Inverse Laplace Transform

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.

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,

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

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,

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

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 ${s}={s}-{1}+{1}={\left({s}-{1}\right)}+\sqrt{{\frac{{2}}{{11}}}}\cdot\sqrt{{\frac{{11}}{{2}}}}$.

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

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.

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,

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