Random Processand Noise

10.2.5 Solved Problems


A stochastic process may involve several related random variables. Common examples include the growth of a bacterial population, an electrical current fluctuating due to thermal noise, or the movement of a gas molecule. Stochastic processes are widely used as mathematical models of systems and phenomena that appear to vary in a random manner. The primary purpose of a Kalman filter is to minimize the effects of observation noise, not process noise. I think the author may be conflating Kalman filtering with Kalman control (where you ARE trying to minimize the effect of process noise). The state does not 'fluctuate' over time, except through the influence of process noise.

In mathematics, the Ornstein–Uhlenbeck process is a stochastic process with applications in financial mathematics and the physical sciences. Its original application in physics was as a model for the velocity of a massive Brownian particle under the influence of friction. It is named after Leonard Ornstein and George Eugene Uhlenbeck. The Ornstein–Uhlenbeck process is a stationary Gauss. The theory of random processes is very elegant and allows for simple analysis of systems with 'noise'. Moreover, even deterministic components such as quantization or rounding errors are modeled as random processes (independent of the actual signal) in order to study system performance. And these approximations are quite good in general. Random signals, also called stochastic signals, contain uncertainty in the parameters that describe them. Because of this uncertainty, mathematical functions cannot be used to precisely describe random signals. Instead, random signals are most often analyzed using statistical techniques that require the treatment of the random parameters of the signal with probability distributions or simple.

Consider a WSS random process $X(t)$ with begin{align} nonumber R_X(tau) =left{ begin{array}{l l} 1- tau & quad -1 leq tau leq 1 & quad 0 & quad text{otherwise} end{array} right. end{align} Find the PSD of $X(t)$, and $E[X(t)^2]$.

  • Solution
    • First, we have begin{align*} E[X(t)^2] &=R_X(0)=1. end{align*} We can write triangular function, $R_X(tau)=Lambda(tau)$, as begin{equation*} R_X(tau)=Pi(tau) ast Pi(tau), end{equation*} where begin{align} nonumber Pi(tau) =left{ begin{array}{l l} 1 & quad -frac{1}{2} leq tau leq frac{1}{2} & quad 0 & quad text{otherwise} end{array} right. end{align} Thus, we conclude begin{align*} S_X(f) &=mathcal{F} {R_X(tau)} &=mathcal{F} {Pi(tau) ast Pi(tau)} &=mathcal{F} {Pi(tau)} cdot mathcal{F} {Pi(tau)} &=big[textrm{sinc} (f)big]^2. end{align*}


Let $X(t)$ be a random process with mean function $mu_X(t)$ and autocorrelation function $R_X(s,t)$ ($X(t)$ is not necessarily a WSS process). Let $Y(t)$ be given by begin{align*} Y(t)&=h(t)ast X(t), end{align*} where $h(t)$ is the impulse response of the system. Show that

  1. $mu_Y(t)=mu_X(t) ast h(t)$.
  2. $R_{XY}(t_1,t_2)=h(t_2) ast R_X(t_1,t_2)=int_{-infty}^{infty} h(alpha) R_X(t_1,t_2-alpha) ; dalpha$.
  • Solution
      1. We have begin{align*} mu_Y(t)=E[Y(t)]&=Eleft[int_{-infty}^{infty} h(alpha)X(t-alpha) ; dalpharight] &=int_{-infty}^{infty} h(alpha)E[X(t-alpha)] ; dalpha &=int_{-infty}^{infty} h(alpha) mu_X(t-alpha) ; dalpha &=mu_X(t) ast h(t). end{align*}
      2. We have begin{align*} R_{XY}(t_1,t_2)=E[X(t_1)Y(t_2)]&=Eleft[X(t_1) int_{-infty}^{infty} h(alpha)X(t_2-alpha) ; dalpharight] &=Eleft[ int_{-infty}^{infty} h(alpha)X(t_1)X(t_2-alpha) ; dalpharight] &=int_{-infty}^{infty} h(alpha)E[X(t_1)X(t_2-alpha)] ; dalpha &=int_{-infty}^{infty} h(alpha) R_X(t_1,t_2-alpha) ; dalpha. end{align*}

Random Process And Noise Pollution


Prove the third part of Theorem 10.2: Let $X(t)$ be a WSS random process and $Y(t)$ be given by begin{align*} Y(t)&=h(t)ast X(t), end{align*} where $h(t)$ is the impulse response of the system. Show that begin{align*} R_{Y}(s,t)=R_Y(s-t)&=int_{-infty}^{infty} int_{-infty}^{infty} h(alpha) h(beta) R_X(s-t-alpha+beta) ; dalpha dbeta. end{align*} Also, show that we can rewrite the above integral as $R_{Y}(tau)=h(tau) ast h(-tau) ast R_X(tau)$.

  • Solution
    • begin{align*} R_{Y}(s,t)&=E[X(s)Y(t)] &=Eleft[ int_{-infty}^{infty} h(alpha)X(s-alpha) ; dalpha int_{-infty}^{infty} h(beta)X(s-beta) ; dbetaright] &= int_{-infty}^{infty} int_{-infty}^{infty} h(alpha) h(beta) E[X(s-alpha)X(t-beta)] ; dalpha ; dbeta &= int_{-infty}^{infty} int_{-infty}^{infty} h(alpha) h(beta) R_X(s-t-alpha+beta) ; dalpha ; dbeta. end{align*} We now compute $h(tau) ast h(-tau) ast R_X(tau)$. First, let $g(tau)=h(tau) ast h(-tau)$. Note that begin{align*} g(tau)&=h(tau) ast h(-tau) &= int_{-infty}^{infty} h(alpha)h(alpha-tau) ; dalpha. end{align*} Thus, we have begin{align*} g(tau) ast R_X(tau)&= int_{-infty}^{infty} g(theta) R_X(theta-tau) ; dtheta &=int_{-infty}^{infty} left[int_{-infty}^{infty} h(alpha)h(alpha-theta) ; dalpha right] R_X(theta-tau) ; dtheta &=int_{-infty}^{infty} int_{-infty}^{infty} h(alpha)h(alpha-theta) R_X(theta-tau) ; dalpha ; dtheta &=int_{-infty}^{infty} int_{-infty}^{infty} h(alpha)h(beta) R_X(alpha-beta-tau) ; dalpha ; dbeta &=int_{-infty}^{infty} int_{-infty}^{infty} h(alpha)h(beta) R_X(tau-alpha+beta) ; dalpha ; dbeta &big(textrm{since } R_X(-tau)=R_X(tau)big). end{align*}

Problem Processand

Let $X(t)$ be a WSS random process. Assuming that $S_X(f)$ is continuous at $f_1$, show that $S_X(f_1) geq 0$.

  • Solution
    • Let $f_1 in mathbb{R}$. Suppose that $X(t)$ goes through an LTI system with the following transfer function begin{align*} H(f)=left{ begin{array}{l l} 1 & quad f_1 lt f lt f_1+Delta & quad 0 & quad text{otherwise} end{array} right. end{align*} where $Delta$ is chosen to be very small. The PSD of $Y(t)$ is given by begin{align*} S_Y(f)=S_X(f) H(f) ^2=left{ begin{array}{l l} S_X(f) & quad f_1 lt f lt f_1+Delta & quad 0 & quad text{otherwise} end{array} right. end{align*} Thus, the power in $Y(t)$ is begin{align*} E[Y(t)^2] & =int_{-infty}^{infty} S_Y(f) ; df &=2int_{f_1}^{f_1+Delta} S_X(f) ; df &approx 2 Delta S_X(f_1). end{align*} Since $E[Y(t)^2] geq 0$, we conclude that $S_X(f_1) geq 0$.

Noise In Communication Process


Let $X(t)$ be a white Gaussian noise with $S_X(f)=frac{N_0}{2}$. Assume that $X(t)$ is input to an LTI system with begin{align*} h(t)=e^{-t}u(t). end{align*} Let $Y(t)$ be the output.

  1. Find $S_Y(f)$.
  2. Find $R_Y(tau)$.
  3. Find $E[Y(t)^2]$.

Random Noise Button

  • Solution
    • First, note that begin{align*} H(f)&=mathcal{F} {h(t)} &=frac{1}{1+j2 pi f}. end{align*}
      1. To find $S_Y(f)$, we can write begin{align*} S_Y(f) &=S_X(f) H(f) ^2 &=frac{N_0/2}{1+(2 pi f)^2}. end{align*}
      2. To find $R_Y(tau)$, we can write begin{align*} R_Y(tau)&=mathcal{F}^{-1}{S_Y(f)} &=frac{N_0}{4}e^{- tau }. end{align*}
      3. We have begin{align*} E[Y(t)^2]&=R_Y(0) &=frac{N_0}{4}. end{align*}

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