NUS/ECE
EE4101
Waveguides
At high frequencies, the loss of electromagnetic waves traveling
along transmission lines due to conductor resistance and radiation
leakage becomes exceedingly large. To alleviate this problem,
hollow waveguides can be used. We will study the rectangular
waveguide as a typical example.
y
x
z
x
z
b
ε, μ
r ε, μ
y
a
Circular waveguide
Rectangular waveguide
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Waveguides
NUS/ECE
EE4101
General Field Expression inside a Waveguide
Transverse directions: (x, y) or (r, φ)
Longitudinal direction: z (propagation direction)
In general
~
E = E ( x , y )e − γz
Then
∂ 2E
2
=
γ
E
2
∂z
Helmholtz’s equations:
∇ 2E + k 2E = 0
∇2H + k 2H = 0
(k = ω
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με )
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Waveguides
NUS/ECE
EE4101
Method of Solution:
Step 1
Express transverse field components Ex, Ey in
terms of longitudinal field component Ez
Step 2
Obtain solution for the longitudinal field
Ez from the wave equation
Step 3
Obtain Ex, Ey from Ez
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Waveguides
NUS/ECE
EE4101
In rectangular coordinates:
∇ 2E + k 2E = 0
⎛ ∂2
∂2
∂2
2⎞
⇒ ⎜⎜ 2 + 2 + 2 + k ⎟⎟E = 0
∂y
∂z
⎝ ∂x
⎠
⎛ 2
∂2
2⎞
⇒ ⎜⎜ ∇ xy + 2 + k ⎟⎟E = 0
∂z
⎝
⎠
⇒ ∇ 2xy + γ 2 + k 2 E = 0
(
)
(
)
(1a)
(
)
(1b)
⇒ ∇ 2xy E + γ 2 + k 2 E = 0
Similarly,
∇ 2xy H + γ 2 + k 2 H = 0
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Waveguides
NUS/ECE
EE4101
∇ × E = − jωμ H
xˆ
∂
∂x
Ex
yˆ
∂
∂y
Ey
zˆ
∂
= − jωμH
∂z
Ez
i = x, y , z
~
∂E z
~
~
+ γE y = − jωμH x
∂y
~
∂
E
~
~
⇒ − γE x − z = − jωμH y
∂x
~
∂E y ∂E~x
~
= − jωμH z
⇒
−
∂y
∂x
∂E z ∂E y
−
= − jωμH x ⇒
∂y
∂z
(2a)
∂E x ∂E z
= − jωμH y
−
∂z
∂x
(2b)
∂E y
∂E x
= − jωμH z
−
∂y
∂x
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Note that:
~
E i ( x , y , z ) = E i ( x , y )e − γz
~
H i ( x , y , z ) = H i ( x , y )e − γz
5
(2c)
Waveguides
NUS/ECE
EE4101
Similarly from
∇ × H = jωε E
We have
~
∂H z
~
~
+ γH y = jωεE x
∂y
(3a)
~
~ ∂H z
~
= jωεE y
− γH x −
∂x
(3b)
~
∂H y
~
∂H x
~
= jωεE z
−
∂y
∂x
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(3c)
Waveguides
NUS/ECE
EE4101
Finally from equation sets in (2) and (3), we have:
~
~
1 ⎛ ∂H z
∂E z ⎞
~
⎟
⎜
− jωε
Hx = − 2
γ
2 ⎜
∂y ⎟⎠
γ + k ⎝ ∂x
~
~
⎞
⎛
1
∂
∂
E
H
~
z
z
⎟
⎜γ
+ jωε
Hy = − 2
2 ⎜
γ + k ⎝ ∂y
∂x ⎟⎠
~
~
⎛
1
∂H z ⎞
∂E z
~
⎟
⎜γ
+ jωμ
Ex = − 2
2 ⎜
∂y ⎟⎠
γ + k ⎝ ∂x
~
~
⎛
1
∂H z ⎞
∂E z
~
⎟
⎜γ
− jωμ
Ey = − 2
2 ⎜
∂x ⎟⎠
γ + k ⎝ ∂y
(4a)
(4b)
(4c)
(4d)
~
Hence, we can solve the scalar Helmholtz’s equations for Ez and
~
Hz, and use the above formulas to determine the other components.
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Waveguides
NUS/ECE
EE4101
Waveguide Mode Classification
It is convenient to first classify waveguide modes as to
whether Ez or Hz exists according to:
TEM:
TE:
TM:
Ez = 0
Ez = 0
Ez ≠ 0
Hz = 0
Hz ≠ 0
Hz = 0
TEM = Transverse ElectroMagnetic
TE = Transverse Electric
TM = Transverse Magnetic
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Waveguides
NUS/ECE
EE4101
(1) TEM Modes:
~
~
Ez = H z = Ez = H z = 0
From the equations in (4), for the existence of non-trivial solutions,
the denominators must be zero also. That is,
γ 2 + k2 = 0
Propagation constants: ∴ γ = ± jk = ± jω με
Phase velocity:
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∴u p =
ω
k
9
=
1
με
Waveguides
NUS/ECE
EE4101
From the equations in (4), the field components take an indefinite
mathematical form of 0/0, whose definite values have to be
determined by boundary conditions. In general, we can write:
∵∇ E = 0
2
xy
∵ ∇2xy H = 0
~
Ex = Ex0 ,
⇒
E x = E x 0 e ± jkz
~
Ey = Ey0 ,
⇒
E y = E y 0 e ± jkz
~
H x = H x0 ,
⇒
H x = H x 0 e ± jkz
~
H y = H y0 ,
⇒
H y = H y 0 e ± jkz
The relations between Ex, Ey, Hx, and Hy can be further obtained
from the equations in (2) and (3), as shown below.
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Waveguides
NUS/ECE
EE4101
Wave impedance:
Z TEM
~
Ex
Ex
jωμ
=
= ~ =
=
Hy Hy
γ
μ
=η
ε
(5a)
Wave impedance = Intrinsic impedance of the medium
~
Ey Ey
jωμ
μ
(5b)
= ~ =−
=−
= − Z TEM
γ
ε
Hx Hx
Combining (5a) & (5b),
H x xˆ + H y yˆ = −
Ey
Z TEM
xˆ +
Ex
yˆ
Z TEM
Therefore
H=
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Z TEM
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zˆ × E
Waveguides
NUS/ECE
EE4101
TEM modes can only exist in two-conductor waveguides such as
two-wire transmission lines, co-axial lines, parallel-plate
waveguides, etc, but not in single-conductor waveguides such as
rectangular waveguides and circular waveguides. This is because
either longitudinal field components or longitudinal currents are
required to support the transverse magnetic field components Hx
and Hy which form close loops in the transverse plane. There are
no longitudinal currents (not longitudinal surface currents) inside
hollow waveguides and hence hollow waveguides cannot support
TEM modes. But they can support TE and TM modes.
(2) TE and TM Modes:
TE and TM modes in general exist in hollow waveguides such as
rectangular waveguides and circular waveguides. They will be
studied in the context of these waveguides.
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Waveguides
NUS/ECE
EE4101
Rectangular Waveguide
(A) TM Modes:
~
Hz = Hz = 0
y
b
x
z
a
We first find the longitudinal field Ez
~
E z ( x, y, z ) = E z ( x, y )e −γ z
From (1a), the equation for the Ez field is:
⎛ ∂2
∂2
2⎞~
⎜⎜ 2 + 2 + h ⎟⎟ E z (x,y ) = 0
∂y
⎠
⎝ ∂x
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h2 = γ2 + k 2
Waveguides
NUS/ECE
EE4101
Let
~
E z ( x, y ) = X (x )Y ( y )
function of x only
function of y only
Then
1 d 2 X (x )
1 d 2Y ( y )
2
+
=
−
h
X ( x ) dx 2
Y ( y ) dy 2
The above equation can be satisfied for all values of x
and y inside the waveguide only when both terms on the
left-hand side being equal to a constant.
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Waveguides
NUS/ECE
Hence let
EE4101
1 d 2Y ( y )
2
=
−
k
y
Y ( y ) dy 2
1 d X (x )
2
=
−
k
x
X ( x ) dx 2
2
where
k x2 + k y2 = h 2
Boundary conditions:
Solutioins:
~
E z (0 ,y ) = 0
~
E z (a,y ) = 0
~
E z ( x,0) = 0
~
E z (x,b ) = 0
X ( x ) = C1 sin k x x
with
Y ( y ) = C2 sin k y y with
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mπ
kx =
a
nπ
ky =
b
C1 and C2 are
constants to be
determined by
the boundary
conditions along
the z direction.
(m = 1, 2, 3, …)
(n = 1, 2, 3, …)
Waveguides
NUS/ECE
EE4101
~
⎛ mπ ⎞ ⎛ nπ ⎞
E z ( x,y ) = E0 sin ⎜
x ⎟ sin ⎜
y⎟
⎝ a ⎠ ⎝ b ⎠
2
⎛ mπ ⎞ ⎛ nπ ⎞
h =k +k =⎜
⎟ +⎜ ⎟
⎝ a ⎠ ⎝ b ⎠
2
2
x
(m = 1, 2, 3,…)
(n = 1, 2, 3, …)
2
2
y
γ ⎛ mπ ⎞
~
⎛ mπ ⎞ ⎛ nπ ⎞
E x ( x,y ) = − 2 ⎜
x ⎟ sin ⎜
y⎟
⎟ E0 cos⎜
h ⎝ a ⎠
⎝ a ⎠ ⎝ b ⎠
γ ⎛ nπ ⎞
~
⎛ mπ ⎞ ⎛ nπ ⎞
E y ( x,y ) = − 2 ⎜ ⎟ E0 sin ⎜
x ⎟ cos⎜
y⎟
h ⎝ b ⎠
⎝ a ⎠ ⎝ b ⎠
jωε ⎛ nπ ⎞
~
⎛ mπ ⎞ ⎛ nπ ⎞
H x ( x,y ) = 2 ⎜ ⎟ E0 sin ⎜
x ⎟ cos⎜
y⎟
h ⎝ b ⎠
⎝ a ⎠ ⎝ b ⎠
jωε ⎛ mπ ⎞
~
⎛ mπ ⎞ ⎛ nπ ⎞
H y ( x,y ) = − 2 ⎜
x ⎟ sin ⎜
y⎟
⎟ E0 cos⎜
h ⎝ a ⎠
⎝ a ⎠ ⎝ b ⎠
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E0 is a contant
equal to C1×C2
and is to be
determined by
the excitation
condition of the
waveguide.
Waveguides
NUS/ECE
EE4101
Every combination of the integers m and n defines a possible TM
mode that may be designated as a TMmn mode. Hence there are
infinite number of TM mode that can exist inside the waveguide.
Propagation constant :
γ = h2 − k 2
⎛ mπ ⎞ ⎛ nπ ⎞
2
= ⎜
⎟ +⎜
⎟ − ω με
⎝ a ⎠ ⎝ b ⎠
2
Note that the cutoff
frequency for a TEM
mode is zero (i.e., DC).
2
⎛ mπ ⎞ ⎛ nπ ⎞
2
= ⎜
⎟ +⎜
⎟ − (2πf ) με
⎝ a ⎠ ⎝ b ⎠
The frequency at which γ = 0 is called the cutoff frequency fc.
1
2π
2
2
(
)
=
λc mn =
1
2
2
⎛ mπ ⎞ ⎛ nπ ⎞
f
με
c
( f c )mn =
+
m
n
π
π
⎛
⎞ ⎛
⎞
⎜
⎟ ⎜
⎟
+
⎜
⎟ ⎜
⎟
2π με ⎝ a ⎠ ⎝ b ⎠
⎝ a ⎠ ⎝ b ⎠
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NUS/ECE
EE4101
(a) When f > fc, the propagation constant is an imaginary number and
the mode can travel inside the waveguide.
⎛ 2πf c με ⎞
⎟
γ = jβ = j k − h = jk 1 − ⎜⎜
⎟
⎝ 2πf με ⎠
2
2
⎛ fc ⎞
= jk 1 − ⎜⎜ ⎟⎟
⎝ f ⎠
∴
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2
⎛ fc ⎞
β = k 1 − ⎜⎜ ⎟⎟
⎝ f ⎠
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2
2
Waveguides
NUS/ECE
EE4101
Guided wavelength:
λg =
λ=
2π
β
=
2π
⎛ fc ⎞
k 1 − ⎜⎜ ⎟⎟
⎝ f ⎠
2
λ
=
⎛ fc ⎞
1− ⎜ ⎟
⎝ f ⎠
2
2π
1
=
k
f με
where λ is the wavelength of a plane wave with a frequency f.
Note that λg > λ .
λ =
2
g
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2
λ
2
⎛λ⎞
1 − ⎜⎜ ⎟⎟
⎝ λc ⎠
2
⎛λ⎞
λ2
1 − ⎜⎜ ⎟⎟ = 2
λg
⎝ λc ⎠
19
1
λ
2
=
1
λ
2
g
+
1
λ2c
Waveguides
NUS/ECE
EE4101
Phase velocity:
up =
ω
=
β
=
up =
ω
⎛f ⎞
k 1 − ⎜⎜ c ⎟⎟
⎝ f ⎠
2
ω
⎛ fc ⎞
ω με 1 − ⎜⎜ ⎟⎟
⎝ f ⎠
u
⎛ fc ⎞
1 − ⎜⎜ ⎟⎟
⎝ f ⎠
2
2
⎛
⎜u =
⎜
⎝
1 ⎞⎟
με ⎟⎠
The phase velocity is frequency dependent.
A rectangular waveguide is a dispersive device.
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Waveguides
NUS/ECE
EE4101
Group velocity:
⎛ fc ⎞
dω
1
ug =
=
= u 1 − ⎜⎜ ⎟⎟
dβ dβ
⎝ f ⎠
dω
2
∴ug < u
Note that:
u pu g = u 2
Wave impedance:
Z TM
Ey
⎛ fc ⎞
Ex
γ
β
=
=−
=
=
= η 1 − ⎜⎜ ⎟⎟
Hy
H x jωε ωε
⎝ f ⎠
2
μ
η=
ε
See animation “Group Velocity”
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Waveguides
NUS/ECE
EE4101
Graphical Interpretation of up and ug
ω
β-ω curve for waveguide TE and TM modes
ωc
u p = slope of this
P
u g = slope at P
β-ω curve for TEM modes
β
straight line
Group velocity ug is the signal propagation velocity if we assume the signal
composed of a narrow band of frequencies centered around f. Phase
velocity up is the speed of a constant-phase point of a particular mode.
Group velocity is also the speed of energy flow inside the waveguide. (See
Ref. 5, Section 8.5, for more details.)
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Waveguides
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EE4101
(b) When f < fc, the propagation constant is a real number and the
mode is non-propagating. The amplitude of the mode becomes
smaller (with the e-αz) along the z direction. This mode is called an
evanescent mode.
k2
γ = α = attenuation constant = h 1 − 2
h
⎛ f ⎞
α = h 1 − ⎜⎜ ⎟⎟
⎝ fc ⎠
E
γ
jα
= x =
=−
= − jη
ωε
H y jωε
2
2
⎛ fc ⎞
⎜⎜ ⎟⎟ − 1 ⇒ imaginary
Z TM
⎝ f ⎠
Note that the energy of an evanescent mode is not lost but only
transferred back to the excitation source. That is, an evanescent mode
is constantly exchanging energy with the excitation source.
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Waveguides
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NUS/ECE
EE4101
Example 1
What are the instantaneous field expressions for the TM11 mode in a
rectangular waveguide of side lenghts a and b? Sketch its field lines.
Solution
With m = 1 & n = 1,
~
⎛π ⎞ ⎛π ⎞
E z ( x,y ) = E0 sin ⎜ x ⎟ sin ⎜ y ⎟
⎝a ⎠ ⎝b ⎠
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γ ⎛π⎞
~
⎛π ⎞ ⎛π
E x (x,y ) = − 2 ⎜ ⎟ E0 cos⎜ x ⎟ sin ⎜
h ⎝a⎠
⎝a ⎠ ⎝b
⎞
y⎟
⎠
γ ⎛π⎞
~
⎛π ⎞ ⎛π
E y ( x,y ) = − 2 ⎜ ⎟ E0 sin ⎜ x ⎟ cos⎜
h ⎝b⎠
⎝a ⎠ ⎝b
⎞
y⎟
⎠
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Waveguides
NUS/ECE
EE4101
~
H z ( x,y ) = 0
jωε ⎛ π ⎞
~
⎛π ⎞ ⎛π ⎞
H x ( x,y ) = 2 ⎜ ⎟ E0 sin ⎜ x ⎟ cos⎜ y ⎟
h ⎝b⎠
⎝a ⎠ ⎝b ⎠
jωε ⎛ π ⎞
~
⎛π ⎞ ⎛π ⎞
H y ( x,y ) = − 2 ⎜ ⎟ E0 cos⎜ x ⎟ sin ⎜ y ⎟
h ⎝a⎠
⎝a ⎠ ⎝b ⎠
For propagation
modes: γ = jβ
~
~
Ei ( x, y, z ) = Ei ( x, y )e −γ z = Ei ( x, y )e − jβ z , i = x, y, z
~
~
−γ z
H i (x, y, z ) = H i (x, y )e = H i ( x, y )e − jβ z , i = x, y, z
Instantaneous field expressions:
{
}
H ( x, y, z; t ) = Re{H ( x, y, z )e },
Ei ( x, y, z; t ) = Re Ei ( x, y, z )e jωt , i = x, y, z
jω t
i
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25
i = x, y , z
Waveguides
NUS/ECE
EE4101
⎛π ⎞ ⎛π
E z ( x,y, z; t ) = E0 sin ⎜ x ⎟ sin ⎜
⎝a ⎠ ⎝b
⎞
y ⎟ cos(ωt − βz )
⎠
β ⎛π⎞
⎛π ⎞ ⎛π ⎞
E x (x,y, z; t ) = 2 ⎜ ⎟ E0 cos⎜ x ⎟ sin ⎜ y ⎟ sin (ωt − β z )
h ⎝a⎠
⎝a ⎠ ⎝b ⎠
β ⎛π⎞
⎛π ⎞ ⎛π ⎞
E y ( x,y, z; t ) = 2 ⎜ ⎟ E0 sin ⎜ x ⎟ cos⎜ y ⎟ sin (ωt − βz )
h ⎝b⎠
⎝a ⎠ ⎝b ⎠
H z ( x,y, z; t ) = 0
ωε
H x (x,y, z; t ) = − 2
h
ωε
H y ( x,y, z; t ) = 2
h
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⎛π⎞
⎛π ⎞ ⎛π ⎞
⎜ ⎟ E0 sin ⎜ x ⎟ cos⎜ y ⎟ sin (ωt − βz )
⎝b⎠
⎝a ⎠ ⎝b ⎠
⎛π⎞
⎛π ⎞ ⎛π ⎞
⎜ ⎟ E0 cos⎜ x ⎟ sin ⎜ y ⎟ sin (ωt − βz )
⎝a ⎠ ⎝b ⎠
⎝a⎠
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Waveguides
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TM11 mode has the lowest cutoff frequency among
all the TM modes. Its field lines are shown below.
Solid lines: E field, dash lines: H field
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Waveguides
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EE4101
~
(B) TE Modes: E z = E z = 0
Using a similar analysis as for the TM modes, we can obtain field
expressions for TE modes as:
(m = 0,1, 2, …)
~
⎛ mπ ⎞ ⎛ nπ ⎞
(n = 0,1, 2, …)
H z ( x, y ) = H 0 cos⎜
x ⎟ cos⎜
y⎟
⎝ a ⎠ ⎝ b ⎠
m & n cannot
jωμ ⎛ nπ ⎞
~
⎛ mπ ⎞ ⎛ nπ ⎞
be both equal
x ⎟ sin ⎜
y⎟
E x ( x, y ) = 2 ⎜
⎟ H 0 cos⎜
to zero
h ⎝ b ⎠
⎝ a ⎠ ⎝ b ⎠
jωμ ⎛ mπ ⎞
~
⎛ mπ ⎞ ⎛ nπ ⎞
x ⎟ cos⎜
y⎟
E y ( x, y ) = − 2 ⎜
⎟ H 0 sin ⎜
H0 is a constant
h ⎝ a ⎠
a
b
⎠
⎝
⎠ ⎝
to be determined
by the excitation
γ ⎛ mπ ⎞
~
⎛ mπ ⎞ ⎛ nπ ⎞
H x ( x, y ) = 2 ⎜
x ⎟ cos⎜
y⎟
condition of the
⎟ H 0 sin ⎜
h ⎝ a ⎠
waveguide.
⎝ a ⎠ ⎝ b ⎠
γ ⎛ nπ ⎞
~
⎛ mπ ⎞ ⎛ nπ ⎞
H y ( x, y ) = 2 ⎜
x ⎟ sin ⎜
y⎟
⎟ H 0 cos⎜
h ⎝ b ⎠
⎝ a ⎠ ⎝ b ⎠
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Waveguides
NUS/ECE
EE4101
Cutoff frequency:
( f c )mn =
⎛ mπ ⎞ ⎛ nπ ⎞
⎟
⎜
⎟ +⎜
⎝ a ⎠ ⎝ b ⎠
2
1
2π με
2
Cutoff wavelength:
(λc )mn =
1
f c με
=
2π
⎛ mπ ⎞ ⎛ nπ ⎞
⎜
⎟ +⎜
⎟
⎝ a ⎠ ⎝ b ⎠
2
2
Propagation constant:
⎛ fc ⎞
β = k 1 − ⎜⎜ ⎟⎟
⎝ f ⎠
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2
Waveguides
NUS/ECE
EE4101
Guided wavelength:
λg =
λ
⎛ fc ⎞
1− ⎜ ⎟
⎝ f ⎠
2
Phased velocity:
up =
u
⎛ fc ⎞
1 − ⎜⎜ ⎟⎟
⎝ f ⎠
2
Group velocity:
⎛ fc ⎞
u g = u 1 − ⎜⎜ ⎟⎟
⎝ f ⎠
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2
Waveguides
NUS/ECE
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Wave impedance:
Z TE =
η
⎛ fc ⎞
1 − ⎜⎜ ⎟⎟
⎝ f ⎠
2
Attenuation constant for evanescent modes:
⎛ f ⎞
γ = α = h 1 − ⎜⎜ ⎟⎟
⎝ fc ⎠
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Note that in TE mode propagation, the lowest order mode is TE10 which
also has the lowest cutoff frequency among all the propation modes in a
rectangular waveguide. The cutoff frequencies of the different modes
are shown below for two cases of waveguide dimensions.
Case 1:
TE 01
b/a=1/2
TE10
TE 20
↓
↓
1
TE11
TM11
↓
f c / (f c )TE
10
3
2
Case 2:
TE 01
b/a=1
TE10
↓
TE11
TM11
↓
TE 20
↓
f c / (f c )TE
10
2
1
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TE 02
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TE10 Mode - Rectangular Waveguide
TE10 is the dominant mode in a rectangular waveguide with lowest
cutoff frequency (when a > b).
(Picture form)
E field: solid lines
H field: dash lines
Surface current
(Schematic form)
TE10
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Field expression of TE10 mode (m = 1 & n = 0):
~
⎛ π ⎞ − jβ z
− jβ z
= H 0 cos⎜ x ⎟e
H z = H z ( x, y )e
⎝a ⎠
~
⎛ 2a ⎞
⎛ π ⎞ − jβ z
− jβ z
= − jη ⎜ ⎟ H 0 sin ⎜ x ⎟e
E y = E y ( x, y )e
⎝a ⎠
⎝ λ ⎠
⎛a⎞
⎛π
H x = H x ( x, y ) e− jβ z = j β ⎜ ⎟ H 0 sin ⎜
⎝π ⎠
⎝a
⎞
x ⎟ e− jβ z
⎠
Ez = Ex = H y = 0
Cutoff frequency:
( f c )TE
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10
=
34
1
2a με
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Cutoff wavelength:
(λc )TE
Propagation constant:
10
= 2a
⎛ λ ⎞
2
β TE = k 1 − ⎜
⎟
⎝ 2a ⎠
10
Guided wavelength:
(λ )
g TE
10
=
λ
⎛ λ ⎞
2
1− ⎜ ⎟
⎝ 2a ⎠
Wave impedance:
Z TE10 =
Hon Tat Hui
η
⎛ λ ⎞
1− ⎜ ⎟
⎝ 2a ⎠
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Excitation of the Rectangular Waveguide
Cross-section at x = a/2
Probe
Coaxial line
Excitation of a rectangular waveguide by a coaxial line.
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A Note on the Propagating Modes inside
the Rectangular Waveguide
Note that in a rectangular waveguide with an excitation source
frequency f = fi, all those TM and TE modes with a cutoff frequency
lower than fi can propagate inside the waveguide. Whether they will
actually appear inside the waveguide depends on the excitation method.
The excitation method, for example the orientation of the coaxial
probe, can be chosen to excite certain modes while suppress other
modes. Those modes with a cutoff frequency higher than fi cannot
propagate inside the waveguide no matter what excitation method
chosen to excite them.
However, in the most general case, an EM wave inside the rectangular
waveguide is a linear combination of all those TE and TM modes
whose cutoff frequencies being lower than the excitation frequency.
Hence the rectangular waveguide is a high-pass filter.
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Example 2
A standard rectangular waveguide WG-16 is to be designed for the Xband (8-12.4 GHz) radar application. The dimensions are a = 2.29 cm
and b = 1.02 cm. If only the lowest mode TE10 mode is to propagate
inside the waveguide and that the operating frequency be at least 25%
above the cutoff frequency of the TE10 mode but no higher than 95% of
the next higher cutoff frequency, what is the allowable operatingfrequency range of this waveguide?
Solution
a = 2.29 cm
( f c )TE
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3 × 108
=
=
= 6.55 × 109
2a με 2 × 0.0229
1
10
b = 1.02 cm
38
(Hz )
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( f c )TE
( f c )TE
( f c )TE
mn
=
1
2π με
⎛ mπ ⎞ ⎛ nπ ⎞
⎟
⎜
⎟ +⎜
⎝ a ⎠ ⎝ b ⎠
2
2
3 × 108
=
=
= 13.10 × 109
a με 0.0229
1
20
m = 2,n =0
3 × 108
=
=
= 14.71× 109
m = 0 , n =1
2b με 2 × 0.0102
1
01
(Hz )
(Hz ) > ( f c )TE
20
Hence the allowable operating-frequency range is:
125%( f c )TE10 ≤ f ≤ 95%( f c )TE 20
That is:
8.19 GHz ≤ f ≤ 12.45 GHz
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References:
1. David K. Cheng, Field and Wave Electromagnetic, AddisonWesley Pub. Co., New York, 1989.
2. David M. Pozar, Microwave Engineering, John Wiley & Sons,
Inc., New Jersey, 2005.
3. Fawwaz T. Ulaby, Applied Electromagnetics, Prentice-Hall, Inc.,
New Jersey, 2007.
4. Robert E. Collin, Field theory of guided waves, IEEE Press, New
York, 1991.
5. J. D. Jackson, Classical Electrodynamics, John Wiley & Sons,
Inc., New York, 1975, Chapter 8, Section 8.5.
6. Joseph A. Edminister, Schaum’s Outline of Theory and Problems
of Electromagnetics, McGraw-Hill, Singapore, 1993.
7. Yung-kuo Lim (Editor), Problems and solutions on
electromagnetism, World Scientific, Singapore, 1993.
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