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

Question
Transformation properties
A function f (t) that has the given Laplace transform F (s).

2020-10-22
Given:
Laplace transform F (s) of a function f (t) defined on an interval $$[0, \infty) - F (s) = \frac{2}{s(s^{2}+16)}$$
Calculation:
Given,
$$F (s) = \frac{2}{s(s^{2}+16)}$$
Partial fraction decomposition,
$$F (s) = \frac{2}{s(s^{2}+16)}$$
Approach:
The
$$=\frac{A}{s}+\frac{Bs+C}{s^{2}+16}$$
$$2=A(s^{2}+16)+s(Bs+C)$$
$$=s^{2}(A+B)+Cs+16A$$
Compare the Left hand side and Right hand side,
$$A = \frac{1}{8}$$
$$B = -\frac{1}{8}$$
$$C = 0$$
Substitute -
$$F (s) = \frac{1}{8s} - \frac{s}{8(s^{2}+16)}$$
Laplace transform,
$$L [f] = \int_{0}^\infty e^{-s}tf(t)dt$$
$$= F (s)$$
$$= \frac{2}{s(s^{2}+16)}$$
$$= \frac{1}{8s} - \frac{s}{8(s^{2}+16)}$$
Inverse Laplace transform-
$$L^{-1} [F] (t) = L^{-1} [ \frac{2}{s(s^{2}+16)}]$$
$$={L}^{ -{{1}}}{\left[\frac{1}{{{8}{s}}}-\frac{s}{{{8}{\left({s}^{2}+{16}\right)}}}\right]}$$
$$=\frac{1}{{8}}{L}^{ -{{1}}}{\left[\frac{1}{{s}}\right]}-\frac{1}{{8}}{L}^{ -{{1}}}{\left[\frac{s}{{{s}^{2}+{4}^{2}}}\right]}$$
$$=\frac{1}{{8}}-\frac{1}{{8}} \cos{{\left({4}{t}\right)}}$$
$${L}^{ -{{1}}}{\left[{F}\right]}{\left({t}\right)}=\frac{1}{{8}}{\left({1}- \cos{{\left({4}{t}\right)}}\right)}$$
$$= f(t)$$
Conclusion:
Hence, a function f (t) that has the given Laplace transform
$${F}{\left({s}\right)}=\frac{2}{{{s}{\left({s}^{2}+{16}\right)}}}{i}{s}\frac{1}{{8}}{\left({1}- \cos{{\left({4}{t}\right)}}\right)}.$$

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