Inverse Laplace Transform of 1 | Inverse Laplace of a constant

The inverse Laplace transform of 1 is the Dirac delta function δ(t). The inverse Laplace transform of a constant is equal to constant times δ(t). In this post, we will learn how to find the inverse Laplace transform of 1, and more generally of a constant.

Dirac delta function

How is Dirac delta function defined? The Dirac delta function is defined as follows:

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

Note that Dirac delta function is also known as the unit impulse function. It satisfies the following property:

$\int_{-\infty}^\infty$ δ(t-a) f(t) dt = f(a) for a≥0 …(I)

We will show that the inverse Laplace transform of 1 is the Dirac delta function.

What is the inverse Laplace transform of 1?

Answer: The inverse Laplace transform of 1 is δ(t) where δ(t) denotes the Dirac delta function.

Proof:

Recall, the definition of the inverse Laplace transform: Let F(s) be the Laplace transform of f(t), that is,

F(s) = L{f(t)} = $\int_0^\infty$ f(t) e^-st dt. …(II)

Then the inverse Laplace transform of F(s) is f(t). In other words,

L^-1{F(s)} = f(t).

Now, to find the inverse Laplace transform of 1, we will put f(t)=e^-st in the above Property (I) that Dirac delta function satisfies.

∴ We have that

$\int_{-\infty}^\infty$ δ(t-a) f(t) dt = f(a) = e^-as

⇒ $\int_{-\infty}^0$ δ(t-a) f(t) dt + $\int_{0}^\infty$ δ(t-a) f(t) dt= e^-as

⇒ $\int_{-\infty}^0$ 0⋅f(t) dt + $\int_{0}^\infty$ δ(t-a) f(t) dt= e^-as. Because, δ(t-a) = 0 for all t ∈ (0, ∞) as a≥0. It follows from the definition of the Dirac delta function.

⇒ 0 + $\int_{0}^\infty$ δ(t-a) f(t) dt= e^-as

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

Thus, the definition (II) of the Laplace transform implies that

L{δ(t-a)} = e^-as …(*)

Put a=0

∴ L{δ(t)} = e⁰ = 1.

This shows that the Laplace transform of δ(t) is 1. Hence,

L^-1{1} = δ(t).

Thus, we have proven that the Laplace transform of 1 is δ(t).

Find the inverse Laplace transform of 1. Summary: The inverse Laplace transform of 1 is the Dirac delta function δ(t).

What is the inverse Laplace transform of constants?

Answer: The inverse Laplace transform of a constant c is cδ(t).

Proof:

We have

L{c δ(t-a)}

= c L{δ(t-a)}

= c e^-as by (*)

Therefore, L{c δ(t-a)} = c e^-as.

Put a=0

∴ L{c δ(t)} = c e⁰ = c.

This shows that the Laplace transform of cδ(t) is c. Hence,

L^-1{c} = c δ(t).

Thus, the inverse Laplace transform of a constant is equal to constant times the Dirac delta function.

Find the inverse Laplace transform of c, where c is a constant. Summary: The inverse Laplace transform of c is c δ(t), where δ(t) is the Dirac delta function.

Also Read:

Laplace transform of 1

Laplace transform of constants

Question: Find the inverse Laplace transform of 5.

Solution:

From above we have that the inverse Laplace transform of c is cδ(t), that is L{c}=cδ(t). Putting c=5, we obtain the inverse Laplace transform of 5 is 5δ(t) where δ(t) is the unit impulse function.

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