Dirac Delta Function

The Heaviside function represents switches from one value to another at some point, but what if we need an instant change to a very big value? This is what the Dirac delta function means.

This is a very strange function. It is zero everywhere except one point, but anyway the integral containing this one point equals $$${1}$$$. The Dirac delta is not a function in the traditional sense. It is instead an example of something called a generalized function or distribution. However, it is very good for modeling instant big changes.

Note that the first and second properties are true for any interval that contains $$${c}$$$ and on which $$${c}$$$ is not an endpoint: $$${\int_{{{c}-\tau}}^{{{c}+\tau}}}\delta{\left({t}-{c}\right)}{d}{t}={1}$$$, and $$${\int_{{{c}-\tau}}^{{{c}+\tau}}}{f{{\left({t}\right)}}}\delta{\left({t}-{c}\right)}{d}{t}={f{{\left({c}\right)}}}$$$, where $$$\tau>{0}$$$.

Now, let's see what the Laplace transform of the Dirac function is (this can be calculated easily using the second property):

$$${L}{\left(\delta{\left({t}-{c}\right)}\right)}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}\delta{\left({t}-{c}\right)}{d}{t}={{e}}^{{-{c}{t}}}$$$, provided $$${c}>{0}$$$.

Let's see another fact about the Dirac function. For this, use the fact that $$${\int_{{-{\infty}}}^{{t}}}\delta{\left(\tau-{c}\right)}{d}\tau={\left\{\begin{array}{c}{1}{\quad\text{if}\quad}{t}>{c}\\{0}{\quad\text{if}\quad}{t}<{c}\\ \end{array}\right.}$$$ and note that this is exactly the definition of the Heaviside function:

$$${\int_{{-{\infty}}}^{{t}}}\delta{\left(\tau-{c}\right)}{d}\tau={u}_{{c}}{\left({t}\right)}$$$

Now, using the fundamental theorem of calculus: $$$\frac{{d}}{{{d}{t}}}{u}_{{c}}{\left({t}\right)}=\delta{\left({t}-{c}\right)}$$$.