Math Methods · Learn
A periodic forcing isn't one frequency, it's a chord. Break it into harmonics, solve for each one separately, then add the answers back up — and discover why one quiet harmonic can shake the whole system.
Take the same damped oscillator from the frequency response lab, but instead of driving it with a single sine wave, drive it with any periodic signal $f(t)$:
If $f(t)$ is periodic, it has a Fourier series: $f(t) = \sum_{n=1}^{\infty} b_n \sin(n\omega_0 t)$, a sum of harmonics of the fundamental frequency $\omega_0$.
The ODE is linear, so the response to a sum of inputs is the sum of the responses. Each harmonic $b_n\sin(n\omega_0 t)$ is just a plain sinusoid at frequency $n\omega_0$, and its steady-state response is read straight off the transfer function $H(i\omega)$ with frequency ratio $r_n = n\omega_0/\omega_n$:
This is Bjorn Poonen's method in §27.2 of his MIT 18.03 notes (linked below): find the Fourier series of the input, solve each harmonic with the usual complex-replacement trick (the Exponential Response Formula), then add up the periodic steady-state pieces.
Every harmonic gets its own point on the Bode curve. Most sit far from $r_n=1$ and pass through close to unchanged. But if the fundamental $\omega_0$ is tuned so that some harmonic $n\omega_0$ lands near $\omega_n$, that single term gets the full resonant boost $\approx 1/(2\zeta)$ — and with light damping, a harmonic that was a small fraction of the input can become the dominant term in the output.
| Situation | What happens |
|---|---|
| All $r_n \ll 1$ | $x(t)$ tracks $f(t)$ almost exactly — every harmonic keeps up. |
| One $r_n \approx 1$ | That harmonic is boosted by $\approx1/(2\zeta)$ and can dominate $x(t)$. |
| All $r_n \gg 1$ | Every harmonic rolls off — $x(t)$ is smaller and smoother than $f(t)$. |
A square wave's harmonics fall off only like $1/n$ (and only odd $n$ appear) — its high harmonics stay loud, so it's more likely some harmonic lands near resonance. A triangle wave's harmonics fall like $1/n^2$ — almost everything is in the fundamental, so resonance mostly matters when $\omega_0$ itself is near $\omega_n$. This is the same orthogonality and convergence behavior from the Fourier series page, now feeding directly into a vibrating system instead of just being decomposed and looked at.
Pick a waveform, then sweep the fundamental frequency $\omega_0$ and watch the spectrum chart — the colored output bars stay close to the gray input bars until one harmonic crosses resonance, where it suddenly towers over the rest. Open the lab →
Bjorn Poonen's free MIT 18.03 lecture notes (PDF) cover this exact method in §27.2, "Solving an ODE using Fourier series" — building directly on Fourier series (§26) and the Exponential Response Formula from earlier sections.
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