Antennas
Simple array of two antennas
Consider two identical parallel Hertzian dipoles separated by a
distance d
iθ
iθ i
2 θ
z
r2
θ2
I2
θ
d 2
© Amanogawa, 2006 – Digital Maestro Series
1
r
r1
θ1
ψ
d 2
I1
x
64
Antennas
The dipole currents have the same amplitude and total phase
difference α
I1 ( t ) = Io cos(ω t + α 2)
I2 ( t ) = Io cos(ω t − α 2)
⇒
phasor
⇒
phasor
I1 = Io e jα 2
I 2 = I o e− jα 2
The electric far−field components at the observation point are
E1 ≈ i θ1
E2 ≈ i θ 2
− jβ r1 + jα 2
µ jβ Io ∆ z e
4π r1
ε
− jβ r2 − jα 2
µ j β Io ∆ z e
4π r2
ε
© Amanogawa, 2006 – Digital Maestro Series
sin θ 1
sin θ 2
65
Antennas
At long distance, we have
r >> d
θ1 ≈ θ 2 ≈ θ
iθ 1 ≈ iθ 2 ≈ i θ
d
r1 ≈ r − cosψ
2
d
r2 ≈ r + cosψ
2
and the field components can be written as
E1 ≈ i θ
µ jβ Io ∆ z e− jβ ( r − ( d 2) cosψ ) + jα 2
sin θ
ε
4π ( r − ( d 2) cosψ )
E2 ≈ i θ
µ jβ Io ∆ z e− jβ ( r + ( d 2) cosψ )− jα 2
sin θ
ε
4π ( r + ( d 2) cosψ )
© Amanogawa, 2006 – Digital Maestro Series
66
Antennas
After applying the approximations, the two components can be
combined to give the total electric field
µ j β Io ∆ z sin θ − jβ r
E = E1 + E2 ≈ i θ
e
ε
4π r
j β ( d 2) cosψ +α 2 )
− j β ( d 2) cosψ +α 2 )
× e (
+e (
(
)
The final result is
E ≈ iθ
µ j β Io ∆ z sin θ − jβ r
 β d cosψ + α 
e
2 cos


ε
4π r
2
field of a Hertzian dipole located
at the center of the array
© Amanogawa, 2006 – Digital Maestro Series
array factor
67
Antennas
The resultant radiation pattern of the electric field is proportional to
sin θ
unit
pattern
 β d cosψ + α 
× cos


2
group pattern
The unit pattern is proportional to the radiation pattern of the
individual antennas, assumed to be identical.
The group pattern is proportional to the radiation pattern the array
would have with isotropic antennas.
• Note: on the x−y plane, ψ coincides with the azimuthal angle ϕ.
Following are examples of two−antenna arrays with specific values
of dipole distance and current phase difference.
© Amanogawa, 2006 – Digital Maestro Series
68
Antennas
d = λ/2
α=0
Broad-side pattern
z
z
z
x
Unit Pattern
x
Group Pattern
y
© Amanogawa, 2006 – Digital Maestro Series
Resultant Pattern
y
y
x
x
x
x
69
Antennas
d = λ /2
α=0
Broad-side pattern
Radiation Pattern for E and H
© Amanogawa, 2006 – Digital Maestro Series
Power Radiation Pattern
70
Antennas
d = λ /2
α = 180
End-fire pattern
Radiation Pattern for E and H
© Amanogawa, 2006 – Digital Maestro Series
Power Radiation Pattern
71
Antennas
d = λ /4
α = 90
Cardioid pattern
Radiation Pattern for E and H
© Amanogawa, 2006 – Digital Maestro Series
Power Radiation Pattern
72
Antennas
d = λ /4
α = −90
Cardioid pattern
Radiation Pattern for E and H
© Amanogawa, 2006 – Digital Maestro Series
Power Radiation Pattern
73
Antennas
d = λ /2
α = 90
Radiation Pattern for E and H
© Amanogawa, 2006 – Digital Maestro Series
Power Radiation Pattern
74
Antennas
d=λ
α=0
Radiation Pattern for E and H
© Amanogawa, 2006 – Digital Maestro Series
Power Radiation Pattern
75
Antennas
d=λ
α = 180
Radiation Pattern for E and H
© Amanogawa, 2006 – Digital Maestro Series
Power Radiation Pattern
76
Antennas
CASE STUDY - 1) A radio broadcast transmitter is located 15 km
West of the city it needs to serve. The FCC standard is to have 25
mV/m electric field strength in the city. How much radiation power
must be provided to a quarter wavelength monopole?
We consider θ=90° for transmission in the plane perpendicular to
the antenna
E = iθ
µ j e− jβ r Imax
 π cos90° 
cos

ε 2π r sin 90° 
2
Imax
E = 120π
= 0.025 V/m
2π × 15, 000
⇒
λ / 2 dipole
Imax = 6.25 A
In a monopole, the lower wire is substituted by the ground. The
equivalent radiation resistance is half that of the corresponding
dipole. Therefore, the total radiated power is half the power
radiated by the half-wavelength dipole, for the same current.
© Amanogawa, 2006 – Digital Maestro Series
77
Antennas
Monopole
P
λ/4
GROUND PLANE
A perfect ground would act like a metal surface, reflecting 100% of
the signal. The ground creates an image of the “missing” wire
delivering to a given point above the ground the same signal as a
complete dipole. The transmission line connected to the antenna
sees only half of the radiation resistance, with total radiated power:
1 2
µ
2 73.07
Ptot = Imax
⋅ 0.193978 / 2 = 0.5 ⋅ 6.25 ⋅
= 713.6W
2
ε
2
Req
© Amanogawa, 2006 – Digital Maestro Series
78
Antennas
2) Improve the design by using a two−antenna array.
A good choice of array parameters is
d = λ /4
phase(antenna B) − phase(antenna A) = α = −90°
which gives a cardioid pattern
d
A
B
15 km
© Amanogawa, 2006 – Digital Maestro Series
79
Antennas
The Poynting vector is given by
P( t ) = i r
µ
2
Imax
ε 8π r sin θ
2 2
2
cos
2π

cos θ 

2
4 cos
2β
( array factor ) 2
λ 2 dipole Poynting vector
= ir
µ
2
Imax
ε 8π r sin θ
2 2
2
cos
2π


d cos ψ + α 

2
cos θ 

2
4 cos
2π
4
cos ψ −
π
4
( array factor ) 2
Note
cosψ = sin θ cos ϕ
R cos ψ
R sin θ cos ϕ
© Amanogawa, 2006 – Digital Maestro Series
θ
ψ
R
ϕ
R sin θ sin ϕ
R sin θ
80
Antennas
The total radiated power is
Only π/2 because it is a monopole
2π
2 π 2
Ptot = r
sin θ dθ
P( t ) dϕ
0
0
∫
π 2
2
Ptot = r ∫
0
∫
2
I max
µ
2  π cos θ 
cos
sin
θ
θ
d


ε 8π 2 r 2 sin 2 θ
 2 
2π
π
2 π
∫0 4cos  4 sin θ cosϕ − 4  dϕ
( array factor )2
© Amanogawa, 2006 – Digital Maestro Series
81
Antennas
The integral over the azimuthal angle ϕ gives
2π
π
2 π
4
cos  sin θ cos ϕ −  dϕ
0
4
4
∫
2π  1 1
π
π 

=4
+ cos  sin θ cos ϕ −   dϕ

0 2 2
2 
2
∫
2π  1 1
π

  dϕ = 4π
=4
+
θ
ϕ
sin
sin
cos


0  2 2
2


∫
For a monopole, we only have the integral
2π
dϕ = 2π
0
∫
© Amanogawa, 2006 – Digital Maestro Series
82
Antennas
In the direction of maximum
θ = 90° & ϕ = 90° ⇒ array factor = 2
The same field strength (25 mV/m) is obtained by applying half the
current of the original monopole to the array elements (also
monopoles)
6.25
Imax =
= 3.125 A
2
The total radiated power is proportional to the square of the current,
and the integral over ϕ gives a factor 4π instead of 2π for the array.
Overall, the total radiated power needed by the array, to produce
the same electric field, is half that of the individual monopole
714
Ptot =
= 357 W
2
© Amanogawa, 2006 – Digital Maestro Series
83
Antennas
For a monopole
Radiated Power = 0.5×1428.3127
= 714.155635 [ W ]
Rad. Resistance = 0.5×73.129616
= 36.564808 [ Ω ]
SAME FEEDING CURRENT
FOR BOTH DIPOLE AND
MONOPOLE
© Amanogawa, 2006 – Digital Maestro Series
84
Antennas
For monopoles
Radiated Power
= 0.5×714.1532
= 357.0766 [ W ]
SAME
FEEDING
CURRENTS FOR
BOTH
DIPOLES
AND MONOPOLES
© Amanogawa, 2006 – Digital Maestro Series
85
Antennas
N-Element Antenna Array
Assume a uniform array of N identical antennas. The elements are
fed by currents with constant amplitude and with phase increasing
by an amount α from one to the other. The spacing d between the
antennas is uniform.
r
1 23
r − ( N − 1) d cosψ
ψ
N
d
I (1) = Io ; I (2) = Io e jα ; … ; I ( N ) = Io e j( N −1)α
© Amanogawa, 2006 – Digital Maestro Series
86
Antennas
The electric field at the observation point (r,ψ) is of the form
E( r, ψ ) = Eo e− jβ r + Eo e− jβ ( r − d cosψ ) e jα + …
… + Eo e− jβ ( r − ( N −1) d cosψ ) e j ( N −1)α
= Eo e− jβ r 1 + e j( β d cosψ +α ) +

… + e j( N −1)( β d cosψ +α ) 

= Eo e
jN ( β d cosψ +α )
−
1
e
− jβ r
N −1
We have used
∑
1 − e j( β d cosψ +α )
e jn( β d cosψ +α ) =
n= 0
© Amanogawa, 2006 – Digital Maestro Series
1 − e jN ( β d cosψ +α )
1 − e j ( β d cosψ +α )
87
Antennas
The magnitude of the electric field is given by
E( r, ψ ) = Eo
1 − e jN ( β d cosψ +α )
1 − e j ( β d cosψ +α )
sin[ N(β d cosψ + α ) / 2]
= Eo
sin[(β d cosψ + α ) / 2]
array factor
We have used
1− e
jx
x jx 2
x
= 2 j sin e
= 2 sin
2
2
© Amanogawa, 2006 – Digital Maestro Series
88
Antennas
We can rewrite
1 sin[ N (β d cosψ + α ) / 2]
E( r, ψ ) = N Eo
N sin[(β d cosψ + α ) / 2]
group pattern
The group pattern has
Maxima when
Nulls
β d cosψ + α = 0, 2π , 4π …
when N ( β d cosψ + α ) = 2 mπ
for m = integer ≠ 0, N , 2 N ,…
© Amanogawa, 2006 – Digital Maestro Series
89
Antennas
Examples
Broadside array (plots on the azimuthal plane, with θ = 90°)
ϕ
2 dipoles
© Amanogawa, 2006 – Digital Maestro Series
4 dipoles
90
Antennas
8 dipoles
© Amanogawa, 2006 – Digital Maestro Series
16 dipoles
91
Antennas
End-fire array (plots on the azimuthal plane, with θ = 90°)
2 dipoles
© Amanogawa, 2006 – Digital Maestro Series
4 dipoles
92
Antennas
8 dipoles
© Amanogawa, 2006 – Digital Maestro Series
16 dipoles
93
Antennas
Cardioid array (plots on the azimuthal plane, with θ = 90°)
2 dipoles
© Amanogawa, 2006 – Digital Maestro Series
4 dipoles
94
Antennas
8 dipoles
© Amanogawa, 2006 – Digital Maestro Series
16 dipoles
95