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Antennas
Amanogawa, 2006 Digital Maestro Series 65
The dipole currents have the same amplitude and total phase
difference
The electric farfield components at the observation point are
( ) ( )
21 1
22 2
cos( 2) I
cos( 2) I
jo o
phasor
jo o
phasor
I t I t I e
I t I t I e
= + == =
j r j o
j r j o
j I z ei
r
j I z ei
r
1
1
2
2
2
1 1
12
2 22
E sin
4
E sin4
+
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Antennas
Amanogawa, 2006 Digital Maestro Series 66
At long distance, we have
and the field components can be written as
1 2
1 2
1 2cos cos2 2
i i i
d dr r
d
r
r
r
+
>>
j r d j oj I z ei
r d
( ( 2)cos ) 2
1E4 ( 2)cos
+ ( )
j r d j oj I z ei
r d
( ( 2)cos ) 2
2
sin
E4 ( 2)cos
+ + ( ) sin
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Antennas
Amanogawa, 2006 Digital Maestro Series 67
After applying the approximations, the two components can be
combined to give the total electric field
The final result is
( ) ( )( )j ro
j d j d
j I z i e
r
e e
1 2
( 2)cos 2 ( 2)cos 2
sinE E E
4
+ += +
+
j ro j I z d i e
rfield of a Hertzian dipole located
at the center of the arraarray
yfactor
sin cosE 2cos
4 2
+
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Antennas
Amanogawa, 2006 Digital Maestro Series 68
The resultant radiation pattern of the electric field is proportional to
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 arraywould have with isotropic antennas.
Note: on thexy plane, coincides with the azimuthal angle .
Following are examples of twoantenna arrays with specific valuesofdipole distance and current phase difference.
d
group patter
unit
patt rn ne
cossin cos
2
+
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Antennas
Amanogawa, 2006 Digital Maestro Series 69
Broad-side/ p2 e0 att rnd =
x x
x
y
x
yUnit Pattern Group Pattern
x
yResultant Pattern
x
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Antennas
Amanogawa, 2006 Digital Maestro Series 70
d Broad-side p/ rn0 t2 at e =Radiation Pattern forEandH Power Radiation Pattern
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Antennas
Amanogawa, 2006 Digital Maestro Series 71
d End-fir/ 2 1 e pattern80 = Radiation Pattern forEandH Power Radiation Pattern
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Antennas
Amanogawa, 2006 Digital Maestro Series 72
d Cardioid/ 4 90 pattern = Radiation Pattern forEandH Power Radiation Pattern
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Antennas
Amanogawa, 2006 Digital Maestro Series 73
d Cardioid patter4 9 n/ 0 = Radiation Pattern forEandH Power Radiation Pattern
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Antennas
Amanogawa, 2006 Digital Maestro Series 74
d / 2 90 = Radiation Pattern forEandH Power Radiation Pattern
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Antennas
Amanogawa, 2006 Digital Maestro Series 75
d 0 = Radiation Pattern forEandH Power Radiation Pattern
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Antennas
Amanogawa, 2006 Digital Maestro Series 76
d 180 =
Radiation Pattern forEandH Power Radiation Pattern
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Antennas
Amanogawa, 2006 Digital Maestro Series 77
CASE STUDY - 1) A radio broadcast transmitter is located 15 kmWest of the city it needs to serve. The FCC standard is to have 25mV/m electric field strength in the city. How much radiation powermust be provided to a quarter wavelength monopole?
We consider =90 for transmission in the plane perpendicular tothe antenna
In a monopole, the lower wire is substituted by the ground. Theequivalent radiation resistance is half that of the correspondingdipole. Therefore, the total radiated power is half the powerradiated by the half-wavelength dipole, for the same current.
max
maxmax
/ 2 dipolecos90
E cos
2 sin90 2
E 120 0.025 V/m 6.25 A2 15,000
j r j e I
i
rI
I
= = =
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Antennas
Amanogawa, 2006 Digital Maestro Series 78
A perfect ground would act like a metal surface, reflecting 100% ofthe signal. The ground creates an image of the missing wiredelivering to a given point above the ground the same signal as acomplete dipole. The transmission line connected to the antennasees only halfof the radiation resistance, with total radiated power:
2 2max
1 73.070.193978 0.5 6.25 713.62
2W
2/
eqR
totP I= = =
P
GROUND PLANE
Monopole
/4
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Antennas
Amanogawa, 2006 Digital Maestro Series 79
2) Improve the design by using a twoantenna array.A good choice of array parameters is
which gives a cardioid pattern
d
phase(antenna B) phase(antenna A) 90
/ 4
= = =
15 km
BA
d
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Antennas
Amanogawa, 2006 Digital Maestro Series 80
The Poynting vectoris given by
( )
( )
r
r
I d P t i
r
Ii
r
2
2
2 dipole Poynting vecto array factor
array fa
22 2max
2 2 2
22 2max
ct
2
r
r
o
2 2
cos cos( ) cos 4 cos
2 28 sin
coscos 4 cos cos
2 48 sin 4
+=
=
R
sinR sin cosR
cosR No
cos sin cos
te
sin sinR
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Antennas
Amanogawa, 2006 Digital Maestro Series 81
The total radiated poweris
2 220 0
sin ( )tot P r d P t d Only /2 because it is a monopole
( )2
22max
2 2 2
array facto
2
r
2 cos2 cos sin0 28 sin
24cos
0 4sin cos
4
I dtotr
P r
d
=
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Antennas
Amanogawa, 2006 Digital Maestro Series 82
The integral over the azimuthal angle gives
For a monopole, we only have the integral
2 2
0
2
0
20
4 cos sin cos4 4
1 14 cos sin cos
2 2 2 2
1 14 sin sin cos 42 2 2
d
d
d
= +
= + =
2
02d =
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Antennas
Amanogawa, 2006 Digital Maestro Series 83
In the direction of maximum
The same field strength (25 mV/m) is obtained by applying half the
current of the original monopole to the array elements (alsomonopoles)
The total radiated poweris proportional to the square of the current,
and the integral over gives a factor 4 instead of 2 for the array.Overall, the total radiated power needed by the array, to producethe same electric field, is half that of the individual monopole
90 & 90 array factor 2 = =
max6.25
3.125 A2
I = =
714357 W
2totP = =
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Antennas
Amanogawa, 2006 Digital Maestro Series 84
For a monopole
Radiated Power= 0.51428.3127= 714.155635 [ W ]
Rad. Resistance = 0.573.129616= 36.564808 [ ]
SAME FEEDING CURRENT
FOR BOTH DIPOLE AND
MONOPOLE
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Antennas
Amanogawa, 2006 Digital Maestro Series 86
N-Element Antenna Array
Assume a uniform array of N identical antennas. The elements arefed by currents with constant amplitude and with phase increasing
by an amount from one to the other. The spacing dbetween theantennas is uniform.
r ( 1) cosr N d
d
1 2 3 N
( 1)(1) ; (2) ; ; ( )
j j N o o oI I I I e I N I e
= = =
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Antennas
Amanogawa, 2006 Digital Maestro Series 87
The electric field at the observation point (r,) is of the form j r j r d j
o o
j r N d j No
j r j d o
j N d
jN d j r
o j d
r E e E e e
E e e
E e e
e
eE e
e
( cos )
( ( 1) cos ) ( 1)
( cos )
( 1)( cos )
( cos )
( cos )
E( , )
1
1
1
+ +
+ +
= + ++
+ + =
We have used
( cos )1( cos )
( cos )0
1
1
jN d N jn d
j dn
ee
e
++ +==
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Antennas
Amanogawa, 2006 Digital Maestro Series 88
The magnitude of the electric field is given by
We have used
ar
( cos )
( cos
ray facto
)
r
1E( , )
1
sin[ ( cos ) / 2]
sin[( cos ) / 2]
jN d
o j d
o
er E
e
N dE
d
++= += +
21 2 sin 2 sin2 2
jx j x x xe j e= =
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Antennas
Amanogawa, 2006 Digital Maestro Series 89
We can rewrite
The group pattern has
group pattern
1 sin[ ( cos ) / 2]
sin[( cosE( ,
) /2]) o
N d
N dr N E
+= +
Maxima when
Nulls when
for
( cos ) 2
cos
integer
0,
0,
2
, 2 ,
, 4
N
N
d
d m
m N
==
+ =
A t
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Antennas
Amanogawa, 2006 Digital Maestro Series 90
Examples
Broadside array (plots on the azimuthal plane, with = 90)
2 dipoles 4 dipoles
A t
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Antennas
Amanogawa, 2006 Digital Maestro Series 91
8 dipoles 16 dipoles
Antennas
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Antennas
Amanogawa, 2006 Digital Maestro Series 92
End-fire array (plots on the azimuthal plane, with = 90)
2 dipoles 4 dipoles
Antennas
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Antennas
Amanogawa, 2006 Digital Maestro Series 93
8 dipoles 16 dipoles
Antennas
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Antennas
Amanogawa, 2006 Digital Maestro Series 94
Cardioid array (plots on the azimuthal plane, with = 90)
2 dipoles 4 dipoles
Antennas
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Antennas
Amanogawa, 2006 Digital Maestro Series 95
8 dipoles 16 dipoles