---
title: "Free space path loss: 6 GHz, 10,500 km"
subject: "Propagation and Antenna"
chapter: 5
key: "semester_6/propagation_and_antenna:fspl-derivation-6ghz-10500km"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/fspl-derivation-6ghz-10500km
site: BCT Engineering
---

# Free space path loss: 6 GHz, 10,500 km

Propagation and Antenna · Chapter 5: Propagation between Antennas

- Asked in: 2071 Magh, 5+3 marks
- Video: [Problems solved in Losses - Satellite Communication - EIRP](https://www.youtube.com/watch?v=pItOMqKiEEI) — 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/fspl-derivation-6ghz-10500km
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-5

## Question

Derive the expression for free space path loss propagation. Calculate free space path loss for 6 GHz frequency, if transmitter and receiver are 10,500 km line of sight apart.

## Short answer

Free space path loss is derived from isotropic antennas: LFS = (4πd/λ)^2, and in dB: LFS(dB) = 20 log(4πd/λ) = 32.44 + 20 log d(km) + 20 log f(MHz). For f=6 GHz, d=10500 km, λ=0.05 m, LFS = 20 log(4π×1.05e7/0.05) = 188.43 dB.

## Answer

#### Derivation of free space path loss

Free space path loss is the ratio of transmitted to received power between two isotropic antennas in free space, caused by spherical spreading of the wave.

1. An isotropic antenna radiating *Pt* produces power density at distance *d*:
   P_d = Pt / (4πd²)
2. An isotropic receiving antenna has effective aperture Ae = λ²/4π.
3. Received power:

```text
Pr = P_d · Ae = Pt/(4πd²) · λ²/(4π) = Pt·(λ/4πd)²
```

4. The path loss is therefore

```text
LFS = Pt/Pr = (4πd/λ)² = (4πdf/c)²

LFS(dB) = 20 log(4πd/λ)
        = 22 + 20 log(d/λ)
        = 32.44 + 20 log d(km) + 20 log f(MHz)
        = 92.44 + 20 log d(km) + 20 log f(GHz)
```

With real antennas of gains Gt and Gr, the received power is Pr = Pt + Gt + Gr − LFS (dB), which is the Friis equation.

#### Numerical: f = 6 GHz, d = 10,500 km

```text
λ = c/f = 3×10⁸ / 6×10⁹ = 0.05 m
4πd/λ = 4π × 1.05×10⁷ / 0.05 = 2.639×10⁹
LFS = 20 log(2.639×10⁹) = 188.43 dB

Check: 92.44 + 20 log 10500 + 20 log 6
     = 92.44 + 80.42 + 15.56 = 188.43 dB
```

**Answer: Free space path loss ≈ 188.4 dB.** (Losses of this size are met on long satellite paths at C-band.)

## Check yourself

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

1. The effective aperture of an isotropic receiving antenna is:
   A. 4π/λ^2
   B. λ/4π
   C. πλ^2
   D. λ^2/4π
   Answer: D. Ae = λ^2/4π for gain 1. (Key point: Isotropic aperture)

2. The free space path loss is:
   A. (λ/4πd)^2
   B. (4πd/λ)^2
   C. 4πd/λ
   D. Pt·Pr
   Answer: B. LFS = Pt/Pr = (4πd/λ)^2. (Key point: Path loss expression)

3. Free space loss with f in GHz and d in km is:
   A. 62.44 + 20 log d
   B. 32.44 + 20 log d + 20 log f
   C. 92.44 + 20 log d + 20 log f
   D. 22 + 20 log f
   Answer: C. The constant is 92.44 when f is in GHz. (Key point: Formula in GHz)

4. LFS at 6 GHz over 10,500 km is about:
   A. 228.4 dB
   B. 168.4 dB
   C. 188.4 dB
   D. 148.4 dB
   Answer: C. 92.44 + 80.42 + 15.56 = 188.43 dB. (Key point: Result)

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