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 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.
FAQs
Answer: The inverse Laplace transform of 1 is the Dirac delta function δ(t).
Answer: The inverse Laplace transform of 2 is 2δ(t) where δ(t) is the unit impulse function.