FIELDS AND WAVES IN
COMMUNICATION
ELECTRONICS
THIRD EDITION
SIMON RAMO
TRW Inc.
IOflN R. WI'IINNERY
University of California, Berkeley
THEODORE VAN DUZER
University of California. Berkeley
@
JOHN WILEY & SONS, INC.
._,4-—
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Library of Congress Catalaging-in-Publicatian Data
E
Ramo, Simon
Fields and
R.
waves
in communication electronics / Simon Ramo,
Whinnery, Theodore
.
E
John
Van Duzer.-—-—3rd ed.
E
cm.
Includes index.
ISBN 978-0—471-58551-0
1.
Electromagnetic
fields.
3. Telecommunication.
Theodore.
I.
2. Electromagnetic waves.
Whinnery, John R. 11. Van Duzer,
III. Title.
QC665. E4R36
E
‘
1993
537———dc20
93434415
crp
Printed in the United States of America
15 14 13 12 11
E
To
our
wives
Virginia
Patricia
Eanice
1.
.r
This book is
an
intermediate-level text
on
electromagnetic fields and waves. It represents
revision of the first two editions of the text, which in turn built upon an earlier volume
by two of the authors.* It assumes an introductory course in field concepts, which can
a
be the lower-division
physics course in
in calculus. Material
on
many
vectors, differential
notation for sinusoids is included in
We have
a
colleges or universities, and a background
equations, Fourier analysis, and complex
form suitable for review
or a
first introduction.
such introductions wherever the material is to be used and have related
given
problems in
them to real
fields and
waves.
Throughout
the book, the derivations and
done in the most direct way possible. Emphasis is placed
analyses
understanding, enhanced by numerous examples in the early chapters.
are
The fundamentals of
electromagnetics,
based
on
on
physical
Maxwell’s brilliant theories, have
since the first version of the text, but
emphases have changed and new
changed
field
coherent
continue
The
of
to
applications
optics for communications and
appear.
information processing continues to grow. New materials of importance to electronic
devices (for example, superconductors) have been developed. Integrated circuit ap—
proaches to guides, resonators, and antennas have grown in importance. All these ev~
olutions are reflected in expanded text and problem material in this edition. Perhaps the
most important change for persons who need to solve field problems is the growing
power of computers. At the simplest level, computers greatly speed numerical evalua~
tion of analytic expressions, easily giving answers over a wide range of parameters.
But there is also a growing library of wholly numerical techniques for finding solutions
to field and wave problems in which complex boundary shapes preclude analytic solutions. We can only give an introduction to this important subject but excellent texts
and reviews are available to carry the interested student farther. It still remains important
to understand the basic laws and to develop strong physical pictures and computer
simulations can substantially add insight, especially in dynamic problems.
The basic order remains that of the second edition. The purpose of beginning with
static fields is not only to develop familiarity with vector field concepts but also to
recognize the fact that a large number of practical time-varying problems (especially
with small devices) can be treated by static techniques (i.e., are quasistatic). The dynamic treatment of Maxwell, with wave examples, follows immediately so that even in
a first term, the student will meet a mix of static, quasistatic, and wave problems. Once
the reader has covered the material of the first three chapters, he or she will find con—
not
'
Whinnery, Fields and Waves in Modern Radio, John Wiley & Sons (first
edition 7944; second edition. 7953). The first edition was prepared with the assistance of
the General Electric Company when the authors were employed in its laboratories.
S. Rama and J. R.
vii
V355
Preface
1
using the text. Material on the elecitromagnetics of circuits (Chap4)
special waveguides (Chapter 9) may be delayed or even omitted and the
later chapters on microwave circuits, materials, and optics can be used in various orders
Selections from among the more advanced sections within a chapter are also possible
without disrupting the basic flow
The authors Wish to thank the following reviewers for their helpful comments: Professor Dennis Nyquist, Michigan State University;
Fred Fontaine, Cooper
siderable flexibility in
ter
or on
=
Professor
Union; Professor Von R. Eshlernann, Stanford University; Professor Paul Weaver, University of Hawaii; Professor Charles Smith, University of; Mississippi; Professor Murray
Black, George Mason University; Professor Donald Dudley, University of Arizona;
Professor B J Rickett, University of California at San D1ego; and Professor Emily Van
Deventer, University of Toronto.
1
gratefully acknowledge the helpful suggestions of students and col~
leagues
Berkeley and users in other universities and in industry. We particularly
thank D. J. Angelakos, C. K. Birdsall, K. K. Mei, J. Fleisichman, M. Khalaf, B. Peters,
and Guochun Liang for their important contributions we also express appreciation to
Doris Simpson, Ruth Dye, Lisa Lloyd-Maffei and Patricra Chen for their careful work
in the preparation of the manuscript.
1
The authors
at
August
1993
Simon Ramo
John
Whinnery
Theodore Van Duzer
CHAPTER fl
1. 1
STATHGNARY ELECTRIC NEEDS
Introduction
BASIC LAWS AND CONCEPTS OF ELECTROSTATICS
1.2
Force Between Electric
1.3
The
Concept
Charges: The Concept of Electric Field
of Electric Flux and Flux
Density: Gauss's Law
of the Use of Gauss's Law
1.4
Examples
1.5
Surface and Volume
1.6
Tubes of Flux:
1.7
Energy Considerations: Conservative Property of
Integrals: Gauss’s
Law in Vector Form
03me
Plotting of Field Lines
Electrostatic Fields
1.8
1.9
Electrostatic Potential;
Equipotentials
Capacitance
DIFFERENTIAL FORMS OF ELECTROSTATIC LAWS
.10 Gradient
NHy—A
.11
The
.12
Laplace‘s
Divergence
an
Electrostatic Field
and Poisson's
.13 Static Fields
.14
of
Equations
Arising from Steady Currents
Boundary Conditions
.15 Direct
in Electrostatics
Integration
Laplace’s Equation: Field
with
Two
Dielectrics
Cylinders
.16 Direct
pt .17
of
Integration of
Equation:
42
Uniqueness of Solutions
45
SPECIAL TECHNIQUES FOR ELECTROSTATIC PROBLEMS
46
Properties
Examples
46
Images
of Two~Dimensional Fields;
1.20 Numerical Solution of the
1.21
Energy of
Graphical
an
Electrostatic
Field
Mapping
Laplace and Poisson Equations
of Information Obtained from Field
ENERGY IN FIELDS
1.22
40
The pn Semiconductor Junction
.18 The Use of
.19
Poisson‘s
Between Coaxial
Maps
50
52
57
59
System
59
Contents
X
sranomwg
cmm'm 2
2.1
1
manaricirtams
70
Introduction
70
STATIC MAGNETIC FIELD LAWS AND CONCEPTS
72
2.2
Concept
of
a
Magnetic
Field
72
Law
2.3
Ampere5
2.4
The Line
2.5
Inductance from Flux
73
of
Integral
Magnetic
1
Field
77
Linkages: External Inductance
81
DIFFERENTLAL FORMS FOR MAGNETOSTATICS AND THE USE
OF POTENTLAL
2.6
The Curl of
2.7
Curl of
2.8
Relation Between Differential and
Vector
Vector Field
84
Field
88
Magnetic
of the Field
2.9
a
2.12 Differential
2.14
of
Magnetic
Equation
Magnetic
93
Loop. Magnetic Dipole
96
Flux
98
Density
for Vector
Magnetic Potential
98
Regions
99
Potential for Current- Free
Boundary Conditions for Static Magnetic F1elds
2.15 Materials with Permanent
Energy of
a
Static
2.17 Inductance from
101
Magnetization
102
MAGNETIC FIELD ENERGY
2.16
90
-
Magnetic Potential
Divergence
213 Scalar
Integral Forms
Equations
2.10 Distant Field of Current
2.11
84
1.
Magnetic
1
106
Field
106
Energy Storage; Internal
Inductance
108
1
CEAPTER S
3.1
MAXWEEE’S EQHA’E‘EQNS
114
Introduction
1
LARGE-SCALE AND DIFFERENTIAL FORMS
1
114
or MAXWELL’S EQUATIONS
3.2
Voltages
Induced
Law for
by Changing Magnetic Fields
3.3
Faraday’s
3.4
Conservation of
Charge
of
Current
Displacement
Physical
3.6
Maxwell' 5
3.7
a
Pictures of
3.5
Maxwell’s
116
Moving System
119
1
Concept
1
121
Displacement Current
Equations1n
Equations
and the
116
in
Differential
Large—Scale
Equation Form;
Form
123
126
128
Xi
Contents
iaxwell's
a.
Equations
for the Time-Periodic Case
EXAMPLES OF USE OF MAXWELL’S EQUATIONS
139
Maxwell’s
139
Equations
and Plane Waves
3.10 Uniform Plane Waves with
3.11 The Wave
Equation
3.19 Power Flow in
Steady-State
Sinusoids
138
Electromagnetic Fields: Poynting's Theorem
Theorem for Phasors
Poynting’s
3.14
Continuity Conditions for
Uniqueness of Solutions
ac
Boundary Conditions
Perfect Conductor for
3.16 Penetration of
3.17 Internal
135
in Three Dimensions
3.13
3.15
199
a
of
a
143
Fields at
a
Boundary:
145
Electromagnetic
Impedance
3.18 Power Loss in
at a
Fields into
a
ac
Fields
Good Conductor
Plane Conductor
3.91
Time~Varying
Integrals
over
158
Fields
Charges
158
and Currents
The Retarded Potentials for the Time~Periodic Case
160
169
THE ELECTROMAGNEHQS
0E CERCHETS
CHAPTER 4
4. 1
as
149
156
POTENTIALS FOR TlME-VARYING FIELDS
3.19 A Possible Set of Potentials for
148
153
Plane Conductor
3.90 The Retarded Potentials
139
171
Introduction
17 1
THE IDEALIZATIONS IN CLASSICAL CIRCUIT THEORY
179
Law
4.9
Kirchhoifs
4.3
Kirchhoif’s Current Law and Multimesh Circuits
177
SKIN EFFECT IN PRACTICAL CONDUCTORS
180
4.4
Voltage
Distribution of
Time~Varying
179
Currents in Conductors of Circular
180
Cross Section
4.5
4.6
of Round Wires
189
CALCULATION OF CIRCUIT ELEMENTS
186
Impedance
186
Self-Inductance Calculations
‘
4.7
Mutual Inductance
189
4.8
Inductance of Practical Coils
193
4.9
Self and Mutual
196
Capacitance
CIRCUITS WHICH ARE NOT SMALL COMPARED
WITH WAVELENGTH
4.10 Distributed Effects and Retardation
198
198
xii
Contents
4.11 Circuit Formulation
Through
the Retarded Potentials
enAMEn s
5.1
200
’
4.12 Circuits with Radiation
205
‘
EEANSMISSIoN tINEs
Introduction
213
213
1
TIME AND SPACE DEPENDENCE or SIGNALS
ON IDEAL TRANSMISSION LINES
214
5.2
Voltage and Current Variations Along
5.3
Relation of Field and Circuit
5.4
Reflection and Transmission at
5.5
Pulse Excitation
5.6
Pulse
on
Analysis
a
Ideal Transmission Line
Resistive
214
218
219
Diseontinuiry
Transmission Lines
1
Forming Line
221
227
I,
SINUSOIDAL WAVES ON IDEAL
5.7
an
for Transmission Lines
TRANSMISSIQN
Reflection and Transmission Coefficients and
Admittance Transformations for Sinusoidal
5.8
Standing Wave
5.9
The Smith Transmission—Line Chart
LINES
Impedance
229
and
Voltages
229
Ratio
233
236
5.10 Some Uses of the Smith Chart
238
\
NONIDEAL TRANSMISSION LINES
245
Transmission Lines with General Forms of DistrIbuted
lmpedances: Lossy
5.12
Eilter~Type
Lines
~
Distributed Circuits. the
(Io-13
RBSONANT TRANSMISSION LINES
5.13
5.14
Purely Standing Wave
Input Impedance
and
on an
Ideal Line
Quality
252
Diagram
‘}
254
254
‘
Factor for Resonant
Transmission Lines
256
=
SPECIAL TOPICS
5.15
Group
and
Energy
245
260
Velocities
260
5.16 Backward Waves
263
‘
5.17 Nonuniform Transmission Lines
CHAPTER 6
6.1
264
‘
PEANE-WAVE PROPAGATION
AND REFLECTION
Introduction
PLANE-WAVE PROPAGATION
274
274
I
l
275
5.2
Uniform Plane Waves in
a
Contents
xiii
Perfect Dielectric
275
6.3
Polarization of Plane Waves
280
6.4
Waves
283
6.5
6.6
in
Dielectrics and Conductors
imperfect
PLANE WAVES NORMALLY INCIDENT ON DISCONTINUITIES
287
Reflection of Normally Incident Plane Waves from Perfect
Conductors
287
Transmission-Line
Analogy
of Wave
Propagation: The Impedance
Concept
289
6.7
Normal Incidence
6.8
Reflection Problems with Several Dielectrics
295
PLANE WAVES OBLIQUELY INCIDENT ON DISCONTINUITIES
300
Incidence at
300
6.9
6.10 Phase
on a
Any Angle
Velocity
6.11 Incidence at
and
Dielectric
on
Perfect Conductors
Impedance
Any Angle
on
292
for Waves at
Oblique Incidence
Dielectrics
306
6.12 Total Reflection
6.13
6.14
Polarizing
Multiple
or
7.1
3 10
Brewster
Angle
Dielectric Boundaries with
CHAPTER 7
303
312
Oblique
Incidence
TWO- AND THREE-DHMENSEGNAE
EGUNDARY VAEEE PROBEEMS
Introduction
313
321
321
THE BASIC DIFFERENTIAL
EQUATIONS
AND NUMERICAL METHODS
322
Laplace, and Poisson Equations
7.2
Roles of Helmholtz.
7.3
Numerical Methods: Method of Moments
324
METHOD OF CONFORMAL TRANSFORMATION
33 1
7.4
Method of Conformal Transformation and Introduction
to
331
Complex-Function Theory
7.5
Properties of Analytic Functions of Complex Variables
7.6
Conformal
7.7
The Schwarz Transformation for General
7.8
Conformal
7.9
Mapping
for
333
336
Laplace‘s Equation
Polygons
345
Wave Problems
348
SEPARATION OF VARIABLES METHOD
351
Mapping for
Laplace’s Equation
in
7.10 Static Field Described
7.11
322
Fourier Series and
Rectangular
by
a
Integral
Coordinates
Single Rectangular
351
Harmonic
353
355
xiv
Contents
7.12 Series of
Rectangular Harmonics for Two- and
ThreeADimensional Static Fields
7.15
Cylindrical
360
Harmonics for Static Fields
365
,
7.14 Bessel Functions
368
%
7.15 Bessel Function Zeros and Formulas
7.16
Expansion
of
Function
a
7.17 Fields Described
7.18
as a
373
Series of Bessel
Functions
by Cylindrical Harmonics
375
377
1
Spherical Harmonics
379
7 19 Product Solutions for the Helmholtz
Equation
in
Equation
1
Rectangular
Coordinates
385
7. 20 Product Solutions for the Helmholtz
in
Cylindrical Coordinates
QE‘HAP'E'ER 8
386
WAVEGEEEBES WE'RE
CONDEEQHNG
QYMNBRECAE.
BGENDAREES
8.1
395
'
Introduction
395
GENERAL FORMULATION FOR GUIDED
WAVES
396
Systems
CYLINDRICAL WAVEGUIDES OF VARIOUS CROSS SECTIONS
Basic
8.3
Waves Guided
8.4
Guided Waves Between Parallel Planes
Equations
and Wave
for Uniform
8.2
Types
by Perfectly Conducting
Parallel IPlates
as
Parallel—Plane
398
398
Superposrtion
of Plane Waves
8.5
396
405
Guiding System
with Losses
407
8.6
Planar Transmission Lines
410
8.7
Rectangular Waveguides
41 7
8.8
The
8.9
TE10 Wave in a Rectangular
Circular Waveguides
8.10
Higher Order
Modes
8.11 Excitation and
on
Guide
423
428
Coaxial Lines
433
of Waves in Guides
435
GENERAL PROPERTIES OF GUIDED WAVES
438
Reception
8.12 General
Properties of TEM Waves
8.13 General
Properties of TM Waves in Cylindrical
Arbitrary Cross Section
on
Multiconductor Lines
Guides of
8.14 General
of
of TE Waves in
Properties
Arbitrary Cross Section
Cylindrical
440
Conducting
442
Conducting
Guides
447
1
Contents
fives
XV
Below and Near Cutoff
449
ispersion of Signals Along Transmission Lines
CHAPTER 9
and
Waveguides
SPECEAE. WAVEGEEEDE TYPES
45 1
462
9.1
Introduction
462
9.2
Dielectric
462
9.3
Parallel-Plane Radial Transmission Lines
464
9.4
Circumferential Modes in Radial Lines: Sectoral Horns
468
Waveguides
Between Inclined Planes
9.5
Duality: Propagation
9.6
Waves Guided
9.7
Ridge Waveguide
474
9.8
The Idealized Helix and Other Slow-Wave Structures
476
9.9
Surface
479
by Conical Systems
472
Guiding
Harmonics
482
RESONANT CAVHTEES
490
9.10 Periodic Structures and
CHAPTER 10
470
Spatial
Introduction
490
RESONATORS OF SIMPLE SHAPE
491
10.2
Fields of
491
10.3
Energy Storage, Losses.
10.4
Other Modes in the
Rectangular
10.5
Circular
Resonator
10.6
Strip Resonators
10.7
Wave Solutions in
10.8
Spherical
10.1
Simple Rectangular
Cylindrical
and
Resonator
Q of
a
Rectangular Resonator
Resonator
493
494
496
500
Spherical
Coordinates
504
508
Resonators
SMALL~GAP CAVITIES AND COUPLING
510
10.9
SmalI~Gap Cavities
510
10.10
Coupling
to Cavities
10.11 Measurement of Resonator
510
O
515
10.12 Resonator Perturbations
518
10.13 Dielectric Resonators
521
XVi
Contents
\
MHCRGWAVE
CHAPTER 1 1
550
NETWORIES
11 1
Introduction
530
11.2
The Networlz Formulation
532
11.5
Conditions for
.
TWO-PORT
Reciprocity
Circuits for
535
JUNCTIONS
WAVEGUIDE
536
Two Port
556
11.4
Equivalent
11.5
Scattering and Transmission Coefficients
11.6
Measurement of Network Parameters
11.7
Cascaded Two Ports
11.8
Examples of Microwave and Optical Filters
l
548
N—PORT WAVEGUIDE JUNCTIONS
}
554
11.9
Circuit and 5~Parameter
11.10 Directional
Couplers
a
559
541
i
545
\_
Representation
and
Hybrid
of N Ports
Networks
554
557
.
FREQUENCY CHARACTERISTICS OF WAVEGUIDE NETWORKS
11.11 Properties of
11.12
Equivalent
a
One-Port
Circuits
Impedance
561
'1
Showing Frequency Characteristics
of One Ports
11.15
Examples
11.14 Circuits
of
564
Cavity Equivalent
Giving Frequency
Circuits
569
Characteristics of N Ports
571
JUNCTION PARAMETERS BY ANALYSIS
11.15
12.1
12 2
573
Quasistatic and Other Methods of Junction
warren 12
Analytsis
RADEA’E‘EGN
573
584
1
Introduction
Some
Types
561
584
of Practical
Radiating Systems
i
586
FIELD AND POWER CALCULATIONS WITH CURRENTS
ASSUMED ON THE ANTENNA
12.3
12.4
Electric and
Magnetic Dipole
Systemization
Currents
of Calculation of
on an
589
Radiators
589
Radiating
Fields
Antenna
and
1
Antenna; Half—Wave
12.5
Long Straight Wire
12.6
Radiation Patterns and Antenna Gain
12.7
Radiation Resistance
12.8
Antennas Above Earth
Dipole
595
i
596
599
*
or
Conducting
Power from
Plane
602
1
603
xvii
Contents
12.9
Traveling Wave
Straight
on a
Wire
606
12.10 V and Rhombic Antennas
607
12.11 Methods of
61 1
Wire Antennas
Feeding
RADIATION FROM FIELDS OVER AN APERTURE
12.12 Fields
as
Sources of Radiation
614
12.13 Plane Wave Sources
of
617
Radiating Apertures Excited by
12.14
Examples
12.15
Electromagnetic
614
Plane Waves
619
Horns
624
12.16 Resonant Slot Antenna
625
12.17 Lenses for
628
Directing
Radiation
ARRAYS OF ELEMENTS
12.18 Radiation
12.19 Linear
630
Intensity with Superposition of Effects
634
Arrays
12.20 Radiation from Diffraction
12.21
Gratings
637
Polynomial Formulation of Arrays and Limitations
Directivity
638
Yagi—Ilda Arrays
641
on
12.22
12.23
12.24
Frequency-Independent
Periodic Arrays
Integrated
Antennas:
Logarithmically
643
Antennas
646
FIELD ANALYSIS OF ANTENNAS
12.25 The Antenna
as a
12.27 Mutual
65 1
Boundary Value Problem
12.26 Direct Calculation of
Impedance
Input Impedance
Between Thin
for Wire Antennas
Dipoles
12.28 Numerical Methods: The Method of Moments
RECEIVING ANTENNAS AND RECIPROCITY
12.29 A
Reciprocity Relations
12.31
Equivalent Circuit of
CHAPTER ES
13.2
65 1
655
659
660
663
663
Transmitting—Receiving System
12.30
13.1
630
666
the
Receiving
Antenna
ELECTROMAGNETEC PRGPERTEES
GE MATEREAES
668
677
Introduction
677
LINEAR ISOTROPIC MEDIA
678
Characteristics of Dielectrics
678
XViii
Contents
and Semiconductors
13.3
Imperfect Conductors
13.4
Perfect Conductors and
13.5
Diamagnetic
and
682
Superconductors
687
Paramagnetic Responses
689
NONLINEAR ISOTROPIC MEDIA
13.6
Materials with Residual
13.7
Nonlinear Dielectrics:
69 1
Magnetization
Application
in
691
Optics
AN ISOT ROPIC MEDIA
13.8
13.9
695
-
699
i
Crystals
Plane-Wave Propagation in Anisotropic Crystals
Representation
of
Anisotropic
Dielectric
699
701
I
13.10 Plane-Wave
13.11
13.12
Electro-Optic
Permeability
13.13 TEM Wave
13.14
Propagation
Faraday
in Ilniaxial
Crystals
705
Effects
707
Matrix for Ferrites
Propagation
7 13
in Ferrites
7 16
Rotation
721
13.15 Ferrite Devices
13.16
13.17
Permittivity
of
725
a
Stationary
Waves
Space-Charge
Magnetic
on a
with Infinite
13.18 TEM Waves on
a
Plasma in
Field
Magnetic
Moving Plasma
a
Field
730
Stationary Plasma
in
a
Finite
Magnetic Field
OPHCS
CHAPTER 14
14.1
Introduction
14.2
Geometrical
742
745
Optics Through Applications
of
Laws
of Reflection
and Refraction
743
Limiting Case
14.3
Geometrical
14.4
Rays
14.5
Ray Matrices for Paraxial Ray Optics
14.6
Guiding
Optics
as
of Wave
Optics
Inhomogeneous Media
of
753
742
RAY OR GEOMETRICAL OPTICS
in
728
Rays by
a
749
752
Periodic Lens
System
756
or
in
Spherical
Mirror Resonators
760
DIELECTRIC OPTICAL WAVEGUIDES
763
14.7
Dielectric Guides of Planar Form
765
Form
14.8
Dielectric Guides of
14.9
Dielectric Guides of Circular Cross Section
14.10
Propagation of
Rectangular
Gaussian Beams in
Graded-Indeii Fibers
767
77 l
775
xix
Contents
14.11
Intermode
Delay and Group Velocity Dispersion
778
14.12 Nonlinear Effects in Fibers: Solitons
14.13
GAUSSIAN BEAMS IN SPACE AND IN OPTICAL RESONATORS
783
PrOpagation of Gaussian Beams
783
in
a
14.14 Transformation of Gaussian Beams
14.15 Gaussian Modes in
14.16
780
Stability
Optical
and Resonant
Homogeneous
by Ray
Medium
Matrix
786
Resonators
Frequencies
of
789
Optical-Resonator
Modes
BASIS FOR OPTICAL INFORMATION PROCESSING
14.17 Fourier
14.18
Transforming Properties
of Lenses
Spatial Filtering
14.19 The
Principle of Holography
APPENDEX fl
APPENBEX 2
APPENDEX 3
APPENBEX 4
APPENDHX 5
ENSEX
793
795
795
798
80 I
QONVERSEON FACE'QRS BETWEEN
SYSTEMS OE EENH'E‘S
813
QOGRDHNA'H‘E SYSTEMS AND VEQEGR
REEA’E‘EONS
815
SKETEH 0? THE DEREVA’E‘EON
GE MAGNE'E‘EQ HERB EAWS
821
QQMPEEX PHASQRS AS HSEB
EN EEECERECAH. £ER£EEWS
824
SQEEE'E‘EON FOR RETARSED
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828
831
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Electric fields have their
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|NTRODUCTION
in electric
charges—electrons
and ions.
Nearly
all real
electric fields vary to some extent with time, but for many problems the time variation
is slow and the field may be considered stationary in time (static) over the interval of
interest. For still other
(called quasistatic) the spatial distribution is
nearly
though the actual fields may vary rapidly with
time. Because of the number of important cases in each of these classes, and because
static field concepts are simple and thus good for reviewing the vector operations needed
to describe fields quantitatively, we start with static electric fields in this chapter and
static magnetic fields in the next. The student approaching the problem in this way
the
same as
important
cases
for static fields
even
must remember that these are
varying
fields
can
give
special cases, and that the interactions between timephenomena, most notably the wave phenomena to
rise to other
be studied later in the text.
beginning the quantitative development of this chapter, let us comment briefly
applications to illustrate the kinds of problems that arise in electrostatics or
quasistatics. Electron and ion guns are good examples of electrostatic problems where
the distribution of fields is of great importance in the design. Electrode shapes are
designed to accelerate particles from a source and focus them into a beam of desired
size and velocity. Electron guns are used in cathode-ray oscilloscopes, in television
tubes, in the microwave traveling wave tubes of radar and satellite communication
systems, in electron microscopes, and for electron~beam lithography used for precision
definition of integrated-circuit device features.
Many electronic circuit elements may have quite rapidly varying currents and voltages and yet at any instant have fields that are well represented by those calculated
from static field equations. This is generally true when the elements are small in comparison with wavelength. The passive capacitive, inductive, and resistive elements are
thus commonly analyzed by such quasistatic laws, up to very high frequencies; so also
Before
on a
are
few
the semiconductor diodes and transistors which constitute the active elements of
electronic circuits.
2
Chapter
Fields
Stationary Electric
1
Transmission lines, including the strip line used in microwave and millimeter-wave
integrated circuits even for frequencies well above 10 GHZ, have properties that can be
calculated
using
the laws for static fields. This is far
nearly exactly,
There
are
electrostatic
static
problem, but
along one
structural
transverse plane satisfy, exactly or
we will see later in the text that for systems having
axis (along the transmission line), the fields in the
frogrncbeing
a
variations
no
:
static field laws.
many other
examples
precipitators
of
used to
application of knowledge
remove
dust and
other
of static field laws. The
solid
particles
from air,
systems
(which must be designed to
xerography, and power switches and transmission
avoid dielectric breakdown) all use static field concepts. Electric fields generated by
body are especially interesting examples. Thus the fields that are detected
by electroencephalography (fields of the brain) and electrocardiography (fields of the
1n the bodyin the same way
heart) are of sufficiently low frequency to be
the human
distributedt
that static fields would be
examples mentioned, the general problemis that of finding the distribution
of fields produced by given sources in a specified mediuin with defined boundaries on
the region of interest. Our approach will be to start with a simple experimental law
(Coulomb’s law) and then transform it into other forms which may be more general or
*
more useful for certain classes of problems
Most readers will have met this material before in physics courses or introductory
electromagnetics courses, so the approach will be that of review with the purposes of
deepening physical understanding and improving familiarity with the needed vector
algebra before turning to the more difficult time-varying problems.
In all the
Basic laws
and garments of Eiectrostatacs
l
1.2
It
was
FORCE BETWEEN ELECTRIC CHARGES: THE
The effect
a
COlexert
thEPT
known from ancient times that electrified bodies
torsion
was
quantified by
Charles A. Coulomb
OF
forces upon
through
brilliant
balance. His experiments showed that like charges repel
opposite charges
attract; that force is
proportional
to the
ELECTRIC FIELD
pioduct
one
another.
experiments using
one
another whereas
of charge
magnitudes;
l
‘
description of Coulomb ’5 experiments and} the groundwork of earlier re
given in R. 3. Elliott, Electromagnetics, McGrgw~Hil/, New York, 7966. For a
detailed account of the history of this and other aspecfsfof electromagnetics, see E. T.
Whittaker, A History of the Theories of Aether and Electricity, Am Inst. Physics New York,
1987 orP F. Matte/0y, Biographical History of Electriclty and Magnetism, Ayer Co. Pub-
An excellent
searchers is
lishers, Salem, NH, 7975
Force Between Electric
1.2
Charges: The Concept of Electric Field
3
that force is
inversely proportional to the square of the distance between charges; and
along the line joining charges. Coulomb’s experiments were done in air,
but later generalizations show that force also depends upon the medium in which
charges are placed. Thus force magnitude may be written
that force acts
f
where ql and q2
representing
are
charge strengths,
the effect of the
defining
a
4142
K
vector 1“ of unit
(1)
7
81‘”
r
is the distance between
medium, and K is
rection information is included
and
=
a
constant
charges, a is a constant
depending upon units. Di-
by writing force as a vector f (denoted here as boldface)
length pointing from one charge directly away from the
other:
=
K
L? t
(2)
81“"
Various systems of units have been used, but that to be used in this text is the
System (SI for the equivalent in French) introduced by Giorgi in 1901.
lntemational
This is
meter—~kilogram—second (mks) system, but the great advantage is that electric
quantities are in the units actually measured: coulombs, volts, amperes, etc. Conversion
factors to the classical systems still used in many references are given in Appendix 1.
a
Thus in the SI system, force in (2) is in newtons (kg-m/sz), q in coulombs, r in meters,
and a in farads/ meter. The constant K is chosen as 1 /477 and the value of a for vacuum
found from
experiment
is
80
=
8.854 X
10"12
F
1
m
—
X
10—9—
36w
(3 )
m
For other materials,
x
8
(4)
81.80
where er is the relative permittivity or dielectric constant of the material and is the
value usually tabulated in handbooks. Here we are considering materials for which a
is
a
scalar
general
independent
media
are
of
strength
and direction of the force and of
discussed in Sec. 1.3 and considered in
more
position. More
Chapter 13.
detail in
Thus in SI units Coulomb’s law is written
f
=
gqu
'J
f'
(5)
4W8)"
Generalizing from the example of two charges, we deduce that a charge placed in
the vicinity of a system of charges will also experience a force. This might be found
by adding vectorially the component forces from the individual charges of the system,
but it is convenient at this time to introduce the concept of an electric field as the force
per unit charge for each point of the region influenced by charges. We may define this
by introducing
a
test
charge Aq
at the
point
of definition small
enough
not to
disturb
4
Chapter
the
charge
Stationary Electric
1
Fields
distribution to be studied. Electric field E is then
f
(6)
=
E
where f is the force
The electric field
given by
acting on the infinitesimal test charge Aq.
arising from a point charge q in a homogeneous
the force law
E
q
x
charges
as seen
.
are
point
to
units
in volts per meter,
as
may be found
__
farads
from the form of
by substituting
volts
coulombs meter
can see
:
in a direction away from the charge,
from
point away
positive charges and toward negative
of electric field magnitude in
in the lower half of Fig. 1.2a. The
seen
_.
We
(7 )
f
r
4'n'er2
Since 1“ is the unit vector directed from the
the electric field vector is
the SI system
dielectric is then
(5):
(meter)2
units in (7):
(V)
meter (m)
(7) that the total
electric
field for
a
system of point
may be found by adding vectorially the fields from the individual charges, as
is illustrated at point P of Fig. 1.20 for the charges q and l- q. In this manner the electric
charges
field vector could be found for any
point in
the
vicinity bf the
two
charges.
An electric
4——
A/\
/
\
l
l
Lower?
FIG. 1.2a Electric fields around two opposite charges.
half of figure shows the separate
fields E + and E_ of the two charges. Upper half shows the vector sum of E+ and E... Construction of E is shown at one point P.
1
1.2
field
Force Between Electric
line is defined
line drawn tangent to the electric field vector at each
as a
5
Charges: The Concept of Electric Field
point
in
space. If the vector is constructed for enough points of the region, the electric field lines
can be drawn in roughly by following the direction of the vectors as illustrated in the
top half of Fig. 1.20. Easier methods of constructing the electric field will be studied
although laborious, demonstrates clearly the
in later sections, but the present method,
meaning of the electric field lines.
For fields
continuous distributions of
charges, the superposition is by
charge. For volume
distributions, the elemental charge rig is p (N where p is charge per unit volume (C/m3)
and (1V is the element of volume. For surface distributions, a surface density pg (C/mz)
is used with elemental surface d8. For filamentary distributions, a linear density p,
(C/m) is used with elemental length dl. An example of a continuous distribution
integration
produced by
of field contributions from the differential elements of
follows.
Example
FIELD
Let
us
radius
in
Fig.
so
the
dE in
OF A
calculate the electric field at
1.2
RING OF CHARGE
points
on
the
axis for
z
a
ring
of
positive charge
of
located in free space concentric with and perpendicular to the z axis, as shown
1.21). The charge p, along the ring is specified in units of coulombs per meter
a
charge in
Fig. 1.2b
axis is dE
cos
a
differential
and is
8 where
length
given by (7)
cos
8
==
is p, dl. The electric field of p, (I! is designated by
a2 + 22. The component along the z
with r2
z/(a2
=
22)“. Note that, by
+
symmetry, the component
on the opposite side
charge
by
perpendicular
of the ring. The total field at points on the axis is thus directed along the axis and is
a dd) we have
the integral of the differential axial components. Taking d1
to the
element
that of the
axis is canceled
=
E
z
i 41mm2
pzaz
0
(14>
+
pm
___
22)3/2
28(a2
+
22)3/2
pl (11
dd,
0
\
\\\\
0
FIG. 1.2b
dEccfl
\EL
2
Electric field of
a
ring
of
charge.
dB
(8)
6
Chapter
1.3
I
Stationary Electric
THE CONCEPT OF ELECTRIC FLUX AND
FLUX; DENSITY: GAuss’s LAW
handling electric field problems to
charges than is electric field E. If we
It is convenient in
directly
related to
D
we
notice from
Eéelds
Eq.
1. 2(7) that D about
material. Moreover, if
we obtain
we
multiply
=
a
introduce another
vector more
define
3E
(1)
point charge
is radial and
the radial component
D by
the
independent of the
of a sphere of
area
radius r,
4177'2D,
i
=
(2)
q
quantity exactly equal to the charge (111 coulombs) so that it may be
thought of as the flux arising from that charge. D may then be thought of as the electric
flux density (C/mz). For historical reasons it is also known as the displacement vector
We thus have
or
a
electric induction.
shaped closed surface as in Fig.
surface surrounding a point charge
whichi the fluid passing surface S of
It18 easy to show (Prob 1. 3b) that for an arbitrarily
1.3a, the normal component of D integrated over the
also
q. The analogy is that of fluid flow in
in
a given time is the same as that passing
1.3b
gives
plane S perpendicular to the flow.
region
depend upon} the shape of the surface used
to monitor it, so long as all surfaces enclose the sameisource. By superposition, the
result can be extended to a system of point charges or a continuous distribution of
Fig.
So fluid flow rate out of
charges, leading
1
does not
a
to the conclusion
electric flux flowing
out
1
of a
closed
I: change enclosed
surface;
(3)
This is Gauss’ 3 law and, although argued here from Coulomb 3 law for simple media,
is found to apply to more general media. ItIS thus a most general and important law.
illustrating its usefulness, let us look more carefully at the role of the medium.
simplified picture showing why the force between charges depends upon the presence of matter is illustrated in Fig. 1.3c. The electron clouds and the nuclei of the atoms
experience Oppositely directed forces as a result of the presence of the isolated charges.
Before
A
FIG. 1.3a
Charge
q and
arbitrary surrounding
surface.
1.3
The
FIG. 1.3b
are
distorted
i i
it
r
Thus the atoms
7
Concept of Electric Flux and Flux Density: Gauss's Law
Flow of
or
a
v
fluid
polarized.
v
it
surfaces S and S i.
through
There is
v
a
shift of the center of symmetry of
the electron cloud with respect to the nucleus in each atom as indicated schematically
in Fig. 1.3c. Similar distortions can occur in molecules, and an equivalent situation
arises in
materials where
some
naturally polarized
molecules have
a
tendency
to be
aligned in the presence of free charges. The directions of the polarization are such for
most materials that the equivalent charge pairs in the atoms or molecules tend to counteract the forces between the two
isolated
charges.
The
magnitude
and the direction of
polarization depend upon the nature of the material.
The above qualitative picture of polarization introduced
the
tified
by giving
a more
by a dielectric may be quanfundamental definition than (1) between D and E:
(4)
D=30E+P
®O®
§
®
@
©® ® @ ®%§@ ®©e @§%Q® QO
FIG. 1.3c
Polarization of the atoms of
a
dielectric
by
a
pair
of
equal positive charges.
8
Chapter-
The first term
gives
!
Stationary Electric
fields
the contribution that would exist if1 the electric field at that
point
in free space and the second measures the effect of the polarization of the material
in
(as
Fig. 1.3a) and is called the electric polarization.
The polarization produced by the electric field in a niaterial depends upon material
were
properties. If the properties do not depend upon position, the material is said to be
homogeneous. Most field problems are solved assuming homogeneity; inhomogeneous
media, exemplified by the earth’s atmosphere, are more difficult to analyze. If the
of the electric field vector, it is
handling a field problem in an
applied steady magnetic field. A
response of the material is the same for all directions
called isotropic. Special techniques are required for
anisotropic medium such as an ionized gas with an
material is called linear if the ratio of the response P to the field E is independent of
amplitude. Nonlinearities are generally not present except for high-amplitude fields. It
possible that the character of a material may be time variable, imposed, for
example, by passing a sound wave through it. Throughout most of this text, the media
will be considered to be homogeneous, isotropic, linear, and time invariant. Exceptions
will be studiedin the final chapters.
I
For isotr0pic, linear material the polarizationis proportional to the field intensity and
is also
we can
write the linear relation
1
P
—-
20er
where the constant Xe is called the electric
as
(5)
susceptibility
Then
(4) becomes the
same
(1),
D
=
80(1
+
XQE
=
SE;
(6)
1 + x,
permittivity, definedin Eq. 1.2(4) is s,
s/s0
Although we will describe a dielectric material largely by its permittivity,
cepts of polarization and susceptibility are in a sense more fundamental and
sidered in more detail in Chapter 13.
and the relative
:
l.4
The
simple
law
can
EXAMPLES OF THE USE
OF
=
the
con-
are con-
GAUSS'S LAW
but
important examples to be discussed in this section show that Gauss’s
strength in a very easy way for problems with certain
kinds of symmetry and given charges. The symmetry givfes the direction of the electric
field directly and ensures that the flux is uniformly distributed. Knowledge of the charge
gives the total flux. Symmetry is then used to get flux density D and hence E
D/e.
be used to find field
=
Example
1.4a
FIELD IN A PLANAR SEMICONDUCTOR DEDLETION LAYER
For the first
intimate
valence
example, we consider a one~dimensional situation where a metal is in
(atomic) contact with a semiconductor. We assume that some of the typically
4 (e.g., silicon) atoms have been replaced by “dopant” atoms of valence 5
Examples of the use of Gauss’s Law
1.4
+
Metal
9
Semiconductor
I
l
Charge-free
region
I
l
l
O
I
A
i
l
l
l
l
l
Depletion
layer
FIG. 1.4a
is the
Model of
region
a
between
(e.g., phosphorus).
metal—semiconductor contact. The
The
and becomes free to
one extra
S x S
d is called
to
which Gauss’s law is
applied
electron in each atom is not needed for atomic bonding
about in the semiconductor.
move
Upon making the metal contact,
forced away from the surface for a distance d. The
depletion region because it is depleted of the free electrons.
it is found that the free electrons
region 0
region
shown dashed.
parallel planes
a
are
Since the
dopant atoms were neutral before losing their extra electrons, they are positively charged when the region is depleted. This can be modeled as in Fig. 1.4a. In the
region x > d the donors are assumed to be completely compensated by free electrons
and it is therefore charge-free. (The abrupt change from compensated to uncompensated
behavior at x
d is a commonly used idealization.) By symmetry, the flux is
x~directed only. The surface used in application of Gauss’s law consists of two infinite
d. This approximation is made because
parallel planes, one at x < d and one at x
the transverse dimensions of the contact are assumed to be much larger than d. With
no applied fields, there is no average movement of the electrons in the compensated
region x Z d so E must be zero there. Thus D is also zero in that region and all the
flux from the charged dopant atoms must terminate on negative charges at the metal
contact. Therefore, Gauss’s law gives, for a unit area,
=
—-
x
—Dx(x)
where
ND
and
e are
nitude of electronic
==
NDe(d
—-
x)
(l)
the volume density of donor ions and the charge per donor (mag
charge), respectively. Then the x component of the electric field is
D
E
It should be clear that the
system; that is, that there
=
—"
NDe(x
—
d)
——
8
simplicity
were no
__
of solution
(2)
8
depended
variations in y and
z.
upon the symmetry of the
'50
Chapter
1
Stationary Electric
Example
FIELD ABOUT
A
Fields
n. 4b
.
LINE CHARGE OR BEIWEEN COAXIAL CYLINDERS
charge in ring form in Ex. 1.2. Let us now find the field E
infinitely
long line of uniformly! distributed charge. The radius
straight,
of the line is negligibly small and can be thought of as the two-dimensional equivalent
of a point charge. Practically, a long thin charged wireI is a good approximation. The
charge, and hence the electric
symmetry of this problem reveals that the force on a
field, can only be radial. Moreover, this electric field will not vary with angle about the
line charge, nor with distance along it. If the strength of the radial electric field is desired
at distance r from the line charge, Gauss’s law may be applied to an imaginary cylin—
drical surface of radius r and any length I. Since the electric field (and hence the electric
flux density D) is radial, there is no normal component Eat the ends of the cylinder and
hence no flux flow through them. However, D is exactly normal to the cylindrical part
of the surface, and does not vary with either angle or distance along the axis, so that
the flux out is the surface area 2771-1 multiplied by the electric flux density Dr. The
charge enclosed is the length l multiplied by the charge per unit length q,. By Gauss’s
law, flux out equals the charge enclosed:
We introduced the line
produced by
a
tesli
27rrlD,
If the dielectric
surrounding
:
lq,
the wire has constant e,
D
Er
=
—r
8
:
—-—ql
(3)
27781‘
Hence, the electric field about the line charge has been
obtained by the
use
of Gauss’ 8
special symmetry of the problem.
The same symmetry applies to the coaxial transmissmn line formed of two coaxial
law and the
.
conducting cylinders of radii a and b with dielectric 8 between them (Fig. 1.41)). Hence
the result (3) applies for radius r between a and b. We use this result to find the
capacitance in Sec. 1.9.
i
‘
I
FIG. 'l.4b
Coaxial line.
1.4
FIG. 1.4a
Spherical
Examples
electrodes
of the use of Gauss’s Law
separated by
Example
two
layers
'I '5
of dielectric materials.
1.4::
FIELD BETWEEN CONCENTRIC SPHERTCAL ELECTRODES WITH Two DIELECTRICS
1.4c shows
formed of two
conducting spheres of radii a and c, with
r
b, in spherical coordinates,2 and a
has
problem
spherical symmetry about the center,
which implies that the electric field will be radial, and independent of the angular
direction about the sphere. If the charge on the inner sphere is Q and that on the outer
sphere is Q, the charge enclosed by an imaginary spherical surface of radius :- selected
anywhere between the two conductors is only that charge Q on the inner sphere. The
flux passing through it is the surface 4772‘2 multiplied by the radial component of the
flux density Dr. Hence, using Gauss’s law,
Figure
a
structure
dielectric 81 extending from r
b to r
c. This
second, 82, from r
one
=
=
to
a
=
=
—-
D,-—-
Q
4777‘?“
(4)
equation for the flux density is the same for either dielectric, since the flux passes
positive charge on the center conductor continuously to the negative charge
on the outer conductor. The electric field has a different value in the two regions,
however, since in each dielectric, D and E are related by the corresponding permittivity:
The
from the
2
E.
=
Er
=
Q
9
b
(5)
< C
( 6)
a < r <
47T811'”
Q
477'82i‘2
b <
‘
I
Note that r is used both for radius from the axis in the circular cylindrical coordinate
system and for radius from the origin in the spherical coordinate system. p is frequently
used for the former but may be confused with charge density, and R is used for the latter,
but is here reserved for distance between source and field points.
12
Chapter
The radial flux
density
1
Stationary Electric
is continuous at the dielectric
Fields
discontinuity at
r
b, but the
=
radial electric field is discontinuous there.
‘
Example
r .4d
FIELDS OF A SPHERICAL REGION OF UNIFORM CHARGE DENSITY
0 to r
a. As in
region of uniform charge density p extending from r
the preceding example, Gauss’s law can be written as (4) where, in this case, Q
{37771-3 p
for r .<_ a and Q
§wa3p for r 2 (I. Then the flux densities for the two regions are
Consider
=
a
z
=2
=
Drxgp
a
3
(8)
I'Za
Dr=3—r§p
I5
(7)
rSa
SURFACE AND VOLUME INTEGRALS: GAuss LAW IN VECTOR FORM
Gauss’ 3 law,
f
given
in words
by Eq.1 3(3),
may be
wntien
figDcosfla’quI
The
symbol 953
is used if the surface18 closed.
The surface
is
integral over a surface and Of course cannot be performed
specified. ItIS in general a double integral. The circle on the
denotes the
until the actual surfaceis
integral Sign
integral
can
also be written in
I
a
still more
Define the unit vector normal to the
employed
given point on
(l)
compact form if vector notation
surface
under consideration, for any
replace Dicos 6 by D fi. This particular product
of the two vectors D and fi denoted bythe dot between the two is known as the dot
product of two vectors, or the scalar product, since it results by definition in a scalar
quantity equal to the product of the two vector magnitudes and the cosine of the angle
between/them. Also, the combination [‘1 (15 is frequentlyIabbreviated further by writing
it dS. Thus the elemental vector dS, representing the element of surface in magnitude
and orientation, has a magnitude equal to the magnitude of the element (15 under con—
sideration and the direction of the outward normal to the surface at that point. The
surface integral in (1) may then be written in any of theI equivalent forms:
the surface,
as
n. Then
~
ngcosedS=fDrfidS=3gD-d8
5
s
All of these say that the normal component of the
general closed surface S.
(2)
ES
vector D is to be integrated
'
a
over
the
1.5
Surface and Volume
Integrals:
'33
Gauss‘s Law In Vector Form
If the charge inside the region is given as a density of charge per unit volume in
coulombs per cubic meter for each point of the region, the total charge inside the region
must be obtained by integrating this density over the volume of the region. This is
analogous to the process of finding the total mass inside a region when the variable
density is given for each point of a region. This process may also be denoted by
a general integral. The symbol ft, is used to denote this, and, as with the surface integral,
the particular volume and the variation of density over that volume must be specified
before the integration can be performed. In the general case, it is performed as a triple
integral.
mass
Gauss’s law may then be written in this notation:
Sin-dszfpdv
(3)
V
S
Although the above may at first appear cryptic, familiarity
mediately reveal that the left side is the net electric flux out
side is the charge within the region.
Example
with the notation will imof the
region
and the
right
1.5
ROUND BEAM OF UNIFORM CHARGE DENSITY
Consider the circular
cylinder of uniform charge density p and infinite length shown in
Fig.
integration is taken in the form of a prism of square cross section.
region
We will demonstrate the validity of (3), utilizing the fact that D has only a radial
component. The right side of (3) is
1.5. A
of
11/2
1
I
V
FIG. 1.5
p dV
2
b/2
2f dy]
Square cylindrical region
—-b/2
of
p dx
=
lbzp
(4)
-—b/?.
integration
in
a
circular
cylinder
of free
charge.
14
Chapter
At any radius,
D,
can
be found
1
as
Stationary Electric
in Sec. 1.4 to be
D:
I‘
(7773);)
__
2777‘
:2
the left side of
integration
6, that cos 6
Thus
b / Zr and that the four sides make
=
on
(3),
(5)
I
2
To do the surface
cos
Fields
i
we! note
equfal
that D
-
d8
=
D,
dx dz
contributions to the integral.
*
b/2
£1)
and from
(6) and (4)
dS
—
we see
4i dzf
that
D
cos
adx 1192p
.——
(6)
b/2
(3) is satisfied. Problem 1.5b contains a similar situcharge density dependent upon radius.
ation, but is somewhat complicated by having the
E
1.6
FIELD LINES
TUBES OF FLUX: PLOTTING OF
isotropic media, the electric field E is in the same :direction as flux density 1). A
charge—free region bounded by E or D lines must their have the same flux flowing
through it for all selected cross sections, since no flux can flow through the sides parallel
with D, and Gauss’s law will show the conservation of flux for this source-free region.
Such a region, called a flux tube, is illustrated in Fig.j 1.6a. Surface S3 follows the
direction of D, so there is no flow through S3. That
floviIing in S1 must then come out
These
if
there
no
tubes
are
are
internal
charges
82
analogous to the flow tubesIn the
fluid analogy usedin Sec 1. 3.
The concept of flux tubesIS especially useful1n making maps of the fields, and will
be utilized later (Sec. 1.19)in a useful graphical field-mapping technique. We showin
the following example how it may be used to obtain field lines in the vicinity of parallel
line charges.
For
Example
a6
FLUX TUBES AND FIELD LINES ABOUT PARALLEL
I
LINES
OF
OPPOSITE CHARGE
flujx
To show how field lines may be found by constructing
tubes, we use, as an example,
of
lines
It
is
obvious from symmetry that
infinitely long, parallel
opposite charge.
the plane in which the two charge lines lie will
D lines and hence can be a
two
contain
boundary
of
a
flux tube. We introduce the
fluxfuncz‘ionf
¢I=LD-dS
(1)
1.5
Tubes of Flux:
plotting
of Field Lines
15
31
FIG. 1.6a
which
Tube of flux.
the flux
crossing some chosen surface. For example, suppose S is the
Fig. 1.6b. First, let the angles or! and a2 shrink to zero so S
also vanishes. Then the flux function will be zero along the plane of the lines of charge.
(The surface S also could be chosen in some other way, making the flux function
different from zero on that plane. The resulting additive constant is arbitrary, so we
take it to be zero.) We get other flux tube boundaries by taking nonzero values of the
angles 011 and a2. First we derive an expression for the flux passing between the 111
0 plane and line L in Fig. 1.6b. Then paths will be formed along which L may be moved
0 surface. Moving L along such
while keeping the same flux between it and the 2/1
a path therefore generates a surface which is the boundary of a flux tube of infinite
length parallel to L. The flux may be divided into the part from the positive line charge
and the part from the negative line charge, since the effects are superposable. The flux
from the positive line goes out radially so that the amount (per unit length) crossing S
is q,(a, /277). The flux passing radially inward toward the negative line charge through
S adds directly to that of the positive line charge and has the magnitude q,(a2/27T). The
total flux per unit length crossing S is
measures
cross-hatched surface in
=
=
(it
FIG. 1.6b
=
Eq-I—(oz1
7r
+
(2)
a2)
Construction of flux tubes about line
charges.
l6
Chapter
FIG. 1.6a
1
Stationary Electric Fields
Tubes of flux between
line charges.
generated by moving L in such a way as to keep (,0 constant is a circular
that
cylinder
passes through the charge lines with its axis in the plane normal to the
line
0
midway between the line charges. Figure 1.6c shows several flux tubes,
[,0
The surface
==
with the values of
1/! indicating
the amount of flux between the
:11
=
0 surface and the
being considered,
a2 is increased from zero to 2'77. Note that as a path is taken
around one of the lines, the flux function goes from O to q,; the total flux per unit length
coming from a line charge is q,. It is clear from this example that the flux function 1,1:
one
as
is not
are
single-valued since
crossed
as
it continues to increase
motion about the line
charge
as a1
continues.
0:2 to the range 0 to 273' to ensure unique values for
Since the boundaries of the flux tubes lie along D
a2 increases; more flux lines
We must therefore limit 011 and
or
4/.
vectors and D
2
8E, they also lie
along E vectors. Thus, by plotting flux tubes, we find the directions of the electric field
vectors surrounding the charges. There is a given
of flux in each tube so the
flux
density D,
and therefore also E, become
large
amount
where the cross section of the tube
becomes small.
1.7
Since
of the
force
CONSERSVATIVE PROPERTY
ENERGY CONSIDERATIONS:
OF ELECTROSTATIC FIELDS
charge placed in the vicinity of other charges experiences a force, movement
charge represents energy exchange. Calculation 10f this requires integration of
components over the path (line integrals). It will lbe found that the electrostatic
a
Energy Considerations: Conservative Property
3.7
system is conservative in that
about
a
closed
path, returning
Consider the force
in the
of
small
on a
net energy is exchanged if
its initial position.
no
to
positive charge Aq
'5 7
of Electrostatic Fields
a test
moved from
charge
infinity
is moved
to a
point
P
vicinity
system of positive charges: (11 at Q1, q2 at Q2, q3 at Q3, and so on
at any point along its path would cause the particle to accelerate
The
force
(Fig. 1.7a).
and move out of the region if unconstrained. A force equal to the negative of that from
a
applied to bring Aq from infinity to its final position.
Aq from q1 in the system is the negative of the force
of the path, multiplied by differential path length:
must then be
surrounding charges
The differential work done
component in the direction
on
dU1
—-F1-dl
:
Or, using the definition of the scalar product, the angle 9 as defined in Fig. 1.7a, and
as stated above, we write the line integral for total work related to (11 as
the force
PQ‘
U1
where
r
._L
:
PQ‘
F,
d1
-
A
cos
6 dl
‘L ————qqimz
=
(1)
is the distance from q1 to the differential path element (ii at each
Since all cos 6 is dr, the integral is simply
point
in the
integration.
U1
47781‘2
no
and
similarly
for contributions from other
PQI
U
=
-J
co
Note that there is
itself.
no
charges,
PQ2
A
flair
’)
471'81‘“
dr
JPQi Aqql
._.
~J
so
PQ3
A ([612
(17‘
9
4778f“
on
that the total work
—f
integral
is
A
—q€3—d)
471'81‘2
component of the work arising from the
test
charge acting
upon
Integrating,
__
A4611
_
4778PQ1
A4612
+
A443
+
+
47TSPQ3
47TSPQ2
El
P
6
dl
/
/
/
/r
/
.q
Q:
//
a.
91
FIG. 1.7a
Integration path
”’23:
for force
on
test
charge.
(2 )
'98
Chapter
1
Stationary Electric Fields
Equation (2) shows that the work done is a function only of final positions and not
path of the charge. This conclusion leads to another: if a charge is taken around
closed
any
path, no net work is done. Mathematically this is written as the closed line
3
integral
of the
l
ngocur—o
(3)
,
general integral signifies that the component of electric field in the direction of
path is to be multiplied by the element of distance along the path and the sum taken
by integration as one moves about the path. The circle through the integral sign signifies
that a closed path is to be considered. As with the designation for a general surface or
volume integral, the actual line integration cannot be performed until there is a specification of a particular path and the variation of E about that path.
In the study of magnetic fields and time—varying electric fields, we shall find corresponding line integrals which are not zero.
This
the
‘
Exampfie
1.7
-'
DEMONSTRATION OF CONSERVATIVE PROPERTY
To illustrate the conservative property and the use of line
integral (3) around the somewhat arbitrary path through
density
p shown in
FIG. 1.7b
to show
Path of
Fig.
1.7b. The
path
is chosen, for
integration (broken line) through electric
conservative property.
integrals,
a
let
uniform
simplicity,
field of
us
take the line
sphere
of
to lie in the
sphere
of uniform
charge
x
=
0
charge
The
plane.
integrand E dl
pr / 3 80 is
E
-
electric field
and
E2
2:
2:
E,.(z/r)
E},
so
19
Electrostatic Potential: Equlpotentlals
1.8
involves electric field components
E), and E2. The radial
Eq. 1.4(7). The components are
E,( y/ r)
found from
py/38O
=
and
E),
E5
’72
02
The
pz/3so.
=
integral (3)
1'72
2
..
The
general
example.
yz
b2
2
+—
b1
C1
Eydy
b1
.2
a
+._.....
becomes
1’!
Cl
36E.dl=£1E_,dz+fblEydy+£Ezdz+f
2.3:. :
380 2
=
[’2
C2
}
z
0
conservative property of electrostatic fields is thus illustrated in this
1.8
ELECTROSTATIC POTENTIAL: EQUIPOTENTIALS
The energy considerations of the preceding section lead directly to an extremely useful
concept for electrostatics—that of potential. The electrostatic potential function is de-
fined
as
the work done per unit charge. Here we start generally and define a potential
points 1 and 2 as the work done on a unit test charge in moving
difference between
from
P1
to
P2.
P2
(13,,o
—
_
(DPl
=
E
-
-
d1
(1)
P:
The conclusion of the
preceding section that the work in moving around any closed
potential function is single-valued; that is, corresponding
to each point of the field there is only one value of potential.
Only a difference of potential has been defined. The potential of any point can be
arbitrarily fixed, and then the potentials of all other points in the field can be found by
application of the definition to give potential differences between all points and the
reference. This reference is quite arbitrary. For example, in certain cases it may be
convenient to define the potential at infinity as zero and then find the corresponding
potentials of all points in the field; for determination of the field between two conduc»
tors, it is more convenient to select the potential of one of these as zero.
If the potential at infinity is taken as zero, it is evident that the potential at the point
P in the system of charges is given by U of Eq. 1.7(2) divided by Aq, so
path
is
zero
shows that this
(D
2
41
This may be written in
a more
versatile form
seen
R,. is
in Fig.
the distance of the ith
1.80.
=
charge
Z
at
43
+
4778PQ2
@(r)
where
£12
+
4778PQ1
+
"'
47TSPQ3
( 2)
as
i-
r;-
from the
(3)
point
of observation at r,
as
23
Chapter
1
Stationary Electric
Fields
yi
oq5
3
qu‘
P
l‘
4:1:
2'
FIG. 1.8a
R.-
=
lr
-
Potential of q1, q2, (13,
ri-i
[(x
=
-
act->2
+
.
.
.
(y
q; is
,
—
found at point P.
yi)ti+
(2
—
210211”
(4)
the
Here x, y, and z are the rectangular coordinates of
point of observation and
the
ith
coordinates
are
the
of
and
charge.3 Generalizing to the case
rectangular
z}
XE, YE,
of
continuously varying charge density,
(Mr)
The
—
p(r’)dV'
(5)
478R
p(r') is charge density at point (x’, y’, z’), and the integral signifies that a summation
should be made similar to that of
potential
is
[V
_
zero
zero
is not at
infinity,
at the desired reference
(2) but continuous over‘ all space. If the reference for
a constant must
potential
position.
(1)0.)
It should be
be added of such value that
_
_
J’V
p<r')dV'
+ C
(6)
4778R
kept in mind that (2)-—(6) were derived assuming that the charges are located
infinite, homogeneous, isotropic medium If conductors or dielectric discontinu—
ities are present, differential equations for the potential (to be given shortly) are used
in
an
for each
region.
We will
is
usually
see
in Sec. 1.10 how the electric field E
easier to find the
it, the field E than to do the
potential by
vector
in electrostatic calculations is
the scalar
can
be found simply
operations
summations discussed
in
in Sec.
from
(Mr).
It
(3) and (5) and, from
1.2. Such convenience
for
introducingithis potential.
In any electrostatic field, there exist surfaces on which ihe potential is a constant, socalled equipotential surfaces. Since
the potential is singlei—valued, surfaces for different
3
one reason
Throughout the text we use primed coordinates to designinte the location of sources and
unprimed coordinates for the point at which their fields tire to be calculated.
1.8
values of
potential
surface in
no
a
do not intersect. The
pictorial representation
three-dimensional field distribution is
variation in
dimension
one
21
Electrostatic Potential: Equipotentlals
dimension and
are
quite
of
more
than
one
such
difficult. For fields that have
therefore called two—dimensional
fields, the third
be used to represent the potential. Figure 1.8b shows such a representation for the potential around a pair of infinitely long, parallel wires at potentials CD1 and
can
height of the surface at any point is the value of the potential. Note that
height or constant potential can be drawn. These equipotentials can
projected onto the x—y plane as in Fig. 1.8c. In such a representation, the equipo-
—CI>1.
The
lines of constant
be
(b)
(C)
FIG. 1.8
the
(b) Plot of a two-dimensional potential distribution using the third dimension to show
same potential system as in (b) plotted onto the x—y plane.
potential. (c) Equipotentials for the
22
Chapter
Stationary Electric
1
tentials look like the contour lines of
energy of
a
unit
charge
relative to
a
and, in fact, measure potential
zero-potential point just as contours meas—
reference altitude, often sea level. It should be
topographic
potential energy relative to some
kept in mind that these lines are actually
cylindrical equipotential surfaces.
boundary
map
selected
a
ure
We will discuss the
Plfelds
traces in the x—y
conditions
on
plane
conductors in
of three—dimensional
some
detail in Sec. 1.14.
point it is sufficient to say that the electric fields inside of a metallic conductor
be considered to be zero in electrostatic systems. Therefore, (1) shows that the
At this
can
conductor is
an
equipotential region.
Exampfie
POTENTIALS AROUND
A
r .Sa
LINE CHARGE AND BETWEEN COAXIAL CYLINDERS
As an example of the relations between potential and electric field, consider first the
problem of the line charge used as an example in Sec.1 .4, with electric field given by
Eq.1.4.(3) By (1) we integrate this from some radius 10 chosen as the reference of zero
potential to radius r:
'
@=*J'ErdI:-JLL=—I—ql—hli‘
2778
r
r
0
Or this
expression
for
0
potential
about
a
line
”3—4—
(I:-
a
27781'
charge
(7)
r0
may be written
lnr + C
(8)
271'8
infinity as the reference of zero potential for the
charge,
by (7)
potential at any finite point would be infinite. As in (6)
the constant is added to shift the position of the zero
potential.
In a similar manner, the potential difference between
coaxial cylinders of Fig. 1. 4b
Note that it is not desirable to select
line
for then
the
theI
may be found:
I
®a-©b:"ffl::q—Irn(2>
1,
27781‘
27119
a
(9)
1. 8b
SPHERICALLY SYMMETRIC CHARGE
Example
POTENTIAL OUTSIDE
We
saw
in
Eq. 1.4(4)
=Q./4m Using
finity,
we see
that the
A
that the flux
density outside a spherically symmetric charge Qrs
and taking the
referen<'g:e potential to be zero at inpotential outside the charge Q is the negative of the integral of
E:
D/so
g’
f‘Er'
f‘ dr from
infinity
to
23
Electrostatic Potential: Equlpotentials
1.8
radius
r:
r
(130‘)
=
-
erI
4723017
Q
._
Example
(10)
_.
9
00
477801‘
1.80
POTENTIAL OF A UNIFORM DISTRIBUTION OF CHARGE
HAVING SPHERICAL SYMMETRY
Consider
(I)
=
O at
a
2‘
volume of
==
00, the
charge density p
potential outside a
that extends from
is
given by (10)
=
r
with
Q
0 to
=
r
=
(1.
érreag’p,
Taking
so
3
(130-)
In
particular,
at
r
z
=
51—3
r 2 a
a
W)
“4"
==
a we
must add to
given
as
E,
=
(12) the integral of the
(Ex. 1.7) and the
pr / 330
is
©(r)
-—
(13(a)
—-f
=
a
So the
(12)
380
Then to get the potential at a point where r S
electric field from a to r. The electric field is
integral
(11)
3802'
potential
at a
radius
r
inside the
@(r)
=
35-1-
380
drl
=
680
charge region
~9— (3a2
—-
680
Jo— (a?-
—
1'2)
(13)
is
r2)
1‘ s a
(14)
Example 1.8d
ELECTRIC DIPOLE
particularly important set of charges is that of two closely spaced point charges of
opposite sign, called an electric dipole.
Assume two charges, having opposite signs to be spaced by a distance 28 as shown
in Fig. 1.8d. The potential at some point a distance r from the origin displaced by an
angle 6 from the line passing from the negative to positive charge can be written as the
sum of the potentials of the individual charges:
A
1
1
c1:
=
i<—~ —)
—
477'8
r+
r
(15)
26
chapter
Stationary Electric
1
fields
-+++4P4HP4¥+4? «1“HHHHHHHH-«Ind-
4|- 1?
4|- ‘Ih’ir
0- 4b
1»
#115 1|- 1? 4P
«1-
1?: 1!- fl- 1|- 4» 15
1P 4?-
1» 4r {1-
1;
1p
19
11-
~a-_----“-—b—v--n-‘—----—‘--’-‘
i
FIG. 'I .9
(a) Parallel-p1ane capacitor with
fnngmgifields (b) Idealization.
charge density p5 on each plate. Since the field E
potential difference (Pa
(1),, is, by Eq. 1.8( 1),
the surface
=
D / s is assumed uniform,
'
the
-
%—%=i
@
8
The total
charge on each plate of areaA is pSA so (1) and 62) give the familiar expression
C
=
--
F
practice, (3)18 modified by the fringing fields, whlcht are increasingly important as
plate spacing to area is increased.
Next consider a capacitor made of coaxial, circular cylindrical electrodes We assume
that fields are only radial and neglect any fringing at the ends if it is of finite length
The charge on each conductor is distributed uniformly in this idealization, as required
by symmetry, with the total charge per unit length being q, The potential difference
found from the field produced by this chargeIS given by Eq. 1. 8(9) The capacitance
per unit length18 thus
In
the ratio of
271's
C:
ln(b/a)
F/In
(4)
a are the radii of larger and smaller
conductors, respectively.
Finally, consider two concentric spherical conductors bf radii a and b, with b > a,
D/s, it is clear that
separated by a dielectric 3. Using symmetry, Gauss’s law, and E
where b and
=
at any radius
Er
where
Q is the charge
on
=
Q
4782‘2
the inner conductor
(equal in magnitude
(5)
and
opposite
in
sign
to
the
(Pa
charge on
-
CIDb
27
Gradient
1.10
the inside of the outer
and this, substituted into
sphere). Integration of (5)
(1), yields
47T8
C
between
spheres gives
47rsab
_
(6)
_
~-(l/a)~--(1/b)—-
b
-—
a
The flux tubes in these three
highly symmetric structures are very simple, being
example and by cylindrically or spherically
radial surfaces in the last two. In Sec. 1.21 we will see a way of finding capacitance
graphically for two-dimensional structures of arbitrary shape in which the flux tubes
have more complex shapes.
Capacitance of an isolated electrode is sometimes calculated; in that case, the flux
from the charge on the electrode terminates at infinity and the potential on the electrode
is taken with respect to an assumed zero at infinity. More extensive considerations of
capacitance are found in Sec. 4.9.
bounded
by parallel
surfaces in the first
mu"its”;fifi’SfiééKfié‘fim‘figzfifimfifi’sfifi”?”Rf”a9353§$§52fl°§3fi§§Vfi¥§§§¥§$€nW ”ifkii‘éwfikfii'i’skfigi’fizfi“3/4”,Kraft?” triflzi‘fifi Wm: 2‘3 3‘
m2: (at)?
‘K‘é‘.
'.
ES 73‘ Lima; 3%?“ 237’. 91.37.31’ai‘1‘23'ifikz» 7% Vin)» "1233424? 61 an:
Eifierentiafi Forms of Etectmstatic laws
GRADIENT
HO
macrosc0pic forms. It is also useful
equivalents in differential forms. Let us start with the relation between
and potential. If the definition of potential difference is applied to two
We have looked at several laws of electrostatics in
to have their
electric field
points
a
distance dl apart,
dCD
=
——E
.
dl
(1)
where d] may be written in terms of its components and the defined unit vectors:
(2)
dlzftdx—l-ydy—l—idz
We
expand
the dot
product:
(M)
Since CD is
a
-(E_\.
=
dx +
E3 dz)
=
—-~
‘
5
+
ax
comparison
of the two
6x
——
6y
dy +
——
dZ
62
expressions,
6(1)
3(1)
6(1)
Ex==--—
6(1)
6(1)
6(1)
a
+
function of x, y, and z, the total differential may also be written
dd)
From
E}, dy
’
E=-—
y
6y
’
E-=-—
‘
62
(3)
28
Chapter
Stationary Electric
1
fields
I
so
6(1)
6(1)
E
2
..
6(1)
(X6x+yay+z 62)
‘—-
“—
4
()
or
E:
~grad<l>
(5)
grad 43, an abbreviation of the gradient of (I), is a vector showing the direction
magnitude of the maximum spatial variation of the scalar function (I), at a point in
where
and
‘-
space
Substituting
back in
(1),
we
have
=
(grad <13)
-
dl
(6)
change in (I) is given by the scalar product of the gradient and the vector (11,
a given element of length (11, the maximum
:value of dCD is obtained when
element is oriented to coincide with the direction ofi the gradient vector. From (6)
Thus the
so
that, for
that
grad (I)IS perpendicular to the equipotentlals because 61(13- 0 for
along
equipotential
The analogy between electrostatic potential and grav1tational potential energy discussedin Sec. 1. 8IS useful for understanding the gradient. It13 easy to see in Fig 1. 8b
that the direction of maximum rate of change of potential is perpendicular to the
equipotentials (which are at constant heights on the potential hill).
If we define a vector operator V (pronounced del)
it is also clear that
d]
-
an
Q
v
then
grad (I)
may be written
as
——
act:
——
+
6x
VCI) if the
VCID—
6
a
2
X
-—
6}:
y
+2
-—
6y
6'
(7)
62g
operation18
intilerpreted
acb
an)
6y
62
as
+ y—- + 2 —‘~
(8)
and
.
E
The
gradient operator
inside front
in circular
=
—~grad<I)
cylindrical
Q:
and
I
+ch
(9)
spherical
coordinates is
given
on
the
cover.
Exampie
1.10
i
ELECTRIC FIELD OF A DIPOLE
I
As
of the
of the
gradient operator, we will find an expression for the
field around an electric dipole. The potential for a dipole is given in Sec. 1.8 in spherical
coordinates so the spherical form of the gradient operator is selected from the inside
an
example
use
1.: r
front
The
Divergence or
29
eteaunsmic Fleld
an
cover.
at?
V<I>= r
—
A
+ 6
61'
and
Substituting Eq. 1.8(16)
Vq)
=
1
a)
+
39
r
the
noting
1%
1‘ Sin
independence
of
E
6
49¢
43,
we
get
qucoseiéfiqsmG
net-3
2'Irer3
Then from (9) the electric field is
'
8
E
:
(9
‘73
1T8"
The second differential form
by
9 + e
2
we
shall consider is that of Gauss’s law.
D dS
L~
.
Mac
right
side is,
Equation 1.5(3)
by inspection, merely
good place
as
and their sources; the
to comment on
dV
Li—
AV~>D
(1)
AV
p. The left side is the outward electric flux
the
div D
a
lim
=
AV
per unit volume, This will be defined
D, Then
This is
(10)
the volume element AV and the limit taken:
lim
The
9)
THE DIVERGENCE OF AN ELECTROSTATIC FIELD
1.11
may be divided
5'"
.1
cos
divergence affiux density, abbreviated div
=
(2)
p
the size scale
implicit
in
our
treatment
of fields
apply to the central set of relations, Maxwell’s
equations,
building. In reality, charge is not infinitely divisible—
the smallest unit is the electron. Thus, the limit of AV in (I) must actually be some
small volume which is still large enough to contain many electrons to average out the
granularity. For our relations to be useful, the limit volume must also be much smaller
than important dimensions in the system. For example, to neglect charge granularity,
the thickness of the depletion layer in the semiconductor in Ex, 1.4a should be much
greater than the linear dimensions of the limit volume, which, in tum, should be much
greater than the average spacing of the dopant atoms. Similarly the permittivity e is an
average representation of atomic or molecular polarization effects such as that shown
in Fig. 1.3c. Therefore, when we refer to the field at a point. we mean that the field is
an average over a volume small compared with the system being analyzed but large
enough to contain many atoms. Analyses can also be made of the fields on a smaller
scale, such as inside an atom, but in that case an average permittivity cannot be used.
In this book, we concentrate on situations where p, 5,, and other quantitmyerages
comments
toward which
we are
also
33
over
Chapter
small volumes. There
are
thin films of current
few atoms thick and semiconductor devices such
ditions
are
must be
not well
satisfied,
as
practical interest which are only a
that of Ex. 1.4a where these cone
calculations using
that-results from
so
Pgtelds
Stationary Electric
I
p and
e as
defined
used with caution.
seeking an understanding of (:2). Consider the infinitesimal
rectangular parallelepiped of dimensions Ax, Ay, Az as shown in Fig.
1.11a. To compute the amount of flux leaving such a volume element as compared with
that entering it, note that the flux passing through any :face of the parallelepiped can
differ from that which passes through the opposite face only if the flux density perpenNow let
volume
us return to
as a
dicular to those faces varies from
faces is small, then to
two faces will
a
simply
distance between faces.
we
to
first
one
If the distance between the two
face to the other.
approximation
be the rate of
the
change
difference in any vector function on the
of the
function
According
calculus,
higher-order differentials are
basis of
to the
the limit, since the
pass
If the vector D at the center x, y,
2
has
a
Dx(x >
Dix 7)
~
—-
2
Dx(x)
+
Ax
-
=
then
exactly
correct
when
zero.
component 12x00, then
Ax
+
with distance times the
this is
w
--
AféDx(x)
2
6x
Ax
anxx
( )
(3)
1
—————
a.
show
In this functional notation, the arguments in parentheses
the points for evaluating
the function Dx. When not included, the point (x, y, 2) Will be understood. The flux
flowing out
Ay Az Dx(x
the y and
2'
the
-
right face is Ay Az Dx(x + Alf/2), and that flowing in the left face is
Ax/Z), leaving a net flow out of Ax Ay ‘ZAz(an/Bx), and similarly for
directions. Thus the net flux flow out of the
AxAyAz
6Dx
6x
parallelepiped is
6D
+
AxAyAz—l
33’
aD+
AxAyAz—i
‘33
y
A2
Dx<x
-
¥)——
2
y
9
(syn)
at
it:
+
Ag‘—>
2
i
Ax
}
j
2
FIG. 1.11:!
Volume element used in div D derivation.
Ara:
The
1.1 l
By
Gauss’s law, this must
p Ax
equal
60.
A
Jr.
D in
expression for div
33
Electrostatic Field
an
A2. So, in the limit,
Ay
6Dy
__
6}:
An
of
Divergence
aD,
+
_-
(4 )
.—
p
__
62
By
rectangular coordinates is
obtained
8D.
6D.
aD~
6x
By
62
by comparing (2)
and
(4):
disz—¢+—i+—‘
If
we
make
that div D
of the vector operator V defined by Eq. l.lO(7) in (4), then (5) indicates
conveniently be written as V D. It should be remembered that V is not
use
can
a true vector
another
(5 )
—
but rather
quantity
in
a
a
operator. It has
vector
defined
manner.
meaning only
Summarizing,
when it is
operating
V-DédivD=&+@+E=p
By
6x
on
(6)
62
divergence is made up of spatial derivatives of the field, so (6) is a partial
equation expressing Gauss’s law for a region of infinitesimal size. The
physical significance of the divergence must be clear. It is, as defined, a description of
the manner in which a field varies at a point. It is the amount of flux per unit volume
emerging from an infinitesimal volume at a point. With this picture in mind, (6) seems
a logical extension of Gauss’s law. In fact (6) can be converted back to the large-scale
form of Gauss’s law through the divergence theorem, which states that the volume
integral of the divergence of any vector F throughout a volume is equal to the surface
integral of that vector flowing out of the surrounding surface,
The
differential
v
Although not a proof,
multiplied by volume
V-FdV:§F‘dS
(7)
5
this is made
plausible by considering Fig.
1.1117. The
element for each elemental cell is the net surface
\
\\
\
\ix
\K
/
\
e
divergence
integral out of
“
_.
~
\
\
\\
\T\F
FIG. 1.1 'l b
Solid divided into subvolumes to illustrate the
divergence
theorem.
32
Chapter
that cell. When summed
by integration,
Stationary Electric Fields
1
all internal contributions cancel since flow out
of one cell goes into another, and only the external surface contribution remains.
D gives
Applications of (7) to (6) with F
=
35D.ds:iv.DdV=ipdV
V
S
whichis the
original
(8)
‘V
Gauss’ 3 law.
expressions for the divergence; and other operations involving
treatment of problems having corresponding
an
as
Let
us,
symmetries.
example, deve10p here the idivergence of D in spherical
coordinates.4 We use the left side of (1) as the definitiojn of div D and apply it to the
differential volume shown in Fig. 1.11c. We will find first the net radial outward flux
from the volume. Both the radial component Dr and the element of area 7'2 d6 sin 6 dqfi
change as we move from r to r + dr. Thus the net flux flow out the top over that in at
It will be useful to have
Vin other coordinate systems for simpler
the bottom is
arr/1r
(r
_—.
To first-order
dr)2
+
6d6dd><
r
r
a ,.
r)
2sin 6d6
sin 9 dr d6
ll
for the 6 and
:5
dqsaalirdr
(145
+ 27' dr sin 6d6
—
divergence
=
d¢
6
sin 6
dd (11)
=
1'
dr d6
4
dip,
+
dqb
6—6
(sin 6 D6)
i
7‘
d6
d7.)
=
i'
(11' d6
dill divided by
(1%
+
the
d¢
3D¢
Ta?
element of volume
Lil/1'45
r2 sin 6dr as d¢
(9)
16(2-2D)+r$111160608(
V-D
cover
a
(D45
is then the total
V-D=
£1ng
5-,. (2-20,)
6
dlil’gb
For the
r2 sin 6 d6 quDr
directions,
d6:«(D92
cit/16
The
—
differentials, this leaves
dill,
Similarly
sin
.......__
rr26
———
’
corresponding expression
in circular
1
In
9
6D)+
16¢
111669!)
cylindrical coordinates,
see
inside front
and Prob. 1.116.
Note that here,
of the
from
as
with other curvilinear coordinate
systems, it is not the scalar product
gradient operator and the vector in spherical coordinates,
the basic definition given by ( l) and (2).
but must be obtained
Laplace’s and Poisson's Equations
1.12
FIG. 1.1'lc
Element of volume in
Example
spherical
33
coordinates.
1.“
UNIFORM SPHERE OF CHARGE
We will show
an example of the application of (6) using a
sphere of charge of uniform
density p having a radius a with divergence written in spherical coordinates. Since the
D
O, the last two terms of
spherical symmetry of the charge region leads to D 9
d,
(9) vanish. We use the value of Dr
rp/ 3 from Eq. 1.4(7) for the region inside the
Charge sphere and obtain
=
=
2
1
V
-
D
=
a
“‘3“[(1‘2)<fl>]
r"
For the
region
outside the
sphere
'
we use
=
3
r
Dr(r)
=
a3
p
p/3r2 from Eq. 1.4(8)
(10)
to show
that
l
V
D
-
N
a
03p
‘261‘[(] (3,2)]
7
_
—
0
(11)
example shows that divergence is zero outside the charge region but equal to
charge density within it. The same result is obtained if one uses divergence expressed
in rectangular coordinates or other coordinates not so natural to the symmetry (Prob.
This
1.1
1d).
1.12
LAPLACE'S
POISSON’S EQUATIONS
preceding sections allow us to derive an important
equation for potential. Differential equations can be applied to problems
general than those solved by symmetry in the first part of the chapter and it is
The differential relations of the two
differential
more
AND
34
Chapter
is
fields
potential as the dependent variable. This is because
specified boundary conditions are often given in
often convenient to work with
potential
Stationary Electric
1
scalar and because the
a
terms of
potentials.
permittivity s is constant throughout the regiOn,
Eq.1.10(5)1n Eq. 1.11(6) with D— 8E yields
If the
the substitution of E from
—’
div(grad<1>)
V
=
-
V61)
=
But, from the equations for gradient and divergence
1.10(7) and 1 11(6)],
so
—
ml‘D
rectangular
1I1
coordinates
[Eqs
a
6ch
az_®
(12(1)
6x2
Ely2
622
( )
that
62—4)
This is
a
differential
density
at
that
point
62(1)
:3ch
6322
62?-
_
6x2
equation relating potential
and is known
Paisson’s
as
v2<1>
__,_J
is
variation
equation.
V2613 (del squared of (1)) is known
V261)
In the
special
case
of
a
9
V
-
as
charge-free region,
(9ch
9
ax~
the
W)
at any point to the
It is often written
—B
=
=
charge
(3)
8
where
(2)
i
Laplacicin of (I).
div(grad (1))
Poisson’s
equation
(4)
reduces to
62(1)
6ch
+——+
ay2
6—22
=0
01'
VZCI) O
(5)
-—
whichIS known
form, V2
can
Laplace’ 3 equation. Although illustrated1n its rectangular coordinate
expressedin cylindrical or spherical coordmates through the relations
as
be
given on the inside front cover.
Any number of possible configurations
of (3)
the conditions
ments
i
and
(5). All
are
of
potential suirfaces
called solutions to these
will
satisfy
the
require—
equfations. It is necessary to know
existing around the boundary of the region to select the particular solution
which applies to a given problem. We will see in Sec. 1.17 a proof of the uniqueness
of a function that satisfies both the differential equation and the boundary conditions.
Quantities other than potential can also be shown to satisfy Laplace’s and Poisson’s
equations, both in other branches of physics and in other parts of electromagnetic field
theory. For example, the magnitudes of the rectangular components of E and the component E: in cylindrical coordinates also satisfy Laplace"s equation.
1.13
Static Fields
from
Arising
Steady
35
Currents
A number of methods exist for solution of two— and three~dimensional
problems with
Laplace’s
equation.
separation
technique a very general
method for solving two-- and three—dimensional problems for a large variety of partial
differential equations including the two of interest here. Conformal transformations of
complex variables yield many useful two-dimensional solutions of Laplace’s equation.
Increasingly important are numerical methods using digital computers. All these methods are elaborated in Chapter 7. Examples for this chapter, after discussion of boundary
conditions, will be limited to one~dimensional examples for which the differential equations may be directly integrated. These show clearly the role of boundary and continuity
or
Poisson’s
of variables
The
is
conditions in the solutions.
1.13
STATIC FIELDS ARISING FROM STEADY CURRENTS
arising from dc potentials applied to conductors are not static becharges producing the currents are in motion, but the resulting steady~state
fields are independent of time. Quite apart from the question of designation, there is a
close relationship to laws and techniques for the fields arising from purely static charges.
We consider ohmic conductors for which current density is proportional to electric
field E through conductivity5 0" Siemens per meter (S / m):
Stationary
cause
currents
the
J
Such
relationship
a
comes
in Sec. 1.10,
be
a
steady
more
a
in
scalar
Chapter
potential
so
stationary
zero
(1)
of time, it is derivable from
independent
J
For
GE
from internal “collisons” and is discussed
13. Since the electric field is
as
=
current,
~0VCI)
=
(2)
continuity requires that the net flow
a buildup or decay of charge
since there cannot be
out of any closed
within the
region
region
in the
state,
jg
J
'
dS
m
0
(3)
s
or
in differential form,
V
Substitution of (2) in (4), with
0
taken
'
as
J
=
(4)
O
constant,
yields
V-Vcbévchzo
Thus
potential
Laplace’s equation as in other static field problems (Sec. 1.12).
boundary condition corresponding to the applied potentials, there is
satisfies
In addition to the
5
The 5/ unit for conductivity is Siemens per meter (8/ m), but older literature sometimes
mho
as
the
(5)
conductivity unit.
uses
36
a
Chapter
1
Stationary Electric
constraint at the boundaries between conductors and
current flow across such
boundaries,
where
n
on
boundaries.
Referring
to
[Fields
insulators since there
(2),
this
requires
can
be
no
that for such
the conductor side,
—=0
(6)
denotes the normal to such boundaries.
between different
conductors
are
considered in the
Continuity relations
following section.
I
1.14
BOUNDARY CONDITIONS IN ELECTROSTATICS
practical field problems involve systemscontainiiig more than one kind of maseen some examples of boundaries between various regions in the
examples of earlier sections, but now we need to develop these systematically to utilize
the differential equations of the preceding sections.
Most
terial. We have
Let
us
consider the relations between normal flux
density
components
across an
arbitrary boundary by using the integral form of Gauss?s law. Consider an imaginary
pillbox bisected as shown in Fig. 1.14a by the interface; between regions 1 and 2. The
thickness of the pillbox is considered to be small enough that the net flux out the sides
vanishes in comparison with that out the flat faces. If we assume the existence of net
surface charge p3 on the boundary, the total flux out of the box must equal p8 AS. By
Gauss’s law,
D n1 AS
FIG. 1.340
—
Boundary
1),,2 AS
:
pSAS
between two different media.
Boundary Conditions
1.14
37
in Blectrostatics
01'
D11]
'—
DnZ
:
(1)
pr
where AS is small
enough to consider D" and p5 to be uniform.
A second relation may be found by taking a line integral about a closed path of length
Al on one side of the boundary and returning on the other side as indicated in Fig.
l.l4a. The sides normal to the
net
boundary are considered to be small enough that their
integral vanish in comparison with those of the sides parallel
By Eq. 1.7(3) any closed line integral of electrostatic field must be zero:
contributions to the
to the surface.
%
E
‘
d].
:
Ell A]
Etz
—
A]
=
O
01‘
Etl
:
(2)
Erz
subscript t denotes components tangential to the surface. The length of the tangenloop is small enough to take E as a constant over the length. Since the
the boundary is negligibly small,
E
of
across
integral
The
tial sides of the
,
CI)1
across
the
boundary. Equations (1)
Consider
(1)
(2),
(3)
or
(1) and (3), form
conditions for the solution of electrostatic field
boundary
and
and
CD2
=
an
interface between two dielectrics with
no
problems.
charge on
a
complete
set of
the surface. From
Eq. 13(1).
81
E211
=
82
(4)
E122
It is clear that the normal component of E changes across the boundary, whereas the
tangential component, according to (2), is unmodified. Therefore the direction of the
resultant E must
Suppose
region 1
that at
makes
change across such a boundary except where either ER or E, are zero.
point at a boundary between two dielectrics the electric field in
an angle 91 with the normal to the boundary. Thus, as seen in Fig.
some
1.14b,
51
01
Enl
Etz
Ed
92
52
FIG. 1 .14!)
dielectrics
Vector relations among electric field components at
(:32
>
8,).
E2
En2
a
point
on a
boundary
between
38
Chapter
Then
and
using (2)
(4)
we see
01
=
62
a
"1—”—
t
'1
t an
—’2
(6)
E122
i
=
tan
82
”1
-
tan
81
us now
<5 )
E11
that
92
Let
{Fields
Stationary Electric
1
consider the
(7)
6,
boundary properties of conductors,
as
exemplified by a piece
of semiconductor with metallic contacts. The currents flowing in both the semiconductor
by the respective conductivities so
developed in Sec. 1.1l3.
At the interface of two contiguous conductors, the: normal component of current
density is continuous across the boundary, because othei'wise there would be a continual
buildup of charges there. The normal component of electric field18 therefore discontinuous and given by
and the metallic contacts
potential drops
occur
related to the fields
are
in them,
as
E111
x
92.
E112
‘
(8)
01
The argument used in dielectric problems for tangential components of electric field
also applies1n the case of a discontinuity of conductivity so that tangential electric field
is continuous
In
across
problems
conductor,
the
boundary.
normally
one
the metal. The electrodes
faces. In this text
or
assumes
with that of the
high compared
electrostatic
conductive material such as a semi»
conductivity of the metallic electrodeis so
other material that negligible potential drops occur in
with metallic electrodes
do
are
on a
assumed to be
normally
conduction problems
we
electric fields at their surfaces
less
that the
assume
perfect conductors
for dc
are
problems
equipotentlals
with
equipotential
and therefore the
tangential
are zero.
Example
BOUNDARY CONDITIONS
1.14
IN A DC
!
CONDUCTION PROBLEM
points
The structure in
sur—
that the electrodes in either
we have made about
Fig. l.14c illustrates some of the
boundary
that a potential difference is provided between the .two metallic
electrodes by an outside source. The electrodes are considered to be perfect conductors
and, therefore, equipotentials. The space directly between them is filled with a layer of
conductive material having conductivity at and permittivity a. The space surrounding
the conductive system is filled with a dielectric having permittivity 31 and 0'
O.
conditions. We
assume
=
1.14
Boundary Conditions
39
in Electrostatics
+
+
Conductor
+
+
(1,0
+
+
+
+
+
+
+
+
\
L
Semiconductor
£1
a. 5
Dielectric
y
J/
(I)
=
5 1
Dielectric
0
Conductor
FIG. 1.140
boundary
Structure
involving
both do conduction and static fields in
a
dielectric to illustrate
conditions.
The normal component of the current density at the conductor—dielectric interface
since the current in the dielectric is zero. This follows formally from (8)
must be zero
using
zero
conductivity
for the dielectric. Therefore, since the electric field is J / 0‘ the
normal component of electric field inside the conductor must be zero. Then the electric
fields inside the conductor at the boundaries with the dielectric are wholly tangential.
Since the electrodes
and the field lines
are
are
assumed to be
perpendicular to
It is clear that since there is
permittivity,
there is also
a
The flux must terminate
to-conductor
flux
on
an
equipotentials,
the
tangential
field there is
zero
the electrode surfaces.
electric field in the conductive medium and it has
density following the same paths as
charges so there will be charges
the current
on
a
density.
the electrode»
boundary.
The electric field in the dielectric region terminates on charges both on the perfectly
conducting electrodes and on the sides of the other conductive medium. On the boundary of the imperfect conductor there is a tangential component and also surface charges,
so the field makes an oblique angle with the conductor surface.
49
Chapter
Stationary Electric
1
fields
LAPLACiEE'S
H5 DIRECT INTEGRATION OF
EQUATION:
FIELD BETWEEN COAXIAL CYLINDERS WITH; Two DIELECTRICS
Laplace’s equation we will take is that of finding the
conducting cylinders of radii a and c (Fig.
potential
1.15), with a dielectric of constant 81 filhng the region between a and b, and a second
"dielectric of constant 82 filling the region between I) and c. The inner conductor is at
potential zero, and the outer at potential V0. Because of: the symmetry of the problem,
the solution could be readily obtained by using Gauss Is law as in Example 1 .,4b but
in the solution by
the primary purpose here is to demonstrate severalii
gprocesses
differential equations.
The geometrical form suggests that the Laplacian VZCI) be expressedin cylindrical
coordinates (see inside front cover), giving for Laplace 3 equation
The first
simple example
of
distribution between two coaxial
2
2
VZQD
=
12-0132)
r
It will be assumed that there is
07‘
61'
no
+
a?
Jae—Cg
dq’)
+
2
0
62‘
variation in the axial (z) direction, and the
angle (1). Equation (1) then reduces to
(1)
cylindrical
symmetry eliminates variations with
la
r
dr
612
dr
_0
'
(2)
Note that in
one
(2) the derivative is written as a total derivative, since there is now only
independent variable in the problem. Equation (2) may be integrated directly:
dd)
7
—
Integrating again,
we
C1
—-
d,
(3)
have
(PlzCllnr-i—Cz
FIG. 1.15
Coaxial
cylinders
with two dielectrics.
(4)
Direct
1.15
Integration
of
4'!
Laplace’s Equation
This has been labeled
(131 because we will consider that the result of (4) is applicable
region (a < r < b). The same differential equation with the same
to the first dielectric
symmetry applies to the second dielectric region, so the same form of solution applies
there also, but the arbitrary constants may be different. 80, for the potential in region
2 (b < r < 6), let us write
(132
The
In
boundary
(a) CD1
3
(b) (132
=
Oatr
continuous
(c) (131
Voatr
(d) Dr1
The
D,.2
application
at
conditions at the
continuity
this
(Dzatr
application
(b)
b,
or
81
(a)
(d®1/dl‘)
to
=
32
(dCIDZ/dr)
applied
(4) yields
=
--C1
in
(6)
a
(5) yields
to
(7)
to
(4) and (5) gives
(8)
Cllnb+C2=C31nb+C4
And condition ((1)
be
there.
C4=VO—~C3lnc
Condition (c)
must
b
=
of condition
of
density
charge—free boundary (Sec. 1.14):
=
r
between the two dielectric
boundary
and the normal component of electric flux
C2
The
are:
c
potential
:
(5)
a
=
are
across
=
=
C4
+
r
conditions at the two conductors
addition, there
media. The
1n
C3
=
applied
(4) and (5) gives
to
81
Any one of the constants,
equations, (6) to (9):
as
C1
C1,
C1
=
82
(9)
C3
may be obtained
by eliminating
between the four
V0
=
ln(b/a)
—
(10)
(81/82)1n(b/c)
remaining constants, C2, C3, and C4, may be obtained from (6), (9), and (7),
respectively. The results are substituted in (4) and (5) to give the potential distribution
in the two dielectric regions:
The
(pl
V0 ln(r/a)
:
1n<b/a)
___
VO[
1
'—
+
“
< r <
b
(11)
(er/82) 1n<c/b)
(81/82) 1n<c/r)
1n(b/a) + (81/82)
1n(c/b)]
b <
'i
< C
(12)
42
Chapter
l
Stationary Electrlcjfilields
be checked that these distributions do satisfy Laplace’s equation and the boundary
continuity conditions of the problem. Only in such simple problems as this will it
be possible to obtain solutions of the differential equation by direct integration, but the
method of applying boundary and continuity conditions to the solutions, however obtained, is well demonstrated by the example.
It
can
and
Ho
DIRECT INTEGRATION OF POISSON’S EQUATION:
THE PN SEMICONDUCTOR JUNCTION
practicial example
The pn semiconductor junction is an important
can be found by direct integration of Poisson’s equation.
fied pn junction. The basic semiconductor is
silicon (or a compound semiconductor such as
typically
Figure
a
in which
properties
a simpli—
1.16a shows
valence 4 material such
galliumjarsenide
as
that behaves much the
region of the figure has been “doped” With valence 5 impurity atoms
such as phosphorus (donors), which although electrically neutral in themselves, have
more electrons than needed for bonding with adjacent silicon atoms and so contribute
electrons which can move relatively freely about the material. The p region of the figure
has valence 3 impurities such as boron (acceptors) which have fewer electrons than
needed for bonding with adjacent silicon atoms. These too are electrically neutral in
same). The
11
themselves, but leave holes that
from atom
move
to atom
with electric fields
or
other
charges. Although the transition between p and 22 regions
must be over some finite region, we assume an idealized model in which it is abrupt—
a step discontinuity.
When the junction is formed, the excess electrons in the n-type region at first diffuse
forces much like small positive
side. The electrons flowing into the
causing them to become negatively
charged. (Remember that they were originally electrically neutral). Likewise, the holes
moving into the n-type side are filled by the excess electrons there. The result is a zone
into the p-type side. The holes diffuse to the n-type
p-type side fill the vacancies in the acceptor bonds,
the
near
junction
in which there is
a
net
negative charge density
in
a
region
on
the
p-type side and a net positive charge density on the n-typje side called a depletion region,
as in the metal—semiconductor junction of Sec. 1.4. The density on the n-type side is
since each of the donor atoms has been
eND
stripped Qf
one
electron. The
density
on
the p-type side is
eNA since each acceptor atom has one additional electron. The
widths of the zones stabilize when the potential arising in the charge regions is sufficient
-
to
prevent further diffusion. Outside the charged regions, the semiconductors are neuequilibrium situation that we shall examine; if a field
tral. No fields exist there in the
did exist it would cause motion of the
The
regions
One
can
of
charge
the flux from the
in
—
a’p
as was
charges and violate the assumption of equilibrium.
to scale) in Fig.
l.16b.
gradient of potential dCI)/dx
—Ex
—-Dx/e
(not
< x <
=
=
done in Ex. 1.4a for the metal-semiconductor contact. We
Ex is zero outside
positive charges in 0
that
just argued
charges
shown
deduce the form of the
from Gauss’s law
have
are
0. This
charge regions (x < ---dp and x > (In). All
<
(1,, must therefore end on the negative
determines the relation
ibetween d" and dp in terms of
the
<
x
[.16
Direct
Integration of Poisson’s Equation: The
pa Semiconductor Junction
43
l-—-l++l
-—-
++
|___l++|
|—---—|++|
P
~--—
n
++
|—~— ++
l'fi +fl
(a)
PM
END
“dp
-
O
eNA
(a)
fix)
b.—
—_w
O
_’dp
dn
((1)
FIG. 1.16
(a)
pn diode
diode. (c) Potential
the
showing regions
gradient. (d) Potential.
of
uncompensated charge. (b) Charge density
in the
values of
NA and ND. The flux density Dx is negative and its magnitude
0 since the charge density is taken to be
linearly from x
wdp to x
O andx
constant. It then falls linearly to zero between x
d". The gradient therefore
takes the form shown in Fig. 1.160. The potential is the integral of the linearly varying
gradient so it has the square—law form in Fig. 1.16d.
Now let us directly integrate Poisson’s equation to get the complete analytic forms.
The boundary conditions on the integration are that the gradient is zero at x
—dp
and at d,,. The potential may be taken arbitrarily to have its zero at x
-dp. Specializing Poisson’s equation 1.12(3) to one dimension and substituting for charge
density the value eNA for the region wdp < x < O, we have
given
increases
=
=
=
=
=
=
-—
(132(1)
eNA
_
-—
dx2
8
1
()
44
Chapter
Ig'lelds
Stationary Electric
1
permittivity 8 is not appreciably affected by the dopant, at least for low-frequency
Integrating (1) from x
dp to an arbitrary x 0 making use of the
zero boundary condition on the gradient at x
dp gives
The
considerations.
=
=
—
=
(1CD
Integrating
a
second time
gradient
and
)
+
(d
(2)
x
0
--dp):
taking ®(-—
CI) (x)
The
i
eN
—A
:
3—
x
-*
828A(36
=
potential evaluated
£12
3
dx
0
==
potential
for
-—dp
<
p) 2:
+ d
—
at x
the
gives
x _<_
O
as
(3 )
are
Ella d
(4)
e
0
eN
42(0)
A
:
28
d-
(5)
boundary conditions for the integration in the region 0 S x S d”.
Poisson’s equation
region differs from (1) in the choice of charge density:
These constitute the
for this
d 2(1)
eN
=
——D_
3
d:c
Integrating (6) with
the
(6)
s
boundary condition (4) gives
dd)
N
_f___12
x
:
_._
dx
+
e
N
L“:
dp
(7)
e
x
Then
using the condition that the total positive charge must equal the
charge and therefore thath NA(dp /dn), (7) can be put in the form
total
negative
—
eNA d”
dd)
=
—
dx
which15
seen to
Integrating (8)
be
1
i‘
(8)
d"!
dn, as expected from
boundary condition (5), we
zero at x—-
with the
-
Gauss’ 3 law
as
discussed above.
find
(I) (x)
The maximum value of the
eNAd2
=
—p
28
potential
<
=
‘
x
1 + 2
“-
=
gdndp
(9)
!
is reached at x
(man)
x2
---2
dp
eN a!2
M)
Note that distance
-
8
—A—”
2? d”:
N
1 + ——4
28
dp in (10) is not immediately known.
ND
Since the
(10)
potential barrier arises
45
Uniqueness of Solutions
1.17
stop the diffusion of the charge carriers, it is expected that diffusion
to
considered. Diffusion
theory
reveals that the
height
k r
Act)
=
of the
potential
must also be
barrier
(10) is
N
3—4459]
(11)
7
e
n;
where k8 is Boltzmann’s constant, NA and ND are acceptor and donor doping densities,
respectively, and ni is the electron density of intrinsic (undoped) silicon, which is about
1.5 X 1010 electrons/cm3 at T
300 K. Once ND is calculated, dp can be found from
=
(10) and
all
quantities
in the field and
potential expressions
are
then known.
UNIQUENESS OF SOLUTIONS
l.l7
potentials governed either by the Laplace or Poisson equation
regions
given potentials on the boundaries are unique. With normal derivatives
of potential (or, equivalently, charges) specified on the boundaries, the potential is
unique to within an additive constant. Here we will prove the theorem for a charge—
free region with potential specified on the boundary. The proofs of the other parts of
the theorem are left as problems.
The usual way to demonstrate uniqueness of a quantity is first to assume the contrary
and then show this assumption to be false. Imagine two possible solutions, (131 and (132.
Since they must both reduce to the given potential along the boundary,
It
can
be shown that the
in
with
cp,
along
the
boundary
surface. Since
--
they
V2CD,
q»,
are
0
(1)
both solutions to
O and
=
=
V2432
=
Laplace’s equation,
O
or
'\72(CI)1
throughout the entire region.
In the divergence theorem, Eq. 1.110),
particular, let it be the quantity
((1)1
‘“
(132)
—
=
O
(2)
F may be any continuous vector
©2)V(®1
—'
quantity.
(132)
Then
[VV
.
[(CDI
From the vector
—_
©3)V(®1
._
(1%)]
dV
2
i [((Dl
"“
©2)V(¢’1
identity
diva/IA)
=
tbdiv
A + A
.
grad t1;
-
(132)]
.
dS
In
46
Chapter
the
equation
I
((131
—
V
may be
(Dame.
expanded
—-
c132)
Electrrclrrems
Stationary
1
to
(W +
f
[V(<I>1
(1)2)12
—
V
£5]? ((1)1
3
The first
over
the
integral must be zero by (2); the
boundary surface There remains
last
dV
integral
—
(1)2)V((I31
must be zero,
"‘
CD2)
'
(18
since (1) holds
i
i
[V [We1
—
@912
W
=
0
gradient of a real scalar is real. Thus its square can only
integral is to be ZCI'O, the gradient itself must be zero:
The
we.
——
a2)
=
0
(3)
be
positive
or zero.
If its
<4)
z
01‘
€
((131
-—
(1)2)
=
constanjt
(5)
I
This constant must
apply
are
even to
the
boundary,
where we know that (1)
is true. The
is then zero, and (I)l
C132 is everywhere zerb, which means that CD1 and (132
identical potential distributions. Hence the proof
uniqueness: Laplace’s equation
constant
~—
solution which satisfies the
of;
boundary conditions of the given region.
If by any method we find a solution to a field problem that fits all boundary conditions
and satisfies Laplace’s equation, we may be sure it is the only one.
can
have
only
one
Spectni Techniques for Efiectmstattc Probiems
1.18
The so-called method of
THE USE OF
lMAgEs
the
with
fields produced by charges in
images is a way of finding
the presence of dielectric6 or conducting boundaries
certain symmetries Here we
concentrate on the more common situations, those With conducting boundaries (But
see
6
Prob. 1.18d.)
W. R
Smyfhe, Static and Dynamic Electricity, Hem/sphere Publishing Co., Washington,
DC 7989
The Else of
mm
FIG. 1.18a
charge
of
Image
a
in
point charge
a
conducting plane.
The field lines shown
Example
1.18a
simplest
case
is that of
a
point charge
A
PLANE
near a
grounded7 conducting plane (Fig.
1.18a). Boundary conditions require that the potential along the plane be
requirement is met if
charge is placed at x
in
m
1
CI)
:
——
4178
=
fl
~——
4178
7
for the
are
q with the conductor.
POINT IMAGE IN
The
47
Images
Use of the term
3
—
r
{[(x
-—d.
zero.
of the
51-,r
—-
d)?-
The
conducting plane an equal and opposite image
Potential at any point P is then given by
place
(1)
+
y2
grounded implies
+
221“”2
a source
-—
for the
[(x
+
(02
+
y2
+
charge that builds
2214/2}
up
on
the
plane.
48
Chapter
This reduces
to
the
required
potential along
zero
fields
Stationary Electric
1
the
pl: mex
0,
=
so
that
(1) gives the
potential for any point to the right of the plane. The exp] ression of course does not
for x < O, for inside the conductor the potential must b DB everywhere zero.
apply
potential other than zero, the value of this constant potential is
(1)
give the expression for potential at any point for x > O.
simply
The charge density on the surface of the conducting plane must equal the normal
flux density at that point. This is easily found by usingI
If the
plane
is at
added to
a
to
GCIJI
psfitheEx:
Substituting (1)
in
(2)
“8—
-
ax
x=0
(2) and performing the indicated differentiation gives
d
p5
g. (d2
=
+
'n'
y2
..
+
22)
3/2
(3)
2
0,
(3) shows that the surface charge density has its peak value at y
with circular contours of equal charge density centered about that point. The density
Analysis
of
decreases
monotonically to zero
studying the
method is in
image
==
=
as
y
and/ or
2
extraction of
the metal—semiconductor surface shown in
go to
Fig.
One
infinity.
electrons
from
a
application
of the
metallic surface
as
in
1.4a.
Example Ltsb
IMAGE OF
If there is
d from it,
——
d. The
A
LINE CHARGE IN A PLANE
charge of strength q, C/m parallel to a conducting plane and distance
proceed as above, placing an image line charge of strength wq, at x
potential at any point x > O is then
a
line
=
we
a
=
..
fl;
271's
1
q,
(x r_*_ d)2
118
(x
1n<_,)=.—. ti
r
Example
IMAGE OF
For
line
A
5-
+y2
(1)2 +y2
I
(4)
1.38::
LINE CHARGE IN A CYLINDER
of strength q, parallel to the axis of a conducting circular cylinder,
from
the axis, the image line charge of strength
q, is placed at radius
rq
r’
where
is
a
the
radius
the
of
The
combination of the
(12/ rq,
cylinder (Fig.El 1.18b).
two line charges can be shown to produce a constant potential along the given cylinder
a
charge
and at radius
—
=
of radius a. Potential outside the
cylinder may be computed from the original line charge
image. (Add q,
cylinder is uncharged.) If the original line charge is
within a hollow cylinder a, the rule for finding the image is the same, and potential
inside may be computed from the line charges.
and its
on
axis if
1.18
FIG. L‘lab
Image
The Ilse of
of line
charge
49
Images
q, in
a
parallel conducting cylinder.
Example 1.18d
IMAGE OF
For
A
POINT CHARGE IN A SPHERE
point charge (1 placed distance r from the center of a conducting sphere of radius
image is a point charge of value (— qa/rq) placed at a distance (612/ rq) from the
center (Fig. 1.186). This combination is found to give zero potential along the spherical
surface of radius a, and may be used to compute potential at any point P outside of
radius 0. (Or, if the original charge is inside, the image is outside, and the pair may be
used to compute potential inside.)
a
a, the
Example
1.182
MULTIPLE IMAGINGS
charge in the vicinity of the intersection of two conducting planes, such as q in
region of A08 of Fig. 1.18d, there might be a temptation to use only one image in
each plane, as 1 and 2 of Fig. 1.18d. Although + (1 at Q and —q at 1 alone would give
constant potential as required along 0A, and + (1 at Q and —q at 2 alone would give
constant potential along OB, the three charges together would give constant potential
For
a
the
Q
\
_ . -b
L
FIG. 1.186
Image of
a
point charge
in
a
sphere.
"U
53
Chapter
Stationary Electric fields
1
A
-
+
4P\q
I
\\
|
l
l
\
\\
z
7K+q
\\
”
//
\
Iq
I
i
/
|\\
I
'
I
\
\\
/|’I
\\
"‘x \ll
\
\
\I
2
I
\\ I
34
\
64—1]
I
i
i
I
\
I
\
B
\
\I
\
/
l
r
X“
/
I
I
\
l/
/
x
I
\
O
I
I
I
\.
,
4——-*—r“\~——_,\
I
I
\\ /
i
I
I
1
\
\
I\\
i
\E
‘
x:
i
—————
.
I
.\
\\
I
,
\\
I
\
I
""N\5
ql‘xl
—q
+q
/
FIG. 15le
Multiple images
of
a
point
or
line
chargegbetween intersecting planes.
I
along
neither 0A
nor
these
OB. It is necessary to image
images in turn, repeating until
or until all further images are too far distant from the region to
coincide
further
images
potential. It is possible to satisfy exactly; the required conditions with a
finite number of images only if the angle A03 is an exact submultiple of 180 degrees,
as in the 45-degree case illustrated by Fig. 1.18d.
influence the
I I9
.
PROPERTIES OF TWO*DIMENSIONAL FIELDS:
Many important electrostatic problems may
the pair of parallel wires of Fig. 1.8b or the
field distribution is the
same
be
GRAPHIC/XL FIELD
considered
as
MAPPING
two-dimensional,
as
in
coaxial system of Fig. 1.15. In these the
in all cross-sectional planes, and although real systems are
the idealization is often
useful one. In the
examples cited above,
analytically, but for cylindrical systems with more
complicated boundaries, numerical techniques may be called for and will be introduced
in the next section. We wish to give first some properties of two~dimensional fields that
can be used to judge the correctness of field maps and can even be used to make useful
pictures of the fields and to obtain approximate values; of such things as capacitance,
conductance, and breakdown voltage. Perhaps the greatest value in making a few such
maps is the feel they give for field behavior.
It has already been established that equipotentials and electric field lines intersect at
right angles, as in the coaxial system of Fig. 1.19a, where field lines are radial and
equipotentials are circles in any given cross—sectional plane. It has also been shown that
the region between two field lines may be considered aafiux tube, and if the amount of
never
infinitely long,
the field distributions could be found
a
Properties
1.19
of Two-Dimensional Fields:
5'5
Graphical Field Mapping
An-—>-
Atll
4) + At?
<1?
(b)
(a)
FIG. 1.19
graphical
flux is
(a) Map
field
of field between coaxial
properly chosen,
ratios, that
conducting cylinders. (b)
Curvilinear
rectangle for
mapping.
the map is made up of small curvilinear figures with equal side
squares.” This also is illustrated in Fig. 1.19a. To. show this
is, “curvilinear
consider
of the curvilinear
rectangles from a general plot, as in
Fig.
adjacent equipotentials, and As the distance
between two adjacent field lines, the magnitude of electric field, assuming a small
square, is approximately ACID/An. The electric flux flowing along a flux tube bounded
by the two adjacent field lines for a unit length perpendicular to the page is then
generally,
more
one
1.1%. If A12 is the distance between two
A¢=DA3=8EA33M
All
or
AS
All!
(1)
_
All
8
ACID
So, if the flux per tube Alp, the potential difference per division ACID, and the permittivity
8 are
constant
throughout the plot,
the side ratio
As/An must also be constant,
as
stated
above.
We
saw
in Sec. 1.14 that
conducting
surfaces
are
equipotentials
in
an
electrostatic
field. Thus, the electric field lines meet the electrodes at right angles.
In applying the principles to the sketching of fields, some schedule such
following
will be
as
the
helpful.
making a number of rough sketches, taking only a minute or so apiece,
starting any plot to be made with care. The use of transparent paper over
basic boundary will speed up this preliminary sketching.
1. Plan
on
before
the
potential difference between electrodes into an equal number
or eight to begin with.
of
sketch
the
Begin
equipotentials in the region where the field is known best, as
for example in some region where it approaches a uniform field. Extend the equi—
potentials according to your best guess throughout the plot. Note that they should
2. Divide the known
of divisions, say four
3.
52
Chapter
‘P
"‘=
1
Stationary Electric
V0
:Eields
I
i
I
._
.J‘
I
3V0
\
\
/
_
T
/
/
X,—
\
_
/
15
V0
fl
‘
a
_________
4
FIG. 1.19::
.
tend to
hug
vicinity
of obtuse
acute
Drawin the
a
plane
angles of the conducting
angles of the boundary.
orthogonal
curvilinear squares, but
be
of fields between
Map
kept paramount,
unity.
even
boun9ary,
spread
out in the
of field lines. As these
are
other than
.
and be
are started, they should form
the
extended,
they
condition of orthogonality should
this
will
result
in some rectangles with ratios
though
set
as
and stepped conductor.
l;
regions with poor side ratios and try to
first guess of equipotentials. Correct them and
able curvilinear squares exist throughout the
Look at the
see
what
was
wrong with the
repeat the procedure until reasonploti
.
regions of low field intensity, there will be large figures, often of five or six
judge the correctness of the plot in this region, these large units should
be subdivided. The subdivisions should be started back away from the region
needing subdivision, and each time a flux tube is divided in half, the potential
divisions in this region must be divided by the same factor. As an example, Fig.
1.190 shows a map made to describe the field between a plane conductor at
potential zero and a stepped plane at potential V0 With a step ratio of ~12:
In
sides. To
1.20
NUMERICAL SOLUTION OF THE LAPLACE
Numerical methods
are
AND
becoming increasingly attractive
to increase. Among the
as
continue
memory
finite differences,8 finite
capacity8
POISSON EQUATIONS
digital computer speed and
methods are those using
powerfulg or method of moments
transformations,
(Sec. 7. 3). Still others will undoubtedly be developed as Computing capabilities continue
to
increase. Here
proach
8
and
some
we
elements,8
Fourier
illustrate the idea
through
the elemental difference
equation
ap-
of its extensions.
L. Co/Iaiz, The Numerical Treatment of Differential
7966, D. Potter, Computational Physics, Wiley, New
Springer-Verlag, New York,
Equations,
7973. L J.
Finite Element
York,
Analysis, 2nd ed” Wiley, New York 7984.
Segerlind, Applied
R. Sorrentino (Ed), Numerical
Methods for Passive Microwave and Millimeter Wave Structures, IEEE Press New York,
7989.
R. W.
Hockneyand J W Eastwood, ComputerSlmulatlon Using Particles, Am. Inst. Physics,
New York, 7988.
Numerical Solution of the
1.20
53
Laplace and Poisson Equations
We consider first the Poisson
equation with potential specified on the boundary. For
simplicity
problem (no variations in z). The internal region
is divided by a grid of mutually orthogonal lines with potential eventually to be determined at each of the grid points. Rectangular coordinates are used and potential at a
point (x, y) is expanded in a Taylor series:
we
take
a
two-dimensional
6430:, y)
1
®x+li,
(
7y) «8613;,
(XY) +/—
+
By adding (1)
and
(2)
and
62CI)(x, y)
~
(Deny)
rearranging,
_
(130:
+
have the
we
(2)
8x
approximation
2430‘, y)
-
-,
L-—(?;—Z)—
2
8x
12, y)
+
12—21)-
—
ax-
12 62(1)
act)
11, y)
-
l
()
9
2
a;
C130:
E 62¢(x, y)
+
(13(x
—-
12, y)
(3)
w
6x2
112
The second
Poisson’s
partial derivative with respect
equation in two dimensions
62(1)
ax"
@(x
be
+
expressed
12, y)
+
in the
(130:
-
approximate
11, y)
+
be obtained in the
62(1)
+
'7
can
to y can
=
6y
same
way. Then
p
_
2
8
form
(DOC,
h)
y +
2
+
where the distance increment h is taken, for
(130w
-
h)
--
4¢(x, y)
=
”mi
(4)
simplicity, to be equal in the two directions.
potential at a given point is the
It is of interest to note that, if Space charge is zero, the
average of the potentials at the surrounding points.
potential is known on boundary points so that a straightforward approach
set of equations such as (4) for the
unknown potentials at the grid points
in terms of the known values on boundary points. This is sometimes done by a matrix
inversion technique, but if memory capacity is a problem, it may be better to use a
method for direct iterative adjustment of grid potentials. This starts from an initial guess
and corrects by bringing in the given values on the boundary through successive passes
through the grid. We illustrate this first by a simple averaging technique.
Note that
is to solve
a
Example
1.20
NUMERICAL SOLUTION OF LAPLACE EQUATION BY SIMPLE AVERAGING
simple averaging, let us find the potentials for the
grid of points
Fig. 1.20a. This is an infinite cylinder of square cross
section with the potentials specified on the entire boundary. The space charge will be
assumed to be zero so we will be solving Laplace’s equation. The broken lines represent
As
an
illustration of iteration with
in the structure in
54
Chapter
1
Stationary Electric
fields
100
"///////////////////////////////%/
l
l
a
a
/
/
I
l
60
FIG. 1.20a
Cylinder
of square
cross
section and
for difference
grid
equation
solution.
grid to be used to approximate the region for the finite-difference solution. The
grid was chosen to simplify the example; a finer grid would be used in most
practical problems. The four unknown potentials designated CI)1 to CD4 are assumed
initially to be the average of the boundary potentials, 65 V. The first calculation is to
find (131 as the average of the four surrounding potentials (80, 100, (132
65, CP3
65). Therefore, in the column labeled Step 1 in Table 1.20, (D1 is given the value 77.50.
Then C132 is found as the average of 100, 20, 77.50, and 65, and this value is put in the
table in Step 1. The procedure is repeated for (133 and CD4 The Step 2 proceeds in the
the
coarse
=
same
Since
potentials converge to
way. It is seen that after several steps the
(4) is approximate, the potentials have converged
would differ less from the correct solution if the
potentials
for the four
Mesll Relaxation
tials leads to
a
points
are
grid
definite values.
to approximate
were
=-
answers.
They
made finer. The correct
also listed in the table.
The above method, in which
final result not far from the correct
successive averaging of the potenpotentials,
is convenient for small
convergence. A more generally
residual for each grid point that
in many cases, has a satisfactory rate of
useful method of calculation is based on a defined
problems and,
measures
the amount
by which
the
potential
there differs from the value dictated
Table 1.20
lterative Calculation for
Example
1.20
Step
1
2
77.50
by
Correct
3
4
5
Potentials
75.2
(I),
(132
CD3
79.07
77.89
77.60
77.51
65.63
63.28
62.70
62.55
62.51
60.5
70.63
68.28
67.70
67.52
67.52
65.4
CI) A
54.06
52.89
52.60
52.51
52.51
50.7
1.20
Numerical Solution of the
the
potentials on the neighboring grid points.
grid is defined by
R“)
came,
=
h)
y +
The residual for the kth pass
<I><")(x,
+
y
h)
-
(13%:
+
where i
k
k
(Ema
12, y)
-
-
4<I3<k_1)(x, y)
+
through
+
the
M)
(5)
122p
.
+
55
Laplace and Poisson Equations
grid point, the potentials at some of the neighboring
generally
already
points
adjusted on that pass through the grid, as was
seen in the above illustration. On each pass (corresponding to Steps 1-—5 in Table 1.20)
and at each grid intersection, one calculates the residual R“) and then the new potential
according to
=
or
-
1
will
since,
at any
have
been
(Pm
(DOC-l)
:
Rae)
+
(6)
97
where Q is called the relaxation factor because it determines the rate at which the
potentials relax toward the correct solution. It is taken in the range 1 S Q S 2. Selection
of Q
1, called simple relaxation, corresponds to the method of averaging illustrated
=
above. When 0
> 1, the procedure is called the method of successive averrelaxation
(SOR). If 0. is fixed, it is usually taken near 2 for problems with many mesh points,
but this
can cause an
initial increase in error,
so
it is often better to start with Q
=
1
and increase it
gradually with each iteration step. One procedure that is found useful
for large grids is the cyclic Chebyshev method in which the following program of
varying Q is used:
0.“)
==
0(2)
:
1
(7)
4
1
'12???“
(8)
'—
where
2
1
=
Tin
for
a
grid
with
22
mesh
points
Q“
in
+
”(cos3 0033)
one
1)
m
n
direction and
at
in the other.
points
1
2
=
( 10)
—,
"
0“”)
(9)
+
4
Q
fining“)
2_
m
op t
1 +
1
"
(11)
nit
When this method is used, the mesh is swept like a checkerboard, with all the red
squares being treated on the first pass with (1“), all the black squares being treated on
the second pass using 0‘2), the reds on the third with 0(3), and so on. This method
requires an additional memory cell in the computer for each mesh point but the speeded
convergence
usually
makes it worthwhile. The convergence
can
be
improved appreci-
56
Chapter
r
Stationary Electric Fields
ably by having a good initial guess for the potentials,
problem
say by using
results from
a
similar
Boundary gonditions In setting up the boundary conditions on a grid, the easiest
situation occurs when potentials are specified on grid pbints. A more difficult problem
is where the normal derivatives are specified. The derivatives are usually zero, as would
be the case at an insulating surface in a conduction problem (Ex. 1.14) or along a line
of symmetry used as an artificial boundary to reduce the required grid size. Examples
of such boundaries are shown in Figs. 1.2019 and 1.206;. In one case the structure has
obvious symmetry in the x—y plane so that a solution need be found for only one-fourth
of the structure. In the other example, the boundary may be taken along the axis of a
cylindrically symmetric system. In the latter case, the difference equations can include
the symmetry (Prob. 1.20e). In the former case, to make the normal derivative of po—
tential zero at a boundary, an imaginary grid point is set up outside the boundary and
its potential is kept the same as at the point symmetrically located just inside the bound—
ary. Sometimes the boundary points do not lie on mesh points; in such cases, linear
interpolation13 used to set the potentials at mesh pomts nearest to the boundary.
0
0 2\\\‘\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
rY/,4’
‘
u““-—_
,-
1
l
I
/?
/
’/\\\\*E\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
(b)
V1
‘Q
3\
x
x
\
x
\
s
_
_____
_
$5.“. .
_
V0
IWI_T__|_I_I~_1_I__I,._E§
___l_l_lwl_l___l_l_! \
_u_|_,1_«_‘1~|_l'__,|j§
1__l~l_l___l_l__l_l §
_______
§
V1
\
FIG. 'l.20b,c Examples in rectangular and cylindrical coordinates where symmetry reduces the
size of the grid for finite-difference solutions.
required
1.2:
1.21
57
Examples of information Obtained from Field Maps
EXAMPLES OF INFORMATION OBTAINED FROM FIELD MAPS
Field maps of the two~dimensional regions, made by either numerical or graphical
techniques, may be used to find field strength within the dielectric region or integral
properties of the systems, such as capacitance per unit length and conductance per unit
length. Field strengths are simply ACID/An in the notation of Sec. 1.19, provided the
divisions are fine enough. Danger of breakdown is obviously greatest near acute angles
where Spacing of equipotentials is smallest, as in the right-angle corner of Fig. 1.1%.
For capacitance, we need to know electric flux density at the conductors, which
corresponds to surface charge density. Values of potential, and not field, are typically
obtained from numerical solutions, but the approximation to a normal derivative at the
boundary is readily calculated. Equipotentials and field lines may be drawn in and then
calculation of capacitance becomes particularly simple. By Gauss’s law, the charge
induced on a conductor is equal to the flux ending there. This is the number of flux
tubes Nf multiplied by the flux per tube. The potential difference between conductors
is the number of potential divisions
Np multiplied by the potential difference per
division. So, for a two—conductor system, the capacitance per unit length is
Q
C:
o,
The ratio
Alp/AC1)
can
--
be obtained from
2
c1),
a
small-square plot
with
NpAcI)
Eq. l.19(1):
_Np
And, for
NM
An
As/An equal
to
unity,
N
C
:2-
8
——f
Np
For
example,
16 flux tubes,
plot of Fig. 1.190, there
capacitance, assuming air dielectric,
in the coaxial line
so
the
F/m
10—9
Cm
3611'
(2)
are
4
potential
divisions and
is
16
><--=35.3><10”12F /m
(3 )
4
‘2
5.2, gives 33.6 X 10‘ F/m, indicating that
Eq. 1.9(4), with b/ a
the map is not perfect.
This same technique can be used to find the conductance between two electrodes
placed in a homogeneous, isotropic, conductive material. The conductivity of the electrode materials must be much greater than that of the surrounding region to ensure that,
when current flows, there is negligible voltage drop in the electrodes and they can be
considered to be equipotential regions. The potential and electric field are related in the
same way as for the casein which there is no conductivity (Sec. 1.13). There is a current
(r E, where (r is conductivity, and current tubes replace the flux tubes of
density J
the dielectric problem. The current in a tube is
Calculation from
=
=
53
Chapter
1
Stationary Electric
AI==JAs
The conductance per unit
length
=
UEAS
=
Using (4)
and
taking AS/An
=
-—
=
cpl
1vp
(4)
as
*
Nf AI
——
cI>2
M» A
0——S
electrodes is defined
between two
1
G
:
iElelds
ACID
iS/m
(5)
1
(6)
1,
G
Nf
S /m
~-----—
2
UN
P
From
(2) and (6)
we see
the useful conclusion that
per unit
electrodes is related to the
thegconductance per
length between the
unit
length
of
electrodes
by
capacitance
example, in transmission-line problems in giving
the conductance per unit length between conductors when the capacitance is known.
Before digital computer methods for solving field problems became so readily avail~
able, the analogy seen above between the field distributibns in conducting and dielectric
media formed the basis for an important means of determining fields in dielectric systems. Electrodes corresponding to those of the dielectric problem are set up in an
electrolytic tank or on conduction paper and equipotentials measuredin the conducting
the ratio
0/8.
This
can
same
be of use, for
system (see Prob. l 21c).
Conducting
region
Dielectric
1
*
a
HT"
FlG. 'I .21
Field map for
conductive electrodes
a
conductive medium
partially
filling
the space between two
highly
1.22
If
in
Energy of an
Electrostatic
59
System
part of the boundary of the conducting region is a nonconducting dielectric, as
Fig. 1.21, current does not flow in the nonconductive region. As pointed out in
a
Sec. 1.14, the normal component of E inside the conductive region must then vanish
at the boundary with the dielectric region. By use of this condition and the fact that E
is
perpendicular to the equipotential electrode surfaces, the field in the conductor can
mapped. Suppose, for example, the conductive material between the electrodes in
100 S/rn. Then
Fig. 1.21 is silicon with the common value of conductivity 0'
be
=
G
=
100 X
(8/23)
=
35
1.22
S/m.
ENERGY OF AN ELECTROSTA'I'IC SYSTEM
an expression for electrostatic energy in terms of
quantities. The result we will obtain can be shown to be true in general; for
simplicity, however, it is shown here for charges in an unbounded region.
The work required to move a charge in the vicinity of a system of charges was
The aim of this section is to derive
field
discussed in Sec. 1.7. The work done must appear
as
energy stored in the system, and
consequently the potential energy of a system of charges may be computed from the
magnitudes and positions of the charges. To do this, let us consider bringing the charges
from infinity to their positions in space. No force is required to bring the first charge
in since no electric field acts on the charge. When the second charge q2 is brought to
a position separated from the location of (11 by a distance R12, an energy
(1142
=
12
is
it
expended, as was shown in
experiences the fields of q1
Sec. 1.7. When the third
and q: and
U 13 + U23
is
expended.
The total energy
=
an
41513
-—
+
47781613
expended
summing
over
the three
charges,
charge
is
4243
may write
(2)
-—
4778R23
to assemble these three
we
brought from infinity,
energy of
(1) and (2).
In
(1)
4178R12
charges
is the
sum
of
63
Chapter
Stationary Electric
1
Eiields
yields the sum of (1) and (2), since by convention i and j are
particles, and each contribution to energy enters twice. In physical
terms, the factor of 2 would result from assuming all dither charges in position when
j has been excluded since the selffinding the energy of the ith charge. The term i
energy of the point charge (i.e., the electron, ion, etc.) does not affect the energy of the
field. Its contribution to the total system energy does (not depend upon the relative
positions of the charges. For 22 charges, the direct extension gives
With the factor
summed
over
31,3,
this
all the
=
1
UE
where the
Eq. 1.8(3)
=
-
II
n
qj
2
i; $214778)?
subscript E indicates energy
for potential, this becomes
stored in electric
Us
we
to a
=
2
'
a5 J
charges
<3 )
and fields.
By
use
of
II
1
Extending (4)
.
’
.
21 (15131
(4)
system with continuously varying charge density p per unit volume,
have
1
UE
=
'
2
f [3(de
(5)
V
i,
The
charge density
p may be
replaced by
the
divergence of D by Eq.
1
Using
the vector
equivalence
l.11(2):
‘
of Prob. 1.11a,
t
UE=lJv-(cpn)dV—-—1—fn-(vqa)dv
2Vi
2V
integral may be replaced by the surface integral of CPD over the closed
surrounding the region, by the divergence theoiem [Eq. l.11(7)]. But, if the
region is to contain all fields, the surface should be taken at infinity. Since (13 dies off
at least as fast as l / r at infinity, D dies off at least as fast as l/rz, and area only
increases as r2, this surface integral approaches zero as the surface approaches infinity.
The first volume
surface
f'V-(CIDD)dV=3g
v
cIJD-dS=O
5,,
Then there remains
l
l
UE=—-fDO(V(I>)dV=-JD-EdV
2
This result
volume dV
seems
V
to say that the energy is
appearing
2
actually in
(6)
V
the electric
to contain the amount of energy
field, each element of
1.22
of
Energy
dUE
an
=
Electrostatic
in
-
63
System
E (W
(7)
The
right answer is obtained if this energy density picture is used. Actually, we know
only that the total energy stored in the system will be correctly computed by the total
integral in (6).
The derivation of (6) was based on a system of charges in an unbounded, linear,
homogeneous region. The same result can be shown if there are surface charges on
conductors in the region. With both volume and surface charges present, (5) becomes
1
1
Ugm~fpci>dV+-fps®d5
2
2
v
(8)
S
The
proof that this leads to (6) is left to Prob. 1.22e. Note that in this equation, and the
Special case of (5), (I) is defined with its reference at infinity, since the last term of
(3) is identified as (I). The advantage of (6) is that it is independent of the reference for
potential.
For a nonlinear medium, the incremental energy when fields are changed (Prob.
1.22f) is
dngfE'dDa’V
Exampie
ENERGY STORED IN
(9)
H.252
A
CAPACITOR
interesting to check these results against a familiar case. Consider a parallel-plate
capacitor of capacitance C and a voltage V between the plates. The energy is known
from circuit theory to be éCVZ, which is commonly obtained by integrating the product
of instantaneous current and instantaneous voltage over the time of charging. The result
may also be obtained by integrating the energy distribution in the field throughout the
volume between plates according to (6). For plates of area A closely spaced so that the
end effects may be neglected, the magnitude of field at every point in the dielectric is
E
distance between plates) and D
eV/d. Stored energy UE given by
V/d (d
(6) becomes simply
It is
==
2
=
This
can
:
be put in terms of
v‘
2
8V
1
I
U5
(volume)(DE)
:
~---«
2
(A
V
d)( > (d)
——
_.
d
capacitance using Eq. 1.9(3):
18A
2(a)
1
UE =--——--V2=—-cv2
2
10
()
62
Chapter
Stationary Electric
1
fields
PROBLEMS
1
1.2a
(i) Compute the force between two charges of l C each, placed 1 m apart in vacuum.
(ii) The esu unit of charge (statcoulomb) is defined as gone that gives a force of 1 dyne
when
placed
1
cm
from
a
like
in
charge
between statcoulornbs and coulombs
vacuum.
given
in
Use
this fact to
Appendix
check the conversion
1.
repulsion
between two electrons to the
1.2!) Calculate the ratio of the electrostatic force of
gravitational force of attraction, assuming that Newton’s law of gravitation holds. The
electron’ 3 chargeis l 602 X 10'19 C, its mass is 9 11 X 10”31 kg, and the
gravitational constant K1s 6. 67 X 10“11 Nmz/kg2
experiment performedin 1785, Coulomb suspended a horizontal rod from its
by a filament with which he could apply a torque to the rod. On one end of the
rod was a charged pith ball In the plane in which the rod could rotate was placed
another, similarly charged, pith ball at the same radiusi By turning the top of the filament he applied successively larger torques to the rod with the amount of torque proportional to the angle turned at the top. With the angle at the top set to 36 degrees, the
angle between the two pith balls was also 36 degrees. Raising the angle at the top to
144 degrees decreased the angular separation of the pith balls to 18 degrees. A further
increase of the angle at the top to 575.5 degrees decreased the angular separation of
the balls to 8.5 degrees. Determine the maximum difference between his measurements and the inverse-square law (For more details
see R. S Elliott. 1)
1.2c In his
center
1.2d Construct the electric field vector for several points in the x-y plane for like charges q
at ((1/2, 0, O) and (— d/ 2 O, 0), and drawin roughly a few electric field lines.
1.2e*
Repeat Prob. 1.2d for charges
tively. Find a point where the
of 2g and-q at
field is zero.
(d/Z, 01, O)
and
(- d/ 2, O, 0), respec-
1.2f Calculate the electric field at points along the axis perpendicular to the center of a disk
of charge of radius a located in free space. The charge ion the disk is a swface charge
ps
C/rn2 uniform
1.3a Show
at any
over
the disk.
by symmetry arguments and the results of Sec. 11.3 that there
point inside a spherical shell of uniform surface Lcharge.
is
no
electric field
geireral
closed surface as in
1.3b Show that the integral of normal flux density over a
Fig. 1.30 with charge q inside gives q. Hint: Relate surface element to element of
solid angle.
1.31: Calculate that electric flux emanating from a point charge q and passing through a
mathematical plane disk of radius (I located a distance d from the charge. The charge
lies on the axis of the disk. Show that in the limit where a/d ——> 00, the flux through
the disk becomes q/Z.
1.43 A coaxial transmission line has
conducting cylinder of radius
c.
an
inner
conducting cylinder of radius
the inner conductor and
q, over
b and dielectric 82 from r :1) to r
c, find the electric field for
b, for b < r < c, and for r > c. Take the conducting cylinders as
—
a
and
an
outer
4, per unit length is uniformly distributed over
the outer. If dielectric 81 extends from r
a to r
Charge
—
=
I
< a,
for
=
a < r <
infinitesimally
thin
Sketch the variation of D and E with radius.
1.4b A
long cylindrical beam of electrons of radius a moving with velocity v,
vo[1 + 6 (1/a)2] has a charge- density radial variation p!
8202/0) ]. Find the
po[l
radial electric field1n terms of the axial velocity vo and the total beam current 10 and
=
=
sketch its variation with radius.
-
63
Problems
1.4c Derive the
for the field about
expression
point charge.
line
a
charge, Eq. 1.4(3), from
the field of
a
1.4d* A
sphere of charge of radius a has uniform density pO except for a spherical cavity
zero charge with radius b, centered at x
0, where d < a and
d, y
0, z
00 < x < 00. Hint: Use
b < a
(2’. Find electric field along the x axis from
super»
position.
of
=
=
——
=
-
1.4e* As in Prob. 1.4d, but
inside the
now
find
an
expression
for electric field for
that the field in the
cavity, showing
a
general point
is constant.
cavity
1.53 A
point charge q is located at the origin of coordinates. Express the electric field vecin its rectangular coordinate components, and evaluate the surface integral for S
chosen as the face perpendicular to the x axis of a cube of side lengths 20 centered on
the charge. Use symmetry to show that Gauss’s law is satisfied.
tor
1.51) Perform the integrations in Eq. 1.5(3) for an infinitely long circular cylindrical ion
beam with p
p0[1 + (r/a)2] using the square prism shown in Fig. 1.5 and plane
ends at z
O and I orthogonal to the axis.
=
=
1.5c If A, B, and C
are
vectors, show that
A‘B
B-A
(A+B)+C
ll
A+(B+C)
A-(B+C) =A‘B+A'C
1.5d Vector A makes
angles
scalar
:21, [3,, y, with the x, y, and z axes respectively, and B makes
with the axes. If 6 is the angle between the vectors, make use of the
angles
B2, y2
product A
522,
cos
°
B to show that
0
=
cos
(21
(:2 +
cos
cos
,8,
B2
cos
+ cos yl cos yz
1.6a Show how the flux function may be used to plot the field from point charges q and
—q distance d apart. Hint: Make use of solid angles and relate these to angle 6 from
the axis
joining charges.
1.6b Plot the field from like
charges
(1 distance d apart
(Prob. 1.2d) by making
use
of the
flux function.
1.6c Plot the field of charges 2:] and ~q distance d apart (Prob. 1.2e)
that not all flux lines terminate at both ends on charges.
1.7a Evaluate
lar
95
F
-
(11 for vectors F
from
(O, 1)
path
Iar path from (O, 0)
to
to
(1, 1)
(O, 1)
to
to
=
it my +
y x2
and F
=
it y
~—
by
9x
use
of flux. Note
about
a
rectangu-
(l, 2) to (0, 2) and back to (0, 1). Repeat for a triangu~
( 1, 1) back to (O, 0). Are either or both nonconser-
vative?
point charge (1 is located at the origin of a system of rectangular coordinates. Evalu1 to x
2, and next
f E (11 in the x—y plane first along the x axis from x
along a rectangular path as follows: along a straight line from the point (1, O) on the x
axis to the point (1, 313); along a straight line from (1, é) to (2, i); along a straight line
1.7b A
ate
=
'
from (2,
%)
1.821 A circular
charge
pS
distance
2
to
=
(2, O).
insulating disk
C/mz.
Find
an
of radius
a
expression
is
charged
uniform surface density of
potential (I) at a point on the axis
with
for electrostatic
a
from the disk.
1.8b* A charge of surface density pS is spread
Find the potential for r < a and for r >
uniformly over a spherical surface of radius 0.
a by integrating contributions from the differ~
64
chapter
1
Stationary Electric Fields
ential elements of charge. Check the results
symmetry of the problem
by making
use
of Gauss’ 5 law and the
.
1. 8c Check the result Eq. 1. 8(8) for the potential about a line charge by
butions from the differential elements of charge Note lthat the problemis
handling properly the infinite limits.
contriintegrating
one of
layer of charge of density p0 lies perpendicular to the z axis and1s infinitely
broadin the x and y directions Using Gauss’ 5 law and Eq. 1. 8(1), find the dependence
d.
of the potential difference across the layer on its
1.8d A flat
thickpess
Stirface
charge densities but with
parallel sheets of charge having equal
opposite sign. The sheets are both of infinite transverse dimension and are spaced by a
distance d. Using Gauss’s law and Eq. 1.8(1), find theielectric fields between and outside the sheets and find the dependence of potential difference between the pair of
1.88 Consider two
sheets
1.8f In
a
on
the
spacing. (This
system of infinite
is called
transverse
a
dipole layer.)
dimension,
a
sheet
__
of charge of pS C/m2 lies be-
tween, and parallel to, two conducting electrodes at zero potential spaced by distance
d. Find the distribution of electric field and potential between the electrodes for arbi—
trary location of the charge sheet. Sketch the results for the
(i) in the
1.8g
center and
Show that all the
are
(ii)
at
equipotential
cylinders whose
cases
where the sheet is
position d/4.
surfaces for two
traces in the
parallel line charges of opposite sign
perpendicular plane are circles as shown in Fig. 1.80.
1.811 A linear quadrupole is formed by two pairs of equal and opposite charges located
along a line such that +q lies at + 5, 2q at the origin, and +q at 5. Find an
approximate expression for the potential at large distances from the origin. Plot an
-
equipotential
—
line.
magnitude of the torque on a dipole in an electric field is the product
magnitude of the dipole moment and the magnitude of the field component
perpendicular to the dipole.
1.8'- Show that the
of
the
1.9 Find the capacitance of the system of two concentric
two different dielectrics used as Ex. 1 .40.
1.10a Find the
gradient
of the scalar function M
eax
=
cos
spherical electrodes containing
i
By cosh 012.
d/2,
0, 0), respectively, find the
—q at (d/2, O, O) and(point charges q and—
potential for any point (x, y, z) and from this derive the electric field. Check the result
by adding vectorially the electric field from the individual charges.
1.1011 For two
1.100 Three
positive charges of equal magnitude q are located at the
and the
triangle. Find the potential at the center of the
charges.
triangle
1.10d For two line charges q, and~
~q, at (d/Z, O) and (—tential for any point (x, y) and from this derive the
comers
force
of
an
on one
equilateral
of the
respectively, find the pod/Z}, 0),field.
electric
1.10e* Find the
expression
for
potential
as
electric field for that region
inside the large sphere.
large sphere of Prob.
outside the
well as for the
region outside
1.10f Find
1 4.d. Also find the
the small sphere but
E and E), in the void1n the sphere of chargein Prob. 1.4d by first finding the
potential. The zeros for the potentials of both large and small spheres should be at
infinity.
i
l
65
Problems
1.11:! Utilize the
rectangular coordinate form
V(l/I{D)
V~(t//A)
11/ and (I)
where
are
1.11c Derive the
V
identity
.13): and F
=
prove the vector
=
i/JVCI)
=
illV~A
equivalences
(1)th
+
+
A-Vil/
any scalar functions and A is any vector function of space.
1.11b Show that the vector
satisfied for [I]
to
expression
for
=
.
M
77:13
A
=
divergence
-
yxyz
+
th
+
111V
+
-:A
(inside back cover) is
iyzz.
in the circular
cylindrical
coordinate system.
1.11d Evaluate the
charge
1.11e Given
divergence of D in Exs. 1.4a, 1.4b, and 1.4c and compare with the known
densities. Evaluate V D for Ex.l.11 using rectangular coordinates.
.
a
vector F
=
3112,
evaluate
sides 2a centered about the
F
é,
origin.
-
dS for S taken
as
the surface of
a
cube and show that the two results
equivalent,
are
as
cube of
integral of V F for this
they should be by the divergence
Then evaluate the volume
«
theorem.
1.11f The width of the
depletion region at a metal—semiconductor contact (Ex. 1.4a) can be
2
using the relation d
(286133 / eN), where C133 is the barrier potential, 8 is
electron charge, and N is the density of dopant ions. Calculate d for ion densities of
1016, 1013, and 1020 crn‘3 assuming a barrier potential of 0.6 V. Comment on the applicability in this calculation of the concept of smoothed~out charge as assumed in
11.7.
using Poisson’s equation.Take a,
calculated
=
=
1.123 Find the
gradient
and
Laplacian
of
a
scalar field
varying
l/r
as
in two dimensions and
in three dimensions. Use the operators in rectangular form and also in
appropriate coordinate system in each case.
1.12b Find the electric field and
expressed
charge density
as
functions of x, y, and
2
if
a more
potential
is
as
CD
=
C sin
out
sin
By
e73
where
7
Val
2
+
32
charge density as functions of x for a space-charge—limited,
potential variation given by CD
V0(x/d)4/3. Find the
density J
pv and note that it is independent of x.
1.12(: Find the electric field and
parallel-plane
diode with
convection current
1.12d
=
=
Laplace’s equation that relative extrema of the electrostatic potential cana charge placed in an electrostatic field cannot be in stable
equilibrium (Earnshaw’s theorem).
Argue
not
from
exist and hence that
potential around a perpendicular intersection of the straight edges of two large
perfectly conducting planes, where the line of intersection is taken to be the axis of a
cylindrical coordinate system, can be shown to be expressible as
1.12e The
(I)
#Arz/B'
sin
§q§
Laplace’s equation in cylindrical coordinates and satzero-potential boundary condition on the planes. Find a generalization to an
arbitrary angle a between planes and verify that it satisfies Laplace’s equation.
Show that this function satisfies
isfies
a
1.13a Which of the
($0:
1.13b
+
yy)(x2
following
+
y2)_ “.7
xx +
may represent steady currents: J
Sketch the form of the two vector fields.
=
yy
or
J
=
Conducting coaxial cylinders of radii a and b have a conducting dielectric with permitivity a, and conductivity 0'1 for the sector 0 < gb < a, and loss-free dielectric 82 for
56
Chapter
1
Stationary Electric
P fields
/i
FIG. P1.13b
the remainder of the dielectric
ance
per unit
Find
regions (Fig. Pl.13b).
capacitance and
conduct—
length
1.143 Sketch the field and current lines for a structure of the formm Fig. 1.14c but with the
dielectric and perfect conductor regions exchanged an d a potential difference applied
between the two perfect conductors. Check all continuity conditions at boundaries.
1.14b A solution to the
(1906, y)
=
99.
problem
+
2
of
1.146
Fig.
can
be
2x
_1x +
1+-—tan
0
y
939
2w
a/Z)
(1 j>tan-l (Lila)
2
+
where
a
is the
on
length
i
‘
--
_
a
conditions
to be
shown:
y
of the conductive
the surface y
=2
+
region.
Show
O and find induced
2’.
1n
a
that this
y2
)72
+
(x
+
(x
satisfies the
—-
+
a/2)2
(1/2)2
boundary
surface charge density along this
1
boundary.
by means of Laplace’ 3 equation the potential distribution between two concenspherical conductors separated by a single dielectric. Theinner conductor of radius
is at potential V0, and the outer conductor of radius 17 is at potential zero.
1.1521 Obtain
tric
a
distribution between two
1.151) Obtain by means of Laplace’s equation the potential
tric spherical conductors with two dielectrics as in Exf, 1.40.
concen-
cylindrical conductors of radii a and b are at potentials zero and V0, re
spectively. There are two dielectrics between the conductors, with the plane through
the axis being the dividing surface. ThatIS, dielectric :81 extends from (15
O to cl)
271" Obtain
distribution from
77, and .92 extends from 4) 7r to <15
potential
the
Laplace’ 3 equation.
1
1.15c Two coaxial
=
—
=
=
1.15d Obtain the electrostatic capacitances for the two conductor systems described1n Sec.
1.15 and inProbs 115a b, andc
1.163 Assume the
ND
=
charge--density profile
1019 cm‘3,
T
and the width of the
of electric field.
=
shown111
300 K, and 8
=
Fig 1.1619
12. Find the
space—charge region (1,,
+
dp.
with
NA
=
1016 cm 3and
of the
potential barrier
jheight
Determine maximum value
1.16b Calculate
(130:) for the metal—semiconductor junction in Ex. 1.4a by integrating Poisequation. Call total barrier height (I),- in this case Find the width of the spacecharge region d, assuming ND constant.
son’s
1.16c* To illustrate the effect of a continuous charge profile in the pn junction example of
Sec. 1.16, consider a charge density in the depletion
region of the form p
=
67
Problems
(ech/a)exp[—— Ir/al]. Find
the electric field and potential as a function of x, and
00 and
sketch p, E, and (1) versus x. Find the potential difference between x
00 and
x
compare with Eq. 1.16(10), taking NA
ND.
=
=
—
=
1.173 Prove that, if
charge density p is given throughout a volume, any solution of Poisson’s
equation 1.122(3) must be the only possible solution provided potential is specified on a
surface surrounding the region.
1.17b Show that the
potential in a charge-free region is uniquely determined, except for an
arbitrary additive constant, by specification of the normal derivatives of potential on
the bounding surfaces.
1.183 Prove that the line
charge and its image as described
give constant potential along a cylindrical
conducting cylinder.
1.180 will
of the
for
a
conducting cylinder
surface at radius
a
in Ex.
in the absence
1.18b Prove that the
point charge and its image as described for the spherical conductor in
gives zero potential along a spherical surface at radius a in the absence of
conducting sphere.
Ex. 1.18d
the
1.18c A
circularly cylindrical electron beam of radius a and uniform charge density p passes
conducting plane that is parallel to the axis of the beam and distance S from the
axis. Find the electric field acting to disperse the beam for the edge near the plane and
for the edge farthest from the plane.
near a
1.18d* For
q(el
d from the plane boundary
q lying in a dielectric 8‘ distance x
second dielectric 52, the given charge plus an image charge
+ 82) placed at x
d with all space filled by a dielectric a, may
point charge
a
between 81 and
—
reg/(el
=
a
=
—
be used to compute the potential for any point x > 0. To find the potential for a point
x < 0, a single charge of value 2(182/(2-31 + 82) is placed at the position of q with all
space filled by dielectric 82. Show that these
lations at a dielectric boundary.
1.18e Find and
plot
of y, when
a
the surface
line
charge
images satisfy
the
required continuity
re-
charge density induced on the conducting plane as a function
(1 above the plane.
lying parallel to the z axis is at x
=
q,
applicability of the image concept for the case of a line charge parallel
vicinity of the intersection of two conducting planes with an angle
270 degrees. (See Fig. 1.18d.)
1.18f Discuss the
to
and in the
AOB
1.18g
Find the
(Do
(a
1.19a
=
when
a
conducting sphere
of radius
a
held at
distance d from the center of the
potential
sphere
d).
<
fields between
Map
tor at
1.19b
potential at all points outside
a point charge q is located a
potential V0,
Map fields
plane. The
d/lz
=
1,
an
as
plane conductor at potential zero and a
Fig. 1.190, but for step ratios a/ b of ti- and ~32
infinite
in
second conduc-
between an infinite flat plane and a cylindrical conductor parallel to the
conductor has diameter d, and its axis is at height 11 above the plane. Take
i.
a rectangular tube of sides
cylinder of radius a, with axis coincident
with the central axis of the rectangular cylinder. Sketch equipotentials and field lines
for the region between conductors, assuming a potential difference V0 between
1.19c The
outer
conductor of
a
two-conductor transmission line is
30 and 50. The inner conductor is
a
circular
conductors.
a and y
parallel conducting planes defined by y
semi—infinite conducting plane of negligible thickness at y
1.19d Two infinite
zero.
A
=
=
—
=
a are
0 and
potential
extending
at
68
Chapter
1
Stationary Electric
Fields
'
yt
0
<1>=o
Va
1
a?
’
°°
:t(L
A
>00
cp=o
FIG.P‘I.19d
from
=
x
0 to
map for the
x
=
region
00
is at
potential V0. (See Fig.
between conductors.
|
P1.l9d.)
Sketch
a
graphical
field
1.19e In Prob. l. 12e take A to be 35 and 7 to be measured 111‘ centimeters and make a plot of
the 100 V equipotential. Construct a graphical field map between the zero and 100- V
equipotentials showing the 25—, 50-, and 75~V equipotentials. Then sketch in the same
equipotentials found from the formula given in Prob. 1.12e to evaluate your field map.
Find the radial distance from the corner where the gradient exceeds the breakdown
fieldin air, 30 kV/cm. What does this suggest about the shape the corner should have
to avoid breakdown?
..
region in Fig. 1.196 into a mesh of squares of sides a/ 2 Terminate the
right a distance a from the corner and on the left a distance 3a/ 2 from
2a and V0= 100 V. Consider the potentials at the left and right
the comer Take b
edges of the above defined grid to be fixed at the values found1n linear variation from
top to bottom Start with all interior grid points at 50 V. Find the potentials at the
mesh points assuming zero space charge and
applying‘ the simple averaging method.
1.20a Subdivide the
region
on
the
=
1.20!)
Repeat Prob.
1. 20a
using
the
cyclic Chebyshev method
1.20c Solve for the potentials at the grid points in the problem in Fig. 1.20a by direct inversion of the set of difference equations expressing Laplace’s equation for all grid
the results with those111 Table 1 20 and discuss differences. Does di~
rect inversion give the exact values of potentials at the grid points? Explain your
points. Compare
answer.
1.20d Set up the difference equation for a
distribution in rectan»
gular coordinates. Consider a cubical box with the following potentials on the various
sides: top, 80 V; right side, 60 V; bottom, 0 V; left side, 100 V; front, 40 V; back, 100
three-dimensionallpotential
grid of the same coarseness as in Fig. 1.20a and assume initial potentials
grid points to be the average of the boundary potentials Calculate the
of corrected potentials by the three-dimensional equivalent of the simple
V. Define
a
for all interior
first set
scheme used for Table 1.20.
i
equation for potential1n cylindrical
O).
(6(13/ 645
1.20e Derive the difference
symmetry assumed
coordinates with axial
=
1.20f* An electron beam accelerated from
i
potential passes normally through a pair of
parallel-wire grids. Model the beam as infinitely broad and without transverse variation. Set up a one-dimensional difference equation for the potential between the grids.
Divide the 5~mm space between grids into five segments. Take both grid potentials to
zero
104 A/m“. Assume 1000 V as a first guess for
Take three steps ofpotential adjustment with space
charge based on the first guess. Recalculate space charge based on the new potentials
and again iterate the potential three times. Repeat recalculations of space charge and
be 1000 V and the beam current to be
all
difference-equation grid points.
i
69
Problems
potentials until the latter differ by no
charge. Use the simple iterative form
more
than 3% between recalculations of space
with (2
1.
==
1.213 Assume that
Fig. 1.19c is full scale, and that V0 is 1000 V. Find the approximate direction of the minimum and maximum electric fields in the figure. Plot a curve of
electric field magnitude along the bottom plane as a function of distance along this
plane, and a curve showing surface charge density induced on this plane as a function
of distance.
1.21b Calculate the
capacitance
per unit
length
from your
plots
for Probs. 1.1% and
c.
1.21c Describe the simplest way to use resistance paper to determine the capacitance per
wire between a grid of parallel round wires and an electrode lying parallel to the grid.
(See Fig. P1.2lc.) Assume the grid to be infinitely long and wide. Defend all decisions
made in the design of the analog.
<—@®®G)GD@®®~-—-—>
FIG. I"! .21
1.22a For
a
given potential
difference
V0
c
between conductors of
the stored energy in the electrostatic field per unit
evaluate the capacitance per unit length.
a coaxial capacitor, evaluate
length. By equating this to %CV2,
1.22b The energy required to increase the separation of a parallel-plate
tance dx is equal to the increase of energy stored. Find the force
per unit cross-sectional
plates
1.220 Discuss in
more
explain why
Eq. 1.222(5).
the
area
assuming
constant
charge
on
capacitor by a dis
acting between the
the plates.
self-energy term in Eq. 1.222(3), and
going to continuous distributions, as in
detail the exclusion of the
problem disappeared
in
1.22d Show the equality of the energies found using Eqs. 122(5) and (6) for
volume of charge of radius a and charge density p C/m3.
a
spherical
an arbitrarily shaped, charged finite conductor embedded in a homogeneousdielectric region of infinite extent that also contains a volume-charge density distribution. Starting from Eq. 122(8) show that (6) results. Make use of the identity in Prob.
1.1121 and consider the dielectric to be bounded by the surface of the conductor and
1.22e Consider
that at
1.22f If
an
infinity.
incremental
charge
distribution is
brought
into
a
field, the incremental energy
may be written
V
Use this
to
nonlinear.
develop Eq. 1.222(9)
for
an
unbounded
region
with
a
medium which may be
2. l
Magnetic
lNTRODUCTlON
effects have many similarities to electric
Magnetic forces were first observed
differences.
materials such
effects, but there
are
also
important
through
lodestone. The compass,
the attraction of iron to natu-
apparently devela profound effect
Europe
Oped
1&190,
In
William
to
thereafter.
1600
Gilbert,
upon navigation
physician Queen Elizabeth I,
a
an
De
book,
published
important
Magnete, presenting rational and thorough summary
of the magnetic effects known to that date, with
discussions of some of the similarities
to and differences from the electric effects then known.
l-Iad discoveries stopped at that
we
could
the
of
point,
immediately adapt
development
ithe preceding chapter to magnetic fields, the two kinds of magnetic “charges” being called north and south poles.
The important difference is that magnetic charges have so far been found only in pairs,
rally occurring magnetic
in China,
not
isolated,
was
so
introduced into
that
we
as
around
A.D.
would be concerned with fields
and had
Efrom dipoles,
as
in Ex. 1.8d.
Discoveries did not stop, however. In 1820, Hans Christian Oersted, during a class
demonstration of an electric battery, observed that the electric current in a wire caused
compass needle to be deflected, thus establishing clearly the first of several
important relationships between electric and magnetic gel‘fects. André~Maiie Ampere
a
nearby
very
quickly
extended the
experiments
and
developed 5» quantitative
Others who contributed both to the
law for the
phe-
practical use of
understanding
within
a
short
were
Jean—B
Felix
Sav art, Joseph
Biot,
electromagnets
very
period
aptiste
Henry, and Michael Faraday. The force produced by magnetic fields (either from permanent magnets or from electromagnets) on electric
currents was also clearly estab—
nomenon.
and to the
through these many experiments. These relationships between electric currents
magnetic fields will constitute the starting point for our development of magnetic
fields in this chapter. The relationships are somewhat more complicated than those of
the preceding chapter, primarily because both the current that acts as the source of field
lished
and
70
2.1
and the current element
must
acting
as a
71
Introduction
probe
to measure it are vectors
whose directions
be introduced into the laws.
As with electric fields, the distributions studied in this
chapter, although called
time-varying phenomena. These “quasistatic” prob
lems are among the most important uses of the laws and, in some cases, are valid for
extremely rapid rates of change. Still we must remember that other phenomena enter—u
and are likely to be important—~when the fields change with time. These are studied in
the following chapter.
Before beginning the detailed development, let us look briefly at a few examples of
important static or quasistatic magnetic field problems. There was the prompt app1i~
cation of Oersted’s observation to useful electromagnets. One of Henry’s early magnets
supported more than a ton of iron, with the current driven only by a small battery.
Electromagnets are now routinely used in loading or unloading scrap iron and many
other applications. The development of practical superconductors in the 19603 has made
possible magnets with high fields in large volumes with additional advantages of stability and light weight. Large currents can be made to flow in the magnet winding since
there is no voltage drOp and no heating. The need to refrigerate is compensated suffi—
ciently for a number of special applications. Superconductors are used extensively in
high—energy physics, where the need is for large volumes of strong field. Fusion research
depends on massive superconductive magnets for containment of the ionized gases of
a plasma. Motors and generators for special applications such as ship propulsion are
being made lighter and smaller by using superconductors.l
Moving charges constitute currents and magnetic fields produce forces on them as
they travel through a vacuum or a semiconductor. Thus magnetic field coils are used
for deflection and focusing of beams of electrons in television picture tubes and electron
microscopes. The magnetic deflection of flowing charge carriers in a semiconductor is
known as the Hall eflect; it is used for measurement of the semiconductor properties
or, with a known semiconductor, may be used as a probe for measurement of magnetic
“static,”
are
applicable
to many
field.
Coils
are
magnetic
used to
fields
can
provide
the inductance needed for
be found from the currents
as
high-frequency
circuits and the
in static calculations when the sizes
compared with wavelength. (However, current distributions are complicated at high frequencies by distributed capacitance in the windings.) Just as we
noted in Sec. 1.1 for electric fields, the distribution of magnetic field in the cross section
of a transmission line is essentially the same as calculated using static field concepts,
even though the fields can actually be varying at billions of times per second.
involved
‘
are
small
More details
on
superconductors
can
be found in Sec. 73.4.
72
Chapter
2
Stationary Magnetic fields
i
t
"
{isflhé‘zy-«in.<15It“"a?Li...Figs!4"?"\AV‘V‘ “‘7é“1e,»§.‘fl$“§7w¢“W"tf‘w‘Z‘is(”WT—VTHT’E‘“?4"1'1‘913QrW‘V'ép’k“)id:\‘l‘jém‘hw'qi'"‘1t
Static
Magnetic
them laws and aoncepts
CONCEPT OF
2.2
A
MAeNt—J'Icf:
FIELD
t
preceding chapter, we use the measurable quantity,
that magnetic forces may arise
either from permanent magnets or from current flow. Since the approach from currents
is more general—and on the whole more important—we start by consideration of the
As with the electrostatic fields of the
force,
to
define
force between
a
field. We noted in Sec. 2.1
magnetic
current
elements. Permanent magnets
may
then be included, at least
conceptually, by considering the effects of these as arising from atomic currents of the
magnetic materials.
The force arising from the interaction of two current elements depends on the magnitude of the currents, the medium, and the distance beitween currents analogously to
the force between electric charges. However, current has direction so the force law
more complicated than
that for charges. Consequently,
proceed by first defining the quantity we will call the magnetic field
and then, in another section, give the law (Ampere’s) that describes how currents contribute to that magnetic field. A vector field quantity B, usually known as the magnetic
flux density, is defined in terms of the force df produced on a small current element of
l
length dl carrying current I such that
between the two currents will be
it is convenient to
,
df=1dlBsinG
where 6 is the
angle
between £11 and B. The direction
(1)
I
relations
of the vectors
are so
along a perpendicularlto the plane containing dl and
B, and has the sense determined by the advance of a right~hand screw if (11 is rotated
into B through the smaller angle (Fig. 2.2). It is convenient to express this information
more compactly through the use of the vector product.
The vector product (also called
cross product) of two vectors (denoted by a cross) is defined as a vector having a
magnitude equal to the product of the magnitudes of the two vectors and the sine of
the angle between them, a direction perpendicular to lthe plane containing the two
vectors, and a sense given by the advance of a right—hand screw if the first is rotated
defined that the vector force df is
FIG. 2.2
Right-th
screw
rule for force
on a
current
element in
a
magnetic field.
into the second
through
the smaller
angle. Relation (1)
df
The
known
quantity
as
the
=
magnetic
as
the
permeability,
may then be written
I dl X B
field vector
H and is related to the vector B defined
medium known
73
Ampére’s Law
2.3
by
or
(2)
magnetic field intensity is denoted
(2) through a constant of the
the force law
[Li
B
=
,U.H
(3)
Many technologically important materials such as iron and ferrite are nonlinear and/or
anisotropic, in which case p. is not a scalar constant, but to keep this introductory
treatment simple, the medium will first be assumed to be homogeneous, isotropic, and
linear. A somewhat more general form of (3) will be given in Sec. 2.3.
In SI units, force is in newtons (N). Current is in amperes (A), and magnetic flux
density B is in tesla (T), which is a weber per square meter or volt second per square
meter and is 104 times the common cgs unit, gauss. Magnetic field H is in amperes per
meter and ,u is in henrys (H) per meter. Conversion factors to other cgs units are in
Appendix I. The value of p. for free space is
,LLO
=
2.3
477 X 10"7
H/m
AMPERE’S LAW
experimentally from a series of ingenious experiments,2 de~
magnetic field vector defined in Sec. 2.2 is calculated from a system
of direct currents. Consider an unbounded, homogeneous, isotropic medium with a
small line element of length (11’ carrying a current 1’ located at a point in space defined
by a vector i" from an arbitrary origin as in Fig. 2.30. The magnitude of the magnetic
field at some other point P in space defined by the vector r from the origin is
deduced
Ampere’s law,
scribes how the
dH(r)
where R
The
2
2
Ir
angle qb
-—
r’
,
==
I’(r’) dZ’ sin qb
477R2
the distance from the current element to the
is that between the direction of the current defined
point of observation.
by (11’ and the vector
description, see J. C. Maxwell, A Treatise on Electricity and Magnetism, 3rd ed,
Chap. 2, Oxford Univ. Press. Oxford, 7892. The law is now more frequently named
after Biot and Savart, but the assignment remains somewhat arbitrary. Following Oerst»
ed ’3 announcement of the effect of currents on permanent magnets in 7820, Ampere
immediate/y announced similar forces of currents on each other. Biot and Savart pre»
sented the first quantitative statement for the special case of a straight wire; Ampere
later followed with his formulation for more general current paths. The form given here
is a derived form borrowing from all that work. For more of the history see E. T. Whittaker,
A History of the Theories of the Aether and Electricity, Am. Inst. Physics, New York, 1987.
or P. F. Mottelay, Bibliographical History of Electricity and Magnetism, Ayer Co. Publishers,
For
a
Part iV,
Salem, NH, 7975.
74
Chapter
2
fields
Stationary Magnetic
dH.
P
l
FIG. 2.3a
Coordinates for calculation of
magnetic field from
current element.
r' from the current element to the point offE observation. The direction of
dH(r) is perpendicular to the plane containing d1 and R, and the sense is determined
by the advance of a right-hand screw if dl is rotated through the smaller angle into the
vector R. Thus, with the current direction shown in Fig. 32.35:, dH at P is outward from
the page. We see then that the cross product can be used to write the vector form of
R
=
r
—-
law:
Arnpere’s
dH(r)
To obtain the total
integrated
over
the
magnetic
path
I'(r’) dl’
=
=
l
directly
(1)
f
‘
elements along
I’(r’) dl’ X
417R3
It is of interest to examine further the relation
field H is
R
477R3
field of the current
PM)
X"
related to the currents, without
a current
path, (1)
is
R
(2)
between
B and H. We
see
that the
regard for the nature of the medium
homogeneously. The force on a current element was seen in
depend upon magnetic flux density. The infidence of the medium in relating
B and H comes about in the following way. The electrdnic orbital and spin motions in
the atoms can be thought of as circulating currents on jwhich a force is exerted by B
and which produce a field M (called magnetization) that adds to H. This is analogous
to the response of a dielectric medium shown in Fig. ll.3c. Then B is related to H as
though there were only free space but with the added field of the atomic currents
as
long
as
it fills all space
Sec. 2.2 to
B
=
mm
+
M)
(3)
f
Magnetization M may have a permanent contribution (to be considered in Sec. 2.15),
but here we neglect this and assume the material isotropic so that M is parallel to H.
2.3
We
can
75
Ampére’s Law
then write
B
==
#00
+
Xm)H
pH
=
=
I-l'r/J'oH
where )(m is called the magnetic susceptibility, p, is the permeability introduced in Sec.
2.2, and ,u, is the relative permeability. Many materials have nonlinear behavior so Xm
and ,u. are, in
general, functions of
the field
strength.
For
diamagnetic
materials Xm <
> 0. Most
O, and for paramagnetic, ferromagnetic, and fenimagnetic materials Xm
materials
commonly considered to be dielectrics or metals have either diarnagnetic or
“'5
so we treat them as free space, taking
paramagnetic behavior and typically I Xm' < 10
p.
,LLO. Ferromagnetic and ferrimagnetic materials usually have Xm and tL/ito much
greater than unity and in some cases are anisotropic, that is, dependent upon direction
of the field. All of these aspects are considered in more detail in Chapter 13.
=
Example
2.3a
FIELD ON AXIs OF CIRCULAR LOOP
application of the law, the magnetic field is computed for a point
loop of wire carrying dc current I (Fig. 2.31)). The element dl’
and
a
is always perpendicular to R. Hence the contribution dH from
magnitude dgb’
As
an
on
the axis of
has
an
example
of the
a
circular
element is
Ia
dH
As
one
integrates
about the
FIG. 2.3b
Magnetic
loop,
2
dqb’
(4)
———————-——
477(02
+
22)
the direction of R
field from element of
a
changes,
and
circular current
so
the direction of
loop (Ex. 2.3a).
76
dH
Chapter
Stationary Magnetic fields
2
changes, generating a conical surface as <1) goes through 277 radians (rad). The radial
components of the various contributions cancel,
and
the axial components add.
'
Using (4)
I
dH__
a
de
Integrating
in
qS
amounts to
==
dH sin 0
=
thus
multiplying by 277;
Ia2
H3
Note that for
a
point
(5)
=
2m2
at the center of the
22W?-
+
loop,
=
2
O,
I
H:
=
(6)
—
261
2:0
g
Exampte 2.3b
FIELD OF
A
FINITE STRAIGHT LINE
OF CURRENT
magnetic field H at a point P a perpendiIsular distance r from the center
length of current I, as shown in Fig. 2. 30. It is easy to see from the righthand rule that thereIS
only an H g component. Its magnitudeIS given by the integral of
over
the
20
(1)
length
Let
of
us
a
find the
finite
a
I sin
9b
We
can see
from
Fig.
2.3!: that sin
45
"
_
4’
which becomes I/ 2771‘ if
Ia]
r/R
andR
dz
.11.
477
=
_
-a
->
(7‘2
00.
+
This
([2
dz
4721?:2
“a
22)”
(r2
=
22)”?
+
Thus,
1
__I_
[(1‘/a)2 + 111/2
same result is found in Ex. 2.4a by
(7)
2771‘
a
different
method.
FIG. 2.3a
Calculation of
magnetic
field of
straight
section of current (Ex.
I
2.3b).
2.4
FIG. 2.3d
Tightly
77
The Line Integral of Magnetic Field
("I
O
wound solenoid of
22
turns per meter
and its
l
representation by
a current
sheet
A/m (Ex. 2.30).
of 22]
Example
2.5c
FIELD IN AN INFINITE SOLENOID
Let
us
here model the
current
long, tightly wound solenoid shown in Fig. 2.3d by an equivalent
magnetic field inside. We assume that though
small helical angle with a cross~sectional plane, we can adequately
sheet to facilitate calculation of the
the wire makes
a
model it with
circumferential current. The current
meter
flowing around the solenoid per
is the number of turns per meter and I is the current in each turn.
differential length of the sheet model, there is a current 111612. We will cala
is 221, where
n
Then, in a
culate, for simplicity, the field on the axis. But one can show, by means that will come
later (see Ex. 2.4d) that the field for an infinitely long solenoid is uniform throughout
the inside of the solenoid. We can adapt (4) for the present calculation by taking 1
in (4) to be 121512. Then the total field on the axis for the infinitely long solenoid is
given by
nIa2 dz
Hz
In
evaluating
the
integral in (8),
infinity
then lets the limits go to
:
one
f—oo 2(a2
first takes
a
symmetrical
=
to
length, it is easy to modify (8)
perform the integrals for fields not
2.4
finite limits
n]
solenoid of finite
2.30) but difficult
(8)
2:3)3/2
as
in
(7) and
with the result
H:
For
+
(9)
to
on
obtain on-axis fields (Prob.
the axis.
THE LINE INTEGRAL OF MAGNETIC FIELD
field may be computed from a given
system of currents, other derived forms of the law may be more easily applied to certain
types of problems. In this and the following sections, some of these forms are presented,
Although Ampere’s
law describes how
magnetic
examples of their application. The sketch of the derivations of these forms, because
are more complex than for the corresponding electrostatic forms, will be left to
Appendix 3.
with
they
78
Chapter
Stationary Magnetic: Fields
2
magnetic field laws derived from Ampere’s law
integral of static magnetic field taken about any given
current enclosed by that path. In the vector notation,
One of the most useful forms of the
is that which states that
closed
path
must
equal
a
the
line
iH-dltLJ-dS=I
(1)
Equation (1) is often referred to as Ampére’s circuittzl law. The sign convention for
right side of (1) is taken so that it is positive if it has the sense of advance
of a right-hand screw rotated in the direction of circulation chosen for the line integration. This is simply a statement of the well-known right-hand rule relating directions
of current and magnetic field.
Equation (1) is rather analogous to Gauss’s law in electrostatics in the sense that it
is an important general relation and is also useful for problem solving if there is sufficient symmetry in the problem. If the product H d1 is constant along some path, H
can be found simply by dividing I by the path length.
current on the
'
Example
2.4a
MAGNETIC FIELD ABOUT
A
LINEICURRENT
important example is that of a long, straight, round} conductor carrying current I. If
integration is made about a circular path of radius r centered on the axis of the wire,
the symmetry reveals that magnetic field is circumferential and does not vary with angle
as one moves about the path. Hence the line integral is
just the product of circumference
and the value of H g. This must equal the current enclosed
An
an
E£H~di=2mfl¢=1
01‘
I
H
as was
Fig.
found
by
a
=
‘6
——
2777‘
A/m
(2)
I
different method in Ex. 2.3b. The
sense
relations
'
are
given
in
2.4a.
Example 2.4b
MAGNETIC FIELD BETWEEN COAXIAL CYLINDERS
A coaxial line
(Fig. 2.4b) carrying current I on the inner conductor and ——I on the outer
(the return current) has the same type of symmetry as the isolated wire, and a circular
path between the two conductors encloses current I so that the result (1) applies directly
,
for the
region
between conductors:
The Line
2.4
79
of Magnetic Field
integral
1%
I’ll-
’
a
I
’
I
(E
‘
I
”‘1
I
I
I
—-—>
:
I
I
l
l
l
‘
\
I
I
\
\.,
/'
(a)
~fi___I
I
HCP
c
I
b
/1:\‘
I
—"
I
___r_ll.....t‘....._.___.._.._...___
a
l
_.._..........
A
i
.
I
U
I
._
i
I
_____
i....._'__...__....__.___.._
\
I
_—-——~.h_‘——-.———.u__._——__
(b)
FIG. 2.4
2.4a and
(a) and (b) Magnetic field about line
b).
current
or a net
a
circular path encloses both the
current of zero. Hence the
cylinders (Exs.
a<r<b
H¢:_
Outside the outer conductor,
and between coaxial
magnetic
field outside is
Example
(3)
going
and return currents,
zero.
2.4a
MAGNETIC FIELD INSIDE A UNIFORM CURRENT
Let
a
us
find the
magnetic
field inside the round inner conductor in
uniform distribution of current. We will
current
by a circle
density is I/ 77122.
enclosed
the current
at radius
r.
The current
apply (2)
but with I
Fig. 2.4b assuming
replaced by 10'), the
The total current in the wire is
[(1‘)
1(a)
=
I and
is
‘7
10-)
and
=
(2-)
I
(4)
—'3
(5)
using (2),
Him
=
—
=
89
Chapter 2
Stationary Magnetic Fields
I
Z)
l
(1
W]
.m—rv‘l
seem/mgeeeeeeegms
‘a\<____..._..74r
[3/
\\
I
______
__________
_l.__.___,z
A
P
B
S®®®®®®®®®®®®®S
FIG. 2.4a
through axis of infinite solenoid
symmetrically spaced elements.
Section
axis from two
for Ex. 2.4d
showing
contributions to H
on
Example 2.4d
MAGNETIC FIELD OF
In Ex. 2.30
A
SOLENOID
magnetic field HZ
carrying a current I A
showed that the
on the
axis of
infinitely long
the integral
relation (1) to Show that the field outside is zero and that inside is uniformly 121. Figure
2.40 shows the section through the solenoid in a plane containing the axis. Let us
consider the integration paths shown by broken lines to be 1 m long in the z direction
for simplicity of notation. Any radial component of H lproduced by a current element
is canceled by that of a symmetrically located element; This is illustrated in Fig. 2.4c
for the fields Ha and Hb from elements a and I) located equal distances from the point
P. Thus, H dl is zero along the sides BD and AE.
Taking the line integral around path ABDEA and setting it equal to the enclosed
current gives
solenoid of
we
12
turns
per meter
is
[n]
.
Now let
an
us use
.
I
E
fiH-dl=n1+fH-d1=nl
(6)
D
since H
the axis is 121. From
(6) the integral from D
to E is zero. Since the
placement
zero.
arbitrary,
path
The line integral around path ABCFA encloses no current so the integral along the
arbitrarily positioned path CF must be equal in magnitude to, and of Opposite sign from,
that along AB. Thus, the internal field is everywhere z—directed and has the value
on
of the outside
DE is
external H must be
H:
=
M
(7)
Note that these symmetry arguments cannot be made for a solenoid of finite length, but
the results given here are reasonably accurate for a
solenoid having a length much
greater than its diameter, except
near
the ends.
2.5
Inductance from Flux
Linkages:
8'5
External Inductance
t
“’3' 39:03”.
FIG. 2.5a
Loop
of wire.
1
shows surface used for calculation of external
Cross—hatching
inductance.
2.5
The
for
important
INDUCTANCE FROM FLUX LINKAGES: EXTERNAL INDUCTANCE
circuit element which describes the effect of
electric circuit is the inductor. It is of
an
primary
concern
magnetic energy storage
dynamic, that is, time—
for
varying, problems, but the inductance calculated from static concepts is often useful up
to very high frequencies. This is the quasistatic use discussed in the introduction to this
chapter. In a manner similar to the capacitance definition of Sec. 1.9, inductance can
be defined in terms of flux linkage by
1
Lz—Jan
(1)
Is
where the surface S must be
in
Fig.
by
the
specified. Consider, for example, the loop of wire shown
produces magnetic flux in the crossnhatched area S bounded
of the flux produced by the current is inside the wire itself. It
2.5a. The current I
loop. Also,
some
is convenient to separate the inductances related to these two components of flux and
call them, respectively, external inductance and internal inductance. Examples of cal—
culations of external inductance for
of
an
simple structures are given
presented in Sec. 2.17.
below and
an
example
internal inductance calculation is
Exampie
2.5a
EXTERNAL INDUCTANCE OF A PARALLEL-PLANE TRANSMISSION LINE
Here
we
find the external inductance for
a
unit
length
of
a
parallel~plane
structure
(Fig.
2.517) which is wide enough compared with the conductor spacing that the fields between
the conductors are, to a reasonable degree of accuracy, those of infinite parallel planes,
as suggested in Fig. 2.5c. Note that the flux tubes (bounded by the field lines) spread
out greatly outside the edges of the conductors. Thus, there is a strong reduction of flux
density B and, therefore, also H. The line integral of H around one of the conductors
82
Chapter
2
Stationary Magnetic Fields
I
_l_
h
T
I
J;
\\\\
Surface for calculation of external inductance of
FIG. 2.5b
has its
h
predominant
contribution from the field
a parallel—plane transmission line.
HO between
the conductors,
I=§H°dlEHowi
where I is the total current in
one
conductor and
w
is
(2)
the conductor width. This result
path in the cross-sectional plane (Fig. 2.5c) between and parallel to the
conductors, so HO can be considered approximately uniform.
The external inductance for a unit length is found by applying (1) to the surface
between the conductors which is shown shaded in Fig. 2.51). Since I is independent of
z and H0 is nearly constant through the space between the conductors and is perpenapplies
to
any
dicular to the shaded surface, ( 1) becomes
1
L
=
-
I
This relation is based
on
the
d
I
flo<w>
—
d
=
H /m
—
“0
neglect of fringing
w
(3)
fields and is most accurate for small
d/w.
FIG. 2.50
Cross-section of
character of
magnetic
parallel-plane
field lines.
transmission line of finite width
i
showing general
2.5
FIG. 2.5d
Inductance from Flux
33
Linkages: External Inductance
Surface for calculation of external inductance of
a
coaxial transmission line.
Exampie 2.5b
EXTERNAL INDUCTANCE OF
For
a
coaxial line
COAXIAL TRANSMISSION LINE
A
pictured in Fig. 2.5d with axial
returning in the outer, the magnetic field
(Ex. 2.4b)
as
conductor and
r
<
b, is
current I
flowing
in the inner
is circumferential and, for
a
<
I
H
For
a
unit
shaded
length the magnetic
in Fig. 2.5d,
::
¢
(4 )
—
2777‘
flux between radii
a
and b is,
by integration
over
the
area
b
1
b
[B‘dS=Ju*d7-=filnS
217
2777‘
a
(5)
a
So, from (1), the inductance per unit length is
L
=
i In b
-
277
a
H/m
(6)
high frequencies, there is not much penetration of fields into conductors as will
in Chapter 3, so this is then the main contribution to inductance. The internal
inductance for low frequencies will be considered in Sec. 2.17.
For
be
seen
84
Stationary Magnetic tPields
Chapter 2
I
Dtfiere tiafl Forms t
:23"
~'~‘u:;x<
‘2r:‘
r‘
«:r—m
a:
uncaéifikauée
.
.2
1..
.735
g
7;.“
a, any»;
925%. mi
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netostatistie
a
and the Rise oi Fotentiai
THE CURL OF A
2.6
VECTOR: FIELD
having to do with line integrals, it will be
operation called cur'l. This is defined in terms of a
line integral taken around an infinitesimal path, divided by the area enclosed by that
path. It is seen to have some similarities to the operatipn of divergence of Sec. 1.11,
which was defined as the surface integral taken about an infinitesimal surface divided
by the volume enclosed by that surface. Unlike the divergence, however, the curl op~
eration results in a vector because the orientation of the surface element about which
To write differential
equation forms
for laws
necessary to make use of the vector
the
taken must be defined. This additional
integral is
complication seems
to be
enough
student.
The student should attempt
make curl a more difficult concept for a beginning
to obtain as much physical significance as possible from the definitions to be given, but
to
at the same
with
time should recognize that full
practice
in its
The curl of
in
a
at
that
a
vector
field is defined
particular direction is found
point,
and
appreciation of the operation will come only
I
use.
finding
as a vector
by orienting
a
the limit of the line
A
[cur1F.=
],
function whose component at a point
small
area normal to the desired direction
integral divided by
the
area:
tF'dil
.
1
A5130
(1)
—T
AS, E
where i denotes a particular direction, AS, is normal to that direction, and the line
integral is taken in the right-hand sense with respect ito the positive 2‘ direction. In
rectangular coordinates, for example, to compute the 2 component of the curl, the small
area AS
Ax Ay is selected in the ray plane to be normal to the z direction (Fig.
2.6a). The right-hand sense of integration about the path with respect to the positive 2
direction is as shown by the arrows of the figure. The line integral is then
=
ffSF
-
(11
=
AyFy
Axe
-
x+Ax
We find Fy at x + Ax and
Fx
at y +
—
AyFy
y+Ay
Ay by
+
Axe
I
truncated
y
Taylor series expansions
l
Fx
E
y+
Ay
+
F_,.
y
an
Ay—
6y
F},
;
3;I
x+
y
Ax
So
6F
6F
8x
6y
+
Fy
3gF-dl§<—y-— x)AxAy
x
6F),
Ax—
ax
(2)
x
2.6
The can of
85
Vector Field
a
Z
ll
p,
(x. y. 2)
F3,
Ay
F1
3:
FIG. 2.6a
Then
using
the definition
Path for line
(1),
we
integral
get
6F
[curl F],
=
"
because the
of
area
in definition of curl.
6F-
,
-—3
«-
8x
—-‘
(3)
6y
expansions (2) become exact in the limit. Similarly, by taking
plane and x—z plane, respectively, we find
6F
a1:~
[curl F1,
=
[curl F],
=
3-:
y
BF
These components may be multiplied
form the vector representing the curl:
aF-
8F
.
cur1F=i——-——’
By
62
by
(4)
a2
6F-
—
62
the
,
t
—
.
--‘-‘~
'
--‘*
(5)
6x
corresponding
6F
31?.
.
+y—-‘——‘
62
8F
8x
6F.
6y
compared with the form of the cross product and the definition
operator V, Eq. 1.10(7), the above can logically be written as
it
a
curlFEVXF=
—
x
Ft
In
unit vectors and added to
+2—y——A
6x
If this form is
vector
the elements
in the y—-z
deriving
y
a
—
By
(6)
of the
2
a
—
62
(7)
Fy F:
curl for other coordinate systems, the variation of line elements with co—
just as the variation of surface elements with coordinates
ordinates must be considered,
86
Chapter
Stationary Magnetic
2
lf‘ields
spherical coordinates was consideredin Sec. 1. 11. (See Appendix 2.) Results for
cylindrical and spherical coordinates are given 0n the inside front cover?
The name curl (or rotation as it is sometimes called) has some physical significance
in the sense that a finite value for the line integral taken in the vicinity of a point is
obtained if the curl is finite. The name should not be associated with the curvature of
in
circular
lines, however, for a field consisting of closed circles may have zero curl nearly
everywhere, and a straight-line field varying in certain ways may have a finite curl. The
following examples illustrate these points.
the field
Exampie
26a
CURL—FREE FIELD WITH CIRCULAR
LINES
FIELD
magnetic field in the region surrounding a current in a long straight round wire
in Eq. 2.4(2) to be H¢
I/ 2771'. If we write this in rectangular coordinates
x2 + yz, we get
x/r, and r2
y/r, cos g!)
using sin Q!)
The
=
was seen
=
=
=
:
_H¢
8111
d)
=
—_27sz
I
Hy
Hz
as can
(6)
(8)
-—
=
H¢
d):
x
compo31ent and
seen
This result is found
for the curl in
VXH=
more
cylindrical
16H
of the curl
=
aHx
6x
6y
2(—
and
naturally
aH
directly
AaH
dz
62
dependence on 2,
Substituting
_
( 11 )
for this
coordinates found inside the
6H
no
are zero.
iproblem using the expression
frbnt cover:
arH
f--[ agb_ ¢]+¢[—’ua—Ii]+2[l( ¢)_16_I£]
r
Since there is
and
(9)
(10)
shows
and
.
Tflm
from Fig. 2. 6b Since there is no 2
immediately that the x and )7 components
(9) into (6) with F: H we obtain
be
(8)
y;
+
0
curl H
r
COS
3’
I
.
Hr
only
an
H ¢ and
91) components
no H and
1H¢
are
there18
no 2
62‘
dependence,
the transverse
does not
-
ones
depend
the
firstI two
corresponding
upon r,
61'
r
we
to
x
2‘
dub
(12)
components vanish. The
and y components. Since
see that
V
X
H: O
as
shown
above.
3
As with the
divergence (footnote 4
basic definition ( i).
cannot
of Chapter 7), one
take the cross product of
a curvilinear coordinate system, but must use the
V and the vector to obtain the curl in
2.6
The Curl of
a
37
Vector Field
L‘1"!
A
¢H¢
FIG. 2.6b
Resolution of
H4, of a
line current into
rectangular components (Ex. 2.6a).
Example 2.6b
FIELD WITH NONVANISHING CURL
The
magnetic field inside a uniform current with circular symmetry was seen in Ex. 2.4c
Ir / 27mg. As in the preceding example, we see that the symmetries
¢(r)
indicate the presence of only the 2 component of the curl in (12). Also, the second term
to be H
in the
2
2
component is
zero.
Thus
1
6(qub)
VXH=2
61‘
r
Example
NONVANISHING CURL
A
theoretically
magnetron has
Fig. 2.6c. We
immediately
IN
see
electron
there
evident
by
velocity
a vector
7m
(13)
I0
2.6c
FIELD OF STRAIGHT PARALLEL VECTORS
stable electron flow in
an
1
=2
a
type of microwave electron tube called
planar
by
2y
vectors straight and parallel. It is
distribution described
function with all
substitution of
v
in
(6)
v
=
that
T
xmeV
FIG. 2.6::
Electron flow in
2
planar magnetron (Ex. 2.60).
a
and is shown in
83
Chapter
curl
ficurl v]x
=
v
there
be
can
AS—
~
Ay
only
are
an x
on
the
A2
that the curl is
lim
=
A 3’)
+
[v‘(y
Ay
a
(6) shows
small
nonzero
because the
gvz(y)]Az
area
(15)
AZ
larger
is
velocity
on one
side of the
loop
i
Example 2.6d
OF THE
GRADIENT
1
OF A
ScALAR
gradient of
show the useful fact that the curl of the
we
and
write for
.-
the other.
CURL
Here
plane
yuz
we can
='
AS——>O
see
in
component of the curl. Then
_
than
(14)
definition of the curl (1) why this result
[Curl le
We
it #10
:
by using the line-integral
and their spatial variations
It is instructive to see,
obtains. All vectors
fields
Stationary Magnetic
2
a
a
scalar is
zero.
If
we
write
Ea
a
F=V§zi£+§j+2é
62
6x
By
and substitute it in
VXF=X
(6),
we
62 E
By
get
62
(92 5
62“
62
+9
By
62 5
§_
62 6x
62
+2—§6x
6x 62
6y
62 ,5
6y
6x
(16)
O. A particpartial derivative operations is arbitrary V X F
X
V
is
field.
The
fact
that
0 follows
the
electrostatic
E:
ularly important example
E:
VII)
E0.
from
either
or
dl
shall
see
in
We
95
immediately
Chapter 3 that
these properties of E do not apply for time-varying fields.
Since the order of the
=
=2
-
2.7
Now let
us use
magnetic
MAGNETIC FIELD
OF
the formulations of the last two
field. The line
of the curl,
CURL
Eq. 2.6(1),
integral
to
of H around
get
95
.
[curl H] ,-
hm
=
95
H
-
(ii is the current
through
the
area
=
lim
~
{11
"
AS- by Eq.
IAS
,
[curl H],.
H
AS;
ASi—eO
But
sections to derive a new relation for
AS, is substituted in the definition
an area
2 4(1)
so
as
—J,.
—
(1)
2.7
This relation holds for all three
corresponding
Curl of
orthogonal components.
unit vectors and added,
we
89
Magnetic Field
get the
vector
If these
are
multiplied by
amnevxn=i
This
the
relation
m
be
thought of as the equivalent of Eq. 2.4(1) for a differential path taken
point. Note that the curl H found in Eq. 2.6(13) is the current density, as
required by (2).
can
around
a
Exampfie
2.7
CURRENT DENSITY AT SUPERCONDUCTOR SURFACE
If
a
sheet of
2H3 parallel to its
superconductive material4 is in a magnetic field H
a penetration of H only a very short distance into the superconductor
Fig. 2.7. The decay of H: with distance is given by
=
surface, there is
as
shown in
it==H¢“V“
where
H0
AS, called
corresponding
is the value at the surface and
of the material. We
can
find the
e)
the
penetration depth, is a property
density using (2) and the
current
H“
Ho
0
FIG. 2.7
Penetration of
magnetic
field into
a
A
0
thick sheet of
)\
4
47x
2h
superconducting
material.
Superconductors include lead, tin, niobium, and numerous other elements, alloys, and
compounds. They have zero dc resistance and other special properties below their crit—
ical temperatures. See, for example, V. Z. Kresin and S. A. Wolf, Fundamentals of Superconductivity, Plenum Press, New York, 7990.
90
Chapter
expansion
in
Eq. 2.6(6):
Jy
Thus, the
fields
Stationary Magnetic
2
I
I
=
[curl
current also is found
HL,
only
=
6H,
Hog
=
near
“ac/A:
/\.—58
"‘3;
the surface.
f
AND
INTEGRAL FORMS
RELATION BETWEEN DIFFERENTIAL
OF THE FIELD
EQUATIONs
2.8
relating magnetic field to current; density was
through the definition of curl. One can‘ proceed in
integral
Stokes’s theorem, which states that for a vector function F,
The differential form
form
derived from the
by using
reverse
§Frdl=[(curlF)'dSEf(V,XF)'dS
S
This theorem is made
areas.
the line
integral
about that
area
tesimal
areas are
summed
over
places
in
Fig. 2.8a, breaking
contribution (V X F) (18 gives
of cuil. If contributions from infiniline integral must disappear for all
For each differential area, the
by
the definition
the surface, the
is first traversed in
boundary
determining
Subdivision of
‘
one: direction
the contribution from
where these contributions do not
FIG. 2.86
(l)
g
plausible by looking at a general sjurface as
it into elemental
internal areas, since a
opposite direction in
S
disappear
arbitrary
are
surface for
an
along
and then later in the
adjacent area. The only
boundary, so that
the outer
proof of Stokes’s
theorem.
Relation Between Differential and
2.8
the result of the summation is then the line
integral Forms of the Field Equations
integral
of the vector around the
9'!
boundary
stated in (1). It is recognized that the process is similar to the transformation from
the differential to the integral form of Gauss’s law through the divergence theorem in
as
Sec. 1.11. Then
writing
Stokes’s theorem for
magnetic field,
we
have
figH-dl=f(VXH)OdS
(2)
5
But, by Eq. 2.7(2), the curl may be replaced by the
current
density:
3gH‘dl=fJ-ds
(3)
s
The
line
right side represents the
integration on the left is
through the surface of which the path for the
(3) is exactly equivalent to Eq. 2.40).
boundary.
flow
current
a
Hence
Example
DEMONSTRATION
Let
us
netic
OF
demonstrate Stokes’s theorem for
wave
in
a
magnetic
field that is part of
certain kind of transmission structure. The field at
of time is described
apply (2)
illustrated. The line
a
electromag—
particular instant
an
by
H
We will
a
2.8a
STOKES'S THEOREM
=
to the area shown in
integral
of
7%
3%
cos
7—75
(4)
a
Fig.
2.8b where the field distribution (4) is
(4) along the broken path is
flW/flflfl
,_,/
/
flfl/fl/fl/fl/ %
FIG. 2.8b
(Ex. 2.8a).
Area for
integration
of field H to demonstrate the
validity of
Stokes’s theorem
92
chapter
2
Stationary Magnetic
0
1
a
fields
0
§H-dl fodx+f Hydy-l-J dex+f
O
O
a
=O+Acosw+O—AcosO§=
where the facts that
The curl of H in
O and
Hx
Hy # f( y)
rectangular coordinates
=
are
1
,
Hydy
(5)
—2A
used.
is
3H
VXH=2——y-=—2A°—Tsinlx
6x
The
integral
of
(6)
over
the surface bounded
a
by
the
(6)
a
brolten line in Fig. 2.819 is
0
mt
7T,
72x
fS(VxH)'dS qu-sm—dx=Acos——
0
=
Since (5) and
(7) give
the
same
a
a
t
I
a
o
~2A
<7)
results, Stokes’s
theorern (l)
is illustrated.
i
l
Example 2.8b
That V
as was
-
V
X F
r-
O
can
done for V X
be
Vz/r
F
PROOFTHATV-V
x
proved by using the
expressions in rectangular coordinates
={U
take a different approach that uses
applies to any surface, we may treat the
let the bounding line shrink to zero so the surface
line integral on the left side of (2) vanishes and we
in Ex. 2.6d. Here
we
Stokes’s theorem. Since Stokes’s theorem
surface shown in
Fig.
2.8a and
becomes
one.
Then the
a
closed
have
i
£(VXF)°dS:O'
Line bounding
surface S
FIG. 2.8a
Surface used in Ex. 2.8b.
(8)
2.9
We may then
apply
the
Vector
divergence
93
Magnetic Potential
theorem
(Sec. 1.11)
to the vector
V
x
F:
%(VXF)‘dS=JV-VXFdV
s
Since
in (8) that the left side is
we saw
integrand
on
the
right
was to
with
zero
an
arbitrary
choice of surface, the
side must vanish,
V
which
(9)
v
be shown. This is
2.9
a
V
-
F
X
O
=
(10)
useful relation in the
study
of
electromagnetic
fields.
VECTOR MAGNETIC POTENTIAL
We introduce here another
potential, which is often used as a conveniently calculated
quantity from which the magnetic field can be found. An integral expression for the
flux density can be obtained from Eq. 2.3(2) by multiplying by it for homogeneous
media:
It is shown in
certain vector
><R
ul’(r’)dl’
_
Appendix 3 that this can be
equivalences. The result gives
B(r)
=
broken into two steps
by making
use
V X A(r)
of
(2)
where
I
I
d]!
The current may be given as a vector density J in current per unit area spread over a
volume V’. Then, since I
J d5, where dS is the differential area element perpendicular
=
to
J and d1 is in the direction of J, dS all forms
to
(3) is
,
A(r)
In both
_
—
a
volume element (IV and the
ILJOJ)
equivalent
dVI
L—MR
(3) and (4), R is the distance from
a
current element of the
integration
to
the
introduced as an intermediate step,
point
computed.
is computed as an integral over the given currents from (3) or (4) and then differentiated
in the manner defined by (2) to yield the magnetic field. Function A is called the
magnetic vector potential. Note that each element of A has the direction of the current
element producing it. It is analogous to the potential function of electrostatics, which
is found in terms of an integral over charges and then differentiated in a certain way to
yield the electric field. The magnetic potential A is different, however, because it is a
at
which A is to be
The function A,
(4)
94
Chapter 2
Stationary Magnetic fields
vector, and does not have the simple physical significance of work done in moving
through the field that electrostatic potential has. Some physical pictures can be formed
but the student should not worry about these until more
has been developed through certain examples.
VECTOR POTENTIAL AND MAGNETIC FIELD OF
Here
we
show that the
magnetic
(2), in that order, is the
The
magnetic
current
vector
element,
as
same as
the function
IA CURRENT ELEMENT
density of a current element found using (3) and
integrand of (l), which expresses Ampére’s law.
exists throughout the Iregion surrounding the given
flux
the
potential A
Fig.
shown in
familiarity with
2.90. From
(3)
we
find
dZ
A=2A,=2”‘[4771‘
(5)
'
origin of coordinates is positioned at the current element. As noted earlier dA
is parallel to the current element producing it. It is most convenient to use spherical
coordinates in this example. From the figure we see that IA,
A2 cos 6 and A a
~AZ
the
The
coordinates
curl in spherical
sin 6.
front cover) reduces to
(from inside
since the
=
B
V x A
=
=
(I)
a
--
-~
Br
'
since
Ag
=
O and
6/34)
=
(IA 6)
I
(M
—r
-'
d
-'
B:V><A=¢(”Jz)3m2
477
(I)
dz
(6)
66
O, by symmetry. Substituting Ar and Ag using (5),
A
Note that d]' X R is
=
sin 6,
r
so
(7)
r
Z
FIG. 2.9a
Vector
potential
AI
in
region surrounding
a
find
6
(7) is equivalent to the integrand in (1).
A
we
current
element.
Vector
2.9
FIG. 2.9b
95
Magnetic Potential
Parallel-wire transmission line.
Example 2.9b
VECTOR POTENTIAL
Let
us
consider
a
AND
FIELD OF
A
PARALLEL-WIRE TRANSMISSION LINE
transmission line of infinite
parallel—wire
length carrying
current I in
conductor and its return in the other distance 2a away. The coordinate system is
set up as in Fig. 2.9b. Since the field quantities do not vary with 2, it is convenient to
one
calculate them in the
2
z
=
—L to
z
m
plane 2
=
O. The conductors will first be taken
L to avoid indeterminacies in the
direction, A by (3) will be in the
as
integrals.
direction also. The contribution
z
extending from
only in the
to A3 from both
Since current is
Wires IS
[L
A,
«-
—L
ll
23
477
The
integrals
#1 dz"
47r\/(x
UL V(x
O
may be
a)2
—-
y2
2’2
+
a)2
+
+
+
Z"?-
I dz”
I (12'
.
--
y2
+
_.I_# {ln[z'
+
L is allowed to
+
271'
—-
as
+
-L
2’2
+
+
0
+
2"2
evaluateds:
A,
Now,
,uJ dz"
f” 477\/(x (1)3 yz
f“ V(x a)2 y2 ]
__
ln[z"
+
V0:
Va
approach infinity,
—-
+
(2)2
(2)2
+
+
y2
y’l
+
+
2’2]
rang
the upper limits of the two terms cancel.
Hence
A‘
5
2
{£1
477
[(x
(x
+
-—
a)~
+
a)2
+
y]
y2
(8)
integrals of this text can be found in standard handbooks such as the CRC Hand»
Chemistry and Physics (any recent edition); M. R. Spiegel, Mathematical Handbook. Schaum’s Outline Series, McGraw-Hill, 7968; or M. Abromowitz and I. A. Stegun
(Eds), Handbook of Mathematical Functions, National Bureau of Standards Applied
Mathematics, Dover, New York, 7964. One of the most complete listings is l. S. Gradshteyn
and l. M. Ryzhik, Table of Integrals, Series, and Products (A. Jeffrey, Trans), Academic
Press, San Diego, CA, 7980.
Most
book of
96
If
Chapter
Stationary Magnetic Fields
2
(2) is then applied, using the expression for curl in rectangular coordinates,
1 BA,
I
pay
277'
ya
Hx:_—~_—
I
1 6A,
Hy
z
...._.__...:
Max
2.10
(x+a)+y
m
__
277‘
(x
ya
2—
(x—a)+y
(.3:
—
7a)
(x-a)~+y‘
DISTANT FIELD OF CURRENT LOOP:
find
(9)
2
+
__
7
we
(st-Fa)
2a)+y"
MAGNETIC
9
(10)
DIPOLE
I
magnetic field on the axis of a loop of current was derived in Ex. 2.3a. Here we
magnetic vector potential and field at locations not restricted to the axis
but distant from the loop. The arrangement to be analyzed is shown in Fig. 2.10. For
any point (r, 6, 915) at which A is to be found, some current elements I dl’ are oriented
such that they produce components of A in directions other than the qS direction. However, by the symmetry of the loop, equal and opposite amounts of such components
exist. As a result A is (f) directed and is independent of the value of qb at which it is to
be found. For convenience, we choose to calculate A at the point (r, 6, 0). The qidirected
The
will find the
contribution of
a
differential element of current is
dA¢
FIG. 2.10
__
[4.] dl’
cos
gb’
(1)
477R
Coordinates for calculation of magn
étic-dipole fields.
Distant Field of Current Loop:
2.10
where R is the distance from the element dl’ to
around the
a
(r, 6, 0). The total is found
as
the
integral
100p
A
where
97
Magnetic Dipole
(11'
pl
__
qS'
cos
__—
—
477
is the radius of the
[.LIa
z
"
()5
loop.
F77
—
R
477
The distance R
qb' dqb’
cos
R
o
can
be
(2)
——
expressed in
terms of the radius
‘
from the
origin
to
(r, 6, O)
as
R2
To get ra cos 4/ we note that
radius line to dl ’. Therefore
r2
=
ra cos
Substituting (4)
into (3) and
l/l
=
assuming
2m
—
is the
{/1
2‘ cos
a?“
+
projection
sin 6
ra
cos
>> a, we
1'
cos
it!
(3)
of
onto the extension of the
r
d)’
(4)
find
1/2
R
CZ
~r<1—-
2
.
-
sm
6cos
,.
¢>’>
01‘
R”1
Utilizing
this
in
expression
14¢
(2),
[-Lla
m
-
4771'
__
~r“‘(1
Ma
we
was
(cos
f
o
an"
45’
sin 6
2
'—
+
in
2 sin 6 cos2
r
#077112)
4717'
r
noted at the outset the result
by substituting (6)
45')
(5)
277
47rr
As
r
find
_
found
gsin6cos
+
applies
Eq. 2.9(2), are
Br
=
Be
2
Bq,
=
qb’)
27173
[1.177612
4771.3
cos
(6)
sin 6
2
to any value of
,quz?’
dqb’
qb.
The components of
B,
6
(7 )
6
(8)
,
Sin
0
(9)
The group of terms [71112 can be given a special significance by comparison of (7)—(9)
with the fields of an electric dipole, Eq. 1.10( 10). The identity of the functional form
of the fields has led to
defining
the
magnitude of the magnetic dipole
m
The
dipole
direction is
along
the 6
=
=
17m2
0 axis in
Fig.
moment as
(10)
2.10 for the direction of 1 shown.
98
Chapter
The vector
potential
can
Stationary Magnetic
2
be written in terms of the
=
in:
partial
derivatives in the
moment In as
V<—>
x
477
where the
magnetic dipole
l
*
A
Plelds
(11)
r
gradient operation
are
with respect to the
point
of
observation of A.
2.1 l
DIVERGENCE OF MAGNETIC
FLEUX
DENSITY
E
given by Eq. 2.9(2) (derived in Appendix 3), the magnetic flux density B can be
expressed as the curl of another vector A when the sources of B are currents. We have
shown in Ex. 2.8b that the divergence of the curl of any Evector is zero. Thus,
As
i
V
A
major
-
B
=
O
(1)
magnetic fields is now apparent. The magdivergence everywhere. That is, when the magnetic field is
no sources of magnetic flux which correspond to the electric
difference between electric and
netic field must have
zero
due to currents, there
are
of electric flux. Fields with
zero d1vergence such as these are con~
charges
ee
often
called
source
sequently
-fi fields.
Magnetic field concepts are often developed from an exact parallel with electric fields
by considering the concept of isolated magnetic poles as sources of magnetic flux,
corresponding to the charges of electrostatics. The result of zero divergence then follows
because such poles have so far been found in nature only as equal and opposite pairs.
Physicists continue to search for isolated magnetic poles;E if they are found, a magnetic
charge density pm will simply be added to the equations giving a finite V B.
as sources
=-
-
i
2.12
The differential
MAGNETIC POTENTIAL
DIFFERENTIAL EQUATION FOR VECTOR
equation
for
magnetic field
in terms of
ciurrent density
was
developed
in Sec. 2.7:
V X H
If the relation for B
as
the curl of vector
a
J
potential
V X V X A
:
sfibstituted,
A is
p.J
(1)
This may be considered a differential equation relating A to current density. It is more
common to write it in a different form utilizing the
of a vector function
Laplacian
definedin rectangular coordinates
as
the vector sum of the Laplacians of the three scalar
components:
VZA
=
It may then be verified that, for
iVZAx
+
rectangular
$472.43,
+
fiVZAZ
coordinates
(2)
Scalar
2.13
99
Magnetic Potential for Current-Free Regions
VxVxA=—V2A+V(V-A)
For other than
so
and
simply
With V
'
A
(3)
rectangular coordinate systems, separation in the form (2) cannot be done
(3) may be taken as the definition of V2 of a vector.
O, as shown in Appendix 3 for statics, (3) and (1) give
=
2A==—ILJ
This is
a
vector
@)
equivalent of the Poisson equation first met in Sec. 1.12.
equations which are exactly of the Poisson form.
It includes
three component scalar
in;.-.'./'/_(.~;;-,’.-;\I:313.-.=
Example
VECTOR POTENTIAL
Let
us
AND
a
A:
From
..;..(
..:,.;
,
:
:
I.
:'..-~:w;~:--:-,
§_.._.:';;._5
.--,-:
-,-.-,-..
§;- .I..-/s--..="r:-:'---.'..7. .'.';
FIELD OF UNIFORM CURRENT DENSITY FLOWING AXIALLY
appropriate form for the vector potential
circularly cylindrical system is
show that the
z—directed current in
=_~-.:/-.:-.;;
2.12
~% ()62
=
+
in
a
uniform flow of
yz)
(5)
(4) and (2),
1
fl.
From this
we see
conductor
carrying
=
—;
that
a
62A“
1
.,
V”Az
"'1;
(5) is the apprOpriate form for
current of constant
62A,
(1;; ELY—i)
+
=
density JO.
2
(6)
Jo
potential in a cylindrical
magnetic field found from
vector
The
(5) is
—J
1
A
A
H=~WXM=~Jmew3
2
m
,u.
In
cylindrical coordinates, this
is
A
J
r
H=¢§
which is the value of
2.I3
m
Eq. 2.4(5).
SCALAR MAGNETIC POTENTIAL FOR CURRENT-FREE REGIONS
In many problems concerned with the finding of magnetic fields, at least a part of the
region is current-free. The curl of the magnetic field vector H is then zero for such
current-free
regions [Eq. 27(2)]. Any
vector with zero curl may be
represented
as
the
'3 30
Chapter
gradient
pomts
of
a
Stationary Magnetic
2
scalar (see Ex. 2.6d). Thus the
magnetic
{fields
iield
as
can
be
expressed
for such
i
H
—vq>,,,
=
(1)
conventionally taken only to complete the analogy with elecpotential applies to bOth cufirent~carrying and current-free
but
it
more
convenient for the latter to use this scalar potential.
is
regions,
usually
Since the divergence of the magnetic flux density B is everywhere zero,
where the minus
is
sign
trostatic fields. The vector
=
v
Thus, for
a
-
homogeneous medium, (13,7,
ave”,
satisfies
VZCID,"
It will be observed from
=
0
=
(2)
Laplace’s equation
0
(3)
(1) that
2
(DmZ
~
(1)1711
=
#J.
H
.
1
Thus, if the path of integration encircles
For if 1 and 2
are
two values of
the
(Pm,
magnetic potential unique,
1,
a
current,
(Pm does
not have a
unique
value.
of integration
ithe point. To make the scalar
we must restrict attention to
iregions which do not entirely
point in space and the path
differing by I, will be assigned to
same
(4)
(in
encloses
a
current
encircle currents. Suitable regions are called “simply
connected” because any two
paths connecting a pair of points in the region form a loop which does not enclose any
exterior points. An example of a simply connected region between coaxial conductors
is shown in Fig. 2.13. The restriction to a simply connected region is not a serious
limitation
once
it is understood.
IN
-—._’
FIG. 2.13
Simply
connected
region between
doaxial cylinders.
Boundary Conditions
2.14
for Static
Magnetic
101
Fields
The
importance of the scalar potential for current-free regions is that it satisfies
Laplace’s equation, for which exist numerous methods of solution. The graphical and
numerical methods given in Chapter 1 for electrostatic fields are directly applicable, as
are the more powerful numerical methods, conformal transformations, and method of
separation of variables to be studied in Chapter 7.
BOUNDARY CONDITIONS FOR STATIC MAGNETIC FIELDS
2.14
The
boundary conditions
bilities
Consider
media
a
as
volume in the
shown in
small
arbitrarily
so
Fig.
interface between two
at an
be found in the
can
same
shape
way
of a
as was
regions
with different permea-
done for static electric fields in Sec. 1.14.
pillbox enclosing
the
boundary
2.14. The surfaces AS of the volume
that the normal flux
density 8,,
are
between the two
considered to be
does not vary across the surface.
so that there is negligible flux
Also, the thickness of the pillbox is vanishingly small
flowing through
the side wall. The net outward flux from the box is
13,,1 AS
where the
sense
of
B"
is
as
=
3,,2
as
or
B nl
=
B n-
,.
shown in the
figure.
magnetic fields may be found by integrating
enclosing the interface plane as shown in Fig. 2.14,
The relation between transverse
magnetic
field H
along
a
line
ng-dl =H,1 ill—H,2 A1=J5Al
where J5 is
a
surface current in amperes per meter width
FIG. 2.14
(1)
Magnetic
fields at
the
(2)
flowing in the direction shown.
boundary between
two
different media.
J 02
The
Chapter
:Flelds
Stationary Magnetic
2
lengths AZ of the sides are arbitrarily small so H, may be considered uniform.
legs of the integration path are effectively reduced to zero length. From (2)
The
other
Htl
There is
a
discontinuity
of the
.-
HtZ
tangential field
i
J3
=
at the
(3)
boundary
between two
regions
any surface current which may exist on the boundary. With direction
equal
information included, where 1‘] is the unit vector normal to the surface,
to
a
><(H1— H2)
J
=
(4)
Although the concept of a surface current is an idealization, it is useful when the
depth of current penetration into a conductor is small, as in the skin effect to be studied
later. In problems involving the scalar magnetic
potential, continuity of H where J
0IS ensured by taking (Pm to be continuous across the boundary. Where surface currents
exist, (4) leads to
«
i
f1 X
may be
as
(VCIJm2
J,
==
with the definition
by combining (3)
seen
VIPml)
-
(5)
ofSCIJm, Eq.
213(1).
S
T
MATERIALS WITH PERMANENT
2.15
MAGNETIZATION
1
Permanent magnets have
a
remnant value of magnetization
[defined in Eq. 23(3)]
when
Magnetic
Chapdiscussed
examples with permanent magnetization M0 and no
true current flow. There are two ways of analyzing such problems: through the scalar
magnetic potential and through the vector potential.
all
ter
applied
fields
13, but here
are
we
materials
removed.
consider
Now
a
scalar
using
potential
as
the definition of
more
detail in
some
use of Scalar Magnetic Potential
H from
in
are
Since current
densrty Jrs
zero, we may derive
in Sec. 2.13:
magnetization from Eq.
l
B
H
=
2.36),
—-
-
M
i
(2)
[Jo
If the
divergence
of
(2) is taken, with V B
-
V2®m
=
:
O
utilized,
we can
write
ifl
(3)
Mo
where
pm
In this formulation
we see
that
we
equivalent magnetic charge density
=
”#0V
have
a
in the
'
M
Poisson
(4)
equation for potential (Pm,
region proportional
to the
with
an
divergence
of
2J5
Materials with Permanent
103
Magnetization
magnetization. For a uniform magnetization, the divergence is zero and CD," satisfies
Laplace’s equation. At the boundaries of the magnet, however, integration of (3) would
show that there is an equivalent magnetic surface charge density pm given by
psm
:
“‘Ofi
‘
M
(5)
The arguments for this are similar to those for surface charge density p5 when there is
a discontinuity in D, as
explained in Sec. 1.14. We will illustrate this through an example
after
giving
formulation
a
use of Vector
in
Eq. 2.9(2)
using
the vector
potential.
Magnetic Potential If we write B
use the definition of magnetization,
as
curl of vector
potential
B==p,O(H+M)=-VXA
we can
A
as
and
take the curl of this
equation, using V
H
X
=
(6)
0 since J
z
0,
to
write
VXVXA=%%
0)
V
(8)
where
Jeq
So
=2
X
M
with
currents
in free space
Eq. 2.132(1), the problem is equivalent to one with internal
proportional to the curl of magnetization. Inside a region of
uniform
magnetization,
the curl is
by comparison
zero
and there
are no
internal currents. At the bound-
ary of the magnetic material, a surface integral of (8) over the area enclosed
used to get Eq. 214(2) and application of Stokes’s theorem give
by the path
%M-dl=JJeq'dS
S
In the
same
way
as
for
Eq. 2.14(4),
this
(JCq)S
gives
:
an
equivalent
M X fl
surface current
(9)
since M
0 outside the magnetic material. So in this formulation the magnet is re—
placed by a system of volume and surface currents from which magnetic field may be
found through use of the vector potential, or directly by using Ampere’s law. Example
2.15b illustrates this procedure.
=
Example
2.15a
UNIFORMLY MAGNETIZED SPHERE
Consider first
direction
as
in
uniform, there
sphere of magnetic material with uniform magnetization MO in the z
Fig. 2.150 using the method with scalar magnetic potential. Since M is
is no volume charge by (4), but if space surrounds the sphere, there is
a
1&4
Chapter
Stationary Magnetic
2
Fields
I
|
i
l
i
\
FIG. 2.15a
the
a
sphere
surface
Sphere
of radius
with uniform
a
magnetic charge density
at
Solutions of
pm
O,
=
r
a
Field lines
(H
or
B) outside
given by
I'LOMO
:
psm
=
magnetization 2M0
shown dashed
(3), in spherical coordinates with
6
COS
a
(10)
variation
corresponding
to
(10)
and
are
(I)
=
__Cr
cos
6
r
< a
a
(11)
’J
r>a
2c056
as can
on
be verified by substitution in the
the inside front
nuity
cover.
The surface
0 in spherical coordinates
expression for V233
magnetic charge given by (10) gives a disconti=
in derivative,
i
,LLOI:
{3(me
BCDmI
Br
Br
i
lad
__
—
,uollglo
cos
6
(12)
from which
f
M
C
Thus for
r <
a,
using (1)
in
spherical
M
=
399—
(13)
coordinates
,
H=~—39[i~cose—esm0]=
ll
1
‘—
N
(14)
Materials with Permanent
2.15
which is
a
uniform field within the
sphere.
A
30‘;
=
> a,
r
3
M
H
For
105
Magnetization
[2t
cos
a + 0 sin
a]
(15)
,.
which
are curves
(shown dashed in Fig. 2.150) similar
to those
outside
a
magnetic
dipole (Sec. 2.10).
Example 2.n5b
ROUND ROD WITH UNIFORM MAGNETIZATION
A circular
cylindrical
direction is shown in
from
a
(8) that there
bar magnet of length I having uniform magnetization in the axial
Fig. 2.15b. Using the second formulation given above, we see
are no
volume currents since V X M
equivalent
surface current at the discontinuous
boundary
JS
=
2'
=
=
0, but there is
a:
(BMO
(16)
problem is then identical to that of the solenoid of length l with current
per unit length given by (16) insofar as the calculation of A (and hence B) is concerned.
As noted in Ex. 2.3c, it is difficult to calculate field lines for an off~axis point, but B
lines will appear somewhat as shown dashed in Fig. 2.151). Lines of magnetic field H
will be of the same form outside the magnet, but will be of different form inside through
We
see
that this
the vector addition H
=
B / #0
M.
—-
/
//
\
/”F‘~‘\\\
\
/"“""‘*\
\
\
\
/
.'
y
;\
———W%——————————>z
"p"
u.“
’A"""+.~
/
\
\
\\1‘
/
\
Cylinder
of radius
shown dashed. (H lines
are
a
of the
/
/
/
\
FIG. 2.1513
"
j
and
same
length
l with
magnetization 2M0. Flux density
magnet.)
form outside the
lines B
106
Chapter
‘"
ENERGY OF
A
Pijelds
1.132=,3$c1'-‘I§’ ufiigiifiiiééigéfifi ’31:” gfréfiéfiriaéfié
f
“R
"i z” '533fi.§~r‘i§i._s
2.16
In
Stationary Magnetic
2
.,
tiiksffif’riizax 42321:}? “136's? a»?
STATIC MAGNETIC FIELD
the energy of a magnetic field, it would appear by analogy with Sec.
should consider the work done in bringing} a group of current elements
considering
1.22 that
we
point of view is correct in principle, but not only is it more
difficult
charges because of the vector nature of currents, but it
also requires consideration of time-varying effects as shojwn in references deriving the
relation from this point of View.6 We will consequently set down the result at this point,
waiting for further discussion until we derive a most important general energy rela~
tionship1n Chapter 3. The general relation for nonlinear materials, correSponding to
together from infinity.
This
to carry out than for
1 22(9) for electric
fields, is
=fH-dBdV
(1)
dUH is the energy added to the system when R} is changed by a differential
(possibly different amounts for different positionslwithin the volume). For linear
materials, H is pr0portional to B so (1) may be integrated over B to give
where
amount
I
UH=-fB-HdV=J’-&H2dV
v2;
(2)
21/
The
analogy
to
Eq. 1.222(6) is apparent, and here also Iwe interpret the energy of a
as actually stored in the fields
produced by those sources. The result
system of sources
is consistent with the inductive circuit energy term,
and will be utilizedin the following section
Exampte
2L1
*
when circuit concepts hold
-
2 tea
ENERGY STORAGE IN SLIPERCONDUCI ING SOLENOID
superconducting
It has been
coils be used to meet
proposed that energy stored in large
in
electric
demand.
coils
chosen
because their zero
peaks
power
Superconducting
iare
dc resistance allows very large currents to be carried with zero power loss (though, of
refrigeration power must be supplied). To be useful such a storage system must
capable of providing about 50 MW for 6 hours, thatiis, storing an energy of about
course,
be
106 MJ. Let
to
6
find the
us assume
required
J. A. Sfrah‘on,
the coil is
coil
properties
a
the field is uniform, and we wish
fieldi from Eq. 2.4(7) is Hz n] so
solenoid and that
and current. The
Electromagnetic Theory, pp.
7 78— 724,
=
McéGraw—Hill, New York, 794 I.
2.16
B:
=
p.121.
The energy from
Energy of a
a
Magnetic
Field
107
(2), for volume V, is
UH
For
Static
“it/«(fille
=
realistic current of 1000 A and flux
20-m
with 1.2 X
length
ising
coil
same
magnitude
is
shape
as
density of 15 T, a coil of 27—m diameter and
104 turns/m would give the required energy. The most prom-
actually
a
toroid but it would have dimensions and currents of the
calculated in this
example.
Example 2.16!)
ENERGY DISSIPATION
IN
HYSTERETIC
MATERIALS
see here how energy loss in hysteretic materials can be explained in terms of
their nonlinear B-H relations. A typical hysteretic relation is shown in Fig. 2.16. We
We will
will
assume an
traversal of the
BH. The energy required for one
isotropic material so that B H
a
H
from
loop by varying
large negative value to a large positive value
‘
=
again can be found from (1). The differential energy is shown as a shaded
hysteresis loop in Fig. 2.16. When the field is decreased, a portion of the
indicated
energy
by the part of the bar outside the loop is returned to the field. The
result of integrating around the loop is that total expended energy per unit volume is
equal to the area of the loop.
and back
bar
on
the
4‘.
’-
dB
FIG. 2.16
Hysteretic,
nonlinear B—H relation.
'3 08
Chapter 2
Stationary Magnetlé:
Fields
INDUCTANCE FROM ENERGY ST ORAGE; INTERNAL lNDUCTANCE
2.]7
was shown in Sec. 2.16 that the magnetic energy may be found by integrating an
fields. From a circuit point
energy density of é-qu throughout the volume of
current flow through the
of View, this is known to be é-le, where I is the
It
significant
instantaneous
inductance.
these two forms
Equating
i112
1
gives
1’5 H2 dV
=
(1)
2
V
given
in Sec. 2.5. It is
linkage method of calculating
for
convenient
especially
problems that would require
consideration of
partial linkages
if done
The form of
inductance
calculating
(1) is useful
-.
alternate to the flux
the
by
-.
-
are of this
g;.",'_7:_.’.‘-,‘71‘;'.'
“'7.7’721-7-1112)-??fifi"3."=':.".2-'.':',:-':7.5‘.'"?.:-‘}'.?X:Z.27)E
REV-22?}???#4321C1?_“-""..,’.F;.'_.'.’:':"".~.
.~—.....---..-. --,;:-.-.-.-
1443:3345 ....._.‘._._.L\.b‘.;... 'aLS'flfihilexi-Limix’in:
.-
.
of flux
method
internal inductance, defined in Sec. 2.5,
,_
':.-7.'"\‘~7‘:~'=7.'--"-":"2'37“;
fr,Tz{7/T‘E(T{:Tx'—2
“571‘7'”?/.'.1’,~_'_."'V"
...._.‘--,;.._(-_‘W_,_..
.:.-5_...-- 2.2:" -"*.-' '",gzi' '.'x-'."v-'</.-:-.1._-:_. """’-'-u§2.'i-A.'-'3';:
501;.“
as an
w)“
.
17")».
-/.
--
-
-.
,~
..
..
..
linkages.
Problems of
nature.
..
.
-
..
,.
,
.1
.
,_
"
.
’
;.
-.-.
-,.-.-:
.
'
'
'
'
--_‘-‘.I’-.'51%".‘3-3?$147714?131-37.?7.7-?":7‘:'-I’.°.‘.!E-’Q'-‘.?’.'-Tz1.7-","77:
L»
T's/1::
MIKE; 2/
1-
/-'
'
.
INTERNAL INDUCTANCE OF CONDUCTORS WTI—I UNIFORM CURRENT
DISTRIBUTION IN A COAXIAL TRANSMISSION LINE
an example of the use of the energy method of inductance calculation, we will find
the internal inductances for the two conductors of a coaxial transmission line under the
As
assumption
uniformly in the conductors. The result for
generally to any straight, round wire with a unimagnetic field in the inner conductor of Fig. 2.4!)
that the current is distributed
the inner conductor
applies
more
form current distribution. The
(EX. 2.4c) is
-
Ir
For
a
unit
length, utilizing (1),
2
.
2
”LI
Ir
2
2
fl!- (277(8)
m
o
L
The
magnetic field
2 77’ d'
'
'
=
57—:
in the outer conductor
=
.
Substituting (5)
in
( 1)
we
=
c2
_
277(c2
a4
'
—
4
b2)
(3)
(4)
(Prob. 2.4a)
--
477514
H/m
I
H O)
#12
z
2
is
x
(7' '>
_
'_
.
(5)
find
“i.
277
c4lnc/b
(62
b2)2
”"-
+
[72—322
4(C2
“
b2)
]
H/m
(6)
1 09
Problems
For
low
frequencies
enough
to assume
the total inductance per unit
uniform current distribution in the conductors,
the coaxial line is the sum of (4), (6), and
length for
Eq. 2.5(6).
PROBLEMS
2.23
Assuming
tor
that each electron
is acted
Eq. 2.2(1).
on
by
force
a
--
constituting
How is the force
2.2b The Hall effect
X
ev
the current in
a
differential
length
B, show that the total force is equal
to
of conduc-
that
given by
the electrons transferred to the structure of the wire?
on
charges in crossed fields within a semiconductor as
important properties of a semiconductor. Consider a
p—type material so that the charge carriers are holes of charge + e. Electric field E0 applied in the x direction causes a current [X
wda'Eo to flow. The magnetic field causes
a buildup of positive charge on the plate at y
O and an equal negative charge on the
top plate because of the velocity uHEO of the holes. The field produced by these charges
on the bottom and top plates EH is exactly of the magnitude to counteract the ev X
B0
force on the holes so that, in steady state, the flow is only in the x direction. Show how
the Hall mobility pH can be determined from measurement of Ix and VH.
shown in
Fig.
uses
motion of
P2.2b to
measure
=
=
FIG. P2.2b
2.2c Show the
following:
AX(B+C)=AXB+AXC
AX(BXC)=B(A‘C)—C(A'B)
A-(BXC)=B'(CXA)=C'(AXB)
2.2d*
when a particle of charge q and mass m is placed in crossed
fields. To demonstrate this, take a uniform electric field E0 in the
y direction and uniform magnetic flux density BO in the x direction. The charge starts at
O with zero velocity. Show that the trajectory can be
the coordinate origin at time t
written in the form (2
Razor)2 + (y R)2 R2, whereR EO/wOBO and too
Cycloidal
motion
electric and
can occur
magnetic
=
—
(180/272. Explain
the motion.
—
=
==
=
119
Chapter
2
Pijelds
Stationary Magnetic
loop of wire is formed by two semicircles,
joined by radial line segments at qb
magnetic field at the origin.
2.3a A
radius 17,
theinner of radius
=
O and
g!)1:
at
a
and the outer of
(Fig
P2 3a) Find the
FIG. P2.3a
2.3b Direct current I0 flows in a square loop of wire
magnetic field on the axis at a point z from the
2.3c
having sides of length
plane ofi the loop.
20. Find the
12 turns per meter by a cona solenoid of finite length L and radius 0
tinuous sheet of circumferential current Find the axial magnetic field at the center of
the solenoid and determine the length for which the fieldIS one-half that of the infinite
Represent
having
solenoid.
2.3d Show that the
an
,
magnetic
field
on
axis of
a
long solenordt at
the endsis half the value for
infinite solenoid.
2.3e An arrangement that can provide a region of relatively uniform fields consists of a pair
of parallel, coaxial loops; the uniform-field region is on the axis midway between the
magnetic field, expressed as a Taylor series exPansion along
point midway between the coils, wilf have zero first, second, and
third derivatives if the loop radii a are equal to the spacmg d of the loops ThisIS the
so-called Helmholtz configuration.
loops.
Show that the axial
the axis about the
field
for b < r < 0,
2.4a For the coaxial line of Fig. 2.4b, find the magnetic
current is distributed uniformly over the conductor cross section.
2.4b A certain kind of electron beam of circular
J0[l
2.4c
-
cross
section contains
a
assuming
current
that
density J
=
(1 /a)4]. Find Hq,,(I) inside the beam.
Express the magnetic field about a long line current in :rectangular coordinate components, taking the wire axis as the z axis, and evaluate 55 H (11 about a square path1n the
x—yplanefrom(- 1,—1) to (1, ——l) to (1, 1) to(" 1i 1) backto (“—1, —l). Also
evaluate the integral about the path from ( 1,1) to
1) to (1,2) to (— 1, 2) back to
( 1, 1). Comment
2.4d A
on
the two results.
(1:,
long thin wire carries a current II in the positive 2
along the axis of a cylindrical coordinate system as shown in Fig. P2.4d. A
rectangular loop of wire lies in
a plane containing the axis. The loop contains the regiOn O _<. 2 S b, R
a/2 _<. r S
R + a/ 2 and carries a current 12 which has the direction of I 1 on the side nearest the
direction
thin
-
l
‘3 '5 I!
Problems
FIG. P2.4d
axis. Find the vector force
on
each side of the
loop
and the
resulting
force
on
the entire
loop.
2.4e* Consider
for
a
stant.
a
round
straight wire carrying a uniform current density J throughout, except
cylindrical void parallel with the wire axis so that the cross section is con-
round
Call the radius of the wire 6, the radius of the hole b, and the distance of the
of the hole from the center of the wire 0. Take 19 < a < c and b < c
(I. Use
center
-—
to find the field H as a function of
radial line
center of the hole for all values of radius from the center of the wire.
superposition
2.4f A demonstration
position along
a
through
the
be given that a thin metal tube can be crushed by magnetic forces
through it. Take the radius of the tube to be 2 cm and the magnetic
field at which failure of the metal occurs as 9 Wb/rnz. (i) What is the maximum current
that could flow axially along the tube before it would be crushed by the magnetic forces
arising from this current? (ii) What is the force per unit area on the surface of the tubing under this condition?
by passing
2.4g
For
the
can
current
an infinitely long cylindrical hollow pipe of any
pipe, magnetic field within the hollow portion is
cross
section
zero.
Show
2.5 A coaxial transmission line with inner conductor of radius
radius b has
r
=
d
a
(with d
b),
a
current
along
and outer conductor of
=
a to
u, extending from r
and air from radius d to I). Find the external inductance per unit
coaxial
<
ferrite of
carrying
why.
cylindrical
permeability
length.
2.6a Find the curl of
2.6b
a
vector
field F
=
fuzz:
By using the rectangular coordinate forms
+
3732322
+
ixzyz.
show that
VX(l,l/F)==t/JVXF~FXV¢I
where F is any vector function and
2.6c Derive the
expression
:11 any
for curl in the
scalar function.
spherical
coordinate system.
Fig. 2.4b, express the magnetic field found in Ex. 2.4b and Prob.
rectangular coordinates and find the curl in the four regions, 2‘ < a, a < r < b,
2.7 For the coaxial line of
2.4a in
b <
r <
c,
r > 6.
Show that V X
theorem.
Comment
Vi/I
E
O
on
the results.
by integrating
over an
arbitrary
surface and
applying
Stokes's
'5 12
Chapter
2.9a Check the results
individual wires,
2311* A square
to
(a, a)
loop
Fields
Stationary Magnetic
2
Eqs. 2 .9(9) and (10) by adding vectorially
using the result of Ex. 2.4a.
the
field from the
magnetic
of thin uvire lies1n the x—y plane extending from ( a -a) to (a, -a)
to (- a, ——a) and carries current Iin that sense of circulation.
( a, a) back
to
Find A and
H_ for
any
point (x,
z)
y,
.-
2.9c* A circular loop of thin wire carries current I. Find A for a point distance: from the
plane of the loop, and radius 1 from the axis, for r /2 << 1. Use this to find the expression for magnetic field on the axis.
2.9d Show that the line integral of vector potential A about a closed
magnetic flux enclosed,
pathis equal
to the
,
§
this
Apply
to
A
-
dl
J
=
B
-
d8
5
long solenoid
find the form of A inside the
of Ex. 2.4d.
infinite single-wire line of current, show that A: as calculated in Ex. 2.9b is in-L < z < L
Then Show that if vector potential is calculated for a finite length-and B calculated from this before letting L approach infinity, the correct value of BIS
2.91% For
an
finite
obtained
2.9f As
an
1
exercise in
having
length.
a
width b
using
carrying
a
{very long thin conducting sheet
current I in the direction of its
direct}
the
with the axis
its
consider
the vector
potential,
uniformly distributed
Show that if the sheet18 assumed to lie1n
xii—«2
centerline, the magnetic field about the strip will be
b
1
Hx
H
=
27Tb
=
y
2.10 Show that the torque
2.1221 Show that V2A
2.121) Use the
=
(12111"1 —/—2——3
——
+.
plane
1111‘1
+
(w2+n2+f
(b/2
small
loop
O for the vector
.,
.,
x)"
+ y-
of current
rectangular coordinate forms
can
around
potential
to prove
Eq.
—;
—-/——i)
y
471'b
-
along
z
given by
b
f
y
1
-—-ln
on a
a
1
1
be; expressed
a
as 7
=
m
X
B.
pair of currents, Eq. 2.9(8).
2.12:6).
2.12c A certain current density is said to produce within itself a vector potential having the
“r
form A
201'
in circular cylindrical coordinates where C13 a constant. Find the
=
divergence
of A, current
density,
and
magnetic field, assuming
2.12:1 We
saw
in Ex. 2.7 that
magnetic
superconductor decays from the surface
m=mrw_
field in
z is parallel to the plane of the
corresponding vector potential A,
where
the
the medium to be free
1
space
a
as
Eperpendicular to the surface. Find
surface and x is
and from it the current
density comparing
with
the result of Ex. 2.7.
2.133 Show whether either of the following
tial, and give the potential function if
F
=2
vector
fields
can
i3y22
+
y6iyz
be obtained from
1
applicable:
+
@23ny
F=fi3y+92x+24§
a
scalar poten~
ll3
Problems
2.13b Find the form of scalar
in
Fig. 2.13,
magnetic potential for the region between conductors as shown
region 0 .<.. d) < 277-, similarly for the region outside the
defined for the
outer conductor.
Current I flows in the inner conductor and the return current in the
outer one.
2.14* Consider the boundary between free space and a plane superconductor with nearby parallel line current I at x
d. It is the nature of a superconductor that when placed in a
weak magnetic field, currents flow in such a way as to eliminate flux inside the super=
conductor
so
that
B,,
at the
surface is zero,
ductor. Show that fields in the
superconductor
netic field
2.15 For the
at .r
with
=
problem
an
image
as
is the
free-space region x
current at .t
=
—d
tangential H
>
O
can
inside the superconby replacing the
be found
carrying current
density J5.
-I. Find the mag
O + and from this the surface current
of
Fig. 2.151), what magnetic charge distribution would be obtained for
equivalent magnetic charges? How would this be modified
inhomogeneous as defined below?
the formulation in terms of
if
magnetization is
M
2.1621 Show that
Eq. 216(1)
=
2M0(1
leads to (2) for linear,
+
k2)
isotropic
materials.
2.16b Assume that the material having the B-—H relation shown in Fig. 2.16 saturates at B
1000 G and estimate graphically the energy per unit volume for one complete traversal
=-
of the
hysteresis 100p.
2.17a Find the external inductance per unit
ergy considerations.
length
for the arrangement of Prob. 2.5 from
en-
2.17b Find the internal inductance per unit length for the parallel-plane transmission line of
Fig. 2.5c if current is assumed of uniform density in each of the conductors.
l
3.1
The laws of static electric and
It has been noted that these
INTRODUCTION
magnetic
are
been studied in Chapters 1 and 2.
predictingéeffects for many time-varying
fields have
useful in
problems, but there are important dynamic effects not described by the static relations,
so other time~varying problems require a more
complete formulation. One important
dynamic effect is the generation of electric fields by time-varying magnetic fields as
expressed through Faraday’s law. A second is the complementary effect whereby time—
varying electric fields produce magnetic fields. This latter effect is expressed through
the concept of displacement current, introduced by Maxwell.
Faraday’s law is well known to us through its importance in transformers, motors,
generators, induction heaters, and similar devices. The effect can be sirnply demonstrated by moving a coil of wire through the field of a strong permanent magnet and
noting
the trace
on an
oscilloscope
connected to the coil
(Fig. 3.1a).
With
readily
available magnets and practical numbers of turns in the coil, movement by hand will
generate millivolts, and such voltages are readily observed on the oscilloscope. An
alternative demonstration utilizes
A switch to turn
on
an
electromagnet with
and off the current in the
its flux
electromfagnet
threading a
buildup
causes
fixed coil.
and
decay
field and generates the voltage to be observed.
The above-described demonstrations and useful devices utilize induced effects in
of the
magnetic
conductors. An
interesting example showing that changing magnetic fields
induce elec-
tric fields in space is that of the betatron. This useful pfarticle accelerator, illustrated in
Fig. 3.1b, accelerates electrons or other charged particles by means of a circumferential
electric field induced
by
a
changing magnetic
field
between poles
tromagnet. The charges are in an evacuated chamber,
law applies in space as well as along conductors.
N and S of
clearly illustrating
that
an
elec-
Faraday’s
dynamic effect, referred to above as a; displacement current effect, is
us through the concept of a capacitance current. We may,
probably
this
think
of
however,
only as current in the conductors to capacitor terminals, supplying
the time rate of change of charge on capacitor plates. 5We shall see that the changing
electric flux in the dielectric between plates contributes to magnetic fields, just as does
conduction current, and acts to complete the current path Displacement currents also
The second
best known to
”4
3.1
'5 '5 5
Introduction
‘9)"‘
eee
FIG. 3.1a
Experimental arrangement to demonstrate the induced voltage predicted by Faraday’s
be moved with sufficient speed by hand to display the induced voltage on a simple
oscilloscope.
law. Coil
can
exist in the
of
moving charges, and so are important in vacuum tubes or solidexample, time-varying effects in the Schottky barrier of
Ex. 1.4a or the pn junction of Sec. 1.16 produce displacement currents in the respective
depletion regions. Effects of these displacement currents must be understood in the
analysis of devices using such junctions.
There is a far-reaching consequence of the fact that changing magnetic flux density
produces a change in electric field and vice versa: it leads to propagation of electro-
vicinity
state electron devices. For
FIG. 3.1b
of
an
Schematic illustration of
electric field induced
by
a
a
betatron, which is used
changing magnetic
field.
to
accelerate electrons
by
means
i 16
Chapter
magnetic
waves.
In
general,
wave
Maxwell’s
3
Equaticjms
when there are two forms of
oneileads to a change of the other.
variation in air (potential energy) in
air molecules (kin:etic energy) that varies both in
phenomena
result
em
ergy, and the presence of a time rate of change of
In a sound wave, for example, an initial pressure
one
location
causes a
motion of the
time and in space. This builds up excess pressure at another position, and the effect
continues. Similarly, changing the magnetic field (or flux density) at one position gen-
change of electric field in both time and space; by Faraday’s law. The subseof electric field produces a change of magnetic field through the displacechange
quent
erates a
interchanges between electric and
magnetic types as the wave progresses
Electromagnetic waves exist in nature in the radiation that takes place when atoms
or molecules change from one energy state to a lower
gone, with frequencies from the
microwave through visible into x-ray regions of the spectrum. (Still lower frequencies
are generated by lightning and other natural
fluctuations.) These natural radiations are
utilized in astronomy and radio astronomy. Telecommunications, navigational guid~
ance, radar, and power transmission depend upon our ability to generate, guide, store,
radiate, receive, and detect electromagnetic waves. This involves many kinds of structures whose properties the designer must be able to predict. The complete set of laws
for time-varying electromagnetic phenomenais known as Maxwell’s equations andIS
central to such predictions.
ment current, and so on In energy terms the energy
I
large-Seek and flatter-entrant
3.2
VOLTAGES INDUCED
BY
Forms of Maxweil’s
Equations
CHANGING: MAGNETIC FIELDS
experimentally that a voltage is iirduced in a conducting circuit
linking that circuit is altered? The voltage is proportional to
magnetic
the time rate of change of magnetic flux linking the circuit. For a circuit of n turns, the
I
induced voltage V can be written
Faraday
discovered
when the
field
V
—-
)1
dt
(1)
where 111m is the magnetic flux linking each turn of the coil. This equation may be used
directly to find the voltage induced in the secondary coil of a transformer, for example,
or to find the voltage induced in a single circuit
becjause of a time-varying current
interacting with the self-inductance of that circuit. For an electric machine, such as a
generator or a motor, the change in flux linkages to be used in (1) may be found from
the movement of the coil of the machine through a spatially variable magnetic field.
3.2
'5 '57
Voltages Induced by Changing Magnetic Fields
Faraday’s experiments
included both
moving systems may be
following section.
approached
stationary
and
moving systems.
The
in several ways and will be discussed
question
more
of
in the
One very important generalization of (1) is to a path in space or other nonconducting
medium. Such an extension is plausible since the resistance of the path does not appear
in the
equation.
Nevertheless the extension should have
it does. Much of the
experimental support
comes
experimental
from the
wave
verification and
behavior to be studied
in the remainder of the book. As described in Sec. 3.1, the betatron1 accelerates
charged
by means of an electric field induced by a changing magnetic
field, as predicted by Faraday’s law. (See Prob. 3.2c.)
Before defining Faraday’s law more precisely, we should be clear about several
definitions. By voltage between two points along a specified path, we mean the negative
line integral of electric field between the points along that path. For static fields, we
saw that the line integral is independent of the path and equal to the potential dzfierezzce
between the two points, but this is not true when there are contributions from Faraday’s
law. When there is a contribution from changing magnetic flux, the voltage about a
closed path is frequently called the electromotivefarce (emf) of that path.
particles
in
a vacuum
emf
E
voltage
about closed
equal, by Faraday’s law, to the time
path. For a circuit which is not moving,
It is
rate
of
(2)
change of magnetic flux through
the
§Etdl=~%:~§at
where the flux
is found
rpm
by evaluating
~§
dl
path
61‘
7-:
E
-
Bids
(3)
S
the normal component of flux
density
B
over
any surface which has the desired path as its boundary, as indicated by the last term in
(3). The negative sign is introduced in the law to agree with the sense relations revealed
by experiment, using the usual right-hand convention in the integrals of (3). Thus, as
in Fig. 3.2a, if the rate of change of flux is positive in the directions shown by the
vertical arrow, the line integral will be positive in the direction shown—opposite to the
conventional right~hand positive direction. If there are several turns, the line integral
of (3) is taken about all of them, and if flux through each is the same, we have the form
first stated in (1).
l
D. W. Kerst and R. Serber,
Phys.
Rev. 60, 53 ( 7947).
Sense of positive
$23!;-
Positive sense
for
Sense
of
.
right-hand
FIG. 3.2a
positive
§E
convention
Sense relations for
Faraday’s
law.
.
d1
'5 '53
Chapter 5
Maxwell’s
Equations
cah
To transform (3) to differential equation form, we
apply Stokes’s theorem (Sec.
the
and
time
of
move
to
the
left
side
(3)
2.8)
differentiation inside the integral:
[(VxEydsz—Ja—Bnds
561‘}
(4)
S
For this
equation
to be valid for an
arbitrary surface,
that
integrands must be equal
the
i
so
63
V
X
E
=
—-—
(5 )
1
at
Faraday’s
law
(4) of
course
reduces to the static
case
when time
derivatives
are zero
in Sec. 1.7, the line
integral of electric field about a closed path is then
zero. For the time-varying field it is not in general zero,
:showing that work can be done
in taking a charge about a closed path in such a fielfd. This work comes from the
changing stored energy of the magnetic field.
and,
as we saw
Example
5.2
AIR BREAKDOWN FROM INDUCED
ELECTROMOTIVE
FORCE
possibility of an ionizing breakdown in air because of electric fields gen~
changing magnetic fields. An axially symmetric electromagnet (Fig. 3.22)) has
Consider the
erated by
FIG. 3.2b
Electromagnet with time-varying
current
producing
a
time~varying magnetic
field.
3.3
radius 020
it.
beyond
linearly
m
and has
Suppose
essentially
Law for
short
as
we can
a
uniform field up to this radius and negligible field
magnetic field from zero to 10 T (tesla)
time
write
27TI‘E
l
7 as possible without such breakdown.
Faraday’s law for a loop of radius r as
A
==
771'2
BB-
strength
as
10'5 V/ m,
3 X
=
7T)
2-295
at
3.3
(6)
7
m.
If
we
take breakdown
then
I‘Bmax
0.2 X 10
2|E¢I
2><3><106
l
2:————:——-—--————--:—
T
Because of
B
—-—i
Electric field is thus maximum at the outer radius of 0.20
of air
'5 '59
Moving System
a
that it is desired to raise
with time in
the axial symmetry
Faraday’s
FARADAY'S LAW FOR
A
3
“S
7
U
MOVING SYSTEM
moving system, one must find the change of magnetic
through the field. A simple and classical example
is that of an elemental ac generator as pictured in Fig. 3.3a. This indicates a single
rectangular loop rotating at constant angular frequency Q in the uniform field BO between the two pole pieces. When the plane of the loop is at angle 4‘) with respect to the
horizontal axis, the magnetic flux passing through it is
For the
flux
use
of
threading
Eq. 3.2(1)
for
circuit
it
a
as
a
moves
t/lm
But
angle qb changes
voltage
is the rate of
FIG. 3.3a
sin
2Boal
(1)
qb
with time and may be written Qt. Thus
dim
And if
=
change
==
280a]
sin Qt
of this flux
Elemental generator with
(2)
(neglecting signs),
rotating loop
between permanent magnets.
129
Chapter
Maxwell’s
3
Equafléns
d_¢m =ZQBOal
(it
V==
fit
cos
(3)
this
basic generator. Now let us
voltage produced by
the
'same
of
view
to
answer. A point of view
point
get
slightly
used effectively by Faraday is that the electric field of the’ moving conductor is generated
Thus
we see
introduce
the sinusoidal
ac
different
a
by its motion through, and
hence‘ cutting”
of, the lines of force. Faraday gave much
to the flux tubes and lines of force. This
physical sigificance
developed rigorously by writing the time derivative
total derivative instead of a partial derivative:
on
point
of view
the right side of Eq.
can
3. 2(3)
iE dl——-——deS
For
a
(4)
moving in space with velocity v, this may be transformed by
developed by Helmholtz,2 which, with V B— O, is
closed path
transformation
be
as a
a vector
r'
63
3gE-dl=~fs|:E—VX(VXB)]
(5)
1
The first term
on
the
right18
the
one we
have
seen
before
The second term
gives
an
of Stokes’ s theorem, may be written as the line
by
X B about the closed path. The result may be interpreted as a motional
added contribution to emf and,
integral of v
electric field given
each
at
use
point
of the circuit
path by
E=v><B
m
In the
example
of
Fig. 3.3a,
the motional field in the upper conductor,
Em:
That in the lower conductor is the
side elements since
Thus the line
3g
E
v
integral
-
dl
==
X
B
(6)
=
(cz.Q)B0
negative
cos
of this.
contribution
gives
loop yields
a
by (6),
1;!)
is
(7)
There is no motional field along the
normal
to the
wires for these sides.
about the
laQBO
cos
01‘
~—
(-lcz().BO
cos
which18 identical to (3).
The differential form of
(pa)
=
ZlaflBO
cos
Dr
(8)
-'
Faraday’ 3 law, Eq. 3. 2(5), may be transformed to a set of
moving coordinates with the same result. This may be dqne by a Galilean transformation
for low velocities and by a Lorentz transformation for relativistic velocities.2 Although
relativity is beyond the scope of this text, it is important to know that Maxwell’s equa~
tions are consistent with the theory of relativity, although Einstein developed that theory
later. Their invariance to Lorentz transformations, in fact, had much to do with the
development of the theory of special relativity.
{
2
See, for example, C. T. Tai, Proc. IEEE 60, 936 ( 7972).
Conservation of Charge and the Concept of Displacement Current
3.4
l
L—
FIG. 33!)
/////
I
'52}
___,v
Hz: Cx
Rectangular loop
of wire
field which varies with distance.
moving through magnetic
..................
Example
3.5
RECTANGULAR LOOP MOVING THROUGH INHOMOGENEOUS FIELD
If
loop of wire is moved through a region of static magnetic field which is a function
position, the flux threading the 100p changes as the loop moves and an emf is generated. Consider the rectangular loop of wire (Fig. 3.3b) translated in the x direction
with velocity 0 through a z-directed static magnetic field which varies linearly with x,
Cx. If the left—hand edge is at x
O at t
vi at time t, and the
0, it is at x
B:
magnetic flux threading the loop is
a
of
=
=
=
Ut+a
JB~dS=bf
S
‘2
2
Ut+a
dexzbci
2
m
The induced emf is then the time rate of
change
=
U!
3:2 (th
+
(9)
a)
of this flux,
d
—§E-dl=-JB'dS=ban
dt
This result
field
v
and
on
can
(10)
5
by finding the motional field in the four sides using (6). The
along the top and bottom. On the left it is -vi
vC(x + a). Thus the integral is
be checked
X B is normal to the wires
the
right side,
—
~§
agreeing
with the result
3.4
Faraday’s
law is but
E- (11
=
va(x
+
(1)
~
vax
=
[3an
(11)
(10).
CONSERVATION OF CHARGE AND THE CONCEPT
OF DISPLACEMENT CURRENT
one
of the fundamental laws for
changing
for the moment that certain of the laws derived for static fields in
fields. Let
Chapters
us assume
1 and 2
can
122
Chapter
Maxwell’s
5
Equaticfms
simply to time—varying fields. We will write the divergence of electric and
magnetic fields in exactly the same form as in statics, with the understanding that all
field and source quantities are functions of time as well as of space. For the curl of
electric field we take the result of Faraday’s law, Eq. 3i2(5). For the curl of magnetic
field, we take for the time being the form from statics, Eq. 2.7(2).
be extended
Vfl=p
m
g
V-BzO
(2)
i
an
VXE——‘é;
(3)
VxH=J
(4)
An elimination among these equations can be made to
charge and current. We would expect this to show that
give
an
equation relating
however p varies with space or
time, total charge is conserved. If current flows out of any volume, the amount of charge
inside must decrease, and if current flows in, charge
increases. Considering a
of current per unit time and
smaller and smaller volume, in the limit the outward
iiiside
flow
per unit volume (which is recognized as the divergence of current density) must give
the negative of the time rate of change of charge per unit volume at that point:
._____a_p
VJ~
at
l
If, however,
we
take the
divergence
of J from
(5)
(4),
vu=vwxm30
which does not agree with the
similar to this, recognized that
continuity argument and (5). Maxwell, by reasoning
(4), borrowed from statics, is not complete for time—
varying fields.
added term
aD/at:
V
J
He
postulated
an
61)
Continuity is now satisfied,
substituting from (1):
as
X
H
=
+
at
may be shown
by taking
a
V
J“
azw
(6)
——
D)"
the
divergence
of
(6) and
8p
:at
(6) contributes to the curl of (magnetic field in the same way
density (motion of charges in conductors) or convection
current density (motion of charges in space). Because
it arises from the displacement
vector D, it has been named the displacement
currenti term. There is an actual timeThe term added to form
as an
actual conduction current
varying displacement
of bound
charges
in
a
material
ment current can be nonzero even in a vacuum. Thus
dielectric,
(6)
but note that
could be written
displace-
3.5
where
Physical Pictures
of
‘23
Displacement Current
conduction
or convection current density in amperes per square meter and
aD/at amperes per square meter.
displacement current density
The displacement current term is important within the dielectric of a capacitor when~
ever the capacitive voltage changes with time. It also always plays a role when moving
charges induce currents in nearby electrodes. Both of these phenomena will be explored
in the following section. Displacement current is negligible for many other lowfrequency problems. For example, it is negligible in comparison with conduction currents in good conductors up to optical frequencies. (This point will be explored more
in Sec. 3.16.) But displacement current becomes important in more and more situations
as the frequency of time-varying phenomena is increased. It is essential, along with the
Faraday law terms for electric field, to the understanding of all electromagnetic wave
phenomena.
Jd
JC
2
=
=
3.5
PHYSICAL PICTURES OF DISPLACEMENT CURRENT
The
displacement current term enables one to explain certain things that would have
proved inconsistent had only conduction or convection current been included in the
magnetic field laws. Consider, for example, the circuit including the ac generator and
the capacitor of Fig. 3.5a. Suppose that it is required to evaluate the line integral of
magnetic field around the loop a~b—c-—d—a. The law from statics states that the result
obtained should be the current enclosed, that is, the current through any surface of
which the loop is a boundary. If we take as the arbitrary surface through which current
is to be evaluated one which cuts the wire A, as does S I, a finite value is clearly obtained
for the line integral. But suppose that the surface selected is one which does not cut the
wire, but instead passes between the plates of the capacitor, as does 52. If conduction
current alone were included, the computation would have indicated no current passing
through this surface and the result would be zero. The path around which the integral
is evaluated is the same in each case, and it would be quite annoying to possess two
different results. It is the displacement current term which appears at this point to
(b)
(a)
FIG. 3.5
Illustrations of how
capacitor; (b)
near a
displacement
moving charge.
current
completes
the circuit:
(a) in
a
circuit with
.524
Chapter
3
Maxwell’s
preserve the continuity of current between the
answer in either case
Equattoris
of the
plates
capacitor, giving
the
same
*
continuity is preserved, consider an ideal parallel-plate capacitor
capacitance C, spacing d, area of each plate A, and applied voltage V0 sin cot. From
circuit theory the charging current is
E
To show how this
of
IC
The field inside the
capacitor
=
has
C
a
=
E;
wCVO
cos
(1)
wt
i
E
magnitude
=
V/ d
so
the
displacement
current
density
IS
‘
Ja
Total
by
=
88E
83;
=
Vo
—
we
cos
d
displacement current flowing between the plates
density of displacement current:
w;
is
(2 )
t
the area of the plate multiplied
the
A
1d
=
A1,
=
m(%->VO
(3)
cos wt
The factor in parentheses is recognized as the electrostatic capacitance for the ideal
parallel-plate capacitor, so (1) and (3) are equal. This value for total displacement
current flowing between the capacitor plates is then exactly the same as the value of
charging current flowing in the leads, calculated by the usual circuit methods above, so
the displacement current does act to complete the circuit, and the same result would be
obtained by the use of either S1 or $2 of Fig. 3 50, as required
Inclusion of displacement current is necessary for a valid discussion of another examplem which a charge region q (Fig 3.511) moves with velocity v. If the line integral
of magnetic field18 to be evaluated about some loop A at a given instant, it should be
possible to set it equal to the current flow for that instant through any surface of which
A is a boundary. If the displacement current term were ignored, we could use any one
of the infinite number of possible surfaces, as S 1, havirig no charge passing through,
and obtain the result zero. If one of the surfaces is selected, as 52, through which charge
is passing at that instant, however, there is a contribution from convection current and
a nonzero result. The apparent inconsistency is resolved
when one notes that the electric
field arising from the moving charge must vary with tithe, and thus will actually give
rise to a displacement current term through both of the surfaces S1 and $2. The sum of
displacement and convection currents for the two surfaces is the same at the given
instant.
..
Z'l;-'_'.i;';';gang-(1.5;.
Example
CLIRRENTS INDUCED
BY A
A”
J.
-'--'.:.-:
,-.:::-..
--:-:
.,
K
5:»......
.:-*_' mi;......
3. 5
SLAB OF CHARGE MOVING
IN A
PLANAR DIODE
planar vacuum diode, as sketched in Fig. 3.5c, the cathode has been pulsed to
produce a slab of charge moving from cathode to anode} The density of charge is taken
In
a
3.5
Physlcal
Plctures of
125
Dlsplacement Current
V0
Slab of charge
FIG. 3.5c
as a
uniform pot Width is
w
moving between parallel platest
and at time t the left-hand
velocity vi We note that the
linearly for x' < x < x' + w,
x’ moving with
edge is at x
electric field EII is independent of x for x < x’, varies
and is again independent of x for x > x’ + w, Its integral
=
is
p
pWI
,
evo=Efld+fw<d—x>—g—E
Then the electric fields for the three
x’ +
w
<
x
< d are,
regions 0
(Prob.
3.5b),
respectively
@LW
__fi
E-“_
<
.r
x’, x’
< x <
1-
d+s
d+w2d1
V
,w
p
<
V0
Pow
—
=v—
d+sd
BE“
8E2
_x
w
=
fi
6x’
=
__
p”
powu
_,
d
d at
gives velocity
1).
(7 )
w
:
at
a
(5)
6
0
2
at
e
and
W
A"+—
a
g.
w,
w
function of time and differentiation with respect to time
Thus displacement current density for the three regions is
But x’ is
+
(4)
BIZ:—Z“+;°[x(gei>+w(E—t)+x]
5‘3
x’
2
pov<d 1)
_
Mai,
-
:
M
(8)
(9)
326
Chapter
To the
displacement
Maxwell’s
5
Equatlolns
density of the region within the charge, given by (8), we
density pov so that the sum of convection and displacement
each region, and this will also be the current per unit area
current
add the convection current
currents is the same for
induced in the
plane
electrodes.
i
MAXWELL'S EQUATIONS IN DIFFERENTIAL EQUATION FORM
3.6
I,
Rewriting the
we
group of equations of Sec. 3.4 with the
displacement current term added,
have
V'DZP
(1)
V-B=0
(2)
613
V
x
E
(3)
=——
at
6D
v x H
This set of
equations, together
z
with certain
J
+—
a:
auxiliary
I
(4)
1
reIlations
and
definitions, is the
electricity
magnetism, governing all electromagfrom
zero
frequencies
phenomena
through the highest-frequency
radio waves (and many phenomena at light frequencies) and in the range of sizes above
atomic size. The equations were first written (not in the above notation) by Maxwell in
1863 and are known as Maxwell’s equations. The material in the sections preceding
this should not be considered a derivation of the laws, for they cannot in any real sense
basic set of
equations
of classical
and
in the range of
netic
be derived from less fundamental laws. Their ultimate
experimental laws, in that they have predicted
justification
correctly,
comes,
and continue to
as
with all
predict,
all
wide range of physical experience.
is a set of differential equations, relating the time and
electromagnetic phenomena
The foregoing set of equations
space rates of change of the various
over a
use
field
of these will be demonstrated in the
quantities at a Ipoint in space and time. The
following chapters Equivalent large-scale
equations will be given in the following section.
The major definitions and auxiliary relations
information
are as
1. For ce Law
and
point of View, merely the definition of the
charge (1 moving with velocity v through an
a magnetic field of flux density B, the forceIS
This
is, from
fields. For
one
a
Definition of Conduction
Current
J
where
o-
is
conductivity
the
(Ohm’s Law)
=
as
in Siemens/meter.
A/mz
For
electric
electric
(5)
f=q[E+va]1£\I
2.
complete
I
follows:
magnetic
field E and
.
that must be added to
a
conductor,
(6)
3.6
3.
Definition of Convection Current
Vp, the current density is
J
4.
127
Maxwell’s Equations in Differential Equation Form
For
=
a
charge density
=
SE
moving
with
velocity
A/m2
pr
(7)
Definition of Permittivity (Dielectric Constant) The
related to the electric field intensity E by the relation
D
p
=
electric flux
density
D is
8r80 E
(8)
12
where 80 is the permittivity of space R“-8.854 X 10““
F/m and a, characterizes
the effect of the atomic and molecular dipoles in the material.
As with static fields
(Sec. 1.3) e, or 8,, is, in the most general case, anisotropic
function of space, time, and the strength of the applied field. But for many
materials it is a scalar constant, and unless specifically noted otherwise, the text
and
a
will be concerned with
rials for which
5.
8
is
a
homogeneous, isotropic, linear,
and time-invariant mate-
scalar constant.
Definition of Permeability
intensity H by
The
magnetic flux density B
is related to the
magnetic
where #0 is the permeability of space
4n X 10‘7 H/m and Mr measures the
effect of the magnetic dipole moments of the atoms constituting the medium (Sec.
2
2.3). In general it and It, are anisotropic and functions of space, time, and magnetic field strength, but unless otherwise noted they will be considered scalar
constants, representing homogeneous, isotropic, linear, and time-invariant
materials.
Example
3.6
NONARBITRARlNESS OF FORMS WHICH SATISFY MAXWELL'S EQUATlONS
Much of the work for the remainder of the text will be in
finding forms that are solutions
interrelationship among electric and magnetic field comequations.
defined
Maxwell’s
ponents
by
equations means that we cannot select arbitrary functions
for any one component. To illustrate the point, let us consider a capacitor formed by
of Maxwell’s
concentric
The
spherical conductors with an ideal dielectric between. As in Ex. 1.4c, Gauss’s
give the electrostatic solution for the dielectric region as
law and symmetry
Dir/313:1“
Q
"J
4777‘“
where Q is the charge on the
might expect the solution
inner
sphere.
Ext
For
a
sinusoidally time—varying charge,
Oasinmt
(10)
one
(11)
123
Chapter
in
EquationsE
Maxwell's
3
coordinates shows that it: is zero,
A check of
V-(eE)
charge-free
dielectric. But consider the Maxwell
spherical
as
expected
for the
equationi
6H
V X E:
electric field of the form
—
“‘
at
zero, so H can
(11) is
The curl of
an
independent
of time. But then the other curl
( 12 )
—
only
be
a
function
equation
6E
V X H“ J +
‘
cannot be satisfied since
but V X H is
are
zero
(13)
—-
(it
for the ideal dielectric,
«SE/at by (11)
is
time-varying
of time. Thus (11), though it- may be a useful quasistatic
is not a true solution of Maxwell’s equations. Proper solutionsin spher-
independent
approximation,
ical form
J is
s
discussedin
Chapter
10.
i
I
5.
3.7
MAXWELL’S EQUATIONS
It is also convenient to have the information of
integral
form
applicable
to
overall
3.2)
for
Eqs. 3.6(1)—3.6(4)
Maxwellis equations
in
large-scale
or
paths
discussion of Faraday’s law (Sec.
of finite size. This is of
regions of space and
started with in the
the type of relation that we
when we derived the differential
course
LARGE-SCALE FORM
IN
expression
from itI The
large-scale equivalents
1
are
§D~dS=fpdV
(I)
ng‘dS=0
(2)
V
S
S
i
at
E-dl=———
B-dS
(its
.
3gH-dlsz-dS+2-J’D‘ds
815!
5
(3)
(4)
(2) are obtained by integrating respectively Eqs. 3.6(1) and 3.6(2)
applying the divergence theorem. Equations (3) and (4) are obtained
by integrating, respectively, Eqs. 3. 6(3) and 3. 6(4) over a surface and applying Stokes’ s
theorem For example, integrating Eq. 3. 6(1),
Equations (1)
over a
and
volume and
fV-DdV=deV
v
and
applying
the
divergence
v
theorem to the left-hand
side, (1) follows. Equation (1) is
3.8
seen to
129
Maxwell’s Equations for the Time-Periodic Case
be the familiar form of Gauss’s law utilized
concerned with fields that
so
much in
Chapter
1. Now that
function of time, the interpretation is that the
electric flux flowing out of any closed surface at a given instant is equal to the charge
enclosed by the surface at that instant.
we are
are a
Equation (2) states that the surface integral of magnetic field or total magnetic flux
flowing out of a closed surface is zero for all values of time, expressing the fact that
magnetic charges have not been found in nature. Of course, the law does not prove that
such charges will never be found; if they are, a term on the right similar to the electric
charge term in (1) will simply be added, and a corresponding magnetic current term
will be added to (3). We will later find situations in which fictitious magnetic charges
and currents will be helpful and may be added to the equations.
Equation (3) is Faraday’s law of induction, stating that the line integral of electric
field about a closed path (electromotive force) is the negative of the time rate of change
of magnetic flux flowing through the path. The law was discussed in some detail in
Sec. 3.2.
is the
generalized Ampere’s law including Maxwell’s displacement
integral of magnetic field about a closed path
is
to
the
total
current (conduction, convection, and dis~
force)
(magnetomotive
equal
the
The
placement) flowing through
path.
physical significance of this complete law
Equation (4)
current
term, and it states that the line
has been discussed in Secs. 3.4—3.5.
3.8
MAXWELL'S EQUATIONS FOR THE TIME~PERIODIC CASE
By far the most important time-varying case is that involving steady-state ac fields
varying sinusoidally in time. Many engineering applications use sinusoidal fields. Other
functions of time, such as the pulses utilized in a digitally coded system, may be considered a superposition of steady-state sinusoids of different frequencies. Fourier
analysis (Fourier series for periodic functions and the Fourier integral for aperiodic
functions) provides the mathematical basis for this superposition. Rather than using real
sinusoidal functions directly, it is found convenient to introduce the complex exponential e1“. Electrical engineers are familiar with the advantages of this approach in the
analysis of ac circuits, and physicists use the complex exponential in a variety of physical problems with sinusoidal behavior. The advantage, which comes from the fact that
derivatives and integrals of ejw’ are proportional to ejw’ so that the function can be
canceled from all equations, is even more important for the vector field problems than
for scalar problems such as the circuit example. It is assumed that the reader has used
this technique before in circuit analysis or other physical problems, but if review is
needed, the use in analysis of a simple electrical circuit may be found in Appendix 4.
Formally, the set of equations 3.6(l)-—3.6(4) is easily changed to the complex form by
replacing a/ar by jw:
V-D
ll
V~BmO
(1)
(2)
139
Chapter
Maxwell’s
3
V X E
Equatiops
—ij
=
(3)
VXH=J+ij§
And the
auxiliary relations, Eqs. 3.6(6)——3.6(9),
J
=
0E
D
=
SE
(4)
remain
conductors;
for
(5)
I
=
(6)
ereoE
’
B
Equations 3.6(5)
and
nonlinear terms in the
=
pH
MOH
=
<7)
i
3.60) should be used with instantaneous values because of the
equations. The constitutive paraineters ,LL and s, are in general
frequency. Materials for which frequency dependence is important are
i
dispeisz've.
It must be recognized that the symbolsin the equations of this article have a different
meaning from the same symbols usedin Sec. 3.6. There they represented the instantaneous values of the indicated vector and scalar quantities. Here they represent the
complex multipliers of 6}”, giving the in—phase and out-of-phase parts with respect to
the chosen reference. The complex scalar quantities are lcommonly referred to as phasors, and by analogy the complex vector multipliers of ej‘fi" may be called vector phasors.
It would seem less confusing to use a different notation for the two kinds of quantities,
but one quickly runs out of symbols. The difference is normally clear from the context,
and when thereIS danger of confusion, we will use
functional notation to denote the
time-varying quantities.
If we wish to obtain the instantaneous values of a given quantity from the complex
functions of
called
value,
we
that the
insert the 61“" and take the real part. For
value of p is
example,
p
where p, and Pi
are
pr +
=
for the scalar p suppose
i
complex
(8)
jp;
real scalars The instantaneous value of p is then
p(t)== Re[(pr
+
jpi)"‘"]
=
pr
COS wt
-
pi sin wt
(9)
Or, alternatively, if p is given in magnitude and phase,
i
=
(plejflp
<10)
I
where
lpl
=
VP?
p?
+
I
6p
The true
a
-
tan
~1
E;
Pr
5,
time-varying form is
p(t)
For
_
vector
quantity,
=
Re[|p|ej(“"+‘9p)]
such
as
=
lplcos(ti)t
E, the complex value
+
6p)
rhay be written
(11)
E
where
E
and
E.
are
E(t)
'53?
Maxwell's Equations for the Time-Periodic Case
5.8
=
13:r
+
jEi
:
E,
(12)
real vectors. Then
=
+
Re[(E,
jEi)ej“”]
cos (or
-~
Ei
Sin wt
(13)
Note that
Er and E,- have the same directions in space only for certain special cases.
they are in the same direction, the vector phasor (12) can be expressed as a vector
“magnitude” and a scalar phase angle, but in the general case, when they are in differ~
ent directions, the six scalar quantities defining the two vectors must be specified
(Prob. 3.8a).
When
Example
3.8
A PHASOR SOLUTION OF MAXWELL'S EQUATIONS
As
an
example
later find to be
of
phasor solutions, let us consider
important as standing waves:
E
DO
:
J
V
Let
us
show that these do
rectangular
COS(CU
,
the
satisfy
-\
the
/
following fields,
which
we
will
(15)
#080X)
#080
phasor forms
of Maxwell’s
coordinate components of (3) and (4), with J
==
0,
equations.
The needed
are
813
—1
6H.
1
-—-‘
=
C!)
——
BB-
—i
=
,LLO 6x
6x
Substitution of (14) and (15)
(16)
——ij2
=
8):
~jcasoEy
gives
p"
00—130 sm(w\/ M0801) j)2cuDo sin(wV/.L080x)
—-
V
=
-
1““0
divergences
(18)
#030
VP“
______0_Pv30
__JDow
cos(wV #08015)
The
(17)
of
V
(14) and (15)
are
6:
(15)
are
6E
0 and ——’
(19)
#080
also found to be
BB-
——‘
So the forms (14) and
M cos(wV/.Loaox)
zero:
,
-:-=
—=_
0
6y
solutions of the phasor Maxwell equations for this
(20)
source-
132
free
Chapter
region.
If
we
wish the
Bz(x, t)
E
y
(x z‘)
=
=
5
time-varying forms,
Re[Bzej“”]
Re[Ey ejw’]
Equationjs
Maxwell’s
=
we
insert ejw‘ and take the real part:
D0 sin(wV M08017)
DO
=
V
sin
(21)
wt
cos(w\/)u.080x)
(22)
cos wt
#030
Exampfies ofi use of Maxwefifi’sr Equations
3.9
MAXWELL'S EQUATIONS
AND
WAVES
PLANE
l
equations still more concrete, let us show how
equations predict the propagation of uniform plane gelectromagnetic waves. Such
waves illustrate the interplay of electric and magnetic effects and are also of great
fundamental and practical importance. Let us begin from the time-varying forms of
Sec. 3.6. We postulate a simple medium with constant, scalar permittivity and permeO, J -'—- O). Maxwell’s equations are
ability and with no free charges and currents (p
To make the information of Maxwell’s
the
=
then
V-D=O
(1)
:
V~B=0
a3
VXE=
——=
~—
a:
VXH=Q=3
6t
For uniform
plane
(2)
aH
“at
(3)
——
!
m
é
at
m
+
waves, we assume variation in
only
z
direction of
direction. Take this
one
rectangular coordinate system. Then la/ax
us start with the two curl equations (3) and (4) in
specialization defined above,
the
a
rectangular
i
6H
VXE=———leadsto
H
at
6E;
._._2___
0 and
=
p”
———-‘5=—
62
0
k
as
0. Let
at
(5 )
6H
6E
i
—
coordinates. With the
6Hx
__
32
a/ay—
=
.U»
——y
at
3H:
‘M
at
( 6)
(7)
133
Maxwell’s Equations and Plane Waves
3.9
6H
____>’_
62
(
6E
VXH23—
6E.
z
8
__.\
62‘
(8)
6E,
leadsto
aH'r-r—e—J
<
6t
62
at
0
GEz
=
8
at
L
(9)
( 10 )
(10) show that the time-varying parts of Hz and E2 are zero. Thus the
entirely transverse to the direction of propagation. The remaining
equations break into two independent sets, with (5) and (9) relating Ey and Hx, and (6)
and (8) relating Ex and
Hy.
The propagation behavior is illustrated by either set. Choose the set with Ex and H
differentiating (6) partially with respect to z and (8) with respect to t:
Equations (7)
fields of the
and
wave are
,
a212,
62Hy
'
-
:2
--
82?—
Substitution of the second
“
62Hy
——
equation
at 32
into the first
yields
62E.
.
=
M8
622
62E.
'
z
’
az a:
8
62‘2
aZEx
(11)
61‘2
important partial differential equation (1 l) is a classical form known as the oneequation, having solutions that demonstrate propagation of a function
(a “wave”) in the z direction with velocity
The
dimensional wave
p4
c:
(12)
H
is
To show this, test
a
solution of the form
z
E_‘.(z, t)
=
z
f1<t ~) f2<r —)
(13)
+
+
-
v
v
Differentiating,
6E.1
I
=
62E?
I!
prime
I
f2
II
1+f2
61:2”:
where the
+
1
at
6E.
____.1
1
:
I
____~
82
Ufl
aZE'X
1
I!
1
+
I
_
Ufz
1
I!
azgzij—Zl+i}—2f2
denotes differentiation of the function with respect to the entire
ar—
gument, and the double prime denotes the corresponding second derivative. Comparison
of the two second derivatives shows that (11) is satisfied by such a solution with 1)
given by (12).
the
as
z
The first term of the solution in
direction with
illustrated in
Fig.
(13) represents
a
function f1
moving
in
To show this consider the function f1 (2) at various times
velocity
3.9. To keep on a constant reference of this wave, we must maintain
v.
134
Maxwell's
5
Chapter
Equatlpns
Ex(t)
FIG. 3.9
A
general
wave
of electric field
versus
distance for three different times.
t
the argument 1
to
keep
on a
2/0 equal to
—
a
implying a velocity dz/ dt
f2 traveling in the negative 2 direction
constant,
=
of
“fwaves ’so that the
name
1
v
=
=
ll
=
c
V
(477
x
second term represents the function
Thus the
v.
*-
with
velocity
“wave
thought
The velocity v defined by (12)13 found
In particular, for free space
as
v. Similarly,
implies a velocity dz/dt
we
must
keep 1‘ + z/v a
in] (12)
constant. This
constant reference of the second term
v.
Theseis moving functions may be
equation”
to be the
velocity
explained.
of light for
the medium
i
10-7
x
8.85419 ><
iii-”rm
#080
(14)
2.9979 x 108
m/s
:
1
(Note that to three significant figures, this is the
3 X 108 m/s, corresponding to 30 taken as 1/3611
remembered value
conveniently
X 10"9 F/m.) This equivalence
between the velocity of light and the predicted velocity of electromagnetic
Maxwell to establish
For
of the
a
medium
plane
electromagnetic phenomenon
with relative permittivity a and relative permeability
wave
light
waves
helped
as an
,ur, the
velocity
is then
(15)
Example
3.9
SINUSOIDAL WAVE
The most
common
and useful
wave
solutionis
one
varying sinusoidally1n space and
time. Consider the function
i
l
Egg, 1)
which is
a
special
differentiations:
case
of
(13).
=
A sin
To
w(z g)
(16)
-—
show that lit satisfies (11),
..
perform
the
Uniform Plane Waves with
3.10
azEx
(1)2
=
672
,us
But since
can
stay
v2
=
on a
(,us)’ 1,
maximum
aZEx
.,
=
61‘2
this is
a
or crest
w<t —)
—
z
,
t
3111 a)
“55A
(l7)
*-
5
2
.
-—,u,8w‘A
SlIl a)
t
(411
=
(18)
-
5
solution. To show that it is
of the function if
0
135
Steady-State Sinusoids
a
“wave,”
we see
that
we
we set
+1):
2
22
=
0,1,2,...
or
(411
vI
-
l)'n'v
+
(19)
———-~
260
so
that the crest does
3.10
move
in the
z
direction with
UNIFORM PLANE WAVES
velocity
v as
time progresses.
STEADY-STATE SINUSOIDS
WITH
To Show the usefulness of the
us
complex phasor approach for steady-state sinusoids, let
important special case. Replacement of Eqs. 3.9(6) and 3.9(8)
complex phasor equivalents, obtained by replacing time derivatives with jw,
continue with this
with the
yields
dEt
.
=
71;;
dHy
~72:
Here
we
amt/11..
<1)
_
==
stEx
have utilized the total derivative with respect to 2 since that is now the
(1) with respect to z and substitution of (2) yield
(2)
only
variable. Differentiation of
dzE‘.
(172'
=
-
wzueEx
(3)
equation, Eq. 3.9(11), but now written in phasor
equation. It could also be obtained by
replacing 62/62‘2 with w2 in Eq. 3.9(11). Solution is in terms of exponentials, as can
be verified by substituting in (3)
This is the
equivalent
form. It is called
a
of the
wave
one~dimensi0nal Helmholtz
_,
cle‘j":
Ex
+
€28ij
(4)
where
k
=
WE
<5)
136
Chapter
Equatidns
Maxwell’s
3
frequently in wave problems. It is a constant of the medium
particular angular frequency a) and is frequently called the wave number. It may
be written in terms of the velocity v defined by Eq. 3.9( 12):
The constant k will be met
for
a
also
k=—
The first term of
(4) is
one
that
its
changes
(6)
phase linearly
with z
becoming
increas-
positivez7direction. This behavior18
as one moves
in the
Re[Exej"”]
Re[cle“jkzej“’3’
lagging
interpretation that the sinusoidIS travehn g in the positive 2 direction
with velocity v, resulting in a phase constant k rad/n1. The second term of (4) is delayed
(becomes more negative) in phase as one moves in the negativez direction and so
represents a negatively traveling wave with the same phase constant.
To show the exact correspondence of this approach with that of Sec. 3.9, let us
convert the phasor form to a time—varying form by tlie rules given in Sec. 3.8. We
multiply the phasor by the exponential ejw’ and take the real part of the product:
ingly negative
or
consistent with the
Ex(z, t)
For
simplicity,
=
=
+
c2521“: em]
(7)
take (31 and 62 to be real. Then
Exe, t)
=
c1
cos(a)t
61
cos
k2)
-—
+ c2
w<t E)
_
1)
cosi(wt
+ 62
cos
l
+
k2)
(8a)
w<t :7)
+
(8b)
v
Following the interpretation of Sec. 3.9, we see two trial sinusoids, the first traveling
in the positive 2 direction with velocity v and the second traveling in the negative 2
direction with the same velocity. The result is then exactly as in Sec. 3.9.
Figure 3.10a shows the sinusoidal variation of Ex with 2 at a particular instant (say
0). This pattern moves to the right with velocity v if it is a positively traveling
wave and to the left if it is negatively traveling. The distance between two planes with
3
the
same
magnitude and direction of E
phase changes by 277
is called
wavelength
distance for which
)1 and18 found
by
the
i
I
!
[smug -z/v)
at t =0
U
FIG. 3.10:1
traveling
Sinusoidal function
plotted
versus
wave, the function progresses in the
distance for
positive
2
ojne instant of time. For
diredtion with
velocity
v.
a
positively
5.10
uniform Plane Waves with
137
Steady-State Sinusoids
A=——=——=—
where
Let
f
us
solution
is
frequency. Figure 3.10!) shows electric field vectors of a sinusoidal
magnetic fields. Returning to the complex forms, we
(4) in the differential equation (1):
also look at the
_l_ dEx
“jaws
3)
Using
(9)
the definition of k from
——
8
H),(z,
t)
=
equivalent
the
[cleflkz
-
czejkz]
(10)
(5),
Hy
The instantaneous
use
k
__.
3;;
dz
wave.
=
$1:
.,_
[Cite—1’“
--
6281“]
(11)
of this is
Re[H),(z)ej“"]
=
J}:
[C1 cos(wt
—-
k2)
-
c2
cos(wt
+
1(2)]
(12)
[.L
V ,u/e for the negatively
positively traveling wave and is
for
problems of wave transtraveling wave. The consequences of these relationships
in
mission and reflection are discussed
Chapter 6.
So
Ex/H), is
V
,u/a
for the
-—
)— r
l°l
_,.
O
4..
<—« —>- «<—<— —>a <—
4(—
0
_;,.
—)—
O
<—
i
‘
l
———>
Extends tOoo
-(—-
O
——>
Vectors showing magnitude and direction of electric field in a sinusoidal, uniform
filling the half-space O S 2 for one instant of time. For a positively traveling wave,
pattern moves to right with velocity 1 / V M8.
FIG. 3.10b
plane
wave
138
Chapter
dimensional
one-
cause
it illustrates
EquatioIns
THE WAVE EQUATION IN THREEI DIMENSIONS
3.l I
The
Maxwell's
5
example
wave
important practical problems
precedingI two
studied in the
sections is
and also because it is
important
be-
useful model for many
Nevertheless we have to be‘concemed with wave behavior
behavior
simply,
a
three dimensions also. To derive the equation governing such phenomena let
specialize to simple media in which a and ,u. are scalar constants and assume no
free charges or convection currents within the region of concern. We may then return
to the special form of Maxwell’s equations given as Eqsl 3.9(1) to 3.9(4). Take the curl
of Eq. 3.9(3), interchanging time and space partial derivatives:
in two
or
still
us
VxVXE=—M§E(V>,<H)
The left side is
magnetic
field
expanded by a vector identity. (See
the right side utilizes Eq. 3. 9(4).
iriside back cover.) The curl of
:
on
__
2
V E
62E
6E
+
V(V
.
source-free dielectric, V D
V E
0 also. Then
For
(1)
a
E)
=
=1
.._
--s--(”)1—
_
__
[rat
O and, if
61‘
a
[L8
,,
(2)
is not a function of space coordinates,
i'
=
V3E=
62E
(3)
‘LL 3——
at2
This is the three-dimensional
wave equation to be
derived. It will be found useful in a
in
to
be
considered
as
the
later,
problems
analysis of propagating modes of
variety
a waveguide, resonant modes of a cavity resonator, or radiating waves from an antenna.
Note that the vector equation breaks into three scalar ;equations, and for rectangular
of
coordinates it separates into three scalar
V2E_
and
we
similarly for By
have the
=
8
=
"
8
622
The
here,
wave
as can
form:
(4)
2
O and a/ay
6/6):
studied
in 1Sec.
equation
62153,.
same
62E,
a:
and E-. Note that if
one
dimensional wave
equations of the
wave
6~Ex
62‘2
‘2
O, V2is just
a
/622 and
3. 9.
(5)
equation applies also to magnetic field forIthe simple medium considered
by taking the curl of Eq. 3.9(4)1 and substituting Eq. 3.9(3) to
be shown
obtain
VZH
In
complex
or
phasor
62H
==
[.LE
gt?
(6)
notation these reduce to three-dimensional Helmholtz equa~
3. 12
Power Flow in
Electromagnetic
(lg/at2
tions, obtained by replacing
phasor
with
-
seen
that the
wave
13?
forms
==
~k2E
(7)
V2H
=
-k2H
(8)
k2
=
(ages
(9)
3.i n
RECTANGULAR Box
FOR A
equation has traveling—wave solutions.
boundary conditions. Consider
It also has
standing»
solutions under proper
Ex
We
Theorem
VZE
Example
wave
Poynting’s
cu2 in (3) and (6):
RESONANT WAVE SOLUTION
We have
Fields:
use
the
.r
=
C
out
lg):
sin
kyy
sin
kzz
(10)
component of (7), expressed in rectangular coordinates:
62E,
Iii-7
Carrying
cos
the
82E,
6y2
6213,.
-
822
”IRE“,
(11)
differentiations,
-
EE
——
kgE
-
TEE
=
-
=
k2
EE
or
k3.
So the
shall
phase
constants
in
k§
+
k3.
(12)
in the three directions must be related
by the condition (12). We
boundary conditions at the
of waves in a rectangular cavity
10 that this relation, combined with
Chapter
conducting walls, gives
see
+
the conditions for
resonance
resonator.
3.12
POWER FLOW
IN
ELECTROMAGNETIC FIELDS: POYNTING'S THEOREM
preceding sections have shown how electromagnetic waves may propagate through
space or a dielectric. We know from experience that such waves can carry energy. The
sun’s rays, which are now known to be electromagnetic waves, warm us. The radio
waves from a distant antenna bring power, admittedly small, to drive the first amplifier
stage of a receiver. For lumped electrical circuits we express power through voltage
The
electromagnetic fields, we can find a similar but more general relationship giving power and energy relationships in terms of the fields. The resulting
theorem, Poynting’s theorem, is one of the most fundamental and useful relationships
of electromagnetic theory.
and current. For
14%
Maxwell’s
chapter 3
We start with the
3.6) and write the
forms (Sec.
time-varying
Equatirims
two curl
equations
of
Maxwell:
an
V x E
=
(l)
i
———
at
VXH=J+%
An
equivalence
of vector
operations (inside
back
m
cover) shows that
H-(VxE)-—E'(VXH):Y;-(EXH)
If
and
products involving (1)
—
(2)
are
now
be
as
indicated;a (3) becomes
an
an
H -—-———E--——E J:
integrated
vi- (E
at
6t
This may
taken
over
(3)
XH )
(4)
the volume of concern:
as
an
at
6
[(H*+E—+EJ>W= fV-(EXHMV
V
From the
divergence theorem (Sec. 1.11), the volume integral
integral of E x H over the boundary
of
div(E
x
H) equals
the surface
[(H-@+EH§2+EJ)W= —f(ExH)-ds
This is the
since
we
at
62‘
V
important Paynting’s theorem and in this form is
so far made no specializations with respect to
have
time-invariant media (5)
a
can
8-H
valid for
general
media
the medium. For linear,
be recast into the form
a
D-E
L[5§< )+5}<'—"2)+E-J]dv__£(ExH)'dS
2
Problem 3.12f shows that
(5)
is
(6)
(6) is consistent with (5) for isotropic media. Equation (6) is
anisotropic media (Prob. 13.80). The term 3E2/ 2 was shown (Sec. 1.22)
to represent the energy storage per unit volume for ant electrostatic field. If this interpretation is extended by definition to any electric field,3 the second term of (6) represents
also valid for
the time rate of increase of the stored energy in the electric fields of the region. Similarly,
MHZ/2 is defined as the density of energy storage foi a magnetic field, the first term
if
represents the time
rate of increase of the stored energy in the magnetic fields of the
region. The third term represents either the ohmic powerI loss if J is a conduction current
density or the power required to accelerate charges if I is a convection current arising
from moving charges. Both of these cases will be illustrated in the examples at the end
of this section. Also, if there is an energy source, E
l is negative for that source and
-
3
For an excellent discussion of the arbitrariness of
Electromagneticlheory. p. 733. McGraw-Hiil.
thesei definitions, refer to J. A. Sircn‘ton.
New
York.
794 i.
3.12
Power Flow in
represents energy flow
out of the
externally.
Thus the term
unit time.
Changing sign,
‘41
Electromagnetic Fields: Poynting’s Theorem
region. All the
right represents
net energy change must be supplied
the energy flow into the volume per
the rate of energy flow out through the enclosing surface is
on
the
W
H
3gP-ds
(7)
E X H
(8)
S
where
P
and is called the
Poynting
=
vector.
it is known from the
proof only that total energy flow out of a region per
given by
integral (6), it is often convenient to think of the
vector P defined by (8) as the vector giving direction and magnitude of energy flow
density at any point in space. Though this step does not follow strictly, it is a most
useful interpretation and one which is justified for the majority of applications. (But
Although
unit time is
see
the total surface
Prob. 3.1221.)
It should be noted that there
the
electromagnetic
we see
field.
that it will be
are cases
Accepting
zero
the
for which there will be
foregoing interpretation
when either E
or
H is
zero or
power flow through
of the Poynting vector,
no
when the two vectors
are
mutually parallel. Thus, for example, there is no power flow in the vicinity of a system
of static charges that has electric field but no magnetic field. Another very important
case
is that of
a
which
perfect conductor,
by definition
must
have
a zero
tangential
component of electric field at its surface. Then P can have no component normal to the
conductor and there can be no power flow into the perfect conductor.
Example
3.12a
OHMIC Loss
To demonstrate the
interpretation
of the theorem, let
us
take the
simple example of a
length,
the resistance per unit
the electric field in the wire is known from Ohm’s law to be
round wire
carrying
direct current
I: (Fig. 3.12). If R is
E,
The
magnetic
field at the surface,
or
at any
H
The
Poynting
vector
P
=
E X H is
=
==
13R
radius
r
outside the wire, is
—‘
(9)
45
everywhere radial,
directed toward the axis:
RI ‘3'
P.I
We then make
an
integration
2‘
over a
*E-H 4’
:
"
cylindrical
(10)
-
2772‘
surface of unit
length
and radius
equal
E42
Chapter
FIG. 3.12
Round wire with
Maxwell’s
3
Equatiojns
Poynting vector directed radiallyinward to supply power for ohmic
Q
losses.
Q
of the wire (there is no flow through the ends of the cylinder since P has no
component normal to the ends). All the flowis through the cylindrical surface, giving
to that
a
power flow inward of amount
W
=
2m'(-P,.)
I;R
=
(11)
We know that this result does represent the correct power flow into the conductor, being
giving the correct density of
dissipated in heat. If we accept the Poynting vector
as
battery or other source of energy as
Qenergy flows through the field
power flow at each point, we must then picture the
setting up the electric and magnetic fields, so that the
and into the wire
surface. The Poynting theorem cannot be considered a
interpretation, for it says only that the total power balance
computed correctly in this manner, but the interpretation is
through its
proof of the correctness of this
for
a
given region
nevertheless
a
will be
useful
one.
-
Example 5.1%
MOVING CHARGES
5
us next consider the example in which J is a convection current For simplicity take
region containing particles of charge value q, mass m, and velocity vp. The convection
current densityIS
Let
a
J
Where
n
is the
density
=
pvp
=
of particles. From the force law
F
and the third term in the
Poynting
=
qE=
theorem
(12)
12qu
m
the acceleration of charges is
dvp
E
(13)
(6)IS
QVE.JdV L—VQ—w (nqv)dV== anaxv)
(14)
Poynting’s Theorem
5.15
.543
for Phasors
which
we recognize to be the rate of
change of the total kinetic energy of the charge
group. In this example we will not try to work out the right side of (6), but the E in
that term is related to the accelerating field, the H is that from the convection current,
and the
Poynting
theorem will
always
be
satisfied.4
Example
5.12s
POYNTING FLOW IN A PLANE WAVE
look at the Poynting theorem applied to the plane electromagnetic wave
preceding sections. The form of a sinusoidally varying wave with Ex and H),
propagating in the positive 2 direction was shown to be
Finally
we
studied in
Ex
EO cos(cor
i
Ji-
H),
The
*-
kz)
E0 cos(wt
Poynting vector is then in the z direction,
flowing in that direction:
-—
(15)
k2)
(16)
which is consistent with
our
interpretation
that power is
P:
By
the
use
of
a
=
Eva
8
=
Note that there is
as
E5 cosz(wt
-
kz)
(17)
this is also
l
l
-
2
+
~
2
cos
2(cot
—-
1(2)
(18)
showing that the wave carries an average power, as
time-varying portion representing the redistribution of stored
maxima and minima of fields pass through a given region.
a
constant term
There is also
energy in space
7
“E5
p.
expected.
l~b
trigonometric identity
P:
J;
=
'
a
3.13
POYNTING'S THEOREM FOR PHASORS
importance of phasors for sinusoidal electromagnetic fields, we need
Poynting theorem in phasor form also. It might seem that we could simply substitute
in the time-varying theorem, Eq. 3.126), replacing 6/6: by jw, but this does not work
since the expression is nonlinear, involving products of the fields. We start with Maxwell’s equations in complex form and derive the complex Poynting theorem by steps
Because of the
the
4
charges move through the surface surrounding the region, the net kinetic energy
transport by the charge stream through the surface is also included. This is actually contained in the third term on the left as shown by L. Tonks, Phys. Rev. 54 863 ( 7938).
If the
144
Maxwell’s
Chapter 3
Equatiojns
tune-varying:E field quantities. The
parallel to those used for the theorem in
equations in complex phasor form are (Sec. 3.8)
V x E
two curl
l
~ij
=
(1)
VXH=J+ij§
Consider the vector
(2)
identity
(3)
V-(EXH*)=H*‘(VXE)-§E‘(VXH*)
where the asterisk denotes the
substituted in this
complex conjugate.
eqiations (1) and (2) may
V
(E
H*)
x
be
;
identity:
~
now
H*
=
(uij)
-
E 0*
—-
—-
we
(4)
i
This
expression
is
volume V and the
integrated through
divergence
theorem utilized:
JV-(EXH*)JV=3€(EXH*)-ds
._f [E°J* +jw(H*'B -E'D*)]dV
s
V
3
=
(5)
:2
v
l
Equation (5) is the general Poynting theorem as it
interpret, consider an isotropic medium in which all
currents
J
2
(TE
so
that a, ,u, and
s are
real scalars.
applies to complex phasors. To
lbsses occur through conduction
Then (5) becomes
.5
éman-ds: —f o-E-Ei‘dV-‘jcuf
s
v
[{nH-H’F—eE-E’fldv
(6)
V5
integral on the right side represents power loss in the conduction
just twice the average power loss. (SeeiAppendix 4.) Thus the real part
of the complex Poynting flow on the left side can be related to this power loss. Or,
interpreting the Poynting vector itself as a density of power flow as in Sec. 3.12,
The first volume
currents and is
Pa
The second volume
=
integral
é-ReGE
on
the
right
War?
H*)
x
of
is proportional to
volume and average
(6)
between average stored magnetic energy in the
energy. Taking into account a factor of-é~ in the energy
averaging
of squares of sinusoids,
Im
f (E
X
we can
H*)
.
S
then
d8
=
interprget
4w(UE!m,
'
where UEav is average stored energy in electric fields
So the imaginary part of the Poynting flow through
reactive power flowing back and forth to
energy in the volume.
(7)
the difference
stored electric
expressions and another ~33- for
the imaginary part of (6) as
~—
UHav)
(8)
and UHav that in magnetic fields.
the surface can be thought of as
supply the instantaneous changes
i
in net stored
3.14
Continuity Conditions
for
Example
AVERAGE POWER
IN
ac
Fields at
8
5.13
UNIFORM PLANE WAVES
To illustrate the average Poynting vector for the planeuwave case, let
expressions for plane waves derived in complex form in Sec. 3.10:
Ex
€12”ij
=
The
complex Poynting
vector
E X H*
=
\/:
[616“1’“
density, by (7),
P2W
take the field
—-jk:
-—
(9)
cze
jkz
]
(10)
is then
p.
and the average power
\/~;—[cle
~—
us
626ij
+
8
_
Hy
145
Boundary
1
8
E
L:
_
-—
cgejk‘jkfelk‘
+
is in the
a:
2
direction and
a
[610,
-—
C§6w1L°]Z
-
C262]
W/m
equal
(11)
to
2
(12)
This
equation states that the average power is simply the average power of the positively
traveling wave minus that of the negatively traveling wave. The cross-product terms of
(l l) contribute only to reactive power, that is, to the interchange of stored energy within
the
wave.
3.14
In the
CONTINUITY CONDITIONS FOR AC FIELDS
UNIOUENESS OF SOLUTIONS
study of static fields, certain boundary
such fields and
were
and
AT A
continuity
BOUNDARY:
conditions
found essential in the solution of the field
were
problems by
stated for
the
use
of
the differential
equations. Similarly, for the use of Maxwell’s equations in differential
we need corresponding boundary and continuity conditions.
Consider first Faraday’s law in large-scale form, Eq. 3.2(3), applied to a path formed
by moving distance Al along one side of the boundary between any two materials and
returning on the other side, an infinitesimal distance into the second medium (Fig.
3.140). The line integral of electric field is
equation form,
%E
Since the
zero
rate
path
is
an
’
dl
:
(Etl
..
Erz) Al
(1)
either side of the boundary, it encloses
changing magnetic flux is zero so long as
density is finite. Consequently,
infinitesimal distance
on
area; therefore the contribution from
of
change of magnetic
(E11
flux
"“
BIZ) A]
z
0
Of
Etl
:
EIZ
(2)
346
Chapter
Maxwell’s Equations
3
fl
:Etl
WW
——"E
2
FIG. 3.14a
Continuity
of
tangential
elecm‘c field
components at
a
dielectric
boundary.
Similarly, the generalized Ampére law in large—scale form, Eq. 3.7(4), may be applied
like path with its two sides on the two sides of the boundary. Again zero area is
enclosed by the path, and, so long as current density and rate of change of electric flux
density are finite, the integral is zero. Thus, as in (2),
to a
Htl
Or in vector form,
by
3.140, (2) and (3)
can
Thus
use
(3)
:
t2
of the unit vector fi normal
be written
tangential components
to: the boundary as shown in Fig.
as
fiX(E1-E2)=OE
(4)
fiX(H1“H2):O
(5)
of electric and
magnetic field
must be
equal
on
the two
sides of any boundary between physically real media. The condition (3) may be modified for an idealized case such as the perfect conductor where the current densities are
allowed to become infinite. This
The
A5
are
integral
case
is discussed
form of Gauss’s law is
separately
Eq. 3.7(1). If
in Sec. 3.15.
two very small
elements of
area
(Fig. 3.141)), one on either side of the boundary between any two
surface charge density p, existing on the boundary, the application of
considered
materials, with
a
Gauss’s law to this elemental volume
AS(D721
gives
““
D112)
PS AS
3
or
Dnl
For
a
"
D122
2
(6)
P:
charge-free boundary,
Dnl
That is, for
=
Dnz
01’
81En1
.—.—..
:92En2
(7)
charge-free boundary, normal components of electric flux density are
a boundary with charges, they are discontinuous by the amount of the
surface charge density.
Since there is no magnetic charge term on the right of Eq. 37(2), 21 development
corresponding to the above shows that always the magnetic flux density is continuous:
a
continuous; for
;
Bnl
:
3712
01'
:u’IHnl
.2:
“2&1?!
(8)
time—varying case, which is of greatest importance to our study, the conditions
tangential compoequations (or their
and
in
these
be
obtained
from
the
two
curl equations
form),
may
equivalent large-scale
For the
normal components are not independent of those given for the
nents. The reason is that the former are derived from the divergence
on
5.14
Continuity Conditions
for
FIG. 3.14b
Diagram showing how discontinuity
boundary is related to surface charge density.
at a
ac
Fields at
8
147
Boundary
in normal components of electric flux
density
in the
time-varying case (Prob. 3.6b). The conditions on tangential components were
large-scale equivalents of the curl equations. Hence, for the ac solutions, it is necessary only to apply the continuity conditions on tangential components
of electric and magnetic fields at a boundary between two media, and the conditions
derived from the
on
components may be used
discontinuous, (6) tells the
normal
out to be
as a
check; if the normal components of D
amount
of surface
charge
that is induced
turn
on
the
boundary.
procedure to prove the uniqueness of solutions of Maxwell’s equaphilosophy in Sec. 1.17. One assumes two possible solutions with the
same given tangential fields on the boundary of the region of interest. The difference
field is formed, and found to satisfy Poynting’s theorem in the form of Eq. 3.12(5).
Stratton5 shows that for linear, isotropic (but possibly inhomogeneous) media, specification of tangential E and H on the boundary and of initial values of all fields at time
zero is sufficient to specify fields uniquely within the region at all later times. The
argument can be extended to anisotropic materials and certain classes of nonlinear
uniqueness
The
tions follows the
materials, but
and
H)
or to
not to
materials that have multivalued relations between D and E (or B
produce oscillations. In steady-state problems we
“active” materials that
not generally concerned with the specifications of initial conditions.
Although the discussion has been given for a region with closed boundaries, uniqueness arguments also apply to open regions extending to infinity, provided certain radiation conditions are satisfied by the fields. These require that the products 2E and 1H
remain finite as r approaches infinity6 and are satisfied by fields arising from real charge
are
and current
important
5
b
sources
to the
contained within
potential
a
finite
region.
The extension to open
formulation of the last part of this
regions
chapter.
J. A. Strafion. Electromagnetic Theory, pp. 486-488, McGraw-Hill, New York, 794 I.
8. Silver, Microwave Antenna Theory and Design, p. 85, IEEE Press, New York, 7984.
is
148
Chapter
3
Maxwell’s Equations
BOUNDARY CONDITIONS AT A PERFECT CONDUCTOR FOR AC FIELDS
3.15
practical problems to treat good conductors (such
as copper and other metals) as though of infinite conductivity when finding the form
of fields outside the conductor. We will study the effect of large but finite conductivity
It is
a
good approximation
in many
do, we will find that all fields and
currents concentrate in a thin region or “skin”
time—varying fields,
and this region approaches zero thickness as the conductivity approaches infinity. Thus,
on
fields within the conductor in Sec. 3.16. When
we
near
the surface for
(infinite conductivity), we find lthat all fields are zero inside
the conductor and any current flow must be only on the surface. The physical pr0perties
for the
perfect
conductor
perfect conductors are discussed in Sec. 13.4. Since the electric field is zero within
the perfect conductor, continuity of tangential electric field at a boundary requires that
the surface tangential electric field be zero just outside the boundary also,
of
E,
and
Eq. 3.114(6) gives
0
=
the normal electric flux
i
density
(1)
as:
I
D"
=
(2)
ps
Furthermore, since magnetic fields also vanish inside the conductor, the
continuity of magnetic flux lines, Eq. 3.14(8), indicates that
statement of
B II =0
(3)
pointed out in the last section, however, the continuity
independent of the condition on tangential E in the time-
at the conductor surface. As was
condition
varying
useful
on
normal B is not
case.
Thus, in the
check
solution, (3) follows from (1), but may sometimes be
ac
alternative
boundary condition.
tangential component of magnetic field is likewise zero inside the perfect conductor but is not in general zero just outside. This discontinuity would appear to violate
the condition of Eq. 3.l4(3), but it will be recalled that a condition for that proof was
that current density remain finite. For the perfect conductor, the finite current J per unit
as a
or as an
The
width is assumed to flow
current
density
on
the surface
is infinite. The
current
as a
discontinuity
in
sheet of
zero
tangential magnetic
thickness,
so
field is found
that
by
a
construction similar to that of Fig. 3.14a. The current enclosed by the path is the current
per unit width J flowing on the surface of the conductor
of the tangential magnetic field at the surface. Then
perpendicular to
the direction
§H-dI=H,d1=J,dz
01'
15
where
sense
15
=
H,
A/m
(4)
;=
is current per unit width, called
relations for
(4)
are
a surface
current density. The direction and
given most conveniently by the vector form of the law below.
To write the relations of
(1)--(4) in
vector
notation,
a
unit vector
1‘1, normal
to the
Penetration of
3.16
Electromagnetic Fields
into a Good Conductor
149
A
FIG. 3.15
Conducting boundary
with the normal unit vector.
conductor at any given point and pointing from the conductor into the
fields exist, is defined (Fig. 3.15). Then conditions (1)—(4) become:
For
region
where
fiXEno
(5)
firBzO
(6)
ps=fi'D
(7)
JszfiXH
(8)
problem, (5) represents the only required boundary condition at a perfect
Equation (6) serves as a check or sometimes as an alternative to (5). Equa~
tions (7) and (8) are used to give the charge and current induced on the conductor by
the presence of the electromagnetic fields.
an ac
conductor.
3.16
PENETRATION OF ELECTROMAGNETIC FIELDS INTO A GOOD CONDUCTOR
equations have been illustrated by showing the wave behavior of electro~
magnetic
good dielectrics. A second extremely important class of materials
used in many electromagnetic problems is that of “good conductors.” Let us examine
the basic behavior of electromagnetic fields in such conductors. The development in
this and the following section will be for steady-state sinusoids using phasor notation,
with the usual understanding that more general time variations may be broken up into
Maxwell’s
fields in
a
series
those
or
continuous distribution of such sinusoids. The conductors of
satisfying
concern are
Ohm’s law,
J
=
GE
(1)
conductivity of the conductor. At optical frequencies metals are
represented by a real constant 0', but the approximation is valid for microwaves
and millimeter waves (Sec. 133). Substitution of (1) into the Maxwell equation 3.8(4)
gives
The constant
a
is the
not well
V X H
=-
(a
+
jw8)E
It is easy to show that the assumption of Ohm’s law implies the absence of
density. Since the divergence of the curl of any vector is zero,
V-VXH=(0”+jwe)V‘E=O
(2)
charge
150
Chapter
where
have assumed
we
Maxwell's Equations
3
homogeneity
V
'
of
and
0'
D
:
p
8.
=
Thus
0
(3)
simple picture of the situation in a conductor is that mobile electrons drift through
positive ions, encountering frequent collisions. On the average, over a vol—
ume large compared with the atomic dimensions but small compared with dimensions
of interest in the system under study, the net charge is zero even though some of the
charges are moving through the element and causing current flow. The net movement
or “drift” in such cases is found proportional to the electric field.
For metals and other good conductors, it is found that displacement current is neg—
ligible in comparison with conduction current for microwave and millimeter-wave frequencies, and in fact is not measurable until frequencies are well into the infrared. For
the present we concentrate on the important cases for which we in (2) is negligible in
comparison with 0'.
Thus, to summarize, the following specializations are appropriate to Maxwell’s equations applied to good conductors, and may in fact be taken as a definition of a good
The
a
lattice of
conductor.
1. Conduction current is
2.
Displacement current
given by
is
Ohm’s law, J
negligible
in
a consequence of (1), the net
conductors.
3. As
To derive the differential
07E.
=2
comparison
charge
With current,
density
is
zero
one << 0.
for
homogeneous
equation which determines the penetration of the fields into
first take the curl of the Maxwell curl equation for electric field, Eq.
3. 8(3), and make use of a vector identity (see inside
cover) and the definition of
permeability to obtain
the conductor
we
back
VxVxE=V(V-E)—V2E=,—jw,u.V><H
Then
using (3)
and
substituting (2)
in
(4) with displacement
VZE
Equations
with forms identical to
and current
(5)
current
(4)
neglected,
we
find
J
=
can
jqu‘E
be found in
(5)
a
similar way for
magnetic
field
density:
VZH
=
jwtw'H
(6)
VJ
=
jquJ
(7)
equations (5)—-(7) for the simple but useful example
plane
depth, with no field variations along the width or length
dimension. This case is frequently taken as that of a conductor filling the half-space
x > O in a rectangular coordinate system with the y—«z plane coinciding with the conductor surface, and is then spoken of as a “semi-infinite solid.” In spite of the infinite
depth requirement, the analysis of this case is of importance to many conductors of
finite extent, and with curved surfaces, because at high frequencies the depth over which
We first consider the differential
of
a
conductor of infinite
Penetration of
3.16
Electromagnetic Fields
into
a
'55?
Good Conductor
significant fields are concentrated is very small. Radii of curvature and conductor depth
may then be taken as infinite in comparison. Moreover, any field variations along the
surface due to curvature, edge effects, or variations along a wavelength are ordinarily
small
so
compared
with the variations into the conductor that
For the uniform field situation shown in
the
z
direction,
we assume no
they may be neglected.
3.16a with the electric field vector in
Fig.
variations with y
or z
and
(5) becomes
'7
.—
jag/.MTEz
=
d1;
72E:
=
(8)
where
4;
72
Since
\/j
j) / Vi (taking
+
(l
=
(9)
jw/.L0'
the root with the
positive sign),
r=(l+j)\/;Tm=l—5u
(10)
where
5
———1——-
-
V
A
complete solution
of
(8) is in
The field will increase to the
The coefficient
x
=
=-
as
exponentials:
+
Cle‘“
impossible
may be written
C1
of
terms
E2
(11)
m
77pr
(12)
C26“
value of
infinity
at x
=
the field at the surface if
00
we
unless
let
E2
C2 is zero.
EO when
=
0. Then
E:
Or, in
terms
of the
quantity
6 defined
E
Since the
equation
where
as
H0
=
2':
by (10)
and
magnetic field and the current density
and
JO
are
the
=
J2
=
magnitudes
(11),
Egg—Vac ”ix/5
the electric field, forms identical to
H,
(13)
E06“?t
are
(14)
governed by
the
same
differential
(14) apply; that is,
Hoe“"‘/5e"j"'/5
Joe"“'/‘Se
of the
(15)
“Ix/5
magnetic
(16)
field and current
density
at
the
surface.
(14)-—(l6) that the magnitudes of the fields and current
decrease exponentially
penetration into the conductor, and 5 has the significance
of the depth at which they have decreased to 1/ 3 (about 36.9%) of their values at the
surface, as indicated in Fig. 3.16a. The quantity 6 is accordingly called the depth of
It is evident from the forms of
with
152
Chapter
Maxwell’s Equations
3
8
Plane solid
FIG. 3.160
illustrating decay
of current into conductor.
10'4
10-—1
00
3
00k
)7
RsCu
O!) O C:
/r
,2§
/
K
10'2
E23
“i 10*3
I
E
§10~*4
“5
‘(~
10—5
10-6
s
g
[
Y—BauCu-O
77 K
L
Q)
E
n
10‘7 g:
Nb 4 K
am +j)
1 0-5
E
m
77 K
(I)
10~6
11
\\
/
/
0L)
%
6/(1 +1) Y—~Ba-—Cu——O
{
17;
‘5;
1‘
‘(x
/
4
=
L
10—8
Nb
4 K
10-9
i
0.001
0.01
0.1
1
10
100
Frequency (GHz)
FIG. 3.16b Skin depth and surface resistance for 00pper at two temperatures and for two superconductors. Note that the skin depth for superconductors is (1 + j) times a real number, so
penetration of fields and current density in Eqs. 3.16(14-16) have only the real exponential decay.
internal
3.17
Impedance
of
a
153
Plane Conductor
penetration, or Skin dept/2. The phases of the current and fields lag behind their surface
values by x/ 5 radians at depth it into the conductor. The penetration depths for copper
at room temperature (300 K) and 77 K are shown in Fig. 3.1617.
Except for ferromagnetic
and ferrite materials, p.
~
#0-
Example
3.16
SKIN DEPTH IN AUDIO TRANSFORMER WITH IRON CORE
An audio
frequency transformer has
,u.
lOOOuO. It is designed to work
highest design frequency. From (11)
=
5
=
=
(30772
Note that this is
but with
a
x
(77
more
relative
15 x
103
x
106)“1/2
><
a core
103
=
made of iron with
x 477 x
0.058
X
10*7
10'"3
than 30 times smaller than for
of
permeability
=
cr
0.5 X 107 and
up to 15 kHz. Let us find the skin
a
m
x
=
107)'-1/2
0.5 X
0.058
depth
at
this
( 17 )
mm
material of the
same
conductivity
unity.
Advantageous electromagnetic behavior can be obtained in circumstances where
cooling to cryogenic temperatures is possible if superconductors are used.7 For reasons
to be explained in Sec. 13.4, the conductivity is complex and frequency dependent, and
8 is constant up to about 100 GHz at (1 + j) times the value of the dc penetration
depth AS. Values of 6 found experimentally for niobium at 4 K and for the oxide
superconductor YBazCu3O7m or simply Y—«Ba—«Cu—O, are shown in Fig. 3.16!) for
comparison with the frequency—dependent values for copper. The oxide superconductor
Y—Ba—Cu—O has an anisotropic crystal structure; it is assumed here that the highly
conducting Cu—O planes are parallel to the surface. The behavior is otherwise more
complicated.
3.17
INTERNAL IMPEDANCE OF A PLANE CONDUCTOR
conductor or superconductor may be looked at as the
it
propagates into the conductor or from the point of
plane
view that induced fields from the time—varying currents tend to counter the applied
The
decay
of fields into
attenuation of
a
good
wave as
a
especially applicable to circuits, in which case we
applied field. Currents (resulting from GE) con—
centrate near this surface and the ratio of surface electric field to current flow gives an
internal impedance for use in circuit problems. By internal, we mean the contribution
fields. The latter
point
of view is
think of the field at the surface
7
as
T. Van Duzer and C. W. Turner,
the
Principles of Superconductive Devices and Circuits, Sec.
by Prentice Hall.)
3. l4, Elsevier, New York, 798 l. (To be reissued
154
to
chapter
impedance
from the fields
sistance term and
ance
contribution
Maxwell’s
3
penetrating
the
Equations
conductor.
gives, in general,
This
a re-
internal inductance, the latter to be added to any external inductarising from the fields outside the conductor.
an
flowing past a unit width on the surface of the plane conductor is
by integrating the current density, Eq. 3.1606), from the surface to the infinite
The total current
found
depth:
J“
=
f
.1,
dx
f
=
o
Joe"(1+1)("‘/5)
o
dx
(1)
,
+
(1
i
The elecuic field at the surface is related to the
0
=
currentidensity
J)
at the surface
by
J
EEO
Internal
impedance
for
a
unit
‘9'
(2)
0'
and unit width is
length
Z
=
A
E20
———
=
1 +
=
defined
as
j
3
—-—
’
With the further definition
2,
Q
R5
+
ij,
(4)
We then have
I
R,
=
wLi
:
—
/—-—7Tf"
:
0'5
(5)
0~
1
With
real, the resistance and internal
frequency.
Equation (5) gives
shows
the dc
0'5
:
(6)
Rs
.
of such
plane conductor are equal
impedance ZS
phase angle of 45 degrees.
another interpretation of depth of penetration 6, for this equation
that the skin-effect resistance of the semi-infinitegplane conductor is the same as
resistance of a plane conductor of depth 6. That is, resistance of this conductor
a"
at any
—
The internal
reactance
thus hasl
a
a
Table 5.1m
Siam Effect Properties of
Depth of Penetration
a (m)
Surface
2.52 x
5.80 x 107
0.0642f‘I/2
O.O826f“1/3
0.127f-1/90.066f“1/?
18 x 107
0.037f“‘/?
Conductivity
o—
(S/rn)
Silver
6.17 x 107
Aluminum
3.72 x 107
(300 K)
(300 K)
Brass (300 K)
Copper (300 K)
Copper (77 K)
Typical Metais
1.57 x 107
i
Resistivity
125 (9)
10‘7f1/2
10'7f1/2
5.01 x 10-7f1/2
2.61 x io-Vfl/2
1.5 x 10-7”2
3.26 x
5.17
internal Impedance of
a
155
Plane Conductor
Table 3.17!)
Shin Effect
Properties
of
Typical Superconductors
Surface
Complex Conductivity
0'
YBaECu3O7_x (77 K)
8.2 x
Niobium (4 K)
5.2 X
2
0"1
—jch(S/rn)
Penetration
A,
106
j20 x 1017f“
106 -—jl75 X 1037f“1
—
=
Depth
6/(1 + j) (m)
250 x 10*9
85 X
10“9
Resistivity
95(0)
40 x
1.0 X
10"253c2
10““25]C2
with exponential decrease in current density is the same as though current were uniformly distributed over a depth 5.
The resistance R5 of the plane conductor for a unit length and unit width is called
the surface resistivity. For a finite area of conductor, the resistance is obtained by
multiplying R, by length, and dividing by width since the width elements are essentially
in parallel. Thus the dimension of RS is ohms or, as it is sometimes called, ohms per
square. Like the depth of penetration 6, RS as defined by (5) is also a useful parameter
in the analyses of conductors of other than plane shape, and may be thought of as a
constant of the material at frequency f.
Superconductors are somewhat different from the good conductor discussed above
in having a complex conductivity with the result that the surface resistance and reactance
terms are not equal. But the definitions in (3) and (4) still apply. Again, p.
#0- They
differ also in that R, increases as f2 rather than as fI/z, as in the case of a good
conductor. Values of depth of penetration (skin depth) and surface resistivity are tabulated for several metals in Table 3.1'7a and are plotted in Fig. 3.16!) as functions of
frequency. Table 3.17b gives experimentally derived data for the complex conductivity,
penetration depth, and surface resistance for two prominent superconductors; the
penetration depth and surface resistance are also plotted in Fig. 3.16!) as functions of
frequency.
z
Example
5.17
APPROXIMATE lNTERNAL IMPEDANCE OF
A
COAXIAL LlNE
The usefulness of this concept for practical problems may now be illustrated by considering the coaxial transmission line of Fig. 3.17. We select as a circuit path one which
follows the outer surface of the inner conductor, AB, traversing radially across to C and
then following the inner surface of the outer conductor CD, returning back radially to
and V013 will arise in part because of the in—
path ABCDA, that is, the inductance external
to the conductors. We consequently call this the external inductance and recognize it
as that found for a coaxial line in Chapter 2. But there is also a voltage contribution
along the path AB due to the internal impedance of the inner conductor and one along
CD arising from internal impedance of the outer conductor.
A. The difference between
voltages VDA
ductance calculated from flux within the
156
Chapter
Maxwell’s
5
Section of coaxial transmission line. Line
FIG. 3.17
relates to
magnetic
Equationis
integral of electric
field about
path ABCD
|
flux associated with external inductance.
large in comparison with skin depth 8, and if
large compared! with 8, both conductors may
good degree of approximation by the planar analysis of this and the
If radii of curvatures
a
and b
are
thickness of the outer tubular conductor is
be treated to
preceding
a
section. Current concentrates
on
the inner surface of the outer conductor,
the outer
surface of the inner conductor and
region of the fields.
the
adjacent to
The inner
conductor, if curvature is negligible, then appears as a: plane of width equal
circumference, 27m. Internal impedance per unit length is then
Zn
The outer conductor, with these
Z51
z
—
27m
its
to
=
Q /m
E
approximations, appearsi as
a
plane
of width
equal
to
its inner circumference, 2771). Its thickness does not enter since it is presumed much
larger than 8, so that fields have died to a negligible value at the outer surface. Internal
impedance per unit length
i
from this part is then
.
i
=
2,:
'-
The
sum
of these two
conductors and
can
gives
22
S
2771;
n /m
the total contribution to
be used in the transmission-line
3.18
POWER Loss IN
A
impedance from
analysis
PLANE
of
cbnductor,
x
direction,
or
into the conductor Utilization of
the total power flowing from the field into the
gives
using Eq 313(7),
P
—
~1-
2Re[Eo
x
5.
CouDUCTOR
To find average power loss per unit area of the plane
Poynting theorem of Sec. 3.13. The field components E-
in the
fields within the
Chapter
we
may
and Hy produce
the
apply
the
power flow
field values at the surface
a
conductor. In complex phasor form,
H*0]: —x——
Re(E-0Hjjo)
(1)
3.18
Power Loss in
FIG. 3.18
Surface of plane conductor
flow per unit width.
a
.357
Plane Conductor
illustrating how magnetic field at surface relates to current
The surface value of
magnetic field can readily be related to the surface current, as
by taking the line integral of magnetic field about some path ABCD of Fig.
3.18 (C and D at infinity). Since magnetic field is in the
y direction for this simple
case, there is no contribution to H (1] along the sides BC and DA; there is no contribution along CD since field is zero at infinity. Hence, for a width w,
can
be
seen
-
-
B
39 H-d12fH-dlr—wHyo
ABCD
(2)
A
integral of magnetic field must be equal to the conduction current enclosed,
displacement current has been shown to be negligible in a good conductor. The
current is just the width w times the current per unit width J3. Then, utilizing (2),
This line
since
«wHyO
=
stz
or
15:
=
(3)
~H),O
This may be written in a vector form which includes the magnitude and
mation of (3) and the fact that J and H are mutually perpendicular,
sense
(4)
JS=fiXH
where fl is
joining
of the
unit vector
a
dielectric
Eq. 3.17(3),
region
form
same
we
as
perpendicular
and H is the
a
obtain for power loss
form that
multiplied by
might
the conductor surface,
magnetic
field
at
pointng into
the ad-
the surface. Note that
(4) is
perfect conductors, Eq. 3.15(8). Then using (1), (3), and
for
WL
This is
to
infor-
=
WL
=
%Re[ZSJSJ:]
=
saga? W/m2
(5)
expected in that it gives loss in terms of resistance
magnitude. An alternate derivation (Prob. 3.18a) is by
have been
square of current
IPLI
158
Chapter
Maxwell’s Equations
3
integration of power loss at each point of the solid Ifrom the known conductivity
density function.
Equation (5) will be found of the greatest usefulness throughout this text for the
computation of power loss in the walls of waveguides, cavity resonators, and other
electromagnetic structures. Although the walls of these structures are not plane solids
of infinite depth, the results of this section may be applied for all practical purposes
whenever the conductor thickness and radii of curvature are much greater than 8, depth
of penetration. This includes most important cases at high frequencies. In these cases
the quantities that are ordinarily known are the fields at the surface of the conductor.
the
and current
=
§otentials for
As
we
heads
A POSSIBLE SET OF POTENTIALS FOR TlME-VARVING FIELDS
3.19
time-varying electromagnetic fields are related to each other and to
sources through the set of differential equations known as Max-
have seen,
the
Time-Varying
and current
charge
equations. It is sometimes convenient to introduce some intermediate functions,
known as potential functions, which are directly related to the sources, and from which
the electric and magnetic fields may be derived. Such functions were found useful for
static fields, and in the case of the electrostatic potential, the potential itself had useful
physical significance. The physical interpretation was less clear in the case of the mag
netic vector potential, but it does provide a useful simplification in the analysis of some
problems. In this and following sections we look for similar potential functions for the
time-varying fields. It turns out that there are many possible sets. We select a commonly
used set known as retarded potentials, which reduce to the potentials used for statics
well’s
in the limit of
no
time variations.
V43 and B
V X A,
might try first to use the forms found for statics,iE
are
functions
of
faced
with
this
time.
We
the
electric
field
problem:
quantities
for time-varying conditions cannot be derived only as the gradient of scalar potential
since this would require that it have zero curl, and it may actually have a nonzero curl
of value
aB/ar; it cannot be derived alone as the curl of a vector potential since this
would require that it have zero divergence, and it may have a finite divergence of value
We
at
=
=
-—
with all
——
p/e.
Since the
static, it
tial, A.
divergence
seems
of
magnetic
that B may still be set
field is
equal
zero
in the?
to the curl
of
general
some
case as
magnetic
it
was
in the
vector
poten-
I
B=VXA
(l)
A Possible Set of Potentials for
3.19
This relation may
written
now
Fields
159
equation 3.6(3)
and the result
Time-Varying
be substituted in the Maxwell
6A
(E 3;)
V X
This
equation
condition that
+
that the curl of
states
permits
a
a
(2)
certain vector
vector to be derived as the
E +
O
=
E
=
quantity is zero.
gradient of a scalar,
But this is the
say (I). That
is,
uvo
62‘
01‘
E
=
-—Vci>
—-
95-
(3)
a:
and
then valid
relationships between fields and potential functions
specializations on the medium have been made to this point. It
is found that the potential functions are most useful, though, for linear, isotropic,
homogeneous media, so in the remaining part of this discussion, we take ,u and s as
scalar constants appropriate to such media. With this specialization we substitute (3) in
Gauss’s law, Eq. 3.6(1), to obtain
Equations (1)
(3)
A and CD. Note that
are
no
6
-V2<D—~—(V‘A)=-p~
a:
(4)
8
Then, substituting B
=
V X A and (3) in
V x V x A
Using
the vector
=
“J
+
Eq. 3.6(4),
we
find
62A
act)
“B[ (61‘) er]
—V
—
—
,
identity
V xVx AEVW-A)
—
V2A
this becomes
V(V
Equations (4)
and
(5)
'
A)
can
-
be
VZA
62A
act)
=
[.LJ
simplified by
—-
iteV<—BT>
further
—
[is
specification
at2
of A. That is, there
(5)
are
any number of vector functions whose curl is the same. One may specify also the
divergence of A according to convenience.8 If the divergence of A is chosen as9
8
Specification of divergence and curl of a vector, with appropriate boundary conditions,
determines the vector uniquely through the Helmholtz theorem. See, for example, R. E.
Collin, Field Theory of Guided Waves, 2nd ed., Appendix A l, lEEE Press, Piscataway, NJ,
7997.
as the Lorentz condition or Lorentz gauge and leads to the symmetry
of (7) and (8). Other useful gauges are the Coulomb and London gauges. See, for example, A. M. Portis, Electromagnetic Fields: Sources and Medla, Wiley, New York, 7978.
This choice is known
See also Prob. 3. 790.
360
Chapter
5
Maxwell’s Equations
6CD
V- A
(4) and (5) then simplify
=
(6)
~—
—
,LLS
at
to
V2¢“#8
32(1)
9=—3
6t“
(7)
-'
8
t
V2A
a 2A
~
its
(8)
—-/.LJ
=
62:2
potentials A and (I), defined in terms of the sources J and p by the differential
equations (7) and (8), may be used to derive the electric and magnetic fields by (l) and
(3). It is easy to see that they do reduce to the corresponding expressions of statics, for
if time derivatives are allowed to go to zero, the set of equations (1), (3), (7), and (8)
Thus the
becomes
Vch
=
f
E
VA: war
which
3.20
are
recognized
as
the
=
-VCI):
(9)
B=VXaA
appropriate expressions
from
Chapters
(10)
1 and 2.
THE RETARDED POTENTIALS AS INTEGRALS OVER CHARGES AND CURRENTS
potential functions A and CI) for time-varying fields are defined in terms of the
charges by the differential equations 3.190) and 3. 19(8). General solutions
of the equation give the potentials as integrals over the charges and currents, as in the
static case. The following discussion applies to the very important case of a region
extending to infinity with a linear, isotr0pic, and homogeneous medium.
From Chapters 1 and 2 the integrals for the static potentials, which may be considered
the solutions of Eqs. 3.19(9) and 3.l9(10), are
The
currents and
®=fpm
v
A
(u
47TSI‘
J
(2)
=
V
4777'
A mathematical development to yield the corresponding integral solutions of the inhomogeneous wave equations 3. 19(7) and 3.19(8) is given in Appendix 5. A plausibility
argument is given here. The solutions are
®mian=ip
(X’,
V
A(x,
y, z,
2:)
=
u
f
V
y ’a
Z”
t
~—
R/ U) dV’
478R
J(x’, y ', z',: -i
477R
_
R / v) dV’
(a
(4)
5.20
The Retarded Potentials
as
Integrals
over
Charges
and Currents
.961
where
=
v
(for free space, v
point (x', y', 2’) and
=
R
In the above, I
——
c
=
field
=
(M8)”2
2.9987 X 108
point (x,
[(x
—-
y,
x’)2
m/s)
(5)
and R is the distance between
source
2),
+
(y
R / v denotes that, for
y')2
—
an
+
(z
-
2:321”2
(6)
evaluation of d) at time t, the value of
R / u should be used. That is, for each element of charge
Charge density p at time t
p (W, the equation says that the contribution to potential is of the same form as in statics,
(1), except that we must recognize a finite time of propagating the effect from the charge
element to the point P at which potential is being computed, distance R away. The
effect travels with velocity v
1/ WL—E: which, as we have seen, is just the velocity
of a simple plane wave through the medium as predicted from the homogeneous wave
equation. Thus, in computing the total contribution to potential (I) at a point P at a given
instant I, we must use the values of charge density from points distance R away at an
R / v, since for a given element it is that effect which just reaches P at
earlier time, t
time t. A similar interpretation applies to the computation of A from currents in (4).
Because of this “retardation” effect, the potentials CI) and A are called the retarded
potentials. Once the phenomenon of wave propagation predicted from Maxwell’s equations is known, this is about the simplest revision of the static formulas (l) and (2) that
could be expected.
-
=
-
Example
5.20
FIELD FROM AN AC CURRENT ELEMENT
simplest examples illustrating the meaning of this retardation, and one that
again in the study of radiating systems, is that of a very short wire carrying
an ac current varying sinusoidally in time between two small spheres on which charges
accumulate (Fig. 3.20). For a filamentary current in a small wire, the differences in
One of the
will be met
L
h
,/
TLfIZ
FIG. 3.20
Retarded
potential
from small current element.
162
Chapter
distance from P to various
points
Maxwell’s
3
of
a
given
Equations
cross
section of the wire
are
unimportant,
that two parts of the volume integral in (4) may be; done by integrating current
density over the cross section to yield the total current in the wire. Thus, for any
so
filamentary
current,
r/v)
—-
A
For the
particular case
(7) is
so
(7)
of Fig. 3.20, current is in the
z
direction
small
h
only; so, by the above,
remaining integration
the
t—i)
Az=-'u——I
47rrz
Finally,
d1
4777‘
compared with r and wavelength,
performed by multiplying the current by I2:
A is also. If h is
of
“J10:
___
8
()
v
if the current in the small element has the form
IO
==
3
substitution in
(8) gives Az
as
i
A,
=
'
From this value of A, the
done when
we return
3.2l
(9)
cos wt
M h]0
.
w<t 3-)
cos
magnetic
to radiation in
5
v
and electric fields
Chapter
(10)
-—
4772'
may be derived.
This will be
12.
THE RETARDED POTENTIALS FOR THE TIME-PERIODIC CASE
electromagnetic quantities of interest are varying sinusoidally in time, in the
complex notation with eja” understood, the set of equations 319(1), 319(3), 320(3),
320(4), and 319(6) becomes
If all
~
=
w/v
=
E
=
y,
2)
A(x,
y,
z)
-
A
__
-~
V x A
~-V<I>
f
_
—-
=
n
(1)
ij
-—
p(x', y’, 2’)?ij
(2)
dV’
<3)
4%,,
J;
J(x’, y’, 2’)e“”"R
477R
—-jw,ue<l>
dV’
(4)
(5)
cum, and R is the distance between source and field points. Note
that the retardation in this
shift in
=
choc,
V
where k
B
case
by the factor 6‘ij and amounts to a
to the distance R from the
according
potential
at which potential is to be computed. (From here
is taken
of each contribution to
phase
contributing element
to the
point
P
care
of
1 63
Problems
will
frequently leave out the functional notation of coordinates, with the under~
standing
potentials are computed for the field point, and the integration is over all
source points.)
For these steady-state sinusoids the relation between A and (I) in (5) fixes (I) uniquely
once A is determined. Thus, it is not necessary to compute the scalar potential (I)
separately. Both E and B may be written in terms of A alone:
on we
that
B
=
E
=
A
==
V
x A
(6)
"112(3) V(v
A)
-
‘18—ij
—-
ij
(7)
W!
(8)
——
477R
v
It is then necessary only to specify the current distribution over the system, to compute
the vector potential A from it by (8), and then find the electric and magnetic fields by
(6) and (7). It may appear that the effects of the charges of the system
out, but of
course
the
charges
to the currents
-
J
=
bution
(1)
to
by means
(4).
left
(9)
~ij
and, in fact, in this steady-state sinusoidal
the distribution of J is
80
case, fixes
but
given.
equivalent
lengthier procedure
computing the charge distribution from the specified current distriof the continuity equation (9), then using the complete set of equations
p uniquely
would be that of
once
being
continuity equation
V
relates the
are
an
PROBLEMS
iCOx sin wt passes through a rectangumagnetic field of the approximate form B
loop in the x~y plane following the path (0, O) to (a, 0) to (a, b) to (O, b) to (O, 0).
Show that if Faraday’s law in microscopic form is used to give E, the macroscopic
form of Faraday’s law is satisfied.
3.221 A
=
lar
3.2b Find the emf around a circular loop in the z
axial magnetic field varying with r, 43, and 2‘
B:
=
Cor sin qb sin
Now find the emf for a circuit consisting of
71'.
Otor
fromr
a,q§
a,qb
=
=
=
0 plane if the loop
approximately as
=
a
is threaded
by
an
wt
half-circle of radius
a
and
a
straight wire
=
use of the electric field produced by a time-varying magnetic field
in space to accelerate charged particles. Suppose that the magnetic field of a betatron
has an axial component in circular cylindrical coordinates of the following form:
3.2c* The betatron makes
B:(r, t)
i
Ctr”
(t
Z
O)
problems denotes ones longer or harder than the average; two asterisks
unusually difficult or lengthy problems.
The asterisk
denote
=
on
364
Chapter
5
Maxwell’s Equations
magnitude and direction at a particular radius a.
velocity
charge q after a time t, assuming that the charge stays in
a path of constant radius a, and calculate the magnetic force on the charge. For what
value(s) of )2 will this magnetic force just balance the centrifugal force (mug/a), where
m is mass of the particle, so that it can remain in the path of constant radius as
Find the induced electric field in
From this find
v on a
assumed?
a long, round wire varies sinusoidally with time. Assume no variation
(coordinate along the wire) or qb (angular coordinate around the wire). Find
magnetic field outside the wire, assuming it related to:current flow at any instant as in
the statics form. (This is called the “quasistatic” approximation.) From the differential
3.2d Current I in
with
z
equation form of Faraday’s law find
generated by the changing magnetic
the electric field in the space outside the wire
field of the form found. Use phasor forms and
take the electric field at the surface of the wire
internal
impedance Z,-
per unit
(2‘
==
0)
Note behavior at
length.
as the
product of current and
infinity. What is unrealistic
about this model?
3.3a To demonstrate
poles
of
a
Faraday’s law
in class,
we
often
move a
coil
by hand through the
permanent magnet and observe the generated voltage
on an
oscilloscope.
One magnet used has a flux density of 0. 1 T and pole pieces about 2 cm in diameter.
Estimate the velocity you can conveniently obtain byhand motion and find how many
turns
you need to produce peak voltages around 10 mV. Sketch the waveform
as the coil is moved through the region between poles.
ex-
pected
3.3!) In the generator of Fig. 3.3a, the poles are reshaped so that magnetic flux density is
inhomogeneous. Take the magnetic field direction as the z direction and the vertical
direction of the
figure
as
the
x
direction. Assume the
variation with x,
quadratic
inhomogeneous field to have a
x2
B:
Find emf
by
use
generated
in the
2
Hm
I
-
g
rotating loop by considering
rate
of
change
of flux, and also
wave in
of the motional electric field in the wires. Plot? the waveform of this
time.
3.3b, it may seem surprising that motional field depends only on the value of
B at the instantaneous position of the conductors, whereas fiux enclosed depends upon
3.3c In Prob.
integration
identical
of field
answers.
throughout the region of inhomogeneous variation, yet both give
Explain why this is so for any arbiirary variation with x.
3.3d In the generator of Fig. 3.30, the rectangular loop is replaced by a circular loop of
radius a, rotated about a line in the plane of the loop and passing through the center.
This axis is normal to B
this
loop
as
as
in
it is rotated with
Fig. 3.3a. Take BO as uniform and find emf generated
angular velocity 0 abodt the defined axis.
in
3.3e The
rectangular 100p of Ex. 3.3 is moved with constant velocity v in the x direction
through an inhomogeneous magnetic field which varies sinusoidally with x,
mt
s1n(L)
'
8.: C
;
Find the induced emf, both from
electric field. Find values for the
rate
of
special
—
change
cases
of flux and
by
use
of the motional
a/L=v_i:, 1,2
3.3f A wire in the form of a rectangular loop with one arm at r
0 extending from y~— O
to y: b and two parallel arms aty
O and y— b
extending from x 01n the +x
direction as in Fig. P3. 3 f has static flux density B01n the z direction. A sliding short at
=
=
=
-
=
3 65
Problems
FIG. P3.3f
x moves
and
3.3g
by
with
velocity
Find the emf induced in the
1).
loop by
rate of
change of flux
the motional method.
A
long, straight wire carries a time—varying current I. A rectangular circuit of length 6
r, 2 plane, with one leg distance r! from the axis and the other at 2'2 as in
Fig. P3.3g. Find the emf induced in the loop.
lies in the
s———z———>1
l
7‘
2T
is
J->I (25)
FIG.
P3.39
density for copper with a field of 0.1 V/m (1 mV/ cm) applied is
What number density of electrons is required to produce the same
3.4 Conduction current
5.8 X 106
A/mz.
density if the electrons have been accelerated in vacuum
potential of 1 kV? What electric field magnitude would be required to pro~
duce the same magnitude of displacement current density in space for sinusoidally
varying waves as follows: (1) a power wave of frequency 60 Hz; (2) a microwave
beam of frequency 3 GHz; (3) a laser beam of wavelength 1.06 pm?
value of convection current
3.521
through
a
Starting
from
Eq. 3.4(7),
prove that for
a
closed surface
s
From this, show that the sum of convection and displacement currents is the same for
both of the surfaces 8, and $2 in Fig. 3.5b. For a spherical capacitor with concentric
conductors of radii
displacement
the leads to the
3.5!) Obtain the
equation.
a
current
and b, with sinusoidal voltage applied between conductors, find
a < r < b and show that it is equal to the charging current in
for
capacitor.
expressions
for electric field,
Eqs. 3.5(4)~—(6),
from the
divergence
166
Chapter 3
Maxwell's Equations
p(x)dx
a: :-<——
I
I
I
I
I
I
I
I
|
I
I
I
I
I
I
I
I
I
I
+3“
3
I
I
|
|
I
I
I
I
I
I
I
|
I
I
I
I
I
I
I
'
”0
d
A
a
I
>
FIG. P3.5c
3.5a In a large class of electron devices, of which the klystron is a good example, current in
the output circuit is induced because of the time-varying conduction current crossing
the output gap. The ac current is superposed on the dc beam and moves across the gap
approximately at the dc velocity 110 of the electrons so that convection current density
in phasor form may be written
Jc(x)
=
10
+
Aerials/”u
Suppose the output gap may be represented by parallel~plane electrodes as in Fig.
P3 .5c. The results of Ex. 3.5 may then be used for the induced current for each elemental slice of length dx, and total induced current for the gap may be obtained by
integrating contributions over the total length d of the gap. Carry out the integration to
find the induced current in the output gap and noticeihow it depends upon transit
angle, aid/v0.
3.63 Check the dimensional
of Eqs.
(1) through (9) of Sec. 3.6.
for continuity of charge islassumed, the two divergence
(2), may be derived from the curl equations, (3) and (4), so far
components of the field are concerned, for regions with finite p and J. This fact
3.6b Show that, if the
equations, 3.6(1)
as ac
consistency
has made it
quite
equation
and
common to
refer to the two curl
equations
alone
as
Maxwell’s
equations.
3.6c Check to
which if any of following could be a field consistent with Maxwell’s
Special condition for p and J is needed, discuss its physical
reasonableness.
see
equations.
(i) B
(ii) E
(iii) E
=
=
=
If
a
220:
f'C / r
f(C/r) cos(wt
—-
wV p.82)
(rectangular coordinates)
(circular cylindrical coordinates)
(circular cylindrical coordinates)
I! 57
Problems
3.7a A
balloon is
conducting spherical
charged with
metric, radially outward propagating electromagnetic
the electric field
happen by finding
radius
at some
r
wave.
>
and its radius
charge Q,
a constant
made to vary sinusoidally in time in some manner from a minimum
maximum value, ram. It might be supposed that this would produce
value, rm“,
a
to a
spherically
sym—
Show that this does not
rm“.
3.7 b A
capacitor formed by two circular parallel plates has an essentially uniform axial
electric field produced by a voltage V0 sin a)! across the plates. Utilize the symmetry to
find the magnetic field at radius r between the plates. Show that the axial electric field
could not be exactly uniform under this time-varying condition.
3.7c
that there
were free magnetic charges of density pm, and that a continuity reEq. 3.4(5) applied to such charges. Find the magnetic current term
that would have to be added to Maxwell’s equations in such a case. Give the units of
pm and of magnetic current density.
Suppose
lation similar to
3.83 Under what conditions can a
magnitude and phase angle,
complex
vector
E
where
E0
is
a
real vector and
3.8b** Consider a case in which the
of magnitude and phase:
Substitute in Maxwell’s
6,;
a
quantity
E be
'0.
E
field vectors
be
can
E
=
1300;,
y,
Z)€j91(x.y.:-)
H
=
H003
y,
z)e 1'91“)“
J
2
J00}
y,
Z)efl93(x.y.z)
p
=
poor,
y,
new-“W
in the
substituting
Re[EOejglej“”]
=
in Maxwell’s
complex form,
3.8c Check to
see
represented by single
(i)
(ii) H
(iii) E
z
=
=
xC e "1"”
V
and separate real and imagiHO,
64. Check the
or,
equations,
V
and 277, and
”8:
3.9(16)
interpret
y,
.
,
2)],
etc.
general time variations, eliminating
equations relating E0,
94.
.
.
.
the time
,
=
plane
wave
a
charge-free region:
a
wt
==
0, 77/4, 77/2,
time. That is,
otherwise. Plot EJr versus
waveshape E,c rectangular in
=
0, l, 2, 3, and
zero
7T/4.
has electric field at
El. versus distance
dispersion. Repeat for a dielectric in
frequency w.
3wt. Sketch
in
wz/v for various times,
traveling wave.
plane wave is excited by
C for I711" < t < (m + 19;).7‘, m
T/4, 3T/4, 5T/4,
distance 2 for t
é’COS
610:,
versus
as a
=
3.9c A uniform
+
.
(rectangular coordinates)
(circular cylindrical coordinates)
(spherical coordinates)
(i)(C/r)e'“j“’
é(C/r)e‘j“’V WW
3.9b A uniform
E,
z) cos[wt
.
for
of
set
"5"
3.9a Plot the sinusoidal solution
37/4,
y,
values
which, if any, of the following could be phasor representations of fields
consistent with Maxwell’s
E
E0(x,
=
equations
variations, and again getting the
vector
EOeJ
__
-
nary parts to obtain the set of differential equations relating ED,
result by using the corresponding instantaneous expressions,
Einst.
a
real scalar?
complex
equations
represented by
for
a
z
cos wt +
0 given as 5,.(0, t)
periods in an ideal dielectric with no
wave velocity at frequency 3w is % that at
=
few
which
=
368
Chapter
Maxwell’s Equations
3
a wave with Ex and
Hy is analyzed. The other set of fields (or the other
polarization as it will be called in Chapter 6) relates E) and Hx. With the same assumptions as to uniformity in the x—y planes, find the Wave solutions for this set in
phasor form.
3.10a In Sec. 3.10
l
that may be considered a uniform plane wave propagates through the
ionosphere and interacts with some charged particles which are moving with a velocity
a tenth the velocity of light in the direction normal to the magnetic field of the wave.
3.10b A radio
wave
Find the ratio of the
magnetic
force of the
the
wave on
charges
to the
electric force.
3.10c Show that Faraday’s law is satisfied in the forward-traveling plane wave with sinuO to z
soidal variations by taking a line integral of electric field from 2
0, x
O,
a to z
a to z
0 back to (O, O) and relating it to the magnetic
x
d, x
d, x
=
=
=
=
=
=
=
=
'
flux
through
that
path.
3.10d Show that the generalized Ampére’s law is satisfied fqr the forward—traveling plane
wave with sinusoidal variations by taking a line integral of magnetic field from 2
O,
btoz
Otoz
btoz
0backto(0,0) andrelatingit
d,y
0,32
d,y
y
=
to
2'
==
=
=
=
displacement current through
that
=
=
path.
3.1121 What are the relations among the constants required for each of the
solution of the three—dimensional Helmholtz equation?
to be a
following
‘
(i) Ex
=
(ii) Ex
2
(iii) Ex
=
kyy sin k22C sinh Kg sin kyy sin kzz
C sinh Kxx sinh Kyy sinh K32
C sin
klx
sin
3.111) Show that the wave equation may be written directly in terms of any of the components of H or E in rectangular coordinates, or for the axial components of H or E in
any coordinate system, but not for other components, such as radial and tangential
components in cylindrical coordinates,
or
any component in
That is,
spherical
coordinates.
.
VZE
=
.Y
['1' a
aZEx
atz
,VZHr
62H
ll. e——:, etc
=
4
atz
but
V2 E
62E
7‘5 pa
3.11c Check to see under what conditions the
tion in circular cylindrical coordinates:
(Note that the
3.123 Describe the
charge Q
vector form of
Poynting
V2
62H
9
1.1.8
523’
is
solution of the Helmholtz equa-
Ff, V‘H¢ ¢
must
following
be used;
see
a
etc.
Prob. 3.11b.)
vector and discuss its
located at the center of
a
small
interpretation for the case of a static point
loop of wire carrying direct current I
.
3.12b
Assuming current density constant over the conductor cross section in the Ex. 3.12a,
find the Poynting vector within the wire and interpret this in terms of the distribution
of dissipation.
3.12c
Interpret
some
some
the
Poynting vector about a parallel-plate capacitor charged from zero to
charge Q. Repeat for an inductor in which current builds up from zero to
final value. Repeat for each of these cases as charge and current is made to definal
1 69
Problems
cay from
given
a
value to
For the
zero.
inductor,
use a
straight
section of wire
as
example.
3.12d Show that the power flow in the uniform plane wave of Ex. 3.12c
of the average energy density and the velocity v of the wave.
3.12e In each of the
following,
plane-wave model
use a
to
estimate the
equals
the
product
quantities for
various
laser systems:
A small helium-neon laser
(i)
1
()t
in diameter. Estimate
mm
=
633 nm) typically produces 1 mW in a beam
of electric and magnetic fields in the laser
strengths
beam.
(ii) It is fairly easy
so
to focus the power for a medium-power CO2 laser ()t
10.6 um)
that there is breakdown in air. Taking breakdown strength as at lower frequen~
cies, about 3
=
106 V/rn, estimate
density in such a laser beam.
(iii) Very high power Nd~glass
(it
pm) have been used in laser—fusion
experiments. Estimate electric field strength at the target for one producing
X
the power
lasers
10.2 kl in 0.9 ns, focused to
3.12f Show that
Eq. 3.12(6)
a
=
1.06
target about 0.5
follows from
Eq. 3.12(5)
mm
in diameter.
for linear,
isotropic,
time-invariant
media.
imaginary part of Eq. 3.13(11) and simplify by letting C,
Ale”!
Azej‘bl where A1, A2, (15,, and 492 are real. Show that the variation with
that on the right side of Eq. 3.l3(8) using time-dependent E and H.
3.133 Find the
C2
=
=
with
3.13b The field
a
large
distance from
a
dipole
radiator has the form, in
A
Ea
:
Find the average power radiated
3.14
Elia (7>e“lk’
a
agrees
spherical coordinates,
.
sin 0
=
through
and
z
large sphere
of radius
1'.
O and 82 filling the halfis filled by two dielectrics, 81 filling the half~space
space x < 0. Determine whether or not there can exist a uniform plane wave with E...
and Hv only and no variations with x or y, propagating in the z direction in this comx >
Space
The propagation factor may be 6““ij with any value of k. Note that
the wave, if it exists, must satisfy the wave equation in each region and the continuity
conditions at the plane between the two regions.
posite-dielectric.
3.1621 Find the variation of
ductor and
3.16b
Repeat
an
average
Poynting
vector for
a
plane
wave
within
a
good
con-
interpret.
Prob. 3.16a for
an
instantaneous
Poynting
vector.
tors of each of these used at 60
conductivity (7, around 107 S/m. For slab conducHz, 1 kHz, and 1 MHz, find the surface resistance of
the two materials if the relative
permeability
3.173 Iron and tin have the
same
order of
of the iron is 500.
magnetic field H for any point .r in the plane conductor in terms of Jo by first
finding the electric field, and then utilizing the appropriate one of Maxwell’s equations
to give H. Show that 15: of Eq. 3.l7(l) is equal to
“Hy at the surface.
3.17b Find the
3.183 The average power loss per unit volume at any point in the conductor is llez/Zoz
Show that Eq. 3.18(5) may be obtained by integrating over the conductor depth to obtain the total power loss per unit area.
3.18b A uniform plane wave of frequency 1 GHz has a power density of 1 MW/m2 and falls
upon an aluminum sheet. It can be shown that upon reflection from a good conductor,
178
Maxwell’s Equations
5
chapter
field at the surface of the conductor is essentially double that in the incident
Estimate the power absorbed in the aluminum per unit area and note it as a
fraction of the incident power.
magnetic
wave.
satisfy the following differential equations
containing charges and currents:
3.193 Show that E and H
medium
V2E
82E
—
”'8
—-
1
=
—
6:2
V p + p“
e
32H
in
a
homogeneous
aJ
—
at
—VXJ
VzH-MSE§=
3.19b A potential function commonly used in elecnomagnetic theory is the Hertz
tential H,
for
a
so
defined that electric and
fields
magnetic
homogeneous medium:
are
derived from it
as
vector po-
follows,
..
H==e-a-V><II
at
E:
59-11
V(V'II)
—
us—,
61"
I
where
V211
and P, the
polarization
vector
2II
—
M8
6—,61*
=
'4'
P
—
a
associated with sources, is
so
defined that
6P
J
=
Show that E and H derived in this
p
_...,
a:
=
manner are
—V=- P
consistent with Maxwell’s
equations.
3.19c An alternative to the Lorentz gauge, which defines V; A by Eq. 319(6), is the Cow
O. Givegthe differential equations relating
lomb gouge which selects it to give V A
(P and A to sources p and J in this case. Discuss problems in use of this apparently
-
=
Note that the equations for Lorentz and Coulomb gauges become identical in the static limit and for charge-free time—varying systems.
simpler gauge.
potentials are generally used only for homogeneous media. Show the
complications in attempting to extend the development to media with p. and a functions of position.
3.19d The retarded
3.20a
with the integral solutions for A and (I), write the integral for the Hertz
11 in terms of the polarization P. (See Prob. 3319b.) Note the relation between
H and A when time variations are as e”.
By analogy
vector
3.20b*
From continuity
of
charge,
find the values of the
charges
that must exist at the ends of
the small current element of Ex. 3.20. Find scalar potential CD from these charges, using Eq. 320(3). Show that Cl) and the A of Eq. 320(8) are related by the Lorentz condition
3.20c
3.19(6).
Using A from Sec 3.20 and (I) from Prob 3.20b, find electric and magnetic fields in
spherical coordinates for the small current element of Ex. 3.20, with sinusoidal current
variation given by Eq. 320(9).
2
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4.l
|NTRODUCTION
Much of the
engineering design and analysis of electromagnetic interactions are done
through
lumped—element circuits. In these, the energy~storage ele~
ments (inductors and capacitors) and the dissipative elements (resistors) are connected
to each other and to sources or active elements within the circuit by conducting paths
of negligible impedance. There may be mutual couplings, either electrical or magnetic,
but in the ideal circuit these couplings are planned and optimized. The advantage of
this approach is that functions are well separated and cause-and-effect relationships
readily understandable. Powerful methods of synthesis, analysis, and computer
optimization of such circuits have consequently been developed.
Most of the individual elements in an electrical circuit are small compared with
wavelength so that fields of the elements are quasistatic; that is, although varying with
time, the electric or magnetic fields -have the spatial forms of static field distributions.
There are important distributed effects in many real circuits, but often they can be
represented by a few properly chosen lumped coupling elements. But in some circuits,
of which the transmission lines are primary examples, the distributed effects are the
major ones and must be considered from the beginning. In some cases in which the
lumped idealizations described above do not strictly apply, lumpednelement models can
nevertheless be deduced and are useful for analysis because of the powerful circuit
methods that have been developed.
We have introduced the lumped-circuit concepts, inductance and capacitance, in our
studies of static fields. We have also seen how the skin effect phenomenon in conductors
changes both resistance and inductance at high frequencies. We now wish to examine
circuits and circuit elements more carefully from the point of view of electromagnetics.
It is easy to see the idealizations required to derive Kirchhoff ’s laws from Maxwell’s
equations. It is also possible to make certain extensions of the concepts when the simplest idealizations do not apply. In particular, introduction of the retardation concepts
shows that circuits may radiate energy when comparable in size with wavelength. The
the mechanism of
amount
of radiated power may be estimated from these extended circuit ideas for
'37?
some
1 72
chapter 4
configurations.
The
Electromagnetlcs of; Circuits
But for certain classes of circuits it
extensions without
a true
field
becomes impossible
We shall look
analysis.
at both types
to make the
of circuits in this
chapter.
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massicai aircuit Theory
The ideafiizaiions 1n
KIRCHHOFF ’s VOLTAGE LAW
4.2
Kirchhoff ’3 two laws
53‘
provide
the basis for classical circuit
theory.
We
with the
begin
voltage
way of reviewing the basic element values of lumped~circuit theory
The law states that for any closed loop of a circuit, the' algebraic sum of the voltages
law
as a
for the individual branches of the
loop'18
I
zero:
f
Zm=0
The basis for this law is
Faraday’s law
for
m
closed
a
path,
written
as
a
~3gE-dl=—JB-dS
at
and the definition of
voltages
(2)
s
between two reference
points
of the
loop,
I)
~fEhl
m
To illustrate the relation between the circuit expressibn (l) and the field expressions
(2) and (3), consider first a single lopp with applied voltage V0(t) and passive resistance,
inductance, and capacitance elements in series (Fig 42(1). A convention for positive
voltage at the source is selected as shown by the plus and minus signs on the voltage
generator, which means by (3) that field of the source is directed from b to a when V0
is
positive A convention for positive current is also chosen, as shown by the arrow on
interpretation of (l) by circuit theory for this basic circuit is then known to be
I(t). The
V00)
—-
RIO)
-
To compare, we break the closed line
several elements:
b
C
an
L
d1
1
-
C
integral of
d
f {(1)
d1
(2): into
a
:
0
(4)
its contributions
over
a
"IE'dl~jE'dl—fE'dl-f.E‘dl=-JB'dS
a
b
c
d5
the
(its
(5)
4.2
192
1 73
Kirchhoff's Voltage Law
b
b
+
Vo(t)
I(t)
C
l
l
I
d
(b)
FIG. 4.2
(a) Series circuit with resistor, inductor, and capacitor. (1)) Detail of inductor.
01'
3
Von)
The
+
+
vcb
+
vdc
Vad
f
=
5;
B
-
d8
(6)
S
(6) is not zero as is the right side of (4), but we recognize it as the
generated by any rate of change of magnetic flux within the path
defined as the circuit. If not entirely negligible, it can be considered as arising from an
inductance of the loop which can be added to the lumped element L, or a mutually
right
side of
contribution to emf
induced
coupling if the
as negligible
consider it
effects
the
are
flux is from
or
added later.) We
now
passive components R, L,
Resistance Element
an
external
included in L
so
examine
source.
that the
the
separately
Thus
we
will from here
on
side of
(6) is zero. (Mutual
three voltage terms related to
right
and C.
The field
expression
to be
applied
to the resistive
material is
the differential form of Ohm’s law,
J
so
that the
voltage Vcb
=
(TE
(7)
is
C
C
Vcb=~fE‘dl=--
b
b
where the
path is
taken
along
some current
flow
J
---d1
path
(8)
or
of the conductor.
Conductivity
a
may vary along this path. At dc or low frequencies, current I is uniformly distributed
over the cross section A of the conductor, which can also
vary with position. Thus
C
Va];
2
I (11
“‘—
"
b
074
=
“IR
(9)
i 74
The
Chapter 4
Electromagneflcs of Circuits
where
6
R
(II
0A
b
This last is the usual dc
at
higher frequencies
or
low-frequency resistance. Thegsituation is more complicated
changing magnetic fields on currents
because of the effect of the
within the conductor. Current distribution
over
the
cross
the conductor must be
and the
particular path along
analysis of Chapter 3, current was
surface impedance. We shall return
of circular
(10)
—
cross
section is then nonuniform,
Specified.
In the
plane skin effect
related to electric field at the surface to define
to this
concept later: in the
chapter
a
for conductors
section.
inductance Element
voltage across the terminals of the inductive element
change of magnetic flux within the inductor, shown in the
a
as
that resistance of the conductor of the coil is negligible,
coil.
first
figure
Assuming
let us take a closed line integral of electric field along the conductor of the coil, returnng
by the path across the terminals (Fig. 42b). Since the contribution along the part of the
path which follows the conductor is zero, all the voltage appears across the terminals:
comes
The
from the time rate of
d
C
~3§E°dl=—f
C(cond.)
By Faraday’s law,
c
E‘dIHJ
E-dl=~f
d(term.)
this is the time rate of
change
a
“J
so
the
at
magnetic flux linkage
the
voltage contributed by
this term,
computing
flux enclosed
aassuming
L
per unit of current
by
the
path,
(Sec. 2.5)
a)
independent of time,
gar-(LI) g
—
(12)
S
“Mn/I
Vcd=
Note that in
B'dS
E'dlz—VdC=‘—g
dam-in.)
as
(ll)
magnetic flux enclosed:
of
"
Inductance L is defined
E‘dl
d(term.)
L
we
add
is
(14)
a
contribution each time
we
follow another turn around the flux. Thus for N turns, the contribution to induced voltage
is
just N times
that of
one
turn,
the calculation of L and will be
provided the same flux links each turn. This enters into
seen specifically when :we find inductance of a coil.
If there is finite resistance in the turns of the coil, the second term of (11) is not zero
multiplied by current; therefore (11) becomes
but is the resistance of the coil, RL,
——-f}€E dlz—RLlfl
=3]st
4.2
Kirchhofi’s
l 75
Voltage Law
01‘
VCd
-
as expected, we simply add
conductivity in the conductors of the
Thus,
RLI + L
511
(15)
dt
another series resistance to take
care
of finite
coil.
fiapacitive Element The ideal capacitor is one in which we store only electric
energy; magnetic fields are negligible so there is no contribution to voltage from changing magnetic fields but only from the charges on plates of the capacitor. The problem
is then quasistatic and voltage is synonymous with potential difference between capac—
itor plates. So, in contrast to the inductor, we can take any path between the terminals
of the capacitor for evaluation of voltage Vda, provided it does not stray into regions
influenced by magnetic fields from other elements. We also take the definition of
capacitance from electrostatics (Sec. 1.9) as the charge on one plate divided by the
potential difference:
C
=
Q
( 16 )
-
V
Thus, from continuity,
I
The last term in
of
(17)
(17) implies
=
a
dQ
——-
d
==
—-
CV
=
C
ch/a
——
17
capacitance which is not changing with time. Integration
with time leads to
Vd”
If
=—-—
( 18 )
Id:
C
capacitor is lossy, there are conduction currents to add to (17),
l/RC in parallel with C;
represented in the circuit as a conductance GC
the value of RC may be calculated from (10) by using conductivity of the dielectric and
area of the capacitor plates.
If the dielectric of the
which
=2
are
induced Voltages from (Miner Parts of the Circuit In addition to voltages
by charges and currents of the circuit path being considered, there may be
induced
particular, if the magnetic field
voltage is produced through
Faraday’s law when this magnetic field changes with time. This coupling is represented
in the circuit by means of a mutual inductor M, as shown in Fig. 4.26. The value of M
is defined as the magnetic flux (1112 linking path 1, divided by the current 12:
induced
from
voltages
one
from other
portions
of the circuit. In
part of the circuit links another part,
an
induced
4'-
M:M17:—1:
_
12
(19)
'5 76
Chapter
4
The
Electromagnetics of Circuits
12
11
12
-—-->
o
+
+
V12
V21
0
((1)
FIG. 4.2
The
(0) Circuit with a mutual inductor. (d) Designation of mutual coupling with negative M.
voltage induced
in the first
path
V12
and the circuit
equation (4)
V0
~—
is then
d9012
:
——
is modified
R11
—-
M
=
dt
to
0%
( 20 )
—“
dt
be
1
dII
div
L ———«—M—:-—
dt
C
dt
J,
(21 )
Idt=0
1
The mutual inductance M may be either positive or negative depending upon the sense
of flux with respect to the defined positive reference for 12. The sign of M is designated
on a
circuit
diagram by
the
shown; those
placing of dots and with sign conventions for currents and
on Fig. 4.2c denote positive M; negative M would be
voltages
designated as in Fig. 4251.
Except for certain materials (to be considered in Chapter 13) there is a reciprocal
relation showing that the same M gives the voltage induced in circuit 2 by time-varying
as
current in circuit 1:
V21
d 11/21
—
d1 1
——
M
dt
All mutual effects to be considered in this
(22)
dt;
chapter have
this
reciprocal relationship.
In summary, we find that if losses in inductor andlcapacitor are ignored, the field
approach, with understandable approximations, leads i:to the definitions for the three
induced
voltage
terms for the
passive
elements used in the circuit
Moreover, the definitions (10), (13), and (16)
these elements. If losses
are
present,
a
are
the lusual
approach, Eq. (4).
quasistatic definitions for
series resistance is added to L and
a
shunt
again as is commonly done in the circuit approach. Coupling between
circuit paths by magnetic flux adds mutual inductanceaelements. We next examine the
Kirchhoff current law and the extension'through this to multimesh circuits.
conductance to C,
4.3
4.3
KIRCHHOFF ’8 CURRENT LAW
AND
The current law of Kirchhoff states that the
junction
is
zero.
l 77
Kirchnofi’s current Law and Multimesh Circuits
Thus, referring
to
algebraic
Fig. 4.30,
MULTIMESH CIRCUITS
sum
of currents
flowing
out
of
a
N
2
n
==
1,,(t)
=
0
(1)
1
It is evident that the idea behind this law is that of
so we
the
its
continuity equation implicit
equivalent:
continuity of current,
equations, Eq. 3.4(5), or
in Maxwell’s
refer to
large-scale
6
%J‘dS=-—~deV
81‘
S
If
(2)
v
apply this to a surface S surrounding the junction, the only conduction current
flowing out of the surface is that in the wires, so the left side of (2) becomes just the
algebraic sum of the currents flowing out of the wires, as in (l). The right side is the
negative time rate of change of charge Q, if any, accumulating at the junction. So (2)
we
may be written
I}; luv):
__dQ____(_t)
(3)
d,
A comparison of (l) and (3) shows an apparent difference, but it is only one of
interpretation. If Q is nonzero, we know that we take care of this in a circuit problem
by adding one or more capacitive branches to yield the capacitive current dQ/dt at the
junction. That is, in interpreting (3), the current terms on the left are taken only as
convection or conduction currents, whereas in (l) displacement or capacitance currents
are included. With this understanding, (1) and (3) are equivalent.
With the two laws, the circuit analysis illustrated in the preceding section can be
extended to circuits with several meshes. As a simple example, consider the low-pass
filter of Fig. 4.31) or 4.3c. Although currents and voltages are taken as time-varying,
in
FIG. 4.30
Current flow from
a
junction.
'3 78
The
Chapter 4
Electromagnetics of aircuits
L1
L2
(b)
(2)
FIG. 4.3
Low~pass
filter:
([7) loop
current
analysis; (c)
node
voltage analysis.
drop the functional notation for simplicity. Figure 4.3b illustrates the standard
12),
utilizing mesh currents I1 and 12. Note that the netzcurrent through C is (11
which automatically satisfies the current law at node b. The voltage law is then written
about each loop as follows:
we
method
-
dll
I
-
1f
"ll—“u“-
The two
equations
are
”-19
dt=0
4
O
(5)
dl
1
“Efaz
I
11)
“'
then solved
dt
‘“
L225:
_
RLI2
Z:
by appropriate means to give I1 and I2 for a given VS.
analysis uses node voltages Va, Vb, and VC as
A second standard method of circuit
shown in
Fig.
4.36. These
are
defined with respect to some reference, here taken
voltage generator, denoted 0. Then Kirchhoff ’s voltage
automatically satisfied, for if we add voltages around the first loop we have
lower terminal of the
VS
+
(Va
—-
VS)
Kirchhoff ’s current law is then
V
Node
a:
-
V
—a-———5
RS
If
Node b:_
L1
+
(Vb
applied
—-
Va)
+
(Os
—
Vb)
a
O
at each of the three nodes as
as
the
law is
(6)
follows:
1
+
~j‘(Va
L1
-
Vb)
If
( V—b
Va) dt+—*
L2
dt
=
O
(7)
dVb
( V-th+C——=O
b
c)
dt
8
()
Node
i-
6:
L2
] (VC
179
Kirchhoff’s Current Law and Multimesh Circuits
4.5
Vb)
~—
KC-
dt +
:
O
(9)
RL
Solution of these
by appropriate means yields the three node voltages in terms of the
no equation for the reference node need be written as it is
given voltage VS.
Note that
contained in the above.
In the above
of
we seem
to be
treating voltage
as a
potential
difference when
we
take
node with respect to the chosen reference, but note that this is only after
voltage
the circuit is defined and we are only breaking up I E d1 into its contributions over
a
.
the various branches. As illustrated in the
whenever there
preceding section,
inductances
we
do have to define the
other elements with contributions to
path carefully
voltage from Faraday’s law.
Finally, a word about sources. The voltage generator most often met in lumpedelement circuit theory is a highly localized one. For example, the electrons and holes
of a semiconductor diode or transistor may induce electric fields between the conducting
electrodes fabricated on the device. The entire device is typically small compared with
wavelength so that the electric field, although time-varying, may be written as the
gradient of a time—varying scalar potential. The integral of electric field at any instant
thus yields an instantaneous potential difference VS between the electrodes, which is the
source voltage (or V,
[ZS if current flows). The induced effects from a modulated
electron stream passing across a klystron gap are similar, as are those from many other
practical devices. There are interesting field problems in the analysis of induced effects
from such devices, but from the point of view of the circuit designer, they are simply
point sources representable by the VS used in the circuits.
A quite different limiting case is that in which the fields driving the circuit are not
localized but are distributed. An important example is that of a receiving antenna with
the fields set down by a distant transmitting antenna. If voltage is taken as the line
integral of electric field along the antenna, applied voltage clearly depends upon the
circuit configuration and orientation with respect to the applied field. Although quite
different from the case with a localized source, it is found that circuit theory is useful
here also. A formulation in terms of the retarded potentials will be applied to this case
are
or
——
in Sec. 4.11.
Current generators are natural to use as sources in place of voltage generators if
emphasis is on the current induced between electrodes of the point source or small—gap
device.
Similarly for the distributed source, if applied magnetic field at the circuit
given, induced current can be calculated and a current representation is
conductor is
natural. One, however, has
theoremsl show that the
two
a
choice in any
of
representations
case
since the Thévenin and Norton
Figs.
4.351 and 4.36
are
equivalent
with
the relations
YS
Thus
‘
an
equivalent
to
Fig.
:
2:13
4.3c is that of
[s
Fig.
:
4.3 f,
(10)
VSYS
utilizing
a current
generator.
8. E. Schwarz and W. G. O/dham, Electrical Engineering: An Introduction. 2nd ed.
Saunders, Fon‘ Won‘h, TX, 1993.
180
Chapter
Electromagnetics of Circuits
The
4
(d)
{e}
(d) Thévenin circuit configuration. (e) Norton circuit form. (f) Equivalent of circuit
(c) using Norton source.
FIG. 4.3
in
’
»
A:——v;
r:
r;
.
‘112“r-~
)Afwr'm't.
(—2:
51:“- 1:
.:;r»
.1
“(45
31/; 'I.
‘
R“3.f}:5&1132552éwswkclg‘mfiflgfiz‘flgflfR““ZKGZZfila-évffi'vfi?)fifiMz’hzsfi‘ngzsfii’fifi'fifigfikfinflmfl nthklfitmvéfk"mgi'lfifi'zigfig‘591i:s‘fiéfic‘axflgfiegig,
..
,
,
.
,
A
.
,
,..
m.
.7.
.9»...
\.
.,
.
v:-
r—r'n
.
~..
Siam Effect in firacticai gonductors
4.4
DISTRIBUTION
To
study
not
uniform,
for
plane
the resistive term at
we
negligible
frequencies high enough
now
wish
that current distribution is
do this for the useful
was
case
done in Sec. 3.16
of round conductors.
good conductor is defined as one for Iwhich displacement
comparison with conduction current so that
V
Faraday’s
to
a
in
so
need to first find the current distribution. This
conductors. We
Recall that
TIME—VARVING CURRENTS IN CONDUCTORS
CIRCULAR CROSS SECTION
OF
OF
law
equation
is
X
H
=
J
=
O‘E
current is
(1)
(in phasor form)
v x E
=
——jmnH,
(2)
4.4
From these two
we
Distribution of
derived the differential
V3.1
We
now
take current in the
(3), expressed
in circular
z
density, Eq. 3.16(7):
jwuch
=
no
(3)
variations with
coordinates
(£21,
for current
equation
direction and
cylindrical
'5 81
Time-Varying Currents
1 dJ-
z or
angle 4). Equation
front cover), is then
(inside
,
(4)
F+;;+T~JZ=O
where
T2
=
*jw/w'
or
1:3
=14” W =j'1/2
T
(5)
where 5 is the useful parameter called “depth of penetration” or “skin depth.” The
differential equation (4) is a Bessel equation. Equations of this type will be studied in
detail in
Chapter 7,
but for the present
J:
For
a
since
study
AJO(Tr)
=
r
+
independent
solutions
as
BHg‘)(Tr)
(6)
=
=
0
=
J:
The
write the two
O is included in the solution, and then it is necessary that B
0. Therefore,
of H 8)(Tr) shows that this is infinite at r
solid wire,
a
we
(7)
AJO(Tr)
=
arbitrary constant A may be evaluated in terms
0E0, with E0 the surface electric field.
of current
density
at
the surface,
which is
J:
Then
=
CEO
at
1'0
(7) becomes
J-—~
E
”0
__
_
J0me)
<8 >
.
10m)
study of the series definitions of the Bessel functions with complex argument shows
that JD is complex. It is convenient to break the complex Bessel function into real and
imaginary parts, using the definitions
A
Ber(v)
.
Be1(U)
43
A
::
real part of
.
Jo(j“/2v)
..__
.
imaginary part of
JOU
1/2
v)
That is,
1004/31»
as
Ber(v)
+
jBei(v)
(9)
182
Chapter
The
4
Electromagnetics of Circuits
Ber(v) and Bei(v) are tabulated in many references.2 Using these definitions and (5),
(8) may be written
J
__
a
‘
0
Ber(\/ir/6)
Ber(\/§ro/5)
+
+
jBeio/Er/a)
j BefiV—irO/a)
( IO )
Fig. 4.4a the magnitude of the ratio of current density to that at the outside of the
plotted as a function of the ratio of radius to outer radius of wire, for different
values of the parameter (7‘0/5). Also, for purposes of the physical picture, these are
interpreted in terms of current distribution for a l-mm-diameter copper wire at different
frequencies by the figures in parentheses.
As an example of the applicability of the plane analysis for curved conductors at
high frequencies where 6 is small compared with radii, we can take the present case of
the round wire. If we are to neglect the curvature and apply the plane analysis, the
coordinate x, distance below the surface, is (rO
r) for a round wire. Then Eq. 3.l6(16)
In
wire is
—
gives
.1.
.-
~—
%
e
—(fo-r)/5
< 11 )
0E0
In
4.427
Fig.
made with
1‘0/8
=
are
plotted
curves
2.39 and
curves
of
lJz/O‘EOI by using this formula,
and
comparisons
are
obtained from the exact formula
2‘0/6
=
(10). This is done for two cases,
7.55. In the latter, the approximate distribution agrees well
with the exact; in the former it does not. Thus, if ratio; of wire radius to 8 is
seems
for
that there should be little
plane
solids. This
point
4.5
will be
in
the wire from the results
analyzing
pursued in impedance
error
large, it
developed
calculations to follow.
IMPEDANCE OF ROUND WIRES
The internal
impedance (resistance and contribution to reactance from magnetic flux
wire) of the round wire is found from total current in the wire and the electric
at
intensity the surface, according to the ideas of Sec. 4.2. Total current may be obtained
from an integration of current density, as for the plane conductor in Sec. 3.17; however,
it may also be found from the magnetic field at the surface, since the line integral of
magnetic field around the outside of the wire must be equal to the total current in the
inside the
Wire:
2
H. B.
Dwight. Tables
Bessel Functions for
7955. M. R.
of Integrals, 3rd ed., MacMi/Ian. New York, 7967. N. W. McLacnlan.
Engineers, 2nd ed., Oxford University Press (Clarendon). New York,
Spiegel. Mathematical Handbook of Formulas
Series, McGraw—Hill, New York. 7968.
and Tables, Schaum’s Outline
’0
0.239
=
5
1-0
(f
=
383
impedance of Round Wires
4.5
103
Hz for
l~mm~diam Cu wire)
1'0
M—
i9-
=
Actual
=
-~—
1a
1:.
Jo
Jo
0.5
I63
:3
2.39
(f
=
105
Hztor
l—mm-diam Cu wire)
:
(f = 106 Hz for
lumm—diam Cu wire)
755
Outer
radius
Wire
axis
Outer
radius
Outer
radius
Wire
axis
R»
(a)
FIG. 4.4
Parallel plane
formula
0-5
'"
’TO
"-
104 HZ for
0.755 (f
l~mm—diam Cu wire)
(a) Current distribution in cylindrical wire for several frequencies. (b) Actual and
distribution in
approximate (parallel-plane formula)
cylindrical
wire.
J0
_.
0E0.
or
277”0H¢ir==ro
Magnetic
field is obtained from the electric field
V
For the round wire with
and
only
r
no
E
X
variations in
derivatives remain,
so
expression
(I),
z or
Maxwell’s
equations:
(2)
the fields
E3
and H 4,, alone
dE,
‘
=
are
present,
(3)
—-—-—
jar/J,
dr
for current
density has already been
through
conductivity 0:
obtained in
Eq. 4.4(8).
Electric field
the
J0 (Tr)
J-
E:
By substituting
by
~jwlLH
==
l
is related to this
(1)
(2) is simply
H 4”
An
I
=
in
z
.2
0
(3) and recalling that T2
__.
‘15
where J ’O(Tr) denotes
=
15a]: J’0(Tr)
__
‘
JO(T70)
-—jw,u,0',
-959 J’O(Tr)
T
jam, J0me)
[d/d(Tr)]JO(Tr).
I
z
(4)
:2
0‘
From
10(Tr0)
(1),
‘27TI'00‘E0 J ’O(Tr0)
T
J0(Tro)
(5)
.384
Chapter
The internal
impedance
The
4
per unit
of: Circuits
Electromagnetics
is defined
length
Z,-
as
13
T] T‘
2,.
Note the
similarity
to internal
in
impedance per square
For low
low-Frequency Expressions
sions of the Bessel functions show that
Then
'
——-——°(,'°)
27rr00‘J 0(Tr0)
=
E200) / I.
(6)
.
Eq. 3.17(3).
frequencies, Tro is small
expanded as
and series expan-
may be
(6)
2
1
2.8
'
The real
or
771%0‘[
1
1+——~
r0
(5)]
+
—-
48
,cop,
7
()
—~
1877
resistive part is
,2
17
Rifz
The first term of this
useful for
term of
(7) corresponds
to a
low-frequency
48
8
that is, for radius
low—frequency
(Li)lf
The
359
(8)
is the dc resistance, and the second is
expression
I‘D/5 as large as unity,
1 +
771‘50‘
—1—
“
equaltto
skin
depth 8.
a
correction
The
imaginary
internal inductance:
Si;
internal inductance is the
H/m
same as
(9)
that found by
energy methods in
Sec. 2.17.
High-Frequency Expressions For high frequencies, the complex argument Tr0 is
large. It may be shown that JO(TrO)/J'0(Tr0) approaches j and the high-frequency
approximation to (6) is
—
(z.1)hf
_
_.
jaw/2 =<1+j>RS
——-—
Vi’lTl‘OU'S
2777.0
o /m
(10)
01'
(Rn
=
(«0me
So resistance and internal reactance
to
the values for
Sec. 3.17 where
a
Rs
plane
:
are
solid of width
Wm
(11)
equal at high frequencies, and both are equal
27m, just as assumed on physical grounds in
Expression for Arbitrary Frequency
Eq. 4.4(9),
27770
(05)”.
it is useful to break into real and
defined in
R5
=
To
interpret (6) for arbitrary frequencies,
imaginary parts using the Ber and Bei functions,
and their derivatives. That is,
Berv +
jBei
v
=
Jo(j“‘/2v)
4.5
185
Impedance of Round Wires
Also let
d
Ber’
+
v
j Bei’
v
M
55
(6)
+
v
j
Bei v)
1“” IOU—”20)
N
Then
(Ber
may be written
‘
R
13
Z.=R+ij,.=-
B
+‘
[erg 1136161]
Vim-0
Ber’ q +
jBei’
q
where
R “‘
.....
—1_
“
7170/1,
_
Vil'o
q
08
5
(r
or
R
R3
:
\/—2_7TI‘O
(”Li
I
“I
B
~
e12
(Ber Q)"
'
B
q]
:|
elf] er
(B31 Q)"
+
Ber q Ber' q + Bei q Bei’ q
RS
:
[
B erq B
7
I
Warm
(Ber 4)"
7
‘I
G351 (1)"
+
Q/m
(12)
Q/m
These
are the expressions for resistance and internal reactance of a round wire at any
frequency in terms of the parameter g, which is W times the ratio of wire radius to
depth of penetration. Curves giving the ratios of these quantities to the do and to the
high—frequency values as functions of I'D/5 are plotted in Figs. 4.5a and 4.5b. A careful
study of these will reveal the ranges of 1-0/5 over which it is permissible to use the
approximate formulas for resistance and reactance.
4'0
r
*F
t
/
3.0
X? /
2.0
¢<~e
L
”u
_-‘
1.0
L—
0
o
B
A
1
659.
FIG. 4.50
2
L
3
g
l
J
4
a
(
t
0
—|
l
5
6
7
Ratio of radius to depth of penetration
Solid~wire skin effect
quantities compared
with dc values.
336
Chapter
4
~21.
FIG. 4.5b
Electromagnetics of Circuits
4
2
0
The
5
10
8
Ratio of radius to depth of
Solid—wire skin effect quantities
12
14
penetration
compared with values from high frequency formulas.
aaicuiatmn of flircunt Eiements
4.6
SELF-INDUCTANCE CALCULATIONS
Chapter 2, was related to field concepts in the first part
examples of inductance calculations for simple configchapter.
urations by the method of flux linkages (Sec. 2.5) and: from an energy point of View
(Sec. 2.17). We now give additional examples of each method.
Self~inductance,
of this
as
defined in
We have shown
Example
4.6a
EXTERNAL INDUCTANCE OF PARALLEL-WIRE TRANSMISSION LINE (APPROXIMATE)
Figure
4.6 shows two
parallel
conductors of radius R with their
distance 2d. Current I flows in the
in the other.
Magnetic
z
direction in the
right~hand
axes
separated by
conductor and returns
field at any point (x, y) is the superposition of that from the two
are far enough apart, the current distribution in either con-
conductors. If conductors
ductor is not much affected by the presence of the other, so that magnetic field from
each conductor may be taken as circumferential about its axis and equal to the current
divided by 211- times radius from the axis. For the y
0 plane passing through the axes
=
'I 87
Self-Inductance Calculations
4.6
FIG. 4.6
Parallel-wire transmission line.
of the two wires, the contribution from both wires is vertical
I
H,x,0
>0 )
The
so
that field, to the
described above, is
approximation
I
+
~
277((1
27m
x)
+
(1)
x)
—~
magnetic flux between the two conductors (used in finding external inductance) is
by integrating over this central plane. For a unit length in the z direction,
then found
“"‘R’
I
LL.—
1,0,"
1
1
dx
+
27f
d +
w(gr—R)
d
x
-—
x
(2)
M1
Inductance per unit
length
2d
rpm
IL
I
277
spacing
+
x)
-
111(d
‘“
x)](3(dI?—)R)
is then
-
R
R
pt
11
n
2d
R
EXTERNAL INDUCTANCE
When
..
2—7; [1n(d
:2
OF
-
R
2d
In
1
R
7T
(3)
Example 4.6b
PARALLEL‘WIRE TRANSMISSION LINE (EXACT)
between conductors is
comparable
with wire radii, current distribution
in the wires is affected and the result obtained above is modified. It
by
a
method of
that the exact
Eq. 1.6( 1)]
and
can be shown either
images or by conformal transformations to be described in Chapter 7
magnetic flux function (11”, [analogous to electric flux function in
scalar magnetic potential CI) for this problem are
In
Wm
(pm
=
,uJ
“'—
477
__L
272'
1H
(x
(x
[tan
-
+
a)2
7
a)'
+
y2
y
-1
__
(x
-
(4)
7
+ y“
a)
tan
y
--I
(x
+
Q]
(5)
138
Chapter
Electromagnetics of Gil-cults
The
4
where
a
Taking
d2
=2
the flux difference atx
Alpm
=
l//m(€ll
d
2'
13,0)
‘*
R2
-
R andx
-—
lullm(—d
“
“2
~d + R (both at )2
d
#1
In
—~———
(6)
R +
--
2
~—d + R
a
-~—
d
477'
R
-—
-
-
(1
In
-d .+ R +
a
a
mmde~Vd2~R2
flld—Rua
__
0),
R, 0)
+
2
-
==
77nd~R+a
7
()
duR—q—VdZ—Rz
7r
By multiplying numerator and denominator by [(d
—-
R)
—-
Va?2
-
R2], this reduces
to
A rpm
=
d
#1
~—-ln
d
—-
1
—
pJ
=
—
R
R
W
1’-
-—
--
._
cosh
77
1
d
(8)
—
R
so
L
=
Ail!
J
I
=
,u,
-
7Tcos
h-I
d
(R)
(9)
—
Examme 4. 6c
PLANE CONDUCTOR
__
INrERNAL INDUCTANCE
As
third
we
assumed. Moreover,
we
basis,
as
WITH
SKIN EFFECT
utilize the energy method and calculate internal inductance.
from that of Ex. 2.17,for which uniform current distribution was
example,
example differs
a
This
OF
phasor forms of the skin effect formulation. The
equation of the circuit forrnsof energy storage to the field
utilize the
in Sec. 2.17, is the
form:
1
..
L12
2
The
[.L
=
H?- dV
—
v
( 10)
2
magnetic field distribution for a semi~infinite conductor with sinusoidally varying
found to be [Eq. 3.l6(15)]
currents was
:
___.
e—(1+j).t/:5
where the coordinate system of Fig. 3.16a13 used. The; current per unit width, J
just the value of Hy at the surface:
Jsz
:2
~Hy(0)
:
USEO
(1
+
j)
(11)
_,
is
(12)
'3 89
Mutual Inductance
4.7
We may now apply (10) to the calculation of L. But first we recognize (10), as written,
is for instantaneous I and H. To use with phasors, we must either convert to instantaneous
forms
procedure
is
or
write the
simpler
equivalent
and
of
(10) for time~average stored energies. The latter
find
we
1
pa
| l
l I
—L12=f—H3dv
4
v
where the factor of fig rather than
is on
of sinusoids.
so
Taking
a
(11) and (12) in (13),
width w,
we
< 13 )
4
each side
from the time average of squares
miss, and a length l, and substituting
comes
that current is
obtain
w
L
_
0’25ZE3
4
7
M r.
w ,1
z
2
0
__
,u
025253
4
2
e
a.
-—_.\/8
t
a"-
or
m
L
&1
__
w
“Zr/5 (It
15
ILL—[
_
7
2w
0
15
L
#5100
2w
so
I
(Liz—.939.
20
w
where relations for skin
depth
2
l
R l
w/.L0'
14205
w
___:__~_=—5
and surface
resistivity
(14)
have been substituted from Secs.
3.16 and 3.17. As found there, the internal reactance per square is equal to surface
resistivity, RS. This is multiplied by length l and divided by width w to give the internal
reactance
of the overall unit.
MUTUAL INDUCTANCE
4.7
The mutual inductance
in
one
was
defined in Sec. 4.2
circuit due to current
approaches
to
its calculation,
flowing
some
as
that
arising from
in another circuit. We
of which may also be
the induced
now
applied
voltage
discuss several
to
calculation of
self-inductance.
Flux Linkages The most direct approach is that from Faraday’s law, finding the
magnetic flux linking one circuit related to current in the other circuit, as in Eq. 4.2( 19).
Thus for two circuits 1 and 2
we
write
M12
2
f3]
'
312
dsl
(1)
.1
where
32
magnetic flux arising from current 12 and integration is over the surface
By reciprocity M21
M12 (for isotropic magnetic materials), so the cal»
is the
of circuit 1.
2
.399
Electromagnetics of Circuits
The
Chapter 4
culation may be made with the inducing current in either circuit. Consider, for example,
the two parallel, coaxial conducting loops pictured in Fig. 4.70. The magnetic field from
a
current
in
one
loop has
been found for
32(0, d)
a
point
[11202
=
enough compared with spacing d,
(1) gives
loop
rrbZBz(O,
—_
be relatively
constant over
-
>
#7703192
d) -._
<3)
*
W
I.
approach
this will
and the relation
The exact formula is found
(2)
W
If loop 2IS small
the second
the‘ axis in EX. 2.3a:
on
by integrating the field
by another method.
over
the
section, but
cross
we
will
the exact calculation
V X A, application of Stokes’s
Magnetic Vector Potential Since B
theorem to ( 1) yields an equivalent expression in terms of the magnetic vector potential:
use of
==
M
=
In (V
X
An'dsI
12
Circuit 1
=
sfiAz-dll
4
()
[2
lfi
d6
dll
(a)
/
/'
/
I
a}
02
’
(zf/
r—-——————-
—!
b1
// ///
55//
b2
F—i‘
L—
//’I
I
////
/
I,
/
/
I/z”’
,IB //
[I //
(2..
C
d
—>l
a
(b)
FIG. 4.7
elements
,
(a) Two circular 100ps. (b)
displaced from
one
another.
Two
(c)
rectangular coupling loops. (c)
Parallel current
4.7
191
Mutual inductance
This form is useful in any problem for which the vector potential is more easily found
than the magnetic field directly. It is especially useful for problems in which the circuit
has
straight-line segments
or can
be
such segments,
approximated by
of the
coupling of square loops pictured in Fig. 4.71). Vector
direction of the current element contributing to it by Eq. 2.9(5),
A from horizontal sides (21 and
mutual inductance only through
2, 02 and [22. Similarly, vertical
as
in the
potential
so
problem
A is in the
the contribution to
horizontal. These sides thus contribute to
is
only
191
integration by (4)
over
in c1 and
the horizontal parts of circuit
contribute to mutual inductance
d1
only by integration over the two parallel (vertical) sides 62 and (12. The basic coupling
element in such a configuration is then that of two parallel but displaced current elements as pictured in Fig. 4.76. The contribution to mutual inductance from such elements
(Prob. 4.7d)
currents
be shown to be
can
+A
Mz—china
+dlna
+A
d+D
c+C
417
(5)
+bln
[29+]?A]+(C+D)~(A+B)}
a—l—
This
as
point
well
as
of view is
for
quite useful for qualitative thinking
quantitative analysis.
about
couplings
Another standard form for calculation of mutual
Neumann’s Form
filamentary circuits follows directly from the above. We
arising from current in circuit 2, assuming that current
neglecting retardation.
in
a
coupling of two
potential A
write the vector
to be in
line filaments and
1. d1
A2
3g [Ul a-TR
2
where R is the distance between current element
dlg
circuit
(6)
and the field
point. Substitution
in
(4) yields
1
[L17 d";
‘
dll
[1.»
477'
47rR
12
This standard form is due to Neumann. Note in
procity
relation
M12
==
M21,
since
integrations
§%
dll
particular
‘
dlr)
( )
R
its illustration of the reci—
about circuits 1 and 2 may be taken in
either order.
Example
4.7a
MUTUAL INDUCTANCE OF COAXIAL LOOPS BY NEUMANN'S FORM
For the coaxial
of
loops
Fig. 4.70,
let
dlI
(16
cos
be any element of circuit 1 and
(112
be any
element of circuit 2. Then
dlI
‘
dl2
=2
dlza
6
(8)
392
Chapter
R
By substituting
6
=
7r
Va??—
=
2gb
—-
The
4
integral-(7)
(61
cos
6
-
b)2
(9)
401)
d2
+
+
(a
b)2
( 10 )
‘
will then be found to become
7r/2
my“V—k
can
+
=
__
which
6)2
(a sin
+
and
k2
the
Electromagnetics of Circuits
be written
(23in2 41—1)qu
[0
V1
—-
( 11 )
k2 sin2 qb
as
=
,n/ZEK— k>K(k) Eats]
(12)
-—
~—
where
77/2
E(k)
V1
=
-—~
k2 sin2 95 dqb
(13)
o
”/2
K(k)
M
=
V1
—
(14)
lc2 sin2 ()5
integrals (13) and (14) are given in tables3 as functions of
complete elliptic integrals of the first and second kinds, respectively.
The definite
called
Example
k and
are
4. 7b
SELF-lNDUCTANCE OF ClRCULAR LOOP THROUGH
IVIUTUAL INDUCTANCE CONCEPTS
Neurnann’s form does not appear useful for the calculation of self-inductances of filamentary current paths, since radius R in (6) becomes zero at some point in the integration
for such filaments. For
pictured in Fig. 4.7d,
culating induced field
a
conductor of finite area,
one
however,
as
in the round
loop
of wire
obtains the external contribution to self-inductance
at the surface of the
conductor, say through the
vector
by calpotential
is nearly
(6). If wire radius a is small compared with loop radius 1', this field
though current were concentrated along the center of the wire. Thus we
conclude that the external inductance of the loop is well approximated by the mutual
A
as
the
3
in
same as
For
R.
example, H. B. Dwight, Tables of Integrals, 3rd ed, Macmillan, New York, 796 l; or M.
Splegel, Mathematical Handbook of Formulas and Tables, Schaum’s Outline Series,
MoGraw—Hi/l, New York, 7968.
(9)
(d)
FIG. 4.7
loops,
one
I93
Inductance of Practical Coils
4.8
((1) Conducting loop for which external self-inductance is to be found. (e) Filamentary
through center of wire and other along inside edge, for which mutual inductance may
be calculated.
inductance between the two filaments of
Fig.
4.7a.
ductance between two concentric circles of radii
L0
=
9
_~
__
,u(2r
4r(r
r
Utilizing (12) for the mutual
a) we then have
and (r
a)l:<l -::>K(k) E00]
-
—-
in~
-
—-
(15)
(2)
-
"
(2r
where
E(k) and K(k)
unity,
and K and E may be
are as
—-
(1)2
defined
by (13) and (14).
approximated by
If
a/r
is very small, It is
nearly
4
K(k)
E
E(k)
.2.
LO
E
ln(fi)
1
so
To find total L, values of internal
4.8
ru[ln<§f) 2]
inductance,
-—
as
(16)
found in Sec. 4.5, must be added.
lNDUCTANCE OF PRACTICAL COILS
study of the inductance of coils at low frequencies involves no new concepts but
only new troubles because of the complications in geometry. Certain special cases are
simple enough for calculation by a straightforward application of previously outlined
methods. For example, for a circular coil of N turns formed into a circular cross section
(Fig. 4.80) we may modify the formula for a circular loop of one turn, Eq. 4.7(16),
A
394
Chapter 4
The
Electromagnetics of Circuits
L;
__..J
l
0000000000000
(6)
(6)
FIG. 4.8
(a) Coil of large radius—to-length ratio. ([7) Solenoidal coil. (6) Solenoidal coil
permeability
on
high-
core.
provided the cross section is small compared with the coil radius. Magnetic field must
be computed on the basis of a current N]; in addition, to compute the total induced
voltage about the coil, N integrations must be made about the loop. Equation 4.7(16)
is thus modified by a factor N 2. The external inductance for this coil is then
'
L0
8R
N~Rn[m<?> 2]
,
=
—-
For the other extreme, the inductance of
computed.
constant,
If the solenoid is
as
long enough,
the
(1)
very long solenoid (Fig. 4.81)) may be
magnetic field on the inside is essentially
a
for the infinite solenoid,
H,
=
-~—
<2)
4.8
195
Inductance of Practical Coils
where N is the total number of turns and l the
length.
The flux
linkage
for N turns is
then N sznHz, and the inductance is
L0
=
'n'p.R2N2
—l—
(3)
For coils of intermediate
frequently
to the
length-to-radius ratio, empirical or semiempirical formulas
Nagaoka formula applies a correction factor F
long solenoid.4 A simple approximate form5 very close to
have to be used. The famous
formula (3) for the
R/l up to 2 or 3 is
this for
=
0
If
a
coil is wound
toroidal
7T].LR2N2
(4)
——————
1 + 0.9R
of
high permeability as shown in Fig. 4.86, the
region, independent of the length of the winding.
is
intensity again given by (2) and the inductance by (3) with l
flux
on a
is restricted to the
core
essentially
magnetic field
2771b: where rO is the mean radius of the toroid.
At higher frequencies the problem becomes more complicated. When turns are rela—
tively close together, the assumption made previously in calculating internal impedance
(other portions of the circuit so far away that circular symmetry of current in the wire
is not disturbed) certainly does not apply. Current elements in neighboring turns will
be near enough to produce nearly as much effect upon current distribution in a given
The
=
turn as the current
ductance
since
core
are
in that turn itself. Values of skin effect resistance and internal in-
then not
changes
as
previously calculated. External
inductance may also be different
symmetrical distribution
in external fields result when current loses its
separation of internal and external
given field line may be sometimes
inside and sometimes outside of the conductor. Finally, distributed capacitances may
be important and further complicate matters (see following section).
Coils utilizing superconductors, which are materials giving zero resistance below
some critical temperature near absolute zero, have become important because one can
obtain with proper design very high values of uniform magnetic fields with them, with—
out the use of iron. They may also be very efficient devices for storage of large energies.
The electromagnetic principles of design are the same as given for other coils, and in
fact, the approximations may be better satisfied by the thin wires typically used in
superconducting magnets. The mechanical forces of the large currents must be consid»
ered in the design, and the transient behavior of a superconductor is very different from
that of an ordinary conductor. Wilson6 gives examples of various coil configurations,
with reference to the background literature.
with respect to the wire axis. In fact, the strict
inductance may not be possible for these coils, for
4
5
‘5
a
E. C. Jordan (Ed). Reference Data for Engineers: Radio, Electronics. Computer, and
Communications, 7th ed, Howard W. Sams, Indianapolis, IN, 1985.
H. A. Wheeler, Proc. IRE 16. 7398 (7928).
M. N. Wilson, Superconducting Magnets. Clarendon Press, Oxford, 7983.
396
Chapter
4.9
The
4
SELF
Electromagnetics ofiCircuits
AND
MUTUAL CAPACITANCE
The concept of electrostatic capacitance between two conductors was introduced in
Chapter 1 as the charge on one of the conductors divided by the potential difference
analysis of Sec. 4.2 this definition was carried over
as a quasistatic concept to give
capacitance term utilized in the analysis of
Little
more need be said about the simple two—
excitation.
circuits with time-varying
is
collect
but
it
useful
to
conductor capacitor,
expressions we have developed for some
between conductors. In the circuit
the usual
of the
common
1. Parallel
capacitive
planes
with
elements.
negligible fringing, A
C
area, d
2-“
=:
spacing:
fl
“‘
"
(1)
a
2. Concentric
spheres
of radii
a
and b
(b
>
a):
47TSCZb
C
(b
3. Coaxial
cylinders
of radii
a
and b
"
(2)
F/m
(3)
_.
(b
>
-
a)
a):
2778
C
--
"
1n<b/a)
4. Parallel
in
cylinders
Chapter 7):
with wires of radius a, with
are
separated by
d (to be derived
i
C
If there
axes
=
—7T8——
1
cosh
(d/ 2a)
"
F/rn
several conductors, the electric flux from
several of the others and induce
one
(4)
conductor may end
on
charge on each of those. Consider, for example, the
multiconductor problem diagramrned in Fig. 4.9a. Suppose conductor 1 is raised to a
positive potential with the other three bodies, 0, 2, and. 3, grounded. The electric flux
from 1 will divide among the other three bodies and induce negative charges on each
of these. The amounts of the separate charges may be used to define capacitances C10,
C12, and C13 in the circuit representation of Fig. 4.919. Similarly, raising conductor 2 to
a nonzero potential and finding induced charges on grounded conductors 0, ,1, and 3
determines C20 and C23 and provides a check on C12. Repetition of the process with
conductor 3 at a nonzero potential gives the remaining element C30 and provides a
check on 6'13 and C23. However, it is usually not possible to measure the individual
charges on the conductors which are tied together. Usually a capacitance current, dQ/dt,
is measured by applying a time-varying voltage, and Q is the sum of the charges on
electrodes connected together. Thus the three measurements described would yield
(C10 + C12 + C13), (C20 + C12 + C23), and (C30 + C13 + C23). Three additional
measurements with linearly independent combinations of V1, V2, and V3 are required to
determine the six elements of the circuit.
197
Self and Mutual Capacitance
4.9
C12
C31
023
Clo
{3
2
3
C20
C30
(17)
Zero potential
electrode
9
((1)
//”~“\\
/
/
/’
/
\
\
/
~~~~~~
1Q
\
\
I
\
~~~~~~~~~
\\
\
wwwwww
\\\
{c}
’u—n—..
I
(l
It}
\\
I
I
l
l
\
l
t
////
\
\\\
\
“
\‘x““““““““
/
a”?
(d)
’’’’’
/’—~_~\\\
I
n—n
/// / /
’//
/
/
\
\
/
2
\
\\
/
/
\\
.a—N
0/
\
l
3
Q
g\\
4
Li
91.
2
13:;
I
l
\
1
l
J
\\
~~~~~
\:\\
‘
~~~~~~~~~
\
\
/
\
‘i
I
(e)
5:3
\
1/ l‘. \L\L
\
r
4)
2
07
FIG. 4.9
(a) Four conducting bodies, one of which is chosen to have zero potential. (1)) The
equivalent circuit for (a). (C) Electrostatic shielding by a grounded sphere. (d) Flux lines between
a pair of conductors without shielding. (e) Partial shielding by a grounded conducting plane.
(f) Ungrouncled nearby conductor increases coupling.
problem is that of decreasing the capacitive coupling between two bodies,
electrostatically shielding them from one another. Consider, for example, the
conductors 1 and 3 of Fig. 4.96. If a grounded conductor 2 is introduced and made to
surround either body 1 or 3 completely, as in Fig. 4.9a, it is evident that a change in
potential of 3 can in no way influence the charge on 1 so that mutual capacitance
0.
C13
More often the added conductor may not completely enclose any body, so that the
capacitive coupling may not be made zero, but may only be reduced from its original
A
common
that is, of
=
value.
Any
finite conductor,
as
2, introduced into the field
acts to
decrease the mutual
T 93
Chapter 4
The
Electromagnetics of Circuits
prior to the introduction of 2, and hence provides some
capacitive coupling between 1 and 3. The reason is that fewer of the
flux lines of the charge on 1 will terminate on 3 with a grounded conductor, as shown
by comparing Figs. 4.9d and 4.9g. However, if 2 is not connected to the ground (the
infinite supply of charge), the effect of the added electrode will be to shorten the flux
lines as seen in Fig. 4.9;“. In terms of the equivalent circuit, the effective capacitance
between 1 and 3 is seen from the equivalent circuit of Fig. 4.91) to be given by C13 in
parallel with C12 and C23 in series:
capacitance C12
from its value
decrease in the
(C13)eff
7‘
C13
+
C12C23
C12
+
C23
generally greater than the value of C13 prior to the introduction of 2 (though
along an equipotential surface of the original field); so, if insulated
from ground, the additional conductor may act to increase the effective capacitive coupling between 1 and 3. It often happens that electrodes, although grounded for direct
current, may be effectively insulated or floating at high frequencies because of imped»
ance in the grounding leads. In such cases the new electrodes do not accomplish their
shielding purposes but may in fact increase capacitive coupling.
This value is
it need not be if 2 lies
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Grunts Which
4.IO
Not
Smahégempared
Wavetength
are
With
DISTRIBUTED EFFECTS AND RETARDATION
generalizations to circuit theory when effects are distributed rather
comparable in size with wavelength so
lumped,
that retardation from one part of the circuit to another must be considered. Considering
first the distributed effects, we recognize that the fields contributing to circuit elements
are always distributed in space and the representation by a lumped element is valid only
when the region is small in comparison with wavelength and when only one type of
energy storage (electric or magnetic) is important for that region. If the electric energy
storage in parts of a primarily inductive element, or magnetic energy in a primarily
capacitive element, becomes important, the approach through classic circuit theory is
to divide into subelements that can be treated as one or the other. For example, suppose
there is electric field (capacitive) coupling between the turns of the inductor of Fig.
4.10a. A first approximation is that of adding a capacitive element across the terminals
of L to represent all the electric energy storage of the element as shown in Fig. 4.1019.
A still better approximation is that of adding a capacitive element between each pair
We
now
than
consider the
and also when circuits become
4.10
Wit
FIG. 4.10
Distributed Effects and Retardation
(a)
(b)
(c)
(d)
199
(a) Coil. (b) Circuit with single capacitance representing electric-field coupling
among turns. (6) Circuit
turn.
of
representation with capacitive coupling shown between each adjacent
((1) Representation with capacitances added between nonadjacent turns.
in
Fig. 4.10c. But there may be coupling between nonadjacent
capacitances can be added as in Fig. 4.10:1. The effect of these at
high frequencies is to bypass some of the turns so that not all turns have the same
current. This last effect would not be at all included in the simpler representation of
Fig. 4.1017. Finally one might go to the limit and consider differential elements of the
coil, attempting to find couplings to all other differential elements, to write and solve
a differential equation for current distribution. This process could be carried out only
for simple configurations, and even the approach through a finite number of lumped
elements as in c or (1 becomes complicated if there are many turns.
Consider next the retardation effect arising from the finite time of propagation of
electromagnetic effects across the circuit. To simplify this discussion, we consider only
sinusoidal excitation so that we can define a wavelength and discuss phase relationships.
More general excitations can of course be broken into a series of sinusoids through
Fourier analysis. Consider, for example, the simple single-100p antenna of Fig. 4.108.
At low frequencies, with diameter d small in comparison with wavelength, the time of
propagation of effects from one part of the loop to another is negligible. Thus, magnetic
field produced by a current element at a point such as A travels to another point such
as B in a negligible part of a cycle and so has negligible phase delay. The induced field
from the time rate of change of the field is then 90 degrees out of phase with current
in B and contributes to the inductive effect we expect for the loop at low frequencies.
adjacent turns,
turns and
as
still other
233
Chapter
4
The
Electromagnetics of Circuits
(e)
FIG. 4.10
at B
(6) Loop
when d is
antenna
comparable
showing phase retardation between
wavelength.
sources at
A and induced fields
with
higher frequencies, with d comparable with wavelength, the finite time of propa—
gation about the circuit must be considered. Current at B may then not be in phase with
current at A, and the magnetic field at B arising from the element at A may not be in
phase with either. The time rate of Change of magnetic field induces an electric field
which may then be not exactly 90 degrees out of phase with 13. If there is an in-phase
At
component, it represents energy transfer, which
turns out to
be
a
contribution to the
energy radiated by this antenna. If current distribution is known, fields throughout the
circuit can be calculated and the contribution to radiated power represented in the circuit
by
so—called radiation resistance. But to find the actual
distribution,
really
boundary-value problem represented by
conducting loop. For
some antennas or other circuits comparable in size with wavelength, it is possible to
make reasonable assumptions about current distribution and extend circuit theory in
this way, but the extension must be done carefully. Additional discussion of this point
will be given in the next section utilizing a retarded potential formulation for circuit
theory.
One important circuit having both distributed and propagation effects is the uniform
transmission line. It turns out that circuit theory can be extended to this case. Agreement
with field solutions is exact for perfectly conducting transmission lines and very good
for lines with losses, as will be seen in Chapter 8. The circuit theory of transmission
lines, to be developed in Chapter 5, is thus of very special importance.
a
needs to solve the
4.11
CIRCUIT FORMULATION THROUGH
current
one
the
THE
RETARDED POTENTIALS
relationships embodied in the retarded potentials of Sec. 3.19 can
provide additional insights into the circuit formulation? for electromagnetic problems,
especially for circuits large in comparison with wavelength. This approach was first
The cause-and~effect
Circuit Formulation
4.“
used
by Carson.7
start
with Ohm’s law in field form for
A
typical
Through
circuit follows
a
a
201
the Retarded Potentials
conductor for all
point along
this
or
part of its path,
so we
path,
E=*
where
cr
moves
about the circuit
(1)
is the
conductivity for the point under consideration and may vary as one
path. We next break up the field into an applied portion, E0,
and an induced portion, E’, the latter arising from the charges and the currents of the
circuit itself. We also write E’ in terms of the retarded potentials of Sec. 3.19
E0+E’=EO-V(P—-a—é==~J—
at
(2)
0'
where A and C1)
are
given
as
integrals
over
the
charges
and currents of the
circuit,
as
defined in
Eqs. 320(3) and 320(4).
The term J / 0' in (2) is indeterminate
nonconducting portions of the path since
insulating portions and 0- is generally undefined within any
localized source. We consequently integrate (2) over conducting portions of the path,
obtaining a cause~and—effect relationship which can be considered the general circuit
equation:
both J and
0- are zero
over
for
A
[ED-dl-fJ-'dl—f§—-dl~fV©-dl==0
at
0'
In
a
conventional circuit, the first term is
the third
an
(3)
applied voltage, the second a resistive term,
a capacitive term. The terms are discussed
inductive term, and the fourth
separately.
Applied Voltage The first term of (3) can be identified as the applied voltage of
circuit theory and is just the integral of applied electric field over the circuit path. In a
circuit such as a receiving antenna (Fig. 4.1141), the applied field is clearly distributed
over the circuit through the mechanism of the incoming electromagnetic wave and the
integration of E0 is about the complete path:
n=fmai
(e
For the localized sources, discussed in Sec. 4.3, for which electric field can be con~
gradient of a scalar potential, the integration of E0 from 2 to 1 about the
sidered the
Fig. 4.1 lb is the negative of that from 1 to 2 of the source since the closed
integral of the gradient is zero. The gap in the capacitor can be ignored in this step
since the localized source produces negligible field there. Thus the source voltage is
circuit of
line
2
1
1 (source)
-(circuit)
7
J. R. Carson. Bell
System Tech. J. 6,
I (1927).
232
Chapter
The
4
Electromagnetlcs of circuits
Incoming E-M radiation
LocaHzed
generator
2
+
4 H
Y?
i
C
(b)
(a) Closed filamentary loop excited by incoming electromagnetic wave. (b) Filamen~
a localized generator (point source). (c) Circular loop of
round wire with circuit path along inner boundary.
FIG. 4.11
tary circuit with capacitor excited by
In this class of
problem, V0 is independent of the circuit path, whereas in the receiving
problem discussed above, V0 depends; very much upon the circuit
antenna class of
configuration
and orientation with respect to the fields.
internal impedance Term The second term in (3) is of exactly the same form as
the ohmic term for the resistor in the circuit example of Sec. 4.2. There we showed that
in the limit of dc this
corresponds
to the
expected resistance term.
Here it is understood
may vary over different parts of the circuit path and the integration brings in the
total resistance of the circuit path. For ac circuits it turns out that this term may also
that
0‘
include
circuit
a
contribution from the inductance internal to the conductor
is taken,
along
for the round wire in Sec. 4.5. Thus for the
which the
important
phasor representations for currents and voltages, this term gives a
complex contribution resulting from internal reactance in addition to the resistance.
That is, if internal impedance per unit length is defined-as the ratio of surface electric
path
sinusoidal
case
with
as was seen
4.1 1
Circuit Formulation
203
the Retarded Potentials
Through
field to the total current in the conductor,
Zi
=
—S
(6)
the term under consideration becomes the total internal
impedance Z,- multiplied by
current I :
fg-dl=fEs'dl:IfZ}dl=IZi
where the
integrals are taken
Fig. 4.1lb.
over
the
conducting portions
(7)
of the circuit from 2 to 3
and 4 to 1 in
Externai inductance Term
the circuit
The third term in
(3) is the inductance
term
and, if
is
properly selected, represents only the contribution from magnetic flux
external to the conductor. Consider, for example, the loop of wire in Fig. 4.11c, and
take the circuit path along the inner surface of the conductor. We will assume that the
integral in the third term of (3) taken over the conducting portions of the circuit differs
negligibly from an integral which would include the small gaps at the source and in
any capacitors included in the circuit. This allows evaluation of that term with closed
integrals. We take the path as stationary so that
path
aA
3g
——
.
(ii
=-
dfig
--
A
-
dl
d:
at
(8)
From Stokes’s theorem,
3gA-dlzf(VXA)~dS
(9)
5
But
V
A
X
(10)
B
=
so
aA
——
.
cl!
(If
=
-
dr
at
B
'
d8
< 11 )
5
integral of (11) is the magnetic flux linking the chosen circuit, exactly as
approach through Faraday’s law in Sec. 3.2. Thus the term may be defined as an
inductance term, as before, recognizing that it is the contribution from flux threading
the chosen circuit path (i.e., the external inductance):
The surface
in the
d1
aA
is:
~—
Thus this
provides
an
alternate way of
d1
-
=
L-—
dr
calculating
( 12 )
inductance:
l
L=7§Audl
(13)
234
The above
neglected.
assumes
Electromagnetics of Circuits
The
Chapter 4
the circuit small
The extensions when this
compared with wavelength so that retardation
assumption is not valid are discussed shortly.
is
fiapacitive Term As with the other terms in (3) we must integrate the WP over the
conducting portions of the circuit, that is, from 2 to 3 and 4 to l in Fig. 4.1119. Here
we assume that“ the fields arising from charges on the capacitor are negligible at the
source, so we may use, as the range of integration, 4 to 3 through the source. Then,
since the integral of the gradient of a scalar completely around a closed path (here
including the capacitor gap) is zero, we may write
4
3
3(gap)
4(circuit)
a lumped capacitor this potential
capacitance C
In
difference is
related
to
the
charge Q through
,
(I) 3
so
that this term is the
capacitance
fl3 4
——
( 15 )
-
C
term of circuit
3
J.4(<:ircuit)
Q
=
theory,
Q
1
C
C
J
VCD-dlz—z—Jdt
( 16 )
Circuits Comparable in Size with Wavelength The formulation in terms of
retarded potentials has been shown to reduce to the usual low-frequency circuit concepts
as
obtained in the earlier formulation
using only fields,
under the
same
assumptions.
The present formulation is attractive in that it appears more readily extendible to large—
dimension circuits, such as an antenna, when retardation effects are important. To i1~
lustrate, consider the circuit of Fig. 4.110, for which
in
a
thin wire. Let
capacitor
also take
so
us assume
that there is
steady~state
only
current is assumed concentrated
that ohmic resistance is
an
and
applied voltage
phasor notation
sinusoids and
a
negligible
and that there is
term from the
potential
no
A. We
for this discussion. Thus,
(3)
becomes
ngo'dl—jwng-dlr-O
and
A, for the filamentary current, by Eq. 321(4), is
A
Substituting (18) in (17)
we
(l7)
=
i
11.13"ij
dl’
( 18 )
477R
and breaking up the
exponential into its sinusoidal components,
obtain
'
fEO-d1~jw3€3€“l(c°s
[(12
-—
477R
‘
”m kR
)dl-dl’r-O
(19)
We
see
that
even
if current I
205
Circuits with Radiation
4.12
were
assumed
values of kR would lead to both real and
entirely in phase about the circuit, finite
imaginary parts of the contribution from this
term. The
imaginary part is the inductive reactance, as found before for this term, except
by the integration of the retardation term. But there is a new term in
phase with I which corresponds to the energy radiated from the circuit, and can be
expressed as current times a radiation resistance.
Although the general modifications for large—dimension circuits are shown by this
approach, it is difficult to carry much further since we really do not know the distribution
of I about the circuit, and cannot find it without a field solution of the problem. In some
antennas it is possible to make reasonable guesses about the current and proceed, but
it is clear that these guesses must eventually be checked through either experiment or
a field analysis. Also, as we have seen, the integration is to be only over conducting
surfaces to avoid the indeterminacy of the second term of (3). For many antennas, the
“gaps” are larger than the conductors, and fields definitely not quasistatic in the open
regions, so this further limits the applicability of this approach. A specific example will
be carried further in the following section.
now
modified
4.12
CIRCUITS
WITH
RADIATION
relationship between field theory and circuit theory,
specific circuits with radiation. As we saw in Sec. 4.11, a circuit that
is not small in comparison with wavelength has retardation of induced fields from one
part of the circuit to the other. The resulting phase changes produce components of
induced field which are in phase with the currents, and an average power flow results.
This power can be shown to be the radiation from the circuit. The phase shifts also
produce some changes in the reactive impedance of the circuit, but this is usually a
higher—order effect. The term we are concerned with is the integration of induced effects,
the second term of Eq. 4.1 1(19):
To conclude this discussion of the
let
us
look at two
_
V.induced
=
We illustrate this with two
.10)
ti
,uI(cos
kR
-
j sin kR)
(1)
477.}?
examples.
Example
4.12a
SMALL CIRCULAR LOOP ANTENNA 0R CIRCUIT
The first
example
is that of a circular
loop,
as
in
Fig. 4.120,
small
enough so that current
may be considered constant about the loop. If I is independent of position, it may be
taken outside the integral (1) so that the induced term may be written
Vinduced
:
(R2
+
ij)I
(2)
296
Chapter
The
4
Electromagnetics of Circuits
(b)
(a)
((2) Circular loop with constant current I with coordinates for calculation of retardation
(b) Straight antenna of finite length with current distribution.
FIG. 4.12
effects.
where
Rr
=
L
The value of CH is
(d)
-
¢')
(ha dd),
wavelength, kR <<
terms of its Taylor series:
27"
my, sin H?
——
sin[(q§
,u
cos
(3)
H?
——-—
dl
-
(11’
477R
——
(h’a dqfi’.
qb’)/ 2].
1 and the sine term in
2”
d1» d1'
477R
and that of dl’ is
and the distance R is 2a
with
=
36%
if;
The
angle
between (11 and dl’ is
Ifthe circuit is small in
comparison
(1) may be replaced by the first
two
k3R3
M l. m Mina.
The first term of
(4) integrates
two terms of the
series. The second term
271'
277
to zero, so we see
R,
z
I f
M
readily
evaluated to
0
The
integrals
are
2477‘
o
__
R,
=
.
length (but
1
_.
sin2<¢ 4’)
2
cos(<b
~—
qb’) dqs drb’
understanding
<5)
give
k341/2
$0“ flu -<—)<ka)4
Thus radiation resistance increases
with the
it is necessary to retain at least
gives
4
__
why
(6)
—
as the fourth power of the ratio of radius to wave
0.05 A, the
that this ratio is always small). For a
=
201
Circuits wllh Radlaflon
“a
value is
R,
If
kR in the
cos
12211
:
(21r
x
0.05)“
for L is likewise
expression
NZHCL
L~LL 47rR|:1
J21?
:
expanded
1.923 n
as a
series,
EL“.
+
--rrcos(q$
7
+
(7)
-1
¢)d¢dd)
41
2!
,
(8)
The first term is
recognized as the Neumann form for inductance of this loop (Sec. 4.7),
remaining terms represent corrections to the inductance because of retardation.
It is seldom necessary to calculate these last-mentioned corrections for circuits properly
considered as lumped~element circuits
and the
Example 4.12b
RADIATION RESISTANCE
As
a
second
current
OF A STRAIGHT ANTENNA BY
example, consider
straight dipole
a
antenna as
CIRCUIT METHODS
shown in
Fig. 4.1217
with
distribution
1(2)
=
(9)
1,..f(:)
where
f(z) is real. But here the conductor does not form a closed circuit, and as ex—
plained earlier. the Carson formulation (Sec. 4.ll) only applies unambiguously over
the surface of conductors. Thus we first find retarded potential A, which has only a 2
component:
/
A.
it];
'
given in
terms
~1w[A
of A
=
=
+
,
c
f(z )e
I'LL"
—‘—-
4742
—;
7
2']
7
d.’
( 10)
by Eq. 321(7):
1
,
E
dz
,
471R
—I
Electric field is
I
—’k.R
1
——
:-
FWV
1
63A
m] —]a)z[A: 1371—2]
,
-
A
+
=
(11)
The portion of E in phase with current causes the radiated power in this picture, and
that clearly comes from the imaginary part of A? The integration of l(z)Ei,,,pmS= over
the antenna
terms
of
a
gives the total
power transferred.
radiation resistance:
or
radiated. and this may be expressed in
I
W
Thus, substituting (10)
It.
_2&l
~
4”
and
=
2
f [Itszgmhrs
(11),
we
m
I R.’
:
—~"'2
<12)
obtain
’
if"
_
f(2)
Lift: ){r—' \
,.
T—u
:1sz
w
298
Chapter
4
The Electromagnetics of circuits
expansions of the sinéklz
z’l terms are often made.
and
cos k2], R, is found to
f(2)
/\/4
dipole [l
be about 73.1 9 in agreement with the value found by a Poynting integration to be
utilized in Sec. 12.7. This method of finding radiation resistances of antennas is called
the induced emf method. It is seldom easier than the Poynting integration but does show
the relationship to circuit theory.
Note that for both examples, we had to assume a form for current distribution to
proceed: This is a clear limitation as it can be done with reasonable confidence only in
specific cases. When that is not possible, field theory must be invoked for the whole
problem.
Note also that until this example, we have neglected any distributed charges along
the interconnecting conductors of the circuit. Here, with current varying as f(z), the
continuity equation requires distributed charges, but these are taken care of by the
V(V A) term in (11), as shown in Sec. 3.21.
In
evaluating these integrals,
series
—
When carried out for the half~wave
=
=
:
’
PROBLEMS
4.23
nonlinear elements, that is, ones for which ,u., e, or a, or some
combination is a function of the field for at least a part of the circuit. Review the formulation of Sec. 4.2 to show this behavior explicitly. Is the general form Eq. 4.2(1)
Many circuits contain
changed
in such cases?
4.2b Some circuits contain
a function of time for
time-varying elements,
at
least
a
for which ,u, a,
or 0', or a
part of the circuit. Discuss these
combination is
in Prob. 4.2a.
cases as
sign of mutual inductance coupling is designated on a circuit diagram by the placing of black dots. With the sign convention for positive voltage and current shown in
Figs. 4.26 and d, the dot location in the former denotes positive M and in the latter,
negative M. Show that either can be represented by a “T~networ as in Fig. P426,
where the upper signs denote Fig. 4.26 and the lower, Fig. 42d.
4.2c The
”
L21|M|
+
+
iIMI
V1
V2
0*
4)
FIG. P4.2c
4.33 It has been
out that the mesh
analysis utilizes Kirchhoff ’s voltage law expliconly implicitly. Show that the current law is satisfied for each
node of the circuit with mesh currents defined by Fig. 4.31). Similarly show that the
voltage law is satisfied by each mesh of Fig. 4.30, with node voltages as shown.
itly but
pointed
the current law
4.3b The generator in the example of Figs. 4.3b and C is-taken as a
series with a source resistance. It can alternatively be taken as
parallel
with
and node
4.3c With
a
a source
equations
conductance
Gs.
voltage generator
a current
in
generator IO in
Make this substitution and write the
new
loop
for the filter.
complex load impedance (admittance)
connected to the
source
terminals
as
in
209
Problems
Figs.
4.351 and 6, show that the two source representations are equivalent in producing
in and voltage across this impedance, when the conditions of Eq. 4.3(10) are
current
sadsfied.
4.3d Show that for fixed
when it is
a
V5 and ZS, the maximum possible power is
“conjugate match” to Z5, that is, ZL
Z“: (or YL
=
delivered
==
the load
to
Yf).
4.4a Make
a power series expansion of
Eq. 4.4(8), retaining up to quadratic tenns, to Show
the variation of magnitude and phase with r to this approximation. Up to about what
1-0/6
will this be
a
reasonable
approximation? (See
4.4b Utilize the
Sec. 7.14.)
of Bessel functions to derive the
asymptotic expansions
expression Eq. 4.4(11). What phase variation
(See Sec. 7.15 or Ref. 2.)
is found in this
4.4c Obtain tables of the Ber and Bei functions and
2.39.
r/rO for I'D/5
plot phase
approximate
approximation?
of current
density
versus
=
4.53 Show that the ratio of very high frequency resistance to dc resistance of a round
conductor of radius 1‘0 and material with depth of penetration 5 can be written
Rm
R0
4.5b
the
Using
approximate formula 4.5(8),
R0 by less than 2%.
from dc resistance
"0
25
find the value of
I‘D/5 below
To what size wire does this
which R differs
correspond
for
copper at 10 kHz? For copper at 1 MHz? For brass at 1 MHz?
4.56“ For
two
z-invariant systems
ductors of the
rent
densities
same
equal
having
the
in
magnitude
at
same
shape
of
cross
section and of
good
con—
distributions will be similar, and cursimilar points, if the applied voltage to the small
material, show that
current
system is 1 /K in magnitude and K2 in frequency that of the large system. Also show
that the characteristic impedance of the small system will be K times that of the large
system under these conditions. Check these conclusions for the case of two round
wires of different radii. K is the ratio of linear dimensions
(K
>
1).
4.63 For the symmetric
R / d from both the
the
parallel-wire line, plot normalized external inductance, irL/p. versus
approximate and exact expressions and note the range over which
approximate formula gives good results.
4.6b Derive the formula for external inductance of the coaxial line in Fig. 2.41) by the en—
ergy method assuming the usual situation of a material with permeability #0 between
the electrodes.
4.6c For
a
parallel-plane
equals
as in Fig. 2.5c, find the dielectric thickness, in
thickness, for which the low-frequency internal inductance
transmission line
terms of the conductor
the external inductance. Take both conductor thicknesses to be the
same.
a solid copper inner conductor of radius 020 cm and a
tubular copper outer conductor of inner radius 1 cm, wall thickness 0.1 cm. Find the
total impedance per unit length of line for a frequency of 3 GHz, including the internal
4.6d A coaxial transmission line has
impedance
4.6e
of both conductors.
a differential length of coaxial transmission line. Take the lines
Fig. 3.17 to be separated by a differential distance dz with 2 positive
to the right. Write Faraday’s law for the loop ABCDA and use the capacitance expression given in Eq. 1.9(4) to show that the equivalent circuit shown in Fig. P4.6e is correct for high frequencies. (Li is internal inductance per square and L.: is external
inductance per unit length.)
Equivalent
circuit for
C -—B and D~A in
2'3 0
Chapter 4
The
Electromagnetics of Circuits
Lib
(—27% Lia)
+
22:0.
dz
Ledz
C
Rsb
V+PA(27:?)
+
RSa)d
z
thedz
27rd
tub/a
0——
T
VT.03
w
B
FIG. P4.6e
encountered problem in microwave circuits is the wire bonding of one
circuit to another as suggested in Fig. P4.6f. The configuration is usually too
complex to fit any simple models, but some useful approximate formulas exist. F. E.
Terman, Radio Engineers Handbook, McGraw-Hill, New York, 1943, gives, for round
l + 51/226], where {f and d are
wires at high frequencies, L
0.20€[ln(4€/d)
length and diameter in millimeters, and L is in nanohenries. Estimate the inductance of
4.6f A
frequently
part of
a
-
=
the 0.5»mm—diameter wire bond in
Note its magnitude compared with
Fig. P4.6f and calculate its reactance at 1.0 GHz.
a typical characteristic impedance of 50 .Q.
l/Emm
W
V4f//[/rf[f/////1
Z
FIG. P4.6f
4.73 A coaxial line, shorted at 2
0, has a rectangular 100p introduced for coupling, lying
in a longitudinal plane with dimensions as shown in Fig. P4.7a. Find the mutual in»
ductance between loop and transmission line assuming (1 <</\ so that field is
essentially independent of z.
==
lrffl/f/flljfffr/T/I'Y/l/////////////a
,
1
r2
T.
fiW/WW/A\
\.
rf/f/fllllljll/llf/fl/f/fl/ff/rfIf/f
FIG. P4.7a
4.7b From tables of the
complete elliptic integrals given in
Eq. 4.7(12) against d/a for b/a
mutual inductance in
4.7c
the references, plot the form of
1 and for 17/0
=
==
—
1,
properties of the complete elliptic integrals for k << 1 and for k
approximate expressions for mutual inductance for these two cases. Inter~
pret physical meaning of these limits and compare with approximate Eq. 4.7(3).
Investigate
the
z
and obtain
4.7d
By integration of Eq. 4.7(4), show
parallel line segments displaced as
that the contribution to mutual inductance from two
shown in
Fig.
4.7c is
as
given by (5).
23 II
Problems
4.7e
Apply Eq. 4.7(5) to the calculation of mutual inductance between
used for coupling between open-wire transmission lines as shown
length of each side is 0.03 m; the separation x is 0.01 m. Assume
which the lines enter are small enough to be ignored.
4.7f Plot
two
in
square loops
4.7b. The
Fig.
that the gaps at
for the circular loop of round wire versus a/r from approximate and “exexpressions and note the range of usefulness of the former. Comment on the validity of the selected mutual approach for a/r approaching unity. (Note that tables of
elliptic functions are required for this comparison.)
LO/ap.
act”
4.7g Suppose
l~mm~diameter copper wire is formed into a single circular loop having a ra~
cm. A voltage generator of 1 V ms and 10 MHZ is connected to an inflni~
dius of 10
tesimal gap in the loop. Find the current flowing in the loop, taking into account internal impedance as well as external inductance. Justify all approximations used.
4.711?“ Check
dipole
4.83 Plot
Eq. 4.7(3) by taking the magnetic field of the small loop as that
integrating flux from this over the area of the larger loop.
of
a
magnetic
and
LO/uR
versus
expression 4.8(4)
add this
curve
R / I from the
expression
for
and compare. (If you have
a
long
access to
solenoid and the
tables for the
empirical
Nagaoka formula,
also.)
ideal infinite solenoid, magnetic flux is uniform everywhere inside the solenoid.
For a finite coil, as pictured in Fig. 4.81), there may be more flux through the central
turns of the coil than those near the ends. Explain how you make a circuit model of the
4.8b For
an
coil in view of these
“partial
flux
linkages.”
circular coil of square cross section, Fig. P4.8c, it has been shown that the larg—
1.5 for a fixed length of wire of chosen
est possible inductance results when R / s
1.7 X 10‘6 RN 2.
size. The value of this inductance for N turns is L
4.8c* For
a
=
=
“‘1
WA
FIG. P4.8c
212
Chapter
4
The
Electromagnetlcs ofE Circuits
1 m of wire with cross section 1 m2, find the values of R, s, N, and
inductance for its maximum according to this rule;
(i) Given
(ii) Repeat for
4.8d
2
m
of the Same wire.
the formula of Prob. 4.8c with that of
Compare
(areaW2
=
Eq. 4.8(1), taking
for the latter
R/1.5.
qualitatively the case of Fig. 4.96 in which twp bodies, 1 and 3, which are
relatively far apart have a grounded conducting plane brought in their vicinity. Give
energy argument to show that C13 is decreased when the plane is added.
4.93 Discuss
4313*
Suppose
a
that the bodies 1 and 3 of
distance d with
a/ d
<<
plane when grounded,
from ground.
4.9a and f are spheres of radii a separated
plane is parallel to the line joining their
by
1. If the added
centers and distance b from it
the
Figs.
an
(a/b
<<
1), find C 13 before and after introduction of
and the effective
CI3
with the
plane present
and insulated
cylindrical conductors of radius 1 cm have their axes 4 cm apart and each axis is
above a parallel ground plane. Make rough graphical field maps and'estimate the
capacitances (per unit length) for this three-conductoriproblem. (Think about how
many plots you need and the best choice of potentials for each.)
4.9c Two
4
cm
4.103 Make
coil
as
wire 1
order-of-magnitude estimate of the capacitance between adjacent turns of a
pictured in Fig. 4.10a if the diameter of the coil is 2 cm, the diameter of the
mm, and the spacing between turns 1 mm. Clearly state your model for the
an
calculation.
4.10b For
a coil as in Fig. 4.100, in the form of a fairly open helix, it is found that the phase
delay of current along the coil is well estimated by assuming propagation along the
wire at the velocity of light in the surrounding dielectric material. For a helix of 100
turns in air, each turn 1 cm in diameter and spaced 1 mm apart on centers (wire
diameter being appreciably smaller than this),
(i) Find the phase difference between current at the end and that at the beginning of
150 MHZ.
the helix for such a traveling wave at f
the
in
the
this
with
difference
retardation term, calculated along a
(ii) Compare
phase
=
direct
path
4.11 We will find
though there
possible.
between the two ends.
waves on an
is
clearly
infinite ideal transmission line that do not radiate
retardation
to
different
points along
the line.
Explain
even
how this is
4.12a What radius do you need to give a radiation resistance of 50 Q from the expression
derived from the small~loop circuit?~Do you think the approximations reasonably
satisfied for this size?
4.12b
Using
the method of Sec. 4.12 derive radiation resistance for
sides d and uniform current assumed about the
4120* As indicated in Ex. 4.12b, make
gral (retain three terms),
A/4.
dipole,l
=
assume
a
series
f(z’)
=
expansion
cos
a
small square
loop
with
loop.
for the sine terms within the inte-
kz’, andaestimate R, for the half~wave
5.1
INTRODUCTION
In
Chapter 3, we saw that the interchange of electric and magnetic energy gives rise to
propagation of electromagnetic waves in space. More specifically, the magnetic
fields that change with time induce electric fields as explained by Faraday’s law, and
the time~varying electric fields induce magnetic fields, as explained by the generalized
Ampere’s law. This interrelationship also occurs along conducting or dielectric boundaries, and can give rise to waves that are guided by such boundaries. These waves are
of paramount importance in guiding electromagnetic energy from a source to a device
or system in which it is to be used. Dielectric guides, hollow-pipe waveguides, and
surface guides are all important for such purposes, but one of the simplest systems to
understand—and one very important in its own right—is the two-conductor transmis—
the
sion line. This system may be considered a distributed circuit and so is useful in estab~
lishing a relation between circuit theory and the more general electromagnetic theory
equations. The concepts of energy propagation, reflections at
discontinuities, standing
traveling waves and the resonance properties of standing
and
waves, phase
group velocity, and the effects of losses upon wave properties are
these transmission-line results to the more general classes of
extended
from
easily
structures.
guiding
A parallel two~wire system is a typical and important example of the transmission
lines to be studied in this chapter. In any transverse plane, electric field lines pass from
one conductor to the other, defining a voltage between conductors for that plane. Mag—
netic field lines surround the conductors, corresponding to current flow in one conductor
and an equal but oppositely directed current flow in the other. Both voltage and current
(and, of course, the fields from which they are derived) are functions of distance along
the line. In the two following sections we set down the transmission~line equations from
distributed-circuit theory, but then discuss its relation to field theory.
Transmission~line effects are not always desirable ones. A cable interconnecting two
high-speed computers may be intended as a direct connection, but will at the very least
introduce a time delay (around 5 ns / m in typical dielectric~filled cable). Moreover, if
the interconnections are not impedance matched at the two ends there will be reflections
of the waves (as we shall see later in the chapter). These “echoes” of pulses representing
expressed
in Maxwell’s
versus
213
2'9 4
Chapter
Transmission Lines
5
complication is dispersion. In a
some degree with frequency,
velocity
so the frequency components which represent the pulse (by Fourier analysis) travel at
different velocities and the pulse distorts as it travels. If dispersion is excessive, the
pulses may be blurred enough so that individual digits cannot be clearly distinguished.
All of these effects occur also in the interconnections of elements in printed circuits
and even in semiconductor integrated circuits, but the close spacings in the last case
limit performance only for extremely short pulses.
Transmission-line analysis is useful, by analogy, in studying a variety of wave phe»
the
could introduce serious errors. A still further
digits
of propagation varies to
real transmission line the
nomena, such
the
as
pr0pagation of acoustic waves and their reflection from materials
properties. An especially interesting analog is that of the pro—
with different acoustic
of
pagation
signals along
a nerve
of the human
body.
Time and Space fiependence of Signafis
an ideal transmission lines
5.2
We
VOLTAGE
AND
CURRENT VARIATIONS ALONG AN IDEAL TRANSMISSION LINE
begin by considering
positely
the transmission line
as a
distributed circuit. In Sec. 2.5
we
inductance per unit length associated with: the flux produced by the opdirected currents in a pair of parallel conductors. When the currents vary with
identified
an
time, there is
a
voltage change along
the line.
Likewise, the distributed capacitance
between the conductors when the
voltage is time-varying
produces displacement
and leads to change in the current flowing along the conductors. The interrelationship
leads to the wave equation for voltage and current along an ideal lossless transmission
current
line.
Figure
5.2 shows
a
representative two-conductor line and the circuit model for a
kept in mind that the external inductance per unit length
It should be
differential
length.
parallel—conductor line is not associated with one conductor or the other. Also, the
circuit model is simply a representation of a differential length of line; there is not a
of
a
one—for-one identification of the two sides of the circuit with the two conductors of the
modeled line.
Consider
length,
‘
a
differential
length of line dz, having the distributed inductance, L per unit
capacitance, C per unit length.1 The length dz then has
and the distributed
Note that,
as
in
Chapter 4,
the
same
symbols are used here for inductance and capaccapacitance. Some texts use I and c
confusion with length and light velocity,
itance per unit length as for total inductance and
for the distributed quantities, but there is then
respectively.
5.2
Voltage
and Current Variations
Along
/
1/
/
y
!
7
l
FIT—dz—d
/
25 5
//
/
L
ideal Transmission Line
/
//
/
l
an
V
fifiz
/
/
//
I
f
FIG. 5.2
Section of
a
representative transmission
line and the
equivalent circuit for a differential
length.
inductance L dz and
equal
capacitance
C dz. The
change
in
voltage across this length is then
change of current. For such
of this inductance and the time rate of
to the
product
length, the voltage change along it at any instant may be written
length multiplied by the rate of change of voltage with respect to length. Then
a
differential
av
voltage change
2
E
as
the
61
dz
:
-(L dz)
5}-
(1)
Note that time and space derivatives are written as partial derivatives, since the reference
point may be changed in space or time in independent fashion.
Similarly,
the
that is shunted
distance is
derivatives
along the line at any instant is merely the current
capacitance. The rate of decrease of current with
capacitance multiplied by time rate of change of voltage. Partial
change
across
in current
the distributed
given by the
are again called
for:
61
current
The
length
change
dz may be canceled in
(1)
=
-—
6V
dz
=
82
and
at
(2)
81
=
-L
-~
at
62
(3 )
8V
61
82
—
(2):
6V
—
-(C dz)
=
-C
—
at
<4)
equations for the analysis of the
they are identical in form with the pairs, Eqs. 3.9(5)
and 39(9) or Eqs. 3.9(6) and 3.9(8), found from Maxwell’s equations for plane electromagnetic waves. As was done there, (3) and (4) can be combined to form a wave
Equations (3)
and
(4)
are
the fundamental differential
ideal transmission line. Note that
21 6
Transmission Lines
Chapter 5
equation for either of the
respect
to distance and
variables. For
example, one
-L
=
7
62V
62]
—C
=
(6)
are
the
same
(6 )
61‘2
at a:
may be substituted
(5)
62 at
62“
Partial derivatives
taken in either order
in
directly
differentiate (3) partially with
821
62V
so
can
to time:
(4) with respect
(assuming continuous functions)
(5):
62V
62V
:
9
L
—
62V
1
=
~77-
:2
62"
U~
(7 )
?
at“
where
v
All real
signals
are
=
(LC)"V2
continuous functions,
as
(8)
required for (7) to apply.
The discontinuous
step waves used later as examples are to be understood as approximations to real signals.
Equations (3) and (4) are known as the telegraphist’s equations, and the differential
equation (7) is the one~dimensiona1 wave equation. A similar equation may be obtained
in terms of current by differentiating (4) with respect to z and (3) with respect to t, and
combining the results:
621
1
321
9)
g=a?
We
saw
in Sec. 3.9 that
an
equation
V(z, z)
of the form
(7) has a solution
Fl(t —) Fz<t 3)
:
"+
+
__
(10)
U
U
z/v) would be seen
F2 are arbitrary functions. A constant value of F10
z/v) represents
by an observer moving in the + z direction with a velocity D, so F1(z‘
a wave traveling in the + z direction with velocity v. Similarly, F20 + z/v)
represents
a wave moving in the —~z direction with velocity v.
To find the current on the line in terms of the functions F1 and F2, substitute the
expression for voltage given by (10) in the transmission—line equation (3):
where
F1
and
—
—
—La—-—1F't-—z- +—1-F’t+-Zat
This
expression
may be
”LU
were
v
integrated partially
1—iFit
If this result
2
v1
—3
v
v
v
with respect to
—-F21‘
(11)
t:
+3
v
substituted in the other transmissionrline
+
f(Z)
equation (4),
12
()
it would be
5.2
Voltage
and Current Variations
Along
found that the function of
integration, f(z), could only be a constant. This
superposed
studying the wave solution, so
will be ignored. Equation (12) may then be written
is
dc solution not of interest in
1
I
=
217
Ideal Transmission Line
an
z
a
possible
the constant
z
Zoi 1< v) 2< ”)1
~—
F
[-
_.
F
-
r+
(13)
—
where
L
Z0
The constant
and is
seen
ZO
from
as
defined
=
—
Q
1
(4)
C
by (14)
(10) and (12)
Lu
=2
is called the characteristic
impedance of the line,
for a single one of
The negative sign for the
wave propagates to the left, and by our
to be the ratio of
voltage to
given instant.
current
the traveling waves at any given point and
negatively traveling wave is expected since the
convention current is positive if flowing to the right.2
.y'.'.l';.:'-'/M5.
Example
CHARACTERISTICS IMPEDANCE
Let
us
find
expressions
AND
is
some
large enough to neglect
Eq. 2.5(6) in (14) we find
typical
where
a
and b
respectively.
are
A
impedance
values. We will
internal inductance.
Z0
2
.:
-L':.~YI..>:,-§y-n\1<2.."'.:~.':-'/XI
1
WAVE VELOCITY FOR A COAXIAL L|NE
for the characteristic
coaxial line and examine
A
5.2
1 Z7
M
271'
and
wave
assume
C from
Using
velocity
for
the conductor
Eq. 1.9(4)
an
ideal
spacing
and L from
\f'g’
(15)
e
the radii of the inner and outer conductors at the dielectric surfaces,
common
dielectric of relative
commercial coaxial cable is
2.26 and the radii
designated RG58/U.
0.406
and b
It has
a
1.48
permittivity
51.6 0.
Substituting these values in (15) and taking it E #0 one finds that ZO
53.5 D. in part because of the
This is slightly below the published normal value 20
neglect of the frequency-dependent internal inductance of the conductors. There is not
much variation of the relative permittivity among the various materials used as the
dielectrics in coaxial lines and since the radius ratio comes in only in a logarithm, one
finds that most commercial coaxial lines have characteristic impedance in a limited
are a
=
mm
=
=
mm.
=
2
20 as defined here is real, it is more logical to call it a “characteristic resistance,”
especially since the concept of impedance implies use with the phasor forms appropri~
ate to steady-state sinusoidal excitation. That is an importantspecial case to be considered later, but even for transmission lines used with pulses or other general signals, it is
common to refer to the defined Z0 as characteristic impedance.
Since
2'5 8
Chapter
range,
usually 50 .0.
S
S
ZO
Transmission Lines
5
80 9.. The
wave
velocity (8) becomes
1
(16)
=
v
g,
which is the
same as
the
velocity
of
plane
waves
in the
same
dielectric
(Sec. 3.9). This
result obtains for all two—conductor transmission lines when the internal inductance and
losses
the
be
can
velocity
5.3
neglected.3
of
light
The
velocity is usually between about 0.5 and 0.7 of
108 m/s, for lines with plastic dielectrics.
wave
in vacuo, 3 X
RELATION OF FIELD AND CIRCUIT ANALYSIS FOR TRANSMISSION LINES
Although we largely utilize the distributed-circuit model» for transmission~line analysis
chapter, let us relate the equations obtained in Sec. 5.2 to field concepts of
Chapter 3. First, let us take the special case of a parallel-plane transmission line, as
indicated in Fig. 5.2, with the conducting planes assumed wide enough in the y direction
so that fringing at the edges is not important. If the planes are also assumed perfectly
conducting, it is clear that a portion of a uniform plane wave with Ex and H), as studied
in Chapter 3, can be placed in the dielectric region between the planes and will satisfy
the boundary condition that electric field enter normally to the perfectly conducting
planes. The Maxwell equations for such a wave [Eqs. 3.9(6) and 39(8)] are
in this
dExCZ, t)
=
“
T
aHy(Z, t)
M
3Hy (2, 2‘)
z
8E.
,\(Z
“-8
62
If
we
define
voltage
as
the line
=
-—
integral
f
of
-
Current for
a
E
(ll
-
positive
by
E between.
—f
2
sense
[(2, t)
With these substituted in
(2)
planes
at
a
given
2,
E)C
dx
i:
-aE_\.(z, t)
(3)
0
width b, with
field
z‘ )
a
1
tangential magnetic
,
a:
2
V(z, t)
<1)
”8—
I
1.
Eqs. 5.2(3)
=
defined for the upper
plane,
is related to the
—bHy(z, t)
and 5 2(4),
we
(4)
find the result identical to
(l) and
(2) if
C
=
b
1°1—
a
These are,
3
respectively,
the
F/m,
capacitance per
It follows that knowledge of either L
or
L
=
—-—
b
unit
H/m
length and inductance
(5)
per unit
length
C for such ideal lines determines the other.
Reflection and Transmission at
5.4
a
Resistive
219
Discontinulty
for such
a system of parallel~p1ane conductors, calculated from static
concepts. The
field and circuit concepts are thus identical in this simple case.
We would also find field and distributed-circuit approaches identical if we applied
them to the coaxial transmission line with ideal conductors
of other
conductors
or to
two~conductor systems
shape
long
perfect. This is because such systems can be
shown to propagate transverse electromagnetic (TEM) waves, for which both electric
and magnetic fields have only transverse components. The absence of an axial magnetic
field
so
means
as
that there
are no
contributions to the line
are
induced transverse electric fields and
integral f
E
no
corresponding
so long as
(1] taken between the two conductors,
.
the
integration paths remain in the transverse plane; thus the voltage between conductors
be taken as uniquely defined for that plane. Similarly the absence of an axial electric
field means that there is no displacement current contribution to 55 H d1 for paths in a
given transverse plane, and if such a closed path surrounds one conductor, the integral
will be just the conduction current flow in that conductor for that plane at that instant
of time. Moreover, the transverse E and H fields can be shown to satisfy Laplace’s
equation in the transverse plane (Prob. 5.3), thus explaining the appropriateness of
using Laplace solutions for the calculation of the L and C of the transmission line.
When the finite resistances of conductors are taken into account, the identity of circuit
and field analysis is no longer an exact one, but has been shown to be a good approximation, for practical transmission lines. This field basis for TEM waves will be developed more in Chapter 8.
can
-
5.4
REFLECTION
Most transmission~line
AND
TRANSMISSION AT A RESlSTlVE DISCONTINUITV
problems
are
concerned with junctions between
a
given line and
other element
impedance,
current
must be
’s
and
Kirchhoff
total
laws,
voltage
discontinuity. By
continuous across the discontinuity. The total voltage in the line may be regarded as
the sum of voltage in a positively traveling wave, equal to V+ at the point of discon»
tinuity, and voltage in a reflected or negatively traveling wave, equal to V“ at the
discontinuity. The sum of V+ and V_ must be VL, the voltage appearing across the
junction:
another of different characteristic
that introduces
a
Similarly,
or
at
the
the
or some
a
V+
line,
load resistance,
sum
point
of
of current in the
discontinuity,
+
V“
=
(1)
VL
positively and negatively traveling waves of the
equal to the current flowing into the junction
must be
load:
I+
The
to the
simplest
form of
equivalent
line of infinite
I...
discontinuity is one
junction, as
transmission line at the
is that of
+
=
(2)
IL
in which
a
shown in
load resistance R L is connected
Fig.
5.4a. Another
5.4b in which the first ideal line is connected to
Fig.
length and characteristic impedance 20L;
here R L
=
ZOL.
a
case
that is
second ideal
Still other forms
229
Chapter
Transmission Lines
5
Z0
VL
.
{a}
11.
-—->
ZOL
VL
20
(b)
(a) Ideal transmission line with
FIG. 5.4
acteristic
Zoz.
as a
impedance ZO
VL
a
resistive load.
second ideal line of infinite
a
(b) Ideal transmission line of char
length and characteristic impedance
load.
produce an effective resistance RL at the junction. In all these cases
By utilizing the relations between voltage and current for the two traveling
found in Sec. 5.2, Eq. (2) becomes
of load circuits
=
with
can
R L IL.
waves as
___-=_...
20
ZO
3
()
RL
By eliminating between (1) and (3), the ratio of voltage in the reflected wave to
the incident wave (reflection coefiicient) and the ratio ofthe voltage on the load
in the incident wave (transmission coefficient) may be fOund:
p91:
T
{:2 YA
V+
The most
interesting,
relations is this: there is
and
no
perhaps
reflected
=
:
ireL—z0
4
———————2RL
+
RL
the most
wave
(5)
20
obvious, conclusion from the foregoing
if the
terminating resistance is exactly equal
impedance of the line. All energy of the incident
transferred to the load and r of (5) is unity.
to
that in
to that
the characteristic
wave
is then
In Sec. 5.7 the definitions of reflection and transmission coefficients will be
for the
case
of sinusoidal
signals
and will include other than
The instantaneous incident power at the load is
W;
=
purely resistive
1+ V+
V2+ /Zo.
=2
given
loads.
The frac—
5.5
Pulse Excitation
on
221
Transmission Lines
tional power reflected is, therefore, the constant
W71"
2
The remainder of the power goes into the load resistor
WTL
_=1__
5.5
Transmission lines
consider
some
or
the second line,
so
0
7
_
PULSE EXCITATION ON TRANSMISSION LINES
increasingly used for digital or pulse-coded information. We
simple examples utilizing the boundary and continuity conditions given
are
above.
Example
5.5a
PULSE ON SHORT-CIRCUITED LINE
Let
us
O and
length
consider
t
=
2‘1/5
a
and
I such that l
in the form of
pulse having a constant value V0 between t
zero otherwise. The pulse is fed into an ideal transmission line of
vi]. We will analyze what happens when the pulse reaches the
signal
=2
a
=
end of the line, which will be taken
as
short-circuited.
Drawing (i) in Fig. 5.50 shows the pulse moving along the line at t
0.3t1. The
arrows connecting the charges on the conductors are electric field vectors. The integral
of the electric field is the voltage between conductors. The current flows in the conductors only where there is voltage, and current continuity is accounted for by displacement currents a aE/at at the leading and trailing edges of the pulse.
At time [I the leading edge of the pulse reaches the end of the line. Drawing (ii) in
Fig. 5.5a shows the pulse shortly after t
1:1. The short circuit requires that the voltage
be zero. To maintain the voltage at zero during the time that the incident pulse is at the
termination I] < t < 2‘2, a negative~z~traveling (reflected) wave having opposite voltage
polarity and equal amplitude is generated as shown in drawing (iii). Note that this result
is predicted by (4) for R L
0. Also, the zero voltage on the load agrees in (5) with
0. Note that the polarity of the current in the reflected wave is the same as in
RL
the incident wave, as could be argued from Eqs. 5.2(10) and 5.2( 12) with the fact that
V +. The total voltage on the line at the time used for drawings (ii) and (iii)
V,
is their superposition; this is shown in drawing (iv), where it is seen that the voltages
are almost completely canceled. At a still later time, the reflected pulse is seen on its
way to the generator [drawing (v)]. At t
22‘1, the pulse will reach the pulse generator.
What happens there depends on the impedance seen looking into the generator. Typically, the characteristic impedance of the transmission line equals the output impedance
-«
=2
z
2
=
-
x
222
Chapter
gefiz'ritor
Transmission Lines
5
HUM
firs-o
_.,.v
t
=
0.3t,
4——
I4(i)
Jr4--—>-
++++
Incident
v
‘
_
wave
*—
11+
(ii)
I
.
—->-
Reflected
wave
v
i
TTT
btl<t<t2
+ + +
T
(iii)
17
Superposition
l<—+
IT
(0)
pulse at the end of a shorted line. The times t, and t2 are those for
trailing edges, respectively, reach the end of the line. Drawings (ii),
(iii), and (iv) are for various instants during the period in which the reflected wave is generated
to maintain the voltage at zero across the short circuit.
FIG. 5.50
which the
Reflection of
leading edges
a
and
of the generator (it is matched). In that case, the reflected pulse is absorbed in the
generator. Otherwise, another reflection takes place.
5.5
Pulse Excitation
No. 1
Computer
No. 2
Z:
Transmission line cable for
223
Transmission Lines
lnterconnectmg cable
Computer
FIG. 5.5b
on
transmitting digital signals
between two computers.
Exampie 5.5b
PULSE REFLECTIONS ON A TRANSMISSION LINE
INTERCONNECTING TWO COMPUTERS
The aim of this
example
controlling reflections for
is to show the
importance
of transmission~line
matching
in
transmission line used to interconnect two computers. Consider the two computers shown in Fig. 5.5b interconnected by a coaxial cable 100 m
long and with a velocity of propagation 2 X 108 m/s, so that there is a time delay of
500
ns
of the
a
for a pulse to propagate from input of the cable to
digitally coded signal, made up of lO-ns pulses
sketched in
Fig.
5.5c. This is sketched
010
versus
30
its output. Consider a portion
with basic spacing 20 us as
distance in
Fig. 5.56:!
90
4o
at
5
ns
before the
an?)
100
(C)
I
I
l
l
l
.
~9~7
«21—19
7|
—3-io
fli
:flfl
~1S—13
~21-19
«3—1
Ar
z(m)
:6)
(9)
FIG. 5.5
(c) A portion of a pulse-coded signal versus time for the computer interconnection of
Fig. 5.51). (d) Voltage versus distance for a time 5 us before leading edge of first pulse reaches
0). (6) Voltage versus distance for reflected signal 110
input of computer No. 2 (defined as 2
ns after instant of sketch (d).
=
224
first
Chapter
pulse edge
Transmission Lines
5
reaches computer No. 2. If input impedance of the second computer
impedance of the line, there is no reflection and the entire
matches the characteristic
signal is accepted. But suppose its input impedance IS 100 Q and
impedance of the cable is 50 0.. By (4), the reflection cdefficient is
RL-ZO
RL+ZO
So the train of
versus
100—50
1
100+50
3
the characteristic
pulses is reflected, at é— amplitude, toward computer No.
Fig. 5.56 at 110 us after time of Fig. 5561. If there
distance in
l
is
as
sketched
impedance
mismatch at the terminals of computer No. 1 additional reflection will take place when
the signal returns there and a spurious signal will be superposed on whatever desired
being sent between the computers at that time. Since the reflected “echo” is of
amplitude than the original signal, differentiation is possible on the basis of
amplitude level, but there is an obvious advantage in matching impedances well enough
that reflected signals are small.
code is
lower
Exampte
SQUARE WAVE
5.5a
ZAPPLIED TO INFINITE LINE
infinite line
O with a square
suddenly charged at t
m. Voltage distribution at t
0 is then
as in Fig. 5.5f. This may be considered a superposition of two such square waves, each
of amplitude VO/Z. One of these moves to the right and the other to the left, each with
3 X 108 m/s) the two partial waves and
1.667 ns (taking 0
velocity v. Thus at 1‘
As
a
wave
third
IN
example,
consider
in distance from
=
2
an
——l
m
to
z
=
=
+1
z
=
=-
V0
V0
'5-
““““““““““
|
-1
>z(m)
1
0
(1‘)
V0
V0
ru—uwa-_—_.—__:-_—_:I____———,
0-1—1
l
~15
“‘4
F5“
I
|
-O.5
O
—2_
'90
P
05
=z(m)
1-5
(g)
FIG. 5.5 Voltage and current distributions for Ex. 5.50. (f) Voltage distribution
1.667 ns.
(g) Traveling waves (dashed) and total voltage versus 2 (solid) at t
=
at t
=
0.
Pulse Excitation
5.5
on
225
Transmission Lines
V0
““““““““““
r
-o.5|
l
“*1
:L’QL
270
L4- v
I
—1.5
"
0.5
o
_________
Adm)
1
I
.l
220
(h)
Zn
39.
2
Ir“
2 5
_
_.
0.5
I
I
39L
__
FIG. 5.5
o
ns.
their
sum are
_________
and
tributions att
rz(m)
.I
current
=
5
(solid)
versus 2 at t
I!
as.
shown in
Fig. 5.5g. Figure 5.511 shows the corresponding current distridifferent sign relations between current and voltage for
negatively traveling waves. Figure 551' shows voltage and current dis-
bution, taking into
positively
2.5
0.5
2
|
(/1) Traveling waves of current (dashed) and total
(1') Voltage (solid) and current (dashed) versus 2 at t
1.667
39—:
220
=
5
account the
ns.
Example 5.5d
PULSE REFLECTIONS WHEN PULSE ls LONGER THAN TRAVEL TIME DOWN THE LINE
Example
5.5b considered reflections for
pulses
much shorter than the travel time down
the transmission line. For many interconnections, especially those within integrated
circuits, the delay time is shorter than pulse width. Reflections because of mismatch
may still cause a problem, causing structure within the
consider the first part of the pulse as a step function of
transmission line
at
may be considered
z
pulse. To illustrate the point,
voltage VO applied to an ideal
0 so high that it
l, with terminating impedances at end 2
open circuit. The front of the wave travels along the line in the
=
an
2
-
(1. Since current must be zero at the
positive 2 direction and reaches the end at I.‘
O, a reflected step wave must be generated with current opposite
open-circuited end 2
to that of the positively traveling wave, starting at the instant of arrival of the latter at
2
0. From Eq. 5.2(13), we know that current in the negatively traveling wave is
V+.
V/ZO, so it follows that V*
As time goes on, the conditions to be met are that the total voltage at the input of
the line (2
I) must be the value VO of the pulse applied there and the current at the
open end must be zero. These conditions can be satisfied by the sum of two square
waves, one traveling in the positive 2 direction and one in the negative 2 direction.
Figure 5.5j shows the voltages of the two individual waves and the superposed, or total,
==
=
==
=
~—
=
--
226
Chapter
_‘_
(UN)
Transmission Lines
5
.1.
(I/v)
NOJ-bU‘ICDVL mwhuot
H
I—‘
O
O
—l
Positively traveling
Negatively traveling
Superposition
FIG. 5.5] Positively and negatively traveling wave components and their superposition to match
boundary conditions on a transmission line with a pulse of voltage V0 applied to the line at z
-l and the line open-circuited at 2
0. The applied pulse length is much longer than the travel
time along the line.
=
==
l, as required by the
voltage. Note that the total voltage is always equal to V0 at z
boundary condition. As discussed in the preceding paragraph, the condition where the
negatively traveling wave has the same polarity of voltage as the positively traveling
wave gives a zero total current. Thus, for either the situation with both waves having
zero value or both with V0, the boundary condition at the open end is satisfied. It is
seen in Fig. 551' that the sum of the two waves meets that requirement. Since the two
waves satisfy the transmission-line equations and their sum satisfies the boundary conditions, they constitute the unique solution.
0 with voltage zero for intervals
Note that a square wave of voltage is produced at z
of 2L/v, interspersed with intervals of the same value with V
2V0.
z
-
z
:
Example
5.52
TRANSMISSION LINE WITH CAPACITIVE TERMINATION
As
a
example, consider the ideal transmission line of Fig. 5.5k
uncharged capacitor. We take the incident wave as the exponential
somewhat different
terminated in
buildup
of
an
Fig. 5.51,
V+ (t)
where, for convenience, time
the
capacitor. By continuity
zero
of
=
Vo[1
is taken
voltage
V+ (t)
as
_
8"”0]
the instant the forward
(1)
wave
arrives at
and current,
+
V_ (t)
=
VC (1‘)
(2)
Pulse
5.6
227
Forming Line
V+(t)
”l
V0
-<---V_
_______________
V+—>
J.
00
J
Z0
|--->-z
z
=
O
(k)
FIG. 5.5
voltage
(1)
(k) Ideal transmission line terminated with
wave
incident
V}. (I)
V
——
-—
these
IC
=
20
equations
(t)
__.__.dV‘_
(t)
—"——
20
By combining
a
capacitor; (1) form
of
forward-traveling
CO.
on
and
using (1)
(I)
___V"
+
(11‘
3
T1
Kg
t
0
we
(1
C0
=
(W (t)
—£—-
(3 )
dt
obtain
__
e—I/To)
__
fie—I/TO
(4)
T0
T1
where T1
C0 Z0 is a time constant set by the capacitor and transmission-line
ance. A solution of this first-order differential equation is
=
V“ (r)
The
arriving
charted,
so
wave
V_ (O)
has
=
VO[1
zero
T
AI er”?!
+
voltage
at
+
imped~
T
(0—J)e*’/To]
--
(5)
the instant of arrival and the
capacitor
is
un-
O and
=
271
A1
=
(6)
‘—
70"”71
Using (2),
the
V60)
It
can
is found to build up in the
voltage
=
be checked that
by substituting
t +
V0|:2
+
2
——1L—
art/TI
(3) is satisfied. The
for t in (5).
2
capacitor
—
as
2
———7°——
ear/TO]
variation of the reflected
wave
(7)
is obtained
Z/U
5.6
PULSE FORMING LINE
One way of forming pulses of a desired length is by charging a transmission line of
l to a dc voltage V0 and then connecting to a resistor as shown in Fig. 5.6a. (In
length
223
Chapter
Transmission Lines
5
o:
++++++++++++++
R
ZO
V0
Voi
V0
Var
R>Z0
R=Z0
(before
discharge)
R<ZO
1/;
2
t
2:1
l
a
g
0
o
2:]
m
61?}
4:1
2‘!
a;
81311?
0
r—L‘S‘I
J4“
i;
“—8“
t
(a)
1
Transmission line of
FIG. 5.6
(0)
length (time delay one way
t1) charged to potential V0
O. (b) Lumped element approximation to (a). (6) Wave
and connected to resistance R at t
shapes of resistor voltage for different relations of R to Z0.
=
=
practice, the line is often approximated by lumped elements, so the lumped-circuit
approximation is shown in Fig. 5.6b.) If the resistance Ris matched to the characteristic
impedance, a pulse of height Vo/ 2 is formed across R for a time 22:], where t1 is oneway prepagation time down the line and the line completely discharged. It may then
be recharged and the process repeated. It may at first seem puzzling that voltage across
the resistor is not just V0 when the switch is closed, but this is because a traveling wave
to the right is excited by the connection. Voltage across the resistor just after closing
the switch is then the sum of dc voltage and the voltage of the positive wave, V+2
VR
The current
traveling
flowing
=
in the resistor is
V0
just
+
the
V+
(1)
negative
of current in the
positively
wave.
V+
az“h=‘7z
0
(2)
Since
VR
2
RI
combination of (l) and
,
1+
be
=
=
Z0, VR
as
==
and
V+
=
Current in the
VO/ 2.
-
wave
started to the
right;
=
=
wave
required.
VO/ 2
3
I, there must then
(2). When this current reaches the open end, 2
with current I
so
that
the
net
current
is
zero at the open
V0/2Z0
VO/ZZO by
reflected
a
end
--
(2) gives
R
R
-————v==——V
V:
Thus if R
229
Reflection and Transmission Coefficients
5.7
_
This
requires
a
in the reflected
voltage
wave
V_
=
-ZOI_
=
-
VO/ 2,
which travels to the left, canceling the remaining voltage on the line and bringing zero
voltage and current to the resistance at t
2r]. From then on all is still. Thus in the
=
of R
right discharges half the voltage initially on the line,
half, yielding a rectangular pulse as shown in Fig. 5.6c.
If R # 20, the wave started to the right upon closing of the switch is other than VO/ 2,
so cancellation of the voltage on the charged line in one round trip does not occur, and
there are further reflections when the wave returns to the input. Figure 5.6a sketches
the form of resistor current for R > Z0 and for R < 20. (See also Prob. 5.6b.)
case
and the
=
wave
Z0,
the
wave to
the
to the left the other
Sinnsmdal Waves
ideal Transmnssnon Lines
on
REFLECTION AND TRANSMISSION COEEHCIENTS AND IMPEDANCE
ADMITTANCE TRANSFORMATIONS FOR SINUSOIDAL VOLTAGES
5.7
AND
preceding discussion has involved little restriction on the type of variation with
voltages applied to the transmission lines. Many practical problems are
concerned with sinusoidal time variations. If a sinusoidal voltage is supplied to a line,
O by
it can be represented at z
The
time of the
=
V(O, t)
The
corresponding
in the
traveling
wave
V+(z, x)
and that
traveling
in the
negative
V_(z, t)
The total
voltage
V(z, r)
is the
=
sum
lV+|
=
2
V
(1)
cos wt
positive
|V+|
=
cos
direction is
2
(0(1‘ >
——
Up
direction is
[VJ
of the two
cos
=
cos[w<t 5-) 64
+
+
UP
traveling
w<t -“->
-
Up
+
waves:
[V_|
cos[w<t > 9p]
+
+
Up
(2)
230
The
Chapter
current, from
corre5ponding
[(2, t)
_m.
Eq. 5.2(13),
t
cos a)
—-
Transmission Lines
5
is
liflfll
Zo
cos
(0
Z0
Up
(3
t +
Up
+
(3)
6p
2/1)) is seen
point on a wave described by F(t
the
2
in
direction.
The
positive
argument of a
by an observer moving with velocity v
for
which
is
constant
is called the
its
so
the
sinusoid is called
velocity
phase,
phase
phase velocity Up.
For sinusoidal time variations, it is useful to rewrite (2) and (3) in phasor form:
In Sec. 5.2
we saw
that
a
constant
-—
a
V
=
I
=
V+e"jfiz
1
V_ejfiz
+
.
.
[V+e"JBz
“—
Zo
(4)
-
VHeJ‘BZ]
(5)
where
B
=
~53
==
wVLC
<6)
p
We may take
complex and
incident
V+ as
equal
waves at z
the reference for
to
=
‘V_|ej‘9p,
O
as
with
phase so that it is real. Then V__ is in general
HP being the phase angle between reflected and
zero
in the instantaneous form
(2).
of the line since
,Bz measures the instanwith respect
voltage (or current) is
are
at
the
two
same
observed to be
separated in 2 such
any
points
The
shortest
distance
between
of
277'.
that ,82 differs by multiples
points of like current
A.
the
a
is
called
or voltage
foregoing reasoning,
wavelength By
is called the
The
quantity ,8
taneous phase at a point
phase
constant
to
z
,B/i
=
0. Moreover,
along the line that
z
=
277
01'
2
s=f=w1£
The
expressions
can now
m
for reflection and transmission coefficients in
be written in
a
special
form for sinusoidal
waves.
Eqs. 5.4(4)
and
5.4(5)
It is convenient to choose
origin of the z coordinate at the discontinuity to be analyzed, as shown in Fig. 5.7
representative discontinuities. It is assumed that a previous analysis gave the
O in the two lower
equivalent value of impedance looking in the + z direction at z
lines. We will see below how this is done. The ratio of total phasor voltage to total
phasor current at any point is the definition of impedance. We set the impedance at
z
O, the load impedance ZL, equal to the ratio (4) to (5). Solving for the ratio V_ /V+,
the
for three
=
=
the reflection coefficient for sinusoidal
waves
is obtained:
<1
p=Z=4+%
w
23'!
Reflection and Transmission Coefficients
5.7
or
20
G
|
‘
0%
z
i
G
201
202
or
a}
I
i'"‘*’3
~1
O
G:
‘1“
Z01
[C
o:
ZL2
Z02
w
—l
0
FIG. 5.7
subject
Representative
analysis.
situations where
a
line of
length
l with
a
discontinuity
at z
=
O is the
of the
Also, (4) and (5)
can
be combined to
::
give
the transmission coefficient:
{if—L— 2.2—2.9...Z0
:
L
+
+
(9)
0.
voltage VL is the total voltage at z
The expressions for power reflected and transmitted at a discontinuity, given for real
functions of time in Eqs. 5.4(6) and 5.4(7), can be adapted to sinusoidal signals of the
complex exponential type. Since power in this case is VI* / 2, and for a single wave in
The load
a
loss-free line this is
=
VV’VIZZ0
=
[VIE/220,
W“
—-f:
=
WT
the fractional reflected power is
| V_
MI
2
'2
,,
=
lpl"
(10)
and the remainder goes into the load:
W
i
W?
—-=
1
-
lplz
(11)
232
Transmission Lines
Chapter 5
expressions for the input impedance
l:
impedance by dividing (4) by (5) for z
Now let
us
find
=
2,.
and admittance at
-—
Z. We find
-
Z0[W
=
(12>
Or, substituting (8),
ZI
ZLcosfil +jZosinBl
ZO cos ,8] + jZL sin? [31
Z0
__
I
By defining admittances Y,expression of the same form for
l/Zi, YL
=
Y-l
l/ZL,
and
Y0
=
1/ZO,
we can
find
an
Yi:
YO
:
=
(13)
[YL
Yo
cos
cos
,Bl
El
Example
+
+
jYO
jYL
sin
sin
Bl]
,3]
5.7
CASCADED THIN-FILM LINES
diagram of Fig. 5.7 represents two thin~film transmission lines in
integrated circuit. The load ZL2 represents a device having a real impedance of 20 D. at the signal frequency, 18 GHz. Line 2, of characteristic impedance 202
2 mm. Line 1, of characteristic impedance Z01
30 Q, has a length 12
20 Q,
has a length 11
1.5 mm. The phase velocities for both lines are the same, 2 X 108
m/s. Let us find the impedance at the input to line 1.
First we must solve for the way the load impedance 21.2 transforms along the line
attached to it. To do this we apply formula (13) to that section of line. For both lines
(6) gives
Suppose
a
the second
microwave
=2
=
=
=
s
=
5’3-
=
(277)(18
2 X
Up
x
109) rad/s
=
108 m/s
566
rad/m
variety we will take angles in degrees here, as both degrees and radians are com~
64.9 degrees and .351
manly used in transmissiondine calculations. Thus [382
48.6 degrees:
For
=
20
cos
64.9°
+
j30
sin 649°
30
cos
649° +
j20
sin 649°
__
36.7 +
Now
we can
find
Zi1
Zil
at the
20
=
18.9
—~
j11.8 n
to line 1
input
(36.7
__
=
+
cos
j 11.8)
486°
j14.6 n
+
using ZQ
cos
as
486° +
j(36.7
+
the load
1'20
Zn:
sin 486°
j 11.8) sin 486°
5.8
233
Standing Wave Ratio
Systems with several lines of different characteristic impedances in cascade can be
analyzed as we have in this example. In each case the analysis starts with the point
farthest from the signal source, transforming the impedance back
successively to the
next discontinuity until the input is reached. In
general, ZO, B, and I will be different
for each section.
5.8
Let
STANDING WAVE RATIO
examine the
phases of the voltages in Eq. 5.7(4). One coefficient, say V+, can
by choice of origin of time. The reflection coefficient, Eq. 5.7(8),
which is a complex number, can be written in the form lplejep so V~ in Eq. 5.7(4) can
be replaced with V+lplejga giving
us
be chosen to be real
=
Let
us
write this
as a
V+efifl
+
V+hkfl%+&’
real function of time with
~
2
to
replaced by l,
the distance from the
end of the line:
V0“, -—1)
=
any
particular instant,
the cosine in the incident
=
2
phases
are
£31)
+
+
V+lp|
cos(wt
-
£31
+
6/)
O, we can readily see that the argument of
say t
(first term), which is its phase, increases with distance
O and that the phase of the reflected wave (second term) decreases. These
Considering
from
V+ cos(wt
=
wave
shown in the top drawings of Fig. 5.8 where it is clear that at some distance
are the same. At 27,, they differ by 77' rad, one having decreased by 77/2
20, the phases
and the other
there
are a
having increased by 7r/2. At 227, they differ by 277' rad, and so on. So
series of locations where the two sinusoids are in phase and another series
of locations where
Where they are in phase, they add
phase, they subtract. At the former
locations the total voltage has its maximum amplitude, and at the latter, its minimum.
Analysis of the sum of the incident and reflected waves given in (1) (see Prob. 5.8f)
shows that the total voltage can also be represented as the sum of a standing wave and
a traveling wave. The total voltage is shown in the lower drawing of Fig. 5.8 for several
particular times in a cycle selected to show the voltage when it has its maximum and
minimum peak values. The broken line shows the voltage amplitude along the trans-
directly
at
each
they are 77' rad
instant, and where
out
7r
of
phase.
rad out of
mission line.
The maximum
voltage
is
Ics=har+WJ
and the minimum, found
a
quarter—wavelength
Vmin
The
standing
wave
Z
ratio is then defined
iv+i
as
(m
from the maximum, is
__
iV—~i
the ratio of the maximum
(3)
voltage amplitude
234
Chapter
Transmission Lines
5
Phase
I
Incident (- Liz)
I
l
:5
I
2
:
|
0
I):
“““““““
t
i221,
“““7
I
,
l
:3-
'
I
>
0
:20
I
I
I
Reflected (0p + I32)
I
I
l
I
I
mmx
2
”mx
W“
I
I
I
\
Vmin /
\
/
,.-
..
V
FIG. 5.8 The upper graph shows the
line with a reflection coefficient p
phases
of the incident
reflected
and
in the lower
+
37r/ 2.
graph, (at,
=
The broken line
to the minimum
waves at t
lplejep. The total voltage: V(z) is shown
~6p/2 +
~0p/2, cot2
*Gp/Z + 7r/2, (013
=
=
=
gives
the
voltage amplitude along
=
0
on a
for selected times
71',
can,
==
~6p/2
the line.
voltage amplitude:
V
s
By substituting (2)
and
(4)
=Y/mi"
(3) and the definition of reflection coefficient, Eq. 5.7(8),
we
find
S
_____IV+I+IV-I:1+IpI
IV+l
It is
seen
that
coefficient p,
standing
giving the
wave
same
ratio is
5.8 is
MI
directly
information
W
Figure
-
as
S
1
(5)
IpI
related to the
this
—
-
quantity.
magnitude
of reflection
The inverse relation is
l
(6)
=
S + 1
3, corresponding to IpI
in
plotted for S
negative sign appearing in the current equation, Eq. 5.7(5), it is evident
Because of the
=
=
that at the
position
235
Standing Wave Ratio
5.8
where the two
traveling
add in the
wave terms
voltage relations,
position is
subtract in the current relation, and Vice versa. The maximum voltage
then a minimum current position. The value of the minimum current is
they
[min
At this
at any
position impedance
point along the line:
is
iV+l “IV—l
___
_‘
(7)
ZO
resistive and has the maximum value it will have
purely
inwwnlz
;%““Ziwn~wmd
At the
position
of the
voltage minimum,
current is a
“)
%S
_
maximum, and impedance is
a
minimum and real:
IV+I
lel
is“ahmtwud“s£9
__
--
Example
“m
__
5.8
SLOTTED-LINE IMPEDANCE MEASUREMENT
A slotted line is
mission line
position
of
a
an
instrument that
movable
containing
voltage minimum
a
or
be used to
can
probe
impedances. It is a trans~
standing wave ratio and the
found. The unknown impedance is
measure
with which the
maximum
be
can
connected to the end of the slotted line.
50 (1
Suppose a measurement on a slotted line of characteristic impedance ZO
reveals a standing wave ratio S
3 and the closest voltage minimum is 0.33A from
the unknown load impedance. Let us see how ZL is deduced. In Fig. 5.8, we see that
at 27, where the minimum is found, the phase of the incident wave,
[32 attains the
=
n
——
value
(6p
+
77)/2,
so
and
277
6p
where
we
=
2,3(O.33/\)
have used
-—
77
==
Eq. 5.7(7). Using
p
2
the
=
T
(0.33)»)
given
S and
0.56110
-
(6)
77'
we
=
1.0 rad
find
[p]
to
be 0.5. Then
235
From
Chapter
Eq. 5.7(8)
ZL
we can
get ZL in
term
50
=
Z0.
Thus
0.56110
[1 0.56110]
=
—~——:—
52.8 + J59.3 0
—-
-
5.9
of p and
1 +
1 + p
0L p]
Z
=
Transmission Lines
5
THE SMITH TRANSMISSION-LINE CHART
Many graphical aids for transmission-line computations have been devised. Of these,
the most generally useful has been one presented by Smithf’ which consists of loci of
constant resistance and reactance plotted on a polar diagram in which radius corresponds
to magnitude of reflection coefficient. The chart enables one to find simply how imped»
ances are transformed along the line or to relate impedance to reflection coefficient or
to standing wave ratio and position of a voltage minimum. By combinations of Oper—
ations, it enables one to understand the behavior of complex impedance~matching techniques and to devise new ones. It is much used in displaying the locus of impedance
of many useful devices as frequency is varied. Although computer programs are avail—
able for transmission-line calculations, its role in diSplaying and understanding match—
ing mechanisms remains useful. This 'chart utilizes the reflection coefficient plane.
Impedance for any point along a transmission line with a passive load then lies within
the unit circle.
circles
and in
Loci of constant resistance are circles
orthogonal
the following
section
give
are
of reflection
Coefficient.
examples of the chart’s use.
begin with Eq. 5.7(12), which gives impedance
define a normalized impedance
some
The discussion of the chart will
terms
and loci of constant reactance
to those of constant resistance. We will first show the basis for this,
We
in
Z.
m=o+méj
(n
0;
a complex variable w equal to the reflection coefficient at the end of the line, shifted
phase to correspond to the input position 1:
and
in
w
Equation 5.7(12)
=
u
+
jv
é pie-2113?
(2)
may then be written
1+
mzl
w
"W
w
or
.
.____l+(u+jv)
7+Jx-————l-(u+jv)
4
P. H. Smith, Electronics 12, 29—37 (7939); 17, 730 (7944).
(4)
This
equation
may be
237
The Smith Transmission-Line Chart
5.9
into real and
separated
as
follows:
(112 + v2)
(1—L£)2+U2
1
I -:
imaginary parts
—
s
(V)
20
\
.«z
,
6
()
,
(1—-u)"+v‘
01'
"'
—-——"——-~+u2—-———1——-—
l+r
1"
(u-i)~+(v--)
,
1
we
then wish to
(8)
:7
x
If
X"
the loci of constant resistance
plot
(7)
_(1+r)2
the
r on
w
plane (u
and
v
serving
rectangular coordinates), (7) shows that they are circles with centers on the
u axis at [r/(l +r), 0] and with radii l/(l + r). The curves for r
0, 335;, l, 2, 00 are
as
=
sketched in
Fig.
circles, with
5.90. From
centers at
(l,
(8),
the
curves
of constant x
1 /x) and with radii
1/le
plotted
on
Circles for
x
the
=
w
O,
plane
1%,
are
:3:
also
1, 1:2,
Fig. 5.9a. Any point on a given transmission line will have some
with
impedance
positive resistance part, and so will correspond to a particular point on
the inside of the unit circle of the w plane. Several uses of the chart will follow. Many
00
are
sketched in
90
{Lo
FIG. 5.9a
6‘0
Basic features of the Smith Chart.
23$
Chapter
5
Transmission Lines
FIG. 5.9b
extensions and combinations of the
chart with
more
divisions is
5.10
In this section
impedance
the
we
show the
and reflection
given
ones to
in
Fig.
SOME USES
use
Smith Chart.
OF THE
or
SMITH CHART
of the Smith chart in
displaying
coefficient, in transferring impedances
line, and in impedance matching. Other
still other extensions
be cited will be obvious to the reader. A
5.91).
uses are
illustrated
the relation between
or
by
combinations will be evident to the reader.
admittances
the
along
problems and
239
Some Uses of the Smith Chart
5.10
To Find Reflection coefficient Given Load Impedance, and Conversely
point within the unit circle of the Smith chart corresponding to a particular position
The
transmission line may of course be located at once if the normalized impedance
corresponding to that position is known. This is done with a reasonable degree of
on a
accuracy by utilizing the orthogonal families of circles giving resistance and reactance
as described above. Thus, point A of Fig. 5.10a is the intersection of the circles r
1
=
and x
from
r
corresponds
Eq. 5.9(2) that [w]
Ipl
to a
=
position
point
on
If the
the
point
so
the
graph
on
as a
with normalized
and from
O circle, which is the outer
=2
Lp,
1 and
=
Eq. 5.9(7) that
the graph. A
of
edge
fraction of the radius to the
r
impedance
M
=
(u2
measure
phase angle
course, reverse the process to find
ZL
if p is
jl.
122)”2
+
=
It is clear
l
on
of the radius to
the
some
gives lPl directly.
O, and Aw
impedance, Z
can be read directly. One can, of
=
O circle thus
the Smith chart is the normalized load
of the reflection coefficient
1 +
=
=-
given.
Example
5.10a
REFLECTION COEFFICIENT FROM LOAD IMPEDANCE
70 .Q is terminated with
Suppose a transmission line of characteristic impedance ZO
1 + jl, shown
a load ZL
70 + j7O 0.. The normalized load impedance is {(0)
1.11 rad so
Aw
as point A in Fig. 5.100. The magnitude of p is 0.45 and Lp
0.456j1'11. The angle may be found by reading the outside wavelength scales,
p
recognizing that a quarter-wave is 7r radians on the chart.
=
==
:
=
=
=
To Transform impedance Along the Line As position I along a loss—free line,
changed, only the phase angle of w changes, as can be
measured relative to the load, is
Eq. 5.9(2) wherein p is a complex number, the reflection coefficient at the
change of position along an ideal line is represented on the chart by movement along circles centered at the origin of the w plane. The angle through which w
changes is proportional to the length of the line and, by Eq. 5.9(2), is just twice the
electrical length of line B]. (Most charts have a scale around the outside calibrated in
fractions of a wavelength, so that the angle need not be computed explicitly. See Fig.
5.10a.) Finally, the direction in which one moves is also defined by Eq. 5.9(2). If one
moves toward the generator (increasing 1), the angle of w becomes increasingly negative, which corresponds to clockwise motion about the chart. Motion toward the load
corresponds to decreasing l and thus corresponds to counterclockwise motion about the
seen
from
load. Thus,
chart.
249
FIG. 5.10a
to
Chapter
Smith chart for
5
impedances.
Transmission Lines
Points A, B', and C and associated broken lines relate
Exs. 5.10a, 5.10b,and 5.100.
Example
5.10
IMPEDANCE TRANSFORMATION
Consider the line and load of Ex. 5.10a for which the; normalized load impedance is
l + jl and is shown at point A in Fig. 5.10a. If the line is a quarter-wave long
(90 electrical degrees),
we move
through
an
angle
of 180
degrees
at constant
the chart toward the generator (clockwise) to point B. The normalized input
is then read as 0.5
j0.5 for point B. If input impedance is given and load
-
desired, the
reverse
of this
procedure
can
obviously
be used.
radius
on
impedance
impedance
5.10
FIG.5.'I 0b
C2,
and
Polar transmission~line chart for admittances. The constructions
01-133
To Find
24‘
Some uses of the Smith Chart
involving points C1,
relate to BK. 5.10d.
Standing
Wave Ratio and Position of
Voltage
Maximum from
3
Given impedance, and Conversely If we wish the standing wave ratio of an
ideal transmission line terminated in a known load impedance, we make use of the
information found in the
maximum
impedance
is real. We
see
preceding
can see
Equation 5.8(8) shows that the location of
voltage and the impedance there
that
S
and
section.
is also the location of maximum
from
Fig.
5 10a that the
.
=
—-——"‘“
=
(1)
gm
point where impedance is
real and maximum
along
constant circle) lies on the right side
any ideal transmission line (represented by a |pl
in
the
horizontal
axis
(u axis). Thus,
following about the circle on the chart
along
=
determined
by the given load impedance, we note its crossing of the right-hand u axis
plane. The value of the normalized resistance of this point is then the standing
wave ratio; the angle moved through to this position from the load impedance fixes the
position of the voltage maximum.
of the
w
242
Chapter
5
Transmission Lines
procedure to determine the load impedance, if standing wave
ratio and position of a voltage maximum are given, is straightforward, as is the extension
to finding position of voltage minimum or finding input impedance in place of load
impedance.
The reversal of this
*
I
‘VW
Example
5.n0c
DETERMINATION OF STANDING WAVE RATIO AND LOCATION
OF VOLTAGE MAXIMUM
Let
us
continue
our
analysis
of the ideal transmission line and load discussed in Exs.
impedance is l + j 1 plotted at point A in Fig.
Moving along the line away from the load (clockwise), one arrives at the pureresistance point C by going 0.088 wavelength. The value of maximum normalized
resistance, which equals the standing wave ratio S, by (1) is read as 2.6.
5.10a and 5.1%. The normalized load
5.1061
Diagram Since admittance transforms along the ideal line
impedance, Eq. 5.7(14), it is evident that exactly the
same chart may be used for transformation of admittances with the same procedure as
for impedances described in the above. Admittance is read for impedance, conductance
for resistance, and susceptance for reactance as seen in Fig. 5.1017. There are differences
Rise
in
as an
exactly
Admittance
the
same manner as
to remember: the
right-hand u
axis
now
a
current maximum instead
of
represents
an
admittance maximum and, there-
voltage maximum; the phase of the reflection
coefficient read as described above and corresponding to a given normalized load admittance is that for current in the reflected wave compared with current in the incident
wave and is therefore different by at from that based on voltages. (See Prob. 5.7d.)
fore,
a
lacs. 5,
.t,-'.\;-.-:..t
t/-\'—'l<-;u
I)
ADMITTANCE ANALYSIS OF VARIABLE SHORTED-STUB TUNER
In this
example we will use the Smith chart for admittances to analyze a mismatched
employs a variable shorted—stub tuner to produce a unity standing wave ratio
in the line leading up to the stub. Figure 5.100 shows the arrangement. Suppose Z0
50 Q in the main line and Z0,
20
70 Q in the stub. Assume ZL
j20 Q. We
2
and
the
will find the appropriate stub location
stub length 15. The admittance
1m
chart rather than the impedance form is used because of the convenience in handling
line that
2
==
=
=
-—
shunt circuits in the admittance formulation.
—
5.10
243
Some uses of the Smith Chart
0
"'
1i
‘"
i
lm
g
or
Z0
ZL
A
g
a
z
Z05
FIG. 5.10c
therefore 8
Variable shorted-stub tuner connected in shunt to
=
1 for
2
<
The load admittance is
admittance is
provide matching
at
-lm
and
—-—l,,,.
YL
=
l/ZL
=
0.025 +
j0.025
S and the characteristic
0.020 S. The normalized load admittance is 1.25 +
Y0
j 1.25;
1/ZO
in
5.10b.
point D1
Fig.
The input admittance of the shorted stub is seen from Eq. 5.7(14) to be purely im»
aginary since Yi
—jYO cot [31. This suggests that it should be placed at a point along
z
the main line
1 and an
1,", where the admittance has a normalized real part g
b
Y
a
that
can
be
canceled
of
constant
imaginary part
by is. Following path
lPl
lwl
toward the generator, we come to the g
1 curve where b
1.13 (point D2) corre—
0.0226 S. That is
(0.020)(1.13)
sponding to an unnormalized susceptance of B
seen to be 0.485}\ from the load. A stub with an input susceptance of —0.0226 S is
connected in shunt to cancel out the imaginary part of the admittance. This moves us
on the admittance chart from D2 to
1 and the line is perfectly matched
D3 where g
for waves approaching the stub location from the generator.
70 0., Yes
0.014 S. The
Now let us find the length 15, of the stub. Since 205
1.61.
normalized input susceptance of the stub must be bis
—-0.0226/ 0.014
The shorted end of the stub has an infinite admittance. That is at C1 on the Smith
admittance chart in Fig. 5.10!) at the right end of the real axis. To transform this admittance to the normalized susceptance
1.61 we move clockwise as shown to point
The
of
be
the
stub
must
0.088/‘t.
length
C2.
The line to the right of the stub appears as a conductance in parallel with a capacitance; addition of the shunt stub provides an inductance which makes the combination
appear as a parallel tuned circuit. It should be clear that this matching technique leaves
a standing wave in the line between the load and the stub as well as in the stub and
provides exact matching only at the one frequency.
=2
=
this is at
=
-
=
=
=~=
=
=
=
=
=
=
2
~—
=
=
-—
244
Chapter 5
Transmission Lines
To Transform Impedance Aiong fiascaded Lines It is often useful to find the
input impedance of cascaded lines of differing characteristics as in Ex. 5.7. The Smith
chart involves
impedance
irnpedances
normalized to the characteristic
transformation in
a
impedance,
requires
by line, back toward
One starts at the load and transforms, line
study of
so a
renormalization for each line.
cascade of lines
the generator.
“Henchman-Lawma- .........
Exampie
5.n0e
IMPEDANCE TRANSFORMATION ALONG CASCADED LINES
For the cascaded ideal lines in
in line 1 in
a
Fig. 5.1001,
let
find the fraction of the power incident
us
that is absorbed in the load. This is
lO—GHz
signal
using Eq. 5.7(11)
given by
a
knowledge
recognizing that the power paSsing the junction
[Pl
at the end of line 1 must be absorbed in the load since we are assuming ideal lossless
in line 1
of
and
lines.
Using the parameters for line 3 given in Fig. 5.10d we find the normalized load
2 cm, so 13
2. The wavelength A3
impedance {L3
ZL/Z03
03/f
0.1)r3.
The load impedance {L3 is marked as point E1 on the Smith chart in Fig. 5.106. We
move along the constant le circle toward the generator by (m. The point E2 is at the
normalized input impedance {,3
0.98
1070.
To transform the impedance along line 2, we must first denormalize 5,3 and then
normalize it to line 2 to get the load impedance gm; thus, 5L2
0.70
Z(,3§,-3/Z02
2
is
marked
The
in
line
is
1.5
This.
cm,
--j0.50.
wavelength
point
v2/f
E3.
A2
so 12
0.2)t2. We move along a constant le circle clockwise from E3 by 0.2/\ to the
0.65 + j0.46.
input of line 2 (marked E4) where {,2
To find the reflection coefficient at the load point for line 1, we renormalize 5.2, so
0.91 + j0.64. This is the point E5. Measuring the distance from
gm
Zozg’,.2/ZO1
to
and
the
0 line, we obtain Jpll
0.32.
origin
dividing by the radius of the r
E5
=
=
=
=
=
=
—
==
=
=
=
=
:
=
=
=
=
Then
Err
=
l
We
-—
lp 1'
2
z
is the fraction of the incident power in line 1 that is
0.90
dissipated in the load.
fibk‘wi—lg‘v-I
201
%
FIG 5.10d
70 n,
2 X
203
Zoz
303
":12
”p3
Zr.
4
VI“;
Cascaded transmission lines with parameters fdr Ex. 5.10e. Z0,
50 Q, Z02
50 Q, ZL
100 0,12
3 “1111,13
2 mm,
1.5 X 108 I'll/S,
UP2
Ups;
=
=
108 m/s.
=
:3:
=
:r.
ll
ll
5.11
Transmission Lines with General Forms of Distributed
FIG. 5.10e
Determination of
Ipl
lmpedances
245
for Ex. 5.106.
£3333F3fifiafifiifiéi9.515:§$§§sfl§kfifmfi§fifmfifiéfi§fifMWQEQ'MMQQF‘fiYMWWR$115533}'@flifimfly‘flfiwWW'fi????t¢§%§§?fiifiéfm%fli€f
‘
;'
s
A’;
.
vita 1&1
x
Nonideai Transmission Lines
5.11
TRANSMISSION LINES
For lines with losses
or
for
WITH
GENERAL FORMS OF DISTRIBUTED IMPEDANCES:
LOSSY LINES
filter-type
may generalize the
Z
impedance per unit length, and
admittance Y per unit length, as shown in
transmission circuits,
distributed series element in the circuit model to
the distributed shunt element to
5.11a. For
Fig.
for voltage
we
an
general
Steady-state sinusoids, using complex notation, the differential equations
a
and current variations with distance
are
then
dV
~
2
—~ZI
1
()
=
-—YV
2
()
dz
d1
dz
Differentiation of (1) and substitution in (2) then
sz
=
6122
’Y 2V
yield
(3 )
246
Chapter
Transmission Lines:
5
I
Z dz
I
+
CH
—>-
—>—
VT
de
0%
TV-b
dV
0
fi
(a)
L dz
R dz
———mmm—ww
«s
C
(121 -;
G dz
(b)
FIG. 5."
(a) Differential length of general transmission line. (b) Lossy line with series
resistance and shunt conductance.
where
=
7
The solution to
(3) may be written in
substitution of the
VZY
terms
of
(4)
exponentials,
be verified
by
expression
:=w&fl¢+vje
From
as can
(1), the corresponding solution for
(a
current is
1
I
==
-
Zn
[VJre‘Vz
-
V_e”z]
(6)
where
Z
23—:
o
7
Z
J;
(7)
.-
The characteristic
impedance Z0 is in general complex, indicating that the voltage
single traveling wave are not in phase. The quantity "y is called the
constant
and is also generally complex,
propagation
and current for
a
v=a+m=V5
so
that if
(5) is written in
terms
V
=
of
a
and
(&
,8,
V+e""ze'jfiz
+
Vueazejfiz
(9)
5.1!
Thus
Transmission Lines with General Forms of Distributed impedances
247
tells the rate of
exponential attenuation of each wave and is correspondingly
,B tells the amount of phase shift per unit
length for each wave and is called the phase constant, as in the loss—free case.
The formula for reflection coefficient derived in Eq. 5.7(8) applies to this case also,
——l in terms of a
remembering that Z0 is complex. To find the input impedance at z
reflection
coefficient
at
z
of
division
V_ /V+
O,
given
(5) by (6) yields
p
a
called the attenuation constant. The constant
=
=
=
V4427”
This may be put in terms of load
Z"
V_,e“7’
+
1 +
Pie—2y]
impedance by substituting Eq. 5.7(8):
__
0
ZL
cosh
2O
cosh
yl +20 sinh yl
”y! + ZL sinh 'yl
(11)
Transmission Line with Series and Shunt Losses
practice
is
one
A very important
in which losses must be considered in the transmission line. In
case
in
general
there may be distributed series resistance in the conductors of the line, and distributed
shunt conductance because of leakage through the dielectric of the line. Distributed
impedance
and admittance
then
are
(Fig. 5.11b)
(12)
Y=G+ij
Z==R+ij,
where L includes both external and internal inductance. These may be used as the values
of Z and Y in (4) and (7) to determine propagation constant and characteristic imped»
ance.
The formula
(10) applies
to
impedance transformations,
and the Smith transmis~
sion-line chart may be utilized with a modification which recognizes that y is complex.
The procedure is as in Sec. 5.10 except that, in moving along the line toward the
generator,
one moves not
‘
to the
exponential
e
along a circle but along a spiral of radius decreasing according
2‘”.
For many important problems, losses are finite but relatively small. If R / (0L << 1
and G/ wC << 1, the following approximations are obtained by retaining up to second—
order terms in the binomial
GV
R
at
+
z
expansions
L/C
(13)
2
z~
0
In
using
the
a)
VLC
l
51+
C
--
4co2LC
+
8th2
302
..
8&1}
R2
0-”-
RG
R“:
(4) and (7), with (12) substituted.
—-———
2VL/C
'8
of
8w2C2
+
( 14 )
8002142
+119. is“):
J
4w2LC
2wC
ZwL
(is)
foregoing approximate formulas, it is often sufficient to retain only first~
case ,8 reduces to its ideal value of 27r/A, a is computed
order correction terms, in which
]
R2
(al ?
8
+
for
Results Lines below)
G2
B
Aproximate
and
Low- s (See
a
Bl]
,8]
8(02C2
j
'[31
4w2LC
cos
5.1m
Table
+
+
Bl
Bl Bl
cos cos
20
alcos
4
]
[31 H]
sin sin
Line
Ideal
jZo jZL
jwVLC
+
+
,8! ,Bl
cos cos
31
tanfil
cot
120
~jZo
yl
71
ZL Z0
for
Zoi
Formulas
Genral
+
[a1
éi
j
jal jal
+
Lines
Transmio
[31 [31
sin sin sin sin
+
RG
]
]
'yl 'yl
ij)
Line
sinh sinh
+
Gen ral ij)(G
Z0 ZL
1mm Rem
+
R+ij G+ij cosh cosh
+
+
71 3/1
V(R
Zotanh
coth
ZO
ZL Z0
2i
Z0
Quantiy consta
,8
consta
Propagtion 7=a+fi Phase
248
impedance
Charcteis
line
Z,
impedance
Input
shorted
of
line
open
of
Impedance Impedance
line
of
+ZLal +Zoatl
dielctri
200:! ZLal
+
in
light
+
of
20 ZL ZL 20
Zo
20
end
input
line
velocity
along equals
from line;
measured
along
line
[32
32
sin
sin
v.
22 — ZL
ZL
jl,-Z0 120
-
,82
Vicos
z—' ’ ZL—‘ZO n+20 1+lpl
lpl
of
line
velocity ideal
Distance Wavelngth Phase
an
for
N<
D
1
,[.c08B
per
a1 a1
a! a!
yz
cosh cosh sinh sinh sinh
Z0 ZL Z0 ZL
+
+
a] a1
+
+
a! a!
[,ZO
—
yz
sinh sinh cosh cosh
cosh
ZL Zn ZL ZO
Vi
20
line
quarte—wv
of
cap itance
z
v
4w—’s'nh
v.
201
,COSV‘
h
ZL—Zo zL+zO 1+lpl 1—|p| condutace,
inductane, quanties quanties
1-
20
resitance,
line
half-wve
of
Impedance Impedance
V(z) [(2)
line
line
along along
Voltage Curent
p
end end
line input
load
of
coefi nt
Reflction
ratio
Standig~wve
Distrbued length Length Denotes Denotes
unit
i
C
G,
R,L,
1
249
L
Subscript Subscript
250
Chapter
Transmission Lines
5
Table 5.11b
Pormnias for Specific Transmission Line
Configurations
I
é—h-é
U
Formulas for
21:
Capacitance 0,
—"'—T
r
{mods/meter
ln
(3)
{2
(rs)
cosh“I
r;
External inductance L.
2::
Conductance G.
Siemens/meter
:-
2__wt”
111:)"111—(22)
r
0
6
______
2:
re
+
a
:1
a
(ii)
u
3
__—___~
(2)
cash“
.1)
m
a
Internal inductance L;,
1 +
1,213:
8&2?9’
“[1
E: __‘/d
rd
1/(a/d)’— 1
r;
b
0—11
='
cosh"
g
me”
t:
2R1!
E: .1
Rats tanceR ' ohms/meter
(b
:3
___._
(~)
(5)
{L cash.4
_l-‘_ In
2:
hemys/meter
a
re
+
(1
_
M5)
I]
b
a
2.3;!
b
1+4
R
beams/meter (for high
frequency)
4
,-
—-
‘9
a
_
Characteristic impedanceat‘
high frequency Z 0. ohms
i In
2:
2. cosh"
([9)
r;
1r
it” [MC ”9]
1 +q
r
(3)
I +
:1
11
411’
,
2
b
16p
(I-q’)
__
601116?)
Zn forairdielectric
1”) cosh
'
Attenuation due to conductor
2.120111{ [2p———
‘(&)=1201n(:{)1(l.+ql)+4
.p
8
“—7
~
it all! >> 1
12019-
3
—
16;)
(1
—
4%)}
b
..i
‘L—
5
220
a,
Attenuation due to dielectrio «4
‘—
Total attenuation dBlmeter
4—'
95.9
,
..
are
=
2
I.(E’.:)
:
e’
x
8585b: + as)
Phase constant for low-log
lines fl
All units above
3.17..
2
u
1"
:7
a
—+
if
>
)t
mks.
4 —-.ie'
permittivity. funds/meter
permeability, henrys/mcter
a
n
==
:4
==
1?
=Vp/1 ohms
for the dielectric
a"
=
3.
=4
A
=-
loss factor of dielectric
U‘lw
akin effect surface mistivity a! conductor. ohms
=
wavelength in dielectric
Curtis, Bell System
Formulas for shielded pair obtained from Green. Leibe. and
Tech. Jam.. 15, pp. 248-284
(April 1938’.
from (13), and Z0 includes a first—order reactive part, given by the last part of (15). Note
that the first-order effects of the two loss factors tend to cancel in ZO whereas they add
in
0:.
Several of the
important formulas
for loss-free, low—loss, and general lines are sum—
properties of lines of several different cross
marized in Table 5.1 la. Formulas for
sectional forms
are
listed in Table 5.11b.
Physical Approximations for low-loss
Lines
The
approximate form (13)
for
attenuation in transmission lines with small losses may be derived from physical ideas.
This approach will be especially useful in estimating attenuation of more general guid-
ing systems
to
be considered later in the text. Let
us
consider the
positively traveling
5.1:
voltage
wave
Transmission Lines with General Forms of Distributed
of
(9) and its corresponding
V
=
I
=
The average power transfer at any
It is assumed here that the
251
current:
V+e"‘”e_jflz
(16)
I+e_"ze_jfiz
(17)
position
WT
lmpedances
=
is then
given by WT
%Re(VI*):
=
%V+I+e“2az
imaginary part
of
Z0
is
(18)
negligible
so
that I + is in
phase
with V +.
The rate of decrease of the average power (18) with distance
the average power loss wL in the line, per unit length:
along
the line must
equal
1
5W]-
~7az
0r
0‘
W’L
:
( 19 )
__
2W7
This is
an important formula relating attenuation constant to power loss per unit length
and average power transfer. By the nature of the development, it applies to the atten»
nation of a traveling wave along any uniform system.
To
apply (19)
to a
transmission line with series resistance R and shunt conductance
first calculate the average power loss per unit length, part of which comes from
the current flow through the resistance and another part from voltage appearing across
G,
we
the shunt conductance. For convenience
W“
=
13R
ViG
+
calculate wL and
=
Vi
by
the
wave
2
at
z
=
1
Wr
So
=
-—
2
1
V+ I +
=
WT
at
z
=
O:
R
—
2
2
The average power transferred
we
-
2
i 25]
G +
*7
( 20 )
O is
V3:
—-
20
( 21 )
(19) gives attenuation in agreement with (13):
1
0‘
=
-
2
[
R
620 +
20]
—
NP/m
( 22 )
The neper (Np) is a unit-free name for attenuation that measures the decay of voltage
amplitude. One neper per meter indicates that the amplitude has decayed to 1/2 of its
incoming value in 1 m. The decibel (dB) is an alternative measure describing the rate
of power decrease
by
the formula 10
longTZ/WTI.5 Attenuation in decibels per meter
is 8.686 times the attenuation in nepers per meter.
5
are often used for ratios of voltage amplitudes using the formula 20 logmvg/ V1.
only correct if the voltages are across identical lmpedances, as in the case of
voltages at points along a transmission line.
Decibels
This is
252
Transmission Lines
Chapter 5
Example
ATTENUATION
IN A
5.1 t
.
THIN-FILM TRANSMISSION LINE
us find the attenuation in an aluminum thin—film parallel~plane transmission line for
signal of 18 GHz. The structure has the form in Fig. 2.50 and flinging fields will be
neglected. The metal thickness 11 is 2.0 ,um with a dielectric thickness d also of 2.0 pm.
The width of the conductors is typical of photolithographically defined lines, w
10
pm. The relative permittivity of the dielectric is 3.8 and it is assumed to be lossless.
1.67 X 10“10 F/m.
From Eq. 1.9(3) the capacitance per unit length is C
sw/d
From Eq. 2.5(3) the external inductance per unit length is L
2.51 X 10‘7
nod/w
Let
a
=
=
=
=
=
H/m.To see how to treat the internal inductance and resistance of the conductors, the
depth of penetration must be compared with the film thickness. From Table 3.17 the
0.0826/\/)-°, so for 18 GHZ, 5 0.616 um.
depth of penetration for aluminum is 3
Thus the aluminum films are 3.2 times the depth of penetration so they are well
approximated by arbitrarily thick layers. Then the internal inductance and resistance
are given by the surface impedance. From Table 3.17 the surface resistivity is RS
3.26 X 10‘7W, so from Eqs. 317(4) and 3.17(6) the surface impedance is
=
=
2
ZS
and the internal
irnpedance
:
10*7\/}(1
3.26 x
per unit
length
+
(23)
j)
for both electrodes is
22
Z.
The characteristic
8.75 x
=
=
T
impedance
is found from
103(1
ZO
+
j)
(24)
V L/ C, where L includes both
=
ternal and internal inductances. The latter is found from
7.74 X 10‘8 H.
n
(24) by dividing by
this to the external inductance: and
w so
the
ex-
Li
=
and
substituting
Adding
44.3 0. Note that if we had neglected internal
capacitance into 20, we find ZO
inductance, Z0 would have been calculated as 38.4 0. The resistance per unit length R
is the real part of (24). Substituting Z0 and R in (22) we get the attenuation constant:
the
=
2
CK:'2?O
From this
we see
5.12
Suppose
sum
that the
wave attenuates
by
Np/cm
about
a
(25)
faCtor of
e
in
a
distance of 1
cm.
FILTER—TYPE DISTRIBUTED CIRCUITS: THE w—B DIAGRAM
the distributed series
inductance and
0.988
capacitance
impedance
in series,
as
of the
general
Fig. 5
shown in
transmission line is formed
.
12a. The
prepagation
by
constant
'y is then
.
'Y
=
1
.
Ja’Cz .1le
+
.
=
ijI
J‘”
L1C2
1
““
or?
'52“
(1)
Filter-Type Distributed
5.12
Circuits: The
253
(0—3 Diagram
w
(jwlCl) dz
jw lez
W
Fir—<1
J'w
T
a?
ng2
/
/
//
\ SiODe
Up(u.‘1)
4‘3
(a)
FIG. 5.12
=
fl
j,
(b)
(a) Filter-type distributed circuit. (b) Its w—B diagram, showing phase velocity.
where
CDC
The
a)
:
(L1C1)_1/2
(2)
interesting characteristic of this system is that for the lower range of frequencies,
is purely real, representing an attenuation without losses in the system
< wt, y
y=a=wfilC2<9§-l),
w-
The attenuation in this circuit
occurs
(3)
co<wc
below the cutoff frequency defined
by (2),
so
that
the system is a distributed high-pass filter. The reactive attenuation which occurs arises
essentially because of continuous reflections in thesystern, and is of the same nature
as
the attenuation in
a
loss~free, lumped-element filter in the attenuating band.
j,8
frequencies above we, the propagation constant 'y is purely imaginary so y
given by (1). It is found useful to plot relations between B and a) with fl on the
abscissa and co on the ordinate. These are called m—B diagrams. Figure 5.12!) shows
the (0—3 relation (1) for the line discussed here. Note that B goes to zero for w
we
and does not exist for w < arc. We saw in (3) that, for this line, there is only attenuation
for a) < me. An important reason for choosing the coordinates of the CO—'B diagram as
done is that the phase velocity at any frequency, which from Eq. 5.7(6) is up
(0/13,
can be seen immediately as the slope of a line to the origin from the curve, as illustrated
in Fig. 5.12b. We see that the phase velocity for the line under consideration is
For
=
and is
=
=
l
up
(02
[ £02]
1
=—__
r—LlCQ
—-
—£
4/2
(4)
strongly for frequencies just above cutoff. Signals with several
frequency components propagating in this range will thus have large dispersion, as will
which is seen to vary
be discussed
more
in Sec. 5.15.
254
Chapter
5
Transmission Lines-
Resonant Eransmnssnon lines
PURELY STANDING WAVE ON AN IDEAL LINE
5.13
important special case of standing waves on a transmission line, introduced in
general in Sec. 5.8, is one in which all the incident energy is reflected. The reflected
00. It is clear from Eq. 5.7(8)
wave has the same amplitude as the incident wave so 8
the
if
of
that Ipl
1 so that |V_I
following conditions exist: (1) short—
|V+| any
circuit load, ZL
00; (3) purely reactive load. The last
O; (2) open-circuit load, ZL
is less obvious than the others but is easily shown (Prob. 5.13b). In each case [VJ
{17+} because the load cannot dissipate power and it must, therefore, be fully reflected.
Suppose that a transmission line, shorted at one end, is excited by a sinusoidal voltage
O. The short
at the other. Let us select the position of the short as the reference, 2:
z
be
must
From
the
condition
at
zero.
that,
0, voltage
always
imposes
Eq. 5.7(4),
An
t
=
=
=
=
=
=
=
V(O)=V++V_:O
If V__
=
-
+
is substituted in
V
=
I
=
V+[e‘j'8"
V
—
1.
Voltage
typical
is
for
always
8152]
standing
5.7(5),
=
.
.
ins—'15:
+
ZO
These results,
and
Eqs. 5.7(4)
e132]
=
~—2jV+
V
2 —+
waves, show the
zero not
only
V=0
at
at
sin
cos
Z0
32
(l)
(2)
[32
following.
the short, but also at
multiples
of
A/ 2
to the
left; that is,
A
—~,Bz=mr
or
2:
-n-2-
32 is an odd multiple of 77/2. These
quarter-wavelength from the short circuit.
Figure 5.13 shows this and also the time evolution of the voltage found by multiplying (l) and (2) by em and taking the real part. Time origin is chosen so that
V+ is real.
3. Current is a maximum at the short circuit and at all points where voltage is zero;
it is zero at all points where voltage is a maximum; Figure 5.13 shows the time
variation of current along the line.
4. Current and voltage are not only displaced in their space patterns, but also are 90
degrees out of time phase, as indicated by the j appearing in (1) and as seen in
Fig. 5.13.
5. The ratio between the maximum current on the linezand the maximum voltage is
ZO, the characteristic impedance of the line.
6. The total energy in any length of line a multiple of a quarter—wavelength long is
2.
Voltage is
are at
a
maximum at all
distances odd
points
multiples of
for which
a
5.13
Purely Standing Wave
on an
255
Ideal Line
t—%—~t
Vlz)
_:t
2
-51
-5
6'
6
wt=0.1r
%
6‘
_5__-_
_.
I
2
l
3.7.
§
“3.1
1
—§
I
rfiz
0
-7r
271'
6
5
1(2)
wt=0
z
3
5
i
0:63
~27r
FIG. 5.13
Time evolution of
voltage
and current
on a
shorted transmission line. The
zeros
and
extrema remain at the same locations.
constant,
merely interchanging between energy in
magnetic field of the currents.
the electric field of the
voltages
and energy in the
magnetic energy of the
voltage is zero everywhere
along the line. Current is given by (2). The energy is calculated for a quarter-wavelength
of the line, assuming VJr to be real.
To check the energy relation
just stated,
currents at a time when the current
pattern is
let
a
us
calculate the
maximum and
256
Chapter
0
L
UM
Transmission Lines
5
0
L
4V2
—3L~cos2 Bzdz
'
=-fnailIzdz=~f
2
2
mm
23
o
2
=
Since
,8
=
2L
1
[-2-
V:
20
+
---
2
277/)L by Eq. 5.7(7),
2,82]
sin
4B
the
foregoing
4/4
is
VzL/‘t
U
=+—
3
The maximum energy stored in the distributed capacitance effect of the line is calquarter-wavelength when the voltage pattern is a maximum and current
culated for the
is
everywhere
is
Voltage
zero.
given by (1).
0
UE
=
SJ
2
M2
-}I/4
o
C
dz
=
~f
2
«um/4
4V3
sin2 32 dz
0
=
By
the definition of
2CV2+
20, (3)
1
z
[—
——
—
2
sin
43
2w]
CVZA
=
—+
(4)
4
W4
may also be written
VZLA
UM
=
—+—-—
VZCA
:
+—
4L/C
4
=UE
(5)
Thus, the maximum energy stored in magnetic fields is exactly equal
in electric fields 90
degrees
later in
phase.
It
can
also be shown that the
to
that stored
sum
of electric
and
magnetic energy at any other part of the cycle is equal to this same value.
Expressions (l) and (2) are also valid for a transmission line with short circuits both
at z
0 and another point where z
n( 77/3), for any integer 12. With some way to
couple energy into a section of line short—circuited at both ends, at a frequency such
that the above criterion on 2 is satisfied (recall that ,B
(0/0), there will be voltages
and currents satisfying (1) and (2). At each such frequency, the line is said to have a
resonance. This idea will be developed further in Sec. 5.14.
=
=
——
=
5.14
INPUT IMPEDANCE AND QUALITY FACTOR
TRANSMISSION LINES
FOR RESONANT
Resonant systems play a very significant role in communication systems for impedance
matching and filtering and we have already seen some aspects of this in Ex. 5.lOd,
where resonant sections of lines
were
used for
matching impedances.
In Sec. 5.13
we
short~circuited ideal lines. In the present section we con—
analyzed standing
sider a low-loss line shorted at either one or both ends so that standing waves similar
waves on
to
those discussed in Sec. 5.13
occur.
The line is
supplied?E by a voltage source connected
5.14
Input Impedance and Quality Factor for Resonant Transmission Lines
257
F“
a
“j
l
t——%
FIG. 5.140
a
a
Resonant transmission lines driven
by voltage
sources at
positions
of maximum
voltage.
at a
voltage maximum in either of the ways shown in Fig. 5.14a. We will find approxexpressions for the resistance seen by the source and for the quality factor Q of
imate
the line, considered
For
as a resonant
ideal line, there
circuit.
points of voltage maximum and zero current at odd
multiples of a quarter—wavelength from the short—circuited ends so the impedance is
infinite there. When losses are present, however, the impedance at these positions is
high but finite, representing the energy dissipated in the losses of the line. Let us find
these losses approximately for a line of n quarter-wavelengths using the expressions
for voltage and current derived for an ideal line, Eqs. 513(1) and 5.130), assuming
that they are not greatly changed by the small losses. The average power dissipated in
an
are
the shunt conductance is then
[IA/4
WG
f
=
(2v+
sin
9
[32)"
dissipated
“/4
WR
:
O
The
input
across
of
resistance
(at
there is
2V+.
4V"G
+
/\
5-
4
4
=
(1)
in the series resistance is
“
a:
R
T”
o
'3
2v+
cos
d2
2
4V3R
n/‘t
4Z2O
—4_
(2)
voltage maximum) must be such that the voltage appearing
produce losses equal to the sum of (l) and (2). The magnitude
a
this resistance will
voltage
dz
2
o
and the average power
7
G
—-
Thus
l(2V+)2
2
Ri
:
nViA
4
G +
5;
Z5
258
Transmission Lines
Chapter 5
01‘
R”
A
general expression
for the
Q
For
a
820
__
mezo
We
see
that
on
=
FL
resonant
system is
0U
to
z
average power loss
(4)
WL
quarter-wavelengths, the stored energy for each
for the ideal line, Eq. 513(5), and the power loss
n
quarter-wavelength is taken as that
is given by the sum of (l) and (2). The
2
(R/ZO)]
stored )
0(energy
a)
resonant transmission line of
Q
+
Q of any
factor
quality
Q
(3)
_
result is
4wOCVin/\
4Vin/\[G + (12/23)]
wOCZO
=
6-20
+
(R/ZO)
(5)
Q is independent of n; this results from the fact that both the stored energy
dissipated are proportional to the length. Thus, Q is a property of the
and the power
line, independent of the number of resonant quarter-wavelengths.
The input resistance for a shorted quarter-wavelength section or at the maximum
voltage point of a line rut/4 (12 even) long shorted at both ends can be rewritten using
(3) and (5) with Eqs. 5.2(8) and 5.2(14) and 5.7(6) and 5?.7(7):
8Q
_
4QZo
(6)
-
,~
nAwoC
n77
supplied to maintain a given voltage level; Q
higher input resistance.
leading
If the frequency and, therefore, A are changed so the distance from the input point
to the short circuit differs from A / 4, the input impedance acquires a reactive component
of first—order importance. With the same amount of frequency change, the voltage and
current patterns do not change much, so the resistive part (6) does not change much.
It will be convenient to complete the analysis in terms of admittance. The susceptance
that arises is in parallel with the conductance equivalentof (6). Let us calculate it for
the lower circuit in Fig. 5.14:2. It consists of two susceptances in parallel, that for the
A/4 section on the left and that of the remaining (12
1))t/ 4 portion on the right. For
each section, the load admittance is infinite so Eq. 527(14) gives, in the lossless
approximation,
The
input resistance measures
increases
as
the power
the losses decrease,
to a
—-
jBi
where
Y0
LettingB
(12
—-
=
1/20
=
w/vp
=
IL and [R are
(00(1 + 6)/vp
and
=
——jY0(cot BIL
1)7T/ 2, the cotangents in
(7)
BZR)
(7)
lengths of the left and right sides of the line.
77/2 and ,BOZR
BO + BOB and taking [SOIL
be approximated for small 5 to give
the
=
can
7T
B.=.-Y
+ cot
—5+
=
(n
—-
2
l )77
8]
==
rm
z
_‘
5Y0
(8)
Input Impedance and Quality Factor for Resonant Transmission Lines
5.14
Then
using (6)
and
lower circuit of
(8),
Fig.
we
have the admittance at the feed
point in
the
259
ml/ 4 line in the
5.14a:
1
MT
Yr
0(55 J5)
.
+
“
‘5‘
(9)
For the upper circuit in Fig. 5.14a the cot BZR terms in (7) and (8) are missing so (9)
describes that circuit with n
1. From this we see that the fractional frequency shift
=
for which the susceptance becomes
circuit sharpness, is
equal
the conductance,
to
a common measure
of
1
51
(10)
~
E
Of
Q
where 2
Af 1
reaches
is the
frequency
C00
a
__
f0
(11)
“
’"
2
Aw,
width between
2Af,
points where
we). Thus Q,
the admittance
magnitude
by (4), is useful
lumped-element circuits.
defined
2 times its value at
resonance
((0
of
frequency
response, as it is for
have Q’s of thousands in the UHF range of
as a measure
of
sharpness
Resonant low-loss transmission lines
=
as
can
frequencies.
Example
5.14
OPEN-ENDED PARALLEL-PLANE TRANSMISSION LINE
Consider
standing
waves
in
an
open-circuited
section of transmission line and the
con-
tribution to Q from radiation at the ends. Radiation loss may be expressed in terms of
load conductances GL at each end, as pictured in Fig. 5.14b, and when radiation is
GL
GL
FlG. 5.14b
Model for
open-ended
transmission line.
269
small, GL
line,
Chapter
<<
Transmission Lines
5
1/ZO, fields in the line are essentially those of a completely open—circuited
V
=
2V+
I
=
—~j
cos
2V+
——
(12)
,82
sin
20
(13)
B2
Power loss from the two end conductances is then
w,
Energy storage
for
a
length
some
O
U
=
2‘
multiple
2V
2
2+)
of
a
<14)
a,
half—wavelength
'J
CV~ A
4311—-
C
-+(2V )- cosqsz
=
dz
=
the
Q from
(15)
2
"Hui/2 2
Using (4),
is
the radiation component is found to be
Q
m0 szi
=
ma"
(16)
=
86L
4ZOGL
There may also be contributions to Q from conductor and dielectric losses,
Losses, when small, add, so reciprocals of Q’s add also:
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A function of time with
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AND
‘3‘
sci
in
(5).
(17)
=——+——+~.
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Topics
ENERGY VELOCITIES
arbitrary wave shape may be expressed as a sum of sinusoidal
by Fourier analysis. If it happens that up is the same for each frequency com~
ponent and there is no attenuation, the component waves will add in proper phase at
each point along the line to reproduce the original wave shape exactly, but delayed by
the time of propagation
z/vp. The velocity up in this case describes the rate at which
the wave moves down the line and could be said to be the velocity of propagation. This
case occurs, for example, in the ideal loss~free transmission line already studied for
which
up is a constant equal to (LC )"1/ 2. If Up changes with frequency, there is said
waves
26'!
Group and Energy Velocities
5.15
to be
dispersion and a signal may change shape as it travels. This causes distortion of
analog signals and limits data rates because of the spreading of pulses in digital signals.
In the nondispersive situation described above, a wave of a given shape propagates
along a line without distortion. The velocity of the group of frequency components is
the same as the phase velocity of any one. In many transmission systems the velocity
of the envelope of the wave such as that in Fig. 5.15a can be different from the phase
velocities of the frequency components, and it is useful to introduce a so—called group
velocity to describe the motion of the group, or envelope of the wave. This is the typical
case when a high~frequency carrier is modulated by a digital or analog
signal.
Let us consider the simplest possible group, a wave having two equal-amplitude
sinusoidal components of slightly different frequency. The voltage at z
O with unityarnplitude components is
=
V0?)
Then for
WI, 2)
=
in which
can
lossless line, the
a
sin[(co0
[3’ is
to
be
-
dw)z‘
From
up
voltage
([30
-
regarded
as a
-
at
(1001‘
any
point
(13):]
-—
+
+
dw)r
(1)
is
sin[(cuO
function of w;
+
sin(au0
+
dw)t
(13 corresponds
—-
to
(,80
+
dry.
Expression (2)
dB)z]
(2)
be put in the form
WI,
for
sin(coO
=
=
=
2
cos[(dw)t
-
(dfi)z] sin(w0t “3305)
(3)
we see that the voltage in this wave group has the form shown in Fig. 5.15b
instant of time. The sinusoid of center frequency moves at the phase velocity
(3)
one
wave
2
coo/BO
but it
cosine term
whereas the
moves at a
a
envelope, described by cos[(dw)t (dB)z], has the form of a
velocity. This is found by keeping the argument of the
——
different
constant:
(4)
”3:
Velocity Ug is called the group velocity and13 shownin Fig. 5.151). Note that vg is the
slope of w—~B curve at the center frequency of the group. Forming the derivative dvpMa)
using up
(0/3, another useful form for group velocity can be derived:
=2
U
P
vg
2:
1
—
(w/vadvp/dw)
Group velocity
FIG. 5.15:1
Envelope
or
group
velocity.
(5 )
262
Chapter
U
FIG. 5.15b
5
Transmission Lines
U
U
Phase and group velocities for
a
group of two sinusoids of
slightly different
frequencies.
For groups more complicated than the two-frequency one considered above, we will
Show later by Fourier analysis that the envelope of a modulated wave retains its shape
long as 218 is constant (i.e., the CO—"fi curve is linear) over the range of frequencies
required to represent the wave. In that case, a pulse such as that illustrated in Fig. 5.15a
would retain its envelope shape and propagate with delay time rd over distance I:
so
l
Td=*“‘: l
vg
If
dB/dw
319.
do)
(6)
over the frequency band of the signal, there is a broadening
envelope. This is known as group dispersion and will be seen in
several later examples.
Group velocity is often referred to as the “velocity of energy travel.” This concept
has validity for many important cases, but is not universally true.6'7 To illustrate the
basis for the concept, let us define here a separate velocity ()3 based on energy flow so
that power transfer is stored energy multiplied by this velocity. That is,
or
is not constant
distortion of the
..
DE
6
7
WT
u
J. A. Sfrafion, Electromagnetic Theory, pp. 330~340, MCGraw—Hill, New York, 194 l.
L. BriI/ouin, Wave Propagation and Group Velocity, Academic Press. New York, l960.
(7)
263
Backward Waves
5.15
WT is average power flow in a single wave and aav is average energy storage per
length. If this definition is applied to the ideal transmission line of Sec. 5.7, we
find that 05
DS),
up. More interesting is the case of the filter-type circuit of Fig.
5.12a, which is a case of normal dispersion (dvp/da) < 0). Here
where
unit
=
2
11*
1
WT ==—-ZII*
O
2
1
“av
:
_
2
CZVV'i‘
2
+
a)“
[C2( 012)]
—‘
[4111*
-——-—€
+
2
2
C1
[1*
2
(O‘C‘f
—
'7
2.
C922
L
1/2
0
L
==-———
7
LIP“
—“——°+1+%II*=—’——
2
L1
=—‘
4
to
SO
1/2
9
0);
(L1C2)—V"[1 “CU—2]
’3
()5
=
(8)
"“
equal to group velocity dw/dB,
and is different from phase velocity.
This is
as can
be found
by differentiating Eq. 312(1),
identity of group velocity and energy velocity can also be shown to be true for
simple waveguides, and it also applies to many other cases of normal dispersion. It
does not usually apply to systems with anomalous dispersion (dvp/dw > 0), including
the simple transmission line with losses. In any event the concept of an energy velocity
is useful only when there is limited dispersion so that the input signal can be recognized
The
at the
output.
5.16
BACKWARD WAVES
phase velocity and group velocity have opposite signs is known as a
Conditions for these may seem unexpected or rare, but they are not.
Consider for instance the distributed system of Fig. 5.16a in which there are series
A
wave
in which
backward
wave.
capacitances
and shunt inductanceswthe dual of the
simple transmission
line of Sec.
5.2. From Sec. 5.11,
"”3 'VZ—YT
_-
This
w—B
__
__
relation is shown in
Fig.
_1_
_L
(ij)(ij)
5.1619. The
up=g==
phase
—-w2
z-
j
and group velocities
LC
(1)
ME
are
(2)
and
do)
v
=
—-
7
2
arVLC
(3)
264
Transmission Lines
Chapter 5
(is)
dz
__..‘
(3-253)
“‘2
(a)
(.0
Slope
z
>
ug
Ogll
/
\
/
Slope
=
up
\
<
0\
(b)
FIG. 5.16 ((1) Equivalent circuit for a transmission line which propagates backward waves.
(1)) wufi relation for line of (a) showing phase and group velocity directions discussed in the
text.
So it is
ward
seen
waves.
that this very simple transmission system satisfies the conditions for back—If energy is made to flow in the positive 2 direction, group Velocity will
direction, as this is a case where 128 does represent energy flow (see Prob.
5.160). However, the phase becomes increasingly negative or “lagging” in the direction
be in this
of
propagation
velocity.
because of the C—L
configuration.
Thus there is
a
negative phase
There are many other filter-type circuits having other combinations of series and
shunt inductances and capacitances on which can exist waves with increasingly lagging
phase in the direction of energy propagation. Also, all periodic circuits (Sec. 9.10)
equal numbers of forward and backward “space harmonics.”
5.17
have
NONUNIFORM TRANSMISSION LINES
varying spacing or size of conductors, as illustrated in Fig.
5.17,
analysis would lead one to consider
with
distance
in the transmission-line equations.
and
as
admittance
varying
impedance
is not this simple, but it is a
so
that
the
formulation
fields
be
distorted
Actually,
may
and
the methods discussed apply
in
a
number
of
important cases,
good approximation
For
a
transmission line with
a
natural extension of the transmission-line
FIG. 5.17
to some wave
problems
265
Nonuniform Transmission Lines
5.17
with
Nonunifomi transmission line.
spatial
variations of the medium (i.e.,
materials). The remainder of this section will consider
cases
inhomogeneous
where such nonuniform
transmission-line
If
line
theory yields a good approximation.
impedance and admittance per unit length vary with distance,
equations corresponding to Eqs. 5.1 1(1) and 511(2) are
dV
2
72—)
(11(2)
~th>1<2>
=—-
denoting
V”
To obtain
a
differential
equation
in
=
—-
(2)
_
"
-
to z,
(1)
Y(Z)V(4)
__
“21;Differentiate (l) with respect
the transmission~
z
[21’
differentiation
+
voltage alone,
by primes:
2’1]
(3)
I may be substituted from
( 1) and 1'
from (2). The result is
Z!
V”
A similar
differential
procedure, starting
equation in I :
I”
'
—-
(E)V’
-—
(ZY)V
O
=
with differentiation of
-
(2;)1’
—
(ZY)I
=
(4)
(2), yields
O
a
second~order
(5)
equations for a uniform
nonzero, representing the nonuniform line
(Sec.
the
be
solved
discussed,
numerically for arbitrary variations of Z and Y
equations may
with distance. A few forms of the variation permit analytic solutions, including the
“radial transmission line,” where either Z or Y is proportional to z, and their product is
constant. Another important case is the “exponential line,” which is taken as the example for this article.
If Z’ and Y
line
are zero,
(4) and (5) reduce,
5.11). When these derivatives
as
are
they should,
to
the
266
chapter
Transmission Lines
5
Exampie
5.17
LINE WITH EXPONENTIALLY VARYING PROPERTIES
Let
us
consider
a
loss-free
Z
These variations
equations
line with Z and Y
exponential
=
Y
ijOqu,
yield constant values
varying
follows:
as
(6)
ijOe"qz
=
of ZY, Z’ /Z, and Y’ / Y so that (4) and (5) become
with constant coefficients,
V”
qV’
-
I" +
+
(11’
+
ngOCOV
wZLOCOI
=
O
(7)
:
O
(8)
exponential propagating form,
These have solutions of the
V
=
I
Voe”7’lz,
(9)
106‘”)?
=
where
to
I
l
:3
1+
We
see the interesting property
frequencies a) < we where
IQ
R /“"‘\ V
\/<§>
IQ
3N5‘£3
(10)
(021.000
(11)
l
N
to
722+
low
IQ
1+
N
of “cutoff”
——
again,
for yl and 3/2
are
purely
real for
’)
agree,
=
(g)
(12)
represented by these real values, like that for the loss-free filter-type
lines, is reactive. This represents no power dissipation but only a continuous reflection
of the wave. For a) > me, however, the values of 7 have both real and imaginary parts,
The attenuation
behavior different from that of the loss-free filters.
which is
a
represent
no
imaginary
The greatest
use
of this type of line is in matching between lines of different charUnlike the resonant matching sections (Prob. 5.7c), this type of
acteristic
impedance.
matching
is insensitive to
frequency.
Note the variation of characteristic
2
Z(Z)
Thus
Again the real parts
approach purely
power dissipation (see Prob. 5.17b). The values of y
values representing phase change only for a) >> we.
.._
.—
__
1(2)
__._.._
Joe"
7’23
_.
VO
__
r0
-
6
“(71-72%
impedance:
_
_._
Z°( 0 )6 q;
ZO can be changed by an appreciable factor if qz is large enough.
( 13 )
The transmission-
267
Problems
line
approximation will become poor, however, if there is too large a change of Z
a wavelength or in a distance
comparable to conductor spacing.
The design of nonuniform matching sections is explored in detail by Elliot.8
and
Y in
PROBLEMS
5.23 Sketch the function
1
z
V(z, t)
=
wz/v for values of
traveling-wave behavior.
tot
versus
5.2b
(i)
Derive
Fig.
cos
=2
w<t 2;)
+
0, 7r/ 2,
7r,
+
N
3cos
H
37r/2
and
2w<t .)
+
explain
I
how this shows
expression for the characteristic impedance of the parallel-plate line in
having a width w and spacing a neglecting the internal inductance of the
an
5.2
conductors. Thin—film transmission lines in
some
computer circuits
can
be mod-
approximately by the parallel-plane line. The line width is usually about
5 pm and the spacing is by means of dielectric of 1~,u.rn thickness and relative
permittivity 2.5 (as is usually true for dielectrics, the relative permeability can be
eled
taken as ~19).
(ii) Calculate the characteristic impedance Z0 and wave velocity 0.
(iii) Suppose the dielectric thickness is halved and find the new values of Z0
A better model for such lines is
in
given
Chapter
and
v.
8.
capacitance per unit length of a parallel-wire line having radii R with distance 2d
between axes is C
we / cosh” 1(d/ R). Find characteristic impedance of a line with
5.2c The
=
air dielectric and
radius is 0.5
5.2d Calculate
spacing
between
1
axes
cm
if
(i) wire radius is
2
mm
and
(ii) wire
mm.
propagation
time
along
the
following transmission
lines
interconnecting
computer elements:
(i)
A thin-film line
(ii) Transmission line
apart, er
2
11) between circuit elements 100 um apart
(er
interconnecting two devices on a silicon computer chip 1
on
GaAs
=
mm
12
(iii) Coaxial cable 100
m
long with
8r
=
2.4, used
to
interconnect computer terminal
and central processor
5.2e A second type of solution to the wave equation, to be studied later in the chapter, is
the standing wave solution. Find under what conditions the following such solution
satisfies Eq. 5.2(7):
V(z, 2‘)
Find the current
[(2, t) corresponding
==
to
Vm
this
cos wt
sin
,82
voltage distribution.
1?. S. Ellioh‘, An Introduction to Guided Waves and Microwave Clrcui’rs,
Hall, Eng/ewood Cliffs, NJ, 1993.
Chap. 8, Prentice
268
5.2f
Chapter 5
Transmission Lines
the cosine and sine in the expression of Prob. 5.2e in terms of complex expo~
Cos x
é-(ejx + 6‘”), and so on, and after multiplying out, show that the
products can be interpreted as traveling waves of the form of Eq. 5 .2( 10).
Expand
nentials,
=
5.2g
Examine the expression for characteristic impedance of a coaxial transmission line,
Eq. 5.2(15),and explain why it is difficult to obtain high characteristic impedances for
such a line without having unreasonable dimensions or very high losses. Plot dc resistance per unit length for such a line versus Z0 if the outer conductor is a tubular copper
conductor of inner radius 1 cm and wall thickness 1 mm, and the inner conductor a
solid cepper cylinder.
5.2h
Repeat Prob. 5.2g
but
plot
ac
resistance at 100 MHz
using
the
approximation
of Ex.
3.17.
5.2 in" Use Eqs. 5 .2(3) and (4) to Show that the spatial rate of change of power flow on an
ideal line is equal to the negative of the time rate of change of the stored energy per
unit length.
simple properties of transverse electromagnetic (TEM) waves, utilize
equations in rectangular coordinates (though the boundaries need not be
rectangular). Take the dielectric as source-free and without losses. Show that if
0 and E:
O,
H;
5.3 To show
some
Maxwell’s
=
=
(i) Propagation
(ii)
must be at the
Both E and H
5.4a Derive
5.4!) Plot
p2
(which
are
velocity
of
light
Eq. 5.4(7) directly from voltage and
and l
p2
-
as
in the dielectric.
transverse) satisfy the Laplace equation in x and y.
current for the load.
functions of RL/ZO and note
region
of reasonable power transfer
to the load.
5.5a
Analyze, as in Ex.5.5a and with drawings like those in Fig. 5.5a, the case of a pulse of
length t1/5 reaching a termination at I
vz‘l with RL
220. Find an expression for
the energy dissipated in the load in terms of the voltage of the incident pulse.
:
=
50 .Q and length I
5.5b A transmission line of characteristic impedance Z01
200 m is
connected to a second line of characteristic impedance Z02
100 Q and infinite
100 V is
length. Velocity of propagation in both lines is 2 X 108 m/s. Voltage V0
==
=
=
=
line 1 at
0. Sketch current versus distance 2 at
t
1.3 us. Calculate power in the incident wave, the reflected wave, and the wave
transmitted into line 2, showing that there is a power balance.
suddenly applied
at
the
input
to
t
=
=
5.5c Plot the reflected
SOQandRL
=
wave
from the terminal of computer No. 2,
as
in
Fig. 5.4g, if Z0
=
10.0.
0 a charge distribution is suddenly placed in the central portion of an infinite
5.5d At t
line as in Ex. 5 .50 except. that the voltage distribution in z is triangular, with maximum
i l m. Find voltage and current distribuO, falling to zero at 2
voltage V0 at z
tions at t
1.667 ns and at t i 5 us, as in Ex. 5.50.
=
=
=
=
5.5e
Repeat Ex.
5.5e but with the transmission line terminated with inductor L.
5.5f The
problem is as in Ex. 5.5e except that the transmission line continues beyond the
capacitor, where it is terminated by its characteristic impedance. Find V_(t), Vc(t), and
V20) in this case, where V20) is the voltage at the input to the continuation transmis~
sion line.
5.63 An ideal open ended line of length l is charged to dc voltage V and shorted at its input
at time t
0. Sketch the current wave shape through the short as a function of time.
=
269
Problems
5.6b A charged cable is connected suddenly to a load resistor equal to its 50-!) characteristic impedance. If its length is 3 m and its phase velocity is one-half the velocity of
light, how long is the pulse in the load? Sketch the waveform at the midpoint of the
line assuming the cable is initially charged to 100 V and the load is 25 0 instead of
50.
5.6c The circuit shown in
Fig.
P5.6c is
a
so~called Blumlein
pulse generator (A.
T. Starr,
Radio and Radar Technique, Pitman & Sons, London, 1953) and has the property that
it produces a voltage pulse equal to the voltage to which the lines are initially charged
by the
source Vc. The resistor RC can be considered essentially infinite. At a time 7
after the switch is closed, a voltage VC appears across the output line terminals 0—0’
and that voltage remains across the terminals 0—0’ for a time 27. The initially charged
lines
of
are
equal length. Analyze the behavior of the
treating the output line as a lumped
scribed behavior,
BC
Vc
circuit to show the above-deresistor
20
RL
=
220.
Zo
o
I
0‘
220
BL
=
220
FIG. P5.6c
5.7a The alternative
approach to derivation of the phasor forms for voltage and current
along a transmission line is to replace a/ar by fan in Eqs. 5.2(3) and (4). Write such
equations and show that Eqs. 5.7(4) and (5) satisfy them.
Find the
special cases of Eq. 5.7(13) for a shorted line, an open line, a half-wave
impedance ZL, and a quarter-wave line with load impedance ZL.
line
with load
When two transmission lines
to
be transmitted from
teristic
one to
impedances. Show
caded lines will
cause
are to
be connected in cascade, a reflection of the wave
occur if they do not have the same charac-
the other will
that
a
quarter-wavelength
the first line to
see
termination and thus eliminate reflection in transfer
where
203 are the characteristic impedances
line, respectively.
Z02
the final
5.7d Derive
an
line inserted between the
cas—
its characteristic
and
impedance ZOl as a
if ,Bgl2
77/ 2 and Z02
VZO,ZO3,
=
=
of the quarter—wave section and
I /I +
expression for a reflection coefficient for current p,
phase from the voltage reflection coefficient by qr rad.
=
_
and show that
it differs in
receiving line of negligible loss is one—third of a wavelength long and has
impedance of 100 Q. The detuned receiver acts as a load of 100 + leO
0. Find the input impedance. Sketch a phasor diagram showing the values of V+, V_,
I +, and 1.. at both the load and input, and check the calculated results for impedance
from this diagram.
5.7e A television
characteristic
5.7f A
strip
transmission line of characteristic impedance 20 Q is used at a frequency of
a load that is a microwave diode with conductance 0.05 S in parallel
10 GHz with
with
a
1-pF capacitor.
The line is
one-eighth wavelength long at
input admittance.
Find reflection coefficient at the load and the
the
design frequency.
279
Chapter
Transmission Lines
5
multiple of quarter-wavelength in length, show
expansion for Eq. 5703) give the following
for
impedance:
expression
input
approximate
5.7g IfZL
<<
Z0
and the line not
that the first-order terms of
a
near a
binomial
Z-z'Z
.10 tan
1
l+ZL
seczfil
multiple of quarter-wavelength in length, show
expansion for Eq. 5.7(13) give the following
for
input
impedance:
approximate expression
5.7h If ZL >>
Z0
and the line not
that the first~order terms of
a
near a
binomial
z,
z
—ij
cot
+
[31
2?.
J
Zr
csc2 31
5.8a An impedance of 100 + j 100 O. is placed as a load on a transmission line of characteristic impedance 50 .0. Find the reflection coefficient in magnitude and phase and the
standing wave ratio of the line.
5.8b
Suppose
that reflection coefficient is
given
in
magnitude
and
phase
as
Iplej9” at the load
0. Find the value of (negative) 2 for which voltage is a maximum. Show that
current is in phase with voltage at this position, so that impedance there is real, as
at
z
=
stated. Calculate the
position
of maximum
voltage
for the numerical values of
Prob. 5.8a.
5.8c A slotted line measurement shows a standing wave ratio of 1.5 with voltage minimum
0.1)t in front of the load. Find magnitude and phase of reflection coefficient at the load
and the input impedance for a length 0.2/\ of the line.
5.8d An ideal transmission line is terminated
tic impedance, RL
in front of the load
=
Can you find
a
20/ 2.
to
by
a
What resistance
eliminate reflections
value for such
a
parallel
resistance with value half the characterisyou put in parallel with the line )t/ 4
the generator side of that resistance?
can
on
resistance if the load resistance is
220?
5.8e* Give two designs for a power splitter consisting of one 504). input line T-connected to
two 50-9. lines with matched terminations, using quarter-wave transformers (see Prob.
5.70) as necessary to ensure unity standing wave ratio at the input at the design fre—
quency and an equal power split. Plot power reflected in the input line as a function of
frequency.
Eq. 5.8(1) can be written in phasor notation in the form of a standing wave
traveling wave. Rewrite as a real function of time and, for the example in Fig.
5.8, calculate V(z) at a value of wt shifted in phase by 77/ 4 rad from wtl.
5.8f Show that
plus
a
5.9 The Smith chart
uses
loci of constant resistance and reactance
on
the reflection coeffi-
cient
plane. Other charts have used loci of reflection coefficient magnitude and phase
plotted on the impedance plane. Show that curves of constant I pl2 are circles on the
impedance plane, and give radii and position of the centers as functions of Iplz. Simi—
larly define the circles corresponding to constant phase of p. Explain the advantages of
the Smith chart.
5.102 A 50-0. line is terminated in a load impedance of 75
j69 9.. The line is 3.5 m long
and is excited by a source of energy at 50 MHZ. Velocity of propagation along the line
is 3 X 108 m/s. Find the input impedance, the reflection coefficient in magnitude and
—-
phase, the value of the standing
using the Smith chart.
5.10b The
standing
mum
wave
ratio
is observed 0.23
the Smith chart.
on an
wave
ratio, and the position of a voltage minimum,
ideal 70~Q line is measured
wavelength
as
a voltage mini~
impedance using
3.2, and
in front of the load. Find the load
273
Problems
5.10c Show that the Smith chart for admittance has the form in
tions corresponding to Eqs. 5.9(l)——(8).
5.10d
Fig. 5.10b, developing
Prob. 5.10b using the chart to determine load admittance. Check to
result is consistent with the impedance found in Prob. 5.10b.
Repeat
see
equa-
if the
5.10e A 70‘!) line is terminated in an impedance of 50 + le 0.. Find the position and
value of a reactance that might be added in series with the line at some point to elimiv
nate reflections of waves incident from the source end. Use the Smith chart.
5.10f
5.10g*
Repeat Prob. 5.10e to determine
placed on the line for matching.
the
position
and value of shunt
A 50-0. transmission line is terminated with
of
main line
<
-z
=
a
the
lengths
of the stubs to
give unity standing
wave
ratio for
0.45 )l.
/2
/1
i
ZO
Z”
FIG. P5.1
antennas
have
impedances
l
>§<
}<—————0.25>.
5.10M" Two
be
a load of ZL
20 + j30. A double-stub
pair of shorted 50-0 transmission lines connected in shunt to the
spaced by 0.25A is located with one stub at 0.2). from the load.
consisting
at points
(See Fig. P5.10g.) Find
timer
to
susceptance
of 100 +
oat—4
09
leO .0
for
a
particular frequency
and
are
transmission lines of characteristic impedance 300 D. By inspection of the
Smith chart, show that it is possible to choose different lengths for the two feed lines
so that when combined in series at the input, the series combination perfectly matches
fed
by
the 300-0. line to which
they are connected. Give the lengths of the two lines in frac—
wavelength. By study of the procedure you have used, state whether or not
this compensation approach will work if the two antennas have arbitrary but equal
tions of
a
impedances.
S.10i* The
problem
parallel.
in
5.10j
is
as
in Prob. 5.10h except that the two transmission lines
A certain coaxial line has
430 and p.
equal to the
an
alternating
dielectric of
vacuum
and
a
are
connected
material with
a
=
Mo and is terminated at the end of a vacuum section by an impedance
characteristic impedance of the vacuum regions. At frequency f0 the die
=
lectric and
a
regions are each A/Z long (A appropriate to each region). Show on
path of impedance variation along the line for Operating frequencies
Also plot the standing wave ratio as a function of distance from the load.
vacuum
Smith chart the
f0
and
5.10k* Show
fo/ 2.
on
the Smith chart
regions of admittance which
cannot be matched
by
the
double-stub arrangement of Prob. 5.10g. Repeat for a spacing of )t/ 8 between stubs.
Repeat for a three~stub arrangement with spacing A/ 8 between stubs.
5.11a Use the formula for
input impedance
of
a
transmission line with losses to check
5.11(22), making approximations consistent with R/wL
<<
1 and
6/ 00C
Eq.
<< 1.
272
5.11b
Chapter
5
Transmission Lines
Defining ac from conductor losses and ad from dielectric losses by identifying the ap~
propriate parts of Eq. 5.11(l3), write the approximate expressions, Eqs. 5 .11(14) and
(15) for [3 and Z0 in terms of ac and ad.
5.11c If the transmission line of Ex. 5.11 is made with thinner films, the currents in the
metal films can be almost uniform. With that approximation for films 0.2 pm thick,
calculate the characteristic impedance and attenuation. Comment on the effects of
thinner films.
using
5.11d* Show that the variation of
complex
power
along
a
lossy transmission line carrying
be written as (d/dz)(-%VI*) + -§-(RII* + GVV*) +
0, where X and B are the imaginary parts of Z and Y, respecBVV*)
(j/2)(XU*
tively. Also, show that a direct tenn-by-term identification with the complex Poynting
theorem can be made by multiplying the above expression by dz and considering the
volume for Eq. 3.13(6) to be of dz thickness and infinite width.
sinusoidal
waves can
=
-
5.11e Calculate for frequencies 1 MHz and 1 GHz the attenuation in decibels per meter for
an air-filled coaxial transmission line with copper conductors using the skin~effect ap~
proximations for high~frequency resistance of Ex. 3.17. The inner conductor is a solid
5.11f
cylinder
of radius 2
radius 1
cm.
mm
and the outer conductor is
a
thick tubular
cylinder
of inner
Repeat Prob. 5.11e if the transmission line is now filled with a polystyrene dielectric.
The equivalent conductances of polystyrene at l Mz and 1 GHZ are, respectively,
(rd
10"8 S/m and 2.8
2
10‘5 S/m.
X
filter-type circuit studied in Sec. 5.12, find expressions for characteristic
impedance Z0. Show that this is real in the propagating region and imaginary in the
attenuating region. What does this signify with respect to power flow in a single
traveling wave?
5.123 For the
5.12b A certain continuous transmission line has an equivalent circuit consisting of series
inductance LI H/m and a shunt element consisting of capacitance C F/ m and
1 /L2C and
inductance L2 H m in parallel. Let a)?
=
-
(i) Obtain expressions for
(ii) Plot 3/2 versus a).
(iii) Replot with
co versus
7, a,and
a,
where
3
a
in
terms
of L1,
is real, and
L2,
versus
0),, and
B,
where
0).
[3
is real.
5.13a Write the instantaneous expressions for voltage and current represented by the complex values in Eqs. 513(1) and (2). Make an integration of total energy, electric plus
magnetic,
for
5.13!) Show that
lpl
sketch
as
in
a
quarter-wavelength
Fig.
pure inductive
1 for
of the line and show that it is
reactive load
independent
of time.
ideal transmission line. Find and
purely
suitably normalized V(z, t) and [(2, t) for a line terminated
reactance equal in magnitude to the characteristic impedance.
2
a
on an
5.13
in
a
5.136 Calculate the instantaneous power flow for a short~circuited line WT(t, z)
V(t, z)I(t, z) and plot the results in a diagram like Fig. 5.13. Discuss this WT(t, z) in
connection with conclusion 6 reached from (1) and (2) in Sec. 5.13.
=
5.13d One of the limitations of energy-storage systems for large energies is the breakdown
of air, unless the system can be evacuated. For air with breakdown strength 3 X 106
V/m,
plate
estimate the maximum energy storage in a resonant air-filled half-wave parallelSpacing between plates is 2 cm and characteristic impedance is 50 .Q. Is
line.
there any advantage with respect to breakdown over the use of a parallel-plate capacitor? Discuss an inductor as an energy~storage system from the same point of view.
273
Problems
5.1421 Plot
Eq. 514(7)
see over
5.14b A
normalized
what range
half-wavelength
(8) is
a
to
Y0
as a
function of
frequency
approximation.
reasonable
near
where
n
=
l and
coaxial transmission line with air dielectric has copper conductors
designed for resonance at 6 GHz. Find the
with the dimensions of Prob. 5.11e, and is
bandwidth 2Af,.
5.153 Is
phase
or
anomalous
5.15b Find the
group velocity the larger for
dispersion (dvp/dw > 0)?
phase
5.15c Consider
a
and group velocities for
normal
a
dispersion
(dvp/dw <
0)? For
transmission line with small losses.
transmission line with very
that series resistance R and shunt
high leakage conductance G per unit length so
capacitance C are negligible. Find phase and group
velocities.
5.15d For Probs. 5.15b and 5.150, show that
not
equal
to
group
5.15e Certain water
an
energy
velocity
waves
of
large amplitude
have
phase
defined
by Eq. 5.15(7)
is
plot up and 12g
versus to
given by up
g / m,
frequency. Determine the ratio
velocities
where g is acceleration due to gravity and a) is angular
of group velocity to phase velocity for such a wave.
5.15f For Prob. 5.12b,
as
velocity.
and show that
vpvg
=
=
l /L1C for all
frequencies.
5.163 Plot the
per unit
w~B diagram for a
length, and a shunt
distributed transmission system with series inductance L,
admittance made up of L2 in parallel with C2, for a unit
length of the system. Is this a backward or forward wave system?
frequency and illustrate phase and group velocity on the plot.
5.16b
Repeat
and CI
Show cutoff
a system made up of a series impedance per unit length of
and the shunt admittance resulting from inductance L2.
Prob. 5.16a for
in
parallel
L1
5.16:: Calculate the velocity of energy propagation, as defined in Sec. 5.15 for the backward
wave line of Fig. 5.16a. Show that it does correspond to group velocity for this line
and discuss the concepts of normal and anomalous dispersion for backward waves.
5.17a Show that Eqs. 517(9) with definitions (10) and (11) do give the solutions of (7) and
(8) with the variations (6). Describe ways in which the exponential variation of L and
C might be achieved, at least approximately, (i) for a parallel-plane transmission line
and
(ii) for
a
coaxial line.
5.17b Show that average power transfer is independent of 2 in the loss-free
considered here for frequncies above cutoff, to > me.
exponential
line
are two values of Eqs. 5.17(10) and 5.17(1 1) representing positively and negatively traveling waves, as expected. Write the complete solutions for V(z) and [(2),
showing both waves, using as constants the voltage amplitudes in positively and nega0. Note the interchange of behavior of positive and
tively traveling waves at z
negative waves if the sign of q is changed, and explain physically.
5.17c There
=
5.17d
Modify the analysis for the exponential line to include losses, retaining constancy of
ZY, Z'/Z, and Y' / Y, and interpret the effect of attenuation constant. Assume RO/XO and
Go/BO
small.
6.1
INTRODUCTION
example of the application of Maxwell’s equations in Chapter 3 was that of
electromagnetic wave propagation in a simple dielectric medium. We now return to the
plane wave example and extend it in this chapter, before considering the more general
guided, resonant, and radiating waves.
Plane waves are good approximations to real waves in many practical situations.
Radio waves at large distances from the transmitter, or from diffracting objects, have
negligible curvature and are well represented by plane waves. Much of optics utilizes
the plane-wave approximation. More complicated electromagnetic wave patterns can
be considered as a superposition of plane waves, so in this sense the plane waves are
basic building blocks for all wave problems. Even when that approach is not followed,
the basic ideas of propagation, reflection, and refraction, which are met simply here,
help the understanding of other wave problems. The methods developed in the preced—
ing chapter on transmission lines will be very valuable for such problems. A large part
of this chapter is concerned with the reflection and refraction phenomena when waves
pass from one medium to another, with examples for both radio waves and light.
The first
274
uniform Plane Waves in
6.2
1:3
3»
,::;:
at: ,',‘
0}
,
275
Perfect Dielectric
‘§<~ 2: wastes, “”4 :3, »i«. 222:» at: we:
..:=
Mane-Wave
(3.2
a
Prepagmion
UNIFORM PLANE WAVES IN A PERFECT DIELECTRlC
The uniform
in
Chapter 3 as the first example of the use of
properties in more detail, restricting attention
to media for which u and e are constants. For a uniform plane wave, variations in two
directions, say x and y, are assumed to be zero, with the remaining (2) direction taken
as the direction of propagation. As in Sec. 3.9, Maxwell’s equations in rectangular
Maxwell’s
plane wave
equations. We
given
was
discuss its
now
coordinates then reduce to
8E
6H
VXH=e-—
VXE=-,u-—
a:
at
In component form these
6E,3
are
6H.t
p“
a:
at
6H),
6E.
.\
z
__
a:
6t
aE.
.t
8
32
at
6H.
.1
(2)
__
“
6Hy
( )
:
8
62
=
.u
(5 )
_
at
BE.
6H.
0
6Ey
( )
-—"
O
(3 )
at
:
8
-—‘
6
()
6t
equations (3) and (6) show that both E2 and H2 are zero,
(static) parts which are not of interest in the wave solution.
magnetic fields of this simple wave are transverse to the direction
As noted in Sec. 3.9, the above
except possibly
for constant
That is, electric and
of
propagation.
In Sec. 3.9
we
showed that combination of the above
the one-dimensional
wave
equation
in
822’
a
and
(4) leads
to
Ex,
6213,.
which has
equations (2)
3213.
2
“8
(7)
at;
general solution
=
f1(; S) f2<t 22;)
(8)
L
(9)
+
—-
+
where
—
u
__
us
light for the medium. The first term of (8) can be interpreted
propagating with velocity v in the positive 2 direction, and the second as a
which is the
as a wave
velocity
of
276
Chapter
with the
propagating
wave
Plane-Wave
6
Propagation and Reflection
in the
velocity
same
negative
By
use
of either
(2)
or
=
f1< U)
t
Ex
—
-
,
(4), magnetic field Hy
Hy
=
n
=
+
Hy+
direction. That is,
z
z
Eh+
2
Hy“
f2( U)
t +
=2
_
was
found
Eh+
=
-
to
( 10 )
—-
be
Eh-
(11)
T]
Ty
where
f5
(12)
8
quantity 77 is thus seen to be the ratio of Ex to Hy in a single traveling wave of this
simple type, and as defined by (12) it may also be considered a constant of the medium,
and will be a useful parameter in the analysis of more complicated waves. It has di—
mensions of ohms and is known as the intrinsic impedance of the medium. For free
The
space
#59
=
no
376.73
=
120770
z
(13)
0
Now
looking
leads to the
at the
wave
remaining two components, E
equation
in
and
Hx,
combination of
(l) and (5)
By
62gm
a
=
‘7
62"
which also has solutions in the form of
Me
2
Ey
( 14)
62‘2
positively
and
negatively traveling
waves as
in
(8). We write this
2
Ey
Either (I)
=
z
f3(t 5) f4(t 5)
+
+
--
=
Ey+
+
Ey_
(15)
(5) then shows that magnetic field is
or
E
E
-—y-+- + —y-..
Hx
=
77
To stress the
results of
relationship
(11) and (16)
13r
If
y+
of electric and
(16)
77
magnetic
fields for the
waves we
write the
as
E
2
“Hy
Ex-
+
x+
:
77’
H
y-
E
2
_Hy
_,
___
”’7
(17)
x”
a number of things. First, relations (17) are sufficient to require that
perpendicular to one another in each of the traveling waves. They
These results show
E and H shall be
also
require
that the value of E at any instant must be 77 times the value of H at that
we note that, if E X H is formed, it points in the positive
instant, for each wave. Finally
6.2
FIG. 6.20
uniform Plane Waves in
Relations between E and H for
a wave
277
Perfect Dielectric
a
propagating
in
positive
2
direction (out of
page).
direction for the
positively traveling parts of (17) and in the negative 2: direction for
negatively traveling part, as expected. These relations are indicated for a positively
traveling wave in Fig. 6.20.
2
the
The energy relations
volume is
are
also of interest. The stored energy in electric fields per unit
8E2
uE
and that in
magnetic fields
2
«~—
8
:
--
(E2
.
2
2
,t
+
E)
y
"
(18 )
is
=
,uin
=
,LL
—
,,
H ‘1 + H
7
“
By (17), rig and uH are equal for a single propagating wave, so the energy density at
each point at each instant is equally divided between electric and magnetic energy. The
Poynting vector for the positive traveling wave is
l
Pz+
:-
E.r+Hy+
"-
Ey+Hx+
:
:7.
(13:34»
+
Eat)
(20)
always in the positive 2: direction except at particular planes where it may be
a given instant. Similarly, the Poynting vector for the negatively traveling wave
is always in the negative 2 direction except Where it is zero. The time-average value of
the Poynting vector must be the same for all planes along the wave since no energy
can be dissipated in the perfect dielectric, but the instantaneous values may be different
at two different planes, depending on whether there is a net instantaneous rate of increase or decrease of stored energy between those planes.
In Sec. 3.10 we also studied the important phasor forms for a plane wave with Ex
and Hy. Extending that
analysis to include the remaining components, we have
and is
zero
for
E_\.(z)
==
77H),(z)
=
Ele‘jkz
Ele"j"'z
+
-
Ezejkz
(21)
E2491":
(22)
278
Chapter 6
Plane-Wave
E),(z)
=
nH...(z)
=
Propagation and Reflection
E3e
--
*sz +
5:453sz
(23)
+
Eiesz
(24)
cow—22
m”
(25)
Egg-fly
where
k
$3
=
=
phase constant for the uniform plane wave, since it gives the change
phase per unit length for each wave component. It may also be considered a constant
of the medium at a particular frequency defined by (25), known as the wave number,
and will be found useful in the analysis of all waves, as will be seen.
The wavelength is defined as the distance the wave propagates in one period. It is
then the value of 2 which causes the phase factor to change by 271':
This constant is the
in
2
kA=27r
k=~il
or
(26)
01‘
2 77
=3
A:
a)
(27)
f
11.8
wavelength, phase velocity, and frequency. The
free-space wavelength
by using the velocity of light in free space in (27)
and is frequently used at the higher frequencies as an alternative to giving the frequency.
It is also common in the optical range of frequencies to utilize a refi‘acz‘ive index n
given by
This is the
common
relation between
is obtained
n
5
—-—.-
~8—
/i
#0 80
=
U
For most materials in the
optical range ,u.
frequency.
To summarize the properties for a single
described as a uniform plane wave:
relative
.
.
QMesJNr-d
.
permittivity
Velocity
There is
=
so
that
n
is
just
the square root of the
for that
of propagation is
no
no,
(28)
electric
or
v
of this
simple type,
field in direction of
propagation.
1
=
magnetic
wave
/\/E.
The electric field is normal to the
magnetic field.
The value of the electric field is 17 times that of the
The direction of
propagation
is
which may be
given by
magnetic
the direction of
field at each instant.
E X
H.
Energy stored in the electric field per unit volume at any instant and any
equal to energy stored in the magnetic field.
7. The instantaneous value of the
E and H
are
given by E2/n
nHz, where
of total electric and magnetic field strengths.
Poynting
the instantaneous values
point is
vector is
=
uniform Plane Waves in
6.2
Example
PROPAGATION OF
If
radio
A
a
279
Perfect Dielectric
6.2
MODULATED WAVE IN A NONDISPERSIVE MEDIUM
of
angular frequency we is amplitude modulated by
angular frequency mm, the resulting function may be written
a
wave
E(t)
Suppose this
2
function at
z
=
==
A[1
O excites
direction. To obtain the form of the
[by
t
-—
z/v
to
+
a
sine
wave
(29)
wmt]cos wot
m cos
of
uniform
plane wave propagating in the positive
propagating wave it is straightforward to replace
a
obtain
E(z, t)
=
A[l
+ m cos
wn,(t §)]cos a)0(t 5)
(30)
-
-
interpretation of this expression is that the entire function propagates in the z
velocity v as illustrated in Fig. 6.2b. All this is correct provided that the
medium is nondispersive (i.e., that v is independent of frequency). If there is dispersion,
an expansion of (29) shows that different frequency components are present and that
each component then propagates at its appropriate I). The result then is that the envelope
moves with a different velocity than the modulated wave as shown in the discussion of
group velocity (Sec. 5.15).
The
direction with
[Enveiope
Modulated
\
wave
/
\
.
/
\
\
(l +
m
(l
\
IA
-
//
m)A
/
\
\
/
\
z
"fl
—_
/
\
\
\
\
/
/
\
/
\
/
\
U
\
/
\
/
l
FIG. 6.2b
time
in
z
t
=
Simple modulated wave having form
a nondispersive medium, envelope
O. In
/
//
\‘~#
of
Eq. (30).
The
and modulated
wave
portion
is
plotted
move
versus 2 at
with
direction.
I/
velocity
v
289
Chapter
6
plane
waves
Propagation and Reflection
POLARIZATION OF PLANE WAVES
6.3
If several
Plane-Wave
have the
same
direction of
propagation, it
is
straightforward
to
superpose these for a linear medium. The orientations of the field vectors in the indi—
vidual waves and the resultant are described by the polarization1 of the waves. In this
concerned primarily with sinusoidal waves of the same frequency.
positively traveling wave only, use phasor representation, and assume
both it and y components of electric field. The general expression for such a
discussion
Let
us
we are
take
there
are
wave
is then
a
E
--=
yEgefhersz
+
(irEl
(1)
E1 and E2 are taken as real and ([1 is the phase angle between x and y components.
corresponding magnetic field is
where
The
1
H
—
s
.
.
(—fizeflt
+
sane-‘1‘“:
(2)
n
The several classes of polarization then
E1
and
depend upon
the
phase
and relative
E2.
Linear
or
Mane Polarization
add at every plane 2' to give an
a with respect to the x axis, as
a
If the two components are in phase, 1/;
electric vector in some fixed direction defined
angle is
pictured
in
i
Fig. 6.3a:
E?
an
=t”-—~=t
an
real and hence the
0, they
by angle
=
—1
E9
3
(i
—
Ex
This
amplitudes
same
for all values of
z
and 2‘. Since E maintains its
direction in space, this polarization is called linear. It is also called plane polarization
since the electric vector defines a plane as it propagates in the z direction. In commu-
FIG. 6.30
‘
Components
of
a
linearly (plane) polarized
wave.
polarization is used in electromagnetics both for this purpose and for the un~
concept of the contribution of atoms and molecules to dielectric properties as
described in Secs. 7.3 and 73.2. Usually, the intended usage is clear from the context.
The term
related
nications
it is
281
Polarization of Plane Waves
6.5
describe
polarization by the plane of the electric
polarization implies that E is vertical. In older optics
texts the magnetic field defined the plane of polarization, but it is now also common in
optics to use electric field.2 To avoid ambiguity in either case it is best to specify
explicitly, “polarized with electric field in the vertical plane.”
vector,
engineering
common to
that the term vertical
so
alrcular Polarization
and
E2
are
and
equal
A second
phase angle
E
The
of E is
magnitude
seen to
be
important special case arises when amplitudes E1
: 77/2. Equation (1) then becomes
(,0
is
=
(i
=
The
sum
WE!
jy)Elej‘”’e"jkz]
=
Remit
2
ED? cos(a)t
i
of the squares of instantaneous
E§(z,
t)
does define the
+
E§(z,
equation
t)
of
=
a
(4)
from the above, and it may be inferred that it
this clearly let us go to the instantaneous forms:
rotates in circular manner, but to see
E(z, t)
ji)Ete'j""
i
k2)
——
E.
and
E%[cosz(mt
-—
I
(5)
5' sin(cot
1(2)]
_.
Ey,
k2)
sin2(cot
+
circle. The instantaneous
angle
—-
a
k2)]
=
E91,
(6)
with respect to the
x
axis is
a
1
tan“
1
Ey(2,
1)
-
__
_.
t a“
In
a
given 2 plane,
_1<:sin(wt kz))
cos(cot
Eta. r)
w
=
k2)
—.
+00 t
—
kz >
7
(i
i cut.
angular velocity with a
2 as pictured in Fig. 6.3b. The
the vector thus rotates with constant
=
a fixed time, the vector traces out a spiral in
propagation may be pictured as the movement of this “corkscrew” in the z direction
with velocity v.
—wt (for z
Note that ill
O) and «,1;
”7T/2 leads to
+7r/2 leads to a
rotation in the opposite direction. The first case is called the left~hand or counterclockwise sense of circular polarization (looking in the direction of propagation) and the
latter, the right~hand or clockwise. The magnetic field for the circularly polarized wave,
i7T/2, is
using E1
E2 and it;
For
2
=
=
=
=
=
E
H
:
——‘
.
(:jft
+
milk:
(8)
77
E2 but it! other
E1
than 0 or : 7r/ 2, the terminus of the electric field traces out an ellipse in a given 2
plane so that the condition of polarization is called elliptical. To see this, again let us
Elliptic Polarization
For the
take instantaneous forms of
no, r)
=
=
2
M. Born and E. Wolf,
general
case, with
E1
¢
E2,
or
=
(1),
Re[(xE1
+
as, cos(a}t
yEgeN)efw‘e‘j“—’]
--
k2)
+
ya, cos(wt
(9)
—
k2 +
to)
Principles of Optics, 6th ed, p. 28, Macmillan, New York, 7980.
282
Chapter
6
Plane-Wave Propagation and Reflection
I
I:
"71
v l\\\‘~/
~
,
vim
it
FIG. 6.3b Circularly polarized wave. Terminus of the electric field vectors forms a spiral of
period equal to the wavelength at any instant of time. This spiral moves in the z direction with
velocity u, so that the vector in a given 2 plane traces out a circle as time progresses.
or
in
a
given
2
plane,
say
2
==
0,
Ex(z, t)
:
E1
Ey(z, t)
2
E2 cos(a)t
cos wt
(10)
+
parametric equations of an ellipse. If 4:
axes of the ellipse are aligned with the x and y axes,
tilted as illustrated in Fig. 6.30.
These
are
the
H
/
/~\
X
l \
\
\/
\/
\y/
/\
/
i
FIG. 6.3a
vectors
'-
/’7
/
/\ \[
ll!)
=
i
77/ 2
but for
the
and minor
the
ellipse
is
7[\;’/- Tirfnjligius
I
I
\
be
ex
0‘\
/MEwas“
7\g/
,
T
\
/
/\\._../X
Elliptically polarized wave. The locus of the terminus of electric
case an ellipse for a given .7. plane as time progresses.
is in each
major
general :11
and
magnetic field
6.4
unpoiarized Wave
there is
a
283
Waves in imperfect Dielectrics and Conductors
We sometimes also
speak
of
an
unpolarized
wave
in which
component in any arbitrary direction for each instant of time. Note that this
to the superposition of waves of different frequency, or of random
superposition of any number of components of the same frequency and
concept applies only
phases,
defined
larized
since
phase
waves
reduces to
when
of
phase between
ionosophere.
of the three
one
have
described above. We may meet unpo»
cases
frequency spectrum (as in sunlight) or a random variation
components, as in the propagation of a radio wave through the
we
a
Example
6.5
LINEARLY POLARIZED WAVE As SUPERPOSITION
OF TWO CIRCULARLY POLARIZED WAVES
as the circularly polarized wave may be looked at as the superposition of two
linearly polarized waves, so may a linearly polarized wave be considered a superposition
of two oppositely circulating circular polarized waves. To show this, let us add righthand and left-hand circularly polarized waves of the same amplitude. Using (4),
Just
E
and
(it
=
+
jy)Ele"fl‘z
magnetic field, using (8),
=
——1(—-ji
+
may!“
7?
The results
are
the
(i
--
jy)Ele_j"z
=
ZKEIe‘ij
is
E
H
+
2E
E
+ ——-1-
(ji
+
We’ll“
expressions
for E and H in
direction,
A
—-—1ye“1’“
77
7?
the x direction. To obtain E in the y
=
we
polarized with electric field in
only to subtract the two circularly
a wave
have
polarized parts.
6.4
WAVES
1N
IMPEREECT DIELECTRICS
AND
CONDUCTORS
propagation will be dealt with extensively
only isotrOpic, linear materials and bring in the effect
of losses on the response of applied fields. The effect of losses can enter through a
response to either the electric or magnetic field, or both. An example of a material in
which the response to the magnetic field leads to losses is the so-called ferrite (or
ferrimagnez‘ic materials), and in this case the permeability must be a complex quantity
,u
,u/
ju". For most materials of interest in wave studies, magnetic response is
and
the permeability is a real constant that differs very little from the penne—
weak
very
of
free
ability
space; this will be assumed throughout this text unless otherwise specified.
Materials
and their effect
in
we
properties
Chapter 13. Here
=
-—
consider
on wave
234
Chapter
Mane-Wave
6
Propagation and Reflection
study of the response of bound electrons in atoms and ions in molecules
shows that the total current density resulting from their motion is
A
J
jcer
=
=
jw(e’
(Sec. 13.2)
jag)E
-
If, in addition, the material contains free electrons or holes, there is
a
conduction current
density
J
=
O’E
At sufficiently high frequencies, a" can be complex (Sec. 13.3), but we will assume that
frequency is low enough to consider it real (satisfactory through the microwave range).
Then the total current density is
J
jw<s'
=
-
jez
—-
19-)E
<1)
(I)
In materials called dielectrics, there are usually few free electrons and any free-electron
current component is included in a” and o~ is taken as zero. On the other hand, in
materials considered conductors such
as
normal metals and semiconductors, the con—
a}; is subsumed in the
duction current dominates and the effect of the bound electrons
conductivity. Although different physically, the two loss terms enter into equations
the same way through the relation 0'
me”.
For dielectric materials with real permeability and complex permittivity,
in
=
V X H
where conduction currents
The
wave
are
=
wave
=
jw(8’
-
number k may be
jk
ll
jc”)E
(2)
included in the loss factor a".
number, Eq. 6.2(25), is complex in this
k
The
jwsE
==
wV,u,(3’
separated into
-
case
and is
ja”)
real and
imaginary parts:
WNW“ 5.0]
where
and
(at was
(3)
According
to Sec.
6.2, the exponential propagation factor for phasor
which becomes, when k is
wave
waves
is
€77“?
complex,
6—122
Thus the
285
Waves in imperfect Dielectrics and Conductors
6.4
attenuates as it
e—aze—-jfiz
2
propagates through the material and the attenuation
upon the dielectric losses and the conduction losses, as would be expected.
The intrinsic impedance, or ratio of electric to magnetic field for a uniform plane
depends
wave, becomes
7’
=
\L \/8[1*J(8/8)]
#4
fl
z
“
,
.
n
(7)
,
An
important parameter appearing in (4) through (7) is the ratio e”/e’: For low-loss
as the examples given in Table 6.4a, this ratio is much less than unity.
We may refer to such a material as an imperfect dielectric. Under these conditions, the
attenuation constant (5) and the phase constant (6) may be approximated by expanding
materials such
both
as
binomial series:
ks"
‘1
<8)
~
to;
7
Mme]
wV
where k
=
tan 6
e”/e’.
=
us’.
It is
Dielectric losses
are
that there is
seen
a
usually
described
by
small increase of the
the “loss
phase
tangent”
and,
constant
Table 6.4a
Properties of Common Dielectrics
at 25°C
104
e'/eO
Materialb
2:
106
f
=
to8
f
=
1010
f
106
=
f
tan 8"
108
=
f
101°
=
Fused quartz (SiOz)
Alumina (96%)
3.8
3.8
3.8
l
2
2
8.8
8.8
8.8
3.3
3.0
14
Alsimag ceramic
MgO
SrTiO3
Polyethylene
Polystyrene
5.7
5.6
5.2
30
16
9.6
9.6
——-
<3
<3
230
230
2
1
2.3
2.3
2.3
<2
2
5
2.6
2.6
2.5
0.7
<1
10
Teflon
2.1
2.1
2.1
<2
<2
5
“tan 8
==
~-
20
-
~-
e"/e'.
”A microwave-absorbing material, Eccosorb, is available with 1.5
<
(sf/so)
<
50 and 0.08
<
(s"/eo)
<
1.
286
Chapter
Propagation and Reflection
Plane-Wave
6
therefore, decrease of the phase velocity when losses
impedance in this case is, from (7),
are
present in
a
dielectric. The
intrinsic
2
'
a:
where
In
a
77’
=
as a
below
frequencies
(8)
0/3’
'
8”
10
+
semiconductor where the losses
V X H
mations
8”
~—
VM/s'.
material such
For radian
1—-
3
—-
=
(jwe’
much
larger
+
0')E
than
are
predominantly conductive,
=jw8’[1-j;:7:|E
cr/e’
the loss term is small and the
(11)
approxi~
may be used with e”/e’ replaced by cr/wa’. For frequencies much
the material may be considered a good conductor. If we make use of the
to
(10)
fact that
0'
——,>>1
£08
in(4),wefind
.
M0“
.
Jk=Jw
.\/——'
71'pr
Tar-(1+1)
1+j
=~——8
(12)
depth of penetration defined by Eq. 3.16(l 1) and used extensively in
Chapter
prepagation factor for the wave shows that the wave decreases in mag
nitude exponentially, and has decreased to l /e of its original value after propagating a
distance equal to depth of penetration of the material. The phase factor correSponds to
a very small phase velocity
where 5 is the
3. The
2
to
=
p
where
c
is the
velocity
of
light
E
(08
=
c
in free space and
8
(13)
7‘;—
A0
is
free—space wavelength.
Since
B/AO is usually very small, this phase velocity is usually much less than the velocity of
light.
Equation (7) gives,
72
for
:
a
good conductor,
FEE =(1+j)/fl° =<1+j)RS
(T
(14)
0'
R5 is the surface resistivity or high-frequency skin effect resistance per square
plane conductor of great depth. Equation (14) shows that electric and magnetic
fields are 45 degrees out of time phase for the wave propagating in a good conductor.
Also, since R5 is very small (0.014 0 for copper at 3 GHZ), the ratio of electric field
to magnetic field in the wave is small.
where
of
a
Reflection of
6.5
Normally
Incident Plane Waves from Perfect Conductors
287
Table 6.4!)
Material Parameters for Microwave
Frequencies and Below
.8...
Conductivity
(S/m)
Material
Copper
5.80 X 107
—-~
Platinum
0.94
107
-~—
Germanium
(lightly doped)
Seawater
X
at which
we’ (Hz)
Frequency
so
=
0'
(Optical)
(Optical)
102
16
1.1 X
4
81
8.9 X 108
2.2 X 105
1011
Fresh water
10"3
81
Silicon
10
12
1.5 X 1010
10‘“2
10'3
30
6.0
(lightly deped)
Representative wet earth
Representative dry earth
X
106
2.6 X 10‘5
7
The above-mentioned results for the
good conductor agree with those found in Chapappeared to be a different analysis. In Sec. 3.16 the assumption that
the conduction current greatly exceeds the displacement current is made at the outset,
with the result that the differential equation is not the wave equation but the so-called
difiusion equation. The assumptions in both approaches are identical, however, so it is
to be expected that the results would be the same.
Table 6.4b lists several materials for which the dominant losses are those resulting
from finite conductivity. The values listed for conductivity and real part of permittivity
are those applicable up through the microwave frequency range. The frequencies at
which displacement currents equal conduction currents are listed and serve to indicate
the range of frequencies in which the two above-mentioned approximations are valid.
ter 3
by using
what
mane waves Normally incident
If
"
a
=
uniform
O,
we
Discontinnities
6.5
REFLECTION OF NORMALLY INCIDENT PLANE WAVES
FROM PERFECT CONDUCTORS
plane
wave
is
normally
know that there must be
incident
some
on a
plane perfect
reflected
boundary conditions cannot be satisfied by
solutions, but will require just enough of the two so
wave.
on
The
the conductor surface is
zero
wave
a
single one of the traveling wave
that the resultant electric field at
for all time. From another
point
Poynting theorem that energy cannot pass the perfect
brought by the incident wave must be returned in a reflected
the
conductor located at
in addition to the incident
of View,
conductor,
wave.
we
so
In this
know from
all energy
simple
case
238
Chapter
the incident and reflected
Plane-Wave
6
waves are
of
Propagation and Reflection
equal amplitudes
pattern whose properties will now be studied.
Let us consider a single—frequency, uniform plane
and
together form
a
standing
wave
axes so
waves
traveling
in both the
positive
E
If Ex
2
O at
z
=
D
as
=
required by
E,r
and
the
E+(e—-”‘z
==
is
negative
E+e“j""
magnetic field to
given by Eq. 6.2(11). Hence
The relation of the
waves
wave
that total electric field lies in the x direction. The
+
2
phasor electric field, including
(Fig. 6.5), is
directions
Exit:
perfect conductor,
—
elk)
=
and select the orientation of
~2jE+
E
==
__
-—E+:
sin kz
(1)
the electric field for the incident and reflected
(2)
Equations (1)
and
(2)
state
that, although total elecuic and magnetic fields for the
waves are still mutually perpendicular in space
they are now in time quadrature. The pattern is a standing
wave since a zero of electric field is always at the conductor surface, and also always
at kz
mr or z
nA/Z. Magnetic field has a maximum at the conductor surface,
and there are other maxima each time there are zeros of electric field. Similarly, zeros
of magnetic field and maxima of electric field are at k2
——(2n + 1)7r/ 2, or 2
-(2n + l)/\/4. This situation is sketched in Fig. 6.5, which shows a typical standing
combination of incident and reflected
and related in
==
magnitude by
=
——~
17,
—
=
wave
pattern such
as was
=
found for the shorted transmission line in Sec. 5.13. At
an
instant in time,
the
field; 90
The average
occurring twice each cycle, all the energy of the line is in
degrees later the energy is stored entirely in the electric field.
Incident
Reflected
a
—->E +3}th
wave-e—
-E+
-
8 met
kt)
+
In)
Standing wave patterns of electric and magnetic fields when a plane wave is reflected
perfect conductor, each shown at instants of time differing by one~quarter of a cycle.
FIG. 6.5
from
wave
magnetic
Transmission-Line
6.6
value of the
fact that
on
289
Analogy of Wave Propagation
vector is zero at every cross-sectional plane; this emphasizes the
the average as much energy is carried away by the reflected wave as is
the incident wave.
Poynting
brought by
These points
are
also shown
E’\.(z, t)
=
HA2, t)
=
by
the instantaneous forms for fields:
Re[-—2jE+
sin
kzejw’]
2E
'
6.6
Re[ 71+ kzejw‘]
wt
(3)
cos wt
(4)
2E
.
COS
2E + sin k2 sin
2
=
——+-
cos
kz
77
TRANSMISSION-LINE ANALOOY OF WAVE PROPAGATION:
THE |MPEDANCE CONCEPT
In the
problem of wave reflections from a perfect conductor, we found all the properties
previously studied for standing waves on an ideal transmission line. The analogy between the plane-wave solutions and the waves along an ideal line is in fact an exact
and complete one. It is desirable to make use of this whether we start with a study of
classical transmission-line theory and then undertake the solution of wave problems or
proceed in the reverse order. In either case the algebraic steps worked out for the
solution of one system need not be repeated in analyzing the other; any aids (such as
the Smith chart or computer programs) developed for one may be used for the other;
any experimental techniques applicable to one system will in general have their counterparts in the other system. We now show the basis for this analogy.
Let us write side by side the equations for the field components in positively and
negatively traveling uniform plane waves and the corresponding expressions found in
Chapter 5 for an ideal transmission line. For simplicity we orient the axes so that the
wave has Ex and
Hy components only:
E42)
=
Hi-(Z)
=
E+e“j":
l
*-
'
k
77
=
__
n
_.
+ E
que
a
8
(1)
V(z)
=
(2)
1(2)
=
*—
EJ’kzl
(3)
B
=
(4)
z0
V+erfflz +V__eJ‘Bz
1
.
.
[E+€“’k:
.ef"z
...
_
(5)
.
.
‘7
'73—O [V+e~1fl
_
—'
V_E’fl‘]
(6)
(ox/[E
<7)
l:
(8)
C
see that if in the field equations we replace Ex by voltage V,
H), by current I,
permeability ,u. by inductance per unit length L, and dielectric constant e by capacitance
per unit length C, we get exactly the transmission-line equations (5) to (8). To complete
the analogy, we must consider the continuity conditions at a discontinuity between two
regions. For the boundary between two dielectrics, we know that total tangential electric
and magnetic field components must be continuous across this boundary. For the case
of normal incidence (other cases will’ be considered separately later), Ex and H1, are the
We
29%
Chapter 6
tangential components,
so
Pierre-Wave Propagation and Reflection
these
continuity conditions are in direct correspondence to
require that total voltage and current be continuous
those of transmission lines which
junction between two transmission lines.
exploit this analogy fully, it is desirable to consider the ratio of electric to magnetic
fields in the wave analysis, analogous to the ratio of voltage to current which is called
impedance and used so extensively in the transmission—line analysis. It is a good idea
to use such ratios in the analysis, quite apart from the transmission—line analogy or the
name given these ratios, but in this case it will be especially useful to make the identification with impedance because of the large body of technique existing under the
heading of “impedance matching” in transmission lines, most of which may be applied
to problems in plane-wave reflections. Credit for properly evaluating the importance
of the wave impedance concept to engineers and making its use clear belongs to
at
the
To
Schelkunoff.3
At any plane 2, we shall define the field or wave
electric field to total magnetic field at that plane:
impedance
as
the ratio of total
(9)
single positively traveling wave this ratio is n at all planes, so that 77, which has
impedance of the medium, might also be thought of as a
characteristic wave impedance for uniform plane waves. For a single negatively travn for all 2. For combinations of positively and negatively
eling wave the ratio (9) is
traveling waves, it varies with z. The input value Z,- distance I in front of a plane at
which the “load” value of this ratio is given as ZL may be found from the corresponding
transmission-line formula, Eq. 5.7(13), taking advantage of the exact analogy. The
intervening dielectric has intrinsic impedance 7;:
For
a
been called the intrinsic
—
z.'
z
7?
[ZL
cos
kl +
ncos kl +
jn
jZL
sin
kl]
sin kl
(10)
argued that in wave problems the primary concern is with reflections and
impedances directly. This is true, but as in the transmission-line case there is
a one~to~one correspondence between reflection coefficient and impedance mismatch
ratio. The analogy may again be invoked to adapt Eqs. 5.7(8) and 5.7(9) to give the
reflection and transmission coefficients for a dielectric medium of intrinsic impedance
7] when it is terminated with some known load value of field impedance ZL:
It may be
not with
P
u
Polite 2m
ZL+77
(11)
+
(12)
3
S. A. Schelkunoff, Bell
Sys’r. Tech. J. 17, 77 (7938).
6.6
We
Transmission-Line
from this that there is
291
Analogy of Wave Propagation
reflection when
77 (i.e., when impedances are
1 when ZL is zero, infinity, or purely
matched). There is complete reflection lpl
imaginary (reactive). Other important uses of formulas (10) to ( 12) will follow in sucsee
no
ZL
=
=
ceeding
sections.
Example
FIELD IMPEDANCE
If
we
calculate
2(2) for the plane
studied in Sec. 6.5,
to be zero
Ex
we
take
ZL
=
we
recognize
(10),
we
normally incident on a plane conductor, as
plane acts as a short circuit since it constrains
zero voltage in the transmission-line analogy. If
wave
to
obtain
2,.
The
same
result is obtained
—-1 from
6.6::
FRONT OF CONDUCTING PLANE
that the
there, corresponding
O in
IN
==
jn
tan
2
==
(13)
the ratio of
by taking
Like the
and
6.5(2).
Eqs. 6.5(1)
line, this is always imaginary (reactive). It is
(2n + l)7r/ 2.
at
kl
phasor electric and magnetic fields
impedance of a shorted transmission
zero at
kl
=
1277'
and infinite at kl
=
Exampie 6.5b
ELIMINATION OF WAVE REFLECTIONS FROM CONDUCTING SURFACES
wave reflections from a plane perfectly conducting surface, it is clear that
coating the conductor with a thin film having conductivity 0 does not help since the
perfect conductor merely shorts this out. As in Fig. 6.6, one can place a sheet with
resistivity 71 0/ square a quarter-wavelength in front of the conductor surface. As seen
from (13), this is at a point of infinite impedance because of wave reflections so that
there is no shunting effect. The wave impinging on the front (if the sheet is thin com-
To eliminate
lncident
wave
Conducfing
sheet
FIG. 6.6
Elimination of reflection from
Perfect
conductor
perfect conductor by placing
'
front of it.
\V
a
conducting
sheet
A/ 4
in
292
Chapter
pared
with its
depth
matched. The needed
Plane-Wave Propagation and Reflection
6
of penetration) then sees wave impedance
conductivity of the sheet is then
7; and is
perfectly
1
=
——
d << 8
n,
ad
(14)
The arrangement just described is not very convenient to make, and is sensitive to
frequency of operation and to angle of incidence. For this reason coatings of ‘anechoic
‘
chambers”
lossy
often
more
material, with
pyramidal taperings
a
uniform
plane
wave
medium with \/
approach to matching, with a porous,
facing the incident wave.
the surface
on
NORMAL INCIDENCE
6.7
If
nonuniform-line
use a
is
incident
normally
ON A
DIELECTRIC
single
on a
V
dielectric
boundary from
a
,uQ/ez
772, the wave reflection and
[1.1/81
transmission may be found from the concepts and equations of Sec. 6.6. Select the
=
771 to one with
direction of the electric field
incident
wave as
the
positive
in that
n2
2
the x direction, and the direction of propagation of the
0 (Fig. 6.7a). The
direction, with the boundary at z
=
infinite in extent,
so that there is no
effectively
there
is
then
the
intrinsic impedjust
region.
impedance
for all planes, and in particular this becomes the known load impedance at the
0. Applying Eq. 6.6(11) to give the reflection coefficient for medium I
right
reflected
plane
2
is assumed to be
medium to the
ance
as
=
wave
The field
=
referred to
z
=
0,
Di 1—
p:
[‘13 1+
The transmission coefficient
the
second dielectric, from
is
giving
Eq. 6.6(12),
“‘
z
E9
~
(1)
of electric field transmitted into the
27],
=-——=—
131+
density
771
712 + 771
amplitude
7:
The fraction of incident power
”'72
(2)
772+771
reflected is
"I
P
4:
131+
=
19:2
E2
2771
2‘01
(ix—1+)
And the fraction of the incident power
density
2
=
Pl+
From
1
-
=
W
(3)
transmitted into the second medium is
W
(4)
(1), we see that there is no reflection if there is a match of impedances, 771
712.
course occur for the trivial case of identical dielectrics, but also for the
This would of
=
6.7
Normal Incidence
on a
293
Dielectric
2:0
/%
Incident
wransma‘tte d
E1+ H1+
wave:
,
//y)§i§"ei E2.H2
—-><—
Reflected
E1_.H1_
wave:
//
i,
Medium 2
Medium 1
(a)
m>n2
-~——
~-—~e\"'E
——
E
7ll<772
""“""
H
”*—‘
—~W\[~T
Mediuml
H
=
E=
T=
Standing
Standing wave
Traveling wave
wave
'
Medium 2
712
771
of magnetic field
of electric field
(6)
FIG. 6.7
Reflection and transmission of
a
plane
wave
from
a
plane boundary
between
two
dielectric media.
they could be made with the same ratio of ,u. to a. This
importance since we do not commonly find high~frequency
dielectric materials with permeability different from that of free space, but it is interesting since we might not intuitively expect a reflectionless transmission in going from
free space to a dielectric with both permittivity and permeability increased by, say, 10
case
of different dielectn'cs if
latter
case
is not
usually
of
times.
a finite value of reflection in the first region, and
magnitude of p is always less than unity. (It approaches
unity as 772/77, approaches zero or infinity.) The reflected wave can then be combined
with a part of the incident wave of equal amplitude to form a standing wave pattern as
in the case of complete reflection studied in Sec. 6.5. The remaining part of the incident
wave can be thought of as a traveling wave carrying the energy that passes on into the
second medium. The combination of the traveling and standing wave parts then pro—
duces a space pattern with maxima and minima, but with the minima not zero in general.
As for corresponding transmission lines, it is convenient to express the ratio of ac
In the
from (1)
general
we can
case
there will be
show that the
294
Chapter 6
Plane-Wave
Propagation and Reflection
amplitude at the electric field maximum to the minimum
quarter-wavelength away) as a standing wave ratio S:
IE..(Z)lmax
__
__
By utilizing (1), it
1 +
z
ac
amplitude (occurring
lpl
a
(5)
may be shown for real 77 that
if
{TL/"71
7
S
=
, > 771
77.
(6)
.
1f
771/772
771 > 772
Since 171 and 112 are both real for perfect dielectrics, p is real and the plane 2
0 must
be a position of a maximum or minimum. It is a maximum of electric field if p is
positive, since reflected and incident waves then add, so it is a maximum of electric
=
field and minimum of
0 is a minimum of
magnetic field if 772 > 171. The plane 2
electric field and a maximum of magnetic field if 771 > 772. These two cases are sketched
in Fig. 6.71).
==
x. {-3 ._-e'-'-.'-.--...¢ t».«..-,-.».-~.-.;
tn;l—'u-AN¥é-'Ii<fiwm-uIn“1'1-L4h,‘.'-u>—’.»’-;,“ALL-;<-fiv-Ql.l"->mv;<-~—~2J~'
52:2
..
..
t
Examp
e 6.7a
REFLECTION FROM QUARTZ AND GERMANIUM AT INFRARED WAVELENGTHS
plane waves normally incident from air onto quartz, and also
germanium,
wavelength of 2 pm, neglecting losses. Equation (1) may
be written in terms of refractive indices, Eq. 6.2(28). Since [1.1
[1.2 and n1
becomes
81/80, n2
V82/80, (1)
Let
us
find reflection of
onto
at a
=
2
2-"
p
’21
=
121
Using
unity.
122
(7)
+ )12
data in Fig. 13.2b, n2 for quartz is about 1.5 at 2 pm and 121 may be taken as
Reflection coefficient p is then ——O.2 and fraction of incident power reflected
p2 is 0.04 or 4%.
or
""
For
germanium,
n2 is 4 and p is found to be —-O.6,
so
that
p2 is 0.36
36%.
Examme 6.7b
REFLECTION FROM
A
GOOD CONDUCTOR
From
Eq. 6.4(14) the characteristic wave impedance of a conductor is seen to be
j)RS, with R5 the surface resistivity defined in Sec. 3.17. Thus for a plane wave
normally incident from a dielectric of intrinsic impedance 7) onto such a conductor, (1)
yields
(1
+
=<1+j)R.
—
n
<1+j>R.+n
z
<1 +j)R./n
1+<1+j>R./n
__1
—-
(8)
6.8
295
Reflection Problems with Several Dielectrics
Examination of values for
conductors from Table 3.17a shows that
typical
very small. Thus
normally
retained gives
a
binomial
of
expansion
(8) with first~order
RS/n
terms
is
only
'
P
z
.___1 +
2 l + R
M
(9)
7)
Then to first order,
7
lpl-
e
1
4R5
--
<10)
77
the fraction of power transmitted into the conductor is approximately 4R
For air
into copper at 100 MHZ, this is 4 X (2 61 X 10 3)/377== 2.76 X 10‘s, or only
0. 0028%
so
/5.n
6.8
We
are next
REFLECTION PROBLEMS WITH SEVERAL DIELECTRICS
interested in
the
considering
of several
case
parallel
dielectric disconti—
pictured in the case
for three dielectric materials in Fig. 6.8a. We might at first be tempted to treat the
problem by considering a series of wave reflections, the incident wave breaking into
one part reflected and one part transmitted at the first plane; of the part transmitted into
region 2 some is transmitted at the second plane and some is reflected back toward the
first plane; of the latter part some is transmitted and some reflected; and so on through
an infinite series of wave reflections. This lengthy procedure can be avoided by considering total quantities at each stage of the discussion, and again the impedance for~
nuities with
a
uniform
wave
incident in
some
material to the left,
as
mulation is useful in the solution.
region to the right has only a single outwardly propagating wave, the wave or
impedance at any plane in this medium is 773, which then becomes the load imped~
If the
field
\\\\
//\\
N\\\\\\
;+////
//
Transmitted
new \‘
FIG. 6.8a
Wave reflections from
other media.
a
system with
a
dielectric medium interfaced between two
296
Chapter
ance to
place
at
z
=
l. The
6.6(10), and since this is
impedance for region 1:
ZL1
Propagation and Reflection
Plane-Wave
6
input impedance
impedance at z
for
the
=
The reflection coefficient in
Z.'2
=2
173
772(7),
region 1,
cos
k2]
k2!
referred to
=
p
cos
region
ZLi
2 is then
given
at once
by Eq.
0, it may also be considered the load
=
“
+
jnz
sin
+
j'n3
sin
z
=
k2!
k2!
(1)
0, is given by Eq. 6.6(11):
7h
(2)
——-——~
+ 771
ZLI
The fraction of power reflected and transmitted is
and
given by Eqs. 6.7(3)
6.7(4),
respectively.
parallel dielectric boundaries, the process is simply
region becoming the load value for the next, until
repeated,
input impedance
one arrives at the region in which reflection is to be computed. It is of course desirable
in many cases to utilize the Smith chart described in Sec. 5.9 in place of (l) to transform
load to input irnpedances and to compute reflection coefficient or standing wave ratio
once the impedance mismatch ratio is known, just as the chart is used in transmission—
If there
are more
than the two
for
the
one
line calculations.
We
now
wish to consider several
special
cases
which
‘-
\>--'rx:*r7-;-.'-;-7-':
Example
-'-:..
are
of
..-':-','..'".~';.‘-=.'/:.
;
importance.
;
__
».
..
:1-;.-.;,
;.-
.,.
-.
-,l
.7;
,,;~---.-./-
.
.
.«_
;-.'<‘..'T'e.\:
6.8a
HALF—WAVE DIELECTRIC WINDOW
input and output dielectrics are the same in Fig. 6.8a (771
773 ), and the intervening dielectric window is some multiple of a half-wavelength referred to medium 2
so that k9}
may then (1) gives
If the
==
=
ZLI
so
that
by (2) reflection
at the
input
=
773
face is
the above will of
1"
(3)
771
zero.
give reflections for frequencies other
a half-wavelength. For
example, a
4 and thickness 0.025 m corresponds to [<21
Corning 707 glass window with 8,.
71' at a frequency of 3 GHz. Let us calculate the reflection at 4 GHz using the Smith
chart. Normalized load impedance for region 2 is 377/ 188.5
2, so we enter the chart
at point A of Fig. 6.8b. We then move on a circle of constant radius 4/3 X A/ 2 or
0.667A toward the generator. This is one complete circuit of the chart plus 0.167A,
0.62
j0.38.
ending at point B where we read normalized input impedance Zig/ZOZ
Renormalizing, to the characteristic impedance of the input region, we multiply by
188.5/377 or g- and find 0.31
j0.19. This is entered at point C, for which<2ve read
A window such
as
than that for which its thickness is
a
course
multiple
of
=
=
2
=
—-—
~
297
Reflection Problems with Several Dielectrics
6.8
try
[I'O
t! D
FIG. 6.8b
from the radial scale
value
can
a
be checked
Smith Chart constructions for Exs. 6.8a and 6.8d.
reflection coefficient
by
'9'
=
numerical calculation
0.55,
so
that
|p|2 is 0.30 or 30%. This
using (1) and (2).
Example 5.8b
ELECTRICALLY THIN WINDOW
If 771
2
773 and
k2]
is
so
small
compared
with
unity
that tan
kg!
m
kg], (1)
becomes
'
ZLI
“
772(771
+
772 +
1.172
kl
2) 771[1 szl<32= 121)]
177119.]
.
z
9
+
"
771
772
(4l
293
Chapter 6
Substituting
in (2),
we see
Plane~Wave
Propagation and Reflection
that
k!
perfeg-p—I)
771
The
magnitude of
(5)
772
reflection coefficient is thus
proportional
to
the electrical
length
of
the dielectric window for small values of kzl; and the fraction of incident power reflected
proportional to the square of this length.
For example, a polystyrene window (a,
2.54) 3 mm thick, with a normally incident
-j0.145 so that 2% of incident power
plane wave at 3 GHz from air, would give p
is
z
:
is reflected.
Example
6.8a
QUARTER-WAVE COATING FOR ELIMINATING REFLECTIONS
Another
important case
is that of
a
quarter~wave coating: placed between
is the
dielectrics. If its intrinsic
impedance
it will eliminate all
reflections for energy
wave
geometric
passing
mean
of those
on
two
different
the two sides,
from the first medium into the
third. To show this, let
7:“
kzl
From
r—-—
=
712
"2"
=
771773
(6)
(1),
ZL1
=
712
w-
“
771713
——
“*
1) 1
773
713
0.
perfect match to dielectric 1, so that p
in
is
for
technique used,
example, coating Optical lenses to decrease the amount
of reflected light, and is exactly analogous to the technique of matching transmission
lines of different characteristic irnpedances by introducing a quarter-wave section hav~
ing characteristic impedance the geometric mean of those on the two sides. In all cases
the matching is perfect only at specific frequencies for which the length is an odd
multiple of a quarter-wavelength, but is approximately correct for bands of frequencies
about these values. Multiple coatings are used to increase the frequency band obtainable
with a specified permissible reflection.4
As a numerical example of this technique, consider the coating needed to eliminate
reflections from a 488-nm~wavelength argon laser beam (taken as a plane wave) in
going from air to fused silica with refractive index 1.46. It follows from (6) that refractive index of the coating should be the geometric mean of that of the air and window,
or about 1.21. Note from Fig. 13.217 that there are few materials with this low index,
but if one is found its thickness should be a quarter-wavelength measured in the coating,
This is
a
=
This
which is about 0.1 pm.
4
C. A. Bolanis, Advanced
Engineering Electromegnetics,
Sec. 5.5,
Wiley, New York, 1989.
299
Reflection Problems with Several Dielectrics
6.8
Example 6.8d
REFLECTION FROM
TWO-PLY DIELECTRIC
A
Now take the
2 mm, 82/80
2.54
two-ply dielectric of Fig. 6.8c in which 0?2
4
3.0
The
707
incident
mm, 83/80
(polystyrene)
(Coming
glass).
d3
normally
wave is at frequency 10 GHz (A0
3 cm). We will do this just on the Smith chart and
enter points on Fig. 6.81). The steps are clear extensions of the above.
Starting at the 3—4 interface,
=
and
=
=
=
=
2L3
203
~—
377
2
(pom tM )
188.5
toward generator,
Moving
d3
A3
—-
We read
ZB/ZO3
ZL"
2;.
==
.__
0.30
=
Moving
0.2 wavelen g th
=
3.0/\/A_l
0.55
(0,55
-—
j0.235
(p oint N)
and renormalize to get load
I254
j0.235)
-
=
T
07
0.438
-
on
2:
region
(point 0)
j0.187
toward generator,
0.20
d,
__-.
A2
Finally
,
2.00
=
~--——
we
read
—————-
———
=
0.49 +
j0.38
.
=
.201
< 0.49
+ J 0.38 )
(p 01111: P)
0.106 wavelen g th
=
3.0/V254
Z,.2/Z02
Zn
Radius to
:
/
and renormalize to air to get load
1
——
2.54
0.31 + J 0.24
Q (as fraction of radius of chart) gives
Ip]
=
0.55
region
so
(pomt Q)
that
over
30% of incident
power is reflected.
CD
@
®
@
“—-—d2—-——>T¢———-——d3—————+~
FIG. 6.8a
A
30
52
composite
window with
5:3
two
1:
-
.
=
on
£0
slab dielectrics and free space
on
either side.
3m
Chapter
Plane-Wave
6
Propagation endinefiectlon
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Mane Waves
incident onfliscontinuities
INCIDENCE AT ANY ANGLE ON PERFECT CONDUCTORS
6.9
We next
fibiiqneiy
remove
the restriction to normal incidence which has been assumed in all the
preceding examples. It is possible and desirable to extend the impedance concept to
apply to this case also, but before doing this we shall consider the reflection of uniform
plane waves incident at an arbitrary angle on a perfect conductor to develop certain
ideas of the behavior at oblique incidence. It is also convenient to separate the discussion
into two cases, polarization with electric field in the plane of incidence and normal to
the plane of incidence. Other cases may be considered a superposition of these two.
The plane of incidence is defined by a normal to the surface on which the wave impinges
and a ray following the direction of‘propagation of the incident wave. That is, it is the
plane of the paper as we have drawn sketches in this chapter. We consider here lossfree media.5 Polarization with E in the plane of incidence may also be referred to as
transverse magnetic (TM) since magnetic field is then transverse. When E is normal to
the plane of incidence it may be called transverse electric (TE).6
Polarization with Electric Field in the Plane of incidence (TM)
In
Fig.
6.9a the ray drawn normal to the incident wavefront makes an angle 6 with the normal
to the conductor. We know that since energy cannot pass into the perfect conductor,
there must be a reflected wave, and we draw its direction of propagation at some
angle 6’. The electric and magnetic fields of both incident and reflected waves
perpendicular to their respective directions of propagation by the properties of
uniform plane waves (Sec. 6.2), so the electric fields may be drawn as shown by E+
and E_. The corresponding magnetic fields H+ and H- are then both normally out of
the paper, so that E X H gives the direction of propagation for each wave. Moreover,
unknown
must
lie
with the
senses as
shown,
E
E+
..
~
H+
If
we
draw
a
z direction in
-
H“
(1)
n
the actual direction of
propagation
for the incident
wave
shown, and a f’ direction so that the reflected wave is traveling in the negative
5’ direction, we know that the phase factors for the two waves may be written as
e’j"; and em” respectively. The sum of incident and reflected waves at any point
as
,
x,
z
(z
<
O)
can
be written
E(x, z)
5
6
For a treatment of waves in free
=
E+e'flԤ
space
+
E_ej"’§'
obliquely incident on the
(2)
plane
surface of a
lossy
medium, see C. A. Balanis, Advanced Engineering Electromagnetics, Sec. 5.4, Wiley,
New York, 7989.
An alternate designation, common in the scientific literature, employs P (for German
“parole/’9 and S (for German “senkrecht,” i.e., perpendicular), respectively, for the
orientations of E in the two
cases.
Incidence at
6.9
Any Angle
on
30.5
Perfect Conductors
(a)
l
l
\
fl
H
E~//
:
E+
/
a
an
\
H
\<’F\/
_
//
/
x
t’
r
z
(b)
FIG. 6.9
in
plane
where
Wave incident at an angle on a dielectric interface. (a) Polarization with electric field
of incidence. (b) Polarization with electric field perpendicular to plane of incidence.
E+
and E_
are
reference values at the
in terms of the
of
g and
rectangular system aligned
5’ from the diagram is
s"
g’
so
that, if these
into their
The
x
are
and
2
Exflt‘, Z)
E43; Z)
=
magnetic
5+
COS
sin
.._E+
1_[y(x,
Z)
The next step is the
z
sin 6 +
-—x
=
lit-(x, O)
:
we
=
waves
of
(4)
(2), and the
two waves broken
__
E_
aiejk(—-xsin6
COS
+zc056)
E, SlI'l glejk(—-.rsin6
_
+zc056)
(5)
(6)
is
of the
Oe“j""5i“0
x
+
Hwejk(-xsin6'+zcos€’)
boundary
for all
must be zero
COS
6’
2 cos
phase factors
H+e—jk(xsin6+:cos«9)
E+
(3)
have
equation can be satisfied for all
equal, and this in turn requires that
This
sin 6’ +
ee—jk(xsir16+zcos«9)
application
0, Ex
6
2 cos
Oe“jk('\”5im9+3C056)
field in the two
which is that, at
=
x
substituted in the
components,
2
ll
origin. We now express all coordinates
with the conductor surface. The conversion
-—
x.
E
..
only
if the
6
6’
==
condition of the
From
cos
(7)
perfect conductor,
(5),
6’e‘flmm6'
phase
=
0
factors in the two terms
(8)
are
(9)
302
Plane-Wave Propagation and Reflection
Chapter 6
That
is, the angle of reflection is equal
it follows that the two
amplitudes
the
to
15,,
If the results
(9) and (10)
are
for field components at any
(8),
(10)
(5), (6), and (7), we have the final expressions
O:
<
z
With‘ this result in
E__
=
substituted in
point
angle of incidence.
equal:
must be
6
Exes, z)
=
~2jE+
cos
Ez(x, z)
=
~2E+
sin 6
nHy(x, z)
=
2E+ cos(kz
sin(kz
cos(kz
cos
(an-“fern"
cos
(11)
6)e‘jmi“‘9
cos
(12)
6)e“j‘“'5i"‘9
(13)
foregoing field has the character of a traveling wave with respect to the x direction,
a standing wave with respect to the z direction. That is, E, is zero for all
time at the conducting plane, and also in parallel planes distance mi in front of the
The
but that of
conductor, where
A
d
2
The
of Ex is
amplitude
ac
conductor.
and
1
(14)
=
z
cos
6
2fV ,ue
6
cos
planes an odd multiple of d/Z in front of the
Ex is zero, are zero where Ex is maximum,
of time phase with respect to Ex. Perhaps the most
maximum in
a
maximum where
E3
H3,
everywhere 90 degrees out
interesting result from this analysis is that the distance between successive maxima and
minima, measured normal to the plane, bec'omes greater as the incidence becomes more
oblique. A superficial survey of the situation might lead one to believe that they would
be at projections of the wavelength in this direction, which would become smaller with
increasing 6. This point will be pursued more in the following section.
and
are
are
Polarization with Electric Field Normal to the Plane of incidence (TE)
polarization (Fig. 6.91)), E+ and E__ are normal to the plane of the paper, and
this
and H__
are
of the two
The
then
waves
=
71Hx(x, z)
=
77H.(x, Z)
=
boundary
=
shown.
in the x,
Eycxa Z)
Proceeding exactly
2
system of coordinates
E+e—jk(xsin6+
-E+
cos
sin
E+
before,
as
:cosfl)
we can
as
Enejk(-.rsin6’+zc086')
+
6e“jk(xsm6+zc°se)
ee~jk(xsin6+zcosfl)
+
by
the
same
+ E_
cos
6’ejk(‘xsm6'+z°°sg')
E__ SlIl Otejk(—-xsin6’+zcosfl’)
perfectly conducting plane is that E), is zero at z
reasoning as before leads to the conclusion that 6
-—E_. The field components, (15)
Ey
77H
=
~2jE+ sin(kz
=
~2E+
x
"qH,
~
=
—-2jE+
cos
6
sin 6
to
cos
(17),
sin(kz
cos
cos
(15)
(16)
(17)
=
m
O
6’
then become
6)e'jmim9
cos(kz
H+
write the components
condition at the
for all x, which
and E +
as
In
(18)
6)e‘j"‘"sm9
(19)
“1mm”
(20)
6)e
Phase
6.10
This set
wave
again
and
Impedance for Waves
shows the behavior of
pattern in the
conducting plane
6.10
Velocity
a
direction, with
2
and at
traveling
zeros
parallel planes
PHASE VELOCITY
wave
of
Ey
at
in the
and
H:
x
direction and
and maxima of
distance nd away, with d
|MPEDANCE FOR WAVES
AND
303
Oblique Incidence
AT
standing
Hx at the
a
given by (14).
OBLIQUE INCIDENCE
Phase
Velocity Let us consider an incident wave, such as that of Sec. 6.9, traveling
velocity v
1N; in a positive direction, which makes angle 0 with a desired
2 direction aligned normally to some
reflecting surface. We saw that it is possible to
the
in
factor
terms
of
the
x and z coordinates:
express
phase
with
=
EOE, Z)
E+e—jk;
=
:
E+e—jk(xsin6+zcost9)
(1)
For many purposes it is desirable to concentrate on the change in phase as one moves
in the x direction, or in the z direction. We may then define the two phase constants for
these directions:
[3,.
=
k sin 6
(2)
B:
=
k
6
(3)
Re[E+ef<w’-flvt“fi=z>]
(4)
cos
(1) in instantaneous form is then
Wave
E(x,
z,
r)
=
keep the instantaneous phase constant as we move in the x direction, we
[31x constant (the last term does not change if we move only in the x
direction), and the velocity required for this is defined as the phase velocity referred to
If
wish to
we
keep
wt
the
direction:
x
——
6):
v
=
a)
—
.
pm
at
BX
(wt—firmzconst
OI
(5)
v,~—-—“——-—“.—-
v
w
6
()
up;_=-= c036
,8:
.
where
We
v
is the
see
velocity
that in both
normal to its
cases
the
wave
front,
phase velocity
normal to the wavefront and will in fact be
violation of relativistic
velocity.
It is the
principles by
velocity
of
a
so
1/V—FE.
is greater than the velocity measured
for any oblique direction. There is no
this result, since
fictitious
point
no
material
object
moves at
this
of intersection of the wavefront and
a
394
Chapter
b
6
Plane-Wave
Propagation and Reflection
_____________
4“"
t“
_....__....\._
2’
b
/_.__
)7!—
/X\
A
/X'\
//
/
/
a/
|
/
/a
/va
______
b
I
/
I
Mediuml
0
l
/
\ l
L
/
04/
0
FIG. 6.10:1
Uniform
plane
wave
moving
line drawn in the selected direction. Thus in
moves to
a’a’ in
a
given
at
angle
6 toward
Fig. 6.10a, if a plane
a
plane.
of constant
phase aa
interval of time, the distance moved normal to the wavefront
is XX ', but the distance moved
by
this constant
phase reference along
the
z
direction is
the greater distance YY’. Since
YY’ =XX’secB
again lead to the result (6) for phase velocity in the z direction.
phase constant B in a particular direction that is reduced by cosine or
sine of the angle between normal to the boundary and normal to the wavefront, whereas
phase velocity is increased by the same factor. The concept of a phase velocity, and
the understanding of why it may be greater than the velocity of light, is essential to the
discussion of guided waves in later chapters, as well as to the remainder of this chapter.
this
picture
would
Thus it is the
Wave Hmpedance In the problems of oblique incidence on a plane boundary between different media, it is also useful to define the wave or field impedance as the
magnetic field components in planes parallel to the boundary. The
continuity of the tangential components of electric and magnetic
fields at a boundary and the consequent equality of the above-defined ratio on the two
sides of the boundary. That is, if the value of this ratio is computed as an input imped—
ance for a region to the right in some manner, it is also the value of load impedance at
that plane for the region to the left, just as in the examples of normal incidence.
Thus, for incident and reflected waves making angle 6 with the normal as in Sec.
6.9, we may define a characteristic wave impedance referred to the z direction in terms
of the components in planes transverse to that direction. From Eqs. 6.9(5) and 6.9(7)
for waves polarized with electric field in the plane of incidence,
ratio of electric to
reason
for this is the
(an
+ and
—
refer, respectively,
=
to
E,
I7:
=
Ex“
“Hy—
=
77
cos
6
incident and reflected wave; the
(7)
sign
of the ratio is
6.10
Phase
chosen for each
waves
wave
polarized with
Velocity
yield
to
a
Impedance for Waves
at
Oblique incidence
positive impedance. From Eqs. 6.9(15)
to the plane of incidence,
and
305
6.9(16) for
electric field normal
(25)“;
we see
and
=
-—
Ey+
_.
Ey_
H.1'
Ha
==
1)
sec
6
(8)
—
that, for the first type of polarization, the characteristic wave impedance is always
as we would expect, since only a component of total electric field lies in
less than n,
magnetic field lies in that plane. In the latter
polarization,
Z: always greater than 77.
The interpretation of the example of the last section from the foregoing point of View
is then that the perfect conductor amounts to a zero impedance or short to the transverse
field component Ex. We would then expect a standing wave pattern in the z direction
with other zeros at multiples of a half—wavelength away, this wavelength being computed from phase velocity in the z direction. This is consistent with the interpretation
of Eq. 6.9(14).
the transverse x—y
the
plane,
reverse
whereas the total
is true and
is
Exampte
6.10
DIFFRACTION ORDERS FROM
It is known that
a
periodic grating
A
of wires, slots,
BRAGG GRATING
or
similar
perturbations reradiates,
or
incident upon it into various directions described as the alifi‘i‘ac~
tion orders for the grating. The directions are defined as those for which phase constants
diffracts,
a
plane
wave
grating match on the two sides, except that multiples of 277 difference may
grating elements and the contributions still add constructively. Consider
a grating as in Fig. 6.10b, with a plane wave incident from the bottom at angle 61 from
the normal. Phase constant in the x direction is k sin 61 so that the phase difference
between induced effects in adjacent grating elements is kd sin 61. For the reradiated
along
the
exist between
ma
[:3
:3::H:H::3E::l[::———>-x
311.5
FIG. 6.10b
Diffraction
grating
with
plane
wave
incident at
an
angle.
396
wave
Chapter
in the top, at
Plane-Wave
6
Propagation and Reflection
angle 62 from the normal, there will then be constructive interference
if
kd sin
61
=
sm62
=
kd sin
62
211277,
i
m==
O, l, 2,
or
.
ma
.
s11161
i—d—
0 (the principal order) but there can be other diffraction orders
Angle 62
61 for m
or lobes provided that Isin 61 : m/‘l/dl ..<. 1, which requires (1 > )t/ 2.
=
=
6.11
Law of Reflection
to
the
wave
INCIDENCE
For
a
AT
uniform
ANY ANGLE ON DIELECTRICS
plane
wave
incident at
angle 61
from the normal
plane boundary between two dielectrics 31 and 82 (Fig. 6.11), there is a reflected
at some angle 6; with the normal and a transmitted (refracted) wave into the
second medium at
some
condition
the
angle 62 with the normal. For either type of polarization,
tangential components of electric and magnetic field at the
continuity
0 must be satisfied for all values of x. As in the argument applied to the
boundary 2
from the perfect conductor, this is possible for all x only if inciof
reflection
problem
on
=
\\
\\
91
\
|
l
l
\
\
\\
\
n1
2/
61
//
/
//
//
/
//
¥
fix
n2
\
\
\
62 \
\
\
l
2
FIG. 6.11
Oblique
incidence
on
boundary
between two
\
\
isotropic
dielectrics.
Incidence at
6.11
dent, reflected, and refracted
x
waves
Any Angle
all have the
on
same
307
Dielectrics
phase
factor with respect to the
direction:
k1
The first
pair
in
sin
61
==
k1
the
of reflection is
angle
of refraction
62
to the
equal
Sneii’s Law of Refraction
angle
6;
=
91
2
k2
sin
62
(l)
(1) gives the result
6;
or
sin
and the
angle
(2)
of incidence.
From the last pair of (1)
angle of incidence 61:
we
find
a
relation between the
(3)
——=—-:—==—
This relation is
index
n
measure
a
familiar
one
in
Optics
and is known
as
Snell’s law. The
refi‘active
is defined to be
of the
phase
unity for free space so its value for any other dielectric is a
velocity of electromagnetic waves in the medium, relative to free
space. It is common to use the refractive index to characterize properties of dielectrics
in the infrared and optical frequency ranges as explained in Sec. 6.2. At microwave
frequencies it is more common to express the velocities in (3) in terms of
permittivity and permeability. For most dielectrics 111/112 may be replaced by (81/82)1/2
and lower
since #1
z
in2
8
#0-
Reflection and Transmission for Polarization with E in Name of incidence
use
To compute the amount of the wave reflected and the amount transmitted, we
the impedance concept as extended for oblique incidence in the last section.
may
To show the
and
validity of this procedure, we write the continuity conditions for
Hy, including both incident and reflected components in region 1:
Following
Sec. 6.10, if
nents for this TM
we
Ex
Ex+
+
E.\‘---
__-—.,
EL?
(4)
Hy,
+
H),_
=
H),2
(5)
define
wave
impedances
in terms of the
tangential
compo
polarization,
221
=
EMF
:
Hy+
_§§;
H),_
Ex;
2,
Equation (5)
total
(6)
<7)
__
Hyz
may be written
€31:
2:!
__
£2:
Zzl
2
£3.
ZL
(8)
308
Chapter
An elimination between
Plane-Wave Propagation and Reflection
6
(4) and (8) results in equations for reflection and transmission
coefficients defined in terms of
tangential
electric field:
hiI‘’0
:
7-
=
z.
(9 )
=
EHL
2..1
+
ZL
22
E.
*2
——————L
=
EM
+
ZL
(10)
Z21
For the present case, in which we assume there is no returning wave in medium 2, the
load impedance ZL is just the characteristic wave impedance for the refracted wave
referred to the
z
direction, obtainable from (3) and Eq. 6.10(7):
'1
..
”2\/1 (i)
v
ZL
And the characteristic
=
772
wave
62
cos
=
vi
z
(11) is applicable
that, for dielectrics with ,u1
=
(11)
for medium 1 referred to the
impedance
Zzi
The second form of
sin2 61
-
COS
"'71
even
z
direction is
(12)
61
when the result for
ZL
is
complex.
Note
pa,
—=
(13)
—z——
32
771
”2
The total fields in
region 1 may then be written as the sum of incident and reflected
and the basic preperties of uniform plane waves. We shall use
utilizing (9)
(denoted H +) of the incident
the boundary:
waves,
This field
again
standing
general reach
wave
wave
Hy+
mH+
Ex
=
Hy
:
E5
=
'th+
fix
2
k1
ele‘fflv‘tew
cos
H+e—jflr‘f!:e_ja6:z
sin
sin
zero.
z
-
+
91
,8:
a
:
peat-’1
k1
traveling
cos
+
<14)
pal-‘35]
(16)
(17)
61
wave
field in the
The ratio of maxima to minima could be
magnitude
to
(15)
direction, but here the minima in the
ratio and would be related to the
parallel
pejfizz]
616“m\"’[ue"m=z
has the character of
field in the
the reference component since it is
wave as
z
x
direction and
a
direction do not in
expressed
as a
of reflection coefficient
by
standing
the usual
expression, Eq. 6.7(5).
Reflection and Transmission for Poiarization with E Normal to Plane of
incidence
ances
For this
and reflection
(TE) polarization, the basic relations (9) and (10) between impedtransmission may also be shown to apply. Note that they were
or
first introduced in connection with transmission-line
waves
in
Chapter Tbut have
now
Incidence at
6.11
found usefulness for many
wave
phenomena. Again
wave
the proper
E—:
=
Z“
Z———L
E"°‘
Ey+
=
wave
+
ZL
Ey+
(18)
Z3]
————‘°ZL
=
+
ZL
impedances
(19)
Zzl
are
obtained from
712
62
sec
n2[1 (5.. )
=
sin2
--
1
Z:I
The total fields in
1
region
E),
=
anx
=
an:
=
Bx
:
are
=
'01
9,]
(20)
61
sec
(21)
(E + denotes the value of Ey+ in the incident wave)
E +e”j/3-“"[e"j/3=: +
-E+
cos
sin
pej5=:]
616“J33r"[e"jfi=Z
Ole—jfir‘[e_jfii‘
E + sin
k1
Eq. 610(8):
—1/2
_
:
to
*
=
7
2L
309
Dielectrics
problems through the impedance concept applied
tangential electric field:
7
polarization,
on
define in terms of
we
p
For this
Any Angle
9,,
,8:
Example
==
k1
+
(22)
pej3=z]
—
(23)
pejfl=:]
(24)
(25)
61
cos
6.1 n
REFLECTION AND TRANSMISSION OF A CIRCULARLY POLARIZED WAVE
AT OBLIQUE |NC|DENCE
at an angle on a dielectric discontinuity as in Fig.
linearly polarized parts (Sec. 6.3) and relations of
60
this section used. To be specific, let us assume such a wave incident at angle 61
3.78. By Snell’s law,
degrees from air onto fused quartz with 8,.
A
circularly polarized wave,
6.11,
incident
be resolved into the two
can
=
=
6»)
“
=
sin—’[J-S: 6,]
sin
=
sin—19162
For the TM component,
wave
impedances
2:}
==
77,
cos
61
2:2
=
772
COS
92
=
sin“1(0.445)
V3.78
82
=
are
377
cos
60°
=
188.5 0
377
"‘
cos
V
3.78
264°
2
174 O,
=
26.4°
319
Chapter
from which
we
Plane-Wave
6
calculate p and
'r
=
(9) and (10)
from
174
p
Propagation and
Reflection
as
188.5
—
——-——
=
--
00 4
.
174 + 188.5
2 x 174
T
0‘96
_
T
174 + 188.5
For the TB component,
Z,1
=
771
sec
61
=
756 Q
Z.2
=
712
sec
62
=
217 D.
p
=
7
=
—O.554
0.446
So the second component is highly reflected, the first part is weakly
reflected and transmitted waves will be elliptically polarized.
TOTAL REFLECTION
6.12
A
study of
the
general
results from Sec. 6.11 shows that there
conditions of incidence of
special
reflected, and both
interest. The first is
one
are
several
that leads to
a
particular
condition of
total reflection. From the basic formula for reflection coefficient, Eq. 6.11(9)
1) if the load impedance
6.11(18), we know that there is complete reflection (I pl
=
is zero,
infinity,
that
is real:
Z21
or
purely imaginary.
'X
lpl
The value of ZL for TM
some
critical
angle
6
=
=
—
To show the last
VX2
z_‘1
’.——L
+
JXL
z
221
XZ
66:
Z?“1
—,———i‘
+
polarization, given by Eq.
6c such that
sin
+
condition, let ZL
H
Cid
=
=
jXL
1
seen to
ZL
and note
<1)
ZEI
6.1 1(11), is
or
become
zero
for
(2)
to
The value of
for TE
polarization, given by Eq. 6.11(20), becomes infinite for this
polarizations, ZL is imaginary for angles of incidence greater
than 0c, so there is total reflection for such angles of incidence.
For loss-free dielectrics having ,LLl
#2, (2) reduces to
same
ZL
condition. For both
=
sin
9c
=
82
*-
31
(3)
It is
seen
that there
when the
81 > .32, or
31 1
Total Reflection
6. 12
real solutions for the critical
are
angle
in this
case
only
when
passes from an optically dense to an optically rarer medium.
6.11(3), we find that the angle of refraction is 77/ 2 for 6
Q.
wave
From Snell’s law, Eq.
and is imaginary for greater
=
angles
of incidence. So from this
point
of View also
we
transfer of energy into the second medium. Although there is no energy
expect
transfer, there are finite values of field in the second region as required by the continuity
no
conditions at the
as
the
phase
Although the
boundary.
,3:
constant
Fields die off exponentially with distance from the
becomes
reflected
wave
boundary
imaginary.
has the
same
amplitude
as
the incident
wave
for
angles
of incidence greater than the critical, it does not in general have the same phase. The
phase relation between E_,.__ and EH for the first type of polarization is also different
from that between
and
for the second type of polarization incident at the same
+
E)”.
E},
angle. Thus,
if the incident
flected
under these conditions is
has both types of
polarization components, the re(Sec. 6.3).
polarized
elliptically
The phenomenon of total reflection is very important at optical frequencies, as it
provides reflection with less loss than from conducting mirrors. The use in total reflect~
ing prisms is a well-known example, and its importance to dielectric waveguides will
be shown in later chapters.
wave
wave
Example
6.12
EVANESCENT DECAY IN SECOND MEDIUM
It
was
noted above that fields
are nonzero
in the second medium under conditions of
boundary (i.e., are evanescent). Let
exponentially
us show this more specifically using expressions already developed. We have so far
considered the propagation factor in the z direction as
from the
total reflection, but die off
e
-j.8;z
(4)
where, for medium 2,
:82
and
using
:2
k2
COS
(5)
62
Snell’s law,
B;
=
Icy/1
-
(5—)
sin2 61
(6)
1
This becomes
(02/01)
sin
61
imaginary, which by (4) represents
>
an
attenuation factor f”, for
1, with
2
a
=
MR?)
1
sin2 .9,
—-
1
Np/m
(7)
3'3 2
Chapter
Plane-Wave
6
and Reflection
Propagation
277
277V SZr/Ao, a will be of order of magnitude
@
for
angles well beyond the critical.
wavelength
Since
:
6.13
Let
us
is incident at
9
be
might
plane
occurs
for
on
no
the dielectric
angle
matching of impedances between the
wave with TM polarization and for a medium with M
this
wave
54.5
dB) per
POLARIZING OR BREWSTER ANGLE
next ask under what conditions there
uniform
Np (or
a
two
==
,uQ,
reflected
wave
boundary.
media, ZL
Eqs.
when the
We know that
=
Zzl.
For the
6.1 MM) and 6.1 1(12)
become
2,
2,,
These two
quantities
equation
has
a
\[1
lfl
cos
-
0,,
equal
=
for
J;32
3331112
(1)
61
32
31
=
sm-l
/ 81
e 9-
(2)
9,
a
particular angle 61
/1 3-1-31112
32
_
:
+ 82
Note that (4) yields real values of
polarization there is always some
incident at this
#51
82
2
6p
such that
(3)
9,,
solution
.
9.,
:
may be made
cos
This
=
tan
1
e
/_2_
81
_
=
tan
n
1(3)
<4)
221
for either 31 > 32 or 82 > 31, and so for TM
angle for which there is no reflection; all energy
GP
angle passes into the second medium.
polarization a study of Eqs. 6.11(20) and 6.11(21) shows that there is no
angle yielding an equality of impedances for materials with different dielectric constants
but like permeabilities. Hence, a wave incident at angle
6p with both polarization corn~
ponents present has some of the second polarization component but none of the first
reflected. The reflected wave at this angle is thus plane polarized with electric field
normal to the plane of incidence, and the angle
99 is correspondingly known as the
polarizing angle. It is also alternatively known as the Brewster angle. Early gas lasers
made use of end windows placed at the Brewster angle to provide for oscillation for
only one of the two polarizations. For the TM polarization, there is no reflection from
the ends of the tube, so an external optical resonator governs the oscillation behavior.
For TE
6.14
313
Multiple Dielectric Boundaries with Oblique Incidence
Output window
/
/
Laser beam
Laser beam
I
spy
FIG. 6.13
Window set at the Brewster
angle
/
eliminate reflection for
to
a
laser beam with TM
polarization.
.................................
Example
LASER LIGHT THROUGH
Let
us
calculate the
A
6.13
WINDOW SET
needed to pass without reflection TM polarized light from a
0.633 urn) using fused quartz with refractive index 222
1.46.
angle
helium-neon laser (A
From (4),
=
=
1.46
6p =tan"1—1
Note that this is the
(Fig. 6.13).
BREWSTER ANGLE
AT THE
=
55.6°
angle between the beam axis and
angle in the glass by Snell’s law,
the normal to the window
We find the
.
62
2
Sin
._.
{174—6 55.6]
.
o
sm
Note that this is the proper Brewster
02
1’
1
=
angle
tan“1
for
going
=
34.4
from
O
glass
to
air:
1
=
34.4°
1.46
So the exit from the window is without reflection also.
6.14
MULTIPLE DIELECTRIC BOUNDARIES WITH OBLIOUE INCIDENCE
regions with parallel boundaries, the problem may be
by successively transforming impedances through the several regions, using the
standard transmission~line formula, Eq. 6.6(10), or a graphical aid such as the Smith
If there
solved
are
several dielectric
3i 4
chapter
chart. For each
Plane-Wave Propagation and Reflection
6
phase constant and characteristic wave impedance must
angle from the normal as well as the properties of the dielectric
ith region, from the concepts of Sec. 6.11, the phase constant is
region
the
include the function of
material. Thus for the
B3,.
and the characteristic
When the
impedance
wave
impedance
=
k;
cos
is
Zzi
=
71,-
cos
6,-
for TM
222'
=
77,-
sec
6,.
for TB
is
finally
(1)
6,.
polarization
(2)
polarization
(3)
transformed to the surface at which it is desired to find
reflection, the reflection coefficient is calculated from the basic reflection formula,
Eq. 6.11(9), and the fraction of the incident power reflected is just the square of its
magnitude. The angles in the several regions are found by successively applying
Snell’s law, starting from the first given angle of incidence.
Example
OBLIQUE INCIDENCE
ON A
6.14
TWO-PLY DIELECTRIC WINDOW
4.0 and consider
2.54, 223
two-ply dielectric of Ex. 6.8d with r22
incidence of plane wave at angle 61
40° from the normal (Fig. 6.14). From Snell’s
40.0°. Note that the
law, angles in the dielectric are 62
18.7°, and 64
233°, 63
exit angle is equal to the entrance angle as expected (Prob. 6.14d).
Wave impedance for the ith region using TE polarization is given by
We return to the
7-
=
:
a
=
221'
n1=1
=
z
=
771/003 6i
n2
92
)7
_________ _._._..
-7wms
FIG. 6.14
Composite
window with
wave
incident
at an
angle.
3'! 5
Problems
492 Q, 222
yielding Z21
2:4
lengths for each region is given by
2
2
ei/Ai
yielding wig/Ag
=
0.097 and
leads to the normalized
ZL3/ZO3
63/)t3
e;
‘2
0, and Z:3
0.189. Use of the above values
=
length
in
wave-
err/110
6i
COS
199 0.. The
=
on a
Smith chart
impedances
2.3/203
=
2.47,
212/202
2
0.37
213/2OI
=
0.20 +
This last value
258.5
=
corresponds
0.48
—-
232/202
j0.25,
——
=
j0.325
0.38 +
jO.3O
j0.158
|p|
to
=2
0.66
=
or
ipl2
0.435. This is somewhat greater
=2
than the value for normal incidence in Ex. 6.8d.
PROBLEMS
6.23 For
an inhomogeneous dielectric, find the differential
equation
Eq. 62(7)) if e is a function of 2 only. Repeat for s a function
for
El. (replacing
only (/2
#0
of 3:
=
in
both cases).
6.2b A
step~function uniform plane wave is generated by suddenly impressing a constant
O at time t
0 and maintaining it thereafter. A perfectly
C at z
Ex
600 m. Sketch total Ex and
conducting plane is placed normal to the z direction at z
electric field
=
=
=
=
nHy
versus 2 att
=
1 us and at!
=
3 [.LS.
6.2c Write the instantaneous forms corresponding to phasor fields given by Eqs.
6.02(21)—(24). Find the instantaneous Poynting vector and show that the average part
is equal to the average Poynting vector of the positively traveling wave minus that for
the
negatively traveling
6.2d For the modulated
number k
wave
wave.
of Ex. 6.2, suppose that the medium is dispersive with
with frequency over the frequency band of interest:
wave
varying linearly
Ak
M
-
k
a:
B:
+
00
Describe
propagation
6.3a Check to show that
zations
appropriate
of the modulated
‘90
wave
and
(2) satisfy
Eqs. 6.3(1)
plane waves.
=
x
equations
under the
speciali-
following special cases of Eq. 6.3(1), identifying the type
2, (/1
7r;
2, [/1
0; (ii) E1
1, E2
1,
2, ip
7r/2; (v) E]
l,E2
l,E2
1, E2
polarization for each: (i) E1
1, ll:
(iii) E1
7r/2; (iv) E1
1,E2
77/4.
1,1!
of
=
case.
Maxwell’s
to
6.3b Sketch the locus of E for the
=
in this
=
=
=
=
==
=
=
=
=
=
=
3'36
Chapter 6
Plane-Wave Propagation and Reflection
6.3c A circularly polarized wave of strength E1 travels in the positive
reflected circularly polarized wave of strength E { returning:
2
direction with the
E=c+nmfm+c+nmwz
Find the average
Poynting
types of terminations
cuss
vector.
at z
=
Repeat for reflected wave ()2
jy)E{ej"z and
O to give the two forms of reflected waves.
-
dis-
arbitrary elliptically polarized wave may be broken up into two oppositely rotating circularly polarized components instead of the two plane-polarized
6.3d Show that any
components.
6.43 For
uniform
a
plane
wave
of
10 GHz
frequency
propagating in polystyrene, calculate
impedance. What are the most
the attenuation constant, phase velocity, and intrinsic
important differences over air dielectric?
6.41) Plot
a curve
showing attenuation
constant in sea water from
104
109 Hz, assuming
to
that the constants given do not vary over this range. Comment on the implications of
the results to the problem of communicating by radio waves through seawater.
6.4c A horn antenna excited at 1 GHz is buried in dry earth. How thick could the earth
covering be if half the power is to reach the surface? Neglect reflections at the surface
in your first calculation and then estimate additional fraction lost
by
the reflection.
6.4d Plot attenuation in nepers per micrometer versus wavelength for nickel and silver
the wavelength range of Fig. 13.3b, using data of that figure.
expression for group velocity of a uniform plane wave propagating in
conductor and compare with phase velocity in the conductor.
6.4e Derive the
good
over
a
6.4f Determine the group velocity for uniform plane waves in a lossy dielectric, asSuming
(i) or and 8’ independent of frequency and (ii) assuming e”/8’ independent of frequency. Use the approximate form for B assuming s"/‘e’ small and compare in each
case
6.4g”?
with
phase velocity.
plane wave in loss-free dielectric 81, impinging upon a second loss~free dielec~
pic 82 at normal incidence, the average Poynting vector in region 2 is just the difference between the average Poynting vectors of incident and reflected waves in region 1.
Show that this is not so if 31 is lossy (even if 82 is taken to be loss free). Explain.
For
a
6.53 Find the instantaneous
Note the
planes
of these planes.
Poynting
for which it is
vector
zero
for
plane 2
for the
standing
wave
for all values of t and comment
Show that the average
Poynting
on
of Sec. 6.5.
the
significance
vector is zero as stated in Sec. 6.5
.
6.51) Evaluate instantaneous values of stored energy in electric field and in magnetic field
between the conductor and plane 2 in the standing wave of Sec. 6.5. Note planes for
which the two forms of energy have the
times) and verify the
statements
same
concerning
maximum values
6.6a Write the formulas for a uniform plane wave with E), and
spondence to voltage and current in the transmission-line
6.6b Consider
tively.
a
lossy ferrite with
(occurring
at
different
stored energy made in Sec. 6.5.
Hx only, and give
equations.
the
corre—
a complex, p.’
je”, respec~
j/i” and 8’
analogy for a plane wave in this material
conductance. Determine expressions for these
both ,u. and
—
~—
Show that the transmission~line
has both series resistance and shunt
elements.
6.6a Reflection and transmission coefficients are given in Eqs. 6.6(11) and (12) in terms of
electric field. Give corresponding expressions in terms of magnetic field and in terms
of power ratios.
31 7
Problems
6.6d A
conducting film of impedance 377 fl/square is placed a quarter-wave in air from a
plane conductor to eliminate wave reflections at 9 GHz. Assume negligible displacement cutrents in the film Plot a curve showing the fraction of incident power reflected
versus
frequency for frequencies from 6 to 18 GHZ.
6.7a A iO—GHZ radar produces
a
still
substantially plane wave which is normally incident upon
magnitude and phase of reflection coefficient and percent of in-
Find the
ocean.
a
cident energy reflected and percent transmitted into the
6.7b For
body of the
sea,
certain dielectric material of
effectively infinite depth, reflections of an incident
from free space are observed to produce a standing wave ratio of 2.7 in the
free space. The face is an electric field minimum Find the dielectric constant.
a
plane
wave
6.7 r: Check the
for power transmitted into
expression
a
good conductor, given
at the end
of
Ex. 6.713, by assuming that magnetic field at the surface is the same as for reflection
from a perfect conductor and computing the conductor losses due to the currents associated with this
magnetic
field.
6.83 Calculate the reflection coefficient and percent of incident energy reflected when a uniform plane wave is normally incident on a plexiglas radome (dielectric window) of
gin.a relative permittivity a,
2.8, with free space on both sides. Frequency
10 cm and 3 cm.
corresponds to free-space wavelength of 20 cm. Repeat for A0
thickness
=
=
6.8VL For
sandwich-type radome consisting of two identical thin sheets (thickness 15 mm.
4) on either side of a thicker foam»type dielectric (thickness
permittivity e,
1.81 cm, relative permi
1.1). calculate the reflection coefficient for waves
ty e,
striking at normal incidence. Take frequency 3 X 109 Hz; repeat for 6 X 109 Hz.
Suggestion: Use the Smith chart.
a
relative
—
2
6.8: What refractive index and what thickness do you need to make a quarterewave antie
10 um? Assume unreflection coating between air and silicon at 10 GI-lz'.7 At A0
=
doped
silicon with losses
negligible.
6.8d For the 10pm design of Prob. 6.8c, plot fraction of incident energy reflected
wavelength
6.8e*
Imagine
from
two
15 pm to A“
=
versus
5 [.Lm.
=
quaner»wave layers of intrinsic impedance 172 and 113 between dielectrics
17, and 174. Show that perfect matching occurs if 772M3
For 71,,
4, 713
3, 712
1.5, 17.
l, calculate reflection coefficient at
10%
below
that
for
with the result for a single
matching.
Compare
frequency
perfect
of intrinsic
impedances
(Th/m)”;
a
)to
=
=
=
:
quarterewave matching coating with T,
6.3m A dielectn‘c window of
:
=
polystyrene (see
2.
Table 6.4a) is made
a
half<wavelength
thick
the dielectric) at log Hz, so that there would be no reflections for normally
incident uniform plane waves from space. neglecting losses in the dielectric. Consider-
(referred
ing
to
the finite losses, compute the reflection coefficient and fraction of incident energy
reflected from the front face Also determine the fraction of the incident energy lost in
the dielectric window.
6.8g
A slab of dielectric of
length l.
constants
5' and e", is backed by aconducting plane
at
I which may be considered perfect. Determine the expression for field impedance
3 X 109 Hz,
at the front face, 2
0. Find value for 5750
4. e”/e'
0.01. f
z
=
=
=
1
1.25
:
cm.
Compare
=
with the value obtained with losses
=
neglected.
6.9a Write instantaneous forms for the field components of Eqs. 6.9(5)—(7) for TM polar-i»
zation. Find the average and instantaneous components of the Poynting vector for both
the
6.9b
,r
and
Repeat
2
directions.
Prob. 6% for the TE
polarization, defined by Eqs. 6.9(18)—(20)r
318
Chapter
Propagation and Reflection
Plane-Wave
6
6.9c For the TM polarization with E, E,, and H, find stored energy in electric fields and
in magnetic fields, for unit area, between planes 2
0 and z
77/ (k cos 6) and
=
==
interpret.
6.10a Show that
a generalization of Eq.
written in the convenient form
6.10(1) for
E(.r,
y,
z)
=
a wave
oblique
to all three axes can
be
E+e‘fl“'
where
k=n,+yk,+2k_,
r
=
xx +
yy
+ 22
6.10b A diffraction grating as in BX. 610 has grating spacing 2 ,um. Find the number and
0.488 pm at (i) norangle of the diffracted orders for an argon laser beam with A0
60 degrees.
mal incidence, 91
O, and (ii) incidence with 61
=
=
=
6.10(: Find the modified relation for diffracted angle in Ex. 6.10 if there is dielectric 81 on
the incident side of the grating and a different 82 on the exit side (,u.
no for both).
=
6.113 Plot
6.111)
incident angle the phase and magnitude of p for a wave incident from air
4.95 for both TM and TE polarizations.
ceramic with 8,.
versus
onto a
=
Repeat Prob. 6.11a with the wave passing from the ceramic into air. (Note
tion is total beyond some angle, as will be explained in Sec. 6.12.)
6.11c Obtain the special forms of p and r for
1) for both polarizations and 32 > 81.
grazing incidence (0— 77/2
—
-
that reflec-
8 with 8 <<
6.11d For both polarizations, give the conditions for which the standing wave pattern in
shows a minimum of tangential electric field at the boundary surface; repeat for a
maximum of tangential E at the surface.
2
expressions for Ex, Hy, and E: in region 2 as functions of x and z for polarization
plane of incidence. Obtain the Poynting vector for each region and demonstrate the power balance. (Take 82 > 81 to avoid the special case to be treated in
6.11e Write
with E in the
Sec.
6.12.)
10. For normal incidence
6.11f A wave passes from air to a medium with a,
10, in
there is no reflection since m
772
170. There is a nonzero reflection for incident
angles other than zero. Plot reflection coefficient versus incident angle for both TM
and TE polarizations.
=
==
=
=
6.123 A microwave transmitter is placed below the surface of a freshwater lake. Neglecting
absorption, find the cone over which you could expect radiation to pass into the air.
6.12b
Using data of Fig. 13.2b, find the critical angle for a wave of wavelength 3 am passing from silica into air and also for silicon into air. For both cases plot phase of reflec—
tion coefficient versus incident angle over the range BC S 6 S 77/ 2 for a wave polarized with E in the plane of incidence.
6.12c In
GaAs laser, the
0.85 pan) is “trapped” or guided along
generated radiation (A0
junction region by total reflection from the adjacent layers. (The mechanism
of dielectric guiding will be explored more in Chapters 8 and 14.) If the junction layer
has refractive index n
3.60, the upper layer n
3.45, and the lower layer n
3.50, find critical angle at upper and lower surfaces.
a
=
the thin
=
6.12d*
Defining 1/1 as
the
phase E_t_ /E.,+
=
and
11/
as
the
phase
=
of Ey_ /Ey+, find
expressions
3'! 9
Problems
for
1,1! and t//’
tween
under conditions of total reflection. Show that the
these two
polarization components,
an
2
ill
—
ill’,
is
phase
given by
1] sin?’ 9
(772/171)[(vg/v,)2
cos
1]
[072/7702
6\/(v2/v,)2 sin2
difference be-
—
<§>
=
2
6.12c
6
6
-
—
1
now of importance in optical communications, guide light by the pheof total reflection (as will be discussed more in Chapter 14). The evanescent
fields outside of the guiding core could cause coupling or “crosstalk” between adja-
Optical fibers,
nomenon
cent
fibers. To estimate the size of this coupling, take a plane model with silica core
refractive index n
1.535 and external cladding of a borosilicate glass having
1.525. The optical signal has free~space wavelength of 0.85 pm. How far away
the next core be placed if field is to be 10"3 the surface value at that plane, if
having
=
12
can
=
(i) incident angle of
89
waves
within the
core
is 85
degrees,
6.12f For several
on
5
and
(ii) if incident angle is
degrees.
applications of total reflection (note Probs. 6.12c
boundary are not very different. If 122
A where A
26.
1, show that 6C
72/2
the two sides of the
<<
a
and
=
e), refractive indices
6) where
221(1
—
=
-
6.133 In Ex. 6.13, it was found that polarizing angle for entrance to the window is also
rect for exit from the window. Prove this for general ratios of 122/221.
6.13b Examine
cor-
Fig. 13.2!) to determine materials which might be suitable as Brewster win—
a C02 laser operating at 10.6 ,u.m. Calculate the appropriate window angle
dows for
for each such material.
6.13c A green ion laser beam,
then passes
Design
window
a
E in the
through
plane
to
0.545 pm, is generated in vacuum, and
operating at A0
1.34.
glass window of refractive index 1.5 into water with n
give zero reflection at the two surfaces for a wave polarized with
=
=
a
of incidence.
6.13d For a,
82 but it, # #2: show that the TB polarization will have an incident angle
with zero reflection like the Brewster angle for TM polarization. Also, what conditions
must be satisfied for such an angle for TE polarization if both 31 s5 82 and pt, # [.LZ?
=
wave in medium 1 of permittivity 81 makes angle 9, with the normal.
Find the prOper length and permittivity of a medium 2 to form a “quarter-wave matching section” to a medium of permittivity 83. Consider both polarizations.
6.1421 An incident
6.14b* A uniform plane wave of free~space wavelength 3 cm is incident from space on a window of permittivity 3 and thickness equal to a half-wavelength referred to the dielec»
tric material so that it gives no reflections for normal incidence. For general angles of
incidence, plot the fraction of incident energy reflected
E in the
6.14c
plane
of incidence and also for
polarization
versus
polarization with
plane of incidence.
6 for
normal to the
of the transmission-line analogies, determine the spacing between a thin film
parallel perfect conductor, and the conductivity properties of that film if reflections are to be perfectly eliminated for a wave incident at an angle 6 from the normal
for the two types of polarization. (See Ex. 6.6b.)
By
use
and
a
6.14d For
a
boundaries between different dielectrics, as in Fig. 6.14, show
angle and angle of incidence upon the first boundary is
Snell’s law utilizing indices of refraction of entrance and exit materials only,
there is no intermediate surface at which total reflection occurs.
series of
parallel
that the relation between exit
given by
provided
6.14e An
optical instrument called an ellipsometer employs a beam of monochromatic light
elliptical polarization incident at an oblique angle on a surface whose properties
with
32%
Chapter
are
6
Plane-Wave
Propagation and Reflection
to be determined. Measurements on the incident and reflected beams
yield
a
value
ratio of reflection coefficients for components in the plane of incidence
and normal to it. Find the expression for the ratio of reflection coefficients that will be
measured by the ellipsometer in terms of incident angle, refractive index of the sub—
of the
complex
supporting the film, and the unknown film thickness and reflective index. Assuming angle and substrate index known, and neglecting losses, explain how the two
strate
unknowns could be determined.
~
‘
/
“3‘
a
.1
“1'
A
.‘3/‘121
_
‘\
‘9‘}:
‘~/
.
“of:
was
‘Tssiz’ét
§3afifisach:
(-.
*
x
h;
‘.“..‘C
7. l
‘
.
.-\
.
i
>
\v
’p
m
«
‘9,
A,
-
i
,,.=
V'
<
2
INTRODUCTION
In the
preceding chapters, numerous special techniques have been presented for solving
dynamic field problems. Before continuing with the important problems of
wave guiding, resonance, and interaction of fields with materials and radiation, it is
necessary to develop some more general and somewhat more powerful techniques of
problem solution. The methods developed in this chapter will usually be illustrated first
through static examples before extending to dynamic problems, and in some cases are
most useful for static or quasistatic problems. Even then such solutions are of use in
certain time-varying problems, as we have seen in the case of circuits and transmission
lines in the preceding chapter.
The approach in this chapter is mostly through the solution of differential equations
subject to boundary conditions. In certain cases the field distributions themselves are
desired, but in other cases (as we saw in the calculation of circuit elements) these
distributions are only steps along the way to other useful parameters.
The most general analytical method to be considered in this chapter is that of sepa—
ration of variables, leading to orthogonal functions which may be superposed to rep—
resent very general field distributions. In developing this method we will spend some
time on the special functions needed for circular cylindrical coordinates (Bessel func»
tions) and for spherical coordinates (Legendre functions). A second powerful analytical
method is that of conformal transformation. Although restricted to two-dimensional
problems and useful primarily (but not exclusively) for solutions of Laplace’s equation,
it is the most convenient way of solving many problems of importance in circuits and
static and
transmission lines.
32?
322
Chapter
Two- and Three-Dimensional
7
Boundary Value
Problems
problems becomes increasingly important with the concomputing power. This is a special field in itself and a rapidly
but we will give some idea of its basis and some elementary approaches
Numerical solution of field
advances in
tinuing
changing one,
to
its
-
use.
'E'he Bast
Bifierential Equations and Name teal Methods
ROLES OF HELMHOLTZ, LAPLAOE, AND POISSON EQUATIONS
7.2
We have
differential
equations—the wave equation, the Helmholtz
equation,
equation—result from Maxwell’s equations with certain
We
shall
generally be concerned with such Special cases, but let us look
specializations.
first at somewhat more general forms. We use the phasor forms, and limit ourselves to
homogeneous, isotropic, and linear media. Starting with the Maxwell equation for curl
E [EL 33(3)],
seen
how
Specific
and the diffusion
V X E
The curl of this is taken and
~jmuH
=
expanded (inside
front
(l)
cover)
VxVxE==—V2E+V(V°E)=—janxH
The
divergence
Eqs. 3.8(1)
and
of E and curl of H
substituted from the Maxwell
are
(2)
equations
3.8(4):
*VZE
+
V(E)
~jwu[J
:
e
jweE]
+
or
VZE
where
k2
2
rogue. By
similar
+
operations
VZH
Equations (3)
and
(4)
eral solutions of these
sz
+
1
=
jqu
on
kZH
+
the curl H
--=
(3)
ng
equation,
we
obtain
—-—V x J
(4)
may be considered inhomogeneous Helmholtz equations. Gendifficult, but usually start from solutions of the corresponding
are
homogeneous equations1
l
J. D. Jackson, Classical
VZE
+
sz
=
0
(5)
VZH
+
k2H
=
0
(6)
Electrodynamics, 2nd ed, Wiley,
New York, 7975.
7.2
Many
so
of the
Roles of Helmholtz, Laplace. and Poisson
Equations
323
problems we are concerned with have no sources except on the boundaries,
equations considered in this chapter are the homogeneous ones, (5)
the Helmholtz
and
(6).
Note that the vector
equations separate simply
V213,.
and
for
+
szx
in
=—
rectangular coordinates,
0
(7)
and
H3. They do not separate so simply for curvilinear
coordinates,
by examining the expansion for V2 of a vector in cylindrical
and spherical coordinates (inside front cover). But for any cylindrical coordinate system,
the axial component of (5) or (6) satisfies a simple Helmholtz equation,
similarly
E), E2, H_ Hy,
1.,
as one can see
VzEZ
similarly for H3.
quasistatic problems,
Laplace equations:
+
sz2
==
0
(8)
and
For
the term in k2 is
negligible
so
that
(5) and (6) reduce
to
V2E
==
0
(9)
V2H
=
O
(10)
These separate into coordinate components as discussed above. However, for quasistatic
or purely static problems it is often more convenient to use the scalar potential functions
defined
by
E
with (I) and
:
:
~Vct>m
(11)
CID,” satisfying Laplace equations,
VZCD
In certain
H
~VCI),
cases we are
taining charges,
=
VZCID,”
O,
case
the Poisson
V2613
0
quasistatic solutions for regions
equation applies to (I) (See. 1.12):
concerned with static
in which
==
:
or
J
(12)
con-
(13)
8
equations govern a large number of important
used for illustration of solution methods in this chapter.
Thus the Laplace, Helmholtz, and Poisson
problems
and will be the
ones
Boundary Conditions As noted in Sec. 1.17, unique solutions of the Laplace or
equation resulted if the function is specified on a boundary surrounding the
of
region interest. Specification of the normal derivative on such a boundary determines
the solution within a constant. Section 3.14 pointed out that unique solutions of the
Helmholtz equation (5) or (6) are obtained by specifying the tangential component of
E or H on the closed boundary, or tangential E on a part of the boundary and tangential
Poisson
H
on
the remainder.
Superposition Since V72 is a linear operator (as are the other operators in Maxwell’s
equations), any two solutions are superposable and the sum is a solution provided that
the medium itself is linear. We have made use of this fact previously, as in the super—
324
Chapter
7
Two- and Three-Dimensional
Boundary Value Problems
V=0
V=0
V=O
V0
2
1
Series of circular
FIG. 7.2
multiples
of
a
cylinders with
71
sectors of
angle
a:
at
the
potential VO
oriented
at
with respect to each other.
position of linearly polarized waves to form a circularly polarized one. We will use the
principle in many future examples. Here we give an example of its use in reasoning to
a simple result.
r-';;/3',1-i:;;-:z>,\\:;.;-:'s-/> .-rL:4:;-zi<-;,:...I.';':\\-/ =,'.:, .21 .............
.-
Example
SOLUTION
BY
7.2
INVERSE APPLICATION OF SUPERPOSITION
interesting example of the use of superposition is the solution for the potential at
a symmetrical structure. For example, consider a homOgeneous dielectric
surrounded by the circular cylinder shown in Fig. 7.2, with a potential VO applied over
a portion of the boundary subtending the angle a and zero potential on the remainder.
27r/n. If the potential at the center were found for 12 different sets
Suppose that a
of boundary conditions as shown in Fig. 7.2, where the only difference between these
is that the section of the boundary to be at potential V0 is rotated by the angle ka, with
k an integer, the sum of the 22 solutions would be the potential at the center of a cylinder
with VO over the entire boundary; this is just V0. Since every problem is identical except
for a rotation by a, which would not affect the potential at the center, the potential at
the center for the original problem must be VO/n. This same technique could be applied
to find the potential at the center point of a square, cube, equilateral polygon, sphere,
and so on, with one portion at a given potential.
An
the center of
=
7.3
NUMERICAL METHODS: METHOD OF MOMENTS
powerful computers has greatly eXpanded our ability to obtain
electromagnetic field problems. The range of use extends from
convenient evaluation of analytic expressions, including ones for which no closed-form
Easy accessibility
accurate
to
solutions for
7.3
325
Numerical Methods: Methods of Moments
solutions exist, to
wholly numerical solutions. Whenever it is possible to find even an
approximate analytic solution, it is useful for seeing parametric dependences to gain
physical insight, but more precise solutions can be obtained numerically.
We consider here only some basic methods. There are other, more specialized techniques, several of which find use in analyzing transmission structures of the kind to be
studied in the following chapters?"3 The choice of method should be based on a tradeoff among accuracy, speed, versatility, and computer memory requirements.
The finite~difierence method was introduced in Sec. 1.20; though simple, it has con»
siderable range of application. A method with similar use, called the finite-element
method, is somewhat more difficult to understand and the programming is more complex, but it has the advantage of adapting well to complex boundary shapes and also
to spatially varying properties of the medium (i.e., permittivity or permeability). In both
of these methods the typical calculation involves a large banded matrix (nonzero ele~
ments only along and near the diagonal). There are well—developed methods for inverting such a sparse matrix, and we saw in Sec. 1.20 an iterative method.
One may use a more computationally efficient approach, called the method of moments, for some problems, especially when integral quantities such as capacitance are
required.4 It is based on an integral equation rather than the differential equation on
which the finite-difference and finite-element methods
are
based.
If
charge is transferred between two conducting bodies in otherwise free space, a
potential difference will exist between them and the charges will become distributed
over the surfaces in such a way that the tangential electric field at the conductor surfaces
is zero. This is analogous to the situation seen in the study of images in Sec. 1.18,
where a point or line charge placed near a conducting surface induces surface charge
on the conducting body and this cancels the tangential electric field of the source charge.
Likewise, if charge is placed on an isolated conducting body, the charges will distribute
themselves on the surface to eliminate the tangential electric field. The method of
moments results in knowledge of the charge distribution on the surfaces and the total
charge for a given potential, and hence the capacitance.
We introduce here a simple way of applying the method of moments to find static
charge distributions and capacitances for two- and three-dimensional electrode systems.
Some structural forms that can be treated are shown in Figs. 7.3a—c. They are shown
with their surfaces subdivided into small elements to prepare for discrete numerical
charge density p5, is assumed to be uniform over each element.
calculations. The surface
The total
the
area.
potential
charge
3
4
as a
point charge
on a
at
calculations. The 2D structures have
the surfaces
2
ascribed to the ith element
We will treat it
are
divided into
strips
of width
3D structure is
psiASi,
where
the center of the element in
no
A1,.
AS,-
making
is
the
variations in the axial direction and
The
charge density
p5,. in this
case
is
R. C. Boonton. Jr., Computational Methods for Electromagnetlcs and Microwaves, Wiley,
New York, 7992.
R. Sorrentino (Ed. ). Numerical Methods for Passive Microwave and Millimeter Wave Struc-
tures, IEEE Press, New York, 7989.
R. F. Harrington, Field Computation by Moment Methods, 1?. E. Krieger, Malabar, FL, 7987;
orig. ed, 1968.
326
(Chapter
7
Two- and Three-Dimensional
Boundary Value
Problems
Z / Z Z /
Z / Z Z Z /
/ Z Z Z Z /
////z7
/;Z/1/17
(a)
A4
(b)
QQQQQ Q Q Q QQ
/
ASi
(c)
of structures suited for evaluation
FiG. 7.3 Examples
by methods of moments. (a) Three»
dimensional parallel-plate capacitor. (b) Round two-dimensional cylinder over a ground plane.
(c) Isolated rod of finite length.
multiplied by All. to give the charge per unit length along the axial direction, which is
represented by a line charge q , in the center of the element for the purposes of calculating
potential. These point and line charges are used to calculate the potentials also at the
centers of the elements.
Three-Dimensional Structures
charges
Potentials
in free space since the conductors
are
are
calculated
accounted for
using the formulas for
by including all charges
on
their surfaces. The
using Eq. 1.8(3)
potential
at the center
of the ith element in the 3D
N
CD“U +
=
I
on
(Di;
is written
is the
potential
at the center
stASj
121‘ 47T€|Fi
(1)
—-——~-—---
_"
l‘j‘
of the ith element
the ith element itself; it must be handled
of (l)
case
as
CDThe term
327
Numerical Methods: Methods of Moments
7.5
resulting
from the
charge
since the terms in the remainder
separately
find (13h- by integrating
over the element. For
clearly singular when i
j. We
we neglect the exact shape of the element, often a square, and
replace it
disk having the area of the element. Thus, with r0
(ASi/a'r)”2
=
are
convenience,
with
a
:
277
ct)“.
f
=
dqs
o
o
One
f
ro
d
Psi—r
0.282 BE
.
.
=
4778)”
8
VASi
(2)
equation of the form (1) is written for each element, thus giving a set of N equations
charges in terms of the given potentials on the electrodes.
in the N unknown
Example
7.53
THREE-DIMENSIONAL CAPACITOR
Let
us
calculate the
To achieve
but here
is
charge
distribution and
capacitance
of
the
structure
in
Fig. 7.351.
accuracy, it is necessary to have many subdivisions of the surfaces,
take a very coarse grid to illustrate the procedures. It is assumed that there
high
we
negligible charge
the outer surfaces of the conductors. The
on
and bottom electrodes
are
taken
as
+ V and
-—
potentials on the top
V, respectively. Multiplying (1), with
(2) substituted, by 47r8/a, the equation for element
1
can
a
a
47T(O.282)pS 1
+
———-—--
‘rl
"‘
be written
rzl
p$2 +
+
--——---
Ir!
“
’8’
p5 8
=
as
4778
——V
a
wa/yyg
a/
a—a—a
((1)
FIG. 7.3d
Coarse subdivisions of
parallel-plate capacitor for
Ex. 7.3a.
(3)
328
Chapter
Two- and Three-Dimensional
7
the other
Writing similar equations for
Boundary Value
subdivisions and
seven
Problems
in matrix form,
casting
have
we
a
r
3.54
11‘1"“ rzl
~——"———
Ira
‘"
r3}
lr1_ r8,
4-"—
__E__
lrl
3.54
-—-—a
“
[1'2
1‘1]
1'3]
“'
[1‘2
V
[7’P52
51
V
V
1'81
"
""
V
..
'
a
V
4773
=
——
a
-—V
(4)
—V
a
L.
Ira
‘
1‘1]
irs
—-V
a
a
"
r2]
The coefficients include
Ira
’13
rjl
-
directly
L-V-
J-Ps8.1
and must be evaluated
The matrix could be entered into
densities found
3.54
1‘3]
“'
geometrically
from
Fig. 7.3d.
inversion program in a computer and the charge
in terms of the potentials on the electrodes. The total charge Q
an
Q/ 2V.
just C
For the purpose of illustration, we will solve the problem by hand, making use of its
symmetry to reduce the computational work. Symmetry dictates that the assumed uniform charge densities satisfy: psl
p55
p58 and p52
ps4
p33
p86
ps7. Since there are just two unknown variables, it is necessary to use only the first
two rows of (4). Substituting values of Ir1
rjl and [r2 rji and using dimensions in
Fig. 7.3d, we obtain
on one
electrode is found and the
capacitance
=
=
is
=
=
——
==
-
=
—-
=
-
-—
([3
J
'
87
1.50
_.
_l_
__
(d/a)
V9
(d/a)2
+
1
——-—-————
V1
+
[
[4
taking 51/5:
(Ci/(2)2]
————————
—-
V4
(cl/a)2
+
1
1
1.50
or
1;]
1
—-
—
—
———-—
V1
1
'
=
54
-
V4
(at/a)2
+
—
(d/a)
(d/a)2]
——
—~
+
[p51]
p52
———1————]
V1
+
(Cl/a)?‘
0.121
Ps1
47reV
1
a
1
__
0.121 1.65
we
sl“ [2:]
0.5, for example,
1.54
Inverting (6),
3
ll
P52
(6)
find
478V
p$1
=
1.071032
=
0.605
(7)
7.3
329
Numerical Methods: Methods of Moments
The total
29.38CZV and the
charge on the top electrode is Q
2c22(p31 + p52)
14.7sa. Application of the method of moments
capacitance is therefore C
Q / 2V
with a fine grid would lead to a more accurate value for capacitance, which would be
Sea.
larger than the fringing-free idealization C
eA/d
4ea2/ (a/ 2)
=
=
=
2
=
Two-Dimensional Structures
==
=
For the 2D case,
we
write, using Eq. 1.8(8),
N
CI).1
The distribution of
=
charge
‘13..n
j2¢i
-
+ CT
2778
is unaffected
by
the constant
(8)
CT and it can be neglected (see
Prob. 7.30).
As in the 3D case, it is necessary to handle (Dig separately. The approach is to integrate
the effect of the surface charge density over an assumed flat strip of width Ali. Thus,
(Dz;
so
2p
Ali/2
-——-S—'f
=
‘
277’s
—E§'- [xlnx
‘
lnxdx
=
--
7T8
0
x]OA’i/2
(9)
that
psi
"
A11'
277's
[n
A1
:|
2
(
)
Example 7.5b
SrRIPLINE CAPACITANCE
us calculate the charge distribution and capacitance per unit length of a twodimensional system of conductors, the so-called stripline configuration, which will be
discussed in Sec. 8.6 and is shown in Fig. 7.38. Here there are three conductors, with
Let
the outer
ones
extending to y
=
and the center conductor carries
infinity,
the surface
charge
conductor. Therefore,
we
d {<
j: 00. The two outer ones are at the same
a
V.
Although
voltage
rapidly with y beyond
decreases
will cut off the outer conductor at
w
(zero) potential
the outer conductors extend to
the
edge of the center
appropriate point
some
>{
LJLW4444444#4i14®-:0
(8)
[‘16. 7.3a
Stripline structure with discretization
for method of moments calculation in Ex. 7.3b.
333
Chapter
Boundary Value Problems
Two- and Three-Dimensional
7
ymax, the suitability of which could be tested by doing the problem twice with different
values of ymax. We will simplify the notation by taking the widths AZ of all segments
to
be the
same.
segmenting is shown in Fig. 7.3e, where it is seen that there are 36 equal
sufficiently fine for illustration, but in practice more divisions
be
this
For
example we have 36 equations of the form (8); 8 have (I)
might
chosen.
V and 28 have CI)
O. The equation for element 1 (on center conductor) is obtained
by multiplying (8), with (10) substituted for CI)“, by ~27Ts/Al:
The chosen
subdivisions. This is
=
=
AZ
(1113‘ 1)psl
_
+
mlrr
’
r2lp52
Or, subtracting psi 1n AZ from each
—
"l'
+
term and
irr
_
lnlrl
summing
2778
r36lp536
‘
=
“K;
the subtracted terms
lrl
rzi
V
(11)
separately,
r36’
"
(1n 2+1 )Psr +1n—-—— P52 +m+1n————— P536
Al
Al
(12)
2778
lnAllpsl
+
on
the set of
in the form of
charge
equations
the
plates is
But the total
+
+ pa +
zero so
9535]
=
“T”
V
the term in brackets vanishes. We
(12) for the N elements
as a
can
write
matrix:
B
‘
T
__
1n
1'6 93
irz
"“
1“
Ir;
—
rol
.—__.:._
.
.
.
1D.
lri
Al
I'll
_
-
Al
rssl
-*
F
Al
1.693
'
-
.
ln
lrz
l"V
-
---——
p51
.
I'36]
““
—-—
”52
Al
_
—
23
V
Al
[1'11
“1n
—
rrl
A1
111
(1‘11
"
rzl
'
1.693
A]
Lp536
“OJ
_
a
(13)
To obtain
4d and a,
numerical result, we take w
4. Inversion of this matrix
the
on
each
element.
The
gives
charge
capacitance for the complete structure
=
a
=
equation
Fig. 7.32 is the sum of the charges on the center conductor, found from (13), divided
by V. Its value is 344p. F/m. The value calculated analytically for an infinitely thin
center conductor is found (Sec. 8.6) to be 346p. F/ m.
in
In the method of moments, the order of the matrix to be inverted is much smaller
or finite-element method, but the main"): is full so that sparse
than in the finite-difference
matrix
techniques cannot be used.
Chapter 12.
will be shown in
An
application
to a
time-varying
radiation
problem
7.4
33'!
Method of Conformal Transformation
Method of conformai Transformation
7.4
METHOD OF CONFORMAL TRANSFORMATION AND INTRODUCTION
TO COMPLEX-FUNCTION THEORY
general mathematical attack for the two—dimensional field distribution problem
theory of functions of a complex variable. The method is in principle the
most general for two-dimensional problems, and the work can be carried out to yield
actual solutions for a wide variety of practical problems. For these reasons, the general
method with some examples will be presented in this and the following sections.
In the theory of complex variables, we use the complex variable Z
x + jy, where
both it and y are real variables. It is convenient to associate any given value of Z with
a point in the x—-y plane (Fig. 7.4a), and to call this plane the complex Z plane. Of
course the coordinates may also be expressed in polar form in terms of r and 6:
A very
utilizes the
=
5
N
-—
sz
yz,
+
6
=tan’1(z)
x
Then
Z
Suppose
that there is
x
now a
+
jy
=-=
different
W
such that W is
is
a
rule
some
specifying
=
corresponding
re”
+
complex
variable W, where
u
function of Z. This
a
j sin 6)
r(cos 6
+
jU
means
=
=
pej‘b
that, for each assigned value of Z, there
value of W. The functional
=
(1)
relationship
is written
(2)
f(Z )
W
W
C’
(b)
(a)
FIG. 7.4
(a) Z plane. (b) W plane.
332
Chapter
Two- and Three-Dimensional
7
If Z is made to vary
moves
about, tracing
out a curve
graph,
the
continuously,
Boundary Value
Next consider
in the
complex Z plane
correspondingly, tracing
usually shown on a separate
corresponding point
out some curve C. The values of W vary
C’. To avoid confusion, the values of W
called the
Problems
are
W
plane (Fig. 7.41:).
complex
small change AZ in Z and the corresponding change AW in
a
derivative of the function will be defined
the usual limit of the ratio
as
W. The
AW/ AZ
as
the
element AZ becomes infinitesimal:
fl:
complex
f(Z
.
+
32:10—— :Alzl’fI-EO
function is said to be
analytic
or
AZ)
"
fCZ)
AZ
AZ
dZ
A
AW
.
regular
(3)
whenever the derivative defined
The derivative may fail to exist at certain isolated (singular)
points where it may be infinite or undetermined, somewhat as in real function theory.
But it would appear that there is another ambiguity with respect to complex variables,
above exists and is
unique.
since AZ may be taken in any arbitrary direction in the Z plane from the original point.
For the derivative to be unique, the ratio AW/ AZ should turn out to be independent of
this direction.
If this
the
AZ
same
=
independence of direction is to result, a
result if Z is changed in the x direction
Ax,
W
(W
flu
a
—-—=—-—=-—
dZ
For
a
necessary condition is that we obtain
alone or in the y direction alone. For
change
63cm
ax
in the y direction, AZ
2
av
=~—+
ax
6a
60
8y
6y
8(1)!)
complex quantities are equal if and only if their real
separately equal. Hence, (4) and (5) yield the same result if
=
6):
imaginary parts
are
(6)
-—
6y
~63
-
~95
6x
(7)
6y
Cauchy-Riemann equations, are then necessary conunique
point and the function f(Z) analytic there. It can
These conditions, known
as
the
dW/dZ to be
they are satisfied,
at a
be shown that, if
direction of the
and
81)
all
arbitrary
(5)
+jv)=————j—
Two
ditions for
4
H
—-—
jax
j Ay,
=——.—=—.———~<u
] 6y
612
jv)
1
6W
(1W
+
change AZ,
the
so
same
they
result for
are
dW/dZ
is obtained for any
also sufficient conditions.
7.5
Pmperties of Analytic Functions of Complex Variables
Exampie
333
7.4
ANALYTICITY OF POWER FUNCTIONS
W
M
+
1'0
=
(x
+
M"
Z2
=
(xi
=
yz)
-
,
A check of the
+
flay
(8)
,
u
=
x“
2)
=
21y
-—
y“
Cauchy—Riemann equations yields
61,!
—*
av
=
6x
92
—
2x
=
By
._
a:
2y
_.
ax
6y
So they are satisfied everywhere in the finite Z plane, and the function is analytic
everywhere there.
Actually, it is not necessary to apply the check when the functional relation is expressed explicitly between Z and W in terms of functions which possess a power-series
expansion about the origin, as ei, sin Z, and so on. The reason is that each term in the
series C,,Z” can be shown to satisfy the Cauchy—Riemann conditions, and consequently
a
series of such terms also satisfies them.
7.5
If
PROPERTIES OF ANALYHC FUNCTIONS OF COMPLEX VARIABLES
Eq. 7.4(6) is differentiated with respect to 3:, Eq. 7.4(7)
resulting equations added, there results
differentiated with respect to
y, and the
6211
62a
-—-
——
6x2
Similarly,
=
O
ayz
(I)
if the order of differentiation is reversed, there results
621)
62v
=
—-
6x2
ayz
O
(2 )
are recognized as Laplace equations in two dimensions. Thus, both the real and
imaginary parts of an analytic function of a complex variable satisfy Laplace’s
equation, and would be suitable for use as the potential functions for two~dimensional
electrostatic problems. The manner in which these are used in specific problems and
the limitations on this usefulness are demonstrated by examples in this and the next
These
the
section.
334
Chapter
7
Two- and Three-Dimensional
Boundary Value Problems
problem in which one of the two parts, it or v, is chosen as the potential function,
proportional to the flux function (Sec. 1.6). To show this, let us
u
that
is
the
potential function in volts for a particular problem. The electric
suppose
field, obtained as the negative gradient of u, yields
For
a
the other becomes
Bu
Ex
By
the
the
x
equation
an
=
=
75;,
for the total differential, the
and y coordinates of dx and
dy
6y
change
in
v
corresponding
to
changes
in
is
6v
(10
(3)
_.......
,
=
av
dx +
—
-
Bx
6y
dy
But, from Cauchy—Riemann conditions, Eqs. 7.4(6) and 7.4(7),
a
—-dv
==
ldx
a
Edy
6x
—
6y
—E dx +
=
3'
Erdy
‘
or
~e
By inspection of Fig. 7.5a, this
curves U
and
v
+
dv, with the
dv
a
dx +
Dx dy
recognized to be just the electric flux dip between
positive direction as shown by the arrow. Then
is
~d¢r
And, except for
flux at 0
0,
—Dy
=
constant that can be set
=
e
dv
equal
to zero
(4)
the
(5)
by choosing the reference for
==
~90
=
av
C/m
(6)
u+dv
4x
FtG. 7.5a
Coordinates for the flux function.
7.5
Similarly,
if
v
335
Properties of Analytic Functions of Complex Variables
is chosen
as
the
function in volts for
potential
some
problem,
flux function in coulombs per meter, with proper choice of the direction for
flux.
an
is the
positive
We have seen that either it or u may be used as a potential function, and then the
other may be used as the flux function, since both satisfy Laplace’s equation. The utility
of the concept, however, hinges on being able to find the analytic function W
f(Z)
=
such that
u
and
v
also
satisfy
the
conditions for the
boundary
Example
ELECTRODES
As
the
example, suppose we desire
given boundary condition is
an
V
If
we
IN
7.5
PARALLEL~PLANE DIODE
the distribution of
=
problem being
considered.
x4/3,
y
=
potentials
in the Z
plane
where
0
(7)
let
W
it is clear that for y
dW/dZ exists and is
=
=
24/3
O, the real part of W is
(8)
u
=
x4/3. Furthermore,
we see
that
O (Prob. 7.4e). Thus it is a suitable potential
unique except at Z
function for this problem; the real part of (8) gives the potential distribution. It is most
convenient for this particular function to express Z in polar coordinates
=
W
=
u
+
jv
=
1'4/3ej49/3
(9)
Thus
Equipotentials,
found
u
=2
v
=
3cos 36
(10)
3sin 36
by setting it equal to a constant, are shown in Fig. 7.5b for u
boundary function (7) has the same form as the potential
1.0:
O and the anode potential unity at x
the cathode at x
=
O and 1. It is of interest that the
in
a
plane
diode with
(I)
Using
on
=
1“”
a plane diode can be truncated and the correct potentials produced
edge by placing electrodes along the equipotential lines as shown in Fig.
procedure is most important in designing electron guns with regular flow
these ideas,
the free
7.517. This
and the result is known
5
=
=
J. R. Pierce, J.
as
the Pierce
gun.5
Appl. Phys. 11, 548 (7940).
336
Chapter
7
Two- and Three-Dimensional
Boundary Value Problems
0 volts
1.0 volts
V=atvJ
Cathode
0 volts
Anode
1.0 volt
.2”
FIG. 7.5b
Focusing
for electron flow in
the electrodes outside the electron flow
7.6
a plane diode.
region.
The upper
portion
of the
figure shows
CONFORMAL MAPPING FOR LAPLACE'S EQUATlON
point of view toward the method in Sec. 7.5 follows if we refer
planes introduced in Sec. 7.4. Since the functional relationship fixes a
of W corresponding to a given value of Z for a given function
A somewhat different
to the Z and W
value
W=f(Z)
point (x, y) in the Z plane yields some point (u, v) in the W plane. As this point
F ( y) in the Z plane, the corresponding point in the W
along some curve x
plane traces out a curve u
F102). If it should move throughout a region in the Z
plane, the corresponding point would move throughout some region in the W plane.
Thus, in general, a point in the Z plane transforms to a point in the W plane, a curve
transforms to a curve, and a region to a region, and the function that accomplishes this
is frequently spoken of as a particular transformation between the Z and W planes.
When the function f(Z ) is analytic, as we have seen, the derivative dW/dZ at a point
is independent of the direction of the change dZ from the point. The derivative may be
written in terms of magnitude and phase:
any
moves
---
=
dW
——
.
.___
M6 J“
(1)
=
Me” dZ
(2)
d2
01‘
W
By the rule for the product of complex quantities, the magnitude of W is M times the
magnitude of dZ, and the angle of W is a: plus the angle of dZ. So the entire infini-
Conformal
7.6
tesimal
337
Mapping for Laplace’s Equation
in the
vicinity of the point W is similar to the infinitesimal region in the
vicinity
point
magnified by a scale factor M and rotated by an angle or.
It is then evident that, if two curves intersect at a given angle in the Z plane, their
transformed curves in the W plane intersect at the same angle, since both are rotated
through the angle a. A transformation with these properties is called a conformal transformation.
In particular, the lines it
constant and the lines 0
constant in the W plane
intersect at right angles, so their transformed curves in the Z plane must also be orthogonal (Fig. 7.6a). We already know that this should be so, since the constant 0 lines
have been shown to represent flux lines when the constant it lines are equipotentials,
and vice versa. From this point of view, the conformal transformation may be thought
of as one that takes a uniform field in the W plane (represented by the equispaced
constant it and constant v lines) and transforms it so that it fits the given boundary
conditions in the Z plane, always keeping the required properties of an electrostatic
region
of the
Z. It is
=
=
field.
Frequently
the transformation is done in steps. That is, the uniform field is trans~
some intermediate complex plane by Z1
f(W), then perhaps into
formed first into
=
finally into a plane
general, there can be any
12(Z2 )
Z3
number of steps. Of course, these functions can be combined into a single transformation, the inverse of which can then be understood on the basis of finding a function
with real or imaginary part satisfying the given boundary conditions as discussed in
a
second intermediate
complex plane 22
boundary conditions
=
in which the
2
g(Z1),
are
and then
satisfied. In
Sec. 7.5.
There
are
knowledge of the required boundary conditions
gives the solution. For help in finding the
of conformal transformations6 which show how one field
few circumstances in which
will lead
transformation that
to the
directly
required form there are
tables
maps into another. The
mapping
ug
U1
UT
functions
U3
given
in the tables may be used
“3
yT
I
”2
ll}
U1
individually
.
US
we
'
"‘
a
r
W
FIG. 7.60
6
For
example,
7952. Also
see
see
R.
A
Z plane
plane
mapping
r
r
of coordinate lines of the W
plane
in the Z
plane.
Dictionary of Conformal Representations, Dover, New York,
Shinzinger and P. A. A. Laura, Conformal Mapping: Methods and
H. Kober,
Applications, Elsevier, Amsterdam, 7997.
338
Chapter
combined in
or
given problem.
a
Two- and Three-Dimensional
7
Boundary Value
Problems
series of steps to transform the uniform field into a field that fits the
examples of the simpler transformations will be given to illustrate
Some
the method.
Example
7.6a
THE POWER FUNCTION: FIELD NEAR A CONDUCTING CORNER
As
a
basic
example,
consider W
expressed
W
It is convenient to
use
the
polar form
W
=
as
some
power:
2"
=
for Z
Z raised to
(3)
[Eq. 7.4(l)]:
(rel-9)”
=
rpejp”
or
u
=
rp
v
=
rp sin
cos
p6
(4)
p6
(5)
conformal-mapping point of View, the field in the W plane is uniform. The
equal potential (say, u equals constant) in the W plane can be mapped
parallel
into the Z plane by setting 1) equal to constant in (5). From the viewpoint of Sec. 7.5
one does not take explicit consideration of the existence of the W plane but simply
recognizes that v is a solution of Laplace’s equation and tries to adjust constants such
that constant 0 lines fit the equipotentials of the given problem. When only one step of
transformation is required, the vieWpoints are wholly equivalent.
If v is chosen as the potential function, the form of one curve of constant 0 (equiO and also at 6
77/p. Thus,
potential) is evident by inspection, for v is zero at 6
if two semi-infinite conducting planes at zero potential intersect at angle a, where
From the
lines of
=
=
77’
p
(6)
=
a
they coincide with this equipotential,
of the
curves
of constant
it
and
boundary conditions are satisfied. The form
v within the angle then give the field
and of constant
configuration near a conducting corner. The field is assumed to result from the presence
an electrode with nonzero potential that either fits one of the constant v lines or is
far enough away that its shape causes no significant deviation of the u and 0 lines in
the region of interest.
The equipotentials in the vicinity of the corner can be plotted by choosing given
values of v, and plotting the polar equation of r versus 6 from (5) with p given by (6).
Similarly, the flux or field lines can be plotted by selecting several values of u and
plotting the curves from (4). The forms of the field, plotted in this manner, for corners
with a
77/4, 77/2, and 377/ 2 are shown in Figs. 7.6b, 7.6a, and 7.6d, respectively.
of
=
7.6
FIG. 7.6b-d
These plots
field map
are
Conformal
Field
near
for
Mapping
conducting
339
Laplace‘s Equation
corners
of 45, 90, and 270
degrees.
of considerable help in judging the correct form of the field in
having
one or more
conducting
a
graphical
boundaries.
Example 7.6b
THE LOGARITHMIC TRANSFORMATION: CIRCULAR CONDUCTING BOUNDARlES
Consider next the
logarithmic
function
(7)
W=C11nZ+C2
The
logarithm of a complex
number is
In Z
=
W
=
readily
111018”)
found if the number is in the
=
1n
2‘
+
1'6
polar form:
(8)
so
Take the constants C 1 and
C2
as
C1(lnr
+
16) +C2
real. Then
u=Cllnr+C2
v
If
u
C16
(10)
as the potential function, we recognize the logarithmic potential
previously for potential about a line charge or a charged cylinder or be-
is to be chosen
forms found
=
(9)
340
Chapter
Two- and Three-Dimensional
7
Boundary Value Problems
av, is then proportional to angle 0,
cylinders. The flux function, 1,11
as it should be for a problem with radial electric field lines.
To evaluate the constants for a particular problem, take a coaxial line with an inner
conductor of radius a at potential zero and an outer conductor of radius [7 at potential
V0. Substituting in (9), we have
tween coaxial
=
0
-—
Cllna-I‘CZ
II
VO=C11nb+C2
Solving,
we
have
C1
so
(7)
can
V0
ln(b/a)
_
""
C2
_
“
_VD
In
a
1n(b/a)
be written
W
..
(I)
—~
__
ln(Z/a)
V0[1n(b/a)]
(11)
or
91/
In the
6
=
foregoing,
u
V0[ln(b/a)]
—-
__
fl
—-
=
8v
“8V06
111(b/a)
V
(12)
C/m
(13)
the reference for the flux function
0. If it is desired to
and its
111(r/a)
_
__
imaginary part
use some
serves to
came
out
automatically at
as complex,
other reference, the constant C 2 is taken
fix the reference
Example
1/!
=
O.
7.6c
THE INVERSE-COSINE TRANSFORMATION:
HYPERBOLIC AND ELLIPTIC CONDUCTING BOUNDARIES
Consider the function
W
=
cos"1 Z
(14)
or
x—I—jy—cosm +jv)
H
II
y
=
cos u
cosh
=cosucoshv
v
—sinusinhv
~jsinusinhv
Conformal
Mapping
for
34'!
Laplace’s Equation
It then follows that
x2
——.,— +
cosh~
v
y2
=
9
,
smhr
x2
y2
.-
~
_
0
COS“ ll
-
1
(15)
1
(16)
v
7
SID" u
Equation (15) for constant 0 represents a set of confocal ellipses with foci at :1, and
[16) for constant it represents a set of confocal hyperbolas orthogonal to the ellipses.
These are plotted in Fig. 7.68. With a proper choice of the region and the function
{either it or v) to serve as the potential function, the foregoing transformation could be
made to give the solution to the following problems:
1. Field around
a
charged elliptic cylinder, including
the
2. Field between two confocal
a
.
flat
strip
conductor
elliptic cylinders or between
extending between the foci
Field between two confocal
and
a
plane
limiting
conductor
an
hyperbolic cylinders or between
extending from the focus to infinity
case
Plot of the transformation
u
+
ft:
=
a
flat
strip
elliptic cylinder
a
cos”'(x
and
hyperbolic cylinder
(e)
FIG. 7.6a
of
+
jy).
342
Chapter
Two- and Three-Dimensional
7
Boundary
Value Problems
4. Field between two semi-infinite
rating
them
(This
5. Field between
an
separated from
it
is
conducting plates, coplanar
of 3.)
limiting
infinite
by
and with
a
gap sepa-
case
a
a
conducting plane and a perpendicular semi-infinite plane
gap
To demonstrate how the result is obtained for
a
particular problem,
consider problem
5, illustrated by Fig. 7.6f. The infinite plane is taken at potential zero, and the perpendicular semi-infinite plane is taken at potential V0. In using the results of the foregoing
general transformation, we must now put in scale factors. To avoid confusion with the
preceding, let us denote the variables for this specific problem by primes:
W’
Clcos“1kZ’
=
+
C2
(17)
C1 is inserted to fix the proper scale of potential, the constant k to fix the
scale of size, and the additive constant C2 to fix the reference for the potential. By
The constant
comparing
with
(14),
Z==kZ
W
['he constants
C1
and
C2
CIW+C2
may be taken
u'
real for this
as
=
+
Clu
problem.
Then
C2
(18)
u'=0
/
/\\
/
/
/
\/
/\
“*‘1\\
I/
/
\
”
/ \
F“7~~g
‘é—La
,
»/
/\
K'4‘,
=
V0
l J/ \/
I
”fit, \ //\\~L
\ /x
/-\\
\
/
L-.v\
-uv”
\<\
\
//
A/
\
\
()9
FIG. 7.6f
Field between
perpendicular planes
with
a
finite gap.
Conformal
7.6
By comparing Figs. 7.6g
and
7.6)”,
O, we want u'
these values in (18) yields
Also, when
u
x
want
we
Z’
to be
and when
V0;
=
u
a
=2
So the transformation with proper scale factors for this
W’
where u’ is the
u’
=
343
Mapping for Laplace’s Equation
+jv’
when Z is
77/2,
u'
problem
2
Z’
77
a
so
k
=
l/a.
O. Substitution of
is
V0[l —cos“1(-—)]
=
unity,
=
~—
(19)
proportional function in volts, and ev’ is the flux function in coulombs
equipotential and flux lines with these scale factors applied are
per meter. A few of the
shown on Fig. 7.6f.
Example 7.6d
PARALLEL CONDUCTING CYLINDERS
Consider next the function
W*Kan—a
Z+a
(20)
1
This may be written in the form
W
K,[ln(Z
=
—-
a) -~ln(Z
+
a)]
By comparing with the logarithmic transformation of Ex. 7.6b which, among other
things, could represent the field about a single line charge, it follows that this expression
a and the other of equal
can represent the field about two line charges, one at Z
it
more
is
a. However,
interesting to show that this
strength but opposite sign at Z
of
form can also yield the field about parallel cylinders
any radius.
Taking K 1 as real,
=
=
u
=
Thus, lines of
constant
logarithm equal
K
J.
(x
2
(x
--1- In
Kll:
u
u
-—
can
2.
__
a),
a)"
+
+
2
y,
(21)
+ y“
y
_.
an
an
(x
-—
(x
a)
be obtained from
(x
-
+
+
(1)]
(
)
(21) by setting the argument of the
to a constant:
(x
y
-1
at)2 + y2
2K2
a)2 +y2
344
Chapter
Two- and Three-Dimensional
7
Boundary Value Problems
i”
Q.
i”
/
k%
\
d
H:
J
(g)
FIG.
Two
7.69
parallel conducting cylinders.
As this may be put in the form
2
[rm]
1
the
curves
of constant
14 are
u-
=4_°I<_
+
(23)
K2
circles with centers at
a(1
+
K?)
K2). If u is taken as the potential function, any one of the
may be replaced by an equipotential conducting cylinder. Thus, if
d (Fig. 7.6g), the values of a
R is the radius of such a conductor with center at x
and radii
(20 VK2)/ (1
circles of constant
—
u
=
and the
particular
value of
K2 (denoted K0)
may be obtained
by setting
,
1
1
KO
"'
_‘
K0
Solving,
a
=
\/—
i
d2 ~-R2
d
d2
(24)
(25)
KO=E+ E—1
The constant
K1 in
inder. Let this be
and
the transformation
VO/Z. Then, by
depends
on
the definition of
the
potential of the conducting cylK2 (= K0 on conducting cylinder)
(25),
V
d2
d
x—-
01‘
K1
V”
=
2
]n[(d/R)
+
V(d2/R2)
=
-—
1]
V0
2 cosh
_1
(d/R)
(26)
The Schwarz Transformation for General
7.7
in (21), the
Substituting
(I)
z
potential
> O with x > O, a
(24). The flux function 1,0
4/
=
~av
<
=
0 if
-
(x
cosh“(d/R)
4
is
K1
av
2
-
111[(x
positive
so
+
the
a)2
a)2
+
+
y2
yz]
negative sign
(27)
must be
chosen in
is
8V0
z
is
point (x, y)
V0
2:
u
For CD
at any
345
Polygons
[tan“1
cosh‘1(d/R)
y
-
(x
+
t an
y
—1
a)
(x
--
a)
:l
(28)
we have not put in the 1eft~hand conducting cylinder
explicitly, the odd symmetry of the potential from (27) will cause this boundary condition to be satisfied also
Although
if the left—hand
of radius R with center at
d is at
potential VO/Z.
If we wish to use the result to obtain the capacitance per unit length of a parallel—
wire line, we obtain the charge on the right~hand conductor from Gauss’s law by finding
the total flux ending on it. In passing once around the conductor, the first term of (28)
changes by 277, and the second by zero. So
cylinder
q=2 77
=
x
8V0
2
—-
-
C/m
cosh“1(d/R)
01‘
q
we
VO
cosh—1(d/R)
C=~=————
A similar
F/ m
(29)
be used to find the external inductance of the
parallel—wire
opposite from the above electric field problem,
with 0 being proportional to the magnetic scalar potential. The result given in Eq. 4.6(9)
line. In that
procedure
case
can
the roles of
u
and
v are
for inductance is
L
=
if
’77
d
cosh-*<-)
R
H/m
(30)
From (29) and (30) we see that LC
#8 as was shown to be the case for other twoconductor lines in Chapter 5. That this is a general result is shown in Sec. 8.12.
=
7.7
THE SCHWARZ TRANSFORMATlON FOR GENERAL POLYGONS
examples in Sec. 7.6 specific functions have been set down, and the electrostatic
problems solvable by these deduced from a study of their properties. In a practical
problem, the reverse procedure is usually required, for the specific equipotential conducting boundaries will be given and it will be desired to find the complex function
useful in solving the problem. The greatest limitation on the method of conformal
transformations is that, for general shaped boundaries, there is no straightforward
In the
346
Chapter
Two- and Three-Dimensional
7
Boundary Value
Problems
Z’ plane
Z plane
P4
P3
(13
a4
P1
lb)
(:1)
FIG. 7.7
2'
plane.
(a) General polygon in Z plane. (1)) Polygon of figure transformed into straight line in
x; is at infinity.
Vertex
procedure by which one can always arrive at the desired transformation if the two—
dimensional physical problem is given. There is such a procedure, however, when the
boundaries consist of straight—line sides with angle intersections.
The Schwarz transformation takes an arbitrary polygon in the Z plane into a series
of segments along the real axis in a 2’ plane as shown in Figs. 7.7a and 7.7b. The
segments correspond to the sides of the polygon.7 The transformation may be found by
integrating the derivative:
(:32
_.
._
:
dZ’
[C(Z'
__
firm/77‘) 1(Z'
__
xé)(“2/7T)
g
1
_
,
.
(Z'
__
x;l)(a,,/vr)
_
1
(1)
Each factor in
vertices
as
(1) may be thought of as straightenng out the boundary at one of the
the transform of Ex. 7.6a did for the single corner. The setting down of (l)
specific problem is usually easy, but the difficulties come in its integration.
Although we have spoken of the figure to be transformed as a polygon, in the practical
application of the method, one or more of the vertices may be at infinity, and part of
the boundary may be at a different potential from the remaining part. Then the real axis
in the Z' plane consists of two parts at different potentials. This latter electrostatic
problem may be solved by a transformation from the Z’ to the W plane, and thus the
transformation from the Z to the W plane is given with the Z' plane only as an inter»
mediate step. Another sort of problem in which the method is useful is that in which a
thin charged wire lies on the interior of a conducting polygon, parallel to the elements
of the polygon. By the Schwarz transformation, the polygon boundary is transformed
to the real axis and the wire corresponds to some point in the upper half of the 2' plane.
This electrostatic problem can be solved by the method of images, and so the original
problem can be solved in this case also.
for
7
a
Formore details see 1?. V. Churchill and J. W. Brown, Complex Variables and Applications,
4271 ed, McGraw~Hfl/, New York, 7984.
The Schwarz Transformation for General
7.7
347
Polygons
Tablet?
2
=
jy; W
b
V0
Z
I
=
b
x
”7’”
-
qr
T
y
Cl‘
x +
a
a
2
g,
=
+
jv, where
[cosh“‘(a~
k
Results for
are
given
some
Flux Function.
=
70:
=
> acosh”‘<
Potential
2"kW—e
—
l—cr~
1+ S
1
S
0
l—
(of
or
+
1)>]
1
=
2
——
kW
:2,
b
u
”+1—22kW
e
[9
=—
u
etw
__
+
t
“1
-
a
—-1
important problems
that have been solved
by the Schwarz technique
in Table 7.7.
-
Example
FRINGING FIELD
IN
..,,_‘_.-’._
—-=
7
,.
v»
7.7
PARALLEL—PLATE CAPACITOR
To illustrate how the concept of a polygon can be applied, consider the parallel~plate
O
capacitor structure in Fig. 7.7c with d3
V0 on the infinite bottom plane and (I)
=
on
the
=
D~C. For the purposes of the Schwarz transformation, the structure may
a polygon with interior angles (21, a2, and
a3 and sides of infinite length.
plane
be considered
Application
of the transformation puts all the boundaries along the real axis as in Fig.
O for x3 > 0. As was shown in EX. 7.6b, a
V0 for xé < O and CI)
7.7d where (I)
2
=
subsequent logarithmic transformation converts such a set of boundary potentials into
a parallel—plane uniform field in the W plane. Combining the two transformations, one
//
//
/
(13:
\\
\\
71’
<1: =0\
/
All
I\
(I):
‘
V"\:
\\
E
621—
\
y:
(0)
x\
g
O
B
Z’
x'2=1
plane
(d)
(6) Edge of parallel-plate capacitor with one plane of infinite extent (equivalent to onesymmetric parallel—plate capacitor). (d) Transformation of the capacitor of (c) into a
single plane.
FIG. 7.7
half of
a
348
Chapter
7
Two- and Three-Dimensional
finds that the flux lines and
in terms of
position x,
potential lines, u and v, respectively,
plane by the relation
h
=
select values of
plotting
out the
and
u
v
field sketched in
7.8
W
(aw/V0
u—
7T
can
Value Problems
are
implicitly defined
y in the Z
Z
One
Boundary
——
1
3V—
-
and calculate the
Fig.
+
0
jar)
(2)
corresponding points
x
and )2, thus
1.90.
CONFORMAL MAPPING FOR WAVE PROBLEMS
that in conformal transformations for statics
complicated boundaries are
simple ones. Conformal transformations of wave problems can similarly
simplify complicated boundaries.8 The transformations can be made only in two dimensions so the fields must be independent of the third dimension. The simplification
of the boundaries is also normally accompanied by increased complexity of the dielectric so this trade-off only occasionally helps.
u + jv
Let us assume that there exists an analytic function W
f(Z)
f(x + jy) which transforms the given boundary shapes in the Z plane to lines of constant
it and v in the W plane. We will first determine the relation between
V391! and Vivilr in
order to transform the scalar Hehnholtz equation, Eq. 7.2(8),
We have
seen
transformed to
=
<92!!!
+
6x2
from the Z
plane
to the
W
plane.
624/
ayg
+
9
Int:
To transform the
~
=
0
=
(1)
derivatives,
we
apply
the chain rule.
First,
n_na+na
(2)
_
an ax
fix
Applying
the chain rule
2
an
—§
=
similarly
for
_
second time leads to
2
2
6x
and
a
av 61
an
—2Bu
2
9
an
+
“-
6):
an
—2
+ 2
60
fix
62
av
dl/Idudv
—
""'
w
an 80 6x Bx
(3)
azu/ayzz
7
7
"
82
95’
8)"
3
2
an
““—
62
=
‘5’
6w
'
a
it
ay
F. E. Borgnis and C. H. Papas,
Springer Var/cg, Berlin, 7958.
+
‘5’
60‘
—
By
in Hondbuch der
+ 2
a2 l” —“
6 av
6y 6y
--
an 60
Physik (3. Fidgge, Ed),
<4)
Vol. 76, p. 358,
7.8
Making
terms
use
of the
of the
right
Conformal
conditions in
Cauchy—Riemann
and 7.4(7) in the second
(4), and adding (3) and (4),
Eqs. 7.4(6)
sides of
(3) and (4) and in the last
a?-
33
term
of
2
45'
8.1"
Note from
Eq. 7.4(4)
+
and
—'By1,D
31’-
‘5'
av“
2
(1W
(12
+
2
au
+
~-
ax
a
—"
8y
(5)
7.4(7) that
2
—
az
45'
611-
=
349
Wave Problems
Mapping for
2
au
=
2
00
+
-—
6x
2
all
=
-—
6x
an
—
+
6x
—-
6y
(6)
Thus
2
dW
v2,
Avil/j
-.=
v2(It)!!!
—
[[2
7
()
quantity le/dWI is a scale factor which relates a differential length lde in the W
plane to the corresponding length lel in the Z plane, as we discussed in Sec. 7.6. The
Helmholtz equation (1) is thus transformed to the W plane giving
The
.—
Visit
Euzo
+
@
general, dZ/dWl is a function of the coordinates so that (8) is equivalent
equation in an inhomogeneous medium.
Boundary conditions in the Z plane consisting of zero values of (II or its normal
derivatives carry over unchanged to the corresponding boundaries in the W plane since
the orthogonality of coordinates is conserved. If a nonzero normal derivative is specified
on a boundary, the scale factor IdW/dZI enters the conversion of the boundary condition
through the relation between gradients in the two planes:
Note that, in
to the Helmholtz
|V..¢|
=
g2:
a’W
Example
lwl
<9)
7.8
CURVED DIELECTRIC WAVEGLHDE
layer of a dielectric material embedded in materials of lower permittivity can serve
guide electromagnetic waves, as will be studied in more detail in Chapter 14. The
phenomenon of total internal reflection analyzed in Sec. 6.12 supplies a qualitative
understanding of dielectric waveguides. Here we see how wave propagation in a curved
layer, as in Fig. 7.8a, can be treated using conformal mapping.
The wave is assumed to be polarized with its electric field in the z direction and E:
is independent of 2:. Then identifying E: with ill, we can write from Eq. 7.2(8), with
A
to
6/6;
=
,
Vi),E:(x,
y)
+
k2E3(x, y)
=
O
(10)
350
Chapter
7
Two- and Three-Dimensional
Boundary Value Problems
53
u=Roln
Ru
“"
(b)
FIG. 7.8
in Z
(a) Curved dielecnic waveguide
into W
plane.
Using
the transformation of Ex. 7 .6b in
W
plane. (1))
slightly
guide transformed
different form,
R20
(11)
ell/R0
(12)
In
R0
3
Curved dielectric
for which
%
=
(8) becomes
Vfiszm,
From
(11) it is easily
seen
Hall/Roam,
+
v)
v)
a
0
(13)
that
1n
u
=
R0
v
=
R06
F2:
(14)
and
Therefore, the edge of the guiding layer
(15)
is the
u
0 line in the W plane and
R0 ln(R0
a)/R0. The result is
that the curved layer in the Z plane becomes the planar region shown in Fig. 7.817. Note
that this layer in the W plane is inhomogeneous. Equation (13) has the usual form of
the Helmholtz equation only if we identify a new wave number k’ by
the other
edge
of the
guide
at
r
=
k’
R0
=
at
—
a)
r
a
/
=
R0
is at
u
=
=
—
2
he exp
R__u
o
(16)
7.9
Since k’ is
a
Laplace’s Equation
function of it,
one cannot
In
351
Rectangular Coordinates
substitute it
directly
for k in the usual
solution. However, the problem is solvable and has been used to
energy from bends in dielectric waveguides.9
study
the
wave
leakage
of
Separation of Variables Method
7.9
LAPLACE'S EQUATION IN RECTANGULAR COORDINATES
One of the most
powerful techniques for solution of linear partial differential equations
separation of variables. This leads to solutions which are products of three
functions (for three—dimensional problems), each function depending upon one coordinate variable only. Such solutions might not seem very general, but they may be
added to form a series which can represent very general functions. Moreover, singleproduct solutions of the wave equation represent modes which can propagate individ~
ually. These are of great practical importance in waveguides and resonant systems and
are studied extensively in following chapters.
As the simplest example of the method of separation of variables, let us first consider
two~dimensional problems in the rectangular coordinates x and y, as we have in the
transformation method of the past section. Laplace’s equation in these coordinates is
is that of
2
2
a?+a?=0
6);“
6y“
We wish to
study product solutions
of the form
(1903 y)
where
point
we see
on
that
we
have
a
(u
=
(2)
X(x)Y(y)
function of x alone times
a
function of y alone. From this
X(x) will be replaced by X and Y( y) by Y. Substituting in (1),
X"Y + XY"
The double
=
we
0
have
(3)
prime denotes the second derivative with respect to the independent variable
sum of functions of one variable only, divide
in the function. Now to separate into the
(3) by (2):
Y”
XII
-—
X
9
+
—-
=
O
Y
M. Herb/um and J. H. Harris, lEEE J. Quantum Electronics 85-] 1, 75 (7975).
(4 )
352
Chapter
Next follows the
Boundary Value
Two- and Three~Dimensional
7
key argument
for this method.
Equation (4)
Problems
is to hold for all values
and y. Since the second term does not contain x, and so cannot vary
with x, the first term cannot vary with x either. A function of x alone which does not
vary with x is a constant. Similarly, the second term must be a constant. Let us denote
Then
the first as K3 and the second as
of the variables
x
Kg.
iii
13%
-+
==
0
(5)
and
X"~sfiX=()
(6)
2
I
T—Kfl=0
recognize that these are in the standard form having real exponentials or hyperbolic
as solutions. Let us write them in hyperbolic form and substitute in (2):
We
functions
<I>(x, y)
It is clear from
Ky
the
must be
same.
(A cosh Kxx
=
(5) that either K i
imaginary
Thus
+ B sinh
(7)
=
@(x, y)
=
K
cosh
+ D
Kyy
sinh
(7)
Kyy)
39‘ must be negative and therefore either Kx or
while the other is real. Furthermore, their
must be
magnitudes
have either of two forms,
can
(13(x, y)
or
KxxXC
+ B sinh
(A cosh Kx
Kx)(C
cos
Ky
+ D’ sin
Ky)
(8)
+ D sinh
Ky)
(9)
or
where, since
{le
=
(A
[Ky
,
cos
Kx + B’ sin
Kx)(C cosh Ky
we
have used the
single symbol
indicate that the constants have
K. The
The choice between
primes
and
are
used to
is dictated
(8)
(9)
changed.
by
boundary conditions. If the potential is required to have repeated zeros
as a function of y, then (8) is used; if repeated zeros are specified for the x variation,
(9) is chosen. If the boundaries extend to infinity in one direction, real exponentials are
used in place of hyperbolic functions. It may be noted from (6) that for Kr
jKy O,
the general solution has the form
the nature of the
=
(Many)
It is
typical
for
product
+
(Alx
:
31)(C1y
solutions that when the
functional forms of the solutions
We will
+
D1)
using
(10)
separation constants go to
see in subsequent sections
change.
boundary conditions.
For the three-dimensional case in rectangular coordinates,
extended. Laplace’s equation is
constants are evaluated
=
zero
the
how the
the
yo
axz
yo
.+-
procedure is simply
yo
0
(11)
X(X)Y()’)Z(Z)
(12)
-+
ayz
the
==
7%}?
Consider solutions of the form
@(x,
y,
Z)
=
Static Field Described
7.10
where each term
variables.
on
the
right side is
Substituting (12)
in (l
1),
by
just
dividing by (I),
we see
one
of the
independent
space
have
X"YZ + XY”Z + XYZ”
and
353
Single Rectangular Harmonic
function of
a
we
a
:
0
that
XII+YII+ZNmO
X
z
Y
( 1 3)
—
We
the
use
argument
same
terms do not vary with x,
used in the two-dimensional
as was
neither
the first. Since it is
can
not vary with x, it must be a constant. Similar
terms.
If
we
let the first term he
a
case.
arguments apply for the second and third
K3, and the third K2,
K§+K§+K°gz0
K3
If the second two
function of x alone and does
the second
(13) becomes
(14)
and differential
written
as
the
(13(x,
equations of the form (6) apply for X, Y, and Z. So the general solution,
product of X, Y, and Z, and sometimes called a rectangular harmonic, is
y,
z)
+ B
[A cosh Krr
=
sinh
er] [C
cosh
Kyy
+ D sinh
Kyy]
,
X
It is clear that at least
least
one
in the
x
of
XX, Ky,
and
+ F smh
[E cosh K52
one
K:
of K7,“, K
must be
i, and K
2
imaginary.
and y directions, the functions of
x
K32]
(15)
be negative for (14) to hold, so at
repeated potential zeros are required
y must be trigonometric functions so
must
If
and
K" are imaginary. There are various other combinations which may be useful.
In some cases it is advantageous to replace the hyperbolic functions by real exponentials
K,
as
and
mentioned earlier for the two-dimensional solutions.
In (15) there appear to be nine constants, to be evaluated using the six possible
boundary conditions, two for each of the three coordinate directions. If, however, one
divides the first bracket
entire
by BDF,
constants. From
so
Let
us see
(14)
we see
that
STATIC FIELD DESCRIBED BY
by D,
A
are
and the third
just
four
SINGLE RECTANGULAR HARMONIC
required to have one of the forms of Sec. 7.9 as
O. The product of
O, C
Eq. 7.9(9) with A
denoted as a single constant C1:
what boundaries would be
solution. Take the
remaining
by F and multiplies the
independent multiplicative
there are only two independent separation constants
equals the number of boundary conditions.
the second
the total number of unknowns
7.10
a
by B,
it becomes clear that there
constants,
special
B’D,
case
may be
q)
of
=
=
CI
sin Kx sinh
Ky
=
(l)
354
Chapter
7
Two- and Three-Dimensional
Boundary Value Problems
O for all x. Hence one boundary can
Similarly, potential is zero along the
rm. Let us confine
O and also at other parallel planes defined by Kx
plane x
attention to the region 0 < Kt < 77 and O < y < 00. The intersecting zero~potential
planes of interest then form a rectangular conducting trough. Let its depth in the x
It is evident from (1) that potential is zero at y
0.
be a zero-potential conducting plane at y
=
=
=
=
direction be
a.
Then Ka
=
'n' or
77‘
K
(2)
=
a
If there is to be
other than
zero.
a
finite field in the
Without
at
which it
as
V0. Then, from (1),
crosses
the
knowing
midplane x
region, there must be some electrode at a potential
shape for the moment, let us take the value of y
a / 2 as y
b, and the potential of the electrode
its
=
=
or,
substituting
in
(1),
we
sm
2
77])
Sinh—
77b
77
'-
Vo :C 1
s inh—==C 1
a
a
have
(I)
_
V0 sinhm/a)
“-
.
sm
1135
sinh(77b/a)
(3)
(1
The potential at any point x, y may be computed from (3). In
that the electrode at potential V0 must take can be found from (3)
particular, the
by setting CI)
form
=
V0,
yielding
sinh
EX
a
:
Sinh(7Tb/a)
sm(mc/a)
(4)
Equation (4) can be plotted to show the form of the electrode. This is done for a value
b/a
é-in Fig. 7.100. Actually, the electrodes should extend to infinity, but if they
are extended a large but finite distance, the solution studied here will represent the
potential very well everywhere except near edges.
19 could be supplied with a sinusoidal
Alternatively, a straight boundary at y
==
=
“*7
U
FIG. 7.10a
Electrodes and
3
potentials for which
a
single
harmonic is the
complete
solution.
Fourier Series and
7.1 1
3
FIG. 7.10b
one
Potential in
a
355
Integral
b
two-dimensional box with
sinusoidal distribution of
a
potential
on
side.
distribution of potential and
in the box. Thus, for
a single harmonic would describe the potential at all points
372* and the boundary potential were
example, if [(0
=
3
and, b)
=
VO
sin
~53
(5)
x
then the harmonic
ct)
shown in
Fig.
V0
sinh(377/a)y
sinh(37rb/a)
=
7. 10b satisfies the
boundary
.
sm
377
—-—
(6 )
-
A
0
conditions and describes the
potential
at
all
points.
7.1 l
FOURIER SERIES AND INTEGRAL
In the
preceding section, we saw that a single-product solution could satisfy only very
special forms of boundary conditions. For more general boundaries a sum of such
solutions must be used. This is one example of situations where Fourier series or integrals are useful in forming solutions for field problems. We provide here a review of
the Fourier tools with the assumption that the reader has already a measure of familiarity
with them.
Fourier Series
dependent
Fourier series
variable x, the
required
used to represent periodic functions. For the inperiodicity is expressed by
are
f(X)
=
f(x
+
L)
(1)
356
Chapter
Two» and Three-Dimensional
7
Boundary Value
Problems
period of the function. We assume that the function can be represented
plus the sum of infinite series of sine and cosine functions of harmonics
fundamental spatial frequency k:
where L is the
by
of
a
a
constant
f(x)
=
620 + a1
cos
kx +
(12C082kx
+ a3 cos 3kx +
2
H
+blsinkx+bzsin2kx+b3sin3kx+..
where the
phase
factor k is related to the
To evaluate the unknown constants in
so~called
orthogonality preperties
period L
(2) for
a
in the usual way:
given
of sinusoids. These
function f(x),
we
make
use
of
are
L/2
j
mkx dx
=
O
m
#
n
(4)
sin 22kt sin mloc dx
=
O
m
#
n
(5)
cos
nkx
cos
-—L/2
L/2
J,
~L/2
and
U2
.
8111
mkx
cos
1210c dx
=
m
#
m
=
n
O
(6)
—L/2
12
However,
L/2
f
L/2
f
cos2 mlcr dx=
-—L/2
sin2 m/Ct' dx
L
=
(7)
-—
2
-~L/2
use of these properties, we multiply each term in (2) by cos Izkx and integrate
period. Every term on the right vanishes because of the properties in (4)—(6)
except the one containing cos nkx; that term gives aflL/ 2 according to (7). Thus,
T0 make
over one
2
an
=
-
L
L/2
f
f(x)
cos
n/cx dx
(8)
mL/Z
Similarly, multiplication of (2) by sin nkx with integration from L / 2 to L / 2 leaves
only the term involving sin nkx on the right-hand side, and its coefficient, by (7), is
__
2
b”
Finally,
to
=
L/2
f(x) sin Izkx dx
~
L
(9)
—L/?.
obtain the constant term (10, every term is integrated directly over
on the right side disappear except that containing
00 so that
a
period
and all the terms
L/2
(10:—
L
This
merely
states that a0 is the average
I
fix) dx
—L /2
of the function f(x).
(10)
7.11
Fourier Series and
357
Integral
For a general function, an infinite number of terms is required in the Fourier series
representation. But often a sufficient degree of approximation to the desired wave shape
is obtained when only a finite number of terms is used. For functions with sharp discontinuities, however, many tenns may be required near the sharp corners, and the
theory of Fourier series shows that the series does not converge to the function in the
neighborhood of the discontinuity (Gibbs phenomenon). The derivative of the series
also does not converge to the derivative of the function, but the integral of the series
always converges to that of the function.
Example
FOURIER SERIES REPRESENTATION OF
A
7.1 in
FUNCTION OVER
A
FINITE INTERVAL
In static field
problems, one commonly has the boundary potential specified over a finite
as along a straight boundary at a constant value of one coordinate. For
the purposes of matching the given boundary potential, it is desirable to express it in a
Fourier series. This can be done even though the function is not periodic, having been
specified only over a finite interval. The point of view is that the interval of length a
may be considered a period or an integral fraction of a period, and a periodic function
defined to agree with the given function over the given interval, repeating itself outside
that interval. A Fourier series may then be written for this periodic function which will
give desired values in the interval, and although it also gives values outside the interval,
that is of no consequence since the original function is not defined there.
The interval is commonly selected as a half-period since the function extended outside the interval may then be made either even or odd, and the corresponding Fourier
series will then have respectively either cosine terms alone or sine terms alone. Figure
7.11a shows by solid. lines some possible examples of functions specified over the
interval, such
interval 0
are
< x < a.
shown
by
Their extensions outside that interval
the broken lines. Note that in
one case
as
either odd
the interval is
or even
L/ 4.
functions
The choice of
as an odd or an even function depends upon
the form used to represent the potential in the problem. Thus, for example, in Eq. 7.lO(3)
the potential is expressed in terms of sin 77x / a and the appropriate series for the rep—
resentation of the boundary potential will be in Sines,
whether to consider the function continued
f(x)
mrx
2
=
b"
sin
where
are
we
have made
found from
positive
use
(9) noting
intervals
are
of the fact that
a
x
(11)
Cl
12:!
L/ 2,
equal. Thus,
with
2
b"
2
—
a
a
2
half~period. The coefficients
integral from the negative and
one
that the contributions to the
L/2,
(I
f
O
J.
3—7—73
dx
fix) sin
a
(12)
358
7
Chapter
Two- and Three-Dimensional
Boundary Value Problems
fix)
fix)
I
,._J
1I
I
g
F
I
+¥>x
|
I
l
I
I;
L
l
L>x
a
I
I
(i)
(iii)
17x)
fix}
(ii)
(iv)
Examples of functions specified
(broken lines).
FIG. 7.11a
I
over a
finite interval
(solid line) and odd and
even
continuations
Suppose,
< a
for
example,
that the
and it is desired to
expand
2
b
specified
it in
a
f(x) =C over the interval
series. Then, (12) yields
function is
sine
0 <
x
a
a
20
Csinfldx=-[—-cos§fl]
=—
(13)
a
and the series is
2C
f(x)=qr
The series
2
77x
3m:
2
577x
231n—-+-srn—+—sm—+--3
a
a
5
a
_
,
(14) has the required value f(x)
represents the dashed
portion
of waveform
.
=
C
(i)
in
over
Fig.
the interval 0 <
(14)
x < a
but also
7.110 outside that interval.
integral In some problems the function of interest is defined over the entire
aperiodic. An example is a square function that is constant in some range
—-a S x S a and zero elsewhere, as shown in Fig. 7.1119. This could be considered the
limiting case of a periodic series of square pulses where the period L goes to infinity.
12k
The spacing of the components (12 + 1)k
277/L from (3) becomes vanishingly
small as the period L approaches infinity, and in the limit the spectrum of component
sinusoidal waves becomes a continuum.10
Fourier
range and is
-—
10
R. Bracewe/I, The Fourier Transform and its
York, 7986.
=
Applications, 2nd rev. ed, McGraw—Hill, New
7.1 1
Fourier Series and
359
Integral
fix)
l
C
l glk)
2C0
7—
r
311'
0
~27rV—7r
WVZTE’
(c')
37r\
FIG. 7.11
(b) Inverse Fourier transform of function in (c). (c) Fourier transform of
function in (1)). Value of g(O)/2Ca is unity.
In the
limiting, aperiodic
case
the series
(2) is replaced by
an
rectangular
integral
°°
1
fOC)
and the function
g(k)
=
in
theory
£0061”
place
of a” and
277-
which takes the
800
The
f
*-
=
f
.
b”
of
(8)—(lO) is given
by“
f(x)e ”’7‘" dx
(16)
integrals shows that for (15) to give the same f(x) that appears
be continuous or have only a finite number of finite disconfinite interval and must be absolutely integrable, that is,
must
L lax)!
The
(15)
of Fourier
(16) the function
tinuities in any
n
dk
«~00
placement of 297 in the pair ( 75)
the literature.
dx <
and ( 76) is
0°
arbitrary and is done in various
(17)
ways in
369
Chapter
Two- and Three-Dimensional
7
Boundary Value Problems
place strong limitations on
These conditions do not
the
utility of the transform pair (1551‘s.
‘
and
(16).
'
U
FOURIER TRANSFORM OF
Using (16)
we
function in
Fig.
find the spectrum of
7.11b:
A
g(k)
RECTANGULAR PULSE
spatial frequency components
for the
rectangular
a
a
‘
Nj/O:
f
=
-.'t1L:-.'.'.'.-
7.1 lb
Example
‘
Ce—Jadx
=
k
4‘:ij
=
2043”]; a)
(18)
important function occurs frequently in practice and is shown in Fig. 7.11c.
f(x) in Fig. 7.11b is called the inverse Fourier transform of that in Fig.
The two are called a transform pair.
This very
The function
7.110.
SERIES OF RECTANGULAR HARMONICS FOR TWO
AND THREE-DIMENSIONAL STATIC FIELDS
7.12
We
in Sec. 7.10 that
saw
ditions
on
product
solutions
(harmonics) satisfying zero boundary con—
two~dimensional rectangular structure can
harmonic
as the expression for the potential
single
three of the four boundaries in
be found. However, the
use
of
a
a
boundary of complicated shape or a simple flat one with a
along it. To solve problems with an arbitrary variation
of potential along a flat boundary on a coordinate line, one may use a sum of harmonics,
each of which satisfies the zero conditions on three boundaries and has a weighting in
the sum such that it equals the given potential at the fourth boundary. Then the given
potential is expanded in a Fourier series of either sines or cosines, chosen to match the
either
requires
a
fourth
sinusoidal variation of potential
functions in the
boundary
and
sum
of harmonics. The harmonic series is evaluated at the fourth
by term, with the Fourier series to evaluate the weighting
procedures are sometimes slightly modified by use of
as
seen in the following examples and problems.
superposition,
compared,
term
coefficients in the former. These
symmetries
and
-.~-'v.--TN .22.???“
.
“iii/‘gygq,
\_
.--,.:<.-:.-
--
.,_
-
-.-.--_
-,-.
,
........
Example
7.1%
TWO-DIMENSIONAL PROBLEM WITH SPECIFIED BOUNDARY POTENTIALS
AS
an
example of
a
problem
solutions of Sec. 7.9, but
two-dimensional
region
can
of
which cannot be solved
be
Fig.
by
means
of
a
7.12a bounded
by using
a
single
one
of the
series of these
by
a
solutions, consider the
O, a
zero~potentia1 plane at y
=
Series of
7.12
36'!
Rectangular Harmonics
(1)::0
we
FIG. 7.120
JL_
b
r‘
Two—dimensional box for Ex. 7.12a.
0, a parallel zero—potential plane at x
a, and a plane
zero-potential plane at x
b. In the ideal problem, the lid is separated from
conducting lid of potential V0 at y
the remainder of the rectangular box by infinitesimal gaps. In a practical problem, it
would only be expected that these gaps should be small compared with the rest of the
=
=
=
box.
In
selecting the proper forms from Sec. 7.9, we will choose the form having sinusoidal
it since potential is zero at x
O and also at x
a, and sinusoids have
0 for
repeated zeros. So the form of Eq 7 9(9) is suitable Moreover, (I): 0 aty
all x of interest, so the function of y must go to zero at y= 0, showing that C
0.
O at x
0. Then CI) is again zero at
O for all y of interest, A
Similarly, since (I)
solutions in
=
=
==
=
=
x
2:
a, so Ka
==
=
=2
mar, or
nm
Kr:
Cl
Denoting
the
of the
product
remaining
constants
B’D
mm
(I)
=
Cm
sin
as
C[71’
we
have
mrry
1nh
sin
a
O, at
Laplace equation and the boundary conditions at y
x
O, and atx
a, but a single term of this form cannot satisfy the boundary condition
b, as the study in Sec 7.10 has shown. A series of such
along the plane lid at y
solutions also satisfies Laplace’ 3 equation and the boundary conditions at y: 0, at
This forms satisfies the
=
=
=
=
x
--
O, and
air 0:
~
2
m7rx
Cm
12le
For the
sum
(1)
the interval 0 <
x
to
give
the
< a, we
required
require
=2
m=l
Cm
sinh
sin
constant
sinm xsinh
fl
(1)
a
a
potential V0 along
the
plane y
=
I)
over
mn-b
——
,
a
O <
x < a
(2)
362
Chapter
But this is
recognized
the interval 0 <
7
Two- and Three-Dimensional
as a
x < a.
Boundary Value
Problems
expansion in Sines of the constant function VO
expansion was carried out in Ex. 7.11a to yield
Fourier
This
Zamsmp—zgi‘,
f(x)=VO=
O<x<a
over
(3)
:1
with
4V
am
Comparison
of
—9,
m
O,
m even
odd
”“7
:
(4)
(3) with (2) shows that
b
Cm
sinh
flf—
Substitution of the results of (5) and (4) in (1)
c1)
:
=
(5)
am
gives
———S@(mW/a) sin 572-773
2 fl
mqr
m
smh(m77b/a)
odd
(6)
a
rapidly convergent except at corners of x ——> O, a, and y ——> b, so it can
be used for reasonably convenient calculation of potential elsewhere.
We note that the evaluation of the constants in the general solution depended upon
the fact that the boundary potentials were specified on surfaces in the coordinate system.
Furthermore, nonzero conditions, potential or normal derivative of potential, must exist
on some part of the boundary to yield a nonzero solution. As will be clarified in the
next example, SUperposition may be used to solve problems where the boundary conThis series is
ditions involve several sides.
Example tub
TWO-DIMENSIONAL PROBLEM REQUIRING SUPERPOSITION
In this
example, we see a boundary potential having a Fourier series expansion which
trigonometric functions and a constant. Matching such a boundary condition requires the superposition of two solutions, one to match the constant and one
for the trigonometric functions. Consider the problem of finding the potentials in the
conducting rectangular solid of infinite extent in the z direction shown in Fig. 7.1%.
The surrounding region contains free space, the potential at y
O is zero, and that
the
b
is
(ID
along
edge y
given by
VOx/a.
This problem requires a solution with repetition in the x direction since the boundary
conditions at the sides x
O, a are the same; the appropriate general form is that in
At
a
x
x
the
O,
Eq. 7.9(9).
component of current density must be zero since no current
includes both
=
=
=
=
=
Series of
7.12
363
Rectangular Harmonics
=0
b
030
FIG. 7.12b
can
Two—dimensional conductive solid embedded in
flow in the free space outside the conductor. Since .1
zero at x
O,
=
nonconductive medium.
a
0E, then Ex
=
must also be
Therefore, 8613/6x, given by
a.
661)
must be zero at
sin Ka
y
=
0
=
requires
0,
=
x
0 if [(0
C
+ 8’ cos
K(~—A sin Kx
=
5;
==
=
Kx)(C cosh Ky
for all y. Since cos Kx
K
mar/a. To match the
1 at
=
a
m7r so
=
0. Thus the
(13,"
“2
cos
x
=
boundary
(7)
Ky)
O, B’
=
0. Also,
condition (I)
=
O at
in the mth harmonic is
potential
Cm
+ D sinh
_m_7r x
31—7:
sinh
a
(8)
y
a
I) should be expanded in a series of cosines
potential on the boundary at y
so that tenn-by-term matching with a series of terms like (8) can be done. The appropriate periodic continuation of the given boundary potential is shown in Fig. 7.11a(iv).
It is seen to have an average value of VO/2 which will be present in the series expansion.
Applying Eqs. 7.1 1(8)—~(10),
The
=
4V0
V
ct>(x,b)=—9—~
2
2
771
‘7
(771 77)“
add
cos-m—Wx
Matching the boundary potential (9) requires both a series
(8) and a separate solution having a constant value at y
form is found from Eq. 7.9( 10). The function
in
a
(131
satisfies the
as
boundary
VO/Zb gives
conditions at
x
=
of harmonics of the form
b. A solution of the latter
(10)
Aly
==
O,
(9)
a
a
and at y
=
0. Evaluation of the constant
V
(plzfl
2b
The solution
in
(11)
involving harmonic terms is found by equating a series
at the boundary y
b, with the series in (9):
(8), evaluated
of terms like that
=
Z
m
mrrb
mvrx
Cm
sinh
cos
a
=
a
——
Z
m
odd
4V
07
(27277)"
marx
cos
——
a
(12)
364
chapter
Two- and Three-Dimensional
7
The result for the second
is
potential
4VO sinh(mqry/a)
("277)2 Sinh(m7rb/a)
@93—
171
The
complete
solution is the
Boundary Value Problems
odd
superposition
(13)
a
(11) and (13), (I)
of
Example
mm:
(I),
2
+
(132.
Lisle
THREE-DIMENSIONAL RECTANGULAR Box WITH POTENTIAL SPECIFIED ON ONE FACE
The method discussed above
consider
a
box with
be extended to three dimensions. As
can
potential
the origin
zero
The box extends from
all faces except
on
of coordinates to
on
x
the side
=
2
=
a, y
==
example,
we
6, where it is
V0.
an
b, and
z
=
c.
The
appropriate general form of the space harmonic is Eq. 7.9(15) with K, and Ky imaginary
and K3 real. To meet the zero-potential boundary conditions at x
O, y
O, and
z
O, the constants A, C, and E must be zero. Also, to satisfy the zero-potential
condition at x
a and y
b, the corresponding separation constants must be 1727/ a
and mr/ b, respectively. Then from Eq. 7.9(14) we get K:
[(flZ’JT/CZ)?‘ + (n77/b)2]1/2.
The general form of the potential must be a doubly infinite sum of the resulting
=
=
=
=
=
=
functions:
(I)
22
=
If
CW,
sin
Ex
(14) is evaluated
at the
sin
boundary
V(x, y)
2
=
22,
=
?y sinh\/(m—W>
+
a
a
In
I?
n
c, the
(14)
2
series becomes
mn-
,
BM
(L21)
Sln
——
1277
,
x 3m
b
a
m
y
(15)
c
(16)
-—
where
2
2
D
=
CW,
31th ("2—”) (335)
+
b
a
The coefficients
Dmn can be evaluated by multiplying (15) by sin(p77x/a) sin(qary/b)
integrating over it from O to a and over y from O to b. Application of the orthogonality conditions Eqs. 7.11(5) and 7.11(7) yields
and
4
Dnm
For the
special
c1)
-—
—
22
'7
where
m
and
case
m
n are
of
=
—
ab
b
a
If
V(x, y)
0
=
V(x, y) sin
Ex
o
V0, (14), (16),
a
and
dx dy
fly
b
odd in the summation.
(17)
(17) give
SI'Inlfi/(ntm/a)2 + our/b)2
sinh\/(nzn‘/cz)2 + (WIT/b)2 c
16V0 Si“(”1”T/0)x SmOW/bb’
””2772
sin
z
(18)
7.13
7. l 3
In
large
a
CYLINDRICAL HARMONICS
class of problems of major
with boundaries
365
Cylindrical Harmonics for Static Fields
interest, the field distribution is desired for regions
the surfaces of
lying along
STATIC FIELDS
FOR
a
cylindrical coordinate system. Examples
the familiar electrostatic electron lenses found in many cathode—ray tubes or certain
coaxial transmission line problems for which static solutions are useful. As has been
are
pointed out in Sec. 7.12, the ability to evaluate the constants in product solutions depends upon having boundaries on coordinate surfaces. Therefore, the fields for this type
of problem are found by separating variables in cylindrical coordinates.
A variety of types of solution are found, depending upon symmetries assumed. In
general, Laplace’s equation in cylindrical coordinates has the form
132d)
,ar +1384)?”
la(_acb)
az©_0
+
Axiai
zero
Symmetry
variations with
with
qb
Longitudinal invariance
Longitudinai invariance
It
was
(13(1‘, qb)
2
In Ex. 1.8a
we saw
that for
r
+
(2)
C2
shown in Prob. 7.9a that the solutions for
zero 2
given by
are
+
(C1)“
=
in
C1
a
variation, called circular harmonics,
n
I
at
and z,
(130-)
Note that, for
1
()
_
"2
rar
Czl'””)(C3
cos
2qu
+
C4
sin
(3)
22¢)
O, axial symmetry exists but (3) breaks down and the solution is
given by (2).
Axial
Symmetry Since
equation (1) becomes
it is assumed that there
32:13
_
61‘“
To solve this
equation,
let
us
try
to
in the differential
(I), Laplace’s
=
9
0
(4)
62‘
find solutions of the
C1303 2)
Substituting
+
61'
r
variations with
63(1)
1 6(1)
...
9
are no
form
(5)
R(I‘)Z(2)
=2
equation (4),
product
we
have
1
R"Z +
*-
R'Z + RZ"
=
O
r
where R” denotes
by dividing by
dzR/drz, Z"
denotes
(I’ZZ/alz2
so on.
The variables
are
separated
_
Z
1 RI
R”
Z"
By
and
RZ:
:
_.
._
R
the standard argument for the method of
+
_...__
r
R
separation of variables,
the left side, which
366
is
a
Chapter
7
Two- and Three-Dimensional
function of 2 alone, and the
right side,
to each other for all values of the
constant. Let this constant be
Boundary Value
which is
variables
r
and
2.
a
function of
r
Problems
alone,
must be
Both sides must then be
equal
equal to a
T2. Two ordinary differential equations then result
as
follows:
1d2
1dR
__+_
R r2
1R
:---:Z"2
6
()
—T2
7
()
dr
IdZZ
-—
Zdzg
Equation (6)
be written
can
as
dZR
ldR
——+——-+T2~R
air2
7dr
-—
Equation (8)
8
()
simplest form of the so~called Bessel equation. A sketch of the
given. One method of finding a solution is to substitute a series and
is the
solution will be
find the conditions
equation.
O
on
the terms of the series for it to be
Thus to solve
R
=
Clo + alr +
a
valid solution of the differential
must be assumed to be a power series in r:
(8), the function R
azr2
+
(131-3
+
-
-
01‘
20 aprP
=
p
(9)
=
Substitution of this function in (8) shows that it is
a
solution if the constants
are as
follows:
0p
(C 1 is
any
arbitrary constant).
°°
“51.2.
is
2
a2,”
:2:
C1(_'1)n1
(T/2)2’"
(m!)2
That is,
2
(-1)m(T;-/2)2m
(sz
(Tr /2)4
Tr
.
=CI[1"<E)+W"“]
‘10)
solution to the differential equation (8). It is easy to show that (10) is convergent
that values may be calculated for any value of the argument (Tr). Such calculations
have been made over a wide range of the values of the argument and the results are
tabulated.
a
so
If
T2 is positive, the function defined by the series is denoted by JO(Tr) and called
zero order. This function is defined by
a
Bessel function of the first kind and of
2
100;)
The
particular
A
=
v
1
—
(5)
+
02/2)“
(202
——
solution (10) may then be written
R
=
<——
_1_>"'(v/2>2'"
3.120 —(—m!)2
.
simply
C1J0(Tr)
as
(11)
7.15
367
Harmonics for Static Fields
Cylindrical
The differential
equation (8) is of second order and so must have a second solution
arbitrary constant. (The sine and cosine constitute the two solutions for
the simple harmonic motion equation.) This solution cannot be obtained by the power»
series method outlined above, since a general study of differential equations would
show that at least one of the two independent solutions of (8) must have a singularity
at r
0. Once one solution is found there is a technique for obtaining a linearly
independent solution for this class of equations12 and several different forms are possible. One form for the second solution (any of which may be called Bessel functions
of second kind, order zero) easily found in tables is
with
a
second
=
2
<yv>i
~;nz=l~W—(1+—2—+§+m+;)
_
7T
2
The constant ln 3/
0.5772
=
.
.
C IJO(Tr) +
=
C3
=
(12)
1
general, then,
(13)
C2N0(Tr)
(8), and the corresponding solution
Z(z)
1
1
is Euler’s constant. In
.
R
is the solution to
1)m(u/2)°"'
(
sinh Tz +
C4
(7) is
to
cosh T2
(14)
It should be noted from (12) that NO(Tr), the second solution to R, becomes infinite at
r
0 is included in the region
0, so it cannot be present in any problem for which 2'
z
=
over
If
a
which the solution
T2 is negative,
solution and T in
(10)
imaginaries disappear,
applies.
let
we
T2
=
may be
and
1'2
jr, where r is real. The series (10) is still
replaced by jr. Since all powers of the series are even,
—-
a new
or
T
=
series is obtained which is also real and convergent.
That is,
2
100”)
3
(2g)
1 +
4
+(12/2) +(12/2)
(21)2
6
(15)
+
(3!)2
Values of J0(jv) may be calculated for various values of u from such a series; these
also tabulated in the references and are usually denoted 10(v). Thus, one solution to
with T
=
jr
(8)
is
R
There must also be
general solution
to
a
=
(8)
The second solution
(3110017) 9 CHOW)
(16)
commonly denoted K007),
7'2 may be written
second solution which is
with T2
=
R
‘2
are
K0 becomes
=
—
Ci10(7'7')
infinite at
r
+
m
so
(17)
C§K0('n')
0 just
as
does
that the
N0,
and
so
is not
required
See, for example, E. T. Whittaker and G. N. Watson, A Course in Modern Analysis, p. 369,
University Press, Cambridge, 7927.
4th ed,
368
Chapter
Two- and Three-Dimensional
7
problems which include the axis r
apply. The solution to the 2 equation (7)
in
=
Z
sin
Cg
=
Summarizing, either of the following
cylindrical coordinates r and z:
As
was
the
O in the range
when
T2
+
72
[C1J0(Tr)
+
C2N0(Tr)][C3
(130‘, z)
=
[C{10(7r)
+
C§K0(rr)][C§
rectangular harmonics,
over
Problems
which the solution is to
7‘2 is
-—
(18)
cos 7'2
forms satisfies
=
with the
=
C1,
(130‘, z)
case
Boundary Value
Laplace’s equation
sinh Tz +
sin
C4
cosh
T2]
in the two
(19)
C,
cos
Its]
(20)
the two forms
are
not
really different
72
+
since (19) includes (20) if T is allowed to become imaginary, but the two separate ways
of writing the solution are useful, as will be demonstrated in later examples. The case
with
no
assumed
symmetries
is discussed in the
7. l 4
following
section.
BESSEL FUNCTIONS
an example of a Bessel function was shown as a solution of the differential
equation 7.13(8) which describes the radial variations in Laplace’s equation for axially
symmetric fields where a product solution is assumed. This is just one of a whole family
of functions which are solutions of the general Bessel differential equation.
In Sec. 7.13
Bessel Functions with Real
For certain
problems, as, for example,
longitudinally split cylinder, it may
be necessary to retain the qb variations in the equation. The solution may be assumed
in product form again,
RF¢Z, where R is a function of 1' alone, Fq, of qb alone, and Z of
2 alone, Z has solutions in
hyperbolic functions as before, and F¢ may also be satisfied
by sinusoids:
Arguments
the solution for field between the two halves of
Z
a
C cosh Tz + D sinh T2
:—
(2)
F¢=Ecosv¢+Fsinv¢
The differential
equation
for R is then
equation
previously:
slightly
(1)
different from the
zero-order Bessel
obtained
1 (1R
d2]?
,+——-—+
dr~
dr
2‘
2
T-—-P7R=O
,,
7'"
(3)
It is apparent at once that Eq. 7 .13(8) is a special case of this more general equation,
that is, v
O. A series solution to the general equation carried through as in Sec. 7.13
=
shows that the function defined
by
the series
°°
mm
is
a
solution to the
(_ 1)m(Tr/2)v+2m
2:3, mlf(v
+
m
+
1)
(4)
equation.
l) is the gamma function of (v + m + l) and, for v integral, is equivalent
the factorial of (v + m). Also for v nonintegral, values of this gamma function are
F(v
to
=
+
m
+
1.0fl
369
Bessel Functions
7.14
l
J00!)
Jllv)
0.5—-
J20!)
l
0
l
u
2
|
4
l
n
6
8
l
I
.L
M
W12
-0.5L
(a)
l
l.O--
Now}
0.5
-1.0
(b)
FIG. 7.14
tabulated. If
(a) Bessel functions of the first kind. (b) Bessel functions of the second kind.
v
is
an
integer
n,
Jn(Tr)
:
(___1)m(Tr/2)n
E
m==0
It
can
Similarly,
COS
If v is
or
N”
for v
m)!
=
Nv(Tr)
‘3
+
(- 1)”J,,. A few of these functions are plotted in Fig.
thatJ_,,
second independent solution13 to the equation is
be shown
a
271102
-m
nonintegral, J..,, is
=
vwfv(Tr)
sin
-
J _v(Tr)
(5)
7.14a.
(6)
v'lr
linearly related to JV, and it is then proper to use either J-”
integral, NV must be used. Equation (6) is indeterminate
integral but is subject to evaluation by usual methods.
as
not
the second solution; forv
373
Chapter
and N _,,
=
complete
solution
Two- and Three-Dimensional
7
As may be noted in
(3) may be written
(*1)"N,,.
to
R
The constant
is known
1:
of first kind, order 1);
for this
chapter
Nu
are cases
It is useful to
keep
in
number of radians of the
as
is
the order of the
a
Fig. 7.14b
+
A1,,(Tr)
=
Boundary Value Problems
these
are
infinite at the
equation. 1,,
is then called
1).
a
Bessel function
Of most interest
integer.
mind that, in the physical problem considered here, 12
sinusoidal variation of the potential per radian of angle
=
1)
A
(7)
BNv(Tr)
Bessel function of second kind, order
in which
origin.
22, an
is the
about
the axis.
Jv(v) and NV (1)) are tabulated in the references.14'15 Some care should
using these references, for there is a wide variation in notation for the
second solution, and not all the functions used are equivalent, since they differ in the
values of arbitrary constants selected for the series. The Nv(v) is chosen here because
it is the form most common in current mathematical physics and also the form most
commonly tabulated. Of course, it is quite proper to use any one of the second solutions
throughout a given problem, since all the differences will be absorbed in the arbitrary
constants of the problem, and the same final numerical result will be obtained; but it is
necessary to be consistent in the use of only one of these throughout any given analysis.
It is of interest to observe the similarity between (3) and the simple harmonic equa~
The functions
be observed in
tion, the solutions of which are sinusoids. The difference between these two differential
equations lies in the term (1 / r)(dR / (17') which produces its major effect as r —-> 0. Note
regions far removed from the axis as, for example, near the outer edge of Fig.
region bounded by surfaces of a cylindrical coordinate system approximates
a cube. For these reasons, it may be expected that, away from the origin, the Bessel
functions are similar to sinusoids. That this is true may be seen in Figs. 7.140 and b.
For large values of the arguments, the Bessel functions approach sinusoids with magnitude decreasing as the square root of radius, as will be seen in the asymptotic forms,
Eqs. 7.15(l) and 7.15(2).
that for
1. 19a, the
Barthel functions
monic
equation
It is sometimes convenient to take solutions to the
in the form of
complex exponentials
simple
har-
rather than sinusoids. That is, the
solution of
(122
.,
,
+ K-Z
O
(8)
Be'jK"
(9)
=
612“
can
be written
as
Z
‘4
’5
=
Ae+sz
+
E. Jahnke. F. Emde, and F. Lose/1, Tables of Higher Functions, 6th ed. revised by F. Lésch.
McGraw—Hill, New York, 7960.
M. Abramowifz and I. A. Sfegun (Eds), Handbook of Mathematical Functions, Dover,
New York, 7964.
7.14
3T5
Bessel Functions
where
6:sz
Since the
Kz i-
cos
j sin
K2
complex exponentials are linear combinations
general solution of (8) as
(10)
of cosine and sine functions,
may also write the
we
Z
or
=
2
A’ejK:
+ B’ sin K2
other combinations.
Similarly,
it is convenient to define
new
Bessel functions which
nations of the
of the
Jv(Tr) and Nv(Tr) functions. By
complex exponential, we write
Ht‘irr)
H£2)(Tr)
These
are
are
Mfr)
=
Jva‘r)
related
Nv(Tr), they
are
—-
both
are
linear combi-
with the definition
(10)
M17)
(11)
mar)
(12)
singular
at
r
z
respectively. Since they
Negative and positive
0.
by
H‘D,(Tr)
H<E),(Tr)
For
+
analogy
called Hankel functions of the first and second kinds,
both contain the function
orders
=
direct
=
==
eJ‘WHg1)(Tz-)
e-J‘WH;2)(Tr)
large values of the argument, these can be approximated by complex exponentials,
magnitude decreasing as square root of radius. For example,
with
H(I)(TI‘)
:
¥r——>uo
_;2._
ej(Tr- 7r/4~Wr/2)
rrTr
asymptotic form suggests that Hankel functions may be useful in wave propagation
problems as the complex exponential is in plane-wave propagation. It is also sometimes
convenient to use Hankel functions as alternate independent solutions in static problems.
Complete solutions of (3) may be written in a variety of ways using combinations of
This
Bessel and Hankel functions.
Bessel and flannel Functions of
T
=
jr,
and
(3)
imaginary Arguments
1 dR
c1212
—,+——-—
a’r~
The solution to
Nv(Tr).
In this
If T is
imaginary,
becomes
r
dr
2
72+v—,R=O
(13)
r~
(3) is valid here if T is replaced by jr in the definitions of Jv(Tr) and
case
Nv(jrr)
is
complex
and
so
requires
two
numbers for each value of
the argument, whereas j”"J,,(j77‘) is always a purely real number. It is convenient to
replace Nv(j'n') by a Hankel function. The quantity 1"" 1H9077) is also purely real and
so requires tabulation of only one value for each value of the argument. If v is not an
integer, j”J,,,,(j'n') is independent of j “’J,,( jrr) and may be used as a second solution.
372
Chapter
Two- and Three-Dimensional
7
Thus, for nonintegral
11
two
possible complete
Boundary Value Problems
solutions
are
R=Auamo+8ynom
no
R=Annma+aflwom
as
and
where powers of j are included in the constants. For 11
12, an
in (14) are not independent but (15) is still a valid solution.
=
It is
common
practice
integer, the two
to denote these solutions as
[$1.01)
=
jzvjtvgv)
(16)
so):§““H9ma
where
v
=
solutions
no
77‘.
As is noted in Sec. 7.15
some
of the formulas
relating
Bessel functions and Hankel
for these modified Bessel functions.
functions must be
Special cases of these
changed
in
and
7.13
for
Sec.
the
axially
symmetric field.
[C(77)
K007)
The forms of Iva-r) and [($77) for v
O, l are shown in Fig. 7.14c. As is suggested
by these curves, the asymptotic forms of the modified Bessel functions are related to
growing and decaying real exponentials, as will be seen in Eqs. 7.15(5) and 7.15(6). It
is also clear from the figure that Kv(17') is singular at the origin.
functions
were seen as
=
A
5
4
......
'”
10(1))
11(0)
3
.—
Kan)
2
l
Ko(v)
l
0
0
1
FIG. 7.140
v
2
3
4
Modified Bessel functions.
7.15
BESSEL FUNCTION ZEROS AND FORMULASM
7.15
The first several
functions
are
zeros
given
373
Bessel Function Zeros and Formulas
of the low~order Bessel functions and of the derivatives of Bessel
in Tables 7.15a and 7.15b,
respectively.
Table 7.158
Zeros of Bessel Functions
10
II
2.405
J._,
3.832
N0
N1
N2
5.136
0.894
2.197
3.384
5.520
7.016
8.417
3.958
5.430
6.794
8.654
10.173
11.620
7.086
8.596
10.023
Table 7.15!)
Zeros of Derivatives of Bessel Functions
1;,
1;
J;
N3
0.000
1.841
3 .054
2.197
3.683
3.832
5.331
6.706
5.430
6.942
8.351
10.173
8.536
9.969
8.596
10.123
11.574
Asymptotlc
5.003
Forms
2
J.,(v)
—->
—-—
77
(fig-co)
~—>
H‘Wv)
3......
—-9
v
—
sin
7TU
2
__
(
___>
__
Jim
(7r/4)
-_
2
WT
-
——
2
>
(2 )
o
(WT/4]
_
—‘
6
JD)
—.
(77/4)
_—
1
-
(3)
9
(WT/J]
(4)
1
L,
(10—9er
(5)
#30-
(6)
u—aoc
u—eoo
j”+1H9)(jv)
are
-
4
.
More extensive tabulations
—-
(1)
—
7w
J.(Jv)-—
y—eoo
WT
~
77
v
-
eflv
2
')
.__1,
-
7w
H(“)(U)
J
~
4
2
N
v
cos
7w
Wm
‘8
N;
N;
=
K
"7; .(v)-->
u—eoo
found in the
sources
given in footnotes
l4 and 15.
374
Chapter
Boundary Value
Two- and Three-Dimensional
Problems
following formulas which may be found by differentiating the
by term, are valid for any of the functions Jv(v), Nv(v),
and H 3WD). Let Rv(v) denote any one of these, and R; denote (d/dv)[R,,(v)].
Berivatives
The
appropriate series,
H $00)),
7
term
R6
=
Rav)
=
Row)
=
vav)
—R1(v)
(7)
1
vR;<v)
vR;(v)
-
(8)
5161(1))
(9)
vaw)
—
+
(10)
-=
~vR,(v>
vR,-1<v)
=
-v"”R,,+1(v)
(11)
v”R,,_1(v)
(12)
d
;—D [v‘vRv(v)]
d
d—U [vvRv(v)]
=
Note that
R‘v( T‘’)
—
._..
i
I
[R (T91
1-4-
—
“-
v
dCTI')
I
[R v(T-)]
le'
For the I and K functions, different forms for the
(13 )
foregoing derivatives must be used.
Eqs. 7.l4(16) and 7.l4(17)
may be obtained from these formulas by substituting
in the preceding expressions. Some of these are
They
v1;(v)
vI;<v)
UK;(U)
urge)
Recurrence Formulas
By
v1,(v)
=
+
v1,+,(v)
(14)
——v1v(v)
=
va(v)
=
+
—
”Iv—1(1))
va+1(v)
(15)
-va(v)
==
recurrence
-
vK,._,(v)
formulas, it is possible
to obtain the values
for Bessel functions of any order, when the values of functions for any two other orders,
differing from the first by integers, are known. For example, subtract (10) from (9).
The result may be written
2
flaw)
As before, Rv may denote J”,
formulas
==
Nv, H 3}),
R,,_,(v)
or
+
R,-,(v)
H S”, but not 1,,
or
K1,.
(16)
For these, the
recurrence
are
2
3311,02)
=
Lara»
=
Kv+1(v)
——
Ana»
(17)
212
—v-K,,(U)
-—
Kv__1(v)
(18)
7.16
Expansion of a Function
integrals Integrals that will
Rv denotes JU, Nv, H9), or Hi2):
fvavflw)
f vvRv_1(v)
I vRv(av)Rv(,Bv)
do
=
dv
=
dv
=
later
problems
given
are
below.
(19)
vvRv(v)
(20)
7—0—3
B
0‘“
dv
solving
-v“"Rv(v)
x
I va;(ozv)
be useful in
375
Series of Bessel Functions
as a
"
(21)
[BRv(av)Rv_1(Bv)
cuRv..1(av)Rv<Bv)],
—-
a
at
B
~1ng [R§(av) Rv_1(ozv)Rv+1(av)]
Nice [133mm <1 ——l,’—,)R§(au)]
=
—-
(22)
2
+
—-
CFU"
7.16
In Sec. 7.11
be
EXPANSION OF
a
expressed
study
over a
was
A
FUNCTION AS
SERIES OF BESSEL FUNCTIONS
made of the method of Fourier series
given region as
the coefficients in such
A
a case
series of sines
a
because of the
or
by
which
cosines. It is
a
function may
possible to evaluate
orthogonality property
of sinusoids. A
of the
study
onality
integrals, Eqs. 7.15(21) and 7.15(22), shows that there are similar orthogw
expressions for Bessel functions. For example, these integrals may be written
for zero-order Bessel functions, and if
and
the mth and
qth
pq
pq, then Eq. 7.15(21) gives
are
roots
of
a
and
J0(v)
=
B
are
taken
O, that is,
as
pm/ a
Jo(p,,,)
=
and
pq/ a,
O and
where pm
J0(pq)
=
0,
pm #
a
I‘
.
210(E"’—’>J0(€1—)
f
a
O
So,
a
function
f(r) may be expressed
z
f(2‘)
=
infinite
2
sum
0
(1)
of zero—order Bessel functions
r
r
7‘
f(")
as an
dr
(I
bIJO(p1;> [7210(P2 Z) b3J0(p3 E)
+
+
+
‘
"
OI'
2
m=l
(2)
bm may be evaluated in a manner similar to that used for Fourier
by multiplying each term of (2) by rJo(pmr/a) and integrating from O to
The coefficients
coefficients
@1062")
376
a.
Chapter
Then
by (1)
Two- and Three-Dimensional
7
all terms
on
the
right disappear except
the mth term:
2
a
a
f0 7f(r)JO< >
M
.
d}
.
f0 bmr[Jo( >:|
Bani
_.
-—
d}
.
a
a
From
Boundary Value Problems
Eq. 7.15(22),
f
o
bmrJ%(p——"‘r>
dr
5
=
2
a
Maps
(3)
or
a
2
p
r
orthogonality relations enabled us to obtain
assumption that the series is a proper representation
of the function to be expanded, but two additional points are required to show that the
representation is valid. The series must of course converge, and the set of orthogonal
functions must be complete, that is, sufficient to represent an arbitrary function over
the interval of concern. These points have been shown for the Bessel series of (2) and
for other orthogonal sets of functions to be used in this text.17
Expansions similar to (2) can be made with Bessel functions of other orders and
In the above,
as
in the Fourier series, the
coefficients of the series under the
types (Prob. 7.16a).
-----
.--:a.;.-..::::=-==-,-.-:-\
E
..
ampfie
7.16
BESSEL FUNCTION EXPANSION FOR CONSTANT IN RANGE O
If the function
f(r) in (4)
is
a
I:m
Using Eq. 7.15(20)
with R
=
constant
=2
V0
in the range 0 <
2V0
—
azfflpm)
J,
v
=
fa
v
< a, we
have
I
d-
(5)
0
0
l, and
pm"
°( )
1 -J
r
< r < A
=
pmr/a,
the
integral
in
(5) becomes
(if l: (were) Thistle]:
=
02
_"
Pm
‘7
(6)
Jl(pnz)
See, for example, E. T. Whittaker and G. N. Watson, A Course in Modern Analysis, 4th
ed, pp. 374-678, University Press, Cambridge, 7927.
Fields Described
7.17
and the series
expansion (2)
by Cylindrical
for the constant
V0
377
Harmonics
is
cc
f“)
or,
using
the values of the
N)
zeros
m;
of JO in Table
2.405r
0.832V0
——Jo
“2.405)
=
2V0
12mm...)
pmr
°< >
7.15a,
0.362VO
116.520)
——
(7)
a
5.520r
——
a
0
a
8
U
8.6542-
0.231110
11(8.654)
Further evaluation of
14 and 15
+
0
a
(8) requires reference to tables in the
Eq. 7.13(11).
sources
given
in footnotes
numerical evaluation of
or
7. l 7
FIELDS DESCRIBED
BY
CYLINDRICAL HARMONICS
We will consider here the two basic types of boundary value problems which exist in
axially symmetric cylindrical systems. These can be understood by reference to Fig.
7.1761. In
one
potential (133
(or both) CD]
ligibly
type both
is
CD2
expand
the
are nonzero.
the
a
and
2
a nonzero
O and either
The gaps between ends and side are considered neg—
potentials will be taken to be independent of
simplicity,
along the surface. In
boundary potentials
second situation,
along
(132, the potentials on the ends, are zero
cylindrical surface. In the second type CD3
and
the
small. For
the coordinate
to
applied
or
(131
to
nonzero
the first type, a Fourier series of sinusoids is used
as was done in the rectangular problems. In the
series of Bessel functions is used
to
expand
the
boundary potential
the radial coordinate.
Nonzero Potential
on fiylindrical Surface
Since the boundary potentials are
axially symmetric, zero-order Bessel functions should be used. The repeated zeros along
the z coordinate dictate the use of sinusoidal functions of z. The potential in Eq. 7.13(20)
is the appropriate form. Certain of the constants can be evaluated immediately. Since
K0(rr) is singular on the axis, C 53 must be identically zero to give a finite potential there.
0. As
The cos 7': equals unity at z
0 but the potential must be zero there so C;
in the problem discussed in Sec. 7.10 the repeated zeros at z
[require that T mar/l.
Therefore the general harmonic which fits all boundary conditions except (I)
V0 at
=
=
=
=
2
‘
=
a
is
chm
:
Am10(fl:-Z)sin<m;rz)
l
Figure 7.1717 shows a sketch of this harmonic for m
we
here
the
is
clear
that
have
potential on
cylinder. It
=
(1)
and with the
the
problem
boundary
expanding the
nonzero
of
378
Chapter
7
Two- and Three-Dimensional
Boundary Value
Problems
FIG. 7.17
(0) Cylinder with conducting boundaries. (1)) One harmonic component for matching
boundary conditions when nonzero potential is applied to cylindrical surface in (a). (c) One
harmonic component for matching boundary conditions when nonzero potential is applied to end
surface in (a).
boundary potential in sinusoids just as in
ing the procedure used there we obtain
(DO, Z)
,
Nonzero Potential
(131
=
CD3
2
O and
7.13, the boundary
on
ever
become
End
rectangular problem of Sec.
59:9 [WWW/Z)
mrr
.
sm
3235
10(m7Ta/l)
In this situation if
we
7.12. Follow-
(2)
1
refer to
Fig. 7.17a,
we see
that
selecting the proper form for the solution from Sec.
(130
V0.
O at r
a for all values of 2 indicates that the
condition that (I)
=
In
=
R function must become
do not
”221d
_
_
the
zero.
zero at r
=
a.
=
Thus,
(The corresponding
we
select the
JO
functions since the
10’s
second solution, NO, does not appear since
potential
must
remain finite
of
10(1))
0, T
the
axis.) The value of T in Eq. 7.13(19) is determined
a for all values of 2. Thus, if
Pm is the mth root
be pm/a. The corresponding solution for Z is in hyperbolic
from the condition that (I)
=
must
on
O at
=
functions. The coefficient of the
at
z
=
379
Spherical Harmonics
7. l8
O for all values of
r.
amplitudes which satisfy the
imposed may be written
=
r
hyperbolic
cosine term must be
zero
since (I) is
zero
of all
cylindrical harmonics with arbitrary
symmetry of the problem and the boundary conditions so
Thus,
a sum
far
@(r, z)
m
One of the harmonics and the
BmJO(-p—”’—'-) Siflh<gfli>
2
=
=
(3)
a
(Z
1
in
Fig. 7.170.
remaining
V0 at r < a.
Here we can use the general technique of expanding the boundary potential in a series
of the same form as that used for the potentials inside the region, as regards the de—
pendence on the coordinate along the boundary. In Ex. 7.16 we expanded a constant
The
over
required boundary potentials
condition is that, at
the range 0 <
constants in
r
in J0 functions
< a
at the
(3). Evaluating (3)
CD0; 1)
so
B,”
”i=1
O at
=
that result
boundary
2
=
1, CI)
=
z
2
1,
==
r
can
we
shown
are
=
a
and (I)
=2
be used here to evaluate the
have
sinh(&i)10(’1§5)
(4)
0
Equations (4) and 7.16(7) must be equivalent for all values of r. Consequently, coef—
ficients of corresponding terms of JO( pmr / a) must be equal. The constant Bm is now
completely determined, and the potential at any point inside the region is
on, 2)
=
E
”1:31
N"
pmJl(pm) Sinh(pnzl/a)
Consider next
Laplace’s equation
about the axis
so
in
spherical
(5)
a
a
coordinates for
that variations with azimuthal
in the two
<p"‘Z>JO<Il—')
SPHERICAL HARMONICS
7.18
equation
sinh
regions
with symmetry
may be neglected. Laplace’s
and 6 then becomes (obtainable
angle qb
remaining spherical
cover)
coordinates
a~(rcp)
a
r
from form of inside front
61‘2
+
——1—
1'
sin 6
5—0
(sin 96—?)
=
69
0
(1)
or
r
1 3ch
as)
52(1)
,+2———+-—
61‘“
Assume
a
product
62'
1
66“
solution
®=R6
as)
——-~=
,,
r
1‘
tan 9 66
(2)
380
Chapter
where R is
a
7
Two- and Three-Dimensional
function of
r
Boundary
Value Problems
alone, and 8 of 6 alone:
1
rR”G + 2R’6
-R6"
+
.R6’
+
r
r
0
=
tan 6
and
r
2R”
From the
previous logic,
for all values of
may be
and
r
expressed
ordinary
=
—
----
(3 )
—
9
G
R
R
9’
G”
ZI‘R'
+
—-—
6
tan
if the two sides of the
6, both sides
equations are to be equal to each other
equal only to a constant. Since the constant
way, let it be m(m + l). The two resulting
be
can
in any nonrestiictive
equations are then
differential
dZR
2'2
(IR
+ 2r—
.,
dze
Equation (4)
has
a
~-
—
+
tan6d6
solution which is
R
+
mo"
+ 1 )e
easily verified
Clr’”
z
DR
m(m
d9
1
+
(Z62
-
dr
dr‘“
==
0
(4)
=-
0
5
U
to be
C2r_("’+1)
+
(6)
functions is not obvious, so, as with the Bessel
equation, a series solution may be assumed. The coefficients of this series must be
determined so that the differential equation (5) is satisfied and the resulting series made
to
A solution to
(5) in
define
function. There is
Bessel
a new
functions,
terms of
Thus, for any
new
departure
one
here from
exact
an
analog with
the
proper selection of the arbitrary constants will
function terminate in a finite number of terms if m is an
for it turns out that
make the series for the
integer.
simple
integer m,
the
a
polynomial
defined
by
”1
d
Pm(cos 6)
is
a
2mm!
solution to the differential
equation;
the solutions
first few values of
polynomials
values of the
are
m are
called
d(cos 6)
tabulated below and
are
P0(cos 6)
=
P1(cos 6)
=
P2(cos 6)
=
P3(cos 6)
=
P4(cos 6)
=
P5(cos 6)
=
6
——
1)’"
(7)
equation (5). The equation is known as Legendre’s
Legendre polynomials of order m. Their forms for the
are
and not infinite series, their values
polynomials
(cos2
2
shown in
can
Fig.
7.180. Since
be calculated
easily
they
are
if desired, but
also tabulated in many references.
1
cos
6
%(3 cos2
6
-%(5 cos3
6
—
1)
(8)
-
%(35 cos“
6
lsL-(63 cos5
6
-
~—
3
cos
6)
30
C082 6
+
70
cos3 6
+ 15 cos
3)
6)
7.18
38'!
Spherical fiarmonics
1.0
>0
--1.0-—-
FIG. 7.18a
Legendre polynomials.
C 1P,,,(cos 6) is only one solution to the second-order
recognized that 8
differential equation (5). There must be a second independent solution, which may be
It is
=
obtained from the first in the
same manner as
this solution becomes infinite for 6
of
spherical coordinates
is
the axis is excluded, the
for Bessel functions, but it turns out that
0.
Consequently it is not needed when the axis
included in the region over which the solution applies. When
second solution must be included. It is typically denoted
=
Q,,(cos 9) and tabulated in the references.”
An orthogonality relation for Legendre polynomials
sinusoids and Bessel functions which led
to
is
quite
the Fourier series and
similar to those for
expansion in
Bessel
functions, respectively.
j
Pm(cos 6)P,,(cos 0)
sin 6 d6
H
0,
m
#
(9)
n
0
2
f
[P,,,(cos 6)]2
sin 6 d6
o
It follows that, if
series of
a function f(6)
Legendre polynomials,
(10)
2m + l
defined between the limits of O to
77'
is written
as a
CG
f(6)
=
2
a,,,P,,,(cos 6),
0 < 0
< rt
(11)
m=0
‘5
W. R. Smyfhe, Static
ington, DC, 7989.
and
Dynamic Electricity, 3rd ed, Hemisphere Publishing Co., Wash»
382
Chapter
7
Two- and Three-Dimensional
the coefficients must be
given by
am
Problems
the formula
+ 1
2
=
Boundary Value
7’
"IT I
f(6)P,,,(cos .9)
sin 6 d6
(12)
O
Exampfie
7.183
HIGH-PERMEABILITY SPHERE IN UNIFORM FIELD
We will examine the field distribution in and around
a sphere of permeability p, a5 #0
placed in an otherwise uniform magnetic field in free space. The uniform
field is disturbed by the sphere as indicated in Fig. 7.1817. The reason for choosing this
example is threefold. It shows, first, an application of spherical harmonics. Second, it
is an example of a situation in which the constants in series solutions for two regions
are evaluated by matching across a boundary. Finally, it is an example of a magnetic
boundary-value problem.
Since there are no currents in the region to be studied, we may use the scalar magnetic
potential introduced in Sec. 2.13. The magnetic intensity is given by
when it is
H
As the
=
MW)”,
(13)
problem is axially symmetric and the axis is included in
Pm(cos 6) are applicable. The series solutions with
the solutions
came, a)
The
=
Z
Pm(cos 6)[C,,,,rm
+
the
region
of interest,
these restrictions
C2mr"('”+1)]
are
(14)
procedure is to write general forms for the potential inside and outside the sphere
across the boundary. Since the potential must remain finite at r
O,
and match these
FIG. 7.18b
=
Sphere
of magnetic material in
an
otherwise uniform
magnetic
field.
the coefficients of the
becomes, for the inside
powers of
negative
region,
cram
383
Spherical Harmonics
7.18
Z
=
vanish for the interior. The series
must
1'
Amr”’Pm(cos a)
(15)
Outside, the potential must be such that it gives a uniform magnetic field H0
potential form which satisfies this condition is
at
infinity.
The
CD,"
That this
gives
uniform field may be
a
H3
Terms of the series
all vanish at
__acI>,,,
__
*
-—H0r
=
cos
6
by noting
seen
1
that dz
ad)",
.1
_
62
(16)
6
cos
=
dr
cos
6
so
H0
_
6r
(17)
(14) having negative powers of
infinity.
r may be added to (16), since
Then the form of the solution outside the sphere is
c1)",
—Hor
=
cos
Z
9 +
Bum(cos arm“)
they
(18)
m
It
was
pointed
out in Sec. 2.14 that
Cl)",
surface currents. Therefore, the terms in
dependence
are
is continuous
across
(15) and (18) having
boundaries without
the
same
form of 6
equated, giving
A0
:
Boa—1
n1
AazBa“2—Ha
1
o
1
.2:
O
m=l
(19)
.
Ama’"
=
Bma”’(m+1)
m
Furthermore, the normal flux density is continuous
Substituting (15)
and
(18)
in
6’.
(20)
the
1
boundary
so
6(1) 171
6(1) I71
”D
at
>
6’.
I~a+
and
equating
terms
(20 )
I~a—-
with the
same
6
dependence,
we
find
B0
#441
=
=
O
—~2 #0 B 1 (1‘3
--
MOI—IO
m
=
m
=
O
(21)
'
umAnflm‘1
From
be
(19)
zero to
=
“11.0071
+
l)B,,,a"(’"+2)
m >
1
O, and that for m > 1, all coefficients must
(21) we see that A0
B0
The only remaining terms are those with
of
conditions.
the
two
sets
satisfy
and
2
=
384
=
m
tuting
Chapter
7
1. These two
the results in
Two- and Three-Dimensional
equations may be solved
(18) gives, for r > a,
to
Boundary Value
give A1
and
B1
Problems
in terms of
H0.
Substi-
3
<1) m
from which H
for
r <
be found
can
a
[.L—po
——---1H‘
[(2M0+M>13 :IOICOS
=
by using (13)
for
r
> a.
6
22
()
Substitution of A1 into (15)
gives,
a,
3
<1)”,
Applying (13),
we
=
<—#—O)H0r
—
6
cos
(23)
+ 1“
2:“0
find the field inside to be
A
H
=
z
3#0
(zflo )
——
+ F»
H0
(24)
It is of interest to observe that the field inside the
Finally, multiplication
of
(24) by
B
p.
gives
A
=
z
the flux
homogeneous sphere
density
3#o
(ZULO/IJ') >
——
+ 1
From
(25)
we see
that for it >> #0 the maximum
B
:
is uniform.
HO
(25)
possible
value of the flux
density is
(26)
23am,
Exampfie 7.18!)
EXPANSION IN SPHERICAL HARMONICS WHEN FIELD Is GIVEN ALONG
AN
AXIS
relatively simple to obtain the field or potential along an axis of symmetry
application of fundamental laws, yet difficult to obtain it at any point off this
axis by the same technique. Once the field is found along an axis of symmetry, expansions in spherical harmonics give its value at any other point. Suppose potential, or any
component of field which satisfies Laplace’s equation, is given for every point along
an axis in such a form that it may be expanded in a power series in z, the distance along
It is often
by
direct
this axis:
(I:
=
axis
If this axis is taken
as
axis may be written for
the axis of
r
Z
0 <
z
<
spherical coordinates,
6
bmz’",
a
(27)
”’30
=
0, the potential off the
< a
on, a)
=
Z
”2:0
bmrum(cos I9)
(28)
7.19
Product Solutions for the Helmholtz Equation in Rectangular Coordinates
385
This is true since it is
a solution of Laplace’s equation and does reduce to the
given
0 where all Pm(cos 6) are unity.
potential (27) for 6
If potential is desired outside of this region, the potential along the axis must be
expanded in a power series good for a < z < 00:
=
c1)
Z cmz-W”,
:
Then (I) at any
point
outside is
(I)
=
> a
z
(29)
"1:1
5:0
given by comparison
Z
with the second series of
c,,,P,,,(cos ray-“0"“),
r>
(14):
(30)
a
m=0
For
wire
example, the magnetic field H:
carrying current I in Sec. 2.3 as
2(a2
The binomial
is
good for
O <
H3,
at any
u) -3/32
(1
+
lul
< 1.
axis
Since
found
22)“2
the axis of
along
2a[1
+
a
circular
loop
of
(22/a3)]3/2
expansion
H3
H:
+
was
3
—_
-
1
_.
_.
15
Lt
+
to
u
2
105
___
_.
8
2
Applied
_
(31), this gives for
I
322
15
20
2a~
8
u
3
+
.
8
22
2
< a
105
48
a
axial component of magnetic field, satisfies
r, 6 with r < a is given by
223
a
Laplace’s equation (Sec. 7.2),
point
1
H:(r, 9)
7.19
2
—~
2a
3
1
-
r2
—
7
ad
‘6CZ“
15
P2(cos 6)
+
r4
—-
8
7
a
P4(cos 6)
+
(32)
PRODUCT SOLUTIONS FOR THE HELMHOLTZ EQUATION
IN RECTANGULAR COORDINATES
technique used in the preceding sections for finding product solutions to Laplace’s
equation will be applied here to the scalar Helmholtz equation. Whereas the single»
product solution for static problems was seen in Sec. 7.10 to be of little value, such
solutions will be seen in the next chapter to be of great importance as waveguide
propagation modes and will be analyzed extensively there.
Let us consider the scalar Helmholtz equation. Here we make the assumption that
the dependent variable depends on 2 in the manner of a wave, as e" 7". The variable (,1;
The
386
Chapter
7
Two- and Three-Dimensional
Boundary Value
Problems
‘
remaining in the equation is, therefore, the coefficient of 60“” 7").
Laplacian explicitly in rectangular coordinates, we have
62
62
—$+f:
6y
“kit!
ax“
where
k3;
solution
=
(,0
72
+
rogue.
Let
us assume
Written with the
(1)
that the solution
can
be written
as
the
product
X(x)Y( y). Substituting this form in (1),
=
X”Y + XY”
=
~k§XY
01‘
Y!)
XII
+
_
_
:
..k2
(2 )
c
Y
X
indicate derivatives. If this equation is to hold for all values of x and y,
"
and y may be changed independently of each other, each of the ratios X /X and
Y”/ Y can be only a constant. There are then several forms for the solutions, depending
The
primes
since
x
upon whether these ratios
negative constant and one
are
taken
positive
as
negative
constants,
positive constants,
negative,
or one
constant. If both are taken as
X'II
_
=
__
k?
.
X
y”
9
?=“a
The solutions to these
kf.
of
and
k;
is
kg.
ordinary differential equations
are
sinusoids, and by (2) the
sum
Thus
II!
=
XY
(3)
where
X
=
Y
=
Acoskxx
+
C
+ D sin
cos
kyy
Bsinkxx
kyy
(4)
a+a=a
Either
kI and ky may be imaginary in which case the corresponding sinusoid
becomes a hyperbolic function. Values of the constants kx and
ky are determined by
conditions on 1,0 at the boundaries in the 35—32 plane. Examples of the application of these
general forms will be seen extensively in the following chapter where the dependent
variable all is identified as E2 or Hz.
or
both of
7.20
In
cylindrical
the
wave
PRODUCT SOLUTIONS FOR THE HELMHOLTZ EQUATION
lN CYLINDRICAL COORDINATES
structures, such
components
coaxial lines or waveguides of circular cross section,
conveniently expressed in terms of cylindrical coordi-
as
are most
Product Solutions for the Helmholtz
7.19
Assuming that
equation becomes
nates.
the
z
dependence
fix
1
+
61'2
where
kg
3/2
=
assumed
an
+
rogue.
product
-
in
Rectangular Coordinates
337
is in the waveform if”, the scalar Helmholtz
an
——-
1
+
——
-—k2 1P
=
'2
6r
1‘
For this
Equation
(1)
partial differential equation,
we
solution and separate variables to obtain two
shall
again substitute
ordinary differential
equations.
Assume
where R is
a
function of
2'
alone and F ¢ is
”
+
RF¢
Separating variables,
we
R’F¢
a
function of
F251?
+
9
--k;RF<25
=
r2
r
alone:
9b
have
7
R”
:8;
+
R
:5
[Ci-,2
+
R
F
45
equation is a function of r alone; the right of qb alone. If both sides
equal for all values of r and qS, both sides must equal a constant. Let this
be v2. There are then the two ordinary differential equations:
The left side of the
to
are
be
constant
““F"
__
2
=
v
==
1)“
(2 )
F
and
R"
22—+
R
R,
i-—-—l— kL’c'r~
R
7
or
1
R”+—R
r
The solution to
to
(3)
I
v2
+(k;~—;>R:O
7
(3)
r~
(2) is in sinusoids. By comparing with Eq. 7. 14(3)
may be written in terms of Bessel functions of order
(p
=
we see
that solutions
v:
(4)
RF¢
where
R
Fa
Either
or
AJv(kcr)
=
+
Ccoqub
BNv(kcr)
+
(5)
Dsinvqb
both of the Bessel functions may be replaced by Hankel functions [Eqs
one desires to look at waves as though propagation were in
7.l4(1 l) and (12)] when
388
Chapter
Two- and Three-Dimensional
7
the radial direction. Thus, for
example,
R
Fqb
If kc is
imaginary,
the
Boundary Value Problems
ll
=
A1H£1)(kc")
+ B 1H
+
Ccoqufi
EEK/Cc")
(6)
Dsinvqb
ordinary Bessel functions
can
be
replaced by
the modified Bessel
functions, Eqs. 7. 14(16) and (17). In the examples in the following chapter, the variable
(/1 will be identified
with
E2
or
Hz.
PROBLEMS
equation satisfied by E, in cylindrical coordinates
region. Repeat for E (,5. Note that these are
7.23 Find the form of differential
charge~free, homogeneous
Laplace equations.
dielectric
none of the spherical components
O.
quasistatic problems in which VZE
7.2b Show that
for
of electric field
for
a
not
satisfy Laplace’s equation
=
7.2c* Show that the
tion
expressed
rectangular component E:
in
Spherical
of electrostatic field satisfies
Laplace’s equation for H, A, and CI)”, in a current-free region
a homogeneous conductor with do currents.
7.2d Derive
Laplace’s
equa-
coordinates.
with static fields
and for J and E in
7.2e Use superposition to find the potential on the axis of an infinite cylinder with
tial specified as (Md)
V0 sin (13/2, for O S (,6 S Zn on the boundary.
a
poten-
=
7.2f A
spherical
O < 6
<
surface is
77/ 2.
potential except for a sector in
potential at the center of the sphere.
at zero
Find the
the
region
0
<
(p
<
7r/ 3,
capacitance of a parallel~plate capacitor with square plates having edge
a/ 2 situated in free space using the method of moments. If
spacing d
you do the calculations by hand, divide each plate into four equal squares. If a computer program is written, run it for several subdivisions of the plates and plot the effect
on capacitance.
7.33 Calculate the
length
7.3b Find
a
a
and
better
=
approximation
to
capacitance of the structure in Ex. 7.3a by subdividFig. 7.3d into four equal parts and repeating the
the
each of the squares shown in
method of moments calculation.
ing
7.3e In
applying the method of moments calculation to two-dimensional problems, the ln r0
in Eq. 1.80) is neglected. As an illustration of the validity of this procedure, find
the potential of two parallel line charges located as follows: +q, at 45
6 and
O, r
R on the qb
0 axis.
O, r
26; take the zero potential point to be r
—-q1 at q5
Apply Eq. 1.8(7) and show that the ln rO terms cancel to arbitrary accuracy as R —-> 00.
How does this explain that the In re terms can be neglected in the two-electrode twodirnensional method of moment problems in which the line charges have a variety of
term
=
=
=
=
=
=
values?
7.3d* Write
a
to find the stripline capacitance as in Ex. 7.3b. Extend the
larger electrodes by one unit of the division in Fig. 7.3e and
capacitance. Then use a subdivision of the electrodes one-half as
example. Compare the results to evaluate the importance of the grid size.
computer program
range included on the
evaluate the effect on
fine
as
in the
339
Problems
7.4a Check
W
by the Cauchy—Riemann equations
CnZ” and a series of such terms,
:
W
=
the
analyticity
2
6.2"
of the
general
power term
"=1
7.4b Check the
following functions by
analytic:
are
7.4c Check the
of the
analyticity
the
Cauchy-Riemann equations
W
:
sinZ
W
2
e2
W
=
2*
W
=
22*
=
following, noting
.r
-
to determine if
they
jy
isolated
points
where the derivatives
may not remain finite:
W==an
W
change AZ
7.4d Take the
Riemann conditions
when the
7.4e If
change is
by following
a
tan Z
general direction Ax + j Ay. Show that, if the Cauchysatisfied, Eq. 7.4(3) yields the same result for the derivative
in any
are
in the
path
ferent values when the
called
ll
branch
x
direction
around
or
point
some
as
the y direction alone.
in the Z
plane,
the variable W takes
on
dif-
Z is reached, the point around which the path is taken is
24/3 along a path of constant
Evaluate W x Z”2 and W
same
point.
O is a branch point for these functions.
origin to show that Z
Discuss the analyticity of these functions at the branch point.
a
=
radius around the
7.5a Plot the
=2
of the
shape
u
=
$0.5
equipotentials
for the V
=
.1'4/3,
y
=
O
boundary
condition used in Ex. 7.5.
7.5b A thin
Use
cylindrical
shell of radius
a
has
a
potential
method similar to that in Ex. 7.5 to find
a
7 .5c Show that if
part of
described
by @(a, 6)
potential function, the field intensity E), is equal
EA. equals the negative of the real part.
is the
u
dW/dZ
and
=
V0
cos
26.
CD036).
to the
imaginary
7.5d Use the results of Prob. 7.5c to find an expression for the slope of equipotential lines
O are normal to the
in terms of dW/dZ. Show that all equipotential lines except it
=
edge in
beam
the electron flow in
Fig.
=
24/3 is
not
analytic
at
Z
=
O,
as
O is a special case.) Him: Write
shown in Prob. 7.4e, and the W
0 to get relations
expression for du in terms of partial derivatives and set du
2
was
2
==
an
existing along
7.6a Plot
a
7.51). (W
0 line at y
=
a
few
77/3
an
equipotential.
equipotentials
377/4.
and flux lines in the
7.6b Evaluate the constant C 1 and
the
potential
zero
at
r
vicinity
of
conducting
corners
of
angles
and
=
C2
in the
function in volts about
a
logarithmic transformation so that it represents
charge of strength q, C/m. Let potential be
line
a.
v is taken as the potential function in the logarithmic transformation, it is
applicable to the region between two semi-infinite conducting planes intersecting at an
0
angle a, but separated by an infinitesimal gap at the origin so that the plane at 6
a at potential V0. Evaluate the
may be placed at potential zero and the plane at 6
7.6c Show that if
=
=2
390
Chapter
Two- and Three-Dimensional
7
C1 and C2, taking the reference for
tion in coulombs per meter.
constants
7.6d Find the form of the
curves
of constant
it
cosh‘I Z, and sinh“1 Z. Do these permit
from the function cos”I Z?
7.6e
Apply
zero
flux at
and constant
one to
solve
v
r
=
a.
Write the flux func-
for the functions sin“1 Z,
in addition to those
problems
cos“1 transformation to item 4 in Ex. 760. Take the right00 at
a to x
plane extending from x
potential V0. Take the left00 at
x
x
from
a
to
extending
potential zero. Evaluplane
the results of the
hand semi—infinite
hand semi-infinite
=
=
=
=
—
-
factors and additive constant.
ate the scale
7 .6?”
Boundary Value Problems
Apply the results of the transformation to item 2 of Ex. 7.6c. Take the elliptic cylindrical conductor of semimajor axis a and semiminor axis 17 at potential V0. The inner
conductor is
a
strip
conductor
extending
between the foci,
x
=
i c, where
c==\/a2—b2
Evaluate all
required
scale factors and constants. Find the total
induced upon the outer
cylinder and
the electrostatic
charge per unit length
capacitance of this two-conductor
system.
7.6g* Modify the derivation in Ex. 7.6d to apply to the problem of parallel cylinders of unequal radius. Take the 1eft~hand cylinder of radius Rl with center at x
d1, the
right-hand cylinder of radius R2 with center at x
d2, and a total difference of poten—
tial VO between cylinders. Find the electrostatic capacitance per unit length in terms of
R1, R2, and ((11 + d2).
=
-
=
7 .6h The
important bilinear
transformation is of the form
(22’ + b
__
cZ' + d
Take a, b, c, and d as real constants, and show that any circle in the 2’ plane is trans«
formed to a circle in the Z plane by this transformation. (Straight lines are considered
circles of infinite radius.)
7.6' Consider the
1.
R, b
—R, c
1, and d
special case of Prob. 7.6h with a
Show that the imaginary axis of the Z’ plane transforms to a circle of radius R, center
at the origin, in the Z plane. Show that a line charge at x’
d and its image at x’
-d in the Z’ plane transform to points in the Z plane at radii rI and r2 with rlr2
R2.
=
=
=
=
=
=
=
Compare
7.73
with the result for
imaging
line
charges
in
a
cylinder (Sec. 1.18).
Explain why factor in the Schwarz transformation may be left out when it
sponds to a point transformed to infinity in the Z’ plane.
a
corre~
7.7b In
Eq. 7.7(2), separate Z into real and imaginary parts. Show that the boundary condition for potential is satisfied along the two conductors. Obtain the asymptotic equa~
tions for large positive it and for large negative u, and interpret the results in terms of
the type of field
7.7c* Work the
approached
in these limits.
of Prob. 7.66
by the Schwarz technique and show that the same
problem of two coplanar semi-infinite plane conductors
separated by a gap 20, with the left-hand conductor at potential zero and the right—
hand conductor at potential V0.
example
result is obtained. This is the
7.7d* For the first
example
would be obtained if
7.7e Plot the
of Table 7.7, find the electrostatic capacitance in excess of what
uniform field existed in both of the parallel~plane regions.
a
V0/ 2 equipotential for Bx.
7.7.
39?
Problems
7.8
Suppose
that the
dielectric
33(2‘)
lower value
as
wave-guiding structure in Fig. 7.8a is bounded on the outside by
which has the value 82 at R0 and then decreases to an appreciably
r is increased. As was seen in Sec. 6.12, waves incident on a
plane
a
boundary between two dielectrics from the higher 8 side can be totally reflected. Find
the limiting rate of decrease of 83 at RO which can permit total reflection of rays approaching the boundary, by studying the variation of the equivalent dielectric constant
in the W plane.
7.93 The so-called circular harmonics
are the
product solutions to Laplace’s equation in the
cylindrical coordinates r and 43. Apply the basic separation of variables
technique to Laplace’s equation in these coordinates to yield two ordinary differential
equations. Show that the r and (1; equations are satisfied respectively by the functions
circular
two
R and
17¢, where
R
C3
Feb
7 .9b An infinite rod of
a
the rod to be of circular
cos
Cgr""
12gb
+
a
cross
uniform
sin
C4
material of relative
magnetic
direction of
to the
pendicular
+
Cir"
2ch
[it lies with its axis perfield in which it is immersed. Take
permeability
magnetic
section with radius
a
and
the expressions
uniformity of the
use
7.9a to find the fields inside and outside the rod. Note the
in Prob.
field in—
side.
7.103 Plot the form of
equipotentials
for (I)
Vo/4,V0/ 2,
=
7 .10b Describe the electrode structure for which the
sin
a
ky
is
a
when y
solution for
potential.
and
for
3VO/4
single rectangular harmonic C1 cosh kx
potential V0 passing through lxl
Take electrodes at
=
(1/2.
=
7.10c Describe the electrode structure and exciting potentials for which the
harmonic (Prob. 7.9a) Cr2 cos 2g!) is a solution.
7.11:! Obtain Fourier series in sines and cosines for the
(i)
A
triangular
V0[(2x/L)
Suppose
1]
defined
from
a
-—a
V0(l
—
2x/L)
from 0 to
=
=
Vox/L
(Vm
cos
for O
kx
circular
functions:
L/ 2
and
for)
=
and also fora < kx<
< x < L
-
V0)
for
a
—-
<
kx
< a,
f(x)
=
77.
given over the interval 0 to a as f(x)
representations yield? Explain how this single
function is
the cosine and sine
represented
=
defined
—7r< kx<
that
L/2
by f(.r)
following periodic
single
to L
by f(.r)
pulse given by f(x)
wave
A sinusoidal
O, for
7.11b
wave
-
(ii) A sawtooth
(iii)
7.100.
Fig.
=
sin
m/ a.
sine term
What do
can
be
in tenns of cosines.
7.11c Find sine and cosine
representations
for the function (3“ defined
over
the interval
O<x<a
for) given by Eq. 7.11(14) in the neighborhood of the discontinuities using
five
sine terms and (ii) ten sine terms and discuss differences from the rectangular
(i)
function being represented.
7.11d Plot
7.11e A
complex form
0 <
x < a
of the Fourier series for
a
function
f(x) defined
is
m
f(.\.‘)
z
Z Cnejlrrnx/a
n=-m
over
the interval
392
Chapter
Two- and Three-Dimensional
7
Show that if this is valid, c" must be
l
=
c”
_
a
given by
a
f
.
f(x)e~}27m:c/a
dx
0
7.11f Find representations for the constant C over the interval 0 < x
form of Prob. 7.11e, and compare the result with Eq. 7.11(14).
7.11g
Find the Fourier
O and
x <
7.123 Obtain
and y
a
=
for)
Problems
Boundary Value
integral representation
for
a
< a
in the
decaying exponential, f(x)
complex
=
Magnum.
O, for
ce‘mforx > 0.
=
series solution for the two-dimensional box
b are at potential zero, and end planes at x
problem
=
a
and
in which sides at y
x
=
—-a are
=
van-5.—
0
,
.3
.,_—m
..
at
potential V0.
7.121) Find the
-
potential
distribution for the box of Prob. 7.12a with the
ditions except that the
should be
V,.
potential
on
the side at y
=
0 should be
same
V1
boundary
and that at y
=
con-
17
-—
7.12c In
a
to x
two-dimensional
=
Obtain
00
a
and
0 and y
problem, parallel planes at y
potential. The one end plane at x
=
are at zero
=
=
b extend from
0 is at
x
=
O
potential V0.
series solution.
fringing that occurs at the open ends of a pair of parallel plates as seen in Fig.
1.9a leads to a modification of the fields between the plates from the ideal uniform
0 to be the ends of the plates, which are at y
distribution. Consider it
O, b. The
analysis of Ex. 7.7 can show that the potential between the ends of the plates may be
Vo[(y/b) + 0.06 sin 277y/ b]. Find the distance
expressed approximately as c13(0, y)
x at which the potential distribution between the plates is linear to within 1%, using
7.12d The
=
=
=
the
analysis
of Prob. 7.12c.
conducting rectangular solid
7.12c A two-dimensional
conductors: at y
0, CD
x
a by a dielectric with
2
=
=
tribution inside the
O;
atx
zero
conducting
=
O, (I)
=
O;
is bounded
at y
conductivity. Find
an
=
on
b, (I)
three sides
=
expression
V0.
».M—_W
of
b-MW
(.w‘
by perfect
It is bounded at
for the
potential
dis—
solid.
a and r
b. The inner (r
a) cylinder
cylinders are located at r
split along
length into two halves which are at different potentials. Potential is
b is at zero
-—V0 for ~7T< 43 < O and V0 forO < 4) < 77. The cylinder atr
potential. Find the potential between the two cylinders.
7 .12f Two concentric
=
=
=
its
is
=
7.12g
plane boundary of a half—space is in strips of width (1 and alterO and the strips to be
V0. Take the boundary to be at y
invariant in the z direction. The origin of the x coordinate lies in the gap between
strips so that the potential is V0 for —a < x < O and V0 for O < x < a. Find the
potential distribution for y 2 O and determine the surface charge density along the y
0 plane. Put the result in closed form (see Collin, footnote 3 of Chap. 8, p. 813)
and plot for ~a < x < a.
The
potential along
V0
nates between
-
the
and
=
—
=
7 .12h* Infinite
atx
=
parallel conducting plates
O, a/2
S
y
S 0,
~00
< z
located at y
O and y
a. A
< 00, is connected to the plate aty
are
=
=
conducting strip
=
(1, thus
“Am-
«awe-
’
n
«an.
"
—
(D:VO
w-asm
,,
e—Qfi»
lag.)
y
L
(D
=
O
as. 7.12h
\
'zaqu-‘L.~i:{,
._-:S.g
‘ags:.’-;="m
Problems
393
introducing additional capacitance between the plates. (See Fig. P7.12}2.) Assume a
potential variation for O _<_ y .<. a/ 2 at x
O, and use superposition of boundary
conditions to find an expression for the capacitance per meter in the z direction added
O.
by the strip at x
linear
=2
=
7.12i Consider
rectangular prism of width at in the x direction
zero potential extending from 2
O to z
cap with the following potential distribution:
has
a
=
O
'
is
Vx,
,0
(3’)
Find the
7.12j*
For
and b in the y direction with
00. At 2
0 the prism
a
all four sides at
potentials within
the
forO
{V0
=
=
=
<x<a/2, ally
fora/2<x<a,ally
prism.
in Ex. 7.120, find the potential distribution if the box is filled with a
homogeneous, isotropic dielectric with permittivity e1 in the bottom half of the box
0
a
box
as
c/ 2
S 2 S
and free space in the remainder.
7.13 Demonstrate that the series
7.163 Write
a
function
f(r) in
Eq. 7.l3(10) does satisfy
terms of
the differential
nth-order Bessel functions
over
equation 7.l3(8).
the range 0 to
a
and
determine the coefficients.
7.16b Determine coefficients for
function
a
zero-order Bessel functions
as
f(r) expressed
fv‘)
2
=
"1-”:
where
7.173 A
p,;,
denotes the mth root of Jc’,(v)
divided into
the range 0 to
over
series of
a as a
follows:
of
l
0
z
magi)
[i.e., 11(0)
O].
=
with
appropriately applied voltages may be used
along the axis with advantageous focusing
properties for electron beams. Suppose the field at the radius a of the cylinder is given
277/12 and p is the period of
approximately by E:(a, 2)
E00 + cos a2), where 0:
the rings. Find the potential variation along the rings (1‘
a) and for r < a. Determine the field on the axis and the period required to have the periodic part of the field
1% of E0.
cylinder
to set up a
a set
nearly uniform
rings
electric field
==
==
=
7.17b Show that the function
(P03 2)
satisfies the
requirement
relative maxima
1 at
of solutions of
.
““e
==
potential
zero
inside the
cylindrical region
of radius
and the
=
‘
cos 1":
Laplace’s equation
potential
cylinder
[/2 to
1/2, it is at potential V0; from 2
2
A1001?)
z
=
with end
problem is as in Prob. 7.17c except that the cylinder
l
b and also from 2
O to z
potential zero from 2
==
7.17c Write the
a, with
=
f(z)
btoz
2:
l
r
=1
end
2
z
=
0
O and
to 2
==
is divided in three parts with
b to z
1. Potential is VO
—
=
for obtaining potential
0 and z
plates at z
z
a cylindrical region of radius
I, provided potential is given as (I)
inside
m
a.
general formula for obtaining potential inside a cylinder of radius
0, is at potential zero, provided that the potential
plane base at z
7.17f Write the
with its
no
b.
general formula
zero-potential
at
—
=
=
=
plates
in two parts. From 2
I, it is at potential ~VO.
a
7.17d The
fromz
that there should be
minima.
or
7.17c Find the series for
=
=
a
which,
is
given
394
Chapter
across
the
Two- and Three-Dimensional
7
plane
surface at
z
=
l,
as
(130‘, 1)
7.17g Find
the
potential
surface at
z
7.183
=
=
r
I. It also has
Apply
the
distribution inside
a, on the end
CI>(r, 1)
separation
==
plate at
V0 for 0
of variables
Boundary Value Problems
a
z
f(r)
==
cylinder
=
with
zero
0 and where
_<. r <
technique
potential
(1/2
on
the
< r < a on
cylindrical
plate
the end
at
a/2.
to
Laplace’s equation
in the three
spherical
0, and 4‘), obtaining the three resulting ordinary differential equations.
Write solutions to the r equation and the ()5 equation.
coordinates,
r,
7.18b Assume a spherical surface split into two thin hemispherical shells with a small gap
between them. Assume a potential V0 on one hemisphere and zero on the other and
find the potential distribution in the surrounding space.
7.18c Write the
tential is
general formulas for obtaining potential for r < a and for r >
given as a general function 1‘09) over a thin spherical shell at r
7.18d For Ex. 7.18b, write the series for
H2
at any
point
r, 6
with
a, when po~
=
a.
r > a.
7.18e A Helmholtz coil is used to obtain very nearly uniform magnetic field over a region
through the use of coils of large radius compared with coil cross sections. Consider
two such coaxial coils, each of radius a, one lying in the plane 2
d and the other in
-d. Take the current for each coil (considered as a single turn) as 1.
the plane 2
=
=
Obtain the series for
H: applicable
to a
region containing
forms for the first three coefficients. Show that if
(other than the
7.19 In
constant
a
=
the
origin, writing specific
2d, the first
nonzero
coefficient
term) is the coefficient of r4.
Eqs. 7.19(3) and (4), let 1,11 be the axial electric field component BE, and simplify by
taking A and C zero in (4). Discuss the forms of solutions and the question of finding
physical boundary conditions for (i) both kx and ky real, (ii) kx real but It), imaginary,
and (iii) both kx and
ky imaginary. For (ii) and (iii) would physical applicability of
solutions be changed if either or both of A and C were nonzero?
8.1
A
waveguide
is
a
structure,
chosen direction with
or
part of
some measure
direction of propagation. If the
INTRODUCTION
structure, that causes
of confinement in the
a
to
planes
transverse to the
a
change direction, within reason—
example, in a transmission line
used to transfer energy from a transmitter to an antenna, the energy follows the path of
the line, at least for paths with only small discontinuities. The guiding of the waves in
all such systems is accomplished by an intimate connection between the fields of the
wave and the currents and charges on the boundaries or by some condition of reflection
at the boundary.
In this chapter we concentrate on cylindrical structures with conducting boundaries.
Multiconductor lines can be used for frequencies from dc up to the millimeter-wave
range. At the highest frequencies, they are often in the form of metallic films on insu»
lating substrates. Hollow conducting cylinders of various cross-sectional shapes
are used in the microwave and millimeter-wave frequency ranges (approximately
l-100 GHz).
Generally, in waveguide analyses we are interested in the distribution of the e1ectro~
magnetic fields; but of greatest importance is the dependence of the propagation con~
stant upon frequency. From the propagation constant one finds wave velocities, phase
variation, and attenuation along the guide and the pulse dispersion properties of the
guide.
Several different types of guides are analyzed in this chapter, including the simple
parallel—plate structure (and some more practical, related forms) and hollow-tube guides
of rectangular and circular cross section. An introduction to means for exciting waves
in waveguides is presented. The chapter concludes with a study of the general properties
of waves in cylindrical waveguides with conducting boundaries.
able limits, the
wave
waveguide
boundaries
propagate in
a wave
is constrained to follow it. For
395
395
Chapter
8
Waveguides
with
Cylindrical Conducting
Boundaries
General Eormutation for fiuided Waves
8.2
BASIC EQUATIONS AND WAVE TYPES FOR UNIFORM SYSTEMS
consider here cylindrical systems with axes taken along the z axis. We also consider
time-harmonic waves with time and distance variations described by 60“” 7‘"), as in the
We
"
study of transmission—line waves. The character of the propagation constant y tells much
about the properties of the wave, such as the degree of attenuation and the phase and
group velocities. The fields in the wave must satisfy the wave equation and the boundary
conditions. We will assume that there is no net charge density in the dielectric and that
cog/ls
any conduction currents are included by allowing permittivity and therefore k2
to be complex. The wave
equations, which reduce to the Helmholtz equations for phasor
fields (Sec. 3.11), are
=
The three-dimensional
V2
may be broken into two parts:
VZE
The last term is the contribution to
is the two-dimensional
term
VfE
:
62E
+
7
62‘
V2 from derivatives in the axial direction. The first
Laplacian in the transverse plane, representing contribuplane. With the assumed propagation function
tions to V2 from derivatives in this
e"
“’2
in the axial direction,
62M
62
The
foregoing
wave
equations
2
:VZE
may then be written
WE
Wu
H
H
~ot+fim
m
_W+Em
m
Equations (1) and (2) are the differential equations that must be satisfied in the dielectric
regions of the transmission lines or guides. The boundary conditions imposed on fields
follow from the configuration and the electrical properties of the boundaries.
The usual procedure is to find two. components of the fields, usually the 2 components
of E and H, that satisfy the wave equations (1) and (2) and the boundary conditions;
then the other field components can be found from these by using Maxwell’s equations.
To facilitate finding the other components, it usually is most convenient to have them
explicitly in terms of the 2 components of E and H.
The curl equations with the assumed functions 60”" 7’” are written below for fields
Basic
8.2
in the dielectric system, assumed here to be linear,
homogeneous,
VXE== ~ja),uH
a___E_.
6y
~in-
—7"
as
“A
=
“(cusEt
(6)
~m—4=ma
6x
a)
6H,
+
6y
'yHy
(4)
=
“1‘60qu
(5)
analysis to
only, by
and y
x
functions in the assumed e‘jw’
'
6H_
.
"MW,
It must be remembered in all
functions of
.
—:
(3)
=
6y
isotropic:
jweE
BE.
r"
an
so on, are
V X H
~JwiLHt-
==
8E,
-
8E.
:1
7E
and
=
.
+
397
Equations and Wave Types for uniform Systems
‘
95—wi='E
””8
ax
a)
ay
follow that these coefficients,
our
agreement
to take care
EX, Hx, By,
of the
z
and
and time
73).
foregoing equations, it is possible to solve for E, E), Hx, or Hy in terms of
and
H:. For example, H, is found by eliminating E), from (3) and (7), and a similar
E:
procedure gives the other components.
From the
1
E"
2“-
E,.x
=
.
6E:
7/
-‘”—'—_~,)‘
7,2
+ k“
6}:
,
+ J wM
6E-
1
'Y'+/€'<
-—--
——i
——
—-—-----:,~
ray
Jwtbax)
——-‘
10
()
8H.
8E-
1
(9)
ay
6H.
‘
+
6H:
~—-—-
an
m=———;MVeey—=
+ k“
6x
'
6y
'y
1
H,"
For
propagating
real if there is
no
:
8E:
——
7/2
8H.
(1608+ 6y)
’
——-————-—,)
+ 16'
ax
+7]
substitution 3/
jB where [3 is
the above with this substitution,
waves, it is convenient to use the
attenuation.
Rewriting
( 12 )
——
=
BH.
‘2’? (‘3‘— Hi)
EX
“3)
'
6H.
B3— War)
if: (-35
(14)
'
6H.
=
(we‘— ta?)
k3; (we
Ely)
"
2:
j
3H.
6E,
63:
”5)
'8
(
)
va— *kEE:
(17)
vm~ ~16in
(18)
—-
u
——
398
Chapter 8
Waveguides with Cylindrical Conducting Boundaries
where
k§§72+k2=k2~—,82
In
studying guided waves along
following types:
(19)
uniform systems, it is
common to
classify
the
wave
solutions into the
nor magnetic field in the direction of propagation. Since electric and magnetic field lines both lie entirely in the transverse
plane, these may be called transverse electromagnetic ( TEM) waves. They are the
usual transmission~line waves along a multiconductor guide.
1. Waves that contain neither electric
magnetic field in the direction of propa—
entirely in transverse planes, they are known
magnetic
waves.
They have also been referred to in the litermagnetic (TM)
2. Waves that contain electric field but
gation.
as transverse
ature as E waves, or waves
3. Waves that contain
gation.
These
referred to
4.
Hybrid
as
waves
are
no
field lies
Since the
of electric type.
magnetic
known
H waves,
field but
or waves
for which
may often be considered
no
as transverse
of
electric field in the direction of propa(TE) waves, and have also been
electric
magnetic type.
conditions
boundary
as a coupling
require
all field components. These
by the boundary.
of TE and TM modes
is not the only way in which the possible wave solutions may be
useful way in that any general field distribution excited in an ideal
guide may be divided into a number (possibly an infinite number) of the above types
with suitable amplitudes and phases. The propagation constants of these tell how the
The
preceding
divided, but is
individual
a
waves
change phase
and
amplitude
as
they
travel down the
guide,
so
that
they may be superposed at any later position and time to give the total resultant field
there. Since it is disadvantageous to have a signal carried by several waves traveling at
different velocities because of the resultant distortion,
signed
so
that
only
one wave can
prepagate
even
waveguides
if many
are
are
normally
de-
excited at the entrance to
guide. As we shall see in Chapter 14, multimode guides are sometimes used in
optical communications; they have also been used in “overmoded” millimeter-wave
systems, but the single-mode guide is the norm.
the
””77
’2'“
(g‘h“if?“<7“YT?"
*‘dmm“
t7?”sf7n*‘Wnth}?:3?Weir"?!lW’fi‘TBKf}“WWW“"T
""T
:6’3‘} (“a—7%"?$2533.Ti???"’WJ?“
“‘9’35a;1?‘» PM
3‘35“fiiflfw
xii-T31W52%iii.
gyiindricai Waveguides at Various cross Sections
8.3
WAVES GUIDED av PERFECTLY CONDUCTING PARALLEL PLATES
simplest wave-guiding systems for analysis is that formed by a slab of
parallel-plane conductors on top and bottom. The fields are assumed to
same as if the plates were of infinite width, which means that any edge effects
One of the
dielectric with
be the
Waves Guided
8.3
or
other variations
ysis
along
of this model. We
by Perfectly Conducting
one transverse
in
coordinate
5 that such
are
399
Parallel Plates
neglected for
a
first-order anal—
system may be considered a twoconductor transmission line, with upper and lower plates acting as the two conductors
of the line. But we shall see that the system also guides waves of other types. Analysis
saw
Chapter
a
of this system helps in the understanding of all the wave types before going on to more
complicated boundaries for the guiding system. We consider the three classes of waves
defined in the
section.
preceding
'E'EM Waves
The transverse
electromagnetic
have neither
waves
EE
nor
HZ.
From
that all transverse components must be zero also unless
Eqs. 8.2(9)——(12)
+
k2
The
O.
72
propagation constant for a TEM wave must then be
we
see
=
That is,
is with the
propagation
velocity
ijk
=
YTEM
of
light
(1)
in the dielectric medium. Since this
argument does not make use of the specific configuration of parallel planes, it applies
to TEM waves in any shape of guide, as will be discussed more later. Moreover, if
7/2 + k2 is zero, we see from Eqs. 8.2(1) and 8.2(2) that both electric and magnetic
fields
satisfy Laplace’s equation
that both have the
so
spatial distribution of two—diparallel~plane
mensional static fields. It is known that the static electric field between
conductors is uniform and normal to the
planes
E.
and
magnetic field,
from
Eq. 8.2(4)
H‘,
l—E
ICU/«b
=
'
that
E:
:1
may write
(2)
E0
=
=
O, is
'wV p‘ 8
==
we
.
10).“
E
=
if?
(3)
E
M
positively traveling waves and the lower sign for negatively
interpreted in terms of voltage and current, we find the same
obtained from the transmission-line analysis.
where the upper
traveling
waves.
results
those
as
with
so
sign
is for
When
TM Waves
Transverse
Eq. 8.2(17).
The transverse
magnetic waves have finite E: but no HZ so we may use
Laplacian is taken as dz/dxg because of the neglect of
the y derivatives:
dzE:
dig
==
-—-kZ:E
<4)
fi=fi+fi
Solution of
(4) is in sinusoids:
E:
Boundary
these
are
requires
=
=
A sin
are
(6)
kcx
next
=
=
as
O.
+ B cos
kcx
0 and a. Since
applied at the conducting planes at x
at
x
O in (6)
0
there.
O
Placing E3
perfectly conducting, E:
O at x
in
The condition E2
a then requires either A
which
O,
conditions
taken
B
m
=
=
=
=
=
490
Chapter
case we
have
Waveguides
8
nothing left,
sin
or
kca
So
a
solution for
E2
kca
==
Cylindrical Conducting Boundaries
with
0, which is satisfied by
=
which satisfies
boundary
A
=
mm:
(8)
81117
.—
to
The transverse field components are now obtained from
O:
0 and a/ay
8.2(12), letting HZ
E"
=
3/ (IE.
----—-”=
kg
H
::
jcos
seen
that there is
guide (in
Let
a
an
=
dE-
=3
—
dx
E
O,
COS
of
Eqs. 8.2(9)
a
jcoea
mm
A
( 10)
CO S
mw
2
y
(9 )
——
0
0
(11)
infinite number of solutions for the various
different field distribution. These solutions
this case, TM
us now
means
mm
ya
—~---A
m'n‘
“—4
kfi
Hx
=
dx
—-—-—
3’
m, each with
E3 by
'2
=
It is
(7)
conditions is
,
E,
1, 2, 3,...
=
m
mar,
are
integral
values of
called the modes of the
modes).
examine the
properties
of the
propagation constant. Solving (5)
for 'y
we
have
2
“y
At
=
VICE
sufficiently high frequencies,
and
we can
rewrite
(12)
—
k2
=\/<Z-7:1-1T)
—-
wzne
the second term in the radicand is
(12)
larger than
the first
as
7=jB=jthbe
(mu/a)2
1—-
'7
(13)
corpse
see that the phase constant approaches that for a plane wave as frequency approaches infinity. Lowering the frequency reduces B and it goes to zero at the frequency
We
mm)
m'n'
wc
:
2:
a
V #8
a
(14)
velocity of light We)“ 1/2 in the given material. We call cuc the cutoff
frequency since pr0pagation takes place only where B is real. The variation of 3 with
w is often
plotted as in Fig. 8.3a for the reasons discussed in Sec. 5.12. It is conveniently
expressed in terms of the cutoff frequency
where
v
is the
2
a)
Waves Guided
8.3
40'!
by Perfectly Conducting Parallel Plates
a,
¢E.Q
FIG. 8.3a
Phase constant and attenuation
as
functions of
frequency for
a
TM
or
TE
waveguide
at
the cutoff
mode.
The cutoff
point
may
usefully
expressed in
be
terms of the
wavelength
frequency
2772)
)LC
z
-—
2a
=
(16)
--
m
coc
That is, cutoff occurs when the spacing between plates is m half-wavelengths, measured
at the velocity of light for the dielectric material. We see the important feature that TM
waves can
propagate above
than the cutoff
a
wavelength);
cutoff frequency (or,
at
lower
frequencies
equivalently,
wavelengths
given by
at
y is real and is
shorter
(1)2
”177'
y:a:~—~1—-<—*>,
a
a)..<_.wc
(17)
a)c
Fig. 8.3a there is attenuation without phase shift for frequencies below the
cutoff frequency of a given mode, phase shift without attenuation for frequencies above
cutoff, and neither attenuation nor phase shift exactly at cutoff. Each of these modes
then acts as a hi gh-pass filter; the attenuation below cutoff, like that for a loss—free filter,
is a reactive attenuation representing reflection but no dissipation for this nondissipative
As
seen
in
system.
For the
propagating regime,
w
> we, we can
define
a
phase velocity
in the usual
way:
v1)
Group velocity
can
=
3-)
B
,—————-———-U
=
be derived from this
1
as
,7
—
in Sec. 5.15:
co2
dw
U
_.___.
g
(18)
(we/0))“
v
1..
<)
__<2
1
(9)
482
Chapter
Waveguides with Cylindrical Conducting Boundaries
8
velocity is always greater than the velocity of light in the medium and group
velocity is always less, both approaching each other and v at frequencies far above
cutoff. A wavelength along the guide, Ag, may also be defined as the distance for which
phase shift increases by 272',
Phase
A
where A is
wavelength
of
a
A
27:
—
g
3
plane
(20)
*-
V1
(we/co)2
--
in the dielectric
wave
A
:
medium,
2
3
(21)
w
The ratio of transverse electric field to transverse magnetic field of a single propagating wave may be defined as a characteristic wave impedance and is useful for certain
types of reflection problems, much as the field impedance for plane waves was found
to be in Chapter 6. For the TM wave, this impedance, from (9) and (10) with 77
=
(.u/ (9)1”,
is
E
s
Z TM =—"=—= 7?
Hy
Note that this ratio is not
cutoff,
so
that
It is real for
Poynting
frequencies
a
a
1—-
(08
function of x and y. It is
calculation will show
no
above cutoff
a
so
that
(02
(w)
—-~"3
22
()
imaginary
for
frequencies
below
average power flow in that regime.
Poynting calculation in that regime
shows finite average power carried by the wave.
A plot of electric field lines in a TM1 mode from the
equations for Ex and E: is shown
Fig. 8.31) (Prob. 8.30). Note that induced charges on top and bottom plates are of the
same sign in a given 2 plane for this wave, in contrast to the sign relations found for
in
Electric field lines
starting on these charges turn and go axially down
guide, ending
charges
sign a distance a half guide wavelength down
the guide. The entire pattern translates along the guide at the phase velocity for a
traveling wave in one direction.
the TEM
wave.
the
of opposite
on
fil -t
1::_.%
”l
,§\
0
kW
1
fl
7r/2
.
it
37/2
11'
fiz———>—
FIG. 8.3b
Electric field lines of
TM1
wave
between
plane
conductors.
m :1
Waves Guided
8.3
TE Waves
by Perfectly Conducting
The transverse electric class of
waves
has
403
Parallel Plates
nonzero
Hz
but
no
E2. Equation
8.2(18) is then utilized:
d2H_
VZH.
==
—-k2
=
2
-
ki=v2+k2=k2~32
The solution will
term
again
is retained since
at the
From
be written in terms of
Ey, proportional
perfectly conducting plane
Eqs. 8.2( 13)
to
0:
Hz
=
B
sinusoids, but this time only the cosine
derivative of Hz with x, must become
to the
=
x
kcx
cos
1'3
=
E),
=
14;.
=
the
conducting plane x
=
zero,
B sin
HI
E
zero
(25)
8.2(16), remembering that E: is
Jé £512.
kg dx
(24)
(26)
kcx
.
.
dH_
Also,
E), must be
some
of
multiple
zero at
qr/a.
1:.” E: ~wa
2
0,
H),
B sin
kcx
(27)
0
=
(28)
=
a, so
kC
is determined from (27)
As with the TM wave, this is identified from
(24)
as
as
the value
of k at cutoff:
”2W
kc
The
propagation
=
27TfCV/.L8
constant from
-£
H
=
m
—-,
a
(24) may
II
l, 2, 3,...
(29)
then be written
a=<m) 1~<3>,
(3-),
a
.y
=
w<wc
(30)
w> avg
(31)
a)C
1—
jfi =jk
to
The forms for attenuation constant in the cutoff range and phase constant in the prop—
agation range are thus exactly the same as for the TM waves (Fig. 8.3a), and by (14)
and
(29) the conditions for cutoff
same
in the
Wave
TE
For
are
the
same
for TE modes
expressions for phase velocity,
propagation range follow from (31) and
or field impedance for the TB wave is
order. The
frequencies
Ev
jCU/l:
Hr
,y
below cutoff this
as
for TM modes of the
group
and
are
same as
velocity,
exactly the
guide wavelength
(18) to (20).
'
wave
7‘]
1/1
impedance
(
is
c/:))2
imaginary,
but for
frequencies
494
Chapter
Waveguides with (Zylindricai Conducting Boundaries
8
FIG. 8.3::
lines of TEI
field
Magnetic
always greater than
always less than n.
above cutoff it is real and
for TM waves, which is
wave
as
n,
between
plane
conductors.
contrasted with the
wave
impedance
The form of the field lines for the first-order TE mode is indicated in
Fig. 8.36. Here
surrounding the y-direction displacement
current. There is no charge induced on the conducting plates and only a y component
of current corresponding to the finite Hz tangential to the plates.
the
magnetic
field lines form closed
curves
1-451'1::-,;':i\4;l.L-;n'::-.;'.: .: ------
Exampre 8.5
INTERPRETATION OF GROUP VELOCITY AS AN ENERGY VELOCITY
Let
us
define
velocity
a
of energy flow in terms of energy stored per unit
length
and
average power flow:
vE
2
WT
( 33 )
~—
u
Take the TM
ing
example. Average power flow
wave as
theorem for
a
width
a
WT
=
w
I
2
Re(E1H3’f)dx
'
2
I
=
Time~average
complex Poynt—
1
-
o
is found from the
w:
£-
lAzfiwe
2
a
"
f
mm“
cos2
)
ma
m7”
dx
energy storage per unit
“'
"
:1
4
a
o
length, including
(34)
7
A?
2
,
mm?"
Bros
both electric and
magnetic
parts, is
a
__
u
Using
the fields from
=2
—-—
4
Azwas
l
g...
8
E
2
+
2
lEzl]
+
[*2
4
W
2
}
dx
.
(35)
(8)-—( 10) this is
{[
[
8
o
...._
IO {4 [lExl
a
Azw
u
14 ,
-
.
sm
2
mm:
+
a
[3202
Bzag
9
,,
”2‘77"
kza2
]
’J
‘7
cos
”1711‘
~—
(1
+
'J
')
“(0'8“0”
——
2
”2 ‘77
.,
'7
COS“
fil’lTX
dx
Guided Waves
8.4
But from (13),
17127r2/ a2
B2
+
is
405
Superposition of Plane Waves
as
just k2,
so
this reduces to
’)
7
’7
A"wask‘a"
2
11
(36)
~-—-———
47722772
and energy
velocity from (33)
is
L
var—9?:
k"
1——
which is
the
93-92
(37)
a)
[1’8
expression (19) for group velocity. The same equivalence
We will say more about the use of these velocities for signal
at the end of the chapter.
exactly
is found for TB
propagation
same as
waves.
GUIDED WAVES BEn/VEEN PARALLEL PLANES AS SUPERPOSITION
OF PLANE WAVES
8.4
The modes of the
parallel-plane guide, studied in the preceding section from the wave
equation, can also be found to be superpositions of plane waves propagating at various
angles. This picture is very helpful in developing a physical feeling for the different
types of modes. The TEM wave is clearly just a portion of a uniform plane wave
polarized with electric field in the x direction and propagating in the z direction:
Ex(x, z)
The
magnetic
field for such
=
Eoe
E
H,,(x, z)
=
(1)
6.2, is
wave, from Sec.
a
‘ka
—9
.
er!“
(2)
'
n
where k
GOV/1;.
=
These
6 from the normal,
waves
only
strike
at
that
are
of the
same
form
as
Eqs. 8.3(2)
8.3(3).
as
we
angle
plane
conducting plane at an angle, tangential electric field must be zero not
plane, but also at other planes which are multiples of a half»wavelength
a
/\
/
/
-
/
\
_
_
\/
Diagram showing
or
\
\\ /74
AL
component of TM
9
\
/
\
TE
\Va/LA
/
\
/
FIG. 8.4a
and
superpose uniform plane waves propagating at an
pictured in Fig. 8.40. It was found in Sec. 6.9 that when
For TM and TE waves,
/
/P
//\ \
\
_
fl
\
/_
Wavefront
\\ //V\Ray direction
A!
ray directions and wavefront direction of
between parallel planes.
waves
z
/
\
a
uniform
plane—wave
406
Chapter
measured at
conditions
Waveguides with Cylindrical Conducting Boundaries
8
phase velocity in the
satisfied if spacing a
are
direction normal to the
between
plates
plane.
That is,
boundary
is
m/‘t
a
(3 )
=
2
cos
6
or
m/‘t
A
2a
AC
w
( )
COS
to
where
277?)
AC
2
mm
(5)
C
The
phase
constant in the
z
direction is
3=ksm6=kV1~00326
or,
(6)
using (4),
2
B=k1—<fl)
m
(I)
exactly the same as found from the detailed analysis in Sec. 8.3. Since ,8 is the
phase and group velocity will be also. We see from Fig. 8.4a that phase velocity
is the velocity of the imaginary point P of intersection of the plane-wave fronts with
the z axis, and is greater than velocity v of the plane wave in the medium, as found in
Sec. 8.3. Group velocity is the component of v in the z direction and is always less than
I). At cutoff (Fig. 8.4b), the waves simply bounce laterally back and forth between
planes without any forward progression so that group velocity is zero and phase velocity
infinite. For frequencies much above cutoff, the angle is very flat (Fig. 8.40) and the
wavefronts are nearly normal to 2, so that both up and vg approach v.
This is
same,
Both TM and TE
wave
types show the behavior described above. TM
waves corre~
spond to the plane waves polarized with electric field in the plane of incidence,
waves, to those polarized with magnetic field in the plane of incidence.
/
.—-> r+—
-.
—
>2
—————‘
//
///
3
,//
“
ET
and TE
\\
\\
'——”Z
Wavetront
9
*1
(b)
(0)
FIG. 8.4 ([7) Special case of (a) when wave is at cutoff. (0) Special case for frequency far above
cutoff so that 6 —> 77/ 2 and wavefront is nearly normal to guide axis.
Parallel-Plane
8.5
Example
DETAILED FIELD EXPRESSIONS FOR
497
Guiding System with Losses
T|\/lm
8.4
WAVE FROM PLANE—WAVE COMPONENTS
To show that this
point of View is exactly equivalent to that of Sec. 8.3, consider the
expressions given by Eqs. 6.9(11)—-(l3) for a wave polarized with electric field in
the plane of incidence striking a conductor at an angle. We rewrite these with the
coordinates x and z primed, since we take different coordinate systems from those in
field
Sec. 6.9:
Extx', 2’)
=
~2jE+
cos
E:.(x’, 2')
=
~2E+
Sin 6
2’)
=
77H),(x’,
To
9
6
2E+ cos(lcz’
sin(kz'
cos(kz’
cos
cos
cos
(8)
memes“
(9)
6)e‘j’°"'sm'9
(10)
z and z’
change to our present coordinate system, let x’
k
6
and
from
cos
(3). Equations (8)-(10)
mrr/a
,B by (6),
2-—
=
=
==
Z'E
E_(z, x)
=
‘
.‘
=
nHV(z, x)
=
we
.
(11)
.
,
(12)
cos
k
-
Also let k sin
a
2E
£42,1-)
——x.
then become
sin(m7n)e—Jfiz
J
(m)efiflz
cos<m)e'fl33
er
ka
‘
If
waft-“tin"
a
2E+
(13)
a
multiplier (2jE +mqr/kaz) equal to a constant A and remember that the
e~le2 has been suppressed in Eqs. 8.3(8)—-(10), we find that
identical to Eqs. 8.3(8)—(10).
next set the
propagation
(ll)—~(l3)
factor
are
8.5
PARALLEL-PLANE GUIDING SYSTEM WITH LOSSES
filling the region between conductors of the parallel—plane guide is
ja" (Sec. 6.4) in the
may replace permittivity by its complex forms 8’
imperfect,
the
with
effect on the propwe
are
concerned
Sec.
Here
8.3.
of
primarily
expressions
have
we
For
TEM
wave
constant.
the
agation
When the dielectric
-
we
”YTEM
If
e”/e’
is small in
comparison
=
with
ja)
V
unity,
#«(3'
a
—
IS")
binomial
expansion
(1)
shows that the real
part of this, which is the attenuation constant, is to first order
(UN
(ad)TEM
E
/
lug! 8”
2
8,
(2)
408
Chapter
For TM and TE
8
waves
Waveguides with Cylindrical Conducting Boundaries
from
Eq. 8.3(12)
2
m,”
=
7TM,TE
[(7)
"
6021148,
je")]
“
1/2
1/2
2
m
2
-
2
II
2
The second
expression
attenuation constant
is obtained
(real part) for
by making
a
binomial
expansion
of the first. The
a)c is then
a) >
m
( “d) W
a
-1
V
_
us’(8”/e')
(4)
“
MW
gives a correction to phase constant, which may be
considering the dispersive properties of the guided wave.
An exact solution is considerably more difficult to obtain when the finite conductivity
of the conducting boundaries must be considered. The approach is to obtain field solutions in both dielectric and conductor regions, with proper continuity conditions applied at the boundary between them. This approach cannot even be carried out for some
geometrical configurations, although it can for this simple system and shows that the
approximations to be used in the following analysis are justified so long as the plane
boundaries are made of much better conducting material than the intervening dielectric
region.1 The expression to be used is that giving attenuation in terms of power loss per
unit length and average power transferred by the mode, Eq. 5.1 1(19). This is an exact
expression, but the approximation comes in by calculating power transfer as that of the
ideal guide, and loss per unit length as that from currents of the ideal guide flowing in
Retention of second-order terms
important
in
the real conductors. For the TEM wave, the power transfer for
(WT)TEM
“’2
a
width
w
is
awEg
(5)
21)
The average power loss per unit area in the plates, if plates are thick
skin depth, is $4?le52 2, so for a unit length and width w, counting both
E
wRS
compared
plates,
with
2
77
Attenuation constant is then calculated
as
:7.
This is the
same as
wL
—_
2
Rs
~—
7
that which would be obtained from the transmission-line formula,
Eq. 5.1102).
‘
precisely, it is required that displacement current in the conductor be negligible
comparison with displacement current in the dielectric, and conduction currentin the
dielectric be small in comparison with conduction current in the conductor.
More
in
Parallel-Plane
8.5
For the TM
of order m, average power transfer from the
w > we is
wave
409
with Losses
Guiding System
Poynting theorem
and
the results of Sec. 8.3 for
C1
(WT)TM
=
w
1
J
‘
5 (EH?)
0
a
w
=
—
2
.
.
aA
mm
<__1_@__
f
wws
______B___f
7722 772
A
The current flow in both upper and lower
dx
a
2A2
w
—-
22722772
0
0
.
cos
mrr
9m
cos~
mm:
e‘Jfl‘) (J
6132)
fldx=(wel3a )g
_
a
a
(12142
wea
.
——
cos
m7r
0
2
dx
(8)
is
plates
msaA
IJszl
z
So average power loss per unit
(WL)TM
iHyix=O
length
for
2 R5
“g
=
iHy|x=a
:
width
a
(9)
:
”277'
w
is
A
I153?
By
a
similar calculation
wL
:
——
=
we
2st8
2RS
naVI —<wc/w)2
find attenuation for
(0:c ) TE
=
approximately
=
[3a
ZWT
(10)
”MT
The attenuation constant from conductor losses is then
( ac ) W
3
sz<W )
=
a
(11)
TB mode to be
2Rs(a)C/a))2
1 ‘(wc/w)?‘
Tia
(12)
'
Curves of normalized attenuation
several
interesting
features of these
versus
frequency
curves.
are
Fig. 8.5. There are
expressions (11) and (12)
shown in
Note first that
3.0
.
a
8
3‘3
TM1
mode
G
g
1.0
TE;
3
mode
E
Q)
0:
O
2.0
1.0
3.0
4.0
///c
FIG. 8.5
Attenuation
curves
of
waves
between
imperfectly conducting planes.
4i 0
Chapter 8
Waveguides
with
Cylindrical Conducting
Boundaries
approach infinity as a) —> me, but the approximations break down at cutoff so that
attenuation is actually finite although high there. The curve for the TM wave starts to
decrease with increasing frequency above cutoff, but reaches a minimum (at w
Sara) and thereafter increases with increasing frequency because of the increase of RS
with frequency. The curve for the TB wave is always lower than that for the TM and,
=
gnoreover, continues
cause
to decrease with
the currents in the conductors
component of field approaches
stantially
11on
to the
axis,
Several different forms of
a
the
y~directed
high frequencies
explained in Sec. 8.4.
zero at
as
8.6
increasing frequency.
are
This behavior arises be-
currents related to
as
H2,
and this
the wavefront becomes sub
PLANAR TRANSMISSION LINES
wave-guiding
dielectric substrate have found
use
structures made from
parallel
metal
strips
in microwave and millimeter~wave circuits
as
on
well
high—speed digital circuits.2 In this section we will examine in some detail three
types, called stripline, microstrip, and coplanar waveguide. Emphasis will be on the
lowest-order mode, which is a TEM wave in the stripline and a quasi~TEM wave in
microstrip and coplanar waveguide.
in
as
stripline consists of a conducting strip lying between, and parallel to,
conducting planes, as shown in Fig. 8.6a. The region between the strip and
The
Stripiine
two wide
the
planes
and
more
is filled with
than
one
a
uniform dielectric. Such
conductor,
can
support
a
a
TEM
structure, with
wave.
If the
a
uniform dielectric
strip
width
w
is much
greater than the spacing d and the two planes are at a common potential, the structure
is roughly approximated by two parallel-plane lines connected in parallel. More precise
results
are
velocity
=
up
is
capacitance per unit length. For a TEM wave, the phase
(#3)" 1/2, and from the transmission-line formalism it is also given by
Then the characteristic impedance is
found from the
Up
=
(LC)'“1/2.
L
20:
VLC
VMS
23:?"
c
(1)
e and [.L the characteristic impedance can be found from
capacitance, which can be determined in a number of ways, as we saw in Chapters
1 and 7. An approximate expression for the characteristic impedance of the stripline,
assuming a zero~thickness strip, has been found by conformal transformation3 to be
Thus, for systems with uniform
the
Z0
2
3
zflJL
4mm)
(2)
T. Itoh (Ed), Planar Transmission Line Structures, IEEE Press, Piscatawoy, NJ, 7987.
Theory of Guided Waves. 2nd ed. Sec. 4.3, IEEE Press, Piscafawoy, NJ,
7997.
R. E. Collin. Field
4'5 1
Planar Transmission Lines
8.6
i
.
FIG. 8.6
where n
2
V
n/s,
k is
1:;-;:-.\".-< 3 <‘,-:-".-"-.............
(a) Stripline. (b) Microstrip.
given by
*1
k-
cos
——
2-d—
h<W
(3)
and
K(k) is the complete elliptic integral of the first kind (see Ex. 4.7a). A
approximate expression for Z0, accurate over the range w/ 2d > 0.56, is4
20
Approximate techniques
Hoffmann.4 The velocity
convenient
m
~
8 ln[2
(4)
exp(7rw/4d)]
used to obtain corrections to
Z0
for t
>
O
discussed
by
depend
(M8)_1/2
propagation
as for all TEM waves. Both velocity of propagation and characteristic impedance,
neglecting loss, are independent of frequency and may be used up to the cutoff fre»
quency of modes between the ground plates, Eq. 8.3(14).
An approximate expression for attenuation resulting from conductor surface resistivity RS is3
__
ac
which is valid if
w >
—
of
does not
lei
[aw/2d
+
1n 2 +
2nd
4d and t <
in footnote 4. Attenuation from
on
1n(8d/m)
7rw/4d
]
(1/5. Approximations
lossy
dielectrics is
are
thickness and is
“Wars/m
(5 )
for other dimensions
exactly
as
in
are
given
Eq. 8.5(2).
Microstrip Probably the most widely used thin-strip line is one formed with the strip
lying on top of an insulator with a conductive backing. A different dielectric (usually
air) is above the insulator and strip (Fig. 8.6b). In such an arrangement, there cannot
be a true TEM wave. The reason is that such a wave, as shown in connection with Eq.
O. The propagation constant 3/ is a single quantity for
8.3(1), requires that y: + k2
2r-
the wave,
this
so
problem
7/2
is
+
k?
and
y?
+
complicated by
k3
cannot
lectrics. As mentioned above for the
microstrip is
4
a
zero
if
strip
k1
#
k2.
Exact solution of
and the two different die-
stripline, the lowest—order approximation for the
guide neglecting fringing. A much better, though
section of parallel-plane
R. K. Hoffmann, Handbook of Microwave
MA, 7987.
both be
the finite width of the
Integrated Circuits, An‘ech House, Norwood,
4i 2
Chapter
8
Cylindrical Conducting Boundaries
Waveguides with
wave is approximately
plane is nearly the same as
that for static fields. This so—called quasiszatic approach employs calculations made
with static fields to determine the transmission-line parameters for propagation of the
lowest-order mode, even though it is not a pure TEM wave. The approximation is very
useful in practical applications but has some limitations that we will mention below.
A common approach to obtaining simple, accurate expressions is to first find the
characteristic impedance ZOO of an electrode structure identical to the one of interest
but with the strip electrode having zero thickness and the dielectric being free space
everywhere.‘ This problem can be solved exactly by the method of conformal mapping,
but the expressions are very complex and more useful results are the various approxi—
mations to the exact expressions. A particularly useful expression is4
still
a
approximate, approach
TEM
wave so
is to consider that the lowest-order
the distribution of fields in the transverse
0.172
w
_1
w
3[d 9(d) ]
Z00 =77m+1.8-“
(w/d)
which is accurate to <0.3% for all
ance
of the actual line, it is corrected
by
>
0.06. Then to get the characteristic impedefiective dielectric constant eeff,
the so-called
which, if filling the entire space, would give the
structure. Since the inductance
same
capacitance
as
that of the actual
is unaffected
by the presence of a dielectric, correction
the correction for the characteristic impedance.
for the
capacitance gives
approximation to the
conformal mapping methods; one
The static
88ff
which is
6
()
always
z
between
effective relative dielectric constant is also found
of several useful
1+(8r‘1)[1+
—
2
unity
and er.
Zo
we
obtain the static
for
a
approximation
1
—_
V1
Applying (6)
:
Zoo/
V
by
approximations is5
lOd/w
+
and
l
(7)
(7) in
(8)
Seff
for the actual characteristic
of insulator dielectric constants
shown in
impedance. The results
Fig. 8.6c. Corrections for
variety
strip thickness have been published.4
Since the phase velocity in a TEM wave is
(the)— 1/2, the ratio c/vP
Q
up
in the quasistatic case. As frequency increases, the longitudinal field components become increasingly important; this can be represented by a frequency-dependent effective dielectric constant 8650:) to express the variation of phase velocity or phase
constant. Approximate expressions for seffq’) and for attenuation from conductor and
are
nonzero
=
=
dielectric losses have been obtained from numerical calculations
or
empirically.4 Some
results follow:
VseffO‘)
5
=
k1:
VS—r
_._
'
8e’ff(0)
=
H. A. Wheeler, IEEE Trans. Microwave
1 +
41745
+
Theory Tech. MTT-25,
Vseffa»
63] (7977).
(9)
8.6
413
Planar Transmission Lines
300
T
\\\
\{1
\\
200
\N\\
\\\
8
100
A"
\
16
20
377d/w
‘V»\
\
2
\\ ‘L\
\
\\
\\
50
\\
40
\\\\\
X‘\Q\Q\:\\
\
30
20
\\\
rem—>1
d
T
10
0.1
0.2
\
\\
\ \
8r
a
,
0.5
1.0
2
4
6
810
w/d
(C)
FIG. 8.6a Approximate characteristic impedance of stripline. The dashed line is the parallelplane approximation for air dielectric. Adapted from H. A. Wheeler, IEEE Trans. Microwave
T/zeozy Tech. MTT-ZS, 631 (1977). © 1977 IEEE.
where
883(0)
is
given by (7)
and
2
1
4dee,
F=-————
—-
w
0.5+
The
and
(10)
1+21n1+g
C
frequency above which the frequency-dependent effective dielectric constant in (9)
(10) is needed is given empirically by4
21 x
fmax
106
(11)
=
+
(w
2d)\/er
+1
frequency where a higher-order mode
relationship of the higher-order
modes to the quasi-TEM mode is analogous to the relationship of the TM and TE
modes to the TEM mode in the parallel-plate guide (Sec. 8.3), but here the higher-order
modes are hybrids of TE and TM components.
Attenuation from losses in conductors with surface resistance R5 is
be used up to
The
quasi—TEM
can
propagate, which is (fc)HEl
mode
can
=
=
ac
near
the
620/ 277061.
The
RSVSBff(0)A/d
(12)
where the parameter A is given for various strip widths and thicknesses in
a dielectric having a loss tangent tan 68 is
Fig.
8.6d.
Attenuation from
__
as
77f
tan
—
c
5,;
/ar(1
+
2
F1)
[I
1
+
—-
F1
+ F1)
——_—_8r(1
‘1
nepers/m
(13)
4'5 4
Chapter
8
Waveguides with Cylindrical Conducting
Boundaries
0.015
E
0.010
.C
\O
U)
E
o.
(D
5
<1 0.005
0
0.1
1.0
10
w/d
((1)
FIG. 8.6d
Factor for calculation of conductor losses in
8.6(12). Both conductors
are
assumed to have the
same
microsuip using the loss formula, Eq.
RS and thicknesses satisfying t > 35.
curves may require adjustment of w to account for nonzero electrode thickness
See footnote 4 from which these data are taken.
Broken parts of the
t.
[l + (lOd/w)]"1/2. The total attenuation can be taken
(12) and (13), although there are also radiation and scattering losses.
where
F1
=2
Of the several different forms of
gopfianar Waveguide
systems in which all conductors
are on one
surface of
a
to be the sum
of
stripline wave-guiding
dielectric substrate, the most
widely used is the coplanar waveguide shown in Fig. 8.66, in which the signal voltage
is applied between the center strip and the grounded outer strips.
As with the microstrip, the fundamental mode of propagation in the coplanar wave~
guide is a quasi-TEM mode. Because the dielectric is not homogeneous in the transverse
plane, the wave cannot be a pure TEM mode. The distribution of electric fields in the
space above the line is the same as in the substrate (assuming negligible thickness of
the strips and infinite substrate thickness). We assume here that air is above the strips
so 3].
1. If the capacitance per unit length for a line with the same conductors but
l
everywhere is C0, in the actual line the capacitance contributions above and
er
=
:
below the substrate surface will be
stant can be defined for the
CO/ 2
quasistatic
and
limit
C
8 eff
and the
phase velocity
is
up
=
:
.........
C08,/ 2.
Then
an
effective dielectric
con-
as
2
CO
(neosefir 1/2.
ar + l
2
(14)
43 5
Planar Transmission Lines
8.6
iii—Q
mm
77777772)
fave—we
Dielectric
1
(2)
FIG. 8.6a
The characteristic
Coplanar waveguide.
for zero-thickness conductors,
impedance
conductors, and infinite-thickness substrate
liptic integrals
of first~order
ZOO
:
and k'
w/a (see Fig. 8.6e)
(15) (<O.24%
error) that
20
much
are
77'
infinitely
in terms of
wide
ground
complete el—
fl
770
4
(1
=
more
Seff
"
(15)
8effK(k)
k2)1/2. Very good approximations
-
convenient in
1n<2\/g>
#70—
=
:
Seff
"
2--
expressed
K(k) and K(k’):
Z0
where k
be
can
for 0
practice
<
w/a
<
to
are
0.173
(16)
and
ZO
These
Vw
1 +
7m"
—
4V8eff
—1
a
[142—4)]
1
for 0.173
<
Vw/a
-—
w/a
< l
(17)
expressions for effective dielectric constant and characteristic impedance are also
accurate to
within several percent if the substrate thickness d is finite but greater than
the total gap width a. More complex relations are required for thinner substrates.4
One reason for interest in the coplanar waveguide is that dispersion is typically less
than in the
microstrip
has been shown to
(9)
coplanar waveguide
for microwave and lower
give
over a
Veefig)
where
F2
=
2fd V
er
--
a
good
very wide range of
.3
=
1/6
I)
6
\fgr
1
eeff(0)
=
G. Hasnian, A. Dienes, and J. 1?.
738 ( 7986).
=
E;
and
frequencies.
A form
nearly identical to
dispersion in the
fit to numerical calculations of the
"'
parameters.6
V
Seff(0)
+13]:ng
is the value in
exp[u ln(w/s)
+
r]
+
Thus,
Veeffm)
(18)
(14). Also,
(19)
Whinnery, IEEE Trans. Microwave Theory Tech. MTT-34,
4'5 6
Chapter
8
Waveguides with Cylindrical Conducting Boundaries
0.1
E
J:
2
a
0.01
O.
(D
5
‘1
0.001
1.0
0.1
0.01
10
w/d
(f)
FIG. 8.6f
formula, Eq. 8.6(12). Conductors all have the
loss
t >
Factor for calculation of conductor losses in
coplanar waveguide using the general
RS and have thicknesses satisfying
same
38. Data taken from footnote 4.
where parameters
u
and
r
depend
substrate thickness
on
u
x
0.54
r
=
0.43
—
according
0.64q
+
0.015q2
0.86q
+
0.54m]2
to
and
-
in which q= ln(w/d).
Attenuation from conductor losses in the
coplanar waveguide can be found using the
general form (12) with parameterA given by the data in Fig. 8.6f.
Attenuation resulting from losses in the dielectric is found using4
77f
__
ad
V
Seff(0)
1
"‘
1/ 3mm)
—
6
1
u
1/8r
]tan
88
nepers/rn
(20)
only if d/a > 1. More accurate expressions
1.
quasi-TEM solution used here applies up to F2
There are several other varieties of strip-type lines. Two prominent types are the slotline waveguide and coplanar strips shown, respectively, in Figs. 8.6g and h. These are
both versions of two~conductor transmission lines. As with the microstrip and 00planar
waveguide, the lowest mode is not TEM but rather quasi-TEM, because of the different
with
eeff(0) given by (14),
which is accurate
exist for thinner substrates.4 The
dielectrics above and below the conductors.
=
8.7
41 7
Rectangular Waveguides
[//////////Z//[/l
J//////Z/fl/////l
Dielectric
(g)
W
W
Dielectric
0?»)
FIG. 8.6
(g) Slot-line waveguide. (12) COplanar-strip waveguide.
8.7
RECTANGULAR WAVEGUIDES
The most
important of the hollow-pipe guides is that of rectangular cross section. As
Fig. 8.7a, a dielectric region of width a and height 17 extends indefinitely in the axial
(z) direction and is closed by conducting boundaries on the four sides. In the ideal
guide, both conductor and dielectric are loss-free. There can be no transverse electromagnetic (TEM) wave inside the hollow pipe since, as was shown in Sec. 8.3, TEM
in
waves
a
have transverse variations like static
region
electric
bounded
(TE)
by
a
waves can
single
fields, and
conductor. Transverse
exist and will be
static fields
no
magnetic (TM)
below.
analyzed
/
/
//
//
/
/
//
//
//
T
/
//
b
y
x
L
if
FIG. 8.7a
a
fl
i 27/
//
j
7!
Coordinate system for
rectangular guide.
can
exist inside
and transverse
4'! 8
Chapter
TM Waves
Waveguides with Cylindrical Conducting Boundaries
8
Transverse
is
equation governing E2
magnetic waves have zero HZ but nonzero E2. The differential
Eq. 8.2(17), here expressed in rectangular coordinates:
WE:
This
to
equation
BZEZ
=2
solved in Sec. 7.19
was
62E;
6y2
+
6x2
—k§Ez
:
(1)
of variables
by separation
procedures and
found
have solutions of the form
E2
(A’ sin lgx+ 3'
=
cos
sin
kxx)(C’
+ D’ cos
kyy
(2)
kyy)
‘
where
ki
The
perfectly conducting boundary
Similarly the ideal boundary at y
k?
=
(3)
O to produce E2
requires 3’
let A’C’ be a new
D’
0.
We
requires
at x
0
=2
k3
+
0
=
==
=
=
0 there.
constant
A and have
E
Axial electric field
(except
A sin
=
must also be zero at
Ez
for the trivial solution A
=
0) if kx
kxa==mm
Similarly,
to make
E3
zero
at y
=
b,
kyb
found from
k6
=
(l)
Since
k3
frequency
is
of
k2
a
—-
B2
given
quency have the
same
4
/,.L8
an
and y
=
b. This
integral multiple
of
can
only
be
l,2,3,...
magnetic
(5)
multiple
wave
so
77:
of
7T:
(6)
with
m
variations in
x
and
n
(3):
—
.‘
is
a
must also be a
1
m.n
::
CHI."
a
(4)
kyy
n:1,2,3,...
So the cutoff condition of the transverse
(designated Tan) is
=
x
m:
kyb=n7n
in y
sin
kxx
/#8
[(mn'>2 <n7r)2]1/2
_______
+
(7 )
_
b
a
Eq. 8.2(19), attenuation for frequencies below the. cutoff
phase constant for frequencies above the cutoff freforms as for the parallel-plane guiding system:
as
in
mode and
2
w
a
=
B
=
kc“
l—
< >
a)c
k
1
—
w<
mom,"
(8)
>
wem.“
<9)
2
(——)
w
,
CO
Phase and group velocities then also have the
8.309)].
,
same
forms
as
before
[Eqs 8.3( 18)
and
The
with
4i 9
Rectangular Waveguides
8.7
remaining field components of the Tan
O and EE from (4):
H3
wave are
found from
Eqs. 8.2(13)——(16)
=
Ex
__
E),
=
Hx
:
—-
1'ka
k2
'Bk,"
—-ch2
jwsk‘,
‘
cos
er
Sin
kyy
(10)
A sin
kAx
cos
kyy
(ll)
_
'
H),
.
A
——
A
Sin
k3;
jwekx
._
—-
—-T
CHI."
kg
(12)
kyy
cos
_
A
[9.1
_
cos
3111
(13)
kyy
where
kx, ky, kc,” and ,B are defined by (5), (6), (7), and (9), respectively, and all fields
multiplied by the propagation terms, ("”33 Plots of electric and magnetic field lines
in the TM11 and TM21 modes are shown in Table 8.7. Note that electric field lines
(shown solid) begin on charges on the guide walls at some fixed 2 plane, turn and go
axially down the guide, and end on charges of opposite sign a half-guide wavelength
down the guide. Magnetic field lines (shown dashed) surround the displacement currents
represented by the changing electric fields as the pattern moves down the guide with
velocity up. The pattern for the TM21 mode is that of two TM11 modes side by side and
of opposite sense.
The attenuation resulting from losses in the conducting walls can be calculated for
the TMM wave following the procedure used to find Eq. 8.5(11):
are
( ac) TM
2R5
__
1
bn
——
[1722(b/a)3
(J‘s/ff Mai/602
222]
n21
+
+
(14)
points before leaving this class of waves. Note from (4) and the
kx
ky given by (5) and (6) that neither m nor :2 can be zero for the
TM wave without its disappearing entirely. The second point is that we have required
as boundary condition that E2 be zero along the perfectly conducting boundary, but
should also be sure that other tangential components of electric field are zero there.
From (10) and the definitions (5) and (6) we can see that Ex is zero as required at
We make two
definitions of
y
=
O and y
and
=
b; from (11)
that
we see
E),
is
zero at x
=
O and
x
=
a.
Thus all
tangential components do satisfy the boundary conditions at the conductors. It
shown (Prob. 8.7h) that imposition of the boundary condition on E2 necessarily
the other
tangential components
of the relations
TE Waves
is from
of E to be
zero on
can
be
causes
the boundaries because of the form
8.2(13)—(l6).
Transverse electric
Eq. 8.2(18), again expressed
VEH,
“
have
waves
in
zero
E3
rectangular
32H.
-
2-
7
8x"
+
62H.
;
6})“
=
and
nonzero
Hz
so
that the start
coordinates:
vkfiHz
(15)
I
'\° .1. Jo
c
0‘. cl. 0‘. #0
‘
o
'-d§
o
o
f. ‘I
q a,
o
o
l’ 0
o
n
o
TE;l
'r‘ 0'. cl. .\o
“a cl
O
ftT \0
o
oh 0‘. o\o
o
II
1'
Rectangulr
Table Types
Wave
.
010'. ‘ol-
“"1 a‘o‘n
8.7 for
0
"_'—j,‘-\_ ,; ‘1. ”}Ia\-—o‘J w-"'¢73‘~ ." ~*OI-’O\-I~»_I
/-‘I§;‘_: 5—.a-u —UFO-v ~‘«& ~—*-/
nln
Guides”
0'
0
0;.
0‘
o
/_-¢.—~ Irlt
,.-—K (*1‘0Iotl \_._/ 1:!
o
‘ufi .I
If. .
o\olo O‘n‘n
a
\o
I
filr
:1
11
1
r1
TE“
(I1
1
r1
l
o
G
o
‘0
.l’o ll 'O
|\
dashed.
of
are
lines
Sum ary
field
I’l
TE“)
TIJ
“MW-"1'~_‘,/
magnetic
ln\
and
‘~>- " s_- 1
solid
‘c
shown
are
l
lines
field
Electri
“
429
3.7
Solution
by
the
separation
H:
z
Rectangular
of variables
sin
k_3sr)(C”
421
of Sec. 7.19
techniques
+ B" cos
(A” sin k_3.r
Wavegtlides
gives
+ D” cos
kyy
(16)
kyy)
where
k Old
=A33
Imposition of boundary conditions in this case is
and
8.2(14)
E3,
find electric field components
we
:
‘
A3
—+
a
(17)
little less direct, but from
Eqs. 8.2( 13)
as
81.1.:
Jwfi
k; 6y
Awakr
H
A2
k 3.1' +8”
(A” sin
cos
k33x)(C”
cos
kyy
-
D" sin
(18)
k3,y)
C
‘
E3.
J
=
'
“’5
k“
C
6H.
——~
(19)
dx
_____,-1
:Jw/Lk
k"
(A"
cos
k31'
-
B" sin k 3r)(C" sin k 3y + D"
cos
k3,y)
C
For
A”
E3.
=
to
0.
be
zero at
Defining
y
B"D"
=
=
O for all x, C
B,
at x
=
=
0, and for
E3,
=
0 at
x
=
O for all y,
B
cos
k_3..1'
b so that
require E3. to be zero at y
k3,];
that [(30 is also a multiple of 77:
(20)
k3,),
cos
must be a
=
multiple
of
77.
E3,
is
zero
a so
kxa
In contrast to the TM waves,
wave’s
=
have then
we
H:
We also
”
vanishing. Although
=
one
we
k,b
3
17277,
=
but not both of
found the
(21)
1277
and 11 may be zero without the
conditions by first calculating elec-
m
boundary
from the way in which E is related to H: that the derivative of H:
conducting boundary must be zero for the tangential electric field to be
zero there, so boundary conditions can be imposed directly on the form (16) without
requiring the explicit forms for E33. and E3.
The forms of transverse electric field' with the derived simplifications to (18) and
tric field,
we can see
normal to the
( 19)
are
E33.
——
'3.a),LL/c
J]?
B
cos
k_3.r
sin k 3y
(22)
CI}! II
k‘
E3,
=
__j_wk—__,u
C)" Il
B sin k 33:
cos
k 3y
(23)
422
Chapter
8
Waveguides
with
Cylindrical Conducting
Boundaries
Corresponding transverse magnetic field components from Eqs. 8.2( 15)
and
8.2( 16)
are
'
H.
.1
Hy
=
_-_~
k.
lg;
sin
kxx
cos
J'Bky B
cos
er
Sln
kyy
(24)
kyy
(25)
.
C!" .II
Since
comparison
of
(21) with (5) and (6) shows that kx
for TM and TE waves, cutoff
frequency
and
and
propagation
ky
have the
same
characteristics for
a
forms
TE”,n
(17) are exactly the same as for the same order TMnm mode. That is,
the expressions (7), (8), and (9) apply here without change. Modes that have different
field distributions but the same cutoff frequencies are said to be degenerate modes.
mode found from
Table 8.7
gives
the field distribution for several different TE modes. Since electric
field is confined to the transverse
plane,
we
find that for each
one
of the TB modes
shown, electric fields begin on charges for a portion of the boundary and end on charges
sign on another portion, in the same x—y plane. Magnetic field lines surround
displacement currents represented by the changing transverse electric fields. For the
TE1 0 mode having no variations in the vertical direction, electric fields go between top
and bottom of the guide in straight lines, and magnetic fields lie entirely in planes
parallel to top and bottom. The TE10 mode is so important that it will be discussed
separately in the following section.
Figure 8.717 shows a line diagram indicating the cutoff frequencies of several of the
lowest-order modes referred to that of the so-called dominant TE10 mode for a guide
with a side ratio b/a
é, which is close to the value used in most practical guides.
Normally, such a guide is designed so that its cutoff frequency for the TED mode is
somewhat (say, 30%) below the operating frequency. In this way only one mode can
propagate so signal distortion caused by multirnode propagation is avoided. Also, by
not being too close to the cutoff frequency, dispersion caused by having different group
velocities for different frequency components of the signal is minimized for the one
propagating mode. Higher-order modes may be excited at the entrance to the guide but
they are below their cutoff frequencies and die away in a short distance from the source.
of opposite
the
=
Tao;
TE10
¢
¢,
L
b
L
O
1
2
”51!
TM“
b
3
Ire/(mm
FIG. 8.7b
Relative cutoff
frequencies
of
waves
in
a
rectangular guide (b/a
=
fir).
8.8
The
TE“,
Wave in
a
423
Rectangular Guide
0.20
TE10 g
0.18
1’10
.-.-.
016
0.14
(dB/m)
Atenuaio
0.12
0.10
0.08
P.
NIH
I!
a
0.06
P. H
NIH
a
0.04
.13.
II
H
0.02
0
5
15
10
25
20
30
Frequency (GHz)
FIG. 8.70
Attenuation due to copper losses in
The attenuation constant for
power loss per unit
length
(a) TE“
as
TEnm ()1 ¢ 0)
Eq. 8.5(11).
2R5
=
bn
1
j-
using
power transfer and
1
+
{(1 Ext)“
+
(ft/f)2
7
22>”
_
f
TE,"0
modes is found
of fixed width.
in
°
And for
rectangular waveguides
f
a
(26)
(b/a)((b/a)m2 + 112)
(bzmz/az) + 122
modes
(a) TEmO
RS
=
C
£977
1
____
(fc/f)2
[1
+
2‘12
a
(JG-C)?
f
(27)
Figure 8.7c shows attenuation versus frequency for TM11 and TE10 modes in rectangular
copper waveguides with various side ratios [3/ a found using (14) and (27), respectively.
It is seen that small b / (2 ratios give large attenuations because of the high ratio of surface
to
cross-sectional
area.
8.8
One of the
THE
TEm
WAVE
IN A
RECTANGULAR GUIDE
simplest of all the waves which may exist inside hollow-pipe waveguides
TEIO wave in the rectangular guide, which is one of the TB modes
is the dominant
424
Chapter
Waveguides
8
preceding section.
following reasons:
studied in the
for the
Cylindrical Conducting Boundaries
with
This mode is of great
engineering importance, partly
frequency is independent of one of the dimensions of the cross section.
Consequently, for a given frequency this dimension may be made small enough
so that the TE10 wave is the only wave which will propagate, and there is no
difficulty with higher-order waves that end effects or discontinuities may cause
1. Cutoff
to
be excited.
2. The
to
polarization
of the field is
bottom of the
definitely fixed, electric field passing from top
polarization may be required for certain
This fixed
guide.
applications.
a given frequency the attenuation due to copper losses
pared with other wave types in guides of comparable size.
3. For
is not excessive
com»
us now rewrite the expressions from the previous section for general TE waves
O
l, n
O, in which case ky
rectangular guides, Eqs. 8.7(20)—(24), setting m
and kC
77/0.
lgE
Let
in
=
=
=
=
=
Hz
=
B
_,
Ey
-
kxx
cos
jwMB
(l)
.
sm
—-
k):
kxx
(2)
'
B
H.
All other components
=
1—f—
sin
a
y
Hz
(3)
This set may be rewritten in
are zero.
a
useful alternate form,
77x
Slfl< >
(—> (3)
'
E
ex
-ZTE Hx
EO
=:
(4)
~—
a
==
'E
1—0
A
(5)
cos
2a
77
a
where
E0
=
jthB
127703
=--
—-———
A
kx
«1/2
2
2TB
77
“'
77
we
1
A
__
"
77
a)
=
A
==
**
f
frequency, wavelength,
(8)
(0V! pas
and wavenumber
are
7T
.
20 V us
—1/2
(7 )
2a
=
1
fa:
2
1
277
v
,u
“‘3
8
Cutoff
(6 )
———
Ac=2m
kc=*
a
(9 )
The
8.8
TE", Wave
Phase and group velocities and
measured
wavelength
along
1
U
U
Ag
v
,
V WV 1
J-
=
=
f
wza)2
—~
2
1
B
the
guide
1
=
g
are
A2
1
=
"
425
Rectangular Guide
in a
-
V 1w
(M)
(10)
A
=
(11)
--———————;
V1 “(A/2a)“
The attenuation
arising from an imperfect dielectric is obtained by replacing a with
equation for 7. Since kC in the TEmO modes is of the same form as in
the parallel-plane modes, Eq. 8.5(3) applies here and. leads to a result equivalent to Eq.
8.5(4):
8’
—-
in the
je”
H
k
ad
by
the
wave
from the
Poynting
Utilizing
the forms
(4),
we
-—
first calculate the power transferred
imperfect, we
=
—
”
a
If
Re
2
0
.
(—E),Hf)
dx
(13)
dy
'
o
have
WT
=
a
Ezb
0
'
ZZTE f
,,
sin"
E dx
==
a
O
Next
(ll/2a)“
theorem:
1
WT
<12)
,,
2
To find attenuation if the conductor is
I
/8
r———8
l
=
E2!)
---9—3
42TH
(l4)
find
approximate losses in the walls by using currents of the ideal mode in
resistivity RS. Current in a conductor is related to the tangential
0 and x
field
at
a so there is current per unit width
magnetic
Hz the side walls x
there.
and
are
Both
tangential at top and bottom surfaces
components Hx
H:
'sz| ‘Hzl
rise
to
surface
current
densities
giving
lHZI. Thus, power loss
IJSZI
IHxI and V“!
we
material of surface
=
=
=
=
=
per unit
length
is
(WL)SIDES
=
(WL)TOP AND BOTTOM
:
2
(
[JR S
2
Ez/‘L2
-Il—O)
l H}-
bR 0
————-S
=
( 15 )
477202
0
2
RS
J0 (Wt-l"
7
“2""
0
RS]
o
E20
Adding
the two contributions
W1.
=
RSE‘S
In)?
—o_
7
a
7)“
4a“
77x
477‘0“
a
07
gooszw-J
dx
E2/\2
+
0,
47712
(16)
2
(15) and (16) and substituting ZTE from (7),
“
2
A2
A2
a
+
E21)2
9771‘
[2%:
70
3R5
2
Zing
dx
sin-—— +
E:2
=
'Hzlz)
+
1
--
9
4a“
+
=
4a
2
RSECZ,
277
2
bAZ
a
+ ———2—
2a
426
Chapter
Waveguides with Cylindrical Conducting Boundaries
8
The attenuation from conductor losses,
ac
1.9!;
2WT
:
Eq.5.9(4),
bag
R 2
:
_ST_I’_E_
is then
a
+
(17)
o
2a~
"rrba
01‘
°
b7)
A
this
study
wave
r—Rs
(M2602 [In-{>1
a
1—
of the field distributions
(1)
to
(3)
or
(18)
2a
(4) and (5) shows the field patterns for
sketched in Table 8.7. First it is noted that
no
field components vary in the
only electric field component is the vertical one E), passing
between top and bottom of the guide. This is a maximum at the center and zero at the
conducting walls, varying as a half-sine curve. The corresponding charges induced by
the electric field lines ending on conductors are (l) charges zero on side walls and (2)
a charge distribution on top and bottom with ,05
8E), on the bottom and sEy on the
top. The magnetic field forms closed paths surrounding the vertical electric displacement currents arising from
Ey, so that there are components HA. and H2. Component H;
is zero at the two side walls and a maximum in the center, following the distribution
of Ey. Component Hz is a maximum at the side walls and zero at the center. Component
Hx corresponds to a longitudinal current flow down the guide in the top, and opposite
in the bottom; Hz corresponds to transverse currents in the top and bottom and vertical
currents on the side walls. These current distributions are sketched in Fig. 8.8a.
This simple wave type is a convenient one to study to strengthen some of our physical
pictures of wave propagation. Electric field is confined to the transverse plane and so
passes between opposite charges of equal density on the top and bottom. The electric
field E), and the transverse magnetic field Hx are maximum at planes a half guide wavelength apait. Halfway between those planes is the maximum rate of change of E), for
the traveling wave and therefore the location of maximum displacement current. The
conduction currents in the metal walls, related to the tangential magnetic field, vary
with position. Displacement currents provide the continuity of total current. The mag—
netic fields surround the electric displacement currents inside the guide and so must
vertical
or
y direction. The
=
have
As
an
axial
as
well
as a
transverse
—
component.
crude way of looking at the problem, one might also think of this mode
formed by starting with a parallel-plate transmission line A of width w to carry
a
fairly
being
the longitudinal current in the center of the guide, and then adding shorted troughs B
of depth 1 on the two sides to close the region, as pictured in Fig. 8.819. Since one would
expect the lengths l to be around a quarter—wavelength to provide a high impedance at
the center, the overall width should be something over a half—wavelength, which we
know to be true for propagation. The picture is only a rough one because the fields in
the two regions are not separated, and propagation is not purely longitudinal in the
center portion or transverse in the side portions.
A third viewpoint follows from that used in studying the higher-order waves between
parallel planes. There it was pointed out that one could visualize the TM and TE waves
8.8
The
TE“,
\\\\
Wave in
a
427
Rectangular Guide
figs >9;
W
‘
(a)
1-4 "wFAF—l—as
B___/
xfl/
—>lwl<——
(b)
xW/WWWWW
/
/
x/
\
/
\
0
\
71/
v
\
/
\
\\
/
.
// vp-u/smfi
\3
-———-——->
Ug=U$ln0'
(Top view)
(6)
FIG. 8.8
(a) Current flow in walls of rectangular guide with TE10 mode. (b) Guide roughly
regions. (c) Path of uniform plane—wave component of
divided into axial- and transverse-current
TE“,
wave
in
in terms of
rectangular guide.
plane waves bouncing
between the two
planes
at such an
angle
that the
interference pattern maintains a zero of electric field tangential to the two planes. Similarly, the TE10 wave in the rectangular guide may be thought of as arising from the
interference between incident and reflected
vector
is vertical, and
bouncing
with the sides that the
zero
component uniform plane
plane
waves,
polarized so that the electric
guide at such an angle
between the two sides of the
electric field is maintained at the two sides. One such
is indicated in Fig. 8.80. As in the result of Sec. 8.4,
exactly A/ 2, the waves travel exactly back and forth across the
guide with no component of propagation in the axial direction. At slightly higher fre)t/Z cos 6, and there is a small propaquencies there is a small angle 0 such that a
gation in the axial direction, a very small group velocity in the axial direction v sin 6,
when the width
a
wave
is
=
428
and
Chapter
a
very
proaches
8
Waveguides with Cylindrical conducting Boundaries
large phase velocity 0/ sin
degrees, so that the wave
frequencies approaching infinity, 6 ap—
guide practically as a plane
6. At
travels down the
90
in space propagating in the axial direction.
All the foregoing points of view explain why the dimension [7 should not enter into
wave
frequency. Since the electric field is always normal to top
and bottom, the placing of these planes plays no part in the boundary condition. However, the dimension b does affect other characteristics of the guide. Small 17 gives a
larger separation between cutoff frequencies of the TEIO and TE01 modes. But it increases attenuation as
shown in Fig. 8.76 and limits power~handling capabilities because
the determination of cutoff
of breakdown-field
limits.
8.9
CIRCULAR WAVEGUIDES
Hollow-pipe waveguides of circular cross section are used in a number of instances,
example, when circular polarization is to be transmitted to certain classes of antennas. Also, as will be shown, the class of TED,z modes (called circular electric) is inter—
esting because of the low attenuation in this class at high frequencies. As before, we
start with ideal dielectric and conducting boundary and make approximate modifications
to these solutions when the materials have small losses. Before treating separately the
TM'and TE classes of waves, it is desirable to have the set of equations 8.2(13)——(l6)
transformed to circular cylindrical coordinates. A straightforward transformation gives
for
(up, 6H,
BE,
j
'
s 513.
j
kil
[
4”
Hr
H
-_—_—
j
rec/2
we
___._
:
j
....—~--
‘5
k3
6E,
_._-.
7
kg
3H,
'8
._
at
r
[we
6E,
—:
+
ar
“marl
6r]
”
6H,
___-
B 6H,
_
r
—:
(3)
(4)
a¢
where
fi=¢+E=E-E
TM
The transverse part of the
circular cylindrical coordinates for this
Waves
shown in
Fig.
m
Laplacian in Eq. 8.2(17) for E2 is expressed in
configuration, with the coordinate system as
8.9a.
1 a
V-j-E,
=
«
—~—
,.
.
BE.
(r—L)
.
1
+
82E‘
-
-k§E.
(6)
Circular
8.9
FIG. 8.9a
Separation
Hollow~pipe
of variables
E:(r, (1'))
where
J”
and
The second
N,,
are
circular
techniques
[A'J”(kcr)
z
cylindrical guide showing coordinate system.
in Sec. 7.20 led to the solution
B'N”(kcr)][C’
+
cos
q’)
so
that
kind, N,,(kcr), is infinite
we
have
just
the
cos
(1)
to
(4) with H:
E¢
where the
prime
D’ sin
+
=
r
variation.
=
AJ,,(kcr)
n
and
_
—
ZTMHd,
simplicity,
A,
Letting A’C’
__
—-
-~
__
—
kind, respectively.
we
be included in
choose the
)qu
cos
origin
(8)
jfi AJ,,(kCI)
.
cos
It
~ZTMH,
(7)
=
,
__
-—
12¢]
so cannot
0, the remaining field components
=
E,
at
22¢
E:
From
12¢
nth-order Bessel functions of first and second
0 for any
the interior solution which includes the axis. Also, for
of
429
Waveguides
jfin
are
72g!)
(9)
mi)
(10)
.
.
EAJHUCCI)
srn
denotes the derivative with respect to the argument and
Z TM
=
( 11 >
—-—
we
The
boundary
E 4, to be
zeros
zero
condition
there. We
imposed by
see
from
(8)
the
that
perfect conductor at r
E: is zero at the boundary
=
a
if
requires E: and
kca is one of the
of the Bessel function,
2
ka=quea=-;T—a=pn,
(12)
C
where
J,,( 13",)
Equation (12)
==
0. We
see
from
(10) that this makes E (:5
zero
there also.
frequency
wavelength for any mode
order. For any 22 there is an infinite number of zeros of J,,(kcr) so there is a doubly
infinite set of modes, denoted TM,,,. Note that the first subscript denotes angular vari~
allows
one to
calculate cutoff
differing from the usual cyclic order of coor~
TM01, TMOZ, and TM11 modes are given in Table 8.9
ations and the second radial variations,
dinates. Field distributions of the
or
‘49
,-——
Wm]
OI...
—--
OI..-
‘
4‘ -«o-
IJ...
.
«Lup.
~-_.—¢
’-* ‘ I’\' -—4t
.‘a1—E r‘ .Iu. 33-1-1 o‘u\‘o
—- §—_~
.
I
l
I
."
n
'\
.
.l".
It.
t
.
.I.
a,
a
.\
‘
a
)
a
'
.‘I‘’l's
'l“
a
,q- :u oh
I
‘\- g}:
I
c
a
W
’-_—. \1‘-‘ d’"“~ ~_-‘—9 I‘D-F }-p.:’ :t=I(”"\’:\~‘_.‘-"
"—_-n~‘tan-.ap: r"'b ~‘_a lpu—dn 5. -.{ 6".-\-¢._n’
1:
s
TEm
u"). :m: .lo\o
I
H-
I
o
t
a}
:)
5-;
1:
i.
:1: 1|.
s
o
ll. 't:
I
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field
Electric
”
8.9
43'!
Circular Waveguides
with
expressions for cutoff frequency and wavelength, and approximate attencylindrical wall has surface resistivity RS. Phase
velocity, group velocity, guide wavelength, and attenuation for frequencies below cutoff
are of the same forms in terms of cutoff frequency as we have seen for the
parallel—
plane and rectangular guides. Wave impedance, when ,8 in terms of cutoff frequency
is substituted in (11), also has the form found for TM modes in the guides studied
together
uation from conductor losses if the
earlier.
TE Waves
expressed
in
The differential
cylindrical
for the
equation
nonzero
H:
of TE waves,
Eq. 8.2(18),
coordinates, is
1
V311:
3
7‘
6H,
(-9
t
1
62H.
72'
ea"
k‘“
2
“3)
The solution for this from Sec. 7.20 is
H303 qb)
Here
the
to
we
BJ,,(kCr)
=
cos
22¢
(14)
have left out the second solution and chosen the cosine variation with
justification as
O, are
(4) with E:
same
for the TM
The
waves.
remaining
qb
with
field components, from (1)
=
E,
Ed,
=
2,511,,
sin
(15)
22¢
Cr
HZTEH,
2
11’2“” BJ,,(kcr)
=
=
J—EZ—Ff 31;,(kcr)
cos
(16)
1245
where
va
In this
requires
the
case
that
13¢,
=
boundary
0 there,
condition
imposed by
,—
along
the
=
guide
0.
:2
we
#80
Again, expressions
are as
perfect
conductor at
2‘
==
(1
or
kca
where J,’,( p,’1,)
the
for
=
2
fl
=
/\ C
phase
(18)
Pin
and group
velocity
and
wavelength
before.
The field distributions of the
TEOl
and
TEll
modes and
in Table 8.9. Note that the field distribution of the
TEII
some
data for these
mode is
quite
a
are
given
bit like that of
TEIO mode in the rectangular guide, with electric field going from top to bottom of
guide, so this is the one that would be primarily excited if a TEIO-mode rectangular
guide were properly tapered and connected to the circular guide. It also has the lowest
cutoff frequency of any mode in a given size circular pipe, as shown by Fig. 8.9b. In
the TEOI mode, the electric field lines do not end on the guide walls, but form closed
the
the
432
Chapter
8
Waveguides with Cylindrical Conducting
TE21 TEollTEm TE41 TE”
FE“
n
t
O
1
L. t L
Mr
21‘
f
c
W
TEn
T7 3
TM21 TMoz
TMu
TMm
Relative cutoff frequencies of
FIG. 8.9b
Boundaries
waves
in
a
circular
guide.
surrounding the axial time-varying magnetic field. This latter wave is especially
interesting as a potential low-loss transmission system at high frequencies and will be
considered more in the example.
circles
Exampie
CIRCULAR ELECTRIC
The field
expressions
for the
mode from the
TE01
Hz
=
13¢
=
kca
:
8.9
MODE
TEm
OVERSIZE GUIDE
1N
general forms (l4)——(16)
BJ0(kcr)
are
(19)
and
The average power transfer
a
by
u
ZTEH,
p61
=
The Bessel
integral
-E
is evaluated
Conduction current in the
so
3.83
the mode, from the
27TI‘
—
WT
Hz,
=
P11
’f
1‘]
T02(¢H)
W
J?“ BJ1(kcr)
=
dr===
.
(20)
(21)
..
Poynting
theorem is
wzngBZ'n'
[0 I‘Jzkcl‘
kzczmolt)
—
dr
22
()
by Eq. 7.15(22):
=
(0251:2an (22
m
[3139011)]
(23)
guide walls is purely circumferential, related to the tangential
length are
that wall losses per unit
R
w.
Attenuation per unit
=
length,
2m
35 IHZIEM
=
waRsBzfécpn)
(24)
in terms of power transfer and loss, is then
ac
_ln_@
——
——
2WT
wzuza
(25 )
8.10
Higher Order Modes
on
433
Coaxial Lines
0.07
0.06
.e/TMOI.
l
0.05
1
c0
3 0.04
l\ l
‘5
g
c
0.03
2
\
0.02
/
\
TE
TE11>\
\<:
01L...—
0.01
\
l
3
J
J
J
o
7
5
9
ll
13
15
Frequency (GHz)
FIG. 8.90
When
2m
Attenuation due to copper losses in circular
is substituted from
(17), this
oz
2
°
waveguides;
diameter
=
2 in.
may be put in the form
[gave/cu)2
anV 1
(we/cor
(26)
—-
is
proportional to the square root of frequency, but the overall expression decreases
frequency. Thus we have the unusual result that attenuation in this mode, for a
given size guide, decreases with increasing frequency, as shown in Fig. 8.96 which
gives a comparison with TE11 and TMOI modes for a 2-in.~diameter guide. The low
attenuation is because the mode fields are very little coupled to the guide walls at high
frequencies. However, other modes may propagate, as shown by Fig. 8.91), so that there
are problems with mode conversion when such guides bend to go around corners.
Practical ways of solving such problems were developed,7 and this system was dem—
onstrated as a low-loss guiding system for millimeter waves, but has been replaced by
optical fiber.
R5
with
8.10
The lowest—order mode
HIGHER ORDER MODES ON COAXIAL LINES
on a
coaxial line is
a
TEM wave; this
in the transmission-line treatment of Ex. 5.2 where
7
we
was
used the
See, for example, 3. E. Miller, Bell System Tech. J. 33, 7209 (1954).
assumed implicitly
capacitance and in—
434
Chapter
8
Waveguides
with
Cylindrical Conducting
Boundaries
ro/rt
(b)
FIG. 8.10
TM
waves
(a) Cross section of
a
coaxial line. (b) Cutoff
wavelength for
some
higher~order
in coaxial lines.
ductance found from static fields. As in the
parallel—plane guide, higher-order (TM and
TE) modes can also exist. Normally, the line is designed in such a way that the cutoff
frequencies of the higher-order moues are well above the operating frequency. Even in
that case, these modes can be of importance near discontinuities.
The general forms useful for the TM and TE modes in circular cylindrical coordinates
are listed in Sec. 7.20. The boundary conditions require that Ez for TM waves be zero
at re and r,- (Fig. 8.1051).
For TM waves,
AnJ12(kcri )
+
anvnaccri)
=
0
An]11(kcrO)
+
BnNnUCerO)
=
0
Of
Nn(kcri)
Jn(kcri)
..
—
Mx(kcr0)
111(kcr0)
(1)
8.11
435
Excitation and Reception of Waves in Guides
For TE waves, the derivative of H: normal to the two conductors must be
inner and
outer
radii.
NI’1(kC) ‘1’ )
r;
1:1(kcri)
Solutions to the transcendental
hence cutoff
analogy
zero at
the
[See discussion following Eq. 8.7(21).]‘Then, in place of (l),
for any
N220%, ‘0)
(2)
Jr,1(kc’.0)
type and any
(2) determine the values of kc and
particular values of 2-,. and r0. By
ri),
l, 2, 3,...
equations (1)
and
frequency
parallel-plane guide, we would expect to find certain modes with a
that the spacing between conductors is of the order of p half-wavelengths.
wave
with the
cutoff such
2
AC
z
:0-
(r0
—
p
=
(3)
This is verified
by Fig. 8.10b for values of I‘D/r,- near unity.
Probably more important is the lowest~order TE wave with circumferential variations.
This is analogous to the TE10 wave of a rectangular waveguide, and physical reasoning
from the analogy leads one to expect cutoff for this wave type when the average cir~
cumference is about equal to wavelength. The field picture of the TEIO mode given in
Sec. 8.8 should make this reasonable. Solution of (2) reveals this simple rule to be
within about 4% accuracy for rO/r, up to 5. In general, for the nth-order TE wave with
circumferential variations,
2
.+
-.
AC~1<’°—-L>,
There are, of course, other TE
of these has
a
n=l,2,3,...
2
12
waves
cutoff about the
(4)
with further radial variations, and the lowest order
same as
the lowest-order TM
wave.
EXCITATION AND RECEPTION OF WAVES IN GUIDES
8.11
exciting or receiving waves in a waveguide are not simple field prob—
we give only a qualitative introduction to the manners of excitation
of fields in various kinds of guides. Approaches to analysis and measurement of these
junctions are given in Chapter 11. Reception of the energy of a wave uses the same
kind of structure as excitation and is just the reverse process. To excite any particular
desired wave, one should study the field pattern and use one of the following concepts.
The
problems
of
lems. In this section
probe or antenna oriented in the direction of electric
placed near a maximum of the electric field of the
exact placing is a matter of impedance matching. Examples are
1. Introduce the excitation in
field. The
probe
a
is most often
mode pattern, but
shown in Figs. 8.11a and b.
2. Introduce the excitation
through a loop
(Fig. 8.1 1c).
oriented in
a
plane normal to the magnetic
field of the mode pattern
3.
Couple
to the
iris, the
two
desired mode from another
guiding systems having
guiding system, by
some common
means
of
field component
a
hole
over
or
the
436
Chapter
8
Waveguides
with
Cylindrical Conducting
Boundaries
(b)
(a)
Soldered connection
Microstrip
AAAAAAAA
vvvvvv
top conductor
Coaxial—
line
j
Substrate
Shielding
box
connector
(e)
(a) Antenna in end of circular guide for excitation of TMOI wave. (b) Antenna in
guide for excitation of the TE10 wave. (0) Loop in end of rectangular guide
for excitation of TEIO wave. (d) Junction between circular guide (TM01 wave) and rectangular
guide (TE10 wave); large-aperture coupling. (6) Coaxial line coupling to microstrip. (f) Excitation
FIG. 8.11
bottom of rectangular
of the
T1320
wave
in
rectangular guide by
extent of the hole. An
is shown in
coupling
Fig.
from
5. For
to resonant
antennas.
example of coupling between waveguides using a large iris
coupling is sometimes done with a small hole as for
cavities (Sec. 10.10).
coaxial line to
one
kind of transmission line into another,
microsuip
shown in
higher-order waves combine as
phasings (Fig. 8.1 1 f).
with proper
oppositely phased
8.11d. The
4. Introduce currents from
a
two
Fig.
in
coupling
sources as are
required,
as
8.11e.
many of the
exciting
8.1 1
6.
Excitation and Reception
of Waves
437
in Guides
Gradually taper a transition between two types of guides,
a rectangular guide to a
TE“ in a circular guide.
as
for
a
TE10
wave
in
Since most of these
exciting methods are in the nature of concentrated sources, they
purely one wave, but all waves that have field components in
a favorable direction for the particular
exciting source. That is, we see that one wave
alone will not suffice to satisfy the boundary conditions of the guide complicated by
the exciting source, so that many higher~order waves must be added for this purpose.
If the guide is large enough, several of these waves will then proceed to propagate.
Most often, however, only one of the excited waves is above cutoff. This will propagate
down the guide and (if absorbed somewhere) will represent a resistive load on the
source, comparable to the radiation resistance of antennas which we shall encounter
further in Chapter 12. The higher~order waves that are excited, if all below cutoff, will
be localized in the neighborhood of the source and will represent purely reactive loads
on the source. For practical application, it is then necessary to add, in the line that feeds
the probe or loop or other exciting means, an arrangement for matching to the load that
has a real part representing the propagating wave and an imaginary part representing
the localized reactive waves. In a practical design, it is important to be concerned that
the match is good over the frequency band of interest.
will not in
general
excite
Example
8.1 i
EXCITATION OF A WAVEGUIDE BY A COAXIAL LINE
Let
us
look in
more
depth
at the structure in
in the center of the broad side of
a
Fig. 8.11b where
waveguide of rectangular
a
coaxial line is inserted
cross
section
to
excite
a
TElO mode. The waveguide is short-circuited at a distance I from the probe to aid in
matching the coaxial line to the waveguide. The fields associated with the probe excite
both the desired TE 10 mode and other higher~order modes. The latter are cutoff and do
not propagate, but they store reactive energy and therefore constitute a reactive com—
ponent of the load on the coaxial line. Proper choice of the size and location of the
probe for a given frequency and guide dimensions makes the standing wave between
the probe and the shorted end contain reactive energy of opposite sign and equal mag—
nitude so that the net reactive component of the input impedance is zero. These ad—
justments are used to make the real part of the load impedance on the coaxial line equal
to its characteristic impedance so that perfect matching is achieved and all the power
is coupled into the guide. Figure 8.11g shows the calculated results for probes of various
radii in a guide of dimensions appropriate for use at about 10 GHz (X band).8 Similar
graphs can be calculated using the methods in the reference of footnote 8 for other
guide sizes and probe radii.
3
R. E. Collin, Field
1997.
Theory
of Guided Waves, 2nd ed. Sec. 7. 7, IEEE Press,
Piscotaway, NJ,
438
Chapter
8
Waveguides
with
Cylindrical Conducting
Boundaries
6O
Ohms
9
10
11
GHz
(g)
Probe input resistance and reactance as a function of frequency for d
0.62 cm,
0.495 cm, guide width 0
2.286 cm, guide height b
1.016 cm. For the thin probe of
radius r
0.5 mm, 1
0.505 cm. Reproduced by permission from R. E. Collin, Field Theory
of Guided Waves, 2nd ed., Sec. 7.1, IEEE Press, Piscataway, NJ, 1991.
FIG.
I
8.119
=
=
=
=
=
=
General Properties of Guided Waves
8.12
GENERAL PROPERTIES OF TEM WAVES
ON
MULTICONDUCTOR LINES
The classical two-conductor transmission system was studied extensively in Chapter 5,
starting from a distributed circuits point of view. We used wave solutions to verify the
of TEM
results for the
special
point
show that TEM
we can
case
waves
waves
between
parallel planes in Sec. 8.3. At this
cylindrical system with
in any two-conductor
General Properties of TEM Waves
8.12
isotropic, homogeneous dielectric, and
by the transmission—line equations.
The
general relations
show that, with
between
loss-free conductors
wave
components
439
Multiconductor Lines
as
are
exactly
those
predicted
expressed by Eqs. 8.2(9)—-(12)
E:
H: zero, all other components must of necessity also be zero,
+ k2 is at the same time zero. Thus, a transverse electromagnetic wave must
and
unless
y2
satisfy
the condition
'y
For
on
:L-jk
=
i1;
=
=
(1)
ijwVMe
perfect dielectric, the propagation constant 7/ is thus a purely imaginary quantity,
signifying that any completely transverse electromagnetic wave must propagate unattenuated and with velocity v, the velocity of light in the dielectric bounded by the guide.
With (1) satisfied, the wave equations, as written in the form of Eqs. 8.2(1) and
a
8.2(2), reduce
to
ViE
These
V3,.H
o,
=
2
0
(2)
Laplace equation written for E and
plane.
entirely in the transverse
E3
H_._
Since
electric
and
fields
both
plane.
magnetic
satisfy Laplace’s equation under static
conditions, the field distribution in the transverse plane is exactly a static distribution
if boundary conditions to be applied to the fields in (2) are the same as those for a static
field distribution. The boundary condition for the TEM wave on a perfect conducting
guide is that electric field at the surface of the conductor can have a normal component
only, which is the same as the condition at a conducting boundary in statics. The line
integral of the electric field between conductors is the same for all paths lying in a
given transverse plane, and may be thought of as corresponding to a potential difference
are
exactly
the form of the two—dimensional
H in the transverse
Since
and
are
between the conductors for that value of
To
study the
character of the
zero, E and H lie
2.
magnetic field,
note
Eqs. 8.2(3)
and
8.2(6) with
zero
E- and H.:
.
Hy
=
Hr
:
'
J—C—"EE
7’
E=
—"
<3)
77
and
T; Ey
1w,“
E,
:
“—3
(4)
TI
[The signs of (3) and (4) are for a positively traveling wave; for a negatively traveling
they are opposite] Study shows that (3) and (4) are conditions that require that
electric and magnetic fields be everywhere normal to each other. In particular, magnetic
field must be tangential to the conducting surfaces since electric field is normal to them.
The magnetic field pattern in the transverse plane then corresponds exactly to that
arising from dc currents flowing entirely on the surfaces of the perfect conductors.
These characteristics Show that a transverse electromagnetic wave may be guided by
two or more conductors, or outside a single conductor, but not inside a closed conwave
449
8
Chapter
FIG. 8.12
Waveguides with Cylindrical Conducting Boundaries
Two-conductor transmission line with
integration paths.
ducting region, since it can have only the distribution of the corresponding two~dirnensional static problem, and no electrostatic field can exist inside a source-free region
completely closed by a conductor (see Prob. 8.12d).
We next may show an exact identity with the ordinary transmission-line equations
for TEM
waves on
the systems that support them. Consider
a
transmission line
con-
sisting of two conductors A and B of any general shape (Fig. 8.12). The voltage between
the two conductors may be found by integrating electric field over any path between
conductors, such as l—O-2 of the figure. It will have the
path is chosen, since E satisfies Laplace’s equation in the
be considered the
plane
are
gradient
of
a
scalar
potential
insofar
same
value
as
no matter
and
which
plane
may
variations in the transverse
transverse
so
concerned.
2
2
V=-JE-dl=—f
1
Differentiating
this
equation
1
with respect to
6V
__
dz
=
__
[2
1
6E"
_
(Exdx-t-Eydy)
(5)
2
dx +
62
as,
__
62
dy
(6 )
But the curl relation,
63
V X E
=
——
at
shows that, if E is zero,
6E
6E
83
-——’—’=—x
and
BB
—*=———y
(7)
8.!2
General Properties of TEM Waves
By substituting (7)
in
(6),
on
441
Multiconductor Lines
have
we
av
ar
ar1(
—=—-—
a2
~Bd+B.d
w)
<8)
H
A
study of Fig. 8.12 reveals that the quantity inside the integral is the magnetic flux
across the path 1—0—2 per unit length in the z direction. According to the usual
definition of inductance, this may be written as the product of inductance L per unit
length and current I, so (8) becomes
flowing
6V
—
—---
a:
62
Equation (9)
is
one
of the differential
tional transmission-line
with current in line A
is
no
as
contribution from
a]
6
=
(L1 )
“L
=
(9 )
--
at
used
equations
as a
starting point
displacement
current
since there is
no
I=f£H-dl=§(1—dex+Hydy)
Differentiating
conven-
E3.)
(10)
with respect to z,
61
—
=
82
From the curl
for
analysis [Eq. 52(3)]. The other may be developed by starting
the integral of magnetic field about a path a~b—c——d—a. (There
fit (
an
—"
dx +
6H,
y)
'
d
62
62
( 11 )
equation,
8D
V X H
=
—
at
it follows that, if Hz
=
0,
6H,
__)
0D.
A
:
in
(11),
we
BD
aH.
—-—'—\
and
2
62
at
62
Substituting (12)
..
--—y
81‘
(12)
have
a]
-—
62
a
=
——
at
56
(D.dy
~
A
D
3,.
dX)
( 13 )
Inspection of Fig. 8.12 shows that this must be the electric displacement flux per unit
length of line crossing from one conductor to the other. Since it corresponds to the
charge per unit length on the conductors, it may be written as the product of capacitance
per unit length and the voltage between lines and (13) becomes
8V
61
-
62
=
-C
——-
61‘
(14)
Equations (9) and (14) are exactly the equations used as a beginning for transmissionanalysis, if losses are neglected (Sec. 5.2). It is seen that these equations may be
line
442
Chapter
Waveguides with Cylindrical Conducting Boundaries
8
exactly from Maxwell’s equations provided the conductors are perfect, and
since fields in the transverse plane satisfy Laplace’s equation, the inductance and ca~
pacitance appearing in the equations are the same as those calculated in statics. This is
of course not the case for the TM and TE waves met in the preceding sections. Waves
on the ideal transmission line have been shown to be TEM waves with phase velocity
2
marl/2. Transmission~line phase velocity is (LC )" 1/ so it follows that inductance
and capacitance per unit length of an ideal transmission line are related by LC
us.
derived
=
Transmission Eines with losses
If the conductor of the transmission line has
finite losses, the above argument does not apply exactly. There must be finite E2 at the
conductor to force the axial currents through the imperfect conductors. In that case
+ k2 of Eqs. 8.2(9)——(12) cannot be zero, and Eqs. 8.2(1) and 8.2(2) do not reduce
Laplace’s equation. But so long as the conductors are reasonably good, the axial
component of electric field is small compared with the transverse component and cor~
rections are small. The usual way of handling the losses through a series resistance in
the transmission-line equations can then be shown by perturbation arguments to be an
y2
to
excellent approximation.9 Losses in the dielectric, however, do not in themselves disturb
the TEM nature of the
since these
cause conduction currents to flow only in
by inclusion of shunt conductance computed
from static concepts and by the wave method with 8 replaced by s’
je” can be shown
to be the same (Prob. 8.120).
In addition to the principal TEM or n‘arzsmz‘ssion-Zine mode on the two-conductor
system, there may propagate higher-order modes as well. The higher-order modes may
be excited at discontinuities in the transmission line and may cause dispersion effects
or radiation. It is difficult to work out the forms of the higher—order modes in open
structures such as the two-wire line, but is straightforward to derive them for the coaxial
transverse
wave
directions. In this
case treatment
._
line
as was
8.13
done in Sec. 8.10.
GENERAL PROPERTIES OF TM WAVES IN CYLINDRICAL CONDUCTING
GUIDES OF ARBITRARY CROSS SECT|ON
In the earlier sections
we
have
seen
several
specific examples
of TM waves; it is the
purpose of the present section to generalize the formulation for any cylindrical structure.
The analysis can be done in a generalized coordinate system10 and might appear more
general,
but for
simplicity
that boundaries may be of
will
rectangular
arbitrary shape.
we
use
coordinates with the
understanding
The Diflerentiai Equation With the assumed propagation constant 60“” 7”), the
finite axial component of electric field for the TM waves must satisfy the wave equation
9
R. E. Collin. Field
Theory of Guided Waves, 2nd ed, Sec.
4. 7, IEEE Press,
Piscatawoy, NJ.
7997.
"3
R. E. Collin, Field
7997.
Theory of Guided Waves, 2nd ed, Sec. 5. 1. IEEE Press, Piscatoway, NJ,
General Properties of TM Waves in
8.15
in the form of
443
Cylindrical Conducting Guides
Eq. 8.2(17):
V2xy E
c
(it)
The value of kc, which is
condition to be
applied
a
to
Boundary Condition
—-k2E-
=
,
i
(72
=
constant
for
(1)
.'.
k2)
+
a
v2
=
+
£02m:
(2)
particular mode, is determined by the boundary
(1).
for
a
Perfectly Conducting Guide
As in the
examples,
the first step in the solution of a practical waveguide problem is to assume that the
waveguide boundaries are perfectly conducting. The appropriate boundary condition is
E:
r
0. It is
easily
shown from the
general relations for the transverse field components
in Sec. 8.2 that
__
E"
m
_
7
+—
k?
BE:
_.
E
—
=
’
6x
,.
Relations
(3)
y
'
6E-
‘
6E:
_._..
k2
'
:
7
+.......
'
6E-
’wf—=
k;
(3) may be written in the
H).
6y
vector
=
k;
(4)
x
form
_
Et
”is”;
=
7
4-? V,E2
(5)
C
where
verse
Et
as
is the transverse part of the electric field vector, and ‘Vt represents the trans—
part of the gradient. By the nature of the gradient, the transverse electric vector
E,
is normal to any line of constant E. It is then normal to the
required, once the boundary is made a curve of constant E2
the
only required boundary
conducting boundary,
0. Thus E2
O is
=
=
condition for solutions of (1).
fintoff Properties of TM Waves Solution of the homogeneous differential equa»
tion (1) subject to the given boundary condition is possible only for discrete values of
the constant
the
It
kc.
These
are
the characteristic values, allowed
values,
or
eigenvalues
of
any one of which determines a particular TM mode for the given guide.
be shown (Prob. 8.13e) that, for any lossless dielectric region which is completely
problem,
can
closed
by perfect conductors,
gation
constant from
the allowed values of
lcC
must be real. Hence the
propa»
(2),
v=
kirk?
(6)
always exhibits cutoff properties. That is, for a particular mode in a perfect dielectric,
kc, and 'y is
y is real for the range of frequencies such that k < kc, y is zero for k
imaginary for k > kc. The cutoff frequency of a given mode is then given by
=
2
ZWfCVas
=
{I
C
=
kc
(7)
444
Chapter
8
Boundaries
Waveguides with Cylindricai Conducting
3
fcz cutoff frequency
21r
kC—E
__L_
2
"’w?
f/fc
FIG. 8.13
and
The
Frequency characteristics of all
(6) may be written in
phase velocity for
and cutoff
terms of
frequencyff>
7=jB=Jk/1—<—>2
>guide
all TM modesm
an
ideal
(D
_.
Up
The group
TE and TM
—
—
——
E
v
1-
E
2
wave
types.
frequency fc:
f>fc
(9)
then has the form
—-I/?.
(10)
f
velocity is
2
d
c1;— v[1
—
vgz
Universal
for attenuation constant,
—-
(%>]
1/2
(11)
phase velocity, and group velocity as funcf/fC
velocity is infinite at cutoff frequency and
Fig.
than
the
of
is always greater
velocity
light in the dielectric; group velocity is zero at
the
is
cutoff and
always less than
velocity of light in the dielectric. As the frequency
increases far beyond cutoff, phase and group velocities both approach the velocity of
light in the dielectric.
tions of
curves
are
shown in
8.13. Phase
8.13
General Properties of TM Waves in
Cylindrical Conducting Guides
445
Magnetic Fields of the Waves Once the distribution of E___. is found by solution
of the differential equation (1) subject to the boundary condition E:
O, the transverse
electric field of a given mode may be found from relation (3) or (5). The transverse
magnetic field may be found from relations (4). By comparing (3) and (4), we see that
x
E.
E.
__i
“.3.
_.
i1-
2
Hr.
Hy
(12)
jwe
These relations show that transverse electric and
are at
that their
may be
the
wave
magnetic fields
magnitudes are related by the quantity y/jcue, which
impedance or field impedance of the mode:
ZTM
7
::
——
:7.
1
7]
jwe
-——
fc
right angles and
thought of as
2
( f)
-—
(13)
,___
8
The wave impedance is imaginary (reactive) for frequencies less than the cutoff
frequency and purely real for frequencies above cutoff, approaching the intrinsic impedance of the dielectric at infinite frequency. This type of behavior is also found in the
study of lumped-element filters, and it emphasizes that the wave can produce no average
power transfer for frequencies below cutoff, where the impedance is imaginary.
The relations between electric and magnetic fields may also be given in the following
vector form, which expresses the properties described above:
H=i
where 2 is the unit vector in the
waves, the lower
sign
for
z
2x13,
‘erx
direction. The upper
negatively traveling
Power Transfer in the Waves
cutoff if the conductor of the
guide
is
(14)
sign
is for
positively traveling
waves.
The power transfer down the guide is zero below
perfect. Above cutoff it may be obtained in terms
of the field components by integrating the axial component of the Poynting vector over
the cross-sectional area. Since it has been shown that transverse components of electn'c
magnetic fields are in phase and normal to each other, the axial component of the
average Poynting vector is one-half the product of the transverse field magnitudes. For
a positively traveling wave,
and
By
use
=
LSEReflZ
Z
1
1
W
><
H’l‘L-
as
=
if |E,HH,lds= :——M-CS Incl2
d5
(15)
of (4), this may be written
WT
z
ZTMw
2
21::
e
2
f
Iv,E_,¢-°—
d5
(16)
446
Chapter
Making
use
Waveguides with Cylindrical Conducting Boundaries
8
of the relation
f
(Prob. 8.13e),
we
[V,EZ|2
d3
f
k3;
=
E3
d5
(17)
obtain
WT
=
me282
21%
f
——
2
E; dS
=
2
f
—
ZTM
-~
2772
f0
f
-
as
2
E;
-
dS
( 1 8)
CS
imperfectly fienducting Boundaries When the conimperfect, an exact solution would require solution of Maxwell’s
both the dielectric and conducting regions. Because this procedure is immost geometrical configurations, we take advantage of the fact that most
Attenuation Due to
boundaries
ducting
equations in
practical for
practical conductors are good enough to cause only a slight modification of the ideal
solution, and the expression wL/ZWT in formula 5.1 1(19) may be used. To compute the
average power loss per unit length, we require the current flow in the guide walls, which
is taken to be the same as that in the ideal guide. By the n X H rule, the current per
unit width in the boundary is equal tothe transverse magnetic field at the boundary and
flows in the axial direction since magnetic field is entirely transverse:
are
WL
=
2
bound
The attenuation constant is then
a0
Js
—'
(11
6
if
=
2
bound
’11,?
dl
(19)
approximately
iv}—
ZWT
2
R
2
..
ngstEnletlz d1
:
2'ZTMICSII-It‘z
nepers/ m
d5
(20)
If desired, the power loss and hence the attenuation constant may be written in terms
of the distribution of E2 only. By use of (4)
Since
E2
is
zero at
all
R
(0282
2
k:
—5
=
L
points along
the
Rw2823g
2k?
S
=
bound
GE.
[an]
—‘
I VE,
2
dl
(21)
'
bound
boundary,
there; E has only the derivative normal
WL
3g
there is
no
tangential derivative of E2
to the conductor:
2
d1
R5
=
f
—-~
2,7721% fc
er
BE.
——e
all
2
dl
(2 2)
An alternative form for the attenuation constant is then
BB
wgmw Mixes]
611
(23,
Attenuation Due to Imperfect Bietectrfic It is noted that the general form for
constant (9) is exactly the same as that for the special case of the parallel-
propagation
General
8.14
Preperties of TE Waves
in
Cylindrical Conducting
Guides
447
plane guide, Eq. 8.3(15). Hence, the modification caused by an imperfect dielectric,
taken into account by replacing jws by jw(e’
je”) or 0' + jws, yields the same form
for attenuation as Eq. 8.5(4):
—
ad
r______k8”/8’
2
______0n nepers/rn
=
7
It is
especially interesting to note that the
imperfect dielectric is the same for all modes
the amount of attenuation is
the
guide
form of the attenuation
produced by an
shapes
though of course
cutoff frequency, which does depend on
of guides,
and all
function of the
and the mode.
GENERAL PROPERTIES OF TE WAVES IN CYLINDRlCAL CONDUCTING
GUIDES OF ARBITRARY CROSS SECTION
8.14
Finally,
a
(24)
7
we
consider
waves
that have
magnetic
field but
no
electric field in the axial
direction. Because of the treatment is similar to that of TM
section, it will be given
more
waves
in the
preceding
briefly.
The Differential Equation The finite
tion in the form of Eq. 8.2(18):
V3H,
H:
=
waves must
satisfy
the
wave
—k§Hz
=
k%
of the
72
+
equa—
(1)
k2
(2)
Boundary conditions for a Perfectly conducting Guide Allowable solutions
to (l) are determined by the single boundary condition that at perfect conductors the
normal derivative of Hz must bezero:
31i_
O
(3) is the required boundary condition, the
Eqs. 8.2(9)—(12) are written:
To Show that
from
E.
"
=
—
_
Hr
Relation
(5)
jwu 6H,
=
Ey
—=
kfi
y
(3)
boundary
at
all
6y
6H:
t
jwu 6Hz
——
k%
__
Hy
2:3?
__
transverse
6x
6Hz
“
“:3;
fields of the
wave
(4 )
<5)
may be written in the vector form
“y
+1;
a
H,
__
—-
k...C
V,H_,
(6)
448
Chapter
Waveguides with Cylindrical Conducting
8
Boundaries
boundary, its transverse gradient has only a comboundary, so by (6), H, does also. Comparison of (4) and (5)
shows that transverse electric and magnetic field components are normal to one aother,
so electric field is normal to the conducting boundary as required.
If H, has
no
normal derivative at the
ponent tangential
to the
Qumff Properties of TE Waves
waves
that
kC
ductors; the
the
exactly
is
same can
same as
It
mentioned in the preceding section on TM
regions completely closed by perfect conwaves. By (2), y then shows cutoff properties
was
real for dielectric
always
be shown for TB
for TM
waves:
y=vg—a
Formulas for attenuation constant below cutoff,
velocities above cutoff then follow
curves
of Fig. 8.13
exactly
m
phase constant, and phase and group
Eqs. 8.13(8)—(11) and the universal
in
as
apply.
Electric Pic“ of the Wave
The electric field is
everywhere transverse and every—
field
magnetic
components. Transverse components of
field may again be related through a field or wave impedance
where normal to the transverse
electric and
magnetic
E)’
r:
=
“a
(8)
ZTE
where, from (4) and (5),
z TE
jam
1224/2
77[ (f)]
1-—
=——=
”y
9
()
-
This
impedance is imaginary for frequencies below cutoff, infinite at cutoff, and purely
frequencies above cutoff, approaching the intrinsic impedance 1) as f / fc becomes large.
real for
Electric field may also be written in the vector form
E
where 2 is the unit vector in the
respectively
to
positively
and
z
as
x
H,)
Average
usual, obtained from the Poynting
waves.
power transfer in the
L
=§fiJW£
=
E
Re[E
x
propagating
range
vector:
1
WT
(10)
direction, and the upper and lower signs apply
negatively traveling
Power Transfer in TE Waves
is,
121,306.
=
H*]
1
-
d8
=
5
[a |E,||H,|
(15
Z
an
449
Waves Below and Near Cutoff
8.15
Or, using (6) and
f
(see Prob. 8.13e),
we
IV,H2I2dS
=
kg
f
H3
45
(12)
obtain
”C(f/f )“
WT ___"E__
:2sz
f H2-
d8
(13’
c
Attenuation Due to
Hmperfectly Conducting Boundaries As with the TEM
true transverse electric wave in most guides with imperfect
conductors, since most (but not all) of the TB modes have axial currents that require a
certain finite axial electric field when conductivity is finite. This axial field is very small
compared with the transverse field, however, so the waves are not renamed.
The axial component of current arises from the transverse component of magnetic
field at the boundary:
mode, there
cannot
be
a
JS;
=
IHA
glthzl g—i
=
The last form follows since it has been shown that the transverse
only
a
current
tangential component a/al
arising from the axial magnetic
at the
boundary.
length
by
:
TE
waves
has the
uation due to
an
=
Hz
lel
=
Rs
—2—
’
r)
|_[H,|~
has
(15)
+
|H,|
2
] d1
(16)
RssttIHzlz + IH,I21dz
nepers/m
zzTEIIHr d5
Imperfect Dielectric
same
of
transverse
the conductor losses is
°
Attenuation Due to
a
is then
wL
The attenuation caused
gradient
There is in addition
field:
letl
The power loss per unit
(14)
=
propagation
constant
of the
for the TM waves, it follows that the form for atten—
dielectric does also. For a reasonably good dielectric, the
form
imperfect
Since the
(17)
as
approximate form, Eq. 8.13(24), may be used.
8.15
The
higher—order waves that may exist in transmission lines and all waves that may
hollow~pipe waveguides are characterized by cutoff frequencies. If the waves
to be used for propagating energy, we are of course interested only in the behavior
exist in
are
WAVES BELOW AND NEAR CUTOFF
459
Chapter
8
Waveguides with Cylindrical Conducting Boundaries
above cutoff. However, the behavior of these reactive
is
important in
1.
at least two
Application
to
practical
waveguide
or
evanescent waves below cutoff
cases:
attenuators
2. Effects of discontinuities in transmission systems
The attenuation properties of these waves below cutoff have been developed in the
previous analyses. It has been found that below the cutoff frequency there is an attenuation only and no phase shift in an ideal guide. The characteristic wave impedance is
a purely imaginary quantity, reemphasizing the fact that no energy can propagate down
the guide. This is not a dissipative attenuation, as is that due to resistance and conductance in transmission systems with propagating waves. It is a purely reactive atten~
nation, analogous to that in a filter section made of reactive elements, when this is in
the cutoff region. The energy is not lost but is reflected back to the source so that the
guide acts as a pure reactance to the source.
The expression for attenuation below cutoff in an ideal guide, Eq. 8.l3(8), may be
written
as
y=a=kcl—<—-
=—1—-—~
I‘tc
fc
As
f is decreased below 3“,,
or
increases from
zero
(1)
fc
toward the constant value
277
a
=
(2)
-—
AC
when
(f/fc)2 <<
important point in the use of waveguide attenuators, since
substantially independent of frequency
far
the
cutoff
is
below
frequency.
operating frequency
1. This is
an
it shows that the amount of this attenuation is
if the
Now let
magnetic
us
look for
a
moment at the relations among the fields of both transverse
below cutoff. If y
for field components of transverse
and transverse electric
substituted in the
expressions
8.13(3) and 8.13(4),
=
waves
'
16E-
2f.
[11:21.'0
__4
Er
w
1...
E
j; 21
“6—;
'
‘
kc 6y
fc
fc
‘
H
y
a
a as given by (l) is
magnetic waves, Eqs.
_L
n
1 E3: ._i_~
f(:
kc
6x
E
y
___.._
__
1__
i
fc
kc
2
6x
E
_1_a__§
kc By
For a given distribution of E2 across the guide section, which is determined once the
guide shape and size and the wave type are specified, it is evident from relations (3)
that, as frequency decreases, f/fc ~> 0, the components of magnetic field approach
zero whereas the transverse components of electric field approach a constant value. We
draw the conclusion that electric fields are dominant in transverse magnetic or E waves
far below cutoff. Similarly, magnetic fields are dominant in transverse electric or H
waves far below cutoff. If the waves are far below cutoff, the dimensions of the guide
8.16
451
Dispersion of Signals Along Transmission Lines and Waveguides
are small compared with wavelength. For any such
region small compared with wave—
length, the wave equation will reduce to Laplace’s equation so that low-frequency
analyses neglecting any tendency toward wave propagation are applicable.
The presence of losses in the guide below cutoff causes the phase constant to change
from the zero value for an ideal guide to a small but finite value, and modifies slightly
the formula for attenuation. These modifications are most important in the immediate
vicinity of cutoff, for with losses there is no longer a sharp transition but a more gradual
change from one region to another. It should be emphasized again that the approximate
formulas developed in previous sections may become extremely inaccurate in this region. For example, the approximate formulas for attenuation caused by conductor or
dielectric losses would yield an infinite value at f
fc. The actual value is large
compared with the minimum attenuation in the pass range since it is approaching the
relatively larger magnitude of attenuation in the cutoff regime, but it is nevertheless
finite. Previous formulas have also shown an infinite value of phase velocity at cutoff,
=
and with losses it too will be finite.
8.16
DISPERSION OF SIGNALS ALONG TRANSMISSION LINES AND WAVEGUIDES
dispersive properties of transmission systems
both vary with frequency. In Chapter 5 we
phase velocity, group velocity,
considered a simple two-frequency group in a dispersive system, but we now wish to
be more general, using the Fourier integral of Sec. 7.11. There are two classes of
problems of concern. One is that of a base-band signal, in which the detailed signal is
of concern. Examples are audio or video signals, or electrical pulses from a computer,
before being placed on other carrier frequencies. The other is that of modulated signals
in which the base-band signal is placed on a high-frequency carrier. For the latter case
we shall consider amplitude modulation and examine the distortion of the envelope.
We have in several instances noted the
when
or
Base-Band
waveform,
Given
Signals
we can
express it
f(z‘), the transform pair
audio
an
as a
Fourier
series of
as
in
Eq.
pulses,
or
similar electrical
7.1 1(15). For
a
time function
may be written
51; L now“
no
£0»)
If each
signal,
integral
=
in... f(t)e“j“”
(1)
do»
dt
(2)
frequency component is delayed in phase by 32 in propagating
(1) gives the delayed function at 2 as
distance
2
along
the transmission system,
fit, 2)
==
517—, f”
g<w>ef<wz>
dc»
(3)
452
Chapter
Waveguides with Cylindrical Conducting Boundaries
8
mj
l
L=Q
T
V
L=5mm
AkL=10mm
—-+ Hos
FtG. 8.16a
Propagation
of
a
5—ps gaussian pulse along
a
microstrip
line.
Strip width
=
0.32
6.9. Reproduced by permission from K. K. Li,
0.4 mm, and 81.
mm, dielectric thickness
G. Arjavalingam, A. Dienes, and J. R. Whinnery, IEEE Trans. MTT-30, 1270 (1982). © 1982
=
=
IEEE.
Now if
”m
n
is
(4).
.51
with
121) independent
f(t. z)
Thus the
original
(3) is
of w,
i-
=
277
]
g<w)ejw<’*-’/vv>
dc»
=
f<t 5-)
—
(5)
Up
“cc
function maintains its
have assumed in many wave
tion, at least to some degree.
we
Transmission lines
shape and propagates at the phase velocity, as
problems. But any dispersion in up modifies the func~
often used for base—band
signals and have some dispersion
through
by skin effect. Some lines, as
the microstrip line of Sec. 8.6, have additional dispersion from the presence of multiple
dielectrics. Figure 8.16a shows the result of a numerical calculation from (3), using the
dispersion relation of Eq. 8.6(18), for the change in shape of a 5-ps gaussian pulse in
propagating along a typical rnicrostrip used with short electrical pulses.11
are
loss terms and internal inductance
”
K. K. Li, G.
Arjavalingam.
as
A. Dienes. and J. R.
affected
Whinnery, IEEE Trans. [WW-30, 7270 (7982).
Dispersion of Signals Along Transmission Lines and Waveguides
8.16
Modulated
amplitude Vc
Signals If the signal (1) is used to amplitude modulate a carrier of
angular frequency we, the resulting modulated wave may be written
and
vm(1‘)
where
is
453
=
Re{Vcej‘”C’[l
mf(r)])
+
(6)
modulation coefficient. In
substituting (l) in (6), we use com for the
frequency
modulating (base—band) signal, and assume that its significant frequency components extend only over a band
(03 S a) S mg:
m
a
of the
—
“’3
Um(t, 0)
Or
letting
w
=
Re{Vcej“’C’[l 5—1]
g(wm)ej‘”m’
+
==
77
dwm]}
(7)
a)c + cum,
Re
=
VCerC’
w°+w3
mVC
.
vm(t, O)
+
.
g(w
277‘
—-
(9,)er
do)
(8)
("c”wB
above we in the integral in (8) correspond to upper sideband terms and
those below we to lower sideband terms. Each frequency component propagates
according to its appropriate phase constant B. Let us expand B as a Taylor series
Frequencies
about me:
B”)
“
So the modulated
UmU’ Z)
:
Mme)
_
signal,
+
(‘0
after
a
éé
”Jam
propagating
+
(Q)
my 5133
-
M
2
we
a
distance
+...
2, is
R6{Vcej[wc’“,3(wc):]
(10)
we
1
X
+
l
277’
f
(9)
we
3(a)
m
)enwmu-:/vg>—<afi,/2>:<d313/dw3)+-~1
them
wa
where
l
—-
=
dflB/dw2
and
higher
Um(z‘, z)
=
terms are
(11)
dw
vg
Now if
a’
“E
we
negligible, (10)
is
interpreted
as
Re{Vcej[“’C’“fl(a’°):][l mf<t 5):”
+
-
(12)
s
envelope propagates without distortion at group velocity vg (though the carrier
generally different phase velocity). But if the higher-order terms are
not negligible, the envelope is distorted and there is said to be group dispersion. For a
gaussian envelope,
so
the
inside
moves at a
f(t)
x
Ce”(2’/T)2
(13)
454
Chapter
8
Waveguides with Cylindrical conducting Boundaries
/W\
i ll
\
”a
41
\U
4
<71!“
_,
t
lllllllhzrg
W
at
t
x
\\
i
\‘iji
[I
i
Illustration of the
FIG. 8.16b
mm
inmflllir
lHlll III | ||
\
,
\
a
J
spread
of the modulated
envelope
of
a
pulse
as
it travels down
a
system with group dispersion.
It
a
can
be shown
width 7" after
(Prob. 8.160) that the
propagating
term
dzfi/ (1602
distance 2, with 7’
the
2
to
spread
to
{1 (-13-;ij ]
1/2
+
z
envelope
given by
2
7'
causes
(14)
The spread of a gaussian envelope, illustrated in Fig. 8.1619, clearly limits data rates as
pulses begin to overlap their neighbors. Although a factor in some waveguide problems
(Prob. 8.16a) the limitation is most important for optical fibers and will be met again
in Chapter 14.
PROBLEMS
will see later, one mode of a rectangular waveguide is a TM wave with Hz
A sin(7rx/ a) sin(71y/ b) with z and t dependence assumed to be e1(“’"fiz).
E5
Find expressions for the transverse field components. At a given plane what are the
phase relations among the transverse components and between them and E2.
8.2a As
=
we
and
8.2b The division into TM and TE classes is not the
as
O
=
noted in Sec. 8.2. Another
electric (LSE) with EI
magnetic (LSM) with Hx
tween E: and Hz for each
=
0 but all
=
way of
classifying guided waves,
employs longitudinal-section
other components present and longitudinal-section
frequently
only
useful division
0 but all other components present. Find the relations be-
of these classes.
8.3a Add induced charges and current flows, with attention to sign, to the pictures of Figs.
8.319 and c for the positively traveling TMI and TE1 waves. Repeat for negatively
traveling
waves.
455
Problems
8.3b Calculate cutoff
1.5
cm
for
TEI, TEE, T133, TMI, TMZ, TM3 waves between planes
a glass dielectric with e'/eO
4. Suppose
GHz is provided at a cross section of the air-filled line and all waves
frequency
apart with air dielectric. Repeat for
excitation at 8
=
excited. Which wave(s) will propagate without attenuation? At what distance from
plane will each of the nonpropagating waves be attenuated to 1 / e of its
are
the excitation
value at the excitation
8.3c The
slope
curve
and
of
for
E:
an
an
plane?
electric field line in the
electric field line of
of the wave, is defined
cos
a
TMl
x~—z
plane is (Ix/dz
=
Ex/E,
wave, obtained from the
Show that the
for Ex
expressions
by
B:
=
mo/a][cos(n:r/a)]“‘
[cos
where 3:0 is the value of x for a given curve at z
0. Plot one or two lines to
the form shown in Fig. 8.3b. [Hint First express fields as real functions of 2.)
2
8.3d
Similarly
wave
to
and
8.3e* Find the
Prob. 8.3c, derive the expression defining magnetic field lines for
one or two lines to verify the form shown in Fig. 8.36.
expression
for electric field lines for
field lines for
8.3f Show that the
also
applies
1.5
give a plot similar
a TE2 wave.
expression
to
8.43 Calculate the
planes
a
TB,
plot
sketch the remainder to
magnetic
verify
TBIII
angle
cm
for energy
to
velocity
a TM2 wave, plot one or
Fig. 8.31). Similarly, plot
derived for
as
TM,"
two
lines, and
and sketch
waves
[Eq. 8.367)]
waves.
6
as
defined in
Fig.
8.40 for ray directions of a TM1 mode between
4, for frequencies of 5, 6, 10, and
8780
apart with glass dielectric,
=
30 GHz.
8.4b Obtain the
uniform
8.4c*
expressions for wave impedance
plane waves reflecting at an angle.
of TM and TE waves,
using
the
of
picture
By suitably changing coordinates as in Ex. 8.4, show that the expressions
6.09(l8)—-(20) for a wave polarized with electric field normal to the plane of incidence
striking a conductor at an angle correspond exactly to the field expressions for a TE,"
wave.
8.5a Find average power transfer and conductor loss for a TB mode between
to verify the expression for attenuation, Eq. 8.5( 12).
parallel planes
8.5b Calculate attenuation in decibels per meter for a TM, wave between copper planes
1.5 cm apart with air dielectric. Frequency is 12 GHz. For the same frequency and
2 X 10'"3 is introduced. Calculate
4, e"/e’
spacing, a glass dielectric with e’/s0
=
=
attenuation from both dielectric and conductor losses.
8.5c Prove that the
losses, is
frequency
of minimum attenuation for
attenuation and calculate for silver conductors 2
m
=
a
TM," mode,
from conductor
VSfC, where fc is cutoff frequency. Give the expression for the minimum
cm
apart and air dielectric for the
l, 2, and 3 modes.
8.5d Show that the transmission-line formula for attenuation constant, Eq. 5.9(7), gives precisely the same result as the approximate wave analysis of Sec. 8.5 for the TEM wave.
8.5e Derive the
ing
a
8.5f* Since
=
E:
approximate
formula for attenuation constant due to dielectric losses
by
wL/2WT.
is
equal
and
opposite
at
top and bottom conductors for TEM
wave
in the
us—
456
Chapter
Waveguides
8
parallel—plane line,
with
Cylindrical Conducting Boundaries
it is reasonable to
assume a
linear variation between the two
values:
2
R
E
=
(1
+13fi<1 i)
—-
(I
77
Find the modification in the distribution for
E. Find the corresponding modification in
Er
the average
Poynting
the
divergence equation for
equations. Describe
of position in the guide.
satisfy
from Maxwell’s
Hy
qualitatively
to
vector as a
function
1 mm, d
2 mm, at
8.6a For a symmetric stripline as in Fig. 8.60 with w
2.7, and
thickness t negligible (but larger than several penetration depths), calculate Z0 and
phase velocity of the TEM mode and the cutoff frequency of the next higher mode.
[Note that tables of elliptic integrals are required]
=
=
8.6b For er
==
1, the lossless microstrip of Fig. 8.6b
can
propagate
=
a true
TEM
wave
at the
capacitance per unit length for a 50-9. line with
such a dielectric and the required w/d for this from Fig. 8.6c. Calculate the difference
due to fringing fields between the capacitance per unit length found above and that
given by the parallel~p1ane approximation, and express this as an equivalent extra
width, Aw/d. Now maintaining w/ d constant, assuming inductance is independent of
8., and transmission—line equations applicable, repeat for other values of a, and plot the
extra equivalent Aw/d due to fringing as a function of 8..
velocity
of
light.
Find inductance and
impedance for a copper microstrip line with an alumina
8
dielectric and air above the line. The dimensions should be w/ d
and d
0.2 mm. Compare the results obtained using the formulas with the graphical
3 GHZ.
O and f
data in Sec. 8.6. Find the fractional change of Zo between f
Calculate the maximum frequency at which the static approximation should be used.
8.6c Calculate the characteristic
(A1203 ceramic)
=
z
=
8.6d
=
stripline with the same materials and substrate thickness d and having the
impedance found in Prob. 8.6a for the microstrip line. Calculate and
compare the attenuations in the microstrip and snipline at 3 GHz assuming conductor
thicknesses of 0.01 mm. Neglect dielectric losses.
Design
a
characteristic
15-0.
with the maximum
possible delay achievable
possible lines. One is to
100 um and alumina (A1203 ceramic) diebe made with cepper conductors with w
lectric and is to be used at room temperature. The other is made with superconducting
100 um and undoped silicon dielectric, having a,
niobium conductors with w
10—5 at 4.2 K, at which temperature the line is to be
11.7 and loss tangent tan 65
10'5 9. for niobium at 4.2 K and the strip thickness to be 5 pm for
used. Take R5
copper and 1 pm for niobium. Find the maximum delay achievable with each of the
8.6e It is desired
with
to
no more
make
a
stripline
than 3 dB attenuation at 10 GHz. Consider two
=
=
=
=
=
lines.
8.6f Consider the coplanar waveguide strip transmission line shown in Fig. 8.6f. Assuming
the line is on an infinitely thick dielectric substrate, the electric fields are distributed
symmetrically above and below the line.
(i) Argue that this leads to an effective dielectric constant gaff
50 0. using a,
(ii) Find the dimensions to give a line with Z0
lowing design formula“
=
=
(e,
=
+
l)/ 2.
3.78 and the fol-
-
w
—~
a
where
Zoo
is the characteristic
.,
=
tanh"
impedance
in 2
77770
(8200
—-—
—-
——
2
)
when the dielectric constant is cr
=
l
457
Prohiems
everywhere,
width
8.6g
and
w
is the width of the
strip
located in the center of
a
gap of
a.
The various
frequency components in a signal (eg. a pulse) propagate at phase velocities determined by the effective dielectric constants at those frequencies As will be
discussed in Sec. 8.16. this variation of velocity leads to dispersion of signals. The
fractional variation of phase velocity with frequency in a coplanar waveguide is lower
at low frequencies than it is in microstrip. Consider the following 50—0 lines with copper conductors and 0r635~mm-thick alumina (A1203 ceramic) substrates The coplanar
line has a strip width w of 0.266 mm and gaps s of 0.117 mm each. The strip width in
the
(i)
line is 0.598
microstrip
min.
Plot the fractional change of phase velocity of the quasi-TEM mode as a function
of frequency in the range 0 < f < 50 GHz for the coplanar guide and 0 < f < 35
GHz for the microstrip. For the microstrip. mark fmx, the limit of applicability of
the static formulation, and also the cutoff frequency of the next higher mode,
c/4dV er
(fJHE,
cZu/Znud. Also mark the cutoff frequency f-rE
next higher mode for the coplanar waveguide.
(ii) Find the fractional change of the phase velocity at the cutoff frequency
higher mode for the coplanar waveguide.
=
=
8.6h
Compare the total attenuation at 3 GHz in nepers/meter for the two lines
Prob. 8.6g and explain the physical reason why the higher one is higher.
8.7 a For
—
l of the
of the next
described in
rectangular waveguide with inner dimensions 3 X 1.5 cm and air dielectric. calfrequencies of the TB“), T520, TEl I, TED, TEZI, TE“. TM, ,, TMZZ
modes. Repeat for a glass dielectric with 575.,
4.'Find lengths to the l/e distances
for the nonpropagating modes excited at 10 GHz.
a
culate the cutoff
=
8.71) Derive the
expression
plot
two such
one or
for
magnetic
lines in the transverse
8.7:: Derive the expression for electric field lines in the
plot
one or two
such
plane
of
a
TM“
transverse
plane of a TEH
guide
is
(neither
this does
8.7f
not
zero)
apply
wave
and
to
in the
m
=
a
TM,,,,,
mode
given by Eq. 8.7(14).
as
8.7e Show that the expression for attenuation because of conductor loss for
m nor :1
and
lines, comparing with Table 8.7, (See approach in Prob. 8.3a.)
8.7d Show that the expression for attenuation because of conductor loss for
in the rectangular
wave
lines, comparing with Table 8.7, (See approach in Prob. 8.3c.)
rectangular guide
O
or I1
=
0
is
as
'I'E,,,,, mode
given by Eq. 8.7(26). Explain why
a
case.
that surface resistivity RE is a function of frequency. find the frequency of
minimum attenuation for a TM”. mode. Show that the expression for attenuation of
TE,,,,, mode must also have a minimum
Recalling
3
the wave types studied so far, those transverse magnetic to the axial direction were
obtained by setting H:
0; those transverse electric to the axial direction were ob—
tained by setting E:
0r For the rectangular waveguide, obtain the lowest»order mode
with H‘,
0 but all other components present. This may be called a wave transverse
8.7g“ Of
s
2
S
magnetic
and TE
waves
to
the
waves
exactly
8.7h* Repeat Prob.
cancelr This is
8r7g
8.7i From the form of
tion
E:
=
0
direction. Show that it may also be obtained by superposing the TM
of just sufficient amounts so that fix from the two
.\‘
given previously
on a
for
a
longitudinal—section
a wave transverse
electric
to the x
Eqs. 8.2(9)7(12). show that for a
perfectly conducting boundary of
tangential component of E also
to
be
zero
wave as
discussed in Prob. 8.2r
direction.
imposition of the condicylindrical guide causes the other
boundary.
TM wave,
a
along that
458
Chapter
8.83 For
tric
also
f
8
with
Waveguides
Cylindrical Conducting
Boundaries
design a rectangular waveguide with COpper conductor and air dielec1.30fc) but
TB“) wave will propagate with a 30% safety factor (f
the wave type with next higher cutoff will be 20% below its cutoff fre—
3 GHz,
=
that the
so
so
that
=
quency. Calculate the attenuation due to copper losses in decibels per meter.
8.81) For Prob. 8.8a, calculate the attenuation in decibels per meter of the three modes with
cutoff frequencies closest to that of the TElo mode, neglecting losses.
8.8e
Design
guide for
guide is to
a
that the
use at
3 GHz with the
be filled with
a
requirements as in
having a permittivity
Prob. 8.83 except
same
dielectric
four times that of
air. Calculate the increase in attenuation due to copper losses alone, assuming that
perfect. Calculate the additional attenuation due to the dielectric, if
the dielectric is
e" c'
=
0.01.
8.8d Find the maximum power that can be carried by a 6~GHz TE10 wave in an air—filled
guide 4 cm wide and 2 cm high, taking the breakdown field in air at that frequency as
2 X
106 V/m.
8.8e The transmission~1ine analogy can be applied to the transverse field components, the
ratios of which are constants over guide cross sections and are given by wave imped~
ances, just as in the case of plane waves in Chapter 6. A rectangular waveguide of
inside dimensions 4 X 2 cm is to propagate a TE10 mode of frequency 5 GHz. A
dielectric of constant a,
3 fills the guide for z > 0, with an air dielectric for z < 0.
Assuming the dielectric-filled part to be matched, find the reflection coefficient at
z
O and the standing wave ratio in the air~filled part.
=
=
8.8f Find the length and dielectric constant of a quarter~wave
between the air and given dielectric of Prob. 8.8e.
8.9a Derive the set of
matching
Eqs. 8.9(1)-(4) by utilizing Maxwell’s equations
assuming propagation as e‘lfi‘.
section to be
in circular
placed
cylindri-
cal coordinates and
8.9b What inner radius do you need for an air-filled round pipe to propagate the TE11
at 6 GHz with operating frequency 20% above the cutoff frequency? What is the
for this mode? Find the attenuation in decibels per meter of the
mode at this frequency, neglecting losses for that calculation.
wavelength
8.9c Show that the
expression
for attenuation from conductor losses of
ac
a
TM,,,
a
TB",
wave
guide
TM01
mode is
Rs
=
1
mlV—fi
(CDC/(D)
....
At what value of
8.9d* Show that the
(OJ/(Dc
is this
expression for
c
a
minimum?
attenuation from conductor losses of
017%
-
(inc/co)2
0’
13:3
"
mode is
”2
8.9a For
a circular air-filled guide with copper conductor, select a radius so that the
TEOI
mode has attenuation of 0.3 dB / km for a frequency of 4 GHz. Estimate the number of
modes (counting only the symmetric ones with n
0) that have cutoff frequencies
==
below the
operating frequency.
8.10 Use the
asymptotic forms of Bessel functions in Eqs. 8.10(1) and (2) for TM and TE
respectively, to show that for large kcr, and ro/r, near unity, the cutoff wave1 modes is approximately twice the spacing between
O, p
length of the n
waves,
=
conductors.
=
459
Problems
8.113 Sketch
using
of mode
examples
for each
a
8.11b Plot fraction of power
function of
sions
are as
couplings by
each of the six methods described in Sec. 8.11
system different from the
frequency
coupled
from
a
one
coaxial line into
from 10 to 11 GHz if
stated in the
utilized in
probe
a
Fig.
8.11 to illustrate it.
waveguide (Fig.
radius is 1.5
mm
8.1 1 g)
as a
and other dimen—
figure caption.
8.123 Demonstrate that, although in a TEM wave E does satisfy Laplace’s equation in the
transverse plane and so may be considered a gradient of a scalar insofar as variations
in the transverse
in all directions
plane
(3:,
are
y, and
concerned, E is
z) are included.
not
the
gradient
of
a
scalar when variations
8.12b Two
perfectly conducting cylinders of arbitrary cross-sectional shapes are parallel and
a dielectric of conductivity 0- and permittivity a. Show that the ratio of
electrostatic capacitance per unit length to dc conductance per unit length is 8/0‘.
separated by
8.12c If the conductors
ity
H:
a,
=
are
perfect
but the dielectric has
show that 3/ must have the
following
conductivity
value for
a
TEM
0 as
permittivs
O,
(E:
as
exist
=
O):
7
=
iijchr
+
jcvefl"2
the distribution of fields may be a static distribution
unlike the case for a lossy conducting boundary.
Explain why
line,
well
wave to
as
in the loss~free
8.12d How many linearly independent TEM waves may exist on a three-conductor transmission line? Describe current relations for a basic set. Complete the proof that there
can be no static field, and hence no TEM wave, inside a single infinite cylindrical
conductor.
8.1321 Show that the circuit of Fig. P8.13a may be used to represent the propagation characteristics of the transverse magnetic wave, if the characteristic wave impedance and
propagation constant are written by analogy with transmission~line results in terms of
an impedance Z1 and an admittance Y , per unit length, and the medium is ,LL,, 8].
ZTM
=
[Z
r——
3;“,
”Y
ZxY1
z
1
Note the
bering
of
similarity
course
between this and the circuits of conventional filter
that all constants in this circuit
are
in
sections,
reality distributed
remem-
constants.
(jig? )dz (j::)dz
0W
W—o
jwmdz
jwuldz
1108le
G
I
v
——o
FIG. P8.'l3a
8.13b Show that all field components for a TM wave may be derived from the axial compo—
nent of the vector potential A. Obtain the expressions relating Ex, Hx, and so on to A2,
the differential equation for A3, and the boundary conditions to be applied at a perfect
conductor. Repeat using the axial component of the Hertz potential defined in
Prob. 3.19b.
466
Chapter
Waveguides with Cylindrical Conducting
8
Boundaries
8.13c Show for a TM wave that the magnetic field distribution in the transverse plane can be
derived from a scalar flux function, and relate this to E5. With transverse electric field
derivable from a scalar potential function and transverse magnetic field derivable from
a scalar flux function, does it follow that both are static—type distributions as in the
TEM Wave?
Explain.
8.13d Show that energy velocity equals group
guide of general cross section.
velocity
for the TM modes in
a
lossless
wave»
.
8.13e* Show that
E2(x, y)
for
a
general
TM
wave
in
a
perfectly conducting guide satisfies
the
equation
—1
k3
[f
=
d8] [f d5]
(Vth)2
E?
5
where
V, represents
the transverse
s
integral is over the cross section
positive for waves in which phase is
and the
gradient
From this argue that k3: is real and
constant over the transverse plane.
the
guide.
of
8.13f* Numerical methods can be used to find the propagation constants for waveguides
of arbitrary cross section. Following the procedures used in solving the Laplace
or Poisson equations in Sec. 1.21 to get a difference equation solution for the scalar
Helmholtz equation V24! + kit/r
0, one finds the residual at the kth step to be
=
With y)
ll/‘Wx. y + 12)
(4
kfilzz)/1,U(k' 1)(x, y).
=
-
-
+
1.0%. y
The
-
change
h)
Woe
+
.QR‘W (4
~
one
+
115%"
-
k, y)
iteration step to
governed by W)
the next in the successive overrelaXation method is
+
12, y)
+
of variable from
=
Wk" 1)
with (1 set to 1.0 for convenience to make
kfihz). Apply
equations
k9; for‘a TMll mode
the
anumerical evaluation of
in a rectangular waveguide. Assume a
rectangular guide with side ratio 1:2. The Helmholtz equation to be solved is Eq. 813(1).
Divide the waveguide into a grid of 18 squares and number the interior points 1-10
left to right, top to bottom. A reasonable initial guess for the product lag—[22
1(2122 can
be formed assuming a one~dimensional variation in the smallest dimension; here take
1632122: 1.1. Start with E2 having 'the following values at the grid points as a first guess:
for points 1, 5, 6, and 10, E:
30; for points 2, 4, 7, and 9, E3
50; for points 3
and 8, E:
70. Use simple relaxation twice to improve the values of E: for the given
kfihz. Then calculate an improved value of kill2 using the relation
=
=
=
=
19/2
°2
2
+
En, y>[E.~
E...
+
+
Es
aw
-
433a, yr]
=
,
ZQW»
where N, E, S, W indicate the points surrounding the grid point at (x, y) and the sum~
are over all grid points. Next make two more steps of relaxation to adjust the
mations
fields to the
Compare
new
ICE/12.
Then
use
the above formula to get
the result with the value of
a
second correction to
kih2 found using differential equations
kihz.
in
Sec. 8.7.
8.142 Derive the equivalent circuitfor
Prob. 8.13a.
a
TB
wave
analogous
to that of a TM wave
given
in
--
8.14b Show that fields
satisfying
Maxwell’s
equations
in
a
homogeneous charge-free,
cur-
463
Problems
rent~free dielectric may be derived from
E
a
~1V
=
vector
X
potential
F:
F
8
1-1:.
V(V-F)-—ij
jw/.L8
(V2 +k2)F=
Obtain
0
expressions for all field components of a TB wave from the axial component F
potential function, and give the differential equation and boundary condi—
:
of the above
tions for F.
8.14c Show that if
derivation of
8.14d Show for
a
one
a
utilizes the
TB wave,
potential function
more
than
one
A instead of the F of Prob. 8.l4b for
component is required.
TE mode that transverse distribution of electric field
scalar flux function. How is this related
8.149. Show that the energy
waveguide of general
velocity equals
cross
to
can
be derived from
a
HS?
the group
velocity
for the TB modes in
a
lossless
section.
in any shape of guide passing from one dielectric material to
frequency the change in cutoff factor may cancel the change in n,
and the wave may pass between the two media without reflection. Identify this condition with the case of incidence at polarizing angle in Sec. 6.13. Determine the require—
ment for a similar situation with TB waves, and show why it is not practical to obtain
8.14f Show for
a
TM
another, that
wave
at one
this.
8.15 A
particular waveguide
attenuator is circular in cross section with radius 1 cm. Plot
attenuation in decibels per meter for the TB“ mode over the frequency range 1—4
GHz. Also plot attenuation of the mode with next nearest cutoff frequency.
8.16a For
a
term
with
hollow—pipe waveguide, with ,8 given by Eq. 8.13(9), find the group dispersion
Find the length of waveguide for which the width of a gaussian pulse
0.85.
7
1 ns is doubled if frequency is 10 GHz and (ac/co
dZB/dwz.
:
=
8.16b Find d 3,8/ (In)2 for a transmission line with series resistance R and shunt conductance G
independent of frequency, where R/wL and G / wC are small compared with unity. ReO and R governed by skin effect. Is the resulting
peat for a coaxial line with G
group dispersion likely to be significant in usual applications?
=
a gaussian function f(r) given by Eq. 8.1603) and find its g(w). Using this
Eq. 8.16(10), show that the envelope broadens with 2 as given by Eq. 8.1604).
8.166“ Start with
in
8.16d“‘ From the solution of Prob. 8.16c find phase (1) at z for the high-frequency
gaussian envelope and find the frequency “chirp,” defined as dq’J/dt.
pulse
with
9.]
INTRODUCTION
preceding chapter dealt with the important special case of waveguides with cylinconducting boundaries. In this chapter, we examine several examples of waveguiding systems of different shapes and properties.
We start with dielectric guides which demonstrate that boundaries other than metal—dielectric boundaries can guide waves in cylindrical systems. Dielectric guides are now
important for optical communication uses and are explored more in Chapter 14. There
follows an examination of radial guiding, both in cylindrical coordinates and in Spher~
ical coordinates. The former is important in certain classes of resonant systems and the
latter in antenna theory. Waveguides of special cross section, used either for impedance
matching or to lower the cutoff frequency for a given transverse dimension, are ana—
lyzed. Finally, classes of waves with phase velocity much slower than the velocity of
light are studied both in uniform systems and in periodic systems. These are important
for such purposes as the interaction with electron beams in traveling-wave tubes and
for confining electromagnetic energy near a surface.
The examples selected are not only important in themselves but illustrate a number
of important principles. The principle of duality shows how the solution of one problem
may sometimes be used for another by interchange of electric and magnetic fields.
Cutoff in a waveguide is shown to correspond to a condition of transverse resonance,
which leads to a variety of approximate and exact techniques for analyzing guides of
irregular shape. Periodic systems show the importance of spatial harmonics for inter~
preting the behavior of such systems.
The
drical
9.2
Dielectric rods, slabs,
dielectric of lower
or
films
DIELECTRIC WAVEGUIDES
can
permittivity.
guide electromagnetic
Guides of this type
462
were
energy if surrounded by a
analyzed by Hondros and
9.2
Dielectric
463
Waveguides
(d)
FIG. 9.2
((1) Rays totally reflected from dielectric boundaries when 81 > 82 and 61 > Be.
(b) Form of electric field distribution versus x in lowest-order mode. (6) Form of field versus )5
in next higher-order mode. (d) Leaky wave when 61 < 6c.
in 1910 and demonstrated by Zahn2 in 1916. They have become very important
guides for light in Optical communication systems, and so will be treated in detail in
Chapter 14. Because of the importance of the principle and some use in other frequency
ranges, we introduce the subject here with some physical pictures and comparisons with
metal—clad guiding systems.
To illustrate dielectric guiding, consider the slab guide of Fig. 9.20 with dielectric
sI surrounded by 82 < 31. In this picture we consider the mode as made up of plane
waves reflecting at an angle from the boundaries between dielectrics, interfering within
the slab to produce a particular mode pattern when conditions are correct.3 From the
concept of total reflection (Sec. 6.12), we know that all energy is reflected from
the interface if angle 61 of the rays (normals to the wavefronts) is greater than critical
angle 6C,
Debyel
as
ac
where
is
we assume
analogous
differences
,u.1
=
(1)
{1,2 The interference within the slab to produce a mode pattern
TE10 mode of rectangular guide (Sec. 8.8). The
.
are
in the
phases
extending
The exponential decay
D. Hondros and P.
3
sin"1(ez/el)1/2
to that described for the
into the dielectric
2
=
of reflections at the boundaries and in the evanescent fields
regions
above and below the dielectric
of fields in the dielectric of
region 2,
guide.
6I > 6C,
when
is
given
Debye, Ann. Phys. 32, 466 (7970).
Phys. 49, 907 (7976).
H. Kogelnik, in Guided Wave Optoelectronics, (T. Tamir, Ed), 2nd ed. Springer-Verlag.
H. Zahn, Ann.
New York, 7990.
464
Chapter
which for #1
by Eq. 6.l2(7),
==
Special Waveguide Types
9
#2,
=
coefficient in the
decay
8
a)
a,
the
gives
3 8:52
[—1
sin2 6,
-
2
1]
x
higher
mode in
Fig.
l8
Cutoff for the dielectric
2
9.26. The
klz
phase
k1
:
sin
as
1/2
(2)
The transverse field distribution in the lowest-order mode is sketched in
that for the next
direction
constant
along
the
Fig. 9.2b,
guide is
and
(3)
91
guide is considered the condition for which '61
=
0,:
at
which
point B
k2. For steeper angles, (91 < Go, there is some energy transmission into the
outer medium on each reflection, leading to leaky waves, as indicated in Fig. 920?.
=
For the
interfering zigzag plane
waves
to
form
mode, the phase lag after reflection
a
from top and bottom and retum to the top must be
~2k1d
where
m
we can
is
cos
2q5
+
01
a
multiple
of 2m
772277
=
(4)
integer and qb is the phase at reflection from medium 2. Using Sec. 6.11
phase at reflection from p for TE and TM4 waves, respectively, taking
an
find
Fitz/123
chE
H
2
tan“1
61]
[Vsinzfi1
[J63- 61) ?/cos 6,]
sz/al/cos
——
(5)
2
2 tan“1
than
sin
——
2
The mode with
m
=
0 in
but for other values of
m
(6)
2
(4) exists for arbitrarily small kid for both TE and TM modes,
there is a minimum kld for cutoff (Prob 9.2d). For the m
=
0 mode with
kid small, ax is also small so that fields extend well into the external
the
mode is only weakly guided (Prob. 9.2e).
i.e.,
region:
physical principle of wave reflection at the interface applies to fibers and
guides of circular cross section, but the detailed development is more
These
will consequently be treated in Chapter 14 by a field analysis,
complicated.
with
a
more
detailed treatment of the planar guides.
together
The
same
other dielectric
9.3
Linearly propagating
a
connection
the two
waves
4
plates
was
are
between
PARALLEL-PLANE RADIAL TRANSMISSION LINES
waves
between
parallel planes
made between the TEM
the conductors of the line. Here
parallel conducting planes
as
in Sec. 6. l i, there is
result of a reversal in
an
we
a
extra
an
er
studied in Secs. 8.3-8.5 and
transmission-line wave, where
analyze radially propagating TEM
and introduce
For TM reflection coefficient defined in terms of
the interface,
were
and
wave
a
transmission-line type of
component parallel to
phase at reflection. But this is a
electric field
in the
spatial direction, compensated for on the next reflection.
9.3
465
Parallel-Plane Radial Transmission Lines
input voltage and load
impedance assumed
uniformly distributed
about circumference
NJ
(0)
(b)
FIG. 9.3
(a) Radial transmission line with input
with
at
input
at outer
radius.
(b) Radial transmission line
inner radius.
formalism to facilitate
impedance-matching calculations (Figs. 9.30 and 1)). Higher—
following section. The wave under consideration has
no field variations either circumferentially or axially. There are then field components
E2 and H¢ only. The component E2, having no variations in the z direction, corresponds
to a total voltage Ezd between plates. The component
H45 corresponds to a total radial
current
in
outward
one plate and inward in the other. This wave is then anal277111,,”
ogous to an ordinary transmission-line wave and thus derives its name, radial transorder modes
are
introduced in the
mission line.
For the
simple
wave
described there
are no
radial field components, and
analysis
may be made by the nonuniform transmission-line theory of Sec. 5.17, allowing L and
C to vary with radius. However, the wave solution for fields may also be obtained
directly from the results of Sec. 7.20. Since there are no qb or z variations, 1) and 7/ may
be set equal to zero. Special linear combinations of Bessel functions have been defined
particularly for this problem,5 but for occasional solution of radial line problems, known
forms of the Bessel functions are satisfactory. The form of Eq. 720(6) in the Hankel
functions is particularly suitable since these can be shown to have the character of
waves traveling radially inward or outward. Since kC
('y2 + k2)”2 by Eq. 8.2(19)
=
and y
=
O has been assumed, then
E
With
v
and 3/ zero, the
=
kc
com.
=
AHg’Ucr)
only remaining
+
BH32>(kz-)
field component in
(1)
Eqs. 8.9(1)—(4) is 1-14,:
'
BE-
1
H4,
The H 5,” terms
the
5
are
___
_--
=
fro/.1. 6r
identified
positively traveling
N. Morouvir‘z, in
=
wave
as
i
7]
the
[AH§1)(kr)
+
BH‘EKIa-n
negatively traveling wave and the H f3) terms as
asymptotic forms that approach complex
because of the
Principles of Microwave Circuits (C. 6. Montgomery,
Chap. 8, McGrow—Hifl, New York, 7947.
E. M. Purcell. Eds).
(2)
R. H. Dicks, and
466
Chapter
exponentials (Sec. 7.15).
It is convenient to utilize the
Special Waveguide Types
9
and
magnitudes
phases
of these
functions,
H81)(v)
H 8%!)
J'HWU)
jH§2)(v)
=
Jo(v)
=
fo(v)
—
-N1(v)
=
=
+
=
jN0(v)
=
+
+
N1(v)
jN0(v)
171(0)
17101)
G0(v)ej9(”)
(3)
Game—1'9“”
(4)
Gl(v)e"”‘”)
(5)
~G1(v)6'j“’(")
(6)
==
=
where
G0(v)
=
01(1))
=
Expressions (l)
and
[J 3(1))
+
We)
+
N%(v)]1/2,
6(v)
=
Nam/2,
W)
=
tan“l[m]
tan“[%]
(7)
10(0)
J
(8)
(2) then become
E2
:
Go(kr)[Aej9(k’)
G
:
95
(,0)
Be“j9("")]
+
'
’1‘
[Aemm
%
_
(9)
-~'
‘1‘
Be mu )1
(10)
where
2000)
.
is
a
characteristic
radially dependent
__
'-
wave
77
6000‘)
(11)
51210.)
impedance.
specification of two
Evaluation of the constants A and B follows from
given
radii. For
example, given Ea
E,
G0
z
a
~
003(9
at rd and Hb at rb, the
*-
GOa 003(9a
“‘
ll/b)
.
90b)
+ J zObe
field values at
fields at any radius
Go sin(0
Goa 003(6a
—--
6)
a
““
W27)
r are
( 12 )
'
H95
G1
2
b
0050;”
-
Glb 005(60
"‘
6a)
+
4’s)
j
E061
51110.0
—-
ZOaGla COS(60
_
$17)
11%)
(13)
ordinary transmission~line equations except for the use
special magnitudes
special radial line quantities versus
kr is given in Fig. 9.34:. This is not very accurate for small values of kr, so the following
approximations are then preferred:
These
are
similar in form to the
of the
and phases. A plot of these
cow)
w
01(1))
z
7371477”)
6(v)
m
my)
z
(14)
1
2
7—7-5
tan—1E; 4123)]
tan"1<wvz)
(15)
2.4
467
Parallel-Plane Radial Transmission Lines
9.3
600
£—
2.0
500
r
V; (scale
L6l-
400
L
300
1.2
A), ohms
8
Scale
0.8
200
*-
(scale '8)
-
0-4
0
L"
100
11/011 degrees,
scale A)
0
0
fl
1
(In degrees,
scale A)
3
2
5
4
(2—?)
FIG. 9.3c
Radial transmission-line
quantities.
where y
More accurate values can be calculated from the definitions (7)
0.5772.
and (8) utilizing tables of J and N, although some tables give magnitude and phase of
=
.
H SE) and H E?)
input
wave
.
directly.
Forms similar to
different radii
.
or
(12) and (13)
two values of H
impedance Zi
:
be derived for two values of
can
E2 specified at
defining an
/H¢,L is given.
The most useful form, however, is that
g.
when load impedance ZL
EzL
Ezi/Hd,i
=
This is
Z1
_
0*
[2L
cos(6i
20L cos(z,bi
—
——
+
jzoL sin(9i
+
91)
sz smog
90L)
-—
—-
60]
sz)
(16)
O) and open (ZL
00) are especially
Special forms of this for output shorted (ZL
simple and are analogous to the corresponding forms for uniform transmission lines.
All of the foregoing relationships are given in terms of fields or wave impedances.
=
:
468
Chapter
Usually current, voltage,
and total
Special Waveguide Types
9
impedance
V
I
—E_,d,
__
Zmfl
d
=
are
(17)
2777-11,,
E,
+
-
desired. The relations
are
m
(H)
(18)
The Sign convention defines higher voltage in the upper plate and outward current in
the upper plate as positive. The upper sign in (18) is for input radius less than that of
the load, ri < rL, and the lower sign for the reverse, ri > rL, since in this case the
convention for positive current would be the opposite of (17).
9.4
CIRCUMFERENTIAL MODES
Many higher—order
modes
section. All those with
z
can
IN
RADIAL LINES: SECTORAL HORNS
exist in the radial transmission lines studied in the last
variations
require
a
spacing between plates greater than
a
half~
wavelength
propagation of energy. For smaller spacing, modes are possible
with circumferential variations but no 2 variations. The field components may be written
for radial
E
=
Hq,
==
H,
z
AvaUcr)
sin
M
“.37. A,Z{,(kr)
(1)
sin
vqb
(2)
'
A
13—”
knr
Zv(kr)
cos
(3)
ms
foregoing equations, 2,, denotes any solution of the ordinary vth order Bessel
equation. For example, to stress the concept of radially propagating waves it may again
In the
be convenient to utilize Hankel functions:
Zva'I‘)
=
H 900‘) +
CvH§2)(/<I‘)
(4)
These circumferential modes may be important as disturbing effects excited by asymmetries in radial lines intended for use with the symmetrical modes studied in the
preceding section. In this case 11 must be an integer since the wave must have the same
0 and 43
value at (I)
277. Waves of the form (1) to (3) may also be supported in a
with
O and 45
wedge—shaped guide
conducting planes at q!)
(no as well as at z
z
=
=
=
=
Circumferential Modes in Radial Lines: Sectoral Horns
9.4
FIG. 9.4a
0, d (Fig. 9.40). The latter
Wedge—shaped guide
case
is
for radiation. In this case, since
E:
important
must
be
——
v
as a
or
sectoral horn.
sectoral
qb
zero at
==
electromagnetic horn6
discussed here
waves
are
interesting
1737—7
in
used
0, qbo,
(5)
$0
The
469
one
respect especially. If we think of the
lowest~order mode (m
l) propagating radially inward in the pie—shaped guide for
it
would
be
Fig. 9.4a,
quite similar to the TE10 mode of the rectangular guide, although
2
modified
by
phenomenon
larly, for the
we
the convergence of the sides. We would consequently expect a cutoff
at such a radius rC that the width rcqio becomes a half—wavelength. Similowest~order circumferential mode in the radial line of
would expect
a
cutoff at such
a
radius that circumference is
one
Figs. 9.30 and b,
wavelength:
A
27n-c
A casual
inspection
=
of
A for radial line,
(l)
to
(3) would
gborc
=
E
for sectoral horn
(6)
not reveal this cutoff since there is no sudden
in the rectangular guide at cutoff. However,
study would reveal that there is a very effective cutoff phenomenon at
about the radius predicted by (6) in that the reactive energy for given power transfer
becomes very great for radii less than this. The radial field impedance for an inward
traveling wave is
change
of mathematical form
a more
detailed
as
there
E~
“‘5‘
This
'
:
was
H (”(kr)
___v’__
=
r"
7
'X
impedance becomes predominantly reactive at a value kr v, which is compatible
Figure 9.417 shows real and imaginary parts of the wave impedances versus
-~
with (6).
krforv
6
R
=
9.
W. L. Barrow and L. J. Chu, Proc. IRE 27, 57 (7939).
470
Chapter
9
Special Waveguide Types
X/R
AMW”
\
1
y A:
SJ 1%
15
Wave resistance and reactance for circumferential mode in radial line for
FIG. 9.4b
9.5
v
=
9.
DUALITY: PROPAGATION BETWEEN lNCLlNED PLANES
Given certain solutions of Maxwell’s
equations, we may obtain other useful ones by
making
important principle of duality. This principle follows from
simple
the symmetry of the field equations for charge-free regions:
use
of the
but
V X E
=
V X H
=
-jpr
jwsE
(1)
(2)
replaced by H, H by -—E, p. by a, and a by ,u, the original
again obtained. It follows that if we are given any solution for such a
dielectric, another may be obtained by interchanging components as stated. It may be
difficult to supply appropriate boundary conditions for the new solution since the magnetic equivalent of the perfect conductor is not known at high frequencies, so the new
solution is not always of practical importance.
One example in which the principle of duality may be utilized to save work in a
practical problem is that of the principal mode in the wedge—shaped dielectric region
between inclined plane conductors (Fig. 9.5). This mode has electric field E «b and mag
netic field HZ. If there are no variations with ()5 or 2, it is evident that the field distriIt is evident that if E is
equations
are
9.5
butions
can
Duality: Propagation
be obtained from those of the radial transmission~line mode
through the foregoing principle of duality. Replacing E by H,
a by ,u in Eqs. 9.3(1) and 9.3(2),
The real
471
Between Inclined Planes
H:
=
Ecfi
z
AH SIX/tr) +
*1.
\/_E_
H
by ~E,
Bng)(kr)
[AHil)(k7‘)
+
u
(Sec. 9.3)
by
8, and
(3)
BHi2)(/C")l
(4)
is that all the derived
expressions 9.3(9)-—(16) may be used without
interchange of quantities as above.
Admittance should be read in place of impedance, and the numerical scale of ZO\/s_r
in ohms (Fig. 9.36) should be divided by (377)2 to give the characteristic admittance
YO/Ve—l. in Siemens. Total admittance is obtained from the field admittance by
advantage
rederivation,
as
may the
curves
of Fig. 9.30, with the
_
Y(0121.1
where the upper
sign
z
.+.
1
—-———
I-(bo
Hz
E¢
is for ri < rL, and the lower for ri
Example
(5)
—
>
rL.
9.5
USE OF INCLINED PLANES FOR IMPEDANCE MATCHING
application of the inclined-plane transmission line might be in impedance matching
parallel-plate transmission lines of different spacings, a'I and at2 (Fig. 9.5). It
is known from practical experience that such transitions, if gradual enough, supply a
good impedance match over a wide band of frequencies (unlike schemes studied in
Prob. 5.7c, which depend upon quarternwavelengths of line). It is seen from Fig. 9.30
that for both lcri and er large (say, greater than 5) the characteristic admittance Y0 is
1,14, 6L
1,01,). If the parallel-plane
nearly 1/77 and 6 and it! are nearly equal (i.e., 6i
One
between
z
FIG. 9.5
Inclined—plane guide.
w
472
line to the
Chapter
right is matched,
Special Waveguide Types
9
its characteristic
admittance is that of
a plane wave,
approximations the input wave
admittance is also approximately 1/77, so that the parallel-plane line to the left is also
nearly matched. This gives some quantitative support to the matching phenomenon
1/77. Equation 9.3(16)
wave
then shows that with the above
mentioned.
9.6
WAVES GUIDED BY CONICAL SYSTEMS
problem of waves guided by conical systems (Fig. 9.6) is important to a basic
understanding of waves along dipole antennas and in certain classes of cavity resona—
tors. In particular, one very important wave propagates along the cones with the velocity
of light, has no field components in the radial direction, and so is analogous to the
transmission-line wave on cylindrical systems. This basic wave is symmetric about the
axis of the guiding cones, so that if the two curl relations of Maxwell’s equations are
written in spherical coordinates with all gb variation eliminated, it is seen that there is
one independent set containing E 9,
11¢, and IE, only:
The
1 6(rE9)
1
1 aEr
'
+ 10);]. H
”b
W
r
66
6
—-
r
_
6r
2'
sin a
as
x
O
l
()
=-
O
(2)
=
O
(3)
'
8111 6H)
(‘
4‘
-
J me Er
1 6 7H
~f—g—j32
I
—
67
jweEg
’///”’-_“\\\
///// //”—~\\\ \\\ \\\
\
\\E “neS\’/
\
\\ \
\\\\\
’/
/
//
/
\
FIG. 9.6
/
//
\\ \
Biconical
guide.
/
Waves Guided
9.6
It
can
be checked
by
substitution that the
by
Conical
473
Systems
following
solution does
satisfy
the three
equations:
5,.
:
"E6
=
0
(4)
,
6
sm
[Ae‘jk’
+
Bejkr]
(5)
3617”]
(6)
1
1H 45
=
sin 9
[Ae‘jk’
—-
These
equations show the now-familiar propagation behavior, the first term representing
traveling radially outward with the velocity of light in the dielectric
material surrounding the cones, the second term representing a radially inward traveling
wave of the same velocity. The ratio of electric to magnetic field E 9 /H is given by 11
g
for the positively traveling wave and by
n for the negatively traveling wave. There
is no field component in the radial direction, which is the direction of propagation.
The above wave looks much like the ordinary transmission-line waves of uniform
cylindrical systems. This resemblance is stressed if we note that the E 9 corresponds to
a voltage difference between the two cones,
a wave
—
7T—
60
“I
V
77"“
E9"
d6
6
27)
In cot
.
6
.
.
8111
00
60
H
90
—77f
=
6
[Ag-17"
+
Bejkr]
(7)
..
-29 [Ae‘lk’
+
BeJA’]
equal-angle cones (Fig. 9.6). This is a voltage which
through the propagation term, e 317". Similarly the azimuthal
independent
field
magnetic
corresponds to a current flow in the cones at 6
60:
where the
case
is
treated is that of
of r, except
=
I
=
=
This current is also
of the
sign
2771'H¢ sin 60
Bejkr]
277[Ae“jk"
independent of radius, except through the propagation term. A study
opposite radial directions in the two cones at
relations shows that it is in
any given radius.
The ratio of voltage to current in
we
and
(8)
-
call characteristic
(8) with
B
=
impedance
in
a
single outward-traveling wave, a quantity which
ordinary transmission line, is obtained from (7)
an
O:
20
__
77 In cot
60/2
(9)
71'
a negatively traveling wave, the ratio of voltage to current is the negative of this
quantity. This value of impedance is a constant, independent of radius, unlike those
ratios defined for a parallel-plane radial transmission line in Sec. 9.3. We can also see
this starting from the familiar concept of ZO as v L/ C, since inductance and capacitance
For
474
Chapter
between
surface
the
path
Special Waveguide Types
per unit radial length are independent of radius. This comes about since
increases proportionally to radius, and distance separating the cones along
cones
area
So far
9
of the electric field also increases
as
conductors
can
proportionally
to radius.
is concerned, the system arising from two ideal coaxial conical
be considered a uniform transmission line. All the familiar formulas for
this
wave
input impedances and voltage and current along the line hold directly, with Z0 given
by (9) and phase constant corresponding to velocity of light in the dielectric:
[3
If the
=
2—;
=
cox/£73
(10)
conducting cones have resistance, there is a departure from uniformity due to this
usually not serious in any practical cases where such conical
resistance term, but this is
systems are used.
Of course, a large number of higher—order waves may exist in this conical system
and in similar systems. These will in general have field components in the radial direction and will not propagate at the velocity of light. We shall consider such general wave
types for spherical coordinates in Sec. 10.7.
9.7
RIDGE WAVEGUIDE
shapes of cylindrical guides that have been utilized, one rather
ridge waveguide, which has a central ridge added to either the top
or bottom, or both, of a rectangular section as in Fig. 9.7a. It is interesting from an
electromagnetic point of view since the cutoff frequency is lowered because of the
capacitive effect at the center, and could in principle be made as low as desired by
decreasing the gap width g sufficiently. Of course, the effective impedance of the guide
also decreases as g is made smaller. One of the important applications is as a nonuniform
transmission system for matching purposes obtained by varying the depth of ridge along
the guide, similar to the approach described in Sec. 9.5.
The calculation of cutoff frequency, which is a very important parameter for any
shape of guide, also illustrates an interesting approach that may be applied to many
guide shapes which cannot be solved exactly. At cutoff, there is no variation in the z
direction (y
0), so one may think of this as a resonant condition for waves propa—
gating only transversely in the given cross section, according to the desired mode. For
example, the TEIO wave in a rectangular guide has a cutoff frequency equal to the
resonant frequency for a plane wave propagating only in the x direction across the
guide, thus corresponding to a half-wavelength in the x direction. A very approximate
calculation of cutoff frequency for the ridge guide might then be made as in Fig. 9.7a
by considering the gap a capacitance and the side sections one-turn solenoidal inductances, and writing the condition for resonance:
Of the miscellaneous
important
one
is the
=
@"I/QLM'1’2-_1€1.21.”er
f__1_
32
2
°_27T
‘27 g
“277
g
Mel/rd
”2
(1)
9.7
475
Ridge Waveguide
0.50
0.75
0.2
0.4
0.6
wa+m
(b)
(a) Cross section of ridge waveguide and approximate equivalent circuit for cutoff
([3) Curves giving cutoff wavelength for a ridge waveguide as in (a). Solid curves:
data from Cohn.7 Dashed curves from Eq. (1).
FIG. 9.7
calculation.
A better
equivalent
circuit for calculation of the transverse
the two sections A and B
considered
resonance
is
one
in which
transmission lines with
a dis~
parallel-plane
between
them.
(This junction effect
continuity capacitance Cd placed
junction
will be discussed in Chapter 11.) Curves of cutoff frequency and a total impedance for
the guide have been calculated in this manner by Cohn.7 Some results are shown in
Fig. 9.71) with comparisons of results from the lumped element approximation ( 1). As
might be expected, agreement is better for the smaller gaps. The technique of finding
cutoff by calculating transverse resonance frequencies—often by numerical meansg—
is valuable for a variety of irregularly shaped guides.
The ridge guide illustrates that guides other than two-conductor transmission lines
can be appreciably smaller than the half-wavelength measure found for the rectangular
and circular guides if the boundaries concentrate energy appropriately. However, the
boundary irregularities generally decrease power-handling capacity.
are
at
7
3
the
S. B. Conn, Proc. IRE 35, 783—788 (7947).
See, for example, R. Bulley, IEEE Trans. Microwave Theory Techniques MIT-18, 7022 ( 7970).
476
Chapter
9
Special Waveguide Types
THE IDEALIZED HELIX AND OTHER SLOW~WAVE fiRUCTURES
9.8
A wire wound in the form of
found useful for
a
antennas9 and
helix
as
(Fig. 9.8a)
makes
a
slow~wave structures in
type of guide that has been
traveling~wave tubes.10 It is
interesting as an example of a general class of structures that possess waves with a
phase velocity along the axis much less than the velocity of light, as contrasted with
most of the waves so far studied, which have phase velocities greater than the velocity
of light. A rough picture suggests that the wave might follow the wire with about the
velocity of light, so that its rate of progress along the axis would correspond to a phase
velocity
up
where
rtr is
mation
over a
can
the
be made
It is rather
pitch angle.
z
0
sin
(l)
1,0
surprising that this represents a good approxiinteresting to find that a useful analysis
wide range of parameters. It is also
by considering
an
The idealization commonly
idealization of the actual helix.
analyzed,10 referred to as the helical sheet, is a cylindrical
surface in which the component of electric field along the direction of 1,1; is assumed to
zero at all points of the sheet (Fig. 9.81)). Moreover, the component of electric field
be
lying in the cylindrical surface normal to the direction of 1,0 is assumed to be continuous
through the surface, as is the component of magnetic field along it: (the latter because
there is to be no current flow normal to the direction of iv). Since the idealization takes
WWW
a
z
¢
lllllllillliilll‘lll
b
a
.
HIHIIHIHHHIIII
(b)
FIG. 9.8
10
(a) Wire helix. (1)) Idealized conducting sheet and curve giving propagation constant.
(c) Section of disk~loaded waveguide.
9
‘0
Chap. 7, McGraw—Hill, New York, 7988.
Traveling—Wave Tubes, Chap. Ill and Appendix II, Van Nostrand. Princeton,
J. D. Kraus, Antennas, 2nd ed”
J. R. Pierce,
NJ, 7950.
these conditions to be the
same over
results for fine-wire helices of small
close
477
The Idealized Helix and Other Slow-Wave Structures
9.8
all the sheet, it would be
pitch angle
or
expected
to
give
best
for multifilar helices with fine wires
together.
To find the
propagation
constant,
general solutions inside and outside
boundary according to continuity conditions
the boundary conditions is a hybrid mode con-
we set
up the
the helical sheet and match them at the
stated above. The
that satisfies
wave
taining both TE and TM components. We assume an unattenuated wave so that y
j,8. Then the separation constant in the general solutions [Eqs. 720(5)] is k3
=
+
y:
=
k2
below the
—-
Bl.
speed
Since k
w/c and B
=
=
light. Therefore, kg
of
<
=
k2
w/vp, B > kfor waves with phase velocities
O and
kC
jr,
==
where
'r
is
a
real
quantity.
Assuming axial symmetry and that the fields must be finite on the axis and zero at
infinite radius, the general solutions can be expressed in terms of modified Bessel
functions
and
[Eqs. 7.1406)
(17)]. The
2
[”133 is understood in all of the
variation
following:
r
E:I
=
< a
> a
r
E:2
A1100”)
=
’T
'1'
'we
'we
Hd’l
:
H:l
=
(2)
A2K0('rr)
J—rr_
[4111(77.)
[—1452
m
H22
2
15¢2
:
3110(7))
“L;— A2K1(Tr)
(4)
BZKOUV‘)
(5)
7'
‘0)
‘w
EM
The idealized
=
boundary
conditions first described
w
+
E“
cos
(I;
=
O
E:2
sin
1,0
+
Egg
cos
ill
=
0
E:l
cos
Egl
sin
ill
2
E:2
cos
([1
cos
it:
=
H:2
sin
ti!
(2)—(7)
ll!
-—
sin
it!
are
substituted in
+
qu1
2
{/1
4, the
(9)
-
-l-
Eda
sin
H¢2
cos
(0
(10)
1/1
(ll)
(8)—(ll). A nonvanishing solution for the field
10(Ta)K0(m)
11(ra)K1('ra)
2
(ka
cot
11/)2
Pierce10 is shown in Fig. 9.8b. It is
approximation (1) gives good results.
A solution of this taken from
>
(8)
that the determinant of the coefficients be zero, which leads to
(m)
cot
(7)
are
sin
amplitude requires
ka
1—;- 82K1(rr)
Efl
H:l
The fields
J—Tfi 3111(7))
(12)
seen
that for
478
Special Waveguide Types
Chapter 9
Some
comments about slow-wave structures are in order. If it is desired to
general
an electric field along the axis, propagating with a phase velocity
light (as in a traveling-wave tube where the phase velocity should be
approximately equal to the beam velocity for efficient interaction with the electrons),
produce
a wave
with
less than that of
then,
as
discussed above,
k2
O:
<
7-?—
-k§
:
32
=
02
It is
e(1 :3)
=
.,.
k2
—
(13)
1/2
(14)
—
consequently necessary in a cylindrically symmetric system that the Bessel function
they may therefore be written as
solutions (Sec. 7.20) have imaginary arguments, and
modified Bessel functions. For a TM wave,
E3
=2
qu
=
(15)
A100”)
jwe
we
*‘B-Er
=
Example
CYLINDRICAL REACTANCE WALL
If
we
r
=
ask about the
a
to
boundary conditions
support slow
waves, we see
(16)
7.411(7))
that
9.8
TO
PRODUCE SLOW WAVES
might be supplied
at a
cylindrical
that, if uniform, it should be of the
surface
nature of a
reactive sheet with
Jjg
_._42
:=
11¢
==
m
W
I 100m)
k 11m)
(17)
The helical sheet studied earlier may be considered as supplying this required reactance
through the interaction with the TE waves and external fields caused by the helical cuts.
The short«circuited sections of radial lines of
also be considered
actance at
ance
r
=
supplying
as
a, and will therefore
X
the
_
"
boundary
of electron space
on
when
the axis that is to act
support
a
slow
waveguide (Fig. 9.8c) may
required uniform re—
to the above
wave
also. The
approximate
react—
charge
Jo(ka)No(kb)
J1<ka)No<kb>
"
-
Jo(kb)No(kd)
Jorkbwlrka)
(18)
the axis of such slow-wave structures is much less than that
B0
on
7’
1, ris substantially equal to ,8. By the nature of the I0 functions
(Up/c)2 <
(Fig. 7.140), the field
on
disk~loaded
structure is
supplied by this
Note that if
a
approximation
an
is
large.
electrons,
will
This is of
as
modify
course
undesirable when it is the field
on
traveling-wave tube. Of course, the presence
the forms of solution somewhat.
in
a
9.9
9.9
Surface
479
Guiding
SURFACE GUIDING
The result of Sec. 9.8 suggests a localization of fields near a surface that possesses a
reactive surface impedance. This concept appeared there in an interior region, but may
are in the external region also. Because of the surface-guiding
principle, the energy is maintained near the surface so that it is not radiated or coupled
seriously to nearby objects. Thus the external region corresponding to Fig. 9.86 would
be as shown in Fig. 9.9a. The proper solutions for the external cylindrical region for
a TM wave are the same as Eqs. 9.8(2)—-(4) for r > a. For E: and
H4” with .2"sz
be useful when the fields
understood,
E:
=
His
=
(1)
AKO(T)‘)
(08
“1—7— AK1(7r)
x
~
7" \ \V
\\
\
(2)
\. \ \ \
7'1,
7¢’/////////////%
\\
i
\ \ k\ \ k\
/
Zr
f
.
/
//'
.//
//
fl/5 M
%\
%//g5,
/
//
(a)
FIG. 9.9
equivalent
(0) Rod with periodic radius variations capable of propagating
of (a). (c) Dielectric~coated conducting surface.
a
slow
wave.
(b) Planar
489
Chapter
The field reactance at
r
=
Special Waveguide Types
9
necessary to support such
a
E2
.
1X
__
a wave
is then
7KO(rr)
,
_
(3)
—-
-
Hg
m; K1(7))
,
r=a
Equation (3) again represents a positive, or inductive, reactance as did Eq. 9.8(17), and
the solutions (1) and (2) die off for large r. Phase velocity is less than the velocity of
light in the external dielectric in order to maintain 7 real, as in Eq. 9.8(13):
7-2
$.15; in “w, mag-a Ridinhilkiivj. :‘Liiiiabfi
=
I32
k2
—
<4)
:' ---------
Exampfle
SURFACE GUIDING BY
9.98
A
REACTANCE SHEET
perhaps more easily visualized for a plane sheet, as pictured in
Fig. 9.9b. Substitution in Eqs. 8.2(13)—(16) easily verifies that the following are solutions of Maxwell’s equations, using the definition (4), again with e'jflz understood:
The above concept is
E
=
Ger“
(5)
'
Hy
=
E
=
*JCUS C
e‘“
(6)
T
.
C
A surface
wave
impedance
defined for this
J.X
Again
TM
we see
wave
that the
(but
see
impedance
Prob.
“’37?”
9.9a),
z
example
Ez
=
—"
Hy
.
x30
can
parallel-plane
'7'
’—
(8)
(1)8
sheet should be inductive for surface
guiding
of this
B
greater than k if the
less
is
than the velocity
Therefore, phase velocity
is to die away with increasing 1:.
of light in the dielectric. The spaces between the
fins
is then
and that the value of
wave
be shorted
(7)
must be
conducting fins may be considered to
impedance at the ends of the
transmission lines and the field
be found from Sec. 8.3
as
E—-‘='t
jnan kd
y
where d is the
height
inductive reactance.
of the fins. It is clear that if kd <
77/ 2,
the surface appears
as an
48'!
Surface fiuiding
9.9
1.x
.=.o-..-,_./..-; t...,-,....:.-.'./ ........
Example 9.9b
SURFACE GUIDING
Another important way of obtaining
DIELECTRIC COATING
av A
surface
impedance sheet is by coating a conductor
Fig. 9.96. The following TM solutions
for region 1 can be verified by substitution in Eqs. 8.2(13)-—~(16). The conductor is
assumed perfect and the argument of the sine is selected to make Ez zero at the conductor
with
a
thin
surface,
x
layer
=
~—
of dielectric,
a
illustrated in
as
d:
E3
D sin
=
kx(x
d)
+
(9)
'
D
H),
=
E
=
«’10—?—
cos
kxn-
+
d)
(10)
'
D
kit
From (9) and
(10)
a
field
Ji
kt
=
impedance
cos
This is inductive
the
B of the
as
required
desired surface
=
x
z
for
—‘
H3,
0
=
(11)
can
'k.
1J—
be found
tan
kxd
as
follows:
(13)
coal
F0
and, if equated to (8), will define
For small thickness, kxd < 1, (13) becomes
guiding
wave.
d)
(12)
E-
jX
+
32
—
at
kx(x
TM
jX~
waves
J'kid
(14)
(1)81
Using (4), (8), (12), and (14), we can show that the phase velocity is less than the
velocity of light in the external dielectric and greater than that in the coating.
The principle is also applicable to lossy dielectrics and/or conductors, for which a
finite but relatively small attenuation in the z direction is obtained. ZenneckII and
Somrnerfeld12 provided the early classical analyses of this phenomenon; Goubaul3 i1—
1ustrated its usefulness as a practical wave-guiding means by using either thin dielectric
coatings or corrugations on round wires; a thorough treatment is given by Collin.14
Surface guiding and dielectric wave guiding are closely related phenomena since one
boundary is a dielectric interface as treated in Sec. 9.2. The relationship will become
clearer with the field analysis of dielectric guides in Chapter 14.
”
J. ZenneCk, Ann.
Phys. 23, 846 (7907).
Phys. Chem. 67, 233 (7899). Described in J. A. STroh‘on, Electromagnetic Theory. p. 527. McGraw~HilL New York, 7946.
G. Goubau, Proc. lRE 39, 679 (7957);.1. Appl. Phys. 21, 7779 (7950).
7?. E. Col/in. Field Theory of Guided Waves. 2nd ed.. Chap. 77. IEEE Press, Piscataway.
A. Sommerfeid. Ann.
.u—l 1:03
NJ. 7997.
482
Chapter
9.“)
9
Special Waveguide Types
PERIODIC STRUCTURES AND SPATIAL HARMONICS
corrugated surfaces used as illustrations of reactance walls in the two preceding
sections are special examples of periodic systems if the spacing between corrugations
is uniform. Periodic systems have interesting properties and important applications and
so will be examined more carefully in this section. We recognize that the treatment as
a smooth reactance wall, used in Secs. 9.8 and 9.9, is only an approximation since the
grooves will cause field disturbances not accounted for by this “smoothed out”
approximation.
Let us begin by consideration of a specific example, the parallel-plane transmission
line with periodic troughs in one plate, as illustrated in Fig. 9.1041. If the troughs are
relatively narrow, they act to waves with z—directed currents in the bottom plate as
shorted transmission lines in series with the conductor. These lines produce values of
O, which are essentially constant over the gap width w. Thus the
E: and Hy at x
for E2 at x
0 is as shown in Fig. 9.101), a phase shift of [Bed
condition
boundary
we will stress propagating waves. The square
allowed
over
each
since
being
period
waves shown neglect higher-order fringing fields at the corners, but are better approx—
imations than the complete smoothing out of the effect (as in Sec. 9.9) and are sufficient
to illustrate the basic preperties of periodic structures.
To fulfill the boundary condition at x
O, we might expect to add solutions of
Maxwell’s equations just as we added solutions of Laplace’s equation in Chapter 7 to
satisfy boundary conditions not satisfied by a single solution.
For the present problem let us take 6/6))
O and consider waves with E, E2, and
Thus
a
of
solutions
Maxwell’s
sum
satisfying
equations and the boundary
H), only.
a may be written by adding waves of the TM form, the
O at x
condition of E2
TEM component being included if the sum includes 72
0:
The
=
=
=
=
=
=
=
E50.) 2)
2
=
A”
sin
K,,(a
—
Joe-'13":
(1)
°°
Ex(x, z)
Hy(x,
z)
E 11in
A"
K
_
—_
=
209% An
cos
cos
K,,(a
Kn(a
—
--
x)e —m
(2)
x)e rfflnz
(3)
,.
where
Ki=wzlw8The
i=k2-Bi
(4)
condition illustrated
by Fig. 9.1017 is a periodic function of 2 if the phase
separately.
expand the resulting periodic function in the comform
the
Fourier
of
series
(Prob. 7.11e), we may write for the boundary condition
plex
boundary
factor is taken out
152(0, 2)
=
e'ffioz
If
we
Z Cue—(j2””Z/d)
[2: ~60
(5)
9.10
Periodic Structures and
483
Spatial Harmonics
rr-—d-—+1
‘1
W.
x
a.
*0
(a)
a
§
Pk
EZ
i
(b)
z
\!
\i\\“‘\
\\
k
k\
k
~9le—
Eoe-‘J'Bo d
HTonVj
It
E0 8‘21fl0d
n.
(C)
(d)
i
(e)
:
¢
i
T
a
l
a
‘
I
l
I
j
:
a
a
l
l
j
a
i
:
4»
i
W
“034:2
T
(f)
to
”coal
fi~—-—>~
(a) Parallel—plane transmission line with periodic short~circuited troughs. (b) Idealized
along the lower plate in the structure of (a). (c) w—fi relation for spatial harmonics
of wave with fundamental traveling in + z direction. ((1) w—B relation for spatial harmonics of
wave with fundamental traveling in —-z direction. (e) Low-frequency equivalent circuit for fun—
damental spatial harmonic of structure in (a). (f) Composite w-B diagram including passband
FIG. 9.10
variations of E:
at
higher frequency.
484
Chapter
Special Waveguide Types
9
where
(1/2
C"
a
f
..
d
—°
7171
By comparing (5)
2
f
_
d
—-d/2
E
=
‘
w/2
1
+j<2mzz/d) dz
E—(O, mining
"iv/2
Eoe+j<2mzz/d)
dz
mTw
srn< )
,
(6)
d
with
(1) evaluated
A”
at x
CI!
=
:—
_
sin
K,,a
:80
+
0,
we can now
identify A"
and
B”:
fl sin(n7rw/d)
sin
7m
(7)
Kna
27m
:8"
A
wave
(8)
7
solution is thus determined for this
problem with the approximations described.
problem compared
We wish first to stress the different role of the TM solutions in this
with that considered
previously.
In earlier sections TM solutions have been considered
“modes” with the inference that each may be excited independently. Here they are
coupled by the periodic boundary condition and must exist in the proper relationship
as
to each other to
satisfy
this
boundary condition.
In this
capacity they
are
known
as
natural extension of the harmonic character of the Fourier series
“spatial harmonics,”
the periodic system in space. We note especially from (8) that determination of B
for any spatial harmonic automatically determines the value of all others. Thus the (0—3
diagram (Sec. 5.12) is periodic in B with repetition at intervals of 277/07 as illustrated
in Figs. 9.100 and d.
Determination of the shape of one period of this plot (say ,80) requires study of the
boundary conditions on two field components. We have already considered E2 and now
choose Hy as the second component for this example. If the solution for the troughs is
well approximated by the shorted-transmission-line behavior, the value of
Hy at x
0 for the shorted lines is given by
O, z
a
to
=
2:
Hy
Thus
one
might equate
this to
]
cot kl
This with (4) and (8) determines
difficult.
special
case
of
,BOd
A
<
at
st
—
cot
x
”2'... 7“an
,80
kl
'0
(3) evaluated
n
For the
:
in
=
,
sm
O, and A" substituted from (7):
mzw
a
cot
principle, although
the
K,,a
general
(9)
solution may be
1, lumped—element approximations may be used. The
fundamental harmonic solution of the
corresponding low~pass filter then gives the char—
acteristics of the fundamental, and from that, the values of ,8 for other spatial harmonics.
The filter corresponding to the line used in the above example is shown in Fig. 9.106.
capacitance and a part of the series inductance represent the fields in the
parallel-plate section between grooves, and the shorted-transmission~line grooves con~
The shunt
tribute
C
=
an
8(d
additional series inductance. The
-—
w)/cz
for
between grooves is
tan
kl
(assuming
a
485
Periodic Structures and Spatial Harmonics
9.10
capacitors
in
Fig.
9.10:? each have the value
length in the y direction. The inductance from the section
w) and that from the shorted grooves is L2
,u.a(d
(nw/w)
unit
L1
2
kl
<
=
—-
77/ 2). Each inductor in
Fig.
9.108 has the value L
=
(L:
L2)/2Although illustrated for a particular example to provide concreteness, the periodic
character of the (Li—,8 diagram and the relation (8) for phase constant of a spatial har—
monic apply to any structure of period d. Other important properties are as follows:
+
1. The group velocities of all spatial harmonics of a given wave are equal. This
would be expected so that the wave would stay together, but may be noted either
from the
w~B diagrams
or
2. Of the infinite number of
with
and
by differentiating (8):
spatial harmonics,
half
are
backward
waves
so on
=
=
backward waves,
various
(Sec. 5.16)
O, 1, 2,
and group velocities in the opposite directions. Thus the n
of Fig. 9.10c are forward waves, whereas 22
l, 2, and
phase
as
is
seen
—
—
so on are
by comparing the signs of w/fi and (Ito/(1,8 for the
portions of the figure. If the phase velocity of the fundamental spatial
0,
negative, the harmonics are as shown in Fig. 9.10a’. Here the n
2, and so on have both phase and group velocities in the negative direction
harmonic is
~—
1,
and
-
are not
with
==
backward waves, whereas the
a
=
1, 2, and
so on are
backward
waves
group velocities. The structure of Fig.
fundamental forward wave, but other periodic structures may have
positive phase
9.10a has
n
velocities and
fundamental backward
negative
waves.
md (m an integer) is the same as at z
plane 2
e
the
multiplication by
phase “130””. This important property is related
to Floquet’s theorem,15 and is often taken as the fundamental starting point for
the study of periodic systems. For the particular example used in this section,
(1) and (8) yield
3. The field distribution at any
=
=
0 except for
7.:
E:(.r, 172d)
H
Z
A"
Sill
Kn(a
-
x)e "jfiolmle “1.27m”:
(10)
)1: —SC
.7:
=
e-J‘flomd
Z
A”
sin
K,,(a
——
x)
=
[mom/3’43
0)
n=—f~
and
similarly
for the other field components.
is cut off where group velocity becomes zero, shown as we in Figs.
9.106 and (1. Above this frequency there is a region of reactive attenuation, typical
4. The
wave
of filters in the
attenuating region. As frequency
propagating waves, as
bands will be found with
‘5
R. E. Collin, Field
7997.
is increased, however, other pass
illustrated in Fig. 9.10f. Each of
Theory of Guided Woves, 2nd ed, Sec. 9.1. IEEE Press. Piscatoway, NJ,
486
Chapter
Special Waveguide Types
9
waves has spatial harmonics, just as the low-pass wave studied. The Pierce
coupled-mode theory16 is especially powerful in giving a picture of the entire
w- B diagram in a structure with relatively small periodic perturbations if the phase
constants of the unperturbed system are known.
these
PROBLEMS
9.23 For
a
dielectric slab
Fig. 9.2a, 2,0
=
,ug, plot the complement of the critical angle of
guide with ,u.1
66, as a function of 82/81. Many dielectric guides have small
the two permittivities, 82 and 81(1
A), where A < 1. For these,
=
7r/2
differences between
show that (/1
W.
—
—
=8
9.2b Demonstrate that when 6
6,: (defined as cutoff for the dielectric slab guides), there
is no decay in medium 2 (outside of slab) and propagation is at the velocity of light in
medium 2. Show that when 9 approaches 77/2 (grazing incidence at the boundary),
=
phase velocity approaches
9.2c
the
velocity
of
light
in medium 1.
the definition of cutoff given in Prob. 9.2b, show that the x component of kI at
cutoff is k1,:
(k?
3W2. Calculate the lowest cutoff frequencies (other than zero)
for TM modes with even and odd symmetry of E: across the slab and similarly for TB
1 mm, s,
modes with even and odd symmetries of H: with d
4.0880, 82
430,
and #1
#2.
Using
=
-
=
=
=
=
9.2d Demonstrate that
Eq. 9.2(4)
has
TM modes in a dielectric slab
increases with m for m > O.
9.2e
a
solution for
guide.
(i) For the dielectric slab mode with
kid
(ii)
arbitrarily
Show that there is
m
=
small
a
kid
for either TE
or
minimum of kid which
0, find the external decay
constant ax for
> 1.
Also find ax for TB modes with
k,d
< l and compare with the result of
part (i).
9.2f If 31 > 82, very little energy is contained in medium 2, and the boundaries appear like
“magnetic shorts”; that is, tangential magnetic field approaches zero at the interface.
Show this from the equations for wave reflections and solve for the lowest~order TE
mode with such conditions.
9.3a For
for
a
a
TMOI wave in a circular waveguide it is desired to insert a blocking impedance
given frequency. To do this, a section of shorted radial line (Fig. P9.3a) is in—
(b)
(a)
FIG. P9.3a, b
“5
A. Yariv and P. Yen,
Optical Waves in Crystals, Sec. 6.4, Wiley, New York, 7984.
487
Problems
serted in the
its outer radius
guide,
impedance looking
a
chosen
so
that with the
into the radial line is infinite
at
the
the radius b is 1.25 times greater than cutoff radius
wave and find the radius a.
9.3b It is sometimes
required
at
guide radius b given, the
given frequency. Suppose that
this frequency for the TM01
break the outer conductor of
to
a
coaxial line for insulation
purposes, without interrupting the rf current flow. This may be accomplished by the
radial line as shown (Fig. P9.3b) in which a is chosen so that with b and the operating
wavelength specified, the radial line has zero input impedance seen from the line. Find
1.
assuming that end effects are negligible and that an/A
the value of 0,
9.3c Find the
=
voltage
at the
radial line at radius b
9.3d
radius
in terms of the coaxial line’s current
a
flowing
into the
(Fig. P9.3b).
Taking the classical transmission-line equations with distributed inductance
itance varying with radius as appropriate to the radial line,
L
=
and capac—
27Tsr
fl
and
C
=
2771'
Show that the
equation for voltage as a function of radius is a Bessel equation, and
voltage and current obtained in this manner are consistent with
that the solutions for
Eqs. 9.3(1)
and
(2).
9.3a Derive forms similar
Repeat
for
Eqs. 9.302)
to
Ha specified
at
and
(13)
if
E0
is
specified
at
rd, and
Eb
at
rb.
re, and Hb at I").
a lossy radial transmission line, find series resistance per unit radial distance (assuming skin-effect condition) for a conductor of surface resistance R5. Also find shunt
conductance per unit radial distance for a dielectric with complex permittivity 3’
js". Add these to the transmission-line equations (Prob. 9.3d) and obtain the differen~
tial equation for voltage.
9.3f For
-
9.43 Sketch lines of
Sec. 9.4.
current
(Take Z"
by judicious
cuts
=
flow for
J"
a
circumferential mode with
for this
without
purpose.) Suggest methods
disturbing the symmetric mode.
v
of
=
l in
a
radial line of
suppressing
this mode
wedge—shaped guide as in Fig. 9.4a may be used to join two waveguides of the same height but different width, both propagating the TEIO mode, and a
good degree of match is obtained by the process so long as the transition is gradual.
-%(Fig. 8.70) and the width aI of the
Suppose the smaller rectangular guide has b/a
1.2. The larger guide has the same height and
smaller guide is such that (fc
f/
)rem
(fC)TEzo f/ 0.8. Evaluate the wave impedances of the TE10 modes in both guides
and compare with those in the sectoral horn at the junctions to the rectangular guides,
using data of Fig. 9.4b for a wedge angle of 20°. Frequency is 10 GHz.
9.4h A section of the
=
=
=
9.53 In the
use
of the inclined
plane
line for
to
=
z
=
approximate standing wave ratio in the line
Compare with that which would exist with a sudden transition, considering
discontinuity of characteristic impedance.
the left.
only
9.5b
=
as discussed in Sec. 9.5, suppose that
2.5, and the dielectric is air. If the line to
matching
2 cm, k)"
1 cm, (12
3 GHz, d1
the right is perfectly matched, obtain the
f
the
Apply the principle of duality to the TEI 1 T1301, and TMOl modes in circular cylindrical waveguides to obtain qualitatively the fields of the “dual” modes. For which of
these might boundary conditions be supplied, allowing changes in the conductor posi—
tion or shape from those of the original mode?
9.5c A
,
wedge—shaped
dielectric
region
is bounded
by conducting planes
at
(p
=
O and
(150,
488
Chapter 9
Special Waveguide Types
2
O and d. Find the field components of the lowest-order mode with
Is this the dual of the mode discussed for sectoral horns in Sec. 9.4?
=
Eg, H,,
and
H2.
application of the mode of Prob. 9.5c to the matching between rectangular
waveguides of different height, both propagating the TE10 mode, describing how to
use the solution of Prob. 9.5c to estimate reflections and the approximations involved.
9.5d Discuss the
kr1 and krz in the inclined planes of Fig. 9.5 are both large, a matched
parallel—plane transmission line with wave impedance 1] at rI gives the same '11 at r2,
so that a parallel~p1ane line at that position would also be matched.
9.5c Show that if
expressions for the voltage, current, and characteristic impedance for the principal waves on a transmission line consisting of two coaxial, common-apex cones of
unequal angles.
9.6a Find
9.6b Write the dual of the mode studied in Sec. 9.6. Can it be
boundaries using perfect conductors?
supported by physical
point of view toward a cylindrical conductor excited by a source across a gap at
0 (Fig. P9.6c) is that it is a nonuniform transmission line propagating effects
away from the gap. (This point of view is used in one theory of cylindrical antennas to
be presented in Sec. 12.25.) That is, an elemental section at 2 may be considered as a
biconical line passing through radius a at 2. Write the expression for characteristic
impedance as a function of z and plot versus 2/0 for air dielectric.
9.6c One
2
==
Exciting
field
0
FIG. P9.6c
9.6d For the biconical line of Fig. 9.6, find a capacitance per unit length at radius 1' based
upon charge per unit length along the cones and voltage as defined in Sec. 9.6. Similarly, find an inductance per unit length based upon current at r and the magnetic flux
between concentric
acteristic
spheres at r
impedance (L/ C )1/2.
and
r
+ dr. Then check
velocity (LC )" 1/
2
and char-
9.6e For
a lossy biconical guide, find series resistance per unit length (assuming skin-effect
conditions) for a conductor of surface resistance RS. Also, find shunt conductance per
unit length for a dielectric with complex permittivity e’
je".
-—
9.7a As
a
demonstration of the
verse
nance
9.7 b For
resonant
of
a
frequency,
technique
of
finding cutoff frequencies by calculating trans—
a TM01 mode in circular guide as the reso-
find the cutoff of
radial line mode.
TB mode in the
small-gap ridge waveguide, the energy stored in electric fields is
in the gap region. If electric field is assumed uniform at E0 over this
region, estimate the stored electric energy uE per unit length for a single propagating
wave in such a guide. If power transfer is
+ uM), and magnetic energy is equal
a
predominantly
vg(uE
electric energy, calculate power transfer for a guide with g
1 mm, 2d
1 cm,
106 V/m, air dielectric, and operating frequency 1.3 times cutoff frequency.
E0
to
=
=
=
9.8a
Assuming (up/c)2
quirements on the
<
type. What should X/n
kaX/n
versus Ba for a slow-wave structure. State the rein order that there may be any slow-wave solution of this
be for ,Ba large?
l, plot
reactance
489
Problems
9.8b Show that
the
a reactance
TM01 type
sheet
kca. Under what conditions might
given guide of this type?
9.8c**
Imagine
be used
might
studied in Sec. 8.9. Plot the
there be
the boundary condition on fast waves of
required value of kaX/1] as a function of
as
a
slow
wave
and
a
series of fast
waves
in
a
transmission line of
spacing 2a in the x direction, for which
with many fine straight parallel slits that lie at an
as in Fig. P9.8c. Assume no variations with y, and
apply
utilized in the helical sheet analysis, obtaining the field components
parallel-plane
a
both upper and lower planes
angle it! from the y direction
are cut
approximations as
for propagation in the z direction, the complete equation determining ,8,
proximate solution of this for ka cot :11 > 1.
and the ap-
FIG. P9.8c
9.9a
Analyze
a
TB surface
plane with no variations in y and show
produce an exponential decay with x.
wave over a
citive reactance is necessary to
9.9b For the TM surface
established
wave
by
the thin dielectric
ductor, find the average power transfer in the
9.9c
uation if the conductor
at x
Repeat Prob. 9.913 with
81
a; ~je’1’with e’{
a
=
=
——
perfect
<
d has
a
z
coating
direction. Find the
surface resistance
conductor and
a
that
a
capa-
on a perfect con~
approximate atten»
Rs.
dielectric with
a
small
lossy part,
8;.
9.9d It is desired to pass an electron beam 1 mm above the structure in Fig. 9.9b. To match
the velocity of the electrons, a wave at 18 GHz should travel at 0.1 the speed of light.
Design
the fins to
produce
the
the field at the beam than at
9.103‘“
Making
the
appropriate impedance
=
x
at .r
=
0. How much weaker is
0?
assumptions that kl < 1 and Bod
Fig. 9.100. Plot the curve
for the structure of
<
1, plot the lowest-frequency passband
for all
,8
and state your
reasoning.
approximations as in Prob. 9.10a, but assume the “troughs” are open
equivalent lumped~element circuit, the propagation constant of the
fundamental, and sketch the (0—3 diagram of this wave. Note the fact that the cutoff
0 and frequencies up to some lower cutoff frequency.
region includes to
9.10b* Take the
same
circuited. Find the
=
low-frequency portion of the (0—3 diagram for a periodic transmission system
composed of parallel LhC circuits in both the series and shunt legs. Note that the fundamental spatial harmonic is a backward wave if the resonant frequency of circuit in
the series leg is lower than that for the shunt leg.
9.10c Find the
9.10d The
example used in Sec. 9.10
was a
those studied in Sec. 9.9 for surface
will represent
nonradiating
or
closed line. Consider
wave
guided
propagation.
open region, such as
regions of the (0—3 plot
an
What
surface waves?
0
9.10e Illustrate Prob. 9.10d by solving a problem with the structure at x
but with the top plate removed so that the region extends to infinity.
=
as
in
Fig. 9.10a
10.]
lNTRODUCTiON
lumped-element type, made by combining separate
small compared with a wavelength, or distributed,
as was seen for sections of transmission line in Chapter 5. In a lumped-element circuit,
a capacitor is used for storage of electric energy and an inductor stores magnetic energy;
at the resonant frequency, there is an exchange of energy between the inductor and the
capacitor every quarter-cycle. There is also an exchange between electric and magnetic
energies every quarter-cycle in the distributed resonant circuit. In this case, however,
the same region is used for both energies, rather than having separate components for
Resonant circuits
capacitors
can
be either the
and inductors, which
are
each type.
Distributed resonant circuits utilize the resonant
properties of standing waves set up
traveling waves on transmission structures
and, hence, are generally of a size comparable with a wavelength. This idea was intro—
duced in Sec. 5.13 for a shorted transmission line with length equal to a multiple of a
quarter~wavelength. Some closed metal cavities can be understood as sections of transmission structures with short-circuited ends and therefore containing standing waves.
Strip-type structures such as microstrip, used in microwave and millimeter-wave cir—
cuits, make light, compact resonant transmission~type structures. A solid dielectriccylinder resonator can be made using a section of a dielectric rod transmission structure
by
interference between forward and
reverse
(Sec. 9.2).
Not all resonators
are
simple enough in shape
to be considered as sections of a wave-
guiding system, and for these, other methods of solving the boundary value problem
are required. We shall see some small—gap cavities that are particularly useful for elec—
devices. Some of these may be considered as capacitively loaded transmission
lines. In others, the electric and magnetic energies are effectively separated so they may
be considered as lumped L-—C circuits, with the one-tum inductor providing the selftron
shielding.
The oscillating
energy is introduced
by a probe or other means of coupling to the
provided by the probe at one of the many resonant
the
seen
by the input probe is real. If the energy coupled in is
frequencies,
impedance
each
than
the
losses
in
cycle, the oscillating waves will increase in amplitude
greater
until the losses just equal the energy supplied. Losses take place in the surfaces of the
metals, in any dielectrics present, and, in open structures, through radiation. If the source
resonant structure. If the energy is
495
10.2
excites the structure at
a
frequency
491
Resonator
Simple Rectangular
Fields of
somewhat off resonance, the
energies
in electric
and magnetic fields do not balance. Some extra energy must be supplied over one part
of the cycle and it is given back to the source during another part of the cycle; thus,
the line acts
as a
ponent representing
is
evident, and
of losses
.3
on
as
exciting source, in addition to a resistive com~
similarity to ordinary tuned—circuit operation
the
circuits,
concept of Q is useful in describing the effect
reactive load
on
the
the small losses. The
with such
bandwidth.
«anagramsmeannaameWanna»mmmmewmsmnwswrmImaerzazerzemmmenaemearfiawammnnzzawmtmeimaawmnrszinanemssmmwmssarncmenfi
Resonators of
10.2
Simpte Shape
FIELDS or SIMPLE RECTANGULAR RESONATOR
For the first mode to be studied in
some
detail,
shall choose that mode in
we
a rectan-
gular conducting box which may be considered the standing wave pattern corresponding
to the TE10 mode in a rectangular guide. As was done in the study of waveguides, the
conducting walls will be taken as perfect, and losses in an actual resonator will be
computed approximately by taking the current flow of the ideal mode as flowing in the
walls of known conductivity.
In the rectangular conducting box of Fig. 10.20, imagine a TE10 waveguide mode
oriented with its electric field in the y direction and propagating in the z direction. The
O and d, as required by the perfect conductors,
E), shall be zero at z
condition that
=
is satisfied if the dimension d is
a
half—guide wavelength. Using Eq. 8.8(11),
/\
A
2
Then the resonant
frequency
WI
(A/Za)2
—-
is
V612
u
f0
+
d2
(1)
m
K
Zaa’V/ie
To obtain the field distribution in the dielectric interior,
pr0pagating
waves
of the form of
E,
=
Ht
2
H.
=
Eqs. 8.8(4)
(E+e"j'83
1
+
and
5.61133)
we
add
positive
and
negative
8.8(5)
33
sin
(2)
a
.
.
————
'
ZTE
(Em—mg
—
EJWZ)
sin
731-—
(3)
a
'
i
7)
)t
<—~>(E+e"mz
2a
.
.
+
E__e”3£)
cos
3
a
(4)
492
Chapter
10
Resonators
(a)
(a) Rectangular cavity. (b) Electric and magnetic fields in rectangular resonator with
mode. Solid lines represent electric field, and dashed lines, magnetic field.
FIG. 1 0.2
TE101
Since
E), must be
wave
from the
also be
E0
=
zero at z
=
O, E“
perfectly conducting
zero at z
=
(1',
so
that
=
-—E+,
B
=
77/61.
would expect, since the reflected
must
equal to the incident wave.
as we
wall should be
Then
E),
(2)—(4)
may be
simplified, letting
~2jE+2
7rx
Ey =E
Hx
'-—-
'
772
5
()
——
osma 31nd
~
fiisin—cos
JnZd
E
)t
W2
d
(6)
‘
z
Hz=j-—°—cos-7E~sin—~
(7)
In studying the foregoing expressions, we find that electric field passes vertically
from top to bottom, entering top and bottom normally and becoming zero at the side
walls as required by the perfect conductors. The magnetic field lines lie in horizontal
displacement current resulting from the time rate
roughly in Fig. 10.217. There are equal and opposite
charges on top and bottom because of the normal electric field ending there. A current
flows between tap and bottom, becoming vertical in the side walls. Here we are re-
xuz
planes
and surround the vertical
of change of Ey. Fields
are
sketched
Energy Storage, Losses.
10.3
minded of
a
493
Q of a Rectangular Resonator
conventional resonant circuit, with the top and bottom acting as capacitor
as the current path between them. In the lumped-element
and the side walls
plates
circuit, electric and magnetic fields
the y direction, and one in the
coordinate system is of course
be described in this
10.3
z
separated,
are
Because the mode studied here has
can
and
one
whereas here
they
half—sine variation in the
direction, it is sometimes known
x
as a
are
intermingled.
direction,
T151101
none
in
mode. The
arbitrary, but some choice must be made. before the mode
manner.
ENERGY STORAGE, LOSSES, AND 9 OF
A
RECTANGULAR RESONATOR
The energy storage and energy loss in the rectangular resonator of the preceding section
are quantities of fundamental interest. Since the total energy passes between electric
magnetic fields, we may calculate it by finding the energy storage in electric fields
the instant when these are a maximum, for magnetic fields are then zero in the
and
at
standing
wave
pattern of the
resonator:
d
8
U
Utilizing Eq. 10.2(5),
(UE)max
=
=
7)-
]; L J0 [33,.“
d
b
*J J J
7
o
o
SE23
(1
and
9
dx
d)?
612
a
s
=
a
that
we see
U
b
o
,
7r:
m
E851n2—31n2—dxdydz
d
a
(1)
2
To obtain
an
If the
to
J5),
=
Left side:
J's),
=
Top:
J5“.
=
conducting
losses
as
8
for power loss in the walls, we utilize the current flow
obtained from the tangential magnetic field at the surface.
~Hx|3£
Back:
“H:'.\~=o
Right
-H:, J5:
Hx
=
walls have surface
side:
Bottom:
resistivity R5,
the
sz
2
J5),
=
J3,
=
Hi":=0
Hit-=0
H3, JSZ
foregoing
=
~Hx
currents will
produce
follows:
R
b
2
+ 2
o
o
b
d
a
wL=—§{2 f f
at: tidy-+2
f: f: [|H,.|2
+
lelgl
dx
ff
o
o
ngli-zodydz
dz}
equation, the first term comes from the front and back, the second from the left
right sides, and the third from top and bottom. Substituting from Eqs. 10.2(6) and
In this
and
2
Fig. 10.2a,
Front:
as
....,,
2
bd
a;
E20
approximation
in the ideal conductors
Referring
d
494
Chapter
Resonators
10
102(7) and evaluating the integrals,
R 5 A2
WI“
The
Q of the
3
8772
0|:a’2
+
—
1
+
——
a2
—
2
d
a
(d (1)]
—
+
(2 )
——
resonator may be defined from the basic definition of
Q
Substituting (1)
a
2
cube,
5:
b
=
=
wOU
WL
(3 )
we
have
219m2
czd(cz:2 + d2)
+
+
d2)”
2b(a3
d, this reduces
to
the
==
0.742
77
chbe
==
71
—-
—-—
6
Eq. 5.14(4):
—-~
(2) with A from Eq. 10.2(1),
and
4RS
Note that for
bd
ab
E?—
R5
+
£13)
expression
77
(5 )
-
R5
377 9., and for a copper conductor at 10 GHz,
0, the Q is about 10,730. Thus we see the very large values of Q for such
resonators as compared with those for lumped circuits (order of a few hundred) or even
with resonant lines (order of a few thousand). In practice, some care must be used if
Q’s of the order of that calculated are to be obtained, since disturbances caused by the
coupling system, surface irregularities, and other perturbations will act to increase the
For
RS
e:
an
air dielectric,
77
z
0.0261
losses. Dielectric losses and radiation from small holes, when present, may be
serious in lowering the Q.
It will be shown in
in the
vicinity
of
Chapter
resonance.
11 that
a
From it
lumped
we can
circuit model
cavity mode
Q, defined in terms
applies
deduce that the
especially
to a
of stored energy and power loss, is also useful in estimating bandwidth of the cavity
as for the lumped resonant circuit. If Af is the distance between points on the
just
response
curve
for which
amplitude
response is down to l
/\/2 of its maximum value
(Sec. 5.14),
-—
Thus, for the foregoing,
bandwidth between
10.4
a
e
Q of 10,000 in
“half-power” points
a
(6)
—-
cavity
resonant at
10 GHz will
yield
a
of 1 MHz.
OTHER MODES IN THE RECTANGULAR RESONATOR
As has been noted, the
particular mode studied for the rectangular box is only one of
possible modes. If we adopt the point of view that a resonant
mode is the standing wave pattern for incident and reflected waveguide modes, any one
of the infinite number of possible waveguide waves might be used, with any integral
an
infinite number of
Other Modes in the
10.4
number of half-waves between
ends. We
shorting
495
Rectangular Resonator
that this
recognize
description
of
a
particular field pattern is not unique, for it depends on the axis chosen to be the “di—
rection of propagation” for the waveguide modes. Thus (see Prob. 10.2b) the simple
mode studied in past sections would be a TE101 mode if the z axis or x axis were
considered the direction of propagation, but it would be a TM110 mode if the vertical
(y)
axis
were
taken
will be chosen
as
as
in
direction. In the following, a coordinate system
and
field
10.2a,
patterns will be obtained by superposing
Fig.
the
incident and reflected
propagation
for various
waves
modes
waveguide
propagating
in the
z
direction.
The
TEmnp
Mode
If
Sec. 8.7), addition of
we
select the
positively
TEnm mode of a rectangular waveguide (see
negatively traveling waves for H3 gives
and
mm
=(Ae’jfi‘
+
Belfl‘)
)2
--—
cos
cos
G
_:'y_
Since the normal component of magnetic field H., must be zero at
then B: —A and Bd= p77 with p an integer Let C-— ~2jA.
z
=
O and
z
=2
(I,
~
-
ll
A(e”JB‘
_
613‘)
—~
)2
mm
.
.
Hz
cos
——
cos
a
73772
(1)
mm
=
C
cos
mry
——
cos
pi)":
,
——
3111
d
b
a
Then, substituting (l) in Eqs. 8.2(13)—(l6), remembering that for the negatively
eling
waves
all terms
HI
=
=
H),
=
E.
=
E),
where
,B
=
135—3: and,
trav-
multiplied by [3 change sign,
(A6
#3
kg
“13‘
mm:
mqr
B
—
eJB)<——>
sin
cos
a
a
~52;- (p777)(1:?) 3253: 51—2-3}- p—ZE
—% (Bi-T) (1%) mz'x 12%?- p732
ngC (3;) ??- n—ZZ L22
_ngC (M) 2:111- 12—272 p—ZE
sin
cos
cos
sin
sin
from
cos
sin
sin
cos
cos
sin
cos
Eq. 8.2(18),
<-—> <>
a
b
12
_’n'y_
b
(2)
(3)
(4)
(5)
496
Chapter
From
(6)
and
Eq. 8.2(19)
we
Resonators
10
find the resonant
frequency:
f0
=
Thin,” Mode
The
modes in
a
31-3
If
271' ,ue
In
=
'
37-7
p—W
+
b
a
1/2
(7)
d
positively
and
may be combined to
rectangular waveguide
E,
+
similar manner,
a
2
2
2
negatively traveling '1‘an
yield
Dsinm—msmmcoslfl
(8)
d
b
a
5, (19—5) (11;) 2153 "—22 233
Ey
~53 <p;><_7r) m—ZZ 12—? sin?
Hx=
fulfil) (n_
—>sin~lz:lflcosn—Z-y—cosP—Z:E
H3,
—jC::D (—
—>cos $ 22—y- 1%
E,
sin
sin
.—..—
sin
:
(9)
(10)
cos
b
(11)
b
sin
2
(12)
cos
a
The
quantity kg
and resonant
frequency f0
are as
in
(6)
and
(7).
General £0mments
We note first that TM and TE modes of the
p have identical resonant
frequencies. Such modes with different field patterns but the
are known as degenerate modes. Other cases of degeneracy
same resonant
frequency
b
may exist as in a cube, (1
TE types have the same resonant
r"—
It is also
=
frequency
order m, n,
d, where orders 112, 121, and 211 of both TM and
frequency.
apparent from (7) that,
box size, the resonant
same
as
the order of
increases. Put
a
mode becomes
it
higher
for
a
given
that to be resonant
differently,
bigger as the order increases.
means
This is to be
given frequency, the box must be made
fit
waves
in
each
It
can
half-sine
are
to
dimension.
be shown
since
more
expected,
a
10.4b
that
increases
at
as
one
and
1040)
Q
(Probs
given frequency
goes to higher
mode orders. This too is logical, since the larger box has a greater volume~to~surface
ratio, and energy is stored in the volume, whereas it is lost on the imperfectly conducting
surface. The high—order modes are consequently useful in “echo boxes” where a high
Q is desired so that the energy will decay at a very slow rate after being excited by a
pulse.
at a
10.5
For
a
circular
ClRCULAR CYLINDRICAL RESONATOR
cylindrical resonator (Fig. 10.5) there is a sirnple mode analogous to that
rectangular box (Sec. 10.2). The vertical electric field has a maxi»
first studied for the
497
Cylindrical Resonator
Circular
10.5
h-a—i
‘3
mxs
/
a
Current
-fi¢
-
/
§‘\\lfl/lfll/ll/l/I/fl/l/l/l/Il ’l/ll/I/YI/ll/I/ll/Illlf/I //:
Electric field
—-——-—
Sections
FIG. 10.5
through
a
Magnetic field
cylindrical cavity with fields
of
TMO10 mode.
Convention is
as
in
Table 8.9.
conducting side walls. A circumferential
magnetic
represented by the time—varying elec~
displacement
tric field. Neither component varies in the axial or circumferential direction. Equal and
opposite charges exist on the two end plates, and a vertical current flows in the side
walls. The mode may be considered a TM01 mode in a circular waveguide operating at
cutoff (to give the constancy with respect to z), or it may be thought of as the standing
wave pattern produced by inward and outward radially propagating waves of the radial
transmission line type (Sec. 9.3). From either point of view we obtain the field
mum at
the center and dies off to
zero at
field surrounds the
the
current
components
E:
=
ab
=
k
'E
1—“
frequency
(2)
Jl<kr)
77
=
E
2.4 05
(3)
=
(2
a
Then the resonant
(l)
EOJO(kr)
is
2.405
k
f0
(4)
=
=
277V us
2770 V [1.8
The energy stored in the cavity at resonance may be found from the energy in the
electric fields at the instant these have their maximum value. Take a and (1, respectively,
498
as
radius and
Chapter
length
of the
cavity:
a
U
d
=
a
SligJ2
9
’7
2771' dr
——
=
are
2
0
This may be
Resonators
10
115(k))
dE5
dr
o
integrated by Eq. 7.1502):
7
U
If the walls
of
are
=
778
dEg
0? mica)
(5)
the power loss may be calculated
imperfect conductors,
approximately:
a
R
WL
=
Zarad
—§
2
152
2
+
2f
o
R
—5
lerlz
2
2
7rr
dr
The first term represents losses on the side wall, the second on top and bottom. The
current per unit width J5, on top and bottom is ng, and JSZ on the side wall is the
value of H g at
r
=
a.
WL
Substituting
=
77R s ad
from
E3
—
2
(2),
we see
12(ka)
l
=
naRsEz0
—2
,,
rJ‘f(kr)
--;
J}(ka)[d
dr
7]
O
integrated by Eq. 7.15(22), recalling
L
The
E3
+ 2
7]
This may also be
for resonance:
that
"
that
+
J0(ka)
=
O is the condition
(5)
a]
Q of the mode may then be obtained as usual from power losses (6) and energy
(5), using (4) for resonant frequency:
stored
on
Tl
P01
(U
QzfiiziflfiTfi
where
P01
:3
2.405
An infinite number of additional modes may be obtained for the cylindrical resonator
by considering others of the possible waveguide modes for circular cylindrical guides
propagating in the axial direction with an integral number of half guide wavelengths
plates. In this manner the standing wave pattern formed by the superposition of incident and reflected waves fulfills the boundary conditions of the conducting
ends. Table 10.5 shows a TE11 mode, a TM01 mode, and a TEOI mode, each with one
half guide wavelength between ends. The resonant wavelengths shown are obtained by
solving the equation
as
between end
A
A.
)k
2
d‘%=rb'hfl
-
1/2
®
Circular
10.5
499
Cylindrical Resonator
Table 10.5
TE111, Cylinder
A
/
/////////
/
0
DO
0
I\°°/\
/
E/o
;
00
O
00
0
00
0
00
/\
A
//////////
li/VVLUUK
A
coo
e
0
l\’°/\
0
I
.'
i/
E
j
on
o
o.
g
0.
°
O.
O
2
/
/
I
///////
Cross section
through AuA
2l
+(m)
2Z‘HJ'I'81‘
;
/
j
1
ejfo:c
.
O
D
j\
0'
/l
//
///
/
A
l
’
/
I
‘
l
A
TMOH, Cylinder
I
1+
c
__
21
(2.610,)
2
zit—mar
Cross section
through A~A
|
A
l
F
I
j
A
TECH, Cylinder
l
///////////////////////
/
____________
/e
/
o
o
a
0\
a
5
55—1—4 5
j’.("'.‘“.: 3.7“")
xii—1°...
j
j
.l
/
{
/
5
/B““o‘-‘o“T’—5—“‘J‘”€\
o(
”“—"—o
o
o——
o
o
o
\
\1...°.....°_|
“.2__°
/
/
01
o
°
1
f0
c
:
+
(1.52:)
2
————_
ZZVWE—r
/
/
j
if///////////{////////////
Cross section
through A-A
<
l
A
I
"
corresponding resonant frequencies are shown in the table. For these modes the
integer p is unity in (8), and cutoff wavelength AC is obtained from Sec. 8.9. Note that
in the designations TEN], TMOH, TED“, the order of subscripts is not in the cyclic
order of coordinates, r, ()5, 2, since it is common in circular waveguides to designate the
qS variation by the first subscript.
Of the foregoing modes, the TE011 is perhaps the most interesting since it has only
circumferential currents in both the cylindrical wall and the end plates. Thus, if a re—
sonator for such a wave is tuned by moving the end plate, one does not need a good
contact between the ends and the cylindrical wall since no current flows between them.
The
509
Chapter
For both of the other modes shown
current does
good
to
flow between the
Resonators
no
(and in fact all except those of type
and its ends
cylinder
so
that any
sliding
TEOmp)
a
finite
contact must be
prevent serious loss.
it would be found that
higher wave orders (those
require larger reson~
ators to be resonant at a given frequency. The Q would become higher because of the
increased volume-to—surface ratio, but the modes would become close together in
frequency relative to the resonant frequency so that it might be difficult to excite one
mode only.
As with the
having
more
rectangular resonator,
variations with any
or
all coordinates r,
qS, 2)
would
STRIP RESONATORS
10.6
Strip-type resonant structures are used in microwave and millimeter-wave circuits as
single resonant elements and as components in filters. Such structures are light and
compact, though more limited in power—handling capability than the box resonators
discussed in the preceding sections. The quality factor Q of this kind of resonator is
limited by losses in the conductors and dielectric, and is also decreased by radiation,
since they are open structures. The examples shown here are of the common microstrip
configuration (Sec. 8.6), with the strips on a dielectric substrate coated on the opposite
side with a metallic ground plane, but similar arrangements can be used with the co~
planar configurations.
Figure 10.6a shows a microstrip resonator equivalent to the resonant transmission
line with short-circuited ends described in Sec. 5.13. To satisfy the boundary conditions
at the ends, the strip length 1 must be
1:
where
Ag
the fields
=
vp/ f
across
=
gaff and
)to/
n
is
mtg/2
an
integer.
(1)
We
assume
negligible
variation of
the
Short circuits in
strip.
microstrip require connections (“vias”) through
the
dielectric,
so a
and convenient arrangement is to leave the ends of the strip open, rather
in the simple model that assumes ideal open circuits at
than shorted. Again, 1
more common
=
nag/2
the ends. However, there
by
an
added
length
Al
so
are
flinging fields
that the
at the
ends and these
resonance condition
z + 2A1
=
can
be
represented
is
rug/2
(2)
The value of Al in the usual situation where the dielectric and
ground plane extend
beyond the end of the strip has been found by numerical methods. Practical calculations
can be made from empirical formulas that
have been fit to the results of the calculations.
The following formula is accurate to about 5% for 0.3 < w/ d < 2 and l < er < 50:1
Al
z
‘22“
‘
0’412
(gaff
(361.,
+
——
O.3)[(w/d) + 0.262]
O.258)[(w/d) + 0.813]
R. K. Hoffmann, Handbook of Microwave
MA, 7987.
(3)
Integrated Circuits, An‘ech House, Norwaod.
Shorts to
5%?
Strip Resonators
10.6
ground plane
all/4;:7r’
:l/
T'
g
\
Ground
plane
(b)
(a)
FIG. 10.6
resonator
where
8.6.
w
ends.
(a) Microstrip resonator with short-circuited
showing capacitively coupled input and output.
is the
Figure
d is the dielectric thickness, and gaff is defined as in Sec.
open—circuit resonator with capacitively coupled input and
strip width,
10.6b shows
(1)) Open-circuited microstrip
an
output connections.
Another
in
Fig.
important form
10.6c. There
circumference is
an
of
strip-type
structure is the
I—
27Travel"
‘“
straight-line
adjacent straight
mean
(4)
"Ag
2
Because of the curvature of the line, radiation is
in
microstrip ring resonator shown
end effects and, if the curvature is not too great, the
integral number of wavelengths at resonance. Thus,
are no
important source of loss than
example
by distributed coupling to an
of microstrip, though capacitive coupling as in Fig. 10.6b can
resonators.
section
a more
Excitation in this
is
be used.
Strip—type
also be made in the form of
rectangular or circular patches,
Note that the rectangular patch
Figs.
width
comparable with wavelength. If the
having
in
the
resonance
circuits,
rectangular patch is as in the rectangular box
lO. 2~— 10.4, except with open-~circuit (zero tangential H) boundary con-
resonators can
normally with open sides as shown in
differs from the strip in Fig. 10.6b by
10.6d and
6.
edges are open
cavity in Secs.
ditions. Only modes with nonzero E._. and no variations in the: direction can be accom~
modated in the strip~type structure. Thus, the modes are the TMnmp types in Eqs.
10.4(8)-(12) with
p
=
0:
E2
2
E0
Hx
=
1-;-
HV
:
k)’
[(3.
=
mw/(a
+
Aa)
and
k),
=2
kyr
cos
(5)
kyy
.
1‘0.“
'
with
cos
E0
k‘.
--;‘—
10).“
mar/(l)
klx
cos
sm
(6)
kvy
'
.
E0
+
Sin
kfr
cos
Ab). Here, Aa
and Ab account for fringing,
and may be estimated from (3). In this case the empirical formula,
limiting case of the end of a very wide strip, w/d —-9 00, isl
Aa/d (or Ab/d)
=
2035/8r
(7)
kyy
+
0.44)
adapted
from the
(8)
502
Chapter
Resonators
10
l
--
Input T
"
"
j;
u
F16. 10.6c
The
resonance
condition
1/
..
Microstrip ring resonator
can
be
Ground p lane
with distributed
adapted from Eq. 10.4(7)
coupling.
with p
2
1
fO
>Output
O:
=
2
7277
mar
+
_
277V “3
a
+ Aa
(9)
b + Ab
rectangular patch resonator, a variety of resonant modes exist
simplest of these is the lowest-order azimuthally
patch
a + Aa.
radial—transmission—line
mode,
assuming an open circuit at r
symmetric
Since a is the radius, the Aa/d here is one-half the value in (8). The fields are described
by Bessel functions as in the piflbox resonator of Fig. 10.5:
As in the
case
in the circular
of the
resonator. The
=
E:
=
H¢
=
(10)
EOJo(k7')
E
1—739- J1(kr)
(11)
id
T
—7/
?\
Ground
fig
plane
\
can
Ground plane
(8)
FIG. 10.6
patch
(d) Rectangular patch
resonator.
resonator
in
microstrip technology. (6)
Circular
microstrip
With
an
open circuit at
condition is
a
Art,
+
503
Strip Resonators
10.6
the
field
tangential
must vanish
H¢
there,
the
so
resonance
1
1
0
P11
O
27rV,LLe
(
A“
+
ZWVMSO
where p11
3.832 is the first root of the Bessel function J1.
More generally, the modes in the circular patch may be considered
a
analytically the
modes
at
cutoff
where
there
is
2
no
variation (B
(Sec.
8.9)
waveguide
TM”1
0), except that here the open-circuit boundary condition at the edge radius must be
applied. From Eqs. 8.9(8)—(10),
same as
=
E2
=
EOJ,,(kCr)
,
H 4,
=
H,
—-J
we
kc
=
the resonant
must use
we
frequency
f0
no
variation in the
Its resonant
frequency
z
or
22¢
(14)
22¢
(15)
f0
=
kC/ZWVTL—g.
The cutoff
instead of pm, which are the roots
the roots of its derivative, pf,1 (see Table
is
pr)!
__
27r(a
The mode with the lowest resonant
dicates
cos
r
frequency f0 where k
kc,
given by Eq. 8.9(12) except that
of the Bessel function in (13),
7.15b). Thus,
cos
Jgalz— EOJ,,(kCr)
:
Resonance is at that
is
(13)
,
k— EOJ,,(kCr)
C
wavenumber
1qu
cos
(16)
Aa) V
+
frequency
is
/.L8
TMIIO,
direction. This mode does
where the third
not
subscript
in~
have azimuthal symmetry.
is
1.841
f0
(17)
_
27r(a
Aa)\/,u,8
+
which is lower than that of the lowest~order
All of the
and
e are
common
asymmetric,
methods of
so one
symmetric mode (12).
patch resonators indicated in Figs. 10.603
expect the asymmetric modes to be preferentially
exciting
should
the
excited.
The factors that contribute to the
are
proportional
to the loss
resonances.
of these resonators
RS,
are
conductor losses, which
dielectric losses, which
are
proportional
of the dielectric, radiation losses, and excitation of spurious
Radiation losses can be minimized by enclosing the structure in a metal
factor tan
58
box, and dielectric losses
of
Q’s
to the surface resistance
Q associated
be combined
are
usually
high values
nearly independent and can
less than those in the conductors. For
with the various loss mechanisms,
they
are
as
Q---<i-}-i+—1-->~l
Qc
Qd
Qt
(18)
504
Chapter
Resonators
10
subscripts refer to conductor, dielectric, and radiation, respectively. Each Q
in (18) is given by Q
wOU/WL. Energy loss WL in the conductors can be calculated
as for the box cavities, i.e., integration of Eq. 3.18(5) over the metal surfaces. The
dielectric energy loss is found by integrating the density, WM
woa”E2/2, where E is
where the
=
=
electric field,
over
Procedures for
U is calculated
the volume of the resonator.
calculating radiation
as
are
found in the
literature.1 Energy storage
Q’s are much lower
100 to 1000 for copper or gold on any of several different
temperature. Strip-type resonators of the same configurations but
than for box cavities,
dielectrics
loss
for the box cavities. With normal metals, the
at room
typically
superconductors have Q’s 10 to 100 times higher at 10 GHz. The advantage
of using superconductors is greatest at microwave (< 20 GHz) frequencies, and decreases at higher frequencies because of the difference of frequency dependencies of
RS shown in Fig. 3.16b.
made with
10.7
WAVE SOLUTIONS IN SPHERICAL COORDINATES
Spherical
resonators are of more intellectual than
introduce
wave
solutions in
spherical
only those solutions
solutions. We shall sketch here
The solutions with
practical interest,
coordinates. In this section
variations
but will
we
serve
develop
with axial symmetry,
involved, but have been
a/a¢
to
such
=
0.
general 4)
given
pletely by Stratton.2 It is found that with axial symmetry the solutions separate into
waves with components Er, E
and those with components H,, H 0,
6., H46
E¢. These are
called TM and TE types, respectively, the spherical surface r constant serving here as
are more
com-
the transverse surface.
0 in
spherical modes with axial symmetry by setting 6/69!)
The
three
coordinates.
curl
equations
spherical
equations containing E,,
Consider then TM
Maxwell’s
E 6, H
4,
2
in
are
a
“a“;(rEa)
1
-
air,
,
=
60
59-01€581“
'
rsinaaa
‘
E
a) ’Ja’gr
a
and
2
U
—
.
’5? (rH¢)
Equations (2)
(1)
“anO-Hqg
:
jtdSO‘Eg)
(3)
(3) may be differentiated and substituted in (1), leading to
an
equation
in H
alone:
4,
62
_
61'20
2
J. A. Strah‘on,
.H9”) +
E
ia[ia
W...
’2
....____.
_
sin a 66
(rqu
sin
9)]
+
k2(rH¢)
Electromagnetic Theory, Chap. VII, McGraw—Hill,
=
0
New York, 794 I.
(4)
To solve this
differential
partial
505
Wave Solutions in Spherical Coordinates
10.7
equation,
we
follow the
solution
product
technique.
Assume
0-11,)
where R is
function of
a
:
alone, and 9 is
r
a
Re
(5)
function of 6 alone. If this is substituted
in (4), the functions of r may be separated from the functions of 6, and these must then
be separately equal to a constant if they are to equal each other for all values of r and
6. For
a
definitely
ulterior motive,
"R"
k2-2——i—dGale
I
R
Thus there
Let
us
label this constant
we
are two
1d<e
__
1101
6)
SM]
sin6d6
ordinary differential equations, one
making the substitutions
+
1):
(
-
in
only
1'
(6 )
+1)
“n”
and
one
in 6
only.
consider that in 9 first,
d
Vl—u2=sin6,
uxcosd,
d
“sine—
du
-——=
d6
Then
2
(1
—-
The differential
in fact
a
L12)
d8
d 8
d
2n
2
__
+
du
equation (7)
[”02
is reminiscent of
1
+
1)
112]
——
—~
l
—
6
O
S
Legendre’s equation (Sec. 7.18)
(7 )
and is
standard form. This form is
dy
2x;
One of the solutions is written3
m2
+
y[iz(n
+
1)
-—
1
_
x2};
=
O
(8)
:P";(x)
by this solution is called an associated Legendre function of
degree 271. These are related to the ordinary Legendre functions
and the function defined
the first
by
the
kind, order
n,
equation
m/2
Pm (x) ( 1__ x2 )
....
As
a
matter
of fact,
d’”P,,(x)
dx’"
(9)
(8) could be derived from the ordinary Legendre equation by this
substitution. A solution to (7) may then be written
2,1"(u)=P,1,(COS
(10)
0)
And, from (9),
P},(cos
3
A second solution,
(Ref. 2).
@300,
6):
—a%0— P(cos
6)
is needed when the axis is not included in the
(11)
region of solution
506
Chapter
Resonators
:0
integral values of n these associated Legendre functions are also polynomials
consisting of a finite number of terms. By differentiations according to (9) in
Eq. 7.123(8), the polynomials of the first few orders are found to be
Thus for
Pcl,(cos
6)
Pi(cos 6)
P911003 6)
P§(cos 6)
=
O
=
sin 6
=
3 sin 6
cos
%sin
6
(5 cos2 6
gsin
6
(7 cos3 6
=
Pi(cos 6)
2
6
(12)
l)
-—
3
-
cos
6)
Other properties of these functions that will be useful to us, and which may be found
a study of the above, are as follows:
from
'n".
P,1,(cos 6) are zero at 6 O and 6
2
if
n
is
even.
are
zero
at
6
6)
77/
P},(cos
P},(cos 6) are a maximum at 6 77/ 2 if n is odd,
1. All
2.
3.
=
=
=
=
is
and the value of this maximum
given by
1
10,,(0)
(-l)_(”""1)/2n!
=
12
odd
(13)
Wit" ;: ii
Legendre functions have orthogonality properties
Legendre polynomials studied previously:
4. The associated
of the
I
0
P}(cos 6)P,1,(cos 6)
sin 6 d6
7’
J
o
[P,1,(cos 9)]2
sin 6 d6
=
=
l #
O,
similar to those
(14)
n
2
+ 1
101—)
(15)
212 + 1
5. The differentiation formula is
§6[P,1,(Cos 6)]
Note that
only
one
=
sin 6
[nPi+1(COS
antenna
analysis
on
R1
the
1)
cos
the
and
6P},(cos 9)]
equation (7)
will not be
axis,
axis, but will be needed in problems such
r
differential
R/W:
on
so
as
(16)
has been
required
the biconical
obtainable from (6), substitute the variable
equation
.
c121?1
air2
By comparing
+
of Sec. 12.25.
To go back to the
=
(n
—
solution for this second-order differential
considered. The other solution becomes infinite
in solutions valid
9)
with
+
lde
——
r
+
dr
Eq. 7.l4(3) it
is
[
seen
(n
9
k~
—-
+
$92
—
r2
R1
=
0
that this is Bessel’s differential
equation
of
Wave Solutions in
10.7
order
%j.
+
n
A
507
Spherical Coordinates
solution may then be written
complete
R1
+
Aan+l/2(kr)
=
Bner+1/2(kr)
(17)
and
R
If It is
an
WR,
=
integer, these half-integral~order Bessel functions reduce simply to algebraic
example, the first few orders are
combinations of sinusoids.4 For
J1/2(x)
=
J3/Z(‘x)
:
/2
E
sin
2
——
x
[ sin
x
cos
..
x
L...
N1/2(x)
=
N3/2(I)
2
i2
——
acosx
2
3:]
{SID-x
cos x
.
——
«—
+
L.
___-x__:|
‘
J5/2(x)
2
3
E
J?
l
-—
L
-—
-3-
‘3
2
,
=
N5/2(x)
Sinx
=
;
L;
3
x]
cos
x
.
smx
-—-
+
(7 1) x]
cos
w
x“
The linear combinations of the J and N functions into Hankel functions
represent
as
found
traveling radially inward or outward,
previously for other Bessel functions:
1. If the
region
infinite
2. If the
at r
of interest includes the
=
region
resent
of interest extends to
radially
a
particular
infinity,
function, H f3; 1 /2
outward
traveling
combination of
:
All
V7-
=
J,,+1/2(kr)
4
——
present since it is
J,,+1/2
-
jN,,+1/2,
must be used to
and
N,,+1/2(kr) required
for any
by combining correctly (17), (10),
(2) and (3), respectively.
rep-
problem
and
(5),
P},(cos 6)Z,,+1/2(kr)
A,P‘(cos 6)
jw8’3/2
__
cannot be
the linear combination of J and N into
Ex—’—"———Z
I
[I1 n+l/2(k‘-k-Z__
I)
6
E,.
conditions will be
wave.
may be denoted as Zn+1 /2(kr), and now
H is determined. E, and E 9 follow from
g
H4,
origin, N,,+1/2
(Sec. 7.14)
boundary
O.
the second Hankel
The
and
waves
(18)
A n 222n
"jaw—V:
7(kr)
lgsin 0
n
[cos
6P,l,(cos 9)
1/2(kr)]
——
P},
+
1
(9)
l(cos 0)]
Special notations for the spherical or half-in tegraI—order Bessel functions have been introduced and are useful if one has much to do with these functions. Thus Stratton, following
Morse (Vibrations and Sound, p. 246, McGraw~HilL New York, 7936) usesjn(x) to denote
(77/ 2X)‘ / 2.1,, + I ,2(x), and similar small letters denote other spherical Bessel and Hankel functions. Still other specialized notations have been used. Because of our limited need for
spherical coordinates, we shall retain the original Bessel function forms so that standard
recurrence
formulas may be used.
503
Resonators
10
Chapter
spherically symmetric TE modes may be obtained by the above and the principle
duality (Sec. 9.5). We then replace E, and E 9 by Hr and H 9, respectively, H96 by
E
and s by ,u:
(,5,
The
of
—
E¢
=
H9
:
BIZ
P},(cos 6)Zn+1/2(kr)
7,—
_B,f ,1,(cos 0) [n.2,+1
flaw)
ij‘B/z
12,2211
2
—
k7)
31:1 [cos 6P,1,(cos 6)
flax-3721
I‘
P},+1(cos 6)]
_
0
10.8
(20)
krz,,_1 ”(100]
SPHERICAL RESONATORS
general discussion of spherical waves from the preceding section will now be
applied to the study of some simple modes in a hollow conducting spherical resonator.
Since the origin is included within the region of the solution, the Bessel functions can
1 in Eq.
only be those of first kind, 1,, + 1/2. For the lowest-order TM mode, let n
10.7(19) and utilize the definitions of Eqs. 10.7(12) and 10.7(18). Letting C
The
=
=
A1(2k/7r)1/2,
we
then have
C sin 6
H¢
Er
=
kr
=
_.
Zan
=-
6
cos
'
mC
[(2:11
cos
—
kr
kzrz
‘
E,
<
sin kr
6
[(
(
kr)
sin kr
_
cos
kr
[0‘2
(1)
[0)
.
1
-
)kr
sin kr +
cos
(2)
kr]
(3)
The mode may be designated TMIOI’ the subscripts here giving variations in the order
1‘, d), and 6. Electric and magnetic field lines are sketched in Fig. 10.80.
To obtain the
resonance
condition,
perfectly conducting shell,
r
=
a.
we
know that E 9 must be
From (3), this
zero at
the radius of the
requires
M
tan
ka
:
l
Roots of this transcendental
equation
found at ka
resonant
m
2.74, giving
a
(4)
—_
-
(ka)2
may be determined
frequency of
numerically
and the first is
1
f0
a»
———————_—
2.29
,uea
(5)
509
Spherical Resonators
10.8
Electric field
-~-—-
Section
Magnetic
field
Section
through axis
through equator
(a)
Section through equator
Section through axis
(b)
FIG. 10.8
mode in
(a) Field patterns for TMW mode in spherical
spherical
resonator.
(b) Field patterns for TE‘01
resonator.
The energy stored at
resonance
may be found from the
peak
energy in
magnetic
fields:
U
=
ff
0
The value of HQ5 is
the
resonance
given by (1),
requirement (4):
The
=
sin 6516 d;-
and the result of the
1
+
k
in conductors of finite
2
f Alidi
o
IH¢I227rr2
—
”R H
=
2
integration
:
9
sin"
[ca]
conductivity
4 R
,,
27m“ sin 6 d6
may be
simplified by
2
fig;— [Ira ——k—(E‘1
approximate dissipation
WL
5"-
C2
2
U
o
,,
(6)
is
% a'C2 sin2
ka
(7)
5'3 0
So the
Chapter
10
Q of this mode is
"n
Q:
[k0 _1+_(ka>2]efl
31112 [ca
2Rs(ka)2
The “dual” of the above mode is the
obtained
fields
Resonators
in
ka
TE“,1 mode,
E 45 for
(l)
(3)
by substituting
Hg,
sketched in Fig. 10.8b. Note that the
O at r
a, requires
by setting 13¢
to
are
obtained
~
(8)
R5
and its field components may be
Hr for E,, and “Hg for E 9. The
resonance
condition for this mode,
=
=
tan
Numerical solution of this
yields
ka
z
ka
ka
2
4.50,
or
1
f0
w
1.395 V p.861
Small flap gavities and
10.9
(9)
—
Qonpimg
SMALL-GAP CAVITIES
Because of their shielded nature and
high Q possibilities, resonant cavities
are
ideal for
in many high-frequency tubes such as klystrons, magnetrons, and microwave
tn'odes. When they are used with an electron stream, it is essential for efficient energy
use
transfer that the electron transit time
sible. If resonators such
across
the active field
those studied in
region be
sections
as
small
as
pos-
used, very short
preceding
cylinders or prisms would be required, and Q would be low and interaction weak.
Certain special shapes are consequently employed which have a small gap in the region
that is to interact with the electron stream. Several examples of useful small-gap cavities
as
were
will follow.
Exampie
nose
FORESHORTENED COAXIAL LINES
Fig. 10% may be considered a coaxial line A terminated in the gap
capacitance (leading to the equivalent circuit of Fig. 10.917) provided that the region
B is small compared with wavelength. The method is particularly useful when the region
B is not uniform, but contains dielectrics or discontinuities, so long as a reasonable
estimate of capacitance can be made.
The structure in
B
5'! l
Small-Gap Cavities
no.9
I
2
'9
'{II/IIIIll/IlllflllllI’llIll/IIIlllll/I/I/I/I/Illllllli.’o
I
\
g
{VII/II’llIII/III’ll/IIIII/llIII/IIIIIIIII/Illll[III/I'l/I/Il/J:
(a)
(b)
_IIIIII’IIIIIIIIIIIIIIIIIIIIII/IIIIIIIIIIIIIIII
'
\
\
WW
III/IIIIIIIIIIIIIIIII
r
/
I
III/IIIIIIIl/IIIIII
h...
1—54
‘<———
ll l ll
\
°
liip/
0
r2——.—l
(d)
(c)
Sphedcal
conducfing
surface
Conical
conducfing
surface
(«2)
FIG. 10.9
(a) Foreshortened coaxial-line resonator. (b) Approximate equivalent circuit for (a).
(c) Foreshortened radial—line resonator. (d) Resonator intermediate between foreshortened coaxial
line and foreshortened radial line. (e) Conical—line resonator.
For resonance, the reactance at any plane should be equal and opposite,
opposite directions. Selecting the plane of the capacitance for this purpose,
looking
in
1
jwoc
Of
Bl
=
t an
"1
1
——
20ch
( 1)
found numerically from (1).
l), the line is practically a quarter~wave length. For larger
values of C, the line is foreshortened from the quarter-wave value and would approach
The resonant
frequency
(ZOwOC
If C is small
zero
length
if
must be
<
ZOwOC approached infinity.
5'! 2
Chapter
Resonators
10
Example rash
FORESHORTENED RADIAL LINES
proportions of the resonator are more as shown in Fig. 1090, it is preferable to
problem as one of a resonant radial transmission line (Sec. 9.3) loaded or
foreshortened by the capacitance of the post or gap. Then, for resonance, the inductive
reactance of the shorted radial line looking outward from radius r1 should be equal in
magnitude to the capacitive reactance of the central post. Using the results and notation
If the
look at the
of Sec. 9.3
we see
that
h
1
3111(6),
_
wC
2m,
01
cos
(a,
—-
—-
62)
62)
01‘
'
62
Once
62
is found,
kr2
:
tan“1[sm
cos
is read from
01 +
( 2 7711/0) CZ01 h)cos
61
(2771'1 / wCth)
-
'
Fig. 9.30,
Example
and resonant
sin
1,01]
(2)
1/11
frequency is
found from k.
10.9c
RESONATORS OF INTERMEDIATE SHAPE
In the coaxial-line resonator of Fig. 10.951 the electric field lines would be
substantially
region far from the gap. In the radial line resonator of Fig. 10.9c the electric
field lines would be substantially axial in the region far from the gap. For a resonator
of the same general type, but with intermediate proportions, the field lines may be
transitional between these extremes as indicated in Fig. 10.9d, and neither of these
approximations may yield good results. Some useful design curves for a range of proportions have been given in the literature.5 Of course, if the capacitive loading at the
center is great enough, the entire resonator will be relatively small compared with
wavelength, and the outer portion may be considered a lumped inductance of value
radial in the
L
m
E
277’
Resonance is
computed from
1n<'—"-)
(3)
2'1
this inductance and the known
capacitance.
Example iced
CONICAL-LINE RESONATOR
A somewhat different form of
at radius
5
a on a
conical line
as
small-gap resonator, formed by placing a spherical short
studied in Sec. 9.6, is shown in Fig. 10.9.9. Since this is
T. Moreno, Microwave Transmission
Design Data,
Artech House, Nam/cod, MA, 7989.
a
uniform line, formula
(1) applies
to
ZO
In the limit of
gap),
this
case as
,8
=
-9
2
cot
7T
0
(4)
zero
becomes
C
kr
cos
__
9
sin e
Q
of the resonator in this
Q
R
sin 10'
.
.
JT]
8111
case
6
In cot
7777
1n cot
r
may be shown to be
—
10.10
(6)
z
d)
limiting
4Rs
(5)
r
C
H
The
k and
6
27- ln
=
well. For the conical line,
capacitance (the two conical tips separated by an infinitesimal
exactly a quarter-wavelength. The field components in this
by forming a standing wave from Eqs. 9.6(5) and 9.6(6), are
the radius
case, obtained
513
Coupling to Cavities
redo
(60/2)
COUPLING
(60 / 2)
+ 0.825
csc
<7)
00
CAVITIES
TO
The types of electromagnetic waves that may exist inside closed conducting cavities
have been discussed without specifically analyzing ways of exciting these oscillations.
excited if the resonator is completely enclosed by conductors.
coupling electromagnetic energy into and out of the resonator must be
introduced from the outside. The most straightforward methods, similar to those discussed in Sec. 8.11 for exciting waves in waveguides, are:
Obviously they
Some
means
cannot be
of
1. Introduction of
a
lines, driven by
conducting probe
an
2. Introduction of
conducting loop
a
3. Introduction of
a
or antenna
in the direction of the electric field
external transmission line
pulsating
with
plane
electron beam
normal to the
magnetic
field lines
passing through a small gap in the
similarly for carriers in solid-
resonator, in the direction of electric field lines, and
state devices
4. Introduction of
being
located
a
common to one
5. In
hole
or
iris between the
cavity
and
field component in the
in the wave mode
so
that
some
driving waveguide, the hole
cavity mode has a direction
a
microstrip or coplanar strip versions, coupling by adjacent strip
examples of Fig. 10.6
lines
as
illus-
trated in the
example, in a velocity modulation device of the klystron type, as in Fig. 10.10a,
input cavity may be excited by a probe, the oscillations in this cavity producing a
voltage across gap g1 and causing a velocity modulation of the electron beam. The
velocity modulation is converted to convection current modulation by a drifting action
For
the
514
Chapter
10
Output
ln put
resonator
resonator
f
\‘
Electron
beam
\
,..-
u.
______
g1
I"“‘
1
Resonators
/
K
\\
g2
/
}
Probe
/
l
_______
Signal
Loop
m
//
‘—-.~wI”
\
Power out
__.
”
\\
\
I
I
"
..
\
//
\‘v’
\
__
i
TE
I
\;
TMmo Cylindrical
resonator
10‘ Rectangular guide
(b)
(a)
Cylindrical
cavity
Coaxial
line
.
(a)
FIG. 10.10
(a) Couplings to the cavities of a velocity modulation tube amplifier. (1)) Section
showing approximate form of magnetic field lines in iris coupling between a guide and cavity.
(c) Magnetic coupling to a cylindrical cavity.
so
that the electron beam may then excite electromagnetic oscillations in the second
by passing through the gap g2. Power may be coupled out of this resonator
resonator
by a coupling loop and a coaxial transmission line. Iris coupling between a TMO10 mode
in a cylindrical cavity and the TE10 mode in a rectangular waveguide is illustrated in
Fig. 10.1019. Here the H¢ of the cavity and the HI of the guide are in the same direction
over
the hole.
rigorous approach to a quantitative analysis of cavity coupling is given in the
following chapter. Some comments and an approximate approach are, however, in order
The
here.
"'::',—,.-;-}
Example
LOOP COUPLING
Let
Fig.
us
concentrate on the
10.100. If
a
IN A
loop coupling
current is
none
CYLINDRICAL CAVITY
to a
TM010 cylindrical mode as sketched in
loop, all wave types will be excited
made to flow in the
Measurement of Resonator
10.11
which have
515
Q
magnetic field threading the loop. The simple TM010 mode is one of these,
resonance, certainly it will be excited most. However, this wave is
known to fit the boundary conditions imposed by the perfectly conducting box alone.
Other waves will have to be superposed to make the electric field zero along the perfectly conducting loop, but these will in general be far from resonance and so will
contribute only a reactive effect. In fact, they may be thought of as producing the selfinductance reactance of the loop, taking into account the presence of the cavity as a
a
and, if it is
near
shield.
The
voltage magnitude
induced in the
M
where
[HI
is
magnetic
field from the
Consider power loss
in terms of magnetic field
area.
WL
This loss
requires
an
as
only
at
r
m1(d
=
the
cavity
mode is
wxxSlHl
(1)
TM010 mode, averaged over the loop, and S is loop
walls, Eq. 10.5(6). This can be written
through Eq. 10.5(2):
that from the
=
a
+
£1)ng %(ka)RS
n2
=
input
loop by
2:
7m(d
+
a)R,|H|2
(2)
resistance
R
:2
v2
L
2W,
(wMS)
__
2
'"
27m(d
+
(1)12,
(3)
impedance from self-inductance of the loop, ij, is added to this, but may
by a frequency shift in the cavity or by a portion of the input line as will
discussed in the following section.
The reactive
be tuned out
be
10.11
The
Q of
a
cavity
MEASUREMENT
OF
RESONATOR Q
has been defined in terms of power loss and energy storage and has
a number of ideal configurations. It has also been noted that the Q
been calculated for
cavity mode, just as for a lumped—element reso~
vicinity of resonance for a single mode,
a lumped—element equivalent circuit such as that of Fig. 10.110 is a good representation.
The excitation means may excite a number of modes, but in general only one is near
resonance. The elements G, L, and C represent the mode near resonance, and jX the
reactive effect of modes far from resonance. Such an equivalent circuit might be suspected to give correct qualitative results, but as will be shown in the next chapter, it
actually gives useful quantitative results also. The equivalent circuit also permits one
to devise ways of measuring Q when it is difficult or impossible to calculate.
Many methods of Q measurement are possible,6 and commercial instruments are
is useful in
nant
6
describing
system. The
T. S.
bandwidth of
reason
Loverghefio. Handbook
MA. 7987.
a
for this is that in the
of Microwave
Testing, Sec. 9.2, An‘ech House, Norwood,
516
Chapter
Resonators
10
G‘
..
Q
i"
I!
——l
‘i
cal“
3‘
ll
jxl.
—-:f_- —_
L
Q
T
._
(b)
(a)
Locus of R
=
X
(a)
FIG. 10.11
waveguide.
(a) Cavity equivalent circuit. (b) Equivalent circuit for cavity with
(c) Locus of impedance on Smith chart for Q measurement.
available which do this
We shall describe
coupling
to a
method
using elemental transmis—
equivalent circuit of Fig. 10.11a
and brings in the importance of the correct coupling. The coupling of the waveguide
to the cavity is illustrated here by the ideal transformer of turns ratio m:1 (Fig. 10.11b).
The guide is assumed to have unity characteristic unpedance for simplicity, so that
terminating impedances are automatically normalized. The input impedance at reference
directly.
a
sion—line measurements which illustrates the use of the
a
is then
9
Z“
By defining Q0
z
vicinity
m
wOC/G, (9%
za
In the
"
=
[JX
=
mz[jX
of resonance,
co
1
.
=
+
G +
l/LC,
and
j(wC
R0
-
l/wL)]
l/G,
=
this is
R°
+
,
1 +
w0(1
+
1(w/w0
8’)
-
(1)
wO/w)QD
]
(2)
where 5’ is small,
7
Z,
a.
ijX
m ~R 0
+ ——,—
l + ZJQOS'
(3)
10.11
Measurement of Resonator
53 7
Q
The series reactance may be removed either by defining a new resonant frequency or
by referring input to a shifted point on the waveguide. The latter is common, and the
reference may be taken as the position where the impedance seen looking toward
cavity is zero when the cavity is tuned off resonance; this is called the “detuned
new
the
short”
position. The detuning is sufficient to make QOB’ > 1 either by detuning the
cavity (by changing coo) or by changing frequency a). Then, by (3), Za
jsz and,
from the impedance transformation formula, Eq. 5.7(13), which for ZO
l is
z
=
Z”
we
find that
Zb
is
zero
t
1
impedance Zb
+
j
tan
jZa
B!
(4)
[31
tan
when
tan
The
+
Zn
at
[3/
~2722X
2
in the
arbitrary frequencies
Z},
=
(5)
neighborhood
of
”22120,,
is then
(6)
——.—“
l +
resonance
ZjQOB
where
R0,,
2
120(1
m4X2)"
+
(7)
and
1714XR0
5=5’-—~
+
2QO(1
(8)
1724X2)
impedance is measured as 5’ is varied either by changing frequency or
cavity. As impedance (6) is of the linear fraction form, it will produce
a circular locus for each value of 1723R0b when plotted on the Smith chart as illustrated
1, passes through
by circles A, B, and C in Fig. 10.11c. Circle A, for which 1722R0b
the origin and is called the condition of critical coupling since it provides a perfect
match to the guide at resonance; circle C with I723R0b < 1 is said to be mzdercoupled;
and circle B with 2212130,) > 1 is overcouplea’. To match the last two, the coupling ratio
1722 would have to be changed. As with the lumped resonant system, the value of Q0
can now be found from the specific value of 6 which reduces impedance magnitude at
reference I) by 1 /\/E of its resonant value. On the Smith chart, this is the point R
X and the corresponding 5 may be denoted 51. At this point the known quantities are
6; corresponding to 51, nsz found from (5), 1722R0b, which is the value of 2,, at reso~
1 in the denominator of (6). Then
nance, and ZQOB1
The locus of
by detuning
the
==
=
==
1
Q0
and (8) may be solved to find
QO
(9)
:
251
in terms of the known
quantities.
518
chapter
Resonators
:0
Q0 thus determined is the “unloaded Q” since it does not account for
guide. A loaded Q which accounts for this is also used and may be found
The value of
the
loading by
Fig. 10.11]:
from
as
i_£:fi_i+1
a’OC
QL
where “external
Q0
am
Qext
Q,” Qext, results from
Qext
=
(00C
(11)
W
It is assumed here that the generator is matched so that the impedance
the guide is its characteristic impedance, taken here as unity.
looking
toward
RESONATOR PERTURBATIONS
lO.l 2
magnetic and electric energies are equal.
cavity walls, this will in general change
one type of energy more than the other, and resonant frequency would then shift by an
amount necessary to again equalize the energies. Perturbation theory7 shows the amount
of frequency shift when a small volume AV is removed from the resonator by pushing
In the condition of resonance, average stored
If a small perturbation is made in one of the
in the boundaries. This may be written
A33
__
mo
IAV (MHZ
IV (,uH2
"‘
+
8E2) dV
8E2) W
AUH
__
"
AUE
(1)
U
energy removed, AUE the electric energy removed, and U
the total stored energy, all time averages. We illustrate with two examples.
where
AUH
is the
magnetic
Example
no.12a
PERTURBATION OF RESONANT PARALLEL-PLANE LINE
We first take
plane
a
simple
case
for which
we can
__
perturb by moving
one
2771) p
parallelunperturbed reso-
end
R. F.
A
770
=
)t
plate
in
=
___r2
(2)
l
by Al,
77v
+
7
Consider the
=
CL’0
we
answer.
Fig. 10.12a, shorted at the two ends. The
with the lowest frequency is at l
)t/ 2 [Eq. 10.6(1)]:
nance
If
check the
transmission line of
p
the
new resonance
is
Al
z
1 +
——-
Harrington, Time~Harmonic Electromagnetic Fields, Chap. 7, McGrow-Hill, New York,
7961.
1472
5'5 9
Resonator Perturbations
10.12
i
T
Ad
I
d
i
_L
T
Lu}
T
(ll—$2
_L
l
(b)
(a)
FIG. 10.12
Parallel-plane transmission line with perturbations: (a) by decreasing length by Al;
(b) by introducing a rectangular indentation at the center of the line.
In using the perturbation formula (1), only magnetic
magnetic field is of the form
110(2)
Total stored energy
[/2
=
2 M 2d
(4)
id
w
d
I
4
is width of the line and d the
=
z
——9-“H
HHS
(5)
4
The energy removed is
spacing.
AU”:
so
3%“
9
I'd-IO
“1/2
w
sin
H0
unperturbed
(twice the average energy in magnetic fields) is
U
where
2
energy is removed. If
w
dAl
(6)
by (1)
Al
Am
——
=--
(7)
—
1
mo
which agrees with (3).
Now suppose the perturbation is at the center of the line where
is removed,
as
shown
by
the dashed outline in
Fig.
10.1217. If
only electric field
unperturbed electric
field is
772
the total energy stored is
[/2
U
—
ZUE
-
')
E9
.
2
214sz Siocoszl
l
[/2
z
=
wlde
E...0
4
(9)
And the electric energy removed is
8
AUE
=
w
Ad A2
E’)°
(10)
529
so
Chapter
Resonators
10
by (1)
Ad AZ
A9
(11)
~_
Id
(00
this, we may use the equivalent circuit of the
line shown in Fig. 10.121). Resonance is given by
To check
wave
AC
capacitively
loaded quarter-
col
w( )
:
__
2
Y000 t
(21113)
(12)
_
where
AC
If
m
=
“’0 + Am and
:2
8W
cool/vp
A
=
2((1
1
1
——
n', to
A_a)~ _ACUP__
_
_.....
Ad
d)
—-
This reduces to (11),
swAzAd
i
Id2
[.18
__
the check of
(13)
d2
(12) gives
1Y0
including
Az Ad
——
first order
._
we
SW
m
1/2
1111/22
3
(14)
w
sign.
Example 10.1%
PERTURBATION ON BOWOM OF CYLINDRICAL CAVITY
For
practical example, imagine a small volume AV taken out of the pillbox
V0 in Fig. 10.12c, along the axis where electric field is maximum
field
magnetic
negligible. The change in energy stored is then
a more
resonator of volume
and
E2
AUE
z
3;” AV
(15)
Vi/I/l/I/l/l/I/I/Ill/Illll/Illlllflllllf’l/I/I/I////////////l/f/fl/M/fl//////I/{
\
\\
\
FIG. 10.126
\
A V».~
9
\
”IIII/11//////////////////////////é////III/x///////////////////////////////A
Small
perturbation
in bottom of circular
cylindrical cavity.
52'!
Dielectric Resonators
10. 1 5
The total energy of the resonator is given
the lowest mode with ka
2.405 is then
by Eq. 10.5(5). Frequency shift
from
(1) for
2
Am
nan—n.“
-~
.—
277's
we
The shift in resonant
RO/ Q. Frequency
1
AV
—-1.85
=
a
( l 6)
——
V0
determines the ratio
frequency
shifts
.955 AV
dEaazJfl/ca)
E%/ U needed for determination of
be measured
accurately, and the perturbation can be made
in the form of a small conducting bead moved by an insulating thread along the axis.
Field can be measured at all points on the axis, and thus its integral found even when
can
field cannot be assumed to be uniform
10.13
We
across
the gap.
DIELECTRIC RESONATORS
in the earlier sections of this
chapter that a section of a hollow metal waveguide
with properties similar to resonant
L—C circuits. Similarly, a section of dielectric waveguide (Sec. 9.2) exhibits resonances
and can be used for the same purposes. Some materials have very high permittivities
and wave energy is therefore strongly confined within the material. Wavelength is small
saw
shorted at the ends constitutes
so
the dielectric resonator
can
a resonant structure
be much smaller than
an
empty hollow metal
structure.
100 and e"/e’
Early work8 employed high~purity TiO2 ceramic material with er
10”. Values of Q (not accounting for losses in supporting structures) then are high
(about 104) as shown in Prob. 10.3d. The difficulty with TiO2 is that its a, has an
intolerably strong temperature dependence of 103 parts per million (ppm) per degree
Celsius, which leads to a resonant frequency dependence of 500 ppm / °C. More recently,
*3
ceramics have been
to
developed9 that can be made with temperature coefficients selected
offset those of the
These have 8r
w
z
supporting
37.5 and
8”/8’
structures,
z
2 X
giving
10“"’
at
a net zero
temperature dependence.
about 10 GHz. Disks of these materials
conveniently be introduced as resonators into microwave-integrated circuits and
beyond 100 GHz is expected.
Exact analysis is possible for a sphere and a toroid, but shapes of greater technical
interest such as rectangular prisms, disks, and rods must be treated approximately. It is
seen that fields at the surface of a region of very high permittivity satisfy approximately
the so-called 0pen~circuit boundary condition for which the normal component of electric field and the tangential component of magnetic field are zero. This is made plausible
by considering reflections of a plane wave in going from a dielectric of high permittivity
(low intrinsic impedance) to one of low permittivity (high intrinsic impedance). One
could calculate the resonant frequency of a dielectric resonator by surrounding it with
a contiguous perfect magnetic conductor to impose the above stated conditions. How-
can
use
3
S. B. Cohn. IEEE Trans. Microwave
9
M. 7?.
Stig/iiz,
Theory Tech. MIT-76,278 (7968).
Microwave J. 24. 79 ( 798 7).
522
Chapter
/I
.I
no
Resonators
\.
\\
Perfect
magnetic
wall
I
I
L
T/
/
\h/
‘m
/
2a
FIG. 10.13 (a) Dielectn'c cylinder in magnetic-wall waveguide
a
mode where L s 2a in dielectric resonator. Fields H outside r
=
by model
in
(a).
Lowest-order
extensions of those given
boundary. ([9)
are
10.13
523
Dielectric Resonators
7000
6000
[(MHz)
5000
4000
3000
0.100
0.400
0.300
0.200
0.500
L(in.)
FIG. 10.130 Experimental data on the lowest resonant frequencies versus length of a dielectric
O. 162 in. After S. B. Cohn, IEEE Trans. Microwave
cylinder of circular cross section. Radius a
Theory Tech. MTT-lfi, 218 (1968). © 1968 IEEE.
=
ever, since the
only approximately satisfied for
perfect magnetic boundary
have been found to give better results. The two lowest—order modes for circular cylin»
drical disks and rods are one with zero electric field along the axis and one with zero
magnetic field along the axis. The model for the former places the solid dielectric
cylinder inside a contiguous infinitely long, magnetic cylindrical waveguide so that the
open—circuit boundary
conditions
are
in the
use
finite-permittivity resonators, modifications
of the
imposed only on the cylindrical surface as in Fig. 10.130.
zero magnetic field along the axis imposes the open—
circuit condition only on the end faces of the dielectric cylinder by means of infinite
parallel magnetic conducting plates. In either case the resonant fields and frequency are
found by setting up propagating-type solutions of the wave equation inside and atten—
uating fields outside the dielectric and matching boundary conditions. The result is that
the mode with zero axial electric field has the lowest resonant frequency for dielectric
cylinders with length less than the diameter. The field distribution for this mode is
shown in Fig. 10.13b. Some experimental results are shown in Fig. 10.13c. The curve
labeled fl is the mode with zero axial electric field and f2 has zero axial magnetic field.
A general treatment of arbitrary shapes of dielectric resonators is given by Van Bladel.10
open—circuit
conditions
are
The model for the mode with
‘0
J. Van Bladel, IEEE Trans. Microwave
Theory Tech. MTT-23, 799 (7975).
524
Chapter
Resonators
10
Exampfie
TE MODE
10.15
DIELECTRIC RESONATOR
IN
example we do the analysis for the case with the circular cylinder of dielectric
a perfectly conducting magnetic waveguide shown in Fig. 10.13a. Use is
made of the concept of duality (Sec. 9.5) in which E is replaced by H, H by -—E,
y, by s, and s by ,u, in known field distributions to find another solution of Maxwell’s
equations that fits boundary conditions dual to those of the given field. The circular
waveguide modes in Sec. 8.9 can be adapted. In particular, the TM mode with an electric
boundary becomes a TB mode with the magnetic boundary. The H 9,.) component in Eq.
8.9(9) becomes the E ¢ component and can be written as
As
an
enclosed in
11(kcr) erlfidz
110901)
L
.
0
where the propagation factor is included for a
d
signifies
dielectric,
the
one
wave
3
(1)
-
in the + z direction and the
subscript
account of waves in both directions in the
dielectric region. Taking
has
E¢
The
[2|
,
“z
J 1 (k I)
2E0—
110201)
L
lzl
Bd2,
cos
_<.
(2)
-
2
frequency is assumed low enough that the waves outside the dielectric (Izl > L/ 2)
off. Choosing the coefficient to ensure continuity of E ¢, across the end of the
are cut
dielectric,
we
may write
Eq15
The
magnetic
=2EO
J 1(k1)
-
2E0
mu
Hr
=
BdL
--—
ijaa“2E
_
e
“0‘“
fl
_.
2
11(P01)
field components
.Bd
H,—— j—-
cos
tangential
110%I)
110701)
to
”2),
J1(kc1.)
‘zt
B"2,
L
COS
3
L
Z
(3)
-
2
the end faces of the dielectric
_.
..
e
are
L
.
sm
I2i
s
(4)
—
2
HAM
7
-
L
7
L/-),
2
2
_
(5
where we have used the relation dual to Eq. 89(9). The upper sign in (5) applies at
the lower at z .~= —L/ 2. Then, equating Hr across the dielectric boundary
L/ 2
z
at
and
=
z
=
:L-L/ 2
,we.
find the determinantal relation for the resonant frequency:
Ba
tan
B—‘Z’L-
—
(6)
an
where
a,
=—-
\/
c028r
62
2
p
—
(7‘?)
<7)
525
Problems
and
__
These calculations
give
resonant
data in the range 0.24 <
L/ 2a
<
3912__ £92
t.)
(.)
frequencies
about 10% lower than the
experimental
0.62.
PROBLEMS
10.23 An
analogy can be made between acoustic resonances in a closed box with fixed
electromagnetic resonances in a box with walls of high conductivity. Develop the comparison further, stressing similarities and differences.
walls and
10.2b Show that the mode described in Sec. 10.2 (resonant condition and field expressions)
would be obtained if one started with the point of view that it was a TE10 mode
propagating in the x direction; similarly consider it a TM 11 mode propagating in the
y direction
exactly
at
cutoff.
10.20. Find the total
charge on the top plate and bottom plate for the resonator of Sec. 10.2.
equivalent capacitance that would give this charge with a voltage equal
to that between top and bottom at the center of the box. Compare this equivalent
capacitance with the parallel—plate capacitance of the top and bottom plates with
fringing neglected.
Determine
an
10.2d Find the total
current
in the side walls of the resonator of Sec. 10.2. Determine
an
equivalent inductance in terms of this current and the magnetic flux linking a vertical
path at the center of the box. What resonant frequency would be given by this
inductance and the equivalent capacitance of Prob. 10.2c? Compare with result of
Eq. 10.2(1).
10.2e
Suppose
that in
wave reactance
Discuss
place
Ey/HJ.
physical
approximation.
of
=
a
perfect conductor
jX is placed
at z
=
d,
a
“reactive wall”
there. Obtain the condition for
ways in which the reactance wall
might
be
giving
a
resonance.
produced,
at least as
an
10.33 Calculate the maximum energy stored in magnetic fields for the simple mode of the
rectangular resonator and show that it is the same as Eq. 10.3(1).
10.3b Convert the phasor forms of electric and magnetic fields of Sec. 10.2 to instantaneous
forms and find stored electric and magnetic energy as functions of time. Show that
the sum is a constant equal to the value of Eq. 10.3(1).
10.3c From the definition of Q in terms of energy storage and power loss, Eq. 10.3(3),
show that the decay of energy in a natural oscillation after excitation is removed is of
the form exp(
Q/ mo. How many periods of oscillation are there in
t/r) where r
=
-—
the l /e time of the
decay?
an imperfect dielectric, show that the Q for any resonant mode is just e’/s”, ne—
glecting losses of the conducting walls. Give the value for the dielectrics of Table
6.4a at 10 GHZ and compare with the value of Q for a copper cube-shaped cavity
given in the text.
10.3d For
526
10.3e
Chapter
10
Resonators
expression 10.3(2) for wall losses if front and back are of one material
resistivity R51, the two sides of another with R52, and top and bottom of
material with R53. Give the special case of this for a cube.
Modify
the
with surface
a
10.3f
third
that a perfect dielectric were available with 8’
530. How would the Q of
dielectric-filled cube compare with that of an air-filled one for the simple mode
studied? Why are they different? (Compare for the same resonant frequency in both
Suppose
=
a
cases.)
10.3g
A square cavity resonator utilizing the simple TEN, mode is to be designed for a
0.3 THz. Height of the cavity
millimeter—wave electron device for Operation at f
must be kept small (0.1 mm) to minimize electron transit time. The dielectric in the
=
copper cavity is vacuum and the cross section is square. Calculate cavity size, the
and bandwidth. Also calculate the shunt resistance R0
(Eob)2/2WL, which is a
Q,
=
measure
of the interaction of the electron beam with the
advantageously shaped cavities, such as that
irnpedances; for example, see Prob. 10.90.
10.4a For
TENp
and
TM,,,,,p
modes in
a
in
cavity. Note that more
Fig. 10.90, have much higher
cubic resonator, consider combinations of the inte»
integer higher than 2. How many dz'fierent resonant frequencies do these
represent? What is the degree of degeneracy (number of modes at a given frequency)
for each frequency? (Note carefully which, if any, integers may be zero.)
gets with
10.4b*
no
By combining incident and reflected waves as was done for the T1310l mode of Sec.
10.2, find electric and magnetic field expressions of the TE111 mode of the rectangu»
lar box. Sketch field distributions in the three coordinate planes. Find the expression
for Q if conductors are imperfect.
10.4c Derive the
expression
increases as m
for
b
d and show that
Q of a TEmmm mode in a cube 0
a given dielectric and resonant
frequency. Explain the
=
=
it
increases for
result.
10.5a Find
and
expressions for resonant frequency in terms of length
TM111 modes of a circular cylindrical cavity.
and radius for the
TM02I
10.5!) A circular resonator has height h and radius 0. Give the lowest resonant frequency
for which a degeneracy between two modes occurs and the designations of these
modes. Make rough sketches of the field patterns. How might a small perturbation be
added to change the resonant frequency of one mode but not of the other?
10.5c Plot
curves of d/A versus a/d for all the significant modes in a circular cylindrical
cavity over the range 0 < d/A < 2, 0 < a/d < 5. Note especially ranges of
operation where there is only one mode over a considerable region of operation.
10.5d* Give the field components and obtain expressions for energy storage, power loss, and
Q for the TM011 mode of a circular cylindrical resonator.
10.5e* Repeat Prob. 10.5d for the
TEO11
mode.
10.6a Show that the resonant frequency of an open~circuited microstrip resonator as in
0.2 mm, d
0.5 m, I
5 mm, and a,
9.8, taking account
Fig. 10.6b with w
of end corrections but using the static value of cam is 11.43 GHz. What fractional
error is incurred if the end correction is neglected? Estimate the error in the lowest
resonant frequency resulting from neglecting dependence of gaff on frequency?
=
=
10.6b* Assume that the conductors in the
2.5 pm thick.
=
resonator in
Prob. 10.6a
=
are
copper and
are