Question

# Find the inverse Laplace transform of the given function.F(s)=\frac{2e^{-2s}}{s^2-4}

Laplace transform
Find the inverse Laplace transform of the given function.
$$\displaystyle{F}{\left({s}\right)}={\frac{{{2}{e}^{{-{2}{s}}}}}{{{s}^{{2}}-{4}}}}$$

2021-09-18

$$\displaystyle{F}{\left({s}\right)}={\frac{{{2}{e}^{{-{2}{s}}}}}{{{s}^{{2}}-{4}}}}$$
$$\displaystyle{\frac{{{2}}}{{{\left({s}-{2}\right)}{\left({s}+{2}\right)}}}}={\frac{{{A}}}{{{s}-{2}}}}+{\frac{{{B}}}{{{s}+{2}}}}$$
$$\displaystyle{2}={A}{\left({s}+{2}\right)}+{B}{\left({s}-{2}\right)}$$
$$A+B=0$$
$$2A-2B=2$$
$$3A=2$$
$$\displaystyle{A}={\frac{{{2}}}{{{3}}}},{B}={\frac{{-{2}}}{{{3}}}}$$
$$\displaystyle{F}{\left({s}\right)}={\left({\frac{{{2}{e}^{{-{2}{s}}}}}{{{3}{\left({s}-{2}\right)}}}}-{\frac{{{2}{e}^{{-{2}{s}}}}}{{{3}{\left({s}+{2}\right)}}}}\right)}$$
By applying Inverse Laplace Transform
$$\displaystyle{f{{\left({t}\right)}}}={\frac{{{2}}}{{{3}}}}{e}^{{{2}{\left({t}-{2}\right)}}}{u}{\left({t}-{2}\right)}-{\frac{{{2}}}{{{3}}}}{e}^{{-{2}{\left({t}-{2}\right)}}}{u}{\left({t}-{2}\right)}$$