Transcript TE wave

Chapter 9 Guided Electromagnetic Waves导行电磁波
Several wave guiding systems, Electromagnetic
waves in rectangular and circular waveguides
Coaxial line，Cavity resonator
1. TEM Wave, TE Wave, and TM Wave
2. Equations for Electromagnetic Waves in Rectangular
Waveguides
3. Characterization of Electromagnetic Waves in Rectangular
Waveguides
4. TE10 Wave in Rectangular Waveguides
5. Group Velocity
6. Circular Waveguides
7. Transmitted Power and Loss in Waveguides
8. Resonant Cavity
9. Coaxial Lines
The electromagnetic waves to be transmitted along a
confined path are called guided electromagnetic waves,
and the systems to transmit the guided electromagnetic
waves are called the wave guiding systems.
Two-wire line, coaxial line, strip line, microstrip, and
metal waveguides are often used in practice.
we will discuss the metal waveguides and the coaxial
line only.
Two-wire
line
Strip line
Coaxial line
Rectangular
waveguide
Microstrip line
Circular
waveguide
Dielectric waveguide,
Fiber optic
1. TEM Wave, TE Wave, and TM Wave
E
E
es
H
es
H
TEM wave
E
TE wave
es
H
TM wave
The wave guiding systems in which an electrostatic field can exist
must be able to transmit TEM wave.
From Maxwell’s equations we can prove that the metal waveguide
cannot transmit TEM wave.
The main properties of several wave guiding systems
Systems
Wave types
EM
Wave band
shielding
Two-wire line
TEM wav
Poor
> 3m
Coaxial line
TEM wave
Good
> 10cm
Strip line
TEM wave
Poor
Centimeter
Microstrip line Quasi-TEM wave
Poor
Centimeter
Rectangular
waveguide
TE or TM wave
Good
Centimeter
Millimeter
Circular
waveguide
TE or TM wave
Good
Centimeter
Millimeter
Fiber optic
TE or TM wave
Poor
Optical wave
The general approach to study the wave guiding systems
Suppose the wave guiding system is infinitely long, and let it be
placed along the z-axis and the propagating direction be along the
positive z-direction. Then the electric and the magnetic field
intensities can be expressed as
E ( x, y, z )  E0 ( x, y )e  jk z z
H ( x , y , z )  H 0 ( x , y ) e  jk z z
where kz is the propagation constant in the z-direction, and they satisfy
the following vector Helmholtz equation:
 2 E  2 E  2 E
2
 2  2  2 k E 0
y
z
 x
 2
2
2
 H   H   H  k 2 H  0
 x 2
y 2
z 2
The above equation includes six components, E x , E y , E z and
H x , H y , H z , in rectangular coordinate system, and they satisfy the
scalar Helmhotz equation.
Based on the boundary conditions of the wave guiding system
and by using the method of separation of variables, we can find
the solutions for these equations.
From Maxwell’s equations, we can find the relationships between
the x-component or the y-component and the z-component as
Ex 
1 
E z
H z 



j
k

j

z
2 
kc 
x
y 
Ey 
1 
E z
H z 



j
k

j

z
2 
kc 
y
x 
1 
E z
H z 


j


j
k
z
2 
kc 
y
x 
1 
E
H z 

H y  2   j z  jk z
kc 
x
y 
Hx 
Where kc2  k 2  k z2 . These relationships are called the representation of
the transverse垂直 components by the longitudinal纵向 components.
We only need to solve the scalar Helmholtz equation for the
longitudinal components, and then from the relationships between the
transverse components and the longitudinal components all transverse
components can be derived.
In the same way, in cylindrical coordinates the z-component can
be expressed in terms of the r-component and –component as
Er  
1 
Ez
 H z 


j
k

j
z
2 
kc 
r
r  
E 
1  k z E z
H z 



j

j

2 
kc 
r 
r 
Hr 
1   E z
H z 


j

j
k
z
2 
kc  r 
r 
H  
1 
E z
k z H z 


j


j
2 
kc 
r
r  
2.
Equations for Electromagnetic Waves in Rectangular
Waveguides
Select the rectangular coordinate system and let the broad side
be placed along the x-axis, the narrow side along the y-axis, and the
propagating direction be along the z-axis.
For TM waves, Hz = 0 , and
according
y
to
the
method
of
longitudinal fields, the component
b
a
z
Ez should first be solved, and from
 ,
x
which the other components can
be derived.
The z-component of the electric field intensity can be written as
E z  E z 0 ( x , y ) e  jk z z
It satisfies the following scalar Helmholtz equation, i.e.
 2 Ez  2 Ez
2


k
Ez  0
c
2
2
x
y
And the amplitude is found to satisfy the same scalar Helmholtz
equation, given by
 2 Ez 0  2 Ez 0

 kc2 Ez 0  0
2
2
x
y
In order to solve the above equation, the method of separation of
variables is used. Let
E z 0 ( x、y)  X ( x)Y ( y)
We obtain
X  Y 

  k c2
X
Y
where X" denotes the second derivative of X with respect to x, and Y"
denotes the second derivative of Y with respect to y.
X  Y 

  k c2
X
Y
The second term on the left side of the above equation is a function
of y only, while the right side is a constant. The only way the equation
can be satisfied is that both terms on the left side are constants.
Now let
X 
  k x2
X
Y 
 k y2
Y
where k x and k y are called the separation constants, and they can be
found by using the boundary conditions.
Obviously
kc2  k x2  k y2
The two equations are second order ordinary differential equations,
and the general solutions, are respectively
X  C1 cos k x x  C 2 sin k x x
Y  C 3 cos k y y  C 4 sin k y y
where all the constants C1 , C2 , C3 , C4 , and k x , k y , depend on the
boundary conditions.
Since the component Ez is parallel to the walls, we have Ez = 0 at
the boundaries x = 0, a and y = 0, b . Using these results we find
kx 
mπ
, m  1,2,3, 
b
ky 
nπ
, n  1,2,3, 
b
And all the field components are
E z  E0 sin
mπ
nπ  jk z z
x sin
ye
a
b
Ex   j
k z E0  mπ   mπ   nπ
x  sin 
 cos
2 
kc  a   a   b

y e  jk z z

Ey   j
k z E0  nπ   mπ   nπ
x  cos
 sin 
2 
kc  b   a   b

y e  jk z z

Hx  j
E0  nπ 
Hy  j
kc2
 mπ   nπ
sin
x  cos
  
 b   a   b
E0  mπ 
kc2

y e  jk z z

 mπ   nπ
cos
x  sin 

 
 a   a   b

y e  jk z z

E z  E0 sin
mπ
nπ  jk z z
x sin
ye
a
b
Ex   j
k z E0  mπ   mπ   nπ
x  sin 
 cos
2 
kc  a   a   b

y e  jk z z

Ey   j
k z E0  nπ   mπ   nπ
x  cos
 sin 
2 
kc  b   a   b

y e  jk z z

Hx  j
E0  nπ 
Hy  j
kc2
 mπ   nπ
sin
x  cos
  
 b   a   b
E0  mπ 
kc2

y e  jk z z

 mπ   nπ
cos
x  sin 

 
 a   a   b

y e  jk z z

(c)
If
mmodes
ormnand
iswith
zero,
then
for TM
wave),
and
all
Ez and
 0 ( Hare
 0called
z the
(e)
(d)
Since
larger
are
multi-valued,
m
pattern
the
of
modes
the
ofishigher
has
(b)
The
plane
zof= the
0n is
a wave
front.nBecause
therelated
amplitude
related
(a) The
phase
electromagnetic
wave
is
tofield
the
variable
components
willalso
bemodes,
zero.
Thus
the
mwith
andless
n are
non-zero
integrals,
and
order
multiple
or
the
forms,
higher
called
and
multiple
that
modes.
A
m
pair
and
of
n
m
are
and
called
n
lead
the
to
modes
a
ztoonly,
the variables
x and
y. Hence, a traveling
x andwhile
y, thethe
TMamplitude
wave is a to
non-uniform
plane
wave.
they
have
clear
physical
meanings.
The
value
ofinstance,
m stands
for
the
number
of
mode,
lower
and
it isor
denoted
the
the
modes.
TMmn
Since
mode.
m
and
n are
TM
not11is
zero,
denotes
andthe
wave
isorder
formed
in lower
theasz-direction,
andboth
aFor
standing
wave
in the
x-the
of
half-cycle
variations
ofis=the
fields
along
thewave
broadwith
side,this
while
n denotes
lowest
pattern
mode
of
the
of
field
TM
wave
for
m
TM
1
,
n
=
in
1
the
,
and
rectangular
the
waveguide.
character
is
11
direction and y-direction.
that
for
the
narrow side.
called
TM
11 wave or mode.
Similarly, we can derive all the components of a TE wave in the
rectangular waveguide, as given by
 mπ   nπ
H z  H 0 cos
x  cos
 a   b

y  e  jk z z

Hx  j
k z H 0  mπ   mπ   nπ   jk z z
x  cos y e

 sin 
kc2  a   a   b 
Hy  j
k z H 0  nπ   mπ   nπ   jk z z
x  sin  y e
  cos
kc2  b   a   b 
Ex  j
H 0  nπ 
Ey   j
kc2
 mπ   nπ   jk z z
x  sin 
y e
  cos
 b   a   b 
H 0  mπ 
kc2
 mπ   nπ   jk z z
x  cos y e

 sin 
a
a

 
  b 
where m, n  0, 1, 2, , but both should not be zero at the same time.
TE wave has the multi-mode characteristics as the TM wave. The
lowest order mode of TE wave is the TE01 or TE10 wave.
3. Characterization of Electromagnetic Waves in Rectangular
Waveguides
Since kc2  k 2  k z2, or k z2  k 2  kc2, if k  kc , then k z  0 . This means
that the propagation of the wave is cut off, and kc is called the cutoff
propagation constant.
2
kc2  k x2  k y2
 mπ   n π 
k 
  
a

  b 
2
2
c
From k  2πf  , we can find the cutoff frequency f c corresponding
to the cutoff propagation constant kc , as given by
2
kc
1
m n
fc 

   
2π  2   a   b 
2
The propagation constant kz can be expressed as
f 
k z   k 1   c 
 f 
2
2



f
k 1   c  ,
 f 

 

2

 fc 
 jk    1,

 f 
f  fc
f  fc
if f  f c , k z is a real number, and the factor e  jk z z stands for the
wave propagating along the positive z-direction.
2
If f  f c , k z is an imaginary number, then e  jk z z  e
 f 
 kz  c  1
 f 
which states that this time-varying electromagnetic field is not
transmitted, but is an evanescent field.
For a given mode and in a given size waveguide,f c
is the lowest
frequency of the mode to be transmitted. In view of this, the waveguide
acts like a high-pass filter.
From k  2π , we can find the cutoff wavelength c corresponding

the cutoff propagation constant kc as
c 
2π
2

2
2
kc
m n
   
 a  b
The cutoff frequency or the cutoff wavelength is related to the
dimensions of the waveguide a, b and the integers m, n . For a given
size of waveguide, different modes have different cutoff frequencies
and cutoff wavelengths. A mode of higher order has a higher cutoff
frequency截止频率, or a shorter cutoff wavelength截止波长.
TE01
The cutoff wavelength of the TE10
wave is 2a, and that of TE20 wave is a.
TE10
TE20
TM11
0
a
2a
c
The left figure gives the distribution of
the cutoff wavelength 截 止 波 长 for a
waveguide with a  2b .
TE01
TE10
TE20
TM11
0
a
2a
Cutoff area
If   c , then the corresponding mode will be cut off. From the
figure we see that if   2a , all modes will be cut off.
If a    2a , then only TE10 wave
exists, while all other modes are cut
off . If   a , then the other modes
will be supported.
c
Hence, if the operating wavelength
工作波长 satisfies the inequality
a    2a
Then the transmission of a single mode is realized, and the TE10 wave is
the single mode to be transmitted.
The transmission of a single mode单模传输 wave is necessary in
practice since it is helpful for coupling energy into or out of the
waveguide.
TE10 wave is usually used, and it is called the dominant mode主模
of the rectangular waveguide矩形波导.
In practice, we usually take a  2b to realize the transmission of the
single mode TE10 in the frequency band频带 a    2a .
To support the TE10 mode the sizes of the rectangular waveguide
should satisfy the following inequality

2
a
b

2
The lower limit for the narrow side depends on the transmitted
power, the allowable attenuation衰减, and the weight per unit length.
In practice, we usually take a  0.7 and b  (0.4 ~ 0.5)a or (0.1 ~ 0.2)a .
As the wavelength is increased, the sizes of the waveguide must be
increased proportionally to ensure the dominant mode主模 is above
cutoff. If the frequency is very low, the wavelength will be very long so
that it may not be convenient for use.
Consequently, metal waveguides are used for microwave bands
above 3GHz.
The phase velocity vp can be found from the phase constant as
vp 
Where v 
1


kz

v
f 
1   c 
 f 
2

v

1   
 c 
2
v
. If the inside of the waveguide is vacuum, then
v
1
 0 0
c
Since the operating frequency f  f c and the operating wavelength
  c , we have vp  c for a vacuum waveguide. Hence, the phase velocity
does not represent the energy velocity 能速in a waveguide.
The phase velocity 相 速 depends on not only the sizes of the
waveguide, the modes, and the properties of the media within the
waveguide, but also the frequency. Hence, an electromagnetic wave will
also experience dispersion色散 in a waveguide.
Based on the relationship between the wavelength and the phase
constant, we find the wavelength of the electromagnetic wave in a
waveguide, g , as
2π


g 


2
2
kz
f 

1   c 
1   
 f 
 c 
where  is the operating wavelength工作波长. The quantity g is called
the guide wavelength波导波长.
Due to f  f c ，
c   ，thus g   .
The ratio of the transverse electric to the transverse magnetic
field intensities as the waveguide impedance of the waveguide. For a
TM wave the waveguide impedance is
Z TM
Ey
Ex


Hy
Hx
2
Z TM

f 
 Z 1   c   Z 1   
 f 
 c 
2
Z


In the same way, we find the waveguide impedance阻抗 of a
TE wave as
Z TE 
Z
f 
1   c 
 f 
2

Z

1   
 c 
2
If f  f c ,  c , then Z TM and Z TE are both imaginary numbers.
This means that the transverse横向 electric field and the transverse
π
横向magnetic field have a phase difference of . Hence, there is no
2
energy flow in the z-direction, and it indicates that the propagation
of the electromagnetic wave is cut off.
Example. The inside of a rectangular metal waveguide is
vacuum, and the cross-section截面 is 25mm10mm. What modes can
be transmitted if an electromagnetic wave of frequency f  10 4 MHz
enters the waveguide? Will the modes be changed if the waveguide is
filled with a perfect dielectric of relative permittivity介电常数  r  4 ?
Solution: Due to the inside is vacuum, the operating wavelength is
c
   30 mm
f
and the cutoff wavelength is c 
2
2
m n
   
 a  b
2

50
m 2  6.25n 2
Then the cutoff wavelength of TE10 wave is c  50mm , that of TE20
wave is c  25mm , and that of TE01 wave is λc  20mm . The cutoff
wavelength of the higher modes will be even shorter. In view of this,
only TE10 wave can be transmitted in this waveguide.
Example. The inside of a rectangular metal waveguide is
vacuum, and the cross-section截面 is 25mm10mm. What modes can
be transmitted if an electromagnetic wave of frequency f  10 4 MHz
enters the waveguide? Will the modes be changed if the waveguide is
filled with a perfect dielectric of relative permittivity介电常数  r  4 ?
If the waveguide is filled with a perfect dielectric of  r  4 , then
the operating wavelength is


 15mm
r
Hence, TE10 and TE20 waves can be transmitted, and some other modes
TE01，TE30，TE11，TM11，TE21，TM21 can exist.
4. TE10 Wave in Rectangular Waveguides
Let m  1, n  0 , we find
Ey   j
Hx  j
H 0  π 
kc2
 π   jk z
  sin  x e z
a a 
k z H 0  π   π   jk z z
 sin  x e
2 
kc  a   a 
π 
H z  H 0 cos x e  jk z z
a 
And H y  E x  E z  0 . The corresponding instantaneous瞬时 values are
E y (r , t )  
H x (r , t ) 
2H 0  π   π 
π
sin
x
sin(

t

k
z

)
  

z
2
kc
2
a a 
2k z H 0
kc2
π
π π 
sin
x
sin(

t

k
z

)
  

z
2
a a 
π 
H z (r , t )  2 H 0 cos x  sin( t  k z z )
a 
The above equations are simplified as
π
π 
E y (r , t )   A sin  x  sin( t  k z z  )
2
a 
π
π 
H x (r , t )  B sin  x  sin( t  k z z  )
2
a 
π 
H z (r , t )  C cos x  sin( t  k z z )
a 
Where A, B, C are positive real numbers.
Ey
Hz
z
Hx
Hz
g
Hx
The right figure gives the distributions
x
of the TE10 wave along the z-direction and
Ey
x-direction at t = 0 .
a
A standing wave驻波 is found in the x -direction, while a traveling
wave is seen in the z -direction.
The amplitude of Hz follows a cosine function, while the amplitudes
of Hx and Ez depend on x with the sine function. But all of them are
independent of the variable y.
The electric and magnetic field lines and the currents of TE10 wave .
z





x
y



a


Electric field lines
Magnetic field lines
g
x
z
b
y
y
The electric currents
on the inner walls
z
x
The modes with higher orders
TE10
TE11
TE20
TE21
TM11
TM21
Electric field lines
Magnetic field lines
Let m = 1, n = 0, we find the cutoff wavelength of TE10 mode as
c  2a
It means that the cutoff wavelength of the TE10 wave is independent
of the narrow side.
The phase velocity相速 and the guide wavelength波导波长 can be
v

found as
vp 
g 
2
2
  
  
1  
 2a 
1  
 2a 
To visualize the physical meaning 物理含义 of the phase
velocity, the energy velocity, as well as the guide wavelength for the
TE10 wave, the expression of electric field intensity Ey is rewritten as
E y  E0 ( e
π
j x
a
e
π
j x
a
) e  jk z z
  π  1 j aπ x - j aπ x 
 sin  x   (e  e ) 
  a  2j



Furthermore, we have
E y  E0e
 jk ( x cos  z sin )
 E0e
 jk (  x cos  z sin )

 
 cos  
 
2a c 

which states that a TE10 wave can be considered as the resultant wave
comprising two uniform plane waves均匀平面波 with the same
propagation constant k .
The propagating directions of the
two plane waves are laid on the xzz
plane. They are parallel to the broad
a
  ②
①
wall, and the two plane waves are
combined into a plane wave taking a
zigzag path between the two narrow
x
walls.
If   c , then   0 . The plane wave will be reflected vertically
between two narrow walls. Hence it cannot propagate in the z-direction
and is cut off.
when the wave loops of the two plane waves meet, a wave loop of
the resultant wave is formed. A wave node of the resultant wave is
formed when the wave nodes of the two plane waves meet.
z
B
a
x

②
A
D ①
C
The bold lines denote the wave loops
of plane wave ①, and the dashed lines
denote that of plane wave ②.
Obviously, the length of the line AB
is equal to the guide wavelength , and
the length of the line AC is equal to the
operating wavelength .
If the inside of the waveguide is vacuum, then the length of the
line AC is equal to the wavelength in vacuum. From the figure, we

find
g 
2


g 




1   
sin 
1  cos 2 
 c 
The space phase of plane wave ① is changed by 2 from A to C,
while that of the resultant wave is changed by 2 over the distance AB.
In view of this, the phase velocity of the resultant wave is greater than
that of the uniform plane wave v,
z
B
a

②
A
D ①
v
vp 
sin 
v
vp 

1   
 c 
C
x
2
From the point of the view of energy, when the energy carried by
plane wave ① arrives at C from A, the movement in z-direction is just
over the distance AD. Hence, the energy velocity is less than the energy
velocity of the uniform plane wave v. From the figure, we find the
energy velocity as
ve  v sin   v  vp

ve  v 1   
 c 
2
Example. The broad side of a rectangular waveguide filled with air
satisfies the condition   a  2 , and the operating frequency is 3GHz. If
the operating frequency is required to be higher than the cutoff
frequency of the TE10 wave by 20% and less than the cutoff frequency of
the TE01 by 20%. Find: (a) The sizes for a and b. (b) The operating
wavelength, the phase velocity, the guide wavelength, and the wave
impedance for the designed waveguide.
Solution: (a) The cutoff wavelength of the TE10 wave is c  2a , and
c
c
the cutoff frequency is f c  
. The cutoff wavelength of TE01 wave
c 2a
c
is c  2b , and the cutoff frequency is f c 
. According to the given
2b
condition, we have
c
c
3  10 9   1.2
3 109   0.8
2b
2a
We find a  0.06m , b  0.04m . Take a  0.06m , b  0.04m.
(b)
The operating wavelength, the phase velocity, the guide
wavelength, and the wave impedance:

vp 
c
 0.1m
f
c
  
2
 5.42  103 m/s
1  
 2a 
g 

  
1  
 2a 
Z TE 1 0 
2
 0.182m
Z
  
1  
 2a 
2
 682Ω
5.
Group Velocity
When the phase velocity is frequency dependent, a single phase
velocity alone cannot account for the speed at which a wave consisting
of multiple frequency components propagates.
As an example, we consider an amplitude-modulated wave to
illustrate the concept of the group velocity.
Suppose an electromagnetic wave propagating in the z-direction
has two components with frequencies close to each other as given by
 A1 ( z, t )  A0 cos(1t  k1 z )

 A2 ( z, t )  A0 cos( 2 t  k 2 z )
with the resultant signal
A  A1  A2  2 A0 cos(Δ t  Δkz) cos(0t  k0 z )
where
1



 0 2 (1  2 )

Δ  1 (   )

1

2
1

k

 0 2 (k1  k 2 )

Δk  1 (k  k )
 0 2  1
A  A1  A2  2 A0 cos(Δ t  Δkz) cos(0t  k0 z )
Since  1 ~  2 , and Δ  0 . Therefore, in a very short time interval,
the first cosine function show little change, but the second cosine function
has large variations. So  0represents the carrier frequency while Δ is the
frequency of the envelope or the modulating frequency.
This is an amplitude-modulated signal with a slower variation in the
amplitude.
If the medium is non-dispersive, the envelope of the amplitude is
moving together with the carrier, both maintaining the sinusoidal
behavior in the movement. Therefore, by the locus of a stationary point
on the envelope, we can find the velocity of the envelope, and this
velocity is called the group velocity, denoted as vg .
Let Δt  Δkz  constant , we find
vg 
dz Δ

d t Δk
For non-dispersive media, the relationship between k and  is
linear, and Δ  d . We obtain
Δk
dk
vg 
d
dk
Let 0t  k0 z  constant , and we find the phase velocity of the
carrier as
vp 
0
k0
Since k    in non-dispersive media, we find the group
velocity as
1
vg 
d  dk 

 
dk  d 
1

 vp
In non-dispersive media, the group velocity is equal to the phase
velocity.
For dispersive media, the relationship between k and  is nonlinear. In this case, for a given operating frequency
,
can be
 0 k (ω)
expanded by Taylor series
 0 around as
1  d 2k 
 dk 
k ( )  k 0  
 (   0 )   2  (  0 ) 2  
2  d  
 d   0
0
For a narrow band signal, take only the first two terms as
approximation, so that
 dk 
k ( )  k 0  
 (   0 )
 d   0
Δ d
Consider vg 
, we have

Δk dk
1
 dk 
 d 
vg  
 

 d 0  dk 0
With a nonlinear relation between the propagation constant k
and  the frequency for a dispersive medium, the phase velocity is
frequency dependent and it is not the same as the group velocity.
Carrier
Envelope
Carrier
It gives the waveforms of the above narrow band signal at three
different moments for the case of vp  2vg .P  is a stationary point on the
envelope, and P is that for the carrier. When the displacement of the
point P is d, the point P  is moved only byd , (d   d ) because the velocity
of the envelope is less.
The actual signal waveform will be modified as it propagates.
dk
d  ω  1 ω dvp
1 
ω dvp 

1

  2
Consider
, we find




dω dω  vp  vp vp dω vp  vp dω 
vp
vg 
 dv p
1
v p d
For the narrow band signal, the above equation becomes
vp
vg 
  dv 
1   p 
vp  d 
0
If the phase velocity vp is independent of frequency,
vg  v p
If
If
dvp
d
dvp
d
dvp
d
 0 , then
 0 , then vg  vp , and the dispersion is called normal dispersion.
 0 , then vg  vp , and it is called abnormal dispersion.
dv
For a rectangular waveguide, p  0 , it is normal dispersive
d
and the group velocity is
2
2

f 
vg  v 1   c   v 1     ve
 f 
 c 
The group velocity is equal to the energy velocity in the
rectangular waveguide, which is the same behavior for all normal
dispersive media.
The phase velocity vp and the group velocity vg in a waveguide
satisfy the following equation
vpvg  v 2
When an electromagnetic wave is propagating in a conductive
medium, abnormal dispersion is observed. In this case, the group
velocity is not equal to the energy velocity, and the above equation
is not valid for this case.
6. Circular Waveguides
The inner radius a is the only dimension to be specified. Select
the cylindrical coordinate system, and let the z-axis be the axis of
the cylinder.
Similar to the rectangular waveguide, the longitudinal components Ez
y
or Hz is first obtained, from which the
transverse components Er , E , Hr , H
a
x
can be derived.
 ,
z
The field intensities in
waveguide can be written as
E ( r ,  , z )  E 0 ( r ,  ) e  jk z z
the
H ( r ,  , z )  H 0 ( r ,  ) e  jk z z
The corresponding longitudinal components are, respectively
E z ( r ,  , z )  E z 0 ( r ,  ) e  jk z z
H z ( r ,  , z )  H z 0 ( r ,  ) e  jk z z
For a TM wave, Hz = 0 . In a source-free region, Ez satisfies the
scalar Helmholtz equation given by
 2 Ez  k 2 Ez  0
Expanding this equation in cylindrical coordinate system, we have
 2 Ez 0 1 E20 1  2 Ez 0
2



k
Ez 0  0
c
2
2
2
r
r r
r 
Using the method of separation of variables is used, and let
E z 0 (r ,  )  R(r ) ( )
Substituting it into the above equation gives
r 2 R rR
 
2 2

 kc r  
R
R

where R  and R  are the second and the first derivatives of the
function R with respect to r, respectively, and   is the second
derivative of the function  with respect to  .
Using the same derivation as before，we obtain the equation
for the function  as
   m 2  0
The general solution is
  A1 cos m  A2 sin m
Since the period of variation of the field with the angle is 2 .
Hence m must be integers so that
m  0,  1,  2
The circular waveguide is symmetrical with respect to the zaxis; thus the plane   0 can be chosen arbitrarily. In this way, we
can always select the plane properly so that the first term cos m
or the second term sin m vanishes.
Therefore, the solution of  can be expressed as
cos m
sin m
  A
We find
d2R
dR
2 2
2
r

r

(
k
r

m
)R  0
c
2
dr
dr
2
Let kc r  x , then the above equation becomes the standard Bessel
equation
d2 R
dR
2
2
x

x

(
x

m
)R  0
2
dx
dx
2
The general solution is
R  BJ m ( x)  CN m ( x)
whereJ m ( x)
is the first kind of Bessel function of order m, and
N m ( x) is the second kind of Bessel function of order m.
If r  0 , x  0 , N m (0)   , But the field should be finite in the
waveguide. Hence the constant C  0 . The solution should then be
R  BJ m (kc r )
Consider all results above, we find the general solution of Ez as
cos m  jk z z
E z  E0 J m (kc r )
e
sin m
And the transverse components are
Er   j
E  j
cos m  jk z z
k z E0
Jm (kc r )
e
kc
sin m
sin m  jk z z
k z mE0
J
(
k
r
)
e
m
c 
2
kc r
 cos m
 sin m  jk z z
J
(
k
r
)
e
m
c 
2
kc r
cos m
cos m  jk z z
E0
H   j
Jm (kc r )
e
kc
sin m
Hr  j
mE0
where Jm (kc r ) is the first derivative of Bessel function J m (kc r ) .
The constant kc depends on the boundary condition.
The components Ez and E are tangential to the inner wall of the
circular waveguide; hence, E z  E  0 at r  a .
2
P


kc2   mn 
We find
 a 
Pmn is the n -th root of the first kind of Bessel function of order m.
The values of Pmn
n
1
2
3
4
0
2.405
5.520
8.654
11.79
1
3.832
7.016
10.17
13.32
2
5.136
8.417
11.62
14.80
m
A pair of m and n corresponds to aPmn , and that corresponds to a
kind of field distribution or a mode. Hence, the electromagnetic waves
have multiple modes in a circular waveguide also.
For the TE wave, Ez = 0. We can use the same approach to find
the component Hz first, and then the other transverse components
can be determined.
TE wave:
cos m  jk z z
H z  H 0 J m (kc r )
e
sin m
cos m  jk z z
kH
H r   j z 0 Jm (kc r )
e
kc
sin m
sin m  jk z z
k z mH 0
H   j 2 J m (kc r )
e
kc r
 cos m
Er  j
mH 0
E  j
kc2 r
H 0
kc
sin m  jk z z
J m (kc r )
e
 cos m
cos m  jk z z
Jm (kc r )
e
sin m
Based on the boundary conditions, we find
 
 Pmn
2


kc  
 a 
.2
 is the root of the first derivative of Bessel function.
Where Pmn
The values of Pmn

n
1
2
3
4
0
3.832
7.016
10.17
13.32
1
1.841
5.332
8.526
11.71
2
3.054
6.705
9.965
13.17
m
As with the rectangular waveguide, if k  kc , then the propagation
constant k z  0 , meaning that the wave is cut off, so that propagation
ceases.
From
k c  2 πf c  
2π
For TM wave, we have
c
fc 
For TE wave, we have
fc 
Pmn
;
2π a 
c 
2π a
Pmn

Pmn
;
2π a 
c 
2π a

Pmn
Cutoff area
The following figure gives the cutoff wavelength of several modes
in a circular waveguide.
The TE11 wave has the longest
TE11
cutoff wavelength, and the next one
TM01
TE21
is the TM01 wave.
TE01
0
a
2a
3a
4a
c
The cutoff wavelengths of the
TE11 and TM01 waves, respectively, as
TE 11 : c  3.41a, TM 01 : c  2.62a
If the operating wavelength  satisfies the following inequality
2.62a    3.41a
The transmission of a single mode (TE11 wave) can be realized,
and the TE11 wave is the dominant mode for the circular waveguide.
If the operating wavelength  is given, to realize the transmission of only the TE11 wave, the radius a must satisfy the following


inequality:
a
3.41
2.62
From the cutoff frequencies or the cutoff wavelengths, the phase
velocity, the group velocity, the guide wavelength and the wave
impedance of each mode can be found using the same equations as
those for the rectangular waveguide.

 







TE11
TE01
TM01















Electric field lines
Magnetic field lines
Example. A circular waveguide of radius a = 5mm is filled with a
perfect dielectric of relative permittivity r = 9 . If it is to be operated
in the dominant TE11 mode, find the permissible frequency range.
Solution: To operate on the mode TE11, the operating wavelength
must satisfy the following inequality
2.62a    3.41a
Hence
max  3.41 5mm  17.1mm
min  2.62  5mm  13.1mm
The corresponding range of the operating frequency is
f max 
f min 
v
 min
v
 max


1
 min  0
1
 max  0
 7634MHz
 5848MHz
7. Transmitted Power and Loss in Waveguides
The longitudinal component of the complex energy flow density
vector is given by the cross product of the transverse components of
the electric and magnetic fields. The integration of the real part over
the cross-sectional area of the waveguide gives the transmitted power.
Take the TE10 wave as an example, we find the transmitted power
as
abE02
P
2Z TE
Where E 0 is the amplitude of the electric field in the middle of
the broad side
If the dielectric strength of the filling dielectric is Eb , then the
maximum transmitted power of the rectangular waveguide is
abEb2
Pb 
4Z TE
In practice, the transmitted power is limited as P   1 ~ 1  Pb for
3 5
safety purpose.
The two major mechanisms for energy loss are imperfect dielectric
and finite conductivity of the waveguide walls.
The effect of the dielectric can be accounted for by introducing the
equivalent permittivity to replace the original one, i.e.

e    j

A vigorous analysis of the waveguide walls is very complicated.
An approximation that retains the magnetic field that would have
existed if the walls were perfectly conducting may be employed.
If we assume the attenuation constant is k  , then the amplitude
of the electric field intensity propagating along the positive zdirection has the form
E  E0 e  k z
The transmitted power can be expressed as
P  P0 e 2 k z
P  P0 e 2 k z
Take the derivative of the above equation with respect to z, we
find the power attenuation per unit length as

P
 2k P
z
Obviously, that is the power loss per unit length, so that Pl1  2k P
Hence, the attenuation constant k is obtained as k   Pl1
2P
To calculate the loss of the walls, we consider a piece of conductor
making up the broad side wall, with unit width and length and a
thickness equal to d .
The current in the conductor is flowing
y
in the z-direction. The resistance of the piece
1
1
of conductor is given by
d
l
1
πf

R



S
1

S d

z

x
where  is the conductivity of the wall.
RS is called the surface resistivity.
The surface resistivities of three types of metal
Metals
RS
Silver
2.52  10 7
Copper
f
2.61  10 7
Aluminum
f
3.26  10 7
f
The surface current density is the current per unit width. Hence
the power loss per unit length and width of the waveguide wall PlS is
PlS  J S2 R S
where the surface current J S  en  H S , and H S is the magnetic field
intensity on the surface of the wall.
Taking the integration of PlS over the inner wall for a section of the
waveguide of unit length, the power loss per unit length of the wall Pl1
can be obtained.
TM1
1
For a given size of rectangular
waveguide, the loss of the TE10 wave
is minimum. For a given width, the
smaller is the narrow wall, the larger
will be the attenuation constant.
The loss of the TE01 wave in a
circular waveguide is minimum in the
higher frequency range. 。
The cutoff wavelength of the TE01
wave is not the largest. In order to realize
the transmission of the single mode TE01,
the modes TM01, TE21 and TE11 have to
be suppressed.
r
For the same cross-section, the perimeter of a rectangle is larger
than that of a circle, and the loss in a circular waveguide is less than
that of the rectangular waveguide.
However, when a TE11 wave is propagating in a circular waveguide,
the fields could be rotated.
An elliptical waveguide
does not result in the rotation
of the fields, and the loss is less
E
also.
In addition, in order to reduce the wall losses, the inner surface
should be polished, and plated with silver or gold.
To prevent
oxidation of the surface, the waveguide may be filed with inert gas.
Example. Calculate the attenuation caused by finite conductivity of
the wall when a TE10 wave is propagating in a rectangular waveguide.
Solution:
When a TE10 wave is
propagating in a rectangular waveguide,
there are x-component and z-component
of the surface currents on the broad sides,
while there is only the y-component on the
narrow sides.
y
Therefore, the power loss per unit length
of the broad wall is
z
x
a
a
2
2

Pla  2  J Sz RS dx   J Sx
RS dx 
 0

0
Where J Sz  e y  H x , J Sx  e y  H z .
The power loss per unit length of the narrow wall is
b
Plb  2 J S y RS dy
2
0
Where J Sy  e x  H z .
Then the total power loss per unit length is
Pl1  Pla  Plb
Based on the transmitted power P and the total power loss
per unit length Pl1 , we find the attenuation constant as
P
k   l1 
2P
 1 2   2 
    
2

    b a  2a  
1  

 2a 
RS
8. Resonant Cavity
In microwave band, the lumped LC tank circuits cannot be used,
we usually employ a transmission line to construct a resonant device,
and it is called cavity resonator.
with the increase of the resonant frequency the inductance and
the capacitance must be reduced.
However, for small L and C,
distributed effects cannot be neglected. The inductance of the lead
wires of capacitors, the distributed capacitances among the coils or
the devices have to be considered. This means that a pure capacitor
or a pure inductor is very difficult to be made at microwave
frequencies.
Furthermore, with the increase in frequency, the radiation effect
of the circuits becomes significant, and the power loss in the dielectric
of the capacitor is more severe as well. All of these will result in the
decrease of the quality factor Q of the lumped tank circuit.
When a metal plate is placed at the end of a waveguide, the electromagnetic wave will be completely reflected, leading to a standing wave.
For a rectangular waveguide operating in the dominant mode,
the closed end corresponds to a wave node for the electric field since
it is tangential to the metal plate. Another wave node for the electric
λ
field appears at a distance g from the closed end. If one more metal
2
plate is placed there, the boundary condition is still satisfied,
y
d
b
z
a
g /2
x
In this way, there is a standing
wave in the cavity formed by the
waveguide walls and the end plates.
Based on the field intensity and the
boundary condition, we find the
equations for the standing waves in
the cavity as follows:
π 
H z  H 0 (e
 e ) cos x 
a 
k aH
π 
H x  j z 0 (e  jk z z  e jk z z ) sin  x 
π
a 
 jk z z
Ey   j
aH 0
π
jk z z
π 
(e jk z z  e  jk z z ) sin  x 
a 
π 
H z  2 jH 0 sin( k z z ) cos x 
a 
k aH
π 
H x  2 j z 0 cos( k z z ) sin  x 
π
a 
2aH 0
π 
Ey  
sin( k z z ) sin  x 
π
a 
There are standing waves of the electric and the magnetic fields
along both the x-direction and the z-direction, but they are out of the
time phase by π . When the electric energy is maximum, the magnetic
2
energy is zero. Conversely, when the magnetic energy is maximum,
the electric energy is zero.
The electromagnetic energy is exchanged between the electric field
and the magnetic field, and this phenomenon is called resonance. So
the cavity is called a resonant cavity, and it is used as a resonant
device in microwave circuits.
For a given cavity, resonance occurs only at certain frequencies.
The particular frequency is called resonant frequency, and the
corresponding wavelength is called the resonant wavelength.
 g 
d  l  ,
 2
If the length of the cavity is
l  1,2,3,
Then the boundary condition can be satisfied, and the resonance will
happen. Hence, the resonant frequency or wavelength of a resonant
cavity is multi-valued, and it is called multiple resonance.
since the guide wavelength is related to the mode, different modes
have different resonant frequencies.
2
2

m
π
n
π
Consider k z2  k 2       , when d  l g , k z d  lπ , k z  lπ ,
2
d
 a   b 
and we find
2
2
 mπ   nπ   lπ 
k 
    
 a   b  d
2
From k 
2π

 2πf  , we find
 mnl 
f mnl 
2
2
2
m n  l 
     
 a  b d 
1
2 
2
2
2
m n  l 
     
 a  b d 
2
The resonant wavelengths and frequencies depend on not only the
sizes of the cavity, but also the mode.
A set of mnl corresponds to a mode. For instance, TE101 stands for
that a rectangular waveguide cavity operating on a TE10 wave, and
the length of the cavity is half of the guide wavelength.
In order to properly design the coupling and the tuning devices of
the cavity, knowledge about the distribution of the fields in the cavity
is required.
The figure gives the distribution of the fields in a rectangular cavity
operating at the TE101 mode.
d
As with other resonant devices,
y
z
a real cavity always has some loss.
a
z
Electric field lines
In order to assess the loss of a
resonant device, the quality
factor Q is usually employed, and
its definition is
 0W
Magnetic field lines
Pl
x
x
b
y
Q
where 0 is the resonant angular frequency, W is the total energy,
Pl is the power loss in the cavity.
The maximum value of the energy density of the electric field in
the cavity is
a 3bd02  2 | H 0 |2
W
2π 2
In order to calculate the power loss of the cavity wall, the same
method for waveguide analysis may be applied. We find the power
loss of a rectangular cavity operating with the TE101 mode as
2a 3b  a 3 d  ad 3  2d 3b
2
Pl 
2
R
|
H
|
S
0
d2
and the Q value is
03  2a 3bd 3
Q 2
4π RS (2a 3b  a 3d  ad 3  2d 3b)
The resonant angle frequency for the TE101 mode is
101  2πf101 
π

2
1 1
   
a d 
2
Therefore, the Q value for the TE101 mode can be expressed as
πZb (a 2  d 2 )3
Q101 
4 RS (2a 3b  a 3d  ad 3  2d 3b)
where Z   .

Since the circular waveguide has less loss,
the Q value of the
cylindrical cavity is higher, and it is more popular than the rectangular
cavity.
The method for calculating the resonant frequency and the Q value
of a cylindrical cavity is the same as that above.
TM wave: f TM 
TE wave: f TE 
QTE
1
2π 
1
2π 
d


2
 Pmn   lπ 

  
 a  d
2
   lπ 
 Pmn

  
 a  d
2
  m 2 
 
1  
  
  Pmn

QTM
2
 lπa 
P 

d
 d 


 2a 
2 π 1  
d 

2
mn

lπa 
 ) 2  

( Pmn




2 3



2
2




2
a
l
π
a
2
a
ml
π
a




2
 )  
 
2π ( Pmn
  1  
d 
d  d  
d  Pmn


2
TE01l modes have higher Q
values, and the maximum
Q value of the TE011 mode
occurs around d  2a.
If  = 3cm, then Q value
will be 104~4104. 。
The approach to increase
the Q value is the same as
that of decreasing the loss of
waveguide wall.
In addition, the volume of the cavity should be as larger as possible
to increase the stored energy, while the area of the walls should be as
small as possible to decrease the loss.
Example. Show that for any mode the resonant wavelength r of
the cavity can be expressed as
c
r 
l  1,2,3, 
2
 l 
1  c 
 2d 
where c is the cutoff wavelength, and d is the length of the cavity.
Consider k z2  k 2  kc2
Solution:
If the length d  l
g
2
, then k z d  lπand k z 
lπ
. Resonance will occur.
d
2π
And due to k  2 π , k c 
, we find
c
r
2
 π   2 π   2 π 
 l       
 d   r   c 
2
2
2
1 1
 l 
   2 2
λr λc
 2d 
Rewriting the equation gives the general formula.
9.
Coaxial Lines
A coaxial line is shown in the figure, with an inner radius a and
an outer radius b.
The electromagnetic wave propagates in the
region between the two conductors, which may be filled with air or
a dielectric.
The coaxial line is a good transmission
y
line in microwave band. It possesses the
electromagnetic shielding function as the
waveguides, but it has a wider frequency
b
x
range of operation.
z
a
Electric field lines
Magnetic field lines
The coaxial line is a typical TEM
wave transmission line a coaxial line,
and the electric field lines are along with
the radial direction , while the magnetic
field lines are a set of circles.
A coaxial line can also be considered as a circular waveguide that
supports TE and TM waves, besides the TEM wave. However, if we
properly design the dimensions according to the operating frequency,
these non-TEM waves can be restrain .
The method for analyzing non-TEM waves in a coaxial line is similar
to that for a circular waveguide. However, the coaxial line has an internal
conductor, the range of the variable r is a  r  b , andr  0 . Hence, the
second kind of Bessel function with the singularity at r = 0 should be the
solution of Bessel equation as well, i.e.
R  BJ m ( x)  CN m ( x)
TE11
TM01
TE10
0
(b - a) (a + b)
c
For TM and TE waves，based on
the boundary conditions the cutoff
propagation constant are found first,
then the cutoff wavelengths can be
obtained.
TM01
TE10
0
TEM wave
TE11
(b - a) (a + b)
c
The TE11 wave has the longest cutoff
wavelength, and it is π(a  b) .
To restrain the non-TEM wave, the
operating wavelength  must satisfy the
following inequality
  π ( a  b)
In other words, the dimensions of the coaxial line should satisfy
the following inequality
 
ab 
π

3
Hence, in order to eliminate the higher order modes in a coaxial
line, the dimensions have to be decreased as the frequency increases.
But small sizes will result in the increase of loss and the restriction of
the transmitted power. For this reason, the coaxial line is usually used
for the frequencies below 3GHz.
However, the operating frequency has no lower limit, and the coaxial
line can also be used to construct a cavity.