Change of Scale Property of Laplace Transforms

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

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/(s²+4), and this is proved by the change of scale property of Laplace transforms.

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