Reflection of electromagnetic waves Lect

8/4/2019 Reflection of electromagnetic waves Lect

http://slidepdf.com/reader/full/reflection-of-electromagnetic-waves-lect 1/27

Notes by: Debbie Prestridge 1

Ht ax

O

X

Y

Et

Hr

E

r

Hi

Ei

Region 2

2 2 2 , ,

Z

Figure 5.1

*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 5Normal 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 waveswhen 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 youtry and make a call on your cell phone and you are downtown amongst all those tallbuildings. 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, youknow 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 one-

dimensional 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

1 1 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:

x

i

m

ze

11

(5.1)

y

i m ze

1

1

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 1

j . *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 andmagnetic fields is

tan

x

y

j

j

e

1

214

1

2

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 thevelocity of light vp < c. The wavelength λ in the conductive medium will be shorter than

the wavelength λ o in free space at the same frequency, λ = 2π/β < λ o. The factor ze

will attenuate the magnitudes of both E and H as they propagate in the +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 medium2. The following are the electric and magnetic fields related to the transmitted wave:

x

t

m

ze

2

2 (5.2)

y

t m ze

2

2

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

2

2

2

x y z

For now, let’s look at the simplest system, that consisting of a plane wave of coordinatez.

d

d z

2

2

2

t1 = 0t2 > t1 e

-αz

Figure 5.2 The electric field associated with a plane wave

propagating along the positive z direction.




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, onlysuperconductors have infinite conductivity and are truly perfect conductors.

Static (time independent)n · D1 = ρs

n · (B1 – B2) = 0

n1= 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 andnonzero 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. Astatic 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 arethose outside the conducting body.

x

t & x

i are tangential to the interface therefore the boundary conditions will require

these fields be equal at z = 0. Equate x

t & x

i , and set z = 0. *Note: (t) is

transmitted wave, (i) incident wave. The result will be

m m1 2

*Note: (m) medium, (+) transmitted wave (5.3)




y

t & y

i 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 incidentwave 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

x

r

m

ze

11

(5.4)

yr m ze

1

1

1

Note: (r) reflected wave, (-) wave traveling in the – z direction

Equation (5.4) is related by

x

r

y

r

1because the reflected wave is traveling in the –

z direction and the Poynting vector E H will be in the -az direction. To satisfy theboundary conditions for the tangential electric field, z = 0. This is important because

the basic model assumes three waves incident, reflected and transmitted:

x

i

x

r z

x

t z 0 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 m m1 1 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,

m m m1 1 2

Therefore,

m m m1

1

1

1

2

2

(5.6)




To solve for m2

, multiply equation (5.6) by 1and add the result to equation (5.5). The

result is:

m

m

2

1

2

1 2

2

(5.7)

The transmission coefficient is the ratio of the amplitudes of the transmitted to theincident fields:

2

2

1 2

(5.8)

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:

m m1 1

2 1

2 1

(5.9)

The reflection coefficient is the ratio of the amplitudes of the reflected and incidentelectric fields given by:

m

m

1

1

2 1

2 1

(5.10)

From equations (5.8) & (5.10), note that the reflection and transmission coefficients are

related by1 .

EXAMPLES:

An H field travels in the a z

direction in free space with a phaseshift constant (β) of

30.0 rad/m and an amplitude of 1

3 A/m. If the field has the direction – ay when t = 0 and

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 complexform. In free space the restriction is unnecessary. Using cosines, then

H (z, t) = 1

3 cos( )t z

For propagation on – z,

x

y

o 120 Or x

t z V m

40cos( )




Thus E z t t z V m( , ) cos( ) 40

Since 30rad m,

2

15m f

c z

3 10

15

4510

8

8

Determine the propagation constant γ for a material having r r

1 8, ,

and 025. S m , if the wave frequency is 1.6MHz.

In this case,

0 25 10

2 1 6 10 8 10 3610 0

12

6 9

9.

.

So that

α = 0

2 9 48 102

f

crad m

r r .

And

j j m9 48 102 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

2

0

j

, 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,




(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 penetratemedia 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 m1 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 theperfect 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 i r

m

j z

x z z z e a

11

m

j z

xe a

1

1

Substituting m m1 1

, for a result of:

( ) tot

m

j z j z

x z e e a

1

1 1

= -2j

21 1

j z t am x sin sin (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 e jωt

Step 2: Take the real part of the resulting expression

( ) Re ( ) tot jwt tot

z e z

=

2

1 1m x

z t a

sin sin (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 itwas obtained by combining two traveling waves of the same frequency

and equal amplitudes of which are propagating in the opposite direction.




Perfect Conductor

σ → ∞

21m

z = 0

2 1m

ωt = π/2

0 < ωt < π/2

λ /4 λ/2

3λ/4

π < ωt <3π/2

ωt = 3π/2

Z

Figure 5.2 shows a variation of the total electric field in equation (5.13) as a functionof z at various time intervals.

From figure 5.2 you can make the following observations:

1. α = 0, indicating that the total field meets the boundary condition at alltimes.

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 thetotal electric field is always zero. This is happening when the two fields

are going through destructive interference process for all values of ωt, alsoknown 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 amplitudeof 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 expressionsfor the traveling and standing waves. For a traveling wave, the electric field is given

by:

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.




( , ) cos( ) z t t am x

1 1

The term ( ) ( ) t z or t z v 1 1

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 aconstant 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 z 1is

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 canbe determined, thus making the value of the field zero. So, from equation (5.12) we

can see that ( ) ( , , ,....

tot z at z n n 0 0 1 2

1

Therefore, z z n 2

1

Or z n

1

2(5.14)

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 inFigure (5.2).

Total Magnetic Field Expression:

( ) ( )

tot i r m j z m j z

y z z e e a

1

1

1

1

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 m1 1

yields:

( )

tot m j z j z

y z e e a

1

1

1 1

= 21

1

1

cos m

y z a

(5.15)




Perfect

Conductor

23

1

m

z = 0

ωt = π

ωt = 0 0 < ωt < π/2

5

4

1

λ 1

3

4

1

11

2

The time-domain magnetic field expression is obtained from equation (5.15) as:

( , ) tot z t 21

11

cos cos m

y y z a t a

(5.16)

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 maximumamplitude 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 arewhere the values of z at cos

1z = 0, therefore,

1z = odd number of

22 1

20 1 2 m m( , , ,... or z = (2m + 1)

1

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 = 14 . 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 fieldsto calculate the time-average Poynting vector Pav (z):

Z

ωt = π/2




P z z zave

( ) Re ( ) ( ) 1

2

1

22 2

1 1

1

1

1Re sin cos j z a z am x

m

y

= 0 (5.18)

The zero value of Pav (z) is obtained because the result of the vector product of is a ( ) ( ) z z

is an imaginary number. This zero value of average power transmitted

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 ( , ) cos( ) ( ). z t t z a V m x 50 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

x

o 120 Or x

t z A m 10

120

3

sin ( )

And ( , ) sin ( ) z t t z a A m x

10

120

3

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 j

f

c j j mo o

2 2 955 10

3 102 0

6

8

1( .

( . )

This result shows that the attenuation factor is α = 0 and the phase-shift constant is

β = 2.0 rad/m.




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 threedifferent media.

The development of these reflections as a function of time can be seen in Figure 5.4b.

x

i

y

i

Region 1




BTravel time

A

i

1

1

1

2

1

3

1

4

2

1

2

1

2

2

2

2

2

3

2

3

3

1

3

2

3

3

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 steady-

state expressions can be obtained.

Time




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 1

1

zi N

N

(Total electric field) (5.19)

i

m

z

xe a

11 (Incident electric field)

( ) ( )

1 1

1

N

m

N z

xe a

(Each reflected wave) (5.20)

And ( )

m

N

1

is the amplitude of the (Nth) reflected wave. All reflected waves are

propagating in the – z direction and therefore, e

z 1

term is common to all the fieldexpressions for these waves.

Substituting equation (5.20) in equation (5.19) for a result of:

( ) ( )

1 1 1

1

1

z e am

z

x m

N

N

(5.21)

( )

m

z

x

z

m

N

xe a e a

N 1 1

1 1

1

The summation ( )

m

N

N 1

1

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 1 11 2

z e a e am

z

x m

z

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.




Region 2

In this region at steady state there will be an infinite number of waves propagatingin the +z direction and an infinite number of other waves propagating in the – z direction.

The steady state expression of the electric field:

( ) x

N N z

N N

2 2

1 1

2 2

2 2

1 1

N z

x

N z

xe a e a

N N

= e a e a z N

x

z N

x

N N

2 2

2 2

1 1

The sum over the amplitudes of the waves propagating in the +z

direction ( )

m

N

N 21

is a complex number m2

, which represents the steady state

amplitude of the wave propagating in the +z direction. Similarly, in the – z direction

m2

= ( )

m

N

N 21

. A general expression for the steady state fields in region 2 is,

therefore, given by:

( )

2 2 2

2 2 z e a e a

m

z

x m

z

x

(5.23)

Region 3In 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 3

1

3 z em

N z

z

N

a

= ez

m

N

z

N

( )

3

3

1

a

(5.24)

=

m

z

xe

3

3 a

This is the steady state amplitude if the transmitted wave in region 3:

( )

m m

N

N 3 3

1




Region 1

(ε1, μ1, σ1)

Region 2

(ε2, μ2, σ2)

Region 3

(ε3, μ3, σ3)

Incident

Wave

Transmitted

(+z)

m z

ye a

2

2

2

m

z

xe a

2

2

m z

ye a

3

3

3

m

ze2

2

m z

ye a2

2

2

Total

Reflected

Wave

m

z

xe a

1

1

m z

ye a1

1

1

Z

i

m

z

xe a

1 1

i m z

ye a

1

1

1

Figure 5.5 shows the summary of the electric & magnetic fields expressions in each

region.

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

Therefore, ( )

( ) 1 20 0

z z z z (5.25)

Substitution: Substituting equations (5.22) & (5.23) in equation (5.25), result is;

m m m m1 1 2 2

(5.26)

Enforce the continuity of the tangential component of the magnetic field at the interfacez = 0, result is:

mz

mz

ye e a z

1

1

1

1

1 1

0

m

zm

z ye e a z

2

2

2

2

2 2

0 (5.27)

OR

m m m m1

1

1

1

2

2

2

2

(5.28)

m

z

xe a3

3

Figure 5.5




In equation (5.27), the amplitude of the magnetic field propagating in the +z direction is

obtained by dividing m1 by

1and the amplitude of the magnetic field propagating in

the – z direction is obtained from m1 1 . The minus sign is to maintain Poynting

vector in the – z direction.

Now apply the boundary conditions in a similar fashion for the tangential electric andmagnetic fields shown in regions 2 & 3 at the interface z = d, result is:

m

d

m

d

m

d e e e

2 2 3

2 2 3

(5.29)

m d m d m d e e e

2

2

2

2

3

3

3 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 m m m1 2 2 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 = ( )

( )

x

y

z

z(5.31)




The following equation can be obtained by substituting the general expression in equation5.22 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 theelectric field by , for a result of:

( )

z

e e

e

m

z

m

z

m z m

1

1

2

2

m

m

z

m

m

z

e

e

(5.32)

Definition: Reflection coefficient is the ratio of the reflected to the incident waves

m m

z

e 2

at an arbitrary location z in any region.

( )

z e

m

m

z

2 (5.33)

d2 d3 d4

r

Hr

O1 O2 O3 O4 O5 Z

Region 1

, 1 1

Region 2

, 2 2

Region 3

, 3 3

Region 4

, 4 4

Region 5

, 5 5

i

Hi

t

Ht

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

and used to illustrate the characteristics of ( ) ( ) z and z .




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

z

z

1

1(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 boundaryconditions). If the interface at z = a separates two media where the electric and

magnetic fields are given by: ( )

1z & ( )

1z Region 1

( ) 2z & ( ) 2

z Region 2

then the boundary conditions at z = a will require that – ( ) ( ) t t a a

1 2

(5.36)

( ) ( ) t t

a a1 2

(5.37)

(“t” is the tangential components of and )

The total electric and magnetic fields are then all tangential to the interfacesseparating 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 –




( )

( ) ( )

1

1

a

a

a

, and

( )

( ) ( )

2

2

a

a

a

The result will be ( ) ( ) 1

a a

(5.38)

Therefore, enforcing the continuity tangential components of the electric and magneticfields across an interface separating two media is equivalent to satisfying the boundarycondition 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:

( )

2

2

2

2 2 z e

m

m

z

(5.39)

In equation (5.39), we substituted the specific characteristic parameters of region 2

such as , , m m

and 2 2 2

. At any other location as for as an example, z΄ within region

2, ( )2 z is given by:

( )

2

2

2

2 2

z e

m

m

z (5.40)

So, from equations (5.39) & (5.40), it is clear that ( )2 z can also be expressed in

terms ( )2

z of by:

( ) ( ) ( )

2

2

22

z e z

z 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 bewithin’ 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

a

a

2

2

(5.42)




At z = a+

just after the interface, reversed in relation, the characteristic impedance of

region 3 is 3, therefore ( ) a

is given by:

( ) ( )

( )

a

a

a

3

3

5.43)

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 ( ) 5 5O = ( )4 4

O

Use a separate origin for each region.

Place the origin of each region at the end except in the case of region 5 whichis semi infinite.

Use equation (5.35) to calculate ( ) O4

in terms of the value of ( ) z at the

point ( ) 4 4O .

Use equation (5.41) to calculate

( ) d 4 , d4 being the thickness (m) of region

4, in terms of the value of ( )4 4O at the origin.

( )4 4

d = ( )4 4

O ed 2 04 4

( )

Calculate the total field impedance ( ) 4 4d in terms of a ( )4 4

d , with a

result of: ( ) 4 4d

( )

( )

4

4 4

4 4

1

1

d

d

This process continues until the values of the total field impedance ( )

1 1O and the reflection coefficient ( )

1 1O in region 1 are obtained.

Calculate the amplitude of the reflected wave in region 2, using ( )1 1O and

the amplitude of the incident wave. In other words from the definition of ( ) z in equation (5.33), we have:

( )

m

m

z z

e1

1

2 1




Evaluate ( )

m

m

z z

e1

1

2 1

at the origin O1, at z = 0 for a result of:

( ) m m

O1 1 1

(5.44)

( )

x

tot

m

z

m

z z e e1 1 1

1 1 is the total electric field in region 1.

m1

was

obtained from equation (5.44)

The total magnetic field in region 1 is given by:

( )

y

m z m z z e e

1

1

1

1

1

1 1

(5.45)

Finally, calculate the amplitudes of the electric and magnetic fields in theother 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 byE (z, t) = 50 cos (2π x 10

8t – 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

f =

2

2 10

2100

8

MHz

1 2 094

.

rad

m

1120

o

o

And the phasor expression for the electric field is

( ) . z e a

j z x

50 2 0945




an expression for relating ( )2 2d to ( ) 2 2O , we employ the

reflection coefficient parameter. Hence,

( )2 2O

2 2 2

2 2 2

( )

( )

O

O

30 60

30 60

1

3

From equation (5.41), we have

( )2 2d = ( )2 2O2 2 2 0 ( )

d e

2

24

3 3

3

1

3

1

3

j je e

The desired value of ( )2 2d is then

2 2

( )d

2

2 2

2 2

1

1

( )

( )

d

d

= 60

11

3

11

3

3

3

j

j

e

e

= 60117 0 29

0 83 0 29

. .

. .

j

j

= 60π (1.15 – j0.75)

(f) From the continuity of the total field impedance at the interface between

regions I & 2, we have

1 1

1 1 1

1 1 1

( )( )

( )

OO

O

=

160 23 141 37

593 77 141 370 35

234 8. .

. ..

. j

je

j

(g) The amplitude of the reflected wave in region1 can now be calculated

from 1 1

( )O and using equation (5.44)

m

m

1

1

1 1

( )O

m1

50 * 0 35 234 8. . je

= 17 5 234 8. . je

V m




(h) The total electric field in region 1 is hence

x1

m j

e z1234 8

11

. ( )

At z = 0, we have

x1 0

( ) 50 1 0 35234 8

.. j

e

Because of the continuity of the electric field at the interface between

regions 1 & 2, we have

x1 0

( ) x2

(-d2)

= m j d e d 2

2 22 21

( )

Hence,

m

j

j j

e

e e

2

234 8

2 2 2 33

50 1 0 3 5

11

3

. .

( )

= 50 1 0 35

1

3

234 8

2 33

.. j

j J

e

e e

= 351712581

.. j

e V m

2. The time average power density in region 1 is given by

Pav1 =1

21 1Re x

=1

21 11

11

1

1

11Re ( ) ( )m

j z x

m j z ye z a x e z a

=1

21

1

2

1

1

2m

z z a

( ( ) )

where we have substituted

m j z ye z a

1

1

1 11

( ) for the magnetic field

in region 1 as given by equation (5.45). Substitutingm1 50

, η1 =120π, and

1 0 35

( ) . z in the average power expression, we obtain

Pav1 = 2.91 a zW/m2



Region 3ε = 16 εo

O3 O2 O1

Region 2ε = 4 εo

μ = μo

Region 1εo, μo

2

2

3d

a z

Z

x

y

Figure 5.7 A plane wave is normally incident on the interfaces between three

dielectric regions

Reflection of electromagnetic waves Lect

Documents

Transcript of Reflection of electromagnetic waves Lect