Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

January 2001

# Part 5

Let a linear discrete parameter shift-invariant system have the following difference equation: $y\left(n\right)=0.7y\left(n-1\right)+x\left(n\right)$ where $x\left(n\right)$ in the input and $y\left(n\right)$ is the output. Now suppose this system has as its input the discrete parameter random process $\mathbf{X}_{n}$ . You may assume that the input process is zero-mean i.i.d.

(a) (5 pts) Is the input wide-sense stationary (show your work)?

(b) (5 pts) Is the output process wide-sense stationary (show your work)?

(c) (5 pts) Find the autocorrelation function of the input process.

(d) (5 pts) Find the autocorrelation function, in closed form, for the output process.

# Solution 1 (retrived from here)

(a)

$E\left[\mathbf{X}_{n}\right]=0.$

$R_{\mathbf{XX}}\left(n+m,\; n\right)$

$\therefore\;\mathbf{X}_{n}\text{ is wide-sense stationary.}$

(b)

$E\left[\mathbf{Y}_{n}\right]=0.7E\left[\mathbf{Y}_{n-1}\right]+E\left[\mathbf{X}_{n}\right]=0.7E\left[\mathbf{Y}_{n-1}\right]=0.7^{2}E\left[\mathbf{Y}_{n-2}\right]=0.7^{n}E\left[\mathbf{Y}_{0}\right]=0.$

$E\left[\mathbf{Y}_{0}\right]=E\left[\sum_{n=-\infty}^{\infty}h\left(0-n\right)\mathbf{X}\left(n\right)\right]=\sum_{n=-\infty}^{\infty}h\left(-n\right)E\left[\mathbf{X}\left(n\right)\right]=0.$

$R_{\mathbf{YY}}\left(n+m,\; n\right)$

$R_{\mathbf{YY}}\left(n+m,\; n\right)$ depends on the time difference $m$ . Thus, $\mathbf{Y}_{n}$ is wide-sense stationary.

(c)

$R_{\mathbf{XX}}\left(n,n+m\right)=R_{\mathbf{X}}\left(m\right)=\sigma_{\mathbf{X}}^{2}\delta\left(m\right).$

(d)

$R_{\mathbf{Y}}\left(m\right)$

$\because\; E\left[\mathbf{X}\left(n\right)\mathbf{Y}\left(m\right)\right]=E\left[\sum_{k=-\infty}^{\infty}h\left(m-k\right)\mathbf{X}\left(n\right)\mathbf{X}\left(k\right)\right]=\sum_{k=-\infty}^{\infty}h\left(m-k\right)\left(\sigma_{\mathbf{X}}^{2}\delta\left(n-k\right)\right).$

Write it here.

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