Illustrated Physics
physics, drawn
Series

Impedance — the series

9 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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.

    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.

    part 1 · waves
  2. What a real coating leaves behind, and over what range. Reflectance against wavelength for a glass surface of index 1.52 in air, uncoated and with quarter-wave layers of 2 different indices, each a quarter of a wave thick at 550 nm. The ideal index is the geometric mean, 1.2329, and the layer made of it takes the reflectance to zero at the design wavelength exactly. No durable solid has that index: magnesium fluoride at 1.38 is the usual compromise and it leaves 1.26% at the design wavelength against 4.26% bare — a reduction of 3.4× rather than a removal. Both curves rise away from the design wavelength, because the thickness is a quarter of a wave only there, and the useful band is wide but not unlimited: the better coating stays under a quarter of the bare reflectance from 415 to 780 nm. The purple cast of a coated lens is that residual — the ends of the visible reflecting while the middle does not.

    The layer that makes a reflection vanish

    A wave meeting a step in impedance reflects, and nothing can be done about the step. Put a third medium between the two, a quarter of a wavelength thick and of exactly the intermediate impedance, and the reflection stops existing — not reduced, cancelled.

    part 2 · waves
  3. A graded junction against a quarter-wave layer. Reflectance against wavelength for four ways of joining a medium of index 1 to one of 1.52. The bare interface reflects 4.26 per cent at every wavelength. A single quarter-wave layer of index √(n₀n_s) takes that to zero at 550 nm exactly and rises symmetrically either side, which is the shape of every single-layer coating and every single-section transformer. The remaining curves are graded junctions of 150 nm and 400 nm, built as 120 thin layers whose index climbs geometrically and put through the same matrix product. They do not have a design wavelength at all. Below a cutoff they are flat and negligible; above it they climb steeply toward the bare value, and the cutoff is set by the length: 398 nm for the 150 nm taper, 1062 nm for the 400 nm taper. Roughly, a taper works for every wavelength shorter than about twice its own optical length, which is the statement that a reflection needs a partner a quarter of a wavelength further in to cancel against. The engineering versions of this are everywhere: the horn on a loudspeaker, the moth's eye, the flared transition between two waveguides, and the graded layer that lets an ultrasound probe reach tissue across a hundredfold impedance step.

    The taper that matches every note

    A quarter-wave layer cancels a reflection at one wavelength and only near it. Spread the same change of impedance over a distance instead, and the reflection vanishes for every wavelength shorter than about twice that distance — not by cancelling one echo against another, but by leaving no step anywhere for an echo to come from.

    part 3 · waves
  4. How close a network gets to a load that stores charge. The fraction of a wave's amplitude reflected from a resistance shunted by a capacitance, with RCωc = 2, against frequency in units of the band edge ωc, for the load alone and for networks of inductors and capacitors whose values, with an ideal transformer at the source, were optimised numerically to keep the reflection low across the band. The bare load holds it to 0.707 across the band; the 1-element network holds it to 0.392 across the band; the 2-element network holds it to 0.320 across the band; the 3-element network holds it to 0.289 across the band. The dashed line is Bode and Fano's floor, exp(−π/RCωc) = 0.208, which no network of any size can go beneath over the whole band; each added element brings the design closer to it, and each buys its flatter band with a reflection that climbs to total just beyond the band edge.

    The mismatch no network can remove

    A quarter-wave layer or a taper can match a resistance to a resistance as well as anyone likes. Put a capacitance across the load and that stops being true for every network that could ever be built from lossless parts: Bode and Fano proved that the total amount of match available is fixed by the load's resistance and capacitance, so a network can only move it about — and a flat match across a band can never be better than e to the minus π over the load's time constant times the band.

    part 4 · waves
  5. The thickness fixes the area, and the design only moves it. How much a metal-backed resistive sheet 7.5 mm from the metal absorbs, measured as ln(1/|Γ|), against wavelength, for sheets of 150, 250, 377, 800 Ω per square. Rozanov's bound says the area under any such curve, over all wavelengths, can be at most 2π² times the thickness: 147.9 mm here. The areas computed are 147.9 mm for 150 Ω, 147.9 mm for 250 Ω, 147.9 mm for 377 Ω, 69.7 mm for 800 Ω. Every sheet up to the impedance of free space meets the bound exactly and only moves its area between wavelengths; the more resistive sheet wastes part of it.

    The absorber whose bandwidth is its thickness

    A metal plate reflects everything, and a surface that must swallow a radar wave instead has to be a match between free space and a short circuit. A resistive sheet a quarter-wave in front of the metal does it perfectly at one frequency. Widening the band takes more sheets and more depth, and a theorem says why that is not a failure of design: for any absorber backed by metal, the total absorption over all wavelengths is fixed by its thickness, and by nothing else except its permeability.

    part 5 · waves
  6. Echoes from a layer that thins to nothing. Synthetic seismic echoes from a layer between two identical rocks — the top reflecting +R, the base −R — for thicknesses of 0, 5, 10, 15, 20, 25, 30, 40, 50, 60, 75, 100 m, sounded with a 30 Hz pulse in rock of 3000 m/s, whose wavelength in the layer is 100 m. Time runs downward; each trace is one thickness. Thick, the layer gives two separate echoes, one from each face, and its thickness can be read from the time between them. As it thins the two echoes overlap and merge. Near 20 m, about a fifth of a wavelength, they interfere constructively and the combined echo is strongest, 1.45 times one face's echo. Thinner still, the combined echo keeps a fixed shape — the pulse's derivative — and simply weakens in proportion to the thickness; nothing in its timing changes.

    The layer too thin to see

    A seismic survey sends a pulse of sound into the ground and listens to the echoes from every boundary where the rock changes. A thick layer returns two echoes, one from its top and one from its base, and the time between them gives its thickness. A thin one returns one echo, and below about a quarter of a wavelength the timing says nothing more: every thin layer looks the same, just weaker as it thins. It has not become invisible. It has become unmeasurable in the one way that was being used, and its thickness has moved into the echo's strength and its spectrum.

    part 6 · waves
  7. A lever that makes air and water meet. The fraction of sound power crossing from air into water through an ideal lossless transformer that multiplies pressure by a factor n, against n on a logarithmic axis. With no transformer, n = 1, it is 0.11 per cent. It rises as the transformer makes the water look less stiff and air look stiffer, reaching everything at n = 60, and falls again beyond, where the water is made to look softer than air. The middle ear's 22 gives 43 per cent in this picture — a gain of 26 dB over no transformer at all. The curve is flat near its top: a lever within a factor of two of the ideal still passes about two-thirds of the power, which is why a mechanism need not be precise to be effective.

    The lever that lets air speak to water

    Sound in air striking a surface of water is almost entirely reflected: 99.9 per cent bounces off, which is why a swimmer underwater hears the world above as a muffled murmur. The inner ear is filled with fluid, and the eardrum faces air, so every land animal has the problem of getting sound across that boundary. The answer is three tiny bones and a difference in area, which together multiply the pressure about twenty-fold and turn a thousandth of the power into nearly half. The middle ear is an impedance transformer, the mechanical twin of the one in every radio, and it works best over just the band where speech carries its meaning.

    part 7 · waves
  8. The mass law: how much sound a limp wall stops. The transmission loss of a wall that only has mass — no stiffness — for sound arriving square-on, against frequency on a logarithmic axis, for surface masses of 10 kg/m², 20 kg/m², 40 kg/m². It is 10 log(1 + (ωm/2ρc)²) decibels. At 500 Hz, 10 kg/m² stops 31.6 dB, 20 kg/m² stops 37.6 dB, 40 kg/m² stops 43.6 dB. Doubling the mass or the frequency adds six decibels, and nothing else about the wall matters: it is the wall's inertia, set against air's characteristic impedance ρc = 413 rayl, that turns the sound back.

    The wall that stops sound by its weight

    A wall between two rooms keeps out sound not by absorbing it but by being hard to shake. The sound's pressure pushes the wall back and forth, the moving wall becomes a loudspeaker on the far side, and how much gets through depends almost entirely on the wall's mass per square metre against the frequency: six decibels more for every doubling of either. Read as impedance, a wall is a mass placed in series with the air, and the same reading explains the frequency at which a stiff pane of glass nearly vanishes and why two thin walls with a gap beat one thick one.

    part 8 · waves
  9. A light panel that blocks one band as a heavy one would. Sound transmission loss at normal incidence against frequency, on a logarithmic frequency axis. Dashed: a limp panel of 2 kg/m², following the mass law, 13.4 dB at 300 Hz. Dotted: the same panel made 3 kg/m² by adding 1 kg/m² of plain mass, 16.8 dB. Solid: the 2 kg/m² panel with the same added mass hung on springs tuned to 300 Hz (damping 0.02). It blocks 35 dB at 300 Hz — the loss the mass law would give a wall 26 kg/m² heavy — and just above, at 367 Hz, falls to 5.2 dB, far below even the bare panel's 15.1: there the panel lets sound through more easily than if the resonators were not there.

    The wall made heavy only at one note

    A wall stops sound by being hard to shake, and how hard a plain wall is to shake is its weight: that is the mass law, and it makes quiet walls heavy. Hang small masses on springs inside a light panel, tuned to one frequency, and near that frequency the panel becomes many times harder to shake than it weighs, because the hidden masses swing against it. A two-kilogram panel then blocks a tone as a wall thirteen times heavier would. The price comes just above the tuning, where the panel's dynamic mass passes through zero and it lets sound through more easily than if the masses were not there.

    part 9 · waves

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