Phasors: the language of waves

The puzzle: adding sound and getting silence¶

Put on a pair of noise-cancelling headphones and switch them on. The roar of the aeroplane engine fades. But the headphones did not block the sound — a soft earpad cannot stop a 40-hertz rumble. They added more sound to it.

That should bother you. Two loudspeakers playing at once are louder than one. Add water to water and you get more water; add sound to sound and — sometimes — you get silence.

The answer is the single most reused idea in this course. It is the reason a thin film of soap shimmers in colours, the reason a diffraction grating splits light, and — much later — the arithmetic underneath quantum mechanics. This module builds it in the simplest possible setting: no waves yet, no space, just things that repeat in time.

Predict before you calculate¶

Commit to an answer for each of these before running anything. Write them down. The point is not to be right — it is to find out which of your intuitions need rebuilding.

Two oscillations of equal amplitude $A$ and the same frequency are added. What resultant amplitudes are possible? Give the largest, the smallest, and the rule in between.
Two oscillations of equal amplitude but frequencies 440 Hz and 441 Hz are added. What does the result sound like? Is it even a single “note”?
One hundred violinists play the same note at the same loudness, but nobody synchronises their bowing, so the phases are random. Is the sound intensity 100 times one violinist — or 10,000 times, or something else?
An oscillation is described by an amplitude and a phase. Can the phase of a single oscillation be measured? Can the phase difference of two?

Explore the model¶

The laboratory notebook for this module puts two (then many) rotating arrows on the screen — complex amplitudes, called phasors — and lets you set each one’s length and starting angle. The physical signal is the horizontal shadow of the arrow tip; superposition is placing arrows tip-to-tail.

Two phasors rotating clockwise at the same frequency (left), placed tip-to-tail. Their sum — the third arrow — also rotates rigidly, and its horizontal projection (right) is again a cosine of the same frequency. Change the phase difference and only the length of the sum arrow changes: constructive at $\delta = 0$ , destructive at $\delta = \pi$ . The animation is generated by the same wavelab.phasors code you can run and read.

Two equal amplitudes at slightly different frequencies. In the rotating frame of the first, the second arrow slowly turns: the pair drifts between agreement and cancellation, and the projection (right) pulses — beats. The envelope repeats at the frequency difference.

Model specification

System: one or more harmonic oscillations of a single angular frequency, each represented by a complex amplitude; the observable is the real signal their sum projects out.
Dynamics: uniform rotation only — every phasor turns as $e^{-\ii\omega t}$ and nothing else happens; there are no equations of motion here.
Boundary: none; phasors neither grow, decay, nor exchange anything.
Ensemble: deterministic, except the random-phase experiment, whose phases are independent uniform draws from a seeded generator.
Ignored: everything physical — what oscillates, what drives it, and every mechanism (damping, forcing, nonlinearity) that would change an amplitude in time.
Valid when: the signals really are single-frequency sinusoids, so a fixed complex amplitude captures each of them completely.
Failure modes: different frequencies summed as if phasors (their relative phase is not constant — that is what beats are), and any nonlinear operation applied to the complex representation instead of the real signal.

Run the laboratory now, then come back. Three things are worth doing before you read on:

Set two equal amplitudes and sweep the phase difference from 0 to $2\pi$ . Watch the sum arrow’s length — and find the two phase differences where the resultant equals one of the originals.
Detune the second frequency by one percent and watch the sum in the rotating frame.
Add one hundred unit phasors with random phases, re-roll the randomness, and watch how far the sum arrow typically reaches.

Derive the result¶

The representation. Euler’s formula, $e^{\ii\theta} = \cos\theta + \ii\sin\theta$ , lets us write any harmonic oscillation as the real part of a rotating complex number. Under the course convention (the phase convention) the rotation is clockwise:

x(t) = A\cos(\omega t + \varphi) = \Real\!\left[\underbrace{A e^{-\ii\varphi}}_{\hat{x}}\; e^{-\ii\omega t}\right].

(1)

The complex number $\hat{x} = A e^{-\ii\varphi}$ — the phasor — freezes the two numbers that distinguish this oscillation from any other of the same frequency: length is amplitude, angle is (minus) phase. The rotation $e^{-\ii\omega t}$ is common to every oscillation of frequency $\omega$ , which is precisely why it can be divided out.

Superposition. Add two oscillations of the same frequency:

A_1\cos(\omega t + \varphi_1) + A_2\cos(\omega t + \varphi_2) = \Real\!\left[(\hat{x}_1 + \hat{x}_2)\, e^{-\ii\omega t}\right].

(2)

The right-hand side is again the real part of one complex number times the rotation — that is, again a harmonic oscillation at $\omega$ . This deserves its box:

The interference law. The squared length of a sum of two arrows is the law of cosines:

A^2 = |\hat{x}_1 + \hat{x}_2|^2 = A_1^2 + A_2^2 + 2 A_1 A_2 \cos\delta, \qquad \delta = \varphi_1 - \varphi_2 .

(3)

Everything the opening puzzle asked is in this one line. Equal amplitudes and $\delta = \pi$ : $A^2 = 2A_1^2 - 2A_1^2 = 0$ — the headphones. Equal amplitudes and $\delta = 0$ : $A = 2A_1$ , and the intensity, which goes as amplitude squared, is four times one source, not two. Where did the missing energy go at $\delta=\pi$ , and where did the extra come from at $\delta=0$ ? Hold that thought — it is a quiz question now and becomes module 23’s central topic, when this same formula reappears wearing its optics costume, $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\delta$ .

Different frequencies: beats. If the frequencies differ, there is no common rotation to divide out, and the phasor picture must be used honestly: in the frame rotating with $\omega_1$ , the second arrow turns at the difference rate. For equal amplitudes the identity

a\cos\omega_1 t + a\cos\omega_2 t = 2a\,\cos\!\Big(\tfrac{\omega_1 - \omega_2}{2}\,t\Big)\cos\!\Big(\tfrac{\omega_1 + \omega_2}{2}\,t\Big)

(4)

says the sum is a fast oscillation at the average frequency inside a slow envelope at half the difference — and since loudness follows the envelope squared, you hear the pulse at the full difference $|\omega_1 - \omega_2|$ : one beat per second for 440 Hz against 441 Hz.

Random phases. Add $N$ unit phasors whose phases are independent and uniform. The expected intensity is

\Big\langle \big|\textstyle\sum_k e^{-\ii\varphi_k}\big|^2 \Big\rangle = \sum_{k,l} \big\langle e^{-\ii(\varphi_k - \varphi_l)} \big\rangle = N ,

(5)

because every cross term averages to zero. The resultant amplitude therefore grows only as $\sqrt{N}$ — a random walk in the complex plane — while $N$ synchronised phasors reach amplitude $N$ and intensity $N^2$ . The hundred unsynchronised violinists are 100 times one violinist; a hundred phase-locked emitters would be 10,000 times. That gap, $N$ versus $N^2$ , is the entire engineering case for the laser.

Verify computationally¶

Deriving a formula and trusting it are different things. The laboratory checks each result numerically, and each check is also a test in the project’s test suite, so the claims on this page cannot silently rot.

1. The representation. wavelab.phasors.real_signal computes $\Real[\hat{x}\,e^{-\ii\omega t}]$ and is compared point-by-point against the direct cosine $A\cos(\omega t + \varphi)$ . Machine-precision agreement — and this is the test that would scream if the phasor sign convention were ever broken.

2. The interference law. The phasor prediction for $(A, \varphi)$ of a two-oscillation sum is compared against a brute-force sum of cosines on a fine time grid. Agreement to 10^-12.

3. The beat identity. The honestly-summed two-frequency signal against the carrier-times-envelope form. Exact.

4. The random walk.

Transfer the idea¶

You have just learned the arithmetic that runs the rest of the course — and a surprising amount of physics beyond it.

Interference (module 23). Replace “two oscillations” by “two paths light can take”: the interference law becomes $I = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\delta$ , with $\delta$ set by the path difference. Young’s double slit is this module plus geometry.
Diffraction (module 29). An aperture is a continuum of sources; the tip-to-tail polygon of arrows becomes an integral — and computing the arrow sum for every direction at once is exactly a Fourier transform.
Coherence (module 27). The violinists: sources with locked phases add like arrows, sources with random phases add like intensities. “Coherent” names the first situation.
AC circuits. Impedance is the phasor idea applied to voltage and current; an engineering course rotates the arrows the other way (the $j$ of the conventions page).
Quantum mechanics. Probability amplitudes add as complex numbers before being squared — Feynman’s “arrows”. The arithmetic you just learned is that arithmetic.

Check your understanding¶

Two identical loudspeakers each produce, at your ear, a pure tone of amplitude A at the same frequency. Which received amplitudes are possible, depending on their phase difference?
A. Anything from 0 up to 2A, set by the phase difference. B. Exactly 2A — two sources of amplitude A always add to 2A. C. Between A and 2A — a second source can only add, never remove. D. Always sqrt(2) A, because uncorrelated signals add in quadrature.
Check your answer
Correct answer: A — Anything from 0 up to 2A, set by the phase difference.
- A. Right. The resultant obeys A_sum^2 = A^2 + A^2 + 2 A^2 cos(delta): 2A when in phase, exactly 0 when opposed, everything in between otherwise. Adding a signal can silence a signal — that is how noise cancellation works.
- B. Only when the phases agree. Amplitudes are not volumes of water; they carry signs moment by moment, and with a half-cycle offset every push from one speaker meets an equal pull from the other.
- C. This is the intuition this module exists to replace. At phase difference pi the two signals cancel completely: the sum of two oscillations can be smaller than either of them, all the way down to zero.
- D. Quadrature addition applies to random-phase (incoherent) sources on average. These two speakers hold a fixed phase relation, so the full range 0 to 2A is available — which value you get is set by delta.
Under the course convention, the oscillation x(t) = 5 cos(omega t + pi/3) is represented by which phasor?
A. 5 e^(-i pi/3), an arrow of length 5 at angle -60 degrees. B. 5 e^(+i pi/3), an arrow of length 5 at angle +60 degrees. C. 5 cos(pi/3) = 2.5, a real number. D. (5, pi/3) — a pair of real numbers; no complex number is needed.
Check your answer
Correct answer: A — 5 e^(-i pi/3), an arrow of length 5 at angle -60 degrees.
- A. Correct. The phasor is A e^(-i phi), so that Re[phasor x e^(-i omega t)] reproduces A cos(omega t + phi). The minus sign in the phasor pairs with the clockwise rotation e^(-i omega t).
- B. That is the engineering convention’s phasor (which rotates counterclockwise as e^(+j omega t)). Under this course’s clockwise rotation it represents 5 cos(omega t - pi/3) — the wrong sign of the phase.
- C. That is the signal’s value at t = 0, one snapshot. The phasor must encode both numbers that characterise the oscillation — amplitude and phase — which takes a complex number, not a real one.
- D. The pair does record the information, but it does not add correctly: summing two oscillations means summing their phasors as complex numbers, and no componentwise rule on (A, phi) pairs does that.
Two same-frequency oscillations have amplitudes 3.0 and 4.0 and a phase difference of a quarter cycle (delta = pi/2). What is the amplitude of their sum?
Check your answer
Correct answer: 5.0 (± 0.1)
A^2 = 3^2 + 4^2 + 2 x 3 x 4 x cos(pi/2) = 9 + 16 + 0 = 25, so A = 5. At a quarter cycle the cross term vanishes and amplitudes add like the sides of a right triangle — the 3-4-5 triangle, drawn with arrows.
A 440 Hz tuning fork sounds together with a string vibrating at 442 Hz, at equal loudness. What do you hear?
A. A tone near 441 Hz whose loudness pulses 2 times per second. B. Two separate steady tones, at 440 Hz and at 442 Hz. C. A steady tone at 882 Hz, the sum of the frequencies. D. A steady tone at 441 Hz of constant loudness.
Check your answer
Correct answer: A — A tone near 441 Hz whose loudness pulses 2 times per second.
- A. Right. The sum is a carrier at the average frequency, 441 Hz, inside a slow envelope; loudness follows the envelope squared and pulses at the full difference, 2 Hz. Musicians tune by slowing these beats to a stop.
- B. The ear cannot separate tones 2 Hz apart — and at any single moment the air carries one summed pressure signal. That sum is a 441 Hz oscillation whose amplitude slowly pulses: beats.
- C. Frequencies do not add when signals superpose — signals add. The sum frequency appears only through nonlinear processing; linear superposition produces the average frequency modulated at the difference.
- D. The carrier is indeed 441 Hz, but its amplitude cannot be constant: the two components drift from agreement to opposition once per second per hertz of detuning, so the loudness must pulse. Constant loudness would need equal frequencies.
One hundred small radio transmitters each deliver unit intensity to a distant receiver. Compare the received intensity when their phases are independent and random with when they are locked together.
A. About 100 when random, exactly 10,000 when locked. B. About 100 in both cases — the total power is fixed by the sources. C. About 10,000 in both cases, since there are 100 sources either way. D. About 10 when random — a random walk of 100 steps typically reaches sqrt(100).
Check your answer
Correct answer: A — About 100 when random, exactly 10,000 when locked.
- A. Right. Random phases: cross terms average away, intensities add, giving N = 100. Locked phases: amplitudes add to N, and intensity is amplitude squared, N^2 = 10,000. That factor of N between them is why a laser outshines a lamp.
- B. Total emitted power is fixed, but locking the phases redirects it: the locked array concentrates emission into narrow directions (this receiver among them) at the expense of others. Locally, intensity can reach N^2.
- C. With random phases the field arrows perform a random walk: typical resultant amplitude sqrt(N), hence intensity N = 100, not N^2. Getting N^2 requires the arrows aligned — coherence.
- D. sqrt(N) = 10 is the typical amplitude, and it is the right random-walk reasoning — but intensity is amplitude squared, so the expected intensity is N = 100.
A student says: “Phase is just a mathematical bookkeeping angle — nothing physical ever depends on it.” What is the correct assessment?
A. Phase differences are measurable physics — they decide interference — while one oscillation’s absolute phase is a clock-setting convention. B. The student is right — only amplitudes are measurable, and phase never affects any observation. C. All phases, including the absolute phase of a single oscillation, are separately measurable. D. Phase matters only when the amplitudes of the two oscillations are exactly equal.
Check your answer
Correct answer: A — Phase differences are measurable physics — they decide interference — while one oscillation’s absolute phase is a clock-setting convention.
- A. Exactly. Whether two waves brighten or silence each other is set by their phase difference, and interferometers measure such differences to fractions of a wavelength. Only the overall phase — when you started the clock — is conventional.
- B. Noise-cancelling headphones falsify this in your ears: the anti-noise differs from the noise only in phase, and the result is silence instead of doubled sound. Phase differences have loudly observable consequences.
- C. Too strong in the other direction. The absolute phase encodes the choice of t = 0; shift your clock and it changes with no physical consequence. What experiments access is phase differences.
- D. The cross term 2 A1 A2 cos(delta) is present for any two nonzero amplitudes; equal amplitudes merely allow the special case of complete cancellation. Interference does not need equality.

Exam-style problems for this module — the ones worth doing with a pen — are collected in the problem set.

Explain it in your own words¶

Answer in a few sentences each. These are the questions that reveal whether the ideas landed; no number will save you.

A friend says: “Noise-cancelling headphones absorb the noise.” Correct the sentence, and explain where the sound’s energy actually goes. (Careful — this is deeper than it looks.)
Why is the sum of two same-frequency cosines always a cosine of that frequency, while the sum of two different-frequency cosines is not a cosine at all?
Explain to a musician why an out-of-tune unison produces a slow pulsing, and what exactly sets the pulsing rate.
One hundred radio antennas broadcast at one frequency. Compare the received intensity when their phases are random with when they are synchronised, and explain the difference without algebra.

Advanced: beyond the simplest picture¶

Where the cancelled energy goes. Superposition never destroys energy; it redistributes it. Two loudspeakers driven in antiphase cancel along the midplane but reinforce elsewhere, and the total radiated power integrates correctly over all directions. The headphone case is sharper: the anti-noise wave changes the radiation load the speaker and the noise source each see, so less acoustic energy is emitted into the ear canal in the first place. Energy bookkeeping in interference is module 23’s subtlest topic, and this paragraph is its trailer.

The rotating-wave idea. Dividing out $e^{-\ii\omega t}$ — working “in the rotating frame” — is a trick with a long career ahead of it: it is how the driven oscillator of module 02 is solved, how magnetic resonance is analysed, and how quantum optics tames fast-oscillating terms (there it is called the rotating-wave approximation).

Negative frequencies. A real cosine is the sum of two complex exponentials, $\cos\omega t = \tfrac12(e^{-\ii\omega t} + e^{+\ii\omega t})$

— one clockwise, one counterclockwise. Keeping only the clockwise half (the “analytic signal”) is what this module quietly did, and it is legitimate exactly because the physical signal is recovered by taking twice the real part. The full two-sided picture returns with the Fourier transform in module 04.