# 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?

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

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}

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.

Your function $$G(s) = \frac{2(s+1)}{s(s^2 + s + 2)}$$ has the partial fraction decomposition $$G(s) = \frac{A}{s} + \frac{Bs + C}{(s^2 + s + 2)}$$The way how I will solve this is to use complex analysis. Your original function can be broken down into three distinct linear factors by solving for the zeros of $$(s^2 + s + 2)$$ using quadratic formula. $$\frac{b\pm\sqrt{b^2 -4ac}}{2a}$$ with a = b = 1 and c = 2. The zeroes are $$s = -\frac{1}{2}+j\frac{\sqrt7}{2}$$ and $$s = -\frac{1}{2}-j\frac{\sqrt7}{2}$$. Your original function can now be written as $$G(s) = \frac{2(s+1)}{s(s-\frac{1}{2}+j\frac{\sqrt7}{2})(s-\frac{1}{2}-j\frac{\sqrt7}{2})}$$ fixing the brackets, we have $$G(s) = \frac{2(s+1)}{s(s-[\frac{1}{2}-j\frac{\sqrt7}{2}])(s-[\frac{1}{2}+j\frac{\sqrt7}{2}])}$$ The new partial fraction decomposition now is $$G(s) = \frac{A}{s}+\frac{B}{[s-(\frac{1}{2}-j\frac{\sqrt7}{2})]}+\frac{C}{[s-(\frac{1}{2}+j\frac{\sqrt7}{2})]}$$ Using cover-up method, if we set $$s = 0$$ we will be able to solve for $$A$$. Thus $$A = \frac{2(s+1)}{[s-(\frac{1}{2}-j\frac{\sqrt7}{2})][s-(\frac{1}{2}+j\frac{\sqrt7}{2})]}$$ $$A = \frac{2}{2} = 1$$ Utilize FOIL method to get this. To solve for B and C, we set $$s = (\frac{1}{2}-j\frac{\sqrt7}{2})$$ and $$s = (\frac{1}{2}+j\frac{\sqrt7}{2})$$ respectively. Since B and C are complex conjugates, C's imaginary part will be the negative of B. Solving for B, $$B = \frac{2(s+1)}{s[s-(\frac{1}{2}+j\frac{\sqrt7}{2})]}$$ $$B = \frac{(1 + j\sqrt7)}{{-j\sqrt7}}$$ $$B = 1 -j\frac{\sqrt7}{7}$$

Because C is the complex conjugate of B, C is therefore $$C = 1 +j\frac{\sqrt7}{7}$$ Thus, your original function is decomposed into $$G(s) = \frac{1}{s} + \frac{1 - j\frac{\sqrt7}{7}}{[s-(\frac{1}{2} - j\frac{\sqrt7}{2})]} + \frac{1 + j\frac{\sqrt7}{7}}{[s-(\frac{1}{2} +j \frac{\sqrt7}{2})]}$$ The inverse Laplace Transform of this is $$g(t) = 1 + (1 - j\frac{\sqrt7}{7})e^{(\frac{1}{2} - j\frac{\sqrt7}{2})t} + (1 + j\frac{\sqrt7}{7})e^{(\frac{1}{2}+j\frac{\sqrt7}{2})t}$$ Do note that this is not yet the final answer. What I will be doing is I will make use of sine and cosine's equivalent complex functions, namely: $$sin(\theta) = \frac{1}{j2}(e^{j\theta} - e^{-j\theta})$$ $$cos(\theta) = \frac{1}{2}(e^{j\theta} + e^{-j\theta})$$ We will then factor out $$e^{\frac{1}{2}t}$$ and distribute $$e^{j\frac{\sqrt7}{2}t}$$ to each terms. Upon doing so, we get

$$g(t) = 1 + e^{\frac{1}{2}t}[(1 - j\frac{\sqrt7}{7})e^{-j\frac{\sqrt7}{2}t}] + [(1 + j\frac{\sqrt7}{7})e^{j\frac{\sqrt7}{2}t}]$$ $$g(t) = 1 + e^{\frac{1}{2}t}[(e^{-j\frac{\sqrt7}{2}t} - j\frac{\sqrt7}{7}e^{-j\frac{\sqrt7}{2}t}) + (e^{j\frac{\sqrt7}{2}t} + j\frac{\sqrt7}{7}e^{j\frac{\sqrt7}{2}t}) ]$$ We will then collect the real and imaginary parts of $$e^{j\frac{\sqrt7}{2}t}$$ and $$e^{-j\frac{\sqrt7}{2}t}$$ because we will be able to get the inverse laplace transform from those. By doing so, we have $$g(t) = 1 + e^{\frac{1}{2}t}[(e^{j\frac{\sqrt7}{2}t} + e^{-j\frac{\sqrt7}{2}t}) - j\frac{\sqrt7}{7}(e^{j\frac{\sqrt7}{2}t} - e^{-j\frac{\sqrt7}{2}t})]$$ $$g(t) = 1 + 2e^{\frac{1}{2}t}cos(\frac{\sqrt7}{2}t) - j\frac{\sqrt7}{7}(j2)e^{\frac{1}{2}t}sin(\frac{\sqrt7}{2}t)$$ $$g(t) = 1 + 2e^{\frac{1}{2}t}cos(\frac{\sqrt7}{2}t) + \frac{2\sqrt7}{7}e^{\frac{1}{2}t}sin(\frac{\sqrt7}{2}t)$$