Stato dei lavori
Ottimizzazione dei wiggler di DAFNE
Simona Bettoni
Outline
 Method to reduce the integrated octupole in the wiggler of DAFNE
 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
Other methods to reduce the integrated octupole
CURVED POLE
Reduction of the octupole around the beam trajectory in the region of the
poles
Proposed by Pantaleo
MOVING MAGNETIC AXIS
Compensation of the integrated octupole in each semiperiod
New method
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
0.05
xT +3 cm
0.04
0.03
0.02
Beam trajectory (xT)
x (m)
0.01
0
-0.01
xT -3 cm
-0.02
-0.03
-0.04
-0.05
-1.5
-1
-0.5
0
0.5
1
1.5
z (m)
By x  xT   b0  b1  x  xT   b2  x  xT   b3  x  xT   b4  x  xT 
2
3
4
bn  bn (z )
The integrated multipoles in periodic magnets
In a displaced system of reference:
B A y x   b A0  b A1  x  b A 2  x 2  b A3  x3  b A 4  x 4
y
x'  x  xT
y’
BT y x'  bT 0  bT 1  x'bT 2  x'2 bT 3  x'3 bT 4  x'4
xT
OA
OT
x
Even multipoles →
j  0,1,2,...
x’
bAk → defined in the reference centered in OA (wiggler axis)
bTk → defined in the reference centered in OT (beam trajectory)
bT 2 j  c2 j b A 2 j  c2 j 1b A 2 j 1  xT  c2 j  2b A 2 j  2  xT  c2 j 3b A 2 j 3  xT ...
2
3
Left-right symmetry of the magnet
Multipoles change sign from a pole to the next one
Sum from a pole to the next one
Odd multipoles →
bT 2 j 1  c2 j 1b A 2 j 1  c2 j  2b A 2 j  2  xT  c2 j 3b A 2 j 3  xT  c2 j  4b A 2 j  4  xT ...
2
3
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
↑
Octupole
bT 3  c4b A4  xT  c6b A6  xT ...
3
WITH THE POLE MODIFICATION
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
Optimization of the pole of the wiggler
Goals
 Reduce as less as possible the magnetic field in the gap
 Maintain the left-right symmetry
FC1-like
FC2-like
FC 1
9
FC 2
Positron trajectory
By (T)
7
5
x (cm)
3
1
-1
-112 -96 -80 -64 -48 -32 -16 0
-3
-5
-7
-9
z (cm)
16
32
48
64
80
96 112
Analysis
0.05
0.04
0.03
0.02
0
0.05
-0.01
0.04
-0.02
0.03
-0.03
-0.04
0.02
-0.05
-1.5
-1
-0.5
0
0.5
z (m)
1
1.5
0.01
x (m)
x (m)
0.01
0
-0.01
-0.02
-0.03
-0.04
-0.05
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
z (m)
0.4
0.6
0.8
1
For each z fit of By vs x in the system of reference perpendicular to the beam trajectory
Cut poles model: analysis perpendicular to s
IFC = 693 A
Perpendicular to z
Cut poles model
Perpendicular to s (Matlab)
300
Perpendicular to s (Tosca)
250
200
150
b3 (T/m^3)
100
50
0
-0.16
-0.12
-0.08
-0.04
-50
0
0.04
0.08
0.12
0.16
-100
-150
-200
z (m)
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!
Sector poles wiggler
IFC = 693 A
Cut the poles in z to have sector poles
Alfa = 0
300
Straight poles
250
Sector poles
200
150
b3 (T/m^3)
100
50
0
-0.16
-0.12
-0.08
-0.04
0
0.04
0.08
0.12
0.16
-50
-100
-150
-200
z (m)
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
Shifted poles solution
$ and field roll-off
Shifted poles model
For the moment shifted the coils with the poles
Cut-shifted poles: the comparison of the field (at the same current = 550 A in FC)
CUT POLES
SHIFTED POLES
Cut-shifted poles: the comparison of the field (at the same current = 550 A in FC)
Cut poles (550 A)
Cut poles (693 A)
Shifted poles (550 A)
2
1.8
By (T)
1.6
1.4
1.2
1
0.8
-0.07
-0.05
-0.03
-0.01
0.01
0.03
0.05
0.07
x (m)
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%
Trajectory optimization
3000
y = -158.43x + 67936
415
Integral of By dz (G.cm)
2000
1000
429
0
-1000
400
410
420
430
430
440
450
460
470
-2000
-3000
452
-4000
460
-5000
-6000
IHC (A)
Determined the best value of the current in HC to minimize the integral of By over z
Trajectory optimization
By integrated over z = 2 G.m
0.015
0.01
x (m)
0.005
0
-1.2
-0.7
-0.2
0.3
-0.005
-0.01
-0.015
z (m)
Exit angle = 8 x 10-2 mrad
x-shift exit-entrance = 0.13 mm
0.8
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
Tools analysis: tracking
1.
Beam enters at several x
0.04
2.
Tosca tracks the trajectory of each beam
0.03
3.
Calculated the x exit-xTR NOM and x’exit in function of the
x-shift at the entrance
0.02
x exit (m)
0.01
0
-0.01
-0.02
-0.03
-0.04
-0.05
-0.04
-0.03
-0.02
-0.01
0
0.01
x enter (m)
0.02
0.03
0.04
0.05
-0.03
-0.02
-0.01
0
0.01
x enter (m)
0.02
0.03
0.04
0.05
-3
8
x 10
6
4
angle exit (rad)
2
0
-2
-4
-6
The curves are only to show the tool
-8
-10
-0.05
-0.04
Axis optimization
200
100
b3 (T/m3)
0
-0.16
-0.12
-0.08
-0.04
0
0.04
0.08
0.12
0.16
-100
-200
-300
-400
Axis
Axis
Axis
Axis
1.15 cm
1 cm
0.85 cm
0.5 cm
z (m)
For the moment used these codes to optimize the position of the axis
Multipoles
Presence of spikes in my analysis
Multipoles
150
Miro passo 2 mm
Miro passo 1 cm
b3 (T/m3)
100
50
0
-1.2
-0.7
-0.2
-50
0.3
0.8
-100
-150
z (m)
Beam trajectory at fixed Dz and parabolic interpolation in z
Spikes
Miro passo 2 mm
175
Miro passo 1 cm
150
Mio con passo 2 mm
125
100
b3 (T/m3)
Integrato
75
50
25
-1.05
-1.2
-1.15 -1
-1.1
-0.95
-1.05
-1 -0.9 -0.95
-0.85
-0.9
-0.85 -0.8
-0.8
0
-0.75
-0.75
-25
-50
-75
-100
Miro passo 2 mm
Miro passo 1 cm
-125
-150
Mio con passo 2 mm
Integrato
z (m)
Spikes: solved problem
Axis optimization
I3 su TUTTO il wiggler (T/m2)
100
50
0.73 cm
0
0.2
0.4
0.6
0.8
1
-50
-100
y = -294.07x + 216.19
-150
Posizione asse (-) (cm)
Minimized I3 calculated in the entire wiggler
1.2
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
Analysis of the results: tracking (±3 cm)
0.05
0.04
0.03
0.02
x (m)
0.01
0
-0.01
-0.02
-0.03
-0.04
-0.05
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
z (m)
0.4
0.6
0.8
1
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
Analysis of the results: tracking: the x exit (±3 cm)
0.03
y = - 0.76*x 2 + 0.93*x - 8e-005
0.02
y = 13*x 3 - 0.76*x 2 + 0.92*x - 8e-005
y = 53*x 4 + 13*x 3 - 0.8*x 2 + 0.92*x - 7.6e-005
y = 1.5e+004*x 5 + 53*x 4 - 2.7*x 3 - 0.8*x 2 + 0.92*x - 7.6e-005
0.01
x exit (m)
0
-0.01
data 1
quadratic
cubic
4th degree
5th degree
-0.02
-0.03
-0.04
-0.04
-0.03
-0.02
-0.01
0
0.01
0.02
0.03
x enter (m)
The fit is satisfactory already for the 3rth-4rth order
Coefficient of the 3rd order term = 13 m-2
Analysis of the results: tracking: the x’ exit (±3 cm)
-3
1.5
x 10
data 1
cubic
4th degree
5th degree
1
0.5
exit angle (rad)
0
-0.5
y = 10*x 3 - 0.64*x 2 - 0.065*x - 2.4e-005
y = 19*x 4 + 10*x 3 - 0.65*x 2 - 0.065*x - 2.3e-005
-1
y = 1.2e+004*x 5 + 19*x 4 - 2.3*x 3 - 0.65*x 2 - 0.062*x - 2.3e-005
-1.5
-2
-2.5
-0.04
-0.03
-0.02
-0.01
0
0.01
0.02
0.03
x enter (m)
The fit is satisfactory for the 3th-4th order
Coefficient of the 3rd order term = 10 rad/m3
Analysis of the results: comparison with the experimental data
(Miro)
Experimental
(Miro)
Experimental
Asse073CorrHC430
Asse073CorrHC430
Asse075CorrHC430
0.08
0.06
0.06
Exit x (m)
0.04
0.04
-0.05
-0.05
-0.04
-0.04
-0.03
-0.03
-0.02
-0.02
0.02
0.02
0.00
-0.010.000.00
-0.02
0.00
-0.01
-0.02
-0.04
0.01
0.01
0.02
0.02
0.03
0.03
0.04
0.04
0.05
0.05
-0.04
-0.06
-0.08
-0.06
x-x0 (m)
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)
Analysis of the results: comparison with the experimental data
Exit angle (rad)
Experimental (Miro)
Asse073CorrHC430
Asse073CorrHC430
Asse075CorrHC430
0.05
0.04
0.04
0.03
0.03
0.02
0.02
0.01
-0.05
-0.05
-0.04
-0.04
-0.03
-0.03
-0.02
-0.02
0.00
0.00
-0.01
0.00
-0.01-0.010.00
0.01
0.01
0.02
0.02
0.03
0.03
0.04
0.04
0.05
0.05
-0.01
-0.02
-0.03
-0.02
x-x0 (m)
I could compare the results only with the results of the experimental map at about 700 A
Analysis of the results: tracking: the y exit
x = x_nom-0.01
6.00E-04
x = x_nom
4.00E-04
x = x_nom+0.01
Exit y (m)
2.00E-04
-3.E-03
-2.E-03
0.00E+00
-1.E-03
0.E+00
-2.00E-04
-4.00E-04
-6.00E-04
Entrance y (m)
1.E-03
2.E-03
3.E-03
Analysis of the results: tracking: the y’ exit
x = x_nom-0.01
3.00E-03
x = x_nom
2.00E-03
x = x_nom+0.01
Exit y' (m)
1.00E-03
-3.E-03
-2.E-03
0.00E+00
-1.E-03
0.E+00
-1.00E-03
-2.00E-03
-3.00E-03
Entrance y (m)
1.E-03
2.E-03
3.E-03
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?)
Di scorta…
Situation in the present configuration (I = 693 A): x exit
4
3
2
y = 727.49x + 307.4x + 6.1081x + 1.0132x + 0.0175
y = 316785x 5 + 7063.2x 4 - 231.14x 3 - 0.7598x 2 + 1.1965x + 0.0183
0.10
6
5
4
3
y = 2E+06x + 357542x + 3602x - 290.86x + 1.028x 2 + 1.2127x + 0.0182
0.08
Exit x (m)
0.06
0.04
0.02
0.00
-0.05
-0.04
-0.03
-0.02
-0.01 -0.02 0.00
-0.04
-0.06
x-x0 (m)
The fit is satisfactory for the
5rth-6rth order
0.01
0.02
0.03
0.04
Sperimentali (conti Miro)
Poly. (Sperimentali (conti Miro))
Poly. (Sperimentali (conti Miro))
Poly. (Sperimentali (conti Miro))
│Coefficient of the 3rd order term │ >200 m-2
Situation in the present configuration (I = 693 A): x’ exit
y = 7E+06x6 + 669739x5 + 2028.2x4 - 592.4x3 - 1.6938x2 + 0.2157x + 0.0004
Experimental (Miro)
0.04
Poly. (Experimental (Miro))
Exit angle (rad)
0.03
0.02
0.01
-0.05
-0.04
-0.03
-0.02
0.00
-0.01
0.00
-0.01
0.01
0.02
0.03
0.04
-0.02
x-x0 (m)
The fit is satisfactory for the 6rth order
│Coefficient of the 3rd order term │ ~600 rad/m3
Trajectory optimization
y = 52.323x + 390.2
460
1.15
Optimized I in HC (A)
450
440
430
0.75
420
410
400
390
0
380
0
0.2
0.4
0.6
0.8
1
1.2
Axis position with respect to the geometric axis (m)
To determine the best value of I in HC for the several axis displacements
1.4
Fine!