The Dirac Delta Function

Background

The Dirac delta function, also called the unit impulse function, is a bit weird. It’s defined by two properties. First, its value is $0$ everywhere except at the impulse point.

\[\delta(t) = \begin{cases} 1 &\quad\text{if } t = 0 \\ 0 &\quad\text{otherwise} \end{cases} \qquad\qquad \delta(t-c) = \begin{cases} 1 &\quad\text{if } t = c \\ 0 &\quad\text{otherwise} \end{cases}\]

Second, So long as the impulse point is within the bounds of integration, the integral will be $1$, otherwise it is $0$. More generally, we write

\[\int_{a}^{b} \delta(t) f(t) \ dt = \begin{cases} f(0) &\quad\text{if } a \leq 0 \leq b \text{ and } a \neq b \\ -f(0) &\quad\text{if } b \leq 0 \leq a \text{ and } a \neq b \\ 0 &\quad\text{otherwise} \end{cases} \qquad\qquad \int_{a}^{b} \delta(t-c) f(t) \ dt = \begin{cases} f(c) &\quad\text{if } a \leq c \leq b \text{ and } a \neq b \\ -f(c) &\quad\text{if } b \leq c \leq a \text{ and } a \neq b \\ 0 &\quad\text{otherwise} \end{cases}\]

You can think about this function defining a rectangle with an infinitesimally small width. Thus, the area under $\delta(t-c) f(t)$ will just be the height of the function at $c$.

The Laplace Transform Proof

Given a function $f(t)$ with Laplace Transform $F(s)$ and any constant $c, d \in \mathbb{R}$.

\[\mathcal{L}\{ \delta(t-c) f(t-d) \} = \int_{0}^{\infty} \delta(t-c) f(t-d) e^{-st} \ dt = \begin{cases} e^{-cs} f(c-d) &\quad\text{if } c \geq 0 \\ 0 &\quad\text{otherwise} \end{cases}\]

Laplace Transforms Series