Download PARALLEL-PLATE WAVEGUIDES

PARALLEL-PLATE WAVEGUIDES
Wave Equation
E + ω2µεE = 0
(1)
∂ 2 Ex
∂2 E x
∂2 Ex
2
2 +
2 +
2 = - ω µεE x
∂x
∂y
∂z
(2a)
∂ 2 Ey
∂2 E y
∂2 Ey
2
2 +
2 +
2 = - ω µεE y
∂x
∂y
∂z
(2b)
∂ 2 Ez
∂ 2 Ez
∂2 Ez
2
+
+
2
2
2 = - ω µεE z
∂x
∂y
∂z
(2c)
PEC
x
µ, ε
z
y
PEC
Transverse Electric (TE) Modes
For a parallel-plate waveguide, the plates are infinite in the y-extent; we need to study the
propagation in the z-direction. The following assumptions are made in the wave equation
⇒
∂
∂
∂
= 0, but
≠ 0 and
≠0
∂y
∂x
∂z
⇒ Assume Ey only
These two conditions define the TE modes and the wave equation is simplified to read
∂ 2 Ey
∂2 E y
2
2 +
2 = - ω µεE y
∂x
∂z
(3)
General solution (forward traveling wave)
[
E y (x,z) = e− jβ zz Ae − jβ x x + Be + j βx x
]
(4)
-2At x = 0, E y = 0 which leads to A + B = 0. Therefore, A = -B = Eo/2j, where Eo is an arbitrary
constant
E y (x,z) = E oe− jβ zz sinβx x
(5)
x=a
x
µ, ε
z
x=0
At x = a, Ey(x, z) = 0. Let a be the distance separating the two PEC plates
E oe − jβz z sinβx a = 0
(6)
β xa = mπ, where m = 1, 2, 3, ...
(7)
mπ
βx = a
(8)
This leads to :
or
Moreover, from the differential equation (3), we get the dispersion relation
β z2 + β x2 = ω2µε = β2,
(9)
which leads to
 mπ 
βz = ω 2µε − 
 a 
2
(10)
where m = 1, 2, 3, ... Since propagation is to take place in the z direction, for the wave to
propagate, we must have βz2 > 0, or
 mπ 
ω 2µε > 
 a 
2
(11)
This leads to the following guidance condition which will insure wave propagation
f>
m
2a µε
(12)
-3The cutoff frequency fc is defined to be at the onset of propagation
fc =
m
2a µε
(13)
The cutoff frequency is the frequency below which the mode associated with the index m will
not propagate in the waveguide. Different modes will have different cutoff frequencies. The
cutoff frequency of a mode is associated with the cutoff wavelength λ c
v 2a
λ c = fc = m
(14)
Each mode is referred to as the TE m mode. From (6), it is obvious that there is no TE0 mode and
the first TE mode is the TE1 mode.
Magnetic Field
From ∇ × E = −jωµH
(15)
we have
−1
H=
jωµ
xˆ
∂
∂x
0
yˆ
0
Ey
zˆ
∂
∂z
(16)
0
which leads to
Hx = −
βz
E oe − jβ z z sinβx x
ωµ
Hz = +
jβx
E e − jβz z cosβx x
ωµ o
(17)
(18)
As can be seen, there is no Hy component, therefore, the TE solution has Ey, Hx and Hz only.
x
θ θ
z
µ, ε
From the dispersion relation, it can be shown that the propagation vector components satisfy the
relations
-4β z = βsinθ, βx = βcosθ
(19)
where θ is the angle of incidence of the propagation vector with the normal to the conductor
plates.
Transverse Magnetic (TM) modes
The magnetic field also satisfies the wave equation:
H + ω2µεH = 0
(20)
∂ 2 Hx
∂2 Hx
∂2 H x
+
+
= - ω 2 µεH x
∂x2
∂y2
∂z2
(21a)
∂ 2 Hy
∂2 Hy
∂2 H y
2
+
+
2
2
2 = -ω µεH y
∂x
∂y
∂z
(21b)
∂ 2 Hz
∂2 Hz
∂2 H z
2
+
+
2
2
2 = - ω µεH z
∂x
∂y
∂z
(21c)
For TM modes, we assume
⇒
∂
∂
∂
= 0, but
≠0 and
≠0
∂y
∂x
∂z
⇒ Assume Hy only
These two conditions define the TM modes and equations (21) are simplified to read
∂ 2 Hy
∂2 Hy
2
2 +
2 = - ω µεH y
∂x
∂z
(22)
General solution (forward traveling wave)
[
H y (x,z) = e − jβ z z Ae − jβ x x + Be + jβx x
From ∇ ×H =jωεE
we get
]
(23)
(24)
-5-
1
E=
jωε
xˆ
∂
∂x
0
yˆ
0
Hy
zˆ
∂
∂z
(25)
0
This leads to
[
]
E x (x,z) =
β z − jβ z z
e
Ae− jβ xx + Be + jβ x x
ωε
E z (x,z) =
βx − jβ z z
e
−Ae − j βx x + Be+ jβ xx
ωε
[
(26)
]
(27)
At x=0, Ez = 0 which leads to A = B = Ho/2 where Ho is an arbitrary constant. This leads to
H y (x,z) = Ho e− jβzz cosβ xx
(28)
E x (x,z) =
βz
H o e− jβ zz cosβ xx
ωε
(29)
E z (x,z) =
jβ x
H e − jβz z sinβ xx
ωε o
(30)
At x =a, Ez = 0 which leads to
β xa = mπ, where m = 0, 1, 2, 3, ...
(31)
This defines the TM modes which have only Hy, Ex and Ez components.
NOTE: THE DISPERSION RELATION, GUIDANCE CONDITION AND CUTOFF EQUATIONS FOR A PARALLEL-PLATE WAVEGUIDE ARE THE SAME FOR TE AND TM
MODES.
Equation (31) defines the TM modes; each mode is referred to as the TMm mode. It can be seen
from (28) that m=0 is a valid choice; it is called the TM0, or transverse electromagnetic or TEM
mode. For this mode βx=0 and,
H y = H oe − jβ z z
Ex =
βz
µ
Ho e − jβz z =
Ho e − jβz z
ωε
ε
Ez = 0
(32)
(33)
(34)
-6where β z = β, and in which there are no x variations of the fields within the waveguide. The
TEM mode has a cutoff frequency at DC and is always present in the waveguide.
x=a
x
E
z
µ, ε
H
x=0
TEM mode
Time-Average Poynting Vector
TE modes
P =
1
Re{E × H *}
2
P =
1
Re yˆ Ey × xˆ H *x + zˆ H*z
2
{
[
(35)
]}
2
2

Eo
1  Eo
2
P = Re zˆ
βz sin βx x + xˆ j
βx cosβx xsin β x x
2  ωµ
ωµ

P = zˆ
Eo 2
βz sin2 βx x
2ωµ
(36)
(37)
(38)
TM modes
P =
1
Re{E × H *}
2
P =
1
Re [ xˆ Ex + zˆ Ez ] × yˆ Hy*
2
{
(39)
}
2
2

Ho
1  H o
2
P = Re zˆ
βz cos βx x − xˆ j
βx sinβx xcos βx x
2  ωε
ωε

P = zˆ
Ho 2
βz cos2 βx x
2ωε
The total time-average power is found by integrating <P> over the area of interest.
(40)
(41)
(42)