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<h2 class="hd hd-2 unit-title">4.1. Applying Laplace transform to solve DEs.</h2>
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<p><b class="bfseries">Objectives</b></p><p>
After completing this lecture the student will be able to </p><ol class="enumerate"><li value="1"><p>
Solve <b class="bfseries"><span style="color:#0000FF">initial value problems</span></b> using Laplace transform methods. </p></li><li value="2"><p>
Employ the method of <b class="bfseries"><span style="color:#0000FF">partial fractions</span></b> to find the <b class="bfseries"><span style="color:#0000FF">inverse Laplace transform</span></b> of [mathjaxinline]Q(s)/P(s)[/mathjaxinline] where [mathjaxinline]\mathrm{deg}Q < \mathrm{deg}P[/mathjaxinline] and [mathjaxinline]P(s)[/mathjaxinline] has only simple roots. </p></li></ol>
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<h2 class="hd hd-2 unit-title">4.2. Application to differential equations.</h2>
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<p>
The [mathjaxinline]t[/mathjaxinline]-derivative rule shows that differentiation in the time domain becomes multiplication by [mathjaxinline]s[/mathjaxinline] in the frequency domain (up to the annoying initial condition term). This leads to the amazing fact that </p><p><b class="bfseries"><span style="color:#0000FF">Fact :</span></b> Laplace transform carries differential equations to algebraic equations. </p><p>
So in using Laplace transform to solve a differential equation, the process is as follows: </p><center><img src="/assets/courseware/v1/e48fa15e0104564e851fec28d1d2f37b/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c3-laplacediagram.png" width="600"/></center><div><br/></div><p>
The missing ingredient here is getting back from the [mathjaxinline]s[/mathjaxinline]-world into the [mathjaxinline]t[/mathjaxinline]-world. We will do this by using the rules and calculations of the Laplace transform to find [mathjaxinline]x(t)[/mathjaxinline] with the desired Laplace transform [mathjaxinline]X(s)[/mathjaxinline]. To make this work, we need to know the following rule. </p><p><b class="bf">Inverse rule.</b> A continuous function [mathjaxinline]f(t)[/mathjaxinline] for [mathjaxinline]t\geq 0[/mathjaxinline] is determined by its Laplace transform [mathjaxinline]F(s)[/mathjaxinline] (if it exists). We use the notation [mathjaxinline]\mathcal{L}^{-1}(F(s);t) = f(t)[/mathjaxinline] or [mathjaxinline]\mathcal{L}^{-1}(F(s)) = f(t)[/mathjaxinline]. </p><p><p><b class="bfseries">Example 2.1 </b> What function of [mathjaxinline]t[/mathjaxinline] has [mathjaxinline]\displaystyle F(s) = \frac{1}{s-5}[/mathjaxinline] as its Laplace transform? </p><p><b class="bfseries"><span style="color:#0000FF">Solution:</span></b> We know that the function [mathjaxinline]\, f(t) = e^{5t}[/mathjaxinline] has [mathjaxinline]\displaystyle F(s) = \frac{1}{s-5}[/mathjaxinline] as its Laplace transform. Every other continuous function that has this Laplace transform agrees with this one for [mathjaxinline]t\geq 0[/mathjaxinline]. We write [mathjaxinline]\displaystyle \mathcal{L}^{-1}\left( \frac{1}{s-5}\right) = e^{5t}[/mathjaxinline] for [mathjaxinline]t\geq 0[/mathjaxinline]. </p></p>
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Inverse Laplace practice 1
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<p>
Use linearity and the Laplace table of transforms in the show/hide box below to find the inverse Laplace transform. </p>
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<div class="hideshowbox">
<h4 onclick="hideshow(this);" style="margin: 0px">Laplace Table<span class="icon-caret-down toggleimage"/></h4>
<div class="hideshowcontent">
<table id="a0000000355" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000356">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle 1[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac1{s}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;0[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.42)</td>
</tr>
<tr id="a0000000357">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle e^{rt}[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac1{s-r}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;\mathrm{Re}\, r[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.43)</td>
</tr>
<tr id="a0000000358">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \cos \omega t[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac{s}{s^2+\omega ^2}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;0[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.44)</td>
</tr>
<tr id="a0000000359">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \sin \omega t[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac{\omega }{s^2+\omega ^2}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;0[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.45)</td>
</tr>
<tr id="a0000000360">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle t[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac1{s^2}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s &gt; 0[/mathjaxinline]
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<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.46)</td>
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<p>
Find a function whose Laplace transform is [mathjaxinline]\displaystyle \frac{3}{s+1}.[/mathjaxinline] </p>
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(Enter your answer as a function of [mathjaxinline]t[/mathjaxinline].) <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_lec4-tab2-problem1_2_1" class="text-input-dynamath capa_inputtype textline">
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Inverse Laplace practice 2
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Use linearity and the Laplace table of transforms in the show/hide box below to find the inverse Laplace transform. </p>
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<h4 onclick="hideshow(this);" style="margin: 0px">Laplace Table<span class="icon-caret-down toggleimage"/></h4>
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<table id="a0000000362" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000363">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle 1[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac1{s}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;0[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.47)</td>
</tr>
<tr id="a0000000364">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle e^{rt}[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac1{s-r}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;\mathrm{Re}\, r[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.48)</td>
</tr>
<tr id="a0000000365">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \cos \omega t[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac{s}{s^2+\omega ^2}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;0[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.49)</td>
</tr>
<tr id="a0000000366">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \sin \omega t[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac{\omega }{s^2+\omega ^2}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;0[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.50)</td>
</tr>
<tr id="a0000000367">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle t[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac1{s^2}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s &gt; 0[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.51)</td>
</tr>
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<p>
Find [mathjaxinline]\displaystyle \mathcal{L}^{-1}\left(\frac{s}{s^2+16}; t\right).[/mathjaxinline] </p>
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Inverse Laplace practice 3
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Use linearity and the Laplace table of transforms in the show/hide box below to find the inverse Laplace transform. </p>
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<h4 onclick="hideshow(this);" style="margin: 0px">Laplace Table<span class="icon-caret-down toggleimage"/></h4>
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<table id="a0000000369" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000370">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle 1[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac1{s}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;0[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.52)</td>
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<tr id="a0000000371">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle e^{rt}[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac1{s-r}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;\mathrm{Re}\, r[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.53)</td>
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<tr id="a0000000372">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \cos \omega t[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac{s}{s^2+\omega ^2}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;0[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.54)</td>
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<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \sin \omega t[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
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<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac{\omega }{s^2+\omega ^2}, \qquad \qquad[/mathjaxinline]
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&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;0[/mathjaxinline]
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<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.55)</td>
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[mathjaxinline]\displaystyle t[/mathjaxinline]
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<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac1{s^2}, \qquad \qquad[/mathjaxinline]
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&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s &gt; 0[/mathjaxinline]
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<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.56)</td>
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<p>
Find [mathjaxinline]\displaystyle \mathcal{L}^{-1}\left(\frac{1}{s^2+16}; t\right).[/mathjaxinline] </p>
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<th class="formulainput" scope="row" rowspan="4">Operators</th>
<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
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<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
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<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
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<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
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<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
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<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
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<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
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<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
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<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
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<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
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Inverse Laplace practice 4
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Use linearity and the Laplace table of transforms in the show/hide box below to find the inverse Laplace transform. </p>
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<h4 onclick="hideshow(this);" style="margin: 0px">Laplace Table<span class="icon-caret-down toggleimage"/></h4>
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<table id="a0000000376" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000377">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle 1[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac1{s}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;0[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.57)</td>
</tr>
<tr id="a0000000378">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle e^{rt}[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac1{s-r}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;\mathrm{Re}\, r[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.58)</td>
</tr>
<tr id="a0000000379">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \cos \omega t[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac{s}{s^2+\omega ^2}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;0[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.59)</td>
</tr>
<tr id="a0000000380">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \sin \omega t[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac{\omega }{s^2+\omega ^2}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s&gt;0[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.60)</td>
</tr>
<tr id="a0000000381">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle t[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \frac1{s^2}, \qquad \qquad[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:center; border:none">
&#160;
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \mathrm{Re}\, s &gt; 0[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.61)</td>
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Find [mathjaxinline]\displaystyle \mathcal{L}^{-1}\left(\frac{1}{s^2} + \frac{1}{s^2+16}; t\right).[/mathjaxinline] </p>
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<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
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Enter <font color="#0078b0">lambda </font> for \(\lambda \)
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Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
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<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
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<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
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<p><h3>Coming next</h3></p><p>
We'll look at a simple example of using this method of Laplace transforms to solve a differential equation next, and we will go through a more systematic approach after we review partial fractions. </p>
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<h2 class="hd hd-2 unit-title">4.3. Example: solving a first order LTI.</h2>
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<p><p><b class="bfseries">Example 3.1 </b> Let's start simple, and solve the first order constant coefficient linear differential equation </p><table id="a0000000383" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\dot x+kx=ky[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
using the Laplace transform. This equation models Newtonian cooling, or the water level in a bay changing under the influence of ocean tides, where [mathjaxinline]\, y \,[/mathjaxinline] is the external temperature, or the water level outside of the bay. We'll assume that the input signal is given by </p><table id="a0000000384" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]y=e^{-t}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
In solving an ODE using Laplace transform, one must specify initial ([mathjaxinline]t=0[/mathjaxinline]) conditions. So here we suppose that [mathjaxinline]\, x(0) \,[/mathjaxinline] is known. </p></p><p><b class="bfseries"><span style="color:#0000FF">Solution:</span></b></p><p><b class="bf">Step 1</b>. Apply [mathjaxinline]\, \mathcal{L}\,[/mathjaxinline] to the differential equation: </p><table id="a0000000385" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax](sX(s)-x(0))+kX(s)=kY(s)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The result is a linear equation for [mathjaxinline]\, X(s)\,[/mathjaxinline], so it is easy to solve. </p><p><b class="bf">Step 2</b>. Solve the equation for [mathjaxinline]\, X(s) \,[/mathjaxinline]. </p><p>
First put all terms not involving [mathjaxinline]X(s)[/mathjaxinline] as a factor on the right: </p><table id="a0000000386" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax](s+k)X(s)=x(0)+kY(s)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none;text-align:right">(3.62)</td></tr></table><p>
Then solve: </p><table id="a0000000387" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]X(s)=\frac{x(0)+kY(s)}{s+k}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p><b class="bf">Step 3</b>. Find the inverse Laplace transform [mathjaxinline]\, \mathcal{L}^{-1}(X(s);t)[/mathjaxinline]. </p><p>
Our input signal is [mathjaxinline]\, y(t)=e^{-t} \,[/mathjaxinline]. We know the Laplace transform of [mathjaxinline]\, y(t) \,[/mathjaxinline] is given by </p><table id="a0000000388" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]Y(s)=\frac{1}{s+1}\, ,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
so </p><table id="a0000000389" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]X(s)=\frac{x(0)+k/(s+1)}{s+k}=\frac{x(0)}{s+k}+\frac{k}{(s+1)(s+k)}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
We want to find a function [mathjaxinline]\, x(t)\,[/mathjaxinline] whose Laplace transform is [mathjaxinline]\, X(s) \,[/mathjaxinline]. By the Linearity rule, we can treat each term separately. The first one is easy; by linearity again, and using the known Laplace transform for an exponential function, we have </p><table id="a0000000390" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x(0)e^{-kt}\rightsquigarrow \frac{x(0)}{s+k}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The second term is trickier, and actually has to be treated in two cases according to whether [mathjaxinline]\, k\neq 1 \,[/mathjaxinline] or [mathjaxinline]\, k=1\,[/mathjaxinline]. For now we'll do the more generic one and suppose that [mathjaxinline]\, k\neq 1 \,[/mathjaxinline]. </p><p>
The relevant technique is now the method of partial fractions, which is important enough to get its own section. For the moment, we'll just recall the simplest case, which is what we need here: if [mathjaxinline]\, k\neq 1 \,[/mathjaxinline], there are unique numbers [mathjaxinline]\, a \,[/mathjaxinline] and [mathjaxinline]\, b\,[/mathjaxinline] such that </p><table id="a0000000391" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{k}{(s+1)(s+k)}=\frac{a}{s+1}+\frac{b}{s+k}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none;text-align:right">(3.63)</td></tr></table><p>
The traditional way to find [mathjaxinline]a[/mathjaxinline] and [mathjaxinline]b[/mathjaxinline] is to cross multiply and set coefficients equal: </p><table id="a0000000392" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{k}{(s+1)(s+k)}=\frac{a(s+k)+b(s+1)}{(s+1)(s+k)}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Another way to do this, that is more fun, is the “cover-up" method. Here's how that works: to find [mathjaxinline]a[/mathjaxinline], multiply the entire equation [mathjaxinline]\displaystyle \, \frac{k}{(s+1)(s+k)}=\frac{a}{s+1}+\frac{b}{s+k} \,[/mathjaxinline] by [mathjaxinline]\, (s+1) \,[/mathjaxinline] and set [mathjaxinline]\, s=-1 \,[/mathjaxinline]. So </p><table id="a0000000393" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{k}{s+k}=a+(s+1)\frac{b}{s+k}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
becomes </p><table id="a0000000394" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{k}{-1+k}=a[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
so [mathjaxinline]\, a=k/(k-1) \,[/mathjaxinline]. Informally, we “covered up" the factor [mathjaxinline]\, s+1 \,[/mathjaxinline] in the denominators, and the term not involving it in the denominator, and get an equation for the coefficient. We can do this calculation without writing by taking the expression [mathjaxinline]\, \displaystyle \frac{k}{(s+1)(s+k)} \,[/mathjaxinline] and “covering up" the [mathjaxinline]\, (s+1) \,[/mathjaxinline] and substituting [mathjaxinline]\, s=-1 \,[/mathjaxinline] in the remaining expression. That is, cover-up gives [mathjaxinline]\, \displaystyle \frac{k}{s+k} \,[/mathjaxinline]. Then, substitution gives [mathjaxinline]\, \displaystyle \frac{k}{-1+k} \,[/mathjaxinline]. </p><p>
Similarly, to find [mathjaxinline]\, b\,[/mathjaxinline], multiply through by [mathjaxinline]\, s+k\,[/mathjaxinline] and set [mathjaxinline]\, s=-k\,[/mathjaxinline]: </p><table id="a0000000395" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{k}{s+1}=(s+k)\frac{a}{s+1}+b[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
becomes </p><table id="a0000000396" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{k}{-k+1}=b[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
so [mathjaxinline]\, b=-k/(k-1)\,[/mathjaxinline]. </p><p>
Putting these back into the equation gives </p><table id="a0000000397" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{k}{(s+1)(s+k)}=\frac{k}{k-1}\left(\frac{1}{s+1}-\frac{1}{s+k}\right)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The payoff is that it's easy to recognize this as the Laplace transform of a function. By linearity and using the known Laplace transform for an exponential function: </p><table id="a0000000398" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{e^{-t}-e^{-kt}}{k-1}\rightsquigarrow \frac{1}{(s+1)(s+k)}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Putting all this together, using linearity as usual, we find that </p><table id="a0000000399" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x(t)=x(0)e^{-kt}+\frac{k}{k-1}(e^{-t}-e^{-kt})\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The presence of [mathjaxinline]\, k-1\,[/mathjaxinline] in the denominator confirms that the case [mathjaxinline]\, k=1\,[/mathjaxinline] has to be treated separately. We will explain how to do that in a later section. </p><p>
Note the structure of this expression: The term [mathjaxinline]\, \displaystyle \frac{k}{k-1}e^{-t} \,[/mathjaxinline] is the “steady state," the exponential response given by the ERF. The two terms involving [mathjaxinline]\, e^{-kt} \,[/mathjaxinline] together form the transient required to produce the given initial condition. One of these terms has the effect of canceling the initial value of the [mathjaxinline]\, \displaystyle \frac{k}{k-1}e^{-t} \,[/mathjaxinline] term, and other imposes the desired initial value. </p>
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<h3 class="hd hd-3 problem-header">Recap</h3><p>
To recap, in this example we did the following: </p><center><img src="/assets/courseware/v1/ef74a9c06a57a0d7774a46a903f14d75/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c3-laplacediagram-v2.png" width="750"/></center><div><br/></div>The inverse Laplace transform involved using partial fractions to write the algebraic equation in a form that we could look up the inverse Laplace function easily because we could recognize it as the Laplace transform of an exponential function. <p>
Of course we could have found this result using the ERF and finding the appropriate transient. The Laplace transform gives us another tool for solving differential equations. More importantly, we will see that it provides a connection to the transfer function of any LTI system. </p>
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<h2 class="hd hd-2 unit-title">4.4. Partial fractions.</h2>
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<p>
We'll begin with a factorization that sticks to real numbers, and give a more systematic factorization involving complex roots later. So to begin let's suppose that the coefficients of both polynomials are real. </p><p><b class="bfseries"><span style="color:#FF7800">Step 1:</span></b> The degree of [mathjaxinline]Q(s)[/mathjaxinline] might be bigger than the degree of [mathjaxinline]P(s)[/mathjaxinline]. In that case, use the division algorithm you learned about in high school algebra to write </p><table id="a0000000400" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{Q(s)}{P(s)}=F(s)+\frac{R(s)}{P(s)}\quad ,\quad \deg R(s)<\deg P(s)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p><b class="bfseries"><span style="color:#FF7800">Step 2:</span></b> The fundamental theorem of algebra assures us that [mathjaxinline]P(s)[/mathjaxinline] factors into a product of linear and quadratic factors: terms of the form [mathjaxinline](s+b)[/mathjaxinline] and [mathjaxinline](s^2+bs+k)[/mathjaxinline] (where the quadratic terms do not have real roots). The "partial fractions" algorithm tells us that if [mathjaxinline]\deg Q(s)<\deg P(s)[/mathjaxinline] and there are not repeated factors in the denominator, then the rational function [mathjaxinline]Q(s)/P(s)[/mathjaxinline] can be written as a sum of terms of the form </p><table id="a0000000401" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{a}{s+b}\quad \text {and}\quad \frac{as+c}{s^2+bs+k}\, ,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
(where the constants [mathjaxinline]a[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], [mathjaxinline]c[/mathjaxinline], [mathjaxinline]k[/mathjaxinline] are all real) and that there is only one such expression. </p><p><p><b class="bfseries">Example 4.1 </b> Find the partial fraction decomposition of </p><table id="a0000000402" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{s^2}{s^2-1}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table></p><p><b class="bfseries"><span style="color:#FF7800">Solution:</span></b> When the degree of the numerator is greater or equal to the degree of the denominator, you must first divide through, and then apply partial fractions decomposition to the remainder. </p><table id="a0000000403" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{s^2}{s^2-1}= 1+\frac{1}{s^2-1}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The denominator [mathjaxinline]s^2-1[/mathjaxinline] factors into a product of linear factors [mathjaxinline]s^2-1=(s-1)(s+1)[/mathjaxinline]. So we want to find a partial fractions decomposition in the form </p><table id="a0000000404" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{1}{(s-1)(s+1)} = \frac{A}{s-1} + \frac{B}{s+1}.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
To find [mathjaxinline]A[/mathjaxinline], multiply through by [mathjaxinline]s-1[/mathjaxinline]: </p><table id="a0000000405" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{1}{(s+1)} = A + (s-1)\frac{B}{s+1},[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
and set [mathjaxinline]s=1[/mathjaxinline]: </p><table id="a0000000406" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{1}{(1+1)} = A + 0\cdot \frac{B}{s+1},[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
which gives us [mathjaxinline]A=1/2[/mathjaxinline]. </p><p>
To find [mathjaxinline]B[/mathjaxinline], multiply through by [mathjaxinline]s+1[/mathjaxinline]: </p><table id="a0000000407" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{1}{(s-1)} = (s+1)\cdot \frac{A}{s-1} +B.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
and set [mathjaxinline]s=-1[/mathjaxinline]: </p><table id="a0000000408" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{1}{(-1-1)} = + 0\cdot \frac{A}{s-1} +B,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
which gives us [mathjaxinline]B=-1/2[/mathjaxinline]. Therefore the partial fractions decomposition is </p><table id="a0000000409" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{s^2}{s^2-1}= 1+\frac{1/2}{s-1}-\frac{1/2}{s+1}.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
We have just described the algorithm for finding these expressions if the roots of the denominator are real and distinct. Here is a video reminding you of that method. </p>
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<h3 class="hd hd-2">Cover-up method in practice</h3>
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Practice with cover-up
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Use the method of cover-up to find the coefficients [mathjaxinline]A[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], and [mathjaxinline]C[/mathjaxinline]: </p>
<p>
[mathjaxinline]\displaystyle \frac{s^2+3s+8}{(s-1)(s-2)(s+5)} = \frac{A}{s-1} + \frac{B}{s-2} + \frac{C}{s+5}[/mathjaxinline] </p>
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(Enter as fractions.) </p>
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<p style="display:inline">[mathjaxinline]A=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]B=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]C=[/mathjaxinline]</p>
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<h3 class="hd hd-3 problem-header">Complex cover-up</h3><p>
Cover-up works equally well with the quadratic factors of the denominator. The method is the same: multiply through by the quadratic expression, and then set [mathjaxinline]s[/mathjaxinline] equal to a root. </p><p><p><b class="bfseries">Example 4.2 </b> Find [mathjaxinline]a[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], and [mathjaxinline]c[/mathjaxinline] so that </p><table id="a0000000416" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{s^2+1}{s^3+2s^2+2s} = \frac{a}{s} + \frac{bs+c}{s^2+2s+2}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p><b class="bfseries"><span style="color:#FF7800">Solution:</span></b> Multiply through by [mathjaxinline]s[/mathjaxinline] and set [mathjaxinline]s=0[/mathjaxinline] to see that [mathjaxinline]a = 1/2[/mathjaxinline]. </p><p>
Then multiply through by [mathjaxinline]s^2+2s+2[/mathjaxinline] and set [mathjaxinline]s[/mathjaxinline] equal to either of the roots of this quadratic, say [mathjaxinline]s=-1+i[/mathjaxinline]. To make things clearer we'll swap sides. </p><table id="a0000000417" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000418"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle bs+c + \frac{a}{s}(s^2+2s+2)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{s^2+1}{s}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000419"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle b(-1+i)+c[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{(-1+i)^2+1}{-1+i}\, .[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
This gives </p><table id="a0000000420" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax](c-b)+bi=\frac{(1-2i-1)+1}{-1+i}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Rationalize the denominator by multiplying numerator and denominator by [mathjaxinline]\overline{-1+i}=-1-i[/mathjaxinline]: </p><table id="a0000000421" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax](c-b)+bi=\frac{(1-2i)(-1-i)}{2}=\frac{-3+i}{2}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
This implies that [mathjaxinline]b=1/2[/mathjaxinline], and then [mathjaxinline]c=b-3/2=-1[/mathjaxinline]. So </p><table id="a0000000422" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{s^2+1}{s^3+2s^2+s} = \frac{1/2}{s} + \frac{(1/2)s-1}{s^2+2s+2}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Note the process we call cover-up: </p><ol class="enumerate"><li value="1"><p>
Write out the proposed partial fractions decomposition. </p></li><li value="2"><p>
Pick one of the linear or quadratic denominators; cover up all the other terms in the partial fractions sum and the chosen factor in the original denominator (and in the chosen term's denominator). </p></li><li value="3"><p>
Set [mathjaxinline]s[/mathjaxinline] equal to a root of the chosen factor and solve. </p></li></ol></p>
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Practice complex cover-up
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Use complex cover up to find the partial fraction decomposition of </p>
<p>
[mathjaxinline]\displaystyle \frac{1}{s^3-1} = \frac{a}{s-1} +\frac{bs+c}{s^2+s+1}[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]a=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]b=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]c=[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">4.5. Partial fractions continued: repeated roots</h2>
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<h3 class="hd hd-3 problem-header">Repeated roots and partial fraction decomposition</h3><p>
If the denominator has repeated factors, you have to allow more complicated terms in the partial fractions expression. For example if [mathjaxinline](s+3)^2[/mathjaxinline] occurs in the denominator, you have to include </p><table id="a0000000428" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{a}{s+3}+\frac{b}{(s+3)^2}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
If [mathjaxinline](s^2+2s+2)^2[/mathjaxinline] occurs, you have to allow </p><table id="a0000000429" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{as+b}{s^2+2s+2}+\frac{cs+d}{(s^2+2s+2)^2}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
You need to include all powers between the first power and the power that occurs in the denominator. </p><p><p><b class="bfseries">Example 5.1 </b> [mathjaxinline]\displaystyle \frac{3s+2}{s(s+1)^2}=\frac{a}{s}+\frac{b}{s+1}+\frac{c}{(s+1)^2}[/mathjaxinline].</p></p><p><b class="bfseries"><span style="color:#FF7800">Solution:</span></b> We can use cover-up to find the constants [mathjaxinline]a[/mathjaxinline] and [mathjaxinline]c[/mathjaxinline], but not [mathjaxinline]b[/mathjaxinline]. </p><p>
Covering up [mathjaxinline]s[/mathjaxinline]: </p><table id="a0000000430" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]a=\frac{2}{(0+1)^2}=2\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Covering up [mathjaxinline](s+1)^2[/mathjaxinline]: </p><table id="a0000000431" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]c=\frac{3(-1)+2}{-1}=1\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
There are many ways to find [mathjaxinline]b[/mathjaxinline]. For example, we can just pick some other value for [mathjaxinline]s[/mathjaxinline], say [mathjaxinline]s=1[/mathjaxinline]: </p><table id="a0000000432" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{2}{1}+\frac{b}{2}+\frac{1}{4}=\frac{5}{4}\, ,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
giving [mathjaxinline]b=-2[/mathjaxinline]. </p>
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Practice cover-up with repeated factors
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Find the constants [mathjaxinline]a[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], and [mathjaxinline]c[/mathjaxinline]: </p>
<p>
[mathjaxinline]\displaystyle \frac{s^2}{(s+1)^2(s-1)} = \frac{a}{s-1}+\frac{b}{s+1} + \frac{c}{(s+1)^2}[/mathjaxinline] </p>
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(Enter as fractions.) </p>
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<p style="display:inline">[mathjaxinline]a=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]c=[/mathjaxinline]</p>
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<h3 class="hd hd-3 problem-header">Complete decomposition</h3><p>
A significant feature of the complex numbers is that any nonzero polynomial factors into a product of linear factors. Each quadratic term we saw above factors as a product [mathjaxinline](s-r)(s-\overline r)[/mathjaxinline]. This lets us write out partial fraction expansion involving only terms of the form </p><table id="a0000000436" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{a}{(s-r)^ k}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
where [mathjaxinline]r[/mathjaxinline] ranges over the roots of [mathjaxinline]P(s)[/mathjaxinline] and [mathjaxinline]k[/mathjaxinline] ranges up to the multiplicity of the root [mathjaxinline]r[/mathjaxinline]. </p><p><p><b class="bfseries">Example 5.2 </b> We can write [mathjaxinline]\displaystyle \frac{s+1}{s^2+1}=\frac{a}{s-i}+\frac{b}{s+i}[/mathjaxinline]. </p></p><p><b class="bfseries"><span style="color:#FF7800">Solution:</span></b> To find [mathjaxinline]a[/mathjaxinline] and [mathjaxinline]b[/mathjaxinline], we can use cover-up. Rewrite as </p><table id="a0000000437" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{s+1}{(s-i)(s+i)}=\frac{a}{s-i}+\frac{b}{s+i}.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Cover up the [mathjaxinline]s-i[/mathjaxinline] term and plug in [mathjaxinline]s=i[/mathjaxinline] to get </p><table id="a0000000438" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{i+1}{2i}=a.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Cover up the [mathjaxinline]s+i[/mathjaxinline] term and plug in [mathjaxinline]s=-i[/mathjaxinline]: </p><table id="a0000000439" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{-i+1}{-2i}=b.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Notice that [mathjaxinline]a[/mathjaxinline] and [mathjaxinline]b[/mathjaxinline] are complex conjugates; this reflects the fact that the coefficients in the original expression are real. </p>
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Practice with a full linear decomposition
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Write [mathjaxinline]\displaystyle \frac{s+2i}{s^2+1} = \frac{a}{s+i} + \frac{b}{s-i}[/mathjaxinline] as a partial fraction with only linear denominators. </p>
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<p style="display:inline">[mathjaxinline]a=[/mathjaxinline]</p>
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<p>Let's get some more practice with calculations.</p>
<p>Recall that our <a href="/courses/course-v1:OCW+18.031+2019_Spring/courseware/week1/lec1/5" target="_blank">automobile suspension</a> (in suitable units!) is modeled by the equation</p>
<table id="a0000000444" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
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<td class="equation" style="width: 80%; border: none;">[mathjax]m\ddot x+b\dot x+kx=b\dot y+ky[/mathjax]</td>
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<p>with</p>
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<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]m=1 \, ,\quad b=2\, ,\quad k=3\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>Suppose that the road we are bouncing over is modeled by</p>
<table id="a0000000446" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]y=4\sin (5t)[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>and that we start from rest initial conditions: [mathjaxinline]x(0) = 0[/mathjaxinline] , and [mathjaxinline]\dot x(0) = 0[/mathjaxinline].</p>
<p>So, using operator notation, our equation is</p>
<table id="a0000000447" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax](D^2+2D+3I)x=(2D+3I)4\sin (5t)[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>Applying Laplace transform (with rest initial conditions) replaces application of [mathjaxinline]D[/mathjaxinline] by multiplication by [mathjaxinline]s[/mathjaxinline]. Using the value of [mathjaxinline]L(\sin (5t);s)[/mathjaxinline] from the table, we find</p>
<table id="a0000000448" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax](s^2+2s+3)X=(2s+3)4\frac{5}{s^2+25}[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>or</p>
<table id="a0000000449" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]X=\frac{20(2s+3)}{(s^2+25)(s^2+2s+3)}\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>We need to find a function that has this as its Laplace transform. It's a rational function – a quotient of a polynomial by a polynomial – but a pretty complicated one, and it certainly doesn't show up in our look-up table!</p>
<p>This is where the technique of partial fractions comes in handy. The degree of the numerator is 1, which is less than the degree of the denominator (which is 4), so partial fractions assures us that we can write the right hand side as a sum of fractions with denominators [mathjaxinline]s^2+25[/mathjaxinline] and [mathjaxinline]s^2+2s+3[/mathjaxinline]. The entries in our table with quadratic denominators have those denominators written as completed squares – i.e. in the form [mathjaxinline](s+a)^2+b[/mathjaxinline] – so let's be smart and start by completing the squares in our case. [mathjaxinline]s^2+25[/mathjaxinline] is already in this form. For the other,</p>
<table id="a0000000450" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]s^2+2s+3=(s+1)^2+2\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>So the partial fractions process guarantees that there are real constants [mathjaxinline]a[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], [mathjaxinline]c[/mathjaxinline], and [mathjaxinline]d[/mathjaxinline] such that</p>
<table id="a0000000451" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]\frac{20(2s+3)}{(s^2+25)(s^2+2s+3)}= \frac{as+b}{s^2+25}+\frac{c(s+1)+d}{(s+1)^2+2}\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>We can discover the values of [mathjaxinline]a[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], [mathjaxinline]c[/mathjaxinline], and [mathjaxinline]d[/mathjaxinline] by cross-multiplying and equating coefficients. But it will be more fun to use <b class="bf">complex coverup</b>.</p>
<p>Recall the process:</p>
<p><b class="bfseries"><span style="color: #0000ff;"> Complex coverup:</span></b></p>
<p><b class="bf">Multiply through by one factor of the denominator and then set [mathjaxinline]s[/mathjaxinline] equal to a root of that factor.</b></p>
<p>So multiply through by [mathjaxinline]s^2+25[/mathjaxinline]:</p>
<table id="a0000000452" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]\frac{20(2s+3)}{s^2+2s+3}= (as+b)+(s^2+25)\frac{c(s+1)+d}{(s+1)^2+2}\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>Now set [mathjaxinline]s=5i[/mathjaxinline] (a root of [mathjaxinline]s^2+25[/mathjaxinline]):</p>
<table id="a0000000453" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]\frac{20(10i+3)}{-25+10i+3}=5ai+b[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>Now it's just a matter of simplifying the left hand side. Multiplying numerator and denominator by the conjugate of the denominator,</p>
<table id="a0000000454" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]b+5ai=20\frac{(3+10i)(-22-10i)}{22^2+10^2} =20\frac{(-66+100)+(-220-30)i}{584}[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>so</p>
<table id="a0000000455" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]b=20\frac{34}{584}=\frac{85}{73}[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>and</p>
<table id="a0000000456" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]a=-4\frac{250}{584}=-\frac{125}{73}\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
</div>
</div>
</div>
</div>
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<h2 class="hd hd-2 unit-title">4.7. Rest initial conditions.</h2>
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<p>Recall that in the formula for the [mathjaxinline]t[/mathjaxinline]-derivative, we found that</p>
<table id="a0000000457" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
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<td class="equation" style="width: 80%; border: none;">[mathjax]f^{(n)}(t)\rightsquigarrow s^ nF(s)-\left(f(0)s^{n-1}+f'(0)s^{n-2}+\cdots +f^{(n-1)}(0)\right)\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>The most important part of the formula for [mathjaxinline]\mathcal{L}(f^{(n)}(t);s)[/mathjaxinline] is [mathjaxinline]s^ nF(s)[/mathjaxinline]. The whole business would be easier without the other terms. This can be arranged by assuming that the function and the relevant derivatives are all zero at [mathjaxinline]t=0[/mathjaxinline]. This is the hypothesis of “rest initial conditions."</p>
<p><b class="bf">Observation:</b> If [mathjaxinline]f(0)=\cdots =f^{(n-1)}(0)=0[/mathjaxinline], then</p>
<table id="a0000000458" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]f^{(n)}(t)\rightsquigarrow s^ nF(s)\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>If we are working with an LTI system that is <b class="bf">stable</b>, the initial conditions don't matter much anyway, and we should feel free to let them be whatever is convenient. In earlier lectures we studied the sinusoidal solution; now, in this section, we will study the solution with rest initial conditions. This is also called the ZSR solution, for “zero state response." See the <a href="/courses/course-v1:OCW+18.031+2019_Spring/courseware/week2/lec4/8" target="_blank">footnote</a> for more on ZSR.</p>
<p><b class="bf">Example.</b> Suppose we want to analyze the automobile suspension system modeled in <a href="/courses/course-v1:OCW+18.031+2019_Spring/courseware/week1/lec1/5" target="_blank">Activity 5 of Lecture 1</a>.</p>
<center><img src="/assets/courseware/v1/5627b85bbc5900d279614ebe43b66d7d/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_ps2-susp3.svg" width="400px" style="margin: 0px 10px 10px 10px;" /></center>
<p>One quarter of the cabin of the car is represented by the mass at the top of the picture. As the car moves along a road, potholes cause the wheels to move up and down; this is represented by vertical motion of the platform at the bottom. This is modeled by the equation</p>
<table id="a0000000459" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]m\ddot x + b \dot x + kx = b\dot y + k y,[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>where the input signal [mathjaxinline]y(t)[/mathjaxinline] is the displacement of the blue platform upwards from some arbitrarily chosen zero position, and the system response [mathjaxinline]x(t)[/mathjaxinline] is the displacement of a point in the car cabin (the orange mass). The modeling assumption is that [mathjaxinline]x=y[/mathjaxinline] when the spring is relaxed, and [mathjaxinline]\dot x=\dot y[/mathjaxinline] when the platform and mass are moving together.</p>
<p>Let's study this using Laplace transform. The rest initial condition assumption on [mathjaxinline]x[/mathjaxinline] is that [mathjaxinline]x(0)=\dot x(0)=0[/mathjaxinline]. Since only the first derivative of [mathjaxinline]y[/mathjaxinline] enters the equation, we need only assume that [mathjaxinline]y(0)=0[/mathjaxinline]. Then if we apply Laplace transform to the equation we get</p>
<table id="a0000000460" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]ms^2X+bsX+kX=bsY +kY[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>or, collecting terms in [mathjaxinline]X[/mathjaxinline] and dividing through by [mathjaxinline]ms^2+bs+k=P(s)[/mathjaxinline],</p>
<table id="a0000000461" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]X(s)=\frac{bs+k}{ms^2+bs+k}Y(s)[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>The factor on the right is precisely the transfer function of the system! This is a general fact:</p>
<p><b class="bfseries">Theorem 7.1 </b> If we have the differential equation</p>
<table id="a0000000462" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]P(D)x=Q(D)y[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>with rest initial conditions, then</p>
<table id="a0000000463" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]P(s)X(s)=Q(s)Y(s).[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>Note that this is equivalent to saying that</p>
<table id="a0000000464" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]X(s)=H(s)Y(s)\,[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>where [mathjaxinline]H(s)[/mathjaxinline] is the <b class="bfseries"><span style="color: #ff7800;">transfer function</span></b> .</p>
<p>This is a fantastic result! In the [mathjaxinline]s[/mathjaxinline]-domain, the system simply multiplies the input signal by the transfer function (subject to rest initial conditions). This is the key observation of this entire course.</p>
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<h2 class="hd hd-2 unit-title">4.8. Footnote.</h2>
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<h3 class="hd hd-3 problem-header"> ZIR/ZSR.</h3><p>
In Section 3 we solved the initial value problem [mathjaxinline]\dot x+kx=ke^{-t}[/mathjaxinline] with arbitrary initial condition [mathjaxinline]x(0)[/mathjaxinline], and got the following result: </p><table id="a0000000465" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x(t)=x(0)e^{-kt}+\frac{k}{k-1}(e^{-t}-e^{-kt})\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Our method dealt separately with the term involving the initial condition [mathjaxinline]x(0)[/mathjaxinline], and so our formula for [mathjaxinline]x(t)[/mathjaxinline] has two parts: the first part gives a transient with the correct initial condition; the second part consists in the steady state solution plus a transient designed to make it vanish at [mathjaxinline]t=0[/mathjaxinline]. </p><p>
This is a standard and useful decomposition of the system response of any LTI system, with given initial conditions: say [mathjaxinline]P(D)x=Q(D)y[/mathjaxinline], with initial values of [mathjaxinline]x[/mathjaxinline] and some of its derivatives given. </p><p>
The <b class="bfseries"><span style="color:#0000FF">zero state response</span></b> or <b class="bfseries"><span style="color:#0000FF">ZSR</span></b> , is the solution to [mathjaxinline]P(D)x=Q(D)y[/mathjaxinline] with rest initial conditions. (The “state" here refers to the initial condition.) </p><p>
The <b class="bfseries"><span style="color:#0000FF">zero input response</span></b> or <b class="bfseries"><span style="color:#0000FF">ZIR</span></b> is the solution to the homogeneous equation [mathjaxinline]P(D)x=0[/mathjaxinline], with the same initial conditions. It is a transient (assuming that the system is stable). </p><p>
The solution to the initial value problem is the sum </p><table id="a0000000466" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x=ZSR+ZIR\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table>
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