1 answer

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

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,

(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).

Answers

Answer:

Given that:

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

x_{(t,s1)}=e^{-t},x_{(t,s2)}=-e^{-t}

X_t=\begin{pmatrix} X_{s1};p_1=0.5\\ X_{s2};p_2=0.5 \end{pmatrix}

Where X_{s1}=e^{-t} and X_{s2}=-e^{-t}

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

.

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