Scaling property - laplace-transform.com

1. What do we mean by scaling?

If $a\in\mathbb{R},a\neq0$ , then we call $f(at)$ the dilation of $f$ by $a$ .

For example, if $f(t)=\sin(t)$ then

$f(2t)=\sin(2t)$

is the dilation of $\sin(t)$ by 2.

2. Laplace transform of a scaled (dilated) function

Proof:

By definition, we have:

$\displaystyle\mathscr{L}\{f(at)\}=\int_{0}^{\infty}e^{-st}f(at)dt$

Changing the variable as follows

$u=at\Rightarrow du=adt$

then

$\displaystyle t=\frac{u}{a},\quad dt=\frac{du}{a}$

and therefore

$\begin{aligned} \mathscr{L}\{f(at)\} & =\int_{0}^{\infty}e^{-s\frac{u}{a}}f(u)\frac{du}{a} \\ & =\frac{1}{a}\int_{0}^{\infty}e^{-s\frac{u}{a}}f(u)du \end{aligned}$

Thus

$\displaystyle \mathscr{L}\{f(at)\}=\frac{1}{a}F\left(\frac{s}{a}\right)$

3. Illustrative Example of laplace transform of sin(2t)

For $f(t)=\sin(t)$ , we know

$\displaystyle\mathscr{L}\{\sin(t)\}=\frac{1}{s^{2}+1^{2}}=\frac{1}{s^{2}+1}=F(s)$

Hence

$\displaystyle \mathscr{L}\{\sin(2t)\}=\frac{1}{2}F\left(\frac{s}{2}\right)$

i.e.,

$\begin{aligned} \mathscr{L}\{\sin(2t)\} & =\frac{1}{2}\cdot\frac{1}{\left(\dfrac{s}{2}\right)^{2}+1} \\ & =\frac{1}{2}\cdot\frac{1}{\dfrac{s^{2}}{4}+1} \\ & =\frac{1}{2}\cdot\frac{1}{\dfrac{s^{2}+4}{4}} \\ & =\frac{1}{\dfrac{s^{2}+4}{2}} \end{aligned}$

$\displaystyle \mathscr{L}\{\sin(2t)\}=\frac{2}{s^{2}+4}$

We reached the end of this short lesson about the Laplace transform of dilated functions. For a full list of Laplace transform properties, check this post!