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      <text>20.2.2.1 The inverse Laplace transform</text>
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      <text><![CDATA[<h2>The inverse Laplace transform</h2>
<p>The Laplace transform takes a causal function \(f(t)\) and transforms it into a function of \(s\), \(F(s)\): </p>
<p>\(\qquad {\cal L}\{f(t)\}\equiv F(s)\)</p> The inverse Laplace transform operator is denoted by \({\cal L}^{-1}\) and involves recovering the original causal function \(f(t)\). That is,

<p><br></p>
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    <h4>Key Point</h4>
    <center><strong>Inverse Laplace Transform</strong></center>
    <p>\(\qquad{\cal L}^{-1}\{F(s)\} = f(t) \qquad\)where \(\qquad{\cal L}\{f(t)\} = F(s)\) </p>
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<p>For example, using standard transforms from Table 1:</p>
<p>\(\qquad {\cal L}^{-1}\left\{{s\over s^2+4}\right\}=\cos2t\,.\,u(t)\) since \({\cal L}\{\cos 2t\,.\,u(t)\}={s\over s^2+4}\).\(\quad\) (Table 1, Rule 6)</p>
<p>Also</p>
<p>\(\qquad {\cal L}^{-1}\left\{{3\over s^2}\right\}=3t\,u(t)\) since \({\cal L}\{3t\,u(t)\}={3\over s^2}\).\(\quad\) (Table 1, Rule 3)</p>
<p>Because the Laplace transform is a linear operator it follows that the inverse Laplace transform is also linear, so if \(c_1,\ c_2\) are constants:</p>

<p><br></p>
<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <center><strong>Linearity Property of Inverse Laplace Transforms</strong>
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            <p>\(\qquad{\cal L}^{-1}\{c_1F(s)+c_2G(s)\} = c_1{\cal L}^{-1}\{F(s)\} + c_2{\cal L}^{-1}\{G(s)\}\) </p>
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<p><br></p>


<p>For example, to find the inverse Laplace transform of \({2\over s^4}-{6\over s^2+4}\) we have</p>
<p>\[\begin{aligned} {\cal L}^{-1}\left\{{2\over s^4}-{6\over s^2+4}\right\}&amp;=\dfrac{2}{6}{\cal L}^{-1}\left\{{6\over s^4}\right\}-3{\cal L}^{-1}\left\{{2\over s^2+4}\right\}\\ &amp;={1\over 3}t^3u(t)-3\sin 2t\,.\, u(t)\text{ (from Table 1)} \end{aligned}\]
</p> Note that the fractions have had to be manipulated slightly in order that the expressions match precisely with the expressions in Table 1.
<br>
<p>Although the inverse Laplace transform can be examined at a deeper mathematical level we shall be content with this simple-minded approach to finding inverse Laplace transforms by using the table of Laplace transforms. However, even this approach is not
    always straightforward and considerable algebraic manipulation is often required before an inverse Laplace transform can be found. Next we consider two standard rearrangements which often occur.</p>
<p></p>
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<p><strong>Inverting through the use of partial fractions</strong></p>

<p>The function </p>
<p>\(\qquad F(s)={1\over (s-1)(s+2)}\)</p> does not appear in our table of transforms and so we cannot, by inspection, write down the inverse Laplace transform. However, by using partial fractions we see that
<p>\(\qquad F(s)={1\over (s-1)(s+2)}={{1\over 3}\over s-1}-{{1\over 3}\over s+2}\)</p> and so, using the linearity property: \[\begin{aligned} {\cal L}^{-1}\left\{{1\over (s-1)(s+2)}\right\}&amp;={\cal L}^{-1}\left\{{{1\over 3}\over s-1}\right\}-{\cal L}^{-1}\left\{{{1\over
3}\over s+2}\right\}\\ &amp;=\tfrac{1}{3}\text{e}^t-\tfrac{1}{3} \text{e}^{-2t} \text{(Table 1, Rule 4)} \end{aligned}\]]]></text>
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