A stationary random process X(t) is applied to an LTI system whose impilse response is $h (t) = e^{- 2 t^{2}}$
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 ______
2.51

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Solution

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) = $\int_{- \infty}^{\infty} h (t) e^{- s t} d t$

$H (0) = \int_{- \infty}^{\infty} h (t) d t = \int_{- \infty}^{\infty} e^{- 2 t^{2}} d t = 2 \int_{0}^{\infty} e^{- 2 t^{2}} d t$

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

So, H(0) = $\frac{2 \sqrt{2}}{4} \int_{0}^{i n f t} u^{- 1 / 2} e^{- u} d u = \frac{1}{\sqrt{2}} \int_{0}^{\infty} u^{(\frac{1}{2} - 1)} e^{- u} d u$

$= \frac{1}{\sqrt{2}} Γ (\frac{1}{2}) = \frac{1}{\sqrt{2}} \sqrt{π} = \sqrt{\frac{π}{2}}$

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

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