Ensino Superior ⇒ Transformada de Laplace Tópico resolvido

[fmatheus]

Encontre f(t):

[tex3]F(s)=\frac{1}{s(s^2 - 2s + 5)}[/tex3]

[poisedom]

[tex3]\frac{1}{s(s^2 - 2s + 5)}=\frac{A}{s}+\frac{Bs+C}{s^2 - 2s + 5}=\frac{As^2 - 2As + 5A+Bs^2+Cs}{s(s^2 - 2s + 5)}=\frac{(A+B)s^2+(-2A+C)s+5A}{s(s^2 - 2s + 5)}[/tex3]
temos então
[tex3]\begin{cases}
A+B=0 \\
-2A+C=0 \\
5A=1
\end{cases}[/tex3] [tex3]\rightarrow A=\frac{1}{5}[/tex3], [tex3]B=-\frac{1}{5}[/tex3] e [tex3]C=\frac{2}{5}[/tex3]

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[tex3]\frac{1}{s(s^2 - 2s + 5)}=\frac{ 1/5}{s}+\frac{(-1/5)s}{s^2 - 2s + 5}+\frac{2/5}{s^2 - 2s + 5}[/tex3]
[tex3]\frac{1}{s(s^2 - 2s + 5)}=\frac{ 1/5}{s}+\frac{(-1/5)s}{(s-1)^2 + 4}+\frac{2/5}{(s-1)^2 + 4}[/tex3]
[tex3]\frac{1}{s(s^2 - 2s + 5)}=\frac{1}{5}\cdot\frac{ 1}{S}+\frac{-1}{5}\cdot\frac{(s-1)+1}{(s-1)^2 + 4}+\frac{2}{5}\cdot\frac{1}{(s-1)^2 + 4}[/tex3]
[tex3]\frac{1}{s(s^2 - 2s + 5)}=\frac{1}{5}\cdot\frac{ 1}{S}+\frac{-1}{5}\cdot\frac{(s-1)}{(s-1)^2 + 4}+\frac{-1}{5}\cdot\frac{1}{(s-1)^2 + 4}+\frac{2}{5}\cdot\frac{1}{(s-1)^2 + 4}[/tex3]
[tex3]\frac{1}{s(s^2 - 2s + 5)}=\frac{1}{5}\cdot\frac{ 1}{S}+\frac{1}{5}\cdot\frac{(s-1)}{(s-1)^2 + 2^2}+\frac{-1}{5}\cdot\frac{1}{(s-1)^2 + 2^2}[/tex3]
[tex3]\frac{1}{s(s^2 - 2s + 5)}=\frac{1}{5}\cdot\frac{ 1}{S}+\frac{1}{5}\cdot\frac{(s-1)}{(s-1)^2 + 2^2}+\frac{-1}{2\cdot5}\cdot\frac{2}{(s-1)^2 + 2^2}[/tex3]
[tex3]\frac{1}{s(s^2 - 2s + 5)}=\frac{1}{5}\cdot\frac{ 1}{S}+\frac{1}{5}\cdot\frac{(s-1)}{(s-1)^2 + 2^2}+\frac{-1}{10}\cdot\frac{2}{(s-1)^2 + 2^2}[/tex3]

então [tex3]f(t)=\frac{1}{5}\cdot 1+\frac{1}{5}e^t\cdot cos(2t)-\frac{1}{10}\cdot e^t\cdot sen(2t)[/tex3]
[tex3]f(t)=\frac{1}{5}+\frac{e^t\cdot cos(2t)}{5}-\frac{e^t\cdot sen(2t)}{10}[/tex3]
[tex3]f(t)=\frac{2+2e^t\cdot cos(2t)-e^t\cdot sen(2t)}{10}[/tex3]