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Antenna Arrays
1 Introduction
Antenna arrays are becoming increasingly important in
wireless communications. Advantages of using antenna
arrays:
1. They can provide the capability of a steerable beam
(radiation direction change) as in smart antennas.
2. They can provide a high gain (array gain) by using
simple antenna elements.
3. They provide a diversity gain in multipath signal
reception.
4. They enable array signal processing.
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2 Far-Field Expression of An Antenna Array
Observation
point P(x,y,z) at
infinity
r1
z
r2
ρ2
ρ1
r0
rN
ri
y
ρi
x
ρN
An arbitrary antenna array
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ρ1, ρ2,…, ρi,…, ρN
= position vectors of the
antenna elements
r1, r2,…, ri,…, rN
= distances of the antenna
elements from the
observation point
r0 = distance of the origin from
the observation point
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The sum of the far fields E radiated from the array is:
N
E = ∑ Ei
(1)
i =1
where Ei is the far field of the ith antena given by:
(2)
Ei = ⎡⎣aˆ θ fθ i (θ ,φ ) + aˆ φ fφi (θ ,φ ) ⎤⎦ wi K i e − jkri
fθ i (θ ,φ ) = the θ component of the radiation pattern
fφi (θ ,φ ) = the φ component of the radiation pattern
wi = the weighting factor of the excitation source
K i = a constant accounting for the path loss
Note that fθ i (θ ,φ ) and fφi (θ ,φ ) are obtained with the ith
(for all i ) antenna element located at the origin.
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For identical antenna elements, the element radiation
patterns are the same and independent of i. Hence,
N
E = ⎡⎣aˆ θ fθ (θ ,φ ) + aˆ φ fφ (θ ,φ ) ⎤⎦ ∑ wi K i e
− jkri
(3)
i =1
Furthermore, we assume that Ki is approximately constant
such that
K1 = K 2 =
= KN = K
(4)
We can also write
ri = r0 − Δr = r0 − ρi ⋅ aˆ r (θ ,φ )
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(5)
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Then
E = Ke
− jkr0
N
⎡⎣aˆ θ fθ (θ ,φ ) + aˆ φ fφ (θ ,φ ) ⎤⎦ ∑ wi e
jkρi ⋅aˆ r (θ ,φ )
i =1
= Ke − jkr0 ⎣⎡aˆ θ fθ (θ ,φ ) + aˆ φ fφ (θ ,φ ) ⎤⎦ f array (θ ,φ )
(6)
The result in (6) is known as the Principle of Pattern
Multiplication, which states that the array pattern is the
product of the antenna element pattern multiplied with an
array factor f array (θ ,φ ) .
Hereafter, we focus on the study of the array factor.
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N
f array (θ ,φ ) = ∑ wi e
jkρi ⋅aˆ r (θ ,φ )
i =1
N
= ∑ wi e jbi
(7)
i =1
Putting in the matrix form, we have
f array (θ ,φ ) = w T b
⎡ w1 ⎤
⎢w ⎥
w=⎢ 2⎥
⎢ ⎥
⎢w ⎥
⎣ N⎦
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(8)
jkρ1⋅aˆ r (θ ,φ )
⎡
⎤
e
⎡e ⎤
⎢ jb2 ⎥ ⎢ jkρ2 ⋅aˆ r (θ ,φ ) ⎥
e ⎥ ⎢e
⎥
⎢
=⎢
b=
⎥
⎢
⎥
⎢ jbN ⎥ ⎢ jkρ N ⋅aˆ (θ ,φ ) ⎥
r
⎥⎦
⎣e ⎦ ⎢⎣e
jb1
6
(9)
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Eq. (8) is an important expression which tells us that the
array radiation pattern (array factor) can be obtained from
a given array element weight vector w and vice versa. If
we consider the weight vector as an N dimensional
function w(xi,yi,zi), i =1, 2, …, N, then the array factor
and the weight function is related by a certain type of
transformation operation ℑ. That is,
f array (θ ,φ ) = ℑ ( w ( x, y, z ) )
(10)
Similarly, the weight function w(x,y,z) can be determined
from a given required array factor function.
w ( x, y, z ) = ℑ−1 ( f array (θ ,φ ) )
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(11)
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3 Uniform Liner Arrays (ULAs)
Dipoles are
parallel to the z
direction
Far field
observation
point
y
r
φ
rN-1
x
Dipole N
Dipole 1
d
An N-element uniform antenna array with an element
separation d and at the plane of θ = π/2
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Uniform linear arrays (ULAs) mean that the array
elements are same as each other and they are aligned along
a straight line with equal element separations. If the
excitation currents have the same amplitude (I = 1) but the
phase difference between adjacent elements is β (the
progressive phase difference), then the weight vector is:
⎡ w1 ⎤ ⎡ I ⎤
⎢ w ⎥ ⎢ Ie jβ ⎥
⎥
⎢ 2⎥ ⎢
j 2β
⎥
w = ⎢ w3 ⎥ = ⎢ Ie
⎥
⎢ ⎥ ⎢
⎥
⎢ ⎥ ⎢
⎢⎣ wN ⎥⎦ ⎢⎣ Ie j( N −1)β ⎥⎦
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(12)
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The array factor (AF) for this array specified on the plane
θ = π/2 is:
N
AF = f array (θ = π 2,φ ) = ∑ wi e
N
jbi
i =1
= 1 + e j ( kd cosφ + β ) + e j 2( kd cosφ + β ) +
= I ∑e
j ( i −1) β
i =1
+ e j ( N −1)( kd cosφ + β )
ψ⎞
⎛
sin ⎜ N ⎟
ψ
N
j
N
−
1
(
)
2
j ( n −1)ψ
⎝
⎠e
2
= ∑e
=
ψ⎞
⎛
n =1
sin ⎜ ⎟
⎝2⎠
where ψ = kd cos φ + β and 0 ≤ φ , β ≤ 2π
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e
j ( i −1) d cos φ
(13)
(14)
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The normalized array factor is:
ψ⎞
⎛
sin ⎜ N ⎟
1
⎝ 2⎠
AFn (ψ ) =
Γ sin ⎛ψ ⎞
⎜ ⎟
⎝2⎠
where Γ is a constant to
make the largest value of
|AFn| equal to one. Note
that Γ is not necessarily
equal to N.
(15)
The relation between |AFn|, Ψ, d, and β is shown graphically
on next page. Note that |AFn| is a period function of Ψ,
which is in turn a function of φ. The angle φ is in the real
space and its range is 0 to 2π. However, Ψ is not in the real
space and its range can be greater than or smaller than 0 to
2π, leading to the problem of grating lobes or not achieving
the maximum values of the |AFn| expression.
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|AFn(Ψ )|
Ψ = kd cosφ +β
kd
β
φ
kdcosφ
The relation between AFn,Ψ, d, and β
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3.1 Properties of the normalized array factor AFn:
1. |AFn| is a periodic function of ψ, with a period of 2π.
This is because
(16)
|AFn(ψ + 2π)| = |AFn(ψ)|.
2. As cos(φ) = cos(-φ), |AFn| is symmetric about the line of
the array, i.e., φ = 0 & π. Hence it is enough to know
|AFn| for 0 ≤ φ ≤ π.
3. The maximum values of |AFn| occur when (see
Supplementary Notes):
ψ 1
= (kd cos φ + β ) = ± mπ , m = 0,1,2,
2 2
−1 ⎡ λ
(− β ± 2mπ ) ⎤⎥ (17)
⇒ φmax = main beam directions = cos ⎢
⎣ 2π d
⎦
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Note that there may be more than one angles φmax
corresponding to the same value of m because cos-1(x) is
a multi-value function. If there are more than one
maximum angles φmax, the second and the subsequent
maximum angles give rise to the phenomenon of grating
lobes. The condition for grating lobes to occur is that d ≥
λ (disregarding the value of β) as shown below:
2nd grating lobe
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1st grating lobe
Main lobe
14
1st grating lobe
2nd grating lobe
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2nd grating lobe 1st grating lobe
1st grating lobe
Main lobe
2nd grating lobe
|AFn(Ψ )|
Ψ
(1) When d ≤ 0.5λ, no
grating lobes can be
formed for whatever
value of β. (2) When d
≥ λ, grating lobe(s) is
(are)
formed
for
whatever value of β.
(3) When 0.5λ <d< λ,
formation of grating
lobes depends on β.
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Visible region
kd
φ
ψ=kdcosφ
15
General conditions to
avoid grating lobes
with β [0,2π] and d
[0.5λ,λ]:
1.For 0 ≤ β < π, the
requirement is:
kd + β ≤ 2π
2. For π ≤ β < 2π, the
requirement is:
kd - β ≤ 0
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4. There are other angles corresponding to the maximum
values for the minor lobes (minor beams) but these angles
cannot be found from the formula in no. 3 above.
5. When β and d are fixed, it is possible that ψ can never be
equal to 2mπ. In that case, the maximum values of |AFn|
cannot be determined by the formula in no. 3.
6. The main beam directions φmax are not related to N. They
are functions of β and d only.
7. The nulls of |AFn| occur when:
nπ
ψ
⎧n = 1,2,3,
=± ,
⎨
N
2
⎩n ≠ N ,2 N ,3 N ,
2n ⎤
−1 ⎡ λ
⇒ φnull = null directions = cos ⎢
(− β ± π ) ⎥ (18)
N ⎦
⎣ 2π d
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Note that there may be more than one angles φnull
corresponding to a single value of n because cos-1(x) is a
multi-value function.
8. The null directions φnull are dependent on N.
9. The larger the number N, the closer is the first null (n = 1)
to the first maximum (m = 0). This means a narrower
main beam and an increase in the directivity or gain of the
array.
10.The angle for the main beam direction (m = 0) can be
controlled by varying β or d.
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Example 1
A uniform linear array consists of 10 half-wave dipoles with
an inter-element separation d = λ/4 and equal current
amplitude. Find the excitation current phase difference β
such that the main beam direction is at 60° (φmax = 60°).
Solutions d = λ/4, θmax = 60°, N = 10
main beam dirction = φmax
⇒
⇒
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−
2
π
λ
⎡
⎤
β
m
2
π
= 60° = cos ⎢ −
±
(
)⎥
⎣ 2π d
⎦
−1
( β ± m2π ) = cos ( 60° ) = 0.5
π
β = − ∓ m2π = −45° + 360° = 315°,
4
18
when m = 1
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Other values of β corresponding to other values of m are
outside the range of 0 ≤ β ≤ 2π and are not included.
⎡ ⎛π
⎤
⎞
sin ⎢5 ⎜ cos φ + β ⎟ ⎥
1
2
⎝
⎠⎦
⎣
AFn =
Γ
⎡1 ⎛ π
⎤
⎞
sin ⎢ ⎜ cos φ + β ⎟ ⎥
⎠⎦
⎣2⎝ 2
Γ = 10
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3.2 Phased (Scanning) Arrays
It was mentioned earlier that by controlling the values of d
or β, the maximum radiation direction of an array can be
arbitrarily pointed to any direction. In practice, the
element separation d is usually fixed while the excitation
current phase β between elements is controlled
electronically. The current amplitudes of the all the
elements are assumed to be the same. This kind of
steerable direction arrays is called uniform phased
scanning arrays. To accomplish this, the excitation
current phase β must be adjusted so that:
ψ = kd cos φ + β φ =φ = 0,
0
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⇒
β = − kd cos φ0
(19)
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For a phased scanning array, the length L = (N-1)d of the
array (where N is the number of elements) can be
determined from the graph on next page. The graph
shows the relation between the array length, the halfpower beamwidth, and the maximum radiation direction.
The half-power beamwidth is an alternative way to
specify the gain of the array. Once the half-power
beamwidth and the maximum radiation direction are
specified, the number
of elements required to design such
y
an array can be calculated.
L
Ant. 1
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Ant. N
d
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Maximum radiation direction
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Example 2
Design a uniform linear phased scanning array whose
maximum radiation direction is in 30° (θ = 30°). The
desired half-power beamwidth is 2° while the element
separation is d = λ/4. Determine the excitation current phase
β, the length of the array L, and the number of elements N in
the array.
Solutions
Since the array is uniform, the current amplitude is same for
all elements. The excitation current phase β is found from:
2π λ
o
o
β = − kd cosθ 0 = −
cos30 = −1.36 rad = −77.94
λ 4
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To find the length L of the array, we use the graph on page
24. From that graph with HPBW=2° and maximum
radiation direction θ =30°:
(L+d)/λ=50
Therefore with d = λ/4,
L = 49.75λ
The number of elements is then:
L
49.75
N = +1 =
+ 1 = 200
d
0.25
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4 Circular Arrays
Δrn = a cosψ n
= a sin θ cos (φ − φn )
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4.1 Advantages of Circular Arrays
1. Unlike linear arrays, circular arrays can provide a 2D
angular scan, both horizontal φ and vertical θ scans.
2. Unlike 2D planar arrays, circular arrays are basically
1D linear arrays but in a circular form.
3. Unlike linear arrays, a circular array can scan
horizontally for 360° with no distortions near the
end-fire directions.
4. Unlike linear arrays, distortions in the array pattern
of a circular array due to mutual coupling effect are
same for each element and this makes it easier to
deal with the mutual coupling effect.
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4.2 Array Factor
For a uniform circular array with N elements and an
equal excitation current amplitude I0 and a current
phase of βn (reference to the central point of the array)
for the nth element (and φn=2πn/N), the array factor is:
AF = e
j[ ka sin θ cos(φ −φ1 )+ β1 ]
+
+e
N
= ∑e
+e
j[ ka sin θ cos(φ −φ2 )+ β 2 ]
j[ ka sin θ cos(φ −φN ) + β N ]
j[ ka sin θ cos(φ −φn )+ β n ]
(20)
n =1
(For a detailed derivation of the AF above, see ref. [1])
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Note that in the above expression, AF is a sum of N
complex exponentials. The magnitude of each complex
exponential is 1. Hence the maximum value of the
magnitude of |AF| is the addition of magnitudes of the
complex exponentials, i.e., N. The maximum radiation
direction (θmax, φmax) is therefore achieved when:
⎧ka sin θ max cos (φmax − 2π N ) + β1 = ±2qπ
⎪ka sin θ cos (φ − 4π N ) + β = ±2qπ
⎪
max
max
2
⎨
⎪
⎪⎩ka sin θ max cos (φmax − 2π ) + β N = ±2qπ
q = 0,1,2,
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(21)
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One excitation method to achieve the above maximum
radiation direction is:
β n = ±2qπ − ka sin θ max cos (φmax − 2π n N ) ,
q is choosen to make β n ∈ [ 0,2π )
n = 1,2,
, N (22)
Hence in a circular array, we first choose a desired
maximum radiation direction (θmax, φmax). Then the
excitation phase βn for each element is determined
according to the above formula. The βn so determined
may not be equally increasing from one element to the
next. This is different from the case of a linear array.
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Example 3
A uniform circular array with a radius a = 0.5λ and the
number of elements N = 8. The maximum radiation
direction of the array factor AF is at (60°, 30°). What should
be the excitation phases βn for the elements?
Solutions
Using the formula for βn , we have:
2π λ π
π 2π ⎞
⎛
β1 = ±2qπ −
sin cos ⎜ −
⎟
λ 2 3
⎝6 8 ⎠
= 2π − 2.63
= 3.66
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2π λ π
π 4π ⎞
⎛
β 2 = ±2qπ −
sin cos ⎜ −
⎟
λ 2 3
⎝6 8 ⎠
= 2π − 1.36
= 4.92
2π λ π
π 6π ⎞
⎛
β 3 = ±2qπ −
sin cos ⎜ −
⎟
λ 2
3
⎝6 8 ⎠
= 0 + 0.70
= 0.70
2π λ π
π 8π ⎞
⎛
β 4 = ±2qπ −
sin cos ⎜ − ⎟
λ 2
3
⎝6 8 ⎠
= 0 + 2.36
= 2.36
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2π λ π
π 10π ⎞
⎛
β 5 = ±2qπ −
sin cos ⎜ −
⎟
λ 2 3
8 ⎠
⎝6
= 0 + 2.63
= 2.63
2π λ π
π 12π ⎞
⎛
β 6 = ±2qπ −
sin cos ⎜ −
⎟
λ 2 3
8 ⎠
⎝6
= 0 + 1.36
= 1.36
2π λ π
π 14π ⎞
⎛
β 7 = ±2qπ −
sin cos ⎜ −
⎟
λ 2 3
8 ⎠
⎝6
= 2π − 0.70
= 5.58
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2π λ π
π 16π ⎞
⎛
β8 = ±2qπ −
sin cos ⎜ −
⎟
λ 2 3
8 ⎠
⎝6
= 2π − 2.36
= 3.92
Vertical pattern (φ = 30°)
Azimuth pattern (θ = 60°)
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Note that in the above discussion on circular arrays, we
have only derived the array factor AF. The array pattern
of any circular array with practical antenna elements
must be obtained by multiplying the array factor with the
actually element radiation pattern. For example, if the
elements are half-wave dipole antennas, then the array
pattern F(θ) is:
cos [(π 2 ) cosθ ]
F (θ ) =
AF
sin θ
cos [(π 2 ) cosθ ] N j[ ka sin θ cos(φ −φn )+ βn ]
=
e
∑
sin θ
n =1
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References:
1. C. A. Balanis, Antenna Theory, Analysis and Design, John Wiley
& Sons, Inc., New Jersey, 2005.
2. W. L. Stutzman and G. A. Thiele, Antenna Theory and Design,
Wiley, New York, 1998.
3. David K. Cheng, Field and Wave Electromagnetic, AddisonWesley Pub. Co., New York, 1989.
4. John D. Kraus, Antennas, McGraw-Hill, New York, 1988.
5. Fawwaz T. Ulaby, Applied Electromagnetics, Prentice-Hall, Inc.,
New Jersey, 2007.
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