This is the square wave.
\begin{equation*}
f(t) = \left\{ \begin{matrix}
0 \amp t \in [2n+1, 2n+2] \\
1 \amp 1 \in (2n, 2n+1)
\end{matrix} \right.
\end{equation*}
Section 5.7 Laplace Transforms of Periodic Functions
Laplace transforms cooperate with periodic functions. I’ve already discussed the sine and cosine transforms, but it is also true for arbitrary, even discontinuous periodic functions. Here are a couple examples of useful discontinuous periodic functions. (Like the unit step function and the \(\delta\)-function, these are both reasonable functions for applied mathematics even though they are discontinuous.)
Example 5.7.1.
Example 5.7.2.
This is the sawtooth wave.
\begin{align*}
f(t) \amp = \{ t-n \amp \amp t \in [n, n+1) \amp \amp
\forall n \in \NN
\end{align*}
For each of these functions, doing an integration over \((0,\infty)\) is problematric. Instead, there is a convenient theorem for Laplace transforms of such waves which only involves the integral over one period.
Proposition 5.7.3.
If \(f(t)\) is bounded periodic with period \(T\text{,}\) then
\begin{equation*}
\calL\{ f(t) \} = \frac{1}{1-e^{-sT}} \int_0^T e^{-st} f(t) dt
\end{equation*}
Example 5.7.4.
Let \(f(t)\) be the square wave, which has period 2. Then I can use \(f\) as a forcing term and solve the following DE with a Laplace transform (I factor \(1-e^{-2s}\) as a difference of squares in this calculation). Assume the initial conditions are \(y(0) = y^\prime(0) = 0\text{.}\)
\begin{align*}
y^\prime + by \amp = f(t)\\
sY + bY \amp = \frac{1}{1-e^{-2s}} \int_0^2 f(t) e^{-st}
dt\\
\amp = \frac{1}{1-e^{-2s}} \int_0^1 e^{-st} dt\\
\amp = \frac{1}{1-e^{-2s}} \left. \frac{e^{-st}}{-s}
\right|_0^1 = \frac{-(e^{-s}-1)}{s(1-e^{-2s})} \\
\amp = \frac{1}{s(1+e^{-s})}\\
Y \amp = \frac{1}{s(s+b)(1-(-e^{-s}))}
\end{align*}
This expression is problematic for an inverse transform. Specficially, the \(\frac{1}{1-e^{-s}}\) term is the problem. I solve the problem by expressing it as a geometric series, which is a very common approach to dealing with period forcing terms.
\begin{align*}
Y \amp = \frac{1}{s(s+b)(1-(-e^{-s})}\\
\amp = \frac{1}{s(s+b)} \left( 1 - e^{-s} + e^{-2s} -
e^{-3s} + \ldots \right)
\end{align*}
I use partial fractions to split up the rational function in front of the series. After doing so, I distribute the result to product two series.
\begin{align*}
\amp = \frac{1}{b} \left( \frac{1}{s} - \frac{1}{s +
b}\right) \left( 1 - e^{-s} + e^{-2s} - e^{-3s} + \ldots
\right) \\
\amp = \frac{1}{b} \left( \frac{1}{s} - \frac{e^{-s}}{s} +
\frac{e^{-2s}}{s} - \frac{e^{-3s}}{s} + \ldots \right)\\
\amp \hspace{1cm} - \frac{1}{b} \left( \frac{1}{s + b} -
\frac{e^{-s}}{s + b} + \frac{e^{-2}}{s + b} -
\frac{e^{-3}}{s + b} + \ldots \right)
\end{align*}
Then I apply the inverse transform. I can work with the series, since each term in the series is a shift of the base term. After transforming both series, I can group them together again and write them in sigma notation.
\begin{align*}
y \amp = \frac{1}{b} \left( 1 - u_1(t) + u_2(t) - u_3(t)
+ \ldots) \right)\\
\amp \hspace{1cm} - \frac{1}{b} \left( e^{-bt} - u_1(t)
e^{-b(t-1)} + u_2(t) e^{-b(t-2)} - u_3(t) e^{-b(t-3)} +
\ldots \right)\\
y \amp = \frac{1}{b} \left( 1-e^{-bt} \right) +
\frac{1}{b} \sum_{n=1}^\infty (-1)^n u_n(t)
(1-e^{-b(t-n)})
\end{align*}
Is this a good solution? It’s not the easiest to work with, since it is expressed as an infinite series (and not even a Taylor series). But it does measure the behaviour. The square wave introduces a new force every two units of time. The infinite series is simply capturing the collective effect of all those forces on the system. All the terms are exponential decay, so the system is still trying to decay to equilibrium, but it keeps getting pulled up again. I could analyze the series more carefully to determine the net effect: whether this function is actually growing and at what rate due to the forces.
