---
title: "Numerical: two-ray vs free-space path loss at 800 MHz"
subject: "Wireless Communication"
chapter: 3
key: "semester_7/wireless_communication:two-ray-path-loss-800mhz"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/two-ray-path-loss-800mhz
site: BCT Engineering
---

# Numerical: two-ray vs free-space path loss at 800 MHz

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

- Asked in: 2070 Magh, 4+4 marks
- Video: [Problem #2 - Two-Ray Propagation Model - Path Loss - Wireless Communication](https://www.youtube.com/watch?v=M2--Wiu4X4M) — 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/two-ray-path-loss-800mhz
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3

## Question

Determine the propagation path loss for signal at 800 MHz, with a transmitting antenna height of 30 m and a receiving antenna height of 2 m, over a distance of 10 km, using two-ray mobile point-to-point propagation model. How is it compared with that of free-space propagation path loss model?

## Short answer

Two-ray model PL = 40 log(10000) − 20 log(30) − 20 log(2) = 124.44 dB. Free-space PL = 110.50 dB. Two-ray predicts ~13.9 dB more loss due to ground reflection causing 1/d⁴ power decay.

## Answer

**Given:** f = 800 MHz, h_t = 30 m, h_r = 2 m, d = 10 km. Antenna gains are taken as unity (0 dB) because they are not given.

#### Path loss by two-ray model
For large distances (d ≫ √(h_t h_r)), the two-ray ground reflection model gives a path loss that does not depend on frequency:
```text
PL(dB) = 40 log d − (10 log G_t + 10 log G_r
                     + 20 log h_t + 20 log h_r)
```
With G_t = G_r = 1:
```text
PL = 40 log(10000) − 20 log(30) − 20 log(2)
   = 160 − 29.54 − 6.02
   = 124.44 dB
```

#### Path loss by free-space model
```text
λ  = c / f = 3 × 10⁸ / 800 × 10⁶ = 0.375 m
PL = 20 log(4π d / λ)
   = 20 log(4π × 10000 / 0.375)
   = 20 log(335103) = 110.50 dB
```
(Same using PL = 32.44 + 20 log f(MHz) + 20 log d(km) = 32.44 + 58.06 + 20 = 110.50 dB.)

#### Comparison
```text
Difference = 124.44 − 110.50 = 13.93 dB
```
| Model | Path loss | Fall-off with distance |
|---|---|---|
| Free space | 110.50 dB | 20 dB/decade (1/d²) |
| Two-ray | 124.44 dB | 40 dB/decade (1/d⁴) |

**Answer:** Two-ray path loss = **124.44 dB**, free-space path loss = **110.50 dB**. The two-ray model predicts about **13.9 dB more loss**, because at large distances the ground-reflected wave nearly cancels the direct wave, so power falls as 1/d⁴ instead of 1/d². The two-ray result is more realistic for a mobile link over flat ground.

## Check yourself

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

1. At large distance, the two-ray path loss formula depends on:
   A. Frequency strongly
   B. d, h_t, h_r and gains, not frequency
   C. Only wavelength
   D. Only f
   Answer: B. PL = 40 log d - (10 log Gt + 10 log Gr + 20 log ht + 20 log hr), which has no frequency term. (Key point: Two-ray frequency dependence)

2. For ht = 30 m, hr = 2 m, d = 10 km, unity gains, two-ray path loss is:
   A. 110.5 dB
   B. 124.4 dB
   C. 160 dB
   D. 95 dB
   Answer: B. 160 - 29.54 - 6.02 = 124.44 dB. (Key point: Two-ray value)

3. How fast does loss rise with distance in the two-ray model?
   A. 20 dB/decade
   B. 10 dB/decade
   C. 40 dB/decade
   D. 80 dB/decade
   Answer: C. Power falls as 1/d^4, i.e. 40 dB per decade, versus 20 dB/decade in free space. (Key point: Fall-off rate)

4. Two-ray loss compared with free space (110.5 dB) is:
   A. 13.9 dB less
   B. 13.9 dB more
   C. Equal
   D. About 3 dB more
   Answer: B. 124.44 - 110.50 = 13.93 dB more, because the reflected wave nearly cancels the direct wave. (Key point: Comparison)

## Related in Wireless Communication

- [Two-ray model advantages and path loss derivation](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Atwo-ray-advantages-path-loss) — Wireless Communication, chapter 3
- [Numerical: Hata path loss for medium city](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Ahata-medium-city-pathloss) — Wireless Communication, chapter 3
- [Numerical: cordless phone free space path loss](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acordless-phone-path-loss) — Wireless Communication, chapter 3, asked 2×
- [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
- [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
- [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

## 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×
- [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
- [Friis, path loss and LOS distance](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Afriis-path-loss-los-distance) — Propagation and Antenna, chapter 5