---
title: "Short note: Power flux density"
subject: "Propagation and Antenna"
chapter: 2
key: "semester_6/propagation_and_antenna:power-flux-density-short-note"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/power-flux-density-short-note
site: BCT Engineering
---

# Short note: Power flux density

Propagation and Antenna · Chapter 2: Antenna Parameters and Arrays

- Asked in: 2080 Chaitra, 3 marks
- Video: [Antenna Radiation Density & Radiated Power in Antenna and Wave Propagation by Engineering Funda](https://www.youtube.com/watch?v=T7lxZRWiQek) — 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/power-flux-density-short-note
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-2

## Question

Write a short note on power flux density (PFD).

## Short answer

Power flux density is the power flowing per unit area normal to propagation, in W/m², and equals the magnitude of the average Poynting vector. For isotropic radiation PFD = Pt/4πd²; with gain, PFD = EIRP/4πd² = E²rms/120π. For EIRP 1 kW at 10 km, PFD = 7.96×10⁻⁷ W/m² and E_rms = 17.3 mV/m.

## Answer

**Power flux density (PFD)** is the power flowing per unit area normal to the direction of propagation, in W/m² (or dBW/m²). It is the magnitude of the average Poynting vector.

For an isotropic radiator, power Pt spreads uniformly over a sphere of area 4πd²:

```text
PFD = Pt / (4π d²)
With antenna gain:  PFD = Pt Gt / (4π d²) = EIRP / (4π d²)
In terms of field:  PFD = E²rms / η0 = E²rms / 120π
```

**Example:** EIRP = 1 kW, d = 10 km:
PFD = 1000/(4π × 10⁸) = 7.96 × 10⁻⁷ W/m² (−61.0 dBW/m²), giving E_rms = √(PFD × 120π) = 17.3 mV/m.

Uses: received power Pr = PFD × Ae; ITU sets PFD limits for satellites to protect terrestrial services. PFD falls as 1/d² (inverse square law).

## Check yourself

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

1. What is power flux density?
   A. Power per unit length, in W/m
   B. Power per unit area normal to the propagation direction, in W/m²
   C. Total radiated power, in W
   D. Field strength in V/m
   Answer: B. It is the magnitude of the average Poynting vector. (Key point: Definition and unit)

2. For EIRP radiated from distance d, what is PFD?
   A. EIRP/(4πd²)
   B. EIRP/(2πd)
   C. EIRP·4πd²
   D. EIRP/d
   Answer: A. Power spreads over a sphere of area 4πd². (Key point: Isotropic formula)

3. If the distance from the transmitter is doubled, the PFD becomes
   A. Half
   B. Double
   C. One quarter
   D. One eighth
   Answer: C. PFD falls as 1/d². (Key point: Inverse square law)

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