In this note I look more carefully at an event with a missing
roll-over word and by using the data and poisson statistics attempt to
estimate the likelihood that this problem is simply due to low rate in
a tank.
First a correction. Nat asked if this roll-over problem only happened
on tanks with three inputs. I said 'no' and pointed out that one of
the cases was a channel with 4 inputs. I checked my notes however and
I noticed that the fourth input to that channel was broken at the
time. So in fact it had only 3 inputs. So it looks as if all of the
examples are verticals that have only have three inputs that are
either plugged in or functional.
So the obvious hypothesis to make is "these tanks don't have a high
enough rate to cause roll-overs to happen every time." So, to test
this theory, what I did was to take one of the channels and calculate
the times between each pulse in the WFD leading up to the muon. By
looking at the distribution we should get an idea of how likely it
would be to miss 328uSec of data.
So here we go:
Times between pulses in nano-seconds:
101125
14475
20665
28935
3545
49795
28060
300
38080
57965
35125
26540
24115
29290
41223
13700
73545
2415
14889
2997
If we assume that the pulses are randomly(that is poisonally[wow is
that a word?]) distributed then the distribution should be an
exponential. Are the pulses random? Well, I'm not sure how good an
approximation it is. Correlated radioactivity and after pulsing could
mean that the pulses are not purely randomly distributed. In this
case measuring an exponential might be(for example) measuring some
nuclear decay constant.
But if you forge ahead, histogram the data, and fit it to the
exponential
f(t) = A*exp[-b*t]
you find that b= .25E-4 (approx a 2.3 usec decay time)
Now you can build a normalized probability distribution p(x) by
p(t) = f(t) / integral(f(t)dt)
where the integral is evaluated from 0 to infinity. The result is:
p(t) = b*exp(-b*t)
Now to calculate the probability that you will measure a time
difference between two pulses of time t_0 or greater:
P(t>t_0) = integral(from t_0 to infinity)p(t)dt
P(t>t_0) = exp[-b*t_0]
So sticking in some numbers:
t_0 prob pulses separated > t0 ns
0 100%
30000 47%
330000 .03%
So the chances of missing a roll-over word(going for 330uSec or longer
with no data) is .03%. So how many roll-over misses do we expect to
see?
Well I didn't examine all of the errors in this run but the 7 events I
wrote about were taken from a selected sample of 1575 well constructed
events. By well constructed I mean there were lip triggers, single
tracks, no errors, etc. They were culled from 6400 raw
events. Of the seven events with errors three of them were due to the
roll-over problem. We need an estimate of how many of these event had
a track which actually went through one of these vertical boxes with
only three of its four inputs used. Instead of doing the work right
now I'll just be extremely generous and say that 1/4 of the event
sample meets this requirement. This results in 393 events. Applying
our .03% estimate we would expect .12 events.
In fact we saw 3 events. This is an order of magnitude difference.
I'm not sure what conclusions to draw given the assumptions I had to
make but it seems plausible to try to look for other explanations.
For example: could something 'bad' happen if one of the inputs has
nothing plugged in to it?
-Chris