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Duct propagation; critical frequency and MUF

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.

Explain the duct propagation mechanism in radio wave propagation. Define critical frequency and MUF with necessary derivations.

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.

   ~~~~~~~~~~~ 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:

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):

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):

f_MUF = fc √(1 + D² / 4h²)

✓ Answer checked by Claude Opus 5.5 (high effort)

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