Finding the inverse Laplace transform

Rubik
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Homework Statement


Find the inverse Laplace transform of the following function
1/[s(s2 + 4)2]


Homework Equations


1/ (s2 + \omega2)2 = (1/ 2\omega3) (sin\omegat -\omegat cos\omegat)


The Attempt at a Solution


L-1 = (1/s) (1/ s2 +22)2
= (1/16) (sin 2t - 2t cos 2t) as the Laplace transform of (1/s) = 1

Is this what I am meant to do?
 
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The inverse Laplace transform of a product isn't the product of the inverse Laplace transforms in general. You can go for the convolution theorem (might be a mess), or use partial fractions to decompose that, which requires you to use a few other inverse identities once you get the decomposition.
 
Right.. how silly of me, so doing it by partial fractions I got 0=B=C and A= 1/16
giving me F(s) = (1/16)(1/s) does that seem a little more on track?
 
Well using partial fractions is definitely on track. But remember that when you have a quadratic term in the denominator, you have to have a linear term in the numerator (plus, it's squared, so there are going to be two.) The form for the expansion should be:

\frac{1}{(s)(s^2+4)^2} = {\frac{A}{s}} + \frac{Bs + C}{s^2 +4} + \frac{Ds + E}{(s^2+4)^2}

Give that a shot.
 
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Okay after that I am now getting A = 1/16, B = -1/16, C = 0, D = -1/4, E = 0.

(1/16)(1/s) - (1/16)(s/[s2 + 4]) - (1/4)(s/(s2 + 4)2
 
That looks good! Now that you have those, you can invert them term-by-term since the inverse Laplace transform is linear, so \mathcal{L}^{-1}(af(t)+bg(t)) = a\mathcal{L}^{-1}(f(t)) + b\mathcal{L}^{-1}(g(t)) for constants a and b.
 
Thank you so much! :D
 
You're welcome! :biggrin:
 
There is a standard result: if f(t) <---> g(s), then integral_{x=0..t} f(x) dx < ---> g(s)/s.

RGV
 
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