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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]]
= [[ECE-QE_CS1-2011|Question 1, August 2011]], Part 1 =
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[[Category:CNSIP]]
 
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[[Category:problem solving]]
:[[ECE-QE_CS1-2011_solusion-1|Part 1]],[[ECE-QE CS1-2011 solusion-2|2]]]
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[[Category:communication networks signal and image processing]]
  
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= [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]] in Communication Networks Signal and Image processing (CS),  Question 1, August 2011=
 
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==Question==
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'''Part 1. ''' 25 pts
  
&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>
 
  
'''<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 optimization problem, }</math></span></font>  
  
===== <math>\color{blue}\text{Solution 1:}</math> =====
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>\text{maximize} -x_{1}^{2}+x_{1}-x_{2}-x_{1}x_{2}</math>  
  
<math> f_{YZ}\left (y,z \right )=\int_{-\infty}^{+\infty}f_{XYZ}\left(x,y,z \right )dx </math>&nbsp;
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>\text{subject to  }   x_{1}\geq0, x_{2}\geq0</math><font color="#ff0000" face="serif" size="4"><br></font>
  
&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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'''<math>\color{blue}\left( \text{i} \right) \text{ Characterize feasible directions at the point } x^{*}=\left[ \begin{array}{c} \frac{1}{2} \\ 0 \end{array} \right]</math>'''<br>  
\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>\color{blue}\left( \text{ii} \right) \text{Write down the second-order necessary condition for } x^{*} \text{. Does the point } x^{*} \text{ satisfy this condition?}</math><br>  
 
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<math> =\frac{3z^{2}}{7\sqrt[]{2\pi}}e^{-zy}
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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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=\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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:'''Click [[ECE-QE_CS1-2011_solusion-1|here]] to view student [[ECE-QE_CS1-2011_solusion-1|answers and discussions]]'''
 
----
 
----
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'''Part 2.'''
  
<math>\color{blue}\text{Solution 2:}</math>
 
  
here put sol.2
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&nbsp;<font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue}  \text{ Use the simplex method to solve the problem, }</math></span></font>
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<math>\color{blue}\left( \text{b} \right) \text{Find}
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class="texhtml">maximize &nbsp; &nbsp; &nbsp; &nbsp;''x''<sub>1</sub> + ''x''<sub>2</sub></span>  
f_{x}\left( x|y,z\right )
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</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;<math>\text{subject to  }  x_{1}-x_{2}\leq2</math><font color="#ff0000" face="serif" size="4"><br></font>'''&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>x_{1}+x_{2}\leq6</math>''' &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;
  
<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math>
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>x_{1},-x_{2}\geq0.</math>  
= \frac{f_{XYZ}\left( x,y,z\right )}{f_{YZ}\left(y,z \right )}
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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>
 
= \frac{e^{-\frac{1}{2}\left(\frac{x-y}{z} \right )^{2}}}{\sqrt[]{2\pi}z}
 
</math>&nbsp;&nbsp;</font>'''
 
  
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:'''Click [[ECE-QE_CS1-2011_solusion-2|here]] to view student [[ECE-QE_CS1-2011_solusion-2|answers and discussions]]'''
 
----
 
----
 
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[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]
<math>\color{blue}\text{Solution 2:}</math><br>
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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>
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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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=\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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----
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<math>\color{blue}\text{Solution 2:}</math><br>
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sol2 here
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----
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<math>\color{blue}\left( \text{d} \right) \text{Find}
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f_{Y}\left(y|z \right )
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</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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=\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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</span></font>
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'''<font face="serif"><math>
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=e^{-zy}z\cdot1_{\left[0,\infty \right )}(y)
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</math>&nbsp;&nbsp;</font>'''
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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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----
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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>
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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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=\frac{f_{XYZ}\left(x,y,z \right )}{f_{Z}(z)}
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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{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>&nbsp;&nbsp;</font>'''
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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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----
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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_CS1-2011_solusion-1|solutions and discussions]]
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*Part 2: [[ECE-QE CS1-2011 solusion-2|solutions and discussions]]
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----
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[[ECE PhD Qualifying Exams|Back to ECE Qualifying Exams (QE) page]]
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[[Category:ECE]] [[Category:QE]] [[Category:Automatic_Control]] [[Category:Problem_solving]]
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Revision as of 09:16, 27 July 2012


ECE Ph.D. Qualifying Exam in Communication Networks Signal and Image processing (CS), Question 1, August 2011


Question

Part 1. 25 pts


 $ \color{blue} \text{ Consider the optimization problem, } $

               $ \text{maximize} -x_{1}^{2}+x_{1}-x_{2}-x_{1}x_{2} $

               $ \text{subject to } x_{1}\geq0, x_{2}\geq0 $

$ \color{blue}\left( \text{i} \right) \text{ Characterize feasible directions at the point } x^{*}=\left[ \begin{array}{c} \frac{1}{2} \\ 0 \end{array} \right] $

$ \color{blue}\left( \text{ii} \right) \text{Write down the second-order necessary condition for } x^{*} \text{. Does the point } x^{*} \text{ satisfy this condition?} $

Click here to view student answers and discussions

Part 2.


 $ \color{blue} \text{ Use the simplex method to solve the problem, } $

               maximize        x1 + x2

               $ \text{subject to } x_{1}-x_{2}\leq2 $
                                        $ x_{1}+x_{2}\leq6 $                                         

                                        $ x_{1},-x_{2}\geq0. $


Click here to view student answers and discussions

Back to ECE Qualifying Exams (QE) page

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