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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>
+
<math>\Rightarrow  
 +
P(Y(t)<=\gamma)
 +
=\int_{-\infty}^{r} \dfrac{1}{\sqrt{2\pi\sigma^2}}e^{-\dfrac{1}{2\sigma^2}(x-\mu)^2}dx
 +
=\dfrac{1}{\sigma}\int_{-\infty}^{r} \dfrac{1}{\sqrt{2\pi}}e^{-\dfrac{(\dfrac{x-\mu}{\sigma})^2}{2}}dx \\
 +
=</math>
 
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Revision as of 08:15, 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 P(Y(t)<=\gamma) =\int_{-\infty}^{r} \dfrac{1}{\sqrt{2\pi\sigma^2}}e^{-\dfrac{1}{2\sigma^2}(x-\mu)^2}dx =\dfrac{1}{\sigma}\int_{-\infty}^{r} \dfrac{1}{\sqrt{2\pi}}e^{-\dfrac{(\dfrac{x-\mu}{\sigma})^2}{2}}dx \\ = $


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