---
title: "Antenna beamwidth (HPBW)"
subject: "Propagation and Antenna"
chapter: 2
key: "semester_6/propagation_and_antenna:beamwidth"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/beamwidth
site: BCT Engineering
---

# Antenna beamwidth (HPBW)

Propagation and Antenna · Chapter 2: Antenna Parameters and Arrays

- Asked in: 2077 Chaitra, 2 marks; 2076 Baisakh, 2 marks; 2070 Bhadra, 2 marks (asked 3 times)
- Video: [Antenna Beamwidth - HPBW and FNBW](https://www.youtube.com/watch?v=aqSGyrhH0AU) — Padmasri Naban
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/beamwidth
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-2

## Question

Explain the beamwidth (half power beamwidth) of an antenna.

## Short answer

HPBW is angular width between half-power (-3 dB) points of main lobe where field is 0.707 of maximum. FNBW ≈ 2×HPBW. Narrow beamwidth gives high directivity: D ≈ 41253/(θE°×θH°). Examples: short dipole 90°, λ/2 dipole 78°.

## Answer

**Beamwidth** is the angular width of the main lobe of the radiation pattern. The **Half Power Beamwidth (HPBW)** is the angle between the two directions in the main lobe where the radiated power falls to **half (−3 dB)** of its maximum, i.e. where the field is 1/√2 = 0.707 of maximum.

```text
        max
       .---.     -3 dB
  0.707/  |  \0.707
      /<-HPBW->\
     /    |     \
  null    |      null
     <---FNBW---->
```

- **FNBW** (first null beamwidth): angle between the first nulls; FNBW ≈ 2 × HPBW.
- Narrow beamwidth means high directivity: D ≈ 41253/(θE° × θH°).
- Examples: short dipole 90°, λ/2 dipole 78°.

## Check yourself

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

1. What is the half power beamwidth?
   A. Angle between the first nulls
   B. Angle between the points where power falls to half (-3 dB) of maximum in the main lobe
   C. Width of the side lobe
   D. Total pattern width of 180°
   Answer: B. HPBW spans the directions where power is half the maximum, i.e. field 0.707 of the maximum. (Key point: HPBW definition)

2. At the half-power points, the field strength equals what fraction of maximum?
   A. 0.5
   B. 0.25
   C. 0.707
   D. 0.1
   Answer: C. Half power corresponds to 1/√2 = 0.707 of the maximum field. (Key point: Field level at HPBW)

3. How are beamwidth and directivity related?
   A. Narrower beamwidth means lower directivity
   B. Beamwidth has no effect on directivity
   C. D ≈ θE × θH
   D. Narrower beamwidth means higher directivity, D ≈ 41253/(θE° θH°)
   Answer: D. A narrower beam concentrates power, giving higher directivity. (Key point: Beamwidth and directivity)

## Related in Propagation and Antenna

- [Gain, efficiency and beamwidth](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Again-efficiency-beamwidth) — Propagation and Antenna, chapter 2
- [Maximum power transfer theorem in antennas](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Amaximum-power-transfer-theorem) — Propagation and Antenna, chapter 1, asked 4×
- [Short note: Effective isotropic radiated power](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Aeirp-short-note) — Propagation and Antenna, chapter 2, asked 4×
- [Received power: 50 km, 10 GHz link](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Areceived-power-50km-10ghz) — Propagation and Antenna, chapter 5, asked 2×
- [Received power: 0.5 km, 1 GHz circuit](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Areceived-power-0-5km-1ghz) — Propagation and Antenna, chapter 5, asked 2×
- [Short note: Power flux density](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Apower-flux-density-short-note) — Propagation and Antenna, chapter 2

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