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
0
