LECTURE 13: LINEAR ARRAY THEORY - McMaster University

[Pages:21]LECTURE 13: LINEAR ARRAY THEORY - PART I (Linear arrays: the two-element array. N-element array with uniform amplitude and spacing. Broad-side array. End-fire array. Phased array.)

1. Introduction

Usually the radiation patterns of single-element antennas are relatively wide, i.e., they have relatively low directivity. In long distance communications, antennas with high directivity are often required. Such antennas are possible to construct by enlarging the dimensions of the radiating aperture (size much larger than ). This approach, however, may lead to the appearance of multiple side lobes. Besides, the antenna is usually large and difficult to fabricate.

Another way to increase the electrical size of an antenna is to construct it as an assembly of radiating elements in a proper electrical and geometrical configuration ? antenna array. Usually, the array elements are identical. This is not necessary but it is practical and simpler for design and fabrication. The individual elements may be of any type (wire dipoles, loops, apertures, etc.)

The total field of an array is a vector superposition of the fields radiated by the individual elements. To provide very directive pattern, it is necessary that the partial fields (generated by the individual elements) interfere constructively in the desired direction and interfere destructively in the remaining space.

There are six factors that impact the overall antenna pattern: a) the shape of the array (linear, circular, spherical, rectangular, etc.), b) the size of the array, c) the relative placement of the elements, d) the excitation amplitude of the individual elements, e) the excitation phase of each element, f) the individual pattern of each element.

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2. Two-element Array

Let us represent the electric fields in the far zone of the array elements in the form

E1

=

M1En1(1,1)

-

e

j

kr1

-

2

r1

^1 ,

(13.1)

E2

=

M 2En2 (2 ,2 )

-

e

j

kr2

+

2

r2

^ 2 .

(13.2)

P

z

1

d 2

d 2 2

r1 r

r2 y

Here, M1, M1 En1, En2 r1, r2

^1 , ^ 2

field magnitudes (do not include the 1/r factor); normalized field patterns; distances to the observation point P; phase difference between the feed of the two array elements; polarization vectors of the far-zone E fields.

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The far-field approximation of the two-element array problem:

P

r1

z

r

d

2

r2 y

d

2

d cos

2

Let us assume that: 1) the array elements are identical, i.e.,

En1( , ) = En2 ( , ) = En ( , ) ,

(13.3)

2) they are oriented in the same way in space (they have identical polarizations), i.e.,

^1 = ^ 2 = ^ ,

(13.4)

3) their excitation is of the same amplitude, i.e.,

M1 = M2 = M .

(13.5)

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Then, the total field can be derived as

E = E1 + E2 ,

( ) E = ^MEn

,

1

e-

jk

r

-

d 2

cos

+

j

2

+

-

e

jk

r

+

d 2

cos

-

j

2

,

r

(13.6)

( ) E

=

^

M r

e-

jkr En

,

e

j

kd 2

cos + 2

+

-

e

j

kd 2

cos + 2

,

E

=

^ M

e- jkr r

En ( ,) ?

2

cos

kd

cos 2

+

.

EF

AF

(13.7)

The total field of the array is equal to the product of the field created by a single element located at the origin (element factor) and the array factor (AF):

AF

=

2 cos

kd

cos 2

+

.

(13.8)

Using the normalized field pattern of a single element, En ( ,) , and the

normalized AF,

AFn

=

cos

kd

cos 2

+

,

(13.9)

the normalized field pattern of the array is expressed as their product:

fn ( ,) = En ( ,) ? AFn ( ,).

(13.10)

The concept expressed by (13.10) is the so-called pattern multiplication rule valid for arrays of identical elements. This rule holds for any array consisting of decoupled identical elements, where the excitation magnitudes, the phase shift between the elements and the displacement between them are not necessarily the same. The total pattern, therefore, can be controlled via the single?element pattern En ( ,) or via the AF. The AF, in general, depends on the:

? number of elements, ? mutual placement, ? relative excitation magnitudes and phases.

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Example 1: An array consists of two horizontal infinitesimal dipoles located at a distance d = / 4 from each other. Find the nulls of the total field in the

elevation plane = ?90 , if the excitation magnitudes are the same and the

phase difference is:

a) = 0 b) = / 2 c) = - / 2

= 0? z

= ?90

8

0

8

= 90? y

= 180?

The element factor En ( ,) = 1 - sin2 sin2 is the same in all three cases producing the same null. For = ?90 , En ( ,) =| cos | and the null is at

1 = / 2.

(13.11)

The AF, which depends on , produces the following results in the 3 cases: a) = 0

AFn

=

cos

kd

cosn 2

=

0

cos

4

cosn

=

0

,

4

cosn

=

(2n

+1) 2

cosn = (2n +1) 2, n = 0, ?1, ?2,....

A solution with a real-valued angle does not exist. In this case, the total field pattern has only 1 null at = 90?, which is due to the element factor.

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Nikolova 2022

Fig. 6.3, p. 255, Balanis

6

b) = / 2

AFn

=

cos

4

cosn

+

4

=

0

4

(cosn

+

1)

=

(2n

+

1)

2

,

cosn + 1 = (2n + 1) 2 cos(n=0) = 1 2 = 0 .

The solution for n = 0 is the only real-valued solution. Thus, the total field

pattern has 2 nulls: at 1 = 90? and at 2 = 0?:

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Fig. 6.4, p. 256, Balanis

7

c) = - / 2

AFn

=

cos

4

cosn

-

4

=

0

4

(cosn

- 1)

=

(2n

+

1)

2

,

cosn -1 = (2n + 1) 2 cos(n=-1) = -1 2 = .

The total field pattern has 2 nulls: at 1 = 90? and at 2 = 180? .

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Fig. 6.4b, p. 257, Balanis

8

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