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<h2 class="hd hd-2 unit-title">9.1. Impulses and impulse response.</h2>
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<h3 class="hd hd-3 problem-header">Objectives</h3><p>
After completing this lecture the student will be able to </p><ol class="enumerate"><li value="1"><p>
incorporate the <b class="bfseries"><span style="color:#0000FF">delta function</span></b> into a description of signals. </p></li><li value="2"><p>
graph <b class="bfseries"><span style="color:#0000FF">generalized functions.</span></b> </p></li><li value="3"><p>
describe and compute the <b class="bfseries"><span style="color:#0000FF">unit impulse response</span></b> of an LTI system using Laplace transform. </p></li><li value="4"><p>
distinguish between <b class="bfseries"><span style="color:#0000FF">pre-</span></b> and <b class="bfseries"><span style="color:#0000FF">post-initial conditions,</span></b> and identify post-initial conditions of the unit impulse response. </p></li></ol>
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<h2 class="hd hd-2 unit-title">9.2. The derivative of the unit step function.</h2>
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<p>
Consider the unit step function </p><table id="a0000000953" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]u(t)=\begin{cases} 0& \quad t<0\\ 1& \quad t>0\end{cases}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
In calculus you would say that the unit step function [mathjaxinline]u(t)[/mathjaxinline] is non-differentiable at [mathjaxinline]t=0[/mathjaxinline]; it's not even continuous there. But we need to talk about its derivative. Here are two examples of how this shows up. </p><ol class="enumerate"><li value="1"><p>
A bank account is modeled by the equation </p><table id="a0000000954" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\dot x - Ix = q(t)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
where [mathjaxinline]I[/mathjaxinline] is the interest rate, and [mathjaxinline]q(t)[/mathjaxinline] is the net rate of deposit. One day you deposit a large amount of money. This total deposit is best modeled by a step function. Thus the deposit rate [mathjaxinline]q(t)[/mathjaxinline] is best modeled by the derivative of this step function. </p></li><li value="2"><p>
Suppose we give a soccer ball a good swift kick. Its velocity increases abruptly from [mathjaxinline]v=0[/mathjaxinline] to a positive value [mathjaxinline]v_0[/mathjaxinline]. The velocity function can be modeled by [mathjaxinline]v(t)=v_0u(t)[/mathjaxinline]. What can we say about the force applied by my foot? </p><p>
Force is the time rate of change of momentum. Momentum is mass times velocity, so it's given by [mathjaxinline]mv(t)=mv_0u(t)[/mathjaxinline]. So the force is the derivative of [mathjaxinline]mv_0u(t)[/mathjaxinline]. </p></li></ol><p>
While the ideal step function has no derivative, our treatment of the step function lets us deal with its derivative. </p><p>
We saw that in reality the unit step function is an idealization of a smooth function that increases in value from 0 to 1 in a very short period of time. Viewed in that light, the derivative of [mathjaxinline]u(t)[/mathjaxinline] is easy to picture. Expanding the [mathjaxinline]t[/mathjaxinline] axis so we can see how [mathjaxinline]u(t)[/mathjaxinline] increases in value from 0 to 1, we get a graph like this: </p><center><img src="/assets/courseware/v1/20adbd95f3ec846aa126359805ab3ec7/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c5_heaviside_zoom.svg" width="200px" style="margin: 0px 10px 10px 10px"/></center><p>
The derivative [mathjaxinline]\dot u(t)[/mathjaxinline] will then have a graph that looks like this: </p><center><img src="/assets/courseware/v1/631ca1c0d4dd6134323f338fd20903fe/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c5_delta_approx.svg" width="200px" style="margin: 0px 10px 10px 10px"/></center><p>
Since [mathjaxinline]u(t)[/mathjaxinline] increases in value by 1, the area under the graph of [mathjaxinline]\dot u(t)[/mathjaxinline] is 1. It's a very narrow tall spike. The same kind of idealization that we used to replace the smooth function of [mathjaxinline]u(t)[/mathjaxinline] with the Heaviside step function leads to the <b class="bf"> Dirac delta function</b> </p><table id="a0000000955" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\delta (t)=\dot u(t)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The delta function is secretly just a smooth function which takes on nonzero values only in a neighborhood of 0 that is too small for us to resolve, and has total integral 1. </p><p>
When we graph it, we will denote the delta function by a “harpoon." </p><center><img src="/assets/courseware/v1/c6c1459ef6ec8af84d11dae591b579e9/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c5_delta.svg" width="200px" style="margin: 0px 10px 10px 10px"/></center><p>
Since we can't picture the area under the harpoon, we'll indicate the area by a number next to it. We can also shift the delta function by [mathjaxinline]a[/mathjaxinline] units to the right and scale it by [mathjaxinline]c[/mathjaxinline], to get </p><table id="a0000000956" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{d}{dt}\, cu(t-a)=c\dot u(t-a)=c\delta (t-a)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Its graph looks like this: </p><center><img src="/assets/courseware/v1/a7fe9887602d41a58a1f30f5941c29a0/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c5_delta_shift.svg" width="200px" style="margin: 0px 10px 10px 10px"/></center>
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<p><b class="bfseries"><span style="color:#FF7800">Food for thought:</span></b> How would you graph [mathjaxinline]-2\delta (t-1)[/mathjaxinline]? </p><p><div class="hideshowbox"><h4 onclick="hideshow(this);" style="margin: 0px">Conventions<span class="icon-caret-down toggleimage"/></h4><div class="hideshowcontent"><p>
While you might decide to draw a upwards arrow at [mathjaxinline]t=1[/mathjaxinline] with a [mathjaxinline]-2[/mathjaxinline] to decorate it indicating a value of [mathjaxinline]-2[/mathjaxinline] for the integral, it is more common to draw a downwards pointing arrow, decorated with a positive [mathjaxinline]2[/mathjaxinline]. </p><center><img src="/assets/courseware/v1/fa94720ceef9a2b3de4f037b4be698e1/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c5_delta_convention.svg" width="200px" style="margin: 0px 10px 10px 10px"/></center><p>
However, note that in the image below, we do not require that the constant [mathjaxinline]c[/mathjaxinline] be positive. </p><center><img src="/assets/courseware/v1/a7fe9887602d41a58a1f30f5941c29a0/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c5_delta_shift.svg" width="200px" style="margin: 0px 10px 10px 10px"/></center><p>
Instead, we allow the possibility that [mathjaxinline]c[/mathjaxinline] be negative, but keep the convention of drawing the delta function pointing upwards. If we know that [mathjaxinline]c[/mathjaxinline] is negative, we instead draw the delta function pointing downwards as is the convention in this course. </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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Practice with delta functions
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Graph [mathjaxinline]\ f(t) = \delta (t) - 3\delta (t-1) + 2u(t-2) - u(t-3)[/mathjaxinline]. </p>
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(Use the line tool, arrow tools, and label tools as needed. Use the class convention that negative delta functions point downwards.) </p>
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( Label the delta functions with the appropriate number by choosing the 'Select' menu item, and double clicking the word 'Value' and typing the number.) </p>
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Derivative of the square wave
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The square wave is defined as </p>
<table id="a0000000957" 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]u(t)-u(t-1)+u(t-2)-u(t-3) + \dotsb[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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Find the derivative of the square wave. </p>
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<text> [mathjaxinline]\delta (t)+\delta (t-1)+\delta (t-2)+\delta (t-3)+\dotsb[/mathjaxinline]</text>
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<text> [mathjaxinline]\displaystyle \frac{\delta (t)}{2}-\frac{\delta (t-1)}{2}+\frac{\delta (t-2)}{2}-\frac{\delta (t-3)}{2}+\dotsb[/mathjaxinline]</text>
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<text> [mathjaxinline]\delta (t)-\delta (t-1)+\delta (t-2)-\delta (t-3)+\dotsb[/mathjaxinline]</text>
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<text> [mathjaxinline]\delta (t)-2\delta (t-1)+2\delta (t-2)-2\delta (t-3)+\dotsb[/mathjaxinline]</text>
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Graph the derivative of the square wave
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Graph the derivative of the square wave. </p>
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(Use the line tool and arrow tools to graph the derivative of the square wave in the window below. Use the class convention that negative delta functions point downwards.) </p>
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(Label the delta functions with the appropriate number by choosing the 'Select' menu item, and double clicking 'Value' and typing the number.) </p>
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<h2 class="hd hd-2 unit-title">9.3. The integral of the delta function.</h2>
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Now, how do we compute the Laplace transform of the delta function? </p><p>
The [mathjaxinline]\delta[/mathjaxinline] function has a special relationship with integration. Fix a number [mathjaxinline]a[/mathjaxinline], and let [mathjaxinline]f(t)[/mathjaxinline] be any function that is continuous at [mathjaxinline]a[/mathjaxinline]. Then, for any numbers [mathjaxinline]b[/mathjaxinline] and [mathjaxinline]c[/mathjaxinline] such that [mathjaxinline]b<a<c[/mathjaxinline], </p><table id="a0000000959" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\int _ b^ c\delta (t-a)f(t)\, dt=f(a)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
To see this just remember that [mathjaxinline]\delta (t-a)[/mathjaxinline] is zero except very near to [mathjaxinline]a[/mathjaxinline], and near to [mathjaxinline]a[/mathjaxinline] it becomes very large with area 1. Its graph looks like a very tall rectangle. Since [mathjaxinline]f(t)[/mathjaxinline] is continuous at [mathjaxinline]a[/mathjaxinline], the values of [mathjaxinline]f(t)[/mathjaxinline] near [mathjaxinline]a[/mathjaxinline] are very near to [mathjaxinline]f(a)[/mathjaxinline]. So the graph of [mathjaxinline]\delta (t-a)f(t)[/mathjaxinline] is also a very narrow rectangle that is [mathjaxinline]f(a)[/mathjaxinline] times taller than the one for the delta function alone. Hence, the area under [mathjaxinline]\delta (t-a)f(t)[/mathjaxinline] is [mathjaxinline]f(a)[/mathjaxinline]. </p><p>
This formula implies other similar formulas. For example </p><table id="a0000000960" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\int _{-\infty }^\infty \delta (t-a)f(t)\, dt=f(a)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
and </p><table id="a0000000961" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\int _{a-}^{a+}\delta (t-a)f(t)\, dt=f(a)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
These formulas reflect the actual practice of measuring the value of a function. When you measure voltage, or current, or position, or temperature, you can't make an <b class="bf"> instantaneous</b> measurement. Your instrument accumulates the effect of the quantity you are measuring over some small time interval – milliseconds if your time scale is in seconds – and averages over that small interval. This averaging process, heavily weighting the values near [mathjaxinline]t=a[/mathjaxinline], is precisely the process of forming the integral </p><table id="a0000000962" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\int _ b^ c \delta (t-a)f(t)\, dt[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
as above! </p><p>
This expression for the integral of [mathjaxinline]\delta (t-a)f(t)[/mathjaxinline] is a characteristic property of the [mathjaxinline]\delta[/mathjaxinline] function, and is the basis for more a sophisticated development of the theory of “distributions", which covers the delta function and much more. </p>
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Compute the integral
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<p style="display:inline">[mathjaxinline]\displaystyle \int _{1}^{\infty } \delta (t)f(t)\, dt =[/mathjaxinline]</p>
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Practice with integrals
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Compute the following integrals. </p>
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<p style="display:inline">[mathjaxinline]\displaystyle \int _{-\infty }^{\infty } \delta (t)\cos (\omega t)\, dt =[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\displaystyle \int _{-\infty }^{\infty } \delta (t)\tan (t+\pi /4)\left(\frac{e^{-3t}}{1+t^2}\right)\, dt =[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\displaystyle \int _{-\infty }^{\infty } \delta (t-\pi )e^{t^2-5}\cos (5t-1)\, dt =[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\displaystyle \int _{-1}^{1} (\delta (t)+\delta (t+1/2)+\delta (t-1/2))(e^{t}t)\, dt =[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\displaystyle \int _{0^{-}}^{5} \delta (t)\arctan (t+1)\, dt =[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\displaystyle \int _{3^{-}}^{3^{+}} t^3 \Big( \delta (t)+\delta (t-1)\cos (\pi t) +\delta (t-2)\cos (2\pi t)+\dotsb +\delta (t-k) \cos (k\pi t) + \dotsb \Big)\, dt =[/mathjaxinline]</p>
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Integration of delta functions.<br/>a) Compute i) [mathjaxinline]\displaystyle {\int _{0^-}^\infty \delta (t)e^{t^2}\, dt}[/mathjaxinline] <br/> (ii) [mathjaxinline]\displaystyle {\int _{0^-}^{\infty } \delta (t-2)e^{t^2\sin (t)\cos (2t)}\, dt}[/mathjaxinline] <br/> (iii) [mathjaxinline]\displaystyle {\int _{0^+}^{\infty } \delta (t) e^{t^2}\, dt}[/mathjaxinline]. </p></li><li value="2"><p>
. Generalized Derivatives.<br/>a) Find the generalized derivative of [mathjaxinline]f(t) = 3u(t) - 2u(t-1)[/mathjaxinline]. Graph [mathjaxinline]f(t)[/mathjaxinline] and [mathjaxinline]f'(t)[/mathjaxinline]. </p><p>
b) Find the generalized derivative of [mathjaxinline]f(t) = \begin{cases} t^2 & \text { for } t < 0\\ e^{-t} & \text { for } t > 0 \end{cases}[/mathjaxinline]. </p></li></ol>
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<h2 class="hd hd-2 unit-title">9.5. The Laplace transform of the delta function.</h2>
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<p>Using the <a href="/courses/course-v1:OCW+18.031+2019_Spring/courseware/week2/lec3/3" target="_blank">integral definition</a>,</p>
<table id="a0000000963" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]\delta (t-a)\rightsquigarrow \int _0^\infty \delta (t-a)e^{-st}\, dt=e^{-as}\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>This is good as long as [mathjaxinline]a>0[/mathjaxinline]. To find the Laplace transform of [mathjaxinline]\delta (t)[/mathjaxinline] itself, though, we can't make sense of the lower limit – it asks us to evaluate [mathjaxinline]\delta (t)[/mathjaxinline] at [mathjaxinline]t=0[/mathjaxinline].</p>
<p>To accommodate this we will make a slight adjustment in our integral definition of the Laplace transform. We want to include the whole of the delta function, so we want to start the integral at a point somewhat to the left of [mathjaxinline]t=0[/mathjaxinline]. This can be formalized in just the same way as we dealt with one-sided values of functions: take a limit from below: say</p>
<table id="a0000000964" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]\int _{0-}^\infty g(t)\, dt=\lim _{a\uparrow 0}\int _ a^\infty g(t)\, dt\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>So our <b class="bfseries"><span style="color: #0000ff;">final definition of the Laplace transform</span></b> is this:</p>
<table id="a0000000965" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]\mathcal{L}(f(t);s)=\int _{0-}^\infty f(t)e^{-st}\, dt\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>With this definition,</p>
<table id="a0000000966" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout: auto;">
<tbody>
<tr>
<td class="equation" style="width: 80%; border: none;">[mathjax]\delta (t)\rightsquigarrow \int _{0-}^\infty \delta (t)e^{-st}\, dt=\lim _{a\uparrow 0}e^{-s0}=1\, .[/mathjax]</td>
<td class="eqnnum" style="width: 20%; border: none;"> </td>
</tr>
</tbody>
</table>
<p>The Laplace transform of the Dirac delta function is the constant function in the frequency domain with value 1. Compare this with [mathjaxinline]\mathcal{L}(1;s)=1/s[/mathjaxinline]: the constant time function produces a pole in the [mathjaxinline]s[/mathjaxinline] domain.</p>
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<h2 class="hd hd-2 unit-title">9.6. Activity: discovering the delta function.</h2>
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Skyscraper function
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The delta function may be approximated by the skyscraper function </p>
<table id="a0000000967" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto">
<tr>
<td class="equation" style="width:80%; border:none">[mathjax]d_ h(t)=\begin{cases} 0&amp; \text {for}\quad t&lt;0\\ 1/h&amp; \text {for}\quad 0&lt;t&lt;h\\ 0&amp; \text {for}\quad t&gt;h\end{cases},[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
</tr>
</table>
<p>
where [mathjaxinline]h[/mathjaxinline] is the width of the skyscraper, and [mathjaxinline]1/h[/mathjaxinline] is the height. </p>
<p>
Sketch the skyscraper function for [mathjaxinline]h=1[/mathjaxinline], [mathjaxinline]h=1/2[/mathjaxinline], and [mathjaxinline]h=1/4[/mathjaxinline]. </p>
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(Verify for yourself that this function approaches the delta function in the limit as [mathjaxinline]h\rightarrow 0[/mathjaxinline].) </p>
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Write in terms of unit step functions
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Write the skyscraper function </p>
<table id="a0000000968" 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]d_ h(t)=\begin{cases} 0&amp; \text {for}\quad t&lt;0\\ 1/h&amp; \text {for}\quad 0&lt;t&lt;h\\ 0&amp; \text {for}\quad t&gt;h\end{cases},[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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in terms of unit step functions. </p>
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<text> [mathjaxinline]u(t)-u(t-h)[/mathjaxinline]</text>
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<text> [mathjaxinline]\displaystyle \frac{u(t)-u(t-h)}{h}[/mathjaxinline]</text>
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Find the Laplace transform
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Find the Laplace transform of the skyscraper function [mathjaxinline]\mathcal{L}(d_ h(t);s)[/mathjaxinline] and [mathjaxinline]\mathcal{L}(d_ h(t-a);s)[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\mathcal{L}(d_ h(t);s)=[/mathjaxinline]</p>
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\(\)
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<p style="display:inline">[mathjaxinline]\mathcal{L}(d_ h(t-a);s)=[/mathjaxinline]</p>
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A limit of Laplace transforms
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(Compare your result with [mathjaxinline]\mathcal{L}(\delta (t-a);s)[/mathjaxinline].) </p>
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<h2 class="hd hd-2 unit-title">9.7. Impulse response.</h2>
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Earlier we studied the response of an LTI system to a unit step input function, with rest initial conditions. Now we will study the response to a unit impulse, that is, to [mathjaxinline]\delta (t)[/mathjaxinline] as input signal. The example to keep in mind is a spring system, in which the mass is struck hard. The equation: </p><table id="a0000000978" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]m\ddot x+b\dot x+kx=p\delta (t)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
We expect its position to be a continuous function of time, but its momentum (and hence its velocity) undergoes a discontinuity. </p><p>
For definiteness, let's take [mathjaxinline]m=2, b=8, k=10[/mathjaxinline], and subject the mass to a unit impulse: </p><table id="a0000000979" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]2\ddot x+8\dot x+10x=\delta (t)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
In this section we'll find the system response, with rest initial conditions, using the Laplace transform. Later we will learn to find this response using the concept of pre- and post-initial conditions. </p><p><b class="bfseries"><span style="color:#FF7800">Solution:</span></b> Apply Laplace transform: </p><table id="a0000000980" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]2s^2X+8sX+10X=1[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
or </p><table id="a0000000981" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]X=\frac{1}{2s^2+8s+10}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Complete the square: </p><table id="a0000000982" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]X=\frac{1/2}{(s+2)^2+1}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
This is the Laplace transform of </p><table id="a0000000983" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\frac{1}{2}e^{-2t}\sin t\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The unit impulse response should vanish for [mathjaxinline]t<0[/mathjaxinline], so it is </p><table id="a0000000984" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]w(t)=\frac{1}{2}u(t)e^{-2t}\sin t\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Here's the graph. This is a typical unit impulse response of an underdamped second order system. </p><center><img src="/assets/courseware/v1/2dcd34992b3648713294713c323758ed/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_c5_impulseresponse_spring.svg" width="550px" style="margin: 0px 10px 10px 10px"/></center>
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<h3 class="hd hd-2">Stopping a pendulum with delta input</h3>
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<h2 class="hd hd-2 unit-title">9.9. Pre- and post-initial conditions.</h2>
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<p>
We've just looked at the impulse response of a spring system in which the mass is struck hard, modeled by the equation: </p><table id="a0000000985" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]m\ddot x+b\dot x+kx=p\delta (t)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
We expect its position to be a continuous function of time, but its momentum (and hence its velocity) undergoes a discontinuity. </p><p>
When the delta function is thought of as an input function it's often called a “unit impulse." The term comes from this physical example: in physics, a impulse is a time integral of force, or a change in momentum. </p><p>
One of the initial conditions for a second order equation like this is the velocity [mathjaxinline]\dot x(0)[/mathjaxinline]. But [mathjaxinline]\dot x[/mathjaxinline] has a discontinuity at [mathjaxinline]t = 0[/mathjaxinline]! Which value shall we take, the one relevant to [mathjaxinline]t<0[/mathjaxinline] (namely [mathjaxinline]\dot x(0)=0[/mathjaxinline]) or the one relevant to [mathjaxinline]t>0[/mathjaxinline] (namely [mathjaxinline]\dot x(0)=p/m[/mathjaxinline])? To clarify this ambiguity, we will talk about “pre-" and “post-initial conditions." </p><p>
When we model an LTI system by the differential equation </p><table id="a0000000986" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P(D)x = Q(D)y[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
a certain number of initial conditions are required to specify a system response [mathjaxinline]x(t)[/mathjaxinline]. If the operator [mathjaxinline]P(D)[/mathjaxinline] is of order exactly [mathjaxinline]n[/mathjaxinline], we will need the [mathjaxinline]n[/mathjaxinline] of them. More precisely, we will need the values </p><table id="a0000000987" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x(0^+),\dot x(0^+),..., x^{(n-1)}(0^+).[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
These are called the "post-initial conditions" of the solution. They may differ from the "pre-initial conditions," </p><table id="a0000000988" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x(0^-),\dot x(0^-),..., x^{(n-1)}(0^-).[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
"Rest initial conditions" really means rest <b class="bf">pre</b>-initial conditions; all these values should be zero. </p><p><p><b class="bfseries">Example 9.1 </b> The pre-initial conditions of [mathjaxinline]\cos (t)[/mathjaxinline] are given by [mathjaxinline]\cos (0^-)=1,\cos '(0^-)=0,\ldots[/mathjaxinline]. Since [mathjaxinline]\cos (t)[/mathjaxinline] is continuous, the post-initial conditions are the same. On the other hand, the pre-initial conditions of [mathjaxinline]u(t)\cos (t)[/mathjaxinline] are all zero, while the post-initial conditions are [mathjaxinline]\cos (0^+)=1,\cos '(0^+)=0,\cos ^{\prime \prime }(0^+)=-1\ldots[/mathjaxinline]. </p></p><p>
This plays into the [mathjaxinline]t[/mathjaxinline]-derivative rule for the Laplace transform. Since we have refined the definition by using [mathjaxinline]0^-[/mathjaxinline] as lower limit, the proper form of the [mathjaxinline]t[/mathjaxinline]-derivative rule is this: </p><p><b class="bf">The [mathjaxinline]t[/mathjaxinline]-derivative rule.</b></p><table id="a0000000989" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f'(t)\rightsquigarrow sF(s)-f(0^-)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Similarly, all the derivatives occurring in the rule for the Laplace transform for higher derivatives of [mathjaxinline]f(t)[/mathjaxinline] are to be evaluated at [mathjaxinline]0^-[/mathjaxinline]. </p><p><p><b class="bfseries">Example 9.2 </b> We check this refined rule on the two functions [mathjaxinline]f(t)=1[/mathjaxinline] and [mathjaxinline]f(t)=u(t)[/mathjaxinline]. These two functions coincide for [mathjaxinline]t>0[/mathjaxinline], and so have the same Laplace transforms, namely [mathjaxinline]1/s[/mathjaxinline]. But they have different pre-initial values, and different derivatives at [mathjaxinline]t=0[/mathjaxinline]. Let's check the rule: </p><p>
With [mathjaxinline]f(t)=1[/mathjaxinline], [mathjaxinline]f(0^-)=1[/mathjaxinline], and [mathjaxinline]f'(t)=0[/mathjaxinline] has Laplace transform [mathjaxinline]\mathcal{L}(f'(t))=0[/mathjaxinline]. Check: </p><table id="a0000000990" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]0=1'(t)\rightsquigarrow s\frac{1}{s}-1=0\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Good! </p><p>
With [mathjaxinline]f(t)=u(t)[/mathjaxinline], [mathjaxinline]f(0^-)=0[/mathjaxinline], and [mathjaxinline]f'(t)=\delta (t)[/mathjaxinline] has Laplace transform [mathjaxinline]\mathcal{L}(f'(t))=1[/mathjaxinline]. Check: </p><table id="a0000000991" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\delta (t)=u'(t)\rightsquigarrow s\frac{1}{s}-0=1\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Also good! </p></p>
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<p><p><b class="bfseries">Example 9.3 </b> Return to the example of a spring system being kicked: For definiteness, let's take [mathjaxinline]m=2, b=8, k=10[/mathjaxinline] as we did two pages ago, and subject it to a unit impulse: </p><table id="a0000000992" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]2\ddot x+8\dot x+10x=\delta (t)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Let's find the displacement of the mass, a second way using the fact that by rest initial conditions, we really mean rest pre-initial conditions. </p></p><p><b class="bfseries"><span style="color:#FF7800">Solution:</span></b> Since there is no force acting for [mathjaxinline]t>0[/mathjaxinline], the solution is a transient for the system. The completion of the square we did shows that the roots of the characteristic polynomial are [mathjaxinline]-2\pm i[/mathjaxinline], so the general transient is </p><table id="a0000000993" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]e^{-2t}(a\cos t+b\sin t)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
We need to figure out the post-initial conditions. The unit impulse increases the momentum by one unit, and hence increases the velocity by [mathjaxinline]1/2[/mathjaxinline]. So immediately after the kick, we have [mathjaxinline]x(0^+)=0[/mathjaxinline] and [mathjaxinline]\dot x(0^+)=1/2[/mathjaxinline]. This implies [mathjaxinline]a = 0[/mathjaxinline] and [mathjaxinline]b = 1/2[/mathjaxinline]; so we get the same solution. </p><table id="a0000000994" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x(t)=\frac{1}{2}e^{-2t}\sin t\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table>
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<h2 class="hd hd-2 unit-title">9.10. Practice with pre- and post- initial conditions</h2>
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<h3 class="hd hd-2">Recitation video: problem setup</h3>
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<h3 class="hd hd-3 problem-header">Problem statement from recitation</h3><p>
Take a moment to work through the problems from the recitation video on your own. Once you are satisfied with your answer, watch the next video to see a worked solution. </p><ol class="enumerate"><li value="1"><p>
Find the unit impulse response for the following systems a) [mathjaxinline]\dot x + 2x = f(t)[/mathjaxinline]. </p><p>
b) [mathjaxinline]2\ddot x + 7\dot x + 3x = f(t)[/mathjaxinline]. </p></li><li value="2"><p>
Find the unit step response to the system in problem (1a). </p></li></ol><p><b class="bfseries"><span style="color:#FF7800">Note on video below:</span></b> In the video below, Lydia finds new pre-initial conditions for homogeneous problems that are consistent with the problems with delta function input. These should be new <b class="bfseries"><span style="color:#0000FF">post-initial conditions</span></b> instead of pre-initial conditions. </p>
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First order system with pre and post initial conditions
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<p>
Let's model a savings account. On your birthday ([mathjaxinline]t=0[/mathjaxinline]), you open an account with $ 500. That day, you receive and immediately deposit $1000. The savings account accrues interest at a continuous rate of [mathjaxinline]I[/mathjaxinline]. The money in your savings account can be modeled by </p>
<table id="a0000000995" 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]\dot x - Ix =1000\dot u(t),[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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where [mathjaxinline]u[/mathjaxinline] is the unit step function. </p>
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In other words, </p>
<table id="a0000000996" 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]\dot x - Ix = 1000\delta (t).[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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<p>
Identify the pre- and post- conditions at [mathjaxinline]t=0[/mathjaxinline] of the differential equation </p>
<table id="a0000000997" 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]\dot x - Ix = 1000\delta (t) ,[/mathjax]</td>
<td class="eqnnum" style="width:20%; border:none">&#160;</td>
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<p>
assuming that the response [mathjaxinline]x[/mathjaxinline] is as smooth as possible. </p>
<p>
<p style="display:inline">[mathjaxinline]x(0^-)=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]x(0^+)=[/mathjaxinline]</p>
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Second order system with pre and post initial conditions
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A unit impulse is applied to a spring/mass/dashpot system modeled by the equation </p>
<table id="a0000000998" 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]2\ddot x + \dot x + kx = 2\delta (t) .[/mathjax]</td>
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<p>
The system is at rest before time [mathjaxinline]0[/mathjaxinline]. Find the pre- and post- initial conditions. </p>
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<p style="display:inline">[mathjaxinline]x(0^-)=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\dot x(0^-)=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]x(0^+)=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\dot x(0^+)=[/mathjaxinline]</p>
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