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    <name>
      <text>20.2.1.1 The Transform and its Inverse</text>
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      <text><![CDATA[<h2>The Laplace transform</h2>

<p>If \(f(t)\) is a <strong>casual function</strong> then the Laplace transform of \(f(t)\) is written \({\cal L}\{f(t)\}\) and defined by:</p>
<p>\(\qquad {\cal L}\{f(t)\}=\int_0^\infty \text{e}^{-st}f(t)\,dt.\)
</p>
<p>Clearly, once the integral is performed and the limits substituted the resulting expression will involve the \(s\) parameter alone since the dependence upon \(t\) is removed in the integration process. This resulting expression in \(s\) is denoted by
    \(F(s)\); its precise form is dependent upon the form taken by \(f(t)\). We now refine a previous Key Point.</p>
<br>
<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <center><strong>The Laplace Transform of a Causal Function</strong></center>
    \[{\cal L}\{f(t)u(t)\}=\int_0^\infty \text{e}^{-st}f(t)u(t)\,dt \equiv F(s)\]
</div>
<br>
<p>To begin, we determine the Laplace transform of some simple causal functions. For example, if we consider the <strong>ramp function</strong> \(f(t)=t.u(t)\) with graph </p>

<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/20-2-fig-1-rampfunction.svg" alt="Workbook 20.2 Figure 11
" width="368" height="164" class="img-responsive atto_image_button_text-bottom">
    <figcaption>Figure 11</figcaption>
</figure>

<p>we find: </p>
<p> \({\cal L}\{t\ u(t)\}=\int_0^\infty \text{e}^{-st}t\ u(t)\,dt\)</p>
<p>\(\qquad =\int_0^\infty \text{e}^{-st}t\ \,dt \qquad\) since in the range of the integral \(u(t)=1\)</p>
<p>\(\qquad =\left[{t\ \text{e}^{-st}\over (-s)}\right]_0^\infty-\int_0^\infty {\text{e}^{-st}\over (-s)}\ \,dt\qquad\) using integration by parts</p>
<p>\(\qquad =\left[{t\text{e}^{-st}\over (-s)}\right]_0^\infty-\left[{\text{e}^{-st}\over (-s)^2}\right]_0^\infty \)</p>
<p>Now we have the difficulty of substituting in the limits of integration. The only problem arises with the upper limit (\(t =\infty\)). We shall always assume that the parameter \(s\) is so chosen that no contribution ever arises from the upper limit (\(t=\infty\)).
    In this particular case we need only demand that \(s\) is real and positive. Using this ‘rule of thumb’: </p>
<p>\({\cal L}\{t\ u(t)\}=[0-0]-\left[0-\left({1\over (-s)^2}\right)\right]\)</p>
<p>\(\qquad\qquad ={1\over s^2}\)
</p>
<p>Thus, if \(f(t)=t\ u(t)\) then \(F(s)=1/s^2\).</p>
<p>A similar, but more tedious, calculation yields the result that if \(f(t)=t^nu(t)\) in which \(n\) is a positive integer then:
</p>
<p>\(\qquad {\cal L}\{t^nu(t)\}={n!\over s^{n+1}}\)</p> [We remember \(n!\equiv n(n-1)(n-2)\dots (3)(2)(1).\)]]]></text>
<file name="20-2-fig-1-rampfunction.svg" path="/" encoding="base64">[encoded data, 8036 characters]</file>
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