# Inverse Laplace Transform -s domain

How can I find the inverse Laplace transforms of the following function?

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

I solved so far. After that, how do I do?

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

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Use: $$\mathcal{LT}_s\left( \sin(\alpha x) \mathrm{e}^{-b x} \right) = \int_0^\infty \sin(\alpha x) \mathrm{e}^{-b x} \mathrm{e}^{-s x} \mathrm{d} x = \frac{\alpha}{(s+b)^2 + \alpha^2}$$ $$\mathcal{LT}_s\left( \cos(\alpha x) \mathrm{e}^{-b x} \right) = \int_0^\infty \cos(\alpha x) \mathrm{e}^{-b x} \mathrm{e}^{-s x} \mathrm{d} x = \frac{b+s}{(s+b)^2 + \alpha^2}$$ Completing the square: $s^2+s+2 = \left(s+\frac{1}{2}\right)^2 + \frac{7}{4}$. Therefore, decompose the image of Laplace transform accordingly: $$\frac{2 (s+1)}{s \left(s^2+s+2\right)}=\frac{1}{s}-\frac{s+\frac{1}{2}}{\left(s+\frac{1}{2}\right) ^2+\frac{7}{4}}+\frac{3}{\sqrt{7}}\frac{\sqrt{7}/2}{\left(\left(s+\frac{1}{2}\right)^2+\frac{7}{4}\right)}$$ Compare with the answer by WolframAlpha.

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To find the inverse Laplace transforms of the function $\ G\left(s\right)=\dfrac{2\left(s+1\right)}{s\left(s^2+s+2\right)}$

You have solved up to partial fraction form of $G\left(s\right)$ i.e

$$G\left(s\right)=\frac{1}{s}+\frac{1}{s^2+s+2}+\frac{s}{s^2+s+2}$$ Now taking the Laplace inverse \begin{align}\mathcal{L^{-1}}\left\{G\left(s\right)\right\}&=\mathcal{L^{-1}}\left\{\frac{1}{s}+\frac{1}{s^2+s+2}+\frac{s}{s^2+s+2}\right\}\\&=\mathcal{L^{-1}}\left\{\frac{1}{s}\right\}+\mathcal{L^{-1}}\left\{\frac{1}{s^2+s+2}\right\}+\mathcal{L^{-1}}\left\{\frac{s}{s^2+s+2}\right\}\\ \end{align} Now the first term $$\mathcal{L^{-1}}\left\{\frac{1}{s}\right\}=1 \\ \qquad$$

Second term is

\begin{align}\mathcal{L^{-1}}\left\{\frac{1}{s^2+s+2}\right\}&=\mathcal{L^{-1}}\left\{\frac{1}{s^2+2\times s\times\frac{1}{2}+\left(\frac{1}{2}\right)^2+2-\left(\frac{1}{2}\right)^2}\right\}\\ &=\mathcal{L^{-1}}\left\{\frac{1}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\ &=\mathcal{L^{-1}}\left\{\frac{2}{\sqrt{7}}\frac{\frac{\sqrt{7}}{2}}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\ &=\frac{2}{\sqrt{7}}\mathcal{L^{-1}}\left\{\frac{\frac{\sqrt{7}}{2}}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\ &=\frac{2}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right)\\ \left[\text{since}\quad \mathcal{L^{-1}}\left\{\frac{b}{\left(s-a\right)^2+b^2}\right\}=e^{at}\sin\left(bt\right)\right] \end{align}

Third term is

\begin{align}\mathcal{L^{-1}}\left\{\frac{s}{s^2+s+2}\right\}&=\mathcal{L^{-1}}\left\{\frac{s}{s^2+2.s.\frac{1}{2}+\left(\frac{1}{2}\right)^2+2-\left(\frac{1}{2}\right)^2}\right\}\\ &=\mathcal{L^{-1}}\left\{\frac{s+\frac{1}{2}-\frac{1}{2}}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\ &=\mathcal{L^{-1}}\left\{\frac{\left(s+\frac{1}{2}\right)}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}-\mathcal{L^{-1}}\left\{\frac{\frac{1}{2}}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\ &=e^{\frac{-1}{2}t}\cos\left(\frac{\sqrt{7}}{2}t\right)-\frac{1}{2}\mathcal{L^{-1}}\left\{\frac{1}{\left(s+\frac{1}{2}\right)^2+\left(\frac{\sqrt{7}}{2}\right)^2}\right\}\\ &=e^{\frac{-1}{2}t}\cos\left(\frac{\sqrt{7}}{2}t\right)-\frac{1}{2}\frac{2}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right) \quad \left(\text{By Second term}\right)\\ &=e^{\frac{-1}{2}t}\cos\left(\frac{\sqrt{7}}{2}t\right)-\frac{1}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right)\\ \qquad \qquad \qquad \left[\text{since}\quad \mathcal{L^{-1}}\left\{\frac{\left(s-a\right)}{\left(s-a\right)^2+b^2}\right\}=e^{at}\cos\left(bt\right)\right]. \end{align} Collecting all three terms, \begin{align}\mathcal{L^{-1}}\left\{G\left(s\right)\right\}&=1+\frac{2}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right)+e^{\frac{-1}{2}t}\cos\left(\frac{\sqrt{7}}{2}t\right)-\frac{1}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right)\\ &=1+\frac{1}{\sqrt{7}}e^{\frac{-1}{2}t}\sin\left(\frac{\sqrt{7}}{2}t\right)+e^{\frac{-1}{2}t}\cos\left(\frac{\sqrt{7}}{2}t\right)\\ \end{align}

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