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<h2 class="hd hd-2 unit-title">5.1. Poles and vibrations activity.</h2>
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Poles of a Laplace transform
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<p>
There is a connection between the poles in the frequency domain and exponents in the time domain. We know that </p>
<ul class="itemize">
<li>
<p>
[mathjaxinline]\displaystyle e^{at}\cos (bt) = \mathrm{Re}\, \left(e^{(a\pm bi)t} \right)[/mathjaxinline] </p>
</li>
<li>
<p>
[mathjaxinline]\displaystyle e^{at}\sin (bt) = \mathrm{Im}\, \left(e^{(a + bi)t} \right)[/mathjaxinline] </p>
</li>
</ul>
<p>
Find the poles of [mathjaxinline]\mathcal{L}\left(e^{-4t}\cos (5t); s\right)[/mathjaxinline]. </p>
<p>
(Enter as a list of complex numbers, separated by commas: e.g. 2-3*i, 2+3*i.) </p>
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Find the poles of [mathjaxinline]\mathcal{L}\left(e^{-4t}\sin (5t); s\right)[/mathjaxinline]. </p>
<p>
(Enter as a list of complex numbers, separated by commas: e.g. 2-3*i, 2+3*i.) </p>
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Mathlet activity
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<p>
The mathlet above displays the graph of the function </p>
<table id="a0000000669" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]x(t) = Ae^{at}\cos (bt) + Be^{ct}\cos (dt).[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
The mathlet starts with [mathjaxinline]B=0[/mathjaxinline]. Play with the applet to get familiar with the controls. </p>
<p>
Now set [mathjaxinline]A=2.0[/mathjaxinline], [mathjaxinline]B=2.0[/mathjaxinline]. (<b class="bfseries"><span style="color:#FF7800">Pro tip:</span></b> Clicking on a hashmark sets the slider at that point.) Please place poles (either using the [mathjaxinline]a[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], [mathjaxinline]c[/mathjaxinline], [mathjaxinline]d[/mathjaxinline] sliders or dragging the poles directly) at [mathjaxinline]-0.4\pm 2i[/mathjaxinline] and [mathjaxinline]-0.1\pm 2i[/mathjaxinline]. </p>
<p>
Which will have a bigger effect on the long-term behavior of [mathjaxinline]x(t)[/mathjaxinline]? </p>
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<text> Moving the poles at [mathjaxinline]-0.4\pm 2i[/mathjaxinline] a little.</text>
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<text> Moving the poles at [mathjaxinline]-0.1\pm 2i[/mathjaxinline] a little.</text>
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Activity: Poles and graphs
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<p>
The following are the pole diagrams of 6 functions of the form </p>
<table id="a0000000670" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]x(t) = Ae^{at}\cos (bt) +Be^{ct}\cos (dt).[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
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<p>
Below that are 6 graphs. Match the graph to the pole diagram. </p>
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<img src="/assets/courseware/v1/c05f8c44e81b1ea87d8406476ffe194b/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_l4b-8-p5.svg" width="300px" style="margin: 0px 10px 10px 10px"/>
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Beats
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Now sum 2 sinusoidal oscillations. Set [mathjaxinline]A=B=2[/mathjaxinline] to get better resolution, and set [mathjaxinline]a=c=0[/mathjaxinline]: there is no decay, both [mathjaxinline]f(t)[/mathjaxinline] and [mathjaxinline]g(t)[/mathjaxinline] are undamped sinusoids. All the poles are on the imaginary axis. </p>
<p>
Observe what happens when the frequencies of the two oscillations are far apart: [mathjaxinline]d=8[/mathjaxinline], [mathjaxinline]b=1[/mathjaxinline]. When the frequencies are close together ([mathjaxinline]d=7[/mathjaxinline] and [mathjaxinline]b=6.5[/mathjaxinline]), you see the phenomenon called beats. You would hear this as &#8220;waawaawaa..." &#8211; the amplitude rises and falls. Let's control the time [mathjaxinline]T[/mathjaxinline] from one maximum amplitude to the next by adjusting the location of the poles. </p>
<iframe style=" display:block; border-left-width: 0px; padding-left: 0px; padding-top: 0px; padding-right: 0px; padding-bottom: 0px; border-right-width: 0px; border-top-width: 0px; border-bottom-width: 0px;" src="https://mathlets1803.surge.sh/polesVibrations.html" width="1100 px" height="640 px"/>
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How would we make [mathjaxinline]T[/mathjaxinline] smaller? <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_rec5-tab1-problem4_2_1">
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Please suggest a formula for [mathjaxinline]T[/mathjaxinline] in terms of [mathjaxinline]b[/mathjaxinline] and [mathjaxinline]d[/mathjaxinline], based on experimentation with the mathlet. </p>
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Growth and decay
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Which of the following functions grows the fastest in amplitude as [mathjaxinline]t\rightarrow \infty[/mathjaxinline]? </p>
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Which of the following functions decays the fastest in amplitude as [mathjaxinline]t\rightarrow \infty[/mathjaxinline]? </p>
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<h3 class="hd hd-3 problem-header">Exponential type practice exercise</h3><p>
Recall from <a href="/courses/course-v1:OCW+18.031+2019_Spring/courseware/laplace/lec3a/8" target="_blank">Lecture 3</a>: a function [mathjaxinline]\, f(t)[/mathjaxinline] is of <b class="bfseries"><span style="color:#0000FF">exponential type [mathjaxinline]k[/mathjaxinline],</span></b> for some real number [mathjaxinline]k[/mathjaxinline], if for some constant [mathjaxinline]C>0[/mathjaxinline] </p><table id="a0000000672" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle \left| f(t) \right| \leq C e^{kt} \qquad \text { for all } t\geq 0.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Match the following. (Pick the smallest available exponential type.) </p><table class="tabular" cellspacing="0" style="table-layout:auto"><tr><td style="text-align:left; border:none">
[mathjaxinline]f(t)[/mathjaxinline] </td><td style="text-align:left; border:none"><b class="bf">region of convergence</b></td><td style="text-align:left; border:none"><b class="bf">exponential type</b></td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]e^ t[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]\mathrm{Re}\, (z) > 0[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]-2[/mathjaxinline] </td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]e^{-t}\sin (4t)[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]\mathrm{Re}\, (z) > 0[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]-1[/mathjaxinline] </td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]t^{30}e^{-t}+1[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]\mathrm{Re}\, (z) > 1[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]0[/mathjaxinline] </td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]\sin (t) + 1[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]\mathrm{Re}\, (z) > -1[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]1[/mathjaxinline] </td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]e^{-2t}+e^{-3t}\cos (t)[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]\mathrm{Re}\, (z) > -2[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]0[/mathjaxinline] </td></tr></table><p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">Solution.<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><table class="tabular" cellspacing="0" style="table-layout:auto"><tr><td style="text-align:left; border:none">
[mathjaxinline]f(t)[/mathjaxinline] </td><td style="text-align:left; border:none"><b class="bf">region of convergence</b></td><td style="text-align:left; border:none"><b class="bf">exponential type</b></td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]e^ t[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]\mathrm{Re}\, (z) > 1[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]1[/mathjaxinline] </td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]e^{-t}\sin (4t)[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]\mathrm{Re}\, (z) > -1[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]-1[/mathjaxinline] </td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]t^{30}e^{-t}+1[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]\mathrm{Re}\, (z) > 0[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]0[/mathjaxinline] </td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]\sin (t) + 1[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]\mathrm{Re}\, (z) > 0[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]0[/mathjaxinline] </td></tr><tr><td style="text-align:left; border:none">
[mathjaxinline]e^{-2t}+e^{-3t}\cos (t)[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]\mathrm{Re}\, (z) > -2[/mathjaxinline] </td><td style="text-align:left; border:none">
[mathjaxinline]-2[/mathjaxinline] </td></tr></table></div><p class="hideshowbottom" onclick="hideshow(this);" style="margin: 0px"><a href="javascript: {return false;}">Show</a></p></div></p><p>
When [mathjaxinline]F(s) = \mathcal{L}(f(t);s)[/mathjaxinline] has a pole at [mathjaxinline]s=r[/mathjaxinline], the value of the integral defining it becomes infinite, and fails to converge. The right-most poles in [mathjaxinline]F(s)[/mathjaxinline] mark the left edge of the region of convergence. </p><SCRIPT src="/assets/courseware/v1/631e447105fca1b243137b21b9ed6f90/asset-v1:OCW+18.031+2019_Spring+type@asset+block/latex2edx.js" type="text/javascript"/><LINK href="/assets/courseware/v1/daf81af0af57b85a105e0ed27b7873a0/asset-v1:OCW+18.031+2019_Spring+type@asset+block/latex2edx.css" rel="stylesheet" type="text/css"/>
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Find the minimum exponential type
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The Laplace transform of a function [mathjaxinline]h_1[/mathjaxinline] is </p>
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<td class="equation" style="width:80%; border:none">[mathjax]\mathcal{L}(h_1) = \frac{s^2-3s+2}{(s^2-6)(s^2+2s+5)}.[/mathjax]</td>
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Without calculating the function, what is the minimum (infimum) exponential type of [mathjaxinline]h_1[/mathjaxinline]? </p>
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Find the minimum exponential type again
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The Laplace transform of a function [mathjaxinline]h_2[/mathjaxinline] is </p>
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<td class="equation" style="width:80%; border:none">[mathjax]\mathcal{L}(h_2) = \frac{s^2-3s+2}{(s^2-4)(s^2+2s+5)}.[/mathjax]</td>
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Without calculating the function, what is the minimum (infimum) exponential type of [mathjaxinline]h_2[/mathjaxinline]? </p>
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<h2 class="hd hd-2 unit-title">5.2. Conclusions on growth and decay.</h2>
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We can make some conclusions from this investigation. Most important: </p><ol class="enumerate"><li value="1"><p><b class="bf">[mathjaxinline]f(t)[/mathjaxinline] decays exponentially to zero exactly when all of the poles of [mathjaxinline]F(s)[/mathjaxinline] have negative real part; that is, they all lie in the left half plane.</b></p><p>
To put this in context in our analysis of LTI systems: the homogeneous solutions to an LTI system are transient when the poles of the transfer function of that system all have negative real part. </p></li><li value="2"><p><b class="bf">The right-most poles of [mathjaxinline]F(s)[/mathjaxinline] determine the long-term behavior of [mathjaxinline]f(t)[/mathjaxinline].</b></p><p>
If the real part of some pole is positive, then [mathjaxinline]f(t)[/mathjaxinline] grows exponentially as [mathjaxinline]t\to \infty[/mathjaxinline]. But in either case, we can be very precise about the rate of growth or decay. </p></li></ol><p><p><b class="bfseries">Example 2.1 </b> What is the exponential type of the function whose Laplace transform has the following pole diagram? </p><center><img src="/assets/courseware/v1/9543c8dac1a985bcb40b977b31a857d3/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c2-polestability-5.svg" width="300px" style="margin: 0px 10px 10px 10px"/></center></p><p><b class="bfseries"><span style="color:#0000FF">Solution:</span></b> The dominant term is [mathjaxinline]Ae^{-t}\cos (2t-\phi )[/mathjaxinline]. This function is of exponential type [mathjaxinline]-1[/mathjaxinline]. </p>
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Identify the minimum exponential type
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What is the minimum (infimum) exponential type of the function whose Laplace transform has the following pole diagram? </p>
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<h3 class="hd hd-3 problem-header">Poles with multiplicity</h3><p>
A further complication is the possibility of a repeated factor in the denominator. This results in a “double pole" (if the factor occurs squared) or higher in the pole diagram. We have seen the effect of this, too: </p><table id="a0000000678" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]t^ n\rightsquigarrow \frac{n!}{s^{n+1}}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
so, by the [mathjaxinline]s[/mathjaxinline]-shift rule, </p><table id="a0000000679" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]t^ ne^{at}\rightsquigarrow \frac{n!}{(s-a)^{n+1}}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
A similar rule holds for repeated complex roots: an [mathjaxinline](n+1)[/mathjaxinline]-fold root produces multiplication of the expected [mathjaxinline]f(t)[/mathjaxinline] by [mathjaxinline]t^ n[/mathjaxinline]. This is a general case of “resonance." </p><p>
The main point is that [mathjaxinline]t^ n e^{at}[/mathjaxinline] is of exponential type [mathjaxinline]b[/mathjaxinline] for any [mathjaxinline]b > a[/mathjaxinline]. Thus [mathjaxinline]\mathcal{L}(t^ ne^{at}; s)[/mathjaxinline] converges for [mathjaxinline]\mathrm{Re\, }(s) > a[/mathjaxinline]. This is exactly the same region of convergence as for [mathjaxinline]e^{at}[/mathjaxinline] which has exponential type [mathjaxinline]b[/mathjaxinline] for any [mathjaxinline]b\geq a[/mathjaxinline]. </p>
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