<?xml version="1.0" encoding="UTF-8"?>
<quiz>
  <question type="description">
    <name>
      <text>20.1.1.4 Casual Functions</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h5>Causal functions<br></h5>

<p>As we have seen above, the Laplace transform involves an integral with limits \(t=0\) and \(t=\infty\). Because of this, the nature of the function being transformed, \(f(t)\), when \(t\) is negative is of no importance. In order to emphasize this we
    shall only consider so-called <strong>causal functions</strong> all of which take the value \(0\) when \(t&lt;0.\)</p>
<p>The simplest causal function is the <strong>Heaviside</strong> or unit step function denoted by \(u(t)\) and defined by:</p>
<p>\(\qquad u(t)= \begin{cases} 1&amp; \text{if}\quad t\ge 0\\ 0&amp; \text{if}\quad t&lt;0 \end{cases}\)</p>
<p>with graph as in the following figure.</p>

<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/figout2.svg" alt="Graph of Heaviside function" width="292" height="109" class="img-responsive atto_image_button_text-bottom">
</figure>

<p>Similarly we can consider other 'step-functions'. For example, from the above definition we deduce</p>

<p>\( u(t-3)= \begin{cases} 1&amp; \text{if}\quad t-3\ge 0\\ 0&amp; \text{if}\quad t-3&lt;0 \end{cases}\)</p>
<p>or rearranging the inequalities:</p>
<p>\( u(t-3) = \begin{cases} 1&amp; \text{if}\quad t\ge 3\\ 0&amp; \text{if}\quad t&lt;3 \end{cases}\)</p>
<p>with graph as in the following Figure:</p>

<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/figout3.svg" alt="Graph of u(t-3)" width="384" height="113" class="img-responsive atto_image_button_text-bottom">
</figure>

<p>The step function has a useful property: multiplying an ordinary function \(f(t)\) by the step function \(u(t)\) changes it into a causal function; e.g. if \(f(t)=\sin t\) then \( \sin t.u(t)\) is causal. This is illustrated in the change bellow:</p>

<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/figout4.svg" alt="Graph of the sine function" width="356" height="107" class="img-responsive atto_image_button_text-bottom">
</figure>

<p><br></p>
<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/figout5.svg" alt="Graph of the function h(t) = sin t.u(t)" width="356" height="107" class="img-responsive atto_image_button_text-bottom">
</figure>

<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <p></p>
    <center><strong>Causal Functions</strong></center>
    <p></p>
    <p>If \(u(t)\) is the unit step function and \(f(t)\) is any function then</p>
    <p></p>
    <center>\(f(t)u(t)\) is a causal function</center>
    <p></p>
</div>

<p>The step function can be used to 'switch on' functions at other values of \(t\) (which we will normally interpret as time). For example \(u(t-1)\) has the value 1 if \(t\ge 1\) and \(0\) otherwise so that \(\sin t.u(t-1)\) is described by the (solid)
    curve in the following figure:</p>

<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/figout6.svg" alt="Graph of sin t.u(t−1)" width="389" height="144" class="img-responsive atto_image_button_text-bottom">
</figure>

<p>The step function can also be used to 'switch-off' signals. For example, the step function \(u(t-1)-u(t-3)\) in Figure 7 has the effect on \(f(t)\) such that \(f(t)\left[u(t-1)-u(t-3)\right]\) (described by the solid curve in Figure 8) switches on at
    \(t=1\) (because then \(u(t-1)-u(t-3)\) takes the value 1), remains 'on' for \(1\le t\le 3\), and then switches 'off' when \(t&gt;3\) (because then \(u(t-1)-u(t-3)=1-1=0\)).</p>

<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/figout7a.svg" alt="Graph of u(t − 1) − u(t − 3)" width="491" height="132" class="img-responsive atto_image_button_text-bottom">
</figure>

<p><br></p>
<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/figout8.svg" alt="Graph of f(t)[u(t − 1) − u(t − 3)]" width="491" height="133" class="img-responsive atto_image_button_text-bottom">
</figure>

<p>If we have an expression \(f(t − a)u(t − a)\) then this is the function \(f(t)\) translated along the \(t\)-axis through a time a. For example \(\sin(t − 2).u(t − 2)\) is simply the causal sine curve \(\sin t.u(t)\) shifted to the right by two units as
    described in the following Figure.</p>

<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/figout7.svg?time=1654638679293" alt="Graph of sin(t − 2).u(t − 2)" width="488" height="184" class="img-responsive atto_image_button_text-bottom">
</figure>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Sketch the curve \(f(t)=e^t (u(t-1)-u(t-2)).\)</p>
<h4 class="HELM_solution">Solution</h4>
<p>You should obtain</p>
<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/figout9.svg" alt="The graph of the above function, which is a curve between x=1 and x=2" width="299" height="168" class="img-responsive atto_image_button_text-bottom">
</figure>

<p>This is botained since, if \(t &lt; 1\) then \(t − 1 &lt; 0\) and \(t − 2 &lt; 0\) and so</p>
<p>\( \quad u(t − 1) = 0, \quad u(t − 2) = 0 \quad \) leading to \(f(t) = 0.\)</p>
<p>Also if \(1 &lt; t &lt; 2\) then \(t − 1 &gt; 0\) and \(t − 2 &lt; 0\) so</p>
<p>\( \quad u(t − 1) = 1 \quad \) and \(\quad u(t − 2) = 0 \quad\) implying \(f(t) = e^t \quad\)for this range of \(t-\)values.</p>
<p>Finally if \(t &gt; 2\) then \(t − 1 &gt; 0\) and \(t − 2 &gt; 0\) and so</p>
<p>\(\quad u(t − 1) = 1, \quad u(t − 2) = 1 \quad \) giving \( \quad f(t) = e^t (1 − 1) = 0.\)</p>]]></text>
<file name="figout2.svg" path="/" encoding="base64">[encoded data, 13576 characters]</file>
<file name="figout3.svg" path="/" encoding="base64">[encoded data, 18744 characters]</file>
<file name="figout4.svg" path="/" encoding="base64">[encoded data, 56248 characters]</file>
<file name="figout5.svg" path="/" encoding="base64">[encoded data, 45972 characters]</file>
<file name="figout6.svg" path="/" encoding="base64">[encoded data, 22764 characters]</file>
<file name="figout7.svg" path="/" encoding="base64">[encoded data, 36552 characters]</file>
<file name="figout7a.svg" path="/" encoding="base64">[encoded data, 20508 characters]</file>
<file name="figout8.svg" path="/" encoding="base64">[encoded data, 44460 characters]</file>
<file name="figout9.svg" path="/" encoding="base64">[encoded data, 27836 characters]</file>
    </questiontext>
    <generalfeedback format="html">
      <text/>
    </generalfeedback>
    <defaultgrade>0</defaultgrade>
    <penalty>0</penalty>
    <hidden>0</hidden>
    <idnumber/>
  </question>
</quiz>
