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Okumura model and median path loss numerical

Short answer

Okumura model gives L50 = LF + Amu - G(hte) - G(hre) - GAREA. Here LF = 125.51 dB, G(hte) = -6.02 dB, G(hre) = 10.46 dB, so median path loss is 125.51 + 43 + 6.02 - 10.46 - 9 = 155.07 dB.

Explain Okumura model of outdoor radio propagation. Determine the median path loss for T-R separation of 50 km, transmit antenna effective height of 100 m and receive antenna effective height of 10 m in a suburban area with correction factor of 9 dB at 900 MHz. Assume median attenuation relative to free space 43 dB.

Okumura model

The Okumura model is an empirical model based on extensive measurements in and around Tokyo. It is one of the most widely used models for macrocell planning in urban areas.

L50(dB) = LF + Amu(f,d) − G(hte) − G(hre) − GAREA
  • LF = free space loss.
  • Amu(f,d) = median attenuation relative to free space in an urban area over quasi-smooth terrain, read from curves (reference hte = 200 m, hre = 3 m).
  • G(hte) = BS antenna height gain = 20 log(hte/200), 30 m < hte < 1000 m.
  • G(hre) = mobile antenna height gain = 10 log(hre/3) for hre ≤ 3 m; 20 log(hre/3) for 3 m < hre < 10 m.
  • GAREA = gain due to environment (suburban, quasi-open, open), from curves.
  • Valid for 150–1920 MHz (extendable to 3 GHz), d = 1–100 km.
  • Merits: simple, accurate for urban/suburban cellular systems (about 10–14 dB standard deviation).
  • Demerits: slow to respond to rapid terrain changes; works better in urban than rural areas; curves are hard to use in software (hence the Hata formula).

Calculation

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

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

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

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

Answer: Median path loss L50 ≈ 155.07 dB.

✓ Answer checked by Claude Opus 5.5 (high effort)

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