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<h2 class="hd hd-2 unit-title">1.1. RLC circuits and the System function.</h2>
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<h3 class="hd hd-3 problem-header">Objectives.</h3><p>
After completing this lecture you will be able to: </p><ol class="enumerate"><li value="1"><p>
model <b class="bfseries"><span style="color:#0000FF">RLC circuits</span></b> using differential equations.</p></li><li value="2"><p>
determine the <b class="bfseries"><span style="color:#0000FF">complex gain</span></b> of an <b class="bfseries"><span style="color:#0000FF">LTI system</span></b> as a function of the input frequency. </p></li><li value="3"><p>
read off information about the <b class="bfseries"><span style="color:#0000FF">gain</span></b> and <b class="bfseries"><span style="color:#0000FF">phase lag</span></b> from the <b class="bfseries"><span style="color:#0000FF">Nyquist plot</span></b> in the complex plane. </p></li><li value="4"><p>
interpret the <b class="bfseries"><span style="color:#0000FF">complex gain</span></b> as the ratio of exponential system response to exponential input, and relate it to the standard form of the modeling differential equation. </p></li><li value="5"><p>
represent the frequency response with <b class="bfseries"><span style="color:#0000FF">Bode plots</span></b> and the <b class="bfseries"><span style="color:#0000FF">Nyquist plot</span></b> , and understand the relationship between these plots. </p></li><li value="6"><p>
find the <b class="bfseries"><span style="color:#0000FF">transfer function</span></b> (or <b class="bfseries"><span style="color:#0000FF">system function</span></b> ) of an LTI system. </p></li></ol>
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<h2 class="hd hd-2 unit-title">1.2. Motivation and course mascot.</h2>
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<p>
In this course, you will learn the mathematical principles underlying the analysis and design of mechanisms and circuits of all types, from radios, to the big tuned mass dampers that stabilize tall buildings such as the “John Hancock" tower, now known as 200 Clarendon St. in Boston. </p><center><img src="/assets/courseware/v1/b57fed446febc478e79aeebe99188404/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_John_Hancock_Tower__200_Clarendon.jpg" width="300"/><br/><font size="2">Photo by Stephen Foskett, Wikimedia Commons. (CC BY-SA) 3.0 </font></center><p>
There's an excellent description of the tuned mass damper <a href="https://en.wikipedia.org/wiki/200_Clarendon_Street" target="_blank">on Wikipedia</a> under the heading Engineering Flaws. </p><blockquote class="quote"> "Two 300-ton weights sit at opposite ends of the 58th floor of the Hancock. Each weight is a box of steel, filled with lead, 17 feet square by 3 feet high. Each weight rests on a steel plate. The plate is covered with lubricant so the weight is free to slide. But the weight is attached to the steel frame of the building by means of springs and shock absorbers. When the Hancock sways, the weight tends to remain still, allowing the floor to slide underneath it. Then, as the springs and shocks take hold, they begin to tug the building back. The effect is like that of a gyroscope, stabilizing the tower. The reason there are two weights, instead of one, is so they can tug in opposite directions when the building twists. The cost of the damper was $3 million. The dampers are free to move a few feet relative to the floor. " – Robert Campbell, architecture critic for The Boston Globe </blockquote>
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<h3 class="hd hd-2">Modeling a tuned mass damper</h3>
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<h3 class="hd hd-3 problem-header">The course mascot</h3><p>
As a model of the tuned mass damper we described, we will be using a beautiful small mechanism we jokingly call the mascot for this course. You may recognize it from the course image. This mechanism was built for use in teaching in the Mechanical Engineering Department here at MIT. We are grateful to Professor David Trumper for letting us use it in this course! And thanks to Jin Young Yoon for supplying the demonstrations and data. </p>
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<h3 class="hd hd-2">Mascot introduction</h3>
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<h3 class="hd hd-3 problem-header">Meet the mascot</h3><center><img src="/assets/courseware/v1/08d0116b4c7adba2eea8f7139d3007c8/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_mascot_diagram.png" width="400"/><br/>A diagram of the mascot<br/><font size="2">Diagram by Professor Dave Trumper, MIT MechE</font><div><br/></div></center><p>
The bearing shaft (along with all attachments) form a lumped mass ([mathjaxinline]m_1[/mathjaxinline]). The slender spring rod behaves as a massless spring with spring constant [mathjaxinline]k_1[/mathjaxinline]. The voice coil actuator serves as a damper with damping constant [mathjaxinline]b_1[/mathjaxinline], as well as providing the driving force. An additional mass [mathjaxinline]m_2[/mathjaxinline] is attached to the bearing shaft by a thin aluminum blade (note in the diagram it is attached horizontally, while in the demo below, it is attached vertically). The aluminum blade can be reasonably modeled as a massless spring with light damping with spring constant [mathjaxinline]k_2[/mathjaxinline] and damping constant [mathjaxinline]b_2[/mathjaxinline]. The LVDT (linear variable displacement transformer) is the mechanism that measures the position of the [mathjaxinline]m_1[/mathjaxinline]. All together, this is a fourth order system— a coupled spring mass system, which can be modeled as below. (This model should seem very familiar.) </p><center><img src="/assets/courseware/v1/3d6022c472ff32bda4afad1c03a716e3/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c8_tmdmodel.svg" width="400px" style="margin: 0px 10px 10px 10px"/><br/>A model of the mascot<br/></center>
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We will keep returning to this system in this course, but in order to illustrate the breadth of application of our methods, we will discuss electrical systems as well. So next we consider a simple series circuit. </p>
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<h2 class="hd hd-2 unit-title">1.3. Simple RLC Circuits.</h2>
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A very simple passive AM radio receiver may be modeled by the following circuit. </p><center><img src="/assets/courseware/v1/0e39487db086dc39ba66db4e3a9cd27a/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_RLC.svg" width="250px" style="margin: 0px 10px 10px 10px"/></center><p>
The power source at left is the antenna, being driven by radio waves. The resistor at top is a speaker. The other components are a capacitor, at bottom, and an inductance coil, at right. </p><p>
We will model this circuit using the same five-step process we used for mechanical systems in Lecture 1. </p><ol class="enumerate"><li value="1"><p><b class="bfseries"><span style="color:#0000FF"> Draw a diagram of the system.</span></b> We've already done Step 1! </p></li><li value="2"><p><b class="bfseries"><span style="color:#0000FF">Identify and give symbols for the parameters of the system.</span></b> The diagram shows symbols for four standard electronic components. The first step in understanding the diagram is to define an orientation; say that current flows clockwise. Since it's a series circuit, the current through any of the wires is the same; write [mathjaxinline]I(t)[/mathjaxinline] for it. Then we can say that the power source, at left, produces a voltage <b class="bf">increase</b> of [mathjaxinline]V(t)[/mathjaxinline] volts at time [mathjaxinline]t[/mathjaxinline]. This voltage increase may vary with time, and may be negative as well as positive. In fact we'll be especially interested in the case in which it is sinusoidal! </p><p>
The meaning of each of the other components is specified by how the voltage <b class="bf">decrease</b> across it is related to the current flowing through it. For us, these relationships <b class="bf">define</b> the components. The impact of each component is determined by the constant appearing in these relationships. </p><table class="tabular" cellspacing="0" style="table-layout:auto"><tr><td style="text-align:left; border:none">
Voltage drop across inductor: </td><td style="text-align:left; border:none">
[mathjaxinline]V_ L(t)=L\dot I(t)[/mathjaxinline]. </td><td style="text-align:left; border:none"> </td></tr><tr><td style="text-align:left; border:none">
Voltage drop across resistor: </td><td style="text-align:left; border:none">
[mathjaxinline]V_ R(t)=RI(t)[/mathjaxinline]. </td><td style="text-align:left; border:none"> </td></tr><tr><td style="text-align:left; border:none">
Voltage drop across capacitor: </td><td style="text-align:left; border:none">
[mathjaxinline]C\dot V_ C(t)=I(t)[/mathjaxinline]. </td><td style="text-align:left; border:none"> </td></tr></table><p>
These components are connected in series, and so their voltage drops are related by Kirchhoff's voltage law: The voltage gain across the power source must equal the sum of the voltage drops across the other components: </p><table id="a0000000104" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]V=V_ L+V_ R+V_ C\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table></li><li value="3"><p><b class="bfseries"><span style="color:#0000FF"> Declare the input signal and the system response.</span></b> The system is being driven by a signal coming in via antenna. This signal causes a voltage difference across the antenna, and it is reasonable to declare this voltage increase [mathjaxinline]V(t)[/mathjaxinline] as the input signal. For system response, we are interested in the loudness of the speaker, which is proportional to the voltage drop across the resistor. So we declare [mathjaxinline]V_ R[/mathjaxinline] to be the system response. </p><p>
This means that we want to set up a differential equation relating [mathjaxinline]V(t)[/mathjaxinline] to [mathjaxinline]V_ R(t)[/mathjaxinline]. </p></li><li value="4"><p><b class="bfseries"><span style="color:#0000FF"> Write down a differential equation relating the input signal and the system response, using Newton's “[mathjaxinline]F=ma[/mathjaxinline]" in the mechanical case or Kirchhoff's laws in the electrical case.</span></b> Because [mathjaxinline]\dot V_ C[/mathjaxinline] appears in the definition of a capacitor, it is natural to differentiate Kirchhoff's voltage law, and rewrite in terms of [mathjaxinline]I[/mathjaxinline]; </p><table id="a0000000105" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\dot V=\dot V_ L+\dot V_ R+\dot V_ C=L\ddot I+R\dot I+(1/C)I\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
To make this into an equation relating the input signal [mathjaxinline]V[/mathjaxinline] to the system response [mathjaxinline]V_ R[/mathjaxinline], we just have to remember that [mathjaxinline]V_ R=RI[/mathjaxinline]. So multiply through by [mathjaxinline]R[/mathjaxinline] and make this substitution, along with its consequences [mathjaxinline]\dot V_ R=R\dot I[/mathjaxinline] and [mathjaxinline]\ddot V_ R=R\ddot I[/mathjaxinline]: </p><table id="a0000000106" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]R\dot V=L\ddot V_ R+R\dot V_ R+(1/C)V_ R\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table></li><li value="5"><p><b class="bfseries"><span style="color:#0000FF"> Rewrite the equation in standard form.</span></b> Input and output are already separated; to put this in standard form we just swap sides: </p><table id="a0000000107" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]L\ddot V_ R+R\dot V_ R+(1/C)V_ R=R\dot V\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table></li></ol><p>
We're done. But before we discuss consequences, recall the equation we discussed in Lecture 1 describing the spring/mass/dashpot system driven through the dashpot: </p><table id="a0000000108" 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\dot y\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
These two equations are formally identical in the way they relate input and system response. This reflects a rough parallel between mechanical and electrical systems, in which </p><center><span><style>
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</style><table id="spring_circuit"><tr><th colspan="2">Mechanical</th><th colspan="2">Electrical</th></tr><tr><td> displacement</td><td> [mathjaxinline]x,y[/mathjaxinline]</td><td> voltage drop, gain</td><td>[mathjaxinline]V_R,V[/mathjaxinline]</td></tr><tr><td> mass</td><td> [mathjaxinline]m[/mathjaxinline]</td><td> inductance</td><td>[mathjaxinline]L[/mathjaxinline]</td></tr><tr><td> damping constant</td><td> [mathjaxinline]b[/mathjaxinline]</td><td> resistance</td><td>[mathjaxinline]R[/mathjaxinline]</td></tr><tr><td> spring constant </td><td> [mathjaxinline]k[/mathjaxinline]</td><td> 1/capacitance</td><td>[mathjaxinline]1/C[/mathjaxinline]</td></tr></table></span></center><p>
Rather than trying to develop this as a formal equivalence, though, we think it's best to focus on the fact that the <b class="bf">mathematics</b> is identical up to change of notation. </p>
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Consider the same RLC circuit. </p>
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Please develop a model for the same series circuit system, but now with system response defined as the voltage drop across the <b class="bf">capacitor</b> [mathjaxinline]V_ C[/mathjaxinline]. As before, the input is [mathjaxinline]V[/mathjaxinline]. </p>
<p>
The differential equation model that you find will take the form: </p>
<table id="a0000000109" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\displaystyle {\color{blue}{a}} \ddot V_ C + {\color{blue}{b}} \dot V_ C+(1/C)V_ C = {\color{blue}{??}}[/mathjax]</td>
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for constants <b class="bfseries"><span style="color:#0000FF">[mathjaxinline]a[/mathjaxinline]</span></b> and <b class="bfseries"><span style="color:#0000FF">[mathjaxinline]b[/mathjaxinline]</span></b> that depend on the parameters of the circuit and an input function <b class="bfseries"><span style="color:#0000FF">[mathjaxinline]??[/mathjaxinline]</span></b> that depends on the parameters in the circuit as well as the voltage drop [mathjaxinline]V[/mathjaxinline] and/or its derivatives [mathjaxinline]\dot V[/mathjaxinline] and [mathjaxinline]\ddot V[/mathjaxinline]. </p>
<p>
Determine these constants in terms of the system parameters [mathjaxinline]R[/mathjaxinline], [mathjaxinline]L[/mathjaxinline], and [mathjaxinline]C[/mathjaxinline], and the voltage [mathjaxinline]V[/mathjaxinline]. Type <b class="bf">dotV</b> for [mathjaxinline]\dot V[/mathjaxinline], and <b class="bf">ddotV</b> for [mathjaxinline]\ddot V[/mathjaxinline]. </p>
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In the systems below, [mathjaxinline]y[/mathjaxinline] is the input, and [mathjaxinline]x[/mathjaxinline] is the response. Which of the mechanical systems below is most like the electrical circuit system where the system response is the voltage drop across the capacitor? </p>
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<text> Spring/mass/dashpot in series driven through a piston at the top. [mathjaxinline]\quad m\ddot x + b\dot x + kx = ky[/mathjaxinline]</text>
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<text> Spring/mass/dashpot in series driven through the dashpot. [mathjaxinline]\quad m\ddot x + b\dot x + kx = b\dot y[/mathjaxinline]</text>
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<text> Spring and dashpot in parallel driven through a piston at the top. [mathjaxinline]\quad m\ddot x + b\dot x + kx = ky+b\dot y[/mathjaxinline]</text>
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<text> Spring and dashpot in parallel driven through a force applied to the mass. [mathjaxinline]\quad m\ddot x + b\dot x + kx = y[/mathjaxinline]</text>
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(Note these appear in the Mathlets: Amplitude and Phase: Second Order <a href="https://mathlets.surge.sh/ampPhaseSecondOrderI" target="_blank">I</a>, <a href="https://mathlets.surge.sh/ampPhaseSecondOrderII" target="_blank">II</a>, <a href="https://mathlets.surge.sh/ampPhaseSecondOrderIII" target="_blank">III</a>, and <a href="https://mathlets.surge.sh/ampPhaseSecondOrderIV" target="_blank">IV</a>.) </p>
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<h2 class="hd hd-2 unit-title">1.4. Frequency response.</h2>
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<h3 class="hd hd-3 problem-header">Circuits as filters.</h3><p>
Let's now study the model of the AM radio receiver from the previous page. </p><center><img src="/assets/courseware/v1/0e39487db086dc39ba66db4e3a9cd27a/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_RLC.svg" width="250px" style="margin: 0px 10px 10px 10px"/></center><p>
The environment is filled with radio waves, electromagnetic oscillations vibrating at different frequencies. For example, </p><ul class="itemize"><li><p>
many cellphones are broadcast at 1900MHz, </p></li><li><p>
medium range AM radio in the US is broadcast at 540-1600kHz, and </p></li><li><p>
FM radio is broadcast at 30-300MHz. </p></li></ul><p>
The input signal to our system is a variable voltage increase across the antenna that is a superposition of the results of all these radio waves. </p><p>
Each AM radio station broadcasts in a small “band" of frequencies centered at a nominal broadcast frequency. Tuning the radio should allow frequencies near this nominal frequency to pass through to the speaker while suppressing all other frequencies. </p><p>
We want to design our radio receiver so that its system response to the frequency of the radio station we want to listen to is much larger than the response to other frequencies. </p><p>
In terms of gain, we can say that we want the gain to be greatest for a specific input frequency – call it [mathjaxinline]\omega _ r[/mathjaxinline]. The range of frequencies that is allowed through with little diminution is called the “pass band" of the receiver. To ensure that we don't receive interference from unwanted stations, it is desirable to try to arrange that the pass band is narrow enough to effectively suppress the signals arising from other sources. </p><center><img src="/assets/courseware/v1/e104bfc4c87e77eff8c0dd0ab3e2effe/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_radiostations_tuned.svg" width="450px" style="margin: 0px 10px 10px 10px"/><br/><font size="2"> Plot of amplitude gain with respect to input frequency (kHz).<br/> The boxes indicate the transmission bands of several <br/> AM radio stations in the Boston area. If we want to tune <br/> to the station centered at 950kHz, we seek a frequency<br/> response profile that looks rather like the blue curve.<br/> </font></center><div><br/></div><p>
A mechanism or circuit designed to suppress all frequencies except those near a specified frequency is called a “filter." A radio receiver suppresses signals with both higher and lower frequencies, and thus it is called a “mid-pass filter," and the range of frequencies it allows through is called the “pass-band." </p>
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<h3 class="hd hd-3 problem-header">LTI systems as filters</h3><p>
Recall that we modeled this AM radio receiver circuit by the differential equation </p><table id="a0000000127" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]L\ddot V_ R+R\dot V_ R+(1/C)V_ R=R\dot V\,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
that is analogous to the associated mechanical system </p><table id="a0000000128" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle m\ddot x + b\dot x + kx = b\dot y.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
In <a href="/courses/course-v1:OCW+18.031+2019_Spring/courseware/review/lec1/3" target="_blank">Lecture 1</a> you made some observations about the mechanical system using the <a href="http://mathlets.org/mathlets/amplitude-and-phase-2nd-order-ii/" target="_blank">Mathlet</a> Amplitude and Phase 2nd Order II. You discovered through the mathlet that the maximum gain of this system (when [mathjaxinline]m=1[/mathjaxinline]) for any fixed values of [mathjaxinline]k[/mathjaxinline] and [mathjaxinline]b[/mathjaxinline], is [mathjaxinline]g=1[/mathjaxinline]; and that this value occurs when [mathjaxinline]\omega =\omega _ r=\sqrt {k/m}[/mathjaxinline], independent of [mathjaxinline]b[/mathjaxinline]. If we think of this system as an AM radio receiver, the resonant peak is at the frequency we are tuning the radio to. The resonant value is given by </p><table id="a0000000129" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\omega _ r = \frac{1}{\sqrt {LC}}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
and is independent of the resistance [mathjaxinline]R[/mathjaxinline]. So this is important information: With inductance fixed, you tune the radio by changing the capacitance. The environment is filled with radio waves at a variety of frequencies. The AM receiver responds weakly to most and relatively strongly only to frequencies at or near the resonant frequency. </p><p>
There's a general lesson here for LTI systems: it's important to be able to visualize the system response to a whole spectrum of different input frequencies. Since we are interested in the sinusoidal system response, we need to specify only two parameters relative to the input frequency to understand this response completely: the gain and the phase lag. These parameters depend upon the input frequency, but not on the input amplitude and not on the phase of the input relative to some standard signal. </p><p>
The two main tools that we have for visualizing the gain and phase lag as functions of the frequency are the <b class="bfseries"><span style="color:#0000FF">Bode Plot</span></b> and the <b class="bfseries"><span style="color:#0000FF">Nyquist Plot</span></b> . There are two Bode plots, one for gain and one for phase. The horizontal axis for each plot is frequency, one plot shows how the gain depends on the frequency of the input signal, and the other shows how the phase depends on the frequency of the input signal. The next problems allow you to explore the Bode plots to understand the frequency response directly. </p>
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Bode Plots
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<p>
Each of the Amplitude and Phase <a href="http://mathlets.org" target="_blank">Mathlets</a> allows visualization of plots of the gain and the phase lag as functions of frequency. To see these two graphs on the Mathlet below, check the box marked &#8220;Bode plots." (<a href="/courses/course-v1:OCW+18.031+2019_Spring/courseware/intro/lec2a/10" target="_blank">Why &#8220;Bode"?</a>) Take a minute now to understand how the Bode plots reflect the behavior of the system. Note that the lower graph plots the phase <b class="bf">gain</b> rather than the <b class="bf">lag</b>, that is, [mathjaxinline]-\phi[/mathjaxinline] rather than [mathjaxinline]\phi[/mathjaxinline]. </p>
<p>
Here are some questions for you to guide your exploration of the Mathlet. Remember, you can change the system parameters [mathjaxinline]k[/mathjaxinline] and [mathjaxinline]b[/mathjaxinline] using the sliders below the graph. Vary them and confirm your hypothesis that [mathjaxinline]\omega _ r=\sqrt k[/mathjaxinline]. </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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(Answer these questions based on your exploration of the Mathlet.) </p>
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What is the phase lag [mathjaxinline]\phi[/mathjaxinline] when [mathjaxinline]\omega =\omega _ r[/mathjaxinline]? <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_lec1-tab4-problem1_2_1" class=" capa_inputtype textline">
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What is the sign of the phase lag [mathjaxinline]\phi[/mathjaxinline] when [mathjaxinline]\omega &lt;\omega _ r[/mathjaxinline]? <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div class="choicegroup capa_inputtype" id="inputtype_lec1-tab4-problem1_3_1">
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How does changing the damping constant [mathjaxinline]b[/mathjaxinline] affect the peak in the Bode plot near the resonant frequency? </p>
<p>
(Choose all that apply.) <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 3" role="group"><div class="choicegroup capa_inputtype" id="inputtype_lec1-tab4-problem1_4_1">
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<input type="checkbox" name="input_lec1-tab4-problem1_4_1[]" id="input_lec1-tab4-problem1_4_1_choice_0" class="field-input input-checkbox" value="choice_0"/><label id="lec1-tab4-problem1_4_1-choice_0-label" for="input_lec1-tab4-problem1_4_1_choice_0" class="response-label field-label label-inline" aria-describedby="status_lec1-tab4-problem1_4_1"> <text><b class="bf">Decreasing</b> [mathjaxinline]b[/mathjaxinline] makes the resonant peak sharper.</text>
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<input type="checkbox" name="input_lec1-tab4-problem1_4_1[]" id="input_lec1-tab4-problem1_4_1_choice_1" class="field-input input-checkbox" value="choice_1"/><label id="lec1-tab4-problem1_4_1-choice_1-label" for="input_lec1-tab4-problem1_4_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_lec1-tab4-problem1_4_1"> <text><b class="bf">Increasing</b> [mathjaxinline]b[/mathjaxinline] makes the resonant peak sharper.</text>
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<input type="checkbox" name="input_lec1-tab4-problem1_4_1[]" id="input_lec1-tab4-problem1_4_1_choice_2" class="field-input input-checkbox" value="choice_2"/><label id="lec1-tab4-problem1_4_1-choice_2-label" for="input_lec1-tab4-problem1_4_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_lec1-tab4-problem1_4_1"> <text><b class="bf">Decreasing</b> [mathjaxinline]b[/mathjaxinline] increases the maximum gain.</text>
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<input type="checkbox" name="input_lec1-tab4-problem1_4_1[]" id="input_lec1-tab4-problem1_4_1_choice_3" class="field-input input-checkbox" value="choice_3"/><label id="lec1-tab4-problem1_4_1-choice_3-label" for="input_lec1-tab4-problem1_4_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_lec1-tab4-problem1_4_1"> <text><b class="bf">Increasing</b> [mathjaxinline]b[/mathjaxinline] increases the maximum gain.</text>
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<input type="checkbox" name="input_lec1-tab4-problem1_4_1[]" id="input_lec1-tab4-problem1_4_1_choice_4" class="field-input input-checkbox" value="choice_4"/><label id="lec1-tab4-problem1_4_1-choice_4-label" for="input_lec1-tab4-problem1_4_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_lec1-tab4-problem1_4_1"> <text>Varying [mathjaxinline]b[/mathjaxinline] does not change the gain plot.</text>
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<h3 class="hd hd-3 problem-header" id="lec1-tab4-problem2-problem-title" aria-describedby="block-v1:OCW+18.031+2019_Spring+type@problem+block@lec1-tab4-problem2-problem-progress" tabindex="-1">
AM radio
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<p>
Suppose you make a homemade AM radio receiver out of an RLC circuit to receive signals from your favorite Boston AM radio station 950. Unfortunately, you aren't getting good reception. </p>
<p>
When you look at the Bode plot for your radio receiver, you understand why your reception is bad. </p>
<center><img src="/assets/courseware/v1/4dd744d1b4f5da07a74f0c903c51ff03/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_radiostations.svg" width="450px" style="margin: 0px 10px 10px 10px"/><br/><b class="bfseries"><span style="color:#0000FF"> Bode plot of gain picking up stations 890, 950, and 1030. :</span></b><br/>Stations are depicted by their broadcast frequency in kHz. <br/>The frequencies band of the broadcast signal is depicted<br/>by the small box around the broadcast frequency. </center>
<div>
<br/>
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<p>
How would you modify the circuit to improve the reception of the station 950 AM? That is, how do you narrow the pass-band? </p>
<p>
(Assume that the inductance is fixed.) </p>
<p>
(Choose all that apply.) </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_lec1-tab4-problem2_2_1">
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<text>By increasing the resistance in the circuit.</text>
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<text>By reducing the resistance in the circuit.</text>
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<text>By increasing the capacitance in the circuit.</text>
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The Bode plot and system response (*)
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(The (*) indicates that this is a difficult problem.) </p>
<p>
Consider the differential equation </p>
<table id="a0000000130" 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">&#160;</td>
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where </p>
<table id="a0000000131" 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]y=\sin \pi t + \sin 2 \pi t + \sin 3\pi t + \dotsb + \sin 10\pi t.[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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The input signal looks like this. </p>
<center><img src="/assets/courseware/v1/166c2bc2371b4e004782b7cee88b8947/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_bodeconceptcheck_input.svg" width="400px" style="margin: 0px 10px 10px 10px"/><br/>Input signal </center>
<p>
Suppose that the Bode plot for the amplitude response for this system looks like this: </p>
<center><img src="/assets/courseware/v1/1d2635be9b5ecde36f0b7f390aa8d25d/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_bodeconceptcheck_bode.svg" width="400px" style="margin: 0px 10px 10px 10px"/><br/>Amplitude response: <br/>the horizontal axis is frequency, <br/>the vertical axis is amplitude. </center>
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Which of the following looks most like the system response? </p>
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<img src="/assets/courseware/v1/b6c8a590e1c1fc6391b0fd6426938fc4/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_bodeconceptcheck_2.svg" width="200px" style="margin: 0px 10px 10px 10px"/>
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<img src="/assets/courseware/v1/495857673b78c43c2b07ee35201684f0/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_bodeconceptcheck_3.svg" width="200px" style="margin: 0px 10px 10px 10px"/>
</td>
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<td style="text-align:center; border:none"><b class="bf">(a)</b> &#8195;[mathjaxinline]3\sin 2\pi t[/mathjaxinline] </td>
<td style="text-align:center; border:none"><b class="bf">(b)</b> &#8195;[mathjaxinline]3\sin 3\pi t[/mathjaxinline] </td>
<td style="text-align:center; border:none"><b class="bf">(c)</b> &#8195;[mathjaxinline]2\sin 5\pi t[/mathjaxinline] </td>
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<td style="text-align:center; border:none">
<img src="/assets/courseware/v1/8ed653c565a79e3dc18c0e5d45670c24/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_bodeconceptcheck_4.svg" width="200px" style="margin: 0px 10px 10px 10px"/>
</td>
<td style="text-align:center; border:none">
<img src="/assets/courseware/v1/8e374d1e3256d2521900a72f01c0b801/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_bodeconceptcheck_5.svg" width="200px" style="margin: 0px 10px 10px 10px"/>
</td>
<td style="text-align:center; border:none">
<img src="/assets/courseware/v1/0ae9c609d8bedbc74148e00082e54d61/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_bodeconceptcheck_6.svg" width="200px" style="margin: 0px 10px 10px 10px"/>
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<td style="text-align:center; border:none"><b class="bf">(d)</b> &#8195;[mathjaxinline]\frac12(\sin \pi t + \sin 3\pi t)[/mathjaxinline] </td>
<td style="text-align:center; border:none"><b class="bf">(e)</b> &#8201; [mathjaxinline]\frac1{10}(\sin \pi t + \sin 3\pi t+\sin 5\pi t)[/mathjaxinline] </td>
<td style="text-align:center; border:none"><b class="bf">(f)</b> &#8201; [mathjaxinline]\frac{1}{10}(\sin \pi t + \dotsb + \sin 10\pi t)[/mathjaxinline] </td>
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<h2 class="hd hd-2 unit-title">1.5. Bode plots of the mascot.</h2>
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<p>
How do you get a Bode plot in the real world? One way, is to hit your system with varying input frequencies, and measure the amplitude and phase lag of the system response. That is exactly what we've done here for our very own 4th order system, the mascot. </p>
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<p>
Here is a reproduction of the same plot using a larger number of input frequencies. (It takes several minutes to cycle through all of these frequencies, which is why we didn't put this in the video.) </p><center><img src="/assets/courseware/v1/1d6ca23228a90445536a0ed86f515e40/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_FR_4thOrder.png" width="600"/></center><p>
Notice that the Bode plot exhibits two peaks, that is two different local maxima, and one local minimum between them. The local minimum corresponds to the case where mass 1 stops moving, but the second mass is still moving quite a bit. By the end of the course, you will gain a better understanding of how to construct a system with this Bode plot. </p><p>
But for now, let's go ahead and take a look at the Bode plot for a system we are much more familiar with, the second order spring mass dashpot. We obtain this system by removing the second mass from the mascot. Take a look at the Bode plot we obtain by hitting the spring mass system with a variety of inputs and measuring the response. </p>
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Here is a reproduction of the Bode plot coming from the same second order system as in the video. </p><center><img src="/assets/courseware/v1/1b80b2cb7f9d492fa769520775256bb1/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_Bode2ndOrder.png" width="400"/></center><div><br/></div><p>
As the frequency increases, the amplitude increases until it reaches a maximum value, at which point it decreases again towards zero. Note this plot looks a bit different from what you've seen on the mathlets because the axes are both in log scale. </p>
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Identify spring constant stiffness from Bode plot
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<p>
We recreated the Bode plot for the same second order system as in the video, but with data taken over a larger number of input frequencies. Then we changed the stiffness of the spring, and made another Bode plot. Then we changed the spring constant one more time, for a third Bode plot. What you see below is three Bode plots for the same second order system, where each plot has a different spring constant [mathjaxinline]k[/mathjaxinline]. </p>
<center>
<img src="/assets/courseware/v1/0390be59b7416f3762a4d92ffcaa9472/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_FR_2ndOrder_Diff_Stiffness.png" width="600"/>
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<p>
Note that we are driving this spring system by a force [mathjaxinline]F[/mathjaxinline] applied to the mass, so the differential equation modeling this system is </p>
<table id="a0000000132" 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 = F.[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
Which Bode plot has the stiffest spring? In other words, for which Bode plot is the spring constant [mathjaxinline]k[/mathjaxinline] the largest? </p>
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<text> The blue Bode plot, which has the smallest practical resonant frequency, and the largest maximum amplitude.</text>
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<input type="radio" name="input_lec1-tab5-problem1_2_1" id="input_lec1-tab5-problem1_2_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="lec1-tab5-problem1_2_1-choice_2-label" for="input_lec1-tab5-problem1_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_lec1-tab5-problem1_2_1">
<text> The red Bode plot, which has the largest practical resonant frequency, and the smallest maximum amplitude.</text>
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<input type="radio" name="input_lec1-tab5-problem1_2_1" id="input_lec1-tab5-problem1_2_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="lec1-tab5-problem1_2_1-choice_3-label" for="input_lec1-tab5-problem1_2_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_lec1-tab5-problem1_2_1">
<text> The green Bode plot, which has the practical resonant frequency and maximum amplitude between the other two.</text>
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<input type="radio" name="input_lec1-tab5-problem1_2_1" id="input_lec1-tab5-problem1_2_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="lec1-tab5-problem1_2_1-choice_4-label" for="input_lec1-tab5-problem1_2_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_lec1-tab5-problem1_2_1">
<text> It cannot be determined from this Bode plot.</text>
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<h2 class="hd hd-2 unit-title">1.6. Activity: Spring systems and gain.</h2>
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In <a href="/courses/course-v1:OCW+18.031+2019_Spring/courseware/review/lec1/3" target="_blank">Lecture 1</a>, we used a mathlet to explore the spring-mass-dashpot system driven through the dashpot. We also discussed the general theory of gain and complex gain. In this activity we will see how this theory works in the specific case of the spring-mass-dashpot system. We start with a summary of gain and complex gain. </p><p><b class="bfseries"><span style="color:#FF7800">Summary:</span></b></p><blockquote class="quote"><ol class="enumerate"><li value="1"><p>
The complex gain is defined as the complex number such that [mathjaxinline]G(\omega )e^{i\omega t}[/mathjaxinline] is the exponential system response to input signal [mathjaxinline]e^{i\omega t}[/mathjaxinline]. </p></li><li value="2"><p>
The gain [mathjaxinline]g(\omega )[/mathjaxinline] is the magnitude of the complex gain [mathjaxinline]G(\omega )[/mathjaxinline]: </p><table id="a0000000140" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\displaystyle g(\omega ) = \left| G(\omega )\right|.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table></li><li value="3"><p>
If the system is modeled by </p><table id="a0000000141" 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>
(with input signal [mathjaxinline]y(t)[/mathjaxinline] and system response [mathjaxinline]x(t)[/mathjaxinline]), then </p><table id="a0000000142" 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></li></ol></blockquote><p>
Now you'll apply this to the spring-mass-dashpot system modeled by the differential equation </p><table id="a0000000143" 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\dot y\, ,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
so </p><table id="a0000000144" 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{bi\omega }{m(i\omega )^2+b(i\omega )+k} =\frac{bi\omega }{(k-m\omega ^2)+bi\omega }\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table>
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<p>
In Lecture 1 you made the following observations about this system using the <a href="https://mathlets.surge.sh/ampPhaseSecondOrderII1803.html" target="_blank">Mathlet</a>. </p><ol class="enumerate"><li value="1"><p>
The maximum gain is 1, independent of the values of [mathjaxinline]m[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], and [mathjaxinline]k[/mathjaxinline]. </p></li><li value="2"><p>
The maximum gain occurs at [mathjaxinline]\omega _ r=\sqrt {k/m}[/mathjaxinline], independent of [mathjaxinline]b[/mathjaxinline]. </p></li><li value="3"><p>
The phase lag at [mathjaxinline]\omega =\omega _ r[/mathjaxinline] is [mathjaxinline]\phi (\omega _ r)=0[/mathjaxinline], while for [mathjaxinline]\omega <\omega _ r[/mathjaxinline] the phase lag is <b class="bf">negative</b>: the system response appears to run <b class="bf">ahead</b> of the input signal. </p></li><li value="4"><p>
As [mathjaxinline]b[/mathjaxinline] decreases, the pass-band narrows. </p></li></ol>
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Checking stability
</h3>
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Your goal is now to verify these observations, using the machinery we have developed in this lecture. </p>
<p>
Recall we are considering the equation </p>
<table id="a0000000145" 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\dot y\, ,[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
where [mathjaxinline]m[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], and [mathjaxinline]k[/mathjaxinline] are positive. </p>
<p>
Verifying that the system is stable guarantees that: (check all that apply) </p>
<p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_lec1-tab6-problem1_2_1">
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<text>all solutions decay to zero as [mathjaxinline]t\rightarrow \infty[/mathjaxinline].</text>
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<input type="checkbox" name="input_lec1-tab6-problem1_2_1[]" id="input_lec1-tab6-problem1_2_1_choice_1" class="field-input input-checkbox" value="choice_1"/><label id="lec1-tab6-problem1_2_1-choice_1-label" for="input_lec1-tab6-problem1_2_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_lec1-tab6-problem1_2_1">
<text>all homogeneous solutions decay to zero as [mathjaxinline]t\rightarrow \infty[/mathjaxinline].</text>
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<input type="checkbox" name="input_lec1-tab6-problem1_2_1[]" id="input_lec1-tab6-problem1_2_1_choice_2" class="field-input input-checkbox" value="choice_2"/><label id="lec1-tab6-problem1_2_1-choice_2-label" for="input_lec1-tab6-problem1_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_lec1-tab6-problem1_2_1">
<text>all solutions are sinusoidal.</text>
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<text>all solutions are asymptotically the same as [mathjaxinline]t\rightarrow \infty[/mathjaxinline].</text>
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<text>all solutions are overdamped and do not oscillate.</text>
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<text>initial conditions don't have a significant long-term effect on the solutions.</text>
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<text>the unforced system always returns to equilibrium.</text>
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<p>
To check stability of [mathjaxinline]P(D)x = Q(D)y[/mathjaxinline], we must verify the following. <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div class="choicegroup capa_inputtype" id="inputtype_lec1-tab6-problem1_3_1">
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<input type="radio" name="input_lec1-tab6-problem1_3_1" id="input_lec1-tab6-problem1_3_1_choice_1" class="field-input input-radio" value="choice_1"/><label id="lec1-tab6-problem1_3_1-choice_1-label" for="input_lec1-tab6-problem1_3_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_lec1-tab6-problem1_3_1"> <text> The real parts of the roots of [mathjaxinline]P[/mathjaxinline] and [mathjaxinline]Q[/mathjaxinline] are negative.</text>
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<input type="radio" name="input_lec1-tab6-problem1_3_1" id="input_lec1-tab6-problem1_3_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="lec1-tab6-problem1_3_1-choice_2-label" for="input_lec1-tab6-problem1_3_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_lec1-tab6-problem1_3_1"> <text> The real parts of the roots of [mathjaxinline]P[/mathjaxinline] are negative.</text>
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Is this system ([mathjaxinline]m\ddot x+b\dot x+kx=b\dot y[/mathjaxinline]) stable? <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 3" role="group"><div class="choicegroup capa_inputtype" id="inputtype_lec1-tab6-problem1_4_1">
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Study the gain
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<p>
Study the complex gain </p>
<table id="a0000000146" 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]G(\omega )=\frac{bi\omega }{(k-m\omega ^2)+bi\omega }\, .[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
Here are two pictures of what the numerator (blue) and denominator (orange) of [mathjaxinline]G(\omega )[/mathjaxinline] might look like for a specific value of [mathjaxinline]\omega[/mathjaxinline]. Right? </p>
<center><img src="/assets/courseware/v1/c9f82c2c2be45da03af162e9d681c5d7/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c2_complexgain.svg" width="600px" style="margin: 0px 10px 10px 10px"/><br/>For fixed parameters [mathjaxinline]m[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], and [mathjaxinline]k[/mathjaxinline], possible numerators<br/>are depicted in blue with potential denominators in orange. </center>
<p>
Decide what the maximum of [mathjaxinline]g(\omega ) = \left| G(\omega )\right|[/mathjaxinline] is. <p style="display:inline">Maximum of [mathjaxinline]g(\omega )[/mathjaxinline]: </p><div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_lec1-tab6-problem2_2_1" class=" capa_inputtype inline textline">
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<p>
What determines the sign of the phase lag? Give an inequality involving the parameters that guarantees that the phase lag is positive. </p>
<p>
(Enter answer in terms of system parameters [mathjaxinline]k[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], and [mathjaxinline]m[/mathjaxinline].) </p>
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<p style="display:inline">Phase lag is positive when [mathjaxinline]\omega[/mathjaxinline]</p>
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<p>
If you fix [mathjaxinline]k[/mathjaxinline], [mathjaxinline]m[/mathjaxinline], and [mathjaxinline]\omega[/mathjaxinline] (with [mathjaxinline]\omega \neq \omega _ r[/mathjaxinline]), what happens to the gain as [mathjaxinline]b[/mathjaxinline] gets small? </p>
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<h4 onclick="hideshow(this);" style="margin: 0px">Hint<span class="icon-caret-down toggleimage"/></h4>
<div class="hideshowcontent">For this it is helpful to divide numerator and denominator of the complex gain by [mathjaxinline]bi\omega[/mathjaxinline]: <table id="a0000000147" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]G(\omega )=\left(1+\frac{k-m\omega ^2}{bi\omega }\right)^{-1}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none">&#160;</td></tr></table> So [mathjaxinline]g(\omega )=|G(\omega )|[/mathjaxinline] is small when <table id="a0000000148" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{|k-m\omega ^2|}{b\omega }[/mathjax]</td><td class="eqnnum" style="width:20%; border:none">&#160;</td></tr></table> is large. </div>
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<text> The gain approaches [mathjaxinline]0[/mathjaxinline].</text>
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<text> The gain approaches [mathjaxinline]1[/mathjaxinline].</text>
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<text> The gain approaches [mathjaxinline]\infty[/mathjaxinline].</text>
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<h2 class="hd hd-2 unit-title">1.7. Resonance.</h2>
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<h3 class="hd hd-3 problem-header">The simple harmonic oscillator</h3><p>
Let's begin by thinking about the example of an undamped spring system, driven through the spring. Let's begin by thinking about the simple harmonic oscillator, which is modeled by the differential equation </p><table id="a0000000152" 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 + kx = f(t).[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Let's look at the response to sinusoidal input to remind ourselves about the definition of pure resonance. </p><p><p><b class="bfseries">Example 7.1 </b> Find the general solution to the DE: [mathjaxinline]m\ddot x + kx = A\cos \omega t[/mathjaxinline]. </p></p><p><b class="bfseries"><span style="color:#FF7800">Answer:</span></b> We do not show all the algebra. You should try it for yourself. </p><p>
Homogeneous solution: </p><table id="a0000000153" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x_ h = c_1\cos (\omega _ nt) + c_2\sin (\omega _ nt),[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
where [mathjaxinline]\omega _ n = \sqrt {k/m}[/mathjaxinline] is the <b class="bfseries"><span style="color:#0000FF">natural frequency</span></b> of the oscillator. </p><p>
Particular solution: </p><table id="a0000000154" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x_ p = \begin{cases} \displaystyle \frac{A\cos \omega t}{|k -m\omega ^2|} \, = \, \frac{A}{m}\frac{\cos \omega t}{|\omega _ n^2 -\omega ^2|} & \text { if $\omega < \omega _ n$}\\ \displaystyle \frac{A\cos (\omega t -\pi )}{|k -m\omega ^2|} \, = \, \frac{-A}{m}\frac{\cos \omega t}{|\omega _ n^2 -\omega ^2|} & \text { if $\omega > \omega _ n$}\\ \displaystyle \frac{A\, t\cos (\omega _ n t-\pi /2)}{2m\omega _ n} \, = \, \frac{A}{m}\frac{t\sin (\omega _ nt)}{2\omega _ n} & \text { if $\omega =\omega _ n$.} \end{cases}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The bottom case ([mathjaxinline]\omega = \omega _ n[/mathjaxinline]) requires complex replacement and the extended ERF. You'll see an alternative approach soon. </p><p>
The general solution (to the inhomogeneous DE) is given by [mathjaxinline]x = x_ p + x_ h[/mathjaxinline]. The simple harmonic oscillator may be simple, but it is not stable. </p><h3 class="hd hd-3 problem-header">Resonance for the simple harmonic oscillator</h3><p>
The solution for [mathjaxinline]x_ p[/mathjaxinline] shows that the gain of the system is given by </p><table id="a0000000155" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]g = g(\omega ) = \frac{1}{m|\omega _ n^2 -\omega ^2|}.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
This is a function of [mathjaxinline]\omega[/mathjaxinline] and the right hand plot below shows its graph. </p><p>
In this case when the input frequency coincides with the natural frequency [mathjaxinline]\omega _ n[/mathjaxinline] the system is in what is called <b class="bfseries"><span style="color:#0000FF">pure resonance</span></b> and [mathjaxinline]\omega _ n[/mathjaxinline] is then called the resonant frequency of the system. When [mathjaxinline]\omega = \omega _ n[/mathjaxinline], we have [mathjaxinline]\displaystyle x_ p = \frac{A\, t\sin \omega _ n t}{2m\omega _ n}[/mathjaxinline]. The left-hand plot below graphs [mathjaxinline]x_ p[/mathjaxinline] in this case. Notice how the response is oscillatory but not periodic. The factor of [mathjaxinline]t[/mathjaxinline] in [mathjaxinline]x_ p[/mathjaxinline] causes the amplitude to keep growing over time. </p><center><img src="/assets/courseware/v1/6280c5c950402d47d2dea96c601af5bd/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_figc2_3.svg" width="400px" style="margin: 0px 10px 10px 10px"/> <img src="/assets/courseware/v1/f69f9f91f266e9d475a0d34ef654a56e/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_figc2_4.svg" width="250px" style="margin: 0px 10px 10px 10px"/><br/> Resonant response ([mathjaxinline]\omega =\omega _ n[/mathjaxinline]) Undamped amplitude response </center>
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<p><h3>Amplitude response and resonance</h3></p><p>
The gain [mathjaxinline]g(\omega )[/mathjaxinline] is a function of [mathjaxinline]\omega[/mathjaxinline]. It tells us the size of the system's response to the given input frequency. The graph of [mathjaxinline]g(\omega )[/mathjaxinline] vs [mathjaxinline]\omega[/mathjaxinline] is one of the Bode plots. In many complex systems, the gain Bode plot can be quite complicated, exhibiting maxima and minima. The maxima occur at frequencies near to some natural frequency of the system, and are called <b class="bfseries"><span style="color:#0000FF">resonant</span></b> frequencies. In second order systems, there is at most one positive resonant frequency, which we will often denote by [mathjaxinline]\omega _ r[/mathjaxinline]. </p><p><p><b class="bfseries">Example 7.2 </b> Consider the spring/mass/dashpot system driven through the spring </p><table id="a0000000156" 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 = kA\cos (\omega t)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
where we consider [mathjaxinline]A\cos (\omega t)[/mathjaxinline] to be the input. </p></p><p>
Find an expression for [mathjaxinline]\omega _ r[/mathjaxinline] when it exists in terms of system parameters. </p><p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">A hint to get started<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><p>
Find a formula for [mathjaxinline]g(\omega )=|G(\omega )|[/mathjaxinline]. Then [mathjaxinline]\omega _ r[/mathjaxinline] is the value where [mathjaxinline]g(\omega )[/mathjaxinline] attains its maximum. </p><p>
The function [mathjaxinline]g(\omega )[/mathjaxinline] is rather complicated; are there simpler functions that have minima or maxima at the same places? </p></div><p class="hideshowbottom" onclick="hideshow(this);" style="margin: 0px"><a href="javascript: {return false;}">Show</a></p></div></p><p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">Full solution<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><p>
The gain is </p><table id="a0000000157" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]g(\omega ) = \left| \frac{Q(i\omega )}{P(i\omega )}\right| = \left|\frac{k}{(k-m\omega ^2)+b\omega i}\right| = \frac{k}{\sqrt {(k-m\omega ^2)^2+b^2\omega ^2 }}.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Finding practical resonance: If it exists the practical resonance frequency [mathjaxinline]\omega _ r[/mathjaxinline] is the frequency where [mathjaxinline]g(\omega )[/mathjaxinline] has a maximum. Now, [mathjaxinline]g(\omega )[/mathjaxinline] has a maximum where the expression under the square root in the denominator </p><table id="a0000000158" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f(\omega ) = (k-m\omega ^2)^2 + b^2\omega ^2[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
has a minimum. Taking the derivative and setting it to 0 we get </p><table id="a0000000159" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f'(\omega ) = -4m\omega (k - m\omega ^2) + 2b^2\omega = 0.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
We find two possible solutions, [mathjaxinline]\omega = 0[/mathjaxinline], and </p><table id="a0000000160" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\omega _ r \, = \, \sqrt {k/m-b^2/2m^2} \, = \, \sqrt {\omega _ n^2 - b^2/2m^2}.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Since the resonant frequency must be real we ignore the case [mathjaxinline]\omega = 0[/mathjaxinline]. The second possibility gives a resonant frequency as long as [mathjaxinline]\omega _ n^2 - b^2/2m^2 > 0[/mathjaxinline]. In conclusion, the resonant frequency for the second order system is </p><table id="a0000000161" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\omega _ r = \begin{cases} \sqrt {k/m-b^2/2m^2} & \text { when }\omega _ n^2 - b^2/2m^2 > 0 \\ \text {no resonant frequency } & \text { otherwise}.\end{cases}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Note that, in the damped case [mathjaxinline]b > 0[/mathjaxinline], when it exists, [mathjaxinline]\omega _ r < \omega _ n[/mathjaxinline]. </p></div><p class="hideshowbottom" onclick="hideshow(this);" style="margin: 0px"><a href="javascript: {return false;}">Show</a></p></div></p><p>
Note that the ratio [mathjaxinline]b/m[/mathjaxinline] has units 1/time; if [mathjaxinline]b/m > \sqrt {2}\omega _ n[/mathjaxinline] then there's no positive resonant frequency. In this case, we say that there is no practical resonance. (Even though the gain is maximized at [mathjaxinline]\omega =0[/mathjaxinline], we do not consider this to be a resonant frequency.) </p><center><img src="/assets/courseware/v1/7d43215f99e5a9c266ff14f85bc6154b/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_figc2_1.svg" width="300px" style="margin: 0px 10px 10px 10px"/> <img src="/assets/courseware/v1/c5ed8b06008c4398604ac33cb78c5c6d/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_figc2_2.svg" width="300px" style="margin: 0px 10px 10px 10px"/><br/>(<b class="bfseries">practical resonance</b>) (no practical resonance)<br/>[mathjaxinline]\omega _ r = \sqrt {\omega _ n^2 - b^2/2m^2}[/mathjaxinline] <br/>. </center><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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<p><b class="bfseries">Example 7.3&#8195;&#8195;</b> Consider the system </p>
<table id="a0000000162" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]\ddot x + b\dot x + kx = k\cos (\omega t)[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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where we consider [mathjaxinline]\cos (\omega t)[/mathjaxinline] to be the input. </p>
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<p>
This system is modeled in the following mathlet. Note that this models a different system that the mathlet we looked at before. This system is driven through the spring at the top. The system we were looking at before (which was analogous to the RLC circuit modeling a passive AM radio receiver) was being driven through the dashpot. </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/ampPhaseSecondOrderI.html" width="1100 px" height="640 px"/>
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<p>
Set [mathjaxinline]k=1[/mathjaxinline] and [mathjaxinline]b=1[/mathjaxinline]. Estimate [mathjaxinline]\omega _ r[/mathjaxinline] the value of the practical resonant frequency if it exists. (Enter decimal answer from mathlet. Enter <b class="bf">none</b> if there is no practical resonance.) <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_lec1-tab7-problem1_2_1" class=" capa_inputtype textline">
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Set [mathjaxinline]k=0.5[/mathjaxinline] and [mathjaxinline]b=1.5[/mathjaxinline]. Find the value of the practical resonant frequency if it exists. (Enter decimal answer from mathlet. Enter <b class="bf">none</b> if there is no practical resonance.) <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div id="inputtype_lec1-tab7-problem1_3_1" class=" capa_inputtype textline">
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<h2 class="hd hd-2 unit-title">1.8. The Nyquist Plot.</h2>
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<p>There is a second checkbox on the Mathlet, labeled “Nyquist Plot." Click it. A new graphing window appears in the lower right corner of the Mathlet. This is a picture of the complex plane, displaying the trajectory of the complex gain, as a function of [mathjaxinline]\omega[/mathjaxinline]. An orange line segment connects the origin to the value [mathjaxinline]G(\omega )[/mathjaxinline] of the complex gain at the selected value of [mathjaxinline]\omega[/mathjaxinline]. Set [mathjaxinline]b=1[/mathjaxinline] and watch what happens to this vector as you move the [mathjaxinline]\omega[/mathjaxinline] slider. The length of the strut is the magnitude of the complex gain, that is, the gain [mathjaxinline]g(\omega )[/mathjaxinline]. You can see it vary with [mathjaxinline]\omega[/mathjaxinline]. The argument of the complex gain is the phase gain [mathjaxinline]-\phi (\omega )[/mathjaxinline]. This curve in the complex plane, called the <b class="bf">Nyquist plot</b>, displays the relationship between the gain and the phase lag.</p>
<p>In this example, the Nyquist plot is actually independent of the system parameters. Check this by varying the values of [mathjaxinline]k[/mathjaxinline] and [mathjaxinline]b[/mathjaxinline] and watching the effect on the curve. This is unusual behavior; most of the time, changing the parameters will change the gain and phase lag profiles, altering the Nyquist plot.</p>
<p><iframe style="display: block; border-width: 0px; padding: 0px;" src="https://mathlets1803.surge.sh/ampPhaseSecondOrderII.html" width="1100 px" height="640 px">
<h3 class="hd hd-3 problem-header">Food for thought.</h3>
<p><b class="bf">1.</b> We have seen that when [mathjaxinline]b[/mathjaxinline] is small the resonant peak is narrow; that is, as soon as [mathjaxinline]\omega[/mathjaxinline] differs much from [mathjaxinline]\omega _ r[/mathjaxinline], the gain becomes very small. Make a prediction about how the Nyquist trajectory will be traversed, based on this observation. That is, as [mathjaxinline]\omega[/mathjaxinline] increases at a steady rate, will [mathjaxinline]G(\omega )[/mathjaxinline] move at steady rate along its trajectory, or will it move faster in some portions than in others? (Check your hypothesis using the mathlet.) </p>
<p><b class="bf">2.</b> Show that the Nyquist plot, for this system, is given by a circle of radius [mathjaxinline]1/2[/mathjaxinline] with center at the complex number [mathjaxinline]1/2[/mathjaxinline] (minus the origin). </p>
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<h4 onclick="hideshow(this);" style="margin: 0px;">Remarks on food for thought.<span class="icon-caret-down toggleimage"></span></h4>
<div class="hideshowcontent"><ol class="enumerate">
<li value="1">
<p>
When [mathjaxinline]b[/mathjaxinline] is small, the gain is very close to zero everywhere except near the resonant peak. Thus we expect that as we vary [mathjaxinline]\omega[/mathjaxinline], the complex gain stays close to zero most of the time, and then traverses very quickly to a large amplitude near resonance. </p>
</li>
<li value="2">
<p>
The Nyquist plot is a plot of the complex gain </p>
<table id="a0000000175" 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{ib\omega }{k-\omega ^2 +ib\omega }.[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
A circle of radius [mathjaxinline]1/2[/mathjaxinline] centered at [mathjaxinline]1/2 + 0i[/mathjaxinline] in the complex plane can be described by the complex equation </p>
<table id="a0000000176" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]|z-1/2| = 1/2.[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
We verify that [mathjaxinline]\displaystyle z=\frac{ib\omega }{k-\omega ^2 +ib\omega }[/mathjaxinline] satisfies the equation for a circle of radius [mathjaxinline]1/2[/mathjaxinline]: </p>
<table id="a0000000177" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout: auto;">
<tr id="a0000000178">
<td style="width: 40%; border: none;"> </td>
<td style="vertical-align: middle; text-align: right; border: none;">
[mathjaxinline]\displaystyle \displaystyle |z-1/2|[/mathjaxinline]
</td>
<td style="vertical-align: middle; text-align: center; border: none;">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align: middle; text-align: left; border: none;">
[mathjaxinline]\displaystyle \left| \frac{ib\omega }{k-\omega ^2 +ib\omega } - \frac12 \right|[/mathjaxinline]
</td>
<td style="width: 40%; border: none;"> </td>
<td style="width: 20%; border: none; text-align: right;" class="eqnnum">(2.16)</td>
</tr>
<tr id="a0000000179">
<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 \left| \frac{ib\omega }{k-\omega ^2 +ib\omega } - \frac12\frac{k-\omega ^2 +ib\omega }{k-\omega ^2 +ib\omega } \right|[/mathjaxinline]
</td>
<td style="width: 40%; border: none;"> </td>
<td style="width: 20%; border: none; text-align: right;" class="eqnnum">(2.17)</td>
</tr>
<tr id="a0000000180">
<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 \left| \frac12\frac{-(k-\omega ^2)+ib\omega }{k-\omega ^2 +ib\omega } \right| = 1/2[/mathjaxinline]
</td>
<td style="width: 40%; border: none;"> </td>
<td style="width: 20%; border: none; text-align: right;" class="eqnnum">(2.18)</td>
</tr>
</table>
</li>
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<h3 class="hd hd-3 problem-header">Worked example.</h3>
<p>Consider a spring/mass/dashpot system being driven through the spring. This is modeled by the equation</p>
<table id="a0000000181" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]\ddot x + \dot x +2x = 2f(t)[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>with input [mathjaxinline]\ f(t)[/mathjaxinline], as in the mathlet below (with b=1 and k=2).</p>
<p><iframe style="display: block; border-width: 0px; padding: 0px;" src="https://mathlets1803.surge.sh/ampPhaseSecondOrderI.html" width="1100 px" height="640 px">
<p>
The Bode plots suggest answers to various questions about the behavior of this system. Let's enumerate some and then verify them by computation. </p>
<ul class="itemize">
<li>
<p>
What is [mathjaxinline]G(0)[/mathjaxinline]? So what are [mathjaxinline]g(0)[/mathjaxinline] and [mathjaxinline]\phi (0)[/mathjaxinline]? </p>
</li>
<li>
<p>
What is the limiting value of [mathjaxinline]G(\omega )[/mathjaxinline] as [mathjaxinline]\omega \rightarrow \infty[/mathjaxinline]? </p>
</li>
<li>
<p>
Does this system exhibit a resonant peak at a positive value of the input frequency? At what value of [mathjaxinline]\omega[/mathjaxinline]? What is the maximal gain? </p>
</li>
</ul>
<div class="hideshowbox">
<h4 onclick="hideshow(this);" style="margin: 0px;">Worked solution.<span class="icon-caret-down toggleimage"></span></h4>
<div class="hideshowcontent">
<p>
The system has complex gain </p>
<table id="a0000000182" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]G = \frac{2}{P(i\omega )} = \frac{2}{2-\omega ^2+i\omega },[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
where [mathjaxinline]P(s) = s^2 + s + 2[/mathjaxinline]. So the gain is given by </p>
<table id="a0000000183" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout: auto;">
<tr id="a0000000184">
<td style="width: 40%; border: none;"> </td>
<td style="vertical-align: middle; text-align: right; border: none;">
[mathjaxinline]\displaystyle \displaystyle g(\omega )[/mathjaxinline]
</td>
<td style="vertical-align: middle; text-align: center; border: none;">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align: middle; text-align: left; border: none;">
[mathjaxinline]\displaystyle \frac{2}{\left|P(i\omega ) \right|}[/mathjaxinline]
</td>
<td style="width: 40%; border: none;"> </td>
<td style="width: 20%; border: none; text-align: right;" class="eqnnum">(2.19)</td>
</tr>
<tr id="a0000000185">
<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 \frac{2}{\left|2-\omega ^2 + i\omega \right|}[/mathjaxinline]
</td>
<td style="width: 40%; border: none;"> </td>
<td style="width: 20%; border: none; text-align: right;" class="eqnnum">(2.20)</td>
</tr>
<tr id="a0000000186">
<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 \frac{2}{\sqrt {(2-\omega ^2)^2 + (\omega )^2}}.[/mathjaxinline]
</td>
<td style="width: 40%; border: none;"> </td>
<td style="width: 20%; border: none; text-align: right;" class="eqnnum">(2.21)</td>
</tr>
</table>
<p>
The phase lag is given by </p>
<table id="a0000000187" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout: auto;">
<tr id="a0000000188">
<td style="width: 40%; border: none;"> </td>
<td style="vertical-align: middle; text-align: right; border: none;">
[mathjaxinline]\displaystyle \displaystyle \phi (\omega )[/mathjaxinline]
</td>
<td style="vertical-align: middle; text-align: center; border: none;">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align: middle; text-align: left; border: none;">
[mathjaxinline]\displaystyle -\mathrm{Arg}\left(\frac{2}{P(i\omega )}\right)[/mathjaxinline]
</td>
<td style="width: 40%; border: none;"> </td>
<td style="width: 20%; border: none; text-align: right;" class="eqnnum">(2.22)</td>
</tr>
<tr id="a0000000189">
<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 \mathrm{Arg}(P(i\omega ))[/mathjaxinline]
</td>
<td style="width: 40%; border: none;"> </td>
<td style="width: 20%; border: none; text-align: right;" class="eqnnum">(2.23)</td>
</tr>
<tr id="a0000000190">
<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 \mathrm{Arg}(2-\omega ^2 + i\omega ).[/mathjaxinline]
</td>
<td style="width: 40%; border: none;"> </td>
<td style="width: 20%; border: none; text-align: right;" class="eqnnum">(2.24)</td>
</tr>
</table>
<p>
Answering the questions posed we find </p>
<ul class="itemize">
<li>
<p>
[mathjaxinline]G(0) = 1[/mathjaxinline]. So [mathjaxinline]g(0) = 1[/mathjaxinline] and [mathjaxinline]\phi (0) = 0.[/mathjaxinline] </p>
</li>
<li>
<p>
As [mathjaxinline]\omega \rightarrow \infty[/mathjaxinline], the complex gain is roughly </p>
<table id="a0000000191" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]G(\omega ) \approx \frac{2}{-\omega ^2}[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
Thus [mathjaxinline]g(\omega ) \rightarrow 0[/mathjaxinline] and [mathjaxinline]\phi (\omega ) \rightarrow \pi[/mathjaxinline] as [mathjaxinline]\omega \rightarrow \infty .[/mathjaxinline] </p>
</li>
<li>
<p>
To complete our understanding of the frequency response of this system, we will look for resonant peaks. The Mathlet suggests that one occurs at about [mathjaxinline]\omega = 1.22[/mathjaxinline]. The function [mathjaxinline]g(\omega )[/mathjaxinline] will be maximal exactly when the denominator [mathjaxinline]|P(i\omega )|[/mathjaxinline] is minimal – or, what is the same, when [mathjaxinline]\omega[/mathjaxinline] minimizes the square: </p>
<table id="a0000000192" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]|P(i\omega )|^2 = (2-\omega ^2)^2 + \omega ^2 .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
To find such [mathjaxinline]\omega[/mathjaxinline], differentiate this function with respect to [mathjaxinline]\omega[/mathjaxinline]: </p>
<table id="a0000000193" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]2(2-\omega ^2)(-2\omega ) + 2\omega = \omega (4 \omega ^2 - 6)[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
The roots of this polynomial are [mathjaxinline]\omega = 0[/mathjaxinline] and [mathjaxinline]\omega = \pm \sqrt {3/2}[/mathjaxinline]. Selecting the positive root gives </p>
<table id="a0000000194" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]\omega _ r = \sqrt {3/2} \sim 1.2247,[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
in good agreement with our observation. The maximal gain is </p>
<table id="a0000000195" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]g(\omega _ r) = g\left(\sqrt {3/2}\right) = 2\sqrt {4/7} \sim 1.5119 ,[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</table>
<p>
again in good agreement. </p>
<p>
Below you can see the Bode plots and Nyquist plot. </p>
<center><img src="/assets/courseware/v1/3ce8b96c3d1ca87c7c1c42fac1fff1cb/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_figc2_6.png" width="325" /> <img src="/assets/courseware/v1/21c3743b5bae020b2d58f6e7265b5879/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_figc2_7.png" width="325" /> <br /> Bode gain plot Bode phase plot (degrees)<br /><br /></center><center><img src="/assets/courseware/v1/188dd4f059b7407cf3a7aaf60d0af595/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_figc2_8.png" width="300" /><br /><b class="bfseries"><span style="color: #0000ff;">Nyquist plot</span></b></center></li>
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<h2 class="hd hd-2 unit-title">1.9. The Transfer Function.</h2>
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<p>
The complex gain [mathjaxinline]G(\omega )[/mathjaxinline] of an LTI system is the factor by which the input signal [mathjaxinline]e^{i\omega t}[/mathjaxinline] gets multiplied. For a system modeled by the differential equation </p><table id="a0000000196" 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)e^{i\omega t},[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
the complex gain is given by the formula [mathjaxinline]Q(i\omega )/P(i\omega )[/mathjaxinline]. </p><p>
Both these facts encourage the following question: what is the exponential system response to a more general exponential input signal, one of the form </p><table id="a0000000197" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]e^{st},\quad \quad s \, \text { a complex constant}\, ?[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
This is a reasonable question. For example, the input signal might be a damped sinusoid, something like [mathjaxinline]e^{-t/10}\cos (\pi t) = \mathrm{Re}\, \left((e^{(-.1+\pi i)t}\right)[/mathjaxinline]. </p><p>
Exactly the same reasoning we used earlier leads to perfectly parallel conclusions. We will use a different letter to denote the multiplicative factor, in order to avoid confusion with the special case given by the complex gain. </p><blockquote class="quote"> The exponential system response of an LTI system to the input signal [mathjaxinline]e^{st}[/mathjaxinline], for [mathjaxinline]s[/mathjaxinline] any complex constant, is of the form <table id="a0000000198" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]H(s)e^{st}.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table> If the LTI system is controlled by the differential equation <table id="a0000000199" 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> then <table id="a0000000200" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]H(s)=\frac{Q(s)}{P(s)}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table> </blockquote><p>
The factor [mathjaxinline]H(s)[/mathjaxinline] is the <b class="bfseries"><span style="color:#0000FF">transfer function</span></b> of the LTI system. You can think of the name as meaning that it <b class="bfseries"><span style="color:#0000FF">transfers</span></b> the input [mathjaxinline]e^{st}[/mathjaxinline] to the response [mathjaxinline]x_ p = H(s)e^{st}[/mathjaxinline]. The transfer function is also called the <b class="bfseries"><span style="color:#0000FF">system function</span></b> . </p><p>
Note that all of the work we have done with complex gain was a study of a special case of the transfer function when [mathjaxinline]s=i\omega[/mathjaxinline]. </p><p>
In particular, </p><table id="a0000000201" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]G(\omega ) = H(i\omega )\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><h3 class="hd hd-3 problem-header">Comments.</h3><ol class="enumerate"><li value="1"><p>
In this class, the transfer function will usually be a quotient of one polynomial by another: a <b class="bf">rational function</b>. In more general LTI systems it may be more complicated. But it is always a function of a <b class="bf">complex number</b> [mathjaxinline]s[/mathjaxinline], taking on <b class="bf">complex values</b>. </p></li><li value="2"><p>
If [mathjaxinline]r[/mathjaxinline] is a root of the denominator, [mathjaxinline]P(r)=0[/mathjaxinline], then [mathjaxinline]H(s)[/mathjaxinline] is not defined. This corresponds to <b class="bf">resonance</b>; the input signal [mathjaxinline]e^{rt}[/mathjaxinline] does not produce an exponential system response at all. We will say more about this situation later. </p></li></ol>
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<p>
What is the transfer function of the system </p>
<table id="a0000000202" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]2\ddot x + 3\dot x + x = 5\dot y + 3y.[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
Here we consider [mathjaxinline]y[/mathjaxinline] the input and [mathjaxinline]x[/mathjaxinline] the response. </p>
<p>
(Enter as a rational function in the variable [mathjaxinline]s[/mathjaxinline].) </p>
<p>
<p style="display:inline">[mathjaxinline]H(s)=[/mathjaxinline]</p>
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Question 2.
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Use the transfer function you found for the system in the previous problem. </p>
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For what values of [mathjaxinline]s[/mathjaxinline] is the transfer function undefined? </p>
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(Enter as a list of points separated by commas: e.g. 2*i, -2*i.) </p>
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<h2 class="hd hd-2 unit-title">1.10. Footnotes.</h2>
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<p><b class="bf">1. Why “Bode"?</b> Bode plots are named after Hendrick Wade Bode, 1905–1982. Bode (rhymes with toady) was one of the fathers of modern control theory. For most of his career he worked at Bell Labs. During World War II he devised the feedback mechanism linking radar data to anti-aircraft fire. After the war he occupied high administrative positions at Bell Labs. In 1967 he retired as Vice President in charge of military development and systems engineering, and took up the position of Gordon McKay Professor at Harvard, a post he held till 1974. </p><p>
Our use of the phrase “Bode plots" in this course is inaccurate in several respects. First of all, in engineering applications one typically has to cope with a wide range of frequencies. To represent them, one plots the log of the frequency horizontally. Similarly, the gain often spans a wide range of values, and to provide clear visualizations of the gain it is sensible to plot [mathjaxinline]\log g(\omega )[/mathjaxinline] rather than [mathjaxinline]g(\omega )[/mathjaxinline]. So typically the gain Bode plot uses a log-log scale, as shown below. </p><center><img src="/assets/courseware/v1/066644fd880ac8162bd2d215e3c55436/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_loglogbode.png" width="450"/><br/>Log-log scale Bode plot.<br/>(Note the vertical axis appears linear, but is plotting log(H(iω)).) <div><br/></div></center><p>
The vertical logarithmic scale is often measured in “decibels." The decibel measure of gain [mathjaxinline]g[/mathjaxinline] is [mathjaxinline]20\log _{10}g[/mathjaxinline] Db. This can also be written [mathjaxinline]10\log _{10}g^2[/mathjaxinline], which is of interest since the power of a sinusoidal signal is proportional to the square of its amplitude. Use of these units originated at Bell Labs in the 1920's. </p><p>
The phase Bode plot uses [mathjaxinline]\log \omega[/mathjaxinline] horizontally, and [mathjaxinline]-\phi[/mathjaxinline] vertically. No need for a log vertically; after all, in a sense [mathjaxinline]\phi[/mathjaxinline] is already a logarithm – in the complex gain it occurs as [mathjaxinline]e^{-i\phi }[/mathjaxinline]. </p><p>
Even these smooth log-log plots are not quite the thing that Bode introduced. He provided quick and efficient ways to sketch these plots, or piecewise linear approximations of them. </p>
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