The previous lesson handed you a number and a warning attached to it. Free-space path loss is what a link would suffer if the only thing between transmitter and receiver were empty space — and it is the best case, the loss you can never beat and almost never achieve. This lesson is about what the world actually does to the wave instead. Three mechanisms account for nearly all of it: reflection, diffraction and scattering. Between them they explain why your phone works in a lift lobby, why a hilltop can be dark while the valley behind it is not, and why millimetre-wave 5G falls over when somebody stands up.
Empty space is a remarkably rare medium. A real path has a ground under it, buildings beside it, leaves and rain in it, and people walking through it. Every one of those boundaries is a place where the electrical properties of the medium change abruptly, and a wave meeting such a boundary does not politely continue — it splits. Part of the energy carries on into the new material, part comes back, and if the boundary is small or rough compared with the wavelength, part goes off in directions that no straight line would predict.
That sounds like nothing but bad news, and for a single clean line-of-sight link it mostly is: every mechanism in this lesson takes energy out of the direct path. But the same mechanisms are the only reason non-line-of-sight coverage exists at all. A base station on a mast cannot see into your kitchen. The signal that reaches you there arrived by bouncing off a neighbour’s wall, bending over your roofline, and scattering off the window frame. The channel that ruins your link budget is also the channel that delivers your service.
Start with the simplest case: a wave striking a large, smooth, flat boundary — a wall, a lake, a stretch of level ground — where “large” and “smooth” both mean compared with the wavelength. The geometry is the one you already know from mirrors, and it is exactly the same law, because light and radio are the same phenomenon at different frequencies (Module 1). Measuring angles from the normal to the surface, the reflected ray leaves at the angle the incident ray arrived at, and the part that penetrates is bent according to Snell’s law:
Geometry tells you where the reflected ray goes; it says nothing about how strong it is. That is the job of the reflection coefficient Γ, the ratio of reflected to incident electric field. Γ is a complex number — it carries a magnitude and a phase shift — and it depends on three things: the material’s permittivity εr (and its conductivity, which makes εr complex), the angle of incidence, and the polarization of the wave relative to the surface. The Fresnel equations give it exactly. Here is the horizontally-polarized case, written in terms of the grazing angle ψ measured up from the surface rather than from the normal:
That grazing limit is the single most consequential line in this section, so read it twice. At shallow angles almost all the field comes back, and it comes back with a minus sign — a 180° phase flip. Both polarizations do this; the flip is not a property of the material but of the geometry. Hold onto it, because in a moment it is going to turn a second, apparently helpful path into a subtraction.
| Surface | Typical εr at UHF | |Γ| at shallow grazing | Note |
|---|---|---|---|
| Metal (sheet, mesh, vehicle body) | — (conductor) | ≈ 1 | Essentially a mirror; almost nothing gets through |
| Concrete | ≈ 5–7 | 0.7–0.9 | Strong reflector and a heavy attenuator in transmission |
| Glass (plain), plasterboard | ≈ 4–7 | 0.7–0.9 | Coated low-emissivity glass behaves far more like metal |
| Dry ground / average earth | ≈ 4–15 | 0.7–0.9 | Wet ground and water are better reflectors still |
One more feature is worth naming even though we will not use it numerically. For vertical polarization there is a particular angle — the Brewster angle — at which the reflected field very nearly vanishes, because at that geometry the boundary charges cannot radiate back the way the wave came. Horizontal polarization has no such angle. It is the reason polarized sunglasses kill glare off a wet road, and at radio frequencies it means the strength of a ground reflection can depend sharply on which polarization you chose — a design lever, not merely a curiosity.
Smooth, large, and flat are all assumptions. Every statement above assumes the surface is far bigger than the wavelength, flat over the region the wave illuminates, and smooth compared with λ. Break the first and you get diffraction round the edges; break the third and you get scattering. Both are coming up, and in a real city all three happen at the same wall.
Now put the grazing phase flip to work. Consider the most common geometry in terrestrial radio: a transmitter at height ht, a receiver at height hr, flat ground between them. The receiver hears two copies — the direct ray, and one that bounced off the ground at a very shallow angle. The bounced path is slightly longer, and for a separation d much larger than either height that extra length has a tidy approximation:
Take a mast at ht = 30 m, a handset at hr = 1.5 m, and f = 900 MHz so that λ = 3×10⁸/9×10⁸ = 0.333 m (M1-L2). At d = 2 km the extra path is Δ = 2 × 30 × 1.5 / 2000 = 90/2000 = 0.045 m, which is 4.5 cm. As a phase that is 360° × 0.045/0.333 = 48.6°, and combining a unit direct ray with a flipped, 48.6°-delayed reflection gives a resultant field of 2 sin(48.6°/2) = 2 × 0.412 = 0.823, i.e. 1.7 dB below free space (M2-L4 for the decibel; M2-L3 for the interference arithmetic). Now the breakpoint:
That exponent change is not a rounding detail. Under free space, doubling the distance costs 10 log₁₀(4) = 6 dB; under a two-ray ground reflection it costs 10 log₁₀(16) = 12 dB. Cell radius predictions built on the free-space law are therefore wildly optimistic, which is precisely why real coverage tools fit a measured exponent instead of assuming one — the log-distance model and its path-loss exponent n are M8-L4’s subject. Notice also that raising either antenna pushes the breakpoint further out, which is the physical reason mast height buys coverage so effectively.
Geometry alone says that behind an obstacle there is a shadow, and that inside the shadow there is nothing. Step behind a building with a radio and geometry is immediately shown to be wrong: the signal weakens, but it does not stop. The mechanism is diffraction, and the cleanest way to see why it must happen is Huygens’ principle: treat every point on a wavefront as a source of a new spherical wavelet, and the wavefront an instant later as the envelope of all those wavelets.
Now block half the wavefront with a screen. The wavelets that survive on the open side still radiate in all directions, including sideways past the screen edge, so their envelope curls into the region behind it. No wave is being bent by a force; energy fills the geometric shadow because the wavelets that were going to cancel it were the very ones you removed. The consequence is a smooth gradient rather than a cliff — strong just outside the shadow, weakening as you move deeper in — and, crucially, the amount of curl scales with λ. Long waves reach deep behind an obstacle; short ones barely reach at all.
Real obstacles are ridges, rooflines and hills, and modelling them exactly is a research problem. The workhorse approximation replaces the obstacle with a perfectly absorbing, infinitely thin half-plane — a knife edge — standing a height h above the straight line from transmitter to receiver, at distances d₁ and d₂ from the two ends. All of that geometry collapses into a single dimensionless number, the Fresnel–Kirchhoff diffraction parameter v:
Turning v into a loss requires the Fresnel integral, which has no closed form, so engineering practice uses a fitted approximation. The one below is the standard ITU-R P.526 / Lee expression, accurate to a few tenths of a decibel and valid for v > −0.7; below that the model simply reports no diffraction loss, because the path is clear enough that the assumption of a single dominant edge stops meaning anything:
Six decibels for an obstacle that is not even blocking anything surprises people, and it is worth understanding rather than memorising. At grazing the edge removes exactly half the wavefront, so half the field — a quarter of the power — arrives, and a quarter of the power is 6 dB down. The Huygens picture pays for itself here: the field at the receiver was never delivered by the geometric ray, it was the sum of contributions from the whole wavefront, and you have deleted half of them.
Take a 10 km link at 2.4 GHz, so λ = 3×10⁸/2.4×10⁹ = 0.125 m, with a ridge at the midpoint: d₁ = d₂ = 5000 m. Suppose the ridge top sits 10 m above the line of sight, so h = +10 m. Then 2(d₁+d₂) = 20000 and λd₁d₂ = 0.125 × 5000 × 5000 = 3.125×10⁶, and the ratio is 20000/3.125×10⁶ = 0.0064, whose square root is 0.08. So:
Run it again with the ridge 5 m below the line of sight, h = −5 m, and you get v = −0.4 and L = 6.9 + 20 log₁₀(√1.25 − 0.5) = 6.9 + 20 log₁₀(0.618) = 6.9 − 4.18 = 2.7 dB. Read that carefully: the path is visually clear and it still loses 2.7 dB. Clearing the line of sight is not the same as clearing the path, and the next section says how much clearance you actually need.
If the receiver is fed by contributions from the whole wavefront, the useful question is: which part of the space around the direct line actually matters? The answer is a family of nested ellipsoids called Fresnel zones, each defined by the extra path length a detour through it would add. The first zone is the set of points adding no more than λ/2, so contributions from within it arrive within half a cycle of the direct ray and reinforce it. Its radius at any point along the path is:
Put our 10 km, 2.4 GHz link into it, with the obstacle at the midpoint: r₁ = √(0.125 × 5000 × 5000 / 10000) = √(0.125 × 2500) = √312.5 = 17.7 m. Seventeen metres of required clearance, on a path where the line of sight might be a hundred metres up. This is why microwave link planning is a terrain exercise and not a ruler-and-map exercise, and why a link that surveys clear can still fail once the crops grow.
The planning rule of thumb is to keep at least 60% of the first Fresnel zone clear of obstruction, which here means 0.6 × 17.7 = 10.6 m. The number is not arbitrary: 10.6 m of clearance is h = −10.6 m, which is v = −10.6 × 1.414/17.7 = −0.85 — past the −0.7 limit of the loss formula, in the region where diffraction loss has effectively gone to zero. The 60% rule is exactly the statement “stay in the region where the knife-edge model reports nothing”, and it is why a link engineer talks about Fresnel clearance rather than about seeing the far tower.
Why 60% and not 100%. Clearing the whole first zone would be ideal but is usually unaffordable in mast height, and it buys very little: by 60% the loss is already negligible, and pushing further mainly moves you into the second zone, whose contributions arrive out of phase and can actually reduce the received field. Clearance beyond about 0.6 r₁ is therefore spent money, not margin.
Specular reflection needs a surface that is smooth on the scale of the wavelength. When it is not — when the surface has bumps comparable to λ — the reflected energy no longer leaves in one direction. Each bump reradiates on its own, and the result sprays over a wide range of angles: scattering. Whether a given surface counts as rough is not an absolute property of the surface but a joint property of surface and frequency, and the usual test is the Rayleigh criterion, a critical roughness height hc:
Those two numbers are the same brick wall judged at two frequencies. Mortar joints and surface texture of a centimetre or so sit comfortably under 9 cm, so at 2.4 GHz the wall is a mirror. The same centimetre is above 7.7 mm at 28 GHz, so the same wall scatters — and a millimetre-wave system loses the strong, predictable specular bounce that a microwave system could rely on. The catalogue of scatterers in a real environment is long:
Rain deserves one honest sentence rather than a hand-wave. Below roughly 10 GHz rain scattering and absorption are usually negligible in a terrestrial link budget; above it they grow steeply with frequency and rain rate, and by the millimetre bands a heavy downpour is a first-order effect that link designers must budget an explicit rain margin for. The numbers belong with the other loss mechanisms in M8-L4, and the millimetre-wave consequences with the 5G material in M11-L3. What matters here is the mechanism: the drops are scatterers, and their size relative to λ is what decides whether they matter.
Assemble the three mechanisms and the behaviour of real radio coverage stops being mysterious. Reflection supplies strong alternative paths and, at grazing incidence, a phase-flipped one that turns the distance law from d⁻² into d⁻⁴. Diffraction leaks energy into shadows, at a cost that the knife-edge model prices in decibels. Scattering redistributes what is left into a large number of weak paths arriving from many directions. Practically:
There is one more consequence, and it is large enough to have its own lesson. If reflection, diffraction and scattering all deliver copies of the same signal by different routes, those copies arrive at different times and with different phases, and they add as vectors — sometimes reinforcing, sometimes cancelling, and differently at every position and every frequency. That is multipath, and its statistics are M8-L3. Module 9 then takes up the other half of the story: even a perfectly predicted signal is only useful relative to the noise and interference it competes with.
A note on what these models are for. None of the formulae here is exact. The knife-edge model replaces a hill with a razor blade; the two-ray model replaces a landscape with a plane; the Rayleigh criterion replaces a texture with one number. They earn their keep because they turn geometry into decibels quickly enough to plan a network with, and because they fail in known, conservative directions. Treat them as instruments with a stated accuracy, not as descriptions of reality.