CoPhased Dipoles

8/6/2019 CoPhased Dipoles

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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


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


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


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


Broad-side/ p2 e0 att rnd =

x x

x

y

x

yUnit Pattern Group Pattern

x

yResultant Pattern

x


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Antennas


d Broad-side p/ rn0 t2 at e =Radiation Pattern forEandH Power Radiation Pattern


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Antennas


d End-fir/ 2 1 e pattern80 = Radiation Pattern forEandH Power Radiation Pattern


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Antennas


d Cardioid/ 4 90 pattern = Radiation Pattern forEandH Power Radiation Pattern


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Antennas


d Cardioid patter4 9 n/ 0 = Radiation Pattern forEandH Power Radiation Pattern


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Antennas


d / 2 90 = Radiation Pattern forEandH Power Radiation Pattern


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Antennas


d 0 = Radiation Pattern forEandH Power Radiation Pattern


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Antennas


d 180 =

Radiation Pattern forEandH Power Radiation Pattern


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Antennas


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


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


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


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


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


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


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


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


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


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


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


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


Examples

Broadside array (plots on the azimuthal plane, with = 90)

2 dipoles 4 dipoles

A t


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Antennas


8 dipoles 16 dipoles

Antennas


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Antennas


End-fire array (plots on the azimuthal plane, with = 90)

2 dipoles 4 dipoles

Antennas


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Antennas



Antennas


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Antennas


Cardioid array (plots on the azimuthal plane, with = 90)

2 dipoles 4 dipoles

Antennas


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Antennas



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CoPhased Dipoles