91Ó°ÊÓ

• For a given transfer function H, the Inverse Laplace Transform takes the output Y(s) and determines what X(s) it is in terms of (s).
• Consider a function $\mathrm{F}\left(\mathrm{s}\right)$, if there is a function $\mathrm{f}\left(\mathrm{t}\right)$that is continuous on $[0,\mathrm{∞})$and satisfies $\mathcal{L}\left\{\mathrm{f}\right\}=\mathrm{F}$then we say that $\mathrm{f}\left(\mathrm{t}\right)$is the inverse Laplace transform of $\mathrm{F}\left(\mathrm{s}\right)$and employ the notation
• $\mathrm{f}={\mathcal{L}}^{−1}\left\{\mathrm{F}\right\}$
  
• ${\mathcal{L}}^{−1}\left\{\frac{\mathrm{n}!}{{(\mathrm{s}−\mathrm{a})}^{\mathrm{n}+1}}\right\}={\mathrm{e}}^{\mathrm{at}}{\mathrm{t}}^{\mathrm{n}},\mathrm{n}=1,2,…$
  

The given function is $\frac{\mathrm{s}−1}{2{\mathrm{s}}^2+\mathrm{s}+6}$

Simplify $\frac{\mathrm{s}−1}{2{\mathrm{s}}^2+\mathrm{s}+6}$ as:

$\begin{array}{c}\frac{\mathrm{s}−1}{2{\mathrm{s}}^2+\mathrm{s}+6}=\frac{\mathrm{s}−1}{2\left({\mathrm{s}}^2+\frac{1}{2}\mathrm{s}+3\right)}\\ =\frac{1}{2}\left(\frac{\mathrm{s}−1}{{\mathrm{s}}^2+\frac{1}{2}\mathrm{s}+3}\right)\\ =\frac{1}{2}\left(\frac{\mathrm{s}−1}{\left({\mathrm{s}}^2+\frac{1}{2}\mathrm{s}+\frac{1}{16}\right)+3−\frac{1}{16}}\right)\\ =\frac{1}{2}\left(\frac{\mathrm{s}−1}{{\left(\mathrm{s}+\frac{1}{4}\right)}^2+{\left(\sqrt{\frac{47}{16}}\right)}^2}\right)\end{array}$

Further simplify the equation as follows:

$\begin{array}{c}\frac{s−1}{2s^2+s+6}=\frac{1}{2}\left(\frac{s+\frac{1}{4}−\frac{5}{4}}{{\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{47}}{4}\right)}^2}\right)\\ =\frac{1}{2}\left(\frac{\left(s+\frac{1}{4}\right)−\frac{5}{4}}{{\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{47}}{4}\right)}^2}\right)\\ =\frac{1}{2}\left(\frac{s+\frac{1}{4}}{{\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{47}}{4}\right)}^2}\right)−\frac{5\sqrt{47}}{94}\left(\frac{\frac{\sqrt{47}}{4}}{{\left(s+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{47}}{4}\right)}^2}\right)\end{array}$

Find the inverse Laplace transform of

$\frac{\mathrm{s}−1}{2{\mathrm{s}}^2+\mathrm{s}+6}=\frac{1}{2}\left(\frac{\mathrm{s}+\frac{1}{4}}{{\left(\mathrm{s}+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{47}}{4}\right)}^2}\right)−\frac{5\sqrt{47}}{94}\left(\frac{\frac{\sqrt{47}}{4}}{{\left(\mathrm{s}+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{47}}{4}\right)}^2}\right)$using ${\mathcal{L}}^{−1}\left\{\frac{\mathrm{b}}{{(\mathrm{s}−\mathrm{a})}^2+{\left(\mathrm{b}\right)}^2}\right\}={\mathrm{e}}^{\mathrm{at}}\mathrm{sinbt}$ and ${\mathcal{L}}^{−1}\left\{\frac{\mathrm{s}−\mathrm{a}}{{(\mathrm{s}−\mathrm{a})}^2+{\left(\mathrm{b}\right)}^2}\right\}={\mathrm{e}}^{\mathrm{at}}\mathrm{cosbt}$as:

$\begin{array}{c}{\mathcal{L}}^{−1}\left\{\frac{\mathrm{s}−1}{2{\mathrm{s}}^2+\mathrm{s}+6}\right\}={\mathcal{L}}^{−1}\left\{\frac{1}{2}\left(\frac{\left(\mathrm{s}+\frac{1}{4}\right)}{{\left(\mathrm{s}+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{47}}{4}\right)}^2}\right)−\frac{5\sqrt{47}}{94}\left(\frac{\frac{\sqrt{47}}{4}}{{\left(\mathrm{s}+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{47}}{4}\right)}^2}\right)\right\}\\ ={\mathcal{L}}^{−1}\left\{\frac{1}{2}\left(\frac{\left(\mathrm{s}+\frac{1}{4}\right)}{{\left(\mathrm{s}+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{47}}{4}\right)}^2}\right)\right\}−{\mathcal{L}}^{−1}\left\{\frac{5\sqrt{47}}{94}\left(\frac{\frac{\sqrt{47}}{4}}{{\left(\mathrm{s}+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{47}}{4}\right)}^2}\right)\right\}\\ =\frac{1}{2}{\mathcal{L}}^{−1}\left\{\frac{\left(\mathrm{s}+\frac{1}{4}\right)}{{\left(\mathrm{s}+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{47}}{4}\right)}^2}\right\}−\frac{5\sqrt{47}}{94}{\mathcal{L}}^{−1}\left\{\frac{\frac{\sqrt{47}}{4}}{{\left(\mathrm{s}+\frac{1}{4}\right)}^2+{\left(\frac{\sqrt{47}}{4}\right)}^2}\right\}\\ =\frac{1}{2}{\mathrm{e}}^{−\frac{1}{4}\mathrm{t}}\cos \frac{\sqrt{47}}{4}\mathrm{t}−\frac{5\sqrt{47}}{94}{\mathrm{e}}^{−\frac{1}{4}\mathrm{t}}\sin \frac{\sqrt{47}}{4}\mathrm{t}\end{array}$

Therefore, the inverse Laplace transform for the given function is

${\mathcal{L}}^{−1}\left\{\frac{\mathrm{s}−1}{2{\mathrm{s}}^2+\mathrm{s}+6}\right\}=\frac{1}{2}{\mathrm{e}}^{−\frac{1}{4}\mathrm{t}}\cos \frac{\sqrt{47}}{4}\mathrm{t}−\frac{5\sqrt{47}}{94}{\mathrm{e}}^{−\frac{1}{4}\mathrm{t}}\sin \frac{\sqrt{47}}{4}\mathrm{t}$