The change of scale property of Laplace transforms is useful to find the Laplace of f(at) if the Laplace of f(t) is known. It states that if L{f(t)} = F(s) then

L{f(at)} = $\dfrac{1}{a} F(\dfrac{s}{a})$.

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## Statement of Change of Scale Property of Laplace

Let the Laplace of f(t) be F(s), that is, L{f(t)} = F(s). Then the Laplace of f(at) is given by the formula:

L{f(at)} = $\dfrac{1}{a} F(\dfrac{s}{a})$. |

## Proof of Change of Scale Property

By the definition of Laplace transforms, we have

L{f(at)} = $\int_0^\infty$ e^{-st} f(at) dt …(∗)

Let us make the following change of variable.

z=at.

So t=z/a.

Differentiating, dt =dz/a.

when t=0, z=0.

when t = ∞, z = ∞.

Thus, from (∗) we get that

L{f(at)} = $\int_0^\infty$ e^{-s(z/a)} f(z) dz/a

⇒ L{f(at)} = $\dfrac{1}{a} \int_0^\infty e^{-\frac{s}{a}z} f(z) dz$

⇒ L{f(at)} = $\dfrac{1}{a} F(\dfrac{s}{a})$.

This is the change of scale property of Laplace transforms, and we have proved it.

**Read Also: **

Laplace Transform: Definition, Table, Formulas, Properties

First shifting property of Laplace transforms

Second shifting property of Laplace transforms

**Question1:** Find the Laplace of sin2t.

**Answer:**

Let f(t)=sint.

So F(s)=L{sint} = $\dfrac{1}{s^2+1}$.

By change of scale property,

L{sin2t} = $\dfrac{1}{2} F(\dfrac{s}{2})$

= $\dfrac{1}{2} \dfrac{1}{(\frac{s}{2})^2+1}$

= $\dfrac{2}{s^2+4}$.

So the Laplace of sin2t is 2/(s^{2}+4), and this is proved by the change of scale property of Laplace transforms.

## FAQs

**Q1: What is the change of scale property in Laplace transforms?**

Answer: The change of scale property in Laplace transforms states that if L{f(t)} = F(s) then L{f(at)} = $\dfrac{1}{a} F(\dfrac{s}{a})$.