EE 543 Theory and Principles of Remote Sensing

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EE 542
Antennas and Propagation for
Wireless Communications
Array Antennas
1
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
2
O. Kilic EE 542
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
3
O. Kilic EE 542
Array Factor
The pattern of an array by neglecting the
patterns of the individual elements; i.e.
assume individual elements are isotropic
4
O. Kilic EE 542
Linear Receive Array
j
A
+
Receiver
O. Kilic EE 542
5
Case 1:
Array Factor for Two Isotropic Sources with Identical
Amplitude and Phase (d = l/2)
P(x,y,z)
z
r1
r
r2
q
(2)
(1)
x
d
(I0,j0)
(I0,j0)
Isotropic sources are assumed for AF calculations. The radiated fields are
uniform over a sphere surrounding the source.
O. Kilic EE 542
6
Radiation from an Isotropic Source
E 
e
 jk r
r
r
7
O. Kilic EE 542
Case 1: Total E Field
E (q , j ) 
I
o
e
 Ioe
jj o
jj o

e
 jk r1
r1
  Ioe
jj o

e
 jk r2
r2
 jk r
 e  jk r

e



r2 
 r1
1
2
where
r1  r1 ;
r2  r2 ;
r1  r 
r2  r 
O. Kilic EE 542
d
2
d
2
ˆ
x
ˆ
x
8
Case 1: Far Field Approximation
In the far field, r>>d or (d/r) <<1
r1   r1 r1 
1
2
2
d
d 
 2
ˆ
 r  2r x


2
4 

1
2
2

ˆ d
r x
d 
d
1 d  

ˆ
 r 1  2 2

 r 1  rˆ x

  
2 
r 2
4r 
r
4  r  


2
1
d

ˆ 
r 1  rˆ x
2
r 

 r 
d
2
co s q
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O. Kilic EE 542
Case 1: Far Field Approximation
Similarly,
r2   r2 r2 
1
2
2
d
d 
 2
ˆ
 r  2r x


2
4 

1
2
2
2

ˆ d
r x
d 
d
1 d  

ˆ
 r 1  2 2

 r 1  rˆ x

  
2 
r 2
4r 
r
4  r  


1
d

ˆ 
r 1  rˆ x
2
r 

 r 
d
2
co s q
Thus, in the far field
r1
r2
r 
r 
d
2
d
2
co s q
co s q
O. Kilic EE 542
10
Case 1: Far Field Geometry
r1
r2
r 
r 
d
2
d
2
co s q
z
co s q
P(x,y,z)
r1
r
r2
q
(2)
(1)
x
d
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.
O. Kilic EE 542
11
Case 1: Total E in the Far Field
E (q , j )  I o e
e
jj o
d


 jk  r  co s q 
2


d


 r  2 co s q 


d
Ioe
 Ioe
jj o
jj o
 2 Ioe
e
 jk r
e
 jk r
r
jj o
e
d


r  2 cos q 


d
 jk
co s q
  j k 2 co s q
2
e
e


r
r

d




d
 jk
co s q 
  jk 2 c o s q
2
 e
e



 jk r
r
 I oe
e
jj o
d


 jk  r  co s q 
2


 d

co s  k
co s q 
 2

12
The slight difference in path length can NOT be neglected for the exponential term!!
O. Kilic EE 542
Case 1: Total E for d=l/2
Note that d=l/2
E (q , j )  2 I o e
 2 Ioe
jj o
e
 jk r
r
jj o
e
 jk r
r
 2 l

co s 
co s q 
 l 4



co s  co s q 
2

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Case 1: Array Factor
The array factor is described as the magnitude of E at a
constant distance r from the antenna (i.e. unit V)
A F (q , j )  A F  r E (q , j )


 2 I o co s  co s q 
2

q
0
/2

3/2
AFn
0
1
0
1
O. Kilic EE 542
Normalized values
14
Case 1: Radiation Pattern
z
q
(2)
(1)
x
(I0,j0)
(I0,j0)
Notice how the two element array is more directive than the single element;
which is an isotropic source.
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O. Kilic EE 542
Case 2:
Array Factor for Two Isotropic Sources with Identical
Amplitude and Opposite Phase
P(x,y,z)
z
r1
r
r2
q
(2)
(1)
x
d
I1  I 2  I 0
(I1,j1)
(I2,j2)
j1  j 0
j2  j0  
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Case 2 – Far Field Geometry
r1
r2
r 
r 
d
2
d
2
co s q
z
co s q
P(x,y,z)
r1
r
r2
q
(2)
(1)
x
d
I1  I 2  I 0
(I1,j1)
(I2,j2)
j1  j 0
j2  j0  
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Case 2: Total E in the Far Field
E (q , j )  I o e
 Ioe
 Ioe
Ioe
jj o
e
 jk r1
 Ioe
r1
jj o
jj o
jj o
j j o  

e
 jk r2
r2
 jk r
 e  jk r

e
j
 e


r2 
 r1
1
2
 jk r
 e  jk r

e



r
r
 1

2
1
e
  2 jI o e
 jk r
r
jj o
2
d
d
 jk
co s q 
  jk 2 co s q
2
e
e



e
 jk r
r
 d

sin  k
co s q 
 2

O. Kilic EE 542
18
Case 2: Radiation Pattern
Note that d=l/2
E (q , j )   j 2 I o e
q AFn
0
1
/2 0
1

3/2 0
jj o
e
 jk r
r


sin  co s q 
2

z
q
(2)
(1)
x
(I0,+j0)
(I0,j0)
Observe how the pattern is rotated compared to Case1 by simply changing the
phase of element 2
O. Kilic EE 542
19
Case 3:
Array Factor for Two Isotropic Sources with Identical
Amplitudes and 90o Phase Shift
Homework: Show that:
E (q , j )  2 I o e
 

j jo  
4

e
 jk r
r


co s  1  co s q  
4

20
O. Kilic EE 542
q
0
/2

3/2
Case 3
AFn
0
cos(/4)
1
cos(/4)
z
q
(2)
(1)
x
(I0,+j0)
(I0,j0)
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O. Kilic EE 542
Generalization to N Equally Spaced
Elements
r
dcosq
dcosq
dcosq
q
d
0
d
1
d
2
3
N-1
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General Case for Linear Array
Total E field:
E (q , j )  I o e


e
jj o
 jk r1
 I1e
r
jj1
e
 jk r2
r

I N  1e
jj N 1
e
 jk rN  1
r
 jk r
r
e
e
 jk r
r
 I o e
N 1

jj o
I ne
 I1e
jj n
e
jj1
e
 jk d co s q

I N  1e
jj N 1
e
 j ( N  1 ) k d co s q

 jn k d co s q
n0
Array Factor:
A F  r E (q , j ) 
N 1

I ne
jj n
e
 jn k d co s q
n0
23
O. Kilic EE 542
Special Case (A)
Equally Spaced Linear Array with Linear
Phase Progression
j n  n
E (q , j ) 
N 1

I ne
 jn  k d co s q  
n0


N 1

I ne
 jn 
n0
w h e re

k d co s q  
Fourier series
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Some Observations
* A F (  )  A F (   2  ) p e rio d ic in  (p e rio d = 2  )
* A F is a fu n ctio n o f q o n ly , n o t  . (R o t a tio n a l sy m m e try )
* * N o te th a t th e e le m e n t p a tte rn ca n b e a fu n ctio n o f  .
* V isib le re g io n : q :  0 - 2 
q :  0 - 2


   kd      kd
   2kd
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Special Case (B)
Uniformly Excited, Equally Spaced Linear Array
with Linear Phase Progression
L e t I o  I1 
E (q , j ) 
e
 jk r
r
 In
N 1
Io  e
jn 
;
    k d co s q
n 0
N 1
A F  r E (q , j )  I o  e
 I o 1  e
jn 
n 0
 Io
1e
1e
jN 
j
 Io
e
e
j
N
2
j

2
N
j
 e
N

2
e 2
e


 j
 j

2
e 2
 e
 j
A F  Io
 j

 e

( N 1) 
j
2
  I e
o


 N 
s in 

 2 
 
s in 

 2 
O. Kilic EE 542
j2
j ( N 1) 

 N 
sin 

2


 
sin 

 2 
26
Observations
A F  Io
 N 
s in 

2


 
s in 

 2 
• 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 0, 2k.
27
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N=4 Case
AF
AF (N=4)
Period: 2
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0

/2
0
0.5
1
1.5
3/2
nulls
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
 (rad)
AF
sin(y/2)
sin(Ny/2)
28
O. Kilic EE 542
More Observations
A F  Io
 N 
s in 

 2 
 
s in 

 2 
For all k values except
when y/2 becomes an
integer multiple of 
• Zeroes (Nulls) @ N/2 = k  ok=2k/N, k=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/N.
• First null at o1=2/N.
• Within one period, N null points  N-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*2/N, twice the width of sidelobes.
• Max value ( = NIo) @  =2k, k=0,1,2, …
O. Kilic EE 542
29
Effect of Increasing N
AF
AF for Different N
12
10
8
6
4
2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
y (rad)
N=3
HW: Regenerate this plot.
N=5
N = 10
30
O. Kilic EE 542
Construction of Polar Plot from AF(y)
• 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. q, j.
• Since linear arrays are rotationally
symmetric wrt j, we are concerned with q
only.
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O. Kilic EE 542
Case 1: Construction of Polar Plot
N = 2, d = l/2,  = 0 (uniform phase)
Using the general representation from Page 24
A F  Io
 N 
sin 

sin 
 
 2 
 Io
 2 I o co s 

 
 
 2 
sin 
sin 


2


 2 
  k d co s q ;
  0, d  l / 2
  co s q
  co s q 
A F  2 I o co s 

2


Compare to page 62
z
r
q
Io, j =0
Io, j =0
l/2
O. Kilic EE 542
x
32
Normalized AF for Case1
f ( ) 
A F ( )
A Fm a x
 
 co s 

 2 
  co s q 
 co s 

2


Normalized AF, N = 2
Period = 2
2.5
f(Y)
2
1.5
1
0.5
0
-7
-6
2
-5
-4
-3

-2
-1
0
y (rad)
O. Kilic EE 542
1
2
3

4
5
6
7
2
33
Normalized AF for Case1 – Polar Plot
Visible range: q: [0-]  : [-kd,kd]
 = kdcosq = cosq
Visually relate q to 
Circle of radius kd 
q1
q2

1
2
; x
  kd
Normalized AF, N = 2
2.5
f(Y)
f()
2
1.5
f1
1
f2
0.5
0
-7
-6
-5
-4

-3
-2
-1
0
O. Kilic EE 542
1
1
2
2

3
34
4
5
6
7
Constructing the Polar Plot
Circle of radius kd 

1
2
; x
  kd
Normalized AF, N = 2
2.5
f(Y)
f()
2
1.5
f1
1
f2
0.5
0
-7
-6
-5
-4

-3
-2
-1
0
O. Kilic EE 542
1
1
2
2

3
35
4
5
6
7
Case 2
N = 2, d = l/2,  = 
Note: AF() same for all N=2
A F  Io
 N 
sin 

2
 


 2 I o co s 

 
 2 
sin 

 2 
  k d co s q  
Value of  different, depends on , d
O. Kilic EE 542
36
Case 2: Polar Format
 = kdcosq + 
Circle of radius kd 
q1
Shifted by 
1
  kd
; x
Normalized AF, N = 2
2.5
f()
2
1.5
f1
1
0.5

0
-7
-6
-5
-4

-3 -2
-1
00
1
2
O. Kilic EE 542
3
4
51
6 2 7
37
Normalized AF for Case 2 – Polar Plot
Circle of radius kd 
2
1
0
; x
Normalized AF, N = 2
f(Y)
f()
2.5
2
f2
1.5
f1
1
0.5
0
-7
-6
-5
-4

-3
-2
-1
O. Kilic
0
EE 542
0
1
2

3
4
1
5
2
6
38
7
Shift by 
  k d co s q  
q : [0,  ]
 : [  k d ,   k d ]
kd
  kd

  kd
q1
1

kdcosq1
39
O. Kilic EE 542
Generalize to Arbitrary N
A F  Io
 N 
s in 

2


 
s in 

 2 
  kd cos q  
Visible Range:
Shift by 
 : [  k d ,   k d ]
  2kd
40
O. Kilic EE 542
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.
41
O. Kilic EE 542
Shift and construct
Observe the dependence of main beam direction on , the phase progression.
Main beam
qpeak
cos(qpeak) = /kd
q1
 - kd
Normalized AF, N = 2
1

f()
6

 + kd
5
4
3
2
f1
1
0
-6
-5
-4
-3
-2
-1
0
1 O. Kilic2 EE 5423
4
5
6

42
Shift and construct
Observe the dependence of main beam direction on , the phase progression.
Main beam
 - kd
Normalized AF, N = 2
f()
6
 + kd
1


5
4
3
2
1
0
-6
-5
-4
-3
-2
-1
0
1 O. Kilic2 EE 5423
4
5
6

43
Array Pattern vs kd
• If kd > 2; i.e. d>l/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
44
O. Kilic EE 542
Grating Lobes
Three main beams.
, x
kd
-kd
N=5
5
4
f(Y)
3
2
1
0
-15
-10
-5
0
5
10
15
, x
Y (rad)
45
O. Kilic EE 542
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 (q , j )  e(q , j )f (q , j )
Normalized
Array Pattern
Normalized
element
pattern
Normalized Array factor
46
O. Kilic EE 542
Validation with Dipoles
• Consider the case of an ideal dipole array
as below.
r
dcosq
dcosq
q
d
I0
d
I1
d
I2
I3
(N-1)d
47
0
O. Kilic EE 542
Sum of the E fields
For the center dipole, assuming z << l
z / 2
ˆI
A  z

 z / 2
e
 jk R
4 R
d z '  z Az  I
e
 jk r
4 r
z
E  E q qˆ  jw  sin q A z
E  sin q A z  e  q

 sin q
Normalized pattern
48
O. Kilic EE 542
Vector Potentials for Each Dipole
e
Az  I0
4 r
0
A z  I1
e
1
Az
m
 jk r
 Im
z
 jk r
4 r
e
e
 jk d co s q
z
 jk r
4 r
e
 jm k d co s q
z
w h e re
Rm
r  m d co s q
49
O. Kilic EE 542
Total Vector Potential
Az 
N 1

Az
m 0


e
 jk r
4 r
e
m
 z  I 0  I 1 e
 jk r
4 r
z
N 1

Ime
jk d co s q

 I N  1e
jk ( N  1 ) d co s q

jk m d co s q
m 0
50
O. Kilic EE 542
Total E field
Eq
jw  sin q A z
 jw  sin q
E q  sin q
N 1

e
 jk r
4 r
Ime
z
N 1

Ime
jk m d co s q
m 0
jk m d co s q
m 0
Array factor
Array pattern
Normalized element pattern
F (q , j )  e(q , j )f (q , j )
O. Kilic EE 542
51
Directivity of Linear Arrays
D (q , j ) 
U (q , j )
Uo

4 U
Pra d
where
U (q , j )  r
Pra d 

2
S av
S av . d s
4
52
O. Kilic EE 542
Radiation Intensity
1
E (q , j )  E o f (q , j )e (q , j )
S a v  rˆ
 rˆ
U (q , j )  r
1
2
1
2
2
E (q , j )
Eo
2
S av 
2
f (q , j )
1
2
r
Eo
2
2
e (q , j )
f (q , j )
2
2
1
r
2
e (q , j )
2
53
O. Kilic EE 542
Total Radiated Power
Pra d 

S av . d s
4


2
Eo
2 r
Eo

2
f (q , j )
2
2
e (q , j ) r rˆ.rˆ sin q d q d j
2
4
2

2 
f (q , j )
2
2
e (q , j ) sin q d q d j
4
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O. Kilic EE 542
Directivity
D (q , j ) 
4  f (q , j )

f (q , j )
2
e (q , j )
2
2
2
e (q , j ) sin q d q d j
4

A 
4  f (q , j )
2
e (q , j )
2
A

f (q , j )
2
2
e (q , j ) sin q d q d j
4
Dm ax 
4
A

4

f (q , j )
2
2
e (q , j ) sin q d q d j
4
55
O. Kilic EE 542
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.
f (q , j )
2


e (q , j )
2
sin ( N  / 2)
2
N sin (  / 2)
1
N

2
N
2
N 1

Using
sin( a  b )  sin a cos b  cos a sin b
( N  m ) co s( m  )
m 1
1
56
O. Kilic EE 542
Directivity:
Isotropic Elements, Linear Phase Progression,
Uniform Spacing, Uniform Amplitude
2
A 

0

A

 kd  
2
d j  f (q ) sin q d q  2 
2
kd
0
 kd  

kd  
2 
1 
f ( )  
d
 kd 
2
f ( ) d 

kd  
2  1

k d  N
2
 kd  
d 

kd  
N
2
N 1
 N
m 1
 m
 kd  

kd  

co s m  d  



Dm ax 
4
N
4
A


4
N
2
N 1
N  m
m 1
m kd

2 co s  m   sin  m k d 
1
1
N

2
N
2
N 1
N  m
m 1
m kd

c o s  m   sin  m k d 
O. Kilic EE 542
57
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.
58
O. Kilic EE 542
Array Factor for Non-Uniform
Excitation
AF 
N 1

Ime
jm 
;
    k d co s q
m 0
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O. Kilic EE 542
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+ejj| = 1Z
60
O. Kilic EE 542
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.
61
O. Kilic EE 542
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
62
O. Kilic EE 542
Rectangular Array Geometry
z
r
q
rmn
y(n)
rmn
dx
j
dy
x(m)
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O. Kilic EE 542
Individual Fields
E m  0 ,n  0 q ,    e q ,   I 00 e
E m n q ,    e q ,   I m ne
jj 0 0
e
 jk r
r
jj m n
e
 ik rm n
rm n
rm n  r  r m n
rm n  r  r m n 
r
2
 2 r .r m n   r m n 
2
r  ˆ.
r r mn
E m n q ,    e q ,   I m ne
jj m n
e
 ik r
r
e
jk . r m n
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O. Kilic EE 542
Total Field
E q ,   
M
N
 
E m n q ,  
m M nN
 e q ,  
e
 ik r
r
M
N
 
I m ne
jj m n
e
jk . r m n
m M nN
Array factor
Element pattern
65
O. Kilic EE 542
Array Factor
Sa 
M
N
 
I m ne
jj m n
e
jk . r m n
m M nN
ˆ  kyy
ˆ  kzz
ˆ
k  kxx
ˆ  ndyy
ˆ
r mn  m d x x
Sa 
M
N
 
I m ne
jj m n
e
jm k x d x
e
jn k y d y
m M nN
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O. Kilic EE 542
Array Factor:
Linear Phase, Uniform Amplitude
j m n  m  x  n y
Imn  I0
S a  I0
M
N
 


e
j m  x  n y
e
jm  k x d x   x
e
jm k x d x
e
jm k y d y
m M nN
 I0
M
N
 





e
jn k y d y   y
e
jn k y d y   y
m M nN
 I0
M

m M
e
jm  k x d x   x

N

nN
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O. Kilic EE 542
Factors of Planar AF
Sa  SxSy
Sx 
N

e
jn  k x d x   x
e
jn k y d y   y

nN
Sy 
N



nN
k x  k sin q co s j
k y  k sin q sin j
68
O. Kilic EE 542
Homework, Problem 1
Show that the Array Factor for two isotropic sources
with identical amplitudes and 90o phase shift is given by
E (q , j )  2 I o e
 

j jo  
4

e
 jk r
r


co s  1  co s q  
4

69
O. Kilic EE 542
HW Problem 2
• Construct by hand after plotting the AF for N=4,  =
/2, d = l/2
• Hint: The AF vs  should look like this:
AF
AF (N=4)
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
 (rad)
AF
70
O. Kilic EE 542
References
• Stutzman, et. al. “Antenna Theory”
provides an excellent discussion on array
antennas!!!
71
O. Kilic EE 542
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