text:
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

 * {| style="width:100%" border="0"

$${e}^{(3)}(t)={A}^{(3)}(t)-{{A}_{2}}^{L(3)} \quad$$
 * style="width:95%" |
 * style="width:95%" |

$$
 *  $$ \displaystyle (Eq. 4.4.1)
 * }

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

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

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

 * {| style="width:100%" border="0"

$${A}^{(3)}(t) = {F}^{(2)}(-t) + {F}^{(2)}(t)\quad$$
 * style="width:95%" |
 * style="width:95%" |

$$
 *  $$ \displaystyle (Eq. 4.4.2)
 * }

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

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

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

 * {| style="width:100%" border="0"

$${{A}_{2}}^{L(3)}= \frac{1}{3}[{F}^{(2)}(-t) + {F}^{(2)}(t)] + \frac{1}{3}[{F}^{(2)}(-t) + {F}^{(2)}(t)] + \frac{1}{3}[{F}^{(2)}(-t) + {F}^{(2)}(t)] + \frac{t}{3}[-{F}^{(3)}(-t) + {F}^{(3)}(t)]$$
 * style="width:95%" |
 * style="width:95%" |

$$
 *  $$ \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

 * {| style="width:100%" border="0"

$$\underline{F} = \underline{I} + \triangle \underline{A} $$
 * style="width:95%" |
 * style="width:95%" |

$$
 *  $$ \displaystyle (Eq. 5.2.1)
 * }

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

 * {| style="width:100%" border="0"

$$LSSM: \quad \underline{x}_{k+1} = \underline{F} \quad \underline{x}_{k}$$
 * style="width:95%" |
 * style="width:95%" |

$$
 *  $$ \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

 * {| style="width:100%" border="0"

$$\lim_{k \to \infty}\underline{x}_{k}=\lim_{k \to \infty} \underline^{k+1}\underline{{x}_{0}} = \underline{\hat x}$$
 * style="width:95%" |
 * style="width:95%" |

$$
 *  $$ \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

 * {| style="width:100%" border="0"

$$ LSSMRN: \quad \underline{{x}_{k+1}} = \underline{F} \quad \underline{x_k} + \underline{G} \quad \underline{w}_{k+1}$$
 * style="width:95%" |
 * style="width:95%" |

$$
 *  $$ \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
 * {| style="width:100%" border="0"

$$\underline{F} = \underline{I} + \triangle \underline{A} $$
 * style="width:95%" |
 * style="width:95%" |

$$
 *
 $$ \displaystyle (Eq. 7.0.1)
 * }

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

 * {| style="width:100%" border="0"

$$LSSM: \quad \underline{x}_{k+1} = \underline{F} \quad \underline{x}_{k}$$
 * style="width:95%" |
 * style="width:95%" |

$$
 *
 $$ \displaystyle (Eq. 7.0.2)
 * }

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

Solution

 * {| style="width:100%" border="0"

$$ LSSMRN: \quad \underline{{x}_{k+1}} = \underline{F} \quad \underline{x_k} + \underline{G} \quad \underline{w}_{k+1}$$
 * style="width:95%" |
 * style="width:95%" |

$$
 *
 $$ \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.
 * {| style="width:100%" border="0"

$$ \left\{ {\begin{array}{*{20}{c}} \dot x_1 \\ \dot x_2 \\ \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 0 & 1 \\  -\frac{k}{m} & -\frac{c}{m} \\ \end{array}} \right] \left\{ {\begin{array}{*{20}{c}} x_1 \\ x_2 \\ \end{array}} \right\} + \left[ {\begin{array}{*{20}{c}} 0 & 0 \\  1 & 0 \\ \end{array}} \right] \left\{ {\begin{array}{*{20}{c}} u \\ 0 \\ \end{array}} \right\} $$   $$ (Eq.7.0.2) \ $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

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

title: User:Egm6341.s11.team3.sungsikkim


text: File:Knew.jpg

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: File:Knew.jpg


text:
User:Eml5526.s11.team6.kurth/HW3

Homework 3

=Problem 3.5=

Refer to lecture slide [[media:Fe1.s11.mtg15.djvu|15-1]] for problem assignment.

Find
Show that the equivalent stiffness of a spring aligned in the x direction for the bar of thickness t with a centered square hole shown above is

where E is the Young's modulus and t is the width of the bar.

Solution
First, subdivide the bar as shown below.

This bar can be equivalently modeled as the following system of springs along the x direction.

where the respective element stiffness is given by equation (2.8) (F&B pg. 14)

By similar geometry and $$\displaystyle (Eq. 3.5.2)$$

Through | Hooke's Law, the system can be further simplified by combining the two parallel springs into one equivalent one as shown below

where

Finally, this system of three springs in series is reduced to a single spring along the x direction with stiff $$k$$ given by

Substituting the values of $$\displaystyle (Eq. 3.5.3)$$ into $$\displaystyle (Eq. 3.5.5)$$

Solved by William Kurth

Given
Strong form:

Weak form:

Find
Consider a trial (candidate) solution and weight functions of the form

Find a solution to the weak form, then check if the equilibrium equation and natural boundary condition in the strong form are satisfied.

Solution
From $$\displaystyle (Eq. 3.10.2)$$ $$w(3) = 0$$ to satisfy the homogeneous essential boundary condition.

Likewise, to satisfy the essential boundary condition of the strong form, $$u(3) = 0.001$$

Next we will evaluate the derivatives of the trial solution and the weight function.

Plugging $$\displaystyle (Eq. 3.10.7)$$ and $$\displaystyle (Eq. 3.10.8)$$ into $$\displaystyle (Eq. 3.10.2)$$ and assuming A and E are uniform in the x-direction (i.e. not functions of x):

For legibility, each term in $$\displaystyle (Eq. 3.10.9)$$ will be evaluated separately.

Now, substituting equations $$\displaystyle (Eq. 3.10.10)$$,$$\displaystyle (Eq. 3.10.11)$$, and $$\displaystyle (Eq. 3.10.12)$$ back into $$\displaystyle (Eq. 3.10.9)$$:

Rearranging and isolating $$\beta_1$$ and $$\beta_2$$:

Equation $$\displaystyle (Eq. 3.10.14)$$ must hold for any arbitrary weight function $$w(x)$$, so long as it satisfies the homogeneous essential boundary condition. It then also follows that it must also hold for any arbitrary $$\beta_1$$ and $$\beta_2$$, leading to the following algebraic equation:

Pre-multiplying both sides $$\displaystyle (Eq. 3.10.15)$$ by the inverse of the coefficient matrix on the left side yields:

Substituting $$\displaystyle (Eq. 3.10.17)$$ back into $$\displaystyle (Eq. 3.10.3)$$ yields the quadratically approximated solution to the weak form $$\displaystyle (Eq. 3.10.2)$$.

Substituting $$\displaystyle (Eq. 3.10.19)$$ back into the equilibrium equation from the strong form $$\displaystyle (Eq. 3.10.1)$$

Therefore the equilibrium equation is not satisfied over the whole interval ]1,3[.

Now using $$\displaystyle (Eq. 3.10.19)$$ to check the natural boundary condition

Therefore, the solution $$\displaystyle u^{\text{quad}}(x)$$ satisfies the natural boundary condition only in the case where $$ A = \frac{40}{3}$$.

Solved by William Kurth.

title: User:Eml5526.s11.team6.kurth/HW3


text:
User talk:Jeromemicheal

willing to have notes of medical surgecalnursing notes on power point please try to do so.

title: User talk:Jeromemicheal


text:
Engineering Experience 4: Design a Small Solar Vehicle/Nl/Groep 6

Team Raton Tijl Crauwels Pieter Cuyvers Pieter Dewaele Korneel Pardaens Christoph Smets Benoît Van Lessen Niels Vertessen Nicholas Ververken


title: File:Fe1.s11.team6.kurth.hw3.fig1.jpg


title: File:Fe1.s11.team6.kurth.hw3.fig3.jpg


title: File:Fe1.s11.team6.kurth.hw3.fig4.jpg


title: File:Fe1.s11.team6.kurth.hw3.fig2.jpg