---
title: "Okumura model and median path loss numerical"
subject: "Wireless Communication"
chapter: 3
key: "semester_7/wireless_communication:okumura-model-median-path-loss"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/okumura-model-median-path-loss
site: BCT Engineering
---

# Okumura model and median path loss numerical

Wireless Communication · Chapter 3: Radio Wave Propagation in Mobile Networks

- Asked in: 2074 Magh, 4+4 marks
- Video: [Outdoor Propagation Model Okumura - Mobile Radio Propagation - Mobile Communication System](https://www.youtube.com/watch?v=GqploKEmD3w) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/wireless_communication/okumura-model-median-path-loss
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3

## Question

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.

## 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.

## Answer

#### 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.

```text
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.

```text
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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The Okumura model is:
   A. A theoretical model from Maxwell's equations
   B. An indoor partition loss model
   C. A fading model
   D. An empirical model from measurements in and around Tokyo, using curves
   Answer: D. It is empirical, built from Tokyo measurements and read from curves. (Key point: Basis of Okumura model)

2. Which is the Okumura equation?
   A. L50 = LF + Amu − G(hte) − G(hre) − GAREA
   B. L50 = LF − Amu − G(hte) − G(hre) − GAREA
   C. L50 = LF + Amu + G(hte) + G(hre) + GAREA
   D. L50 = Amu − LF
   Answer: A. Free-space loss plus median attenuation, minus the height and area gains. (Key point: Okumura equation)

3. LF for 50 km at 900 MHz is:
   A. 105.5 dB
   B. 145.5 dB
   C. 125.51 dB
   D. 115.5 dB
   Answer: C. 20 log(4π x 50000/0.3333) = 125.51 dB. (Key point: Free-space loss)

4. L50 for the given data is:
   A. 167.1 dB
   B. 155.07 dB
   C. 143.1 dB
   D. 130.2 dB
   Answer: B. 125.51 + 43 + 6.02 − 10.46 − 9 = 155.07 dB. (Key point: Median path loss)

## Related in Wireless Communication

- [Numerical: Okumura median path loss (unity gain)](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aokumura-median-loss-unity-gain) — Wireless Communication, chapter 3, asked 3×
- [Numerical: 10 km link feasibility with Okumura](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aokumura-link-feasibility) — Wireless Communication, chapter 3
- [Numerical: Okumura path loss, EIRP and received power](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aokumura-eirp-received-power) — Wireless Communication, chapter 3, asked 3×
- [Numerical: Okumura T-R separation for 167 dB](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aokumura-separation-distance) — Wireless Communication, chapter 3
- [Numerical: two-ray vs free-space path loss at 800 MHz](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Atwo-ray-path-loss-800mhz) — Wireless Communication, chapter 3
- [Numerical: Hata path loss for medium city](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Ahata-medium-city-pathloss) — Wireless Communication, chapter 3

## Related in other subjects

- [Path loss in radio wave propagation](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Apath-loss-derivation) — Propagation and Antenna, chapter 5, asked 2×
- [Free space path loss: 6 GHz, 10,500 km](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Afspl-derivation-6ghz-10500km) — Propagation and Antenna, chapter 5
- [Short note: Effective antenna height](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Aeffective-antenna-height-short-note) — Propagation and Antenna, chapter 5
- [10 GHz LOS link loss and Friis derivation](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Alos-link-10ghz-100km-friis) — Propagation and Antenna, chapter 5