Convolution Integral

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

Definition. Let f(t) and g(t) are functions on `[0,oo)` then the convolution integral of f(t) and g(t) is `(f**g)(t)=int_0^t f(t-tau)g(tau)dtau` .

One property of convolution integral is that `(f**g)(t)=(g**f)(t)`


`(f**g)(t)=int_0^t f(t-tau)g(tau)dtau`

Now use change of variable: let `t-tau=u` (or `tau=t-u` ) then `dtau=-du` . Since `tau` is changing from 0 to t then u is changing from `t-0=t` to `t-t=0` .

And integral can be rewritten as follows:

`int_0^t f(t-tau)g(tau)dtau=int_t^0 f(u)g(t-u)*(-du)=int_0^t f(u)g(t-u)du=(g**f)(t)`

This completes the proof.

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

`L((f**g)(t))=L(int_0^tf(t-tau)g(tau)dtau)=int_0^oo e^(-st) int_0^t f(t-tau)g(tau)dtaudt=int_0^oo int_0^t e^(-st)f(t-tau)g(tau)dtaudt`

Now, reverse order of integration: `int_0^oo int_tau^oo e^(-st)f(t-tau)g(tau)dtdtau` .

Next step is to make substitution `t-tau=a` (or `t=tau+a` ) then `dt=d/(dt)(t-tau)*da=da` . Since t is changing from `tau` to `oo` then a is changing from `tau-tau=0` to `oo-tau=oo`

This gives `int_0^oo int_tau^oo e^(-st)f(t-tau)g(tau)dtdtau=int_0^oo int_0^oo e^(-s(tau+a))f(a)g(tau)dadtau=int_0^oo (e^(-s tau)g(tau) int_0^oo e^(-as)f(a)da)dtau=int_0^oo e^(-sa)f(a) da int_0^oo e^(-s tau)g(tau)dtau=F(s)G(s)`

This gives that `L((f**g)(t))=F(s)G(s)` . Taking inverse Laplace trnasform of both sides gives `L^-1(F(s)G(s))=(f**g)(t)`

Very nice! This means that if you want to find inverse Laplace transform of `H(s)=F(s)G(s)` then you can use convolution integral.

Convolution Theorem. `L^(-1)(F(s)G(s))=(f**g)(t)` where `f(t)=L^(-1)(F(s))` and `g(t)=L^(-1)(G(s))`.

Let's work a couple of examples.

Example 1. Calculate `L^-1(s/(s^2+1)^2)`

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

Note, that `s/(s^2+1)^2=s/(s^2+1)1/(s^2+1)` . Also it is known that `L^(-1)(s/(s^2+1))=cos(t)` and `L^-1(1/(s^2+1))=sin(t)`



`=int_0^t cos(t-tau)sin(tau)dtau=1/2 int_0^t (sin(tau-(t-tau))+sin(tau+(t-tau)))dtau=`

`=1/2 int_0^t (sin(2tau-t)+sin(t))dtau=1/2(-1/2cos(2tau-t)+tau sin(t))|_0^t=1/2(-1/2cos(2t-t)+t sin(t))-1/2(-1/2cos(2*0-t)+0* cos(t))=-1/4cos(t)+1/4cos(-t)+tsin(t)=-1/4cos(t)+1/4cos(t)+tsin(t)=tsin(t)`

Example 2. Calculate `L^-1(1/(s(s-2)))`

Since `1/(s(s-2))=1/s 1/(s-2)` ,`L^-1(1/s)=1` and `L^-1(1/(s-2))=e^(2t)` then

`L^-1(1/(s(s-2)))=(1**e^(2t))=int_0^t 1*e^(2tau)dtau=1/2e^(2tau)|_0^t=1/2e^(2t)-1/2e^(2*0)=1/2(e^(2t)-1)`