ITS Track based Alignment

By

Bjørn S. NilsenThe Ohio State University

Contents

• Quick overview– What are the mis-alignments– Track model assumptions– Re-alignment form

• Results from perfect tracking/detectors

Mis-Alignment overview

• There are 2198 ITS Si detector

• Each detectors can be mis-positioned in all of 6 different ways.– Collection of detectors may be

effected in similar/correlated ways, whole ladders…

– SDD also has an additional possible mis-alignment, “Jitter”, or better t0

• ITS must also be aligned with respect to the rest of ALICE, TPC.

Track model

• Not dependent on a specific track model– Must return point of intersection with an

arbitrary plane, including error matrix– Must return direction of track at that point– Should be refit excluding points from a specific

detector, quickly

• with no ambiguity as to which points belong to track i.

€

tμi s;M{xν

i0L xνij L }( )

Re-Aligning

• For ITS is the thin detector/plane direction in the detector (local) coordinate system.

• Where i is the index for different tracks, and j is the index for different detectors. Together they define a RecPoint.

€

χ 2 = tμij (s j;M{xξ

ik≠ j L }) − (ΔRμτj 0 xτ

ij − ΔTμj )( )

1

σ μνij

( )2

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟tν

ij (s j ;M{xξik≠ j L }) − (ΔRντ

j 0 xνij − ΔTν

j( )

ij

∑

€

0 xμij ≡0Rμτ

j l xτij −0Tμ

j f xμij = ΔRμτ

j 0 xτij − ΔTμ

j

€

s j ( f ) n j = ΔRμτ

j 0Rτνj )

y νj , f Tμ

j = ΔRμτj 0Tτ

j + ΔTμj )

€

) y

€

∂χ 2

∂ΔTηl

= 0ηl

€

∂χ 2

∂ΔRηυl

= 0ηυl

Perfect Tracks & Detectors3000 events with 5 tracks each

____

__T

rack

s___

___

____

__D

etec

tors

____

_

Miss-alignedDet=885 x=0.1cm z=1.0E-7cm

3000 events with 5 tracks each

____

__D

etec

tors

____

___

____

Tra

cks_

____

_

Det=885

Det=885

After Re-Alignment__

____

Det

ecto

rs__

____

____

_Tra

cks_

____

_

Perfect — Re-Aligned__

____

Det

ecto

rs__

___

____

__T

rack

s___

___

1E-15

Numbers

€

0R =

1 0 0

0 −1 0

0 0 −1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟€

0T = 0 −38.5684 1.9754( )

€

χ 2 j= 885= 2.0081E −10

Perfect Alignment

€

mR =

1 0 0

0 −1 0

0 0 −1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟€

mT = −0.1 −38.5684 1.9753999( )

€

χ 2 j= 885= 244.451

Mis-Alignment

€

rR =

1 −8.59E − 20 −4.96E − 9

6.08E − 7 −1 1.27E − 6

−7.46E − 9 −2.54E − 21 −1

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟€

rT = −3.35432E − 9 −38.5684026 1.975399991( )

€

χ 2 j= 885= 2.3579E −11

Re-Alignment

Realignment Results1E-8 to 1cm in x & z

1 0

0

0

0

0

0-1

-1

-38.57

0

1.975

Conclusion

• In plane displacements realignment good

• Progress is under way

• Must take into account track angles

Additional Slides

SDD “Jitter”Better t0

• ALICE SDD readout clock synchronous with Beam.– Jitter poor choice of word– t0 fix for all events– t0 may differ by detector

(half detector)

• Test Beam SDD readout clock asynchronous with Beam– Jitter varies event by event– Must be measured

Solution for In-plane

Equation for Full

€

) Q μν

ij ≡ΔRμτ

j Qτνij

ΔRρςj Qςρ

ij

Wμνij ≡ δμν +

) Q μν

ij

Yμνij ≡ δμν −

) Q μν

ij

Qμνij ≡0 )

n μj cos(θν

i )

χ2 After Re-aligning

Z

X

1.0 0.1 0.01 1.0E-3 1.0E-4 1.0E-5 1.0E-6 1.0E-7 1.0E-8 0.0

1.0 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.76e-11

0.1 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11

0.01 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11

1.0E-3 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11

1.0E-4 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11

1.0E-5 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11

1.0E-6 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11

1.0E-7 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11

1.0E-8 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11 2.36e-11

Differ at the 0.1% level