Lecture 5 - USPASuspas.fnal.gov/materials/10MIT/Lecture6.pdf · Lecture 5 Wave guides and...

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

Lecture 5

Wave guides and Resonators

A.Nassiri

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

Dispersion

First, consider a very simple sinusoidal electromagnetic wave propagating along along the x-axis, with electric field

( )kxtAE −ω= cos

kv

vx

tAE

p

p

ω=

−ω= cos

vp is the phase velocity. The phase velocity is the velocity at which the crest of the sinusoidal wave move through space. If one moves the observation point from 0 to x, the wave will arrive

Later and tp is referred to as the phase delay.p

p vx

t =


Dispersion

In vacuum (free space) the wave vector is given by k = ω/c and the phase velocity is just the vacuum velocity of light. In a material the wave vector is given by k = ωn/c so that the phase velocity is the vacuum velocity of light divided by the refractive index n.

Lets consider the a superposition of two waves at slightly different frequency, that is:

( ) ( )( ) ( )kxtkxtA

xktAxktAE

∆−ω∆−ω=−ω+−ω=

coscoscoscos

22211

I have defined four new quantities as:

kkkkkk

∆−=ω∆−ω=ω∆+=ω∆+ω=ω

22

11 ,,


Dispersion

0

2

→∆∂ω∂

→∆

ω∆=

−ω∆

−ω=

kkk

v

where

vx

tvx

tAE

g

gp

for

coscos


Dispersion

vg is the group velocity and vp is the phase velocity. The crests of the wave still move at the phase velocity but the modulations move at the group velocity. As before the group delay is defined by: tg = x/vg

There is another, slightly different, definition of the phase andgroup delay. When a wave travels over a certain distance x, throughsome medium, it will accumulate phase. In complex notation, theoutput field is related by the input field by

kx

eEE jinout

=φ= φ

Referring to these equations, it can be seen that the phase andgroup delay can also be expressed as:

ω∂φ∂

=ωφ

= gp tt ,


Dispersion

How fast do signals travel?

When an electromagnetic wave travels through a medium, itaccumulates phase. Naturally, the question arises what the velocityis at which a signal propagates. This in not the phase velocity. Thephase velocity is the velocity of the waves or, put in other words, itis the velocity at which the zero crossings of the field propagate.Thus, the phase velocity at frequency ω0 is the velocity at whichthe wave E(t,x)∝ cos(ω0t-kx) propagates. A perfect cosine repeatsitself every 2π radians and cannot therefore transport anyinformation. Information is, for example, a series of bits that canonly be encoded on a electromagnetic wave by modulating it. Themodulations propagates with the group velocity. Right ?


Dispersion

Unfortunately, things are not that simple. Consider a wave guide, consisting of a hollow metal tube filled with air.

Ld

It can be shown that an electromagnetic wave traveling through this wave guide will accumulate the phase

. Assumingthat the electric field vector iseverywhere perpendicular to the metal-air interface (TE mode).

. xc=1.841.Using φ=kL,we can write:

( ) 12

−

ωωω

=ωφc

cc

L

dcxc

c2

=ω


Dispersion

( ) 22cck ω−=ω

0 10

1

( )12102 ×πω

( )12102 ×πck

Cut-off


Dispersion

The frequency scale ~1THz (λ~1mm)

The “light line” corresponds to light propagating in free space

At very large k, the dispersion curve becomes the light line

Thus, the wavelength of the light (k=2π/λ) becomes much shorter than the diameter of the wave guide.

When the wavelength becomes of the order of the diameter of the wave guide (k→0), the dispersion curve flattens. At k=0, the frequency has finite value, therefore the phase velocity, which is ω/k, is infinite! An infinite velocity means the electromagnetic wave travels the distance L through the wave guide in zero time.

On the other hand, the group velocity, given by is zero at k=0!Thus, in a wave guide at cut-off the phase velocity is infinite and the group velocity is zero. If signals travel with the group velocity then at cut-off signals do not travel at all. This a good example where the group velocity does not represent the signal velocity.

kv g ∂

ω∂=


Dispersion

What happens to electromagnetic waves that have frequencies below cutoff frequency?These waves are in a “forbidden region,” much like the bad gap in a semiconductor.

( ) 12

−

ωωω

=ωφc

cc

L ( )2

1

ωω

−ω

=ωc

cc

ik

Therefore, below cutoff the electromagnetic wave is no longer a propagating wave with well defined wavelength. Instead, the wave decays exponentially and smoothly. Such wave is called evanescent wave. Sine k is imaginary, ik is a purely real number.


Dispersion for Waveguide

βz

ω

ωc

Slope = group velocity = vg

Parallel Plate Wave guide

Slope=vpz

βz1 βz2 βz3

ω1

ω2

ω3

} Slope=vpz

µε=

1Slope


The cavity resonator is obtained from a section of rectangular wave guide, closed by two additional metal plates. We assume again perfectly conducting walls and loss-less dielectric.

d

bdpβ

bnβ

amβ

z

y

x

π=

π=

π=

Rectangular Waveguide Resonator (Recap)

a


The addition of a new set of plates introduces a condition forstanding waves in the z-direction which leads to the definition of oscillation frequencies

222

21

+

+

µε=

dp

bn

am

fc

The high-pass behavior of the rectangular wave guide is modified into a very narrow pass-band behavior, since cut-off frequencies of the wave guide are transformed into oscillation frequencies of the resonator.



In the wave guide, each mode isassociated with a band of frequencieslarger than the cut-off frequency.

In the resonator, resonant modes canonly exist in correspondence of discreteresonance frequencies.

0 fc1 fc2 f fc10 fc2 f



The cavity resonator will have modes indicated as

TEmnp TMmnp

The values of the index corresponds to periodicity (number of sine orcosine waves) in three direction. Using z-direction as the reference forthe definition of transverse electric or magnetic fields, the allowedindices are

===

===

,...,,,,...,,,,...,,,

,...,,,,...,,,,...,,,

321032103210

321032103210

pnm

TMpnm

TE

With only one zero index m or n allowed

The mode with lowest resonance frequency is called dominant mode. In case a ≥ d>b the dominant mode is the TE101.



Note that a TM cavity mode, with magnetic field transverse to the z-direction, it is possible to have the third index equal zero. This is because themagnetic field is going to be parallel to the third set of plates, and it cantherefore be uniform in the third direction, with no periodicity.

The electric field components will have the following form that satisfies theboundary conditions for perfectly conducting walls.

π

π

π

= zdp

ybn

xa

mE xx sinsincosE

π

π

π

= zdp

ybn

xa

mE yy sincossinE

π

π

π

= zdp

ybn

xa

mE zy cossinsinE



The amplitudes of the electric field components also must satisfy the divergence condition which, in absence of charge is

00 =

π

+

π

+

π

⇒=⋅∇ zyx Edp

Ebn

Ea

mE

The magnetic field intensities are obtained from Ampere’s law:

π

π

π

ωµβ−β

= zdp

ybn

xa

mj

EEH zyyz

x coscossin

π

π

π

ωµβ−β

= zdp

ybn

xa

mj

EEH xzzx

y cossincos

π

π

π

ωµβ−β

= zdp

ybn

xa

mj

EEH yxxy

z sincoscos



Similar considerations for modes and indices can be made if the other axesare used as a reference for the transverse field, leading to analogousresonant field configurations.

A cavity resonator can be coupled to a wave guide through a small opening.When the input frequency resonates with the cavity, electromagneticradiation enters the resonator and a lowering in the output is detected. Byusing carefully tuned cavities, this scheme can be used for frequencymeasurements.



Example of resonant cavity excited by using coaxial cables.

The termination of the inner conductor of the cable acts like an elementary dipole (left) or an elementary loop (right) antenna.

Excitation with a dipole antenna Excitation with a loop antenna



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

=

=

cossin

sincos

coscos

2

2

2

2

4

4

0

Field Expressions For TE Modes – Rec. WG

g

yx

EH

η=

g

xy

EH

η±=

m,n = 0,1,2,… but not both zero


22

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



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

=

=

cossin

sincos

coscos

2

2

0

2

2

g

yx

EH

η=

g

xy

EH

η±=

m,n = 1,2,3,…



22

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


Rectangular Cavity Resonator

Add 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 oscillations at frequencies foe which the foregoing condition, that is

d=l λg/2, l=1,2,3,… is satisfied.

a

bdx

y z


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 Resonator

Substituting 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 Q

The 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

strPWQ ω=π=

cycleper dissipatedEnergy storedenergy Maximum2

mestr WWW +=Consider the TE101 mode:

+

µ=

π

π

πωµε

=ε

= ∫∫∫∫

116

44

2

22

2

0

2

00

222

daHabdW

dxdydzdzsin

axsinHadvEW

e

abd

v ye

2 and 11

02

22

22101

2 adxaxsin

da

a=

π

+

µεπ

=ω=ω ∫


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

sincoscossin

+

µ= 1

16 2

22

daHabd

Wm

Note that the the time-average electric and magnetic energies are preciselyequal. 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 theaverage over a period, it is shared equally between the electric and magneticforms. The total time-average stored energy is

+

µ=+= 1

8 2

22

daHabd

WWW mestr


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=

+++

==

∫∫∫∫∫∫

∫

==

bottomtop

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

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

str

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


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


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.


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⋅=⋅=

**


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 cavity

We 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

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 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 z

zeekH φ

ρρ=ψρ

2,,


( )( )

( )( )( )( ) ρ

ρ

−ρ

ρ

ρ

ρ

ρ

ρ

ρ

ρ

jkn

jkn

n

n

ekH

ekH

kρsinkN

kρcoskJ

toanalogous

toanalogous

toanalogous

toanalogous

2

1

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:

( )ρρkJ n ( )ρρkN n

( )( )ρρkH n1 ( )( )ρρkH n

2

( ) ( )

( ) ( ) ( )( )αρ−−π

=αρΚ

αρ−=αρ

+ 212 n

nn

nn

n

Hj

jJjI

( )( ) toanalogous

toanalogous αρ−

αρ

αρΚ

αρ

e

eI

n

n

Bessel functions


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,,,md

zqcosmcosmsin

axJ mn

nTMmnq

32103213210 where ===

π

φφ

ρ

=ψ

,...,,,...;q,,,...;n,,,md

zqsinmcosmsin

axJ mn

nTEmnq

3213213210 where ===

π

φφ

ρ′

=ψ



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)



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 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

+−=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

πφρ

da

xJxma

axJ

xdHa

dzddEEWW

a mnn

mn

mnn

mn

d ae

4

2

2

02

22

2

2222

0

2

0 022

( )( )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



Where is the loss tangent of the dielectric. This is the same as the resultof 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

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,

δtan



Cavity wave guide mode patterns


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=

Lecture 5 - USPASuspas.fnal.gov/materials/10MIT/Lecture6.pdf · Lecture 5 Wave guides and...

Documents

Transcript of Lecture 5 - USPASuspas.fnal.gov/materials/10MIT/Lecture6.pdf · Lecture 5 Wave guides and...