From an array of dual pol antennas and knowing the array peak gain, and the power in W of each component in the array, what is the correct way to calculate EIRP?

In linear terms is it just the simple sum of all the components multiplied by the gain (linear); do I need to take into account the phase and the possibly different orientations of my antennas?

  • \$\begingroup\$ Presumably, when the gain calculation or measurement was made, the phase and orientation of the antennas was known. Similarly with the individual power measurements. If the measurements are of a different configuration, ( phase or orientation) then they are not germane to computing anything. \$\endgroup\$
    – user69795
    Jan 9 at 21:37
  • 1
    \$\begingroup\$ Yes, it's the sum of the linear components (in voltage!). You'd need array directivity which should already include phase and orientations of your elements. Is "array peak gain" that you've indicated dBi or dBiC? \$\endgroup\$
    – Jason
    Jan 9 at 21:54
  • \$\begingroup\$ @Jason Yes peak gain is in dBi \$\endgroup\$ Jan 10 at 8:34

1 Answer 1


The "simple" equation is:

$$ EIRP_{dBmiC} = Gain_{Linear} + 10log_{10}(N) + P_{device} + 3 dB$$ Note: power in log.

The equation above assumes the array gain is identical for both polarizations and they are perfectly 90 degrees out of phase. 3 dB comes from linear->circular polarization conversion.

For a more rigorous equation, you must combine the EIRP components as vectors.

Circular polarization from linear components: $$ \hat{e}_{\frac{l}{r}hcp}=\frac{1}{\sqrt{2}}\left [ \vec{\theta } \mp j\vec{\phi }\right ] $$

Or, in your case: $$ \vec{eirp}_{CP}=\frac{1}{\sqrt{2}}\left [ \vec{eirp}_{L1} \pm j\vec{eirp}_{L2}\right ] $$ Note: eirp is voltage/field vector.

For a simple 2-element, dual polarized element with 0 dBm at each port you would expect:

$$ EIRP_{dBmiC} = EIRP_{dBmi} + 3 dB = 3 dB + 3 dB + 0 dBm + 3 dB = 9 dBmiC$$

  • 3 dB from array gain, per polarization (estimate)
  • 3 dB from conducted power (10logN), per polarization
  • 0 dBm output power, per element
  • 3 dB from linear polarization -> circular polarization

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