Random Processand Noise
2021年6月12日Download here: http://gg.gg/uyhw9
Gtalk windows download free. If you want to talk with your friends, both you must have Google talk installed, and once you have installed it, you can start talking as long as you want.for FREE! Google Talk is compatible with Apple iChat, Gaim, Adium, Trillian Pro and Psi. Google Talk enables us to quickly and easily talk or send instant messages to our friends for free.
Yes, many DSP texts (as well as Wikipedia’s definition of a discrete-time white noise process) and many people with much higher reputation than me on dsp.SE say that uncorrelatedness suffices for defining a white noise process, and in the case of white Gaussian noise it does because Gaussianity brings in the jointly Gaussian property: a. Random Process and Noise Autumn 2018 Course No.: ECE 704 T Instructor Shahid M Shah, email id: FIRST NAME AT ece.iisc.ernet.in, FIRST NAME DOT nit AT gmail.com Credits 4 Location AB-VI-101 Lecture Hours Announcement.
Looking for Army games to download for free? Here are the top free Army games for PC for 2020, including Zombie Derby: Pixel Survival, Royal Adventure, Arkheim - Realms at War, and more. Free military games for pc windows 7 without graphic card. We collected 98 of the best free online army games. These games include browser games for both your computer and mobile devices, as well as apps for your Android and iOS phones and tablets. They include new army games such as Battalion Commander and top army games such as Forward Assault Remix, War Clicks.
*Because noise simulation spreads over so many diverse disciplines, it is difficult to find a notation agreeable to all participants. This paper, therefore, uses common engineer- ing notation and methodology for the stochastic process and system theory discussions. This notation is consistent with that used in 9, 14, 33, 52, 62.
*Random nosebleeds are always a nuisance. Anticoagulants, like heparin or warfarin, which slow your body’s clot-making process, and antiplatelet drugs, like aspirin, which make it harder for.
*Why this Video is Important? In this video the effect of noise on the system is discussed and how the performance of any digital communication system is degr.10.1.5 Gaussian Random Processes
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 vectors. As we saw before, random variables $X_1$, $X_2$,.., $X_n$ are said to be jointly normal if, for all $a_1$,$a_2$,.., $a_n$ $in mathbb{R}$, the random variablebegin{align}%label{} a_1X_1+a_2X_2+..+a_nX_nend{align}is a normal random variable. Also, a random vectorbegin{equation}nonumber textbf{X} = begin{bmatrix} X_1 %[5pt] X_2 %[5pt] . [-10pt] . [-10pt] . [5pt] X_nend{bmatrix}end{equation}is said to be normal or GaussianRandom Noise Button if the random variables $X_1$, $X_2$,.., $X_n$ are jointly normal. An important property of jointly normal random variables is that their joint PDF is completely determined by their mean and covariance matrices. More specifically, for a normal random vector X with mean $mathbf{m}$ and covariance matrix C, the PDF is given bybegin{align*} f_{mathbf{X}}(mathbf{x})=frac{1}{(2pi)^{frac{n}{2}} sqrt{dettextbf{C}}} exp left{-frac{1}{2} (textbf{x}-textbf{m})^T mathbf{C}^{-1}(textbf{x}-textbf{m}) right}.end{align*}Now, let us define Gaussian random processes.A random process $big{X(t), t in J big}$ is said to be a Gaussian (normal) random process if, for allbegin{align}%label{} & t_1,t_2, dots, t_n in J,end{align}the random variables $X(t_1)$, $X(t_2)$,.., $X(t_n)$ are jointly normal.
Example
Let $X(t)$ be a zero-mean WSS Gaussian process with $R_X(tau)=e^{-tau^2}$, for all $tau in mathbb{R}$.
* Find $Pbig(X(1) lt 1big)$.
* Find $Pbig(X(1)+X(2) lt 1big)$.
*Solution
*
* $X(1)$ is a normal random variable with mean $E[X(1)]=0$ and variance begin{align*}%label{} textrm{Var}big(X(1)big)&=E[X(1)^2] &=R_X(0)=1. end{align*} Thus, begin{align*}%label{} Pbig(X(1) lt 1big)&=Phi left(frac{1-0}{1} right) &=Phi(1) approx 0.84 end{align*}
* Let $Y=X(1)+X(2)$. Then, $Y$ is a normal random variable. We have begin{align*}%label{} EY &=E[X(1)]+E[X(2)] &=0; end{align*} begin{align*}%label{} textrm{Var}(Y) &=textrm{Var}big(X(1)big)+textrm{Var}big(X(2)big)+2 textrm{Cov}big(X(1),X(2)big). end{align*} Note that begin{align*}%label{} textrm{Var}big(X(1)big)&=E[X(1)^2]-E[X(1)]^2 &=R_X(0)- mu_X^2 &=1-0=1=textrm{Var}big(X(2)big); end{align*} begin{align*}%label{} textrm{Cov}big(X(1),X(2)big)&=E[X(1)X(2)]-E[X(1)]E[X(2)] &=R_X(-1)-mu_X^2 &=e^{-1} -0=frac{1}{e}. end{align*} Therefore, begin{align*}%label{} textrm{Var}(Y) &=2+frac{2}{e}. end{align*} We conclude $Y sim N(0,2+frac{2}{e})$. Thus, begin{align*}%label{} Pbig(Y lt 1big)&=Phi left(frac{1-0}{sqrt{2+frac{2}{e}}} right) &=Phi(0.6046) approx 0.73 end{align*}
An important property of normal random processes is that wide-sense stationarity and strict-sense stationarity are equivalent for these processes. More specifically, we can state the following theorem.
Theorem Consider the Gaussian random processes $big{X(t), t in mathbb{R}big}$. If $X(t)$ is WSS, then $X(t)$ is a stationary process.
*Proof
*We need to show that, for all $t_1,t_2,cdots, t_r in mathbb{R}$ and all $Delta in mathbb{R}$, the joint CDF of begin{align}%label{} X(t_1), X(t_2), cdots, X(t_r)end{align} is the same as the joint CDF of begin{align}%label{} X(t_1+Delta), X(t_2+Delta), cdots, X(t_r+Delta).end{align}Since these random variables are jointly Gaussian, it suffices to show that the mean vectors and the covariance matrices are the same. To see this, note that $X(t)$ is a WSS process, sobegin{align}%label{} mu_X(t_i)=mu_X(t_j)=mu_X, quad textrm{for all }i,j,end{align} andbegin{align}%label{} C_X(t_i+Delta,t_j+Delta)=C_X(t_i,t_j)=C_X(t_i-t_j), quad textrm{for all }i,j.end{align}From the above, we conclude that the mean vector and the covariance matrix ofbegin{align}%label{} X(t_1), X(t_2), cdots, X(t_r)end{align} is the same as the mean vector and the covariance matrix ofbegin{align}%label{} X(t_1+Delta), X(t_2+Delta), cdots, X(t_r+Delta).end{align}
Similarly, we can define jointly Gaussian random processes.Two random processes $big{X(t), t in J big}$ and $big{Y(t), t in J’ big}$ are said to be jointly Gaussian (normal), if for allbegin{align}%label{} & t_1,t_2, dots, t_m in J & quad quad textrm{and} & t’_1,t’_2, dots, t’_n in J’,end{align}the random variablesbegin{align}%label{} & X(t_1), X(t_2), cdots, X(t_m), Y(t’_1), Y(t’_2), cdots, Y(t’_n)end{align}are jointly normal.Note that from the properties of jointly normal random variables, we can conclude that if two jointly Gaussian random processes $X(t)$ and $Y(t)$ are uncorrelated, i.e.,begin{align*}%label{} C_{XY}(t_1,t_2)=0, quad textrm{for all }t_1,t_2,end{align*}then $X(t)$ and $Y(t)$ are two independent random processes.
Random Process And Noise Ordinance10.2.5 Solved ProblemsProblem
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
Shokuhou Misaki appears briefly in the arc, taking control of a mind of a college-aged girl wearing a miniskirt Santa Claus costume. After a short banter with Mikoto, Misaki tells Mikoto to combine forces as the two Level 5s of Tokiwadai Middle School in order for them to. This was the name of an apparition who appeared in the last episode of Divergence Eve. There, she asked Misaki if she was truly sure she wanted to proceed. She makes an appearance in the first episode, as the object of Lyar’s pursuit. The daughter of Ryoko and Tenchi, has appeared in a picture with her half-sister, Ryo-ohki’s daughter. In one particular Doujinshi, Tenchi Muyo -if- a pregnant Ryoko, Ayeka, and Ryo-ohki are shown.
* $mu_Y(t)=mu_X(t) ast h(t)$.
* $R_{XY}(t_1,t_2)=h(t_2) ast R_X(t_1,t_2)=int_{-infty}^{infty} h(alpha) R_X(t_1,t_2-alpha) ; dalpha$.
*Solution
*
* 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*}
* 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
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
Let $X(t)$ be a WSS random process. Assuming that $S_X(f)$ is continuous at $f_1$, show that $S_X(f_1) geq 0$. Random Noise Generator
*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 $S_X(f_1) geq 0$.
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. Random Noise Maker
* Find $S_Y(f)$.
* Find $R_Y(tau)$.
* Find $E[Y(t)^2]$.
*Solution
* First, note that begin{align*} H(f)&=mathcal{F} {h(t)} &=frac{1}{1+j2 pi f}. end{align*}
* 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*}
* 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*}
* We have begin{align*} E[Y(t)^2]&=R_Y(0) &=frac{N_0}{4}. end{align*}Random Process And Noise
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Gtalk windows download free. If you want to talk with your friends, both you must have Google talk installed, and once you have installed it, you can start talking as long as you want.for FREE! Google Talk is compatible with Apple iChat, Gaim, Adium, Trillian Pro and Psi. Google Talk enables us to quickly and easily talk or send instant messages to our friends for free.
Yes, many DSP texts (as well as Wikipedia’s definition of a discrete-time white noise process) and many people with much higher reputation than me on dsp.SE say that uncorrelatedness suffices for defining a white noise process, and in the case of white Gaussian noise it does because Gaussianity brings in the jointly Gaussian property: a. Random Process and Noise Autumn 2018 Course No.: ECE 704 T Instructor Shahid M Shah, email id: FIRST NAME AT ece.iisc.ernet.in, FIRST NAME DOT nit AT gmail.com Credits 4 Location AB-VI-101 Lecture Hours Announcement.
Looking for Army games to download for free? Here are the top free Army games for PC for 2020, including Zombie Derby: Pixel Survival, Royal Adventure, Arkheim - Realms at War, and more. Free military games for pc windows 7 without graphic card. We collected 98 of the best free online army games. These games include browser games for both your computer and mobile devices, as well as apps for your Android and iOS phones and tablets. They include new army games such as Battalion Commander and top army games such as Forward Assault Remix, War Clicks.
*Because noise simulation spreads over so many diverse disciplines, it is difficult to find a notation agreeable to all participants. This paper, therefore, uses common engineer- ing notation and methodology for the stochastic process and system theory discussions. This notation is consistent with that used in 9, 14, 33, 52, 62.
*Random nosebleeds are always a nuisance. Anticoagulants, like heparin or warfarin, which slow your body’s clot-making process, and antiplatelet drugs, like aspirin, which make it harder for.
*Why this Video is Important? In this video the effect of noise on the system is discussed and how the performance of any digital communication system is degr.10.1.5 Gaussian Random Processes
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 vectors. As we saw before, random variables $X_1$, $X_2$,.., $X_n$ are said to be jointly normal if, for all $a_1$,$a_2$,.., $a_n$ $in mathbb{R}$, the random variablebegin{align}%label{} a_1X_1+a_2X_2+..+a_nX_nend{align}is a normal random variable. Also, a random vectorbegin{equation}nonumber textbf{X} = begin{bmatrix} X_1 %[5pt] X_2 %[5pt] . [-10pt] . [-10pt] . [5pt] X_nend{bmatrix}end{equation}is said to be normal or GaussianRandom Noise Button if the random variables $X_1$, $X_2$,.., $X_n$ are jointly normal. An important property of jointly normal random variables is that their joint PDF is completely determined by their mean and covariance matrices. More specifically, for a normal random vector X with mean $mathbf{m}$ and covariance matrix C, the PDF is given bybegin{align*} f_{mathbf{X}}(mathbf{x})=frac{1}{(2pi)^{frac{n}{2}} sqrt{dettextbf{C}}} exp left{-frac{1}{2} (textbf{x}-textbf{m})^T mathbf{C}^{-1}(textbf{x}-textbf{m}) right}.end{align*}Now, let us define Gaussian random processes.A random process $big{X(t), t in J big}$ is said to be a Gaussian (normal) random process if, for allbegin{align}%label{} & t_1,t_2, dots, t_n in J,end{align}the random variables $X(t_1)$, $X(t_2)$,.., $X(t_n)$ are jointly normal.
Example
Let $X(t)$ be a zero-mean WSS Gaussian process with $R_X(tau)=e^{-tau^2}$, for all $tau in mathbb{R}$.
* Find $Pbig(X(1) lt 1big)$.
* Find $Pbig(X(1)+X(2) lt 1big)$.
*Solution
*
* $X(1)$ is a normal random variable with mean $E[X(1)]=0$ and variance begin{align*}%label{} textrm{Var}big(X(1)big)&=E[X(1)^2] &=R_X(0)=1. end{align*} Thus, begin{align*}%label{} Pbig(X(1) lt 1big)&=Phi left(frac{1-0}{1} right) &=Phi(1) approx 0.84 end{align*}
* Let $Y=X(1)+X(2)$. Then, $Y$ is a normal random variable. We have begin{align*}%label{} EY &=E[X(1)]+E[X(2)] &=0; end{align*} begin{align*}%label{} textrm{Var}(Y) &=textrm{Var}big(X(1)big)+textrm{Var}big(X(2)big)+2 textrm{Cov}big(X(1),X(2)big). end{align*} Note that begin{align*}%label{} textrm{Var}big(X(1)big)&=E[X(1)^2]-E[X(1)]^2 &=R_X(0)- mu_X^2 &=1-0=1=textrm{Var}big(X(2)big); end{align*} begin{align*}%label{} textrm{Cov}big(X(1),X(2)big)&=E[X(1)X(2)]-E[X(1)]E[X(2)] &=R_X(-1)-mu_X^2 &=e^{-1} -0=frac{1}{e}. end{align*} Therefore, begin{align*}%label{} textrm{Var}(Y) &=2+frac{2}{e}. end{align*} We conclude $Y sim N(0,2+frac{2}{e})$. Thus, begin{align*}%label{} Pbig(Y lt 1big)&=Phi left(frac{1-0}{sqrt{2+frac{2}{e}}} right) &=Phi(0.6046) approx 0.73 end{align*}
An important property of normal random processes is that wide-sense stationarity and strict-sense stationarity are equivalent for these processes. More specifically, we can state the following theorem.
Theorem Consider the Gaussian random processes $big{X(t), t in mathbb{R}big}$. If $X(t)$ is WSS, then $X(t)$ is a stationary process.
*Proof
*We need to show that, for all $t_1,t_2,cdots, t_r in mathbb{R}$ and all $Delta in mathbb{R}$, the joint CDF of begin{align}%label{} X(t_1), X(t_2), cdots, X(t_r)end{align} is the same as the joint CDF of begin{align}%label{} X(t_1+Delta), X(t_2+Delta), cdots, X(t_r+Delta).end{align}Since these random variables are jointly Gaussian, it suffices to show that the mean vectors and the covariance matrices are the same. To see this, note that $X(t)$ is a WSS process, sobegin{align}%label{} mu_X(t_i)=mu_X(t_j)=mu_X, quad textrm{for all }i,j,end{align} andbegin{align}%label{} C_X(t_i+Delta,t_j+Delta)=C_X(t_i,t_j)=C_X(t_i-t_j), quad textrm{for all }i,j.end{align}From the above, we conclude that the mean vector and the covariance matrix ofbegin{align}%label{} X(t_1), X(t_2), cdots, X(t_r)end{align} is the same as the mean vector and the covariance matrix ofbegin{align}%label{} X(t_1+Delta), X(t_2+Delta), cdots, X(t_r+Delta).end{align}
Similarly, we can define jointly Gaussian random processes.Two random processes $big{X(t), t in J big}$ and $big{Y(t), t in J’ big}$ are said to be jointly Gaussian (normal), if for allbegin{align}%label{} & t_1,t_2, dots, t_m in J & quad quad textrm{and} & t’_1,t’_2, dots, t’_n in J’,end{align}the random variablesbegin{align}%label{} & X(t_1), X(t_2), cdots, X(t_m), Y(t’_1), Y(t’_2), cdots, Y(t’_n)end{align}are jointly normal.Note that from the properties of jointly normal random variables, we can conclude that if two jointly Gaussian random processes $X(t)$ and $Y(t)$ are uncorrelated, i.e.,begin{align*}%label{} C_{XY}(t_1,t_2)=0, quad textrm{for all }t_1,t_2,end{align*}then $X(t)$ and $Y(t)$ are two independent random processes.
Random Process And Noise Ordinance10.2.5 Solved ProblemsProblem
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
Shokuhou Misaki appears briefly in the arc, taking control of a mind of a college-aged girl wearing a miniskirt Santa Claus costume. After a short banter with Mikoto, Misaki tells Mikoto to combine forces as the two Level 5s of Tokiwadai Middle School in order for them to. This was the name of an apparition who appeared in the last episode of Divergence Eve. There, she asked Misaki if she was truly sure she wanted to proceed. She makes an appearance in the first episode, as the object of Lyar’s pursuit. The daughter of Ryoko and Tenchi, has appeared in a picture with her half-sister, Ryo-ohki’s daughter. In one particular Doujinshi, Tenchi Muyo -if- a pregnant Ryoko, Ayeka, and Ryo-ohki are shown.
* $mu_Y(t)=mu_X(t) ast h(t)$.
* $R_{XY}(t_1,t_2)=h(t_2) ast R_X(t_1,t_2)=int_{-infty}^{infty} h(alpha) R_X(t_1,t_2-alpha) ; dalpha$.
*Solution
*
* 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*}
* 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
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
Let $X(t)$ be a WSS random process. Assuming that $S_X(f)$ is continuous at $f_1$, show that $S_X(f_1) geq 0$. Random Noise Generator
*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 $S_X(f_1) geq 0$.
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. Random Noise Maker
* Find $S_Y(f)$.
* Find $R_Y(tau)$.
* Find $E[Y(t)^2]$.
*Solution
* First, note that begin{align*} H(f)&=mathcal{F} {h(t)} &=frac{1}{1+j2 pi f}. end{align*}
* 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*}
* 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*}
* We have begin{align*} E[Y(t)^2]&=R_Y(0) &=frac{N_0}{4}. end{align*}Random Process And Noise
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