# Inverse Laplace Transform

As has been seen previously, the formula for the Laplace transform is ${F}{\left({s}\right)}={L}{\left({f{{\left({t}\right)}}}\right)}$; in other words, we are given a function ${f{{\left({t}\right)}}}$ and we need to find ${F}{\left({s}\right)}$. The inverse Laplace transform is the operation of finding ${f{{\left({t}\right)}}}$ given ${F}{\left({s}\right)}$.

The operation of the Laplace transform is quite straightforward: we need to calculate the imporper integral and we are done. Finding the inverse Laplace transform is more tedious and complicated. The table of Laplace transforms is very useful for this.

Property. Linearity of the inverse Laplace transform: ${{L}}^{{-{{1}}}}{\left({a}{F}{\left({s}\right)}+{b}{G}{\left({s}\right)}\right)}={a}{{L}}^{{-{{1}}}}{\left({F}{\left({s}\right)}\right)}+{b}{{L}}^{{-{{1}}}}{\left({G}{\left({s}\right)}\right)}={a}{f{{\left({t}\right)}}}+{b}{g{{\left({t}\right)}}}$, where ${a}$ and ${b}$ are some constants.

Note that there can be more than 2 terms, and linearity will still hold.

Proof.

From the linear property of the Laplace transform, we have that ${a}{F}{\left({s}\right)}+{b}{G}{\left({s}\right)}={L}{\left({a}{f{{\left({t}\right)}}}+{b}{f{{\left({t}\right)}}}\right)}$. Taking the inverse Laplace transform of both sides yields

${{L}}^{{-{{1}}}}{\left({a}{F}{\left({s}\right)}+{b}{G}{\left({s}\right)}\right)}={{L}}^{{-{{1}}}}{\left({L}{\left({a}{f{{\left({t}\right)}}}+{b}{g{{\left({t}\right)}}}\right)}\right)}$,

or

${{L}}^{{-{{1}}}}{\left({a}{F}{\left({s}\right)}+{b}{G}{\left({s}\right)}\right)}={a}{f{{\left({t}\right)}}}+{b}{g{{\left({t}\right)}}}$, since ${{L}}^{{-{{1}}}}{\left({L}{\left({f{{\left({t}\right)}}}\right)}\right)}={f{{\left({t}\right)}}}$.

This completes the proof.

Let's go through a couple of examples.

Example 1. Calculate ${{L}}^{{-{{1}}}}{\left(\frac{{5}}{{s}}+\frac{{2}}{{{s}-{2}}}+\frac{{5}}{{{{s}}^{{2}}+{1}}}+\frac{{1}}{{{s}+{3}}}\right)}$.

${{L}}^{{-{{1}}}}{\left(\frac{{5}}{{s}}+\frac{{2}}{{{s}-{2}}}+\frac{{5}}{{{{s}}^{{2}}+{1}}}+\frac{{1}}{{{s}+{3}}}\right)}={5}{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{s}}\right)}+{2}{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{s}-{2}}}\right)}+{5}{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{{s}}^{{2}}+{1}}}+\frac{{1}}{{{s}-{\left(-{3}\right)}}}\right)}={5}\cdot{\left({1}\right)}+{2}{{e}}^{{{2}{t}}}+{5}{\sin{{\left({t}\right)}}}+{{e}}^{{-{3}{t}}}={5}+{2}{{e}}^{{{2}{t}}}+{5}{\sin{{\left({t}\right)}}}+{{e}}^{{-{3}{t}}}$

The above inverse Laplace transforms can be seen in the table of Laplace transforms.

Now, let's see another example.

Example 2. Calculate ${{L}}^{{-{{1}}}}{\left(\frac{{{s}+{1}}}{{{{s}}^{{2}}+{3}}}+\frac{{1}}{{{5}{s}-{7}}}\right)}$.

${{L}}^{{-{{1}}}}{\left(\frac{{{s}+{1}}}{{{{s}}^{{2}}+{3}}}\right)}={{L}}^{{-{{1}}}}{\left(\frac{{s}}{{{{s}}^{{2}}+{3}}}\right)}+{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{{s}}^{{2}}+{3}}}\right)}+{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{5}{\left({s}-\frac{{7}}{{5}}\right)}}}\right)}$.

On this stage, look at the denominator of the first two terms. Since $\frac{{1}}{{{{s}}^{{2}}+{3}}}=\frac{{1}}{{{{s}}^{{2}}+{{\left(\sqrt{{{3}}}\right)}}^{{2}}}}$, it seems that the first term should be ${\cos{{\left(\sqrt{{{3}}}{t}\right)}}}$ and the seond term is ${\cos{{\left(\sqrt{{{3}}}{t}\right)}}}$. However, $\sqrt{{{3}}}$ is missing in the numerator for ${\sin{{\left(\sqrt{{{3}}}{t}\right)}}}$. So,

${{L}}^{{-{{1}}}}{\left(\frac{{s}}{{{{s}}^{{2}}+{3}}}\right)}+{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{{s}}^{{2}}+{3}}}\right)}+{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{5}{\left({s}-\frac{{7}}{{5}}\right)}}}\right)}={\cos{{\left(\sqrt{{{3}}}{t}\right)}}}+{{L}}^{{-{{1}}}}{\left(\frac{\sqrt{{{3}}}}{\sqrt{{{3}}}}\frac{{1}}{{{{s}}^{{2}}+{3}}}\right)}+\frac{{1}}{{5}}{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{s}-\frac{{7}}{{5}}}}\right)}={\cos{{\left(\sqrt{{{3}}}{t}\right)}}}+\frac{{1}}{\sqrt{{{3}}}}{{L}}^{{-{{1}}}}{\left(\frac{{\sqrt{{{3}}}}}{{{{s}}^{{2}}+{3}}}\right)}+\frac{{1}}{{5}}{{e}}^{{\frac{{7}}{{5}}{t}}}={\cos{{\left(\sqrt{{{3}}}{t}\right)}}}+\frac{{1}}{\sqrt{{{3}}}}{\sin{{\left(\sqrt{{{3}}}{t}\right)}}}+\frac{{1}}{{5}}{{e}}^{{\frac{{7}}{{5}}{t}}}$

The denominator is the key part in taking the inverse transform. Identify what function are you dealing with based on the denominator. If there is more than one possibility, examine the numerator. Then correct the numerator to get the form you need.

Example 3. Calculate ${{L}}^{{-{{1}}}}{\left(\frac{{{7}{s}-{5}}}{{{{s}}^{{2}}+{2}{s}+{5}}}\right)}$.

Again, examine the denominator: ${{s}}^{{2}}+{2}{s}+{5}={{s}}^{{2}}+{2}{s}+{1}+{4}={{\left({s}+{1}\right)}}^{{2}}+{{2}}^{{2}}={{\left({s}-{\left(-{1}\right)}\right)}}^{{2}}+{{2}}^{{2}}$.

Seems like we are dealing with either ${{e}}^{{-{t}}}{\cos{{\left({2}{t}\right)}}}$ or ${{e}}^{{-{t}}}{\cos{{\left({2}{t}\right)}}}$. Now, take a look at the numerator. It is ${7}{s}-{5}$, or ${7}{\left({s}-\frac{{5}}{{7}}\right)}$. For ${{e}}^{{-{t}}}{\cos{{\left({2}{t}\right)}}}$, we need ${s}+{1}$ in the numerator, and for ${{e}}^{{-{t}}}{\sin{{\left({2}{t}\right)}}}$, we need ${2}$ in the numerator; so, rewrite the numerator as ${7}{\left({s}+{1}-{1}-\frac{{5}}{{7}}\right)}={7}{\left({s}+{1}-\frac{{12}}{{7}}\right)}={7}{\left({s}+{1}\right)}-{12}={7}{\left({s}+{1}\right)}-{2}\cdot{6}$.

Thus,

${{L}}^{{-{{1}}}}{\left(\frac{{{7}{s}-{5}}}{{{{s}}^{{2}}+{2}{s}+{5}}}\right)}={{L}}^{{-{{1}}}}{\left(\frac{{{7}{\left({s}+{1}\right)}}}{{{{\left({s}+{1}\right)}}^{{2}}+{{2}}^{{2}}}}\right)}-{{L}}^{{-{{1}}}}{\left(\frac{{{2}\cdot{6}}}{{{\left({s}+{1}\right)}+{{2}}^{{2}}}}\right)}={7}{{L}}^{{-{{1}}}}{\left(\frac{{{s}+{1}}}{{{{\left({s}+{1}\right)}}^{{2}}+{{2}}^{{2}}}}\right)}-{6}{{L}}^{{-{{1}}}}{\left(\frac{{2}}{{{\left({s}+{1}\right)}+{{2}}^{{2}}}}\right)}={7}{{e}}^{{-{t}}}{\cos{{\left({2}{t}\right)}}}-{6}{{e}}^{{-{t}}}{\sin{{\left({2}{t}\right)}}}={{e}}^{{-{t}}}{\left({7}{\cos{{\left({2}{t}\right)}}}-{6}{\sin{{\left({2}{t}\right)}}}\right)}$

Let's proceed to the next example.

Example 4. Calculate ${{L}}^{{-{{1}}}}{\left(\frac{{s}}{{{2}{{s}}^{{2}}-{4}{s}-{9}}}\right)}$.

Examine the denominator: ${2}{{s}}^{{2}}-{4}{s}-{9}={2}{\left({{s}}^{{2}}-{2}{s}-\frac{{9}}{{2}}\right)}={2}{\left({{s}}^{{2}}-{2}{s}+{1}-{1}-\frac{{9}}{{2}}\right)}={2}{\left({{\left({s}-{1}\right)}}^{{2}}-\frac{{11}}{{2}}\right)}$.

So, we are dealing with ${{e}}^{{t}}{\sinh{{\left(\sqrt{{\frac{{11}}{{2}}}}{t}\right)}}}$ and ${{e}}^{{t}}{\cosh{{\left(\sqrt{{\frac{{11}}{{2}}}}{t}\right)}}}$. For the term involving ${\cosh{}}$, we need ${s}-{1}$ in the numerator, and for the term involving ${\sinh{}}$, we need $\sqrt{{\frac{{11}}{{2}}}}$; so, rewrite the numerator as ${s}={s}-{1}+{1}={\left({s}-{1}\right)}+\sqrt{{\frac{{2}}{{11}}}}\cdot\sqrt{{\frac{{11}}{{2}}}}$.

${{L}}^{{-{{1}}}}{\left(\frac{{s}}{{{2}{{s}}^{{2}}-{4}{s}-{9}}}\right)}={{L}}^{{-{{1}}}}{\left(\frac{{{\left({s}-{1}\right)}+\sqrt{{\frac{{2}}{{11}}}}\cdot\sqrt{{\frac{{11}}{{2}}}}}}{{{2}{\left({{\left({s}-{1}\right)}}^{{2}}-{{\left(\sqrt{{\frac{{11}}{{2}}}}\right)}}^{{2}}\right)}}}\right)}=\frac{{1}}{{2}}{{L}}^{{-{{1}}}}{\left(\frac{{{s}-{1}}}{{{{\left({s}-{1}\right)}}^{{2}}+{{\left(\sqrt{{\frac{{11}}{{2}}}}\right)}}^{{2}}}}\right)}+\sqrt{{\frac{{1}}{{22}}}}{{L}}^{{-{{1}}}}{\left(\frac{{\sqrt{{\frac{{11}}{{2}}}}}}{{{{\left({s}-{1}\right)}}^{{2}}-{{\left(\sqrt{{\frac{{11}}{{2}}}}\right)}}^{{2}}}}\right)}=\frac{{1}}{{2}}{{e}}^{{t}}{\cosh{{\left(\sqrt{{\frac{{11}}{{2}}}}{t}\right)}}}+\sqrt{{\frac{{1}}{{22}}}}{{e}}^{{t}}{\sinh{{\left(\sqrt{{\frac{{11}}{{2}}}}{t}\right)}}}=\frac{{{e}}^{{t}}}{{22}}{\left(\sqrt{{{22}}}{\sinh{{\left(\sqrt{{\frac{{11}}{{2}}}}{t}\right)}}}+{11}{\cosh{{\left(\sqrt{{\frac{{11}}{{2}}}}{t}\right)}}}\right)}$

The next examples are those that you will be dealing with most often. They require the knowledge of partial fraction decomposition.

Example 5. Calculate ${{L}}^{{-{{1}}}}{\left(\frac{{{s}+{2}}}{{{\left({s}+{3}\right)}{\left({s}-{1}\right)}}}\right)}$.

Looking at the denominator, we see that it has the terms ${s}+{3}$ and ${s}-{1}$, and it looks like that inverse transform will contain ${{e}}^{{-{3}{t}}}$ and ${{e}}^{{t}}$.

However, to do that, we need to split the "product" into sum.

To do this, assume that $\frac{{{s}+{2}}}{{{\left({s}+{3}\right)}{\left({s}-{1}\right)}}}=\frac{{A}}{{{s}+{3}}}+\frac{{B}}{{{s}-{1}}}$, where ${A}$ and ${B}$ are unknown constants.

Then,

$\frac{{A}}{{{s}+{3}}}+\frac{{B}}{{{s}-{1}}}=\frac{{{A}{\left({s}-{1}\right)}+{B}{\left({s}+{3}\right)}}}{{{\left({s}-{1}\right)}{\left({s}+{3}\right)}}}=\frac{{{s}{\left({A}+{B}\right)}-{A}+{3}{B}}}{{{\left({s}-{1}\right)}{\left({s}+{3}\right)}}}$.

And this must equal $\frac{{{s}+{2}}}{{{\left({s}+{3}\right)}{\left({s}-{1}\right)}}}$; so, we obtain following system:

${\left\{\begin{array}{c}{A}+{B}={1}\\-{A}+{3}{B}={2}\\ \end{array}\right.}$.

It has the solution ${A}=\frac{{1}}{{4}}$ ${B}=\frac{{3}}{{4}}$ (for steps, see system of linear equations calculator).

So, $\frac{{{s}+{2}}}{{{\left({s}+{3}\right)}{\left({s}-{1}\right)}}}=\frac{{1}}{{4}}\frac{{1}}{{{s}+{3}}}+\frac{{3}}{{4}}\frac{{1}}{{{s}-{1}}}$.

This is called partial fraction decomposition.

Finally,

${{L}}^{{-{{1}}}}{\left(\frac{{{s}+{2}}}{{{\left({s}+{3}\right)}{\left({s}-{1}\right)}}}\right)}=\frac{{1}}{{4}}{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{s}+{3}}}\right)}+\frac{{3}}{{4}}{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{s}-{1}}}\right)}=\frac{{1}}{{4}}{{e}}^{{-{3}{t}}}+\frac{{3}}{{4}}{{e}}^{{t}}$.

Note that we could write the denominator as ${\left({s}+{3}\right)}{\left({s}-{1}\right)}={{s}}^{{2}}+{2}{s}-{3}={{s}}^{{2}}+{2}{s}+{1}-{4}={{\left({s}+{1}\right)}}^{{2}}-{4}={{\left({s}+{1}\right)}}^{{2}}-{{2}}^{{2}}$ and solve it like in the previous example.

However, there are problems where this approach wouldn't work.

In general, if you have in the denominator ${{\left({a}_{{n}}{{x}}^{{n}}+{a}_{{{n}-{1}}}{{x}}^{{{n}-{1}}}+\ldots+{a}_{{0}}\right)}}^{{k}}$, the corresponding term in partial fraction decomposition is $\frac{{{A}_{{{n}-{1}}}{{x}}^{{{n}-{1}}}+{A}_{{{n}-{2}}}{{x}}^{{{n}-{2}}}+\ldots+{A}_{{0}}}}{{{a}_{{n}}{{x}}^{{n}}+{a}_{{{n}-{1}}}{{x}}^{{{n}-{1}}}+\ldots+{a}_{{0}}}}+\frac{{{A}_{{{n}-{1}}}{{x}}^{{{n}-{1}}}+{A}_{{{n}-{2}}}{{x}}^{{{n}-{2}}}+\ldots+{A}_{{0}}}}{{{\left({a}_{{n}}{{x}}^{{n}}+{a}_{{{n}-{1}}}{{x}}^{{{n}-{1}}}+\ldots+{a}_{{0}}\right)}}^{{2}}}+\ldots+\frac{{{A}_{{{n}-{1}}}{{x}}^{{{n}-{1}}}+{A}_{{{n}-{2}}}{{x}}^{{{n}-{2}}}+\ldots+{A}_{{0}}}}{{{\left({a}_{{n}}{{x}}^{{n}}+{a}_{{{n}-{1}}}{{x}}^{{{n}-{1}}}+\ldots+{a}_{{0}}\right)}}^{{k}}}$

In particular,

1. ${a}{x}+{b}\to\frac{{A}}{{{a}{x}+{b}}}$
2. ${a}{{x}}^{{2}}+{b}{x}+{c}\to\frac{{{A}{x}+{B}}}{{{a}{{x}}^{{2}}+{b}{x}+{c}}}$
3. ${{\left({a}{x}+{b}\right)}}^{{2}}\to\frac{{A}}{{{a}{x}+{b}}}+\frac{{B}}{{{\left({a}{x}+{b}\right)}}^{{2}}}$
4. ${{\left({a}{{x}}^{{2}}+{b}{x}+{c}\right)}}^{{2}}\to\frac{{{A}{x}+{B}}}{{{a}{{x}}^{{2}}+{b}{x}+{c}}}+\frac{{{C}{x}+{D}}}{{{\left({a}{{x}}^{{2}}+{b}{x}+{c}\right)}}^{{2}}}$

Example 6. Calculate ${{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{{s}}^{{2}}{{\left({s}+{1}\right)}}^{{2}}}}\right)}$.

The partial fraction decomposition is

$\frac{{1}}{{{{s}}^{{2}}{{\left({s}+{1}\right)}}^{{2}}}}=\frac{{A}}{{s}}+\frac{{B}}{{{s}}^{{2}}}+\frac{{C}}{{{s}+{1}}}+\frac{{D}}{{{\left({s}+{1}\right)}}^{{2}}}$.

$\frac{{A}}{{s}}+\frac{{B}}{{{s}}^{{2}}}+\frac{{C}}{{{s}+{1}}}+\frac{{D}}{{{\left({s}+{1}\right)}}^{{2}}}=\frac{{{A}{s}{{\left({s}+{1}\right)}}^{{2}}+{B}{{\left({s}+{1}\right)}}^{{2}}+{C}{{s}}^{{2}}{\left({s}+{1}\right)}+{D}{{s}}^{{2}}}}{{{{s}}^{{2}}{{\left({s}+{1}\right)}}^{{2}}}}=\frac{{{A}{{s}}^{{3}}+{2}{A}{{s}}^{{2}}+{A}{s}+{B}{{s}}^{{2}}+{2}{B}{s}+{B}+{C}{{s}}^{{3}}+{C}{{s}}^{{2}}+{D}{{s}}^{{2}}}}{{{{s}}^{{2}}{{\left({s}+{1}\right)}}^{{2}}}}=$

$=\frac{{{{s}}^{{3}}{\left({A}+{C}\right)}+{{s}}^{{2}}{\left({2}{A}+{B}+{C}+{D}\right)}+{s}{\left({A}+{2}{B}\right)}+{B}}}{{{\left({{s}}^{{2}}{\left({s}+{1}\right)}\right)}}^{{2}}}$

It should equal $\frac{{1}}{{{{s}}^{{2}}{{\left({s}+{1}\right)}}^{{2}}}}$; so, we have following system:

${\left\{\begin{array}{c}{A}+{C}={0}\\{2}{A}+{B}+{C}+{D}={0}\\{A}+{2}{B}={0}\\{B}={1}\\ \end{array}\right.}$

From the last equation, ${B}={1}$; from the third equation, ${A}=-{2}$; from the first equation, ${C}={2}$; and from the second equation,${D}={1}$. So,

$\frac{{1}}{{{{s}}^{{2}}{{\left({s}+{1}\right)}}^{{2}}}}=-\frac{{2}}{{s}}+\frac{{1}}{{{s}}^{{2}}}+\frac{{2}}{{{s}+{1}}}+\frac{{1}}{{{\left({s}+{1}\right)}}^{{2}}}$.

Therefore,

${{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{{s}}^{{2}}{{\left({s}+{1}\right)}}^{{2}}}}\right)}=-{2}{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{s}}\right)}+{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{s}}^{{2}}}\right)}+{2}{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{s}+{1}}}\right)}+{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{\left({s}+{1}\right)}}^{{2}}}\right)}=-{2}\cdot{1}+{t}+{2}{{e}}^{{-{t}}}+{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{\left({s}+{1}\right)}}^{{2}}}\right)}$

To find ${{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{\left({s}+{1}\right)}}^{{2}}}\right)}$, note that $-\frac{{d}}{{{d}{s}}}\frac{{1}}{{{s}+{1}}}=\frac{{1}}{{{\left({s}+{1}\right)}}^{{2}}}$. Now, using property 3 of the Laplace transform and the fact that ${L}{\left({{e}}^{{-{{t}}}}\right)}=\frac{{1}}{{{s}+{1}}}$, we obtain that

${L}{\left({t}{{e}}^{{-{{t}}}}\right)}=-\frac{{d}}{{{d}{s}}}\frac{{1}}{{{s}+{1}}}=\frac{{1}}{{{\left({s}+{1}\right)}}^{{2}}}$.

Taking the inverse transform of both sides yields

${{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{\left({s}+{1}\right)}}^{{2}}}\right)}={t}{{e}}^{{-{{t}}}}$.

Finally,

${{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{{s}}^{{2}}}}{{\left({s}+{1}\right)}}^{{2}}\right)}=-{2}+{t}+{2}{{e}}^{{-{{t}}}}+{t}{{e}}^{{-{{t}}}}$.

As can be seen, to successfully take the inverse Laplace transform, you should identify the denominator and master partial fraction decomposition. These are two essential parts, and only a lot of practice can improve your skills.