Friis, path loss and LOS distance

Friis transmission equation

Consider a transmitter of power Pt feeding an antenna of gain Gt, and a receiving antenna of gain Gr (effective aperture Aer) at distance d in free space. Assume far field, matched polarisation, matched loads and no obstacles.

1. An isotropic source spreads Pt uniformly over a sphere of area 4πd², so the power density at distance d is P_iso = Pt / (4πd²) (W/m²)
2. A directive antenna with gain Gt increases the density in its main direction by Gt: P_d = Pt·Gt / (4πd²)
3. The receiving antenna collects the power falling on its effective aperture: Pr = P_d · Aer = Pt·Gt·Aer / (4πd²)
4. Effective aperture and gain are related by Aer = Gr·λ² / (4π). Substituting:

Pr = Pt·Gt·(Gr·λ²/4π) / (4πd²)

Pr = Pt·Gt·Gr·λ² / (4πd)²

Pr/Pt = Gt·Gr·(λ / 4πd)²

This is the Friis transmission equation. In terms of the two effective apertures (using Gt = 4πAet/λ²) it can also be written as Pr/Pt = Aet·Aer / (λ²d²).

In decibels:

Pr(dBW) = Pt(dBW) + Gt(dB) + Gr(dB) − 20 log(4πd/λ)

Path loss in free space

With isotropic antennas (Gt = Gr = 1):

LFS = Pt/Pr = (4πd/λ)²
LFS(dB) = 20 log(4πd/λ) = 22 + 20 log(d/λ)
        = 32.44 + 20 log d(km) + 20 log f(MHz)

Line-of-sight distance d = 3.57(√ht + √hr) km

Because of the curvature of the earth, a straight ray from an antenna of height h touches the earth at the radio horizon.

          Tx
          |\   ht
          | \
          |  \  d1
   -------+---*  <- tangent point
     \    |   /
    R \   |  /  R
       \  | /
         O  (centre of earth)

From the right-angled triangle (radius R to the tangent point is perpendicular to the ray):

(R + ht)² = R² + d1²
d1² = 2R·ht + ht²
d1 ≈ √(2R·ht)          (ht ≪ R)

With R = 6370 km = 6.37×10⁶ m and h in metres:

d1 = √(2 × 6.37×10⁶ × ht) m
   = 3569√ht m
   = 3.57√ht km

Similarly for the receiving antenna, d2 = 3.57√hr km. The total LOS distance is

d = d1 + d2 = 3.57(√ht + √hr) km   (ht, hr in m)

This is the optical (geometric) horizon. Allowing for standard atmospheric refraction with effective earth radius kR (k = 4/3), the radio horizon becomes d = 4.12(√ht + √hr) km.