---
title: "Two-ray model path and phase difference"
subject: "Wireless Communication"
chapter: 3
key: "semester_7/wireless_communication:two-ray-path-phase-difference"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/two-ray-path-phase-difference
site: BCT Engineering
---

# Two-ray model path and phase difference

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

- Asked in: 2081 Bhadra, 6 marks; 2071 Bhadra, 8 marks (asked 2 times)
- Video: [Ground Reflection (2 Ray) Model - Mobile Radio Propagation - Mobile Communication System](https://www.youtube.com/watch?v=Ul38Tp1gTwk) — Ekeeda
- 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-phase-difference
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3

## Question

Derive an expression for path difference and phase difference between line of sight (LOS) path and ground reflected (GR) path in two-ray model using simple geometrical relations.

## Short answer

Using the image method, LOS path ≈ d + (ht−hr)²/(2d), ground-reflected ≈ d + (ht+hr)²/(2d). Path difference Δ ≈ 2 ht hr / d. Phase difference θΔ = 2πΔ/λ ≈ 4π ht hr/(λ d).

## Answer

The **two-ray ground reflection model** assumes the received field is the sum of a line-of-sight (LOS) wave and a wave reflected from the ground. The path difference between them decides whether they add or cancel.

```text
  Tx
   |\  d' (LOS)
   | \______________________  Rx
ht |  \                  __/|
   |   \  d'' (GR)    __/   | hr
   |    \          __/      |
 --+-----\______/-----------+----- ground
          reflection point
   |<---------- d ---------->|

 Image method: image of Tx at depth ht below ground
```

#### Path lengths (method of images)
Replace the ground-reflected path by a straight line from the image of the transmitter (at −ht) to the receiver.

```text
LOS:  d'  = √[d² + (ht − hr)²]
GR :  d'' = √[d² + (ht + hr)²]
```

#### Path difference
Using binomial expansion √(1 + x) ≈ 1 + x/2 for d ≫ ht + hr:

```text
d'  = d√[1 + ((ht−hr)/d)²] ≈ d + (ht − hr)²/(2d)
d'' = d√[1 + ((ht+hr)/d)²] ≈ d + (ht + hr)²/(2d)

Δ = d'' − d'
  = [(ht + hr)² − (ht − hr)²] / (2d)
  = 4ht·hr / (2d)
Δ ≈ 2ht·hr / d
```

#### Phase difference and time delay
A path difference of one wavelength corresponds to 2π radians:

```text
θΔ = 2πΔ/λ = Δωc/c ≈ 4π·ht·hr/(λd)

τd = Δ/c = θΔ/(2πfc)
```

#### Use in received field
With ground reflection coefficient Γ ≈ −1, the total field is

```text
|ETOT| ≈ 2E0d0/d · sin(θΔ/2)
```
For large d, θΔ/2 < 0.3 rad, so sin(θΔ/2) ≈ θΔ/2 and

```text
ETOT ≈ (2E0d0/d)(2πht·hr/(λd)) = k/d²
Pr = Pt·Gt·Gr·ht²·hr² / d⁴
```
So at large distance received power falls by 40 dB/decade, and is independent of frequency.

## Check yourself

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

1. In the two-ray derivation, the ground-reflected path is replaced by:
   A. A curved path
   B. A line from the receiver to its image
   C. A line from the image of the transmitter below ground to the receiver
   D. A line along the ground
   Answer: C. The image of Tx at depth ht gives d'' = sqrt(d^2 + (ht+hr)^2). (Key point: Method of images)

2. For d much larger than ht+hr, the path difference is approximately:
   A. (ht+hr)/d
   B. 2*ht*hr/d
   C. 4*ht*hr/d^2
   D. ht*hr/d
   Answer: B. Delta = [(ht+hr)^2 - (ht-hr)^2]/(2d) = 2 ht hr / d. (Key point: Path difference)

3. The phase difference between the two paths is:
   A. pi*ht*hr/(lambda*d)
   B. ht hr/(lambda d)
   C. 4 pi ht hr/(lambda d)
   D. 2 pi d/lambda
   Answer: C. theta = 2 pi Delta/lambda = 4 pi ht hr/(lambda d). (Key point: Phase difference)

4. At large distance in the two-ray model, received power falls with distance as:
   A. 1/d^3
   B. 1/d^4 (40 dB/decade)
   C. 1/d^2
   D. 1/d
   Answer: B. Pr = Pt Gt Gr ht^2 hr^2 / d^4, independent of frequency. (Key point: Large distance result)

## Related in Wireless Communication

- [Scattering and two-ray ground reflection model](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Ascattering-two-ray-ground-reflection) — Wireless Communication, chapter 3
- [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: 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
- [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
- [Diffraction and Fresnel zone phase difference](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Adiffraction-fresnel-zone) — Wireless Communication, chapter 3
- [Log-distance path loss model](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Alog-distance-path-loss) — Wireless Communication, chapter 3

## Related in other subjects

- [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
- [Inductance of a single-phase two-wire line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Asingle-phase-inductance-derivation) — Power System Analysis I, chapter 4
- [10 GHz LOS link loss and Friis derivation](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Alos-link-10ghz-100km-friis) — Propagation and Antenna, chapter 5
- [Phong illumination model expression](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Aphong-illumination-model-derive) — Computer Graphics, chapter 8, asked 8×