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 $2 π$ , let $s q (t)$ equal $π / 4$ on the first half-period and $minus π / 4$ on the second. (The odd-looking amplitude $π / 4$ 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,

$s q (t) = \frac{π}{4} (1 - 2 u_{π} (t) + 2 u_{2 π} (t) - 2 u_{3 π} (t) + \dots),$

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

$ℒ {s q} (s) = \frac{π}{4 s} (1 - 2 e^{- π s} + 2 e^{- 2 π s} - \dots) = \frac{π}{4 s} \cdot \frac{1 - e^{- π s}}{1 + e^{- π s}},$

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:

$s q (t) = \sin t + \frac{\sin 3 t}{3} + \frac{\sin 5 t}{5} + \frac{\sin 7 t}{7} + \dots,$

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 plots the first three partial sums: $S sub 1 equals sin t$ , then $S sub 2$ adds $\frac{1}{3} \sin 3 t$ , then $S sub 3$ adds $\frac{1}{5} \sin 5 t$ .

Three blue waves of increasing darkness drawn over the same two periods: a dashed pale sine wave, a dashed medium curve whose crests have begun to flatten, and a solid dark curve with flatter tops, steeper crossings, and a small ripple. Toggling the curves' visibility shows each added odd harmonic pushing the shape closer to a square wave of height about 0.785. — The first three Fourier partial sums of the square wave of height $π / 4 approximately equals 0.785$ , drawn over nearly two periods: the pale dashed $S sub 1 equals sin t$ , the medium dashed $S_{2} = S_{1} + \frac{1}{3} \sin 3 t$ , and the solid dark $S_{3} = S_{2} + \frac{1}{5} \sin 5 t$ , its tops flatter and its crossings steeper. Toggle each curve's visibility to replay the convergence one harmonic at a time.

Explore

Observe. Hide $S sub 2$ and $S sub 3$ . Alone, $S sub 1$ 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 $π / 4 approximately equals 0.785$ it is aiming for.
Predict. At the center of the plateau, $t equals π / 2$ , the next term contributes $\frac{1}{3} \sin (3 π / 2) = - \frac{1}{3}$ . Will revealing $S sub 2$ raise or lower the middle of the crest? Decide, then toggle it on.
Verify. Read the three center heights at $t equals π / 2$ : they are $1$ , then $1 - \frac{1}{3} \approx 0.67$ , then $1 - \frac{1}{3} + \frac{1}{5} \approx 0.87$ — the partial sums of the alternating series $1 - \frac{1}{3} + \frac{1}{5} - \dots$ , whose limit is exactly $π / 4$ . The plateau height is that famous series, straddled from above and below.
Observe. Count the ripples riding the solid curve's plateau: three crests, at $t equals π / 6$ , $π / 2$ , and $5 π / 6$ . Each harmonic added contributes one more ripple and shrinks them all.
Predict, then verify. The solid curve's tallest crests are the ones nearest the jumps — at $t equals π / 6$ and $5 π / 6$ , height $\frac{1}{2} + \frac{1}{3} + \frac{1}{10} = \frac{14}{15} \approx 0.93$ — not the one at the center. Predict where the tallest crest of $S sub 4$ would sit, then check your reasoning against the pattern from $S sub 1$ to $S sub 3$ . (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,

$x^{″} + ω_{0}^{2} x = s q (t),$

and each harmonic $\frac{1}{n} \sin n t$ produces its own steady response $\frac{1}{n (ω_{0}^{2} - n^{2})} \sin n t$ , exactly as in the beats section — provided $ω sub 0 not equal to n$ . The denominators tell the story: the response is dominated by whichever harmonic sits closest to the natural frequency. And if $ω sub 0$ 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 $1, 3, 5, 7, …$ 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 $x^{″} + 9 x = s q (t)$ has natural frequency $3$ , so its natural period $2 π / 3$ is one third of the forcing period — no match. Show that it resonates anyway.

Feed the Fourier series through the equation term by term. The fundamental $\sin t$ yields the bounded response $\frac{1}{9 - 1} \sin t = \frac{1}{8} \sin t$ . The fifth harmonic $\frac{1}{5} \sin 5 t$ yields $\frac{1 / 5}{9 - 25} \sin 5 t = - \frac{1}{80} \sin 5 t$ , tiny. But the third harmonic $\frac{1}{3} \sin 3 t$ drives the oscillator exactly at its natural frequency, and the resonant repair from Chapter 2 — multiply by $t$ — gives the response $- \frac{1}{18} t \cos 3 t$ , growing without bound. (Check: for $x^{″} + 9 x = F \sin 3 t$ the resonant solution is $- \frac{F}{6} t \cos 3 t$ ; here $F = \frac{1}{3}$ .) 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.