---
title: "Numerical: Okumura median path loss (unity gain)"
subject: "Wireless Communication"
chapter: 3
key: "semester_7/wireless_communication:okumura-median-loss-unity-gain"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/okumura-median-loss-unity-gain
site: BCT Engineering
---

# Numerical: Okumura median path loss (unity gain)

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

- Asked in: 2082 Baisakh, 8 marks
- Video: [Wireless mobile communication, Lecture no 14, Okumura Model Curve frequency vs distance, attenuation](https://www.youtube.com/watch?v=DkLFbCiF0xA) — EC Video
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/wireless_communication/okumura-median-loss-unity-gain
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3

## Question

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]

## 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.

## Answer

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

#### Step 4: Received power
```text
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).**

## Check yourself

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

1. LF at 50 km, 900 MHz is:
   A. 145.5 dB
   B. 125.51 dB
   C. 105.5 dB
   D. 95.5 dB
   Answer: B. 20 log(4 pi 50000/0.3333) = 125.51 dB. (Key point: Free space loss)

2. For hre = 10 m, G(hre) is:
   A. 20 dB
   B. 10.46 dB
   C. -6.02 dB
   D. 6.02 dB
   Answer: B. 20 log(10/3) = 10.46 dB. (Key point: G(hre))

3. L50 equals:
   A. 171.5 dB
   B. 155.07 dB
   C. 125.51 dB
   D. 143.0 dB
   Answer: B. 125.51 + 43 + 6.02 - 10.46 - 9 = 155.07. (Key point: Median loss)

4. With 60 dBm EIRP and unity gain receiver, Pr is:
   A. -85.07 dBm
   B. -155.07 dBm
   C. -65.07 dBm
   D. -95.07 dBm
   Answer: D. 60 - 155.07 + 0 = -95.07 dBm. (Key point: Received power)

## Close variants set in other exams

- [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×

## Related in Wireless Communication

- [Okumura model and median path loss numerical](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aokumura-model-median-path-loss) — Wireless Communication, chapter 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

## Related in other subjects

- [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