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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. The entropy power of a random signal is defined to be the power of band-limited white noise having the same differential entropy rate and bandwidth as the original noise. If a random signal has an entropy H (bits), the power of band-limited white noise having the same entropy rate is given.

Noise is most meaningfulif it is put in perspectiveby comparingto signal strength. In X-ray radiography,the relevant comparison is to contrast. Artifacts It is important to distinguish random noise effects from image artifacts. Although artifacts can appear random, such as the.
Here, we will briefly introduce normal (Gaussian) random processes. We will discuss some examples of Gaussian processes in more detail later on. Many important practical random processes are subclasses of normal random processes. First, let us remember a few facts about Gaussian random.
How quickly our random signal or processes changes with respect to the time function 2. Whether our process has a periodic component and what the expected frequency might be As was mentioned above, the autocorrelation function is simply the expected aluev of a product. Assume we have a pair of random ariablesv from the same process, X 1 = X.

10.2.5 Solved Problems

Problem

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*}

Problem

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

$mu_Y(t)=mu_X(t) ast h(t)$.
$R_{XY}(t_1,t_2)=h(t_2) ast R_X(t_1,t_2)=int_{-infty}^{infty} h(alpha) R_X(t_1,t_2-alpha) ; dalpha$.

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*}

Problem

Airline tycoon deluxe download. 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. Change oem boot logo windows 10. 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

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 $S_X(f_1) geq 0$.

Random Process And Noise

Problem

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.

Find $S_Y(f)$.
Find $R_Y(tau)$.
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*}

10.2.5 Solved Problems

Random Process And Noise

Random Noise Button

Random Process And Noise Cancelling

Random Process And Noise Pollution

Random Processand Noise