Wireless 101
M08 · L02
Module 8 · Lesson 2

What the World Does to the Wave

Free-space path loss was the best case — the loss you never beat and almost never reach. Three mechanisms explain the difference: reflection, diffraction, scattering.

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Wireless 101
M08 · L02
Reflection

Angle In, Angle Out

Geometry says where the ray goes; Γ says how strong it is — set by polarization, angle and permittivity. Metal gives |Γ| ≈ 1, concrete and glass 0.7–0.9 at shallow angles.

Fresnel coefficient, grazing angle ψ
\Gamma_{\perp} = \frac{\sin\psi - \sqrt{\varepsilon_r - \cos^2\psi}}{\sin\psi + \sqrt{\varepsilon_r - \cos^2\psi}}

Let ψ → 0 and it collapses to Γ → −1 for any material: near-total reflection with a 180° phase flip.

02 / 11
Wireless 101
M08 · L02
Two rays

The Ground Subtracts

Path difference and breakpoint
\Delta \approx \tfrac{2 h_t h_r}{d} \;\Big|\; d_{bp} = \tfrac{4 h_t h_r}{\lambda}

Direct plus ground-bounce. The extra path shrinks as 1/d, so the rays converge in phase — but the bounce flipped 180°, so converging in phase means cancelling.

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Wireless 101
M08 · L02
900 MHz, 30 m mast, 1.5 m handset

Power Falls as d⁻⁴

Beyond the breakpoint the cancellation is monotonic and the exponent changes. Doubling distance now costs 12 dB, not 6 — and that is why mast height buys coverage.

Δ at 2 km
4.5 cm
Breakpoint
540 m
Per doubling
12 dB
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Wireless 101
M08 · L02
Diffraction — Huygens

Why the Shadow Is Not Dark

Every point on a wavefront is a secondary source. Block half of them and the survivors still radiate sideways, so their envelope curls into the shadow. All the geometry reduces to one number:

Fresnel–Kirchhoff parameter
v = h\sqrt{\tfrac{2(d_1+d_2)}{\lambda d_1 d_2}} = \tfrac{h\sqrt{2}}{r_1}
05 / 11
Wireless 101
M08 · L02
Try it — 10 km link

Knife Edge and Fresnel Clearance

Move the obstacle above or below the line of sight and watch v, the loss and the 60% verdict.

Height h +10 m
Position 50% · 5.0 km
Freq 2.40 GHz
v 0.80 loss 12.6 dB r1 17.7 m clear −57% · FAIL <60%
Interactive: put walls and edges on a map and watch reflection, shadow and coverage — AI-generated
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Wireless 101
M08 · L02
Worked example — 10 km, 2.4 GHz

A 10 m Ridge at Midpath

  • λ = 0.125 m, d₁ = d₂ = 5000 m, r₁ = 17.7 m
  • v = 10√2/17.7 = 0.8
  • √(0.7²+1) + 0.7 = 1.221 + 0.7 = 1.921
  • L = 6.9 + 20log₁₀(1.921) = 12.6 dB
  • At v = 0, grazing, L = 6.0 dB — never zero
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Wireless 101
M08 · L02
Fresnel zones

Keep 60% Clear

r₁ = √(λd₁d₂/(d₁+d₂)) is an ellipsoid with the antennas at its foci. Clear 60% of it and v reaches −0.85, past the −0.7 limit where the model reports no loss at all.

r₁ midpath
17.7 m
60% of it
10.6 m
That is v
−0.85
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Wireless 101
M08 · L02
Scattering

When a Mirror Stops Being One

Rayleigh roughness criterion
h_c = \frac{\lambda}{8\sin\theta}

At 10° grazing, hc is 9.0 cm at 2.4 GHz but 7.7 mm at 28 GHz. The same brick wall is a mirror in one band and a scatterer in the other — and rain joins in above about 10 GHz.

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Wireless 101
Knowledge Check

Check what stuck

Four questions from this lesson. Answer to see why — the explanation appears whether you were right or wrong. Nothing is scored or saved.

Question 1 of 0
Score 0/0

10 / 11
Wireless 101
M08 · L02
Recap

What you learned

  • Grazing reflection: Γ → −1, a 180° flip
  • Past the breakpoint, power falls as d⁻⁴
  • Huygens is why shadows are lit at all
  • Knife edge: v = 0 costs 6 dB, v = 0.8 costs 12.6 dB
  • Keep 60% of the first Fresnel zone clear
Up next in Module 8
Multipath & Fading →
All lessons Read article
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