In this topic, you study the Table of Inverse Laplace Transforms.
Definition: If Laplace transform $L[f(t)]=F(s)$ then inverse Laplace transform ${L^{ – 1}}[F(s)] = f(t)$. Using above property, the inverse Laplace transform of standard forms are
| \[{L^{ – 1}}[F(s)]\] | \[f(t)\] |
| \[{L^{ – 1}}\left[ {\frac{1}{s}} \right]\] | \[u(t)\] |
| \[{L^{ – 1}}\left[ {\frac{1}{{{s^2}}}} \right]\] | \[t \] |
| \[{L^{ – 1}}\left[ {\frac{{n!}}{{{s^{n + 1}}}}} \right]\] | \[{t^n} \] |
| \[{L^{ – 1}}\left[ {\frac{1}{{s + a}}} \right]\] | \[{e^{ – at}}\] |
| \[{L^{ – 1}}\left[ {\frac{1}{{s – a}}} \right]\] | \[{e^{ at}}\] |
| \[{L^{ – 1}}\left[ {\frac{a}{{{s^2} + {a^2}}}} \right]\] | \[\sin at\] |
| \[{L^{ – 1}}\left[ {\frac{s}{{{s^2} + {a^2}}}} \right]\] | \[\cos at\] |
| \[{L^{ – 1}}\left[ {\frac{a}{{{s^2} – {a^2}}}} \right]\] | \[\sinh at\] |
| \[{L^{ – 1}}\left[ {\frac{s}{{{s^2} – {a^2}}}} \right]\] | \[\cosh at\] |
Note:- Defined for $t$ ≥ 0, $f(t)$ = 0, for $t$ < 0.