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Utah State University
ECE 6010
Stochastic Processes
Homework # 8 Solutions

  1. Suppose $ \{X_t, t \geq 0 \} $ is a Wiener process. Define a process $ \{Y_t, t \geq 0 \}$ by $ Y_{t} = X_{t+D} - X_{t}$ for a fixed positive number $ D$ .
    1. Find the mean and autocorrelation functions of $ \{Y_{t} \}$ .
      Mean :

      $\displaystyle \mu_{Y}(t) = E[Y_{t}] = E[X_{t+D}]-E[X_{t}] = \mu(t+D) - \mu(t) = \mu D$

      Autocorrelation :

      $\displaystyle R_{Y}(t,s)$ $\displaystyle =$ $\displaystyle E[(X_{t+D} - X_{t})(X_{s+D} - X_{s})]$  
        $\displaystyle =$ $\displaystyle E[X_{t+D}X_{s+D}] - E[X_{t+D} X_{s}] - E[X_{t}X_{s+D}] + E[ X_{t}X_{s}]$  
        $\displaystyle =$ $\displaystyle [\sigma^{2} \min(t+D,s+D) + \mu^{2} (t+D)(s+D)] - [\sigma^2 \min(t+D,s) + \mu^2 (t+D)s]$  
          $\displaystyle -[ \sigma^{2} \min(t,s+D) + \mu^2 (S+D)t ]+ [\sigma^2 \min(t,s) + \mu^2 st]$  
        $\displaystyle =$ $\displaystyle \sigma^2 [ \min(t+D,s+D) - \min(t+D,s) - \min(t,s+D) + \min(t,s)]$  
          $\displaystyle + \mu^2 [ ts +Dt + sD + D^2 - st - sD - st -Dt + st]$  
        $\displaystyle =$ \begin{displaymath}\left\{ \begin{array}{ll} \sigma^2[s+D - s - \min(t,s+D) +s] ... ...- t - \min(t,s+D) +t] + \mu D^2 & t < s \\ \end{array} \right.\end{displaymath}< s \\ \end{array} \right.\end{displaymath}" border="0" height="60" width="380" />  
        $\displaystyle =$ \begin{displaymath}\left\{ \begin{array}{ll} \mu^2 D^2 & t \geq s , t-s-D \g... ... + \mu^2 D^2 & t < s , t-s+D \geq 0 \\ \end{array} \right.\end{displaymath}< s , t-s+D \geq 0 \\ \end{array} \right.\end{displaymath}" border="0" height="104" width="359" />  
        $\displaystyle =$ \begin{displaymath}\left\{ \begin{array}{ll} \mu^2 D^2 & \vert t-s\vert \geq D \... ...s\vert] + \mu^2 D^2 & \vert t-s\vert <D \\ \end{array} \right.\end{displaymath}  

      So $ R_{Y}(t,s) = \sigma^2 ax(0,D-\vert t-s\vert) + \mu^2 D^2$ .

    2. Show that $ \{Y_{t} \}$ is a stationary and find its spectrum.
      $ \mu_{y}(t)$ is a constant and $ R_{Y}(t,s) = R_{Y}(t-s) \ \Rightarrow$ W.S.S.

      Since Gaussian too $ \Rightarrow$ Strictly Stationary.

      $\displaystyle R_{Y}(\tau) = \mu^2 D^2 + \left[ \mathrm{rect}\left(\frac{2\tau}{D}\right) * \mathrm{rect}\left(\frac{2\tau}{D}\right) \right] \sigma^2 D $

      So,

      $\displaystyle S_{Y}(\omega) = \mathcal{F}\{R_{Y}(\tau)\} = 2 \pi \mu^2 D^2 \delta(\omega) + \sigma^2 D \frac{ \sin^{2}(\omega D/2)}{(\omega D/2)^{2}} $

  2. Suppose $ \{X_{t} \}$ and $ \{Y_{t} \}$ are zero mean and individually and jointly W.S.S. Show that the mean-square error associated with the noncausal Wiener filter for estimation of $ X_{t}$ from $ \{Y_{t}, t \in \mathbb{R} \}$ is

    $\displaystyle \frac{1}{2 \pi} \int_{-\infty}^{\infty} \left[ S_{X}(\omega) - \frac{\vert S_{XY}(\omega)\vert^{2}} {S_{Y}(\omega)} \right] d\omega. $


    $\displaystyle MSE$ $\displaystyle =$ $\displaystyle E[(X_{t} - \hat{X}_{t})^2]$  
      $\displaystyle =$ $\displaystyle E[(X_{t} - \hat{X}_{t})X_{t}] - \underbrace{ E[(X_{t} - \hat{X}_{t})\hat{X}_{t}]}_{0}$  
      $\displaystyle =$ $\displaystyle E[X_{t}^2] - E[X_{t} \hat{X}_{t}]$  
        $\displaystyle \quad \quad (E[(X_{t} -\hat{X}_{t})\hat{X}_{t}] = 0 \ \Rightarrow E[X_{t} \hat{X}_{t}] = E[\hat{X}_{t}^2])$  
      $\displaystyle =$ $\displaystyle E[X_{t}^2] - E[\hat{X}_{t}^2]$  
      $\displaystyle =$ $\displaystyle R_{X}(0) - R_{\hat{X}}(0)$  
      $\displaystyle =$ $\displaystyle \frac{1}{2 \pi} \int_{-\infty}^{\infty} (S_{X}(\omega) - S_{\hat{X}}(\omega) ) d\omega$  
      $\displaystyle =$ $\displaystyle \frac{1}{2 \pi} \int_{-\infty}^{\infty} (S_{X}(\omega) - \vert H_{0}(\omega)\vert^2 S_{Y}(\omega) ) d\omega$  
      $\displaystyle =$ $\displaystyle \frac{1}{2 \pi} \int_{-\infty}^{\infty} (S_{X}(\omega) - \frac{\vert S_{XY}(\omega)\vert^2}{\vert S_{Y}(\omega)\vert^2} S_{Y}(\omega) ) \ d\omega$  
      $\displaystyle =$ $\displaystyle \frac{1}{2 \pi} \int_{-\infty}^{\infty} (S_{X}(\omega) - \frac{\vert S_{XY}(\omega)\vert^2}{S_{Y}(\omega)}) \ d\omega$  

  3. Suppose $ Y_t = S_t+N_t$ for $ t \in \mathbb{R}$ , where $ \{S_{t} \}$ and $ \{N_{t} \}$ are zero-mean, W.S.S., and orthogonal. Suppose that we wish to estimate $ X_t = \int_{-\infty}^{\infty} k(t- \tau) S_{\tau} d\tau $ , with an estimate of the form $ \hat{X}_t = \int_{-\infty}^{\infty} h(t- \tau)Y_{\tau} d\tau $ , where $ k$ and $ h$ are impulse responses of linear time-invariant systems. show that

    $\displaystyle E[(X_{t} - \hat{X}_{t})^{2}] = \frac{1}{2 \pi} \int_{-\infty}^{\i... ...t^{2} S_{S}(\omega) + \vert H(\omega)\vert^{2} S_{N}(\omega) \right] d\omega $

    where $ K$ and $ H$ are the transfer functions of $ k$ and $ h$ , respectively, and $ S_{S}$ and $ S_{N}$ are the power spectral densities of $ \{S_{t} \}$ and $ \{N_{t} \}$ . (Note the case that $ k(t) = \delta(t-\lambda)$ for some fixed $ \lambda \in \mathbb{R}$ .)


    $\displaystyle E[(X_t - \hat{X}_t)^2]$ $\displaystyle =$ $\displaystyle E \left[ \left(\int_{-\infty}^{\infty} k(t- \tau) S_{\tau} d\tau - \int_{-\infty}^{\infty} h(t- \tau)Y_{\tau} \ d\tau \right)^2 \right]$  
      $\displaystyle =$ $\displaystyle E \left[ \left( \int_{-\infty}^{\infty} k(t- \tau) S_{\tau} d\t... ... d\tau - \int_{-\infty}^{\infty} h(t- \tau)N_{\tau} \ d\tau \right)^2 \right]$  
      $\displaystyle =$ $\displaystyle E \left[ \left( \int_{-\infty}^{\infty} (k(t- \tau) - h(t- \tau))... ... d\tau - \int_{-\infty}^{\infty} h(t- \tau)N_{\tau} \ d\tau \right)^2 \right]$  
      $\displaystyle =$ $\displaystyle E \left[ \left( \int_{-\infty}^{\infty} (k(t- \tau) - h(t- \tau))... ...ft[ \left(\int_{-\infty}^{\infty} h(t- \tau)N_{\tau} \ d\tau \right)^2 \right]$  
      $\displaystyle =$ $\displaystyle R_{W}(0) + R_{Z}(0)$  
        \begin{displaymath}\quad \quad \quad \left( \begin{array}{ll} \mathrm{Where} & W... ...infty}^{\infty} h(t- \tau)N_{\tau} \ d\tau \end{array} \right)\end{displaymath}  
      $\displaystyle =$ $\displaystyle \frac{1}{2 \pi} \int_{-\infty}^{\infty} S_{W}(\omega) \ d\omega + \frac{1}{2 \pi} \int_{-\infty}^{\infty} S_{Z}(\omega) \ d\omega$  
      $\displaystyle =$ $\displaystyle \frac{1}{2 \pi} \int_{-\infty}^{\infty} \left[ \vert K(\omega) - ... ...rt^{2} S_{S}(\omega) + \vert H(\omega)\vert^{2} S_{N}(\omega) \right] d\omega$  

  4. Consider the situation of the previous problem with $ k(t) = \delta(t-\lambda)$ ,

    $\displaystyle S_{S}(\omega) = \frac{A^2}{\alpha^2 + \omega^2} \quad \quad S_{N}(\omega) = \frac{N_{0}}{2} $

    1. Find the noncausal Wiener filter for estimating $ X_{t}$ from $ \{Y_{t}, t \in \mathbb{R} \}$ . Find the corresponding mean-square error.

      When $ k(t) = \delta(t-\lambda)$ ,

      $\displaystyle X_{t} = \int_{-\infty}^{\infty} \delta(t-\lambda -\tau) S_{\tau} d \tau = S_{t-\lambda} $

      and

      $\displaystyle K(\omega) = \mathcal{F}\{\delta(t-\lambda)\} = e^{-j\lambda \omega}$


      $\displaystyle R_{Y}(t,u)$ $\displaystyle =$ $\displaystyle E[(S_{t}+N_{t})(S_{u}+N_{u})]$  
        $\displaystyle =$ $\displaystyle E[S_{t}S_{u} + S_{t}N_{u} + N_{t}S_{u} + N_{t}N_{u}]$  
        $\displaystyle =$ $\displaystyle R_{S}(\tau) + R_{SN}(\tau) + R_{SN}(-\tau) + R_{N}(\tau)$  
      $\displaystyle S_{Y}(\omega)$ $\displaystyle =$ $\displaystyle S_{S}(\omega) + S_{N}(\omega) + S_{SN}(\omega) + S_{SN}(-\omega)$  
        $\displaystyle =$ $\displaystyle S_{S}(\omega) + S_{N}(\omega) + 2 \ \mathrm{Re}[S_{SN}(\omega)]$  
             $\displaystyle \mbox{ (Since $S \perp N$, $S_{SN} = 0$)}$  
        $\displaystyle =$ $\displaystyle S_{S}(\omega) + S_{N}(\omega)$  


      $\displaystyle R_{XY}(t,u)$ $\displaystyle =$ $\displaystyle E[X_{t} Y_{u}]$  
        $\displaystyle =$ $\displaystyle E[S_{t-\lambda} S_{u}] + E[S_{t-\lambda} N_{u}]$  
        $\displaystyle =$ $\displaystyle R_{s}(t-\lambda - u) + R_{SN}(t- \lambda - u)$  
      $\displaystyle R_{XY}(\tau)$ $\displaystyle =$ $\displaystyle R_{s}(\tau - u) + R_{SN}(\tau - u)$  
      $\displaystyle S_{XY}(\omega)$ $\displaystyle =$ $\displaystyle e^{-j \omega \lambda}S_{S}(\omega) + e^{-j \omega \lambda}S_{SN}(\omega)$  
             $\displaystyle \mbox{ (Since $S \perp N$, $R_{SN} = S_{SN} = 0$)}$  
        $\displaystyle =$ $\displaystyle e^{-j \omega \lambda}S_{S}(\omega)$  


      $\displaystyle H_{0}(\omega)$ $\displaystyle =$ $\displaystyle \frac{ S_{XY}(\omega)}{S_{Y}(\omega)}$  
        $\displaystyle =$ $\displaystyle \frac{e^{-j \omega \lambda}S_{S}(\omega)}{S_{S}(\omega) + S_{N}(\omega)}$  
        $\displaystyle =$ $\displaystyle \frac{2 A^2}{N_{0}} \frac{1}{\omega^2 + \beta^2} e^{-j \omega \lambda} \quad \mbox{ (where $ \beta^2 = \alpha^2 + \frac{2A^2}{N_{0}}$)}$  

      $\displaystyle h_{0}(t) = \mathcal{F}^{-1}\{H_{0}(\omega) \} = \frac{A^2}{\beta N_{0}} e^{-\beta \vert t - \lambda \vert} $

      From the previous problem
      $\displaystyle MSE$ $\displaystyle =$ $\displaystyle \frac{1}{2 \pi} \int_{-\infty}^{\infty} \left[ \vert K(\omega) - ... ...rt^{2} S_{S}(\omega) + \vert H(\omega)\vert^{2} S_{N}(\omega) \right] d\omega$  
        $\displaystyle =$ $\displaystyle \frac{1}{2 \pi} \int_{-\infty}^{\infty} \left\vert e^{-j \omega \... ...da}S_{S}(\omega)}{S_{S}(\omega) + S_{N}(\omega)} \right\vert^2 S_{N} d \omega$  
        $\displaystyle =$ $\displaystyle \frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{S_{N}^2 (\omega)}{(... ...} + \frac{S_{S}^2 (\omega)}{(S_{S}(\omega) + S_{N}(\omega))^2} S_{N} d \omega$  
        $\displaystyle =$ $\displaystyle \frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{S_{N}^2 (\omega) S_... ... S_{S}^2 (\omega) S_{N}(\omega)}{(S_{S}(\omega) + S_{N}(\omega))^2} \ d \omega$  
        $\displaystyle =$ $\displaystyle \frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{S_{S}(\omega) S_{N}(\omega)}{S_{S}(\omega) + S_{N}(\omega)} d \omega$  

    2. Find the causal Wiener filter for estimating $ X_{t}$ from $ \{ Y_{\tau}, \tau \leq t \}$ . Consider $ \lambda <0$<0$" align="middle" border="0" height="32" width="46" /> and $ \lambda \geq 0$ .

      We have

      $\displaystyle H_0(\omega) = \frac{1}{S_Y^+(\omega)}\left[ \frac{S_{XY}(\omega)}{S_Y^-(\omega)}\right]_+ $

      where

      $\displaystyle S_Y(\omega) = S_S(\omega) + S_N(\omega) = \frac{A^2}{\alpha^2 + \... ...(\omega - [\alpha^2 + 2A^2/N_0]^{1/2})}{(\omega + i\alpha) (\omega - i\alpha)} $

      so

      $\displaystyle S_Y^+(\omega) = \sqrt{N_0/2} \frac{\omega - i[\alpha^2 + 2A^2/N_0... ...= \sqrt{N_0/2} \frac{\omega + i[\alpha^2 + 2A^2/N_0]^{1/2}}{ \omega + i\alpha} $

      Now

      $\displaystyle H_2(\omega) \defeq \frac{S_{XY}(\omega)}{S_Y^-(\omega)} = \frac{A... ...omega} \right] e^{-i\omega \lambda} \qquad \text{(partial fraction expansion)} $

      Taking the inverse Laplace transform,

      $\displaystyle h_2(t) = \frac{A^2}{\alpha+\beta}\left[ e^{-\alpha(t-\lambda)} u(t-\lambda) + e^{-\beta(-t+\lambda)}u(-t+\lambda)\right] $


      \begin{picture}(0,0)% \includegraphics{hw10fig4.ps}% \end{picture}
      @font
      picture(4149,1632)(1939,-2731) (3751,-2686)(0,0)[b] $ \lambda$ % (2326,-1636)(0,0)[b] \bgroup\color[rgb]{0,0,0}$ A^2/(\alpha+\beta)$\egroup % (4351,-1936)(0,0)[lb] \bgroup\color[rgb]{0,0,0}$ e^{-alpha(t-\lambda)}$\egroup % (3301,-1936)(0,0)[rb] \bgroup\color[rgb]{0,0,0}$ e^{-beta(-t+\lambda)}$\egroup % (1951,-2686)(0,0)[rb] \bgroup\color[rgb]{0,0,0}$ e^{-\beta \lambda}$\egroup %

      Let

      $\displaystyle H_2(\omega) = \frac{e^{-i \omega \lambda} S_S(\omega)}{S_S(\omega) + S_N(\omega)} $

      (the thing whose $ [ ]_+$ we need to compute. Then

      $\displaystyle H_2(\omega) = e^{-i\omega \lambda} \frac{2A^2}{N_0} \frac{1}{(2A^... ...omega^2} = e^{-i\omega \lambda} \frac{2A^2}{N_0} \frac{1}{\beta^2 + \omega^2}, $

      where $ \beta^2 = 2A^2/N_0 + \alpha^2$ . Then

      \begin{displaymath}h_2(t) = \frac{2A^2}{2N_0 \beta} e^{-\beta\vert t-\lambda\ver... ...\beta} e^{\beta t} e^{-\beta \lambda} & t < \lambda \end{cases}\end{displaymath}< \lambda \end{cases}\end{displaymath}" border="0" height="70" width="346" />

      For $ \lambda > 0$ 0$" align="middle" border="0" height="32" width="46" /> we have delay, so that the filter performs smoothing.

      $\displaystyle h_1(t) = [h_t(2)]_+ = e^{-\alpha(t-\lambda)} u(t-\lambda) + e^{-\beta(-t+\lambda)}[u(-t-\lambda) - u(-t)] \frac{A^2}{\alpha+\beta} $

      Transforming

      $\displaystyle H_1(\omega) = H_2(\omega) - \frac{A^2}{\alpha+\beta} e^{-\beta \lambda} \frac{1}{b-i\omega} \qquad \text{(throw away the noncausal part)} $

      Then

      $\displaystyle H_0(\omega) = \frac{1}{\sqrt{N_0/2}}\frac{\omega-i\alpha}{\omega ... ...ega) - \frac{A^2}{\alpha+\beta} e^{-\beta \lambda} \frac{1}{b-i\omega} \right] $


Copyright 2008, Todd Moon. Cite/attribute Resource. admin. (2006, June 13). Homework Solutions. Retrieved November 23, 2009, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Stochastic_Processes/hw10sol.html. This work is licensed under a Creative Commons License. Creative Commons License
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