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Path loss in radio wave propagation
Short answer
Path loss is Pt/Pr; free-space loss with isotropic antennas is LFS=(4πd/λ)^2, so LFS(dB)=20log(4πd/λ)=32.44+20log d(km)+20log f(MHz). With antenna gains, L(dB)=20log(4πd/λ)-Gt-Gr. Over a plane reflecting earth loss becomes 40 dB per decade and independent of frequency.
Derive the expression for the path loss in case of radio wave propagation.
Path loss is the reduction in power of a radio signal as it travels from the transmitting antenna to the receiving antenna, expressed as the ratio of transmitted power to received power (usually in dB). In free space it is caused only by the spherical spreading of the wave; over real paths, ground reflection, obstacles and the atmosphere add further loss.
Basic idea: spreading of power
.-~~~-. sphere of area 4πd²
/ \
Tx ( * ------- )--> Rx (aperture Ae)
\ /
'-...-.
An isotropic antenna radiating Pt watts spreads it uniformly over a sphere of radius d:
P_iso = Pt / (4πd²) W/m²
Effect of antenna gains
A transmitting antenna of gain Gt concentrates the power in its main beam:
P_d = Pt·Gt / (4πd²)
The receiving antenna extracts power in proportion to its effective aperture:
Pr = P_d · Ae , with Ae = Gr·λ² / 4π
Pr = Pt·Gt·Gr·λ² / (4πd)² (Friis equation)
In dB form:
Pr = Pt + Gt + Gr − 20 log(4πd/λ) (dBW, dB)
Path loss (with antenna gains)
The overall transmission loss between the transmitter output and the receiver input is
L = Pt/Pr = (4πd/λ)² / (Gt·Gr)
L(dB) = 20 log(4πd/λ) − Gt(dB) − Gr(dB)
Free space (basic) path loss
With isotropic antennas (Gt = Gr = 1) we get the basic free space loss, which depends only on distance and wavelength:
LFS = (4πd/λ)²
LFS(dB) = 20 log(4π) + 20 log d − 20 log λ
= 21.98 + 20 log(d/λ)
Putting λ = c/f = 3×10⁸/f:
LFS = 20 log(4π/3×10⁸) + 20 log d(m) + 20 log f(Hz)
= −147.56 + 20 log d(m) + 20 log f(Hz)
With d in km and f in MHz:
LFS = 32.44 + 20 log d(km) + 20 log f(MHz) dB
Path loss over a flat reflecting earth (two-ray model)
For a real LOS link above ground, the receiver gets a direct ray and a ground-reflected ray. For a perfectly reflecting ground (Γ ≈ −1) and d ≫ ht, hr, the path difference is Δ = 2ht·hr/d, and the received power becomes
Pr = Pt·Gt·Gr · (ht·hr)² / d⁴
L(dB) = 40 log d − 20 log ht − 20 log hr − Gt − Gr
So over a plane earth the loss rises at 40 dB per decade (12 dB per doubling of distance), compared with 20 dB per decade in free space, and it no longer depends on frequency.
Factors that add to path loss in practice
- Obstacles and diffraction (knife-edge loss, Fresnel zone blockage)
- Atmospheric absorption by oxygen and water vapour (above ~10 GHz)
- Rain attenuation (important above ~10 GHz)
- Multipath fading and shadowing in mobile links
Summary
| Model | Pr depends on | Loss slope |
|---|---|---|
| Free space | 1/d², λ² | 20 dB/decade |
| Plane earth (two-ray) | 1/d⁴, (ht·hr)² | 40 dB/decade |
Example: f = 6 GHz, d = 10,500 km gives LFS = 20 log(4π × 1.05×10⁷/0.05) = 188.4 dB.
✓ Answer checked by Claude Opus 5.5 (high effort)
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