Questions tagged [inverse-laplace]

Definition: If $~\mathcal L\{f(t)\}=F(s)~$, then $~f(t)~$ is the inverse Laplace transform of $~F(s)~$, the inverse being written as:$$f(t)=\mathcal L^{-1}\{F(s)\}$$

Properties :

• Linearity Property

$~\mathcal{L}^{{\left.-{1}\right.}}{\left\lbrace{a}\ {G}_{{1}}{\left({s}\right)}+{b}\ {G}_{{2}}{\left({s}\right)}\right\rbrace}={a}\ {{g}_{{1}}{\left({t}\right)}}+{b}\ {{g}_{{2}}{\left({t}\right)}}~$

• Shifting Property

If $~\mathcal{L}^{{\left.-{1}\right.}}{G}{\left({s}\right)}= g{{\left({t}\right)}}~$, then $~\mathcal{L}^{{\left.-{1}\right.}}{G}{\left({s}-{a}\right)}={e}^{{{a}{t}}} g{{\left({t}\right)}}~$

• If $~\mathcal{L}^{{\left.-{1}\right.}}{G}{\left({s}\right)}= g{{\left({t}\right)}}~$, then $~\mathcal{L}^{{\left.-{1}\right.}}{\left\lbrace\frac{{{G}{\left({s}\right)}}}{{s}}\right\rbrace}={\int_{{0}}^{{t}}} g{{\left({t}\right)}}{\left.{d}{t}\right.}~$

• If $~\mathcal{L}^{{\left.-{1}\right.}}{G}{\left({s}\right)}= g{{\left({t}\right)}}~$,then $~\mathcal{L}^{{\left.-{1}\right.}}{\left\lbrace{e}^{{-{a}{s}}}{G}{\left({s}\right)}\right\rbrace}={u}{\left({t}-{a}\right)}\cdot g{{\left({t}-{a}\right)}}~$

Note: The inverse can generally be obtained by using standard transforms. Often $~F(s)~$ is the ratio of two polynomials and cannot be readily identified with a standard transform. However, the use of partial fractions can often convert such an expression into simple fraction terms which can then be identified with standard transforms.

References:

• "Integral Transforms and Their Applications" by Dambaru Bhatta and Lokenath Debnath
• https://en.wikipedia.org/wiki/Inverse_Laplace_transform