lecture 33 antenna arrays and phase arrayscourses.cit.cornell.edu/ece303/lectures/lecture33.pdf ·...

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ECE 303 – Fall 2007 – Farhan Rana – Cornell University

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


Two Hertzian Dipoles – A Two Element Array

y

z

( ) ( )03

00 ˆ hrdIzrJrrrr

−= δ

x

0hr

1hr0J

r1Jr

( ) ( )13

11 ˆ hrdIzrJrrrr

−= δ

( ) ( ) [ ]( )

( ) ( )φθφθ

θπ

ηθ

θπ

ηθ

,,,

sin4

ˆ

sin4

ˆ

10

10

.ˆ

0

1.ˆ

0

00

.ˆ1

.ˆ0

Fr

eIIe

IIe

rdIkj

eIeIerkdjrE

hrkjhrkjrkjo

hrkjhrkjrkjoff

Ε=

⎥⎦

⎤⎢⎣

⎡+=

+=

−

−

r

rr

rr

rr

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:

Consider first an array of just two Hertzian dipoles:

Element Factor Array Factor

Each antenna in the array is an “element” of the array

• 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

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An N-Element Antenna Array - IConsider 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

Let the position vector of the m-th antenna be mhr

y

x

mhr

z

mI

0I1I

1−NIzOne can write the E-field in the far-field as a superposition of the E-fields produced by all the elements in the array:

( ) ( )

( ) ( )φθφθ

φθ

,,,

,, 1

0

.ˆ

0

Fr

eIIrrE

N

m

hrkjmff m

Ε=

∑Ε=−

=r

rrr r


An N-Element Antenna Array - II

( ) ( )

( ) ( )φθφθ ,,,

1

0

.ˆ

0

Fr

eIIrrE

N

m

hrkjmff m

Ε=

∑Ε=−

=r

rrrr r

( )( )

( ) ( )222 ,, ofpart angular

4

ˆ.,, φθφθ

πφθ Fr,

rP

rtrSG ⎥⎦

⎤⎢⎣⎡ Ε∝=

rrr

( ) ( )( )max,

,,φθ

φθφθG

Gp =

Gain:

Pattern:

y

x

mhr

z

mI

0I1I

1−NIz

3


An N-Element Hertzian Dipole Phase Array

x

z

y

N-element Hertzian dipole array

a

αjeα2je( )α1−Nje

• Current magnitude is the same for all the dipoles

• Current phase difference between successive dipoles is α

1

αj

m

m eI

I=+1

( ) ( ) ( )

( ) ( ) ( )( )21

0

cossin2

1

0

cossin.ˆ1

0

.ˆ

0

,

, 0

∑=⇒

∑=∑=

−

=

+

−

=

−

=

N

m

makj

N

m

makjmjhrkjN

m

hrkjm

eF

eeeeIIF m

αφθ

φθα

φθ

φθrr

( ) ( )( )

( ) ( )( )⎥⎦⎤

⎢⎣⎡ +

⎥⎦⎤

⎢⎣⎡ +

=φθα

φθα

cossin21sin

cossin2

sin

2

2

ka

kaN

Array Factor:

Element Factor:

( ) ( ) rkjo er

dIkjr −=Ε θπ

ηθφθ sin4

ˆ ,, 0r


x

z

y

N-element Hertzian dipole phase array antennaa


1

( ) ( )( ) ( )( )

( ) ( )( )⎥⎦⎤

⎢⎣⎡ +

⎥⎦⎤

⎢⎣⎡ +

=φθα

φθαθφθ

cossin21sin

cossin2

sinsin1,

2

22

2ka

kaN

Np

Pattern:

Coming from the element factor

array factor

( ) ( ) ( )222 ,sin1, φθθφθ F

Np =

An N-Element Hertzian Dipole Phase Array

NdIkP o 2012π

η=

( )( )

( ) ( )222 ,sin

23

4

ˆ.,, φθθ

πφθ F

NrP

rtrSG ==

rr

Coming from the element factor

array factor

4


An N-Element Hertzian Dipole Phase Array: Maxima

Lets look at radiation in the x-y plane:

φ

a

( )φcosa

y

x

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:

( ){ KK,3,2,1,0

2cos=

±=+n

nak παφ

( )φπθ ,2=p

φ

2

4

20

π

πα

=

−=

=

ka

N

( )2,2 φπθ =F

2

4

20

π

πα

=

−=

=

ka

N

(degrees) φ

nulls

smallmaxima(side lobes)

bigmaxima(main lobes)

main lobes

side lobes


An N-Element Hertzian Dipole Phase Array: Maxima

φ

a

( )φcosa

y

x

Condition for a big maximum in the x-yplane:

( ){ KK,3,2,1,0

2cos=

±=+n

nak παφ

At a big maximum the value of the array factor is:

( )( )( )

( )( )

2

2

22max

cos21sin

cos2

sin,2

N

ka

kaN

F

=

⎥⎦⎤

⎢⎣⎡ +

⎥⎦⎤

⎢⎣⎡ +

==φα

φαφπθ

And the value of the antenna gain is:

( )2,2 φπθ =F

2

4

20

π

πα

=

−=

=

ka

N

(degrees) φ

nulls



( ) ( ) ( )

NG

FN

G

23,

2

,sin23,

max

22

=⎟⎠⎞

⎜⎝⎛ =⇒

=

φπθ

φθθφθ

5


An N-Element Hertzian Dipole Phase Array: Nulls

φ

a

y

x

Radiation in the x-y plane:

( )( )( )

( )( )⎥⎦⎤

⎢⎣⎡ +

⎥⎦⎤

⎢⎣⎡ +

==φα

φαφπθ

cos21sin

cos2

sin,2

2

22

ka

kaN

F

The array factor will give a null in the radiation pattern provided:

( )φcosa

( )

{ KK,3,2,,0

2cos

NNNnN

nak

≠

±=+παφ

( )2,2 φπθ =F

2

4

20

π

πα

=

−=

=

ka

N

(degrees) φ

nulls




An N-Element Hertzian Dipole Phase Array: Width of Big Maxima

( )2,2 φπθ =F

2

4

20

π

πα

=

−=

=

ka

N

φ

φ

a

y

x

φ∆Angular width of a lobe associated with a big maxima:

For a big maxima: ( ) {

,3,2,1,022cos KK±±±===+ nN

nNnak ππαφ

At the nulls nearest to the above maxima we must have:

122

cosN

nNak ±=+⎟

⎠⎞

⎜⎝⎛ ∆

+ παφφ

Which can be solved for ∆φ – and for N large (N >> 1) one gets approximately: ( )

sin14

φπφ

kaN≈∆

6


x

z

y

aka = π


Phase difference between successive elements is α

N = 2 N = 10

( )φθ ,p ( )φθ ,p

1


An N-Element Hertzian Dipole Phase Array: Examples


x

z

y

aka = π / 2


Phase difference between successive elements is α

N = 2 N = 10

( )φθ ,p ( )φθ ,p

1


An N-Element Hertzian Dipole Phase Array: Examples

7


An (N+1)-Element Hertzian Dipole Binomial Array

x

z

y

(N+1)-element Hertzian dipole array

a • Current phase is the same for all the dipoles

• Currents magnitudes in the dipoles follow a binomial distribution

( )!!!

0 mNmN

IIm

−=

( ) ( )( ) ( )

( ) ( ) ( ) ( ) ( )⎥⎦⎤

⎢⎣⎡=+=⇒

∑−

=∑===

φθφθ

φθ

φθ

φθ

cossin2

cos21,

!!!,

222cossin2

0

cossin.ˆ

0

.ˆ

00

akeF

emNm

NeeIIF

NNNakj

N

m

makjhrkjN

m

hrkjm mrr

Array Factor:

Element Factor:

( ) ( ) rkjo er

dIkjr −=Ε θπ

ηθφθ sin4

ˆ ,, 0r

π

λ

=⇒

=

ka

a2


Lets look at radiation in the x-y plane:

φ

a

( )φcosay

x

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:

( ){

( ) n

nnak

πφπ

πφ

±=⇒

=±=

cos2

,3,2,1,0 2cos

KK

An (N+1)-Element Hertzian Dipole Binomial Array: Maxima and Nulls

( ) ( )⎥⎦⎤

⎢⎣⎡== φπφπθ cos2

cos2,2 222 NNF

( ) ( )⎥⎦⎤

⎢⎣⎡== φπφπθ cos2

cos,2 2Np

N+1=2

N+1=40

N+1=2N+1=40

( )φπθ ,2=p ( )φπθ ,2=p

Note:

• No side lobes• Two nulls at φ = 0,π

(degrees) φ

(degrees) φ

8


Phased Array TRack to Intercept Of Target

Alaska

Applications of Antenna Arrays - I


Communication satellite with two phase array antenna panels MRI system with phase array

RF receivers

Applications of Antenna Arrays - II

Jicamarca facility (Peru) with an array of half-wave dipoles

SETI

lecture 33 antenna arrays and phase arrayscourses.cit.cornell.edu/ece303/lectures/lecture33.pdf ·...

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