Question

Determine the inverse Laplace transform of the given function.

$\frac{\mathbf{2}\mathbf{s}\mathbf{-}\mathbf{1}}{{\mathbf{s}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{s}\mathbf{+}\mathbf{6}}$

Short answer

The solution is

${\mathcal{L}}^{-1}\left\{\frac{2(s-2)+3}{{(s-2)}^2+{\left(\sqrt{2}\right)}^2}\right\}\left(t\right)=2e^{2t}\cos \sqrt{2}t+\frac{3}{\sqrt{2}}{e}^{2t}\sin \sqrt{2}t$

Step-by-step solution

Given Information

The given function value in s domain is$\frac{2s-1}{s^2-4s+6}$

Use partial fractions

Factorize the denominator of the given function as

$\begin{array}{c}{s}^2-4s+6=s^2-4s+4+2\\ ={\left(s-1\right)}^2+{\left(\sqrt{2}\right)}^2\end{array}$

The function becomes.$\frac{2s-1}{{(s-2)}^2+{\left(\sqrt{2}\right)}^2}$Decompose the function as:

$\begin{array}{c}\frac{2s-1}{{(s-2)}^2+{\left(\sqrt{2}\right)}^2}=\frac{2(s-2)+3}{{(s-2)}^2+{\left(\sqrt{2}\right)}^2}\\ =\frac{2(s-2)}{{(s-2)}^2+{\left(\sqrt{2}\right)}^2}+\frac{3}{\sqrt{2}}\frac{\sqrt{2}}{{(s-2)}^2+{\left(\sqrt{2}\right)}^2}\end{array}$

Take inverse Laplace transform using ${\mathcal{L}}^{-1}\left\{\frac{{\displaystyle s-a}}{{\displaystyle {(s-a)}^2+b^2}}\right\}\left(t\right)=e^{at}\cos bt$and ${\mathcal{L}}^{-1}\left\{\frac{{\displaystyle b}}{{\displaystyle {(s-a)}^2+b^2}}\right\}\left(t\right)=e^{at}\sin bt$as:

${\mathcal{L}}^{-1}\left\{\frac{{\displaystyle 2(s-2)+3}}{{\displaystyle {(s-2)}^2+{\left(\sqrt{2}\right)}^2}}\right\}\left(t\right)=2e^{2t}\cos \sqrt{2}t+\frac{3}{\sqrt{2}}{e}^{2t}\sin \sqrt{2}t$

Therefore,the required inverse Laplace transform is:

${\mathcal{L}}^{-1}\left\{\frac{{\displaystyle 2(s-2)+3}}{{\displaystyle {(s-2)}^2+{\left(\sqrt{2}\right)}^2}}\right\}\left(t\right)=2e^{2t}\cos \sqrt{2}t+\frac{3}{\sqrt{2}}{e}^{2t}\sin \sqrt{2}t$