10 GHz LOS link loss and Friis derivation

Path loss and received power

Given: f = 10 GHz, Gt = Gr = 25 dB, d = 100 km. The transmitted power is not given, so the received power is found in terms of Pt and also for an assumed Pt = 1 W (0 dBW).

Wavelength:

λ = c/f = 3×10⁸ / 10¹⁰ = 0.03 m

Free space path loss:

4πd/λ   = 4π × 10⁵ / 0.03 = 4.189×10⁷
LFS(dB) = 20 log(4.189×10⁷) = 152.44 dB
Check:  32.44 + 20 log 100 + 20 log 10000
      = 32.44 + 40 + 80 = 152.44 dB

Net link loss including antenna gains:

L = LFS − Gt − Gr = 152.44 − 25 − 25 = 102.44 dB

Received power:

Pr(dBW) = Pt(dBW) + 25 + 25 − 152.44 = Pt(dBW) − 102.44

For Pt = 1 W:
Pr = −102.44 dBW = −72.44 dBm ≈ 5.7×10⁻¹¹ W

Answer: Path loss = 152.44 dB (102.44 dB after the 50 dB of antenna gain); Pr = Pt − 102.44 dB, i.e. −102.44 dBW ≈ 57 pW for Pt = 1 W.

Derivation of the 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)