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The resulting thermal noise is given by
![\begin{displaymath}1 \sigma = \frac{2 k T_{sys}}{\eta A \sqrt{\Delta \nu \times t}} (\mathrm{K})
\end{displaymath}](img886.gif) |
(10.13) |
where k is the Boltzmann's constant, A is the geometric collecting area of the
telescope,
the global efficiency factor (including decorrelation,
quantization, etc...),
the bandwidth in use and t the integration
time. The resulting phase noise is inversely proportional to the signal to noise
ratio.
S.Guilloteau
2000-01-19