Fourier Series and Periodic Forcing

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The idea

A Fourier series expresses a periodic waveform as sinusoidal harmonics. A linear system responds to each harmonic separately, so its steady periodic response can be assembled from frequency responses. Higher harmonics can have very different amplitudes and phases after filtering.

Visual guide

VISUAL GUIDEMore harmonics sharpen the transition but retain overshoot

First harmonic
Seven odd harmonics
Target square wave

Odd sine harmonics approximate the square wave. The seven-term sum is closer away from jumps than the first harmonic, but overshoots near each transition. At a jump the series converges to the midpoint value zero.

Method and assumptions

For period $2\pi$ , write $f(t)\sim a_0/2+\sum_{n\ge1}(a_n\cos nt+b_n\sin nt)$ , with coefficients obtained by integrating over one period and dividing by $\pi$ . Under standard piecewise smoothness conditions, the series converges to the midpoint of the two one-sided limits at a jump.

Worked example: an odd square wave

Take $f=1$ on $(0,\pi)$ and $f=-1$ on $(-\pi,0)$ , periodically extended. Oddness makes $a_n=0$ . Integration gives $b_n=4/(\pi n)$ for odd $n$ and zero for even $n$ . Its first terms are $(4/\pi)(\sin t+\sin3t/3+\cdots)$ .

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

Why can a linear system’s responses to Fourier harmonics be added?

All systems have constant coefficients.

Linearity permits superposition, with suitable convergence.

Every harmonic has the same response.

Hint 1 · Find a starting point

Distinguish linearity from other properties of a model.

Hint 2 · Take the next step

A sum of inputs leads to a sum of zero-state responses when the operation and limits are valid.

Show the reasoning

Answer: Linearity permits superposition, with suitable convergence.

Superposition is the reason; each harmonic can still have a different amplitude and phase response.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: filtering one harmonic

For $y'+y=B\sin nt$ , seek $a\sin nt+b\cos nt$ . Matching gives $a=B/(1+n^2)$ and $b=-nB/(1+n^2)$ . The amplitude is $B/\sqrt{1+n^2}$ , so high-frequency harmonics are reduced more strongly.

Interpreting the result

A finite Fourier sum overshoots near a jump; adding terms narrows the affected region but does not eliminate the limiting Gibbs overshoot. For an undamped oscillator, a resonant harmonic needs separate treatment because no bounded steady periodic response exists.

Practice

Which coefficients vanish for an even input?
What value does the square-wave series take at a jump from $-1$ to $1$ ?
What is the output amplitude for $y'+y=\sin3t$ ?

Show worked solutions

The sine coefficients $b_n$ vanish.
The midpoint is $0$ .
$1/\sqrt{10}$ .

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

MAKE IT YOURS

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Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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