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Received power: 15.5 km, 3.4 GHz, dish receiver

Short answer

Using λ = 0.08824 m, receiving dish effective aperture Ae = ηπD^2/4 = 24.13 m², so Gr = 4πAe/λ² = 45.9 dB. Applying the Friis equation, maximum received power is Pr = Pt Gt Gr (λ/(4πd))^2 = 6.32×10⁻⁴ W (0.632 mW, −1.99 dBm).

What is the maximum power that can be received over a distance of 15.5 km line-of-sight free space with a 3.4 GHz frequency consisting of the transmitting antenna gain of 25 dB and receiving antenna diameter 6.4 m with 75% antenna efficiency? Where transmitted power is 250 W.

Maximum power is received when the antennas are matched and aligned in free space, so the Friis equation applies directly. The receiving gain is found from the dish size and efficiency.

Given

  • d = 15.5 km, f = 3.4 GHz, Pt = 250 W
  • Gt = 25 dB = 10^2.5 = 316.23
  • Receiving dish D = 6.4 m, η = 0.75

Step 1: Wavelength

λ = c/f = 3×10⁸ / 3.4×10⁹ = 0.08824 m

Step 2: Effective aperture and gain of the receiving dish

Ae = η·πD²/4 = 0.75 × π × 6.4² / 4 = 24.13 m²
Gr = 4π·Ae/λ² = 4π × 24.13 / 0.08824²
   = 3.894×10⁴  = 45.90 dB

Step 3: Received power (Friis, aperture form)

Pr = Pt·Gt·Ae / (4πd²)
   = 250 × 316.23 × 24.13 / (4π × (15500)²)
   = 1.9075×10⁶ / 3.0191×10⁹
   = 6.32×10⁻⁴ W

Check in dB

LFS = 20 log(4πd/λ) = 126.88 dB
Pt  = 10 log 250    = 23.98 dBW
Pr  = 23.98 + 25 + 45.90 − 126.88 = −31.99 dBW
    = −1.99 dBm ≈ 0.632 mW

Answer: Maximum received power ≈ 6.32×10⁻⁴ W = 0.632 mW (−31.99 dBW, −1.99 dBm).

✓ Answer checked by Claude Opus 5.5 (high effort)

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