0
\$\begingroup\$

in this paper the following statement is written:

An area A is allotted to each element in the infinite array. This is the maximum area available to each element, and is usually greater than the physical size of the actual element. It is natural to assume that the maximum gain obtainable from an element in the array is related to the area A by the well-known gain formula for apertures large compared with a wavelength, because the entire array is indeed large. Furthermore, since the effective area of an element should be proportional to its projected area in the direction of interest, the element gain should have a cos theta variation with angle. Based on this intuitive reasoning, the maximum element gain would be

enter image description here

This relation gives a fundamental upper limit to the gain obtainable in an element of an infinite planar array. It also implies that the ideal shape of the gain pattern of such an element would approach the cos theta variation. However, there are factors not contained in (1) [the previous equation] which must be considered in any objective analysis of element gain.

enter image description here

Well, this reasoning is not intuitive for me. Precisely:

  1. For hypothesis, mutual coupling is absent. Mutual coupling is a real cause of anisotropic radiation pattern but the author takes it into account in the following part of the article by adding an active reflection coefficient multiplicative term. For what concerns my doubts, mutual coupling does not exist.

  2. For hypothesis, the planar array is infinite. So, there is no reason for why different radiating elements should have different patterns, as they are surrounded each one by the same environment.

Provided these hypotheses, looks like the author says that:

Even if the isolated single element pattern is isotropic, it would become anisotropic in a phased array.

This result comes from the known relationship:

$$D(\theta, \phi) = \frac{4*\pi *A_e}{\lambda^2} $$

This relationship is fine. What I do not understand is why the author does consider the geometric area of the cell (with its corresponding cosine projection term) where the single element is put, instead of the effective area of the single element.

If the radiating element is isotropic, its effective area will be:

$$A_e = \frac{\lambda^2}{4*\pi} $$

This obviously will say that the radiating element inside the array is still isotropic.

My question could be rephrased like that: "Why does the effective area of each radiating element in a phased array equal its available cell and not its isolated effective area?" The space between the single element and the cell A is empty: why should it catch power?

\$\endgroup\$
4
  • \$\begingroup\$ the individual element remains unchanged. It's that the sum of all the wavefields have a directionality. \$\endgroup\$ Commented Apr 2, 2022 at 11:33
  • \$\begingroup\$ @MarcusMüller The directionality is embodied in the array factor. But the author adds an additional directional contribution, that is the non-isotropic cos(theta) pattern shown above. \$\endgroup\$
    – Kinka-Byo
    Commented Apr 2, 2022 at 11:37
  • \$\begingroup\$ Real world elements I've seen are never isotropic, and ones used in arrays always have a roll off that is cos^n(theta), even when measured in isolation. The exact roll off factor (n) does change when the element is embedded in a array. That's one of the reason real world arrays may have several rows or columns of dummy elements around the edges of the active part of the array. \$\endgroup\$
    – SteveSh
    Commented Apr 2, 2022 at 12:51
  • \$\begingroup\$ @Kinka-Byo I guess a real infinite array would have the same area facing you no matter which angle you were standing at (that area being infinity) but a finite-area one will have an angle that varies as cos(theta). I think you are right that this doesn't really square with the hypotheses \$\endgroup\$ Commented Apr 19, 2022 at 16:44

2 Answers 2

0
\$\begingroup\$

A while ago I've played with overlapping circular wave patterns. In Inkscape, I've made a few groups of concentric circles, with smeared edges and some alpha thrown in.

Radiators spaced λ/2:

lambda/2

Radiators spaced 1 λ:

1 lambda

Radiators spaced 2 λ:

2 lambda

This is just the basic idea. There are things I don't understand, such as, why the typical spacing of a reflector turns out to be about λ/6 :-)

\$\endgroup\$
1
  • \$\begingroup\$ This is interesting, but not really relevant to the question posted. This is showing visually how the contributions from all the elements combine in the far field as a function of angle off of boresight, theta and element spacing. For a spacing of lambda/2. you see the single main beam at 0 deg, where all the wave fronts line up. For a spacing of lambda, you get a main beam at 0 deg and two other main beams (grating lobes) at + and - 90 deg (not shown on the graphic). This has little to do with the element pattern and the effective area of the element. \$\endgroup\$
    – SteveSh
    Commented Apr 2, 2022 at 15:25
0
\$\begingroup\$

In answer to you last question regarding effective area, I offer this qualitative explanation.

Think about how an element behaves when emitting (transmitting) RF energy. The RF energy leaves the element and you can measure or model the e-field and h-field at any point around the element, anywhere within the effective area of the element, and beyond.

Since elements are usually always reciprocal, the same thing happens when the element is receiving energy. Hence the effective area of an element applies to receive as well as transmit.

To analyze this correctly, you need to use field solvers like HFSS or CST Microwave. This is above my pay grade!

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.