# A function f (t) that has the given Laplace transform F (s).

A function f (t) that has the given Laplace transform F (s).
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Given:
Laplace transform F (s) of a function f (t) defined on an interval $\left[0,\mathrm{\infty }\right)-F\left(s\right)=\frac{2}{s\left({s}^{2}+16\right)}$
Calculation:
Given,
$F\left(s\right)=\frac{2}{s\left({s}^{2}+16\right)}$
Partial fraction decomposition,
$F\left(s\right)=\frac{2}{s\left({s}^{2}+16\right)}$
Approach:
The
$=\frac{A}{s}+\frac{Bs+C}{{s}^{2}+16}$
$2=A\left({s}^{2}+16\right)+s\left(Bs+C\right)$
$={s}^{2}\left(A+B\right)+Cs+16A$
Compare the Left hand side and Right hand side,
$A=\frac{1}{8}$
$B=-\frac{1}{8}$
$C=0$
Substitute -
$F\left(s\right)=\frac{1}{8s}-\frac{s}{8\left({s}^{2}+16\right)}$
Laplace transform,
$L\left[f\right]={\int }_{0}^{\mathrm{\infty }}{e}^{-s}tf\left(t\right)dt$
$=F\left(s\right)$
$=\frac{2}{s\left({s}^{2}+16\right)}$
$=\frac{1}{8s}-\frac{s}{8\left({s}^{2}+16\right)}$
Inverse Laplace transform-
${L}^{-1}\left[F\right]\left(t\right)={L}^{-1}\left[\frac{2}{s\left({s}^{2}+16\right)}\right]$
$={L}^{-1}\left[\frac{1}{8s}-\frac{s}{8\left({s}^{2}+16\right)}\right]$
$=\frac{1}{8}{L}^{-1}\left[\frac{1}{s}\right]-\frac{1}{8}{L}^{-1}\left[\frac{s}{{s}^{2}+{4}^{2}}\right]$
$=\frac{1}{8}-\frac{1}{8}\mathrm{cos}\left(4t\right)$
${L}^{-1}\left[F\right]\left(t\right)=\frac{1}{8}\left(1-\mathrm{cos}\left(4t\right)\right)$
$=f\left(t\right)$
Conclusion:
Hence, a function f (t) that has the given Laplace transform
$F\left(s\right)=\frac{2}{s\left({s}^{2}+16\right)}is\frac{1}{8}\left(1-\mathrm{cos}\left(4t\right)\right).$