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Analysis Test Beam Pixel TPC

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Presentation on theme: "Analysis Test Beam Pixel TPC"— Presentation transcript:

1 Analysis Test Beam Pixel TPC
Peter Kluit, Michael Lupberger, Jan Timmermans Nikhef/PixelTPC meeting 3 December

2 Introduction Analysis of the pixel testbeam data
Description of the setup 10 octoboards consisting of 2 x 4 chips Geometry y along radius of the TPC: x orthongonal z from time measurement 25 nsec clock Start by a zero B field run: straight tracks Run _ offset z is 6 cm Typically track crosses 2 x 10 Chips Cannot align all chips in an octoboard because only 2/8 are illuminated by the beam

3 Chip layout in the test beam

4 Chip alignment in xy All tracks should go through chips 10 and 153
Require > 50 hits in each chip Select clean and clear track: On the full track 80% of all hits in the event should be used Measure the rotations of chips 10 and 153 Fit phi angle in chip 10 and 153 separately Compare to measured phi angle from fit to chips Create new geometry file for rotated chips 10 and 153 Preciser alignment procedure for chips 10 and 153 Apply a stricter track selection > 100 hits with cuts on d0 and z0 Select zone in x measure residual in xy plane (d0) per pixel Fit the residual vs y (y=outward radius of TPC) ypixel = 0 to 256 Impose that shift = 0 (nominal position)

5 Chip 10 residuals Note the correlation with x due to beam profile
The plots investigate possible a periodic structure in y Residuals flat in y and no clear periodic structure Fitted slope in y = mm/pixel

6 Chip 153 residuals Residuals flat in y and no clear periodic structure. Some deformation at the edge. Fitted slope = mm/pixel

7 Then aligning all Chips
From this point onwards it is rather easy: just fit per chip the residuals in pixel x and pixel y These form the corrections to the individual chips This gives the first iteration for the corrections Finally, a second iteration is done using the refit of the full track Here examples of the residuals after this procedure A total of 5000 events were used

8 Chip 1 residuals Residuals flat in y and no clear periodic structure

9 Chip 150 residuals Residuals flat in y and no clear periodic structure

10 Corrections per chip So shifts are of the order of 250 microns and rotations of 0.6 micron per pixel (or 150 microns over full chip 256)

11 Hit residuals The xy plane looks perfect with a sigma per hit of 0.86 mm The z residuals have a non zero mean, a long tail and an rms of 4 mm The fit with exponential convoluted gaussian gives a sigma gaus of 1.7 mm and a tau of 3.4 mm

12 Sagitta straight line measurement
Now we are set to measure the sagitta (and thus momentum) resolution of the long track We use the B = 0 field so straight line run with all the alignment constants defined before Here, a method is used where each track is split in three pieces. The first quarter called Inner part; the second half Middle and the last quarter the Outer part. The sagitta can be calculated from fits: Fit Inner & Outer parts -> track IO Fit the Middle part -> track M; calculate the position in middle of hits pos M Fit Inner part -> track I; and Fit Outer part -> track O Sagitta = distance of track IO to pos M Compare phi angle IO with phi M (and phi I and phi O) The sagitta and delta phi plots will tell us how well we do

13 Sagitta and pull Clean tracks 498 tracks
abs (d0 + 8) < 2 mm and abs(z0-74)<10 mm abs(phiI,phiO,phiIO )<0.005 |sagitta|< 1 mm nhits < 4000 and nhits > 0.95 nhits all 498 tracks The sagitta resolution is fitted 65 microns. The pull is sagitta / error where error = 0.75/sqrt(nHits/4) ( <nHits> = 3000) The fitted pull sigma is 2.2

14 Sagitta angle and pull Using clean tracks
The sagitta angular resolution is fitted 0.6 mrad. The pull is angle / error where error = 0.75/sqrt(nHits/8)/(3*L/16) The fitted pull sigma is 1.5

15 Aligning the z coordinate
Same procedure as for XY start with chips 10 and 153 Start with aligning (measure rotation) chip 10 and 153 To obtain a preciser prediction only hits were fitted to a track if the z residuals were smaller than 2 mm There is more structure in the z coordinate (time): saw tooth structure over 96 pixels plus odd/even modulations in y and x Chip 10 Chip 153

16 Chip 1 Aligning the Z coordinate
Local y shift and rotation Local x rotation periodicity Break points and amplitude of the saw tooth vary from chip to chip As well as magnitude of odd/even structure Local y Saw tooth

17 Chip 150 Aligning the Z coordinate
Sawtooth structure likely due to the clock distribution over the chip 200 ps delay per pixel.

18 Chips 1& 150 after aligning the Z coordinate
Residuals after applying the alignment corrections for shift, rotation y and rotation x, saw tooth y and odd/even Chip 150 Chip 1

19 Concerning the z resolution
The expected resolution has three components: Gaussian smearing due to diffusion 0.6 mm Clock binning 25 nsec about 2 mm Clock time walk parametrized by an exponential mean 3.4 mm Observation: the chip 1 fit shows a repetitive peak structure with a 2 mm shifts. In chip 153 the peak is more diluted. The question is whether we can measure the gaussian (sigma) component in the full distribution. The two fits yield 0.5 and 1 mm respectively. What happens is that for chip 1 the rise of the binned clock close to the gaussian diffusion is fitted. In the second fit the bins do not align that well so part of the smearing coming from the clock is included in the chip153 result. Still I think that we can conclude that the data are compatible with the expected resolution. It is however difficult to extract a precise number for the gaussian diffusion component.

20 Summary & further steps
Concerning the z resolution: this is basically understood. Current issues will be resolved in the new TimePix chip. The resolution in xy is a puzzle. The single hit resolution is understood but …the sagitta plots show a resolution of 65 microns. This is quite precise, but still a factor 2.2 too large. This is very important because we expect a 1/sqrt(N) dependence, where N is here 3000! This affects directly the momentum resolution. In Whistler I did quite a few studies to try to pin down the problem. One of the points is that the deformations at the edges of the Chips have to be removed .This makes things better but we are still not there Already inside one Chip the resolution is a factor 2 too large We made plots and saw deformations as a function of z. The residuals have offsets as a function of z. Very very puzzling… This has to be investigated further

21 Further steps Concerning dEdx
Below a plot of the energy loss per mm for the selected tracks No special analysis cuts: nor adjustments for inefficiencies (dead pixels/edges). This gives 6.1 hits per mm (not correcting for dead space between Chips). Resolution 14% for 3000 hits. Can be further improved with truncated mean method to 10% As Michael showed! From Michaels talk 20 sept 2015


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