| Line 28: | Line 28: | ||
b)<br> | b)<br> | ||
| − | <math>f_x(\omega)=\dfrac{\ | + | <math>f_x(\omega)=\dfrac{\partial F_x(\omega)}{\partial\omega}</math><br> |
<math>f_x(\omega)= | <math>f_x(\omega)= | ||
\begin{cases} | \begin{cases} | ||
| Line 38: | Line 38: | ||
c)<br> | c)<br> | ||
| − | <math>X(\omega)\bar=\int_{-\infty}^{\infty} \ | + | <math>X(\omega)\bar=\int_{-\infty}^{\infty} \omega f_x(\omega) dx =\int_{0}^{h} -\dfrac{2}{h^2}(\omega)^2 +\dfrac{2}{h}\omega d\omega</math> |
---- | ---- | ||
Revision as of 22:15, 18 February 2019
Communication Signal (CS)
Question 1: Random Variable
August 2016 Problem 1
Solution
a)
$ F_x(\omega)= \begin{cases} 0 & \omega<0 \\ \dfrac{1}{2}hb-\dfrac{1}{2}hb(\dfrac{h-\omega}{h})^2=\dfrac{2\omega}{h}-\dfrac{w^2}{h^2} & 0<=\omega<h \\ 1 & \omega>=h \end{cases} $
b)
$ f_x(\omega)=\dfrac{\partial F_x(\omega)}{\partial\omega} $
$ f_x(\omega)= \begin{cases} 0 & \omega<0 \\ \dfrac{-2}{h^2}\omega+\dfrac{2}{h} & 0<=\omega<h \\ 0 & \omega>=h \end{cases} $
c)
$ X(\omega)\bar=\int_{-\infty}^{\infty} \omega f_x(\omega) dx =\int_{0}^{h} -\dfrac{2}{h^2}(\omega)^2 +\dfrac{2}{h}\omega d\omega $
