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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, 2020 at 13:25

3 Answers 3

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In the textbook starting case of an antenna composed of N identical elements, with some spacing and individual pattern, all fed in phase with equal power, then the directivity, gain and effective aperture increase with $N$.
This array may or may not meet your definition of "phased array" - it depends on what other information you have about the array.

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

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 neglects any deliberately applied taper (feeding the edge elements with less power than the centre ones to control sidelobes), and ignores the real-world effects of coupling between elements, overlaping element apertures.

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  • $\begingroup$ 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, 2020 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, 2020 at 20:25
  • $\begingroup$ You should include efficiency too, but in an ideal antenna $\eta=1$ so Gain = Directivity. $\endgroup$
    – tomnexus
    Jul 14, 2020 at 20:26
  • $\begingroup$ OK! I get it. thanks again. $\endgroup$
    – tyrela
    Jul 15, 2020 at 2:05
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The previous answers are correct only in the limit for a sparse array, where the elements are spaced at greater than a wavelength apart. N is the "array gain" (for an un-tapered array) which is not the same thing as the "gain" of the array. Array gain is the improvement in SNR from the array, caused by the fact that the signal is added coherently while the noise is added non-coherently. The directivity of a planar array is the product of N and the directivity of an element only in the limit as element spacing becomes much larger than lambda. The general calculation for a planar array is more complicated. See Van Trees, "Optimum Array Processing" section 4.1.1.2.

Let us be clear on some definitions, referencing IEEE Std 145-2013, "IEEE Standard for Definitions of Terms for Antennas" and D.K. Cheng, "Field and Wave Electromagnetics". "Array gain" is actually not a sanctioned term, but is nonetheless often used in practice. "Directivity" is always greater than "gain" by a factor called "radiation efficiency." "Gain" is always in turn greater than "realized gain" by an "impedance mismatch" factor. To make matters even more confusing, in the past "gain" was often used synonymously with "directivity." The terms "power gain" and "directive gain" are both deprecated by IEEE.

However, to answer your specific question, the ratio of effective aperture to directivity is a universal constant, $\frac{\lambda^2}{4\pi}$. Effective aperture is also always less than or equal to the physical aperture, which in this case is $N\times dy\times dx$, where dy and dx are the spacing between elements in the x and y dimension. Thus, $A_e= N\times dx\times dy\times \eta$, where $\eta$ is a factor called the "illumination efficiency" that accounts for tapering of the array and spacing of the array. In the case of a sparse array, where element spacing $>\lambda$, $\eta$ is reduced because the array is not uniformly illuminated.

There is a physically intuitive reason for this relationship; essentially there are a limited number of photons per unit area to be captured by the individual antennas. Placing two high gain antennas very close to each other (less than a wavelength) does not buy twice the gain, for example. This is why the physical aperture size must be taken into account.

Let's assume a 16 x 16 un-tapered standard rectangular array (which means that elements are spaced at $\lambda \over 2$.) The array gain is $10log_{10}(N) = 10log_{10}(256) = 24.1$dB. If the array were tapered, this value would go down. The directivity, assuming isotropic elements, is 25.9dBi Van Trees. Now assume elements with 9.0dBi directivity. The directivity is not 33.1dBi, but rather is only 29.2dBi.[4] The reason for this is that the the effective aperture of the individual elements limits their directivity. So, $D = A_e \times \frac{4\pi}{\lambda^2} = N \times dx \times dy \times \eta \times \frac{4\pi}{\lambda^2} = N \times \frac{\lambda}{2} \times \frac{\lambda}{2} \times \frac{4\pi}{\lambda^2} = N\pi$.
Note, in this case $\eta = 1$ because the array is un-tapered. Why the slight difference from $10log_{10}(N\pi)=$ 29.05 dBi? The elements around the edge of the array aren't as limited in their effective aperture as are the majority of elements.

Now let's move the array elements to $\lambda$ spacing. From the above formula, we expect the directivity to peak at $D = A_e \times \frac{4\pi}{\lambda^2} = N \times dx \times dy \times \eta \times \frac{4\pi}{\lambda^2} = N \times \lambda \times \lambda \times \frac{4\pi}{\lambda^2} = 4N\pi$. The actual result is 34.6380 dBi, just shy of the ideal 35.0745 dBi we expected.[4]. Why the difference from the ideal? If the spacing in the x and y dimensions is $\lambda$, then the spacing along the diagonals is $\lambda \sqrt{2}$, thus creating tiny regions in the overall array where photons are missed, leading to $\eta < 1$.

Now go to $10 \lambda$ spacing. The result now should converge to N times the element gain, or $10log_{10}(N)$ + 9 dBi = 33.1 dBi. The actual result is in fact, 33.1 dBi.[4]

[4]: MATLAB Phased Array System Toolbox, https://www.mathworks.com/products/phased-array.html

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  • $\begingroup$ Hello Jonathan, and welcome to this site! :-) $\endgroup$
    – Mike Waters
    Mar 17 at 22:39
  • $\begingroup$ So basically, directivity is not array gain, am I getting it right? $\endgroup$
    – tyrela
    Mar 19 at 6:39
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    $\begingroup$ Hi JW and welcome, your input will be super useful to this stack site. What you say is correct but it's for real-world phased arrays. For eg. two antennas a reasonable distance apart do give double the gain&directivity (ignore coupling&overlap). The spirit of the question is "N isotropic radiators" so I started by assuming they're well spaced out. In the limit I think it's N, but of course lots of real-world effects start to happen in real arrays. If I'm wrong about ideal N please explain, and especially if it's more (not considering weird coupling stuff). I can't open that book page. $\endgroup$
    – tomnexus
    Mar 20 at 19:59
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    $\begingroup$ Hi Tom and thank you for the welcome. I see that you are a long-time contributor and work on large radio telescope arrays. I think that leads to the difference in our thinking, as I interpreted the question in terms of a planar phased array and not a sparse array such as what you work on. My formulas technically still apply, but as the array becomes more sparse (elements move away from each other by a factor of many wavelengths, the illumination of the "array" is no longer unity and therefore the directivity goes down. I need to put some thought into the sparse situation, frankly. $\endgroup$ Mar 21 at 17:02
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    $\begingroup$ Thanks JW. I'm an antenna guy, so was thinking more along the lines of Lossless Passive Splitter -> Several Antennas, neglecting all coupling etc. This I think is the way it would have been treated in Kraus or Balanis. Your answer brings some rigour to the answer, better covering a real phased array today. $\endgroup$
    – tomnexus
    Mar 21 at 23:44
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First, your premise that for a planar phased array "all the elements are same isotropic antennas" is not correct. By definition, a the elements of a planar phased array are planar antenna elements, which are not isotropic: they have a unidirectional directivity perpendicular to the plane of the antenna.

But let's say that $G$ is the gain of each planar (e.g. patch) element in the orthogonal, or "broadside", direction ($\theta_0$, $\phi_0$). You can prove quite simply that your formula is not universally correct by looking at the case $N=2$ and assuming that the two elements are fed $180^\circ$ out of phase. In that case, the fields from each element will cancel each other out and $A_e(\theta_0, \phi_0)$ = 0.

As the other answers allude, the directivity - and hence capture cross-section - of an antenna array as a function of direction is quite complicated. If, however, the antenna elements are in-phase, then the electric fields (V/m) will add "constructively" as $N$ in the broadside direction, which implies the power density (W/m$^2$) will add as $N^2$. So with the in-phase condition (and assuming a loss-less combiner), your relation would be: $$A_e(\theta_0,\phi_0) = \frac{\lambda}{4\pi}GN^2$$

Off the broadside direction the relation would not be so simple, and would depend on the phasing of and distance between each element.

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    $\begingroup$ Fair enough about the 180 degrees, but that's a strange case (and there would be another beam some angle away, depending on the spacing). But your second example is strange in that it also assumes double the input power which is not wrong but needs to be disclosed... you can't compare an array of one element with 1 W total power, to an array of two with 2 W power, etc. If you split the power fairly then it's only N. But we all need to better define the basic terms and assumptions here before jumping in! $\endgroup$
    – tomnexus
    Mar 22 at 22:49
  • $\begingroup$ Total power would be power captured over all angles. $\endgroup$
    – AG5CI
    Mar 22 at 23:28
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    $\begingroup$ I mean input power... We know two antennas under ideal conditions give 3 dB more gain, not 6 dB. The power divider splits the power N ways so you need to divide the gain or power density or aperture in your eqn by N. $\endgroup$
    – tomnexus
    Mar 23 at 3:32

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