Question

A stationary random process X(t) is applied to an LTI system whose impilse response is $h\left(t\right)=e^{-2t^2}$
The output process of this LTI system is Y(t). If the mean value of the process X(t) is 2, then the mean value of the process Y(t) will be ______

• A. 2.51

Answer 1

The correct option is A 2.51
As X(t) is a stationary process, Y(t) will be also a stationary process.

E[Y(t)] = H[0] E[X(t)]

H(0) = H(s) at s=0

H(s) = ${\int }_{-\infty }^{\infty }h\left(t\right)e^{-st}dt$

$H\left(0\right)={\int }_{-\infty }^{\infty }h\left(t\right)dt={\int }_{-\infty }^{\infty }{e}^{-2t^2}dt=2{\int }_0^{\infty }{e}^{-2t^2}dt$

Let, $u=2t^2\Rightarrow du=4tdt\Rightarrow dt=\frac{\sqrt{2}}{4}{u}^{-1/2}du$

So, H(0) = $\frac{2\sqrt{2}}{4}{\int }_0^{inft}{u}^{-1/2}{e}^{-u}du=\frac{1}{\sqrt{2}}{\int }_0^{\infty }{u}^{(\frac{1}{2}-1)}{e}^{-u}du$

$=\frac{1}{\sqrt{2}}\Gamma \left(\frac{1}{2}\right)=\frac{1}{\sqrt{2}}\sqrt{\pi }=\sqrt{\frac{\pi }{2}}$

$E\left[Y\right(t\left)\right]=H\left(0\right)E\left[X\right(t\left)\right]=2\sqrt{\frac{\pi }{2}}=\sqrt{2\pi }=2.51$