Waves & Oscillations · Upper Division

Fourier Analysis

Any periodic function — however complicated — can be decomposed into a sum of sines and cosines. This single idea underlies signal processing, quantum mechanics, optics, and partial differential equations.

PrerequisitesWaveproperties(Ch.8)⋅Calculus(Ch.22)⋅LinearalgebrahelpsbutisnotrequiredWave properties (Ch. 8) \cdot Calculus (Ch. 22) \cdot Linear algebra helps but is not required
Learning Goals
  • Compute Fourier series coefficients using orthogonality integrals.
  • Identify the Gibbs phenomenon and explain why the overshoot persists at discontinuities.
  • Define the Fourier transform and apply it to compute spectra of simple functions.
  • State and apply Parseval's theorem to relate energy in position and frequency space.
  • Derive the Heisenberg uncertainty relation from the Fourier uncertainty principle.

F.1 The Fourier Series

Let f(x) be a periodic function with period 2L. The Fourier series of f is the decomposition into harmonics:

f(x)=a02+∑n=1∞[ancos⁡(nπxL)+bnsin⁡(nπxL)]f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left[a_n\cos\left(\frac{n\pi x}{L}\right)+b_n\sin\left(\frac{n\pi x}{L}\right)\right](F.1)

The coefficients aₙ and bₙ are found by exploiting the orthogonality of sines and cosines: ∫(−L to L) sin(mπx/L) sin(nπx/L) dx = 0 unless m = n. Multiplying both sides by the appropriate trig function and integrating isolates each coefficient:

an=1L∫−LLf(x)cos⁡(nπxL) dxbn=1L∫−LLf(x)sin⁡(nπxL) dxa_n=\frac{1}{L}\int_{-L}^{L}f(x)\cos\left(\frac{n\pi x}{L}\right)\,dx \qquad b_n=\frac{1}{L}\int_{-L}^{L}f(x)\sin\left(\frac{n\pi x}{L}\right)\,dx(F.2)
Example F.1 — Fourier Series of a Square Wave

f(x)=+1for0<x<L,−1for−L<x<0.FindtheFouriercoefficientsf(x) = +1 for 0 < x < L, -1 for -L < x < 0. Find the Fourier coefficients.

Symmetry:fisodd→allan=0(cosinetermsvanishf is odd \to all a_{n} = 0 (cosine terms vanish
bₙ:bn=(1/L)∫(−LtoL)f(x)sin(n\pix/L)dx=(2/L)∫(0toL)sin(n\pix/L)dxb_{n} = (1/L)\int(-L to L) f(x)sin(n\pix/L)dx = (2/L)\int(0 to L) sin(n\pix/L)dx
Integrate:= (2/L)[−L/(nπ)cos(n\pix/L)]0L=(2/nπ)(1−cosnπ2/L)[-L/(n\pi) cos(n\pix/L)]_{0}^L = (2/n\pi)(1 - cos n\pi
Result:bn=4/(nπ)foroddn,0forevennb_{n} = 4/(n\pi) for odd n, 0 for even n
Series:f(x)=(4/π)[sin(\pix/L)+sin(3\pix/L)/3+sin(5\pix/L)/5+⋯f(x) = (4/\pi)[sin(\pix/L) + sin(3\pix/L)/3 + sin(5\pix/L)/5 + \cdots
Figure F.1. Fourier series approximation. Each slider adds more harmonics to the sum (blue curve). Individual harmonics are shown as faint colored lines. The target function is the dim gray curve. Notice the Gibbs phenomenon — the overshoot near discontinuities never disappears, it just moves closer to the jump.

F.2 The Gibbs Phenomenon

When a Fourier series approximates a function with a jump discontinuity, the partial sums overshoot by approximately 9% of the jump height near the discontinuity — and this overshoot persists no matter how many terms are added; it just becomes narrower. This is theGibbs phenomenon, discovered in 1899.

It reflects a deep truth: a uniformly convergent series cannot converge to a discontinuous function. The Fourier series converges in the mean square sense (minimum total error), not pointwise at the discontinuity. This distinction matters in signal processing (ringing artifacts in audio and image compression) and in solving PDEs with discontinuous boundary conditions.

F.3 The Fourier Transform

For non-periodic functions, we take the period L → ∞ and the discrete sum becomes an integral. The Fourier transform decomposes f(x) into a continuous spectrum of frequencies:

F^(k)=∫−∞∞f(x)e−ikx dx(forward transform)\hat F(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx \qquad \text{(forward transform)}(F.3)
f(x)=12π∫−∞∞F^(k)eikx dk(inverse transform)f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat F(k)e^{ikx}\,dk \qquad \text{(inverse transform)}(F.4)

The variable k is the spatial frequency (wavenumber). |F̂(k)|² is the power spectrum— how much of each frequency is present. The Fourier transform turns convolutions into multiplications, differential equations into algebraic equations, and is the mathematical backbone of every filter, antenna, and spectrometer.

Theorem F.1 — Parseval's Theorem
The total energy is the same whether computed in position space or frequency space:∫−∞∞∣f(x)∣2dx=(1/2π)∫−∞∞∣F^(k)∣2dk\int_{-}\infty^\infty |f(x)|^{2} dx = (1/2\pi) \int_{-}\infty^\infty |F̂(k)|^{2} dkThis means no information is lost in the transform — it is a unitary operation.

F.4 Connection to Physics

The Fourier transform is everywhere in physics, often wearing disguises:

Quantum mechanics. The position wavefunction ψ(x) and momentum wavefunction φ(p) are Fourier transforms of each other: φ(p) = (1/√2πℏ)∫ψ(x)e^(−ipx/ℏ)dx. The Heisenberg uncertainty relation Δx·Δp ≥ ℏ/2 is a direct consequence of the Fourier uncertainty principle: a narrow ψ(x) requires a broad φ(p) and vice versa.

Diffraction.The far-field diffraction pattern of an aperture is exactly the Fourier transform of the aperture's transmission function. The single-slit sinc² pattern from Chapter 18 is the squared magnitude of the Fourier transform of a rectangular function.

Signal processing. The discrete Fourier transform (DFT), computed efficiently by the Fast Fourier Transform (FFT) algorithm (Cooley–Tukey, 1965), underlies MP3 audio compression, JPEG image compression, radar, MRI, and Wi-Fi (OFDM).

Definition F.2 — Common Traps
  • Orthogonality sets coefficients: guessing harmonic amplitudes usually misses normalization factors.
  • Gibbs overshoot does not vanish: adding terms narrows the ringing but leaves the overshoot height near a jump.
  • Transform conventions vary: factorsof2πmovebetweentheforwardandinversetransformsbyconventionfactors of 2\pi move between the forward and inverse transforms by convention
  • Narrow in x means broad in k: localization and bandwidth are reciprocal, not independently adjustable.
Exercises — F.1–F.4 Fourier Analysis
0.
ComputetheRMSvalueofthesawtoothwavef(x)=x/πon−π<x<πusingthedefinition(Compute the RMS value of the sawtooth wave f(x) = x/\pi on -\pi < x < \pi using the definition (1/2π)\intf(x)2dx,andverifyusingParseval′stheorem1/2\pi)\intf(x)^{2}dx, and verify using Parseval's theorem.
Straightforward
1.FindtheFouriersineseriesoff(x)=x/πon−π<x<π(thesawtoothwaveFind the Fourier sine series of f(x) = x/\pi on -\pi < x < \pi (the sawtooth wave.
Straightforward
2.FindtheFouriertransformF^(k)off(x)=e−a∣x∣foraFind the Fourier transform F̂(k) of f(x) = e^{-a|x|} for a > 0. How does the width of f in x relate to the width of F̂ in k?
Intermediate
3.State the convolutiontheoremforFouriertransforms.ShowhowitcanbeusedtosolvetheODEy′′+y=olution theorem for Fourier transforms. Show how it can be used to solve the ODE y'' + y =
Intermediate
4.DerivetheHeisenberguncertaintyrelation\Deltax\Deltap≥ℏ/2fromtheFourieruncertaintyprincipDerive the Heisenberg uncertainty relation \Deltax\Deltap \ge \hbar/2 from the Fourier uncertainty principle. Which wavefunction shape achieves the minimum uncertainty?
Challenging
Key Takeaways
  • Fourierseries:anyperiodicf(x)=∑[ancos+bnsin];coefficientsfoundbyorthogonalityFourier series: any periodic f(x) = \sum[a_{n}cos + b_{n}sin]; coefficients found by orthogonality integrals.
  • Squarewave:onlyoddharmonics,bn=4/(nπ).AddingtermsreduceserrorbutGibbsovershoSquare wave: only odd harmonics, b_{n} = 4/(n\pi). Adding terms reduces error but Gibbs overshoot persists.
  • Fouriertransform:extendsseriestonon−periodicf;F^(k)=\intf(x)e−ikxdxFourier transform: extends series to non-periodic f; F̂(k) = \intf(x)e^{-ikx}dx.
  • Parseval's theorem: energy is preserved by the transform (it's unitary).
  • Narrow in x ↔ broad in k: the Fourier uncertainty principle, which implies Heisenberg.
  • Diffractionpattern=∣FTofaperture∣2.Convolutioninspace=multiplicationinfrequencDiffraction pattern = |FT of aperture|^{2}. Convolution in space = multiplication in frequency.