Time alignment studies

M. Palutan, A. Sarti, A. Sciubba10/11/08

disclaimer: 1) we’ve not set up an alternative time alignment procedure with respectto what presented by Giacomo/Giovanni/Maddalena 2) we report on a couple of issues (mostly related to hardware) which could beuseful to better understand timing

1

Introduction

• We use Giacomo/Giovanni software tools to reconstruct cosmic muon tracks • Data sample: 33878-33892 + 34063-34120 (september) + mask 133 noisy channels (as Firenze analysis)

• From a sample of 700k events we selected ~80k FORWARD tracks, with 3 or 4 associated hits (we reject events with more than one track)

• We don’t apply any correction to account for track projectivity; the expected muon time is evaluated as the average of the track hit times

We focus on two aspects of the time alignment procedure

1) The use of tX and tY instead of their average

2) The effect of the TDC non-linearities

2

Time residuals before alignment: ODE 109

M3R4: 4 chambers

4 padY

24 padX

ODE channel

ODE channel

M3R4:with IB

tX-t0

tY-t0

(ns)

(ns)<tX-t0><tY-t0>

tX-t0 = Rx

tY-t0 = RY

let’s define

3

Time residuals before alignment: ODE 60

M2R2: 1 chamber

4 padY

48 padX

M2R2: no IB

ODE channel

ODE channel

(ns)

(ns)

Rx

RY

an average alignment could introduce triggerinefficiency

4

Time residuals: M2R2 vs M2R3

M2R2: 1 chamber

4 padY

24 padX

M2R2: no IB M2R3: with IB

<RX> = -0.66 ns

<RY> = 3.85 ns

<RX> = -0.67 ns

<RY> = -0.47 ns

M2R3: 2 chambers

4 padY

48 padX

(ns) (ns)

Rx

RY

Rx

RY

5

Time residuals before alignment: ODE 149

M5R4: 4 chambers

4 padY

6 padX

M5R4: 4 chambers

M5R4:with IB

ODE channel

ODE channel

(ns)

(ns)Rx

RY

the alignment is correct only on average

6

Time alignment

ODE channel

ODE channel

ODE channel

a) Each ODE channel is shifted by the average time residual

b) We don’t correct channel with 1 entry

warning: the same data are used both for evaluating the correction and to check the quality of the procedure

<Rx><RY>

7

Time residuals before/after alignment

M2 M3 M4 M5

beforealignment

after alignment

<RY> ns<RX> ns

RMS(RX) nsRMS(R) ns

pad time:t=(tX+tY)/2R = t-t0

M2 M3 M4 M5

M2 M3 M4 M5 M2 M3 M4 M5

8

PAD time residuals before/after alignment

beforeRMS = 7.7 ns

afterRMS = 6.8 ns

R (ns)

R (ns)

(X+Y)/2all stations

9

PAD time residuals after alignment

M3R4Rx

RY

(ns)

Dt = (tX-tY)/2

(ns) 10

Time spectra vs TDC range

To investigate if tails are due to bad time measurement, a symmetric cut is applied: we eliminate TDC bins 14-15 on one side and bins 0-1 on the other side, to avoid systematic displacement of the time average

The aim is to use only the center of the TDC range, where time is better determined

Time alignment should be insensitive to this cut, but for some reduction in statistics

11

Time spectra vs TDC range

Dt Dt

tX

tY

tX

tY

(ns) (ns)

(ns) (ns)

1 < binX AND binY < 14 binX OR binY = 1, 14, 15

time of theother

12

tY vs tX

tYtX

tYtX

nsns

1 < binX AND binY < 14 binX OR binY = 1, 14, 15

13

An example: ODE 106 (M3R4)

tX

tY

RX

RY

RX

RY

all hits

1 < binX AND binY < 14

14

Time alignment without bin 1, 14, 15

RX

RY

all hits

1 < binX AND binY < 14

RX

RY

ns

ns M3R4

padXpadY

15

Time alignment without bin 1, 14, 15

before align.RMS = 6.9 ns

(X+Y)/2all stations

all hits

1 < binX AND binY < 14

R (ns)

R (ns)

after align.RMS = 5.8 ns

before align.RMS = 7.7 nsafter align.RMS = 6.8 ns

= 3.7 ns

16

€

7.72 −6.82 = 3.6 ns

€

6.92 −5.82

Comparison of alignment constants

K0 – K1 (ns) K1 vs K0 (ns)

17

Let’s compare the alignment constants evaluated without the TDC cut (K0) with the ones evaluated using the TDC cut (K1)

When applying the TDC cut, the number of emptyODE channels increases -> K1=0

Conclusions

18

1) The use of tX and tY in the alignment procedure is more directly related to the hardware misalignments

2) In some steps of the alignment procedure, it could be useful to use only the central region of the TDC range