Appendix A: Formulation of the final data flow

In this Appendix we will discuss how to account for the distribution of the acquisition parameters between the different detectors in the computation of the overall compression rate. Since the formalism is simpler we will develop expressions for $\mbox{$\eta_{{\rm c}}$ } = 1/\mbox{$C_{{\rm r}}$ }$ instead of $C_{{\rm r}}$.

We have pointed out in Sect. 5.2 that the compression efficiency is a random variable, whose distribution is a function of all those parameters which are relevant to fix the statistical distribution of the input signal. In our case: $\nu$, ${{\rm VOT}}$, ${\rm AFO}$, $N_{{\rm circ}}$ are the relevant parameters, so that the conditioned probability to have a compression efficiency in the range $\mbox{$\eta_{{\rm c}}$ }$, $\mbox{$\eta_{{\rm c}}$ } + {\rm d}\mbox{$\eta_{{\rm c}}$ }$ is:

$$% {\mathcal P}_{\nu,N_{{\rm Circ}}}\left(\mbox{$\eta_{{\rm c}... ...box{${{\rm VOT}}$ } \right) {\rm d} \mbox{$\eta_{{\rm c}}$ }.$$ (A1)

This probability may be obtained by our Monte-Carlo simulations for different combinations of ${\rm AFO}$, ${{\rm VOT}}$, $N_{{\rm circ}}$ and $\nu$. Then the averaged compression efficiency is:

$$% \overline{\eta}_{{\rm {c}},\nu,N_{{\rm Circ}}}(\mbox{${\rm AFO}... ...eta_{{\rm c}}$ }\, \mbox{$\eta_{{\rm c}}$ }\, {\mathcal P}_{\nu,N_{{\rm Circ}}}$$

$$\times \left(\mbox{$\eta_{{\rm c}}$ } \vert \mbox{${\rm AFO}$ }, \mbox{${{\rm VOT}}$ }\right).$$ (A2)

Of course we assumed that for any $\nu$, ${{\rm VOT}}$, ${\rm AFO}$, $N_{{\rm circ}}$ the probability distribution is integrable and normalized to 1, while the integration limits 0, $+\infty$ are to be intended as formal. There are several detectors for any frequency channel, each one having its own ${\rm AFO}$ and ${{\rm VOT}}$, so distributions of ${\rm AFO}$ and ${{\rm VOT}}$ values may be guessed among the different detectors. Assuming they are integrable and normalized to 1 as well it is possible to compute the most probable $\mbox{$\overline{\eta}_{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}}$ as

 

$$% \mbox{$\overline{\eta}_{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}} = \... ...{\rm max}}} {\rm d}\mbox{${\rm AFO}$ } \, {\mathcal P}_\nu(\mbox{${\rm AFO}$ })$$

$$\times \int_{{\rm VOT}_{{\rm min}}}^{{\rm VOT}_{{\rm max}}} {\rm ... ...\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}}(\mbox{${\rm AFO}$ }, \mbox{${{\rm VOT}}$ }).$$ (A3)

With this definition the final overall compression efficiency is:

$$% \mbox{$\overline{\eta}_{{\rm {c}}}$ }_{,N_{{\rm Circ}}} = \... ...u \mbox{$\overline{\eta}_{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}}$$ (A4)

where $f_\nu$ is the partition function for the data flow through the different detectors, if $n_{{\rm dtc},\nu}$ is the number of detectors for the frequency channel $\nu$ (see Table 1), $n_{{\rm dtc}} = \sum_{\nu = 30, 44, 70, 100~{\rm GHz}} n_{{\rm dtc},\nu} = 112$, is the total number of detectors and if the number of samples for frequency is a constant, then:

$$% f_\nu = \frac{n_{{\rm dtc},\nu}}{112},$$ (A5)

so that for $\nu = 30$, 44, 70 and 100 GHz respectively: $f_\nu = 0.0714$, 0.1071, 0.2143 and 0.6071, finally the expect data rate for each set of 60 circles is:

$$% \overline{R}_{N_{{\rm Circ}}} = 16 \, {\rm bits} \; \times \; 60 \, {\rm circles} \; \times \; 8640 \, {\rm samples}$$

$$\times \; 112 \, {\rm detectors} \; \times \; \mbox{$\overline{\eta}_{{\rm {c}}}$ }^{N_{{\rm Circ}}}.$$ (A6)

Presently there are no data to know in advance the distribution of ${{\rm VOT}}$ and ${\rm AFO}$ values between the different detectors. For this reason in this work we assumed simply flat distributions, identical for each frequency for such parameters. More over, the ${\rm AFO}$ contribution is negligible, so that the variance introduced by this parameter is neglected. From (9) we assumed that the compression efficiency is approximately a linear function of $\ln \mbox{${{\rm VOT}}$ }$ or:

$$% \mbox{$\overline{\eta}_{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}}... ...{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}} \ln \mbox{${{\rm VOT}}$ }$$ (A7)

where $\mbox{$\dot{\overline{\eta}}_{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}}$ is the first derivative of $\mbox{$\overline{\eta}_{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}}(\mbox{${{\rm VOT}}$ })$ with respect to $\ln \mbox{${{\rm VOT}}$ }$ computed for $\mbox{${{\rm VOT}}$ } = 1$ V/K, $\mbox{$\overline{\eta}_{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}},1} \equiv \mbox{$\ov... ...eta}_{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}}(\mbox{${{\rm VOT}}$ } = 1\;{\rm V/K})$. As an example, at 30 GHz for arith-n1 the full signal compression rate is $\mbox{$\overline{\eta}_{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}}(\mbox{${{\rm VOT}}$ }) \approx 0.3534 + 0.287 \times \ln \mbox{${{\rm VOT}}$ }({\rm K/V})$ with one interpolation error less than $0.2\%$. With these approximations Eq. (A3) becomes

 

$$% \mbox{$\overline{\eta}_{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}} \ap... ...rm Circ}},1}+ \mbox{$\dot{\overline{\eta}}_{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}}$$

$$\times \int_{ 0.5\;{\rm V/K}}^{1.5\;{\rm V/K}} {\rm d}\mbox{${{\rm VOT}}$ } { \frac{ \ln \mbox{${{\rm VOT}}$ }}{1.0 \;{\rm V/K}}}$$ (A8)

and after integration we obtain the final formula

$$% \mbox{$\overline{\eta}_{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}}... ...ox{$\dot{\overline{\eta}}_{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}}$$ (A9)

for the case in the previous example: $\mbox{$\overline{\eta}_{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}=2} \approx 0.34$ which is equivalent to $\mbox{$C_{{\rm r}}$ } \approx 2.94$.

To understand the influence of the error in the ${{\rm VOT}}$ determination over the distribution on the final predictions the computation is made for a truncated (i.e. zero outside the ${{\rm VOT}}$ range of interest) normal distribution of ${{\rm VOT}}$. The rms for the ${{\rm VOT}}$ distribution is chosen in the ${{\rm VOT}}$ range [0.5, 1.5] V/K we obtain respectively $\mbox{$\overline{\eta}_{{\rm {c}}}$ }_{,\nu,N_{{\rm Circ}}} \approx 0.35$, 0.34, 0.34; which corresponds to $\mbox{$C_{{\rm r}}$ } \approx 2.86$, 2.91, 2.92 respectively. Similar results are obtained with a quadratic ${{\rm VOT}}$ distribution. In conclusion these predictions are robust against the shape of the ${{\rm VOT}}$ distribution, at least for distributions which are symmetric around the nominal $\mbox{${{\rm VOT}}$ } = 1$ V/K value.