The Laplace transform of delta function is equal to L{δ(t)} = 1. Here we find the Laplace of Dirac delta function, that is, unit impulse function.

The Dirac delta function is also known as the unit impulse function. Its Laplace transform formula is given by

L{δ(t)} = 1. |

## Laplace Transform of Dirac Delta Function

Recall that the Dirac delta function is defined by

$\delta(t) = \begin{cases} 0, & \text{if } t\neq 0 \\ \infty, & \text{if } t=0. \end{cases}$

It satisfies the following property

$\int_{-\infty}^\infty$ δ(t) f(t) dt = f(0)

Put f(t)=e^{-st}, so that we have

$\int_{-\infty}^\infty$ δ(t) e^{-st} dt = e^{-s ⋅0}

⇒ $\int_{-\infty}^0$ δ(t) e^{-st} dt + $\int_{0}^\infty$ δ(t) e^{-st} dt = 1

⇒ 0 + $\int_{0}^\infty$ δ(t) e^{-st} dt = 1, by the above definition of δ(t).

⇒ L{δ(t)} = 1, by the definition of Laplace transforms of δ(t).

So the Laplace transform of delta function, δ(t), is equal to 1.

**Summary:**

L{δ(t)} = 1. |

Have You Read These?

Laplace transform of unit step function

## FAQs

**Q1: What is the Laplace transform of delta function δ(t)?**

Answer: The Laplace transform of the delta function is equal to 1.

**Q2: What is L{δ(t)}?**

Answer: L{δ(t)} = 1.