Download C. 8

8. Wave Guides and Cavities
8A. Wave Guides
Boundary Conditions at Perfect Conductors
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•
•
•
Suppose we have a region bounded by a conductor
We want to consider oscillating fields in the non-conducting region

For oscillating fields, changing B would imply non-zero E
E   B
t
But E must vanish in the conductor Ec  Bc  0  Dc  Hc
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•
On the surface of the conductors,  and J are present
Which appear in two of Maxwell’s equations
Therefore, D and H|| need not be continuous
But E|| and B must be continuous
So the correct boundary conditions must be
E  S   0, B  S   0
D  

H  J  D
t
Cylindrical Wave Guides
• Consider now a hollow infinite cylinder of arbitrary cross-section
– We’ll make it along the z-direction
• We want to find solutions moving along the cylinder
D   E , B  H
• Assume it is filled with a linear material:
• We will use complex notation
D  0  B
– Time-dependence will look like e-it

• Maxwell’s equations with no sources
E   B
t
• Use linearity plus time dependence

H  D
  E  0    B ,   E  iB ,   B  i E
t
• Take curl of either of the last two equations:
2
 2 E       E   i  B    E
2
 2 B       B   i  E    B
2
2

E


E0
• Use the double cross-product rule    V     V    V
• And therefore
2B   2B  0
2
Transverse and Longitudinal Dependence
• Since we have translation dependence along z, it
makes sense to look for
 ikz it
E

E
x
,
y
e
ikz


solutions that go like e
– Moving in the z direction B  B  x, y  e  ikz it
2E   2E  0
2B   2B  0



• Divide any derivative into
  t  zˆ , t  xˆ  yˆ
z
x
y
transverse and longitudinal parts
• Then we have, for example
t2 E    2  k 2  E  0
t2 B    2  k 2  B  0
E  Et  E z , E z  Ez zˆ , Et  Ex xˆ  E y yˆ .
• Also break up fields into
longitudinal and transverse parts:
• We now want to write our Maxwell’s equations broken up this way
Breaking Up Some Maxwell’s Equations
  t  zˆ



, t  xˆ  yˆ
z
x
y
E  Et  E z , E z  Ez zˆ , Et  Ex xˆ  E y yˆ .
• Let’s look at some Maxwell’s equations:
  E  iB ,   B  i E
   E t  iBt    zˆ  Et   t   Ez zˆ   iBt  zˆ   Et  t Ez  zˆ  iBt
z
z
   E  z  i Bz  zˆ     E   i B
t
t z
z
• But we assume fields look like
E  E  x, y  e  ikz it
B  B  x, y  e  ikz it
• Therefore the last equation becomes
• We similarly have
zˆ   ikEt  t Ez   iBt
zˆ   ikBt  t Bz   iEt
TEM Modes
• Can we find solutions with Ez = Bz = 0?
zˆ   ikEt  t Ez   iBt
• Such modes are called TEM modes
zˆ   ikBt  t Bz   iEt
– Because both E and B are transverse
• Then we
kzˆ  Et  Bt and  kzˆ  Bt  Et
would have
2
2



zˆ  Bt
ˆ
ˆ
ˆ
k z   z  Et   kz  Bt
• Multiplying the first by kzˆ 
k 2 Et   2 Et
• But the left side is
• We would therefore have k   
– Implies phase velocity equal to free waves
• Recall also zˆ   t  Et z  i Bz  0  t  Et  0
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 E  0  t  Et  0
And also
This tells us finding Et is a 2D electrostatic problem
Recall also that potential is constant on surfaces
Only get non-trivial solutions if there are at least two conducting surfaces
Sample Problem 8.1
A coaxial cable has inner radius a and outer radius b, and is filled with a material with
electric permittivity and magnetic susceptibility . Find exact electric and magnetic
field solutions for TEM modes.
• We will start by finding the electric
t  Et  0  t  Et
field, which is transverse and satisfies
• Since it has no curl, it is derivable from a potential
    0 n 2
• Potential must be constant on the inner and outer surfaces
Et  E    ρˆ
• Symmetry implies that Et must be radial:
1 
• No divergence
C

E

C


E


0



E


E   ,    ρˆ
t
tells us:
 

C  ikz it
• Put back in the z- and t-dependence
E  ρˆ e
• Where:

k   
• The magnetic
C   ikz it
Bt  kzˆ  Et
Bt    zˆ  Et
field is then
B
φˆ e

• Take real part to get actual fields
All We Need is the Z-Direction of the Fields
zˆ   ikEt  t Ez   iBt
• Now consider non-TEM modes
zˆ   ikBt  t Bz   iEt
– Ez or Bz are non-zero
• Multiply second equation by i and substitute the first one
zˆ   ikzˆ   ikEt  t Ez   it Bz    2Et
• For a transverse vector,
• We therefore have
• Solve for Et:
• Also recall
zˆ   zˆ  At   At
k 2Et  ik t Ez  i zˆ  t Bz   2Et
Et 
Bt 
i
  k
2
2
 kt Ez  zˆ t Bz 
2
 kt Bz  zˆ t Ez 
i
  k
2
• Normally you have just Bz or Ez
– Modes with Ez= 0 are called TE modes (transverse electric)
– Modes with Bz = 0 are called TM modes (transverse magnetic)
Finding Non-TEM modes
Et 
i
  k
2
 k t Ez   zˆ t Bz 
2 
Bt 
i
  k
• All modes of E and B satisfy
t2    2  k 2   0
• Let us define
 2   2  k 2
t2   2  0
• Then our equations become
Et  i 2   k t Ez   zˆ   t Bz 
Bt  i 2   k t Bz   zˆ   t Ez 
• We must also satisfy our boundary conditions
E
S
 0, B 
S
0
2
2
 kt Bz  zˆ t Ez 
TE Modes
 2  k 2   2
Et  i 2   k t Ez   zˆ   t Bz 
     0
2
t
2
Bt  i 2   k t Bz   zˆ   t Ez 
Case 1: TE modes (transverse electric)
• Search for solutions with Ez = 0, so everything comes from Bz
t2 Bz   2 Bz  0
• The transverse fields are then
Bt  ik 2t Bz , Et  i 2 zˆ  t Bz
• We must also satisfy boundary conditions E
• These imply tBz must be parallel
to the walls of the cylinder

Bz
n
S
S
 0  B
S
0
• Solve the eigenvalue equation subject to the boundary conditions
TM modes
 2  k 2   2
     0
2
t
2
Et  i 2   k t Ez   zˆ   t Bz 
Bt  i 2   k t Bz   zˆ   t Ez 
Case 2: TM modes (transverse magnetic)
• Search for solutions with Bz = 0, so everything comes from Ez
t2 Ez   2 Ez  0
• The transverse fields are then
Et  ik 2t Ez , Bt  i 2 zˆ  t Ez
• We must also satisfy the boundary conditions
E S  0  B S
• These imply that Ez must vanish on the walls
Ez S  0
• Solve the eigenvalue equation subject to the boundary conditions
Sample Problem 8.2 (1)
A hollow cylindrical waveguide has circular cross-section of
radius a. Find the relationship between the frequency and wave
number for the lowest frequency modes for the TE and TM modes.
• The frequencies are given by
 2  k 2   2
• The 2 values are eigenvalues of the equation
 t2   2   0     0 n2
– Where  is Bz (TE) or Ez (TM) modes
• Makes sense to work in
 2 1  1  2
2




 0
cylindrical coordinates
2
2
2

   
• Rotational symmetry
im


,


R

e




implies solutions of the form
• Substitute in:
d 2 R 1 dR m2
2


R


R0
• Let x  
2
2
d
 d 
• Then we have
d 2 R 1 dR  m2 

 1  2  R  0
2
dx
x dx 
x 
Sample Problem 8.2 (2)
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•
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•
A hollow cylindrical waveguide has circular cross-section of
radius a. Find the relationship between the frequency and wave
number for the lowest frequency modes for the TE and TM modes.
2
2


d
R
1
dR
m
2
2
2
x  
  k  

 1  2  R  0
2
dx
x dx 
x 
This is Bessel’s Equation
Solutions are Bessel functions R  x   J m  x   J m  
    0 n 2
Recall that  represents Bz or Ez

Bz S  0 or Ez S  0
And we have boundary conditions
n
We therefore must have
TE modes: J m  a   0,
TM modes: J m  a   0
• Let xmn be the roots of Jm and let ymn the roots of its derivative
• Then the formula for
2
2
y
x
2
2
mn
the frequencies will be: TE:  2  k 2  mn ,
TM:


k

a2
a2
Sample Problem 8.2 (3)
A hollow cylindrical waveguide has circular cross-section of
radius a. Find the relationship between the frequency and wave
number for the lowest frequency modes for the TE and TM modes.
2
2
y
x
TE:  2  k 2  mn2 ,
TM:  2  k 2  mn2
a
a
• Maple is happy to find roots of Bessel’s equation
> for m from 0 to 3 do
evalf(BesselJZeros(m,1..3)) end do;
x01  2.4048, x11  3.8317,
    0 n 2
x21  5.1356, x02  5.5201, x31  6.3802,
• With a little coaxing we can also get it to find the y’s
> for m from 0 to 3 do seq(fsolve(diff(BesselJ(m,x),x),
x=(n+m/2-3/4)*Pi),n=1..3) end do;
y11  1.8412,
y21  3.0542,
y01  3.8317,
y31  4.2012,
y41  5.3175,
Comments on Modes
2
2
y
x
TE:  2  k 2  mn2 ,
TM:  2  k 2  mn2
a
a
y11  1.8412, y21  3.0542, y01  3.8317, y31  4.2012, y41  5.3175,
x01  2.4048, x11  3.8317, x21  5.1356, x02  5.5201, x31  6.3802,
 ynm a TE modes
• Note that for each mode, there
min  
is a minimum frequency
 xnm a TM modes
• If you are at a limited , only a finite number of modes are possible
– Usually a good thing
– Ideally, want only one mode
• The lowest mode is usually a TE mode
• Note that the lowest modes often have m > 1
– Actually represent two modes because modes can be eim
8B. Rectangular Wave Guides
Working out the Modes
• Let’s now consider a rectangular wave guide
– Dimension a  b with a  b
 2  2
2



 0
• We will work in Cartesian coordinates
2
2
x
y
• Boundary conditions:

Bz S  0 or Ez S  0
n
  mx 
  ny 
• For TE modes, we
Bz  x, y   B0 cos 
cos 


want waves of the form
 a 
 b 
• For TM modes, we
  mx    ny 
want waves of the form Ez  x, y   E0 sin 
 sin 

 a   b 
• In each case, we have
2
2


m
n
2
2
   2  2 
b 
a
b
a
Restrictions on Modes
  mx    ny 
  mx 
  ny 
Ez  x, y   E0 sin 
Bz  x, y   B0 cos 
 sin 

 cos 

a
b
a
b








2
2


m
n
2
2
2
2
2
   2  2 


k


b 
a
• For the TM modes (Ez), we must have m > 0 and n > 0
• For the TE modes (Bz), we can’t have m = 0 and n = 0
– Since then Bz has no transverse derivative
• Hence TE modes have m > 0 OR n > 0
• Degeneracy between TE mode and TM mode if both positive

• Lowest frequency mode: TE mode
 10 
a
with (m,n) = (1,0)
• Second lowest modes: modes with   2 ,   
b
20
01
a
b
a
(m,n) = (2,0) or (0,1)
• If a  2b, then  factor of two difference between lowest and next lowest modes
• Normally make a  2b
The TE10 mode
• We can work out all the fields for the TE10 mode
– Put back in the z and t dependence
• We then work out all the other pieces using
Bt  ik 2t Bz , Et  i 2 zˆ  t Bz
  x   ikz it
Bz  B0 cos 
e
 a 
• We therefore have
Bt 
  x   ikz it
B0 xˆ sin 
e

 a 
iak
  x   ikz it
B0 yˆ sin 
e

 a 
• The factors of i represent a phase shift between the
various modes
Et 
ia
b
a
8C. Cavities
Cylindrical Cavities
• Let’s now cap off the cylindrical cavity, make the ends conducting
• Let it be along z with z from 0 to d
• On the end, we must have E  0  B 
S
S
• This means at z = 0 and z = d, we have Et
• We have up until now used modes with
Bz
0,c
 0  Bz
0,c
or Ez   x, y  eikz it
• We no longer want eikz, instead we want
• We will have to take linear combinations
of our previous solutions
sin  kz  
1
2i
cos  kz  
1
2
d
ikz
 ikz
e

e


e
ikz
 e ikz 
TE Modes in Cavities
• For TE modes, we had t2 Bz   2 Bz  0
 2  k 2   2
Bz  Bz  x, y  eikz it
Bt  ik 2t Bz , Et  i 2 zˆ  t Bz
• But we also need
Bz  0  Bz  d   0
ikz
 ikz
• We therefore must have Bz to be like sin  kz   21i  e  e 
• To get it to vanish at z = d, must have
p
k
, p  1, 2,3,
• So Bz is given by
d
Bz  Bz  x, y  sin  kz  eit
d
• To get the transverse components, take same linear combinations:
Bt   ik t Bz  x, y 
2
1
2i
e
Et  i zˆ  t Bz  x, y 
2
ikz
1
2i
e
e
ikz
 ikz
e
e
Bt   2kt Bz  x, y  cos  kz  eit
 it
 ikz
e
 it
Et  i 2zˆ t Bz  x, y  sin  kz  eit
TM Modes in Cavities
• For TM modes, we had t2 Ez   2 Ez  0
Ez  Ez  x, y  eikz it
•
•
•
•
•
 2  k 2   2
Et  ik 2t Ez , Bt  i 2 zˆ  t Bz
But we also need
Et  0  Et  d   0
We have to make the two contributions to Et cancel at z = 0, d
This can be done if Ez is like
cos  kz   12  eikz  e ikz 
To get it to work at z = d, must have
p
So Ez is given by
k
, p  0,1, 2,
d
it
Ez  Ez  x, y  cos  kz  e
d
• To get the transverse components, take same linear combinations:
Et   ik t Ez  x, y 
2
1
2
e
ikz
Bt  i zˆ  t Ez  x, y 
2
1
2
e
e
 ikz
ikz
e
e
Et   2kt Ez  x, y  sin  kz  eit
 it
 ikz
e
 it
Bt  i 2 zˆ t Ez  x, y  cos  kz  eit
Rules for Cavities Summarized
 2  k 2   2 k   p
• For both TE and TM modes, we have
d
– k > 0 for TE, k  0 for TM
2
2
t      0
• We have to solve the equation
–  represents Bz for TA and Ez for TM 
 S 0  S 0
• Boundary conditions for TE:
n
• Boundary conditions for TM:
d
Bz    x, y  sin  kz  eit
• For TM modes, we then have
Bt   2kt  x, y  cos  kz  eit
Et  i 2zˆ t  x, y  sin  kz  eit
• For TM modes, we then have
Ez    x, y  cos  kz  eit
Et   2 kt  x, y  sin  kz  eit
Bt  i 2 zˆ t Ez  x, y  cos  kz  eit
Box-Shaped Cavities
• Consider a cavity of dimensions a  b  d
• Frequency is given be
 2  k 2   2
• Where k is given by
p
– p = 1, 2, 3, … for TE
k
d
– p = 0, 1, 2, … for TM
d
2 2
2 2
• In both cases, we had

m

n
2
 mn  2  2
– m > 0 OR n > 0 for TE
a
b
– m > 0 AND n > 0 for TM
2
2
2
m
n
p
• Therefore, our frequencies are:
2
mnp

 2 2
2
a
b
c
• At least two of m, n, p must be non-zero
– If m = 0 or n = 0, it is a TE mode
– If p = 0, it is a TM modes
– If all three are non-zero, it can be a TE or TM mode
b
a
Sample Problem 8.2 (1)
A conducting box-shaped cavity has dimensions a  b  d. Write
explicitly the form of all fields for the TE1,0,1 mode, and find the
energy in electric and magnetic fields as a function of time.
• For TE modes, we first solve  2   2  0 
t

– Where  is Bz
n

x


• The 1,0 solution is
  B0 cos  
 a 
– I arbitrarily
2
included 2/2

2

 2
•  is given by
a
• Next find Bz: Bz    x, y  sin  kz  e
• Where k is given by
• So we have
  x    z  it
Bz  B0 cos 
 sin 
e
 a   d 
it
k
S
0
d
    0 n 2

d
b
a
Sample Problem 8.2 (2)
. . . Write explicitly the form of all fields for the TE modes, and
find the energy in electric and magnetic fields as a function of time.
x 
  B0 cos  
 a 
• Now let’s get Bt:
 
2
2
a2

k
d
Bt  k 2t  x, y  cos  kz  eit
a2 
x 
  z  it
a
x 
  z   it
Bt  B0 2  t cos 
cos
e
ˆ
  B0 x sin 
cos 
e

 


 d
a
d


 
d
 a 
 d 
• And finally, let’s go for Et:
Et  i 2zˆ t  x, y  sin  kz  eit
ia 2
i a
  x    z  it
  x    z   it
ˆ
Et   2 B0 z t cos 
B0 yˆ sin 
sin 
e
 sin   e  




 a   d 
 a   d 
• We will also need the
2
2
2
2
2


frequencies, which are given by   k  
 2  2  2
a
d
Sample Problem 8.2 (3)
. . . find the energy in electric and magnetic fields as a function of time.
i a
  x    z   it
E
B0 yˆ sin 
 sin 
e

 a   d 
 2 
2
a
2

2
d2
a
x  z 
• Keep in mind that we need to take
ˆ
E
B0 y sin 
 sin 
 sin t 

 a   d 
the real part of these expressions
2 2
• The electric energy

a 2 2 x  2 z  2
2
1
uE  2  E 
B0 sin 
density is:
 sin   sin t 
2
2
 a 
 d 
• Total electric energy is
a2  d 2 2
2 x 
2 z 
2
U E   u E dx dy dz 
B
sin
sin
sin
t  dx dy dz
0


 
2
2 d
 a 
 c 
a2  d 2 2 a d
2
UE 
B
b
sin
t 
0
2
2 d
2 2
UE 
a  a2  d 2  b
8 d
B02 sin 2 t 
Sample Problem 8.2 (4)
. . . find the energy in electric and magnetic fields as a function of time.
a
  x    z  it
x 
  z   it
Bz  B0 cos 
Bt   B0 xˆ sin 
 sin 
e
 cos 
e
d
 a   d 
 a 
 d 
• Again, take the real part
• Now let’s go for the magnetic energy density:
2
1 2

1

x

z
a
 2 
2
2
2 x
2   z 
2
uB 
B 
B
cos
sin

sin
cos
cos
t 
0 


 


 
2
2
2 
 a 
 d  d
 a 
 d 
U B   uB dx dy dz
1 2  2   x  2   z  a2
2 x 
2   z 
2

B0  cos 
sin

sin
cos
cos
t  dx dy dz

 


 
2
2
 a 
 c  d
 a 
 c 

1 2  a d a2 a d  2

B0  b  2 b  cos t 
2  2 2 d 2 2 
UB 
a  a2  d 2  b
8 d
B02 cos2 t 
Sample Problem 8.2 (5)
A conducting box-shaped cavity has dimensions a  b  d. Write
explicitly the form of all fields for the TE1,0,1 mode, and find the
energy in electric and magnetic fields as a function of time.
UE 
a  a2  d 2  b
8 d
B02 sin 2 t 
• It is pretty easy to see that
UB 
a  a2  d 2  b
2 d
UE UB 
B02 cos2 t  d
a  a2  d 2  b
2 d
    0 n 2
B02
b
• Of course, this is just conservation of energy
a