Energy Calibration of BESIII EMC  ‘Digi’-calibration Bhabha calibration  0 calibration Radiative Bhabha calibration  ‘Cluster’-calibration.

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Presentation transcript:

Energy Calibration of BESIII EMC  ‘Digi’-calibration Bhabha calibration  0 calibration Radiative Bhabha calibration  ‘Cluster’-calibration

Digi-calibration  Converting ADC counts into MeV PED i - pedestal value with respect to ADC i, e i - the electronic gain constant , c i - the energy conversion constant. obtained from that cosmic calibration , before counter installation into the container obtained from online Pressure of structure Geometry of calorimeter Radiation damage Crystal non-uniformities etc. To achieve more accurate c i, counter-by-counter calibration is executed using Bhabha events.

 Energy losses in material in front of the Emc On average reduced deposited energy Energy losses fluctuate Reduced energy resolution  Need known energy processes over the entire physics range to calibrate clusters   0 decays and Radiative Bhabha events can be used ‘cluster’ calibration

EMC calibration flow chart ADC Cosmic const.(ci) & gi g i =g b i / a  i gbigbi aiai

By minimizing the  2, matrix equation is extracted. Q is matrix with order 6272, is sparse. All g i are decided simultaneously by inverting matrix equation. Sparse matrix package (SLAP etc.), solving the matrix equation.. E e - electron or positron energy from kinematic f ( E e, ,  ) - Shower leakage correction obtained from MC E k exp - expected (deposited ) energy Index k - shower range around maximum energy counter Find constants g i minimizing: Bhabha calibration

Gamma-Gamma Calibration gi Intermediate Data (matrix, vector) Bhabha calibration const. Bhabha calibration flow chart

Algorithm EmcSelectBhabhaEmcCollectBhabhaEmcChiCalib EmcBhabhaEvent e +, e - EmcBhaCalibData stores matrix, vector, the number of hits/Xtal and other stuff Fill inRead outFill inRead out SLAP Algebra package solve large sparse symmetric and nonsymmetric positive definite linear systems, Ax = b, using preconditioned iterative methods EmcBhaMatChan holds the intermediate data (matrix, vector) to be stored in the spatial DB, represents one channel = crystal (one row in the matrix + vector element + number of direct and indirect hits) EmcLSSMatrix a Large Sparse Symmetric Matrix, Per row only the nonzero elements are stored together with their column indices. (the storiage of the row index is a tribute to SLAP which needs it) Solve linear equations Calibration constants are be Saved in the condition database check Class diagram for Bhabha calibration

 0 calibration

Energy correction function Energy loss/leakage is shower depth dependent  Energy dependency as polynomial in lnE Material in front of the Emc / tilt of the crystal varies with polar angle  angular dependency as polynomial in cos 

Iterative procedure to obtain the coefficients 1.Select photon candidates and calibrate their energies with the old calibration function 2.Build all possible two-photon combinations 3.Histogram the 2-photon-mass in bins of E 1, E 2 (2 entries per combination) and in bins ((cos  1 +cos  2 )/2 )(1 entry per combination) 4.Fit the function fNov(m)+poly(m) to the histogramsfNov(m) 5.Use the ‘peak’-masses m 0 from the fits and plot ln(m  0 /m 0 ) as function of lnE or cos  6.Fit a 3 rd order polynomial in lnE to ln(m  0 /m 0 ) (lnE) and 2 nd order polynomial in cos  to ln(m  0 /m 0 ) (cos  ) 7.Add the coefficients of the polynomials to the old coefficients 8.Repeat from step 1 untill coefficients don’t change anymore

‘Novosibirsk’ function A, , and m 0 correspond to the height, width and mean of a Gaussian distribution t is a measure of the asymmetric tail. the normalization for an tail t,  is the full width at half maximum divided by, and the most probable value is reached for m = m 0.

Iterative procedure to obtain the coefficients 1.Select photon candidates and calibrate their energies with the old calibration function 2.Build all possible two-photon combinations 3.Histogram the 2-photon-mass in bins of E 1, E 2 (2 entries per combination) and in bins ((cos  1 +cos  2 )/2 )(1 entry per combination) 4.Fit the function fNov(m)+poly(m) to the histograms 5.Use the ‘peak’-masses m 0 from the fits and plot ln(m  0 /m 0 ) as function of lnE or cos  6.Fit a 3 rd order polynomial in lnE to ln(m  0 /m 0 ) (lnE) and 2 nd order polynomial in cos  to ln(m  0 /m 0 ) (cos  )ln(m  0 /m 0 ) (lnE) ln(m  0 /m 0 ) (cos  ) 7.Add the coefficients of the polynomials to the old coefficients 8.Repeat from step 1 untill coefficients don’t change anymore

Why ln(m  0 /m 0 )? Fit to lnE yields a 0-3 Fit to (cos  1 +cos  2 )/2 yields a 4, 5

Iterative procedure to obtain the coefficients 1.Select photon candidates and calibrate their energies with the old calibration function 2.Build all possible two-photon combinations 3.Histogram the 2-photon-mass in bins of E 1, E 2 (2 entries per combination) and in bins ((cos  1 +cos  2 )/2 )(1 entry per combination) 4.Fit the function fNov(m)+poly(m) to the histograms 5.Use the ‘peak’-masses m 0 from the fits and plot ln(m  0 /m 0 ) as function of lnE or cos  6.Fit a 3 rd order polynomial in lnE to ln(m  0 /m 0 ) (lnE) and 2 nd order polynomial in cos  to ln(m  0 /m 0 ) (cos  ) 7.Add the coefficients of the polynomials to the old coefficients 8.Repeat from step 1 untill coefficients don’t change anymore

谢 谢 !谢 谢 !