let's consider a simple linear uniform spaced array: enter image description here

We know that, from elementary array theory, the total pattern (for instance in terms of E-field) may be evaluated, in case no phase shift is applied to the excitations, as:

$\vec{E}_{total} = \vec{E}_{single\,element} \cdot \sum_{n=1}^{M} e^{j\cdot (n-1)\cdot k \cdot d \cdot (\theta)}$

where the last term is the array factor. $\vec{E}_{single\,element}$ is the single element pattern when it's isolated in free space.

Obviously it is an approximation, since in real life all the array elements won't have the same radiation pattern, because of their different position inside the array and mutual coupling effects. For instance, an element which is at the edge of the array will have around himm less elements than one at the centre, and so will interact with less elements. To take into accounts this effects, I'll call single elements pattern not simply as a unique quantity $\vec{E}_{single\,element}$, but as:

$\vec{E}_{n}$, with n = 1,2,3,...,M

If I simulate an antenna array with a simulator, I've the possibility to see both the total pattern $\vec{E}_{total}$ and all the single elements patterns $\vec{E}_{n}$ evaluated inside the array. Now, if I want to check my simulation of the overall array from single patterns $\vec{E}_{n}$, should I use:

  1. This:

$\vec{E}_{total} = \sum_{n=1}^{M}\vec{E}_{n}\cdot e^{j\cdot (n-1)\cdot k \cdot d \cdot (\theta)}$


  1. This:

$\vec{E}_{total} = \sum_{n=1}^{M}\vec{E}_{n}$

My attempts: in my attemps, it seems that 2) is true. But I'm asking this question because I'm not sure of I from a theory point of view, and because if it's wrong, I have probably done a mistake in my code for evaluating the total field.


The first is still correct, and can be evaluated at any position.

The second is true only at broadside, when $\theta=0$.

In your attempts, do you find that $\vec{E}_{total}$ is the same in all directions? Then you are doing something else wrong, or $d \ll {2\pi\over{\lambda}}$
Or are you accidentally including the $e^{-j{Phase}}$ in the individual element pattern? If you are simulating the elements one by one, with the others' excitation at $0 V$ then of course they already include the whole array term. Perhaps start by considering $\vec{E}_{n} = 1$.

As a hint for getting clean array patterns without mutual coupling: use very short dipoles as elements, instead of resonant dipoles. The radiated power scales down but you can just normalise this, but they don't interact with each other very much.

  • $\begingroup$ I'm including the e^-jphase in the individual element pattern, is it wrong? I thought that the electric field in the formula is the complex electric field. Moreover, each element is simulated one by one with others' excitation at 0V. Why do they include the array term in this case? $\endgroup$ – Kinka-Byo Jan 8 at 16:15
  • 1
    $\begingroup$ Well you have to include one or the other. But the En in the first equation is the element pattern when it is at the origin. By simulating each element in turn, in its actual position, you are already fully accounting for the array pattern. Then by superposition you can just add them up, so the second equation applies. $\endgroup$ – tomnexus Jan 8 at 16:43
  • $\begingroup$ Perfect, now it's explained why my results are correct works only with 2, thank you very much. Instead, in case I've the element pattern in the origin, should I consider both the term e^-jPhase and the array term? $\endgroup$ – Kinka-Byo Jan 8 at 17:09
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    $\begingroup$ Yes if the element is simulated at the origin (usually just by simulating a single element, but if you want to include coupling you could move the array along for each simulation) then you need to put back the phase term. But I don't see the point of doing it in two parts - either simulate the whole thing or use the array formula to simplify your life. $\endgroup$ – tomnexus Jan 9 at 21:22

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