# Inverse Laplace transform - using the table

I am trying to find the inverse Laplace transform $(g(t))$ of

$$G(s) = \frac{2s}{(s+1)^2+4}$$

I know about the inverse transforms $e^{a t}\cos(\omega t)$ and $\mathrm{e}^{at}\sin(\omega t)$ however I am trying to get the inverse transforms without these, as these are not on the table of standard transforms for our course.

I was also attempting to use $\mathrm{e}^{at} f(t) \leftrightarrow F(s-a)$, but I'm not sure how to go about this, either.

Any help would be appreciated. :)

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For some basic information about writing math at this site see e.g. here, here, here and here. – Jayesh Badwaik Sep 2 '12 at 3:26
Some MSE users tried to improve your post using TeX (for better readability). Please check whether these edits did not unintentionally change the meaning of your post. – Jayesh Badwaik Sep 2 '12 at 3:27
if I were you then I would work on seeing how the transforms of $e^{at}\cos(t)$ and $e^{at}\sin(t)$ arise from the shift theorem. – James S. Cook Sep 2 '12 at 3:28
sorry about the maths writing format. the edit was fine, thanks for that. Also, the shift theorem helped a lot, thanks for pointing me in that direction! :) – user1393689 Sep 2 '12 at 6:13
If you don't mind a little complex arithmetic then you can also derive both shifted sine and cosine from $\mathcal{L}^{-1}(\frac{1}{s-c}) = e^{ict}$ with $c=a+ib$. Just multiply by $\frac{s-(a-ib)}{s-(a-ib)}$ and see how the denominator goes to $(s-a)^2+b^2$ whereas the numerator has $(s-a)$ and $ib$. Since $e^{(a+ib)t} = e^{at}\cos(bt)+ie^{at}\sin(bt)$ equating real and imaginary parts reveals both the transform for sine and cosine at once. – James S. Cook Sep 2 '12 at 17:49

If you have $(s+1)^2$ in the denominator then make it appear in the numerator by adding zero:

$$G(s) = \frac{2s}{(s+1)^2+4} = \frac{2(s+1-1)}{(s+1)^2+4} = 2 \frac{s+1}{(s+1)^2+4}-\frac{2}{(s+1)^2+4}$$

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The inverse Laplace transform is defined by,

$$f(x) = \frac{1}{2\pi i}\int_C F(s) {\rm e}^{sx} ds \,$$

where $F(s)$ is the Laplace transform of $f(x)$, and $C$ is the Bromwich contour.

$$f(x) = \frac{1}{2\pi i} \int_C \frac{{2s\,\rm e}^{sx}}{((s+1)^2+4)} ds = \frac{1}{2\pi i} \int_C \frac{{2s\,\rm e}^{sx}}{(s^2+2s+5)} ds$$

$$= \frac{1}{2 \pi i} \int_C \frac{{2s\,\rm e}^{sx}}{(s+1+2i)(s+1-2i)} ds\,.$$

The integrand has two simple poles at $s=-(1+2i)$ and $s=-1+2i$ which implies,

$$f(x) = \lim_{s\rightarrow -(1+2i)} \frac{{2s\,\rm e}^{sx}}{(s+1-2i)} + \lim_{s\rightarrow (-1+2i)} \frac{{2s\,\rm e}^{sx}}{(s+1+2i)}$$

$$= {{\rm e}^{-x}} \left( 2\,\cos \left( 2\,x \right) -\sin \left( 2\,x \right) \right)$$

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