---
title: "Scattering and two-ray ground reflection model"
subject: "Wireless Communication"
chapter: 3
key: "semester_7/wireless_communication:scattering-two-ray-ground-reflection"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/scattering-two-ray-ground-reflection
site: BCT Engineering
---

# Scattering and two-ray ground reflection model

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

- Asked in: 2074 Bhadra, 2+8 marks
- Video: [Two Ray Ground Reflection Model In Wireless Communication](https://www.youtube.com/watch?v=fajmNSr3PdM) — ECIT Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/wireless_communication/scattering-two-ray-ground-reflection
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3

## Question

What is known as scattering? Derive an expression for two ray ground reflected model.

## Short answer

Scattering reradiates energy in many directions when radio waves hit objects comparable to wavelength or rough surfaces. In the two-ray model, total field magnitude is 2(E0d0/d)sin(θΔ/2), and at large distances received power becomes Pt Gt Gr ht^2 hr^2 / d^4, giving 40 dB/decade loss independent of frequency.

## Answer

#### Scattering
**Scattering** occurs when a radio wave meets an object whose size is comparable to or smaller than the wavelength, or a rough surface (foliage, street signs, lamp posts, rough ground), and the energy is re-radiated in many directions. It explains why received signal is often stronger than reflection and diffraction models predict. A surface is treated as rough when its height variation exceeds the critical height hc = λ/(8 sin θi) (Rayleigh criterion).

#### Two-ray ground reflection model
```text
  Tx
   |\  LOS path d'
ht | \_______________________ Rx
   |  \                    _/|
   |   \  reflected d''  _/  | hr
 --+----\____________/-------+-- ground
   |<----------- d --------->|
```

**1. Field expressions.** With E0 the field at reference d0, the LOS and ground-reflected fields are:

```text
E_LOS = (E0d0/d')·cos[ωc(t − d'/c)]
E_g   = Γ·(E0d0/d'')·cos[ωc(t − d''/c)]
```
For grazing incidence, Γ ≈ −1.

**2. Path difference** (method of images):

```text
d'  = √[(ht − hr)² + d²] ≈ d + (ht − hr)²/(2d)
d'' = √[(ht + hr)² + d²] ≈ d + (ht + hr)²/(2d)
Δ = d'' − d' ≈ 2ht·hr/d
```

**3. Phase difference and delay:**

```text
θΔ = 2πΔ/λ = Δ·ωc/c
τd = Δ/c
```

**4. Total field.** Taking d' ≈ d'' ≈ d in the amplitudes and Γ = −1:

```text
|ETOT| = (E0d0/d)·|1 − e^(−jθΔ)|
       = (E0d0/d)·√(2 − 2cosθΔ)
       = 2(E0d0/d)·sin(θΔ/2)
```

**5. Large-distance approximation.** When θΔ/2 < 0.3 rad (i.e. d > 20πht·hr/(3λ) ≈ 20ht·hr/λ), sin(θΔ/2) ≈ θΔ/2:

```text
|ETOT| ≈ (2E0d0/d)·(2π·ht·hr/(λd)) ≈ k/d²   V/m
```

**6. Received power and path loss.** Since power ∝ |E|²:

```text
Pr = Pt·Gt·Gr·ht²·hr² / d⁴

PL(dB) = 40 log d − (10 log Gt + 10 log Gr
                     + 20 log ht + 20 log hr)
```

So at large distances received power falls at **40 dB/decade**, compared with 20 dB/decade in free space, and the path loss is independent of frequency.

## Check yourself

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

1. Scattering occurs when a wave meets:
   A. A very large smooth surface
   B. An object much larger than the wavelength with a smooth face
   C. An object comparable to or smaller than λ, or a rough surface, re-radiating in many directions
   D. A vacuum
   Answer: C. Foliage, signs and rough ground scatter energy in many directions. (Key point: Meaning of scattering)

2. At grazing incidence the ground reflection coefficient is taken as:
   A. +1
   B. 0
   C. +0.5
   D. −1
   Answer: D. Γ ≈ −1, so the reflected wave subtracts from the direct wave. (Key point: Reflection coefficient)

3. The approximate path difference d'' − d' is:
   A. 2 ht hr / d
   B. (ht + hr)/d
   C. ht hr / d
   D. d/(2 ht hr)
   Answer: A. Expanding both square roots gives Δ ≈ 2 ht hr / d. (Key point: Path difference)

4. When θΔ/2 < 0.3 rad, sin(θΔ/2) is replaced by θΔ/2, giving:
   A. |E| proportional to 1/d
   B. |E| proportional to 1/d^2, so Pr = Pt Gt Gr ht^2 hr^2/d^4
   C. |E| independent of d
   D. |E| proportional to d
   Answer: B. |ETOT| ≈ k/d^2 and power goes as the field squared, i.e. 1/d^4. (Key point: Approximation for large d)

## Related in Wireless Communication

- [Two-ray model path and phase difference](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Atwo-ray-path-phase-difference) — Wireless Communication, chapter 3, asked 2×
- [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: 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
- [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

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