### 3 The delta function (or impulse function)

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.

Figure 14

This can be represented formally using step functions; it switches on at $t=d$ and switches off at $t=d+a$ and has amplitude $b$ :

$\phantom{\rule{2em}{0ex}}f\left(t\right)=b\left[u\left(t-d\right)-u\left(t-\left\{d+a\right\}\right)\right]$

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.

Figure 15

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 $a\to 0$ , the ‘function’ becomes a ‘spike’ at $t=d$ . The resulting function is called a delta function (or impulse function ) and denoted by $\delta \left(t-d\right)$ . This notation is used because, in a very obvious sense, the delta function described here is ‘located’ at $t=d$ . Thus the delta function $\delta \left(t-1\right)$ is ‘located’ at $t=1$ whilst the delta function $\delta \left(t\right)$ is ‘located’ at $t=0$ .

If we were defining an ordinary function we would write

$\phantom{\rule{2em}{0ex}}\delta \left(t-d\right)=\underset{a\to 0}{lim}\frac{1}{a}\left[u\left(t-d\right)-u\left(t-\left\{d+a\right\}\right)\right]$

However, this limit does not exist. The important property of the delta function relates to its integral:

$\begin{array}{llll}\hfill {\int }_{-\infty }^{\infty }\delta \left(t-d\right)\phantom{\rule{0.3em}{0ex}}dt=\underset{a\to 0}{lim}{\int }_{-\infty }^{\infty }\frac{1}{a}\left[u\left(t-d\right)-u\left(t-\left\{d+a\right\}\right)\right]\phantom{\rule{0.3em}{0ex}}dt& =\underset{a\to 0}{lim}{\int }_{d}^{d+a}\frac{1}{a}\phantom{\rule{0.3em}{0ex}}dt\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\underset{a\to 0}{lim}\left[\frac{d+a}{a}-\frac{d}{a}\right]=1\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

which is what we expect since the area under each of the limiting curves is equal to $1$ .

A more technical discussion obtains the more general result:

##### Key Point 11

Sifting Property of the Delta Function

${\int }_{-\infty }^{\infty }f\left(t\right)\delta \left(t-d\right)\phantom{\rule{0.3em}{0ex}}dt=f\left(d\right)$

This is called the sifting property of the delta function as it sifts out the value $f\left(d\right)$ from the function $f\left(t\right)$ . 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

$\phantom{\rule{2em}{0ex}}{\int }_{\alpha }^{\beta }f\left(t\right)\delta \left(t-d\right)\phantom{\rule{0.3em}{0ex}}dt=f\left(d\right)$

Thus, as long as the delta function is ‘located’ within the range of the integral the sifting property holds. For example,

$\phantom{\rule{2em}{0ex}}{\int }_{1}^{2}sint\phantom{\rule{1em}{0ex}}\delta \left(t-1.1\right)\phantom{\rule{0.3em}{0ex}}dt=sin1.1=0.8112\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}{\int }_{0}^{\infty }{\text{e}}^{-t}\delta \left(t-1\right)\phantom{\rule{0.3em}{0ex}}dt={\text{e}}^{-1}=0.3679$

Write expressions for delta functions located at $t=-1.7$ and at $t=2.3$

$\delta \left(t+1.7\right)$ and $\delta \left(t-2.3\right)$

Evaluate the integral ${\int }_{-1}^{3}\left(sint\phantom{\rule{1em}{0ex}}\delta \left(t+2\right)-cost\phantom{\rule{1em}{0ex}}\delta \left(t\right)\right)\phantom{\rule{0.3em}{0ex}}dt$

You should obtain the value $-1$ since the first delta function, $\delta \left(t+2\right)$ , is located outside the range of integration and thus

$\phantom{\rule{2em}{0ex}}{\int }_{-1}^{3}\left(sint\phantom{\rule{1em}{0ex}}\delta \left(t+2\right)-cost\phantom{\rule{1em}{0ex}}\delta \left(t\right)\right)\phantom{\rule{0.3em}{0ex}}dt={\int }_{-1}^{3}-cost\phantom{\rule{1em}{0ex}}\delta \left(t\right)\phantom{\rule{0.3em}{0ex}}dt=-cos0=-1$

#### 3.1 The Laplace transform of the delta function

Here we consider $\mathsc{L}\left\{\delta \left(t-d\right)\right\}$ . From the definition of the Laplace transform:

$\phantom{\rule{2em}{0ex}}\mathsc{L}\left\{\delta \left(t-d\right)\right\}={\int }_{0}^{\infty }{\text{e}}^{-st}\delta \left(t-d\right)\phantom{\rule{0.3em}{0ex}}dt={\text{e}}^{-sd}$

by the sifting property of the delta function. Thus

##### Key Point 12

Laplace Transform of the Sifting Function

##### Exercise

Find the Laplace transforms of   $3\delta \left(t-3\right).$

$3{e}^{-3s}$