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I have a question: Find the Laplace transform of $$ \frac{1}{s^2(s^2+4)}. $$ Now I also do have its solution. I wanted to know why do we Laplace inverse integral method (Please correct me if the name of the method is wrong) and not just simply take its Laplace inverse (i.e use linearity principle and find the Laplace of $\frac{1}{s^2}$ and $\frac{1}{s^2+4}$ separately and multiply them)?

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3 Answers 3

up vote 5 down vote accepted

Use partial fractions: $$ \frac{1}{s^2(s^2+4)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs+D}{s^2 + 4} $$ Find $A$, $B$, $C$, $D$ and then apply linearity.

Linearity says that you can find inverse Laplace transforms of the terms in a sum and then add them, but it doesn't say you can find inverse Laplace transforms of the factors in a product and then multiply them.

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I solved the question using partial fractions and I got D=-4 and A=B=C=0. Taking laplace inverse gave the answer -2sin2t. It was different from the answer that the teacher got from integration. –  Fahad Uddin Oct 8 '11 at 16:52
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Dear Michael: You can write directly $$\frac{1}{s^2\,(s^2+4)}=\frac{A}{s^2}+\frac{B}{s^2 + 4}\quad.$$ –  Pierre-Yves Gaillard Oct 8 '11 at 18:57
    
Dear @Akito: You may want to check your computation. –  Pierre-Yves Gaillard Oct 8 '11 at 18:58
    
@Akito: If the numbers you got are right, that would meant that $\dfrac{1}{s^2(s^2+4)}=\dfrac{-4}{s^2+4}$. That is certainly wrong, since $\dfrac{1}{s^2(s^2+4)}$ obviously has a vertical asymptote at $s=0$ and $\dfrac{-4}{s^2+4}$ does not. –  Michael Hardy Oct 8 '11 at 21:15
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@Pierre-Yves: I hadn't actually noticed that. Generally a repeated linear factor requires multiple terms like that, but if one can write every factor as a function of $s^2$, with no odd-degree terms, then you can say $s^2$ appears only once as a factor. –  Michael Hardy Oct 9 '11 at 1:06

Partial Fractions
$$ F(s)=\frac{1}{s^2(s^2+4)} = \frac{K_{11}}{s^2} + \frac{K_{12}}{s} + \frac{k_2}{s^2 + 4} $$ $$ K_{11}= F(s)\cdot s^2 = \frac{1}{s^2+4}\text{ (when $s=0$) }\rightarrow k_{11}=\frac{1}{4} $$ $$ k_{12}=\frac{d}{ds}[ F(s)\cdot s^2]\text{ (when $s=0$) }\rightarrow k_{12}= 0 $$ $$ k_2=F(s)\cdot(s^2+4)\text{ (when $s=2i$) }\rightarrow k_2=\frac{-1}{4} $$ So : $$ f(t)=\mathcal{L^{-1}}\left[\frac{\frac{1}{4}}{s^2} + \frac{\frac{-1}{4}}{s^2 + 4}\right] $$ $$ f(t)=\frac{t}{4}-\frac{\sin(2t)}{8} $$

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Because that's not how linearity works. The definition of a map $T: V \to W$ being linear is that $T(a_1v_1 + a_2v_2) = a_1T(v_1) + a_2T(v_2)$. Nowhere does it speak anything about $T(v_1v_2)$ - in fact "$v_1v_2$" doesn't even have a meaningful definition for most vector spaces.

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@Akito: Therefore, if you want to exploit the linearity, you'll need the old standby, partial fraction decomposition... –  J. M. Oct 8 '11 at 13:44
    
@J.M.: Does the linearity principle apply on the inverse laplace too? –  Fahad Uddin Oct 8 '11 at 14:09
    
@Akito: Certainly. –  J. M. Oct 8 '11 at 14:11

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