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Find the inverse Laplace transform of the giveb function by using the convolution theorem.

$F(x) = \frac{s}{(s+1)(s^2+4)}$

If I use partial fractions I get: $\frac{s+4}{5(s^2+4)} - \frac{1}{5(x+1)}$

which gives me Laplace inverses:

$\frac{1}{5}(\cos2t + \sin2t) -\frac{1}{5} e^{-t}$

But the answer is: $f(t) = \int^t_0 e^{-(t -\tau)}\cos(2\tau) d\tau$

How did they get that?

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    @tomcuchta I have the inverses so to put it in convolution form then all I have to do is plug it in accordingly but even if I do that my inverses are not correct, since the correct answer above doesn't represent the inverse I gave.2012-12-08

1 Answers 1

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Related techniques (I), (II). Using the fact about the Laplace transform $L$ that

$ L(f*g)=L(f)L(g)=F(s)G(s)\implies (f*g)(t)=L^{-1}(F(s)G(s)) .$

In our case, given $ H(s)=\frac{1}{(s+1)}\frac{s}{(s^2+4)}$

$F(s)=\frac{1}{s+1}\implies f(t)=e^{-t},\quad G(s)=\frac{s}{s^2+4}\implies g(t)=\cos(2t).$

Now, you use the convolution as

$ h(t) = \int_{0}^{t} e^{-(t-\tau)}\cos(2\tau) d\tau . $

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    @BabakSorouh thanks for clearing that up!2012-12-08