## 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

then

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}

So

\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!