ECE440 MW Chapter_2_EM Plane Wave Propagation

7/23/2019 ECE440 MW Chapter_2_EM Plane Wave Propagation

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Prof Fawzy IbrahimECE440 Ch2_EM Plane Wave1 of 53

ECE440 (EET416) Microwave Engineering

Chapter 2

Electromagnetic Plane Wave Propagation

Prof. Fawzy Ibrahim

Electronics and Communication Department

Misr International University (MIU)


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Chapter Contents

2.1 Introduction

2.2 Plane Wave Equation

2.3 Plane Wave Solution in a Lossless Medium

2.3.1 Classification of wave solution

2.3.2 Solution of Wave Equation

2.3.3 Basic plane wave parameters2.4 Plane Wave in a general Lossy Medium

2.5 Wave Propagation in Good Conductors

2.6 Poyntings Theorem and Wave Power

2.6.1 General EMFs2.6.2 Sinusoidal EMFs

2.7 Reflection of Uniform Plane Wave


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2.1 Introduction Our first application of Maxwell's equations will be in relation to

electromagnetic wave propagation. The existence of EM waves, predicted byMaxwell's equations, was first investigated by Heinrich Hertz. After severalcalculations and experiments Hertz succeeded in generating and detectingradio waves, which are sometimes called Hertzian waves in his honor.

In general, waves are means of transporting energy or information. Typicalexamples of electromagnetic waves include radio waves, TV signals, radarbeams and light rays.

All forms of electromagnetic energy have three fundamental characteristics:1. They all travel at high velocity (velocity of light, c =3x108 m/s) in free space.2. While travelling, they assume the properties of waves.3. They radiate outward from a source.

This chapter is concerned with the application of Maxwells equation to theproblem of electromagnetic wave propagation.

The uniform plane wave represents the simplest case, while it is appropriatefor an introduction, it is of great practical important.


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2.1 Introduction (Continued)

In this chapter, our major goals are to:1. Solve Maxwell's equations and determine the wave characteristics andparameters in the following media:

a. Free space ( = 0, = o = 4 x 10-7 H/m and = o= 8.854 x10

-12 =(1/36)x10-9 F/m)

b. Lossless dielectrics ( = 0, = ro and = ro)

c. Lossy dielectrics ( 0, = ro, = ro ) and > )

where is the angular frequency of the wave.

2. Investigate the power carried by a wave.

3. Understand the reflection and transmission of plane wave between twodifferent Media.

Chapter Objectives


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2.2 Plane Wave Equation When considering electromagnetic waves, in free space, it is called sourceless

(v = 0 , J = 0 a n d = 0). Under these conditions, Maxwells equations in point

form may be written in terms of and only as:

0.

vE

(2.3)t

E

t

DJHx

(2.1)t

H

t

BEx

(2.2)

0. H

(2.4) Most of our work will be involved with fields having a sinusoidal or harmonic

time dependence. In this case phasor notation is very convenient and so all field quantities will be

assumed to be complex vectors with an implied eit time dependence. As an example if a sinusoidal electric field in the x-direction has the form:

xatzyxAtzyxE )cos(),,(),,,(

(2.5)

Where A is the real amplitude, is the radian frequency, = 2f, f is the wavefrequency and is the phase reference of the wave, at t = 0, it has the phasorform:

x

j

s aezyxAzyxE ),,(),,(

(2.6)


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Example 2.1

Express Ey(z, t) =0.5 cos (2x109t 30z + 60o) V/m as a phasor.

Solution:

Step #1: Write the exponential notation or equation as:

Ey(z, t) = Re[0.5 e j(2x109t 3 0 z + 6 0o) ]

Step #2: drop Re and suppress ej2x109t to obtaining the phasor:

Eys(z) = 0.5 e(- j 30z + j 60o)

Note that Ey is real, but Eys is in general complex. Note also that a mixed nomenclature is commonly used for the angle; that is,

30z is in radians, while 30o is in degrees.

2.2 Plane Wave Equation (Continued)

We will assume cosine-based phasor, so the conversation form phasor quantities to real time-varying quantities is accomplished by multiplying thephasor by eit and taking the real part:

]),,(Re[),,,( tjs ezyxEtzyxE

(2.7)


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Note: Using phasor notation we get the following relations:

Ej

dtE

1

HjHjEx o

EjtE

(2.8) (2.9)

In source free, linear isotropic, homogenous region, applying Eq.(2.8), theMaxwells equations [(2.1) and (2.2)] in the Phasor form are:

EjHx o

(2.10)

(2.11)


EEEEExx

222 0).(

HxjExx o

Equations (2.10) and (2.11) constitute two equations for the two unknowns

and as follows:

1. Take the Curl operator of both sides of (2.10), we have:

2. Apply the curl identity described in Eq.(A.5) we have:

(2.12)

(2.13)

0.

vE

0. H

ExxEE

).(2


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022 HH oo

022 EE oo


From (2.11) and (2.13) substitute (2.12) we have

022 EkE o

(2.14)

An identical Equation for can be derived in the same manner:

022 HkH o

or

or(2.15)

Equations (2.14) and (2.15) are called the wave equations or Helmholtz

equations The constant is calledthe wave number or propagation constant

In the free space ; its unit is rad/m.

so ko is real and ko = 2/o = / c, where c is the speed of light.

oook

o

oooc

f

ck

22

2

2

2

2

2

22 .

z

V

y

V

x

VVV

The Laplacian operator, 2 is defined in Equ (A.6) as:

(A.6)


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It is now possible to separate the solutions of the wave equation in acharge free region into three basic types of fields. These are:1. Transverse Electro-Magnetic wave (TEM wave):

It is characterized by the condition that both the electric and magnetic fieldvectors lie in a plane perpendicular to the direction of propagation, i.e.having no components in the direction of propagation.

2. Transverse Electr ic wave (TE, or H-wave):

It is characterized by having an electric field which is entirely in a planetransverse to the (assumed) direction of propagation. Only the magneticfield has a component in the direction of propagation and hence thiswave type is also known as H-waves. For TE waves, it is possible toexpress all field components in terms of the axial magnetic fieldcomponent.

3. Transverse Magnet ic wave (TM, or E-wave)It is characterized by having a magnetic field which in entirely in a planetransverse to the (assumed) direction of propagation. Only the electric field

has a component in the direction of propagation, and hence this wavetype is also known as E-wave. For TM-wave, it is possible to express allfield components in term of the axial electric field component.

2.3 Plane Wave Solution in a Lossless Medium

2.3.1 Classification of wave solution


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A basic plane wave solution to the above equations can be found by

considering an electric field with only an x component and uniform (novariation) in the x and y, then we get:

zjkzjk

xoo eEeEzE

)(

2.3 Plane Wave Solution in a Lossless Medium (continued)


0)(

)(

zEthenazEE xxx

The Helmholtz or wave equation (2.14) is reduced to:

(2.16)and 0)(

y

zEx

0)(

)( 22

2

zEkz

zE

xo

x

(2.17)

The solution to Eq.(4.17) will be forward and backward propagating waveshaving the general form:

(2.18)

Where E+ and E- are arbitrary constants.

The above solution is for time harmonic case at frequency . In time domain,this result is written as:

])(Re[),( tjxx ezEtzE

)cos()cos(),( zktEzktEtzE oox (2.19)


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Example 2.2

Given the complex amplitude of the electric field of a uniform plane wave,E+ = [100 x + 2030o y] V/m, construct the phasor and real instantaneousfields if the wave is known to propagate in the forward z direction in freespace and has frequency of 10 MHz.

Solution

We begin by constructing the general Phasor expression, Eq. (2.18):(z) = [100x + 20ej30

o y] e-jk0z

Where ko =/ c = 2 f/c = (2 x 107)/(3 x 108) = 0.21 rad/m.

The real instantaneous form is then found through the rule expressed in Eq.(2.7):

(z, t) = Re[100e j0.21z e j2x107 x + 20e j30o

e j0.21z e j2 x107 y]

= Re[100e j(2 x 107t 0.21z) x + 20ej(2 x 107t 0.21z 30o) y]

= 100 cos (2 x107t 0.21z) x + 20 cos (2 x107t 0.21z +30o) y




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Note the following characteristics of the forward wave in eq.(2.19):

1. It is time harmonic because we assumed time dependence ejt to arrive .2. E+ is called the amplitude of the wave and has the same units as E.3. (t - k0 z) is the phase (in radians) of the wave; it depends on time t and space

variable z.4. is the angular frequency (in radians/second) and k0 is the phase constant or

wave number (in radians/meter).

Due to the variation of E with both time t and space variable z, we may plot Eas a function of t by keeping z constant and vice versa. The plots of E(z, t = constant) and E(t, z = constant) are shown in Fig. 2.3(a) and (b),respectively.

From Fig. 2.3(a), we observe that the wave takes distance to repeat itself andhence is called the wavelength (in meters).

From Fig. 2.3(b), the wave takes time T to repeat itself; consequently T isknown as the period (in seconds). Since it takes time T for the wave to travel

distance at the speed v, we expect that:

2.3.3 Basic plane wave parameters


)cos(),( zktEtzE ox

f

vTv

p

p


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Fig. 2.3 Plot of : (a) with constant t and (b) with constant z.)sin(),( zktEtzE ox

But T = l/f, where f is the

frequency of the wave in (Hz).

Because of this fixedrelationship betweenwavelength and frequency,

one can identify the position ofa radio station within its bandby either the frequency or thewavelength. Usually thefrequency is preferred.

Which shows that for everywavelength of distancetraveled, a wave undergoes aphase change of 2 radians.

fvf

vTv p

p

p

Tf /22

ook /2


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1. Phase velocity, vp To maintain a fixed point on the wave, the phase is constant, therefore:

phase = (t - ko z) = constant (2.20)

The velocity of the wave is called the phase velocity, because it is the velocityat which a fixed phase point on the wave travel and is given by dz/dt:

00)( dt

dz

kzktdt

d

oo

ooooo

pkdt

dzv

1



(2.21)

For free space, the phase velocity is given by:

cmx

xxx

voo

p

sec/10310

36

1104

11 897

Where c is the speed of light in the free space.


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Fig. 2.4 Plot of : (a) at t=0; (b) at t=T/4 (c) at t=T/2.P moves along +z direction with velocity vp.

)sin(),( zktEtzE ox

In summary, we note the following:1. A wave is a function of both time

and space.2. Though time t = 0 is arbitrarily

selected as a reference for thewave, a wave is without beginning

or end.3. A negative sign in (t ko z) is

associated with a wave propagatingin the +z direction (forward travelingor positive-going wave) whereas apositive sign indicates that a wave istraveling in the -z direction(backward traveling or negativegoing wave).

1. Phase velocity, vp


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2. The Wavelength,

The wavelength o, is defined as the distance between two successivemaximum (or minima, or any other reference points) on the wave, at a fixedinstant of time, thus:

[(t koz ) (t ko (z + o) ] = 2 ko o = 2So, o = 2/ ko = 2 Vp / = Vp / f For free space = c / f (2.22)

3. The Wave Impedance or medium intrinsic impedance, In section 2.3.2, one of wave equations is solved [(2.14) for electric field or

(2.15) for the magnetic field ]. In general, whenever or is known, the other filed vector can be readily

found by using one of Maxwells curl equations. Thus applying (2.10) to the electric field obtained (2.18) gives:

2.3.3 Basic plane wave parameters (continued)


sos HjEx

Hjaaz

zEa

zE

zyx

aaa

EEE

zyx

aaa

ECurl ozyx

x

x

zyx

zyx

zyx

0

)(0

00)(


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Substitute using Eq.(2.18) for Exs, we have:

yox Hjdz

zdE

)(



])()[(1 zjk

o

zjk

oyoo eEjkeEjk

jH

][ zjkzjko oo eEeEk

][)( 0 zjkzjkyoo eEeE

kzH

)]cos()cos([),( zktEzktEk

tzH ooo

oy

In real instantaneous form, (2.23) becomes:

(2.23)

where E+ and E- are assumed real.

(2.24)

3. The Wave Impedance or medium intrinsic impedance,

Which is greatly simplified for a single Ex component varying only with z andHx = Hz = 0, we get:

So


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In general, Eqs.(2.23) and (2.24) show that the electric and magnetic fieldamplitudes of the forward propagating wave in free space are related through



yoy

o

oy

ox HHH

kE

0

We also find the backward propagating wave amplitudes are related through

(2.25)

yoyyx

HHHk

E

0

(2.26)

Where is called wave impedance for the plane wave and the intrinsicimpedance for the medium, for free space, the intrinsic impedance is given by:

377120

1014436/10

104 229

7

0

x

x

k o

ooo

3. The Wave Impedance or medium intrinsic impedance,


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Example 2.3

A uniform plane wave in free space its electric field intensity is given by:

V/m. Find:

a) The propagation constant, ko. b) The wavelength, .

c) The magnetic field intensity, at the point P(0.1, 1.5, 0.4) at t = 8 ns.

Solution a) The propagation constant, ko is given by:

b) The wavelength, is given by:

c) The magnetic field intensity, is given by (2.25) as:

At the point P(0.1, 1.5, 0.4) at t = 8 ns, we have:

zo ayktxE )102cos(8.0 8

||H

mradx

x

ck ooo /094.2

103102

8

8

mxf

c33.0

103

108

8

H

xx

o

o

a

ytx

a

yktx

H 120

)094.2102cos(8.0

)102cos(8.0 88

mmAaxxxxH o

x /78.3|108cos|2.12)5.1094.2108102cos(12.2|| 98

mmAaytx x /)094.2102cos(12.2 8




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Fig.2.5 (a) Plot of E and H as functions of z at t = 0; (b) plot of E and H atz = 0. The arrows indicate instantaneous values.

The first term in (2.19) represents a wave traveling the +ve z direction andthe second term represents a wave traveling the - ve z direction

Some feeling for the way in which the fields vary in space may be obtainedfrom Fig. 2.5.




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Fig.2.5 (c) Relation between E and H field vectors and the direction of propagation, z.




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Example 2.4 The electric field in free space is given by:

(a) Find the direction of wave propagation.

(b) Calculate ko and the time it takes to travel a distance of /2.

(c) Sketch the wave at t = 0, T/4, and T/2.

Solution

(a) From the positive sign in (t + kox), we infer that the wave is propagatingalong the ve x direction (- x).

(b) In free space, vp = c. From (2.21), we have:

Since the wave is traveling at the speed of light c, we have:

mVaxktE yo /)10cos(50 8

mradxc

kk

cv oo

p /333.03

1

103

108

8

nsxxc

k

ctct o 42.31

1032

6

2

/2

22 811




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Example 2.4 Solution:

(c) As discussed before:

Ey at t = 0, T/4 and T/2 is plotted against x as shown in Fig. 2.6. Notice that apoint P (arbitrarily selected) on the wave moves along - x as t increases.

This shows that the wave travels along the ve x direction (- x).

mVxkxktEtat ooy /)cos(50)cos(500

mVxkxk

xkxktETtat

oo

ooy

/)sin(50)2/cos(50

)4

2cos(50)cos(504/

mVxkxk

xkxktETtat

oo

ooy

/)cos(50)cos(50

)2

2cos(50)cos(502/




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Fig.2.6 The wave travels along the ve x direction (- x).

Example 2.4 Solution:

2.3.3 Basic plane wave

parameters (continued)



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The analytical treatment of the uniform plane wave is extended when the wavepropagates in dielectric of conductivity, , permittivity, and permeability, .

The medium is assumed to be homogenous (having constant and withposition) and isotropic (in which and are invariant with the field orientation).

In this case ( = 0 a n d J 0)The wave equations or Helmholtz equations are:

ss HjEx

2.4 Plane Wave in a general Lossy Medium

sss EjEHx

(2.27)

The resulting wave equation for E then becomes:

0)1(22

ss EjE

022 ss EE

(2.28)

(2.29b)or

t

EJ

t

DJHx

t

H

t

BEx

or

or

(2.29a)


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Where is the wave number or propagation constant of the medium. Inthis case is a function of the material properties, as described by , and

. is a complex value os it is referred to as complex propagation constantfor the medium defined by:

zzxs eEeEzE

)(

2.4 Plane Wave in a general Lossy Medium (continued)

zjz

xs eEeEzE )()(

jjj 1

If we again assume an electric field with only an x-component and uniform inx and y, the wave equation (2.29) has the phasor solutions in the form:

The positive traveling wave in phasor form after substituting for is:

(2.30)

(2.32)

(2.31)

)cos(),( zteEtzE zx

Multiplying (2.32) by ej and taking the real part yields a form of the field in

time domain that can be visualized as:(2.33a)

The magnetic field intensity is given by:

)cos(),( zteE

tzH zy

(2.33b)


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From eqn. (2.33), it is seen that as and waves travel in a conductingmedium, its amplitude is attenuated by the factor e-az. The distance ,

through which the wave amplitude decreases by a factor e-1 (about 37%) iscalled skin depth or penetration depth of the medium.


E

H

Fig.2.7 Skin depth

1 eEeE

1

The skin depth is a measureof the depth to which anelectromagnetic wave canpenetrate the medium.


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][1200 r

r

r

r

k

j

Notes:1. The uniform plane wave that propagates in the forward z-direction with

phase constant () and an exponential decay factor e-z

.2. The rate of decay with distance is called the attenuation constant or attenuation coefficient (). The attenuation coefficient is measured in nepersper meter (Np/m).If the region for z > 0 is a lossless dielectric, then = 0 or


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Example 2.5 A plane wave propagating in lossless dielectric medium (r, r = 1and = 0) has an electric field given as Ex= E

+ cos (1.51x 010 t 61.6 z). For this wave, determine the:

(a) Wavelength, . (b) Phase velocity, vp.(c) Dielectric constant of the medium,r. (d) Intrinsic or Wave impedance,

Solution

(a) By comparison with (2.19) we identify = 1.51 x 1010 rad /sec and k= 61.6 m-1

= + j = j 61.6 m-1, then = 0 and k = = 61.6 which gives the wavelength (2.35) as:

m

k

102.0

6.61

222


(b) The phase velocity can be found from (2.34):

sec/1045.26.61

1051.1 810

mxx

kp

50.1)1045.2100.3()( 28

8

2 xxcgivesccv

p

r

rrr

p

8.3075.1

377

r

o

k

j

(c) This is slower than the speed of light by a factor of 1.225. The dielectricconstant,rof the medium can be found as:

(c) The intrinsic or wave impedance, is given by (2.36):


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Many problems of practical interest involve loss or attenuation due to good(but not perfect) conductors.

If metals can be categorized as good conductors. From Eq.(2.30) thecomplex propagation constant

is given by:

2.5 Wave Propagation in Good Conductors (lossy Medium)

jjj 1

for , the one can be neglected with respect to / and we get

jjjj

but

therefore

fjf

j

jjjjj

)1(2

2)1(

2)1(

2)1(

ooandj 451901901

)1(2

1)45sin45(cos451 0 jjj oo


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Regardless of the parameter and of the conductor or the frequency of

applied field, and are equal. If we again assume only an Ex componenttraveling in the +z direction the Eq.(2.33) becomes:

2.5 Wave Propagation in Good Conductors (continued)

)cos(),( zteEtzE zx )cos(

fzteE

fz (2.44)

Definition: The depth of penetration (s) is the distance traveled inside themedium at which the amplitude of the wave attenuates e (e 2.7) times from

its initial level as shown in Fig.2.2. This is also called skin depth or skin effect. From Eq.(2.44) the field amplitude is given by:

zfz

x eEeEzE )(

1x |(z)E|

eEeE f

s

(2.45)

At z = 0 |Ex(z)| = E+ and

at z = s |Ex(z)| = E+/ e, therefore:

Fig.2.2

211

fs

The skin depth or depth of penetration s is:

(2.46)

f (2.44)Therefore:

)cos(),( zteEtzE zx


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/2/2/ 2 spV

ffs 2

122

2

Notes:

1. The uniform plane wave that propagates in the forward z-direction with

phase constant and attenuation constant or coefficient () as:

2. The phase velocity

3. The wave length

(2.47)

(2.48a)

(2.48b)

4. The intrinsic impedance

j

f

(2.48c)


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mxx

xAls

77

3 1014.810816.3

11003.5 a) For Aluminum:

mxx

xCos

7

7

3 1060.610813.5

11003.5

mxx

xGls

77

3 1086.710098.4

11003.5

mxx

xSls

7

7

3 1040.610173.6

11003.5

b) For Copper:

c) For Gold:

d) For Silver:

11003.5

1

)104)(10(

112 3710

x

xf os

Example 2.6 Compute the skin depth (s) at a frequency of 10 GHz of thefollowing materials with conductivity, as indicated:

a) Aluminum (Al = 3.816x107

S/m). b) Copper (Co = 5.818x107

S/m).c) Gold (Gl = 4.098x107 S/m). d) Silver (Sl = 6.173x107 S/m).Solution: Eq. (2.46) gives the skin depths as:


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Example 2.7A plane wave propagating in the sea water medium (r=81, r = 1and = 4 S/m) Consider an incident wave of frequency 1 MHz, find the:a) Skin depth,s. b) Wavelength, .c) Phase velocity, vp . d) Intrinsic or Wave impedance, .

Solution

(a) The loss tangent is given by:

ms

6.122


(c) The phase velocity can be found from (2.35):

sec/106.1)25.0(102 66 mxxp

1890)1085.8)(81)(102(

4

' 126

xx

987.0987.0)44()44(

)44(1041012 76j

jxj

jxxxxxj

Sea water is therefore a good conductor at 1 MHz, the skin, s of Eq. (2.46) is:

(d) The intrinsic or wave impedance, is given by (2.49):

(b) the wavelength as:

cmmxxf os 2525.0)4()104)(10(

1121176

in free space the wavelength 0 = 300 m

in free space v = c

44 jj


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Example 2.8A plane wave is given by:

Determine the following:

a) The propagation constant and the wave parameters (Vp, , and s)b) The magnetic field, associated with the wave in both phasor and time domain

representations.

Solution

a)

b)


mVaztetzE xz /)410cos(5.0),( 94

mradjThen /)44(4

222)5050(

)44(

7104910

)44(

71049104

j

ejj

xx

j

xxjj

ms

25.04

11

mm 57.124

22

sec/8

105.24

910

mxp

mVxazt

zetzE /)4910cos(45.0),(

xa

zj

e

z

eEzsE )(

mVxa

zj

e

z

ezsE /

44

5.0)(

ya

zje

ze

Ez

sH )(

mA/m

)44(425.2

4222

5,0)(

ya

jzje

ze

ya

zje

ze

je

zs

H

myazt

zetzH /mA)

44910cos(425.2),(


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In 1884 an English physicist, John H. Poynting developed a power theorem

for the electromagnetic field. This theorem determined the power flow associated with an electromagneticwave which can be transported from one point (where a transmitter islocated) to another point (with a receiver) by means of EM waves.

The Poynting vector in watts per square meter (W/m2); that is defined as:

It represents the instantaneous power density vector associated with theEM field at a given point. The integration of the Poynting vector over anyclosed surface gives the net power flowing out of that surface.

2.6 Poyntings Theorem and Wave Power

(2.49)

2.6.1 General EMFs

)),,,(),,,((),,,( tzyxHxtzyxEtzyxP


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Notes:

1. The Poynting vector2. Direction of : in other words is in the direction of the

wave propagation.

3. Magnitude of : power density (power/area or W/m2)

4. The theorem states that the electromagnetic power that enters a volumethrough its boundary is equal to the rate of increase of stored energy in the

volume and the electromagnetic power lost in the volume .2.6.2 Sinusoidal EMFs

For sinusoidal EMFs if we assume that the electric and magnetic fieldintensities have the forms:

then

The instantaneous Poynting vector or instantaneous power density vector is:

2.6 Poyntings Theorem and Wave Power (continued)

)( HxEP

HPandEP

P

P

P

2.)(

m

W

m

A

m

VHxEP

2.6.1 General EMFs

x

z azteEtzE )cos(),(

yz azteEtzH )cos(),(

(2.50a)

(2.50b)

z

zaztzte

EHxEP )cos()cos()( 2

2

(2.51a)


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Using the trigonometric identity: Cos A Cos B = [Cos (A+B) +Cos (A-B)],

Eqn.(2.51a) becomes:

To determine the time-average Poynting vector or power density in[W/m2], which is of more practical value than the instantaneous Poyntingvector , we integrate eq. (2.51b) over the period T = 2/; that is:

Eqn.(2.52) can be derived using vector analysis as follows:


(2.51b)

(2.52)

2.6.2 Sinusoidal EMFs

z

zazte

EtzP )]cos()22[cos(

2),( 2

2

)(zPdav

),( tzP

T

z

z

T

dav dtazteE

TdttzP

TzP

0

22

0

)]cos()22[cos(2

1),(

1)(

z

z

z

z

dav aeE

aeE

zP )cos(2

)]cos(0[2

)( 22

22

]Re[)cos(),( )( ztjzxz eeEazteEtzE

]Re[)cos(),( )(

ztjzyz

eeE

azteE

tzH

(2.53a)

(2.53b)


39/53


The time-average power density or Poynting vector, Eqn.(2.51), is

equivalent to :

The total time-average power, crossing a given surface S is given by:

We should note the difference between , , and

is the instantaneous Poynting vector or instantaneous powerdensity vector in [W/m2] and it is time varying. .

is the time average of the Poynting vector or time average powerdensity vector also in [W/m2] and it is time invariant.

is the total time average power through a surface in watts; it is a scalarand time invariant.


(2.54)

(2.55)

2.6.2 Sinusoidal EMFs

z

z

ssdav aeE

HxEHxEzP )cos(2

]Re[2

1]Re[

2

1)( 2

2**

)(zPdav

S

tav sdHxEP

).Re(

2

1 *

),,,( tzyxP

),,( zyxPdav

tavP),,,( tzyxP

),,( zyxPdav

tavP

tavP


40/53


2.6 Poyntings Theorem and Wave Power (continued)2.6.2 Sinusoidal EMFs

Example 2.9 In a nonmagnetic medium (o and ) if the electric field intensity of

a plane wave is given by:Determine and calculate the following:a) rand .b) The time-average power density of the wave, .c) The total time-average power, crossing an area, A = 100 cm2 of plane

2x + y = 5 .

Solution a) Since = 0 and ( /c, the medium is not free space but alossless medium.

= k = 0.8, = 2 x107, =o (nonmagnetic), = ro

and

b) The time-average power density of the wave is given by:

mVaxtxtxE z /)8.0102sin(4),( 7

12

102

1038.07

8

xxc

ck rroro

59.14r

][7.9812120

1200

xkrr

r

),,( zyxPdav

22

* /817.982

16

2]Re[

2

1)( mmWaa

xa

EHxExP xxxdav

tavP


41/53


2.6 Poyntings Theorem and Wave Power (continued)2.6.2 Sinusoidal EMFs

Example 2.9 Solution

c) The normal to plane 2x + y = 5 is given by:Hence the total time-average power is: 5

2 yxn

aaa

Wx

aaxax

aAxSsdHxEP

yxx

n

S

tav

5.7245

10162

52)10100.()1081(

).().Re(2

1

5

43

*


42/53


We consider the phenomenon of reflection which occurs when a uniformplane wave is incident on the boundary between regions composed of two

different materials. We study the reflection of a plane wave normally incident from free spaceonto a surface of a conducting half-space. The geometry is shown in Fig.2.3where the lossy half space z > 0 is characterized by parameters , and.

With no loss of generality, we assume that the incident plane wave has anelectric field vector oriental along the x-axis and is propagating along thepositive z-axis.

2.7 Reflection of Uniform Plane Wave

Fig. 2.3 Plane wave reflection from a lossy medium, normal incidence


43/53


The incident fields can be then written in the phasor form, for z < 0 as:

2.7 Reflection of Uniform Plane Wave (continued)

x

zjk

is aeEzE

o

)(

y

zk

ois

aeEzH 1

)( 0

(2.53) (2.57)

Where, for free space Ko is the wave number, o is the wave impedance andE+ is an arbitrary amplitude.

A reflected wave may exist in the region z < 0, with the form:

x

zjk

rs aeEzE o )(

)()( 0 yzjk

o

rs aeEzH

(2.58) (2.59)

Where is called the reflection coefficient of the reflected electric field.

Note the sign in the exponential terms of (2.56 and 2.57) has been chosenas negative, to represent waves traveling in the +ve z direction of propagation.

On the other hand, the sign in the exponential term of (2.58 and 2.59) has

been chosen as +ve, to represent waves traveling in the ve z direction ofpropagation. The transmitted fields for z > 0 in the lossy medium can bewritten in the phasor form as:

x

z

ts aeTEzE )(

y

z

ts aeTE

zH )(

(2.60) (2.61)


44/53


Where T is the transmission coefficient of the transmitted electric field an isthe intrinsic impedance of lossy medium.

From (2.42) the intrinsic impedance is given by:

And from (2.30) complex propagation constant or wave number is given by:

)()()( zEzEzE Tsrsis

jjj /1


(2.62)

We now have a boundary value problem, where the general form of fields areknown via (2.66) (2.61) on either side of the material discontinuity at z = 0.

The two unknown constants and T are found by applying two boundaryconditions on Ex and Hy a t z = 0 .

Since these tangential field components must be continuous at z = 0, for theelectric field, we have the following equation:

(2.63)

From Equations (2.60), (2.61) and (2.62) substitute in (2.63), we get:

(2.64)

/j

x

zjk

is aeEzE o )(

x

zjk

rs aeEzE o )(

x

z

ts aeTEzE )(

T1


45/53


Similarly for the magnetic field, from Equations (2.57), (2.59) and (2.61) weget:


0

1(2.65)

o

T

21

o

o

Solving Equation (2.64) and (2.65) for the reflection and transmission

coefficients, we have:

(2.66)

(2.67)

This is the general solution for reflection and transmission of a normallyincident wave at the interface of lossy material, where is the intrinsicimpedance of the material.

and T are complex numbers.

y

zk

o

is aeEzH 1

)( 0

)()( 0 yzjk

o

rs aeEzH

y

z

ts aeTE

zH )(

)()()( zHzHzH Tsrsis


46/53


Example 2.10

Consider a plane wave normally incident on a half-space of copper ( =

5.813x107

S/m, r= 1 and r= 1). If wave frequency, f = 1 GHz, compute:(a) The propagation constant or wave number,..(b) The intrinsic impedance, .(c) The skin depth, s for the conductor.(d) The reflection coefficient, .(e) The transmission coefficient, T.Solution:

(a) The complex propagation constant for the medium is given by (2.30) as:


5

5779

10)79.479.4(

1079.410813.510410

xjSo

xxxxxxf

j Where and are given by (2.43) as:

(b) From (2.36) The intrinsic impedance, is calculated as:

3

5

79

10)239.8239.8(10)79.479.4(

104102xj

xj

xxxj


47/53


Example 2.10 Solution (continued)

o

o

o

xj

xjxj

xj

99.179110239.8008.377

10239.899.37637710)239.8239.8(

37710)239.8239.8(

3

3

3

3


(e) The transmission coefficient, T is calculated from (2.67) as:

o

o

x

xjxjT

4510181.6

10239.8008.37710)239.8239.8(22

5

3

3

(c) The copper is a good conductor, then the skin, s Eq. (2.46) is:

(d) The reflection coefficient, is calculated from (2.66) as:

mxf o

s

610088.2121


48/53


Example 2.11

Consider a plane wave normally incident from free space on a half-space of

a medium 2 with parameters (r = 3.5 and r=2.5) as shown in Fig. 2.4. If the wave frequency, f = 1 GHz and the incident electric field in the phasorform, for z < 0 is:

where E+ = 10 mV/m, find the phasor and instantaneous values of theincident wave ( , ), the reflected wave ( , ) and the transmitted wave( , ).

Solution:

The incident magnetic field intensity in phasor form is:

The instantaneous incident magnetic field intensity is:


x

zjk

is aeEzE o )(

iE

iH

Fig. 2.4 Plane wave reflection.

rE

rH

iE

iH

]/[94.20

5.26377

10

0

1)( 00 mA

ya

zje

ya

zjke

ya

zjkeEz

isH

sradx

x

ck ooo /94.20

103

1028

9

]/[)94.20102cos(5.26),( 9 mAyaztxtz

isH

44610)36/1(5.2

1045.39

7

xx

xx

j

jj

084.0

377446

377446

0

0


49/53


Example 2.11 Solut ion:

The reflected electric field intensity in instantaneous and phasor forms are:

The reflected magnetic field intensity in instantaneous and phasor forms are:

The transmitted electric field intensity in instantaneous and phasor forms are:

The transmitted magnetic field intensity in instantaneous and phasor forms are:


]/[27.110377

084.0)( 000 mA

ya

zjke

ya

zjkxex

ya

zjkeE

o

zrs

H

1084.1377446

)446(22

0

T

]/[94.61

84.1094.61

)10)(084.1()( mmVzj

exa

zje

xa

jkzeTE

xaz

tsE

]/[94.61

3.24446

84.10)( mA

zje

ya

zke

ya

zjke

TE

yaz

tsH

]/[94.20

84.0)10)(084.0()( 00 mmVxazj

exazjk

exazjk

eEzrsE

94.615.25.394.20 xkk rro

]/[)94.20102cos(84.0),( 9 mmVxaztxtz

rsE

]/[)94.20102cos(27.1),( 9

mAyaztxtzrsH


50/53


Example 2.12

Consider a plane wave normally incident from a medium 1 with parameters(

1,

1and

1) on a half-space of a medium 2 with parameters (

2,

2and

2) as shown in Fig. 2.6. For 1 = 100 , 2 = 300 and The incidentelectric field in the phasor form, for z < 0 is:

If E+ = 100 V/m calculate the values for:

(a) The incident wave and the magnitude for the average incident power

density.(b) The reflected wave and the average

reflected power density.

(c) The transmitted wave and the average

transmitted power density.


Fig. 2.5 Plane wave reflection from a medium 1, normal incidence on medium 2.

x

zjk

is aeEzE )( 1


51/53



(a) The incident magnetic field in the phasor form, for z < 0 is calculated from

(2.66) as:

The magnitude for the average incident power density is given by.

(b) The reflection coefficient, is calculated from (2.66) as:

The reflected wave that exists in the region z < 0, is calculated from (2.66) as :


5.0100300

100300

12

12

]/[100

100

1)( 111

1

mAaeaeaeEzH yzjk

y

zjk

y

zjk

is

]/[50)1)(100(2

1

)1

()(Re

2

1)Re(

2

1

2

1

* 11

mWaa

aeExaeEHxEP

zz

y

zjk

x

zjk

iiidav

]/[50)100)(5.0()( 111 mVaeaeaeEzE xzjk

x

zjk

x

zjk

rs


52/53



The reflected magnetic field in the phasor form, for z < 0 is calculated from

(2.69) as:

The magnitude for the average reflected power density is given by.

(c) The transmission coefficient, T is calculated from (2.67) as

The transmitted wave that exists in the region z > 0, is calculated from (2.68) as:


]/)[(5.12)5.0)(50(2

1)(

2

1)Re(

2

1 2** mWaaHEHxEP zzrrrrrdav

]/[)(5.0)(100

50)()( 111

1

mAaeaeaeEzH yzjk

y

zjk

y

zjk

rs

5.1400

600

100300

)300(22

12

2

T

]/[150)100)(5.1()( 222 mVaeaeaeTEzE xzjk

x

zjk

x

zjk

ts


53/53



The transmitted magnetic field in the phasor form, for z > 0 is calculated

from (2.59) as:

The magnitude for the average transmitted power density is given by.

We may check and confirm the power conservation requirement:

Then:


]/[5.37)5.0)(150(2

1

2

1)Re(

2

1 2* mWaaaHEHxEPzzztttststdav

]/[5.0300

150)( 222

1

mAaeaeaeTE

zH yzjk

y

zjk

y

zjk

ts

tdavrdavidav PPP

]/[5.37]/[5.12]/[50 222 mWmWmW

ECE440 MW Chapter_2_EM Plane Wave Propagation

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Transcript of ECE440 MW Chapter_2_EM Plane Wave Propagation