Lecture5.pdf

Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 1
Waveguides
A. Nassiri
Lecture 5

Waveguides are used to transfer electromagnetic power efficiently from one point
in space to another.
Waveguides
x
y
z
Coaxial line
Two-wire line
Microstrip line
Rectangular waveguide Dielectric waveguide

Waveguides
In practice, the choice of structure is dictated by: (a) the desired operating frequency
band, (b) the amount of power to be transferred, and (c) the amount of transmission
losses that can be tolerated.
Coaxial cables are widely used to connect RF components. Their operation is practical
for frequencies below 3 GHz. Above that the losses are too excessive. For example, the
attenuation might be 3 dB per 100 m at 100 MHz, but 10 dB/100 m at 1 GHz, and 50
dB/100 m at 10 GHz. Their power rating is typically of the order of one kilowatt at 100
MHz, but only 200 W at 2 GHz, being limited primarily because of the heating of the
coaxial conductors and of the dielectric between the conductors (dielectric voltage
breakdown is usually a secondary factor.)
Another issue is the single-mode operation of the line. At higher frequencies, in order to
prevent higher modes from being launched, the diameters of the coaxial conductors
must be reduced, diminishing the amount of power that can be transmitted. Two-wire
lines are not used at microwave frequencies because they are not shielded and can
radiate. One typical use is for connecting indoor antennas to TV sets. Microstrip lines
are used widely in microwave integrated circuits.

Waveguides
In a waveguide system, we are looking for solutions of Maxwell’s equations that are
propagating along the guiding direction (the z direction) and are confined in the near
vicinity of the guiding structure. Thus, the electric and magnetic fields are assumed to
have the form:
( ) ( ) z
j
t
j
e
y
x
E
t
z
y
x
E β
−
ω
= ,
;
,
,
( ) ( ) z
j
t
j
e
y
x
H
t
z
y
x
H β
−
ω
= ,
;
,
,
Where β is the propagation wave number along the guide direction. The
corresponding wavelength, called the guide wavelength, is denoted by λg=2π/β .
The precise relationship between ω and β depends on the type of waveguide structure
and the particular propagating mode. Because the fields are confined in the
transverse directions (the x, y directions,) they cannot be uniform (except in very
simple structures) and will have a non-trivial dependence on the transverse
coordinates x and y. Next, we derive the equations for the phasor amplitudes E (x, y)
and H (x, y).

Because of the preferential role played by the guiding direction z, it proves convenient to
decompose Maxwell’s equations into components that are longitudinal, that is, along the z-
direction, and components that are transverse, along the x, y directions. Thus, we decompose:
Waveguides
( ) ( ) ( ) ( ) ( ) ( )
y
x
E
z
y
x
E
y
x
E
z
y
x
E
y
y
x
E
x
y
x
E z
T
al
longitudin
z
transverse
y
x ,
ˆ
,
,
ˆ
,
ˆ
,
ˆ
, +
≡
+
+
=







 


 

In a similar fashion we may decompose the gradient operator:
z
j
z
z
y
x T
z
T
z
y
x ˆ
ˆ
ˆ
ˆ
ˆ β
−
∇
=
∂
+
∇
=
∂
+
∂
+
∂
=
∇
Where we made the replacement ∂z  -jβ because of the assumed z-dependence. Introducing
these decompositions into the source-free Maxwell’s equation we have:
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
( ) ( ) 0
0
0
0
=
+
⋅
β
−
∇
=
⋅
∇
=
+
⋅
β
−
∇
=
⋅
∇
+
ωε
=
+
×
β
−
∇
ωε
=
×
∇
+
ωµ
−
=
+
×
β
−
∇
ωµ
−
=
×
∇
z
T
T
z
T
T
z
T
z
T
T
z
T
z
T
T
H
z
H
z
j
H
E
z
E
z
j
E
E
z
E
j
H
z
H
z
j
E
j
H
H
z
H
j
E
z
E
z
j
H
j
E
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ

Transverse and Longitudinal Components
( )
( )E
E
z
E
E
xy
xy
z
xy




2
2
2
2
2
2
2
2
γ
+
∇
=








∂
∂
+
∇
=
∇
+
∇
=
∇
The wave equations
become now
( )
( )
2 2 2
2 2 2
0
0
xy
xy
E k E
H k H
γ
γ
∇ + + =
∇ + + =
 �

�
 �

�

Solution strategy
We still have (seemingly) six simultaneous equations to solve.
In fact, the 6 are NOT independent. This looks complicated!
Adopt a strategy of expressing the transverse fields (the Ex,Ey,
Hx,Hy components in terms of the longitudinal components Ez
and Hz only. If we can do this we only need find Ez and Hz from
the wave equations….Too easy eh!
The first step can be carried out directly from the two curl
equations from the original Maxwell’s eqns. Writing these out:

First step
( )
( )
( )
3
2
1
z
x
y
y
z
x
x
y
z
H
j
y
E
x
E
H
j
x
E
E
H
j
E
y
E
ωµ
ωµ
γ
ωµ
γ
−
=
∂
∂
+
∂
∂
−
=
∂
∂
−
−
−
=
+
∂
∂
( )
( )
( )
6
5
4
z
x
y
y
z
x
x
y
z
E
j
y
H
x
H
E
j
x
H
H
E
j
H
y
H
ωε
ωε
γ
ωε
γ
=
∂
∂
+
∂
∂
=
∂
∂
−
−
=
+
∂
∂
All replaced by -γ. All fields are functions of x and y only.
z
∂
∂

Result
Now, manipulate to express the transverse in terms of the
longitudinal. E.g. From (1) and (5) eliminate Ey
x
z
x
z
H
j
x
H
H
j
y
E
ωµ
γ
ωε
γ
−
=






∂
∂
−
−
+
∂
∂
transverse
longitudinal
2
2
2
2
where
1
k
k
y
E
j
x
H
k
H
c
z
z
c
x
+
γ
=








∂
∂
ωε
−
∂
∂
γ
−
=
kc is an eigenvalue
(to be discussed)

The other components








∂
∂
−
∂
∂
−
=








∂
∂
+
∂
∂
−
=








∂
∂
+
∂
∂
−
=








∂
∂
−
∂
∂
−
=
x
H
j
y
E
k
E
y
H
j
x
E
k
E
x
E
j
y
H
k
H
y
E
j
x
H
k
H
z
z
c
y
z
z
c
x
z
z
c
y
z
z
c
x
ωµ
γ
ωµ
γ
ωε
γ
ωε
γ
2
2
2
2
1
1
1
1
So find solutions for
Ez and Hz and then use
these 4 eqns to find all
the transverse components
We only need to find
Ez and Hz now!

Wave type classification
It is convenient to to classify as to whether Ez or Hz exists
according to:
TEM: Ez = 0 Hz = 0
TE: Ez = 0 Hz ≠ 0
TM Ez ≠ 0 Hz = 0
We will first see how TM wave types propagate in waveguide
Then we will infer the properties of TE waves.

The TE modes of a parallel plate wave guide are preserved if
perfectly conducting walls are added perpendicularly to the electric
field.
On the other hand, TM modes of a parallel wave guide disappear if
perfectly conducting walls are added perpendicularly to the
magnetic field.
E
H
The added metal plate does not
disturb normal electric field and
tangent magnetic field.
H
E
The magnetic field cannot be
normal and the electric field
cannot be tangent to a perfectly
conducting plate.

TM waves (Hz=0)
y
E
k
E
x
E
k
E
x
E
k
j
H
y
E
k
j
H
z
c
y
z
c
x
z
c
y
z
c
x
∂
∂
−
=
∂
∂
−
=
∂
∂
−
=
∂
∂
=
2
2
2
2
γ
γ
ωε
ωε
Longitudinal: 2nd order PDE for
Ez. we defer solution until we have
defined a geometry plus b/c.
Transverse solutions once
Ez is found
( ) 0
2
2
2
=
+
γ
+
∇ z
z
xy E
k
E

Further Simplification
The two E-components can be combined. If we use the notation:
y
y
x
x
E
k
y
E
x
E
E xy
z
xy
c
y
x
t
ˆ
ˆ
where
ˆ
ˆ 2
∂
∂
+
∂
∂
=
∇
∇
−
=
+
=
γ

TM
x
y
y
x
TM
z
xy
c
t
Z
E
z
H
j
H
E
H
E
Z
E
k
E

 ×
=
Ω
=
−
=
=
∇
−
=
ˆ
2
ωε
γ
γ

Eigenvalues
We will discover that in closed systems, solutions are possible
only for discrete values of kc. There may be an infinity of values
for kc, but solutions are not possible for all kc. Thus kc are
known as eigenvalues. Each eigenvalue will determine the
properties of a particular TM mode. The eigenvalues will be
geometry dependent.
Assume for the moment we have determined an appropriate
value for kc, we now wish to determine the propagation
conditions for a particular mode.

We have the following propagation vector components for the modes
in a rectangular wave guide
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2





 π
−





 π
−
µε
ω
=
β
β
−
β
−
µε
ω
=








λ
π
=








λ
π
=
β
π
=
β
π
=
β
β
+
β
+
β
=
µε
ω
=
a
n
a
m
a
n
a
m
β
z
y
x
g
z
z
y
x
z
y
x
;
At the cut-off, we have
( )
2
2
2
2
2
0 




 π
−





 π
−
µε
π
=
=
β
a
n
a
m
fc
z

Operating bandwidth
All waveguide systems are operated in a frequency range that ensures that only the lowest
mode can propagate. If several modes can propagate simultaneously, one has no control
over which modes will actually be carrying the transmitted signal. This may cause undue
amounts of dispersion, distortion, and erratic operation.
A mode with cutoff frequency ωc will propagate only if its frequency is ω≥ ωc, or λ <
λc. If ω< ωc, the wave will attenuate exponentially along the guide direction. This
follows from the ω,β relationship
2
2
2
2
2
2
2
2
c
c c
c
ω
−
ω
=
β
⇒
β
+
ω
=
ω
If ω≥ ωc, the wavenumber β is real-valued and the wave will propagate. But if ω< ωc,
β becomes imaginary, say, β = -jα, and the wave will attenuate in the z-direction, with
a penetration depth δ= 1/α:
z
z
j
e
e α
−
β
−
=

Operating bandwidth
If the frequency ω is greater than the cutoff frequencies of several modes, then all of
these modes can propagate. Conversely, if ω is less than all cutoff frequencies, then
none of the modes can propagate.
If we arrange the cutoff frequencies in increasing order, ωc1< ωc2 <ωc3 < · · · , then, to
ensure single-mode operation, the frequency must be restricted to the interval ωc1
<ω<ωc2, so that only the lowest mode will propagate. This interval defines the
operating bandwidth of the guide.
This applies to all waveguide systems, not just hollow conducting waveguides. For
example, in coaxial cables the lowest mode is the TEM mode having no cutoff
frequency, ωc1 = 0. However, TE and TM modes with non-zero cutoff frequencies do
exist and place an upper limit on the usable bandwidth of the TEM mode. Similarly, in
optical fibers, the lowest mode has no cutoff, and the single-mode bandwidth is
determined by the next cutoff frequency.

The cut-off frequencies for all modes are
2
2
2
2
2
2
1






+






=
λ






+






µε
=
a
n
a
m
a
n
a
m
f
c
c
With cut-off wavelengths
With indices
TE modes m=0,1,2,3,… TM modes m=1,2,3,…
n=0,1,2,3,… n=1,2,3,…
(but m=n=0 not allowed)

Cut-off
Since the wave propagates according to e±γz. Then propagation
ceases when γ = 0.
2
2
2
2
implies
0
then
since c
c
c k
k =
=
−
= µε
ω
γ
µε
ω
γ
µε
π
2
c
c
k
f =
Or
Cut-off
frequency

Write γ in terms of fc
It is usual, now to write γ in terms of the cut-off frequency.
This allows us to physically interpret the result.
2
2
2
2
2
2
2
2
1
c
c
c
c
c
c
f
f
k
k
k
k −
=
−
=
−
=
ω
ω
µε
ω
γ
This part from
the definition
see slide 8/5.
Substitute for µε
from definition of fc
Recall similarity
of this result with
β for an ionized gas.
see slide 7/12
2
2
2
k
kc −
=
γ

Conditions for Propagation
There are two possibilities here:
1 c
f
f > γ is imaginary
2
2
2
2
2
2
1
1
with
f
f
jk
k
k
jk
k
k
j
j
c
c
c
−
=
−
=
−
=
= β
γ
We conclude that if the operational frequency is above
cut-off then the wave is propagating with the form e-jβz
this is a special case of
the result in the previous slide
This says now that
γ becomes jβ with
2
2
c
k
k −
=
β

Different wavelengths
The corresponding wavelength inside the guide is
λ
λ
β
π
λ >
−
=
=
2
2
1
2
f
fc
g
g for guide
This is the “free space”
wavelength
The free space
wavelength may
be written alternatively
k
π
λ
2
=
Now if we introduce a cut-off wavelength
λ=v/fc where v is the corresponding velocity
(=c, in air) in an unbounded
medium. We can derive: 2
2
2
1
1
1
c
g λ
λ
λ
+
=

Dispersion in waveguides
The previous relationship showed that β was a function of
frequency i.e. waveguides are dispersive. Hence we expect the
phase velocity to also be a function of frequency. In fact:
v
v
f
f
v
v
g
c
p >
=
−
=
=
λ
λ
β
ω
2
2
1
So, as expected the phase velocity is always higher than in an
unbounded medium (fast wave) and is frequency dependent.
So we conclude waveguides are dispersive.
This can be > c!

Group velocity
This is similar to as discussed previously.
v
v
f
f
v
v
g
c
g <
=
−
=
∂
∂
=
λ
λ
ω
β 2
2
1
1
So the group velocity is always less than in an unbounded
medium. And if the medium is free space then vgvp=v2=c2
which is also as previously discussed. Finally, recall that
the energy transport velocity is the group velocity.

Dispersion in waveguides
The previous relationship showed that β was a function of
frequency i.e. waveguides are dispersive. Hence we expect the
phase velocity to also be a function of frequency. In fact:
v
v
f
f
v
v
g
c
p >
=
−
=
=
λ
λ
β
ω
2
2
1
So, as expected the phase velocity is always higher than in an
unbounded medium (fast wave) and is frequency dependent.
So we conclude waveguides are dispersive.
This can be > c!

Wave Impedance
Wave impedance can also be written in terms of the radical:
2
2
1
f
fc
−
In particular:
2
2
2
2
1
1
f
f
Z
f
f
k
j
c
TM
c
−
=
=
−
= η
ωε
ωε
γ
The factor k/ωε can be
shown to be:
air)
(if
377
0
Ω
=
=η
ε
µ
For f>fc Then the impedance is
real and less than the surrounding
medium dielectric

Evanescent waves
2 c
f
f < γ is real
2
2
2
2
1
1
with
f
f
k
k
k
k c
c
−
=
−
=
=α
γ
We conclude that the propagation is of the form e-αz i.e.
the wave is attenuating or is evanescent as it propagates in
the +z direction. This is happening for frequencies below
the cut-off frequency. At f=fc the wave is said to be cut-off.
Finally, note that there is no loss mechanism contributing
to the attenuation.

Impedance for evanescent waves
A similar derivation to that for the propagating case produces:
2
2
1
c
c
TM
f
f
k
j
Z −
−
=
ωε
This says that for TM waves, the wave impedance is capacitive
and that no power flow occurs if the frequency is below cut-off.
Thus evanescent waves are associated with reactive power only.

TE Waves
A completely parallel treatment can be made for the case of
TE propagation, Ez = 0,Hz ≠ 0. We only give the parallel
results.
( )
( )
H
z
Z
E
f
f
j
Z
H
k
H
H
k
H
TE
c
TE
z
xy
c
TE
t
z
z
xy


×
−
=
−
=
=
∇
−
=
=
+
+
∇
ˆ
)
(
2
2
2
2
2
2
1
0
η
γ
ωµ
γ
γ

Dispersion
For propagating modes (γ =jβ), we may graph the variation of
β with frequency (for either TM or TE) and this determines the
dispersion characteristic.
2
2
2
2
1
ely
alternativ
or
medium
unbounded
in the
velocity
the
is
where
1
ω
ω
β
ω
β
c
c
v
v
f
f
k
−
=
−
=
This is more useful form
for plotting
Note that vp>v
vg<v
vpvg= v2
Equation of red plot

Wave Impedance
f/fc
Normalized frequency
Normalized
wave impedance
Z (ohms)
1
1
Evanescent
here
ZTE/η
ZTM/η

Dispersion for Waveguide
TEM
slope=ω/β
β
ω
ωc
Slope = group velocity = vg
Propagating TE and TM modes
Slope=phase velocity=vp

Rectangular waveguide
Convention always says that a is the long side.
X
Y
a
b
ε, µ
Assume perfectly conducting walls and perfect dielectric filling the wave guide.
b
a

TM waves
TM waves have Ez ≠ 0. We write Ez(x,y,z) as Ez(x,y)e-γz.
The wave equation to solve is then
( ) 0
2
2
2
2
2
=








+
∂
∂
+
∂
∂
y
x
E
k
y
x
z
c ,
Plus some boundary conditions on the walls of the
waveguide. The standard method of solving this PDE is to
use separation of variables. I.e..
( ) ( ) ( )
y
Y
x
X
y
x
Ez =
,

Possible Solutions
If we substitute into the original equation we get two more
equations. But this time we have full derivatives and we can
easily write solutions.
2
2
2
2
2
2
2
2
2
with
0
and
0
c
y
x
y
x
k
k
k
Y
k
dy
Y
d
X
k
dx
X
d
=
+
=
+
=
+
Mathematics tells us that the solutions depend on the sign of kx
2
kx
2 kx Appropriate X(x)
0 0 A0x+B0
+ k A1sin kx+B1cos kx; C1ejkx +D1e-jkx
- jk A2sinh kx +B2cosh kx; C2ekx +D2e-kx

Boundary conditions
Boundary conditions say that the tangential components of Ez
vanish on the walls of the guide :
( )
( )
( )
( )
walls.
bottom
and
top
0
,
0
0
,
E
walls.
hand
right
and
left
0
,
0
,
0
z



=
=



=
=
b
x
E
x
y
a
E
y
E
z
z
z
We choose the sin/cos form (why?) and directly write:
( ) ( )( )
y
k
B
y
k
A
x
k
B
x
k
A
y
x
E y
y
x
x
z cos
sin
cos
sin
, 2
2
1
1 +
+
=

Final solution
Using the boundary conditions, we find:
X(x) must be in the form sinkxx
Y(y) must be in the form sinkyy
,.....
,
,
, 3
2
1
and
integer
n
m,
with =







=
=
n
m
b
n
k
a
m
k
y
x
π
π
( )
2
2
2
0






+






=
=
b
n
a
m
k
b
y
n
a
x
m
E
y
x
E
c
z
π
π
π
π
sin
sin
,
We can only get discrete
values of kc -eigenvalues!
This satisfies all the
boundary conditions
Do not start from 0

Mode numbers (m,n)
The m,n numbers will give different solutions for Ez (as well as
all the other transverse components. Each m,n combination will
correspond to a mode which will satisfy all boundary and wave
equations. Notice how the modes depend on the geometry (a,b)!
We usually refer to the modes as TMmn or TEmn eg TM2,3
Thus each mode will specify a unique field distribution in the
guide. We now have a formula for the parameter kc once
we specify the mode numbers.
The concept of a mode is fundamental to many E/M problems.

From previous formulas, we have directly upon using the value kc
2
2
2
2
2
2
1






+






=






+






=
b
n
a
m
b
n
a
m
f
c
c
λ
µε
Note this!

TE Modes
For TE modes, we have Ez = 0, Hz ≠ 0 as before.
( ) 0
2
2
2
2
2
=








+
∂
∂
+
∂
∂
y
x
H
k
y
x
z
c ,







=
=
⇒
∂
∂
=
=
⇒
∂
∂







=
=
⇒
∂
∂
=
=
⇒
∂
∂
=
=
=
=
b
E
y
H
E
y
H
a
x
E
x
H
x
E
x
H
x
y
z
x
y
z
y
x
z
y
x
z
y
at
0
0
y
at
0
at
0
0
at
0
0
0
0
0
Boundary
conditions
Boundary conditions
for Hz (longitudinal)
are equivalently
expressed in terms of
Ex and Ey (transverse)

TEmn Results
The expressions for fc and λc are identical to the TM case.
But this time we have that the TE dominant mode (ie. the
TE mode with the lowest cut-off frequency) is TE10 This
mode has an even lower cut-off frequency than TM11 and
is said to be the Dominant Mode for a rectangular
waveguide.
( )
( )
( ) a
a
v
a
f
b
y
n
a
x
m
H
y
x
H
TE
c
TE
c
z
2
2
2
1
cos
cos
,
10
10
0
=
=
=
=
λ
µε
π
π
This is provided
we label the large
side ‘a’ and associate
this side with the
mode number ‘m’

View of TE10 mode for waveguide.
TE10
H field
E field

For mono-mode (or single-mode) operation, only the fundamental TE10 mode
should be propagating over the frequency band of interest.
The mono-mode bandwidth depends on the cut-off frequency of the second
propagating mode. We have two possible modes to consider, TE01 and TE20.
( )
( ) ( )
10
20
01
2
1
2
1
TE
f
a
TE
f
b
TE
f
c
c
c
=
µε
=
µε
=

( ) ( ) ( )
µε
=
=
=
⇒
=
a
TE
f
TE
f
TE
f
a
b c
c
c
1
2
2
10
20
01
If
( ) ( ) ( )
20
01
10
2
TE
f
TE
f
TE
f
a
b
a c
c
c <
<
⇒
>
>
If
f
Mono-mode Bandwidth
0 fc(TE10) fc(TE20)
f
Mono-mode Bandwidth
0 fc(TE10) fc(TE20)
fc(TE01)

( ) ( )
01
20
2
TE
f
TE
f
a
b c
c <
⇒
<
If
f
Mono-mode Bandwidth
0 fc(TE10) fc(TE20)
f
Useful Bandwidth
0 fc(TE10) fc(TE20) fc(TE01)
fc(TE01)
In practice , a safety margin of about 20% is considered, so that the useful
bandwidth is less than the maximum mono-mode bandwidth. This is
necessary to make sure that the first mode (TE10) is well above cut-off, and
the second mode (TE01 or TE20) is strongly evanescent.
Safety margin

If a=b (square wave guide) ⇒ ( ) ( )
20
10 TE
f
TE
f c
c =
f
0 fc(TE10) fc(TE20)
fc(TE01) fc(TE02)
In the case of perfectly square wave guide, TEm0 and TE0n
modes with m=n are are degenerate with the same cut-off
frequency.
Except for orthogonal field orientation, all other properties
of the degenerate modes are the same.

Example – Design an air-filled rectangular wave guide for the following
operation conditions:
a. 10 GHz in the middle of the frequency band (single mode operation)
b. b=a/2
The fundamental mode is the TE10 with cut-off frequency
( ) Hz
a
m
a
c
a
TE
fc
2
10
3
2
2
1 8
10
sec
/
×
≈
=
ε
µ
=


For b=a/2, TE01 and TE20 have the same cut-off frequency
( )
( ) Hz
a
m
a
c
a
TE
f
Hz
a
m
a
c
a
c
b
c
b
TE
f
c
c
sec
/
sec
/
8
20
8
01
10
3
1
10
3
2
2
2
2
1
×
≈
=
ε
µ
=
×
≈
=
=
=
ε
µ
=





The operation frequency can be expressed in terms of the cut-off frequencies
( ) ( ) ( )
( ) ( )





 ×
+
×
=
×
⇒
=
+
=
−
+
=
a
a
GHz
TE
f
TE
f
TE
f
TE
f
TE
f
f
c
c
c
c
c
8
8
9
01
10
01
10
10
10
3
2
10
3
2
1
10
0
10
0
10
2
2
.
.
cm
a
b
cm
a 125
1
2
25
2 .
. =
=
=
⇒

An example
We consider an air filled guide, so εr=1. The internal size of
the guide is 0.9 x 0.4 inches (waveguides come in standard sizes).
The cut-off frequency of the dominant mode:
( )
( ) GHz
k
f
a
k
b
n
a
m
k
c
TE
c
TE
c
c
56
6
2
10
3
43
137
2
43
137
10.16mm
4
0.
b
22.86mm;
9
0.
a
8
0
0
2
2
10
10
.
.
.
=
×
×
=
=
=
=
=
′
′
=
=
′
′
=






+






=
π
ε
µ
π
π
π
π

An example
The next few modes are:
( )
( )
( )
,11
01
20,
10,
is
mode
of
order
ascending
the
86
.
274
21
.
309
38
.
338
20
01
11





=
=
=
c
c
c
k
k
k
The next cuff-off frequency after TE10 will then be
( ) GHz
f TE
c 12
13
2
10
3
86
274 8
20
.
.
=
×
×
=
π
So for single mode operation we must operate the guide within
the frequency range of 6.56<f <13.12GHz.

It is not good to operate too close to cut-off for the reason that
the wall losses increases very quickly as the frequency approaches
cut-off. A good guideline is to operate between 1.25fc and 1.9fc.
This then would restrict the single mode operation to 8.2 to 12.5
GHz.
The propagation coefficient for the next higher mode is:
( ) ( ) 2
2
20
2
2
20
20
1
c
c
c
f
f
k
k
k −
=
−
=
γ
Specify an operating frequency f, half way in the original
range of TE10 i.e.. 9.84GHz.
An example

( )
.
evanescent
strongly
very
is
TE
ie.
dB/m
1581
8.7
x
181.8
dB
in
or
8
181
So
(Real)
8
181
12
13
84
9
1
86
274
20
2
20
=
=
=






−
=
m
Np /
.
.
.
.
.
α
γ
In comparison for TE10:
( ) ( )
rad/m
153.64
So
)
(Imaginary
64
153
56
6
84
9
1
43
137
1
2
2
10
2
10
10
=
=






−
=
−
=
β
γ j
f
f
k
c
c .
.
.
.
All further higher order modes will be cut-off with higher rates
of attenuation.
An example

Field Patterns
Field patterns for the TE10 mode in rectangular wave guide

The simple arrangement below can be used to excite TE10 in a
rectangular wave guide.
The inner conductor of the coaxial cable behaves like a dipole
antenna and it creates a maximum electric field in the middle of
the cross-section.
TE01
Closed end

Waveguide Cavity Resonator
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
a
b
d
p
β
b
n
β
a
m
β
z
y
x
π
=
π
=
π
=

The addition of a new set of plates introduces a condition for
standing waves in the z-direction which leads to the definition of
oscillation frequencies
2
2
2
2
1






+






+






µε
=
d
p
b
n
a
m
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 is
associated with a band of frequencies
larger the cut-off frequency.
In the resonator, resonant modes can
only exist in correspondence of discrete
resonance frequencies.
0 fc1 fc2 f fc1
0 fc2 f

The cavity resonator will have modes indicated as
TEmnp TMmnp
The values of the index corresponds to periodicity (number of sine or
cosine waves) in three direction. Using z-direction as the reference for
the definition of transverse electric or magnetic fields, the allowed
indices are





=
=
=





=
=
=
,...
,
,
,
,...
,
,
,
,...
,
,
,
,...
,
,
,
,...
,
,
,
,...
,
,
,
3
2
1
0
3
2
1
0
3
2
1
0
3
2
1
0
3
2
1
0
3
2
1
0
p
n
m
TM
p
n
m
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 the
magnetic field is going to be parallel to the third set of plates, and it can
therefore be uniform in the third direction, with no periodicity.
The electric field components will have the following form that satisfies the
boundary conditions for perfectly conducting walls.





 π





 π





 π
= z
d
p
y
b
n
x
a
m
E x
x sin
sin
cos
E





 π





 π





 π
= z
d
p
y
b
n
x
a
m
E y
y sin
cos
sin
E





 π





 π





 π
= z
d
p
y
b
n
x
a
m
E z
y cos
sin
sin
E

The amplitudes of the electric field components also must satisfy the
divergence condition which, in absence of charge is
0
0 =





 π
+





 π
+





 π
⇒
=
⋅
∇ z
y
x E
d
p
E
b
n
E
a
m
E

The magnetic field intensities are obtained from Ampere’s law:





 π





 π





 π
ωµ
β
−
β
= z
d
p
y
b
n
x
a
m
j
E
E
H z
y
y
z
x cos
cos
sin





 π





 π





 π
ωµ
β
−
β
= z
d
p
y
b
n
x
a
m
j
E
E
H x
z
z
x
y cos
sin
cos





 π





 π





 π
ωµ
β
−
β
= z
d
p
y
b
n
x
a
m
j
E
E
H y
x
x
y
z sin
cos
cos

Similar considerations for modes and indices can be made if the other axes
are used as a reference for the transverse field, leading to analogous
resonant field configurations.
A cavity resonator can be coupled to a wave guide through a small opening.
When the input frequency resonates with the cavity, electromagnetic
radiation enters the resonator and a lowering in the output is detected. By
using carefully tuned cavities, this scheme can be used for frequency
measurements.

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

Characteristics of some standard air-filled rectangular waveguides.
Here are some standard air-filled rectangular waveguides with their naming
designations, inner side dimensions a, b in inches, cutoff frequencies in GHz,
minimum and maximum recommended operating frequencies in GHz, power
ratings, and attenuations in dB/m (the power ratings and attenuations are
representative over each operating band.) We have chosen one example from
each microwave band.
Waveguides

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