---
title: "Numerical: coverage range of a 150 W base station"
subject: "Wireless Communication"
chapter: 3
key: "semester_7/wireless_communication:coverage-range-path-loss-exponent"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/coverage-range-path-loss-exponent
site: BCT Engineering
---

# Numerical: coverage range of a 150 W base station

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

- Asked in: 2070 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/wireless_communication/coverage-range-path-loss-exponent
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3

## Question

Determine the radio coverage range of a base station that transmits a RF signal at 150 W, given the receiver threshold level is −104 dBm. Assume that the path loss at the first meter is 15 dB in a mobile radio propagation condition. (Path loss exponent = 4)

## Short answer

Convert 150 W to 51.76 dBm; max allowed path loss = 155.76 dB. Using log-distance model PL(d) = 15 + 40 log(d) gives log(d) = 3.519, d = 10^3.519 = 3304 m. Coverage range ≈ 3.30 km.

## Answer

**Given:** P_t = 150 W, receiver threshold = −104 dBm, path loss at 1 m (d₀) = 15 dB, path loss exponent n = 4.

**Step 1 – Transmit power in dBm**
```text
P_t = 10 log(150 W / 1 mW) = 10 log(150000)
    = 51.76 dBm
```

**Step 2 – Maximum allowed path loss**
The signal can be received while P_r ≥ −104 dBm, so:
```text
PL_max = P_t − P_r(min) = 51.76 − (−104)
       = 155.76 dB
```

**Step 3 – Log-distance path loss model**
```text
PL(d) = PL(d₀) + 10 n log(d/d₀),   d₀ = 1 m
155.76 = 15 + 10 × 4 × log(d)
40 log(d) = 140.76
log(d) = 3.519
d = 10^3.519 = 3304 m
```

**Answer:** The radio coverage range of the base station is about **3.30 km** (radius ≈ 3304 m).

## Check yourself

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

1. 150 W in dBm is approximately:
   A. 21.8 dBm
   B. 51.8 dBm
   C. 41.8 dBm
   D. 15 dBm
   Answer: B. 10 log(150000) = 51.76 dBm. (Key point: Transmit power in dBm)

2. Maximum allowed path loss with threshold -104 dBm is:
   A. 51.76 dB
   B. 104 dB
   C. 155.76 dB
   D. -52.24 dB
   Answer: C. PL_max = Pt - Pr(min) = 51.76 + 104 = 155.76 dB. (Key point: Maximum path loss)

3. With PL(d0) = 15 dB, n = 4 and d0 = 1 m, which equation gives d?
   A. 155.76 = 15 + 40 log d
   B. 155.76 = 15 + 4 log d
   C. 155.76 = 15 x 40 log d
   D. 155.76 = 15 + 20 log d
   Answer: A. 10n log(d/d0) = 40 log d, so log d = 140.76/40 = 3.519. (Key point: Solving for d)

4. The coverage range is about:
   A. 330 m
   B. 33 km
   C. 3.3 km
   D. 330 km
   Answer: C. d = 10^3.519 = 3304 m. (Key point: Coverage range)

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