---
title: "10 GHz LOS link loss and Friis derivation"
subject: "Propagation and Antenna"
chapter: 5
key: "semester_6/propagation_and_antenna:los-link-10ghz-100km-friis"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/los-link-10ghz-100km-friis
site: BCT Engineering
---

# 10 GHz LOS link loss and Friis derivation

Propagation and Antenna · Chapter 5: Propagation between Antennas

- Asked in: 2078 Chaitra, 5+3 marks
- Video: [Examples of Friis Transmission Formula | Antenna Parameters | Engineering Funda](https://www.youtube.com/watch?v=Q0C9ZMcrwn4) — 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/los-link-10ghz-100km-friis
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-5

## Question

A free space LOS (Line of Sight) microwave link operating at 10 GHz consists of a transmit and receive antenna each having gain of 25 dB. Calculate the path loss of the link and received power for 100 km distance. Derive Friis transmission equation.

## Short answer

Friis equation: Pr=PtGtGrλ²/(4πd)². At 10 GHz λ=0.03 m, d=100 km, LFS=152.44 dB; net loss after 25+25 dB gains is 102.44 dB. Thus Pr=Pt-102.44 dB, i.e. -102.44 dBW≈57 pW for Pt=1 W.

## Answer

#### Path loss and received power

**Given:** f = 10 GHz, Gt = Gr = 25 dB, d = 100 km. The transmitted power is not given, so the received power is found in terms of Pt and also for an assumed **Pt = 1 W (0 dBW)**.

**Wavelength:**

```text
λ = c/f = 3×10⁸ / 10¹⁰ = 0.03 m
```

**Free space path loss:**

```text
4πd/λ   = 4π × 10⁵ / 0.03 = 4.189×10⁷
LFS(dB) = 20 log(4.189×10⁷) = 152.44 dB
Check:  32.44 + 20 log 100 + 20 log 10000
      = 32.44 + 40 + 80 = 152.44 dB
```

**Net link loss including antenna gains:**

```text
L = LFS − Gt − Gr = 152.44 − 25 − 25 = 102.44 dB
```

**Received power:**

```text
Pr(dBW) = Pt(dBW) + 25 + 25 − 152.44 = Pt(dBW) − 102.44

For Pt = 1 W:
Pr = −102.44 dBW = −72.44 dBm ≈ 5.7×10⁻¹¹ W
```

**Answer: Path loss = 152.44 dB (102.44 dB after the 50 dB of antenna gain); Pr = Pt − 102.44 dB, i.e. −102.44 dBW ≈ 57 pW for Pt = 1 W.**

#### Derivation of the 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)
```

## Check yourself

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

1. For f = 10 GHz, what is the wavelength used in the path loss calculation?
   A. 0.1 m
   B. 0.3 m
   C. 0.003 m
   D. 0.03 m
   Answer: D. λ = c/f = 3×10^8 / 10^10 = 0.03 m. (Key point: Wavelength)

2. What is the free space path loss at 10 GHz over 100 km?
   A. 132.44 dB
   B. 202.44 dB
   C. 102.44 dB
   D. 152.44 dB
   Answer: D. LFS = 32.44 + 20 log 100 + 20 log 10000 = 152.44 dB. (Key point: Free space path loss)

3. With Gt = Gr = 25 dB, how is the received power related to Pt?
   A. Pr = Pt − 102.44 dB
   B. Pr = Pt + 102.44 dB
   C. Pr = Pt − 202.44 dB
   D. Pr = Pt − 152.44 dB
   Answer: A. Gains are added and loss subtracted: 25 + 25 − 152.44 = −102.44 dB. (Key point: Received power)

4. In deriving the Friis equation, which step introduces the receiving gain Gr?
   A. Dividing Pt by the sphere area 4πd^2
   B. Multiplying the power density by Gt
   C. Replacing the effective aperture Aer by Gr·λ^2/4π
   D. Squaring the ratio λ/d
   Answer: C. 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)

## Related in Propagation and Antenna

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- [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
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