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Friis equation and free space path loss
Short answer
Friis transmission equation: Pr = Pt*Gt*Gr*(lambda/(4*pi*d))^2, derived from power density of isotropic radiator, antenna gain, and effective aperture. Free space path loss LFS = (4*pi*d/lambda)^2 = 32.44 + 20log(d_km) + 20log(f_MHz) dB for isotropic antennas.
How do you get Friis transmission equation and path loss in case of free space wave propagation?
The Friis transmission equation gives the power received by one antenna from another in free space in terms of the transmitted power, the two antenna gains, the wavelength and the distance. The path loss follows from it by setting both gains to unity.
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.
- 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)
- A directive antenna with gain Gt increases the density in its main direction by Gt: P_d = Pt·Gt / (4πd²)
- The receiving antenna collects the power falling on its effective aperture: Pr = P_d · Aer = Pt·Gt·Aer / (4πd²)
- 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
The free space path loss LFS is the ratio of transmitted to received power when both antennas are isotropic (Gt = Gr = 1), i.e. the loss due to spreading of the wave alone:
LFS = Pt/Pr = (4πd/λ)²
LFS(dB) = 20 log(4πd/λ)
= 20 log(4π) + 20 log(d/λ)
= 21.98 + 20 log(d/λ)
≈ 22 + 20 log(d/λ) dB
Using λ = c/f, with d in km and f in MHz:
LFS(dB) = 32.44 + 20 log d(km) + 20 log f(MHz)
(With f in GHz the constant becomes 92.44.) The loss grows by 6 dB each time the distance or the frequency is doubled.
✓ Answer checked by Claude Opus 5.5 (high effort)
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