--- title: "Critical frequency; MUF and skip distance (flat earth)" subject: "Propagation and Antenna" chapter: 4 key: "semester_6/propagation_and_antenna:critical-frequency-skip-muf-flat" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/critical-frequency-skip-muf-flat site: BCT Engineering --- # Critical frequency; MUF and skip distance (flat earth) Propagation and Antenna · Chapter 4: Propagation and Radio Frequency Spectrum - Asked in: 2081 Chaitra, 2+6 marks; 2080 Chaitra, 2+7 marks (asked 2 times) - Video: [Sky Wave Propagation 2 | Critical Frequency I MUF I Skip Distance | Optimum Frequency IAntenna Hindi](https://www.youtube.com/watch?v=VqgcHBtYorM) — Dopamine - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/critical-frequency-skip-muf-flat - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-4 ## Question Define the critical frequency (fcr). Derive an expression for the relation between skip distance and maximum usable frequency (fmuf) in SW propagation assuming that the Earth is flat. ## Short answer Critical frequency fcr = 9√Nmax is the maximum vertical-incidence frequency reflected by a layer. For flat earth, with virtual height h, MUF = fcr √(1 + D²/4h²) and skip distance D = 2h √[(fMUF/fcr)² − 1]. ## Answer #### Critical frequency The **critical frequency (fcr)** of an ionospheric layer is the highest frequency that is reflected back to earth when the wave is sent **vertically upward**. Higher frequencies at vertical incidence pass through the layer. It depends on the maximum electron density of the layer: ```text fcr = 9 √Nmax (fcr in Hz, Nmax in electrons/m³) ``` Typical values: E ≈ 3–4 MHz, F2 ≈ 5–12 MHz. #### Relation between skip distance and MUF (flat earth) **Skip distance (D)** is the shortest distance from the transmitter at which a sky wave of a given frequency returns to earth. The **maximum usable frequency (f_MUF)** is the highest frequency that returns to earth for a given distance (angle of incidence). Assumptions: earth and ionosphere are flat; reflection takes place at virtual height *h*; the path is symmetric (both antennas on ground). ```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] ``` **Result** ```text f_MUF = fcr √(1 + D²/4h²) D = 2h √[(f_MUF/fcr)² − 1] ``` - For f ≤ fcr, D = 0 (no skip zone). - As frequency rises above fcr, the skip distance grows. **Example:** h = 300 km, fcr = 8 MHz, f = 16 MHz: D = 2 × 300 × √(4 − 1) = 1039 km. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the critical frequency? A. The highest frequency reflected at vertical incidence, fcr = 9√Nmax B. The lowest frequency absorbed by D layer C. The highest frequency at any incidence D. The frequency at which skip distance is largest Answer: A. Above fcr, a vertically sent wave passes through the layer. (Key point: Definition of critical frequency) 2. At the point of reflection, what is the refraction angle, and what does Snell's law give? A. 90°; sin θi = n = √(1 − 81Nmax/f²), so f = 9√Nmax / cos θi B. 0°; sin θi = 1 C. 45°; f = fc D. 90°; sin θi = 81N/f² Answer: A. The ray turns horizontal, giving cos²θi = 81Nmax/f², so f = fc sec θi. (Key point: Condition at reflection) 3. From the flat-earth geometry, what is sec θi? A. √(1 + D²/4h²) B. √(1 + D²/h²) C. D/2h D. √(1 − D²/4h²) Answer: A. tan θi = (D/2)/h, so sec θi = √(1 + tan²θi) = √(1 + D²/4h²). (Key point: Flat-earth geometry) 4. What is the skip distance in terms of MUF? A. D = 2h √[(fMUF/fcr)² − 1] B. D = h √[(fMUF/fcr)² + 1] C. D = 2h (fMUF/fcr) D. D = h/2 √[1 − (fMUF/fcr)²] Answer: A. Solving fMUF/fcr = √(1 + D²/4h²) gives D = 2h√[(fMUF/fcr)² − 1]. Example: h = 300 km, fcr = 8 MHz, f = 16 MHz gives 1039 km. 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