Transcript Waveguide - Portal UniMAP

Chapter 2
Waveguide
Chapter Outlines
Chapter 2








Waveguide
Rectangular Waveguide Fundamentals
Waveguide Field Equations
Parallel Plate Waveguide
Rectangular Waveguide Modes
Cylindrical Waveguide Fundamentals
Cylindrical Waveguide Modes
Resonant Cavity
Dielectric Waveguide
2
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
3
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) :
• TLine can only support TEM wave whereas waveguide
can support many possible field configurations.
4
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 operate above dc (f =0) to a very high
frequency, but waveguide can operate only above cutoff
frequency and therefore acts as a high pass filter.
5
Introduction (Cont’d..)
The most common waveguide types:
6
1.1 Rectangular Waveguide
Fundamentals
A cross section of rectangular waveguide is shown below:
• 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
7
Rectangular Waveguide
Fundamentals (Cont’d..)
The interior dimensions are a x b, where the longer side is a.
‘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
8
Rectangular Waveguide
Fundamentals (Cont’d..)
Waveguide can support TE and TM modes, where:

In TE Modes, the electric field is transverse to the
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.
9
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.
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
10
Rectangular Waveguide
Fundamentals (Cont’d..)
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
For conventional rectangular waveguide filled with air,
where a = 2b, the dominant or lowest order mode is TE10
with cutoff frequency fc10 = c/2a
11
Rectangular Waveguide
Fundamentals (Cont’d..)
The relative cutoff frequencies for the first 12 modes of
conventional rectangular waveguide filled with air,
Location
of
modes
relative
to
the
dominant TE10 mode in
standard
rectangular
waveguide where a=2b.
12
Rectangular Waveguide
Fundamentals (Cont’d..)
13
Example 1
Calculate the cutoff frequency for
the first four modes of WR284
waveguide.
14
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 
1
2 
2
2
1
c
1 0
c
where u p 
      fc10 

2a
a b
15
Solution to Example 1(Cont’d..)
Then, we have:
3 108 m
100 cm
s
fc10 
 2.08 GHz
27.214 cm  1 m
This agrees with the cutoff frequency cited in table. Then :
3 108 m
c
100 cm
s
fc01 

 4.41 GHz
2b 23.404 cm  1 m
c
fc20   4.16 GHz
a
not same with fc10 since a ≠ 2b
16
Solution to Example 1(Cont’d..)
and for the fourth mode,
3 108 m
2
2
1
1

 
 100 cm
s
fc11 

 

2
 7.214 cm   3.404 cm  1 m
 4.87 GHz
17
Rectangular Waveguide
Fundamentals (Cont’d..)
The field pattern for two modes where E only varies in
the x direction, since n=0, the field is constant in the y
direction.
The field patterns and
associated field
intensities in a cross
section of rectangular
waveguide for (a) TE10
and (b) TE20. Solid
lines indicate electric
field; dashed lines are
the magnetic field.
18
1.2 Waveguide Field Equations
Beginning with Maxwell’s equations, develop the time
harmonic field equations for rectangular waveguide. For
simplicity, consider the guide filled with lossless, charge
free media and the walls to be perfect conductors.
  ES  0    D  0
  HS  0    B  0
  E S   jH S
  H S  jES
 E S   E S  0
2
2
 2 H S   2 H S  0
19
Waveguide Field Equations (Cont’d..)
For conventional rectangular waveguide,
components in Cartesian coordinates are:
the
field
E S  E xsa x  E ysa y  E zs a z
H S  H xsa x  H ysa y  H zs a z
Inserting these equations into previous Maxwell’s
equation..
20
Waveguide Field Equations (Cont’d..)
 E zs E ys 
 E ys E xs 
 E xs E zs 




 y  z a x   z  x a y   x  y a z   j o H xsa x  H ysa y  H zs a z





 H zs H ys 
 H ys H xs 
H xs H zs 





 y  z a x   z  x a y   x  y a z  j o E xsa x  E ysa y  E zs a z







From previous divergence equations,
E xs E ys E zs


0
x
y
z
H xs H ys H zs


0
x
y
z
21
Waveguide Field Equations (Cont’d..)
From previous wave equation,
 2E S
x 2
 2H S
x 2

 2E S

 2H S
y 2
y 2

 2E S

 2H S
z 2
z 2
  2 E S
  2 H S
22
Waveguide Field Equations (Cont’d..)
From the first expanded Maxwell’s equations,
H zs H ys

 j E xs
y
z
H xs H zs

 j E ys
z
x
H ys H xs

 j E xz
x
y
E zs E ys

  j H xs
y
z
E xs E zs

  j H ys
z
x
E ys E xs

  j H zs
x
y
23
Waveguide Field Equations (Cont’d..)
Then, consider the fields only propagate in the z direction,
in harmonic fields:
ES  E0 e jt e z
The partial derivative with respect to z is:
ES
jt z
 Eo e e
z
 2 ES
2
jt z
  Eo e e
2
z
Substitute these into the expanded Maxwell’s equations
24
Waveguide Field Equations (Cont’d..)
Those equations can be reduced to:
H zs
 H ys  j E xs
y
H zs
 H xs 
 j E ys
x
H ys H xs

 j E zs
x
y
E zs
 E ys   j H xs
y
E zs
 E xs 
  j H ys
x
E ys E xs

  j H zs
x
y
25
Waveguide Field Equations (Cont’d..)
So, using these equations, we can find expression for the
four transverse components (Ex, Ey, Hx, Hy) in terms of z
directed components (Ez and Hz), where:
E zs
H zs 
1 
  

E ys 
 j
y
x 
k c2 
(1)
E zs
H zs 
1 
 j

H xs 

2 
y
x 
kc 
(2)
26
Waveguide Field Equations (Cont’d..)
And also..
E xs
H zs
1  E zs
 

 j
y
k c2  x



(3)
H ys
E zs
H zs
1 
 j


2 
x
y
kc 



(4)
With,
kc     
2
2
2
Try to derive these four equations on your own!
27
Waveguide Field Equations (Cont’d..)
So, these four important equations will be used to find
the transverse components for TM and TE mode, where:
• for TM mode, Hz=0, then use these four equations to
find the transverse components.
• for TE mode, Ez=0, then use these four equations to
find the transverse components.
28
Waveguide Field Equations (Cont’d..)
• for TM mode, Hz=0, equation (1) to (4) can be reduced to:
E zs 
1 
  
 (5)
E ys 
y 
k c2 
H xs
E zs 
1 
 j
 (7)

y 
k c2 
E xs
1  E zs 




2
k c  x 
H ys
E zs 
1 

 j

2
x 
kc 
(6)
(8)
29
Waveguide Field Equations (Cont’d..)
• for TE mode, Ez=0, equation (1) to (4) can be reduced to:
E ys
H zs 
1 

 j
 (9)
2
x 
kc 
H xs
H zs 
1 
(11) H ys   1   H zs





x 
k c2 
k c2  y
H zs
1 
E xs    j
y
k c2 

 (10)




(12)
30
1.3 Parallel Plate Waveguide
The parallel plate waveguide is the simplest type of
guide that can support TM and TE Modes, and TEM as
well because it’s formed from two plates as shown.
The width W is assumed to be
much greater than the separation
d, so that fringing fields and any
x variation can be ignored. A
material with permittivity εr and
permeability µr is assumed to fill
the region between the two
plates.
31
Parallel Plate Waveguide (Cont’d..)
By considering the boundary
condition, the magnitude E will
change with y, E=0 at y=0 and at
y=a but maximum in the middle. y=a
Magnitude E will not change with
x since x is infinity (no boundary),
so the value will constant . The E
will along the propagation
i.e. y=0
+z direction. Thus,
z
x

0
x
32
Parallel Plate Waveguide Modes
(Cont’d..)
• For TM Mode
For TM Mode, the magnetic field has its components
transverse or normal to the direction of wave propagation,
where Hz=0 and a nonzero Ez fields. So,

2
2
E z  kc E z
0

 2 E zs
x 2

 2 E zs
y 2

 2 E zs
z 2
 k c2 E zs  0
Since the field is going to +z direction, it can be reduced to only:
 2 E zs
x 2

 2 E zs
y 2
 k c2 E zs  0
33
Parallel Plate Waveguide TM Modes
(Cont’d..)

 0 , solution for this is:
With
x
E zs  A sin kc y  B cos kc y
At y=0 Ezs =0 , thus
Ez 
k c2
 A cos kc y  B sin kc y 
j
E zs  0  A sin kc (0)  B cos kc (0)
Hence B=0
At y=a Ez =0
E zs  0  A sin kc (a)
This valid if
kc a  n
or
n
kc 
a
n=1,2,3,4….
34
Parallel Plate Waveguide TM Modes
(Cont’d..)
Thus with
e
 j z
,
ny   jz

E zs  A sin
e
a 

Then, from equation (5) to (8), since d/dx=0, so only (5) and
(7) remain, therefore:
E ys  
H xs
  E zs 


k c2  y 
E zs 
1 
 j


2 
y 
kc 

j 
ny   jz
E ys  
A cos
e
kc 
a 

j 
ny   jz
H xs 
A cos
e
kc 
a 
35
Parallel Plate Waveguide Modes
(Cont’d..)
• For TE Mode
For TE Mode, the electric field has its components
transverse or normal to the direction of wave propagation,
where Ez=0 and a nonzero Hz fields. So,

2
2
H z  kc H z
0

 2 H zs
x 2

 2 H zs
y 2

 2 H zs
z 2
 k c2 H zs  0
Since the field is gong to +z direction, it can be reduced to only:
 2 H zs
x 2

 2 H zs
y 2
 k c2 H zs  0
36
Parallel Plate Waveguide TE Modes
(Cont’d..)
With

 0 , solution for this is:
x
H zs  A sin kc y  B cos kc y
But, the boundary are that Ex=0 at y=0,a. So that from (10),
E xs 
 j
A cos kc y  B sin kc y 
kc
At y=0 Ex =0 , thus
E xs
 j
A cos kc 0  B sin kc 0
0
kc
Hence A=0
37
Parallel Plate Waveguide TE Modes
(Cont’d..)
At y=a Ex =0
E xs  0  B sin kc (a)
This valid if
kc a  n
or
kc 
n
a
n=1,2,3,4….
Thus with e  jz , and get back the solution for Hz,
ny   jz

H zs  B cos
e
a 

38
Parallel Plate Waveguide TE Modes
(Cont’d..)
Then, from equation (9) to (12), since d/dx=0, so only (10) and
(12) remain, therefore:
H zs
1 
E xs    j
y
k c2 



1  H zs 


H ys  
2
k c  y 
where
j  ny   jz
B sin
 Exs 
e
kc 
a 
j  ny   jz
 H ys 
B sin
e
kc 
a 
j  
39
1.4 Rectangular Waveguide Modes
• For TM Mode
For TM Mode, the magnetic field has its components transverse
or normal to the direction of wave propagation. At the walls of
the waveguide, the tangential components of the E field must be
continuous, that is:
E zs  0 at y  0 and y  b
E zs  0 at x  0 and x  a
40
Rectangular Waveguide TM Modes
(Cont’d..)
For TM mode, Hz=0, then lets:
E zs  XYe z
Where X is a function of x and Y is a function of y
Then the wave equation becomes:
Y
d2X
dx 2
X
d 2Y
dy 2
  2 XY   2  XY
41
Rectangular Waveguide TM Modes
(Cont’d..)
We divide the equation by XY
1 d 2 X 1 d 2Y
2


k
c 0
2
2
X dx
Y dy
1 d 2Y
2

k
 0 and then
Let’s
y
Y dy 2
Transitional Page
or
1 d2X
 k x 2  k y 2  k c2
X dx 2
1 d2X
 kx2  0
X dx 2
and
2
kc
2
 kx  k y
2
42
Rectangular Waveguide TM Modes
(Cont’d..)
Solution for both equations:
X  C1 cos k x x  C2 sin k x x Y  D1 cosk y y  D2 sin k y y
And the whole expression becomes:


Ezs  C1 cosk x x  C2 sin k x x D1 cosk y y  D2 sin k y y e z
43
Rectangular Waveguide TM Modes
(Cont’d..)
We know that the tangential electric fields at the walls of the
waveguide must be zero. Then by applying the boundary
conditions:
Applying boundary condition at x=0 where Ez=0

Transitional Page
Ezs  0  C1 cos0  C2 sin 0 D1 cosk y y  D2 sin k y y

So C1=0
Applying boundary condition at y=0 where Ez=0
E zs  0  C2 sin k x xD1 cos 0  D2 sin 0
So D1=0
44
Rectangular Waveguide TM Modes
(Cont’d..)
Then the equation reduced to




Ezs  C2 D2 sin k x x sin k y y e z  E0 sin k x x sin k y y e z
Applying the other boundary condition x=a where Ez =0


Ezs  0  E0 sin k x a sin k y y e z
E0 and sin kyy cannot be zero
This means that sin kxa = mπ = 0 or k x 
m
where m=0,1,2,3,4…
a
45
Rectangular Waveguide TM Modes
(Cont’d..)
Applying the other boundary condition y=b where Ez =0


E zs  0  E0 sin k x x  sin k y b e z
But E0 and sin kxx cannot be zero,
Transitional Page
This means that sin kyb = 0 or k y 
So,
n
b
where n =0,1,2,3,4…
mx 
ny   jz

E zs  E0  sin
 sin
e
a 
b 

46
Rectangular Waveguide TM Modes
(Cont’d..)
Then, substitute Hzs=0 and
  j
we have:
E xs
 j m
mx 
ny   jz


E0  cos
sin
e



a 
b 

k c2 a
E ys
 j n
mx 
ny   jz


E0  sin
cos
e



a 
b 

k c2 b
47
Rectangular Waveguide TM Modes
(Cont’d..)
and also..
H xs
j n
mx 
ny   jz


E0  sin
cos
e



a 
b 

k c2 b
 j m
mx 
ny   jz
 Page
Transitional
H ys 
E0  cos
 sin
e
kc2
a

a 
b 
Denote as TMmn ,the field vanish for TM00 , TM10 and
TM01. The lowest mode is TM11
48
Rectangular Waveguide TM Modes
(Cont’d..)
Useful relations to be remember,
2
 m   n 
 

 a   b 
2
Wave number, k c2  k x2  k y2  
The propagation constant is:
2
2
 m   n 
2
  


k
 

 a   b 
where k  

49
Rectangular Waveguide TM Modes
(Cont’d..)
Where we have 3 possibilities depending on k (or ω), m and n:
• Cutoff mode :
2
 m   n 
k 2   2   
 

 a   b 
  0 or     0
2
Transitional
Page
At this time, ω is called cutoff angular frequency:
c 
1

2
 m   n 

 

 a   b 
2
50
Rectangular Waveguide TM Modes
(Cont’d..)
• Evanescent mode :
2
 m   n 
k 2   2   
 

 a   b 
   or   0
2
We have no propagation at all. These non propagating or
attenuating modes are said to be evanescent.
• Propagating mode :
2
 m   n 
2
2
k     
 

 a   b 
  j or   0
2
51
Rectangular Waveguide TM Modes
(Cont’d..)
Where the phase constant becomes:
2
 n 
2  m 
  k 
 

 a   b 
2
For each mode (combination of m and n) thus has a cutoff
frequency fcmn given by:
Transitional Page
fc mn 
kc
2 

1
2 
2
 m   n 

 

 a   b 
2
The cutoff frequency is the operating frequency below which
attenuation occurs and above which propagation takes place.
52
Rectangular Waveguide TM Modes
(Cont’d..)
The cutoff wavelength,
c 
2
2
m n
   
 a  a
2
The phase constant,
 fc 
    1   
 f 
2
The intrinsic wave impedance,
TM
 fc 
 ' 1  
 f 
2
Where,  ' 


Intrinsic impedance
in the medium
53
Rectangular Waveguide Modes
(Cont’d..)
• For TE Mode
For TE Mode, the electric field has its components transverse or
normal to the direction of wave propagation. At the walls of the
waveguide, the tangential components of the E field must be
continuous, that is:
E xs  0 at y  0 and y  b
E ys  0 at x  0 and x  a
54
Rectangular Waveguide TE Modes
(Cont’d..)
For TE mode, Ez=0, then lets:
H zs  XYe z
Where X is a function of x and Y is a function of y
Transitional Page
From previous derivation, the wave equation becomes:


H zs  C1 cosk x x  C2 sin k x x D1 cosk y y  D2 sin k y y e z
55
Rectangular Waveguide TE Modes
(Cont’d..)
Hz cannot impose boundary condition since it is not zero at
boundary, so we determine Ex and Ey as follow :
From, E xs
E xs
H zs
1  E zs
 

 j
y
k c2  x
H zs
1 
 j

2 
y
kc 
E xs 
j
kc2

 it becomes:


 Then reduced to:

C1 cos k x x  C2 sin k x x D1 sin k y y  D2 cos k y y e z
56
Rectangular Waveguide TE Modes
(Cont’d..)
From,
E ys
E ys
H zs
1  E zs


 j
2
x
k c  y
H zs 
1 

j


x 
k c2 



it becomes:
Then reduced to:
Transitional Page
E ys  
j
kc2
C1 sin k x x  C2 cos k x xD1 cos k y y  D2 sin k y y e z
57
Rectangular Waveguide TE Modes
(Cont’d..)
Then by applying the boundary conditions:
Applying boundary condition at y=0 where Ex=0
E xs  0 
j
kc2
C1 cos k x x  C2 sin k x xD1 sin k y 0  D2 cos k y 0e z
So D2=0, then applying boundary condition at x=0 where Ey=0
E ys  0  
j
kc2
C1 sin k x 0  C2 cos k x 0D1 cos k y y  D2 sin k y y e z
So C2=0
58
Rectangular Waveguide TE Modes
(Cont’d..)
Applying the other boundary condition x=a where Ey =0
E ys  0  
j
kc2
C1 sin k x (a)D1 cos k y y e z
m
This means that sin kxa = 0 or k x 
a
where m=0,1,2,3,4…
Transitional Page
Applying the other boundary condition y=b where Ex =0
E xs  0 
j
kc2
C1 cos k x xD1 sin k y (b)e z
n
This means that sin kyb = 0 or k y 
b
where n=0,1,2,3,4…
59
Rectangular Waveguide TE Modes
(Cont’d..)
So,
So,
mx 
ny   jz

H zs  H 0  cos
 cos
e
a 
b 

Then, substitute Ez=0 and
  j
we have:
E xs
j n
mx 
ny   jz


H 0  cos
 sin
e
2 b
a 
b 

kc
E ys
 j m
mx 
ny   jz


H 0  sin
cos
e



a
a 
b 

k c2
60
Rectangular Waveguide TE Modes
(Cont’d..)
and also..
H xs
j m
mx 
ny   jz


H 0  sin
cos
e



a 
b 

kc2 a
j n
mx 
ny   jz
 Page
Transitional
H ys 
H 0  cos
 sin
e
k c2 b

a 
b 
Denote as TEmn ,the field vanish for TE00. The lowest
mode is TE10 for a>b and TE01 for b>a.
61
Rectangular Waveguide Modes
(Cont’d..)
From equations for the TM and TE modes, we can obtain the
field patterns..
For example, the dominant TE10 mode, where m = 1 and n = 0,
so the Ex, Ey, Hx, Hy and Hz equations becomes,
x   jz

H zs  H 0  cos e
a 

So then in the time domain, H z  Re H zs e jt


x 

H z  H 0  cos  cost  z 
a 

62
Rectangular Waveguide Modes
(Cont’d..)
Similarly..
H ys  0
E ys
E xs  0
 j 
 x   jz

H 0  sin e
2
a 

kc a
 a
 x 
Ey 
H 0  sin  sin t  z 

a 

H xs
j 
 x   jz

H  sin e
2 a 0
a 

kc
a
 x 
Hx  
H 0  sin  sin t  z 

a 

63
Rectangular Waveguide Modes
(Cont’d..)
So, for TE10 mode, the variation of the E and H fields with x
in an xy plane, say plane cos (ωt-βz)=1 for Hz and plane sin
(ωt-βz)=1 for Ey and Hx
64
Rectangular Waveguide Modes
(Cont’d..)
The corresponding field lines:
65
Rectangular
Waveguide
Modes
Field
lines for some of the
lower order modes
of a rectangular
waveguide :
(Cont’d..)
66
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 medium
characterized by σ=0, µr=1, ε=4ε0. Calculate the
cutoff frequencies of the modes.
67
Solution to Example 2
The cutoff frequency is given by:
fc mn 
2
2
1
 m   n 

 
 
2 
 a   b 
1
2 
2
m n
   
 a  b
2
With a = 2.5b or a/b = 2.5,
up 
Or,
1


c
r  r
c

2
So,
c
fc mn 
4a
m2 
a2
b2
n2
fc m n  3 m 2  6.25 n 2 GHz
68
Solution to Example 2 (Cont’d..)
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 TE 01 (m  0, n  1)  fc 01  32.5  7.5 GHz
TE 02 (m  0, n  2)  fc 02  35  15 GHz
TE 03 (m  0, n  3)  fc 03  37.5  22 .5 GHz
Thus, for fcmn < 15.1 GHz, the maximum n = 2.
69
Solution to Example 2 (Cont’d..)
Then, fix n and increase m,
For TE10 (m  1, n  0)  fc10  3 GHz
TE 20  fc 20  6 GHz
TE 30  fc 30  9 GHz
TE 40  fc 40  12 GHz
TE 50  fc 50  15 GHz
TE 60  fc 60  18 GHz
Thus, for fcmn < 15.1 GHz, the maximum m = 5.
70
Solution to Example 2 (Cont’d..)
We know the maximum value of m and n, so try other
possible combinations in between the maximum values.
For TE11, TM 11 (degenerate modes )  fc11  3 7.25  8.078 GHz
TE 21, TM 21  fc 21  3 10 .25  9.6 GHz
TE 31, TM 31  fc 31  3 15 .25  11 .72 GHz
TE 41, TM 41  fc 41  3 22 .25  14 .14 GHz
TE12 , TM 12  fc12  3 26  15 .3 GHz
71
Solution to Example 2 (Cont’d..)
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:
72
Example 3
In a rectangular waveguide with dimension a=1.5 cm,
b=0.8 cm, σ=0, µ=µ0, ε=4ε0, find:


A
 x   3y 
11
H x  2 sin  cos
 sin   10 t  z
m
 a   b 
• The mode of operation
• The cutoff frequency
• The phase constant
• The propagation constant
• The intrinsic wave impedance
73
Solution to Example 3
We could find that the given expression is in instantaneous field
expression form which obtained from the phasor forms by using:

E  Re E S e jt


H  Re H S e jt

From the given expression we could find that m=1, and n=3. That
is the guide is operating at TM13 or TE13.
Suppose that after this, we choose TM13 mode.
74
Solution to Example 3 (Cont’d..)
Where for air filled waveguide,
2
c m n
fc m n 
   
2  a  b
2
Hence, for m=1 and n=3, the cutoff frequency is:
fcmn
c

4

1

 1.5 10  2


2

 
3
 
  0.8 10 2
 

2


  28.57 GHz


75
Solution to Example 3 (Cont’d..)
The phase constant,
2
 r
 fc 
    1    
c
 f 
 fc 
1  
 f 
2
Where,
  2f  

100
11 or
f 
 50 GHz
10
2
  10 (2)
11
3  10 8
2
 28 .57 
1 
 1718 .81 rad / m

 50 
76
Solution to Example 3 (Cont’d..)
The propagation constant,
    j
but   0
Because it’s in propagating mode,
So,
  j
,   j1718 .81 / m
The intrinsic wave impedance,
2
TM
f 
377
 ' 1  c  
r
 f 
2
 28 .57 
1 
 154 .7 

 50 
77
1.5 Cylindrical Waveguide
Fundamentals
A hollow metal tube of circular cross section also supports TE
and TM waveguide modes as shown.
Transitional Page
78
Cylindrical Waveguide
Fundamentals (Cont’d..)
For cylindrical waveguide, the field components is in
cylindrical coordinates which are:
E S  E s a   Es a  E zs a z
H S  H s a   Hs a  H zs a z
Inserting these equations into previous Maxwell’s
equation,
 E S   ES  0
2
2
 2 H S   2 H S  0
79
Cylindrical Waveguide
Fundamentals (Cont’d..)
By using the same method of derivation for rectangular
waveguide (starting from slide 21), we could get four equations
of Eρ, Eφ, Hρ and Hφ in terms of Ez and Hz
j  E zs  H zs

E s  

2
 
k c  



Transitional
Page
(13)
H zs 
 j   E zs


Es 
 
2   
 
kc 
(15)
H zs 
j   E zs


H s 

2   
 
kc 
(14)
E zs  H zs
j 

H s  


2

 
kc 
Try this!!!!



(16)
80
1.6 Cylindrical Waveguide Modes
• For TM Mode
For the TM modes of the circular waveguide, we must solve Ez
from the wave equation in cylindrical coordinates, then through
a very long and difficult derivation, we could get the transverse
fields as:
 j
 A sin n  B cos n J n ' kc  e  jz
E s 
kc
 jn
Es  2  A cos n  B sin n J n kc  e  jz
kc 
81
Cylindrical Waveguide TM Modes
(Cont’d..)
and also..
H s
jn
 jz




 2
A cos n  B sin n J n kc  e
kc 
Transitional Page
 j
 A sin n  B cos n J n ' kc  e  jz
Hs 
kc
82
Cylindrical Waveguide TM Modes
(Cont’d..)
Some useful parameters for TM mode:
ZTM 
E
H

p
kc  nm
 E
H


Wave impedance for TM modes
k
Cutoff wavenumber
a
 nm  k  k c  k   pnm a 
2
fc nm 
kc
2 
2

2
pnm
2a 
2
Propagation constant
of the TMnm modes
Cutoff frequency
83
Cylindrical Waveguide TM Modes
(Cont’d..)
Values of pnm for TM Modes of a Circular Waveguide
n
pn1
pn2
pn3
0
2.405
5.520
8.654
Transitional
Page7.016
1
3.832
10.174
2
5.135
8.417
11.620
pnm is the roots of Jn(x) which recognized as solution for
Bessel’s differential equation.
84
Cylindrical Waveguide
Modes (Cont’d..)
• For TE Mode
For the TE modes of the circular waveguide, we must solve Hz
from the wave equation in cylindrical coordinates, after which
we could get the transverse fields as:
E s 
 j n
kc 2 
 A cos n  B sin n J n kc  e  jz
 j
 A sin n  B cos n J n ' kc  e  jz
Es 
kc
85
Cylindrical Waveguide TE Modes
(Cont’d..)
and also..
 j
 A sin n  B cos n J n ' kc  e  jz
H s 
kc
Transitional Page
Hs 
 jn
kc 2 
 A cos n  B sin n J n kc  e  jz
86
Cylindrical Waveguide TE Modes
(Cont’d..)
Some useful parameters for TE mode:
ZTE 
kc 
E
H
 E

p'nm
H

k

Wave impedance for TE modes
Cutoff wavenumber
a
 nm  k  k c  k   p'nm a 
2
fc nm 
kc
2 
2
2

p'nm
2a 
2
Propagation constant
of the TEnm modes
Cutoff frequency
87
Cylindrical Waveguide TE Modes
(Cont’d..)
Values of p’nm for TE Modes of a Circular Waveguide
n
p'n1
p’n2
p’n3
0
3.832
7.016
10.174
Transitional
Page5.331
1
1.841
8.536
2
3.054
6.706
9.970
p’nm is the roots of Jn(x) which recognized as solution for
Bessel’s differential equation.
88
Field lines for some of the lower order modes of a cylindrical
waveguide :
89
1.7 Resonant Cavity
The length of resonator, d is made multiple of waveguide
wavelength, i.e.
dp
a
g
p =1,2,3……
2
b
d=p/2
Resonator wavelength can be
calculated as :
1
2r

1
2c

1
2g

2


where c
kc
1
2c

1
 
a
2d 2
p
d=p2
90
Resonant Cavity (Cont’d..)
And kc for rectangular waveguide
2
k c2
 m   n 

 

 a   b 
2
For cylindrical wave guide
p nm
TM mode k c 
a
TE mode
p ' nm
kc 
a
91
Example 4
A cylindrical resonator has a radius of 5cm
which is used to measure frequency from 8GHz
to 12GHz at TE11 mode . What is the required
length, d for tuning those frequency in that
particular mode.
92
Solution to Example 4
First we calculate the cutoff wavelength
2
2
c 
 1.841  0.1706 m
kc
0.05
First frequency wavelength at 8 GHz,
8
c
3 10
r1 

 0.0375 m
f r1 8 10 9
Second frequency wavelength at 12 GHz,
r 2 
c
fr2

3 10 8
12 10
9
 0.025 m
93
Solution to Example 4 (Cont’d..)
Calculate the length of wave guide

p
d 
2

1
1
2r

1
2c





For first frequency at 8 GHz,
1
d1 
2
1
1
0.03752

1
0.17062
 0.0192 m
For second frequency at 12 GHz,
1
d2 
2
1
1
0.0252

1
0.17062
 0.0126 m
94
Solution to Example 4 (Cont’d..)
So, the cavity need to have length, d in this range in
order to make the cavity operates at resonant
frequencies between 8GHz to 12 GHz:
0.0126 m  d  0.0192 m
95
1.8 Dielectric Waveguide
Air filled wave guide
g 
 o2
    s 
chnm
fc 
2 a
c
fc 
2
  2
 g   r  o2  k c2
 k c2
2
2
fc 
2
c
2

    s 
2
o
m n
   
 a  b
1
Dielectric filled wave guide
1
o
2
2
fc 
2  r
o
chnm
2 a  r
c
2 r
  2
2
 m  n
   
 a  b
1
c 2

r
o 2
2
96
Chapter 2
Waveguide
End