Part 1 · Oscillators and waves

1.6 Fourier series and the elementary wave

Source: Postorino, week 2 slides, pp. 1–5, and week 1 slides, pp. 24–27

Objectives

Theorem. A periodic function is a sum of sines and cosines of the multiples of its frequency; the coefficients are integrals over one period.
Recipe. Compute $A_{n}$ , $B_{n}$ (or $C_{n}$ ) for a given periodic function, using parity to skip half the work.
Picture. Any wave is a sum of elementary waves $sin (k x - ω t + φ)$ ; study one, then add them up.
Vocabulary. Wave vector, angular frequency, harmonics, orthogonality, dispersion relation $ω = k v$ .

Week 2, pp. 1–5, reworks week 1, pp. 24–27. This section follows week 2 and cites both.

Periodicity

Definition 1.6.1Wave vector and angular frequencyslides w2, p. 1

A function periodic in space with period $λ$ , $f (x + λ) = f (x)$ , has wave vector (vettore d’onda) $k = 2 π / λ$ .
A function periodic in time with period $T$ , $h (t + T) = h (t)$ , has angular frequency (pulsazione) $ω = 2 π / T$ .

$k$ counts radians of phase per unit length, $ω$ radians per unit time. The ordinary frequency is $ν = 1/ T = ω /2 π$ .

The Fourier series

Theorem 1.6.2Fourier seriesslides w2, p. 1–2

A function $f (x)$ of period $λ$ and wave vector $k = 2 π / λ$ can be written as an infinite series of sines and cosines: $f (x) = \frac{A _{0}}{2} + n = 1 \sum \infty [A_{n} cos (nk x) + B_{n} sin (nk x)],$ $A_{n} = \frac{2}{λ} \int_{- λ /2}^{+ λ /2} f (x) cos (nk x) d x, B_{n} = \frac{2}{λ} \int_{- λ /2}^{+ λ /2} f (x) sin (nk x) d x .$ In time, for $h (t)$ of period $T$ and $ω = 2 π / T$ , the same with $k x \to ω t$ and $λ \to T$ : $h (t) = \frac{A _{0}}{2} + n = 1 \sum \infty [A_{n} cos (nω t) + B_{n} sin (nω t)], A_{n} = \frac{2}{T} \int_{- T /2}^{+ T /2} h (t) cos (nω t) d t, B_{n} = \frac{2}{T} \int_{- T /2}^{+ T /2} h (t) sin (nω t) d t .$

The terms $n = 1, 2, 3, \dots$ are the harmonics: waves whose wavelength is $λ$ , $λ /2$ , $λ /3$ , … so that each repeats a whole number of times in one period.

Key observation 1.6.3The constant term is the meanslides w2, p. 2

Since $cos 0 = 1$ and $sin 0 = 0$ , the $n = 0$ term is the mean over the period (valor medio sul periodo): $\frac{A _{0}}{2} = \frac{1}{λ} \int_{- λ /2}^{+ λ /2} f (x) d x, B_{0} = 0.$

WarningSlide slip: the constant term

Week 2 writes $f = \frac{A _{0}}{2} + \sum_{n = 0}^{\infty} \dots$ with $A_{0} = \frac{1}{λ} \int f$ . Taken literally that counts the mean one and a half times: the $n = 0$ term of the sum adds $A_{0}$ to the $\frac{A _{0}}{2}$ in front. Week 1 writes $\sum_{n = 0}^{\infty}$ with $A_{0} = \frac{1}{λ} \int f$ and no $\frac{A _{0}}{2}$ , which is consistent. So is the form above: the sum starts at $n = 1$ and $A_{0}$ uses the same $\frac{2}{λ}$ formula as every other $A_{n}$ , so that $\frac{A _{0}}{2}$ is the mean. It also matches the table below ( $C_{0} = A_{0} /2$ ). Week 2’s intro line calls the time function $g (t)$ while the formulas use $h (t)$ : the same function.

Why the coefficients are these integralsbeyond the lectures

Over one period the functions $cos (nk x)$ , $sin (nk x)$ are orthogonal: $\int_{- λ /2}^{λ /2} cos (nk x) cos (mk x) d x = \frac{λ}{2} δ_{nm} (n, m \geq 1), \int sin (nk x) sin (mk x) d x = \frac{λ}{2} δ_{nm}, \int cos (nk x) sin (mk x) d x = 0,$ with $δ_{nm} = 1$ if $n = m$ and 0 otherwise. They follow from the product-to-sum formulas, e.g. $cos a cos b = \frac{1}{2} [cos (a - b) + cos (a + b)]$ , and the fact that a cosine of a non-zero whole number of periods integrates to zero.

Multiply the series by $cos (mk x)$ and integrate over a period. Every term dies except $A_{m} \cdot \frac{λ}{2}$ , so $A_{m} = \frac{2}{λ} \int f cos (mk x) d x$ . The same with $sin (mk x)$ gives $B_{m}$ . Integrating the series itself kills every term except the constant: $\int f = \frac{A _{0}}{2} λ$ .

Recipe 1.6.4Computing a Fourier seriesbeyond the lectures

Find the period $λ$ (or $T$ ), and $k = 2 π / λ$ .
Use parity: if $f$ is even, $f sin$ is odd and all $B_{n} = 0$ ; if $f$ is odd, all $A_{n} = 0$ , including the mean.
Compute the remaining integrals over one period, splitting it where $f$ changes formula.
Check $A_{0} /2$ is the mean, and that the partial sums look like $f$ .

Example 1.6.5The square wavebeyond the lectures

$f (t) = + 1$ for $0 < t < T /2$ , $- 1$ for $- T /2 < t < 0$ , repeated.

$f$ is odd, so $A_{n} = 0$ for all $n$ , including the mean.
$B_{n} = \frac{2}{T} \cdot 2 \int_{0}^{T /2} sin (nω t) d t = \frac{4}{T} \cdot \frac{1 - c o s ( nπ )}{nω} = \frac{2}{nπ} (1 - (- 1)^{n})$ .
So $B_{n} = \frac{4}{nπ}$ for odd $n$ and 0 for even $n$ : $f (t) = \frac{4}{π} [sin ω t + \frac{1}{3} sin 3 ω t + \frac{1}{5} sin 5 ω t + \dots] .$

Loading the demo…

Figure 1.6.1 The partial sum of a Fourier series, with

N

harmonics, over the function it converges to. On the right, the bar chart of the coefficients

A_{n}

and

B_{n}

. Change the shape and add harmonics: a smooth function needs few, a jump needs many, and near a jump the partial sums overshoot by about 9 % however many you add (the Gibbs phenomenon).

A site for playing with Fourier series, given on the slides: falstad.com/fourier.

Complex form

Proposition 1.6.6Complex Fourier seriesslides w2, p. 2–3

$h (t) = n = - \infty \sum + \infty C_{n} e^{inω t}, C_{n} = \frac{1}{T} \int_{0}^{T} h (t) e^{- inω t} d t, C_{n} \in C;$ $f (x) = n = - \infty \sum + \infty C_{n} e^{ink x}, C_{n} = \frac{1}{λ} \int_{- λ /2}^{+ λ /2} f (x) e^{- ink x} d x .$

Any interval of one full period gives the same integral, so $\int_{0}^{T}$ and $\int_{- T /2}^{T /2}$ are interchangeable. One sum over positive and negative $n$ replaces the separate cosine and sine sums. Orthogonality becomes a single statement: $\frac{1}{T} \int_{0}^{T} e^{i (n - m) ω t} d t = δ_{nm}$ beyond the lectures.

Proposition 1.6.7Real and complex coefficientsslides w2, p. 3

$C_{n}$ from $A_{n}, B_{n}$	$A_{n}, B_{n}$ from $C_{n}$
$C_{0} = A_{0} /2$	$A_{0} = 2 C_{0}$
$C_{n} = (A_{n} - i B_{n}) /2 (n > 0)$	$A_{n} = C_{n} + C_{- n} (n > 0)$
$C_{- n} = (A_{n} + i B_{n}) /2 (n > 0)$	$B_{n} = i (C_{n} - C_{- n}) (n > 0)$

If the function is real, the coefficients have conjugate symmetry: $C_{- n} = C_{n}^{*}$ .

Proofbeyond the lectures

Use $cos θ = \frac{e ^{i θ} + e ^{- i θ}}{2}$ and $sin θ = \frac{e ^{i θ} - e ^{- i θ}}{2 i} = - i \frac{e ^{i θ} - e ^{- i θ}}{2}$ (Exercise 1.2.2). Then $A_{n} cos θ + B_{n} sin θ = \frac{A _{n} - i B _{n}}{2} e^{i θ} + \frac{A _{n} + i B _{n}}{2} e^{- i θ} .$ Read off the coefficients of $e^{inω t}$ and $e^{- inω t}$ . Adding the two gives $A_{n}$ ; subtracting gives $C_{n} - C_{- n} = - i B_{n}$ , so $B_{n} = i (C_{n} - C_{- n})$ . For real $f$ , $A_{n}$ and $B_{n}$ are real, so $C_{- n}$ is the conjugate of $C_{n}$ .

From the series to elementary waves

The most general disturbance is $Y (x, t) = f (x - v t) + g (x + v t)$ (Theorem 1.5.3), with $f$ and $g$ practically arbitrary. Studying waves in general is very complicated, because those functions can be almost anything.

But if $f$ and $g$ are periodic, the Fourier series applies (the Fourier transform will relax this). Combining the general form of the solution, the superposition principle and the Fourier series, $Y$ can be rewritten as an infinite sum of waves, the Fourier components, each a sine or cosine of $(x - v t)$ , or of $(x + v t)$ for $g$ . Each depends on $x$ and $t$ only through $x \mp v t$ , so by Theorem 1.5.3 it solves the same wave equation. (Week 2 writes “ $(x - v t)$ or $(x - v t)$ ”; the second is meant to be $(x + v t)$ .)

Key observation 1.6.8One component at a timeslides w2, p. 4

Without loss of generality, one can study what happens to a single sin/cos component, then add the components back up with the right Fourier coefficients.

The elementary wave

Definition 1.6.9Elementary waveEssentialslides w2, p. 5

The elementary wave is the building block of any (periodic) wave: $sin [k x - ω t + φ], k = \frac{2 π}{λ}, ω = \frac{2 π}{T},$ or $cos [k x - ω t + φ]$ , or in complex form $e^{i [k x - ω t + φ]}$ . Its wave vector and angular frequency are linked by the dispersion relation $ω = k v .$

Where it comes from. A Fourier component is $sin [(x - v t) \cdot const + φ]$ . Setting the constant equal to the wave vector $k$ meets three necessary conditions:

the argument of the sine is dimensionless: $k = 2 π / λ$ has dimensions $[L^{- 1}]$ ;
the wave has spatial period $λ$ ;
the wave has time period $T$ .

So $sin [k (x - v t)] = sin [k x - ω t]$ , which defines $ω = k v$ .

Proposition 1.6.10Speed of the elementary waveslides w2, p. 5

$v = \frac{ω}{k} = \frac{2 π / T}{2 π / λ} = \frac{λ}{T} = ν λ .$ In one period $T$ the wave moves forward by exactly one wavelength.

In terms of $T$ and $λ$ the elementary wave is $sin [\frac{2 π}{λ} x - \frac{2 π}{T} t]$ : $x \to x + λ$ or $t \to t + T$ shifts the argument by $2 π$ . The dispersion relation links the spatial and temporal periodicities; as the course will show, it also links energy to momentum (§5.5 Franck–Hertz, Stern–Gerlach and wave–particle duality).

Loading the demo…

Figure 1.6.2 The elementary wave

sin (k x - ω t + φ)

, with

its snapshot at t = 0

. Set

λ

T

and

φ

, then press Play and follow the

crest

: after one period it has moved one wavelength. Recipe: two snapshots and a speed fix the wave.

N.B. Sine and cosine differ only by a phase, and the elementary wave is defined up to an arbitrary phase $φ$ . From now on waves are written $sin [k x - ω t + φ]$ , $cos [k x - ω t + φ]$ or $e^{i [k x - ω t + φ]}$ ; the complex form is the complex substitution applied to waves.

Key observation 1.6.11Why we see sinusoidal wavesslides w2, p. 5

We are used to seeing sinusoidal waves because resonance is usually present. Out of all the frequencies that make up a generic wave, resonance selects only a few, as on a vibrating string. Why is explained with the Dirac delta.

RemarkA missing phase on week 1

Week 1, p. 27, writes the elementary wave as $sin [(x - v t) \cdot cost +]$ , with an empty slot after the $+$ and a missing-glyph box. The slot is presumably $φ$ , as week 2 has it.

Exercises

Exercise 1.6.1beyond the lectures

Show that $Y = A sin (k x - ω t)$ solves the wave equation only if $ω = k v$ .

Solution

$\partial_{x}^{2} Y = - k^{2} Y$ and $\partial_{t}^{2} Y = - ω^{2} Y$ . The equation needs $- k^{2} Y = - \frac{ω ^{2}}{v ^{2}} Y$ , so $ω^{2} = k^{2} v^{2}$ : $ω = k v$ for a wave moving towards $+ x$ .

Exercise 1.6.2beyond the lectures

Give the complex coefficients of the square wave of Example 1.6.5.

Solution

$C_{n} = (A_{n} - i B_{n}) /2 = - i B_{n} /2 = - \frac{2 i}{nπ}$ for odd $n > 0$ , $C_{- n} = C_{n}^{*} = + \frac{2 i}{nπ}$ , and $C_{n} = 0$ for even $n$ (including $C_{0}$ , the zero mean).

Exercise 1.6.3beyond the lectures

Find the Fourier series of the triangle wave $f (x) = ∣ x ∣$ for $- λ /2 < x < λ /2$ , repeated.

Solution

$f$ is even: $B_{n} = 0$ . Mean: $\frac{A _{0}}{2} = \frac{1}{λ} \cdot 2 \int_{0}^{λ /2} x d x = \frac{λ}{4}$ . $A_{n} = \frac{4}{λ} \int_{0}^{λ /2} x cos (nk x) d x = \frac{4}{λ} \cdot \frac{c o s ( nπ ) - 1}{( nk ) ^{2}} = - \frac{2 λ}{n ^{2} π ^{2}}$ for odd $n$ , 0 for even $n$ (integrate by parts; $nkλ /2 = nπ$ ). So $f = \frac{λ}{4} - \frac{2 λ}{π ^{2}} [cos k x + \frac{1}{9} cos 3 k x + \frac{1}{25} cos 5 k x + \dots]$ . The coefficients fall as $1/ n^{2}$ , faster than the square wave’s $1/ n$ : no jumps, quicker convergence.

Exercise 1.6.4beyond the lectures

Sound in air travels at 343 m/s. Find $λ$ , $k$ and $ω$ for the note A at 440 Hz.

Solution

$λ = v / ν = 343/440 = 0.78$ m; $k = 2 π / λ = 8.06$ m⁻¹; $ω = 2 π ν = 2765$ rad/s. Check: $ω / k = 343$ m/s.

Exercise 1.6.5beyond the lectures

Write $3 cos (k x - ω t) + 4 sin (k x - ω t)$ as a single elementary wave $A cos (k x - ω t + φ)$ .

Solution

It is $Re {(3 - 4 i) e^{i (k x - ω t)}}$ , since $Re {- 4 i e^{i θ}} = 4 sin θ$ . $3 - 4 i = 5 e^{- i 0.927}$ , so $A = 5$ , $φ = - 0.927$ rad $= - 53.1°$ .