View source for ECE PhD QE CNSIP Jan 2001 Problem1.5 - Rhea

[[Category:ECE]]
[[Category:QE]]
[[Category:CNSIP]]
[[Category:problem solving]]
[[Category:random variables]]

==Question from [[ECE_PhD_QE_CNSIP_Jan_2001_Problem1|ECE QE January 2001]]== 

Let a linear discrete parameter shift-invariant system have the following difference equation: <math class="inline">y\left(n\right)=0.7y\left(n-1\right)+x\left(n\right)</math> where <math class="inline">x\left(n\right)</math>  in the input and <math class="inline">y\left(n\right)</math>  is the output. Now suppose this system has as its input the discrete parameter random process <math class="inline">\mathbf{X}_{n}</math> . 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.

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==Share and discuss your solutions below.==
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=Solution 1 (retrived from [[ECE600_QE_2000_August|here]])=
'''(a)''' 

<math class="inline">E\left[\mathbf{X}_{n}\right]=0.</math> 

<math class="inline">R_{\mathbf{XX}}\left(n+m,\; n\right)</math> 

<math class="inline">\therefore\;\mathbf{X}_{n}\text{ is wide-sense stationary.}</math> 

'''(b)''' 

<math class="inline">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.</math> 

<math class="inline">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.</math> 

<math class="inline">R_{\mathbf{YY}}\left(n+m,\; n\right)</math> 

<math class="inline">R_{\mathbf{YY}}\left(n+m,\; n\right)</math>  depends on the time difference <math class="inline">m</math> . Thus, <math class="inline">\mathbf{Y}_{n}</math>  is wide-sense stationary. 

'''(c) '''

<math class="inline">R_{\mathbf{XX}}\left(n,n+m\right)=R_{\mathbf{X}}\left(m\right)=\sigma_{\mathbf{X}}^{2}\delta\left(m\right).</math> 

'''(d) '''

<math class="inline">R_{\mathbf{Y}}\left(m\right)</math> 

<math class="inline">\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).</math> 

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==Solution 2==
Write it here.
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[[ECE_PhD_QE_CNSIP_Jan_2001_Problem1|Back to QE CS question 1, January 2001]]

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