A(β) = C m e jmβ (1.3) A(β)e jmβ dβ. (1.4) A d ne jmβn. (1.5)

Transcription

1 Chapter 1 Circular Arrays 1.1 Circular Ring Arrays Radiated far-field of a continuous circular ring array 1 (Fig. 1.1) is given by, ˆ F (θ,φ) = R A(β)G e (θ,φ β)e sinθcos(φ β)] } jkr dβ. (1.1) For discrete circular ring array A(β) = A d (β) = N above equation becomes where F (θ,φ) = R N A d n δ(β β n ) ], the A d ng e (θ,φ β n )e jkr sinθcos(φ βn)]}. (1.2) Since A(β) is a periodic function with a period of, A(β) = C m e jmβ (1.3) C m = 1 ˆ For discrete arrays, the above equation becomes C m = 1 A(β)e jmβ dβ. (1.4) N A d ne jmβn. (1.5) 1 In this article, the word circular array is defined as a circular arrangement of antennas in which orientation of each individual array element is a function of the angular position β. Their orientation is such that EP(β) = G e(θ,φ β) (see (1.1)). 1

2 (a) A(β) in xy domain (b) A(β) in β domain Figure 1.1: Continuous array excitation A(β) in both xy and β domains (a) Discrete array starting at β = (b) Discrete array starting at β = π/2 Figure 1.2: Discrete uniformly spaced arrays with different starting positions Two different types of uniformly spaced discrete arrays are shown in Fig If the array starts at β =, If the array starts at β = β /2, then C m = C m+n. C m = C m+n. 2

3 1.2 Phase Mode Theory Now, substituting (1.3) in (1.1) gives F (θ,φ) = R C mˆ G e (θ,φ β)e jkr sinθcos(φ β) e jmβ] dβ }} pattern resulting from the mth excitation phase mode (1.6) So, entire far-field can be decomposed into individual phase modes as shown in the above equation. Let us further explore the far-field generated by the mth order phase mode in the next subsection mth Order Phase Mode Excitation Like A(β), G e (θ,φ) also can be written 2 in-terms of a Fourier series as shown below: where G e (θ,φ) = D l (θ)e jlφ (1.7) From 1.7, we get D l (θ) = 1 ˆ G e (θ,φ)e jlφ dφ. (1.8) G e θ,(φ β)] = D l (θ)e jl(φ β). (1.9) From (1.9) and (1.6), far-field generated by the mth order phase mode is given as F m (θ,φ) = R C mˆ = R C m = R C m = R C m e jmφ G e (θ,φ β)e jkr sinθcos(φ β) e jmβ] dβ D l (θ)e jlφˆ e jkr sinθcos(φ β) e j(m l)β] } dβ } D l (θ)j (m l) e jmφ J m l (k R sinθ) j (m l) D l (θ)j m l (k R sinθ) 2 because G e(θ,φ) is a periodic function in φ with a period of } (1.1) 3

4 Thus, total far-field generated by the circular ring array is given by F (θ,φ) = = R F m (θ,φ) C m e jmφ ] } j (m l) D l (θ)j m l (k R sinθ) 1.3 Circular Disc Arrays (with separable distribution) (1.11) A general theory for analyzing circular ring arrays was provided in the previous section. Such a general theory does not exist for the case of circular disc arrays!. However, if the disc array excitation is separable (i.e., A(ρ, β) = A ρ (ρ) A β (β)), a simple theory can then be provided. For a circular disc array with separable distribution, radiated far-filed is given by ˆ ˆ ] F (θ,φ) = ρa ρ (ρ) A β (β)g e (θ,φ β)e jkρsinθcos(φ β) dβ dρ (1.12) From the circular ring array theory of the previous section, it can be proved that where ˆ A β (β)g e (θ,φ β)e jkρsinθcos(φ β) dβ = C β m ejmφ C β m = 1 j (m l) D l (θ)j m l (k ρsinθ)] } (1.13) ˆ Substituting (1.13) in (1.12) gives ˆ F (θ,φ) = Cme β jmφ = = C β m ejmφ C β me jmφ A β (β)e jmβ dβ. (1.14) j (m l) D l (θ)j m l (k ρsinθ)] } ρa ρ (ρ)dρ j (m l) D l (θ) ˆ ]} } A ρ (ρ)j m l (k ρsinθ)ρdρ ]} } j (m l) D l (θ) A Hankel ρ,(m l) (k sinθ) (1.15) 4

5 where 3 A Hankel ρ,(m l) (k sinθ) = ˆ A ρ (ρ)j m l (k ρsinθ)ρdρ. 1.4 Examples Circular Ring Antenna with Uniform Excitation For a circular ring antenna of radius R (which physically resembles Fig. 1.1.(a)), fundamental array element 4 is an infinitesimal current source oriented along the Y-direction (assuming the current is flowing in counter-clockwise direction). Element pattern E φ corresponding to J e = δ(x)δ(y)δ(z)]ŷ is given as G E φ e (θ,φ) = jke jkr ηcosφ 4πr Since the current distribution is assumed to be uniform, A(β) = I. Therefore, from (1.4) I for m = C m = for m. (1.16) Similarly, from (1.8) D l (θ) = 1 2 ( jηke jkr ) for l = ±1 4πr for l ±1. (1.17) Substituting C m and D l values in (1.11) gives E φ (θ,φ) = I R j l D l (θ)j l (k R sinθ) ] l= 1,1 = J 1 (k R sinθ) I R kηe jkr. 2r The above equation is nothing but Eq. 5-54b (p-248, 1]). Following a similar procedure, it can be proved that E θ = Discrete Circular Ring Array In this section, we will analyze a discrete circular ring array (shown in Fig. 1.2) with uniform distribution. All elements of the array are assumed to be isotropic elements and the array starts at β = as shown in Fig. 1.2.(a). So, A(β) = N δ(β β n) and β n = (n 1) N. 3 from the Hankel transform properties 4 here, fundamental array element is related to the current direction at β = 5

6 Therefore, from (1.5) C m = 1 = 1 = 1 N A d n e jmβn N e (n 1)( jm N ) ( 1 e j2mπ 1 e j2mπ N Since we assumed that the array is made up of isotropic elements,g e (θ,φ) = 1. So, 1 for l = D l (θ) = for l. ). Therefore, substituting C m and D l values in (1.11) gives F (θ,φ) = = R F m (θ,φ) ( 1 e j2mπ 1 e j2mπ N ) j m J m (k R sinθ)]e jmφ } Circular Uniform Distribution Aperture on Ground Plane (1.18) For more details regarding this radiating source, see (P. 688, 1]). The electric field in the circular aperture is given as E = E ŷ, ρ a. So, the equivalent magnetic current (including the ground plane effect) is given as J m = 2E ˆx. In this example, the current J m is uni-directional. However, in this article, till now it is assumed that EP(β) = G e (θ,φ β). Let us not worry!. We can simply separate the entire analysis into array factor and element pattern evaluations. When we say array-factor, we are talking about an array of isotropic elements (since, element pattern is already separated). So, A ρ (ρ) = 1, ρ a A β (β) = 2E. Element pattern of an isotropic element is given as G e (θ,φ) = 1. 6

7 So, from (1.15), we get because AF (θ,φ) = 4πE A Hankel ρ, (k sinθ) (1.19) Cm β = 2E for m = for m, and 1 for l = D l = for l. From the identity d dx xm J m (x)] = x m J m 1 (x), Hankel transform mentioned in (1.19) is given by A Hankel ρ, (k sinθ) = ˆ a J (k sinθρ)ρdρ = a k ρ J 1 (k ρ a). where k ρ = k sinθ. So, finally we obtain the far-field using the pattern multiplication technique and is given as E = 4πE a J 1(k ρ a) jke jkr sinφ k ρ 4πr ˆθ +cosθcosφ ˆφ ]. 7

8 Bibliography 1] C. A. Balanis, Antenna Theory: Analysis and Design. Hoboken, NJ: John Willey, ,