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<h2 class="hd hd-2 unit-title">11.1. Lecture 7: Convolution.</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>
Describe the response of an LTI system to any input signal, with rest initial conditions, as a <b class="bfseries"><span style="color:#0000FF">convolution integral.</span></b> </p></li><li value="2"><p>
Compute the convolution of two signals using the Laplace transform. </p></li><li value="3"><p>
Carry out formal calculations of convolutions, including the possibility of generalized functions as signals. </p></li><li value="4"><p>
Explain the significance of the terms in the convolution integral as they occur in expressing the solution to an LTI system. </p></li></ol>
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<h2 class="hd hd-2 unit-title">11.2. Why convolution?</h2>
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<p>
We have learned to determine the system response of LTI systems by using the Laplace transform to convert the problem to a simple algebraic relation in the frequency domain. This is effective, and this method reveals important features of the system response, especially about its long term behavior. </p><p>
This method depends on the observation that the system function [mathjaxinline]H(s)[/mathjaxinline] is the Laplace transform of the unit impulse response (or “weight function") [mathjaxinline]h(t)[/mathjaxinline]. This implies that the Laplace transform of the system response (with rest initial conditions) to an input signal [mathjaxinline]f(t)[/mathjaxinline] is simply the product </p><table id="a0000001066" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]X(s)=H(s)F(s)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
In this lecture we will analyze the relationship between the unit impulse response, the input signal, and the system response (with rest initial conditions) directly in the time domain. The result can be written as </p><table id="a0000001067" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x(t)=h(t)*f(t)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
where the asterisk denotes a new “product operation" on signals, known as <b class="bf">convolution</b>. </p><p>
So one way to define convolution is by means of the Laplace transform: </p><table id="a0000001068" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\mathcal{L}(h(t)*f(t);s)=H(s)F(s)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Here [mathjaxinline]h(t)[/mathjaxinline] and [mathjaxinline]f(t)[/mathjaxinline] can be pretty much any pair of signals. </p>
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Using this equation as the definition of convolution, find the convolution [mathjaxinline]1*1[/mathjaxinline]. Here [mathjaxinline]1[/mathjaxinline] denotes the constant function with value [mathjaxinline]1[/mathjaxinline]. </p>
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(Enter as a function of [mathjaxinline]t[/mathjaxinline]. We've multiplied by [mathjaxinline]u(t)[/mathjaxinline] for you.) </p>
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<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div id="formulaequationinput_lec11-tab2-problem1_2_1" class="inputtype formulaequationinput">
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<span class="trailing_text" id="trailing_text_lec11-tab2-problem1_2_1">[mathjaxinline]\cdot u(t)[/mathjaxinline]</span>
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<h2 class="hd hd-2 unit-title">11.3. Convolution.</h2>
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The integral definition of convolution is </p><table id="a0000001070" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f(t)*h(t)=\int _0^ t f(\tau )h(t-\tau )\, d\tau \, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
(Some motivation for this formula can be found on the last tab of this lecture.) </p>
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Use the integral definition of convolution to compute [mathjaxinline]t^3*t[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]t^3*t=[/mathjaxinline]</p>
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<span class="trailing_text" id="trailing_text_lec11-tab3-problem1_2_1">[mathjaxinline]\cdot u(t)[/mathjaxinline]</span>
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<font color="#0078b0">2/3</font>
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<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
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<th class="formulainput" scope="row" rowspan="4">Operators</th>
<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
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<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
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<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
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<tr class="formulainput">
<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
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<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
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<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
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<th class="formulainput" scope="row">Basic functions</th>
<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
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<th class="formulainput" scope="row" rowspan="3">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
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<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
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<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
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Practice problem 2
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Use the integral definition of convolution to compute [mathjaxinline]1*u(t-1)[/mathjaxinline]. </p>
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(Enter your answer as a function of [mathjaxinline]t[/mathjaxinline] times a unit step function. You only need to enter the argument of the unit step function in the box provided.) </p>
<p>
<p style="display:inline">[mathjaxinline]\displaystyle \left. \phantom{\int } 1*u(t-1)= \right[[/mathjaxinline]</p>
<div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="formulaequationinput_lec11-tab3-problem2_2_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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<span class="trailing_text" id="trailing_text_lec11-tab3-problem2_2_1">[mathjaxinline]\displaystyle \left] \phantom{\int } \right.[/mathjaxinline]</span>
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<p style="display:inline">[mathjaxinline]\left. \phantom{\int } \cdot u\right([/mathjaxinline]</p>
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<span class="trailing_text" id="trailing_text_lec11-tab3-problem2_3_1">[mathjaxinline]\displaystyle \left) \phantom{\int }\right.[/mathjaxinline]</span>
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<th class="formulainput" scope="row" rowspan="3">Numbers</th>
<td class="formulainput">Integers</td>
<td class="formulainput">
<font color="#0078b0">2520</font>
</td>
</tr>
<tr class="formulainput">
<td class="formulainput">Fractions</td>
<td class="formulainput">
<font color="#0078b0">2/3</font>
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<td class="formulainput">Decimals </td>
<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="4">Operators</th>
<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Basic functions</th>
<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="3">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
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<h2 class="hd hd-2 unit-title">11.4. Farm run-off and convolution.</h2>
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To repeat, convolution may also be defined as the following integral: </p><table id="a0000001081" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f(t)*h(t)=\int _0^ t f(\tau ) h(t-\tau ) \, d\tau \, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
We confess that this formula is at first rather hard to understand. The following story may help elucidate it. </p><p>
Phosphate fertilizer is applied seasonally to the ground in a certain farm. Some of the fertilizer runs off into a lake, where it undergoes natural decay (by chemical process or being carried away by an outflow from the lake). </p><p>
We can model this situation as follows: Write [mathjaxinline]f(t)[/mathjaxinline] for the rate at which phosphate is entering the lake at time [mathjaxinline]t[/mathjaxinline] (in kg/year). Write [mathjaxinline]x(t)[/mathjaxinline] for the phosphate load in the lake (in kg). Write [mathjaxinline]a[/mathjaxinline] for the rate of natural decay of the phosphate in the lake. We then have a simple linear model for this system: </p><table id="a0000001082" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\dot x+ax=f(t)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The input signal is [mathjaxinline]f(t)[/mathjaxinline]; the system response is [mathjaxinline]x(t)[/mathjaxinline]. Suppose that at [mathjaxinline]t=0[/mathjaxinline] there is no phosphate in the lake at all; this is rest initial conditions. </p><p>
Let's fix a moment in time, after the system has run for a while and try to evaluate [mathjaxinline]x[/mathjaxinline] at that instant. We will denote the fixed time we are looking at by [mathjaxinline]t[/mathjaxinline]. This is dangerous notation, because we really want [mathjaxinline]t[/mathjaxinline] to be "fixed" for the moment. We will determine the amount of phosphate, [mathjaxinline]x(t)[/mathjaxinline], in the lake at time [mathjaxinline]t[/mathjaxinline] using the principle of superposition. </p><p>
The phosphate added to the lake at each moment between time 0 and time [mathjaxinline]t[/mathjaxinline] makes a contribution to the value of [mathjaxinline]x(t)[/mathjaxinline]. We need another symbol for these intermediate times; let's use the Greek letter [mathjaxinline]\tau[/mathjaxinline]. </p><p>
Begin by thinking about the contribution to [mathjaxinline]x(t)[/mathjaxinline] made by the phosphate added to the lake right at the beginning – say between time [mathjaxinline]0[/mathjaxinline] and time [mathjaxinline]\Delta \tau[/mathjaxinline], where [mathjaxinline]\Delta \tau[/mathjaxinline] is very small. The amount of phosphate added in that time interval is close to [mathjaxinline]f(0)\Delta \tau[/mathjaxinline]: the rate of deposition at time 0, times the amount of time elapsed. But not all this phosphate is still present at the later time [mathjaxinline]t[/mathjaxinline]. It has decayed according to the natural decay process. So at time [mathjaxinline]t[/mathjaxinline] this first time interval contributes </p><table id="a0000001083" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f(0)\Delta \tau \cdot e^{-at}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
kilograms to the value of [mathjaxinline]x(t)[/mathjaxinline]. </p><p>
Now think about a later small time interval, still between 0 and [mathjaxinline]t[/mathjaxinline]; say between time [mathjaxinline]\tau[/mathjaxinline] and time [mathjaxinline]\tau +\Delta \tau[/mathjaxinline]. The value of the input signal is [mathjaxinline]f(\tau )[/mathjaxinline], so in this time interval some [mathjaxinline]f(\tau )\Delta \tau[/mathjaxinline] kg is added to the lake. But not all of this is still present at the later time [mathjaxinline]t[/mathjaxinline]. How much time has elapsed between time [mathjaxinline]\tau[/mathjaxinline] and the later time [mathjaxinline]t[/mathjaxinline]? Answer: [mathjaxinline]t-\tau[/mathjaxinline]. So the phosphate added around time [mathjaxinline]\tau[/mathjaxinline] has been reduced by the factor [mathjaxinline]e^{-a(t-\tau )}[/mathjaxinline]; the contribution to the value [mathjaxinline]x(t)[/mathjaxinline] is </p><table id="a0000001084" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f(\tau )\Delta \tau \cdot e^{-a(t-\tau )}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
kilograms. </p><p>
The value [mathjaxinline]x(t)[/mathjaxinline] is the sum of these small contributions. For [mathjaxinline]\Delta \tau[/mathjaxinline] sufficiently small, this sum becomes indistinguishable from an integral; the only notational change is to replace [mathjaxinline]\Delta \tau[/mathjaxinline] by [mathjaxinline]d\tau[/mathjaxinline] (and place it at the end of the product. The integral runs from [mathjaxinline]\tau =0[/mathjaxinline] to [mathjaxinline]\tau =t[/mathjaxinline]: </p><table id="a0000001085" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x(t)=\int _0^ t f(\tau )e^{-a(t-\tau )}\, d\tau \, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
This is the <b class="bf">convolution integral</b>. </p><p>
We can recognize the function [mathjaxinline]e^{-a\tau }[/mathjaxinline] as the unit impulse response for the differential operator in our system, [mathjaxinline]D+aI[/mathjaxinline]: </p><table id="a0000001086" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]h(\tau )=u(\tau )e^{-a\tau } .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
In the integral, [mathjaxinline]t>\tau[/mathjaxinline], so we never evaluate [mathjaxinline]e^{-a\tau }[/mathjaxinline] at negative values of [mathjaxinline]\tau[/mathjaxinline], and so the step function [mathjaxinline]u(\tau )[/mathjaxinline] has no impact. </p>
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<h2 class="hd hd-2 unit-title">11.5. Mathlet and general statement.</h2>
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Mathlet and general statement.
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<p>
We hope this story helps to justify and explain the terms in the convolution integral. To visualize this process more clearly, please use the mathlet <b class="bf">Convolution: Accumulation</b>. </p>
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<p>
Select as Signal [mathjaxinline]f(t)=1+\cos (bt)[/mathjaxinline] and as weight function [mathjaxinline]g(t)=e^{-at}[/mathjaxinline]. Recall that the weight function is the unit impulse response. </p>
<p>
(The mathlet writes [mathjaxinline]g(t)[/mathjaxinline] for what we are writing [mathjaxinline]h(t)[/mathjaxinline]. Also, it takes [mathjaxinline]a=\ln (2)[/mathjaxinline] and [mathjaxinline]b=2[/mathjaxinline].) </p>
<p>
This replicates the farm run-off scenario. </p>
<p>
The lower window shows a graph of the input signal; you can see the seasonal variation in the rate of run-off. </p>
<p>
Please click on the time slider at [mathjaxinline]t=3[/mathjaxinline]. Then position the cursor in either window near the vertical line [mathjaxinline]t=3[/mathjaxinline]. This is near to [mathjaxinline]t=\pi[/mathjaxinline], which is the period of the oscillation, so the value of the input signal [mathjaxinline]f(t)[/mathjaxinline] there is about 2. </p>
<p>
Now depress the mouse key. Blocks of color appear in both windows. </p>
<p><b class="bf">Look at the bottom window.</b> The colored block in the bottom window represents the area under the graph of what function? </p>
<p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_lec11-tab5-problem1_2_1">
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<text> The weight function: [mathjaxinline]g(t)[/mathjaxinline]</text>
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<text> The weight function delayed by [mathjaxinline]3[/mathjaxinline]: [mathjaxinline]g(t-3)[/mathjaxinline]</text>
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<text> The weight function [mathjaxinline]g(t)[/mathjaxinline] times the value of the signal at [mathjaxinline]t=3[/mathjaxinline]: [mathjaxinline]f(3)g(t)[/mathjaxinline]</text>
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<text> The weight function delayed by [mathjaxinline]3[/mathjaxinline] and multiplied by the value of the signal at [mathjaxinline]t=3[/mathjaxinline]: [mathjaxinline]f(3)g(t-3)[/mathjaxinline]</text>
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Now please listen to the demo associated with this Mathlet. (Towards the end, there's a voice-over error. I said that the operator involved is [mathjaxinline]d/dt+ax[/mathjaxinline], but of course I meant [mathjaxinline]d/dt+aI[/mathjaxinline].) </p>
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<p>
The process we described actually applies verbatim to any LTI system! The time invariance tells us that the system response to a delayed delta function is just the correspondingly delayed impulse response; and linearity lets us apply superposition to get the result. Let us restate the conclusion: </p><blockquote class="quote"> The system response (with rest initial conditions) by an LTI system to a signal [mathjaxinline]f(t)[/mathjaxinline] is given by convolving with the unit impulse response: <table id="a0000001087" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x(t)=f(t)*w(t)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table> </blockquote><h3 class="hd hd-3 problem-header"> Another perspective.</h3><p>
The signal [mathjaxinline]f(t)[/mathjaxinline] can be written as an integral of scaled delta functions: </p><table id="a0000001088" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f(t)=\int _{0}^{t}f(\tau )\delta (t-\tau )\, d\tau .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
(Recall that [mathjaxinline]t[/mathjaxinline] is the variable, not [mathjaxinline]\tau[/mathjaxinline].) </p><p>
By the superposition principle, then, the system response (with rest initial conditions) is the corresponding integral of the system responses to the constituent signals, [mathjaxinline]f(\tau )\delta (t-\tau )\, d\tau[/mathjaxinline]. If [mathjaxinline]h(t)[/mathjaxinline] is the system response to [mathjaxinline]\delta (t)[/mathjaxinline], then, by time invariance, the system response to the delayed delta function [mathjaxinline]\delta (t-\tau )[/mathjaxinline] is [mathjaxinline]h(t-\tau )[/mathjaxinline]; so the response to [mathjaxinline]f(\tau )\delta (t-\tau )\, d\tau[/mathjaxinline] is </p><table id="a0000001089" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f(\tau )h(t-\tau )\, d\tau[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Then, by linearity (aka superposition) the system response to the input signal [mathjaxinline]f(t)[/mathjaxinline] is </p><table id="a0000001090" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]x(t)=\int _0^ tf(\tau )h(t-\tau )\, d\tau \, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table>
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The mathlet <b class="bf">Convolution: Accumulation</b> suggests that the differential equation [mathjaxinline]\dot x+ax=1+\cos (bt)[/mathjaxinline] has a particular solution of the form [mathjaxinline]A+B\cos (bt-\phi )[/mathjaxinline]. What are [mathjaxinline]A[/mathjaxinline] and [mathjaxinline]B[/mathjaxinline]? </p>
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(Express in terms of [mathjaxinline]a[/mathjaxinline] and [mathjaxinline]b[/mathjaxinline].) </p>
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<p style="display:inline">[mathjaxinline]A=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]B=[/mathjaxinline]</p>
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<h2 class="hd hd-2 unit-title">11.6. Proof of convolution formula.</h2>
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Laplace transform takes convolution to the ordinary product: </p><table id="a0000001097" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f(t)*h(t)\rightsquigarrow F(s)H(s)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
This follows from what we have done above; but let's check it directly. For this we need to return to the definition of Laplace transform. </p><table id="a0000001098" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f(t)*h(t) \rightsquigarrow \int _0^\infty (f(t)*h(t))e^{-st}\, dt =\int _0^\infty \left[ \int _0^ t f(\tau )h(t-\tau )\, d\tau \right] e^{-st}\, dt \, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
To get the Laplace transforms of [mathjaxinline]f(t)[/mathjaxinline] and [mathjaxinline]h(t)[/mathjaxinline] involved, we should reverse the order of integration: </p><table id="a0000001099" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\cdots = \int _0^\infty \left[ \int _\tau ^\infty f(\tau )h(t-\tau )e^{-st}\, dt\right] \, d\tau \, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The expression [mathjaxinline]h(t-\tau )[/mathjaxinline] suggests that we change variables; say [mathjaxinline]u=t-\tau[/mathjaxinline]. In the inner integral, [mathjaxinline]\tau[/mathjaxinline] is constant, so [mathjaxinline]du=dt[/mathjaxinline]. The limits on [mathjaxinline]u[/mathjaxinline] are [mathjaxinline]\tau[/mathjaxinline] less than the limits on [mathjaxinline]t[/mathjaxinline], so we have </p><table id="a0000001100" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\cdots = \int _0^\infty \left[ \int _0^\infty f(\tau )h(u)e^{-s(u+\tau )}\, du \right] \, d\tau \, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Now pull the factors depending only on [mathjaxinline]\tau[/mathjaxinline] outside the integral over [mathjaxinline]u[/mathjaxinline], remembering that [mathjaxinline]e^{-s(u+\tau )}=e^{-su}e^{-s\tau }[/mathjaxinline]: </p><table id="a0000001101" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\cdots = \int _0^\infty f(\tau )e^{-s\tau }\left[\int _0^\infty h(u)e^{-su}\, du\right]\, d\tau \, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
This is indeed just the product of the integrals defining [mathjaxinline]F(s)[/mathjaxinline] and [mathjaxinline]H(s)[/mathjaxinline]. </p>
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<p><b class="bfseries"><span style="color:#FF7800">Note on video below:</span></b> In the video below 4:00 there is a misprint on the board. The integral whose limits appear to be from infinity to infinity [mathjaxinline]\displaystyle \int _{\infty }^{\infty }[/mathjaxinline] should really be zero to infinity [mathjaxinline]\displaystyle \int _{0}^{\infty }[/mathjaxinline]. </p>
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<h2 class="hd hd-2 unit-title">11.7. Properties.</h2>
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The convolution is a new kind of <b class="bf">“product"</b> of two signals. It enjoys the following properties for functions [mathjaxinline]f(t)[/mathjaxinline], [mathjaxinline]g(t)[/mathjaxinline], and constants [mathjaxinline]a[/mathjaxinline], [mathjaxinline]b[/mathjaxinline]. </p><ul class="itemize"><li><p>
Bi-linearity: </p><table id="a0000001102" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax][af(t)+bg(t)]*h(t)=a[f(t)*h(t)]+b[g(t)*h(t)][/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
and </p><table id="a0000001103" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f(t)*[ag(t)+bh(t)]=a[f(t)*g(t)]+b[f(t)*h(t)]\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table></li><li><p>
Associativity: </p><table id="a0000001104" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax][f(t)*g(t)]*h(t)=f(t)*[g(t)*h(t)][/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table></li><li><p>
Commutativity: </p><table id="a0000001105" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f(t)*h(t)=h(t)*f(t)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table></li><li><p>
Unit: The delta function serves as the unit for convolution: </p><table id="a0000001106" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\delta (t)*f(t)=f(t)=f(t)*\delta (t)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table></li></ul><p>
These properties can be checked by direct manipulation of the convolution integral. But it's easier to check that they hold after application of Laplace transform, where they are obvious; in the frequency domain they are just the properties of the usual multiplication of functions. </p>
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Write [mathjaxinline]t*t^{10}[/mathjaxinline] in two ways using the integral definition of convolution, and compute one of them. </p>
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<p style="display:inline">[mathjaxinline]t*t^{10}=[/mathjaxinline]</p>
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<span class="trailing_text" id="trailing_text_lec11-tab7-problem1_2_1">[mathjaxinline]\cdot u(t)[/mathjaxinline]</span>
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<h2 class="hd hd-2 unit-title">11.8. Block diagrams: back to the time domain.</h2>
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The formalism of the frequency domain lets us represent both systems and signals in the same language. A signal is represented by its Laplace transform. A system is represented by its transfer function, or, what is the same thing, the Laplace transform of its unit impulse response. </p><p>
The convolution product allows us to do the same in the time domain. We have just seen that if you have an LTI system with system unit impulse response [mathjaxinline]h(t)[/mathjaxinline], then a general input signal [mathjaxinline]f(t)[/mathjaxinline] produces [mathjaxinline]f(t)*h(t)[/mathjaxinline] as system response. So, working entirely in the time domain, we can write </p><center><img src="/assets/courseware/v1/ddf0ecddb361d6fd4a1aeac1bf51cf48/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_l7-10-block1.png" width="260"/></center><div><br/></div><p>
In particular we can take [mathjaxinline]f(t)=\delta (t)[/mathjaxinline]. Then by definition the output is [mathjaxinline]h(t)[/mathjaxinline]; that is, </p><table id="a0000001112" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\delta (t)*h(t)=h(t)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
How does a cascade work out? </p><center><img src="/assets/courseware/v1/f7fcfa7f79907c063c264dae379d41d0/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_l7-10-block2.png" width="360"/></center><div><br/></div>To determine the answer, we just need to determine the unit impulse response of the cascaded system. Since [mathjaxinline]\delta (t)*g(t)=g(t)[/mathjaxinline], <center><img src="/assets/courseware/v1/c5938f83c307f03e28a661ec85a06c13/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_l7-10-block3.png" width="390"/></center><div><br/></div>That is, the unit impulse response of the cascade is the convolution of the two constituent unit impulse responses: <center><img src="/assets/courseware/v1/0a974f2e57a7d588ab1a89769af4c10c/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_l7-10-block4.png" width="630"/></center><div><br/></div><p>
This perspective makes associativity obvious. Each of the 5 block diagrams below represent the same system. The first translates to [mathjaxinline](f*g)*h[/mathjaxinline] and the last to [mathjaxinline]f*(g*h)[/mathjaxinline]. </p><center><img src="/assets/courseware/v1/3bbfd8467781acfc7aaa8acfb3c092dd/asset-v1:OCW+18.031+2019_Spring+type@asset+block/images_l7-10-block5.png" width="580"/></center><div><br/></div>But commutativity seems more surprising: if you feed signal [mathjaxinline]f(t)[/mathjaxinline] to a system with unit impulse response [mathjaxinline]g(t)[/mathjaxinline], the outcome is the same as if you were to feed [mathjaxinline]g(t)[/mathjaxinline] to a system with unit impulse response [mathjaxinline]f(t)[/mathjaxinline]! Commutativity is one of the miraculous outcomes of the LTI hypothesis.
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<h2 class="hd hd-2 unit-title">11.9. Convolution and generalized functions.</h2>
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The delta function [mathjaxinline]\delta (t)[/mathjaxinline] serves as the unit for the convolution product. To accommodate the delta function in the convolution integral, we should refine it to read </p><table id="a0000001113" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]f(t)*h(t)=\int _{0^-}^{t^+}f(\tau )h(t-\tau )\, d\tau \, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Then, indeed, </p><table id="a0000001114" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\delta (t)*h(t)=\int _{0^-}^{t^+}\delta (\tau )h(t-\tau )\, d\tau =h(t-\tau )|_{\tau =0} =h(t)\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
More generally, </p><table id="a0000001115" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\delta (t-a)*h(t)=\int _{0^-}^{t^+}\delta (\tau -a)h(t-\tau )\, d\tau = h(t-\tau )|_{\tau =a}=h(t-a)[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
– that is, convolving with [mathjaxinline]\delta (t-a)[/mathjaxinline] (for [mathjaxinline]a\geq 0[/mathjaxinline]) delays a signal by [mathjaxinline]a[/mathjaxinline] units; it shifts the graph to the right by [mathjaxinline]a[/mathjaxinline] units. In particular, </p><table id="a0000001116" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\delta (t-a)*\delta (t-b)=\delta (t-(a+b))\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
These formulas let us convolve generalized functions as well as piecewise continuous functions. </p>
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<p>Compute the convolution \((1+\delta(t-1)) * (1-\delta(t-1)) \). </p>
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<tr class="formulainput">
<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Basic functions</th>
<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="3">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
</tr>
</tbody>
</table>
</div>
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<h2 class="hd hd-2 unit-title">11.10. Practice problems</h2>
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<h3 class="hd hd-3 problem-header" id="lec11-tab10-problem1-problem-title" aria-describedby="block-v1:OCW+18.031+2019_Spring+type@problem+block@lec11-tab10-problem1-problem-progress" tabindex="-1">
1 (a)
</h3>
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<p>
Calculate [mathjaxinline]t*t^2[/mathjaxinline] and [mathjaxinline]e^ t*e^ t[/mathjaxinline] using the Laplace transform. </p>
<p>
(The [mathjaxinline]u(t)[/mathjaxinline] is provided for you. Do not enter it into the answer box.) </p>
<p>
<p style="display:inline">[mathjaxinline]t*t^2=[/mathjaxinline]</p>
<div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="formulaequationinput_lec11-tab10-problem1_2_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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<input type="text" name="input_lec11-tab10-problem1_2_1" id="input_lec11-tab10-problem1_2_1" data-input-id="lec11-tab10-problem1_2_1" value="" aria-describedby="trailing_text_lec11-tab10-problem1_2_1 status_lec11-tab10-problem1_2_1" size="20"/>
<span class="trailing_text" id="trailing_text_lec11-tab10-problem1_2_1">[mathjaxinline]\cdot u(t)[/mathjaxinline]</span>
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\(\)
<img src="/static/images/spinner.bc34f953403f.gif" class="loading" alt="Loading"/>
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<div class="script_placeholder" data-src="/static/js/capa/src/formula_equation_preview.b1967ab28c31.js"/>
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<p>
<p style="display:inline">[mathjaxinline]e^ t*e^ t=[/mathjaxinline]</p>
<div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="formulaequationinput_lec11-tab10-problem1_3_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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<span class="trailing_text" id="trailing_text_lec11-tab10-problem1_3_1">[mathjaxinline]\cdot u(t)[/mathjaxinline]</span>
<span class="status unanswered" id="status_lec11-tab10-problem1_3_1" data-tooltip="Not yet answered.">
<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
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\(\)
<img src="/static/images/spinner.bc34f953403f.gif" class="loading" alt="Loading"/>
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<div class="formulainput">
<table class="formulainput">
<tbody>
<tr class="fiptitle">
<th class="formulainput" scope="col">Allowable Entries</th>
<th class="formulainput" scope="col">Descriptions</th>
<th class="formulainput" scope="col">Example Entries</th>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="3">Numbers</th>
<td class="formulainput">Integers</td>
<td class="formulainput">
<font color="#0078b0">2520</font>
</td>
</tr>
<tr class="formulainput">
<td class="formulainput">Fractions</td>
<td class="formulainput">
<font color="#0078b0">2/3</font>
</td>
</tr>
<tr class="formulainput">
<td class="formulainput">Decimals </td>
<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="4">Operators</th>
<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Basic functions</th>
<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="3">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
</tr>
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1(b)
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<p>
Calculate [mathjaxinline]e^{at}*e^{bt}[/mathjaxinline] for [mathjaxinline]a\neq b[/mathjaxinline] using the Laplace transform. </p>
<p>
<p style="display:inline">[mathjaxinline]e^{at}*e^{bt}=[/mathjaxinline]</p>
<div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="formulaequationinput_lec11-tab10-problem2_2_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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<span class="trailing_text" id="trailing_text_lec11-tab10-problem2_2_1">[mathjaxinline]\cdot u(t)[/mathjaxinline]</span>
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<div class="formulainput">
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</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="3">Numbers</th>
<td class="formulainput">Integers</td>
<td class="formulainput">
<font color="#0078b0">2520</font>
</td>
</tr>
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<td class="formulainput">Fractions</td>
<td class="formulainput">
<font color="#0078b0">2/3</font>
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<td class="formulainput">Decimals </td>
<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="4">Operators</th>
<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Basic functions</th>
<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="3">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
</tr>
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<h3 class="hd hd-3 problem-header" id="lec11-tab10-problem3-problem-title" aria-describedby="block-v1:OCW+18.031+2019_Spring+type@problem+block@lec11-tab10-problem3-problem-progress" tabindex="-1">
1(c)
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Take the limit in part (b) as [mathjaxinline]b\to a[/mathjaxinline] to find [mathjaxinline]e^{at}*e^{at}[/mathjaxinline]. (Check against the case [mathjaxinline]a=1[/mathjaxinline] in part (a).) </p>
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<p style="display:inline">[mathjaxinline]e^{at}*e^{at}=[/mathjaxinline]</p>
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<span class="trailing_text" id="trailing_text_lec11-tab10-problem3_2_1">[mathjaxinline]\cdot u(t)[/mathjaxinline]</span>
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<th class="formulainput" scope="col">Example Entries</th>
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<th class="formulainput" scope="row" rowspan="3">Numbers</th>
<td class="formulainput">Integers</td>
<td class="formulainput">
<font color="#0078b0">2520</font>
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<td class="formulainput">Fractions</td>
<td class="formulainput">
<font color="#0078b0">2/3</font>
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<td class="formulainput">Decimals </td>
<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="4">Operators</th>
<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Basic functions</th>
<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="3">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
</tr>
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(d)
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Find [mathjaxinline]\cos t* \cos t[/mathjaxinline] using [mathjaxinline]\cos t = (e^{it} + e^{-it})/2[/mathjaxinline] and the formulas from parts (b) and (c) for complex values of [mathjaxinline]a[/mathjaxinline] and [mathjaxinline]b[/mathjaxinline]. (Note your answer must be a real-valued function.) </p>
<p>
<p style="display:inline">[mathjaxinline]\cos t* \cos t=[/mathjaxinline]</p>
<div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="formulaequationinput_lec11-tab10-problem4_2_1" class="inputtype formulaequationinput" style="display:inline-block;vertical-align:top">
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<span class="trailing_text" id="trailing_text_lec11-tab10-problem4_2_1">[mathjaxinline]\cdot u(t)[/mathjaxinline]</span>
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<th class="formulainput" scope="col">Descriptions</th>
<th class="formulainput" scope="col">Example Entries</th>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="3">Numbers</th>
<td class="formulainput">Integers</td>
<td class="formulainput">
<font color="#0078b0">2520</font>
</td>
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<td class="formulainput">
<font color="#0078b0">2/3</font>
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<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
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<th class="formulainput" scope="row" rowspan="4">Operators</th>
<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
</td>
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<h2 class="hd hd-2 unit-title">11.11. (Optional) Motivation for convolution.</h2>
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<h3 class="hd hd-2">Motivation with power series</h3>
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Let [mathjaxinline]f(t)[/mathjaxinline] and [mathjaxinline]g(t)[/mathjaxinline] be two functions with Laplace transforms </p><table id="a0000001126" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001127"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle f(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle F(s)[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(7.1)</td></tr><tr id="a0000001128"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle g(t)[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle \rightsquigarrow[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle G(s).[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(7.2)</td></tr></table><p>
A natural question one might ask is if there is a formula for the Laplace transform of [mathjaxinline]f(t)g(t)[/mathjaxinline] in terms of [mathjaxinline]F(s)[/mathjaxinline] and [mathjaxinline]G(s)[/mathjaxinline]. No such formula exists. Instead, one can ask if the function [mathjaxinline]F(s)G(s)[/mathjaxinline] is the Laplace transform of a function that can be built out of [mathjaxinline]f(t)[/mathjaxinline] and [mathjaxinline]g(t)[/mathjaxinline]. This is possible, and the answer is the convolution. </p><p><p><b class="bfseries">Definition 11.1 </b> The convolution of [mathjaxinline]f(t)[/mathjaxinline] and [mathjaxinline]g(t)[/mathjaxinline], written as [mathjaxinline]f(t)*g(t)[/mathjaxinline], is the function of [mathjaxinline]t[/mathjaxinline] whose Laplace transform is [mathjaxinline]F(s)G(s)[/mathjaxinline]. </p></p><h3 class="hd hd-3 problem-header">Power series motivation</h3><p>
The power series </p><table id="a0000001129" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]F(x) = \sum _{n=0}^{\infty } a_ n x^ n[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
is the discrete analogue of the Laplace transform. You can see this by writing [mathjaxinline]a_ n = a(n)[/mathjaxinline] as a function of [mathjaxinline]n[/mathjaxinline], and writing [mathjaxinline]x= e^{-s}[/mathjaxinline]. Then we can think of </p><table id="a0000001130" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]F(x) = \sum _{n=0}^{\infty } a(n) e^{-sn}[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
as the discrete version of the Laplace transform of [mathjaxinline]a(n)[/mathjaxinline]. </p><p>
Write two power series </p><table id="a0000001131" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000001132"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle F(x)[/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 \sum _{n=0}^{\infty } a_ nx^ n[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(7.3)</td></tr><tr id="a0000001133"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle G(x)[/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 \sum _{n=0}^{\infty } b_ nx^ n.[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none;text-align:right" class="eqnnum">(7.4)</td></tr></table><p>
Write the product as a power series </p><table id="a0000001134" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]F(x)G(x) = \sum _{n=0}^{\infty } c_ n x^ n .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The question analogous to convolution in power series is to find the formula for [mathjaxinline]c_ n[/mathjaxinline] in terms of the [mathjaxinline]a_ i[/mathjaxinline]'s and [mathjaxinline]b_ j[/mathjaxinline]'s. </p>
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Find the formula for [mathjaxinline]c_0[/mathjaxinline], [mathjaxinline]c_1[/mathjaxinline] and [mathjaxinline]c_2[/mathjaxinline]. Generalize the formula to find a formula for [mathjaxinline]c(n)[/mathjaxinline]. </p>
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(Type a_1 for [mathjaxinline]a_1[/mathjaxinline], b_1 for [mathjaxinline]b_1[/mathjaxinline] etc.) </p>
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<p style="display:inline">[mathjaxinline]c_0=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]c_1=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]c_2=[/mathjaxinline]</p>
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<th class="formulainput" scope="col">Allowable Entries</th>
<th class="formulainput" scope="col">Descriptions</th>
<th class="formulainput" scope="col">Example Entries</th>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="3">Numbers</th>
<td class="formulainput">Integers</td>
<td class="formulainput">
<font color="#0078b0">2520</font>
</td>
</tr>
<tr class="formulainput">
<td class="formulainput">Fractions</td>
<td class="formulainput">
<font color="#0078b0">2/3</font>
</td>
</tr>
<tr class="formulainput">
<td class="formulainput">Decimals </td>
<td class="formulainput"><font color="#0078b0">3.14</font>, <font color="#0078b0">.98</font></td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="4">Operators</th>
<td class="formulainput">+ - * / (add, subtract, multiply, divide)</td>
<td class="formulainput">Enter <font color="#0078b0"> (x+2*y)/(x-1)</font> for \( \displaystyle \frac{x+2y}{x-1} \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">^ (raise to a power)</td>
<td class="formulainput">Enter <font color="#0078b0"> x^(n+1) </font> for \( x^{n+1} \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">_ (add a subscript)</td>
<td class="formulainput">Enter <font color="#0078b0"> v_0 </font> for \( v_0 \) </td>
</tr>
<tr class="formulainput">
<td class="formulainput">Use ( ) to clarify order of operations</td>
<td class="formulainput"> Enter <font color="#0078b0">(2+3)*2 </font> for 10 <br/>
Enter <font color="#0078b0"> 2+3*2 </font> for 8 </td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Greek letters</th>
<td class="formulainput">Enter (english) name of letter</td>
<td class="formulainput">Enter <font color="#0078b0">alpha </font> for \( \alpha \)<br/>
Enter <font color="#0078b0">lambda </font> for \(\lambda \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Mathematical <br/> constants</th>
<td class="formulainput">e, pi</td>
<td class="formulainput">Enter <font color="#0078b0">e^x </font> for \( e^x \)<br/>
Enter <font color="#0078b0">2*pi </font> for \( 2\pi \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row">Basic functions</th>
<td class="formulainput">abs, ln, log, log_2, sqrt</td>
<td class="formulainput">Enter <font color="#0078b0">abs(x+y) </font> for \( \left|x+y \right| \)<br/>
Enter <font color="#0078b0">sqrt(x^2-y) </font> for \( \sqrt{x^2-y} \)
</td>
</tr>
<tr class="formulainput">
<th class="formulainput" scope="row" rowspan="3">Trigonometric <br/> functions</th>
<td class="formulainput">sin, cos, tan, sec, csc, cot</td>
<td class="formulainput">Enter <font color="#0078b0">sin(4*x+y)^2 </font> for \(\sin^2(4x+y) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput">arcsin, arccos, arctan, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">arctan(x^2/3) </font> for \(\tan^{-1}\left(\frac{x^2}{3}\right) \)</td>
</tr>
<tr class="formulainput">
<td class="formulainput"> sinh, cosh, arcsinh, etc.</td>
<td class="formulainput">Enter <font color="#0078b0">cosh(4*x+y) </font> for \(\cosh(4x+y) \)</td>
</tr>
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