Laplace transform

Problem 01
Find the Laplace transform of   $f(t) = \dfrac{\sin t}{t}$.
 

Solution 01
$\mathcal{L} (\sin t) = \dfrac{1}{s^2 + 1}$
 

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \int_s^\infty \dfrac{du}{u^2 + 1}$

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \big[ \arctan u \big]_s^\infty$

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \lim_{a \to \infty}\big[ \arctan u \big]_s^a$

Problem 03
Find the Laplace transform of   $f(t) = t^2 \cos 3t$.
 

Solution 03
$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s)$
 

$\mathcal{L} (\cos 3t) = \dfrac{s}{s^2 + 3^2}$

$\mathcal{L} (\cos 3t) = \dfrac{s}{s^2 + 9}$
 

$\mathcal{L} (t^2 \cos 3t) = (-1)^2 \dfrac{d^2}{ds^2} \left[ \dfrac{s}{s^2 + 9} \right]$

$\mathcal{L} (t^2 \cos 3t) = \dfrac{d}{ds} \left[ \dfrac{(s^2 + 9)(1) - s(2s)}{(s^2 + 9)^2} \right]$

Problem 02
Find the Laplace transform of   $f(t) = t \sin 3t$.
 

Solution 02
$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s)$
 

$\mathcal{L} (\sin 3t) = \dfrac{3}{s^2 + 3^2}$

$\mathcal{L} (\sin 3t) = \dfrac{3}{s^2 + 9}$
 

$\mathcal{L} (t \sin 3t) = (-1)^1 \dfrac{d}{ds} \left[ \dfrac{3}{s^2 + 9} \right]$

$\mathcal{L} (t \sin 3t) = -\left[ \dfrac{-3(2s)}{(s^2 + 9)^2} \right]$

$\mathcal{L} (t \sin 3t) = -\dfrac{-6s}{(s^2 + 9)^2}$

Problem 01
Find the Laplace transform of   $f(t) = t \cos 2t$.
 

Solution 01
$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s)$
 

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

$\mathcal{L} (\cos 2t) = \dfrac{s}{s^2 + 4}$
 

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

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

$\mathcal{L} (t \cos 2t) = -\dfrac{s^2 + 4 - 2s^2}{(s^2 + 4)^2}$

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

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