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      <text>20.2.3.1 The first shift theorem</text>
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      <text><![CDATA[<h2> The first shift theorem</h2>

<p>The first and second shift theorems enable an even wider range of Laplace transforms to be easily obtained than the transforms we have already found. They also enable a significantly wider range of inverse transforms to be found. Here we introduce the
    first shift theorem. If \(f(t)\) is a causal function with Laplace transform \(F(s)\), i.e. \({\cal L}\{f(t)\}=F(s)\), then as we shall see, the Laplace transform of \(\text{e}^{-at}f(t)\), where \(a\) is a given constant, can easily be found in terms
    of \(F(s)\).</p>
<p>Using the definition of the Laplace transform: </p>
<p>\[\begin{aligned} {\cal L}\{\text{e}^{-at}f(t)\}&amp;= \int_0^{\infty}\text{e}^{-st} \biggl[\text{e}^{-at}f(t)\biggr]\,dt\\ &amp;=\int_0^{\infty}\text{e}^{-(s+a)t}f(t)\,dt\end{aligned}\]
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<p>But if
</p>
<p>\(\qquad F(s)={\cal L}\{f(t)\}=\int_0^{\infty}\text{e}^{-st}f(t)\,dt\) </p>then simply replacing ‘\(s\)’ by ‘\(s+a\)’ on both sides gives:
<p>\(\qquad F(s+a)=\int_0^{\infty}\text{e}^{-(s+a)t}f(t)\,dt\)</p>
That is, the parameter \(s\) is shifted to the value \(s+a\).
<p>We have then the statement of the <strong>first shift theorem</strong>:</p>
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<div class="HELM_keypoint">
    <h4>Key Point</h4>

    <center><strong>First Shift Theorem</strong></center>

    <p>If \(\qquad{\cal L}\{f(t)\} = F(s) \qquad \)then \(\qquad {\cal L}\{\text{e}^{-at}f(t)\} = F(s+a)\) </p>
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<p></p>



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<p>For example, we already know (from Table 1) that
</p>
<p>\(\qquad {\cal L}\{t^3u(t)\}={6\over {s^4}}\)</p>
and so, by the first shift theorem:
<p>\(\qquad {\cal L}\{\text{e}^{-2t}t^3u(t)\}={6\over {(s+2)^4}}\)</p>]]></text>
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