Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2004

4. (35 pts.)

Assume that $\mathbf{X}\left(t\right)$ is a zero-mean, continuous-time, Gaussian white noise process with autocorrelation function $R_{\mathbf{XX}}\left(t_{1},t_{2}\right)=\frac{N_{0}}{2}\delta\left(t_{1}-t_{2}\right).$ Let $\mathbf{Y}\left(t\right)$ be a new random process defined as the output of a linear time-invariant system with impulse response $h\left(t\right)=\frac{1}{T}e^{-t/T}\cdot u\left(t\right),$ where $u\left(t\right)$ is the unit step function and $T>0$ .

(a)

What is the mean of $\mathbf{Y\left(t\right)}$ ?

$E\left[\mathbf{Y}\left(t\right)\right]=E\left[\int_{-\infty}^{\infty}h\left(\tau\right)\mathbf{X}\left(t-\tau\right)d\tau\right]=\int_{-\infty}^{\infty}h\left(\tau\right)E\left[\mathbf{X}\left(t-\tau\right)\right]d\tau=\int_{-\infty}^{\infty}h\left(\tau\right)\cdot0d\tau=0.$

(b)

What is the autocorrelation function of $\mathbf{Y}\left(t\right)$ ?

$S_{\mathbf{XX}}\left(\omega\right)=\int_{-\infty}^{\infty}\frac{N_{0}}{2}\delta\left(\tau\right)e^{-i\omega\tau}d\tau=\frac{N_{0}}{2}.$

Let $\alpha=\frac{1}{T}$ .

$H\left(\omega\right)=\int_{-\infty}^{\infty}h\left(t\right)e^{-i\omega t}dt=\int_{0}^{\infty}\alpha e^{-\alpha t}\cdot e^{-i\omega t}dt=\alpha\int_{0}^{\infty}e^{-\left(\alpha+i\omega\right)t}dt=\alpha\frac{e^{-\left(\alpha+i\omega\right)t}}{-\left(\alpha+i\omega\right)}\biggl|_{0}^{\infty}=\frac{\alpha}{\alpha+i\omega}.$

$S_{\mathbf{YY}}\left(\omega\right)=S_{\mathbf{XX}}\left(\omega\right)\left|H\left(\omega\right)\right|^{2}=S_{\mathbf{XX}}\left(\omega\right)H\left(\omega\right)H^{*}\left(\omega\right)=\frac{N_{0}}{2}\cdot\frac{\alpha}{\alpha+i\omega}\cdot\frac{\alpha}{\alpha-i\omega}=\frac{\alpha^{2}N_{0}}{2\left(\alpha^{2}+\omega^{2}\right)}.$

$S_{\mathbf{YY}}\left(\omega\right)=\frac{\alpha^{2}N_{0}}{2\left(\alpha^{2}+\omega^{2}\right)}=\left(\frac{\alpha N_{0}}{4}\right)\frac{2\alpha}{\alpha^{2}+\omega^{2}}\leftrightarrow\left(\frac{\alpha N_{0}}{4}\right)e^{-\alpha\left|\tau\right|}=R_{\mathbf{YY}}\left(\tau\right).$

$\because e^{-\alpha\left|\tau\right|}\leftrightarrow\frac{2\alpha}{\alpha^{2}+\omega^{2}}\text{ (on the table given)}.$

$\therefore R_{\mathbf{YY}}\left(\tau\right)=\left(\frac{\alpha N_{0}}{4}\right)e^{-\alpha\left|\tau\right|}=\left(\frac{N_{0}}{4T}\right)e^{-\frac{\left|\tau\right|}{T}}.$

(c)

Write an expression for the $n$ -th order characteristic function of $\mathbf{Y}\left(t\right)$ sampled at time $t_{1},t_{2},\cdots,t_{n}$ . Simplify as much as possible.

(d)

Write an expression for the second-order pdf $f_{\mathbf{Y}\left(t_{1}\right)\mathbf{Y}\left(t_{2}\right)}\left(y_{1},y_{2}\right)$ of $\mathbf{Y}\left(t\right)$ . simplify as much as possible.

$\mathbf{Y}\left(t\right)$ is a WSS Gaussian random process with $E\left[\mathbf{Y}\left(t\right)\right]=0 , \sigma_{\mathbf{Y}\left(t\right)}^{2}=R_{\mathbf{YY}}\left(0\right)=\frac{N_{0}}{4}$ .

$r_{\mathbf{Y}\left(t_{1}\right)\mathbf{Y}\left(t_{2}\right)}=r\left(t_{1}-t_{2}\right)=\frac{C_{\mathbf{YY}}\left(t_{1}-t_{2}\right)}{\sqrt{\sigma_{\mathbf{Y}\left(t_{1}\right)}^{2}\sigma_{\mathbf{Y}\left(t_{2}\right)}^{2}}}=\frac{R_{\mathbf{YY}}\left(t_{1}-t_{2}\right)}{R_{\mathbf{YY}}\left(0\right)}=e^{-\alpha\left|t_{1}-t_{2}\right|}.$

$f_{\mathbf{Y}\left(t_{1}\right)\mathbf{Y}\left(t_{2}\right)}\left(y_{1},y_{2}\right)=\frac{1}{2\pi\sigma_{\mathbf{Y}\left(t_{1}\right)}\sigma_{\mathbf{Y}\left(t_{2}\right)}\sqrt{1-r^{2}}}\exp\left\{ \frac{-1}{2\left(1-r^{2}\right)}\left[\frac{y_{1}^{2}}{\sigma_{\mathbf{Y}\left(t_{1}\right)}^{2}}-\frac{2ry_{1}y_{2}}{\sigma_{\mathbf{Y}\left(t_{1}\right)}\sigma_{\mathbf{Y}\left(t_{2}\right)}}+\frac{y_{2}^{2}}{\sigma_{\mathbf{Y}\left(t_{2}\right)}^{2}}\right]\right\} $$=\frac{1}{2\pi\frac{N_{0}}{4}\sqrt{1-e^{-2\alpha\left|t_{1}-t_{2}\right|}}}\exp\left\{ \frac{-1}{2\left(1-e^{-2\alpha\left|t_{1}-t_{2}\right|}\right)}\left[\frac{y_{1}^{2}}{N_{0}/4}-\frac{2y_{1}y_{2}e^{-\alpha\left|t_{1}-t_{2}\right|}}{N_{0}/4}+\frac{y_{2}^{2}}{N_{0}/4}\right]\right\}$$ =\frac{2}{\pi N_{0}\sqrt{1-e^{-2\alpha\left|t_{1}-t_{2}\right|}}}\exp\left\{ \frac{-2}{N_{0}\left(1-e^{-2\alpha\left|t_{1}-t_{2}\right|}\right)}\left[y_{1}^{2}-2y_{1}y_{2}e^{-\alpha\left|t_{1}-t_{2}\right|}+y_{2}^{2}\right]\right\}$ .

(e)

Find the minium mean-square estimate of $\mathbf{Y}\left(t_{2}\right)$ given that $\mathbf{Y}\left(t_{1}\right)=y_{1}$ . Simplify your answer as much as possible.

$\widehat{y_{2}}_{MMS}\left(y_{1}\right)=E\left[\mathbf{Y}\left(t_{2}\right)|\mathbf{Y}\left(t_{1}\right)=y_{1}\right]=\int_{-\infty}^{\infty}y_{2}\cdot f_{\mathbf{Y}\left(t_{2}\right)}\left(y_{2}|\mathbf{Y}\left(t_{1}\right)=y_{1}\right)dy_{2}$

$\text{where }f_{\mathbf{Y}\left(t_{2}\right)}\left(y_{2}|\mathbf{Y}\left(t_{1}\right)=y_{1}\right)=\frac{f_{\mathbf{Y}\left(t_{1}\right)\mathbf{Y}\left(t_{2}\right)}\left(y_{1,}y_{2}\right)}{f_{\mathbf{Y}\left(t_{1}\right)}\left(y_{1}\right)}.$

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