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)\le\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 \\ =\dfrac{1}{\sigma}\int_{-\infty}^{\dfrac{r-\mu}{\sigma}} \dfrac{1}{\sqrt{2\pi}}e^{-\dfrac{z^2}{2}}dz=\dfrac{1}{\sigma}\Phi(\dfrac{r-\mu}{\sigma}) $
$ \sigma^2=c_1^2\sigma_x^2+c_2^2\sigma_x^2-2c_1c_2R_{xx}(\tau) $ $ \mu=(c_1-c_2)\mu_x $
$ P(Y(t)\le r)=\dfrac{1}{\sqrt{c_1^2\sigma_x^2+c_2^2\sigma_x^2-2c_1c_2R_{xx}(\tau)}}\Phi(\dfrac{r-(c_1-c_2)\mu_x}{\sqrt{c_1^2\sigma_x^2+c_2^2\sigma_x^2-2c_1c_2R_{xx}(\tau)}}) $


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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva