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<h2 class="hd hd-2 unit-title">2.1. Pole diagrams.</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>
determine the <b class="bfseries"><span style="color:#0000FF">pole/zero diagram</span></b> of the system function of an LTI system, and use it to sketch both the modulus of the system function and the <b class="bfseries"><span style="color:#0000FF">amplitude response curve</span></b> . </p></li><li value="2"><p>
identify the resonant frequencies, stability, and gain of an LTI system from the pole diagram of its transfer function. </p></li><li value="3"><p>
use the location of <b class="bfseries"><span style="color:#0000FF">poles</span></b> and <b class="bfseries"><span style="color:#0000FF">zeros</span></b> to determine the amplitude response. </p></li></ol><h3 class="hd hd-3 problem-header">Higher Goal.</h3> Confront higher order systems without fear, and appreciate the potential complexity of their frequency response.
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<h2 class="hd hd-2 unit-title">2.2. Poles of the transfer function.</h2>
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<p>The transfer function formalism makes our notation neater, and broadens its scope: [mathjaxinline]s[/mathjaxinline] rather than [mathjaxinline]i\omega[/mathjaxinline], and it gives information about more general system responses. But this generalization of [mathjaxinline]G(\omega )[/mathjaxinline] is extremely useful in understanding the complex gain itself, as well.</p>
<p>The transfer function is a fairly complicated gadget. You can't really graph it! It takes a complex number as input, so its graph lies over the complex plane; and it produces a complex number as output, which needs another two dimensions to represent. So we can't graph it in three dimensions.</p>
<p>But the most useful part of the complex gain is its magnitude – the gain. So instead of graphing [mathjaxinline]H(s)[/mathjaxinline], we will graph [mathjaxinline]|H(s)|[/mathjaxinline]. This is now a surface lying over the complex plane. Let's do some examples.</p>
<p>For a start, suppose our system is an undamped spring system, driven through the spring. This system is modeled by the equation</p>
<table id="a0000000208" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
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<td class="equation" style="width: 80%; border: none;">[mathjax]m\ddot x+kx=ky[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>This is the harmonic oscillator with natural frequency</p>
<table id="a0000000209" 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]\omega _ n=\sqrt {k/m}[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>It's expressed in standard form [mathjaxinline]P(D)x=Q(D)y[/mathjaxinline], with</p>
<table id="a0000000210" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
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<td class="equation" style="width: 80%; border: none;">[mathjax]P(s)=ms^2+k,\quad \quad Q(s)=k.[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
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<p>So, its transfer function is</p>
<table id="a0000000211" 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]H(s)=\frac{k}{ms^2+k}=\frac{k/m}{s^2+k/m}= \frac{\omega _ n^2}{s^2+\omega _ n^2}\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
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<p>Let's think first of the graph of</p>
<table id="a0000000212" 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]\displaystyle g(\omega )=|G(\omega )|=|H(i\omega )|=\frac{\omega _ n^2}{|(i\omega )^2+\omega _ n^2|} =\frac{\omega _ n^2}{|\omega _ n^2-\omega ^2|}\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
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<p>When [mathjaxinline]\omega =\pm \omega _ n[/mathjaxinline], the denominator vanishes and [mathjaxinline]G(\omega )[/mathjaxinline] isn't defined. This is resonance! When [mathjaxinline]\omega[/mathjaxinline] is near to [mathjaxinline]\pm \omega _ n[/mathjaxinline] but not quite equal to it, the value of the gain is large; the sinusoidal system response has amplitude much larger than the amplitude of the input signal. As [mathjaxinline]|\omega |[/mathjaxinline] gets large, the gain falls off to zero. The graph looks like this:</p>
<center><img src="/assets/courseware/v1/47a4ef21b2cceae6007dc09d9476a908/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_figc2_hm_poles.svg" width="450px" style="margin: 0px 10px 10px 10px;" /><br />The resonance occurs at input frequency [mathjaxinline]\omega _ n[/mathjaxinline].</center>
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<p>This is the amplitude response curve of the system. Note that the frequency enters here only through its square. We lose nothing, and gain quite a bit, by formally allowing [mathjaxinline]\omega[/mathjaxinline] to take on negative values as well as positive ones.</p>
<p>Now let's think about the absolute value of the full transfer function,</p>
<table id="a0000000213" 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]|H(s)|=\frac{\omega _ n^2}{|s^2+\omega _ n^2|}[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
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<p>The polynomial [mathjaxinline]s^2+\omega _ n^2[/mathjaxinline] has two roots, [mathjaxinline]s=\pm i \omega _ n[/mathjaxinline]. At these values of [mathjaxinline]s[/mathjaxinline], the transfer function is not defined. When [mathjaxinline]s[/mathjaxinline] is near to [mathjaxinline]\pm i \omega _ n[/mathjaxinline], the value of [mathjaxinline]|H(s)|[/mathjaxinline] is large; when [mathjaxinline]|s|[/mathjaxinline] gets big, [mathjaxinline]|H(s)|[/mathjaxinline] falls off to small values. The graph looks like this:</p>
<center><img src="/assets/courseware/v1/26798b05637d9747645c12abb4786440/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c2_3d_poles_1.png" width="400" /><br />Graph of [mathjaxinline]\left|H(s)\right|[/mathjaxinline] with purely imaginary poles<br /><span size="2" style="font-size: small;"> The blue curve lies over the imaginary axis. The blue curve on the surface is precisely the amplitude response curve.<br /> We have also outlined a box in the plane above the imaginary axis for emphasis. The green curve lies over the real axis.<br /> It represents the gain of the system if it is fed a <b> real</b> exponential input.</span></center>
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<p>If you slice the graph of [mathjaxinline]H(s)[/mathjaxinline] with a plane above the imaginary axis, you get the graph of the gain, which is the amplitude response curve – the graph of [mathjaxinline]g(\omega )=|H(i\omega )|[/mathjaxinline].</p>
<p>I think of this graph as like a tent. The canvas is draped over two tall poles, one at [mathjaxinline]i\omega _ n[/mathjaxinline], the other at [mathjaxinline]-i\omega _ n[/mathjaxinline]. Do you agree?</p>
<p>The points [mathjaxinline]\pm i\omega _ n[/mathjaxinline] are called <b class="bf">poles</b>, perhaps because they lie at the base of the tent-poles. The most important features of the graph of [mathjaxinline]|H(s)|[/mathjaxinline] – and so of the gain curve – are captured by the location of the poles. A diagram of the complex plane with the location of the poles marked by <b class="bfseries"><span style="color: #ff7800;">X</span></b> 's is called a <b class="bfseries"><span style="color: #0000ff;">pole diagram</span></b> of [mathjaxinline]H(s)[/mathjaxinline].</p>
<center><img src="/assets/courseware/v1/ee52d432caf1eda50df6e1f495bc47c5/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_figc2_firstpolediagram.svg" width="450px" style="margin: 0px 10px 10px 10px;" /></center>
<p>In our cases the poles of [mathjaxinline]H(s)[/mathjaxinline] are exactly the roots of the denominator, [mathjaxinline]P(s)[/mathjaxinline] (assuming [mathjaxinline]P(s)[/mathjaxinline] and [mathjaxinline]Q(s)[/mathjaxinline] have no common factors).</p>
<p>Now, what happens if we introduce some damping? Let's set [mathjaxinline]m=1[/mathjaxinline] to save on symbols. So</p>
<table id="a0000000214" 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]\ddot x+b\dot x+kx=ky\, ,[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<table id="a0000000215" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
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<td class="equation" style="width: 80%; border: none;">[mathjax]P(s)=s^2+bs+k,\quad \quad Q(s)=k\, ,[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
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<p>and</p>
<table id="a0000000216" 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]H(s)=\frac{Q(s)}{P(s)} = \frac{k}{s^2+bs+k}\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>The roots of [mathjaxinline]P(s)[/mathjaxinline] can be found by the quadratic formula, or, if you prefer, by completing the square:</p>
<table id="a0000000217" 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]s^2+bs+k=(s+(b/2))^2+(k-(b^2/4))\, ,[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
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<p>so [mathjaxinline]s[/mathjaxinline] is a root if [mathjaxinline](s+(b/2))^2=-(k-(b^2/4))[/mathjaxinline]. With [mathjaxinline]b[/mathjaxinline] small (smaller than [mathjaxinline]2\sqrt k)[/mathjaxinline], the system is underdamped; the right hand side is negative; and the roots are</p>
<table id="a0000000218" 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]r=-(b/2)\pm i\omega _ d\, ,[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>where</p>
<table id="a0000000219" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]\omega _ d=\sqrt {k-(b^2/4)}[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>is the <b class="bf">damped angular frequency</b>. The roots have also acquired a real part that is negative. The damping has pushed them off the imaginary axis!</p>
<p>Because of damping there is no pure resonance. Engineers call the frequency that produces the largest gain (at least if [mathjaxinline]b[/mathjaxinline] is small relative to [mathjaxinline]k[/mathjaxinline]) “practical resonance". The amplitude response curve is:</p>
<center><img src="/assets/courseware/v1/f78920bb2be16eb062ce7791ba970e23/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_figc2_damped.svg" width="400px" style="margin: 0px 10px 10px 10px;" /></center>
<p>The transfer function is now</p>
<table id="a0000000220" 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]H(s)=\frac{k}{s^2+bs+k}\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>This function has poles at the roots of the denominator, that is, at [mathjaxinline]-(b/2)\pm i\omega _ d[/mathjaxinline]. The pole diagram is this:</p>
<center><img src="/assets/courseware/v1/872b9d54b7298e6ce54527171e8fc71d/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_figc2_secondpolediagram.svg" width="450px" style="margin: 0px 10px 10px 10px;" /></center>
<p>The graph of [mathjaxinline]|H(s)|[/mathjaxinline] exhibits two infinitely high peaks over these two points.</p>
<center><img src="/assets/courseware/v1/7430e7a5f2cff10b006ca37537d74202/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c2_3d_poles_2.png" width="400" /><br />Graph of [mathjaxinline]\left|H(s)\right|[/mathjaxinline], poles no longer along imaginary axis</center>
<div></div>
<p>These points are no longer on the imaginary axis, but, if [mathjaxinline]b[/mathjaxinline] is small relative to [mathjaxinline]k[/mathjaxinline], they are not far off. So if you walk this terrain on a path over the imaginary axis, starting from [mathjaxinline]i\omega[/mathjaxinline] for [mathjaxinline]\omega[/mathjaxinline] larger than the resonant frequency, you will climb up the shoulder of the mountain near [mathjaxinline]-(b/2)+i\omega _ d[/mathjaxinline]; then descend to a valley that reaches a relative minimum at [mathjaxinline]\omega =0[/mathjaxinline]; and then goes through a symmetrical elevation gain and loss for negative [mathjaxinline]\omega[/mathjaxinline].</p>
<p>This hiking imagery is very useful. It explains how poles influence the gain curve.</p>
<p>Play with the mathlet below which plots [mathjaxinline]|H(s)|[/mathjaxinline] to get a better understanding of what is happening. (By clicking on and dragging the large graph on the left, you can change the orientation of the graph of [mathjaxinline]|H(s)|[/mathjaxinline]. Click the top or side button to reset the orientation.)</p>
<p><iframe style="display: block; border-width: 0px; padding: 0px;" src="https://mathlets1803.surge.sh/ampRespPoleDiagram.html" width="1100 px" height="640 px"></iframe></p>
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<h2 class="hd hd-2 unit-title">2.3. Poles, amplitude response, connection to ERF.</h2>
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<h3 class="hd hd-3 problem-header">What we know</h3><p>
We've briefly introduced the idea of poles and the transfer function. We need to tie these ideas back to what we already know, and develop the pole diagram. </p><p>
If we consider an LTI system </p><table id="a0000000221" 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;text-align:right">(2.27)</td></tr></table><p>
where [mathjaxinline]y(t)[/mathjaxinline] is the input and [mathjaxinline]x(t)[/mathjaxinline] is the system response: </p><ul class="itemize"><li><p>
The <b class="bfseries"><span style="color:#0000FF">stability</span></b> of the system is determined by the roots of the characteristic polynomial [mathjaxinline]P(r)[/mathjaxinline]. </p></li><li><p>
The <b class="bfseries"><span style="color:#0000FF">amplitude response</span></b> of the system to a sinusoidal input of frequency [mathjaxinline]\omega[/mathjaxinline] is determined by the characteristic polynomial [mathjaxinline]P[/mathjaxinline] and the polynomial [mathjaxinline]Q[/mathjaxinline] via the equation </p><table id="a0000000222" 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|.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table></li><li><p>
The <b class="bfseries"><span style="color:#0000FF">transfer function</span></b> is [mathjaxinline]H(s) = Q(s)/P(s)[/mathjaxinline]. </p></li></ul><p>
We see that the roots of [mathjaxinline]P[/mathjaxinline] give us important information about the system. The roots of [mathjaxinline]P(s)[/mathjaxinline] are the poles of [mathjaxinline]H(s)[/mathjaxinline], so all of this information is present in the pole diagram. In summary, the roots of [mathjaxinline]P[/mathjaxinline] occur in the denominator of the transfer function. These are called <b class="bfseries"><span style="color:#FF7800">poles</span></b> of the transfer function. </p><p>
Here we will learn about <b class="bfseries"><span style="color:#0000FF">poles</span></b> and the <b class="bfseries"><span style="color:#0000FF">pole diagram</span></b> of an LTI system. The pole diagram ties together stability, amplitude response, and transfer function in one diagram in the complex [mathjaxinline]s[/mathjaxinline]-plane. </p><blockquote class="quote"> The <b class="bfseries"><span style="color:#0000FF">pole diagram</span></b> of the transfer function <ol class="enumerate"><li value="1"><p>
shows whether the system is stable; </p></li><li value="2"><p>
shows whether the unforced system is oscillatory; and </p></li><li value="3"><p>
gives a rough picture of the amplitude response and (practical) resonances of the system. </p></li></ol> </blockquote><p>
For these reasons the pole diagram is a standard tool used by engineers in understanding and designing systems. </p><p>
Since every LTI system has a transfer function, everything we learn here will apply to all such systems. </p>
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<h2 class="hd hd-2 unit-title">2.4. Defining poles.</h2>
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<h3 class="hd hd-3 problem-header">Rational Functions.</h3><p><p><b class="bfseries">Definition 4.1 </b> A <b class="bfseries"><span style="color:#FF7800">rational function</span></b> is a ratio of polynomials [mathjaxinline]Q(s)/P(s)[/mathjaxinline], where [mathjaxinline]P[/mathjaxinline] is not the zero function. </p></p><p><p><b class="bfseries">Example 4.2 </b> The following are all rational functions. </p><p>
[mathjaxinline]\displaystyle \frac{s^2 + 1}{s^3+3s+1}[/mathjaxinline], [mathjaxinline]\displaystyle \frac{1}{ms^2+bs+k}[/mathjaxinline], [mathjaxinline]\displaystyle \frac{s^5 + s^3 + s+1}{s^3}[/mathjaxinline]. </p></p><p>
If the numerator [mathjaxinline]Q(s)[/mathjaxinline] and the denominator [mathjaxinline]P(s)[/mathjaxinline] have no roots in common, then the rational function [mathjaxinline]Q(s)/P(s)[/mathjaxinline] is in <b class="bfseries"><span style="color:#0000FF">reduced form</span></b> . </p><p><p><b class="bfseries">Example 4.3 </b> The three functions in the example above are all in reduced form. </p></p><p><p><b class="bfseries">Example 4.4 </b> [mathjaxinline](s-2)/(s^2-4)[/mathjaxinline] is not in reduced form, because [mathjaxinline]s=2[/mathjaxinline] is a root of both numerator and denominator. We can rewrite this in reduced form as </p><table id="a0000000223" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{s-2}{s^2-4} = \frac{s-2}{(s-2)(s+2)} = \frac{1}{s+2}.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table></p><h3 class="hd hd-3 problem-header">Poles.</h3><p>
For a rational function in reduced form the <b class="bfseries"><span style="color:#0000FF">poles</span></b> are the values of [mathjaxinline]s[/mathjaxinline] where the denominator is equal to zero; or, in other words, the points where the rational function is not defined. We allow the poles to be <b class="bf">complex numbers</b> here. </p><p><p><b class="bfseries">Remark 4.5 </b> For the system [mathjaxinline]P(D)x = Q(D)y[/mathjaxinline] with transfer function [mathjaxinline]H(s) = Q(s)/P(s)[/mathjaxinline], the poles of the transfer function are among the roots of [mathjaxinline]P(s)[/mathjaxinline]. That is, they are among the characteristic roots used to solve the homogeneous equation [mathjaxinline]P(D)x = 0[/mathjaxinline]. </p></p>
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4.1 Find the poles.
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For each rational function below, find the poles. </p>
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(Type <b class="bf">i</b> for the complex number [mathjaxinline]i[/mathjaxinline]. Enter answers as a list of poles separated by commas, e.g. 2+3*i, 2-3*i, 3. If there are no poles, type None.) </p>
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<p style="display:inline">(a) &#8195;[mathjaxinline]\displaystyle \frac{1}{s^2+8s+7}[/mathjaxinline] &#8195;</p>
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<p style="display:inline">(b) &#8195;[mathjaxinline]\displaystyle \frac{s-2}{s^2-4}[/mathjaxinline] &#8195;</p>
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<p style="display:inline">(c) &#8195;[mathjaxinline]\displaystyle \frac{1}{s^2+4}[/mathjaxinline] &#8195;</p>
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<p style="display:inline">(d) &#8195;[mathjaxinline]s^2+1[/mathjaxinline]&#8195;</p>
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<p style="display:inline">(e) &#8195;[mathjaxinline]\displaystyle \frac{1}{(s^2+8s+7)(s^2+4)}[/mathjaxinline] &#8195;</p>
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<p style="display:inline">(f) &#8195;[mathjaxinline]\displaystyle \frac{3s}{(s^2+6s+10)(s-2)}[/mathjaxinline]&#8195;</p>
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4.2(a) Find the poles of W
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4.2(b) Sketch for real s
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Recall [mathjaxinline]W(s) = 1/(s-2) + 1/(s+3)[/mathjaxinline]. </p>
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Sketch the graph of [mathjaxinline]|W(s)|[/mathjaxinline] for <b class="bfseries"><span style="color:#FF7800">real</span></b> [mathjaxinline]s[/mathjaxinline]. </p>
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(Indicate the location of any vertical asymptotes using the "Vertical asymptote" tool. Sketch the function paying attention to regions where the function is increasing and decreasing, and where the function is [mathjaxinline]0[/mathjaxinline].) </p>
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<h3 class="hd hd-3 problem-header">4.2(c) Graphing [mathjaxinline]|W(s)|[/mathjaxinline]</h3><p>
Try sketching the graph of [mathjaxinline]|W(s)|[/mathjaxinline] for <b class="bfseries"><span style="color:#FF7800">complex</span></b> [mathjaxinline]s[/mathjaxinline] on a piece of paper. </p>
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<h2 class="hd hd-2 unit-title">2.5. Pole diagrams.</h2>
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We now have pole diagrams for functions. This allows us to define the pole diagram for an LTI system using the system function. </p><p><p><b class="bfseries">Definition 5.1 </b> The <b class="bfseries"><span style="color:#0000FF">pole diagram of a function</span></b> [mathjaxinline]F(s)[/mathjaxinline] is simply the complex [mathjaxinline]s[/mathjaxinline]-plane with an <b class="bfseries"><span style="color:#FF7800">X</span></b> marking the location of each pole of [mathjaxinline]F(s)[/mathjaxinline]. </p></p><p><p><b class="bfseries">Example 5.2 </b> The function [mathjaxinline]\displaystyle F(s) = \frac{1}{s+2}[/mathjaxinline] has a pole at [mathjaxinline]s=-2[/mathjaxinline]. The pole diagram is a copy of the complex plane with an <b class="bfseries"><span style="color:#FF7800">X</span></b> at the location [mathjaxinline]s=-2[/mathjaxinline]. </p><center><img src="/assets/courseware/v1/92c8ad49b1437bd4454d70af0d791e94/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c2-polediag-1.svg" width="400px" style="margin: 0px 10px 10px 10px"/><br/>Pole diagram for [mathjaxinline]\displaystyle F(s) = \frac{1}{s+2}[/mathjaxinline] </center></p><p><p><b class="bfseries">Example 5.3 </b> The function [mathjaxinline]\displaystyle F(s)[/mathjaxinline] = [mathjaxinline]\frac{1}{s^2+4}[/mathjaxinline] has poles at [mathjaxinline]s=\pm 2i[/mathjaxinline]. The pole diagram is a copy of the complex plane with an <b class="bfseries"><span style="color:#FF7800">X</span></b> at the location [mathjaxinline]s=-2i[/mathjaxinline] and [mathjaxinline]s=2i[/mathjaxinline]. </p><center><img src="/assets/courseware/v1/9b6c4999d94f044202253882eac4f6d4/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c2-polediag-3.svg" width="400px" style="margin: 0px 10px 10px 10px"/><br/>Pole diagram for [mathjaxinline]\displaystyle F(s) = \frac{1}{s^2+4}[/mathjaxinline] </center></p><p><p><b class="bfseries">Example 5.4 </b> Draw the pole diagram for [mathjaxinline]\displaystyle F(s) = \frac{s}{s^2+6s+10}[/mathjaxinline]. </p><p>
Note that the poles of [mathjaxinline]F[/mathjaxinline] are where the denominator is zero. We complete the square to find the roots: </p><table id="a0000000224" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000225"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle 0[/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 s^2+6s+10[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.28)</td></tr><tr id="a0000000226"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle (s+3)^2 - 9 + 10[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.29)</td></tr><tr id="a0000000227"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
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[mathjaxinline]\displaystyle =[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle (s+3)^2 +1[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.30)</td></tr><tr id="a0000000228"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
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[mathjaxinline]\displaystyle \Longrightarrow s=-3\pm i[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(2.31)</td></tr></table><p>
Thus the pole diagram is as follows. </p><center><img src="/assets/courseware/v1/bbb53f7b6a840aea1e1ffd8773eb953b/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c2-polediag-4.svg" width="400px" style="margin: 0px 10px 10px 10px"/><br/>Pole diagram for [mathjaxinline]\displaystyle F(s) = \frac{s}{s^2+6s+10}[/mathjaxinline] </center></p>
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Draw pole diagrams (a)
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Draw the pole diagrams for the following function. </p>
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[mathjaxinline]\displaystyle F(s) = \frac{1}{s-2}[/mathjaxinline] </p>
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Draw pole diagrams (b)
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Draw the pole diagrams for the following function. </p>
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[mathjaxinline]\displaystyle G(s) = \frac{1}{((s+3)^2+1)(s+2)(s+4)}[/mathjaxinline] </p>
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Draw pole diagrams (c)
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Draw the pole diagrams for the following function. </p>
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[mathjaxinline]\displaystyle H(s) = \frac{3s}{(s^2+6s+10)(s-2)}[/mathjaxinline] </p>
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<h2 class="hd hd-2 unit-title">2.6. Pole diagram for an LTI system.</h2>
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<p><h3>The pole diagram for an LTI system</h3></p><p><p><b class="bfseries">Definition 6.1 </b> The <b class="bfseries"><span style="color:#0000FF">pole diagram for an LTI system</span></b> is defined to be the pole diagram of its transfer function.</p></p><p>
Try solving the problem(s) yourself before looking at the worked solution. </p><p><p><b class="bfseries">Example 6.2 </b> Give the pole diagram for the system </p><table id="a0000000230" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\ddot x + 8\dot x + 7x = f(t) ,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
where we take [mathjaxinline]f(t)[/mathjaxinline] to be the input and [mathjaxinline]x(t)[/mathjaxinline] the output.</p></p><p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">Worked solution<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><p>
The transfer function for this system is [mathjaxinline]\displaystyle {W(s) = \frac{1}{s^2+8s+7} = \frac{1}{(s+1)(s+7)}}[/mathjaxinline]. Therefore, the poles are [mathjaxinline]s=-1, \, -7[/mathjaxinline] and the pole diagram is<br/></p><center><img src="/assets/courseware/v1/b17eab38d9ad318a759f7ff3bc97c5af/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c2-polediag-7.svg" width="300px" style="margin: 0px 10px 10px 10px"/><br/>Pole diagram for the system in [mathjaxinline]\ddot x + 8\dot x + 7x = f(t)[/mathjaxinline] </center></div><p class="hideshowbottom" onclick="hideshow(this);" style="margin: 0px"><a href="javascript: {return false;}">Show</a></p></div></p><p><p><b class="bfseries">Example 6.3 </b> Give the pole diagram for the system </p><table id="a0000000231" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\ddot x + 4\dot x + 6x = \dot y ,[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
where we consider [mathjaxinline]y(t)[/mathjaxinline] to be the input and [mathjaxinline]x(t)[/mathjaxinline] to be the output. </p></p><p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">Worked solution<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><p>
The system is [mathjaxinline](D^2+4D+6)x = Dy[/mathjaxinline], so the transfer function is [mathjaxinline]W(s) = Q(s)/P(s) = s/(s^2+4s+6)[/mathjaxinline]. This has poles at the roots of the denominator, i.e. [mathjaxinline]s=-2 \pm \sqrt 2\, i[/mathjaxinline]. </p><center><img src="/assets/courseware/v1/bece5a9bf6ea7c6d57ed23ac71b052ed/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c2-polediag-8.svg" width="400px" style="margin: 0px 10px 10px 10px"/><br/>Pole diagram for the system in [mathjaxinline]\ddot x + 4\dot x + 6x = \dot y[/mathjaxinline] </center></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">2.7. Poles and stability.</h2>
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<p>
The pole diagram gives us information about the <b class="bfseries"><span style="color:#0000FF">stability</span></b> of a system. </p><p>
Recall that the LTI system </p><table id="a0000000232" 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;text-align:right">(2.32)</td></tr></table><p>
has an associated homogeneous equation </p><table id="a0000000233" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(D)x \, = \, 0[/mathjax]</td><td class="eqnnum" style="width:20%; border:none;text-align:right">(2.33)</td></tr></table><p>
The system is stable exactly when all the roots of [mathjaxinline]P[/mathjaxinline] have negative real part. Stability can be stated equivalently in terms of poles. </p><p>
Suppose that the LTI system is modeled by the differential equation [mathjaxinline]P(D)x=Q(D)y[/mathjaxinline], where [mathjaxinline]y[/mathjaxinline] is the input signal and [mathjaxinline]x[/mathjaxinline] the system response. The system function is then </p><table id="a0000000234" 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><p>
Then the poles of [mathjaxinline]H(s)[/mathjaxinline] are among the zeros of [mathjaxinline]P(s)[/mathjaxinline], and are exactly the roots of [mathjaxinline]P(s)[/mathjaxinline] as long as [mathjaxinline]Q(s)[/mathjaxinline] doesn't share any roots with [mathjaxinline]P(s)[/mathjaxinline]. We will make the general assumption that [mathjaxinline]P(s)[/mathjaxinline] and [mathjaxinline]Q(s)[/mathjaxinline] have no roots in common. (The cases where this is not true can be handled, but the complications distract from the main goal.) </p><ul class="itemize"><li><p>
The system is stable if all its poles have <b class="bfseries"><span style="color:#0000FF">negative real part</span></b> . </p></li><li><p>
Equivalently, the system is stable if all its poles lie strictly in the left half of the complex plane, i.e. [mathjaxinline]\mathrm{Re}(s) < 0[/mathjaxinline]. </p></li></ul><p>
This gives us a visual criterion for identifying whether or not a system is stable from the pole diagram, as illustrated in the following problem. </p>
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Identify stable LTI systems
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Each of the following six graphs is the pole diagram of an LTI system. Identify which systems are stable. (Choose all that are stable. Note that the two columns are graded independently.) </p>
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<h2 class="hd hd-2 unit-title">2.8. Poles and amplitude response.</h2>
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In this section we will explore how the pole diagram gives us a sense of the amplitude response, that is of the gain [mathjaxinline]g(\omega )[/mathjaxinline]. We will see that the pole diagram gives us a useful graphical tool for spotting resonant or near-resonant frequencies of LTI systems. </p><p>
We will go into more detail below. In brief we will make the following argument. </p><ol class="enumerate"><li value="1"><p>
We have seen that the graph of [mathjaxinline]|H(s)|[/mathjaxinline] is large near the poles of [mathjaxinline]H[/mathjaxinline]. </p></li><li value="2"><p>
Thus, the pole diagram provides a crude graph of [mathjaxinline]|H(s)|[/mathjaxinline]: roughly speaking, [mathjaxinline]|H(s)|[/mathjaxinline] will be large for values of [mathjaxinline]s[/mathjaxinline] near the poles. </p></li><li value="3"><p>
The gain of a system with transfer function [mathjaxinline]H(s)[/mathjaxinline] is given by [mathjaxinline]g(\omega ) = |H(i\omega )|[/mathjaxinline]. </p></li><li value="4"><p>
Thus, by restricting our attention to the imaginary axis [mathjaxinline]s = i\omega[/mathjaxinline], the pole diagram also gives us a crude sense of the gain [mathjaxinline]g(\omega )[/mathjaxinline]. </p></li></ol><p>
We will illustrate the above through a series of problems that get you to think about what the pole diagram means geometrically for the graph of [mathjaxinline]H(s)[/mathjaxinline]. </p>
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Where is amplitude largest?
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The figure below shows the pole diagram of a function [mathjaxinline]H(s)[/mathjaxinline]. At which of the points [mathjaxinline]A[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]C[/mathjaxinline] on the diagram would you guess [mathjaxinline]|H(s)|[/mathjaxinline] is largest? </p>
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Largest amplitude on imaginary axis
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The pole diagram of a function [mathjaxinline]H(s)[/mathjaxinline] is shown in the figure below. At what point&#8201; [mathjaxinline]s[/mathjaxinline] on the positive imaginary axis would you guess that [mathjaxinline]|H(s)|[/mathjaxinline] is largest? </p>
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<h2 class="hd hd-2 unit-title">2.9. Amplitude response and the transfer function.</h2>
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<p>
Recall that for the system [mathjaxinline]P(D) x \, = \, Q(D)y[/mathjaxinline], the transfer function is [mathjaxinline]H(s) = \frac{Q(s)}{P(s)}.[/mathjaxinline] The gain is given by </p><table id="a0000000235" 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;text-align:right">(2.34)</td></tr></table><p>
which is the ratio of amplitudes: response over input. </p><p>
Note: Rephrasing this in graphical terms: we can graph the magnitude of the system function [mathjaxinline]|H(s)|[/mathjaxinline] as a surface over the [mathjaxinline]s[/mathjaxinline]-plane. The amplitude response of the system [mathjaxinline]g(\omega ) = |H(i\omega )|[/mathjaxinline] is given by the part of the system function graph that lies above the imaginary axis. </p>
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Amplitude response and the pole diagram
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In the mathlet below, you see a large pole diagram on the left and a smaller Bode gain plot on the right. (There is another picture of the same pole diagram below the Bode plot.) This is the pole diagram of the transfer function for the spring/mass/dashpot system driven through the spring (as in the mathlet <a href="https://mathlets.surge.sh/ampPhaseSecondOrderI" target="_blank">Amplitude and Phase: Second Order I</a>). </p>
<p>
Take out a piece of paper, and make a sketch of the graph of the absolute value [mathjaxinline]|H(s)|[/mathjaxinline] of a transfer function that has this pole diagram. </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/ampRespPoleDiagram.html" width="1100 px" height="640 px"/>
<p>
The pole diagram of the left is actually a view from above of the graph of [mathjaxinline]|H(s)|[/mathjaxinline]. If you grab the diagram in the window, you can rotate it to see the 3D surface of the magnitude of the transfer function. How was your sketch? </p>
<p>
Note that you can get back to the pole diagram view by clicking the &#8220;Top" button, and see the amplitude response curve more clearly by clicking the &#8220;Side" button. </p>
<p>
Use the mathlet above to answer the following questions. </p>
<ol class="enumerate">
<li value="1">
<p>
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<input type="radio" name="input_lec2-tab9-problem1_2_1" id="input_lec2-tab9-problem1_2_1_choice_1" class="field-input input-radio" value="choice_1"/><label id="lec2-tab9-problem1_2_1-choice_1-label" for="input_lec2-tab9-problem1_2_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_lec2-tab9-problem1_2_1"> <text> The poles lie on the imaginary axis.</text>
</label>
</div>
<div class="field">
<input type="radio" name="input_lec2-tab9-problem1_2_1" id="input_lec2-tab9-problem1_2_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="lec2-tab9-problem1_2_1-choice_2-label" for="input_lec2-tab9-problem1_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_lec2-tab9-problem1_2_1"> <text> The poles are close to the imaginary axis.</text>
</label>
</div>
<div class="field">
<input type="radio" name="input_lec2-tab9-problem1_2_1" id="input_lec2-tab9-problem1_2_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="lec2-tab9-problem1_2_1-choice_3-label" for="input_lec2-tab9-problem1_2_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_lec2-tab9-problem1_2_1"> <text> The poles lie on the real axis.</text>
</label>
</div>
<div class="field">
<input type="radio" name="input_lec2-tab9-problem1_2_1" id="input_lec2-tab9-problem1_2_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="lec2-tab9-problem1_2_1-choice_4-label" for="input_lec2-tab9-problem1_2_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_lec2-tab9-problem1_2_1"> <text> The poles are close to the real axis.</text>
</label>
</div>
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<input type="radio" name="input_lec2-tab9-problem1_2_1" id="input_lec2-tab9-problem1_2_1_choice_5" class="field-input input-radio" value="choice_5"/><label id="lec2-tab9-problem1_2_1-choice_5-label" for="input_lec2-tab9-problem1_2_1_choice_5" class="response-label field-label label-inline" aria-describedby="status_lec2-tab9-problem1_2_1"> <text> The poles could be anywhere.</text>
</label>
</div>
<span id="answer_lec2-tab9-problem1_2_1"/>
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<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
</span>
</div>
</div></div> </p>
</li>
<li value="2">
<p>
Fix [mathjaxinline]k=0.5[/mathjaxinline], and move the damping constant [mathjaxinline]b[/mathjaxinline] from [mathjaxinline]0[/mathjaxinline] to its maximum value. What happens to the poles during that process? <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div class="choicegroup capa_inputtype" id="inputtype_lec2-tab9-problem1_3_1">
<fieldset aria-describedby="status_lec2-tab9-problem1_3_1">
<div class="field">
<input type="radio" name="input_lec2-tab9-problem1_3_1" id="input_lec2-tab9-problem1_3_1_choice_1" class="field-input input-radio" value="choice_1"/><label id="lec2-tab9-problem1_3_1-choice_1-label" for="input_lec2-tab9-problem1_3_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_lec2-tab9-problem1_3_1"> <text> The poles slide along a vertical line.</text>
</label>
</div>
<div class="field">
<input type="radio" name="input_lec2-tab9-problem1_3_1" id="input_lec2-tab9-problem1_3_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="lec2-tab9-problem1_3_1-choice_2-label" for="input_lec2-tab9-problem1_3_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_lec2-tab9-problem1_3_1"> <text> The poles traverse curves converging to the real axis, then move apart along the real axis.</text>
</label>
</div>
<div class="field">
<input type="radio" name="input_lec2-tab9-problem1_3_1" id="input_lec2-tab9-problem1_3_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="lec2-tab9-problem1_3_1-choice_3-label" for="input_lec2-tab9-problem1_3_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_lec2-tab9-problem1_3_1"> <text> The poles traverse curves converging to the imaginary axis, then move apart along the imaginary axis.</text>
</label>
</div>
<div class="field">
<input type="radio" name="input_lec2-tab9-problem1_3_1" id="input_lec2-tab9-problem1_3_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="lec2-tab9-problem1_3_1-choice_4-label" for="input_lec2-tab9-problem1_3_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_lec2-tab9-problem1_3_1"> <text> The poles don't move at all.</text>
</label>
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<li value="3">
<p>
(Keep [mathjaxinline]k=0.5[/mathjaxinline] fixed.) Now execute the same variation of the damping term (start at [mathjaxinline]b=0[/mathjaxinline] and move [mathjaxinline]b[/mathjaxinline] to its maximum value), but observe what happens to the amplitude of the system response, as represented by the graph at upper right. Which best describes the behavior? (Choose all that apply.) </p>
<p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 3" role="group"><div class="choicegroup capa_inputtype" id="inputtype_lec2-tab9-problem1_4_1">
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<text>As the damping constant increases, the maximum gain decreases (or stays the same).</text>
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<text>There is always a positive frequency at which the gain is larger than anywhere else.</text>
</label>
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<text>For any positive input frequency, when you increase the damping constant you decrease the gain.</text>
</label>
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<text>For large damping term, largest gain occurs at [mathjaxinline]\omega = 0[/mathjaxinline].</text>
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<li value="4">
<p>
(Keep [mathjaxinline]k=0.5[/mathjaxinline] fixed.) Finally, we'll compare observations of the gain curve with observations about the position of the poles. Still with reference to the behavior when [mathjaxinline]b[/mathjaxinline] changes: It appears that as you increase the damping coefficient [mathjaxinline]b[/mathjaxinline], two things happen at about the same time: </p>
<ul class="itemize">
<li>
<p>
the resonant peak moves to [mathjaxinline]\omega = 0[/mathjaxinline] ; and </p>
</li>
<li>
<p>
the poles collide on the real axis. </p>
</li>
</ul>
<p>
It's hard to see from the Mathlet which comes first; maybe they occur for the same value of [mathjaxinline]b[/mathjaxinline]. Make a computation to decide. </p>
<p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 4" role="group"><div class="choicegroup capa_inputtype" id="inputtype_lec2-tab9-problem1_5_1">
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<text> The resonant peak moves to [mathjaxinline]\omega =0[/mathjaxinline] first.</text>
</label>
</div>
<div class="field">
<input type="radio" name="input_lec2-tab9-problem1_5_1" id="input_lec2-tab9-problem1_5_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="lec2-tab9-problem1_5_1-choice_2-label" for="input_lec2-tab9-problem1_5_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_lec2-tab9-problem1_5_1">
<text> The poles collide on the real axis first.</text>
</label>
</div>
<div class="field">
<input type="radio" name="input_lec2-tab9-problem1_5_1" id="input_lec2-tab9-problem1_5_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="lec2-tab9-problem1_5_1-choice_3-label" for="input_lec2-tab9-problem1_5_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_lec2-tab9-problem1_5_1">
<text> The poles collide to real axis exactly when the resonant peak moves to [mathjaxinline]\omega =0[/mathjaxinline].</text>
</label>
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<h3 class="hd hd-3 problem-header" id="lec2-tab9-problem2-problem-title" aria-describedby="block-v1:OCW+18.031+2019_Spring+type@problem+block@lec2-tab9-problem2-problem-progress" tabindex="-1">
Stability
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<p>
Consider the pole diagram of the transfer function for the spring/mass/dashpot system driven through the spring (as in the mathlet above). </p>
<p>
For [mathjaxinline]b&gt;0[/mathjaxinline] and [mathjaxinline]k&gt;0[/mathjaxinline], there is some choice of [mathjaxinline]b[/mathjaxinline] and [mathjaxinline]k[/mathjaxinline] such that the poles of the transfer function lie in the right half plane. </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_lec2-tab9-problem2_2_1">
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<p>
Consider the following pole diagrams for linear time invariant systems of the form [mathjaxinline]P(D)y = f[/mathjaxinline].<br/></p><table class="tabular" cellspacing="0" style="table-layout:auto"><tr><td style="text-align:left; border:none">
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[mathjaxinline]\rightarrow[/mathjaxinline] List all the stable systems. <br/>[mathjaxinline]\rightarrow[/mathjaxinline] Choose the stable system where the transient has the fastest decay.<br/>[mathjaxinline]\rightarrow[/mathjaxinline] Choose the stable system where the transient decays as fast as possible without oscillation.<br/></p>
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<h3 class="hd hd-2">Problem 1 solution, problem 2 set up</h3>
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<p>
Consider the following pole diagrams for linear time invariant systems of the form [mathjaxinline]P(D)y = f[/mathjaxinline]. </p><center><img src="/assets/courseware/v1/66d98497cf82f36a48c4cfb23808e33a/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c2-recitationG.svg" width="450px" style="margin: 0px 10px 10px 10px"/></center><p>
Suppose the input is [mathjaxinline]f(t) = F_0\cos (\omega t)[/mathjaxinline] with [mathjaxinline]\omega[/mathjaxinline] an integer between 1 and 5. </p><p>
Which value of [mathjaxinline]\omega[/mathjaxinline] causes the biggest response from the system? </p>
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<h3 class="hd hd-2">Problem 2 solution</h3>
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<h2 class="hd hd-2 unit-title">2.11. Zeros of the transfer function.</h2>
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The “zeros" of a rational function are the values of [mathjaxinline]s[/mathjaxinline] for which [mathjaxinline]H(s)=0[/mathjaxinline]. If [mathjaxinline]H(s) = Q(s)/P(s)[/mathjaxinline] is in reduced form, the zeros are exactly the roots of [mathjaxinline]Q(s)[/mathjaxinline]. If an input frequency is near a zero then we would expect the gain to be small. Thus the <b class="bfseries"><span style="color:#0000FF">zeros of the system</span></b> also give us important information about the system. It is common practice to mark the zeros on the pole diagram with a circle. </p><p><p><b class="bfseries">Example 11.1 </b> The pole-zero diagram below shows zeros at [mathjaxinline]-0.3 \pm 2i[/mathjaxinline], [mathjaxinline]-0.3\pm 3i[/mathjaxinline] and poles at [mathjaxinline]-0.6\pm 1.8i[/mathjaxinline], [mathjaxinline]-0.6\pm 3.2i[/mathjaxinline]. One system function with these zeros and poles is </p><table id="a0000000243" 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{((s+.3)^2+4)((s+.3)^2+9)}{((s+.6)^2+1.8^2)((s+.6)^2+3.2^2)}.[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><center><img src="/assets/courseware/v1/285259690a446e8210c4ab0a9eda0370/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c2-poles-max-3.svg" width="400px" style="margin: 0px 10px 10px 10px"/></center><p>
The Bode gain plot for this system is shown below. </p><center><img src="/assets/courseware/v1/ff28bd5fa38b88a29d79e032e3fa4d2e/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_rfig-polediag-2.svg" width="250px" style="margin: 0px 10px 10px 10px"/><br/><font size="2">Note that when [mathjaxinline]\omega [/mathjaxinline] ( not [mathjaxinline]i\omega [/mathjaxinline]) is near 2 or 3, the gain is small.</font><div><br/></div></center><p>
Note that when [mathjaxinline]i\omega[/mathjaxinline] is near the zeros the gain is small. The poles were chosen to raise the gain as soon [mathjaxinline]\omega[/mathjaxinline] is outside the range [mathjaxinline][2, 3][/mathjaxinline]. Because the number of zeros equals the number of poles the gain goes asymptotically to a non-zero value (in this case 1) as [mathjaxinline]\omega[/mathjaxinline] gets large. The net result is a <b class="bfseries"><span style="color:#0000FF">band-stop</span></b> filter: any input with frequency in the frequency band [2,3] is suppressed because the gain is near 0. Other frequencies are passed through because the gain is near 1. </p></p><p><p><b class="bfseries">Remark 11.2 </b></p><ol class="enumerate"><li value="1"><p>
The location of zeros has nothing to do with stability. </p></li><li value="2"><p>
A <b class="bf">rational</b> transfer function is determined by its pole-zero diagram up to a constant. (For this we are assuming that there are no repeated poles or zeros.) </p></li></ol></p>
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Consider the spring/mass/dashpot system with [mathjaxinline]m=1[/mathjaxinline], [mathjaxinline]b=2[/mathjaxinline], [mathjaxinline]k=2[/mathjaxinline], </p>
<table id="a0000000244" 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]\ddot x + 2\dot x + 2x = 2 \dot y, \qquad y = \cos \omega t[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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Find the pole-zero diagram and use it to get an idea of the sketch of [mathjaxinline]g(\omega )[/mathjaxinline]. </p>
<p>
Which best illustrates [mathjaxinline]g(\omega )[/mathjaxinline]? </p>
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Using the pole-zero diagram, what can you infer about the value of the resonant frequency [mathjaxinline]\omega _ r[/mathjaxinline]? </p>
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<text> [mathjaxinline]\omega _ r &lt; 1[/mathjaxinline]</text>
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<text> [mathjaxinline]\omega _ r =1[/mathjaxinline]</text>
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<text> [mathjaxinline]\omega _ r &gt;1[/mathjaxinline]</text>
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A question of resonance
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Consider the system [mathjaxinline]m\ddot x + b\dot x + kx = b\dot y[/mathjaxinline], where [mathjaxinline]y[/mathjaxinline] is the input and [mathjaxinline]x[/mathjaxinline] is the output. True or false: for any (positive) values of [mathjaxinline]m[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], and [mathjaxinline]k[/mathjaxinline] the system has a positive resonant frequency [mathjaxinline]\omega _ r[/mathjaxinline]. </p>
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(Hint: Use the Nyquist plot.) </p>
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<text> False</text>
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