Question

Can someone please explain how to solve the problem below? I keep getting the answer incorrect.

(13 points) The random process X(t) consists of the following two sample functions which are equally likely: x(t,sı)=e", x(t,82)=-et Determine the mean and autocorrelation function of X(t), and also determine whether X(t) is wide sense stationary. (Note: no credit will be awarded for correct guesses without justification).

Answer 1

Answer:

Given that:

The random process X(t) consists of the following two sample functions which are equally likely:

-2t,81) = e = e, (t,52) = -6

Xt X 31, p1 = 0.5 X32:p2 = 0.5

Where $X_{s1}=e^{-t}$ and

X 2 = -e

Mean of $X_{t}$ , $\mu _{t}=E(X_t)$

$=\frac{1}{2}X_{s1}+\frac{1}{2}X_{s2}$

$=\frac{1}{2}(e^{-t})+\frac{1}{2}(-e^{-t})$

$=0$

Variance of $X_t$ , $\sigma _t^2=E[(X_t-\mu _t)^2]$

$=E(X_t^2)$

$=\frac{1}{2}X_{s1}^2+\frac{1}{2}X_{s2}^2$

$=\frac{1}{2}(e^{-2t})+\frac{1}{2}(-e^{-2t})$

$=e^{-2t}$

Autocorrelation function of $X_t$ , $R_X(t_1,t_2)=\frac{E[(X_{t1}-\mu _{t1})(X_{t2}-\mu _{t2})]}{\sigma _{t1}\sigma _{t2}}$

$=\frac{E[X_{t1}X_{t2}]}{e^{-2t_1}e^{-2t_2}}$

$=\frac{0.5\int_{-\infty }^{\infty }x_1x_2X_{s1}dx_1dx_2+0.5\int_{-\infty }^{\infty }x_1x_2X_{s2}dx_1dx_2}{e^{-2(t_1+t_2)}}$

$=\frac{0.5\int_{-\infty }^{\infty }x_1x_2(e^{-t})dx_1dx_2+0.5\int_{-\infty }^{\infty }x_1x_2(-e^{-t})dx_1dx_2}{e^{-2(t_1+t_2)}}$

$=0$

Covariance of $X_t=E[(X_{t1}-\mu _{t1})(X_{t2}-\mu _{t2})]$

$=0$

Since,Mean and covariance of $X_t$ are independent of t

$X_t$ is WSS