--- title: "MUF and critical frequency relation" subject: "Propagation and Antenna" chapter: 4 key: "semester_6/propagation_and_antenna:muf-critical-frequency-relation" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/muf-critical-frequency-relation site: BCT Engineering --- # MUF and critical frequency relation Propagation and Antenna · Chapter 4: Propagation and Radio Frequency Spectrum - Asked in: 2072 Magh, 2+4 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-critical-frequency-relation - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-4 ## Question Define maximum usable frequency. Derive the relation between the critical frequency and maximum usable frequency. ## Short answer MUF is the highest frequency returned by the ionosphere for a given oblique incidence. It relates to critical frequency fc=9√Nmax by the secant law: f_MUF = fc sec θi, where θi is the incidence angle, so MUF always exceeds fc and increases with path distance. ## Answer #### Maximum usable frequency The **maximum usable frequency (MUF)** is the highest frequency that is returned to earth by the ionosphere for a given angle of incidence (that is, for a given distance between transmitter and receiver). #### Relation between critical frequency and MUF **Critical frequency (fc)** is the highest frequency reflected at vertical incidence. ```text ~~~~~~~ ionised layer ~~~~~~~ \ normal / θi \ | / \ | / Tx ___________________\_|__/_______ Rx ``` The refractive index of a layer with electron density N (per m³) at frequency f (Hz) is ```text n = √(1 − 81N / f²) ``` Using Snell's law, a ray entering from free space (n = 1) at angle θi is bent until its angle of refraction is 90° (ray horizontal), where it starts to return: ```text n = √(1 − 81N / f²) (refractive index) Snell's law at return point: sin θi = n sin² θi = 1 − 81Nmax / f² cos² θi = 81Nmax / f² f = 9√Nmax / cos θi Vertical incidence (θi = 0): fc = 9√Nmax Hence f_MUF = fc sec θi (secant law) ``` So **f_MUF = fc sec θi = fc / cos θi**. This is the **secant law**. Since sec θi ≥ 1, the MUF is always higher than the critical frequency, and it increases as the path becomes longer (more oblique). For flat earth with virtual height h and distance D: sec θi = √(1 + D²/4h²), so f_MUF = fc√(1 + D²/4h²). **Example:** fc = 6 MHz, θi = 60°: f_MUF = 6 × sec 60° = 6 × 2 = 12 MHz. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The MUF is: A. The highest frequency reflected at vertical incidence B. The highest frequency returned to earth for a given angle of incidence C. The lowest frequency absorbed by the D layer D. The frequency with maximum absorption Answer: B. MUF is the highest returned frequency for a given incidence angle (distance); fc is the vertical-incidence value. (Key point: MUF definition) 2. At the return point of the ray, which condition is used? A. sin(theta_i) = n B. cos(theta_i) = n C. tan(theta_i) = n D. n = 0 at all angles Answer: A. Snell's law with refraction angle 90 gives sin theta_i = n = sqrt(1 - 81Nmax/f^2). (Key point: Derivation step) 3. What is the resulting relation? A. f_MUF = fc cos(theta_i) B. f_MUF = fc sec(theta_i) C. f_MUF = fc sin(theta_i) D. f_MUF = fc tan(theta_i) Answer: B. cos^2 theta = 81Nmax/f^2 gives f = 9sqrt(Nmax)/cos theta = fc sec theta. (Key point: Secant law) 4. If fc = 6 MHz and theta_i = 60 degrees, the MUF is: A. 3 MHz B. 6 MHz C. 12 MHz D. 10.4 MHz Answer: C. sec 60 = 2, so MUF = 12 MHz. 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