### 2 The inverse Laplace transform

The Laplace transform takes a causal function $f\left(t\right)$ and transforms it into a function of $s$ , $F\left(s\right)$ :

$\phantom{\rule{2em}{0ex}}\mathsc{L}\left\{f\left(t\right)\right\}\equiv F\left(s\right)$

The inverse Laplace transform operator is denoted by ${\mathsc{L}}^{-1}$ and involves recovering the original causal function $f\left(t\right)$ . That is,

##### Key Point 5

Inverse Laplace Transform

For example, using standard transforms from Table 1:

${\mathsc{L}}^{-1}\left\{\frac{s}{{s}^{2}+4}\right\}=cos2t\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{0.3em}{0ex}}u\left(t\right)$ since $\mathsc{L}\left\{cos2t\phantom{\rule{0.3em}{0ex}}.\phantom{\rule{0.3em}{0ex}}u\left(t\right)\right\}=\frac{s}{{s}^{2}+4}$ .

Also

$\phantom{\rule{2em}{0ex}}{\mathsc{L}}^{-1}\left\{\frac{3}{{s}^{2}}\right\}=3t\phantom{\rule{0.3em}{0ex}}u\left(t\right)$ since $\mathsc{L}\left\{3t\phantom{\rule{0.3em}{0ex}}u\left(t\right)\right\}=\frac{3}{{s}^{2}}$ .

Because the Laplace transform is a linear operator it follows that the inverse Laplace transform is also linear, so if ${c}_{1},\phantom{\rule{1em}{0ex}}{c}_{2}$ are constants:

##### Key Point 6

Linearity Property of Inverse Laplace Transforms

${\mathsc{L}}^{-1}\left\{{c}_{1}F\left(s\right)+{c}_{2}G\left(s\right)\right\}={c}_{1}{\mathsc{L}}^{-1}\left\{F\left(s\right)\right\}+{c}_{2}{\mathsc{L}}^{-1}\left\{G\left(s\right)\right\}$

For example, to find the inverse Laplace transform of $\frac{2}{{s}^{4}}-\frac{6}{{s}^{2}+4}$ we have

Note that the fractions have had to be manipulated slightly in order that the expressions match precisely with the expressions in Table 1.

Although the inverse Laplace transform can be examined at a deeper mathematical level we shall be content with this simple-minded approach to finding inverse Laplace transforms by using the table of Laplace transforms. However, even this approach is not always straightforward and considerable algebraic manipulation is often required before an inverse Laplace transform can be found. Next we consider two standard rearrangements which often occur.

#### 2.1 Inverting through the use of partial fractions

The function

$\phantom{\rule{2em}{0ex}}F\left(s\right)=\frac{1}{\left(s-1\right)\left(s+2\right)}$

does not appear in our table of transforms and so we cannot, by inspection, write down the inverse Laplace transform. However, by using partial fractions we see that

$\phantom{\rule{2em}{0ex}}F\left(s\right)=\frac{1}{\left(s-1\right)\left(s+2\right)}=\frac{\frac{1}{3}}{s-1}-\frac{\frac{1}{3}}{s+2}$

and so, using the linearity property:

Find the inverse Laplace transform of $\frac{3}{\left(s-1\right)\left({s}^{2}+1\right)}$ .

Begin by using partial fractions to write the given expression in a more suitable form:

$\phantom{\rule{2em}{0ex}}\frac{3}{\left(s-1\right)\left({s}^{2}+1\right)}=\frac{\frac{3}{2}}{s-1}-\frac{\frac{3}{2}s+\frac{3}{2}}{{s}^{2}+1}$

Now continue to obtain the inverse: