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<h2 class="hd hd-2 unit-title">12.1. Lecture 8: Return of the Mascot.</h2>
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<h3 class="hd hd-3 problem-header">Objectives.</h3><p>
Apply the results of this course to study a genuine mechanical system in the frequency domain. </p>
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<h2 class="hd hd-2 unit-title">12.2. Reintroducing the Mascot.</h2>
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<p>
A tuned mass damper is a system of coupled damped oscillators in which one oscillator is regarded as primary and the second as a control or secondary oscillator. If tuned properly the maximum amplitude of the primary oscillator in response to a periodic driving force will be lowered and much of the energy will be absorbed by the secondary oscillator. </p><p>
This kind of system is used for example in tall buildings to limit the swaying of the building in the wind. People are sensitive to this swaying, so by adding a tuned mass damper the building sways less and the damper, which no one can feel, vibrates instead. Another application is to stabilize laboratory tables supporting experiments that are sensitive to vibrations. </p><p>
The 18.03Lx mascot is an example of such a system. The figure below represents an idealized version of it. </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"/></center><p>
The first mass [mathjaxinline]m_1[/mathjaxinline] is attached on one side to a wall by a spring and damper and on the other side it is attached to a second mass [mathjaxinline]m_2[/mathjaxinline] by another spring and damper. The spring and damping constants [mathjaxinline]k_1[/mathjaxinline], [mathjaxinline]k_2[/mathjaxinline], [mathjaxinline]b_1[/mathjaxinline], [mathjaxinline]b_2[/mathjaxinline] are indicated on the figure. A force [mathjaxinline]f(t)[/mathjaxinline] pushes on the first mass. The absolute positions of the masses are given by [mathjaxinline]x_1[/mathjaxinline] and [mathjaxinline]x_2[/mathjaxinline], arranged so that the spring between them is relaxed when [mathjaxinline]x_1=x_2[/mathjaxinline] and the spring connecting the first mass to the wall is relaxed when [mathjaxinline]x_1=0[/mathjaxinline]. </p><p>
We will regard first mass as the building, or the table – it's being shaken by some force, and we wish to control the amplitude of its resulting oscillation – so the system response of interest is [mathjaxinline]x_1[/mathjaxinline]. The second mass is presumably smaller, and the behavior of [mathjaxinline]x_2[/mathjaxinline] is of only secondary interest. </p><p>
In this Lecture, we will work out the system function of this fairly complicated system. To accomplish this goal, we will watch it in action and record its system response. Finally, we match the resulting Bode plot in order to discover the pole/zero diagram of the system and hence information about the system parameters. </p>
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<h2 class="hd hd-2 unit-title">12.3. The differential equations.</h2>
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<p>
Newton's law [mathjaxinline]f=ma[/mathjaxinline] and the usual assumptions about linear damping and spring force lead to the following differential equations governing the motion of the system. </p><table id="a0000001152" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\begin{array}{llllllllllll} m_1\ddot{x}_1 & +& b_1\dot{x}_1 & +& k_1x_1 & -& b_2(\dot{x}_2-\dot{x}_1) & -& k_2(x_2-x_1) & =& f(t)\\ m_2\ddot{x}_2 & +& & & & +& b_2(\dot{x}_2-\dot{x}_1) & +& k_2(x_2-x_1) & =& 0 \end{array}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Let's rearrange these equations to put all the [mathjaxinline]x_1[/mathjaxinline]'s on one side and all the [mathjaxinline]x_2[/mathjaxinline]'s on the other: </p><table id="a0000001153" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001154"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle (m_1\ddot{x}_1 + b_1\dot{x}_1+k_1x_1)+(b_2\dot{x}_1 + k_2x_1)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle = f(t)+b_2\dot{x}_2+k_2x_2[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000001155"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle m_2\ddot{x}_2 + b_2\dot{x}_2 + k_2x_2[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle = b_2\dot{x}_1 + k_2x_1.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
We can simplify the notation, and clarify the structure of these equations, by using operator notation. Define polynomials </p><table id="a0000001156" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001157"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle P_1(s)[/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 m_1s^2 + b_1s + k_1[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000001158"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle P_2(s)[/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 m_2s^2 + b_2s + k_2[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000001159"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle Q_2(s)[/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 b_2s + k_2[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
Thus [mathjaxinline]P_1(s)[/mathjaxinline] is the characteristic polynomial of the first oscillator, [mathjaxinline]P_2(s)[/mathjaxinline] is the characteristic polynomial of the second oscillator, and [mathjaxinline]Q_2(s)[/mathjaxinline] is a first order polynomial reflecting the components connecting the two systems. Our system of equations becomes </p><table id="a0000001160" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001161"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle (P_1(D)+Q_2(D))x_1[/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 f(t) + Q_2(D)x_2[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(8.1)</td></tr><tr id="a0000001162"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle P_2(D)x_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 Q_2(D)x_1[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(8.2)</td></tr></table>
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<h2 class="hd hd-2 unit-title">12.4. The system function.</h2>
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<p>
When we transform this system to the frequency domain we get </p><table id="a0000001163" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="eqn8p3"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle (P_1(s)+Q_2(s))X_1(s)[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle F(s) + Q_2(s)X_2(s)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(8.3)</td></tr><tr id="a0000001165"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle P_2(s)X_2(s)[/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 Q_2(s)X_1(s) \notag[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
(For simplicity, from now on we'll write [mathjaxinline]P_1[/mathjaxinline], [mathjaxinline]X_1[/mathjaxinline]. etc instead of [mathjaxinline]P_1(s)[/mathjaxinline], [mathjaxinline]X_1(s)[/mathjaxinline], etc.) </p><p>
This is a pair of linear equations relating [mathjaxinline]X_1[/mathjaxinline], [mathjaxinline]X_2[/mathjaxinline], and [mathjaxinline]F[/mathjaxinline]. We're interested in [mathjaxinline]X_1[/mathjaxinline], so let's begin by isolating [mathjaxinline]X_1[/mathjaxinline] on the left of the first equation: </p><table id="a0000001166" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]X_1=\frac{1}{P_1+Q_2}\left(F+Q_2X_2\right)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none;text-align:right">(8.4)</td></tr></table><p>
According to the second equation, </p><table id="a0000001167" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr id="eqn8p5"><td class="equation" style="width:80%; border:none">[mathjax] X_2 = \frac{Q_2}{P_2}X_1.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none;text-align:right">(8.5)</td></tr></table><p>
Thus [mathjaxinline]X_1[/mathjaxinline] arises from summing [mathjaxinline]F[/mathjaxinline] with a multiple of [mathjaxinline]X_2[/mathjaxinline], and [mathjaxinline]X_2[/mathjaxinline] is in turn a multiple of [mathjaxinline]X_1[/mathjaxinline]. We can represent this recursion directly with a closed loop block diagram: </p><center><img src="/assets/courseware/v1/ae9f8e19c5026930738bb8109e9d63f3/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_l8-4-block.png" width="430"/></center><p>
The bottom branch shows [mathjaxinline]\displaystyle X_2 = \frac{Q_2}{P_2}\cdot X_1[/mathjaxinline]. This is then scaled by [mathjaxinline]Q_2[/mathjaxinline] and fed back into the top branch, which shows </p><table id="a0000001168" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]X_1 = \frac{1}{P_1+Q_2}\cdot (F+Q_2X_2).[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
These are exactly the equations above. </p><p>
Altogether the bottom part of the loop multiplies [mathjaxinline]X_1[/mathjaxinline] by [mathjaxinline]\frac{Q_2^2}{P_2}[/mathjaxinline] and feeds this back into the top branch. Notice that this feedback is entirely determined by the parameters of the secondary oscillator. This acts like a "feedback control system", but strictly speaking it is not one. Roughly, such control systems are characterized by the following feedback loop, </p><center><img src="/assets/courseware/v1/31bc03d284e2df2e03e5a692fb817c50/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_l8-4-block2.png" width="360"/></center><p>
with the requirement that the “plant" parameters on the top branch are independent of the “control" parameters on the bottom branch. </p><p>
In the tuned mass damper, we are trying to moderate the oscillations of the primary mass by tuning the parameters in the secondary system. But the secondary system parameters in [mathjaxinline]Q_2[/mathjaxinline] enter into the system function along the top branch. However, in practice the secondary mass damping term [mathjaxinline]b_2[/mathjaxinline] is very small, and if we can make the spring constant [mathjaxinline]k_2[/mathjaxinline] small relative to the quantities occurring in the primary system, then the model is essentially an example of feedback control. </p><p>
Let's find the transfer function. We use the equation for [mathjaxinline]X_2[/mathjaxinline] to eliminate the [mathjaxinline]X_2[/mathjaxinline] from the equation for [mathjaxinline]X_1[/mathjaxinline]: </p><table id="a0000001169" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001170"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \displaystyle X_2[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{Q_2}{P_2}X_1[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000001171"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle X_1[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{1}{P_1+Q_2}\left(F+Q_2 \frac{Q_2}{P_2}X_1 \right)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000001172"><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{1}{P_1+Q_2}\left(F+\frac{Q_2^2}{P_2}X_1 \right)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000001173"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle X_1\left(1-\frac{Q^2_2}{(P_1+Q_2)P_2}\right)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{1}{P_1+Q_2}F[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000001174"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle X_1\left(\frac{(P_1+Q_2)P_2 - Q^2_2}{(P_1+Q_2)P_2}\right)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \frac{1}{P_1+Q_2}F[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000001175"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle X_1[/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{P_2}{P_1P_2+Q_2(P_2-Q_2)}F[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
Then using the fact that [mathjaxinline]P_2-Q_2=m_2s^2[/mathjaxinline], we find </p><table id="a0000001176" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]X_1 = \frac{P_2}{P_1P_2 + m_2s^2Q_2} F[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The transfer function is thus </p><table id="a0000001177" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]H_1(s)=\frac{P_2}{P_1P_2 + m_2s^2Q_2}.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p><b class="bf">Question:</b> Obtain this transfer function using Black's formula. </p><p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">Check your answer.<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><p>
We essentially re-derived Black's formula in finding the transfer function above. Still, it is worth seeing that Black's formula would have covered this situation without having to reiterate the logic. </p><p>
For the following closed loop system with positive feedback </p><center><img src="/assets/courseware/v1/1ddf80101b2f766173a6f0cce985631c/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_l8-4-block3.png" width="360"/></center><p>
Black's formula says the closed loop transfer function is [mathjaxinline]\displaystyle \frac{G(s)}{1-K(s)G(s)}.[/mathjaxinline] In our case, </p><table id="a0000001178" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]G(s) = \frac{1}{P_1+Q_2} \quad \text { and } \quad K(s) = \frac{Q_2^2}{P_2}.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Putting these into Black's formula gives the same formula for the transfer function that we found above. </p></div><p class="hideshowbottom" onclick="hideshow(this);" style="margin: 0px"><a href="javascript: {return false;}">Show</a></p></div></p><SCRIPT src="/assets/courseware/v1/631e447105fca1b243137b21b9ed6f90/asset-v1:OCW+18.031+2019_Spring+type@asset+block/latex2edx.js" type="text/javascript"/><LINK href="/assets/courseware/v1/daf81af0af57b85a105e0ed27b7873a0/asset-v1:OCW+18.031+2019_Spring+type@asset+block/latex2edx.css" rel="stylesheet" type="text/css"/>
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<h2 class="hd hd-2 unit-title">12.5. Activity: The secondary oscillator.</h2>
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Now suppose that we consider [mathjaxinline]x_2[/mathjaxinline] to be the system response of our Mascot system. Make a closed loop block diagram that produces that system. (There is more than one way to choose [mathjaxinline]A[/mathjaxinline] and [mathjaxinline]B[/mathjaxinline].) </p>
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Identify [mathjaxinline]A[/mathjaxinline] and [mathjaxinline]B[/mathjaxinline] in terms of [mathjaxinline]P_1[/mathjaxinline], [mathjaxinline]P_2[/mathjaxinline] and [mathjaxinline]Q_2[/mathjaxinline]. </p>
<p>
(Type <b class="bf">P_1</b> for [mathjaxinline]P_1(s)[/mathjaxinline], <b class="bf">P_2</b> for [mathjaxinline]P_2(s)[/mathjaxinline], and <b class="bf">Q_2</b> for [mathjaxinline]Q_2(s)[/mathjaxinline]; omit the (s)es.) </p>
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What is the transfer function for the feedback loop you described in the problem above? </p>
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(Type <b class="bf">P_1</b> for [mathjaxinline]P_1(s)[/mathjaxinline], <b class="bf">P_2</b> for [mathjaxinline]P_2(s)[/mathjaxinline], and <b class="bf">Q_2</b> for [mathjaxinline]Q_2(s)[/mathjaxinline]; i.e. omit the (s).) </p>
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<p style="display:inline">[mathjaxinline]H_2=[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">12.6. Measured frequency response of the full mascot system.</h2>
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<h3 class="hd hd-2">Review the mascot setup.</h3>
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<p>
First let's recall the simple system with no secondary mass attached to it. </p><center><img src="/assets/courseware/v1/3dc5ebce2d522c4141a86490b28ce9fa/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_mascot_diagram_2ndorder.png" width="400"/><br/>A diagram of a spring mass dashpot driven through the mass<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. The LVDT (linear variable displacement transformer) is the mechanism that measures the position of the [mathjaxinline]m_1[/mathjaxinline]. This is a second order system modeled by the equation </p><table id="a0000001186" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]m_1\ddot x + b_1 \dot x + k_1x = f(t).[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Typical of a physical apparatus, our system does not exactly match the ideal system just described. The main difference is that the input force [mathjaxinline]f(t)[/mathjaxinline] is actually the response of the voice coil to an input voltage. For now we will ignore this issue. In a later section we will examine it more closely. </p><p>
We videotaped this system responding to a sequence of different input signals [mathjaxinline]f(t) = \sin (\omega t)[/mathjaxinline] as [mathjaxinline]\omega[/mathjaxinline] increased from [mathjaxinline]2\pi[/mathjaxinline] to [mathjaxinline]200\pi[/mathjaxinline] radians per second, i.e. from 1 to 100 hertz. </p>
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<h3 class="hd hd-2">Bode plot of system with no secondary mass</h3>
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<p><b class="bfseries">Comment on the Bode plot:</b></p><p>
Here is a reproduction of the Bode plot coming from the same 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>
We attempted to normalize the record of the magnitude of the low frequency system response to 1, but we missed, as you can see, and set it at about 0.82 instead. </p><p>
We recreated the Bode plot for the same 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/v1/4f4a06d2101771eafce4322b70b3cf15/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_TunedMassDamper_Plots-1.png" width="600"/></center><div><br/></div><p>
This is a classic second order Bode plot up to about 15 hertz. As mentioned above, the physical apparatus is slightly more complicated than the ideal system. We will look at this in more detail in section 8.8. </p><p>
Now let's look at the behavior of the Mascot when we add a secondary mass, to produce a fourth order system. </p>
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<h3 class="hd hd-2">Bode plot of the Mascot system with both masses</h3>
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<p>
As you see, this is more complicated! Here's a cleaned up and re-normalized version of the Bode plot, again created by sweeping out more input frequencies. </p><center><img src="/assets/courseware/v1/6f8fdd1ee8f127bf18ecaec0cb2c36b0/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_TunedMassDamper_Plots-2a.png" width="600"/></center><div><br/></div><p>
The data has a noticeable dip around 2.75 Hz. You can see that the cart underneath the mascot starts vibrating, quite dramatically at around minute mark 0:51 in the video above. </p><p>
We now know that to see the transfer function of a system, such as the cart, it is enough to know the unit impulse response. So here you see we've hit the cart with an impulse, and you can see the response curve on the computer behind the cart. </p><p>
(The video below has no audio, except for the sound of the impulse.) </p>
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<h3 class="hd hd-2">Cart dynamics</h3>
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<p>
We recorded the position of the cart. Here's a graph of the impulse response: </p><center><img src="/assets/courseware/v1/99cb9ace99ef41c23ff654c1bbc23726/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_cartDynamics.png" width="600"/></center><div><br/></div><p>
This is a complex mix of transients! The cart itself is represented by a high order LTI differential operator. The medium term behavior looks rather simple, though: It looks like a single damped sinusoid. </p><p><b class="bfseries"><span style="color:#FF7800">Note:</span></b> Due to the calibration of the measuring devices the 0 point is reported as slightly less than -0.3. </p>
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Please select from the following pole diagrams the one most likely to represent the cart. </p>
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(Remember, the imaginary part of a pole represents an angular pseudo-frequency, measured in radians per second, not a frequency, measured in Hertz.) </p>
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<h2 class="hd hd-2 unit-title">12.7. Interpretation.</h2>
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<p>
We will try to understand various behaviors we saw in the demonstration. Recall the system </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"/></center><p>
and its system function </p><table id="a0000001187" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]H_1(s)=\frac{P_2}{P_1P_2 + m_2s^2Q_2}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
with </p><table id="a0000001188" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001189"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle P_1(s)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle m_1s^2 + b_1s + k_1[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000001190"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle P_2(s)[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle m_2s^2 + b_2s + k_2[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr><tr id="a0000001191"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle Q_2(s)[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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[mathjaxinline]\displaystyle b_2s + k_2[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
This formula contains several interesting pieces of information. </p><ol class="enumerate"><li value="1"><p>
When the fourth order system was driven at a certain frequency, the mass seemed to just stop. The force continued to be applied to it, and the secondary mass was vibrating rapidly, but the shaft itself was completely stationary. What was happening? The <b class="bf">gain</b> at that frequency – say [mathjaxinline]\omega[/mathjaxinline] – was apparently zero. This only happens when [mathjaxinline]H_1(i\omega )=0[/mathjaxinline]. </p><p>
From the formula for [mathjaxinline]H_1[/mathjaxinline], you see that the zeros of the transfer function for this closed loop system occur precisely at the poles of the transfer function for the secondary oscillator – namely, at the roots of the characteristic polynomial [mathjaxinline]P_2(s)=m_2s^2+b_2s+k_2[/mathjaxinline]. </p><p>
Suppose that the secondary oscillator is lightly damped; in fact, for simplicity, suppose that it is <b class="bf">undamped</b>: [mathjaxinline]b_2=0[/mathjaxinline]. Then these roots are [mathjaxinline]\pm i\omega _2[/mathjaxinline], where [mathjaxinline]\omega _2=\sqrt {k_2/m_2}[/mathjaxinline] is the natural frequency of the secondary oscillator. That is, </p><table id="a0000001192" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\omega =\omega _2\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
In the real situation of a tuned mass damper, the main natural frequency of a building might be, say, [mathjaxinline]\omega _ n[/mathjaxinline]. The trick is then to adjust the parameters [mathjaxinline]m_2[/mathjaxinline] and [mathjaxinline]k_2[/mathjaxinline] of the secondary oscillator so that [mathjaxinline]\omega _2=\omega _ n[/mathjaxinline], because then the mass in the primary oscillator will be <b class="bf">stationary</b>. This is ideal! </p><p>
In the real situation of a tuned mass damper, the practical resonant frequency of a building might be, say, [mathjaxinline]\omega _ r[/mathjaxinline]. This is the input frequency that would cause the largest oscillation. The trick to damping out these oscillations is then to adjust the parameters [mathjaxinline]m_2[/mathjaxinline] and [mathjaxinline]k_2[/mathjaxinline] of the secondary oscillator so that [mathjaxinline]\omega _2 = \omega _ r[/mathjaxinline], i.e. we tune the system so the zero of [mathjaxinline]P_2[/mathjaxinline] is at the resonant frequency for [mathjaxinline]P_1[/mathjaxinline]. </p><p>
A peculiar feature of this result is that it appears that it can be made to function no matter what the secondary mass is. </p><p><b class="bf">Question</b>: What's the downside to using very small mass for the secondary oscillator? Why do actual tuned mass dampers weigh hundreds of tons? </p><p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">Answer<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><p>
We'll continue with the approximation that [mathjaxinline]b_2 = 0[/mathjaxinline], and think about the angular frequency of interest, [mathjaxinline]\omega _ r = \sqrt {k_2/m_2}[/mathjaxinline]. Then [mathjaxinline]P_2(i\omega _ r) = 0[/mathjaxinline], so </p><table id="a0000001193" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]H_2(s) = \frac{Q_2}{P_1P_2 + m_2s^2Q_2}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
evaluates at [mathjaxinline]s=i\omega _ r[/mathjaxinline] to </p><table id="a0000001194" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]H_2(i\omega _ r) = \frac{Q_2(i\omega _ r)}{m_2(i\omega _ r)^2Q_2(i\omega _ r)} = -\frac{1}{m_2\omega _ r^2}.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The fact that this is real and negative indicates that at this resonant point the secondary mass is oscillating in “anti-phase" with a phase lag of 180 degrees relative to the input signal. This makes sense! </p><p>
The expression for [mathjaxinline]H_2(i\omega _ r)[/mathjaxinline] also shows that if [mathjaxinline]m_2[/mathjaxinline] is small then the gain [mathjaxinline]|H_2(i\omega _ r)| = 1/(m_2\omega _ r^2)[/mathjaxinline] is large. So large that the amplitude of oscillation of the secondary mass will send it right through the wall of the building! </p></div><p class="hideshowbottom" onclick="hideshow(this);" style="margin: 0px"><a href="javascript: {return false;}">Show</a></p></div></p></li><li value="2"><p>
The transients of this system are of the form [mathjaxinline]e^{rt}[/mathjaxinline] where [mathjaxinline]r[/mathjaxinline] is a pole of [mathjaxinline]H_1(s)[/mathjaxinline] (assuming all these poles are simple). For almost all choices of system parameters, the roots of the numerator will differ from the roots of the denominator. (Actually, you can check that they definitely differ as long as [mathjaxinline]m_2\neq 0[/mathjaxinline] and [mathjaxinline]k_2\neq 0[/mathjaxinline].) In that case, the poles of [mathjaxinline]H_1(s)[/mathjaxinline] are precisely the roots of the denominator, [mathjaxinline]P_1(s)P_2(s)+m_2s^2Q_2(s)[/mathjaxinline]. This is a fourth degree polynomial (as long as [mathjaxinline]m_1\neq 0[/mathjaxinline] as well!) so there are four independent transients. This makes sense, since there are four degrees of freedom in choosing initial conditions: you can set the position and the velocity of each of the two masses. This is a <b class="bf">fourth order</b> system. </p></li><li value="3"><p>
It is always interesting to check out what happens for large frequencies. When [mathjaxinline]|s|[/mathjaxinline] is very large </p><table id="a0000001195" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]H_1(s) \approx \frac{m_2s^2}{m_1s^2m_2s^2} = \frac{1}{m_1}s^{-2}.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Three things to note: </p><ul class="itemize"><li><p>
The gain at high frequency [mathjaxinline]\omega[/mathjaxinline] is proportional to [mathjaxinline]\omega ^{-2}[/mathjaxinline]. This reflects the fact that the degree of the denominator in [mathjaxinline]H_1(s)[/mathjaxinline] is the degree of the numerator plus 2. </p></li><li><p>
The phase lag is about [mathjaxinline]180^\circ[/mathjaxinline], because with [mathjaxinline]s=i\omega[/mathjaxinline], [mathjaxinline]s^{-2}[/mathjaxinline] is real and negative. The mass is moving against the impressed force. </p></li><li><p>
At high frequencies the system is approximately the same as that of the primary mass without the secondary oscillator attached. </p></li></ul></li></ol><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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<h2 class="hd hd-2 unit-title">12.8. Conclusions of mascot analysis.</h2>
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<p>
To conclude this analysis of the Mascot, let us now be a little more careful about the model. We've already made the point that the system response is actually the voltage step across the LVDT sensor, rather than the position of the shaft. If we are more realistic about the input signal, it is also a voltage, the voltage applied to the voice coil actuator. The voice coil actually has its own dynamics, and provides most of the damping in the primary spring system. </p><p>
When we account for the input signal as a voltage, we end up with a system function with a degree 5 polynomial in the denominator, and with a degree 2 polynomial in the numerator, as before. </p><p>
As a result, when [mathjaxinline]\omega[/mathjaxinline] is large, the gain becomes proportional to [mathjaxinline]\omega ^{-3}[/mathjaxinline]. This power law is visible in right side of the log-log Bode plot where the graph straightens out into a straight line for large [mathjaxinline]\omega[/mathjaxinline]. </p><center><img src="/assets/courseware/v1/6f8fdd1ee8f127bf18ecaec0cb2c36b0/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_TunedMassDamper_Plots-2a.png" width="600"/></center><div><br/></div><p>
In the plots below, the green circles indicate recorded data from our frequency sweep with the mascot. The red line labeled Model (Mech+Elec) shows the gain from a model that models the voice coil actuator as an order 1 subsystem used to transform an input voltage to a mechanical force. The black line labeled Model (Mech) shows the gain from a model that considers the mechanical force as the input to the system. </p><center><img src="/assets/courseware/v1/d594c13c7e94406471de99bd7e225747/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_TunedMassDamper_Plots-3.png" width="600"/><br/>Measured frequency: response of the Mascot system </center><div><br/></div>
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In a real world tuned mass damper in a tall building, there is very little damping. However, in our mascot, there is a large amount of damping coming from the voice coil. If you are curious about where the damping is coming from, check out the demonstration in this video. </p>
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<h3 class="hd hd-2">Damping effect of voice coil</h3>
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