Lecture 25
Guided Waves in Parallel Plate Metal Waveguides
In this lecture you will learn:
• Parallel plate metal waveguides
•TE and TM guided modes in waveguides
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Parallel Plate Metal Waveguides
W
d
z
• Consider a parallel plate waveguide (shown above)
• We have studied such structures in the context of transmission lines
• We know that they can guide TEM waves (Transverse Electric and Magnetic)
in which both the electric and magnetic fields point in direction perpendicular
to the propagation direction
• But these structures can guide more than just the TEM waves that we have
considered so far ………….
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
1
Basic Wave Equations
Consider a parallel plate waveguide:
x
ε
d
µo
z
The electric field of any guided wave will satisfy the complex wave equations:
r r
r r
∇ 2 E (r ) = −ω 2 µo ε E (r )
r r
r r
∇ 2 H (r ) = −ω 2 µo ε H (r )
r r
r r
∇ × E (r ) = − j ω µo H (r )
r r
r r
∇ × H (r ) = j ω ε E (r )
We look for solutions of the equation,
r r
r r
∇ 2 E (r ) = −ω 2 µo ε E (r )
where the z-dependence is that of a wave going in the z-direction, and where the
E-field is pointing in the y-direction:
r r
E (r ) = yˆ F ( x ) e − j k z z
Some unknown function of “x”
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
TE Guided Modes - I
x
ε
d
µo
z
r r
The assumed solution form: E (r ) = yˆ F ( x ) e − j k z z
represents a TE guided wave (Transverse Electric) since the direction of E-field is
transverse to the direction of wave propagation
Plugging the assumed solution into the equation gives:
r r
r r
∇ 2 E (r ) = −ω 2 µo ε E (r )
⇒
r r
⎛ ∂2
∂ 2 ⎞⎟ r r
2
⎜
( )
( )
⎜ ∂z 2 + ∂x 2 ⎟ E r = −ω µo ε E r
⎠
⎝
⇒
−
∂ 2F ( x )
∂x 2
(
)
= ω 2 µo ε − k z2 F ( x )
Perfect metal boundary conditions ⇒ F ( x = 0 ) = F ( x = d ) = 0
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
2
Need to solve: −
∂ 2F ( x )
∂x 2
TE Guided Modes - II
(
)
= ω 2 µo ε − k z2 F ( x )
With boundary conditions ⇒
F ( x = 0) = F ( x = d ) = 0
Solution is: F ( x ) = Eo sin(k x x )
Automatically satisfies the
boundary condition: F ( x = 0 ) = 0
But the value of kx cannot be arbitrary – boundary condition at x = d dictates that:
kx =
mπ
d
where : m = 1, 2, 3, KK
⎛ mπ ⎞
x⎟
⎠
⎝ d
Solution becomes: F ( x ) = Eo sin⎜
⎛ mπ ⎞ − j k z z
x⎟ e
⎠
⎝ d
r r
And: E (r ) = yˆ Eo sin⎜
{ m = 1, 2, 3, KK
x
d
Ey
ε
m=2
m=1
Ey
µo
z
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
TE Guided Modes - III
r r
⎛ mπ ⎞ − j k z z
E (r ) = yˆ Eo sin⎜
x⎟ e
⎠
⎝ d
{ m = 1, 2, 3, KK
x
E-field: m=1 mode
d
E
z
π
kz
x
E-field: m=2 mode
d
E
z
π
kz
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
3
TE Guided Modes – Dispersion Relation
x
d
Ey
ε
m=2
m=1
Ey
µo
z
r r
⎛ m π ⎞ − j kz z
E (r ) = yˆ Eo sin⎜
x⎟ e
⎠
⎝ d
{ m = 1, 2, 3, KK
Different “m” values correspond to different TE modes – labeled as TEm modes
The equation: −
∂ 2F ( x )
∂x 2
(
)
= ω 2 µo ε − k z2 F ( x )
implies:
k z2 + k x2 = ω 2 µo ε
2
⇒
⎛m π ⎞
2
k z2 + ⎜
⎟ = ω µo ε
⎝ d ⎠
⇒
⎛m π ⎞
k z = ω 2 µo ε − ⎜
⎟
⎝ d ⎠
2
Dispersion relation for
TEm guided mode
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
TE Guided Modes –xCut-off Frequency
d
Ey
ε
m=2
m=1
Ey
⎛m π ⎞
k z = ω 2 µo ε − ⎜
⎟
⎝ d ⎠
2
z
Dispersion relation for
TEm guided mode
For the TEm mode, if the frequency ω is
less than:
kz
plane wave dispersion
relation: k z = ω µo ε
1 ⎛m π ⎞
⎜
⎟
µo ε ⎝ d ⎠
TE1 mode
dispersion relation
Then kz becomes entirely imaginary
and the mode does not propagate (but
decays exponentially with distance)
TE2 mode
dispersion relation
⇒ Cut-off frequency for TEm mode:
ωm =
⎛m π ⎞
⎜
⎟
µo ε ⎝ d ⎠
1
µo
ω
ω1
ω2
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
4
TE Guided Modesx– Magnetic Field
d
Ey
ε
m=2
m=1
µo
Ey
z
r r
⎛ m π ⎞ − j kz z
E (r ) = yˆ Eo sin⎜
x⎟ e
⎠
⎝ d
{ m = 1, 2, 3, KK
r r
r r
∇ × E (r ) = − j ω µo H (r )
Magnetic field is given by the equation:
r r
jE ⎡ m π
⎛ m π ⎞⎤ − j k z z
⎛ mπ ⎞ ˆ
H (r ) = o ⎢ zˆ
cos⎜
x ⎟ + x j k z sin⎜
x ⎟⎥ e
ω µo ⎣ d
⎠⎦
⎝ d
⎠
⎝ d
Note that the perfect metal boundary condition for the magnetic field is
automatically satisfied i.e:
r
r
H x (r ) x = 0 = H x (r ) x = d = 0
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
TE Guided Modes – Field Profiles
The E-field and H-field lines for the TE1 mode are shown below:
x
d
H
H
E
H
E
z
π
kz
r r
⎛ m π ⎞ − j kz z
E (r ) = yˆ Eo sin⎜
x⎟ e
⎠
⎝ d
{ m = 1, 2, 3, KK
r r
jE ⎡ m π
⎛ m π ⎞⎤ − j k z z
⎛ mπ ⎞ ˆ
H (r ) = o ⎢ zˆ
cos⎜
x ⎟ + x j k z sin⎜
x ⎟⎥ e
ω µo ⎣ d
⎠⎦
⎝ d
⎠
⎝ d
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
5
TE Guided Modes – Another Perspective - I
Consider TE-wave reflection off a perfect metal:
ε
x
Ei
r
ki
Hi
µo
r
kr
Er
E
z
Hr
r
k i = − k x xˆ + k z zˆ
r
k r = k x xˆ + k z zˆ
k z2 + k x2 = ω 2 µo ε
r r
E (r )
r r
E (r )
⇒
x >0
= yˆ E i e − j (− k x
x + kz z )
+ yˆ ΓE i e − j (k x
[
= yˆ E i e − j ( − k x x + k z z ) − e − j (k x
r r
E (r )
= yˆ 2 j E i sin(k x x ) e − j k z z
x >0
x +kz z )
x +kz z )
Γ = −1
]
x >0
Notice the “sine” variation of the
y-component of the E-field
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
TE Guided Modes – Another Perspective - II
If another top metal plate is placed at one of the nodes of the “sine” function
then this additional metal plate will not disturb the field
r r
E (r )
x >0
= yˆ 2 j Ei sin(k x x ) e − j k z z
x
Ei
Hi
r
ki
r
kr
Er
Hr
r
k i = − k x xˆ + k z zˆ
Ei
Hi
ε
r
ki
Ey
µo
z
r
k r = k x xˆ + k z zˆ
This is exactly what guided TE modes are – TE-waves bouncing back and fourth
between two metal plates and propagating in the z-direction !
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
6
TM Guided Modes - I
Consider TM-wave reflection off a perfect metal:
x
Ei
Hi
ε
r
ki
µo
r
kr
Er
Hy
Hr
z
r
k r = k x xˆ + k z zˆ
r
k i = − k x xˆ + k z zˆ
k z2 + k x2 = ω 2 µo ε
r r
H (r )
r r
H (r )
⇒
x >0
= yˆ Hi e − j (− k x
x >0
= yˆ Hi e − j (− k x
[
r r
H (r )
x >0
x + kz z )
+ yˆ ΓTM Hi e − j (k x
x + kz z )
+ e − j (k x
x +kz z )
x +kz z )
ΓTM = +1
]
= yˆ 2 Hi cos(k x x ) e − j k z z
Notice the “cosine” variation of
the y-component of the H-field
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
TM Guided Modes - II
If another top metal plate is placed at the maximum points of the “cosine”
function then this additional metal plate will not disturb the field
r r
H (r )
x >0
= yˆ 2 Hi cos(k x x ) e − j k z z
x
Ei
Hi
r
ki
r
k i = − k x xˆ + k z zˆ
Er
Ei
r
kr
Hr
Hi
ε
r
ki
Hy
µo
z
r
k r = k x xˆ + k z zˆ
This is exactly what guided TM modes are – TM-waves bouncing back and fourth
between two metal plates and propagating in the z-direction !
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
7
TM Guided Modes –xBasic Equations - I
ε
d
µo
z
r r
r r
Need to solve the equation: ∇ 2 H (r ) = −ω 2 µo ε H (r )
r r
H (r ) = yˆ G ( x ) e − j k z z
Assume the solution form:
It represents a TM guided wave (Transverse Magnetic) since the direction of
H-field is transverse to the direction of wave propagation
Plugging the assumed solution into the equation gives:
r r
r r
∇ 2 H (r ) = −ω 2 µo ε H (r )
⇒
r r
⎛ ∂2
∂ 2 ⎞⎟ r r
2
⎜
( )
( )
+
⎜ ∂z 2 ∂x 2 ⎟ H r = −ω µo ε H r
⎠
⎝
⇒
−
∂ 2G ( x )
∂x 2
(
)
= ω 2 µo ε − k z2 G ( x )
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
TM Guided Modes – xBasic Equations - II
d
m=1
Hy
Need to solve: −
∂ 2G ( x )
∂x 2
π
d
µo
z
(
)
= ω 2 µo ε − k z2 G ( x )
Motivation for this is obtained from the
TM-wave reflection analysis discussed
earlier
Solution is: G ( x ) = Ho cos(k x x )
kx = m
ε
Hy
m=2
where : m = 0, 1, 2, 3, KK
⎛ mπ ⎞
x⎟
⎝ d
⎠
Solution becomes: G ( x ) = Ho cos⎜
r r
⎛ m π ⎞ − j kz z
x⎟ e
⎝ d
⎠
And: H (r ) = yˆ Ho cos⎜
{ m = 0, 1, 2, 3, KK
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
8
TM Guided Modes – Electric Field
x
d
m=1
Hy
Hy
m=2
ε
µo
z
r r
⎛ m π ⎞ − j kz z
H (r ) = yˆ Ho cos⎜
x⎟ e
⎠
⎝ d
{ m = 0, 1, 2, 3, KK
r r
r r
Electric field is given by the equation: ∇ × H (r ) = j ω ε E (r )
r r
jH ⎡
mπ
⎛ mπ ⎞ ˆ
⎛ m π ⎞⎤ − j kz z
E (r ) = − o ⎢ − zˆ
sin⎜
x ⎟ + x j k z cos⎜
x ⎟⎥ e
ωε ⎣
d
⎝ d
⎠
⎝ d
⎠⎦
Note that the perfect metal boundary condition for the electric field is
automatically satisfied, i.e.:
r
r
E z (r ) x = 0 = E z (r ) x = d = 0
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
TM Guided Modes – Dispersion Relation
x
d
m=1
Hy
ε
Hy
m=2
µo
z
r r
⎛ m π ⎞ − j kz z
H (r ) = yˆ Ho cos⎜
x⎟ e
⎝ d
⎠
{ m = 0, 1, 2, 3, KK
Different “m” values correspond to different TM modes – labeled as TMm modes
The equation: −
∂ 2G ( x )
∂x 2
(
)
= ω 2 µo ε − k z2 G ( x )
implies:
k z2 + k x2 = ω 2 µo ε
2
⇒
⎛m π ⎞
2
k z2 + ⎜
⎟ = ω µo ε
⎝ d ⎠
⇒
⎛m π ⎞
k z = ω 2 µo ε − ⎜
⎟
⎝ d ⎠
2
Dispersion relation for
TMm waveguide mode
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
9
TM Guided Modes –x Cut-off Frequency
Hy
m=1
d
Hy
m=2
⎛m π ⎞
k z = ω 2 µo ε − ⎜
⎟
⎝ d ⎠
ε
µo
z
2
Dispersion relation for
TMm guided mode
kz
For the TMm mode, if the frequency ω is
less than:
1 ⎛m π ⎞
⎜
⎟
µo ε ⎝ d ⎠
kz = ω
TM0 mode
dispersion relation
µo ε
TM1 mode
dispersion relation
Then kz becomes entirely imaginary
and the mode does not propagate (but
decays exponentially with distance)
TM2 mode
dispersion relation
⇒ Cut-off frequency for TMm mode:
ωm =
ω
⎛m π ⎞
⎜
⎟
µo ε ⎝ d ⎠
1
ω0
ω1
ω2
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
TM Guided Modes – Field Profiles
The E-field and H-field lines for the TM1 mode are shown below:
x
d
E
H
z
π
kz
r r
⎛ m π ⎞ − j kz z
H (r ) = yˆ Ho cos⎜
x⎟ e
⎠
⎝ d
{ m = 0, 1, 2, 3, KK
r r
jH ⎡
mπ
⎛ mπ ⎞ ˆ
⎛ m π ⎞⎤ − j kz z
E (r ) = − o ⎢ − zˆ
sin⎜
x ⎟ + x j k z cos⎜
x ⎟⎥ e
ωε ⎣
d
⎝ d
⎠
⎝ d
⎠⎦
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
10
TM0 Guided Mode – Field Profiles
The E-field and H-field for the TM0 mode are:
r r
H (r ) = yˆ Ho e − j k z z
Note that fields are not a function of “x”
r r
k
E (r ) = xˆ z Ho e − j k z z
ωε
The E-field and H-field lines for the TM0 mode are shown below:
x
E
d
z
H
π
kz
The TM0 mode is just the TEM mode that we worked with when dealing with
transmission lines !
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
11