EELE 3332 – Electromagnetic II

Chapter 10

Electromagnetic Wave

Propagation

Islamic University of Gaza Electrical Engineering Department

Dr. Talal Skaik

2012 1

•A changing magnetic field produces an electric field, and a

changing electric field produces a magnetic field.

•Accelerating charges produce electromagnetic waves.

•The existence of EM waves, predicted by Maxwell’s equations, was

first investigated by Heinrich Hertz (sometimes called Hertzian

waves).

In general, waves are means of transporting energy or

information.

Typical examples of EM waves include radio waves, TV signals,

radar,… 2

Electromagnetic wave propagation

3

Longitudinal Waves: Vibration is parallel to the direction of propagation. Sound and pressure waves are longitudinal. Transverse Waves: the motion of the matter particles is perpendicular to the direction of propagation of the wave itself.

Transverse waves Longitudinal waves

Mechanical waves

Transverse waves

Electromagnetic waves

WAVES

4

Longitudinal Wave

(in spring)

Transverse Wave

(in string)

• A material medium is necessary for the transmission for mechanical waves. Mechanical waves cannot travel through vacuum.

• Disturbance is transmitted from one layer to the next through the medium.

Mechanical Waves

• Electromagnetic waves travel through empty space!!

• All electromagnetic waves travels at c = 3108 m/s in vacuum (speed of light).

• Electric field, magnetic field, and direction of travel are mutually perpendicular. 5

Electromagnetic Waves

•Material medium is not essential for propagation. Em waves travel through vacuum. •Disturbance of electric and magnetic fields travelling through space. •All electromagnetic waves are transverse waves.

Goal in this chapter: Solve Maxwell’s equations and describe EM

wave motion in the following media:

6

Electromagnetic waves

0 0

0 0

0 0

0 0

1. Free Space ( =0, = , = )

2. Lossless Dielectrics ( =0, = , = or )

3. Lossy Dielectrics ( 0, = , = )

4. Good Conductors ( , = , = or )

where is the angular fre

r r

r r

r

quency of the wave

The wave has the following characteristics:

It varies with both time and space.

It is time harmonic.

The amplitude of the wave is A.

The phase of the wave (in radians) is the term (ωt-βz), depends

on time t and space variable z.

ω is the angular frequency in (radians/second).

β is the phase constant, or wave number in (radians per meter).

7

Electromagnetic waves

sin( )E A t z

8

sin( )E A t z

Wave takes distance λ to repeat itself.

λ is called the wavelength (in meters)

Wave takes time T to repeat itself.

T is called the Period (in seconds)

Since it takes time for the wave to travel distance at the speed ,

1( speed of the wave, depends on the medium )

But 1 / , where

T u

uT

u u

T f f

is the of the wave in Hertz (Hz).

Since 2 , =

frequency

u f

f andu

2 2

2 (rad/m)

This shows that for every wavelength of distance traveled,

a wave undergoes phase change of 2 radians.

f

f

9

10

Consider a fixed point P on

the wave.

Sketch E=A sin(ωt-βz) at

times t=0, T/4, and T/2.

It is evident that as the wave

advances with time, point P

moves along +z direction.

So, the wave E=A sin(ωt-βz) is

travelling with a velocity u in

the +z direction.

11

Point P is a point of constant phase, therefore

Notes:

A sin(ωt - βz) is wave propagating in +z direction (forward travelling,

or positive-going wave)

A sin(ωt + βz) is wave propagating in -z direction (backward travelling,

or negative going wave)

constantt z

dzz u

dt

12

13

8

8

( ) The wave is propagating along - direction.

10 1(b) in free space , = rad/m

3 10 3

If T is the period of the wave, it takes T seconds to travel

a distance at speed c. Hence to travel a

xa a

u cu c

8

distance /2 will take:

1 21 31.42 ns.

2 2 2 10

Tt

f

14

An Electric field in free space is given by

(a) Find the direction of wave propagation.

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

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

Example 10.1

8

y50cos(10 ) a V/mE t x

15

( ) At t=0, 50cos

1 2At t= = = ,

4 4 4

2 50cos .

4

50cos / 2 50sin

1 2At t= = = ,

2 2 2

2 50cos .

2

50cos 50cos

y

y

y

c E x

T

f

E x

x x

T

f

E x

x

Notice that the wave travels along x

x

a

Example 10.1 8 y50cos(10 ) a V/mE t x

v

Consider a linear, isotropic, homogeneous, LOSSY dielectric medium

that is charge free ( =0). Maxwell's equations in phasor form are:

E 0 (1)

H 0

s

s

(2)

E H (3)

H E (4)

Taking the curl of both sides of equation (3) gives:

E H (3)

Since A (

s s

s s

s s

j

j

j

2A) A

( E )s

2

2 2 2

E E

E E 0 , where

s s

s s

j j

j j

16

10.3 Wave Propagation in Lossy Dielectrics

A lossy dielectric is a medium in which an EM wave, as it propagates, loses power owing to imperfect dielectric. (partially conducting medium

with σ≠0)

2 2 2

2 2

propagation consta

E E 0 where

is called the of the medium.

By similar procedure, it can be shown that for the H-Field, H H 0

nt

s s

s s

j j

2 2

2 2

22

2 2 2

2 2 2

E E 0

H H 0

is a complex quantity, let

(Vector Wave Equations)

Sinc

2

,

e

2

s s

s s

j

j j j j j

j j

2

2

(solve for )

1 12

1 12

and

17

Wave Propagation in Lossy Dielectrics

18

• α is attenuation constant (Np/m): defines the rate of decay of the

wave in the medium. measured in Nepers per meter (Np/m).

• β is phase constant (rad/m) : is a measure of the phase shift per

unit length in radians per meter. (also called wave number)

• α=0 for lossless medium (σ=0)

• An attenuation of 1 neper indicates a reduction of

e-1 of the original value.

• (1 Np=20 log10e=8.686 dB).

Wave Propagation in Lossy Dielectrics

2

u

z s

s

2 2 2

2

2

Assume the wave propagates along a and E has only x component

, then E E ( ) a

Since E E 0 ( E Vector Laplacian )

( )Hence

xs x

s s s

xs

z

E z

x

2

2

( )xsE z

y

22

2

22

2

' '

0 0

( )( ) 0

( ) ( ) 0

This is a scalar wave equation, a differential equation with solution:

( ) (where and are constants)

( econd part is zero

xsxs

xsxs

z z

xs o o

E zE z

z

E zor E z

z

E z E e E e E E

S

z

( )

0

0

since we assumed wave traveling along +a ).

E( , ) Re ( ) a Re a

E( , ) cos( ) a

j t z j t z

xs x x

z

x

z t E z e E e e

z t E e t z

19

Wave Propagation in Lossy Dielectrics

A Sketch of |E| at times t=0 and t=Δt is shown

Notice E has only x-component and it is travelling in the +z direction. 20

0E( , ) cos( ) az

xz t E e t z

Wave Propagation in Lossy Dielectrics

21

( ) 00 0

H( , ) can be obtained as:

H( , ) Re a ,

where is complex quantity known as , in ohms

of the medium.

intrinsic imped

/

a

1

nce

z j t z

y

j

z t

Ez t H e e H

je

j

with

1/42

( )0

0

, tan2

H= Re a

H( , ) cos( ) a

z j t z

yj

z

y

Ee e

e

Eor z t e t z

• Notice that E and H are out of phase by Ѳη at any instant of time.

Thus , E leads H (or H lags E) by Ѳη.

• The ratio of the magnitude of the conduction current density Jc to

that of the displacement current density Jd in a lossy medium is

Where tanѲ is known as the loss tangent and Ѳ is the loss angle of the

medium. 22

0

0

E( , ) cos( ) a

H( , ) cos( ) a

z

x

z

y

z t E e t z

Ez t e t z

s

s

J Etan

J E

tan

cs

ds j

or

s

Since tan and tan2 2

H E 1 E = E

where , 1 , ' ''

complex permitt

' , ''

is ca ille he i vd t

s s c s

c c

c

jSince j j j

j j j with

of the medium.

Notice that the ratio of '' to ' is the loss tangent of the medium.

'' tan =

t

'

y

23

tanѲ is used to determine how lossy the medium is:

Good (lossless or perfect) dielectric if tanѲ is vary small (σωε)

• Behaviour of a medium depends not only on parameters σ, ε, and µ,

but also on the frequency. A medium regarded as a good conductor

at low frequencies may be a good dielectric at high frequencies.

24

10.4 Plane waves in lossless dielectrics

In a lossless dielectric, σ

25

10.5 Plane waves in free space

In free space

0 00, = , =

0 0

0 0

8

This may be regarded as special case of section 10.4.

0,

1 2 , =

where c=3 10 m/s, is the speed of light in a vaccum.

c

u c

o00

0

0

0 120 377

is called the intrinsic impedance of free space.

Plots of E and H (a) as functions of z at t 0; and (b) at z 0. The arrows indicate instantaneous values. 26

Plane waves in free space

0 x

0

0

If E= cos( t- z) a

then H= cos( t- z) a

H cos( t- z) a

In general, if a , a , and a are

unit vectors along the E field,

H field, and the direction of

ave propagation:

a a

y

y

E H k

k E

E

H

E

a

a a a

a a a

H

k H E

E H k

27

Wave representation

Plane Wave and its representation Circular Wave and its representation

28

Plane waves in free space

29

Plane waves in free space

Both E and H fields (or EM waves) are everywhere normal to the

direction of wave propagation, ak.

They form an EM wave and have no electric or magnetic field

components along the direction of propagation. Such a wave is

called a transverse electromagnetic (TEM) wave.

A combination of E and H is called a uniform plane wave

because E (or H) has the same magnitude throughout any

transverse plane, defined by z=constant.

Uniform plane waves serve as approximations to practical waves

such as those from a radio antenna a distance sufficiently far from

radiating sources.

21

= = (1 )22

2

2 2 , =

Also, Since and

jj j j j j

f

u

j

j

j

o o45 Thus E leads H by 45 .

30

Plane Waves in Good Conductors

A perfect or good conductor is one in which σ>>ωε

0 0, = , = r

Therefore, as the wave travels in a conducting medium, it

amplitude is attenuated by a factor e-αz.

The distance δ, through which the wave amplitude decreases to a

factor e-1 (about 37% of the original value) is called skin depth or penetration depth of the medium.

31

0 x

o0

If E cos( ) a

H cos( 45 ) a

z

z

y

E e t z

Ethen e t z

Plane Waves in Good Conductors

1

0 0

1 (general)

E e E e

The skin depth is a measure of the depth to which an EM

wave can penetrate the medium.

32

Plane Waves in Good Conductors

Skin depth

illustration

/

0

Since for good conductors =

1 1

1Since = E cos( ) z x

f

f

zE e t a

33

Plane Waves in Good Conductors Skin Depth in Copper

Frequency (Hz) 10 60 100 500 104 108 1010

Skin depth (mm) 20.8 8.6 6.6 2.99 0.66 6.6 x 10-3 6.6 x 10-4

For copper, σ=5.8x107 S/m, µ= µ0, 66.1/ (in mm)f

The skin depth decreases with increasing frequency. Thus, E and H can

hardly propagate through good conductors.

The fields and currents are confined to a very

thin layer (the skin) of the conductor surface.

For a wire of radius a, it is a good

approximation at high frequencies to assume

that all of the current flows in the circular ring of

thickness δ. (as shown)

34

35

Skin effect : is the tendency for high-frequency currents to flow

on the surface of a conductor.

* The effective conductor cross section decreases and the

conductor resistance increases.

It is used to advantage in many applications:

Since the skin depth in silver is very small, silver plating is often

used to reduce the material cost of waveguide components. (e.g

silver-plating on brass).

Hollow conductors are used instead of solid conductors in

outdoor television antennas, and thus saving weight and cost.

Plane Waves in Good Conductors

The is given by:

The (in ) is given by:

1 (real part of for a good c

dc

s

lR

S

fR

s

dc resistance

surface or skin resistance R

onductor)

The is calculated by using the dc formula with ,

where is the width:

For a conductor of radius , 2 . So,

sac

S w

w

l R lR

w w

a w a

acac resistance R

2

2 2 2

At high frequencies, is far greater than

ac

dc

ac dc

lR a aa f

lR

a

R R

36

Plane Waves in Good Conductors

37

0A lossy dielectric has an intrinsic impedance of 200 30 at

a particular radian frequency . If, at that frequency, the plane

wave propagating through the dielectric has the magnetic field

component:

1 10 cos A/m

2

find E and . Determine the skin depth and wave polarization.

x

ye t x

H a

Example 10.2

/6 /6

The given wave travels along so that

; , so

10,so 200 30 200 2000

Except for the amplitude and phase difference, and always

hav

x

k x H y

E k H x y z

E z

o j joo o

o

EH e E e

H

a

a a a a

a a a a a a

a a

E H

/6e the same form. Hence

Re 2000

2 cos kV/m2 6

j x j t

E

x

z

e e e

xe t

E a

E a38

Example 10.2 - Solution

2 2

1/22

2

Knowing that =1/2 , we need to determine . Since

1 1 and 1 12 2

1 1

1 1

But

o

1/2

z

tan 2 =tan 60 3 . Hence

2 1 1 1 0.2887 Np/m

2 1 3 3 2 3

1= 2 3

The wave has an component; hence it is polarized along the z-direction. E

39

Example 10.2 - Solution

40

r

x y

r

In a lossless medium for which = 60 , =1 and

= 0.1 cos ( t z) + 0.5 sin ( t z) A/m,

calculate , , and .

H a a

E

Example 10.3

8 8

In this case , =0 , =0 , and =1, so

120 120 120= / 2 4

60

24

1 3 10 1.5 10 rad/s

2 2

o rr r

o r r

o o r r

or

c c

cor

Solution

41

From the given field, can be calculated in two ways: using the

techniques (based on Maxwell's equations) developed in this chapter

or directly using Maxwell's equations as in Chapter 9.

To

H E

Method 1 :

1 2

1 x 2 y

use the techniques developed in this chapter, we let

where

= 0.1 cos ( t z) and = 0.5 sin ( t z)

and the corresponding electric field

H = H + H

H a H a

1 2

1 2

1 1o E 2 2o E

where

=E cos ( t z) and = E sin ( t z)

E = E + E

E a E a

Example 10.3 - Solution

1 1

x y

1

E k H z x y

Notice that although has components along and , it has no component

along the direction of propagation; it is therefore a TEM wave.

For

H a a

E

a a a a a a

2 2

1o 1o

1 y

2

E k H z y x

60 (0.1) 6

Hence

= 6 cos

For

E H

E t z

a

E

a a a a a a

2o 2o

2 x

1 2

8

x

60 (0.5) 30

Hence

= 30 sin

Adding and gives ; that is,

=94.25 sin (1.5 10 t z) + 18.85 cos (1.5

E H

E t z

a

E E E

E a8

y10 t z) V/m a 42

Example 10.3 - Solution

We may apply Maxwell's equations directly.

1

where 0.

But

( ) ( ) 0

x y z

y

x x

dtt

a a a

x x z z

z z

Method 2 :

EH = E E H

HH a

H H

x y

x y

x y

y

= 0.5 cos ( t z) 0.1 sin ( t z)

Hence

1 0.5 0.1sin ( t z) sin ( t z)

= 94.25 sin ( t z) + 18.85 cos ( t z) V/m

x

x

z

dt

Ha

a a

E H a a

a a43

Example 10.3 - Solution

44

Example 10.4

8

y

r r

A uniform plane wave propagating in a medium has

E 2 sin(10 ) V/m. If the medium is characterized by

= 1, = 20, and = 3 mhos/m, find , , and .

ze t z

a

H

98

We need to determine the loss tangent to be able to tell whether the medium is

a lossy dielectric or a good conductor.

3 3393 1

1010 1

36

showing that

Solution

1/27 8

the medium may be regarded as a good conductor at the frequency

4 10 20(10 )(3)of operation. Hence, = = 61.4

2 2

61.4 Np/m , 61.4 rad/m

45

1/27 8

3

3 61.4 8

4 10 20(10 ) 800Also, | |

3 2

tan 2 3393 454

Hence

sin4

and

32 69.1 10

| | 800

Thus 69.1 10 sin 10

o

z

o H

H k E z y x

oo

z

H e t z

EH

e t

H a

a a a a a a

H 61.42 mA/m4

xz

a

Example 10.4 - Solution

46

o xA plane wave E = cos ( t z) is incident on a good

conductor at z = 0. Find the current density in the conductor.

E a

Example 10.5

2 2

2 2

Since the current density = , we expect to satisfy the wave equation

( E E 0), that is,

0

Also the incident has only an x-compone

s s

s s

Solution

J E J

J J

E

x x

22

2

nt and varies with z.

Hence = J (z, t) and

0

which is an ordinary differential equation with solution

sx sx

dJ J

dz

J

J a

z z

sx Ae Be

47

Example 10.5 - Solution

The constant must be zero because is finite as z .

But in a good conductor , >> so that =1/ .

1Hence

= 1

an

d

sx

sx

J

J A

B

jj j

e

(1 )/

(1 )/

or

(0)

where (0) is the current density on the conductor surface.

z j

z j

sx sx

sx

J J Ae

J

48

For the copper coaxial cable shown, let a = 2 mm, b = 6 mm, and t = 1 mm.

Calculate the resistance of 2 m length of the cable at dc and at 100 MHz.

Example 10.6

2 7 3 2

2 2 2

Let

where and are the resistances of the inner and outer conductors.

At dc,

22.744 m

5.8 10 (2 10 )

[ ] 2

o i

o i

i

o

R = R + R

R R

RS a

RS b t b t bt

Solution :

7 6

dc

2 0.8429 m

5.8 10 1 12 10

Hence R 2.744 0.8429 3.587 m

49

8 7

3 7

At f=100 MHz ,

( )2 2

2 10 4 10 0.41

2 2 10 5.8 10

Since = 6.6 m