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Subsection6.4.1ConvolutionApplying the Convolution Theorem¶ permalink

If \(f\) and \(g\) are two piecewise continuous exponentially bounded functions, then we define the *convolution product* of \(f\) and \(g\) to be
\begin{equation*}
(f*g)(t) = \int_0^t f(t - \tau) g(\tau) \, d\tau = \int_0^t f(\tau) g(t - \tau) \, d\tau.
\end{equation*}
The convolution product has many properties similar to those of ordinary multiplication.

- Commutivity: \(f*g = g*f\text{.}\)
- Distribution: \(f*(g + h) = f*g + f*h\text{.}\)
- Associativity: \(f*(g*h) = (f*g)*h\text{.}\)
- \(0 *f = f*0 = 0\text{.}\)

All of these properties can be proven using the definition of convolution and the properties of integration.

There is, however, no multiplicative identity. In other words, \(f*1 \neq f\text{.}\) For example, suppose that \(f(t) = \cos t\text{.}\) Then
\begin{equation*}
(f*1)(t) = \int_0^t \cos(t - \tau) \cdot 1 \, d\tau = \sin t.
\end{equation*}
Also, it may not be the case that \(f*f\) is a nonnegative function.

One important property of the Laplace transform is how convolution products behave.

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Theorem6.7

Let \(f\) and \(g\) be two piecewise continuous exponentially bounded functions, and suppose that \({\mathcal L}(f)(s) = F(s)\) and \({\mathcal L}(g)(s) = G(s)\) for \(s \geq a \gt 0\text{.}\) Then
\begin{equation*}
F(s) G(s) = {\mathcal L}(f*g)(s)
\end{equation*}
for \(s \gt a\text{.}\)

If
\begin{align*}
F(s) & = \int_0^\infty e^{-s\xi} f(\xi) \, d\xi\\
G(s) & = \int_0^\infty e^{-s\tau} g(\tau) \, d\tau,
\end{align*}
then
\begin{align*}
F(s) G(s) & = \int_0^\infty e^{-s\xi} f(\xi) \, d\xi \int_0^\infty e^{-s\tau} g(\tau) \, d\tau\\
& = \int_0^\infty g(\tau) \left( \int_\tau^\infty e^{-st} f(t - \tau) \, dt \right) \, d\tau,
\end{align*}
where \(\xi = t - \tau\) is the change of variable. Reversing the order of integration, we have
\begin{equation*}
\int_0^\infty g(\tau) \left( \int_\tau^\infty e^{-st} f(t - \tau) \, dt \right) \, d\tau = \int_0^\infty e^{-st} \left( \int_0^t f(t - \tau) g(\tau) \, d\tau \right) \, dt.
\end{equation*}
However, this last expression is just the Laplace transform of \(f*g\text{.}\)

Let us use the Convolution Theorem (Theorem 6.7) to calculate the inverse Laplace transform of
\begin{equation*}
H(s)= \frac{a}{s^2(s^2 + a^2)} = \frac{1}{s^2} \cdot \frac{a}{s^2 + a^2}.
\end{equation*}
instead of using partial fractions. The inverse Laplace transform of \(1/s^2\) is \(t\text{,}\) and the inverse Laplace transform of \(a/(s^2 + a^2)\text{.}\) is \(\sin at\text{.}\) Using the Convolution Theorem, the inverse Laplace transform of \(H(s)\) is
\begin{equation*}
h(t) = \int_0^t (t - \tau) \sin a \tau \, d \tau = \frac{at - \sin at}{a^2}.
\end{equation*}

We can also use the Convolution Theorem to solve initial value problems. Consider the IVP
\begin{align*}
y'' + 4y & = g(t)\\
y(0) & = 3\\
y'(0) & = -1.
\end{align*}
Taking the Laplace transform of both sides of the differential equation and applying the initial conditions, we obtain
\begin{equation*}
s^2 Y(s) - 3s + 1 + 4Y(s) = G(s),
\end{equation*}
where \(G(s)\) is the Laplace transform of \(g(t)\text{.}\) Solving for \(Y(s)\text{,}\) we have
\begin{equation*}
Y(s) = \frac{3s-1}{s^2 + 4} + \frac{G(s)}{s^2 + 4} = 3\frac{s}{s^2 + 4} - \frac{1}{2} \frac{2}{s^2 + 4} + \frac{1}{2} \frac{2}{s^2 + 4} G(s).
\end{equation*}
The last term corresponds to the forcing term of our differential equation. Taking the inverse Laplace transform of both sides and applying the Convolution Theorem, we get
\begin{equation*}
y = 3 \cos 2t - \frac{2}{2} \sin 2t + \frac{1}{2} \int_0^t \sin 2(t - \tau) g(\tau) \, d \tau.
\end{equation*}

It is possible to write a solution for the initial value problem
\begin{align*}
ay'' + by' + cy & = g(t)\\
y(0) & = y_0\\
y'(0) & = y_1
\end{align*}
using the Convolution Theorem. Taking the Laplace transform of both sides of the differential equation and using the initial conditions, we have
\begin{equation*}
(as^2 + bs + c)Y(s) - (as + b)y_0 - ay_1 = G(s)
\end{equation*}
or
\begin{equation*}
Y(s) = \Phi(s) + \Psi(s),
\end{equation*}
where
\begin{align*}
\Phi(s) & = \frac{(as + b)y_0 + ay_1 }{as^2 + bs + c}\\
\Psi(s) & = \frac{G(s)}{as^2 + bs + c}.
\end{align*}
Therefore,
\begin{equation*}
y = \phi(t) + \psi(t),
\end{equation*}
where \(\phi(t) = {\mathcal L}^{-1}( \Phi(s))\) and \(\psi(t) = {\mathcal L}^{-1}( \Psi(s))\text{.}\) Observe that \(\phi(t)\) is the solution to the initial value problem
\begin{align*}
ay'' + by' + cy & = 0\\
y(0) & = y_0\\
y'(0) & = y_1.
\end{align*}
Once we have values for \(a\text{,}\) \(b\text{,}\) and \(c\text{,}\) the function \(\phi(t)\) is easy to find. To find \(\psi(t)\text{,}\) we first write
\begin{equation*}
\Psi(s) = \frac{1}{as^2 + bs + c} G(s) = H(s) G(s),
\end{equation*}
where \(H(s) = 1/(as^2 + bs + c)\text{.}\) If \(h(t)\) is the inverse Laplace transform of \(H(s)\text{,}\) then
\begin{equation*}
\psi(t) = {\mathcal L}^{-1}(H(s)G(s)) = \int_0^t h(t - \tau) g(\tau) \, d\tau.
\end{equation*}

Let us consider the case where \(G(s) = 1\text{.}\) Consequently, \(g(t) = \delta(t)\) and \(\psi(s) = H(s)\text{.}\) This means that \(h(t)\) is a solution to the initial value problem
\begin{align*}
ay'' + by' + cy & = \delta(t)\\
y(0) & = y_0\\
y'(0) & = y_1.
\end{align*}
For this reason, \(h(t)\) is sometimes called the *impulse response* of the system.