Dirac Delta Function

Represents a very strong force given over a very short period of time

$$\delta (t)=\{\begin{array}{ll}0& t\ne 0\\ \infty & t=0\end{array}$$

It having a value of infinity makes it technically a distribution not a function, but we can still use it as a function

Properties

$${\int }_{-\infty }^{\infty }\delta (t)dt=1$$ $${\int }_{-\infty }^{\infty }f(t)\delta (t)dt=f(0)$$

can be shifted by a using $\delta (t-a)$

$${\int }_{-\infty }^{\infty }f(t)\delta (t-a)dt=f(a)$$

Laplace Transform

• $\mathcal{L}\{\delta (t-a)\}=e^{-as}\text{ for }a>0$
• $\mathcal{L}\{f(t)\delta (t-a)\}=f(a)e^{-as}\text{ for }a>0$
• This means we will get an Unit Step Function in our inverse transform

Relationship with Unit Step Function

$${\int }_{-\infty }^t\delta (x-a)dx=\{\begin{array}{ll}0& t<a\\ 1& t>a\end{array}=u(t-a)$$ $$\delta (t-a)=u^{\prime }(t-a)$$

The Dirac Delta Function is the weak derivative of the Unit Step Function