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      <text>20.3.1.1 The second shift theorem</text>
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      <text><![CDATA[<h2>The second shift theorem</h2>
<p>The second shift theorem is similar to the first except that, in this case, it is the time-variable that is shifted not the \(s\)-variable. Consider a causal function \(f(t)u(t)\) which is shifted to the right by amount \(a\), that is, the function \(f(t-a)u(t-a)\)
    where \(a&gt;0\). The following figure illustrates the two causal functions.</p>
<p><img src="@@PLUGINFILE@@/20-3-fig-13-shifted_graph.svg" alt="Two graphs, showing a casual function and the same function, shifted to the right by a." width="643" height="204" class="img-responsive atto_image_button_text-bottom"><br></p>
<p>The Laplace transform of the shifted function is easily obtained: \[\begin{align}{\cal L}\{f(t-a)u(t-a)\}&amp;= \int_0^{\infty}\text{e}^{-st}f(t-a)u(t-a)\,dt \\ &amp;=\int_a^{\infty}\text{e}^{-st}f(t-a)\,dt\end{align}\]
</p>
<p>(Note the change in the lower limit from \(0\) to \(a\) resulting from the step function switching on at \(t=a\)). We can re-organise this integral by making the substitution \(x=t-a\). Then \(\displaystyle \mathrm{d}t=\mathrm{d}x\) and when \(t=a, x=0\)
    and when \(t=\infty\) then \(x=\infty\).</p>
<p>Therefore \[\begin{align} \int_a^{\infty}\text{e}^{-st}f(t-a)\,dt &amp;=\int_0^{\infty}\text{e}^{-s(x+a)}f(x)\,dx\\ &amp;=\text{e}^{-sa}\int_0^{\infty}\text{e}^{-sx}f(x)\,dx\end{align}\]</p>
<p>The final integral is simply the Laplace transform of \(f(x)\), which we know is \(F(s)\) and so, finally, we have the statement of the second shift theorem:</p><br>
<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <center>
        <p><strong>Second Shift Theorem</strong></p>
        <p>If \({\cal L}\{f(t)\}=F(s)\) then \({\cal L}\{f(t-a)u(t-a)\}=\text{e}^{-sa}F(s)\)</p>
    </center>
</div>
<br>
<p>Obviously, this theorem has its uses in finding the Laplace transform of time-shifted causal functions but it is also of considerable use in finding inverse Laplace transforms since, using the inverse formulation of the theorem of the previous Key Point
    we get:</p><br>
<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <center>
        <p><strong>Inverse Second Shift Theorem</strong></p>
        <p>If \({\cal L}^{-1}\{F(s)\}=f(t)\) then \({\cal L}^{-1}\{\text{e}^{-sa}F(s)\}=f(t-a)u(t-a)\)</p>
    </center>
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<hr>
<h4 class="HELM_example">Example</h4>
<p>Find the inverse Laplace transform of  \(\displaystyle {\text{e}^{-3s}\over {s^2}}\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>You should obtain \((t-3)u(t-3)\) for the following reasons. We know that the inverse Laplace transform of \(1/s^2\) is \(t.u(t)\) (from the table of common Laplace transforms) and so, using the second shift theorem (with \(a=3\)), we have \[ {\cal L}^{-1}\biggl\{\text{e}^{-3s}{1\over
    {s^2}}\biggr\}=(t-3)u(t-3).\]
</p>
<p>This function is graphed in the following figure:</p>
<center><img src="@@PLUGINFILE@@/20-3-fig-ex-example_func.svg" alt="Graph of the answer to the exercise. Shows a line leaving the x-axis at an angle of 45 degrees at x = 3/" width="356" height="177" class="img-responsive atto_image_button_text-bottom"></center>]]></text>
<file name="20-3-fig-13-shifted_graph.svg" path="/" encoding="base64">[encoded data, 21820 characters]</file>
<file name="20-3-fig-ex-example_func.svg" path="/" encoding="base64">[encoded data, 8956 characters]</file>
<file name="fig4-1.svg" path="/" encoding="base64">[encoded data, 8956 characters]</file>
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