Simple array of two antennas Consider two identical ...amanogawa.com/archive/docs/antennas2.pdf · Simple array of two antennas Consider two identical parallel Hertzian dipoles separated

Antennas

© Amanogawa, 2006 – Digital Maestro Series 64

Simple array of two antennas

Consider two identical parallel Hertzian dipoles separated by a distance d

I1I2

z

x

1θ2θ θ

r1r 2 r

d 2 d 2

iθ

i1θ

i2θ

ψ

Antennas


The dipole currents have the same amplitude and total phase difference α

The electric far−field components at the observation point are

( )

( )

21 1

22 2

cos( 2) I

cos( 2) I

jo o

phasorj

o ophasor

I t I t I e

I t I t I e

α

α

ω α

ω α −

= + ⇒ =

= − ⇒ =

j r jo

j r jo

j I z eir

j I z eir

1

1

2

2

2

1 11

2

2 22

E sin4

E sin4

β α

θ

β α

θ

µ β θε π

µ β θε π

− +

− −

∆≈

∆≈

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

θ θ θ

θ θ θ

ψ ψ

≈ ≈

≈ ≈

≈ − ≈ +

>>

j r d joj I z ei

r d

( ( 2)cos ) 21E

4 ( 2)cos

β ψ αθ

µ βε π ψ

− − +∆≈

−( )j r d j

oj I z eir d

( ( 2)cos ) 22

sin

E4 ( 2)cos

β ψ αθ

θ

µ βε π ψ

− + −∆≈

+( ) sinθ

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 zi er

e e

1 2

( 2)cos 2 ( 2)cos 2

sinE E E4

βθ

β ψ α β ψ α

µ β θε π

−

+ − +

∆= + ≈

× +

j roj I z di er

field of a Hertzian dipole locatedat the center of the arra

array y

factor

sin cosE 2 cos4 2

βθ

µ β θ β ψ αε π

−∆ + ≈

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 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.

d

group patterunit

patt rn ne

cossin cos2

β ψ αθ + ×

Antennas


Broad-side / p2 e0 att rnd λ α= =

z

x x

z

x

y

x

yUnit Pattern Group Pattern

x

yResultant Pattern

x

z

Antennas


d Broad-side p/ rn0 t2 at eλ α= =Radiation Pattern for E and H Power Radiation Pattern

Antennas


d End-fir/ 2 1 e pattern80λ α= =Radiation Pattern for E and H Power Radiation Pattern

Antennas


d Cardioid/ 4 90 patternλ α= =Radiation Pattern for E and H Power Radiation Pattern

Antennas


d Cardioid patter4 9 n/ 0λ α= = −Radiation Pattern for E and H Power Radiation Pattern

Antennas


d / 2 90λ α= =Radiation Pattern for E and H Power Radiation Pattern

Antennas


d 0λ α= =Radiation Pattern for E and H Power Radiation Pattern

Antennas


d 180λ α= =Radiation Pattern for E and H Power Radiation Pattern

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

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.

max

maxmax

/ 2 dipolecos90E cos2 sin90 2

E 120 0.025 V/m 6.25 A2 15, 000

j rj e Iir

I I

βθ λµ π

ε π

ππ

− ° = °

= =×

⇒=

Antennas


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:

2 2max

1 73.070.193978 0.5 6.25 713.622

W2

/

eqR

totP I µε

= ⋅ = ⋅ ⋅ =

P

GROUND PLANE

Monopole

λ/4

Antennas


2) Improve the design by using a two−antenna array.

A good choice of array parameters is

which gives a cardioid pattern

dphase(antenna B) phase(antenna A) 90

/ 4 α

λ− = = − °

=

15 km

BA

d

Antennas


The Poynting vector is given by

( )

( )

r

r

I dP 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

λ

µ π θ β ψ α

ε π θ

µ π θ πψ

ε π θ

π

+=

−

=

θ

ϕψ R

sinR θsin cosR θ ϕ

cosR ψ

Nocos sin cos

teψ θ ϕ=

sin sinR θ ϕ

Antennas


The total radiated power is

2 220 0

sin ( )totP r d P t dπ πθ θ ϕ= ∫ ∫

Only π/2 because it is a monopole

( )2

22max

2 2 2

array facto

2

r

2 cos2 cos sin0 28 sin2

4cos0 4sin cos

4

I dtotr

P r

d

π µ π θ θ θε π θ

π ππ θ ϕ ϕ

−

= ∫

∫

Antennas


The integral over the azimuthal angle ϕ gives

For a monopole, we only have the integral

2 20

20

20

4 cos sin cos4 4

1 14 cos sin cos2 2 2 21 14 sin sin cos 42 2 2

d

d

d

π

π

π

π πθ ϕ ϕ

π πθ ϕ ϕ

π θ ϕ ϕ π

− = + − = + =

∫

∫

∫

20 2dπ ϕ π=∫

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 (also monopoles)

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

90 & 90 array factor 2θ ϕ= ° = °⇒ =

max6.25 3.125 A

2I = =

714 357 W2totP = =

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

Antennas


For monopoles Radiated Power = 0.5×714.1532 = 357.0766 [ W ]

SAME FEEDING CURRENTS FOR BOTH DIPOLES AND MONOPOLES

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) cosr N d ψ− −

d

ψ

1 2 3 N

( 1)(1) ; (2) ; ; ( )j j No o oI I I I e I N I eα α−= = =…

Antennas


The electric field at the observation point (r,ψ) is of the form

j r j r d jo o

j r N d j Noj r j d

o

j N d

jN dj r

o j d

r E e E e e

E e e

E e e

e

eE ee

( cos )

( ( 1) cos ) ( 1)

( cos )

( 1)( cos )

( cos )

( cos )

E( , )

1

11

β β ψ α

β ψ α

β β ψ α

β ψ α

β ψ αβ

β ψ α

ψ − − −

− − − −

− +

− +

+−

+

= + +

+

= + ++

−=

−

…

…

…

We have used( cos )1

( cos )( cos )

0

1 1

jN dNjn d

j dn

eee

β ψ αβ ψ α

β ψ α

+−+

+=

−=

−∑

Antennas


The magnitude of the electric field is given by We have used

ar

( cos )

( cos

ray facto

)

r

1E( , )1sin[ ( cos ) / 2]sin[( cos ) / 2]

jN do j d

o

er EeN dE

d

β ψ α

β ψ αψ

β ψ αβ ψ α

+

+−

=−

+=

+

21 2 sin 2 sin2 2

jx j xx xe j e− = =

Antennas


We can rewrite The group pattern has

group pattern

1 sin[ ( cos ) / 2]sin[( cos

E( ,) / 2]

) oN d

N dr N E β ψψ α

β ψ α+

=+

Maxima when

Nulls when

for

( cos ) 2

cos

integer

0,

0,

2

, 2 ,

, 4

NN

d

d mm N

β ψ α

ψ

π π

β α π+ == ≠

+ = …

…

Antennas


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

2 dipoles 4 dipoles

ϕ

Antennas


8 dipoles 16 dipoles

Antennas


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

2 dipoles 4 dipoles

Antennas



Antennas


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

2 dipoles 4 dipoles

Antennas



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Simple array of two antennas Consider two identical ...amanogawa.com/archive/docs/antennas2.pdf · Simple array of two antennas Consider two identical parallel Hertzian dipoles separated