Diffraction and Fresnel zone phase difference

Diffraction is the bending of radio waves around the edges of obstacles (hills, building edges) so that the signal reaches the shadow region behind them, even without line of sight. It is explained by Huygens' principle: every point on a wavefront acts as a source of secondary wavelets, which combine behind the obstacle. Diffraction loss depends on how far the obstacle blocks the Fresnel zones.

Fresnel zone geometry (knife-edge)

A transmitter and receiver are separated by an obstacle of effective height h above the LOS line, at distances d1 and d2.

                 |\  diffracted path
                 | \
        ________ h  \________
  Tx  /  α  β     |    γ     \  Rx
   o---------------+----------o    LOS line
   |<---- d1 ----->|<-- d2 -->|
   ht              obstacle   hr

Assume h ≪ d1, d2 and h ≫ λ.

Path difference

Length of the diffracted path (two straight segments over the edge):

√(d1² + h²) + √(d2² + h²)
 ≈ d1 + h²/(2d1) + d2 + h²/(2d2)   (binomial approx.)

Path difference over the direct path (d1 + d2):

Δ ≈ (h²/2)·(1/d1 + 1/d2)
Δ ≈ (h²/2)·(d1 + d2)/(d1·d2)

Phase difference

φ = 2πΔ/λ = (2π/λ)·(h²/2)·(d1 + d2)/(d1·d2)

Using small angles, the angle α ≈ h(d1 + d2)/(d1·d2). Defining the Fresnel–Kirchhoff diffraction parameter:

v = h·√[ 2(d1 + d2) / (λ·d1·d2) ]
  = α·√[ 2d1d2 / (λ(d1 + d2)) ]

the phase difference becomes:

φ = (π/2)·v²

Fresnel zones

Fresnel zones are regions where the path difference is nλ/2 (phase difference nπ). The radius of the n-th zone at the obstacle is:

rn = √[ nλ·d1·d2 / (d1 + d2) ]

• Odd zones add to the direct signal, even zones cancel it.
• If the obstacle does not block the first Fresnel zone (about 55% clearance is enough), diffraction loss is very small. As blockage increases (v increases), diffraction loss increases; at v = 0 (edge on the LOS line) the loss is 6 dB.