Waves

What happens where the medium changes

Two conditions at a join — the displacement is continuous, and so is the transverse force — fix the reflected and transmitted amplitudes completely. What decides them is one quantity, the impedance, and not the stiffness or the density separately: two quite different media with the same impedance are, to a wave, the same medium.

Assumes: The medium decides the speed, and the source only decides the note · A wave is a shape that travels, and nothing else does

A pulse travelling along a rope reaches a knot where the rope becomes a heavier one. Part of it goes on and part comes back, upside down. Every element of that sentence — how much goes on, how much comes back, and why one of them is inverted — follows from two requirements at the knot, and neither of them is about waves.

Reflected upside down. A pulse arriving at a join where the impedance rises by a factor of 3, drawn at three moments. The amplitudes are read off the marched wave: the reflected pulse is -0.500 of the incident one and the transmitted pulse is 0.500, against (1−Z₂/Z₁)/(1+Z₂/Z₁) = -0.500 and 2/(1+Z₂/Z₁) = 0.500 from the two matching conditions. The reflection is inverted, which is the same fact as a pulse on a string flipping when it reaches a wall: a wall is a medium of infinite impedance, and the inversion is what keeps the displacement at the join equal to zero. Note that the transmitted amplitude exceeds one where the second medium is lighter, and that this is not a violation of anything: amplitude is not energy.
Fig. 1 A pulse arriving at a join where the impedance rises threefold, marched through in three snapshots. The amplitudes are read off the drawn wave, not substituted: the reflected pulse is −0.500 of the incident one and the transmitted pulse is 0.500, against the matching conditions’ (1Z2/Z1)/(1+Z2/Z1)(1-Z_2/Z_1)/(1+Z_2/Z_1) and 2/(1+Z2/Z1)2/(1+Z_2/Z_1). Note that the transmitted pulse is narrower — it is travelling more slowly, so the same duration occupies less length.

The two requirements are these. The rope does not break, so the displacement on one side of the knot equals the displacement on the other. And the knot has no mass, so the transverse forces on it must cancel — the slope times the tension has to match across the join. Two conditions, two unknowns, and the problem is closed.

What comes out is one number

Solving the pair gives amplitude coefficients

r=Z1Z2Z1+Z2,t=2Z1Z1+Z2,r = \frac{Z_1-Z_2}{Z_1+Z_2},\qquad t = \frac{2Z_1}{Z_1+Z_2},

where Z=TμZ=\sqrt{T\mu} for a string — the geometric mean of the tension and the mass per unit length. Nothing else survives.

Everything depends on one ratio. The reflected and transmitted amplitudes at a join, against the ratio of the two impedances. Nothing else appears: not the tension, not the density, not the frequency, not the shape of the wave. The reflection changes sign at a ratio of one — that is the inversion at a heavier medium — and it is the only place either curve does anything abrupt. At 0.333 the coefficients are 0.500 and 1.500; at 3 the coefficients are -0.500 and 0.500. A join at ratio one reflects nothing at all, which is what matching a load means and why an anti-reflection coating, an impedance-matching horn and a transformer are the same idea.
Fig. 2 The two coefficients against the ratio of the impedances. Neither the tension nor the density appears on its own, and neither does the frequency or the shape of the wave. The reflection changes sign at a ratio of one, which is the inversion at a heavier medium; the transmission exceeds one wherever the second medium is lighter. At the two marked ratios, which are reciprocals, the coefficients are 0.500 and 1.500 against −0.500 and 0.500.

That a single combination decides everything is worth taking seriously, because it is not obvious. The wave speed is T/μ\sqrt{T/\mu} and the impedance is Tμ\sqrt{T\mu}; both are built from the same two quantities and they are independent of each other. A medium can be made faster without changing its impedance, by raising the tension and lowering the density together — and if that is done at a join, a wave crosses it with no reflection at all while changing speed and wavelength.

Everything depends on one ratio. The reflected and transmitted amplitudes at a join, against the ratio of the two impedances. Nothing else appears: not the tension, not the density, not the frequency, not the shape of the wave. The reflection changes sign at a ratio of one — that is the inversion at a heavier medium — and it is the only place either curve does anything abrupt. At 0.1 the coefficients are 0.818 and 1.818; at 1 the coefficients are 0.000 and 1.000; at 4 the coefficients are -0.600 and 0.400. A join at ratio one reflects nothing at all, which is what matching a load means and why an anti-reflection coating, an impedance-matching horn and a transformer are the same idea.
Fig. 3 The same two curves marked at a tenfold mismatch in each direction and at the matched point. At a ratio of 0.1 the transmitted amplitude is 1.818 and the reflection is 0.818; at 4 they are 0.400 and −0.600; at 1 they are 1 and 0. The curves are smooth through the matched point and nothing special happens there to the amplitudes — what happens is that one of them passes through zero, which is a fact about a sign rather than about a discontinuity.

The other combination of the same two quantities is the speed, T/μ\sqrt{T/\mu}, which decides how fast the wave goes and how long its wavelength is. Speed and impedance are independent, and that independence is what makes matching possible at all: a boundary can change one without changing the other. Everything about whether a wave crosses is impedance; everything about what it looks like once across is speed.

The sign, and what it is doing there

The minus sign for Z2>Z1Z_2>Z_1 is the most visible feature of the whole subject and the one most often stated as a separate rule.

It is a consequence of the first matching condition. At the join the displacement is whatever it is on both sides; if the second medium is very heavy it barely moves, so the sum of the incident and reflected displacements there must be nearly zero, so the reflected pulse must be nearly the negative of the incident one. In the limit of infinite impedance — a wall — the join cannot move at all, the sum is exactly zero, and r=1r=-1.

Reflected the same way up. A pulse arriving at a join where the impedance falls by a factor of 3.33, drawn at three moments. The amplitudes are read off the marched wave: the reflected pulse is 0.538 of the incident one and the transmitted pulse is 1.538, against (1−Z₂/Z₁)/(1+Z₂/Z₁) = 0.538 and 2/(1+Z₂/Z₁) = 1.538 from the two matching conditions. The reflection keeps its sign, which is what a free end does. Note that the transmitted amplitude exceeds one where the second medium is lighter, and that this is not a violation of anything: amplitude is not energy.
Fig. 4 The same join with the ratio the other way up: the second medium is lighter, so it moves more than the first and the reflection keeps its sign. At Z2/Z1=0.3Z_2/Z_1=0.3 the reflected pulse is 0.538 and the transmitted one is 1.538, drawn wider because the wave is travelling faster on the far side. The limiting case here is a free end, where the reflection is +1 and the end of the rope flicks to twice the incident amplitude.

Both limits are what makes a standing wave possible. A string clamped at both ends has r=1r=-1 at each, so a wave sent along it returns inverted, returns again upright, and interferes with itself. Only the wavelengths for which that interference is constructive survive.

Total reflection at both ends produces a standing wave, and the boundary condition doing it is the r=1r = -1 case stated as a constraint rather than as a coefficient: the displacement at a clamped end is zero, so the allowed wavelengths are the ones fitting a whole number of half-cycles between the two ends. Change one end to a free one — r=+1r = +1 — and the allowed set changes to odd quarter-wavelengths, which is the difference between a flute and a clarinet.

The energy, and the factor that gets dropped

The coefficients above are amplitude ratios, and a transmitted amplitude of 1.6 looks alarming until the energies are done properly.

Where the energy goes, and the weighting that is left out. Reflected and transmitted power at 3 joins, as fractions of the incident power. The reflected fraction is r²; the transmitted fraction is not t² but (Z₂/Z₁)t², because the power a wave carries depends on the medium as well as on the amplitude. With that factor the two add to exactly one at every ratio — 0.360 + 0.640, 0.000 + 1.000, 0.360 + 0.640 — and without it they add to 2.920, 1.000, 0.520, which is the arithmetic that makes a transmitted amplitude greater than one look impossible when it is not.
Fig. 5 Reflected and transmitted power at three joins, as fractions of what arrives. The reflected fraction is r2r^2; the transmitted fraction is not t2t^2 but (Z2/Z1)t2(Z_2/Z_1)t^2, because the power a wave carries depends on the medium as well as on the amplitude. With the factor the two add to exactly one at every ratio, to twelve decimal places. Without it they add to 2.920, 1.000 and 0.520 — one of them impossible, one accidentally right, and one apparently a loss.

The reason for the weighting is that power in a wave is 12Zω2A2\tfrac12 Z\omega^2A^2: for the same amplitude and frequency, a high-impedance medium carries more. So a large amplitude in a light medium and a small one in a heavy medium can carry the same power, and the conversion between the two at a join is exactly what the transmission coefficient describes. A wave crossing into a lighter medium gets bigger and does not get more energetic.

That accounting also explains why the reflected fraction is symmetric under swapping the two media — r2r^2 is unchanged by Z1Z2Z_1\leftrightarrow Z_2 — while the amplitudes are not. A boundary reflects the same proportion of energy whichever way a wave crosses it, which is a statement of reciprocity and is why a window that lets 96 per cent of the light in lets 96 per cent of it out.

Where the energy goes, and the weighting that is left out. Reflected and transmitted power at 3 joins, as fractions of the incident power. The reflected fraction is r²; the transmitted fraction is not t² but (Z₂/Z₁)t², because the power a wave carries depends on the medium as well as on the amplitude. With that factor the two add to exactly one at every ratio — 0.111 + 0.889, 0.111 + 0.889, 0.640 + 0.360 — and without it they add to 1.889, 0.556, 0.680, which is the arithmetic that makes a transmitted amplitude greater than one look impossible when it is not.
Fig. 6 The energy accounting at three more joins, including a ninefold mismatch where 64% of the power comes straight back. The pair at 0.5 and 2 are reciprocals and reflect the identical 11.1%, which is the symmetry just described drawn rather than asserted. Every bar sums to exactly one, which is the figure’s own check that the impedance weighting has been applied.

Matching, which is the whole of the engineering

Since the reflection vanishes at Z2=Z1Z_2=Z_1 and grows as the ratio departs from one, the design problem for anything that has to get a wave from one medium into another is always the same: reduce the ratio, or hide it.

A single quarter-wave layer of impedance Z1Z2\sqrt{Z_1Z_2} between the two makes the two reflections it produces cancel, because the second is delayed by half a cycle relative to the first and is of equal size. That is the anti-reflection coating on a lens, the matching layer on an ultrasound probe, the quarter-wave transformer in a waveguide, and the graded layer on a radar-absorbing surface. All four are the same calculation with different words.

A gradual taper does the same job over a band of frequencies rather than at one: if the impedance changes slowly compared with a wavelength, the wave has no boundary to reflect from at all. A horn on a brass instrument is exactly this, matching the high impedance of a narrow tube to the low impedance of open air; so is the flare of a loudspeaker, so is an acoustic ceiling wedge, so is the impedance-graded coating on the inside of an anechoic chamber.

The failure case is as instructive as the success. An ultrasound scan will not work through air, because the impedance ratio between soft tissue and air is about 3,400 — so 99.9 per cent of the energy is reflected at the first millimetre of air anywhere in the path. That is why the gel exists, and it is the entire reason for it: not lubrication, not conduction, but a layer whose impedance is close to tissue’s, filling the gap so the mismatch never occurs.

The same number, read the other way, is what makes the scan work. An echo comes back from every place the impedance changes, and its size says by how much — so the image is a map of impedance discontinuities rather than of density, or of stiffness, or of anything a pathologist would name. Bone and air both show up as bright reflectors and as shadows behind, for the same reason and with opposite signs, and a soft-tissue boundary with a mismatch of a per cent returns a ten-thousandth of the energy and is visible only because the instrument is patient.

Where the two waves overlap they add, and if their amplitudes are unequal the cancellation is incomplete: where they are exactly out of step the sum is the difference of the amplitudes rather than zero. So a partial reflection produces a partial standing wave, with maxima and minima whose ratio measures the reflection coefficient directly. That ratio has a name in every branch of the subject and is the standard way of measuring a mismatch without taking anything apart.

What impedance is a ratio of

The essay has given four different formulas for an impedance — Tμ\sqrt{T\mu} for a string, ρc\rho c for sound, μ/ε\sqrt{\mu/\varepsilon} for light, L/C\sqrt{L/C} for a cable — and has said they are the same quantity without saying what quantity that is. It is a ratio, it is the same ratio in all four cases, and saying so makes the matching conditions look inevitable rather than convenient.

Take the string. At any point, the transverse force one side exerts on the other is the tension times the slope, Ty/x-T\,\partial y/\partial x, and the transverse velocity of the material there is y/t\partial y/\partial t. For a wave travelling in one direction those two derivatives are not independent: a shape moving at speed cc satisfies y/x=(1/c)y/t\partial y/\partial x = -(1/c)\,\partial y/\partial t. Substituting,

F=Tcv=Tμ  v.F = \frac{T}{c}\,v = \sqrt{T\mu}\;v.

So the impedance is the force required per unit velocity — how hard the wave has to push to get the medium moving. That is why it is the geometric mean of the two material constants rather than their ratio, and that is why it is called an impedance, which is a word borrowed from circuits for exactly the same quantity.

Every case on the list is that ratio with different names for the two factors. In sound it is pressure over particle velocity. On a cable it is voltage over current, which is where the word came from. For light it is electric field over magnetic field, up to the constants. In each pair, one member is an effort — a force, a pressure, a voltage, a field — and the other is a flow, and the product of the two is the power the wave carries.

Once that is seen, the two matching conditions stop being facts about knots. They are the statement that the effort and the flow are each continuous across a join, and they are continuous for reasons that are about bookkeeping rather than about physics: a discontinuity in the flow would mean material accumulating at a point of no volume, and a discontinuity in the effort would mean a finite force on a massless element. Both are always true, in every medium, for the same reason.

That is the real content of the claim that all these subjects share one calculation. It is not that four wave equations happen to have similar solutions. It is that a wave always carries an effort and a flow, the impedance is their ratio, and continuity of both at a join is a constraint that writes itself.

A join with mass in it

The second matching condition assumed the knot has no mass, and that assumption is what makes the coefficients independent of frequency. It is worth relaxing, because the frequency dependence it produces is the thing every real boundary has and no figure on this page shows.

Put a bead of mass mm at the junction of two identical strings. The displacement is still continuous — the bead is attached to both — but the forces no longer cancel, because now there is something for the net force to accelerate. The second condition becomes Newton’s law applied to the bead: the difference of the two transverse forces equals the bead’s mass times its acceleration.

Working that through for a sinusoidal wave gives a transmission amplitude

t=11+iωm/2Z,t = \frac{1}{1 + i\omega m/2Z},

which is complex, so the transmitted wave is now shifted in phase as well as reduced in size — something no massless join ever does. Its magnitude falls from one at low frequency toward zero at high, passing through 1/21/\sqrt2 at ω=2Z/m\omega = 2Z/m.

That is a first-order low-pass filter, corner frequency and all, arrived at by putting a bead on a string. Low frequencies do not notice the bead, because over one cycle it has time to be dragged along by the string on either side. High frequencies see an object too heavy to be moved in the time available, and reflect off it as they would off a wall.

The general statement is worth extracting, because it explains a remark made in passing above. A boundary is characterised by a length or a time as well as by an impedance ratio, and whether it counts as sharp depends on the wavelength. A bead is sharp at low frequency and a wall at high. A layer is invisible when it is thin compared with a wavelength and a mirror when it is a quarter of one. A taper is a boundary at low frequency and no boundary at all at high.

This is the same reasoning as the ultrasound gel and the brass bell, and it is why the frequency-independent coefficients of the first half of this essay are the exception rather than the rule. They hold in the limit where the join has no structure of its own — no mass, no thickness, no stiffness — which is a good description of a great many boundaries and of no real one.

Why the boundary is where the physics is

It is worth asking why so much of wave physics happens at joins rather than in the middle of things. The answer is that a uniform medium does very little: a wave crosses it, arriving as it left. Everything interesting — reflection, refraction, standing waves, resonance, the existence of modes, the trapping of light in a fibre, the confinement of an electron in an atom — happens where the medium stops being uniform.

That gives the two matching conditions a status they do not obviously deserve. They are not a special topic within waves; they are the only place where the solutions of a wave equation acquire any structure. In the interior, any wave of any shape at any wavelength is allowed; at a boundary, a condition is imposed, and imposing conditions at two boundaries at once is what turns a continuum of possibilities into a discrete set.

Every quantised system in physics is that argument. A note on a string, a colour in a hydrogen spectrum, a band in a solid and the energy of an electron in a quantum dot are all continuous problems made discrete by the requirement that a wave match something at the edges. The two conditions in this essay are the simplest instance, and the fact that they are simple enough to solve on a rope is why the rope keeps being drawn.

The same algebra, elsewhere

The pair of matching conditions is not about ropes. Any wave equation with a discontinuity in its coefficients gives the same structure, with the impedance built from whatever the local equivalents of stiffness and inertia are.

For sound, Z=ρcZ=\rho c: air is 413 rayl and water is 1.5 million, so the reflection coefficient at the surface of a lake is 0.9995 and essentially no airborne sound gets into it. For light, the impedance is μ/ε\sqrt{\mu/\varepsilon} and the coefficients reduce to the Fresnel formulae at normal incidence, giving the familiar 4 per cent reflection at a glass surface. For a transmission line it is L/C\sqrt{L/C}, and a cable terminated in anything but its own characteristic impedance sends a reflection back down itself — which is why coaxial cable is sold as 50 or 75 ohms and why a laboratory instrument has a terminator hanging off a spare socket.

The list is worth reading as a single fact rather than four analogies. In each case the wave equation is 2y/t2=(1/ρκ)2y/x2\partial^2 y/\partial t^2 = (1/\rho\kappa)\,\partial^2 y/\partial x^2 with two material constants, one of which acts as an inertia and one as a compliance; the speed is the square root of their ratio and the impedance is the square root of their product. Which physical quantity plays which role changes; the algebra does not. That is why an acoustician, an optician and a radio engineer can borrow each other’s results directly, and why the word impedance — coined for electrical circuits — ended up attached to a property of a rope.

The quantum version has the same algebra with a probability amplitude in place of a displacement, and the two conditions become the continuity of the wavefunction and of its derivative. What is different is what plays the part of the impedance: it is set by the local momentum, so a step in potential energy reflects part of a particle even when the particle has more than enough energy to cross it. That has no classical counterpart at all, and it is the same formula as a rope with a knot in it.

A mismatch used deliberately

Matching is one use of the coefficients; the opposite is the other, and it is just as deliberate.

A crystal resonator, a laser cavity and an organ pipe all work by maximising the mismatch at their ends, so that almost nothing leaks out and the wave makes thousands of round trips before the energy is gone. The quality of such a resonator is essentially a count of those trips, and the count is set by how close the reflection coefficient is to one. A laser mirror is specified to a reflectivity of 99.98 per cent for exactly this reason: the difference between that and 99.9 is a factor of five in how long the light stays inside.

What a small leak buys is a sharp response, and the sharpening as the loss falls is the whole design space of a resonator. A resonator’s linewidth and its round-trip loss are the same quantity in two languages, and the mismatch at the ends is usually the dominant term in that loss — so a designer choosing a reflectivity is choosing a linewidth, whether or not the calculation is done that way round.

The same reasoning explains why a musical instrument is a compromise rather than an optimum. A pipe that reflected perfectly at its open end would ring beautifully and be inaudible; one that reflected nothing would radiate everything and would not have modes at all. What is wanted is a mismatch large enough to sustain a standing wave and small enough to leak a usable fraction, and the flare of a bell is a tunable knob on precisely that trade.

There is one more use, which is measurement. Send a pulse down a cable and watch what comes back: the timing of the reflection gives the distance to the fault, and the sign and size of it says what kind of fault. A positive reflection is an open circuit, a negative one a short, and a partial one a change of impedance such as a crushed cable or a badly made joint. The technique is called time-domain reflectometry, it is how a break in a buried cable is found without digging it up, and it is the figure at the top of this page used as an instrument.

What the picture cannot show

Everything above is at normal incidence. At an angle the matching conditions have to be applied to components, the impedance acquires a factor of cosθ\cos\theta or 1/cosθ1/\cos\theta depending on the polarisation, and the two polarisations then behave differently — which is where Brewster’s angle comes from and which a one-dimensional drawing cannot contain.

The figures also draw a sharp join. A boundary is only sharp compared with a wavelength, and the same physical interface is abrupt for one frequency and gradual for another; that frequency dependence is what makes a taper work and is invisible in a snapshot. Nor is any absorption drawn: real joins turn a fraction of the energy into heat, and the accounting that closes to twelve decimal places above closes to rather less in a laboratory.

And the pulse is drawn as a scalar displacement, which is right for a rope and not for light, sound in a solid, or anything with more than one polarisation. A solid supports both compression and shear waves with different impedances, and a wave arriving at an interface generally produces four outgoing waves rather than two — mode conversion, which is a nuisance in ultrasonic testing and the reason seismology can see the Earth’s liquid core.

The ladder from here

Later rungs on this anchor: oblique incidence and the Fresnel coefficients derived the same way; the quarter-wave transformer and multilayer stacks, where the reflections from many boundaries are summed; the transmission-line equations, where the same algebra becomes an engineering discipline; mode conversion at a solid interface; and the impedance of free space, 377 ohms, which is what an antenna has to be matched to and which is a property of the vacuum rather than of anything material.

The neighbouring ladders are what sets a wave’s speed, which is the other combination of the same two properties, standing waves, which is what total reflection at two ends produces, and tunnelling, which is this calculation with a wavefunction in it.

Part 1 of 4

This essay is one argument about Impedance. The others:

What links here

Essays that reach for this one mid-argument — the half of a link its own author cannot write down.

What this makes readable

Essays that declare this one a prerequisite.

The objects named here

The third axis, after the field and the reading path: the things themselves, and every essay that touches each one.

Boundary conditionsEnergy conservationImpedanceImpedance matchingPhase inversionReflection coefficientStanding waveTransmission coefficient