---
title: "Numerical: Okumura path loss, EIRP and received power"
subject: "Wireless Communication"
chapter: 3
key: "semester_7/wireless_communication:okumura-eirp-received-power"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/okumura-eirp-received-power
site: BCT Engineering
---

# Numerical: Okumura path loss, EIRP and received power

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

- Asked in: 2082 Bhadra, 5 marks; 2080 Bhadra, 4+2+2 marks (asked 2 times)
- 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-eirp-received-power
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3

## Question

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.)

## 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.

## 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, EIRP = 1 kW, Gr = 10 dB.

#### Step 1: Free space path loss
```text
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
```text
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
```text
L50 = 125.51 + 43 − (−6.02) − 10.46 − 9
    = 155.07 dB
```

#### Step 4: EIRP in dBm
```text
EIRP = 10 log(1000 W / 1 mW) = 10 log(10⁶) = 60 dBm
```

#### Step 5: Received power
```text
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).**

## Check yourself

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

1. Which expression gives the free space path loss used in Okumura's model?
   A. 20 log(4 pi d / lambda)
   B. 10 log(4 pi d / lambda)
   C. 40 log(d)
   D. 20 log(d)
   Answer: A. LF = 20 log(4 pi d/lambda) = 125.51 dB here. (Key point: Free space loss)

2. G(hte) for hte = 100 m is -6.02 dB. How does it enter L50?
   A. Subtracted, so it adds 6.02 dB to the loss
   B. Ignored
   C. Multiplied by Amu
   D. Added, so it reduces loss by 6.02 dB
   Answer: A. L50 has - G(hte); subtracting -6.02 adds 6.02 dB. (Key point: Height gain sign)

3. What is 1 kW EIRP in dBm?
   A. 30 dBm
   B. 40 dBm
   C. 90 dBm
   D. 60 dBm
   Answer: D. 10 log(1000 W / 1 mW) = 10 log(10^6) = 60 dBm. (Key point: EIRP in dBm)

4. Using L50 = 155.07 dB, EIRP = 60 dBm and Gr = 10 dB, the received power is:
   A. -105.07 dBm
   B. -85.07 dBm
   C. -75.07 dBm
   D. -95.07 dBm
   Answer: B. Pr = 60 - 155.07 + 10 = -85.07 dBm. (Key point: Received power)

## Close variants set in other exams

- [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×

## 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
- [Numerical: 50 W in dBm/dBW and received power](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Atransmit-power-dbm-free-space) — Wireless Communication, chapter 3
- [Numerical: λ/4 monopole and two-ray received power](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Amonopole-two-ray-received-power) — Wireless Communication, chapter 3
- [Numerical: received power with cable and mismatch loss](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Areceived-power-coax-mismatch) — Wireless Communication, chapter 3
- [Numerical: maximum cell radius using Hata model](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Ahata-max-cell-radius) — Wireless Communication, chapter 3

## Related in other subjects

- [AM transmitter 10 kW/11.8 kW: modulation index and power](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-10kw-11-8kw) — Communication Systems, chapter 2
- [Received power: 50 km, 10 GHz link](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Areceived-power-50km-10ghz) — 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
- [Received power: 100 MHz, 0.75 km](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Areceived-power-100mhz-0-75km) — Propagation and Antenna, chapter 5