--- title: "MUF, critical frequency and skip (curved earth)" subject: "Propagation and Antenna" chapter: 4 key: "semester_6/propagation_and_antenna:muf-fcr-skip-curved-earth" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/muf-fcr-skip-curved-earth site: BCT Engineering --- # MUF, critical frequency and skip (curved earth) Propagation and Antenna · Chapter 4: Propagation and Radio Frequency Spectrum - Asked in: 2075 Bhadra, 2+8 marks - 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-fcr-skip-curved-earth - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-4 ## Question What is Maximum Usable Frequency (MUF)? Derive the expression of MUF, critical frequency (fcr) and skip distance assuming curved earth. ## Short answer MUF is the highest frequency reflected between two points for a given angle of incidence. f_MUF = fcr sec φ with fcr = 9√Nmax (critical frequency). For curved earth, tan φ = R sin(D/2R)/(R+h−R cos(D/2R)), and skip distance D = (2R/k)[1−√(1−2k²h/R)] with k = √((f/fcr)²−1). ## Answer #### Maximum Usable Frequency (MUF) The **MUF** is the highest frequency that is returned to earth by the ionosphere between two given points (for a given angle of incidence). Higher frequencies pass through the layer. It is related to the critical frequency fcr (highest frequency reflected at vertical incidence) by the secant law: **f_MUF = fcr sec φ**. In practice the optimum working frequency OWF ≈ 0.85 MUF is used. #### Derivation for curved earth **Assumptions:** spherical earth of radius R; layer acting as a thin reflector at virtual height h; both antennas on the ground; reflection point P above the mid-point of the path. First, the secant law. The refractive index of the layer is n = √(1 − 81N/f²). By Snell's law, a ray entering at angle φ returns where n = sin φ: ```text 1 − 81Nmax/f² = sin² φ 81Nmax/f² = cos² φ f = 9√Nmax / cos φ = fcr sec φ, fcr = 9√Nmax ``` Then the curved-earth geometry: ```text P (reflection point, height h) /|\ / | \ φ = angle of incidence at P /φ | \ ~~~~~~~~/~~~|~~~\~~~~~~~ ionosphere T / | \ R (on earth surface) \. | ./ `. |θ .' θ = D / (2R) `. | .' O (earth centre) OT = R, OP = R + h, arc TR = D ``` Step 1: Drop a perpendicular from T onto OP. Its foot Q lies at distance R cos θ from O, and TQ = R sin θ. In triangle TQP: ```text QP = (R + h) − R cos θ tan φ = TQ / QP = R sin θ / (R + h − R cos θ) ``` Step 2: Secant law for the ionosphere (from n = √(1 − 81N/f²) and Snell's law): f_MUF = fc sec φ, with fc = 9√Nmax. So: ```text f_MUF = fc √(1 + tan² φ) f_MUF = fc √{ 1 + [ R sin(D/2R) / (R + h − R cos(D/2R)) ]² } ``` Step 3: For D small compared with R, sin θ ≈ θ = D/2R and cos θ ≈ 1 − θ²/2: ```text tan φ ≈ (D/2) / (h + D²/8R) f_MUF ≈ fc √{ 1 + [ (D/2) / (h + D²/8R) ]² } ``` Step 4: Skip distance for a given frequency f. Let k = √[(f/fc)² − 1] = tan φ: ```text D/2 = k (h + D²/8R) (k/8R) D² − D/2 + k h = 0 D = (2R/k) [ 1 − √(1 − 2k²h/R) ] (smaller root) ``` Check: as R → ∞, √(1 − x) ≈ 1 − x/2, so D → 2kh = 2h√[(f/fc)² − 1], the flat-earth result. Step 5: Limit due to curvature. The largest angle φ occurs when the ray leaves the transmitter along the horizon (tangent at T): ```text sin φmax = R / (R + h) f_MUF(max) = fc sec φmax D_max (one hop) = 2R cos⁻¹[R / (R + h)] ``` For h = 300 km, R = 6370 km: sin φmax = 0.955, φmax = 72.8°, sec φmax = 3.37, D_max ≈ 3836 km. So single-hop MUF cannot exceed about 3.4 fc. #### Summary ```text fcr = 9 √Nmax f_MUF = fcr sec φ, tan φ = R sin(D/2R) / (R + h − R cos(D/2R)) D = (2R/k)[1 − √(1 − 2k²h/R)], k = √[(f/fcr)² − 1] ``` **Example:** fcr = 8 MHz, h = 300 km, f = 16 MHz: k = √3 = 1.732; D = (2 × 6370/1.732)[1 − √(1 − 2 × 3 × 300/6370)] ≈ 1125 km (the flat-earth formula gives 1039 km). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Using Snell's law n = sin(phi), the secant law is obtained as: A. f = 9 sqrt(Nmax)/cos(phi) = fcr sec(phi) B. f = 9 sqrt(Nmax) cos(phi) C. f = fcr tan(phi) D. f = 81 Nmax sec(phi) Answer: A. 1 - 81Nmax/f^2 = sin^2(phi) gives 81Nmax/f^2 = cos^2(phi). (Key point: Secant law derivation) 2. The optimum working frequency used in practice is about: A. 1.5 MUF B. 0.85 MUF C. 0.85 fcr D. MUF exactly Answer: B. It allows a margin for ionospheric variations. (Key point: OWF) 3. For curved earth, with k = sqrt[(f/fcr)^2 - 1], the skip distance is: A. D = 2kh B. D = (2R/k)[1 + sqrt(1 - 2k^2 h/R)] C. D = (2R/k)[1 - sqrt(1 - 2k^2 h/R)] D. D = kR/2h Answer: C. It is the smaller root of (k/8R)D^2 - D/2 + kh = 0. (Key point: Skip distance equation) 4. In the example (fcr = 8 MHz, h = 300 km, f = 16 MHz) the curved-earth skip distance compared with flat earth is: A. Smaller B. Equal C. Zero D. Larger (about 1125 km vs 1039 km) Answer: D. Curvature reduces the incidence angle for a given D, so the skip distance is longer. 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