User:Egm6341.s11.team3.sungsikkim

= 4.4 Dereivative of $${e}^{(3)}(t)$$ =

Given
$$e(t) = A(t)-{{A}_{2}}^{L}(t)$$

$$A(t) = \int_{-t}^{+t}F(x)dx = \int_{-t}^{k}F(x)dx + \int_{k}^{+t}F(x)dx$$

$${{A}_{2}}^{L}(t) = \frac{t}{3}[F(-t) + 4F(0) + F(t)]$$

Find
$$ Prove \quad $$

$${e}^{(3)}(t)=-\frac{t}{3}[{F}^{(3)}(t) - {F}^{(3)}(-t)] \quad$$

Solution

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$${e}^{(3)}(t)={A}^{(3)}(t)-{{A}_{2}}^{L(3)} \quad$$
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$$
 *  $$ \displaystyle (Eq. 4.4.1)
 * }

$$ Derivative\quad of \quad A(t) \quad $$

$${A}^{(1)}(t) = F(-t) + F(t) \quad$$

$${A}^{(2)}(t) = -{F}^{(1)}(-t) + {F}^{(1)}(t)\quad$$


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$${A}^{(3)}(t) = {F}^{(2)}(-t) + {F}^{(2)}(t)\quad$$
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$$
 *  $$ \displaystyle (Eq. 4.4.2)
 * }

$$Derivative \quad of\quad {{A}_{2}}^{L} \quad$$

$${{A}_{2}}^{L(1)}= \frac{1}{3}[F(-t) + 4F(0) + F(t)] + \frac{t}{3}[-{F}^{(1)}(-t) + {F}^{(1)}(t)] $$

$${{A}_{2}}^{L(2)}= \frac{1}{3}[-{F}^{(1)}(-t) + {F}^{(1)}(t)] + \frac{1}{3}[-{F}^{(1)}(-t) + {F}^{(1)}(t) + \frac{t}{3}[{F}^{(2)}(-t) + {F}^{(2)}(t)]$$


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$${{A}_{2}}^{L(3)}= \frac{1}{3}[{F}^{(2)}(-t) + {F}^{(2)}(t)] + \frac{1}{3}[{F}^{(2)}(-t) + {F}^{(2)}(t)] + \frac{1}{3}[{F}^{(2)}(-t) + {F}^{(2)}(t)] + \frac{t}{3}[-{F}^{(3)}(-t) + {F}^{(3)}(t)]$$
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$$
 *  $$ \displaystyle (Eq. 4.4.3)
 * }

$$Substitute \quad (Eq.4.4.2)\quad and \quad(Eq.4.4.3)\quad into\quad (Eq.4.4.1)\quad$$

$${e}^{(3)}(t)=-\frac{t}{3}[{F}^{(3)}(t) - {F}^{(3)}(-t)]$$

$$
 *  $$ \displaystyle

= 5.2 Linear State Space Model =

= 5.2.1 =

Given

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$$\underline{F} = \underline{I} + \triangle \underline{A} $$
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$$
 *  $$ \displaystyle (Eq. 5.2.1)
 * }

$$\underline{I} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\quad \underline{A} = \begin{pmatrix} -0.2& 1 \\ -1 & -0.2 \end{pmatrix}, \quad \triangle = 0.02 \quad \underline{x_0}= \dbinom{{{x}_{0}}^{1}}{{{x}_{0}}^{2}} $$


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$$LSSM: \quad \underline{x}_{k+1} = \underline{F} \quad \underline{x}_{k}$$
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$$
 *  $$ \displaystyle (Eq. 5.2.2)
 * }

Find
$$ Run \quad LSSM \quad and \quad plot \quad({x}_{j}, j=0,1,...)$$

Solution
$$ By \quad (Eq 5.2.1) \quad $$

$$ \underline{F}=\begin{pmatrix} 0.9960 & 0.200 \\ -0.200 & 0.9960 \end{pmatrix}$$

$$ By \quad (Eq 5.2.2) \quad build \quad code \quad$$

Result is



Given

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$$\lim_{k \to \infty}\underline{x}_{k}=\lim_{k \to \infty} \underline^{k+1}\underline{{x}_{0}} = \underline{\hat x}$$
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$$
 *  $$ \displaystyle (Eq. 5.2.3)
 * }

Find
Find Equilibrium   Point

plot $$\underline{\hat x} $$ as big red dot

$$\underline{{x}_{0}} $$as big blue dot

Solution
By (Eq5.2.3)

Result is



Given

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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. 5.2.4)
 * }

$$\underline{G}= \dbinom{1}{1} \alpha\ $$

where $$\alpha\ = 0.5, 1, 2$$

Find
Plot $$ ( \underline{{x}_{j}}, \quad j=0,1,2,...)$$

Solution
By Eq(5.2.4)

Result is when $$\alpha\ = 0.5$$



when $$\alpha\ = 1$$



when $$\alpha\ = 2$$



Given
Cauchy Random Noise 
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$$\underline{F} = \underline{I} + \triangle \underline{A} $$
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$$
 * <p style="text-align:right"> $$ \displaystyle (Eq. 7.0.1)
 * }

$$\underline{I} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\quad \underline{A} = \begin{pmatrix} -0.2& 1 \\ -1 & -0.2 \end{pmatrix}, \quad \triangle = 0.02 \quad \underline{x_0}= \dbinom{{{x}_{0}}^{1}}{{{x}_{0}}^{2}} $$

<span id="(1)">
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$$LSSM: \quad \underline{x}_{k+1} = \underline{F} \quad \underline{x}_{k}$$
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$$
 * <p style="text-align:right"> $$ \displaystyle (Eq. 7.0.2)
 * }

Find
Plot $$ ( \underline{{x}_{j}}, \quad j=0,1,2,...)$$ in single-slit diffraction

Solution
<span id="(1)">
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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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$$
 * <p style="text-align:right"> $$ \displaystyle (Eq. 7.0.1)
 * }

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. <span id="(1)">
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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 $$x_k$$

Given
Cauchy Random Noise

Find
Plot $$ ( \underline{{x}_{j}}, \quad j=0,1,2,...)$$ in single-slit diffraction

Solution
by (Eq5.2.4) with Cauchy Random Noise

Result is