Referring to the image below, what would the inverse Laplace transform be? I can't seem to find any tables that include this case.
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$\begingroup$ lpsa.swarthmore.edu/LaplaceXform/InvLaplace/… You know how to perform partial fraction expansion ? $\endgroup$ – Ben Sep 12 at 3:21
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$\begingroup$ This table has it all, you just have to fill in the numbers. $\endgroup$ – Matt L. Sep 12 at 5:35
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1$\begingroup$ @Ben: Partial fractions don't really help here. You can't split a double pole into two single poles. $\endgroup$ – Matt L. Sep 12 at 9:17
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$\begingroup$ Yeah you're right... sorry $\endgroup$ – Ben Sep 12 at 11:20
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This is quite straightforward to solve. Either just use this table where you can directly find the corresponding result, or "derive" it yourself with very basic knowledge of the Laplace transform.
You should know that
$$\mathcal{L}\{u(t)\}=\frac{1}{s}\tag{1}$$
Integration is equivalent to multiplication with $1/s$, so integrating $(1)$ gives
$$\mathcal{L}\{t\cdot u(t)\}=\frac{1}{s^2}\tag{2}$$
Replacing $s$ by $s+a$ is equivalent to multiplication with $e^{-at}$. Consequently, we have
$$\mathcal{L}^{-1}\left\{\frac{2}{(s+2)^2}\right\}=2\,t\,e^{-2t} u(t)\tag{3}$$
