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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:random variables]]
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[[Category:probability]]
  
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= [[ECE-QE_CS1-2011|Question 1, 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_CS1-2011_solusion-1|Part 1]],[[ECE-QE CS1-2011 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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----
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Question 1: Probability and Random Processes
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</font size>
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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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===== <math>\color{blue}\text{Solution 1:}</math>  =====
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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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&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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\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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August 2011
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</center>
 
----
 
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<math>\color{blue}\text{Solution 2:}</math>
 
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<math>d\in\Re^{2}, d\neq0 \text{ is a feasible direction at } x^{*}</math>&nbsp;<br>
 
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>\text{ if } \exists \alpha_{0} \text{ that } \left[ \begin{array}{c} \frac{1}{2} \\ 0 \end{array} \right] + \alpha\left[ \begin{array}{c} d_{1} \\ d_{2} \end{array} \right] \in\Omega \text{ for all } 0\leq\alpha\leq\alpha_{0}</math>
 
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'''<math>\because \begin{Bmatrix}x\in\Omega: x_{1}\geq0, x_{2}\geq0\end{Bmatrix}</math>'''
 
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<br> <math>\therefore d=
 
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\left[ \begin{array}{c} d_{1} \\ d_{2} \end{array} \right], d_{1}\in\Re, d_{2}\geq0</math>
 
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----
 
----
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==Question==
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'''Part 1. ''' 25 pts
  
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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>\color{blue}\text{Solution 1:}</math>  
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&nbsp;<font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue}\text{ Let } \mathbf{X}\text{, }\mathbf{Y}\text{, and } \mathbf{Z} \text{ 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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<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><math>\text{Let } f\left(x\right)=x_{1}^{2}-x_{1}+x_{2}+x_{1}x_{2} \text{ , } g_{1}\left(x\right)=-x_{1} \text{ , } g_{2}\left(x\right)=-x_{2}</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;">
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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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</span></font>  
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'''<font face="serif"><math>\text{It is equivalent to minimize } f\left(x\right) \text{,  }</math>&nbsp;&nbsp;</font>'''
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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>  
  
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'''<font face="serif"></font>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>\text{ subject to } g_{1}(x)\leq0, g_{2}(x)\leq0</math>'''
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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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<font color="#ff0000" style="font-size: 17px;">'''<math>\left\{\begin{matrix}
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<math>\color{blue}\left( \text{d} \right) \text{Find }  
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l\left(x,\mu \right) = \nabla f(x)+\mu_{1}\nabla g_{1}(x)+ \mu_{2}\nabla g_{2}(x) \\ =\left( \begin{array}{c} 2x_{1}-1+x_{2} \\ 1+x_{1} \end{array} \right) + \left( \begin{array}{c} -\mu_{1} \\ 0 \end{array} \right) +\left( \begin{array}{c} 0 \\ -\mu_{2} \end{array} \right) =0\\
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f_{Y}\left(y|z \right ).
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-\mu_{1}x_{1}-\mu_{2}x_{2} = 0 \\
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</math><br>  
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x_{1} = \frac{1}{2},x_{2} = 0
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\end{matrix}\right.</math>'''</font><br><math>\Rightarrow \mu_{1}=0 , \mu_{2}=3/2</math>&nbsp; &nbsp;
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<math>\therefore x^{*} \text{ satisfies FONC}</math>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;
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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{green} \text{There exist } \mu \text{ which make point } x^{*} \text{ satisfies FONC.}</math>
 
  
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<math>\text{SONC: } L(x^{*},\mu^{*}) = \nabla l(x^{*},\mu^{*})=\left( \begin{array}{cc} 2 & 1 \\ 1 & 0 \end{array} \right)</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.''' 25 pts
  
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<font color="#ff0000">'''<math>T(x^{*},\mu^{*}): \begin{cases} y^{T}\nabla g_{1}(x)=0 \\ y^{T}\nabla g_{2}(x)=0 \end{cases} : \begin{cases} y^{T}\left( \begin{array}{c} -1 \\ 0 \end{array} \right)=0 \\ y^{T}\left( \begin{array}{c} 0 \\-1 \end{array} \right)=0 \end{cases} \Rightarrow y=\left( \begin{array}{c} 0 \\0 \end{array} \right)</math><br>'''</font>
 
  
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<math>\color{green} \text {Here not using formal set expression. }</math>&nbsp;&nbsp;<math>\color{red} T\left( x^{* },\mu^{* } \right) \text{ should be } T\left( x^{* } \right)</math>  
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&nbsp;<font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue} \text{Show that if a continuous-time Gaussian random process } \mathbf{X}(t) \text{ is wide-sense stationary, it is also strict-sense stationary.}
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</math></span></font>  
  
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<math>\text{The SONC condition is for all } y\in T \left(x^{*},\mu^{*} \right) , y^{T}L\left(x^{*},\mu^{*} \right)y \geq 0</math>
 
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<math>y^{T}L\left(x^{*},\mu^{*} \right)y =0 \geq 0 \text{. So } x^{*} \text{satisfies SONC.}</math><br>
 
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<math>\color{red} \text{For SONC, } T\left( x^{* } \right)= \left \{ y\in\Re^{n}: Dh\left( x^{*} \right)y=0, Dg_{j}\left( x^{*} \right)y=0, j\in J\left( x^{*} \right)  \right \}</math>
 
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>\color{red}  J\left(x^{*}\right)= \left \{  j:g_{j}\left(x^{*}\right)=0 \right \}</math>
 
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<math>\color{red} \text{For SOSC, }  \tilde{T}\left( x^{* },\mu^{*} \right)= \left \{ y: Dh\left( x^{*} \right)y=0, Dg_{i}\left( x^{*} \right)y=0, i\in \tilde{J}\left( x^{*},\mu^{*} \right)  \right \}</math>
 
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>\color{red} \tilde{J}\left ( x^{\ast },\mu ^{\ast } \right )= \left \{ i:g_{i}\left ( x^{\ast } \right ) = 0,\mu_{i}^{\ast }> 0\right \}</math><br>
 
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<math>\color{red} \tilde{J}\left ( x^{\ast },\mu ^{\ast } \right ) \subset 
 
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J\left(x^{*}\right)</math>, &nbsp; &nbsp;&nbsp;<math>\color{red} T\left( x^{* } \right) \subset \tilde{T}\left( x^{* },\mu^{*} \right)</math>
 
  
 +
:'''Click [[ECE-QE_CS1-2011_solusion-2|here]] to view student [[ECE-QE_CS1-2011_solusion-2|answers and discussions]]'''
 
----
 
----
 +
'''Part 3.''' 25 pts
  
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<math>\color{blue}\text{Solution 2:}</math><br>
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Show that the sum of two jointly distributed Gaussian random variables that are not necessarily statistically independent is a Gaussian random variable.
  
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<math>\text{The problem is equivalent to  min} f\left(x_{1},x_{2}\right) = x_{1}^{2}-x_{1}+x_{2}+x_{1}x_{2}</math>&nbsp;&nbsp;
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:'''Click [[ECE-QE_CS1-2011_solusion-3|here]] to view student [[ECE-QE_CS1-2011_solusion-3|answers and discussions]]'''
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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; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>\text{subject to  }  x_{1}\leq0, x_{2}\leq0</math>
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<math>Df\left ( x \right )=\left ( \nabla f\left ( x \right ) \right )^{T} = \left [ \frac{\partial f}{\partial x_{1}}\left ( x \right ),\frac{\partial f}{\partial x_{2}}\left ( x \right ) \right ]=\left [ 2x_{1}-1+x_{2},1+x_{1} \right ]</math><br>
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<math>F\left ( x \right ) =D^{2}f\left ( x \right )=\begin{bmatrix}
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\frac{\partial^{2} f}{\partial x_{1}^{2}}\left ( x \right ) & \frac{\partial^{2} f}{\partial x_{2}\partial x_{1}}\left ( x \right )\\
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\frac{\partial^{2} f}{\partial x_{1}\partial x_{2}}\left ( x \right ) & \frac{\partial^{2} f}{\partial x_{2}^{2}}\left ( x \right )
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\end{bmatrix}=\left [ \begin{array}{cc} 2 & 1 \\ 1 & 0 \end{array} \right ]</math>
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<math>\text{SONC for local minimizer } x^{*}=\begin{bmatrix} \frac{1}{2}\\0 \end{bmatrix}</math>
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>d^{T} \nabla f\left ( x^{*} \right )=0  \cdots \left ( 1 \right )</math>&nbsp; &nbsp; &nbsp;
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>d^{T} F\left ( x^{*} \right )d\geq 0  \cdots \left ( 2\right )</math><br>
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<math>\text{For (1), } \begin{bmatrix} d_{1} & d_{2} \end{bmatrix}\begin{bmatrix} 0\\ \frac{3}{2}\end{bmatrix} =0 \Rightarrow  d_{1}\in\Re, d_{2}=0</math><br>
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<math>\text{For (2), } F\left ( x \right ) = \begin{bmatrix} 2 &1 \\ 1 &0\end{bmatrix}>0</math>&nbsp; &nbsp; &nbsp; &nbsp;<math>\color{green}  A=\begin{bmatrix} a &b \\ c &d\end{bmatrix} \text{ is positive definite when } a>0 \text{ and } ac-b^{2}>0</math><br>
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<math>\therefore \text{ for all } d\in\Re^{n}, d^{T}F\left ( x^{*} \right )d\geq 0</math>
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<font face="serif"><math>\text{The point } x^{*}=\begin{bmatrix} \frac{1}{2}\\0 \end{bmatrix} \text{ satisfies SONC for local minimizer.}</math><br></font>
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----
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<font face="serif"></font><math>\color{blue}\text{Related Problem: For function }</math>
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'''Part 4.''' 25 pts
  
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&nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>f\left( x_{1},x_{2}  \right) =\frac{1}{3} x_{1}^{3} + \frac{1}{3} x_{2}^{3} -x_{1}x_{2}</math>
 
  
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<math>\color{blue} \text{Find point(s) that satisfy FONC and check if they are strict local minimizers.}</math>  
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Assume that <math>\mathbf{X}(t)</math> is a zero-mean continuous-time Gaussian white noise process with autocorrelation function
  
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<math>\color{blue}\text{Solution:}</math>  
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>R_{\mathbf{XX}}(t_1,t_2)=\delta(t_1-t_2).
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</math>
  
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<math>\text{Applying FONC gives } \nabla f\left ( x \right )=\begin{bmatrix}
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Let <math>\mathbf{Y}(t)</math> be a new random process ontained by passing <math>\mathbf{X}(t)</math> through a linear time-invariant system with impulse response <math>h(t)</math> whose Fourier transform <math>H(\omega)</math> has the ideal low-pass characteristic
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x_{1}^{2}-x_{2}\\
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x_{2}^{2}-x_{1}
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\end{bmatrix}=0</math>  
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&nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>\Rightarrow x^{\left ( 1 \right )}=\begin{bmatrix}
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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>H(\omega) =
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0\\
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\begin{cases}  
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0
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1, & \mbox{if } |\omega|\leq\Omega,\\
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\end{bmatrix} \text{ and }x^{\left ( 2 \right )}=\begin{bmatrix}
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0, & \mbox{elsewhere,}  
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1\\
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\end{cases}
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1
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</math>
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\end{bmatrix}</math>  
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<math>\text{The Hessian matrix: } F\left ( x \right )=\begin{bmatrix}
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where <math>\Omega>0</math>.
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2x_{1} & -1\\
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-1 & 2x_{2}
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\end{bmatrix}</math>  
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&nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>\text{The matrix } F\left ( x^{\left ( 1 \right )} \right )=\begin{bmatrix}
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a) Find the mean of <math>\mathbf{Y}(t)</math>.
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0 & -1\\
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-1 & 0
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\end{bmatrix} \text{ is indefinite. The point is not a minimizer.}</math>  
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&nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>\text{The matrix } F\left ( x^{\left ( 2\right )} \right )=\begin{bmatrix}
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b) Find the autocorrelation function of <math>\mathbf{Y}(t)</math>.
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0 & -1\\
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-1 & 0
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\end{bmatrix} \text{ is positive definite. }</math>  
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<math>\therefore x^{\left ( 2 \right )}=\begin{bmatrix}
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c) Find the joint pdf of <math>\mathbf{Y}(t_1)</math> and <math>\mathbf{Y}(t_2)</math> for any two arbitrary sample time <math>t_1</math> and <math>t_2</math>.
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1\\
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1
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\end{bmatrix} \text{ satisfies SOSC to be a strict local minimizer.}</math>  
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----
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d) What is the minimum time difference <math>t_1-t_2</math> such that <math>\mathbf{Y}(t_1)</math> and <math>\mathbf{Y}(t_2)</math> are statistically independent?
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Automatic Control (AC)- Question 3, August 2011
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Go to
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*Part 1: [[ECE-QE_AC3-2011_solusion-1|solutions and discussions]]
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*Part 2: [[ECE-QE AC3-2011 solusion-2|solutions and discussions]]
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*Part 3: [[ECE-QE AC3-2011 solusion-3|solutions and discussions]]
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*Part 4: [[ECE-QE AC3-2011 solusion-4|solutions and discussions]]
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*Part 5: [[ECE-QE AC3-2011 solusion-5|solutions and discussions]]
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:'''Click [[ECE-QE_CS1-2011_solusion-4|here]] to view student [[ECE-QE_CS1-2011_solusion-4|answers and discussions]]'''
 
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[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]
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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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Latest revision as of 15:40, 30 March 2015


ECE Ph.D. Qualifying Exam

Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2011

Question

Part 1. 25 pts


 $ \color{blue}\text{ Let } \mathbf{X}\text{, }\mathbf{Y}\text{, and } \mathbf{Z} \text{ 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) $

$ \color{blue}\left( \text{a} \right) \text{Find the joint probability density function } f_{YZ}(y,z). $

$ \color{blue}\left( \text{b} \right) \text{Find } f_{x}\left( x|y,z\right ). $

$ \color{blue}\left( \text{c} \right) \text{Find } f_{Z}\left( z\right ). $

$ \color{blue}\left( \text{d} \right) \text{Find } f_{Y}\left(y|z \right ). $

$ \color{blue}\left( \text{e} \right) \text{Find } f_{XY}\left(x,y|z \right ). $


Click here to view student answers and discussions

Part 2. 25 pts


 $ \color{blue} \text{Show that if a continuous-time Gaussian random process } \mathbf{X}(t) \text{ is wide-sense stationary, it is also strict-sense stationary.} $


Click here to view student answers and discussions

Part 3. 25 pts

Show that the sum of two jointly distributed Gaussian random variables that are not necessarily statistically independent is a Gaussian random variable.

Click here to view student answers and discussions

Part 4. 25 pts


Assume that $ \mathbf{X}(t) $ is a zero-mean continuous-time Gaussian white noise process with autocorrelation function

                $ R_{\mathbf{XX}}(t_1,t_2)=\delta(t_1-t_2). $

Let $ \mathbf{Y}(t) $ be a new random process ontained by passing $ \mathbf{X}(t) $ through a linear time-invariant system with impulse response $ h(t) $ whose Fourier transform $ H(\omega) $ has the ideal low-pass characteristic

               $ H(\omega) = \begin{cases} 1, & \mbox{if } |\omega|\leq\Omega,\\ 0, & \mbox{elsewhere,} \end{cases} $

where $ \Omega>0 $.

a) Find the mean of $ \mathbf{Y}(t) $.

b) Find the autocorrelation function of $ \mathbf{Y}(t) $.

c) Find the joint pdf of $ \mathbf{Y}(t_1) $ and $ \mathbf{Y}(t_2) $ for any two arbitrary sample time $ t_1 $ and $ t_2 $.

d) What is the minimum time difference $ t_1-t_2 $ such that $ \mathbf{Y}(t_1) $ and $ \mathbf{Y}(t_2) $ are statistically independent?

Click here to view student answers and discussions

Back to ECE Qualifying Exams (QE) page

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