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<h2 class="hd hd-2 unit-title">1.1. Review differential equations.</h2>
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<h3 class="hd hd-3 problem-header">Objectives.</h3><p>
After completing this lecture the student will be able to </p><ol class="enumerate"><li value="1"><p>
find a basis of solutions to a homogeneous linear constant coefficient ODE, in terms of exponentials and sinusoids, and determine whether the equation is stable or not; </p></li><li value="2"><p>
find a particular solution to a linear constant coefficient ODE with right hand side made up of exponentials and sinusoids; </p></li><li value="3"><p>
use the principle of superposition to find the general solution in terms of these first two procedures; </p></li><li value="4"><p>
model mechanical systems using differential equations, using the language of input signals and system response and the standard form [mathjaxinline]P(D)x=Q(D)y[/mathjaxinline] in terms of characteristic polynomials. </p></li><li value="5"><p>
determine the complex gain of a system and extract from it the gain and phase lag, for any given input frequency. </p></li></ol><p>
This lecture is a review, so many explanations are very brief. In particular, we assume that you understand complex numbers. </p>
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<h2 class="hd hd-2 unit-title">1.2. Motivation.</h2>
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<p>In this introductory lecture we will review some of the key ideas from <em><a href="https://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/" target="_blank">18.03 Differential Equations</a></em>. You'll refresh your skill at modeling systems using ordinary differential equations and the input/output model, at finding solutions of these equations, and in interpreting these solutions in terms of the original system. The key word is "complex gain."</p>
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<h2 class="hd hd-2 unit-title">1.3. A mass/spring/dashpot system.</h2>
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<p>To gain an appreciation for the kind of question we are discussing, study the Mathlet below. It represents a mass/spring/dashpot system. We are interested in how the orange mass moves. The spring is attached to a wall, and the far end of the dashpot is being moved.</p>
<p>In analyzing any system, we have to specify the <b class="bfseries"><span style="color: #0000ff;">input signal</span></b> and the output signal or system response. Here we declare the position of the dashpot piston to be the input signal [mathjaxinline]y(t)[/mathjaxinline], and the position of the mass, [mathjaxinline]x(t)[/mathjaxinline], to be the system response. The applet specifies that the input signal is sinusoidal; indeed it is given by [mathjaxinline]\cos (\omega t)[/mathjaxinline], with amplitude 1 and angular frequency [mathjaxinline]\omega[/mathjaxinline]. What you observe is that the motion of the mass, the system response, is <b class="bf">also sinusoidal</b>, with the <b class="bf">same period</b>. This sinusoidal system response is the "steady state response."</p>
<p>We can adjust the angular frequency [mathjaxinline]\omega[/mathjaxinline] using a slider. The Mathlet reminds us that the period is given by [mathjaxinline]2\pi /\omega[/mathjaxinline], so that large [mathjaxinline]\omega[/mathjaxinline] means small period, and [mathjaxinline]\omega =1[/mathjaxinline] means period [mathjaxinline]2\pi[/mathjaxinline]. The angular frequency is measured in radians per second. The angular frequency determines the <b class="bf">frequency</b>; it is [mathjaxinline]\omega /2\pi[/mathjaxinline], measured in cycles per second, or <b class="bf">Hertz</b>. We will always prefer the angular frequency because it makes our formulas neater.</p>
<p>There are three "system parameters": the mass [mathjaxinline]m[/mathjaxinline], the damping constant [mathjaxinline]b[/mathjaxinline], and the spring constant [mathjaxinline]k[/mathjaxinline]. The mass is fixed at [mathjaxinline]m=1[/mathjaxinline], but the other two can be adjusted using the sliders. Spend a few minutes now playing with this Javascript Mathlet. You can animate it with the [[mathjaxinline]>>[/mathjaxinline]] key under the graphing window.</p>
<p><iframe style="display: block; border-width: 0px; padding: 0px;" src="https://mathlets1803.surge.sh/ampPhaseSecondOrderII.html" width="1100 px" height="640 px"></iframe></p>
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Activity 1: Input and Output
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Use the Mathlet to answer the following questions. </p>
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What does the blue curve represent? <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_lec0-tab3-problem1_2_1">
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<input type="radio" name="input_lec0-tab3-problem1_3_1" id="input_lec0-tab3-problem1_3_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="lec0-tab3-problem1_3_1-choice_2-label" for="input_lec0-tab3-problem1_3_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_lec0-tab3-problem1_3_1"> <text> The system response.</text>
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Activity 2: Gain
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The <b class="bf">gain</b> of an LTI system is measured separately for each input frequency [mathjaxinline]\omega[/mathjaxinline]: </p>
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<td class="equation" style="width:80%; border:none">[mathjax]\hbox{gain}=\frac{\hbox{amplitude of system response}}{\hbox{amplitude of input signal}}\, ,[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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where the input signal is a sinusoid of frequency [mathjaxinline]\omega[/mathjaxinline]. In this Mathlet, the amplitude of the input signal is fixed at 1, so the gain equals the amplitude of the sinusoidal system response. </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/ampPhaseSecondOrderII.html" width="1100 px" height="640 px"/>
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Take [mathjaxinline]b=1.0[/mathjaxinline], [mathjaxinline]k=2.0[/mathjaxinline], and [mathjaxinline]\omega =1.00[/mathjaxinline], and measure the gain. (Note: you can set the sliders to certain values by clicking on the hashmarks.) <p style="display:inline">gain=</p> <div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_lec0-tab3-problem2_2_1" class=" capa_inputtype inline textline">
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Stay with these values of [mathjaxinline]b[/mathjaxinline] and [mathjaxinline]k[/mathjaxinline], but vary [mathjaxinline]\omega[/mathjaxinline]. What is the maximum gain you observe for this system? </p>
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<p style="display:inline">Maximum gain=</p>
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<p>
Now pick some other value of [mathjaxinline]b[/mathjaxinline] and [mathjaxinline]k[/mathjaxinline], and vary the input frequency [mathjaxinline]\omega[/mathjaxinline]. What is the maximum gain in these cases? (Can you formulate a general conjecture?) </p>
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<p style="display:inline">Maximum gain=</p>
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Activity 3: Resonant frequency
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The angular frequency at which the gain is maximal is the "resonant angular frequency" [mathjaxinline]\omega _ r[/mathjaxinline]. </p>
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Set [mathjaxinline]b=1.0[/mathjaxinline] and [mathjaxinline]k=1.0[/mathjaxinline]. Measure the value of [mathjaxinline]\omega _ r[/mathjaxinline]. <p style="display:inline">[mathjaxinline]\omega _ r=[/mathjaxinline]</p> <div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_lec0-tab3-problem3_2_1" class=" capa_inputtype inline textline">
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Same question for [mathjaxinline]b=1.0[/mathjaxinline] and [mathjaxinline]k=2.0[/mathjaxinline]. <p style="display:inline">[mathjaxinline]\omega _ r=[/mathjaxinline]</p> <div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="inputtype_lec0-tab3-problem3_3_1" class=" capa_inputtype inline textline">
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<p>
Same question for [mathjaxinline]b=1.0[/mathjaxinline] and [mathjaxinline]k=4.0[/mathjaxinline]. <p style="display:inline">[mathjaxinline]\omega _ r=[/mathjaxinline]</p> <div class="inline" tabindex="-1" aria-label="Question 3" role="group"><div id="inputtype_lec0-tab3-problem3_4_1" class=" capa_inputtype inline textline">
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Please suggest a formula for [mathjaxinline]\omega _ r[/mathjaxinline] in terms of [mathjaxinline]k[/mathjaxinline], for this value of [mathjaxinline]b=1.0[/mathjaxinline]. <p style="display:inline">[mathjaxinline]\omega _ r=[/mathjaxinline]</p> <div class="inline" tabindex="-1" aria-label="Question 4" role="group"><div id="formulaequationinput_lec0-tab3-problem3_5_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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Now select at least one other value for [mathjaxinline]b[/mathjaxinline], and make the same measurements. Based on these experiments, please suggest a formula for [mathjaxinline]\omega _ r[/mathjaxinline] that might be valid for all values of [mathjaxinline]b[/mathjaxinline] and [mathjaxinline]k[/mathjaxinline]. <p style="display:inline">[mathjaxinline]\omega _ r=[/mathjaxinline]</p> <div class="inline" tabindex="-1" aria-label="Question 5" role="group"><div id="formulaequationinput_lec0-tab3-problem3_6_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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<h3 class="hd hd-3 problem-header">Conclusions and what's next</h3><p>
In the rest of this lecture we will remind you how we model systems like the one in the above activities. This allows us to answer questions about gain, and many other analogous questions, by computation. Some of your discoveries about the system in these will be addressed in Lecture 2. In Lecture 2, we see that this example is a mechanical analogue of an extremely simple AM radio receiver, and the analysis we carry out determines important features of the radio, such as the frequency it is tuned to and how sharp the tuning is. </p><p><p><b class="bfseries">Remark 3.1 </b> Notice that the phase lag becomes zero at the resonant frequency. We'll be able to explain why. (This is a special feature of this system.) </p></p><p><b class="bfseries"><span style="color:#FF7800">Note about notation:</span></b> Almost always, in this course, the independent variable is time, denoted [mathjaxinline]t[/mathjaxinline]. Functions of time might be denoted [mathjaxinline]x(t)[/mathjaxinline], [mathjaxinline]y(t)[/mathjaxinline], [mathjaxinline]z(t)[/mathjaxinline], [mathjaxinline]f(t)[/mathjaxinline]; or, for short, [mathjaxinline]x,y,z,f[/mathjaxinline]. As you just saw in the examples we use the notation </p><table id="a0000000004" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\dot x=\frac{dx}{dt}\quad ,\quad \ddot x=\frac{d^2x}{dt^2}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table>
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<h2 class="hd hd-2 unit-title">1.4. Modeling the system.</h2>
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<p>An important part of <em><a href="https://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/" target="_blank">18.03 Differential Equations</a></em>, and of this course, is the process of modeling a mechanical or electrical system using ordinary differential equations. The basic process has the following steps.</p>
<ol class="enumerate">
<li value="1">
<p><b class="bf">Draw a diagram of the system.</b></p>
</li>
<li value="2">
<p><b class="bf">Identify and give symbols for the parameters of the system with units.</b></p>
</li>
<li value="3">
<p><b class="bf">Declare the input signal and the system response.</b></p>
</li>
<li value="4">
<p><b class="bf">Decide what you are looking for: a solution satisfying specific initial conditions or a steady state solution.</b></p>
</li>
<li value="5">
<p><b class="bf">Write down a differential equation relating the input signal and the system response</b>, e.g. using Newton's "[mathjaxinline]F=ma[/mathjaxinline]" in the mechanical case or Kirchhoff's laws in the electrical case.</p>
</li>
<li value="6">
<p><b class="bf">Rewrite the equation in standard form.</b></p>
</li>
</ol>
<p>In our case:</p>
<ol class="enumerate">
<li value="1">
<p>Here is a diagram of the system we are studying.</p>
<center><img src="/assets/courseware/v1/f751f2526727cbcaa5439b2eb4ff232d/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c1_smd_1.svg" width="150px" style="margin: 0px 10px 10px 10px;" /></center></li>
<li value="2">
<p>We have already named the relevant system parameters, on the diagram: [mathjaxinline]m[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], and [mathjaxinline]k[/mathjaxinline]. Note we now allow [mathjaxinline]m[/mathjaxinline] to take a value other than [mathjaxinline]1.0[/mathjaxinline].</p>
</li>
<li value="3">
<p>The input signal is given by the position of the dashpot piston; we will write [mathjaxinline]y(t)[/mathjaxinline] or just [mathjaxinline]y[/mathjaxinline] for this function of time. We have to declare also which is the positive direction for [mathjaxinline]y[/mathjaxinline]: It's value increases when the piston moves up. The system response we are interested in is the position of the mass, which we will write [mathjaxinline]x(t)[/mathjaxinline]. It's value also increases as the mass moves up. We also declare [mathjaxinline]x=0[/mathjaxinline] to be the position at which the spring is relaxed. Thus when [mathjaxinline]x>0[/mathjaxinline] the spring is compressed and exerts a force pointing down, while if [mathjaxinline]x<0[/mathjaxinline] it is extended and exerts a force pointing up.</p>
<p>These declarations are arbitrary – we could instead make [mathjaxinline]y[/mathjaxinline] become positive when the piston moves <b class="bf">down</b>, or say [mathjaxinline]x=1[/mathjaxinline] when the spring is relaxed. We would get the same analysis, but expressed somewhat differently. Specifying that the spring is relaxed when [mathjaxinline]x=0[/mathjaxinline] simplifies the expressions enough to make it a very good idea.</p>
</li>
<li value="4">
<p>We will assume that we are working in the linear regime. That is, we assume first that the force of the spring on the mass is proportional to [mathjaxinline]x[/mathjaxinline], i.e. [mathjaxinline]x[/mathjaxinline] is small enough that Hooke's law is valid. Second we assume that the force exerted by the dashpot on the mass is proportional to the speed so [mathjaxinline]\dot x[/mathjaxinline] and [mathjaxinline]\dot y[/mathjaxinline] are small enough that the damping effect of the dashpot is linear. Thus "Hooke's law" tells us – the force exerted by the spring is proportional to [mathjaxinline]x[/mathjaxinline]. The force exerted on the mass by the dashpot is proportional to the speed at which the plunger is moving through the piston. The two proportionality constants are [mathjaxinline]k[/mathjaxinline] and [mathjaxinline]b[/mathjaxinline], so Newton's law gives us</p>
<table id="a0000000005" 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=F=-kx-b(\dot x-\dot y)\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
</li>
<li value="5">
<p>"Standard form" means: terms coming from the input signal on the right, and terms coming from the system response on the left. So we have:</p>
<table id="a0000000006" 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\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
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</li>
</ol>
<p>This is the differential equation relating the input signal [mathjaxinline]y[/mathjaxinline] and the system response [mathjaxinline]x[/mathjaxinline]. It is valid whether the input signal is sinusoidal or not, and it allows for non-sinusoidal system responses.</p>
<ul class="itemize">
<li>
<p>The model does reflect the assumptions that we could use linear force laws.</p>
</li>
<li>
<p>In this modeling process we did not need to assume [mathjaxinline]m[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], and [mathjaxinline]k[/mathjaxinline] were constant. But we will observe this system over a small enough time scale that we may assume that [mathjaxinline]m[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], and [mathjaxinline]k[/mathjaxinline] are effectively constants.</p>
</li>
</ul>
<p>Putting these two assumptions together we get a <b class="bf">linear ordinary differential equation</b> with <b class="bf">constant coefficients</b>.</p>
<p>A system exhibiting both these hypotheses is called a <b class="bfseries"><span style="color: #0000ff;">Linear time invariant</span></b> or <b class="bfseries"><span style="color: #0000ff;">LTI</span></b> system. This entire course revolves around understanding the behavior of LTI systems, based on an analysis of solutions of the corresponding "LTI" differential equations</p>
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<h2 class="hd hd-2 unit-title">1.5. Operator notation.</h2>
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It is very useful to write differential equations with constant coefficients, like the ones we have just seen, using "operator notation." In this notation, [mathjaxinline]D[/mathjaxinline] stands for Differentiation with respect to time: </p><table id="a0000000007" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]D=\frac{d}{dt}\, ,\quad Dx=\frac{dx}{dt}=\dot x\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
We'll also write [mathjaxinline]I[/mathjaxinline] for the Identity operator, so [mathjaxinline]Ix=x[/mathjaxinline]. Then </p><table id="a0000000008" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]D^2x=\frac{d}{dt}\frac{d}{dt}x=\frac{d^2x}{dt^2}=\ddot x\, ,\quad (5D^3-3I)y=5\frac{d^3y}{dt^3}-3y\, ,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
and so on. Any polynomial </p><table id="a0000000009" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(s)=a_ ns^ n+\cdots +a_1s+a_0[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
determines a linear combination of powers of [mathjaxinline]D[/mathjaxinline], a <b class="bf">differential operator</b> [mathjaxinline]P(D)[/mathjaxinline] with </p><table id="a0000000010" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(D)x=a_ nD^ nx+\cdots +a_1Dx+a_0x= a_ n\frac{d^ nx}{dt^ n}+\cdots +a_1\frac{dx}{dt}+a_0x\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
In these terms, the kind of differential equation we are looking at, relating an input signal [mathjaxinline]y[/mathjaxinline] with the system response [mathjaxinline]x[/mathjaxinline], has the general form </p><table id="a0000000011" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><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></table><p>
The polynomials [mathjaxinline]P[/mathjaxinline] and [mathjaxinline]Q[/mathjaxinline] are the "characteristic polynomials" of the system. Thus, in the "driving through the dashpot" example, </p><table id="a0000000012" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(s)=ms^2+bs+k\, ,\quad Q(s)=bs \, ,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
and in the automobile suspension example, </p><table id="a0000000013" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(s)=ms^2+bs+k\, ,\quad Q(s)=bs+k \, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The operator notation enforces the two key assumptions we are making about the systems we are studying: </p><ol class="enumerate"><li value="1"><p>
They are <b class="bf">linear</b>. This has profound implications about the structure of the collection of system responses, generally captured by the statement of "superposition." For example, if [mathjaxinline]x_1[/mathjaxinline] is a system response to the input signal [mathjaxinline]y_1[/mathjaxinline], and [mathjaxinline]x_2[/mathjaxinline] is a system response to the input signal [mathjaxinline]y_2[/mathjaxinline], then the input signal [mathjaxinline]a_1y_1+a_2y_2[/mathjaxinline] (where [mathjaxinline]a_1[/mathjaxinline] and [mathjaxinline]a_2[/mathjaxinline] are constants) has as a response the linear combination [mathjaxinline]a_1x_1+a_2x_2[/mathjaxinline]. </p></li><li value="2"><p>
They are <b class="bf">time-invariant</b>. This again has a deep impact on the behavior of system responses. For example, if [mathjaxinline]x(t)[/mathjaxinline] is a system response to the input signal [mathjaxinline]y(t)[/mathjaxinline], then the delayed input signal [mathjaxinline]y(t-t_0)[/mathjaxinline] (where [mathjaxinline]t_0[/mathjaxinline] is a constant) has as a system response the delayed function [mathjaxinline]x(t-t_0)[/mathjaxinline]. </p></li></ol><p>
A system exhibiting both these hypotheses is called an <b class="bf">LTI system</b>. This entire course revolves around understanding the behavior of LTI systems, based on an analysis of solutions of the corresponding "LTI" differential equations. </p>
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Using operator notation, we can write the differential equation </p>
<table id="a0000000014" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \frac{d^4x}{dt^4}+2\frac{d^2x}{dt^2}+x=0.[/mathjaxinline]
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<td style="width:20%; border:none" class="eqnnum">&#160;</td>
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in the form </p>
<table id="a0000000016" 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]P\left(D\right)x=0.[/mathjax]</td>
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Find the operator [mathjaxinline]P\left(D\right)[/mathjaxinline]. </p>
<p><b class="bfseries"><span style="color:#FF7800">Tip :</span></b> Type <b class="bf">D^n</b> for [mathjaxinline]D^ n[/mathjaxinline]. Type <b class="bf">I</b> for the operator [mathjaxinline]D^0[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]P\left(D\right)=[/mathjaxinline]</p>
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Question 2.
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Using operator notation, we can write the differential equation that models the wheel of a car </p>
<table id="a0000000021" 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>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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in the form </p>
<table id="a0000000022" 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]P(D)x = Q(D)y.[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
Find [mathjaxinline]P(D)[/mathjaxinline] and [mathjaxinline]Q(D)[/mathjaxinline]. </p>
<p><b class="bfseries"><span style="color:#FF7800">Tip :</span></b> Type <b class="bf">D^n</b> for [mathjaxinline]D^ n[/mathjaxinline]. Type <b class="bf">I</b> for the operator [mathjaxinline]D^0[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]P(D)=[/mathjaxinline]</p>
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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/>
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<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
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<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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<h2 class="hd hd-2 unit-title">1.6. Sinusoids.</h2>
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Typical input signals of interest will be sinusoidal functions. They are more important than they may seem at first. For one thing, Fourier analysis shows that <b class="bf">any</b> periodic input signal can be expressed as a combination of sinusoids. So by superposition understanding the system response to sinusoids lets us understand the system response to general periodic signals. </p><p>
In fact, as we'll see, the most important system response to a sinusoidal input signal is almost always again sinusoidal, of the same period. So let's recall the languages surrounding sinusoidal functions. Suppose we have two sinusoids of the same period, as pictured below. </p><center><img src="/assets/courseware/v1/2da64272ed050b40aef2373afa83d2ee/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_sinusoids.png" width="500"/></center><div><br/></div><p>
Think of the blue one is the input signal and the orange one the system response. We can completely describe the system response to this input signal with just two numbers: </p><p>
(1) The <b class="bf">gain</b> [mathjaxinline]g[/mathjaxinline] is the ratio of the output amplitude to the input amplitude. </p><p>
(2) The <b class="bf">phase lag</b> [mathjaxinline]\phi[/mathjaxinline] is the number of radians the output signal is behind the input signal. </p><p>
Suppose for example (our "standard example") that the input signal is </p><table id="a0000000029" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]y(t)=A\cos (\omega t)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The sinusoidal system response can then be written </p><table id="a0000000030" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x_ p(t)=gA\cos (\omega t-\phi )\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
We can also write this in terms of the <b class="bf">time lag</b> </p><table id="a0000000031" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]t_0=\frac{\phi }{\omega }=\frac{\phi }{2\pi }P[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
where [mathjaxinline]P=2\pi /\omega[/mathjaxinline] is the period: [mathjaxinline]x_ p(t)=gy(t-t_0)[/mathjaxinline]. </p><p>
By linearity and time invariance, it follows that if the input signal [mathjaxinline]y(t)[/mathjaxinline] is <b class="bf">any</b> sinusoid of period [mathjaxinline]P[/mathjaxinline], the sinusoidal system response is </p><table id="a0000000032" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x_ p(t)=gy(t-t_0)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
for some gain [mathjaxinline]g\geq 0[/mathjaxinline] and some time lag [mathjaxinline]t_0[/mathjaxinline] between [mathjaxinline]0[/mathjaxinline] and [mathjaxinline]P[/mathjaxinline]. </p><p>
So a great deal of energy goes into understanding the gain of a system. The phase lag can also be important, but generally plays a secondary role. </p>
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The input of an LTI system is [mathjaxinline]3\sin (4t)[/mathjaxinline]. The gain is [mathjaxinline]g=1.6[/mathjaxinline], and the phase lag is [mathjaxinline]\phi = 0.4[/mathjaxinline] at that input frequency. </p>
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What is the steady state output signal at this frequency? <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div id="formulaequationinput_lec0-tab6-problem1_2_1" class="inputtype formulaequationinput">
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The input of an LTI system is [mathjaxinline]\cos (5t)[/mathjaxinline] and the output is </p>
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<td class="equation" style="width:80%; border:none">[mathjax]\cos (5t)-\sqrt {3}\sin (5t).[/mathjax]</td>
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Find the gain and phase lag at this input frequency. </p>
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<p style="display:inline">[mathjaxinline]g=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\phi =[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">1.7. The complex exponential.</h2>
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Part of the reason why sinusoids are important in the study of differential equations is their differentiation rules: </p><table id="a0000000038" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000039"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle D\cos (\omega t)=[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle -\omega \sin (\omega t)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000000040"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle D\sin (\omega t)=[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \omega \cos (\omega t)\, .[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
These are fairly simple, but not as simple as the exponential differentiation formula </p><table id="a0000000041" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]De^{ct}=ce^{ct}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
In fact the trigonometric differentiation formulas are contained in the exponential one, provided we let [mathjaxinline]c[/mathjaxinline] be a <b class="bf">complex</b> number. We may use <b class="bf">Euler's formula</b> </p><table id="a0000000042" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]e^{i\theta }=\cos (\theta )+i\sin (\theta )[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
as the <b class="bf">definition</b> of [mathjaxinline]e^{i\theta }[/mathjaxinline] for [mathjaxinline]\theta[/mathjaxinline] real. In the complex plane this is the point on the unit circle making an angle of [mathjaxinline]\theta[/mathjaxinline] with the positive real axis. Setting [mathjaxinline]\theta =\omega t[/mathjaxinline], we get a complex-valued function of the real parameter [mathjaxinline]t[/mathjaxinline]. It traces out the unit circle in the complex plane, traversed counterclockwise. We can find its derivative, the velocity: </p><table id="a0000000043" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]De^{i\omega t}=-\omega \sin (\omega t)+i\omega \cos (\omega t) =i\omega (\cos (\omega t)+i\sin (\omega t))=i\omega e^{i\omega t}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
You can check that the exponential rule </p><table id="a0000000044" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]e^{i\alpha }e^{i\beta }=e^{i(\alpha +\beta )}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
is equivalent to the addition laws for cosine and sine. </p><p>
Again enforcing the exponential law, we can define [mathjaxinline]e^ z[/mathjaxinline] for any complex number [mathjaxinline]z[/mathjaxinline]: writing [mathjaxinline]z=a+bi[/mathjaxinline] with [mathjaxinline]a[/mathjaxinline] and [mathjaxinline]b[/mathjaxinline] real, </p><table id="a0000000045" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]e^ z=e^{a+bi}=e^ a(\cos b+i\sin b)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
This is the point in the complex plane with polar coordinates </p><table id="a0000000046" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]|e^ z|=e^ a\, ,\quad \mathrm{arg}(e^ z)=b\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
It follows that </p><table id="a0000000047" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]De^{zt}=ze^{zt}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
for any complex constant [mathjaxinline]z[/mathjaxinline]. In fancy language, you can say that the exponential function [mathjaxinline]e^{zt}[/mathjaxinline] is an <b class="bf">eigenfunction</b> with eigenvalue [mathjaxinline]z[/mathjaxinline] for the differentiation operator [mathjaxinline]D[/mathjaxinline]. </p><p>
If we apply [mathjaxinline]D[/mathjaxinline] twice we find </p><table id="a0000000048" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]D^2e^{zt}=D(ze^{zt})=z^2e^{zt}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
and so on; so if we have a polynomial [mathjaxinline]P(z)[/mathjaxinline], we can compute the effect on [mathjaxinline]e^{zt}[/mathjaxinline] of the differential operator [mathjaxinline]P(D)[/mathjaxinline]: </p><table id="a0000000049" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(D)e^{zt}=P(z)e^{zt}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
This is without doubt the most important formula in the course! </p>
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Geometry of complex exponentials.
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As the real number [mathjaxinline]t[/mathjaxinline] tends to [mathjaxinline]+\infty[/mathjaxinline], how does the value of [mathjaxinline]e^{(-1+2i)t}[/mathjaxinline] move in the complex plane? </p>
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Possible answers (select one): ("Radially" means along part of a ray emanating from the origin.) </p>
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<text> Does not move.</text>
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<text> Clockwise along a circle.</text>
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<text> Counterclockwise along a circle.</text>
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<text> Spirals clockwise as it moves inward.</text>
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<text> Spirals counterclockwise as it moves inward.</text>
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<text> Spirals clockwise as it moves outward.</text>
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<text> Spirals counterclockwise as it moves outward.</text>
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<text> Along part of a hyperbola.</text>
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<text> Along part of a parabola.</text>
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Review argument.
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Let [mathjaxinline]\alpha[/mathjaxinline] be the complex root of [mathjaxinline]z^2+2z+2[/mathjaxinline] with positive imaginary part. What is the value of [mathjaxinline]\mathrm{arg}(\alpha )[/mathjaxinline] in the interval [mathjaxinline](-\pi ,\pi ][/mathjaxinline]? </p>
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<p style="display:inline">[mathjaxinline]\mathrm{arg}(\alpha )=[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">1.8. Solution of homogeneous equations.</h2>
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Our first application of the complex exponential is to solving <b class="bf">homogeneous</b> LTI equations: differential equations of the form </p><table id="a0000000051" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(D)x=0\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
For what constants [mathjaxinline]r[/mathjaxinline] is [mathjaxinline]e^{rt}[/mathjaxinline] a solution to this equation? Using [mathjaxinline]P(D)e^{rt}=P(r)e^{rt}[/mathjaxinline], and the fact that [mathjaxinline]e^{rt}[/mathjaxinline] is never zero (its magnitude is [mathjaxinline]e^{at}[/mathjaxinline] where [mathjaxinline]a[/mathjaxinline] is the real part of [mathjaxinline]r[/mathjaxinline]), we find that [mathjaxinline]e^{rt}[/mathjaxinline] is a solution exactly when </p><table id="a0000000052" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(r)=0\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p><b class="bf">The roots of the characteristic polynomial provide exponential solutions to the homogeneous equation.</b></p><p><b class="bf">Example 1:</b> [mathjaxinline]\dot x+kx=0[/mathjaxinline]. The characteristic polynomial is [mathjaxinline]P(r)=r+k[/mathjaxinline]; it has one root, namely [mathjaxinline]r=-k[/mathjaxinline]; so [mathjaxinline]e^{-kt}[/mathjaxinline] is the one and only exponential solution. By superposition, any constant multiple of [mathjaxinline]e^{-kt}[/mathjaxinline] is again a solution; and they exhaust the set of solutions. </p><p><b class="bf">Example 2:</b> [mathjaxinline]\ddot x+\omega _ n^2x=0[/mathjaxinline]. This is the <b class="bf">harmonic oscillator</b>, and you know very well that [mathjaxinline]\cos (\omega _ n t)[/mathjaxinline] and [mathjaxinline]\sin (\omega _ n t)[/mathjaxinline] are solutions. By superposition, any linear combination of these two functions is again a solution, and any sinusoid of angular frequency [mathjaxinline]\omega _ n[/mathjaxinline] is such a linear combination. So the general real solution to [mathjaxinline]\ddot x+\omega _ n^2x=0[/mathjaxinline] is the general sinusoid of angular frequency [mathjaxinline]\omega _ n[/mathjaxinline]: [mathjaxinline]A\cos (\omega _ nt-\phi )[/mathjaxinline]. </p><p>
But let's get to this result using the roots of the characteristic polynomial, [mathjaxinline]P(s)=s^2+\omega _ n^2[/mathjaxinline]. They are [mathjaxinline]\pm i\omega _ n[/mathjaxinline], so we have two exponential solutions, [mathjaxinline]e^{i\omega _ nt}[/mathjaxinline] and [mathjaxinline]e^{-i\omega _ nt}[/mathjaxinline]. Actually, this follows from what we just did using Euler's formula, since it tells us that these two functions are linear combinations of [mathjaxinline]\cos (\omega _ n t)[/mathjaxinline] and [mathjaxinline]\sin (\omega _ n t)[/mathjaxinline]. </p><p>
Conversely, by superposition, any linear combination of [mathjaxinline]e^{i\omega _ n t}[/mathjaxinline] and [mathjaxinline]e^{-i\omega _ n t}[/mathjaxinline] is again a solution; and since we are now talking about complex-valued functions we mean any linear combination with <b class="bf">complex</b> coefficients. These two functions are complex conjugates of each other. For any complex number, say [mathjaxinline]z[/mathjaxinline], </p><table id="a0000000053" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\mathrm{Re}\, z=\frac{z+\overline z}{2}\, ,\quad \mathrm{Im}\, z=\frac{z-\overline z}{2i}\, ,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
so, with [mathjaxinline]z=e^{i\theta }[/mathjaxinline], we get the "inverse Euler formulas" </p><table id="a0000000054" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\cos (\theta )=\frac{e^{i\theta }+e^{-i\theta }}{2}\, ,\quad \sin (\theta )=\frac{e^{i\theta }-e^{-i\theta }}{2i}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
With [mathjaxinline]\theta =\omega _ n t[/mathjaxinline], we find again that [mathjaxinline]\cos (\omega _ n t)[/mathjaxinline] and [mathjaxinline]\sin (\omega _ n t)[/mathjaxinline] are solutions. </p>
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Question 1.
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Find a basis of real solutions to [mathjaxinline]\ddot x+5\dot x+4x=0[/mathjaxinline]. </p>
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<text>[mathjaxinline]e^{-t}[/mathjaxinline]</text>
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<text>[mathjaxinline]e^{-2t}[/mathjaxinline]</text>
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<text>[mathjaxinline]e^{-4t}[/mathjaxinline]</text>
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<text>[mathjaxinline]\cos (2t)[/mathjaxinline]</text>
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<text>[mathjaxinline]\sin (2t)[/mathjaxinline]</text>
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<text>[mathjaxinline]e^{-t}\cos (\sqrt {3}t)[/mathjaxinline]</text>
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<text>[mathjaxinline]e^{-t}\sin (\sqrt {3}t)[/mathjaxinline]</text>
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Question 2.
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Find a basis of real solutions to [mathjaxinline]\ddot x+2\dot x+4x=0[/mathjaxinline]. </p>
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(Check all that apply.) </p>
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<text>[mathjaxinline]e^{-t}[/mathjaxinline]</text>
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<text>[mathjaxinline]e^{-2t}[/mathjaxinline]</text>
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<text>[mathjaxinline]e^{-4t}[/mathjaxinline]</text>
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<text>[mathjaxinline]\cos (2t)[/mathjaxinline]</text>
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<text>[mathjaxinline]\sin (2t)[/mathjaxinline]</text>
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<text>[mathjaxinline]e^{-t}\cos (\sqrt {3}t)[/mathjaxinline]</text>
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<text>[mathjaxinline]e^{-t}\sin (\sqrt {3}t)[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">1.9. The exponential response formula.</h2>
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We now address the inhomogeneous situation, </p><table id="a0000000057" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(D)x=f(t)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
In our applications, [mathjaxinline]f(t)[/mathjaxinline] will arise as [mathjaxinline]Q(D)y[/mathjaxinline] where [mathjaxinline]y[/mathjaxinline] is the input signal. We will be mainly interested in the case of sinusoidal input signal, which we will treat by considering complex exponential input signals [mathjaxinline]e^{i\omega t}[/mathjaxinline]. According to the formula </p><table id="a0000000058" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]Q(D)e^{rt}=Q(r)e^{rt}\, ,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
the resulting right hand side [mathjaxinline]f(t)[/mathjaxinline] will also be exponential, or a constant (perhaps complex) multiple of an exponential function. Generally [mathjaxinline]r[/mathjaxinline] will be purely imaginary, [mathjaxinline]r=i\omega[/mathjaxinline], but for the moment let's let it be any complex constant. Because of linearity, we might as well take the multiplicative constant to be 1. </p><p>
So we want to find a solution of </p><table id="a0000000059" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(D)x=e^{rt}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Just as in the homogeneous case, let's try for an exponential solution. But now instead of leaving the exponential constant unknown and solving for it, as we did before, we will try for a multiple of the <b class="bf">same</b> exponential function [mathjaxinline]e^{rt}[/mathjaxinline]. What is unknown is not the exponential constant but rather the multiplicative constant out front. So plug in [mathjaxinline]x=Ae^{rt}[/mathjaxinline], with the same [mathjaxinline]r[/mathjaxinline] but unknown [mathjaxinline]A[/mathjaxinline]: </p><table id="a0000000060" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]e^{rt}=P(D)x=P(D)(Ae^{rt})=AP(D)e^{rt}=AP(r)e^{rt}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
For this to work, we must have [mathjaxinline]AP(r)=1[/mathjaxinline], or [mathjaxinline]A=1/P(r)[/mathjaxinline]. We have discovered the </p><p><b class="bf">Exponential Response Formula (ERF):</b> A solution to [mathjaxinline]P(D)x=e^{rt}[/mathjaxinline] is given by </p><table id="a0000000061" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x_ p(t)=\frac{e^{rt}}{P(r)}\, ,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
provided that [mathjaxinline]P(r)\neq 0[/mathjaxinline]. </p><p><b class="bf">Example.</b> Find a solution to the equation </p><table id="a0000000062" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\ddot x+2\dot x+2x=3e^{-3t}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The characteristic polynomial is [mathjaxinline]P(r)=r^2+2r+2[/mathjaxinline], and the exponential constant is [mathjaxinline]r=-3[/mathjaxinline]. So [mathjaxinline]P(-3)=(-3)^2+2(-3)+2=5[/mathjaxinline], and </p><table id="a0000000063" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x_ p=\frac{3e^{-3t}}{5}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Check it! </p>
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<h2 class="hd hd-2 unit-title">1.10. Complex replacement.</h2>
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The ERF merges beautifully with the theory of complex exponentials. This is because of the way you differentiate a complex-valued function of time, such as [mathjaxinline]e^{rt}[/mathjaxinline] (for [mathjaxinline]r[/mathjaxinline] complex): You differentiate real and imaginary parts separately; they form the real and imaginary parts of the derivative. So if the coefficients of a polynomial [mathjaxinline]P(r)[/mathjaxinline] are real, then </p><table id="a0000000064" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\mathrm{Re}\, \left(P(D)f(t)\right)=P(D)\mathrm{Re}\, \left( f(t) \right)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The consequence of this is that if we have a linear differential equation </p><table id="a0000000065" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(D)x=\cos (\omega t)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
with constant real coefficients, then its solutions are the real parts of solutions of a <b class="bf">different</b> equation, a "complex replacement": </p><table id="a0000000066" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(D)z=e^{i\omega t}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table>
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<h2 class="hd hd-2 unit-title">1.11. Transients and Stability.</h2>
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In the previous activity, it turned out that all solutions to the corresponding homogeneous equation die off exponentially. So any two solutions to the original equation become asymptotic as time gets large – and quickly. So for many practical purposes we can pick just one solution to the inhomogeneous equation, and use it. There is one particularly nice solution, and it happens to be the one we obtained using the ERF and Complex Replacement: </p><table id="a0000000067" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x_ p=\frac{1}{\sqrt {85}}\cos (3t-\phi )\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The use of the word "particular" is a kind of pun; it is just one of many, but it is also a particularly nice one! </p><p>
Anyway, when the solutions to the homogeneous equation die off exponentially like this the homogeneous solutions are called <b class="bf">transients</b>, and the equation is termed <b class="bf">stable</b>. </p><p>
The solutions to the homogeneous equation are essential if you have an initial condition you have to meet. But even if you are only interested in the long term behavior, your strategy will be completely different depending on whether the homogeneous solutions die off or blow up. In the latter case, they will likely swamp any particular solution; in the stable case we have steady state solution to which all solutions become asymptotic, independent of initial conditions. (Note that the particular solution does of course have its own initial conditions; but they are not particularly important!) </p><p>
Why did the solutions to the homogeneous equation die off exponentially in this case? Where did the exponential constant [mathjaxinline]-1[/mathjaxinline] come from, in [mathjaxinline]e^{-t}\cos t[/mathjaxinline] and [mathjaxinline]e^{-t}\sin t[/mathjaxinline]? </p><p>
The answer: it is the <b class="bf">real part</b> of both roots of the characteristic polynomial. Remember, with [mathjaxinline]r=a+bi[/mathjaxinline], </p><table id="a0000000068" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]e^{rt}=e^{(a+bi)t}=e^{at}(\cos (bt)+i\sin (bt))[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
so the growth or decay of real part (or any part) of [mathjaxinline]e^{rt}[/mathjaxinline] depends only on the real part of [mathjaxinline]r[/mathjaxinline]. If [mathjaxinline]a>0[/mathjaxinline], it will blow up exponentially; if [mathjaxinline]a<0[/mathjaxinline], it will decay exponentially. In the marginal case [mathjaxinline]a=0[/mathjaxinline], you get pure oscillation. </p><p>
For equations of higher degree, the characteristic polynomial [mathjaxinline]P(s)[/mathjaxinline] will have more roots. If the real part of <b class="bf">any</b> root is positive, then almost all solutions to the homogeneous equation [mathjaxinline]P(D)x=0[/mathjaxinline] will tend exponentially to infinity. If <b class="bf">all</b> the roots have <b class="bf">negative</b> real part, then <b class="bf">all</b> solutions will tend exponentially to zero. </p><p>
So this is the stability criterion: All roots of [mathjaxinline]P(r)[/mathjaxinline] must have negative real part. </p>
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<h3 class="hd hd-3 problem-header" id="lec0-tab11-problem1-problem-title" aria-describedby="block-v1:OCW+18.031+2019_Spring+type@problem+block@lec0-tab11-problem1-problem-progress" tabindex="-1">
Question 1.
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<p>
For the spring-mass-dashpot system modeled earlier </p>
<table id="a0000000069" 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,[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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<p>
the constants [mathjaxinline]m[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], and [mathjaxinline]k[/mathjaxinline] are all positive. </p>
<p>
Is this system stable or unstable? </p>
<p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_lec0-tab11-problem1_2_1">
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<text> unstable</text>
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Question 2.
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Is the system </p>
<table id="a0000000070" 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]x^{(4)}+4x=\dot y + 2y[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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stable or unstable? </p>
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<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_lec0-tab11-problem2_2_1">
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<text> stable</text>
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<h2 class="hd hd-2 unit-title">1.12. Complex Gain.</h2>
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<p>
The combination of complex replacement and ERF efficiently delivers critical information about the sinusoidal solutions of equations of the form </p><table id="a0000000074" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(D)x=\hbox{(a sinusoid)}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
or, more generally, of the form </p><table id="a0000000075" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(D)x=Q(D)\hbox{(a sinusoid)}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
This method leads to sinusoidal solutions of the form </p><table id="a0000000076" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x_ p(t)=\mathrm{Re}\, (Ge^{i\omega t})[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
where [mathjaxinline]\omega[/mathjaxinline] is the angular frequency of the input sinusoid and [mathjaxinline]G[/mathjaxinline] is some complex constant. </p><p>
The smart way to find the amplitude of this sinusoid is to express [mathjaxinline]G[/mathjaxinline] in polar form: say </p><table id="a0000000077" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]G=|G|e^{-i\phi }\, ,\quad \hbox{so}\quad \phi =-\mathrm{arg}(G)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Then </p><table id="a0000000078" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]z(t)=Ge^{i\omega t}=|G|e^{-i\phi }e^{i\omega t}=|G|e^{i(\omega t-\phi )}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The real part of this is </p><table id="a0000000079" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]|G|\cos (\omega t-\phi )\, ,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
which has amplitude [mathjaxinline]|G|[/mathjaxinline]. </p><p>
As we saw in <a href="/courses/course-v1:OCW+18.031+2019_Spring/courseware/review/lec1/11" target="_blank">this previous example</a>, the complex replacement gave the solution </p><table id="a0000000080" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]z_ p(t) = \frac{e^{3it}}{P(3i)} = \frac{e^{3it}}{-7+6i}, \qquad x_ p(t) = \frac{1}{\sqrt {85}} \cos (3t-\phi ).[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
In more detail, the method replaces a sinusoidal input signal by a complex exponential signal [mathjaxinline]e^{i\omega t}[/mathjaxinline]. So we have the equation </p><table id="a0000000081" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(D)z=Q(D)e^{i\omega t}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
But </p><table id="a0000000082" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]Q(D)e^{i\omega t}=Q(i\omega )e^{i\omega t}\, ,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
and we are looking for solutions of the form </p><table id="a0000000083" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]z_ p=G(\omega )e^{i\omega t}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
(where [mathjaxinline]G(\omega )[/mathjaxinline] is a complex number to be determined, depending on [mathjaxinline]\omega[/mathjaxinline] and the system parameter). Making the substitution, </p><table id="a0000000084" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(D)z_ p=P(D)G(\omega )e^{i\omega t}=G(\omega )P(i\omega )e^{i\omega t}\, ,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
so </p><table id="a0000000085" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]G(\omega )P(i\omega )e^{i\omega t}=Q(i\omega )e^{i\omega t}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Canceling the exponential and dividing through, we find </p><table id="a0000000086" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]G(\omega )=\frac{Q(i\omega )}{P(i\omega )}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
This complex number is the <b class="bf">complex gain</b>. It contains both the gain of the system and the phase lag: </p><table id="a0000000087" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]g(\omega )=|G(\omega )|\quad ,\quad \phi (\omega )=-\mathrm{arg}(G(\omega ))\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
(The <b class="bf">negative</b> of the argument, because we are talking about the phase <b class="bf">lag</b> rather than the phase <b class="bf">gain</b>.) </p>
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<h2 class="hd hd-2 unit-title">1.13. Footnotes.</h2>
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<p><b class="bf">1. Question:</b> Both the mass and the amplitude of the sinusoidal input signal are fixed at the value 1 (in appropriate units) in this Mathlet (and others). Isn't this a serious limitation?</p>
<p><iframe style="display: block; border-width: 0px; padding: 0px;" src="https://mathlets1803.surge.sh/ampPhaseSecondOrderII.html" width="1100 px" height="640 px">
<p><b class="bf">Answer:</b> No. </p>
<p>
First, the Mathlet displays the behavior of the system with input amplitude 1; but by linearity an input amplitude of [mathjaxinline]A[/mathjaxinline] simply multiplies the system response by [mathjaxinline]A[/mathjaxinline]. </p>
<p>
Suppose we are interested in a more general case, with mass [mathjaxinline]m[/mathjaxinline] and input signal amplitude [mathjaxinline]A[/mathjaxinline]. The equation is then </p>
<table id="a0000000088" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]m\ddot x+b\dot x+kx=b\frac{d}{dt}A\cos (\omega t)\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
Divide through by [mathjaxinline]m[/mathjaxinline]. We arrive at the equation with [mathjaxinline]m=1[/mathjaxinline], but with [mathjaxinline]b[/mathjaxinline], [mathjaxinline]k[/mathjaxinline], and [mathjaxinline]A[/mathjaxinline] replaced by [mathjaxinline]b/m[/mathjaxinline], [mathjaxinline]k/m[/mathjaxinline], and [mathjaxinline]A/m[/mathjaxinline]. So you can use these modified values for the system parameter in the Mathlet, and then multiply the displayed system response by [mathjaxinline]A/m[/mathjaxinline]. </p>
<p><b class="bf">2. Question:</b> What replaces the ERF if [mathjaxinline]P(r)=0[/mathjaxinline]? </p>
<p><b class="bf">Answer:</b></p>
<p><b class="bfseries"><span style="color: #0000ff;"> Modified ERF:</span></b> If [mathjaxinline]P(r)=\cdots =P^{(k-1)}(r)=0[/mathjaxinline] but [mathjaxinline]P^{(k)}(r)\neq 0[/mathjaxinline], then </p>
<table id="a0000000089" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]x_ p(t)=\frac{t^ ke^{rt}}{P^{(k)}(r)}[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
is a solution to </p>
<table id="a0000000090" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]P(D)x=e^{rt}\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p><b class="bfseries">Example 13.1 </b> Use the generalized ERF to find a particular solution to the system [mathjaxinline]\ddot{x}+x=e^{it}[/mathjaxinline]. </p>
<p>
Solution: The characteristic polynomial of the differential equation [mathjaxinline]\ddot{x}+x=e^{it}[/mathjaxinline] is [mathjaxinline]P\left(r\right)=r^2+1[/mathjaxinline], with roots [mathjaxinline]\pm i[/mathjaxinline]. Since [mathjaxinline]i[/mathjaxinline] is a root of the characteristic polynomial, the <b class="bfseries"><span style="color: #0000ff;">Generalized ERF</span></b> tells us to find the smallest integer [mathjaxinline]s[/mathjaxinline] for which [mathjaxinline]P^{\left(s\right)}\left(i\right)\neq 0[/mathjaxinline]. In this case, [mathjaxinline]P^{\prime }\left(r\right)=2r[/mathjaxinline], so [mathjaxinline]P^{\prime }\left(i\right)=2i[/mathjaxinline], which is not zero, so [mathjaxinline]s=1[/mathjaxinline]. The Generalized ERF tells us that </p>
<table id="a0000000091" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout: auto;">
<tr id="a0000000092">
<td style="width: 40%; border: none;"> </td>
<td style="vertical-align: middle; text-align: right; border: none;">
[mathjaxinline]\displaystyle x_ p\left(t\right)=\frac{1}{P^{\prime }\left(i\right)}t^{1}e^{it}=\frac{1}{2i}te^{it}[/mathjaxinline]
</td>
<td style="width: 40%; border: none;"> </td>
<td style="width: 20%; border: none;" class="eqnnum"> </td>
</tr>
</table>
<p>
is a particular solution to the inhomogeneous ODE. Since [mathjaxinline]\frac1i=e^{-i\pi /2}[/mathjaxinline], we can write [mathjaxinline]x_ p[/mathjaxinline] as </p>
<table id="a0000000093" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout: auto;">
<tr id="a0000000094">
<td style="width: 40%; border: none;"> </td>
<td style="vertical-align: middle; text-align: right; border: none;">
[mathjaxinline]\displaystyle x_ p\left(t\right)=\frac{1}{2}te^{it-i\pi /2}.[/mathjaxinline]
</td>
<td style="width: 40%; border: none;"> </td>
<td style="width: 20%; border: none;" class="eqnnum"> </td>
</tr>
</table>
<div class="hideshowbox">
<h4 onclick="hideshow(this);" style="margin: 0px;">Walk through the proof of generalized ERF<span class="icon-caret-down toggleimage"></span></h4>
<div class="hideshowcontent">
<p>
Return to the key equation </p>
<table id="a0000000095" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]P(D)e^{rt}=P(r)e^{rt}[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
The trick is to regard [mathjaxinline]r[/mathjaxinline] as a variable. So there are two variables in the equation. Differentiate with respect to [mathjaxinline]r[/mathjaxinline], holding [mathjaxinline]t[/mathjaxinline] constant. Remember, [mathjaxinline]D[/mathjaxinline] differentiates with respect to [mathjaxinline]t[/mathjaxinline], and </p>
<table id="a0000000096" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]\frac{d}{dr}\frac{d}{dt}=\frac{d}{dt}\frac{d}{dr}\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
So the left hand side gives </p>
<table id="a0000000097" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]\frac{d}{dr}P(D)e^{rt}=P(D)\frac{d}{dr}e^{rt}=P(D)te^{rt}\, ,[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
while, by the product rule, the right hand side gives </p>
<table id="a0000000098" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]\frac{d}{dr}P(r)e^{rt}=P'(r)e^{rt}+P(r)te^{rt}\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
Set these equal and and remember that we are supposing that [mathjaxinline]P(r)=0[/mathjaxinline]: </p>
<table id="a0000000099" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]P(D)(te^{rt})=P'(r)e^{rt}\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
That is: If [mathjaxinline]P(r)=0[/mathjaxinline] but [mathjaxinline]P'(r)\neq 0[/mathjaxinline], </p>
<table id="a0000000100" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]x_ p(t)=\frac{te^{rt}}{P'(r)}[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
is a solution to the equation </p>
<table id="a0000000101" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]P(D)x=e^{rt}\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
This process may be continued, to give: </p>
<p><b class="bf">Modified ERF:</b> If [mathjaxinline]P(r)=\cdots =P^{(k-1)}(r)=0[/mathjaxinline] but [mathjaxinline]P^{(k)}(r)\neq 0[/mathjaxinline], then </p>
<table id="a0000000102" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]x_ p(t)=\frac{t^ ke^{rt}}{P^{(k)}(r)}[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
is a solution to </p>
<table id="a0000000103" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]P(D)x=e^{rt}\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
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