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Free space path loss: 6 GHz, 10,500 km

Short answer

Free space path loss is derived from isotropic antennas: LFS = (4πd/λ)^2, and in dB: LFS(dB) = 20 log(4πd/λ) = 32.44 + 20 log d(km) + 20 log f(MHz). For f=6 GHz, d=10500 km, λ=0.05 m, LFS = 20 log(4π×1.05e7/0.05) = 188.43 dB.

Derive the expression for free space path loss propagation. Calculate free space path loss for 6 GHz frequency, if transmitter and receiver are 10,500 km line of sight apart.

Derivation of free space path loss

Free space path loss is the ratio of transmitted to received power between two isotropic antennas in free space, caused by spherical spreading of the wave.

  1. An isotropic antenna radiating Pt produces power density at distance d: P_d = Pt / (4πd²)
  2. An isotropic receiving antenna has effective aperture Ae = λ²/4π.
  3. Received power:
Pr = P_d · Ae = Pt/(4πd²) · λ²/(4π) = Pt·(λ/4πd)²
  1. The path loss is therefore
LFS = Pt/Pr = (4πd/λ)² = (4πdf/c)²

LFS(dB) = 20 log(4πd/λ)
        = 22 + 20 log(d/λ)
        = 32.44 + 20 log d(km) + 20 log f(MHz)
        = 92.44 + 20 log d(km) + 20 log f(GHz)

With real antennas of gains Gt and Gr, the received power is Pr = Pt + Gt + Gr − LFS (dB), which is the Friis equation.

Numerical: f = 6 GHz, d = 10,500 km

λ = c/f = 3×10⁸ / 6×10⁹ = 0.05 m
4πd/λ = 4π × 1.05×10⁷ / 0.05 = 2.639×10⁹
LFS = 20 log(2.639×10⁹) = 188.43 dB

Check: 92.44 + 20 log 10500 + 20 log 6
     = 92.44 + 80.42 + 15.56 = 188.43 dB

Answer: Free space path loss ≈ 188.4 dB. (Losses of this size are met on long satellite paths at C-band.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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