Convolution Integral

To find the inverse Laplace transform, partial fraction decomposition is very useful, but sometimes it can be very difficult to find the partial fraction decomposition, so there are cases where the inverse Laplace transform can be found with the help of the convolution integral.

Definition. Let ${f{{\left({t}\right)}}}$ and ${g{{\left({t}\right)}}}$ are functions on ${\left[{0},\infty\right)}$; then, the convolution integral of ${f{{\left({t}\right)}}}$ and ${g{{\left({t}\right)}}}$ is ${\left({f{\star}}{g}\right)}{\left({t}\right)}={\int_{{0}}^{{t}}}{f{{\left({t}-\tau\right)}}}{g{{\left(\tau\right)}}}{d}\tau$.

One property of the convolution integral is that ${\left({f{\star}}{g}\right)}{\left({t}\right)}={\left({g{\star}}{f}\right)}{\left({t}\right)}$.

Proof.

${\left({f{\star}}{g}\right)}{\left({t}\right)}={\int_{{0}}^{{t}}}{f{{\left({t}-\tau\right)}}}{g{{\left(\tau\right)}}}{d}\tau$

Now, use the change of variable: let ${t}-\tau={u}$ (or $\tau={t}-{u}$ ); then, ${d}\tau=-{d}{u}$. Since $\tau$ is changing from ${0}$ to ${t}$, ${u}$ is changing from ${t}-{0}={t}$ to ${t}-{t}={0}$.

And the integral can be rewritten as follows:

${\int_{{0}}^{{t}}}{f{{\left({t}-\tau\right)}}}{g{{\left(\tau\right)}}}{d}\tau={\int_{{t}}^{{0}}}{f{{\left({u}\right)}}}{g{{\left({t}-{u}\right)}}}\cdot{\left(-{d}{u}\right)}={\int_{{0}}^{{t}}}{f{{\left({u}\right)}}}{g{{\left({t}-{u}\right)}}}{d}{u}={\left({g{\star}}{f}\right)}{\left({t}\right)}$

This completes the proof.

Now, let's find the Laplace transform of the convolution integral:

${L}{\left({\left({f{\star}}{g}\right)}{\left({t}\right)}\right)}={L}{\left({\int_{{0}}^{{t}}}{f{{\left({t}-\tau\right)}}}{g{{\left(\tau\right)}}}{d}\tau\right)}={\int_{{0}}^{\infty}}{{e}}^{{-{s}{t}}}{\int_{{0}}^{{t}}}{f{{\left({t}-\tau\right)}}}{g{{\left(\tau\right)}}}{d}\tau{d}{t}={\int_{{0}}^{\infty}}{\int_{{0}}^{{t}}}{{e}}^{{-{s}{t}}}{f{{\left({t}-\tau\right)}}}{g{{\left(\tau\right)}}}{d}\tau{d}{t}$

Now, reverse the order of integration: ${\int_{{0}}^{\infty}}{\int_{\tau}^{\infty}}{{e}}^{{-{s}{t}}}{f{{\left({t}-\tau\right)}}}{g{{\left(\tau\right)}}}{d}{t}{d}\tau$.

The next step is to make the substitution ${t}-\tau={a}$ (or ${t}=\tau+{a}$); then, ${d}{t}=\frac{{d}}{{{d}{t}}}{\left({t}-\tau\right)}\cdot{d}{a}={d}{a}$. Since ${t}$ is changing from $\tau$ to $\infty$, we have that ${a}$ is changing from $\tau-\tau={0}$ to $\infty-\tau=\infty$.

This gives ${\int_{{0}}^{\infty}}{\int_{\tau}^{\infty}}{{e}}^{{-{s}{t}}}{f{{\left({t}-\tau\right)}}}{g{{\left(\tau\right)}}}{d}{t}{d}\tau={\int_{{0}}^{\infty}}{\int_{{0}}^{\infty}}{{e}}^{{-{s}{\left(\tau+{a}\right)}}}{f{{\left({a}\right)}}}{g{{\left(\tau\right)}}}{d}{a}{d}\tau={\int_{{0}}^{\infty}}{\left({{e}}^{{-{s}\tau}}{g{{\left(\tau\right)}}}{\int_{{0}}^{\infty}}{{e}}^{{-{a}{s}}}{f{{\left({a}\right)}}}{d}{a}\right)}{d}\tau={\int_{{0}}^{\infty}}{{e}}^{{-{s}{a}}}{f{{\left({a}\right)}}}{d}{a}{\int_{{0}}^{\infty}}{{e}}^{{-{s}\tau}}{g{{\left(\tau\right)}}}{d}\tau={F}{\left({s}\right)}{G}{\left({s}\right)}$

So, ${L}{\left({\left({f{\star}}{g}\right)}{\left({t}\right)}\right)}={F}{\left({s}\right)}{G}{\left({s}\right)}$. Taking the inverse Laplace trnasform of both sides gives ${{L}}^{{-{{1}}}}{\left({F}{\left({s}\right)}{G}{\left({s}\right)}\right)}={\left({f{\star}}{g}\right)}{\left({t}\right)}$.

Very nice! This means that, if you want to find the inverse Laplace transform of ${H}{\left({s}\right)}={F}{\left({s}\right)}{G}{\left({s}\right)}$, you can use the convolution integral.

Convolution Theorem. ${{L}}^{{-{1}}}{\left({F}{\left({s}\right)}{G}{\left({s}\right)}\right)}={\left({f{\star}}{g}\right)}{\left({t}\right)}$, where ${f{{\left({t}\right)}}}={{L}}^{{-{1}}}{\left({F}{\left({s}\right)}\right)}$ and ${g{{\left({t}\right)}}}={{L}}^{{-{1}}}{\left({G}{\left({s}\right)}\right)}$.

Let's work a couple of examples.

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

We, of course, can use partial fraction decomposition to find the inverse transform, but it is much easier to calculate the inverse transform with the help of the convolution integral.

Note that $\frac{{s}}{{{\left({{s}}^{{2}}+{1}\right)}}^{{2}}}=\frac{{s}}{{{{s}}^{{2}}+{1}}}\frac{{1}}{{{{s}}^{{2}}+{1}}}$. Also, it is known that ${{L}}^{{-{1}}}{\left(\frac{{s}}{{{{s}}^{{2}}+{1}}}\right)}={\cos{{\left({t}\right)}}}$ and ${{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{{s}}^{{2}}+{1}}}\right)}={\sin{{\left({t}\right)}}}$.

So,

${{L}}^{{-{{1}}}}{\left(\frac{{s}}{{{\left({{s}}^{{2}}+{1}\right)}}^{{2}}}\right)}={\left({\cos{{\left({t}\right)}}}\star{\sin{{\left({t}\right)}}}\right)}=$

$={\int_{{0}}^{{t}}}{\cos{{\left({t}-\tau\right)}}}{\sin{{\left(\tau\right)}}}{d}\tau=\frac{{1}}{{2}}{\int_{{0}}^{{t}}}{\left({\sin{{\left(\tau-{\left({t}-\tau\right)}\right)}}}+{\sin{{\left(\tau+{\left({t}-\tau\right)}\right)}}}\right)}{d}\tau=$

$=\frac{{1}}{{2}}{\int_{{0}}^{{t}}}{\left({\sin{{\left({2}\tau-{t}\right)}}}+{\sin{{\left({t}\right)}}}\right)}{d}\tau=\frac{{1}}{{2}}{\left(-\frac{{1}}{{2}}{\cos{{\left({2}\tau-{t}\right)}}}+\tau{\sin{{\left({t}\right)}}}\right)}{{\mid}_{{0}}^{{t}}}=\frac{{1}}{{2}}{\left(-\frac{{1}}{{2}}{\cos{{\left({2}{t}-{t}\right)}}}+{t}{\sin{{\left({t}\right)}}}\right)}-\frac{{1}}{{2}}{\left(-\frac{{1}}{{2}}{\cos{{\left({2}\cdot{0}-{t}\right)}}}+{0}\cdot{\cos{{\left({t}\right)}}}\right)}=-\frac{{1}}{{4}}{\cos{{\left({t}\right)}}}+\frac{{1}}{{4}}{\cos{{\left(-{t}\right)}}}+{t}{\sin{{\left({t}\right)}}}=-\frac{{1}}{{4}}{\cos{{\left({t}\right)}}}+\frac{{1}}{{4}}{\cos{{\left({t}\right)}}}+{t}{\sin{{\left({t}\right)}}}={t}{\sin{{\left({t}\right)}}}$

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

Since $\frac{{1}}{{{s}{\left({s}-{2}\right)}}}=\frac{{1}}{{s}}\frac{{1}}{{{s}-{2}}}$, ${{L}}^{{-{{1}}}}{\left(\frac{{1}}{{s}}\right)}={1}$ and ${{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{s}-{2}}}\right)}={{e}}^{{{2}{t}}}$, we have that

${{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{s}{\left({s}-{2}\right)}}}\right)}={\left({1}\star{{e}}^{{{2}{t}}}\right)}={\int_{{0}}^{{t}}}{1}\cdot{{e}}^{{{2}\tau}}{d}\tau=\frac{{1}}{{2}}{{e}}^{{{2}\tau}}{{\mid}_{{0}}^{{t}}}=\frac{{1}}{{2}}{{e}}^{{{2}{t}}}-\frac{{1}}{{2}}{{e}}^{{{2}\cdot{0}}}=\frac{{1}}{{2}}{\left({{e}}^{{{2}{t}}}-{1}\right)}$