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      <text>20.2.1.3 The Transform and its Inverse</text>
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      <text><![CDATA[<p>As a second example, we consider the decaying exponential \(f(t)=\text{e}^{-at}u(t)\) where \(a\) is a positive constant. This function has graph:</p>

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    <img src="@@PLUGINFILE@@/20-2-fig-2-decayingexp.svg" alt="Workbook 20.2 figure 2" width="304" height="140" class="img-responsive atto_image_button_text-bottom">
    <figcaption>Figure 12</figcaption>
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<p>In this case, \[\begin{aligned} {\cal L}\{\text{e}^{-at}u(t)\} &amp;=\int_0^\infty \text{e}^{-st}\text{e}^{-at}\,dt\\ &amp;=\int_0^\infty \text{e}^{-(s+a)t}\,dt\\ &amp;=\left[{\text{e}^{-(s+a)t}\over -(s+a)}\right]_0^\infty={1\over s+a} \text{(zero contribution
    from the upper limit)} \end{aligned}\]</p>
<p>Therefore, if \(f(t)=\text{e}^{-at}u(t)\) then \(F(s) = {1\over s+a}\).</p>
<p>Following this approach we can develop a table of Laplace transforms which records, for each causal function \(f(t)\) listed, its corresponding transform function \(F(s)\). Table 1 gives a limited table of transforms.<br></p>

<p><br></p>
<h4>The linearity property of the Laplace transformation</h4>

<p>If \(f(t)\) and \(g(t)\) are causal functions and \(c_1\), \(c_2\) are constants then </p>

<p>\[\begin{aligned} {\cal L}\{c_1f(t)+c_2g(t)\}&amp;=\int_0^\infty \text{e}^{-st}[c_1f(t)+c_2g(t)]\,dt\\ &amp;=c_1\int_0^\infty \text{e}^{-st}f(t)\,dt+c_2\int_0^\infty \text{e}^{-st}g(t)\,dt\\ &amp;=c_1{\cal L}\{f(t)\}+c_2{\cal L}\{g(t)\}\end{aligned}\]</p>

<p></p>
<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <center><strong>Linearity Property of the Laplace Transform</strong>
    </center>
    <p>\(\qquad{\cal L}\{c_1f(t)+c_2g(t)\}=c_1{\cal L}\{f(t)\}+c_2{\cal L}\{g(t)\}\)</p>
</div>
<p></p>

<center>
    <br><strong>Table 1.</strong> Table of Laplace Transforms
    <table style="width:70%;">
        <tbody>
            <tr>
                <th>Rule</th>
                <th>Causal function</th>
                <th>Laplace transform</th>
            </tr>

            <tr>
                <td>1\(\qquad\)</td>
                <td>\(f(t)\)</td>
                <td>\(\displaystyle F(s)\)</td>
            </tr>
            <tr>
                <td>2\(\qquad\)</td>
                <td>\(u(t)\)</td>
                <td>\(\displaystyle{1\over s}\)</td>
            </tr>
            <tr>
                <td>3\(\qquad\)</td>
                <td>\(t^nu(t)\) </td>
                <td>\(\displaystyle{n!\over s^{n+1}}\)</td>
            </tr>
            <tr>
                <td>4\(\qquad\)</td>
                <td>\(\text{e}^{-at}u(t)\) </td>
                <td> \(\displaystyle{1\over s+a}\)</td>
            </tr>
            <tr>
                <td>5\(\qquad\)</td>
                <td>\(\sin at\,.\,u(t)\)</td>
                <td> \(\displaystyle{a\over s^2+a^2}\)</td>
            </tr>
            <tr>
                <td>6\(\qquad\)</td>
                <td>\(\cos at\,.\,u(t)\)</td>
                <td>\(\displaystyle{s\over s^2+a^2}\)</td>
            </tr>
            <tr>
                <td>7\(\qquad\)</td>
                <td>\(\text{e}^{-at}\sin bt\,.\,u(t)\)</td>
                <td> \(\displaystyle{b\over (s+a)^2+b^2}\)</td>
            </tr>
            <tr>
                <td>8\(\qquad\)</td>
                <td>\(\text{e}^{-at}\cos bt\,u(t)\) </td>
                <td>\(\displaystyle{s+a\over (s+a)^2+b^2}\)</td>
            </tr>
        </tbody>
    </table>
</center>
<br><br>

<p>That is, the Laplace transform of a linear sum of causal functions is a linear sum of Laplace transforms. For example, </p>
<p> \[\begin{aligned} {\cal L}\{2\cos t\,.\,u(t)-3t^2u(t)\} &amp;=2{\cal L}\{\cos t\,.\, u(t)\}-3{\cal L}\{t^2u(t)\}\\ &amp;=2\left({s\over s^2+1}\right)-3\left({2\over s^3}\right) \end{aligned}\]</p>]]></text>
<file name="20-2-fig-2-decayingexp.svg" path="/" encoding="base64">[encoded data, 9808 characters]</file>
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