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LECTURE 18: PLANAR ARRAYS, CIRCULAR ARRAYS

1. Planar arraysPlanar arrays are more versatile; they provide more

symmetrical patterns with lower side lobes, much higherdirectivity (narrow main beam). They can be used to scan the mainbeam 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

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The AF of a linear array of M elements along the x-axis is:

( )( )1 sin cos1 1

1

x

Mj m kd

x mm

AF I e θ φ β− +

== ∑ (18.1)

where sin cos cos xθ φ γ= is the directional cosine with respect tothe x-axis. It is assumed that all elements are equispaced with aninterval of xd and a progressive shift xβ . 1mI denotes theexcitation amplitude of the element at the point with coordinates:

( 1) xx m d= − , 0y = . In the figure above, this is the element of them-th row and the 1st column of the array matrix.

If N such arrays are placed next to each other in the ydirection, a rectangular array will be formed. We shall assumeagain that they are equispaced at a distance of yd and there is a

progressive phase shift along each row of yβ . It will be also

assumed that the normalized current distribution along each of thex-directed array is the same but the absolute values correspond to afactor of 1 ( 1,..., )nI n N= . Then, the AF of the entire array will be:

( )( ) ( )( )1 sin cos1 sin cos1 1

1 1

y yx x

N Nj n kdj m kd

n mn m

AF I I e eθ φ βθ φ β − +− +

= =

=

∑ ∑ (18.2)

or

M Nx yAF S S= ⋅ , (18.3)

where:

( )( )1 sin cos1 1

1

x

M

Mj m kd

x x mm

S AF I e θ φ β− +

== = ∑ , and

( )( )1 sin sin1 1

1

y y

N

Nj n kd

y y nn

S AF I eθ φ β− +

== =∑

In the array factors above:ˆ ˆsin cos cos

ˆ ˆsin sin cosx

y

x r

y r

θ φ γθ φ γ

= ⋅ == ⋅ =

(18.4)

The pattern of a rectangular array is the product of the array factorsof the linear arrays in the x and y directions.

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For a uniform planar (rectangular) array 1 1 0m nI I I= = , for allm and n, i.e., all elements have the same excitation amplitudes.

( )( ) ( )( )1 sin sin1 sin cos0

1 1

y yx x

N Nj n kdj m kd

n n

AF I e eθ φ βθ φ β − +− +

= == ∑ ∑ (18.5)

The normalized array factor can be obtained as:

sinsin21 12

( , )sin sin

2 2

yx

nx y

NMAF

M N

ψψ

θ φψ ψ

=

, (18.6)

where:sin cos

sin sinx x x

y y y

kd

kd

ψ θ φ βψ θ φ β

= += +

The major lobe (principal maximum) and grating lobes of theterms:

sin1 2

sin2

M

x

xx

MS

M

ψ

ψ

=

(18.7)

sin21

sin2

N

y

yy

N

SN

ψ

ψ

=

(18.8)

are located at angles such that:sin cos 2 , 0,1,x m m xkd m mθ φ β π+ = ± = … (18.9)sin sin 2 , 0,1,y n n ykd n nθ φ β π+ = ± = … (18.10)

The principal maxima correspond to 0, 0m n= = .

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In general, xβ and yβ are independent from each other. But, if it is

required that the main beams ofMxS and

NyS intersect (which is

usually the case), then the common main beam is in the direction:

0θ θ= and 0φ φ= , 0m n= = (18.11)If the principal maximum is specified by 0 0( , )θ φ , then theprogressive phases xβ and yβ must satisfy:

0 0sin cosx xkdβ θ φ= − (18.12)

0 0sin siny ykdβ θ φ= − (18.13)

When xβ and yβ are specified, the direction of the main beam can

be found by simultaneously solving (18.12) and (18.13):

0tan y x

x y

d

d

βφ

β= (18.14)

22

0sin yx

x ykd kd

ββθ

= ± + (18.15)

The grating lobes can be located by substituting (18.12) and(18.13) in (18.9) and (18.10):

0 0

0 0

sin sintan

sin cos

ymn

x

nd

md

λθ φφ λθ φ

±=

±(18.16)

0 00 0sin sinsin cos

sincos sin

yxmn

mn mn

nmdd

λλ θ φθ φθ

φ φ

±±= = (18.17)

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To avoid grating lobes, the spacing between the elements must beless than λ ( yd λ< and yd λ< ). In order a true grating lobe to

occur, both equations (18.16) and (18.17) must have a real solution( , )mn mnθ φ .

3-D pattern of a 5-element square planar uniform array withoutgrating lobes ( / 4d λ= , 0x yβ β= = ):

Fig. 6.24, pp.313 Balanis

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3-D pattern of a 5-element square planar uniform array withoutgrating lobes ( / 2d λ= , 0x yβ β= = ):

Fig. 6.25, pp.314, Balanis

Notice the considerable decrease in the beamwidth as the spacingis increased from / 4λ to / 2λ .

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1.2 The beamwidth of a planar array

x

y

z

hφ

hθ

0θ

0φ

hφ

A simple procedure, proposed by R.S. Elliot1 will be outlined.It is based on the use of the beamwidths of the linear arraysbuilding the planar array.

For a large array, whose maximum is near the broad side, theelevation plane HPBW is approximately:

2 2 2 2 20 0 0

1

cos cos sinh

x y

θθ θ φ θ φ− −

= +

(18.18)

1 “Beamwidth and directivity of large scanning arrays”, The Microwave Journal, Jan. 1964, pp.74-82

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where: 0 0( , )θ φ specify the main-beam direction;

xθ is the HPBW of a linear broadside array whosenumber of elements M and amplitude distributionis the same as that of the x-axis linear arraysbuilding the planar array;

yθ is the HPBW of a linear BSA whose number of

elements N and amplitude distribution is the sameas those of the y-axis linear arrays building theplanar array.

The HPBW in the plane, which is orthogonal to the 0φ φ=plane and contains the maximum, is:

2 2 2 20 0

1

sin cosh

x y

φθ φ θ φ− −=

+(18.19)

For a square array ( )M N= with amplitude distributions along the xand y axes of the same type, equations (18.18) and (18.19) reduceto:

0 0cos cosyx

h

θθθθ θ

= = (18.20)

h x yφ θ θ= = (18.21)

From (18.20), it is obvious that the HPBW in the elevation planevery much depends on the elevation angle 0θ of the main beam.The HPBW in the azimuthal plane hφ does not depend on theelevation angle 0θ .

The beam solid angle of the planar array can be approximatedby:

A h hθ φΩ = (18.22)or

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2 22 2 2 2 2

0 0 0 0 02 2cos sin cos sin cos

x yA

y x

x y

θ θ

θ θθ φ φ φ φθ θ

Ω =

+ +

(18.23)

1.3 DirectivityThe general expression for the calculation of the directivity of

an array is:2

0 00 2

20 0

0 0

| ( , ) |4

| ( , ) | sin

AFD

AF d dπ π

θ φπθ φ θ θ φ

=

∫ ∫(18.24)

For large planar arrays, which are nearly broadside, (18.24)reduces to:

0 0cosx yD D Dπ θ= (18.25)

where xD is the directivity of the respective linear BSA, x-axis;

yD is the directivity of the respective linear BSA, y-axis.

One can also use the array solid beam angle AΩ in (18.23) tocalculate the approximate directivity of a nearly broadside planararray:

2

2

0[ ] [deg ]

32400

A Sr A

Dπ

Ω Ω(18.26)

Remember:1) The main beam direction is controlled through the phase

shifts, xβ and yβ .

2) The beamwidth and side-lobe levels are controlled throughthe amplitude distribution.

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2. Circular array

2.1 Array factorThe normalized field can be written as:

( )1

, ,njkRN

nnn

eE r a

Rθ φ

−

==∑ (18.27)

where:2 2 2 cosn nR r a ar ψ= + − (18.28)

For r a , (18.28) reduces to:

( )ˆ ˆcosnn nR r a r a a rρψ− − ⋅ (18.29)

In rectangular coordinate system:ˆ ˆ ˆcos sin

ˆ ˆ ˆ ˆsin cos sin sin cosn n na x y

r x y z

ρ φ φθ φ θ φ θ

= +

= + +

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Therefore:( )sin cos cos sin sinn n nR r a θ φ φ φ φ= − + (18.30)

Finally, nR is approximated in the phase terms as:

( )sin cosn nR r a θ φ φ= − − (18.31)

For the amplitude term, the approximation1 1

, alln

nR r

(18.32)

is made.Assuming the approximations (18.31) and (18.32) are valid,

the far-zone array field is reduced to:

( ) ( )sin cos

1

, , n

jkr Njka

nn

eE r a e

rθ φ φθ φ

−−

== ∑ (18.33)

where: na is the excitation coefficient (amplitude and phase);2

n nN

πφ = is the angular position of the n-th element.

In general, the excitation coefficient can be represented as:nj

na I e α= , (18.34)where nI is the amplitude term, and nα is the phase of theexcitation of the n-th element relative to a chosen array element ofzero phase.

( ) ( )sin cos

1

, , n n

jkr Nj ka

nn

eE r I e

rθ φ φ αθ φ

−− +

=⇒ = ∑ (18.35)

The AF is obtained as:

( )sin cos

1

( , ) n n

Nj ka

nn

AF I eθ φ φ αθ φ − +

==∑ (18.36)

Expression (18.36) represents the AF of a circular array of Nequispaced elements. The maximum of the AF occurs when all thephase terms in (18.36) equal unity, or:

( )sin cos 2 , 0, 1, 2, alln nka m m nθ φ φ α π− + = = ± ± (18.37)

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The principal maximum ( 0m = ) is defined by the direction

0 0( , )θ φ , for which:

( )0 0sin cos , 1,2,...,n nka n Nα θ φ φ= − − = (18.38)

If a circular array is required to have maximum radiation in thedirection 0 0( , )θ φ , then the phases of its excitations will have tofulfil (18.38). The AF of such an array is:

( ) [ ]0 0sin cos( ) sin cos( )

1

, n n

Njka

nn

AF I e θ φ φ θ φ φθ φ − − −

==∑ (18.39)

0(cos cos )

1

( , ) n n

Njka

nn

AF I e ψ ψθ φ −

==∑ (18.40)

Here:

[ ]1cos sin cos( )n nψ θ φ φ−= − is the angle between r and ˆn

aρ ;

[ ]10 0 0cos sin cos( )

n nψ θ φ φ−= − is the angle between ˆn

aρ and maxr

pointing in the direction ofmaximum radiation.

As the radius of the array a becomes very large as compared toλ , the directivity of the uniform circular array ( 0, allnI I n= )approaches the value of N.

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Uniform circular array 3-D pattern (N=10,2

10ka aπλ

= = )