Definition of Inverse Laplace Transform

From $\mathcal{L} \left\{ f(t) \right\} = F(s)$, the value f(t) is called the inverse Laplace transform of F(s). In symbol,
 

$\mathcal{L}^{-1}\left\{ F(s) \right\} = f(t)$

where $\mathcal{L}^{-1}$ is called the inverse Laplace transform operator.
 

To find the inverse transform, express F(s) into partial fractions which will, then, be recognizable as one of the following standard forms.
 

Table of Inverse Laplace Transform

  1.   $\mathcal{L}^{-1}\left\{ \dfrac{1}{s} \right\} = 1$
     
  2.   $\mathcal{L}^{-1}\left\{ \dfrac{1}{s - a} \right\} = e^{at}$
     
  3.   $\mathcal{L}^{-1}\left\{ \dfrac{1}{s^n} \right\} = \dfrac{t^{n - 1}}{(n - 1)!}; ~ ~ n = 1, ~ 2, ~ 3, ...$
     
  4.   $\mathcal{L}^{-1}\left\{ \dfrac{1}{(s - a)^n} \right\} = \dfrac{e^{at}t^{n - 1}}{(n - 1)!}$
     
  5.   $\mathcal{L}^{-1}\left\{ \dfrac{1}{s^2 + b^2} \right\} = \dfrac{1}{b}\sin bt$
     
  6.   $\mathcal{L}^{-1}\left\{ \dfrac{s}{s^2 + b^2} \right\} = \cos bt$
     
  7.   $\mathcal{L}^{-1}\left\{ \dfrac{1}{s^2 - a^2} \right\} = \dfrac{1}{a}\sinh at$
     
  8.   $\mathcal{L}^{-1}\left\{ \dfrac{s}{s^2 - a^2} \right\} = \cosh at$
     
  9.   $\mathcal{L}^{-1}\left\{ \dfrac{1}{(s - a)^2 + b^2} \right\} = \dfrac{1}{b}e^{at}\sin bt$
     
  10.   $\mathcal{L}^{-1}\left\{ \dfrac{s - a}{(s - a)^2 + b^2} \right\} = e^{at}\cos bt$
     
  11.   $\mathcal{L}^{-1}\left\{ \dfrac{s}{(s^2 + b^2)^2} \right\} = \dfrac{1}{2b}t \sin bt$
     
  12.   $\mathcal{L}^{-1}\left\{ \dfrac{1}{(s^2 + b^2)^2} \right\} = \dfrac{1}{2b^3}(\sin bt - bt\cos bt)$

 

Theorems on Inverse Laplace Transformation

Theorem 1
If a and b are constants,
$\mathcal{L}^{-1}\left\{ a~f(s) + b~g(s) \right\} = a\mathcal{L}^{-1}\left\{ f(s) \right\} + b\mathcal{L}^{-1}\left\{ g(s) \right\}$
 

Theorem 2
$\mathcal{L}^{-1}\left\{ f(s) \right\} = e^{-at} \mathcal{L}^{-1}\left\{ f(s - a) \right\}$
 

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