Rectangular_waveguide_description and equation

Notes 20
Rectangular Waveguides
1
ECE 3317
Applied Electromagnetic Waves
Prof. David R. Jackson
Fall 2023

Rectangular Waveguide
 We assume that the boundary is a perfect electric conductor (PEC).
2
Cross section
x
y
a
b ,
ε µ
No TEMz mode can exist!

Rectangular Waveguide (cont.)
3
Why is there no TEMz mode?
z
k k ω µε
= =
TEMz mode:
b
x
y
a
E
0
z =
( )
2 2
, 2
( , ):
m n
z
m n
m n k k
a b
π π
   
= − −
   
   
Rectangular waveguide mode
( )
,
m n
z
k k
≠
( ) ( )
, 0,0
m n ≠

 Two types of modes can exist independently:
TMz: Ez only
TEz: Hz only
4

 We analyze the problem to solve for Ez or Hz (all other fields come from these).
TMz: Ez only
TEz: Hz only
5
Cross section
x
y
a
b ,
ε µ

TMz Modes
0, 0
z z
H E
= ≠
2 2
0
z z
E k E
∇ + =(Helmholtz equation)
0
z
E = on boundary (PEC walls)
( ) ( )
0
, , , z
jk z
z z
E x y z E x y e−
=
Guided-wave assumption:
2 2 2
2
2 2 2
0
z z z
z
E E E
k E
x y z
 
∂ ∂ ∂
+ + + =
 
∂ ∂ ∂
 
2 2
2 2
2 2
0
z z
z z z
E E
k E k E
x y
 
∂ ∂
+ − + =
 
∂ ∂
 
6

TMz Modes (cont.)
We solve the above equation by using the method of separation of variables.
( )
2 2
2 2
2 2
0
z z
z z
E E
k k E
x y
∂ ∂
+ + − =
∂ ∂
Define: 2 2 2
c z
k k k
≡ −
2 2
2
2 2
0
z z
c z
E E
k E
x y
∂ ∂
+ + =
∂ ∂
Note that kc is an
unknown at this point.
We then have:
2 2
2
0 0
0
2 2
0
z z
c z
E E
k E
x y
∂ ∂
+ + =
∂ ∂
Dividing by the exp(-j kz z) term, we have:
7
Please see Appendix A for the solution.

TMz Modes (cont.)
8
Solution from separation of variables method:
( )
( )
,
, , sin sin
m n
z
jk z
z mn
m x n y
E x y z A e
a b
π π −
   
=    
   
( )
2 2
, 2 2 2
m n
z c
m n
k k k k
a b
π π
   
= − = − −
   
   
1,2,
1,2,
m
n
=
=


Note: If either m or n is zero, the entire field is zero.
0 r r
k k
ω µε µ ε
= =
0 0 0
0
2
k
c
ω π
ω µ ε
λ
= = =
2 2
2
c
m n
k
a b
π π
   
= +
   
   
TMmn mode

Cutoff Frequency for Lossless Waveguide
Set ( )
,
0
m n
z
k =
,
2 2
TM
2 m n
c
c
f f
m n
k f
a b
π π
π µε
=
   
= = +
   
   
We start with
9
TMz Modes (cont.)
( )
2 2
, 2
m n
z
m n
k k
a b
π π
   
= − −
   
   
,
2 2
TM
2
m n d
c
c m n
f
a b
π π
π
   
= +
   
   
( )
d
r
c
c
ε
= nonmagnetic material
This defines the cutoff frequency.
Note:
Cutoff frequency only has
a clear meaning in the
lossless case (k is real).

Summary of TMz Solution: TMmn mode
( )
( )
,
, , sin sin
m n
z
jk z
z mn
m x n y
E x y z A e
a b
π π −
   
=    
   
1,2,
1,2,
m
n
=
=


10
Note: If either m or n is zero, the entire field is zero.
TMz Modes (cont.)
,
2 2
TM
2
m n d
c
c m n
f
a b
π π
π
   
= +
   
   
( )
2 2
, 2 2 2
m n
z c
m n
k k k k
a b
π π
   
= − = − −
   
   
(lossless waveguide)

TEz Modes
( )
2 2
2 2
2 2
0
z z
z z
H H
k k H
x y
∂ ∂
+ + − =
∂ ∂
We now start with
0, 0
z z
E H
= ≠
11
( ) ( )
0
, , , z
jk z
z z
H x y z H x y e−
=
Guided-wave assumption:
2 2
2
0 0
0
2 2
0
z z
c z
H H
k H
x y
∂ ∂
+ + =
∂ ∂
Please see Appendix B for the solution.
Define: 2 2 2
c z
k k k
≡ −

Summary of TEz Solution: TEmn mode
( )
( )
,
, , cos cos
m n
z
jk z
z mn
m x n y
H x y z A e
a b
π π −
   
=    
   
( )
2 2
, 2
m n
z
m n
k k
a b
π π
   
= − −
   
   
0,1,2
0,1,2
m
n
=
=


( ) ( )
, 0,0
m n ≠
Note: Same formula for cutoff frequency as the TMz case!
12
TEz Modes (cont.)
,
2 2
TE
2
m n d
c
c m n
f
a b
π π
π
   
= +
   
   

Summary for Both Modes
( )
( )
,
, , cos cos
m n
z
jk z
z mn
m x n y
H x y z A e
a b
π π −
   
=    
   
( )
2 2
, 2
m n
z
m n
k k
a b
π π
   
= − −
   
   
( ) ( )
0,1,2,
0,1,2,
, 0,0
m
n
m n
=
=
≠


same formula for both modes
( )
( )
,
, , sin sin
m n
z
jk z
z mn
m x n y
E x y z A e
a b
π π −
   
=    
   
TMmn
TEmn
( )
2 2
,
2
m n d
c
c m n
f
a b
π π
π
   
= +
   
   
same formula for both modes
1,2,
1,2,
m
n
=
=


TMz TEz
13
d
r
c
c
ε
=

Field Plots
14
Color denotes magnitude, arrows show direction of electric field.

Wavenumber
2 2
z c
k k k
= −
2 2
c
m n
k
a b
π π
   
= +
   
   
TMz or TEz mode: with
z
k β
=
Note: The (m,n) notation is suppressed here on kz.
15
Above cutoff:
2 2
2 m n
k
a b
π π
β
   
= − −
   
   
z
k jα
= −
Below cutoff:
2 2
2
m n
k
a b
π π
α
   
= + −
   
   
Lossless waveguide:
( )
:
z z
k k j
β α
= −
Recall the general formula for

Wavenumber Plot
β
f
c
k
c
f
α
2
k f
β π µε
= =
2 2
2
, c
m n
k f f
a b
π π
β
   
= − − >
   
   
2 2
2
, c
m n
k f f
a b
π π
α
   
= + − <
   
   
16
“Light line”
( )
2 2
,
2
m n d
c c
c m n
f f
a b
π π
π
   
= = +
   
   
d
r
c
c
ε
=
2 2
c
m n
k
a b
π π
   
+
   
   

Guided Wavelength
Recall: The guided wavelength λg is the distance z that it
takes for the wave to repeat itself.
2
g
π
λ
β
=
(This assumes that we are above the cutoff frequency – otherwise
guided wavelength makes no sense.)
17
( )
2
1 /
d
g
c
f f
λ
λ =
−
After some algebra (see next slide):
0
d
r
λ
λ
ε
 
=
 
 
 
( )
g d
λ λ
>
Note :

Guided Wavelength (cont.)
2 2 2 2
2
2 2 2
g
c
k k
m n
k
a b
π π π
λ
β π π
= = =
−
   
− −
   
   
18
( ) ( ) ( )
2 2 2 2
2
2 2 2
2
1 / 1 /
1 /
d
g
c c c
c
d
k k k k k k k
k k
π π π λ
λ
π
λ
= = = =
− − −
−
2
k f
ω µε π µε
= =
2
c c c
k f
ω µε π µε
= =
( )
2
1 /
d
g
c
f f
λ
λ =
−
Derivation of wavelength formula (lossless waveguide):
/ /
c c
k k f f
=

Dominant Mode
The "dominant" mode is the one with the lowest cutoff frequency.
2 2
2
d
c
c m n
f
a b
π π
π
   
= +
   
   
Assume b < a
Lowest TMz mode: TM11
Lowest TEz mode: TE10
( ) ( )
0,1,2,
0,1,2,
, 0,0
m
n
m n
=
=
≠


1,2,
1,2,
m
n
=
=


TMz
TEz
The dominant mode is the TE10 mode.
19
x
y
a
b ,
ε µ
d
r
c
c
ε
=

Dominant Mode (cont.)
Summary (TE10 Mode)
( ) 10
, , cos z
jk z
z
x
H x y z A e
a
π −
 
=  
 
2
2
z
k k
a
π
 
= − 
 
2
d
c
c
f
a
=
2
2
, c
k f f
a
π
β
 
=
− >
 
 
2
2
, c
k f f
a
π
α
 
= − <
 
 
20
d
r
c
c
ε
=
0 r
k k ε
=

Fields of the Dominant TE10 Mode
( ) 10
, , cos z
jk z
z
x
H x y z A e
a
π −
 
=  
 
Find the other fields from these equations (Appendix A of Notes 19):
2 2 2 2
z z z
y
z z
j H jk E
E
k k x k k y
ωµ
   
∂ ∂
= −
   
− ∂ − ∂
   
2 2 2 2
z z z
x
z z
j E jk H
H
k k y k k x
ωε
   
∂ ∂
= −
   
− ∂ − ∂
   
2 2 2 2
z z z
x
z z
j H jk E
E
k k y k k x
ωµ
   
− ∂ ∂
= −
   
− ∂ − ∂
   
2 2 2 2
z z z
y
z z
j E jk H
H
k k x k k y
ωε
   
− ∂ ∂
= −
   
− ∂ − ∂
   
21

Summary of fields for TE10 mode:
10 10
2 2
z
j
E A
k k a
ωµ π
  
= −
  
−  
 
where
22
( ) 10
, , sin z
jk z
y
x
E x y z E e
a
π −
 
=  
 
( ) 10
, , sin z
jk z
z
x
k x
H x y z E e
a
π
ωµ
−
   
= −  
 
 
 
( ) 10
, , cos z
jk z
z
x
H x y z A e
a
π −
 
=  
 

23
b
x
y
a
E
0
z =
TE10 Mode
x
y
x
y
E
H
a
b
Length of arrows denotes magnitude of field Color denotes magnitude of field
Spacing between arrows denotes magnitude of field

24
TE10 Mode
3D View

What is the mode with the next highest cutoff frequency?
2 2
2
d
c
c m n
f
a b
π π
π
   
+
   
   
Assume b < a / 2
Then the next highest is the TE20 mode.
( ) ( )
2,0 1,0
2
c c
f f
=
25
( )
2
0,1 1
2 2
d d
c
c c
f
b b
π
π
   
= =
   
   
( )
2
2,0 2 1
2 2 / 2
d d
c
c c
f
a a
π
π
   
= =
   
   
x
y
a
b ,
ε µ
fc
TE10 TE20
Useful operating region
TE01
d
r
c
c
ε
=
A 2:1 operating band!
( )
1,0
2
d
c
c
f
a
=

What is the mode with the next highest cutoff frequency?
Assume b > a / 2
Then the next highest is the TE01 mode.
26
( )
2
0,1 1
2 2
d d
c
c c
f
b b
π
π
   
= =
   
   
( )
2
2,0 2 1
2 2 / 2
d d
c
c c
f
a a
π
π
   
= =
   
   
x
y
a
b ,
ε µ
The useable bandwidth is now lower than before.
fc
TE10 TE20
Useful operating region TE01
d
r
c
c
ε
=

Power flow in lossless waveguide (f > fc):
27
[ ]
2
10 W
4
z
ab
P E
β
ωµ
 
=  
 
( ) 10
, , sin z
jk z
y
x
E x y z E e
a
π −
 
=  
 
 Make b larger to get more power flow for a given value of a.
 Keep b smaller than a/2 to get maximum bandwidth.
(watts flowing down the waveguide)
(The derivation is omitted, but please see the formula box above.)
The optimum dimension for b is a/2
(gives maximum power flow without sacrificing bandwidth).
( )
( )
*
0 0
*
0 0
*
0
1
ˆ
Re
2
1
Re
2
1
Re
2
b a
z
b a
y x
a
y x
P E H z dxdy
E H dxdy
b E H dx
= × ⋅
= −
= −
∫∫
∫∫
∫
Note: Above cutoff, there is only watts flowing (no vars). Below cutoff there is no watts flowing (only vars).

Plane wave interpretation of TE10 mode
28
( ) ( )
10 10
10
, , sin sin
2
z z
x x
z
jk z jk z
y x
jk x jk x
jk z
x
E x y z E e E k x e
a
e e
E e
j
π − −
−
−
 
=
 
 
 
−
=  
 
sin
2
jz jz
e e
z
j
−
 
−
=  
 
:
Note
( ) 10 10
, , x x
z z
jk x jk x
jk z jk z
y
E x y z E e e E e e
− − −
′ ′′
= + ( )
( )
10 10
10 10
/ 2
/ 2
E E j
E E j
′ ≡−
 
 
 
′′ ≡+
 
x
k
a
π
 
≡
 
 
PW #1 PW #2
z
x
PW #2
PW #1
(E, H) θ
2
2
/
tan x
z
k a
k
k
a
π
θ
π
= =
 
−  
 

Losses in Waveguide (f > fc)
29
( )
2
2
0
2
1 [np/m]
1 /
r
s c
c
c
R f
b
b a f
f f
ε
α
η
 
 
+
 
 
 
 
−  
( )
2
2
0 1 tan
z r d d
k k j j
a
π
ε δ β α
 
= − − =
−
 
 
( )
2
2
0
Im 1 tan
d r d
k j
a
π
α ε δ
 
=
− − − 
 
Dielectric loss:
Conductor loss:
( )
: 1 tan
rc r j
ε ε δ
= −
Recall
d c
α α α
≈ +
(This is derived in ECE 5317.)
Note:
If we are below cutoff,
attenuation is mainly due to
evanescence, so we don’t
worry about conductor and
dielectric loss then.

Example
Standard X-band* waveguide:
a = 0.900 inches (2.286 cm)
b = 0.400 inches (1.016 cm)
( )
1,0
2
c
c
f
a
=
( )
1,0
6.56 [GHz]
c
f =
( )
2,0
13.11 [GHz]
c
f =
Find the single-mode operating frequency region
for air-filled X-band waveguide.
Hence, we have:
Use
30
Note: b < a / 2
6.56 13.11 [GHz]
f
< <
* X-band: from 8.0 to 12 GHz.
X-band waveguide

Example (cont.)
( )
1,0
6.56 [GHz]
c
f =
0 0 0 0
2 / 2 /
k k f c
ω µ ε π π λ
= = = =
At 9.00 GHz: β = 129.13 [rad/m]
At 5.00 GHz: α = 88.91 [nepers/m]
Recall:
/ 137.43 [rad/m]
c
k a
π
= =
31
: 188.62 [rad/m]
: 104.79 [rad/m]
k
k
=
=
At 9.0 GHz
At 5.0 GHz
2
2
, c
k f f
a
π
β
 
=
− >
 
 
2
2
, c
k f f
a
π
α
 
= − <
 
 
 Find the phase constant of the TE10 mode at 9.00 GHz.
 Find the attenuation in dB/m at 5.00 GHz
X-band waveguide

Example (cont.)
772 dB/m
=
Attenuation
This is a very rapid attenuation!
dB/m 8.68589α
=
32
At 5.0 GHz:
[ ]
88.91 nepers/m
α =
Recall:
X-band waveguide

Waveguide Components
33
Straight sections Flexible waveguides Waveguide bends
Waveguide adapters Waveguide couplers
https://www.pasternack.com
Waveguide terminations

Waveguide Modes in Transmission Lines
34
 A transmission line normally operates in the TEMz mode,
where the two conductors have equal and opposite currents.
 At high frequencies, waveguide modes can also propagate on
transmission lines.
 This is undesirable, and it limits the high-frequency range of
operation for the transmission line.

Waveguide Modes in Coax
35
Dominant waveguide mode in coax (derivation omitted):
11
TE 1 1
1 /
c
r
c
f
b a
a π
ε
  
≈   
+
  
4
4
0.035 inches 8.89 10 [m]
0.116 inches 29.46 10 [m]
2.2
r
a
b
ε
−
−
= = ×
= = ×
=
Example: RG 142 coax
0 4
/ 3.3 4
1 8. [ ]
Z
b a =
= ⇒ Ω
11
TE
16.8 [GHz]
c
f ≈
TE11 mode:
r
ε a
b
z
0
0 ln
2 r
L b
Z
C a
η
π ε
 
= =  
 
This coax cannot be used above 16.8 [GHz]
Note:
In this notation, the “11”
subscript refers to the
angular and radial variation.

Appendix A: TMz Modes
We solve the above equation by using the method of separation of variables.
( )
2 2
2 2
2 2
0
z z
z z
E E
k k E
x y
∂ ∂
+ + − =
∂ ∂
Define:
2 2 2
c z
k k k
≡ −
2 2
2
2 2
0
z z
c z
E E
k E
x y
∂ ∂
+ + =
∂ ∂
Note that kc is an
unknown at this point.
We then have:
We assume: ( ) ( ) ( )
0 ,
z
E x y X x Y y
=
2 2
2
0 0
0
2 2
0
z z
c z
E E
k E
x y
∂ ∂
+ + =
∂ ∂
Dividing by the exp(-jkzz) term, we have:
36
We want to solve:

Hence 2
c
X Y X Y k XY
′′ ′′
+ =
−
2
c
X Y
k
X Y
′′ ′′
+ =
−
Hence 2
c
X Y
k
X Y
′′ ′′
=
− −
Divide by XY :
( ) ( ) ( )
0 ,
z
E x y X x Y y
=
( ) ( )
F x G y
=
This has the form
Both sides of the equation
must be a constant!
2 2
2
0 0
0
2 2
0
z z
c z
E E
k E
x y
∂ ∂
+ + =
∂ ∂
37
Appendix A (cont.)

Denote
2
c
X Y
k
X Y
′′ ′′
=
− − =
constant
2
x
X
k
X
′′
=
− =
constant
(1)
(0
( )
) 0
( 0 2
)
X
X a
=
=
( ) sin( ) cos( )
x x
X x A k x B k x
= +
Boundary conditions:
General solution:
0 ( ) sin( )
x
B X x A k x
= ⇒ =
sin( ) 0
x
k a =
(1)
(2)
38
x
y
a
b ,
ε µ
Appendix A (cont.)

sin( ) 0
x
k a =
This gives us the following result:
, 1,2
x
k a m m
π
= = 
( ) sin
m x
X x A
a
π
 
=  
 
Hence
Now we turn our attention to the Y (y) function.
39
x
m
k
a
π
=
From the last slide:
Appendix A (cont.)

We have
2 2
c x
X Y
k k
X Y
′′ ′′
=
− − =
−
2 2
x c
Y
k k
Y
′′
= −
Hence
2 2 2
y c x
k k k
= −
Then we have
2
y
Y
k
Y
′′
= −
Denote
( ) sin( ) cos( )
y y
Y y C k y D k y
= +
General solution:
40
Appendix A (cont.)

( ) sin( ) cos( )
y y
Y y C k y D k y
= +
(3)
(0
( )
) 0
( 0 4
)
Y
Y b
=
=
0 ( ) sin( )
y
D Y y C k y
= ⇒ =
sin( ) 0
y
k b =
Equation (4) gives us the following result: , 1,2
y
k b n n
π
= = 
(3)
(4)
41
y
n
k
b
π
=
x
y
a
b ,
ε µ
Appendix A (cont.)

( ) sin
n y
Y y C
b
π
 
=  
 
The Y(y) function is then
( ) ( )
0 , ( ) sin sin
z
m x n y
E x y X x Y y AC
a b
π π
   
= =    
   
Therefore, we have
( )
0 , sin sin
z mn
m x n y
E x y A
a b
π π
   
=    
   
( )
( )
,
, , sin sin
m n
z
jk z
z mn
m x n y
E x y z A e
a b
π π −
   
=    
   
New notation:
The Ez field inside the waveguide thus has the following form:
42
Appendix A (cont.)

Recall that
2 2 2
y c x
k k k
= −
2 2 2
c x y
k k k
= +
Hence,
2 2
2
c
m n
k
a b
π π
   
= +
   
   
Therefore, the solution for kc is given by
Next, recall that 2 2 2
c z
k k k
= −
2 2 2
z c
k k k
= −
Hence
43
Appendix A (cont.)
( )
2 2
, 2
m n
z z
m n
k k k
a b
π π
   
= = − −
   
   

Appendix B: TEz Modes
( )
2 2
2 2
2 2
0
z z
z z
H H
k k H
x y
∂ ∂
+ + − =
∂ ∂
We now start with
Using the separation of variables method again, we have
( ) ( ) ( )
0 ,
z
H x y X x Y y
=
( ) sin( ) cos( )
y y
Y y C k y D k y
= +
( ) sin( ) cos( )
x x
X x A k x B k x
= +
2 2 2
c x y
k k k
= + 2 2 2
z c
k k k
= −
where
and
0, 0
z z
E H
= ≠
44

Appendix B (cont.)
( ,0) 0
( , ) 0
x
x
E x
E x b
=
=
(0, ) 0
( , ) 0
y
y
E y
E a y
=
=
( )
0 , cos cos
z mn
m x n y
H x y A
a b
π π
   
=    
   
The result is
2 2 2 2
z z z
x
z z
H jk E
j
E
k k y k k x
ωµ
   
∂ ∂
−
= −
   
− ∂ − ∂
   
2 2 2 2
z z z
y
z z
H jk E
j
E
k k x k k y
ωµ
   
∂ ∂
= −
   
− ∂ − ∂
   
This can be shown by using the following equations:
0, 0,
z
H
y b
y
∂
= =
∂
0, 0,
z
H
x a
x
∂
= =
∂
45
x
y
a
b ,
ε µ

( )
( )
,
, , cos cos
m n
z
jk z
z mn
m x n y
H x y z A e
a b
π π −
   
=    
   
( )
2 2
, 2
m n
z
m n
k k
a b
π π
   
= − −
   
   
0,1,2
0,1,2
m
n
=
=


Note: The (0,0) TEz mode is not valid, since it violates the magnetic Gauss law:
( ) ( )
, 0,0
m n ≠
( ) 00
ˆ
, , jkz
H x y z z A e−
= ( )
, , 0
H x y z
∇⋅ ≠
Same formula for cutoff
frequency as the TEz case!
46
Appendix B (cont.)
The Hz field inside the waveguide thus has the following form:

Rectangular_waveguide_description and equation

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