---
title: "Numerical: received power, E-field, flux density, voltage"
subject: "Wireless Communication"
chapter: 3
key: "semester_7/wireless_communication:free-space-received-power-e-field"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/free-space-received-power-e-field
site: BCT Engineering
---

# Numerical: received power, E-field, flux density, voltage

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

- Asked in: 2080 Baisakh, 8 marks; 2078 Chaitra, 2+2+2+2 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/wireless_communication/free-space-received-power-e-field
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3

## Question

Assume free space propagation, a receiver is located 10 km away from a 50 W transmitter. The carrier frequency is 900 MHz, antenna gain at transmitter and receiver are 1 and 2 respectively. Find:
i) The power received at the receiver
ii) The magnitude of E-field at the receiver antenna
iii) The power flux density
iv) The rms voltage applied to the receiver input.
The receiver antenna has a purely real impedance of 50 Ω and is matched to the receiver.

## Short answer

Use Friis equation with lambda=0.333 m: received power is 7.04e-10 W (-61.5 dBm), E-field magnitude 3.87 mV/m, power flux density 3.98e-8 W/m2, and rms voltage at receiver input is 0.375 mV (0.188 mV across matched terminals).

## Answer

**Friis free space equation:** Pr = Pt·Gt·Gr·λ² / [(4π)²d²]

Given: Pt = 50 W, d = 10 km, f = 900 MHz, Gt = 1, Gr = 2, R = 50 Ω (matched).

```text
λ = c/f = 3×10⁸ / 900×10⁶ = 0.3333 m
```

#### i) Received power
```text
Pr = 50 × 1 × 2 × (0.3333)² / [(4π)² × (10⁴)²]
   = 7.04 × 10⁻¹⁰ W
   = 10 log(7.04×10⁻¹⁰) = −91.53 dBW = −61.53 dBm
```

#### ii) Magnitude of E-field
Effective aperture of the receiving antenna:

```text
Ae = Gr·λ²/(4π) = 2 × 0.1111 / 12.566 = 0.01768 m²
```
Since Pr = |E|²·Ae / 120π:

```text
|E| = √(Pr × 120π / Ae)
    = √(7.04×10⁻¹⁰ × 376.99 / 0.01768)
    = 3.87 × 10⁻³ V/m = 3.87 mV/m
```

#### iii) Power flux density
```text
Pd = |E|²/120π = Pr/Ae
   = 7.04×10⁻¹⁰ / 0.01768
   = 3.98 × 10⁻⁸ W/m²
```

#### iv) RMS voltage at the receiver input
The antenna is modelled as a voltage source V (its open-circuit voltage) in series with Rant = 50 Ω. With a matched receiver, V divides equally between the antenna resistance and the receiver, so Pr = (V/2)²/Rant = V²/(4Rant):

```text
V = √(4 × Rant × Pr) = √(4 × 50 × 7.04×10⁻¹⁰)
  = 3.75 × 10⁻⁴ V = 0.375 mV
```
This V = 0.375 mV is the value the standard textbook (Rappaport) gives as the rms voltage applied to the receiver input. The voltage actually measured across the matched receiver terminals is V/2 ≈ 0.188 mV.

**Answer: Pr = 7.04×10⁻¹⁰ W (−61.5 dBm); |E| = 3.87 mV/m; Pd = 3.98×10⁻⁸ W/m²; Vrms = 0.375 mV (open-circuit antenna voltage; 0.188 mV across the receiver terminals).**

## Check yourself

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

1. The Friis free space equation is:
   A. Pr = Pt Gt Gr / d^4
   B. Pr = Pt Gt Gr lambda^2 / ((4 pi)^2 d^2)
   C. Pr = Pt lambda / d
   D. Pr = Pt Gt Gr lambda / (4 pi d)
   Answer: B. Received power falls as 1/d^2. (Key point: Friis equation)

2. For Pt = 50 W, Gt = 1, Gr = 2, d = 10 km, 900 MHz, Pr is about:
   A. 3.5 x 10^-8 W
   B. 7.04 x 10^-6 W
   C. 7.04 x 10^-12 W
   D. 7.04 x 10^-10 W
   Answer: D. Pr = 50*2*0.1111/(157.9*10^8) = 7.04e-10 W (-61.5 dBm). (Key point: Received power value)

3. The E-field is found from:
   A. Pr = |E| 120 pi
   B. Pr = |E|^2 / Ae
   C. Pr = |E|^2 Ae / (120 pi)
   D. Pr = |E| Ae
   Answer: C. Pr = |E|^2 Ae/120pi with Ae = Gr lambda^2/4pi gives 3.87 mV/m. (Key point: E-field)

4. Power flux density can be found as:
   A. Pr x Gr
   B. Pr / Ae
   C. Pr / Gr
   D. Pr x Ae
   Answer: B. Pd = Pr/Ae = 3.98e-8 W/m^2. (Key point: Power flux density)

5. For a matched antenna of 50 ohm, the rms voltage is found from:
   A. V = Pr*Rant
   B. V = sqrt(Rant*Pr)
   C. V = Pr/Rant
   D. V = sqrt(4*Rant*Pr)
   Answer: D. Pr = V^2/(4 Rant), so V = 0.375 mV; the receiver input sees half. (Key point: RMS voltage)

## Related in Wireless Communication

- [Numerical: free-space received power (165 W, 325 MHz)](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Afree-space-received-power-165w) — Wireless Communication, chapter 3
- [Numerical: 50 W in dBm/dBW and received power](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Atransmit-power-dbm-free-space) — 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: received power with cable and mismatch loss](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Areceived-power-coax-mismatch) — Wireless Communication, chapter 3
- [Numerical: Okumura path loss, EIRP and received power](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aokumura-eirp-received-power) — Wireless Communication, chapter 3, asked 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

## Related in other subjects

- [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: 15.5 km, 3.4 GHz, dish receiver](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Amax-received-power-15-5km) — Propagation and Antenna, chapter 5
- [Received power: 100 MHz, 0.75 km](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Areceived-power-100mhz-0-75km) — Propagation and Antenna, chapter 5
- [Free space path loss: 6 GHz, 10,500 km](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Afspl-derivation-6ghz-10500km) — Propagation and Antenna, chapter 5