---
title: "Numerical: Okumura T-R separation for 167 dB"
subject: "Wireless Communication"
chapter: 3
key: "semester_7/wireless_communication:okumura-separation-distance"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/okumura-separation-distance
site: BCT Engineering
---

# Numerical: Okumura T-R separation for 167 dB

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

- Asked in: 2077 Chaitra, 8 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-separation-distance
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3

## Question

Employing the Okumura model compute the transmitter and receiver separation distance if median loss is 167 dB when the carrier frequency is 2.1 GHz. Assume height of transmitting antenna is 40 m, height of receiving antenna is 2 m, for a large city. [Amu = 34 dB, Garea = 0 dB]

## Short answer

Using Okumura: G(hte)=20 log(40/200)= -13.98 dB, G(hre)=10 log(2/3)= -1.76 dB; required LF = L50 - Amu + G(hte) + G(hre) = 167-34-13.98-1.76=117.26 dB. From LF=20log(4πd/λ) with λ=0.1429 m, solve for d ≈ 8292 m (8.29 km).

## Answer

**Okumura model:** L50 = LF + Amu(f,d) − G(hte) − G(hre) − GAREA

Given: L50 = 167 dB, fc = 2.1 GHz, hte = 40 m, hre = 2 m, large city, Amu = 34 dB, GAREA = 0 dB.

#### Step 1: Antenna height gain factors
```text
G(hte) = 20 log(hte/200) = 20 log(40/200) = −13.98 dB
G(hre) = 10 log(hre/3)   = 10 log(2/3)    = −1.76 dB
         (hre ≤ 3 m, so 10 log is used)
```

#### Step 2: Free space loss required
```text
LF = L50 − Amu + G(hte) + G(hre) + GAREA
   = 167 − 34 + (−13.98) + (−1.76) + 0
   = 117.26 dB
```

#### Step 3: Distance from free space loss
```text
λ  = c/f = 3×10⁸ / 2.1×10⁹ = 0.1429 m
LF = 20 log(4πd/λ)

4πd/λ = 10^(117.26/20) = 7.294 × 10⁵
d = 7.294×10⁵ × λ/(4π)
  = 7.294×10⁵ × 0.011368
  = 8292 m
```

**Answer: T–R separation d ≈ 8.29 km.**

Note: 2.1 GHz is slightly above Okumura's basic 1920 MHz range, but the model is commonly extrapolated up to 3 GHz; Amu is taken as given.

## Check yourself

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

1. Which rearrangement gives the required free-space loss?
   A. LF = L50 + Amu + G(hte) + G(hre) + GAREA
   B. LF = L50 − Amu + G(hte) + G(hre) + GAREA
   C. LF = L50 − Amu − G(hte) − G(hre) − GAREA
   D. LF = L50 + Amu − G(hte) − G(hre)
   Answer: B. From L50 = LF + Amu − G(hte) − G(hre) − GAREA, LF = L50 − Amu + G(hte) + G(hre) + GAREA. (Key point: Okumura equation)

2. For hre = 2 m, which formula is used for G(hre)?
   A. 20 log(hre/3)
   B. 20 log(hre/200)
   C. 10 log(hre/3)
   D. 10 log(hre/200)
   Answer: C. For hre ≤ 3 m, G(hre) = 10 log(hre/3) = −1.76 dB. (Key point: Mobile height gain choice)

3. What free-space loss is required?
   A. 117.26 dB
   B. 152.7 dB
   C. 137.3 dB
   D. 181.7 dB
   Answer: A. LF = 167 − 34 − 13.98 − 1.76 = 117.26 dB. (Key point: Value of LF)

4. The T-R separation obtained is about:
   A. 829 m
   B. 82.9 km
   C. 8.29 km
   D. 0.83 km
   Answer: C. 4πd/λ = 7.294x10^5, so d = 7.294x10^5 x 0.011368 = 8292 m. (Key point: Distance)

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