a Matlab package for phased array beam shape inspection
a Matlab package for phased array beam shape inspection
a Matlab package for phased array beam shape inspection
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20 4 A PHASED ARRAY<br />
From Eq. (57) we Eq. (58) we get antenna’s “power integral” PI as<br />
PI = 1 ∫<br />
G E |AF| 2 dΩ . (59)<br />
4π<br />
4π<br />
The <strong>array</strong> factor can always be evaluated numerically directly from the defining sum<br />
Eq. (41); or by computing the 2D Fourier trans<strong>for</strong>m Eq. (42) by FFT; or, as we mostly<br />
do in this <strong>package</strong>, where we take the excitation field as constant, even from the closed<br />
expression Eq. (50). For many purposes, like when inspecting pointing accuracy, this<br />
is already enough, <strong>for</strong> only the relative gain, that is, the gain relative to the maximum<br />
level, the <strong>beam</strong> center, may be of interest. But when the attainable signal levels are of<br />
concern, the absolutely normalized gain of Eq. (58) is required. The problem then is to<br />
evaluate the power integral.<br />
4.8. The power integral <strong>for</strong> an <strong>array</strong> with isotropic elements<br />
For the <strong>array</strong> with isotropic elements, G E = 1, the power integral over the full solid<br />
angle can readily be evaluated. The squared magnitude of the <strong>array</strong> factor is<br />
|AF| 2 = ∑ a m a m ′e i2πu 0D(m−m ′) e i2πuD(m−m′ )<br />
(60)<br />
The only non-trivial step is to evaluate the integral<br />
∫<br />
I = e i2πuD(m−m′) dΩ(û), (61)<br />
4π<br />
Noting that the orientation of the coordinate system does not matter when integrating<br />
over the full solid angle, we will orient the z-axis of our 3D spherical coordinate system<br />
along the direction of the vector D(m − m ′ ), so that θ is the polar angle of û = (u, u z )<br />
from that direction, and hence the integrand in I becomes<br />
e i2π cos θ‖D(m−m′ )‖ sin θdθdφ .<br />
The integral can be immediately evaluated over θ (and trivially,φ), to give<br />
The power integral thus is<br />
I = 4π sinc(2π‖D(m − m ′ )‖) . (62)<br />
PI = ∑ a m a m ′ sinc(2π‖D(m − m ′ )‖) e i2πu 0D(m−m ′) . (63)<br />
It may be observed that G = |AF| 2 /PI, with AF taken from Eq. (60) and PI from<br />
Eq. (63), has the expected special cases. For instance, when D x = 0 and D y = 0, we<br />
have only one, isotropic, element, so we expect G → 1 when D x → 0, D x → 0, and this<br />
clearly happens. For the case of two elements along the x-axis, a 00 = 1 and a 01 = ±1,<br />
with D x = D, D y = 0, the absolutely normalized gain is<br />
G =<br />
1 ± cos[2πD(u x − u 0 x)]<br />
1 ± sinc(2πD) · cos 2πDu 0 , (64)<br />
x<br />
where u x = sin θ, u 0 x = sin θ 0 . Here, as in general, the normalization depends both on<br />
the <strong>array</strong> spacing, the excitation pattern, and the <strong>beam</strong> direction.