Question

determine a transformada de Laplace inversa de f (s)= 1/(s-1)(s+3)(s+5)

Answer 1

Vamos fazer uma quebra em frações parciais:

$F(s)=\dfrac{1}{(s-1)(s+3)(s+5)}\\\\ F(s)=\dfrac{A}{s-1}+\dfrac{B}{s+3}+\dfrac{C}{s+5}\\\\ F(s)=\dfrac{A(s+3)(s+5)+B(s-1)(s+5)+C(s-1)(s+3)}{(s-1)(s+3)(s+5)}\\\\ F(s)=\dfrac{A(s^2+8s+15)+B(s^2+4s-5)+C(s^2+2s-3)}{(s-1)(s+3)(s+5)}\\\\ F(s)=\dfrac{(A+B+C)s^2+(8A+4B+2C)s+(15A-5B-3C)}{(s-1)(s+3)(s+5)}$

$\Longrightarrow \begin{cases}A+B+C=0\to C=-A-B~~~(i)\\ 8A+4B+2C=0\to 4A+2B+C=0~~~(ii)\\ 15A-5B-3C=1~~~(iii)\end{cases}\\\\\\ (i)\to(ii):\\\\ 4A+2B+(-A-B)=0\to 3A+B=0\to \begin{matrix}B=-3A\\C=2A\end{matrix}~~~(iv)\\\\\\ (iv)\to (iii):\\\\ 15A-5\cdot(-3A)-3\cdot (2A)=1\\\\ 15A+15A-6A=1\\\\ 24A=1\to \boxed{A=1/24}~~~\boxed{B=-1/8}~~~\boxed{C=1/12}$

Voltando à expressão original, temos:

$F(s)=\dfrac{(1/24)}{s-1}+\dfrac{(-1/8)}{s+3}+\dfrac{(1/12)}{s+5}\\\\ F(s)=\dfrac{1}{24}\cdot\dfrac{1}{s-1}-\dfrac{1}{8}\cdot\dfrac{1}{s+3}+\dfrac{1}{12}\cdot\dfrac{1}{s+5}$

Aplicando a Inversa da Transformada de Laplace:

$\mathcal{L}^{-1}\{F(s)\}=\mathcal{L}^{-1}\left\{\dfrac{1}{24}\cdot\dfrac{1}{s-1}-\dfrac{1}{8}\cdot\dfrac{1}{s+3}+\dfrac{1}{12}\cdot\dfrac{1}{s+5}\right\}\\\\ \mathcal{L}\{F(s)\}=\mathcal{L}^{-1}\left\{\dfrac{1}{24}\cdot\dfrac{1}{s-1}\right\}-\mathcal{L}^{-1}\left\{\dfrac{1}{8}\cdot\dfrac{1}{s+3}\right\}+\mathcal{L}^{-1}\left\{\dfrac{1}{12}\cdot\dfrac{1}{s+5}\right\}$

$\mathcal{L}^{-1}\{F(s)\}=\dfrac{1}{24}\cdot\mathcal{L}^{-1}\left\{\dfrac{1}{s-1}\right\}-\dfrac{1}{8}\cdot\mathcal{L}^{-1}\left\{\dfrac{1}{s+3}\right\} +\dfrac{1}{12}\cdot\mathcal{L}^{-1}\left\{\dfrac{1}{s+5}\right\}\\\\ \mathcal{L}^{-1}\{F(s)\}=\dfrac{e^{-(-1)t}}{24}\cdot\mathcal{L}^{-1}\left\{\dfrac{1}{s}\right\}-\dfrac{e^{-3t}}{8}\cdot\mathcal{L}^{-1}\left\{\dfrac{1}{s}\right\}+\dfrac{e^{-5t}}{12}\cdot\mathcal{L}^{-1}\left\{\dfrac{1}{s}\right\}$

$\mathcal{L}^{-1}\{F(s)\}=\dfrac{e^{t}}{24}\cdot1-\dfrac{e^{-3t}}{8}\cdot1+\dfrac{e^{-5t}}{12}\cdot1\\\\ \boxed{\mathcal{L}^{-1}\{F(s)\}=\dfrac{e^{t}}{24}-\dfrac{e^{-3t}}{8}+\dfrac{e^{-5t}}{12}}$