ECE 3317
Prof. David R. Jackson
Spring 2013
Notes 20
Rectangular Waveguides
1
Rectangular Waveguide
Rectangular Waveguide
y
b
ε, µ
x
a
Cross section
 We assume that the boundary is a perfect electric conductor (PEC).
 We analyze the problem to solve for Ez or Hz (all other fields come from these).
TMz: Ez only
TEz: Hz only
2
TMz Modes
Hz
TMz =
0, E z ≠ 0
∇2E z + k 2E z =
0 (Helmholtz equation)
E z = 0 on boundary
Guided-wave assumption:
(PEC walls)
E z ( x, y, z ) = E z 0 ( x, y ) e − jkz z
 ∂2E z ∂2E z ∂2E z
+ 2
 2 +
2
∂y
∂z
 ∂x
 2
0
 + k Ez =

 ∂2E z ∂2E z
 2
2
− kz E z  + k E z =
0
 2 +
2
∂y
 ∂x

3
TMz Modes (cont.)
∂2E z ∂2E z
2
2
+
+
k
−
k
0
(
z ) Ez =
2
2
∂x
∂y
Define:
We then have:
2
2
2
k=
k
−
k
c
z
∂2E z ∂2E z
2
Note that kc is an
E
0
k
+
+
=
c
z
unknown at this point.
∂x 2
∂y 2
Dividing by the exp(-j kz z) term, we have:
∂2E z0 ∂2E z0
2
+
+
k
0
c Ez0 =
2
2
∂x
∂y
We solve the above equation by using the method of separation of variables.
We assume:
E z 0 ( x, y ) = X ( x ) Y ( y )
4
TMz Modes (cont.)
∂2E z0 ∂2E z0
2
+
+
k
0
c Ez0 =
2
2
∂x
∂y
where
E z 0 ( x, y ) = X ( x ) Y ( y )
Hence,
X ′′Y + XY ′′ =
−kc2 XY
Divide by XY :
X ′′ Y ′′
+
=
−kc2
X
Y
Hence
This has the form
X ′′
Y ′′
2
=
− kc −
X
Y
F ( x) = G ( y)
Both sides of the
equation must be a
constant!
5
TMz Modes (cont.)
X ′′
Y ′′
2
=
− kc −
=
constant
X
Y
Denote
X ′′
=
−k x2 =
constant
X
General solution:
=
X ( x)
Boundary conditions:
(1)
(2)
A sin(k x x) + B cos(k x x)
X (0) = 0 (1)
X ( a ) = 0 (2)
B=
0 ⇒ X ( x) =
A sin(k x x)
sin(k x a ) = 0
6
TMz Modes (cont.)
The second boundary condition results in
sin(k x a ) = 0
This gives us the following result:
k x a m=
π , m 1, 2
=
kx =
Hence
mπ
a
 mπ x 
X ( x) = A sin 

 a 
Now we turn our attention to the Y (y) function.
7
TMz Modes (cont.)
We have
X ′′
Y ′′
2
=
− kc −
=
−k x2
X
Y
Hence
Denote
Y ′′
= k x2 − kc2
Y
2
2
2
k=
k
−
k
y
c
x
Then we have
Y ′′
= −k y2
Y
General solution:
=
Y ( y)
C sin(k y y ) + D cos(k y y )
8
TMz Modes (cont.)
=
Y ( y ) C sin(k y y ) + D cos(k y y )
Boundary conditions:
(3)
(4)
Y (0) = 0 (3)
Y (b) = 0 (4)
D=
0 ⇒ Y ( y) =
C sin(k y y )
sin(k y b) = 0
Equation (4) gives us the following result:
k y b n=
=
π , n 1, 2
ky =
nπ
b
9
TMz Modes (cont.)
Hence
 nπ y 
Y ( y ) = C sin 

b


Therefore, we have
 mπ x   nπ y 
E z 0 ( x, y ) X=
=
( x ) Y ( y ) AC sin 
 sin 

 a   b 
New notation:
 mπ x   nπ y 
E z 0 ( x, y ) = Amn sin 
 sin 

a
b

 

The TMz field inside the waveguide thus has the following form:
 mπ x   nπ y  − jkz( m ,n) z
E z ( x, y, z ) = Amn sin 
 sin 
e
 a   b 
10
TMz Modes (cont.)
Recall that
Hence
2
2
2
k=
k
−
k
y
c
x
2
2
2
k=
k
+
k
c
x
y
Therefore the solution for kc is given by
 mπ   nπ 
kc2 
=
 +

a
b

 

2
Next, recall that
Hence
2
2
2
2
k=
k
−
k
c
z
2
2
2
k=
k
−
k
z
c
11
TMz Modes (cont.)
Summary of TMz Solution for (m,n) Mode (Hz = 0)
 mπ x   nπ y  − jkz( m ,n) z
E z ( x, y, z ) = Amn sin 
 sin 
e
 a   b 
k
( m,n )
z
 mπ   nπ 
=k 2 − 
 −

 a   b 
2
2
m = 1, 2,
n = 1, 2,
Note: If either m or n is zero, the entire field is zero.
12
TMz Modes (cont.)
Cutoff frequency
We start with
m,n )
k z( =
where =
kc( )
m,n
Set
k z( m ,n ) = 0
(
k 2 − kc( m ,n )
)
 mπ   nπ 

 +

a
b

 

2
Note: The number kc is the
value of k for which the
wavenumber kz is zero.
2
2
k = kc( m ,n )
ω µε = kc( m,n )
2π f µε = kc( m ,n )
13
TMz Modes (cont.)
Hence
2π f µε = kc( m ,n )
 mπ   nπ 

 +

 a   b 
2
2π=
f µε
which gives us
2
The cutoff frequency fc of the TMm,n mode is then
f c m ,n
=
TM
 mπ   nπ 

 +

 a   b 
2
1
2π µε
2
This may be written as
cd
f
=
2π
TM m ,n
c
 mπ   nπ 

 +

 a   b 
2
2
cd =
1
µε
14
TEz Modes
Ez
TEz=
0, H z ≠ 0
We now start with
∂2H z ∂2H z
2
2
k
k
0
+
+
−
(
z ) Hz =
2
2
∂x
∂y
Using the separation of variables method again, we have
H z 0 ( x, y ) = X ( x ) Y ( y )
where
=
X ( x)
A sin(k x x) + B cos(k x x)
=
Y ( y ) C sin(k y y ) + D cos(k y y )
and
2
2
2
k=
k
+
k
c
x
y
2
2
2
k=
k
−
k
z
c
15
TEz Modes (cont.)
y
Boundary conditions:
E x ( x, 0) = 0
E y (0, y ) = 0
E x ( x, b ) = 0
E y ( a, y ) = 0
b
ε, µ
x
a
The result is
cross section
 mπ x 
 nπ y 
H z 0 ( x, y ) = Amn cos 
 cos 

 a 
 b 
This can be shown by using the following equations:
 − jωµ  ∂H z  jk z  ∂Ez
=
− 2
Ex  2
2 
2 
−
∂
−
k
k
y
k
k


z 
z  ∂x
∂H z
0,=
=
y 0, b
∂y
 jωµ  ∂H z  jk z  ∂Ez
=
− 2
Ey  2
2 
2 
−
∂
k
k
x

 k − k z  ∂y
z 
∂H z
0,=
=
x 0, a
∂x
16
TEz Modes (cont.)
Summary of TEz Solution for (m,n) Mode (Ez = 0)
 mπ x 
 nπ y  − jk z( m ,n) z
H z ( x, y, z ) = Amn cos 
 cos 
e
 a 
 b 
k z( m ,n )
 mπ   nπ 
2
=k − 
 −

a

  b 
2
m = 0,1, 2
n = 0,1, 2
2
Same formula for cutoff
frequency as the TEz case!
( m, n ) ≠ ( 0, 0 )
Note: The (0,0) TEz mode is not valid, since it violates the magnetic Gauss law:
H ( x, y, z ) = zˆ A00 e − jkz
∇ ⋅ H ( x, y , z ) ≠ 0
17
Summary
TMz
 mπ x   nπ y  − jkz( m ,n) z
E z ( x, y, z ) = Amn sin 
 sin 
e
 a   b 
TEz
 mπ x 
 nπ y  − jk z( m ,n) z
H z ( x, y, z ) = Amn cos 
 cos 
e
 a 
 b 
k z( m ,n )
 mπ   nπ 
=k 2 − 
 −

a
b

 

cd
f
=
2π
( m,n )
c
TMz
2
 mπ   nπ 

 +

a

  b 
m = 1, 2,
n = 1, 2,
2
TEz
2
Same formula for both cases
2
Same formula for both cases
m = 0,1, 2,
n = 0,1, 2,
( m, n ) ≠ ( 0, 0 )
18
Wavenumber
General formula for the wavenumber
 mπ   nπ 

 +

a

  b 
2
TMz or TEz mode: =
kz
k 2 − kc2
with
kc
=
2
Note: The (m,n) notation is suppressed here.
Above cutoff:
k z k 1 − ( kc / k )
=
2
= k 1 − ( ωc / ω )
= k 1 − ( fc / f )
Hence
Recall:
k = ω µε
2
kc = ωc µε
2
=
kz k 1 − ( fc / f )
cd
fc
=
2π
2
 mπ   nπ 

 +

 a   b 
2
2
19
Wavenumber (cont.)
Below cutoff:
k z =k 2 − kc2 =
− j kc2 − k 2
Hence
kz =
− j kc2 − k 2
=
− jkc 1 − ( k / kc )
2
=
− jkc 1 − (ω / ωc )
=
− jkc 1 − ( f / f c )
Hence
kz =
− jkc 1 − ( f /
)
2
2
cd
f
=
fc
c
2π
2
 mπ   nπ 

 +

 a   b 
2
2
20
Wavenumber (cont.)
Recal that
k z= β − jα
Hence we have
β=
k 1 − ( fc / f ) ,
f > fc
α=
kc 1 − ( f / f c ) ,
f < fc
2
2
cd
fc
=
2π
 mπ   nπ 

 +

a

  b 
2
2
21
Wavenumber Plot
General behavior of the wavenumber
β
kc
α
“Light line”
β =k
f
fc
β=
k 1 − ( fc / f ) ,
2
α=
kc 1 − ( f / f c ) ,
2
f > fc
f < fc
cd
fc
=
2π
 mπ   nπ 

 +

 a   b 
2
2
22
Dominant Mode
The "dominant" mode is the one with the lowest cutoff frequency.
Assume b < a
y
cd
fc
=
2π
 mπ   nπ 

 +

 a   b 
2
2
b
ε, µ
x
a
Cross section
Lowest TMz mode: TM11
Lowest TEz mode: TE10
TEz
TMz
The dominant mode is the TE10 mode.
m = 1, 2,
n = 1, 2,
m = 0,1, 2,
n = 0,1, 2,
( m, n ) ≠ ( 0, 0 )
23
Dominant Mode (cont.)
Formulas for the dominant TE10 mode
 π x  − jkz z
H z ( x, y, z ) = A10 cos 
e
 a 
=
kz
π 
2
k2 − 
a
fc =
At the cutoff frequency:
f = fc
cd
2a
β=
k 1 − ( fc / f ) ,
2
α=
kc 1 − ( f / f c ) ,
2
=
λ c=
cd / f c
d / f
f > fc
λ = 2a
f < fc
24
Dominant Mode (cont.)
What is the mode with the next highest cutoff frequency?
fc
=
cd
2π
 mπ   nπ 

 +

 a   b 
2
cd  2π 
cd  2 
=
=
fc


 
2π  a 
2 a
2
( 2,0 )
2
y
cd  π 
cd  1 
=
=
fc
 
 
2π  b 
2 b
2
( 0,1)
Assume b < a / 2
The next highest is the TE20 mode.
(1,0 )
( 2,0 )
fc
ε, µ
b
x
= 2 fc
a
Cross section
useful operating region
TE10
TE20
TE01
fc
25
Dominant Mode (cont.)
Fields of the dominant TE10 mode
 π x  − jkz z
H z ( x, y, z ) = A10 cos 
e
 a 
Find the other fields from these equations:
 − jωµ  ∂H z  jk z  ∂E z
=
− 2
Ex  2
2 
2 
 k − k z  ∂y  k − k z  ∂x
 jωµ  ∂H z  jk z  ∂E z
=
Ey  2
− 2
2 
2 
k
−
k
∂
x
k
−
k


z 
z  ∂y
 jωε  ∂E z  jk z  ∂H z
=
− 2
Hx  2
2 
2 
−
∂
−
k
k
y
k
k


z 
z  ∂x
 − jωε  ∂E z  jk z  ∂H z
=
− 2
Hy  2
2 
2 
−
∂
−
k
k
x
k
k


z 
z  ∂y
26
Dominant Mode (cont.)
From these, we find the other fields to be:
 π x  − jkz z
E ( x, y, z ) = yˆ E10 sin 
e
 a 
 kz 
 π x  − jkz z
H ( x, y, z ) = − xˆ 
e
 E10 sin 
 a 
 ωµ 
 jωµ   π 
=
−  A10
E10  2
where
2 
 k − kz   a 
y
E
H
b
a
x
27
Dominant Mode (cont.)
Wave impedance:
 ωµ 
= −

Hx
k
 z 

ωµ
= −
 k 1− f / f 2
( c )

Ey

η
= −
 1− f / f 2
( c )









η=
Define the wave impedance:
Z TE

E y ωµ
1
≡ − = = η
 1− f / f 2
H x kz
( c )





η0
εr
Note: This is the
same formula as for a
TEz plane wave!
28
Example
Standard X-band* waveguide (air-filled):
a = 0.900 inches (2.286 cm)
b = 0.400 inches (1.016 cm)
Note: b < a / 2.
Find the single-mode operating frequency region.
Use
(1,0 )
fc
c
=
2a
Hence, we have
f c(1,0) = 6.56 [GHz]
f c( 2,0) = 13.11 [GHz]
* X-band: from 8.0 to 12 GHz.
29
Example (cont.)
Standard X-band* waveguide (air-filled):
a = 0.900 inches (2.286 cm)
b = 0.400 inches (1.016 cm)
 Find the phase constant of the TE10 mode at 9.00 GHz.
 Find the attenuation in dB/m at 5.00 GHz
Recall:
β=
k 1 − ( fc / f ) ,
2
α=
kc 1 − ( f / f c ) ,
2
f c(1,0) = 6.56 [GHz]
f > fc
f < fc
=
k ω=
µ0ε 0 2=
π f / c 2π / λ0
At 9.00 [GHz]: k =188.62 [rad/m]
=
kc π=
/ a 137.43 [rad/m]
At 9.00 GHz: β = 129.13 [rad/m]
At 5.00 GHz: α = 88.91 [nepers/m]
30
Example (cont.)
At 5.00 [GHz]: α = 88.91 [nepers/m]
 π x  −α z
E y ( x, y, z ) = A10 sin 
e
 a 
Therefore,
dB/m = −20 log10  e −α (1) 
dB/m = 772
A very rapid attenuation!
Note: We could have also used
=
dB/m 8.68589
=
[nepers/m] 8.68589 α
31
Guide Wavelength
The guide wavelength λg is the distance z that it takes for the wave to repeat itself.
λg =
2π
β
(This assumes that we are above the cutoff frequency.)
From this we have
2π
2π
=
λg =
2
2
 2π 
k 1 − ( fc / f )
−
f
f
1
/
( c )


 λ 
Hence we have the result
λg =
λ
1 − ( fc / f )
2
λ0
λ=
εr
32
Phase and Group Velocity
Recall that the phase velocity is given by
ω
vp =
β
Hence
vp
ω
ω
=
=
2
2
ω µε 1 − ( f c / f )
k 1 − ( fc / f )
We then have
vp =
µε 1 − ( f c / f )
2
cd
1 − ( fc / f )
For a hollow waveguide (cd = c):
Hence:
1
vp =
vp > c !
2
c
1 − ( fc / f )
2
(This does not violate relativity.)
33
Phase and Group Velocity (cont.)
The group velocity is the velocity at which a pulse travels on a structure.
The group velocity is given by
1
dω
vg =
=
dβ dβ
dω
Vi ( t )
(The derivation of this is omitted.)
A pulse consists of a "group" of frequencies
(according to the Fourier transform).
t
vg
Vi (t)
+-
waveguiding system
34
Phase and Group Velocity (cont.)
If the phase velocity is a function of frequency, the pulse will be distorting as
it travels down the system.
vg
Vi (t)
+-
waveguiding system
A pulse will get distorted in a rectangular waveguide!
vp ( f ) =
cd
1 − ( fc / f )
2
35
Phase and Group Velocity (cont.)
To calculate the group velocity for a waveguide, we use
β=
k 2 − kc2 =
ω 2 µε − kc2
Hence we have
µε
dβ 1 2
ωµε
ωµε
2 −1/2
2
=
ω
µε
−
k
ωµε
=
=
=
(
)
(
c )
2
dω 2
ω 2 µε − kc2 ω µε 1 − ( kc / k ) 2
1 − ( kc / k )
We then have the following final result:
For a hollow waveguide:
=
vg cd 1 − ( f c / f )
2
vg < c
36
Phase and Group Velocity (cont.)
For a lossless transmission line or a lossless plane wave (TEMz waves):
γ =α + j β = ( R + jω L )( G + jωC )
β= k= ω µε
= jω LC
= jω µε
= jk
We then have
ω
ω
v=
=
=
p
β ω µε
=
vg
1
=
dβ
dω
1
= cd
µε
(a constant)
1
= cd
µε
Hence we have
v=
v=
cd
p
g
For a lossless transmission
line or a lossless plane wave
line there is no distortion,
since the phase velocity is
constant with frequency.
37
Plane Wave Interpretation of Dominant Mode
Consider the electric field of the dominant TE10 mode:
 π x  − jkz z
E ( x, y, z ) = yˆ E10 sin 
e
 a 
 e + jk x x − e − jk x x  − jkz z
= yˆ E10 
e
2j


where
kx = π / a
Separating the terms, we have:
#1
#2
 E10  + jkx x − jkz z
 − E10  − jk x x − jk z z
=
+ yˆ 
e
e
E ( x, y, z ) yˆ 
e
e
 2j 
 2j 
This form is the sum of two plane waves.
=
k1 zˆ k z − xˆ k x
=
k2 zˆ k z + xˆ k x
38
Plane Wave Interpretation (cont.)
x
Picture (top view):
k2
a
θ
TE10 mode
k1
z
θ
tan=
kx
=
kz
π /a
 fc 
k 1−  
 f 
2
=
k1 zˆ k z − xˆ k x
=
k2 zˆ k z + xˆ k x
At the cutoff frequency, the angle θ is 90o. At high frequencies (well above
cutoff) the angle θ approaches zero.
39
Plane Wave Interpretation (cont.)
Picture of two plane waves
λ
cd
Crests of waves
x
k1
λg
a
z
vp
k2
Waves cancel out
cd
The two plane waves add to give an electric field that is zero
on the side walls of the waveguide (x = 0 and x = a).
40
Waveguide Modes in
Transmission Lines
 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 limits the high-frequency range of operation
for the transmission line.
41
Waveguide Modes in
Transmission Lines (cont.)
Consider an ideal parallel-plate transmission line (neglect fringing)
(This is an approximate model of a microstrip line.)
y
H
PEC
Assume w > h
E
PMC
h
x
w
Rectangular “slice of plane wave” (the region inside the waveguide)
A “perfect magnetic conductor” (PMC) is assumed on the side walls.
42
Waveguide Modes in
Transmission Lines (cont.)
Some comments about PMC
A PMC is dual to a PEC:
PEC
PMC
Et = 0
Ht = 0
E = nˆ E n
H = nˆ H n
E
PEC
H
PMC
43
Waveguide Modes in
Transmission Lines (cont.)
From a separation of variables solution, we have the following results for the
TMz and TEz waveguide modes:
y
Assume w > h
PEC
PMC
h
x
w
TMz
 mπ x   nπ y  − jk z( m ,n) z
E z ( x, y, z ) = Amn cos 
 sin 
e
 w   h 
TEz
 mπ x 
 nπ y  − jk z( m ,n) z
H z ( x, y, z ) = Amn sin 
 cos 
e
 w 
 h 
( m,n )
kz
 mπ   nπ 
=k − 
 −

w

  h 
2
2
2
44
Waveguide Modes in
Transmission Lines (cont.)
The dominant mode is TE10
 π x  − jkz(1,0) z
H z ( x, y, z ) = A10 sin 
e
 w 
(1,0 )
k z=
π 
k2 − 
 w
2
y
PEC
PMC
h
w
x
The cutoff frequency of the TE10 mode is
cd  π 
cd
=
fc =
 
2π  w 
2w
2
45
Waveguide Modes in
Transmission Lines (cont.)
y
Example
w
h
x
Microstrip line
Assume:
fc =
ε r = 2.2 (Rogers 5880 Duroid)
cd
2w
=
w 2=
h 3.15[mm]
3 ×108 / 2.2
fc ≈
2 ( 3.15 ×10−3 )
=
Z 0 65.4 [Ω]
f c ≈ 3.2 ×1010 [Hz]
h = 1.575 [mm] (62 [mils])
fc ≈ 32 [GHz]
46
Waveguide Modes in
Transmission Lines (cont.)
Dominant waveguide mode in coax (derivation omitted)
TE11 mode:
 1  1 
fc ≈
 

a εr  π   1 + b / a 
c
Example: RG 142 coax
=
a 0.035 inches
= 8.89 × 10−4 [m]
=
b 0.116 inches
= 29.46 × 10−4 [m]
ε r = 2.2
b / a = 3.31 ⇒ Z 0 = 48.4[Ω]
f c ≈ 16.8 [GHz]
εr
a
b
47

デaⅥ 。。し

Introduction to the Smith Chart

A NOVEL MULTI-FOLD HAIRPIN LINE MICROSTRIP BANDPASS

Document

Machine Translation with Source-Predicted Target