0
\$\begingroup\$

This question is related to the following link and Longley-Rice Propagation Model

https://www.itu.int/dms_pubrec/itu-r/rec/p/R-REC-P.453-11-201507-S!!PDF-E.pdf

In the Longley-Rice Propagation Model the surface refractive index is taken into account and it also discussed how it is dependence on height. The question is: how does it affect different frequencies? I could not find any literature explaining its characteristic related to frequency? Any suggestion how to understand the effect of earth surface refractive affect RF propagation,?

\$\endgroup\$
4
\$\begingroup\$

The paper you reference is quite thorough in giving ranges of temperature and height over which its approximations are valid, but it is totally silent on frequency. Therefore, we can assume that the refractive index is independent of frequency, at least over the commercial communication range that is the ITU's remit.

However, we know that long wave (100s kHz) and medium wave (~1MHz) signals can operate over the horizon, while VHF (~100MHz) and above operates more or less line of sight, so what's going on?

The clue is in the ratio of the signal's wavelength to the rate at which refractive index changes with height, see equation 11. A radio signal, to the extent that it propagates 'quasi-optically', does so over a 'beam' that is a few wavelengths wide, see Fresnel Zone on wikipedia for a proper definition of 'a few'. A long wavelength signal therefore experiences the medium as having a varying index across the beam width, and bends. A short wavelength signal sees a medium with negligible variation over a few wavelengths, and travels in straight lines.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ May I ask where is wavelength in 11? There is only height. I do understand your comment about bend though. \$\endgroup\$ – Creator Nov 9 '17 at 0:30
  • \$\begingroup\$ 11 is only about height. The sentence parses as 'ratio of (signal wavelength) to (index versus height, see 11)', which taken at face value is nonsense. To have a ratio, we need length to length, so what I meant and expressed poorly was the ratio of (the Fresnel height, which depends on wavelength), to a (height over which signifiicant deltaN occurs, which can be deduced from 11). \$\endgroup\$ – Neil_UK Nov 9 '17 at 5:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.