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<h2 class="hd hd-2 unit-title">3.1. Laplace transform of a damped sinusoid.</h2>
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Activity
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(We encourage discussing this problem on the forum. It is worth more points than a regular problem.) </p>
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Damped sinusoids are important signals! Compute [mathjaxinline]\mathcal{L}(e^{at}\cos (\omega t);s)[/mathjaxinline] where [mathjaxinline]a[/mathjaxinline] is real by the method used on the previous page. Don't forget to find the region of convergence. </p>
<p>
<p style="display:inline">[mathjaxinline]\mathcal{L}(e^{at}\cos (\omega t);s)=[/mathjaxinline]</p>
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Region of convergence: <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div class="choicegroup capa_inputtype" id="inputtype_rec3-tab1-problem1_3_1">
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<input type="radio" name="input_rec3-tab1-problem1_3_1" id="input_rec3-tab1-problem1_3_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="rec3-tab1-problem1_3_1-choice_2-label" for="input_rec3-tab1-problem1_3_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_rec3-tab1-problem1_3_1"> <text> [mathjaxinline]|s|&gt;0[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">3.2. Laplace transform of higher derivatives.</h2>
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<p>
We can go on to compute the effect of Laplace transform on the second derivative. To make the calculation clearer, let's write [mathjaxinline]g(t)=f'(t)[/mathjaxinline] and [mathjaxinline]G(s)[/mathjaxinline] for its Laplace transform; so we know that [mathjaxinline]G(s)=sF(s)-f(0)[/mathjaxinline]. Then </p><table id="a0000000336" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000337"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle f^{\prime \prime }(t)=g'(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:left; border:none">
[mathjaxinline]\displaystyle \, sG(s)-g(0)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000338"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
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[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \, s(sF(s)-f(0))-g(0)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000339"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</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 \, s^2F(s)-f(0)s-f'(0)\, .[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
This process continues; we find that [mathjaxinline]\mathcal{L}(f^{(n)}(t);s)[/mathjaxinline] depends upon [mathjaxinline]f(0),f'(0),\ldots ,f^{(n-1)}(0)[/mathjaxinline]: </p><table id="a0000000340" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><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></table>
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Find the Laplace Transform
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Suppose that the Laplace transform of a function [mathjaxinline]x(t)[/mathjaxinline] is [mathjaxinline]X(s)[/mathjaxinline]. Moreover, suppose that </p>
<table id="a0000000341" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000342">
<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle x(0)[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle 2[/mathjaxinline]
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<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.33)</td>
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<tr id="a0000000343">
<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \dot x(0)[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle -1[/mathjaxinline]
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<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.34)</td>
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<p>
Use linearity and the [mathjaxinline]t[/mathjaxinline]-derivative formula to find the Laplace transform of the function: </p>
<table id="a0000000344" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000345">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle y(t)[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \ddot x + \dot x +2 x.[/mathjaxinline]
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<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none;text-align:right" class="eqnnum">(3.35)</td>
</tr>
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<p>
(Type <b class="bf">X</b> for [mathjaxinline]X(s)[/mathjaxinline]. Note that the answer is case sensitive.) </p>
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<p style="display:inline">[mathjaxinline]\mathcal{L}\left(y(t); s\right)=[/mathjaxinline]</p>
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