Lesson 26 - Convolution

In this lesson, we introduce the method of convolution. This method is to be used when you have a composition of piecewise functions and need to take the Laplace transform of those functions.

Definition

The convolution $f\ast g$ of piecewise continuous functions $f$ and $g$ is defined for $t\ge 0$ as follows:

Properties of Convolution

1. $(f\ast g)(t)=(g\ast f)(t)$
2. $(f\ast (g+h))(t)=(f\ast g)(t)+(f\ast h)(t)$
3. $((f\ast g)\ast h)(t)=(f\ast (g\ast h))(t)$
4. $(f\ast 0)(t)=0$

Theorem

$\mathcal{L}[(f\ast g)(t)]=F(s)G(s)$, where $F(s)=\mathcal{L}[f(t)],$ $G(s)=\mathcal{L}[g(t)]$

Example

Use convolution to find: ${\mathcal{L}}^{-1}\left[\frac{1}{(s^2+1{)}^2}\right]$

${\mathcal{L}}^{-1}\left[\frac{1}{(s^2+2{)}^2}\right]={\mathcal{L}}^{-1}[\frac{1}{s^2+1}\cdot \frac{1}{s^2+1}]$

Identify $f(t)$ and $g(t)$. In this case, $f(t)=g(t)=\frac{1}{s^2+1}$.

Next, we take the Laplace transform of $f(t)$ and $g(t)$, which will be the same since the functions are the same.

$\mathcal{L}[f(t)]=\mathcal{L}[g(t)]=\sin t$

Now let's plug this into our definition above.

Now we can utilize a trig identity in order to solve this integral.

Recall $\sin A\sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)]$. Let's say $A=v$, and $B=t-v$.

Now we can perform the integration.

Recall $\sin (-t)=-\sin t$.

So,

Try for Yourself

1. ${\mathcal{L}}^{-1}\left[\frac{1}{s(s^2+1)}\right]$
2. ${\mathcal{L}}^{-1}\left[\frac{s}{(s^2+1{)}^2}\right]$
3. ${\mathcal{L}}^{-1}\left[\frac{s+1}{(s^2+1{)}^2}\right]$

Next Lesson: Lesson 27 - The Dirac Delta Function

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