Question

Determine Determine a transformada de Laplace inversa da função
S+3
F (s)= ------------
S^2 - s - 2

Answer 1

A função é $F(s)=\dfrac{s+3}{s^2-s-2}$. Tentemos separar a fração em soma de fraçoes mais simples

$F(s)=\dfrac{s+3}{s^2-s-2}\\ \\ \\</p><p>F(s)=\dfrac{s+3}{(s-2)(s+1)}\\ \\ \\</p><p>\text{Seja }~~\dfrac{s+3}{(s-2)(s+1)} = \dfrac{A}{s-2}+\dfrac{B}{s+1}\\ \\ \\</p><p>s+3=(A+B)s+(A-2B)\to A+B =1 \wedge A-2B=3\\ \\ \\</p><p>A=\dfrac{5}{3}\wedge B=-\dfrac{2}{3}\\ \\ \\</p><p>\text{Ent\~ao }F(s)=\dfrac{5/3}{s-2}-\dfrac{2/3}{s+1}$

Agora vamos encontrar a transformada de Laplace inversa

$\mathcal{L}^{-1}\{F(s)\}_t=\mathcal{L}^{-1}\left\{\dfrac{5/3}{s-2}-\dfrac{2/3}{s+1}\right\}\\ \\ \\</p><p>\mathcal{L}^{-1}\{F(s)\}_t=\mathcal{L}^{-1}\left\{\dfrac{5/3}{s-2}\right\}-\mathcal{L}^{-1}\left\{\dfrac{2/3}{s+1}\right\}\\ \\ \\</p><p>\mathcal{L}^{-1}\{F(s)\}_t=\dfrac{5}{3}\mathcal{L}^{-1}\left\{\dfrac{1}{s-2}\right\}-\dfrac{2}{3}\mathcal{L}^{-1}\left\{\dfrac{1}{s+1}\right\}\\ \\ \\</p><p>\boxed{\boxed{\mathcal{L}^{-1}\{F(s)\}_t=\dfrac{5}{3}e^{2t}-\dfrac{2}{3}e^{-t}}}$