ch3: EM waves in matter

date: May 17 2026

the question driving this chapter: how do walls effect EM transmission and SNR?

1. The three fates at a boundary

A radio wave travelling through one medium hits a boundary with another. Exactly three things happen, in proportions set by the materials, the angle of arrival, and the frequency.

Reflection. Some of the wave bounces back. The fraction is set by the impedance mismatch between the two media (Section 3).
Transmission, with a bend. The part that does not reflect enters the second medium, propagating in a slightly different direction. The new direction is set by Snell’s law (Section 4).
Absorption inside the second medium. Once inside, the wave loses energy to the material at a rate set by complex permittivity. The wave decays exponentially with depth (Section 5).

2. Free space and the $R^4$ problem

Before the wave reaches the wall, it travels through air. Even in a perfect vacuum, the wave gets weaker with distance, not because the air absorbs anything, but because the energy spreads over an expanding surface.

One-way propagation: $1/R^2$

An isotropic source of power $P_t$ radiating in all directions spreads its energy over a sphere of radius $R$ and area $4\pi R^2$ . The power density at distance $R$ is

S(R) \;=\; \frac{P_t}{4\pi R^2}

A real antenna concentrates energy in a beam, multiplying this density by the antenna gain $G$ , but the $1/R^2$ falloff is unchanged. The total received power at a receiver of effective area $A_{rx}$ is the Friis transmission equation:

P_r \;=\; \frac{P_t \, G_t \, G_r \, \lambda^2}{(4\pi R)^2}

The wavelength squared shows up because effective area of an antenna scales as $\lambda^2$ : at higher frequency, the same physical antenna has smaller effective area, so it captures less incident energy.

Two-way (radar) propagation: $1/R^4$

A radar listens to its own echo. The wave first travels distance $R$ to the target, illuminating it with power density $P_t G_t / (4\pi R^2)$ . The target re-radiates a fraction of that as if it were a tiny isotropic re-radiator with effective cross-section $\sigma$ (the radar cross-section). The re-radiated wave then travels another $R$ back, suffering another $1/R^2$ spreading loss before it reaches the antenna of effective area $G_r \lambda^2 / (4\pi)$ .

P_r \;=\; \frac{P_t \, G_t \, G_r \, \lambda^2 \, \sigma}{(4\pi)^3 \, R^4}

The $R^4$ in the denominator is the central fact of radar design. Doubling the target range cuts the received power by 16. Tripling it cuts by 81.

3. Reflection at a boundary

An electromagnetic wave is a coupled oscillation of E and B fields. The ratio of $|E|$ (eletric field) to $|H|$ (magnetic field) is the wave impedance of the medium. In free space, $\eta_0 \approx 377$ Ω. In a dielectric with relative permittivity $\epsilon_r$ ,

\eta \;=\; \frac{\eta_0}{\sqrt{\epsilon_r}}

When a wave going through medium 1 reaches medium 2, the field has to match boundary conditions on both sides simultaneously, and a single incoming wave cannot do that alone. So part of it reflects back. At normal incidence, the reflection coefficient (field amplitude) is

r \;=\; \frac{\eta_2 - \eta_1}{\eta_2 + \eta_1}

The power reflection coefficient is $|r|^2$ and the power transmission coefficient is $1 - |r|^2$ (for lossless materials). For a wave going from air ( $\epsilon_r = 1$ ) into brick ( $\epsilon_r \approx 3.75$ ):

r \;=\; \frac{1/\sqrt{3.75} - 1}{1/\sqrt{3.75} + 1} \;\approx\; -0.32, \qquad |r|^2 \;\approx\; 0.10

So about 10% of the power reflects at the front surface; 90% enters the brick. The minus sign on $r$ means the reflected E-field is inverted relative to the incident one (the same thing that happens when a string wave reflects off a fixed end).

At an angle: Fresnel coefficients

At oblique incidence, polarization starts to matter. Fix the plane of incidence: the plane containing the incoming ray and the surface normal. Polarization splits into two cases relative to it:

TE (s-polarization): E points out of the plane of incidence. It lies flat along the surface regardless of angle.
TM (p-polarization): E sits inside the plane of incidence. It tilts with the ray, so part of it runs along the surface and part of it sticks out.

The boundary only cares about the tangential component of E. For TE that component is the whole field and doesn’t change with angle; for TM the share between tangential and normal does change with angle. So the two polarizations reflect by different amounts, and the gap widens as the incidence gets more oblique:

r_\text{TE} \;=\; \frac{\eta_2 \cos\theta_i - \eta_1 \cos\theta_t}{\eta_2 \cos\theta_i + \eta_1 \cos\theta_t}, \qquad r_\text{TM} \;=\; \frac{\eta_2 \cos\theta_t - \eta_1 \cos\theta_i}{\eta_2 \cos\theta_t + \eta_1 \cos\theta_i}

Both reduce to the normal-incidence formula at $\theta_i = 0$ , and both go to $|r| = 1$ at grazing. In between, TE climbs monotonically while TM dips to zero at one specific angle before climbing back.

Brewster angle

That zero falls where the numerator of $r_\text{TM}$ vanishes. Solving with Snell’s law:

\tan \theta_B \;=\; \frac{n_2}{n_1} \;=\; \sqrt{\frac{\epsilon_{r,2}}{\epsilon_{r,1}}}

At $\theta_B$ the reflected and refracted rays would be exactly $90°$ apart. The reflected wave would have to be radiated by dipoles in the second medium oscillating along the direction of the reflected ray, and a dipole cannot radiate along its own axis. So nothing reflects, and the entire TM wave passes through. TE has no such zero because its E-field is perpendicular to both rays at any angle.

The practical upshot: an unpolarized wave hitting a dielectric near $\theta_B$ reflects almost purely as TE. Polarizing sunglasses exploit this to cut horizontal glare off roads and water. For radar, oblique reflections off walls, ground, and vehicle panels rotate the polarization of the return, which is what dual-polarized weather and SAR radars use as a discriminator on top of amplitude.

Fresnel geometry: rays and polarization

Snap to θ_B and watch the TM arrows on the transmitted ray line up along the reflected ray direction.

Material Polarization Angle of incidence (deg) 55°

Teal dots are TE (E out of page) and warm arrows are TM (E in page); marker size scales with field amplitude.

Fresnel coefficients vs angle of incidence

Same setup as above, plotted quantitatively: power reflected for TE and TM as a function of incidence angle.

Material Angle of incidence (deg) 30° θ_B = —

|r|² vs incidence angle; TE climbs monotonically while TM dips to zero at θ_B.

4. Refraction and beam bending

The fraction of the wave that does not reflect enters the second medium, but at a different angle. The matching condition at the boundary forces the tangential component of the wavevector to be continuous, which gives Snell’s law:

n_1 \sin\theta_1 \;=\; n_2 \sin\theta_2, \qquad n \;=\; \sqrt{\epsilon_r}

Going from a less dense medium (air) to a denser one (brick, $n \approx 1.94$ ), the beam bends toward the normal. At $\theta_1 = 60°$ into brick, $\theta_2 = \arcsin(\sin 60° / 1.94) \approx 26.6°$ . Inside the wall the beam travels almost straight ahead even when the incident wave comes in at a steep angle.

What this does to through-wall radar

The bend has two practical effects:

The path length inside the wall is longer than the wall thickness by a factor of $1/\cos\theta_2$ . For a 20 cm wall at $\theta_2 = 27°$ , the actual traversal is about 22 cm. Modest at moderate angles, but noticeable at grazing incidence.
The beam exits the back of the wall offset laterally from where it entered. For a thick wall this offset can be enough to point the radar at the wrong room. Not usually a problem at normal-ish incidence.

Total internal reflection (not normally relevant for radar)

When going from high- $n$ to low- $n$ (e.g. brick back to air) at angles above the critical angle $\theta_c = \arcsin(n_2/n_1)$ , the wave cannot leave the dense medium and reflects totally. For brick-to-air, $\theta_c \approx 31°$ . This matters for waves bouncing around inside a wall but rarely for the direct transmission path.

5. Absorption inside a material

Once the wave is inside the second medium, it propagates and loses energy to the material. Real dielectrics have a small imaginary part to their permittivity:

\epsilon \;=\; \epsilon' - j\epsilon'' \;=\; \epsilon_0 \epsilon_r' \left(1 - j \tan\delta\right),

where $\tan\delta = \epsilon''/\epsilon'$ is the loss tangent. The wavenumber in the medium becomes complex:

k \;=\; \beta - j\alpha, \qquad \alpha \;\approx\; \frac{\omega}{c}\sqrt{\epsilon_r'} \cdot \frac{\tan\delta}{2} \quad \text{(for } \tan\delta \ll 1\text{)}

The field inside the medium decays as $E(z) = E_0 \, e^{-\alpha z} e^{-j\beta z}$ , so power decays as $e^{-2\alpha z}$ . The skin depth $\delta_s = 1/(2\alpha)$ used in the table below is the depth at which the power falls to $1/e$ of its surface value (field to $1/\sqrt{e}$ ). The field-amplitude $1/e$ depth is twice that, $1/\alpha$ . Both conventions appear in the literature; what matters is that the same one is used consistently.

From loss tangent to dB per meter

In a low-loss dielectric, the conductivity-equivalent form is

\alpha \;\approx\; \frac{\eta_0 \, \sigma}{2 \sqrt{\epsilon_r'}}, \qquad \eta_0 \approx 377\,\Omega,

where $\sigma$ is the effective conductivity (S/m) at the operating frequency. Loss in dB/m is $8.686 \, \alpha$ . ITU-R Recommendation P.2040 publishes empirical fits of $\sigma(f)$ for common building materials.

material	$\epsilon_r'$	$\sigma$ at 2.4 GHz (S/m)	loss at 2.4 GHz (dB/m)	skin depth at 2.4 GHz
Drywall (plasterboard)	2.94	0.022	~22	~20 cm
Wood (dry)	1.99	0.012	~14	~30 cm
Brick	3.75	0.038	~32	~14 cm
Concrete	5.31	0.063	~46	~10 cm
Glass	6.27	0.014	~9	~48 cm
Fresh water	~80	~1.5	~310	~1.4 cm
Wet brick / wet wood	↑ 2–3x	↑ 5–10x	↑ 3–5x	↓ 3–5x

Approximate one-way attenuation in homogeneous slabs of common materials. Real walls add scattering loss from mortar joints, cavities, and rebar; published in-situ measurements typically run 1.5x to 3x higher than the bulk numbers.

Why water is so bad and metal is opaque

Water has both high $\epsilon_r$ (so the wave is reflected strongly at the air-water surface) and high $\tan\delta$ at GHz frequencies (so any wave that gets in is absorbed quickly). A skin depth of ~1 cm at 2.4 GHz means a few centimeters of water blocks essentially all of an RF signal. Humans are mostly water; this is why a body returns a measurable reflection (high $\epsilon$ mismatch with air) and also why imaging through a body fails (high absorption inside).

Metal has $\sigma$ on the order of $10^7$ S/m; its skin depth at 2.4 GHz is around 1.3 μm. Any practical metal sheet is opaque. A wire mesh acts as a metal sheet to wavelengths longer than the grid spacing and a transparent screen to wavelengths much shorter (a microwave-oven door is opaque at 2.45 GHz but transparent to visible light).

6. Putting it together: the wall budget

The total one-way loss through a homogeneous slab in air is the sum of three contributions:

Reflection at the front surface (some power bounces off and never enters);
Absorption through the bulk material (depends on thickness and frequency);
Reflection at the back surface (some power that reaches the rear surface bounces back inside the wall instead of exiting).

To first order (ignoring the small interference fringes from internal reflections),

T \;\approx\; (1 - R)^2 \, e^{-2 \alpha d}, \qquad R = \left|\frac{n_2 - 1}{n_2 + 1}\right|^2

where $d$ is the wall thickness and $\alpha$ depends on frequency through the loss tangent. Taking $-10 \log_{10}$ of both sides splits the loss in dB cleanly into a reflection term and an absorption term.

Wall transmission vs frequency

For a chosen material and thickness, the plot shows total one-way loss in dB as a function of frequency from 0.3 GHz to 30 GHz, decomposed into the reflection component (two surfaces) and the absorption component (bulk).

Material Thickness (cm) 20 cm Frequency marker (GHz) 2.4 GHz

Stacked decomposition of one-way wall loss vs frequency. Reflection loss (teal) is mostly frequency-independent and dominates below ~1 GHz. Absorption loss (warm) rises with frequency and dominates above a few GHz. The crossover frequency depends on material and thickness.

The 1 to 8 GHz observation

Most published through-wall radar work sits in the 1 to 8 GHz range. Below 1 GHz the wavelengths are too large for compact arrays and angular resolution suffers. Above 8 GHz the absorption climbs steeply for common materials. The interval in between is where the bulk absorption is still under ~10 dB per 20 cm wall and the wavelength is still small enough (a few cm) for reasonable resolution.

next: ch4: how current becomes a wave →