Line 22: Line 22:
 
<math>R_{x(t)x(t+\tau)}=R_{xx}(\tau)</math><br>
 
<math>R_{x(t)x(t+\tau)}=R_{xx}(\tau)</math><br>
 
Such that<br>
 
Such that<br>
<math>Y(t)=c_1X(t)-c_2X(t-\tau)</math> <math>\sim N((c_1-c_2)\mu_x),(c_1^2+c_2^2)\sigma_x^2-2c_1c_2R_{xx}(\tau))</math><br>
+
<math>Y(t)=c_1X(t)-c_2X(t-\tau)</math> <math>\sim N((c_1-c_2)\mu_x),(c_1^2+c_2^2)\sigma_x^2-2c_1c_2R_{xx}(\tau))</math><br>
 +
<math>\Rightarrow </math>
 
----
 
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[[ECE-QE_CS1-2016|Back to QE CS question 1, August 2016]]
 
[[ECE-QE_CS1-2016|Back to QE CS question 1, August 2016]]
  
 
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Revision as of 00:02, 19 February 2019


ECE Ph.D. Qualifying Exam

Communication Signal (CS)

Question 1: Random Variable

August 2016 Problem 4


Solution

Since $ X(t) $ is a wide sense Gaussian Process $ \Rightarrow X(t) $ is SSS.
$ Y(t) $ is a combination of two Gaussian distribution.
$ R_{x(t)x(t+\tau)}=R_{xx}(\tau) $
Such that
$ Y(t)=c_1X(t)-c_2X(t-\tau) $ $ \sim N((c_1-c_2)\mu_x),(c_1^2+c_2^2)\sigma_x^2-2c_1c_2R_{xx}(\tau)) $
$ \Rightarrow $


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