ECE Ph.D. Qualifying Exam

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.

Share and discuss your solutions below.

Solution 1 (retrived from here)


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

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

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


$ 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.


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


$ 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). $

Solution 2

Write it here.

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