# Plane Wave Propagation: Why Study It?

Urban and Civil

Nov 16, 2013 (4 years and 7 months ago)

79 views

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

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

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

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 right-handed 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.

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

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

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

).

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

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
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.

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

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

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
    