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=\int_{-\infty}^{r} \dfrac{1}{\sqrt{2\pi\sigma^2}}e^{-\dfrac{1}{2\sigma^2}(x-\mu)^2}dx | =\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}^{r} \dfrac{1}{\sqrt{2\pi}}e^{-\dfrac{(\dfrac{x-\mu}{\sigma})^2}{2}}dx \\ | ||
| − | =</math> | + | =\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})</math><br> |
| + | <math>\sigma^2=c_1^2\sigma_x^2+c_2^2\sigma_x^2-2c_1c_2R_{xx}(\tau)</math> <math>\mu=(c_1-c_2)\mu_x</math><br> | ||
| + | <math>P(Y(t)<=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)}})</math><br> | ||
---- | ---- | ||
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Revision as of 09:21, 19 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)) $
$ \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 \\ =\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)<=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)}}) $
