Question

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

Laplace transform
ANSWERED
asked 2021-09-17
Find the inverse Laplace transform of the given function.
\(\displaystyle{F}{\left({s}\right)}={\frac{{{2}{e}^{{-{2}{s}}}}}{{{s}^{{2}}-{4}}}}\)

Expert Answers (1)

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
\(\displaystyle{f{{\left({t}\right)}}}={\frac{{{2}}}{{{3}}}}{e}^{{{2}{\left({t}-{2}\right)}}}{u}{\left({t}-{2}\right)}-{\frac{{{2}}}{{{3}}}}{e}^{{-{2}{\left({t}-{2}\right)}}}{u}{\left({t}-{2}\right)}\)

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