Linear Imperfections equations of motion with imperfections: smooth approximation orbit correction...

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Skew Multipoles: Example Skew Quadrupole normal quadrupole:  clockwise rotation by 45 o  skew quadrupole 3 p+p+ p+p+

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Linear Imperfections

equations of motion with imperfections: smooth approximation

orbit correction for the un-coupled case

transfer matrices with coupling: element and one-turn

what we have left out (coupling)

perturbation treatment: driven oscillators and resonances

1

sources for linear imperfections

CAS, Darmstadt 2009 Oliver Brüning CERN BE-ABP

sources for linear imperfections:

ygB

xxgB

x

y

⋅−=

Δ+⋅−= )~(xxx Δ+=~

2

Sources for Linear Field Errors

-magnetic field errors: b0, b1, a0, a1

-powering errors for dipole and quadrupole magnets

-energy errors in the particles change in normalized strength

-feed-down errors from quadrupole and sextupole magnets

example: feed down from a quadrupole field

dipole + quadrupole field component

-roll errors for dipole and quadrupole magnets

p+

Skew Multipoles: Example Skew Quadrupolenormal quadrupole: clockwise rotation by 45o skew quadrupole

By = g ⋅ x ⇒ Fx = −q ⋅v ⋅g ⋅ xBx = g ⋅ y ⇒ Fy = +q ⋅v ⋅g ⋅ y xNvqFxNB

yNvqFyNB

yx

xy

⋅⋅⋅−=⇒⋅−=

⋅⋅⋅−=⇒⋅+=

xB

g y

∂∂

= ⎟⎟⎠⎞

⎜⎜⎝⎛

∂∂

−∂∂

=xB

yB

N xy

21

3

p+ p+

Sources for Linear Field Errors

-magnet positioning in the tunnel transverse position +/- 0.1 mmroll error +/- 0.5 mrad

-tunnel movements:slow driftscivilizationmoonseasonscivil engineering

-closed orbit errors beam offset inside magnetic elements

-energy error: dispersion orbit

sources for feed down and roll errors:

4

Equation of Motion ISmooth approximation for Hills equation:

perturbation of Hills equation:

in the following the force term will be the Lorenz force of a charged particle in a magnetic field:

)/()),(),(()()( 202

2 pvssysxFswswdsd ⋅=⋅+ω

0)()()(22 =⋅+ sωsKsω

dsd

)sin()( 00 φω +⋅⋅= sAsω

(constant -function and phase advance along the storage ring)

0)()( 202

2 =⋅+ sωsωdsd ω

K(s) = const

LQ /2 00 ⋅= πω(Q is the number of oscillations during one revolution)

BvqFrr

×⋅=5

yxw ,=

Equation of Motion Iperturbation for dipole field errors:

perturbation of Hills equation:⎪⎩

⎪⎨⎧

⋅−⋅−

−=⋅+

)()()()(

1

1

02

022

sysxk

ksxsx

dsd

κω

pB

pvF yΔ

−=⋅

6

perturbation for quadrupole field errors:

ypg

pvF

xpg

pvF

y

x

⋅Δ

+=⋅

⋅Δ

−=⋅

normalized multipole gradients:

]/[][3.00 cGeVp

TBk Δ⋅=

]/[]/[3.01 cGeVp

mTN⋅=κ

]/[]/[3.01 cGeVp

mTgk ⋅=

Coupling I: Identical Coupled Oscillators

π mode:

ω mode:

kk 20 +=πω0k=ωω

fundamental modes for identical coupled oscillators:

0)()( 22

22

2 =⋅+ tqtqdtd

πω0)()( 12

12

2 =⋅+ tqtqdtd

ωω

yxtq +=)(1 yxtq −=)(2

7

weak coupling (k << k0):

description of motion in unperturbed ‘x’ and ‘y’ coordinates

degenerate mode frequencies

k0k

0kk

0k

0k

solution by decomposition into ‘Eigenmodes’:

distributed coupling: )()()( 12

22 sysxsx xds

d ⋅−=⋅+ κω

)()()( 12

22 sxsysy yds

d ⋅−=⋅+ κω

8

Coupling II: Equation of Motion in Accelerator

0)()( 22222

2 =⋅+ sqsqdtd ω0)()( 1

2112

2 =⋅+ sqsqdtd ω

ybxasq ⋅+⋅=)(1 ydxcsq ⋅+⋅=)(2

With orthogonal condition: 0=⋅+⋅ dca

take second derivative of q1 and q2:

2

1

22

1

222

1

22

1

22

21

2;

21

2 ⎟⎟⎠⎞

⎜⎜⎝⎛ −

+−−

=⎟⎟⎠⎞

⎜⎜⎝⎛ −

++−

=κωω

κωω

κωω

κωω yxyxyxyx ca

9

Coupling II: Equation of Motion in Accelerator

0)()( 22222

2 =⋅+ sqsqdtd ω0)()( 1

2112

2 =⋅+ sqsqdtd ω

use Orthogonal condition for calculating a,b,c,d (set b=1=d)

expressions for ω1 and ω2 as functions of a, b, c, d, ωx, ωy

with:

ω1,22 = 1

2 ⋅ ωx2 +ωy

2( ) ± Ω

2222

21 ⎟⎟⎠⎞

⎜⎜⎝⎛ −

+=Ω yx ωωκ

yields:

very different unperturbed frequencies:

10

Coupled Oscillators Case Study: Case 1

expansion of the square root:

12

2

1

22

>>⎟⎟⎠⎞

⎜⎜⎝⎛ −

κωω yx

xyx

x ωωω

κωω ≈−

+= 22

21

1 yyx

y ωωω

κωω ≈−

−= 22

21

2

‘nearly’ uncoupled oscillators

ω1,22 = 1

2 ⋅ ωx2 +ωy

2( ) ± 12 ⋅ ωx

2 −ωy2( ) ⋅

2κ1

ωx2 −ωy

2( )

⎝ ⎜ ⎜

⎠ ⎟ ⎟

2

+1

εε 2111 +≈+

1;1;1;1 =−≈=≈ dca

almost equal frequencies:

11

Coupled Oscillators Case Study: Case 2

Ω±= ~02,1 ωω

keep only linear terms in Δ

Δ−= 21

0ωωyΔ+= 21

0ωωx

20

221

20

22,1 ωκωω ⋅Δ+±=

expansion of the square rootfor small coupling and Δ:

20

22 2ωωω ≈+ yx Δ≈− 022 2ωωω yx

20

2

40

21

02,1 1ωω

κωω Δ+±⋅=

2

20

21

21~ Δ+⋅=Ω

ωκ

with:

( )yx ωωω +=21

0

measurement of coupling strength:

12

Coupled Oscillators Case Study: Case 2

measure the difference in the Eigenmode frequencies while bringing the unperturbed tunes together:

220

21

02,1 21

Δ+⋅±=ωκωω

the minimum separation yields the coupling strength!!

ω1,2

ωx

ωy

0ωκ

Δ

Coupled Oscillators Case Study: Case 2

with

initial oscillation only in horizontal plane:

x(s) = A ⋅cos ˜ Ω ⋅s( ) ⋅cos 12 ω1 +ω2[ ] ⋅ s+ 1

2 [φ1 + φ2]( )

yxtq −=)(1

yxtq +=)(2

13

0)0(;0)0(;0)0(;)0( =′==′= yyxAx

)sin( 111 φω +⋅⋅= sAq )sin( 222 φω +⋅⋅= sAqand

and

y(s) = −A ⋅sin ˜ Ω ⋅s( ) ⋅sin 12 ω1 +ω2[ ] ⋅ s+ 1

2 [φ1 + φ2]( )

ω1,2 = 12 ⋅ ωx +ωy( ) ± ˜ Ω

modulation of the amplitudes

sum rules for sin and cos functions:

Beating of the Transverse Motion: Case I

modulation of the oscillation amplitude:

frequencies can not be distinguished and energy can be exchanged between the two oscillators

two almost identical harmonic oscillators with weak coupling:

π-mode and ω=mode frequencies are approximately identical!

x

s

14Ω~

( )yx ωωω +=21

0

Driven Oscillators

large number of driving frequencies!

equation of motion driven un-damped oscillators:

)(

..

21 2

2

2 )()()( klmLyx smlsk

mlkklmwds

dwds

d eWswswQsw φωω π

ωω +⋅⋅+⋅+⋅⋅− ∑=++

15

Perturbation treatment:substitute the solutions of the homogeneous equation of motion:

into the right-hand side of the perturbed Hills equation andexpress the ‘s’ dependence of the multipole terms by their Fourierseries (the perturbations must be periodic with one revolution!)

)sin()( 00 φω +⋅⋅= sAsω

Driven Oscillators

general solution: )()()( swswsw sttr +=

)cos()()()()( 02

01

02

2 φωωω +⋅⋅=⋅+⋅⋅+ − ssΩsωsωQsω dsd

dsd

16

single resonance approximation:

consider only one perturbation frequency (choose ):

without damping the transient solution is just the HO solution

)sin()( 00 φω +⋅⋅= sasωtr

0ωω ≈

Lyx mlk πωωω 2++=

Driven Oscillators

resonance condition:

0ωω =n

17

)](cos[)()( 20

ωaωωω

−⋅⋅= sΩsωststationary solution:

where ‘ω’ is the driving angular frequency!and W(ω) can become large for certain frequencies!

22

001

1)(

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=

+−ω

ω

ω

ωω

Qnn

nΩΩ

justification for single resonance approximation:

all perturbation terms with: 0ωω ≠n de-phase with the transient

no net energy transfer from perturbation to oscillation (averaging)!

Resonances and Perturbation Treatment

nQLn LQ =⏐⏐⏐⏐ →⏐⋅= ⋅=0

/20

00/2 πωπω

example single dipole perturbation:

resonance condition:

∑∞

−∞=

⋅⋅⋅−=⋅+ ⋅n

Lsnlkswsw Ldsd )/2cos()()( 1

02

022 πω

)()(00 ssk

pvsF

L −⋅−=⋅

d

avoid integer tunes!

(see general CAS school for more details) 18€

ΔCO(s) =β (s)

2sin(πQ)⋅ Δk0(t) ⋅ β (t) ⋅cos(|φ(t) −φ(s) | −πQ)∫ dt

Fourier series of periodic d-function

Resonances and Perturbation Treatment integer resonance for dipole perturbations:

19 dipole perturbations add up on consecutive turns! Instability

assume:

Q = integer

Resonances and Perturbation Treatment integer resonance for dipole perturbations:

20

dipole perturbations compensate on consecutive turns! stability

assume:

Q = integer/2

Resonances and Perturbation Treatment Orbit ‘kink’ for single perturbation (SPS with 90o Q = n.62):

21

)|)()(cos(|)sin(2)()(

)( 000 Qsskl

Qss

sCO πφφπ

−−⋅Δ⋅=Δ

)()( 00 εε −′Δ=+′Δ sOCsOC

Resonances and Perturbation Treatment

2/22 0

/20

00nQLn LQ =⏐⏐⏐⏐ →⏐⋅=⋅ ⋅= πωπω

example single quadrupole perturbation:

resonance condition:

avoid half integer tunes!

(see general CAS school for more details) 22

xkpvsF

⋅−=⋅ 1)( )cos()( 0,00 φω +⋅⋅= sAsω xwith:

∑∞

−∞=

±⋅±⋅⋅−=⋅+n

xx sLnAswsw Llk

dsd )]/2cos([)()( 0,0

120,2

2

2 φωπω

dtQstttkQs

s∫ −−⋅⋅Δ⋅=

Δ)2|)()(|2cos()()(

)2sin(21

)()(

10

πφφπ

Resonances and Perturbation Treatment half integer resonance for quadrupole perturbations:

23

quadrupole perturbations add up on consecutive turns! Instability

assume:Q = integer + 0.5

feed down error:

ybvqFybB yx ⋅⋅⋅+=⇒⋅= 11

Resonances and Perturbation Treatment

nQQLn yxLQ

yx =±⏐⏐⏐ →⏐⋅=± ⋅= /20,0, /2 πωπωω

example single skew quadrupole perturbation:

resonance condition:

avoid sum and difference resonances!

24

ypvsFx ⋅−=

⋅ 1)( κ )cos()( 0,00 φω +⋅⋅= sAsy ywith:

∑∞

−∞=

±⋅±⋅⋅−=⋅+n

yx sLnAsxsx Ll

dsd )]/2cos([)()( 0,0

120,2

2

2 φωπω κ

difference resonance stable with energy exchange sum resonance instability as for externally driven dipole

coupling with:

25

drive and response oscillation de-phase quickly no energy transfer between motion in ‘x’ and ‘y’ plane

small amplitude of ‘stationary’ solution:

no damping of oscillation in ‘x’ plane due to coupling

coupling is weak tune measurement in one plane will show both tunes in both planes but

with unequal amplitudes

tune measurement is possible for both planes

yxyx QQorQQ <<>>

2)(2])(1[

1

00

20)(

ωω

ωω

ω

Q

ΩΩ+−

⋅=

Resonances and Perturbation Treatment: Case 1

coupling with:

26

drive and response oscillation remain in phase and energy can be exchanged between motion in ‘x’ and ‘y’ plane:

large amplitude of ‘stationary’ solution:

damping of oscillation in ‘x’ plane and growth of oscillation amplitude in ‘y’ plane

‘x’ and ‘y’ motion exchange role of driving force!

each plane oscillates on average with:

Impossible to separate tune in ‘x’ and ‘y’ plane!

yx QQ ≈

( )yx QQ +21

2)(2])(1[

1

00

20)(

ωω

ωω

ω

Q

ΩΩ+−

⋅=

Resonances and Perturbation Treatment: Case 2

Exact Solution for Transport in Skew Quadrupolecoupled equation of motion: 01 =⋅+′′ yx κ

01 =⋅+′′ xy κand

27

can be solved by linear combinations of ‘x’ and ‘y’:

0)()( 1 =+⋅+′′+ yxyx κ 0)()( 1 =−⋅−′′− yxyx κand

solution as for focusing and defocusing quadrupole

transport matrix for ‘x-y’ and ‘x+y’ coordinates for κ1 > 0:

iniend yxyx

ll

ll

yxyx

⎟⎟⎠⎞

⎜⎜⎝⎛

′−′−

⋅⎟⎟⎟

⎜⎜⎜

=⎟⎟⎠⎞

⎜⎜⎝⎛

′−′−

)cos()sin(

)sin()cos(

111

1

11

κκκκκ

κ

iniend yxyx

ll

ll

yxyx

⎟⎟⎠⎞

⎜⎜⎝⎛

′+′+

⋅⎟⎟⎟

⎜⎜⎜

=⎟⎟⎠⎞

⎜⎜⎝⎛

′+′+

)cosh()sinh(

)sinh()cosh(

111

1

11

κκκκκ

κ

Transport Map with Couplingtransport map for skew quadrupole:

transport map for linear elements without coupling:

inisqend zMz rr⋅=

with:

28

⎟⎟⎟⎟⎟

⎜⎜⎜⎜⎜

′=

yyxx

zr

⎟⎟⎟⎟⎟

⎜⎜⎜⎜⎜

−−

−−=

adcbbadccbaddcba

M sq

11

11

κκ

κκand

inilend zMz rr⋅=

⎟⎟⎟⎟⎟

⎜⎜⎜⎜⎜

=

4443

3433

2221

1211

0000

0000

mmmm

mmmm

M lwith

Transport Map with Couplingcoefficients for the transport map for skew quadrupole:

with:

29

NsNsNsNsN

NsNsNsNsN

dcba

2/)]sinh()[sin(2/)]cosh()[cos(

2/)]sinh()[sin(2/)]cosh()[cos(

−−

++

====

One-Turn Map with Couplingone-turn map around the whole ring:

)()()( 000 szsTLsz rr⋅=+

30

∏=i

iMT

⎟⎟⎠⎞

⎜⎜⎝⎛

=NmnM

T

T is a symplectic 4x4 matrix

with:

with: being 2x2 matrices 16 parameters in total

nmNM ,,,

notation:

STSTt =⋅⋅ with: ⎟⎟⎟⎟⎟

⎜⎜⎜⎜⎜

−=

0100100000010010

S

determines n*(n-1)/2 parameters for a n x n matrix

Parametrization of One-Turn Map with Couplinguncoupled system: parameterization by Courant-Snyder variables

31

)sin()cos( μμ ⋅+⋅= JIT

⎟⎟⎠⎞

⎜⎜⎝⎛

=1001

I ⎟⎟⎠⎞

⎜⎜⎝⎛

−−=

aγa

J

with:

21+=

T is a 2 x 2 matrix 4 parameters

T is symplectic determines 1 parameter

3 independent parameters

Parametrization of One-Turn Map with Couplingrotated coordinate system:

using a linear combination of the horizontal and vertical position vectors the matrix can be put in ‘symplectic rotation’ form

or:

32

⎟⎟⎠⎞

⎜⎜⎝⎛ −⋅⎟⎟⎠⎞

⎜⎜⎝⎛⋅⎟⎟⎠⎞

⎜⎜⎝⎛−

=−−

)cos()sin()sin()cos(

00

)cos()sin()sin()cos( 1

2

11

φφφφ

φφφφ

IDDI

AA

IDDIT

1−⋅⋅= RURT with:

is a symplectic 2x2 matrice 3 independent parameters

total of 10 independent parameters for the One-Turn map

D

2,1);sin()cos( =⋅+⋅= iJIA iiiii μμ

One-Turn Map with Couplingrotated coordinate system:

33

x

y

x~y~

rotated coordinate system:

new Twiss functions and phase advances for the rotated coordinates

)sin()cos( iiii JIA μμ ⋅+⋅= ⎟⎟⎠⎞

⎜⎜⎝⎛

−−=

ii

iiiJ αγ

βα

cos(μ1) − cos(μ2) = 12 Tr M −N( )[ ]

2+ det(m + n+)

with:

( ))(

det)2tan(

21 NMTr

nm−+−

=+

φ

⎟⎟⎠⎞

⎜⎜⎝⎛−

−⏐⏐⏐⏐ →⏐⎟⎟⎠

⎞⎜⎜⎝⎛ −

+

acbd

dcba matrixadjoint

Summary One-Turn Map with Coupling

coupling changes the orientation of the beam ellipse along the ring

34

a global coupling correction is required for a restoration of the uncoupled tune values (can not be done by QF and QD adjustments)

coupling changes the Twiss functions and tune values in thehorizontal and vertical planes

a local coupling correction is required for a restoration of the uncoupled oscillation planes ( mixing of horizontal and vertical kicker elements and correction dipoles)

What We Have Left Out

dispersion beat:

-beat: skew quadrupole perturbations generate -beatsimilar to normal quadrupole perturbations

35

integer tune split and super symmetry

skew quadrupole perturbations generate vertical dispersion

the (1,-1) coupling resonance in storage rings with supersymmetry can be strongly suppressed by an integer tune split

general definition of the coupling coefficients:

dsesss srssiyxr

Lyx ))()((1,

2

)()()(21 ⋅+±

± ⋅⋅⋅= ∫πφφκ

πκ⏐⏐ →⏐= = ω

ωκκ /11

Orbit Correctiondeflection angle:

36

)(]/[][3.0

iy

i sXcGeVp

lTB′Δ=

⋅Δ⋅−=q

))(sin()()( iii sssZ φφq −⋅⋅⋅=Δtrajectory response:

))(cos()(/)( iii sssZ φφq −⋅⋅=′Δ

closed orbit bump

compensate the trajectory response with additional dipolefields further down-stream ‘closure’ of the perturbationwithin one turn

Orbit Correction3 corrector bump:

37

12

113

23

133 )cos(

)tan()sin( q

φφφq ⋅⋅⎟⎟⎠

⎞⎜⎜⎝⎛

Δ−ΔΔ

= −−

closure

sensitive to BPM errors; large number of correctors

123

13

2

12 )sin(

)sin( qφφ

q ⋅ΔΔ

⋅−=−

limits

SVD Algorithm Ilinear relation between corrector setting and BPM reading:

38

global correction:

),...,,( 21 mcccCOR = vector of corrector strengths

CORABPM ⋅=

problem A is normally not invertible (it is normally not even a square matrix)!

),...,,( 21 nbbbBPM = vector of all BPM data

A being a n x m matrix

BPMACOR ⋅= −1

solution minimize the norm: CORABPM ⋅−

SVD Algorithm IIsolution:

39

singular value decomposition (SVD):

find a matrix B such that

pm

i

pixx

/1

⎟⎠⎞⎜

⎝⎛

= ∑

any matrix can be written as:

where O1 and O2 are orthogonal matrices and D is diagonal

CORBABPM ⋅⋅−

attains a minimum with B being a m x n matrix and:

21 ODOA ⋅⋅=

tOO =−1

SVD Algorithm IIIdiagonal form:

40

define a pseudo inverse matrix:

⎟⎟⎟

⎜⎜⎜

⎛==⋅

10

011 O)kDD

⎟⎟⎟⎟⎟

⎜⎜⎜⎜⎜

=

0000

00

0

00

0

0 22

11

LMLL

MLL

LL

O

L

LOM

kk

D

σ

σσ

),min( mnk ≤

⎟⎟⎟⎟⎟⎟⎟⎟⎟

⎜⎜⎜⎜⎜⎜⎜⎜⎜

=

0

0/10

0

0

00

0/10

0

00

0/1

11

11

11

M

M

LML

O

LMLL

M

M)σ

σσ

D

1k being the k x k unit matrix

SVD Algorithm IVcorrection matrix:

41

main properties:

define the ‘correction’ matrix:

( ) ( ) ktt ODOODOBA 11221 =⋅⋅⋅⋅⋅=⋅

)

SVD allows you to adjust k corrector magnets

tt ODOB 12 ⋅⋅=)

if k = m = n one obtains a zero orbit (using all correctors)

for m = n SVD minimizes the norm (using all correctors)

the algorithm is not stable if D has small Eigenvalues can be used to find redundant correctors!

),min( mnk ≤

Harmonic FilteringUnperturbed solution (smooth approximation):

42

orbit perturbation

spectrum peaks around Q = n small number of relevant terms!

periodicity:

022 =⋅⋅+′′ xQx Lπ sQi LeAsx ⋅⋅⋅⋅=

π2

)(

)(22 sFxQx L =⋅⋅+′′ π

sni

nn

LefsF ⋅⋅⋅⋅=∑π2

)( sni

nn

LedsCO ⋅⋅⋅⋅=∑π2

)(

( ) ( )2222 QQfd

LL

nn ππ −

=

Most Effective Correctorthe orbit error is dominated by a few large perturbations:

43

brut force:

minimize the norm: CORBABPM ⋅⋅−

using only a small set of corrector magnets

select all possible corrector combinations time consuming but god result

selective: use one corrector at the time + keep most effective much faster but has a finite chance to miss best solution and can generate π bumps

MICADO: selective + cross correlation between orbitresidues and remaining correcotr magnets

Example for Measured & Corrected Orbit Data LEP:

44

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