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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.

  • 1
    The inverse Laplace transform of $s/(s^2 + a^2)$ is $\cos(at)$. If $3 = a^2$, what do you suppose $a$ is?2012-06-28
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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 $\frac{4s+3}{s^2+3} = {\frac {2-i\sqrt {3}/2}{s-i\sqrt {3}}}+{\frac {2+i\sqrt {3}/2}{ s+i\sqrt {3}}}$