Class Wiki - User contributions [en]

Laplace transforms: Under-damped Mass-Spring System on an Incline

2010-09-27T22:42:05Z

Brandon.plubell:

[[User:Brandon.plubell|Brandon.plubell]] 03:26, 9 November 2009 (UTC)

=Part 1 - Use Laplace Transformations=
By Brandon Plubell

==Problem Statement==
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

==Initial Conditions and Values==
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

==Force Equations==
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

==Laplace Transform==
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

==Inverse Laplace Transform==
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

==Equation of Motion==
Putting it all back together again gives,

<math>
x(t) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Media:BP_BodePlotOctave-1.zip|Octave Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

=Part 2 - Final and Initial Value Theorems=

==Initial Value Theorem==
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

==Final Value Theorem==
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

=Part 3 - Bode Plot=
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

=Part 4 - Breakpoints and Asymptotes on Bode Plot=
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

=Part 5 - Convolution=
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point). Questions left: How could the result be adjusted to account for initial conditions?

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

(Laplace in blue solid, Convolution in red dotted)

=Part 6 - State Equation=

Choose the state variable to be <math> x\frac{}{} </math> and <math> \dot{x} </math>, then following the example from class, the state equation is

<math>
\begin{bmatrix}
\dot{x} \\
\ddot{x}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 \\
\frac{-k}{m} & \frac{-\lambda}{m}
\end{bmatrix}
\,
\begin{bmatrix}
x \\
\dot{x}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
1
\end{bmatrix}
\, g \, \sin{\theta}
</math>

=Appendix A=

==Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-12-13T23:25:12Z

Brandon.plubell:

=Setup State Space Equation=

==Problem Statement==
Find an input function such that the lower mass, <math>m_1</math>, is stationary in the steady state. Find the equation of motion for the upper mass, <math>m_2</math>.

The use of one spring between the masses is just a simplification of a multi-spring system, so the possibility of being off-kilter is neglected and just the vertical forces are considered.

[[Image:BP_Setup-2.jpg|right|Problem Setup]]

==Initial Conditions and Values==

<math>m_1 = 140 kg \frac{}{}</math>

<math>k_1 = 108000 \frac{N}{m}</math>

<math>m_2 = 80 kg \frac{}{}</math>

<math>k_2 = 80000 \frac{N}{m}</math>

==Force Equations==

[[Image:BP_FBD2-2.jpg|right|FBD for m1]]

[[Image:BP_FBD1-2.jpg|right|FBD for m1]]

Sum of the forces in the x direction yields

For <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_2} = m_2 \ddot{x}_2
</math>

<math>
\Rightarrow\ m_2 \ddot{x}_2=F_{s_2} - m_2 g
</math>

Since
<math>
F_{s_2} = k_2(L_2+x_1-x_2)
</math>

<math>
\Rightarrow\ \ddot{x}_2=-g + \frac{k_2}{m_2} \, L_2 + \frac{k_2}{m_2} \, x_1 - \frac{k_2}{m_2} \, x_2
</math>

And for <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_1} = m_1 \ddot{x}_1
</math>

<math>
\Rightarrow\ m_1 \ddot{x}_1=F_{s_1} + F - m_1 g - F_{s_2}
</math>

Since
<math>
F_{s_1} = k_1(L_1-x_1)
</math>

Where
<math>F \frac{}{} = F(t)</math> is the input force

<math>
\Rightarrow\ \ddot{x}_1 = - g +\frac{1}{m_1} \, F + \frac{k_1}{m_1} \, L_1 - \frac{k_1}{m_1} \, x_1
-\frac{k_2}{m_1} \, L_2 - \frac{k_2}{m_1} \, x_1 + \frac{k_2}{m_1} \, x_2
</math>

==State Space Equation==
The general form of the state equation is

<math>
\underline{\dot{x}}(t) = \widehat{A} \, \underline{x}(t) + \widehat{C} \, \underline{u}(t)
</math>

Where <math>\widehat{M}</math> denotes a matrix and <math>\underline{v}</math> denotes a vector.

Let <math>x_1 \frac{}{}</math>, <math>\dot{x}_1 \frac{}{}</math>, <math>x_2 \frac{}{}</math>, and <math>\dot{x_2} \frac{}{}</math> be the state variables, then

<math>

\begin{bmatrix}
\dot{x}_1 \\
\ddot{x}_1 \\
\dot{x}_2 \\
\ddot{x}_2
\end{bmatrix}

=

</math>
<math>

\begin{bmatrix}
0 & 1 & 0 & 0 \\
\frac{-k_1}{m_1}-\frac{k_2}{m_1} & 0 & \frac{k_2}{m_1} & 0 \\
0 & 0 & 0 & 1 \\
\frac{k_2}{m_2} & 0 & \frac{-k_2}{m_2} & 0
\end{bmatrix}

\begin{bmatrix}
x_1 \\
\dot{x}_1 \\
x_2 \\
\dot{x}_2
\end{bmatrix}

+

\begin{bmatrix}
0 & 0 & 0 & 0 \\
-1 & \frac{k_1}{m_1} & \frac{-k_2}{m_1} & \frac{1}{m_1} \\
0 & 0 & 0 & 0 \\
-1 & 0 & \frac{k_2}{m_2} & 0
\end{bmatrix}

\begin{bmatrix}
g \\
L_1 \\
L_2 \\
F
\end{bmatrix}

</math>

=Solve Using Laplace Transform Method=

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-20T07:56:07Z

Brandon.plubell:

=Setup State Space Equation=

==Problem Statement==
Find an input function such that the lower mass, <math>m_1</math>, is stationary in the steady state. Find the equation of motion for the upper mass, <math>m_2</math>.

The use of one spring between the masses is just a simplification of a multi-spring system, so the possibility of being off-kilter is neglected and just the vertical forces are considered.

[[Image:BP_Setup-2.jpg|right|Problem Setup]]

==Initial Conditions and Values==

<math>m_1 = 140 kg \frac{}{}</math>

<math>k_1 = 108000 \frac{N}{m}</math>

<math>m_2 = 80 kg \frac{}{}</math>

<math>k_2 = 80000 \frac{N}{m}</math>

Let the initial conditions be zero for the time being.

==Force Equations==

[[Image:BP_FBD2-2.jpg|right|FBD for m1]]

[[Image:BP_FBD1-2.jpg|right|FBD for m1]]

Sum of the forces in the x direction yields

For <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_2} = m_2 \ddot{x}_2
</math>

<math>
\Rightarrow\ m_2 \ddot{x}_2=F_{s_2} - m_2 g
</math>

Since
<math>
F_{s_2} = k_2(L_2+x_1-x_2)
</math>

<math>
\Rightarrow\ \ddot{x}_2=-g + \frac{k_2}{m_2} \, L_2 + \frac{k_2}{m_2} \, x_1 - \frac{k_2}{m_2} \, x_2
</math>

And for <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_1} = m_1 \ddot{x}_1
</math>

<math>
\Rightarrow\ m_1 \ddot{x}_1=F_{s_1} + F - m_1 g - F_{s_2}
</math>

Since
<math>
F_{s_1} = k_1(L_1-x_1)
</math>

Where
<math>F \frac{}{} = F(t)</math> is the input force

<math>
\Rightarrow\ \ddot{x}_1 = - g +\frac{1}{m_1} \, F + \frac{k_1}{m_1} \, L_1 - \frac{k_1}{m_1} \, x_1
-\frac{k_2}{m_1} \, L_2 - \frac{k_2}{m_1} \, x_1 + \frac{k_2}{m_1} \, x_2
</math>

==State Space Equation==
The general form of the state equation is

<math>
\underline{\dot{x}}(t) = \widehat{A} \, \underline{x}(t) + \widehat{C} \, \underline{u}(t)
</math>

Where <math>\widehat{M}</math> denotes a matrix and <math>\underline{v}</math> denotes a vector.

Let <math>x_1 \frac{}{}</math>, <math>\dot{x}_1 \frac{}{}</math>, <math>x_2 \frac{}{}</math>, and <math>\dot{x_2} \frac{}{}</math> be the state variables, then

<math>

\begin{bmatrix}
\dot{x}_1 \\
\ddot{x}_1 \\
\dot{x}_2 \\
\ddot{x}_2
\end{bmatrix}

=

</math>
<math>

\begin{bmatrix}
0 & 1 & 0 & 0 \\
\frac{-k_1}{m_1}-\frac{k_2}{m_1} & 0 & \frac{k_2}{m_1} & 0 \\
0 & 0 & 0 & 1 \\
\frac{k_2}{m_2} & 0 & \frac{-k_2}{m_2} & 0
\end{bmatrix}

\begin{bmatrix}
x_1 \\
\dot{x}_1 \\
x_2 \\
\dot{x}_2
\end{bmatrix}

+

\begin{bmatrix}
0 & 0 & 0 & 0 \\
-1 & \frac{k_1}{m_1} & \frac{-k_2}{m_1} & \frac{1}{m_1} \\
0 & 0 & 0 & 0 \\
-1 & 0 & \frac{k_2}{m_2} & 0
\end{bmatrix}

\begin{bmatrix}
g \\
L_1 \\
L_2 \\
F
\end{bmatrix}

</math>

=Solve Using Laplace Transform Method=

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-20T07:53:57Z

Brandon.plubell:

=Setup State Space Equation=

==Problem Statement==
Find an input function such that the lower mass, <math>m_1</math>, is stationary in the steady state. Find the equation of motion for the upper mass, <math>m_2</math>.

The use of one spring between the masses is just a simplification of a multi-spring system, so the possibility of being off-kilter is neglected and just the vertical forces are considered.

[[Image:BP_Setup-2.jpg|right|Problem Setup]]

==Initial Conditions and Values==

<math>m_1 = 140 kg \frac{}{}</math>

<math>k_1 = 108000 \frac{N}{m}</math>

<math>m_2 = 80 kg \frac{}{}</math>

<math>k_2 = 80000 \frac{N}{m}</math>

Let the initial conditions be zero for the time being.

==Force Equations==

[[Image:BP_FBD2-2.jpg|right|FBD for m1]]

[[Image:BP_FBD1-2.jpg|right|FBD for m1]]

Sum of the forces in the x direction yields

For <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_2} = m_2 \ddot{x}_2
</math>

<math>
\Rightarrow\ m_2 \ddot{x}_2=F_{s_2} - m_2 g
</math>

Since
<math>
F_{s_2} = k_2(L_2+x_1-x_2)
</math>

<math>
\Rightarrow\ \ddot{x}_2=-g + \frac{k_2}{m_2} \, L_2 + \frac{k_2}{m_2} \, x_1 - \frac{k_2}{m_2} \, x_2
</math>

And for <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_1} = m_1 \ddot{x}_1
</math>

<math>
\Rightarrow\ m_1 \ddot{x}_1=F_{s_1} + F - m_1 g - F_{s_2}
</math>

Since
<math>
F_{s_1} = k_1(L_1-x_1)
</math>

Where
<math>F \frac{}{} = F(t)</math> is the input force

<math>
\Rightarrow\ \ddot{x}_1 = - g +\frac{1}{m_1} \, F + \frac{k_1}{m_1} \, L_1 - \frac{k_1}{m_1} \, x_1
-\frac{k_2}{m_1} \, L_2 - \frac{k_2}{m_1} \, x_1 + \frac{k_2}{m_1} \, x_2
</math>

==State Space Equation==
The general form of the state equation is

<math>
\underline{\dot{x}}(t) = \widehat{A} \, \underline{x}(t) + \widehat{C} \, \underline{u}(t)
</math>

Where <math>\widehat{M}</math> denotes a matrix and <math>\underline{v}</math> denotes a vector.

Let <math>x_1 \frac{}{}</math>, <math>\dot{x}_1 \frac{}{}</math>, <math>x_2 \frac{}{}</math>, and <math>\dot{x_2} \frac{}{}</math> be the state variables, then

<math>

\begin{bmatrix}
\dot{x}_1 \\
\ddot{x}_1 \\
\dot{x}_2 \\
\ddot{x}_2
\end{bmatrix}

=

</math>
<math>

\begin{bmatrix}
0 & 1 & 0 & 0 \\
\frac{-k_1}{m_1} & 0 & \frac{k_2}{m_1} & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & \frac{-k_2}{m_2} & 0
\end{bmatrix}

\begin{bmatrix}
x_1 \\
\dot{x}_1 \\
x_2 \\
\dot{x}_2
\end{bmatrix}

+

\begin{bmatrix}
0 & 0 & 0 & 0 \\
-1 & \frac{k_1}{m_1} & \frac{-k_2}{m_1} & \frac{1}{m_1} \\
0 & 0 & 0 & 0 \\
-1 & 0 & \frac{k_2}{m_2} & 0
\end{bmatrix}

\begin{bmatrix}
g \\
L_1 \\
L_2 \\
F
\end{bmatrix}

</math>

=Solve Using Laplace Transform Method=

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-20T07:43:56Z

Brandon.plubell:

=Setup State Space Equation=

==Problem Statement==
Find an input function such that the lower mass, <math>m_1</math>, is stationary in the steady state. Find the equation of motion for the upper mass, <math>m_2</math>.

The use of one spring between the masses is just a simplification of a multi-spring system, so the possibility of being off-kilter is neglected and just the vertical forces are considered.

[[Image:BP_Setup-2.jpg|right|Problem Setup]]

==Initial Conditions and Values==

<math>m_1 = 140 kg \frac{}{}</math>

<math>k_1 = 108000 \frac{N}{m}</math>

<math>m_2 = 80 kg \frac{}{}</math>

<math>k_2 = 80000 \frac{N}{m}</math>

Let the initial conditions be zero for the time being.

==Force Equations==

[[Image:BP_FBD2-2.jpg|right|FBD for m1]]

[[Image:BP_FBD1-2.jpg|right|FBD for m1]]

Sum of the forces in the x direction yields

For <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_2} = m_2 \ddot{x}_2
</math>

<math>
\Rightarrow\ m_2 \ddot{x}_2=F_{s_2} - m_2 g
</math>

Since
<math>
F_s = -k \, x
</math>

<math>
\Rightarrow\ \ddot{x}_2=\frac{-k_2}{m_2} \, x_2 - g
</math>

And for <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_1} = m_1 \ddot{x}_1
</math>

<math>
\Rightarrow\ m_1 \ddot{x}_1=F_{s_1} + F - m_1 g - F_{s_2}
</math>

Where
<math>F \frac{}{} = F(t)</math> is the input force

<math>
\Rightarrow\ \ddot{x}_1 = \frac{-k_1}{m_1} \, x_1 + \frac{k_2}{m_1} \, x_2 - g + \frac{F}{m_1}
</math>

==State Space Equation==
The general form of the state equation is

<math>
\underline{\dot{x}}(t) = \widehat{A} \, \underline{x}(t) + \widehat{C} \, \underline{u}(t)
</math>

Where <math>\widehat{M}</math> denotes a matrix and <math>\underline{v}</math> denotes a vector.

Let <math>x_1 \frac{}{}</math>, <math>\dot{x}_1 \frac{}{}</math>, <math>x_2 \frac{}{}</math>, and <math>\dot{x_2} \frac{}{}</math> be the state variables, then

<math>

\begin{bmatrix}
\dot{x}_1 \\
\ddot{x}_1 \\
\dot{x}_2 \\
\ddot{x}_2
\end{bmatrix}

=

</math>
<math>

\begin{bmatrix}
0 & 1 & 0 & 0 \\
\frac{-k_1}{m_1} & 0 & \frac{k_2}{m_1} & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & \frac{-k_2}{m_2} & 0
\end{bmatrix}

\begin{bmatrix}
x_1 \\
\dot{x}_1 \\
x_2 \\
\dot{x}_2
\end{bmatrix}

+

\begin{bmatrix}
0 & 0 & 0 & 0 \\
-1 & \frac{1}{m_1} & 0 & 0 \\
0 & 0 & 0 & 0 \\
-1 & 0 & 0 & 0
\end{bmatrix}

\begin{bmatrix}
g \\
F \\
0 \\
0
\end{bmatrix}

</math>

=Solve Using Laplace Transform Method=

Class Wiki:About

2009-11-19T06:27:00Z

Brandon.plubell:

The purpose of ClassWiki is to allow students and the instructor to create documentation that my assist other students.

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-19T03:26:07Z

Brandon.plubell:

=Setup State Space Equation=

==Problem Statement==
Find an input function such that the lower mass, <math>m_1</math>, is stationary in the steady state. Find the equation of motion for the upper mass, <math>m_2</math>.

The use of one spring between the masses is just a simplification of a multi-spring system, so the possibility of being off-kilter is neglected and just the vertical forces are considered.

[[Image:BP_Setup-2.jpg|right|Problem Setup]]

==Initial Conditions and Values==

<math>m_1 = 140 kg \frac{}{}</math>

<math>k_1 = 108000 \frac{N}{m}</math>

<math>m_2 = 80 kg \frac{}{}</math>

<math>k_2 = 80000 \frac{N}{m}</math>

Let the initial conditions be zero for the time being.

==Force Equations==

[[Image:BP_FBD2-2.jpg|right|FBD for m2]]

[[Image:BP_FBD1-2.jpg|right|FBD for m1]]

Sum of the forces in the x direction yields

For <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_2} = m_2 \ddot{x}_2
</math>

<math>
\Rightarrow\ m_2 \ddot{x}_2=F_{s_2} - m_2 g
</math>

Since
<math>
F_s = -k \, x
</math>

<math>
\Rightarrow\ \ddot{x}_2=\frac{-k_2}{m_2} \, x_2 - g
</math>

And for <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_1} = m_1 \ddot{x}_1
</math>

<math>
\Rightarrow\ m_1 \ddot{x}_1=F_{s_1} + F - m_1 g - F_{s_2}
</math>

Where
<math>F \frac{}{} = F(t)</math> is the input force

<math>
\Rightarrow\ \ddot{x}_1 = \frac{-k_1}{m_1} \, x_1 + \frac{k_2}{m_1} \, x_2 - g + \frac{F}{m_1}
</math>

==State Space Equation==
The general form of the state equation is

<math>
\underline{\dot{x}}(t) = \widehat{A} \, \underline{x}(t) + \widehat{C} \, \underline{u}(t)
</math>

Where <math>\widehat{M}</math> denotes a matrix and <math>\underline{v}</math> denotes a vector.

Let <math>x_1 \frac{}{}</math>, <math>\dot{x}_1 \frac{}{}</math>, <math>x_2 \frac{}{}</math>, and <math>\dot{x_2} \frac{}{}</math> be the state variables, then

<math>

\begin{bmatrix}
\dot{x}_1 \\
\ddot{x}_1 \\
\dot{x}_2 \\
\ddot{x}_2
\end{bmatrix}

=

</math>
<math>

\begin{bmatrix}
0 & 1 & 0 & 0 \\
\frac{-k_1}{m_1} & 0 & \frac{k_2}{m_1} & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & \frac{-k_2}{m_2} & 0
\end{bmatrix}

\begin{bmatrix}
x_1 \\
\dot{x}_1 \\
x_2 \\
\dot{x}_2
\end{bmatrix}

+

\begin{bmatrix}
0 & 0 & 0 & 0 \\
-1 & \frac{1}{m_1} & 0 & 0 \\
0 & 0 & 0 & 0 \\
-1 & 0 & 0 & 0
\end{bmatrix}

\begin{bmatrix}
g \\
F \\
0 \\
0
\end{bmatrix}

</math>

=Solve Using Laplace Transform Method=

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-19T03:25:41Z

Brandon.plubell:

=Setup State Space Equation=

==Problem Statement==
Find an input function such that the lower mass, <math>m_1</math>, is stationary in the steady state. Find the equation of motion for the upper mass, <math>m_2</math>.

The use of one spring between the masses is just a simplification of a multi-spring system, so the possibility of being off-kilter is neglected and just the vertical forces are considered.

[[Image:BP_Setup-2.jpg|right|Problem Setup]]

==Initial Conditions and Values==

<math>m_1 = 140 kg \frac{}{}</math>

<math>k_1 = 108000 \frac{N}{m}</math>

<math>m_2 = 80 kg \frac{}{}</math>

<math>k_2 = 80000 \frac{N}{m}</math>

Let the initial conditions be zero for the time being.

==Force Equations==

[[Image:BP_FBD2-2.jpg|right|FBD for m2]]

[[Image:BP_FBD1-2.jpg|right|FBD for m1]]

Sum of the forces in the x direction yields

For <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_2} = m_2 \ddot{x}_2
</math>

<math>
\Rightarrow\ m_2 \ddot{x}_2=F_{s_2} - m_2 g
</math>

Since
<math>
F_s = -k \, x
</math>

<math>
\Rightarrow\ \ddot{x}_2=\frac{-k_2}{m_2} \, x_2 - g
</math>

And for <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_1} = m_1 \ddot{x}_1
</math>

<math>
\Rightarrow\ m_1 \ddot{x}_1=F_{s_1} + F - m_1 g - F_{s_2}
</math>

Where
<math>F \frac{}{} = F(t)</math> is the input force

<math>
\Rightarrow\ \ddot{x}_1 = \frac{-k_1}{m_1} \, x_1 + \frac{k_2}{m_1} \, x_2 - g + \frac{F}{m_1}
</math>

==State Space Equation==
The general form of the state equation is

<math>
\underline{\dot{x}}(t) = \widehat{A} \, \underline{x}(t) + \widehat{C} \, \underline{u}(t)
</math>

Where <math>\widehat{M}</math> denotes a matrix and <math>\underline{v}</math> denotes a vector.

Let <math>x_1 \frac{}{}</math>, <math>\dot{x}_1 \frac{}{}</math>, <math>x_2 \frac{}{}</math>, and <math>\dot{x_2} \frac{}{}</math> be the state variables, then

<math>

\begin{bmatrix}
\dot{x}_1 \\
\ddot{x}_1 \\
\dot{x}_2 \\
\ddot{x}_2
\end{bmatrix}

=

</math>
<math>

\begin{bmatrix}
0 & 1 & 0 & 0 \\
\frac{-k_1}{m_1} & 0 & \frac{k_2}{m_1} & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & \frac{-k_2}{m_2} & 0
\end{bmatrix}

\begin{bmatrix}
x_1 \\
\dot{x}_1 \\
x_2 \\
\dot{x}_2
\end{bmatrix}

+

\begin{bmatrix}
0 & 0 & 0 & 0 \\
-1 & \frac{1}{m_1} & 0 & 0 \\
0 & 0 & 0 & 0 \\
-1 & 0 & 0 & 0
\end{bmatrix}

\begin{bmatrix}
g \\
F \\
0 \\
0
\end{bmatrix}

</math>

=Solve Using Laplace Transform Method=

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-11-19T03:25:21Z

Brandon.plubell:

[[User:Brandon.plubell|Brandon.plubell]] 03:26, 9 November 2009 (UTC)

=Part 1 - Use Laplace Transformations=

==Problem Statement==
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

==Initial Conditions and Values==
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

==Force Equations==
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

==Laplace Transform==
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

==Inverse Laplace Transform==
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

==Equation of Motion==
Putting it all back together again gives,

<math>
x(t) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Media:BP_BodePlotOctave-1.zip|Octave Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

=Part 2 - Final and Initial Value Theorems=

==Initial Value Theorem==
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

==Final Value Theorem==
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

=Part 3 - Bode Plot=
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

=Part 4 - Breakpoints and Asymptotes on Bode Plot=
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

=Part 5 - Convolution=
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point). Questions left: How could the result be adjusted to account for initial conditions?

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

(Laplace in blue solid, Convolution in red dotted)

=Part 6 - State Equation=

Choose the state variable to be <math> x\frac{}{} </math> and <math> \dot{x} </math>, then following the example from class, the state equation is

<math>
\begin{bmatrix}
\dot{x} \\
\ddot{x}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 \\
\frac{-k}{m} & \frac{-\lambda}{m}
\end{bmatrix}
\,
\begin{bmatrix}
x \\
\dot{x}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
1
\end{bmatrix}
\, g \, \sin{\theta}
</math>

=Appendix A=

==Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-11-19T03:24:57Z

Brandon.plubell: /* Part 6 - State Equation */

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-19T03:24:22Z

Brandon.plubell: /* State Space Equation */

=Setup State Space Equation=

==Problem Statement==
Find an input function such that the lower mass, <math>m_1</math>, is stationary in the steady state. Find the equation of motion for the upper mass, <math>m_2</math>.

The use of one spring between the masses is just a simplification of a multi-spring system, so the possibility of being off-kilter is neglected and just the vertical forces are considered.

[[Image:BP_Setup-2.jpg|right|Problem Setup]]

==Initial Conditions and Values==

<math>m_1 = 140 kg \frac{}{}</math>

<math>k_1 = 108000 \frac{N}{m}</math>

<math>m_2 = 80 kg \frac{}{}</math>

<math>k_2 = 80000 \frac{N}{m}</math>

Let the initial conditions be zero for the time being.

==Force Equations==

[[Image:BP_FBD2-2.jpg|right|FBD for m2]]

[[Image:BP_FBD1-2.jpg|right|FBD for m1]]

Sum of the forces in the x direction yields

For <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_2} = m_2 \ddot{x}_2
</math>

<math>
\Rightarrow\ m_2 \ddot{x}_2=F_{s_2} - m_2 g
</math>

Since
<math>
F_s = -k \, x
</math>

<math>
\Rightarrow\ \ddot{x}_2=\frac{-k_2}{m_2} \, x_2 - g
</math>

And for <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_1} = m_1 \ddot{x}_1
</math>

<math>
\Rightarrow\ m_1 \ddot{x}_1=F_{s_1} + F - m_1 g - F_{s_2}
</math>

Where
<math>F \frac{}{} = F(t)</math> is the input force

<math>
\Rightarrow\ \ddot{x}_1 = \frac{-k_1}{m_1} \, x_1 + \frac{k_2}{m_1} \, x_2 - g + \frac{F}{m_1}
</math>

==State Space Equation==
The general form of the state equation is

<math>
\underline{\dot{x}}(t) = \widehat{A} \, \underline{x}(t) + \widehat{C} \, \underline{u}(t)
</math>

Where <math>\widehat{M}</math> denotes a matrix and <math>\underline{v}</math> denotes a vector.

Let <math>x_1 \frac{}{}</math>, <math>\dot{x}_1 \frac{}{}</math>, <math>x_2 \frac{}{}</math>, and <math>\dot{x_2} \frac{}{}</math> be the state variables, then

<math>

\begin{bmatrix}
\dot{x}_1 \\
\ddot{x}_1 \\
\dot{x}_2 \\
\ddot{x}_2
\end{bmatrix}

=

</math>
<math>

\begin{bmatrix}
0 & 1 & 0 & 0 \\
\frac{-k_1}{m_1} & 0 & \frac{k_2}{m_1} & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & \frac{-k_2}{m_2} & 0
\end{bmatrix}

\begin{bmatrix}
x_1 \\
\dot{x}_1 \\
x_2 \\
\dot{x}_2
\end{bmatrix}

+

\begin{bmatrix}
0 & 0 & 0 & 0 \\
-1 & \frac{1}{m_1} & 0 & 0 \\
0 & 0 & 0 & 0 \\
-1 & 0 & 0 & 0
\end{bmatrix}

\begin{bmatrix}
g \\
F \\
0 \\
0
\end{bmatrix}

</math>

=Solve Using Laplace Transform Method=

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-19T03:21:48Z

Brandon.plubell: /* State Space Equation */

=Setup State Space Equation=

==Problem Statement==
Find an input function such that the lower mass, <math>m_1</math>, is stationary in the steady state. Find the equation of motion for the upper mass, <math>m_2</math>.

The use of one spring between the masses is just a simplification of a multi-spring system, so the possibility of being off-kilter is neglected and just the vertical forces are considered.

[[Image:BP_Setup-2.jpg|right|Problem Setup]]

==Initial Conditions and Values==

<math>m_1 = 140 kg \frac{}{}</math>

<math>k_1 = 108000 \frac{N}{m}</math>

<math>m_2 = 80 kg \frac{}{}</math>

<math>k_2 = 80000 \frac{N}{m}</math>

Let the initial conditions be zero for the time being.

==Force Equations==

[[Image:BP_FBD2-2.jpg|right|FBD for m2]]

[[Image:BP_FBD1-2.jpg|right|FBD for m1]]

Sum of the forces in the x direction yields

For <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_2} = m_2 \ddot{x}_2
</math>

<math>
\Rightarrow\ m_2 \ddot{x}_2=F_{s_2} - m_2 g
</math>

Since
<math>
F_s = -k \, x
</math>

<math>
\Rightarrow\ \ddot{x}_2=\frac{-k_2}{m_2} \, x_2 - g
</math>

And for <math>m_1 \frac{}{}</math>

<math>
+ \uparrow \sum F_{x_1} = m_1 \ddot{x}_1
</math>

<math>
\Rightarrow\ m_1 \ddot{x}_1=F_{s_1} + F - m_1 g - F_{s_2}
</math>

Where
<math>F \frac{}{} = F(t)</math> is the input force

<math>
\Rightarrow\ \ddot{x}_1 = \frac{-k_1}{m_1} \, x_1 + \frac{k_2}{m_1} \, x_2 - g + \frac{F}{m_1}
</math>

==State Space Equation==
The general form of the state equation is

<math>
\underline{\dot{x}}(t) = \widehat{A} \, \underline{x}(t) + \widehat{C} \, \underline{u}(t)
</math>

Where <math>\widehat{M}</math> denotes a matrix and <math>\underline{v}</math> denotes a vector.

Let <math>x_1 \frac{}{}</math>, <math>\dot{x}_1 \frac{}{}</math>, <math>x_2 \frac{}{}</math>, and <math>\dot{x_2} \frac{}{}</math> be the state variables, then

<math>
\begin{bmatrix}
& & & \\
& & & \\
& & & \\
& & &
\end{bmatrix}

\begin{bmatrix}
\dot{x}_1 \\
\ddot{x}_1 \\
\dot{x}_2 \\
\ddot{x}_2
\end{bmatrix}

=

</math>
<math>

\begin{bmatrix}
0 & 1 & 0 & 0 \\
\frac{-k_1}{m_1} & 0 & \frac{k_2}{m_1} & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & \frac{-k_2}{m_2} & 0
\end{bmatrix}

\begin{bmatrix}
x_1 \\
\dot{x}_1 \\
x_2 \\
\dot{x}_2
\end{bmatrix}

</math>

=Solve Using Laplace Transform Method=

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-19T01:21:50Z

Brandon.plubell: /* Force Equations */

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-19T01:21:23Z

Brandon.plubell: /* Force Equations */

File:BP FBD2-2.jpg

2009-11-19T01:02:58Z

Brandon.plubell: uploaded a new version of "Image:BP FBD2-2.jpg"

File:BP FBD1-2.jpg

2009-11-19T01:02:18Z

Brandon.plubell: uploaded a new version of "Image:BP FBD1-2.jpg"

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-19T00:48:54Z

Brandon.plubell: /* Force Equations */

File:BP FBD2-2.jpg

2009-11-19T00:47:52Z

Brandon.plubell:

File:BP FBD1-2.jpg

2009-11-19T00:47:36Z

Brandon.plubell:

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-19T00:47:05Z

Brandon.plubell:

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-19T00:24:18Z

Brandon.plubell: /* Initial Conditions and Values */

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-19T00:22:35Z

Brandon.plubell: /* Problem Statement */

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-19T00:21:13Z

Brandon.plubell:

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-19T00:20:27Z

Brandon.plubell: /* Problem Statement */

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-19T00:20:00Z

Brandon.plubell: /* Problem Statement */

File:BP Setup-2.jpg

2009-11-19T00:19:18Z

Brandon.plubell:

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-18T05:33:27Z

Brandon.plubell: /* Initial Conditions and Values */

Fall 2009

2009-11-18T04:54:55Z

Brandon.plubell:

====HW # 5====
Put a link here to an example problem you made up and solved using Laplace Transforms, of the complete solution of a mechanical system or a circuit that is described by linear ordinary differential equations with constant coefficients. Make sure you start with the physical system, and end up with the time response of the system.

Use your wiki page to explain the problem and solution to one of your classmates. Have him or her certify that they have checked it for errors, by listing that on the page. Then go to [http://moodle.wallawalla.edu Moodle] and put a link to your page in the in-box for HW #5.

[[Laplace transforms:Series RLC circuit]]

[[Laplace transforms:Mass-Spring Oscillator]]

[[Laplace transforms:DC Motor circuit]]

[[Laplace transforms: Simple Electrical Network]]

[[Laplace transforms: R series with RC parallel circuit]]

[[Laplace transforms: Critically Damped Motion ]]

[[Laplace transforms: Under-damped Mass-Spring System on an Incline]]

[[Laplace transforms: Critically Damped Spring Mass system]]

====HW #12====
Coupled Oscillator Problem

[[Coupled Oscillator: Hellie]]

[[Coupled Oscillator: Coupled Mass-Spring System with Input]]

==2009-2010 Contributors==

[[Ben Henry]]

[http://fweb.wallawalla.edu/class-wiki/index.php/Fall_2009/JonathanS Jonathan Schreven]

[http://fweb/class-wiki/index.php/Laplace_transforms:DC_Motor_circuit/ Kendrick Mensink]

[http://fweb/class-wiki/index.php/Laplace_transforms:_Critically_Damped_Motion Mark Bernet]

Fall 2009

2009-11-18T04:54:43Z

Brandon.plubell:

====HW # 5====
Put a link here to an example problem you made up and solved using Laplace Transforms, of the complete solution of a mechanical system or a circuit that is described by linear ordinary differential equations with constant coefficients. Make sure you start with the physical system, and end up with the time response of the system.

Use your wiki page to explain the problem and solution to one of your classmates. Have him or her certify that they have checked it for errors, by listing that on the page. Then go to [http://moodle.wallawalla.edu Moodle] and put a link to your page in the in-box for HW #5.

[[Laplace transforms:Series RLC circuit]]

[[Laplace transforms:Mass-Spring Oscillator]]

[[Laplace transforms:DC Motor circuit]]

[[Laplace Transforms: Simple Electrical Network]]

[[Laplace Transforms: R series with RC parallel circuit]]

[[Laplace Transforms: Critically Damped Motion ]]

[[Laplace Transforms: Under-damped Mass-Spring System on an Incline]]

[[Laplace Transforms: Critically Damped Spring Mass system]]

====HW #12====
Coupled Oscillator Problem

[[Coupled Oscillator: Hellie]]

[[Coupled Oscillator: Coupled Mass-Spring System with Input]]

==2009-2010 Contributors==

[[Ben Henry]]

[http://fweb.wallawalla.edu/class-wiki/index.php/Fall_2009/JonathanS Jonathan Schreven]

[http://fweb/class-wiki/index.php/Laplace_transforms:DC_Motor_circuit/ Kendrick Mensink]

[http://fweb/class-wiki/index.php/Laplace_transforms:_Critically_Damped_Motion Mark Bernet]

Fall 2009

2009-11-18T04:54:24Z

Brandon.plubell:

====HW # 5====
Put a link here to an example problem you made up and solved using Laplace Transforms, of the complete solution of a mechanical system or a circuit that is described by linear ordinary differential equations with constant coefficients. Make sure you start with the physical system, and end up with the time response of the system.

Use your wiki page to explain the problem and solution to one of your classmates. Have him or her certify that they have checked it for errors, by listing that on the page. Then go to [http://moodle.wallawalla.edu Moodle] and put a link to your page in the in-box for HW #5.

[[Laplace Transforms: Series RLC circuit]]

[[Laplace Transforms: Mass-Spring Oscillator]]

[[Laplace Transforms:DC Motor circuit]]

[[Laplace Transforms: Simple Electrical Network]]

[[Laplace Transforms: R series with RC parallel circuit]]

[[Laplace Transforms: Critically Damped Motion ]]

[[Laplace Transforms: Under-damped Mass-Spring System on an Incline]]

[[Laplace Transforms: Critically Damped Spring Mass system]]

====HW #12====
Coupled Oscillator Problem

[[Coupled Oscillator: Hellie]]

[[Coupled Oscillator: Coupled Mass-Spring System with Input]]

==2009-2010 Contributors==

[[Ben Henry]]

[http://fweb.wallawalla.edu/class-wiki/index.php/Fall_2009/JonathanS Jonathan Schreven]

[http://fweb/class-wiki/index.php/Laplace_transforms:DC_Motor_circuit/ Kendrick Mensink]

[http://fweb/class-wiki/index.php/Laplace_transforms:_Critically_Damped_Motion Mark Bernet]

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-18T04:52:25Z

Brandon.plubell: /* Problem Statement */

Coupled Oscillator: Coupled Mass-Spring System with Input

2009-11-18T04:51:01Z

Brandon.plubell: New page: =Setup State Space Equation= ==Problem Statement== Find an input function such that the lower mass, <math>m_1</math>, is stationary in the steady state. Find the equation of motion for t...

=Setup State Space Equation=

==Problem Statement==
Find an input function such that the lower mass, <math>m_1</math>, is stationary in the steady state. Find the equation of motion for the upper mass, <math>m_2</math>.

==Initial Conditions and Values==

==Force Equations==

==State Space Equation==

=Solve Using Laplace Transform Method=

=Solve Using Eigenvalue-Eigenvector Method=

Fall 2009

2009-11-18T04:46:21Z

Brandon.plubell:

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-11-09T20:50:55Z

Brandon.plubell: /* Part 6 - State Equation */

[[User:Brandon.plubell|Brandon.plubell]] 03:26, 9 November 2009 (UTC)

=Part 1 - Use Laplace Transformations=

==Problem Statement==
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

==Initial Conditions and Values==
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

==Force Equations==
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

==Laplace Transform==
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

==Inverse Laplace Transform==
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

==Equation of Motion==
Putting it all back together again gives,

<math>
x(t) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Media:BP_BodePlotOctave-1.zip|Octave Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

=Part 2 - Final and Initial Value Theorems=

==Initial Value Theorem==
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

==Final Value Theorem==
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

=Part 3 - Bode Plot=
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

=Part 4 - Breakpoints and Asymptotes on Bode Plot=
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

=Part 5 - Convolution=
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point). Questions left: How could the result be adjusted to account for initial conditions?

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

(Laplace in blue solid, Convolution in red dotted)

=Part 6 - State Equation=

Choose the state variable to be <math> x\frac{}{} </math> and <math> \dot{x} </math>, then following the example from class, the state equation is

<math>
\begin{bmatrix}
\dot{x} \\
\ddot{x}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 \\
\frac{-k}{m} & \frac{-\lambda}{m}
\end{bmatrix}
\,
\begin{bmatrix}
x \\
\dot{x}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
1
\end{bmatrix}
\, g \, \sin{\theta}
</math>

<math>
x=

</math>

=Appendix A=

==Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-11-09T20:50:46Z

Brandon.plubell: /* Equation of Motion */

[[User:Brandon.plubell|Brandon.plubell]] 03:26, 9 November 2009 (UTC)

=Part 1 - Use Laplace Transformations=

==Problem Statement==
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

==Initial Conditions and Values==
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

==Force Equations==
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

==Laplace Transform==
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

==Inverse Laplace Transform==
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

==Equation of Motion==
Putting it all back together again gives,

<math>
x(t) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Media:BP_BodePlotOctave-1.zip|Octave Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

=Part 2 - Final and Initial Value Theorems=

==Initial Value Theorem==
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

==Final Value Theorem==
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

=Part 3 - Bode Plot=
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

=Part 4 - Breakpoints and Asymptotes on Bode Plot=
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

=Part 5 - Convolution=
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point). Questions left: How could the result be adjusted to account for initial conditions?

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

(Laplace in blue solid, Convolution in red dotted)

=Part 6 - State Equation=

Choose the state variable to be <math> x\frac{}{} </math> and <math> \dot{x} </math>, then following the example from class, the state equation is

<math>
\begin{bmatrix}
\dot{x} \\
\ddot{x}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 \\
\frac{-k}{m} & \frac{-\lambda}{m}
\end{bmatrix}
\,
\begin{bmatrix}
x \\
\dot{x}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
1
\end{bmatrix}
\, g \, \sin{\theta}
</math>

<math>
y=

</math>

=Appendix A=

==Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-11-09T20:48:43Z

Brandon.plubell: /* Part 6 - State Equation */

[[User:Brandon.plubell|Brandon.plubell]] 03:26, 9 November 2009 (UTC)

=Part 1 - Use Laplace Transformations=

==Problem Statement==
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

==Initial Conditions and Values==
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

==Force Equations==
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

==Laplace Transform==
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

==Inverse Laplace Transform==
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

==Equation of Motion==
Putting it all back together again gives,

<math>
x(0) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Media:BP_BodePlotOctave-1.zip|Octave Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

=Part 2 - Final and Initial Value Theorems=

==Initial Value Theorem==
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

==Final Value Theorem==
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

=Part 3 - Bode Plot=
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

=Part 4 - Breakpoints and Asymptotes on Bode Plot=
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

=Part 5 - Convolution=
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point). Questions left: How could the result be adjusted to account for initial conditions?

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

(Laplace in blue solid, Convolution in red dotted)

=Part 6 - State Equation=

Choose the state variable to be <math> x\frac{}{} </math> and <math> \dot{x} </math>, then following the example from class, the state equation is

<math>
\begin{bmatrix}
\dot{x} \\
\ddot{x}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 \\
\frac{-k}{m} & \frac{-\lambda}{m}
\end{bmatrix}
\,
\begin{bmatrix}
x \\
\dot{x}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
1
\end{bmatrix}
\, g \, \sin{\theta}
</math>

<math>
y=

</math>

=Appendix A=

==Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-11-09T03:26:03Z

Brandon.plubell:

[[User:Brandon.plubell|Brandon.plubell]] 03:26, 9 November 2009 (UTC)

=Part 1 - Use Laplace Transformations=

==Problem Statement==
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

==Initial Conditions and Values==
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

==Force Equations==
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

==Laplace Transform==
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

==Inverse Laplace Transform==
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

==Equation of Motion==
Putting it all back together again gives,

<math>
x(0) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Media:BP_BodePlotOctave-1.zip|Octave Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

=Part 2 - Final and Initial Value Theorems=

==Initial Value Theorem==
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

==Final Value Theorem==
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

=Part 3 - Bode Plot=
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

=Part 4 - Breakpoints and Asymptotes on Bode Plot=
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

=Part 5 - Convolution=
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point). Questions left: How could the result be adjusted to account for initial conditions?

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

(Laplace in blue solid, Convolution in red dotted)

=Part 6 - State Equation=

Choose the state variable to be <math> \dot{x} </math> and <math> \ddot{x} </math>, then following the example from class, the state equation is

<math>
\begin{bmatrix}
\dot{x} \\
\ddot{x}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 \\
\frac{-k}{m} & \frac{-\lambda}{m}
\end{bmatrix}
\,
\begin{bmatrix}
x \\
\dot{x}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
1
\end{bmatrix}
\, g \, \sin{\theta}
</math>

=Appendix A=

==Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-11-09T03:25:31Z

Brandon.plubell: /* Part 6 - State Equations */

[[User:Brandon.plubell|Brandon.plubell]] 05:54, 29 October 2009 (UTC)

=Part 1 - Use Laplace Transformations=

==Problem Statement==
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

==Initial Conditions and Values==
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

==Force Equations==
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

==Laplace Transform==
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

==Inverse Laplace Transform==
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

==Equation of Motion==
Putting it all back together again gives,

<math>
x(0) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Media:BP_BodePlotOctave-1.zip|Octave Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

=Part 2 - Final and Initial Value Theorems=

==Initial Value Theorem==
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

==Final Value Theorem==
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

=Part 3 - Bode Plot=
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

=Part 4 - Breakpoints and Asymptotes on Bode Plot=
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

=Part 5 - Convolution=
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point). Questions left: How could the result be adjusted to account for initial conditions?

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

(Laplace in blue solid, Convolution in red dotted)

=Part 6 - State Equation=

Choose the state variable to be <math> \dot{x} </math> and <math> \ddot{x} </math>, then following the example from class, the state equation is

<math>
\begin{bmatrix}
\dot{x} \\
\ddot{x}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 \\
\frac{-k}{m} & \frac{-\lambda}{m}
\end{bmatrix}
\,
\begin{bmatrix}
x \\
\dot{x}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
1
\end{bmatrix}
\, g \, \sin{\theta}
</math>

=Appendix A=

==Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-11-09T03:25:21Z

Brandon.plubell:

[[User:Brandon.plubell|Brandon.plubell]] 05:54, 29 October 2009 (UTC)

=Part 1 - Use Laplace Transformations=

==Problem Statement==
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

==Initial Conditions and Values==
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

==Force Equations==
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

==Laplace Transform==
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

==Inverse Laplace Transform==
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

==Equation of Motion==
Putting it all back together again gives,

<math>
x(0) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Media:BP_BodePlotOctave-1.zip|Octave Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

=Part 2 - Final and Initial Value Theorems=

==Initial Value Theorem==
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

==Final Value Theorem==
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

=Part 3 - Bode Plot=
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

=Part 4 - Breakpoints and Asymptotes on Bode Plot=
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

=Part 5 - Convolution=
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point). Questions left: How could the result be adjusted to account for initial conditions?

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

(Laplace in blue solid, Convolution in red dotted)

=Part 6 - State Equations=

Choose the state variable to be <math> \dot{x} </math> and <math> \ddot{x} </math>, then following the example from class, the state equation is

<math>
\begin{bmatrix}
\dot{x} \\
\ddot{x}
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 \\
\frac{-k}{m} & \frac{-\lambda}{m}
\end{bmatrix}
\,
\begin{bmatrix}
x \\
\dot{x}
\end{bmatrix}
+
\begin{bmatrix}
0 \\
1
\end{bmatrix}
\, g \, \sin{\theta}
</math>

=Appendix A=

==Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-10-30T05:14:03Z

Brandon.plubell: /* Equation of Motion */

[[User:Brandon.plubell|Brandon.plubell]] 05:54, 29 October 2009 (UTC)

=Part 1 - Use Laplace Transformations=

==Problem Statement==
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

==Initial Conditions and Values==
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

==Force Equations==
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

==Laplace Transform==
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

==Inverse Laplace Transform==
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

==Equation of Motion==
Putting it all back together again gives,

<math>
x(0) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Media:BP_BodePlotOctave-1.zip|Octave Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

=Part 2 - Final and Initial Value Theorems=

==Initial Value Theorem==
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

==Final Value Theorem==
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

=Part 3 - Bode Plot=
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

=Part 4 - Breakpoints and Asymptotes on Bode Plot=
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

=Part 5 - Convolution=
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point). Questions left: How could the result be adjusted to account for initial conditions?

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

(Laplace in blue solid, Convolution in red dotted)

=Appendix A=

==Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

File:BP BodePlotOctave-1.zip

2009-10-30T05:13:35Z

Brandon.plubell:

File:BP BodePlot-1.zip

2009-10-30T05:13:00Z

Brandon.plubell: uploaded a new version of "Image:BP BodePlot-1.zip"

Matlab (and Octave) script for Bode and position plots.

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-10-30T05:10:36Z

Brandon.plubell: /* Part 5 - Convolution */

[[User:Brandon.plubell|Brandon.plubell]] 05:54, 29 October 2009 (UTC)

=Part 1 - Use Laplace Transformations=

==Problem Statement==
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

==Initial Conditions and Values==
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

==Force Equations==
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

==Laplace Transform==
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

==Inverse Laplace Transform==
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

==Equation of Motion==
Putting it all back together again gives,

<math>
x(0) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

=Part 2 - Final and Initial Value Theorems=

==Initial Value Theorem==
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

==Final Value Theorem==
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

=Part 3 - Bode Plot=
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

=Part 4 - Breakpoints and Asymptotes on Bode Plot=
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

=Part 5 - Convolution=
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point). Questions left: How could the result be adjusted to account for initial conditions?

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

(Laplace in blue solid, Convolution in red dotted)

=Appendix A=

==Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-10-30T05:10:19Z

Brandon.plubell: /* Part 5 - Convolution */

[[User:Brandon.plubell|Brandon.plubell]] 05:54, 29 October 2009 (UTC)

=Part 1 - Use Laplace Transformations=

==Problem Statement==
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

==Initial Conditions and Values==
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

==Force Equations==
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

==Laplace Transform==
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

==Inverse Laplace Transform==
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

==Equation of Motion==
Putting it all back together again gives,

<math>
x(0) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

=Part 2 - Final and Initial Value Theorems=

==Initial Value Theorem==
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

==Final Value Theorem==
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

=Part 3 - Bode Plot=
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

=Part 4 - Breakpoints and Asymptotes on Bode Plot=
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

=Part 5 - Convolution=
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point). Questions left: How could the result be adjusted to account for initial conditions?

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]
(Laplace in blue solid, Convolution in red dotted)

=Appendix A=

==Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-10-29T05:54:11Z

Brandon.plubell:

[[User:Brandon.plubell|Brandon.plubell]] 05:54, 29 October 2009 (UTC)

=Part 1 - Use Laplace Transformations=

==Problem Statement==
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

==Initial Conditions and Values==
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

==Force Equations==
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

==Laplace Transform==
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

==Inverse Laplace Transform==
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

==Equation of Motion==
Putting it all back together again gives,

<math>
x(0) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

=Part 2 - Final and Initial Value Theorems=

==Initial Value Theorem==
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

==Final Value Theorem==
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

=Part 3 - Bode Plot=
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

=Part 4 - Breakpoints and Asymptotes on Bode Plot=
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

=Part 5 - Convolution=
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point). Questions left: How could the result be adjusted to account for initial conditions?

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

=Appendix A=

==Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-10-29T05:53:50Z

Brandon.plubell: /* Appendix A - Poles */

[[User:Brandon.plubell|Brandon.plubell]] 05:44, 26 October 2009 (UTC)

=Part 1 - Use Laplace Transformations=

==Problem Statement==
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

==Initial Conditions and Values==
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

==Force Equations==
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

==Laplace Transform==
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

==Inverse Laplace Transform==
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

==Equation of Motion==
Putting it all back together again gives,

<math>
x(0) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

=Part 2 - Final and Initial Value Theorems=

==Initial Value Theorem==
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

==Final Value Theorem==
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

=Part 3 - Bode Plot=
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

=Part 4 - Breakpoints and Asymptotes on Bode Plot=
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

=Part 5 - Convolution=
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point). Questions left: How could the result be adjusted to account for initial conditions?

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

=Appendix A=

==Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-10-29T05:52:03Z

Brandon.plubell:

[[User:Brandon.plubell|Brandon.plubell]] 05:44, 26 October 2009 (UTC)

=Part 1 - Use Laplace Transformations=

==Problem Statement==
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

==Initial Conditions and Values==
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

==Force Equations==
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

==Laplace Transform==
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

==Inverse Laplace Transform==
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

==Equation of Motion==
Putting it all back together again gives,

<math>
x(0) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

=Part 2 - Final and Initial Value Theorems=

==Initial Value Theorem==
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

==Final Value Theorem==
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

=Part 3 - Bode Plot=
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

=Part 4 - Breakpoints and Asymptotes on Bode Plot=
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

=Part 5 - Convolution=
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point). Questions left: How could the result be adjusted to account for initial conditions?

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

=Appendix A - Poles=
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-10-29T04:48:24Z

Brandon.plubell: /* Part 5 - Convolution */

[[User:Brandon.plubell|Brandon.plubell]] 05:44, 26 October 2009 (UTC)

=Under-Damped Mass-Spring System on an Incline=

==Part 1 - Use Laplace Transformations==

===Problem Statement===
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

===Initial Conditions and Values===
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

===Force Equations===
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

===Laplace Transform===
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

===Inverse Laplace Transform===
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

===Equation of Motion===
Putting it all back together again gives,

<math>
x(0) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

==Part 2 - Final and Initial Value Theorems==

===Initial Value Theorem===
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

===Final Value Theorem===
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

==Part 3 - Bode Plot==
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

==Part 4 - Breakpoints and Asymptotes on Bode Plot==
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

==Part 5 - Convolution==
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point). Questions left: How could the result be adjusted to account for initial conditions?

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

==Appendix A - Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-10-29T04:47:35Z

Brandon.plubell: /* Part 4 - Breakpoints and Asymptotes on Bode Plot */

[[User:Brandon.plubell|Brandon.plubell]] 05:44, 26 October 2009 (UTC)

=Under-Damped Mass-Spring System on an Incline=

==Part 1 - Use Laplace Transformations==

===Problem Statement===
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

===Initial Conditions and Values===
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

===Force Equations===
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

===Laplace Transform===
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

===Inverse Laplace Transform===
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

===Equation of Motion===
Putting it all back together again gives,

<math>
x(0) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

==Part 2 - Final and Initial Value Theorems==

===Initial Value Theorem===
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

===Final Value Theorem===
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

==Part 3 - Bode Plot==
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

==Part 4 - Breakpoints and Asymptotes on Bode Plot==
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint. It looks as though this is also the max of the Bode Plot and possibly the resonant frequency.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

==Part 5 - Convolution==
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point).

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

==Appendix A - Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]

Laplace transforms: Under-damped Mass-Spring System on an Incline

2009-10-29T04:46:59Z

Brandon.plubell: /* Part 4 - Breakpoints and Asymptotes on Bode Plot */

[[User:Brandon.plubell|Brandon.plubell]] 05:44, 26 October 2009 (UTC)

=Under-Damped Mass-Spring System on an Incline=

==Part 1 - Use Laplace Transformations==

===Problem Statement===
Find the equation of motion for the mass in the system subjected to the forces shown in the free body diagram. The inclined surface is coated in 1mm of SAE 30 oil.

[[Image:BP_Setup-1.jpg|right|Problem Setup]]

===Initial Conditions and Values===
* A is the area of the box in contact with the surface
* g is the gravitational acceleration field constant
* bt is the thickness of the fluid covering the inclined surface
* μ is the viscosity constant of the fluid
* m is the mass of the box
* k is the spring constant

<math>
A = \frac{1}{4} m^2
</math>

<math>
g = 9.81 \frac{m}{s^2}
</math>

<math>
b_t = 1 mm \frac{}{}
</math>

<math>
\mu = 0.06 \frac{N \cdot s}{m^2}
</math>

<math>
m = 45 kg \frac{}{}
</math>

<math>
k = 200 \frac{N}{m}
</math>

<math>
\theta = 30^{\circ} \frac{}{}
</math>

Let the initial conditions be

<math>
x(0) = -0.5 m \frac{}{}
</math>

<math>
\dot{x}(0) = 0 \frac{m}{s}
</math>

===Force Equations===
[[Image:BP_FBD-1.jpg|right|Free Body Diagram]]

The sum of the forces in the x direction yields the equation

<math>
+ \swarrow \sum F_x = m\ddot{x}
</math>

<math>
\Rightarrow\ m\ddot{x}=F_s + F_f + mg \sin \theta
</math>

Where

<math>
F_s=-k\,x
</math>

<math>
F_f=-\frac{\mu \, A}{b_t} \, \dot{x}
</math>

To make the algebra easier, let

<math>
\lambda=\frac{\mu \, A}{b_t}
</math>

Then, from the sum of forces equation

<math>
m\,\ddot{x} + \lambda\,\dot{x}+k\,x=mg \sin \theta
</math>

<math>
\Rightarrow\ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x=g \sin \theta
</math>

===Laplace Transform===
<math>
\mathcal{L} \left\{ \ddot{x} + \frac{\lambda}{m}\,\dot{x}+\frac{k}{m}\,x \right\} = \mathcal{L} \left\{ g \sin \theta \right\}
</math>

<math>
\Rightarrow\ s^2\,X(s) - s\,x(0) - \dot{x}(0) + \frac{\lambda}{m}\,s\,X(s) - \frac{\lambda}{m}\,x(0) + \frac{k}{m}\,X(s) = g \sin \theta \, \left(\frac{1}{s}\right)
</math>

<math>
\Rightarrow\ X(s) \left(s^2 + \frac{\lambda}{m}\,s + \frac{k}{m} \right) = g \sin \theta \, \left( \frac {1}{s} \right) + s\,x(0) + \dot{x}(0) + \frac{\lambda}{m}\, x(0)
</math>

If we let <math>x(0)\text{ and }\dot{x}(0)</math> be 0 and rearrange the equation,

<math>
\frac{X(s)}{X_{in}(s)} = \frac{X(s)}{g\sin\theta \left( \frac{1}{s} \right)} = H(s)
</math>

<math>
\Rightarrow H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

The above is the transfer function that will be used in the Bode plot and can provide valuable information about the system.

<math>
X(s)=g\sin\theta \left( \frac{1}{s} \right) \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+x(0) \, \left( \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
+\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right)
</math>

===Inverse Laplace Transform===
Since the Laplace Transform is a linear transform, we need only find three inverse transforms. All of the these have complex roots, since <math> {\left( \frac{\lambda}{m} \right)}^2 < 4 \, \frac{\lambda}{m} </math>. Because I am not yet comfortable finding the inverse with complex roots by hand, I used a laplace transform program for the TI-89.

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s \left( s^2 + \frac{\lambda}{m} \, s + \frac{k}{m} \right)} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40}
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{s}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

<math>
\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

===Equation of Motion===
Putting it all back together again gives,

<math>
x(0) =
g \, \sin {\theta} \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{-9}{40} \cos {\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{3 \, \sqrt{159}}{2120} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] + \frac{9}{40} \right)
</math>

<math>
+ \,
x(0) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} - \frac{\sqrt{159}}{159} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

<math>
+
\left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, \left( e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right] \right)
</math>

It is useful to have the equation in the form given above because <math> x(0) \text{ , } \dot{x}(0) \text{ , } \theta \text{ , and even } g </math> can be varied and still give accurate results. The Matlab (or Octave) script below can be edited as described. Take note! <math> \lambda \text{ (and all that depend on it), } m \text{ , and } k </math> cannot be altered (else the inverse Laplace is false)!

[[Media:BP_BodePlot-1.zip|Matlab Script]]

[[Image:BP_Position-1.jpg|Position of Mass from Laplace Method]]

==Part 2 - Final and Initial Value Theorems==

===Initial Value Theorem===
As was derived in class, there are two theorems that relate the initial and final values (in this case positions) of the output functions in the t domain with the output function in the s domain. In a case such as this, in which the initial values are given, the initial value theorem is just a check.

<math>
\lim_{s \to \infty}{s \, X(s)} = x(0)
</math>

Taking the limit of <math> s \, X(s) </math> gives

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (0) + \left( x(0) \right) \, (1) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(0)
</math>

<math>
\Rightarrow \
x(0) = x(0) \frac{}{}
</math>

===Final Value Theorem===
The Final Value Theorem is a very useful tool that will show what the final value of the output function (as <math> t \to \infty </math>), which in this case is the final position of the block. Notice that it is not the unstretched length of the spring (else <math> x(\infty) = 0 </math>). It is also of interest to note that only the input function comes into play here, as all the others go to zero, and is not dependent on the initial position or velocity.

<math>
\lim_{s \to 0}{s \, X(s)} = x(\infty)
</math>

<math>
\Rightarrow \
\left( g \sin{\theta} \right) \, (\frac{m}{k}) + \left( x(0) \right) \, (0) + \left( \dot{x}(0) + \frac{\lambda}{m} \, x(0) \right) \, (0) = x(\infty)
</math>

<math>
\Rightarrow \
x(\infty) = \frac{m \, g}{k} \sin{\theta} = 1.104 m
</math>

Which can be seen in the plot in the section [[#Equation of Motion|Equation of Motion]].

==Part 3 - Bode Plot==
The bode plot shows useful information about the system we are analyzing. It has only to do with the transfer function, which means that it does not change based upon the input. However, it can show what a given frequency of a harmonic input will do to the output. For my example, it can be seen that at about <math> 2 \, rad/s </math> there is a rise in the magnitude of the transfer function. If it were hit with a corresponding frequency by an input function, it could have very larg oscillations.

[[Image:BP_BodePlot-1.jpg]]

==Part 4 - Breakpoints and Asymptotes on Bode Plot==
From the transfer function in the [[#Laplace Transform|Laplace Transform]] section,

<math>
H(s)=\frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}}
</math>

it can be seen that there are no zeros (nothing in the numerator that would make the function go to zero), but there is a place in the denominator that would exhibit deviant behavior. That is when the <math>s^2</math> and <math>k/m</math> are on the same order of magnitude. That is one stops dominating and the other starts. This point can be visually observed by finding the intersection of the asymptotes in the Bode Plot. Where they intersect is (roughly) a breakpoint.

<math>
\sqrt{\frac{k}{m}} = 2.108
</math>

==Part 5 - Convolution==
The convolution is a equation that relates the output to the input and transfer function. As derived in class, it is

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {x(t_0) \, h(t-t_0) \, dt_0}
</math>

Where <math>h(t)</math> is the inverse laplace of the transfer function.

<math>
h(t)=\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + \frac{\lambda}{m} \, s + \frac{k}{m}} \right\} =
e^{\frac{-1}{6} \, t} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, t}{6} \right)} \right]
</math>

So

<math>
x(t)=
x_{in}(t)*h(t) =
\int_{0}^{t} {\left( g \, \sin{\theta} \right) \,
e^{\frac{-1}{6} \, \left( t-t_0 \right)} \, \left[ \frac{2 \, \sqrt{159}}{53} \, \sin {\left( \frac{\sqrt{159} \, \left( t-t_0 \right)}{6} \right)} \right]
\, dt_0}
</math>

To solve the integral, one must do two integration by parts, or alternatively plug it into a calculator, which yields

<math>
x(t)= \frac{-3}{2120} \, g \, \sin{\theta} \, e^{\frac{-1}{6}t} \left( 159 \cos{\left( \frac{\sqrt{159} \, t}{6} \right)} + \sqrt{159} \sin{\left( \frac{\sqrt{159} \, t}{6} \right)} - 159 \, e^{\frac{1}{6}t} \right)
</math>

As can be seen in the plot, the Convolution method, as executed, resulted in the same results as the Laplace methods, just without any initial conditions (starts at 0 and has a smaller amplitude, but finishes at the same point).

[[Image:BP_PositionBoth-1.jpg|Both Laplace and Convolution Methods of EOM]]

==Appendix A - Poles==
If one puts the transfer function from the [[#Laplace Transform|Laplace Transform]] section, it can be seen that the poles (roots of the denominator) will have both real and imaginary components, which is observable by the quadratic formula

<math>
s = 0 = \frac{- \lambda}{2 \, m} \, \pm \, \frac{1}{2} \, \sqrt{\frac{\lambda^2}{m^2} - \frac{4 \, k}{m}}
</math>

Given
<math>
\frac{4 \, k}{m} = \frac{160}{9}
</math>
and
<math>
\frac{\lambda^2}{m^2} = \frac{1}{9}
</math>

<math>
\Rightarrow \ s = 0 = \frac{-\lambda}{2 \, m} \, \pm \, \frac{1}{2} \, j \, \sqrt{\frac{4 \, k}{m} - \frac{\lambda^2}{m^2}}
= \frac{-1}{6} \, \pm \, j \, \frac{\sqrt{159}}{6} = = -0.1667 \, \pm \, j \, 2.102
</math>

Note the (important) switching of the terms in the square root when the <math>j</math> was taken out front, since it is of course <math>\sqrt{1}</math>. Below is a nice plot (built in Matlab function, like the bode plot) which the plots poles on the imaginary and real axes. The closer the poles get to the the imaginary axis (i.e., the smaller the real values get), the closer to destructive behavior at a certain frequency.

[[image:BP_Poles-1.jpg]]