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\begin{cases} | \begin{cases} | ||
0 & \omega<0 \\ | 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 | + | \dfrac{1}{2}hb-\dfrac{1}{2}hb(\dfrac{h-\omega}{h})^2=\dfrac{2\omega}{h}-\dfrac{w^2}{h^2} & 0\le\omega<h \\ |
| − | 1 & \omega | + | 1 & \omega\ge h |
| + | \end{cases} | ||
| + | </math><br> | ||
| + | b)<br> | ||
| + | <math>f_x(\omega)=\dfrac{\partial F_x(\omega)}{\partial\omega}</math><br> | ||
| + | <math>f_x(\omega)= | ||
| + | \begin{cases} | ||
| + | 0 & \omega<0 \\ | ||
| + | \dfrac{-2}{h^2}\omega+\dfrac{2}{h} & 0\le\omega<h \\ | ||
| + | 0 & \omega\ge h | ||
\end{cases} | \end{cases} | ||
| − | </math> | + | </math><br> |
| + | |||
| + | c)<br> | ||
| + | <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 =\dfrac{1}{3}h</math><br> | ||
| + | |||
| + | d)<br> | ||
| + | <math>P(x>\dfrac{h}{3})=\int_{\dfrac{h}{3}}^{+\infty} \omega f_x(\omega)dx = \int_{\dfrac{h}{3}}^{h} -\dfrac{2}{h^2}\omega+\dfrac{2}{h}d\omega =\dfrac{4}{9}</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]] | ||
Latest revision as of 15:32, 19 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\le\omega<h \\ 1 & \omega\ge 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\le\omega<h \\ 0 & \omega\ge 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 =\dfrac{1}{3}h $
d)
$ P(x>\dfrac{h}{3})=\int_{\dfrac{h}{3}}^{+\infty} \omega f_x(\omega)dx = \int_{\dfrac{h}{3}}^{h} -\dfrac{2}{h^2}\omega+\dfrac{2}{h}d\omega =\dfrac{4}{9} $
