ivan bazarov, single hom two-pass analysis, srf mtg, 4 june 2003 1 chess / lepp single hom, two-pass...

Ivan Bazarov, Single HOM two-pass analysis, SRF mtg, 4 June 2003 1CHESS / LEPPCHESS / LEPP

Single HOM, two-pass analysis

y

z

y

x

E B

injectedbeam

2nd passdeflected

beam

TM110

1

10

100

1000

10000

100000

4.1 4.2 4.3 4.4 4.5 4.6 4.7

recirculation time [ns]

thre

sho

ld [

mA

]

theory

simulation

Motivation:

explain the plot produced by BBU code bi

R/Q = 100 OhmQ = 10000m12 = 10–6 m(c/eV) = 2 2 GHzt0 = (1.3 GHz)–1


Frequency of the instability• Dipole mode is excited by first current moment thru interaction with

longitudinal field of the mode

• Infinite number of bunches with finite number of passes (as opposed to finite number of bunches with infinite number of passes for BBU in storage rings)

• Potentially, any frequency can be present in FT of the current moment for infinite delta-function current train

• Instability occurs with frequency close to that of HOM, where impedance is maximal

xI

t

0t

0n0)1( n 0)1( nFT

)sin()/(2

1)( 2 tekQRtW Q

t

')'()'( '' dtetWiZ ti

2

2

'1)/(

2)'()'(

Q

ikQR

ixIV


Getting the master equation

t

dttxtIttWtV ')'()'()'()( )2()2(

t

rr dtttVmc

ettIttWtV ')'()'()'()( 12)1(

m

mtttItI )()( 000)1(

Sample solution: tieVtV 0)(

Summing geometric series 1)cos(2

)sin(1

02

0

t

tKe ti

0

0

20012 ,2

)/( tiQ

t

eekQRtIm

c

eK


Perturbative approach

Solve the master equation for instability frequency treating K as small parameter:

...2210 KK

The frequency up to the first order in K:

Kt

ee

Qi

t

n Q

t

tir

r

0

2

0 22

2

Requiring Im() = 0 yields famous

Q

t

tQkmQRe

cI r

rth 2

exp)sin()/(

2

12

)1(

Problem 1: Half solution is missing

Problem 2: Unphysical exponent


4.2 10 -94.3 10 -94.4 10 -94.5 10 -94.6 10 -94.7 10 -9

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1

Second order perturbative term

The second order term is found to be:

2222

0

20

0

2

0 )1(4

)]1([

22

20

0

Keet

ettieK

t

ee

Qi

t

ntiti

tir

Q

t

Q

t

ti

r

rr

r

Im() = 0 yields quadratic equation for the threshold current

)]2(sin[)1()]2(sin[

)2sin()cot()2()(sin)1()(cos))(sin(sin(

)/(

4

00

0022

02

12

)2(

0

00

rtt

r

rrrQt

rQt

rQ

t

rth

tttt

tttttttte

mtQkQRe

cI

r

r

1st order2nd order

rt

thI

Observation: Clearly, the other half of the solution is not a 2nd order effect


Complex current approach

Solving master equation directly for current gives the following:

)cos(2)sin()/(

20

22)(2

00120

0

0

0

0

0

0

teeeeeettmkQRe

cI tiQ

t

tiQ

t

ttiQ

t

r

Re(I0)

Im(I0)

solution space


Max and min currents

-750 -500 -250 250 500 750

-750

-500

-250

250

500

750

)sin()/(

])cos(1[4

0120

0

tmtkQRe

tcIout

-0.04 -0.02 0.02 0.04

-0.04

-0.02

0.02

0.04

12)/(

2

mQkQRe

cI in

)1)cos((2...max ,)cos()cos(2... 0001

20

ttt

Q

t


Obtaining complete first order solution

In the following limit (HOM damping is small of timescale of t0)

(instability frequency shift is small compared to bunching frequency, or as seen later, equivalent

to

number of bunches in recirculating loop >> 1)

Solving Im(I0) = 0, yields the threshold and instability frequency

12

0 Q

t

1)( 00 tt

Qie

mkQRe

cI rtti

2)/(

4 )(

120

0

)cot(2

,)sin()/(

2

120

r

r

tQ

tQkmQRe

cI

Note: It’s sin(tr) not sin(tr)Unphysical exponent is gone.


4.2 10 -94.3 10 -94.4 10 -94.5 10 -94.6 10 -94.7 10 -9

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1

Linearized solutions for instability frequency

Transcendental equation for instability frequency can be linearized for two important cases:

)cot(2 rtQ

rrt ,2look at solutions closest to

r

r

tQ

tΔωQ

/2 ,12

r

r

r t

t

tΔωQ

2 ,12

2

120 21

)/(

2

Q

QkmQRe

cI

rt0rt0rrr ttt


Comparison with tracking

10

100

1000

10000

100000

4.05 4.1 4.15 4.2 4.25 4.3 4.35 4.4 4.45 4.5 4.55 4.6 4.65 4.7


thre

sh

old

[m

A]

bi code

solution 1

solution 2

-1.5

-1

-0.5

0

0.5

1

1.5

4.05 4.15 4.25 4.35 4.45 4.55 4.65

recirculation tim e [ns]

Del

ta o

meg

a [G

Hz]

solution 1

solution 2


-0.1 -0.05 0.05 0.1

-0.1

-0.05

0.05

0.1

-40 -20 20 40

-40

-20

20

40

zoomed in

Solving Im(I0)=0 numerically

with increasing tr


Comparison of tracking with numeric solution of Im(I0)=0

10

100

1000

10000

100000

1000000

3.9E-09 4E-09 4.1E-09 4.2E-09 4.3E-09 4.4E-09 4.5E-09 4.6E-09 4.7E-09 4.8E-09

recirculation time [s]

thre

shol

d [m

A]

bi code

solution 1

solution 2

-1

-0.5

0

0.5

1

1.5

2

3.9E-09

4.0E-09

4.1E-09

4.2E-09

4.3E-09

4.4E-09

4.5E-09

4.6E-09

4.7E-09

4.8E-09

recirculation time [s]

Del

ta o

meg

a [G

Hz]

solution 1

solution 2


Large tr (r << /2Q)

0

20

40

60

80

100

120

140

160

180

200

4030.1 4030.2 4030.3 4030.4 4030.5 4030.6


thre

sho

ld [

mA

]

complex current method

1st order perturbation in Omega

tr does not matter as opposed to small accelerators case, threshold is approx. given by Iin

-1.5

-1

-0.5

0

0.5

1

1.5

4030.1 4030.2 4030.3 4030.4 4030.5 4030.6


De

lta

om

eg

a [

MH

z]


A word on quad HOM BBU

ede xzWeqdsFx )( : dipole

teetyxeeqe yyxxyxzWeqdsFxee 00

2220

20 2)(2 : quad

coupling term

Wake functions are identical in the form, except for the loss factor difference

44,

22,

or

or

bkk

bkk

qloss

dloss

In the approximation that alignment error of cavity transverse position dominates and causes dipole-like BBU (b is beam pipe radius)

2

2

,,2b

II dthqth , i.e. ~ 2 orders of magnitude bigger

ivan bazarov, single hom two-pass analysis, srf mtg, 4 june 2003 1 chess / lepp single hom, two-pass...

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