TAU - status December 14. Lab setup 9/7/2011TAU - status report2 Each RO line is X100 Attenuated and...
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Transcript of TAU - status December 14. Lab setup 9/7/2011TAU - status report2 Each RO line is X100 Attenuated and...
TAU - status
TAU - status report 2
Each RO line is X100 Attenuated and 100Ω terminated
10 μm position precision of collimated source
10 RO lines
HV line quenched with 20MΩ connected to a 1MΩ /100MΩ voltage divider
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Source sits here
1cm thick, Delrin collimator, with a 3 mm slit opening
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All night measurement
• collimated Source was left on the same position (above RO channel 7) for ~14 hours.
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Each histogram shows the ratio between each channel and the channel beneath the source (7).
1 2 3 4
5 6 7 8
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• Each full turn of the micrometer moves the source by 1mm• Started the scan when the micrometer read is 60mm• Made 16 measurements each of 5minutes and 1mm apart. • CF4 gas, 600 Torr, 1316V .• Actual voltage on the panel is ~4/5 of the HV applied due
to the voltage divider.9/7/2011
TAU - status report 79/7/2011
49mm 50mm 51mm 52mm
53mm 54mm 55mm 56mm
60mm 59mm 58mm 57mm
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Source position measurement
• points in this graph are the Gaussian fit results
• Errors are σ/√(n-1)• Numbers in the x-axis
have no global meaning, simply the read of the micrometer and not the distance from any real point in space…
• The electrode pitch in this panel (10 electrodes per inch) is the slope of this line: 0.4 electrodes per mm…
XY signal processing
• Measuring the pulse as a function of HV line number.
85 24105 55
HV line Area Width at bottom
Width at 60%
24 46.7 77 bins 21 bins 1.36
55 51.5 75 bins 33 bins 1.2
85 47.7 80 bins 64 bins 0.995
105 48.3 82 bins 57 bins 1.05
XY signal processing
• Cutting out the noise. Flipping the signal to the positive side. Trying to discriminate between the two pulses (table above).
• None of the quantities above seems to give good separation between the two pulses.
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Last minute plots …
• Seems that the r/o pulse can be parameterized according an electric damped oscillator
cos( )tA e t
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r/o pulse fit
If we supposed R=100 ohms, the
fit results gives C=15pF
• Measuring the signal at stripe #1, we expect to get the signal shape with a factor 2. This should include the negative side of the bi-polar signal.
• Fitting the two pulses with the shape extracted from the above measurement. Allowing the location of the two signals to float.