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Optimal number of ON per OFF measurements

This section is just a reformulation of the original demonstration by Ball (1976).

Let's assume that we are measuring \ensuremath{n_\ensuremath{\mathrm{on/off}}} independent on-positions for a single off. The same integration time ( \ensuremath{t_\ensuremath{\mathrm{on}}}) is spent on each on-position and the off integration time is

\ensuremath{t_\ensuremath{\mathrm{off}}}= \alpha \, \ensuremath{t_\ensuremath{\mathrm{on}}},
\end{displaymath} (53)

where $\alpha$ can be varied. Using eq. [*] and $\ensuremath{t_\ensuremath{\mathrm{onoff}}}
= \ensuremath{n_\ensuremath{\mathrm{...
..._\ensuremath{\mathrm{on/off}}}+\alpha)\,\ensuremath{t_\ensuremath{\mathrm{on}}}$, it can be shown than
\ensuremath{t_\ensuremath{\mathrm{onoff}}}= \frac{\ensurema...
...ensuremath{n_\ensuremath{\mathrm{on/off}}}}{\alpha} \right) }.
\end{displaymath} (54)

Differenciating with respect to $\alpha$, we obtain
\end{displaymath} (55)

Setting the result to zero then gives that the minimum elapsed time to reach a given rms noise is obtained for
\alpha = \sqrt{\ensuremath{n_\ensuremath{\mathrm{on/off}}}}...
...\mathrm{on/off}}}} \, \ensuremath{t_\ensuremath{\mathrm{on}}}.
\end{displaymath} (56)

Gildas manager 2014-07-01