errors in beam based alignment measurements

F. Wileke, HERA Seminar Zeuthen, 7-9 January 2002

1

Errors in Beam Based Alignment Measurements

Beam Based Alignment Formulae

Error by thin lens treatment

Error in the Analysis by Optical Error


2

X 0 1 M 01 d 0

M 0 d 0

Beam Based Alignment Formulae

Closed Orbit

Revolution matrix closed orbit vector x,x‘

2sd kd h s( )d

d

2=0

d(0)=d´(0)=0

Closed orbit after a change of a quadrupole strengths k

X = 1 M( ) 1 d M = M 0 kM kd

dd =d 0 k

d kd

d

(L)


3

Comments for Slide 2:

The change of the closed orbit by a change in a quadrupole strength is expanded up to first order in the quadrupole change. The center of changed quadrupole is taken at the beginning of the lattice. As Georg Hoffstaetter has found out, it is very advantageous for the treatment of beam based alignment to consider the magnet offset in the middle (longitudinal middle of the quadrupole.

The vector d is the solution of the orbit for starting values x=x’=0. H are the inhomogeneities.

The term (1-M)^-1 makes the solution ring-periodic.


4

M0

k*dG/dk

lattice

½quad

Start:

Middle oft the quad

Gh-1

Half quadrupole matrix

0

k L

21

sin k L

k L

k L

2 k( ) 1sin k L

k L

0

Gh-1

½quad

x0

kMd

d= M 0 G h

1k

Gd

d G h

1

1 1 M 01 M 0

ka x 0p

b x 0

1

1 M 01 d 0

k 1 M 01

kd 0

d

d


5

Comments slide 4:

In order to express the effect of a change quadrupole up to first order as a function of quadrupole offset and angle in the middle, one starts with the offset in the middle of the quadrupole, transforms it back to the beginning by half an inverse quadrupole matrix. Then one transforms the offset vector at the beginning through the whole quadrupole for the original and changed value. The difference is expressed by the difference matrix d/dkM. The resulting orbit change is transformed back to the middle of the quadrupole. Since we are limiting ourselves to first order in dk, the transformation back kann be performed with the unchange quadruple half matrix.

The matrix G(l/2)-1(d/dkM)G(L/2)-1 has a antidiagonal structure which will ease the analysis and which will clearly separate angle and offset effects.

From the final expression one can see that offset produce cosine like orbits wrt the quadrupole middle and the angle offset produces sine-like orbits wrt to the middle of the quad


6

x = 1 1 M 01 M 0

ka x 0p

b x 0

1

x 0 k 1 M 0

1k

d 0d

d

x = 1 M 01 M 0

ka x 0p

b x 0

x 0

x s( ) = T 0 s( ) 1 M 01 M 0

ka x 0p

b x 0

x 0

1 M 01 M 0

0

b

a

0

x 0

x p0


7

Comments slide 6:

If the quadrupole strength is changed, the inhomogeneity vector d is also changing via the misalignment of this quadrupole. We find that if the offset and angle are taken in the middle, the change of d can be expressed by a similar antidiagonal matrix. Then we can absorb the d/dk d term and express the closed orbit change as a function of the sum of original orbit and misalignment in the test quadrupole


8

Error by Thin Lens TreatmentThin lens matrix

Thin lens differential matrix

0

k L

0

0

C k L( )

K k( )k

k S k L( )

S k L( )

K k( )

C k L( )

1

k L0

1

Thick lens matrix

Thick lens differential matrix

0

k L

21

sin k L

k L

k L

2 k( ) 1sin k L

k L

0

x meas = T f x f

x c = T c1 T f

x f

x meas = T c x c 2

1sin k L

k L

0

0

0

R k L( )2

1S k L( )

K k( ) LS k L( )

K k( ) L1

S k L( )

K k( ) L1

Relative

Pos Error


9

Comments slide 8:

The first error considered is the error made by simplifying the analysis to a thinn lens treatment.

The expressions relate the thick lens analysis of the orbit change caused by the quadrupole change with a thin lens analysis. Again, taking the offset and angle in the middle, the analysis simplifies due to the antidiagonal structure of the difference matrix.

We see that the coefficient between the “true”offset obtained by thick lens analysis and the “false”offset obtained by thin lens offset has a simple analytical form which can be easily evaluated


10

Errors for the various quadrupoles

GO v k 0.14 L 3.2 1 R k L( ) 38.423 %

GO h k 0.14 L 3.2 1 R k L( ) 16.355 %

GG h k 0.001 L 1.3 1 R k L( ) 6.536 %

GI h k 0.25 L 1.88 1 R k L( ) 8.17 %

GI v k 0.25 L 1.88 1 R k L( ) 22.546 %

GJ h k 0.18 L 1.88 1 R k L( ) 20.673 %

GJ h k 0.18 L 1.88 1 R k L( ) 10.326 %

QL14 h k 0.058 L 1.03 1 R k L( ) 0.225 %

QL16 h k 0.112 L 1.03 1 R k L( ) 0.255 %

Errors by thin lens treatment


11

Comments 10th Slide:

This slide shows the errors made by thin lens treatment for the quadrupoles in the interaction regions


12

Error due to optics error in a quad (closed bump method)

kLx0

k2L2x0

kLx0

This point implies large angle error

dkL =0.5/m*200m*1%=1

x

x+x´

First corrector


13


Here we consider the effect of a gradient error of a quadrupole which is between the test quadrupole and the bpms or the compensating kicks. In the IR, the The quadrupole are long, strong and the optics has a large beta. K*L* can be easily 100 so that dkl=1 is possible. If the analysis point is at a distance of from the test quad, and the error quad at a distance /2, the diagram shows that the errors can be quite large. Instead of the yellow arrow which is calculated as an effect of the quadrupole offset without knowledge of the quadrupole error one has the

Red arrow as the “true”effect. From this one would calculate a very large angle offset of the quadrupole which is not there.


14

Undisturbed transformation from komensating kicks to middle of test quad

C u

1 t sin t 1

1

tcos t 1 t sin t 1

2 t sin t 2

2

tcos t 2 t sin t 2

disturbed transformation

Disturbing kick

q S 1 q 1 q 1 S 2 q 2 q 2

k q L q

Z c

S 1 t 1 t

S p 1 t 1 t t

S 2 t 2 t

S p 2 t 2 t t

S q t q t

S p q t q t t

1

2

q


15

Comments 14th slide: The analysis considered is to compensate the closed orbit distortion by two orbit correctors. The two kicks, the in-between-optics and the test quadrupole difference matrix allow to calculate the test quadrupole offset.

The two matrices describe the transformation from the kicks to the beam offset in the test quad middle for the undistorted and the distorted transformation

Comments 16th Slide: The distorted transformation is expressed by the sum of two transformations, one containing the gradient error. The result using the distorted and the undistorted transfromation is expressed by the matrix product in the red box.


16

C 1

S q t q t

S p q t q t t

0

0

S 1 q 1 q

0

S 2 q 2 q

0

C p C u k q L q C 1

transformation expressed by a 2x2 Matrix C1

Now we express the coordinates evaluted assuming an unperturped transformationthus the false result Zfby the coordinates evaluated by a perturbed transformationthus the correct result Zc

Z f C u C p1 Z c

C C u C p1


17

The same can be done for the magnet offsets

X c M k k L( )1

C p

1

2

X c0.027

0.086

X f M k k L( )1

C u

1

2

X f0.026

0.085

X f M k k L( )1

C u C p

1 M k k L( ) X c X f

0.026

0.085

M k k L( )

0

k L

21

S k L( )

K k( ) L

k L

2 k1

S k L( )

K k( ) L

0


18


Finally, we do the same for the quadrupole offset calculation by multiplying the distorted and undistorted transformation with with the differential matrix respectively. The red box contains now the bba result with and without taking into account the quadrupole error between test quad and analysing kicks


19

Error matrix for magnet offssetskL and q are from the error quad, kL, t t are taken from the test quad

kL q

1

2sin 2 qt

t sin qt2

k t sin qt

2 k L sin k L

k L sin k L

cos qt t sin qt 2

k L sin k L

k L sin k L t k

1

2sin 2 qt

t sin qt2

This potentially small denominator enhances the angle error

(infinite for thin lens)


20

Comments Slide 19:

The matrix which relate the correct and the false result has a simple analytic form. The second row describes the errors of the angle. It can be very large for a short test quadrupole

Comments rest of the slides: Erros of analysis for the test quadrupole GO and assuming any other quadrupole as a source of error is performed. An error of 1% is assumed.

The errors of the offset can become as large as 20%

And the errors of the slope can be as large as 100%


21

Test GO horizontal

k 0.14146 L 3.2 k 1 % k t 19.22 t 3.308 2 t 12.5

measured magnet-beam offset X f.0095

.0025

----------------------------------------------------------------------------------------------------------

GI L q 63 q 3.303 2 kL 1.88 0.2459 1 % qt t q

X c M k k L( )1

C m1 qt kL q t t M k k L( ) X f X c

9.436 10 3

2.578 10 3

X c0X f0

X c0

0.679 %X c1

X f1

X c1

3.044 %

----------------------------------------------------------------------------------------------------------


22

GJ L q 44 q 3.296 2 kL 1.88 0.1324 1 % qt t q

X c M k k L( )1


9.496 10 3

2.641 10 3

X c0X f0

X c0

0.046 %X c1

X f1

X c1

5.335 %

------------------------------------------------------------------------------------------------------------------

QL14 L q 33.2 q 3.205 2 kL 1.033 0.055 1 % qt t q

X c M k k L( )1


0.01

3.85 10 3

X c0X f0

X c0

6.29 %X c1

X f1

X c1

35.072 %

------------------------------------------------------------------------------------------------------------------

QL14 L q 24 q 2.950 2 kL 1.033 0.112 1 % qt t q

X c M k k L( )1

C m1 qt kL q t t M k k L( ) X f

X c

0.011

5.724 10 3X c0

X f0

X c0

16.022 %X c1

X f1

X c1

56.326 %


23

Test GO vertical

k .14146 L 3.2 k 1 % k t 28.66 t 2.420 2 t 1.5

measure magnet -beam offset X f.002

.00015

------------------------------------------------------------------------------------------------------------------

GI L q 29 q 2.407 2 kL 1.88 0.2459 1 % qt t q

X c M k k L( )1


1.975 10 3

2.091 10 4

X c0X f0

X c0

1.28 %X c1

X f1

X c1

28.25 %

------------------------------------------------------------------------------------------------------------------ q 50 q 2.399 2 kL 1.88 0.1324 1 % qt t q


24

X c0X f0

X c0

1.28 %X c1

X f1

X c1

28.25 %

------------------------------------------------------------------------------------------------------------------ q 50 q 2.399 2 kL 1.88 0.1324 1 % qt t qGJ L

X c M k k L( )1


1.96 10 3

2.91 10 4

X c0X f0

X c0

2.028 %X c1

X f1

X c1

48.461 %

------------------------------------------------------------------------------------------------------------------

QL14 L q 30.2 q 2.265 2 kL 1.033 0.055 1 % qt t q

X c M k k L( )1


1.949 10 3

9.064 10 4

X c0X f0

X c0

2.639 %X c1

X f1

X c1

83.451 %

------------------------------------------------------------------------------------------------------------------

QL16 L q 84 q 2.225 2 kL 1.033 0.112 1 % qt t q

X c M k k L( )1

C m1 qt kL q t t M k k L( ) X f

X c1.679 10 3

5.69 10 3

X c0X f0

X c0

19.15 %X c1

X f1

X c1

97.364 %

------------------------------------------------------------------------------------------------------------------


25

Errors of Beam Based Alignment Results due to Statistical and Systematical Errors of the Correctors

Beam based alignment measurements are performed by chaging the test quadrupole (index"t") by an amount k and compensatingthe corresponding orbit by two corrector coils indices "1" and "2". Let the optical paramenters for the undisturbed optics (k=0) be

t 0 1 1.2

2 2 0.89

t 200 1 40 2 40

t 10

The test quadrupole parameterrs are: k .25 L 2 k 5 % k

The matrux Cu transports corrector kicks to the center of the test magnet

C u 1 2 t 1 2 t t

1 t sin t 1

1

tcos t 1 t sin t 1

2 t sin t 2

2

tcos t 2 t sin t 2

The matrix M relates the offset and angular misalignment of the magnet taken in its longitudinal and the difference orbit (x,x')T, which is artificially transformed back into the middle of the magnet magnet

M k k L( )

0

k L

21

sin k L

k L

k L

2 k( )1

sin k L

k L

0


26

The matrix A, the product of Cu and M-1, relates the compensating kicks (1,2)T and mid magnet offsets in

A k k L t 1 2 t 1 2 t 2 k

k k

1

tcos t 1 t sin t 1

k L sin k L

k 1 t sin t 1

k L sin k L

2

t

cos t 2 t sin t 2

k L sin k L

k 2 t sin t 2

k L sin k L

to illustrate this, an example is calculated. X is the vector with an assumed misalignment of our test magnet. Given such a misalignment,

the kick vector Z0 =A-1X should be the result of the measurement

x 10 3 xp 10 4 Xx

xpZ 0 C u 1 2 t 1 2 t t 1 M k k L( ) X Z 0

In order to reproduce the misalignment of that size , the measurements has to be precice to the level of 10m.

i 0 1 20 j 0 1 20

Real measured kicks however contain errors, systematical and statistical ones. Lets assume a statictical error of

and systematical errors of

Then the result of the measurement may be

1.0 10 5

1 1 10 5 2 2 10 5

Zi j Z 0

rnd 1.( ) 0.5( ) 1

rnd 1.( ) 0.5( ) 2


27

Zi j( ) 1

105

Zi j 0

104

Using these measured kicks the inverse transformation A-1 is used to calculate the corresponding misalignment

Xi j A k k L t 1 2 t 1 2 t Z

i j

Xi j 1

.0001

Xi j 0

0.001


28

One recognizes that the transformation A tends to enhance the effect of the error of the 's. This is due to the fact, that the effect of the angle misalignment on is small. Therefore, a relatively small error of produces a large error in the angle. The result on the offset, however is not independent of the the angle-component of the kick because of coupling be the presence of a large inside the test quadrupole. This can be seen more easily by reansforming the 's into a base which is in (anti) phase and orthogonal to the test quad

pi j

1 Zi j 0

cos t 1 2 Z

i j 1

cos t 2

ai j

1 Zi j 0

sin t 1 2 Z

i j 1

sin t 2

Yi j

pi j

ai j

The transformation A then simplyfies to As A s2 k

k k

1

t

k L sin k L

0

1

t

t

k L sin k L

k t

k L sin k L

We now recognize that the t couples the offset value to the value of the orthogonal kick a whereas the angle offset depends only on a

Xi j 1

.0001

Xi j 0

0.001

Xi j A k k L t

2 t 1 1 t Y

i j


29

The following fix is proposed: The values for .a should be small since the effect of the angle is small. Therefore the measurement value of .a is dominated by the measurement error. It is therefore proposed to set .a to zero.

Yci j

Yi j 0

0

Xi j A k k L t

2 t 1 1 t Yc

i j

We know recalculate the offset values:

Xi j 1

.0001

Xi j 0

0.001

The error on x by neglecting the effect of the x' ís

x t

2k L2 sin k L

2

k t k L sin k L

2

xp

x

x This is small as long as x.p is small


30

Conclusion:

Errors due to thin lens treatment are in the order of 20%

Errors of the same order of magnitude are produced by a

Gradient error of 1% in the analysis. Angle errors can become quite large.

Facit: I see no reason why not to trust BBA measurement

To a level of 20% or better!

errors in beam based alignment measurements

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