Topics

Wave-guide theory:


SPH 407: Waveguide


Topics

Wave-guide properties:






wavelength,
velocity of propagation,
impedance cut-off,
reflection,
and refraction.
wave-guide transmission,
phase and group velocities, wave-guide
equation.
Guided wave systems.
Transverse electric, transverse magnetic,
transverse electromagnetic modes.
Topics




Simple transmission techniques
applied to wave-guide below and near
cut-off.
Guided waves in layered medium.
Boundary conditions.
Dielectric rod and slab guides
Field concept in relation to lumped
circuit elements.
Propagation Mode of EM Wave
Transverse Electromagnetic (TEM)
• The electric field, E and the magnetic field, H are
oriented transverse to the direction of propagation
of wave.
• Exists in plane waves and transmission lines (2
conductors).
• No cut-off frequency.
y
Ey
Direction of
Travel
x
Hz
Hz
z
1
Transverse Electric (TE)
Transverse Magnetic (TM)
• The electric field, E is transverse to the direction of
propagation of wave and the magnetic field, H has
components transverse and in the direction of the
wave.
• Exists in waveguide modes.
• The magnetic field, H is transverse to the direction
of propagation of wave and the electric field, E has
components transverse and in the direction of the
wave.
• Exists in waveguide modes.
y
y
Ey
Hx
Ey
Direction of
travel
x
Hy
z
E
Ex
H
Direction of
travel
x
Hz
z
Waveguide
What is a waveguide?

Consider the “light pipe analogy” illustrated in
Figure 8.0A. A flashlight serves as our “r-f
source,” which given that light is also an
electromagnetic wave is not all that
unreasonable. The source radiates into freespace, and spreads out as a function of distance.
The intensity per unit area at the destination – a
wall falls off as a function of distance (D)
according to the inverse square law (1/D2).
What is a waveguide?

Now consider the transmission scheme in Figure
8.0B, the light wave still propagates over a
distance D, but is now confined to the interior of
a mirrored pipe. Almost all of the energy
coupled to the input end is delivered to the
output end, where the intensity is practically
undiminished. The light pipe analogy may not be
the best way to explain the operation of
waveguides, but rather a neat summary on a
simple level.
2
Waveguide
Introduction
WAVEGUIDE  any structure that supports propagation
of a wave. In general usage:

The term waveguide refers to constructs that only
support non TEM mode propagation, name in the TE and
TM Mode.
It
also unable to support wave propagation below a
certain frequency, or cutoff frequency
Introduction (Cont’d..)
A waveguide is another means of guiding the EM energy
from one point to another (same as transmission line).
Some differences between waveguide and transmission line
(TLine) :
Introduction (Cont’d..)
• At microwave frequencies (3 to 300 GHz), TLine
becomes inefficient due to skin effect and dielectric
losses,
but
waveguides
are
used
at
microwave
frequencies to obtain larger bandwidth and lower signal
attenuation.
• TLine can only support TEM wave whereas waveguide
• TLine can operate above dc (f =0) to a very high
can support many possible field configurations.
frequency, but waveguide can operate only above cutoff
frequency and therefore acts as a high pass filter.
Properties and characteristics of waveguide:
Properties and characteristics of waveguide:
1. The conducting walls of the guide confine the
electromagnetic fields and thereby guide the
electromagnetic wave through multiple
reflections .
2. when the waves travel longitudinally down the
guide, the plane waves are reflected from wall to
wall .the process results in a component of
either electric or
magnetic fields in the direction of propagation of
the resultant wave.
3. TEM waves cannot propagate through the
waveguide since it requires an axial conductor
for axial current flow .
4. when the wavelength inside the waveguide
differs from that outside the guide, the velocity of
wave propagation inside the waveguide must
also be different from that through free space.
5. if one end of the waveguide is closed using a
shorting plate and allowed a wave to propagate
from other end, then there will be complete
reflection of the waves
resulting in standing waves.
3
Introduction (Cont’d..)
Types of waveguides:
(1)rectangular
waveguide
(2)cylindrical
waveguide
(3)elliptical
waveguide
(4)parallel waveguide
Rectangular Waveguides
Waveguide components
Rectangular waveguide
Waveguide bends
More waveguides
Waveguide to coax adapter
E-tee
4
Uses


High frequencies
High power
Need to find the fields components
of the em wave inside the
waveguide

Can operate only above certain
frequencies



To reduce attenuation loss


Rectangular WG

Acts as a High-pass filter
Ez Hz Ex Hx Ey Hy
We’ll find that waveguides don’t
support TEM waves
Normally circular or rectangular

We will assume lossless rectangular
Rectangular Waveguide
Fundamentals
Rectangular Waveguide
Fundamentals (Cont’d..)
A cross section of rectangular waveguide is shown below:
The interior dimensions are a x b, where the longer side is a.
• Propagation is in the +z direction
or out of page.
• Conducting walls  brass, copper
or aluminium.
• Chosen to be thick enough for
mechanical rigidity and several skin
depths over the frequency of interest.
• The inside wall  electroplated
with silver or gold to improve
performance
Rectangular Waveguide
Fundamentals (Cont’d..)
Waveguide can support TE and TM modes, where:

In TE Modes, the electric field is transverse to the
‘a’ dimension:
Determines the frequency range of the dominant, or lowest
order, the mode of propagation.

Usually operates in lowest propagating mode, since higher
order  higher attenuation + difficult to extract from guide.

‘b’ dimension:
 Affects
 Also

attenuation, smaller b has higher attenuation.
sets the max power capacity
Usually half of the ‘a’ dimension
Rectangular Waveguide
Fundamentals (Cont’d..)
The order of the mode refers to the field configuration in
the guide and is given by ‘m’ and ‘n’ integer subscripts, as
TEmn and TMmn.
direction of propagation. Some magnetic field component
in the direction of propagation.

In TM Modes, the magnetic field is transverse and an
electric field component must be in the propagation
direction.
The ‘m’ subscript corresponds to the number of half wave
variations of the field in x direction
The ‘n’ subscript corresponds to the number of half wave
variations of the field in y direction
5
Rectangular Waveguide
Fundamentals (Cont’d..)
Maxwell’s equations (no sources)


  H  iE



  E  iH
In conjunction with the guide dimensions, m and n
determine the cutoff frequency for a particular mode.
fcmn 
1
2 
2
m n
   
 a  b
2

i

j

k

x

y

z
Hx

i
For conventional rectangular waveguide filled with air,
where a = 2b, the dominant or lowest order mode is TE10
with cutoff frequency fc10 = c/2a



 i ( E x i  E y j  E z k )
Hy Hz


j k

x

y

z
Ex
Ey
Ez



 i ( H x i  H y j  H z k )
Rectangular Waveguides:
Fields inside
Maxwell’s equations (no sources)
i E x  Hyz  ik z H y

i E y  ik z H x  Hxz

H
i EZ  xy  Hyx


E z
 iH x  y  ik z E y


E z
 iH y  ik z E x  x


E y
E
 iH z  x  yx
expansion
Ex 
(k
 
)


E

H
 E y  2i 2 (k z yz   xz )
k k z


E z
H z
i
 H x  k 2  k z2 ( y  k z x )

H  i ( Exz  k z Hyz )

 y k 2  k z2
i
k 2  k z2
E z
z x
H z
y
Using phasors & assuming waveguide
filled with lossless dielectric material
and walls of perfect conductor, the
wave inside should obey…
Helmholz’s
equations:
2 E  k 2 E  0
2 H  k 2 H  0
where
k 2   2  c
By Substituting HZ and EZ =0, We’ll find that waveguides don’t support TEM
waves
Then applying on the z-component…
 2 Ez  k 2 Ez  0
 2 Ez  2 Ez  2 Ez


 k 2 Ez  0
x 2
y 2
z 2
Solving by method of Separation of Variables :
E z ( x, y, z )  X ( x)Y ( y ) Z ( z )
from where we obtain :
X '' Y '' Z ''


 k 2
X
Y
Z
Fields inside the waveguide
X '' Y '' Z ''


 k 2
X
Y
Z
 k x2  k y2   2   k 2
h 2   2  k 2  k x2  k y2
which results in the expressions :
X ''  k x2 X  0
X(x)  c1 cos k x x  c2 sin k x x
Y ''  k y2Y  0
Y(y)  c3 cos k y y  c4 sin k y y
Z  Z  0
Z ( z )  c5ez  c6 e z
''
2
6
Substituting
Ez ( x, y, z)  X ( x)Y ( y)Z ( z)
X(x)  c1 cos k x x  c2 sin k x x
Y(y)  c3 cos k y y  c4 sin k y y
Other components
From Faraday and Ampere Laws we can find the
remaining four components:
Z ( z )  c5ez  c6 e z

E z  c1 cos k x x  c2 sin k x x c3 cos k y y  c4 sin k y y  c5ez  c6 e z
 E z j H z
 2
h 2 x
h
y
 E z j H z
Ey   2
 2
h y
h
x
j E z  H z
Hx  2

h y h 2 x
j E z  H z
Hy   2

h x h 2 y
where
Ex  

If only looking at the wave traveling in  z - direction :
E z   A1 cos k x x  A2 sin k x x A3 cos k y y  A4 sin k y y e z
Similarly for the magnetic field,
H z  B1 cos k x x  B2 sin k x x B3 cos k y y  B4 sin k y y e z
*So once we know
Ez and Hz, we can
find all the other
fields.
h 2   2  k 2  k x2  k y2
Modes of propagation
From these equations we can conclude:
 TEM (Ez=Hz=0) can’t propagate.

TE (Ez=0) transverse electric


TM (Hz=0) transverse magnetic, Ez exists


In TE mode, the electric lines of flux are
perpendicular to the axis of the waveguide
In TM mode, the magnetic lines of flux are
perpendicular to the axis of the waveguide.
HE hybrid modes in which all components
exists
TM Mode
Ez   A1 cos k x x  A2 sin k x x A3 cos k y y  A4 sin k y y e z
 Boundary
E z  0 at y  0 ,b
conditions: E z  0 at x  0 ,a
From these, we conclude:
X(x) is in the form of sin kxx,
where kx=mp/a, m=1,2,3,…
Y(y) is in the form of sin kyy,
where ky=np/b, n=1,2,3,…
So the solution for Ez(x,y,z) is
Ez  A2 A4 sin k x x sin k y y e  jz
TM Mode

Substituting
 mp
E z  Eo sin 
 a
where
  np   j  z
x  sin 
y e
  b 
 mp   np 
2
2
h2  
 
  k
 a   b 
2
2
7
TMmn
TM modes
 mp   np   jz
Ez  Eo sin
x  sin
y e
 a   b 
Hz  0


Other components are
Ex  
 E z

  mp 
 mpx   npy  z
Ex   2 
 Eo cos
 sin
e
h  a 
 a   b 
  np 
m
p
x
n
p
y

 
 z
Ey   2 
 Eo sin
 cos
e
h  b 
 a   b 
j  np 
 mpx   npy  z
Hx  2 
 Eo sin
 cos
e
h  b 
 a   b 
h 2 x
 E z
Ey   2
h y
j E z
Hx  2
h y
j E z
Hy   2
h x
The m and n represent the mode of propagation
and indicates the number of variations of the
field in the x and y directions
Note that for the TM mode, if n or m is zero, all
fields are zero.
j  mp 
 mpx   npy  z

 Eo cos
 sin
e
h2  a 
 a   b 
Hy  
Cutoff frequency
Eigen values
We define
h2   2  u2
  h 2  u2
From
Where βu is the phase constant k
The cutoff frequency exists when  = 0 or h 2   u2  c2 
Solutions for several WG problems will exist only for real
numbers of h or “eigen values” of the boundary value
problems, each eigen value determines the characteristic of
the particular TM mode.
or
k
2
x
Hz.
  h 1 (
f 2
) .
fc
We can write
Evanescent:
TM Cutoff

h
2p 
fc 
 mp   np 
When  2   
 

 a   b 
The cutoff frequency occurs when
 mp   np 
2
When c   
 

 a   b 
2
or f c 
1
2p
1

 k y2   k 2
2
2
2

2
x
 mp   np 
2
 
 
   
 a   b 
 mp   np 
2
 
 
   
 a   b 
2
k

 k y2   k 2
2
   and   0
2
then     j  0
 mp   np 

 

 a   b 
2
2
2

Means no propagation, everything is attenuated
8
Propagation:
Cutoff
k

2
x
 k y2   k 2
attenuation
Propagation
of mode mn
 mp   np 
2
 
 
   
 a   b 
2
2
fc,mn

This is the case we are interested since is when the wave is allowed to
travel through the guide.

The cutoff frequency is the frequency below which attenuation
occurs and above which propagation takes place. (High Pass)
2
f c mn 
 mp   np 
When  2   
 

 a   b 
2
2

  j and   0
u'  m   n 
   
2  a  b
2
The phase constant becomes
f 
 mp   np 
 
   ' 1   c 
 a   b 
 f 
2
2
2
   2   
Phase velocity and impedance

Wave in the dielectric
medium
The phase velocity is defined as
up 

Summary of TM modes

'

2p


 '   / u'   
up
 '  / 
Ey
f 
E
 x 
 ' 1  c 
Hy
Hx
f 
u'   /  '  f  1 / 
2
 '  u' / f
Related example of how fields look:
Parallel plate waveguide - TM modes
Ez
 fc 

f 
2
TM   ' 1  
up 

 fc 

f 
2
/
 ' 1 
 
'
 f 
1  c 
 f 
2
H z  B1 cos k x x  B2 sin k x x B3 cos k y y  B4 sin k y y e z
E x  0 at y  0 ,b
 Boundary
conditions: E y  0 at x  0 ,a
m=1
0
a x
From these, we conclude:
X(x) is in the form of cos kxx,
where kx=mp/a, m=0,1,2,3,…
Y(y) is in the form of cos kyy,
where ky=np/b, n=0,1,2,3,…
So the solution for Ez(x,y,z) is
H z  B1B3 cos k x x cos k y y e  jz
m=3
z
2
TE Mode
 m px  e jt   z 
 A sin

a 

Ez
m=2
f 
   ' 1  c 
f 
f
And the intrinsic impedance of the mode
is
TM
Inside the waveguide
a x
9
TE Mode

Substituting
 mpx   np   jz
H z  H o cos
y e
 cos
 a   b 
where again
 mp   np 
h2  
 

 a   b 
2

TEmn
  np   jz
x  cos
y e
  b 
Other components are
j H z
h 2 y
j H z
Ey   2
h
x
 H z
Hx   2
h x
 H z
Hy   2
h y
Ex  
Note that n and m cannot be both zero
because the fields will all be zero.
Cutoff
 mp
H z  H o cos
 a
Ez  0

2
j  np 
 mpx   npy  z

 H o cos
 sin
e
h2  b 
 a   b 
j  mp 
 mpx   npy  z
Ey   2 
 H o sin
 cos
e
h  a 
 a   b 
Ex 
j   mp 
 mpx   npy  z

 H o sin
 cos
e
h2  a 
 a   b 
j   np 
m
p
x
n
p
y

 
 z
Hy  2 
 H o cos
 sin
e
h  b 
 a   b 
Hx 

The cutoff frequency is the same expression as for the TM mode
2
f c mn 

u'  m   n 
   
2  a  b
2
But the lowest attainable frequencies are lowest because here n or
m can be zero.
attenuation
Propagation
of mode mn
fc,mn
10
Dominant Mode



Summary of TE modes
Wave in the dielectric
medium
The dominant mode is the mode with
lowest cutoff frequency.
It’s always TE10
The order of the next modes change
depending on the dimensions of the
guide.
Inside the waveguide
f 
 '   / u'   
TE 
 '  / 
/
 
'
 f 
1  c 
 f 
2
Consider a length of air-filled copper X-band
waveguide, with dimensions a=2.286cm,
b=1.016cm operating at 10GHz. Find the
cutoff frequencies of all possible propagating
modes.
Solution:
 From the formula for the cut-off frequency

TE
0
2
Example:
Wave impedance varies with
frequency and mode
’
 fc 

f 
2
 ' 1 
 '  u' / f

'
f 
1  c 
f 

up 
u'   /  '  f  1 / 
Variation of wave impedance
2
   ' 1  c 
f 
2
TM
f c mn 
u'  m   n 
   
2  a  b
2
fc,mn
Group velocity, ug
Example
An air-filled 5-by 2-cm waveguide has
Ez  20 sin40px sin50py e
 jz
at 15GHz
 What mode is being propagated?
 Find 
 Determine Ey/Ex

V/m
Is the velocity at which
the energy travels.
ug 

f 
1
 u' 1   c 
 / 
f 
2
 rad/s   m 
 rad/m    s 

  
Ey  
j  p 
 mpx  z
  H o sin
e
h2  a 
 a 
It is always less than u’
u p u g  u '
2
11
Example 1
Group Velocity
Calculate the cutoff frequency for

As frequency is increased,
the group velocity increases.
the first four modes of WR284
waveguide.
Rectangular Waveguide
Fundamentals (Cont’d..)
Solution to Example 1
From table, the guide dimensions are a=2.840 inches and
b=1.340 inches. Converting to metric units:
a  7.214 cm
b  3.404 cm
Therefore,
fc10 
Solution
Then, we have:
1
2 
2
2
1
c
1 0
c
where u p 
      fc10 

2a
a b
Solution 1(Cont’d..)
and for the fourth mode,
3 108 m
s 100 cm  2.08 GHz
fc10 
27.214 cm  1 m
This agrees with the cutoff frequency cited in table. Then :
fc01 
3 108 m 100 cm
c
s

 4.41 GHz
2b 23.404 cm  1 m
fc20 
c
 4.16 GHz
a
3 108 m
2
2
1
1
 
 100 cm
s 
 

2
 7.214 cm   3.404 cm  1 m
 4.87 GHz
fc11 
not same with fc 10 since a ≠ 2b
12
Solution to Example 2
Example 2
A rectangular waveguide with dimension a=2.5
cm, and b=1 cm is to operate below 15.1 GHz.
How
many
TE
and
TM
modes
can
the
waveguide transmit if the guide is filled with
The cutoff frequency is given by:
fcmn 
up 
Or,
Solution to Example 2
Since we are looking for cutoff freq below 15.1 GHz, a
systematic way is to fix m or n and increase the other
until fcmn is greater than 15.1 GHz. So, by fixing m and
increasing n,
For TE01 (m  0, n  1)  fc01  32.5  7.5 GHz
TE02 (m  0, n  2)  fc02  35  15 GHz
TE03 (m  0, n  3)  fc03  37.5  22.5 GHz
Thus, for fcmn < 15.1 GHz, the maximum n = 2.
Solution to Example 2
We know the maximum value of m and n, so try other
possible combinations in between the maximum values.
2
2p 
2
m n
   
 a  b
2
With a = 2.5b or a/b = 2.5,
medium characterized by σ=0, µr=1, ε=4ε0.
Calculate the cutoff frequencies of the modes.
2
1
 mp   np 

 
 
2 
 a   b 
1
1


c
r r

c
2
So,
fcmn 
c
4a
2
m2  a 2 n 2
b
fcmn  3 m2  6.25n 2 GHz
Solution to Example 2
Then, fix n and increase m,
For TE10 (m  1, n  0)  fc10  3 GHz
TE20  fc20  6 GHz
TE30  fc30  9 GHz
TE40  fc40  12 GHz
TE50  fc50  15 GHz
TE60  fc60  18 GHz
Thus, for fc mn < 15.1 GHz, the maximum m = 5.
Solution to Example 2
Those modes whose cutoff freq are less or equal to 15.1
GHz will be transmitted, that is 11 TE modes and 4 TM
modes, as illustrated below:
For TE11 , TM 11 (degeneratemodes)  fc11  3 7.25  8.078 GHz
TE21 , TM 21  fc21  3 10.25  9.6 GHz
TE31, TM 31  fc31  3 15.25  11.72 GHz
TE41 , TM 41  fc41  3 22.25  14.14 GHz
TE12 , TM 12  fc12  3 26  15.3 GHz
13
Solution to Example 3
Example 3
In a rectangular waveguide with dimension a=1.5 cm,
b=0.8 cm, σ=0, µ=µ0, ε=4ε0, find:

We could find that the given expression is in instantaneous field
expression form which obtained from the phasor forms by using:


A
 px   3py 
11
H x  2 sin   cos
 sin p  10 t  z
m
 a   b 
• The mode of operation
• The cutoff frequency
• The phase constant
• The propagation constant
• The intrinsic wave impedance
2

is the guide is operating at TM13 or TE13.
Suppose that after this, we choose TM13 mode.
Solution to Example 3
The phase constant,
Where for air filled waveguide,
c m n
   
2  a  b

H  Re H S e jt
From the given expression we could find that m=1, and n=3. That
Solution to Example 3
fcmn 

E  Re E S e jt
2
f 
2
    1   c  
 f 
Hence, for m=1 and n=3, the cutoff frequency is:
 r
c
f 
1  c 
 f 
2
Where,
fcmn
c

4

1

1
.
5

102


2

 
3
  
0
.
8

102
 

2


  28.57 GHz

  2pf  p 1011

Solution to Example 3
The propagation constant,
    j
but   0
p 1011(2)
3 10
8
or f 
100
 50 GHz
2
2
 28.57 
1 
  1718.81 rad / m
 50 
Power transmission

The average Poynting vector for the waveguide
fields is
1
1
*
*
*
2
So,
2
Ex  E y
2
  j

,   j1718.81 / m
The intrinsic wave impedance,
f 
2
377
TM   ' 1   c  
r
 f 

2
2

[W/m2]
zˆ
where  = TE or TM depending on the mode
2
 28.57 
1 
  154.7 
 50 

Pave  ReE  H   Re E x H y  E y H x
Because it’s in propagating mode,
Pave   Pave  dS 
a
b
 
x 0 y 0
Ex  E y
2
2
2
dydx
[W]
14
Attenuation in Lossy waveguide

When dielectric inside guide is lossy, and walls
are not perfect conductors, power is lost as it
travels along guide.
Attenuation for TE10

Dielectric attenuation, Np/m
d  
Pave  Po e 2z



The loss power is
PL  
dPave
 2Pave
dz

Where c+d are the attenuation due to ohmic
(conduction) and dielectric losses
Usually c >> d
Waveguide Cavities




 '
f 
2 1   c 
 f 
Dielectric
conductivity!
2
Conductor attenuation, Np/m
c  
2


 0.5  b  f c ,10  
 f  

a


 f c ,10  


b ' 1  
 f 
2 Rs
2
Cavity TM Mode to z
Cavities, or resonators, are used for storing
energy
Used in klystron tubes, band-pass filters and
frequency meters
It’s equivalent to a RLC circuit at high
frequency
Their shape is that of a cavity, either cylindrical
or cubical.
Solving by Separation of Variables :
E z ( x, y, z )  X ( x)Y ( y ) Z ( z )
from where we obtain :
X(x)  c1 cos k x x  c2 sin k x x
Y(y)  c3 cos k y y  c4 sin k y y
Z ( z )  c5 cos k z z  c6 sin k z z
where k 2  k x2  k y2  k z
TMmnp Boundary Conditions
2
Resonant frequency
E z  0 at y  0,b
From these, we conclude:
kx=mp/a
ky=np/b
kz=pp/c
m= 1,2,3…………
n= 1,2,3…………
p= 0,1,2,3…………..
where c is the dimension in z-axis
 mpx   npy   ppz 
E z  Eo sin
 sin
 cos

 a   b   c 
where
E z  0 at x  0,a

E y  E x  0, at z  0,c
The resonant frequency is the same
for TM or TE modes, except that the
lowest-order TM is TM111 and the
lowest-order in TE is TE101.
2
c
fr 
2
u'  m   n   p 
     
2  a  b  c 
2
2
2
  mp   np   pp 

 
 

2
a

b 
c
2

 mp   np   pp 
2
k2  
 
 
   
 a   b   c 
2
2
2
15
TEmnp Boundary Conditions
Cavity TE Mode to z
Solving by Separation of Variables :
H z ( x, y, z )  X ( x)Y ( y ) Z ( z )
from where we obtain :
X(x)  c1 cos k x x  c2 sin k x x
H z  0 at z  0,c
From these, we conclude:
kx=mp/a
ky=np/b
kz=pp/c
where c is the dimension in z-axis
E y  0 at x  0,a
 mpx   npy   npz 
H z  H o cos
 cos
 sin

 a   b   b 
c
E x  0, at y  0,b
Y(y)  c3 cos k y y  c4 sin k y y
Z ( z )  c5 cos k z z  c6 sin k z z
where k 2  k x2  k y2  k z
2
Quality Factor, Q



Quality Factor, Q
The cavity has walls with finite
conductivity and is therefore losing
stored energy.
The quality factor, Q, characterized the
loss and also the bandwidth of the
cavity resonator.
Dielectric cavities are used for
resonators, amplifiers and oscillators at
microwave frequencies.

Is defined as
Q  2π
 2p
Time average energystored
loss energy per cycle of oscillation
W
PL
For the dominant mode TE 101
QTE101
a

 c 2 abc

3
 2b a  c 3  ac a 2  c 2
 
2


where


1
pf101 o c
Example
For a cavity of dimensions; 3cm x 2cm x 7cm filled with
air and made of copper (c=5.8 x 107)
 Find the resonant frequency and the quality factor
2
2
2
for the dominant mode.
3 1010  1   1   0 
f r110 
         9GHz
2
Answer:
 3  2  7 
3 1010  1   0   1 
         5.44GHz
2
 3  2  7 
2
fr 

QTE101 
2
2
1
(5.44 109 ) o c
3
2
Summary.
 1.6 106

 72 3  2  7
 2  233  7 3   3  732  7 2 
 568,378
16
Maxwell’s equations (no sources)


  H  iE



  E  iH
 E x  2i 2 (k z Exz   Hyz )
k k z

 E y  2i 2 (k z Eyz   Hxz )
k k z


E z
H z
i
 H x  k 2  k z2 ( y  k z x )

E z
i
H 
( x  k z Hyz )

 y k 2  k z2
 E x  2i 2 (k z Exz   Hyz )
k k z

 E y  2i 2 (k z Eyz   Hxz )
k k z


E z
H z
i
 H x  k 2  k z2 ( y  k z x )

E z
i
H 
( x  k z Hyz )
 y k 2  k 2
TM: HZ= 0, to solve Ez
to solve EZ、HZ
TE: Ez= 0, to solve Hz
z
TE
TEmn mode
Helmholtz equation
E x  i k 2k 2 ( nbp ) H mn cos(map x) sin( nbp y )e ik z z
E y  i k 2k 2 ( map ) H mn sin( map x) cos(nbp y )e ikz z
x  0, a : E y  0, E z  0  Hxz
z

i

j

k

x

y

z
Hx

i
i E x  Hyz  ik z H y

i E y  ik z H x  Hxz

H
i EZ  xy  Hyx


E z

i

H

x  y  ik z E y



i

H


ik z E x  Exz
y



E y
E x

i

H


z
x  y



 i ( E x i  E y j  E z k )
Hy Hz


j k

x

y

z
Ex
Ey
Ez



 i ( H x i  H y j  H z k )
y  0, b : E x  0, E z  0 
H x  i k 2kzk 2 ( map ) H mn sin( map x) cos(nbp y )e ik z z
z
H z
x
H z
x
H z
y
H y  i k 2kzk 2 ( nbp ) H mn cos(map x) sin( nbp y )e ikz z
z
H z  H mn cos( x) cos( y )e
mp
a
np
b
ik z z
H z  H mn cos(map x) cos(nbp y)eikz z
Fields
E y  i
mp
a
( ) H mn sin(
mp
a
np
b
Propagation constant:   k z  k 2  k x2  k y2  k 2  (
mp 2 np 2
) ( )
a
b
x) cos( y )e
f c mn 
Cutoff frequency:
H x  i k 2 k 2 ( map ) H mn sin( map x) cos(nbp y )e
kz
ik z z
z
Wave impedance:
H y  i k 2kzk 2 ( nbp ) H mn cos(map x) sin( nbp y )e ik z z
ZTE 
 Kx 
y b
y 0
0
Separate variable
 D2  0
H z ( x, y, z )  X ( x)Y ( y)e ikz Z
mp
, m  0,1,2,...
a
np
 Ky 
, n  0,1,2,...
b
General solution
Separated PDE
H z ( x, y )  X ( x)Y ( y )
 2 X ( x)
 k x2 X ( x)  0
2 x
 2Y ( y )
 k y2Y ( y )  0
2
 y
(C2 cos k y y  D2 sin k y y )
Amn = C1C2
fields
E x  i k 2kzk 2 ( map ) Emn cos(map x) sin( nbp y )e  ik z z
k x2  k y2
k
1


2p 
2p 
2p 
(
mp 2 np 2
) ( )
a
b
E y  i k 2  k 2 ( nbp ) Emn sin( map x) cos(nbp y )e
Kz
Propagation constant:   k z  k 2  k x2  k y2  k 2  (
ik z z
mp 2 np 2
) ( )
a
b
H x  i k 2k 2 ( nbp ) Emn sin( map x) cos(nbp y )e ik z z
E y k
Ex


Hy
Hx 
f c mn 
Cutoff frequency:
z
E z  Emn sin( map x) sin( nbp y )e ik z z
z
Wave impedance:
ZTM 
H y  i k 2k 2 ( map ) Emn cos(map x) sin( nbp y )e ik z z
k x2  k y2
k
1


2p 
2p 
2p 
(
mp 2 np 2
) ( )
a
b
E y 
Ex


Hy
Hx
k
z
z
H z  H mn cos(map x) cos(nbp y )e ik z z
xa
H z
y
z
ik z z
Ez  0
 D1  0
0
TMmn mode
z

k 2  k z2
x 0
x 0, x  a
H z
y y  0 , y b
 (C1 cos k x x  D1 sin k x x)
TEmn mode
E x  i K 2K 2 ( nbp ) H mn cos(map x) sin( nbp y )e ik z z
2 H Z  k 2 H Z  0
Boundary conditions
z
expansion
Phase velocity:
vp 




k2 (
mp 2 n p 2
) ( )
a
b


k

TE10 mode (the fundamental mode)
E y  i pa H10 sin( pa x)e ikz
H x  i pa ( map ) H10 sin( pa x) cos(nbp y )e ikz
H z  H10 cos(pa x)e ikz
Ex  Ez  H y  0
1

 v p _ dielec
Hz  0
Phase velocity:
vp 




k2 (
mp 2 n p 2
) ( )
a
b


k

1

 v p _ dielec
TM11 mode (the lowest mode)
E x  i akmp2 E11 cos(pa x) sin( pb y )e  ikz
c
E y  i bknp2 E11 sin( pa x) cos(pb y )e ikz
c
E z  E11 sin( pa x) sin( pb y )e ikz
H x  i bkn2p E11 sin( pa x) cos(nbp y )e ikz
c
H y  i akm2p E11 cos(pa x) sin( nbp y )e ikz
c
Hz  0
kc  (
mp 2 np 2
) ( )
a
b
Use Cylindrical coordinates
Circular Waveguides
17
From Maxwell Eqs. we can derive
for Ez and Hz, all other components:
TE Modes: (Ez=0)
Solving the wave equation for Hz:
Use Separation of Variables:
TEM waves
not supported
In cylindrical coordinates:
where
And we have assumed waves travel to +z , so
Since they are both equal to a constant Kc, we can
separate by variables:
This is Bessel’s Equation!
Since Hz must be periodic:
This is general solution for Bessel’s
Equation!
Jn=Bessel function of the 1st kind
kf must be an integer
Yn=Bessel function of the 2nd Kind
n is the order
Bessel Function of the 1st Kind
Bessel’s Functions
Bessel functions
are the radial part
of the modes of
vibration of a
circular drum and
circular antennas!
Similar to Sine and Cosine but
amplitude goes down w/argument
Jn
Yn
n is the order
18
So we are left with:
Substituting
The Cutoff frequency is:
We need to satisfy:
derivative
Therefore, we need:
Note we have A and
B, which depend
on excited power.
TE11 is the dominant mode
Due to symmetry of guide, we can rotate the axis of
the coordinate system so that either A or B are zero:
TM Modes: (Hz=0)
Solving the wave equation for Ez:
Ñ2 Ez + k 2 Ez = 0
Use Separation of Variables:e (r, f ) = R(r )P(f )
z
In cylindrical coordinates:
Following similar procedure as for TE, now for TM we
obtain:
The Propagation Constant
Bessel
The cutoff frequency:
19
TM fields and impedance
Modes of Propagation
Dominant Mode
20
Conventional sizes
d=2a
Advantages
• Circular polarization waves and virtually any other
type of polarization can be propagated thru it.
• Circular waveguides offer implementation
advantages over rectangular waveguide in that
installation is much simpler when forming runs for
turns and offsets.
• Manufacturing is generally simpler, too.
21