LECTURE 18: PLANAR ARRAYS, CIRCULAR ARRAYS 1. Planar arrays Planar arrays are more versatile; they provide more symmetrical patterns with lower side lobes, much higher directivity (narrow main beam). They can be used to scan the main beam toward any point in space. Applications – tracking radars, remote sensing, communications, etc. 1.1 The array factor of a rectangular planar array Fig. 6.23(b), pp.310, Balanis 1 The AF of a linear array of M elements along the x-axis is: M j m −1 kd sin θ cos φ + β x ) AFx1 = ∑ I m1e ( )( (18.1) m =1 where sin θ cos φ = cos γ x is the directional cosine with respect to the x-axis. It is assumed that all elements are equispaced with an interval of d x and a progressive shift β x . I m1 denotes the excitation amplitude of the element at the point with coordinates: x = (m − 1)d x , y = 0 . In the figure above, this is the element of the m-th row and the 1st column of the array matrix. If N such arrays are placed next to each other in the y direction, a rectangular array will be formed. We shall assume again that they are equispaced at a distance of d y and there is a progressive phase shift along each row of β y . It will be also assumed that the normalized current distribution along each of the x-directed array is the same but the absolute values correspond to a factor of I1n (n = 1,..., N ) . Then, the AF of the entire array will be: N j m −1 kd sin θ cos φ + β x ) j ( n −1)( kd y sin θ cos φ + β y ) AF = ∑ I1n ∑ I m1e ( )( x (18.2) e n =1 m =1 or AF = S xM ⋅ S yN , (18.3) where: N M j m −1 kd sin θ cos φ + β x ) S xM = AFx1 = ∑ I m1e ( )( , and m =1 N S yN = AF1 y = ∑ I1n e j ( n −1)( kd y sin θ sin φ + β y ) n =1 In the array factors above: sin θ cos φ = xˆ ⋅ rˆ = cos γ x sin θ sin φ = yˆ ⋅ rˆ = cos γ y (18.4) The pattern of a rectangular array is the product of the array factors of the linear arrays in the x and y directions. 2 For a uniform planar (rectangular) array I m1 = I1n = I 0 , for all m and n, i.e., all elements have the same excitation amplitudes. N N j ( n −1)( kd y sin θ sin φ + β y ) j ( m −1)( kd x sin θ cos φ + β x ) AF = I 0 ∑ e (18.5) ∑e n =1 n =1 The normalized array factor can be obtained as: ψ x sin N ψ y 1 sin M 2 1 2 , AFn (θ ,φ ) = ψ ψ M N x y sin sin 2 2 (18.6) where: ψ x = kd x sin θ cos φ + β x ψ y = kd y sinθ sin φ + β y The major lobe (principal maximum) and grating lobes of the terms: ψ sin M x 1 2 (18.7) S xM = ψ M sin x 2 ψy sin N 2 1 (18.8) S yN = N ψ y sin 2 are located at angles such that: kd x sin θ m cos φm + β x = ±2mπ , m = 0,1,… (18.9) kd y sin θ n sin φn + β y = ±2nπ , n = 0,1,… (18.10) The principal maxima correspond to m = 0, n = 0 . 3 In general, β x and β y are independent from each other. But, if it is required that the main beams of S xM and S yN intersect (which is usually the case), then the common main beam is in the direction: θ = θ 0 and φ = φ0 , m = n = 0 (18.11) If the principal maximum is specified by (θ 0 ,φ0 ) , then the progressive phases β x and β y must satisfy: β x = − kd x sin θ 0 cos φ0 (18.12) β y = −kd y sin θ 0 sin φ0 (18.13) When β x and β y are specified, the direction of the main beam can be found by simultaneously solving (18.12) and (18.13): β ydx tan φ0 = (18.14) β xd y 2 βx βy sin θ 0 = ± (18.15) + kd kd x y The grating lobes can be located by substituting (18.12) and (18.13) in (18.9) and (18.10): sin θ 0 sin φ0 ± nλ dy (18.16) tan φmn = m λ sin θ 0 cos φ0 ± dx sin θ 0 sin φ0 ± nλ sin θ 0 cos φ0 ± mλ dy d x sin θ mn = = (18.17) cos φmn sin φmn 2 4 To avoid grating lobes, the spacing between the elements must be less than λ ( d y < λ and d y < λ ). In order a true grating lobe to occur, both equations (18.16) and (18.17) must have a real solution (θ mn ,φmn ) . 3-D pattern of a 5-element square planar uniform array without grating lobes ( d = λ / 4 , β x = β y = 0 ): Fig. 6.24, pp.313 Balanis 5 3-D pattern of a 5-element square planar uniform array without grating lobes ( d = λ / 2 , β x = β y = 0 ): Fig. 6.25, pp.314, Balanis Notice the considerable decrease in the beamwidth as the spacing is increased from λ / 4 to λ / 2 . 6 1.2 The beamwidth of a planar array θh φh z θ0 y φ0 x φh A simple procedure, proposed by R.S. Elliot1 will be outlined. It is based on the use of the beamwidths of the linear arrays building the planar array. For a large array, whose maximum is near the broad side, the elevation plane HPBW is approximately: θh = 1 1 cos θ 0 2 θ x−2 cos 2 φ0 + θ y−2 sin 2 φ0 (18.18) “Beamwidth and directivity of large scanning arrays”, The Microwave Journal, Jan. 1964, pp.74-82 7 where: (θ 0 ,φ0 ) specify the main-beam direction; θx is the HPBW of a linear broadside array whose number of elements M and amplitude distribution is the same as that of the x-axis linear arrays building the planar array; θy is the HPBW of a linear BSA whose number of elements N and amplitude distribution is the same as those of the y-axis linear arrays building the planar array. The HPBW in the plane, which is orthogonal to the φ = φ0 plane and contains the maximum, is: 1 φh = (18.19) θ x−2 sin 2 φ0 + θ y−2 cos 2 φ0 For a square array ( M = N ) with amplitude distributions along the x and y axes of the same type, equations (18.18) and (18.19) reduce to: θy θx = cosθ 0 cosθ 0 φh = θ x = θ y θh = (18.20) (18.21) From (18.20), it is obvious that the HPBW in the elevation plane very much depends on the elevation angle θ 0 of the main beam. The HPBW in the azimuthal plane φh does not depend on the elevation angle θ 0 . The beam solid angle of the planar array can be approximated by: Ω A = θ hφh (18.22) or 8 θx θ y ΩA = cos θ 0 2 2 sin φ0 + θ y2 θ x2 2 θ x2 cos φ0 sin φ0 + 2 cos 2 φ0 θy (18.23) 2 1.3 Directivity The general expression for the calculation of the directivity of an array is: | AF (θ 0 ,φ0 ) |2 D0 = 4π 2π π (18.24) ∫ 2 φ | AF ( θ , ) | sin θ dθ dφ 0 0 ∫ 0 0 For large planar arrays, which are nearly broadside, (18.24) reduces to: (18.25) D0 = π Dx D y cosθ 0 where Dx is the directivity of the respective linear BSA, x-axis; Dy is the directivity of the respective linear BSA, y-axis. One can also use the array solid beam angle Ω A in (18.23) to calculate the approximate directivity of a nearly broadside planar array: π2 32400 D0 (18.26) Ω A[ Sr ] Ω A[deg 2 ] Remember: 1) The main beam direction is controlled through the phase shifts, β x and β y . 2) The beamwidth and side-lobe levels are controlled through the amplitude distribution. 9 2. Circular array 2.1 Array factor The normalized field can be written as: N e− jkRn E ( r ,θ ,φ ) = ∑ an Rn n =1 where: For r Rn = r 2 + a 2 − 2ar cosψ n a , (18.28) reduces to: Rn r − a cosψ n r − a aˆ ρn ⋅ rˆ ( (18.27) (18.28) ) (18.29) In rectangular coordinate system: aˆ ρn = xˆ cos φn + yˆ sin φn rˆ = xˆ sin θ cos φ + yˆ sin θ sin φ + zˆ cosθ 10 Therefore: Rn = r − a sin θ ( cos φn cos φ + sin φn sin φ ) (18.30) Finally, Rn is approximated in the phase terms as: (18.31) Rn = r − a sin θ cos (φ − φn ) For the amplitude term, the approximation 1 1 , all n (18.32) Rn r is made. Assuming the approximations (18.31) and (18.32) are valid, the far-zone array field is reduced to: e− jkr N jka sin θ cos(φ −φn ) E ( r ,θ ,φ ) = an e (18.33) ∑ r n =1 where: an is the excitation coefficient (amplitude and phase); 2π φn = n is the angular position of the n-th element. N In general, the excitation coefficient can be represented as: a = I n e jα n , (18.34) where I n is the amplitude term, and α n is the phase of the excitation of the n-th element relative to a chosen array element of zero phase. e− jkr N j ka sin θ cos(φ −φn ) +α n I ne ⇒ E ( r ,θ ,φ ) = (18.35) ∑ r n =1 The AF is obtained as: N AF (θ ,φ ) = ∑ I n e j ka sin θ cos(φ −φn ) +α n (18.36) n =1 Expression (18.36) represents the AF of a circular array of N equispaced elements. The maximum of the AF occurs when all the phase terms in (18.36) equal unity, or: ka sin θ cos (φ − φn ) + α n = 2mπ , m = 0, ±1, ±2, all n (18.37) 11 The principal maximum ( m = 0 ) is defined by the direction (θ 0 ,φ0 ) , for which: α n = −ka sin θ 0 cos (φ0 − φn ) , n = 1, 2,..., N (18.38) If a circular array is required to have maximum radiation in the direction (θ 0 ,φ0 ) , then the phases of its excitations will have to fulfil (18.38). The AF of such an array is: N jka sin θ cos(φ −φn ) −sin θ 0 cos(φ0 −φn ) ] AF (θ ,φ ) = ∑ I n e [ (18.39) n =1 N AF (θ ,φ ) = ∑ I n e jka (cosψ n −cosψ 0 n ) (18.40) n =1 Here: ψ n = cos −1 [sin θ cos(φ − φn ) ] is the angle between r̂ and aˆ ρn ; ψ 0 = cos −1 [sin θ 0 cos(φ0 − φn ) ] is the angle between aˆ ρ and r̂max n n pointing in the direction of maximum radiation. As the radius of the array a becomes very large as compared to λ , the directivity of the uniform circular array ( I n = I 0 , all n ) approaches the value of N. 12 Uniform circular array 3-D pattern (N=10, ka = 2π λ a = 10 ) 13