(New page: = ECE Ph.D. Qualifying Exam in "Communication, Networks, Signal, and Image Processing" (CS) = = Question 1, August 2011, Part 1 = :[[ECE...)
 
Line 7: Line 7:
 
----
 
----
  
−
&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;<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.}
 +
</math></span></font>  
  
−
'''<math>\color{blue}\left( \text{a} \right) \text{ Find the joint probability density function } f_{YZ}(y,z).</math>'''<br>
 
  
 
===== <math>\color{blue}\text{Solution 1:}</math>  =====
 
===== <math>\color{blue}\text{Solution 1:}</math>  =====
  
−
<math> f_{YZ}\left (y,z \right )=\int_{-\infty}^{+\infty}f_{XYZ}\left(x,y,z \right )dx </math>&nbsp;
 
−
 
−
&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}
 
−
\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)}
 
−
\left(y \right )\cdot1_{\left [1,2 \right ]}\left(z \right )
 
−
</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 35: Line 19:
  
 
here put sol.2
 
here put sol.2
−
----
 
−
 
−
<math>\color{blue}\left( \text{b} \right) \text{Find}
 
−
f_{x}\left( x|y,z\right )
 
−
</math><br>
 
−
 
−
<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>'''
 
−
 
−
----
 
−
 
−
<math>\color{blue}\text{Solution 2:}</math><br>
 
−
 
−
sol2 here
 
−
----
 
−
 
−
<math>\color{blue}\left( \text{c} \right) \text{Find}
 
−
f_{Z}\left( z\right )
 
−
</math><br>
 
−
 
−
<math>\color{blue}\text{Solution 1:}</math>
 
−
 
−
<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>
 
−
=\frac{3z^{2}}{7}\cdot1_{\left[1,2 \right ]}(z)
 
−
</math>&nbsp;&nbsp;</font>'''
 
−
 
−
----
 
−
 
−
<math>\color{blue}\text{Solution 2:}</math><br>
 
−
 
−
sol2 here
 
−
----
 
−
 
−
<math>\color{blue}\left( \text{d} \right) \text{Find}
 
−
f_{Y}\left(y|z \right )
 
−
</math><br>
 
−
 
−
<math>\color{blue}\text{Solution 1:}</math>
 
−
 
−
<font color="#ff0000"><span style="font-size: 17px;">'''<font face="serif"></font><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>
 
−
=e^{-zy}z\cdot1_{\left[0,\infty \right )}(y)
 
−
</math>&nbsp;&nbsp;</font>'''
 
−
 
−
----
 
−
 
−
<math>\color{blue}\text{Solution 2:}</math><br>
 
−
 
−
sol2 here
 
−
----
 
−
<math>\color{blue}\left( \text{e} \right) \text{Find}
 
−
f_{XY}\left(x,y|z \right )
 
−
</math><br>
 
−
 
−
<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_{Z}(z)}
 
−
</math>'''</span></font><font color="#ff0000"><span style="font-size: 17px;">
 
−
</span></font>
 
−
 
−
'''<font face="serif"><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>'''
 
−
 
−
----
 
−
 
−
<math>\color{blue}\text{Solution 2:}</math><br>
 
−
 
−
sol2 here
 
 
----
 
----
  

Revision as of 12:27, 27 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{Show that if a continuous-time Gaussian random process } \mathbf{X}(t) \text{ is wide-sense stationary, it is also strict-sense stationary.} $


$ \color{blue}\text{Solution 1:} $

$ \color{blue}\text{Solution 2:} $

here put sol.2

"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_CS1-2011_solusion-2&oldid=52085"
Shortcuts
Help Main Wiki Page Random Page Special Pages Log in
Search

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang