Chapter 4:

RF/Microwave interaction and beam loading in SRF cavity

4.1 RF field in SRF cavity

4.2 Beam loading

4.3 Dynamic detuning (microphonics, Lorentz force detuning, etc)

4.4 Basics on RF control

-develop equivalent circuit for rf system, cavity and beam

-develop equations for steady state and transient

-develop concept for the LLRF control

RF circuit modeling

RF components

RF source; klystrons are the most popular devices for f>300MHz.

tetrode, solid state amplifier for low power and/or low frequency

RF transmission; Waveguides or coaxial cables

Circulator; usually used as an isolator with matched load to protect RF source

Power coupler; feed RF power to a cavity

Cavity; electro-magnetic energy storage device

RF control; control cavity field and phase

Fundamental Power Coupler

HOM

Coupler

(HOMB)

HOM

Coupler

(HOMA)

Field

Probe

High Voltage

Power Supply

Klystron

Waveguide

Network

Waveguide/couplers

Low-level rf

EPICS user interface

Control/monitoring

Transmitter

HVPS

High power RF circuit

Low-level RF circuit

Main RF

Amplifier

Transmission

line

LO

LO

LO

I&Q

or

A&P

I&Q (or A&P)

Timing system &

Synchronization

transmitter

AND…

Protection system

(machine & personal)

Diagnostics &

user interfaces

Load

Resonance detection

& control

~

Main RF

Amplifier

Transmission

line

First, main high power RF circuit and cavity responses without beam.

Equivalent circuit (will use effective quantities for the modeling)

Dummy

Load

Le Ce re

Va

~ Beam, Ib

Z0 1:n

gI

Due to the circulator, this is not an exactly equivalent for generator current.

So introduced

gI twice of equivalent generator current.

Power coupler

Circulator

Transmission

line

(with beam later on)

Covert the model to the cavity side

~ Le Ce re

Va

~ Beam, Ib

n

II

g

g

Zext’=

n2Z0T2

Le Ce re

Va

ΩTR

P

V

P

TV

P

TLEr

W 2R

V

R

V

r

TV

r

V

2r

VP

2

sh

c

2

a

c

2

0

c

2

0sh

2

0

sh

2

0

sh

2

0

sh

2

a

e

2

ac

defined in Chap. 2

L C R

V0

~ L C R

V0

n

II

g

g

Zext=

n2Z0

Remember that equivalent circuit

parameters are defined by

references (V0, Va).

Coupling between cavity and the transmission line through a coupler, b

β

rZ,

β

RZ

Zn

R

Z

R

TZ

RT

Z

rβ e

extext

0

2

ext

2

ext

2

ext

e

Coupling factor b

,U/PωQ ex0ex c00 U/PωQ

impedanceshuntloadedeffective:β1

rr

r

β

r

1

r

1 eL

eeL

QLoaded:β1

QQ

Q

1

Q

1

Q

1)PU/(PωQ 0

L

0exL

cex0L

As it should be, coupling factor and Q’s are not function of particle velocity.

It is function of coupler geometry at a given mode (field profile).

That means Qex will be different when there’s a field tilt (field flatness).

impedanceshuntloaded:β1

RR

R

β

R

1

R

1L

L Similarly we can define

Governing equation for RF field in a cavity

~ Le Ce

Va

gI rL

IrL ILe ICe

geLrC LeIIII

eaL

Lar

aeC

dt/L

/r

C

e

L

e

VI

VI

VI

g

e

a

ee

a

eL

aga

e

a

L

aeC

1

CL

1

Cr

1

L

1

r

1C IVVVIVVV

If we use the equivalent circuit with V0,

rL should be replaced with RL

We can eliminate non-practical parameters (Ce, Le, C, L) using the relations:

L0

0

L

e0

LLe0

L

2

a

2

ae

0

cex

0L CRω

Lω

R

Lω

rrCω

r

V

2

1

VC2

1

ωPP

UωQ

g

L

L0a

2

0a

L

0a

Q

rωω

Q

ωIVVV

g

L

L0a

2

0a

L

0a

Q

rωω

Q

ωIVVV

Steady state solution with RF only

Generator current is the only source generator induced voltage Vg=Va

Particular solution in steady state of second order differential equation

ψ)ti(ω

aa eV(t) Vtiω

gg eI(t) Iat

angledetuning:ψ

1 when δ,2Qf

Δf2Q

ω

ωω2Q

ω

ω

ω

ωQtanψ

0)Δfif,eIr(eIψtan1

reI

ω

ω

ω

ωQ1

r(t)

L

0

L

0

0L

0

0L

tiω

gL

ψ)ti(ω

g2

Lψ)ti(ω

g2

0

02

L

La

V

Typical damped driven oscillator equation

Phasor representation

In general, fields can be expressed as

phase:θωtamplitude,:A,Ae )ti( A

To have the total voltage we need to add/subtract generator current/voltage

and beam current/induced voltage. Linear superposition works from the

linearity of Maxwell’s equations. But one should take the relative phase into

account.

If we choose a frame of reference that is rotating at a frequency , the phasor

will be stationary in time.

References can be arbitrary but it is convenient to have:

Reference frequency : operating frequency (rf source frequency) since all

other fields are around operating frequency.

Reference phase: beam arrives at the electrical center of cavity zero phase

(or real axis). How can we represent ‘beam’ at the reference frequency?

1

x

ψcosψtan1

1x

2

tiωiψ

gL

ψ)ti(ω

g2

La eecosψIreI

ψtan1

r(t)

V

common rotating term

relative phase

change due to

detuning

Amplitude

decrease

due to

detuning

iψshiψeiψ

L

g

atot ecosψ

β)2(1

recosψ

β1

recosψr

(t)

(t)Z

I

V

Total impedance of the equivalent model including detuning without beam

0.00E+00

2.00E-01

4.00E-01

6.00E-01

8.00E-01

1.00E+00

1.20E+00

-6000 -4000 -2000 0 2000 4000 6000-1.00E+02

-8.00E+01

-6.00E+01

-4.00E+01

-2.00E+01

0.00E+00

2.00E+01

4.00E+01

6.00E+01

8.00E+01

1.00E+02

-6000 -4000 -2000 0 2000 4000 6000

Va/(rLIg)

f0-f

Ex) QL=7105, f=805 MHz

plot the normalized Va and the detuning angle as a function of cavity detuning

f0-f

Bandwidth at -3bB: 10log10(P/Pref) for power, 20log10(V/Vref) for voltage

20log10(1/sqrt(2))=-3.01 =/4

Half width at -3dB: 1/2= 0/(2QL)=2575 Hz in this example

(time constant of loaded cavity)=1/1/2=2QL/0=277s

f, f/f=QL

RF power without beam loading

As mentioned,

‘due to the circulator, this is not an exactly equivalent for generator current.

So introduced Ig* twice of equivalent generator current.’

To calculate forward power in the transmission line (waveguide or coaxial cable)

8β

rI

β

r

2

I

2

I

2

1P

isgenerator thefrom lineion transmissin thepower forward averaged time theSo,

cavity in the voltageinducedgenerator with confused bet Don'

power. forward actual toscorrespond thisβ),/(r/2)(:Voltage

/2:current Forward

e

2

gegg

g

g

egfor

gfor

V

IV

II

β1

rr,ecosψIreI

ψtan1

rwith e

L

iψ

gL

iψ

g2

La

V

c

2

a0

0

2

ashshe

22

c2

e

22

a

e

2

g

g

a

P

VQ

Uω

Vr,r2r ψ)tan(1

4β

β)(1P

ψcos

1

2r

1

4β

β)(1V

8β

rIP

:is Vget toneeded power' Forward' calculatecan One

This is a useful equation when Pc & b are well defined, as for normal conducting cavity.

loading) beamwithout ( ag VV

forV

refV

phasein,β2

1β

resonanceon voltageinducedgenerator :1β

rr

2β

r

rg,

for

ge

gLrg,

ge

for

V

V

IIV

IV

loading) beamwithout ( ag VV

rg,V forV

refV

rg,V

b>1 b<1

In superconducting cavity, more practical parameters are QL, r/Q, Va (or V0), f1/2

since Q0 is much bigger than Qex, Pc is not well-defined, etc.

ψcos

1

(r/Q)Q

1

4

1V

ψcos

1

r

1

8

1Vψ)tan(1

r

β)(1

8β

β)(1V

8β

rIP

:is Vget toneededpower Forward

2

L

2

a2

L

2

a

2

e

2

a

e

2

g

g

a

relation. same themakes Q ingcorrespond and P ofset Any

)2Q

f(f ,

f

Δf

f

Δf2Q tanψearlier, defined asAnd

QQ

r

2

1r

β1

r

Q

r2

Q

1β

1β

2r

Q

2r

Q

r

Uω

P

P

V

Uω

V

Q

r

this.)of validity check the ,assumption thisuses one(when β.1β 1β

xx

L

01/2

1/20

L

LLe

L

L

0

e

0

e

0

shc

c

2

a

2

a

2

1/2L

2

o

2

1/2L

2

22

o

2

1/2L

2

ag

f

Δf1

(R/Q)Q

V

4

1

f

Δf1

Q(R/Q)T

TV

4

1

f

Δf1

(r/Q)Q

V

4

1P

cosψ(r/Q)QP2cosψ(r/Q)QP2cosψIrIψtan1

rxxLggLg

2

La

V

Don’t be confused with other passive couplings.

Quiz) when we measure Ea through field probe, how?

gV

resonanceonissystemthewhen

fieldcavityinducedGeneratorr gLrg, IV forV

refV

When b >>1

reffor

gL

ge

ge

for2

r

1)2(β

r

2β

r

VV

IIIV

iψ

gLag ecosψIrloading) beamwithout ( VV

HOMEWORK 4-1

for f=805 MHz, Ea=10MV/m, L=0.68m, r/Q=279, and

QL=7105, QL=1106, QL=2106,

1. Plot required forward power as a function of detuning (-500 Hz~500 Hz)

using spreadsheet 4_1.xlsx

2. If Q0=11010, What is cavity wall loss, Pc? What does that mean?

Equivalent beam in a RF circuit model

Micro-pulse Bunch spacing=Tb

RF frequency/n

When we say ‘Beam current’, it is an time averaged DC current.

Ex) Ib0=40 mA CW beam at bunch spacing 402.5 MHz

Tb=1/402.5 MHz~2.5ns,

Q (charge per bunch)=Ib0 (C/s) x Tb (s)=0.04 x 2.5e-9 = 100 pC

Temporal distribution of beam can be described by a Gaussian distribution with

standard deviation t

2t

2

2σ

t

t

bunch eσπ2

Q(t)I

t

Ibunch(t)

Ipeak

If t is 1.0 degree of 402.5 MHz (7 ps),

Ipeak~5.7 A t

Ibunch(t)

Ib0

b

b

2

t

2

b

2

b0

2

t

2

b

2

b

tpeakn

1n

bn0

2

t

2

t

bunch

T

2πω

3,2,1,0,n),2

σωnexp(2I)

2

σωnexp(

T

σ2π2Ia

,t)cos(nωa2

a)

2σ

texp(-

σπ2

Q(t)I

Fourier decomposition from –Tb/2 to Tb/2

1)2

σωnexp(, σ

nω

1If

2

t

2

b

2

t

b

Ex) Bunch spacing=402.5 MHz, Operating RF frequency=805 MHz:

So the Fourier component (n=1, 2, 3,…) of bunched beam at operating frequency

is simply 2Ib0.

0

0.2

0.4

0.6

0.8

1

1.2

0.4025 2.415 4.4275 6.44 8.4525 10.465 12.4775 14.49 16.5025 18.515

f (GHz)

Fo

uri

er

Co

mp

on

en

t E

xp

on

en

tial T

erm

Va

Steady state with beam loading

~ Le Ce re ~ Beam, Ib

n

II

g

g

Zext’=

n2Z0T2

- We added ‘beam’ as a current source like the RF generator. The beam

energy effect is in the effective quantities.

- Beam current at the operating frequency is

beamofcurrentDC:I,2I b0b0b I

- The cavity voltage is the sum of generator induced voltage and beam

induced voltage in a cavity.

bgabg

L

L0a

2

0a

L

0a )(

Q

rωω

Q

ωVVVIIVVV

-As mentioned earlier, set the reference phase for beam phase at the center of

electric center beam induced image current sits on negative real axis.

-Beam induced voltage has the same form as generator induced voltage.

ψ)i(π

bL

ψ)i(π

b2

Lb

iπ

bb

ecosψIreIψtan1

r

eIcurrent,inducedbeam

V

I

bIbLr b, r IV

bV

0

L

0

0L

0

0L

f

Δf2Q

ω

ωω2Q

ω

ω

ω

ωQtanψ

In this example is positive,

Which means the resonance

frequency is higher than

generator frequency.

bIbLr b, r IV

bV

Generator induced voltage is about same as before for the RF only case

except relative phase of generator current.

Let’s start with arbitrary phase first.

ψ)i(θ

gLg

iθ

gg

ecosψIr

eI

V

I

gLrg, r IV

gV

aV

The black circle is rotating around the origin

when generator phase (RF phase) changes.

The real component of Va is for acceleration: cosaV

bV

bIbLr b, r IV

bV

gV

bVaV

gLrg, r IV

-To get required accelerating voltage Va and synchronous phase for a beam

current Ib at a fixed cavity detuning () and Loaded Q (QL),

the generator current Ig (RF power and phase) is uniquely determined.

bIbLr b, r IV

bV

gV

bVaV

gLrg, r IV

How about forward voltage (Vfor) and reflected voltage (Vref) of the system with b>>1

forVrefV

bga

refforrefforreffora . and from calculateddirectly becan PandP :

VVV

VVVVV

Generator power

2

a

Lg

2

a

Lg

L

2

ag sinV

rItanψcos

V

rI1

r

1

8β

β)(1VP

For SRF cavities where b>>1, using the relations

2

b

L

1/2

2

b

L

L

2

ag

ab0b

a

b0ab0

b

b0bLL

tanQ

Q

f

Δf

Q

Q1

4(r/Q)Q

VP

cosVIP,cosV

(r/Q)I

ωU

cosVI

Q

1,2II,Q

Q

r

2

1r

Optimum cavity detuning (opt) and Loaded Q (QL),

cosIVP Δfat &QQ If

detuningoptimum:tan2Q

fΔftan

Q

Q-

f

Δf If

boagoptbL

b

0opt

b

L

1/2

No reflected power

statesteady in PPPP balancePower refbcg

HOMEWORK 4-2, Play with the spreadsheet

Ex) Using parameters in the table, (at particle beta=0.61) calculate Va, Qb, and

optimum f, and

generate table of required RF power as a function of f (-1000 Hz ~ 1300 Hz)

for QL1=3e5, QL2=7e5, QL3=1e6.

r/Q(at b=0.61)= 279 Ohm

TTF (at b=0.61)= 0.68

Ib0= 0.04 A

Syn Phase= -15 degree

E0= 15 MV/m

Length= 0.6816 m

QL1= 3.00E+05

QL2= 7.00E+05

QL3= 1.00E+06

f= 8.05E+08 Hz

0

100

200

300

400

500

600

700

-1500 -1000 -500 0 500 1000 1500

delta f (f0-f) in Hz

RF

po

we

r

QL3

QL2

QL1 fopt

Pb

(4_2.xlsx)

change parameters in blue

HOMEWORK 4-3, Play with the spreadsheet

Ex) Using parameters in the table and TTF data,

generate (r/Q), Qb, RF power required, optimum f, required RF power at

optimum f as a function of particle velocity from b=0.5 to b=0.8.

r/Q(at beta=0.61)= 279 Ohm

Ib0= 0.04 A

Syn Phase= -15 degree

E0= 15 MV/m

Length= 0.6816 m

QL= 7.00E+05

delta f= -300 Hz

f= 8.05E+08 Hz

Transit time factor

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.45 0.55 0.65 0.75 0.85

beta

T (from SF)

T(fit)

TTF=1540.22897*b^6 - 6951.52591*b^5 + 12957.19469*b^4 - 12724.40926*b^3 + 6910.19655*b^2 - 1956.643*b + 224.88436

(4_3.xlsx); change parameters in blue

0

50

100

150

200

250

300

350

0.45 0.55 0.65 0.75 0.85

beta

r/Q

(O

hm

)

0.0E+00

5.0E+05

1.0E+06

1.5E+06

2.0E+06

2.5E+06

0.45 0.55 0.65 0.75 0.85

beta

Qb

0

20

40

60

80

100

120

140

160

180

200

0.45 0.55 0.65 0.75 0.85

beta

Op

t. d

etu

nin

g (

Hz)

0

50

100

150

200

250

300

350

0.45 0.55 0.65 0.75 0.85

beta

Po

wer

(kW

)

Required RF

at -300 Hz detuning

Required RF

at optimum detuning

Transient behavior without beam

g

L

L0a

2

0a

L

0a

Q

rωω

Q

ωIVVV

2t/τ

L

2

refref

t/τ

Lgrefareffor

LggLforrg,

t/τ

Lg

t/τ

gLa

tiω

gga

tiω

gg0

)e2(1P2r

(t)V(t)P

)e2(1(r/Q)QP(t)V

(r/Q)QP/2IrVV2

1

)e(1(r/Q)QP2)e(1Ir(t)V

eI RFon turn & 0(0) 0,at t 1)

axis. realon are all ,eIset weIf phase.in are All ω.ω case, resonanceOn

g

VVV

IV

I

Steady state

term

Transient

term

02

L

01

1/20

L

ψ)ti(ω

gL

tiω

2

tiω

1τ

t

a

ω4Q

11ωfrequency) resonance (systemω,

ω

1

ω

2Qconstant) (time τwhere

ecosψIr)ece(ce(t)V 11

General Solution:

initial condition problem

)/τt2(t2/τt

grefemit

a

0

a

0

2

a

0

2

a

ref

c1

brefcg

1

)/τt(

1aa

i/τt

Lg1a1

11

1

1

e)e(14P)P(P

dt

(t)dV

(r/Q)ω

(t)2VdU/dt

(r/Q)ω

VU

Uω

V

Q

r

example in thisdU/dt (t)P

1)(β negligible is P and at t off turnedis RF example, In this

dU/dtPPPP :balancepower General

t t,)e(t(t)

RF off turn & )ee(1(r/Q)QP2)(t ,tat t 2)

tVV

V

The direction of this power after RF off is same as that of reflected power. This

power is a release of stored energy from cavity. The mechanism is different.

This power is called ‘emitted power’.

Some useful information can be taken just from decay curves of the Va and

Pemit after RF off.

shortlyit at look will weamount; detuning

QPemitor Vaeither ofdecay at fit with

property).cavity :(r/QUω

V

Q

r using calculated becan Va U,know when we

,energy)(storedUPemit(t)dt

L

0

2

a

t1

(transient1.xls); change parameters in blue

Pg= 60 kW

r/Q= 279 Ohm

QL= 7.00E+05

t1= 0.0015 s

f= 8.05E+08 Hz

Ex)

0

1

2

3

4

5

6

7

8

0 0.001 0.002 0.003

Time (s)

Va

(M

V)

0

50

100

150

200

250

0 0.001 0.002 0.003

Time (s)

P (

kW

)

Forward P

Reflected P

Emitted P

Area=

stored energy

)/τt2(t 1e

)/τt(t 1e

tiω

tiω

aa

e(t)I(t)

,e(t)V(t)

I

V

Transient behavior with detuning & beam loading

-For general expressions including time-varying detuning and beam loading,

the system equation will be developed.

-Using finite difference method, cavity behaviors will be explored.

-First, separate out fast rotating terms and build up model in vector space (real

& imaginary)

bg

L

L0a

2

0a

L

0a where,

Q

rωω

Q

ωIIIIVVV

Complex envelope

(slowly varying)

RF term

(fast oscillation)

)/(2Qωω,ωωωΔ where

)IiωI(ω2rV)Δω()iωω(Δ2ωV)iω2(ωV

L01/20

1/2a

2

1/2a1/2a

L

IωrV)iωωi(Δ-VV2iω

11/2a1/2aa L

IωrV)iωωi(Δ-V 1/2a1/2a L

arrange and above,equation in the iIII,iVVVinsert iraiara

i1/2ai1/2arai

r1/2aiar1/2ar

IωrVωωVΔdt

dV:Imaginary

IωrωVΔVωdt

dV : termReal

L

L

System Equation for SRF cavity: First order ordinary differential equation.

Initial condition problem

This equation set describes SRF cavity field in complex space.

Using following relations sets of power and voltage can be calculated.

reffora

rirr

fifr

aiar

b0b

L

2

for

gL

for

bg

LL0L01/2

power reflected & phase votage,reflectedV,V

power forward & phase votage,forwardV,V

phase and tagecavity volV,V

2I,r2

P,2

r

:source driving

input too.an becan in time ωΔ yingslowly var

(r/Q)2

Qr ω,-ωωΔ ),/(2Qωω :constantInput

VVV

IV

IV

III

g

ti,1/2t12,tai,1/2t11,tar,t22,

tr,1/2t12,tai,t11,tar,1/2t21,

ti,1/2tai,1/2tar,t12,

tr,1/2tai,tar,1/2t11,

t22,t12,tai,Δttai,

t21,t11,tar,Δttar,

Iωr)K(Vω)Kω(VΔtK

Iωr)Kω(VΔ)K(VωtK

)IωrVωωVΔt(K

)IωrωVΔVωt(K

)/2K(KVV

)/2K(KVV

(FDM) method difference finite using way simple One

L

L

L

L

(filling_test_FDM.xls); change parameters in blue

Ex) RF only RF power 60000

Phase1 0

F 8.05E+08

w 5.06E+09

QL 7.00E+05

r/Q 279

Cl 0.6816

rL 97650000

w1/2 3.61E+03

detuning end of pulse 300

slope Hz/ms 0

filling time us 3.00E+03

dt 2.00E-06

0.0E+00

1.0E+06

2.0E+06

3.0E+06

4.0E+06

5.0E+06

6.0E+06

-4.0E+06 -2.0E+06 0.0E+00 2.0E+06 4.0E+06 6.0E+06

0.00E+00

1.00E+06

2.00E+06

3.00E+06

4.00E+06

5.00E+06

6.00E+06

7.00E+06

0 1000 2000 3000 4000

forV

refV

loading)beamnoV(or gaVaV

0.00E+00

5.00E+00

1.00E+01

1.50E+01

2.00E+01

2.50E+01

3.00E+01

0 1000 2000 3000 4000

aofPhase V

0.00E+00

1.00E+06

2.00E+06

3.00E+06

4.00E+06

5.00E+06

6.00E+06

7.00E+06

8.00E+06

-2.00E+06 0.00E+00 2.00E+06 4.00E+06 6.00E+06 8.00E+06

HOMEWORK 4-4)

Generate these three plots Va at various

detuning using filling_test_FDM.xlsx (RF only)

0.E+00

1.E+06

2.E+06

3.E+06

4.E+06

5.E+06

6.E+06

7.E+06

8.E+06

0 500 1000 1500 2000 2500 3000

amp

litu

de

of V

a

Time (us)

0

20

40

60

80

100

120

0 500 1000 1500 2000 2500 3000

Ph

ase

of V

a

Time (us)

Hz0Δf

Hz300Δf

Hz600Δf

Hz0021Δf

Hz0036Δf

Hz0036Δf

Hz0021Δf

Hz600Δf

Hz300Δf

Hz0Δf

Hz0Δf

Hz300Δf

Hz600Δf

Hz0036Δf

Hz0024Δf

Hz0015Δf

Hz900Δf

aiV

arV

-2.00E+02

-1.50E+02

-1.00E+02

-5.00E+01

0.00E+00

5.00E+01

1.00E+02

1.50E+02

2.00E+02

0 1000 2000 3000 4000

Ex) pulsed operation. RF pulse length 1500 s, detuning -300 Hz

other parameters are same as in the previous example

0.00E+00

1.00E+06

2.00E+06

3.00E+06

4.00E+06

5.00E+06

6.00E+06

7.00E+06

0 1000 2000 3000 4000

0.00E+00

4.00E+04

8.00E+04

1.20E+05

1.60E+05

2.00E+05

0 1000 2000 3000 4000

(filling_test_FDM.xls)

change parameters in blue

aV

aofPhase V

Reflected power

d/dt=

0.00E+00

4.00E+04

8.00E+04

0 1000 2000 3000 4000

-1.60E+02

-1.40E+02

-1.20E+02

-1.00E+02

-8.00E+01

-6.00E+01

-4.00E+01

-2.00E+01

0.00E+00

0 1000 2000 3000 4000

0.00E+00

1.00E+06

2.00E+06

3.00E+06

4.00E+06

5.00E+06

6.00E+06

0 1000 2000 3000 4000

(pulse_beam_test_FDM.xls)

change parameters in blue

Ex) open loop with beam loading

RF power (W) 60000

Vfor phase (degree) 2

f (Hz) 8.05E+08

w 5.06E+09

QL 7.00E+05

r/Q (Ohm) 279

Cl (m) 0.6816

rL 97650000

w1/2 3.61E+03

rL*w1/2 3.53E+11

detuning end of pulse -300

slope Hz/ms 0

rf pulse length us 2.00E+03

Ib0 (A) 2.50E-02

beam enters at t (us) 500

bPhase (degree) 0

dt 2.00E-06 -6.00E-02

-5.00E-02

-4.00E-02

-3.00E-02

-2.00E-02

-1.00E-02

0.00E+00

0 1000 2000 3000 4000

aV

aofPhase V

Reflected power

-Ib=-2Ib0

d/dt=

RF control

To have a beam with a required quality (emittance, energy spread, etc.) cavity

field amplitude and phase should be maintained within a certain (machine

specific) ranges.

For example, <1% in amplitude and <1 degree in phase are typical values.

In modern digital LLRF systems, feedback control is an essential part and feed

forward control becomes more popular.

LLRF control system should provide required cavity field stability against;

Beam loading (transient including jitter and/or CW)

Beam fluctuations

HPRF droop/ripple (mainly from HVPS)

Dynamic cavity detuning (Lorentz force detuning, microphonics)

Loop delay of RF control system

Reference RF phase/amplitude fluctuations

Electron loading conditions in a cavity and/or at around a power coupler

Any changes that affects characteristics of control system

matching condition in HPRF transmission line

thermal drift (electronic board, cables, etc.)

HVPS

Field detection and control: digital I/Q mostly in modern control system

this uses conceptually same as the phasor relations we learned

Main RF

Amplifier

Transmission

line

LO

I&Q

transmitter

Load

Field Reference

RF

IF LPF

+

- errors controller

I: in-phase (corresponds real component),

Q: quadrant (corresponds imaginary component)

-3.00E+06

-2.00E+06

-1.00E+06

0.00E+00

1.00E+06

2.00E+06

3.00E+06

4.00E+06

5.00E+06

6.00E+06

7.00E+06

8.00E+06

0.00E+00 1.00E+03 2.00E+03 3.00E+03 4.00E+03

-3.00E+06

-2.00E+06

-1.00E+06

0.00E+00

1.00E+06

2.00E+06

3.00E+06

4.00E+06

5.00E+06

6.00E+06

7.00E+06

8.00E+06

0.00E+00 1.00E+03 2.00E+03 3.00E+03 4.00E+03

Ex) Feedback control example:

The real world has many complex practical issues. One can develop a

conceptual understandings of the rf control from this example.

(Feed_back_test_FDM.xls) change parameters in blue

Reference cavity field I

Reference cavity field Q

cavity field I

cavity field Q

Parameters used

In this example

Constant detuning

For reference

During filling

Feedback gains

f (Hz) 8.05E+08

w 5.06E+09

QL 7.00E+05

r/Q (Ohm) 279

Cl (m) 0.6816

Vcav MV 6952320

filling time (s) 4.00E-04

(1-exp(-1)) -1 1.5819767

rL 97650000

w1/2 3.61E+03

rL*w1/2 3.53E+11

detuning end of pulse 168

slope Hz/ms 0

rf pulse length us 1.30E+03

Ib0 (A) 2.60E-02

beam enters at t (us) 700

bPhase (degree) -15

proportional gain K 20

integral gain Ki 1

dt 2.00E-06

-5.00E+01

0.00E+00

5.00E+01

1.00E+02

1.50E+02

2.00E+02

0.00E+00 1.00E+03 2.00E+03 3.00E+03 4.00E+03

0.00E+00

5.00E+04

1.00E+05

1.50E+05

2.00E+05

2.50E+05

3.00E+05

3.50E+05

0.00E+00 1.00E+03 2.00E+03 3.00E+03 4.00E+03

0.00E+00

1.00E+06

2.00E+06

3.00E+06

4.00E+06

5.00E+06

6.00E+06

7.00E+06

8.00E+06

0.00E+00 1.00E+03 2.00E+03 3.00E+03 4.00E+03

-4.00E+01

-2.00E+01

0.00E+00

2.00E+01

4.00E+01

6.00E+01

8.00E+01

1.00E+02

0 1000 2000 3000 4000

Va amplitude

Va phase

Forward power

Reflected/emitted

power

Forward power

phase

Forward power

phase

RF control for constant field/phase:

One can say that it forces the system into steady state condition quickly and stably

0.00E+00

1.00E+06

2.00E+06

3.00E+06

4.00E+06

5.00E+06

6.00E+06

7.00E+06

8.00E+06

0.00E+00 1.00E+03 2.00E+03 3.00E+03 4.00E+03

Va amplitude

There are still errors with feedback control from many other sources.

the errors are repetitive, one can generate error tables including loop delay

information and apply to the next pulses. feed forward

Cavity detuning due to the Lorentz force (dynamic) Vibrations, resonances and damping

-Vibration source; RF pulse repetitive hammering by radiation pressure

with frequencies of repetition rate and harmonics

-Mechanical resonance frequencies (wn)

determined by the equivalent mass of each mechanical mode

& equivalent stiffness of the system

-Resonance; source term hits around the mechanical resonance frequency

-Damping; determined by the whole system

energy transfer; sound wave radiation, internal/structural damping,

helium, heat dissipation, transfer to other system thru propagation

-Vibration amplitude; determined by the relation between the damping and the

resonance.

generally amplitude is smaller at higher frequency and higher damping.

This will add initial frequency off-set.

-2000

-1500

-1000

-500

0

500

1000

-50 -40 -30 -20 -10 0 10 20 30 40 50

Rad

iati

on

Pre

ssu

re (

Pa)

axial coordinate (cm)

Radiation Pressure

Impulsive Forcing Function

(time domain)

Transformed Forcing Function

(Frequency Domain)

Radiation Pressure in time and frequency domain

(vibration source)

0 1000 2000 3000 Time (sec)

Cavit

y F

ield

Time

Cavity displacement

or cavity detuning

Cavity field Corresponds static KL

After transient disappear

CW

Pulsed mode

0.5 1 1.5 2

-2?10-9

-1?10 -9

1?10-9

2?10 -9

Static deformation

Dynamic vibrations. Reaches steady state

after transient period.

Mechanical modes of the system

lQ

l

VkQ

l

l

lcavllll

l

ll

modemechanical offactor Quality:

modemechanical offrequency resonance:

222

Resonance system

Mode, damping, & modal mass finding strongly

depends on boundary conditions and whole

mechanical system details.

Mode frequencies & Q of the mode can have

large spread. large error

l:kl

l

ltotal

modeofcoeff.detuningdynamic

,

Vibration due to LF

Measured transfer function by Lorentz force

Once mechanical responses (amplitude and phase) of one selected cavity are

measured, quite accurate prediction or reconstruction is possible for that

cavity.

But due to the sensitivity of mechanical mode characteristics, there is large

scattering. Sometimes unpredicted mode could be found.

-200

0

200

400

600

800

1000

0 500 1000 1500

Time (us)

Dy

na

mic

de

tun

ing

(H

z)

-200

-100

0

100

200

300

400

500

600

0 500 1000 1500

Time (us)

Dy

na

mic

de

tun

ing

(H

z)

Medium beta cavity (installed cavity)

KLDyn: 3~4 Hz/(MV/m)2

17 MV/m

High beta cavity (installed cavity)

KLDyn: 1~2 Hz/(MV/m)2

16.5 MV/m

Ex) SNS cavities

filling flattop

filling flattop

Unpredicted 1.6kHz component sits on nominal low frequency response in

medium beta cavities.

-200

0

200

400

600

800

1000

0 500 1000 1500

Time (us)

Dy

na

mic

de

tun

ing

(H

z)

-200

-100

0

100

200

300

400

500

600

0 500 1000 1500

Time (us)

Dy

na

mic

de

tun

ing

(H

z)

Medium beta cavity (installed cavity)

KL: 3~4 Hz/(MV/m)2

17 MV/m

High beta cavity (installed cavity)

KL: 1~2 Hz/(MV/m)2

16.5 MV/m

Observed detuning agrees with expectations

-200

0

200

400

600

800

1000

0 500 1000 1500

Time (us)

Dy

na

mic

De

tun

ing

(H

z)

15Hz

30Hz

60Hz

filling flattop

-400

-200

0

200

400

600

800

1000

0 300 600 900 1200 1500

Time (us)

Dyn

am

ic D

etu

nin

g (

Hz)

0.0E+00

2.0E+06

4.0E+06

6.0E+06

8.0E+06

1.0E+07

1.2E+07

1.4E+07

1.6E+07

1.8E+07

Ea

cc (

MV

/m)

Dynamic detuning

Eacc

In this example the accelerating gradient is 12.7 MV/m. (high beta cavity)

The 1.6 kHz components shows

resonances at higher repetition rate

in some of medium beta cavities

some cavities show bigger resonance phenomena

repetition rate dependent

Static driving forces are proportional to ‘square of cavity field’ (Lorentz force).

Dynamic responses could be quite different depending modal mass, modal

boundary conditions, driving force spectrums.

Low Qex and high beam loading structure, not a big issue (will need some extra

RF power).

Very important in pulsed machine especially in pulsed high Qex structure.

Generate steady state vibration pattern. Repetitive from pulse to pulse.

Counter vibration (compensation) for the biggest frequency components can

correct quite efficiently. Demonstrations have been done using piezo-electric

actuators.

Vibration due to PT

An input voltage is applied to the piezoelectric

actuator device which make the piezo stack

shrink/expand

piezo Piezo-stack

Piezo-voltage input

Shrink/expand

LFD compensation

Since the forcing mechanisms are

different between LFD and piezo

detuning system responses are not

same.

Building a virtual cavity for dynamic detuning

(complete set of modeling)

tdeIReVVt tdj

L

tdjt

t

t

0

~

2/1

~

0

~~~0

2 2 2mL,m L,m m L,m m L,m

m

L,m

RF 0 mod

SST L 0 mod mod mod

modmod

1/ 2

mod m2

L,m L,m

mod mod m

k VQ

k 0

P P (1 sin t)

V (t) R I (1 cos sin( t ))2

tan

1 cos cos(t) k V

sin( t / 2)

2 2mP,m P,m m P,m m P,m P

m

P osc osc

osc 1/ 2i osc

osc osc

j cos

n n n

n

n

n

i

k VQ

(t) sin t

; ; t

v( ) e P ( )cos cos(n )

P ( ) polynomes of

ntan

General RF eq.

of Cavity field

Dynamic detuning

Due to LF

Dynamic detuning

Due to Piezo tuner

Virtual Cavity

(analytic basis;

very fast and

provide general view

Verification of Virtual Cavity

Many useful and practical tools

; optimization of compensation for LFD

microphonics compensation study in high Qex cavity

ponderomotive oscillation study for RF system…

Ex. LFD compensation (optimization study)

LF only contains harmonics of the repetition rate.

Ideally, perfect compensation is possible.

Decompose the LFD into harmonics find corresponding piezo signal components

(I) Piezo. Tuner input voltage

(II) Generated detuning

(including transient)

(III) Detuning generated by the

Piezo. Tuner in SST

(IV) Its sum with the initially

Targeted portion of the

Lorentz Detuning

(I)

(II)

(III) (IV) < few Hz

Seems to be complex to apply for the real system

Practically applicable and straightforward compensation

scheme (simple harmonic compensation)

-Only the Lorentz

Detuning during the RF

turn-on transient and

the beam pulse is tried

to be compensated

-A single harmonic

seems sufficient to

obtain a satisfying

compensation

-The Piezo. Tuner input

voltage contains only

this harmonic

Simple waveform

Ex. Analysis with dynamic detuning (feed_back_test_FDM_dyn_detuning.xls)

Microphonics

There are always mechanical vibrations from environments.

These vibrations can shake cavities.

Responses of cavities are function of dynamic/modal characteristics of the

system (not only by cavity mechanical properties).

Qex is normally higher in low beam current machines. Cavity bandwidths are

getting narrower as Qex’s get higher.

If HPRF does not have enough margin, a cavity field may not reach its

operating setpoint.

Examples of vibration sources

helium pressure fluctuations

pumps (water, mechanical vacuum)

ground vibration including ocean waves (1/7 Hz)

traffics

etc.

Ex) Microphonics measurement at JLab

0

200

400

600

800

1000

1200

1400

1600

1800

2000

-10.3 -8.9 -7.5 -6.2 -4.8 -3.4 -2.0 -0.7 0.7 2.1 3.5 4.9 6.2 7.6 9.0 10.4 11.7

Occu

ren

ces

Frequency Deviation (Hz)

Beam current in proposed CW superconducting linacs is < several mA.

Qb is mid 107 range

If loaded Q QL is 5x107 (for RF efficiency), 1/2= 0/(2QL) will be in comparable

ranges of microphonics.

f=650 MHz f1/2=6.5 Hz

f=80.5 MHz f1/2=0.8 Hz

If 1/2 is too small,

a few bandwidth of cavity frequency variation will cause large cavity phase

variations, and/or cavity field can not be kept at operating point.

It may need to make stiff cavities to push mechanical frequencies as high as

possible so they don’t couple to the low frequency mechanical noise that has the

largest amplitudes. But required force of mechanical tuner should be in a

reasonable range.

Amplifier may need to have a certain amount margin that can cover detuned

cavity operation.

Active feedback/feed forward control may need to be used.