07 Plane Electromagnetic Waves

7/27/2019 07 Plane Electromagnetic Waves

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Plane Electromagnetic

Waves

Chapter 7


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Introduction

7 - 1


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Introduction

In a source-free non-conducting simple medium,Maxwells equations are combined to give the

source-free wave equation

The solutions of wave equations represent waves.

Main concern of this chapter is the behavior of

waves that have a 1-D spatial dependence, i.e.,plane waves.

7-1

22

2 2

10

pu t

=

E

E


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Uniform Plane Waves

wavefront : virtual surface in space where thefields have the same phase everywhere on this

surfacealso called equiphase surface

plane wave : a special solution of wave equation

(Maxwells equations) with E assuming the samedirection and phase on infinite planesperpendicularto the propagation direction

in other words, on an infinite plane wavefront, E alsohas the same direction

uniform plane wave : E also has the same

magnitude on the infinite plane wavefront

7-1


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Approximation of Plane Waves

Strictly speaking, uniform plane waves do notexist in practice because a source ofinfinite

extent would be required to create them.However in real life situations, if we are far away

enough from any source, the wavefront becomes

almost spherical, and a very small portion of thesurface of a huge sphere is very close to a plane.

e. g., the ground of the earth

The characteristics of uniform plane waves areparticularly simple, the study of them is offundamental and practical importance.

7-1


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Plane Waves in Lossless

Media

7 - 2

wave equation Doppler effect transverse electromagnetic waves polarization of plane waves


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Plane Waves in Lossless Media (1)7-2

homogeneous vector Helmholtzs equation in free space

2 2 0k + =E E

2

p

ku

= = = : free-space wavenumber

in Cartesian coordinates,

2 2 22

2 2 20

x

y

z

E

k Ex y z

E

+ + + =

consider a UPW characterized by a uniform Ex over planes z,

0 and 0

x y

= =


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Plane Waves in Lossless Media (2)

: arbitrary constants to be determined by BC

2 2 22

2 2 20xk E

x y z

+ + + =

0 , 0x y = =

22

20x x

d Ek E

dz

+ =

0 0( ) ( ) ( ) jkz jkzx x xE z E z E z E e E e+ + = + = +

0 0,E E+

7-2


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Traveling Waves

( )0 0( , ) ( ) cos( )

j t j t kz

x xE z t E z e E e E t kz

+ + + + = = = e e

traveling wavepropagating in+z direction

using cosine reference

traveling wavepropagating in-z direction

0 :jkz

E e

= 2 / k

E0+ up

0

Ex+(z)

z4

2

3

4

7-2

t = 00 cosE kz+ t = /2

0 sinE kz+ t = /

0 cosE kz+


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Phase Velocity

for a particular phase point

cos( )t kz C =

(a constant phase)t kz C =

1p

dzu

dt k

= = = : phase velocity

0dt kdz =

7-2


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Magnetic Field H of a UPWj = E H

0 0 ( )

( ) 0 0

x y z

x x y y z z

x

j H H Hz

E z

+ + +

+

= = + +

a a a

E a a a

( )1, , 0x

x y z

E zH H H

j z

++ + += = =

7-2


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Intrinsic Impedance

0

( )1( ) ( )jkzx

y x

E zH E e jkE z

j z z

++ + + = = =

1( ) ( ) ( )y x x

kH z E z E z

+ + += =

= : intrinsic impedance

0( , ) ( , ) [ ( ) ] cos( )j ty y y y yE

z t H z t H z e t kz

++ += = = H a a ae

0 0 0/ 120 377 ( ) =

7-2

in free space,


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Properties of Uniform Plane Waves

For a UPW, the ratio of the magnitudes ofE andH is the intrinsic impedance of the medium.

E H propagation direction

| | / | | = E H

7-2


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Example 7-1

A uniform plane wave with E = axEx propagates in a lossless simplemedium (r= 4, r= 1, = 0) in the +z-direction. Assume that Ex issinusoidal with a frequency 100 MHz and has a maximum value of +10-4

V/m at t = 0 and z = 1/8 m.

a) Write the instantaneous expression forE

for any t and z.b) Write the instantaneous expression forH.

c) Determine the location where Ex is a positive maximum when t = 10-8 s.

Solution :

find k first,

a) using cost as reference,

8

8

2 10 44

3 10 3r rk

c

= = = =

( )4 8( , ) 10 cos 2 10x x xz t E t kz = = +E a a


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Example 7-1

a) E reaches a maximum when the argument of cosine function is zero:

b) instantaneous expression forH is

82 10 0t kz + =

0 , 1/8t z= =

4 13 8 6

kz = = =

4 8 4 84 4 1

( , ) 10 cos 2 10 10 cos 2 103 6 3 8x xz t t z t z

= + =

E a a

0, 60xy y y

r

EH

= = = = =H a a

4810 4 1

( , ) cos 2 1060 3 8yz t t z

= H a


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Example 7-1

c) E reaches a maximum when the argument of cosine function is 2n

8 8 4 12 10 (10 ) 23 8

mz n

=

13 3, 0,1,2, ; 0

8 2m mz n n z= = >

13

8mz n=

2 3

2k

= =

x

y z

3m

2=

4 4 1( ,0) 10 cos ( )

3 8

xz z= E a

(0, ) ( ,0) /y xt E z =H a

1 m8

O


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Doppler Effect (3)

time elapsed at R, t', corresponding to t at T

2 2 1/ 22 0 0

2 1 /20

0 0

0

0

1[ 2 ( )cos ( ) ]

2[1 ( )cos ( ) ]

(1 cos )

t t r r u t u t c

r u tt u t

c r r

r u tt

c r

= + +

= + +

+

7-2.1

2 20( )u t r

1/ 2

2

1(1 ) 1 ( )

2

1 1( 1)( )

2 2

x x

x

= +

+ +

2 1 (1 cos )u

t t t t c

= =

ift represents a period of the time-harmonic source, i.e., t = 1 / f1

(1 cos )

1 cos

f uf f

ut c

c

= = +

2( / ) 1u c


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Applications of Doppler Effect

The Doppler effect is the basis of operation of theradarused by police to check the speed of a

moving vehicle.The frequency shift of the received wave reflected by amoving vehicle is proportional to the speed of thevehicle.

7-2.1


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Transverse Electromagnetic Waves

A uniform plane wave with the E and H vectorstransverse (perpendicular) to the propagation

direction is called a transverse electromagnetic(TEM) wave.

For a UPW propagating in the z-direction, the phasorfield quantities are functions of only the distance z

along a single coordinate axis.

A UPW does not always propagate along acoordinate axis. It is necessary to consider a

more general case where a UPW propagatesalong an arbitrary direction.

7-2.2


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UPW Propagating Along Arbitrary Direction

7-2.2

a a ax x y y z z kk k k k = + + =k a

x y zx y z= + +R a a a

define a propagation direction vector

position vector of an arbitrary point Q in space

any point Q on this plane has the property:

constantk

OP = =a R

je k R represents a plane wave

has a plane as an equiphase surfacej

e k R

(x, y, z)Q

P

R

ak

yz

x

O

plane of constant phase


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Properties of an Arbitrary UPW (1)7-2.2

in a charge-free region,

0 =E

0 0 0( ) ( ) ( ) 0j j j

e e e = + =k R k R k RE E E

0

0 ( ) 0je =k R

E

7 2 2


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Properties of an Arbitrary UPW (2)7-2.2

E0 is transverseto propagationdirection!

( )

( )

( )

( )

x y z

x y z

k

j k x k y k zj

x y z

j k x k y k zx x y y z z

jk

k

e ex y z

j k k k e

jk e

+ +

+ +

= + +

= + +

=

k R

a R

a a a

a a a

a

0 ( ) 0

j

e

=

k RE

0( ) 0kjk

kjk e =a RE a

0 0k =a E

7 2 2


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H Field of an Arbitrary UPW7-2.2

1( ) ( )

j= H R E R

1 1( ) ( ) ( )k

= = H R k E R a E R

( ) ( )j = E R H R

0

1

( ) ( )kjk

k e

= a R

H R a E

E H ak

7 2 3


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Polarization of Plane Waves

polarization : time-varying behaviorof the E fieldvector at a given point in space looking along thepropagation direction

time-varying behavior means the possible changing ofthe direction ofE, or the magnitude ofE, orboth

separate description of the H field vector is not

necessary since it is definitely related to that ofE

three types of polarization of plane waves

elliptical polarization

linear polarization

circular polarization

7-2.3

7 2 3


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Linear Polarization7-2.3

( ) or ( )x x y yE z E z= =E a E a

for a UPW propagating in the +z-direction,


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7-2 3


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Elliptical Polarization (2)7 2.3

the sum of two linearly polarized waves in bothspace and time quadrature is elliptically polarized

1 2 10 20(0, ) (0, ) (0, ) cos sinx y x yt E t E t E t E t = + = +E a a a a

1 10 1 10(0, ) cos cos (0, ) /E t E t t E t E = =

2 20 2 20(0, ) sin sin (0, ) /E t E t t E t E = =

2 2

1 2

10 20

(0, ) (0, )1

E t E t

E E

+ =

2 2

cos sin 1t t + =2 2

2 21x y

a b+ =

equation foran ellipse

7-2.3


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Circular Polarization7 2.3

1 2 10 20(0, ) (0, ) (0, ) cos sinx y x yt E t E t E t E t = + = +E a a a a

2 2

1 2

10 20

(0, ) (0, )1

E t E t

E E

+ =

2 2 2

x y r+ =

equation fora circle

ifE10 = E20

2 2 21 2 0(0, ) (0, )E t E t E+ =

the angle E makes with the x-axis at z = 0 :

1 2

1

(0, )tan(0, )

E t tE t

= = E rotates with anangular velocity ina counterclockwisedirection!

right-handedcircularpolarization,RHCP

x

y

O E10

E(0, t)

7-2.3


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Left-Handed Circular Polarization

1 2

1

(0, )tan(0, )

E tt

E t

= =

E rotates with anangular velocity ina clockwise direction!

left-handed

circularpolarization,LHCP

10 20( )jkz jkz

x yz E e ejE = +E a a

10 20(0, ) cos sinx yt E t E t =E a a

x

y

O

E10E(0, t)

7-2.3


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Linear Polarization

1 2 10

1 20

(0, )tan

(0, )

E t E

E t E = =

10 20( )jkz jkz

x yz E e E e = +E a a

10 20(0, ) ( )cosx yt E E t = +E a ano phase difference

linear

polarization,LP

x

y

OE

10

E20

P2

P1

1 10

20

tanE

E

7-2.3


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Conditions for Elliptical Polarization

If the E field vector has

two spatially orthogonal components (e. g., Ex and Ey)with unequal amplitudes E10 E20,

the phase difference between the two components isnot zero or an integral multiple of/2,

then E is elliptically polarized with the principalaxes not coincide with the coordinate axes.

Wave Polarizations of Some Practical7-2.3


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Wave Polarizations of Some Practical

Applications

AM broadcast : LP with the E field perpendicularto the ground

TV signals : LP withE

field parallel to the groundFM broadcast : CP

therefore, orientation of an FM receiving antenna is not

critical as long as it lies in a plane normal to thedirection of the signal


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Example 7-2

An LP plane wave can be resolved into RHCP + LHCP

Solution :

for an LP plane wave propagating in the +z-direction,

0 0

0 0

( )

2 2

( ) ( )2 2

( ) ( )

jkz jkz

x x

jkz jkz

x y x y

rc lc

E Ez e e

E Ej e j e

z z

= +

= + +

= +

E a a

a a a a

E E

0 0( ) ( ) , ( ) ( )2 2

jkz jkz

rc x y lc x y

E Ez j e z j e = = +E a a E a a

0( )jkz

xz E e=E a


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Plane Waves in Lossy

Media

7 - 3

complex propagation constant low-loss dielectrics

good conductors skin effect plasma

7-3


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Fields in Simple Conducting Medium

If the simple medium is conducting ( 0), acurrent J = E will flow, Amperes law becomes

The other three equations are unchanged.

Therefore, all the previous equations for non-

conducting media will apply to conducting mediaif is replaced by c.

( ) cj j j jj

= + = + = + =

H E J E E E

c j

= : complex permittivity

,c c c

k k =

7-3


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Fields in Lossy Dielectrics (1)

When an external time-varying E field is appliedto materials, small displacements of boundcharges result, giving rise to a volume density ofpolarization P.

This polarization vector will vary with the samefrequency as the applied field.

As the frequency increases, the inertia of thecharged particles tends to prevent the particledisplacements from keeping in phase with the

field changes, leading to a frictional dampingmechanism that causes power losses.

7-3


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Fields in Lossy Dielectrics (2)

The phenomenon ofout-of-phase polarizationcan be characterized by a complex electricsusceptibility and hence a complex permittivity.

If in addition, the material has an appreciableamount of free charge carriers such as theelectrons in a conductor, where will also beohmic losses.

Forlow-loss media, damping losses are very smalland the real part ofc is usually written as .

An equivalent conductivity can be defined as

( ) ( ) ( )c j =

=

C f f7-3


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Classification of Materials

good conductor : >>

good insulation :


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Example 7-3

A sinusoidal E of amplitude 250 V/m and frequency 1 GHz exists in alossy dielectric medium that has relative permittivity of 2.5 and a losstangent 0.001. Find the average power dissipated in the medium percubic meter.

Solution :

first find the effective conductivity of the lossy medium,

average power dissipated per unit volume is

0

tan 0.001c

r

= =

99 4100.001(2 10 )( )2.5 1.39 10

36

= =

2

4 2 3

1 1

2 2

1

(1.39 10 ) 250 4.34 (W/m )2

p JE E

= =

= =

Pl W i L M di (1)7-3


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Plane Waves in Lossy Media (1)

Helmholtzs equation in lossless medium:

2 2 0ck + =E E

2 2 0k + =E E

Helmholtzs equation in lossy medium:

,

c cjk j = =

c ck =

in order to conform to transmission line notation, we define

: complex propagation constant

is complex

1/ 2

1j jj

= + = +

1/ 2

1j j j

= + =

or

Pl W i L M di (2)7-3


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Plane Waves in Lossy Media (2)

2 2 0ck + =E E

c cjk j = =

2 2 0 =E E

0

z

x x xE E e

= =E a a

UPW solution propagatingin +z-direction with Ex

0=z j z

xE E e e

j = + : attenuation constant : phase constant

general expressions ofand in terms ofandconstitutive parameters , , and are very involved

L L Di l t i (1)7-3.1


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Low-Loss Dielectrics (1)

a low-loss dielectric has a non-zero equivalent conductivity such that,

or / 1

21

12 8

j j j

= + +

( )

(1 / )

cj j j j

j j

= + = =

=

binomial expansion

2'' 1 ''

, ' 12 ' 8 '

+

Low Loss Dielectrics (2)7-3.1


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Low-Loss Dielectrics (2)

intrinsic impedance of a low-loss dielectric

E and H fields are not in time phase

1/ 2

1 1

' 2

c j j

= +

phase velocity in a low-loss dielectric

21 1

1 8pu

=

Good Conductors (1)7-3.2


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Good Conductors (1)

/ 1

( / )

(1 / )

1(1 )

2

cj j j j

j j jj

jj j f

= + = =

=

+= = = +

f = =

/ 2 1/ 2 / 4

( ) (1 ) / 2j j

j e e j

= = = +

Good Conductors (2)7-3.2


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Good Conductors (2)

intrinsic impedance of a good conductor

(1 ) (1 )cc

j fj j

= = + = +

phase velocity in a low-loss dielectric

2pu

=

considercopperas an example,

7 75.8 10 , 4 10 , 3 MHzf = = =

4720 m/s , 0.24 mm , 2.262 10p

p

uu

f

= = = = =

Skin Effect7-3.2


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Skin Effect

Skin effect : A high-frequency EM wave isattenuated very rapidly as it propagates in a goodconductor. The fields exist only in a very thinlayer (skin) close to the conductor surface.

The distance through which the amplitude of atraveling plane wave decreases by a factor ofe-1

( 0.368) is called the skin depth.

e. g., Cu @ 3 MHz, 0.038 mm

1e e

=1 1

f

= = =

1

2

= =

Skin Depths of Various Materials


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Skin Depths of Various Materials

Iron (

0.25 (m)32 (m)4Seawater

0.000160.0050.651.00107Iron (r103)

0.00270.08410.923.54

107

Aluminum

0.00250.07910.144.10107Gold

0.00210.0668.535.80107Copper

0.0020 (mm)0.064 (mm)8.27 (mm)6.17107Silver

1 (GHz)1 (MHz)f= 60 (Hz)(S/m)Material

Example 7-4 (1)


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p ( )

A UPW propagating in the +z-direction in seawater is

E=

ax100cos(10

7t)at z = 0. For seawaterr= 72, r= 1, and = 4.

a) Determine , , c, up, and .

b) Find the distance at which the amplitude ofE is 1% of its value at z = 0.

c) Write the expressions forE(z, t) and H(z, t) at z = 0.8 (m) at functions oft.

Solution :

a) attenuation constant,

phase constant,

7 610 , 5 102

f

= = =

7 90

4200 1

110 10 72

36r

= = =

6 75 10 (4 10 )4 8.89f = = =

8.89f = =

Example 7-4 (2)


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a) intrinsic impedance,

phase velocity,

wave length,

skin depth,

b) distancez

1 at which |E

(z

1)| = 1% |E

(z

= 0)|

6 7/ 4(5 10 )(4 10 )(1 ) (1 )

4j

c

fj j e

= + = + =

7

610 3.53 108.89

pu = = =

2 20.707

8.89

= = =

1 10.112

8.89

= = =

1 10.01 1/ 0.01 100z ze e = = =

1

1 4.605

ln100 0.518 (m)8.89z = = =

Example 7-4 (3)


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c) in phasor notation,

instantaneous expression forE,

at z = 0.8 m,

instantaneous expression forH,

( ) 100 z j zxz e e =E a

( )( , ) [ ( ) ] [ 100 ] 100 cos( )j t z j t z zx xz t z e e e e t z = = = E E a ae e

0.8 7 7(0.8, ) 100 cos(10 0.8 ) 0.082cos(10 7.11)x xt e t t = = E a a

( ) ( )( ) ( , ) j tx x

y y

c c

E z E zH z H z t e

= =

e

0.8 0.8 7.11

1.61/ 4 / 4100 0.082(0.8) 0.026

j j

jy j j

e e eH ee e

= = =

7(0.8, ) 0.026cos(10 1.61)yt t= H a

Ionized Gases8-3.3


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Ionized Gases

In the earths upper atmosphere, roughly from 50~ 500 km, there exist layers of ionized gasescalled the ionosphere.

consists offree electrons and positive ions that areproduced when the ultraviolet radiation from the sun isabsorbed by the atoms and molecules in the upper

atmosphere the charged particles tend to be trapped by the earths

magnetic field

The altitude and character of the ionized layersdepend both on the nature of the solar radiationand on the composition of the atmosphere.


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Group Velocity

7 - 4

dispersion wave packet

distortion group velocity

Review of Phase Velocity7-4


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Review of Phase Velocity

Phase velocity is the velocity of propagation of anequiphase wavefront.

For plane waves in a lossless medium,

is a constant independent of frequency.

In some cases, is not a linear function ofand

waves of different frequencies will propagate withdifferent phase velocities.

e. g., wave propagation in a lossy dielectric, or in a

waveguide

pu

=

= 1/pu =

Dispersion7-4


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p

All information-bearing signals consist of a bandof frequencies, normally centered around a highcarrierfrequency. Different component

frequencies travel with different phase velocitieswill cause a distortion in the signal wave shape.

The phenomenon of signal distortion caused by adependence of the phase velocity on frequencyis called dispersion.

a lossy dielectric medium is a dispersive medium

The various frequency components form a wavepacket. Group velocity is the velocity ofpropagation of the wave-packet envelop.

Group Velocity (1)7-4


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p y ( )

consider the simplest case of a wave packet consists of two travelingwaves with equal amplitude and slightly different angular frequencies,

0 0 0

0 0 0

0 00

( , ) cos[( ) ( ) ]

cos[co cos

( ) ( ) ]2 ( )s( ) t z

E z t E t z

E tE t z

z

= + +

+

=

for the wave inside the envelop,

0

0

p

dzu

dt

= =

0 0t z C =0 E(z, t)

z

up

ug

Group Velocity (2)7-4


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p y ( )

for the envelop,

t z C =

1

/g

dzu

dt

= = =

1

/

gu

d d

=

0

0 0 0coscos) 2 )( ) (( , tt t zE z zE =

0 E(z, t)

z

up

ug

Catagorization of Dispersion7-4


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g p

no dispersion

normal dispersion

anomalous dispersion

2

1 p

p p p

dud d

d d u u u d

= =

1

p

gp

p

uu

du

u d

=

0 ,pg p

duu u

d= =

0 ,pg p

duu u

d< >


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Flow of ElectromagneticPower and

the Poynting Vector

7 - 5

power flow Poynting vector

Poyntings theorem instantaneous and average power

densities

Flow of Electromagnetic Power7-5


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t

=

B

Et

= +

D

H J

vector identity,

( ) ( ) ( ) = E H H E E H

( )t t

=

B D

E H H E E J

in a simple medium , , = = =D E B H J E

2( ) 1 ( ) 1

2 2

H

t t t t

= = =

B H H HH H

2( ) 1 ( ) 1

2 2E

t t t t

= = =

D E E EE E

2

( ) E = =E J E E

Poyntings Theorem (1)7-5


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2 2 21 1( ) E H2 2

Et

= +

E H

2 2 21 1( )2 2S V V

d E H dv E dvt

= +

E H s

( )t t

= B D

E H H E E J

storedelectricenergy

storedmagneticenergy

ohmic powerdissipation

time-rateof change

powerdecrease inside V

powerleavingV through itssurface S

conservationof powertake volume integral



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2 2 21 1( )2 2S V V

d E H dv E dvt

= + E H s

: electric energy density

= E HP

power per unit area

: Poynting vector

( )e mS V V

d w w dv p dvt

= + +

s P

2 *1 1

2 2ew E = = E E

: magnetic energy density

: ohmic power density

2 *1 12 2

mw H = = H H

2 2 * */ /p E J = = = = E E J J



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( )e mS V V

d w w dv p dvt

= + +

s P

total power flows into a closed surface equals thesum of the rate of increase ofstored electric and

magnetic energies and the ohmic powerdissipatedwithin the enclosed volume

About Poynting Vector7-5


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The power relations in Poyntings theorempertain to the total power flow across a closedsurface obtained by the surface integral ofEH.

The definition of the Poynting vector as thepower density vector at every point on thesurface is an arbitrary concept, although useful.

The Poynting vectorP is in a direction normal to

both E and H.

Example 7-5


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2 2z z

I I

b b = = =J

J a E a

Find the Poynting vector on the surface of a long, straight conductingwire (of radius b and conductivity ) that carries a direct current I. VerifyPoyntings theorem.

Solution :

b

direct current

current in the wire is uniformlydistributed over its cross section

2

I

b

=H a

on the wire surface,

2 2

2 3 2 3( ) 2 2z r

I I

b b

= = = E H a a aP

22 2

2 3 22

2r

S S

Id d b I I R

b b

= = = =

s a s

P P

/R S=

Instantaneous Expressions for Fields7-5.1


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( )0( ) ( )

j z

x x xz E z E e += =E a a

( )0 0

( , ) [ ( ) ]

[ ] cos( )

j t

z j t z zx x

z t z e

E e e E e t z

=

= =

E E

a a

e

e

( )0( ) ( )j zz

y y y

c

Ez H z e e

+= =H a a

0( , ) [ ( ) ] cos( )j t zyc

Ez t z e e t z

= = H H ae

j

c c e

=

Instantaneous Power Density8-5.1


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220

220

( , ) ( , ) ( , ) [ ( ) ] [ ( ) ]

cos( )cos( )

[cos cos(2 2 )]2

j t j t

z

z

c

z

z

c

z t z t z t z e z e

Ee t z t z

Ee t z

= =

=

= +

E H E H

a

a

e eP

note :

* 2

( , ) [ ( ) ] [ ( ) ]

1 [ ( ) ( ) ( ) ( ) ]2

[ ( ) ( ) ]

j t j t

j t

j t

z t z e z e

z z z z e

z z e

=

= +

E H

E H E H

E H

e e

e

e

P* *

* *

* *

*

( ) ( )

1 1( ) ( )

2 2

1 [( )4

( )]

1( )

2

= + +

= +

+ +

= +

A B

A A B B

A B A B

A B A B

A B A B

e e

e

*( ) ( ) / 2 = +A A Ae

*( ) ( ) / 2 = +B B Be

Average Power Density8-5.1


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2

20av

01( )= ( , ) cos

2

Tz

z

c

Ez z t dt eT

= aP P

2 20( , ) a [cos cos(2 2 )]2

z

z

c

Ez t e t z

= + P

* 21

( , ) [ ( ) ( ) ( ) ( ) ]2

j t

z t z z z z e

= + E H E HeP

*

av

1( ) [ ( ) ( )]

2z z z= E HeP

*av

1[ ]

2

= E HeP

Example 7-6 (1)

The far field of a short vertical current element Idl located at the origin of


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The far field of a short vertical current element Idl located at the origin of

a spherical coordinate system in free space is

a) Write the expression instantaneous Poynting vector.

b) Find the total average power radiated by the current element.

Solution :

a) instantaneous Poynting vector

60( , ) ( , ) sin j R

IdR E R j e

R

= =

E a a

0

( , )( , ) sin2

j RE R IdR j eR

= =

H a a

22 2

2

2

( , ; ) [ ( , ) ] [ ( , ) ]

( )30 sin sin ( )

15 sin [1 cos2( )]

j t j t

R

R t R e R e

Idt R

R

Idt R

R

=

=

=

E H

a a

a

e eP

Example 7-6 (2)

b) total average radiated power


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b) total average radiated power

2

2av ( , ) 15 sinR

IdR

R

=

a

P

av av

22

2 2

0 0

2

2 2

( , )

15 sin sin

40

SR d

Id

R d dR

dI

=

=

=

s

P P


73/120

Normal Incidence ofPlane Waves at Plane

Boundaries

7 - 6

reflection coefficient transmission coefficient

standing wave standing wave ratio

Incidence onto Different Medium7-6


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EM waves propagate in an environment full ofdifferent media in real situations.

An EM wave traveling in one medium impingeson another medium with a different intrinsicimpedance will undergo a reflection.

unless the second medium is a PEC, a part of the

incident power is transmitted into the second medium

Both medium 1 and 2 are assumed to be lossless.

Normal Incidence at Dielectric Boundary (1)

7-6


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Ei

Hi ani

Er

Hr

anr

x

y zincidentwave

reflectedwave

z = 0

medium 2medium 1(1 , 1) (2 , 2)

Et

Ht ant

transmitted

wave

10( )

j z

i x iz E e=E a

incident wave :

reflected wave :

10

1( )

j zi

i y

E

z e

=H a

transmitted wave :

unknownamplitude

10( )

j z

r x rz E e=E a

10

1 1

1( ) ( ) ( ) j zrr z r y Ez z e

= = H a E a

20( ) j zt x tz E e =E a

20

2 2

1( ) ( ) j ztt z t y

Ez z e

= =H a E a

unknownamplitude

Normal Incidence at Dielectric Boundary (2)

7-6


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BC at the interface z = 0 :

1tan 2 tan( 0) ( 0)z z= = =E E

1tan 2 tan

( 0) ( 0)z z= = =H H

(0) (0) (0)i r t+ =E E E

(0) (0) (0)i r t+ =H H H

0 0 0i r tE E E+ =

00 0

1 2

1( ) ti r

EE E

=

Et

Ht ant

transmittedwave

Ei

Hi ani

Er

Hr

anr

x

yz

incidentwave

reflectedwave

z = 0

medium 2medium 1

(1 , 1) (2 , 2)

Reflection & Transmission Coefficients7-6


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0 2 1

0 2 1

r

i

E

E

= =

+

0 2

0 2 1

2t

i

E

E

= =

+

: reflection coefficient

: transmission coefficient

1 + =

2 1 20 0 0 0

2 1 2 1

2,

r i t iE E E E

= =

+ +0 0 0i r tE E E+ =

00 0

1 2

1( ) ti r

EE E

=

Observations7-6


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can be either positive or negative, dependingon whether2 is greater or less than 1.

The definitions for and apply even when themedia are lossy, i.e., even when 1 and 2 arecomplex. and are therefore complex.a complex means a phase shift is introduced at the

interface upon reflection

Standing Waves7-6


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1 1 1 121 0 0( ) ( ) (1 )

j z j z j z j z

x i x iz E e e E e e = + = + E a a

total field in medium 1,

1 max 0 1 min 0| ( ) | (1 | |) , | ( ) | (1 | |)i iz E z E= + = E E

standing wave

max

min

1

1

ES

E

+ = =

1

1

S

S

=

+

: 1 ~ 1

S : 1 ~

: standing-wave ratio, SWR

Field Expressions in Both Media7-6


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medium 1 :

medium 2 :

1 1 1 12

1 0 0( ) ( ) (1 )

j z j z j z j z

x i x iz E e e E e e

= + = + E a a

1 1 1 120 01

1 1

( ) ( ) (1 )j z j z j z j zi iy yE E

z e e e e

= = H a a

20( )

j z

t x iz E e

=E a

20

2

( ) j zt y iz E e

=H a

Example 7-7 (1)

A UPW in a lossless medium with intrinsic impedance 1 is incident


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normally onto another lossless medium with intrinsic impedance 2through a plane boundary.

a) Obtain the expressions for the time-average power densities in bothmedia.

b) Find the standing-wave ratio in medium 1 if2 = 21.

Solution :

a) time-average power density,

*av

1( )2= E HeP

1 1

1 1

22 20

av1

1

22 2 20

1

2 22 20 0

1

1 1

(1 )(1 )2

(1 ) ( )2

(1 ) 2 sin 2 (1 )2 2

j z j ziz

j z j ziz

i iz z

Ee e

Ee e

E Ej z

= +

= +

= + =

a

a

a a

e

e

e

P

1 121 0( ) (1 )

j z j z

x iz E e e = + E a

1 1201

1

( ) (1 )j z j ziyEz e e

= H a

Example 7-7 (2)

a)2

2

0( )j z

t x iz E e =E a


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since both media are lossless, the power flow in medium 1 must beequal to that in medium 2, i.e.,

b) if2 = 21,

20av2

22i

zE

= aP 2

0

2

( ) j zt y iz E e

=H a

av1 av2=P P

2 21

21

=

2 1

2 1

13

= =+

1 1 / 3 21 1 / 3

S += =

Normal Incidence on a Good Conductor (1)

7-6.1


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Ei

Hi ani

Er

Hr

anr

x

y z

incidentwave

reflectedwave

z = 0

medium 2medium 1(1 = 0) (2 = )

PEC

if medium 2 is a PEC =

2 (1 ) 0j

= + =

2 1 2

2 1 2 1

21 , 0

= = = =

+ +

0 0 0 0 0, 0r i i t iE E E E E= = = =

E2 = 0 , H2 = 0

Normal Incidence at PEC Boundary (2)7-6.1


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Ei

Hi ani

Er

Hranr

x

y zincidentwave

reflectedwave

z = 0

medium 2medium 1

(1 = 0) (2 = )

PEC

10( )

j z

i x iz E e=E a

incident wave :

reflected wave :

10

1( )

j zi

i y

E

z e

=H a

1

0( )j z

r x iz E e+= E a

10

1 1

( )( ) j zr ir y yz Ez e

+= =EH a a

total E field in medium 1 :

1 0 1( ) ( ) ( ) 2 sini r x iz z z j E z= + = E E E a

01 1

1

( ) ( ) ( ) 2 cosii r yE

z z z z

= + =H H H a

Instantaneous Field Expression in Medium 1

7-6.1

x


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2006/10/17 85

1 1 0 1( , ) [ ( ) ] 2 sin sinj t x iz t z e E z t = =E E ae

01 1 1

1

( , ) [ ( ) ] 2 cos cosj t iyE

z t z e z t

= =H H ae

2

4

3

4

z = 0

PEC

z

z

standingwave

t = t = 3/4 t = /4 t = 0t = /2

t = 5/4t = 3/2

Notes About Standing Wave7-6.1


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E1 vanishes on the conducting boundary(Er0 = Ei0) as well as at points that are multiplesof/2 from the boundary.

H1 is a maximum on the conducting boundary(Hr0 = Hi0 = Ei0 /1).

The standing waves ofE1 and H1 are in timequadrature (90phase difference) and are shiftedin space by a quarter wavelength.

Example 7-8 (1)

A y-polarized UPW (Ei, Hi) at 100 MHz propagates in air in the +x

di ti d i i ll PEC t 0 A i th


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direction and impinges normally on a PEC at x = 0. Assuming theamplitude ofEi is 6 mV/m, write the phasor and instantaneousexpressions for

a) Ei and Hi or the incident wave;

b) Erand Hror the incident wave;c) E1 and H1 of the total wave in air.

Solution :

a) incident wave,

88 0

1 0 1 080

2 10 2

2 2 10 , , 1203 10 3f k c

= = = = = = = = =

3 2 /3

( ) 6 10j x

i yx e

= E a4

2 /3

1

1 10( ) ( )

2j x

i x i zx x e

= =H a E a

Example 7-8 (2)

a) incident wave,


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b) reflected wave,

( )3 8( , ) [ ( ) ] 6 10 cos 2 10 2 / 3j ti i yx t x e t x = = E E ae

( )4

810( , ) cos 2 10 2 / 32

i zx t t x

= H a

3 2 / 3( ) 6 10 j xr yx e= E a

42 /3

1

1 10

( ) ( ) ( ) 2

j x

r x r zx x e

= =H a E a

( )

3 8( , ) [ ( ) ] 6 10 cos 2 10 2 / 3j tr r y

x t x e t x = = +E E ae

( )4

810( , ) cos 2 10 2 / 32

r zx t t x

= +H a

Example 7-8 (3)

c) total field,3 2( ) ( ) ( ) 12 10 i E E E


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31 2( ) ( ) ( ) 12 10 sin

3i r yx x x x = + =

E E E a

4

1

10 2( ) ( ) ( ) cos

3i r zx x x x

= + =

H H H a

3 81 1

2( , ) [ ( , ) ] 12 10 sin sin(2 10 )

3j t

yx t x t e x t

= =

E E ae

4

81 10 2( , ) cos cos(2 10 )3

zx t x t

= H a


90/120

Oblique Incidence ofPlane Waves at Plane

Boundaries

7 - 7

Snells law of reflection and refraction total reflection

surface wave reflection and transmission coefficients Brewsters angle

Oblique Incidence (1)7-7

x

reflected since both the incident and reflected


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ani

anr

O z

incident

wave

reflectedwave

z = 0

medium 2medium 1

r

i

O'

(1 , 1) (2 , 2)

antA'

A B

refractedwave

t

wavefront

OA AO =

since both the incident and reflectedwaves propagate in medium 1 with thesame phase velocity up1, the distancesmust be equal

sin sinr iOO OO =

r i =Snells lawof reflection

Oblique Incidence (2)x

reflected the time it takes for the transmitted

7-7


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ani

anr

O z

incident

wave

reflectedwave

z = 0

medium 2medium 1

r

i

O'

(1 , 1) (2 , 2)

antA'

A B

refractedwave

t

wavefront

the time it takes for the transmittedwave to travel from O to B equals thetime for the incident wave to travel fromA to O'

Snells lawof refraction

2 1p p

OB AO

u u=

2

1

sin

sin

pt

pi

uOB OO

uAO OO

= =

2 1 1

1 2 2

sin

sin

pt

i p

u n

u n

= = =

Snells Law of Refraction

i u

7-7


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1 1 1 2

2 2 2 1

sin

sint r

i r

n

n

= = = =

p

cn

u=2 1 1

1 2 2

sin

sin

pt

i p

u n

u n

= = = : refraction index,

for non-magnetic media 1 2 0 = =

if medium 1 is free space 1 11 , 1r n = =

2

22

sin 1 1

sin 120t

i rn

= = =

Note : Snells laws are independent of wave polarizations!

Total Reflection (1)

When the wave in medium 1 is incident on a less

7-7.1


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When the wave in medium 1 is incident on a lessdense medium 2, i.e., 1 > 2,

Since t increases with i, an interesting situationarises when

t= /2, at which angle the refracted

wave will glaze along the interface.

Furtherincrease in i would result in no refracted

wave and this is the total reflection (t = /2)The incident angle for which total reflection

occurs is called the critical angle.

1

2sin sint i

= sin sint i > t i >

Critical Angle7-7.1


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1

2

sin sint i

=

2

1

sin c

=

/ 2i c t = =

1 12 2

1 1sin sincn

n

= =

Total Reflection (2)7-7.1


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ani

anr

x

yz

incidentwave

reflectedwave

z = 0

c

c

medium 2medium 1(1 , 1) (2 , 2)

antsurface

wave

c = / 2

mathematically, what ifi > c, 2

1

sin sini c

> =

1 1 2

2 2 1sin sin 1t i

= > >

2 21

2

cos 1 sin sin 1t t ij

= =

Fields in Medium 2 Under Total Reflection

i +

7-7.1


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2 2sin cos, t tj x j zt t e e E H

22 2 1 2 2 2 1 2( / )sin 1 , / sini x i = =

( )22 sin cos, t tntj x zj

t t e e + =a RE H

sin cosnt x t z t = +a a a

sin 1t >

21 2cos ( / )sin 1t ij =

2 2, xz j xt t e e E H

Surface Wave

For i > c an evanescent wave exists along the

7-7.1


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Fori > c, an evanescent wave exists along theinterface (in the x-direction), which is attenuatedexponentially in medium 2 in the normal direction

(z-direction).This wave is tightly bound to the interface and is

called a surface wave. This is a non-uniform

plane wave.No power is transmitted into medium 2 under

total reflection situation.

Example 7-9

The permittivity of water is 1.750. An isotropic light source at a

distance d under water yields an illuminated circular area of a radius 5m.


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yDetermine d.

Solution :

refraction index of water is

radius of illuminated area, O'P, corresponds to the critical angle,

1.75 1.32wn =

1 1 01 1sin sin 49.21.32

c

wn

= = =

0

54.32tan tan49.2c

O P

d

= = =

Example 7-10

A dielectric rod or fiber of a transparent material can be used to guide

light or EM waves under the conditions of total internal reflection.D t i th i i di l t i t t f th idi di


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gDetermine the minimum dielectric constant of the guiding medium sothat a wave incident on one end at any angle will be confined within therod until it emerges from the other end.

Solution :1 must be greater than or equal to c for total internal reflection,

1sin sin c cos sint c

1

1sin sint i

r

=

2 0

1 1 1

1 11 sin ir r

=

1 / 2 t = 2

1

sin c

=

21 1 sin 2r i + =

Ionized Gases

In the earths upper atmosphere, roughly from 50

7-7.2


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In the earth s upper atmosphere, roughly from 50~ 500 km, there exist layers of ionized gasescalled the ionosphere.

consists offree electrons and positive ions that areproduced when the ultraviolet radiation from the sun isabsorbed by the atoms and molecules in the upperatmosphere

the charged particles tend to be trapped by the earthsmagnetic field

The altitude and character of the ionized layersdepend both on the nature of the solar radiationand on the composition of the atmosphere.

Ionosphere

Ionized gas with equal electron and ion densities

7-7.2


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Ionized gas with equal electron and ion densitiesare called plasma.

The ionosphere plays an important role in the

propagation of EM waves and affectstelecommunication.

Because the electrons are much lighterthan the

positive ions, they are accelerated more by the Efield of EM wave passing through the ionosphere.

Wave Propagations in Ionosphere (1)

Effect of the ionosphere on wave propagation

7-7.2


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Effect of the ionosphere on wave propagationcan be studied on the basis of an effective p,

2 2

0 02 2= 1 1p p

p

f

f

=

: plasma angular frequency

2

0

p

Ne

m

=

e : electron charge , m : electron mass

N : numbers of electrons per unit volume

2 00 2

1 ( / ) ,1 ( / )

p p

p

j f ff f

= =

Wave Propagations in Ionosphere (3)

0 as f fp. When becomes 0, D (which

7-7.2


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, (depends on free charges only) is 0 even when E(which depends on both free and polarization

charges) is not.When f< fp,

indicating an attenuation without propagation. fpis also referred to as the cutoff frequency.

When f> fp, is purely imaginary and EM waves

will propagate unattenuated in the ionosphere.

f< fp 1 < fp / f 1 (fp / f)2 < 0 is purely real

Plasma Frequency of the Ionosphere

319.7 10m = 19

7-7.2


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2

0

19

2 2

p

p

Nef N

m

= =

191.602 10e = 12

0 8.854 10=

10 3 12 310 /m ~ 10 /mN

lowestlayer

highestlayer

0.9 MHz ~ 9 MHzp

f

N at a given altitude varieswith the time of the day, theseason, and other factors.

Effect of Ionosphere on Communication

For communication with a satellite or a spacet ti b d th i h t

7-7.2


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pstation beyond the ionosphere we must usefrequencies much higherthan 9 MHz to ensure

wave penetration through the layer with thelargest N at any angle of incidence.

Signals with frequencies lower than 0.9 MHz can

not penetrate into even the lowest layer.Signals with frequencies between 0.9 ~ 9 MHz

will penetrate partially into the lower ionospheric

layers but will eventually be turned back where Nis large.

Example 7-11

When a spacecraft reenters the earths atmosphere, its speed and

temperature ionize the surrounding atoms and molecules and create aplasma It has been estimated that the electron density is in the


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2006/10/17 107

plasma. It has been estimated that the electron density is in theneighborhood of 2 108 /cm3. Discuss the plasmas effect on frequencyusage and the mission controllers on earth.

Solution :8 3 14 32 10 /cm 2 10 /mN = =

14 79 2 10 12.7 10 Hz 127 MHzpf = = =

radio communication cannot be established forf< 127 MHz

Perpendicular Polarization (1)7-7.3

sin cos = +a a aincident wave :


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2006/10/17 108

Ei

Hi

ani

Er

Hranr

x

yz

incidentwave

reflectedwave

r

i

z = 0

medium 2medium 1(1 , 1) (2 , 2)

Et

Ht

ant

t

transmittedwave

sin cosni x i z i = +a a a

reflected wave :

1 1( sin cos )0 0( , )

ni i ij j x z

i y i y ix z E e E e += =a RE a a

1

1

( sin cos )0

1

( , ) [ ( , )] /

( cos sin ) i i

i ni i

j x zix i z i

x z x z

Ee

+

=

= +

H a E

a a

sin cosnr x r z r = a a a

1 1( sin cos )0 0( , )

nr r r j j x z

r y r y r x z E e E e = =a RE a a

1

1

( sin cos )0

1

( , ) [ ( , )] /

( cos sin ) r r

r nr r

j x zrx r z r

x z x z

Ee

=

= +

H a E

a a

Perpendicular Polarization (2)

sin cos = +a a atransmitted wave :

7-7.3


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2006/10/17 109

Ei

Hi

ani

Er

Hranr

x

yz

incidentwave

reflectedwave

r

i

z = 0

medium 2medium 1(1 , 1) (2 , 2)

Et

Ht

ant

t

transmittedwave

sin cosnt x t z t = +a a a

2 2 ( sin cos )0 0( , )

nt t t j j x z

t y t y t x z E e E e += =a RE a a

2

2

( sin cos )0

2

( , ) [ ( , )] /

( cos sin ) t t

t nt t

j x ztx t z t

x z x z

Ee

+

=

= +

H a E

a a

Field Expressions in Both Media

incident wave :( i )

7-7.3


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reflected wave :

transmitted wave :

1( sin cos )0( , )

i ij x z

i y ix z E e +=E a

1( sin cos )0

1( , ) ( cos sin )i ij x zi

i x i z i

E

x z e

+

= +H a a

1( sin cos )0( , )

r rj x z

r y rx z E e =E a

1( sin cos )0

1

( , ) ( cos sin ) r rj x zrr x r z r E

x z e

= +H a a

2 ( sin cos )0( , )

t tj x z

t y tx z E e +=E a

2 ( sin cos )0

2

( , ) ( cos sin ) t tj x ztt x t z t E

x z e

+= +H a a

Boundary Conditions at the Interface

1tan 2 tan( 0) ( 0)z z= = =E E( 0) ( 0)H H

7-7.3


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1tan 2 tan( 0) ( 0)z z= = =H H

( ,0) ( ,0) ( ,0)

( ,0) ( ,0) ( ,0)

iy ry ty

ix rx tx

E x E x E x

H x H x H x

+ =

+ =

1 1 2

1 1 2

sin sin sin0 0 0

sin sin sin00 0

1 2

1 ( cos cos ) cos

i r t

i r t

j x j x j x

i r t

j x j x j xti i r r t

E e E e E e

EE e E e e

+ =

+ =

all three exponential factors involving xmust be equal ! (phase matching)

Snells Law

1 1 2sin sin sini r tx x x = =

7-7.3


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1 1 2i r t

1 1

2 2

sin,

sint

r i

i

n

n

= = = : Snells law

Reflection & Transmission Coefficients

0 0 0

01i r t

t

E E EE

+ =

7-7.3


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00 0

1 2

1( )cos costi r i t

EE E

=

0 2 2

0 2 1 2 1

2 cos 2( / cos )

cos cos ( / cos ) ( / cos )t i t

i i t t i

E

E

= = =

+ +

0 2 1 2 1

0 2 1 2 1

cos cos ( / cos ) ( / cos )

cos cos ( / cos ) ( / cos )r i t t i

i i t t i

E

E

= = =

+ +

1 + =

: Fresnels equations

Parallel Polarization7-7.4

sin cosni x i z i = +a a aincident wave :

( sin cos )j +


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Ei

Hi

ani

Er

Hr

anr

x

yzincident

wave

reflectedwave

r

i

z = 0

medium 2medium 1(1 , 1) (2 , 2)

Et

Ht

ant

t

transmittedwave reflected wave :

sin cosnr x r z r = a a a

1( sin cos )0( , ) ( cos sin )

i ij x z

i i x i z ix z E e

+= E a a

1( sin cos )0

1

( , ) i ij x zii yE

x z e

+=H a

1( sin cos )

0( , ) ( cos sin ) r rj x z

r r x r z r

x z E e

= +E a a

1( sin cos )0

1

( , ) r rj x zrr yE

x z e

= H a

transmitted wave :

sin cosnt x t z t = +a a a2 ( sin cos )

0( , ) ( cos sin )t tj x z

t t x t z t x z E e += E a a

2 ( sin cos )0

2

( , ) t tj x ztt yE

x z e

+=H a

Reflection & Transmission Coefficients

0 0 0( )cos cosi r i t tE E E + =matching BC at

z= 0,

7-7.4


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0 0 0

0 0 0

1 2

( )

1 1( )

i r i t t

i r tE E E

=

0 2 1

0 2 1

cos cos

cos cos

r t i

i t i

E

E

= =

+

0 2

0 2 1

2 cos

cos cost i

i t i

E

E

= =

+

cos1

cost

i

+ =

Brewster Angle of No Reflection7-7.5

2 1cos coscos cos

t i

=+


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2 1cos cost i +

i = B|| such that || = 0

22 21

2

2

cos 1 sin 1 sint t in

n

= =

1 2sin / sint in n =

i B =

for non-magnetic media 1 2 1 2 0, = =

2 1cos cost B =

2 2 1 1 2

21 2

1 /sin

1 ( / )B

=

1 12 2

1 11 2

1sin tan tan

1 (or

/ )B B

n

n

= = = +

Brewster Angle of No Reflection

2 1cos coscos cos

i t

=+

7-7.5


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2 1cos cosi t +

2 1cos cosB t =

i = B such that = 0

22 21

2

2

cos 1 sin 1 sint t in

n

= =

1 2sin / sint in n =

2 1 2 2 1

21 2

1 /sin

1 ( / )B

=

i B =

for magnetic media

1 2

1sin

1 ( / )B

=

+

1 2 1 2, =

Application of Brewster Angle

Because B and B|| are different, it is possible toseparate these two types of polarization in an

8-10.3


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separate these two types of polarization in anunpolarized wave.

When an unpolarized wave such as random lightis incident upon a boundary at the Brewsterangle B||, only the component with perpendicular

polarization will be reflected.Brewster angle is therefore also called polarizing angle

Based on this principle, quartz windows set at the

Brewster angle at the ends of a laser tube areused to control the polarization of an emitted lightbeam.

Example 7-13 (1)

The dielectric constant of pure water is 80.

a) Determine the Brewster angle for parallel polarization B|| and thecorresponding angle of transmission.

b) A l ith di l l i ti i i id t f i t


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b) A plane wave with perpendicular polarization is incident from air on watersurface at i = B||. Find the reflection and transmission coefficients.

Solution :a) Brewster angle for parallel polarization,

1 1 0

2

1 1sin sin 81.0

1 (1/ ) 1 (1/80)

B

= = =

+ +

1 1 1 0

2 2

sin 1 1sin sin sin 6.381 81

Bt

= = = = +

Example 7-13 (2)

b) reflection and transmission coefficients,

1 2

2

377377 , 40.1

= = =


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2

81 , 6.38i t

= =

1 2/ cos 2410 , / cos 40.4i t = =

40.4 2410 2 40.40.967 , 0.033

40.4 2410 40.4 2410

= = = =

+ +

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07 Plane Electromagnetic Waves