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

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kt foo<br />

N --T j ° I_(_)+K(-_)]__I°<br />

k T<br />

(El 2)<br />

Thus the ratio <strong>of</strong> the peak output pulse power to the<br />

noise power is<br />

S<br />

P<br />

--=<br />

N<br />

2x/_-<br />

kTom 2<br />

(El3)<br />

Now om 2 = o + 6o2/o 3 <strong>and</strong> has a minimum when<br />

O = (3)1/4_ 1/2 (El4)<br />

At this value, m 2 = 4/3 so we find from (El3)<br />

= P P<br />

_ °Pt kT(_,/2) = kTk3.3.3.3._//23/2 _,/2 )<br />

(El5)<br />

The factor<br />

23/2<br />

/3 - 33/4 - 1.24... (El6)<br />

degrades the output signal-to-noise ratio to the value<br />

that would exist with the same filter, <strong>and</strong> a nonsweeping<br />

signal <strong>of</strong> 0.937 dB less power.<br />

198

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