---
title: "Short note: Fresnel zones"
subject: "Propagation and Antenna"
chapter: 5
key: "semester_6/propagation_and_antenna:fresnel-zones-short-note"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/fresnel-zones-short-note
site: BCT Engineering
---

# Short note: Fresnel zones

Propagation and Antenna · Chapter 5: Propagation between Antennas

- Asked in: 2078 Chaitra, 3 marks
- Video: [LoRa/LoRaWAN tutorial 7: Fresnel Zone](https://www.youtube.com/watch?v=HWOivbJjw7s) — Mobilefish.com
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/fresnel-zones-short-note
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-5

## Question

Write a short note on Fresnel zones.

## Short answer

Fresnel zones are ellipsoids where the via-boundary path exceeds the direct path by nλ/2; radius rn=√(nλd1d2/(d1+d2)). Adjacent zones cancel, and most energy is carried by the first zone, so at least 60% of r1 must be kept clear; grazing blockage gives about 6 dB loss.

## Answer

**Fresnel zones** are a series of ellipsoids around the direct line between a transmitter and receiver, such that a path via any point on the boundary of the *n*th zone is longer than the direct path by nλ/2.

```text
          _..--~~~~~~~~--.._     1st Fresnel zone
   Tx o<-------- d1 --->|<--- d2 ------->o Rx
          ~~--..______..--~~
                        ^ radius r1 here
```

#### Radius of the nth zone

At a point that is d1 from the transmitter and d2 from the receiver:

```text
rn = √( n·λ·d1·d2 / (d1 + d2) )
```

The first zone is largest at mid-path: r1(max) = ½√(λd), where d = d1 + d2.

#### Significance

- Secondary wavelets from adjacent zones arrive with a phase difference of 180°, so odd zones add and even zones subtract. Most of the energy reaching the receiver travels inside the **first Fresnel zone**.
- If an obstacle blocks part of the first zone, diffraction loss appears even when the optical LOS is clear. At grazing (obstacle touching LOS) the loss is about 6 dB.
- **Design rule:** keep at least 60% of the first Fresnel zone (0.6r1) clear of terrain and buildings. This sets tower heights for microwave links, with earth bulge (k = 4/3) also included.

**Example:** For d = 20 km at 6 GHz (λ = 0.05 m), r1 at mid-path = ½√(0.05 × 20000) = 15.8 m.

## Check yourself

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

1. A path through a point on the boundary of the nth Fresnel zone is longer than the direct path by:
   A. n^2λ
   B. nλ
   C. nλ/2
   D. λ/n
   Answer: C. Zone boundaries are defined by path excess of nλ/2. (Key point: Fresnel zone definition)

2. Which expression gives the radius of the nth Fresnel zone?
   A. sqrt(nλ·d1·d2/(d1+d2))
   B. sqrt(λ·d1·d2)
   C. sqrt(nλ(d1+d2)/(d1·d2))
   D. nλd1d2/(d1+d2)
   Answer: A. rn = sqrt(nλd1d2/(d1+d2)), largest at mid-path. (Key point: Zone radius)

3. What clearance of the first Fresnel zone is the design rule for microwave links?
   A. At least 10%
   B. Exactly 100% with no margin
   C. At least 6 dB
   D. At least 60%
   Answer: D. Keeping 0.6r1 clear avoids significant diffraction loss. (Key point: Design rule)

4. Why do odd and even zones act differently on the received signal?
   A. Even zones are blocked by the earth
   B. Adjacent zones differ by 180° in phase, so odd add and even subtract
   C. Odd zones have longer wavelength
   D. Even zones carry no energy
   Answer: B. Wavelets from adjacent zones are 180° out of phase. (Key point: Zone phase)