User:Egm6341.s11.team3/sub7a

= Cauchy Distribution =

Given
Cauchy Random Noise 
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$$\underline{F} = \underline{I} + \triangle \underline{A} $$ $$
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 *  $$ \displaystyle (Eq. 7.0.1)
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$$\underline{I} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\quad \underline{A} = \begin{pmatrix} -0.2& 1 \\ -1 & -0.2 \end{pmatrix}, \quad \triangle = 0.02 \quad \underline{x_0}= \dbinom{{{x}_{0}}^{1}}{{{x}_{0}}^{2}} $$


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$$LSSM: \quad \underline{x}_{k+1} = \underline{F} \quad \underline{x}_{k}$$ $$
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 *  $$ \displaystyle (Eq. 7.0.2)
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Find
Plot $$ ( \underline{{x}_{j}}, \quad j=0,1,2,...)$$ in single-slit diffraction

Solution

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$$ LSSMRN: \quad \underline{{x}_{k+1}} = \underline{F} \quad \underline{x_k} + \underline{G} \quad \underline{w}_{k+1}$$ $$
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 *  $$ \displaystyle (Eq. 7.0.1)
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Result is



Given
An ideal mass-spring-damper system with mass m, spring constant k and viscous damper of damping coefficient c is subject to a force F=u. 
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$$ \left\{ {\begin{array}{*{20}{c}} \dot x_1 \\ \dot x_2 \\ \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 0 & 1 \\  -\frac{k}{m} & -\frac{c}{m} \\ \end{array}} \right] \left\{ {\begin{array}{*{20}{c}} x_1 \\ x_2 \\ \end{array}} \right\} + \left[ {\begin{array}{*{20}{c}} 0 & 0 \\  1 & 0 \\ \end{array}} \right] \left\{ {\begin{array}{*{20}{c}} u \\ 0 \\ \end{array}} \right\} $$   $$ (Eq.7.0.2) \ $$
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Finding
Let $$k=1,/ m=1/2,/ x_0=[0.8,-0.4]^T$$ For $$u=0.5$$ Cauchy noise, and $$c=\frac{3}{2}c_{cr}$$, plot $$x_k$$

Solution
Egm6341.s11.team3.sungsikkim 03:29, 21 April 2011 (UTC)

= =

Given: Verhulst Equation with initial conditions
Verhulst equation:

$$ \frac{d}{dt}x(t) = rx(1 - \frac{x}{x_{max}}) \ $$

$$ x_{max} = 15 \ $$ and $$ r = 1.4 \ $$

with initial condition of $$ x_0 \ $$

for time $$ t \in [0, 10] \ $$

Case 1:

$$ x_0 = 3 < \frac{1}{2} x_{max} \ $$

Case 2:

$$ x_0 = 9 > \frac{1}{2} x_{max} \ $$

Find: By using Hermite - Simpson rule, integrate given Verhulst equation
From given Prof. Vu Quoc's notes for

$$ z_{i+ 1} = z_i + \frac{h/2}{3} \left ( f_i + 4f_{i+\frac{1}{2}}[g(z_i,z_{i+1})],t_{i+\frac{1}{2}} + f_{i+1} \right ) \ $$

integrate this function

Solution
Main reference pages were meeting 38 and 39 notes

From given Prof. Vu Quoc's notes for

$$ z_{i+ 1} = z_i + \frac{h/2}{3} \left ( f_i + 4f_{i+\frac{1}{2}}[g(z_i,z_{i+1})],t_{i+\frac{1}{2}} + f_{i+1} \right ) \ $$

,where

$$ f_i = f(z_i, f_i) \ $$

$$ f_{i+1} = f(z_{i+1}, t_{i+1}) \ $$

$$ f_{i+\frac{1}{2}} = f[g(z_i,z_{i+1},t_{i+\frac{1}{2} })] = f(z_{i+\frac{1}{2}}, t_{i+\frac{1}{2}}) \ $$

,where $$ t_{i+\frac{1}{2}} = t_i + \frac{h}{2} \ $$ and as it was proven in HW 6.6 as well as given by Prof. Vu Quoc's

$$ z_{i+ \frac{1}{2}} = \frac{1}{2}(z_i + z_{i+1}) + \frac{h}{8}(f_i - f_{i+1}) \ $$

,note that $$ h \ $$ is the time step

Also specifying the equation to be of a form $$ F(z_{i+ 1})= 0 \ $$

$$ F(z_{i+ 1})= -z_{i+ 1} + z_i + \frac{h/2}{3} \left ( f_i + 4f([g(z_i,z_{i+1})],t_{i+\frac{1}{2}}) + f_{i+1} \right ) = 0\ $$

and, finally, as mentioned in

$$ z_{i+1}^{(k+1)} = z_{i+1}^{(k)} - \left [ \frac{d}{dz}F(z_{i+1}^{(k)}) \right ]^{-1} F(z_{i+1}^{(k)}) \ $$

in our case, for a given function, we have first order derivative. Therefore

$$ z_{i+1}^{(1)} = z_{i+1} - \left [ \frac{d}{dz}F(z_{i+1}) \right ]^{-1} F(z_{i+1}) \ $$

Then check the condition for convergence as it is shown in

$$ z_{i+1}^{(1)} - z_{i+1} ?< Tolerance \ $$

If condition is not met, choose $$ z_{i+1}^{(1)} = z_{i+2} \ $$ and iterate again to get $$ z_{i+2}^{(1)} \ $$ until the tolerance is met

Egm6341.s11.team3.rakesh 02:24, 21 April 2011 (UTC)

Egm6341.s11.team3.JA 17:00, 21 April 2011 (UTC)

Solution Part 2:
All of the necessary information is given in the problem statement. Therefore, all that is left to do is perform the calculations using the proposed methods. The same MatLAB code from Solution Part 1 may be used, with the exception of the file main.m. This file will be replaced by mainB.m as shown below, where the only difference between these files is that mainB.m uses the angle of attack governed by Equation (7.2.2.B), and mainB does not calculate J or ode45.

Below are the new resulting state variable plots, followed by the new M-File that produced them:



Egm6341.s11.team3.russo 03:50, 20 April 2011 (UTC)