Electromagnetic Fields
Normal Incidence on an Interface
Consider a planar interface between two unbounded media, and a
uniform plane wave with normal incidence on the interface.
x
Medium 1
Medium 2
Interface
{x,y}-plane
ε1 = εr1 εo
µ1 = µr1 µo
σ1
ε2 = εr2 εo
µ2 = µr2 µo
σ2
Incident wave
Transmitted wave
Reflected wave
y
0
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z
67
Electromagnetic Fields
Because of the medium discontinuity, the
experiences a partial reflection at the interface.
incident
wave
In medium 2, only a forward transmitted wave exists
The total fields at the interface must satisfy the boundary
conditions for electromagnetic fields. Without loss of generality,
we assume the following orientation for the electromagnetic fields
of the waves
Ex
Propagation
z
Hy
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68
Electromagnetic Fields
Recalling the solution for Helmholtz equation, the phasor fields in
the medium 1 can be written as
E1 ( z) =
E1+ ( z) exp(−γ1 z) + E1− exp( γ1 z)
H1 ( z) =
H1+ ( z) exp(−γ1 z) + H1− exp( γ1 z)
(
1
E1+ ( z) exp(−γ1 z) − E1− exp( γ1 z)
=
η1
Total
Field
γ1 =
Incident wave
jωµ1 (σ1 + jωε1 )
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)
Reflected wave
jωµ1
η1 =
σ1 + jωε1
69
Electromagnetic Fields
The forward transmitted wave in medium 2 is given by
E2 x ( z) = E+2 ( z) exp(−γ 2 z)
H2 y ( z) = H+2 ( z) exp(−γ 2 z)
1 +
=
E2 ( z) exp(−γ 2 z)
η2
Total
Field
γ2 =
Transmitted wave
jωµ 2 (σ 2 + jωε 2 )
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jωµ 2
η2 =
σ 2 + jωε 2
70
Electromagnetic Fields
Both fields are parallel to the interface. The boundary conditions
indicate that the total fields are continuous at the interface. Note
that we are assuming a finite conductivity, therefore no surface
current exists and the tangent magnetic field is also continuous.
The interface is located at z = 0 so all exponentials are equal to 1:
⇒ E1+ + E1− = E2+
E1 x ( z = 0) = E 2 x ( z = 0)
(
)
1
1 +
+
−
H1 y ( z = 0) = H2 y ( z = 0) ⇒
E1 − E1 =
E2
η1
η2
+
Assuming that the amplitude E1 of the incident wave is known, we
−
+
have two unknowns E1 and E2 . In order to obtain a general result,
it is convenient to solve the equations above in terms of reflection
−
+
+
+
coefficient (E1 /E1 ) and transmission coefficient (E2 /E1 ).
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71
Electromagnetic Fields
Reflection Coefficient
E1−
η2 − η1
ΓE =
=
+ η +η
E1
2
1
This is similar to the voltage load reflection coefficient found for a
transmission line, if one considers the following analogy
⇔ Transmission Line
⇔ Z0 Characteristic Impedance
Medium 1
η1
Medium 2 ⇔ Load
η2 ⇔ ZR Load Impedance
For the magnetic field
ΓH =
H1−
H1+
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=
− E1− / η1
E1+ / η1
=−
E1−
E1+
= −Γ E
72
Electromagnetic Fields
Transmission Coefficient
τE =
E2+
E1+
=
E1+ + E1−
E1+
=1+
2 η2
= 1 + ΓE =
η2 + η1
E1−
E1+
For the magnetic field we have
τH =
H+2
H1+
=
H1+ + H1−
H1+
=1+
H1−
H1+
=1−
E1−
E1+
= 1 − ΓE
NOTE: The reflection and transmission coefficients for the fields are
in general complex quantities.
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73
Electromagnetic Fields
Special cases
Matched Impedances
η1 = η2
⇒ Γ E = 0 and τ E = 1
In this case we have total transmission into medium 2 and no
reflection.
Medium 2 = Perfect Conductor
σ 2 → ∞ ⇒ η2 = 0 ⇒ Γ E = −1 and τ E = 0
The wave experiences total reflection, consistent with the fact that
the fields must be zero inside the perfect conducting medium. This
case is analogous to a line with a short circuit load. The total
electric fields at the interface is
E1+ + E1− = E1+ + Γ E E1+ = E1+ − E1+ = E2+ = 0
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74
Electromagnetic Fields
Perfect Dielectric Media
µ1
σ1 = 0 ⇒ η1 =
= Real
ε1
µ2
σ 2 = 0 ⇒ η2 =
= Real
ε2
Usually,
µ1 = µ 2 = µ o
µ o / ε 2 − µ o / ε1 1 − ε 2 / ε1
⇒ ΓE =
=
= Real
µ o / ε 2 + µ o / ε1 1 + ε 2 / ε1
1 − ε 2 / ε1
2
τE = 1 + Γ = 1 +
=
= Real
1 + ε 2 / ε1 1 + ε 2 / ε1
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75
Electromagnetic Fields
Power flow
Assuming dielectric media (no-loss) for simplicity, the time-average
power associated with the incident wave and the reflected wave is
{
}
+2
1 E1
*
1
P( t ) in = Re E × H =
2
2 η1
+2
1 E1
P( t ) refl =
2 η1
ΓE
2
The power reflection coefficient is
P( t ) refl
2
ΓP = = ΓE
P( t ) in
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76
Electromagnetic Fields
The time-average power flow transmitted in medium 2 is
+2
1 E2
Also
P( t ) trans =
2 η2
+2
1 E1
P( t ) trans = P( t ) in − P( t ) refl =
2 η1
(
1 − ΓE
2
)
since power flow normal to the interface must be continuous
⇒
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+2
1 E2
2 η2
=
+2
1 E1
2 η1
(1 − Γ )
E
2
77
Electromagnetic Fields
The power transmission coefficient is
P( t ) trans
2 η1
τP = = τE
P( t ) in
η2
P( t ) in − P( t ) refl
2
=
= 1 − ΓE
P( t ) in
Note that, as a consequence of power conservation, from the
results above one gets
Γ P + τP = Γ E 2 + 1 − Γ E 2 = 1
NOTE: the reflection and transmission coefficients for the timeaverage power flow are always real.
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78
Electromagnetic Fields
Plane Waves in Arbitrary Directions
For a uniform plane wave with general orientation, the direction of
propagation is identified by the propagation vector, normal to the
phase planes
β = β x iˆx + β y iˆy + β z iˆz
Considering the position vector
r = x iˆx + y iˆy + z iˆz
we have the scalar dot product
β ⋅ r = β xiˆx + β yiˆy + β z iˆz ⋅ x iˆx + y iˆy + z iˆz =
= β x x + β y y + β zz
(
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)(
)
79
Electromagnetic Fields
The electromagnetic fields of a plane wave propagating along a
general direction are on the phase plane perpendicular to the
propagation vector and can be written as
E(r, t ) = Eo cos ( ω t − β ⋅ r + ϕ o )
= Eo cos ( ω t − β x x − β y y − β z z + ϕ o )
H (r, t ) = Ho cos (ω t − β ⋅ r + ϕ o )
= Ho cos (ω t − β x x − β y y − β z z + ϕ o )
We have assumed propagation in a dielectric by giving the same
phase to the fields. In addition
Eo
Ho =
η
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µ
η=
ε
(dielectric)
80
Electromagnetic Fields
The fields are perpendicular to each other and to the propagation
vector according to the right hand rule
E( x, y, z)
x
y
H ( x, y, z)
z
, P
The propagation vector is also parallel to the Poynting vector.
E(r, t ) × H (r, t ) = P( t ) ∝ β
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81
Electromagnetic Fields
The orthogonality of the vectors can be expressed mathematically
by the following dot products
Eo ⋅ Ho = Eo ⋅ β = Ho ⋅ β = 0
and cross products
Eo
Ho = iβ ×
η
We have
Eo = − iβ × ηHo
E
E
Eo
β
β
Ho = iβ × o = × o =
×
η β
µ / ε ω µε
µ/ε
β × Eo
Ho =
ωµ
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82
Electromagnetic Fields
β
µ β
µ Eo = − iβ × ηHo = − ×
Ho = −
Ho
×
β
ε
ε
ω µε
β × Ho
Eo = −
ωε
Since the propagation vector is related to the wavelength and the
phase velocity as
2π
β=
=
λ
ω
vp
for each direction corresponding to components of the propagation
vector we can define apparent wavelengths and apparent velocities
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83
Electromagnetic Fields
2π
λx =
βx
2π
λy =
βy
2π
λz =
βz
ω
v px =
βx
ω
v py =
βy
ω
v pz =
βz
The apparent quantities are related to the actual ones as
2
2
β
β 2z
1
1
1
1
β
βx
y
=
=
+
+
=
+
+
2
2
2
2
2
2
2
4π
4π
4π
4π
λ
λ x λ y λ 2z
2
2 β2 β2
1 β
1
1
1
βx
y
z
=
=
+
+
=
+
+
2
2
2
2
2
2
2
vp ω
v px v py v2pz
ω
ω
ω
2
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84
Electromagnetic Fields
Phase planes
r
2
r3
r1
x
x iˆx
An apparent wavelength is greater than the actual one, since it is
measured along a direction which is not normal to the parallel
phase planes.
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85
Electromagnetic Fields
An apparent velocity is greater than the actual velocity since it
seemingly connects longer distances during the same time. With
reference to the previous figure
(r 2 − r1 ) = v p t1
(r 3− r 1 ) = v px ix t1
Since
(r 2 − r1 ) <
(r 3− r 1 )
⇒
v p < v px
The apparent velocity always exceeds the phase velocity. However,
we will see later that relativity laws are not violated.
If one considers a direction parallel to the phase planes, the
apparent velocity is even infinite!
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86
Electromagnetic Fields
Phasor notation
E(r, t ) = Eo cos ( ω t − β x x − β y y − β z z + ϕ o )
jϕ o jω t − jβ x x − jβ y y − jβ z z
e
e
= Re Eo e e e
{
phasor ⇒
}
jϕ o − jβ x x − jβ y y − jβ z z
e
e
E(r) = Eo e e
H (r, t ) = Ho cos (ω t − β x x − β y y − β z z + ϕ o )
jϕ o jω t − jβ x x − jβ y y − jβ z z
e
e
= Re Hoe e e
{
phasor ⇒
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}
jϕ o − jβ x x − jβ y y − jβ z z
e
e
H(r) = Ho e e
87
Electromagnetic Fields
Incidence on Perfect Conductor
Consider first normal incidence at an interface between a dielectric
and a perfect conductor. Total reflection occurs, as in a shortcircuited transmission line.
x
Medium 1
Medium 2
Interface
{x,y}-plane
ε1 = εr1 εo
µ1 = µr1 µo
Perfect
Conductor
σ2→∞
Incident wave
E0
H0
Reflected wave
E 1.0
y
0
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z
88
Electromagnetic Fields
Because of interference between incident and reflected wave, there
is a standing wave in medium 1.
Medium 2
Medium 1
x
ε1 = εr1 εo
µ1 = µr1 µo
E
Perfect
Conductor
σ2→∞
2 Eo
2 Eo
H
E0
H0
y
2 / © Amanogawa, 2001 – Digital Maestro Series
/ 0
z
89
Electromagnetic Fields
Consider now incidence at an angle. We choose an electric field
perpendicular to the plane of incidence.
E×
Medium 1
ε1 = εr1 εo
µ1 = µr1 µo
Medium 2
H
x
x
E
E0
H0
z
H
y
0
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Perfect
Conductor
σ2→∞
z
90
Electromagnetic Fields
Only the normal component, corresponding to β z is reflected.
Medium 2
Medium 1
x
ε1 = εr1 εo
µ1 = µr1 µo
E
Perfect
Conductor
σ2→∞
2 Eo
H
2 Eo
E0
H0
y
2 / z Note: β z
/ z z / 2
0
z
< β ⇒ λz > λ
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91
Electromagnetic Fields
β=
2π
; βz =
2π
=
2π
λ
λz
λ
First maximum
λz
λ
=
zmax =
4
4 cos θ
First minimum
λz
λ
=
zmin =
2
2 cos θ
cos θ
Examples:
θ = 45 ⇒ zmax ≈ 0.35λ
θ = 15 ⇒ zmax ≈ 0.259λ
θ=0
⇒ zmax ≈ 0.25λ
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E×
H
x
Medium 2
Perfect
Conductor
σ2→∞
x
E
E0
H0
z
H
y
0
z
92
Electromagnetic Fields
If we place a second perfect conductor interface, parallel to the
previous one, the wave is guided along the x-direction by reflection.
Perfect
Conductor
σ2→∞
E0
H0
H
E
×
E
x
Perfect
Conductor
σ2→∞
E0
H0
H
y
0
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z
93
Electromagnetic Fields
Parallel Plate Waveguide
w
a
x
0
a
z
y
Assume uniform waves along the y-direction
Assume no fringing effects
w a
y
0
Propagation along the z-direction
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94
Electromagnetic Fields
Maxwell’s equations
∇ × E = − jω µ H
⇓
 iˆx iˆy iˆz 


∂
∂
∂ 

det
∂ x ∂ y ∂ z


 E x E y E z 
© Amanogawa, 2001 – Digital Maestro Series
∂
∂
E z − E y = − jωµH x (1)
∂y
∂z
∂
∂
E x − E z = − jωµH y (2)
⇒
∂z
∂x
∂
∂
E y − E x = − jω µH z (3)
∂x
∂y
95
Electromagnetic Fields
∇ × H = jω ε E
⇓
 iˆx
iˆy
iˆz 


∂
∂
∂ 

det
∂ x ∂ y ∂ z


 H x H y H z 
© Amanogawa, 2001 – Digital Maestro Series
∂
∂
H z − H y = jωεE x (4)
∂y
∂z
∂
∂
H x − H z = jωεE y (5)
⇒
∂z
∂x
∂
∂
H y − H x = jωεE z (6)
∂x
∂y
96
Electromagnetic Fields
From (1) & (2) & (5)
∂
(1)
∂z
∂2
∂
E y = jωµ H x
⇒
∂z
∂ z2
∂
∂2
∂
(3) ⇒
E y = − jω µ H z
2
∂x
∂x
∂x
∂2
∂2
∂
 ∂

Ey +
E y = jωµ  H x − H z  = −ω2µ ε E y
∂x
 ∂z

∂ z2
∂ x2
From (5)
⇓
jωε E y
Wave equation for Transverse Electric (TE) modes
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97
Electromagnetic Fields
From (4) & (6) & (2)
∂
(4)
∂z
∂2
∂
H y = − jωε E x
⇒
∂z
∂ z2
∂
∂2
∂
(6) ⇒
H y = jωε E z
2
∂x
∂x
∂x
∂2
∂2
∂
 ∂

Hy +
H y = − jωε  E x − E z  = −ω2µ ε H y
∂x 
∂z
∂ z2
∂ x2
From (2)
⇓
− jωµ H y
Wave equation for Transverse Magnetic (TM) modes
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98
Electromagnetic Fields
Transverse Electric (TE) modes
E
H
H
×
E
Ey 0
Boundary Conditions
x 0
x a
This solution satisfies the boundary conditions:
E y = Eo sin (β x x ) e
− jβ z z
(
x
© Amanogawa, 2001 – Digital Maestro Series
)
Eo − jβ x x
− jβ z
e
= j
− e jβ x x e z
2
z
99
Electromagnetic Fields
We have
β2 =
4π2
λ2
= β 2x + β 2z = ω2µ ε
and from boundary conditions at the conductor plates
x = 0) E y = 0
x = a) sin (β x a ) = 0 ⇒ β x a = m π
m = 1, 2, 3
mπ
β x = β cos θ =
a
2
2 1 / 2
  mλ 
 mπ 
β z = β sin θ = ω µ ε − 
 = ω µ ε  1 − 
 
a
2
 a 

 

2
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100
Electromagnetic Fields
For each possible index m we have a mode of propagation. Modes
are labeled TE10 , TE20 , TE30 , ….
The first index gives the periodicity (number of half sinusoidal
oscillations) between the plates, along the x-direction. The second
index is zero to indicate uniform solution along the y-direction.
Note that the solution m = 0 (or mode TE00) is not acceptable,
because it would require a field configuration with uniform electric
field tangent to the metal plates. This is an unphysical boundary
condition, which is possible only for the case of trivial solution of
zero field everywhere.
H
E
© Amanogawa, 2001 – Digital Maestro Series
TE00 m 0
x 0 & z
Unphysical !!!
101
Electromagnetic Fields
A mode can propagate only if the frequency is sufficiently high, so
that βz > 0.
We have the cut-off condition when
mπ 2 π
2a
⇒ λc =
β = βx =
=
a
m
λc
1
2
2 2

m
m
π
λ



c
⇒ β z = ω 2µ ε −
= ω µ ε 1 −
=0

 a 
  2a  
vp
m
Cut - off frequency for mode m
fc =
=
λ c 2 a µε
Exactly at cut-off the wave would bounce between the plates,
without propagation along the wave guide axis.
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102
Electromagnetic Fields
When the frequency is below the cut-off value
2a
f < fc ⇒ λ > λ c =
m
1
 mπ 
2
βz = ± ω µ ε − 

 a 
2

 2
  mλ  2 
= ±ω µ ε  1 − 
 
2a  
 


>1 
1
  mλ  2
 2
− 1
= ± j ω µε

  2a 



⇒ β z = − jα ⇒ e− j( − jα ) z = e−αz
The mode attenuates entering the guide as an evanescent wave.
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103
Electromagnetic Fields
Transverse Magnetic (TM) modes
E
H
H
E
The magnetic field can be tangent to the conductor plates. In fact, it
is maximum at the plates, since the reflection coefficient is ΓH = 1.
The solution is of the form:
H y = Ho cos (β x x ) e
− jβ z z
(
x
© Amanogawa, 2001 – Digital Maestro Series
)
Ho − jβ x x
− jβ z
e
=
+ e jβ x x e z
2
z
104
Electromagnetic Fields
At the metal plates
x = 0) H y = H o
x = a) cos (β x a ) = 1 ⇒ β x a = m π
m = 0, 1, 2, 3
Modes are labeled TM00 , TM10 , TM20 , TM30 , …
Note that the solution m = 0 (or mode TM00) is acceptable, because
the magnetic field can be uniform and tangent to the metal plates.
E
H
© Amanogawa, 2001 – Digital Maestro Series
TM00 m 0
x 0 & z
Physical !!!
105
Electromagnetic Fields
The TM00 mode is like a portion of a uniform plane wave sliding
between the plates of the waveguide.
Both the electric and the magnetic field are transverse (normal to
the guide axis) therefore this mode is usually known as Transverse
Electro Magnetic mode (TEM). For this mode we have
2π
2π
β = βz =
⇒ βx =
= 0 ⇒ λc → ∞
λ
λc
vp
= 0 Cut - off frequency for TEM mode
fc =
λc
The TEM mode is the fundamental mode. It can propagate at any
frequency.
All other TM modes have the same cut-off frequency condition as
the TE modes with identical indices.
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106
Electromagnetic Fields
The apparent wavelength along the guide axis is also called the
guide wavelength
2π
2π
λg = λz =
=
β z β sin θ
mπ 2 π 2π λ
Since : β x = β cos θ =
=
=
a λc λ λc
fc
λ
cos θ =
=
λc f
λg =
⇒
λ
2
1 − (λ / λ c )
© Amanogawa, 2001 – Digital Maestro Series
=
 λ 
sin θ = 1 −  
 λc 
λ
2
2
1 − ( fc / f )
107
Electromagnetic Fields
There is a corresponding apparent velocity along the guide axis, or
guide phase velocity
ω
ω
v pz =
=
β z β sin θ
v pz =
vp
2
1 − (λ / λ c )
=
vp
2
1 − ( fc / f )
The expressions for guide wavelength and guide velocity are also
identical for TE and TM modes.
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108
Electromagnetic Fields
Consider a TE wave with electric field amplitude Eo.
amplitude of the magnetic field is
The total
Eo
Ho =
η
The magnetic field has two components with amplitude
Eo
Eo λ
H x = Ho sin θ =
sin θ =
η
η λg
2π
λ
since : λ g =
=
β sin θ sin θ
Eo
Eo λ
H z = Ho cos θ =
cos θ =
η
η λc
λ
2a
since : λ c =
=
m cos θ
© Amanogawa, 2001 – Digital Maestro Series
109
Electromagnetic Fields
Consider a TM wave with magnetic field amplitude Ho. The total
amplitude of the electric field is
Eo = ηHo
The electric field has two components with amplitude
Ex
Ez
λ
= Eo sin θ = η H0 sin θ = η Ho
λg
2π
λ
since : λ g =
=
β sin θ sin θ
λ
= Eo cos θ = η Ho cos θ = η Ho
λc
2a
λ
since : λ c =
=
m cos θ
© Amanogawa, 2001 – Digital Maestro Series
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Electromagnetic Fields
The x−component of the magnetic field for the TE wave is
associated with the wave moving along the z−direction (axis of the
waveguide). The guide impedance for the TE modes is defined as
ηg
TE
=η
λg
λ
=η
1
2
1 − (λ / λ c )
=η
1
2
1 − ( fc / f )
The x−component of the electric field for the TM wave is associated
with the wave moving along the z−direction (axis of the waveguide).
The guide impedance for the TM modes is defined as
λ
2
2
ηg
=η
= η 1 − ( λ / λ c ) = η 1 − ( fc / f )
TM
λg
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Electromagnetic Fields
If there is a discontinuity along the guide axis (e.g., a change in
dielectric medium), one can use transmission line theory to analyze
the mode behavior individually in terms of transmission and
reflection. Sections of the guide can be replaced by a transmission
line, with the guide impedance as the characteristic impedance.
Note that the guide impedance is a function of frequency for all
modes, except for the fundamental TEM mode
fc (TEM ) = 0 ⇒ ηg
TEM
2
= η 1 − (0 / f ) = η
The reflection coefficient at a discontinuity is of the usual form
Γ=
η g2 − η g1
η g2 + η g1
The power reflection coefficient is again |Γ|2 and the power
transmission coefficient is 1−|Γ|2.
© Amanogawa, 2001 – Digital Maestro Series
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Electromagnetic Fields
The phasor fields for TE modes are summarized as follows
Electric Field: a single transverse component
− jβ z z ˆ
 mπ  − jβ z z ˆ
i y = Eo sin 
x e
iy
E = Eo sin (β x x ) e
 a 
Magnetic Field: two components, obtained from Faraday’s law:
∂
∂
ˆ
E y iˆz = − jωµ ( H x iˆx + H z iˆz )
∇ × E y = − E y ix +
∂z
∂x
βz
 mπ  − jβ z z ˆ
x e
ix
sin 
⇒ H = − Eo
ωµ  a 
βx
 mπ  − jβ z z ˆ
x e
iz
cos 
+ jEo
ωµ
 a 
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Electromagnetic Fields
The following relationships are useful to introduce the medium
impedance in the TE field expressions above
βz
ω µε sin θ
=
=
ωµ
ωµ
ε λ
λ
1
=
=
µ λ g η λ g ηg
TE
β x ω µε cos θ
=
=
ωµ
ωµ
ε λ
λ
=
µ λc η λc
Note once again that there is no allowed solution for m = 0 in the
case of TE modes. The first allowed TE mode is the TE10.
© Amanogawa, 2001 – Digital Maestro Series
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Electromagnetic Fields
The phasor fields for TM modes are summarized as follows
Magnetic Field: a single transverse component
− jβ z z ˆ
 mπ  − jβ z z ˆ
i y = H o cos 
x e
iy
H y = H o cos (β x x ) e
 a 
Electric Field: two components, obtained from Ampere’s law:
∂
∂
ˆ
∇ × H y = − H y ix + H y iˆz = jωε ( Ex iˆx + Ez iˆz )
∂z
∂x
βz
 mπ  − jβ z z ˆ
x e
ix
cos 
⇒ E = Ho
ωε
 a 
βx
 mπ  − jβ z z ˆ
x e
iz
sin 
+ jHo
ωε  a 
© Amanogawa, 2001 – Digital Maestro Series
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Electromagnetic Fields
The following relationships are useful to introduce the medium
impedance in the TM field expressions above
βz
ω µε sin θ
=
=
ωε
ωε
µ λ
λ
=η
= ηg
TM
ε λg
λg
β x ω µε cos θ
=
=
ωε
ωε
µ λ
λ
=η
λc
ε λc
The field expressions simplified for the TEM mode resemble a
uniform plane wave propagating along the axis of the guide
− jβ z
H y = Ho e z iˆy
− jβ z
− jβ z
E x = ηHo e z iˆx = Eo e z iˆx
Remember, the TM00 or TEM mode is the fundamental mode.
© Amanogawa, 2001 – Digital Maestro Series
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Electromagnetic Fields
Wave Dispersion
A plane wave by itself does not carry information. For transmission
of information it is necessary to have a frequency spectrum of finite
size, as obtained by modulation of a wave, for instance.
Information does not travel at the guide phase velocity, but it
propagates according to the group velocity
guide phase velocity
group velocity
ω
v pz =
βz
dω
vg =
dβ z
To illustrate the nature of the group velocity, consider the simple
case of an amplitude modulated signal (assume ω >> ∆ω)
E y ( t ) = Eo (1 + m cos ( ∆ω ⋅ t )) cos ( ωo t )
© Amanogawa, 2001 – Digital Maestro Series
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Electromagnetic Fields
This signal has three components
E y ( t ) = Eo cos (ωo t ) + m Eo cos (ωo t ) cos ( ∆ω ⋅ t )
=
Eo cos ( ωo t )
m
+ Eo cos ( ωo t − ∆ω ⋅ t )
2
m
+ Eo cos ( ωo t + ∆ω ⋅ t )
2
ωo−∆ω
© Amanogawa, 2001 – Digital Maestro Series
ωo
ωo+∆ω
ω
118
Electromagnetic Fields
The line at angular frequency ωo is the carrier. The modulation
information is contained in the two side frequency lines at ωo−∆ω
and ωo+∆ω.
Now, consider an amplitude modulated wave propagating in a
parallel plate wave guide. The z−components of the propagation
factor depend on frequency and are different for the two side
frequencies. In general, we have
β z1 = β zm − ∆β z
ωo+∆ω
ωo−∆ω
β z2 = β zm + ∆β z
ω
ωm
z z
z2
z1 zm
© Amanogawa, 2001 – Digital Maestro Series
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Electromagnetic Fields
The dispersion relation β (ω ) is approximately linear when ∆ω << ωo
ωo+∆ω
ωm ≈ ωo
ωo−∆ω
ω
z z
z1
z2
zm z ( o )
z
Under this assumption, we can write
E y ( z, t ) = Eo cos ( ωo t − β z z )
m
+ Eo cos ( ωo − ∆ω ) t − (β z − ∆β z ) z 
2
m
+ Eo cos ( ωo + ∆ω ) t − (β z + ∆β z ) z 
2
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Electromagnetic Fields
E y ( z, t ) = Eo cos ( ωo t − β z z )
m
+ Eo cos  (ωo t − β z z ) − ( ∆ω ⋅ t − ∆β z z )
2
m
+ Eo cos  (ωo t − β z z ) + ( ∆ω ⋅ t − ∆β z z )
2
= Eo cos (ωo t − β z z )
+ mEo cos ( ωo t − β z z ) cos ( ∆ω ⋅ t − ∆β z z )
= Eo 1 + m cos ( ∆ω ⋅ t − ∆β z z ) cos ( ωo t − β z z )
modulated amplitude
The modulation envelope travels at the group velocity
vg = ∆ω / ∆β z
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Electromagnetic Fields
vg 15.0
MODULATION ENVELOPE
z
10.0
5.0
v pz 0.0
z
-5.0
-10.0
-15.0
0.00
0.50
© Amanogawa, 2001 – Digital Maestro Series
1.50
2.00
3.00
CARRIER
4.00
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Electromagnetic Fields
For the parallel plate wave guide
2 1 / 2
1/ 2
  λ 
  f 2 
βz = ω µ ε 1 −   
= ω µ ε 1 −  c  
  f  
  λc  




vp
vp
ω
v pz =
=
=
2
2
βz
1 − (λ / λ c )
1 − ( fc / f )
dω
2
2
vg =
= v p 1 − ( λ / λ c ) = v p 1 − ( fc / f )
dβ z
⇒ v pz ⋅ vg = v2p
Since v pz ≥ v p
⇒ vg ≤ v p
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Electromagnetic Fields
Information travels at the group velocity, which is always less than
the corresponding phase velocity in the given medium.
The group and phase velocities for each mode propagating in the
wave guide are frequency−dependent. This means that frequency
components of a broadband signal travel at different speed and
change their phase relationship as they propagate along the wave
guide. The group and phase velocities of the modes are also
mode−dependent. This means that if a signal is distributed over a
number of different modes, the components spread out over time
during propagation.
This phenomenon is called dispersion. Wave guides are in general
dispersive media.
Note: For the fundamental TEM mode in parallel plate wave guide
fc = 0 ⇒ v pz = v p = vg
© Amanogawa, 2001 – Digital Maestro Series
⇒ no dispersion
124
Electromagnetic Fields
ω
Slope
vg
ω2
Slope
ω1
ωc
vp Slope
at cutoff
vg 0
v pz 1
v pz
z1
z2
z
Dispersion diagram
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Electromagnetic Fields
The power flow follows the Poynting vector, with the same direction
as the propagation vector. The group velocity accounts for the
effective motion of the power flow in the direction parallel to the
axis of the wave guide.
2 L sin vg t
P
L
L
2L vp t
2 L ⋅ sin θ 2 L
∆t =
=
⇒ vg = v p sin θ
vg
vp
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Electromagnetic Fields
The guide phase velocity corresponds to the apparent motion
illustrated by the following diagrams
P
L / sin v pz t / 2
L vp t / 2
L
P
L / sin v pz t / 2
L
© Amanogawa, 2001 – Digital Maestro Series
L vp t / 2
127
Electromagnetic Fields
Therefore, we obtain for the guide phase velocity
vp
2L
2L
∆t =
=
⇒ v pz =
v pz ⋅ sin θ v p
sin θ
From the results above, we have again
v pz ≥ v p
vg ≤ v p
vp
v p sin θ = v2p
v pz ⋅ vg =
sin θ
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