13.
1
Plane Wave Propagation: Why Study It?
Two kinds of propagation:
•
bounded: in waveguides, transmission lines
•
unbounded: antennas, flashlights, stars
Unbounded propagation:
Far from the sources, the wave
appear planar
Ulaby Figure 7

3
13.
2
Plane Wave Propagation: Why Study It?
Two kinds of propagation:
•
bounded: in waveguides, transmission lines
•
unbounded: antennas, stars
Bounded straight

line propagation:
The electric field
and magnetic field
can be treated as transversely modulated
plane waves.
Ulaby Figures 7

1 and 7

2
0 0
(,,;) (,)cos
x y z t x y t kz
E E
0 0
(,,;) (,)cos
x y z t x y t kz
H H
The plane wave approximation allows us to separate material effects
effects (nonlinearity, dispersion, birefringence) from geometric effects
13.
3
Maxwell’s Equations: Phasor Domain
Phasor Domain Fields:
•
Phasors are essentially Fourier transforms at a single frequency
•
Any time domain behavior can be described by adding up phasors
Reminder:
In this section of his book, Ulaby uses
k
and the designation
wavenumber
, instead of
b
and the designation
phase constant
.
He uses tildes (~), not hats (^) to indicate phasors
(,,;) Re (,,)exp( )
x y z t x y z j t
E E
13.
4
Maxwell’s Equations: Phasor Domain
•
We have used
D
=
e
E
and
B
=
m
H
•
In the phasor domain, we can easily generalize
Ampere’s Law
Gauss’s Law of
Magnetics
Faraday’s Law
Gauss’s Law
Phasor Domain
Time Domain
Name of Law
V
D
0
B
t
B
E
t
D
H J
V
/
e
E
j
m
E H
0
H
j
e
H J E
( ),( )
e e m m
13.
5
Maxwell’s Equations: Phasor Domain
Complex permittivity
When Ohm’s law holds,
J
=
s
E
, we may write Faraday’s law
Charge

Free Medium
In a charge

free medium the phasor domain equations become
with ,/
e e e s
c c
( )
,where
j j j j
j j
s
e s e e
e e e e
H J E E E
E
c
0,
0,
j
j
m
e
E E H
H H E
13.
6
Maxwell’s Equations: Phasor Domain
Wave Equations
Combining Faraday’s and Ampere’s laws:
Using the vector relation
Gauss’s law
we obtain the wave equation
Calculating , we find similarly
2
c c
( ) ( ) ( )
j j j
m m e me
E H E E
2
( ) ( ),
E E E
2 2
c
0, and the definition
me
E
2 2
0
E E
( )
H
2 2
0
H H
We have the same wave equation for
This is analogous to in transmission lines
This analogy is no coincidence!!
Transmission lines can be analyzed using EM waves
E H
and
V I
and
13.
7
Lossless Plane

Wave Propagation
Dispersion relation and wave equation:
s
= 0
We have:
and the wave equation becomes
Plane

Wave Properties
(1)
By definition, a plane wave only varies in one direction, which we
choose to the be the
z

direction. Hence,
Our wave equation becomes
2 2 2
c
so that
k k
e e me me
2 2
0
k
E E
2
2
2
0
k
z
E
E
(,,) ( ) and 0
x y z z
x y
E E
E E
13.
8
Lossless Plane

Wave Propagation
Plane

Wave Properties
(2)
.
Proof:
From the
z

components of Ampere’s and
Faraday’s laws,
Plane

wave fields are always orthogonal to the direction of
propagation!
(3)
The
x

and
y

components of are uncoupled.
The wave equations for these components becomes
There are two independent solutions, one with
and the other with
0
z z
E H
E
0 and 0
y y
x x
z z
H E
H E
j E j H
x y x y
e e
2
2
2 2
2 2
0 and 0
y
x
x y
d E
d E
k E k E
dz dz
0
y
E
0
x
E
13.
9
Lossless Plane

Wave Propagation
Plane

Wave Properties
Focusing on the solution, we have
(4)
The general solution to the wave equation has forward

and backward

propagating components
(just like in a transmission line)
(5)
Writing Faraday’s law implies
The analogy with transmission lines is
ˆ
,
x
E
E x
0 0
( ) ( ) ( ) exp( ) exp( )
x x x x x
E z E z E z E jk z E jk z
0
y
E
0 0
( ) 0,( ) ( ) ( )
exp( ) exp( )
x y y y
x x
H z H z H z H z
k k
E jk z E jk z
m m
( ) ( ) ( ) ( ) ( ) ( )
x x x
V z V z V z E z E z E z
0 0
( ) ( ) ( ) ( )
( ) ( ),
x x
y
V z V z E z E z
I z H z
Z Z k
m
13.
10
Lossless Plane

Wave Propagation
Plane

Wave Properties
This is a characteristic property of the medium (not the wave)
Focusing on forward

going waves, we have
(6)
Plane waves are TEM waves (transverse electromagnetic waves),
waves in which the electric and magnetic fields are
orthogonal to each other and the direction of propagation:
Our solution generalizes to:
0
0
( )
ˆ ˆ ˆ ˆ
( ) ( ) exp( ),( ) exp( )
x x
x x
E z E
z E z E jkz z jkz
E x x H y y
( ) intrinsic impedance
k
m m
e
Ulaby Figure 7

4
ˆ
(,,) is a righthanded system
E H k
1
ˆ ˆ
,
H k E E k H
13.
11
Lossless Plane

Wave Propagation
Plane

Wave Properties
(7)
Plane waves (like all waves) are characterized by an amplitude and a
phase. We may write
The electric and magnetic fields are
in phase
(8)
In a vacuum
0 0
exp( ),
x x
E E j
0
ˆ
so that (,) Re ( )exp( ) cos( ),
x
z t z j t E t kz
E E x
0
ˆ
and (,) Re ( ) exp( ) cos( ),
x
E
z t z j t t kz
H H y
0
8
0 p
0
0 0
1
377,3 10 m/s
u
k
m
e
me
The observation that electromagnetic waves propagate at c led
led Maxwell to propose the light is made of electromagnetic waves!
Ulaby 2001
13.
12
Electromagnetic Plane Wave in Air: Ulaby Example 7

1
Question:
The electric field of a 1

MHz plane wave traveling in the
+
z

direction in air points along the
x

direction. If the peak value of
E
is 1.2
p
mV/m and
E
is a maximum at
t
= 0 and
z
= 50 m, obtain expressions for
E
(
z
,
t
)
and
H
(
z
,
t
), and then plot these variations as a function of
z
at
t
= 0.
Answer:
At
f
= 1 MHz, the wavelength in air is given by
so that the wavenumber is
k
= (2
p
/ 300) ~ 0.0209 rad/m. From the general
expression for
E
(
z
,
t
), we have
0
6
ˆ
(,) cos( )
2
ˆ
1.2 cos 2 10 mV/m.
300
x
z t E t kz
t z
p
p p
E x
x
8
6
3 10
300 m
1 10
c
f
Lossless Plane

Wave Propagation
13.
13
Electromagnetic Plane Wave in Air:
Ulaby
Example 7

1
Answer (continued):
The cosine is a maximum when its argument is 0
(or other multiples of 2
p
), so we have
and
which becomes at
t
= 0:
2
ˆ
(,0) 1.2 cos mV/m
300 3
2
ˆ
(,0) 10cos A/m
300 3
z t z
z t z
p p
p
p p
m
E x
H y
2 50
0 or
300 3
p p
Lossless Plane

Wave Propagation
6
6
0
2
ˆ
(,) 1.2 cos 2 10 mV/m
300 3
(,) (,) 2
ˆ
(,) 10cos 2 10 A/m
120 ( ) 300 3
z t t z
z t z t
z t t z
p p
p p
p p
p m
p
E x
E E
H y
13.
14
Electromagnetic Plane Wave in Air: Ulaby Example 7

1
Answer (continued):
We show the plot below.
Note that
E
and
H
are in phase.
Lossless Plane

Wave Propagation
Ulaby Figure 7

5
13.
15
Tech Brief 13: RFID Systems
History
1973: Two patents issued
–
one for active RFID with rewritable memory to Mario
Cardullo
, second for passive RFID system for keyless entry to Charles Walton.
Overview
System consists of a reader (transceiver) and a tag (transponder). Tag broadcasts
information about its identity when polled by the reader.
Active vs. Passive
The reader must generate a strong enough signal to generate the current
necessary in a passive tag to transmit the message. Thus, passive devices are
limited in range. However, active devices are significantly more expensive to
fabricate. Devices with higher frequency of operation generally transmit greater
distances, but are also more expensive.
13.
16
Tech Brief 13: RFID Systems
13.
17
Tech Brief 13: RFID Systems
13.
18
Polarization
Two independent solutions
Forward

propagating and at the same
Each independent solution is referred to as a different
polarization
The time variation of the amplitude and phase that an observer sees depends
on the phase difference between the
x

and
y

components, as well as their
amplitudes. We may write
( )
( )
ˆ ˆ ˆ ˆ
( ) ( ) ( );( )
y
x
x y
E z
E z
z E z E z z
E x y H x y
Ulaby Figure 7

6
( ) exp( )
( ) exp( ) exp( )
x x
y y
E z a jk z
E z a j jk z
13.
19
Polarization
Two independent solutions
The vector sum yields
with the corresponding time

dependent field
At a fixed point in
z
, the tip of the vector
E
(
z
,
t
) can trace out a line, a circle,
or an ellipse on the
x

y
plane
In general, light is elliptically polarized
Field magnitude and phase
ˆ ˆ
( ) exp exp( )
x y
z a a j jkz
E x y
ˆ ˆ
(,) Re ( )exp( ) cos( ) cos( )
x y
z t z j t a t kz a t kz
E E x y
Ulaby 2001
1/2
2 2 2 2
1
(,) cos ( ) cos ( )
(,)
(,) tan
(,)
x y
y
x
z t a t k z a t k z
E z t
z t
E z t
E
13.
20
Polarization
Linear Polarization
For convenience, we will observe the polarization state at
z
= 0
Linear polarization corresponds to
= 0 or
=
p
.
We have
so that
Since
is constant the tip of
the
E

field vector is a
sinusoidally

varying straight line
ˆ ˆ
(0,) ( ) cos
x y
t a a t
E x y
Ulaby Figure 7

7
1/2
2 2
1
(0,) cos
(0,) tan/
x y
y x
t a a t
t a a
E
13.
21
Polarization
Circular Polarization
This case corresponds to
Left circular polarization (LCP):
= +
p
/ 2
Right circular polarization (RCP):
=
–
p
/ 2
Field variation
At
z
= 0, we now find
(0,) constant
(0,)
t a
t t
E
,/2
x y
a a
p
ˆ ˆ ˆ ˆ
( ) exp(/2) exp( ) ( )exp( )
ˆ ˆ
(,) cos( ) sin( )
z a j jkz a j jkz
z t a t kz a t kz
p
E x y x y
E x y
13.
22
Polarization
Circular Polarization
Left circular polarization:
Left

hand screw
points in the direction of propagation
Right circular polarization:
Right

hand
screw points in the direction of propagation
Ulaby Figure 7

8
Ulaby Figure 7

9
13.
23
Polarization
Elliptical Polarization
The general case is somewhat complicated
The polarization ellipse is specified by two angles:
—
= rotation angle
—
c
= ellipticity angle
These are related to
a
x
,
a
y
,
by
0
0
0
tan 2 (tan 2 ) cos
sin 2 (sin 2 )sin
with tan/
y x
a a
c
Ulaby Figure 7

11
13.
24
Polarization
Elliptical Polarization
Ulaby Figure 7

12
13.
25
Right

Hand Circular Polarization State: Ulaby Example 7

2
Question:
A right

circularly

polarized plane wave with electric field
modulus of 3 mV/m is traveling in the +
y

direction in a dielectric medium with
If the wave frequency is 100 MHz, obtain
expressions
E
(
y
,
t
) and
H
(
y
,
t
).
Answer:
Since the wave is traveling in the +
y

direction, its field
components are in the
x

and
z

directions. When a right

circularly polarized
wave travels in the +
z

direction, the
y

component is retarded with respect to the
x

component by a phase
p
/
2. By the cyclic permutation rule,
x

component must
be retarded with respect to the
z

component by a phase
p
/
2 in this case. We thus
conclude,
ˆ ˆ ˆ ˆ
( ) exp(/2) exp( )
ˆ ˆ
3( ) exp( ) mV/m
1 3
ˆ ˆ ˆ
( ) ( ) ( ) exp( ) mA/m
x z
y E E a j jk y
j jk y
y y j jk y
p
E x z x z
x z
H y E z x
Polarization
0 0
4, , and 0.
e e m m s
13.
26
Right

Hand Circular Polarization State:
Ulaby
Example 7

2
Answer (continued):
With we have
The time

domain fields are then given by
ˆ ˆ
(,) Re 3( ) exp( ) exp( )
ˆ ˆ
3 sin( ) cos( ) mV/m
3
ˆ ˆ
(,) Re ( ) exp( ) exp( )
1
ˆ ˆ
cos( ) sin( ) mA/m
20
y t j jk y j t
t k y t k y
y t j jk y j t
t k y t k y
p
E x z
x z
H z x
x z
Polarization
8
r
8
0
r
2 10 2
4
rad/m
3
3 10
120
60
4
k
c
p
e
p
p
p
e
8 1
2 2 10 s,
f
p p
13.
27
Tech Brief 14: Liquid Crystal Display (LCD)
Physical principal
•
Liquid crystals are neither solid nor liquid, but hybrid of both
•
Twisted
nematic
LCs have rod shaped molecules that tend to assume a twisted
spiral shape when sandwiched between two glass substrates with orthogonal
grooving orientations
•
Twisted spiral LCs act as a polarizer since incident light tends to follow the
orientation of the spiral
•
The LC layer is sandwiched between orthogonal polarization filters. The LC
layer rotates the light by 90º when it passes through, allowing light to exit from
the other polarization filter
•
The twisted
nematic
LC spiral untwists under the influence of an electric field
•
Degree of untwisting is proportional to field strength
•
Under an applied voltage, light cannot pass the second filter
Discovery: 1880s by a botanist
Frierich
Reinitzer
Uses:
Digital clocks, cell phones, desktop and laptop
computers, televisions, etc.
13.
28
Tech Brief 14: Liquid Crystal Display (LCD)
13.
29
Tech Brief 14: Liquid Crystal Display (LCD)
13.
30
Lossy Propagation
Attenuation Constant and Phase Constant
Starting from the wave equation,
with
and defining
—
a
= attenuation coefficient;
b
= phase constant
we find
Equating real and imaginary parts and solving for
a
and
b
2 2 2
c
( ) [,/]
j
me me e e ee s
2 2
0
E E
j
a b
2 2 2 2 2
( ) ( ) 2
j j j
a b a b ab me me
1/2 1/2
2 2
1 1,1 1
2 2
me e me e
a b
e e
13.
31
Lossy Propagation
Solution (+
z

direction)
with
The electric and magnetic fields are no longer in phase!
This is analogous to what happens in lossy transmission lines with the
voltage and the current.
0 0
0
c c
ˆ ˆ ˆ
( ) ( ) exp( ) exp( ) exp( )
( )
ˆ ˆ ˆ
( ) ( ) exp( ) exp( )
x x x
x x
y
z E z E z E z j z
E z E
z H z z j z
a b
a b
E x x x
H y y y
1/2
c
c
1
j
m m e
e e e
13.
32
Lossy Propagation
Solution (+
z

direction)
The electric and magnetic fields attenuate
The corresponding attenuation length is called
the skin depth
s
Two important limits
•
low

loss dielectric:
•
good conductor:
0
0
c
( ) exp( ),( ) exp( )
x
x x y
E
E z E z H z z
a a
s
1/
a
/1 ( in practice less than 1/100 )
e e
/1 ( in practice greater than 100 )
e e
13.
33
Lossy Propagation
Low

Loss Dielectric
From the expression for
,
so that
From the expression for
c
,
,
2 2
e m s m
a b me me
e e
1/2
1 1
2
j j j j
e e
me me
e e
1/2
c
1 1 1
2 2
j j j
m e m e m s m
e e e e e e e
13.
34
Lossy Propagation
Good Conductor
From the expression for
,
so that
From the expression for
c
,
,
2
f f
ms
a p ms b a p ms
1
2 2
2
j
j j
ms ms ms
me
c
(1 ) (1 )
f
j j j
m p m a
e s s
Some consequences
•
The
E

field and
H

field are 45
o
out of phase
•
A good conductor shorts out the
E

field, but not the
H

field
As
s
,
c
0
E / H
0
Ulaby 2001
13.
35
Plane Wave in Seawater: Ulaby Example 7

4
Question:
A plane wave travels in the +
z

direction downward in sea water,
with the
x

y
plane at the sea surface and
z
= 0 denoting a point just below the sea
surface. The constitutive parameters are
If the magnetic field at
z
= 0 is given by
(a) Obtain expressions
E
(
z
,
t
) and
H
(
z
,
t
), and (b) determine the depth at which
the amplitude of
E
has fallen to 1% of its value at
z
= 0.
Answer:
Seawater is a good conductor at
f
= 1 kHz, so we have
In phasor form, we have
0
0
c
ˆ
( ) exp( ) exp( )
ˆ
( ) exp( ) exp( )
x
x
z E z j z
E
z z j z
a b
a b
E x
H y
Lossy Propagation
0 0
80, , and 4 S/m.
e e m m s
3
ˆ
(0,) 100cos 2 10 15 mA/m,
t t
p
H y
3 7 1
c
10 4 10 4 0.126 m
(1 )/0.044exp(/4)
f
j j
b a p ms p p
as p
13.
36
Plane Wave in Seawater: Ulaby Example 7

4
Answer (continued):
We may write the initial field as
so that in the time domain
Comparing to the initial conditions, we have
0 0
3
0 0
0
0
3
0
ˆ
(,) Re exp( ) exp( ) exp( ) exp( )
ˆ
exp( 0.126 ) cos 2 10 0.126
ˆ
(,) Re exp( ) exp( ) exp( ) exp( )
0.44exp(/4)
ˆ
22.5 exp( 0.126 ) cos 2 10 0.126
x
x
x
x
z t E j z j z j t
E z t z
E
z t j z j z j t
j
E z t z
a b
p
a b
p
p
E x
x
H y
y
0
45
Lossy Propagation
0 0 0
exp( )
x x
E E j
3
0 0
0 0
22.5 100 10 4.44 mV/m
45 15 60
x x
E E
13.
37
Plane Wave in Seawater: Ulaby Example 7

4
Answer (continued):
Our final result is
The depth at which the amplitude of
E
has decreased to 1% of its initial value at
z
= 0 is obtained from
3
3
0
ˆ
(,) 4.44exp( 0.126 ) cos 2 10 0.126 60 mV/m
ˆ
(,) 100exp( 0.126 ) cos 2 10 0.126 15 mA/m
z t z t z
z t z t z
p
p
E x
H y
Lossy Propagation
ln(0.01)
0.01 exp( 0.126 ) 36.5 m
0.126
z z
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