- 2071 Magh · 4 marks
Fresnel zone and knife edge diffraction
Short answer
Fresnel zones are ellipsoids with path difference nλ/2; blocking the first zone increases diffraction loss. Knife-edge diffraction occurs when a sharp obstacle obstructs the path; loss depends on parameter v = h√[2(d1+d2)/(λ d1 d2)] (h is edge height above LOS), and equals ~6 dB when v=0.
Explain Fresnel zone and knife edge diffraction.
Fresnel zones
Fresnel zones are concentric ellipsoids around the direct Tx–Rx line. A path through any point on the boundary of the nth zone is longer than the direct path by nλ/2.
_..--~~~~~~~~--.._ 1st zone
Tx o<------- d1 ------>|<----- d2 ---->o Rx
~~--..______..--~~
Radius of the nth zone at distances d1 and d2 from the ends:
rn = √( n·λ·d1·d2 / (d1 + d2) )
- Waves from successive zones differ in phase by 180°, so most useful energy travels inside the first zone.
- Blocking part of the first zone causes diffraction loss even if the optical path is clear. Links are designed to keep at least 0.6r1 clear.
Knife-edge diffraction
When a single sharp obstacle (ridge, building edge) blocks the path, the wave bends over the edge into the shadow region (Huygens' principle: each point on the wavefront above the edge acts as a new source). The obstruction is measured by the Fresnel–Kirchhoff parameter
v = h·√( 2(d1 + d2) / (λ·d1·d2) )
where h is the height of the edge above the LOS line (negative if below).
| Condition | v | Diffraction loss |
|---|---|---|
| Path well clear | v ≤ −1 | ≈ 0 dB |
| Edge touches LOS | v = 0 | 6 dB |
| Edge blocks LOS | v = 1 | ≈ 14 dB |
| Deep shadow | v > 2.4 | ≈ 20 log(v/0.225) dB |
The two ideas are linked: v = √2·h/r1, so the loss depends on how much of the first Fresnel zone the edge covers.
✓ Answer checked by Claude Opus 5.5 (high effort)
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