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Is the following fraction (actually a Laplace transform) a kind of partial fraction?

$$\frac{4s+3}{{s^2}+3}$$

Can this be solved this way? $$\frac{A}{s}+\frac{B}{s+{\frac{3}{s}}}$$

If not can you please tell me how to find inverse transform?

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If you want to keep everything real, this is already decomposed into partial fractions. For the inverse Laplace transform, just split it as $$ \frac{4s+3}{s^2+3} = 4 \frac{s}{s^2+3} + 3\frac{1}{s^2+3}. $$ You should be able to invert each term separately.

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  • $\begingroup$ Ok, if the first one had 9 instead of 3 I would say it is $\cos(3t)$ but what is it in this case? $\endgroup$
    – Sean87
    Jun 28 '12 at 19:07
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    $\begingroup$ The inverse Laplace transform of $s/(s^2 + a^2)$ is $\cos(at)$. If $3 = a^2$, what do you suppose $a$ is? $\endgroup$ Jun 28 '12 at 19:09
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If you don't need to keep it real, the roots of $s^2 + 3$ are $\pm \sqrt{3} i$, and the partial fraction decomposition is $$\frac{4s+3}{s^2+3} = {\frac {2-i\sqrt {3}/2}{s-i\sqrt {3}}}+{\frac {2+i\sqrt {3}/2}{ s+i\sqrt {3}}}$$

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