Electromagnetic Waves and Transmission Lines - Machac

7/18/2019 Electromagnetic Waves and Transmission Lines - Machac

http://slidepdf.com/reader/full/electromagnetic-waves-and-transmission-lines-machac-56d6ca7ad7af8 1/191

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CONTENTS

1. INTRODUCTION .............................................................................................. 61.1 Description of an electromagnetic field ………………….….............................. 8

1.2 Wave equation ………………...........................................................…............... 13

1.3 Potentials ………….........................................................................…................. 14

2. ELECTROMAGNETIC WAVES IN FREE SPACE .................................… 172.1 Solution of the wave equation ....................................…..................................... 17

2.2 Propagation of a plane electromagnetic wave …………………………………. 18

2.3 Wave polarization ……………………………………………………………… 28

2.4 Cylindrical and spherical waves ……………………………………………….. 32

2.5 Problems ……………………………………………………………………….. 34

3. WAVES ON A PLANE BOUNDARY ………………………………………. 353.1 Perpendicular incidence of a plane wave to a plane boundary ………………… 35

3.2 Perpendicular incidence of a plane wave to a layered medium ………………... 42

3.3 Oblique incidence of a plane electromagnetic wave to a plane boundary ……... 45

3.4 Problems ……………………………………………………………………….. 54

4. SOLUTION OF MAXWELL EQUATIONS AT VERY HIGH

FREQUENCIES ………………………………………………………………. 56

5. GUIDED WAVES …………………………………………………………….. 61

6. TEM WAVES ON A TRANSMISSION LINE ………………………………. 656.1 Parameters of a TEM wave …………………………………………………….. 65

6.2 Transformation of the impedance along the line ………………………………. 71

6.2.1 An infinitely long line ………………………………………………………….. 71

6.2.2 A line of finite length …………………………………………………………… 72

6.2.3 A line terminated by a short cut and by an open end ……….………………….. 736.3 Smith chart ……………………………………………………………………… 76

6.4 Problems ……………………………………………...………………………… 87

7. WAVEGUIDES WITH METALLIC WALLS ……………………………... 88

7.1 Parallel plate waveguide ……………………………………………………….. 88

7.2 Waveguide with a rectangular cross-section …………………………………… 94

7.3 Waveguide with a circular cross-section ………………………………………. 103

7.4 Problems ………………………………………………………………………... 107

8. DIELECTRIC WAVEGUIDES ……………………………………………… 1088.1 Dielectric layers ………………………………………………………………… 109

8.2 Dielectric cylinders ……………………………………………………………... 114

8.3 Problems ………………………………………………………………………... 115

9. RESONATORS ……………………………………………………………….. 1169.1 Cavity resonators ……………………………………………………………….. 116

9.2 Problems ………………………………………………………………………... 120

10. RADIATION ………………………………………………………………….. 12110.1 Elementary electric dipoles …………………………………………………….. 121

10.2 Elementary magnetic dipoles …………………………………………………... 126

10.3 Radiation of sources with dimensions comparable with the wavelength ………. 129

10.4 Antenna parameters …………………………………………………………….. 133

10.5 Antenna arrays ………………………………………………………………….. 13410.6 Receiving antennas ……………………………………………………………... 137



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10.7 Problems ……………………………………………………………………….. 140

11. WAVE PROPAGATION IN NON-ISOTROPIC MEDIA ………………… 14111.1 Tensor of permeability of a magnetized ferrite ………………………………… 142

11.2 Longitudinal propagation of a plane electromagnetic wave in a magnetized

ferrite …………………………………………………………………………… 146

11.3 Transversal propagation of a plane electromagnetic wave in a magnetized

ferrite …………………………………………………………………………… 15111.4 Applications of non-reciprocal devices ………………………………………… 154

11.5 Problems ………………………………………………………………………... 155

12. APPLICATIONS OF ELECTROMAGNETIC FIELDS ………………….. 156

12.1 Introduction to microwave technology …………………………………………. 156

12.2 Antennas ………………………………………………………………………... 162

12.2.1 Wire antennas …………………………………………………………………... 164

12.2.2 Aperture antennas ………………………………………………………………. 165

12.2.3 Broadband antennas …………………………………………………………….. 167

12.2.4 Planar antennas …………………………………………………………………. 168

12.3 Propagation of electromagnetic waves in the atmosphere ……………………… 169

12.4 Optoelectronic ………………………………………………………………….. 173

12.4.1 Optical waveguides …………………………………………………………….. 173

12.4.2 Optical detectors ………………………………………………………………. 176

12.4.3 Optical amplifiers and sources …………………………………………………. 177

12.4.4 Optical modulators and sensors ………………………………………………… 178

13. MATHEMATICAL APPENDIX …………………………………………….. 180

14. BASIC PROBLEMS ………………………………………………………….. 191

15 LIST OF RECOMMENDED LITERATURE ………………………………. 194



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1. INTRODUCTION

This textbook is aimed at students of the Faculty of Electrical Engineering, Czech

Technical University taking a course in Waves and Transmission Lines. The textbook builds

on basic knowledge of time varying electromagnetic fields gained from courses in physics

and electromagnetic field theory. The textbook introduces all the basic knowledge that an

electrical engineer specializing in radio engineering and telecommunications should have andthat is necessary for further courses such as microwave engineering, antennas and

propagation, optical communications, etc. Sequential mastering of wave theory contributes to

the final objective of university studies, which is to enable graduates to do creative work.

The course in Waves and Transmission Lines studies the theory and applications of

classical electrodynamics. It is based on Maxwell’s equations. This course provides a basis

for understanding the behavior of all high frequency electric circuits and transmission lines,

starting from those applied to transmitting and receiving electric energy, processing signals,

microwave circuits, optical fibers, and antennas. The main applications lie in wireless

communications, radio engineering and optical systems.

The text follows the classical approach to macroscopic electrodynamics. All quantities

are assumed to be averaged over the material, which by its nature has a microscopic structureconsisting of atoms. This confines the description of electromagnetic effects using

macroscopic theory on the high frequency side, as the wavelength must be much longer than

the dimensions of the atoms and molecules. This boundary lies in the range of ultraviolet

light. Nevertheless, the spectrum of frequencies in which electromagnetic effects can be

treated using this macroscopic theory is really huge – over 17 orders. And this whole

spectrum really is used in a variety of different applications. Modern communication systems

use electromagnetic waves with ever shorter wavelengths. The spectrum of electromagnetic

waves is shown in Fig. 1.1.

First, we review the basic relations from electromagnetic field theory and introduce

potentials describing a time varying electromagnetic field. The concept of a plane

electromagnetic wave is carefully reviewed. In addition, a cylindrical wave together with aspherical wave are briefly introduced. The behaviour of a plane electromagnetic wave on the

boundary between two different materials is studied in detail. Here we will start treating the

incidence of a wave perpendicular to the plane boundary and to a layered medium. Oblique

incidence is studied in general, and then special effects such as total transmission and total

reflection are treated. Specific aspects of solving Maxwell equations at very high frequencies

are discussed separately.

Transmission lines are designed to transmit guided waves. After introducing the

general properties of guided waves the TEM wave is treated, and the transformation of

impedances along a line is described. The Smith chart, a very effective graphical tool for

analysis and design of high frequency circuits, is described and its basic applications are

explained through particular problems. Waves propagating along waveguides with metallic

walls of rectangular and circular cross-sections are studied. Dielectric waveguides are treated

separately. They form the basis of optical fibers. Cavity resonators, unlike low frequency L-C

resonant circuits, are able to resonate on an infinite row of resonant frequencies. Several kinds

of such resonators are analyzed in the text.

Attention is paid to problems of radiation of electromagnetic waves. This covers

antenna theory. Elementary sources of an electromagnetic field, such as an electric dipole and

a magnetic dipole, are studied and compared. Then radiation from sources with dimensions

comparable with the wavelength is described. The basic antenna parameters are defined. The

basic idea of antenna arrays is built up. Finally, receiving antennas are dealt with, and the

effective antenna length and effective antenna surface are derived.



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Non-isotropic materials are introduced, and the tensor of permeability of a magnetized

ferrite is calculated. The propagation of a plane electromagnetic wave in this ferrite material

homogeneously filling an unbounded space is studied, in particular when the wave propagates

both in the direction parallel with the magnetizing field and in the direction perpendicular to

the magnetizing field. Some devices using non-isotropic materials are mentioned.

The applications of electromagnetic fields in particular branches of electrical

engineering are briefly introduced. An introduction to microwave technology is given. Here

scattering parameters are introduced. The paragraph on antennas represents a continuation of

Chapter 10, introducing basic types of antennas. Particular mechanisms of wave propagation

in the atmosphere are explained. The transmission formula and radar equations are derived.

The basics of optoelectronics are presented. This involves a characterization of opticalwaveguides, detection and optical detectors, optical amplifiers, and the sources of optical

103

106

109

1012

1015

1018

106

104

103

102

10

10

0

10-2

10-3

10-6

10-9

frequency

(Hz)

wavelength

(m)

classification applications

phone, audio

AM radio

long waves

medium waves

short waves

very short waves

decimeter waves

centimeter waves

millimeter waves

quasi optical waves

ultraviolet radiation

infrared radiation

visible light

nm

360

460

560

660

760

red

yellowgreen blue

violet

argon laser 490 nm

He-Ne laser 630 nm

radar, space investigation

radar, satellite commun.

radar, TV, navigation

TV, FM radio, services

radio, services

micr owaves

The spectrum of electromagnetic waves.

Fig. 1.1



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radiation, namely lasers, and finally modulators of an optical beam and a short review of

optical fiber sensors.

The book has a mathematical appendix summarizing the necessary knowledge of

mathematics. Basic problems in the form of questions are summarized at the end of the text.

They help the students in preparing for their examinations. A list of suggested literature is

given.

The textbook treats time varying electromagnetic fields. Harmonic dependence on

time is assumed throughout most of the text. Such fields are described using symbolic

complex quantities called phasors, and time dependence in the form t jω e is assumed. These

phasors are not marked by special symbols. When we need to emphasize an instantaneousvalue, we will mark it by showing dependence on time, e.g., E (t). Vectors will be marked by

bold characters, e.g., E.

1.1 Description of an electromagnetic fieldThe sources of an electromagnetic field are electric charge Q [C] and electric current I

[A], which is nothing else than the flow of charge. Charge is often distributed continuously in

a space, on a surface, or along a curve. It is convenient in this case to define the correspondingcharge densities. The charge volume density is defined as the charge amount stored in a unit

volume

V

q ρ

V ∆

∆=

→∆ 0lim [C/m

3] . (1.1)

The charge surface density is defined by analogy

S

qσ

S ∆

∆=

→∆ 0

lim [C/m2] . (1.2)

The charge linear density is defined as the charge stored along a line or a curve of unitlength

l

q

l ∆

∆=

→∆ 0limτ [C/m] , (1.3)

Electric current is created by a moving charge. This is defined as the passing charge

per time interval t ∆

t

q I

t ∆

∆=

→∆ 0lim [A=C/s] . (1.4)

It is useful to define current densities, which are vector quantities, as it is necessary to definethe current flow direction. Current density is defined as

0iJS

I

S ∆

∆=

→∆ 0lim [A/m

2] , (1.5)

where i0 is a unit vector describing the current flow direction. It is sometimes useful to use the

abstraction of a surface current passing along a surface, see Fig. 1.2. The linear density of



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this current is defined as

0iK I

x ∆

∆=

→∆ 0lim [A/m] . (1.6)

The surface current and its density are an

abstraction used to simplify the mathematicaldescription of the current passing a conductor at a

high frequency. Due to the skin effect, currentflows through only a very thin layer under the

material surface. In fact the surface current

defined by (1.6) represents the finite current passing a cross-section of zero value. This

requires infinite material conductivity. Another widely used abstraction is a current filament representing a conductor of negligible cross-

section (e.g., a line) carrying a finite current.

The total electric current crossing a closed surface is related to the charge accumulated

inside the volume surrounded by this surface by the continuity equation in an integral form

∫∫ =+⋅

S

Q jd 0ω SJ , (1.7)

or in a differential form

0div =+ ρ ω jJ . (1.8)

where Q is the total charge accumulated in volume V with boundary S . The current density is

related by Ohm’s law in the differential form to an electric field

J=σE , (1. 9)

where σ is conductivity in S/m. It should be noted that in spite of the movement of the free

electrons, a conductor passed by an electric current stays electrically neutral, as the charge of the electrons is compensated by the positive charge of the charged atomic lattice.

The electric field is described by the vector of electric field intensity E, the unit of

which is V/m. The magnetic field is described by the vector of magnetic field intensity H, theunit of which is A/m. These vectors are related to the induction vectors by material relations

HB µ = , (1.10)

ED ε = , (1.11)

where r ε ε ε 0= and r µ µ µ 0= are permittivity and permeability, respectively, and

π ε 36/10 90

−= F/m and 70 104 −= π µ H/m are the permittivity and permeability of a vacuum,

respectively. Vectors E and H are related by the set of Maxwell’s equations. Their differential form reads

( ) S j JEH ++= ωε σ rot , (1.12)

∆

K S

i0 I

Fig. 1.2



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HE ωµ j−=rot , (1.13)

( ) 0div ρ ε =E , (1.14)

( ) 0div =H µ , (1.15)

where JS is a current supplied by an external independent source, ρ 0 is the volume density of afree charge supplied by an external independent source. The differential form of Maxwell’sequations is valid only at those points where the field quantities

are continuous and are continuously differentiated functions of position. They are not valid, for example, on a boundary between

two different materials where the material parameters change

step-wise, Fig. 1.3. For this reason we have to appendcorresponding boundary conditions to these equations. We

suppose that the boundary between two materials contains a free

electric charge with density σ 0, and electric current K passingalong this boundary. The boundary conditions in vector form can be expressed

( ) 02211 σ ε ε =−⋅ EEn ,

( ) 021 =−× EEn ,

( ) 02211 =−⋅ HHn µ µ , (1.16)

( ) K HHn =−× 21 ,

( ) 021 =−⋅ JJn ,

n is the unit vector normal to the boundary. A scalar form using the normal and tangentialcomponents of vectors is

σ ε ε =− 2211 nn E E , (1.17)

21 t t E E = , (1.18)

2211 nn H H µ µ = , (1.19)

K H H t t =− 21 , (1.20)

21 nn J J = . (1.21)

Specially on the surface of an ideal conductor with conductivity ∞→σ , and since theelectric and magnetic fields are zero inside this conductor, we have

011 σ ε =n E , 01 =t E , 01 =n H , K H t =1 . (1.22)

The first Maxwell equation (1.12) has three terms on its right hand side. Term jω E

1

2

nε 1 µ 1 σ 1

ε 2 µ 2 σ 2

Fig. 1.3



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represents the displacement current, term σ E represents the conducting current which causesconducting losses in the material, see (1.9), and JS is the current supplied by an internal

source. Equation (1.12) can be simplified by introducing a complex permittivity

( )

S c

S er S r S r

j

j j j j j

JE

JEJEJEEH

+

=+−=+

−=++=

ε ωε

δ ε ωε ωε

σ ε ωε σ ε ωε

0

00

00 tg1rot

where complex permittivity ε c is defined

( ) '''tg10

ε ε δ ε ωε

σ ε ε j j j er r c −=−=−= , (1.23)

and term

r e ε ωε

σ δ

0tg = , (1.24)

is called the loss factor, or thea loss tangent, and δ e is the loss angle. This loss factor isfrequently used to define the losses in a material in spite of the fact that it is frequency

dependent. The reason is that this loss factor, similarly as the imaginary part of permittivity

ε ’’, contains in practice not only the conducting losses, but also polarization and another kinds

of losses. Similarly we can introduce complex permeability µ c representing all kinds of magnetic losses

( ) '''tg1 µ µ δ µ µ j j mr c −=−= . (1.25)

In material relations (1.9) – (1.11) conductivity σ , permittivity ε and permeability µ represent the electric and magnetic properties of a material. In a linear material these

parameters do not depend on field quantities, while in a nonlinear material they depend on E

or H. These parameters can depend on space coordinates, which is the case of a non-

homogeneous material. In a homogeneous material σ , ε , and µ do not depend on

coordinates. An isotropic material has parameters that are constant in all directions, ε , µ , σ are scalar quantities. Non-isotropic materials possess different behaviour in different

directions. Their permittivity, permeability, or conductivity are tensor quantities that can be

expressed by matrices. An example is provided by magnetized ferrite or magnetized plasma.

E.g., equation (1.11) for magnetized plasma can be rewritten into

ED ε = , (1.26)

which gives three particular scalar equations

z xz y xy x xx x E E E D ε ε ε ++= ,

z yz y yy x yx y E E E D ε ε ε ++= , (1.27)

z zz y zy x zx z E E E D ε ε ε ++= .



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Vector D has a different direction from vector E. Permittivity is then not a scalar quantity, but

a tensor quantity

=

zz zy zx

yz yy yx

xz xy xx

ε ε ε

ε ε ε

ε ε ε

ε . (1.28)

Similarly we can express the permeability and conductivity of an anisotropic material.

The density of power transmitted by an electromagnetic wave is described by the

complex Poynting vector. This vector is defined as

QSHES jav +=×= *

2

1, (1.29)

where Sav is the average value of the transmitted power, which represents the density of active

power

[ ]*Re2

1HES ×=av . (1.30)

Q is the density of reactive power

[ ]*Im2

1HEQ ×= . (1.31)

Poynting’s theorem represents the balance of power in an electromagnetic system in volume

V . It can be read by dividing power into active and reactive power

R J S P P P += , (1.32)

RavemavS QW W Q +−= ω 2 , (1.33)

where P S and QS are the average values of the active and reactive power supplied by an

external source. The active power supplied by an external source is

∫∫∫

⋅=V

S S

dV P *Re2

1JE , (1.34)

JS is the current supplied by this source. The active power is partly lost in materials, power P J ,

and partly radiated outside of our volume, power P R. These quantities are

∫∫∫=

V

J dV E P 2

2

1σ , (1.35)

∫∫∫∫ ⋅×=⋅=

S S

av R d d P SHESS*Re , (1.36)



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where S is the surface surrounding investigated volume V . Similarly, for reactive power we

get

∫∫∫ ⋅−=

V

S S dV Q *Im2

1JE . (1.37)

This power covers the difference between the average values of the energy of an electric field,W eav, and the energy of a magnetic field, W mav, and is partly radiated outside, Q R. These values

are

∫∫ ⋅×=

S

R d Q SHE*Im

2

1, (1.38)

∫∫∫

=

V

mav dV H

W 4

2 µ

, ∫∫∫

=

V

eav dV E

W 4

2ε

. (1.39)

1.2 Wave equationThe solution of an electromagnetic field by Maxwell’s equations (1.12) and (1.13)

involves solving six scalar equations for three components of the electric field and for three

components of the magnetic field. We can reduce this number of equations by extracting one

of the two vectors from (1.12) and (1.13), as vectors E and H are mutually dependent. Let us

start with equation (1.12). Let us apply operator rot to this equation

( ) S j JEH rotrotrotrot ++= ωε σ .

Inserting for doubly applied operator rotation rotrotH HH ∆−= divgrad (13.81) we get

( ) S j JEHH rotrotdivgrad ++=∆− ωε σ .

rotE can be expressed from Maxwell’s second equation (1.13) and Maxwell’s fourth equation

(1.15) states that divH=0. Now we have

( ) S j j JHH rot++−=∆− ωε σ µ ω .

The term containing material parameters and a frequency in front of H on the right hand sideis usually denoted

( ) ωµσ µε ω ωε σ µ ω j j jk −=+−= 22 , (1.40)

where k is a complex propagation constant. Using this simplification we get the wave

equation for the vector of magnetic field intensity in the form

S k JHH rot2 −=+∆ . (1.41)



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The derivation of the equation for E is similar. Let us apply operator rotation to (1.13),

and we get

HE rotrotrot ωµ j−= .

Using (13.81), expressing divE from Maxwell’s third equation (1.14), and using the

propagation constant k (1.40) we get

S jk JEE ωµ ε

ρ +=+∆ 02 grad . (1.42)

Wave equations (1.41) and (1.42) are basic equations which describe an electromagnetic field

in a space with general sources supplying current JS and charge ρ 0. These equations are,

however, not suitable for solving the field in areas with sources, as their right hand sides are

rather complicated functions of source quantities JS and ρ 0. Equations for potentials are

preferably applied to solve these problems, which, e.g., involve radiation of antennas.

1.3 PotentialsStationary or static fields are described by simplified Maxwell’s equations

0rot =E , (1.43)

0div =B . (1.44)

The potentials for these fields can then be simply defined. The scalar potential is defined by

ϕ grad−=E , (1.45)

and the vector potential by

AB rot= . (1.46)

These potentials are solutions of Poisson’s equations

ε

ρ ϕ 0−=∆ , (1.47)

S JA µ −=∆ . (1.48)

Equation (1.48) holds, supposing that the Lorentz calibration condition in the form

0div =A (1.49)

is laid to the vector potential. The solution of these equations can be obtained by applying the

method of superposition



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∫∫∫= dV r

0

4

1 ρ

πε ϕ , (1.50)

∫∫∫= dV r

S JA

π

µ

4. (1.51)

In a non-stationary field we have again equation (1.44) and the vector potential is,therefore, defined by (1.46). An electric field is a different story. It is now described instead

of (1.43) by Maxwell’s equation (1.13), which can be rewritten into

ABE rotrot ω ω j j −=−= ,

( ) 0rot =+ AE ω j .

This last equation tells us that the rotation of the vector in brackets equals zero, consequently

this vector can be expressed by a scalar potential. The electric field is now expressed by both

the scalar potential and the vector potential

AE ω ϕ j−−= grad . (1.52)

Let us now derive equations analogous to (1.47) and (1.48) which describe the

distribution of the vector and scalar potentials of a time varying electromagnetic field. First

we insert (1.46) and (1.52) into Maxwell’s first equation (1.12) and we get

( )( ) S j j JAA µ ω ϕ µσ ωµε +−−+= gradrotrot ,

( ) ( ) S j j JAAA µ ωµσ µε ω ϕ µσ ωµε +−++−=∆+2

graddivgrad ,

( )[ ] S jk JAAA µ ϕ µσ ωµε −++=+∆ divgrad2 .

This equation can be simplified supposing that the argument of the gradient is zero, which is

the Lorentz calibration condition for a time varying electromagnetic field

( )ϕ µσ ωµε +−= jAdiv . (1.53)

Finally the wave equation for the vector potential reads

S k JAA µ −=+∆ 2 . (1.54)

The equation for the scalar potential is derived in a similar way. We insert (1.52) into

Maxwell’s third equation (1.14)

ε

ρ ω ϕ 0divgraddiv =−− A j .

Adiv is expressed by calibration condition (1.53), and we have



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( )ε

ρ ϕ µσ ωµε ω ϕ 0−=+−∆ j j ,

which gives the final form of the wave equation for the scalar potential

ε

ρ ϕ ϕ 02 −=+∆ k . (1.55)

Equations (1.54) and (1.55) are more suitable for solving an electromagnetic field, as the

terms on their right hand sides are simple functions of source quantities JS and ρ0. They arefrequently used to analyze and design antennas. Note that Poisson’s equations (1.47) and

(1.48) represent limiting cases of general wave equations (1.54) and (1.55), assuming that the

frequency is decreased to zero, which is equivalent to decreasing propagation constant k . Due

to this, the solution of wave equations can be expected in a form corresponding to (1.50) and

(1.51). Let us first solve equation (1.55) for the scalar potential in the time domain. A lossless

medium is assumed to simplify understanding. Now the equivalent to (1.55) in the time

domain is

ε

ρ ϕ µε ϕ 0

2

2

−=∂

∂+∆

t .

The variations of an electromagnetic field excited at distance R from a source are delayed by

the time which is necessary for the wave to travel at speed v along this distance ,r r R −= ,

Fig. 1.4

v

R

t =∆ . (1.56)

Consequently we get

( )( )

V d R

v

Rt

V d R

t t t

V V

∫∫∫∫∫∫

−

=∆−

=0

0

4

1

4

1 ρ

πε

ρ

πε ϕ

Applying phasors we get, assuming that k =ω/v

( )

V d R

V

v Rt jt j

∫∫∫−

=/

0 e

4

1e

ω ω ρ

πε ϕ .

The final form of the solution is

V d R

V

jkR

∫∫∫−

=e

4

1 0 ρ

πε ϕ . (1.57)

Similarly for the vector potential we have

dV ρ 0

R

r

0

r’

Fig. 1.4

ϕ



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V d R

V

jkRS

∫∫∫−

=e

4

JA

π

µ . (1.58)

Again we see that (1.57) and (1.58) pass to (1.50) and (1.51) when the frequency is reduced to

zero, and, consequently, the propagation constant goes to zero, which causes a transition

1e →− jkr

.Equations (1.57) and (1.58) are frequently used to calculate the electromagnetic field

excited by sources, e.g., by antennas in space or probes in waveguides. The electric field and

the magnetic field are then calculated from the potentials, using (1.46) and (1.52). Prior to

performing this calculation we usually have to determine the distribution of the electric

charge and the electric current. This task is often not simple, see section 10.

2. ELECTROMAGNETIC WAVES IN FREE SPACE

2.1 Solution of the wave equationLet us solve wave equation (1.42) for an electric field. We will look for a solution

which describes a plane electromagnetic wave propagating in an infinite space filled by a

homogeneous material. Wave fronts, which are the surfaces of a constant field phase, are - in

the case of a plane wave - planes perpendicular to the direction of propagation. Such a wave

can be excited by a source located at infinity, which excites a field with a constant amplitude

and phase within the whole infinite plane. So this wave is only an abstraction. We will study

its behaviour and propagation, as this provides a basis for understanding all phenomena

connected with the propagation of electromagnetic waves. Most real waves can be expressed

as the superposition of a series of plane electromagnetic waves.Let us assume a free space filled by a homogeneous material with parameters ε , µ , and

σ . The wave equation is solved as a homogeneous equation in the form (1.42), i. e., withoutany source. This means that the solution describes the possible particular waves, called modes

or eigen waves, which can propagate in our space. We rewrite equation (1.42) into a form

describing the i-th coordinate of the electric field

02

2

2

2

2

2

2

=+∂

∂+

∂

∂+

∂

∂i

iii E k z

E

y

E

x

E , i= x, y, z (2.1)

This equation can be solved by the method of separation of variables. E i is expected in theform

( ) ( ) ( ) z Z yY x X E i = . (2.2)

Introducing (2.2) into (2.1) we get

02''''''

=+++ k Z

Z

Y

Y

X

X , (2.3)

where



book - 2

18

2''

2''

2''

,, z y x k Z

Z k

Y

Y k

X

X −=−=−= (2.4)

and the propagation constant (1.40) can be written in the form

22222

z y x k k k jk ++=−= ωµσ µε ω . (2.5)

Equation (2.3) is then separated into three equations. The equation for the x-coordinate is

02'' =+ X k X x , (2.6)

with its solution in the form of the two propagating plane waves

x-jk x jk x x B A X ee += . (2.7)

Combining electric field E i (2.2) we get

( ) ( ) z -jk z jk y jk y jk x-jk x jk

i z z y y x x F E DC B A E eeeeee

-+++= . (2.8)

Constants A to F can be determined from the boundary conditions. Multiplying the brackets in

(2.8) we get eight terms of the form( z k yk xk j z y x ±±±

e . We confine the solution of the waveequation to one of the terms representing the particular plane wave with vector complexamplitude E0

( z k yk xk j j z y x ++−⋅− == ee 00 EEErk , (2.9)

where the scalar product rk ⋅ represents the projection of vector r into the direction of k .

These vectors are

000 zyxr z y x ++= ,

000 zyxk z y x k k k ++= , ωµσ µε ω jk k k k z y x −=++= 22222 .

(2.9) describes the plane electromagnetic wave propagating in the direction determined by

vector k .

2.2 Propagation of a plane electromagnetic waveA plane electromagnetic wave propagating in a general direction in an unbounded

space filled by a homogeneous material is described by the electric field intensity (2.9)

( ) 0eeeeeee 0000

φ j j j j j j rβrαrβrαrαβrk EEEEE

⋅−⋅−⋅−⋅−⋅−−⋅− ==== , (2.10)

where the generally complex propagation vector can be rewritten into a real part and animaginary part

αβk j−= . (2.11)



book - 2

19

The instantaneous value of an electric field is

( ) ( )00 sineeIm, φ ω ω +⋅−== ⋅−rβEErE

rα t t t j . (2.12)

Term rαE ⋅−e0 represents the dependence of the wave

amplitude on position r . From the definition of the

propagation constant (2.5) it follows that both its real part β and its imaginary part α are positive numbers.

Exponential function rα⋅−e then decreases with increasing

value of its exponent. The amplitude therefore decreases

and the wave is attenuated in the case of non-zero α .Vector αααα is therefore known as an attenuation constant

(vector), as it describes the measure of the wave attenuation. The wave described by (2.10)with a negative exponent therefore propagates in the positive direction – in the direction

defined by the propagation vector. The argument of the sinus function0

φ ω +⋅− rβt

determines the dependence of the field phase on time and coordinates. For this reason ββββ iscalled a phase constant (vector).

The surfaces of a constant phase – wave fronts – are planes determined by the

condition .cos const r ==⋅ ϕ β rβ We can see from Fig. 2.1 that the surface of a constant

phase is a plane perpendicular to phase vector ββββ. The surfaces of a constant amplitude are

determined by .const =⋅rα , and they represent planes perpendicular to attenuation vector αααα.

The name plane wave is derived from the shape of these surfaces. A uniform wave has

surfaces of constant amplitude and of constant phase that are identical, as vectors αααα and ββββ are parallel. A non-uniform wave has surfaces of the constant phase different from surfaces of

the constant amplitude, as vectors αααα and ββββ are not parallel.The phase velocity is the velocity of the propagation of planes of a constant phase. Let

us follow the propagation of a plane with a constant phase φ which in a time increment ∆t

moves by a distance ∆r

( ) ( ) r t r r t t r t ∆−∆=⇒∆+−∆+=−= β ω β ω β ω φ 0 ,

from this we can define the phase velocity

β

ω =

∆

∆=

t

r v . (2.13)

Group velocity represents the velocity with which the wave transmits energy, or the velocity

of the propagation of planes of a constant amplitude. We will derive the group velocity as thevelocity of the propagation of the amplitude of the superposition of two waves with

frequencies ω and ω +d ω and with corresponding phase constants β and β +d β and with

amplitudes equal to one. The superposition of these waves is

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

+−

+

−=

=+−++−=+=

z d

t d z d t d

z d t d z t t z E t z E t z E

22sin

2cos2

sinsin,,, 21

β β

ω ω

β ω

β β ω ω β ω

.

ββββϕ

r cos ϕ

constant phase

Fig. 2.1



book - 2

20

The amplitude of the resulting field is described by function cosine. In a time increment ∆t it

moves by a distance ∆ z

( ) ( ) z d t d z z d t t d z d t d ∆−∆=⇒∆+−∆+=−= β ω β ω β ω φ 0

from this we can define the group velocity

ω

β β

ω

d

d d

d

t

z v g

1==

∆

∆= . (2.14)

This velocity must in any case be lower than the velocity of light in a vacuum c.

As we will see later, the phase and the group velocities are not functions of frequency

in a lossless material – an ideal dielectric. This is not the case of a lossy material with nonzero

conductivity, where waves with different frequencies propagate with different velocities. Each

signal transmitting any information is represented by the spectrum of frequencies, and

consequently each component of this signal propagates with a different velocity. The result is

that the signal is distorted by passing the lossy material. This distortion is called dispersion

and the material is called dispersive. So only the ideal dielectric is a non-dispersive material

in which a passing signal is not distorted.

The wavelength is defined as the least distance, measured in the direction of

propagation, of two points with the same phase

( ) ( )[ ] π βλ λ β ω β ω 2sinsin =⇒−−=− z t z t ,

from this we can get

f v

v

===ω π

β π λ 22 , (2.15)

where f =ω /2π is the frequency.

In the preceding section we solved the wave equation for an electric field. The relation

between the electric field and the magnetic field can be derived from Maxwell’s equations.

We insert the solutions of the wave equation in the form

rk rk HHEE ⋅−⋅− == j j e,e 00 ,

into Maxwell’s second equation (1.13). The rotation of vector E is (….)

( ) ( ) rk rk rk rk rk Ek Ek EEEE

⋅−⋅−⋅−⋅−⋅− ×−=×−=×+== j j j j j j je eeegradroterotrot 00000 ,

as the rotation of constant vector E0 is zero. Inserting into (1.13) we get

rk rk Ek H

⋅−⋅− ×−=− j j j j ee 00ωµ

00 Ek H ×= j jωµ . (2.16)



book - 2

21

Inserting into Maxwell’s first equation (1.12) we get

( ) 00 Hk E ×−=+ j jωε σ . (2.17)

It is evident from (2.16) and (2.17) that vectors E, H and k represent a right handed rotating

set of three vectors, Fig. 2.2. Vectors E and H are perpendicular to each other and both are

perpendicular to the direction of propagation determined by vector k . For this reason a plane electromagnetic wave propagating in an

unbounded space filled by a homogeneous material is called a

transversal electromagnetic wave, abbreviated as TEM. We

rewrite vector k as k =k n, where vector n is a unit vector

determining the propagation direction. Now using the form of k

(2.5) we get from (2.17)

nHHn

E ×+

=+

×−= 0

00

ωε σ

ωµ

ωε σ j

j

j

jk .

The second root in this equation has the unit [Ω] and therefore it is known as the wave

impedance of the space

ωε σ

ωµ ωµ

ωε σ j

j

k j

k Z w

+==

+= , (2.18)

it is generally a complex number which defines the ratio of E over H including the phase shift

between an electric field and a magnetic field. Consequently, the relations between an electric

field and a magnetic field are

nHE ×= 00 w Z , (2.19)

00

1EnH ×=

w Z (2.20)

Let us now simplify the description of a plane electromagnetic wave by considering

that it propagates in the z direction. So we have 0zk k = and r=z . Vectors E and H can then

be written as

0000 , yxHyxE y x y x H H E E +=+= .

Inserting these vectors into (2.19) we get

0000 xyyx yw xw y x H Z H Z E E +−=+ ,

using this equation we can define the wave impedance as

x

y

y

xw

H

E

H

E Z −== . (2.21)

k

E

H

Fig. 2.2



book - 2

22

The complex propagation constant k was defined by (2.5). The phase constant and the

attenuation constant can be calculated by inserting the values of the material parameters and

frequency into (2.5). Alternatively α and β can be obtained from (2.5) in the form

+

+= 11

2

12

ωε

σ µε ω β , (2.22)

−

+= 11

2

12

ωε

σ µε ω α . (2.23)

The power transmitted by a plane electromagnetic wave is defined by the average

value of Poynting’s vector

( )

=

××

=×=×=−−−−

*

*0002*

00*

Ree2

1

eeeeRe2

1

Re2

1

w

z z j z z j z av

Z

EzE

HEHESα β α β α

( ) ( )[ ] 0*

2

02*0000

*00*

2 Ree2

11Ree

2

1zEzEzEE

=

⋅−⋅= −−

w

z

w

z

Z

E

Z

α α

02

2

0cose

2zS z

z

w

av Z

E ϕ α −= , (2.24)

where ϕ z is the argument of wave impedance z jww e Z Z

ϕ = (2.18). The power is transmitted

in the z direction, i.e., in the direction of the wave propagation.

Let us now discuss the propagation of plane electromagnetic waves in materials with

limit parameters. We will at the same time show the distribution of the electromagnetic field

of this wave in dependence on the time and space coordinates.

An ideal dielectric is a material with zero conductivity. The propagation constant and

the wave impedance are real in this material

0,222 ==⇒== α µε ω β β µε ω k , (2.25)

r r

w Z ε

π

ε ε

µ

ε

µ 120

0

0 === . (2.26)

This means that the wave propagates with a constant amplitude, i.e., without losses, and such

a material is therefore called a lossless material. The electric field and the magnetic field are

in phase, as Z w is a real number. The electric field and the magnetic field and their

instantaneous values are, assuming E parallel with the x axis and H parallel with the y axis

and zero starting phase,

000000 ee,e yyHxEz j

w

z j z j

Z

E H E

β β β −−−

=== ,



book - 2

23

( ) ( ) ( ) ( ) 00

00 sin,,sin, yHxE z t Z

E t z z t E t z

w

β ω β ω −=−= .

These functions are plotted in Fig. 2.3 in dependence on time t for constant z and in

dependence on the z coordinate for constant t.

A good conductor is a material in which we have ωε <<σ . The formulas for the

propagation constant (2.5) and the wave impedance (2.18) can under this condition be

simplified to

( ) j j jk jk −=−=−=⇒−= 12

ωµσ ωµσ α β ωµσ , (2.27)

2

ωµσ α β == , (2.28)

( )4

,e2

1 4π

ϕ σ

ωµ

σ

ωµ

σ

ωµ π

==+== z

j

w j j

Z . (2.29)

In a well conducting material the electric and magnetic fields have a mutual phase shift equal

to π /4, i.e., 45°. Generally the phase shift between E and H lies between 0° for a lossless

material and 45° for a good conductor.

Phase constant β and attenuation constant α have high values (2.28), and the wave is

rapidly attenuated. A material is sometimes characterized by penetration depth δ . This

quantity determines the length at which the wave amplitude decreases to 1/e (e=2.718281 isthe basis of the natural logarithm), i.e., to a value of 36.8 % of its starting value. The value of

the penetration depth can be derived

ωµσ α δ αδ 21

ee 1 ==⇒= −− . (2.30)

____________________________

Example 2.1: Determine the penetration depth of a plane electromagnetic wave into copper

for frequencies of 50 Hz, 10 kHz, and 100 MHz.

Applying formula (2.30) we getδ = 9.35 mm for 50 Hz

=0 m

t

E

ω t=π

E

Fig. 2.3



book - 2

24

δ = 0.66 mm for 10 kHz

δ = 6.6 µm for 100 MHz

At high frequencies the field does not penetrate at all into the conducting material.

____________________________

The electric field and the magnetic field and their instantaneous values are, in the caseof a lossy material

00

00 eee,ee yHxE z j z j z

w

z j z

Z

E E

ϕ β α β α −−−−− == ,

( ) ( ) ( ) ( ) 00

00 sine,,sine, yHxE z z

w

z z t Z

E t z z t E t z ϕ β ω β ω α α −−=−= −− .

These functions are plotted in Fig. 2.4 in dependence on time t for constant z and in

dependence on the z coordinate for constant t , and a general lossy material is assumed. Thedependence on the z coordinate shows the attenuation of the wave according to function exp(-

α z ).

A dielectric material with non-zero conductivity σ which fulfils the condition

σ <<ωε (2.31)

can be called a lossy dielectric. The plane electromagnetic wave propagating in such a

material can be treated as a wave propagating in the dielectric only in the case of propagationover a short distance, for which its attenuation can be neglected. A wave propagating over a

long distance can be substantially attenuated, even in the case of a low attenuation constant.

Under condition (2.31) we can accept the following simplification for the propagation

constant

−≈−=−=

ωε

σ µε ω

ωε

σ µε ω ωµσ µε ω j j jk

2

1112 ,

ε

µ σ µε ω

2

1 jk −= , (2.32)

=0 m

t

E

H

ω t=π

E

H

Fig. 2.4

ϕ z / ω exp(- z )



book - 2

25

r ε

πσ

ε

µ σ α µε ω β

60

2, === . (2.33)

We see that the phase constant is the same as the phase constant of the wave propagating in a

dielectric, but we have non-zero attenuation. Other parameters Z w, v, λ are determined using

the same formulas as in the case of a wave propagating in a dielectric.

________________________________

Example 2.2: Calculate the instantaneous value of voltage received by a frame antenna

located according to Fig. 2.5, due to the propagation of an electromagnetic wave

0

84

0

82

310sin10.33.1,

310sin10.5 yHxE

−=

−= −− z

t z

t

We will calculate the received voltage in two ways. Firstly, according to the definition

of voltage. A closed loop is represented by an antenna perimeter, Fig. 2.5.

( ) ( )

−+−=+−=⋅+⋅=⋅= −

∫∫∫ 310sin10sin10.5 882

21

2

2

1

1

π π π π t t E E d d d t u

c

lElElE

We can rewrite this result to the final form

( )

+−=

310sin157.0 8 π

t t u .

The second way must result in the same

formula. We apply Faraday’s induction law

( )dt

d t u

φ −=

The magnetic flux passing the antenna area is

( )

−

−=

=

−==⋅=

−

−

∫∫∫∫

t z

t

dz z

t dz H d

S

8811

0

84

0

10cos3

10cos10.157

310sin10.33.1

π π

µπ µπ µ φ SH

( ) ( )

+−=

−

−=−=

310sin157.010sin

310sin157.0 888 π π φ

t t t dt

d t u .

The results are identical. Tilting the antenna into plane y-z , voltage u=0, as vectors d S and H

are perpendicular. In this way we can determine the direction of the wave propagation, i.e. the

k

E

H

π

π

E 1 E 2

d l1 d l2

c

Fig. 2.5



book - 2

26

direction in which the transmitter is located. In this example we have the antenna dimensions

comparable with the wavelength, which is π π

β

π λ 6

31

22=== .

____________________________________

Example 2.3: Calculate the effective value of the voltage induced in a frame antenna, see Fig.

2.6, due to the propagating plane electromagnetic wave defined in Example 2.2.In this case the antenna dimensions are much lower than the wavelength. We can

therefore simplify the calculation of the magnetic flux. This can be done by supposing

( )a z j z j +−− ≈ β β ee , i.e. 1e ≈− a j β

,

i.e. 1<<a β .

The magnetic flux is

ψ φ cosS B= ,

where ψ is the angle between vectors B and

d S. In our case this angle is 0°. The inducedvoltage is

µV18.122

,, 0

00 ===−=−=−=H S U

U H S jU dt

dH S

dt

d u ef

ωµ µ ω µ

φ .

This value was obtained using the wave parameters from the previous example.

_______________________________________

Example 2.4: A plane electromagnetic wave propagates in the direction of the positive z axis

in air. It is incident perpendicularly to an ideally conducting plane located at z =0. Determine

the field distribution, which is the superposition of the incident and reflected waves supposing

that an electric field reaches at point z = -1 m and at time t = 0 s its maximum value

( ) V/m1000,1 max ===−= E t z E . The frequency is 100 MHz. Calculate the power

transmitted by this wave.

An incident wave propagating in the positive z direction is

00ee xE

ϕ j jkz

iiE −= ,

00ee yH

ϕ j jkz

iiH −= .

A reflected wave which propagates in the negative z direction is

00 ee xEϕ j jkz

r r E = , 00 ee yHϕ j jkz

r r H = .

The resulting field is the superposition of the incident and reflected waves

0000 eeee xxEEEϕ ϕ j jkz

r j jkz

ir i E E +=+= −

0000 eeee yyHHH ϕ ϕ j jkz r

j jkz ir i H H +=+= −

k

E

Ha

a

Fig. 2.6

a=0.1 m



book - 2

27

The relation between the amplitudes of the incident and reflected waves can be determined

from the boundary conditions. The electric field is parallel to the conducting plane, and due to

(1.22) its value must therefore be zero

( ) ( ) ( ) r ir i E E z z z 00000 −=⇒=+=== EEE .

The relation between the amplitudes of a magneticfield follows from the orientation of the vectors of the

plane wave. The vectors of both the incident wave and

the reflected wave are mutually perpendicular, Fig.

2.7. Consequently we have

r i H H 00 = .

Now we can write the field distributions of both phasors and instantaneous values

( ) ( ) ( )

( )

0

2/

00000 sin2esin2eee xxxE

π ϕ ϕ ϕ −−

=−=−=

j

i

j

i

j jkz jkz

i ekz E kz jE E

( ) ( ) ( ) 00 sinsin2, xE t kz E t z i ω −=

( ) ( ) 0000 ecos2eee yyHϕ ϕ j

i j jkz jkz

i kz H H =+= −

( ) ( ) ( ) 00 sincos2, yH t kz H t z i ω =

We now have a new kind of

electromagnetic wave. The

dependence on the z coordinateand time is separated. This

indicates that it is not a

propagating wave but a standing

wave. The distribution of the

electric field and the magnetic

field of a standing wave is plotted

in Fig. 2.8. The electric field is

shifted by 90 deg corresponding to

the magnetic field. It is evident

from Fig. 2.8 that using an excited

standing wave we can measure the

wavelength, as the distance

between two adjacent minima or

maxima is equal to one half of the

wavelength.

The power transmitted by

the standing wave is determined

by Poynting’s vector

( ) ( ) ( ) ( )( ) 0ecos2sin2ReRe 0002/

0 =×=×= −yxHES

ϕ π ϕ ji

jiav kz H ekz E

k i

Ei

Hi

Er Hr

k r

Fig. 2.7

( ) z E

( ) z H

−λ /4

−λ /4−λ /2

-λ/2

-3 λ /4

-3 λ /4 −λ

−λ

Fig. 2.8



book - 2

28

The power transmitted by the standing wave is zero, which is why it is called standing. This is

a natural result, as the incident wave transmits some power and no power is lost in the ideally

conducting wall. Consequently, the whole power is reflected back in the form of a reflected

wave. The total power, which is the vector sum of these two, must be zero.

The numerical results are:

1m3

2−==

π ω

ck , Ω= π 120w Z ,

at t = 0 s and z = -1 m we have

( )2

-,V/m8.57sin3

2sin2100 00max

π ϕ ϕ

π ==⇒

−== ii E E E

Consequently we have

( ) ( ) 02

sinsin6.115, xE

−=π

ω t kz t z

( ) ( ) 02

sincos6.30, yH

−=π

ω t kz t z

_________________________________________

2.3 Wave polarization

We showed that vectors E and H are mutually perpendicular and lie in a transversal plane perpendicular to the direction in which the wave propagates. Their actual position in

this plane can, however, be quite general, and it can vary with time. The wave polarization

determines the way in which the end point of vector E moves in the transversal plane. Let us

find an equation which gives a general description of the behaviour of vector E in the plane

perpendicular to the direction of wave propagation. Let us suppose that this direction is

identical with the z axis and that we have a lossless material. Vector E is

( ) 00 sinsin yxE y ym x xm kz t E kz t E φ ω φ ω +−++−= .

Let us simplify this formula by choosing the position k z x

φ = . Then we have

( )t E E xm x ω sin= , (2.34)

( )φ ω += t E E ym y sin , (2.35)

where x y φ φ φ −= . We eliminate time from (2.34) and (2.35). From (2.34) we get

( ) xm

x

E

E t =ω sin , ( )

2

1cos

−=

xm

x

E

E t ω ,



book - 2

29

and (2.35) gives

( ) ( ) ( ) φ ω φ ω φ ω sincoscossinsin t t t E

E

ym

y+=+= .

Inserting for ( )t ω sin and ( )t ω cos we get

φ φ sin1cos

2

−+=

xm

x

xm

x

ym

y

E

E

E

E

E

E

This formula gives the equation

φ φ 2

22

sincos2 =

+−

xm

x

ym

y

xm

x

ym

y

E

E

E

E

E

E

E

E . (2.36)

Assuming general values E xm, E ym and φ this equationrepresents the equation of an ellipse, Fig. 2.9, in an

arbitrary position in the coordinate system E x and E y.

Consequently the end point of vector E moves along an

ellipse, and the wave is an elliptically polarized wave.

The position and the shape of this ellipse depend on

values E xm, E ym and φ .A special case is π φ n= , in which we have

.0sinand,1cos =±= φ φ (2.36) then gives the equation

xm

ym

x y E

E E E ±= , (2.37)

which is the equation of a line, Fig. 2.10, and our wave is a linearly polarized wave. Another

special case is ( )2

12π

φ += n , in which 1sinand0cos ±== φ φ . (2.36) then gives the

equation

1

22

=

+

ym

y

xm

x

E

E

E

E , (2.38)

E x

E y

Fig. 2.9

E x

E y

Fig. 2.10

E x

E y

E x

E y

E x

E y E ym=0 E xm=0 n odd n even



book - 2

30

of an ellipse in the basic position, Fig.

2.11. This wave has an elliptic

polarization. Assuming further equality

of amplitudes E xm=E ym this elliptically

polarized wave becomes a wave with a

circular polarization, Fig. 2.12.

In the case of a wave with an

elliptical or circular polarization we can

distinguish two types of polarization

determined by the direction of vector E

rotation. If the angle difference φ is within the interval (0,π ),vector E rotates to the left, when observing the wave in the

direction of propagation, and the wave has a left-handed

polarization, Fig. 2.13a. If the angle difference φ is within the

interval (π ,2π ), vector E rotates to the right, when observing the

wave in the direction of propagation, and the wave has a right-

handed polarization, Fig. 2.13b.

Circularly polarized wavesare widely used in technical

applications, e.g., transmission of

a signal in communications, etc. It

can be shown that the

superposition of two circularly

polarized waves, one rotating to

the left, the second rotating to the

right, gives a wave with an

arbitrary polarization. This is

important, e.g., when treating

propagation of waves in non-isotropic materials, e.g., in a magnetized ferrite. The circularly polarized wave can be

represented by an instantaneous value

( ) ( )kz t E kz t E E kz t E E y x −±=

±−=−= ω π

ω ω cos2

sin,sin 000 ,

( ) ( ) ( )[ ]000 cossin, yxE kz t kz t E t z −±−= ω ω ,

and consequently by

( ) jkz000 e−±= yxE j E .

A survey of basic types of polarization is given in Tab. 2.1.

________________________________

Example 2.5: Decompose the generally elliptical polarized wave into the superposition of the

two circularly polarized waves, one with left-handed polarization, the second with right-

handed polarization.

We start solving a reciprocal problem in which we combine two circularly polarized

waves with opposite orientation of the electric field vector rotation. The waves with left-handed and right-handed polarization are

Fig. 2.11

E x

E y E xm> E ym

E x

E y

E xm< E ym

E x

E y

Fig. 2.12

E xm=E ym

⊗

left hand polarization

⊗

right hand polarization

Fig. 2.13

a b( )π φ ,0∈ ( )π π φ 2,∈



book - 2

31

φ (deg) 0 45 90 135 180 225 270 315

E xm=0

E xm< E ym

E xm= E ym

E xm> E ym

E ym=0

left-handed right-handed

Tab. 2.1

( ) jkz

L j A −+= e00 yxE , ( ) jkz

R j B −−= e00 yxE .

The superposition of these waves is

( ) ( )[ ] jkz

R L j A B B A −−++=+= e00 yxEEE .

In the case A= B the resulting wave is linearly polarized, for A> B it is elliptically polarized

right-handed, for A< B elliptically polarized left-handed. Let us reverse this procedure. Wehave a wave with general elliptical polarization

( ) jkz jDC −+= e00 yxE .

Comparing this formula with the previous formula we get the amplitudes of the particular

waves with circular polarization

2,

2

DC B

DC A

+=

−= .

_________________________________________

Example 2.6: Calculate the power transmitted by a wave with generally elliptical polarization.

A wave with elliptical polarization can be written in the form

ϕ j jkz jkz B A −−− += eee 00 yxE .

Poynting’s vector is



book - 2

32

( ) ( )[ ]

( ) ( )[ ]

[ ] 022

0*

0*

00

*0

*

Re

2

1

eeeeeeRe2

1

Re2

1Re

2

1

z

xyyx

EzEHES

B A

Z

B A B A Z

Z

w

j jkz jkz j jkz jkz

w

w

av

+

=−×+

=××=×=

−−− ϕ ϕ

_________________________________________

2.4 Cylindrical and spherical wavesWe will solve the wave equation for a vector potential in a cylindrical coordinate

system in the free space filled by a lossless material without a source. We will assume that the

field does not depend on α and z. As a result, we obtain a wave which propagates from the z

axis to infinity with surfaces of a constant amplitude and a constant phase in the shape of

cylinders. Such a wave is an abstraction as a plane wave. The excitation of a cylindrical wave

supposes an infinitely long line source with a passing current of a constant amplitude and a

constant phase along this source.

The homogeneous wave equation for the z component of a vector potential A = A(r )z0

is (1.54)

02 =+∆ Ak A .

As the vector potential does not depend on α and z , we have

01 2

2

2

=+∂

∂+

∂

∂ Ak

r

A

r r

A. (2.39)

This is a Bessel-type equation with a solution in the form of the sum of Bessel functions of

zero order of the first and second kind, see mathematical appendix,

( ) ( ) ( ) ( )kr H C kr H C kr Y C kr J C A 20

,2

10

,10201 +=+= , (2.40)

where Hankel functions of the zero order and of the first and second kind are defined by

Bessel functions, see mathematical appendix (13.44) and (13.45). It is evident from their

asymptotical expression (13.46) and (13.47) that function 20 H represents a wave propagating

as kr increases toward infinity, whereas 10 H represents a wave propagating from infinity

towards the z axis. Assuming only a wave propagating from the z axis in an area of high kr ,

i.e., far distant from the axis, we have finally the cylindrical wave described by the form

r

eC A

jkr −

= . (2.41)

Why does the amplitude of this wave decrease as function r /1 , as we have no

losses? The decrease in the amplitude is caused by the spread of power to space. Let us

calculate the density of the power transmitted by this wave



book - 2

33

0

2 1

2rS

r Z

D

w

av = ,

where D is a constant and Z w is the wave impedance. The power transmitted through the

surface of a cylinder with radius r and length l is

l Z

Drl S P

w

av π π 22

22

== . (2.42)

This power is constant, so we have no losses and it confirms why the field amplitude

decreases as r /1 .

For high kr and a small space angle we can approximate this wave by a plane wave.

Let us now solve the wave equation in a spherical coordinate system, assuming that

the vector potential depends only on distance from the origin r . We will solve a homogeneous

equation, as we have a source-less space. The solution corresponds to a spherical wave. Such

a wave can be excited by an elemental omni-directional source placed at the origin. We then

solve the equation

011 22

2=+

∂

∂

∂ Ak

r

Ar

r r . (2.43)

The solution of this equation corresponding to a wave propagating from the origin towards

infinity is

( )kr H r

C A 2

2/1= , (2.44)

using the formula for a Hankel function of order ½ and of the second kind (13.48), we get

r C A

jkr −

=e

. (2.45)

The amplitude of this spherical wave decreases due to the spread of power as 1/r . This wave,

for high kr and within a small space angle, can be approximated by a plane wave.

____________________________

Example 2.7: An ideal omnidirectional antenna transmits a spherical wave. What power mustthe antenna transmit to get transmitted power density S av=1 mW/m2 at distance r 1=1 km. What

power density corresponds to this at distance r 2=1 m? Calculate the corresponding electric

field amplitude at r 2=1 m. Assume a lossless material.

The total transmitted power is

( ) W1256041000 211 ==== ∫∫ r S dS S r P avav π .

The power density at r 2 is



book - 2

34

( ) 2

22 W/m10004

12

===r

P r S av

π

As the power density is

( ) z

w

av Z E S ϕ cos

21

2

=

we get

( )V/m1.868

cos

2==

z

wav Z S E

ϕ

_________________________________________

2.5 Problems2.1 Write the solution of the wave equation describing a plane electromagnetic wave

propagating in the free space with parameters ε r =8, µ r =1, σ =40 S/m. The frequency is

40 GHz.

( ) 12 m20403140 −−=−= j jk ωµσ µε ω

z j z 31402040

0 ee −−= EE

( ) ( )ϕ +−= − z t t z z 314010.5.25sine, 102040

0EE

2.2 The plane electromagnetic wave propagates in a free space filled by a material with

parameters ε r =4, µ r =1, σ =0. The frequency is 10 MHz. The electric field intensity has thedirection of the x axis and has a positive maximum equal to 10 V/m at z =0 and t =0. Calculate

the electric and magnetic field and their instantaneous values and Poynting’s vector at z =6 m

and t =10-8 s.

V/me10ee10 94.042.02/ j z j j

x E −− == π

( ) V/m94.52

42.01014.3sin10, 7 −=

+−⋅=π

z t t z E x

A/me053.0ee053.0 94.042.02/ j z j j

y H −− == π

( ) A/m0315.02

42.01014.3sin053.0, 7 −=

+−⋅=π

z t t z H y

20 W/m26.0 zS =av

2.3 A plane electromagnetic wave propagates in a free space filled by a material with

parameters ε r =80, µ r =1, σ =0.002 S/m. The frequency is 500 kHz. The electric field intensity

has the direction of the x axis and has a positive maximum equal to 50 V/m at z =0 and t =0.

Calculate the electric and magnetic field and their instantaneous values and Poynting’s vector. z j j z

x E 101.02/039.0 eee50 −−= π

( )

+−= −

2

101.0sine50, 7039.0 π ω z t t z E z

x



book - 2

35

z j j z

y H 101.0)366.02/(039.0 eee37.1 −−−= π

( )

−+−= − 366.0

21012.0sine37.1, 7039.0 π

ω z t t z H z

y

20

078.0 W/me32 zS z av

−=

2.4 Two plane electromagnetic waves propagate in the same direction in a lossy material.The first wave has the frequency 1 GHz and the corresponding attenuation constant is β 1=78

m-1. The second wave has the frequency 3 GHz and the attenuation constant is β 2=83 m-1. The

amplitudes of these waves are equal at z =0. Determine the distance z in which E 1=10 E 2.

mm46.010ln

12

=−

= β β

z

2.5 A plane electromagnetic wave propagates in a free space filled by a material with

parameters ε r =2, µ r =1, σ =0.01 S/m. The frequency is 9 MHz. Calculate the electric and

magnetic field amplitudes at the point at which S av=10 W/m2.

V/m8.48cos

2 == z

wav Z S

ϕ E

A/m58.0==w Z

EH

2.6 To measure a receiving antenna we need a plane electromagnetic wave. To measure with

a sufficiently low error we can accept a change in field amplitude of 1% within a region 1 m

in length. The wave is excited by an ideal omnidirectional antenna transmitting a cylindrical

wave. Calculate the distance from the transmitting antenna at which we have to measure.

r =99 m

3. WAVES ON A PLANE BOUNDARY

3.1 Perpendicular incidence of a plane wave to a plane boundaryLet us solve the problem of the perpendicular incidence of a plane electromagnetic

wave on a plane boundary between two different materials, Fig. 3.1. An incident wave carries

some power. Part of this power is reflected back at the boundary in the form of a reflected

wave. In the first material, this wave is superimposed on the incident wave. Part of the power is transmitted to the second material in the form of a transmitted wave. The incident wave is

described by the quantities marked by index i, the reflected wave is marked by index r , and

the wave transmitted to the second material is marked by index t . We assume that the

orientation of the vectors is according to Fig. 3.1. Our task is to determine the relations

between the amplitudes of particular waves. The propagation vectors of these waves are

01zk k i = ,

02zk k t = , (3.1)

01zk k r −= ,



book - 3

36

where k 1 and k 2 are the propagation

constants of a plane wave in the first

and second materials, respectively. The

total field in the first material is the

superposition of the incident and

reflected waves

z jk rx

z jk ixr i E E 11 ee 001 xxEEE +=+=

− ,

( )

( ) z jk rx

z jk ix

z jk rx

z jk ixr i

E E Z

E E k

11

11

ee1

ee

01

00001

11

−=

=×−×=+=

−

−

y

xzxzHΗHωµ

.

In the second material we have only the transmitted wave

z -jk

txt E 2

e02 xEE == ,

z -jk tx

z -jk txt E

Z E

k 22 e

1e 0

20

2

22 yyHH ===

ωµ

Both the electric field and the magnetic field are parallel to the boundary placed at z =

0. These fields must fulfill the boundary conditions for the tangential components (1.18) and

(1.121). So we have at z = 0

txrxix E E E =+ , (3.2)

( ) txrxix E k

E E k

2

2

1

1

µ µ =− . (3.3)

This set of equations has the solution

2112

2112

k k

k k E E ixrx

µ µ

µ µ

+

−= , (3.4)

Ei Et

Er

Hi Ht

Hr

k i k t

k r

ε 2, µ 2, σ 2 ε 1, µ 1, σ 1

Fig. 3.1

0



book - 3

37

2112

122

k k

k E E ixtx

µ µ

µ

+= . (3.5)

Now we can define the reflection coefficient as

12

12

2112

2112

Z Z

Z Z

k k

k k

E

E

R ix

rx

+

−

=+

−

== µ µ

µ µ

, (3.6)

and the transmission coefficient as

12

2

2112

12 22

Z Z

Z

k k

k

E

E T

ix

tx

+=

+==

µ µ

µ , (3.7)

where Z 1 and Z 2 are the wave impedances of the two materials (2.18). Values R and T are

generally complex. Inserting R and T from (3.6) and (3.7) into the boundary condition (3.2)

we get the relation between R and T in the form

RT += 1 . (3.8)

Similarly as in (3.5) and (3.6), we can define the reflection and transmission coefficients for a

magnetic field. The reader can compare the relations for the reflection and transmission

coefficients for a plane electromagnetic wave, which is perpendicularly incident on the plane

boundary of two different

materials, with the incidence

of a wave in the

transmission lines, see Fig.

3.2.

For a lossless

dielectric, the wave

impedance is defined by (2.26), and we have

21

21

21

21

12

12

nn

nn

Z Z

Z Z R

r r

r r

+

−=

+

−=

+

−=

ε ε

ε ε , (3.9)

21

1

21

1

12

2 222

nn

n

Z Z

Z T

r r

r

+=

+=

+=

ε ε

ε , (3.10)

where n is the refractive index

r n ε = . (3.11)

When ε 1<ε 2 we have Z 1> Z 2 and consequently R<0, and the orientation of the electric field

vectors is according to Fig. 3.3a. The electric field is reflected out of phase. When ε 1>ε 2 we

have Z 1< Z 2 and consequently R>0 and the orientation of the electric field vectors is according

to Fig. 3.3b. Then electric field is reflected in phase.

Z 1 Z 212

12

Z Z

Z Z R

+

−=

12

22

Z Z

Z T

+=

Fig. 3.2



book - 3

38

Let us now calculate the power

transmitted by the field in the first

material and the power carried by the

transmitted wave in the second material,

assuming that there are no losses in the

two materials. In the first material we

get

( ) ( ) ( )[ ]

[ ]

( ) [ ]

( )2

02*2***

0*

02*2***

*0

*000

*111

1

eeRe

2

1Re

2

1

eeRe2

1

eeeeRe2

1Re

2

1

11

11

1111

R

H E H E H R E R H E

H E H E H E H E

H H E E

avi

z jk iyrx

z jk ryixiyixiyix

z jk iyrx

z jk ryixryrxiyix

z jk ry

z jk iy

z jk rx

z jk ixav

−=

=+−−+=

=+−−=

=−×+=×=

−

−

−−

S

zz

z

yyxxHES

(3.12)

This is a logical result. The total power transmitted in the first material is equal to the power

transmitted by the incident wave reduced by the power transmitted by the reflected wave. The

power transmitted in the second material must be equal to the power transmitted in the first

material, as no power can be lost in the boundary itself.

Let us now assume that the second material is a perfect conductor with σ 2 infinite

and consequently Z 2=0. There is a zero field in such a material, i.e., no field penetrates into

this material. We get R = -1 and T = 0. The field in the first material is the superposition of the

incident and reflected waves. This was already solved in Example 2.4. The total field has thecharacter of a standing wave described, see Example 2.4, by the phasors of the electric field

and the magnetic field

( ) 00 sin2 xE kz jE i−= , (3.13)

( ) 00 cos2 yH kz H i= , (3.14)

The instantaneous values of these fields are, see Example 2.4,

( ) ( ) ( ) 00 cossin2, xE t kz E t z i ω −= , (3.15)

( ) ( ) ( ) 00 sincos2, yH t kz H t z i ω = . (3.16)

This wave does not transmit any power. The distribution of its amplitudes is plotted in Fig.

2.8. As was mentioned in Example 2.4, the excitation of this standing wave is frequently used

to measure the wavelength, or consequently the frequency.

Let us now turn our attention again to the perpendicular incidence of a plane

electromagnetic wave to the plane boundary between two dielectric materials. Let us

analyze in greater detail the field in the first material, which is the superposition of the

incident and reflected waves, Fig. 3.1. We suppose that ε 1>ε 2 and consequently R>0. The x-component of an electric field is

E i E t k i

E r

k t

k r

a

E i E t k i

E r

k t

k r

b

Fig. 3.3



book - 3

39

( ) z jk z jk z jk z jk z jk rxix x e R R E E R E E E E 11111 1eeee 0001

−−−−++=+=+= ,

( ) ( ) z jk x e R z k R E E 11cos2 101

−−+= , (3.17)

Similarly for a magnetic field we get

( ) ( ) z jk y e R z k jR H H 11sin2 101

−−+−= . (3.18)

In the case ε 1<ε 2 the formulae for an electric field and for a magnetic field are exchanged. Thefirst terms in field distributions (3.17) and (3.18) represent a standing wave with amplitude

2 RE 0 and 2 RH 0, whereas the second terms represent a wave traveling toward the boundary

with amplitude E 0(1- R) and H 0(1- R) and continuing in the second material as the transmitted

wave. Only the traveling part transmits power, it therefore verifies (3.12), and we get

( )2001 1

2

1 R H E S av −= . (3.19)

The distribution of the amplitude of an

electric field of this standing wave is

plotted in Fig. 3.4. We can identify the

minima and maxima of the amplitude,

with their mutual distance equal to λ /2.

The same distribution of a voltage wave

and a current wave on a transmission line

can be measured, assuming that this wave

is incident to the junction between two

lines with different wave impedances,Fig. 3.2.

Standing waves are described by a

standing wave ratio p, abbreviated as

SWR, defined by

R

R

E

E p

−

+==

1

1

min

max . (3.20)

This SWR is very often used in practical applications, as it can be measured simply.

Consequently, the measurement of a reflection coefficient or of an impedance is transposed tothe measurement of SWR.

As a last case, we will study the perpendicular incidence of a plane wave to a well

conducting material with conductivity σ . The case of the ideal conductor has been already

solved in Example 2.4, where the standing wave with p = ∞ was introduced. We assume that

the first material is air. So the wave impedance and the propagation constant for the two

materials are

001 µ ε ω =k ,0

01

ε

µ = Z , ( )

21

22

σ ωµ jk −= , ( )

σ

ωµ

21 2

2 j Z +=

E

E max=E 0(1+ R)

E min

=E 0(1- R)

λ /2

Fig. 3.4



book - 3

40

The transmission coefficient is

( ) j Z Z

Z T

−+

=+

=

12

1

22

2

12

2

ωε

σ . (3.21)

The transmitted wave can be expressed

( )

z jk ixt e

j

E 2

12

1

2

2

0−

−+

=

ωε

σ xE , (3.22)

This intensity enables us to calculate the

total current passing a strip of the second

material of width equal to l , Fig. 3.5, as it

determines the current density (1.9). We

get

tym

tym z jk

txm

tx x

H l

H jk

l Z dz e E l

dz E l dz J l I

=

===

===

∫

∫∫

∞−

∞∞

2

2

0

00

22

22

σ σ

σ

The result is very simple

t tym K l H l I ==2 , (3.23)

the total current is proportional to the value of the magnetic field just below the boundary,

which can be substituted by current density K . The power lost in the strip, Fig. 3.5, of length

d is

( ) ( ) 222*2

*2

2

1

22

1Re

2

1Re

2

1tymhf tymtxm H S H l d I E d I U P ρ

σ

ωµ ==== , (3.24)

where S =ld is the surface of the strip, and ρ hf is the high frequency resistance of the surface of a well conducting material

σ

ωµ ρ

2

2=hf . (3.25)

Formula (3.24) assumes homogeneous distribution of the magnetic field along the boundary.

The more general form is

Ei Et J2

Hi Ht

k i k t

ε 2, µ 2, σ 2 ε 0, µ 0

Fig. 3.5

l



book - 3

41

∫∫=

S

t hf dS H P 2

2

1 ρ . (3.26)

So to calculate the power lost in a material we have to know only the distribution of the

tangent component of the magnetic field just at the surface of this material.

______________________________ Example 3.1: A plane electromagnetic wave with amplitude E im=300 V/m is incident from

the air perpendicular to the surface of a conducting material with parameters ε r =80, µ = µ 0, σ =5S/m, and the frequency is 500 kHz. Calculate:

a) The phasors and the instantaneous values of the reflected and transmitted waves.

b) The power passing the boundary.

c) The penetration depth and the wavelengths in the air and in the material.

d) The distance at which the field amplitude decreases to 1/100 of its amplitude at the

boundary.

The wave impedances and propagation constants are

π 1201 = Z ,4/0

2 e888.0 π

σ ωε

ωµ j

j

j Z ≅

+= ,

1001 m0105.0 −== µ ε ω k , ( ) 1

02 m1 −−=−= π σ ωµ j jk .

The reflection and transmission coefficients are

112

12 −≅+

−=

Z Z

Z Z R , 4/

12

2 e0047.02 π j

Z Z

Z T =

+= .

The phasors and instantaneous values of the electric and magnetic fields are

z ji

0105.00 e300 −

= xE , z ji

0105.00 e796.0 −

= yH ,

z j jr

0105.00 e300 π

exE = , z jr

0105.00 e796.0yH = ,

4/0 e410.1 π π π j z j z

t eexE−−

= , z j z t

π π −−= ee596.10yH ,

( ) ( ) z t t z i 0105.0cos300, 0 −= ω xE , ( ) ( ) z t t z i 0105.0cos796.0, 0 −= ω yH ,

( ) ( )π ω ++= z t t z r 0105.0cos300, 0xE , ( ) ( ) z t t z r 0105.0cos796.0, 0 += ω yH ,

( ) ( )4/cos41.1, 0 π π ω π +−=

− z t et z z t xE , ( ) ( ) z t et z z

t π ω π −=

−cos596.1, 0yH .

The particular power densities are

( )2

01

2

0

*

mW/4.1192

1

Re2

1

zzHES ==×= Z

E iiiiav ,



book - 3

42

( ) 20

* mW/4.119Re2

1zHES −=×= r r rav ,

( ) ( ) 20

2

2

02

2

0* mW/8.0cos

2

1Re

2

1Re

2

1zzzHES ==

=×= z

t t t t tav

Z

E

Z

E ϕ .

The penetration depth is

m318.01

2

== β

δ .

The distance z 100 at which E ( z 100)= E ( z =0)/100 is calculated

( )m488.1

100ln01.0

2

1002 ==⇒=−

β

β z e

z .

______________________________________

3.2 Perpendicular incidence of a plane wave to a layered mediumLet us assume a plane electromagnetic wave incident from a half space filled by

material perpendicular to the system of layers of different materials, Fig. 3.6a. Our task is

to describe the field distribution in all layers and the field penetrating through this structure.

This problem can be effectively solved using transmission matrices. We derive separately thetransmission matrix describing the behaviour of the waves on the boundary between two

adjacent materials, Fig. 3.6b, and the transmission matrix of a single layer, Fig. 3.6c.An electromagnetic wave with the complex amplitude of an electric field a1 is incident

from the left to the boundary between two different materials I and II, Fig. 3.6b. This wave is

partially reflected and partially transmitted through the boundary. Similarly, a wave withamplitude b2 propagating from the right is reflected and transmitted. T 12 is the transmission

coefficient from material I to material II. T 21 is the transmission coefficient from material II tomaterial I. R12 represents the reflection from the left, R21 is the reflection coefficient from the

side of material II. The wave traveling from the boundary to the left has amplitude b1, Fig.

3.6b

2121121 bab T R += . (3.27)

I

k 1 k 2 k 3 k 4 . . .

II

d 2 d 3

a1

b1

a1

b1

T 12 a1

R12 a1

R21 b2 a2

b2T 21 b2

a b

I II

d 2

k 2

c

Fig. 3.6

a1

II

b2

2212

d jk e−= aa

2221

d jk e−

= bb



book - 3

43

Amplitude a2 of the wave propagating from the boundary to the right is, Fig. 3.6b

1122212 aba T R += . (3.28)

From (3.27) and (3.28) we get, using R12 = - R21 and T 12T 21 – R12 R21 = 1,

=

2

2

12

12

121

1

111

ba

ba

R R

T . (3.29)

So the transmission matrix describing the boundary between the two materials is

[ ]

=

1

11

12

12

12III,

R

R

T A . (3.30)

Now we derive the transmission matrix of only a single layer, Fig. 3.6c. In this layer

we have one wave propagating to the left with amplitude b2 and one wave propagating to theright with amplitude a1. These two waves passing the layer of thickness d 2 can be expressed

22

12d jk e−= aa , 22

21d jk e−= bb ,

we consequently get the transmission matrix

=

−

2

2

1

1

22

22

e0

0e

b

a

b

a

d jk

d jk

, (3.31)

So the transmission matrix describing the behaviour of the waves in the single layer is

[ ]

=

−22

22

e0

0e2 d jk

d jk

A . (3.32)

The total transmission matrix of the layered structure, Fig. 3.6a, is the product of

particular matrices of type (3.30) and (3.31)

=

−−

N

N

d jk

d jk

d jk

d jk

R

R

T R

R

T b

a

b

a...

e0

0e

1

11

e0

0e

1

11

33

33

22

22

23

23

2312

12

121

1. (3.33)

Structures of this kind, Fig. 3.6a, are frequently used namely in optical wavelength

technology as an antireflection coating which can provide even frequency selective behaviour.

These structures can serve as band pass filters, band stop filters, etc. They can be designed ina similar way as microwave filters composed of sections of transmission lines of different

wave impedances.

_______________________________ Example 3.2: Design an inter-layer placed between two different dielectric materials to getzero reflection, a so called antireflection layer.



book - 3

44

The structure is shown in Fig. 3.7. There is adielectric layer of thickness d placed between two dielectric

layers. Our task is to design the thickness and the permittivity of this inter-layer to get zero reflection, i.e., to

get b1 = 0. The transmission matrix of our structure is

according to (3.33) the product of three matrices

=

− 01

11

e0

0e

1

11 3

23

23

2312

12

121

1

2

2 a

b

a

R

R

T R

R

T d jk

d jk

From this set of equations we get

( ) 323122312

122 ee

1aa

d jk d jk R R

T T

−+= ,

( ) 323122312

1

22

ee

1

ab

d jk d jk

R RT T

−

+= .

The reflection coefficient is

( ) d jk

d jk

d jk d jk

d jk d jk

R R

R R

R R

R R R

2

2

22

22

22312

22312

2312

2312

1

1

e1

e

ee

ee

−

−

−

−

+

+=

+

+==

a

b.

Inserting for particular reflection coefficients R12 and R23 from (3.9) we get

( )( ) ( )( )( )( ) ( )( ) d jk

d jk

nnnnnnnn

nnnnnnnn R2

2

232213221

221323221

e

e−

−

−−+++

+−++−= .

For the next procedure we assume two possible simplifications relating refractive indicesn1>n2>n3 or n1<n2<n3. Firstly, we have to get the real and minimum value of the reflection

coefficient. This supposes

π π md k d jk 221e 22 2 +=⇒−=

− ,

where m is an integer. As k 2 = 2π /λ 2, we get the value of the antireflection layer thickness

24

22 λ λ md += . (3.34)

The minimum inter-layer thickness is equal to a quarter of the wavelength of a wave in the

second material. Now from the condition R = 0 we get

( )( ) ( )( ) 021323221 =+−−+− nnnnnnnn ,

which gives us the condition of permittivity

ε 1 ε 2 ε 3

d

a1

b1

a3

Fig. 3.7

b3=0



book - 3

45

312 r r r ε ε ε = or 312 nnn = . (3.35)

Results (3.34) and (3.35) are parameters of the well known quarter-wavelength transformer

frequently used in microwave engineering.

__________________________________

3.3 Oblique incidence of a plane electromagnetic wave to a plane boundaryIn the case of the oblique incidence of a plane electromagnetic wave to a plane

boundary between two different materials

and we have to treat two cases

separately. We have to differentiate the two

cases of polarization of an incident wave.

We assume a wave with a linear

polarization. In the first case, vector E is

perpendicular to the plane of incidence,

which is defined by normal vector n

together with the propagation vector of an

incident wave k i, Fig. 3.7. This case is

marked as the incidence of a wave with

horizontal or perpendicular polarization.

This notification follows from the theory of

wave propagation above the earth ground

and, e.g., their incidence to the ionosphere.

The second case is called the incidence of a

wave with vertical or parallel

polarization, as vector E is parallel to the

plane of incidence.Horizontal polarization. In this

case we assume the orientation of all

vectors as defined in Fig. 3.8. Our task is to determine the relation between the amplitudes of

the electric fields of all three waves. To do this we have to force an electric field and a

magnetic field to fulfill the boundary conditions for tangential components at plane x = 0

(1.18) and (1.20), assuming that no current passes along the boundary. Let us first express the

propagation vectors, Fig. 3.8, in their components

( ) ( ) 0101

00

sincos zx

zxk

ii

iz ixi

k k

k k

ϕ ϕ +−=

=+−=, (3.36)

( ) ( ) 0101 sincos zxk r r r k k ϕ ϕ += , (3.37)

( ) ( ) 0202 sincos zxk t t t k k ϕ ϕ +−= , (3.38)

So the electric and magnetic fields of the incident wave can be expressed in the form

( ) ( ) z jk x jk i

r jii

iii E E ϕ ϕ sincos

000011 eee

−⋅−== yyE

k , (3.39)

n

ϕ i ϕ r

ϕ t

Ei

Er

Et

Hi

Hr

Ht

k i

k r

k t

k iz k ix

ϕ i

Fig. 3.8



book - 3

46

( ) ( )( ) ( ) ( ) z jk x jk ii

iiii

ii

Z

E ϕ ϕ ϕ ϕ ωµ

sincos00

1

0

1

11 eesincos−

−−=×

= xzEk

H , (3.40)

For the reflected and transmitted waves we have

( ) ( ) z jk x jk r

r jr r

r r r E E ϕ ϕ sincos

000011 eee

−−⋅−−=−= yyE

k , (3.41)

( ) ( ) z jk x jk t

r jt t

t t t E E ϕ ϕ sincos

000022 eee

−⋅−== yyE

k , (3.42)

( ) ( )( ) ( ) ( ) z jk x jk r r

r r r r

r r

Z

E ϕ ϕ ϕ ϕ ωµ

sincos00

1

0

1

11 eesincos −−+−=

×= xz

Ek H , (3.43)

( ) ( )( ) ( ) ( ) z jk x jk t t

t t t t

t t

Z

E ϕ ϕ ϕ ϕ ωµ

sincos00

2

0

2

22 eesincos−

−−=×

= xzEk

H , (3.44)

The boundary conditions at x = 0 are

( ) ( ) ( ) z E z E z E tyryiy ,0,0,0 =− , (3.45)

( ) ( ) ( ) z H z H z H tz rz iz ,0,0,0 =+ (3.46)

Inserting into (3.45) the expressions for the electric field from (3.39) and similar expressions

for Er and Et (3.41) and (3.42) we get

( ) ( ) ( ) z jk

t

z jk

r

z jk

i

t r i E E E ϕ ϕ ϕ sin

0

sin

0

sin

0

211 eee−−−

=− . (3.47)

This condition must be fulfilled for each coordinate z . This means that the exponents in all

terms must be equal. This results in

( ) ( ) ( )t r i k k k ϕ ϕ ϕ sinsinsin 211 ==

This equation defines Snell’s first and second laws

r i ϕ ϕ = , (3.48)

( ) ( )t i k k ϕ ϕ sinsin 21 = . (3.49)

Snell’s first law tells us that the angle of incidence equals the angle of reflection. Snell’s

second law can be used to calculate the angle of transmission. Boundary condition (3.47) now

reduces to

000 t r i E E E =− . (3.50)

Similarly, inserting (3.40) and analogous formulas for Hr and Ht (3.43) and (3.44) into (3.46),

and applying Snell’s laws, we get





book - 3

48

( ) ( ) z jk x jk t

t t t

Z

E ϕ ϕ sincos

2

00

22 ee−

= yH . (3.59)

Field distributions (3.54) to (3.59) are inserted into boundary conditions (1.18) and (1.20)

assuming K = 0. We get using Snell’s laws the set of equations

( )( ) ( ) 000 coscos t t r ii E E E ϕ ϕ =− , (3.60)

( )

2

0

1

00

Z

E

Z

E E t r i=

+. (3.61)

The reflection and transmission coefficients can be calculated from these two equations

( ) ( )

( ) ( )

( ) ( )

( ) ( )t i

t i

t i

t i

k k

k k

Z Z

Z Z R

ϕ ε ϕ ε

ϕ ε ϕ ε

ϕ ϕ

ϕ ϕ

coscos

coscos

coscos

coscos

2112

2112

21

21

+

−=

+

−= , (3.62)

( )

( ) ( )

( )

( ) ( )t i

i

t i

i

k k

k

Z Z

Z T

ϕ ε ϕ ε

ϕ ε

ϕ ϕ

ϕ

coscos

cos2

coscos

cos2

2112

21

21

2

+=

+= . (3.63)

Angle ϕ t can be expressed in formulas (3.52), (3.53) and (3.62), (3.63) by the angle of

incidence

( ) ( )( )

( )it

t t k k k

k k k k ϕ

ϕ ϕ ϕ 22

122

2

2

12

222 sin

sin1sin1cos −=

−=−= . (3.64)

The reflection and transmission coefficients in the case of both vertical and horizontal

polarization depend not only on the parameters of the material but also on the angle of

incidence. In the case of materials with nonzero conductivity these coefficients are complex

numbers. In many cases the two materials are dielectrics. Then using (3.64) and µ 1 = µ 2 = µ 0 we can express the reflection coefficients in the form

( ) ( )

( ) ( )ii

ii

R

ϕ ε

ε ϕ

ε

ε

ϕ ε

ε ϕ

ε

ε

2

2

1

2

1

2

2

1

2

1

sin1cos

sin1cos

−+

−+−

=⊥ , (3.65)

( ) ( )

( ) ( )ii

ii

R

ϕ ε

ε ϕ

ε

ε

ϕ ε

ε ϕ

ε

ε

2

2

1

1

2

2

2

1

1

2

sin1cos

sin1cos

−+

−−

= . (3.66)

Reflection coefficients (3.65) and (3.66) are plotted in Fig. 3.10 assuming that ε 1/ε 2 = 3, and

ε 1/ε 2 = 1/3. In the following text we will discuss some special cases of the oblique incidence

of a plane electromagnetic wave to a plane boundary between two different materials.



book - 4

49

Fig. 3.10

______________________________ Example 3.3: A plane linearly polarized electromagneticwave with amplitude E i0 = 10 V/m is incident to a plane

boundary between two dielectric materials with permittivitiesε r 1=1 and ε r 2=8. Its frequency is 30 MHz. The angle of incidence is ϕ i = 30°. The electric field contained with the

plane of incidence angle α 0 = 45°. Calculate the electric andmagnetic fields and the transmitted power of the reflectedwave and the transmitted wave. Calculate power transmittedthrough the boundary.

An incident wave must be decomposed to two waves,

the first one polarized perpendicular to the plane of incidence, while the second wave has a parallel polarization,see Fig. 3.11, where the wave is observed in its direction of propagation. The two particular waves have amplitudes

( ) V/m07.7sin 00 ==⊥ α ii E E

( ) V/m07.7cos 00 == α ii E E

The angle of transmission is

( ) °=⇒=

= 2.10177.0sinarcsin

2

1t it

k

k ϕ ϕ ϕ

The amplitudes of the electric field of the particular wave with perpendicular and parallel polarization are

( ) ( )

( ) ( )V/m75.353.0

coscoscoscos

1221

1221 −=−=+

−= ⊥⊥⊥ i

it

it ir E

k k

k k E E

ϕ µ ϕ µ

ϕ µ ϕ µ

ϕi

(°)

R

0 30 60 900

0.2

0.4

0.6

0.8

1

ε1 /ε

2=1/3

R⊥

R

-

+ ϕB

-

ϕi

(°)

R

0 30 60 900

0.2

0.4

0.6

0.8

1

ε1 /ε

2=3

R⊥

R

- +

+

ϕB

c

E i

the plane of

incidence

E i

E i0

0

Fig. 3.11

k r

H i0



book - 4

50

( )

( ) ( )V/m32.347.0

coscoscos2

1221

12 ==+

= ⊥⊥⊥ iit

iit E

k k

k E E

ϕ µ ϕ µ

ϕ µ

( ) ( )

( ) ( )V/m05.343.0

coscoscoscos

2112

2112 ==+

−= i

t i

t iir E

k k

k k E E

ϕ ε ϕ ε

ϕ ε ϕ ε

( )

( ) ( )V/m54.35.0

coscoscos2

2112

21 ==+

= it i

iit E

k k

k E E

ϕ ε ϕ ε

ϕ ε

Now we can draw diagramsshowing the orientation of theelectric field vectors in areflected wave and in atransmitted wave, Fig. 3.12.The amplitudes of the reflectedand transmitted waves and theangles of their declination fromthe plane of incidence, Fig.3.12, are

V/m8.4220 =+= ⊥ r r r E E E

°−== ⊥ 51arctgr

r r E

E α

V/m9.4220 =+= ⊥ t t t E E E

°== ⊥ 43arctgt

t t E

E α

To calculate the amplitudes of a magnetic field we have to calculate the wave impedances

Ω==Ω== 133,12020

02

0

01

r

Z Z

ε ε

µ π

ε

µ ,

A/m0367.0,A/m0127.0,A/m0265.0 000 === t r i H H H

The densities of the active power transmitted by particular waves are

200 W/m1325.0

21

== iiiav H E S ,

2

00W/m0304.0

2

1==r r ravH E S ,

E r

plane of incidence

E r

E r 0

α r

Fig. 3.12

k r

H r 0

E t

plane of incidence

E t E t 0

α t

k t

H t 0

(a) (b)

transmitted wavereflected wave



book - 4

51

200 W/m09.0

21

== t t tav H E S .

To calculate the power transmitted through the boundary we have to decompose these powersto components parallel and perpendicular to the boundary, Fig. 3.13. The power transmittedthrough the boundary is

( ) ( ) ( ) 2W/m0886.0coscoscos ==−= t taviraviiav xav S S S S ϕ ϕ ϕ ,

The power transmitted through the boundary must be of the same value when calculated from both sides. The

power transmitted along the boundary is

( ) ( ) 21 W/m08145.0sinsin =+= iraviiav z S S S ϕ ϕ ,

( ) 22 W/m0156.0sin == t tav z S S ϕ .

The power transmitted in the direction parallel to the boundary has different values in the two materials. ____________________________________

Total transmission. Fig. 3.10 shows that the reflection coefficient for the vertical polarization equals zero at a certain angle. This angle is called Brewster’s angle ϕ B. Thismeans that there is no reflection at Brewster’s angle, and the whole power is transmitted tothe second material. Putting (3.66) equal to zero, we get

( )21

2sinε ε

ε ϕ

+= B , or ( )

1

2tgε

ε ϕ = B . (3.67)

This angle is sometimes called Brewster’s polarization angle. This name is given because anincident wave with an elliptic polarization when it is incident under this angle makes areflection of only the linearly polarized wave, as there is no reflection of the component withan electric field parallel to the plane of incidence. This effect is used in polarizing elements.

Total reflection. The plot in Fig. 3.10 drawn for the case ε 1>ε 2 shows that the

reflection coefficients for the waves of the two kinds of polarization are equal to one startingfrom a certain angle. This angle is called the critical angle ϕ c. The value of this angle followsfrom Snell’s law (3.49). In the case ε 1<ε 2 we have so called refraction to a normal, Fig.3.14a, in which ϕ i>ϕ t . In the case ε 1>ε 2 we have so called refraction from a normal, Fig.3.14b, at which ϕ i<ϕ t . Now angle ϕ t can be equal to 90° assuming that ϕ i = ϕ c. From (3.49)we get for ε 1>ε 2

ϕ i

ϕ t

Fig. 3.13

Siav Srav

Stav



book - 4

52

( )1

2sinε

ε ϕ =c . (3.68)

At angles ϕ i > ϕ c the refraction angle has a complex value and 1= R , Fig. 3.10. This meansthat the whole power is reflected back. This of course does not mean that there is no field inthe second material.

Let us now investigate the field distribution created after total reflection on the boundary between two dielectric materials with permittivities ε 1>ε 2 and assuming ϕ i > ϕ c. Weassume only a wave with a horizontal polarization to simplify the derived formulas. Theresults are also valid in the case of a wave with a vertical polarization. Angle ϕ t has a complexvalue, and from (3.49) we have

( ) ( )it ϕ ε ε ϕ sinsin21= > 1 , ( ) ( ) ( )t t t j ϕ ϕ ϕ cossin1cos 2 =−= .

Now the field in the second material is, Fig. 3.8 and (3.42),

( ) ( )[ ] ( ) ( )t2t2tt2 sincos00

sincosx-00 eee ϕ ϕ ϕ ϕ z jk xk

i z jk

it E T E T −

⊥+−

⊥ == yyE . (3.69)

This formula describes a so-called surface wave. The amplitude of this wave decreasesexponentially in the direction into the second material and propagates in the direction of axis z along the boundary with the propagation constant

( ) ( )it z k k k ϕ ϕ sinsin 122 == . (3.70)

The phase velocity of this wave is

( ) ( ) 2122 sinsin pit z

pz vk k k

v <===ϕ

ω

ϕ

ω ω . (3.71)

ε 2

ε 1

ϕ i

ϕ t

Fig. 3.14a

ε 2

ε 1

ϕ i

ϕ t

b



book - 4

53

Due to this the surface wave is called aslow wave in the second material. This isthe case of a non-uniformelectromagnetic wave, as planes of aconstant amplitude are parallel with the

boundary, and at the same time planes of a constant phase are perpendicular to the

boundary. The field distribution is plottedin Fig. 3.15.

In the case of total reflection wehave 1= R and consequently

( )ψ j R exp= The field in the firstmaterial is the superposition of theincident and reflected waves. Thesewaves are (3.39) and (3.41)

( ) ( )[ ]ii z x jk

ii E

ϕ ϕ sincos

001

e+−−

= yE ,( ) ( )[ ]ii z x jk

r r E ϕ ϕ sincos

001e +−

−= yE ,

their superposition is

( ) ( ) ( )

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

( )[ ] ( )i

iii

iii

z jk jii

z jk j xk j xk ji

z jk x jk x jk ir i

ee xk jE

ee E

e R E

ϕ ψ

ϕ ψ ψ ϕ ψ ϕ

ϕ ϕ ϕ

ψ ϕ sin2/100

sin2/2/cos2/cos00

sincoscos001

1

111

111

2/cossin2

ee

ee

−

−−−−

−−⊥

−=

=−=

=−=+=

y

y

yEEE

. (3.72)

The magnetic field can be derived in a similar way using distributions (3.40) and (3.43)

( ) ( )[ ]

( ) ( )[ ] ( )i z jk jii

iii

r i

ee xk

xk Z

E

ϕ ψ ψ ϕ ϕ

ψ ϕ ϕ

sin2/10

101

01

12/coscoscos

2/cossinsin2

−−−

−−−=+=

z

xHHH

. (3.73)

(3.72) and (3.73) describe a wave which propagates in the z direction and has the character of

a standing wave in the x direction. The distribution of the electric field is plotted in Fig. 3.15.The phase velocity of this wave

( ) 11

11 sin k v

k k v p

i z pz

ω

ϕ

ω ω =>== . (3.74)

For this reason, the surface wave in the first material is known as a fast wave.Later we will explain the principle of a dielectric waveguide on the basis of the total

reflection and of this surface wave.

E y

( )i xk ϕ

π λ cos2

1=

v pz >v1 fast wave

v pz <v2 slow wave

Fig. 3.15



book - 4

54

The oblique incidence of a plane wave to the surface of a lossy material. Let uscalculate the distribution of the electric field in the second lossy material, assuming horizontal

polarization. The transmitted wave is (3.42)

( ) ( )[ ]t t z x jk it E T

ϕ ϕ sincos00

2e +−−⊥= yE .

The second material is a lossy material with nonzero conductivity, consequently its phaseconstant is a complex number k 2 = β 2 – jα 2. From Snell’s law (3.49) it follows that sin(ϕ t ) is acomplex number and cos(ϕ t ) = a + jb. Inserting these values into the formula for an electricfield we get

( ) ( ) ( )[ ]t z k ba x jbait E T

ϕ α β β α sinx00

12222 ee ++−−−⊥= yE . (3.75)

This formula describes a non-uniformelectromagnetic wave. The planes of aconstant amplitude are parallel with

the boundary, Fig. 3.16, whereas the planes of a constant phase aredetermined by the equation

const t z k ba x =−+

ϕ α β sin122 .

This gives the angle under which thewave propagates in the secondmaterial, Fig. 3.16,

( )( )ba

k iT

22

1 sintg

α β

ϕ ϕ

+−= . (3.76)

Increasing the conductivity of the second material causes a rise of α 2 and β 2, which meansthat angle ϕ T decreases to zero. Finally a wave in a well conducting material propagates

perpendicular to the boundary, independently of the angle of incidence, and is a uniform wavewhich is of course attenuated very fast.

3.4 Problems3.1 A plane electromagnetic wave is incident from the air to the plane surface of a dielectric.A reflection causes a standing wave in the air with standing wave ratio p=2.7. Calculate the

permittivity ε r of the dielectric and reflection coefficient R.There are two solutions: R=0.46, ε r =0.138

R=-0.46, ε r =7.2

3.2 What permittivity has a dielectric, the surface of which reflects at most 1% of the energyof a wave incident perpendicular.

0.67<ε r <1.5

ε 2, µ 2, σ 2

ε 0, µ 0

ϕ i

ϕ T

Fig. 3.16

plane of a constantamplitude

plane of a constant phase



book - 4

55

3.3 A plane electromagnetic wave is incident from the air to the plane surface of sea water with parameters ε r =81 and σ =5 S/m. The frequency is 10 MHz, and the amplitude of the waveis E im=100 V/m. Calculate then electric field intensity at depth 1 m under the surface.

( ) m/V00175.01 =t E

3.4 Design an antireflection layer on the surface of a silicon photodiode for light with

wavelength 0.8 µm. The permittivity of silicon is 12.d =(0.11+m0.22) µmε r 2=3.46

3.5 Calculate the complex amplitude of a wave passing the layered structure from Fig. 3.6.The amplitude of the incident wave is E i=10 V/m, the parameters of the structure are: ε r 1=1,ε r 2=4, ε r 3=6, d =8 cm, and the frequency is 10 GHz.

a3=6e- j2π /3 V/m

3.6 Calculate the average value of Poynting’s vectors representing the power transmitted by

the surface wave described by (3.69) and (3.72).( )t

i zav

E T k S ϕ

ωµ sin

2 0

20

22

2⊥

=

( ) ( )

−=

2cossinsin

21

2

1

20

1r

iii

zav xk Z

E S

ψ ϕ ϕ

There is no active power transmitted in the x direction.

3.7 Calculate the slant of an output glass window with permittivity ε r =2.13, as a wave with a parallel polarization passes through it without

losses, Fig. 3.17.The angle must be equal to Brewster’s angle.ϕ = ϕ B = 55.6°

3.8 A plane electromagnetic wave is incident to the plane boundary between two dielectricswith permittivities ε r 1=2.53, ε r 2=1. Calculate the minimum angle of incidence at which totalreflection takes place.

ϕ ic = 33° 3.9 A plane electromagnetic wave with perpendicular polarization is incident from the air tothe plane surface of a material with parameters ε r 2=1.5, σ 2=2.10-5 S/m. The frequency is f = 2MHz, and the angle of incidence is ϕ i = 60°. Calculate the direction of propagation of a wavetransmitted into the second material, its attenuation and phase constants and phase velocity.

ψ T = 44° v p2 = 2.42.108 m/s

0519.02 = β m-1

0382.02 =α m-1

ϕ ε r

Fig. 3.17



book - 5

56

3.10 Calculate the angle of propagation of the wave transmitted into copper after the

incidence of a plane electromagnetic wave from the air under angle ϕ i = 45°. The frequency is

2 MHz the, and conductivity of copper is σ = 5.8.107 S/m.

ψ T = 0°

3.11 Calculate the phase velocity of the surface wave which is excited by the total reflection

of a plane electromagnetic wave that is incident from a dielectric with permittivity ε r 1=4 to its boundary with the air. The angle of incidence is ϕ i = 45°, and the frequency is 10 MHz.

v pz = 2.16.108 m/s

4. SOLUTION OF MAXWELL EQUATIONS AT VERY HIGH

FREQUENCIES

We will assume the propagation of an electromagnetic wave in a generally

nonhomogeneous lossless medium. The medium will be described by the distribution of therefractive index ( ) ( ) ( )rr r n z y xn ε ==,, . We assume a very high frequency, so that the

wavelength is very short, much shorter than the distances at which the refractive index varies

significantly. We assume the distribution of the electric field of a wave propagating in our

medium in a form analogous to a plane electromagnetic wave

( ) ( )rrEE

φ 0 jk m e

−= , (4.1)

where vector function Em(r) determines the distribution of the field amplitude and function

φ (r) determines the distribution of the phase, and k 0 is the phase constant in the vacuum

000 ε µ ω =k . The equation Em(r) = const defines the planes of a constant amplitude, and the

equation φ (r ) = const defines planes of a constant phase, known as wave-fronts. These two

sets of planes are generally different. Nevertheless, we will show that a propagating wave has

locally the same properties as a plane TEM wave propagating in a homogeneous space.

Let us insert (4.1) and a similar formula for the magnetic field into Maxwell’s

equations (1.12), (1.13) and (1.14)

( ) φ φ ωε 00 eerot

jk m

jk m j

−−= EH , (4.2)

( ) φ φ ωµ 00 eerot jk

m jk

m j−−

−= HE , (4.3)

( ) 00 =− φ

ε jk

mediv E . (4.4)

Using formula (13.78) from the mathematical appendix describing the rotation applied to the

product of two functions, we rewrite (4.2)

( ) φ φ φ ωε φ 000 eerote 0

jk mm

jk m

jk j grad jk

−−−=×− EHH .

Reducing the exponential function we get



book - 5

57

( ) mmm j grad jk EHH ωε φ =×− 0rot , (4.5)

and similarly from equation (4.3)

( ) mmm j grad jk HEE ωµ φ −=×− 0rot . (4.6)

Using formula (13.70) for the divergence of the product of two functions, we rewrite (4.4) to

( ) ( ) 0divgrad 00 =+⋅−−

m jk jk

m ee EEφ φ

ε ε

( ) ( )m jk jk jk

m ee jk e EE divgradgrad 0000

φ φ φ ε φ ε ε

−−−−=−⋅ .

Finally we have

( )[ ] ( )mm jk EE divgradgrad 0 ε φ ε ε −=−⋅ . (4.7)

Rewriting (4.5), (4.6) and (4.7) we get

( )mmr m jk

HEH rot1

grad00

0=+× ε

µ

ε φ , (4.8)

( )mmm jk

EHE rot1

grad00

0=−×

µ

ε φ , (4.9)

( )

+=⋅ mmm

grad

jk EEE div1

grad0 ε

ε φ . (4.10)

Now we use the assumption that we treat the field at very high frequencies. This means that

∞→ f , and consequently ∞→0k and 01

0

→ jk

. We can now set the right hand sides of

(4.8), (4.9) and (4.10) equal to zero.

0grad0

0=+× mr m EH ε

µ

ε φ , (4.11)

0grad0

0=−× mm HE

µ

ε φ , (4.12)

0grad =⋅ mEφ . (4.13)

As φ (r ) = const determines the wave-fronts, and vector gradφ is perpendicular to these wave-

fronts we have a very important result. According to (4.13) vectors gradφ and Em are

perpendicular. This means that vector Em is tangent to the wave-front, consequently it is

perpendicular to local direction of the wave propagation.

Inserting now from (4.12) for Hm to (4.11) we have



book - 5

58

( ) mr m EE ε µ

ε φ φ

µ

ε

0

0

0

0 gradgrad −=×× .

Rewriting the double vector product (13.11) we have

( ) ( ) mr mm EEE ε φ φ φ φ −=⋅−⋅ gradgradgradgrad .

Owing to (4.13) we have

22grad nr == ε φ , (4.14)

this equation couples the wave-fronts with the distribution of the relative permittivity. We are

thus able to determine the distribution of phase φ solving the partial differential equation

2222

n z y x

=

∂

∂+

∂

∂+

∂

∂ φ φ φ . (4.15)

The mutual relations between the electric and magnetic fields follow from (4.11) and

(4.12)

mm HE ×= φ ε

ε µ grad-

00, (4.16)

mm EH ×= φ

µ

ε µ grad

00. (4.17)

Using (4.14), vector gradφ can be expressed as

00gradgrad ll n== φ φ , (4.18)

where unit vector l0 determines the local direction of the wave

propagation. (4.16) and (4.17) tell us that vectors Em and Hm are

mutually perpendicular and are perpendicular to the direction of

the wave propagation at each point, Fig. 4.1. This means that a

wave propagating at very high frequencies in a generally

nonhomogeneous medium behaves locally as a TEM wave

propagating in a free space.

The power transmitted by this wave is defined by

Poynting’s vector. Locally this vector again corresponds to the

Poynting vector of a plane TEM electromagnetic wave

[ ] 0

2

2

1Re

2

1lEHES mmmav

µ

ε =×= . (4.19)

The power is transmitted in the l0 direction, i.e., perpendicular to the wave-fronts.

Em

Hm

gradφ = nl0

Fig. 4.1

r



book - 5

59

The row of vectors l0 represents a ray. Let us derive the equation describing the ray.

We will start with equation (4.18), where,

according to Fig. 4.2, l0 = dr/d s. Deriving

(4.18) over s we have

( ) ( )[ ]rr

r φ grad

d

d

d

d

d

d

s s

n

s

=

Let us now pay attention to the right hand

side part of this equation

0d

dgrad

d

d j

j jii

i

x x s

x

sx∑∑

∂

∂

∂

∂=

φ φ ,

where xi is the i-th coordinate, i.e., x, y or z .

Changing the order of the summation we get

( )[ ]

( ) ( )[ ]rllrx

rrxxx

nn x

s x s

x

x x x x s

x

j j

j

j j

ji

i i j j

j j

j jii

i

grad

d

dgrad

d

d

d

d

000

000

=⋅∂

∂=

=⋅∂

∂=

∂

∂

∂

∂=

∂

∂

∂

∂

∑

∑∑∑∑∑ φ φ φ

Consequently we get

( ) ( )[ ]rr

r n s

n s

gradd

d

d

d=

, (4.20)

which is the equation describing the trajectory of the ray. The changes in the direction of the

rays are caused by changes in the refraction index. Solving (4.20) we are able to find the rays

and to represent the wave by rays. In the case of a homogeneous material we have n = const ,

consequently grad(n) = 0 and

( ) 0d

d

d

d=

sn

s

rr => bar += s

where a and b are constant vectors. The ray

represents a straight line.According to (4.19) the power transmitted by

the wave propagates along rays. From this we can

estimate the value of the field amplitude. Let us take

an area dS 1 on the wave-front φ 1. The wave transmits

power P 1 through this area, Fig. 4.3. The rays

starting at the perimeter of area dS 1 circumscribe a

tube of constant power and mark on wave-front φ 2 area dS 2 through which the same power P 1 passes.

This power is

ray

d sdr

r

r+dr

φ

φ +dφ

0

s

l0

Fig. 4.2

φ 1

dS 1

φ 2

dS 2

P 1

P 1

Fig. 4.3

a tube of

constant

power



book - 5

60

2

2

21

2

11 d2

1d

2

1S S P mm EE

µ

ε

µ

ε == .

From this equation we get

2

112

d

d

S

S

E E mm = . (4.21)

The amplitude of the electric field is higher where the area is smaller, which corresponds to

the higher density of the rays.

This technique using rays to describe waves is known as the geometric optic. The

waves are described by a set of rays which propagate independently. Their trajectories are

described by (4.20). If they fall on any boundary, they behave according to Snell’s laws.

________________________________

Example 4.1: Determine the trajectory of the ray which propagates at the plane y = 0 in a

dielectric layer, see Fig. 4.4. The boundary conditions are x( z =0) = 0, ( ) 00dd ϑ ϑ ≈= tg z x . The

refractive index depends on the x component

( ) ( )20 1 xk n xn −= , where constant k << 1 .

This is known as the parabolic distribution of

the refractive index. Assume that the ray

changes its direction very little in the relation

to the z axis, so we can put s = z . Such a ray

is known as a par-axial ray.The trajectory of the ray is described

by (4.20). This equation can be generally

solved only numerically. Assuming a par-

axial ray we have s ≈ , and (4.20) now reads

( )( )

0d

d

d

d

d

dx

r

x

xn

z xn

z =

, r = xx0 + z z0 .

( )( )

000 d

d

d

d

d

dxzx

x

xn

z

x xn

z =

+

as 0d

d=

nwe get

( )( ) x

xn

z

x xn

z d

d

d

d

d

d=

, => ( ) xnk

x xn 02

2

2d

d−= =>

( )20

0

2

2

1

2

d

d

kxn

xnk

z

x

−−=

Now we neglect kx2 comparing to 1 and we get

ϑ 0

Fig. 4.4



book - 5

61

xk z

x2

d

d2

2

−= .

The solution of this equation is

( ) ( )qz Bqz A x cossin += , k q 2= .

Applying the boundary conditions we get

( ) 00 = x => B = 0

( )qz q A z

xcos

d

d= , 0

0d

dϑ =

= z z

x=>

k A

2

0ϑ = ,

( ) z k

k

x 2sin

2

0ϑ = .

A part of the ray trajectory is shown in

Fig. 4.5. The parameters of the

trajectory are

k z p

2

π = ,

20

p z z = ,

k x

2

0max

ϑ =

The ray is coupled to the area around the z axis, and it bends and returns back. For x < 0 the

ray has a symmetric shape. __________________________________

5. GUIDED WAVES

Up to now we have studied electromagnetic waves propagating in a free space filled

by a homogeneous or non-homogeneous material. This is very important for the theory of

wave propagation in the atmosphere, and is used in the design of communication systems, particularly the channels represented by transmitting and receiving antennas, and the space

between them. On the other hand, there are waves the existence of which is based on the presence of a boundary between different materials. These boundaries can be of various

shapes, depending on the required behaviour and the application, and they form a

transmission line. We have studied the surface wave excited due to the incidence of a planeelectromagnetic wave on a plane boundary of two dielectric materials under an angle greater

than the critical angle. This surface wave is one from the examples of guided waves.A wide variety of transmission lines are used. Their geometry depends on many

factors. The most important are the technology of the system or the circuit in which the line is

applied, the frequency band and the transmitted power. Transmission lines can be generallycategorized into three groups according to the form of the transmitted wave. In the first group,

there are transmission lines which are able to transmit a TEM wave. Later we will see thatsuch a transmission line must be able to transmit DC current, so it must consist of at least two

ϑ 0

Fig. 4.5

p0

max



book - 6

62

separate conductors. Examples of these transmission lines are shown in Fig. 5.1a. The two

transmission lines shown in Fig. 5.1b consist of two conductors, and are thus able to conduct

DC current. These are planar lines with conducting strips located on a dielectric substrate. The

electric field and the magnetic field have to fulfill boundary conditions on the surfaces of this

substrate. Due to this, the longitudinal components of the fields are always present. The

propagating wave is not a TEM wave. At low frequencies this wave can be treated as a TEM

wave. Consequently it is known as a quasi TEM wave. The lines shown in Fig. 5.1c do not

guide a TEM wave. The first two lines in Fig. 5.1c are known as waveguides as they guide a

wave along their hollow center.

To simplify the analysis of transmission lines we will assume lines without losses, i.e.,

dielectric materials with zero conductivity, and metals with infinite conductivity. The lines

two-conductor

transmission lineco-axial

transmission line parallel-plate

waveguide

strip-line

Fig. 5.1a Transmission lines with a TEM wave

waveguide with a rectangular

cross-section

waveguide with a circular

cross-sectionoptical fibre

Fig. 5.1c Transmission lines without a TEM wave

dielectric

microstrip line coplanar waveguide

Fig. 5.1b Transmission lines with a quasi-TEM wave



book - 6

63

will be assumed to be infinitely long and longitudinally homogeneous. The particular

materials will be homogeneous. The lines will be directed in the positive z-axis direction,

which is thus equal to the direction of the wave propagation.

The field distribution of a wave propagating along the transmission line in the z-axis

direction will be calculated by solving the homogeneous wave equation (2.1). Therefore we

will get only the waves, that can propagate along the line. These waves are known as the

eigenmodes, and represent particular solutions of the wave equation. Let us apply the method

of separation of variables for solving the wave equation. The longitudinal component of

electric field E z is assumed in the form

( ) ( ) ( ) z P y x E z y x E z z ,,, 0= . (5.1)

This form of the field distribution is inserted into the wave equation (2.1) and is rewritten to

0d

d1 2

2

2

0

0=++

∆k

z

P

P E

E

z

z T (5.2)

where T ∆ represents Laplace’s operator calculated using derivatives over the transversal

coordinates x and y. The first and second terms on the left hand side of (5.2) must to be equal

to constants k p and k z , which represent the transversal propagation constant and the

longitudinal propagation constant. In this way we decompose equation (5.2) into two

equations for E 0 z and P

002

0 =+∆ z p z T E k E , (5.3)

0'' 2=+ P k P z . (5.4)

The propagation constants k p and k z are coupled by

222 z p k k k += . (5.5)

Equation (5.4) has the solution describing the phase variation of the wave propagating along

the line in the form

( ) z jk z e z P −= . (5.6)

We omitted here the wave propagating in the negative z direction. All components of theelectric and magnetic fields have the same character as E z has. Consequently the derivatives

of these components over z can be expressed simply. So for the i-th component of the electric

field we have

i z i E jk

E −=

d

d. (5.7)

The character of the field described by (5.1) enables us to divide the modes

propagating along the transmission line into two groups, and the modes in each group can be

treated separately. These modes are the transversal electric (TE) modes and the transversalmagnetic (TM) modes. The TE modes have E z =0, and their field contains components: E x,



book - 6

64

E y, H x, H y, H z , while in complement the TM modes have H z =0, and their field contains

components: E x, E y, E z , H x, H y. The field distribution and the propagation constants of these

modes are determined by solving the wave equation (5.3) for the longitudinal components of

the electric and magnetic fields, and the transversal components are derived from the

longitudinal components.

We start from Maxwell’s first and second equations (1.12) and (1.13), putting σ =0 and

J s=0. These two vector equations can be expressed in the scalar form

x y z z E j H jk

y

H ε ω =+

∂

∂, (5.8)

y z

x z E j H

H jk ε ω =∂

∂−− , (5.9)

z x y

E j y

H

x

H ε ω =

∂

∂+

∂

∂, (5.10)

x y z z H j E jk

y

E µ ω −=+

∂

∂, (5.11)

y z

x z H j x

E E jk µ ω −=

∂

∂−− , (5.12)

z x y

H j y

E

x

E µ ω −=

∂

∂+

∂

∂. (5.13)

From (5.12) we express H y and insert it into (5.8). Now (5.8) contains only E x, E z , and H z and

in this way we get the dependence E x on the longitudinal components of the electric and

magnetic fields. Similarly we get the other transversal components. So we have

∂

∂+

∂

∂−=

x

E k

y

H

k

j E z z z

p

x µ ω

µ ω 2

, (5.14)

∂

∂+

∂

∂−−=

y

E k

x

H

k

j E z z z

p

y µ ω

µ ω 2

, (5.15)

∂

∂−

∂

∂=

x

H k

y

E

k

j H z z z

p

xε ω

ε ω 2

, (5.16)

∂

∂+

∂

∂−=

y

H k

x

E

k

j H z z z

p

yε ω

ε ω 2

. (5.17)

Now putting E z =0 we will get the transversal components E x, E y, H x, H y as functions of

H z and we have TE modes. Conversely, putting H z =0 we will get the transversal components E x, E y, H x, H y as functions of E z and we have TM modes.



book - 6

65

Let us now turn our attention to (5.5). Using ε µ ω =k we get the longitudinal

propagation constant

22 p z k k −= ε µ ω . (5.18)

Three cases can be now distinguished. At high frequencies we have

22 pk >ε µ ω

and k z is a real number, so the corresponding mode represents a wave propagating along the

transmission line. At low frequencies we have

22 pk <ε µ ω

and k z is an imaginary number, so the corresponding mode represents an evanescent wave

which does not propagate along the transmission line and its amplitude exponentiallydecreases. The boundary point determines the so called cut-off frequency of the mode

ε µ π 2

p

c

k f = . (5.19)

This means that the transmission line behaves

as a high-pass filter, as it transmits the mode

starting from the cut-off frequency, Fig. 5.2.

Note that TEM modes have, as will be shown,

k p=0 and, consequently, f c=0, and they can propagate from zero frequency.

Each transmission line has its dominant

mode. This is the mode with the lowest

possible cut-off frequency, i.e., with the lowest k p. It is desired to operate the transmission line

in the frequency band of the single mode operation. This is the frequency band at which

only the dominant mode propagates. This band is limited from above by the cut-off frequency

of the nearest higher mode. When the two modes can propagate simultaneously, the

transmitted signal is coupled to these two modes. As they have different propagation

constants (5.18), they have different phase velocities and the field of these modes arrives at

the output port with a different delay. This results in distortion of the transmitted signal.

The particular transmission lines and the modes propagating along them will bestudied in the following sections.

6. TEM WAVES ON A TRANSMISSION LINE

6.1 Parameters of a TEM waveLet us study a transmission line with a general cross-section, homogeneous along its

infinite length, located in the unbounded homogeneous space parallel to the z -axis. Let theline have perfect conductors with infinite conductivity. We will study the propagation of a

0 c

the mode does not

propagate

the mode propagates

Fig. 5.2



book - 6

66

TEM (transversal electromagnetic) wave along this transmission line. The TEM wave has

field components only in the transversal plane and E z =0, H z =0. On an infinitely long

homogeneous line we assume dependence on the longitudinal coordinate in the form z jk z −e .

Thus we have

( ) ( ) z jk T

z y x z y x −= e,,, EE , ( ) ( ) z jk

T z y x z y x −

= e,,, HH . (6.1)

The transversal field components have the form 0000 , yxHyxE y xT y xT H H E E +=+= .

Inserting these forms into wave equations (1.41) and (1.43) in the space without sources we

get

02=+∆ T pT T k EE

02=+∆ T pT T k HH

where T ∆ represents Laplace’s operator applied according to transversal coordinates x and y,

and 22222 z z p k k k k −=−= µε ω is a transversal propagation constant. As both the electric and

magnetic fields have zero longitudinal components, these fields have lines of E and H lying

only in the transversal planes and therefore they must have a constant phase in these planes.

This determines k p = 0 ant thus k z = k . As a result we have

0=∆ T T H , (6.2)

0=∆ T T E , (6.3)

Equations (6.2) and (6.3) are Laplace equations. The electric and magnetic field transversal

components are solutions of the Laplace equation, and this electromagnetic field therefore hasthe character of a stationary field, of course except for its wave character in the longitudinal

direction. As the field distribution in the transversal plane is the same as the distribution of thestationary field, the line must be able to conduct a DC current. Such a transmission line must

consist of at least two conductors to transmit the DC current. Generally, a TEM wave can

propagate only along a transmission line which consists of at least two conductors.As the field has the character of a stationary field we can unambiguously define the

voltage and the current at any point on the transmission line. The voltage is defined by the

integral along an arbitrary path c1 lying in the transversal plane

( ).1

,

const z c

T d t z u

=

⋅= ∫ sE . (6.4)

This voltage depends only on the z -coordinate and time t . The electric current is determined

by the integral along the closed path c2 which surrounds one of the conductors (Ampere’slaw), and lies in the transversal plane

( ).2

,

const z c

T d t z i

=

∫ ⋅= lH . (6.5)



book - 6

67

This current depends only on the z -coordinate and time t . Now we can express the electric field and the magnetic field using the voltage and the

current

( ) ( ) ( )t z u y xt z y x T T ,,,,, eE = , (6.6)

( ) ( ) ( )t z i y xt z y x T T ,,,,, hH = . (6.7)

Vector function eT ( x, y) represents the distribution of the electric field when DC voltage 1 V isconnected to the line terminals. Inserting (6.6) into (6.4) we get

( ) 1,

1

=⋅∫ se d y xc

T . (6.8)

Vector function hT ( x, y) represents the distribution of the magnetic field when DC current 1 A passes the transmission line. This function is normalized to one, as follows from the insertion

of (6.7) to (6.5),

( ) 1,

2

=⋅∫c

T d y x lh . (6.9)

The problem of calculating the field distribution is thus divided into two parts. Thecalculation of functions eT ( x, y) and hT ( x, y) can be done according to techniques known from

the electrostatic field and the stationary magnetic field, solving (6.2) and (6.3). The reader istherefore able to solve this part of the problem. The calculation of voltage u( z ,t ) and currenti( z ,t ) is known from circuit theory. This is described by the well known telegraph equations.

Assuming a steady harmonic state, these equations are

( ) I L j RU ω +−=∂∂ , (6.10)

( )U C jG I

ω +−=∂

∂. (6.11)

R, L are line series resistance and inductivity per unit length, G and C are line parallel

conductance and capacity per unit length. These primary parameters of a transmission line

can be calculated by methods known from electromagnetic field theory. We can eliminateeither the voltage or the current from the couple of these equations to get one equation of the

second order

( )( ) 02

2

=++−∂

∂U C jG L j R

z

U ω ω , (6.12)

( )( ) 02

2

=++−∂

∂ I C jG L j R

z

I ω ω . (6.13)

These equations are equivalent to the wave equation. The propagation of the TEM wave along

the line is characterized by secondary parameters. These are propagation constant γ ,

characteristic impedance Z C , phase velocity v, and wavelength λ .



book - 6

68

We start by writing the solution of the telegraph equation for voltage U (6.12)

02

2

2

=−∂

∂U

z

U γ , (6.14)

where the propagation constant is written from (6.12)

( )( )C jG L j R ω ω γ ++= . (6.15)

Here we take the notation of the propagation constant, which is known from circuit theory.

The lossless transmission line has R=0 and G=0 and the propagation constant will be

β ω ω γ j LC j LC ==−=2 , (6.16)

where β is the phase constant in the sense of (2.11) valid for a plane electromagnetic wave propagating in a homogeneous space filled by a lossless material. Note that in most of the

formulas valid for the TEM wave propagating along the transmission line we can substitute L

for µ and C for ε , and we get formulas valid for the plane electromagnetic wave propagating

in free space.The solution of (6.14) and the corresponding equation valid for the electric current are,

assuming a single wave propagating on an infinitely long line in the positive z direction,

z U U γ −= e0 , (6.17)

z I I γ −= e0 , (6.18)

where U 0 and I 0 are wave amplitudes. Inserting into (6.12) for the voltage and the current weget

( ) z z I L j RU γ γ ω γ −−+−=− ee 00 ,

and from this formula we get, using (6.15), the characteristic impedance of the transmissionline, which is defined as the ratio of the voltage amplitude and the current amplitude,

C jG

L j R L j R

I

U Z C

ω

ω

γ

ω

+

+=

+==

0

0 . (6.19)

A lossless line has

C

L Z C = . (6.20)

The voltage and the current are in phase on a lossless line, as the characteristic impedance is a

real number. We can again compare (6.20) with the characteristic impedance of a losslessmaterial for a plane electromagnetic wave (2.26).



book - 6

69

The phase velocity is defined and can be determined in the same way as the phasevelocity of a plane electromagnetic wave in free space (2.13). Assuming a lossless line we can

get

LC v

1==

β

ω . (6.21)

This velocity determines the velocity of the propagation of a constant phase along the line. Inthe case of a lossless line it also determines the velocity of the energy transmitted along the

line, i.e., the group velocity.

The wavelength has again the same meaning as the wavelength of a planeelectromagnetic wave propagating in an unbounded space (2.15)

β

π λ

2==

f

v. (6.22)

___________________________ Example 6.1: Calculate the secondary parameters of the co-axial transmission line shown inFig. 6.1, assuming lossless materials. The line

parameters are 2r 1=0.46 mm, 2r 2=1.5 mm,

ε r =2. Calculate the distribution of the electricfield and the magnetic field in the cross-section of this transmission line.

From the theory of electromagneticfield we know the line capacity per unit

length

1

2ln

2

r

r C πε = ,

and the line inductivity

1

20 ln2 r

r L

π

µ = .

The characteristic impedance is, according to

(6.20),

Ω13.50ln60

ln

2

ln2

1

2

1

2

0

1

20

====r

r

r

r

r

r

C

L Z

r r C

ε ε πε

π

µ

It is evident that the given parameters correspond to the standard 50 Ω co-axial cable. The

phase constant (6.19) is

r 1

r 2

r 3

τ

-τ

Fig. 6.1

ε r



book - 6

70

r k

r

r r

r LC ε µε ω

πε

π

µ ω ω β 0

1

21

20

ln

2ln

2==== ,

where k 0 is the phase constant of the wave propagating in a vacuum. The wavelength (6.22) is

r r k ε

λ

ε

π

β

π λ 0

0

22=== ,

where λ 0 is the wavelength of a plane electromagnetic wave in a vacuum.The distribution of the electric field and the magnetic field in the cross-section of this

transmission line is known from electromagnetic field theory

( )12ln2 r r r

U

r E r ==

πε

τ ,

r

I H

π α

2= .

Note that, using the formula for characteristic impedance Z C of this transmission line, ratio

α H E r gives a value equal to the characteristic impedance for a plane electromagnetic wave

propagating in the space filled by a dielectric with permittivity ε r . ________________________________

The results of the above example can be generalized. The TEM wave propagates along

a transmission line with a homogeneous dielectric material between the conductors in the

same way as a plane electromagnetic wave propagating in the unbounded space filled by the

same homogeneous material. This refers to the propagation constant, the wavelength, and the

phase velocity. The characteristic impedance is different from the characteristic impedance of

the material for a plane wave. The characteristic impedance of a transmission line is defined

as the ratio of the amplitudes of the voltage and of the current. So it must depend not only on

the material parameters but also on the line transversal dimensions. The characteristic

impedance of the material for a plane wave is defined as the ratio of the field amplitudes, so it

depends only on the material parameters.

______________________________

Example 6.2: Calculate the secondary parameters of the two-conductor transmission line

(twin-lead transmission line) from Fig. 6.2, assuming lossless materials.

In electromagnetic field theory we calculated the line capacity and inductivity per unit

length

r

aC

ln

πε = .

r

a L ln0

π

µ = .

The characteristic impedance (6.20) is

Fig. 6.2

a

r r



book - 6

71

r

a

r

a

r

a

C

L Z

r

C ln120

ln

ln0

ε πε π

µ

=== .

The phase constant and the wavelength are

r r k LC ε ε ε µ ω ω β 000 === ,r ε

λ λ 0= .

_____________________________

The results of the above example validate the remarks made to Example 6.1.

The TEM wave is not the only wave that can propagate along the above mentioned

lines. So called waveguide modes, treated in Chapter 7, can propagate along these lines at

sufficiently high frequencies above their cut-off frequency. These modes are undesired, and

the lines must be designed to prevent the propagation of waveguide modes.

6.2 Transformation of the impedance along the line

6.2.1 An infinitely long lineThe general solution of the telegraph equations (6.10) and (6.11) on an infinitely long

transmission line consists of two waves, one propagating to the right (denoted by subscript +),

and the second propagating to the left (denoted by subscript -)

z z

U U U γ γ

ee −

−

+ += , (6.23)

z

C

z

C

z z

Z

U

Z

U I I I γ γ γ γ eeee −−+−

−+ −=+= , (6.24)

assuming the current amplitudes C C Z U I Z U I −−++ −== , , as the current flowing to the

left has the opposite orientation. The impedance at any point along the line is

( ) z z

z z

C

U U

U U Z

I

U z Z

γ γ

γ γ

ee

ee

−

−

+

−−

+

−

+== , (6.25)

On an infinitely long line with the only one source connected at −∞= , we have only one

wave propagating to the right, and the above formulas read z U U γ −

+= e , (6.26)

z I I γ −+= e , (6.27)

The impedance at any point along the line is



book - 6

72

( ) C Z I

U z Z ==

+

+ . (6.28)

The impedance at any point on an infinitely long transmission line is thus equal to the

characteristic impedance. This gives us the recipe for realizing an infinitely long line. A line

of finite length is terminated by an impedance equal to Z C . This ensures no reflection at the

line termination, and the line behaves as an infinitely long line.

6.2.2 A line of finite lengthLet us take a transmission line of finite length l , terminated at the end by impedance

Z L. The voltage ( )l U U =2 and current ( )l I I =2 at the end are known. Their ratio is

2

2

I

U Z L = . (6.29)

To simplify this problem we use a new coordinate s=l-z measured from the line end, Fig. 6.3. As ds=-dz , we can

rewrite equations (6.12) and (6.13) to the form

( )( ) 02

2

2

=−∂

∂ sU

s

sU γ , (6.30)

( )( ) 02

2

2

=−∂

∂ s I

s

s I γ . (6.31)

The solution of these equations is

s s U U U γ γ −−+ += ee , (6.32)

( ) s s

C

U U Z

I γ γ −−+ −= ee

1. (6.33)

For z=l we now have s=0 and

( )−+−+ −=+= U U Z

I U U U C 1, 22 .

From these formulas we get the amplitudes of the wave traveling to the right and the wave

traveling to the left as functions of the voltage and the current at the line end

( )222

1 I Z U U C +=+ , (6.34)

( )22

2

1 I Z U U C −=− . (6.35)

0 l

s l-z 0

Fig. 6.3

U 2, I 2



book - 6

73

Inserting (6.34) and (6.35) into (6.32) and (6.33) we get the voltage and the current at any

point on the transmission line

( ) ( ) ( ) s I Z sU sU C γ γ sinhcosh 22 += , (6.36)

( ) ( ) ( ) s Z

U s I s I

C

γ γ sinhcosh 2

2

+= . (6.37)

The impedance at any point along the transmission line is

( ) ( )

( ) ( )

( )( ) s Z Z

s Z Z Z

s Z

U s I

s I Z sU

I

U Z

LC

C LC

C

C

γ

γ

γ γ

γ γ

tgh

tgh

sinhcosh

sinhcosh

22

22

+

+=

+

+== , (6.38)

Formula (6.38) is the basic formula used in the analysis and design of high frequency circuits.

It determines how the terminating impedance is transformed to any point along the line.

The propagation constant of a lossless line is determined by (6.16), which is a purelyimaginary number. The characteristic impedance is a real number (6.20). Equations (6.36) and

(6.37) can now read (13.35) and (13.36)

( ) ( ) ( ) s I jZ sU sU C β β sincos 22 += , (6.39)

( ) ( ) ( ) s Z

U j s I s I

C

β β sincos 22 += . (6.40)

The impedance at any point along the line is

( )

( ) s jZ Z

s jZ Z Z

I

U Z

LC

C LC

β

β

tg

tg

+

+== . (6.41)

The two formulas (6.38) and (6.39) confirm the fact stated at the end of the preceding

paragraph. Terminating the line by an impedance equal to characteristic impedance Z C , we get

the impedance at any point equal to the characteristic impedance. The line behaves as an

infinitely long line.

6.2.3 A line terminated by a short cut or by an open endA line terminated by a short cut or by an open end is exposed to total reflection, and

thus a reflected wave has the same amplitude as the wave incident to the termination. As a

result we get a standing wave. This effect is analogous to the perpendicular incidence of a

plane electromagnetic wave to the surface of a well conducting material or a material with

infinite permeability – a perfect magnetic material.

An open end termination represents an infinite impedance, and the current passing

through this impedance 02 = I . Formulas (6.39) and (6.40) are now, using λ π β /2= (2.15),

( )

== sU sU U

λ

π β

2coscos 22 , (6.42)



book - 6

74

( )

== s

Z

U j s

Z

U j I

C C λ

π β

2sinsin 22 . (6.43)

These voltage and current

distributions are plotted in Fig.

6.4. Formulas (6.42) and (6.43)

describe a standing wave on atransmission line and are

analogous to formulas describing

the standing wave after the

perpendicular incidence of a

plane electromagnetic wave to the

surface of a perfect magnetic

material. The active power

transmitted by this wave is zero,

which gives the wave its name.

Using a probe sliding along the

line we can detect the distributionof the voltage determined by

(6.42) with stable maxima and

minima. The distance between the

two adjacent maxima or minima

is equal to a half wavelength. In

this way we can measure the

wavelength.

The impedance at any point on

the line is

−==

λ π

s jZ

I

U Z C 2gcot .

(6.44)

A lossless line terminated by an open

end behaves as a reactance with a

value from minus to plus infinity.

This reactance X ( s) is plotted in Fig.

6.5. Fig. 6.5 shows the character of

the impedance. It changes from the

capacitive character to an impedanceof a series resonance circuit, to an

inductive character, and the

impedance of a parallel resonant

circuit. Very close to the line end,

where s<<λ , the impedance is

Cs j s LC C

L j

sC

L j Z

ω ω β

111=−=−≈ , (6.45)

( ) su

( ) si

s

s

λ /4

λ /4λ /2

λ /2

3 λ /4

3 λ /4 λ

λ

Fig. 6.4

( s)

Fig. 6.5

λ /4

λ /2

3λ /4

λ



book - 6

75

and the line behaves here as a capacitor with a capacity proportional to the line capacity per

unit length.

The short cut termination represents zero impedance, and the voltage at the line end

is equal to zero. Formulas (6.39) and (6.40) are now

( )

==

λ

π β s

I Z j s I Z jU C C 2sinsin 22 (6.46)

( )

==

λ π β

s I s I I 2coscos 22 .

(6.47)

These voltage and current

distributions are plotted in Fig.

6.6. Formulas (6.46) and (6.47)

describe a standing wave which

does not transport any active power. The reader can compare

these formulas with (3.13) and

(3.14), which describe the

distribution of a standing wave

created by the perpendicular

incidence of a plane

electromagnetic wave on the

surface of a perfect conductor.

The impedance at any

point on the line is

==

λ π

s jZ

I

U Z C 2tg .

(6.48)

This function is plotted in Fig. 6.7.

Choosing a suitable line length, we

can realize a line stub with an

arbitrary input reactance value. Fig.

6.7 shows the character of this

impedance. Very close to the end of a

line ( s<<λ ) we have the impedance

Ls j s LC C

L j

s jZ

s jZ Z C C

ω ω

λ π

λ π

==

=≈

= 22tg

,

(6.49)

and the line terminated by a short cut

behaves as an inductor with itsinductivity proportional to the line inductivity per unit length.

( s)

Fig. 6.7

λ /4

λ /2

3λ /4

λ

( ) su

( ) si

s

s

λ /4

λ /4λ /2 3 λ /4

3 λ /4 λ

λ

Fig. 6.6



book - 61

76

A line stub with a short cut is applied more frequently than a stub with an open end.

The reason is that the former radiates less energy than the latter. The radiation of energy

causes a non zero real part of the impedance, as it represents losses.

6.3 Smith chart

When designing microwave circuits the designer is interested in the values of voltagesU , currents I , impedances Z and reflection coefficient ρ . It would be a tedious work to

calculate these quantities, specially for a lossy transmission line, and transform them along a

transmission line without a computer and modern CAD tools. There are some graphical tools

that can be used to simplify this task. One of them was introduced by Smith in 1939. It is a

chart that enables us to make a simple transformation of the impedance along a transmission

line and to recalculate the impedance to a reflection coefficient, and vice versa. It is a very

useful tool for designing microwave circuits. It is even used in acoustics. As we work with

values of voltage and current, this tool is applicable in the case of transmission lines with a

TEM wave, where they are uniquely defined. The Smith chart can even be applied in the case

of lines with no TEM wave, e.g., waveguides. Here we can use an appropriate scaling of

propagating waves to some hypothetic TEM waves described by voltages and currents.Let us take a TEM transmission line, Fig. 6.8, fed from a generator and terminated by

a load, with length l

and characteristic

impedance Z C. The

position along this

line is determined by

the z coordinate

measured from the

terminals of the

generator. The

propagation constantis assumed in the

notation used in the

theory of electric circuits (6.15)

β α γ j jk +== , (6.50)

where α is the attenuation constant and β is the phase constant. At point z we have an

impedance Z = U ( z )/ I ( z ), this impedance causes a reflection with a reflection coefficient

( ) ( )( ) C

C

Z z Z Z z Z z

+−= ρ . (6.51)

The reflection coefficient is determined in the same way as the reflection coefficient for the

TEM wave incident perpendicular to the boundary between two materials (3.6). From (6.51)

we get

( )( )( ) z

z Z z Z C

ρ

ρ

−

+=

1

1. (6.52)

The standing wave ratio is defined in the same way as (3.20)

generator load

1

2

l

d

Z C

Fig. 6.8



book - 61

77

( )( )

( ) z

z z p

ρ

ρ

−

+=

1

1. (6.53)

The transformation of the impedance along the transmission line is described by

(6.38). Using (6.38) we can recalculate the impedance from point z 1 to z 2 and back

( ) ( )( )

( ) ( )( ) ( )d z Z Z

d Z z Z Z

z I

z U z Z

C

C C

γ

γ

tgh

tgh

2

2

1

11

+

+== , (6.54)

( )( )( )

( ) ( )

( ) ( )d z Z Z

d Z z Z Z

z I

z U z Z

C

C C

γ

γ

tgh

tgh

1

1

2

22

−

−== , (6.55)

where d = z 2 – z 1, Fig. 6.8. To define the reflection coefficient we have to decompose the

voltage at any point to a wave propagating to the right, i. e., in the positive z direction, and a

wave propagating to the left

( ) z U z U γ −+ = e0 , ( ) z U z U γ e0=− .

The transformation of the voltage along the line is

( ) ( ) ( )12e12 z z z U z U −−++ = γ , ( ) ( ) ( )12e12

z z z U z U −−− = γ , (6.56)

From (6.56) we have the transformation of the reflection coefficient

( )( )

( )( ) d z

z U

z U z γ ρ ρ 2

1

2

22 e==

+

−

, (6.57)

( )( )

( )( ) d z

z U

z U z γ ρ ρ 2

2

1

11 e−

+

−

== . (6.58)

To get a universal tool for analyzing of microwave circuits we use the normalized

impedances

C Z

Z z = . (6.59)

Consequently (6.51), (6.52) and (6.54) have the form

( )1

1

+

−=

z z ρ , (6.60)

ρ

ρ

−

+=

1

1 z , (6.61)



book - 61

78

( )( ) ( )

( ) ( )d z z

d z z z z

γ

γ

tgh1

tgh

2

21

+

+= . (6.62)

Now we have everything necessary for deriving the Smith chart. First we show the

way to make a graphic representation in a complex plane of the reflection coefficient and its

transformation. The reflection

coefficient is generally a complexnumber with an amplitude lower

than 1. It is plotted into a complex

plane u, jv, Fig. 6.9. The lines of

constant amplitude are circles with

their center at the origin, and the

lines of constant phase are radial

lines, Fig. 6.9. The reflection

coefficient is transformed

according to (6.58). For a lossless

line it is γ = j β . The transformation of the reflection coefficient is controlled by

( ) ( ) d j z z β ρ ρ 221 e−= . (6.63)

Now we have to distinguish between the two

directions of the shift along a line. In the case of d >

0 it is z 2 > z 1, z 1 is closer to the generator and the

shift is toward the generator. As d > 0 the phase

in (6.63) decreases and the shift toward the

generator corresponds to the rotation of the

reflection coefficient in the complex plane to the

right, Fig. 6.10. In the case of d < 0 it is z 2 < z 1, z 1 iscloser to the load and the shift is toward the load.

As d < 0 the phase in (6.63) increases and the shift

toward the load corresponds to the rotation of the

reflection coefficient in the complex plane to the

left, Fig. 6.10. According to (6.63), the measure of the shift is equal to the product β d . The

trip around the circumference of the whole complex plane represents angle 2π and, as β =2π /λ , this corresponds to

λ

π

λ

π β π

d d d 4

2222 === => 5.0=

λ

d .

The angle in the complex plane is not defined in degrees, but in the relative distance

measured in the number of wavelengths in the range from 0 to 0.5, see Fig. 6.10. We have

to distinguish between the direction toward the load and the direction toward the generator. In

the case of a lossless line the transformation is performed along a circle, as the modulus of the

reflection coefficient stays unchanged. In the case of a lossy line the modulus of the reflection

coefficient decreases exponentially, and the transformation is therefore performed along a

spiral.

The complex normalized impedance (6.59) and its complex conjugate value are

u

v ρ | ρ |

ϕ

u

v

0

| ρ |=10.7

0.2ϕ =0

90°

180°

-90°

-180°

Fi . 6.9

u

v

0

βλ =const

0.250

0.5

Fi . 6.10

0.125

0.375

shift toward

the generator

shift toward

the load



book - 61

79

ρ

ρ

−

+=+=

1

1 jxr z , (6.64)

*

**

1

1

ρ

ρ

−

+=−= jxr z . (6.65)

By adding and subtracting these two equations we get the real r and imaginary x parts of the

normalized impedance

( )

−

++

−

+=+=

*

**

1

1

1

1

2

1

2

1

ρ

ρ

ρ

ρ z z r , (6.66)

( )

−

+−

−

+=−=

*

**

1

1

1

1

2

1

2

1

ρ

ρ

ρ

ρ z z jx . (6.67)

From (6.66) and (6.67) we can derive equations which determine images of the lines of the

constant values of the real part of normalized impedance r and of the imaginary part of

normalized impedance x in the complex plane of the reflection coefficient. The goal is to

convert (6.66) and (6.67) into the equation of a general circle in the complex plane ρ = u + jv

( ) ( ) ( ) 0222** =−+++−−− Rnm jnm jnm ρ ρ ρ ρ , (6.68)

where R is the radius of a circle, m and jn are the coordinates of

the center of the circle, see Fig. 6.11. After some manipulations,

(6.66) can be rewritten into

01

1

11

** =+

−+

+−

+−

r

r

r

r

r

r ρ ρ ρ ρ . (6.69)

Comparing (6.69) with (6.68) we get the coordinates of the

center of the circle (6.69) and its radius

1+=

r

r m , n = 0 ,

1

1

+=

r R . (6.70)

Now we are able totransform the lines

of the constant

values of the real

part of the

normalized

impedance from the

complex plane z = r

+ jx, where these

lines are lines

parallel with the

imaginary axis, into

u

v

m

n

R

Fig. 6.11

r

x

r =0 r =1 r =2

u

v r =0

r =1

r =2

r =∞

Fig. 6.12





book - 61

81

the unit radius in the plane ρ = u + jv. The final version of the Smith chart is shown in Fig.

6.14.

Each point in the Smith chart corresponds to an impedance and at the same time to the

reflection coefficient (6.60). The scale of the modulus of the reflection coefficient is shown

below the chart. The corresponding angle must be measured using a protractor, or the scale of

the angles is shown in some versions of the chart along its perimeter. The scale of the

standing wave ratio (6.53), shown in some versions of the chart, is a nonlinear rating from 1

to ∞ . The transformation of the reflection coefficient is performed as explained in the text

referring to Fig. 6.10. The scale of the relative distance is shown along the perimeter of the

chart, its range being between 0 and 0.5. There are the two scales of the length. The first

corresponds to the orientation toward the generator, the second toward the load. The scale of

the real part of the normalized impedance is on the horizontal axis of the chart. The scale of

the imaginary part of the normalized impedance is shown along the perimeter of the chart.

Fig. 6.14

In the following examples we show some basic operations with the Smith chart. Toshow the advantages of applying the Smith chart we choose cases where analytical calculation

= 0.4+ 0.7

=0.588



book - 61

82

can be used.

_____________________________________

Example 6.3: Show the normalized impedance z = 0.4+ j07 in the Smith chart.

This impedance is shown in Fig. 6.14 together with the corresponding value of the

reflection coefficient

( ) °=+−=+−= 410588.057.0143.01

1 je j

z z ρ .

_____________________________________

Example 6.4: The lossless transmission line is terminated by a short. Calculate the

normalized impedance at distance l = 3λ /8 from the short.

As z ( z 2) = 0, we can apply (6.48) and we get

( ) ( ) j j jl j z z −=

=

== π λ

λ

π β

4

3tg

8

32tgtg1 .

In the Smith chart we use the following technique, Fig. 6.15. The short is a zero

impedance. From this point we move in the direction toward the generator along the circle of

a constant reflection coefficient by the normalized length l /λ = 3/8 = 0.375. At this distance

we read the normalized impedance – j.

_____________________________________

Fig. 6.15



book - 61

83

Example 6.5: The lossless transmission line is terminated by an open end termination.

Calculate the normalized impedance at distance l = λ /4 from the short.

The normalized impedance at any point along the open end terminated transmission

line is given by (6.44), so we get

( ) ( ) 0

2

cotgcotg1 =

−=−=

π β jl j z z .

The open end transforms as a short. In the Smith chart we perform the transformation

according to Fig. 6.16. The open end termination represents the infinite normalized

impedance. From this point we move in the direction toward the generator along the circle of

a constant reflection coefficient by the normalized length l /λ = 1/8 = 0.25. At this distance we

read the normalized impedance 0.

_____________________________________

Example 6.6: The lossless transmission line is terminated by the normalized impedance z L =

0.8+ j. Calculate the normalized impedance at distance l = 0.2λ from the end of the line.

Using (6.62) applied to the lossless line and the product β l we get

π λ λ

π β 4.02.0

2==l

( )( )

( )

( )

( ) ( )0061.1807.0

4.0tg8.01

4.0tg8.0

tg1

tg

1 j

j j

j j

l z j

l j z z z

L

L

−=++

++

=+

+

= π

π

β

β

Fig. 6.16



book - 611

84

In the Smith chart, Fig. 6.17, we first determine the point corresponding to the loading

impedance z L = (0.8+ j). We draw a radial line of the constant value of β l through this point

and read the value of the relative distance 0.15 on the perimeter of the chart. From this point

we move toward the generator by the relative distance 0.2, i.e., to the point 0.15+0.2 = 0.35.

Here we read on the circle representing the same value of the modulus of the reflection

coefficient as at point z L the normalized impedance z ( z 1) = 0.8 – j, which is approximately the

same value as calculated.

___________________________________

Example 6.7: The lossy transmission line with attenuation α ’ = 0.5 dB is terminated bynormalized impedance z L = 0.8+ j2.2. Calculate the normalized impedance and the standing

wave ratio at distance l = 20 m from the end of the line. The wavelength is 0.44 m.

From the attenuation in dB we determine attenuation constant α and from thewavelength phase constant β

( ) ( )elog20elog20' α α α −== − =>( )

05756.0elog20

'=−=α m-1, 27.14

2==

λ

π β m-1

Using (6.62) we get

( )( )( )

156.002.1tgh1

tgh1 j

l z

l z z z

L

L +=+

+=

γ

γ

Fig. 6.17

0.8+ j

0.8- j

l /λ =0.2



book - 611

85

In the Smith chart, Fig. 6.18, we first determine the point corresponding to the loading

impedance z L = (0.8+ j2.2). We draw the radial line of the constant value of β l through this

point and read the value of the relative distance 0.188 on the perimeter of the chart. From this

point we move toward the generator by the relative distance l /λ = 20/λ = 45.454 so we get to

the point 0.188+45.454 = 45.642. This means that we go round the chart 91 times and stop at

the point 0.142. This is the position of the impedance we are looking for. Now we have to

determine the value of the reflection coefficient. The reflection coefficient transformsaccording to (6.58) so we get

1.022

21 ρ ρ ρ α == − l e

In the Smith chart in Fig. 6.18 we read the normalized impedance z ( z 1) = 1.02+ j0.15, and

using (6.60) and (6.53) we get the standing wave ratio p( z 1) = 1.16.

_______________________________________

The normalized impedance could be calculated simply by using transformation

formulas (644), (6.48) and (6.62) in examples 6.4 to 6.7. The Smith chart however simplifies

the work. In example 6.8 we perform impedance matching, i.e., we make corrections in a

circuit to reduce the reflection at the given frequency to zero. This cannot be done simply, as

it includes the need to solve a transcendent equation. The Smith chart is a very efficient tool

in such a case.

___________________________________

Fig. 6.18

zL=0.8+ j2.2

0.188

0.142

| ρ 2|| ρ 1|

zL=1.02+ j0.15



book - 611

86

Example 6.8: The lossless transmission line with

characteristic impedance Z C = 300 Ω is terminated by

impedance Z L = (420- j180) Ω. Match this impedance by

the stub of the same line terminated by a short end

connected in series to the line so as to get a zero

reflection coefficient, Fig. 6.19. The wavelength is 5 m.

First we must determine distance l 1 at which weconnect the stub. The loading impedance normalized to

the characteristic impedance z L = Z L/ Z C = 1.4- j0.6 must

be transformed within distance l 1 to impedance z 1 =

1+ jx. Here we connect the stub with input impedance z 2

= 0- jx. The resulting impedance at this point is then z = z 1+ z 2 = 1, and therefore the reflection

coefficient is zero.

The solution of this problem in the Smith chart is simple, see Fig. 6.20. Impedance z L

is transformed along the circle of the constant value of the reflection coefficient to the point at

which the impedance lies on the circle where the real part of the normalized impedance is 1.

This gives z 1 = 1- j0.6, Fig. 6.20. Corresponding length l 1/λ = 0.043. Using the given value of

λ we have l 1 = 0.215 m.

Let us now form the stub. At its input it must have the normalized impedance equal to

0+ j0.6. This stub is terminated by a short that is zero impedance. The length of this stub l 2 is

determined by transforming zero impedance to the point corresponding to the impedance

0+ j0.6. From the Smith chart in Fig. 6.20 it follows that the necessary length is l 2/λ = 0.086,

Z C

Z C

l 1

l 2

Z L

Fig. 6.19

Fig. 6.20

z L

l 1/λ =0.043

z 1

0+ j0.6

l 2/λ =0.086



book - 611

87

which gives l 2 = 0.43 m. The second solution can be obtained by transforming z L to the point

with impedance z 1 = 1+ j0.6. Then the stub must have input impedance 0- j0.6. This solution is

not optimal as the corresponding lengths l 1 and l 2 are longer than those obtained above.

_______________________________________

6.4 Problems

6.1 A longitudinally homogeneous transmission line transmits a TEM wave with frequency f =1 kHz. The amplitude of the voltage on the input port is U m=60 V. The end terminals of the

line are connected to an impedance equal to the line characteristic impedance. The line

parameters are R=4 Ω/km, G=0.5 µS/km, L=2 mH/km, C =6000 pF/km. Calculate the phasor

of the voltage on the line end.

U 2=-24.75+ j33.93

6.2 A transmission line is 50 km long, and its parameters are R=6.46 Ω/km, G=1 µS/km, L=4

mH/km, C =12.2 nF/km. The line is loaded by a resistor R L=1000 Ω. The voltage on the load

is U 2=10 V. The frequency is 0.8 kHz. Calculate the voltage and the current on the input of the line.

V892.85921.1e033.9 748.11 jU j +−==

mA20569e77.20 848.11 j I j +−==

6.3 Calculate the input impedance of a lossless transmission line of the length l =0.4 m. The

characteristic impedance is 75 Ω, and the frequency is 0.6 GHz. The line has a shortcut at the

end.

Z 1=230.8 Ω, L1=0.12 µH

6.4 Calculate the input impedance of a lossless transmission line of length l =0.4 m. Thecharacteristic impedance is 75 Ω, and the frequency is 0.6 GHz. The line has an open end.

Z 1=-24.4 Ω, C 1=21.76 pF

6.5 A lossless very short transmission line (l =125 mm) has the input impedance at frequency

f =60 MHz Z 1S = j8850 Ω when terminated by a short circuit, and Z 1O=- j35.5 Ω when terminated

by an open circuit. Calculate the inductivity per unit length and the capacity per unit length.

L=75 µH/m, C =0.56 pF/m

6.6 Match the transmission line with characteristic impedance

50 Ω terminated by a series combination of the resistor withresistivity R = 100 Ω and a capacitor with capacity C = 1.5 pF,

Fig. 6.21, at frequency 5 GHz, which corresponds to the

wavelength on the line λ = 42.5 mm.

The short circuited stub of

the same transmission line of

length l 2 = 4.6 mm must be

connected in series at distance l 1 = 3 mm from the load,

Fig.6.19.

R

C

Z C λ

Fig. 6.21



book - 7

88

7. WAVEGUIDES WITH METALLIC WALLS

In this section we will study transmission lines with metallic walls. Such transmission

lines are known as waveguides. The field distribution is determined by solving the wave

equation for the longitudinal components of the electric and magnetic fields, as described in

section 5. Our task is to solve equation (5.3) with the proper boundary conditions. The

tangential component of the electric field has to be equal to zero on the metallic walls. Thistechnique will be directly applied to a parallel plate waveguide and to a waveguide with the

rectangular cross-section. The field distribution is described in a rectangular coordinate

system in the case of these two lines. In the case of a waveguide with a circular cross-section

we have to use a circular coordinate system. The parallel plate waveguide consists of the two

separate conducting plates, so it can conduct DC current and it can guide the TEM mode.

Together with this TEM mode we will describe the modes known as the waveguide modes.

These modes have nonzero cut-off frequency. Waveguides with rectangular and circular

cross-sections consist of only one conductor, so they are not able to conduct the TEM mode.

7.1 Parallel plate waveguideThe cross-section of the parallel plate

waveguide is shown in Fig. 7.1. It is created by

the two parallel infinitely wide conducting plates

which are located at a distance a. The space

between these plates is filled by an ideal

dielectric material.

We will describe separately the field of

the TE and TM modes. The calculation

procedure is simplified by the fact that the field

does not depend on the y coordinate. So we put

0=∂∂ y in (5.14) to (5.17).

The TE modes thus have E y, H x and H z

components. Setting E z =0 in (5.14) to (5.17) we get

x

H

k

j E z

p

y∂

∂=

2

µ ω , (7.1)

x

H

k

k j H z

p

z x

∂

∂−=

2. (7.2)

Function H 0 z representing the transversal field distribution analogous to (5.1) depends only on

the x coordinate, so we can put

x p k k = , (7.3)

and the H 0 z component is determined by solving the wave equation

0d

d0

20 =+ z x z H k

H . (7.4)

ε , µ

a

∞=σ

∞=σ

Fig. 7.1



book - 7

89

The solution is in the form

( ) ( ) xk B xk A H x x z cossin0 += .

The boundary conditions cannot be fulfilled by this component. We have to use E 0 y, which istangent to both conducting plates. At x=0 we have E 0 y(0)=0, Fig. 7.1, and, consequently, A=0.

The second boundary is at x=a, where E 0 y(a)=0 and sin(k xa)=0 and we get the propagationconstant in the x direction k x

a

mk xm

π = . (7.5)

Consequently, the solution of the wave equation (7.4) is

= x

a

m B H z

π cos0 . (7.6)

(7.1) and (7.2) now have the form

−= x

a

m B

k

j E

xm

y

π µ ω sin0 , (7.7)

= x

a

m B

k

k j H

xm

zm x

π sin0 . (7.8)

To remove k xm from the denominators of (7.7) and (7.8) we introduce a new constant

xk j B µ ω −=' and we get the field distribution of the TE modes

z jk y

zm xa

m B E

−

= esin'

π , (7.9)

z jk zm x

zm xa

m B

k H

−

= esin'

π

µ ω , (7.10)

z jk xm z

zm xa

m B

k j H

−

= ecos'

π

µ ω . (7.11)

The TM modes have nonzero field components E x, E z and H y. The transversal

components are from (5.14) to (5.17)

x

E

k

k j E z

p

zm x

∂

∂−=

2, (7.12)

x

E

k

j H z

p

y∂

∂−=

2

ε ω . (7.13)



book - 7

90

The equation for determining the transversal distribution of the longitudinal component of the

electric field is

0d

d0

20 =+ z x z E k

x

E . (7.14)

The boundary conditions are met directly by E z , which is tangent to both plates. Applying asimilar procedure as for the TE modes we get the field distribution

z jk y

zm xa

mC H

−

= ecos'

π , (7.15)

z jk zm x

zm xa

mC

k E

−

= ecos'

π

ε ω , (7.16)

z jk xm

z

zm

xa

m

C

jk

E

−

= esin'

π

ε ω . (7.17)

The propagation constant in the x-direction has the form (7.5) as in the case of the TE modes,

and (7.3) is again valid.

Modal number m determines the

form of the field distribution in the

waveguide cross section. This is shown in

Fig. 7.2. m determines how many half-

periods of the sinus function across the

space between the plates the distribution

has. Fig. 7.2 shows the distribution of E z for the TM modes, or E y for the TE modes.

The longitudinal propagation

constant k z is given by (5.18). Using (7.3)

and (7.5) we have for the m-th mode

22

−=

a

mk zm

π µε ω . (7.18)

Now we have from (5.19) the cut-off frequency and the cut-off wavelength of both the TE and

TM modes with modal number m in the parallel plate waveguide

µε a

m f cm

2= , (7.19)

m

acm

2=λ . (7.20)

The parallel plate waveguide behaves like the high-pass filter, as the modes can propagate

starting from their cut-off frequency.

From the longitudinal propagation constant we can determine the parametersdescribing the propagation of the modes along the parallel plate waveguide. These are the

a

E z , E y

m=0 m=1 m=2 m=3

Fig. 7.2



book - 7

91

wavelength, the phase velocity, and the group velocity. The phase velocity of the m-th mode

is

22

222

11

1

1

2

−

=

−

=

=

−

=

−

==

cm

cm

zm

pm

v

f

f

v

a

mmk v

λ

λ

µε ω

π µε

π µε ω

ω ω

. (7.21)

Similarly we can get formulas for the group velocity, the wavelength and the longitudinal

propagation constant

22

11

−

=

−

==

cm

cm

pm

gm

f

f f

v

λ

λ

λ λ λ , (7.22)

22

11d

d

−=

−==

cm

cm

zm

gm v f

f v

k v

λ

λ ω , (7.23)

22

11

−=

−=

cm

cm zm k

f

f k k

λ

λ . (7.24)

The plots of these functions

are shown in Fig. 7.3. It is

seen from Fig. 7.3 that as

the frequency approaches

from above the cut-off

frequency, the

corresponding mode stops

propagating as its

propagation constant and

group velocity tend to zero.

At the same time its phasevelocity and wavelength

tend to infinity. Below the

cut-off frequency these

quantities are imaginary

numbers and the mode is

the evanescent mode, the amplitude of which decreases exponentially along the line, and this

mode does not propagate.

The field distributions of the TE and TM modes are different, so their wave

impedances are different. They are defined as the ratio of the transversal components of the

electric and magnetic fields. Using (7.9) and (7.10) we get for the TE modes

1

v pm/v

λ gm/λ

v gm/v

k zm/k

m=1

m=1

m=2

m=2

c1 c2 c3

Fig. 7.3



book - 7

92

2

0TE

1

−

==−=

f

f

Z

k H

E Z

cm zm x

y

m

ωµ . (7.25)

For the TM modes we get from (7.15) and (7.16)

2

0TM 1

−===

f

f Z

k

H

E Z cm zm

y

xm

ωµ . (7.26)

Z 0 is the wave impedance of the free space.

Let us now look for the dominant mode of the parallel plate waveguide. This is the

mode with the lowest possible modal number m. It follows from the distribution of the

electromagnetic field of the TE modes, (7.9), (7.10), (7.11) that these modes cannot have m =

0, as this case gives a zero field. Consequently, the lowest TE mode is the mode TE 1. For the

TM modes we can allow m=0, as from (7.15-17) we get

jkz y C H −= e' , (7.27)

jkz x C

k E −= e'

ε ω , (7.28)

0= z E . (7.29)

This TM0 mode has only the transversal components of the electric and magnetic fields.

Therefore it is the TEM wave. Its cut-off frequency is zero and this mode can propagate fromzero frequency and k z 0 = k . This confirms the fact that the parallel plate waveguide consists of

two separate conductors and is therefore able to conduct the DC current. So the dominant

mode of the parallel plate waveguide is the TEM mode, which can propagate from zero

frequency.

The field distribution of the two lowest TE modes on the parallel plate waveguide is

shown in Fig. 7.4. Fig. 7.5 shows the field distribution of the three lowest TM modes.

H

E

TE1 TE2Fig. 7.4



book - 7

93

The explanation of wave propagation based on geometrical optics is shown in Fig. 7.6.

The wave is coupled between the plates under angle Θ and bounces up and down due to

reflections on the conducting

plates. The wavelength in the

direction along the line is the

projection of the free space

wavelength, so it must be longer.

We have

Θ=

cos

λ λ g ,

consequently

2

1cos

−=Θ

f

f cm . (7.30)

This angle of establishment determines the conditions of the wave propagation. It corresponds

to the relative frequency distance of the wave from the cut-off frequency. The lower this angleis, the further the wave is from the cut-off. Θ = 90°corresponds to the cut-off frequency and it

is obvious that the wave does not propagate.

__________________________

Example 7.1: Determine the modes which can propagate in the parallel plate waveguide at

the frequency 10 GHz. The distance between the plates is 40 mm and the space is filled by air.

The mode can propagate if the wavelength at 10 GHz is lower than the cut-off

wavelength of this mode. The wavelength at 10 GHz is

30== f

c

λ mm

E

H

TM0 = TEM

E

H

Fig. 7.5

TM1 TM2

Θ

λ g

λ

Fig. 7.6



book - 7

94

The cut-off wavelength of the m-th mode is determined by (7.20). So we have

m=0 λ c0 = ∞ the mode can propagate

m=1 λ c1 = 80 mm the mode can propagate

m=2 λ c2 = 40 mm the mode can propagate

m=3 λ c1 = 26.7 mm the mode cannot propagate

The parallel plate waveguide can transmit the TM0, TM1, TE1, TM2 and TE2 modes at thefrequency 10 GHz.

________________________________

Example 7.2: Determine the distance between the two conducting planes to get the parallel

plate waveguide along which the only TE1 mode propagates at 10 GHz and the mode TE2 is

attenuated by –120 dB/m. Do not consider the TM modes.

The attenuation of the TE2 mode is

dB E

e E d jk z

120log20

0

02

−=

−

, d = 1 m

6102 −−=d jk z e , - jk z2d = -13.8 , k z2 = - j13.8

As k z2 is an imaginary number the TE2 mode does not propagate and is an evanescent mode.

From the determined value of k z2 we will calculate the transversal propagation constant k y2

( ) mma jk k aa

mk z y 292108.13

2 2222

22 =⇒=−−=−=== µε ω

π π

The cut-off wavelength of the TE1 mode is

λ c1 = 2a = 58 mm .

The TE1 mode can propagate.

___________________________________

7.2 Waveguide with a rectangular cross-sectionThe parallel plate waveguide is not suitable for practical applications, as it is open

from the sides and it can radiate energy to the

sides. To get a practical line we have toconfine it by two conducting planes from

sides. Thus we get a waveguide of rectangular

cross-section. This waveguide is completely

shielded. It is shown in Fig. 7.7. The inner

dimensions are a in the direction of the x axis,

and b in the direction of the y axis. We will

assume an infinitely long waveguide with

ideally conducting walls filled by a lossless

dielectric.

The process of solving the wave

equation is shown in chapter 5. In the case of

a

b

0

Fig. 7.7



book - 7

95

the TM modes we solve the wave equation (5.3) for E 0 z and the transversal components of the

electric and magnetic fields are calculated from (5.14-17) putting H 0 z = 0.

For the TE modes we determine H 0 z from the equation analogous to (5.3) and put E 0 z

= 0. The longitudinal component of the magnetic field is

( ) y jk y jk x jk x jk z

y y x x DC B A H −−

++= eeee0 , (7.31)

where the separation constants, which represent the propagation constants in the x and y

directions, are coupled with k p by

222 y x p k k k += . (7.32)

Unknown constants A, B, C , D and the propagation constants must be determined applying

the boundary conditions. To meet these boundary conditions we have to know the transversal

components of the electric field. From (5.14) and (5.15) we get

( )( ) y jk y jk x jk x jk

p

y z

p

x y y x x DC B A

k k

y H

k j E −− −+=

∂∂−= eeee

2

0

20 µ ω ωµ

( )( ) y jk y jk x jk x jk

p

x z

p

y y y x x DC B A

k

k

x

H

k

j E

−−+−−=

∂

∂= eeee

2

0

20

µ ω ωµ .

The boundary conditions are

y = 0, E x = 0 => C – D = 0 => ( ) yk jC C y

y jk y jk y y sin2ee =−−

,

x = 0, E y = 0 => A – B = 0 => ( ) ( ) xk jA A x x jk x jk x x sin2ee =−

−,

x = a, E y = 0 => ( ) 0sin =ak x =>a

mk x

π = , (7.33)

y = b, E x = 0 => 0sin =bk y =>b

nk y

π = , (7.34)

Now setting a new constant M = 4 AC , which determines the field amplitude, we have the field

distribution

z jk z

zmn yb

n x

a

m M H

−

= ecoscos

π π , (7.35)

z jk

p

zmn x

zmn yb

n x

a

m M

a

m

k

jk H

−

= ecossin

2

π π π , (7.36)

z jk

p

zmn

y

zmn

yb

n

xa

m

M b

n

k

jk

H

−

−=

esincos2

π π π

, (7.37)



book - 7

96

z jk

p

x zmn y

b

n x

a

m M

b

n

k

j E

−

= esincos

2

π π π ωµ , (7.38)

z jk

p

y zmn y

b

n x

a

m M

a

m

k

j E

−

−= ecossin

2

π π π ωµ , (7.39)

where

22

2

+

=

b

n

a

mk p

π π , (7.40)

A similar procedure can be followed in the case of TM modes. Now we solve the

wave equation (5.3). Its solution is

( ) y jk y jk x jk x jk z

y y x x DC B A E −−

++= eeee0 . (7.41)

The longitudinal component of the electric field is tangent to all walls of the waveguide. This

simplifies the solution. The boundary conditions are

x = 0, E z = 0 => A + B = 0

=> ( ) ( ) yk xk M E y x z sinsin0 =

y = 0, E z = 0 => C + D = 0

x = a, E z = 0 => ( ) 0sin =ak x =>a

mk x

π = , (7.42)

y = b, E z = 0 => 0sin =bk y =>b

nk y

π = . (7.43)

The particular components of the electric and magnetic field are derived from (7.41) using

(5.14-7)

z jk z

zmn yb

n x

a

m M E

−

= esinsin

π π , (7.44)

z jk

p

zmn x

zmn yb

n x

a

m M

a

m

k

jk E

−

−= esincos2

π π π , (7.45)

z jk

p

zmn y

zmn yb

n x

a

m M

b

n

k

jk E

−

−= ecossin

2

π π π , (7.46)

z jk

p

x zmn y

b

n x

a

m M

b

n

k

j H

−

= ecossin

2

π π π ωε . (7.47)



book - 7

97

z jk

p

y zmn y

b

n x

a

m M

a

m

k

j H

−

−= esincos

2

π π π ωε . (7.48)

k p is determined by (7.40) and k zmn by (5.18).

Phase constants k p, k x, k y, k z are equal for the TE and TM modes. This means that all

parameters describing the propagation of these modes along the waveguide with a rectangular

cross-section are equal for the TE and TM modes. Each mode is described by the two modalnumbers m and n. They determine the field distribution in the corresponding direction, as

shown in Fig. 7.2. The longitudinal propagation constant is, as follows from (5.18),

22222

−

−=−=

b

n

a

mk k p z

π π µε ω µε ω . (7.49)

From the condition of zero value of k z we get the cut-off frequency (5.19) of the mode with

modal numbers m and n

22

2

1

+

=

b

n

a

m f cmn

π π

µε π . (7.50)

The cut-off wavelength is

22

2

+

==

b

n

a

m f

v

cmn

cmn

π π

π λ . (7.51)

The reader can now compare relations (7.49), (7.50) and (7.51) valid for a waveguidewith a rectangular cross-section with those valid for a parallel plate waveguide (7.18), (7.19)

and (7.20). Instead of mπ /a we now have (7.40). It follows that the same relations can be

derived for the wavelength along the waveguide, the phase velocity, the group velocity and

the longitudinal propagation constant as (7.21), (7.22), (7.23) and (7.24). The wave

impedance must be defined separately for the TE and TM modes, as their field distribution is

different. It is defined as the ratio of the transversal components

x

y

y

xTM TE

H

E

H

E Z −==, .

These quantities are described as in the parallel plate waveguide by (7.25) and (7.26).

____________________________

Example 7.3: Determine the modes which can propagate along a rectangular waveguide with

the internal dimensions a = 40 mm, b = 25 mm at the frequency f = 10 GHz.

The propagating modes must have f < f cmn, which corresponds to the condition k 2 > k p2.

This gives us

22

22

+

=>

b

n

a

mk p

π π µε ω ,



book - 7

98

and results in the equation

775.15162.6681.43 22 ⋅+⋅> nm

This inequality is fulfilled for the modes: TE10, TE01, TE20, TE11, TM11, TE21, TM21

________________________________ Example 7.4: Calculate the phase and group velocities, the wavelength, the longitudinal

propagation constant and the wave impedance of the TE10 and TM12 modes in a rectangular

waveguide filled by air with the internal dimensions a = 22 mm, b = 10 mm. The frequency is

f = 10 GHz.

First we calculate the cut-off frequencies of these two modes

GHz82.62

110 ==

µε a f c ,

GHz76.3022

122

12 =

+

=

ba f c π π

µε π .

The TE10 mode can propagate, but the TM12 mode cannot as its cut-off frequency is greater

than 10 GHz. Therefore the required parameters will be computed only for the TE 10 mode.

We will first calculate the term

73.01

2

10 =

−

f

f c

Consequently we have

810 101.4

73.0⋅==

cv p m/s

810 1019.273.0 ⋅=⋅= cv g m/s

4173.0

10 ==λ

λ g mm

57.15273.010 =⋅= µε ω z k m-1

2.51673.0

0TE10 ==

Z Z Ω

____________________________________

The most important mode is the dominant mode. This is the mode with the

combination of the lowest possible modal numbers m and n and consequently with the lowest

cut-off frequency and the simplest field distribution. From the distribution of the field of the

TM modes (7.44-48) it follows that it is not possible to set m = 0 and n = 0 as the field is zero.



book - 7

99

In the case of the TE modes it is possible, see (7.35-39), to put m = 0 or n = 0, but not

simultaneously. So we have two possible combinations of m and n (1,0) or (0,1). Which cut-

off frequency is lower depends on the relation between the dimensions of the waveguide.

Accepting the standard notation a > b, we get as the dominant mode the mode TE10. From

(7.50) and (7.51) we will get its cut-off wavelength and its cut-off frequency

ac

210=λ , (7.52)

a f c

1

2

110

µε = . (7.53)

Formula (7.52) tells us that the waveguide transmits the TE10 mode starting from the

frequency at which a = λ /2. The field distribution of the dominant TE10 mode follows from

(7.35-39)

z jk y

z x

a

M a j

E −

−= esin

π

π

µ ω , (7.54)

z jk z

z xa

M H −

= ecos

π , (7.55)

z jk z x

z xa

M a jk

H −

= esin

π

π . (7.56)

There is only one transversal component of the electric field and one transversal component

of the magnetic field. The field does not depend on the y coordinate. The longitudinal

propagation constant is from (7.49)

22

−=

ak z

π µε ω . (7.57)

The electric field of this mode has its maximum at the center of the wider dimension,

so it can be simply excited by a probe located at the

position of the field maximum, Fig. 7.8.

Tangent components of the magnetic field create

an electric current passing along the surface of the

waveguide walls in the direction perpendicular to this

field. This means that the H x component creates currents

flowing along the waveguide. This must be taken into

account when parts of the waveguide are mounted

together. Their flanges must be well fitted to allow these

currents to pass, otherwise the increased resistance

between flanges raises the losses.

____________________________

Example 7.5: Determine the frequency band of the single-mode operation of the rectangular

waveguide with internal dimensions a = 22.86 mm, b = 10.16 mm. The waveguide is filled byair.

E y

Fig. 7.8

coaxial probe



book - 7

100

The frequency band is limited from below by the cut-off frequency of the dominant

mode TE10. The cut-off wavelength of this mode is (7.52)

72.45210 == acλ mm .

The cut-off frequency is

56.610

10 ==c

c

c f

λ GHz .

The frequency band of the single-mode propagation is limited from above by the cut-off

frequency of the nearest higher mode. In the rectangular waveguide it could be TE01 or TE20

modes, depending on the ratio between the waveguide dimensions. The cut-off wavelengths

of these two modes are

32.20201 == bcλ mm, 86.2220 == acλ mm.

In our case a > 2b and consequently λ c20 > λ c01. The nearest higher mode is therefore the TE20

mode. The cut-off frequency corresponding to λ c20 is

12.1320

20 ==c

c

c f

λ GHz.

The frequency band of the single-mode operation is from 6.56 to 13.12 GHz.

____________________________________

Example 7.6: Design the dimensions of the rectangular waveguide a, b as it transmits at thefrequency 10 GHz the only dominant mode TE10 and the two nearest higher modes TE01 and

TE20 are equally attenuated.

The two modes TE01 and TE20 will be equally attenuated if their longitudinal

propagation constants are equal

k z 01 = k z 20 => 220

2201

2 p p k k k k −=− =>

ak

bk p p

π π 22001 ===

Consequently we have a relation that assures equal attenuation of the two modes

a = 2b

The condition for propagation of the dominant mode TE10 is

a

c f f c

210 => => 15

2=>

f

ca mm

The condition for the attenuation of the two higher modes is

b

c f f f cc

22001 ==< => 15

2

=<

f

cb mm, a = 2b => a < 30 mm



book - 7

101

The waveguide has the dimensions a = 2b and 15 < a < 30 mm.

_____________________________________

The power transmitted by the rectangular waveguide is calculated from the Poynting

vector

( ) ( ) 0TE10

2

0TE10

*

0**

2

1Re2

1Re2

1Re2

1 zzzHES Z

E

Z

E E H E y y y

x yav =

−=−=×= .

The total active transmitted power is inserted for E y (7.54)

∫∫ ∫∫∫

==⋅=

am

a b

y

S

av xa

x

Z

b E y x E

Z P

0

2

TE10

2

0 0

2

TE10

dsin2

dd2

1d

π SS ,

TE10

2

4 Z

ab E

P m

= . (7.58)

The maximum amplitude of the electric field is from (7.54) π ωµ /a M E m = .

_________________________________

Example 7.7: Calculate the maximum power that can be transmitted by the dominant mode

TE10 in a rectangular waveguide with the internal dimensions a = 22.86 mm, b = 10.16 mm at

the frequencies 5 GHz and 10 GHz. The waveguide is filled by air with the electric strength

E p = 30 kV/cm.

According to Example 7.5 the cut-off frequency of the dominant TE10 mode in this

waveguide is 6.56 GHz. This means that this mode does not propagate at 5 GHz. So we willcalculate the power only at 10 GHz. The wave impedance of this mode is

Ω=

−

= 500

1

2

10

0TE10

f

f

Z Z

c

,

consequently the transmitted power (7.58) allows the maximum value of the electric field to

be equal to the electric strength

MW05.14 TE

10

2

== Z

ab E P

p.

___________________________________

Example 7.8: A waveguide with a rectangular cross-section transmits the TM11 mode at the

frequency 6 GHz. The waveguide is filled by air and its internal dimensions are a = 71 mm

and b = 35.5 mm. Calculate the magnitude of the total current passing the waveguide walls in

the longitudinal direction and the magnitude of the displacement current passing the

waveguide cross-section in the longitudinal direction. The field distribution is described by

the formulas



book - 7

102

( ) ( ) z j z b ya x j E 2.78e/sin/sin3000 −= π π ,

( ) ( ) z j x b ya x E 2.78e/sin/cos1060 −−= π π ,

( ) ( ) z j y b ya x E 2.78e/cos/sin2120 −−= π π ,

( ) ( ) z j x b ya x H 2.78e/cos/sin04.9 −= π π ,

( ) ( ) z j y b ya x H 2.78e/sin/cos52.4 −−= π π .

The conducting current passing along the surface of the waveguide walls can be calculated

from the magnetic field boundary condition (1.16)

K Hn =× .

The current directions are shown in Fig. 7.9.In the upper and bottom walls we have

( )a x K z /sin04.91 π = A/m.

In the left and right walls we have

( )b y K z /sin52.42 π = A/m.

The total current passing the upper wall is

( ) 409.02

04.9d/sin04.9d

00

11 ==== ∫∫ a xa x x K I

aa

z

π π A

The total current passing the left wall is

( ) 102.02

52.4d/sin52.4d

00

22 ==== ∫∫ b yb y y K I

bb

z

π π A

The total current is

022.122 21 =+= I I I A

The density of the displacement current is

( ) ( )b ya x E J z Dz /sin/sin10000 π π ωε ==

The total displacement current passing the waveguide cross-section is

a

b

0

Fig. 7.9

n H y

K z 2

K z 1

H x

n

nn

K z 2

K z 1

H x

H y



book - 7

103

( ) ( ) 022.14

1000dd/sin/sin1000dd

02

00 0

==== ∫ ∫∫ ∫a ba b

Dz D

ab y xb ya x y x J I

π π π A

The results show that the current continuity is preserved. The total value of the

conducting current passing the waveguide walls equals the value of the displacement current

passing the waveguide cross-section.

____________________________________

All previous results were derived assuming that all materials are without losses. This is

an ideal case. There are two sources of losses in a real waveguide. These are the finite

conductivity of the waveguide walls and the finite conductivity of the dielectric filling the

waveguide. Other losses are caused by the roughness of the waveguide wall and nonperfect

connections between the flanges. All these losses increase with the frequency. They are about

0.1 dB/m for the frequency band from 8 to 12.4 GHz, which is the so called X band.

It is not possible to design a rectangular waveguide with arbitrary dimensions. The

dimensions of these waveguides are determined by international standards depending on the

frequency band. An example is the waveguide for the X band. Its internal dimensions are a =22.86 mm, b = 10.16 mm. The waveguides are at present used at lower frequencies only in the

case of the need to transmit high power. Planar transmission lines such as the microstrip line,

Fig. 5.1b, are used for low power applications instead of waveguides. Metallic waveguides

are very massy and their production is very expensive, so they are not suitable for mass

production. At high frequency bands, above 50 GHz, waveguides are still used, as they have

lower losses than planar transmission lines. Planar transmission lines are open transmission

lines and lose energy due to radiation at high frequencies.

7.3 Waveguide with a circular cross-section

A waveguide with a circular cross-section, Fig. 7.10, is used only in some specialapplications. An example is a rotating joint which transmits an electromagnetic wave to the

feeder of a rotating radar antenna. The problem of these waveguides is that due to their

geometry they do not keep the plane of the polarization of the transmitted mode when the

waveguide is long. The frequency band of the single mode operation of a circular waveguide

is narrower than the same band of a rectangular waveguide.

In this section we will show only the basic

properties of a waveguide with a circular cross-section,

Fig. 7.10. The internal radius of this waveguide is r o, and

the longitudinal axis is z . We will assume an infinitely

long waveguide with ideally conducting walls filled by a

lossless dielectric. Even in a waveguide of this geometrywe can get results similar to those obtained in section 5.

We have to use the cylindrical coordinate system. The

field in a circular waveguide can be separated into TE and

TM modes, which are treated separately. We will see that

unlike in the rectangular waveguide the longitudinal

propagation constants and the cut-off frequencies are

different for the TE modes and for the TM modes.

First we will treat TM modes. The field

distribution of these modes is determined by solving the wave equation (5.3) written for the

longitudinal component of the electric field (5.1). The cylindrical coordinate system must be

used here, see the mathematical appendix.

r o

Fig. 7.10



book - 7

104

( ) ( ) z jk z z

z r E z r E −= e,,, 0 α α , (7.59)

where the transversal field distribution is obtained by solving the wave equation (5.3)

expressed in the cylindrical coordinate system (13.86)

0

11

0

2

2

02

2

0

2

02

=+∂

∂

+∂

∂

+∂

∂

z p

z z z

E k

E

r r

E

r r

E

α . (7.60)

The propagation constants k p, k z and k are coupled by (5.5). We will solve (7.60) by the

method of separation of variables. Function E 0 z is assumed in the form

( ) ( ) ( )α α Α⋅= r Rr E z ,0 . (7.61)

Inserting (7.61) into (7.60) and after some manipulation we will get the equation

0

''''' 222 =+Α

Α++

pk r R

R

r R

R

r . (7.62)

Term ΑΑ /'' must be independent of r and α , and must be equal to constant m, which gives

us the equation for function Α

0'' 2 =Α+Α m . (7.63)

The solution of (7.63) is obtained using a proper origin for reading angle α in the simple form

( )α mC cos=Α . (7.64)

The electric field must be unique, therefore ( ) ( )π α α m E E z z 200 += . It follows from this that

m must be an integer number. Inserting solution (7.64) into (7.62) we will get

0d

d1

d

d2

22

2

2

=

−++ R

r

mk

r

R

r r

R p . (7.65)

This equation is Bessesl’s equation of the m-th order. The solution is in the form of the sum of

the Bessel functions of the m-th order J m and Y m, see the mathematical appendix,

r k Y Br k J A R pm pm += . (7.66)

To determine constants A and B we have to apply the boundary conditions. The tangent

component of the electric field on the surface of the waveguide wall must be zero and the

field must have a finite value at the waveguide center. The second condition gives us B = 0, as

Bessel function Y m has an infinite value at the origin, see the mathematical appendix.

Component E z is tangent to the waveguide wall, so the first boundary condition reads

( ) 0== or r R => 0=o pm r k J . (7.67)



book - 7

105

The product k pr o must be equal to the n-th zero point of the Bessel function of the m-th order

α mn. These values are listed in Table 13.1 in the mathematical appendix. Consequently the

transversal propagation constant of the mode with modal numbers m and n is

o

mn pmn

r k

α =TM . (7.68)

From (5.18) and (5.19) we get the longitudinal propagation constant and the cut-off

frequency of the TM modes in the circular waveguide

2

222TM

−=−=

o

mn pmn zmn

r k k

α µε ω µε ω , (7.69)

o

mncmn

r f

α

µε π 2

1TM = . (7.70)

According to Table 13.1 we have the order of the lowest modes TM01, TM11, TM21,

TM02 and TM31, as the lowest numbers α mn are α 01 = 2.40482, α 11 = 3.83171, α 21 = 5.13562,

α 02 = 5.52007, α 312 = 6.38016.

The field distribution of particular modes can be determined in a similar way as the

distribution of TM modes in a rectangular waveguide. The cylindrical coordinate system must

be applied. Longitudinal component E z has the form, using (7.64) and (7.66),

( ) ( ) z jk pm z

zmnemr k J E E −

= α cos0 , (7.71)

where E 0 is an unknown amplitude. The transversal components are E r , E α , H r and H α . Thefield distribution of the lowest modes is shown in Fig. 7.11.

TE11 TM01 TE01 TM11

Fig. 7.11

The field distribution of the TE modes is determined by solving the wave equation

written for the longitudinal component of the magnetic field. Using the same procedure as in

the case of the TM modes we will get the distribution of H z

( ) ( ) z jk pm z

zmnemr k J H H −

= α cos0 , (7.72)

electric field magnetic field



book - 7

106

where H 0 is an amplitude. The TE modes have the E α component of the electric field tangent

to the waveguide wall. This component is proportional to r H z ∂∂ / , so the boundary condition

is

0=∂

∂

= or r

z

r

H => ( ) 0' =o pm r k J . (7.73)

The product k pr o must be equal to the n-th zero point of the derivative of the Bessel function

of the m-th order 'mnα . These values are listed in Table 13.2 in the mathematical appendix.

Consequently the transversal propagation constant of the mode with modal numbers m and n

is

o

mn pmn

r k

'TE α

= . (7.74)

From (5.18) and (5.19) we get the longitudinal propagation constant and the cut-off

frequency of the TE modes in a circular waveguide

2'

222TE

−=−=

o

mn pmn zmn

r k k

α µε ω µε ω , (7.69)

o

mncmn

r f

'TE

2

1 α

µε π = . (7.70)

According to Table 13.2 we have the order of the lowest modes TE11, TE21, TE01, TE31

and TE41, as the lowest numbers '

mn

α are '

11

α = 1.84118, '

21

α = 3.05424, '

01

α = 3.83170, '

31

α

= 4.20119, '41α = 5.31755. The transversal propagation constants and the cut-off frequencies

of the TE modes are different from those of the TM modes. The field distribution of the

lowest modes over the waveguide cross-section is shown in Fig. 7.11.

For the circular waveguide the same relations as (7.21), (7.22), (7.23) and (7,24) can

be derived for the wavelength along the waveguide, the phase velocity, the group velocity,

and the longitudinal propagation constant, of course separately for the TE and TM modes.

The wavec impedance is defined as the ratio of the transversal components

r

r TM TE

H

E

H

E Z α

α

−==, .

The dominant mode of the circular waveguide is the TE11 mode, as its cut-off

frequency has the lowest value. The value of the corresponding parameter '11α = 1.841. The

cut-off frequency of this mode is

or f

841.1

2

1TEc11

µε π = . (7.71)

The cut-off wavelength of the dominant TE11 mode is



book - 7

107

or or

cmn r r

ε ε π

λ 4.3841.1

2TE == . (7.72)

___________________________________

Example 7.9: Determine the frequency band of the single mode propagation of a circular

waveguide filled by a dielectric with relative permittivity ε r with radius r o = 8.2 mm.

The dominant mode of the circular waveguide is the TE11 mode with the cut-off wavelength (7.72). The nearest higher mode is the TM01 mode with the cut-off wavelength

or or

c

c r r

f

vε

ε π λ 6.2

40482.2

2

TM01

MT01 === .

For the waveguide radius r o = 8.2 mm we get the band of the single mode operation

21.32 < λ < 27.88 mm

10.76 < f < 14.07 GHz ____________________________________

7.4 Problems

7.1 Calculate for the modes propagating between the two parallel plates from Example 7

the wavelength, the phase velocity, the group velocity, the longitudinal propagation constant

and the wave impedance. The frequency is 10 GHz.

mode v pm

(m/s)

v gm

(m/s)

λ gm

(mm)

k zm

(m

-1

)

Z TM

(Ω)

Z TE

(Ω)0th 3 108 3.108 30 209 376.6 -

1st 3.24 108 2.78 108 32.4 193.7 349.1 406.3

2nd 4.55 108 1.98 108 45.5 137.9 248.6 570.6

7.2 Calculate the frequency band of the propagation of only the TE1 mode in a parallel

plate waveguide filled by air. The distance of the plates is 40 mm. Do not consider the TM

modes.

f c1 = 3.75 < f < f c2 = 7.5 GHz

7.3 How does the result of Problem 7.2 change if we replace the air by a dielectric with

permittivity ε r = 9? f c1 = 1.25 < f < f c2 = 2.5 GHz

7.4 Determine the points at the cross-section of the rectangular waveguide at which the

magnetic field of modes TE10, TE01, TE20, TE11 and TM11 is zero.

7.5 Determine the points at the cross-section of the rectangular waveguide at which the

electric field of modes TE10, TE01, TE20, TE11 and TM11 is zero.

7.6 The maximum value of the amplitude of the electric field in the rectangular waveguide

is E m = 600 V/m. The dominant mode TE10 propagates in this waveguide at the frequency 10



book - 8

108

GHz. The waveguide is filled by air and its internal dimensions are a = 22.86 mm and b =

10.16 mm. Calculate the amplitudes of all field components.

E ym = -600 V/m

H xm = 1.19 A/m

H zm = 1.04 A/m

7.7 Calculate the electric field intensity at the point with the coordinates x = 5.5 mm and y

= 5.1 mm in the cross section of a rectangular waveguide filled by air. The waveguide

transmits the dominant mode TE10 at the frequency 15 GHz. The transmitted power is 10 W.

The internal dimensions of the waveguide are a = 22.9 mm, b = 10.2 mm. z j

y j E 282e575 −−= V/m

7.8 Determine which modes can propagate along a circular waveguide filled by air with

radius r o = 5.2 mm at frequency f = 30 GHz.

TE11, TM01, TE21

7.9 Calculate for the mode TE11 propagating along the waveguide from problem 7.8 at

frequency of 30 GHz the longitudinal propagation constant and the wavelength.TE11 z k = 518.9 m-1

TE11 g λ = 12.1 mm

8. DIELECTRIC WAVEGUIDES

In the past, dielectric waveguides were used in microwave systems mainly as

dielectric antennas. The wave was guided by a dielectric rod, the cross-section of which wastapered and the wave was gradually radiated out into the space. Dielectric waveguides are

nowadays frequently applied in optical communication systems as optical fibers and in the

form of planar optical waveguides in the circuits of optical integration systems.

The operation of dielectric waveguides is

based on total reflection of the wave on the boundary

between the two dielectric materials treated in

paragraph 3, see Fig. 3.15. The wave is totally

reflected back if it is incident to the boundary of the

two dielectric materials from the side of the material

with greater permittivity at an angle greater than the

critical angle. The field penetrates into the secondmaterial as an evanescent wave with the amplitude

decreasing exponentially perpendicular to the

boundary. Perpendicular to the boundary the wave in

the first material has the character of a standing wave.

The wave propagates as a surface wave in the

direction along the boundary, Fig. 3.15. Let us now

imagine that we confine the material with greater

permittivity to a layer of finite thickness 2a. We

couple into this layer the wave as it is incident to the

boundary at an angle greater than the critical angle. It

is reflected back and is incident to the opposite

E y

v pz >v1

fast wave

v pz <v2

slow wave

Fig. 8.1

v pz <v2

slow wave

ε 1

ε 2

ε 2

ε 1 > ε 2

standing wave

evanescent wave

evanescent wave



book - 8

109

boundary and bounces up and down. According to Fig. 8.1 the wave is in this way coupled to

the layer, but it also penetrates into the surrounding material, where its amplitude decreases

exponentially. The field distribution across the layer depends on the frequency and on the

thickness of the layer.

Thus any dielectric structure with a core made from a material with permittivity

greater than the permittivity of the surrounding material can behave as a dielectric waveguide.

The guiding of the wave is due to the total internal reflection on the surface of the core.

8.1 Dielectric layersA dielectric layer is the dielectric

guiding structure which can be treated most

simply. Let us assume a dielectric

homogeneous layer that is infinitely long and

wide, and its thickness is 2a, see Fig. 8.2. We

assume that the field does not depend on the y

coordinate, therefore 0=∂

∂

y, and that the

wave propagates in the z axis direction. Thus

we can treat the TE and TM modes

separately, similarly as in the parallel plate

waveguide. According to (5.14-17) the TE modes have non-zero field components E y, H x and

H z , and the TM modes have components H y,

E x and E z .

Let us first study the TE modes. The

mode propagates in the dielectric layer due

to the internal total reflections, see Fig. 8.3.

Its electric field has only the E y component,

and the magnetic field must thus have the H x

and H z components. The wave equation can

by now solved for the E y component of the

electric field. The solution obtained

similarly as in paragraph 5 by the method of

separation of variables can be written in the

particular regionsc, d,e, Fig. 8.3,

( ) ( )[ ] z jk x x y

z xk B xk A E −+= ecossin 111 , x < a , (8.1)

z jk z jk y

z xC E −= ee 22 , x < -a , (8.2)

z jk z jk y

z x D E −−= ee 2

3 , x > a , (8.3)

where the separation constants are coupled by

( ) ( ) 221

2

001

2

0121 z xr r k k k k +=== µ ε ω ε ε , (8.4)

( ) ( )22

2

2

002

2

02

2

3

2

2 z xr r k k k k k +==== µ ε ω ε ε . (8.5)

a

-a

0ε 1

ε 2

ε 2

c

d

e

Fig. 8.2

a

-a

0 ε 1

ε 2

ε 2

c

d

e

Fig. 8.3

H

E

TE modes



book - 8

110

Constants k x1 and k x2 are the propagation constants in the x direction in the dielectric layer and

in the surrounding material, respectively.

We can simplify the treatment of the field by dividing the TE modes into two groups.

In the first group we have symmetric, or even modes, which satisfy the condition

( ) ( ) x E x E y y −= . (8.6)

For these modes we put in (8.1) A = 0 and they are described by the cosine function. In the

second group we have anti-symmetric, or odd modes which satisfy the condition

( ) ( ) x E x E y y −−= . (8.7)

For these modes we put in (8.1) B = 0 and they are described by the sinus function.

The E y component of the even TE modes in the dielectric layer is

( ) z jk x y

z xk B E −= ecos 11 . (8.8)

The magnetic field is related to the electric field by Maxwell’s second equation (1.13).

Assuming E x = E z = 0 we will get

x

E

j H

y

z ∂

∂=

ωµ

1, (8.9)

and the z component of the magnetic field in our three particular areas is

( ) z jk

x

x

z

z xk B j

k H −−= esin

1

1

1 ωµ , x < a , (8.10)

z jk z jk x z

z xC k

H −= ee 222

ωµ , x < -a , (8.11)

z jk z jk x z

z x Dk

H −−−= ee 223

ωµ , x > a . (8.12)

The field must fulfill the boundary conditions on the layer surfaces at x = a and x = -a.

Consequently at x = a and for arbitrary z we have

( ) ( )a E a E y y 31 = , ( ) ( )a H a H z z 31 = ,

( ) a jk x

xe Dak B 21cos

−= , (8.13)

( ) a jk x x

x xe Dk

ak B j

k 22

11 sin

−−=−ωµ ωµ

.

( ) a jk x x x

xe Dk ak B jk 2211 sin

−=− . (8.14)



book - 8

111

Dividing (8.14) by (8.13) we get

( ) ( ) ( )ak jak ak x x x 211 tg = . (8.15)

This equation couples propagation constants k x1 and k x2, but to determine them we need to get

another equation. This equation is derived by inserting from (8.4) for k z into (8.5)

( ) 20021

22

21 ω µ ε ε ε r r x x k k −=− . (8.16)

Now to remove the imaginary unit from (8.15) we introduce a new variable

2'2 x x k jk = . (8.17)

Consequently we have the two equations known as dispersion equations

( ) ( ) ( )ak ak ak x x x'211 tg = , (8.18)

( ) ( ) ( ) 220021

2'2

2

1 aak ak r r x x ω µ ε ε ε −=+ . (8.19)

The E y component of the odd TE modes is

( ) z jk x y

z xk A E −= esin 11 . (8.20)

Applying exactly the same procedure for calculating the distribution of the corresponding

magnetic field and fulfilling the boundary conditions we will get the dispersion equation for

the odd TE modes in the form

( ) ( ) ( )ak ak ak x x x'211 cotg =− . (8.21)

The second dispersion equation remains in the

form (8.19).

The electric field distribution of TE

modes in the plane transversal to the propagation direction is described by (8.8) or

(8.20) in the layer itself and by (8.2) and (8.3) in

the surrounding material. This distribution is

plotted for the three lowest TE modes in Fig.

8.4. The modes are denoted by numbers which

determine the number of zeros of the E y

function across the layer. So we have, Fig. 8.4,

the series of modes: even TE0, odd TE1, even

TE2, odd TE3, etc.

Solving equations (8.18) and (8.19) or

(8.21) and (8.19) we determine the propagation

constants k x1 and k x2, and the longitudinal propagation constant is then calculated from (8.4)

or (8.5). In this way we can get the dispersion characteristic of the given TE mode

propagating along our dielectric layer. The dispersion characteristic represents the dependenceof the propagation constant on the frequency. Dispersion equations (8.18) and (8.19) cannot

2a

E y

m=0 m=1 m=2

Fig. 8.4

c

d

eTE0

evenTE2

even

TE1

odd



book - 8

112

be solved analytically. The solution must be

performed using a proper numerical

method. This problem can be overcome by

a simple graphical technique, which gives a

clear physical insight of the effect of

guiding the waves along the dielectric

layer. In the coordinate system

( ) ( )ak ak x x'21 and , equation (8.19)

represents a circle with its center at the

origin. The radius of this circle is

( ) 0021 µ ε ε ε ω r r a R −= .

(8.22)

The radius is proportional to the frequency,

the layer thickness and the second root of

the difference between the permittivities.The solution of the set of equations (8.18)

and (8.19) or (8.21) and (8.19) can be

obtained graphically as an intersection of

the circle (8.19) with the curve represented

by (8.18) or (8.21), Fig. 8.5.

From Fig. 8.5 we see that we always have one solution of the dispersion equations,

which is represented by the intersection of the circle (8.19) with the first branch of the tangent

function on the left hand side of (8.18). This means that the TE0 mode can propagate under

any circumstances at any layer at any frequency. This is the dominant mode of this dielectric

waveguide. The nearest higher mode is the TE1 mode. So the upper frequency of the

frequency band of the single mode propagation is determined by the cut-off frequency of theTE1 mode. The solutions of the dispersion equations lying in the lower half plane of the plane

( ) ( )ak ak x x'21 and determined by ( )ak x

'2 < 0 are not physical, as they correspond to the field

(8.2) and (8.3) which increases exponentially in the surrounding areas d and e.

Consequently, the cut-off frequency of the TE1 mode follows from the condition R = π /2,

which gives the cut-off frequency of this mode

( ) 0021

1

1

4

1

µ ε ε ε r r

ca

f −

= . (8.23)

The cut-off frequency of any mode with the modal number m > 0 is determined by thecondition R = mπ /2, so generally we have

( ) 0021

1

4 µ ε ε ε r r

cma

m f

−= . (8.24)

Fig. 8.5 shows the plots of the dispersion equations (8.18) for modes TE 0, TE2 and

TE4, (8.21) for modes TE1 and TE3. There are the three propagating modes TE0, TE1 and TE2

at the given frequency. The corresponding solutions of the dispersion equations are marked in

Fig. 8.5 by M0, M1 and M2. The solution marked M2non is non-physical.

R

π /2 π 3π /2 k x1a

k x2a‘

TE0 TE1 TE2 TE3

M0

M1

M2

M0non

TE4

Fig. 8.5



book - 8

113

_______________________________________

Example 8.1: Calculate the longitudinal propagation constant of the TE 0 mode propagating

along a dielectric layer 5 mm in thickness with the relative permittivity ε r = 2.6 at the

frequency f = 62.5 GHz. The layer is surrounded by air.

According to (8.22) we have R = 4.135. The corresponding dispersion equations are

plotted in Fig. 8.5. Our solution is marked by M0. From Fig. 8.5 we can read

( ) ( ) 93.3,29.1'

21 == ak ak x x . From these values we have k x1 = 516 m-1

, and (8.4) gives us k z =2042 m-1.

___________________________________

The same procedure can be followed

for the TM modes. These modes propagate

in the dielectric layer due to the internal total

reflections, see Fig. 8.6. Their magnetic field

has only the H y component and the electric

field must thus have the E x and E z

components. The wave equation can now be

solved for the H y component of the magneticfield. The solution obtained similarly as in

paragraph 5 by the method of separation of

variables can be written in the particular regions c, d,e, Fig. 8.6,

( ) ( )[ ] z jk x x y

z xk B xk A H −+= ecossin 111 , x < a , (8.25)

z jk z jk y

z xC H −= ee 22 , x < -a , (8.26)

z jk z jk

y

z x

D H

−−

= ee2

3 , x > a , (8.27)

where the separation constants are coupled by (8.4) and (8.5). The distribution of the H y

component of the magnetic field of the TM modes is the same as the distribution of the

electric field shown in Fig. 8.4. The rest of the process is the same as in the case of the TE

modes. From the H y component of the magnetic field we calculate the distribution of the E z

electric field by

x

H

j E

y

z ∂

∂−=

ωε

1. (8.28)

We divide the field into even and odd modes and we meet the boundary conditions at x = a,

which are H y1(a) = H y3(a) and E z 1(a) = E z 3(a). As a result we obtain the dispersion equations

for the even TM modes

( ) ( ) ( )ak ak ak x x x'211

1

2 tg =ε

ε , (8.29)

and for the odd TM modes

a

-a

0ε 1

ε 2

ε 2

c

d

e

Fig. 8.6

E

H

TM modes



book - 8

114

( ) ( ) ( )ak ak ak x x x'211

1

2 cotg =−ε

ε . (8.30)

The equation of the circle (8.19) remains

unchanged. The cut-off frequencies of the

TM modes are then defined by (8.24) as for

the TE modes. The dominant mode is theTM0 mode, which can propagate at any

frequency. The frequency band of the single

mode operation is confined from above by

the cut-off frequency of the TM1 mode

(8.23). The only difference between the

solutions of the dispersion equations is their

shift caused by the term ε 2 /ε 1, Fig. 8.7.

The dispersion characteristic of the

modes propagating along the dielectric layer

can be obtained by repeating the procedure

shown in Example 8.1 for the TE modes,and similarly for the TM modes. The

dispersion characteristic of the TM0 and TE1

modes propagating along the dielectric layer

defined in Example 8.1 is plotted in Fig. 8.8.

Note the cut-off frequency of the TE1 mode.

8.2 Dielectric cylindersThe dielectric cylinder, the cross-

section of which is shown in Fig. 8.9,

represents the dielectric waveguide

widely used in practical applications.

Optical fiber is based on it. In the case

of optical fiber the surrounding material

with relative permittivity ε r 2 does not fill

the whole space, but creates a cover

layer, see Fig. 5.1c. The cylindrical

dielectric waveguide is able to transmit

the wave due to the internal reflection,

assuming that ε 1 > ε 2.The field distribution can be determined

similarly as in the case of a cylindrical

waveguide, using the wave equation in the

cylindrical coordinate system. However, we have

to assume that the field penetrates into the

surrounding material d. The field cannot be

divided into TE and TM modes due to the

boundary conditions on the surface of the

dielectric cylinder. All modes propagating alongthe dielectric cylinder have nonzero components

R

π /2 π 3π /2 k x1a

k x2a‘

TM0

TM1

TM2

TM3

M0

M1

M2

M0non

TM4

Fig. 8.7

TM0

TE1

Fig. 8.8

c1

r 0ε r 1

ε r 2

Fig. 8.9

d c



book - 8

115

E z and H z . Only for the cylindrically symmetric modes with the modal number m = 0 do we

have the modes TE0n and TM0n.

The longitudinal components of the electric and magnetic fields are assumed in the

form

( ) ( ) 0101 ,e,,, r r r E z r E z jk z z

z <= −α α , (8.31)

( ) ( ) 0202 ,e,,, r r r E z r E z jk z z

z >= −α α . (8.32)

( ) ( ) 0101 ,e,,, r r r H z r H z jk z z

z <= −α α , (8.33)

( ) ( ) 0202 ,e,,, r r r H z r H z jk z z

z >= −α α . (8.34)

Functions E 0 z 1, E 0 z 2, H 0 z 1 and H 0 z 2 are solutions of the wave equation (7.60) and a similar

equation for the magnetic field. Applying the same procedure as in the case of the metallic

waveguide we will obtain the solution

( ) 011 ,ee r r r k J A E z jk jm pm z

z <= −α , (8.35)

( ) 022 ,ee r r r jk K B E z jk jm pm z

z >= −α , (8.36)

( ) 011 ,ee r r r k J C H z jk jm pm z

z <= −α , (8.37)

( ) 022 ,ee r r r jk K D H z jk jm pm z

z >= −α . (8.38)

Where K m is the modified Bessel function of the second kind, see the mathematical appendix.

This function describes the evanescent character of the field outside the cylinder. The electric

and magnetic fields (8.35-38) together with E α 1, E α 2, H α 1 and H α 2 have to meet the boundaryconditions on the cylinder surface as they are tangent to this surface. From these boundary

conditions we can determine the constants B, C and D and the dispersion equations

necessary to determine the propagation constants of particular modes.

8.3 Problems8.1 Determine the frequency band of the single mode operation and the cut-off frequencies of

four modes in a plexiglass slab 5 mm in thickness with relative permittivity equal to 2.6. Theslab is infinitely wide and long and is surrounded by air.

f c1 = 23.7 GHz

f c2 = 47.4 GHz

f c3 = 71.1 GHz

f c4 = 94.8 GHz

the single mode operation band is 0 < f < 23.7 GHz

8.2 Determine the maximum thickness of a dielectric layer with relative permittivity ε r = 2.28

which ensures propagation of the TE0 mode alone at the wavelength λ 0 = 1.3 µm. The

dielectric layer is placed between two glass layers with the relative permittivity ε r = 2.26.µm6.4≤h



book - 9

116

8.3 Calculate the phase velocity of the TE0 mode propagating along the dielectric layer from

the example 8.1 at the frequency f = 62.5 GHz.

v pTE0 = 2.53 108 m/s

8.4 Determine the field distribution and dispersion equations of the TE and TM modes

propagating along a dielectric layer of thickness a placed on an ideally conducting plane, Fig.

8.10.

The structure in Fig. 8.10 is one half of the structure shown in Fig. 8.2, with the plane

of symmetry located at x = 0. Only the

even TM modes and the odd TE modes

can exist on this structure due to its

symmetry. Apart from this, everything else

is the same as was explained for the

dielectric layer. The dominant mode is the

TM0 mode, and the TE0 mode does not

exist.

9. RESONATORS

9.1 Cavity resonatorsMicrowave resonators are used in microwave circuits similarly as standard resonant

LC circuits at lower frequencies. At high frequencies we have to protect the electromagnetic

field from radiation outside the resonator. For this reason, resonators are very often used as

cavity resonators. These resonators are formed by the volume filled by a dielectric material,

mostly air. This volume is shielded by a suitable metallic coating. The electromagnetic field is

coupled inside, using a suitable probe. We will treat lossless resonators, assuming that the

dielectric material has zero conductivity and on the other hand the metal coating has infinite

conductivity.

Let us first determine the resonant frequency of a general cavity resonator. We use the

power balance for the reactive power described by equation (1.33). Assuming a shielded

lossless structure without any radiation, the power balance for the reactive power is

( ) ( )∫∫∫ ∫∫∫ −=⋅

V V

dV dV 22*

2

1Im

2

1HEEJ µ ε ω .

Disconnecting the source we have J = 0 and we get the self-oscillation of the field in thevolume. The condition for self-oscillation is determined by the equality of energies W e = W m,

which is expressed by

∫∫∫∫∫∫ =

V V

dV dV 22

HE µ ε .

The electric field can be determined from Maxwell’s first equation (1.12) assuming σ = 0 and

no external current. Then we have

a

0ε 1

ε 2

c

d

Fig. 8.10

∞→σ



book - 9

117

( ) 2

22

2rot

1HE

ε ω = .

Finally we get the resonant frequency of our cavity

( ) ( )

∫∫∫

∫∫∫

∫∫∫

∫∫∫==

V

V

V

V

dV

dV

dV

dV

2

2

2

2

20

rot

1

rot

1

E

E

H

H

µε µε ω . (9.1)

It follows from (9.1) that the resonant frequency of the cavity depends on the material

parameters µ and ε and on the cavity geometry that is hidden in the integrals. This is commonwith standard LC resonant circuits. However, (9.1) shows that the resonant frequency

depends on the distribution of the electric and magnetic fields . Consequently as we can

excite an infinite number of particular modes in the cavity, we have an infinite number of

resonant frequencies corresponding to these modes. Some modes are known as degenerated.

These modes have a different field distribution but an equal resonant frequency.Let us first determine the resonant frequency of

the cavity resonator formed by a segment of any

waveguide terminated at both ends by ideally

conducting planes. The planes are perpendicular to the

longitudinal axis of the waveguide. The length of the

segment is d , Fig. 9.1, so the conducting planes are

located at z = 0 and z = d . It is reasonable to assume that

the field distribution of the mode excited in this cavity

originates from the distribution of the modes existing in

the original waveguide forming the cavity. These

modes have transversal and longitudinal propagation constants k pmn and k zmn (5.5) whichdepend on two transversal modal numbers m and n. The distribution of the transversal

components of its electric field is ET 0( x, y) as the function of the two transversal coordinates x,

y. The electric field in our cavity is now the superposition of the two waves traveling in the

positive z direction and in the negative z direction

( ) ( )( ) z jk z jk T T

zmn zmn B A y x z vu ee,,, 0 +=−

EE .

This field must fulfill the two boundary conditions at z = 0 and z = d . At z = 0 we have

( ) 00=+

B AT E => A = - B .

The field distribution is now

( ) ( )( ) ( ) ( ) z k jBvu B Bvu z vu zmnT z jk z jk

T T zmn zmn sin2,ee,,, 00 EEE =+−=

− . (9.2)

At z = d we have

( ) 0sin =d k z .

d 0d

Fig. 9.1



book - 9

118

d

pk zmnp

π = , p = 1, 2, 3, … . (9.3)

(9.3) tells us that we have the set of discrete values of the longitudinal propagation constantsof the modes oscillating in the resonator. These constants depend on modal numbers m, n and

p. Now using (5.5) and (2.25) we have the set of resonant frequencies

2

2

00

02

1

+=

d

pk f pmn

r

mnp

π

ε ε µ π . (9.4)

The particular value of the resonant frequency depends on transversal propagation

constant k pmn, which is determined by the type of waveguide from which the resonator iscomposed. In the case of a waveguide with a rectangular cross section with dimensions a

and b, k pmn is determined by (7.40) and the resonant frequency is

222

00

02

1

+

+

=

d

p

b

n

a

m f

r

mnp

π π π

ε ε µ π . (9.5)

This formula is valid for both the TE and TM modes.

_____________________________________ Example 9.1: Design a resonator tuned to the resonant frequency 15 GHz for the mode TE101.The resonator is formed by the segment of the rectangular waveguide with the dimensions a =

30 mm, b = 15 mm. The resonator is filled with air.

The dominant mode of the rectangular waveguide has the simplest field distribution,so we design the resonator for this mode. Thus for mode TE 101 we have the resonant

frequency (9.5)

220101

11

2 d a

c f += ,

where c is the speed of light in a vacuum. From the above formula we can determine the

necessary length d

6.10

12

1

2

2

0101

=

−

=

ac f

d mm.

The resonator must be 10.6 mm long.

___________________________________

In the case of a cylindrical resonator with radius r o we have to distinguish between

the TE and TM modes, as their transversal propagation constants are different. For TE

modes, k pmn is determined by (7.74), and the resonant frequency is



book - 9

119

22'

00

02

1

+

=

d

p

r f

o

mn

r

mnp

π α

ε ε µ π . (9.6)

For TM modes, k pmn is determined by (7.68), and the resonant frequency is

22

00

02

1

+

=

d

p

r f

o

mn

r

mnp

π α

ε ε µ π . (9.7)

The dominant mode of the cylindrical waveguide is the TE11 mode, so the resonators are

mostly designed to work with this mode, where 841.1'11 =α .

_____________________________________

Example 9.2: Design a cylindrical resonator tuned to the resonant frequency 14 GHz for the

mode TE111. The resonator is filled with air and its radius is r o = 8.2 mm.

Applying the same procedure as in example 9.1 we get the resonator length

67.16

22

'11

2

0101

=

−

=

or c

f

d

α π

π mm.

The resonator must be 16.67 mm long.

___________________________________

A similar procedure can be applied to determine the resonant frequency of a dielectric

resonator of either the circular or rectangular cross section terminated by conducting planes.

The conducting planes must exceed the cross section of the resonator, as the field is notconfined to the dielectric, but it penetrates into the surrounding air. The resonant frequency isthen determined by (9.4), where the proper value of the transversal propagation constant must

be used.

Cavity resonators are used in microwave technology at frequencies above about 5

GHz. At lower frequencies the transversal dimensions of the waveguide, a segment of which

forms the resonator, are too big. For this reason, these resonators become impractical at such

frequencies. We can use either dielectric resonators or ferrite resonators. Another possibility

is to form the resonator from a TEM transmission line. This line can be left either short

circuited or open at its ends. A short circuited termination is more convenient, as it does not

radiate. A TEM transmission line has k pmn = 0, consequently the resonant frequency is

determined directly by the number of periods of the standing wave along the line segment. So

we have from (9.4)

r

pd

p f

ε ε µ 00

02

= . (9.8)

In the case of a real resonator with some losses we define the quality factor of the

resonator. The internal quality factor determined by the losses of the resonator itself is defined

by



book - 9

120

L P

W Q 0ω = , (9.9)

where ω 0 is the resonant frequency, W is the averaged value of the total energy of the

electromagnetic field inside the cavity, P L is the power lost in the resonator. The losses in the

dielectric material filling the cavity can usually be neglected, namely in the case of air.

Consequently the power is lost in the cavity walls. The stored energy is

∫∫∫==+=

V

eavmaveav dV W W W W 2

22 E

ε . (9.10)

The power lost in the cavity walls can be calculated from the known distribution of the

magnetic field on these walls using (3.26).

Dielectric resonators are used namely in microwave integrated systems. They are

significantly smaller than cavity resonators, which is their main advantage. These resonators

are usually designed as cylinders made of a dielectric material with high permittivity. This

significantly reduces the dimensions. The disadvantage is that the field penetrates into thesurrounding space, which may increase the losses and reduce the quality factor. On the other

hand the electromagnetic field can be simply coupled into such a dielectric resonator by the

proximity effect. The resonator is placed in the vicinity of a microstrip line and the field is

directly coupled into its volume. Due to the field penetration outside the dielectric resonator,

its resonant frequency can in most cases be determined by solving the field distribution

numerically. The resonant frequency can be approximately determined by (9.4), as mentioned

above, when the dielectric resonator is terminated by conducting planes.

Ferrite resonators are produced from mono-crystals of yttrium iron garnet (YIG) in

the shape of small spheres. The resonant frequency of these resonators depends on the

stationary magnetic field H0 applied to this material. It is, as will be shown in chapter 11,

00 H γ ω = , (9.11)

where γ is gyromagnetic ratio, see Paragraph 11.1. These resonators can be simply tuned by

changing the applied stationary magnetic field.

9.2 Problems9.1 Calculate the resonant frequency of the cavity resonator formed by a segment of the

waveguide with a rectangular cross section with the dimensions a = 25 mm, b = 15 mm, d =

10 mm for the modes TE101 and TE111. The cavity is filled with air. f 0101 = 18.14 GHz

f 0111 = 19 GHz

9.2 Calculate the necessary length d of a cavity resonator of rectangular shape to get the

resonator tuned to 10 GHz for the mode TE101. The cavity is filled with air. The transversal

dimensions are a = 40 mm, b = 15 mm.

d = 16.2 mm

9.3 Calculate the resonant frequency of a resonator with a circular cross section filled with air

for the mode TE111. The radius of the cavity is r o = 8.2 mm, the length is d = 25 mm.

f 0111 = 12.27 GHz



book - 9

121

9.4 Calculate the resonant frequency and the quality factor of a resonator with a rectangular

cross section filled with lossless air for the TE101 mode. The dimensions of the cavity are a =

22.86 mm, b = 10.16 mm, d = 22.86 mm. The wall conductivity is σ = 5.7 . 107 S/m.

f 0101 = 9.28 GHz

Q = 7758

10. RADIATION

Radiation is usually understood as the transmission of electromagnetic waves in a free

space from a source of finite dimensions supplying finite power. We assume an infinite space

filled by a homogeneous and isotropic dielectric material without any losses. We have already

studied the propagation of waves in a free space, or along transmission lines. We did not take

into account the source of the electromagnetic field. For example, a plane electromagnetic

wave with surfaces of constant amplitude and phase in the shape of an infinite plane must be

excited by a hypothetical source with infinite dimensions that must radiate infinite power. In

the case of transmission lines, we just studied the field in the view of possible waves or modes, known as eigen modes, which can propagate along this line. We did not investigate

whether these modes would really propagate, assuming that the line were connected to a real

source supplying finite power.

Radiation is used for the transmission of signals namely in telecommunications,

radars, sensors and navigation. Radiation also occurs as a parasitic effect that represents

losses, cross-talk and parasitic interference. For example, radiation from open transmission

lines and electric circuits.

In the first section we showed that electric current and electric charges are sources of

an electromagnetic field. However, electric current is the flow of the charge. These two

quantities are coupled by a continuity equation (1.8). This means that we can consider only

the electric current as the source. The real source is therefore a conducting body of finitedimensions, and the electric current passes through its volume or at high frequencies only

along its surface. From the known distribution of this current we can calculate the field

distribution by solving the wave equation. The wave equations written directly for field

vectors (1.41) and (1.42) have rather complicated functions of the source current and charge.

Therefore they are not suitable for solving our problem. To remove this inconvenience we

have introduced vector and scalar potentials, which represent the electric and magnetic fields

by (1.46) and (1.52). Inserting these formulas into Maxwell’s equations we have derived wave

equations for potentials with simple functions of the source current and charge (1.54) and

(1.55). Their solution is in the form of the superposition of spherical waves (1.57) and (1.58).

In this textbook we will calculate the electromagnetic field distribution only in the

cases of simple sources – antennas, where we know in advance the distribution of the electric

current. First we will treat simple elementary electric and magnetic dipoles. An electric dipole

is represented by a very short straight conductor. The magnetic dipole is represented by a

conducting loop of infinitely small dimensions. We will show how to calculate the field

radiated by sources of dimensions comparable with the wavelength. We will define the

parameters of antennas. Finally, we will mention antenna arrays and receiving antennas.

10.1 Elementary electric dipoleAn elementary electric dipole consists of two point electric charges q and -q placed at

distance d . These charges harmonically change their values. The dipole moment is



book - 10

122

ω j

d I Qd p == , (10.1)

as the charge is an integral of the electric current. So the dipole

can be represented by a straight conductor with passing current

I , Fig. 10.1. The conductor must be much shorter than the

wavelength, as the amplitude and phase of the current areconstant along the conductor. The cross-section of the

conductor is negligible. Then the vector potential is simply

calculated using (1.58)

V d r

V

jkr

∫∫∫−

=e

40

JzA

π

µ ,

Current density J has according to Fig. 10.1 the direction determined by unit vector z0, and is

constant within the conductor volume. The integral can be omitted, as J is constant within the

volume of the dipole and we get

r d I

jkr −

=e

40

π

µ zA . (10.2)

The magnetic field is calculated from the known distribution of the vector potential using

(1.46) and (1.12), assuming that the space is filled with a non-conducting material. First we

transform the vector potential from the Cartesian coordinate system to the spherical

coordinate system. Using Fig. 10.2 and using the fact that the field does not depend on angle

α we get

( ) ( ) 0000 sincos ϑ ϑ ϑ ϑ ϑ z z r A A A A +=+= rrA . (10.3)

The magnetic field can be obtained using (1.46), taking into account

that Aϕ = 0 and 0/ =∂∂ , and using the expression of the rotation

operator in the spherical coordinate system (13.89)

( ) ( )

∂

∂−

∂

∂==

ϑ µ µ ϑ

r ArA

r r

0rot1 α

AH . (10.4)

Inserting (10.3) into (10.4) we get

( )( ) 02

2

sine1

4αH ϑ

π jkr

kr kr

j Id k −

+= . (10.5)

The electric field follows from Maxwell’s first equation (1.12) using (13.89)

( )( )

( )[ ] ( )

∂

∂−

∂

∂== α α

ϑ ϑ

ϑ ϑ ωε ωε rH

r r H

r j j

00 sinsin

1rot

1 rHE . (10.6)

Inserting (10.5) into (10.6) we get the components of the electric field vector

d

dS

r

ϑ

J

A

Fig. 10.1

ϑ

ϑ

r

z ϑ

Fig. 10.2



book - 10

123

( ) ( )

( )ϑ λ

ωµ cose

132

jkr r

kr kr

j Id j E −

+−= , (10.7)

( ) ( )( )ϑ

λ

ωµ ϑ sine

11

2 32

jkr

kr kr

j

kr

Id j E −

−−= , (10.8)

The vector of the magnetic field (10.5) has only one component

( )( )ϑ

π α sine

1

4 2

2 jkr

kr kr

j Id k H −

+= . (10.9)

The electric field has two components, E r and E ϑ , and the magnetic field has only one

component H α . This is natural, as the lines of vector H must be of the shape of a circle drawn

around the conductor carrying an electric current.

Let us now discuss the character of formulas (10.7) to (10.9). The first fraction of

these formulas characterizes the source by its current and length, and the frequency and

material filling the space. The second term in rectangular brackets defines the dependence of

the radiated field on the distance in the form of the normalized distance r/ λ as k = 2π /λ . The

third term determines the change of the field phase with distance. The last term describes the

dependence of the radiated field on angle ϑ .The dependence on distance contains three

powers of product kr or relative distance r /λ :

( ) ( ) ( )32

111

kr and

kr kr . These functions are

plotted in Fig. 10.3. They differ by the speed of

their increase when the argument tends to zero

and by the speed of their decrease when the

argument increases to infinity. Consequently, we

can distinguish three areas around the dipole at

which the field has a particular behaviour. The

well known near field zone, which is very close

to the dipole, is defined by condition r /λ << 1.

Here field components with the term 1/(kr )3

prevail. This term is present in the formulas describing the electric field (10.7) and (10.8).

Assuming r /λ << 1, we can simplify (10.7) and (10.8) to

( )( )ϑ

λ

ωµ cos

13

kr j

Id E r = ,

(10.10)

( )( )ϑ

λ

ωµ ϑ sin

1

2 3kr j

Id E = .

Inserting I = jω q into (10.10) we get the well-known distribution of the electrostatic fieldexcited by a dipole. For this reason, this zone is called the static zone. The magnetic field is

of negligible value in this zone.

0

2

4

6

8

0 1 2 3

kr

Fig. 10.3

( )kr

1

( )2

1

kr

( )3

1

kr



book - 10

124

At slightly longer distances, term 1/(kr )2 starts to have an important value, and the

magnetic field must be taken into consideration. In this area it can be expressed as

( )( )

( )202

2

4

sinsine

1

4 r

Id

kr

Id k H jkr

π

ϑ ϑ

π α ≈= − , (10.11)

which, according to the Biot-Savart law, represents the magnetic field excited by the elementof a conductor with electric current I , d in length. This area is known as the inductive zone.

The most important zone is the far-field zone. This zone is defined by r >> λ . Only

terms 1/(kr ) are of importance in the formulas describing the field distribution. The field

contains only two components

( )ϑ ε

µ

λ ϑ sin

e

2 r Id

j E

jkr −

= , (10.12)

( )ϑ λ α sin

e

2 r Id

j

H

jkr −

= . (10.13)

These components of the electric and magnetic fields are in phase, are mutually

perpendicular, and are perpendicular to the direction of propagation, which is determined by

unit vector r0. Their ratio

w Z H

E ==

ε

µ

α

ϑ .

equals the wave impedance of the free space. Thus the field has the character of a TEM plane

wave propagating in the free space. In fact the field is represented by a spherical wave

modified in its direction of propagation by function sin(ϑ ). Only components E ϑ and H α

participate in the transmission of power from the dipole. For this reason, the area is known as

the radiation zone.

Function sin(ϑ ) determines the radiation pattern of the dipole. The radiation pattern

is generally a function dependent on two angles, ϑ and α , and is defined by

( )( )

max

,,,

ϑ

ϑ α ϑ α ϑ

E

const r E F

== , (10.14)

The radiation pattern is usually plotted as its cuts in

characteristic planes. For our

dipole ( ) ( )ϑ α ϑ sin, = F , and

this radiation pattern is shown

in Fig. 10.4. The radiation

pattern is omnidirectional in the

horizontal plane ϑ = 90°.The power radiated by

the dipole to the whole space is

calculated using Poynting’svector, calculated in the far field zone

α = 0 ϑ ϑ = 90°

Fig. 10.4



book - 10

125

( ) ( ) ( ) ( )

( ) 02

22

2

00**

sin8

sine2

sine2

Re2

1Re

2

1Re

2

1

2

r

rrHES

ϑ λ ε

µ

ϑ λ

ϑ ε

µ

λ α ϑ

r

d I

r

d I j

r

d I j H E jkr jkr

av

=

=

−==×= −−

(10.15)

The total power is calculated integrating (10.15) over the surface of a sphere with radius r

much greater than the wavelength. An element of this surface in the spherical coordinate

system is dS = r 2sin(ϑ )d α d ϑ

( )( ) ( )

3

4

4sin

4sin

sin

8 2

22

0

3

2

222

2

0 02

2

2

22

π λ ε

µ ϑ ϑ π

λ ε

µ α ϑ ϑ

ϑ

λ ε

µ π ππ

d I d

d I d d r

r

d I P === ∫∫ ∫

2

2

2

222

2

22 4040

3

===

λ

π

ε λ ε

π

λ

π

ε

µ d I

d I d I P

r r

. (10.16)

The dipole radiating this power represents a load for the source feeding it. So we can

substitute the dipole by a radiating resistor with resistance Rr . This is a resistor in which the

same power is consumed. The power is

2

2

1 I R P r = ,

so we get

2

2

2

2

280

3

22

===

λ π

λ

π

ε

µ d d

I

P Rr . (10.17)

As d << λ , the value of Rr is very small and it is practically impossible to match this

elementary dipole to any generator.

____________________________________

Example 10.1: Calculate the power radiated by a dipole radiating into a space filled with air,

and also the radiation resistor of this dipole. The dipole is d = 2 m in length. The passing

current has amplitude 3 A and frequency 1 MHz.

The radiating power is

16.04040

2

2

2

2 =

=

=

c

df I

d I P

r r

π

ε λ

π

ε W

From the radiated power and the passing current we get

Ω== 035.02

2 I

P Rr



book - 10

126

____________________________________

Example 10.2: What power must be radiated by the electric dipole to get the value of an

electric field E m = 25 mV/m at the distance from the antenna r = 100 km determined by angle

ϑ = 90°. The antenna is much shorter than the wavelength, which is 450 m. The space is

filled with air.

We are at a distance r=105 >> λ = 450, so we are in the far field zone. The value of the

electric field is here, assuming ϑ = 90°,

r Id E E m

1

2

1

0

0

ε

µ

λ ϑ == ,

and from this formula we get the product Id characterizing the antenna

58642

0

0

==

ε

µ

λ r E Id m Am .

The radiated power is (10.16)

5.6940

2

=

=

λ

π Id P kW.

________________________________________

10.2 Elementary magnetic dipole

An elementary magnetic dipole is represented by acurrent loop with radius a and passing current I . The magnetic

dipole is shown in Fig. 10.5. The radius of this loop is much

lower than the wavelength, and consequently we can assume

that the current has a constant amplitude and phase along the

loop. The moment of this dipole is

Sm d I = , (10.18)

where 2adS π = . The excited field has rotational symmetry, so we can look for the field

distribution only in the plane xz . The current densities in the loop at any position x can be

decomposed into two components. The x components have opposite directions, so the field

excited by them vanishes. The radiated field is thus excited by only the y components of the

current density, Fig. 10.6a. Using (1.58), we get the vector potential describing the field

excited at point P, Fig. 10.6b, in the form

( )( )

Q

V QP

jkr

V d r

Q P

QP

∫∫∫−

=e

4

JA

π

µ

I d S

a

Fig. 10.5



book - 10

127

where α d aS dV cQ = , Sc is the cross-section of the loop conductor. The vector potential has

the direction of the exciting current. This is the y component in the plane xz , but in the whole

space it is the α component. Taking into account only the y component J y = J cos(α) of the

current density and assuming that the integral of J over cross-section S c equals current I ,

which is constant along the loop, we get

( ) ( ) ( )

( ) ( )∫∫

∫∫∫∫ ∫∫

−−

−−

==

===

π π

π π

α

α α

π

µ α α

π

µ

α α π

µ α α

π

µ

0

2

0

2

0

2

0

cose

2cos

e

4

cose

4cos

e

4

d r

I ad r

I a

dS J d r

ad adS J r

P A

QP

jkr

QP

jkr

S

c

QP

jkr

S

c

QP

jkr

QP QP

c

QP

c

QP

(10.19)

The result of the particular integration in (10.19) is precise. The final integral cannot be

simply solved, this can be done only numerically. To get an analytical result, we have to

simplify the expression for distance r QP , Fig. 10.6b. We assume r >> a.

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

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

−=−=−≈

≈−+=+−+=

=+−+=+=

α ϑ α ϑ α ϑ

α ϑ ϑ α ϑ ϑ

ϑ α

cossin1cossin21cossin2

cossin2coscossin2sin

coscos2

2

2222222

22222

r

ar

r

ar ar r

ar r ar ar r a

r OX aOX a XP QX r QP

( ) ( )α ϑ cossinar r QP −= . (10.20)

Assuming that ( ) ( )[ ] 1cossincos ≈α ϑ ka and ( ) ( )[ ] ( ) ( )α ϑ α ϑ cossincossinsin kaka ≈ as a << λ

we can simplify

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

( ) ( )[ ]α ϑ

α ϑ α ϑ α ϑ

cossin1e

cossinsincossincoseeee cossin

jka

ka jka

jkr

jkr jka jkr jkr QP

+=

≈+==

−

−−−

,

J

α

α

J y = J cos(α)Q

Q J x = J sin(α)

J y

J x

r

P

ϑ

r cos(ϑ )

r sin(ϑ )

r QP

Fig. 10.6

a b

a

acos( ) X

O



book - 10

128

( ) ( )( ) ( ) ( ) ( )α ϑ α ϑ

α ϑ

cossin1

cossin11

cossin1

112r

a

r r

a

r

r

ar

r QP

+=

+≈

−

= .

Now we can simply express (10.19) as

( ) ( )

( ) ( )[ ] ( ) ( ) ( )∫

∫

++=

==

−

−

π

π

α

α α α ϑ α ϑ π

µ

α α π

µ

02

0

coscossin1

cossin1e2

cose

2

d r

a

r jka I a

d r

I a P A

jkr

QP

jkr QP

After some manipulations and after integration we get

( ) ( )ϑ µ

α sine1

42

2 jkr

r r

jk I a P A −

+= . (10.21)

From this distribution of the vector potential we calculate the magnetic field using

( ) µ AH rot= , and from the magnetic field we calculate the electric field using

( ) ( )ωε jHE rot= . The procedure is the same as that applied in the case of the electric dipole.

Operator rot must be used in the spherical coordinate system (13.89). As a result we obtain

two components of the magnetic field H r and H ϑ and one component of the electric field E α

( )

( ) ( )( )ϑ

λ

π cose

132

2 jkr

r kaka

j I ka H −

+= , (10.22)

( )

( ) ( )( )ϑ

λ

π ϑ sine

11

2 32

2 jkr

kr kr

j

kr

I ka H −

−−−= , (10.23)

( )

( )( )ϑ

ωµ α sine

1

4 2

2 jkr

kr

j

kr

I ka E −

−= . (10.24)

Comparing these formulas with (10.7), (10.8) and (10.9) we can see that the field

excited by the current loop is dual to the field excited by the electric dipole. For this reason

we will not repeat here the discussion from the previous section. It is obvious that, due to this

duality, the field excited by the magnetic dipole can be obtained from the formulas (10.7),

(10.8) and (10.9), using the transformation: H is exchanged by – E, E is exchanged by H, ε isexchanged by µ , and the dipole moment p = Id /( jω ) is exchanged by µ m = µπ a2 I .

_________________________________

Example 10.3: Compare the power radiated by the electric dipole to the power radiated by

the magnetic dipole. Assume the same current and the same length of the conductor. This

means that the length of electric dipole d equals the perimeter of the current loop, i.e., d =

2π a.

The power radiated by the electric dipole is determined by (10.16). The power radiated by the magnetic dipole can be calculated in the same way that (10.16) was derived. Another



book - 10

129

way is to use the above mentioned duality and transformation. To use this transformation we

have to modify (10.16) to

3

42

2

22

123 c

pd I P e

πε

ω

λ

π

ε

µ == .

Inserting here µ m instead of p, µ instead of ε and using d = 2π a we get the power radiated bythe magnetic dipole

( )3

2224

3

224

1212 c

a I

c

m P m

π

µ π ω

πµ

µ ω == .

Now we can express the ratio of these powers

2

4224

3

3

222 12

12

==

aa I

c

c

d I

P

P

m

e

π

λ

µ π ω

π

πε

ω .

As λ >> a, the power radiated by an electric dipole is much greater than the power radiated bya magnetic dipole. For this reason, only an electric dipole is used as a transmitting antenna.

The radiation of the magnetic dipole has very low efficiency. This antenna is used only rarely

and only as a receiving antenna.

_______________________________________

10.3 Radiation of sources with dimensions comparable with the wavelengthIn previous sections we calculated the field radiated by electric and magnetic dipoles.

We assumed that these dipoles have dimensions much smaller than the wavelength, and the

current density is therefore constant within the volume of the dipole. This assumption

simplified the calculation of the integral in (1.58). In the case of sources with dimensions

comparable with the wavelength we cannot neglect the current variation, and we have to

consider that distance R in (1.58) varies within the volume of the source and is different from

the radius vector, which defines the position of the point at which the field is calculated.

Practical antenna problems are very complicated, as we do not know the current distribution

in advance. This distribution however depends on the radiated field via the boundary

conditions valid on the antenna surface. The radiated field is determined by the current

distribution. This vicious circle must be overcome by a proper numerical method, whichenables us to calculate the unknown current distribution applying the proper boundary

conditions. Such a problem is however beyond the scope of this textbook.

Let us turn our attention to the calculation of the radiated field in the far field zone,

and let us assume the known distribution of a current passing our source. We are looking for

the field at point P in the far field zone determined by radius vector r, Fig.10.7. The source

has volume V . Let us put an element of the current passing the source volume at point Q

defined by radius vector rQ, Fig. 10.7. The dimensions of the source are comparable with the

wavelength. Assuming that point P is in the far field zone, vectors r and rQP can be

considered to be parallel. Thus we can express the distance between points Q and P as



book - 10

130

( )α cosQQP r r r −≈ ,

(10.25)

and we can assume an even

coarser approximation in the

denominator of the expression

for a vector potential

r r QP

11≈ .

Using (1.58) we get the vector

potential at point P

( )

( )

( )

V d r V d r V

jkr jkr

V QP

jkr

QQP

∫∫∫∫∫∫

−−

≈=α

π

µ

π

µ cos

eQ

e

4

eQ

4 J

J

A,

Let us denote a radiation vector

( )( )

V d

V

jkr Q

∫∫∫=α cos

eQJN , (10.26)

which defines the properties of the source. Then the vector potential is

NAr

jkr −

=e

4π

µ . (10.27)

The vector of the electric field is calculated from the vector and scalar potentials by (1.52).

The scalar potential is coupled with the vector potential by the Lorentz calibration condition

(1.53). Inserting this condition into (1.52) we get

( )AAE divgradωµε

ω j

j −−= . (10.28)

The vector potential (10.27) is inserted into (10.28), taking into account that radiation vector

N does not depend on r . We get

( ) [ ] LL ++

+−= −

200

1e

4

1

r N N j

r

jkr αE α ϑ ϑ π

µ ω , (10.29)

N ϑ and N α are the components of N in the spherical coordinate system. We are looking for thefield in the far field zone. In this zone only functions with the dependence on distance r of the

order of –1 can be taken into account, and the value of the terms of higher orders can be

neglected. This means that only the first term in (10.29) can be taken. This gives us a

physically reasonable result. The electric field has the form of a spherical wave with a zero

longitudinal component and two nonzero transversal components

Fig. 10.7

P

r

0

rQ

rQP α

r Qcos(α )

Q

J

V



book - 101

131

0=r E ,

ϑ ϑ ϑ λ π

µ ω N

r

Z j N

r j E

jkr jkr −−

−=−=e

2

e

4, (10.30)

α α α

λ π

µ ω N

r

Z j N

r

j E jkr jkr −−

−=−=e

2

e

4

.

As this spherical wave can be locally approximated by a plane TEM wave, we can determine

the magnetic field using the wave impedance of the free space (2.26)

w Z

E H ϑ

α = ,w Z

E H α

ϑ −= , (10.31)

The field in the far field zone has locally the character of a TEM wave with elliptical

polarization, because the particular transversal components E ϑ , and E α , have generally

different amplitudes and phase.

_________________________________

Example 10.4: Calculate the electric field radiated by a

symmetric dipole of finite length, Fig. 10.8, located at

the origin of the coordinate system. Current passing

dipole conductors is in the form of a standing wave.

Let us first calculate the radiation vector, which

has the direction of the passing current. Using (10.26)

we get

( )( ) ( )

∫∫ ∫∫−−

== L

L

jkz L

L S

jkr

z dz z I dSdz J Q ϑ α cos0

cos

0 ee zzN

The dipole is represented by a transmission line with an

open end termination at both ends. So the current must

fulfill here the boundary condition I (- L) = I ( L) = 0, which gives a current distribution in the

form of a standing wave

( )[ ] z Lk I I m −= sin .

The ϑ component of the radiation vector is N ϑ = - N z sin(ϑ). The electric field has the only

component

( )

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

−++=

==−=

∫ ∫−

−

−−

0

0

coscos sinsinsine

2

sine

2

e

2

L

L jkz jkz

jkr

z

jkr jkr

dz e z Lk dz e z Lk r

Z j

N r

Z j N

r

Z j E

ϑ ϑ

ϑ ϑ

ϑ λ

ϑ λ λ

After some rearrangements, this formula reads

I ( z )

L

L

r

r QP P

Q

Fig. 10.8

ϑ



book - 101

132

( )[ ] ( )

( )ϑ

ϑ ϑ

sin

coscoscose60

kLkL

r I j E

jkr

m

−=

−

. (10.32)

The dipole radiation pattern is

( )

( )[ ] ( )

( )ϑ

ϑ

α ϑ sin

coscoscos

,

kLkL

F

−

= . (10.33)

For a very short dipole ( L << λ ) ( ) ( )2

2

11cos kLkL −≈ , ( )[ ] ( ) ( )ϑ ϑ 22

cos2

11coscos kLkL −≈ ,

and the radiation pattern is described by the simplified formula

( )( )

( )ϑ α ϑ sin2

,2

kL F = ,

which is the radiation pattern of the elemental electric dipole. Fig. 10.9 shows the radiation

patterns of dipoles with particular lengths: 4/,0 λ =→ L L (kL = π /2), L = λ /2 (kL = π ), L =

λ 3/4 (kL = π 3/2), . L = λ (kL = 2π ), L = 2λ (kL = 4π ). Increasing further the dipole length we

finally get the radiation pattern with two main beams directed along the conductors. This tells

us that the power is transmitted only along the conductors.

_______________________________________

L → 0 L = λ /4 L = λ /2

L = λ 3/4 L = λ

L = 2λ

Fig. 10.9

ϑ



book - 101

133

The power radiated by our source into the whole space is calculated from Poynting’s

vector

( ) ( ) 0

22

0**

02402

1Re

2

1rrrHES ravav S

E E H E H E =

+=−=×=

π

α ϑ ϑ α α ϑ . (10.34)

Integrating (10.34) over the surface of a sphere with radius r >> λ we get the radiated power

( )∫ ∫=

π π

α ϑ ϑ 0

2

0

2 sin d d r S P ravr . (10.35)

Quantity W = S ravr 2 is known as the radiation intensity. It represents the power radiated into

a unit space angle.

10.4 Antenna parametersWe have already spoken about the antenna radiation pattern (10.14). The radiation

pattern determines the dependence of the radiated field on directional angles ϑ and α . It isusually normalized to the maximum of the radiated field. It is generally a complex function.

Most often we use the amplitude radiation pattern, but the phase radiation pattern can be

also applied. The power radiation pattern corresponds to the second power of (10.14). The

amplitude radiation pattern is expressed in absolute units or in dB. Fig. 10.10 shows the

radiation pattern of

an antenna plotted

in dB normalized to

0. It consists of the

main lobe and anumber of side

lobes. We are in

most cases

interested in the

main lobe, at which

the maximum of

power is radiated.

The main lobe is

determined by its

width, usually

defined as the widthmeasured at the

level –3 dB below

the main maximum

of the electric field

amplitude, i.e., at

one half of the maximum power – the full width at half power (FWHP). We are also

interested in the direction of the main lobe. Next we define the side lobe level (SLL),

usually in dB, which determines the separation of the maximum lobe of the side lobes from

the level of the field in the main lobe.

The radiation pattern defines the antenna directivity. The maximum directivity is

defined as the ratio of the maximum of the radiation intensity and its average value

-3 5

-3 0

-2 5

-2 0

-1 5

-1 0

-5

0

- 1 8 0 - 1 5 0 - 1 2 0 - 9 0 -6 0 - 3 0 0 3 0 6 0 9 0 1 2 0 1 5 0 1 8 0

αααα (° )

F

( d B )

ϑ = 0°

Fig. 10.10

FWHP

SLL

-3 dB



book - 101

134

( ) ( )

( ) ( )∫ ∫

∫ ∫∫ ∫

=

=====

π π

π π π π

α ϑ ϑ α ϑ

π

α ϑ ϑ

π

α ϑ ϑ

π

π

0

2

0

2

0

2

0 max0

2

0

2

max22

maxmaxmax

sin,

4

sin

4

sin

4

4

d d F

d d S

S d d S r

S r

P

r S

W

W D

rav

ravrav

rav

r

rav

av

, (10.36)

The directivity in the general direction then is

( )2

max ,α ϑ F D D = . (10.37)

The antenna directivity evaluates only the directional distribution of the energy flow. It does

not define the efficiency of the radiation η . This efficiency is included into an antenna gain.

The antenna gain incorporates the efficiency of the conversion of the power supplied from the

source feeding the antenna P s to radiated power P r . The gain is defined as

maxmaxmax 4

4

D P

W

P

W G

r s

η

η

π

π

=== . (10.38)

Then antenna input impedance is defined as

A

A A

I

U Z = , (10.39)

where U A and I A are the voltage and the current at the antenna terminals. It is usually difficult

to determine the current distribution over the antenna that defines its input current. In many

cases we thus substitute the real part of the antenna input impedance by the radiating

resistance (10.17). It is

( )∫ ∫==

π π

α ϑ ϑ 0

2

02

2

2sin

22d d S

I

r

I

P R r

A A

r r . (10.40)

10.5 Antenna arraysWe have studied up to now only single radiators. The field radiated by these radiators

is calculated as the superposition of the spherical waves coming out from partial elements of

the radiator. These particular waves depend on the local amplitude and phase of the current

passing this element. The possible ways of modifying the antenna radiation pattern and gain

are therefore limited, as we cannot simply change the current distribution over the antenna

body. This is the reason why antenna arrays are used. Such an array consists of a number of

partial radiating elements. Each element has its own feeding source. Now we have huge

possibilities to change the array radiation pattern and its gain. This can be done by changing

the amplitudes and phases of the feeding currents and changing the positions of theseelements. Antenna arrays can be one dimensional, two dimensional or three dimensional.



book - 102

135

Generally, when we increase the number of particular elements we reduce the width of the

radiation pattern.

Let us show the tremendous possibilities of antenna arrays through the following

simple case, Fig. 10.11. There are two dipoles of total length λ /2 located in parallel with the z

axis at distance d symmetrically to the origin, as shown in Fig. 10.11. These dipoles are fed

by different currents I 1 and I 2. The field radiated by a particular dipole was calculated in

Example 10.4 (10.32). The field radiated by the couple of dipoles is the superposition of the

particular fields (10.32), where kL = π /2

( )

( )

+

=+=−−

2

2

1

121

21 ee

sin

cos2

cos

60r

I r

I j E E E jkr jkr

ϑ

ϑ π

ϑ ϑ ϑ ,

assuming that the mutual coupling

between the dipoles is neglected.

Looking for the field in the far field

zone, we can express distances r 1 andr 2, using r as in (10.20)

( ) ( )α ϑ cossin2

1

d r r += ,

( ) ( )α ϑ cossin2

2

d r r −= .

Consequently we get the field

radiated by the couple of dipoles in

the form

( )( ) ( ) ( ) ( )

( ) ( )α ϑ ϑ ϑ α ϑ α ϑ

ϑ ,e

60eee

60 21

cossin2

1

cossin2

11 F r

F j I I r

F j E jkr d

jk d

jk jkr −−−

=

+= ,

(10.41)

where F 1 is the radiation pattern of one dipole and F 2 is a function known as the array factor.

Expression (10.41) can be used to describe the radiated field in the far field zone for any arrayof equal radiators. We get the radiation pattern of such an array in the simple form of a

product

( ) ( ) ( )α ϑ α ϑ α ϑ ,,, 21 F F F = . (10.42)

Let us return to our problem. (10.41) can be simplified assuming currents of the same

amplitude I 0 and symmetric phase

ψ ψ j j e I I e I I −== 0201 , .

Now the array factor is

λ /2 I 1 I 2

d

r

P

r1 r2

α

ϑ

Fig. 10.11

0



book - 102

136

( ) ( ) ( )

−= ψ α ϑ α ϑ cossin

2cos2, 02

kd I F . (10.43)

This function is plotted for several values of d and ψ in Fig. 10.12 in the plane xy, i.e., for

2π ϑ = which gives F 1 = 1. Fig. 10.12 thus shows the radiation pattern of the couple of

dipoles in the plane xy. It is evident that we have really tremendous possibilities in changing

the shape of the array radiation pattern even in the case of two radiators.

______________________________ Example 10.5: Calculate the

radiation pattern of the array of N

dipoles λ /2 in length in the plane

xy. The dipoles are parallel with

the z axis. Their centers are

equidistantly located on the x

axis with separation d = λ /2, Fig.

10.13. The dipoles are fed by

equal current I 0.

The dipoles radiate in the

xy plane homogeneously in all

-10 1 2 m

mkd cos( )

r rm

α

d

Fig. 10.13

d = λ /8 d = λ /4 d = λ 3/8 d = λ /2 d = λ 3/4 d = λ

2ψ = 0°

2ψ = 45°

2ψ = 90°

2ψ = 135°

2ψ = 180°

Fig. 10.12



book - 102

137

directions, so F 1(π /2,α ) = 1. Using Fig. 10.13 and details of calculation from the derivation of

(10.41), we get

( ) ( )α α ϑ 20

1

0

cos0

1

0

0

ee

ee F

r I A

r I A

r I A E

jkr N

m

jkmd jkr N

m m

jkr m −−

=

−−

=

−

=== ∑∑ .

Substituting ( )α cose jkmd q = we can read the expression for the array factor normalized to 1 as

a geometric sequence of complex terms qm

( )

( )

( )

( )

=

=−

−=

−

−=+++=

−

−−

α π

α π

α

α

cos4

sin

cos4

sin1

cos2

1sin

cos2

sin11

1

111

12121

222

1

322

N

N

kd

kd N

N qq

qqq

N q

q

N qqq

N N

F N N

N N N

L

The array factors

computed for N =

1,2,3,4,10 are plotted in

Fig. 10.14. These

functions are normalized

by dividing their values

by N . For N = 1 the

radiation goeshomogeneously in all

directions – one dipole.

Increasing the number of

radiators, we increase the

amplitude of the main

lobe of F 2 at α = 0, andside lobes appear, as was

mentioned above. The

width of the main beam

decreases with the

number of dipoles.

_______________________________________

10.6 Receiving antennasIn the preceding paragraphs we studied the particular sources of an electromagnetic

field which are known as transmitting antennas. Here we show how to calculate the power

received by an antenna. We define the effective length of an antenna and its equivalent circuit.

Let us assume an antenna irradiated by a plane TEM wave, Fig. 10.15. Impedance Z p

represents the input impedance of a receiver. The wave described by vectors Ei, Hi and by propagation vector k i is incident to the conducting surface of the antenna. Let us assume that

0

0.2

0.4

0.6

0.8

1

1.2

0 30 60 90 120 150 180

αααα (°)

F 2

= 1

= 2

= 3

= 4

= 10

Fig. 10.14



book - 102

138

this wave is not distorted by this incidence. Consequently it does not fulfill the boundary

conditions on the antenna surface. To meet these conditions and to account for the distortion

of the field we have to add the field known as scattered, with vectors Er and Hr . Now the total

field

r i EEE += ,

r i HHH += ,

must fulfill the boundary condition, and it models the

precise distortion of the field by the antenna. Note

that the scattered field must vanish far from the

antenna, as the field here must correspond only to the

incident wave. We will calculate the power received

by the antenna and fed to the input of the receiver as

the total power which crosses closed surface S , fully

covering the antenna. Let us assume that this surface

closely copies the antenna conducting surface. This power is

( ) ( ) ( )[ ]

( )[ ] ( ) ( )∫∫∫∫∫∫

∫∫∫∫

⋅×+⋅×+⋅×+=

=⋅+×+=⋅×=

S

r r

S

r i

S

ir i

S

r ir i

S

d d d

d d P

SHESHESHEE

SHHEESHE

***

***

2

1

2

1

2

1

2

1

2

1

, (10.44)

In the first integral we can reorder vectors in the mixed product and we can get

( ) ( ) ( )[ ] 0*** =⋅+×=⋅×+=⋅×+ dS dS d ir iir iir i HEEnnHEESHEE

due to the fact that ( ) 0=+× r i EEn , as this vector product determines the tangent component

of the total electric field to the conducting surface. The last integral in (10.44) represents the

power radiated by the antenna back in the form of a scattered field. This power can be

represented as

( ) 2*

2

1

2

1 A A

S

r r I Z d =⋅×∫∫ SHE ,

where Z A is the antenna input impedance and I A is the antenna input current. The second

integral in (10.44) can be rewritten

( ) ( )

**0

****

2

1

2

1

2

1

2

1

2

1

2

1

Aeff i

d

Ai

S

Ai

S

r i

S

r i

S

r i

I dl i

dS dS dS d

dEiE

K EHnEnHESHE

⋅−=⋅−=

=⋅−−=×⋅−=⋅×=⋅×

∫

∫∫∫∫∫∫∫∫,

Z p

S

Ei

Hi

k i

n

Fig. 10.15



book - 102

139

where i0 is a unit vector in the direction along the antenna, I A is the current passing along the

antenna, d is its length and deff represents the antenna effective length. This quantity is

defined as the length which, multiplied by the value of the incident electric field, offers the

internal voltage received by the antenna. The received power is

*2*

2

1

2

1

2

1 Aeff i A A A A I Z I I U P dE ⋅−== , (10.45)

and the voltage induced on the antenna terminals is

eff i A A A I Z U dE ⋅−= . (10.46)

(10.45) and (10.46) define the receiving antenna equivalent circuit, Fig. 10.16. The input

impedance of antenna Z A represents the back radiation which in the receiving antenna is

considered as loss.

Similarly as the antenna effective length we can define the antenna effective

aperture by

eff av R AS P = , (10.46)

where P R is the power supplied by the antenna

to its load, and S av is the magnitude of

Poynting’s vector of the incident plane

electromagnetic wave.

_____________________________________

Example 10.6: Calculate the effective aperture

of the elementary electric dipole d in length,located in air. Assume that the dipole is

matched to the load, Fig. 10.16.

According to Fig. 10.16, the current supplied by the receiving antenna to the input

impedance of the receiver and the corresponding power are

p A

A Z Z

U I

+= , r A R R I P

2

2

1= ,

where Rr is radiating resistance of the dipole (10.17), which represents the real part of antenna

impedance Z A, and U = E incd is the voltage received by the antenna. Assuming the impedancematching *

A p Z Z = and the current is

r

A R

U I

2= .

Then the received power is

r

inc

r

R R

d E

R

U P

842

1222

== .

Z A

Z peff i dE ⋅U A

I A

Fig. 10.16



book - 102

140

From (10.46), inserting for the received power, for Poynting’s vector π 2402

2

0

2incinc

av

E

Z

E S == and

for the radiating resistance (10.17), we get the effective aperture

2

2

2

2

2

22

8

3

80

30240

8

λ

λ π

π π =

==

d

d

E R

d E A

incr

inceff .

Using (10.36) we can calculate the directivity of the elementary electric dipole

( ) ( ) ( )2

3

sin2

4

sin,

4

0

3

0

2

0

2max ===

∫∫ ∫π π π

ϑ ϑ π

π

α ϑ ϑ α ϑ

π

d d d F

D .

Combining this result for the directivity with the formula for the effective aperture we get

π

λ

4

2

D Aeff = .

Although this formula has been derived for the elementary electric dipole, it can be applied

for the wide class of antennas.

_________________________________________

10.7 Problems10.1 A short conductor of length l radiates an electromagnetic wave at the frequency f = 100kHz, Fig. 10.17. At the distance r 1 = 1000 m

from this antenna we can measure the electric

field amplitude E m = 10 V/m. There is a

rectangular conducting loop with the surface S =

0.5 m2 located at the distance r 2 = 1200 m from

the antenna. Calculate the angle δ at which the

loop receives the maximum voltage, and also the

value of this voltage.

δ = 0° U = 0.0105 V

10.2 Calculate the average value of the power density radiated by an electric dipole at the

distance r = 20 km in the directions determined by ϑ = 0°, 45°, 90°, 135°, 180°. The

amplitude of the passing current is I m = 50 A, the frequency is f = 500 kHz.

ϑ = 0°, 180° 0=avS W/m2

ϑ = 45°, 135° 81004.2 −⋅=avS W/m2

ϑ = 90° 81008.4 −⋅=avS W/m2

10.3 A conductor 50 km in length conducts a current with the amplitude I m = 100 A. The

frequency is f = 50 Hz. Calculate the power radiated into the whole space.

l I r 1

δ

S

r 2

r 1

I

S

Fig. 10.17



book - 11

141

P = 274 W

10.4 Calculate the power radiated to the whole space by a circular loop of radius a = 5 cm.

The passing current has the amplitude I m = 5 A, and the frequency is f = 500 kHz. Calculate

the radiating resistance of this radiator.

P = 0.22 10-10 W

Rr

= 1.76 10-12 Ω

11. WAVE PROPAGATION IN NON-ISOTROPIC MEDIA

We have studied the propagation of electromagnetic waves in isotropic materials in

previous chapters. The relations between the field vectors in these materials are described by

(1.9), (1.10) and (1.11), where conductivity σ , permeability µ , and permittivity ε are scalar

quantities. In non-isotropic materials these formulas have the form (1.26)

ED ε = , (11.1)

HB µ = . (11.2)

Vectors D, E, and B, H are not parallel (1.27). Tensors of permittivity (1.28) and

permeability can be expressed

=

zz zy zx

yz yy yx

xz xy xx

ε ε ε

ε ε ε

ε ε ε

ε , (11.3)

=

zz zy zx

yz yy yx

xz xy xx

µ µ µ

µ µ µ

µ µ µ

µ . (11.4)

Taking into account the non-isotropy of the materials we are able to clarify a number

of effects occurring in nature. The number of elements applied in microwave technology is

based on the effect of propagation in a non-isotropic material. Such elements are called non-

reciprocal, as a wave propagates through them differently in different directions. Non-

isotropic behaviour of materials can be either natural or induced. Some mono-crystals in

which, e.g., birefringence (double refraction) known from optics, can be observed are

materials with natural non-isotropic behaviour. Plasma (ionized gas) and ferrite material are

examples with induced non-isotropic behaviour. The non-isotropy is induced in these

materials by applying an external magnetic field. Plasma occurs in the upper layers of then

atmosphere, and the non-isotropy is induced by the earth’s magnetic field. We will pay

attention to ferrite materials, e.g., they are widely used in microwave technology in

production of non-reciprocal devices, as isolators, gyrators, hybrids, phase shifters, etc., or

ferrite resonators. Ferrites are mixtures of oxides of ferromagnetic materials (Fe, Ni, Cd, Li,

Mg) sintered at high temperatures. Ferrites must have very low conductivity (10-4 – 10-6 S/m)

to allow an electromagnetic field to propagate through them and to interact with them.



book - 11

142

11.1 Tensor of permeability of a magnetized ferriteAny medium consists of a system of atoms. Each atom consists of a positively charged

core and a number of electrons bearing a negative charge. These electrons move along certain

orbital paths, and at the same time spin around their axes. This movement and rotation of the

charged electrons represents the flow of an electric current. The flowing current excites a

magnetic field which is perpendicular to the plane of the current loop. Consequently we can

define the orbital magnetic moment and the spin magnetic moment of the electron – mo and m sp. The electron has a certain mass. Its movement causes the orbital mechanical

moment Lo and the spin mechanical moment L sp. These moments are coupled by therelations

oom

eLm

20 µ −= , (11.5)

sp sp spm

eLLm γ µ −=−= 0 , (11.6)

where e and m are the charge and the mass of the electron, and me0 µ γ = is known as the

gyromagnetic constant. The total moment of the whole

atom is the sum of these particular moments. The

moments of the core are negligible in comparison with

the moment of the electron. It has been experimentally

proved that the orbital moments are of much lower

value than the spin moments. Therefore we will treat

only the spin moments.

Let us now study the behaviour of an electron

with magnetic spin moment m sp and with mechanical

spin moment L sp located in the external magnetic field parallel to the z axis Ho = H oz0 which is not parallel to

m sp. Under the external field each single magnetic

dipole in the medium rotates. This movement is known

as precession. The end points of vectors m sp and L sp

move along a circle, Fig. 11.1. Torque T affecting the

dipole is

o spo spm

eHLHmT ×−=×= 0 µ .

It is known from mechanics that the change of L sp due to T is

o spdt

d HmT

L×==

sp.

Inserting for L sp from (11.6) we get

o spo spm

e

dt

d HmHm

m×−=×−= γ µ 0

sp. (11.7)

This vector relation can be rewritten into three scalar equations

electron

m sp

L sp

Ho

Fig. 11.1



book - 11

143

0=+ spyo

spxm H

dt

dmγ , (11.8)

0=− spxo

spym H

dt

dmγ , (11.9)

0=dt

dm spz . (11.10)

The solution of (11.8) and (11.9) describes the precession movement of the end point of m sp

( )t mm o spx ω cos∞= , (11.11)

( )t mm o spy ω sin∞= , (11.12)

where oo H γ ω = is known as the frequency of free precession – the Larmor precession

frequency. This frequency is proportional to the magnetizing field. In the case of a ferrite

without losses the precession is not damped, and it would last to infinity. In a real lossy

medium, the end point of m sp moves along a spiral and it finally approaches the direction of

the external magnetic field. Instead of the magnetic dipole we can follow the vector of

magnetization as for a unit volume including N particles with the magnetic moments

organized by the external magnetic field into the same direction, and the magnetization is M

= N m sp. Equation (11.7) now reads

odt

d

HM

M

×−= γ . (11.13)


( )t M M o x ω cos∞= , (11.14)

( )t M M o y ω sin∞= . (11.15)

Let us assume that the ferrite material is exposed to the superposition of a DC

magnetic field Ho and a high frequency field Hm, the amplitude of which is H m << H o. The

directions of these fields are different

t jmo

t jmoc H ω ω ee 0 HzHHH +=+= . (11.16)

The total magnetic field thus has a time varying value and a time varying direction.

Consequently the precession movement of the magnetization does not have a constant

direction but will follow the varying magnetic field, and its frequency will be equal to the

frequency of the varying magnetic field. This is know as forced precession. Similarly as the

magnetic field we can express the magnetization

t jmo

t jmoc M ω ω ee 0 MzMMM +=+= , (11.17)



book - 11

144

where M m << M o. Now we insert the total fields (11.16) and (11.17) into (11.13) and we get

( )( )t j

omt j

om

oot j

mot j

mmt j

omt j

mo

cct j

m

e H e M

eeee

j

ω ω

ω ω ω ω

ω

γ

γ

γ ω

00

2

e

zMHz

HMHMHMHMHM

HMM

×+×−≈

≈×+×+×+×+×−=

=×−=

momom H M j MzHzM ×+×−= 00 γ γ ω .

This vector equation can be expressed using vector components in the following three

equations

myomy M mx M H M j ω ω ω −= ,

mxomx M my M H M j ω ω ω +−= , (11.18)

0=mz M jω ,

where o M M γ ω = is the magnetizing constant. Solving the set of equations (11.18) we get

my

o

M mx

o

M omx H j H M

2222 ω ω

ω ω

ω ω

ω ω

−−

−−= ,

my

o

M omx

o

M mx H H j M

2222 ω ω

ω ω

ω ω

ω ω

−−

−= , (11.19)

0=mz M .

The magnetic induction of the high frequency field is

( )mmm MHB += 0 µ .

Particular components of Bm using (11.19) are read

my

o

M mx

o

M omx H j H B 220220 1 ω ω

ω ω µ ω ω

ω ω µ −−

−−= ,

my

o

M omx

o

M mx H H j B

−−+

−=

220220 1ω ω

ω ω µ

ω ω

ω ω µ , (11.20)

mz mz H B 0 µ = .

Consequently we have the tensor of permeability



book - 11

145

−

=

000

0

0

µ

µ µ

µ µ

µ a

a

j

j

, (11.21)

where

−−=

220 1o

M o

ω ω

ω ω µ µ , (11.22)

220

o

M a

ω ω

ω ω µ µ

−= . (11.23)

It is important to remember that the tensor of permeability (11.21) describes the behaviour of

the ferrite material under magnetization by the field (11.16). The DC field has the direction

parallel to the z axis and the tensor of permeability (11.21) couples together the induction and

the intensity of the high frequency magnetic field

mm HB µ = . (11.24)

We noted that non-isotropic behaviour is also displayed by plasma magnetized by a

DC magnetic field directed parallel to the z axis. In this case we can express the tensor of

permittivity as

−

=

000

0

0

ε

ε ε

ε ε

ε a

a

j

j

, (11.25)

and the vectors of the high frequency field are coupled by

mm ED ε = . (11.26)

_______________________________________

Example 11.1: Calculate magnetic field H o to magnetize a ferrite to magnetic resonance at

the frequency f r = 15 GHz. The mass of an electron is m = 9.109.10-31 kg, the charge of an

electron is e = 1.602.10-19 C.

The gyromagnetic constant is

50 1021.2 ⋅==m

e µ γ m/C .

The frequency of the free precession is

oo H γ ω = => 51026.4 ⋅==γ

ω oo H A/m

_______________________________________



book - 11

146

Example 11.2: Calculate the tensor of permeability of a ferrite material with ω ο = 1.105 1010 s-1 and ω M = 1.547 1010 s-1 (see Problem 11.1) at the frequency f = 300 MHz.

Applying formulas (11.22) and (11.23) we get

0220 44.21 µ ω ω

ω ω µ µ =

−−=

o

M o ,

0220 246.0 µ ω ω

ω ω µ µ =

−=

o

M a ,

and the tensor of permeability (11.21) is

−

=

−

=

100

044.2246.0

0246.044.2

00

0

0

0

0

j

j

j

j

a

a

µ

µ

µ µ

µ µ

µ

_________________________________________

11.2 Longitudinal propagation of a plane electromagnetic wave in a

magnetized ferriteLet us assume an unbound space

filled homogeneously by a ferrite

material considered to be lossless. This

material is magnetized in the z direction

by a static magnetic field Ho = H oz0. A

uniform plane electromagnetic wave propagates in this space in the direction

of the z axis, i.e., in the direction parallel

to the magnetizing field, Fig. 11.2. Both

the electric field and the magnetic field of

this wave do not depend on the x and y

coordinates. Assuming 0=∂

∂

xand

0=∂

∂

ywe can rewrite Maxwell’s first and second equations (1.12) and (1.13) into

x

y E j

H ωε =

∂

∂− , (11.27)

y x E j

H ωε =

∂

∂, (11.28)

z E jωε =0 , (11.29)

( ) ya x

y

H j H j

E

µ µ ω −−=∂

∂

− , (11.30)

Ho

k


Fig. 11.2



book - 11

147

( ) y xa x H H j j

E µ µ ω +−=

∂

∂, (11.31)

z H j 00 ωµ −= , (11.32)

Equations (11.29) and (11.32) tell us that the propagating wave is a transversal

electromagnetic wave, as the longitudinal field components are zero. We are looking for thesolution to above equations in the form of a plane TEM wave

z jk mx x

z H H −= e , (11.33)

z jk my y

z H H −= e , (11.34)

z jk my xy

z jk mx x

z z H Z E E −− == ee , (11.35)

z jk mx yx z jk my y z z H Z E E −− −== ee , (11.36)

where the wave impedances are defined

my

mx xy

H

E Z = , (11.37)

mx

my

yx H

E Z −= , (11.37)

and can be determined together with propagation constant k z by solving equations (11.27) –

(11.32). We insert the form of field (11.33), (11.34), (11.35) and (11.36) into (11.27) –

(11.32)

z jk my xy

z jk mx

z jk my z

z z z H Z j E j H jk −−− −=−= eee ωε ωε ,

my xymy z H Z j H jk ωε −== , (11.38)

z jk mx yx

z jk my

z jk mx z

z z z H Z j E j H jk −−− =−=− eee ωε ωε ,

mx yxmx z H Z j H jk ωε =− . (11.39)

Similarly from (11.30) and (11.31) we get

( )myamx yxmx z H j H Z H k µ µ ω −= , (11.40)

mymxa xymy z H H j Z H k µ µ ω += . (11.41)

From (11.38) and (11.39) we have the wave impedances



book - 11

148

ωε

z yx xy

k Z Z == , (11.42)

From (11.40) and (11.41) we exclude the wave impedance by (11.42). We get the set of two

equations for H mx and H my

( ) 0222 =+− myamx z H j H k µ ε ω µε ω , (11.43)

( ) 0222 =−+− my z mxa H k H j µε ω µ ε ω . (11.44)

Propagation constant k z can be determined from the condition of solubility of the set of the

equations (11.43) and (11.44), which requires that the determinant of the system matrix is

equal to zero

( ) ( ) 022222 =+− a z jk µ ε ω µε ω .

This equation has two solutions defining the propagation constant of a wave propagating

through the space filled by a ferrite material

( )a R zLk µ µ ε ω ±=, . (11.45)

Inserting the propagation constants (11.45) into (11.43) and (11.44) we get

mxLmyL H j H = , (11.46)

mxRmyR H j H −= . (11.47)

The propagating wave can thus be decomposed as the superposition of two partial

electromagnetic waves. These waves can be expressed

( ) z jk mxL L

zL H j −+= e00 yxH , (11.48)

( ) z jk mxR R

zR H j−−= e00 yxH , (11.49)

These two waves are circularly polarized waves. The wave denoted by index L is a circularly

polarized wave with left-handed rotation. The wave denoted by index R is a circularly

polarized wave with right-handed rotation. The wave propagating in the ferrite consequently

behaves as the superposition of the two circularly polarized waves with left-handed and right-

handed polarization. Their propagation constants (11.45) are different, and so are the phase

velocities and the wave impedances

( )a zL

Lk

v µ µ ε

ω

+==

1, (11.50)

( )a zR R k v µ µ ε

ω

−==

1

, (11.51)



book - 11

149

ε

µ µ

ωε

a zL L

k Z

+== , (11.52)

ε

µ µ

ωε

a zR R

k Z

−== . (11.53)

We substitute for the term ( )a µ µ ± , where µ is defined by (11.22) and µ a by (11.23),

the effective permeability µ ef

±±=±=

o

M aef

ω ω

ω µ µ µ µ 10 .

(11.54)

The dependence of

relative effective permeability

0 µ µ ef and of normalized

phase velocity cv on

normalized frequency oω ω is

plotted in Fig. 11.3. There is a

frequency band in which the

permeability of the wave with

right-handed polarization is

negative. The corresponding

phase velocity (11.51) together

with the wave impedance

(11.53) are imaginary in thisfrequency range. This means that

the wave does not propagate.

There is a sharp resonance in the

plots in Fig. 11.3 at ω = ω o. It is

known as ferromagnetic

resonance. At this frequency,

equal to Larmor precession

frequency ω o, vector Hm performs the precession

movement, which is not damped

in the case of a lossless ferrite.

The wave with right-handed

polarization is intensively

attenuated at this resonance. The propagation of the wave with left-handed polarization is not

influenced by the ferromagnetic resonance, as the direction of the magnetic field vector

rotation is opposite to the precession movement. It follows from this that in practical

applications of the ferrite material we have to choose a frequency fairly different from the

Larmor precession frequency ω o. This removes the strong wave attenuation together with the

severe dependence of all wave parameters on frequency. Note that all this concerns the case

of a wave propagating in the direction parallel to DC magnetizing field Ho.

The phase velocities of the partial waves with right-handed and left-handed polarization (11.50) and (11.51) are different. Let us now assume that a plane TEM wave with

-10

-8

-6

-4

-2

0

2

4

6

8

10

0

1

2

3

ω ωω ω /ω ωω ω o

ω ωω ω /ω ωω ω o

1

1

µµµµef /µµµµ0

µ efL

µ efR

v/c

v L

v R

v R

µ efR

Fig. 11.3



book - 11

150

linear polarization propagates in the ferrite material. This wave is the superposition of the two

above defined waves which can be expressed using (11.48) and (11.49)

( ) ( ) mRmL L R H j H j 0000 yxyxHHH −++=+= .

Passing the distance d along the z axis, the two particular waves change their phase

differently, as they have different phase velocities and the wave can consequently beexpressed assuming that the amplitudes are equal H mL = H mR = H mx

( )

( ) ( )[ ]

−−

−=

=−−+=

=+=

−

−−−−

−−

d k k

d k k

H

j H

d

zL zR zL zRd jk mx

d jk d jk d jk d jk mx

d jk R

d jk L

a

zL zR zL zR

zR zL

2sin

2cose2

eeee

ee

00

00

yx

yx

HHH

, (11.55)

where

2

zR zLa

k k k

+= .

Equation (11.55) shows that the x component and the y component of the magnetic field of

the propagating wave stay in phase. This means that the polarization remains linear. The plane

of the polarization is twisted by angle ψ

( )

−== d

k k

H

H zL zR

x

y

2tgtg ψ => d K d

k k F

zL zR =−

=2

ψ , (11.56)

where K F is the Faraday constant.

This effect is known as the Faraday effect. The polarization plane of the wave with

linear polarization propagating in the ferrite material in the direction parallel with the DC

magnetizing field is gradually rotated. The measure of this rotation depends on frequency f ,

magnetizing field H o and the parameters of the ferrite.

________________________________________

Example 11.3: Calculate the Faraday constant determining the rotation of the polarization

plane of the wave propagating in the ferrite material from Example 11.2 in parallel with the

magnetizing field at the frequency 300 MHz. Calculate the distance through which the

polarization plane is rotated by 90°.Using (11.45), we get the propagation constants of the wave with left-handed

polarization and the wave with right-handed polarization

( ) 28.10=+= a zLk µ µ ε ω m-1,

( ) 29.9=−= a zRk µ µ ε ω m-1.

µ and µ a calculated in Example 11.2 were used. Using (11.56) we get



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151

495.02

−=−

= zL zR F

k k K rad/m K F = -28.5 °/m .

The polarization plane is rotated by 90° through the distance 3.16 m.

________________________________________

11.3 Transversal propagation of a plane electromagnetic wave in a

magnetized ferriteLet us assume that a plane

electromagnetic wave propagates in a

ferrite material perpendicular to a DC

magnetizing field Ho = H oz0. Let us say in

the direction of the x axis, Fig. 11.4. Both

the electric field and the magnetic field of

this wave do not depend on the y and z

coordinates. Assuming 0=∂

∂

y and

0=∂

∂

z we can rewrite Maxwell’s first and

second equations (1.12) and (1.13)

similarly as in the case of longitudinal propagation into

0= x E , (11.57)

y z E j

H ωε =

∂

∂− , (11.58)

z

y E j

H ωε =

∂

∂, (11.59)

ya x H j H j µ µ ω −−=0 , (11.60)

( ) y xa z H H j j

E µ µ ω +=

∂

∂, (11.61)

z

y H j

x

E 0 µ ω −=

∂

∂, (11.62)

Assuming dependence on the x coordinate in the form x jk x−

e which describes a plane

electromagnetic wave propagating in the x direction, the equations (11.57) to (11.62) read

0=mx E , (11.63)

mymz x E H k ωε = , (11.64)

Ho

k


Fig. 11.4



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152

mz my x E H k ωε −= , (11.65)

myamx H j H µ µ = , (11.66)

( )mymxamz x H H j E k µ µ ω +=− , (11.67)

mz my x H E k 0 µ ω = , (11.68)

The set of equations (11.63) to (11.68) can be divided into two sets. The equations

(11.64) and (11.68) define the system of E my and H mz . The equations (11.65), (11.66) and

(11.67) define the system of E mz , H mx and H my.

Dividing equations (11.64) and (11.68) we get the propagation constant

ε µ ω 0== xo x k k . (11.69)

Using the propagation constant we can define the phase velocity

ε µ

ω

0

1==

xo

ok

v . (11.70)

The wave impedance is

ε

µ 0==mz

my

o H

E Z . (11.71)

Equations (11.69), (11.70) and (11.71) define the propagation constant, the phase velocity andthe wave impedance of a standard plane TEM wave. This wave is known as an ordinary

wave. The ordinary wave propagates perpendicularly to the DC magnetizing field and its

magnetic field is parallel with Ho. The propagation of the ordinary wave does not depend on

Ho.

The second set of equations describes the wave propagating perpendicular to Ho. It has

the longitudinal component of magnetic field H mx shifted by π /2 to transversal component

H my. Electric field E mz is parallel to Ho. This wave is known as an extraordinary wave. From

(11.65), (11.66) and (11.67) we get the propagation constant, the phase velocity and the wave

impedance of the extraordinary wave

µ

µ µ ε ω

22a

xek −

= , (11.72)

µ

µ µ ε

ω

22

1

amx

ek

v−

== , (11.73)

εµ

µ µ

ωε

22a xe

my

mz

e

k

H

E Z

−==−= . (11.74)



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153

These quantities depend on Ho. At µ = 0 we have ve = 0 and the extraordinary wave does not

propagate. This happens at the frequency

( ) M oot ω ω ω ω += . (11.75)

This effect is transversal ferromagnetic resonance. The extraordinary wave can be

described by the field components

x jk my y

xe H H −= e , (11.76)

x jk my

a x jk mx x

xe xe H j H H −− == ee

µ

µ , (11.77)

x jk mye

x jk mz z

xe xe H Z H E −− −== ee . (11.78)

The plane TEM wave with linear polarization propagating perpendicularly to the DC magnetizing

field can be decomposed into ordinary and

extraordinary waves, Fig. 11.5. These waves have

different phase velocities and therefore they have

different phases. The ratio of the electric fields of

these waves is

( ) ψ j

mz

my xk k j

mz

my

ze

yoe

E

E e

E

E

E

E xo xe == −

.

This shows that the phase between the two

components of the electric field varies with the x

coordinate, and the TEM wave which is the

superposition of the ordinary wave and the

extraordinary wave changes its polarization as the

wave propagates, Fig. 11.6. The wave has linear

polarization at the points at which

( ) π ψ n xk k xo xe =−= . The character of the wave

polarization changes regularly with x from linear

polarization, to elliptical polarization, circular polarization, again to elliptical polarization,

etc., Fig. 11.6.

________________________________________

Example 11.4: A plate TEM wave is incident from the air at the angle ϑ i = 45° to the surface

of the ferrite material from Example 10.2, the magnetizing field is parallel to the ferrite

surface, Fig. 11.7, and the ferrite permittivity is ε r = 10. The frequency is f = 300 MHz. The

wave has arbitrary orientation of vector E to the plane of incidence. Calculate the difference

between refraction angles of an ordinary wave and of an extraordinary wave.

Ho

k xo

Fig. 11.5

H zo

E yo

H ye

E ze

k xe

H xe

ordinary

wave

extra-

ordinary

wave



book - 11

154

The refracted wave propagates in the ferrite perpendicular to the magnetizing field.

The incident wave must be expressed as the superposition of the wave with horizontal

polarization and of the wave with vertical polarization. The wave with horizontal polarization

has vector E perpendicular to the plane of incidence, i.e., parallel with Ho. It represents an extraordinary wave in the

ferrite. The wave with vertical polarization has vector E

parallel to the plane of incidence, i.e., perpendicular to Ho. It

represents an ordinary wave in the ferrite. The propagation

constants of these waves (11.69) (10.72) are

83.190 == ε µ ω ok m-1 ,

7.9

22

=

−

= µ

µ µ

ε ω a

ek m-1

.

From Snell’s law (3.49) we get refraction angles ϑ to = 12.9° and ϑ te = 27°. The difference

between these angles is 14.1°. _________________________________________

11.4 Applications of non-reciprocal devicesThere are devices working with a longitudinally magnetized ferrite which make use of

the Faraday effect, and devices with a transversally magnetized ferrite. The latter have simpler

construction and are more universal.Among nonreciprocal devices, we can mention a gyrator marked in circuits as shown

in Fig. 11.8a. The gyrator

shifts the phase of a

transmitted wave in one

direction by 180° and there

is no phase shift in the

return direction. The

insulator, Fig. 11.8b,

transmits a wave only in

one direction and does not

transmit in the return

Ho

ψ = 0

E

45°

0<ψ <π/2 ψ =π/2 π/2<ψ <π ψ =π π<ψ <3π/2 ψ =3π/2 3π/2<ψ <2π ψ =2π

45° 45°

Fig. 11.6

π

gyrator insulator

1

23

circulator

a b cFig. 11.8

ϑ i

ϑ o

ϑ e

Ho

ferrite

Fig. 10.7



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155

direction. The circulator, Fig. 11.8c, transmits a wave only in the direction of an arrow from

gate 1 to gate 2, from gate 2 to gate 3, and from gate 3 to gate 1. Controlling circuits are

steered by changing the electric current in the winding of an electromagnet magnetizing a

ferrite. Ferrite resonators were mentioned in Chapter 9.

Fig. 11.9 shows a sketch of an insulator with a longitudinally magnetized ferrite

working with the use of the Faraday effect. The wave in the form of the TE 10 mode in a

rectangular waveguide propagates from the left. It is transformed to the TE11 mode of the

waveguide with a circular cross-section. Its electric field, Fig. 11.9, is perpendicular to the

resistive plate and is not attenuated by this plate. This mode travels in the ferrite and its plane

of polarization is rotated by 45°, as its electric field is again perpendicular to the second

resistive plate and is transformed to the TE10 mode of an output rectangular waveguide. The

rotation of the polarization plane of the wave traveling back is in the opposite direction, so the

electric field is parallel to the resistive plate and, moreover, the TE01 mode which cannot

propagate is excited in the output waveguide. The wave is thus highly attenuated.

11.5 Problems

11.1 Calculate the frequency of the ferromagnetic resonance and the magnetizing constant of

a ferrite magnetized by the field H o = 5 104 A/m. The magnetization is M o = 7 104 A/m.

ω ο = 1.105 1010 s-1 f r = 1.76 GHz

ω M = 1.547 1010 s-1

11.2 Recalculate the tensor of permeability of the ferrite material from Example 11.2 at the

frequency 3 GHz.

−

=

100

0266.025.1

025.1266.0

0 j

j

µ µ

Ho

TE10 TE10

TE10TE01

TE11 TE11

TE11 TE11

Fig. 11.9



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156

11.3 Calculate for the ferrite material from Example 11.2 the propagation constants of the

waves with left-handed polarization and with right-handed polarization propagating in parallel

with a magnetizing field at the frequency 3 GHz.

k L = 77.2 m-1

k R = j62 m-1

The wave with right-handed polarization does not

propagate, as k R is a purely imaginary number.

12. APPLICATIONS OF ELECTROMAGNETIC FIELDS

Knowledge of electromagnetic field theory is applied in the analysis and design of

systems in all branches of high frequency technology. These are namely: microwave

technology, antennas, propagation of electromagnetic waves, and optical communications. In

particular paragraphs we will briefly introduce these branches.

All these particular branches contribute to the design of, e.g., the communication link

shown in Fig. 12.1. Microwave technology covers the design of feeders, the output high

frequency part of a transmitter, and the input high frequency part of a receiver. Antennas are

treated usually separately, as their analysis, optimization and design are very specific.

Similarly we treat the propagation of waves in the atmosphere above ground in a special way.

In a communication system antennas can be substituted by proper converters to excite a light

signal in an optical fiber. This problem belongs to optoelectronic technology. Proper

transmission lines, e.g., coaxial cable, were used in older communication links.

Special measurement techniques using special measuring devices and systems are

applied to verify all steps in the development process of all high frequency circuits andsystems. It is beyond the scope of this textbook to treat these techniques. In the following

paragraphs we briefly introduce problems connected with microwave technology, antennas,

propagation of waves and optoelectronic technology.

12.1 Introduction to microwave technologyThe spectrum of electromagnetic waves is shown in Fig. 1.1 together with typical

applications. Up to the frequency of about 1 GHz most circuits are constructed using lumped

parameter circuit components. In the frequency range from 1 up to 100 GHz, lumped circuit

elements are usually replaced by transmission line and waveguide components. Thus the term

microwave technology refers generally to the engineering and design of information handling

transmitter

(generator)receiver

feeder feeder

transmitting

antenna

receiving

antenna

propagating wave

(guiding medium)

Fig. 12.1



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157

systems in the frequency range from 1 to 100 GHz, corresponding to wavelengths as long as

30 cm and as short as 3 mm. At shorter wavelengths we have what can be called optical or

quasi-optical technology since many of the techniques used here are derived from classical

optical techniques. The characteristic feature of microwave technology is the short

wavelengths involved, these being of the same order of magnitude as the circuit elements and

devices employed. Conventional low frequency circuit analysis based on Kirchhoff’s laws

and voltage current concepts cannot be applied. It is necessary instead to carry out the

analysis in terms of a description of electric and magnetic fields.

There is no distinct frequency boundary at which lumped parameter circuit elements

must be replaced by distributed circuit elements. With modern technological processes it is

possible to construct printed circuit inductors and on chip capacitors that are so small that

they retain their lumped parameter characteristics at frequencies as high as 10 GHz or even

higher. Likewise, optical components, such as parabolic reflectors and lenses, are used to

focus waves with wavelengths as long as 1 m or more.

The first development of microwave technology was started with the development of

radars, as the need for high resolution radar capable of detecting small targets is coupled with

the need to raise the frequency. This arises predominantly from the need to have antennas of

sufficiently small dimensions that will radiate essentially all the transmitted power into anarrow pencil-like beam.

In recent years microwave frequencies have also come into widespread use in

communication links. This follows from the demand to increase the amount and the speed of

transmitted information. Satellite and mobile communication systems are based on

microwave technology. At the present time most communication systems use the

transmission of digital signals.

Waveguides periodically loaded with shunt susceptance elements support slow waves

having velocities much less than the velocity of light, and are used in linear accelerators.

These produce high energy beams of charged particles for use in atomic and nuclear research.

The slow traveling electromagnetic waves interact very efficiently with charged-particle

beams having the same velocity, and thereby give up energy to the beam. Another possibilityis for the energy in an electron beam to be given up to an electromagnetic wave, with

resultant amplification. This latter device is the traveling-wave tube.

Sensitive microwave receivers are used in radio astronomy to detect and study the

electromagnetic radiation from the sun and a number of stars that emit radiation in this band.

Microwave radiometers are also used to map atmospheric temperature profiles, moisture

conditions in soils and crops, and for other remote sensing applications.

Molecular, atomic, and nuclear systems exhibit various resonance phenomena under

the action of periodic forces arising from an applied electromagnetic field. Many of these

resonances occur in the microwave range. We have thus a very powerful experimental probe

for studying the basic properties of materials. Out of this research on materials have come

many useful devices employing ferrites, see Chapter 11.The development of the laser, a generator of essentially monochromatic (single-

frequency) light waves, together with semiconductor technology, has stimulated great interest

in the possibilities of developing communication systems at optical wavelengths.

There are plenty of the applications of microwaves in industrial processes –

microwave ovens, drying of materials, manufacturing wood and paper products, material

curing. Microwave radiation has also found some application for medical hyperthermia or

localized heating of tumors.

At frequencies where the wavelength is several orders of magnitude larger than the

greatest dimensions of the circuit or system being examined, conventional circuit elements

such as capacitors, inductors, resistors, transistors or diodes are the basic building blocks for

the information transmitting, receiving, and processing circuits used. The description or



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158

analysis of such circuits may be adequately carried out in terms of loop currents and node

voltages without consideration of propagation effects. The time delay between cause and

effect at different points in these circuits is so small compared with the period of the applied

signal as to be negligible. It might be noted here that an electromagnetic wave propagates a

distance of one wavelength in a time interval equal to one period of a sinusoidally time

varying applied signal. As a consequence, when the distances involved are short compared

with a wavelength, the time delay is not significant. As the frequency is raised to a point

where the wavelength is no longer large compared with the circuit dimensions, propagation

effects can no longer be ignored. A further effect is the great relative increase in the

impedance of the connecting leads, terminals, etc., and the effect of distributed capacitance

and inductance. In addition, currents circulating in unshielded circuits comparable in size

with a wavelength are very effective in radiating electromagnetic waves. The net effect of all

this is to make most conventional low frequency circuit elements and circuits hopelessly

inadequate at microwave frequencies. As is well known, lumped circuit elements do not

behave in the desired manner at high frequencies. For example, a coil of wire may be an

excellent inductor at 1 MHz, but at 50 MHz it may be an equally good capacitor because of

the predominating effect of interturn capacitance. Even though practical low frequency

resistors, inductors and capacitors do not function in the desired manner at microwavefrequencies, this does not mean that such energy dissipating and storage elements cannot be

constructed at microwave frequencies. On the contrary, there are many equivalent inductive

and capacitive devices for use at microwave frequencies. Their geometrical form is quite

different, but they can be and are used for much the same purposes, such as impedance

matching, resonant circuits, etc.

For low power applications, microwave tubes have been completely replaced by

transistors, diodes, and negative resistance diodes. However, for high power applications

microwave tubes are still necessary.

One of the essential requirements in a microwave circuit is the ability to transfer signal

power from one point to another without radiation loss. This requires the transport of

electromagnetic energy in the form of a propagating wave. A variety of such structures have been developed that can guide electromagnetic waves from one point to another without

radiation loss. The simplest guiding structure, from an analysis point of view, is the

transmission line. Several of these, such as the open two-conductor line, the coaxial line, the

parallel plate line, and the strip line, illustrated in Fig. 5.1b, are in common use at the lower

microwave frequencies.

At the higher microwave frequencies hollow-pipe waveguides, as illustrated in Fig.

5.1c, may be used. Today, these waveguides are used predominantly only in special systems,

where high power signals are transmitted, or where extremely low losses are required.

The development of semiconductor high frequency active devices, such as bipolar

transistors, field-effect transistors, and diodes, has had a dramatic impact on the microwave

engineering field. With the availability of microwave transistors, the focus on waveguidesand waveguide components has changed to a focus on planar transmission line structures,

such as microstrip lines and coplanar waveguides. These structures, shown in Fig. 5.1b, can

be manufactured using printed circuit techniques. They are compatible with microwave

semiconductor devices, and can be miniaturized, they are cheap, and are thus suitable for

mass production. Microwave circuits are usually designed as hybrid integrated microwave

circuits. In hybrid circuit construction the transmission lines and transmission line

components, such as matching elements, are manufactured first and then the semiconductor

devices are soldered into place. The current trend is toward the use of monolithic microwave

integrated circuits (MMIC) in which both the transmission line circuits and the active devices

are fabricated on a single chip.



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159

A unique property of the transmission line is that a satisfactory analysis of its

properties may by carried out by treating it as a network with distributed parameters and

solving for the voltage and current waves that may propagate along the line, see Chapter 6. In

the case of planar transmission lines this can be done at lower microwave frequencies only.

At higher frequencies they must be, similarly as hollow-pipe waveguides, treated as

electromagnetic boundary value problems, and a solution for the electromagnetic fields must

be determined, see Chapter 7. The reason is that it is not possible to define a unique voltage

and current that have the same significance as at low frequencies. However using a proper

normalization the graphical means of waveguide analysis, such as the Smith chart, see

Chapter 6.3, are commonly used.

Associated with transmission lines and waveguides are a number of interesting

problems related to methods of exciting fields in guides and methods of coupling energy out.

Signals from generators and measuring devices are usually carried with the use of coaxial

cables. A number of connectors have been produced which are compatible to connecting

planar transmission lines by soldering them directly into their layout, Fig. 12.2. These lines

can be simple coupled using the proximity effect, see the microstrip line coupler in Fig. 12.3

with ports 1 to 4. Three

basic coupling methods

are used in the case of waveguides, Fig. 12.4:

probe coupling, see Fig.

12.4a and Fig. 7.8, loop

coupling, see Fig. 12.4b,

and aperture coupling

between adjacent guides,

see Fig. 12.4c. These coupling devices are actually small antennas that radiate into a

waveguide.

Inductive and capacitive elements take a variety of forms at microwave frequencies.

Perhaps the simplest are short circuited sections of a transmission line and a waveguide, see

Chapter 6.2.3. These exhibit a range of susceptance values from minus to plus infinity,depending on the length of the line, and hence may act as either inductive or capacitive

elements. They may be connected as either series or shunt elements, as illustrated in Fig. 12.5.

They are commonly referred to as stubs and are widely used as impedance matching

elements, see Example 6.8. In a rectangular waveguide thin conducting windows, or

diaphragms, as illustrated in Fig.

12.6, also act as shunt susceptive

elements. Their inductive, Fig. 12.6a,

or capacitive, Fig. 12.6b, nature

depends on whether there is more

magnetic energy or electric energy

stored in the local fringing fields.

Fig. 12.2

21

3 4

Fig. 12.3

a b

aperture

c

Fig. 12.4

a - series stub b – shunt stub

Fig. 12.5



book - 12

160

Inductors and capacitors are designed namely

for microwave integrated circuits in forms resembling

lumped parameter elements. The example is an

interdigital capacitor designed for the coplanar

waveguide illustrated in Fig. 12.7. An example of an

inductor is a spiral inductor designed for a microstrip

line technology, shown in Fig. 12.8.

Resonant circuits are used both at low

frequencies and at microwave frequencies to control the frequency of an oscillator and for

frequency filtering. At low frequencies this function is performed by an inductor and a

capacitor in a series or parallel combination. Resonance occurs when there are equal average

amounts of electric and magnetic energy stored. This energy oscillates back and forth between

the magnetic field around the inductor and the

electric field between the capacitor plates. At

microwave frequencies the LC circuit may be

replaced by a closed conducting enclosure, or

cavity, see Chapter 9.1. The electric and

magnetic energy is stored in the field within thecavity. At an infinite number of the resonant

frequencies (9.1) there are equal average

amounts of electric and magnetic energy stored

in the cavity volume. In the vicinity of any one

resonant frequency, the input impedance to the

cavity has the same properties as for a

conventional LC resonant circuit. One

significant feature worth noting is the very much

larger Q values that may be obtained, these

being often in excess of 104, as compared with

those obtainable from low frequency LC circuits. Dielectric andferrite resonators are mentioned in Chapter 9.1. Resonators can be

formed even using the planar technology of printed transmission

lines. An example is a patch resonator, frequently used as a

microstrip patch antenna, Fig. 12.9. The Q factor of these

resonators is rather low, due to their radiation. Fig. 12.9 shows a

rectangular patch resonator, the resonators of various different

shapes are even used.

When a number of microwave devices are connected by

means of sections of transmission lines or waveguides, we obtain a

microwave circuit. An analysis of the behaviour of such circuits is

carried out either in terms of the equivalent transmission

line voltage and current

waves or in terms of the

amplitudes of the propagating

waves. The first approach

leads to an equivalent

impedance description, and

the second emphasizes the

wave nature of the fields and

results in a scattering matrix

formulation.

a b

Fig. 12.6

Fig. 12.8

metallization slot

Fig. 12.7

dielectric substrate

conducting

patchfeeding

microstrip

Fig. 12.9 backward metallization



book - 12

161

Particular microwave circuits perform a variety of particular tasks in complex systems

or subsystems. We can include among these circuits: oscillators, amplifiers, modulators,

demodulators, mixers, filters, switches, multiplexers, demultiplexers, etc. It is beyond the

scope of this textbook to treat these circuits in detail.

Scattering parameters:A microwave circuit can be generally

represented by an N-port, i.e., a circuit with N

ports, Fig. 12.10. The internal structure of this

circuit may be rather complex and it can be

very difficult to find the solution of the

Maxwell equations describing the field. Let

us try to describe such a circuit. Let us

assume that each port is connected by a

transmission line to the rest of a system.

There is a wave, which is incident to this

circuit port propagating along this line toward

the port, and there is a wave reflected from

the port and propagating back. These wavesare determined by amplitudes +

nU and −

nU .

Let us normalize these amplitudes by

Cn

nn

Z

U a

+

= , (12.1)

Cn

nn

Z

U b

−

= , (12.2)

where Z Cn is the characteristic impedance of the line connected to the n-th port. The voltage

and the current at the n-th port are

( )nnCnnnn ba Z U U U +=+=−+ , (12.3)

( ) ( )nn

Cn

nn

Cn

n ba Z

U U Z

I −=−=−+ 11

. (12.4)

The power at the n-th port is

[ ] ( )[ ] 22**22*

2

1

2

1Re

2

1Re

2

1nnnnnnnnnnnn babaabba I U P P P −=−+−==−=

−+ .

So the normalization was done to be able to express simply the power transmitted by the

incident wave and the power transmitted by the reflected wave

2

2

1nn a P =

+ , (12.5)

U 1

U 2

.

.

.

U n

U N

.

.

.

an

bnb1

b2

b N

a1

a2

a N

Fig. 12.10



book - 12

162

2

2

1nn b P =

− . (12.6)

It is reasonable to assume that all incident waves contribute to each reflected wave. In

the case of a linear circuit, this dependence must be linear. It reads

N nN jnjnnnn aS aS aS aS aS b ++++++= ......332211 , n = 1, 2, 3, …., N . (12.7)

In the matrix form we have

=

N NN N N

N

N

N a

a

a

S S S

S S S

S S S

b

b

b

.

.

....

.

.

.

2

1

21

22221

11211

2

1

. (12.8)

Matrix [ ]S is known as a scattering matrix. Particular elements of the scattering matrix aredefined as

j

iij

a

bS = for 0=k a until jk ≠ . (12.9)

These elements represent the transmission from the j-th port to the i-th port. The main

diagonal elements S ii represent the reflection coefficient at the i-th port. This assumes that all

other ports are matched and no energy comes to them.

The elements of a scattering matrix are generally complex numbers and can be

displayed on a Smith chart as functions of frequency. Their amplitudes are often expressed indB. The scattering matrix can be simply transformed to a transition matrix or to a impedance

matrix.

12.2 AntennasAn antenna is an element that transforms energy from an electric circuit to energy

transmitted by a radiated wave into space, or vice versa. The basic theory of wave radiation,

antennas, and antenna arrays was presented in Chapter 10 together with definitions of the

basic antenna parameters. These are: radiation pattern, input impedance, antenna directivity

and gain, effective length and area of the receiving antenna. All these parameters depend on

frequency. An antenna is an integral part of communication links, navigation systems, radars,telescopes, and various sensors.

Antennas are often designed as metallic bodies. The basic problem in this case is to

determine the distribution of the electric current on the surface of the antenna. Once we know

the distribution of this current we can relatively simply calculate the field radiated by an

antenna using formulas (10.26) to (10.30) derived in Chapter 10.

Let us assume an antenna as a perfectly conducting body with surface S , Fig. 12.11,

located in an unbounded space filled by homogeneous lossless material with wave impedance

Z 0 and propagation constant k . This antenna is irradiated by a plane electromagnetic wave

described by vector Ei. This field induces an electric current with density K on the surface of

an antenna. This current produces electric field E s radiated, or scattered, back by an antenna.

Consequently the total field is the sum of these particular fields



book - 12

163

E = Ei + E s . (12.10)

Let us assume for the sake of simplicity,

the problem that irradiating field Ei

remains unchanged by the incidence to

the surface of an antenna, and so the

disturbance of a field is caused purely by

scattered field Es. The total field must

fulfill the boundary condition on the

antenna surface (1.22)

0=×En ,

i s EnEn ×−=× . (12.11)

The scattered field is determined by

(10.28)

( )AAE divgradωµε

ω j j s −−= , (12.12)

where vector potential A is determined by the surface equivalent of (1.58), Fig. 12.12

( )( )

∫∫−

=

S

jkR

dS R

r r

e'

4

K A

π

µ , (12.13)

where R = r – r’. Inserting for vector potential (12.13) into (12.12) and consequently for the

scattered field into (12.11) we get an integral equation for the unknown distribution of electric

current K on the antenna surface

i

S

jkR

S

jkR

dS R

jdS

R j En

K K n ×−=

−−× ∫∫∫∫

−− e

4divgrad

e

4 π

µ

ωµε π

µ ω .

We can exchange the order of derivatives and integration in the second term as we integrate

according to r ’ and the derivatives included in operators gradient and divergence are

calculated according to r . So we have

i

S

jkR jkR

dS R R j j En

K K

n ×=

+× ∫∫

−− e

divgrad

1e

4 ωµε ω π

µ . (12.14)

This integro-differential equation cannot be solved analytically. A proper numerical method

must be used to solve (12.14). Simplified methods are used in practice to determine the

distribution of current on the surface of antennas. These methods depend on the kind of

antenna.

The solution of the antenna problem is rather difficult. Therefore we often divide this

problem into two parts. Solving an inner antenna problem we in fact solve equation (12.14).

The aim is to determine the distribution of currents or charges on the antenna body or on

some equivalent surfaces. This problem must often be simplified. In the second step we solve

the outer antenna problem, which calculates of the field excited by the source currents. If

0

r’

r

R

S

Fig. 12.11

n

K k

Ei

Hi



book - 12

164

we would like to determine the field in the radiating (far field) zone, we can use methods

explained in Chapter 10. The behaviour of transmitting and receiving antennas is similar. A

radiation pattern for transmission is equal to a radiation pattern for receiving, and the same is

valid for an antenna input impedance. Therefore we analyze antennas mostly as transmitting.

In the case of a receiving antenna we determine in addition the effective length or aperture

and possibly the noise temperature.

It is very difficult to list various kinds of antennas, as there is a great variety of them

and it is difficult to categorize them in some cases. We did not mention aperture antennas in

Chapter 10. The wave passes through an aperture, e.g., through an open-ended waveguide.

Examples of this are horn, reflector and slot antennas. A variety of wire antennas are used

in particular applications. Ultra wide band antennas must by applied in modern radars and

communication systems. Special types of antennas are lens and traveling wave antennas.

Planar antennas are widely used, as they are compatible with planar microwave circuits and

microwave integrated circuits. There is a huge variety of these antennas. All kinds of antennas

can be combined to create an antenna array, see Chapter 10.5. In the following text we briefly

introduce particular types of antennas.

12.2.1 Wire antennasDipole antennas as examples of wire antennas are widely

used. They were analyzed in Example 10.4. Greater directivity and

lower side lobe levels than in the case of a straight dipole can be

gained using the vee dipole, Fig. 12.12. It is possible to find the

angle γ to get maximum directivity.

An extremely practical wire antenna is the folded dipole. It

consists of two parallel dipoles connected at the ends forming a

narrow wire loop, as shown in Fig. 12.13, with dimension d much

smaller than L. For L = λ /2 the input impedance of this dipole is

four times higher than the impedance of a straight dipole. This

antenna is suitable for TV and FM radio receiving, as it is wellmatched to 300 Ω twin-lead transmission line.

We saw in Chapter 10.5 that array antennas can be used to increase

directivity. The arrays we have examined had all elements active,

requiring a direct connection to each element by a feed network. The feed

networks for arrays are considerably simplified

if only a few elements are fed directly. Such an

array is referred to as a parasitic array. The

elements that are not directly driven, called

parasites, receive their excitation by near-field

coupling from the driven elements. A parasitic

linear array of parallel dipoles is known as a Yagi-Uda antenna.

These antennas are simple and have a relatively high gain. The

basic unit consists of three elements one is driven and two are

parasitic, a director and a reflector. More than one director is

usually used. Fig. 12.14 shows a Yagi-Uda antenna with five

directors. The current induced in closely spaced parasitic elements

has the opposite phase to the current of the driven element. From the radiation patterns shown

at the bottom of Fig. 10.12 it is clear that such an array radiates endfire. Lengthening one

parasitic element known as the reflector, the radiation dual endfire beam is changed to a more

desirable single endfire beam. This is more pronounced in a parasitic element shorter than the

driven element, known as a director. An example of the radiation pattern of a Yagi-Uda

L

γ

Fig. 12.12

L

d

Fig. 12.13

driven elementdirectors

reflector

Fig. 12.14



book - 12

165

antenna with 5 directors is shown in Fig. 12.15. The radiation pattern was measured in the

plane of antenna symmetry perpendicular to

the antenna elements.

12.2.2 Aperture antennas

Aperture antennas can be solved using

the equivalence principle. Let

electromagnetic sources be contained in

volume V bounded by surface S with outward

normal n, Fig. 12.16a. The fields E and H

exterior to S can be found by removing the

sources in V and placing the following surface

current densities on S , see Fig. 12.16b,

t HnK ×= , (12.15)

nEM ×= t , (12.16)

where K is current surface

density on S , M is

equivalent magnetic current

density on S , Et and Ht are

the electric and magnetic

fields produced by the

original sources on surface

S . Thus, with a knowledge

of the tangential fields over

a surface due to the original

sources, we can find the fields everywhere external to the surface through the use of

equivalent surface current densities K and M. These densities define vector potentials, see

(1.58),

∫∫−

=

S

jkr

dS r

eK A

π

µ

4, (12.17)

∫∫−

=

S

jkr

dS r

eMF

π

µ

4, (12.18)

where F is the electric vector potential, which is a quantity analogical to magnetic vector

potential A. The magnetic and electric vector potentials determine the field vectors, see

(10.28),

( ) ( )FAAE rot1

divgradε ωµε

ω −−−= j

j , (12.19)

( ) ( )AFFH rot1

divgrad µ ωµε

ω +−−= j

j . (12.20)

Fig. 12.15

V

sources

E, H nEt

Ht S

V

no sources

zero field

E, H nM

K S

a b

Fig. 12.16



book - 12

166

Equations (12.19) and (12.20) follow from the symmetry of Maxwell’s equation gained by

introducing equivalent current densities (12.15) and (12.16). The calculation of the field in the

radiating zone follows the technique described in Chapter 10.3.

The gain of the aperture antenna can be estimated by, see Example 10.6 where the

homogeneous field distribution over the antenna aperture was assumed, not exactly fulfilled

in most aperture antennas,

eff AG2

4

λ

π = , (12.21)

where Aeff = η Aeffm is the effective aperture area of an antenna, Aeffm being the maximum

effective aperture (10.46) area, and η is the

efficiency of radiation.

A typical example of aperture antennas

is a pyramidal horn antenna, shown in Fig.

12.17. This antenna is fed from the

rectangular waveguide. The pyramidal horn

antenna is flared in both the E - and H -planes.This configuration will lead to narrow

beamwidths in both principal planes, thus

forming a pencil beam. The gain can be

simply evaluated by (12.21), where Aeff = ηΑΒ . A horn antenna can also be designed on the

basis of a waveguide of circular cross-section.

In long-distance radio communication and high-resolution radar applications, antennas

with high gain are required. Reflector antennas are

perhaps the most widely used high gain antennas,

which routinely achieve gains far in excess of 30 dB

in the microwave region. The simplest reflector

antenna consists of two components: a large (relativeto the wavelength) reflecting surface and a much

smaller feeding antenna. The most prominent

example is the parabolic reflector antenna, see its

cross section in Fig. 12.18. The reflector has a

paraboloid of revolution shape. A feeder is placed at

the focal point. For large reflectors (a>>λ )geometrical optics principles can be applied and

radiation from the antenna is analyzed by the ray

tracing method, see Chapter 4. The gain can be

simply evaluated by

(12.21), where Aeffm =

4π a2, Fig. 12.18. Theradiation pattern of these

antennas is very narrow –

units of degrees.

There are various

versions of slot antennas.

Radiating slots can be

made in the walls of a

waveguide or can be

designed in a planar version, Fig. 12.19.

E

Fig. 12.17

B

2a

focal point –

feeding point

aperture

plain

Fig. 12.18

reflector

radiated ray


microstrip

feeding the

slot from a

back side

Fig. 12.19

metallization

radiating slot



book - 12

167

12.2.3 Broadband antennasIn many applications an antenna must operate effectively over a wide range of

frequencies. An antenna with wide bandwidth is referred to as a broadband, or an ultra wide

band (UWB), antenna. The term broadband is a relative measure of bandwidth and varies with

circumstances. Let f U and f L be the upper and lower frequencies of operation for which

satisfactory performance is obtained. The center frequency is f C . Then the bandwidth as a

percentage of the center frequency is

100×−

=

C

LU

f

f f BW . (12.22)

Resonant antennas have small bandwidths, while, antennas that have traveling waves on them

operate over wider frequency ranges.

It is frequently desirable to have the radiation pattern and input impedance of an

antenna remaining constant over a very

wide range of frequencies, say 10:1 or

higher. An antenna of this type is referred

to as a frequency independent antenna.This is, for example, the case of an

infinite biconical antenna, see Fig.

12.20. This antenna must however be

truncated, forming a finite biconical

antenna and most of its broadband behaviour

disappears. The condition of frequency

independence is based on a shape determined fully

by angles. The concept of angle emphasis has been

exploited in recent years and has led to a family of

wide bandwidth antennas. These are spiral

antennas and log-periodic antennas.Planar spiral antennas can be simply

constructed using printed circuit techniques. Their

radiation pattern is bi-directional. To obtain

radiation in one direction, the spiral antenna is

placed on a conical surface. The shape of a spiral

antenna is determined by the equation of the spiral

in the polar coordinate system

ϕ ar r e0= , (12.23)

where r 0 is the radius for ϕ = 0 and a is a constant giving the flare rate of the spiral. This

curve can be used to make the angular antenna shown in Fig. 12.21, which is referred as a

planar spiral antenna. The four edges of the metal each have an equation for their curves of

the form (12.23), ϕ ar r e01 = , ( )2/02 e π ϕ −

=ar r , ( )π ϕ −

=ar r e03 , ( )π π ϕ −−

=2/

04 ear r . Maximal

radius R determines a lower frequency 4/ L R λ = and the distance of input ports determines

the upper frequency.

A log-periodic antenna is an antenna having a structural geometry such that itsimpedance and radiation characteristics repeat periodically as the logarithm of the frequency.

There are several configurations of this antenna. Fig 12.22a shows a planar log-periodic toothed trapezoid antenna, Fig. 12.22b shows a log-periodic trapezoid wire antenna, Fig.

12.22c shows log-periodic dipole array geometry. The wide band operation of these

Fig. 12.20

Fig. 12.21



book - 12

168

antennas is obtained due their linearly varied dimensions. Each part of the antenna is active atdifferent frequency band.

12.2.4 Planar antennasPlanar antennas

are designed in a planar

technology. Due to thisthey are compatible with

planar transmission linesand microwave circuits

based on planar

technology. They can bedirectly integrated into

these circuits and can

even be a part of monolithic microwave

integrated circuits.A planar

microstrip patch

antenna is shown in Fig.

12.9. The drawback of

this antenna is its rather narrow band. A number of modifications of a microstrip patchantenna have been designed to widen its frequency band, to get the antenna radiating a

circularly polarized wave, to get dual frequency operation.The simplest way to widen the microstrip patch antenna frequency band is to use a

thicker substrate, or possibly to divide this substrate into two layers, a thin dielectric substrate

and a thick air (foam) layer. This approach is very often combined with an aperture coupling.This antenna is sketched in Fig. 12.23. The thick substrate reduces the antenna quality factor

and in this way widens the frequency band. At the same time, the slot through which energy iscoupled from the feeding microstrip line to the patch, together with the open ended microstrip

line stub terminating the line, is tuned to a frequency slightly different from the patch resonant

frequency. This offset can considerably increase the antenna frequency band.There are a number of other types of planar antennas. Some from them were

introduced above. An electric dipole alone or as a part of an Yagi-Uda antenna can bedesigned in planar geometry. Fig. 12.19 shows a slot antenna. Broad band antennas are often

designed in planar technology, see the spiral antenna in Fig. 12.21, or the planar log-periodictoothed trapezoid antenna shown in Fig. 12.22a.

a b c

Fig. 12.22


microstrip

feeding theslot from

back side

Fig. 12.23

metallization

coupling slot

radiating

patch


air spacer



book – 12.1

169

12.3 Propagation of electromagnetic waves in the atmosphereThe problem of propagation of electromagnetic waves deals with the part of a

communication link shown in Fig 12.1 between the transmitting and receiving antennas. It

includes propagation of electromagnetic waves in the space above the Earth. Wave

propagation is influenced mainly by two factors. The first factor includes the parameters of

the medium itself in which the wave propagates. The second factor covers the geometry of the

whole scene. Here we have to assume the actual profile and cover of the ground, includingvegetation. The atmosphere changes its parameters – complex permeability – depending on

altitude. This causes, e.g., the bending of wave rays in lower parts of the atmosphere (thetroposphere), and wave reflection in the ionosphere. The atmosphere is a time varying

medium. The time changes are slow, and are caused by alternating seasons of the year, and by

changes between day and night. Rapid variation can be caused by changing weather, above all

by hydro-meteors present in the atmosphere – rain, snow, fog, water vapour. They cause

attenuation and scattering of waves. The propagation of an electromagnetic wave is thus not

the matter of a single ray simulating more or less straight propagation of a wave. The wave

received by the antenna is in most cases the sum of the particular waves caused by particular

reflections. Some of these effects are frequency very selective, some are not and are active in

wide frequency bands. Some of these effects vary rapidly with time, some are stable.We can distinguish several basic types of propagation: surface wave, direct wave,

reflected and scattered wave, space wave, tropospheric wave, ionospheric wave, Fig. 12.24.

The ground surface is a boundary between two electrically different materials. Such a

boundary is able to guide a surface wave, see Chapter 8. The surface wave follows the bent

Earth surface. Its polarization is vertical. This kind of connection can be made via long

distances namely in the case of long waves, Fig. 1.1.

A direct wave represents a connection over short distances between places with direct

visibility and in the case of satellite communication. This can happen only at very high

frequencies. In practical cases a wave does not propagate in a straight direction. Due to space

variations of refractive index of the atmosphere, the trajectories of rays representing the wave

Earth

surface wave

direct wave

reflected wave

Fig. 12.24

troposphere up to 10 km

ionosphere 50 - 400 km

tropospheric wave

ionospheric wave



book – 12.1

170

propagation are bent, see Chapter 4.

In the case of a wave transmitted from elevated positions we have to consider not only

the direct wave, but the contribution of waves reflected from the ground or scattered from

various obstacles. These are the reflected wave and the scattered wave. These waves are

superimposed on the direct wave. This interference can cause unpredictable losses of the

received signal.

The simultaneous presence of the direct and reflected wave, the complex sum of these

waves, creates the total field. This, is known as propagation by the space wave. It is typified

namely by the direct visibility between the transmitter and the receiver, ensured by elevated

antennas at frequencies above 30 GHz. In this case we have so called multi-path

propagation, as there could be a number of reflected and scattered waves approaching the

receiving antenna at the same time.

The propagation of a tropospheric wave is used in communication over very long

distances (thousands of kilometers) for short waves. Propagation of this type uses scattering

of a wave on non-homogeneities in the troposphere that have different refractive indices. Only

a very small part of the transmitted power reaches the receiver, but a connection can be made

at distances far beyond the optical horizon.

An electromagnetic wave can propagate to long distances due to reflections from theionosphere. The layer of ionized air causes intensive continuous bending of the trajectories of

the rays as finally the rays are returned toward the Earth’s surface. Waves with a wavelength

longer than 10 m can be propagated by an ionospheric wave.

At present communication over long distances runs via satellites in the microwave

frequency ranges – satellite communication links.

Let us now discuss the power balance of a communication channel between two

antennas located at distance r in free an unbounded space filled by a lossless material

assuming that r is in the far field zone of the two antennas. In this case, we can assume

propagation of a spherical wave with its amplitude modified by angular dependence defined

by the transmitting antenna radiation pattern – antenna directivity DT . Assuming dependence

of the radiated field on distance in the form (2.49) we get the average value of Poynting’svector at the position of the receiving antenna

22 44 r

G P

r

D P S T T T r av

π π == , (12.24)

where P r is the power radiated by the transmitting antenna, and P T is the total power supplied

to the transmitting antenna. The power received by the receiving antenna is

Reffmav R AS P η = , (12.25)

where Aeffm is the receiving antenna maximum effective aperture (10.46) and η R is the

efficiency of the receiving antenna. The effective aperture can be simply calculated for an

elementary electric dipole. It has the form

π

λ

4

2

D Aeffm = , (12.26)

valid for any antenna. Using (10.38), (12.26) and (12.24) we get from (12.25)



book – 12.1

171

π

λ

π η

44

2

2

RT T Reffmav R

G

r

G P AS P == .

Consequently we get the power received by the receiving antenna

T RT R P GGr P

2

4

=

π

λ . (12.27)

This term determines the received power assuming that all kinds of losses are neglected. In

spite of this fact the received power is lower than the transmitted power. The reason is that the

transmitted power is spread into the whole space, see Chapter 2.4, and consequently the field

amplitude decreases as 1/r and Poynting’s vector as 1/r 2. The term ( )24 r π λ is known as the

free space loss.

The transmission formula (12.27) assumes that antennas are targeted mutually in the

directions of the maximum of their radiation. Using (10.38) we can introduce antenna

directivities with their dependences on angles ϑ and ϕ

( ) ( ) R R RT T T RT T R D Dr

P P ϕ ϑ ϕ ϑ η η π

λ ,,

4

2

= , (12.28)

where index R denotes the receiving antenna and index T denotes the transmitting antenna.

Equation (12.28) is known as the Friis transmission formula. It is applicable in the case of

the general directions of the

antennas defined by angles

R RT T ϕ ϑ ϕ ϑ ,,, , Fig. 12.25. In

practical cases of propagationall kinds of losses includingthe polarization mismatch

must be included in (12.28).

Moreover, the multi-path propagation must be taken into

account and particular wavesmust be added with their

proper phase and

attenuation.Radar can be

assumed as a special case of a communication channel.

An electromagnetic wave is

transmitted by a transmitter,reflected back by a target

and received by a receiver,Fig. 12.26. The time delay

between transmission and

reception of the pulses is proportional to the distance of the target. We will calculate the power received by a radar receiver. To simplify this problem we assume that the transmitting

and receiving antennas are pointed such that the radiating pattern maxima are directed toward

the target. The power density incident on the target is then

transmitting antenna receiving antenna

ϑ T , ϕ Τ

ϑ R, ϕ R

r

Fig. 12.25

transmitter

receiver

target

r

Fig. 12.26



book – 12.1

172

2224 r

A P G

r

P S

effT T

T T

incλ π

== , (12.29)

where the gain has been expressed using the transmitting antenna effective aperture (12.21).The power intercepted by the target is proportional to the incident power density (12.29), so

incinc S P σ = , (12.30)

where the proportionality constant σ is the radar cross section and is the equivalent area of thetarget as if the target reradiated the incident power isotropically. Although power P inc is not

really scattered isotropically, the receiver samples the scattered power in only one direction

and we are only concerned about that direction and assume the target scatters isotropically.Because P inc appears to be scattered isotropically, the power density arriving at the receiver is

24 r

P S inc R

π = . (12.31)

The power available at the receiver is then

ReffR R S A P = . (12.32)

Combining the above four equations gives

242 44 λ π

σ

π

σ

r

A A P

r

S A P

effReffT

T inc

effR R == , (12.33)

which is referred to as the radar equation. Usually the transmitting and receiving antennasare identical, that is, AeffT = AeffR and GT = G R = G. Using (12.21) we can rewrite (12.33) in a

convenient form as

( ) 43

22

4 r

G P P T R

π

σ λ = . (12.34)

The combination of (12.30) and (12.31) actually forms the definition of the radar cross

section

inc

R

S

S r 24π σ = , (12.35)

which is the ratio of 4π times the radiation intensity, r 2S R, in the receiver direction to the

incident power density from the transmitter direction.

The radar equation enables us to determine the range of the radar supposing we know

the transmitted power, the antenna parameters, the least power that must be received by the

receiver, and the target radar cross section. Equation (12.33) neglects the attenuation of a

wave by the atmosphere.



book 122

173

12.4 OptoelectronicsWith increasing demands to transmit more and more of information, transmission

channels must use wider and wider frequency bands. This is achieved by increasing the

transmission frequency. Huge possibilities are offered by using optical beams as a medium for

transmitting information. Here we are in the frequency band of the order 10 14 Hz. This gives

us an extremely wide frequency band, which can be used to transmit a signal, i.e., the great

information capacity of an optical link. Optical signals can be exploited when we master thespecial technology. This concerns generation, modulation, transmission, demodulation and

detection of optical signals.Based on the knowledge gained in the preceding text, in this chapter we will introduce

basic problems of optoelectronics. We start with optical waveguides, and will continue with

coupling the optical signal to them, followed by optical detectors, optical amplifiers, lasers,

optical modulators and, finally, optical sensors.

12.4.1 Optical waveguidesA system designer must be aware of the fundamental bandwidth limitations in optical

waveguides. The most prominent limitation is dispersion, which represents the dependence

of the relative propagation constant k /k 0 on frequency, where k 0 is the phase constant in avacuum. Dispersion distorts the shape of the pulses transmitted through the waveguide. The

pulses are spread in time when traveling along the waveguide, so that they even cannot be

recognized at the output. Temporal spreading effectively establishes the maximum data rate

for a communication link. There are three types of dispersion: material dispersion, modal

dispersion, and waveguide dispersion. In material dispersion, different wavelengths of light

travel at different velocities. Consequently, the pulse effectively spreads out (or disperses) in

time and space. Modal dispersion arises in waveguides with more than one propagating

mode. Each allowed mode in the waveguide travels at a different group velocity. The pulse

energy in a waveguide is distributed among the various allowed modes, either through the

initial excitation, or through mode coupling that occurs within the waveguide. The modes

arrive at the end of the waveguide slightly delayed relative to each other. This effectivelyspreads the temporal duration of the pulse, which again limits the bandwidth. Waveguide

dispersion is a more subtle effect. The propagation constant depends on the wavelength, so

even within a single mode different wavelengths propagate at slightly different speeds.

Waveguide dispersion is usually smaller than material and modal dispersion. However, in the

vicinity of the so called zero dispersion point for materials, waveguide dispersion can be the

dominant effect in a single mode system. Waveguide dispersion can be used to cancel

material dispersion, allowing the design of special dispersion shifted waveguides.

In Chapter 8 we studied basic dielectric waveguides: the dielectric layer – the

dielectric slab waveguide and the dielectric cylinder. In these two waveguides the core has a

constant permittivity. Such waveguides are known as step-index waveguides. A very common

parameter for characterizing waveguides is the numerical aperture NA. This concept is based on ray tracing and refraction, so it is applicable to multimode waveguides. The

geometry of this problem is shown in Fig. 12.27. The numerical aperture is defined as the

sinus of the maximum angle θ max

under which the ray is coupled into

the waveguide. Let us assume a fiber

with core refraction index n1,

refraction index of the surrounding

material n2, and this fiber faces a

medium with n = 1. The ray will be

guided if it strikes the fiber interfaceat an angle greater than the critical

fiber core n1

fiber cladding n2

fiber cladding n2n = 1

cθ θ ≥

maxθ θ ≤inc

Fig. 12.27



book 122

174

angle (3.68). Applying the Snell laws we get

( ) ( ) ( )

( ) 22

21

21

221

21

11max

1sin1

cos90sinsin

nnnnnn

nn NA

c

cc

−=−=−=

==−==

θ

θ θ θ . (12.36)

The numerical aperture is a useful parameter for large core multimode waveguides, namely

for describing the coupling of light into the fiber. The light to be coupled into the waveguide

must be focused in such a way that all incident rays lie within this angle θ max.

There are two ways to significantly reduce modal dispersion in a waveguide: by using

only single mode waveguides, or by using a graded-index waveguide. The single mode

waveguide appears to be the simplest, but it is not always a practical solution. It is very

difficult to couple light into single mode waveguides, as their core has small dimensions. The

second method for reducing modal dispersion is to use graded-index waveguides. These

waveguides can be made with relatively large dimensions, easing the coupling and alignment

problems common in single mode devices, and they can dramatically reduce modal

dispersion.

In a graded-index waveguide the permittivity is asmooth function of position. At the waveguide center it has

the maximum value, and it decreases smoothly with

distance away from the central axis, see the sketch in Fig.

12.28. These waveguides can be analyzed as has been

described in Chapter 4. The path of the ray propagating in

the graded-index waveguide has been determined in

Example 4.1. It is described by a sinus function.

The modal dispersion arose due to the path

differences between the high order rays that followed a

long zigzag down the waveguide and the low order rays

that traveled straight. Fig. 12.29a illustrates this case for two extreme modes. In the graded-index structure, a ray

traveling near the axis will

spend more time in a high

index material, Fig.

12.29b, and will travel

more slowly than will a

ray that is farther from the

axis. However, rays far

from the axis follow a

longer sinusoidal path.

Through optimaladjustment of the index gradient, it is possible to minimize the difference in the group delay

between the extreme rays. This will reduce modal dispersion and effectively increase the

information capacity of the waveguide.

Dispersion of the signal can sometimes be compensated or eliminated through clever

design, but attenuation simply leads to a loss of signal. Eventually the energy in the signal

becomes so weak that it cannot be distinguished with sufficient reliability from the noise

always present in the system. Attenuation therefore determines the maximum distance at

which an optical link can be operated without amplification. Attenuation arises from several

different physical effects. In an optical waveguide, one must consider: intrinsic material

absorption, absorption due to impurities, Rayleigh scattering , bending and waveguide

scattering losses, and microbending loss. In terms of priority, intrinsic material absorption

0

n( x)

Fig. 12.28

step index

a

graded index

b

higher order mode

low order mode

higher order mode

low order mode

Fig. 12.29



book 122

175

and Rayleigh scattering are the most serious causes of power loss for long distance systems.

Impurity absorption has become less of a problem as improved material processing techniques

have been developed. Total optical attenuation is characterized

z inout e P P α −

= , (12.37)

where P out and P in are the output and input powers of the optical wave, respectively, and α isthe attenuation coefficient. The attenuation coefficient depends strongly on the wavelength

of the light and the material system involved.

The intrinsic material absorption arises from the atomic, molecular, and vibrational

transitions. These transitions can absorb electromagnetic energy from the applied field and

store it in the excited state. This energy is eventually dissipated through emission of a photon

or through the creation of lattice vibrations, and represents a loss to the electromagnetic field.

Sometimes the attenuation due to impurities is more severe than the attenuation of the basic

material itself. One of the problems of manufacturing optical glass fibers for long distance

communication systems is to provide material with sufficient purity. These losses have been

represented by complex permittivity (1.23) in Chapter 1. Similarly we can introduce the

complex index of refraction (3.11)

n = n’ – jn’’ . (12.38)

The imaginary part of n can lead to attenuation or gain, depending on its sign. In the case of a

passive medium it leads to attenuation and its value is positive. Inserting for wave

propagating constant k = k 0n = k 0(n’-n’’) we can describe the wave traveling along the

waveguide

( ) ( ) z n jk z nk z jnn jk jkz ee E e E e E z E '''

0'''

00000 −−−−−

=== (12.39)

Rayleigh scattering is a fundamentally different attenuation mechanism. Instead of

light being absorbed and converted into stored energy within a medium, it is simply scattered

away from its original direction. This is the scattering of light off the random density

fluctuations that exist in a dielectric material. These losses are fundamental and cannot be

compensated or eliminated.

The present attenuation value of 0.2 dB/km in fused silica is the fundamental limit of

the performance of such glass fibers.

Optical power can be lost through leakage due to fiber bending. If the fiber is bent, the

spatial mode is not appreciably changed in shape compared to the straight fiber. However, the

plane wavefronts associated with the mode are now pivoted about the center of curvature of

the bend. To keep up with the mode, the phase front on the outside of the bend must travel alittle faster than the phase front in the core. At some critical distance from the core of the

fiber, the phase front will have to travel faster than the local speed of light, c/nclad . Since this

is not possible, the field beyond this critical radius breaks away and enters a radiating mode.

The power that breaks away is a loss to the waveguide. Microbending loss is caused by

putting an optical fiber in a cable and wrapping it about the central cord. The degree of

attenuation depends on the specific cabling geometry.

There can be a significant loss in optical connections due to misalignment or due to

mismatch between the two devices. Misalignment between a source and a single mode

waveguide in the case of dimensions less than 1 µm can cause a coupling loss exceeding

1 dB. Coupling problems are exaggerated by the small dimensions of the typical optical

waveguides and sources, which makes alignment a critical and challenging task. The



book 122

176

calculation of the coupling between two optical waveguides is based on a modal description

of the waveguides. The couple depends on alignment, dimension differences, and geometric

shapes. In the calculation process we have to include the forward waves of the waveguides,

the reflected waves and the radiation modes originating from both the reflected wave and the

transmitted field.

The crucial problem is to couple the light from a source into a waveguide. This

coupling is defined by the radiation pattern of the light source, which is defined by the angular

dependence of brightness B(ϕ ,θ ). Brightness is defined as the optical power radiated into a

unit solid angle per unit surface area. It is specified in watts per square centimeter per

steradian. Consider the case

in Fig. 12.30, showing an

optical source end-fire

coupling onto the end of a

waveguide. The source

emits light into a cone,

which partially overlaps the

numerical aperture of the

waveguide. Any lightfalling outside either the

numerical aperture or the

physical core dimensions obviously will not couple to the waveguide. The total power

coupled to the waveguide is, assuming waveguide circular symmetry, given by

( ) ( )∫ ∫ ∫ ∫

=

min max

0

2

0

2

0 0

sin,

r

S d dr r d d B P

π π θ

ϕ ϕ θ θ θ ϕ (12.40)

where the subscript S refers to the source, r min is the smaller radius of either the fiber core or

the source. Brightness B is here integrated over the acceptance solid angle of the fiber. Themaximum acceptance angle θ max is defined through the numerical aperture (12.36).

12.4.2 Optical detectorsThe minimum received power necessary to achieve the desired quality of information

in a communication link is established by noise. There are many ways to detect light. Any

process which converts optical energy into another useful form of energy can be considered to

be a detector. For optoelectronic systems, the most useful detectors are those that convert

optical energy directly into electric current or voltage. There are two fundamental classes of

detector: quantum, or photon, detectors, which respond to the number of photons that are

absorbed, and thermal detectors, which respond to the energy that is absorbed.

The quantum detector measures the number of photons received per second N , whichis proportional to the power of the optical signal P

hc

P

hf

P N

λ η η == , (12.41)

where η is the quantum efficiency of the detector, defined as the probability that a free

electron or hole is generated per absorbed photon. h is Planck’s constant, hf is the energy of

the photon. The number of received photons increases with the wavelength up to the threshold

at which the energy of the photon is not sufficient to excite a free electron or hole.

Fig. 12.30

fiber

acceptance

angle

core

cladding

lost power

optical source

source radiation

pattern



book 122

177

An ideal thermal detector responds to the total power, independent of the wavelength.

Real detectors have spectral response curves which are limited only by the spectral absorption

of their coating. The absorbed optical power is converted to heat. In general, thermal detectors

have a much slower response time than most quantum detectors. Optical communication

systems almost exclusively use quantum detectors.

The quality of a signal is degraded by noise, i.e., random fluctuations that cannot be

distinguished from the signal. There are many sources of noise. Shot noise is a fundamental

noise that exists in all optical detection processes. Its origin is in the randomness of the arrival

of photons in the signal. It is fundamental and, consequently, there is no way to avoid it. The

absolute noise value is rarely a concern – normally we are worried about the relative noise

compared to the signal. The significant parameter is the signal-to-noise ratio - S / N .

Four basic parameters characterize the performance of an optical detector. These are:

responsivity, spectral response, detectivity, and time response. Responsivity tells us how

much signal is obtained per unit optical power. It is measured in A/W or in V/w. Detectivity

tells us the minimum detectable power required to achieve S / N = 1. The quantum detectors

used in optical communication systems work on the basis of the internal photoeffect. In this

case a photon creates a free charge carrier in the material. The creation of excess carriers leads

to a change in conductivity, p-n junction voltage, or junction currents. There are three major categories of optoelectronic detectors: photoconductors, pin diodes, and avalanche

photodiodes.

Photoconductive detectors employ semiconductor materials with a bandgap suited to

the wavelength of the light that is being detected. A photon with sufficient energy can lift an

electron from the valence band to the conducting band, and thus increases the conductivity. p-

n junction detectors are very often used in optical systems. A pair electron-hole is generated

by a photon in the depletion layer. It is separated by an internal field and contributes to the

current excited in this way by the incident light. Adding an intrinsic layer between the p- and

n-layers we get a pin photodiode. The intrinsic layer widens the depleted layer and in this

way reduces the diode capacitance and shortens the diode time response. An avalanche

photodiode uses the internal multiplication effect to achieve higher detectivity than pin photodiodes. The structure of this diode is designed so that the extra electron-hole pairs

excited by the detected light are accelerated through the depletion layer by the intense electric

field as they are able to generate extra free electron-hole pairs. These subsequent electrons

and holes are themselves accelerated, and can lead to further pair creation. As a result of this

impact ionization, the total current in the photodiode is greater than would be produced by

photoionization alone.

12.4.3 Optical amplifiers and sourcesA photon with sufficient energy absorbed in a material can excite an atom from its

basic energetic state to the upper excited stage. Spontaneous emission of radiation occurs

when an excited atom spontaneously relaxes back to the lower basic state, and in the processemits a single photon. This effect of the spontaneous emission of light radiation is used in

standard light emitting diodes. The electromagnetic field of the incident optical wave can

induce a transition between the upper and lower state, and the atom gives up one quantum of

energy to the field. This is known as stimulated emission of radiation. Optical amplifiers and

lasers are based on the stimulated emission of radiation. Let us take a monochromatic optical

beam that propagates through a material in the z direction. Its electrical field then varies as

( ) ( ) gz e E z E 0= , (12.42)

where g is the optical gain. This quantity is proportional to the difference between the populations of atoms in the lower basic state N 1 and the atoms in the upper excited state N 2



book 122

178

( )12 N N g −= σ , (12.43)

where σ is a constant. From (12.43) it follows that for N 2 > N 1 we have g > 0. This condition

is known as the inversion population of states, and the optical beam is amplified due to the

stimulated emission. The inversion population can be obtained by adding the necessary

energy from outside, depending on the kind of active material – pumping the laser or the

amplifier.The simplest optical amplifier compatible with optical communication links using

optical fibers is the erbium-doped fiber optical amplifier. This is a standard optical fiber with

the core doped by erbium. Erbium ions in the glass lattice assure appropriate energy levels.

This active material is pumped optically by

irradiating, e.g., by a tungsten bulb. The energy of the

photons of the pumping light must be higher than the

energy of the photons of the amplified light. The

pumping assures transition of atoms from the basic

energy state E 0 to state E 3, see Fig. 12.31, and the

stimulated emission then goes from state E 2 to state

E 1. The spontaneous emission in the optical amplifier is the source of the noise added to the amplified

signal.

Lasers are optical oscillators. The acronym laser stands for light amplification by the

stimulated emission of radiation. Like other oscillators, the laser is an optical amplifier with

positive feedback. Amplification is assured by the gain, which is established through

population inversion due to pumping an active medium. Positive feedback is accomplished in

the optical frequency range using mirrors appropriately aligned to resonate one, or a few,

cavity modes. Part of the generated power is in this way selectively returned back to the input

of the amplifier.

In optoelectronics, lasers tend to be based on waveguide structures which are already

one-dimensional in nature. The spatial confinement of the oscillating mode is provided by thestructure of the device. The longitudinal mode selection and feedback is accomplished by an

open resonator known as the Fabry-Perot resonator, Fig. 12.32. It consists of two mirrors with

reflection coefficients R1 and R2, separated by

distance d . Only light with an integer number of

half-wavelengths in the cavity will be resonant in

the structure. The field incident to the mirrors is

partially reflected, assuring the feedback, and

partially transmitted, providing the laser output.

The wave reflected back is amplified, which

maintains the oscillations in the cavity.

Modern optoelectronics is based on the application of semiconductor injection lasers.

These lasers are compatible with driving electronic circuits. They are efficient, small and

cheap. The light emitted by these lasers can be relatively simply coupled to optical fibers. The

inversion population of states is achieved by injecting free charge carriers across two

heterojunctions into a narrow active layer. For this reason, these lasers are known as double

heterostructure semiconductor lasers. Semiconductor physics is beyond the scope of this

course.

12.4.4 Optical modulators and sensors

There are two common methods for encoding a signal onto an optical beam: either

directly modulate the optical source, or externally modulate a continuous wave optical source.Direct modulation is the most widespread method of modulation today, but it introduces

pump stimulated

emission

E 0

E 3 E 2

E 1

Fig. 12.31

d

mirror mirror

Fig. 12.32

R1 R2



book 122

179

demanding constraints on semiconductor lasers. For example, it is difficult to directly

modulate a semiconductor laser at frequencies above a few GHz. Furthermore, it is difficult to

maintain single mode operation of these pulsed lasers. Multiple mode lasers have a larger

spectral bandwidth, which leads to increased pulse spreading due to dispersion. External

modulators offer several advantages over direct modulation. First, one can use a relatively

simple and inexpensive continuous wave laser as the primary optical source. Second, since a

modulator can encode information based on a number of externally controlled effects, it is not

compromised by the need to maintain population inversion or single mode control.

The simplest electrooptic modulator is based on the linear electrooptic effect, or

Pockel’s effect. This represents the change in the refractive index due to an externally applied

electric field. Devices which directly modulate the phase, the intensity, or the polarization of

the light are designed with the use of the Pockel’s effect. The phase modulator is simple. The

external electric field changes the refractive index, and in this way the phase velocity of the

wave propagating through the modulator. This causes changes in the output signal. The phase

modulation introduced by the electrooptic effect can be used to create intensity modulation

via changes in polarization and through interferometric effects. Polarization modulation can

be achieved using the differential retardation between two orthogonal polarizations of the

optical wave. To convert polarization modulation into intensity modulation, it is necessary torun the output through a linear polarizer. Phase modulation can be converted into intensity

modulation through constructive interference between two waves. The Fabry-Perot

interferometer and the Mach-Zender interferometer represent two examples for converting

phase modulation into intensity modulation. Fig. 12.33 shows a schematic Mach-Zender

interferometer. The single mode waveguide input is split into two single mode waveguides by

a 3 dB Y junction. The split beams

travel different paths, and then

recombine at another Y junction. The

relative phase difference of the two

beams can be electrooptically

controlled by applying a voltage tothe center electrode in the structure

shown in Fig. 12.33. Because the

change in refractive index n depends

on the direction of the applied electric

field, index n increases in one arm

and decreases in the other arm. This differential change in index is used to alter the relative

phase of the recombining fields and thus the amplitude of the output wave.

Like electrooptic modulators, acoustooptic modulators control the transmission of

light by local changes in the refractive index of the transmission medium. The modulation

occurs by means of a traveling sound wave which induces stress related modifications of the

local refraction index. Optical wave interaction can be produced by either bulk acoustic wavestraveling in the volume of the material, or by surface acoustic waves which propagate on the

surface within approximately one acoustic wavelength of the surface. Surface acoustic wave

devices are well suited to integrated optics applications because the energy of the acoustic

field is concentrated in the region of the optical waveguide.

Anything that can perturb the optical beam in a fiber can be exploited to make a

sensor. Common interactions are length and refractive index modification through stress,

strain, or temperature. The designer must know how to make the fiber interact selectively with

the measured quantity of interest. An optical sensor can be based on changes in intensity,

polarization, phase, wavelength, and direction of propagation. In extrinsic sensors the light

leaves the fiber and is modulated prior to being recoupled onto a fiber. In an intrinsic sensor

the light is modulated inside the fiber.

3 dB coupler

input

output

E

V

Fig. 12.33



book 122

180

Three basic effects are used to intensity modulate the optical signal: mechanical, based

on movement of an aperture or object which interferes with the optical path; refractive, where

the cladding layer of the fiber is modified in such a way as to cause the guided light to leave

the core; and absorptive, where the optical signal is absorbed by a substance. The critical

weakness of the intensity sensor is that the system has no way of knowing if a change in

intensity is in response to a change of the measured quantity, or if it is due to some other

cause.

This weakness can be reduced by using a phase modulation index based, e.g., on the

Mach-Zender interferometer, see Fig. 12.33. The measured physical quantity changes the

phase of the optical wave in one arm of the interferometer. This causes a change in the output

signal.

13. MATHEMATICAL APPENDIX

This section summarizes the basic knowledge of mathematics necessary to understand

the text.

Computation with vectors:A vector quantity is written as a linear combination of basic vectors x0, y0, z0

000 zyxA z y x A A A ++= . (13.1)

We add vectors by adding their coordinates

( ) ( ) 000 zyxBA z z y y x x B A B A B A +++++=+ . (13.2)

The modulus of a vector is

222

z y x A A A A ++==A . (13.3)

The scalar product of two vectors is

( ) ABBA ⋅=++==⋅ z z y y x x B A B A B A AB α cos , (13.4)

where αααα is the angle between the two vectors A and B. It follows from definition (13.4) that

the scalar product of two vectors equals zero for perpendicular vectors. The scalar

product is linear

( ) CABACBA ⋅+⋅=+⋅ , (13.5)

From (13.4) we can determine the angle between vectors

AB

BA ⋅= arccosα . (13.6)



book-13

181

According to Fig. 13.1 the scalar product determines the projection

of vector A in the direction of vector B. Let n be a unit vector

normal to a surface, then it follows from Fig. 13.1 that the

component of A normal to the surface is defined by

An ⋅=n A . (13.7)

The vector product of two vectors is defined

( ) ABiBA ×−==× n AB α sin , (13.8)

where in is a unit vector perpendicular to the plane in which the two vectors lie. The vector

product can be calculated in the rectangular coordinate system as

( ) ( ) ( ) 000

000

zyx

zyx

BA x y y x z x x z y z z y

z y x

z y x B A B A B A B A B A B A

B B B

A A A −+−+−==× . (13.9)

If the vector product of two nonzero vectors equals zero, it

follows from (13.8) that these two vectors are parallel. From Fig.

13.1 it is clear that the value of the vector product determines the

projection of vector A into the direction perpendicular to vector

B. Therefore using the vector product we can compute the

tangential component of a vector to the plane determined by unit

normal vector n

An×=t A . (13.10)

The value of the vector product (13.8) equals the area of

the parallelogram in Fig. 13.2. The vector product is linear

as the scalar product (see (13.5)).

The double vector product is defined as

( ) ( ) ( )BACCABCBA ⋅−⋅=×× . (13.11)

The mixed product is defined as

( ) ( ) ( )

z y x

z y x

z y x

C C C

B B B

A A A

=×⋅=×⋅=×⋅ BACACBCBA . (13.12)

The value of the mixed product equals the volume of the parallelepiped shown in Fig. 13.3.

Complex numbers:

The basic knowledge is, see Fig. 13.4 defining particular quantities,

B

A

•

cos(α )

Fig. 13.1

sin( )

B

Fig. 13.3

C

B

α

BA×

Fig. 13.2



book-13

182

=+==+=

x

yarctg y x jy x z j ϕ ρ ρ ϕ ,,e 22 , (13.13)

( ) ( )ϕ ϕ ϕ sincose j j += . (13.14)

Goniometric functions:

Here is a list of basic formulas

( ) ( ) 1cossin 22 =+ α α , (13.15)

( ) ( )[ ]α α 2cos12

1sin 2 −= , (13.16)

( ) ( )[ ]α α 2cos12

1cos2 += , (13.17)

( ) ( ) ( ) ( ) ( ) β α β α β α sincoscossinsin ±=± , (13.18)

( ) ( ) ( ) ( ) ( ) β α β α β α sinsincoscoscos m=± , (13.19)

( ) ( )

±=±

2cos

2sin2sinsin

β α β α β α

m, (13.20)

( ) ( )

−

+=+

2cos

2cos2coscos

β α β α β α , (13.21)

( ) ( )

−

+−=−

2sin

2sin2coscos

β α β α β α , (13.22)

( ) ( ) ( )

=+=+=+

A

B B AC t C t Bt A arctg,,sincossin 22 ϕ ϕ ω ω ω , (13.23)

( ) ( ) jx jx

j x −−= ee

2

1sin , (13.24)

( ) ( ) jx jx x −+= ee2

1cos . (13.25)

Hyperbolic functions:

( ) ( ) x x x −−= ee2

1sinh , (13.26)

( ) ( ) x x x −+= ee2

1cosh . (13.27)

ρ

ϕ

y

Fig. 13.4



book-13

183

( ) ( ) 1sinhcosh 22 += x x , (13.28)

( ) ( ) ( ) ( ) ( ) y x y x y x sinhcoshcoshsinhsinh ±=± , (13.29)

( ) ( ) ( ) ( ) ( ) y x y x y x sinhsinhcoshcoshcosh ±=± . (13.30)

From definitions of hyperbolic functions (13.26) and from the definitions of goniometricfunctions (13.24) and (13.25) it follows

( ) ( ) ( ) ( ) x jx x j jx coshcos,sinhsin == , (13.31)

( ) ( ) ( ) ( ) ( ) y x j y x jy x sinhcoscoshsinsin +=+ , (13.32)

( ) ( ) ( ) ( ) ( ) y x j y x jy x sinhsincoshcoscos −=+ , (13.33)

( )

( ) ( )

( ) ( ) y x

y j x

jy x 2cosh2cos

2sinh2sin

tg +

+

=+ , (13.34)

( ) ( ) ( ) ( ) ( ) y x j y x jy x sincoshcossinhsinh +=+ , (13.35)

( ) ( ) ( ) ( ) ( ) y x j y x jy x sinsinhcoscoshcosh +=+ . (13.36)

Bessel’s functions:

A linear combination of Bessel’s functions J n( x) and Y n( x) of the n-th order is a

solution of the differential equation

( ) 0''' 222 =−++ yn x y x y x , (13.37)

( ) ( ) xY C x J C y nn 21 += . (13.38)

Functions J n( x) are defined by the sum

( )( )

( )∑∞

=

+

++Γ

−

=0

2

1!

21

k

k n

k

nk nk

x

x J . (13.39)

Functions Y n( x) follow from J n( x) by the formula

( )( ) ( ) ( )

( )π

π

n

x J n x J xY nn

nsin

cos −−= . (13.40)

The integrals of Bessel’s functions are

( ) ( ) ( ) ( ) x J xdx x J x x J xdx x J x n

n

n

n

n

n

n

n −+

−− −== ∫∫ 11 , . (13.41)



book-13

184

Plots of these functions of the three lowest orders are shown in Fig. 13.5 and Fig. 13.6.

Fig. 13.5 Fig. 13.6

To calculate the propagation constants of modes propagating in a waveguide with a

circular cross-section, see paragraph 7.3, we need to know the zero points α mn of the Bessel

function J m and the zero points 'mnα of the derivative of this function. The zero points α mn of

the Bessel function J m are listed in Tab. 13.1. The zero points 'mnα of the derivative '

m J are

listed in Tab. 13.2.

m 0 1 2 3 4 5

n = 1 2.40482 3.83171 5.13562 6.38016 7.58834 8.77148

n = 2 5.52007 7.01559 8.41724 9.76102 11.06471 12.33860

n = 3 8.65372 10.17347 11.61984 13.01530 14.37254 15.70017

n = 4 11.79153 13.32369 14.79595 16.22347 17.61597 18.98013

n = 5 14.93091 16.47063 17.95980 19.40942 20.82693 22.21780

Tab. 13.1 - α mn

m 0 1 2 3 4 5

n = 1 3.83170 1.84118 3.05424 4.20119 5.31755 6.41562

n = 2 7.01558 5.33144 6.70613 8.01524 9.28240 10.51986

n = 3 10.17346 8.53632 9.96947 11.34592 12.68191 13.98719

n = 4 13.32369 11.70600 13.17037 14.58585 15.96411 17.31284

n = 5 16.47063 14.86359 16.34752 17.78875 19.19603 20.57551

Tab. 13.2 - 'mnα

The asymptotic formulas of Bessel’s functions J n(x) and Y n(x) for x increasing to infinity are

( )

+−=

22

1cos

2 π

π n x

x x J n , (13.42)

( )

+−=

22

1sin

2 π

π n x

x xY n , (13.43)

Hankel’s functions (Bessel’s functions of the 3rd kind):

x

J 0 ,

J 1 ,

J 2

0 2 4 6 8 10

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1J

0

J1

J2

x

Y 0 ,

Y 1 ,

Y 2

0 2 4 6 8 10

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4 Y0

Y1

Y2



book-13

185

The solution of equation (13.37) is any linear combination of Bessel’s functions J n andY n. This enables us to introduce Hankel’s functions by

( ) ( ) x jY x J x H nnn +=)(1 , (13.44)

( ) ( ) x jY x J x H nnn −=)(2 . (13.45)

Their approximate values for a high argument are

( ) ( )2/4/1 e2 n x j

n x

x H π π

π −−≈ , (13.46)

( ) ( )2/4/2 e2 n x j

n x

x H π π

π −−−≈ . (13.47)

It is evident from (13.44) and (13.45) that the relations between Hankel’s functions and

Bessel’s functions J n and Y n are similar to the relations between the exponential function of a

complex argument and goniometric functions. Bessel’s functions J n and Y n describe solutions

of the wave equation cylindrical as standing waves, whereas Hankel’s functions describe a

traveling wave.

Hankel’s functions of the order ½ are defined by

( ) jx

x j x H e

212/1

π −= , (13.48)

( )jx

x j x H −

= e22

2/1π . (13.48)

Modified Bessel’s functions:Modified Bessel’s functions are Bessel’s functions of an imaginary argument.

Inserting into (13.37) x = jz we get

( ) 0''' 222 =+−+ yn z zy y z . (13.49)


( ) ( ) z K B z I A y nn += . (13.50)

The modified Bessel’s function of the first kind is defined

( ) ( ) jz J j z I nn

n−= , (13.51)

which gives a real value for real z . Similarly

( ) ( ) jz H j z K n

n

n

11

2

π +

= (13.52)



book-13

186

is the modified Bessel’s function of the second kind, again real for real z . This function can be

defined by function I n

( )( ) ( )

( )π π

n

z I z I z K nn

nsin2

−= − . (13.53)

Their approximate values for a high argument are

( ) z

e z I

z

nπ 2

= , (13.54)

( ) z n e

z z K −=

2

π . (13.55)

Plots of modified Bessel’s functions of the two lowest orders are shown in Fig. 13.7 and Fig.

13.8.

Fig. 13.7 Fig. 13.8

Coordinate systems:We use the three basic coordinate systems: rectangular, cylindrical and spherical.

In the rectangular coordinate system each point is determined by three coordinates

x, y, z , which represent the corresponding segments on the axes, see Fig. 13.9. Elemental

displacements in the directions of the axes are dx, dy and dz . The volume element is

dV =dx.dy.dz .

The cylindrical coordinates are r , α , z , see Fig. 13.10. r represents the perpendicular

distance from the axis z , α is the angle measured in the plane x y starting from the positive xdirection, and z has the same meaning as in the rectangular system. The cylindrical

coordinates are bound with the rectangular coordinates by the relations

( )α cosr x = ,

( )α sinr y = , (13.56)

= .

Elemental displacements in the directions of the axes are dr , rd α and dz . The volume element

is dV=dr.r.d α .dz . If the integrated function depends only on r , we can use the volume element

x

I 0 ,

I 1

0 1 2 3 4 5 60

5

10

15

20

25

30

I1

I0

x

K 0 ,

K 1

0 1 2 3 4 5 6

0

1

2

3

4

K1

K0



book-13

187

in the form dV=2π rldr , which is the volume between two coaxial cylinders with radii r and

r+dr of length l .

The spherical coordinates are r , α , ϑ , see Fig. 13.11. r represents the distance from

the origin, α is the angle measured in the plane x y starting from the positive x direction, and

ϑ is the angle measured from the positive z direction. The spherical coordinates are bound

with the rectangular coordinates by the relations

( ) ( )ϑ α sincosr x = ,

( ) ( )ϑ α sinsinr y = , (13.57)

( )ϑ cosr z = .

Elemental displacements in the directions of the axes are dr , r sin(ϑ )d α and rd ϑ . The volumeelement is dV =r 2.sin(ϑ ).dr.d α .d ϑ . If the integrated function depends only on r , we can use the

volume element in the form dV =4π r 2dr , which is the volume between two concentric spheres

with radii r and r+dr .

The flux of a vector (a surface integral):Let J be the distribution of electric current. The total current passing surface S can be

calculated in the following way. We discretize the

surface, Fig. 13.12, to elementary surfaces dS

determined by vectors d S=dS n, where n is the unit

normal vector. The current passing the elemental

surface dS is

SJnJ d dS dS J dI n ⋅=⋅== .

The total current is the summation of dI , which is a

surface integral

∫∫ ⋅=S

d I SJ . (13.58)

In this way we have presented the physical meaning of the surface integral from the scalar product of a vector and the vector of a surface element. In the case of a closed surface we

J

n

d S

S

dS

Fig. 13.12

(r ,ϑ , )

Fig. 13.11

r

( x, y, z )

Fig. 13.9

z0

y0 x0 (r ,α , z )

Fig. 13.10

r

z0 αααα0

r0

αααα0

r0

ϑϑϑϑ0



book-13

188

have the notation

∫∫ ⋅=S

d I SJ .

Path integral:

We demonstrate the physical meaning of thisintegral by calculating the work performed by force F

along path c. We divide the path into elemental arcs of

length dl. The work performed along an arch is, see Fig.

13.13,

lFtF d dl dl F dW t ⋅=⋅== 0 .

The work performed by force F along the whole path c is the summation of dW , which is an

integral

∫ ⋅=c

d W lF . (13.59)

In the case of a closed path the integral is known as the circulation of vector F

∫ ⋅=c

d W lF . (13.60)

Gradient of a scalar function:

The gradient of a scalar function is a vector which determines the direction in which a

function value increases at maximum speed, and the modulus of this vector determines themagnitude of this speed. This vector is perpendicular to the planes of the constant function

values. In the rectangular coordinate system the gradient is defined by

000grad zyx z y x ∂

∂+

∂

∂+

∂

∂=

ϕ ϕ ϕ ϕ . (13.61)

The increase of a scalar function corresponding to a shift by d r can be determined neglecting

differentials of higher orders

( ) ( )

( ) ...,,

,,

+∂

∂+

∂

∂+

∂

∂+=

=+++=+

dz z

dy y

dx x

z y x

dz z dy ydx xd

ϕ ϕ ϕ ϕ

ϕ ϕ rr

,

( ) ( )

( )Θ=⋅=∂

∂+

∂

∂+

∂

∂≈

≈−+++=

cosgradgrad

,,,,

dr d dz z

dy y

dx x

z y xdz z dy ydx xd

ϕ ϕ ϕ ϕ ϕ

ϕ ϕ ϕ

r

So the increase of a function is greatest in the direction of gradϕ . The gradient is defined by

partial derivatives, so it is a linear operator

dl

t0d l

c

F

Fig. 13.13



book-13

189

( ) 22112211 gradgradgrad ϕ λ ϕ λ ϕ λ ϕ λ +=+ , (13.62)

where λ 1 and λ 2 are two scalar constants. For the product of two functions we have

( ) ϕ ψ ψ ϕ ϕψ gradgradgrad += . (13.63)

In the literature we can find the symbolic vector ∇ , known as Hamilton’s “nabla” operator. Inthe rectangular coordinate system this operator is defined as

000 zyx z y x ∂

∂+

∂

∂+

∂

∂=∇ . (13.64)

Using this operator the gradient can be expressed as

ϕ ϕ ∇=grad . (13.65)

Divergence of a vector function:The divergence of vector function J is defined as the volume density of this vector

quantity flux flowing out of a point, so it characterizes the sources of a vector,

⋅= ∫∫→

S V

d V

SJJ1

limdiv0

, (13.66)

where S is the boundary of volume V . In the case of J being a current density, divJ

corresponds to the current flowing out of a unit volume. The Gauss-Ostrogradsky theorem

follows from this

∫∫∫∫∫ =⋅V S

dV dS JJ div . (13.67)

In the rectangular coordinate system the divergence can be calculated as

z

J

y

J

x

J z y x

∂

∂+

∂

∂+

∂

∂=Jdiv . (13.68)

This operator is a linear operator, as it is defined by partial derivatives,

( ) 2121 divdivdiv JJJJ µ λ µ λ +=+ , (13.69)

where λ and µ are scalar constants. The divergence applied to the product of scalar function f

and vector function J is

( ) JJJ divgraddiv f f f +⋅= . (13.70)

The divergence applied to the vector product of two vectors is

( ) BAABBA rotrotdiv ⋅−⋅=× . (13.71)



book-13

190

The divergence applied to the gradient of a scalar function defines Laplace’s operator

( )2

2

2

2

2

2

graddiv z y x ∂

∂+

∂

∂+

∂

∂=∆=

ϕ ϕ ϕ ϕ ϕ . (13.72)

The divergence can be expressed using Hamilton’s operator

JJ ⋅∇=div . (13.73)

Rotation of a vector function:

The rotation of a vector function is defined as the surface density of this vector

circulation

⋅= ∫→

cS

d S

lFrF1

limmaxrot0

0 , (13.74)

where c is a curve that describes the boundary of surface S , and r0 is a unit vector. This vector

determines the direction of vector d S and is selected to obtain the maximum value of

⋅∫→

cS

d S

lF1

lim0

. As the rotation represents the surface density of a circulation, Stokes’

theorem follows from the definition of the rotation

∫∫∫ ⋅=⋅S c

d d SFlF rot . (13.75)

In the rectangular coordinate system the operator rotation is defined as a symbolicdeterminant

z y x F F F

z y x ∂

∂

∂

∂

∂

∂=

000

rot

zyx

F . (13.76)

This operator is again a linear operator, as it is defined by partial derivatives,

( ) 2121 rotrotrot FFFF µ λ µ λ +=+ . (13.77)

The rotation applied to the product of a scalar function and a vector function is

( ) FFF ×+= f f f gradrotrot . (13.78)

There are mathematical identities

( ) 0gradrot = f , (13.79)

( ) 0rotdiv =F , (13.80)



book-13

191

( ) ( ) FFF2divgradrotrot ∇−= . (13.81)

In the rectangular coordinate system we have ∆=∇2 The rotation can be expressed by

Hamilton’s operator as`

FF ×∇=rot , (13.82)

which follows from (13.63) and (13.9).

Differential operators in the cylindrical coordinate system:

000

1grad zαr

z r r ∂

∂+

∂

∂+

∂

∂=

ϕ

α

ϕ ϕ ϕ , (13.83)

( ) z

F F

r F r

r r

z r

∂

∂+

∂

∂+

∂

∂=

α α 11

div F , (13.84)

( ) 000 11rot zαrF

∂

∂−∂∂+

∂

∂−∂∂+

∂

∂−∂∂=

α α α

α r z r z F F r r r r

F z

F z

F F r

, (13.85)

2

2

2

2

2

11

z r r r

r r ∂

∂+

∂

∂+

∂

∂

∂

∂=∆

ϕ

α

ϕ ϕ ϕ . (13.86)

Differential operators in the spherical coordinate system:

000

1

sin

1grad ϑ

ϑ∂

∂+

∂

∂

ϑ+

∂

∂=

ϕ

α

ϕ ϕ ϕ

r r r αr , (13.87)

( ) ( )ϑϑϑ∂

∂

ϑ+

∂

∂

ϑ+

∂

∂= F

r

F

r F r

r r r sin

sin

1

sin

11div 2

2 α α F , (13.88)

( ) ( )

( ) 0

00

1

sin

1

1sin

sin

1rot

ϑ

∂

∂−

∂

∂

ϑ+

+

ϑ∂

∂−

∂

∂+

∂

∂−ϑ

ϑ∂

∂

ϑ= ϑ

ϑ

α

α

α

α

F r r r

F

r

F F r

r r

F F

r

r

r αrF

, (13.89)

ϑ∂

∂ϑ

ϑ∂

∂

ϑ+

∂

∂

ϑ+

∂

∂

∂

∂=∆

ϕ

α

ϕ ϕ ϕ sin

sin

1

sin

1122

2

22

2

2 r r r r

r r . (13.90)

14. BASIC PROBLEMS

This section lists the basic problems collected throughout the text. The ability to answer them

verifies the student’s knowledge and readiness to sit for the exam.

1 Maxwell’s equations in a differential form for a time-harmonic field expressed by



book-13

192

phasors.

2 Maxwell’s equations in an integral form for time-harmonic field expressed by phasors.

3 Boundary conditions of tangential components of E, H in a nonstationary field.

4 Boundary conditions of normal components of E, H in a nonstationary field.

5 Electromagnetic field on a surface of an ideal conductor.

6 Poynting’s vector. Definition and expression using field vectors.

7 Energy balance of active power. Physical meaning of particular items.

8 Energy balance of reactive power. Physical meaning of particular items.

9 The wave equation for E or H in a general material outside of a source region. Time-

harmonic field. Expression using phasors.

10 Continuity equation for harmonic field

11 Description of vectors E and H by potentials A and ϕ in time varying field.

12 Wave equations for potentials, sources assumed.

13 General solution of wave equations for potentials, harmonic field is assumed.

14 The expression of E of a time-harmonic plane wave in a general material. Phasor and

instantaneous value. Meaning of particular items.

16 What is a surface of constant amplitude and constant phase in a plane wave? What are a

uniform wave and a nonuniform wave?17 What is a phase velocity? How is it defined?

18 What is a group velocity? How is it defined?

19 Draw the orientation of E, H and k in a plane electromagnetic wave. What is the

relation of these vectors?

20 What is the wave impedance of a general material? How is it defined?

21 The expressions for k, vf and Z in an ideal dielectric.

22 The expressions for k, vf and Z in a good conductor.

23 What is a penetration depth? How is it defined?

24 Active power transmitted by a plane electromagnetic wave through a surface 1 m2.

25 What is the polarization of an electromagnetic wave? Which kinds of polarization of an

electromagnetic wave are there?26 Under which conditions can two linearly polarized waves create a wave with linear,

circular and elliptical polarization?

27 Equation of eikonal.

28 Differential equation describing the ray.

29 Coefficient of reflection for electric field in a perpendicular incidence, general

materials.

30 Coefficient of transmission for electric field in a perpendicular incidence, general

materials.

31 Coefficients of reflection and transmission for a perpendicular incidence, lossless

dielectrics.

32 What is the standing wave?33 What is the standing wave ratio, its relation to R?

34 What is λ /4 transformer?

35 Snell’s laws.

36 What is the Brewster’s polarization angle, expression?

37 Draw a plot of R=f(ϕi) for ε1<ε2 for both horizontal and vertical polarization.

38 Draw a plot of R=f(ϕi) for ε1>ε2 for both horizontal and vertical polarization.

39 What is the total reflection, the condition for it?

40 What is the surface wave?

41 What is the character of field in the second material in the case of the total reflection?

42 Write a formula describing the spherical wave.43 Write a formula describing the cylindrical wave.



book-13

193

44 What transmission line can transmit TEM wave?

45 Cut-off frequency of a mode in a parallel plate waveguide.

46 Cut-off wavelength of a mode in a parallel plate waveguide.

47 Sketch the field distribution of the TEM wave propagating in a parallel plate

waveguide.

48 Phase velocity of a wave propagating between two parallel plates.

49 Wavelength of a wave propagating between two parallel plates.

50 Propagation constant of a wave propagating between two parallel plates.

51 Wave impedance of a TM and TE waves propagating between two parallel plates.

52 What mode is dominant in the rectangular waveguide?

53 Cut-off frequency of a mode in the rectangular waveguide.

54 Cut-off wavelength of a mode in the rectangular waveguide.

55 Cut-off frequency of the dominant mode in the rectangular waveguide.

56 Cut-off wavelength of the dominant mode in the rectangular waveguide.

57 Phase velocity of a wave propagating in a rectangular waveguide.

58 Wavelength of a wave propagating in a rectangular waveguide.

59 Propagation constant of a wave propagating in a rectangular waveguide.

60 Wave impedance of a TM and TE waves propagating in a rectangular waveguide.61 Power carried by the dominant mode in a rectangular waveguide.

62 What transmission line can transmit the TEM wave?

63 Cut-off frequency of a TM mode in the circular waveguide.

64 Cut-off frequency of a TE mode in the circular waveguide.

65 What mode is dominant in the circular waveguide?

66 Phase velocity of a wave propagating in a circular waveguide.

67 Wavelength of a wave propagating in a circular waveguide.

68 Propagation constant of a wave propagating in a circular waveguide.

69 Wave impedance of a TM and TE waves propagating in a circular waveguide.

70 Explain the guiding of waves in a dielectric waveguide.

71 Draw the plot of Ey field as a function of the transversal coordinate in the dielectriclayer for the first three TE modes.

72 Which parameters determine the resonant frequency of a cavity resonator?

73 Resonant frequency of a rectangular cavity resonator.

74 The telegraph equations for time-harmonic u and i on a TEM transmission line. The

meaning of particular items.

75 The characteristic impedance of a transmission line - the lossy and lossless case.

Expression and the definition.

76 The input impedance of a lossy and a lossless line of length L terminated by an

impedance ZL.

77 The input impedance of a lossless line of length L with a short and open end.

78 Draw lines of constant amplitude of a reflection coefficient in the Smith Chart.79 Draw lines of constant phase of a reflection coefficient in the Smith Chart.

80 Draw lines of constant real part or a normalized impedance in the Smith Chart.

81 Draw lines of constant imaginary part or a normalized impedance in the Smith Chart.

82 Characteristic regions around the radiating elementary electric dipole.

83 Which components of electric and magnetic field carry power from the radiating

electric dipole?

84 Draw radiation patterns of the radiating elementary electric dipole.

85 What is the radiating resistance of the dipole?

86 What is the antenna directivity?

87 What is the antenna gain?

88 Relation between E and D in anisotropic material.



89 Relation between B and H in anisotropic material.

90 Tensor of permeability of magnetized ferrite.

91 Tensor of permittivity of a magnetized plasma.

92 What is the ferromagnetic resonance?

93 What is the Faraday effect?

94 What is the ordinary wave?

95 What is the extraordinary wave?

14. LIST OF RECOMMENDED LITERATURE

C. A. Balanis: Advanced Engineering Electromagnetics, John Wiley & Sons, Inc., New York,

USA, 1989.

J. D. Jackson: Classical Electrodynamics, 3rd ed., John Wiley & Sons, Inc., New York, USA,

1998.

L. B. Felsen, N. Marcuvitz: Radiation and scattering of waves, IEEE Press, Piscataway, N. J.,1994.

R. E. COLIN: Field Theory of Guided Waves, 2nd ed., IEEE Press, Piscataway, N. J., 1991.

d

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Electromagnetic Waves and Transmission Lines - Machac