Plane Wave Propagation: Why Study It?

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16 Νοε 2013 (πριν από 3 χρόνια και 11 μήνες)

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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 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.



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
    