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: $\frac{{d}}{{{d}{t}}}{u}_{{c}}{\left({t}\right)}=\delta{\left({t}-{c}\right)}$.