Question

$X\left(t\right)$ is a $WSS$ random process with an average power of $2W$, another random process $Y\left(t\right)$ is defined as,
$Y\left(t\right)=X\left(t\right)cos({\omega }_{o}t+\theta )$
Where, $\theta$ is a random variable which is uniformly distributed in the interval $[0,2\pi ]$ and ${\omega }_0$ is a constant. If $X\left(t\right)$ and $\theta$ are statistically independent, then the average power of the random process $Y\left(t\right)$ will be

• A. $4W$
• B. $2W$
• C. zero
• D. $1W$

Answer 1

The correct option is D $1W$

$Y\left(t\right)=X\left(t\right)cos({\omega }_{o}t+\theta )$



ACF of $Y\left(t\right)$, $R_Y\left(\tau \right)=E\left[Y\right(t\left)Y\right(t+\tau \left)\right]$

$=E\left[X\right(t\left)cos\right({\omega }_{o}t+\theta \left)X\right(t+\tau \left)cos\right({\omega }_{o}t+\theta +{\omega }_o\tau \left)\right]$

$=E\left[X\right(t\left)X\right(t+\tau \left)\right]E\left[cos\right({\omega }_{o}t+\theta \left)cos\right({\omega }_{o}t+\theta +{\omega }_0\tau \left)\right]$

$=R_X\left(\tau \right)[\frac{1}{2}cos{\omega }_0\tau +\frac{1}{2}E[cos(2{\omega }_{o}t+2\theta +{\omega }_0\tau )\left]\right]$

$E\left[cos\right(2{\omega }_{0}t+{\omega }_0\tau +2\theta \left)\right]=E\left[cos\right(\alpha +2\theta \left)\right]={\int }_{-\infty }^{\infty }cos(\alpha +2\theta )f_{\theta }\left(\theta \right)d\theta$

$={\int }_0^{2\pi }cos(\alpha +2\theta )\frac{1}{2\pi }d\theta =0$
So, $R_Y\left(\tau \right)=\frac{1}{2}{R}_X\left(\tau \right)cos\left({\omega }_0\tau \right)$
Given that, the average power of $X\left(t\right)=R_X\left(0\right)=2W$
Average power of $Y\left(t\right)=R_Y\left(0\right)=\frac{1}{2}{R}_X\left(0\right)=\frac{1}{2}\left(2\right)=1W$