3
$\begingroup$

I have a theoretical question about EIRP.

EIRP is calculated by summing the power of the radio (the intentional radiator, IR) plus the gain of the antenna (dBd/dBi) minus the sum of the losses. The calculated value is equivalent to the power which a isotropic radiator would need to provide the same power density as the original antenna in the direction of maximum gain.

Is this statement true:

"Since antennas are passive elements it would not be possible to measure more power on any given theoretical client than the power (Watts (Energy/Time)) provided by the I.R. The gain of the antenna increases the power intensity, applying the same amount of power to a smaller area but does not change the total radiated power."

$\endgroup$
3
  • $\begingroup$ When you say I.R do you mean $I^2R$? $\endgroup$
    – rclocher3
    Oct 8 at 19:12
  • $\begingroup$ No, it is IR from "intentional radiator". $\endgroup$
    – 377ohms
    Oct 8 at 22:36
  • $\begingroup$ Thanks for clarifying that. It's always best practice to explain obscure acronyms here, and I hope that you don't mind that I edited your question to do so. (Another best practice is to delete comments after they are no longer useful, so we should delete these.) $\endgroup$
    – rclocher3
    Oct 11 at 16:38
3
$\begingroup$

I think what you're asking is: If a radio+antenna system transmits X Watts, then regardless of the configuration of EITHER the transmitting antenna OR the receiving antenna, the power measured at the receiving antenna can never be MORE than X Watts.

And that would be correct.

An isotropic radiator of 50 Watts would spread that power out evenly on the surface area of a sphere (${4\pi{r^2}}$) so that 50 watts diminishes rather quickly with distance from the radiator.

An antenna with gain will reshape that energy distribution into something with greater density in some areas and less (or none) in others. But no practical antennal will EVER put all 50 Watts into a single direction and zero power in all other directions.

This means that there is already far less power arriving at the receiving antenna.

And the receiving isotropic antenna will be sensitive to energy from all directions equally. Designing it for gain in a direction has the practical effect of making it appear "larger" in one direction and "smaller" in others, but there is no way to make it appear SO large that it will capture ALL of a transmitted signal. The measured gain for the receiving antenna is in terms of dBi again, NOT dBm.

$\endgroup$
1
  • 2
    $\begingroup$ Hi webmarc, thanks for your reply. I mean if the power (Watts) that the radio transmits is the maximum power that can be measured by any receiver (Watts) regardless the gain of the transmitting or receiving antenna. What may increase is the intensity (Watts/m²). Considering that antennas are passive devices and does not add energy to the system. Example: IR: (100mW) , Antenna (3dBi), EIRP: (23dBm/200mW) it would not be possible to have a client receiving more than 100mW. $\endgroup$
    – 377ohms
    Oct 8 at 14:45
3
$\begingroup$

That’s just conservation of energy.

In the one dimensional case, 100 mW would “isotropically” radiate 50 mW backwards and 50 mW forwards. A one dimensional 3 dB directional gain would just move that energy around to radiating all 100 mW forwards. If you put a set of receive antennas both in front and behind that 1D transmitting antenna, they would capture the same sum total amount in both cases (minus system losses).

The 3D case just moves that ratio-ing around to fractions of a sphere or steradians.

$\endgroup$
1
  • $\begingroup$ Yes, that is exactly what I thought due to conservation of energy. Thanks. $\endgroup$
    – 377ohms
    Oct 8 at 18:15
2
$\begingroup$

The following statement:

The calculated value is equivalent to the power which a isotropic radiator would use to provide the same power of the system.

is somewhat confusing.

The calculated value for EIRP is equivalent to the power an isotropic radiator would need to provide the same power density as the original antenna in the direction of maximum gain.

This needed power is greater than the power fed to the antenna: the isotropic radiator distributes power equally in all directions, while the original antenna concentrates power in the direction of maximum gain.

$\endgroup$
1
  • 1
    $\begingroup$ Yes, that is right regarding the EIRP definition, thanks. $\endgroup$
    – 377ohms
    Oct 8 at 15:57

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.