Antenna gain reciprocity violation of conservation of energy?

In summary, the problem is that the amount of energy that is available for free is limited by the amount of energy that is being lost in the process.
  • #36
Quandry said:
Simple answer is that the gain is relative to an isotropic radiator. So the transmit gain is never more (or even as much) as the available power and the receive gain is also relative to that received by an isotropic receive aerial.
I don't see how this is the "simple answer"!
I think we are all agreed that conservation of energy is not at fault! And I hope we all agree that reciprocity is founded on solid physics. So the OP question becomes, why does the Friis equation not work here. That has nothing to do with the units you use to measure gain - doesn't even have to be in any sort of decibels - so long as you know what you are doing and convert to isotropic levels for the values you put in the equation.

As far as I can see, several people have now explained the approximations and assumptions in the Friis equation, which IMO boil down to, the antennae must be much further apart. Neither the Free Space Loss nor the calculated antenna gain will be realistic at short range. And the range at which the Friis equation works, increases with the gain of the antennae.

For an antenna, it is not the units of gain nor what it is relative to that is at fault. It is the concept of an antenna having a definite gain which is fixed from here to eternity and unaffected by its surroundings (I have in mind here that the other antenna is part of the surroundings and has a diminishing influence as it moves further away.)
 
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  • #37
Merlin3189 said:
I don't see how this is the "simple answer"
My answer was to the original question asked by and4bj which was to do with conservation of energy. S(he) had specifically endeavoured to take near field effects out of the equation.
Harald Friis developed his equation using gains relative to isotropic antennas. The equation undergoes many transformations when applied to specific situations but specifically requires the distance between antennas to be a large number of wavelengths so in ande4bj's question the simple answer is that the Friis equation works. But the real simple answer remains that gain is relative to a standard antenna.
 
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  • #38
Quandry said:
But the real simple answer remains that gain is relative to a standard antenna.
True.
Antenna “gain” is not power amplification. It is the effective aperture or capture cross-section, expressed in dB to make computation of the energy budget easier. Since dB is a Log(ratio), the effective aperture must be referenced to something, usually the aperture of either an isotropic radiator, dBi, or a dipole element, dBd.

Antenna gain is only specified in the far field where the spherical wavefronts can be treated as being flat. It is not surprising that something will break if the far field cross-section is misapplied, in the near field, as power amplification.
 
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