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Numerical: Okumura path loss, EIRP and received power

Short answer

Using Okumura’s model: free-space loss = 125.51 dB, add Amu(43 dB), subtract G(hte) = -6.02 dB, G(hre) = 10.46 dB, GAREA = 9 dB → L50 = 155.07 dB. EIRP = 60 dBm; received power = EIRP – L50 + Gr = -85.07 dBm.

Determine the mean path loss using Okumura's model for d = 50 km, hte = 100 m, hre = 10 m in a suburban environment. If the base station transmitter radiates an EIRP of 1 kW at a carrier frequency of 900 MHz, find EIRP (dBm) and the power at the receiver where the gain of receiving antenna is 10 dB. (Given: Amu(f,d) = 43 dB, GArea(f) = 9 dB for a 50 km distance and 900 MHz frequency of operation.)

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

Given: d = 50 km, hte = 100 m, hre = 10 m, f = 900 MHz, Amu = 43 dB, GAREA = 9 dB, EIRP = 1 kW, Gr = 10 dB.

Step 1: Free space path loss

λ  = c/f = 3×10⁸ / 900×10⁶ = 0.3333 m
LF = 10 log[(4π)²d²/λ²] = 20 log(4πd/λ)
   = 20 log(4π × 50 000 / 0.3333)
   = 125.51 dB

Step 2: Antenna height gains

G(hte) = 20 log(100/200) = −6.02 dB
G(hre) = 20 log(10/3)    = 10.46 dB   (3 m < hre < 10 m)

Step 3: Mean (median) path loss

L50 = 125.51 + 43 − (−6.02) − 10.46 − 9
    = 155.07 dB

Step 4: EIRP in dBm

EIRP = 10 log(1000 W / 1 mW) = 10 log(10⁶) = 60 dBm

Step 5: Received power

Pr = EIRP − L50 + Gr
   = 60 − 155.07 + 10
   = −85.07 dBm

Answer: L50 = 155.07 dB, EIRP = 60 dBm, Pr = −85.07 dBm (about 3.1 pW).

✓ Answer checked by Claude Opus 5.5 (high effort)

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