Download Chapter5_Final.doc

*Note: The information below can be referenced to: Iskander, M., Electromagnetic
Fields and Waves, Waveland Press, Prospect Heights, IL, 1992, ISBN: 1-57766-115-X.
Edminister, J., Electromagnetics (Schaum’s Outline), McGraw-Hill, New York, NY,
1993, ISBN: 0-07-018993-5. Hecht, E., Optics (Schaum’s Outlines), McGraw-Hill, New
York, NY, ISBN: 0-07-027730-3
Chapter 5
Normal Incidence Plane Wave Reflection and Transmission at Plane at Plane
Boundaries
5.1 Introduction
Why is there a need to study the reflection and transmission properties of plane waves
when incident on boundaries between regions of different electric properties? Perhaps
you had no idea that we experience this topic daily in our lives. For instance, when you
try and make a call on your cell phone and you are downtown amongst all those tall
buildings. Will you always have great reception? When the hot sun penetrates your
window it can quickly heat up your room, but maybe you have blinds, curtains, or tinted
material to prevent some of that intense heat. For those of you that wear glasses, you
know what happens when you get your picture taken; that annoying glare from those
glasses. What about a light ray on the surface of a mirror? A reflection can be seen and
some of that ray will penetrate the glass.
This chapter focuses on the reflection and transmission properties related to onedimensional problems that have normal-incident plane waves converging on infinite
plane interfaces that will separate two or more different media.
The Figure 5.1 is the fundamental concept. It illustrates the geometry of the positive z
propagating plane wave that is normally incident on a plane interface between regions 1
& 2.
Region 1
E
i
Hi
Region 2
 ,  , 
1
t
E
1
1
X
Er
O
Ht
 ,  , 
2
2
2
ax
Y
Z
Hr
Figure 5.1
Notes by: Debbie Prestridge
1
5.2 Normal Incidence Plane Wave Reflection and Transmissions at Plane
Boundary Between Two Conductive Media
The electric and magnetic fields related to the incident wave are given by the following:

 ix   m 1e  1 z

 i   m1 e 1 z

y
1
(5.1)
* Note: (i) incident, (m1) medium 1, (γ1) propagation constant in region 1, (η1)
wave impedance in region 1, (z) direction of propagating wave
The complex propagation constant in region 1 is   1  j 1 . *Note: α and β are the
real and imaginary parts respectively. The propagation constant γ is that square root of γ2
whose real and imaginary parts are positive:
γ = α + jβ
With
 
 
2

 

1


1



  
2 

 
 
2

 
1     1
  
2 

The wave impedance as defined in chapter 2 as the ratio between the electric and
magnetic fields is

1
  
j tan 1 

 x


  
2
  

e
1


2
4


y
  j      

  1    
 

The wave impedance η in a conductive medium is a complex number meaning that the
electric and magnetic fields are not in phase. The phase velocity will be less than the
velocity of light vp < c. The wavelength λ in the conductive medium will be shorter than
the wavelength λo in free space at the same frequency, λ = 2π/β < λo. The factor e  z
will attenuate the magnitudes of both E and H as they propagate in the +z direction.
Notes by: Debbie Prestridge
2
t1 = 0
t2 > t 1
e-αz
Figure 5.2 The electric field associated with a plane wave
propagating along the positive z direction.
What happens when this wave hits the boundary?
Some of the energy related to the incident wave will transmit across the boundary surface
at z = 0 in region 2, therefore providing a transmitted wave in the +z direction in medium
2. The following are the electric and magnetic fields related to the transmitted wave:

 tx   m 2 e  2 z
 m 2 2 z
t

y 
e
 2
(5.2)
* Note: (t) transmitted wave
Recall Maxwell’s equations:
      j 
    j
 ·E=0
 ·H=0
The Wave equation for H:
 2   2  2


  2
 2 x2  y2  z2
For now, let’s look at the simplest system, that consisting of a plane wave of coordinate
z.
d 2
2
2  
dz
Notes by: Debbie Prestridge
3
Therefore, according to the wave equation as noted above, equations (5.1) & (5.2) satisfy
Maxwell’s equations.

If the amplitude of the transmitted wave  m2
is unknown then boundary conditions at the
interface z = 0 separating the two media must be satisfied.
Good conductors are often treated as if they were perfect conductors. Metallic
conductors such as copper have a high conductivity σ = 6 * 107 S/m, however, only
superconductors have infinite conductivity and are truly perfect conductors.
 Static (time independent)
n · D1 = ρs
n · (B1 – B2) = 0
n  1 = 0
n  ( 1   2 ) = 0
* Subscripts denote the conducting medium.
Characteristics of static cases:
1. Electrostatic field inside a good conducting medium is zero. Free charge
can exist on the surface of a conductor, thus making the normal
component of D discontinuous being zero inside the conductor and
nonzero outside. The tangential component of E just inside the conductor
must be zero even if the surface is charged.
2. The electric and magnetic fields in the static case are independent. A
static magnetic field can therefore exist inside a metallic body, even
though an E field cannot. The normal component of B and the tangential
components of H are therefore continuous across the interface.
For time-varying fields, the boundary conditions for good (perfect)
conductors are:

Time –varying fields (time dependent)
n ·D = 0
n·B=0
n = 0
n   = Js
The subscripts have been deleted because in this case the only nonvanishing fields are
those outside the conducting body.
 tx &  ix are tangential to the interface therefore the boundary conditions will require
these fields be equal at z = 0. Equate  t &  i , and set z = 0. *Note: (t) is
x
x
transmitted wave, (i) incident wave. The result will be
 m 1   m 2
*Note: (m) medium, (+) transmitted wave
Notes by: Debbie Prestridge
(5.3)
4
 ty &  iy are also tangential to the interface, so by applying the same procedure as
above you will notice that it is impossible to satisfy the magnetic field boundary
conditions if  1   2 .
We can then, include a reflected wave in region 1 traveling away from the interface,
or in other words in the –z direction. Only part of the energy related to the incident
wave will be transmitted to region 2 because of the process the incident fields must
encounter prior to crossing the boundary. The fields left behind during this process
will in fact be the reflected wave.
The electric and magnetic fields related to the reflected wave are

 rx   m 1e  1 z
(5.4)
  m 1 1 z
 ry 
e
 1
Note: (r) reflected wave, (-) wave traveling in the –z direction
 rx
Equation (5.4) is related by  r    1 because the reflected wave is traveling in the –

y
z direction and the Poynting vector E  H will be in the -az direction. To satisfy the
boundary conditions for the tangential electric field, z = 0. This is important because
the basic model assumes three waves  incident, reflected and transmitted:

i
x
  rx

z0
  tx
z0
This can be simplified, by adding the E field transmitted to the E field reflected of
medium 1 with a result equal to the E field transmitted wave of medium 2:
 m 1   m 1   m 2
(5.5)
*Note: We can model the system as three waves  incident, reflected and transmitted.
Boundary conditions must be met for the E field as well as the H field. Waves have both

E & H fields -       .
Similarly, enforcing the continuity of the tangential magnetic field at z = 0,
  
  


m1
m1
m2
Therefore,
 m 1  m 1  m 2


 1  1  2
Notes by: Debbie Prestridge
(5.6)
5

To solve for  m2
, multiply equation (5.6) by  1 and add the result to equation (5.5). The
result is:
 m 2
2 2
 








 m1
1
2
(5.7)
The transmission coefficient  is the ratio of the amplitudes of the transmitted to the
incident fields:
2 2
 
(5.8)
 1   2
The amplitude of the reflected wave can be solved for by multiplying equation (5.6)
by 2 and subtracting the result from equation (5.5) for a result of:
   1
(5.9)
 m 1   m 1 2
 2   1
The reflection coefficient  is the ratio of the amplitudes of the reflected and incident
electric fields given by:
 m 1  2   1
    
 m1  2   1
(5.10)
From equations (5.8) & (5.10), note that the reflection and transmission coefficients are
related by 1    .
EXAMPLES:
An H field travels in the  a z direction in free space with a phaseshift constant (β) of
1
30.0 rad/m and an amplitude of
A/m. If the field has the direction –ay when t = 0 and
3
z = 0, write suitable expressions for E and H. Determine the frequency and wavelength.
In a medium of conductivity σ, the intrinsic impedance η, which relates E and H,
would be complex, and so the phase of E and H would have to be written in complex
form. In free space the restriction is unnecessary. Using cosines, then
1
cos( t   z )
H (z, t) = 
3
For propagation on –z,
x
  o   120 
y
Notes by: Debbie Prestridge
Or
 
 x   40cos( t   z) V m
6
E ( z, t )  40 cos( t  z)V m
Thus
Since   30 rad m,
 
2



15
m
f 
c


3  10 8

15

45

 10 8 z
Determine the propagation constant γ for a material having r  1,  r  8,
and   0.25 S m , if the wave frequency is 1.6MHz.
In this case,
0.2510 12


 10 9  0
6
9
  2  16
. 10  8 10 36 
So that
α=0
      2 f
r r
c
 9.48 10 2 rad m
And     j  j 9.48 10 2 m1 . The material behaves like a perfect dielectric at the
given frequency. Conductivity of the order 1  S m indicates that the material is more
like an insulator than a conductor.
5.3 Normal Incidence Plane-Wave Reflection at Perfectly conducting Plane
Special case (analysis of material presented in section 5.2)
Assumptions-(region 2) perfect conductor  2 → ∞, wave impedance
 2 
2

2 j 2

 0 , as σ2→ ∞,
(5.11)
To simplify the standing wave analysis, assume that region 1 is a perfect dielectric σ1 = 0.
Using substitution: take equation (5.11) in the reflection and transmission
coefficient expressions in equations (5.8) and (5.10) in order to obtain
  0,
   1
The zero value of the transmission coefficient simply means that the amplitude of the

transmitted field in region 2 is  m2
= 0. This can be explained in terms of the following:

The depth of penetration parameter is zero in a perfectly conducting region,
Notes by: Debbie Prestridge
7

(Chapter 3, p. 241). Therefore, there will be no transmitted wave in a perfectly
conducting region, because of the inability of time-varying fields to penetrate
media with conductivities converging toward infinity.
Only the incident and reflected fields will be present in region 1.

For   1,


The amplitude of the reflected wave is  m 1    m 1 . The reflected wave is
therefore equal in amplitude and is opposite in phase to the incident wave. This
simply means that the entire incident energy wave is reflected back by the
perfect conductor.
The combination of the two fields meets the boundary condition at the surface of
the perfect conductor.
This can be illustrated by examining the expression for the total electric field  tot ( z)
in region 1, which is assumed to be a perfect dielectric (i.e., α1 = 0)
 tot ( z)   i ( z)   r ( z)   m 1 e  j1 z a x   m 1 e  j1 z a x
Substituting  m 1    m 1 , for a result of:


 j z
 tot ( z)   m 1 e 1  e j1 z a x
= -2j  2 j  m 1 sin   1 z sin  t a x
(5.12)
Note: The total electric field is zero at the perfectly conducting surface (z = 0)
meeting the boundary condition.
To study the propagation characteristics of the compound wave in front of the perfect
conductor, we must obtain the real-time form of the electric field.
Step 1: Multiply the complex form of the field in equation (5.12) by ejωt
Step 2: Take the real part of the resulting expression

 tot ( z )  Re e jwt  tot ( z )

= 2 m 1 sin  1 z sin  t a x
(5.13)

In equation (5.13) the amplitude of the electric field was assumed real  m1
. Our
objective is then to complete the following step:
Step 3: Show that the total field in region 1 is not a traveling wave, although it
was obtained by combining two traveling waves of the same frequency
and equal amplitudes of which are propagating in the opposite direction.
Notes by: Debbie Prestridge
8
Figure 5.2 shows a variation of the total electric field in equation (5.13) as a function
of z at various time intervals.
ωt = 3π/2
Perfect Conductor
σ→∞
π < ωt <3π/2

2  m1
3λ/4
λ /4
λ/2

z=0
Z

 2  m 1
0 < ωt < π/2
ωt = π/2
Figure 5.2b The variation of the total electric field in front of the
perfect conductor as a function of z and at various time intervals ωt.
From figure 5.2 you can make the following observations:
1. α = 0, indicating that the total field meets the boundary condition at all
times.
2. The total electric field has maximum amplitude twice that of the incident
wave. The maximum amplitude occurs at z = λ/4, at z = 3λ/4, etc., when
ωt = π/2, ωt = 3π/2, etc. happening when both the incident and reflected
waves constructively interfere.
3. When z = λ/2, z = λ, z = 3λ/2, etc., in front of the perfect conductor the
total electric field is always zero. This is happening when the two fields
are going through destructive interference process for all values of ωt, also
known as null locations.
4. The occurrence of the null and constructive interference locations do not
change with time. The only thing that changes with time is the amplitude
of the total field at nonnull locations. Therefore, the wave resulting from
the interference of the two waves is called “standing waves”.
It should also be emphasized that the difference between the electric field expressions
for the traveling and standing waves. For a traveling wave, the electric field is given
by:
Notes by: Debbie Prestridge
9
 ( z, t )   m 1 cos( t  1) a x
The term ( t   1 z ) or  (t  z v1 ) emphasizes the coupling between the location
as a function of time of a specific point (constant phase) propagating along the wave.
It also indicates with an increase in t, z should also increase in order to maintain a
constant value of (t –z/v1), and it characterizes a specific point on the wave. This
means that a wave with an electric field expression which includes cos ( t   1 z ) is
a propagating wave in the positive z direction. The time t and location z variables are
uncoupled in equation (5.13), or in other words, the electric field distribution as a
function of z in front of the perfect conductor follows a sin   1 z  variation, with the
locations of the field nulls being those values of z at which sin   1 z  = 0.
The sin (ωt) term modifies the amplitude of the field allowing a variation of a
function of time located at the nonzero field locations as illustrated in Figure 5.2.
By finding the values of   1 z  the permanent locations of the electric field nulls can
be determined, thus making the value of the field zero. So, from equation (5.12) we
can see that
 tot ( z)  0 at 1 z  n (n  0,  1,  2,....
2
z
z  n
Therefore,
1

zn 1
Or
(5.14)
2
This simply shows that  tot ( z)  0 is zero at the boundary z = 0, and at every half
wavelength distance away from the boundary in region 1 which is illustrated in
Figure (5.2).
Total Magnetic Field Expression:
  m 1  j 1 z  m 1 j z 
tot
i
r



   ( z)   ( z)  
e

e 1  ay
1
  1

The minus sign in the reflected magnetic field expression is simply because for a –z
propagating wave the amplitude of the reflected magnetic field is related to that of the
reflected electric field by (   1 ). Substituting  m 1    m 1 yields:
 m 1  j 1 z
tot

 ( z) 
e
 e j1 z  a y


1
= 2
Notes by: Debbie Prestridge
 m 1
1
cos  1 z a y
(5.15)
10
The time-domain magnetic field expression is obtained from equation (5.15) as:

 tot ( z , t )  2  m1 cos  z a cos t a

(5.16)
1
y
y
1
0 < ωt < π/2
ωt = 0
5 1
3 1
4
4
1

Perfect
Conductor

2

Z
m3

1
2
1
λ1
z=0
ωt = π/2
ωt = π
Figure 5.3 The magnetic field distribution in front of a perfect conductor as a
function of time.
Equation 5.16 is also a standing wave as illustrated in Figure 5.3, with the maximum
amplitude of the magnetic field occurring at the perfect conductor interface (z = 0)
where the total electric field is zero. The location of nulls in the magnetic field are
where the values of z at cos  1 z = 0, therefore,



 1 z = odd number of  2m  1
(m  0,  1,  2,... or z = (2m + 1) 1
2
2
4
The magnetic field distribution in front of a perfectly conducting boundary is
illustrated in Figure 5.3, where we can observe that its first null occurs at
z =  1 4 . This is the location of the maximum electric field (see Figure 5.2). By
comparison, equations (5.13) & (5.16) shows that the electric and magnetic fields of a
standing wave are 90° out of time phase due to the sin(ωt) term and cos(ωt),
respectively. This will result in a zero average power transmitting in either direction
of the standing wave. This can be illustrated by using the complex forms of the fields
to calculate the time-average Poynting vector Pav (z):
Notes by: Debbie Prestridge
11


1
 ( z)
Re  ( z )  
2

 m 1
1 

 Re  2 j  m1 sin  1 z a x  2
cos  1 z a y 
2 
1

=0
(5.18)
Pave ( z) 
The zero value of Pav (z) is obtained because the result of the vector product of is a
  ( z) is an imaginary number. This zero value of average power transmitted
 ( z)  
by this wave is yet another reason for calling the total wave in front of the perfect
conductor a “standing wave.”
Examples:
In free space  ( z, t )  50 cos( t   z) a x (V m). Obtain H (z, t).
Examination of the phase, ωt-βz, shows that the direction of propagation is +z, since
E x H must also be in the +z direction, H must have the direction –ax. Consequently,
y
10 3
Or  x 
  o  120 
sin ( t   z) A m
x
120
 ( z, t ) 
And
10 3
sin ( t   z) a x A m
120
For the wave of the problem above determine the propagation constant γ, given that
the frequency is f = 95.5MHz.
In general, γ = j   (   j   ) . In free space, σ = 0, so that
  j  o  o  j
2 f
2 (955
.  10 6
j
 j (2.0) m 1
c
3  10 8
This result shows that the attenuation factor is α = 0 and the phase-shift constant is
β = 2.0 rad/m.
Notes by: Debbie Prestridge
12
5.4
Reflection and Transmission at Multiple Interfaces
Obtain expressions for the electric & magnetic fields in each region shown in Figure 5.4a.
Figure 5.4a A Plane wave normally incident on two interfaces separating three
different media.
Region 1
 ix
 iy
The development of these reflections as a function of time can be seen in Figure 5.4b.
Notes by: Debbie Prestridge
13

i
A

 
1
2
 1
1
B


 
1
3
1
2

 2
2
2
1

 2
3

 2
2

Travel time
Time

 3
2
  3
1

 3
3

 3
2

4
1
Figure 5.4b
Note: Subscripts indicate the region’s number and the superscripts indicate whether the
wave is transmitted in the +z direction or reflected in the –z direction. The superscript 1,
2 and 3, reversed in order, is included to indicate the number of the reflection or
transmission process. Also the plane waves are normally incident at the multiple
interfaces. The reflection process is shown at an angle with respect to the interface in
order to illustrate the development of the multiple reflections as a function of time. So
the travel time of a typical wave  2 (1) between two interfaces is measured by the vertical
distance between the points A & B.
By summing all the fields that resulted from the multiple reflection process the steadystate expressions can be obtained.
Notes by: Debbie Prestridge
14
The following are the steady-state expressions:
Region 1
In this region at steady state, we have the incident wave propagating in the +z
direction and an infinite number of reflected waves  1 (1) ,  1 ( 2) ,..., and so on propagating
in the –z direction. The total electric field in this region is:

 
 1 ( z)   i 
(Total electric field)
( N )
1
(5.19)
N 1

 i   m 1 e  1 z a x
(Incident electric field)

 1 ( N )   m (1N ) e  1 z a x
(Each reflected wave)
(5.20)

And  m(1N ) is the amplitude of the (Nth) reflected wave. All reflected waves are

propagating in the –z direction and therefore, e  1 z term is common to all the field
expressions for these waves.
Substituting equation (5.20) in equation (5.19) for a result of:

 1 ( z)   m 1 e  1 z a x 

 
N1
( N )
m1
(5.21)


 m 1 e  1 z a x  e  1 z

The summation
 
N 1
( N )
m1

 
N 1
( N )
m1
ax
over the complex amplitudes of the infinite number of the

reflected waves is just another complex  m1
representing the steady-state amplitude of
the overall reflected wave. Equation (5.21) will reduce to:


 1 ( z)   m 1 e  1 z a x   m 1 e  2 z a x
(5.22)
Equation (5.22) represents an expression for the overall fields in region 1 at steady-state.
This means after a period sufficiently long for the development of an infinitely large
number of reflections.
Notes by: Debbie Prestridge
15
Region 2
In this region at steady state there will be an infinite number of waves propagating
in the +z direction and an infinite number of other waves propagating in the –z direction.
The steady state expression of the electric field:

 
 x ( z ) 

N 1

N
2


 
N 1
N
2
  2 N e 2 z a x 
N 1
=e
2 z

 
N 1

  2 N a x  e2 z
N 1
e 2 z a x
N
2


 
N 1
N
2
ax
The sum over the amplitudes of the waves propagating in the +z

direction 
  ( N ) is a complex number   , which represents the steady state
N 1
m2
m2
amplitude of the wave propagating in the +z direction. Similarly, in the –z direction

  = 
 ( N ) . A general expression for the steady state fields in region 2 is,
m2
N 1
m2
therefore, given by:


 2 ( z)   m 2 e  2 z a x   m 2 e  2 z a x
(5.23)
Region 3
In region 3 we can see all the transmitted waves in this region are propagating in
the +z direction. The steady state expression for the electric field in this region is:
 3 ( z) 

 
N 1
= e
( N )
m3
e  3 z a z


3 z
 
N 1
( N )
m3
az
(5.24)
= 

m3
e
 3 z
ax
This is the steady state amplitude if the transmitted wave in region 3:
 m3 
Notes by: Debbie Prestridge

 
N 1
( N )
m3
16
Figure 5.5 shows the summary of the electric & magnetic fields expressions in each
region.
Region 1
(ε1, μ1, σ1)
Region 2
(ε2, μ2, σ2)
 i   m 1 e  1 z a x


 m 2 e  2 z a x
Incident
 m 1 1 z Wave
i

 
e ay
 1
Region 3
(ε3, μ3, σ3)

 m 3 e  3 z a x
Transmitted
(+z)
 m2 2 z
e ay
2
 m3 3 z
e ay
 3

 m 2 e  2 z

 m 1e  1 z a x
Z
Total

Reflected  m1 e 1 z  a
y
Wave
 1


 m 2 2 z
e
 ay
 2


Figure 5.5
Solve for the unknown amplitudes of the reflected and transmitted waves in each region

in terms of the known amplitude of the incident electric field  m1
from the previously
determined steady-state expressions.
First enforce the boundary conditions at the interfaces z = 0 & z = d.
At z = 0, regions 1 & 2 have a tangential electric fields that are continuous
 1 ( z) z  0   2 ( z) z  0
Therefore,
(5.25)
Substitution:
Substituting equations (5.22) & (5.23) in equation (5.25), result is;
 m 1   m 1   m 2   m 2
(5.26)
Enforce the continuity of the tangential component of the magnetic field at the interface
z = 0, result is:
  m 1  z  m 1  z 
  
   


e 1 
e1 
a y   m 2 e  2 z  m 2 e  2 z 
ay
 1
2
  1
 z0
  2
 z0
OR
 m 1  m 1  m 2
 m2



 1
 1  2
 2
Notes by: Debbie Prestridge
(5.27)
(5.28)
17
In equation (5.27), the amplitude of the magnetic field propagating in the +z direction is

obtained by dividing  m1
by  1 and the amplitude of the magnetic field propagating in
the –z direction is obtained from     . The minus sign is to maintain Poynting

m1
1

vector in the –z direction.
Now apply the boundary conditions in a similar fashion for the tangential electric and
magnetic fields shown in regions 2 & 3 at the interface z = d, result is:



 m 2 e  2 d   m 2 e  2d   m 3 e  3d
(5.29)
 m 2 3d  m 2 2 d  m 3 3d
e

e 
e
 2
 2
 3
(5.30)
5.5 Reflection Coefficient and Total Field Impedance Solution Procedure
Equations (5.26, 5.28, 5.29, and 5.30) are four equations in the four
unknowns  m 1 ,  m 2 ,  m 2 ,  m 3 . These four equations can be solved using a well known
method for solving simultaneous equations. This way would be rather tedious;
fortunately there is a simpler way for solving multiple-interface problems.
The interest is not in solving these equations, but to emphasize that every additional
region that we introduce between regions 1 & 3 adds two unknowns to our system of
equations, for the amplitudes of the +z and –z traveling waves.
A new parameter must be introduced called “the total field impedance,”  ( z ) at any
location z within anyone of the dielectric media illustrated in Figure 5.6.
Definition: Total field impedance is the ratio between the total electric field and the total
magnetic field in a given region.

 ( z )
  z =  x
 y ( z)
Notes by: Debbie Prestridge
(5.31)
18
Region 1
Region 2
1 , 1
Region 3
2 , 2
Region 5
Region 4
5 , 5
4 , 4
3 , 3
i
t
Hi
Ht
O1
O2
O4 O5
O3
Z
r
Hr
d3
d2
d4
Figure 5.6 Illustration of the multiple region problem and the various waves in
each region. The interface between regions 1 and 2 was assumed to be at z = a
 ( z) and  ( z) .
and used to illustrate the characteristics of 
The following equation can be obtained by substituting the general expression in equation
5.22 for the total electric field in any region containing reflections. Remember the
amplitude of the magnetic field component propagates in the +z direction. This relates to
the amplitude of the electric field by  . The amplitude of the magnetic field component
propagates in the –z direction and is related to the corresponding component of the
electric field by   , for a result of:


 m e  z   m e  z

 ( z)   
 m  z  m
e



  
1   m e 2 z
m
 
  
1   m e 2 z
m
(5.32)
Definition: Reflection coefficient is the ratio of the reflected to the incident waves

 m  m e 2 z at an arbitrary location z in any region.
  
 ( z)   m e 2 z
m
Notes by: Debbie Prestridge
(5.33)
19
Substitution: Substituting equation (5.33) in equation (5.32), the expression for the total
field impedance at an arbitrary location z within one specific region reduces to:
 ( z)  
1  ( z)
1  ( z)
(5.34)
Definition: The characteristic impedance  is the ratio between the electric and
magnetic fields of a plane wave propagating in the +z direction.
*If the plane wave is propagating in the –z direction the ratio will become   .
Notice that region 4 of Figure 5.6 is absence of a reflected wave therefore it is clear from
equation (5.33) that  ( z ) will be equal to zero and  ( z ) in equation (5.34) will be equal
to  . Also, notice that  is calculated specifically at the interface between two media and
at the origin of the coordinate system, meaning z = 0.  ( z ) , is calculated at an arbitrary
location z within any one of the multiple regions illustrated in Figure 5.6. Don’t
substitute any one region by the simplified expression of equation (5.10).
The  ( z ) can be expressed in terms  ( z ) . So, from equation 5.34 it can be shown that
 ( z)  
 ( z)  
 ( z)  
(5.35)
The following are properties of the quantities of  ( z ) and  ( z ) :
1. Most important characteristic of  ( z ) is its continuity at an interface separating
two different media. (See Chapter 3, equation (3.56) & (3.57) under boundary
conditions). If the interface at z = a separates two media where the electric and
magnetic fields are given by:
 1 ( z ) &  1 ( z )
Region 1
 ( z ) &  ( z )
Region 2
2
2
then the boundary conditions at z = a will require that –
 t 1 (a  )   t 2 (a  )
 (a  )  
 (a  )

t1
t2
(5.36)
(5.37)
(“t” is the tangential components of  and  )
The total electric and magnetic fields are then all tangential to the interfaces
separating the various dielectric media. So for applications of interest we may
replace the tangential components in equations (5.36) & (5.37) by the total fields.
Therefore, by dividing equations 5.36 by 5.37 and taking into account that –
Notes by: Debbie Prestridge
20
 1 (a  )  
 (a  )   (a ) , and

1
 2 (a  )  
 (a  )   (a )

2
The result will be
(5.38)
 1 (a  )   (a  )
Therefore, enforcing the continuity tangential components of the electric and magnetic
fields across an interface separating two media is equivalent to satisfying the boundary
condition on the total field impedance as given in equation (5.38). The importance of this
will be clarified in later material.
2. The following are two important characteristics of  ( z ) :
(a) The calculation of  ( z  ) at z΄ from its value  ( z ) at z, where both z΄ and z are
within a specific one of the regions illustrated in Figure 5.6.
To obtain the relationship between  ( z ) &  ( z  ) , just consider equation (5.33)
that will provide the values of  ( z ) within any one region in Figure 5.6.
Example:
Region 2 will give a result of:
 

(5.39)
 2 ( z)  m 2 e 2 2 z
 m2
In equation (5.39), we substituted the specific characteristic parameters of region 2
such as  m 2 ,  m 2 , and 2 . At any other location as for as an example, z΄ within region
2,  ( z ) is given by:
2
 

 2 ( z )  m 2 e 2 2 z 
 m2
(5.40)
So, from equations (5.39) & (5.40), it is clear that  2 ( z ) can also be expressed in
terms  ( z ) of by:
2

 2 ( z )  e 2 2 ( z   z )  2 ( z )
(5.41)
Equation (5.41) presents an important relation that allows us to calculate the
reflection coefficient at z΄ in terms if its value at z. Note that z΄ and z should be
within’ the same region such as region 2.
(b) The other important characteristic of  ( z ) is related to its discontinuity at the
interfaces between the different media.
Example:

Assume the interface between regions 2 & 3 of Figure 5.6 are at z = a,  ( a  ) in
region 2 just before the interface (i.e. at z = a-) is given by:
 (a  )   2
(5.42)
 (a  )   
 (a )  
2
Notes by: Debbie Prestridge
21
At z = a+ just after the interface, reversed in relation, the characteristic impedance of
region 3 is  3 , therefore  ( a  ) is given by:
 (a  )   3
5.43)
 (a  )   
 (a )  
3
Because the total field impedance is continuous across the interfaces between the
various,  (a  )   (a  ) , as described in part 1. The reflection coefficients in the
previous two equations are discontinuous across the interface. This is due to the
different values of 2 and 3 . The reflection coefficient obviously is not an
appropriate quantity used to relate field quantities across an interface between two
different media. It is best to use  ( z ) when making transitions across interfaces
between different dielectric media.
The following is the systematic solution procedure for solving the reflection and
transmission at multiple interfaces:
 Satisfy boundary conditions on the tangential components of the electric and
magnetic fields. *Enforce the continuity of  ( z ) at the
interface  (O ) =  (O )
5



4
4
Use a separate origin for each region.
Place the origin of each region at the end except in the case of region 5 which
is semi infinite.
Use equation (5.35) to calculate  (O4 ) in terms of the value of  ( z ) at the
point  (O ) .
4

5
4
Use equation (5.41) to calculate  ( d 4 ) , d4 being the thickness (m) of region
4, in terms of the value of  ( O ) at the origin.
4


4
2 (  d  0)
 4 ( d 4 ) =  4 ( O4 ) e 4 4
Calculate the total field impedance  4 ( d 4 ) in terms of a  4 ( d 4 ) , with a
1   4 ( d 4 )
 4 ( d 4 )   4
result of:
1   4 ( d 4 )
This process continues until the values of the total field impedance
 1 ( O1 ) and the reflection coefficient  1 (O1 ) in region 1 are obtained.
Calculate the amplitude of the reflected wave in region 2, using  (O ) and
1
1
the amplitude of the incident wave. In other words from the definition of
 ( z ) in equation (5.33), we have:

  
 m 1   ( z ) 2m11z 
e


Notes by: Debbie Prestridge
22


 m 1 



Evaluate  m1   ( z ) 21 z  at the origin O1, at z = 0 for a result of:
e



 m1   (O1 )  m 1
(5.44)

  1 z    e 1 z is the total electric field in region 1.   was
 tot
x 1 ( z )   m1 e
m1
m1
obtained from equation (5.44)

The total magnetic field in region 1 is given by:
 m 1 1 z  m 1 1 z

 y1 ( z ) 
e

e
 1
 1

(5.45)
Finally, calculate the amplitudes of the electric and magnetic fields in the
other regions. Do this by matching the boundary conditions at interfaces with
known field on one side and unknown field on the other side.
EXAMPLE:
A uniform plane wave propagating in free space is normally incident on the multiple
dielectric media shown in Figure 5.7. If the x-polarized electric field associated with this
wave is given by
E (z, t) = 50 cos (2π x 108t – 2.094z) ax
Determine the following:
1. Magnitude of electric field tranmitted to region 2.
2. Time-average power density in region 1.
Solution
1. The solution procedure is summarized as follows:
(a) From the given frequency of propagation and the electrical properties of each
region, calculate the characteristic parameters  and  in each region. From
the given electric field expression, these parameters in region 1 are given by
 2 108
f=

 100 MHz
2
2
 1  2.094 rad m
 1 
o
 120 
o
And the phasor expression for the electric field is

 ( z)  50 e j 2.0945z a x
Notes by: Debbie Prestridge
23
In region 2,
2 
o
4 o
 60 
α2 = 0
Because region 2 is lossless-that is, σ2 = 0
 2    o(4  o)  2
 o o  2 
1
= 4.1488 rad m
In region 3,
3 
o
 30 
16 o
α2 = 0
 3 = 4   = 8.376 rad m
1
(b) To avoid a cumbersome process of carrying phase constants throughout the
calculations, we set a separate origin in each region. We routinely locate
origins at the end of each region, except in region 3 where we set its origin
at the beginning as shown in Figure 5.7.
(c) We start the calculations from the simplest region-that is, region 3-which is
free from any reflections.


 1  3 ( O3 )
 3(O3)  3
1   (O )
3

3

Because 3(O3) = 0, the total field impedance  3(O3) is, hence, equal to the
characteristic impedance η3
 3(O3) = η3 = 30π Ω
(d) Apply the continuity of the total field impedance at the interface between
regions 2 & 3. Hence,
 2 (O2 ) =  3 ( O3) = η3 = 30π Ω
(e) To relate the various parameters of the three regions to those of the incident
plane wave, we need to continue moving toward region 1 ultimately to
calculate  1 ( O1) in this region to do so; we need to calculate
 2 ( d 2) from its value at the origin  2 (O2 ) . Because we do not have
Notes by: Debbie Prestridge
24
an expression for relating  2 ( d 2) to  2 (O2 ) , we employ the
reflection coefficient parameter. Hence,

1
 2 (O2)  2 30  60
 2 (O2 ) 



30  60
3
 2 ( O2 )   2
From equation (5.41), we have

 2 (  d 2) =  2 (O2) e2  2( d 20)
1 4 1 
  2
 2      e j 3  e j 3
 3
3
3
The desired value of  2 (  d 2) is then


 2(  d 2)  

2
1   2 (  d 2)

1   2 (  d2)
1 
1  e j 3
3
= 60
1  j
1 e 3
3
117
.  j 0.29
= 60
0.83  j 0.29
= 60π (1.15 – j0.75)
(f) From the continuity of the total field impedance at the interface between
regions I & 2, we have


1(O1) 
 1 (O1)  1
=
 160.23  j14137
.

 0.35 e j 234.8
593.77  j14137
.
 1 (O1)  1
(g) The amplitude of the reflected wave in region1 can now be calculated

from 1(O1) and using equation (5.44)

 m 1

 m 1


 1(O1)



 50 * 0.35 e j 234.8
 m1
= 17.5 e
Notes by: Debbie Prestridge
j 234.8
V

m
25
(h) The total electric field in region 1 is hence





 x1   m 1 e j 234.8 1  1 ( z ) 


At z = 0, we have



 x1(0)  50 1  0.35 e j 234.8

Because of the continuity of the electric field at the interface between
regions 1 & 2, we have


 x1(0) 
 x 2 (-d2)


=  m 2 e j 2 (  d 2 ) 1   2  d 2 


Hence,



m2

=


50 1  0.35 e j 234.8

 1 
e j 2 2 (  2 3) 1  e j 3 
 3



50 1  0.35 e j 234.8

1 
e j 2 3 e J 3
3
 j125.81 V
= 3517
. e
m
2. The time average power density in region 1 is given by
1   
Pav1 = Re 1 x 1

2 





1     j z 


m1 j  z 




1
1

= Re  m1 e
1  1( z)  a x x
e 1  1( z)  a y

1




2



2

2
1  m 1
=
(1  1 ( z) ) a z
2 1

where we have substituted  

 m1
 

e j 1z 1  1( z)  a y for the magnetic field
1




 50, η1 =120π, and
in region 1 as given by equation (5.45). Substituting  m1

1( z)  0.35 in the average power expression, we obtain
Pav1 = 2.91 a z W/m2
Notes by: Debbie Prestridge
26
Region 1
εo, μo
Region 3
ε = 16 εo
Region 2
ε = 4 εo
μ = μo


x
O2
O1
O3


y
Z
az
d
2


2
3
Figure 5.7 A plane wave is normally incident on the interfaces between three
dielectric regions
Notes by: Debbie Prestridge
27