--- title: "Free space propagation and Friis derivation" subject: "Propagation and Antenna" chapter: 5 key: "semester_6/propagation_and_antenna:free-space-friis-complete" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/free-space-friis-complete site: BCT Engineering --- # Free space propagation and Friis derivation Propagation and Antenna · Chapter 5: Propagation between Antennas - Asked in: 2080 Asoj, 2+4+2 marks; 2075 Baisakh, 2+8 marks (asked 2 times) - 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/free-space-friis-complete - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-5 ## Question What is free space propagation (communication)? Derive the mathematical expression of Friis free space equation. Also, find the expression of free space path loss. ## Short answer Free space propagation is radio communication with clear line-of-sight and no reflections, refraction, or obstacles; only spherical spreading reduces power density as 1/d^2. Friis equation: Pr = Pt*Gt*Gr*(lambda/(4*pi*d))^2. Free space path loss LFS = (4*pi*d/lambda)^2 = 32.44 + 20log(d_km) + 20log(f_MHz) dB. ## Answer **Free space propagation** is radio communication between a transmitter and receiver that have a clear, unobstructed line-of-sight path, with no reflection, refraction, absorption or diffraction from the earth, atmosphere or objects. The only effect on the signal is the spreading of energy over a larger and larger sphere, so the power density falls as 1/d². Satellite links and terrestrial microwave LOS links (with a clear first Fresnel zone) are close to free space. #### Assumptions of the free space model - Antennas are in each other's far field. - The medium is lossless, homogeneous and isotropic (εr = μr = 1). - Polarisations are matched and the antennas are matched to their loads. - No obstacles in the first Fresnel zone, and no ground reflection. #### Derivation of the Friis free space 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) ``` This shows that the received power falls with the **square of distance** (6 dB per doubling of *d*) and, for fixed gains, with the square of frequency. #### Free space path loss The **free space path loss** *LFS* is the ratio of transmitted to received power when both antennas are isotropic (Gt = Gr = 1), i.e. the loss due to spreading of the wave alone: ```text LFS = Pt/Pr = (4πd/λ)² LFS(dB) = 20 log(4πd/λ) = 20 log(4π) + 20 log(d/λ) = 21.98 + 20 log(d/λ) ≈ 22 + 20 log(d/λ) dB ``` Using λ = c/f, with *d* in km and *f* in MHz: ```text LFS(dB) = 32.44 + 20 log d(km) + 20 log f(MHz) ``` (With *f* in GHz the constant becomes 92.44.) The loss grows by 6 dB each time the distance or the frequency is doubled. **Example:** At f = 10 GHz (λ = 0.03 m) and d = 50 km, LFS = 20 log(4π × 50000/0.03) = 146.4 dB. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Free space propagation means: A. Propagation with ground reflection B. A clear unobstructed LOS path with only spreading loss C. Propagation via the ionosphere D. Propagation in a waveguide Answer: B. No reflection, refraction, absorption or diffraction; only spherical spreading. (Key point: Free space definition) 2. Which relation between gain and effective aperture is used in the derivation? A. Aer = Gr lambda^2/(4 pi) B. Aer = Gr/lambda C. Aer = 4 pi Gr/lambda D. Aer = Gr lambda Answer: A. Substituting into Pr = P_d Aer produces the Friis equation. (Key point: Effective aperture) 3. How does received power change when distance doubles? A. Falls 3 dB B. Falls 6 dB C. Falls 12 dB D. Unchanged Answer: B. Pr is proportional to 1/d^2, so doubling d loses 6 dB. (Key point: Distance dependence) 4. With d in km and f in MHz, LFS in dB is: A. 92.44 + 20 log d + 20 log f B. 32.44 + 20 log d + 20 log f C. 22 + 20 log d D. 32.44 + 40 log d Answer: B. The constant is 32.44 for km and MHz (92.44 for GHz). (Key point: LFS formula) ## 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× - [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 - [Friis, path loss and LOS distance](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Afriis-path-loss-los-distance) — Propagation and Antenna, chapter 5 - [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 - [Transmission loss and Friis equation](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Atransmission-loss-friis) — Propagation and Antenna, chapter 5 - [Path loss in radio wave propagation](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Apath-loss-derivation) — Propagation and Antenna, chapter 5, asked 2× ## Related in other subjects - [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 - [Numerical: cordless phone free space path loss](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acordless-phone-path-loss) — Wireless Communication, chapter 3, asked 2× - [Two-ray model advantages and path loss derivation](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Atwo-ray-advantages-path-loss) — Wireless Communication, chapter 3 - [Okumura model and median path loss numerical](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aokumura-model-median-path-loss) — Wireless Communication, chapter 3