3.2 Square-Wave Forcing

Throw the switch on and off forever and you get the simplest periodic forcing there is: the square wave. With period $2\pi$ , let $\mathrm{sq}(t)$ equal $\pi/4$ on the first half-period and $-\pi/4$ on the second. (The odd-looking amplitude $\pi/4$ is chosen to make every coefficient below a clean unit fraction.) In the language of the last section it is an infinite train of switches,

$\mathrm{sq}(t) = \frac{\pi}{4} \left( 1 - 2u_{\pi}(t) + 2u_{2\pi}(t) - 2u_{3\pi}(t) + \cdots \right),$

and its transform costs one geometric series: each delayed switch contributes its exponential tag, and

$\mathcal{L}\{\mathrm{sq}\}(s) = \frac{\pi}{4s} \left( 1 - 2e^{-\pi s} + 2e^{-2\pi s} - \cdots \right) = \frac{\pi}{4s} \cdot \frac{1 - e^{-\pi s}}{1 + e^{-\pi s}},$

a closed form paying for infinitely many switches at once. That is the time ledger: the wave described by when it flips.

There is a second ledger — not when the wave switches but which frequencies it contains:

$\mathrm{sq}(t) = \sin t + \frac{\sin 3t}{3} + \frac{\sin 5t}{5} + \frac{\sin 7t}{7} + \cdots,$

its Fourier series. Only odd multiples of the fundamental frequency $1$ appear, and the $n$ th harmonic arrives with amplitude $1/n$ . The claim looks implausible — smooth waves summing to a jump? — and deserves to be watched happening. The figure steps through the first three partial sums: $S_1 = \sin t$ , then $S_2$ adds $\tfrac{1}{3}\sin 3t$ , then $S_3$ adds $\tfrac{1}{5}\sin 5t$ .

A grey square wave on a time axis, flat at height about 0.785 and flipping sign at time pi, 2 pi and 3 pi, with an amber wave drawn over it. At the opening setting, with the slider k at 1, the amber wave is a plain sine wave of height 1: round-topped, overshooting the flat top of the square wave in the middle of each half-period. Moving k to 2 and then 3 adds the third and then the fifth harmonic: the amber tops flatten toward the square wave, the middle of each top dipping below it at k equal to 2 and rising back above it at k equal to 3, the crossings steepen, and small ripples appear along each top. — The square wave of height $\pi/4 \approx 0.785$ (grey) and one of its Fourier partial sums (amber), drawn over nearly two periods with time $t$ along the horizontal axis. The slider $k$ sets how many harmonics the amber curve sums: $k = 1$ draws $S_1 = \sin t$ , $k = 2$ draws $S_2 = S_1 + \tfrac{1}{3}\sin 3t$ , and $k = 3$ draws $S_3 = S_2 + \tfrac{1}{5}\sin 5t$ , its tops flatter and its crossings steeper. Drag $k$ from $1$ to $3$ to replay the convergence one harmonic at a time.

Explore

Observe. Start at $k = 1$ , where the amber curve is $S_1$ alone. It is a plain sine of height $1$ — the right period and the right sign pattern already, but round-topped and, at its crest, taller than the plateau $\pi/4 \approx 0.785$ it is aiming for.
Predict. At the center of the plateau, $t = \pi/2$ , the next term contributes $\tfrac{1}{3}\sin(3\pi/2) = -\tfrac{1}{3}$ . Will moving to $S_2$ raise or lower the middle of the crest? Decide, then drag $k$ to $2$ .
Verify. Step $k$ through $1$ , $2$ , $3$ and watch the middle of the crest, at $t = \pi/2$ , against the grey plateau: above it, below it, above it again. The three center heights are $1$ , then $1 - \tfrac{1}{3} \approx 0.67$ , then $1 - \tfrac{1}{3} + \tfrac{1}{5} \approx 0.87$ — the partial sums of the alternating series $1 - \tfrac{1}{3} + \tfrac{1}{5} - \cdots$ , whose limit is exactly $\pi/4$ . The plateau height is that famous series, straddled from above and below.
Observe. At $k = 3$ , count the ripples riding the amber curve's plateau: three crests, at $t = \pi/6$ , $\pi/2$ , and $5\pi/6$ . Each harmonic added contributes one more ripple and shrinks them all.
Predict, then verify. At $k = 3$ the tallest crests are the ones nearest the jumps — at $t = \pi/6$ and $5\pi/6$ , height $\tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{10} = \tfrac{14}{15} \approx 0.93$ — not the one at the center. Predict where the tallest crest of $S_4$ would sit, then check your reasoning against the pattern from $S_1$ to $S_3$ . (That stubborn spike beside each jump is the Gibbs phenomenon: it narrows as terms are added but never shrinks below about $9\%$ of the jump.)

Connect

Why take the wave apart at all? Because for a linear equation, superposition turns the series into a term-by-term to-do list. Drive an undamped oscillator with the square wave,

$x'' + \omega_0^2\, x = \mathrm{sq}(t),$

and each harmonic $\tfrac{1}{n}\sin nt$ produces its own steady response $\frac{1}{n\,(\omega_0^2 - n^2)}\sin nt$ , exactly as in the beats section — provided $\omega_0 \ne n$ . The denominators tell the story: the response is dominated by whichever harmonic sits closest to the natural frequency. And if $\omega_0$ equals an odd integer, that single term resonates and grows without bound while every other term stays politely small.

This is the treacherous fact about periodic switching. A pure sinusoid resonates at exactly one frequency; a square wave carries the frequencies $1, 3, 5, 7, \ldots$ all at once and can resonate a system tuned to any of them. Matching the forcing period to the natural period is the resonance you expect. A square wave that switches three or five times slower than the system's natural oscillation excites a resonance you may not expect — the energy arrives through an overtone. Switched machinery must be designed against the whole ladder of frequencies, not a single rung.

Example 1 A hidden resonance

The system $x'' + 9x = \mathrm{sq}(t)$ has natural frequency $3$ , so its natural period $2\pi/3$ is one third of the forcing period — no match. Show that it resonates anyway.

Show solution

Feed the Fourier series through the equation term by term. The fundamental $\sin t$ yields the bounded response $\frac{1}{9 - 1}\sin t = \tfrac{1}{8}\sin t$ . The fifth harmonic $\tfrac{1}{5}\sin 5t$ yields $\frac{1/5}{9 - 25}\sin 5t = -\tfrac{1}{80}\sin 5t$ , tiny. But the third harmonic $\tfrac{1}{3}\sin 3t$ drives the oscillator exactly at its natural frequency, and the resonant repair from Chapter 2 — multiply by $t$ — gives the response $-\tfrac{1}{18}\, t \cos 3t$ , growing without bound. (Check: for $x'' + 9x = F\sin 3t$ the resonant solution is $-\frac{F}{6}\, t\cos 3t$ ; here $F = \tfrac{1}{3}$ .) The square wave never looks like a frequency- $3$ signal, but one of its harmonics is one, and that is all resonance requires.

An original work of XYZ Homework, built around interactive XYZ 3D figures, following the standard first-course sequence (first-order equations through systems and the phase plane). Distinct from the LibreTexts-sourced Differential Equations for Engineers (Lebl) edition, which has its own attribution. License: CC-BY-NC-SA-4.0.