Wireless, RF, inverse fourth power law vs inverse square law

[FrankJ777]

When, in wireless communications, does the inverse fourth power-law become relevant? My understanding is that is that what cause the average signal power to degrade to the forth power is cancellation from self reflections. So by my way of thinking, an LOS point to point system, like a microwave; if the receiving antenna is is mounted several Fresnel zones from the ground, that Friis equation with inverse power would apply. However, in the case of mobile communications, where a handheld receiver is used, and might be close to the ground, it would seem that reflections are more likely to degrade the signal; so the inverse fourth power would be more appropriate to use. Does anyone know if this is that case?
Thanks

 

[tech99]

It is a very complicated and long topic, but yes, this is approximately correct. Of course, mobile communication is also subject to large variations due to multiple reflections from buildings etc and also blocking by hills. At VHF frequencies, it is unusual to have sufficient height to obtain Friis conditions, even with "high" towers.
For frequencies above about 30MHz, with high antennas, the average path loss is independent of polarization, but at MF and below the vertical polarization propagates better due to the surface wave, and over sea water, the inverse square law will apply.
For very low antennas at VHF, the surface wave may also be dominant, which favours vertical polarization.
I am sorry I cannot reply fully in a small space.

 

[skeptic2]

I remember seeing a paper that suggested using an exponent of -2 (inverse square law) the first kilometer and gradually increasing to -4, 10 km or more from the base antenna. The logic is that at 1 km or less propagation is essentially free space. As the distance increases, losses due to terrain and clutter reduce the propagation.

Another possibility is to use the HATA path loss formula. That formula can be found at
https://www.google.com/search?q=axonn rf pasth loss & transmission distance calculations&cad=h

 

[FrankJ777]

and over sea water, the inverse square law will apply

Why would inverse square apply over water? My though is that sea water is a good reflective surface, so there would be more destructive interference, hence the inverse 4th law would apply.

 

[tech99]

Why would inverse square apply over water? My though is that sea water is a good reflective surface, so there would be more destructive interference, hence the inverse 4th law would apply.

For vertical polarization over a good conductor, propagation is predominantly by the surface wave, so that the cancellation effect does not occur and losses can be small.

 

[tech99]

Another possibility is to use the HATA path loss formula. That formula can be found at
https://www.google.com/search?q=axonn rf pasth loss & transmission distance calculations&cad=h[/QUOTE]

The Hata formulas are based on an empirical model, using graphs, produced by Okumura, as a result of many tests in the Tokyo area. It caters for the situation where there is a high central antenna well above the surrounding clutter from buildings, with the mobile embedded in the clutter. It presents curves for urban, suburban and open terrain. The curves are reproduced on the Wiki page, which is quite useful, and it is interesting to compare them with the 1/d^2 and 1/d^4 models. These may also be expressed as 6dB and 12dB per octave of distance respectively.
https://en.wikipedia.org/wiki/Okumura_model