In a previous lesson, we proved that the Laplace transform \mathscr{L} is a linear transformation, i.e.,
\mathscr{L}\{\alpha f(t)+g(t)\}=\alpha\mathscr{L}\{f(t)\}+\mathscr{L}\{g(t)\}
with \alpha\in\mathbb{R}.
But, in this lesson, we will answer the following question:
1. Is the laplace transform linear with respect to the product?
We may wonder if the product between two functions has the same property:
\mathscr{L}\{f(t)\cdot g(t)\}=\mathscr{L}\{f(t)\}\cdot\mathscr{L}\{g(t)\}
But a very simple example show us that the equality above is not valid: if
f(t)=t,\quad g(t)=t^{2}
then the Laplace transforms of f and g are, respectively,
\displaystyle F(s)=\frac{1}{s^{2}},\qquad G(s)=\frac{2}{s^{3}}
Let h(t)=f(t)\cdot g(t). So,
h(t)=t^{3}
and as its Laplace Transform is
\displaystyle H(s)=\frac{6}{s^{4}}
It is clear we are able to conclude that
\displaystyle F(s)\cdot G(s)=\frac{2}{s^{4}}\neq\frac{6}{s^{4}}=H(s)
Thus, the Laplace transform does not have a linearity property in the product of functions.
This leads us to another type of products, known as convolution!
2. What is Convolution?
The following definition provides us a way to combine f(t) and g(t) in a new function h(t) such that
H(s)=F(s)\cdot G(s)
Illustrative example: convolution of t and exp(t)
If f(t)=t and g(t)=e^{t}, then
f(\tau)=\tau,\qquad g(t-\tau)=e^{t-\tau}
and integration by parts gives us
\begin{aligned} (f\ast g)(t) & = \int_{0}^{t}f(t)g(t-\tau)d\tau \\ & = \int_{0}^{t}\tau e^{t-\tau}d\tau \\ & = -t+e^{t}-1 \end{aligned}
From the last example, we can conclude (f\ast g) is a function in the variable t. Our goal, however, is not to evaluate the convolution itself but to use the theorem below.
3. Convolution theorem Laplace
Proof:
We have
\displaystyle F(s)=\int_{0}^{\infty}e^{-s\tau}f(\tau)d\tau
\displaystyle G(s)=\int_{0}^{\infty}e^{-s\kappa}f(\kappa)d\kappa
Then
\begin{aligned} F(s)G(s) & = \left(\int_{0}^{\infty}e^{-s\tau}f(\tau)d\tau\right)\left(\int_{0}^{\infty}e^{-s\kappa}f(\kappa)d\kappa\right) \\ & = \int_{0}^{\infty}\int_{0}^{\infty}e^{-s\tau}e^{-s\kappa}f(\tau)g(\kappa)d\tau d\kappa \\ & = \int_{0}^{\infty}\int_{0}^{\infty}e^{-s(\tau+\kappa)}f(\tau)g(\kappa)d\tau d\kappa \\ & = \int_{0}^{\infty}f(\tau)d\tau\int_{0}^{\infty}e^{-s(\tau+\kappa)}g(\kappa)d\kappa \end{aligned}
If t=\tau+\kappa so dt=d\kappa, and also \kappa=t-\tau
In the boundaries, when \kappa=0, we get
t=\tau+0\Rightarrow t=\tau
and as \kappa\to\infty, it is clear t\to\infty. Hence
\displaystyle F(s)G(s)=\int_{0}^{\infty}f(\tau)d\tau\int_{\tau}^{\infty}e^{-st}g(t-\tau)d\tau
We are able to rewrite the integral above as follows:
\displaystyle F(s)G(s)=\int_{0}^{\infty}e^{-st}\left\{\int_{0}^{t}f(\tau)g(t-\tau)d\tau\right\}dt
By the definition of Laplace Transform, we conclude that
F(s)G(s)=\mathscr{L}\{f\ast g\}
This ends the proof.
Illustrative example of convolution use with Laplace transform
If f(t)=\cos(t) and g(t)=\sin(t) then their Laplace transforms are, respectively,
\displaystyle F(s)=\frac{s}{s^{2}+1},\qquad G(s)=\frac{1}{s^{2}+1}
As
\displaystyle (f\ast g)(t)=\int_{0}^{t}f(t)g(t-\tau)d\tau=\int_{0}^{\infty}\cos(t)\sin(t-\tau)d\tau
Thus, by the last theorem:
\displaystyle\mathscr{L}\left\{\int_{0}^{\infty}\cos(t)\sin(t-\tau)d\tau\right\}=F(s)G(s)=\frac{s}{s^{2}+1}\cdot\frac{1}{s^{2}+1}
4. How to fit the convolution with the inverse laplace transform
The convolution is more useful when we apply its inverse Laplace transform due to the fact we usually face the product of two or more functions when we solve differential equations.
For example, to evaluate
\displaystyle \mathscr{L}^{-1}\left\{\frac{1}{s(s^{2}+1)}\right\}
observe that
\displaystyle \frac{1}{s(s^{2}+1)}=\frac{1}{s}\cdot\frac{1}{s^{2}+1}
and as
\displaystyle F(s)=\mathscr{L}\{1\}=\frac{1}{s},\qquad G(s)=\mathscr{L}\{\sin(t)\}=\frac{1}{s^{2}+1}
we have
\displaystyle\mathscr{L}^{-1}\left\{\frac{1}{s(s^{2}+1)}\right\}=\mathscr{L}^{-1}\{F(s)G(s)\}=(f\ast g)(t)
where f(t)=1 and g(t)=\sin(t). Therefore
\displaystyle\mathscr{L}^{-1}\left\{\frac{1}{s(s^{2}+1)}\right\}=\int_{0}^{t}f(t)g(t-\tau)d\tau=\int_{0}^{t}1\sin(t-\tau)d\tau
In the last integral above, we change the variables
\left\{\begin{array}{l} u=t-\tau \\ du=-d\tau \end{array}\right.\Rightarrow\left\{\begin{array}{l} \tau=0\to u=t \\ \tau=t\to u=0 \end{array}\right.
Hence
\begin{aligned} \int_{0}^{t}1\sin(t-\tau)d\tau & = -\int_{t}^{0}\sin(u)du \\ & = \int_{0}^{t}\sin(u)du \\ & = -\left.\cos(u)\right|_{u=0}^{u=t} \\ & = -\cos(t)+\cos(0) \end{aligned}
And since \cos(0)=1, we may conclude
\displaystyle\mathscr{L}^{-1}\left\{\frac{1}{s(s^{2}+1)}\right\}=-\cos(t)+1
We reached the end of this short lesson about the laplace transform of convolution. For a full list of Laplace transform properties, check this post!
