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Utah State University
ECE 6010
Stochastic Processes
Homework # 9 Solutions

Suppose $\{ X_{t},t \in \mathbb{R} \}$ is a ramdom process with power spectral density

$\displaystyle S_{X} (\omega) = \frac{1}{(1+ \omega^{2})^{2}}.$

Find the autocorrelation function of $X_{t}$ .

$\displaystyle R_{X}(\tau)$ $\displaystyle =$ $\displaystyle \mathcal{F}^{-1} \{S_{X}(\omega) \} = \mathcal{F}^{-1} \left\{ \f... ...ega^{2})} \right\} * \mathcal{F}^{-1} \left\{ \frac{1}{(1+\omega^{2})} \right\}$

$\displaystyle =$ $\displaystyle \frac{1}{2} e^{-\vert\tau\vert} * \frac{1}{2} e^{-\vert\tau\vert}$

$\displaystyle =$ $\displaystyle \frac{1}{4} \int_{-\infty}^{\infty} e^{-\vert t\vert} e^{-\vert\tau - t\vert} dt$

$\displaystyle =$ $\displaystyle \frac{1}{4} \int_{-\infty}^{0} e^{t} e^{t-\tau} dt + \frac{1}{4... ...{t-\tau} dt + \frac{1}{4} \int_{\tau}^{\infty} e^{-t} e^{\tau -t} dt$ $\displaystyle \mbox{ (for $\tau \geq 0$)}$

$\displaystyle =$ $\displaystyle \frac{1}{4} \left[ \int_{-\infty}^{0} e^{2t - \tau} dt + \int_{0}^{\tau} e^{-\tau} dt + \int_{\tau}^{\infty} e^{t - 2\tau} dt \right]$

$\displaystyle =$ $\displaystyle \frac{1}{4} \left[ \frac{1}{2} e^{-\tau} + \tau e^{-\tau} + \frac{1}{2} e^{-\tau} \right]$

$\displaystyle =$ $\displaystyle \frac{1}{4} \left( \tau e^{-\tau} + e^{-\tau} \right) \$ $\displaystyle \mbox{ (for $\tau \geq 0$)}$

Similarly,

$\displaystyle R_{X}(\tau) = \frac{1}{4} \left( - \tau e^{\tau} + e^{\tau} \right)$ $\displaystyle \mbox{ (for $\tau < 0$)}$ < 0$)}$" align="middle" border="0" height="31" width="83" /> $\displaystyle$

Therefore,

$\displaystyle R_{X}(\tau) = \frac{1}{4} e^{-\vert\tau\vert} ( \vert\tau \vert +1)$
Suppose that $\omega$ is a random variable with p.d.f. $f_{\omega}$ and $\theta$ is a random variable independent of $\omega$ uniformly distributed in $(-\pi,\pi)$ . Define a random process by $X_{t} = a \cos(\omega t + \theta), t\in \mathbb{R}$ where is a constant. Find the power spectral density of $\{X_{t} \}$ .

$\displaystyle E[X_{t_{1}} X_{t_{2}} ]$ $\displaystyle =$ $\displaystyle E \{ a^{2} \cos(\omega t_{1} + \theta) \cos(\omega t_{2} + \theta)$

$\displaystyle =$ $\displaystyle \frac{1}{2} a^{2} E \{ \cos(\omega t_{1}-\omega t_{2}) - \cos(\omega t_{1} + \omega t_{2} + 2\theta) \}$

$\displaystyle =$ $\displaystyle \frac{1}{2} a^{2} \left [ E \{ \cos(\omega t_{1}-\omega t_{2}) \} - \underbrace{E \{ \cos(\omega t_{1} + \omega t_{2} + 2\theta) \}}_{0} \right]$

$\displaystyle =$ $\displaystyle \frac{1}{2} a^{2} E \{ \cos(\omega t_{1}-\omega t_{2}) \}$

$\displaystyle =$ $\displaystyle \frac{1}{2} a^{2} \int_{-\infty}^{\infty} \cos(\tau \omega) f_{\o... ...frac{ e^{j \omega \tau} + e^{-j \omega \tau} } {2} f_{\omega}(\omega) d\omega$

$\displaystyle =$ $\displaystyle \frac{1}{4} a^{2} 2\pi [ \mathcal{F}^{-1} \{ f_{\omega}(\omega) \} + \mathcal{F}^{-1} \{ f_{\omega}(-\omega) \}]$

$\displaystyle =$ $\displaystyle \frac{\pi a^{2}}{2} [ \mathcal{F}^{-1} \{ f_{\omega}(\omega) \} + \mathcal{F}^{-1} \{ f_{\omega}(-\omega) \}]$

Therefore,

$\displaystyle S_{X}(\omega) = \frac{\pi a^{2}}{2} [f_{\omega}(\omega) + f_{\omega}(-\omega)]$
Suppose events occur randomly in $T = [0,\infty)$ in the following way:
1. The numbers of events in nonoverlapping intervals are independent of one another.
2. exactly one event in $(t,t+\Delta t)) = \lambda(t) \Delta t + o(\Delta t)$ , where $\lambda(t)$ is a continuous nonnegative function on $[0,\infty)$ .
3. more than one event in an interval of length $\Delta t) = o(\Delta t).$
Define a random process $\{X_{t}, t \in T \}$ by $X_{0} = 0$ and $x_{t}$ i s the number of events occurring in .
1. For $t> s \geq 0$ s \geq 0$" align="middle" border="0" height="28" width="67" /> , show that $(X_{t} - X_{s})$ is a Poisson random variable with parameter $\int_{z}^{t} \lambda (x) \ dx$ .
  
  k events in $[s,t]) = p_{k}(t,s)$ for $t> s \geq 0$ s \geq 0$" align="middle" border="0" height="28" width="67" /> and $k \geq 0$ .
  
  $\displaystyle p_{k}(t+\Delta t,s)$ $\displaystyle =$ $\displaystyle p_{k}(t,s) (1-\lambda(t) \Delta t + o(\Delta t)) + p_{k-1}(t,s)(\lambda(t) \Delta t + o(\Delta t))+ p_{< k-1}(t,s) o(\Delta t)$ < k-1}(t,s) o(\Delta t)$" align="middle" border="0" height="31" width="531" />
  
  $\displaystyle =$ $\displaystyle p_{k}(t,s) (1-\lambda(t) \Delta t + o(\Delta t)) + p_{k-1}(t,s)(\lambda(t) \Delta t + o(\Delta t))+ o(\Delta t)$
  
  $\displaystyle \frac{d}{dt} p_{k}(t,s) = \lim_{\Delta t \rightarrow 0} \frac{ p_... ... - p_{k}(t,s)}{\Delta t} = - p_{k}(t,s) \lambda (t) + p_{k-1}(t,s) \lambda (t)$
  
  If $p_{k}(t,s) = \frac{e^{-\int_{s}^{t} \lambda(q) dq} \left(\int_{s}^{t} \lambda(q) dq \right)^{k} }{k!}$ satisfies the above equation then we can say that $X_{t}-X_{s}$ is a Poisson random variable with parameter $\int_{s}^{t} \lambda (q) dq$ .
  Now, let $I = \int_{s}^{t} \lambda (q) dq$ and $I' = \lambda (t)$
  Therefore,
  
  $\displaystyle \frac{d}{dt} p_{k}(t,s)$ $\displaystyle =$ $\displaystyle \frac{ -I' e^{I} I^{k} + e^{I} k I^{k-1} I'}{k!}$
  
  $\displaystyle =$ $\displaystyle - \frac{I' e^{I} I^{k}}{k!} + \frac{ e^{I} I^{k-1} I'}{(k-1)!}$
  
  $\displaystyle =$ $\displaystyle \frac{ e^{I} I^{k}}{k!} \lambda (t) + \frac{ e^{I} I^{k-1} }{(k-1)!}\lambda (t)$
  
  $\displaystyle =$ $\displaystyle - p_{k}(t,s) \lambda (t) + p_{k-1}(t,s) \lambda (t)$
  
  So that $p_{k}(t,s)$ does satisfy. And $X_{t}-X_{s}$ is Poisson.
2. Find the mean and autocorrelation functions of $\{X_{t} \}$ .
  
  We have, $E[X_{t}-X_{s}] = \sum_{k=0}^{\infty} k p_{k}(t,s)$ . So,
  
  $\displaystyle E[X_{t}]$ $\displaystyle =$ $\displaystyle E[X_{t}-X_{0}] = \sum_{k=0}^{\infty} k p_{k}(t,0)$
  
  $\displaystyle =$ $\displaystyle \sum_{k=0}^{\infty} k \frac{e^{-\int_{0}^{t} \lambda(q) dq} \left(\int_{0}^{t} \lambda(q) dq \right)^{k} }{k!}$
  
  $\displaystyle =$ $\displaystyle \int_{0}^{t} \lambda(q) dq$
  
  Now,
  
  $\displaystyle E[X_{t}X_{s}]$ $\displaystyle =$ $\displaystyle E[X_{t}(X_{t}+X_{s}-X_{t})] = E[X_{t}^2] + E[X_{t}(X_{s}-X_{t})]$
  
  $\displaystyle =$ $\displaystyle E[X_{t}^2] + E[X_{t}Y_t] = E[X_{t}^2] + E[X_{t}]E[Y_t]$
  
  $\displaystyle =$ $\displaystyle \lambda_{X}^{2} + \lambda_{X} + \lambda_{X} \lambda_{Y} = \lambda_{X}( \lambda_{X}+ 1 +\lambda_{Y} )$
  
  $\displaystyle =$ $\displaystyle \int_{0}^{t} \lambda(q) dq \left[ \int_{0}^{t} \lambda(q) dq + 1 + \int_{t}^{s} \lambda(q) dq \right]$
  
  So,
  
  $\displaystyle R_X(t,s) = \int_{0}^{t} \lambda(q) dq \left[ \int_{0}^{s} \lambda(q) dq + 1 \right]$ $\displaystyle \mbox{ (for $s > t$)}$ t$)}$" align="middle" border="0" height="31" width="80" /> $\displaystyle$
  
  and
  
  $\displaystyle R_X(t,s) = \int_{0}^{s} \lambda(q) dq \left[ \int_{0}^{t} \lambda(q) dq + 1 \right]$ $\displaystyle \mbox{ (for $t > s$)}$ s$)}$" align="middle" border="0" height="31" width="80" /> $\displaystyle$
Suppose that $\{X_t, t \in \mathbb{R} \}$ is a w.s.s., zero-mean, Gaussian random process with auto-correlation function $R_{X}(\tau), \tau \in \mathbb{R}$ and power spectral density $S_X(\omega), \omega \in \mathbb{R}$ . Define the random process $\{Y_{t}, t \in \mathbb{R} \}$ by $Y_{t} = (X_{t})^{2}, t \in \mathbb{R}$ . find the mean, autocorrelation, and powerspectral density of $\{Y_{t}, t \in \mathbb{R} \}$ .

Mean :

$\displaystyle \mu_{y}(t) = E[X_{t}^{2}] = R_{X}(0) = \sigma^{2}$

Autocorrelation :

$\displaystyle R_{Y}(t,s)$ $\displaystyle =$ $\displaystyle E[Y_{t}Y_{s}] = E[X_{t}^{2} X_{s}^{2}]$

$\displaystyle =$ $\displaystyle \left. \frac{\partial ^{4}}{\partial u^{2} \partial v^{2}} \Phi_{X_{t}X_{s}}(u,v) \right\vert _{u=v=0} \frac{1}{i^{4}}$

$\displaystyle \vdots$

$\displaystyle =$ $\displaystyle R_{X}^{2}(0) + 2 R_{X}^{2} (\tau) \quad \quad (\tau = t-s)$

PSD :

$\displaystyle S_{Y}(\omega) = \mathcal{F} \{R_{Y}(\tau) \} = 2 \pi R_{X}^{2}(0) \delta(\omega) - 2 (S_{X}(\omega) *S_{X}(\omega))$
Suppose and are independent random variables with and $\mathrm{var}(U) = \mathrm{var}(V) = 1$ . Define random processes by

$\displaystyle X_t = U \cos t + V \sin t \quad Y_{t} = U \sin t + V \cos t, \quad t \in \mathbb{R}.$

Find the autocorrelation and cross-correlation functions of $\{ X_{t},t \in \mathbb{R} \}$ and $\{Y_{t}, t \in \mathbb{R} \}$ . Are $\{X_{t} \}$ and $\{Y_{t} \}$ jointly wide sense stationary? Are they individually wide sense stationary?

$\displaystyle R_{X}(t,s)$ $\displaystyle =$ $\displaystyle E[X_{t}X_{s}] = E[(U \cos t + V \sin t)(U \cos s + V \sin s) ]$

$\displaystyle =$ $\displaystyle E[U^{2} \cos t \cos s + UV (\cos t \sin s + \sin t \cos s) + V^{2} \sin t \sin s ]$

$\displaystyle =$ $\displaystyle \cos t \cos s E[U^{2}] + E[U] E[V] (\cos t \sin s + \sin t \cos s) + E[V^{2}] \sin t \sin s$

$\displaystyle =$ $\displaystyle \cos t \cos s + \sin t \sin s$

$\displaystyle =$ $\displaystyle \cos(t-s)$

Similarly,

$\displaystyle R_{Y}(t,s) = cos(t-s)$

$\displaystyle R_{XY}(t,s)$ $\displaystyle =$ $\displaystyle E[X_{t}Y_{s}] = E[(U \cos t + V \sin t)(U \sin s + V \cos s) ]$

$\displaystyle =$ $\displaystyle E[ U^{2} \cos t \sin s + UV ( \cos t \cos s + \sin t \sin s) + V^{2} \sin t \cos s]$

$\displaystyle =$ $\displaystyle E[ U^{2}] \cos t \sin s + E[UV] ( \cos t \cos s + \sin t \sin s) E[ V^{2}] \sin t \cos s$

$\displaystyle =$ $\displaystyle \cos t \sin s + \sin t \cos s$

$\displaystyle =$ $\displaystyle \sin (t+s)$

$\displaystyle \mu_{X}(t) = 0 \quad \quad \mu_{Y}(t)$

So, $\{X_{t} \}$ and $\{Y_{t} \}$ are individually WSS, but not jointly WSS.

Copyright 2008, Todd Moon. Cite/attribute Resource. admin. (2006, June 13). Homework Solutions. Retrieved October 12, 2008, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Stochastic_Processes/hw9sol.html. This work is licensed under a Creative Commons License.

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Utah State University
ECE 6010
Stochastic Processes
Homework # 9 Solutions

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Utah State University ECE 6010 Stochastic Processes Homework # 9 Solutions

Utah State University
ECE 6010
Stochastic Processes
Homework # 9 Solutions