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

3's = unitcostof signal planelectronics
3'j = unitcostof junction
plane electronics
3'i = unit cost of image plane electronics
3"c = unit cost per cable
nj = number of junctions
N = number of cables
For a 1000 element circular array in a hexagonal lattice,
the maximum delay across the array at the corner of the
image field is about 22 cycles. At an IF frequency of 1
GHz this is about 4.4 m of cable. Each receiver in the
image plane must accept x/4-ff[rr or about 34 inputs,
which, because of the necessary connectors, requires a
unit about 6-1/2 in. wide by 7-1/2 in. high. There are 88
such units across the field, so the image plane will be
about 15 m across and 13-1/2 m high. lfwe separate the
planes by 10 m, the longest cables will be 17 m and the
shortest 12.6 m. Let us take the average length to be 15
m. We estimate the cost of cable at 15 cents/m, the cost
of connectors at $1.50, and the cost of cutting to length
and installing at $2.25 for an average cost 3'c = $6.
Taking the cost of the signal plane units at 3'i = $800
the cost of the repeaters at 3,i = $800 and the cost of the
image plane receiver detectors at $500, we derive the
costs given in Table 11-3.
TABLE 11-3
Antenna Array Signal Junction Image
Shape Lattice Plane Pian._e Plane Cabfing Total
$ Million
Square Square 0.8 1.6 2. 1.14 5.54
Circular Square 0.8 1.8 2. 1.24 5.84
Circular Hexagonal 0.8 2.6 2. 1.43 6.83
In spite of the maze of interconnections, the cabling
is not the major cost. To get really accurate costs we
need to refine our estimates of the electronics costs,
which can only be done by designing the units.
The major operational disadvantage of the delay line
imager is that the size of the useful field is inversely proportional
to the RF frequency. If the field is proper at 1
GHz, only the central third (one-ninth of the area) is
useful at 3 GHz. Unless we raise the surface density of
the image points in the center of the field we will lose
90% of the detail.
Optical Imaging
A radiative imager using light could be made if there
were some neat way to modulate the amplitude and
phase of the coherent light transmitted at a small spot in
a plate. We could them map the array as an array of such
spots and modulate each in accordance with the amplitude
and phase of the RF signal received by each
antenna. But in addition to requiring techniques unavailable
to us, such a method requires close tolerances and is
afflicted with lateral chromatic aberration. Using light as
the imaging radiation is equivalent to making f0 in
equation (52) very large and negative. The focal length is
then proportional to fr and the relative variation over
the band is
AF
B
= -- (63)
P /c
To keep AF/F < 0.01 with B = 100 MHz requires
fc > 10 GHz.
In a radiative imaging system, where the path delays
change continuously with position across the signal and
image planes, we can avoid the chromatic aberration
only by using radiation at the original frequency. This
leaves us with acoustic waves and microwaves to
consider.
Acoustic
Imaging
Piezoelectric transducers having only a few decibels
of conversion loss over the frequency band from 0.6 to
1.8 GHz have been built using thin films of zinc oxide.
Units to cover the band from 1 to 3 GHz appear possible
with further development. Let us therefore examine the
feasibility of acoustic imaging in this frequency region.
Since the waves are generated in a solid a'nedium such
as crystalline lithium niobate or sapphire, focusing lenses
are probably not realizable. Instead we must either curve
the signal and image array surfaces, or use RF delay to
generate coverging wavefronts, or both. Figure 11-18
shows two concave arrays whose vertices are separated
by a distance £. With a solid medium this distance _ is
fixed, and the most economical solution is not to use RF
delays but simply to make the radius of curvature of the
signal array rs equal to £. This places the center of
curvature Cs at the vertex of the image array Vi. On-axis
radiation will therefore focus at Vi. Off-axis radiation
will focus on a spherical surface whose radius of
curvature ri = _/2. (That is, Ci will be located midway
between Vs and Vi.) The image surface is not plane but
is part of a Rowland sphere: a Rowland circle in both
directions.
147