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Antenna Power Loss calculation

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kae_jolie

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Is it fair to say:

Antenna Power Loss = Power Accepted - Power Radiated ?

Basically Power Loss could have been absorbed by other objects around antenna and hence not radiated and measured by receiving antenna (or same antenna if S11).
 

Basically, yes. Metal or dielectrics in the vicinity can cause loss and you can lump that in with the radiation efficiency. Through reciprocity, it's the same for transmit or receive. Power accepted is 1-|S11|^2, which in an array would be active-S11 to account for mutual coupling terms.
 

Thanks ColonelTigh.

I see in some papers S11 used in lieu of antenna power loss, I mean they assume non-radiated power (power loss) to be equivalent to power reflected back to source from antenna (S11). I think this is inherently wrong. S11 deals primarily with circuit impedance in power source, whereas non-radiated power or power loss could be due to adjacent dielectrics in the vicinity. Correct?
 

Hello kae_jolie,

You are right, the total loss is the sum of the mismatch loss (-10*log{1-|S11|^2} ) and the ohmic/dielectric (heat) loss (in dB).

Radiation efficiency = (radiated power) / (actual input power) = (radiated power) / (Pinc*(1-|S11|^2) ).

Antenna efficiency = (radiated power) / Pinc

Note that names for the various phenomena may vary, so this can be confusing.
 
WimRFP,
Thanks for reply. To rephrase what I mean using your terms, I believe (Power Accepted - Power Radiated) = Dielectric Loss only because mismatch loss has been accounted for already in Power Accepted term which is equal to 1-|S11|^2

Which brings me to another question: Which S11 to use? The S11 reading at solution frequency or total S11 in some bandwidth?

Thanks.
 

(Power Accepted - Power Radiated) = Dielectric Loss only because mismatch loss has been accounted for already in Power Accepted term which is equal to 1-|S11|^2

¡OK!

The S11 thing is tricky, it depends on the spectral power distribution of the TX signal, and the actual power amplifier. Many power amplifiers do not behave as a true linear source with 50 Ohms output impedance. So the actual mismatch loss (with respect to delivered power in a real 50 Ohms load), depends on your amplifier.

If you have a setup with a forward and reverse coupler between output of amplifier and load, you may see variation in forward power when changing load impedance.

Maybe you can give some more details about the actual application to give a more specific answer.
 

If you're only concerned about determining antenna efficiency, resistive and dielectric losses are fairly constant over a "narrow" bandwidth. You could determine S11 at the antenna input port at one frequency and use that to determine the efficiency per WimRFP's equations.

If you have a matching network between your input port and actual antenna, this calculation would include resistive/dielectric losses of the matching network with losses of the antenna itself.
 

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