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I was wondering how you would be able to solve the inverse laplace of

$$\mathcal{L}^{-1}\left\{\frac{s}{RCs+1}\right\}\left(s\right)\tag{1}$$

where $R$ and $C$ are constants?

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Write $$\frac{s}{RCs + 1} = \frac{1}{RC}\frac{RCs}{RCs + 1} = \frac{1}{RC}\left(1 - \frac{1}{RCs + 1}\right) = \frac{1}{RC}\left(1 - \frac{1}{RC}\frac{1}{s + \frac{1}{RC}}\right)$$ Since the inverse Laplace of $1$ is $\delta(t)$ and the inverse Laplace of $\frac{1}{s + \frac{1}{RC}}$ is $e^{-t/(RC)}$, we have that the inverse Laplace of $\frac{s}{RCs + 1}$ is $$\frac{1}{RC}\left(\delta(t) - \frac{1}{RC}e^{-\frac{t}{RC}}\right) = \frac{1}{RC}\delta(t) - \left(\frac{1}{RC}\right)^2 e^{-\frac{t}{RC}}$$

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  • $\begingroup$ you give incredible answers! $\endgroup$ – jm324354 Feb 5 '15 at 4:56
  • $\begingroup$ @bd1251252 thank you! $\endgroup$ – kobe Feb 5 '15 at 4:56

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