For a $N$-element planar phased array, all the elements are same isotropic antennas. Then, what is the effective area of this array? Is it $$A_e=\frac{\lambda^2}{4\pi}GN$$ where $G$ is the directivity of a fixed angle $(\theta,\phi)$.

Meanwhile, as the antennas are isotropic, is the beamforming cause the $G$ times change in effective area?

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    $\begingroup$ Hello and welcome to ham.stackexchange.com! $\endgroup$ – rclocher3 Jul 14 '20 at 13:25

Yes, that's correct.

If the elements are isotropic then as you've written it, $G=1$

And the effective aperture increases by $N$.

Of course this new aperture is only effective in the direction of the peak(s) of the response, elsewhere it will be less. And this assumes that the elements are far enough apart that there's no coupling between them.

  • $\begingroup$ Really thank for your answer! So, does $N$ equal to the maximum directivity (When mainlobe is steered at the direction of signal ) of the array? Then, for any fixed angle $(\theta,\phi)$, the result is $\frac{\lambda^2}{4 \pi}G(\theta,\phi)$? ($G(\theta,\phi)$ is the directivity of the array for the angle $(\theta,\phi)$) $\endgroup$ – tyrela Jul 14 '20 at 17:44
  • $\begingroup$ Yes. The formula in general relates effective aperture to gain. It's what connects the idea of gain-over-isotrope with capture-of-power-density. It's only really useful for this purpose, unless the antenna is an actual aperture like a horn or dish, don't read too much more into it. $\endgroup$ – tomnexus Jul 14 '20 at 20:25
  • $\begingroup$ You should include efficiency too, but in an ideal antenna $\eta=1$ so Gain = Directivity. $\endgroup$ – tomnexus Jul 14 '20 at 20:26
  • $\begingroup$ OK! I get it. thanks again. $\endgroup$ – tyrela Jul 15 '20 at 2:05

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