--- title: "MUF and skip distance (flat earth)" subject: "Propagation and Antenna" chapter: 4 key: "semester_6/propagation_and_antenna:muf-skip-distance-flat" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/muf-skip-distance-flat site: BCT Engineering --- # MUF and skip distance (flat earth) Propagation and Antenna · Chapter 4: Propagation and Radio Frequency Spectrum - Asked in: 2078 Chaitra, 8 marks; 2076 Bhadra, 7 marks (asked 2 times) - Video: [Maximum Usable Frequency (MUF) (Definition, Basics, Derivation, Formula & Calculation) Explained](https://www.youtube.com/watch?v=0Houq2d-Jtk) — 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/muf-skip-distance-flat - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-4 ## Question Define and derive the formula for maximum usable frequency and skip distance assuming earth is perfectly flat. ## Short answer MUF is the highest frequency returned for a given path. For flat earth: MUF = fcr sqrt(1 + D^2/(4h^2)), where D is path length and h virtual height; skip distance D = 2h sqrt((MUF/fcr)^2 - 1). ## Answer #### Definitions - **Maximum usable frequency (MUF):** the highest frequency that is returned to earth by the ionosphere for a given angle of incidence (that is, for a given distance between two points). Frequencies above the MUF pass through the layer. - **Skip distance (D):** the minimum distance from the transmitter at which a sky wave of a given frequency returns to earth. - **Critical frequency (fc):** the MUF for vertical incidence, fc = 9√Nmax. #### Derivation (earth assumed perfectly flat) Assumptions: flat earth and flat ionospheric layer; sharp reflection at virtual height *h*; transmitter and receiver at ground level; symmetric path. ```text ionosphere (virtual height h) ~~~~~~~~~~~~~~~~~~ P ~~~~~~~~~~~~~~~~~~ / | \ /θi| \ / |h \ / | \ flat earth T-----M-----R <--- D ----> TM = MR = D/2, θi = angle of incidence ``` Refractive index of an ionised layer (electron density N per m³, frequency f in Hz): ```text n = √(1 − 81N / f²) ``` By Snell's law, a wave entering at angle θi turns back where the ray becomes horizontal (refraction angle 90°): ```text sin θi = n = √(1 − 81Nmax / f²) cos² θi = 81Nmax / f² f = 9√Nmax / cos θi ``` For vertical incidence (θi = 0) this gives the critical frequency fc = 9√Nmax, so ```text f_MUF = fc sec θi (secant law) ``` From the flat-earth geometry (triangle TMP): ```text tan θi = (D/2) / h sec θi = √(1 + tan² θi) = √(1 + D² / 4h²) f_MUF = fc √(1 + D² / 4h²) = fc √(4h² + D²) / (2h) ``` Solving for the skip distance: ```text (f_MUF / fc)² = 1 + D² / 4h² D = 2h √[(f_MUF / fc)² − 1] ``` #### Final formulas ```text f_MUF = fc sec θi = fc √(1 + D² / 4h²) D = 2h √[(f_MUF / fc)² − 1] ``` **Remarks** - MUF increases with distance (larger θi), so long links can use higher frequencies. - The flat-earth result is good for distances up to about 500 km; for longer paths the earth's curvature limits θi, and single-hop MUF reaches about 3–4 × fc. - In practice the optimum working frequency OWF ≈ 0.85 MUF is used, to allow for ionospheric changes. **Example:** fc = 7 MHz, h = 300 km, D = 1000 km: f_MUF = 7 × √(1 + 1000²/(4 × 300²)) = 7 × √3.778 = 13.6 MHz. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The maximum usable frequency is: A. The lowest frequency absorbed by the D layer B. The highest frequency returned to earth by the ionosphere for a given angle of incidence (distance) C. The highest frequency reflected only at vertical incidence D. The frequency that gives zero skip distance Answer: B. Frequencies above the MUF pass through the layer; the vertical-incidence case is specifically the critical frequency. (Key point: Definition of MUF) 2. The secant law for the MUF states: A. f_MUF = fc cos(theta_i) B. f_MUF = fc tan(theta_i) C. f_MUF = fc sec(theta_i) D. f_MUF = fc / sec(theta_i) Answer: C. From cos^2(theta_i) = 81Nmax/f^2, f = 9 sqrt(Nmax)/cos(theta_i) = fc sec(theta_i). (Key point: Secant law) 3. In the flat-earth result f_MUF = fc sqrt(1 + D^2/4h^2), what happens to the MUF as the distance D increases? A. It increases, because the angle of incidence becomes larger B. It decreases, because the path gets longer C. It stays equal to fc D. It becomes independent of h only Answer: A. Larger D means larger theta_i and larger sec(theta_i), so longer links can use higher frequencies. (Key point: MUF vs distance) 4. For fc = 7 MHz, h = 300 km and D = 1000 km, the MUF is about: A. 7 MHz B. 23.3 MHz C. 9.5 MHz D. 13.6 MHz Answer: D. f_MUF = 7 x sqrt(1 + 1000^2/(4 x 300^2)) = 7 x sqrt(3.778), about 13.6 MHz. (Key point: Worked example) ## Related in Propagation and Antenna - [Critical frequency; MUF and skip distance (flat earth)](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Acritical-frequency-skip-muf-flat) — Propagation and Antenna, chapter 4, asked 2× - [MUF, critical frequency and skip (curved earth)](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Amuf-fcr-skip-curved-earth) — Propagation and Antenna, chapter 4 - [Skip distance and MUF (curved earth)](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Askip-muf-curved-earth) — Propagation and Antenna, chapter 4 - [Critical frequency and skip distance (flat earth)](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Acritical-frequency-skip-distance-flat) — Propagation and Antenna, chapter 4, asked 2× - [Skip distance derivation (flat earth)](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Askip-distance-flat-earth) — Propagation and Antenna, chapter 4, asked 2× - [Refractive index, critical frequency and MUF](https://www.bctengineer.com/question.md?key=semester_6%2Fpropagation_and_antenna%3Arefractive-index-fc-muf) — Propagation and Antenna, chapter 4