Laplace transforms: Critically Damped Motion: Difference between revisions

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==Using the Laplace Transform to solve a spring mass system that is critically damped==
=Using the Laplace Transform to solve a spring mass system that is critically damped=


===Problem Statement===
==Problem Statement==
An 8 pound weight is attached to a spring with a spring constant k of 4 lb/ft.
An 8 pound weight is attached to a spring with a spring constant k of 4 lb/ft.
The spring is stretched 2 ft and rests at its equilibrium position.
The spring is stretched 2 ft and rests at its equilibrium position.
It is then released from rest with an initial upward velocity of 3 ft/s.
It is then released from rest with an initial upward velocity of 3 ft/s.
The system contains a damping force of 2 times the initial velocity.

==Solution==


===Things we know===


<math>m=\frac{8}{32}=\frac1 4 slugs</math>

<math>\text {k=4}\,</math>

<math>\text {C=2}\,</math>
<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-3</math>



<math>\text {Standard equation: }\,</math>
<math>m\frac{d^2x}{dt^2}+C\frac{dx}{dt}+khx=0</math>

===Solving the problem===
<math>\text {Therefore the equation representing this system is.}\,</math>

<math>\frac1 4 \frac{d^2x}{dt^2}=-4x-2\frac{dx}{dt}</math>

<math>\text {Now we put the equation in standard form}\,</math>

<math>\frac{d^2x}{dt^2}+8\frac{dx}{dt}+16x=0</math>


<math>\text {Now that we have the equation written in standard form we need to send}\,</math>
<math>\text {it through the Laplace Transform.}\,</math>

<math>\mathcal{L}[\frac{d^2x}{dt^2}+8\frac{dx}{dt}+16x]</math><br /><br />

<math>\text {And we get the equation (after some substitution and simplification)}.\,</math>

<math>\mathbf s^2 {X}(s)+8\mathbf s{X}(s)+16\mathbf{X}(s)=-3</math><br /><br />

<math>\mathbf {X}(s)(s^2+8s+16)=-3</math><br /><br />


<math>\mathbf {X}(s)=-\frac{3}{(s+4)^2} </math><br /><br />

<math>\text {Now that we have completed the Laplace Transform}\,</math>
<math>\text {and solved for X(s) we must so an inverse Laplace Transform. }\,</math>

<math>\mathcal{L}^{-1}[-\frac{3}{(s+4)^2}]</math><br /><br />

<math>\text {and we get, by using the table of transforms on p. 515 of the text bokk}\,</math>

<math>\mathbf {x}(t)=-3te^{-4t}</math><br /><br />

<math>\text {So there you have it the equation of a Critically Damped spring mass system.}\,</math>

==Apply the Initial and Final Value Theorems to find the initial and final values==

:Initial Value Theorem

::<math>\lim_{s\rightarrow \infty} sF(s)=f(0)\,</math>

:Final Value Theorem

::<math>\lim_{s\rightarrow 0} sF(s)=f(\infty)\,</math>


===Applying this to our problem===

<math>\text {The Initial Value Theorem}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf {sX}(s)=-\frac{3}{(s+4)^2}\,</math>

<math>\lim_{s\rightarrow \infty} \mathbf s{X}(s)=-\frac{3}{(\infty+4)^2}=0\,</math>


<math>\text {So as you can see the value for the initial position will be 0. }\,</math>

<math>\text {Which makes sense because the system is initially in equilibrium. }\,</math>

<math>\text {The Final Value Theorem}\,</math>

<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{3}{(s+4)^2}\,</math>


<math>\lim_{s\rightarrow 0} \mathbf s{X}(s)=-\frac{3}{(0+4)^2}=-\frac 3 {16}\,</math>

<math>\text {This shows the final value to be}\,</math>
<math>-\frac{3}{16}ft</math>

<math>\text {Which appears to mean the system will be below equilibrium after a long time. }\,</math>


==Bode Plot of the transfer function==

===Transfer Function===

<math>\mathbf {X}(s)=-\frac{3}{(s+4)^2} </math><br /><br />


===Bode Plot===

<math>\text {This plot is done using the control toolbox in MatLab. }\,</math>

[[Image:bode.jpg|700px|thumb|left|Fig (1)]]




==Break Points and Asymptotes==

<math>\text {A break point is defined by a place in the bode plot where a change occurs.}\,</math>

<math>\text {To find your break points you must start with a transfer function. }\,</math>

<math>\text {Transfer Function: }\,</math>


<math>\mathbf {X}(s)=-\frac{3}{(s+4)^2} </math><br /><br />


<math>\text {A break point is located at any value where s = what is being added to it. }\,</math>
<math>\text {So for this transfer function its at s=4 (that is also the asymptotes location). }\,</math>

=Convolution=
<math>\text {The convolution equation is as follows: }\,</math>

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

<math>\text {It does basically the same thing as the Laplace Transform. }\,</math>
<math>\text {To start we must inverse transform our transfer function }\,</math>


<math>\mathbf {X}(s)=-\frac{3}{(s+4)^2} </math><br /><br />

<math>\text {Which once more yields: }\,</math>

<math>\mathbf {x}(t)=-3te^{-4t}</math><br /><br />

<math>\text {Then we put this into the convolution integral: }\,</math>

<math>
x(t)=x_{in}(t) * h(t) = \int_{0}^{t} {-3(t-t_0)e^{-4t-t_0} \, dt_0}
</math>

<math>\text {Which once more yeilds: }\,</math>


<math>\mathbf {x}(t)=(-cte^{-4t})</math><br /><br />


<math>\text {Not exactly the same but remember initial conditions arnt used}\,</math>



=State Space=

<math>\text {Using state equatons is just another way to solve a system modeled by an ODE }\,</math>

<math>\text {First we need to add an applied force so u(t)=2N }\,</math>

<math>m=\frac{8}{32}=\frac1 4 slugs</math>

<math>\text {k=4}\,</math>

<math>\text {C=2}\,</math>
<math>\text {x(0)=0}\,</math>

<math>\dot{x}(0)=-3</math>

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


<math>\begin{bmatrix} \dot{x} \\ \ddot{x} \end{bmatrix}=\begin{bmatrix} 0 & 1 \\ -k/m & -C/m \end{bmatrix} \begin{bmatrix} x \\ \dot{x} \end{bmatrix} + \begin{bmatrix} 0 \\ 1/m \end{bmatrix}u(t)</math>









---

Written By: Mark Bernet


Error Checked By: Greg Peterson

Latest revision as of 16:58, 18 November 2009

Using the Laplace Transform to solve a spring mass system that is critically damped

Problem Statement

An 8 pound weight is attached to a spring with a spring constant k of 4 lb/ft. The spring is stretched 2 ft and rests at its equilibrium position. It is then released from rest with an initial upward velocity of 3 ft/s. The system contains a damping force of 2 times the initial velocity.

Solution

Things we know


Solving the problem















Apply the Initial and Final Value Theorems to find the initial and final values

Initial Value Theorem
Final Value Theorem


Applying this to our problem




Bode Plot of the transfer function

Transfer Function




Bode Plot

Fig (1)



     ==Break Points and Asymptotes==                                                                                                                                                      






Convolution











State Space






---

Written By: Mark Bernet


Error Checked By: Greg Peterson