Lecture 33
Antenna Arrays and Phase Arrays
In this lecture you will learn:
• Antenna arrays
• Gain and radiation pattern for antenna arrays
• Antenna beam steering with phase array antennas
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Two Hertzian Dipoles – A Two Element Array
z
Consider first an array of just two Hertzian dipoles:
(
(
r r
r r
J0 (r ) = zˆ I0 d δ 3 r − h0
r r
r r
J1(r ) = zˆ I1 d δ 3 r − h1
)
)
r
J0
r r
h0 h1
r
J1
y
x
Each antenna in the array is an “element” of the array
One can write the E-field in the far-field as a superposition of the E-fields produced
by all the elements in the array:
[
r
r
r r
j ηo kd
ˆ
ˆ
Eff (r ) = θˆ
sin(θ ) e − j k r I0 e j k r .h0 + I1 e j k r .h1
4π r
]
r
r
⎡ I
⎤
I
j ηo k I0d
ˆ
ˆ
= θˆ
sin(θ ) e − j k r ⎢ 0 e j k r .h0 + 1 e j k r .h1 ⎥
4π r
I
I
⎣ 0
⎦
0
r Element Factor
= Ε(r ,θ , φ ) F (θ , φ )
Array Factor
• The “element factor” is just the E-field produced by the first element if it were
sitting at the origin
• The “array factor” captures all the interference effects
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
1
An N-Element Antenna Array - I
Consider an N-element antenna array where the elements are:
- all identical (all loops, or all Hertzian dipoles, or all Half-wave dipoles, etc)
- all oriented in the same way
- but with possibly different current phasors
Let the current phasor of the m-th antenna be Im
r
Let the position vector of the m-th antenna be hm
One can write the E-field in the far-field as a superposition of the
E-fields produced by all the elements in the array:
z
r
hm
r
N −1 I
r r
r
m e j k rˆ .hm
Eff (r ) = Ε (r ,θ , φ ) ∑
m = 0 I0
r
= Ε (r ,θ , φ ) F (θ , φ )
IN −1
Im
y
I1
z I0
x
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
An N-Element Antenna Array - II
r r
r r N −1 Im j k rˆ.hr
m
Eff (r ) = Ε(r ) ∑
e
m = 0 I0
r
= Ε(r ,θ , φ ) F (θ , φ )
Gain: G (θ , φ ) =
Pattern:
r r
S (r , t ) .rˆ
P 4π r
p(θ , φ ) =
2
r
2
2
∝ ⎡angular part of Ε(r,θ , φ ) ⎤ F (θ , φ )
⎥⎦
⎢⎣
G (θ , φ )
G (θ , φ ) max
z
r
hm
IN −1
Im
y
I1
z I0
x
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
2
An N-Element Hertzian Dipole Phase Array
z
N-element Hertzian dipole array
• Current magnitude is the same for all
the dipoles
a
e jα
e j 2α
1
• Current phase difference between
successive dipoles is α
y
e j (N −1)α
x
I m +1
= ejα
Im
Element Factor:
r
j ηo k I0d
Ε(r ,θ , φ ) = θˆ
sin(θ ) e − j k r
4π r
Array Factor:
r
N −1
Im j k rˆ.hrm
ˆ
e
= e j k r .h0 ∑
I0
m =0
N −1
F (θ , φ ) = ∑
m =0
⇒
N −1
F (θ , φ ) = ∑
2
m =0
(e
e jα m e j k a m sin(θ ) cos(φ )
)
j k a sin(θ ) cos (φ )+α m
2
⎡N
⎤
sin2 ⎢ (α + ka sin(θ ) cos(φ ))⎥
2
⎣
⎦
=
⎤
2⎡1
sin ⎢ (α + ka sin(θ ) cos(φ ))⎥
⎦
⎣2
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
An N-Element Hertzian Dipole Phase Array
z
N-element Hertzian dipole phase array antenna
a
P=
e j 2α
x
e
e
jα
1
ηo
12π
y
G (θ , φ ) =
j (N − 1)α
2
k I0 d N
r r
S (r , t ) .rˆ
P 4π r 2
=
3
2
sin2 (θ ) F (θ , φ )
2N
Coming from the
element factor
Pattern:
p(θ , φ ) =
1
N2
sin2 (θ ) F (θ , φ )
array factor
2
⎡N
⎤
sin2 ⎢ (α + ka sin(θ ) cos(φ ))⎥
2
⎣
⎦
2
p(θ , φ ) = 2 sin (θ )
⎤
2⎡1
N
sin ⎢ (α + ka sin(θ ) cos(φ ))⎥
⎦
⎣2
1
Coming from the
element factor
array factor
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
3
An N-Element Hertzian Dipole Phase Array: Maxima
y
Lets look at radiation in the x-y plane:
a cos(φ )
Radiation going in the φ direction from
adjacent elements will add in-phase,
and one will have a big maximum in the
radiation pattern, provided:
φ
k a cos(φ ) + α = ±2π n
{ n = 0,1,2,3,KK
x
a
F (θ = π 2 , φ )
p (θ = π 2 , φ )
ka =
big
maxima
(main lobes)
N = 20
N = 20
α=−
2
side
lobes
main
lobes
π
4
φ
α=−
ka =
π
π
4
π
nulls
small
maxima
(side lobes)
2
2
φ (degrees)
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
An N-Element Hertzian Dipole Phase Array: Maxima
y
Condition for a big maximum in the x-y
plane:
a cos(φ )
k a cos(φ ) + α = ±2π n
{ n = 0,1,2,3,KK
φ
x
a
F (θ = π 2 , φ )
N = 20
α=−
ka =
2
F (θ = π 2 , φ )
big
maxima
(main lobes)
π
4
π
At a big maximum the value of the array
factor is:
nulls
2
φ (degrees)
small
maxima
(side lobes)
2
max
⎡N
⎤
sin2 ⎢ (α + ka cos(φ ))⎥
2
⎣
⎦
=
⎤
2⎡1
sin ⎢ (α + ka cos(φ ))⎥
⎣2
⎦
= N2
And the value of the antenna gain is:
G (θ , φ ) =
3
2
sin2 (θ ) F (θ , φ )
2N
π ⎞
3
⎛
⇒ G⎜ θ = ,φ ⎟
= N
2
2
⎝
⎠max
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
4
An N-Element Hertzian Dipole Phase Array: Nulls
y
Radiation in the x-y plane:
a cos(φ )
The array factor will give a null in the
radiation pattern provided:
2π n
N
{ n ≠ 0, N ,2N ,3N ,KK
k a cos(φ ) + α = ±
φ
x
a
F (θ = π 2 , φ )
F (θ = π 2 , φ )
2
⎡N
⎤
sin2 ⎢ (α + ka cos(φ ))⎥
2
⎣
⎦
=
⎤
2⎡1
sin ⎢ (α + ka cos(φ ))⎥
⎣2
⎦
2
big
maxima
(main lobes)
N = 20
α=−
ka =
π
small
maxima
(side lobes)
4
π
nulls
2
φ (degrees)
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
An N-Element Hertzian Dipole Phase Array: Width of Big Maxima
y
F (θ = π 2 , φ )
2
N = 20
φ
a
α=−
x
ka =
π
4
π
2
φ
Angular width of a lobe associated with a big maxima:
∆φ
nN
For a big maxima: k a cos(φ ) + α = 2π n = 2π
N
{ n = 0,±1,±2,±3,KK
At the nulls nearest to the above maxima we must have:
nN ± 1
∆φ ⎞
⎛
k a cos⎜ φ +
⎟ + α = 2π
2 ⎠
N
⎝
Which can be solved for ∆φ – and for N large (N >> 1)
one gets approximately:
∆φ ≈
4π
1
N ka sin(φ )
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
5
An N-Element Hertzian Dipole Phase Array: Examples
z
N-element Hertzian dipole array
a
ka = π
e j 2α
p (θ , φ )
x
Phase difference between
successive elements is α
e jα
1
y
e j (N −1)α
p (θ , φ )
N=2
N = 10
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
An N-Element Hertzian Dipole Phase Array: Examples
z
N-element Hertzian dipole array
a
ka = π / 2
e j 2α
p (θ , φ )
x
e jα
Phase difference between
successive elements is α
1
y
e j (N −1)α
N=2
p (θ , φ )
N = 10
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
6
An (N+1)-Element Hertzian Dipole Binomial Array
a=
⇒
z
λ
(N+1)-element Hertzian dipole array
2
ka = π
• Current phase is the same for all the
dipoles
a
• Currents magnitudes in the dipoles
follow a binomial distribution
y
Im
N!
=
I0 m! (N − m )!
x
Element Factor:
r
j ηo k I0d
Ε(r ,θ , φ ) = θˆ
sin(θ ) e − j k r
4π r
Array Factor:
r
N
F (θ , φ ) = ∑
m =0
⇒
N
Im j k rˆ.hrm
ˆ
e
= e j k r .h0 ∑
I0
m =0
F (θ , φ ) = 1 + e j k a sin(θ ) cos(φ )
2
2N
N!
e j k a m sin(θ ) cos(φ )
m! (N − m )!
⎡k
⎤
= 22N cos2N ⎢ a sin(θ ) cos(φ )⎥
⎣2
⎦
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
An (N+1)-Element Hertzian Dipole Binomial Array: Maxima and Nulls
a cos(φ )
y
Lets look at radiation in the x-y plane:
Radiation going in the φ direction from
adjacent elements will add in-phase,
and one will have a big maximum in the
radiation pattern, provided:
φ
x
a
2N ⎡ π
⎤
F (θ = π 2 , φ ) = 2 cos ⎢ cos(φ )⎥
⎣2
⎦
⎤
2N ⎡ π
p(θ = π 2 , φ ) = cos ⎢ cos(φ )⎥
⎦
⎣2
2
2N
k a cos(φ ) = ±2π n
⇒
π
2
{ n = 0,1,2,3,KK
cos(φ ) = ±π n
p (θ = π 2 , φ )
p (θ = π 2 , φ )
N+1=40
Note:
N+1=2
N+1=40
N+1=2
φ (degrees)
• No side lobes
• Two nulls
at φ = 0,π
φ (degrees)
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
7
Applications of Antenna Arrays - I
Phased Array TRack to Intercept
Of Target
Alaska
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Applications of Antenna Arrays - II
Communication satellite
with two phase array
antenna panels
MRI system with phase array
RF receivers
Jicamarca facility (Peru) with
an array of half-wave dipoles
SETI
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
8