Etude de bruit de fond induit par les muons dans l'expérience ...

tel-00724955, version 1 - 23 Aug 2012
C
152 Definition of the bolometer timing
and
C.3 An example for Run 8
t 0 H + ∆tH · 0.5 + T C
= t max
H − ∆tH · fH + ∆tH · 0.5 + t p
H − tmax H
= t p
H
(C.4)
The event chosen as example for Run 8 is the 3 rd event of the run of Jan 2 th 2008,
which correspond to runntp = ia02a005, and runANA = 120205.
From the data analysis file, we have the time difference of the beginning of the
run and of the beginning of the ionization interval, which is for Run 8 the time of
the beginning of the ionization interval:
T H = t 0 I = 15019.483 ms (C.5)
T H is already corrected by the factor 1.008. We also have the time of the beginning
of the ionization pulse relative to the center of the ionization interval:
T I = 0.68573 · 1.008 ms (C.6)
From the data acquisition, i.e. the ntp file, which contain the same information
than the corresponding raw data files, we have the time of the maximum of the
filtered heat pulse:
t max
H
Thus, it is clear that T H and t max
H
= 1493100 unit of the 10 µs clock
= 14931.000 · 1.008 ms
= 15050.448 ms (C.7)
are different and have a different signification,
despite a misleading name in the bolometer analysis ANA. T H is the beginning of
the ionization interval and tmax H is the maximum of the heat pulse.
The time of the maximum of the heat pulse in ANA can be evaluated if we add
the pre-trigger fI = 0.75 of the sample window to the time of the beginning of the
ionization window:
max
tH ANA = t0I + ∆tI · fI · 1.008 (C.8)
= T H + ∆tI · fI · 1.008
= 15019.483 ms + 40.96 · 0.75 · 1.008 ms
Comparing Equations C.7 and C.9, we have tmax
H
= 15050.449 ms (C.9)
ANA
= tmax H , we find back the ntp
time stamp from the time of the analysis. We are now certain that the time used in
the analysis and the time used in the ntp is the same and is the one coming from the
device which dispatch the clock in 10 µs-beat, and not some arrangements from tPC. ∗
∗ About tPC, the real time read in the ntp file is tPC = 1199263394.124085, then tPC − t0 PC =
1199263394.124085 − 1199263379.193085 = 14931.000 ms. Therefore tPC − t0 PC = tmax H · 10−5 still
holds as it is defined in Table C.1.