Emittance Test

Design

New deisign: We changed to use just one magnet.

File:Emittance Test Optim File.txt

HRRL had steer magnets on itself. We don't need external steer magnets. But Dr. Chouffani said that we might need them after quads, if the quads are not on axis. He will think about it.

Beam Line design on 27th, Jul. 2010:

Optim File on 27th, Jul. 2010:

File:Emittance Test Optim File 27th Jul 2010.txt

File:HRRL Emittance.pdf

YAG (Yttrium Aluminum Garnet) Christal

Faraday Cup

Old Beam Line

Beam Line Parts

Length of the pole face of Quad2Tin z direction is 8 cm.

Constructed Beam Line for Emittance Test

Vacuum pressure on the beam line is less than [math] 10^{-8} [/math] torr (on 2010-July-7th).

Leveling Survey

Downstream of the beam up by 2.6 mm.

How to do Leveling Survey and Alignment with Theodolite on e- Beam Line

A. Leveling Beam Pipe

1. Put the theodolite as high as the center of the beam line at upstream. Turn theodolite up and down. If the up angle is equal to down angle from the center of the pipe, theodolite is exactly as high as beam pipe. If not, adjust theodolite, until up and down angles are equal.

2. Once first step finished, then rotate theodolite horizontally to down stream. Turn it up and down, if up and down angles are not equal, adjust the stand till they are equal.

B. How to Make Sure the Quad is Perpendicular to the Beam Line

If the distances between quads and end of cavity at any given points are equal, then quad and end of cavity are parallel. Pipe is perpendicular to the end of the cavity, thus, magnet would also be perpendicular to the pipe.

I will find at lest three random points quads and three corresponding points at the end of the cavity, make sure they are equal, then I can say, magnet is perpendicular to the beam line.

I am planning to stick a metal ruler two, so two faces are parallel.

C. Centering the Quad Magnet with Theodolite

Vertical Centering

1. Finish the step 1 in A.

2. Then turn up the theodolite from the upper edge of the pipe to the magnet hole, call this angle upper angle. Do the same at lower part, call it lower angle. If the this two angles are not equal adjust the stand till they are equal.

Horizantal Centering

We know the diameter of the pipe [math] d_p [/math]. So, we can calculate from the pipe to the edge of the quad as:

We need to make sure this distance is same when measure it horizontally, so that our quad is centered on the pipe. I am going to use a ruler to ...(thinking)

July 2010 Run

Initial Accelerator Settings

E = 10 MeV [math]\Rightarrow[/math] Control Voltage= 4.09 V

Pulse Width <= 50 ns

Average Beam Current = least amount needed to light up YaG crystal, start at 1 mA and Rep rate of 100 Hz.

At the same time we want to do radiation test on HRRL. This is for predicting the radiation after shifting cavity to new position. Radiation test run plan is at: [1]

7/26/10

Alignment check

1.) Survey the Quad to be sure it is centered on the beam pipe

2.) survey the flag to determine if target is centered on the beam pipe

Transmission tuning

1.) Using the quads and accelerator coils, focus the beam on the YaG crystal to as fine a spot as possible. Monitor the FC during the focusing.

changed control Voltage from

Control Voltage= 4.9 V [math]\Rightarrow[/math] E = 16 MeV

to

4.09 V [math]\Rightarrow[/math] E = 10 MeV

Tune the 4 accelerator coils for max transmission.

We want to maximize the current reach the Farady Cup. This can be determined by the ratio on Channels of FC and Gun Current.

Gun current is the current on the pick up coil, which is input current for cavity. Current on Farady cup is the current reach the end of the beam line (FC is at the end of the beam line). We want to maximize the the ratio of Farady cup current to Gun current (FCC/GC), so there are more electrons will make it to the target. Thus, we enlarge our positron yield.

Basic principle: Fix three coils, change one till it maximized.

Coil 1 upstream left/right (mA)	Coil 2 upstream UP/down (mA)	Coil 3 downstream up/down	Coil 4 downstream left right	Scope Picture	Gun Current	FC current	Transmission [math]\equiv[/math] Gun / FC
	10				15 +/- 0.65 nV s	-5.4[math]\pm[/math]1.8 nV s	36 [math]\pm[/math] 12 %

The HRRL went down after 2 hours of operation.

7/27/10

No HRRL beam time today. Chad tested the Thermatron module and believes it is not the problem. Chad will continue debugging tomorrow.

Filament

BMP format:

Wiki won't allow me to upload with bmp extension. So, I used png extension to upload. After download files, you will find they are still with bmp extension.

7/28/10

The HRRL came back around 3pm today. A capacitor in the bank was found to be the problem.

Maximize Transmission

Maximized setup according to the accelerator operator.

Beam Energy: 16 MeV

Rep Rate: 95 Hz

Pulse Width: 100 ns

Coil 1 upstream left/right (A)	Coil 2 upstream UP/down (A)	Coil 3 downstream up/down (A)	Coil 4 downstream left right (A)	Scope Picture	Gun Current	FC current	Ratio	Spot Size
0.1	0.5	3.1	0.7

Reversed the polarities on coil # 3 and coil # 4. Rep rate: 3Hz

Coil 1 upstream left/right (A)	Coil 2 upstream UP/down (A)	Coil 3 downstream up/down (A)	Coil 4 downstream left right (A)	Scope Picture	Gun Current	FC current	Ratio	Spot Size
1.2	1.2	0.0	3

After image manipulation in GIMP to remove spot from the filament, Arne obtained reasonable projections for horizontal and vertical beam sizes ( < 2 mm). The quad was off, low rep rate to avoid saturation, 10MeV. This was the spot for the initial setup on Wednesday.

Electron beam deflection due to Earth's B-field.

It appears the the image from electrons hitting the YaG crystal is offset (to the right) compare to the image from the filament which appears near the center of the cross hairs.

Is the earths Magnetic field causing the electrons to move this far beam right?

[math]B_{Earth}[/math] = 0.00005 Tesla ( [math]\frac{ \mbox{kg}}{\mbox{C}\cdot \mbox{s}}[/math]) = 0.5 Gauss

E_{electron} = 10 MeV = Pc [math]\Rightarrow[/math] P = 10 MeV/c

Estimation using Classical Lorentz Force (forget about angle of B and V)

[math]\vec{F} = q \vec{v} \times \vec{B}[/math]

assuming [math]v[/math] and [math]B[/math] are orthogonal then we get the cyclotron radius

[math]F= qvB = m\frac{v^2}{r} \Rightarrow r = \frac{qB}{mv}[/math]

[math]r = \frac{(1.6 \times 10^{-19} \mbox{C}) \times ( 5 \times 10^{-5}\frac{\mbox{kg}}{\mbox{C} \cdot \mbox{s}})}{mv}[/math]

[math]mv = P = 10 \frac{ \mbox{MeV}}{\mbox{c}} = 5.6 \times 10^{-21} \frac{\mbox{kg} \cdot \mbox{m} }{\mbox{s}} [/math]

[math]\Rightarrow r = 700 \mbox{m}[/math]

Now some Geometry.

Assuming the orbit is a circle with a 700 m radius, then the question becomes how far does the electron move to maintain the orbit if it travels 1 m in a straight line.

let

[math]r[/math]= radius of circle

[math]l[/math]= distance along the beam line

[math]d[/math] = deflection due to gravity

Note

[math]l[/math] is a tangent to the orbit

one can construct a right angle triangle with side of length d, the [math]l[/math] as the base and [math]r+d[/math] as the hypotenuse.

Then Pythagorean's theorem gives

[math]r^2 + l^2 = (r+d)^2[/math]

Giving

[math]d^2+ (2r)d +l^2 =0[/math]

using quadratic formula you get

[math]2d = -(2r)/2 \pm \sqrt{(2r)^2 + 4\frac{l^2}{r^2}}/2[/math]

since l/r is very small (1/700) we can approximate this as

[math]d = -r \pm r \sqrt{1+ \frac{l^2}{r^2}} \approx \frac{l^2}{r} = \frac{l^2}\frac{P}{qB}= \frac{qBl^2}{P}[/math]

for 10 MeV and a 1 m long beam pipe

[math]d = (l\mbox {m})^2 \frac{(1.6 \times 10^{-19} \mbox{C}) \times ( 5 \times 10^{-5}\frac{\mbox{kg}}{\mbox{C} \cdot \mbox{s}})}{ 5.6 \times 10^{-21} \frac{\mbox{kg} \cdot \mbox{m} }{\mbox{s}} } = l^2 \times 1.42 \mbox{mm}[/math] where [math]l[/math] is in units of m

Radiation Footprint Measurements

Spent the remaining beam time doing radiation footprint measurements.

HRRL_radiation_tests#Run_Plan

7/29/10

Today we first tune up the beam for max transmission.

If the above calculation is correct we can expect the electron beam to move at least 2 mm away from the center of the FC. We could test it the Earth's B-field is the cause of the deflection by watch the deflection as a function of the beam energy. If we had enough [math]\mu[/math]-metal we could wrap the beam pipe in it and see if the deflection goes away.

Quad Scan

First Scan

Beam current

[math]\frac{535.4 \pm 23.4 \times 10^{-12} \;\mbox{V} \cdot \mbox{s} }{50 \Omega} = 10.71 \pm 0.478 \times 10^{-12}\mbox{ C}[/math]

[math]\frac{10.71 \pm 0.478 \times 10^{-12}\mbox{ C}}{105 \times 10^{-9}\; \mbox{s}} = 102 \pm 4.46 \mu \mbox{A peak current}[/math]

Second Scan (Polarity Changed)

PS Output Current to Shunt Resistance Current Conversion

PS negative connected to Quad positive.

Shunt resistance: 50 AMP, 50 MV. (PR 0050, QECO)

PS Reading (Amps)	Shunt Resistance Voltage (mV)	Shunt Resistance Current (Amps) (Calculated)
0	0	0
1	1	1
2	2.1	2.1
3	3.2	3.2
4	4.3	4.3
5	5.4	5.4
6	6.4	6.4
7	7.4	7.4
8	8.5	8.5
9	9.6	9.6
10	10.7	10.7
11	11.7	11.7
12	12.8	12.8
13	13.9	13.9
14	14.9	14.9
15	16	16
16	17	17
17	18.1	18.1
18	19.2	19.2
19	20.3	20.3
20	21.3	21.3
21	22.4	22.4
22	23.5	23.5

Radiation Footprint Measurements

spent noon to 3pm on radiation measurements below

HRRL_radiation_tests#29_Jul_2010

Earth's B-field effect

It the above calculation on the effects of the Earth's Magnetic field on a the electron beam is correct then we should see the amount of deflection depend inversely on the electron's momentum (1/P).

So lets take a few pictures at the energy exterma to see how far the electron spot moves away from the filament image on the crystal.

E beam (V gun)	YaG picture	Beam Spot posiiton (arb offset)
5	100 px
10 (4.09 V)		-31.8 [math]\pm[/math] .1 mm
16 (4.9 V)		-32.0 [math]\pm[/math] 0.1 mm

We did not see the beam spot move when the energy was changed from 10 MeV to 16 MeV, but going to lower energy definitely moved the beam spot farther away from the center filament image that is located on the YaG crystal cross hairs.

Maximize Transmission

Change the linac coils to determine a setting which delivers the maximum current to the FC and has a reasonable spot size on the YaG viewer.

Coil 1 upstream left/right (A)	Coil 2 upstream UP/down (A)	Coil 3 downstream up/down	Coil 4 downstream left right	Scope Picture	Gun Current	FC current	Ratio	YaG Image
0	0.6	1.8	2.7

Coil 1 upstream left/right (A)	Coil 2 upstream UP/down (A)	Coil 3 downstream up/down	Coil 4 downstream left right	Scope Picture	Gun Current	FC current	Ratio	YaG Image
0	0.6	1.8	2.7	100px				100px

a.) Does a small beam spot size on the crystal give maximum FC signal?

b.) Is there a point where the spot size changes and you can't resolve a change in the FC output?

c.) Remove the YaG target and determine if above observations are unchanged.

Pickup Coil Calibration

2.) After tuning the beam to deliver the Maximum beam current to the FC, measure the FC and green pickup coil output as a function of current by integrating the pulse with an oscilloscope.

FC output	Green Pickup Coil output	Spot on Crystal	Scope
(mA)	(mA)
	0.5
	1
	1.5
	2
	2.5
	3
	3.5
	4
	4.5
	5
	5.5
	6

a.) does the FC output change with rep rate? It should not but maybe there is loss at high rates?

Quad scan

Tune the beam so a change in the quad B-field does not cause a deflection.

First attempt to measure the spot size change as a function of Quad current

Quad Current	Beam Spot RMS (0 degrees= Left/right)	Beam Spot RMS (90 degrees= Up/Down)
(Amps)	(mm)	(mm)

Shall we do both polarities?

Imaging

Background subtraction

The filament produces at least a 2 mm diameter spot on the YaG crystal. Beam dominates the image.

7/30/10

At 4pm yesterday we attempted to measure the emittance at high currents of 10 mA, previous measurements were at 0.3 mA. The beam spot became larger than the 1" YaG target holder. This would make emittance measurements impossible.

In order to measure the emittance above 1 mA we will need to insert a doublet as close to the accelerator entrance as possible. Since this would not be done by today we decided to cancel today's beam time and work on the next emittance measurement.

The YaG camera saturated at about 0.6 mA of beam current (our measurements were made at about 0.3 mA). Perhaps we should use a thin Aluminum target instead of the YaG crystal for high current measurements.

We were able to Trigger the GigE camera. We should use this next time.

Imaging scratch area

After image manipulation in GIMP to remove spot from filament. Obtained reasonable projections for horizontal and vertical beam sizes. The quad was off, low rep rate to avoid saturation, 10MeV. This was the spot for the initial setup on Wednesday.

Data Analysis

Root Code: File:Pixle.C

The cut is changed from circular to elliptical on Sep. 23rd, 2010.

Rotating the Image

Images need to be rotated -6 degree, so cross at center of the crystal is perpendicular to the image. Here are the rotated images

1st Scan

2nd Scan

Polarity was switched from 1st scan.

Fits for Beam Spot Projection

1st Scan

2nd Scan

Coil Current vs Beam Spot Projection Gaussian Fit's Sigma

1st Scan

Coil Current Reading on PS (Amps)	Coil Current Reading From Shunt Resistance (Amps)	X Projection Sigma (pixels)	Y Projection Sigma (pixels)
0	0	30.28	30.69
1	1	31.6	29.81
1.4		32.3	29.37
2	2.1	32.93	2864
3	3.2	34.45	27.85
4	4.3	35.51	27.3
5	5.4	36.28	27.06
6	6.4	37.22	26.52
7	7.4	38.14	26.8
8	8.5	38.89	26.97
9	9.6	38.59	28.29
10	10.7	39.97	27.44
11	11.7	39.8	27.72
12	12.8	40.35	27.67
13	13.9	40.06	27.67
14	14.9	40.2	27.76
15	16	40.26	27.96
16	17	40.84	27.81
17	18.1	41.41	27.93
18	19.2	41.55	28.12
19	20.3	41.41	28.18
20	21.3	42.25	28.58

2nd Scan

Coil Current Reading on PS (Amps)	Coil Current Reading From Shunt Resistance (Amps)	X Projection Sigma (pixels)	Y Projection Sigma (pixels)
0	0	31.68	33.79
1	1	30.51	34.77
2	2.1	29.72	36.33
3	3.2	28.81	37.6
4	4.3	28.11	38.3
5	5.4	27.99	38.45
6	6.4	27.78	39.73
7	7.4	28.07	39.32
8	8.5	27.91	39.31
9	9.6	28.38	41.08
10	10.7	28.62	41.87
11	11.7	28.66	41.52
12	12.8	28.91	42.31
13	13.9	28.8	42.16
14	14.9	28.95	42.41
15	16	29.06	42.58
16	17	29.27	42.95
17	18.1	29.15	43.26
18	19.2	29.01	42.98
19	20.3	29.05	42.61
20	21.3	28.93	43.63
22	23.5	29.08	44.39

Multi-Fits for Beam Spot Projection

Multi-Fits 1st Scan

Naming the photos: Rotated_HRRL_Emit_Test_Quad_Scan_First_20Amp_Multi_Fit7 -> RHETQSF20AMF7

Coild Current (Amps)	1st Fit	2nd Fit	3rd Fit	4th Fit	5th Fit	6th Fit	7th Fit	Mean and Error
0	[math]{\sigma}_{x,y}= 21.4,22.5 [/math]	[math]{\sigma}_{x,y}= 25.5,23.2 [/math]	[math]{\sigma}_{x,y}= 22.9,24.5 [/math]	[math]{\sigma}_{x,y}= 28.2,25.0 [/math]	[math]{\sigma}_{x,y}= 26.1,25.7 [/math]	[math]{\sigma}_{x,y}= 26.6,28.2 [/math]	[math]{\sigma}_{x,y}= 26.3,27.3 [/math]	[math]{\sigma}_{x,y}= 25.3\pm2.2 , 25.2 \pm1.9 [/math]
1	[math]{\sigma}_{x,y}= 22.3,21.5 [/math]	[math]{\sigma}_{x,y}= 25.1,21.8 [/math]	[math]{\sigma}_{x,y}= 22.9,24.5 [/math]	[math]{\sigma}_{x,y}= 25.5,22.6 [/math]	[math]{\sigma}_{x,y}= 28.7,28.2 [/math]	[math]{\sigma}_{x,y}= 33.3,26.3 [/math]	[math]{\sigma}_{x,y}= 35.0,28.4 [/math]	[math]{\sigma}_{x,y}= 27.5\pm4.6 , 24.8 \pm 2.7 [/math]
2	[math]{\sigma}_{x,y}= 21.2,20.2 [/math]	[math]{\sigma}_{x,y}= 22.8,20.7 [/math]	[math]{\sigma}_{x,y}= 27.9,21.2 [/math]	[math]{\sigma}_{x,y}= 35.0,27.1 [/math]	[math]{\sigma}_{x,y}= 36.6,26.6 [/math]	[math]{\sigma}_{x,y}= 36.6,26.0 [/math]	[math]{\sigma}_{x,y}= 40.0,24.6 [/math]	[math]{\sigma}_{x,y}= 31.4 \pm 6.9 , 23.8 \pm 2.8 [/math]
3	[math]{\sigma}_{x,y}= 28.7,19.6 [/math]	[math]{\sigma}_{x,y}= 30.4,20.1 [/math]	[math]{\sigma}_{x,y}= 29.8,20.0 [/math]	[math]{\sigma}_{x,y}= 38.8,13.3 [/math]	[math]{\sigma}_{x,y}= 39.5,26.8 [/math]	[math]{\sigma}_{x,y}= 38.8,27.4 [/math]	[math]{\sigma}_{x,y}= 41.0,25.6 [/math]	[math]{\sigma}_{x,y}= 35.3 \pm 5.0 , 21.8 \pm 4.7 [/math]
4	[math]{\sigma}_{x,y}= 28.2,17.6 [/math]	[math]{\sigma}_{x,y}= 30.2,17.6 [/math]	[math]{\sigma}_{x,y}= 29.9,18.1 [/math]	[math]{\sigma}_{x,y}= 37.1,29.8 [/math]	[math]{\sigma}_{x,y}= 39.4,18.5 [/math]	[math]{\sigma}_{x,y}= 40.3,19.1 [/math]	[math]{\sigma}_{x,y}= 42.9,19.2 [/math]	[math]{\sigma}_{x,y}= 35.4 \pm 5.5 , 20.0 \pm 4.1 [/math]
5	[math]{\sigma}_{x,y}= 33.0,17.0 [/math]	[math]{\sigma}_{x,y}= 33.5,16.6 [/math]	[math]{\sigma}_{x,y}= 31.1,17.2 [/math]	[math]{\sigma}_{x,y}= 37.0,17.9 [/math]	[math]{\sigma}_{x,y}= 40.0,18.2 [/math]	[math]{\sigma}_{x,y}= 41.8,18.5 [/math]	[math]{\sigma}_{x,y}= 42.5,18.0 [/math]	[math]{\sigma}_{x,y}= 37.0 \pm 4.2 , 17.6 \pm 0.6 [/math]
6	[math]{\sigma}_{x,y}= 31.8,15.4 [/math]	[math]{\sigma}_{x,y}= 34.7,15.6 [/math]	[math]{\sigma}_{x,y}= 33.4,16.3 [/math]	[math]{\sigma}_{x,y}= 41.0,16.4 [/math]	[math]{\sigma}_{x,y}= 41.4,16.2 [/math]	[math]{\sigma}_{x,y}= 43.2,16.8 [/math]	[math]{\sigma}_{x,y}= 44.8,16.2 [/math]	[math]{\sigma}_{x,y}= 38.6 \pm 4.8 , 16.1 \pm 0.4 [/math]
7	[math]{\sigma}_{x,y}= 36.4,14.8 [/math]	[math]{\sigma}_{x,y}= 39.3,15.3 [/math]	[math]{\sigma}_{x,y}= 40.7,14.9 [/math]	[math]{\sigma}_{x,y}= 43.6,15.1 [/math]	[math]{\sigma}_{x,y}= 43.0,15.5 [/math]	[math]{\sigma}_{x,y}= 44.2,15.9 [/math]	[math]{\sigma}_{x,y}= 49.3,16.1 [/math]	[math]{\sigma}_{x,y}= 42.4 \pm 3.8 , 15.4 \pm 0.5 [/math]
8	[math]{\sigma}_{x,y}= 37.9,14.7 [/math]	[math]{\sigma}_{x,y}= 43.0,14.7 [/math]	[math]{\sigma}_{x,y}= 40.0,14.0 [/math]	[math]{\sigma}_{x,y}= 44.0,15.7 [/math]	[math]{\sigma}_{x,y}= 44.0,15.5 [/math]	[math]{\sigma}_{x,y}= 43.9,15.3 [/math]	[math]{\sigma}_{x,y}= 44.1,15.0 [/math]	[math]{\sigma}_{x,y}= 42.4 \pm 2.3 , 15.0 \pm 0.5 [/math]
9	[math]{\sigma}_{x,y}= 41.0,12.5 [/math]	[math]{\sigma}_{x,y}= 39.0,13.0 [/math]	[math]{\sigma}_{x,y}= 41.0,12.9 [/math]	[math]{\sigma}_{x,y}= 43.0,14.4 [/math]	[math]{\sigma}_{x,y}= 45.0,14.7 [/math]	[math]{\sigma}_{x,y}= 44.9,15.0 [/math]	[math]{\sigma}_{x,y}= 46.0,16.0 [/math]	[math]{\sigma}_{x,y}= 42.8 \pm 2.4 , 14.1 \pm 1.2 [/math]
10	[math]{\sigma}_{x,y}= 41.0,14.0 [/math]	[math]{\sigma}_{x,y}= 45.7,13.7 [/math]	[math]{\sigma}_{x,y}= 41.5,14.2 [/math]	[math]{\sigma}_{x,y}= 46.2,13.6 [/math]	[math]{\sigma}_{x,y}= 46.0,14.7 [/math]	[math]{\sigma}_{x,y}= 45.0,14.2 [/math]	[math]{\sigma}_{x,y}= 46.9,14.1 [/math]	[math]{\sigma}_{x,y}= 44.6 \pm 2.2 , 14.4 \pm 0.3 [/math]
11	[math]{\sigma}_{x,y}= 45.4,13.2 [/math]	[math]{\sigma}_{x,y}= 47.2,12.0 [/math]	[math]{\sigma}_{x,y}= 46.4,12.3 [/math]	[math]{\sigma}_{x,y}= 46.4,13.2 [/math]	[math]{\sigma}_{x,y}= 46.7,14.5 [/math]	[math]{\sigma}_{x,y}= 47.0,14.3 [/math]	[math]{\sigma}_{x,y}= 47.2,14.6 [/math]	[math]{\sigma}_{x,y}= 46.6 \pm 0.6 , 13.5 \pm 1.0 [/math]
12	[math]{\sigma}_{x,y}= 44.5,11.6 [/math]	[math]{\sigma}_{x,y}= 44.9,12.0 [/math]	[math]{\sigma}_{x,y}= 46.4,11.9 [/math]	[math]{\sigma}_{x,y}= 48.6,11.8 [/math]	[math]{\sigma}_{x,y}= 49.1,11.6 [/math]	[math]{\sigma}_{x,y}= 47.8,13.0 [/math]	[math]{\sigma}_{x,y}= 50.0,12.7 [/math]	[math]{\sigma}_{x,y}= 47.3 \pm 2.0 , 12.1 \pm 0.5 [/math]
13	[math]{\sigma}_{x,y}= 48.0,11.8 [/math]	[math]{\sigma}_{x,y}= 50.0,11.6 [/math]	[math]{\sigma}_{x,y}= 50.2,12.1 [/math]	[math]{\sigma}_{x,y}= 49.2,11.0 [/math]	[math]{\sigma}_{x,y}= 50.6,11.2 [/math]	[math]{\sigma}_{x,y}= 50.0,12.0 [/math]	[math]{\sigma}_{x,y}= 51.6,11.6 [/math]	[math]{\sigma}_{x,y}= 49.9 \pm 1.0 , 11.6 \pm 0.4 [/math]
14	[math]{\sigma}_{x,y}= 51.0,11.8 [/math]	[math]{\sigma}_{x,y}= 51.0,11.2 [/math]	[math]{\sigma}_{x,y}= 51.3,11.2 [/math]	[math]{\sigma}_{x,y}= 50.7,11.6 [/math]	[math]{\sigma}_{x,y}= 52.2,11.7 [/math]	[math]{\sigma}_{x,y}= 51.8,11.6 [/math]	[math]{\sigma}_{x,y}= 52.7,11.4 [/math]	[math]{\sigma}_{x,y}= 51.5 \pm 0.7 , 11.5 \pm 0.2 [/math]
15	[math]{\sigma}_{x,y}= 52.3,12.0 [/math]	[math]{\sigma}_{x,y}= 50.0,11.6 [/math]	[math]{\sigma}_{x,y}= 51.0,11.3 [/math]	[math]{\sigma}_{x,y}= 50.6,11.6 [/math]	[math]{\sigma}_{x,y}= 53.0,11.5 [/math]	[math]{\sigma}_{x,y}= 53.2,11.9 [/math]	[math]{\sigma}_{x,y}= 52.7,12.0 [/math]	[math]{\sigma}_{x,y}= 51.8 \pm 1.2 , 11.7 \pm 0.3 [/math]
16	[math]{\sigma}_{x,y}= 53.4,12.5 [/math]	[math]{\sigma}_{x,y}= 54.0,12.0 [/math]	[math]{\sigma}_{x,y}= 54.3,11.8 [/math]	[math]{\sigma}_{x,y}= 51.5,11.5 [/math]	[math]{\sigma}_{x,y}= 53.3,12.7 [/math]	[math]{\sigma}_{x,y}= 52.1,12.8 [/math]	[math]{\sigma}_{x,y}= 55.0,12.5 [/math]	[math]{\sigma}_{x,y}= 53.4 \pm 1.1 , 12.3 \pm 0.5 [/math]
17	[math]{\sigma}_{x,y}= 55.0,13.4 [/math]	[math]{\sigma}_{x,y}= 56.5,13.8 [/math]	[math]{\sigma}_{x,y}= 56.8,13.9 [/math]	[math]{\sigma}_{x,y}= 55.9,12.9 [/math]	[math]{\sigma}_{x,y}= 56.0,11.9 [/math]	[math]{\sigma}_{x,y}= 57.4,12.2 [/math]	[math]{\sigma}_{x,y}= 58.0,13.3 [/math]	[math]{\sigma}_{x,y}= 56.5 \pm 0.9 , 13.1 \pm 0.7 [/math]
18	[math]{\sigma}_{x,y}= 59.0,14.0 [/math]	[math]{\sigma}_{x,y}= 56.0,13.2 [/math]	[math]{\sigma}_{x,y}= 56.0,13.4 [/math]	[math]{\sigma}_{x,y}= 55.0,13.9 [/math]	[math]{\sigma}_{x,y}= 56.1,14.1 [/math]	[math]{\sigma}_{x,y}= 53.5,13.4 [/math]	[math]{\sigma}_{x,y}= 59.0,13.0 [/math]	[math]{\sigma}_{x,y}= 56.4 \pm 1.9 , 13.6 \pm 0.4 [/math]
19	[math]{\sigma}_{x,y}= 60.7,16.1 [/math]	[math]{\sigma}_{x,y}= 53.0,15.0 [/math]	[math]{\sigma}_{x,y}= 60.0,15.7 [/math]	[math]{\sigma}_{x,y}= 55.7,14.1 [/math]	[math]{\sigma}_{x,y}= 58.8,15.0 [/math]	[math]{\sigma}_{x,y}= 53.0,14.0 [/math]	[math]{\sigma}_{x,y}= 54.5,15.0 [/math]	[math]{\sigma}_{x,y}= 56.5 \pm 3.0 , 15.0 \pm 0.7 [/math]
20	[math]{\sigma}_{x,y}= 62.0,18.9 [/math]	[math]{\sigma}_{x,y}= 60.0,19.0 [/math]	[math]{\sigma}_{x,y}= 59.0,19.3 [/math]	[math]{\sigma}_{x,y}= 56.0,17.0 [/math]	[math]{\sigma}_{x,y}= 55.6,17.2 [/math]	[math]{\sigma}_{x,y}= 56.0,17.0 [/math]	[math]{\sigma}_{x,y}= 55.7,17.3 [/math]	[math]{\sigma}_{x,y}= 57.8 \pm 2.4 , 18.0 \pm 1.0 [/math]

Multi-Fits 2nd Scan

Naming the photos: Rotated_HRRL_Emit_Test_Quad_Scan_Second_22Amp_Multi_Fit7 -> RHETQSS22AMF7

Coild Current (Amps)	1st Fit	2nd Fit	3rd Fit	4th Fit	5th Fit	6th Fit	7th Fit	Mean and Error
0	[math]{\sigma}_{x,y}=27.8,29.2 [/math]	[math]{\sigma}_{x,y}=28.4,30.5 [/math]	[math]{\sigma}_{x,y}=28.5,32.1 [/math]	[math]{\sigma}_{x,y}=29.8,37.4 [/math]	[math]{\sigma}_{x,y}=33.7,38.5 [/math]	[math]{\sigma}_{x,y}=33.4,38.6 [/math]	[math]{\sigma}_{x,y}=36.3,38.0 [/math]	[math]{\sigma}_{x,y}=31.1 \pm 3.1,34.9\pm3.8 [/math]
1	[math]{\sigma}_{x,y}=22.6,28.5 [/math]	[math]{\sigma}_{x,y}=24.5,30.7 [/math]	[math]{\sigma}_{x,y}=23.1,29.9 [/math]	[math]{\sigma}_{x,y}=27.4,41.0 [/math]	[math]{\sigma}_{x,y}=27.1,38.6 [/math]	[math]{\sigma}_{x,y}=26.5,39.8 [/math]	[math]{\sigma}_{x,y}=27.0,39.7 [/math]	[math]{\sigma}_{x,y}= 25.5 \pm 1.9 , 35.5 \pm 5.1 [/math]
2	[math]{\sigma}_{x,y}=23.5,32.9 [/math]	[math]{\sigma}_{x,y}=24.2,33.5 [/math]	[math]{\sigma}_{x,y}=22.7,34.7 [/math]	[math]{\sigma}_{x,y}=26.3,41.0 [/math]	[math]{\sigma}_{x,y}=26.8,45.8 [/math]	[math]{\sigma}_{x,y}=26.1,44.1 [/math]	[math]{\sigma}_{x,y}=26.5,48.9 [/math]	[math]{\sigma}_{x,y}= 25.2 \pm 1.5 , 40.1 \pm 6.0 [/math]
3	[math]{\sigma}_{x,y}=22.6,37.0 [/math]	[math]{\sigma}_{x,y}=22.9,36.4 [/math]	[math]{\sigma}_{x,y}=21.7,37.6 [/math]	[math]{\sigma}_{x,y}=23.8,39.5 [/math]	[math]{\sigma}_{x,y}=23.8,42.3 [/math]	[math]{\sigma}_{x,y}=24.1,42.3 [/math]	[math]{\sigma}_{x,y}=25.4,42.3 [/math]	[math]{\sigma}_{x,y}=23.5 \pm 1.1 , 39.6\pm 2.5 [/math]
4	[math]{\sigma}_{x,y}=19.9,38.8 [/math]	[math]{\sigma}_{x,y}=19.8,38.1 [/math]	[math]{\sigma}_{x,y}=19.1,39.3 [/math]	[math]{\sigma}_{x,y}=20.5,41.3 [/math]	[math]{\sigma}_{x,y}=20.8,42.2 [/math]	[math]{\sigma}_{x,y}=21.1,43.3 [/math]	[math]{\sigma}_{x,y}=21.4,44.4 [/math]	[math]{\sigma}_{x,y}= 20.4\pm 0.8 , 41.1\pm 2.2 [/math]
5	[math]{\sigma}_{x,y}=18.2,43.3 [/math]	[math]{\sigma}_{x,y}=18.0,45.9 [/math]	[math]{\sigma}_{x,y}=17.9,48.1 [/math]	[math]{\sigma}_{x,y}=18.2,47.9 [/math]	[math]{\sigma}_{x,y}=18.2,40.9 [/math]	[math]{\sigma}_{x,y}=17.0,48.5 [/math]	[math]{\sigma}_{x,y}=18.4,48.0 [/math]	[math]{\sigma}_{x,y}= 18.0 \pm 0.4 , 46.1 \pm 2.7 [/math]
6	[math]{\sigma}_{x,y}=16.4,47.0 [/math]	[math]{\sigma}_{x,y}=16.2,48.4 [/math]	[math]{\sigma}_{x,y}=15.7,50.6 [/math]	[math]{\sigma}_{x,y}=16.6,50.5 [/math]	[math]{\sigma}_{x,y}=16.5,46.9 [/math]	[math]{\sigma}_{x,y}=16.0,50.6 [/math]	[math]{\sigma}_{x,y}=16.3,50.6 [/math]	[math]{\sigma}_{x,y}= 16.2 \pm 0.3 , 49.2\pm 1.6 [/math]
7	[math]{\sigma}_{x,y}=13.1,44.5 [/math]	[math]{\sigma}_{x,y}=12.8,46.8 [/math]	[math]{\sigma}_{x,y}=13.2,49.3 [/math]	[math]{\sigma}_{x,y}=13.3,49.6 [/math]	[math]{\sigma}_{x,y}=13.7,45.2 [/math]	[math]{\sigma}_{x,y}=13.9,49.7 [/math]	[math]{\sigma}_{x,y}=12.8,49.6 [/math]	[math]{\sigma}_{x,y}= 13.3\pm0.4 , 47.8\pm 2.1 [/math]
8	[math]{\sigma}_{x,y}=12.3,49.0 [/math]	[math]{\sigma}_{x,y}=12.4,46.7 [/math]	[math]{\sigma}_{x,y}=11.7,49.3 [/math]	[math]{\sigma}_{x,y}=11.6,49.2 [/math]	[math]{\sigma}_{x,y}=13.1,45.0 [/math]	[math]{\sigma}_{x,y}=13.0,50.0 [/math]	[math]{\sigma}_{x,y}=12.3,49.6 [/math]	[math]{\sigma}_{x,y}= 12.3\pm 0.5 , 48.4 \pm 1.7 [/math]
9	[math]{\sigma}_{x,y}=13.3,50.4 [/math]	[math]{\sigma}_{x,y}=13.2,52.4 [/math]	[math]{\sigma}_{x,y}=13.1,55.7 [/math]	[math]{\sigma}_{x,y}=13.5,54.8 [/math]	[math]{\sigma}_{x,y}=13.3,50.0 [/math]	[math]{\sigma}_{x,y}=13.5,53.4 [/math]	[math]{\sigma}_{x,y}=13.2,54.5 [/math]	[math]{\sigma}_{x,y}= 13.3 \pm 0.1 ,53.0 \pm 2.0 [/math]
10	[math]{\sigma}_{x,y}=12.6,53.8 [/math]	[math]{\sigma}_{x,y}=13.0,54.9 [/math]	[math]{\sigma}_{x,y}=12.4,58.5 [/math]	[math]{\sigma}_{x,y}=12.9,57.7 [/math]	[math]{\sigma}_{x,y}=12.7,52.7 [/math]	[math]{\sigma}_{x,y}=12.8,55.0 [/math]	[math]{\sigma}_{x,y}=12.0,56.6 [/math]	[math]{\sigma}_{x,y}= 12.6 \pm 0.3 , 55.6 \pm 1.9 [/math]
11	[math]{\sigma}_{x,y}=10.1,52.2 [/math]	[math]{\sigma}_{x,y}=10.2,52.4 [/math]	[math]{\sigma}_{x,y}=9.8,58.3 [/math]	[math]{\sigma}_{x,y}=9.7,56.9 [/math]	[math]{\sigma}_{x,y}=10.9,51.3 [/math]	[math]{\sigma}_{x,y}=10.0,54.5 [/math]	[math]{\sigma}_{x,y}=9.7,55.4 [/math]	[math]{\sigma}_{x,y}= 10.1 \pm 0.4 ,54.4 \pm 2.4 [/math]
12	[math]{\sigma}_{x,y}=11.2,55.9 [/math]	[math]{\sigma}_{x,y}=10.8,55.0 [/math]	[math]{\sigma}_{x,y}=11.0,60.1 [/math]	[math]{\sigma}_{x,y}=10.5,59.8 [/math]	[math]{\sigma}_{x,y}=11.1,56.2 [/math]	[math]{\sigma}_{x,y}=10.4,56.4 [/math]	[math]{\sigma}_{x,y}=10.4,58.0 [/math]	[math]{\sigma}_{x,y}= 10.8\pm 0.9, 57.3\pm1.8 [/math]
13	[math]{\sigma}_{x,y}=8.9,54.8 [/math]	[math]{\sigma}_{x,y}=9.1,56.2 [/math]	[math]{\sigma}_{x,y}=8.7,60.4 [/math]	[math]{\sigma}_{x,y}=8.7,59.6 [/math]	[math]{\sigma}_{x,y}=9.1,53.5 [/math]	[math]{\sigma}_{x,y}=8.8,56.3 [/math]	[math]{\sigma}_{x,y}=8.7,57.1 [/math]	[math]{\sigma}_{x,y}= 8.9 \pm 0.2 , 56.8 \pm 2.3 [/math]
14	[math]{\sigma}_{x,y}=10.5,57.4 [/math]	[math]{\sigma}_{x,y}=10.9,56.0 [/math]	File:RHETQSS14AMF3.png [math]{\sigma}_{x,y}=10.7,61.0 [/math]	[math]{\sigma}_{x,y}=10.7,60.7 [/math]	[math]{\sigma}_{x,y}=10.6,54.9 [/math]	[math]{\sigma}_{x,y}=10.9,60.6 [/math]	[math]{\sigma}_{x,y}=10.3,58.7 [/math]	[math]{\sigma}_{x,y}= 10.7 \pm 0.2 , 58.5\pm 2.3 [/math]
15	[math]{\sigma}_{x,y}=10.0,57.4 [/math]	[math]{\sigma}_{x,y}=10.8,55.2 [/math]	[math]{\sigma}_{x,y}=10.7,60.3 [/math]	[math]{\sigma}_{x,y}=10.5,58.7 [/math]	[math]{\sigma}_{x,y}=10.7,55.0 [/math]	[math]{\sigma}_{x,y}=10.3,57.4 [/math]	[math]{\sigma}_{x,y}=10.2,58.7 [/math]	[math]{\sigma}_{x,y}= 10.5 \pm 0.3 , 57.5 \pm 1.8 [/math]
16	[math]{\sigma}_{x,y}=10.5,60.6 [/math]	[math]{\sigma}_{x,y}=11.2,55.8 [/math]	[math]{\sigma}_{x,y}=11.5,63.4 [/math]	[math]{\sigma}_{x,y}=11.1,62.3 [/math]	[math]{\sigma}_{x,y}=12.0,56.8 [/math]	[math]{\sigma}_{x,y}=11.4,60.6 [/math]	[math]{\sigma}_{x,y}=11.3,56.0 [/math]	[math]{\sigma}_{x,y}= 11.3 \pm 0.4 , 59.4 \pm 2.9 [/math]
17	[math]{\sigma}_{x,y}= 12.2,57.6[/math]	[math]{\sigma}_{x,y}=13.6,58.7 [/math]	[math]{\sigma}_{x,y}=12.1,63.4 [/math]	[math]{\sigma}_{x,y}=11.2,57.4 [/math]	[math]{\sigma}_{x,y}=12.0,57.6 [/math]	[math]{\sigma}_{x,y}=12.0,58.7 [/math]	[math]{\sigma}_{x,y}=12.4,64.0 [/math]	[math]{\sigma}_{x,y}= 12.2 \pm 0.7 , 59.6 \pm 2.6 [/math]
18	[math]{\sigma}_{x,y}=11.4,59.7 [/math]	[math]{\sigma}_{x,y}=12.0,55.7 [/math]	[math]{\sigma}_{x,y}=13.2,61.4 [/math]	[math]{\sigma}_{x,y}=13.3,59.0 [/math]	[math]{\sigma}_{x,y}=12.5,58.5 [/math]	[math]{\sigma}_{x,y}=12.3,60.2 [/math]	[math]{\sigma}_{x,y}=11.9,58.3 [/math]	[math]{\sigma}_{x,y}= 12.4 \pm 0.6 , 59.0\pm 1.7 [/math]
19	[math]{\sigma}_{x,y}=19.99,59.0 [/math]	[math]{\sigma}_{x,y}=13.5,52.6 [/math]	[math]{\sigma}_{x,y}= 11.5,62.0[/math]	[math]{\sigma}_{x,y}=12.2 [/math]	[math]{\sigma}_{x,y}=11.86,59.6 [/math]	[math]{\sigma}_{x,y}=12.2,57.4 [/math]	[math]{\sigma}_{x,y}=13.0,55.0 [/math]	[math]{\sigma}_{x,y}= 13.5 \pm 2.7 , 57.8 \pm 2.9 [/math]
20	[math]{\sigma}_{x,y}= 13.29,60.73[/math]	[math]{\sigma}_{x,y}= 17.2,56.5[/math]	[math]{\sigma}_{x,y}= 13.3,59.8[/math]	[math]{\sigma}_{x,y}= 12.0,60.68[/math]	[math]{\sigma}_{x,y}= 12.84,63.72[/math]	[math]{\sigma}_{x,y}=14.0,59.3 [/math]	[math]{\sigma}_{x,y}= 13.7,61.2[/math]	[math]{\sigma}_{x,y}= 13.8 \pm 1.5 , 60.3 \pm 2.0 [/math]
22	[math]{\sigma}_{x,y}=19,61[/math]	[math]{\sigma}_{x,y}=12,55.7[/math]	[math]{\sigma}_{x,y}=13.2,61.7[/math]	[math]{\sigma}_{x,y}=13.3,59.0[/math]	[math]{\sigma}_{x,y}=11.3,62.4[/math]	[math]{\sigma}_{x,y}=12.3,60.2[/math]	[math]{\sigma}_{x,y}=11.9,58.3[/math]	[math]{\sigma}_{x,y}= 13.3 \pm 2.4 , 59.8 \pm 2.1 [/math]

Current -vs- Sigma Data Files

First Scan:

File:HRRL Emitt Current vs Sigma 1stScan x.txt

File:HRRL Emitt Current vs Sigma 1stScan y.txt

Second Scan:

File:HRRL Emitt Current vs Sigma 2ndScan x.txt

File:HRRL Emitt Current vs Sigma 2ndScan y.txt

Current vs Sigma:

X projections:

File:HRRL Emitt Current vs Sigma Projections x.txt

Fit:

P-value of this fit is less than 0.000001.

Y projections:

File:HRRL Emitt Current vs Sigma Projections y.txt

Fit:

P-value of this fit is less than 0.000001.

Next step: 
calibrate: mm/pxiel 
Parabola: All the x projections makeup one parabola, and y projections makeup one parabola. Put them in together.

Converting pixels to length

Diameter of circle inside crystal is 10 mm.

Convert the pictures to a histogram and measure pixels that way

Figures	Horizontal Diameter (pixels)	Vertical Diameter (pixels)
	91	59
	90	59
	90	59
	90	60
	90	59
	91	60
	90	60
	Average Horizontal Diameter (pixels)	Average Vertical Diameter (pixels)
	[math] 90.29 \pm 0.49 [/math]	[math] 59.43 \pm 0.53 [/math]
	Horizontal mm/pixels	Vertical mm/pixels
	[math] 0.1108 \pm 0.0006 [/math]	[math] 0.1683 \pm 0.0015 [/math]

Sigma (in mm) vs Dipole Coil Current (on Shunt Resistance)

X Projection

The fit is:

[math] \sigma_x = (2.9098 \pm 0.0073) + (0.21775 \pm 0.00062)I + (0.006696 \pm 0.000047)I^{2} [/math]

p-value is less than 0.000001

For a parabola [math] y = a + bx + cx^{2} [/math], the [math] x [/math] corresponding to the minimum [math] y_{min} [/math] is:

[math] x_{min} = - \frac{b}{2c} [/math].

So, [math] I_{min} = -16.259707[/math]

and

[math] \Delta I_{min} = \sqrt{ \frac{(\Delta b)^2}{4c^2} + \frac{ b^2 (\Delta c)^2}{4c^4} } [/math].

[math] \Delta I_{x,min} = \sqrt{ \frac{(0.00062)^2}{4 \times (0.006696)^2} + \frac{ (0.21775)^2 \times (0.000047)^2}{4 \times (0.006696)^4} } = 0.123161 = 0.12[/math].

[math] I_{min} = -16.26 \pm 0.12 [/math].

[math] \sigma_{x,min} = 2.93671995816274 [/math]

[math] \sigma_{x,min, up} = 2.94409703114234 [/math]

[math] \sigma_{x,min, dn} = 2.92934288518313 [/math]

[math] \sigma_{x,min,up} - \sigma_{x,min} = 0.007377072979601 [/math]

[math] \sigma_{x,min} - \sigma_{x,min,dn}= 0.00737707297960144 [/math]

So,

[math] \sigma_{x,min} = 2.9367 \pm 0.0074 [/math]

Y Projection

The fit is:

[math] \sigma_{y} = (4.691 \pm 0.018) + (-0.3729 \pm 0.0013)I + (0.01289 \pm 0.00008)I^{2} [/math]

p-value is less than 0.000001

[math] x_{min} = - \frac{b}{2c} [/math].

So, [math] I_{min} = 14.464701[/math]

[math] \Delta I_{min} = \sqrt{ \frac{(\Delta b)^2}{4c^2} + \frac{ b^2 (\Delta c)^2}{4c^4} } [/math].

[math] \Delta I_{min} = \sqrt{ \frac{(0.0013)^2}{4 \times (0.01289) ^2} + \frac{ (-0.3729)^2 \times (0.00008)^2}{4 \times (0.01289)^4} } = 0.102966 [/math]

[math] I_{min} = 14.46 \pm 0.10 [/math].

[math] \sigma_{y,min} = 4.65274050365148 [/math]

[math] \sigma_{y,min, up} = 4.67087520809017 [/math]

[math] \sigma_{y,min, dn} = 4.63487351175864 [/math]

[math] \sigma_{y,min, up} - \sigma_{y,min} = 0.0181347044386939 [/math]

[math] \sigma_{y,min} - \sigma_{y,min, dn}= 0.0178669918928378 [/math]

So,

[math] \sigma_{y,min} = 4.653 \pm 0.018 [/math]

Conclusions

For X projection: [math] I_{x,min} = -16.26 \pm 0.12[/math], and [math] \sigma_{x,min} = 2.9367 \pm 0.0074 [/math]

For X projection: [math] I_{y,min} = 14.46 \pm 0.10 [/math], and [math] \sigma_{y,min} = 4.653 \pm 0.018 [/math]

Emittance

Quadrupole Strength: [math] k = 0.2998 \frac{g[T/m]}{p[GeV]} [/math]

g here is the gradient of the qudrupole. p is the momentum of the e- beam.

In our case, energy of the beam is 0.01 GeV and gap between dipole center and pole face is 1 inch.

[math] k = 0.2998 \frac{\frac{B(I)}{(1 inch)\times\frac{1 m}{1 inch}}[T/m]}{0.01 GeV} [/math]

[math]B(I) = (3.6 \pm 1.3) \times 10^{-3} + (1945 \pm 2) \times 10^{-5} I [kG] = (3.6 \pm 1.3) \times 10^{-4} + (1945 \pm 2) \times 10^{-6} I [T] [/math]

So,

[math] k = 0.2998 \frac{ (3.6 \pm 1.3) \times 10^{-4} + (1945 \pm 2) \times 10^{-6} I [T] }{0.01 \times 0.0254 [GeV][m]} [/math]

[math] k = (0.42 \pm 0.16) + (2.2956 \pm 0.0024) \times I[/math]

According to File:Emit Meas DAFNE Linac e Beam.pdf, for the case of quadruple scanning one quadrupole and screen, the emittance related to the quadrupole strength.

Transfer Matrix for 1uaruople \tilde{Q} and screen (drift space?) \tilde{S} consist the transfer matrix of the whole system as:

[math] \tilde{M} = \tilde{S}\tilde{Q} [/math]

[math] \tilde{Q} = \begin{bmatrix} 1 & 0 \\ k=\pm\frac{1}{f} & 1 \end{bmatrix}. [/math]

[math] \tilde{S} = \begin{bmatrix} S_{11} & S_{12} \\ S_{21} & S_{22} \end{bmatrix}. [/math]

[math] \sigma_{screen11} = Ak^2 - 2ABk + (C + AB^2) [/math]

[math] \epsilon = \frac{\sqrt{AC}}{S^2_{12}} [/math]

[math] S_{12} [/math] is the distance between quadrupole and screen.

If I have a fit:

[math] \sigma_{screen11} = xk^2 + yk + z [/math]

Then, [math] A=x, B=\frac{y}{-2A}, C=z-AB^2[/math]

Two fits are:

Dr. Kim's Suggestion in his email

from	Yujong Kim <yjkim@isu.edu>
to	Sadiq Setiniyaz <sadik82@gmail.com>
cc	Dr Forest <tforest@athena.physics.isu.edu>
date	Fri, Oct 8, 2010 at 6:44 AM
subject	Re: Quad Scan Parameters

Dear Sadiq,

Many thanks,

I re-checked your QM scan fitting plot (final plots on emittance mewasurement).

Your fitting range too wide (-20 < k < 50).

Please fit again with a narrower range (+10 < k < 50).
Then, you can get much beautiful results.
Normally, we fit for only one k sign (positive or negative).
Yujong Kim

Here are results after doing Dr. Kim's suggested fit.

Note: I also corrected the distance S12 to the distance between quad center and screen (I originally did from the end of the quad to the screen, this is wrong).

For X projection

Minimum ChiSqrd is: 129.749

The fit is:

sig^2 = (7.82463e-06+-2e-08) + (4.23945e-07+-1e-09)*k + (6.82167e-09+-1e-10)*k*k

Probability of obtaining data at least as incompatible with distribution as the data actually observed is 0.000000 A = 7.82463e-06, B = -0.0270904, C = 1.07925e-09, S12=0.64. Geometrical Emittance: 2.24354e-07 (m*rad), Normalized Emittance: 4.39048e-06 (m*rad).

Data file contains: Number-of-points B-Field-Strength Error-on-B-Field-Strength Sigma^2 Error-on-Sigma^2

File:HRRL Emit Parbola Fit Xaxis as BfieldStrength.txt

For Y projection

Minimum ChiSqrd is: 25.3174

The fit is:

sig^2 = (1.71755e-05+-8e-08) + (-8.38108e-07+-3e-09)*k + (1.3229e-08+-1e-10)*k*k

Probability of obtaining data at least as incompatible with distribution as the data actually observed is 0.000000 A = 1.71755e-05, B = 0.0243983, C = 3.00479e-09, S12=0.64. Geometrical Emittance: 5.54629e-07 (m*rad), Normalized Emittance: 1.08538e-05 (m*rad).

Data file contains: Number-of-points B-Field-Strength Error-on-B-Field-Strength Sigma Error-on-Sigma

File:HRRL Emit Parbola Fit Yaxis as BfieldStrength.txt

Data ROOTSYS Fit Script

File:ROOTSCRIPT for HRRL Emittace.C

Emittance Results for HRRL at 10 MeV and 102 micro Amp peak

Measurements of the beam spot size as a function of the quadrupole magnetic field used to calculate the beam emittance. The fits are [math]\sigma^2_x = (7.824\pm 0.02) + (0.424 \pm 0.001)k + (6.8 \pm 0.1 \times 10^{-9} )k^2[/math] and [math]\sigma^2_y =17.17 \pm 0.08) - (0.084 \pm 0.003)k + (13.2 \pm 0.1e-10 \times 10^{-9}) k^2[/math] with a[math] \chi^2/\nu[/math] of 8.1 and 1.7 respectively. The respective P-values for this fit are 0.05 and 10^{-20}.

File:Emmitance HRRL 1mA 10-14-10.xmgrace.txt

Automatic Script to plot RMS

ROOT

How does the fit look when using the script below.

Working on it.

You should do background subtraction instead of cutting out the spot.

 2 bugs remain till Spe 24th. 1. entries; 2. Sub->Draw();

Running Parabola_Fit_for_AutoRMS.C will run AutoRMS.C and do the fits.

File:AutoRMS.C

File:Parabola Fit for AutoRMS.C

This 2 programs and pictures for data can be downloaded at: [2]

MATLAB

These MATLAB scripts will do Gaussian fits for beam spots and get rms beam size on x and projections (by Gauss_Fit.m). These rms values will be passed on (to Emit_Parabola_Fit_kl_YProjection.m and Emit_Parabola_Fit_kl_YProjection.m) to fit parabola. Parabola fit will be plotted,and emittances, normalized emittances, beta functions, alpha functions will given at the end.

File:Guass Fit.txt

File:Emit Parabola Fit kl XProjection.txt

File:Emit Parabola Fit kl YProjection.txt

File:Gauss.txt

File:Gausschisq.txt

File:Gaussfit.txt

File:Plot Beam Spot.txt

File:Plot Gauss Fit.txt

File:ReadMe.txt

Alternative Analysis from the East Cost

Image Analysis

Emittance/TWISS parameter extraction

New Fits with K1L

No Fringe Field Effect

We want to do new fits with K1L. K1 is quadrupole strength and L is the thickness of the quadrupole. We will do it for the data we got previously. They are the following two data sets.

At this step, we will ignore fringe field effect, we want to consider it later.

need uncertainties in plots below

The uncertainties are difficult to believe because they are so small (< 0.1% !)

Redo you plots so they look like the ones below in terms of line boldness and quality.  Identify the fit parameters and their error as well as a measurement of the goodness of fit.

https://wiki.iac.isu.edu/index.php/File:Emittance_HRRL_1mA.png

Data files contains:

Number-of-points B-Field-Strength Error-on-B-Field-Strength Sigma^2 Error-on-Sigma^2

File:HRRL Emit Parbola Fit Xaxis as BfieldStrength.txt

File:HRRL Emit Parbola Fit Yaxis as BfieldStrength.txt

X-Projection:
Fit Parameters by MATLAB:
y = 0.00000784 + (0.00000288)*x + 0.00000031*x*x
Fit Parameters by Matrix Inversion:
y = (0.00000781+-0.00000027) + (0.00000282+-0.00000014)*x + (0.00000030+-0.00000002)*x*x
P-value = 7.6238e-21

Y-Projection:
Fit Parameters by MATLAB:
y = 0.00001766 + (-0.00000599)*x + 0.00000066*x*x
Fit Parameters by Matrix Inversion:
y = (0.00001717+-0.00000084) + (-0.00000558+-0.00000041)*x + (0.00000059+-0.00000005)*x*x
P-value =  0.020964

geometrical emittance:    [math]\epsilon_{x}  =   1.5   \pm  1.4  ~ mm*mrad[/math]     [math]\epsilon_{y} =  3.7 \pm 3.9  ~ mm*mrad [/math]
normalized emittance:     [math]\epsilon_{n,x}  =  29    \pm   27 ~ mm*mrad[/math]     [math]\epsilon_{n,y} = 72.0 \pm  77.0 ~ mm*mrad [/math]
                          [math]\beta_{x} = 0.49  \pm 0.23  [/math]                    [math]\beta_{y} = 0.39 \pm 0.16  [/math]
                          [math]\alpha_{x} = -1.53   \pm  0.72 [/math]                 [math]\alpha_{y} = 2.5  \pm 1.0 [/math] 

X Projection:

Y Projection:

Corrections for vertical calibration

We made error when we are doing vertical calibration. When we calculate vertical calibration, we need to multiply the diameter of the circle by [math] cos45^{o} [/math].

This is the reason that we got dramatically different results in vertical and horizontal emittances.

Errors in Converting pixels to length for Vertical

	Average Horizontal Diameter (pixels)	Average Vertical Diameter (pixels)
	[math] 90.29 \pm 0.49 [/math]	[math] 59.43 \pm 0.53 [/math]
	Horizontal mm/pixels	Vertical mm/pixels
	[math] 0.1108 \pm 0.0006 [/math]	[math] 0.1683 \pm 0.0015 [/math]
		[math] V_{cal} [/math] need to be multiplied by [math] cos45^{o} [/math]
		[math] 0.1190\pm 0.0011 [/math]

New Fits with K1L

Considering fringe field effect with "Cheap and Dirty" way

x-projection:

emit=2.2 +- 1.3 mm*mrad, emit_norm=42.4 +- 25.4 mm*mrad

beta=0.72 +- 0.31, alpha=-1.23 +-0.56

//K1*L(1/m)   er K1*L    sgima^2(mm)   er sigma^2

Media:2010_Jul_Emit_refit_data_x.txt

y-projection:

emit=2.6 +- 2.0 mm*mrad, emit_norm=50.5 +- 38.3 mm*mrad

beta=0.54 +- 0.22, alpha=2.68 +-1.13

//K1*L(1/m)   er K1*L    sgima^2(mm)   er sigma^2

Media:2010_Jul_Emit_refit_data_y.txt

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