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Comparison of the full calibration scheme

As described in sections 2.2.2, the actual calibration of data is a two-steps process. In the first step, \ensuremath{T_\ensuremath{\mathrm{sky}}\ifthenelse{\equal{}{}}{}{^\ensuremath{\mathrm{}}}} is simply deduced from Eq. 11 and the ATM model is then just used to associate to this \ensuremath{T_\ensuremath{\mathrm{sky}}\ifthenelse{\equal{}{}}{}{^\ensuremath{\mathrm{}}}} value, a value of the total opacity of the atmosphere at signal and image frequencies. In the second step, the \ensuremath{T_\ensuremath{\mathrm{cal}}\ifthenelse{\equal{}{}}{}{^\ensuremath{\mathrm{}}}} value is computed with Eq. 17 and then used on the data.

Hence, in Eq. 17, the value of the term $(\ensuremath{T_\ensuremath{\mathrm{hot}}\ifthenelse{\equal{}{}}{}{^\ensuremath{...
...T_\ensuremath{\mathrm{sky}}\ifthenelse{\equal{}{}}{}{^\ensuremath{\mathrm{}}}})$ is independent of the ATM model and/or of the coupling factors, while the value of the term $\ensuremath{\displaystyle\left[ 1+\ensuremath{G_\ensuremath{\mathrm{im}}} \righ...
...}}\,\exp\ensuremath{\displaystyle\left( -\ensuremath{a}\tau \right) } \right] }$ is directly dependent of the ATM modeling and/or of the coupling factors. Two different situations arise: 1) the measured coupling parameters are at best known only to a given accuracy and at worst biased and 2) the model is inaccurate. To model both situations, we will assume that \ensuremath{\tau\ifthenelse{\equal{true}{}}{}{_\ensuremath{\mathrm{true}}}}, \ensuremath{G_\ensuremath{\mathrm{im}}^\ensuremath{\mathrm{true}}}, \ensuremath{F_\ensuremath{\mathrm{eff}}^\ensuremath{\mathrm{true}}} and \ensuremath{T_\ensuremath{\mathrm{cal}}\ifthenelse{\equal{true}{}}{}{^\ensuremath{\mathrm{true}}}} are the actual values of those different parameters while \ensuremath{\tau\ifthenelse{\equal{mod}{}}{}{_\ensuremath{\mathrm{mod}}}}, \ensuremath{G_\ensuremath{\mathrm{im}}^\ensuremath{\mathrm{meas}}}, \ensuremath{F_\ensuremath{\mathrm{eff}}^\ensuremath{\mathrm{meas}}}, and \ensuremath{T_\ensuremath{\mathrm{cal}}\ifthenelse{\equal{meas}{}}{}{^\ensuremath{\mathrm{meas}}}} are the modeled or measured values. It is easy to show that the relative error on \ensuremath{T_\ensuremath{\mathrm{cal}}\ifthenelse{\equal{}{}}{}{^\ensuremath{\mathrm{}}}} is

\begin{displaymath}
\frac{\ensuremath{T_\ensuremath{\mathrm{cal}}\ifthenelse{...
...e{\equal{true}{}}{}{_\ensuremath{\mathrm{true}}}}) \right] }-1
\end{displaymath} (18)

Thus the accuracy of \ensuremath{T_\ensuremath{\mathrm{cal}}\ifthenelse{\equal{}{}}{}{^\ensuremath{\mathrm{}}}} directly depends on the accuracy of the coupling factors and on the modeled opacity. It must be noted that 1) this equation assumes that the load and data calibration are made at the same airmass (i.e. elevation) 2) \ensuremath{G_\ensuremath{\mathrm{im}}} is close to zero for the EMIR receivers, implying that the \ensuremath{F_\ensuremath{\mathrm{eff}}} is one of the key parameter to get high calibration accuracy and 3) it is better to underestimate the opacity than to overestimate it.

Fig. 1 compares the values of opacities, \ensuremath{T_\ensuremath{\mathrm{atm}}\ifthenelse{\equal{}{}}{}{^\ensuremath{\mathrm{}}}} and \ensuremath{T_\ensuremath{\mathrm{emi}}\ifthenelse{\equal{}{}}{}{^\ensuremath{\mathrm{}}}} as a function of precipitable for both version of ATM. The comparison is made at two different frequencies, which have a very different opacity behavior:

115.3 GHz
The opacity is dominated by the dry continuum due to the transparency of water at this frequency and to the presence of a close oxygen line. It is known that at 3mm under submillimeter weather conditions, the measured sky emission is lower than the modeled one with ATM 1985. This facts comes from the uncertainty on the knowledge of the \ensuremath{F_\ensuremath{\mathrm{eff}}}, an incorrect modeling of the dry continuum with ATM 1985, or a combination of both. Anyway, this is why ATM 1985 authorizes negative amount of water vapor. ATM 2009 makes the situation even worse because the total opacity and thus emission are larger with ATM 2009 than with ATM 1985. This comes mainly from the dry continuum opacity being larger with ATM 2009 than with ATM 1985... On the other hand, the same amount of atmospheric emission is reached with significantly less water vapor in ATM 2009.
230.5 GHz
The opacity is dominated by its wet component, the dry continuum being almost negligeable. This is the reason why \ensuremath{T_\ensuremath{\mathrm{atm}}\ifthenelse{\equal{}{}}{}{^\ensuremath{\mathrm{}}}} sharply increases when the amount of water vapor goes to zero with ATM 2009. It is unclear why such an effect is not seen with ATM 1985. Here also, the total opacity and thus emission are larger with ATM 2009 than with ATM 1985, and the wet opacity increases slowlier with ATM 2009 than with ATM 1985. But the effects are much less pronounced at 230.5 than at 115.3 GHz. The consequence is that the amount of precipitable vapor (and thus of opacity) needed to reproduce the emission of the sky is similar in both ATM versions.


next up previous contents
Next: Recommendations Up: Comparison of ATM 1985 Previous: Comparison at constant amount   Contents
Gildas manager 2014-07-01