Question

A random process $X\left(t\right)$ has a periodic sample function as shown in the figure below:



Where $A,T$ and to $\le T$ are constants but $ϵ$ is random variable uniformly distrubuted on the interval $(0,T)$, then the pdf is equal to

• A. $\frac{t_o}{T}\delta \left(x\right)+\frac{T}{t_o}\delta (x-A)$
• B. $\left(\frac{T-t_o}{T}\right)\delta \left(x\right)+\frac{t_o}{T}\delta (x-A)$
• C. $\frac{t_o}{T}\delta \left(x\right)+\left(\frac{T-t_o}{T}\right)\delta (x-A)$
• D. $\left(\frac{T-t_o}{t_0}\right)\delta \left(x\right)+\left(\frac{T-t_o}{T}\right)\delta (x-A)$

Answer 1

The correct option is B $\left(\frac{T-t_o}{T}\right)\delta \left(x\right)+\frac{t_o}{T}\delta (x-A)$

The procedure are indentical, let $ϵ=e$,
The above value should be non-negative, this will happen only for $\left(\frac{T-t_o}{T}\right)$ and also since x can have only two values A and 0, thus
Here, $f_x(x|ϵ=e)=PX\le x|ϵ=e$
$=\left(\frac{T-t_o}{T}\right)\mu \left(x\right)+\frac{t_o}{T}\mu (x-A)$
Because 'x' can have only value of zero and A.
Thus, $f_X(x|ϵ=e)=\left(\frac{T-t_o}{T}\right)\delta \left(x\right)+\frac{t_o}{T}\delta \left(x\right)$
$f_X\left(x\right)={\int }_{-\infty }^{\infty }{f}_x(x|ϵ=e)f_e\left(e\right)de$
$\left(\frac{T-t_o}{T}\right)\delta \left(x\right)+\frac{t_o}{T}\delta (x-A)$