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(t){f{{\left({t}\right)}}} and g(t){g{{\left({t}\right)}}} are functions on [0,){\left[{0},\infty\right)}; then, the convolution integral of f(t){f{{\left({t}\right)}}} and g(t){g{{\left({t}\right)}}} is (fg)(t)=0tf(tτ)g(τ)dτ{\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 (fg)(t)=(gf)(t){\left({f{\star}}{g}\right)}{\left({t}\right)}={\left({g{\star}}{f}\right)}{\left({t}\right)}.

Proof.

(fg)(t)=0tf(tτ)g(τ)dτ{\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 tτ=u{t}-\tau={u} (or τ=tu\tau={t}-{u} ); then, dτ=du{d}\tau=-{d}{u}. Since τ\tau is changing from 0{0} to t{t}, u{u} is changing from t0=t{t}-{0}={t} to tt=0{t}-{t}={0}.

And the integral can be rewritten as follows:

0tf(tτ)g(τ)dτ=t0f(u)g(tu)(du)=0tf(u)g(tu)du=(gf)(t){\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((fg)(t))=L(0tf(tτ)g(τ)dτ)=0est0tf(tτ)g(τ)dτdt=00testf(tτ)g(τ)dτdt{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: 0τestf(tτ)g(τ)dtdτ{\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 tτ=a{t}-\tau={a} (or t=τ+a{t}=\tau+{a}); then, dt=ddt(tτ)da=da{d}{t}=\frac{{d}}{{{d}{t}}}{\left({t}-\tau\right)}\cdot{d}{a}={d}{a}. Since t{t} is changing from τ\tau to \infty, we have that a{a} is changing from ττ=0\tau-\tau={0} to τ=\infty-\tau=\infty.

This gives 0τestf(tτ)g(τ)dtdτ=00es(τ+a)f(a)g(τ)dadτ=0(esτg(τ)0easf(a)da)dτ=0esaf(a)da0esτg(τ)dτ=F(s)G(s){\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((fg)(t))=F(s)G(s){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 L1(F(s)G(s))=(fg)(t){{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(s)=F(s)G(s){H}{\left({s}\right)}={F}{\left({s}\right)}{G}{\left({s}\right)}, you can use the convolution integral.

Convolution Theorem. L1(F(s)G(s))=(fg)(t){{L}}^{{-{1}}}{\left({F}{\left({s}\right)}{G}{\left({s}\right)}\right)}={\left({f{\star}}{g}\right)}{\left({t}\right)}, where f(t)=L1(F(s)){f{{\left({t}\right)}}}={{L}}^{{-{1}}}{\left({F}{\left({s}\right)}\right)} and g(t)=L1(G(s)){g{{\left({t}\right)}}}={{L}}^{{-{1}}}{\left({G}{\left({s}\right)}\right)}.

Let's work a couple of examples.

Example 1. Calculate L1(s(s2+1)2){{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 s(s2+1)2=ss2+11s2+1\frac{{s}}{{{\left({{s}}^{{2}}+{1}\right)}}^{{2}}}=\frac{{s}}{{{{s}}^{{2}}+{1}}}\frac{{1}}{{{{s}}^{{2}}+{1}}}. Also, it is known that L1(ss2+1)=cos(t){{L}}^{{-{1}}}{\left(\frac{{s}}{{{{s}}^{{2}}+{1}}}\right)}={\cos{{\left({t}\right)}}} and L1(1s2+1)=sin(t){{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{{s}}^{{2}}+{1}}}\right)}={\sin{{\left({t}\right)}}}.

So,

L1(s(s2+1)2)=(cos(t)sin(t))={{L}}^{{-{{1}}}}{\left(\frac{{s}}{{{\left({{s}}^{{2}}+{1}\right)}}^{{2}}}\right)}={\left({\cos{{\left({t}\right)}}}\star{\sin{{\left({t}\right)}}}\right)}=

=0tcos(tτ)sin(τ)dτ=120t(sin(τ(tτ))+sin(τ+(tτ)))dτ=={\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=

=120t(sin(2τt)+sin(t))dτ=12(12cos(2τt)+τsin(t))0t=12(12cos(2tt)+tsin(t))12(12cos(20t)+0cos(t))=14cos(t)+14cos(t)+tsin(t)=14cos(t)+14cos(t)+tsin(t)=tsin(t)=\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 L1(1s(s2)){{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{s}{\left({s}-{2}\right)}}}\right)}.

Since 1s(s2)=1s1s2\frac{{1}}{{{s}{\left({s}-{2}\right)}}}=\frac{{1}}{{s}}\frac{{1}}{{{s}-{2}}}, L1(1s)=1{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{s}}\right)}={1} and L1(1s2)=e2t{{L}}^{{-{{1}}}}{\left(\frac{{1}}{{{s}-{2}}}\right)}={{e}}^{{{2}{t}}}, we have that

L1(1s(s2))=(1e2t)=0t1e2τdτ=12e2τ0t=12e2t12e20=12(e2t1){{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)}