a Matlab package for phased array beam shape inspection

More documents

Recommendations

Info

20 4 A PHASED ARRAY From Eq. (57) we Eq. (58) we get antenna’s “power integral” PI as PI = 1 ∫ G E |AF| 2 dΩ . (59) 4π 4π The array factor can always be evaluated numerically directly from the defining sum Eq. (41); or by computing the 2D Fourier transform Eq. (42) by FFT; or, as we mostly do in this package, where we take the excitation field as constant, even from the closed expression Eq. (50). For many purposes, like when inspecting pointing accuracy, this is already enough, for only the relative gain, that is, the gain relative to the maximum level, the beam center, may be of interest. But when the attainable signal levels are of concern, the absolutely normalized gain of Eq. (58) is required. The problem then is to evaluate the power integral. 4.8. The power integral for an array with isotropic elements For the array with isotropic elements, G E = 1, the power integral over the full solid angle can readily be evaluated. The squared magnitude of the array factor is |AF| 2 = ∑ a m a m ′e i2πu 0D(m−m ′) e i2πuD(m−m′ ) (60) The only non-trivial step is to evaluate the integral ∫ I = e i2πuD(m−m′) dΩ(û), (61) 4π Noting that the orientation of the coordinate system does not matter when integrating over the full solid angle, we will orient the z-axis of our 3D spherical coordinate system along the direction of the vector D(m − m ′ ), so that θ is the polar angle of û = (u, u z ) from that direction, and hence the integrand in I becomes e i2π cos θ‖D(m−m′ )‖ sin θdθdφ . The integral can be immediately evaluated over θ (and trivially,φ), to give The power integral thus is I = 4π sinc(2π‖D(m − m ′ )‖) . (62) PI = ∑ a m a m ′ sinc(2π‖D(m − m ′ )‖) e i2πu 0D(m−m ′) . (63) It may be observed that G = |AF| 2 /PI, with AF taken from Eq. (60) and PI from Eq. (63), has the expected special cases. For instance, when D x = 0 and D y = 0, we have only one, isotropic, element, so we expect G → 1 when D x → 0, D x → 0, and this clearly happens. For the case of two elements along the x-axis, a 00 = 1 and a 01 = ±1, with D x = D, D y = 0, the absolutely normalized gain is G = 1 ± cos[2πD(u x − u 0 x)] 1 ± sinc(2πD) · cos 2πDu 0 , (64) x where u x = sin θ, u 0 x = sin θ 0 . Here, as in general, the normalization depends both on the array spacing, the excitation pattern, and the beam direction.
4.9 The power integral for an array with directional elements 21 4.9. The power integral for an array with directional elements When the element gain G E depends on direction, the power integral cannot in general evaluated in a closed form. However, for arrays with reasonably large number of elements both in the x- and y-directions, an approximation can be found as follows. For such an array, the array factor resembles a sum of delta functions, that is, AF has narrow beams around the widely separated grating directions, and pretty much zero elsewhere. Denoting by ∆Ω g the solid angle around the grating direction u g that contain the essential part of the AF in that direction, we write the power integral as the sum of integrals over the support regions, ∫ 4π · PI = G E |AF| 2 dΩ ≈ ∑ ∫ G E |AF| 2 dΩ . (65) 4π g ∆Ω g We assume that the element gain is wide compared to the size ∆Ω g , so that we can take G E to be constant in that region, and take it out of the integral, and use Eq. (42) to replace AF by the 2-D Fourier transform of the excitation amplitude-sequence, 4π · PI ≈ ∑ ∫ G E (u g ) |ã((u − u 0 )D)| 2 dΩ(û) . (66) g ∆Ω g The differential surface element, on the surface of unit sphere, corresponding the solid angle dΩ, is equal of the size of the projection d 2 u to xy-plane of that surface element, divided by the cosine of the polar angle θ g of û at the element, dS g (û) = d2 u cos θ g . (67) This corresponds to change of variables from the spherical coordinates on the surface of the sphere to the cartesian coordinates on the xy-plane. The precise boundary of the area of integration does not matter, as it anyway is assumed to be in the 0-region of the integrand. We make the substitution Eq. (67) to Eq. (66). Then we also expand the area of integration to cover the whole grating zone Z g hosting the direction u g ; this just adds some more 0-region to the area, for there is at most one grating beam in a grating zone. The power integral becomes 4π · PI = ∑ g G E (u g ) cos θ g ∫ Z g |ã((u − u 0 )D)| 2 d 2 u . (68) Finally, we use Parseval’s theorem for 2-D sequencies, Eq. (56), to get ( ) ∑ ∑ G E (u g ) 1 |am | 2 PI = . (69) cos θ g g D x D y 4π 4.10. Array gain for an array with directional elements Consider an array with the excitation amplitudes all unity, and directional elements with power gain G E . From Eq. (50), Eq. (58) and Eq. (69), the absolutely normalized gain is G(û) = G E(û)(M x M y ) 2 [diric()diric()] 2 4πD x D y M x M y ∑g = G E (û) ∑ G E (u g) g cos θ g G E (u g) cos θ g 4πA λ 2 · [diric()diric()] 2 , (70)
Page 1 and 2: E3ANT A Matlab Package for Phased A
Page 3 and 4: List of Figures 1. Beam steering. .
Page 5 and 6: 5 2. Time steering and phase steeri
Page 7 and 8: 2.2 Monochromatic signals—time-st
Page 9 and 10: 3.1 Transmission 9 lobes have equal
Page 11 and 12: 3.2 Reception 11 current I goes via
Page 13 and 14: 3.2 Reception 13 equations. But fir
Page 15 and 16: 15 “losing sensitivity” (due) t
Page 17 and 18: 4.2 Array factor as 2-D discrete Fo
Page 19: 4.6 Parseval’s theorem 19 closed
Page 23 and 24: 4.12 Computing the antenna directiv
Page 25 and 26: 4.14 2-D beam cross section 25 or m
Page 27 and 28: 5.1 Timing accuracy versus pointing
Page 29 and 30: 6.1 Method 29 excitation amplitudes
Page 31 and 32: 31 electric field and the scatterin
Page 33 and 34: 33 Table 1: Predicting SNR for the
Page 35 and 36: 35 s(t, r; û) E p û û 0 0 R m
Page 37 and 38: 37 D=1.0; θ 0 = 60, θ g = −8, 6
Page 39 and 40: 39 (a) (b) (c) (d) Figure 5: Random
Page 41 and 42: 41 A + Z 2 I A Z 1 B + Z 3 V V (B)
Page 43 and 44: 43 E 0 E m û d m P xy ∆ m P u Fi
Page 45 and 46: 45 80 60 40 20 θ 0 (°) 0 −20
Page 47 and 48: 47 10 Beam width (°) 9 8 7 6 5 4 M
Page 49 and 50: 49 Y X Figure 17: A 3D view of an a
Page 51 and 52: 51 45 40 Directivity (dBi) 35 30 25
Page 53 and 54: 53 ARRAY 50×20 1.50×1.50 δx=0.0
Page 55 and 56: 55 ARRAY 50×20 1.50×1.50 δx=0.0
Page 57 and 58: 57 ∆X z E p χ P Q ∆Z z r 2 R 2
Page 59 and 60: 0.02 59 12×4 1.4×1.4 φ 0 =0.0°
Page 61 and 62: 61 V eff for a long pulse (km 3 ) V
Page 63 and 64: 63 V eff for a long pulse (km 3 ) V
Page 65 and 66: 65 B. Pointing geometry The functio
Page 67 and 68: 67 KIRUNA Lat=67.86 Lon=20.44 Hgt=0
Page 69: 69 C. Matlab functions There are tw

a Matlab package for phased array beam shape inspection

Create successful ePaper yourself

Delete template?

Save as template?