Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2006

3 (15 points)

Let $\mathbf{Y}(t)$ be the output of linear system with impulse response $h\left(t\right)$ and input $\mathbf{X}\left(t\right)+\mathbf{N}\left(t\right)$ , where $\mathbf{X}\left(t\right)$ and $\mathbf{N}\left(t\right)$ are jointly wide-sense stationary independent random processes. If $\mathbf{Z}\left(t\right)=\mathbf{X}\left(t\right)-\mathbf{Y}\left(t\right)$ , find the power spectral density $S_{\mathbf{Z}}\left(\omega\right)$ in terms of $S_{\mathbf{X}}\left(\omega\right) , S_{\mathbf{N}}\left(\omega\right) , m_{\mathbf{X}}=E\left[\mathbf{X}\right]$ , and $m_{\mathbf{Y}}=E\left[\mathbf{Y}\right]$ .

Solution

Let $\mathbf{M}\left(t\right)=\mathbf{X}\left(t\right)+\mathbf{N}\left(t\right)$ . Since $\mathbf{X}\left(t\right)$ and $\mathbf{N}\left(t\right)$ are jointly wide-sense stationary. $\mathbf{M}\left(t\right)$ is also a wide-sense stationary random process.

$\mathbf{Y}\left(t\right)=\mathbf{M}\left(t\right)*h\left(t\right).$

$R_{\mathbf{Y}}\left(\tau\right)=\left(R_{\mathbf{M}}*h*\tilde{h}\right)\left(\tau\right)\text{ where }\left(\tilde{h}\left(t\right)=h\left(-t\right)\right).$

$R_{\mathbf{M}}\left(\tau\right)=E\left[\mathbf{M}\left(t\right)\mathbf{M}\left(t+\tau\right)\right] $$=E\left[\mathbf{X}\left(t\right)\mathbf{X}\left(t+\tau\right)\right]+E\left[\mathbf{X}\left(t\right)\right]E\left[\mathbf{N}\left(t+\tau\right)\right]+E\left[\mathbf{X}\left(t+\tau\right)\right]E\left[\mathbf{N}\left(t\right)\right]+E\left[\mathbf{N}\left(t\right)\mathbf{N}\left(t+\tau\right)\right]$$ =R_{\mathbf{X}}\left(\tau\right)+2m_{\mathbf{X}}m_{\mathbf{N}}+R_{\mathbf{N}}\left(\tau\right)$

$R_{\mathbf{XY}}\left(\tau\right)=E\left[\mathbf{X}\left(t\right)\mathbf{Y}\left(t+\tau\right)\right] $$=E\left[\mathbf{X}\left(t\right)\int_{-\infty}^{\infty}\left(\mathbf{X}\left(t+\tau-\alpha\right)+\mathbf{N}\left(t+\tau-\alpha\right)\right)h\left(\alpha\right)d\alpha\right]$$ =\int_{-\infty}^{\infty}\left(R_{\mathbf{X}}\left(\tau-\alpha\right)+E\left[\mathbf{X}\left(t\right)\right]E\left[\mathbf{N}\left(t+\tau-\alpha\right)\right]\right)h\left(\alpha\right)d\alpha $$=R_{\mathbf{X}}\left(\tau\right)*h\left(\tau\right)+m_{\mathbf{X}}m_{\mathbf{N}}*h\left(\tau\right). R_{\mathbf{Z}}\left(\tau\right)=E\left[\mathbf{Z}\left(t\right)\mathbf{Z}\left(t+\tau\right)\right]=E\left[\left(\mathbf{X}\left(t\right)-\mathbf{Y}\left(t\right)\right)\left(\mathbf{X}\left(t+\tau\right)-\mathbf{Y}\left(t+\tau\right)\right)\right]$$ =R_{\mathbf{X}}\left(\tau\right)-R_{\mathbf{YX}}\left(\tau\right)-R_{\mathbf{XY}}\left(\tau\right)+R_{\mathbf{YY}}\left(\tau\right).$

$S_{\mathbf{Z}}\left(\omega\right)=S_{\mathbf{X}}\left(\omega\right)-S_{\mathbf{YX}}\left(\omega\right)-S_{\mathbf{XY}}\left(\omega\right)+S_{\mathbf{Y}}\left(\omega\right)=S_{\mathbf{X}}\left(\omega\right)-S_{\mathbf{XY}}^{*}\left(\omega\right)-S_{\mathbf{XY}}\left(\omega\right)+S_{\mathbf{Y}}\left(\omega\right) $$=S_{\mathbf{X}}\left(\omega\right)-2\Re\left\{ S_{\mathbf{XY}}\left(\omega\right)\right\} +S_{\mathbf{M}}\left(\omega\right)\Bigl|H\left(\omega\right)\Bigr|^{2}$$ =S_{\mathbf{X}}\left(\omega\right)-2\Re\left\{ S_{\mathbf{X}}\left(\omega\right)H\left(\omega\right)+2\pi m_{\mathbf{X}}m_{\mathbf{N}}\delta\left(\omega\right)H\left(\omega\right)\right\} +\left\{ S_{\mathbf{X}}\left(\omega\right)+2\pi m_{\mathbf{X}}m_{\mathbf{N}}\delta\left(\omega\right)+S_{\mathbf{N}}\left(\omega\right)\right\} \Bigl|H\left(\omega\right)\Bigr|^{2} $$=S_{\mathbf{X}}\left(\omega\right)-2\Re\left\{ S_{\mathbf{X}}\left(\omega\right)H\left(\omega\right)+2\pi m_{\mathbf{X}}\left(m_{\mathbf{Y}}-m_{\mathbf{X}}H\left(0\right)\right)\delta\left(\omega\right)\right\} +$$ \left\{ S_{\mathbf{X}}\left(\omega\right)+S_{\mathbf{N}}\left(\omega\right)\right\} \Bigl|H\left(\omega\right)\Bigr|^{2}+2\pi m_{\mathbf{X}}\left(m_{\mathbf{Y}}-m_{\mathbf{X}}H\left(0\right)\right)H\left(0\right)\delta\left(\omega\right).$

$\because m_{\mathbf{Y}}=m_{\mathbf{M}}*h\left(t\right)=\int_{-\infty}^{\infty}\left(m_{\mathbf{X}}+m_{\mathbf{N}}\right)h\left(t\right)dt=\left(m_{\mathbf{X}}+m_{\mathbf{N}}\right)H\left(0\right)\Rightarrow m_{\mathbf{N}}H\left(0\right)=m_{\mathbf{Y}}-m_{\mathbf{X}}H\left(0\right).$

## Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood