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      <text>20.2.1.4 The linearity property of the Laplace transformation</text>
    </name>
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      <text><![CDATA[<h4 class="HELM_exercise">Exercise</h4>
<p>Obtain the Laplace transform of the hyperbolic function \(\sinh at\).</p>
<ol class="HELM_parts">
    <li>Begin by expressing \(\sinh at\) in terms of exponential functions:
        <p>[[input:ans1]] [[validation:ans1]] [[feedback:prt1]]</p>
    </li>
    <li>Now use the linearity property (Key Point above) to obtain the Laplace transform of the causal function \(\sinh at.u(t)\):
        <p>[[input:ans2]] [[validation:ans2]] [[feedback:prt2]]</p>
    </li>
    
</ol>
<p></p>]]></text>
    </questiontext>
    <generalfeedback format="html">
      <text><![CDATA[<ol class="HELM_parts">
    <li>\(\sinh at = \dfrac{1}{2}(\text{e}^{at}-\text{e}^{-at}) \)</li>
    <li>You should obtain \(a/(s^2 − a^2)\) since\[\begin{aligned} {\cal L}\{\dfrac{\text{e}^{at}-\text{e}^{-at}}{2}u(t)\} &amp;=\dfrac{1}{2}{\cal L}\{\text{e}^{at}u(t)\}-\dfrac{1}{2}{\cal L}\{\text{e}^{-at}u(t)\}\\ &amp;=\dfrac{1}{2}\left[\dfrac{1}{s-a}\right]-\dfrac{1}{2}\left[\dfrac{1}{s+a}\right]\\
        &amp;=\dfrac{1}{2}\left[\dfrac{2a}{(s-a)(s+a)}\right]\left[\dfrac{1}{s+a}\right]= \dfrac{a}{s^2-a^2}\end{aligned}\]
    </li>
    <li>You should obtain \(s/(s^2 − a^2)\) since\[\begin{aligned} {\cal L}\{\dfrac{\text{e}^{at}+\text{e}^{-at}}{2}u(t)\} &amp;=\dfrac{1}{2}{\cal L}\{\text{e}^{at}u(t)\}+\dfrac{1}{2}{\cal L}\{\text{e}^{-at}u(t)\}\\ &amp;=\dfrac{1}{2}\left[\dfrac{1}{s-a}\right]+\dfrac{1}{2}\left[\dfrac{1}{s+a}\right]\\
        &amp;=\dfrac{1}{2}\left[\dfrac{2s}{(s-a)(s+a)}\right]\left[\dfrac{1}{s+a}\right]= \dfrac{s}{s^2-a^2}\end{aligned}\]
    </li>
    <li>You should obtain \({\cal L}\{u(t-a)\}= \displaystyle \int_a^\infty \displaystyle \text{e}^{-st} \, \mathrm{d}t \) (note the lower limit is \(a\)) since:
        <p>\({\cal L}\{u(t-a)\} = \displaystyle \int_0^\infty \displaystyle \text{e}^{-st}u(t-a) \, dt = \displaystyle \int_0^a \text{e}^{-st}u(t-a) \, dt + \displaystyle \int_a^\infty \displaystyle \text{e}^{-st}u(t-a) \, dt\) </p>
        <p>In the first integral \(0 &lt; t &lt; a\) and so \((t − a) &lt; 0\), therefore \(u(t − a) = 0\).</p>
        <p>In the second integral \(a &lt; t &lt; \infty\) and so \((t − a) &gt; 0\), therefore \(u(t − a) = 1\). Hence </p>
        <p>\(\qquad {\cal L}\{u(t-a)\} = 0 + \displaystyle \int_a^\infty \text{e}^{-st} \, dt \)
        </p>
    </li>
    <li>\(\qquad {\cal L}\{u(t-a)\} = \displaystyle \int_0^\infty \text{e}^{-st} \, dt = \left[\dfrac{\text{e}^{-st}}{(-s)} \right]_a^\infty = \dfrac{\text{e}^{-sa}}{s} \)</li>
</ol>]]></text>
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    <stackversion>
      <text>2023010400</text>
    </stackversion>
    <questionvariables>
      <text>ta1: 1/2*(e^(a*t)-e^(-a*t));
ta2: a/(s^2-a^2);
ta3: s/(s^2-a^2);
ta4: inf;
ta5: a;
ta6: e^(-s*t);
ta7: e^(-s*a)/s;</text>
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      <text><![CDATA[<span style="font-size: 1.5em; color:green;"><i class="fa fa-check"></i></span> Correct answer, well done.]]></text>
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      <text><![CDATA[<span style="font-size: 1.5em; color:red;"><i class="fa fa-times"></i></span> Incorrect answer.]]></text>
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