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

and the required calculation time is
To _ 2(Nlog2N) 10-6 SeC (5)
Now let us determine what sort of bandwidths may
be efficiently analyzed on a real time basis with a
hard-wired fast Fourier transformer. For this let
T = recording time necessary to achieve the desired
frequency resolution (1/T) in the spectrum
B = bandwidth handled per analyzer
N = number of data points per recording of signal
3BT (to prevent aliasing)
We assume that, for a given channel, two FFTs are used
alternately to achieve real-time operation. One analyzes
while the other is recording. It is clear that if To > T the
analyzing unit will fall behind, while if To < T it will be
idle some of the time. Thus, for most efficient operation
T = To and we have
or
T = 2(Nlog2N) 10--6
T = 2(3BTlog23BT) 10--6
dimensional Fourier transform of the complex amplitude
distribution over the front focal plane (ref.7).
(Note: the front and back focal planes are each one focal
length from the corresponding principal plane and are
not the object and image planes.) The intensity distribution
over the back focal plane is thus the two
dimensional power spectrum of the front focal plane
distribution. Because of these relationships and, because
lenses have enormous information transmission capacity,
coherent optical systems are widely used to obtain two
dimensional power spectra. Not so well known is the
fact that this large information rate of lenses can be used
efficiently to obtain the power spectra of one dimensional
signals (refs. 8,9).
The signal to be analyzed is recorded in a raster scan
on a strip of photographic film or other suitable medium
so that after development the transmittance of the film
is directly proportional to the signal amplitude. Adc
bias or offset is added to the signal (or in the light
modulator) to avoid negative amplitudes. Figure 11-4
shows a laser beam recorder in which the laser beam is
first modulated by the biased signal and then deflected
horizontally by a sawtooth waveform. Blanking signals
are applied to the modulator during flyback. A lens
brings the modulated and deflected beam to a sharp
focus on a strip of film moving downward at a constant
rate. Each line of the raster thus represents a time
segment of the input signal.
5X10 s = 3BIog23BT (6)
If T = l sec (to achieve 1 Hz resolution) then B = 11,094
Hz. Since 2N log2N=106 we f'md from equation (4) an
estimated equipment cost of $250,000 to analyze an
l 1,000 Hz band. The total system cost to analyze two
100 MHz bands would therefore be on the order of $4.5
billion.
Fast as the Cooley-Tukey algorithm may be, it is no
match for the data rate of the Cyclops system. The
above cost, comparable to the cost of the array itself,
plus the expected downtime and maintenance cost of
the 18,000 FFT computers required, cause us to reject
this approach in favor of the inherently more powerful
and less expensive method of optical spectrum analysis
described in the next section.
THE OPTICAL SPECTRUM ANALYZER
it is commonly known that the complex amplitude
distribution over the back focal plane of a lens is the two
ASER
A /MODULATOR FILM
MOT,ON
S,GNAL zLENS r"-'
'OSD-C
DEFLECTION//
v_
SIGNAL / \_._
RECORDED
Figure 11.4. Raster scan recorder.
RASTER /
After exposure, the film is run through a rapid
processor and into the optical spectrum analyzer. The
processor introduces a fixed delay into the system but
otherwise the analysis is done in real time. Figure I 1-5
shows the developed film being pulled through the gate
of the spectrum analyzer where it is illuminated with
H
126