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Numerical: Okumura median path loss (unity gain)

Short answer

Using Okumura model L50 = LF + Amu - G(hte) - G(hre) - GAREA, with free-space loss 125.51 dB, height gains -6.02 and 10.46 dB, median path loss is 155.07 dB. With EIRP 1 kW and unity gain receiver, received power is -95.07 dBm.

Find the median 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 the power at the receiver (assume a unity gain receiving antenna). [Assume Amu = 43 dB, GAREA = 9 dB]

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 (suburban), EIRP = 1 kW, Gr = 1 (0 dB).

Step 1: Free space path loss

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

Step 2: Height gain factors

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

Step 3: Median path loss

L50 = LF + Amu − G(hte) − G(hre) − GAREA
    = 125.51 + 43 + 6.02 − 10.46 − 9
    = 155.07 dB

Step 4: Received power

EIRP = 1 kW = 10 log(1000/0.001) = 60 dBm

Pr = EIRP − L50 + Gr
   = 60 − 155.07 + 0
   = −95.07 dBm

Answer: Median path loss L50 = 155.07 dB; received power Pr = −95.07 dBm (about 3.1×10⁻¹³ W).

✓ Answer checked by Claude Opus 5.5 (high effort)

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