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Two-ray model advantages and path loss derivation

Short answer

Two-ray model adds ground reflection, predicting 40 dB/decade loss (∝ 1/d⁴) for large d, independent of frequency, and showing antenna height effects. Path loss: Pr = Pt Gt Gr ht² hr² / d⁴, or in dB PL = 40 log d - 20 log(ht hr) - 10 log(Gt Gr).

What are the advantages of two ray propagation model over free space path loss model? Derive the equation of path loss using two ray model with appropriate diagram.

Advantages of the two-ray model over free space model

  • Free space assumes only a direct path; in real mobile links there is almost always a ground-reflected wave as well. The two-ray model includes it.
  • It is accurate for predicting large-scale signal strength over distances of several km for systems with tall towers (above about 50 m), and also for LOS microcells in urban areas.
  • It shows that beyond a break distance power falls as 1/d⁴ (40 dB/decade), which matches measurements better than the 1/d² of free space.
  • It shows the effect of antenna heights (loss decreases as ht and hr increase) and the alternating peaks and nulls close to the transmitter.
  • Path loss becomes independent of frequency at large d.

Derivation of path loss

  Tx
   |\  E_LOS (d')
ht | \_______________________ Rx
   |  \                    _/|
   |   \  E_g (d'')     _/   | hr
 --+----\____________/-------+-- ground
   |<----------- d --------->|

Path difference (method of images, image of Tx at −ht):

d'  = √[d² + (ht − hr)²],  d'' = √[d² + (ht + hr)²]
Δ = d'' − d' ≈ 2ht·hr/d   (for d ≫ ht + hr)
θΔ = 2πΔ/λ = 4π·ht·hr/(λd)

Total field. With free space field E0 at reference d0, LOS field E0d0/d', and ground reflection coefficient Γ ≈ −1 (grazing incidence):

ETOT = E_LOS + E_g
|ETOT| = (E0d0/d)·|1 − e^(−jθΔ)|
       = (E0d0/d)·√[(1 − cosθΔ)² + sin²θΔ]
       = (E0d0/d)·√(2 − 2cosθΔ)
       = 2(E0d0/d)·sin(θΔ/2)

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

|ETOT| ≈ 2(E0d0/d)·(2π·ht·hr/(λd)) = k/d²

Received power. Since Pr ∝ |E|², and using the Friis equation at the reference:

Pr = Pt·Gt·Gr·ht²·hr² / d⁴

Path loss in dB:

PL(dB) = 40 log d − (10 log Gt + 10 log Gr
                     + 20 log ht + 20 log hr)

Thus at large distances, path loss rises 40 dB per decade of distance, decreases by 6 dB when either antenna height doubles, and does not depend on frequency.

✓ Answer checked by Claude Opus 5.5 (high effort)

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