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Numerical: Okumura T-R separation for 167 dB

Short answer

Using Okumura: G(hte)=20 log(40/200)= -13.98 dB, G(hre)=10 log(2/3)= -1.76 dB; required LF = L50 - Amu + G(hte) + G(hre) = 167-34-13.98-1.76=117.26 dB. From LF=20log(4πd/λ) with λ=0.1429 m, solve for d ≈ 8292 m (8.29 km).

Employing the Okumura model compute the transmitter and receiver separation distance if median loss is 167 dB when the carrier frequency is 2.1 GHz. Assume height of transmitting antenna is 40 m, height of receiving antenna is 2 m, for a large city. [Amu = 34 dB, Garea = 0 dB]

Okumura model: L50 = LF + Amu(f,d) − G(hte) − G(hre) − GAREA

Given: L50 = 167 dB, fc = 2.1 GHz, hte = 40 m, hre = 2 m, large city, Amu = 34 dB, GAREA = 0 dB.

Step 1: Antenna height gain factors

G(hte) = 20 log(hte/200) = 20 log(40/200) = −13.98 dB
G(hre) = 10 log(hre/3)   = 10 log(2/3)    = −1.76 dB
         (hre ≤ 3 m, so 10 log is used)

Step 2: Free space loss required

LF = L50 − Amu + G(hte) + G(hre) + GAREA
   = 167 − 34 + (−13.98) + (−1.76) + 0
   = 117.26 dB

Step 3: Distance from free space loss

λ  = c/f = 3×10⁸ / 2.1×10⁹ = 0.1429 m
LF = 20 log(4πd/λ)

4πd/λ = 10^(117.26/20) = 7.294 × 10⁵
d = 7.294×10⁵ × λ/(4π)
  = 7.294×10⁵ × 0.011368
  = 8292 m

Answer: T–R separation d ≈ 8.29 km.

Note: 2.1 GHz is slightly above Okumura's basic 1920 MHz range, but the model is commonly extrapolated up to 3 GHz; Amu is taken as given.

✓ Answer checked by Claude Opus 5.5 (high effort)

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