Question

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

Answer 1

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
\({F}{\left({s}\right)}=\frac{2}{{{s}{\left({s}^{2}+{16}\right)}}}{i}{s}\frac{1}{{8}}{\left({1}- \cos{{\left({4}{t}\right)}}\right)}.\)