comparison of different chamber configurations for the high luminosity upgrade of M2R2 G. Martellotti - LNF - 13/03/2015 Roma1 + Alessia

1)Bi-poissonian distribution of hit number N The average number of hits per event is = μ ω μ = average number of interactions per BC (1.7 in 2012 data) ω = average number of hits per interaction in the region under study The distribution of N is the convolution of 2 Poissonian distributions and in the first one P(0) is eliminated to take in to account that BCs with 0 interactions are not acquired 2) CLUSTER SIZE is taken into account (presently only CSX is considered) FURTHER STEPS that should be done -Realistic hit spatial correlation (‘’Alessia’’) -Correct efficiency calculation SIMULATION

mu =1.7 At the same rate (per second) in the acquired events I have more hits in the TS (0 interactions not acquired) mu =7.6 At the same rate I have less hits in the TS - The ghost fraction depends only on the hit number. M2R2 (as it is now) at 40 MHz Rate of particle hits (KHz/cm2) Occupancy per logical pad Effect of bi-poissonian (and zero interactions probability) Particle hits Reco hits

CLUSTER simulation M2R1 PadLog=0.63x3.1 STRIPWIR=0.63x25 STRIPCAT=3.8x3.1 M2R2 PadLog=1.25x6.25 STRIPWIR=1.25x25 STRIPCAT=15.0x6.25  4 x CATphys=7.5x3.1 M2R3 PadLog=2.5x12.5 STRIPX=2.5x50 STRIPY=15.0x12.5 M2R4 PadLog=5.0x25.0 For the moment in the simulation I assumed Clusters only along X

No CS simulated. Assumed uncorrelated hits N part = N hit CS for particles in the TS simulated Not considered contributions from contiguous TS contributions from particles hitting contiguous TS 32% CLSX=1 48% CLSX=2 20% CLSX=3 - NO CLSY Rate of particle hits (KHz/cm2) Occupancy per logical pad CS assumed Increasing CS reduces the rate of particles in the TS and ghost fraction Increasing CS increases the contribution of hits and ghosts due to contiguous TS M2R2 (as it is now) at 40 MHz

M2R2 Can we assume that, at these low rates, ghost fraction is negligible ? Rates of reconstructed hits in 2012 ( by Giacomo) Rates at high luminosity are extrapolated from 2012 data

32% CLSX=1 48% CLSX=2 20% CLSX=3 - NO CLSY Estimate of GHOST % in 2012 Particle hits / Reco hits = 0.97 for minimum = 0.90 for average = 0.85 for maximum M2R2 Also at low luminosity (2012) the simulated ghost fraction is not negligible The ghost fraction on data appears to be significantly larger due to the large spatial correlations (See ‘’Alessia’’) min aver max Assuming Cluster Size: Rate of reco hits (KHz/cm2) Occupancy per logical pad Particle hits Reco hits

Hit number distribution Comparison data - simulation DATA 1 PV plot is zero suppressed - Zeros are ~ 93% M2R1 Simulation 1 PV Simulation 4 PV Zeros <<93% ALESSIA

Average hit number as a function of nPV measured in 4 crowded chambers of M2R1 The hits found for nPV=7.6 correspond to ~ 2 Pad /TS PV number ALESSIA

PARTICLE RATE at luminosity = 2x10 33 Minimum Average Maximum M2R The ghost fraction in 2012 is larger  The rates of particle-hits extrapolated at high luminosity should be even smaller  The ghost fraction even larger Absolute values are still quite uncertain. But this is not so critical in the comparison of different configurations Rates ghost subtracted (simulation)

11 Q1-M2-R2 Assembling Output Connector X Y not in scale comparison of different M2R2 configurations

M2R2 as it is NOW TS =375 cm log.ch. = 12 wir(X) * 4 cath(Y) Area wire pad = 31.3 cm 2 Area cathode logical channel = 93.6 cm 2 composed of 4 phys. Channels 2 Ored on FE, 2 on IB Occupancy per logical pad y0 y3 x0 x11 Trigger Sector Logical cath. channel = 4 Phys. Ch. wire channel GHOST FRACTION at 80 kHz/cm 2 ~28% = 80 readout channels per chamber from FE (64 from IB) A Rate of particle hits (KHz/cm2)

M2R2 HALF TS =375/2 cm log.ch. = 6 wir(X) * 4 cat(Y) Area wire pad = 31.3 cm 2 Area cathode logical pad (1/2) = 46.8 cm 2 Cathode logical channels composed of 2 vertical phys. chann. SAME LOGICAL PAD AS NOW Occupancy per logical pad y0 y3 x0 x5 Trigger Sector Logical cath. channel = 2 phys. ch. wire channel GHOSTS FRACTION at 80 kHz/cm 2 reduced to 16% = 80 readout channels per chamber Same CARIOCA Inefficiency Reduced DIALOG ineff. w.r.t. configuration A B Rate of particle hits (KHz/cm2)

M2R2 TS =375/2 cm log. Ch - X, Y = 3 wir, 8 cat Area wire pad (x2) = 62.6 cm 2 Area cathode pad (1/4) = 23.4 cm 2 LOGICAL PAD area same as now, shape changed. y0 y3 x0 x5 Trigger Sector Logical cath. channel Logical wire channel = 2 phys ch. Occupancy per logical pad GHOST FRACTION reduced Assuming the same CS  15.5% Reducing CS (guess)  14.2% Same CARIOCA Inefficiency DIALOG inefficiency further reduced = 88 readout channels per chamber C Rate of particle hits (KHz/cm2)

Solution C versus solution B ( 88 vs 80 readout channels) Similar inefficiency CS and Ghost fraction (slightly) reduced Better shape  possible improvement of FOIs for medium-high momentum muons FOI can be reduced by a factor 2 reducing MisID without loosing in MuID efficiency B C Matteo

M2R2 TS =375/2 cm 2 half area - 48 log. Ch. X, Y = 6 wir, 8 cat Area wire pad = 31.3 cm 2 Area cathode pad (1/4) = 23.4 cm 2 LOGICAL PAD area 1/2 (and the occupancy per logical pad also 1/2) y0 y3 x0 x5 Trigger Sector Logical cath. channel = 1 phys. Ch. wire channel Occupancy per logical pad GHOST FRACTION slightly higher 19% at 80 kHz/cm = 112 readout channels per chamber Same CARIOCA Inefficiency DIALOG inefficiency further reduced D Rate of particle hits (KHz/cm2)

Reading only all physical cathode pads NO GHOSTS - Too high occupancy per pad  8.25% - Too large X side to build up a convenient FOI for high momentum muons 64 readout channels per chamber

P M2R2 Cathode Pad Detector 192 pads per chamber 1) Pad (=present logical pad) = 1.25 x 6.25 = 7.8 cm 2 (48 x 4 pads/chamber) 2) Pad (=solution D - 2X, 1/2Y) = 2.5 x 3.12 = 7.8 cm 2 (24 x 8 pads/chamber) No ghosts at all Much higher efficiency The second solution is better (FOIs for MuID) 1) 2)  CONCLUSION: a cathode pad detector is certainly convenient

A new topic

AND/O R Two different definitions: MC : % of penetrating tracks Data : reconstructed hits (crossings) AND/OR We cannot assume that all the tracks not giving the AND are ‘’single gap’’ tracks They can be ‘’large angle’’ tracks  Big effect on pad efficiency evaluation Revise efficiency calculations

In the DIALOG the AND is first performed between the physical channels of the two layers and only after the OR of adjacent channels  the AND/OR measures the correlation between physical channels, neither logical channels nor logical pads DIALOG logical channel generation map

We have the OR of two layers (pad of both layers must be inefficient to have an inefficiency)  the inefficiency is essentially due to the fraction of correlated hits due to penetrating tracks If R OR is the rate in the Giacomo table, the rate on single layer is R 1L = R OR (1+fc)/2 where fc is the fraction of correlated hits Layer inefficiency is Ineff 1L ~ δ eff *R 1L where δ eff is the effective dead time The OR inefficiency is fc* Ineff 1L OR inefficiency ~ fc* Ineff 1L Now to build up the logical pad we perform the AND of X, Y stripes AND inefficiency = ineff X + ineff Y + ineff X *ineff Y The ‘’hit correlation’’ measured from the AND/OR ratio for logical pads cannot be assumed as correalation for logical channels X and Y (neither for logical channel composed of the OR of 2 or more adjacent physical channels) Large angle tracks don’t giving AND because they are hitting a different physical channel X, can have correlated hits in the logical channel Y

 The inefficiency has been significantly underevaluated in regions where logical pads are the crossing of physical X,Y channels (M23R12) or/and in regions where logical pads are the OR of adjacent physical pads AND/OR ratio must be evaluated on data separately for X and Y channels. -The difference between the two measurements will give us an idea of the impact of ‘’large angle tracks’’ When beam is on, we have to measure AND/OR switching off one gap

SPARES

Penetrating tracks cross 4 gaps. After crossing the first gap, they can hit next gap in the same pad (small angle-pad centered tracks) or not (large angle-peripheral tracks). - Here pad is the logical pad or the logical channel (X,Y) in M23R12 Lets call: S = fraction of single-gap tracks in each gap P = fraction of penetrating tracks crossing 4 gaps F = fraction of P crossing one pad in a gap and a different pad in the contiguous gap The hit rate on the first gap is R G = K(P + S). FP tracks of this gap will give a new hit in the next gap  The OR rate of 2 contiguous gaps is R 2G = K(P(1-F)+2FP+2S) = K(P+FP+2S) When going from gap 2 to gap 3, by definition of F, again a fraction F of P tracks (not necessarily the same tracks previously jumping pad) gives a new hit in the third gap  The OR rate of 3 contiguous gaps is R 3G = K(P(1-F)+3FP+3S) = K(P+2FP+3S)  The rate of the OR of 4 gaps is R 4G = K(P(1-F)+4FP+4S) = K(P+3FP+4S)

If the logical pad is composed by 1 physical pad (N ph =1) the rate measured with the And is R And = K(1-F)P. Let’s call Corr = R And /R OR  The rate on the single FE to calculate the effect of CARIOCA dead time is R 1FE = R 2G = K(P + FP + 2S) = R OR (1+ Corr)/2  The inefficiency of a straight μ Inef μ =Inef PAD (Corr+Inef PAD (1-Corr)) ~ Inef PAD Corr If the logical pad is composed by N ph physical pads R 1FE > R OR (1+ Corr)/2 (the AND measures only the penetrating tracks hitting in the next gap the same physical pad, not the same logical pad) To evaluate correctly the dead time inefficiency, we should know the % of inclined tracks. For M23R12, it is also crucial to know separately Corr for X and Y while presently correlations were measured for the logical pad. The small correlation measured is probably due to the small wire strip size, while correlations for cathodes could be much higher (and their inefficiency much higher)  % of single gap tracks S is much smaller - inefficiency is underevaluated

2 Pys/L OR 2 Pys/L AND Layer 1Layer 2 X X X X X X X X XX XX ORAND OR/AND of the 2 layers OR =10, AND=2 f = AND/OR = 2/10 OR/AND of the 2 layers OR =8, AND=2 f = AND/OR = 2/8 CHAMBER 1phys.ch.=1log.ch. 2phys.ch.=1log.ch. Hit rate = R OR R OR =10 R 1FE = 6 = R OR (1+f)/2 Hit rate = R OR R OR =8 R 1FE = 6 ≠ R OR (1+f)/2 = 5 This is because the AND is always performed on the single Phys. chann.  Inclined tracks can contribute to 1 hit in the OR and not to the AND  to evaluate correctly the dead time ineffic. I should know the % of inclined tracks Separate readout of all phys channels in a layer 2 phys channels are Ored in each layer