EE 542
Antennas & Propagation for
Wireless Communications
Topic 3 - Basic EM Theory and
Plane Waves
1
Outline
•
•
EM Theory Concepts
Maxwell’s Equations
–
–
–
–
•
Notation
Differential Form
Integral Form
Phasor Form
Wave Equation and Solution (lossless,
unbounded, homogeneous medium)
–
–
Derivation of Wave Equation
Solution to the Wave Equation – Separation of Variables
–
Plane waves
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EM Theory Concept
The fundamental concept of em theory is that a
current at a point in space is capable of
inducing potential and hence currents at
another point far away.
E, H
J
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Introduction to EM Theory
• The existence of propagating em waves
can be predicted as a direct consequence
of Maxwell’s equations.
• These equations satisfy the relationship
between the vector electric field, E and
vector magnetic field, H in time and
space in a given medium.
• Both E and H are vector functions of
space and time; i.e. E (x,y,z;t), H (x,y,z;t.)
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What is an Electromagnetic Field?
• The electric and magnetic fields were originally
introduced by means of the force equation.
• In Coulomb’s experiments forces acting between
localized charges were observed.
• There, it is found useful to introduce E as the
force per unit charge.
• Similarly, in Ampere’s experiments the mutual
forces of current carrying loops were studied.
• B is defined as force per unit current.
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Why not use just force?
• Although E and B appear as convenient replacements
for forces produced by distributions of charge and
current, they have other important aspects.
• First, their introduction decouples conceptually the
sources from the test bodies experiencing em forces.
• If the fields E and B from two source distributions are the
same at a given point in space, the force acting on a test
charge will be the same regardless of how different the
sources are.
• This gives E and B meaning in their own right.
• Also, em fields can exist in regions of space where there
are no sources.
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Maxwell’s Equations
• Maxwell's equations give expressions for electric and
magnetic fields everywhere in space provided that all
charge and current sources are defined.
• They represent one of the most elegant and concise
ways to state the fundamentals of electricity and
magnetism.
• These set of equations describe the relationship
between the electric and magnetic fields and sources
in the medium.
• Because of their concise statement, they embody a
high level of mathematical sophistication.
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Notation:
(Time and Position Dependent Field Vectors)
E (x,y,z;t)
Electric field intensity (Volts/m)
H (x,y,z;t)
Magnetic field intensity (Amperes/m)
D (x,y,z;t)
Electric flux density (Coulombs/m2)
B (x,y,z;t)
Magnetic flux density (Webers/m2, Tesla)
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Notation: Sources and Medium
J (x,y,z;t)
Electric current density (Amperes/m2)
Jd (x,y,z;t)
Displacement current density (Amperes/m2)
re
Electric charge density (Coulombs/m3)
e
er
Permittivity of the medium (Farad/m)
Relative permittivity (with respect to free space eo)
m
er
Permeability of the medium (Henry/m)
Relative permittivity (with respect to free space mo)
s
Conductivity of the medium (Siemens/m)
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Maxwell’s Equations –
Physical Laws
• Faraday’s Law  Changes in magnetic field
induce voltage.
• Ampere’s Law  Allows us to write all the
possible ways that electric currents can make
magnetic field. Magnetic field in space around
an electric current is proportional to the current
source.
• Gauss’ Law for Electricity The electric flux
out of any closed surface is proportional to the
total charge enclosed within the surface.
• Gauss’ Law for Magnetism The net
magnetic flux out of any closed surface is zero.
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Differential Form of Maxwell’s
Equations
Faraday’s Law:
B
 E = t
(1)
Ampere’s Law:
D
 H = J 
;
t
Jd
D
;
t
J = J s  Jc
(2)
Gauss’ Law:
 D = r v
(3)
 B = 0
(4)
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Constitutive Relations
Constitutive relations provide information
about the environment in which
electromagnetic fields occur; e.g. free
space, water, etc.
permittivity
D = eE
9
10
eo 
36
7
m o  4 10
B = mH
(5)
permeability
(6)
Free space values.
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Time Harmonic Representation Phasor Form
• In a source free ( J s  r  0 ) and lossless
(s  0  J  0 ) medium characterized by
permeability m and permittivity e, Maxwell’s
equations can be written as:
H
  E = -m
c
t
E
 H = e
;
t
 D = 0
 B = 0
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Examples of del Operations
• The following examples will show how to
take divergence and curl of vector
functions
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Example 1
Let
Find:
ˆ sin x;   sin(x )sin(2y )
Ay

a) the curl of A   A


b) the divergence of A  A
c) the gradient of 

  
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Solution 1
 

 
ˆ
ˆ
ˆ   y
ˆ sin x 
 A  
x
y
z
y
z 
 x
ˆcos x
z
 

 
ˆ
ˆ
ˆ  y
ˆ sin x   0
 A
x
y
z
y
z 
 x
  sin(x)sin(2y )
 sin( x)sin(2y )
 sin( x)sin(2y ) 
ˆ
ˆ
ˆ
  
x
y
z
x
y
z


ˆ cos(x )sin(2y )  y
ˆ2 sin( x)cos(2y )
x
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Example 2
Calculate the magnetic field for the electric field
given below. Is this electric field realizable?
ˆ  cos( wt  o )
E( x, y, z; t )   xˆ 5 xy  yz
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Solution
H
  E = -m
t
ˆ  cos( wt  o ) 
    xˆ 5 xy  yz
 

 
ˆ  cos( wt  o ) 
  xˆ 
yˆ  zˆ    xˆ 5 xy  yz
y
z 
 x
  5 xy 
z
 cos( wt  o ){
 xˆ  xˆ    xˆ  yˆ  
x
x
  5 xy 
z
 yˆ  xˆ    yˆ  yˆ  
y
y
  5 xy 
z
 zˆ  xˆ    zˆ  yˆ }
z
y
H
 5 x cos( wt  o ) zˆ = - m
t
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Solution continued
H
5
 x cos( wt  o ) zˆ
t
m
5x
H =
zˆ  cos( wt  o )
m
5x

zˆ sin( wt  o )
wm
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Solution continued
To be realizable, the fields must satisfy Maxwell’s equations!
E
 H = e
t
?
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Solution Continued

 5x
zˆsin( wt  o ) 
 H =   

 wm
5
yˆ sin( wt  o )

wm
ˆ  cos( wt  o ) 
  xˆ 5 xy  yz
E
e
e
t
t
ˆ 
  we sin( wt  o )  xˆ 5 xy  yz
5
ˆ 
yˆ sin( wt  o )   we sin( wt  o )  xˆ 5 xy  yz
wm
These fields are NOT realizable. They do not form em fields.
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Time Harmonic Fields
• We will now assume time harmonic fields; i.e.
fields at a single frequency.
• We will assume that all field vectors vary
sinusoidally with time, at an angular frequency w;
i.e.
E( x, y, z; t )  eˆ Eo ( x, y, z )cos( wt  o )
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Time Harmonics and Phasor
Notation
Using Euler’s identity
e j (wt  )  cos(wt   )  j sin(wt   )
The time harmonic fields can be written as
E( x, y, z; t )  Re eˆ Eo ( x, y, z )e
j  wt o 

 Re eˆ Eo ( x, y, z )e e 
jo
Phasor notation
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jwt
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Phasor Form
E ( x, y, z; t )  Re  E ( x, y, z )e jwt 
H ( x, y, z; t )  Re  H ( x, y, z )e
jwt

where
jo
E ( x, y, z ) eˆ Eo ( x, y, z )e
H ( x, y, z ) hˆ H ( x, y, z )e j
o
o
Information on
amplitude,
direction and
phase
Note that the E and H vectors are now complex and are known as phasors
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Time Harmonic Fields in Maxwell’s
Equations
With the phasor notation, the time derivative in
Maxwell’s equations becomes a factor of jw:

 jw
t
E ( x, y, z; t )  Re E ( x, y, z )e
jwt

E ( x, y, z; t )

jwt
 Re  E ( x, y, z )e 
t
t
 Re  jwE ( x, y, z )e
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jwt

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Maxwell’s Equations in Phasor
Form (1)
E
 H = e
t
E ( x, y, z; t )  Re E ( x, y, z )e jwt 
H ( x, y, z; t )  Re H ( x, y, z )e jwt 

 
  H = Re   H ( x, y, z ) e
E

t

 Re E ( x, y, z )e jwt
t
jwt
  Re

 jwE ( x, y, z )  e jwt

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Maxwell’s Equations in Phasor
Form (2)
E
 H = e
t

  H = Re    H ( x, y, z )  e
E

t
 Re  E ( x, y, z )e jwt 
jwt



t
Re    H ( x, y, z )  e   e Re  jwE ( x, y, z )  e 
 Re  jwE ( x, y, z )  e jwt
jwt
jwt
  H ( x, y, z )  jwe E ( x, y, z )
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Phasor Form of Maxwell’s
Equations (3)
Maxwell’s equations can thus be written in phasor
form as:
H
  E = -m
t
E
 H = e
;
t
 D = 0
 B = 0
  E = - jwm H
  H = jwe E
D= 0
B= 0
Phasor form is dependent on position
only. Time dependence is removed.
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Examples on Phasor Form
Determine the phasor form of the following
sinusoidal functions:
a)
b)
c)
d)
f(x,t)=(5x+3) cos(wt + 30)
g(x,z,t) = (3x+z) sin(wt)
h(y,z,t) = (2y+5)4z sin(wt + 45)
V(t) = 0.5 cos(kz-wt)
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Solutions
a)
f (x, t )  Re F (x)e jwt 
 5x  3 cos(wt  30)

 5x  3 Re e
j wt  30 
 Re 5x  3 e e
j 30

jwt

F (x)  5x  3 e j 30  5x  3  cos 30  j sin30 
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Solutions
b)
g(x, z, t )  (3x  z ) sin(wt )
 (3x  z ) cos(wt  90)

 (3x  z )Re e

 Re (3x  z )e
 Re (3x  z )e



j wt  90 
j wt  90
 j 90
e
jwt

G(x, z )  (3x  z )e  j 90   j(3x  z )
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Solution
c)
h(y , z, t )  (2y  5)4 z sin(wt  45)
 (2y  5)4 z cos(wt  45  90)
 (2y  5)4 z cos(wt  45)
 (2y  5)4 z Re e j (wt  45)
 Re (2y  5)4 ze  j 45e j (wt )
H(y , z )  (2y  5)4 ze  j 45
 (2y  5)4 z  cos 45  j sin 45 
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Solution
d)
v(t )  0.5 cos(kz  wt )
 0.5Re e
V  0.5e
jkz  jwt
  Re 0.5e
jkz
e
 jwt

jkz
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Example
• Find the phasor notation of the following
vector:

C (t ) 
E(t )
t
where


ˆ  8 cos wt  sin wt  z
ˆ
E(t )  3 cos(wt )  4 sin(wt ) x
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Solution
E
ˆ
C (t ) 
 3w sin wt  4w cos wt  x
t
ˆ
8 w sin(wt )  w cos wt  z
Using
sin wt  Re   je jwt 
cos wt  Re  e jwt 
ˆ
C (t )  Re  3w   je jwt   4w  e jwt   x
ˆ
 Re 8w   je jwt   8w  e jwt   z
ˆ   8wj  8w  z
ˆ
C  3wj  4w  x
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Example
• Show that the following electric field
satisfies Maxwell’s equations.
ˆ oe
E  xE
 jkz
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ˆ oe
E  xE
B
Solution
 jkz
w em  jkz
1
k
ˆ o e  jkz  yE
ˆ o
  E  yE
e
jw
w
w
ˆ o em e  jkz
 yE
H 
?
E
1
m
ˆ
By
Eo

e  jkz ;
 
m
e
 Eo  jkz 
1
1
ˆ e 
H 
  y
jwe
jwe
 

1 Eo  jkz
ˆ)
  jk( x
e
jwe 

k
w me
ˆ oe  jkz  xE
ˆ oe  jkz
xE
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The Wave Equation (1)
If we take the curl of Maxwell’s first equation:
    E   jwm   H
Using the vector identity:


    A     A  2 A
And assuming a source free, i.e. qe  J  0 and lossless;
i.e.
s  0 medium:
 E  0
 H = J  Jd ;
Jd
D
;
t
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J = J s  Jc  0
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The Wave Equation (2)
 E (r )  w me E (r )
2
2
Define k, which will be known as wave number:
k
2
w me
2
 E ( x, y, z )  k E ( x, y, z )  0
2
2
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Wave Equation in Cartesian
Coordinates
 E ( x, y, z )  k E ( x, y, z )  0
2
2
where
E ( x, y, z )  xˆ Ex ( x, y, z )  yˆ E y ( x, y , z )  zˆ E z ( x, y , z )
and
2
2
2



2
  2 2 2
x y z
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Laplacian
 2 E ( x, y, z )   2  xˆ Ex ( x, y, z )  yˆ E y ( x, y, z )  zˆ E z ( x, y, z ) 
 xˆ  2 Ex ( x, y, z )  yˆ  2 E y ( x, y, z )  zˆ  2 E z ( x, y, z )
  2 Ex  2 Ex  2 Ex 
 xˆ  2 
 2 
2
y
z 
 x
  2 Ey  2 Ey  2 Ey 
 yˆ  2 

2
2 
y
z 
 x
  2 Ez  2 Ez  2 Ez 
 zˆ  2 
 2 
2
y
z 
 x
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Scalar Form of Maxwell’s
Equations
Let the electric field vary with x only.
E
2
 k E
2
x
E  xˆ f ( x)  yˆ g ( x)  zˆ h( x)
2
and consider only one component of the field; i.e. f(x).
 f
2
 k x f
2
x
2
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Possible Solutions to the Scalar
Wave Equation
f ( x)  A1e
 jkx
 A2e
 jkx
 traveling wave
Energy is transported from one
point to the other
or
f ( x)  C1 cos(kx)  D1 sin(kx)  standing wave
Standing wave solutions are appropriate for bounded propagation such as wave
guides.
When waves travel in unbounded medium, traveling wave solution is more
appropriate.
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The Traveling Wave
• The phasor form of the fields is a mathematical
representation.
• The measurable fields are represented in the time domain.
Let the solution to the -component of the electric field be:
E  Ae
1
 jkx
A1  A1 e
;
j
Traveling in +x direction
Then
E  ( x, y, z; t )  Re E ( x, y, z )e
 Re  Ae
1
 jkx
e
jwt
jwt

 xˆ 
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A1 cos  wt  kx   
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Traveling Wave
As time increases, the wave moves along +x direction
1.5
wt = PI/2
wt = 0
E(x,y,z;t)
1
0.5
0
-0.5
-1
-1.5
-4
-3
-2
-1
0
1
2
3
4
kx
E ( x, y, z; t )  cos  wt  kx   
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Standing Wave
E  C1 cos(kx); C1  C1 e j
Then, in time domain:
E  ( x, y, z; t )  Re E ( x, y, z )e jwt 
 ReC1 cos(kx)e jwt  xˆ  C1 cos  wt    cos(kx)
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Standing Wave
Stationary nulls and peaks in space as time passes.
1.5
E(x,y,z;t)
1
0.5
0
-0.5
-1
-1.5
-4
-3
-2
-1
0
1
2
3
4
kx
wt = 0
wt = PI/4
wt = PI/2
wt = 3PI/4
wt = -PI/2
wt = 5PI/8
wt = PI
wt = 3PI/8
47
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To summarize
• We have shown that Maxwell’s equations
describe how electromagnetic energy
travels in a medium
• The E and H fields satisfy the “wave
equation”.
• The solution to the wave equation can be
in various forms, depending on the
medium characteristics
48
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The Plane Wave Concept
• Plane waves constitute a special set of E
and H field components such that E and H
are always perpendicular to each other
and to the direction of propagation.
• A special case of plane waves is uniform
plane waves where E and H have a
constant magnitude in the plane that
contains them.
49
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Plane Wave Characteristics
E ( x, y, z; t )  A1 cos  wt  kx    eˆ
amplitude
Frequency
(rad/sec)
phase
polarization
Wave number, depends on
the medium characteristics
Direction of
propagation
E ( x, y, z; t )  A1 cos   eˆ;   wt  kz  
amplitude
phase
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50
Plane Waves in Phasor Form
E ( x, y, z; t )  A1 cos  wt  kx    eˆ
E ( x, y, z )  A1 e e
j
 E0e
Complex
amplitude
 jkx
 jkx
eˆ
eˆ
polarization
Position dependence
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Example 1
Assume that the E field lies along the x-axis (i.e. xpolarized) and is traveling along the z-direction.
wave number
ˆ oe
E  xE
 jkz
We derive the solution for the H field from the E field using
Maxwell’s equation #1:
1
Eo  jkz
H
  E  yˆ e
o
 - jwm 
k2
w2 me
H
Intrinsic impedance;
377 W for free space
E
Note the I = V/R analogy in circuit theory.
O. Kilic
52
EE542
Example 1 (2 of 4)
z
direction of
propagation
y
x
E, H plane
E ( x, y , z )  E ( z )
H ( x, y , z )  H ( z )
E and H fields are not functions of x and y, because they lie on x-y plane
O. Kilic EE542
53
Example 1 (3 of 4)
In time domain:
ˆ Eo cos(wt  kz   )
E x
H  yˆ
Eo
o
cos(wt  kz   )
phase term
*** The constant phase term  is the angle of the complex number Eo
54
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Example 1 (4 of 4)
Wavelength: period in space
kl = 2
 w me
55
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Velocity of Propagation (1/3)
• We observe that the fields progress with time.
• Imagine that we ride along with the wave.
• At what velocity shall we move in order to keep
up with the wave???
56
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Velocity of Propagation (2/3)
ˆ Eo cos(wt  kz  )
E x
E field as a function of different times
1
0.8
0.6
0.4
Ex
0.2
0
-0.2
-0.4
Constant phase
points
-0.6
wt  kz  a  constant
-0.8
dz
d  wt  a 
v  -1
dt
dt  k 
-4
-3
w
v 
k
-2
0 kz
-1
wt = 0
wtz= PI/2
1
2
3
4
wt = PI
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Velocity of Propagation (3/3)
In free space:
dz w
v 

dt
k
v fs
k  w me
eo
w
v 

k
1
mo
me
c
c
1
moe o
109
F/m
36
4  107 H/m
3  108 m/s
Note that the velocity is independent of the frequency of the wave, but a function of the medium properties.
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Example 2
A uniform em wave is traveling at an angle 
with respect to the z-axis as shown below.
The E field is in the y-direction. What is
the direction of the H field?
x
k
E

z
y
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59
Solution: Example 2
x
The direction of propagation is the unit vector
k
E
ˆ z
ˆ cos   x
ˆ sin
k
The E field is along y

z
y
Because E, H and the direction of propagation are perpendicular to each other, H
lies on x-z plane. It should be in the direction parallel to:


ˆ // k
ˆ y
ˆ z
ˆ sin  x
ˆ cos 
h
60
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Example 3
Write the expression for an x-polarized electric
field that propagates in +z direction at a
frequency of 3 GHz in free space with unit
amplitude and 60o phase.
+ z-direction
E ( x, y, z; t )  A1 cos  wt  kz    eˆ
W = 2f =
=1
x
2*3*109
w moe o
60o
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Solution 3
+ z-direction
E ( x, y, z; t )  A1 cos  wt  kz    eˆ
W = 2f =
=1
x
2*3*109
w moe o
60o
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Example 4
If the electric field intensity of a uniform plane wave
in a dielectric medium where e = eoer and m = mo is
given by:
ˆ
E  377cos(10 t  5y)z
9
Determine:
• The direction of propagation and
frequency
• The velocity
• The dielectric constant (i.e. permittivity)
• The wavelength
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Solution: Example 4 (1/2)
1. +y direction; w = 2f = 109
2. Velocity:
3. Permittivity:
v 
1
em
w 109
v 

 2  108 m/s
k
5

v  2  10 
8
1
e oe r mo

3  108
er
c
er

3  108
er
9
 e r   2.25
4
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64
Solution: Example 4 (2/2)
4. Wavelength:
2 2
l

k
5
m
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Example 5
Assume that a plane wave propagates along
+z-direction in a boundless and a source
free, dielectric medium. If the electric field
is given by:
ˆ  E oe
E  E x (z )x
 jkz
ˆ
x
Calculate the magnetic field, H.
66
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Example 5 - observations
• Note that the phasor form is being used in
the notation; i.e. time dependence is
suppressed.
• We observe that the direction of
propagation is along +z-axis.
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Solution: Example 5 (1/2)
k
H
Eo

ˆ
y
Eo
m
e
ˆ
y
E
Intrinsic impedance, I = V/R
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Solution: Example 5 (2/2)
• E, H and the direction of propagation are
orthogonal to each other.
• Amplitudes of E and H are related to each
other through the intrinsic impedance of
the medium.
• Note that the free space intrinsic
impedance is 377W.
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Example 6
Sketch the motion of the tip of the vector A(t)
as a function of time.
ˆ  jy
ˆ
Ax
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Solution: Example 6 (1/2)
A(t )  Re  Ae jwt 
ˆ
 Re xe
jwt
ˆ
 jye
jwt



j  wt   
 jwt
ˆ
ˆ  2 
 Re  xe
 ye


ˆ cos(wt )  y
ˆ cos(wt 
x

2
)
ˆ cos(wt )  y
ˆ sin(wt )
x
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Solution: Example 6 (2/2)
y
wt = 90o
wt = 180o
x
wt = 0
wt = 270o
The vector A(t) rotates clockwise wrt z-axis. The tip traces a
circle of radius equal to unity with angular frequency w.
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Polarization
• The alignment of the electric field vector of
a plane wave relative to the direction of
propagation defines the polarization.
• Three types:
– Linear
– Circular
– Elliptical (most general form)
Polarization is the locus of the tip of the electric field at a given point
as a function of time.
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Linear Polarization
y
E
x
• Electric field oscillates
along a straight line
as a function of time
• Example: wire
antennas
y
E
x
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Example 7
ˆ o cos(wt  kz)
E (z;t )  xE
For z = 0 (any position value is fine)
ˆ o cos(wt )
E (0;t )  xE
y
Eo
- Eo
x
t=
t=0
Linear Polarization: The tip of the E field always stays on xaxis. It oscillates between ±Eo
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75
Example 8
ˆ cos(wt  kz)  2y
ˆ cos(wt  kz)
E (z, t)  x
Exo=1
Eyo=2
Let z = 0 (any position is fine)
ˆ cos(wt )  2y
ˆ cos(wt )
E (0, t )  x
y
t=0
2
Linear Polarization
x
1
t = /2
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76
Circular Polarization
y
RHCP
x
y
LHCP
x
• Electric field traces a
circle as a function of
time.
• Generated by two
linear components
that are 90o out of
phase.
• Most satellite
antennas are
circularly polarized.
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O. Kilic EE542
Example 7
ˆ cos(wt  kz)  y
ˆ sin(wt  kz)
E (z;t )  x
Exo=1
Eyo=1
Let z= 0
cos(wt  kz  2 )
ˆ cos(wt )  y
ˆ sin(wt )
E (0;t )  x
RHCP
y
t=/2w
t=/w
x
t=0
t=3/2w
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Elliptical Polarization
y
RH
x
y
LH
• This is the most
general form
• Linear and circular
cases are special
forms of elliptical
polarization
• Example: log spiral
antennas
x
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Example 8
ˆ cos(wt  kz  a )  yb
ˆ cos(wt  kz  b )
E (z;t )  xa
Ey
Ex
Linear when
a  b or a  b  
b
Ey   E x
a
Circular when
a  b  
and a  b
2
Ey2  E x2  a2
Elliptical if no special condition is met.
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Example 9
ˆ  j 4y
ˆ e
E   3x
 j 0.5 z
V/m
Determine the polarization of this wave.
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Solution: Example 9 (1/2)
Note that the field is given in phasor form. We would like to
see the trace of the tip of the E field as a function of time.
Therefore we need to convert the phasor form to time
domain.

E (z; t )  Re Ee

jwt

ˆ  j 4y
ˆ e
 Re 3x
 j 0.5 z
e
jwt
;
j e
j 2
ˆ cos(wt  0.5z )  4y
ˆ cos(wt  0.5z  2 )
 3x
ˆ cos(wt  0.5z )  4y
ˆ sin(wt  0.5z )
 3x
E x (z; t ) = 3 cos(wt  0.5z )
E y (z; t ) = 4 sin(wt  0.5z )
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82
Solution: Example 9 (2/2)
Let z=0
E x (0; t ) = 3 cos(0.5z)
E y (0; t ) = 4 sin(0.5z)
E (0; t )
2
x
9

E (0; t )
2
y
16
1
Elliptical polarization
E x  Ey
83
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Example 10
Find the polarization of the following fields:
a)
E (r )   jxˆ  yˆ  e
b)
E (r )   (1  j ) yˆ  (1  j ) zˆ  e
c)
E (r )   (2  j ) xˆ  (3  j ) zˆ  e
 jkz
 jkx
 jky
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Solution: Example 10 (1/4)
a)
E (r )   jxˆ  yˆ  e
 jkz
ˆ
 xe
 j  kz  2 
ˆ
 ye
 jkz
E (r ; t )   xˆ sin( wt  kz )  yˆ cos(wt  kz )
Observe that orthogonal components
have same amplitude but 90o phase
difference.
Circular Polarization
y
t=0
Let kz=0
z
t=/2w
x
t=3/2w
RHCP
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t=/w
EE542
85
Solution: Example 10 (2/4)
b)
E (r )   (1  j ) yˆ  (1  j ) zˆ  e  jkx
j 4
1  j  2e ; 1  j  2e
 j 4
E (r ; t )  yˆ 2 cos( wt  kx  4 )  zˆ 2 cos( wt  kx  4 )
Observe that orthogonal components
have same amplitude but 90o phase
difference.
Circular Polarization
z
t=+/4w
Let kx=0
t=3/4w
x
t=5/4w
O. Kilic EE542
y
t=-/4w
RHCP
86
Solution: Example 10 (3/4)
E (r )   (2  j ) xˆ  (3  j ) zˆ  e  jky
c)
2  j  5e j ; 3  j  10e  j
1
1  1 
  tan   ;   tan  
2
 3
1
E (r ; t )  xˆ 5 cos( wt  ky   )  zˆ 10 cos( wt  ky   )
Observe that orthogonal components
have different amplitudes and are out of
phase.
x
Elliptical Polarization
t=-/w
Left Hand
Let ky=0
z
y
t=+/w
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Solution: Example 10 (4/4)
d)
E (r )   jxˆ  j 2 yˆ  e jkz
je
j 2
E (r ; t )   xˆ sin( wt  kz )  yˆ 2sin( wt  kz )
   xˆ  yˆ 2  sin( wt  kz )
Observe that orthogonal
components are in phase.
Linear Polarization
y
x
z
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Coherence and Polarization
• In the definition of linear, circular and elliptical
polarization, we considered only completely
polarized plane waves.
• Natural radiation received by an anatenna
operating at a frequency w, with a narrow
bandwidth, Dw would be quasi-monochromatic
plane wave.
• The received signal can be treated as a single
frequency plane wave whose amplitude and
phase are slowly varying functions of time.
89
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Quasi-Monochromatic Waves
ˆE x (t )e
E (z;t )  x
 jkz  jx (t )
o
ˆE y (t )e
y
 jkz  jy (t )
o
amplitude and phase are slowly varying functions of time
90
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Degree of Coherence
r xy
E
Ex E
2
x
*
y
Ey
2

1
2
where <….> denotes the time average.
1T
 lim 
T 
T 0
dt
91
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Degree of Coherence – Plane Waves
Ex  Ex e
 jkz
Ey  Ey e
 jkz
o
o
e
 j x
e
 j y
 x , y are constant. Thus:
r xy  1
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Unpolarized Waves
• An em wave can be unpolarized. For
example sunlight or lamp light. Other
terminology: randomly polarized,
incoherent. A wave containing many
linearly polarized waves with the
polarization randomly oriented in space.
• A wave can also be partially polarized;
such as sky light or light reflected from the
surface of an object; i.e. glare.
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Poynting Vector
• As we have seen, a uniform plane wave
carries em power.
• The power density is obtained from the
Poynting vector.
• The direction of the Poynting vector is in
the direction of wave propagation.
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Poynting Vector
S  EH
W/m
*
2
1
*
S  Re E  H
2
where
denotes time average


95
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Example 11
Calculate the time average power density for
the em wave if the electric field is given by:
ˆ oe
E  xE
 jkz
96
O. Kilic EE542
Solution: Example 11 (1/2)
H


1

ˆ  E;
z
1
ˆ x
ˆ E e
z


 jkz
o
Eo

ˆ
ye
 jkz
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Solution: Example 11 (2/2)
1
S  Re E  H *
2
2

Eo 
1
ˆy
ˆ
 Re  x

2
 

2
1
ˆ
z
Eo
2


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Plane Waves in Lossy Media
• Finite conductivity, s results in loss
• Ohm’s Law applies:
Jc  s E
Conduction current
Conductivity, Siemens/m
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Complex Permittivity
From Ampere’s Law in phasor form:
  H  J  jwe E;
where J  Js  Jc
 Js  s E  jwe E
s

 Js  jw  e  j  E
w

 Js  jwe E
s

e  e  j 
w

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Wave Equation for Lossy Media
 E  w me E  0
2
2
Wave number:
Loss tangent, d
k  w me
s 

 w me 1  j

w
e


k   j
Phase constant
1
Attenuation constant
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2
101
Example 12 (1/2)
Plane wave propagation in lossy media:
complex number
ˆ oe
E  xE
ˆ
Hy
Eo

 jkz
  j
k  w me
;
e  jkz ;
  m
e
  e j
ˆ oe   z e  j  z
E  xE
ˆ
Hy
Eo

e
 z
e
 j z
ˆ
y
Eo

e  z e
 j   z  
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Example 12 (2/2)
ˆ oe
E (z;t )  xE
H (z;t )  yˆ
Eo

 z
e
cos(wt   z )
 z
cos(wt   z   )
attenuation
propagation
Plane wave is traveling along +z-direction and dissipating as it moves.
103
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Field Attenuation in Lossy Medium
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Attenuation and Skin Depth
Attenuation coefficient, , depends on the
conductivity, permittivity and frequency.
s 

k  w me 1  j

w
e


k   j
1
2
Skin depth, d is a measure of how far em wave can
penetrate a lossy medium
d 
O. Kilic EE542
1

105
Lossy Media
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Example 13
• Calculate the attenuation rate and skin
depth of earth for a uniform plane wave of
10 MHz. Assume the following properties
for earth:
m = mo
e = 4eo
s = 10-4
107
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Solution: Example 13
First we check if we can use approximate relations.
s
104 36 109
2
Slightly conducting



4.5

10
1
we 2 107
4
s m s mo s


  120  30  104  0.0094
2 e 2 4e o 4
d
1

 106.1 m
108
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References
• http://www.glenbrook.k12.il.us/GBSSCI/PH
YS/Class/waves/u10l1b.html
• Applied Electromagnetism, Liang Chi
Shen, Jin Au Kong, PWS
109
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Homework Assignments
• Due 9/25/08
110
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Homework 3.1
The magnetic field of a uniform plane wave traveling in free
space is given by
ˆ oe
H  xH
 jkz
1. What is the direction of propagation?
2. What is the wave number, k in terms of permittivity, eo
and permeability, mo?
3. Determine the electric field, E.
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Homework 3.2
• Find the polarization state of the following plane
wave:
E (r )   jxˆ  j 2 yˆ  e
jkz
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Homework 3.3
How far must a plane wave of frequency 60
GHz propagate in order for the phase of
the wave to be retarded by 180o in a
lossless medium with mr =1 and er = 3.5?
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O. Kilic EE542
Solution Homework 3.1
ˆ oe
H  xH
 jkz
1. What is the direction of propagation? Ans: -z
2. What is the wave number, k in terms of permittivity, eo
and permeability, mo?
Ans: free space  k  w
moe o
3. Determine the electric field, E.
E
ˆ  Hoo  e
E y
 jkz
;
o 
mo
eo
O. Kilic EE542
H
k
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Solution: Homework 3.2
E (r )   jxˆ  j 2 yˆ  e jkz
je
j 2
E (r ; t )   xˆ sin( wt  kz )  yˆ 2sin( wt  kz )
   xˆ  yˆ 2  sin( wt  kz )
Observe that orthogonal
components are in phase.
Linear Polarization
y
x
z
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O. Kilic EE542
Solution 3.3 (1/2)
Wavelength: period in space
kl = 2
 w me
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Solution 3.3 (2/2)
kz  



z  

k w me
2 f 3.5moe o
1
3 * 108


9
2f 3.5 moe o 2 * 60 * 10 3.5
1

m
400 3.5
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O. Kilic EE542