--- title: "Duct propagation; critical frequency and MUF" subject: "Propagation and Antenna" chapter: 4 key: "semester_6/propagation_and_antenna:duct-critical-frequency-muf" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/duct-critical-frequency-muf site: BCT Engineering --- # Duct propagation; critical frequency and MUF Propagation and Antenna · Chapter 4: Propagation and Radio Frequency Spectrum - Asked in: 2071 Magh, 4+4 marks - Video: [Duct Propagation in Antenna and Wave Propagation | Simplified | Super Refraction | AWP Module 6](https://www.youtube.com/watch?v=4XGi7HdfjBA) — THE BACKBENCH ENGINEERING COMMUNITY - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/duct-critical-frequency-muf - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-4 ## Question Explain the duct propagation mechanism in radio wave propagation. Define critical frequency and MUF with necessary derivations. ## Short answer Duct propagation traps waves in a layer where modified refractive index M decreases with height, enabling long range. Critical frequency fc = 9√Nmax is derived from setting refractive index zero at vertical incidence. MUF = fc secθi from Snell's law for oblique incidence, giving the highest usable frequency for a path. ## Answer #### Duct propagation mechanism **Duct propagation** occurs when a layer of the lower atmosphere has a refractive index that falls very rapidly with height, so that VHF/UHF/microwave rays bend down more than the earth curves and get trapped in the layer (the **duct**), travelling far beyond the horizon. ```text ~~~~~~~~~~~ top of duct ~~~~~~~~~~~ /\ /\ /\ /\ /\ / \ / \ / \ / \ / \ Tx===\=/====\=/====\=/====\=/====Rx ``` - **Cause:** temperature inversion and sharp decrease of humidity with height (common over sea and on calm evenings). - **Condition:** modified refractive index M = (n − 1 + h/R) × 10⁶ decreases with height (dM/dh < 0, i.e. dN/dh < −157 per km). - The trapped wave is repeatedly bent down and reflected from the ground, like a wave in a waveguide; only wavelengths small compared with duct thickness propagate. - Effects: ranges of hundreds of km at microwave frequencies, but also interference and radar false echoes; not reliable because it depends on weather. #### Critical frequency The **critical frequency (fc)** is the highest frequency reflected by a layer at vertical incidence. From n = √(1 − 81N/f²), the wave turns back at vertical incidence where n = 0: ```text 0 = 1 − 81Nmax / fc² fc = 9 √Nmax (Hz, N per m³) ``` #### Maximum usable frequency (MUF) The **MUF** is the highest frequency returned to earth for a given angle of incidence θi (given distance). A ray turns back where sin θi = n (Snell's law): ```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) ``` For flat earth with virtual height h and distance D, sec θi = √(1 + D²/4h²): ```text f_MUF = fc √(1 + D² / 4h²) ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Duct propagation occurs when: A. dN/dh is greater than -40 per km B. dN/dh is less than -157 per km (dM/dh < 0) C. The ionosphere has an F2 layer D. Frequency is below 3 MHz Answer: B. Rays curve more than the earth when dN/dh < -157 per km, trapping the wave. (Key point: Duct condition) 2. Which causes a duct? A. Temperature inversion and sharp fall of humidity with height B. Strong rain C. Sunspot activity D. Ground conductivity Answer: A. Inversions and rapid humidity decrease, common over sea and calm evenings, give the sharp index gradient. (Key point: Duct causes) 3. How is fc obtained? A. Setting n = 0 at vertical incidence at Nmax: fc = 9 sqrt(Nmax) B. Setting n = 1 C. Setting sin theta = 0.5 D. Setting N = 0 Answer: A. 1 - 81Nmax/fc^2 = 0 gives fc = 9 sqrt(Nmax). (Key point: Critical frequency) 4. For flat earth with virtual height h and distance D, f_MUF is: A. fc sqrt(1 + D^2/4h^2) B. fc (1 + D/2h) C. fc D/h D. fc sqrt(1 - D^2/4h^2) Answer: A. sec theta = sqrt(1 + D^2/4h^2) from the geometry, so MUF = fc sec theta. 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