A. Nassiri -ANL · 2015. 3. 5. · f a a 2 12 2 2 2 1 1 1 2 101 101 1 101 ≈. λ = = µε = −...
Transcript of A. Nassiri -ANL · 2015. 3. 5. · f a a 2 12 2 2 2 1 1 1 2 101 101 1 101 ≈. λ = = µε = −...
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010
Resonators
A. Nassiri -ANL
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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.
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 3
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
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 4
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.
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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⋅=⋅=
**
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 6
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
18 2
22
dvEPand
dvEWW
vydielectric
vymstr
2
2
2
22
∫∫
ε′′ω=
ε′==
The total quality factor due todielectric losses is
cd QQQ111
+=
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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.
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 8
Cylindrical Wave Functions
x
y
z ρ
φ
z
The Helmholtz equation in cylindrical coordinates is
011 222
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 222
2
2
2 =+∂+
φ∂
Φ
Φρ+
ρρ
ρρk
z
ZdZ
dddR
dd
R
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 9
Cylindrical Wave Functions
22
21zk
z
ZZ
−=∂
∂
( ) 01 22222
=ρ−+φ∂
ΦΦ
+ρ
ρρ
ρ zkkd
ddR
dd
R
22
21n−=
φ∂
Φ∂Φ
( ) 02222 =ρ−+−ρ
ρρ
ρ zkknddR
dd
R
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( )[ ]
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
Cylindrical Wave Functions
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Cylindrical Wave Functions
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 nnnnn 21 ,,,~( )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.
( )ρρkJ n ( )ρρkN n( )( )ρρkH n1
( )( )ρρkH n2
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Cylindrical Wave Functions
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 z φρ=ψ ρρ ,,
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Bessel functions of 1st kind
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Bessel functions of 2nd kind
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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 zz eekJ φρρ=ψ ρ ,,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,,
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 16
( )( )
( )( )( )( ) ρ
ρ
−ρ
ρ
ρ
ρ
ρ
ρ
ρ
ρ
jkn
jkn
n
n
ekH
ekH
kρkNkρkJ
toous analog
togous analo
toogous anal
tologous ana
2
1
sin
cos
Bessel functions
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 17
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 n2
( ) ( )
( ) ( ) ( )( )αρ−−π=αρΚ
αρ−=αρ
+ 212 n
nn
nn
n
Hj
jJjI
( )( ) tologous ana
tologous anaαρ−
αρ
αρΚ
αρ
e
eI
n
n
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 18
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.
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 19
Circular Cavity Resonators
,...,,,,...;q,,,...;n,,,m
dzq
mm
ax
J mnnTMmnq
32103213210 ===
π
φφ
ρ=ψ
where
coscossin
,...,,,...;q,,,...;n,,,m
dzq
mm
ax
J mnnTEmnq
3213213210 ===
π
φφ
ρ′=ψ
where
sincossin
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 20
Circular Cavity Resonators
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
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 21
Circular Cavity Resonators
d/a TM010 TE111 TM110 TM011 TE211 TM111TE011
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
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 22
Circular Cavity Resonators
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
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 23
Circular Cavity Resonators
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=β
ax mn
mn
While the propagation constant of the TMnm mode is2
2
−κ=β
ax mn
mn
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 24
Circular Cavity Resonators
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 mnrr
mnq
22
2
π+
εµπ=
dq
axc
f nmrr
mnq
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 25
Circular Cavity Resonators
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
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 26
Circular Cavity Resonators
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
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 27
Circular Cavity Resonators
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
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 28
Circular Cavity Resonators
( )
( )( )
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
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 29
Cavity wave guide mode patterns
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 30
Cylindrical Cavity mode patterns
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 31
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 :
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 32
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
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 33
Circular TE011 mode
On the curved surface of the cylinder we have Jsϕdue to Hz given by
( ) aratdz
Hdz
JHJ s =
π−≈
π=
ϕsin.sin. 40308323
On the flat end surfaces we have Jsϕ due to Hr given by
azandzata
rJ
daH
J s ==
π±=
φ 08323
8323 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.
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 34
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
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 35
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
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 36
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 ≈+δ
λ= Q
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 37
Excitation of the TE011 mode in a circular cavity via coupling from a TE01mode in a rectangular wave guide.
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 38
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=
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 39
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
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 40
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
mncmn
mn
( )m 021
80127680210
..
..≈
−
×≅λ
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 41
(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 134
22211
.≈+
=λ
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 42
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.≈
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 43
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.
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 44
Dielectric Wave guide
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
0222
12 =ψ
−
ωε+∇ gs k
cIn the vacuum region we have
0222
2 =ψ
−
ω+∇ gs k
c
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 45
Dielectric Wave guide
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=
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 46
Dielectric Wave guide
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.
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 47
Dielectric Wave guide
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
0222
12 =ψ
−
ωε+∇ gs k
c
22
2
12
gs kc
k −ω
ε=
( )arforkrdrd
rdrd
r s
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 48
Dielectric Wave guide
We can write
02222
2 =ψ
−+ tkrdr
dr
drd
r
This can be rewritten
rkzwherezdzd
zdzd
z t==ψ
−+ 022
22
This type of modified Bessel’s equation,whose most generalform is
( ) 02222
2 =ψ
+−+ mz
dzd
zdzd
z
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 49
Dielectric Wave guide
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!
!
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 50
Dielectric Wave guide
( ) ,
+
π≅
zo
ze
zIz
m11
2
( ) .
+
π≅ −
zoe
zzK zm
112
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
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 51
Dielectric Wave guide
The behavior of I0(z) and K0(z)
I0(z)
K0(z)
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 52
Dielectric Wave guide
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
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 53
Dielectric Wave guide
We can now write
,r
Hk
kiH z
s
gr ∂
∂= 2 ,r
Ek
E zg ∂
∂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 ≥
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 54
Dielectric Wave guide
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
−=′−=′
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 55
Dielectric Wave guide
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
( )( )
( )( ) 0
11 =+akKk
akKakJk
akJ
tt
t
ss
s
where ( ) .22
122 1
ckk st
ω−ε=+
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 56
Dielectric Wave guide
Graphical solution of the dispersion relation
( )( )akJk
akJ
ss
s
1
( )( )akKk
akK
tt
t
1
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 57
Dielectric Wave guide
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 1stroot of J0(z)=0.
The cutoff frequency for the TE0p mode is given by
a
cj pp 11
00 −ε
=ω
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 58
Dielectric Wave guide
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.
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 59
Dielectric Wave guide
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
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 60
Dielectric Wave guide
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 111 ε=−
Again, eliminating constant A between the two equations gives the dispersion relation
( )( )
( )( ) 0
111 =+ε akKk
akKakJk
akJ
tt
t
ss
s
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Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 61
Dielectric Wave guide
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.
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