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Two-ray model path and phase difference

Short answer

Using the image method, LOS path ≈ d + (ht−hr)²/(2d), ground-reflected ≈ d + (ht+hr)²/(2d). Path difference Δ ≈ 2 ht hr / d. Phase difference θΔ = 2πΔ/λ ≈ 4π ht hr/(λ d).

Derive an expression for path difference and phase difference between line of sight (LOS) path and ground reflected (GR) path in two-ray model using simple geometrical relations.

The two-ray ground reflection model assumes the received field is the sum of a line-of-sight (LOS) wave and a wave reflected from the ground. The path difference between them decides whether they add or cancel.

  Tx
   |\  d' (LOS)
   | \______________________  Rx
ht |  \                  __/|
   |   \  d'' (GR)    __/   | hr
   |    \          __/      |
 --+-----\______/-----------+----- ground
          reflection point
   |<---------- d ---------->|

 Image method: image of Tx at depth ht below ground

Path lengths (method of images)

Replace the ground-reflected path by a straight line from the image of the transmitter (at −ht) to the receiver.

LOS:  d'  = √[d² + (ht − hr)²]
GR :  d'' = √[d² + (ht + hr)²]

Path difference

Using binomial expansion √(1 + x) ≈ 1 + x/2 for d ≫ ht + hr:

d'  = d√[1 + ((ht−hr)/d)²] ≈ d + (ht − hr)²/(2d)
d'' = d√[1 + ((ht+hr)/d)²] ≈ d + (ht + hr)²/(2d)

Δ = d'' − d'
  = [(ht + hr)² − (ht − hr)²] / (2d)
  = 4ht·hr / (2d)
Δ ≈ 2ht·hr / d

Phase difference and time delay

A path difference of one wavelength corresponds to 2π radians:

θΔ = 2πΔ/λ = Δωc/c ≈ 4π·ht·hr/(λd)

τd = Δ/c = θΔ/(2πfc)

Use in received field

With ground reflection coefficient Γ ≈ −1, the total field is

|ETOT| ≈ 2E0d0/d · sin(θΔ/2)

For large d, θΔ/2 < 0.3 rad, so sin(θΔ/2) ≈ θΔ/2 and

ETOT ≈ (2E0d0/d)(2πht·hr/(λd)) = k/d²
Pr = Pt·Gt·Gr·ht²·hr² / d⁴

So at large distance received power falls by 40 dB/decade, and is independent of frequency.

✓ Answer checked by Claude Opus 5.5 (high effort)

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