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Mod note: Please don't tinker with SIZE tags. Things are perfectly readable without them.
Find x(t) = L-1[(4e-4s - 3)/(s2 + 6s + 25)].
L(x(t)) = ∞∫∞x(t)e-stdt
L-1(x(s)) = (1/2π)(σ - ∞j)∫(σ + ∞j)[x(s)est]ds, "But you want to avoid this integral."
Laplace Table:
f(t), t > 0 ::::::::::::::::::::::::::::::::::::::::::::::: F(s)
_______________________________________
σ(t) :::::::::::::::::::::::::::::::::::::::::::::::::::::: 1
u(t) :::::::::::::::::::::::::::::::::::::::::::::::::::::: (1/s)
t :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: (1/s2)
e-at :::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s + a)
eat :::::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s - a)
te-at ::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s+a)2
sin(Bt) ::::::::::::::::::::::::::::::::::::::::::::::: B/(s2 + B2)
cos(Bt) ::::::::::::::::::::::::::::::::::::::::::::::: s/(s2 + B2)
_______________________________________________________
Properties:
L[tf(t)] = -dF(s)/ds
L[x(at)] = (1/|a|) X (s/a), (unsure what this means at the moment though)
L[0∫tf(T)dT] = F(s)/s
Other properties
It seems this problem is asking for doing the inverse Laplace transform on the equation
(4e-4s - 3)/(s2 + 6s + 25) but it seems I can't just do the integral. From what I hear "Laplace tables" are pretty normal to use, but I seem to be missing how they are supposed to help here, do they apply to Inverse Laplace transforms too?
Homework Statement
Find x(t) = L-1[(4e-4s - 3)/(s2 + 6s + 25)].
Homework Equations
L(x(t)) = ∞∫∞x(t)e-stdt
L-1(x(s)) = (1/2π)(σ - ∞j)∫(σ + ∞j)[x(s)est]ds, "But you want to avoid this integral."
Laplace Table:
f(t), t > 0 ::::::::::::::::::::::::::::::::::::::::::::::: F(s)
_______________________________________
σ(t) :::::::::::::::::::::::::::::::::::::::::::::::::::::: 1
u(t) :::::::::::::::::::::::::::::::::::::::::::::::::::::: (1/s)
t :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: (1/s2)
e-at :::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s + a)
eat :::::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s - a)
te-at ::::::::::::::::::::::::::::::::::::::::::::::::::::: 1/(s+a)2
sin(Bt) ::::::::::::::::::::::::::::::::::::::::::::::: B/(s2 + B2)
cos(Bt) ::::::::::::::::::::::::::::::::::::::::::::::: s/(s2 + B2)
_______________________________________________________
Properties:
L[tf(t)] = -dF(s)/ds
L[x(at)] = (1/|a|) X (s/a), (unsure what this means at the moment though)
L[0∫tf(T)dT] = F(s)/s
Other properties
The Attempt at a Solution
It seems this problem is asking for doing the inverse Laplace transform on the equation
(4e-4s - 3)/(s2 + 6s + 25) but it seems I can't just do the integral. From what I hear "Laplace tables" are pretty normal to use, but I seem to be missing how they are supposed to help here, do they apply to Inverse Laplace transforms too?
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