---
title: "Short note: Effective antenna height"
subject: "Propagation and Antenna"
chapter: 5
key: "semester_6/propagation_and_antenna:effective-antenna-height-short-note"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/effective-antenna-height-short-note
site: BCT Engineering
---

# Short note: Effective antenna height

Propagation and Antenna · Chapter 5: Propagation between Antennas

- Asked in: 2078 Chaitra, 3 marks
- Video: [Effective Height of Antenna (Definition, Basics, Physical Meaning, Formula & Calculation) Explained](https://www.youtube.com/watch?v=5rmMBBeSug0) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/effective-antenna-height-short-note
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-5

## Question

Write a short note on effective antenna height.

## Short answer

Effective height is the equivalent height of a uniform-current antenna that gives the same maximum radiated field, defined by heI0=∫I(z)dz or he=Voc/E. It is less than physical height; for a short monopole he=h/2, for a quarter-wave monopole he=λ/2π. It sets ground-wave field strength and is used in LOS/plane-earth calculations.

## Answer

**Effective height** (effective length) of an antenna is the height of an imaginary antenna carrying a **uniform** current equal to the actual feed-point current that would radiate the same field in the direction of maximum radiation. It converts the real, non-uniform current distribution into an equivalent uniform one.

#### Definition

```text
h_e · I0 = ∫₀^h I(z) dz      (transmitting)

h_e = V_oc / E               (receiving)
```

where I0 is the base current, I(z) the current along the antenna, V_oc the open-circuit voltage induced and E the incident field. The effective height is always less than the physical height because current tapers to zero at the tip.

#### Typical values

| Antenna | Physical height | Effective height |
|---|---|---|
| Short monopole (triangular current) | h | h/2 |
| Quarter-wave monopole | λ/4 | λ/2π ≈ 0.64h |
| Half-wave dipole (effective length) | λ/2 | λ/π |
| Top-loaded monopole | h | close to h |

#### Use in propagation

- Ground wave field strength from a vertical antenna: E = 120π·h_e·I/(λd) V/m. A larger h_e gives a stronger field, which is why MF broadcast towers use **top loading** to make the current nearly uniform.
- In space-wave (LOS) links, "effective antenna height" also means the height of the antenna above the average (effective) ground level of the reflecting area, not above the local foot of the tower. This height is used in d = 4.12(√ht + √hr) km and in the plane-earth path loss formula.

## Check yourself

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

1. The effective height of a transmitting antenna is the height of an imaginary antenna that:
   A. Radiates in all directions equally
   B. Carries a uniform current equal to the feed-point current and radiates the same field
   C. Has the same physical length but zero loss
   D. Is placed at the average ground level
   Answer: B. It replaces the real non-uniform current by an equivalent uniform one giving the same maximum-direction field. (Key point: Definition)

2. Why is the effective height of a real antenna less than its physical height?
   A. The wavelength is too long
   B. Feeder losses reduce the height
   C. Ground absorbs part of the field
   D. Current tapers towards zero at the tip
   Answer: D. Because current falls to zero at the tip, the equivalent uniform-current length is shorter. (Key point: Physical vs effective)

3. What is the effective length of a half-wave dipole?
   A. λ/4
   B. λ/π
   C. h/2
   D. λ/2
   Answer: B. A half-wave dipole has effective length λ/π. (Key point: Typical value)

4. Why do MF broadcast towers use top loading?
   A. To make current nearly uniform and raise the effective height
   B. To reduce the physical height of the tower
   C. To lower the frequency
   D. To make the radiation pattern omnidirectional in elevation only
   Answer: A. Top loading evens out the current so h_e approaches h, giving stronger field E = 120π·h_e·I/(λd). (Key point: Top loading)

## Related in Propagation and Antenna

- [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×
- [Virtual height of a layer](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Avirtual-height) — Propagation and Antenna, chapter 4, asked 3×
- [MUF range numerical: 400 km reflection height](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Amuf-range-numerical-400km) — Propagation and Antenna, chapter 4

## Related in other subjects

- [Effective height, thickness and slenderness ratio](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Aeffective-height-length-thickness-slenderness) — Design of Timber and Masonry Structures, chapter 5