A. Nassiri -ANL - USPASuspas.fnal.gov/materials/10MIT/Lecture12.pdf · Massachusetts Institute of...

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010

Resonators

A. Nassiri -ANL

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 2

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 thesurrounding metallicwalls, where µm is thepermeability of themetallic walls.


Air-filled cubical cavity

We 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= −

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


Observations

Previous 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

2

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) inpractice act to increase losses and reduce Q.


Observations

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

ε ′′−ε′=ε jEJeff ε ′′ω=

*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⋅=⋅=

**


Dielectric Losses

Using the expression for Ey for the TE101 mode, we have

ε ′′ε′

=ω=

+

µω

ε′ε ′′

=

d

strd

dielectric

PW

Q

d

aabdHP

1

8 2

22

dvEPand

dvEWW

vydielectric

vymstr

2

2

2

22

∫∫

ε′′ω=

ε′==

The total quality factor due todielectric losses is

cd QQQ111

+=

Teflon-filled cavityWe found that an air-filled cubical shape cavity resonating at 10 GHz has a Qc of11,000, 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≅= =

22

22

12101

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

Or times lower than that of the air-filled cavity. The quality factor Qd dueto 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 factorof the resonator.


Cylindrical 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

ZZ

−=∂

∂

( ) 01 2222

2=ρ−+

φ∂

ΦΦ

+ρ

ρρ

ρ zkkd

ddR

dd

R

22

21n−=

φ∂

Φ∂Φ

( ) 02222 =ρ−+−ρ

ρρ

ρ zkknddR

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

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.


( )( )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 kind


Bessel functions of 2nd kind


Bessel functions

The 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

zz

eekH φρρ=ψ

ρ

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 functions


Bessel functions

The and functions represent cylindricalstanding 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:


( )( )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 Resonators

As in the case of rectangular cavities, a circular cavity resonator can beconstructed by closing a section of a circular wave guide at both ends withconducting walls.

ϕ

z

x

r

a

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



,...,,,,...;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



d/a TM010 TE111 TM110 TM011 TE211 TM111

TE011

TE112 TM210 TM020

0.00.501.002.003.004.00∞

1.001.001.001.001.131.201.31

∞2.721.501.001.001.001.00

1.591.591.591.591.801.912.08

∞2.801.631.191.241.271.31

∞2.901.801.421.521.571.66

∞3.062.051.721.871.962.08

∞5.272.721.501.321.201.00

2.132.132.132.132.412.562.78

2.292.292.292.292.603.003.00

antdo

mnq

r

r

f

f

min

for the circular cavity of radius a and length d



mn

0 1 2 3 4 5

1234

2.4055.5208.65411.792

3.8327.01610.17313.324

5.1368.41711.62014.796

6.3809.76113.015

7.58811.06514.372

8.77112.339

Ordered zeros Xmn of Jn(X)

mn

0 1 2 3 4 5

1234

3.8327.01610.17313.324

1.8415.3318.53611.706

3.0546.7069.96913.170

4.2018.01511.346

5.3179.28212.682

6.41610.52013.987

Ordered zeros X`mn J`

n(X)

mn

0 1 2 3 4 5

1234

2.4055.5208.65411.792

3.8327.01610.17313.324

5.1368.41711.62014.796

6.3809.76113.015

7.58811.06514.372

8.77112.339



Cylindrical cavities are often used for microwave frequency meters. The cavity isconstructed with movable top wall to allow mechanical tuning of the resonantfrequency, 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 modecan be written as

( ) ( ) [ ]zjzjt

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

22

′

−κ=βa

x mnmn

While the propagation constant of the TMnm mode is2

2

−κ=β

ax mn

mn



Now 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 integernumber 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



Then the dominant TE mode is the TE111 mode, while the dominant TM mode is theTM110 mode. The fields of the TEnml 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

+−=εµ=η jAH 2 and



Since the time-average stored electric and magnetic energies are equal, the totalstored 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

( )( )mnn

mnmn

xJxm

xdHa ′

′

−′

πηεκ= 2

2

2

2422

18



The 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

( )( )

( ) ( )

′−

′β

+

′

β+

′

−

′

ηκ=

ω=

2

2222

2

2

2

3

112

1

4

mnmnmn

mn

smncc

x

mx

a

x

amad

xm

Rx

adaPW

Q



( )

( )( )

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

xJx

m

x

Ha

da

xJ

ax

Jx

ma

x

dHa

dvEEdvEJP

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

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

δtan


Cavity wave guide mode patterns


Cylindrical Cavity mode patterns


Circular TE011 mode

π

ρωµ−

=

π

ρπ

=

π

ρ

=

φ

ρ

dz

aJ

daHj

E

dz

aJ

daH

H

dz

aJHH z

sin..

cos..

sin.

83238323

83238323

8323

1

1

( ) ( )ρ′−=ρ hJhJNOTE 1 :


Circular TE011 mode

The electric field lines form closed circular loops centered around the cylinder axis.

The electric field lines are threaded with closed loops of magnetic filed lines in the radial planes.

No surface charges appear on any of the cavity walls, since the normal electric field is zero everywhere on the walls

However surface currents do flow in the walls due to tangential magnetic fields.

HnJ s ×= ˆ


Circular TE011 mode

On the curved surface of the cylinder we have Jsϕdue to Hz given by

( ) aratdz

Hdz

JHJ s =

π

−≈

π

=ϕ

sin.sin. 40308323

On the flat end surfaces we have Jsϕ due to Hr given by

azandzata

rJ

daH

J s ==

π

±=φ

083238323 1

..

It is interesting to note that the surface currents are entirelycircumferential. No surface current flows between the flat walls andthe curved walls.Hence, if one end of the cavity is mounted on micrometers and

moved to change the length of the cavity, the TE011can still be fullysupported, since the current flow is not interrupted.


Circular TE011 mode

Movable construction of the end faces also suppress other modes,particularly TM111, which has the same resonant frequency butlower Q.

The currents that are required to support TE111 are interrupted bythe space between the movable ends and the side walls.

As in the case of of the rectangular cavity resonator, the resonantfrequency of the TE011mode can be found by substituting theexpressions for any one of the field components into the waveequation.

201

2

0111

+

π

µε=ω

ax

d


Circular TE011 mode

The resonant free-space wavelength of the cavity corresponding tothe resonant frequency is given by

201

2011011

1

2

π+

εµ==λ

ax

d

fc rr

For the most general case of Temnp mode, the resonant free-spacewavelength λmnp is given by

22

2

π+

εµ==λ

ax

dpf

c

mn

rr

mnpmnp


Circular TE011 mode

Using the expression derived for the Q of the circular Temnp andusing x01=3.832, we have

( ) ( )( )

++

+δ

λ≈ 3

22

216801216801216801610

da

dadaQ

.

...

X-band circular cavity: d=2a such that its TE011 resonates at 10GHz. What is Q?

cma

aa

Hzsm

fc 981

832321

21010

103229

18

011011 .

.=→

π+

=×

−×≈=λ

−

mf

7104261 −×≈σµπ

=δ .

!.. 3079216801610≈+

δλ

= Q


Excitation of the TE011 mode in a circular cavity via coupling from a TE01mode in a rectangular wave guide.


Loop or Probe Coupling

For 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 coupler terminated on a resistive load R for a mode with stored energy W is

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=


A WR-1500 rectangular air wave guide has inner dimensions38.1cm ×19.05 cm. Find (a) the cutoff wavelength for the dominantmode; (b) the phase velocity, guide wavelength and waveimpedance for the dominant mode at a wavelength of 0.8 times thecutoff wavelength;(c) the modes that will propagate in the waveguide at a wavelength of 30 cm.

22222222

anbm

ab

bn

amf

v

mnmn

c

pc

+=

+

==λ

So the cutoff wavelength of the dominant TE10 mode is

cmac 276210

.==λ

Problem 1


The phase velocity, guide wavelength and the wave impedance can be calculates as

2222 1

1

−µε

=

π

−

π

−µεω

ω=

βω

=

f

fbn

am

vmn

mn

cmnp

( )18

2

8105

81103 −−×≅

−

×≅ smv p

.

2222 1

22

−

λ=

π

−

π

−µεω

π=

βπ

=λ

f

fbn

am

mncmnmn

( )m 021

80127680210 .

...

≈−

×≅λ


(c) At λair=40 cm, only the dominate mode TE10 mode propagates since the next higher mode TE20 (or TE11) has λc20= λc10/2 = 38.1 cm < 40 cm.

142

1222

10

10 −µε

µ=

−

η=

fa

fa

ff

Zc

TE

( )Ω≈

−≈ 3628

801377

210.

.TEZ

(d) At λair=30 cm, the propagating modes are TE10, TE20, and TM20(λc20=38.1 cm), TE01(λc01=38.1 cm), and TE11 and TM11

cmba

abc 1342

2211.≈

+=λ


Problem 2

Design a rectangular cavity resonator that will resonant in the TE101mode at 10GHz and resonant in the TM110 mode at 20 GHz.Assuming a=2b. The resonant frequencies of the TE101 and TM110modes are given by

cmaGHzaa

f

GHzda

f

dp

bn

am

mnp

677120412103

10112103

1

22

10

110

22

10

101

2

22

2

22

2

22

.≈→≈+×

≈

≈+×

≈

π+

π+

πµε

=ω

cmab 83902 .≈= from f101 we find cmd 3543.≈


Dielectric Wave guide

We have shown that it is possible to propagate electromagnetic wave s down a hollow conductor. However, other types of guiding structures are also possible.

The general requirement for a guide of electromagnetic waves is that there be a flow of energy along the axis of the guiding structure but not perpendicular to it.

This implies that the electromagnetic fields are appreciable only in the immediate neighborhood of the guiding structure.

Consider an axisymmetric tube of arbitrary cross section made of some dielectric material and surrounded by a vacuum. This structure can serve as a wave guide provided that dielectric constant of the material is sufficiently large.



The boundary conditions satisfied by the electromagnetic fields are significantly different to those of a conventional wave guide. The transverse fields are governed by two equations; one for the region inside the dielectric, and the other for the vacuum regions.

Inside the dielectric we have

022

2

12 =ψ

−

ωε+∇ gs k

cIn the vacuum region we have

022

22 =ψ

−

ω+∇ gs k

c



Here, stands for either Ez or Hz, ε1 is the relativepermittivity of the dielectric material, and kg is the guide propagationconstant.

( ) zikgeyx ,ψ

The guide propagation constant must be the same both insideand outside dielectric in order to satisfy the electromagneticboundary conditions at all points on the surface of the tube.

Inside the dielectric the transverse Laplacian must be negative,so that the constant

is positive. Outside the cylinder the requirement of no transverseflow of energy can only be satisfied if the fields fall offexponentially (instead of oscillating).

22

2

12

gs kc

k −ω

ε=



Thus2

222

ckk gt

ω−=

The oscillatory solutions (inside) must be matched to theexponentially solutions (outside). The boundary conditions are thecontinuity of and on thesurface of the tube.

normal B and D tangential E and H

The boundary conditions are far more complicated than those ina conventional wave guide. For this reason, the normal modescannot usually be classified as either pure TE or TM modes.

In general, the normal modes possess both electric and magneticfield components in the transverse plane. However, for the specialcase of a cylindrical tube of dielectric material the normal modescan have either pure TE or pure TM characteristics.



Consider a dielectric cylinder of radius a and dielectric constantε1. For the sake of simplicity, let us only search for normal modeswhose electromagnetic fields have no azimuthal variation. We canwrite

022

2

12 =ψ

−

ωε+∇ gs k

c

22

2

12

gs kc

k −ω

ε=

( )arforkrdrd

rdrd

r s <=ψ

++ 022

2

22

The general solution to this equation issome linear combination of the Besselfunctions J0(ksr) and Y0(ksr). However,since Y0(ksr) is “badly” behaved at theorigin(r=0) the physical solution is ( )rkJ s∝ψ



We can write

0222

22 =ψ

−+ tkr

drd

rdrd

r

This can be rewritten

rkzwherezdzd

zdzd

z t==ψ

−+ 02

2

22

This type of modified Bessel’s equation,whose most generalform is

( ) 0222

22 =ψ

+−+ mz

dzd

zdzd

z



The two linearly independent solutions are denoted Im(z) andKm(z). The asymptotic behavior of these solutions at small isas follows:

z

( ) ( )( )∑

∞

=+

=

0

2 42

k

km

m mkkzz

zI!!

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )[ ] ( )( )∑

∑∞

=

+∞

=

−

+++ψ++ψ

−+

−+−−−

=

0

2

12

0

41122

11

214122

1

k

mmm

mmk

k

m

m

kmkz

kmkz

zIzzkkmz

zK

!!

ln!

!



( ) ,

+

π≅

zo

ze

zIz

m11

2

( ) .

+

π≅ −

zoe

zzK z

m11

2

Hence Im(z) is well behaved in the limit whereasKm(z) is badly behaved. The asymptotic behavior at large is

,0→zz

Hence,Im(z) is badly behaved in the limit whereasKm(z) is well behaved.

,∞→z



The behavior of I0(z) and K0(z)

I0(z)

K0(z)



Is it clear that the physical solution (I.e., the one which

Decays as is∞→r ( ).rkK t∝ψThe physical solution is

( )

( ).

,

arrkKA

arrkJ

t

t

>=ψ

≤=ψ

for

andfor

A is an arbitrary constant, and stands for either Ez or Hz.

( ) zikgerψ



We can now write

,r

Hk

kiH z

s

gr ∂

∂= 2 ,

rE

kE z

g ∂∂ωµ

−=θ

θεωε= H

kE

r

gr

arfor ≤

There are analogous set of relationships for . The fact that the field components form two groups;(Hr,Eθ), which depends on Hz and (Hθ,Er), which depend on Ez; means that the normal modes takes the form of either pure TE modes or pure TM modes.

arfor ≥



For a TM mode (Ez=0) we find that

( )rkJH sz =

( )rkJk

kiH s

s

gr 1−=

( )rkJk

iE ss

1ωµ

=θ

andarfor ,≤

( )rkAKH tz =

( )rkKk

kiAH t

t

gr 1=

( )rkKk

iAE tt

1ωµ

−=θ

arfor ≥

Here we have used ( ) ( )( ) ( ).

,zKzK

zJzJ

1

1

−=′−=′



The boundary conditions require Hz,Hr, and Eθ to be continuous across r = a. Thus, it follows that

( ) ( ),akJakAK st =

( ) ( ) .s

s

t

t

kakJ

kakK

A 11 =−

Eliminating the arbitrary constant A, will yield

( )( )

( )( ) 011 =+

akKkakK

akJkakJ

tt

t

ss

s

where ( ) .2

2

122 1

ckk st

ω−ε=+



Graphical solution of the dispersion relation

( )( )akJk

akJ

ss

s

1

( )( )akKk

akK

tt

t

1



Since the first root of occurs at z=2.4048 the condition condition for the existence of propagating modes can be written

( )zJ

ac

140482

101 −ε

=ω>ω.

In other words, the mode frequency must lie above the cutofffrequency ω01 for the TE01 mode (here, the 0 corresponds to thenumber of nodes in the azimuthal direction, and 1 refers to the 1st

root of J0(z)=0.

The cutoff frequency for the TE0p mode is given by

a

cj pp 11

00 −ε

=ω



At the cutoff frequency for a particular kt=0, which implies that

ω−= 2

222

ckk gtc

kgω

=

The mode propagates along the guide at the velocity of light invacuum. Immediately below this cutoff frequency the system nolonger acts as a guide but as an antenna, with energy being radiatedradially. For the frequencies well above the cutoff, kt and kg are ofthe same order of magnitude, and are large compared to ks. Thisimplies that the fields do not extend appreciably outside thedielectric cylinder.



For the TM mode (Hz=0) we find that

( )

( )

( )rkJk

kiE

rkJk

iH

rkAJE

ss

gr

ss

sz

1

11

−=

εωε−=

=

θ

arfor ≤

( )

( )

( )

arfor

rkKk

kiAE

rkKk

iAH

rkAKE

tt

gr

tt

tz

>

=

ωε=

=

θ

1

1



The boundary conditions require Ez, Hθ, and Dr to be continuousacross r = a. Thus, it follows that

( ) ( ),akJakAK st =

( ) ( ) .s

s

t

t

kakJ

kakK

A 11

1 ε=−

Again, eliminating constant A between the two equations gives the dispersion relation

( )( )

( )( ) 011

1 =+εakKk

akKakJk

akJ

tt

t

ss

s



It is clear from this dispersion relation that the cutoff frequencyFor the TM0p mode is exactly the same as that for the TE0p mode.

It is also clear that in the limit the propagation constantsare determined by the roots of J1(ksa) ≈ 0. However, this is exactlyThe same as the determining equation for TE modes in a metallicWave guide of circular cross section (filled with dielectric of relativePermittivity ε1).

11 >>ε

Modes with azimuthal dependence (i.e., m > 0) have longitudinalComponents of both E and H. This makes the math somewhat moreComplicated. However, the basic results are the same as for m=0Modes: for frequencies well above the cutoff frequency the modesare localized in the immediate vicinity of the cylinder.

A. Nassiri -ANL - USPASuspas.fnal.gov/materials/10MIT/Lecture12.pdf · Massachusetts Institute of...

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