WAVEGUIDES
x
z
y
µ, ε
Maxwell's Equation
∇ 2E + ω 2µεE = 0
(A)
∇ 2H + ω 2µεH = 0
(B)
For a waveguide with arbitrary cross section as shown in the above figure, we assume a plane
wave solution and as a first trial, we set E z = 0. This defines the TE modes.
∂H
From ∇ × E = −µ ∂t , we have
∂H x
∂Ez ∂Ey
=
-µ
∂y
∂z
∂t ⇒ +jβ zEy = -jωµH x
(1)
∂H y
∂Ex ∂Ez
=
-µ
∂z
∂x
∂t ⇒ -jβ zEx = -jωµH y
(2)
∂H z
∂Ey ∂Ex
∂Ey ∂Ex
=
-µ
⇒
∂x
∂y
∂t
∂x - ∂y = -jωµH z
(3)
From ∇ × E = − jωµH, we get
jωεE =
xˆ
yˆ
zˆ
∂
∂x
Hx
∂
∂x
Hy
∂
∂z
Hz
∂H z
∂H z ∂H y
=
jωεE
⇒
x
∂y
∂z
∂y + jβ zHy = jωεEx
(4)
∂H z
∂H x ∂H z
∂z - ∂x = jωεEy ⇒ -jβ zHx - ∂x = jωεEy
(5)
- 2∂H y ∂H x
∂x - ∂y = 0
(6)
We want to express all quantities in terms of H z.
From (2), we have
βE
Hy = z x
ωµ
(7)
∂H z
2 Ex
∂y + jβ z ωµ = jωεEx
(8)
in (4)
Solving for E x
Ex =
∂H z
β z2-ω2µε ∂y
Hx =
-β zEy
ωµ
jωµ
(9)
From (1)
(10)
in (5)
β z2Ey ∂Hz
- ∂x = jωεEy
+j
ωµ
(11)
Ey =
∂H z
β z2-ω2µε ∂x
(12)
Hx =
∂H z
β z2-ω2µε ∂x
(13)
Hy =
∂H z
β z2-ω2µε ∂y
(14)
so that
-jωµ
jβ z
Ez = 0
jβ z
(15)
- 3Combining solutions for E x and E y into (3) gives
∂2H z ∂2H z
+
= [β z2-ω2µε]H z
∂x 2
∂y 2
(16)
RECTANGULAR WAVEGUIDES
y
x
µ, ε
z
b
a
If the cross section of the waveguide is a rectangle, we have a rectangular waveguide and the
boundary conditions are such that the tangential electric field is zero on all the PEC walls.
TE Modes
The general solution for TE modes with Ez=0 is obtained from (16)
[
][
H z = e − jβ z z Ae − jβ x x + Be + jβ x x Ce − jβ y y + De + jβ y y
Ey =
Ex =
]
(18)
jβ yωµ
]
(19)
β2z
− ω µε
2
[
][
(17)
β xωµ
e − jβ z z − Ae − jβ x x + Be + jβ x x Ce − jβ y y + De + jβ y y
2
− ω µε
β2z
[
]
][
e − jβ z z Ae − jβ x x + Be + jβ x x − Ce − jβ y y + De + jβ y y
At y=0, Ex=0 which leads to C=D
At x=0 Ey=0 which leads to A=B
H z = H o e − jβ z z cos β x x cos β y y
Ey =
Ex =
β2z
β xωµ
H o e − jβ z z sin β x x cos β y y
2
− ω µε
jβ yωµ
β2z
− ω µε
2
H o e − jβ z z cos β x x sin β y y
(20)
(21)
(22)
- 4mπ
At x=a, Ey=0; this leads to β x = a
nπ
At y=b, Ex=0; this leads to β y = b
The dispersion relation is obtained by placing (20) in (16)
β2x + β2y + β2z = ω 2 µε
(23)
 mπ  +  nπ  + β2 = ω 2 µε
z
 a 
 b
(24)
mπ  2  nπ  2
βz = ω 2 µε − 
−
 a 
 b
(25)
2
2
and
The guidance condition is
mπ  2  nπ  2
ω 2 µε > 
+
 a 
 b
(26)
or
f > f c where fc is the cutoff frequency of the TE mn mode given by the relation
m 2  n 2
1

fc =
+
 b
2 µε  a 
(27)
The TEmn mode will not propagate unless f is greater than fc. Obviously, different modes will
have different cutoff frequencies.
TM Modes
The transverse magnetic modes for a general waveguide are obtained by assuming Hz =0. By
duality with the TE modes, we have
∂2Ez ∂2Ez
+ 2 =[β z2-ω2µε]Ez
∂x 2
∂y
(28)
- 5with general solution
[
][
E z = e − jβ z z Ae − jβ x x + Be + jβ x x Ce − jβ y y + De + jβ y y
]
(29)
The boundary conditions are
At x=0, Ez=0 which leads to A=-B
At y=0, Ez=0 which leads to C=-D
At x=a, Ez=0 which leads to β x =
mπ
a
nπ
At y=b, Ez=0 which leads to β y = b
so that the generating equation for the TM mn modes is
E z = E o e − jβ z z sin
mπx
nπy
sin
a
b
(30)
NOTE: THE DISPERSION RELATION, GUIDANCE CONDITION AND CUTOFF EQUATIONS FOR A RECTANGULAR WAVEGUIDE ARE THE SAME FOR TE AND TM
MODES.
For additional information on the field equations see Rao, page 539 Table 9.1.
There is no TE 00 mode
There are no TM m0 or TM 0n modes
The first TE mode is the TE 10 mode
The first TM mode is the TM11 mode
Impedance of a Waveguide
For a TE mode, we define the transverse impedance as
ηgTE =
-Ey Ex ωµ
Hx = Hy = βz
- 6From the relationship for β z and using
fc2 =
1  m  2  n  2 
+

 b 
4 π 2 µε  a 

(31)
we get
ηgTE =
η
(32)
fc2
1− 2
f
µ
. Analogously, for TM modes, it can be shown that
ε
where η is the intrinsic impedance η =
ηgTM
fc2
f2
= η 1−
(33)
Power Flow in a Rectangular Waveguide (TE 10)
The time-average Poynting vector for the TE 10 mode in a rectangular waveguide is given by
2
E
βz
πx
1
P = Re[E × H *] = zˆ o
sin 2
2
2 ωµ
a
Power = ∫
0 ∫0
2
2
βz
πx
dxdy
sin 2
a
2 ωµ
a b Eo
(34)
(35)
2
E βzab
E ab
Power = o
= o
4 ωµ
4 ηgTE10
(36)
Therefore the time-average power flow in a waveguide is proportional to its cross-section area.