Uniform plane waves

Far from the TLR-1 antenna, the wave it launches looks locally like the simplest solution of Maxwell's equations: a uniform plane wave, in which the electric and magnetic fields are perpendicular to each other and to the direction of travel and are the same everywhere on each plane perpendicular to that direction. Real waves spread out from their sources, but a few wavelengths from an antenna the spreading is so gradual that over any small region the wave is plane. Almost everything engineers need to know about propagation, from how fast a signal travels and how much it weakens in concrete to how deep currents penetrate into a metal enclosure and how much power a person standing near an antenna absorbs, can be learned from plane waves. This lesson develops them in lossless and lossy media, introduces the Poynting vector for power flow and describes polarization.

The plane wave in a lossless medium

Take the phasor wave equation of lesson 6 in a source-free lossless medium and look for a field that depends only on z. Let E point along x. The equation reduces to d²E_x/dz² + β²E_x = 0, with β = ω√(με). Its solutions are E_x = E⁺e^{−jβz} + E⁻e^{+jβz}: a wave travelling in +z and one travelling in −z. In the time domain the forward wave is E⁺ cos(ωt − βz), a cosine pattern sliding along z at the phase velocity v_p = ω/β = 1/√(με). The wavelength is λ = 2π/β = v_p/f. The phase constant β, in radians per metre, is also called the wavenumber k.

Faraday's law then fixes the magnetic field: H points along y, in phase with E, with magnitude E/η. The ratio η = √(μ/ε) is the intrinsic impedance of the medium, measured in ohms. In free space η₀ = √(μ₀/ε₀) ≈ 376.7 Ω, often rounded to 377 Ω or written 120π. In a dielectric with ε_r, η = η₀/√ε_r, so in polyethylene it is about 251 Ω and in FR-4 about 180 Ω. The directions of E, H and propagation form a right-handed set: E × H points in the direction of travel. A wave travelling in −z has its H reversed relative to E. These relations, E and H perpendicular, in phase and in the ratio η, are the plane-wave rules you will use constantly.

Power flow: the Poynting vector

Electromagnetic waves carry energy. Poynting's theorem, derived by combining Maxwell's curl equations, shows that the rate at which energy flows out of a closed surface equals the rate of decrease of stored field energy inside minus the power dissipated in it. The power flow per unit area is the Poynting vector S = E × H, in watts per square metre. For sinusoidal fields, the time-average Poynting vector is ½Re(E × H*) with peak phasors, or Re(E × H*) with rms phasors. For a plane wave in a lossless medium this is |E|²/η, using rms values. Power density is what radiation exposure limits restrict and what a receiving antenna collects, so lessons 11 and 12 use it constantly.

Poynting's theorem has a surprising consequence even in circuits. In a coaxial cable carrying DC power, the energy does not flow inside the copper; it flows through the dielectric between the conductors, carried by the radial electric field and the circling magnetic field, whose cross product points along the cable. The copper guides the energy and dissipates a little of it as heat, but the energy itself travels in the fields. The same is true on a printed-circuit board: the signal energy travels in the dielectric between a trace and its ground plane, which is why lesson 11 cares so much about what lies under a trace.

Waves in lossy media

In a conducting medium, ∇×H = (σ + jωε)E, and the wave equation acquires a complex propagation constant γ = α + jβ = √(jωμ(σ + jωε)). The wave becomes E⁺e^{−αz}e^{−jβz}: it still oscillates with phase constant β but its amplitude decays by the attenuation constant α, in nepers per metre. Multiply α by 8.686 to get decibels per metre. The intrinsic impedance becomes complex, η = √(jωμ/(σ + jωε)), so H lags E in phase. The depth at which the amplitude falls to 1/e, about 37 percent, is 1/α.

Two limits make γ easy. In a low-loss dielectric, σ ≪ ωε, the attenuation is α ≈ (σ/2)√(μ/ε) = (σ/2)η and β is almost unchanged. If the loss is described by a loss tangent instead, α ≈ (β tan δ)/2. In a good conductor, σ ≫ ωε, α and β become equal, both √(πfμσ), and the wave decays by a factor of e in a distance δ = 1/√(πfμσ), the skin depth. At radio frequencies δ is tiny in metals: about 2.2 µm in copper at 915 MHz. Currents on metal surfaces flow only in this thin layer, which is why radio-frequency conductors behave as though their resistance per square were the surface resistance R_s = 1/(σδ), and why silver plating a few micrometres thick improves a connector even though the bulk metal is brass.

Skin depth and shielding

Skin depth decides whether a metal enclosure shields. A wave striking a conducting wall is mostly reflected, as lesson 8 explains, and what enters decays by about 8.7 dB per skin depth. A sheet several skin depths thick therefore adds tens of decibels of absorption loss on top of the reflection loss. At 915 MHz, the 1.5 mm aluminium of the TLR-1 enclosure is hundreds of skin depths thick, so the metal itself is effectively opaque; in practice the shielding is limited by seams, gaps and openings, which lesson 8 also addresses. At the loop detector's 71 kHz, the skin depth in aluminium is about 0.3 mm, so the same enclosure still absorbs well. At 60 Hz, the skin depth in aluminium is about 11 mm and the enclosure barely affects power-frequency magnetic fields, which is why hum from a nearby luminaire cable must be fought with twisted pairs and balanced inputs rather than with thin metal.

import math, cmath
eps0, mu0, c = 8.854e-12, 4e-7 * math.pi, 2.998e8
eta0 = math.sqrt(mu0 / eps0)
print(f"free space: eta0 = {eta0:.1f} ohm, lambda at 915 MHz = {c / 915e6 * 100:.1f} cm")
# 1. Skin depth at three frequencies and surface resistance at 915 MHz
print("metal              60 Hz       71.2 kHz     915 MHz     Rs at 915 MHz")
for name, sig, mur in (("copper", 5.8e7, 1), ("aluminum", 3.5e7, 1), ("stainless (304)", 1.4e6, 1), ("mild steel", 6e6, 200)):
    ds = [1 / math.sqrt(math.pi * f * mu0 * mur * sig) for f in (60, 71.2e3, 915e6)]
    print(f"{name:16s} {ds[0]*1e3:8.3f} mm {ds[1]*1e3:8.4f} mm {ds[2]*1e6:8.3f} um   {1 / (sig * ds[2]) * 1e3:6.2f} mohm")
# 2. Exact propagation constant at 915 MHz (effective conductivity includes dielectric loss)
w = 2 * math.pi * 915e6
print("medium            alpha (dB/m)   lambda (cm)   |eta| (ohm)")
for name, er, sig in (("dry concrete", 5, 5e-3), ("typical concrete", 6, 3e-2), ("wet soil", 15, 5e-2), ("fresh water", 80, 0.2), ("seawater", 81, 4.0)):
    gam = cmath.sqrt(1j * w * mu0 * (sig + 1j * w * er * eps0))
    eta = cmath.sqrt(1j * w * mu0 / (sig + 1j * w * er * eps0))
    print(f"{name:16s} {gam.real * 8.686:10.2f}     {2 * math.pi / gam.imag * 100:9.2f}     {abs(eta):9.1f}")
# 3. Power density near the TLR-1 cabinet antenna and the general-public reference levels
eirp = 10 ** ((20 + 6 - 2) / 10) / 1000          # 20 dBm, 6 dBi panel, 2 dB cable loss -> W
for r in (0.2, 0.5, 2.0):
    S = eirp / (4 * math.pi * r * r)
    print(f"r = {r:3.1f} m: S = {S:7.4f} W/m^2, E = {math.sqrt(S * eta0):5.2f} V/m rms")
f = 915
print(f"reference levels at {f} MHz: Safety Code 6 {0.02619 * f**0.6834:.2f} W/m^2, FCC general population {f / 1500 * 10:.2f} W/m^2")

free space: eta0 = 376.7 ohm, lambda at 915 MHz = 32.8 cm
metal              60 Hz       71.2 kHz     915 MHz     Rs at 915 MHz
copper              8.532 mm   0.2477 mm    2.185 um     7.89 mohm
aluminum           10.983 mm   0.3188 mm    2.812 um    10.16 mohm
stainless (304)    54.914 mm   1.5941 mm   14.062 um    50.80 mohm
mild steel          1.876 mm   0.0544 mm    0.480 um   347.00 mohm
medium            alpha (dB/m)   lambda (cm)   |eta| (ohm)
dry concrete           3.66         14.65         168.5
typical concrete      20.01         13.36         153.4
wet soil              21.11          8.46          97.2
fresh water           36.57          3.66          42.1
seawater             664.76          3.33          35.5
r = 0.2 m: S =  0.4997 W/m^2, E = 13.72 V/m rms
r = 0.5 m: S =  0.0800 W/m^2, E =  5.49 V/m rms
r = 2.0 m: S =  0.0050 W/m^2, E =  1.37 V/m rms
reference levels at 915 MHz: Safety Code 6 2.77 W/m^2, FCC general population 6.10 W/m^2

The skin-depth table confirms the worked example and adds two lessons. Stainless steel, a poor conductor, has a skin depth about six times that of copper and a surface resistance about six times higher, which is why stainless hardware near an antenna is lossy. Mild steel has a much smaller skin depth than its conductivity alone would suggest, because its permeability concentrates the current further; its high surface resistance is why a steel pole or a rebar cage absorbs some radio energy. The propagation table uses effective conductivities that include the dielectric relaxation loss at 915 MHz, so they are much larger than the direct-current values in lesson 6. Concrete attenuates the TLR-1 signal by several to around 20 dB per metre depending on its moisture, wet soil similarly, fresh water more, and seawater stops it within centimetres. Those numbers explain why the radio works across an intersection but not from inside a buried vault.

Exposure to radio-frequency fields

Health Canada's Safety Code 6 sets reference levels for human exposure, and ISED's RSS-102 applies them to radio equipment. For the general public between 300 MHz and 6 GHz, the reference level for power density rises with frequency, 0.02619 f^0.6834 W/m² with f in MHz, which is about 2.8 W/m² at 915 MHz, averaged over 6 minutes. The power density near the TLR-1 cabinet antenna, with 24 dBm of effective isotropic radiated power from a 20 dBm transmitter, a 6 dBi antenna and 2 dB of cable loss, is about 0.5 W/m² at 20 cm and 0.08 W/m² at 0.5 m, and the antenna sits on a mast arm several metres above the sidewalk. The far-field formula overestimates the density very close to an antenna, and the regulator's evaluation methods take that into account, but even this simple estimate shows a large margin. The US limit, for comparison, is about 6.1 W/m² at the same frequency.

Polarization

The polarization of a wave describes the path traced by the tip of its electric field vector at a fixed point. If E always points along one line, the wave is linearly polarized; a vertical whip antenna radiates vertical linear polarization. If E has two perpendicular components of equal amplitude 90° out of phase, the tip traces a circle and the wave is circularly polarized, right-hand or left-hand depending on the direction of rotation relative to the direction of travel. In general the components have unequal amplitudes or an arbitrary phase and the tip traces an ellipse. Satellite navigation signals use circular polarization so that the receiver's orientation and the rotation of polarization by the ionosphere do not matter.

Polarization matters because a receiving antenna responds only to the field component that matches its own polarization. A vertical receiving whip picks up a vertically polarized wave fully and a horizontally polarized wave hardly at all; the polarization loss factor is cos²ψ, where ψ is the angle between the polarizations, so a 45° mismatch costs 3 dB and a 90° mismatch, in theory, everything. A linearly polarized antenna receiving a circularly polarized wave loses 3 dB whatever its orientation. For the TLR-1, Amara specifies vertical polarization at both ends of every link. Reflections from building walls and vehicles partly rotate the polarization, which lesson 8 explains, so in a cluttered intersection the polarization mismatch loss is a few decibels rather than the large values of a clean test range.