Question

determine the inverse Laplace transform of the given function. F(s)=\frac{2s+1}{s^2+16}

Laplace transform
ANSWERED
asked 2021-09-18
determine the inverse Laplace transform of the given function.
\(\displaystyle{F}{\left({s}\right)}={\frac{{{2}{s}+{1}}}{{{s}^{{2}}+{16}}}}\)

Expert Answers (1)

2021-09-19

Step 1 Find the inverse Laplace transform of \(\displaystyle{F}{\left({s}\right)}={\frac{{{2}{s}+{1}}}{{{s}^{{2}}+{16}}}}\)
Step 2
\(L^{-1}\left\{\frac{2s+1}{s^2+16}\right\}=L^{-1}\left\{\frac{2s}{s^2+16}\right\}+L^{-1}\left\{\frac{1}{s^2+16}\right\}\)
\(=2L^{-1}\left\{\frac{s}{s^2+16}\right\}+L^{-1}\left\{\frac{1}{s^2+16}\right\}\)
\(=2(\cos 4t)+\frac{1}{4}\sin 4t \left[ \because L^{-1}\left\{\frac{s}{s^2+a^2}\right\}=\cos at \text{ and } L^{-1}\left\{\frac{a}{s^2+a^2}\right\}=\sin at \right]\)
\(\displaystyle={2}{\cos{{4}}}{t}+{\frac{{{1}}}{{{4}}}}{\sin{{4}}}{t}\)

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