Download Chapter 1 - EM Waves

Chapter 1
Electromagnetic
Waves
WHY STUDY ??
• In ancient time
– Why do paper clip get attracted to rod rubbed with silk
– What causes lightening
– Why do colors get separated when light beam passes
through a prism
• Modern days
– How we receive TV/radio signals
– Why is radio signal good in some corner of the room but
not other
– Why do cell phones have signal fluctuation
2
Chapter Outlines
Chapter 1
Electromagnetic Waves
 Faraday’s Law
 Transformer and Motional EMFs
 Displacement Current
 Maxwell’s Equations
 Lossless TEM Waves
 EM Wave Fundamental and Equations
 EM Wave Propagation in Different Media
 EM Wave Reflection and Transmission at
Normal or Oblique Incidence
3
Introduction
In your previous experience in studying electromagnetic,
you have learned about and experimented with
electrostatics and magnetostatics … concentrating on
static, or time invariant electromagnetic fields (EM
Fields).
Henceforth, we shall examine situations where electric
and magnetic fields are dynamic or time varying !!
4
Introduction (Cont’d..)
Where :
• In static EM Fields, electric and magnetic fields are
independent each other, but in dynamic field both are
interdependent.
• Time varying EM Fields, represented by E(x,y,z,t) and
H(x,y,z,t) are of more practical value than static EM
Fields.
• In time varying fields, it usually due to accelerated
charges or time varying currents.
5
Introduction (Cont’d..)
In summary:

Stationary charges  electrostatic fields

Steady currents  magnetostatic fields
Time varying currents  electromagnetic fields or
waves

6
1.1 Faraday’s Law
According to faraday’s experiment, a static magnetic
field produces no current flow, but a time varying field
produces an induced voltage called electromotive force or
emf in a closed circuit, which causes a flow of current.
Faraday’s Law – the induced emf, Vemf in volts, in any
closed circuit is equal to the time rate of change of the
magnetic flux linkage by the circuit.
Faraday’s Law (Cont’d..)
Where,

Vemf  
t
The negative sign is a consequence of Len’z Law. If we
consider a single loop, Faraday’s Law can be written as:


Vemf  
   B  dS
t
t
Faraday’s Law (Cont’d..)
(a)
An increasing magnetic field out
of the page induces a current in
(a) or an emf in (b). (c) The
distributed resistance in a
continuous conductive loop can be
modeled as lumped resistor Rdist
in series with a perfectly
conductive loop.
(b)
(c)
Faraday’s Law (Cont’d..)
Generating emf requires a time varying magnetic flux
linking the circuit. This occurs if the magnetic field
changes with time ‘transformer emf’’ or if the surface
containing the flux changes with time ‘motional emf’.
The emf is measured around the closed path enclosing
the area through which the flux is passing, can be
written as:

Vemf   E  dL    B  dS
t
It is clear that in time varying situation, both E and B
are present and interrelated.
Example 1
Consider the rectangular loop moving with velocity u=uyay
in the field from an infinite length line current on the z
axis. Assume the loop has a distributed resistance Rdist.
Find an expression for the current in the loop including its
direction.
11
Solution to Example 1
First calculate the flux through the loop at an instant time,
0 I
B
a Remember ? a  al  a 
2
Where, a
unit vector along the line current
l
a
So,
unit vector perpendicular from the line
current to the field point
a  al  a   a z  a y  a x
0 I
0 I
B
a  
ax
2
2y
Solution to Example 1 (Cont’d..)
Arbitrarily choose dS in the +ax direction,
dS  dydza x
So the flux can be easily calculated as:
0 I
   B  dS  
dydz

2y
0 I

2
ya

y
b
dy
dz

y
0
0 Ib
ln  y  a   ln  y 

2
Solution to Example 1 (Cont’d..)
Then, we want to find how this flux changes with time,
0 Ib d
d
ln  y  a   ln  y 

dt
2 dt
By chain rule,
0 Ib  1
d
1  dy

 

dt
2   y  a  y  dt
By considering uy=dy/dt,
0 Iabu y
d

dt 2y y  a 
Solution to Example 1 (Cont’d..)
Our emf is negative of this, where:
Vemf  
0 Iabu y
2y y  a 
Since we considered dS in the +ax direction, our emf is taken
counterclockwise circulation. But since the emf is negative,
our induced current is apparently going in the clockwise
direction with value of:
I ind 
0 Iabu y
2y  y  a Rdist
1.2 Transformer and motional emf
The variation of flux with time as in previous equation
maybe caused in three ways:

By having a stationary loop in a time varying B field.
(transformer emf)

By having a time varying loop area in a static B field.
(motional emf)

By having a time varying loop area in a time varying
B field.
Transformer and motional emf
(Cont’d..)
• Stationary Loop in Time Varying B Field
This is the case where a stationary conducting loop is in
a time varying magnetic B field. The equation becomes:
B
Vemf   E  dL   
 dS
t
By applying Stokes Theorem in
the middle term, we obtain:
B
Vemf     E   dS     dS
t
Transformer and motional emf
(Cont’d..)
This leads us to the point or differential form of Faraday’s
Law,
B
E  
t
Based on this equation, the time varying electric field is
not conservative, or not equal to zero. The work done in
taking a charge about a closed path in a time varying
electric field, for example, is due to the energy from the
time varying magnetic field.
Transformer and motional emf
(Cont’d..)
• Moving Loop in static B Field
When a conducting loop is moving in a static B field, an
emf is induced in the loop. Recall that the force on a
charge moving with uniform velocity in magnetic field,
Fm  Qu B
So then, we define the
motional electric field Em,
Fm
Em 
 uB
Q
Transformer and motional emf
(Cont’d..)
If we consider a conducting loop moving with uniform
velocity u as consisting of a large number of free
electrons, the emf induced in the loop is:
Vemf   Em  dL   u  B  dL
L
• Moving Loop in Time Varying Field
The total emf would be:
B
Vemf   Em  dL   
 dS   u  B  dL
t
S
L
Example 2
The loop shown is inside a uniform magnetic field
B = 50 ax mWb/m2 . If side DC of the loop cuts the flux
lines at the frequency of 50Hz and the loop lies in the yz
plane at time t = 0, find the induced emf at t = 1 ms.
21
Solution to Example 2
Since the B field is time invariant, the induced emf is
motional, that is:
Vemf   Em  dL   u  B  dL
L
Where,
dL  dL DC  dza z
u
dL moving loop
dt

d
dt
a  a
  4 cm,   2f  100
Solution to Example 2 (Cont’d..)
As u and dL is in cylindrical coordinates, transform B
field into cylindrical coordinate (Chapter 1 in
Electromagnetic Theory !! ):

B  B0a x  B0 cos a  - sin a

Where B0 = 0.05 , therefore:
a
uB 
0
B0 cos 
a

 B0 sin 
az
0   B0 cos a z
0
Solution to Example 2 (Cont’d..)
And
u  B   dL   B0 cos dz  0.04100 0.05cos dz
 0.2 cos dz
0.03
Vemf 
  0.2 cos dz  6 cos 
mV
z 0
To determine  recall that,

at
d
dt
   t  C
t  0,   
2
C is constant
because the loop is in the yz plane!
Solution to Example 2 (Cont’d..)
Hence,
  t 

2
Therefore,


Vemf  6 cos   6 cos t  
2

 6 sin 100t  mV
So that at t = 1 ms,
Vemf  6 sin 100 (0.001)  5.825 mV
1.3 Displacement Current
We recall from Ampere’s Circuital Law for static field,
  H  Jc
‘c’ subscript is used to identify it as a conduction current
density, which related to electric field Ohm’s Law by:
J c  E
But divergence of curl of a vector is identically zero,
    H  0    J
Displacement Current (Cont’d..)
The current continuity equation,
 v
  Jc  
t
We see that the static form of Ampere’s Law is clearly
invalid for time varying fields since it violates the law of
current continuity, and it was resolved by Maxwell
introduction which what we called displacement current
density,
  H  Jc  Jd
Where Jd is the rate of change
of the electric flux density,
D
Jd 
t
Displacement Current (Cont’d..)
The insertion of Jd was one of the major contribution of
Maxwell. Without Jd term, electromagnetic wave
propagation (e.g. radio or TV waves) would be
impossible. At low frequencies, Jd is usually neglected
compared with Jc. But at radio frequencies, the two
terms are comparable.
Therefore,
D
  H  Jc 
t
Displacement Current (Cont’d..)
By applying the divergence of curl, rearrange, integrate
and apply Stoke’s Theorem, we can get the integral
form of Ampere’s circuital Law:

 H  dL   J c  dS  t  D  dS  ic  id
Do you really understand this displacement current??
Only formula and formula…????
Displacement Current (Cont’d..)
To have clear understanding of displacement current,
consider the simple capacitor circuit of figure below.
A sinusoidal voltage source is applied to the capacitor, and
from circuit theory we know the voltage is related to the
current by the capacitance. i(t) here is the conduction current.
Displacement Current (Cont’d..)
Consider the loop surrounding the plane surface S1. By static
form of Ampere’s Law, the circulation of H must be equal to
the current that cuts through the surface. But, the same
current must pass through S2 that passes between the plates
of capacitor.
Displacement Current (Cont’d..)
But, there is no conduction current passes through an ideal
capacitor, (where J=0, due to σ=0 for an ideal dielectric ) flows
through S2. This is contradictory in view of the fact that the
same closed path as S1 is used.
But to resolve this conflict, the current passing through S2
must be entirely a displacement current, where it needs to be
included in Ampere’s Circuital Law.

 H  dL   J d  dS  t
S2
Q
 D  dS  t  I   J  dS
S2
S1
So we obtain the same current for either surface though it is
conduction current in S1 and displacement current in S2.
Displacement Current (Cont’d..)
Jc
The ratio of conduction current magnitude to the displacement
current magnitude is called loss tangent, where it is used to
Jd
measure the quality of the dielectric
good
dielectric
will
have
Ji
very low loss tangent.
Other example for physical meaning:
Jc
i
Jd
Ji
mi
Ji = current source
Jc= conducted current through resistor
Jd=displacement current through dielectric
material
i
md
mi= magnetic current source
md=displacement magnetic current
1.4 Maxwell Equations
Below is the generalized forms of Maxwell Equations:
Maxwell Equations
Point or
Differential Form
Gauss’s Law
  D  v
Gauss’s Law for
Magnetic Field
B  0
Faraday’s Law
Ampere’s Circuital
Law
B
E  
t
D
  H  Jc 
t
Integral Form
 D  dS  Qenc
 B  dS  0

 E  dL   t  B  dS

 H  dL   J c  dS  t  D  dS
Maxwell Equations (Cont’d..)
It is worthwhile to mention other equations that go
hand in hand with Maxwell’s equations.
Lorent’z Force Equation
F  qE  u  B
Constitutive Relations
v
J  
t
Current Continuity Equation
D  E
B  H
J  E
Maxwell Equations (Cont’d..)
Circuit - Field relations:
Field Relation
Circuit Relation
Jc   E
1
i  V  GV
R
BH
  Li
H
md  
dt
i
VL  L
t
D  E
qe  C V
e  
E
dt
Vc
ic  C
t
1.5 Lossless TEM Waves
Let’s use Maxwell’s equations to study the relationship
between the electric and magnetic field components of an
electromagnetic wave.
Consider an x-polarized wave propagating in the +z direction
in some ideal medium characterized by µ and ε, with σ = 0.
Electromagnetic wave as x-polarized means that the E field
vector is always pointing in the x or –x direction. Choose σ = 0 to
make medium lossless for simplicity, as given by:
Ez, t   E0 cost  z a x
Lossless TEM Waves (Cont’d..)
A plot of the equation E(z,0) = E0cos(z)ax at 10 MHz in
free space
with E0 = 1 V/m.
Fundamentals of Electromagnetics
With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Lossless TEM Waves (Cont’d..)
Upon application of Maxwell equations, we would also
find the magnetic field propagates in the +z direction, but
the field is always normal or perpendicular to the electric
field vector  the wave is said to propagate in a
transverse electromagnetic wave mode, or TEM.
TEM Waves has no E field or H field components along
the direction of propagation.
We can apply Faraday’s Law,   E  
B
H
 
t
t
to the
propagating electric field equation previously.
Solve the right hand side and the left hand side to get H !!
Lossless TEM Waves (Cont’d..)
Where,
ax
ay


E 
x
y
E0 cost  z  0
az

z
0
 

   E0 cost  z a y  E0 sin t  z a y
 z

H
 
 E0 sin t  z a y
t
E0
 dH    sin t  z a y dt

E0
H
cost  z a y  C

Lossless TEM Waves (Cont’d..)
We could see that the time varying E is the only source
of H, if no conduction current given that can also
generate H  Thus C must be zero.
The amplitudes of E and
H are related by Maxwell
equations.
Plot of the equation H(z,0)
= (E0/) cos(–z)ay at 10
MHz in free space with
E0 = 1 V/m along with the
lighter plot of E(z,0).
Lossless TEM Waves (Cont’d..)
We can apply Ampere’s Circuital Law,   H  J c 
propagating magnetic field equation previously.
D to the
t
By considering lossless characteristic, taking the curl
of H, equating the equation and then integrate it, we
could get :
   
-Solution-
And propagation velocity relation:

up 

to get
Try
this!!!
up 
1

Example 3
Suppose in free space that:
E(z,t) = 5.0 e-2zt ax V/m.

Is the wave lossless?

Find H(z,t).
43
Solution to Example 3
Since the wave has an attenuation term (e-2zt) it is clearly
not lossless.
To find H,
ax
H
  E   o
 
x
t
5e2 zt
ay
az

    5e2 zt  a  10te2 zt a
y
y
y
z
z
0
0
Therefore,
10t 2 zt
10
dH 
e dta y , H =  te 2 zt dta y
 o
o
Solution to Example 3 (Cont’d..)
This integral is solved by parts
where we let
  udv  uv   vdu 
u  t and dv  e
2 zt
dt.
We arrive at:
 10t 2 zt
10 2 zt 
A
H
e 
e ay
2
4 o z
m
 2 o z

1.6 EM Wave Fundamental
and equations
In free space, the constitutive parameters are σ = 0, µr = 1,
εr = 1, so the Ampere’s Law and Faraday’s Law equations
become :
B
H
E  
   E   0
t
t
D
E
  H  Jc 
   H  0
t
t
If there is some point in space a source of time varying E field,
a H field is induced in the surrounding region. As this H field
also changing with time, it in turn induces an E field. Energy
is pass back and forth between E and H fields as they radiate
away from the source at the speed of light.
EM Wave Fundamental and
equations (Cont’d..)
The EM waves radiates spherically, but at a remote distance
away from the source they resemble uniform plane wave. In a
uniform plane wave, the E and H fields are orthogonal, or
transverse to the direction of propagation ( to propagate in
TEM mode ).
EM Wave Fundamental and
equations (Cont’d..)
We will briefly review some of the fundamental features
of waves before employing them in the study of
electromagnetic. Consider time harmonic waves,
represented by sine waves, rather than transient waves
(pulses or step functions), generally as:
Ez, t   E0e z cost  z   a x
The E field is a function of position (z) and time (t). It is
always pointing in +x or –x direction  x-polarized
wave.
EM Wave Fundamental and
equations (Cont’d..)
Where,
E0
E0 e



Initial amplitude at z = 0
z
exponential terms  attenuation
Angular frequency
Phase constant
Phase shift
And important relations:
  2f
2 1
T

 f

2

dz 
up 
  f
dt 
EM Wave Fundamental and
equations (Cont’d..)
E(0,t) = Exax = E0cos(t)ax.
E(0,t) =Exax = E0cos(t + )ax.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
E(z, 0) = E0cos(–z)ax.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
Copyright © 2005 by John Wiley & Sons. All rights reserved.
E(z, 0) = E0e–zcos (–z)ax.
EM Wave Fundamental and
equations (Cont’d..)
Use Maxwell’s equations to derive formulas governing EM
wave propagation.
Consider that the medium is free of any charge, where:
D  0
And linear, isotropic, homogeneous, and time invariant
(simple media), whereby the Maxwell’s equation can be
rewritten as :
E
H
  H  E  
,   E  
t
t
E  0 , H  0
EM Wave Fundamental and
equations (Cont’d..)
Take curl of both sides of Faraday’s Law,     E       

Consider position derivative acting on a time derivative in a
homogeneous material,
    E    

  H 
t
Exchange the Faraday’s Law for the curl of H

E 
E
 2E
    E   E  
 
  
t 
t 
t
t 2
Invoking a vector identity,     A    A   2 A
H 

t 
EM Wave Fundamental and
equations (Cont’d..)
So now we have :
2

E

E
2
  E   E   
 
t
t 2
But, medium is charge free, so divergence of E is zero,
2

E

E
2
 E  
 
t
t 2
Helmholtz Wave
Equation for E.
This can be broken up into three vector equations ( Ex, Ey and Ez)
Task 1
By starting with Maxwell’s equations for
simple and charge free media, derive the
Helmholtz
Wave
Equation
for
magnetic
field, H
Solve this! Please submit the solution after class!
-Solution54
EM Wave Fundamental and
equations (Cont’d..)
But, with interest on Helmholtz equations for time
harmonic fields, the previous Helmholtz wave equation
becomes :
 E s  j   j E s
2
because
E s
 jE s
t
Generally written in the form:
 2E s   2E s  0
Where the propagation constant, γ, with real part is attenuation
and an imaginary part is phase constant,
  j (  j )    j
EM Wave Fundamental and
equations (Cont’d..)
The value of
and β in terms of the material’s
constitutive parameters:


2
 
 

 
1 
  1

2 

  




2
 
 

 
1 
  1

2 

  


EM Wave Fundamental and
equations (Cont’d..)
The Helmholtz wave
magnetic fields, H :
equation
for
time
harmonic
 Hs   Hs  0
2
2
With loss of generality, we assume that x polarized traveling
in the +az direction, where Es has only an x component with
function of z, since for plane wave that the fields do not vary
in transverse direction, where in this case is xy plane. Then :
E s  E xs ( z )a x
EM Wave Fundamental and
equations (Cont’d..)
By substituting this equation into the Helmholtz wave equation
for E field, it becomes:
2 -  2 E xs z   0
Hence,
 2 E xs ( z )
x 2

 2 E xs ( z )
y 2

 2 E xs ( z )
z 2
  2 E xs ( z )  0
With first and second term is zero :
 2 E xs ( z )
z 2
2



2
2
  E xs ( z )  0  
   E xs ( z )  0
2
 z

EM Wave Fundamental and
equations (Cont’d..)
This is a scalar wave equation, a linear homogeneous
differential equation. A possible solution for this equation is :
2

E

Esx

z
sx

z
If we let, Esx  Ae
then,
 Ae ,
 2 Aez
z
z 2
 2

2
So, this equation,  2    E xs ( z )  0 becomes:
 z

2   2  0, or         0
Which has two solutions,
(1)     0,     , Esx  Aez , or Esx  E0 e z
(2)     0,     , Esx  Aez , or Esx  E0ez
EM Wave Fundamental and
equations (Cont’d..)
The general solution is linear superposition of these two:
E xs ( z )  E0 e z  E0 e z
For the E0 represents the E field amplitude of +z traveling
wave at z=0, by reinserting the vector, multiply by time factor e jt
apply Euler’s identity and use the propagation constant to get :
Ez, t   E0 e z cost  z a x
EM Wave Fundamental and
equations (Cont’d..)
For the E0 represents the E field amplitude of negative z
traveling wave at z=0, and similarly by previous approach to get :
Ez, t   E0 e z cost  z a x
The general instantaneous solution is superposition of these two
solutions above.


E s  E0 e z  E0 e z a x
 z
Ez, t   E0 e cost  z a x
or generally as :
 z
 E0 e
cost  z a x
EM Wave Fundamental and
equations (Cont’d..)
The magnetic field can be found by applying Faraday’s Law :
  E s   jH s
Evaluating the curl of E to find:


  Es  - E0 ez  E0ez a y
We can solve for H ,
 E0 z E0 z 
Hs  
e

e a y
 j

j



If we assume that the wave propagates along +az and Hs has only an
y component, as what we had assume for Es
H s  H ys ( z)a y
EM Wave Fundamental and
equations (Cont’d..)
We would have been led to the expression,


H s  H 0 ez  H 0ez a y
By making comparison, we can find a relationship between E0


and H 0 , or E0 and H 0

E0
H 0

j

intrinsic impedance (in ohms)
j

   n   e j n
  j
 
E0
H 0
tan 2 


EM Wave Fundamental and
equations (Cont’d..)
By plugged in the equation of intrinsic impedance into equation of
magnetic field, we could get:
E0 z
Hz, t  
e
cost  z   n a y

Where E and H are out of phase by θn at any instant of time due to
the complex intrinsic impedance of the medium. Thus at any time,
E leads H or H lags E by θn . The ratio of magnitude of conduction
current density to displacement current density in a lossy medium
is:
Jc
Jd
E s



 tan 
jE s
j
loss tangent
EM Wave Fundamental and
equations (Cont’d..)
Tan δ is used to determine
how lossy a medium is, i.e.
• Good (lossless or perfect)
dielectric if
tan   1,   
• Good conductor if
tan   1 ,   
EM Wave Fundamental and
equations (Cont’d..)
Basically, we can use Fleming’s Left Hand Rule to
determine the E, H and propagation direction ;
Propagation direction
(first finger)
H direction (second
finger)
E direction (thumb)
EM Wave Fundamental and
equations (Cont’d..)
By knowing the EM wave’s direction of propagation, given
as unit vector ap, is the same as the cross product of Es
with unit vector, aE and Hs with unit vector aH :
ES  H S
aP 
ES  H S
And also, a pair of simple formulas can be derived:
HS 
1
a P  ES

E S  a P  H S
In addition to
properties of
cross product,
a P  aE  a H
a H  a P  aE
 aE  a P  a H
EM Wave Fundamental and
equations (Cont’d..)
Representation of waves; in (a), the wave travels in the ap = +az
direction and has Es = E0+ e–z ax and Hs = (E0+/)e–z ay. In (b),
the wave travels in the ap = –az direction and has Es = E0–ez ax
along with Hs = –(E0–/)ez ay.
Example 4
Suppose in free space :
H(x,t) = 100 cos(2π x 107t – βx + π/4) az mA/m.
Find E(x,t).
69
Solution to Example 4
We could find:
H s  0.100e  j x e j a z , a P  a x ,
   4 
E s   a P  H s  120 a x  0.100e  j  x e j a z  12 e  j  x e j a y
So then,
E  12 cos t   x    a y
Since free space is stated,
2
2


 2 30 rad m
 c f
2


V


7
and then E  12 cos  2 x10 t 
x  ay
30
4 m

Example 5
Suppose:
E (x,y,t) = 5 cos(π x 106t – 3.0x + 2.0y) az V/m.
Find :

H (x,y,t).

The direction of propagation, ap
71
Solution to Example 5
We could find:
E s  5e  j 3 x e j 2 y a z
Assume nonmagnetic material and therefore have:
 Es   jHs  j10e j 3x e j 2 ya x  j15e j 3x e j 2 ya y
Hs 
j10  j 3 x j 2 y
j15  j 3 x j 2 y
e e ax 
e e a y  2.53e j 3 x e j 2 y a x  3.8e  j 3 x e j 2 y a y
 jo
 j
So that,
H( x, y, t )  2.53cos  x106 t  3x  2 y  a x  3.80 cos  x106 t  3x  2 y  a y
A
m
Solution to Example 5 (Cont’d..)
The direction of propagation :
Es  H s
aP 
Es  H s
Where,
Es  Hs  19e j 6 x e j 4 ya x 12.65e j 6 x e j 4 y a y
And with the exponential terms canceling in the top and
bottom of the equation for ap, we have:
a p  0.83e

 j6x j4 y 
e ax  0.55e e a y
 j6x j4 y
1.7 EM Wave Propagation In
Different Media
We will consider time harmonic field propagating in
different types of media : Lossless, Charge-Free
 Dielectrics
 Conductors
• Lossless, Charge - Free
Charge free, ρv=0, medium has zero conductivity, σ=0. This is
the case where waves traveling in vacuum or free space (free
of any charges). Perfect dielectric is also considered as
lossless media.
EM Wave Propagation In
Different Media (Cont’d..)
Evaluate the propagation constant,
  j (  j )    j
So,

Where,
j (0  j ) 
j 2 2   j     j
  0 ,    
Since   0 , the signal does not attenuate as it travels
 lossless medium.
The propagation velocity, u p 

1



EM Wave Propagation In
Different Media (Cont’d..)
Evaluate the intrinsic impedance,
j


0  j


For lossless materials, E and H are always in phase. Again,

 r 0

 r 0
r
0
r
 0  120 
intrinsic impedance of free space
Example 6
In a lossless, nonmagnetic material with :
εr = 16, and H = 100 cos(ωt – 10y) az mA/m.
Determine :

The propagation velocity

The angular frequency

The instantaneous expression for the electric
field intensity.
77
Solution to Example 6
The propagation velocity:

c
3x108
m
up  

 0.75x108

s
r
16
The angular frequency:
  u p    0.75 x108  10   7.5 x108
rad
s
From given H field :
mA
H( y, t )  100 cos  7.5 x10 t  10 y  a z
m
8
Solution to Example 6 (Cont’d..)
So, the time harmonic H field is:
H s  0.100e j y a z ,
Where, Es  a P  H s 
120
r
a y  0.100e j y a z  3 e j y a x
Finally, the instantaneous expression for E field is:
V
E( y, t )  9.4 cos  7.5 x10 t  10 y  a x
m
8
EM Wave Propagation In
Different Media (Cont’d..)
• Dielectric
Treating a dielectric as lossless is often a good
approximation, but all dielectrics are to some degree lossy
 finite conductivity, polarization loss etc. With finite
conductivity, the E field gives rise to conduction current
density results in power dissipation.
Thus, it will give a complex permittivity, complex
propagation constant with attenuation constant greater
than zero. The intrinsic impedance is also complex,
resulting a phase difference between E and H fields.
EM Wave Propagation In
Different Media (Cont’d..)
• Conductor
In any decent conductor, the loss tangent, σ/ωε>>1 or σ>>ωε
so that σ ≈ ∞, so that:
,   

2
 


 
1 
  1

2 
  



where interior
brackets becomes:
Therefore,
  

2
 f


EM Wave Propagation In
Different Media (Cont’d..)
The intrinsic impedance approximated by:

By considering
j

  j
j

1  j leading to:
j
2

 j 45

(1  j ) 
e
2

0
and also..

2
up  



2
f
E leads H by 450
EM Wave Propagation In
Different Media (Cont’d..)
A wave from air (free space)
penetrates rapidly in a good
conductor, with wavelength clearly
much shorter.
A large attenuation means the fields
cannot penetrate far into the conductor.
In a good conductor, the large attenuation means the
penetration depth can be quite small, confining the fields near
the surface or skin, of the conductor  skin depth.
1
1
 

f
Example 7
In a nonmagnetic material,
E(z,t) = 10 e-200z cos(2π x 109t - 200z) ax mV/m.
Find H(z,t)
84
Solution to Example 7
Since
have:
= β, the media is a good metal, with µr = 1 we

2
 200 
2
S
   f o , or  

 10.13
9
7
 f o  1x10  4 x10 
m
We could also find the intrinsic impedance,
 j 45
j 45
  2 e  28e 

Solution to Example 7 (Cont’d)
So, to calculate H,
Es  10e
 z  j  z
e
ax , Hs 
1

a P  Es 
1

a z 10e
 z  j  z
e
ax 
10

e  z e  j  z a y
The instantaneous expression for the magnetic field intensity.
H( z, t )  360e
200 z
mA
cos  2 x10 t  200 z  45  a y
m
9
1.8 EM Wave Reflection &
Transmission at Normal Incidence
What happens when a EM wave is incident on a different
medium? E.g. Light wave incident with mirror, most of it
gets reflected but a portion gets transmitted (rapidly
attenuating in the silver backing of the mirror.
Consider a plane wave that are normally incident which
means the planar boundary separating the two media is
perpendicular to the wave’s propagation direction.
Generally, consider a time harmonic x-polarized electric
field incident from medium 1 (µr1, εr1, σr1) to medium 2 (µr2,
εr2, σr2)
EM Wave Reflection & Transmission
at Normal Incidence (Cont’d..)
With incident field:
Ei ( z, t )  E0i e  1 z cost  1z a x
EM Wave Reflection & Transmission
at Normal Incidence (Cont’d..)
We have the following sets of equations:
E is  E0i e 1z e  j1z a x
H is 
E0i
1
e 1z e  j1z a y
E rs  E0r e1 z e j1z a x
H 
r
s
E
r
0
1
Reflected Fields
e1z e j1z a y
E ts  E0t e  2 z e  j 2 z a x
H 
t
s
E0t
2
e
 2 z  j 2 z
e
Incident Fields
E0i , E0r , E0t The E field
intensities at z=0
Transmitted Fields
ay
EM Wave Reflection & Transmission
at Normal Incidence (Cont’d..)
The boundary conditions:
Et1  Et 2
a 21  H1  H 2   K  Ht1  Ht 2
Applying these boundary conditions at z=0 to get:
r  2  1 i
E0 
E0  E0i ,
 2  1
 2  1 E0r


 2  1 E0i
t
E
2

2

2 E i  E i ,  
2  0
E0t 
0
0
 2  1
 2  1 E0i
Try this!!
Reflection
Coefficient
Transmission
Coefficient
EM Wave Reflection & Transmission
at Normal Incidence (Cont’d..)
By comparison,
 1  
Consider a special case when medium 1 is a perfect dielectric
(lossless,σ1=0) and medium 2 is a perfect conductor (σ2= ∞).
For this case, 2  0,   1,   0 showing that the wave
is totally reflected  fields in perfect conductor must vanish, so
there can be no transmitted wave, E2 = 0.
The totally reflected wave combines with the incident wave to
form a standing wave  it stands and does not travel, it
consists of two traveling waves Ei and Er of equal amplitudes
but in opposite directions.
EM Wave Reflection & Transmission
at Normal Incidence (Cont’d..)
Standing wave pattern for an incident wave in a lossless
medium reflecting off a second medium at z=0 where  = 0.5.
Emax 1  
SWR 

Emin 1  
Example 8
A uniform planar waves is normally incident from
media 1 (z < 0, σ = 0, µr = 1.0, εr = 4.0) to media 2
(z > 0, σ = 0, µr = 8.0, εr = 2.0).
Calculate
the
reflection
and
transmission
coefficients seen by this wave.
93
Solution to Example 8
The reflection coefficient ;
2  1
 120

8

; 1 

 60, 2 
 120
 240
2  1


2
4
This leads to:
240  60 3

  0.60
240  60 5
and the transmission coefficient,
  1    1.60
Example 9
Suppose media 1 (z < 0) is air and media 2 (z > 0)
has εr = 16.
The transmitted magnetic field
intensity is known to be:
Ht = 12 cos (ωt - β2z) ay mA/m.
Determine the instantaneous value of the incident
electric field.
95
Solution to Example 9
We know that,
t
E
mA
mA
 j 2 z
t
o  j 2 z
H s  12e
ay

e
ay
m
2
m
From transmitted H field, we could find the transmitted
E field,
Eot
mA t
V
V
 j 2 z
t
2  30, so
 12 , Eo  0.36 , and  Es  1.13e a x
2
m
m
m
Solution to Example 9 (Cont’d..)
Since we know the relation between transmitted E field
and incident E field,
2  1
3
2
E   E  1    E ;  
  ,   1  
2  1
5
5
t
o
Eoi 
i
o
Eot

Thus,
i
o
 2.83, so  Eis  2.83e j1z a x
V
 E( z, t )  2.83cos t  1 z  a x .
m
1.9 EM Wave Reflection &
Transmission at Oblique Incidence
A uniform plane waves
traveling
in the ai
direction is obliquely
incident from medium 1
onto medium 2.
Plane of incidence  plane
containing both a normal to
the
boundary
and
the
incident’s wave propagation.
In figure, the propagation direction is ai and the normal is az, so the
plane incidence is the x z plane. The angle of incidence, reflection
and transmission is the angle that makes the field a normal to the
boundary.
EM Wave Reflection & Transmission
at Oblique Incidence (Cont’d..)
When EM Wave in plane wave form obliquely incident on the
boundary, it can be decomposed into:
Perpendicular Polarization, or transverse electric (TE)
polarization  The E Field is perpendicular or transverse to

the plane of incidence.
Parallel Polarization, or transverse magnetic (TM)
polarization  The E Field is parallel to the plane of

incidence, but the H Field is transverse.
We need to decompose into its TE and TM components separately, and
once the reflected an the transmitted fields for each polarization
determined, it can be recombined for final answer.
EM Wave Reflection & Transmission
at Oblique Incidence (Cont’d..)
• TE Polarization
EM Wave Reflection & Transmission
at Oblique Incidence (Cont’d..)
For TE polarization, the fields are summarized as follows:
Eis  E0i e  j1  x sin  i  z cos i a y
H is
Incident Fields
E0i  j1  x sin  i  z cos i 
 cos ia x  sin ia z 

e
1
E rs  E0r e  j1  x sin  r  z cos r a y
H rs 
E0r
1
e  j1  x sin  r  z cos r  cos  r a x  sin  r a z 
Ets  E0t e  j 2  x sin  t  z cos t a y
H ts 
E0t  j 2  x sin  t  z cos t 
 cos 
e
2
Reflected Fields
Transmitted Fields
ta x
 sin  t a z 
EM Wave Reflection & Transmission
at Oblique Incidence (Cont’d..)
By applying Snell’s Law,
i   r
1 sin  t

 2 sin  i
Try to solve this!!!
We could get:
r  2 cos  i  1 cos t i
E0 
E0  TE E0i
 2 cos i  1 cos t
t
E0 
2 2 cos i
i
i
E0   TE E0
1 cos t   2 cos i
 TE 1  TE
EM Wave Reflection & Transmission
at Oblique Incidence (Cont’d..)
• TM Polarization
EM Wave Reflection & Transmission
at Oblique Incidence (Cont’d..)
By similar geometric arguments, TM fields are:
Eis  E0i e  j1  x sin  i  z cos i  cos i a x  sin i a z 
Incident Fields
i
E
H is  0 e  j1  x sin  i  z cos i a y
1
E rs  E0r e  j1  x sin  r  z cos r  cos  r a x  sin  r a z 
H rs  
E0r
1
e  j1  x sin  r  z cos r a y
Ets  E0t e  j 2  x sin  t  z cos t  cos t a x  sin t a z 
H ts 
E0t  j 2  x sin  t  z cos t 
e
a
2
y
Reflected Fields
Transmitted Fields
EM Wave Reflection & Transmission
at Oblique Incidence (Cont’d..)
By applying Snell’s Law and employing the boundary
conditions, we could get the following expressions
relating the field amplitudes:
r  2 cos  t  1 cos  i i
i
E0 
E0  TM E0
 2 cos t  1 cos i
E0t 
2 2 cos i
E0i   TM E0i
1 cos i   2 cos t
cos i
 TM  1  TM 
cos t
EM Wave Reflection & Transmission
at Oblique Incidence (Cont’d..)
For TM polarizations, there exists an incidence angle at
which all of the wave is transmitted into the second
medium  Brewster Angle, θi = θBA , where:
sin  BA 
 22 (22 12 )
2 2
2 2
2 1 1  2
sin  BA 
1
r
1
r
1
2
When a randomly polarized wave such as light is incident on
a material at the Brewster angle, the TM polarized portion is
totally transmitted but a TE component is partially reflected.
Example 10
A 100 MHz TE polarized wave with amplitude
1.0 V/m is obliquely incident from air (z < 0)
onto a slab of lossless, nonmagnetic material
with εr = 25 (z > 0). The angle of incidence is
40. Calculate:
(a) the angle of transmission,
(b) the reflection and transmission coefficients,
(c) the incident, reflected and transmitted for E
fields.
107
Solution to Example 10
(a)
1 

c

2 100 x106 
3x108
 r
rad
rad
 2.09
, 2 
 10.45
.
m
c
m
1
1
1
1

 ; sin t  sin 40 ; t  7.4
2
r2 5
5
(b)
120
1  120; 2 
  24
25
2 cos i  1 cos t
TE 
 0.732;  TE  1  TE  0.268
2 cos i  1 cos t
Solution to Example 10 (Cont’d)
(c)
For incident field:
E  1e
i
s

 j 2.09 x sin 40  z cos 40
a  1e j1.34 x e j1.60 z a V
y
y
m
Thus,
V
E ( z, t )  1cos t  1.34 x  1.60 z  a y
m
i
For reflected field:
Eor  TE Eoi  0.732
Solution to Example 10 (Cont’d)
Leading to:
E   0.732  e
r
s
 j1.34 x  j1.60 z
Thus,
e
V
ay
m
V
E ( z , t )  0.732 cos t  1.34 x  1.60 z  a y
m
r
Finally for transmitted field:
Eot   TE Eoi  0.268
Solution to Example 10 (Cont’d)
To get:
E  0.268e
t
s
 j  2  x sin t  z cost 
a y  0.268e
 j1.35 x  j10.4 z
e
Therefore,
V
E ( z , t )  0.268 cost  1.35 x  10.4 z a y
m
r
V
ay
m
Electromagnetic
Waves
End