Solutions to Tutorial 3 – Antenna Arrays
Q.1)
 2    
kd  
  
   4  2
  2m  kdcosm , m  0,1,2,
Q.2)
(i)
AF  1  e jkd cos   2 sin(
For d = /4,
AF  2 sin(
 , in x  z plane
 
 , in x  y plane
kd cos 
),
2
kd cos 

)  2 sin( cos  )
2
4


|AF| is maximum at  = 0 or  and is equal to 2 sin( ) . Hence,   2 sin( ) and
4
4
kd cos 
sin(
)
2
AF 

n
sin( )
4
90
1
120
60
0.8
0.6
150
30
0.4
0.2

180
0
330
210
300
240
270
1
(ii)
x-z plane


Normalized element pattern Fn    sin    ,
2

90
1
60
120
0.8
0.6
30
150
0.4
/2 is added to change the
starting point of  from the
top to the horizontal.

0.2
180
0
210
330
240
300
270
Array pattern,
90
90
1
120
60
0.6
60
0.8
0.6
0.6
150
30
30
0.4
0.2
30
150
0.4
0.4
0.2
180
0

330
210
300
Array factor |AFn|
0.2
180
0
330
210
300
240
270
270

Element pattern |Fn()|
2
1
120
0.8
150
240
90
1
120
60
0.8
=
180
0
210
330
300
240
270
= Array pattern (normalized)
x-y plane
Normalized element patter Fn    1 ,
90
1
60
120
0.8
0.6
150
30
0.4

0.2
180
0
330
210
240
300
270
Array pattern,
90
90
1
120
60
0.6
60
0.8
0.6
0.6
30
150
0.4
30
150
30
0.4
0.4
0.2
0.2
0.2
180
0

330
210
300
Array factor |AFn|
180
0
330
210
240
270
=
180
0
210
330
240
300
300
270
270

1
120
60
0.8
150
240
90
1
120
0.8
Element pattern |Fn()|
= Array pattern (normalized)
Q.3)
(a)
d = /2,   
Array factor with the phase reference point (the origin of the coordinate system) at antenna
#1 is:
AF  1  e j kd cos     1  e j (cos 1)
Shift the origin (the phase reference point) to the centre of the array. The array factor
becomes:
1
1


 j  kd cos    
j  kd cos    
 j  cos  1
j  cos  1


e 2
e 2
e 2
 2 cos   cos   1 
AF  e 2
2

The largest value of |AF| is 2. Hence the normalized array factor is:
AFn 
AF


 cos   cos   1 
2
2

(b)
Dipoles #1 and #3 constitute an array with separation = 0.5   0.5   0.707 and fed
by currents with a phase difference . By part (a), the array factor for this array is:
2
3
2
AF13 
1   j 2 
e
2
2 cos  4  1
e

j  2 cos  4  1
2



  cos   2 cos    4   1 
2


Note that we need to add /4 to  to account for the fact that this array is turned an angle of
/4 rad clockwise with respect to the array in part (a).
Similarly,
dipoles
#2
and
#4
constitute
another
array
with
separation
=
0.5   0.5 
 0.707 and fed by currents with a phase difference . By part (a), the
array factor for this array is:

j  2 cos  3 4  1 
1   j   2 cos 3 41


AF24  e 2 
 e 2
  cos   2 cos   3 4   1 
2
2


2
2
Note that we have added 3/4 to  to account for the fact that this array is turned an angle of
3/4 rad clockwise with respect to the array in part (a).
Hence total AF for the four-dipole array is:
AF  AF13  AF24e j 2




 cos   2 cos    4   1   j cos   2 cos   3 4   1 
2

2

Note that the feeding phase of the array formed by dipoles #2 and #4 is /2 leading the
feeding phase of the array formed by dipoles #1 and #3. Hence we need to add an extra
feeding phase of /2 to AF24.
4