Laplace transforms: Critically Damped Spring Mass system

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

Problem Statement

An 98 Newton weight is attached to a spring with a spring constant k of 40 N/m. The spring is stretched 4 m and rests at its equilibrium position. It is then released from rest with an initial upward velocity of 2 m/s. The system contains a damping force of 40 times the initial velocity.

Solution

Given

$m={\frac {98}{9.81}}kg$

${\text{Spring Constant k=40}}N/m\,$

${\text{Damping Constant C=40}}\,$

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

${\dot {x}}(0)=-4$

${\text{Standard equation: }}\,$

$m{\frac {d^{2}x}{dt^{2}}}+C{\frac {dx}{dt}}+khx=0$

Solving the problem

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

${\frac {98}{9.8}}{\frac {d^{2}x}{dt^{2}}}=-40x-40{\frac {dx}{dt}}$

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

${\frac {d^{2}x}{dt^{2}}}+{\frac {40}{10}}{\frac {dx}{dt}}+{\frac {40}{10}}x=0$

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

${\mathcal {L}}[{\frac {d^{2}x}{dt^{2}}}+{\frac {40}{10}}{\frac {dx}{dt}}+{\frac {20}{5}}x]$

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

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

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

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

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

${\mathcal {L}}^{-1}[-{\frac {4}{(s+2)^{2}}}]$

${\text{and we get}}\,$

$\mathbf {x} (t)=-4te^{-2t}$

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

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

Initial Value Theorem

$\lim _{s\rightarrow \infty }sF(s)=f(0)\,$

Final Value Theorem

$\lim _{s\rightarrow 0}sF(s)=f(\infty )\,$

Applying this to our problem

${\text{The Initial Value Theorem}}\,$

$\lim _{s\rightarrow \infty }\mathbf {sX} (s)=-{\frac {4}{(s+2)^{2}}}\,$

$\lim _{s\rightarrow \infty }\mathbf {s} {X}(s)=-{\frac {4}{(\infty +2)^{2}}}=0\,$

${\text{So as you can see the value for the initial position will be 0.}}\,$ ${\text{Because the infinity in the denominator always makes the function tend toward zero.}}\,$

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

${\text{The Final Value Theorem}}\,$

$\lim _{s\rightarrow 0}\mathbf {s} {X}(s)=-{\frac {4}{(s+2)^{2}}}\,$

$\lim _{s\rightarrow 0}\mathbf {s} {X}(s)=-{\frac {4}{(0+2)^{2}}}=-{\frac {4}{4}}\,$

${\text{This shows the final value to be}}\,$ $-{\frac {4}{4}}m$

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

Bode Plot of the transfer function

Transfer Function

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

Bode Plot

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

Fig (1)

Break Points

${\text{Find the Break points using the transfer function.}}\,$

Transfer fucntion

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

${\text{The equation above contains break points but only in the denominator.}}\,$

${\text{There is only the variable s in the denominator so only those types of break point exist}}\,$

${\text{The break points are asymtotes at the point -2 which occurs twice in this particular equation}}\,$

Convolution

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

$x(t)=x_{in}(t)*h(t)=\int _{0}^{t}{x(t_{0})\,h(t-t_{0})\,dt_{0}}$

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

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

${\text{Which once more yields: }}\,$

$\mathbf {x} (t)=-4te^{-2t}$

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

$x(t)=x_{in}(t)*h(t)=\int _{0}^{t}{-4(t-t_{0})e^{-2t-t_{0}}\,dt_{0}}$

${\text{Which once more yeilds: }}\,$

$\mathbf {x} (t)=(-cte^{-2t})$

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

State Space

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

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

$m={\frac {98.1}{9.81}}=10kg$

${\text{k=40}}\,$

${\text{C=40}}\,$

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

${\dot {x}}(0)=-4$

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

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

Created by Greg Peterson

Checked by Mark Bernet