
Go to  
 
Go to  
  
−
*Part 1: [[ECE-QE_CS1-2011_solusion-1|solutions and discussions]]  
+
*Part 1: [[ECE-QE_CS5-2011_solusion-1|solutions and discussions]]  
−
*Part 2: [[ECE-QE CS1-2011 solusion-2|solutions and discussions]]  
+
*Part 2: [[ECE-QE CS5-2011 solusion-2|solutions and discussions]]  
  
 
----
 
----

Revision as of 12:01, 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:} $

$ \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:} $


$ \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:} $


$ \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

Go to

Back to ECE Qualifying Exams (QE) page

Retrieved from "https://www.projectrhea.org/rhea/index.php?title=ECE-QE_CS5-2011_solusion-2&oldid=52102"
Shortcuts
Help Main Wiki Page Random Page Special Pages Log in
Search

Alumni Liaison

To all math majors: "Mathematics is a wonderfully rich subject."

Dr. Paul Garrett