(New page: = ECE Ph.D. Qualifying Exam in "Communication, Networks, Signal, and Image Processing" (CS) = = Question 5, August 2011, Part 1 = :[[ECE...)
 
 
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= [[ECE PhD Qualifying Exams|ECE Ph.D. Qualifying Exam]] in "Communication, Networks, Signal, and Image Processing" (CS)  =
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[[Category:ECE]]
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[[Category:QE]]
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[[Category:CNSIP]]
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[[Category:problem solving]]
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[[Category:image processing]]
  
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= [[ECE-QE_CS5-2011|Question 5, August 2011]], Part 1 =
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<center>
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<font size= 4>
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[[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]]
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</font size>
  
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:[[ECE-QE_CS5-2011_solusion-1|Part 1]],[[ECE-QE CS5-2011 solusion-2|2]]]
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<font size= 4>
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Communication, Networking, Signal and Image Processing (CS)
  
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Question 5: Image Processing
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</font size>
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August 2011
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</center>
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==Question==
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'''Part 1. ''' 50 pts
  
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&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>
 
  
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'''<math>\color{blue}\left( \text{a} \right) \text{ Find the joint probability density function } f_{YZ}(y,z).</math>'''<br>  
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&nbsp;<font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue}\text{Consider the following discrete space system with input } x(m,n) \text{ and output } y(m,n).
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</math></span></font>  
  
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===== <math>\color{blue}\text{Solution 1:}</math> =====
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>\color{blue}
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y(m,n) = \sum_{k=-\infty}^{\infty}{\sum_{l=-\infty}^{\infty}{x(m-k,n-l)h(k,l)}}.
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</math><br>
  
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<math> f_{YZ}\left (y,z \right )=\int_{-\infty}^{+\infty}f_{XYZ}\left(x,y,z \right )dx </math>&nbsp;
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<math>\color{blue}
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\text{For parts a) and b) let}
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</math><br>
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>\color{blue}
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h(m,n)=sinc(mT,nT), \text{where} T\leq1.
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</math><br>
  
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&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)}
 
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\left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right )</math><br>
 
  
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<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>
 
  
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<math> =\frac{3z^{2}}{7\sqrt[]{2\pi}}e^{-zy}
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<math>\color{blue}\text{a) Calculate the frequency response, }H \left( e^{j\mu},e^{j\nu} \right).</math><br>  
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\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)}
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\left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right )
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</math>
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<math>
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<math>\color{blue}\text{b) Sketch the frequency response for } |\mu| < 2\pi \text{ and } |\nu| < 2\pi \text{ when } T = \frac{1}{2}  
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=\frac{3z^{2}}{7}e^{-zy}\cdot 1_{[0,\infty)}
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\left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right )
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</math>
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----
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<math>\color{blue}\text{Solution 2:}</math>
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here put sol.2
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----
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<math>\color{blue}\left( \text{b} \right) \text{Find}  
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f_{x}\left( x|y,z\right )
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</math><br>  
 
</math><br>  
  
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<math>\color{blue}\text{Solution 1:}</math>
 
  
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<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math>
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<math>\color{blue}
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= \frac{f_{XYZ}\left( x,y,z\right )}{f_{YZ}\left(y,z \right )}
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\text{For parts c), d), and e) let}
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</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;">
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</math><br>
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</span></font>  
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>\color{blue}
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h(m,n)=sinc\left( \frac{(n+m)T}{\sqrt[]{2}},\frac{(n-m)T}{\sqrt[]{2}} \right)
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</math><br>
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<math>\color{blue}
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\text{where } T\leq1.
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</math><br>
  
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'''<font face="serif"><math>
 
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= \frac{e^{-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2}}}{\sqrt[]{2\pi}z}
 
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</math>&nbsp;&nbsp;</font>'''
 
  
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----
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<math>\color{blue}\text{c) Calculate the frequency response, }H \left( e^{j\mu},e^{j\nu} \right).</math><br>
  
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<math>\color{blue}\text{Solution 2:}</math><br>
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<math>\color{blue}\text{d) Sketch the frequency response for } |\mu| < 2\pi \text{ and } |\nu| < 2\pi \text{ when } T = \frac{1}{2}  
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sol2 here
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----
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<math>\color{blue}\left( \text{c} \right) \text{Find}  
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f_{Z}\left( z\right )
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</math><br>  
 
</math><br>  
  
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<math>\color{blue}\text{Solution 1:}</math>
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<math>\color{blue}\text{e) Calculate } y(m,n) \text{ when } x(m,n)=1.</math><br>
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<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math>
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=\int_{0}^{+\infty}{f_{YZ}\left(y,z \right )dy}
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</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;">
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</span></font>  
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'''<font face="serif"><math>
 
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=\frac{3z^{2}}{7}\cdot1_{\left[1,2 \right ]}(z)
 
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</math>&nbsp;&nbsp;</font>'''
 
  
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:'''Click [[ECE-QE_CS5-2011_solusion-1|here]] to view student [[ECE-QE_CS5-2011_solusion-1|answers and discussions]]'''
 
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'''Part 2.''' 50 pts
  
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<math>\color{blue}\text{Solution 2:}</math><br>
 
  
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sol2 here
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&nbsp;<font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue}\text{Consider an image } f(x,y) \text{ with a forward projection}
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----
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</math></span></font>
  
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<math>\color{blue}\left( \text{d} \right) \text{Find}
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>\color{blue}
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f_{Y}\left(y|z \right )
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p_{\theta}(r) = \mathcal{FP}\left \{ f(x,y) \right \}
 
</math><br>  
 
</math><br>  
  
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<math>\color{blue}\text{Solution 1:}</math>  
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>\color{blue}  
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= \int_{-\infty}^{\infty}{f \left ( r cos(\theta) - z sin(\theta),r sin(\theta) + z cos(\theta) \right )dz}.
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</math>
  
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<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math>
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<math>\color{blue}
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=\frac{f_{YZ}\left(y,z \right )}{f_{Z}(z)}</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;">
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\text{Let } F(\mu,\nu) \text{ be the continuous-time Fourier transform of } f(x,y) \text{ given by}
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</span></font>  
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</math><br>
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&nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; &nbsp;<math>\color{blue}
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F(u,v) = \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{f(x,y)e^{-j2\pi(ux,vy)}dx}dy}
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</math><br>
  
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'''<font face="serif"><math>
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<math>\color{blue}
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=e^{-zy}z\cdot1_{\left[0,\infty \right )}(y)
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\text{and let } P_{\theta}(\rho) \text{ be the continuous-time Fourier transform of } p_{\theta}(r) \text{ given by}
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</math>&nbsp;&nbsp;</font>'''
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</math><br>
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&nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; &nbsp;<math>\color{blue}
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P_{\theta}(\rho)  = \int_{-\infty}^{\infty}{p_{\theta}(r)e^{-j2\pi(\rho r)}dr}.
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</math><br>
  
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----
 
  
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<math>\color{blue}\text{Solution 2:}</math><br>
 
  
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sol2 here
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<math>\color{blue}\text{a) Calculate the forward projection }p_{\theta}(r) \text{, for } f(x,y) = \delta(x,y).
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----
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<math>\color{blue}\left( \text{e} \right) \text{Find}  
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f_{XY}\left(x,y|z \right )
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</math><br>  
 
</math><br>  
  
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<math>\color{blue}\text{Solution 1:}</math>  
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<math>\color{blue}\text{b) Calculate the forward projection }p_{\theta}(r) \text{, for } f(x,y) = \delta(x-1,y-1).
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</math><br>  
  
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<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math>
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<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).
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=\frac{f_{XYZ}\left(x,y,z \right )}{f_{Z}(z)}
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</math><br>  
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</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;">
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</span></font>  
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'''<font face="serif"><math>
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<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).
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=\frac{e^{-zy}}{\sqrt[]{2\pi}}e^{-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2}}\cdot1_{\left[0,\infty \right )}(y)
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</math><br>  
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</math>&nbsp;&nbsp;</font>'''
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<math>\color{blue}\text{e) Describe in precise detail, the steps required to perform filtered back projection (FBP) reconstruction of } f(x,y).
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</math><br>
  
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<math>\color{blue}\text{Solution 2:}</math><br>
 
  
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sol2 here
 
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"Communication, Networks, Signal, and Image Processing" (CS)- Question 1, August 2011
 
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Go to
 
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*Part 1: [[ECE-QE_CS5-2011_solusion-1|solutions and discussions]]
 
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*Part 2: [[ECE-QE CS5-2011 solusion-2|solutions and discussions]]
 
  
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:'''Click [[ECE-QE_CS5-2011_solusion-2|here]] to view student [[ECE-QE_CS5-2011_solusion-2|answers and discussions]]'''
 
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[[ECE PhD Qualifying Exams|Back to ECE Qualifying Exams (QE) page]]
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Latest revision as of 10:25, 13 September 2013


ECE Ph.D. Qualifying Exam

Communication, Networking, Signal and Image Processing (CS)

Question 5: Image Processing

August 2011

Question

Part 1. 50 pts


 $ \color{blue}\text{Consider the following discrete space system with input } x(m,n) \text{ and output } y(m,n). $

                $ \color{blue} y(m,n) = \sum_{k=-\infty}^{\infty}{\sum_{l=-\infty}^{\infty}{x(m-k,n-l)h(k,l)}}. $

$ \color{blue} \text{For parts a) and b) let} $
                $ \color{blue} h(m,n)=sinc(mT,nT), \text{where} T\leq1. $


$ \color{blue}\text{a) Calculate the frequency response, }H \left( e^{j\mu},e^{j\nu} \right). $

$ \color{blue}\text{b) Sketch the frequency response for } |\mu| < 2\pi \text{ and } |\nu| < 2\pi \text{ when } T = \frac{1}{2} $


$ \color{blue} \text{For parts c), d), and e) let} $
                $ \color{blue} h(m,n)=sinc\left( \frac{(n+m)T}{\sqrt[]{2}},\frac{(n-m)T}{\sqrt[]{2}} \right) $
$ \color{blue} \text{where } T\leq1. $


$ \color{blue}\text{c) Calculate the frequency response, }H \left( e^{j\mu},e^{j\nu} \right). $

$ \color{blue}\text{d) Sketch the frequency response for } |\mu| < 2\pi \text{ and } |\nu| < 2\pi \text{ when } T = \frac{1}{2} $

$ \color{blue}\text{e) Calculate } y(m,n) \text{ when } x(m,n)=1. $


Click here to view student answers and discussions

Part 2. 50 pts


 $ \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{b) Calculate the forward projection }p_{\theta}(r) \text{, for } f(x,y) = \delta(x-1,y-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{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{e) Describe in precise detail, the steps required to perform filtered back projection (FBP) reconstruction of } f(x,y). $


Click here to view student answers and discussions

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