Stato dei lavori
description
Transcript of Stato dei lavori
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Stato dei lavori
Ottimizzazione dei wiggler di DANE
Simona Bettoni
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Outline Method to reduce the integrated octupole in the wiggler of DANE
Analysis tools at disposal:
→ Multipolar analysis: In (also vs x shift at the entrance)
→ Tracking: x (y) and x’ (y’) vs x (y) shift at the entrance (tools
Tosca+Matlab)
Shifted poles & cut poles models
Axis optimization
Analysis of the results:
→ Multipolar analysis
→ Tracking
→ Comparison with the experimental data at disposal
In the future
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Other methods to reduce the integrated octupole
CURVED POLE
MOVING MAGNETIC AXIS
New method
Reduction of the octupole around the beam trajectory in the region of the poles
Compensation of the integrated octupole in each semiperiod
Proposed by Pantaleo
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Multipolar expansion of the field with respect to the beam trajectory
1. Determination of the beam trajectory starting from the measured data
2. Fit of By between -3 cm and +3 cm by a 4º order polynomial in x
centered in xT(z) = xT
-1.5 -1 -0.5 0 0.5 1 1.5-0.05
-0.04
-0.03
-0.02
-0.01
0
0.01
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0.05
z (m)
x (m
)
443
32
210 TTTTTy xxbxxbxxbxxbbxxB
xT +3 cm
xT -3 cm
Beam trajectory (xT)
)(zbb nn
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The integrated multipoles in periodic magnets
44
33
2210 ''''' xbxbxbxbbxB TTTTT
yT
44
33
2210 xbxbxbxbbxB AAAAA
yA
Even
multipoles
→
Odd
multipoles
→
...34242
232322222121212 Tj
AjTj
AjTj
Ajj
Ajj
T xbcxbcxbcbcb
...33232
222221212222 Tj
AjTj
AjTj
Ajj
Ajj
T xbcxbcxbcbcb
In a displaced system of reference:
bAk → defined in the reference centered in OA (wiggler axis)
bTk → defined in the reference centered in OT
(beam trajectory)x’
y’
x
y
OA O T
xT
Left-right symmetry of the
magnet Multipoles change sign from a pole to
the next one Sum from a pole to the next one
,...2,1,0j
Txxx '
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Method to reduce the integrated octupole: displacement of the magnetic field
WITHOUT POLE MODIFICATION
In each semiperiod the particle trajectory is always on one side with respect the magnetic axis
In each semiperiod the particle travels on both sides with respect to the magnetic axis
Opportunely choosing the B axis is in principle possible to make zero the integrated octupole in each semiperiod
WITH THE POLE MODIFICATION
...366443 TA
TAT xbcxbcb Octupo
le
↑
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Optimization of the pole of the wiggler
FC1-like FC2-like
Goals Reduce as less as possible the magnetic
field in the gap
Maintain the left-right symmetry
-9
-7
-5
-3
-1
1
3
5
7
9
-112 -96 -80 -64 -48 -32 -16 0 16 32 48 64 80 96 112
z (cm)
x (c
m)
Positron trajectory
By (T)FC 2FC 1
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Analysis
For each z fit of By vs x in the system of reference perpendicular to the beam trajectory
-1.5 -1 -0.5 0 0.5 1 1.5-0.05
-0.04
-0.03
-0.02
-0.01
0
0.01
0.02
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0.04
0.05
z (m)
x (m
)
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-0.05
-0.04
-0.03
-0.02
-0.01
0
0.01
0.02
0.03
0.04
0.05
z (m)
x (m
)
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Cut poles model: analysis perpendicular to s
I3 calculated over the entire wiggler varies of more than a factor 2 if the
analysis is performed perpendicular to s and not to z!
Cut poles model
-200
-150
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-50
0
50
100
150
200
250
300
-0.16 -0.12 -0.08 -0.04 0 0.04 0.08 0.12 0.16
z (m)
b3 (T
/m̂3)
Perpendicular to z
Perpendicular to s (Matlab)
Perpendicular to s (Tosca)
IFC = 693 A
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Sector poles wiggler
Cut the poles in z to have sector poles
I3 calculated over the entire wiggler perpendicular to z is 9.09 T/m3 with respect to 4.13 T/m3 of the analysis perpendicular to z
-200
-150
-100
-50
0
50
100
150
200
250
300
-0.16 -0.12 -0.08 -0.04 0 0.04 0.08 0.12 0.16
z (m)
b3 (T
/m̂3)
Alfa = 0
Straight poles
Sector poles
IFC = 693 A
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Shifted poles solution
$ and field roll-off
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Shifted poles model
For the moment shifted the coils with the poles
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Cut-shifted poles: the comparison of the field (at the same current = 550 A in FC)
SHIFTED POLES
CUT POLES
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Cut-shifted poles: the comparison of the field (at the same current = 550 A in FC)
0.8
1
1.2
1.4
1.6
1.8
2
-0.07 -0.05 -0.03 -0.01 0.01 0.03 0.05 0.07x (m)
By
(T)
Cut poles (550 A)Cut poles (693 A)Shifted poles (550 A)
With the shifted poles solution, the field roll-off is improved, therefore the shims can be eliminated maintaining more or less the same dependence of the solution on the x-shift at the entrance.
Shim thick in cut poles solution = 1.15 mm x 2 = 2.3 mm/37 mm = 6 % gap
By(z = 0, x = 0)SHIFTED POLES = By(z = 0, x = 0)CUT POLES+7.6%
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Trajectory optimization
Determined the best value of the current in HC to minimize the integral of By over z
430
415
460
452
429
y = -158.43x + 67936
-6000
-5000
-4000
-3000
-2000
-1000
0
1000
2000
3000
400 410 420 430 440 450 460 470
IHC (A)
Inte
gral
of B
y dz
(G
.cm
)
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Trajectory optimization
Exit angle = 8 x 10-2 mrad
x-shift exit-entrance = 0.13 mm
-0.015
-0.01
-0.005
0
0.005
0.01
0.015
-1.2 -0.7 -0.2 0.3 0.8
z (m)
x (m
)
By integrated over z = 2 G.m
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Tools analysis: multipoles with Tosca & Matlab
TOSCA
1. Determination of the best beam trajectory (tracking Tosca)
2. For each z found By in the points on a line of ±3 cm around (xTR, 0, zTR,) and perpendicular to the trajectory
3. Fit of the By at each point of the line (Tosca) at steps of 1 mm (fit Matlab)
MATLAB
1. Determination of the best beam trajectory (tracking Tosca/0 the integral of By over z)
2. For each found z points on a line of ±3 cm around (xTR, 0, zTR,) and perpendicular to the trajectory
3. Fit of the By at each point of the line at steps of 1 mm interpolated by Matlab
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Tools analysis: tracking
1. Beam enters at several x
2. Tosca tracks the trajectory of each beam
3. Calculated the x exit-xTR NOM and x’exit in function of the x-shift at the entrance
-0.05 -0.04 -0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04 0.05-10
-8
-6
-4
-2
0
2
4
6
8x 10
-3
x enter (m)
angl
e ex
it (r
ad)
-0.05 -0.04 -0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04 0.05-0.04
-0.03
-0.02
-0.01
0
0.01
0.02
0.03
0.04
x enter (m)
x ex
it (m
)
The curves are only to show the tool
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-400
-300
-200
-100
0
100
200
-0.16 -0.12 -0.08 -0.04 0 0.04 0.08 0.12 0.16
z (m)
b3
(T/m
3)
Axis 1.15 cmAxis 1 cmAxis 0.85 cmAxis 0.5 cm
Axis optimization
For the moment used these codes to optimize the position of the axis
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Multipoles
Presence of spikes in my analysis
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Multipoles
Beam trajectory at fixed z and parabolic interpolation in z
-150
-100
-50
0
50
100
150
-1.2 -0.7 -0.2 0.3 0.8
z (m)
b3 (T/m
3)Miro passo 2 mm
Miro passo 1 cm
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Spikes
Spikes: solved problem
-150
-125
-100
-75
-50
-25
0
25
50
75
100
125
150
175
-1.2 -1.15 -1.1 -1.05 -1 -0.95 -0.9 -0.85 -0.8 -0.75
z (m)
b3 (T
/m3)
Miro passo 2 mm
Miro passo 1 cm
Mio con passo 2 mm
Integrato
-150
-125
-100
-75
-50
-25
0
25
50
75
100
125
150
175
-1.05 -1 -0.95 -0.9 -0.85 -0.8 -0.75
z (m)
b3 (T
/m3)
Miro passo 2 mm
Miro passo 1 cm
Mio con passo 2 mm
Integrato
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Axis optimization
Minimized I3 calculated in the entire wiggler
y = -294.07x + 216.19
-150
-100
-50
0
50
100
0.2 0.4 0.6 0.8 1 1.2
Posizione asse (-) (cm)
I3 s
u T
UT
TO
il
wig
gle
r (T
/m2)
0.73 cm
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Multipolar analysis: to summarize
Multipolar analysis (entire wiggler)I0 (T.m) I1 (T) I2 (T/m) I3 (T/m2) I4 (T/m3)
Tosca (2 mm step)
-1.17E-04 2.09 -1.13 0.13 87.8
Tosca (1 cm step)
4.6E-05 2.10 -1.25 -0.98 211
Miro (2 mm step)
1.87E-04 2.09 -1.13 -1.01 95.0
Miro (1 cm step)
1.08E-04 2.08 -1.14 -1.32 101
To do the first optimization I used this technique
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Analysis of the results: tracking (±3 cm)
Beam enters from x = xTR NOM-3 cm to x = xTR NOM+3 cm at steps of 1 mm, where xTR NOM is the position of entrance of the nominal trajectory
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-0.05
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-0.01
0
0.01
0.02
0.03
0.04
0.05
z (m)
x (m
)
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-0.04 -0.03 -0.02 -0.01 0 0.01 0.02 0.03-0.04
-0.03
-0.02
-0.01
0
0.01
0.02
0.03
x enter (m)
x ex
it (m
)
y = - 0.76*x2 + 0.93*x - 8e-005
y = 13*x3 - 0.76*x2 + 0.92*x - 8e-005
y = 53*x4 + 13*x3 - 0.8*x2 + 0.92*x - 7.6e-005
y = 1.5e+004*x5 + 53*x4 - 2.7*x3 - 0.8*x2 + 0.92*x - 7.6e-005
data 1
quadratic
cubic 4th degree
5th degree
Analysis of the results: tracking: the x exit (±3 cm)
The fit is satisfactory already for the 3rth-4rth order
Coefficient of the 3rd order term = 13 m-2
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-0.04 -0.03 -0.02 -0.01 0 0.01 0.02 0.03-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5x 10
-3
x enter (m)
exit
angl
e (r
ad)
y = 10*x3 - 0.64*x2 - 0.065*x - 2.4e-005
y = 19*x4 + 10*x3 - 0.65*x2 - 0.065*x - 2.3e-005
y = 1.2e+004*x5 + 19*x4 - 2.3*x3 - 0.65*x2 - 0.062*x - 2.3e-005
data 1
cubic 4th degree
5th degree
Analysis of the results: tracking: the x’ exit (±3 cm)
The fit is satisfactory for the 3th-4th order
Coefficient of the 3rd order term = 10 rad/m3
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Analysis of the results: comparison with the experimental data
I could compare the results only with the results of the experimental map at about 700 A
Ho riscalato curva di Miro x_exit = x_exitMIRO-x_exitMIRO(xENTR = 0)
-0.08
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0.08
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x-x0 (m)
Exi
t x (m
)Experimental (Miro)
Asse073CorrHC430
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-0.04
-0.02
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0.02
0.04
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0.08
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x-x0 (m)
Exi
t x (m
)Experimental (Miro)Asse073CorrHC430Asse075CorrHC430
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Analysis of the results: comparison with the experimental data
I could compare the results only with the results of the experimental map at about 700 A
-0.03
-0.02
-0.01
0.00
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0.04
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-0.05 -0.04 -0.03 -0.02 -0.01 0.00 0.01 0.02 0.03 0.04 0.05
x-x0 (m)
Exi
t ang
le (ra
d)Experimental (Miro)
Asse073CorrHC430
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
-0.05 -0.04 -0.03 -0.02 -0.01 0.00 0.01 0.02 0.03 0.04 0.05
x-x0 (m)
Exi
t ang
le (ra
d)Experimental (Miro)Asse073CorrHC430Asse075CorrHC430
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Analysis of the results: tracking: the y exit
-6.00E-04
-4.00E-04
-2.00E-04
0.00E+00
2.00E-04
4.00E-04
6.00E-04
-3.E-03 -2.E-03 -1.E-03 0.E+00 1.E-03 2.E-03 3.E-03
Entrance y (m)
Exi
t y (m
)
x = x_nom-0.01
x = x_nom
x = x_nom+0.01
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Analysis of the results: tracking: the y’ exit
-3.00E-03
-2.00E-03
-1.00E-03
0.00E+00
1.00E-03
2.00E-03
3.00E-03
-3.E-03 -2.E-03 -1.E-03 0.E+00 1.E-03 2.E-03 3.E-03
Entrance y (m)
Exi
t y' (
m)
x = x_nom-0.01
x = x_nom
x = x_nom+0.01
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Conclusions
Shifted poles - cut poles solution comparison:
The field roll-off is improved no shim increased BPEAK
Cheaper
At present:
improved the linearity zone of x and x’ with respect to the field map at
dipsosal
In the future:
Shifted poles solution analysis:
Analysis of the field maps by Dragt, Mitchell and Venturini (the map considered the
best one by us, one with the poles more centered and one with the poles more
shifted)
Measurement of the field map of the wiggler at I = 550 A to have a real
comparison with the results of the simulation (at LNF, at ENEA?)
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Di scorta…
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Situation in the present configuration (I = 693 A): x exit
y = 727.49x4 + 307.4x3 + 6.1081x2 + 1.0132x + 0.0175
y = 2E+06x6 + 357542x5 + 3602x4 - 290.86x3 + 1.028x2 + 1.2127x + 0.0182
y = 316785x5 + 7063.2x4 - 231.14x3 - 0.7598x2 + 1.1965x + 0.0183
-0.06
-0.04
-0.02
0.00
0.02
0.04
0.06
0.08
0.10
-0.05 -0.04 -0.03 -0.02 -0.01 0.00 0.01 0.02 0.03 0.04
x-x0 (m)
Exi
t x
(m)
Sperimentali (conti Miro)
Poly. (Sperimentali (conti Miro))
Poly. (Sperimentali (conti Miro))
Poly. (Sperimentali (conti Miro))
The fit is satisfactory for the 5rth-6rth order
│Coefficient of the 3rd order term │ >200 m-2
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Situation in the present configuration (I = 693 A): x’ exit
The fit is satisfactory for the 6rth order
│Coefficient of the 3rd order term │ ~600 rad/m3
y = 7E+06x6 + 669739x5 + 2028.2x4 - 592.4x3 - 1.6938x2 + 0.2157x + 0.0004
-0.02
-0.01
0.00
0.01
0.02
0.03
0.04
-0.05 -0.04 -0.03 -0.02 -0.01 0.00 0.01 0.02 0.03 0.04
x-x0 (m)
Exi
t ang
le (ra
d)
Experimental (Miro)
Poly. (Experimental (Miro))
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Trajectory optimization
To determine the best value of I in HC for the several axis displacements
0.75
0
1.15
y = 52.323x + 390.2
380
390
400
410
420
430
440
450
460
0 0.2 0.4 0.6 0.8 1 1.2 1.4
Axis position with respect to the geometric axis (m)
Opt
imiz
ed I in
HC (A)
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Fine!