Chapter 1

Circular Arrays

1.1 Circular Ring Arrays

Radiated far-field of a continuous circular ring array1 (Fig. 1.1) is given by,

F (θ, φ) = R0

{ˆ 2π

0

[

A (β)Ge (θ, φ− β) ejk0R0 sin θ cos(φ−β)]

dβ

}

. (1.1)

For discrete circular ring array A (β) = Ad (β) =∑N

n=1

[Ad

nδ (β − βn)], the

above equation becomes

F (θ, φ) = R0

{N∑

n=1

[

AdnGe (θ, φ− βn) e

jk0R0 sin θ cos(φ−βn)]}

. (1.2)

Since A (β) is a periodic function with a period of 2π,

A (β) =

∞∑

m=−∞

Cmejmβ (1.3)

where

Cm =1

2π

ˆ 2π

0

A (β) e−jmβdβ. (1.4)

For discrete arrays, the above equation becomes

Cm =1

2π

N∑

n=1

Adne

−jmβn . (1.5)

1In this article, the word circular array is defined as a circular arrangement of antennasin which orientation of each individual array element is a function of the angular position β.Their orientation is such that EP (β) = Ge (θ, φ− β) (see (1.1)).

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(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 = 0 (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.1.2. If the array starts at β = 0,

Cm = Cm+N .

If the array starts at β = β0/2, then

Cm = −Cm+N .

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1.2 Phase Mode Theory

Now, substituting (1.3) in (1.1) gives

F (θ, φ) = R0

∞∑

m=−∞

Cm

ˆ 2π

0

[

Ge (θ, φ− β) ejk0R0 sin θ cos(φ−β)ejmβ]

dβ

︸ ︷︷ ︸

pattern resulting from the mth excitation phase mode

(1.6)

So, entire far-field can be decomposed into individual phase modes as shown inthe above equation. Let us further explore the far-field generated by the mthorder phase mode in the next subsection.

1.2.1 mth Order Phase Mode Excitation

Like A (β), Ge (θ, φ) also can be written2 in-terms of a Fourier series as shownbelow:

Ge (θ, φ) =∞∑

l=−∞

Dl (θ) ejlφ (1.7)

where

Dl (θ) =1

2π

ˆ 2π

0

Ge (θ, φ) e−jlφdφ. (1.8)

From 1.7, we get

Ge [θ, (φ− β)] =∞∑

l=−∞

Dl (θ) ejl(φ−β). (1.9)

From (1.9) and (1.6), far-field generated by the mth order phase mode is givenas

Fm (θ, φ) = R0Cm

ˆ 2π

0

[

Ge (θ, φ− β) ejk0R0 sin θ cos(φ−β)ejmβ]

dβ

= R0Cm

∞∑

l=−∞

{

Dl (θ) ejlφ

ˆ 2π

0

[

ejk0R0 sin θ cos(φ−β)ej(m−l)β]

dβ

}

= R0Cm

∞∑

l=−∞

{

Dl (θ) 2πj(m−l)ejmφJm−l (k0R0 sin θ)

}

= R0Cmejmφ

{∞∑

l=−∞

2πj(m−l)Dl (θ)Jm−l (k0R0 sin θ)

}

(1.10)

2because Ge (θ, φ) is a periodic function in φ with a period of 2π

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Thus, total far-field generated by the circular ring array is given by

F (θ, φ) =

∞∑

m=−∞

Fm (θ, φ)

= R0

∞∑

m=−∞

{

Cmejmφ

∞∑

l=−∞

[

2πj(m−l)Dl (θ)Jm−l (k0R0 sin θ)]}

(1.11)

1.3 Circular Disc Arrays (with separable distribution)

A general theory for analyzing circular ring arrays was provided in the previoussection. Such a general theory does not exist for the case of circular discarrays!. However, if the disc array excitation is separable (i.e., A (ρ, β) =Aρ (ρ) × Aβ (β)), a simple theory can then be provided. For a circular discarray with separable distribution, radiated far-filed is given by

F (θ, φ) =

ˆ

∞

0

ρAρ (ρ)

[ˆ 2π

0

Aβ (β)Ge (θ, φ − β) ejk0ρ sin θ cos(φ−β)dβ

]

dρ

(1.12)

From the circular ring array theory of the previous section, it can be provedthat

ˆ 2π

0

Aβ (β)Ge (θ, φ− β) ejk0ρ sin θ cos(φ−β)dβ =

∞∑

m=−∞

{

Cβmejmφ

∞∑

l=−∞

[

2πj(m−l)Dl (θ)Jm−l (k0ρ sin θ)]}

(1.13)

where

Cβm =

1

2π

ˆ 2π

0

Aβ (β) e−jmβdβ. (1.14)

Substituting (1.13) in (1.12) gives

F (θ, φ) =

ˆ

∞

0

∞∑

m=−∞

{

Cβmejmφ

∞∑

l=−∞

[

2πj(m−l)Dl (θ) Jm−l (k0ρ sin θ)]}

ρAρ (ρ) dρ

=

∞∑

m=−∞

{

Cβmejmφ

∞∑

l=−∞

{

2πj(m−l)Dl (θ)

[ˆ

∞

0

Aρ (ρ)Jm−l (k0ρ sin θ) ρdρ

]}}

=

∞∑

m=−∞

{

Cβmejmφ

∞∑

l=−∞

{

2πj(m−l)Dl (θ)[

AHankelρ,(m−l) (k0 sin θ)

]}}

(1.15)

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where3

AHankelρ,(m−l) (k0 sin θ) =

ˆ

∞

0

Aρ (ρ) Jm−l (k0ρ sin θ) ρdρ.

1.4 Examples

1.4.1 Circular Ring Antenna with Uniform Excitation

For a circular ring antenna of radius R0 (which physically resembles Fig.1.1.(a)), fundamental array element4 is an infinitesimal current source orientedalong the Y-direction (assuming the current is flowing in counter-clockwise di-

rection). Element pattern Eφ corresponding to ~Je = [δ (x) δ (y) δ (z)] y is givenas

GEφe (θ, φ) = −

jke−jkr

4πrη cosφ

Since the current distribution is assumed to be uniform, A (β) = I0. Therefore,from (1.4)

Cm =

{I0 for m = 00 for m 6= 0

. (1.16)

Similarly, from (1.8)

Dl (θ) =

{

− 12

(jηke−jkr

4πr

)

for l = ±1

0 for l 6= ±1. (1.17)

Substituting Cm and Dl values in (1.11) gives

Eφ (θ, φ) = I0R0

∑

l=−1,1

[2πj−lDl (θ)J−l (k0R0 sin θ)

]

= J1 (k0R0 sin θ)I0R0kηe

−jkr

2r.

The above equation is nothing but Eq. 5-54b (p-248, [1]). Following a similarprocedure, it can be proved that Eθ = 0.

1.4.2 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 beisotropic elements and the array starts at β = 0 as shown in Fig. 1.2.(a). So,

A (β) =∑N

n=1 δ (β − βn) and βn = (n− 1) 2πN

.

3from the Hankel transform properties4here, fundamental array element is related to the current direction at β = 0

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Therefore, from (1.5)

Cm =1

2π

N∑

n=1

Adne

−jmβn

=1

2π

N∑

n=1

e(n−1)(−jm 2πN )

=1

2π

(1− e−j2mπ

1 − e−j2mπ

N

)

.

Since we assumed that the array is made up of isotropic elements, Ge (θ, φ) =1. So,

Dl (θ) =

{1 for l = 00 for l 6= 0

.

Therefore, substituting Cm and Dl values in (1.11) gives

F (θ, φ) =∞∑

m=−∞

Fm (θ, φ)

= R0

∞∑

m=−∞

{(1− e−j2mπ

1− e−j2mπ

N

)

[jmJm (k0R0 sin θ)] ejmφ

}

(1.18)

1.4.3 Circular Uniform Distribution Aperture on Ground

Plane

For more details regarding this radiating source, see (P. 688, [1]). The electricfield in the circular aperture is given as

~E = E0y, ρ ≤ a.

So, the equivalent magnetic current (including the ground plane effect) is givenas

~Jm = 2E0x.

In this example, the current ~Jm is uni-directional. However, in this article,till now it is assumed that EP (β) = Ge (θ, φ− β). Let us not worry!. We cansimply separate the entire analysis into array factor and element pattern eval-uations. When we say array-factor, we are talking about an array of isotropicelements (since, element pattern is already separated). So,

Aρ (ρ) = 1, ρ ≤ a

Aβ (β) = 2E0.

Element pattern of an isotropic element is given as

Ge (θ, φ) = 1.

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So, from (1.15), we get

AF (θ, φ) = 4πE0AHankelρ,0 (k0 sin θ) (1.19)

because

Cβm =

{2E0 for m = 00 for m 6= 0

, and

Dl =

{1 for l = 00 for l 6= 0

.

From the identity ddx

[xmJm (x)] = xmJm−1 (x), Hankel transform mentionedin (1.19) is given by

AHankelρ,0 (k0 sin θ) =

ˆ a

0

J0 (k0 sin θρ) ρdρ

=a

kρJ1 (kρa) .

where kρ = k0 sin θ.So, finally we obtain the far-field using the pattern multiplication technique

and is given as

~E = 4πE0aJ1 (kρa)

kρ

jke−jkr

4πr

[

sinφ θ + cos θ cosφ φ]

.

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Bibliography

[1] C. A. Balanis, Antenna Theory: Analysis and Design. Hoboken, NJ: JohnWilley, 2005. 1.4.1, 1.4.3

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