The table of inverse Laplace transformations is very useful if you want to compute inverse Laplace of functions. In this post, we will provide here the inverse Laplace table.

Inverse Laplace Table

Function f(t)	Inverse Laplace L-1{f(t)}
1	δ(t), Dirac delta function
1/s	1
1/s2	t
1/s3	t2/2
1/s4	t3/6
1/sn	tn-1/(n-1)!, n=1, 2, 3, …
$\dfrac{1}{s-a}$	eat
$\dfrac{1}{s+a}$	e-at
$\dfrac{1}{(s-a)^n}$	$\dfrac{e^{at} t^{n-1}}{(n-1)!}$
$\dfrac{1}{s^2+a^2}$	$\dfrac{1}{a}$ sin at
$\dfrac{1}{s^2-a^2}$	$\dfrac{1}{a}$ sinh at
$\dfrac{s}{s^2+a^2}$	cos at
$\dfrac{s}{s^2-a^2}$	cosh at
$\dfrac{1}{(s-a)^2+b^2}$	$\dfrac{1}{b}$ eat sin bt
$\dfrac{s-a}{(s-a)^2+b^2}$	eat cos bt
$\dfrac{s}{(s^2+a^2)^2}$	$\dfrac{1}{2a}$ t sin at
$\dfrac{1}{(s^2+a^2)^2}$	$\dfrac{1}{2a^3}$ (sin at – at cos at)

How to Use Inverse Laplace Table

Example 1: Find the inverse Laplace of $\dfrac{s^2-2s+3}{s^3}$.

Solution:

At first, we will break the given function into parts as follows:

$\dfrac{s^2-2s+3}{s^3}$ = $\dfrac{s^2}{s^3}- \dfrac{2s}{s^3}+\dfrac{3}{s^3}$ = $\dfrac{1}{s}- 2\dfrac{1}{s^2}+\dfrac{3}{s^3}$.

So the inverse Laplace of the given function will be

= L^-1(1/s) – 2 L^-1(1/s²) + 3 L^-1(1/s³).

Now, using the above table of inverse Laplace, the above

= 1 – 2t + 3 t²/2

= (2 – 4t + 3 t²)/2.

Read More Inverse Laplace

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