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MUF and critical frequency relation

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.

Define maximum usable frequency. Derive the relation between the critical frequency and maximum usable frequency.

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.

                 ~~~~~~~ ionised layer ~~~~~~~
                      \  normal  /
                   θi  \   |    /
                        \  |   /
   Tx ___________________\_|__/_______ Rx

The refractive index of a layer with electron density N (per m³) at frequency f (Hz) is

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:

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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