Project Cyclops, A Design... - Department of Earth and Planetary ...
Project Cyclops, A Design... - Department of Earth and Planetary ...
Project Cyclops, A Design... - Department of Earth and Planetary ...
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introduced<strong>and</strong>theorder<strong>of</strong> theintegrating <strong>and</strong>averagingoperations<br />
exchanged togive:<br />
"qn2"f_£g(t)g(t+x)_Ink(t)nk(t+x) ]<br />
dt dx<br />
(C8)<br />
The average over k <strong>of</strong> nk(t) nk(t + x) is the<br />
autocorrelation, A(x), <strong>of</strong> the noise <strong>and</strong> is independent <strong>of</strong><br />
t. Thus,<br />
qn 2 = A(x) (t)g(t + x)d dx<br />
(C9)<br />
In the presence <strong>of</strong> white noise the best gate is the<br />
signal itself. From,equation (C2) we see that what we<br />
should do is cross correlate the received signal with the<br />
expected waveform <strong>of</strong> the noise-free signal. In function<br />
space this amounts to finding the projection <strong>of</strong> s(t) on<br />
f(t).<br />
From equations (C6) <strong>and</strong> (CI1) <strong>and</strong> using the<br />
optimum gate equation (CI2), we obtain the output<br />
signal-to-noise power ratio<br />
= -- = 2 dv (C14)<br />
(S) q$ f°°lF(p)[2<br />
N- opt qn 2 J__oo _u)<br />
If if(v) is constant over the signal spectrum, this ratio<br />
is expressed as<br />
=L_A(x)_°°lG(v)]2ei21rVxdvldx<br />
(lO)<br />
opt<br />
(C15)<br />
where equation (CI0) follows from equation (9) by<br />
applying Parseval's theorem to the second integral. Now<br />
A(x) is the transform <strong>of</strong> the power spectrum <strong>of</strong> the<br />
noise, while the quantity in brackets in equation (C I 0) is<br />
the inverse transform <strong>of</strong> IG(v)l 2. Applying Parseval's<br />
theorem once more equation (CI0) becomes<br />
-- ¢(v)<br />
%7: = _ IG(v)l2 av (c_1)<br />
where if(u) is the one-sided power spectral density <strong>of</strong> the<br />
noise. Since our integrals extend from --_ to +_o in<br />
frequency, we must use the two sided density if(v)/2.<br />
The problem may now be stated as follows: Find the<br />
G(v) that minimizes qn 2 as given by equation (Cll)<br />
under the constraint that qs as given by equation (C6)<br />
is held constant. This is a straightforward problem in the<br />
calculus <strong>of</strong> variations <strong>and</strong> yields the result<br />
G(v) = m --<br />
_(v)<br />
(C12)<br />
where m is an arbitrary real constant. If the<br />
noise is "white" that is, if Pn(V) = constant, then<br />
(;(v) = constant X F(v) <strong>and</strong><br />
where E is the pulse energy, <strong>and</strong>, for thermal noise,<br />
= kT.<br />
If, prior to gating, the received signal <strong>and</strong> noise are<br />
first passed through a filter having the transmission K(v),<br />
then in all subsequent expressions F(u) is replaced by<br />
F(u) K(u) <strong>and</strong> _b(v) by _b(u) [K(u)I 2. Equation (El 2) in<br />
particular then becomes<br />
F(v)<br />
G(v) K(v) = m -- (CI6)<br />
_(v)<br />
<strong>and</strong> we observe that it is only the product <strong>of</strong> the gate<br />
spectrum <strong>and</strong> the conjugate <strong>of</strong> the filter transmission<br />
that is specified. Deficiencies in either can be compensated<br />
for by the the other. A case <strong>of</strong> particular interest<br />
occurs if we set G(v) = constant. This corresponds to a<br />
gate that is a 6-function at t = 0. We then have<br />
<strong>and</strong> for white<br />
noise<br />
F(v)<br />
K(v) = constant --<br />
¢(v)<br />
(el 7)<br />
K(v) = constant X F(v) (c 18)<br />
g(t) = constant × J(t) (C13) As given by equation (C 17) or (CI8) K(v) is called a<br />
matched filter <strong>and</strong> gives the highest ratio <strong>of</strong> peak signal<br />
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