Lecture 8 - U.S. Particle Accelerator Schooluspas.fnal.gov/materials/03UCSB/Lecture08.pdf ·...

Microwave Physics and Techniques UCSB –June 20031

Lecture 8Resonators

A. Nassiri - ANL

June 19, 2003


zjcy

zjcx

zjz

z

z

z

z

eybn

xa

mA

bm

jE

eybn

xa

mA

an

jE

eybn

xa

mAH

E

β

β

β

π

ππ

ωµπ

λ−=

π

ππ

ωµπ

λ=

π

π

=

=

m

m

m

cossin

sincos

coscos

2

2

2

2

4

4

0

Field Expressions For TE Modes – Rec. WGField Expressions For TE Modes – Rec. WG

g

yx

EH

η= m

g

xy

EH

η±=

m,n = 0,1,2,… but not both zero


Field Expressions For TE Modes – Rec. WGField Expressions For TE Modes – Rec. WG22

21

+

µε=

bn

am

fc 222

+

=λ

bn

am

c

( ) ( )22 11 ffcc

g−

λ=

λλ−

λ=λ

( ) ( )22 1

1

1

1

ccpz

ff λλ−µε=

−µε=ν

( ) ( )22 11 ccg

ff λλ−

εµ=

−

εµ=η


Field Expressions For TM Modes – Rec. WGField Expressions For TM Modes – Rec. WG

zj

g

cy

zj

g

cx

zjz

z

z

z

z

eybn

xa

mA

bn

jE

eybn

xa

mA

am

jE

eybn

xa

mAE

H

β

β

β

π

ππ

πλλ

=

π

ππ

πλλ

=

π

π

=

=

m

m

m

m

m

cossin

sincos

coscos

2

2

0

2

2

g

yx

EH

η= m

g

xy

EH

η±=

m,n = 1,2,3,…


Field Expressions For TM Modes – Rec. WGField Expressions For TM Modes – Rec. WG22

21

+

µε=

bn

am

fc 222

+

=λ

bn

am

c

( ) ( )22 11 ffcc

g−

λ=

λλ−

λ=λ

( ) ( )22 1

1

1

1

ccpz

ff λλ−µε=

−µε=ν

( ) ( )22 11 ccg ff λλ−εµ

=−εµ

=η


Determine the lowest four cutoff frequencies of the dominant mode for three cases of rectangular wave guide dimensions b/a=1, b/a=1/2, and b/a =1/3. Given a=3 cm, determine the propagating mode(s) for f=9 GHz for each of the three cases.

The expression for the cutoff wavelength for the TEm n mode where m=0,1,2,3,.. and n=0,1,2,3,.. But not both m and n equal to zero and for TMmn mode where m=1,2,3,.. And n=1,2,3,.. is given by

22

22

1

+

=λ

bn

am

c

The corresponding expression for the cutoff frequency is 2

222

21

221

+

µε=

+

µε=

λν

=ba

nmab

na

mf

c

pc


The cutoff frequency of the dominant mode TE10 is . Hence µεa21

[ ]2

2

10

+=

ba

nmf

f

TEc

c

TE10

TE01

TM11

TE11

TE20

TE02

TM21

TM12

TE21

TE12

[ ]10TEc

cf

f1=ab

1 2 3 4 5 2 5

21

=ab

TE10

TE01

TE20

TE11

TM11

TE21

TM21

[ ]10TEc

cf

f1 2 3 4 5 5 8

31

=ab

TE10 TE20

TE30

TE01

TM11

TE11

[ ]10TEc

cf

f

1 2 3 4 5 10


Hence for a signal of frequency f=9GHz, all the modes for which is less that 1.8 propagate. These modes are:

TE10,TE01,TM11,TE11 for b/a=1

TE10 for b/a=1/2

TE10 for b/a=1/3

[ ]10TEc

cf

f

So for b/a≤1/2, the second lowest cutoff frequency which corresponds to that of the TE20 mode is twice of the cutoff frequency of the dominant TE10 . For this reason, the dimension b of the a rectangular wave guide is generally chosen to be less than or equal to a/2 in order to achieve single-mode transmission over a complete octave( factor of two) range of frequencies.


Rectangular Cavity ResonatorRectangular Cavity ResonatorAdd two perfectly conducting walls in z-plane separated by a

distance d.For B.C’s to be satisfied, d must be equal to an integer

multiple of λg/2 from the wall.Such structure is known as a cavity resonator and is the

counterpart of the low-frequency lumped parameter resonant circuit at microwave frequencies, since it supports oscillationsat frequencies foe which the foregoing condition, that is

d=l λg/2, l=1,2,3,… is satisfied.

a

bdx

y z


Rectangular Cavity ResonatorRectangular Cavity ResonatorSubstituting for and rearranging, we obtainλg

( )2

22

2

211

1

2

=

λ−

λ

λλ−

λ=

dp

pd

c

c

Substituting for givesλc222

2 2221

+

+

=

λ dp

bn

am 222

222

1

+

+

=λ

dp

bn

am

222

2221

+

+

µε=

λν

=dp

bn

am

f posc


Quality Factor QQuality Factor QThe quality factor is in general a measure of the ability of a resonator to store energy in relation to time-average power dissipation. Specifically, the Q of a resonator is defined as

wall

strPW

Q oω=π=per cycle d dissipateEnergystored energyMaximum2

mestr WWW +=Consider the TE101 mode:

+

µ=

π

π

πωµε

=ε

= ∫∫∫∫

116

44

2

22

2

0

2

00

222

d

aHabdW

dxdydzdz

ax

Ha

dvEW

e

abd

v ye

o

o sinsin

211

02

22

22101

2 adx

ax

da

a=

π

+

µεπ

=ω=ω ∫ sin and


The time average stored magnetic energy can be found as

dxdydzdz

ax

dz

ax

d

aH

dvHHW

abd

zv xm

π

π

+

π

πµ

=

+µ

=

∫∫∫

∫

222

0

22

2

00

2

22

4

4

sincoscossino

+

µ= 1

16 2

22

d

aHabdW m

o

Note that the the time-average electric and magnetic energies are precisely equal. This should be true in general simply follows from the complex Poyting’stheorem. Physically, the fact that energy cycles between being purely electric, partly electric and partly magnetic, and purely magnetic storage, such that on the average over a period, it is shared equally between the electric and magnetic forms. The total time-average stored energy is

+

µ=+= 1

8 2

22

d

aHabdWWW mestr

o


We now need to evaluate the power dissipated in the cavity walls. This dissipation will be due to the surface currents on each of the six walls as induced by the tangential magnetic fields, that is . Note that the

power dissipation is given by and that

is the surface resistance.

HnJ s ×= ˆ

ss RJ 2

21

tanHJ s =

σfR ms µπ=

+++

==

∫∫∫∫∫∫

∫

==

4444 34444 21444 3444 21444 3444 21bottomtop

a

xz

d

leftright

b

xz

d

front,back

a

zx

bs

wallswall

dxdzHHdydzHdxdyHR

dsHR

P

,,

tan

0

22

00

20

00

20

0

2

2222

2


After completing the integration steps, we obtain:

+

+

+

= 121

4 3

3

2

222

d

adb

d

adadHR

P swall

o

Therefore the quality factor Q, is

+

+

+

+

πµ=

ω=

121

1

3

3

2

2

2

2

101

d

adb

d

ada

d

a

DRabf

PW

Qswall

stro

Substituting for f101, gives

+

+

+

+

ηπ=

121

1

23

3

2

2

23

2

2

2

d

adb

d

ada

d

a

dRb

Qs


For a cubical resonator with a = b = d, we have

( )

( )[ ]21101

221011

101

33

112

121

−

−

σµπ=δδµ

µ=

πµ=

+

µε=µε=

mms

cube fa

Raf

Q

dafaf

Skin depth of the surrounding metallic walls, where µm is the permeability of the metallic walls.


Air-filled cubical cavityAir-filled cubical cavityWe consider an air-filled cubical cavity designed to be resonant in TE101 mode at 10 GHz (free space wavelength λ=3cm)with silver-plated surfaces (σ=6.14×107S-m-1, µm= µ0.. Find the quality factor.

( ) cmf

cf

aaf 122222

1121101101

1101 .≈

λ==

εµ=⇒µε= −

oo

At 10GHz, the skin depth for the silver is given by

( ) mµ≈×××π×××π=δ−− 6420101461041010

21779 ..

and the quality factor is

0001164203

1223

,.

.≅

µ×≅

δ=

mcma

Q


ObservationsObservationsPrevious example showed that very large quality factors can be achieved with

normal conducting metallic resonant cavities. The Q evaluated for a cubical cavity is in fact representative of cavities of other simple shapes. Slightly higher Q values may be possible in resonators with other simple shapes, such as an elongated cylinder or a sphere, but the Q values are generally on the order of magnitude of the volume-to-surface ratio divided by the skin depth.

( )

cavity

cavity

s

ts

v

wall

m

wall

strS

V

dsHR

dvHf

PW

PW

Qδ

≅

µπ

=ω

=ω=

∫∫ 2

2

22

22

2o

oo

Where Scavityis the cavity surface enclosing the cavity volume Vcavity.

Although very large Q values are possible in cavity resonators, disturbances caused by the coupling system (loop or aperture coupling), surface irregularities, and other perturbations (e.g. dents on the walls) in practice act to increase losses and reduce Q.


ObservationsObservationsDielectric losses and radiation losses from small holes may be especially important in reducing Q. The resonant frequency of a cavity may also vary due to the presence of a coupling connection. It may also vary with changing temperature due to dimensional variations (as determined by the thermal expansion coefficient). In addition, for an air-filled cavity, if the cavity is not sealed, there are changes in the resonant frequency because of the varying dielectric constant of air with changing temperature and humidity.Additional losses in a cavity occur due to the fact that at microwave frequencies for which resonant cavities are used most dielectrics have a complex dielectric constant . A dielectric material with complex permittivity draws an effective current , leading to losses that occur effectively due to

ε ′′−ε′=ε jEJeff ε ′′ω= o

*effJE ⋅

The power dissipated in the dielectric filling is

dydxdzE

dvEEdvJEP

a b dy

vv

effdielectric

2

0 0 02

21

21

∫ ∫ ∫

∫∫ε ′′ω

=

ε ′′ω⋅=⋅=

o

**


Dielectric LossesDielectric LossesUsing the expression for Ey for the TE101 mode, we have

ε ′′ε′

=ω=

+

µω

ε′ε ′′

=

d

strd

dielectric

PW

Q

d

aabdHP

o

oo 1

8 2

22

dvEPand

dvEWW

vydielectric

vymstr

2

2

2

22

∫∫

ε ′′ω=

ε′==

o

The total quality factor due to dielectric losses is

cd QQQ111

+=


Teflon-filled cavityTeflon-filled cavityWe found that an air-filled cubical shape cavity resonating at 10 GHz has a Qc of 11,000, for silver-plated walls. Now consider a Teflon-filled cavity, with ε= ε0(2.05-j0.0006). Find the total quality factor Q of this cavity.

[ ]rrr

da fc

aa

c

aff

ε′=⇒

εµ=

ε′µ≅= =

oo 22

22

12101

µr=1 for Teflon. This shows that the the cavity is smaller, or a=b=d=1.48 cm. Thus we have

Or times lower than that of the air-filled cavity. The quality factor Qd due to the dielectric losses is given by

rε′

76843

≅δ

=a

Qc

rε′

2365≅+

=cd

cdQQ

QQQ

Thus, the presence of the Teflon dielectric substantially reduces the quality factor of the resonator.


Cylindrical Wave FunctionsCylindrical Wave Functions

x

y

z ρ

φ

z

The Helmholtz equation in cylindrical coordinates is

011 22

2

2

2

2 =ψ+∂

ψ∂+

φ∂

ψ∂

ρ+

ρ∂ψ∂

ρρ∂∂

ρk

z

The method of separation of variables gives the solution of the form

0111 22

2

2

2

2 =+∂

+φ∂

Φ

Φρ+

ρρ

ρρk

z

ZdZ

dddR

dd

R



22

21zk

z

ZdZ

−=∂

( ) 01 2222

2=ρ−+

φ∂

ΦΦ

+ρ

ρρ

ρzkk

dddR

dd

R

22

21n

d−=

φ∂

ΦΦ

( ) 02222 =ρ−+−ρ

ρρ

ρzkkn

ddR

dd

R


( )[ ]

0

0

0

22

2

22

2

22

222

=+∂

=Φ+φ∂

ψ

=−ρ+ρ

ρρ

ρ

=+

ρ

ρ

ZKz

Zd

nd

RkkddR

dd

R

kkk

k

z

z

z

p satisfy toDefine




These are harmonic equations. Any solution to the harmonic equation we call harmonic functions and here is denoted by h(nφ)and h(kzz). Commonly used cylindrical harmonic functions are:

Where is the Bessel function of the first kind,

Is the Bessel function of the second kind, is the Hankel

function of the first kind, and is the Hankel function of the second kind.

( ) ( ) ( ) ( ) ( )ρρρρρ ρρρρρ kHkHkNkJkB nnnnn21 ,,,~

( )ρρkJ n ( )ρρkN n

( )ρρkH n1

( )ρρkH n2



Any two of these are linearly independent.

A constant times a harmonic function is still a harmonic function

Sum of harmonic functions is still a harmonic function

We can write the solution as :

( ) ( ) ( )zkhnhkB znknk zφρ=ψ ρρ ,,


Bessel functions of 1st kindBessel functions of 1st kind


Bessel functions of 2nd kindBessel functions of 2nd kind


Bessel functionsBessel functionsThe are nonsingular at ρ=0. Therefore, if a field is finite at

ρ=0, must be and the wave functions are

( )ρρkJ n

( )ρρkBn ( )ρρkJ n

( ) zjkjnnknk z

zeekJ φ

ρρ=ψρ ,,

The are the only solutions which vanish for

large ρ. They represent outward-traveling waves if kρ is

real. Thus must be if there are

no sources at ρ→∞. The wave functions are

( )( )ρρkH n2

( )ρρkBn( )( )ρρkH n2

( )( ) zjkjnnknk z

zeekH φ

ρρ=ψρ

2,,


( )( )

( )( )( )( ) ρ

ρ

−ρ

ρ

ρ

ρ

ρ

ρ

ρ

ρ

jkn

jkn

n

n

ekH

ekH

kρkN

kρkJ

toous analog

togous analo

toogous anal

tologous ana

2

1

sin

cos

Bessel functionsBessel functions


Bessel functionsBessel functions

The and functions represent cylindrical standing waves for real k as do the sinusoidal functions. The

and functions represent traveling waves for

real k as do the exponential functions. When k is imaginary (k = -jα)it is conventional to use the modified Bessel functions:

( )ρρkJ n ( )ρρkN n

( )( )ρρkH n1 ( )( )ρρkH n

2

( ) ( )

( ) ( ) ( )( )αρ−−π

=αρΚ

αρ−=αρ

+ 212 n

nn

nn

n

Hj

jJjI

( )( ) tologous ana

tologous anaαρ−

αρ

αρΚ

αρ

e

eI

n

n


Circular Cavity ResonatorsCircular Cavity ResonatorsAs in the case of rectangular cavities, a circular cavity resonator can be constructed by closing a section of a circular wave guide at both ends with conducting walls.

ϕ

z

x

r

a

dThe resonator mode in an actual case depends on the way the cavity is excited and the application for which it is used. Here we consider TE011mode, which has particularly high Q.


Circular Cavity ResonatorsCircular Cavity Resonators

,...,,,,...;q,,,...;n,,,m

dzq

mm

ax

J mnn

TMmnq

32103213210 ===

π

φφ

ρ

=ψ

where

coscossin

,...,,,...;q,,,...;n,,,m

dzq

mm

ax

J mnn

TEmnq

3213213210 ===

π

φφ

ρ′

=ψ

where

sincossin



The separation constant equation becomes

222

222

kdq

ax

kdq

ax

mn

mn

=

π

+

′

=

π

+

For the TM and TE modes,

respectively. Setting , we can solve for the resonant frequencies

µεπ= fk 2



2.132.132.132.132.412.562.78

TM210

∞2.901.801.421.521.571.66

TE211

1.591.591.591.591.801.912.08

TM110

2.292.292.292.292.603.003.00

∞5.272.721.501.321.201.00

∞3.062.051.721.871.962.08

∞2.801.631.191.241.271.31

∞2.721.501.001.001.001.00

1.001.001.001.001.131.201.31

0.00.501.002.003.004.00∞

TM020TE112TM111

TE011

TM011TE111TM010d/a

antdo

mnq

r

r

f

f

min

for the circular cavity of radius a and length d



8.77112.339

7.58811.06514.372

6.3809.76113.015

5.1368.41711.62014.796

3.8327.01610.17313.324

2.4055.5208.65411.792

1234

543210mn

Ordered zeros Xmn of Jn(X)

6.41610.52013.987

5.3179.28212.682

4.2018.01511.346

3.0546.7069.96913.170

1.8415.3318.53611.706

3.8327.01610.17313.324

1234

543210mn

Ordered zeros X`mn J`

n(X)


Circular Cavity ResonatorsCircular Cavity ResonatorsCylindrical cavities are often used for microwave frequency meters. The cavity is constructed with movable top wall to allow mechanical tuning of the resonant frequency, and the cavity is loosely coupled to a wave guide with a small aperture.

The transverse electric fields (Eρ, Eφ) of the TEmn or TMmn circular wave guide mode can be written as

( ) ( ) [ ]zjzjt

mnmn eAeAzE β−β−+ +φρ=φρ E ,,,The propagation constant of the TEnm mode is

22

′

−κ=βa

xmnmn

While the propagation constant of the TMnm mode is 2

2

−κ=β

axmn

mn


Circular Cavity ResonatorsCircular Cavity ResonatorsNow in order to have Et =0 at z=0, d, we must have A+= -A-, and A+sin βnmd=0 or

βmnd =lπ, for l=0,1,2,3,…, which implies that the wave guide must be an integer number of half-guide wavelengths long. Thus, the resonant frequency of the TEmnlmode is

And for TMnml mode is

22

2

π

+

′

εµπ=

dq

axc

f mn

rrmnq

22

2

π

+

εµπ=

dq

axc

f nm

rrmnq


Circular Cavity ResonatorsCircular Cavity ResonatorsThen the dominant TE mode is the TE111 mode, while the dominant TM mode is the TM110 mode. The fields of the TMnml mode can be written as

( )

( )

0

2

2

2

2

=

π

ρ′

′′

κη=

π

ρ′

ρ′κη

=

π

ρ′

ρ′β−

=

π

ρ′

′′

β=

πφ

ρ′

=

φ

ρ

φ

ρ

z

mnn

mn

mnn

mn

mnn

mn

mnn

mn

mnnz

E

dzq

mφa

xJ

xaHj

E

dzq

mφa

xJ

xmHaj

E

dzq

mφa

xJ

xmHa

H

dzq

mφa

xJ

xaH

H

dzq

ma

xJHH

sincos

sinsin

cossin

coscos

sincos

o

o

o

o

o

+−=εµ=η jAH 2o and


Circular Cavity ResonatorsCircular Cavity ResonatorsSince the time-average stored electric and magnetic energies are equal, the total stored energy is

( )

( )ρρ

ρ′

′

+

ρ′

′′πηεκ

=

φρρ+ε

==

∫

∫ ∫ ∫

=ρ

π

φρ

da

xJ

xma

ax

Jx

dHa

dzddEEWW

amn

nmn

mnn

mn

d a

e

0

22

22

2222

0

2

0 0

22

4

22

o

( )( )mnn

mnmn

xJxm

xdHa ′

′

−′

πηεκ= 2

2

2

2422

18

o


Circular Cavity ResonatorsCircular Cavity ResonatorsThe power loss in the conducting walls is

( ) ( )[ ]( ) ( )[ ]

( )( ) ( )

′−

′

β+

′β

+′π=

φρρ=+=+

φ=ρ+=ρ==

∫ ∫

∫ ∫∫π

=φ =ρφρ

=

π

=φφ

2

2222

222

2

0 0

22

0

2

0

222

1122

002

22

mnnmmnmnn

s

a

d

zz

s

St

sc

xm

xa

xamda

xJHR

ddzHzH

dzadaHaHR

dsHR

P

o

( )( )

( ) ( )

′−

′β

+

′

β+

′

−

′

ηκ=

ω=

2

2222

2

2

2

3

112

1

4

mnmnmn

mn

smncc

x

mxa

x

amad

xm

Rx

adaPW

Q



Where is the loss tangent of the dielectric. This is the same as the result of Qd for the rectangular cavity.

( )

( )( )

δ=

ε ′′ε

=ω

=

′

′

−′

ηκε ′′ω=

ρρ

ρ′

′+

ρ′

ρ′′

πηκε ′′ω=

+

ε ′′ω=⋅=

∫

∫∫

=ρ

φρ

tan

*

1

18

4

221

22

2

2422

022

2

2

2222

22

dd

mnnmnmn

amn

nmn

nmnmn

vvd

PW

Q

xJxm

x

Ha

da

xJ

ax

Jxma

x

dHa

dvEEdvEJP

o

o

To compute the Q due to dielectric loss, we must compute the power dissipated in the dielectric. Thus,

δtan


Cavity wave guide mode patternsCavity wave guide mode patterns


Cylindrical Cavity mode patternsCylindrical Cavity mode patterns


Loop or Probe CouplingLoop or Probe CouplingFor a probe coupler the electric flux arriving on the probe tip furnishes the current induced by a cavity mode:

I = ωε SEwhere E is the electric field from a mode averaged over probe tip and S is the antenna area. The external Q of this simple couplerterminated on a resistive load R for a mode with stored energy Wis

In the same way for a loop coupler the magnetic flux going through the loop furnishes the voltage induced in the loop by a cavity mode: V= ωµ SH

2222

ESR

WQext

ωε=

Lecture 8 - U.S. Particle Accelerator Schooluspas.fnal.gov/materials/03UCSB/Lecture08.pdf ·...

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