The absorber whose bandwidth is its thickness

The mismatch no network can remove ends with a limit that belongs to the load rather than to the design: a resistance with a capacitance across it carries a fixed allowance of match, and every network built to feed it does no more than choose where the allowance is spent. That essay’s load is given and its network is free. The problem here turns it round. The load is the most unforgiving one there is — a sheet of metal, which reflects every wave that reaches it — and the thing to be designed is a coating that stops it reflecting. The coating is free to be made of anything at all, and the limit turns out to be its thickness.

This is the problem of the radar absorber, the anechoic chamber and the coating that keeps a microwave oven’s door from leaking, and at every frequency from radio to millimetre waves it has the same form. A plane wave in air meets a metal surface, and the metal — far below its plasma frequency, where the frequency below which nothing gets in shows a conductor refusing a wave, and penetrated only a skin depth, as how far a field gets into metal computes — is a short circuit: the reflected wave is equal and opposite to the incident one. In the language of what happens where the medium changes, the impedance steps from free space’s 376.7 ohms to zero, and a step to zero reflects everything. An absorber is whatever has to be put in front of the metal so that the wave arriving from free space sees 376.7 ohms instead.

A sheet where the field is #

The simplest absorber is a thin sheet of resistive material — a carbon-loaded film, a coating of resistive ink — with a resistance of 376.7 ohms per square, held some distance in front of the metal. The sheet’s resistance per square is what matters because a plane wave driving a current across a square of it sees the same resistance whatever the size of the square, and 376.7 ohms is exactly the impedance of free space.

Where the sheet is placed decides everything. The bare metal sets up a standing wave in front of it, with a node of electric field at the surface — the field has to vanish on a perfect conductor — and maxima a quarter of a wavelength out. A resistive sheet laid directly on the metal sits at the node, has no field across it, carries no current and absorbs nothing; painting a resistive film straight onto the metal does nothing at all. Move it a quarter-wave out and it sits at the maximum. There, seen from the front, the quarter-wave of air backed by metal behaves as an open circuit — the short at the metal transformed through a quarter-wave line into its opposite, as the quarter-wave layer of the layer that makes a reflection vanish transforms one impedance into another. The wave arriving sees the sheet’s 376.7 ohms in parallel with nothing, which is 376.7 ohms, a perfect match. Every bit of the wave’s energy is dissipated in the sheet.

This is the Salisbury screen, patented by Winfield Salisbury in 1952 and one of the first radar absorbers. It is perfect at one frequency and at its odd multiples, and nowhere else.

One quarter-wave, one narrow band #

Away from 10 gigahertz the spacing is no longer a quarter-wave, the air gap stops looking like an open circuit, and some of the wave is reflected. At 20 gigahertz the gap is half a wavelength, the sheet is back at a node, and the screen reflects everything, as if it were not there. The useful band — where less than one per cent of the power comes back, −20 decibels — runs from 8.74 to 11.26 gigahertz, about ±13 per cent around the design frequency. Between the notches, what stands in front of the screen is a partial standing wave, the pattern of the node that is not standing still, whose minima fill in as the reflection grows; an engineer measuring an absorber with a probe in front of it reads exactly that ratio of maxima to minima.

A sheet of the wrong resistance does worse everywhere. At the design frequency the sheet alone terminates the line, so its reflection is the reflection of a resistance $R$ against 376.7 ohms, $(R - Z_0)/(R + Z_0)$: a third of the amplitude for a sheet of half or twice the impedance of free space. That makes the curves flatter, and whether flatter is worse depends on what is asked of the absorber — which is the first sign of the thing this essay is about.

The thickness fixes the area #

A narrow band invites the obvious question: what material, placed in the same 7.5 millimetres, would absorb over a wider one? Konstantin Rozanov answered it in 2000, and the answer is that none can do better in total. For any flat layer backed by metal, however it is built — sheets, foams, graded mixtures, lossy dielectrics — the reflection coefficient at normal incidence obeys

$$\int_0^\infty \ln\frac{1}{|\Gamma(\lambda)|}\,d\lambda \;\le\; 2\pi^2 \sum_i \mu_{s,i}\, d_i ,$$

where $d_i$ are the thicknesses of its layers and $\mu_{s,i}$ their static permeabilities, which are one for anything non-magnetic. The left side is the total absorption, measured on a logarithmic scale of reflection and integrated over wavelength; the right side is the thickness.

The drawing is the whole content of the result. The four curves have very different shapes: the matched sheet shoots up to total absorption at a wavelength four times its spacing, the 150-ohm sheet absorbs a little everywhere and never much. Their areas, computed numerically across all wavelengths, are the same to five figures — 147.9 millimetres, exactly $2\pi^2$ times 7.5 millimetres. Changing the sheet changes where the absorption goes and cannot change how much of it there is. The sheet more resistive than free space falls short, with less than half the allowance; it wastes the rest, and the reason is a zero of its reflection coefficient hidden off the real frequency axis, which the theorem’s proof charges against the total.

The bound comes from the same place as the Bode–Fano limit, and as the Kramers–Kronig relations of the answer that cannot come first: the reflection coefficient is the response of a causal system, so it is analytic in half of the complex frequency plane, and an integral of its logarithm around that half-plane is fixed by how it behaves at its edge. At very long wavelengths a thin metal-backed layer looks like a small inductance in front of a short — its reflection departs from −1 by an amount proportional to its thickness and permeability — and that departure, which nothing inside the layer can change, sets the value of the integral.

Two consequences follow at once. The weight is in wavelength, not frequency, so low frequencies are expensive: a band from 1 to 2 gigahertz spans 15 centimetres of wavelength, and the band from 10 to 20 gigahertz only 1.5. And the reflection is measured logarithmically, so a deeper absorption costs in proportion to its depth in decibels: holding a band at −40 dB costs twice what −20 dB does.

More sheets, a wider band, a thicker absorber #

The classical way to widen the band is to add sheets — a Jaumann absorber, named after the German engineer credited with it in the Second World War — each a quarter-wave from the next, with resistances rising outwards so that the wave meets a gentle gradient of loss rather than one abrupt sheet, as in the taper that matches every note.

With resistances optimised to hold the reflection under −20 dB over the widest possible band, the two-sheet design covers 5.7 to 14.3 gigahertz and the three-sheet design 4 to 16 — a band three times as wide as its centre frequency is high, against a quarter for the single sheet. The price is visible in the legend. Each sheet brings another quarter-wave of depth, and the three-sheet absorber is 22.5 millimetres thick. Every one of the three designs meets Rozanov’s bound exactly: their integrated absorptions are 147.9, 295.9 and 443.8 millimetres, $2\pi^2$ times 7.5, 15 and 22.5. The extra sheets did not find extra absorption. They bought it, with thickness, and then spread it more usefully.

The ripple in the band is a choice. The optimiser was told to hold the worst reflection at −20 dB, and it did so by letting the reflection rise to exactly that level between the notches rather than wasting absorption on notches deeper than the requirement. Absorption spent at −35 dB in one place is absorption not available to hold −20 dB somewhere else.

How close each design comes #

The bound gives a minimum thickness for any band. If an absorber must hold its reflection below a level $|\Gamma_0|$ from wavelength $\lambda_{\min}$ to $\lambda_{\max}$, that alone uses up $\ln(1/|\Gamma_0|)(\lambda_{\max} - \lambda_{\min})$ of the integral, so its thickness can be no less than that divided by $2\pi^2\mu_s$.

The single Salisbury screen is more than eight times thicker than the bound demands for its band, because it spends most of its allowance where nobody asked for it — on a deep notch at the centre, on the flanks outside the band, and on long wavelengths where it absorbs a little. The three-sheet absorber is three and a half times the bound, because more of its allowance falls inside its band. No design reaches the bound itself, which would require a reflection exactly at −20 dB inside the band and exactly total outside it, a response no finite structure has. What the comparison measures is how much of a fixed budget each design wastes, and it is the right question to ask of any absorber before asking about its materials.

The loophole is permeability #

The bound has one quantity in it besides thickness: static permeability. A non-magnetic absorber is stuck with $\mu_s = 1$, and a magnetic one is not.

The ferrite tiles that line the walls of electromagnetic test chambers are about six millimetres thick and absorb from a few tens of megahertz upwards, where the wavelength is ten metres. A resistive sheet at the same depth absorbs nothing until twelve gigahertz. The difference is the permeability: at 1,500, it multiplies the tile’s thickness in the bound by 1,500, and the tile has 186.5 metres of wavelength to spend where the sheet has 12 centimetres.

The mechanism is worth seeing, because it is not simply that a magnetic material is lossy. A ferrite’s permeability relaxes — it stays near its static value up to a relaxation frequency, a few megahertz here, and falls off above it with a large imaginary part, exactly as the dielectric constant of water relaxes in the constant that depends on how fast it is asked. Well above the relaxation, the thin tile on metal looks like an inductance whose inductance is falling in proportion to frequency, and an inductive impedance $j\omega L$ with $L \propto 1/(j\omega)$ is a pure resistance, the same at every frequency. With a tile thickness, a static permeability and a relaxation frequency chosen so that this resistance equals 376.7 ohms, the tile matches free space over the whole range from well above the relaxation up to where its thickness becomes a sizeable fraction of a wavelength inside it. The flat match that a resistive sheet achieves only at a quarter-wave, the ferrite achieves by having its permeability fall in exactly the right way.

There is a simpler way to see why magnetism is the loophole, and it is in the first figure. At the surface of the metal the electric field has a node, which is why a resistive sheet there does nothing. The magnetic field has the opposite pattern: the surface currents that cancel the electric field double the magnetic one, so the magnetic field is at its maximum right on the metal. Anything that dissipates energy through the electric field has to stand off a quarter-wave to find any field to work on; anything that dissipates it through the magnetic field can lie flat on the metal and find the strongest field there is. A thin non-magnetic layer on metal sits in almost no electric field at long wavelengths, and its absorption falls with the square of thickness over wavelength. A thin magnetic layer sits in the full magnetic field, and its absorption depends on thickness times permeability — which is exactly the combination that appears on the right-hand side of the bound.

The same argument explains the ferrite’s upper edge. Above a few hundred megahertz its permeability has fallen so far that the 6.3 millimetres are no longer electrically thin, the tile stops being a simple resistance, and absorption at those frequencies has to come from something else — which is why chamber walls put pyramids of carbon-loaded foam on top of the ferrite tiles, adding the depth the higher frequencies need.

Where the bound holds #

Normal incidence, flat layers. The integral is stated for a plane wave arriving head-on at an infinite flat stack. At oblique incidence the bound takes a different form for each polarisation and depends on the angle, and for curved or finite objects there is no simple bound at all, which is part of why shaping matters as much as coating in reducing a radar reflection.

Passive, linear, time-invariant materials. The proof uses the causality and passivity of the layer’s response. An absorber with active elements, or one whose properties are switched in time — the same exception the Bode–Fano limit leaves open — is not bound by it.

Static permeability as a material constant. The ferrite’s $\mu_s$ enters as the permeability at zero frequency. Materials with large $\mu_s$ are ferrimagnets and ferromagnets whose permeability relaxes, and Snoek’s law ties the static permeability to the relaxation frequency, roughly holding their product fixed; a magnetic absorber cannot have both an enormous $\mu_s$ and a relaxation at high frequency, and that is the real limit on how thin one can be.

Metal backing. A layer without metal behind it can transmit instead of reflecting, and the absorber problem becomes a different one.

Seams, edges, and the heat nobody draws #

Every curve here is a reflection coefficient for a plane wave on an infinite sheet. Real absorbers are panels with seams and edges, and a seam between two perfect panels is a slot that scatters; radar reflections from aircraft are dominated by edges, cavities and specular flat surfaces that no coating fully treats. The figures also show only where the energy does not go — back towards the source — and not where it does: into heat in the resistive sheets and the ferrite, negligible on a radar target, but in a chamber used to test transmitters at high power the absorber warms, and how much power it can take is part of its specification.

Still open: the thinnest absorber that can be built #

The bound is a ceiling that designs approach but do not reach, and the gap has become a research field. Metamaterial absorbers use resonant structures much smaller than a wavelength to put the absorption where it is wanted with little waste, and the best of them are reported to come within a small factor of the thickness bound over bandwidths of an octave or more. Whether time-modulated and active absorbers can go below it for signals that are not known in advance, and how much power the modulation must cost, is open, as is whether materials can be found that break Snoek’s trade between high static permeability and high relaxation frequency.

The habit worth carrying away is to find the budget before spending effort on the design. An absorber backed by metal has a fixed allowance of absorption, set by its thickness and its permeability; every sheet, gradient and material chosen after that decides only where the allowance goes. Asking for a thinner absorber is asking either for less absorption or for more permeability, and there is no third answer.