---
title: "Numerical: cordless phone free space path loss"
subject: "Wireless Communication"
chapter: 3
key: "semester_7/wireless_communication:cordless-phone-path-loss"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/cordless-phone-path-loss
site: BCT Engineering
---

# Numerical: cordless phone free space path loss

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

- Asked in: 2077 Chaitra, 6 marks
- Video: [Free Space Propagation Model - Wireless Communication](https://www.youtube.com/watch?v=vEbfiBaskVM) — Padmasri Naban
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/wireless_communication/cordless-phone-path-loss
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3

## Question

Given a cordless phone operating at 52000 kHz frequency which has a range of 50 m. Determine the free space path loss.

## Short answer

Free space path loss is found using PL(dB)=20 log(4πd/λ). With f=52 MHz, λ=c/f=5.769 m, d=50 m, PL = 20 log(108.91) = 40.7 dB.

## Answer

Free space path loss (with unity gain antennas) is given by the Friis equation:

```text
PL(dB) = 10 log(Pt/Pr) = 20 log(4πd/λ)
```
Given: f = 52 000 kHz = 52 MHz, d = 50 m (range of the phone).

#### Step 1: Wavelength
```text
λ = c/f = 3×10⁸ / 52×10⁶ = 5.769 m
```

#### Step 2: Check far field
The phone's antenna is small (much less than λ), so the far field begins within a few wavelengths; 50 m is about 8.7λ, so the Friis equation can be used.

#### Step 3: Path loss
```text
4πd/λ = 4π × 50 / 5.769 = 108.91

PL = 20 log(108.91) = 40.74 dB
```

**Check using the practical form:**

```text
PL = 32.44 + 20 log f(MHz) + 20 log d(km)
   = 32.44 + 20 log 52 + 20 log 0.05
   = 32.44 + 34.32 − 26.02
   = 40.74 dB
```

**Answer: Free space path loss at the 50 m range ≈ 40.7 dB** (the received power is about 1/11 860 of the transmitted power for unity-gain antennas).

## Check yourself

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

1. With unity-gain antennas, free-space path loss is:
   A. 20 log(λ/(4πd))
   B. 10 log(4πd/λ)
   C. 20 log(4πd/λ)
   D. 40 log(4πd/λ)
   Answer: C. PL = 10 log(Pt/Pr) = 20 log(4πd/λ). (Key point: Free-space loss formula)

2. What is the wavelength at 52 000 kHz?
   A. 5.769 m
   B. 0.577 m
   C. 57.7 m
   D. 1.73 m
   Answer: A. λ = 3x10^8 / 52x10^6 = 5.769 m. (Key point: Wavelength)

3. Why can the Friis equation be used at 50 m?
   A. 50 m is less than one wavelength
   B. The antenna is small, so the far field starts within a few wavelengths and 50 m is about 8.7λ
   C. Friis is valid at any distance
   D. The phone uses high gain antennas
   Answer: B. Small antenna means a short far-field distance; 50 m is about 8.7λ. (Key point: Far-field validity)

4. The free-space path loss is about:
   A. 32.4 dB
   B. 50.7 dB
   C. 26.0 dB
   D. 40.7 dB
   Answer: D. 20 log(108.91) = 40.74 dB, confirmed by 32.44 + 34.32 − 26.02. (Key point: Result)

## Close variants set in other exams

- [Doppler effect; cordless phone path loss](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Adoppler-effect-cordless-phone) — Wireless Communication, chapter 3, asked 2×

## Related in Wireless Communication

- [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
- [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: 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: free-space received power (165 W, 325 MHz)](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Afree-space-received-power-165w) — 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
- [Friis equation and free space path loss](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Afriis-path-loss-free-space) — Propagation and Antenna, chapter 5, asked 2×
- [Free space loss (LFS)](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Afree-space-loss) — Propagation and Antenna, chapter 5