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      <text>20.3.4.1 The delta function</text>
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      <text><![CDATA[<h2>
    The delta function (or impulse function)
</h2>
<p>There is often a need for considering the effect on a system (modelled by a differential equation) by a forcing function which acts for a very short time interval. For example, how does the current in a circuit behave if the voltage is switched on and
    then very shortly afterwards switched off? How does a cantilevered beam vibrate if it is hit with a hammer (providing a force which acts over a very short time interval)? Both of these engineering ‘systems’ can be modelled by a differential equation.
    There are many ways the ‘kick’ or ‘impulse’ to the system can be modelled. The function we have in mind could have the graphical representation (when \(a\) is small) shown in Figure 14.</p><center><img src="@@PLUGINFILE@@/20-3-fig-14-step_function.svg" alt="Graph of a step function that 'switches' on at x=d and off at x=d+a, with amplitude b." width="588" height="159" class="img-responsive atto_image_button_text-bottom"></center><p></p>
<p>This can be represented formally using step functions; it switches on at \(t=d\) and switches off at \(t=d+a\) and has amplitude \(b\): \[\qquad\displaystyle f(t)=b[u(t-d)-u(t-\{d+a\})]\]</p>
<p>The effect on the system is related to the area under the curve rather than just the amplitude \(b\). Our aim is to reduce the time interval over which the forcing function acts (i.e. reduce \(a\)) whilst at the same time keeping the total effect (i.e.
    the area under the curve) a constant. To do this we shall take \(b=1/a\) so that the area is always equal to 1. Reducing the value of \(a\) then gives the sequence of inputs shown in Figure 15.&nbsp;</p><center><img src="@@PLUGINFILE@@/20-3-fig-15-limit_delta_function.svg" alt="Graph showing step function of width a and height 1/a, for decreasing values of a." width="463" height="293" class="img-responsive atto_image_button_text-bottom"></center><p></p>
<p>As the value of \(a\) decreases the height of the rectangle increases (to ensure the value of the area under the curve is fixed at value 1) until, in the limit as \(a\rightarrow 0\), the 'function' becomes a 'spike’ at \(t=d\). The resulting function
    is called a <strong>delta function</strong> (or <strong>impulse function</strong>) and denoted by \(\delta (t-d)\). This notation is used because, in a very obvious sense, the delta function described here is ‘located’ at \(t=d\). Thus the delta function
    \(\delta (t-1)\) is ‘located’ at \(t=1\) whilst the delta function \(\delta (t)\) is 'located’ at \(t=0\).</p>
<p>If we were defining an ordinary function we would write \(\qquad\displaystyle \delta (t-d)=\lim_{a\rightarrow 0}{1\over a}[u(t-d)-u(t-\{d+a\})]\)</p>
<p>However, this limit does not exist. The important property of the delta function relates to its integral: \[\begin{align*} \int_{-\infty}^{\infty}\delta (t-d)\,dt &amp;=\lim_{a\rightarrow0}\int_{-\infty}^{\infty}{1\over a}[u(t-d)-u(t-\{d+a\})]\,dt \\
    &amp;=\lim_{a\rightarrow0}\int_d^{d+a}{1\over a}\,dt\\ &amp;=\lim_{a\rightarrow0}\biggl[{d+a\over a}-{d\over a}\biggr]=1\qquad \end{align*}\] which is what we expect since the area under each of the limiting curves is equal to \(1\).</p>
<p>A more technical discussion obtains the more general result:</p>
<div class="HELM_keypoint">
    <h4>Key Point</h4>
    <p></p>
    <center><strong>Sifting Property of the Delta Function</strong></center>
    <p></p>
    <p>\[\int_{-\infty}^{\infty}f(t)\delta (t-d)\,dt=f(d)\]</p>
</div>
<br>
<p>This is called the <strong>sifting property</strong> of the delta function as it sifts out the value \(f(d)\) from the function \(f(t)\). Although the integral here ranges from \(t=-\infty\) to \(t=+\infty\) in fact the same result is obtained for any range
    if the range of the integral includes the point \(t=d\). That is, if \(\alpha\le d\le\beta\) then \[\qquad\int_{\alpha}^{\beta}f(t)\delta (t-d)\,dt=f(d)\]</p>
<p>Thus, as long as the delta function is 'located' within the range of the integral the sifting property holds. For example, \[\qquad \int_1^2\sin t\ \delta (t-1.1)\,dt=\sin 1.1=0.8912\qquad\qquad \int_0^{\infty}\text{e}^{-t}\delta (t-1)\,dt=\text{e}^{-1}=0.3679
    \]
</p>]]></text>
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<file name="20-3-fig-15-limit_delta_function.svg" path="/" encoding="base64">[encoded data, 17824 characters]</file>
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