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Unit 3 · Lesson 3-1
The Fourier Series
2h 10m1300 XP
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The Fourier Series

Complex exponentials as eigenfunctions, the synthesis and analysis equations, orthogonality, and the coefficients of square, sawtooth and triangle waves

1872 — a machine that predicts the sea
The tide at a port is not one rhythm but several. The Moon pulls one bulge of water round the Earth every 12.42 hours, the Sun another every 12 hours, and the tilts and ellipses of their orbits add rhythms of about 12.66, 23.93 and 25.82 hours. Each rhythm is a sinusoid with its own amplitude and phase at that port; the tide is their sum. Computing that sum by hand, hour by hour for a year of tide tables, was laborious and full of errors. In 1872 William Thomson — later Lord Kelvin — conceived a machine to do it. A wire runs over a row of pulleys; each pulley is driven up and down by a crank turning at the rate of one tidal rhythm, with its amplitude and phase set for the port; the end of the wire adds all the motions and a pen draws the predicted tide. The first machine, built in 1872–73, summed ten components; larger ones followed in 1876 and 1879, and Thomson's symbols for the tidal constituents are still in use (Tide-predicting machine). It was a sum of sinusoids in brass. This lesson is about writing signals that way — and about why linear systems love it.
Unit 2 ended with a question: is there an input that an LTI system passes through unchanged in shape? There is, and Lesson 1-2 introduced it. Feed an LTI system the complex exponential ejωte^{j\omega t} and the output is the same complex exponential, multiplied by a complex number that depends on ω\omega:
ejωt  ⟶  H(jω) ejωt,H(jω)=∫−∞∞h(τ) e−jωτ dτ.e^{j\omega t} \;\longrightarrow\; H(j\omega)\,e^{j\omega t}, \qquad H(j\omega) = \int_{-\infty}^{\infty} h(\tau)\,e^{-j\omega\tau}\,d\tau .
In the language of linear algebra, complex exponentials are the eigenfunctions of LTI systems, and H(jω)H(j\omega) — the frequency response — holds the eigenvalues. The magnitude ∣H∣|H| scales the sinusoid, the angle ∠H\angle H shifts its phase. No other kind of signal survives every LTI system unchanged in shape.
Derivation
Why exponentials pass through unchanged
Put x(t)=ejωtx(t) = e^{j\omega t} into the convolution integral, using the commuted form y=h∗xy = h * x:
y(t)=∫−∞∞h(τ) x(t−τ) dτ=∫−∞∞h(τ) ejω(t−τ) dτ.y(t) = \int_{-\infty}^{\infty} h(\tau)\,x(t - \tau)\,d\tau = \int_{-\infty}^{\infty} h(\tau)\,e^{j\omega(t - \tau)}\,d\tau .
The factor ejωte^{j\omega t} does not depend on τ\tau, so it comes out of the integral:
y(t)=ejωt∫−∞∞h(τ) e−jωτ dτ=H(jω) ejωt.y(t) = e^{j\omega t}\int_{-\infty}^{\infty} h(\tau)\,e^{-j\omega\tau}\,d\tau = H(j\omega)\,e^{j\omega t}.
The integral is just a number for each ω\omega (it exists when hh is absolutely integrable — a stable system). The same argument in discrete time gives ejΩn→H(ejΩ) ejΩne^{j\Omega n} \to H(e^{j\Omega})\,e^{j\Omega n} with H(ejΩ)=∑kh[k] e−jΩkH(e^{j\Omega}) = \sum_k h[k]\,e^{-j\Omega k}.
💡The payoff
If an input can be written as a sum of complex exponentials, x(t)=∑kak ejkω0t,x(t) = \sum_k a_k\,e^{jk\omega_0 t}, then by linearity the output is y(t)=∑kak H(jkω0) ejkω0t.y(t) = \sum_k a_k\,H(jk\omega_0)\,e^{jk\omega_0 t}. Each term is simply multiplied by the frequency response at its own frequency. Convolution — an integral for every instant — becomes one multiplication per frequency. Lesson 3-3 cashes this in.
Which signals can be written that way? Fourier's claim, made for heat flow in 1807, was: every periodic signal of practical interest. A signal with period TT can only contain frequencies whose cycles fit a whole number of times into TT — the harmonics kω0k\omega_0 of the fundamental ω0=2π/T\omega_0 = 2\pi/T. So a periodic signal is a sum of harmonically related complex exponentials, a Fourier series.
x(t)=∑k=−∞∞ak ejkω0t,ak=1T∫Tx(t) e−jkω0t dt,ω0=2πTx(t) = \sum_{k=-\infty}^{\infty} a_k\,e^{jk\omega_0 t}, \qquad a_k = \frac{1}{T}\int_{T} x(t)\,e^{-jk\omega_0 t}\,dt, \qquad \omega_0 = \frac{2\pi}{T}

The synthesis equation (left) builds the signal from its Fourier coefficients; the analysis equation (right) finds them. The integral runs over any one period.

Derivation
The analysis equation from orthogonality
Two harmonics are orthogonal over a period: for integers mm and kk,
∫0Tejmω0t e−jkω0t dt=∫0Tej(m−k)ω0t dt={T,m=k0,m≠k\int_0^T e^{jm\omega_0 t}\,e^{-jk\omega_0 t}\,dt = \int_0^T e^{j(m-k)\omega_0 t}\,dt = \begin{cases} T, & m = k \\ 0, & m \ne k \end{cases}
because for m≠km \ne k the integrand turns round the unit circle a whole number m−km - k of times and its integral cancels. Now multiply the synthesis equation by e−jkω0te^{-jk\omega_0 t} and integrate over one period:
∫0Tx(t) e−jkω0t dt=∑mam∫0Tej(m−k)ω0t dt=ak T.\int_0^T x(t)\,e^{-jk\omega_0 t}\,dt = \sum_m a_m \int_0^T e^{j(m - k)\omega_0 t}\,dt = a_k\,T .
Every term but m=km = k vanishes. Dividing by TT gives the analysis equation. Each coefficient measures how much the signal looks like one harmonic — a projection, exactly as a vector's components are found by dot products with orthogonal unit vectors.
📐The square wave
Let x(t)=+1x(t) = +1 for 0<t<T/20 < t < T/2 and −1-1 for T/2<t<TT/2 < t < T. Its average is zero, so a0=0a_0 = 0. For k≠0k \ne 0: ak=1T[∫0T/2e−jkω0t dt−∫T/2Te−jkω0t dt]=1−e−jkπjkπ={2jkπ,k odd0,k evena_k = \frac1T\left[\int_0^{T/2} e^{-jk\omega_0 t}\,dt - \int_{T/2}^{T} e^{-jk\omega_0 t}\,dt\right] = \frac{1 - e^{-jk\pi}}{jk\pi} = \begin{cases} \dfrac{2}{jk\pi}, & k \text{ odd} \\[4pt] 0, & k \text{ even}\end{cases} Pairing kk with −k-k turns each pair into a sine: x(t)=4π(sin⁡ω0t+sin⁡3ω0t3+sin⁡5ω0t5+⋯ ).x(t) = \frac{4}{\pi}\left(\sin\omega_0 t + \frac{\sin 3\omega_0 t}{3} + \frac{\sin 5\omega_0 t}{5} + \cdots\right). Only odd harmonics, falling as 1/k1/k. A square wave at 1 kHz is a 1 kHz sine of amplitude 1.27, plus 3 kHz at a third of that, 5 kHz at a fifth, and so on.
💡Real signals and the trigonometric form
For a real signal the coefficients come in conjugate pairs, a−k=ak∗a_{-k} = a_k^*, and the pairs combine into real sinusoids: x(t)=a0+∑k=1∞2∣ak∣cos⁡(kω0t+∠ak).x(t) = a_0 + \sum_{k=1}^{\infty} 2|a_k|\cos\left(k\omega_0 t + \angle a_k\right). a0a_0 is the average (the DC value). 2∣ak∣2|a_k| is the amplitude of the kkth harmonic and ∠ak\angle a_k its phase. Even signals have real coefficients (cosines only), odd signals imaginary ones (sines only) — the square wave above is odd.
📐Smoothness decides how fast the coefficients fall
  • The square wave jumps: ∣ak∣∝1/k|a_k| \propto 1/k.
  • The sawtooth, which ramps and then jumps, also falls as 1/k1/k, but has every harmonic, even and odd.
  • The triangle wave has no jumps, only corners: ∣ak∣∝1/k2|a_k| \propto 1/k^2, odd harmonics only — 8π2(sin⁡ω0t−sin⁡3ω0t9+sin⁡5ω0t25−⋯ )\tfrac{8}{\pi^2}\left(\sin\omega_0 t - \tfrac{\sin 3\omega_0 t}{9} + \tfrac{\sin 5\omega_0 t}{25} - \cdots\right).
  • A signal with a continuous first derivative falls as 1/k31/k^3, and so on.
Each extra degree of smoothness costs the high harmonics a factor of kk. That is why a jump needs so many terms, and why sharp edges make wideband interference.
Profound study of nature is the most fertile source of mathematical discoveries.
– Joseph Fourier, Théorie analytique de la chaleur (1822), chapter 1, quoted on Wikiquote
🕰Heat first, signals later
Fourier presented his memoir on the propagation of heat in solid bodies to the Institut national in Paris on 21 December 1807, and published Théorie analytique de la chaleur in 1822 (Fourier series). The claim that arbitrary functions — even discontinuous ones — could be built from sines was controversial, and making it precise occupied mathematicians for the rest of the century. Engineers took it up because it worked; Lesson 3-2 is about the places where it works only nearly.
⚠Tides are not periodic
The tidal rhythms of 12.42 h and 12 h have no common period — their ratio is not a ratio of small integers, which is Lesson 1-1's condition — so the tide never exactly repeats. It is still a sum of sinusoids, and the tide machine used exactly that. A Fourier series is the special case in which every frequency is a multiple of one fundamental.
💡Where you will use this
Lesson 3-2 asks in what sense the series converges; Lesson 3-3 passes periodic signals through systems harmonic by harmonic. Letting the period grow gives the Fourier transform (Unit 4); sampling the frequency axis gives the DFT (Lesson 6-2). Harmonic content is how power engineers, acousticians and vibration analysts describe periodic signals every day.
📚Further reading
  • Oppenheim and Willsky, Signals and Systems, sections 3.1–3.5 — eigenfunctions of LTI systems and the continuous-time Fourier series.
  • T. W. Körner, Fourier Analysis (Cambridge University Press, 1988) — the mathematics and the history, including Kelvin's tide machines.
  • Tide-predicting machine on Wikipedia.
↓But there is a problem
Ten pulleys predicted the tides because tides are smooth. A square wave is not: its coefficients fall only as 1/k1/k, and any finite number of terms leaves something out. Add more terms and you would expect the error to vanish everywhere. It does not. Near every jump the partial sums overshoot — and the overshoot does not shrink, however many terms you add. In 1898 a machine for summing Fourier series drew it, and it took a famous physicist two letters to Nature to explain it correctly.