---
title: "Friis, path loss and LOS distance"
subject: "Propagation and Antenna"
chapter: 5
key: "semester_6/propagation_and_antenna:friis-path-loss-los-distance"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/friis-path-loss-los-distance
site: BCT Engineering
---

# Friis, path loss and LOS distance

Propagation and Antenna · Chapter 5: Propagation between Antennas

- Asked in: 2076 Bhadra, 4+3 marks
- Video: [Friis Transmission Formula (Basics, Formula & Derivation) Explained](https://www.youtube.com/watch?v=Op7bBz6qbwY) — 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/friis-path-loss-los-distance
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-5

## Question

Derive Friis transmission equation and path loss in case of free space propagation and find out distance of line of sight as: d = 3.57(√ht + √hr) km.

## Short answer

Friis equation is Pr=PtGtGrλ²/(4πd)²; free-space loss is LFS=(4πd/λ)²=32.44+20log d(km)+20log f(MHz). From the right triangle (R+h)²=R²+d² with R=6370 km, the radio horizon is d=√(2Rh)=3.57√h km; with Tx and Rx heights, d=3.57(√ht+√hr) km (4.12 with k=4/3 refraction).

## Answer

#### Friis transmission equation

Consider a transmitter of power *Pt* feeding an antenna of gain *Gt*, and a receiving antenna of gain *Gr* (effective aperture *Aer*) at distance *d* in free space. Assume far field, matched polarisation, matched loads and no obstacles.

1. An isotropic source spreads *Pt* uniformly over a sphere of area 4πd², so the power density at distance *d* is
   P_iso = Pt / (4πd²)  (W/m²)
2. A directive antenna with gain *Gt* increases the density in its main direction by *Gt*:
   P_d = Pt·Gt / (4πd²)
3. The receiving antenna collects the power falling on its effective aperture:
   Pr = P_d · Aer = Pt·Gt·Aer / (4πd²)
4. Effective aperture and gain are related by Aer = Gr·λ² / (4π). Substituting:

```text
Pr = Pt·Gt·(Gr·λ²/4π) / (4πd²)

Pr = Pt·Gt·Gr·λ² / (4πd)²

Pr/Pt = Gt·Gr·(λ / 4πd)²
```

This is the **Friis transmission equation**. In terms of the two effective apertures (using Gt = 4πAet/λ²) it can also be written as Pr/Pt = Aet·Aer / (λ²d²).

In decibels:

```text
Pr(dBW) = Pt(dBW) + Gt(dB) + Gr(dB) − 20 log(4πd/λ)
```

#### Path loss in free space

With isotropic antennas (Gt = Gr = 1):

```text
LFS = Pt/Pr = (4πd/λ)²
LFS(dB) = 20 log(4πd/λ) = 22 + 20 log(d/λ)
        = 32.44 + 20 log d(km) + 20 log f(MHz)
```

#### Line-of-sight distance d = 3.57(√ht + √hr) km

Because of the curvature of the earth, a straight ray from an antenna of height *h* touches the earth at the radio horizon.

```text
          Tx
          |\   ht
          | \
          |  \  d1
   -------+---*  <- tangent point
     \    |   /
    R \   |  /  R
       \  | /
         O  (centre of earth)
```

From the right-angled triangle (radius R to the tangent point is perpendicular to the ray):

```text
(R + ht)² = R² + d1²
d1² = 2R·ht + ht²
d1 ≈ √(2R·ht)          (ht ≪ R)
```

With R = 6370 km = 6.37×10⁶ m and *h* in metres:

```text
d1 = √(2 × 6.37×10⁶ × ht) m
   = 3569√ht m
   = 3.57√ht km
```

Similarly for the receiving antenna, d2 = 3.57√hr km. The total LOS distance is

```text
d = d1 + d2 = 3.57(√ht + √hr) km   (ht, hr in m)
```

This is the optical (geometric) horizon. Allowing for standard atmospheric refraction with effective earth radius kR (k = 4/3), the radio horizon becomes d = 4.12(√ht + √hr) km.

## Check yourself

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

1. What gives the power density 1/(4πd^2) at distance d from an isotropic source?
   A. Power absorbed by the atmosphere
   B. Power spread uniformly over a sphere of area 4πd^2
   C. Power spread over a circle of area πd^2
   D. Power concentrated in a narrow beam
   Answer: B. An isotropic source radiates equally in all directions, so Pt is divided by the sphere area 4πd^2. (Key point: Isotropic spreading)

2. In deriving the Friis equation, which step introduces the receiving gain Gr?
   A. Replacing the effective aperture Aer by Gr·λ^2/4π
   B. Squaring the ratio λ/d
   C. Dividing Pt by the sphere area 4πd^2
   D. Multiplying the power density by Gt
   Answer: A. The received power is density × Aer, and Aer = Gr·λ^2/(4π) links aperture to gain, which brings Gr into the equation. (Key point: Friis steps)

3. With isotropic antennas the free space loss is:
   A. 4πd/λ
   B. (4πd/λ)^2
   C. Gt·Gr(4πd/λ)^2
   D. (λ/4πd)^2
   Answer: B. LFS = Pt/Pr = (4πd/λ)^2 when Gt = Gr = 1. (Key point: Free space loss)

4. How does the geometric horizon distance d1 = 3.57 sqrt(h) km follow from the geometry?
   A. From (R+h)^2 = R^2 + d1^2 with h much less than R, so d1 = sqrt(2Rh)
   B. From d1 = R·h
   C. From Snell's law at the ground
   D. From d1 = h/R
   Answer: A. Right-angled triangle gives d1^2 = 2Rh + h^2 ≈ 2Rh; with R = 6370 km this is 3.57 sqrt(h). (Key point: Radio horizon)

5. With standard refraction (k = 4/3) the radio horizon coefficient becomes:
   A. 5.0
   B. 3.57
   C. 4.12
   D. 2.83
   Answer: C. Using kR = 4/3 R gives d = 4.12(sqrt ht + sqrt hr) km. (Key point: Effective earth radius)

## Related in Propagation and Antenna

- [Friis equation and free space path loss](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Afriis-path-loss-free-space) — Propagation and Antenna, chapter 5, asked 2×
- [10 GHz LOS link loss and Friis derivation](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Alos-link-10ghz-100km-friis) — 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
- [Free space propagation and Friis derivation](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Afree-space-friis-complete) — Propagation and Antenna, chapter 5, asked 2×
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- [Friis equation and LFS = 22 + 20 log(d/λ)](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Afriis-free-space-loss-22-20log) — Propagation and Antenna, chapter 5

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