Page 195 - Satellite Communications, Fourth Edition

Antennas 175

Combining this with Eq. (6.36) gives

2 s
( cos cos ) (6.40)
0
l
This can be substituted into Eq. (6.38) to give the AF as a function of
relative to maximum.

Example 6.2 A dipole array has 2 elements equispaced at 0.25 wavelength. The
AF is required to have a maximum along the positive axis of the array. Plot the
magnitude of the AF as a function of .
Solution Given data are N 2; s 0.25l; 0 0.

From Eq. (6.40):

( cos 1)
2
The AF is

1
AF 2 a e jn 2
n 0
|1 cos jsin |

22(1 cos )

When 0, 5 0 , and hence the AF is 2 (as expected for two elements in the
0 direction). When , 52p , and hence the AF is zero in this direc-
tion. The plot on polar graph paper is shown below.

90
120 2 60
1.5
150 1 30
0.5
|AF(f)| 180 0 0

210 330

240 300
270
f