Difference between revisions of "ECE PhD QE CNSIP 2004 Problem1.4" - Rhea

Latest revision as of 00:38, 10 March 2015

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)}. $