Transcript Crystal-to-crystal calibration of PWO calorimeter

Crystal-to-crystal calibration of
PWO calorimeter
U. Of Minnesota group
Datao Gong, Jingzhi Zhang and
Yuichi Kubota
Purposes
• We need to calibrate crystals to get MeV/ADC count
using physics events. This can be broken into two steps.
– Obtain crystal-to-crystal constants (constants describing the
ratios of constants among crystals)
• This may not give absolute accuracy at energies away from the
typical energies of the calibration source if linearity of the system is
not perfect, but should be very good description of relative gains as
long as the non-linearity is uniform across crystals.
– Study groups of crystals together to get the best absolute
constants at different energies using other physics process
involving photons/electrons of appropriate energies
• The constants will vary over time, and this has to be
tracked.
– Over “short terms,” we need to rely on the laser-based
monitoring system.
– Over longer terms, it would be great to be able to monitor timedependent variation using physics events such as π0, η, ψ and 
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Calibration w/ Single  and 0
• Single  from sources such as Z→μμ or
momentum measured electrons can be
used for calibration.
• From kinematics, one can come up with a
good estimate of the photon energy w/o
calorimeter.
• We use a 2 minimization technique to
obtain constants.
• 0 calibration is a relatively straightforward
extension of the same technique.
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Calibration w/ Single γ
• We form 2 by:
 
2
k
( Ek   Ci Ai( k ) ) 2
i

2
k
• where k runs through events, and i runs
through crystals. Ek is the energy estimated by
an alternative way (tracking) for event k, Ci is a
constant for crystal i, Ai(k )is the ADC pulse
height for event k, crystal i and k is the energy
resolution of the photon candidate k which
should take into account both the calorimeter
resolution and energy estimates.
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Find 2 minimum
• We can find the minimum for 2 by
(  2 )
0
C j
• This results in N linear equations:
2( Ek   Ci Ai( k ) ) A(j k )
i
0
k
2
k
• This will reduce to a set of linear equations for
Cj’s.
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Find 2 minimum II
• The linear equation for Cj’s in matrix form is:
M ijC j  Ri
• where
M ij  
k
Ai( k ) A(j k )
 k2
• and
Ri  
k
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Ek Ai( k )
 k2
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Inverting large matrices
• If we were to calibrate all 80,000 crystals, we will need to
deal with a 80k by 80k matrix.
– This requires 24 GB of memory (single precision)
• However, note that Mij is non-zero only when crystals i
and j are included in at least a photon, i.e. neighbors, i ~
j or Mij is “nearly” diagonal.
– This makes storing of Mij more manageable (~10 MB) and
inverting possible. (24 neighbor crystals)
• CLEO & Belle use a FORTRAN routine MA28 to do the
inversion.
– Actually, it produces only the solution.
• OPAL uses simple iterating method to find the solution.
(A.A. Poblaguev of Yale wrote a nice note about this
method.)
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Iteration method
• The linear equation is still:
M ijC j  Ri
• Suppose we have a good set of constants,
Cip’s, (from test beam or previous calibration),
• Now we try to find better ones, Cin, from new
data: Cin = Cip + i
• or
Mij (C jp   j )  Ri
M ij j  Ri  M ijC jp  Ri '
• Note that the matrix to be inverted does not
change but Rj’ should
be smaller than Rj.
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Iteration method II
• Starting with
M ij j  Ri  M ijC jp  Ri '
• and ignoring the off-diagonal elements,
M  ~ M 
1
1
ii
ii
or i  Ri' / Mii .
• 2nd time around, we Cip + i as Cip and repeat
the whole procedure.
– The convergence should be better if one chooses
those photons which deposit most of their energies
in one crystal, reducing off-diagonal elements.
• Our experience shows that this is not needed.
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Our strategies
• Jingzhi works with a small number of
crystals, not dealing with large matrix and
study physics of our approach.
– 30 by 30 crystal array spanning  from 0 to
0.525 and  from 0 to 0.525
• Datao works with issues associated with
dealing with large matrices, and also
iteration method to find solutions of linear
equations a la OPAL.
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Procedures
• Produced 30k Single photon MC events
– Pt = 10 GeV and 5 GeV.
• Use both 3-by-3 crystal array and BasicCluster
to identify photon candidates.
10 GeV
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Fraction of energy captured in the cluster
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Energy resolution, etc.
• Average energy fraction in 3-by-3 array:
– 92.6% at 10 GeV
– 91.6% at 5 GeV
• Resolution:
– 1.8% at 10 GeV
– 3.2% at 5 GeV
• Naively, we expect the precision of the
constants to be resolution/sqrt(N), where
N is the number of photons/crystal.
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Energy fraction at low energies
• Detected
energy fraction
is smaller at
energies below
6 GeV.
• Cause is still
under
investigation.
• BasicCluster a
bit higher
overall
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3-by-3 array
5%
~93%
80%
fractions
resolution
BasicCluster
5%
96%
80%
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1 GeV
10 GeV
1 GeV
10 GeV 13
Photon results (matrix inversion)
• The constants averages
1.078.
• This is to compensate the
energy loss of ~7%
• Width of distribution (error
in constant) = 0.65%, when
(naively) expected
1.8/sqrt(18k/900) = 0.40%.
• The difference is
presumably because
photon energies are
shared among crystals.*
• 10 GeV Pt
• 30k events generated
• 18k reconstructed (> 88%
of energy detected)
* Next slide for more details
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Constant - 1
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Energy sharing and constant
precision
• Using toy MC, where the energy sharing can
be changed freely, we study their relation.
Relative
constant
precision
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50%
Fraction of energy
the center crystal
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gets
100%
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Energy fraction of the center
crystals
• The center crystals
get between 30 and
90% of the cluster
energy, and
• Typically it is 60-70%.
• From the previous
plot, deterioration
factor would be
between 1.5 and 2.
Single 0 MC
0%
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Fraction of energy
the center crystal
gets
100%
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Qualitative understanding of errors
• Once the Mij is inverted and the solutions, Ci0’s,
are found, errors on the constants can be
calculated.
• It turns out that 2 can be re-written in terms of
possible deviations from the solutions, Ci = Ci0 +
i:
2
2
  min  MijCiC j
• So the contour represented by
M ijCiC j  1
• shows the “1-” limit of the constants in the Ndimensional space.
• M-1 is the error matrix.
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Graphical understanding
• M-1 is the error matrix.
• With energy sharing
among crystals, Mii is
smaller than w/o energy
Ai( k ) A(j k )
sharing. Note that: M ij   2
k
k
-1
• As a result, (M11) would
be larger.
• Off-diagonals make the
error contour tilted, and
(M-1)11 larger than (M11)-1
• This explains qualitatively
why the errors in
constants are larger than
our naïve expectation.
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M ijCiC j  1
2
1= (M-1)11
1
1= (M11)-1
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At 5 GeV
• The constants
averages 1.087.
• This is to compensate
the energy loss of ~8%
• Width of distribution =
1.4%,
• naively expectation:
3.2/sqrt(20k/900) =
0.68%.
• Larger deviation from
expectation than 10
GeV – see next slide?
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• 5 GeV Pt
• 30k events generated
• 20k reconstructed
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More details
• Constants larger at the supermodule boundaries,
presumably to compensate additional energy losses.
• More finer structures at 5 GeV (every 5º). Are these
real?
10 GeV
5 GeV
Supermodule boundaries
Constant -
Constant -
1
1
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=10º
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=30º
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Iteration method a la OPAL
• DaTao wrote a code to use the iteration method to solve
the linear equations (OPAL method)
• It produces virtually the same constants as the matrix
inversion method.
• Once the solution Ci’s are obtained, one can check their
validity by calculating
ΔR2≡|MijCj – Ri|2 which should be zero.
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0 calibration
• Used 5 GeV Pt MC events.
– E > 1 GeV
– At p0 > 10 GeV, many photon
pairs overlap.
– Pt of photons ~ 2.5 GeV, and
fraction of energy detected is
smaller.
– As a result, the π0 mass peak
(118.6 MeV) is lower than 135
MeV.
– Opening angle resolution is
not negligible.
– Angle measurements may
need improvement.
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=8.0%
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0 calibration basics
• For the two photons to form a 0,
4 E1 E2 sin 2
2
~ m2 0


(k )
2



4 sin
C
A
C
A
~
m
 i  
j j 
0
2  i 1
 j 2





  Ci Ai( k )   C j A(j k )  ~ m 2 0 /  4 sin 2 k   mk

 i
 j

2

 1
 2

2
 k 
k
(k )
i
• So 2 can be defined as
2  
k
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

2
(k )
(k )
(mk    Ci Ai   C j A j )
 i 1
 j 2

 k2
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0 calibration basics II
 
2
k


2
(k )
(k )
(mk    Ci Ai   C j A j )
 i 1
 j 2

 k2
• Since this is not quadratic in Ci’s, minimum
cannot be found in a single step using matrix
inversion.
• Assume that we have preliminary constants,
Cip, and we seek the next approximation,
Cin = Cip + i
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0 calibration basics III
2  


2
(k )
(k )
(mk    Cip   i Ai   C jp   j A j )
 i 1
 j 2

 k2
k



2
(k )
(k )
(mk   E1p    i Ai  E2 p    j A j )
i 1
j 2


 ,
 k2
k
where E1p   Cip Ai( k ) and E2 p 
i 1
(k )
C
A
 jp j ,
j 2
energy estimateof photonsbased on t hepreviouscalibration.
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0 calibration basics III
• Taking only 1st order for i,
2  
k


(k) ( k )
(k )
(k )
(k)
(k )
mk  E1p E2 p  E2 p   i Ai  E1p   j A j 
i 1
j 2


2
 k2
• Compared to the single photon calibration case,
(k )
(k ) (k )
mk  E1(k)
E
acts
the
role
of
E
,
and
acts
E
p
2p
k
2 p Ai
the role of Ai(k )
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0 calibration basics III
(  2 )
• Using
0
 l
• Similar matrix equation results, where
M ij  
k
E
M ijC j  Ri
(k )
np

Ai( k ) Em( kp) A(j k )

, whereif i   , n  2 and if i  
1
2
k
2
, n  1,
and if j   1 , m  2 and if j   2 , m  1.
Ri  
k
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m
k

(k )
(k ) (k )
 E1(k)
E
E
p
2p
2 p Ai
 k2
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0 calibration results
• Used, for now, true opening
angles to avoid issues
associated with photon
direction determination.
• Deviation from 1 is as
expected; 135/119 = 1.13
• Errors ~ Width = 3.4%.
• Naïve expectation =
8.0%/sqrt(27) = 1.5%.
• More deviation from
expectation than single
photon – expected from
more energy sharing among
two photons.
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Iteration
• Since in the current case, we approximated the
2 to the linear term of i, we need to iterate the
procedure to get better results.
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QCD Background under peak
• Ptjet < 15 GeV/c.
• p > 1 GeV
• 0.1 0’s/event
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Background effect
• Added background
(event mixing)
• 32886 real 0’s
• 24372 random
background pairs
• Results:
– Mean -1.87±0.13
• -2.00±0.13 w/ no bkg
– σ=3.40±0.11
• 3.37±0.11 w/ no bkg
• Virtually no effect
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0 rate estimate (naïve!)
• L1 trigger is 100 kHz.
• Most of them contain ~20 QCD events.
• We will reconstruct ~2 0’s (each event) or
200 K 0’s/sec.
• If we want calibration accuracy of 0.3%,
we need ~200*30 0’s/crystal or 5×108.
• This implies that we need to collect
2.5×103 s or 1 hour.
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Performance
Matrix inversion
(2/3 barrel)
Iteration
(full barrel)
20
10
19
1033
6
30
Photon ToyMC
Photon MC (GEANT)
0 ToyMC
Iteration method is much faster to solve than matrix inversion.
Need minutes to run over events (30k, small range sample) in ExRoot
For 6k 0 /xtal in full detector range, the evt skimming time is huge!
EventSkim
EventSkim
….
Equation Solver
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EventSkim
Two separate modules
can solve it.
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To do (partial list)
• More effects of background (random
combinations)
– Worsening of errors.
– Biasing of constants.
– ηdependent background levels in QCD events.
• Study of photon direction reconstruction
– Both bias and resolution
• Systematic effects (biases) due to materials, etc.
• Errors in constants vs. resolution/cluster.
– The off-diagonals which increase errors in the
constants should help us when resolution/cluster is
calculated from error/crystal compared to naïve
expectation.
• Expand it to η, J/,  events
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Summary
• We have tools to do single photon/electron
calibration as well as 0 calibration.
– We can calibrate the entire detector in one shot.
• Memory and CPU time are both manageable.
• We understand how it works to the first order.
• Typical 5 GeV Pt 0 sample, we will need 6,000
0’s/crystal to get ~0.3% error on the constants.
(5108 0’s total)
• May take as little as 1 hour to collect enough
0’s.
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