1

EE 542Antennas and Propagation for

Wireless Communications

Array Antennas

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Array Antennas

• An antenna made up of an array of individual antennas

• Motivations to use array antennas:– High gain more directive pattern– Steerability of the main beam

• Linear array: elements arranged on a line• 2-D planar arrays: rectangular, square, circular,

…• Conformal arrays: non-planar, conform to

surface such as aircraft

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Radiation Pattern for Arrays

Depends on:

• The type of the individual elements

• Their orientation

• Their position in space

• The amplitude and phase of the current feeding them

• The total number of elements

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Array Factor

The pattern of an array by neglecting the patterns of the individual elements; i.e. assume individual elements are isotropic

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Linear Receive Array

+

Receiver

A

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Case 1:

Array Factor for Two Isotropic Sources with Identical Amplitude and Phase (d = /2)

d

P(x,y,z)

x

z

(1) (2)

r

r1

r2

(I0,0) (I0,0)

Isotropic sources are assumed for AF calculations. The radiated fields are uniform over a sphere surrounding the source.

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Radiation from an Isotropic Source

jkreE

r

r

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Case 1: Total E Field

2

2

1

1

2

2

1

1

( , ) o o

o

rjk jkj j

o o

jk jkrj

o

r

re eE I e I e

e e

r

I e

r

rr

1 1 1

2 2 2

ˆ; 2

ˆ; 2

dr r r r x

dr r r r x

where

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Case 1: Far Field Approximation

In the far field, r>>d or (d/r) <<1

1

21

2

22

1 1 1

22

2 2

ˆ22 4

ˆ 1ˆ ˆ1 2 1

2 4 4

1ˆ ˆ1

2

cos2

d dr r r r r x

r x d d d dr r r x

r r r r

dr r x

r

dr

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Case 1: Far Field Approximation

1

21

2

22

2 2 2

22

2 2

ˆ22 4

ˆ 1ˆ ˆ1 2 1

2 4 4

1ˆ ˆ1

2

cos2

d dr r r r r x

r x d d d dr r r x

r r r r

dr r x

r

dr

Similarly,

Thus, in the far field

1

2

cos2

cos2

dr r

dr r

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Case 1: Far Field Geometry

d

P(x,y,z)

x

z

(1) (2)

r

r1

r2

dcos

1

2

cos2

cos2

dr r

dr r

If the observation point r is much larger than the separation d, the vectors r1, r and r2 can be assumed to be approximately parallel. The path lengths from the sources to the observation point are slightly different.

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Case 1: Total E in the Far Field

c

cos

os

cos2 2

cos2

co

2

s

s

2

co2

( , )

2 cos cos

ccos o2

2

s2

o o

o

o

o

d

dj

jk r jk r

j j

o o

j jkro

jkr d djk jkj

o

jk

d

djk

rj

o

k

e eE I e I e

r r

e eI e e

r r

eI e e e

r

e dI e k

d

r

d

The slight difference in path length can NOT be neglected for the exponential term!!

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Case 1: Total E for d=/2

Note that d=2

2( , ) 2 cos cos

4

2 cos cos2

o

o

jkrj

o

jkrj

o

eE I e

r

eI e

r

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Case 1: Array Factor

( , ) ( , )

2 cos cos2o

AF AF r E

I

The array factor is described as the magnitude of E at a constant distance r from the antenna (i.e. unit V)

AFn

0 0

/2 1

0

/2 1

Normalized values

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Case 1: Radiation Pattern

x

z

(1) (2)

(I0,0) (I0,0)

Notice how the two element array is more directive than the single element; which is an isotropic source.

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Case 2:

Array Factor for Two Isotropic Sources with Identical Amplitude and Opposite Phase

d

P(x,y,z)

x

z

(1) (2)

r

r1

r2

(I1,1) (I2,2)

1 2 0

1 0

2 0

I I I

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Case 2 – Far Field Geometry

d

P(x,y,z)

x

z

(1) (2)

r

r1

r2

dcos

1

2

cos2

cos2

dr r

dr r

(I1,1) (I2,2)1 2 0

1 0

2 0

I I I

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Case 2: Total E in the Far Field

1 2

1

1

2

2

1 2

1 2

cos cos2 2

1 2

( , )

2 sin cos2

oo

o

o

o

o

jk jkjj

o o

jkr jkrj j

o

jkr jkrj

o

jkr d djk jkj

o

jkrj

r r

o

e eE I e I e

e eI e e

r r

e eI e

r r

eI e e e

r

e djI

r

k

r

er

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Case 2: Radiation PatternNote that d=2

( , ) 2 sin cos2

o

jkrj

o

eE j I e

r

x

z

(1) (2)

(I0,0) (I0,+0)

Observe how the pattern is rotated compared to Case1 by simply changing the phase of element 2

AFn

0 1

/2 0

1

/2 0

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Case 3:

Array Factor for Two Isotropic Sources with Identical Amplitudes and 90o Phase Shift

Homework: Show that:

4( , ) 2 cos 1 cos4

ojkrj

o

eE I e

r

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Case 3 AFn

0 0

/2 cos(/4)

1

/2 cos(/4)

x

z

(1) (2)

(I0,0) (I0,+0)

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Generalization to N Equally Spaced Elements

0 1 2 3 N-1d d d

dcos

dcos

dcos

r

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General Case for Linear Array

1 2 1

1 1

1 1

1 1

cos ( 1) cos1 1

1cos

0

( , )N

o N

o N

n

jkr jkr jkrj j j

o N

jkrj j jjkd j N kd

o N

jkr Nj jnkd

nn

e e eE I e I e I e

r r re

I e I e e I e er

eI e e

r

1cos

0

( , ) n

Nj jnkd

nn

AF r E I e e

Total E field:

Array Factor:

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Special Case (A)

1 1

cos

0 0

( , )

where

cos

n

N Njn kd jn

n nn n

n

E I e I e

kd

Fourier series

Equally Spaced Linear Array with Linear Phase Progression

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Some Observations

* ( ) ( 2 ) periodic in (period = 2 )

* AF is a function of only, not . (Rotational symmetry)

** Note that the element pattern can be a function o

: 0 -

f .

* Visible region:

:

2

0 - 2

AF AF

2

kd kd

kd

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Special Case (B)

1

1

0

12 ( 1)

0

( 1)2 2 22

2 2 2

Let

( , ) ; cos

( , ) 1

sin1 21 sin

2

o n

jkr Njn

on

Njn j j j N

o on

N N Nj j jjN N

j

o o oj j j j

I I I

eE I e kd

r

AF r E I e I e e e

Ne e e e

I I I ee e e e

Uniformly Excited, Equally Spaced Linear Array with Linear Phase Progression

sin2

sin2

o

N

AF I

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Observations

• AF similar to the sinc function (i.e. sinx/x) with a major difference:

• Sidelobes do not die off for increasing values because the denominator is a sine function, and does not increase beyond a value of 1.

• AF is periodic with 2• Maximum value (=Io) occurs at k

sin2

sin2

o

N

AF I

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N=4 Case

AF (N=4)

00.5

11.5

22.5

33.5

44.5

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5

(rad)

AF

AF sin(y/2) sin(Ny/2)

Period:

nulls/2

/2

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More Observations

• Zeroes (Nulls) @ N/2 = kk=2kk=0,1,2, …

• This implies that as N increases there are more sidelobes (i.e. more secondary null points) in one period.

• Sidelobe widths are 2

• First null at1=2

• Within one period, N null pointsN-2 sidelobes (Because we discard k = N case, which corresponds to the second peak. Also 2 nulls create one sidelobe.)

• This implies that as N increases, the main beam narrows.

• Main lobe width is 2*2twice the width of sidelobes

• Max value ( = NIo) @ =2k, k=0,1,2, …

sin2

sin2

o

N

AF I

For all k values except when /2 becomes an integer multiple of

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Effect of Increasing N

AF for Different N

02468

1012

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5

y (rad)

AF

N = 3 N = 5 N = 10

HW: Regenerate this plot.

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Construction of Polar Plot from AF()

• The angle is not a physical quantity.

• We are more interested in observing the AF as a function of angles in real space; i.e. .

• Since linear arrays are rotationally symmetric wrt, we are concerned with only.

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Case 1: Construction of Polar PlotN = 2, d = /2, = 0 (uniform phase)

Using the general representation from Page 24

/2

x

z

Io, =0

r

Io, =0

sinsin2 2 cos

2sin sin2 2

cos ; 0, / 2

cos

cos2 cos

2

o o o

o

N

AF I I I

kd d

AF I

Compare to page 62

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Normalized AF, N = 2

0

0.5

1

1.5

2

2.5

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

y (rad)

f(Y

)

Normalized AF for Case1

max

( )( ) cos

2

coscos

2

AFf

AF

Period = 2

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Normalized AF for Case1 – Polar PlotVisible range: : [0-] : [-kd,kd]

= kdcos = cos

0

0.5

1

1.5

2

2.5

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

f(Y

)

kd

f(

)

f

x

Circle of radius kd

f

kd

Visually relate to

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Constructing the Polar Plot

0

0.5

1

1.5

2

2.5

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

f(Y

)

kd

f(

)

f

x

Circle of radius kd

f

f

f

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Case 2

N = 2, d = /2, =

sin2 2 cos

2sin2

cos

o o

N

AF I I

kd

Note: AF() same for all N=2

Value of different, depends on , d

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Case 2: Polar Format

= kdcos+

0

0.5

1

1.5

2

2.5

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

f(Y

)

kd

f()

f

x

0 2

Shifted by

Circle of radius kd

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Normalized AF for Case 2 – Polar Plot

0

0.5

1

1.5

2

2.5

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

f(Y

)

f(

)

f

x

Circle of radius kd

f

f

f

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Shift by

kdcos

kdkd

: [0, ]

: [ ,

co

]

skd

kd kd

kd

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Generalize to Arbitrary N

sin2

sin2

cos

o

N

AF I

kd

Shift by : [ , ]

2

kd kd

kd

Visible Range:

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General Rule

• AF plot with respect to is identical for all cases with identical N.

• The polar plot is determined by shifting the unit circle by , the linear phase progression amount.

• Visible range is always the 2kd range centered around that point.

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Shift and construct

0

1

2

3

4

5

6

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

f(Y

)

f()

f

Observe the dependence of main beam direction on , the phase progression.

Main beam

+ kd - kd

peak

cos(peak) = /kd

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Shift and construct

0

1

2

3

4

5

6

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

f(Y

)

f()

Observe the dependence of main beam direction on , the phase progression.

Main beam

+ kd - kd

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Array Pattern vs kd

• If kd > 2; i.e. d>/2 multiple peaks can occur in the visible range. These are known as grating lobes, and are often undesirable.

• Why??– Will cause reduced directivity as power will be

shared among all peaks– Likely to cause interference

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Grating Lobes

N=5

0

1

2

3

4

5

-15 -10 -5 0 5 10 15

f(Y

)

Three main beams.

kd, x

, x

-kd

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Pattern Multiplication

• So far only isotropic elements were considered.• Actual arrays are made up of nearly identical

antennas• AF still plays a major role in the pattern

( , ) ( , ) ( , )F e f Normalized Array Pattern

Normalized element pattern

Normalized Array factor

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Validation with Dipoles

• Consider the case of an ideal dipole array as below.

I0

(N-1)d

d d d

dcos

dcos

r

I1 I2 I3

0

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Sum of the E fields

/ 2

/ 2

ˆ '4 4

ˆ sin

sin sin

jkR jkrz

zz

z

z

e eA zI dz zA I z

R r

E E jw A

E A e

For the center dipole, assuming z <<

Normalized pattern

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Vector Potentials for Each Dipole

0

1

0

cos1

cos

4

4

4where

cos

m

jkr

z

jkrjkd

z

jkrjmkd

z m

m

eA I z

re

A I e zr

eA I e z

r

R r md

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Total Vector Potential

1

0

cos ( 1) cos0 1 1

1cos

0

4

4

m

N

z zm

jkrjkd jk N d

N

jkr Njkmd

mm

A A

ez I I e I e

re

z I er

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Total E field

1cos

0

1cos

0

sin

sin4

sin

z

jkr Njkmd

m

Njkmd

m

m

m

E jw A

ejw z I e

E

r

I e

Array pattern

Normalized element pattern

Array factor

( , ) ( , ) ( , )F e f

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Directivity of Linear Arrays

2

4

( , ) 4( , )

( , )

.

o rad

av

rad av

U UD

U P

U r S

P S ds

where

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Radiation Intensity

2

2 2 2

2

2 2 22

1( , ) ( , ) ( , )

1ˆ ( , )

2

1 1ˆ ( , ) ( , )

2

1( , ) ( , ) ( , )

2

o

av

o

av o

E E f er

S r E

r E f er

U r S E f e

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Total Radiated Power

4

22 2 2

24

22 2

4

.

ˆ ˆ( , ) ( , ) . sin2

( , ) ( , ) sin2

rad av

o

o

P S ds

Ef e r r r d d

r

Ef e d d

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Directivity2 2

2 2

4

2 2

2 2

4

2 2max

4

4 ( , ) ( , )( , )

( , ) ( , ) sin

4 ( , ) ( , )

( , ) ( , ) sin

4 4

( , ) ( , ) sin

A

A

A

f eD

f e d d

f e

f e d d

Df e d d

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Directivity for Arrays with Isotropic Elements

• Easier to calculate• Represents an approximate solution for elements

with broad patterns• Uniform amplitude and equal spacing will be

assumed.

22

1

21

2

sin( / 2)( , )

sin( / 2)

1 2( ) cos( )

( , ) 1

N

m

Nf

N

N m mN N

e

sin( ) sin cos cos sina b a b a b Using

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Directivity: Isotropic Elements, Linear Phase Progression,

Uniform Spacing, Uniform Amplitude

22 2

0 0

2

1

21

1

21

max 1

21

1( ) sin 2 ( )

2( )

2 1 2cos

4 42cos sin

4 11 2

c

kd

Akd

kd

kd

kd kdN

Amkd kd

N

m

NA

m

d f d f dkd

f dkd

d N m m dkd N N

N mm mkd

N N mkd

DN m

N N mkd

os sinm mkd

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Non-Uniformly Excited Linear Arrays

• We have seen the effects of phase shifting on the beam direction.

• We can also shape the beam and control the level of sidelobes by adjusting the amplitude of the currents in an array.

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Array Factor for Non-Uniform Excitation

1

0

; cosN

jmm

m

AF I e kd

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Can we eliminate the sidelobes???• Yes!

• First consider the 1x2 element array as in case 1 we studied.

• Recall that the AF did not have any sidelobes

AF = |1+ej| =

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Binomial Series Coefficients

• If the amplitudes are equal to the coefficients of the binomial series, no sidelobes.

• Consider the array factor, which is the square of Case 1:

• AF = (1+Z)(1+Z)=1 + 2Z + Z2

• This corresponds to a three element array with current amplitudes in the ratio of 1:2:1

• Since this array factor is simply the square of an array factor with no sidelobes there are no sidelobes.

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2-Dimensional Arrays

• The elements lie on a plane instead of a line.

• Many geometric shapes are possible; circle, square, rectangle, hexagon, etc.

• Will consider rectangular arrays

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Rectangular Array Geometry

dx

x(m)

y(n)

dy

mn

r

rmn

z

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Individual Fields

00

0, 0 00

22

.

, ,

, ,

2 .

.̂

, ,

mn

mn

mn mn

jkrj

m n

ikrj

mn mn

mn

mn mn

mn mn mn mn

mn

ikrj jk

mn mn

eE e I e

r

eE e I e

r

r r

r r r r

r r

eE e I e e

r

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Total Field

.

, ,

, mn mn

M N

mnm M n N

ikr M Nj jk

mnm M n N

E E

er

I e ee

Element patternArray factor

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Array Factor

.

ˆ ˆ ˆ

ˆ ˆ

mn mn

y ymn x x

M Nj jk

a mnm M n N

x y z

mn x y

M Njnk dj jmk d

a mnm M n N

S I e e

k k x k y k z

md x nd y

S I e e e

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Array Factor:Linear Phase, Uniform Amplitude

0

0

0

0

x y y yx x

y y yx x x

y y yx x x

mn x y

mn

M Nj m n jmk djmk d

am M n N

M Njn k djm k d

m M n N

M Njn k djm k d

m M n N

m n

I I

S I e e e

I e e

I e e

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Factors of Planar AF

sin cos

sin sin

x x x

y y y

a x y

Njn k d

xn N

Njn k d

yn N

x

y

S S S

S e

S e

k k

k k

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Homework, Problem 1Show that the Array Factor for two isotropic sources with identical amplitudes and 90o phase shift is given by

4( , ) 2 cos 1 cos4

ojkrj

o

eE I e

r

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HW Problem 2

• Construct by hand after plotting the AF for N=4, = /2, d = /2

• Hint: The AF vs should look like this:

AF (N=4)

00.5

11.5

22.5

33.5

44.5

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5

(rad)

AF

AF

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References

• Stutzman, et. al. “Antenna Theory” provides an excellent discussion on array antennas!!!