Laplace Transforms: Vertical Motion of a Coupled Spring System: Difference between revisions
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By '''Jimmy Apablaza''' |
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This problem is described in Exercise 14, Section 7.6 (page 323) of the ''A first Course in Differential Equations'' textbook, 8ED (ISBN 0-534-41878-3). |
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__TOC__ |
__TOC__ |
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[[Image:Figure1.gif|thumb|Figure 1. Coupled Spring System.]] |
[[Image:Figure1.gif|thumb|Figure 1. Coupled Spring System.]] |
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= Problem Statement = |
= Problem Statement = |
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Derive the system of differential equations describing the straight-line vertical motion of the coupled spring shown in Figure 1. Use Laplace transform to solve the system when <math>k_1=k_2=k_3=1</math>, <math>m_1=m_2=1</math>, and <math>x_1(0)=0</math>, <math>x'1(0)=-1</math>, <math>x_2(0)=0</math>, and <math>x'_2(0)=1</math>. |
Derive the system of differential equations describing the straight-line vertical motion of the coupled spring shown in Figure 1. Use Laplace transform to solve the system when <math>k_1=k_2=k_3=1 \frac{}{}</math>, <math>m_1=m_2=1 \frac{}{}</math>, and <math>x_1(0)=0 \frac{}{}</math>, <math>x'1(0)=-1 \frac{}{}</math>, <math>x_2(0)=0 \frac{}{}</math>, and <math>x'_2(0)=1 \frac{}{}</math>. |
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= Solution = |
= Solution = |
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At positions <math>x_1</math> and <math>x_2</math>, the masses <math>m_1</math> and <math>m_2</math> are in equilibrium. Thus, the motion equations for <math>m_1</math> and <math>m_2</math> are, |
At positions <math>x_1 \frac{}{}</math> and <math>x_2 \frac{}{}</math>, the masses <math>m_1 \frac{}{}</math> and <math>m_2 \frac{}{}</math> are in equilibrium. Thus, the motion equations for <math>m_1 \frac{}{}</math> and <math>m_2 \frac{}{}</math> are, |
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:<math>m_1\ddot{x}_1=-k_1x_1+k_2(x_2-x_1)</math> |
:<math>m_1\ddot{x}_1=-k_1x_1+k_2(x_2-x_1)</math> |
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where <math>m_1x''_1</math> and <math>m_2x''_2</math> represent the [http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion Newton's Second Law of Motion] and <math>-k_1x_1+k_2(x_2-x_1)</math> and <math>-k_2(x_2-x_1)-k_3x_2</math> represent the net forces acting in the masses. |
where <math>m_1x''_1 \frac{}{}</math> and <math>m_2x''_2 \frac{}{}</math> represent the [http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion Newton's Second Law of Motion] and <math>-k_1x_1+k_2(x_2-x_1) \frac{}{}</math> and <math>-k_2(x_2-x_1)-k_3x_2 \frac{}{}</math> represent the net forces acting in the masses. |
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== Laplace Transform == |
== Laplace Transform == |
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Applying the Laplace Transform to the motion equations and plugging the values of <math>k_1</math>, <math>k_2</math>, <math>k_3</math>, <math>m_1</math>, <math>m_2</math>, <math>x_1(0)</math>, <math>x'1(0)</math>, <math>x_2(0)</math>, and <math>x'_2(0)</math> for this systems, we obtain, |
Applying the Laplace Transform to the motion equations and plugging the values of <math>k_1 \frac{}{}</math>, <math>k_2 \frac{}{}</math>, <math>k_3 \frac{}{}</math>, <math>m_1 \frac{}{}</math>, <math>m_2 \frac{}{}</math>, <math>x_1(0) \frac{}{}</math>, <math>x'1(0) \frac{}{}</math>, <math>x_2(0) \frac{}{}</math>, and <math>x'_2(0) \frac{}{}</math> for this systems, we obtain, |
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:<math>\mathcal{L}[m_1\ddot{x}_1+k_1x_1-k_2(x_2-x_1)]=m_1[s^2X_1(s)-sx_1(0)-\dot{x}_1(0)]+k_1X_1(s)-k_2(X_2(s)-X_1(s))=0</math> |
:<math>\mathcal{L}[m_1\ddot{x}_1+k_1x_1-k_2(x_2-x_1)]=m_1[s^2X_1(s)-sx_1(0)-\dot{x}_1(0)]+k_1X_1(s)-k_2(X_2(s)-X_1(s))=0</math> |
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::<math>X_2(s)=\dfrac{X_1(s)+1}{(s^2+2)}</math> |
::<math>X_2(s)=\dfrac{X_1(s)+1}{(s^2+2)}</math> |
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Finally, solving for <math>X_1(s)</math> and <math>X_2(s)</math> yields, |
Finally, solving for <math>X_1(s) \frac{}{}</math> and <math>X_2(s) \frac{}{}</math> yields, |
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:<math>X_1(s)=\dfrac{-1}{s^2+3}</math> |
:<math>X_1(s)=\dfrac{-1}{s^2+3}</math> |
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== Inverse Laplace Transform == |
== Inverse Laplace Transform == |
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[[Image:mass_spring_displacement_plot.png|thumb|Figure 2. Coupled Spring System |
[[Image:mass_spring_displacement_plot.png|thumb|Figure 2. Coupled Spring System Motion.]] |
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First, we recognize that |
First, we recognize that |
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:<math>x_2=(t)\mathcal{L}^{-1}[X_2(s)] = \mathcal{L}^{-1} \left [ \dfrac{1}{s^2+3} \right ] = \dfrac{1}{\sqrt{3}} \mathcal{L}^{-1} \left [ \dfrac{\sqrt{3}}{s^2+3} \right ] = \dfrac{1}{\sqrt{3}} sin(\sqrt{3}t)</math> |
:<math>x_2=(t)\mathcal{L}^{-1}[X_2(s)] = \mathcal{L}^{-1} \left [ \dfrac{1}{s^2+3} \right ] = \dfrac{1}{\sqrt{3}} \mathcal{L}^{-1} \left [ \dfrac{\sqrt{3}}{s^2+3} \right ] = \dfrac{1}{\sqrt{3}} sin(\sqrt{3}t)</math> |
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A plot of the |
A plot of the oscillatory motion is shown on Figure 2. |
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== Initial-Value & Final-Value Theorem == |
== Initial-Value & Final-Value Theorem == |
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The initial-value and final-value theorem can be useful the finding the behavior of a |
The initial-value and final-value theorem can be useful the finding the behavior of a function at small and large times respectively. By definition, the Initial-Value Theorem is, |
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<math>\lim_{s\rightarrow \infty} sX(s)=x(0)\,</math> |
<math>\lim_{s\rightarrow \infty} sX(s)=x(0)\,</math> |
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<math>H_2(s)=\dfrac{1}{s^2+3}</math> |
<math>H_2(s)=\dfrac{1}{s^2+3}</math> |
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can be easily done using a program like Octave or MATLAB. The code is displayed below. From Figure 3 we may notice that the Amplitude vs. Frequency plot for both functions overlaps. The peak amplitude occurs at |
can be easily done using a program like Octave or MATLAB. The code is displayed below. From Figure 3 we may notice that the Amplitude vs. Frequency plot for both functions overlaps. The peak amplitude occurs at <math>\sqrt{3} \approx 1.73 \frac{}{}</math> seconds, as well as the phase switching as shown in the Phase vs. Frequency plot. |
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h1=tf([-1],[1 0 3]); |
h1=tf([-1],[1 0 3]); |
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== Magnitude Frequency Response == |
== Magnitude Frequency Response == |
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Considering the Transfer Functions <math>H_1(s)</math> and <math>H_2(s)</math> described in the Bode Plot section, we notice that there are no values for the ''s''-variable that make <math>H_1(s)</math> and <math>H_2(s)</math> equal zero. |
Considering the Transfer Functions <math>H_1(s) \frac{}{}</math> and <math>H_2(s) \frac{}{}</math> described in the Bode Plot section, we notice that there are no values for the ''s''-variable that make <math>H_1(s) \frac{}{}</math> and <math>H_2(s) \frac{}{}</math> equal zero. |
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From the Bode Plot, we notice that the break point occurs at ≈ 1.73 seconds, so the |
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== Convolution == |
== Convolution == |
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By definition, the convolution of two functions is, |
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<math>y(t)=f(t)*h(t)=\int_{-\infty}^{\infty}{f(\tau)h(t-\tau)d\tau}=\int_{\infty}^{-\infty}{f(t-\tau)h(\tau)d\tau}</math> |
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where <math>h_1 \frac{}{}</math> refers to the inverse Laplace Transform. Assuming that <math>\tau \frac{}{}</math> is a dummy variable of integration, and <math>f(t) \frac{}{}</math> is the impulse function <math>( f_1(t)=f_2(t)=\delta(t) \frac{}{} )</math>, the convolution for our system is, |
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:<math>y_1(t)=\int_{0}^{t}{\delta(t-\tau)h_1(\tau)d\tau}=-\dfrac{1}{\sqrt{3}}\int_{0}^{t}{\delta(t-\tau)sin(\sqrt{3}\tau)d\tau}=\left [ -\dfrac{1}{\sqrt{3}} sin(\sqrt{3}\tau) \right ]_{\tau=t}=-\dfrac{1}{\sqrt{3}} sin(\sqrt{3}t)</math> |
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:<math>y_2(t)=\int_{0}^{t}{\delta(t-\tau)h_2(\tau)d\tau}=\dfrac{1}{\sqrt{3}}\int_{0}^{t}{\delta(t-\tau)sin(\sqrt{3}\tau)d\tau}=\left [ \dfrac{1}{\sqrt{3}} sin(\sqrt{3}\tau) \right ]_{\tau=t}=\dfrac{1}{\sqrt{3}} sin(\sqrt{3}t)</math> |
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which is the same solution yields by the inverse Laplace Transform |
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== State Equation Model == |
== State Equation Model == |
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By definition, the the state equation is stated as |
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<math> |
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\underline{\dot{x}} = \widehat{A} \, \underline{x} + \widehat{B} \, \underline{u} |
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</math> |
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Now, consider the motion equations described in the Solution section, |
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<math>m_1\ddot{x}_1+k_1x_1-k_2(x_2-x_1)=m_1\ddot{x}_1+k_1x_1-k_2x_2+k_2x_1=0</math> |
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<math>m_2\ddot{x}_2+k_2(x_2-x_1)-k_3x_2=m_2\ddot{x}_2+k_2x_2-k_2x_1-k_3x_2=0</math> |
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Solving for <math>\ddot{x}_1</math> and <math>\ddot{x}_2</math> yields, |
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:<math>\ddot{x}_1=-\dfrac{k_1}{m_1}x_1+\dfrac{k_2}{m_1}x_2-\dfrac{k_2}{m_1}x_1</math> |
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:<math>\ddot{x}_2=\dfrac{-k_2}{m_2}x_2+\dfrac{k_2}{m_2}x_1+\dfrac{k_3}{m_2}x_2</math> |
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Finally, we 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. Thus, |
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<math> |
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\begin{bmatrix} |
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\dot{x}_1 \\ |
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\ddot{x}_1 \\ |
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\dot{x}_2 \\ |
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\ddot{x}_2 |
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\end{bmatrix} |
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= |
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\begin{bmatrix} |
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0 & 1 & 0 & 0 \\ |
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-\frac{1}{m_1}(k_1+k_2) & 0 & \frac{k_2}{m_1} & 0 \\ |
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0 & 0 & 0 & 1 \\ |
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\frac{k_2}{m_2} & 0 & \frac{1}{m_2}(k_3-k_2) & 0 |
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\end{bmatrix} |
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\begin{bmatrix} |
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= References = |
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x_1 \\ |
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\dot{x}_1 \\ |
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x_2 \\ |
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\dot{x}_2 |
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\end{bmatrix} |
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</math> |
Latest revision as of 10:37, 4 December 2009
By Jimmy Apablaza
This problem is described in Exercise 14, Section 7.6 (page 323) of the A first Course in Differential Equations textbook, 8ED (ISBN 0-534-41878-3).
Problem Statement
Derive the system of differential equations describing the straight-line vertical motion of the coupled spring shown in Figure 1. Use Laplace transform to solve the system when , , and , , , and .
Solution
At positions and , the masses and are in equilibrium. Thus, the motion equations for and are,
- ∴
- ∴
where and represent the Newton's Second Law of Motion and and represent the net forces acting in the masses.
Laplace Transform
Applying the Laplace Transform to the motion equations and plugging the values of , , , , , , , , and for this systems, we obtain,
Finally, solving for and yields,
Inverse Laplace Transform
First, we recognize that
On the other hand, we identify that , and so . Hence, we fix the expression by multiplying and dividing by ,
A plot of the oscillatory motion is shown on Figure 2.
Initial-Value & Final-Value Theorem
The initial-value and final-value theorem can be useful the finding the behavior of a function at small and large times respectively. By definition, the Initial-Value Theorem is,
and the Final-Value Theorem is,
Thus, applying both theorems to our the Laplace Transforms,
Bode Plot
The Bode plot for the Transfer Functions
and
can be easily done using a program like Octave or MATLAB. The code is displayed below. From Figure 3 we may notice that the Amplitude vs. Frequency plot for both functions overlaps. The peak amplitude occurs at seconds, as well as the phase switching as shown in the Phase vs. Frequency plot.
h1=tf([-1],[1 0 3]); h2=tf([1],[1 0 3]); bode(h1,'b',h2,'-.r'); legend('H_1(s)','H_2(s)') grid on;
Magnitude Frequency Response
Considering the Transfer Functions and described in the Bode Plot section, we notice that there are no values for the s-variable that make and equal zero.
From the Bode Plot, we notice that the break point occurs at ≈ 1.73 seconds, so the
Convolution
By definition, the convolution of two functions is,
where refers to the inverse Laplace Transform. Assuming that is a dummy variable of integration, and is the impulse function , the convolution for our system is,
which is the same solution yields by the inverse Laplace Transform
State Equation Model
By definition, the the state equation is stated as
Now, consider the motion equations described in the Solution section,
Solving for and yields,
Finally, we let , , , and be the state variables. Thus,