Project Cyclops, A Design... - Department of Earth and Planetary ...

azimuth _) has a delay (2a/c)sin 0 greater than the far
point (at azimuth _b+ 7r). For radiation arriving at zero
elevation (0 = n/2). The delay difference is 2a/c, the
propagation time across the array (Fig. 10-10). The delay
plane is the tilted ellipse, and the delay for an element at
A is ra = z_ + 2a/c while that tbr an element at B is
T b = T_.
= rmi n -7"0 = _'_ + 7"2
a
r
= (r_ + _) (1 - _-) (15)
and a variable part having a delay range
T
A
2O
rl+-- C-
2r
_'v = _'max - rmin = (16)
¢
vl
For an array 10 km in diameter, ale = 16.67/asec and
r_ _ 20 /asec. Thus, the fixed delay needed for the
center element is 36.67 /.tsec and the variable delay
needed for an element at the rim is 33.3/asec. The total
delay (rl + _'2 + rv) needed for any IF line is thus about
37 #sec and must be accurate to -+1/8 nsec, an accuracy
of about 3.4 parts per million. Clearly the delay system
is going to require stabilization of some sort to achieve
this accuracy.
SIGNAL
The average value of zf is one third the sum of the
altitudes of the lower two cones:
1
rf =-_ (r_ +-_-ac) (17)
while the average value of r v is
Figure 10-10. IF delay requirements.
2 2a 4a
7u = 7" c - 3c (18)
As 4_ is varied from 0 to 2rr, the tilted delay plane
rotates around the central axis and is always tangent to
two cones having their common vertex at P. If we
arrange to pivot the delay plane at P, then for greater
elevation angles it will always lie between these two
cones. The required delay for any element is never
greater than that defined by the upper cone and never
less than that defined by the lower cone of the pair. For
an antenna element at azimuth a and radius r from the
center
of the array
The total amount of delay required for n antennas
therefore is
n 5a
T = ff (r£ + "c-') (19)
Taking r£ = a/c gives the minimum possible total delay
for a fully steerable array
r = 7"_+ a + r sin 0 cos(_ - a)
e e
a-r
e
Thus the delay compensation can conveniently consist
of two parts-a
fixed part
a+t
c
_a3)
2a
T = n -- (20)
e
If the delay is obtained with additional transmission
line the total length needed is vT, where v is the velocity
(14) of propagation. Thus Lmi n = 2ha(v/c). Since the total
length of line used in the IF transmission system is about
(2/3) na (and 2/3 < vie < 1), we see that about two to
three times as much line would be needed for the delay
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