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

[[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
* b<sub>t</sub> is the thickness of the fluid covering the inclined surface
* &mu; 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]]