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<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>\ | + | <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> |
---- | ---- | ||
[[ECE-QE_CS1-2016|Back to QE CS question 1, August 2016]] | [[ECE-QE_CS1-2016|Back to QE CS question 1, August 2016]] | ||
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]] | ||
Revision as of 23:01, 18 February 2019
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)) $
