Take the Inverse Laplace Transform of the function.F(s)=\frac{1}{s(s^2+2s+2)}

foass77W 2021-09-07 Answered
Take the Inverse Laplace Transform of the function.
F(s)=1s(s2+2s+2)
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Expert Answer

Khribechy
Answered 2021-09-08 Author has 100 answers

F(s)=1s(s2+2s+2)=1s((s+1)2+(1)2)=1s1((s+1)2+(1)2)
F(s)=L{1}L{etsin(t)}
L1{F(s)}=f(t)=1(etsint)
f(t)=0teusin(u)1du=(ek2[1sin(u)1cos(u)])0t
=(e42[sin(u)+cos(u)])0t
=12[et(sin(t)+cos(t))1]
f(t)=12[1et(sin(t)+cos(t))]

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