Laplace transforms: Critically Damped Motion

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

$m=\frac{8}{32}=\frac{1}{4}slugs$

$\text{k=4}$

$\text{C=2}$

$\text{x(0)=0}$

$\dot{x}(0)=-3$

$\text{Standard equation: }$ $m\frac{d^{2}x}{d{t}^2}+C\frac{dx}{dt}+khx=0$

Solving the problem

$\text{Therefore the equation representing this system is.}$

$\frac{1}{4}\frac{d^{2}x}{d{t}^2}=-4x-2\frac{dx}{dt}$

$\text{Now we put the equation in standard form}$

$\frac{d^{2}x}{d{t}^2}+8\frac{dx}{dt}+16x=0$

$\text{Now that we have the equation written in standard form we need to send}$ $\text{it through the Laplace Transform.}$

${ℒ}[\frac{d^{2}x}{d{t}^2}+8\frac{dx}{dt}+16x]$

$\text{And we get the equation (after some substitution and simplification)}.$

${\mathbf{s}}^{2}X(s)+8\mathbf{s}X(s)+16\mathbf{X}(s)=-3$

$\mathbf{X}(s)(s^2+8s+16)=-3$

$\mathbf{X}(s)=-\frac{3}{(s+4{)}^2}$

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

${{ℒ}}^{-1}[-\frac{3}{(s+4{)}^2}]$

$\text{and we get, by using the table of transforms on p. 515 of the text bokk}$

$\mathbf{x}(t)=-3t{e}^{-4t}$

$\text{So there you have it the equation of a Critically Damped spring mass system.}$

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

Initial Value Theorem

$\underset{s\to \mathrm{\infty }}{\lim }sF(s)=f(0)$

Final Value Theorem

$\underset{s\to 0}{\lim }sF(s)=f(\mathrm{\infty })$

Applying this to our problem

$\text{The Initial Value Theorem}$

$\underset{s\to \mathrm{\infty }}{\lim }\mathbf{s}\mathbf{X}(s)=-\frac{3}{(s+4{)}^2}$

$\underset{s\to \mathrm{\infty }}{\lim }\mathbf{s}X(s)=-\frac{3}{(\mathrm{\infty }+4{)}^2}=0$

$\text{So as you can see the value for the initial position will be 0. }$

$\text{Which makes sense because the system is initially in equilibrium. }$

$\text{The Final Value Theorem}$

$\underset{s\to 0}{\lim }\mathbf{s}X(s)=-\frac{3}{(s+4{)}^2}$

$\underset{s\to 0}{\lim }\mathbf{s}X(s)=-\frac{3}{(0+4{)}^2}=-\frac{3}{16}$

$\text{This shows the final value to be}$ $-\frac{3}{16}ft$

$\text{Which appears to mean the system will be below equilibrium after a long time. }$

Bode Plot of the transfer function

Transfer Function

$\mathbf{X}(s)=-\frac{3}{(s+4{)}^2}$

Bode Plot

$\text{This plot is done using the control toolbox in MatLab. }$

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Fig (1)

     ==Break Points and Asymptotes==                                                                                                                                                      

$\text{A break point is defined by a place in the bode plot where a change occurs.}$

$\text{To find your break points you must start with a transfer function. }$

$\text{Transfer Function: }$

$\mathbf{X}(s)=-\frac{3}{(s+4{)}^2}$

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

Convolution

$\text{The convolution equation is as follows: }$

$x(t)=x_{in}(t)*h(t)=\underset{0}{\overset{t}{\int }}x(t_0)h(t-t_0)d{t}_0$

$\text{It does basically the same thing as the Laplace Transform. }$ $\text{To start we must inverse transform our transfer function }$

$\mathbf{X}(s)=-\frac{3}{(s+4{)}^2}$

$\text{Which once more yields: }$

$\mathbf{x}(t)=-3t{e}^{-4t}$

$\text{Then we put this into the convolution integral: }$

$x(t)=x_{in}(t)*h(t)=\underset{0}{\overset{t}{\int }}-3(t-t_0)e^{-4t-t_0}d{t}_0$

$\text{Which once more yeilds: }$

$\mathbf{x}(t)=(-ct{e}^{-4t})$

$\text{Not exactly the same but remember initial conditions arnt used}$

State Space

$\text{Using state equatons is just another way to solve a system modeled by an ODE }$

$\text{First we need to add an applied force so u(t)=2N }$

$m=\frac{8}{32}=\frac{1}{4}slugs$

$\text{k=4}$

$\text{C=2}$

$\text{x(0)=0}$

$\dot{x}(0)=-3$

$\ddot{x}(0)=0$

$\left[\begin{array}{c}\dot{x}\\ \ddot{x}\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -k/m& -C/m\end{array}\right]\left[\begin{array}{c}x\\ dotx\end{array}\right]+\left[\begin{array}{c}0\\ 1/m\end{array}\right]u(t)$

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Written By: Mark Bernet

Error Checked By: Greg Peterson