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Knife edge diffraction

Short answer

Knife-edge diffraction describes wave bending around a sharp obstacle, allowing energy into the shadow region. The diffraction parameter v = h*sqrt(2(d1+d2)/(lambda*d1*d2)) quantifies obstruction; v=0 (grazing) gives 6 dB loss. Diffraction loss Gd is obtained from Fresnel integral approximations; for clear path, first Fresnel zone should be 60% clear.

Describe (write a short note on) Fresnel knife edge diffraction phenomenon.

Knife-edge diffraction is the bending of a radio wave around a single sharp obstacle (a hill ridge, building edge) that blocks part of the line-of-sight path. It lets some energy reach a receiver in the "shadow" region, and the amount depends on how much of the first Fresnel zone is blocked.

Geometry

                     /\  obstacle edge
                    /  \
   Tx o------------h----\-------------o Rx
        <--- d1 ---->|<----- d2 ----->
   h = height of edge above the LOS line

Huygens' principle

Every point on the wavefront above the edge acts as a secondary source. These secondary wavelets add up (with different phases) behind the obstacle, so the field in the shadow is weak but not zero.

Fresnel–Kirchhoff diffraction parameter

The obstruction is described by the dimensionless parameter

v = h · √( 2(d1 + d2) / (λ·d1·d2) )
  • v < 0: edge is below the LOS line (path clear)
  • v = 0: edge just touches the LOS line
  • v > 0: edge blocks the LOS line (receiver in shadow)

Diffraction loss

The received field is E = F(v)·E0, where E0 is the free-space field and F(v) is the complex Fresnel integral. The diffraction gain Gd = 20 log|F(v)| is read from a chart or approximated (Lee):

Range of vGd (dB)
v ≤ −10
−1 ≤ v ≤ 020 log(0.5 − 0.62v)
0 ≤ v ≤ 120 log(0.5·e^(−0.95v))
1 ≤ v ≤ 2.420 log(0.4 − √(0.1184 − (0.38 − 0.1v)²))
v > 2.420 log(0.225/v)

Key points

  • When the edge just grazes the LOS (v = 0), the field is halved: a loss of 6 dB.
  • For a clear path (v strongly negative) the field oscillates around E0 (±1 dB).
  • Loss rises quickly in the shadow, roughly 20 log(0.225/v) for large v.
  • For a negligible loss the first Fresnel zone should be at least about 60% clear of obstacles; this is used to fix tower heights in microwave link design.

✓ Answer checked by Claude Opus 5.5 (high effort)

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