change of scale

Problem 03
Supposed that the Laplace transform of a certain function   $f(t)$   is   $\dfrac{s^2 - s + 1}{(2s + 1)^2 (s - 1)}$,   find the Laplace transform of   $f(2t)$.
 

Solution 03
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then by change of scale property,   $\mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a} F \left( \dfrac{s}{a} \right)$
 

$\mathcal{L} \left\{ f(t) \right\} = \dfrac{s^2 - s + 1}{(2s + 1)^2 (s - 1)}$

Problem 02
Given that   $\mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \left( \dfrac{1}{s} \right)$,   find   $\mathcal{L} \left( \dfrac{\sin 3t}{t} \right)$.
 

Solution 02
$\mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a} F \left( \dfrac{s}{a} \right)$
 

$\mathcal{L} \left( \dfrac{\sin 3t}{t} \right) = 3\mathcal{L} \left( \dfrac{\sin 3t}{3t} \right)$,   thus,   $a = 3$

Problem 01
Find the Laplace transform of   $f(t) = \cos 4t$   using the change of scale property.
 

Solution 01
$\mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a} F \left( \dfrac{s}{a} \right)$
 

$\mathcal{L} (\cos t) = \dfrac{s}{s^2 + 1}$
 

Thus,
$\mathcal{L} (\cos 4t) = \dfrac{1}{4} \left[ \dfrac{\dfrac{s}{4}}{\left( \dfrac{s}{4} \right)^2 + 1} \right]$

$\mathcal{L} (\cos 4t) = \dfrac{1}{4} \left[ \dfrac{\dfrac{s}{4}}{\dfrac{s^2}{16} + 1} \right]$