Hyo J. Eorn
Electromagnetic Wave Theory for Boundary-Value Problems An Advanced Course on Analytical Methods
With III Figures
Springer
Professor Hyo
J. Eorn
Korea Advanced Institute of Science and Technology Dept. of Electrical Engineering and Computer Science 373-1, Guseong-dong, Yuseong-gu Daejeon, 305-701 Korea e-mail:
[email protected] ISBN 3-540-21266-3 Springer-Verlag Berlin Heidelberg New York
Library of Congress Control Number:
2004102306
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Preface
In this era of reliance on communication and information technologies, electromagnetic wave engineering has become increasingly important. Electromagnetic wave theory finds practical applications in mobile telecommunication and optical fiber transmission. This book is written as a text for the twosemester graduate course on electromagnetic wave theory offered at the Korea Advanced Institute of Science and Technology (KAIST), Daejeon, Korea. The prerequisites for this course are undergraduate courses in advanced calculus and in electromagnetism. In particular, some mathematical background in differential equations and complex variable analysis would be helpful. To understand scattering, propagation, and radiation of electromagnetic waves, boundary-value problems must be solved using electromagnetic wave theory based on Maxwell's equations. The motivation for writing this text is to present solutions to canonical boundary-value problems, thereby helping students better understand electromagnetic wave theory in addition to enhancing their analytic skills. Selected topics are not only representative for electromagnetic boundary-value problems, but also instructive in radio wave communication applications. This text has the following unique attributes: •
• • •
Each topic deals with canonical boundary-value problems that can be solved using basic analytic skills in Fourier series, Fourier transform, and residue calculus. No numerical computations are presented. Necessary mathematical formulas and derivations are included in the text. The technique of Fourier transform and mode matching is utilized to derive rigorous solutions to selected problems in Sections 3.6, 4.4, 7.4, 10.3, 10.4, 10.5, 10.6, and 10.7.
Fundamental electromagnetic issues in communication engineering are presented in Chapters 1 through 11. Chapter 1 introduces Maxwell's equations and related basics including the constitutive relations, boundary conditions, powers, and potential concept. Chapter 2 deals with uniform plane waves, polarization, Gaussian beam, plane wave propagation across plane boundaries,
VIII
Preface
and transmission line theory. Chapter 3 analyzes electromagnetic waves that propagate along a metallic rectangular waveguide, dielectric slab waveguide, circular waveguide, and shielded stripline. Chapter 4 investigates resonance in rectangular, circular, and spherical cavities as well as wave coupling using a groove guide coupler. Chapter 5 discusses wave propagation in anisotropic media including uniaxial and ferrite materials. Chapter 6 introduces some theorems that are useful for scattering computation. Relevant scattering examples are introduced to illustrate these theorems. Chapter 7 considers wave scattering from a circular cylinder, sphere, step, slit, and circular aperture. Chapter 8 introduces Sturm-Liouville problems and free-space Green's functions. Chapter 9 illustrates applications of Green's functions dealing with radiation from currents in waveguides. Chapter 10 presents antenna fundamentals and antenna radiation in terms of the dipole antenna, loop antenna, aperture antenna, groove-backed antenna, slit array antenna, slotted coaxial line antenna, and flanged coaxial line antenna. Chapter 11 considers half-space radiation problems using dipole and line sources. The selected topics can be covered in two semesters. The following sections are covered in the first semester: Chapter 1, Sections 1.1 through 1.6 Chapter 2, Sections 2.1 through 2.6 Chapter 3, Sections 3.1 through 3.5 Chapter 4, Sections 4.1 through 4.2 Chapter 6, Sections 6.1 through 6.6 Chapter 7, Section 7.1 Chapter 8, Sections 8.1 through 8.4 Chapter 9, Section 9.1 Chapter 10, Sections 10.1 through 10.2 The remaining topics are more advanced and, therefore, are recommended for second semester course material. Throughout the text, the exp( -iwt) time factor is suppressed and the International System of Units (SI) is implicitly assumed. Vectors are denoted - with overlines, such as A, B, etc. My sincere appreciation goes to KAIST for providing me with a superb academic environment in which I dared to write this text. The many valuable comments, criticisms, and creative ideas received from students at KAIST have been very much appreciated. My thanks go to Professors Jong K. Park, Yong H. Cho, Sang W. Nam, Young K. Cho, and Sang Y. Shin for their reading of the manuscript and helpful comments. Excellent assistance from my graduate students in drawing figures is greatly acknowledged. I gratefully acknowledge support from the Radio Education Research Center of Information and Communications University, Daejeon, Korea. I also wish to express my gratitude for the patience and support of my family during the writing of this text. Finally, any comments regarding the text would be gratefully received. Daejeon, Korea
Hya J. Earn •
Contents
Notations
xv
Electromagnetic Basics .......... 1.1 Maxwell's Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Maxwell's Equations in Differential Equation Form. . . . 1.1.2 Continuity Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1.1.3 Maxwell's Equations in Integral Form. . . . . . . . . . . . . . . . 1.1.4 Maxwell's Equations in Time-Harmonic Form. . . . . . . . . 1.2 Constitutive Relations 1.2.1 Material Media 1.2.2 Conduction Current and Loss Tangent . . . . . . . . . . . . . . . 1.3 Boundary Conditions 1.3.1 Boundary Conditions Between Two Media. . . . . . . . . . . . 1.3.2 Boundary Conditions with Perfect Electric Conductor.. 1.4 Poynting Vector. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1.4.1 Complex Poynting Vector 1.4.2 Time-Average Poynting Vector. . . . . . . . . . . . . . . . . . . . .. 1.5 Vector and Scalar Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1.5.1 Potential Representations 1.5.2 Free-Space Solution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1.6 Static Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1.6.1 Electrostatic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1.6.2 Magnetostatic Fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Problems for Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..
1 1 1 2 3 5 6 6 7 8 8 10 10 11 12 13 13 16 18 18 19 20
Plane Wave Propagation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.1 Uniform Plane Wave. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.1.1 Propagation in Lossless Medium. . . . . . . . . . . . . . . . . . . .. 2.1.2 Propagation in Lossy Medium
21 21 21 24
2.2
Polarization............................................ 26
2.3
Gaussian Beam
30
X
Contents 2.3.1 Line Source. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.3.2 Point Source. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.4 Reflection at Plane Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.4.1 TE Wave (Perpendicular Polarization) . . . . . . . . . . . . . .. 2.4.2 TM Wave (Parallel Polarization) . . . . .. 2.5 Infinitely Long Transmission Lines. . . . . . . . . . . . . . . . . . . . . . . .. 2.5.1 Coaxial Line. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.5.2 Voltage and Current. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.6 Terminated Transmission Lines. . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.6.1 Reflection Coefficient, Impedance, and Power. . . . . . . .. 2.6.2 Voltage Standing-Wave Ratio. . . . . . . . . . . . . . . . . . . . . .. 2.6.3 Cascaded Lines Problems for Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..
30 33 34 34 37 39 39 41 42 44 44 45 47
Waveguides. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3.1 Cylindrical Waveguides. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3.2 Rectangular Waveguide , 3.2.1 TM Wave : . . . . . .. 3.2.2 TE Wave. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3.3 Dielectric Slab Waveguide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3.3.1 TM Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3.3.2 TE Wave. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3.4 Circular Waveguide. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3.4.1 TM Wave .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3.4.2 Power Delivery by TM mn Mode. . . . . . . . . . . . . . . . . . . .. 3.4.3 TE Wave. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3.5 Circular Dielectric Waveguide 3.6 Shielded Stripline. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3.6.1 Field Representations. . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3.6.2 Boundary Conditions , 3.6.3 Residue Calculus for Problems for Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..
49 49 50 51 53 55 55 57 58 58 60 63 63 67 69 70 75 80
Ji ............................
Cavity Resonators and Coupler. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 83
4.1
Rectangular Cavity Resonator 4.1.1 TE Mode 4.1.2 TM Mode. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4.1.3 Quality Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4.2 Circular Cavity Resonator. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4.2.1 TM Mode. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4.2.2 Quality Factor for TM olO Mode. . . . . . . . . . . . . . . . . . . .. 4.2.3 TE Mode. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4.3 Spherical Cavity Resonator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4.3.1 TM Mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4.3.2 Quality Factor for TM oll Mode . . . . . . . . . . . . . . . . . . . ..
83 83 84 85 88 88 90 91 93 93 96
Contents 4.3.3 TE Mode. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4.4 Groove Guide Coupler 4.4.1 Field Analysis 4.4.2 Residue Calculus for I} and 12 . • . . . . . . . . . . . . . . . . . . . . Problems for Chapter 4
XI 99 100 100 104 109
Propagation in Anisotropic Media 5.1 Propagation in Anisotropic Media 5.1.1 Dispersion Relation 5.1.2 Uniaxial Medium 5.2 Propagation in Ferrites 5.2.1 Magnetized Ferrite 5.2.2 Transversely Magnetized Ferrite 5.2.3 Longitudinally Magnetized Ferrite 5.3 Propagation Along Ferrite-Filled Parallel-Plate Waveguide Problems for Chapter 5
111 111 111 113 116 116 118 119 120 123
Electromagnetic Theorems 6.1 Uniqueness Theorem 6.2 Image Method 6.2.1 Image Method Using Infinite Planes 6.2.2 Current Above Perfect Electric Conducting Plane 6.3 Equivalence Principle 6.3.1 Love's Equivalence Principle 6.3.2 Transmission Through Circular Aperture 604 Induction Theorem 604.1 Equivalence Based on Induction Theorem 604.2 Scattering from Conducting Rectangular Plate 6.5 Duality Theorem 6.6 Reciprocity Theorem 6.6.1 Lorentz Reciprocity Theorem 6.6.2 Reciprocity for Antennas Problems for Chapter 6
125 125 127 127 130 131 131 132 136 136 137 139 141 142 143 144
Wave Scattering . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Dielectric Circular Cylinder 7.1.1 TM Scattering 7.1.2 TE Scattering 7.1.3 Electrostatic Fields 7.2 Dielectric Sphere 7.2.1 Electromagnetic Case 7.2.2 Electrostatic Case 7.3 Step in Parallel-Plate Waveguide 704 Slit in Conducting Plane 704.1 Field Analysis
147 147 147 150 151 154 154 159 161 164 164
XII
Contents 7.4.2 Far Field and Transmission Coefficient 7.4.3 Residue Calculus for I(Ieo) 7.4.4 Thin Slit Within High-Frequency Limit 7.5 Circular Aperture: Electrostatic Case Problems for Chapter 7
Green's Functions: Fundamentals 8.1 Delta Function and Sturm-Liouville Equation 8.1.1 Delta Function 8.1.2 Sturm-Liouville Equation 8.2 One-Dimensional Green's Function 8.2.1 Free Space Approach 1 Approach 2 8.2.2 Half Space Approach 1 Approach 2 Approach 3 Approach 4 8.2.3 Closed Space Approach 1 Approach 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Approach 3 8.3 Two-Dimensional Green's Function 8.3.1 Approach 1 8.3.2 Approach 2 8.3.3 Approach 3 8.4 Three-Dimensional Green's Function 8.4.1 Approach 1 8.4.2 Approach 2 8.4.3 Approach 3 Problems for Chapter 8 Green's Functions: Applications 9.1 Currents in Free Space 9.1.1 Radiation from Sheet Current 9.1.2 Radiation from Shell Current Approach 1 Approach 2 Approach 3 9.2 Line Current in Rectangular Waveguide 9.2.1 Radiation in Parallel-Plate Waveguide Approach 1 Approach 2 . . . . . . . . . 9.2.2 Radiation in Shorted Parallel-Plate Waveguide
168 170 173 176 179 181 181 181 183 186 186 187 188
189 190 191 191
192 193 193 194
195 196 197 199 202 204 205 207 209 212 213 213 213 215 215 216
217 218 219 219 220 222
Contents
Approach 1 Approach 2
9.2.3 Radiation in Rectangular Waveguide 9.3 Line Current in Circular Waveguide 9.3.1 Approach 1 9.3.2 Approach 2 9.4 Sheet Current in Parallel-Plate Waveguide Problems for Chapter 9
XIII
223 223
224 225 226 228 229 232
Antenna Radiation 10.1 Antenna Fundamentals 10.2 Wire Antennas 10.2.1 Dipole Antenna 10.2.2 Circular Loop Antenna 10.3 Aperture Antenna 10.4 Groove-Backed Antenna 10.4.1 Approach 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4.2 Approach 2 10.5 Slit Array Antenna 10.5.1 Field Analysis 10.5.2 Reflection, Transmission, and Far Field 10.6 Slotted Coaxial Line Antenna 10.6.1 Field Analysis 10.6.2 Reflection, Transmission, and Far Field 10.6.3 Residue Calculus for It 10.7 Flanged Coaxial Line Antenna 10.7.1 Field Representations 10.7.2 Boundary Conditions Problems for Chapter 10
235 235 238 238 240 242 245 246 250 251 251 255 257 257 262 263 267 267 270 275
Radiation Above Half Space 11.1 Electric Line Source 11.2 Vertical Electric Dipole 11.3 Horiwntal Electric Dipole Problems for Chapter 11
277 277 281 284 290
References
293
Coordinates and Vector Formulas A.1 Coordinate Relations A.2 Differential Operators A.2.1 Rectangular Coordinates (x, y, z) A.2.2 Cylindrical Coordinates (p, ¢, z) " A.2.3 Spherical Coordinates (r, 0, ¢) A.3 Vector Formulas
295 295 297 297 297 298 298
Bessel Functions B.1 Bessel Functions and Modified Bessel Functions B.1.1 Limiting Forms for Small and Large Arguments B.1.2 Wronskian............................... B.1.3 Generating Function B.1.4 Hankel Functions B.1.5 Recurrence Formulas B.1.6 Modified Bessel Functions B.2 Spherical Bessel Functions
299 299 301 302 302 302 302 303 303
Residue Theorem
305
Associated Legendre Functions
307
Transforms and Series
309
Index
311
Notations
A -B -D E F
g(r; r') H
H~\)
H;;)O •
t
lmO 1m (.) J Jc Ji JmO • JmO
lmO Js k k
KmO M
Ms •
n
NmO P Pav P~O
PEe
magnetic vector potential magnetic flux density electric flux density electric field intensity electric vector potential Green's function magnetic field intensity Hankel function of the first kind of order m Hankel function of the second kind of order m imaginary unit (i = A) modified Bessel function of the first kind of order m imaginary part of (.) electric current density conduction current density impressed electric current density Bessel function of the first kind of order m spherical Bessel function of the first kind of order m spherical Bessel function of the first kind of order m surface electric current density wavenumber wave vector modified Bessel function of the second kind of order m magnetic current density, magnetic polarization vector surface magnetic current density normal unit vector Bessel function of the second kind of order m electric polarization vector time-average power associated Legendre function of the first kind perfect electric conductor
PMC Q~(.)
ReO Res 0 Sav TE TEM TM & &(.)
&mn € €()
-€
em TJ ()
>. fL fLo fL
Pe Pm a
rP rPe rPm Xe
Xm W
\7 \72
(·t
perfect magnetic conductor associated Legendre function of the second kind real part of (.) residue of 0 time-average Poynting vector transverse electric transverse electromagnetic transverse magnetic skin depth delta function Kronecker delta permittivity permittivity of free space (vacuum) tensor permittivity eo= 2 and Cl = e2 = ... = 1 intrinsic impedance of medium angles wavelength permeability permeability of free space (vacuum) tensor permeability electric charge density magnetic charge density conductivity angles electric scalar potential magnetic scalar potential electric susceptibility magnetic susceptibility angular frequency del operator Laplacian operator complex conjugate of (.)
1
Electromagnetic Basics
1.1 Maxwell's Equations Most basic laws in electromagnetic theory were established in the 19th century by a variety of talented scientists, including Oersted, Ampere, and Faraday. Later, James Clerk Maxwell unified these basic laws into fundamental equations, which macroscopically state the relations between electromagnetic fields and their sources (current and charge). These fundamental equations are called Maxwell's equations. Let us first consider Maxwell's equations in a differential equation form. 1.1.1 Maxwell's Equations in Differential Equation Form
Maxwell's equations are written in terms of time-varying quantities as t"7
-E(- ) __ 8B(r, t) r, t 8t
(Faraday's law)
(1.1)
(Ampere's law)
(1.2)
V . D(r, t) = Pe(r, t)
(Gauss' law)
(1.3)
V·B(r,t)=O
(Magnetic Gauss' law)
(1.4)
v X
875(r, t) V x H(r,t) = 8t +J(r,t)
where the time-varying quantities, functions of position vector r and time t, are E(r, t) = electric field intensity or electric field (volts/meter) H(r, t) = magnetic field intensity or magnetic field (amperes/meter) D(r, t) = electric flux density (coulombs/meter2 ) 2 B(r, t) = magnetic flux density (webers/meter ) 2 J(r, t) = electric current density (amperes/meter )
2
1 Electromagnetic Basics
Pe (r, t) = electric charge density (coulombs/meter 3 )
.
8DZ' t) , which is called
Maxwell revised Ampere's law by adding the term
the electric displacement current density. He noted that the addition of the electric displacement current density was necessary to make Maxwell's equations a consistent set. He also predicted the existence of electromagnetic waves, which was later experimentally verified by Heinrich Rudolf Hertz. Maxwell's equations are thus considered empirical in that they were formulated and verified by experimental observations. Maxwell's equations are instrumental in understanding macroscopic electromagnetic phenomena encountered in wave scattering, radiation, and propagation. Maxwell's equations (1.1) through (1.4) are in an asymmetric form due to the absence of magnetic sources. However, from a mathematical viewpoint, it is convenient to transform (1.1) through (1.4) into a symmetric form, which can be achieved by introducing the fictitious magnetic sources M(r, t) and Pm(r, t). Consequently, Maxwell's equations can be written in a symmetric form as
8B(r t) '1 x E(r, t) = 8t' - M(r, t)
(1.5)
8D(r t) 'lxH(r,t)= 8t' +J(r,t)
(1.6)
Peer, t)
(1. 7)
'1. B(r, t) = Pm(f, t)
(1.8)
'1. D(r, t) =
where 2 M(r, t) = magnetic current density (volts/meter ) Pm(r, t) = magnetic charge density (webers/meter3 )
.
The hypothetical assumption of M and Pm is purely for mathematical convenience. This means that as long as the field produced by the equivalent sources M and Pm is identical with the field produced by the real sources J and Pe, the assumption of M and Pm is justified. 1.1.2 Continuity Equations
The assumption of charge conservation stipulates that the current and charge densities be conserved since the sum of them can be neither destroyed nor created. Charge conservation is mathematically stated in terms of the continuity equations given as
1.1 Maxwell's Equations
\7 . J(r, t)
+ ape(r, t)
= 0
8t
r7 •
v
M(-) 8Pm(r, t) = 0 r,t+ at .
3
(1.9) (1.10)
The continuity equations state that the creation of the current J (or M) should be at the expense of a decrease in the charge density Pe (or Pm). It is important to note that Gauss' laws (1.7) and (1.8) are derivable from (1.5), (1.6), (1.9), and (1.10). For instance, applying the divergence to (1.5) • gJves
y. [\7 xE(r,t)l = -%t [\7.B(r,t)] -\7·M(r,t). .
(1.11)
o Substituting (1.10) into (1.11) yields (1.12) Equation (1.12) is integrated with respect to time, and setting the integration constant equal to zero produces (1.8). Similarly, applying the divergence to (1.6) results in (1.7). This implies that the two Gauss' laws are dependent equations, whereas the remaining two Maxwell's curl equations, (1.5) and (1.6), are independent. In general, (1.5) and (1.6) must be solved to determine unique and complete solutions to electromagnetic boundary-value problems.
1.1.3 Maxwell's Equations in Integral Form It is sometimes useful to transform Maxwell's equations into an integral form.
In view of Fig. 1.1, Stokes' theorem is given by
s
\7xA·dS=
-
c
A·dl
(1.13)
where dS is the differential surface vector and dl is the differential line vector encircling the surface S. Applying Stokes' theorem to (1.5) yields E(r, t) . d1 = S
C
8B(r, t) at
M(-) . dS . + r,t
(1.14)
Similarly, applying Stokes' theorem to (1.6) yields -
c
-
H(r,t) ·dl =
s
815(1', t) 8t
J(-) . dS . + r,t
(1.15)
4
1 Electromagnetic Basics
dl
ds
·•-••••• .... •••••• ..... ... ......... . - . ••••••••••• •
• ••••••
• ••••••• • •••••••• • •••••••• • •••••••• • •••••••• • •••••••••• • ••••••••••• • •••••••••• • ••••••••••• • • •••••••••••• • •••••••••••••• • ••••••••••••••••••
s
..•.•••••••••••••••••••••••••••••••••••• . .. .. ... .... .. . ... .. .. ........ ... ..... • • • • • •• • • • • • • • •• • • • ••• •• • • • • • •• • • • • • • • • • • • • • • • •• • • • • • • • • • •
Fig. 1.1. Surface S encircled by line C.
In view of Fig. 1.2, the divergence theorem gives V·Adv=
A·as
v
(1.16)
s
where S is the closed surface surrounding the volume V, and as is the differential surface vector pointing outward from the surface. Applying the divergence theorems to (1.7) and (1.8), respectively, gives
s
D(r, t) . as =
v
B(r, t) . as = s
PeCr, t) dv
(1.17)
Pm(r, t) dv .
(1.18)
v
Equations (1.14), (1.15), (1.17), and (1.18) constitute Maxwell's equations in an integral form.
ds
v
..... .. • ••••• • ••••• • •••••
S • ••••• •
••
• •• • • • •• • • •
........... .. .. . .. .,.. .. . . .. ..... . _ . . ............ .............. . . .... .... . .......... ...... .... ...•........ ... .. .... .. .... . . .. . ... ... . •.•.. • •.•... • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• •• • ............... . .............. . ....... . ...,. .. . . ............. .. .. .. .. ...... .. .. . •
•
•
Fig. 1.2. Volume V surrounded by surface S.
1.1 Maxwell's Equations
5
1.1.4 Maxwell's Equations in Time-Harmonic Form
In most wireless communication channels, information is transmitted by a sinusoidal (time-harmonic) electromagnetic wave with a carrier frequency. Let us assume that a time-varying (instantaneous) field E(r, t) with a carrier angular frequency w is represented as
E(r, t) = Eo(r) cos(wt + ¢)
(1.19)
where Eo(r) is the amplitude vector of E(r, t), and ¢ is its phase. The timevarying field E(r, t) is a real function ofthe position vector r and time t. To effectively represent the time-varying field E(r, t) in a time-harmonic case, the phasor field E(r) = Eo(r)e-i ••••• .'
~
1------';
dz
..
'
Fig. 2.1. Traveling wave at velocity u along +z-direction.
z
wave propagation
- - - -......- - - y
x
Fig. 2.2. Uniform plane wave propagating along +z direction. Let us consider a three-dimensional case using (2.8). The plane wave solution to (2.8) is assumed to be (2.14)
-
Eo
exp(ik·r)
where the wave vector is k = xk x + yk y + zk z and the position vector is
r = xx + yy + zz. This represents a uniform plane wave that propagates in the k direction. Substituting (2.14) into (2.8) yields the dispersion relation (2.15)
24
2 Plane Wave Propagation
-
211" where the wavenumber is k = Ikj = A and A is the wavelength. Substituting
(2.14) into (2.3) gives (2.16) k· Eo
which results in (2.17)
k·E(r)=O.
Equation (2.17) implies that the wave vector k and the electric field vector E(r) are perpendicular to each other. Furthermore, (2.14) and (2.1) yields -
1 -
H(r) =
WJ.L
k x E(r)
(2.18)
indicating that the three vectors, E(r), H(r), and k, are mutually perpendicular. A uniform plane wave is, therefore, referred to as a TEM (transverse electromagnetic) wave since there are nO electromagnetic field components in the k direction. For a lossless medium, the time-average Poynting vector Sav •
IS
S av = =
~
Re (E x H*)
~IEoI2 2
k wJ.L
(2.19)
which indicates that the real power flows along the k direction. 2.1.2 Propagation in Lossy Medium
Consider a lossy medium where the complex medium permittivity
€e
is (2.20)
Then, the solution to the Helmholtz equation
J2
dz 2
2
+ W J.L€e
Ex (z) = 0
(2.21)
•
IS
(2.22) where k = w.jJ.L€e. Expression (2.22) represents a uniform plane wave that propagates in a lossy medium with complex medium permittivity €e. Consider two different lossy media - good dielectric and good conducting media.
2.1 Uniform Plane Wave
•
25
For the good dielectric medium, a «1 WE
(2.23)
•
For the good conductor,
a WE
»
1
= (1
wJ.La
+ i)
2
(2.24)
•
Then, a wave that propagates within the good conductor is written as
E",(z) = E;t exp (i - 1)
wJ.La
(2.25)
2 z
which shows the wave decays exponentially as it propagates in the lossy medium. As the wave propagates over the distance z =
wJ.La
, the field
amplitude decreases by a factor of e- ~ 0.368. This propagating distance is known as the skin depth J of a good conductor 1
J=
•
2
--. wJ.La
(2.26)
The rate of wave attenuation depends upon the size of the skin depth 7 J. For a good conductor such as copper (a = 5.80 x 10 (siemens/m) ) at 100 (MHz), the skin depth is J = 6.6 X 10- 3 (mm), which is indeed very small compared with the wavelength. The field that exists within the interior of good conductors is, therefore, considered approximately as a null field at 100 (MHz). This means that good conductors like copper can be regarded as PEe (perfect electric conductors) at 100 (MHz) and even higher frequencies. Let us find the magnetic field within good conductors. Substituting E",(z) into (2.27) •
gIves
26
2 Plane Wave Propagation
H (z) =
y =
1 8Ex (z) iW/-l 8z
k E+e ikz
w/-l
.
x
(2.28)
The wave impedance Z in good conductors is given by
=
W/-l
k
.
W/-l
= (1- z)
20'
(2.29)
indicating that the phase difference between electric and magnetic fields 7r in good conductors is 4 (radians).
2.2 Polarization As an electromagnetic wave propagates, the tip of its electric field traces a certain curve on a plane perpendicular to the direction of wave propagation. The trace of the electric field tip is described in terms of the polarization. Let us consider the one-dimensional plane wave propagation of (2.14) with the assumption k x = ky = O. Assume that the electric field phasor is given by
E(z) = Eoe
ikz
(2.30)
where E 1 and E 2 are the magnitudes of Ex and E y , respectively. Then, the time-varying form of E(z) is
E(z, t) = Re [E(z)e-
iwt
]
(2.31 ) where
Ex(z, t) = E 1 cos(wt - kz + Bx )
(2.32)
Ey(z, t) = E 2 cos(wt - kz + By) .
(2.33)
2.2 Polarization
27
Let
= cos(wt - kz
+ Ox)
(2.34)
v = Ey(z, t) -
~
= cos(wt - kz
+ Oy)
.
(2.35)
Then, the relation
(2.36) results in (2.37)
It is possible to transform (2.37) into the equation of an ellipse using coordinates based on the relation
~-1J
Ex(z, t) = Ef. cos'l/J - ETJ sin'l/J
(2.38)
Ey(z, t) = Ef. sin 'l/J + ETJ cos'l/J .
(2.39)
The corresponding polarization ellipse is shown in Fig. 2.3. Substituting Ex(z, t) and Ey(z, t) into (2.37) yields
AEl
+ BE; + CEf.ETJ
2
= sin (Oy - Ox)
(2.40)
(0 _ 0 ) sin 2'l/J cos y x El~
(2.41)
(0 _ 0 ) sin 2'l/J + cos y x E 1 ~
(2.42)
where 2
2
cos 'l/J A - E;
+
2
B _ sin 'l/J - E;
sin 'l/J _
E? 2
+
_ sin 2'l/J CE;
cos 'l/J
E?
+
sin 2'l/J _ 2
E?
(0 _ 0 ) cos 2'l/J cos y x El~ .
(2.43)
28
2 Plane Wave Propagation
Ell
\,
,,
, ""'I'"" "" , "" , "" ""
"" "" ""
~,--'--'--- ' - - - - - - - - +
""
"" ""
E%
Fig. 2.3. Polarization ellipse.
•
Elliptic polarization To obtain the equation of an ellipse, it is necessary to eliminate the cross term E~ E'1 (C = 0) by choosing the rotation angle 'I/J appropriately. The condition (C = 0) gives tan 2'I/J = tan 20 cos( Oy - Ox) where tan 0
(2.44)
E2 • = E . Thus, we obtam 1
(2.45) which represents elliptic polarization where the tip of the electric field traces a tilted ellipse as the wave propagates along the z-direction.
•
Linear polarization nn nn . When Oy - Ox = 0 or n, then 'I/J = 0 + 2 or - 0 + 2 ' where n IS an integer. For instance, when Oy - Ox = 0 and 'I/J = 0, A is shown to be 0 and (2.45) reduces to
E'1 = 0
(2.46)
where the tip of the electric field traces a line on the ECaxis and the wave is said to have linear polarization. Its graphical representation is shown in Fig. 2.4.
•
Circular polarization n When Oy - Ox = ± 2 and E 1 = E 2 , (2.45) becomes
2.2 Polarization
29
E~
Ell
\
\ \ \ \
\ --------..~\-'--=-------.
Ex
\
\ \ \ \
Fig. 2.4. Linear polarization.
,/
,
E~
,/ ,/ ,/
,/
\ \
\
,/ ,/
\
,/
'If ----;---,/-7~\-'--=--i_----. Ex ,/ ,/
\
\ \
,/ ,/ ,/ ,/
,/
Fig. 2.5. Circular polarization.
+
=1
(2.47)
which represents a circle on the ECEf) plane; thus, the wave has circular polarization. Its graphical representation is shown in Fig. 2.5. When Oy - Ox = - ;, the polarization is called right-hand circular polarization, since the electric field behaves as a right-handed screw that advances in the +z-direction. Similarly, when Oy - Ox = ;, (2.47) represents left-hand circular polarization.
30
•
2 Plane Wave Propagation
Unpolarized The intensity of natural sun light is completely random in any direction perpendicular to the direction of wave propagation. Therefore, the trace of the electric field tip is random, and thus irregular. This type of wave is called completely unpolarized.
2.3 Gaussian Beam Consider the problem of laser beam propagation when a laser beam illuminates a large-sized target that is not far away from the source. The amplitude shape of a laser beam is commonly Gaussian due to the Gaussian field distribution across a laser source aperture. This section considers the representations of a Gaussian beam that emanates from line and point sources. The cases of line and point sources correspond to two- and three-dimensional problems, respectively. Note that some relevant discussion is available in [1] and [2, pp. 160-165]. 2.3.1 Line Source
Consider a laser beam E(r) = yE(x, z) that propagates along the +z-direction from the source aperture E(x,O), as shown in Fig. 2.6. The line source is infinitely long in the y-direction and the field is assumed to be uniform with respect to y
~
= 0 . The Gaussian beam E(x, z) for z
> 0 is expressed in
terms of the inverse Fourier transform as x
Gaussian field E(x, z)
z E (x, 0)
source aperture
Fig. 2.6. Gaussian wave propagating along z-direction.
2.3 Gaussian Beam
31
<Xl
1
(2.48)
E(x,z) = 271" -<Xl
The expression E(x, z) should satisfy the Helmholtz equation. Substituting E(x, z) into (2.8) yields
rP 2) ( d 2 +1\, E((,z) = z
.
(2.49)
0
where I\, = Jk 2 - (2 and k = wVfi€. The solution for the beam E((, z) is assumed to propagate along the +z-direction. Therefore -. E((,z) = Eo(()e'''Z . (2.50)
-
Substituting E((,z) into E(x,z) gives <Xl
1 E(x, z) = 271"
(2.51 ) -<Xl
where
-
<Xl
E(x, O)e-i(:Z: dx .
E o(() =
(2.52)
-<Xl
The aim is to express E(x, z) in a mathematically tractable form based on approximations. In practice, the source aperture E(x,O) is modeled as approximately Gaussian in amplitude as 2
E(x, 0) = exp
x - (3'5
+ i¢
(2.53)
where (30 is the beamwidth. To estimate the phase variation ¢ at the source aperture, let us consider the line source placed at z = - Zo, as shown in Fig. 2.7. The circular cylindrical phase front arriving at z = 0 from z = -Zo is
Assuming Zo
•
(2.54)
•
(2.55)
» lxi, then ·k 2 .k z x z Zo + 2 ZQ
Therefore, the source aperture E(x, 0) can be rewritten as 2
E(x,O) = A exp
where A
=
e
ikzo
x - (32
(2.56)
ik .. . and (32 = (3'5 - 2za· SubstItutmg E(x,O) mto (2.52) and 1
1
using the integral formula for Re (P)
>0
32
2 Plane Wave Propagation
x \
\
\ circular phase front \ I
line source
• -z o
, I
I
I
I
Fig. 2.7. Line source placed at z = -zoo
Hence, the Gaussian beam is written as 1 E(x, z) = 21T
00
v:rrA(3exp
(2.59)
-00
The expression (2.59) is still difficult to analytically evaluate unless further approximation is made. Assume that the Gaussian beam is strongly collimated in the z-direction so that its transverse wave vector component «() is much smaller than the longitudinal one (1\;); thus I\;
=
y!k 2 _
(2
~ k _ (2
(2.60)
2k .
Substituting (2.60) into E(x, z) and performing integration yields
A
2
x E(x, z) = ";1 + iZ exp ikz - (32(1 + iZ)
(2.61)
where Z = ; 2zk. Note that E(x, z) represents an approximate form for the Gaussian beam that propagates along the +z-direction. When (30 and ikz E(x, z) reduces to the uniform plane wave Ae as it should.
Zo
---t
00,
2.3 Gaussian Beam
33
2.3.2 Point Source
Gaussian beam propagation from a point source can be similarly analyzed as in the line source case. Consider the laser beam E(r) = yE(x, y, z) that propagates along the +z-direction from the source aperture E(x, y, 0). The Gaussian beam E(x, y, z) for z > 0 is expressed in terms of the inverse Fourier transform 00
1
00
E«(,1J, z)e
E(x, y, z) = (211")2 -00
i
«(Z+71Y)
d( d1J .
(2.62)
-00
The solution for the beam E(x, y, z) propagating along the +z direction is 00
1
00
(2.63)
E(x, y, z) = (211")2 -00
where
K.
=
Jk
2 -
(2 -
1J2
-00
and 00
00
E(x, y, O)e- i «(Z+71Y) dx dy . -00
(2.64)
-00
The source aperture E(x, y, 0) is modeled as approximately Gaussian in amplitude with the beamwidth 130, thereby yielding
E(x, y, 0) = exp -
(x2
+ y2) (35
. + up .
(2.65)
Assume that the point source placed at z = - Zo produces quadratic phase variation ¢ over the source aperture at z = O. The source aperture is, therefore, given by
E(x,y,O) = Aexp where A
=e
ikzo
and ;2
= ;5
(2.66)
- ;:0' Substituting E(x, y, 0) into (2.64) gives (2.67)
Assume that the Gaussian beam is strongly collimated in the z-direction to yield
(2.68)
34
2 Plane Wave Propagation
-
Substituting E o((, TJ) and (2.68) into (2.63) finally gives the approximate Gaussian beam expression for a point source A E(x,y,z) = (l+iZ)exp
(2.69)
2z
where Z = (32 k' When (30 and Zo -+ 00, E(x, y, z) reduces to the uniform plane wave Ae
ikz
.
2.4 Reflection at Plane Boundary When a uniform plane wave impinges on a plane boundary, the reflection and transmission of an incident wave occurs at the boundary. The reflected and transmitted waves naturally take the form of uniform plane waves. This section investigates the reflection and transmission across a plane boundary separating two lossless and isotropic media. TE and TM waves are analyzed to obtain their reflection and transmission coefficients based on the boundary conditions.
2.4.1 TE Wave (Perpendicular Polarization) Consider a TE wave that is incident on a plane boundary between two lossless, dielectric media, as shown in Fig. 2.8. For simplicity, it is assumed that the incident electric field has only a y-component and the incident wave vector lies on the x-z plane (plane of incidence). The incident wave, therefore, is referred to as a TE (transverse electric to the plane of incidence) wave. The electric field vector is perpendicular to the plane of incidence and the TE wave thus
z •
E'y region
x
en
region (II)
at
Fig. 2.8. TE wave incident on plane boundary.
2.4 Reflection at Plane Boundary
35
corresponds to the case of perpendicular polarization. The wavenumbers in regions (I) and (II) are k1 (= W-/f.ll€t) and k 2 (= WVf.l2€2), respectively. The total electric field in region (I) consists of the incident and reflected waves as (2.70) (2.71 ) where RTE is the reflection coefficient for a TE wave. The phase terms are given by k zi = k1 sinOi , k zi = k I COSOi, k zr = k I sinOr, and k zr = k I cos Or. The transmitted wave in region (II) is E~(x, z) = TTE exp(ikztx - ikztz)
(2.72)
where k zt = k 2 sin fh, k zt = k 2 cos Ot, and TT E is the transmission coefficient for a TE wave. Let us enforce the boundary conditions for the tangential field continuity at z = O. The tangential E y continuity at z = 0 (2.73) •
reqmres exp(ikzix)
+ RTE exp(ikzrx)
=
TTE
exp(ikztx) .
(2.74)
Equation (2.74) must be satisfied irrespective of x, thereby yielding the phasematching condition (2.75) which results in sin Oi = sin Or (law of reflection) and k I sin Oi - k 2 sin Ot (Snell's law) simultaneously. Hence, (2.74) becomes 1 + RTE =
TTE·
(2.76)
The accompanying Hz and Hz components can be obtained by substituting (2.70) through (2.72) into Faraday's law \l x E = iWf.lll. Since i BEy h . I H . . 0 Hz = B' t e tangentla z contlllmty at z = Wf.l Z (2.77) •
gIves
k zi (R TE- 1) _ - k zt ,." .lTE· f.lI f.l2
(2.78)
36
2 Plane Wave Propagation
Solving (2.76) and (2.78) for the reflection and transmission coefficients gives
R
where Z2 =
'T/2
()
1
Z1 =
- Z2 - Z1 TE - Z2 + Z1
( 7) 2. 9
2Z2 TTE = Z2 + Z1
(2.80)
1]1
()
J1.2
'T/2 =
1
1
and
1]1
=
J1.1
.
cos t cos i 1:2 1:1 Let us evaluate the time-average power densities entering and leaving the boundary at z = 0 for lossless media, where both Z1 and Z2 are real. The incident time-average power density is given in terms of the Poynting vector as Si =
~Re (~x Jt.)
.(-z)
1
2Z1
(2.81 )
•
Similarly, the reflected and transmitted power densities are
(2.82)
(2.83) A law of power conservation (Sr
+ St
= Si) across the boundary can be easily
proved as Sr
+ St
=
1
2 (2.84)
Let us consider the special case when 1:1 > 1:2 and J1.2 = angle is
-
1-
-
• 2 ()
sm
i.
J1.1.
The transmitted
(2.85)
2.4 Reflection at Plane Boundary
When the incident angle (}i is greater than the critical angle (}e = sin-
37
1
, 1"1
the term cos(h (thus Z2) becomes purely imaginary, thereby yielding IRTEI = 1, Sr = Si, and St = 0. This means that the total internal reflection occurs when a wave impinges on a less dense medium (1"1 > 1"2) at an incident angle greater than the critical angle ((}i > (}e). The condition St = 0 should not be construed as a null field in region (II). The transmitted field in region (II) is evanescent, and it decays exponentially away from the boundary at z = 0.
2.4.2 TM Wave (Parallel Polarization) Consider a TM (transverse magnetic to the plane of incidence) wave that impinges on the plane boundary at z = 0, as shown in Fig. 2.9. A TM wave has no magnetic field components in the plane of incidence. The electric field vector is parallel to the plane of incidence and the TM wave thus corresponds to the case of parallel polarization. In regions (I) (k~ = Wv'J.L1f1) and (II) (k 2 = WVJ.L2f2) the magnetic fields are (2.86) (2.87) H~(x, z) =
TTM
exp(ik:z:tx - ikztz)
(2.88)
where k:z: i = k 1 sin (}i, k zi = k 1 cos (}i, k:z: r = k 1 sin (}r, k:z: t = k 2 sin (}t, k zr = k 1 cos (}r, and k zt = k 2 cos (}t. Since the tangential Hy(x, 0) must be continuous, the continuity condition
z HIy
region (I) = = = = = = x :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::.; .', ...y::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
region (II)
et
Fig. 2.9. TM wave incident on plane boundary.
38
2 Plane Wave Propagation
(2.89) results in 1 +RTM =
TTM
(2.90)
with the conditions sin Oi = sin Or (law of reflection) and k 1 sin Oi = ~ sin 0t (Snell's law). Ampere's law, \l x H -iWEE, produces the tangential electric field, i 8Hy (). • Ez = 8 ' Hence, E z x,O contmmty
=
WE
Z
(2.91) yields
k zi (RTM
1) -_
-
E1
-
kzt'T' .l.TM·
(2.92)
E2
Solving (2.90) and (2.92) for the reflection and transmission coefficients for a TM wave gives
R
- Y2-Yj
™ - Y2 + Y1
TTM
2Y2 = Y2 + Y1
(2.93) (2.94)
1 , Y1 = 1 , 'T/2 = J.l2, and "II = J.l1 . Note that 'T/2 cos Ot "II cos Oi f2 f1 R TM and TTM are also obtained from (2.79) and (2.80) based on the duality · h . . c h ' 1 d 1 t heorem, wh1C permIts mter angmg "II -+ - an "12 -+ -. "II 'T/2 Let us consider the special case when the reflection coefficient RTM becomes zero for nonmagnetic media (J.l1 = J.l2 = J.lo and E1 f- E2)' The condition Y2 - Y1 = 0 gives
where Y2
=
(2.95) Solving (2.95) and Snell's law
(2.96) for Oi gives II
ui
=
'-1
SIn
= t an
-1
(2.97)
2.5 Infinitely Long Transmission Lines
39
which indicates that the total transmission occurs when a TM wave impinges on the half space at the incident angle ()i. The angle (2.97) is referred to as the Brewster angle. Let us consider the other special case when 101 > 102 and J.L2 = J.Ll = J.Lo. Like the previous TE wave case, the total internal reflection occurs when a TM wave impinges on a less dense medium (101 > 102) at an incident angle greater than the critical angle
()c
= sin-
1 •
2.5 Infinitely Long Transmission Lines Transmission lines are commonly used to transfer electromagnetic energy from one location to another. Parallel wires, parallel plates, and coaxial lines are all typical examples of transmission lines. Transmission lines consist of two conductors that guide TEM (transverse electromagnetic) waves in a low-frequency regime where their field components are all transverse to the wave propagation direction. 2.5.1 Coaxial Line
This section investigates the behavior of wave propagation using an infinitely long coaxial line, as shown in Fig. 2.10. For a practical coaxial line with an inner conductor radius a and outer conductor radius b, the size is usually small (b - a) « A and its annular interior region is filled with dielectric material of permittivity 10 and permeability J.L. For simplicity, an ideal coaxial line is assumed where the dielectric material is lossless and the inner and outer conductors are PEC material. An input voltage source is applied between the inner and outer conductors at z = 0 to send a TEM wave along the z-direction. The TEM wave consists of E p and Hq, components that propagate along the z-direction with the propagation constant k (= wVJif.) p
. . . -- .....H$ I I '..: I P \ I
\,
J
d}/C
b
I
-:a;;+W~~I---------
---
Fig. 2.10. Coaxial line lying along z-axis.
z
40
2 Plane Wave Propagation
(2.98)
H",(p, z) = H",(p)e
ikz
(2.99)
.
Ampere's law \l x
-H = -iw€E - +J
(2.100)
reduces to Ampere's circuital law \lxH=J
(2.101)
for a TEM wave since the electric current density J that flows through the conductors is such that liw€E « Applying Stokes' theorem to Ampere's circuital law gives
JI.
c
H",(p,z)dl = I(p,z).
Substituting the current I(p,z) = Ite
ikz
into (2.102) gives
1+
H (p,z) =
0
'"
(2.102)
27rp
e
ikz
(2.103)
.
Faraday's law \l x E = iWJ.LH
(2.104)
is rewritten as 8Ep (p, z)
8z
=
.
~WJ.L
H (
'" p, z
)
(2.105)
which results in J.L
-H",(p, z) €
-
(2.106)
The time-average power carried by a coaxial line is
J.L -In €
b -
a
•
(2.107)
2.5 Infinitely Long Transmission Lines
41
2.5.2 Voltage and Current As a TEM wave propagates on a coaxial line, the direction of E p and H¢ remains unchanged as if they were scalars. It is thus expedient to represent the wave propagation on transmission lines in terms of scalars such as the voltage and current. Let us reinterpret the transmission line problem by introducing [+(z) and V+(z) where V+(z) and [+(z) are the incident (forward) voltage and current that propagate along the +z-direction, as shown in Fig. 2.11. The incident waves, [+(z) and V+(z), are (2.108) b
V+(z) =
Ep(p, z) dp a J.L
E
,
ln
.
b a
V+ o ikz = V+e 0
(2.109)
.
The time-average power, which is delivered by the voltage and current, is
J.L
-In E
b a
(2.110)
which is identical with (2.107). The characteristic impedance of a transmission line Zo is defined as the ratio ofthe voltage V+(z) to the current [+(z). The characteristic impedance of a coaxial line is, therefore
V+ 0 Z0= + [0
- - t: In ~ 1
21T
a
E
•
(2.111)
Consider the reflected waves, V-(z) and [-(z), as shown in Fig. 2.11. Assuming
r(z) =
[oe-
ikz
(2.112)
and substituting [- (z) into Ampere's circuital law (2.102) gives
r0
H¢(p, z) = 2
1Tp
.kz
e-·
.
(2.113)
42
2 Plane Wave Propagation
---v +(z), [+ (z) y- (Z), [- (Z)
4
z •
Fig. 2.11. Incident [V+(z), I+(z)] and reflected [v-(z), r(z)] waves.
Substituting H¢(p, z) into Faraday's law (2.105) yields
Ep(p,z) = -
/-L
[0
to
21rp
e -ikz
.
(2.114)
Let us define the reflected voltage V - (z) as
V- (z)
b
=
Ep(p, z) dp a
r0 21r '-
/-LIn to
b
-
e
-ikz
"
'V'
11,-
o
- 11,- -ikz
=
0
e
.
(2.115)
Hence, the ratio yields V-(z) _ Vo= -Zo. [-(z) [0
(2.116)
2.6 Terminated Transmission Lines Consider a terminated, lossless transmission line that is excited by the source Vs with internal impedance Zs at z = 0, as shown in Fig. 2.12. Termination with a load Zl at z = l generates reflected waves, thereby resulting in standing waves due to the interference between incident and reflected waves. The total standing wave voltage V (z) consists of the incident V+(z) and reflected V- (z) components V(z) = V+(z)
+ V-(z) (2.117)
2.6 Terminated Transmission Lines
43
I (z)
Zo
+ Zs
• V+(z) V(z)
+
Vs
~
Zl
V- (z)
z
V(z), l(z)
z=O
z =l
Fig. 2.12. Transmission line terminated with load Z/.
Similarly, the current I(z) is
(2.118) Since 10+
=
11,+ 0
Zo
,
=_
and 10-
11,0
Zo '
I(z) = 1 (Vo+e ikz _ Vo-e-ikz) . Zo
(2.119)
It is possible to solve (2.117) and (2.119) for the two unknown voltage amplitudes Vo+ and Vo-' Two boundary conditions must be enforced to determine Vo+ and Vo-' Applying Kirchhoff's voltage law at z = l and 0 results in
V(l)
(2.120)
Zl = I(l) VB = I(O)Zs
+ V(O)
.
(2.121)
Solving (2.120) and (2.121) for Vo+ and Vo- gives
11,+ = o (Zo 11, o =
r.
Ie
Zo v: + ZB) + Il(Zo _ Zs)e2ikl s
2ikl
(2.122)
Zo v: (Zo + Zs) + Il(Zo _ Zs)e2ikl s 'V
11,+ o (2.123) where
rl
Zl - Zo Zl + Zo
is the voltage reflection coefficient at the load z = l.
44
2 Plane Wave Propagation
2.6.1 Reflection Coefficient, Impedance, and Power
The voltage reflection coefficient F(z) at z is
_ V-(z) F(z) = V+(z) _ ZI - Zo 2ik(l-z) _
Z1+ Z 0 " ~
e
-
T' .I.
Ie
2ik(l-z)
.
(2.124)
y
FI
The impedance Z(z) at z is defined as
Z( ) = V(z) = Z 1 + F(z) 01-F(z) . z - fez)
(2.125)
Let us evaluate the time-average power delivery to the load Pav when the load voltage and current are
(2.126)
(2.127)
The result is Pav =
~ Re
[V (l)r (l)] 2
= 1V0+1 Re (1 r,* 2Z . I o
+ r,I -InI 1
2)
(2.128) -
To achieve a maximum power delivery to the load, the reflection coefficient II. must be zero, leading to the relation ZI = Zo, which is called a matched condition. 2.6.2 Voltage Standing-Wave Ratio
Investigating a voltage wave pattern on a transmission line is instructive. The total voltage on a transmission line is
2.6 Terminated Transmission Lines
=
V/ eikz [1 + llei2k(l-z)]
.
45
(2.129)
Assume that the reflection coefficient at the load is (2.130) The magnitude of V(z) then becomes
I
lV(z)1 = iVo+ 1 + 111le i2k (l-z)-ill .
(2.131 )
The maximum of lV(z)1 occurs when 2k(l- zt} - () = 2mr (n: integer) as
lV(z)lmax = iVo+
11 + 11111
(2.132)
while the minimum occurs when 2k(l - Z2) - () = (2n - 1)71" (n: integer) as (2.133) The distance d = Z2 transmission line is
-
Zl
between the minimum and maximum points on a
kd = 71"
2
d=
~ 4
•
(2.134)
The ratio of the maximum to minimum voltage magnitudes is called the voltage standing-wave ratio (VSWR) and is expressed as VSWR = lV(z)lmax
lV(z)lmin
1+ 1111 1-1111·
(2.135)
2.6.3 Cascaded Lines
Consider a two-stage transmission line, as shown in Fig. 2.13 (a). Note that Zl and Z2 are the characteristic impedances of cascaded transmission lines, respectively. At z = Za on the second transmission line (Z2), the voltage reflection coefficient Fa and impedance Za are
_ Zl r.a -
(2.136)
Z - Z 1 +Fa a - 2 1 _ Fa .
(2.137)
Z2 Zl + Z2
46
2 Plane Wave Propagation
Zs
vs
+ -
I-
l1
Z1
, I• I I I I I
-I
l2
original problem (a)
Zs
vs
+ -
equivalent problem (b)
Fig. 2.13. Two-stage transmission line.
At Z = Zb on the second transmission line (Z2), the voltage reflection coefficient and impedance Zb are
n
(2.138)
Z_Zl+ b2 1
n' n
(2.139)
Therefore, the original problem of a two-stage transmission line is simplified to a single-stage transmission line with the equivalent impedance Zb, as shown in Fig. 2.13 (b). The geometry in Fig. 2.13 (b) is identical with that of the single-stage transmission line in Fig. 2.12. The solution to the problem in Fig. 2.12 is thus available from (2.122) and (2.123). It is often desirable to eliminate the reflected wave on Z1 by appropriately choosing Z2. No reflection on Z1 requires the matched condition (2.140) To realize the matched condition, the length l2 = • • mgm
~
is chosen, thereby result-
Problems for Chapter 2
Zl - Z2
-
Zl
47
(2.141)
+ Z2
_ Z? Z b - Zl .
(2.142)
Assume that the load impedance Zl is real. Equating (2.140) with (2.142) • gIVes (2.143) The inserted ~-IOng transmission line with Z2 = ";ZlZI is called the quarterwave transformer.
Problems for Chapter 2 1. Derive (2.40) and (2.44). 2. Derive (2.61) and (2.69).
•
z
HIy
region (I) ·... ........ ....... .. .. .. .. .. .. .......... .. .. "" ..."" ..."" ..."" ..."" ..."" ..."" ...=.."" ..."-"" ..."-"" ..."" ...-. .............................................. .y • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •• ••
•••
•• • • • • • • •
•
• • • • • • ••
• •••••••••••••
x
® H'; d region (II) · ... ................ .. ... .. .. . . . . . .. .... .. ... .. ·. .. ........ ............. ...... ................ .
"......
. .
. .
region (III)
Fig. 2.14. TM wave impinging on dielectric slab of thickness d.
3. Consider a TM wave that impinges on a dielectric slab, as shown in Fig. 2.14. In regions (I), (II), and (III), the total magnetic field representations are given by
48
2 Plane Wave Propagation
H~(x, z) =
Rexp(ik",rx + ikzrz)
H;(x, z) = exp(ik"'dX) [A exp(ikzdZ) H~(x,z) = Texp(ik",tx - ikztz) .
(2.145)
+ B exp( -ikzdz)]
(2.146) (2.147)
To determine the unknown coefficients, A, B, R, and T, the four boundary conditions for the field continuities, Hy(x,O), Hy(x,-d), E",(x,O), and E",(x, -d), must be enforced. Derive A, B, R, and T using the boundary conditions. 4. In circuit analysis, a lossless coaxial line is modeled as a transmission line with the shunt capacitances C and series inductances L. Given a coaxial line of inner and outer radii a and b, filled with dielectric of J.l and i, derive its shunt capacitance and series inductance per unit length. 5. Consider the terminated transmission line, as shown in Fig. 2.12. Derive · ffi . ]-(z) t h e current re fl ectlOn coe Clent ] + (z) . •
3
Waveguides
3.1 Cylindrical Waveguides The understanding of wave propagation along cylindrical waveguides is of practical interest in microwave and optical communications. Typical cylindrical waveguides include rectangular waveguides, optical fibers, and microstrip lines. This section investigates electromagnetic waves that can propagate along an infinitely long waveguide, as shown in Fig. 3.1. Wave propagation is conveniently analyzed in terms of the magnetic and electric vector potentials A and F. The vector potentials in a source-free cylindrical waveguide satisfy the Helmholtz equations
(3.1 ) (3.2) where k (= wVJIf) is the wavenumber. To understand the wave propagation behavior, the Helmholtz equations must be solved for A and F subject to the boundary conditions imposed at the waveguide boundary.
cylindrical waveguide z -'it~T=-I---------
Ez
p
Fig. 3.1. Infinitely long waveguide that lies along z-direction .
•
50
3 Waveguides
Let us discuss the possible types of waves that can propagate along a waveguide. The propagating waves are conveniently represented in terms of the TE and TM waves. The TE wave refers to the wave whose electric field components are transverse to the wave propagation direction z (Ez = 0). Similarly, the TM wave refers to the wave whose magnetic field components are transverse to the wave propagation direction z (Hz = 0). The description of TM and TE waves is possible using the longitudinal components of A and F, Az(x,y,z) and Fz(x,y,z). •
When F = zFz(x, y, z) and A = 0, then Ez(x, y, z) = 0; this type of wave is called a TE wave (transverse electric to the wave propagation direction z). A TE wave satisfies the Helmholtz equation (3.3)
The explicit field expressions for a TE wave using rectangular and cylindrical coordinates are shown in Tables 3.1 and 3.2, respectively. •
When A = zAz(x,y,z) and F = 0, Hz(x,y,z) = 0 and a TM wave (transverse magnetic to the z-direction) is obtained. A TM wave satisfies the Helmholtz equation (3.4)
A hollow waveguide surrounded with metallic walls can support either TM or TE wave propagation. •
When the wavenumber in the z-direction is k, the propagating term takes the form of e±ikz and the longitudinal field components become zero Ez(x,y,z) = Hz(x,y,z) = 0, as seen in Tables 3.1 and 3.2. This type of wave is referred to as a TEM wave. Transmission lines (coaxial line, paired wire, etc.) consisting of two conducting terminals can support TEM wave propagation.
•
A dielectric waveguide or a partially filled waveguide with metallic walls can support the propagation of a hybrid wave, which is a combination of TM and TE waves. Both Az(x,y,z) and Fz(x,y,z) must be used for a hybrid wave analysis.
3.2 Rectangular Waveguide A rectangular waveguide with conducting walls is commonly used to transmit power in microwave communication and antenna applications. The field analysis for a rectangular waveguide is a canonical boundary-value problem. This section investigates the behavior of TM and TE wave propagation along a rectangular waveguide filled with homogeneous dielectric material (J1., i).
3.2 Rectangular Waveguide
51
Table 3.1. Field representations using rectangular coordinates (x, y, z) Fields
TM wave t
{)2A.
W J.I.€
{)x () Z
•
E",
t
{)2A.
WJ.I.€
{)y {)z
•
Ey
/ {)2
•
t
E.
W J.I.€
\
TE wave 1 {)F•
--€
1 {)F• € {)x
\
() 2 Z
+k
2
A.
0
/
•
t
{)2 F.
WJ.I.€
{)x {)z
•
1 {)A.
-
H",
J.I.
{)y
1
{)A.
t
{)2 F.
J.I.
{)x
WJ.I.€
{)y {)z
--
Hy
•
H.
{)y
t
0
W J.I.€
I
{)2 ()Z2
2'
+k
F.
3.2.1 TM Wave Consider TM wave propagation along a rectangular waveguide with perfect conducting walls, as shown in Fig. 3.2. A TM wave is assumed to propagate along the z-direction with the propagation constant k z . The vector potentials associated with the TM wave are given by Fz(x,y,z) = 0 and y
b
z (a> b)
x
Fig. 3.2. Rectangular waveguide with cross section a x b (a > b).
52
3 Waveguides
Table 3.2. Field representations using cylindrical coordinates (p, c/J, z)
Fields
TM wave
TE wave
•
Ep
z o2A. wltf op oz
1 of• -fp oc/J
•
1 of• f op
E",
02A. wltfP oc/J oz
z
•
2 /0 Z o 2 Z wlt f
~
•
E.
}
D
A.
loA. ItP oc/J
z 02F. wltf op oz
loA. -It op
02F. wltfP oc/J oz
Hp
H",
+k
2
•
•
z
2 / 0 Z o 2 Z wlt f
~
•
D
H.
+k
2
}
F.
Az(x, y, z) = Az(x, y) exp(ikzz). To determine the propagation constant k z , 2 2 it is necessary to solve the Helmholtz equation (\7 + k ) Az(x, y, z) = 0 subject to the boundary conditions. The Helmholtz equation can be rewritten as (3.5) Based on the separation of variables technique, Az(x, y) takes the form of
(3.6)
Az(x,y) = X(x)Y(y) .
Substituting Az(x, y) into the Helmholtz equation (3.5) and dividing by X(x)Y(y) yields 1
J2 X(x)
X(x)
dx2
+
J2y(y) Y(y) dy 2 1
2
+ k - k z = O.
(3.7)
J2y(y) To satisfy (3.7), the terms X(x) dx 2 and Y(y) d 2 must be equal to y constants independent of x and y, respectively. Let the separation constants 2 2 be -kx and _k y as 1
J2 X(x)
2
1
3.2 Rectangular Waveguide
53
J2X(x) X(x) dx 2
(3.8)
J2y(y) Y(y) dy 2
(3.9)
1
1
Then, the dispersion relation is given by
+ k; + k;
k;
= k
2
(3.10)
.
Since E z (hence A z ) must vanish at the boundaries x = 0, a and y = 0, b, the vector potential AAx, y, z) is chosen as
Az(x,y,z) = sink",x sinkyY exp(ikzz) where k", =
":7r, k y =
n;, and
(3.11)
mand n are integers (m, n > 1). The remain-
ing field components are available by substituting Az(x, y, z) into Table 3.1 in Section 3.1. The dispersion relation yields the propagation constant for the TM mn mode (3.12)
while its cutoff wavenumber is given by k c =
m7r)2 (n7r)2 ( a + b . When the
operating frequency is high enough to satisfy the condition k > k c , the TM mn mode can propagate. When k < kc , the TM mn mode cannot propagate and becomes attenuated. Note that the lowest-order TM wave is the TM l1 mode.
3.2.2 TE Wave The analysis of TE wave propagation within a rectangular waveguide is somewhat similar to the analysis of TM waves. Let us start with the assumption Az(x, y, z) = 0 and Fz(x, y, z) = Fz(x, y) exp(ikzz), where the vector potential Fz (x, y) satisfies the Helmholtz equation
2 8
8x 2
2 8 2 2 + 8 +k -k z Fz(x,y) =0. y
2
(3.13)
Solving the Helmholtz equation is possible using the separation of variables technique. The tangential electric fields must vanish on the metallic surfaces, implying that the normal derivatives of Fz(x,y) must be zero,
8FAx,y) 8y of
y=O,b
= 0 and 8Fz (x,y) 8x
= O. Hence, Fz(x,y,z) takes the form "'=O,a
54
3 Waveguides (3.14)
m7r a'
n7rb'
.mtegers
where kz = ky = and m and n are (m > 0, n > 0, m + n :j:. 0). The propagation constant k z and cutoff wavenumber kc for the TE mn mode are (3.15)
7r C: f + (nb7rf .
kc =
When a fields are
(3.16)
> b, the lowest-order TE wave is the TE IO mode, where its nonzero (3.17)
Hz =
-ikza
7r
. H o sm
7r
· 0sm
•
Ey = where H o =
i Wj.Lf
tWj.La H
(7r)2 - and k z =
k2
a
(7rX) (. ) a exp tkzz
(7rX) ('k) a exp t
(7r)2
-
-
(3.18) (3.19)
zZ
.
a
Let us consider a rectangular waveguide of dimension a = 2b, where the lowest propagating wave is TE IO and the next higher wave is TE20 or TEol . If
. . (7r) the operatmg frequency IS chosen as a
f}, f3) to guarantee wave guidance within region (II). The field in region (I) must be evanescent in the +y-direction for y > 0 and propagate along the z-direction. Hence, A;(y,z) is written as (3.25)
where
01
= Jk~
-
W2J1.1f1.
d
The wave in region (II) experiences reflection at y = ± 2' thereby exhibiting standing wave characteristics along the y-direction. The vector potential in region (II) is
A;I (y, z) = (A~ cos kyy + A~ sin kyy) exp(ikzz) (3.26)
+ k z2
k 22 = -
h k were W 2 J1.2 f 2· Similarly, since the field in region (III) must be evanescent along the -ydirection for y < 0, 2 y
--
(3.27)
-
where 03 = Jk~ W2J1.3f3' To represent the vector potentials, A; (y, z), A;I (y, z), and A; II (y, z), four unknown coefficients, A 2 , 'IjJ, A 3 , and k z , are introduced, respectively, whereas A} is considered a known coefficient. The aim is to determine the four unknown coefficients utilizing the four boundary conditions for the E z and Hz field continuities at y =
±~.
The E z and Hz field components are obtained by substituting A;(y,z), A;I(y,z), and A;II(y,Z) into Table 3.1 in Section 3.1. The E z continuities at y = ± ~ are (3.28)
d -- z
2'
l
•
Since E z = - WJ1.f
•
(3.29)
2
8 2 8z 2 + k A z , (3.28) and (3.29) can be rewritten as (3.30)
(3.31)
3.3 Dielectric Slab Waveguide
continuities at y =
Similarly, the Hz
±~
57
yield
(3.32)
ky d a3 . - - A 2 sm -ky 2 + 1/J = A 3 exp( -a3 d / 2 ) . M2 M3
(3.33)
Taking the ratio of (3.30) and (3.32) results in
al
d
+ 1/J -
ky
.
(3.34)
•
(3.35)
tan a ± tan{3 tan (a ± {3) = ..,------'---:: 1 =f tan a tan {3
(3.36)
-
tan
€l
kY 2
€2
Similarly, (3.31) and (3.33) gives a3
d
+ 1/J - Y2
--=- tan -k €3
ky
Using the formula
it is possible to eliminate 1/J in (3.34) and (3.35). The result yields the dispersion relation
a a tan(kyd) - €2 l = €2 3 1 + tan(kyd) €} k y €3ky When region (III) is perfectly conducting (PEe, relation reduces to
€3
a €2 l . €l k y
-+
00),
(3.37) the dispersion
(3.38) The transcendental relation (3.37) can be solved for kz numerically. In general more than one solution exist.
3.3.2 TE Wave The analysis of TE wave propagation is quite analogous to that of TM wave propagation. The TE wave (E z = 0 and Hz f:- 0) is assumed to propagate along a slab waveguide with the propagation constant k z . The associated vector potential F z (y, z) takes the form of
Ff (y, z) = B 1 exp( -alY) exp(ikzz)
(3.39) (3.40) (3.41)
58
3 Waveguides
Enforcing the boundary conditions for the Ex and Hz continuities at y = ± ~ results in the dispersion relation for the TE wave •
(3.42)
Note that the dispersion relation (3.42) can also be directly obtained from (3.37) based on the duality theorem (Section 6.5), which permits interchanging 10 1,2,3 ~
J.l1,2,3·
3.4 Circular Waveguide In addition to rectangular waveguides, circular conducting tubes, known as circular waveguides, are also popular guiding structures for microwave applications. This section investigates wave propagation within a circular waveguide that is surrounded with a conducting wall. Both TM and TE waves are considered in the following subsections.
3.4.1 TM Wave
Consider a TM wave (Hz = 0 and E z =j:. 0) that propagates inside a circular waveguide, as shown in Fig. 3.4. Assume that the circular waveguide is filled with dielectric material of (J.l,€). When the wave propagates along the zdirection with the propagation constant k z , the vector potential A(p, ¢, z) takes the form of (3.43)
The corresponding Helmholtz equation drical coordinates is rewritten as
(\7 + k 2
2
)
A z (p, ¢, z) = 0 using cylin-
y
a Z + - - - -W- d - - ' - - - f - - - - - - - - - - -
x
PEC
Fig. 3.4. Circular waveguide with radius a.
3.4 Circular Waveguide
1 8 p 8p
8 p8p
59
(3.44)
or (3.45)
2 where k~ = k - k;. Based on the separation of variables technique, let
Az(p,¢) = R(p)p(¢) .
(3.46)
Substituting Az(p, ¢) into the Helmholtz equation and dividing by R(p)p(¢) yields
d
dR(p) - p dp pR(p) dp 1
1
1
+ (Ji p(¢)
~p(¢)
d¢2
2
+ kp =
0.
(3.47)
Substituting 1
~p(¢)
p(¢)
d¢2
2
(3.48)
=-m
into (3.47) results in the following two ordinary differential equations ~
2
d¢2 +m 1 d
-
pdp
d pdp
p(¢) = 0
(3.49)
R(p) = O.
(3.50)
2
m + k~ - p2
The solution to (3.49) is sinm¢
p(¢) =
(3.51 )
•
cosm¢ Since the field is periodic in the azimuthal direction with a 27T-periodicity, the parameter m should be an integer. Equation (3.50) is known as Bessel's equation (see Appendix B) and its solution is represented as the linear combination of the Bessel functions Jm(kpp), Nm(kpp), Hg),(2)(k pp), .... Among these Bessel functions, the appropriate choice for R(p) is Jm(kpp), since Nm(kpp) and Hg),(2) (kpp) become infinite at p = O. Hence, A z (p, ¢, z) is given by sinm¢
exp(ikzz) . cosm¢
(3.52)
60
3 Waveguides
The explicit field components are obtained by substituting Az(p, ¢, z) into Table 3.2 in Section 3.1. Since E z and E", must vanish at p = a, Jm(kpmna) = 0, where kpmna is the nth root of the mth order Bessel function. The TM wave with the eigenvalue k pmn is called the TM mn mode, where the roots of Jm(kpmna) = 0 are tabulated in Table 3.3. The propagation constant for the TM mn mode is • gIVen as (3.53)
In order for the TM mn mode to propagate, the wavenumber must be higher than k pmn k (= w...fiii.)
> k pmn
(3.54)
where k pmn is the cutoff wavenumber for the TM mn mode. The lowest cutoff . 2.405 ( ) wavenumber IS k po1 ~ TM Ol mode . a Table 3.3. Values of kpmna m=O
m=l
m=2
m=3
n=l
2.405
3.832
5.136
6.380
n=2
5.520
7.016
8.417
9.761
n=3
8.654
10.173
11.620
13.015
3.4.2 Power Delivery by TMm.n Mode
Let us evaluate the power that is delivered by the TM mn mode when k Since
> k pmn . (3.55)
the transverse field components are
•
3.4 Circular Waveguide
E _
61
2
8 A z (p, ¢, z)
i
p - WJ-Li
8p 8z
z = - kpk J:n(kpp)(A sin m¢+ Bcosm¢) exp(ikzz) WJLi 2 E _ i 8 A z (p,¢,z) cf> WJ-Lip 8¢ 8z
(3.56)
z = - mk Jm(kpp)(Acosm¢ - Bsinm¢) exp(ikzz) WJ-Lip
(3.57)
H _ 1 8A z (p, ¢, z) p - J-Lp 8¢ = m Jm(kpp)(Acosm¢ - B sin m¢) exp(ikzz) J-Lp
(3.58)
H __ .!. 8A z (p,¢,z) cf> J-L 8p =
-~J:n(kpp)(Asinm¢ + Bcosm¢)
exp(ikzz)
J-L
where J:r,(kpp) denotes differentiation with respect to the argument
(3.59)
d~(k~;r).
The total time-average power Pa1J is given by a
1
P a1J
2"
= 2 Re
X
0
=
0
a
~Re 2
(E 1l*) .zpd¢dp
2"
(EpH; - Ecf>H;) pd¢dp . 0
(3.60)
0
Let us first consider a case in which m yield
i
O. The ¢ integrations for m
2" ~m2k E H* d¢ = z J2 (k p) o cf> p WJ-L2 ip2 m p
(IAI + IBI 2
2
)
•
i
0
(3.62)
Therefore, P a1J becomes
pdp.
(3.63)
62
3 VVavegtrides
The evaluation of Pav is tedious but straightforward. Let kpp = u, then
o
=- I
(3.64)
.
Using recurrence formulas for the Bessel function
J;"(u) =
~ [-Jm+1(u) + Jm-1(u)]
(3.65) (3.66)
•
gIves
:>:1[2 2 ] 2 Jm+1(u) + Jm_1(u) udu
1=
(3.67)
o where kpa = x. Based on the formula
J~(u)udu
2
=~
[J~(u) - Jm- 1(u)Jm+1(u)]
(3.68)
I is integrated to
x2
4 [J;'+l (x) -
1=
Jm(x) Jm+2 (x)
+J~_l (x) - Jm- 2 (x)Jm(x)] .
(3.69)
Since Jm(x) = 0, 2
1= ;-
[J~+l (x) + J~_l (x)] .
(3.70)
Further simplification is possible using the relations
Jm+1(x)
= -Jm-1(x) = -J;"(x).
(3.71)
The final result for m = 1,2,3,· .. is 2
1= P av
;-J;" 2(X)
(3.72)
2 2 = (IAI + IBI )
~k~ J.L
(k pa)2 J;" 2(kpa) .
(3.73)
f.
Similarly, it can be shown that for m = 0 P av
2 1fkz = IB 1 2 2 WJ.L
(
f.
)2 12( ) kpa J o kpa .
(3.74)
3.5 Circular Dielectric Waveguide
63
3.4.3 TE Wave The analysis of TE wave (Hz t= 0 and E z = 0) propagation inside a metallic circular waveguide is similar to the analysis of TM wave propagation. The solution to the Helmholtz equation 1 0 pop
p
0
(3.75)
op
•
IS
sinm¢ (3.76) cosm¢ The explicit field components are obtained by substituting F z (p, ¢, z) into Table 3.2 in Section 3.1. Since E¢ vanishes at p = a, J:n(k~mna) = 0, where k~mna is the nth root of the derivative of the mth order Bessel function. The roots of J:n(k~mna) = 0 are tabulated in Table 3.4. Note that the propagation constant is given by k z = Jk 2 - (k~mn)2. The lowest cutoff wavenumber of the TEmn mode is
k~ll ~
1.841, indicating that the lowest propagating a mode among TE and TM waves is the TEll mode. Table 3.4. Values of k~mna
m=O
m=1
m=2
m=3
n=1
3.832
1.841
3.054
4.201
n=2
7.016
5.331
6.706
8.015
n=3
10.173
8.536
9.969
11.346
3.5 Circular Dielectric Waveguide Consider electromagnetic wave guidance by an infinitely long circular dielectric waveguide, as shown in Fig. 3.5. A circular waveguide (/-Ll , €}) is embedded in a background medium (/-L, 10), thus constituting a model for an optical fiber with a core and cladding. It is assumed that the circular dielectric waveguide is denser than the background medium (€} > 10). Let us represent the propagating waves in regions (/-L}, €d and (/-L,€) based on the boundary conditions. The
64
3 Waveguides
y (~, e)
a z
--~
"7''---'
,....7--'--+-----------
Fig. 3.5. Circular dielectric waveguide with radius a.
boundary conditions require that the four tangential electric and magnetic field (E,p, E z , H,p, and Hz) continuities must be simultaneously satisfied at p = a. The four boundary conditions are available by assuming that both the E z and Hz components are nonzero. Therefore, the guided wave is a hybrid mode that is a combination of TE (Hz i- 0) and TM (E z i- 0) waves. Inside the waveguide (p < a), both E z (equivalently A z ) and Hz (equivalently F z ) satisfy the Helmholtz equations
= O.
(3.77)
Hence, A~ and F[ take the form of (3.78) (3.79)
where the dispersion relation is
2 k z
k2 + p
2 k 2 = 1 = W ILl £1
.
(3.80)
The choice of the azimuthal field variation, cos m¢ and sin m¢, in A z and Fz facilitates the enforcement of the boundary conditions later on. Outside the waveguide (p > a), the fields are chosen as evanescent types that vanish at infinjty p -+ 00. The vector potentials satisfy the Helmholtz equations
-0 .
(3.81 )
3.5 Circular Dielectric Waveguide
65
The corresponding A~I and pII are assumed to be (3.82) (3.83) where K m is the modified Bessel function of the second kind and the dispersion relation is (3.84) In order for the wave to be properly guided by the dielectric waveguide, the parameters k z , k p , and Cl p must all be real numbers. The boundary conditions and field components must be used to find the propagation constant k z • The field components in the respective media for P < a and P > a are straightforwardly available from Table 3.2 in Section 3.1. The condition of E,p continuity at P = a then gives
(3.85) where the prime denotes differentiation with respect to the argument as J' (kpa) = dJm(kpp) . m d(kpp) p=a Similarly, matching the remaining boundary conditions for the E z , H,p, and Hz continuities at P = a produces the other three equations for A, B, C, and D. A set of the four simultaneous equations can be written in a matrix form p
k J' (k a) £1
m
p
o
o
-~J' (k a) fJ.1 m p
o
o
66
3 Waveguides A
B = 0.
•
(3.86)
c D The dispersion relation is obtained by setting the determinant to zero as p
J
o where kpa
= Ul, Ci.pa -
-K
o mk z U2,
-
_
= p,
J
_
= Ul
W The dispersion relation can be rewritten as p
-
p
--
-K
U~ -
o
o
P
p
to
-J
p
J:"(Ul) _ K:"(U2) J ( )' and K = U2 K ( ). m Ul m U2
K
/-l
-
-J
J
-/-ll
p --
(3.88)
K
or (3.89)
or
(3.90)
3.6 Shielded Stripline
67
In principle, the propagation constant k z can be determined by solving the dispersion relation (3.90). Let us introduce the normalized frequency V as (3.91) An alternative way of determining k z is to solve the simultaneous equations (3.90) and (3.91) graphically. Plotting (3.90) and (3.91) on the UI-U2 plane and identifying intersections of two curves yields the solution k z . Note that (3.91) represents a circle of radius V on the Ui-U2 plane. For each index m, there are a number of solutions k z = k zmn . Once k zmn is determined, the ratios of the field amplitudes A, B, C, and D can subsequently be obtained from (3.86). The hybrid waves are designated as HE mn or EH mn modes. The HE mn mode refers to when the Hz component is dominant, while the EH mn mode refers to when the E z component is dominant. Table 3.5. Mode cutoff frequencies Modes
Jm(Ul) = 0
Ul
HEll
Jl (Ul) = 0
Ul = 0
TEO! I TMO! , HE21
JO(Ul)=0
Ul = 2.4048
Table 3.5 shows that the HEll mode is the lowest order with zero cutoff frequency. Single-mode HEll propagation along a dielectric circular waveguide is possible if 0 < Ul < 2.4048. Considering the constraint (3.91), the radius of circle V on the UI-U2 plane must be V < 2.4048 to guarantee single-mode HEn propagation. Hence, the condition for single-mode propagation is written as
o < Jkr - k a < 2.4048. 2
(3.92)
Consider a practioal single-mode silica fiber that operates at 1.55 (J.lm) , where the refractive indices for the core and cladding are nl = 1.46 and n = 1.46 0.002, respectively. The condition (3.92) shows that the core radius a must be a < 7.77 (J.lm) to support single-mode propagation.
3.6 Shielded Stripline Due to recent advances in integrated-circuit technologies, planar transmission lines such as striplines are now commonly used in microwave circuits. This section investigates a shielded stripline, as shown in Fig. 3.6. Regions (I), (II),
68
3 Waveguides y
PEC/. /.
•
,
I, I, I ,
'I 'I ' !
original problem (a)
,
I
!,
PEC
wJ-//fl'fl'fl'A0'fl'fl'/Lf//fl'fl'/A.mfl'fl'fl'/fl'////fl'/fl'fl'fl'//fl'//L
I ,I , I,
' I' I
, I, I
I '(I) I ' I,
~ ~ ~ ~ ~[~ ~ ~ ~ ~ ~ ~ ~ I'l""'~11 - e • q ml 2TJ tan (TJb q ) (---ta~ bq 11=0 ell ((~ - a~) ((~ -
L
•
Consider I q when n
(3.150) af) .
> p. The complex (-plane is shown in Fig.
3.9. Note 2
that the function
f (() contains simple poles at ±(v = ±
k2 -
V1r bq
-
k~,
which are the solutions to the equation TJ tan(TJb q ) = O. Thus 00
f(O d( +
f(() d( = 21ri r2
L Res f(() . v=O (-(.
(3.151)
Since
r2
f(() d( = I q
(3.152)
f(() d( = 0
(3.153)
3.6 Shielded Stripline
79
1m (~)
~v X
X X
x'"
rl x, ,
,
Fig. 3.9. Complex (-plane with simple poles.
I q is given by 00
I q = 21Ti
L
v=o
Res f(C)
(=(v
(3.154) •
Consider I q when n < p. Let us choose a semicircular path in the lower half-plane and perform residue calculus. The result is 00
I q = -21Ti
L
v=o
Res f(C)
(--(v
(3.155) The expressions (3.150), (3.154), and (3.155) can be combined into
(3.156)
80
3 Waveguides
where (3.157) Substituting (3.156) into (3.138) gives ()()
(3.158) Summing series over n in
Ji yields
()()
n=-oo
(3.159)
(3.160)
(3.161)
•
Therefore 2 [1 ±
()()
L
(±l)nAn = (±l)P
ei(v T - ei(v a
1 =f
=f ei(v(T-a)]
ei(v T
•
(3.162)
n=-oo
Substituting (3.162) into (3.158) gives the final series expression for
•
Ji as
(3.163)
Problems for Chapter 3 1. The TE mn mode propagates along a rectangular waveguide with dimensions (a x b), as shown in Fig. 3.2. Compute the power delivered by the TE mn mode.
Problems for Chapter 3
81
2. Consider the dielectric slab waveguide shown in Fig. 3.3. The dielectric medium in region (III) is replaced with a perfect magnetic conductor (PMC). Determine the dispersion relation of a TE wave. 3. Construct (3.86) using the boundary conditions for the field continuities. 4. Derive (3.134) from (3.132) and (3.133). 5. Derive (3.155) using residue calculus. 6. Consider the shielded stripline shown in Fig. 3.6. Assume TM wave propagation in the z-direction (Hz = 0 and E z i 0). Following the TE wave analysis in Section 3.6, derive the dispersion relation for a TM wave.
4
Cavity Resonators and Coupler
4.1 Rectangular Cavity Resonator Resonant cavities are basic microwave components that store electromagnetic energy. Microwave resonant cavities are known to have large quality factors. A rectangular cavity resonator is relatively easy to analyze, yet it provides physical insight into the resonance mechanism. This section investigates wave resonance in a rectangular resonant cavity that is surrounded by electric walls. 4.1.1 TE Mode
Consider a rectangular cavity resonator that is surrounded by perfectly conducting walls, as shown in Fig. 4.1. A resonant cavity is filled with a dielectric medium with permeability J..t and permittivity E. A rectangular cavity resonator of dimensions (a x b x d) is viewed as a rectangular waveguide of dimensions (a x b) with perfectly conducting walls placed at z = 0 and d. z
y
d
b
a
, Fig. 4.1. Rectangular cavity resonator surrounded by perfectly conducting walls.
84
4 Cavity Resonators and Coupler
Assume that the TE mn mode propagates along the ±z-direction with A(x,y,z) = 0 and F(x,y,z) = zFz(x,y,z). The Helmholtz equation for Fz(x, y, z) is given by
(4.1)
, Let
n1r (Ok Fz(x, y, z) = cos a x cos b y Ae' zZ m1r
ok
+ Be-'
zZ
)
(4.2)
where
1r
C:
f
+ (nb1rf + k;
= k
2
(4.3)
and A and B are unknown amplitudes. Since the tangential electric fields at z = 0 and d must vanish,
Exl z=O,d
= _~ 8Fz (x,y,z) to 8y
=0
(4.4)
-0 .
(4.5)
z=O,d
E y z=O,d
= ~ 8Fz (x,y,z) to 8x z=O,d
Hence, the unknown coefficients A and B are chosen accordingly to yield m1r
n1r
(4.6)
Fz(x,y,z) = Gcos a x cos b y sinkzz
rnr d'
where k z (m,n) = 0,1,2,"" (m > 0, n ~ 0, m+n f:. 0), and p = 1,2,3, .... The wave associated with F z (x, y, z) is called the rectangular cavity TE mnp mode, whereas its field components can be obtained by substituting Fz(x,y,z) into Table 3.1 in Section 3.1. 4.1.2 TM Mode
Similarly, it is possible to obtain the vector potentials for the rectangular cavity TM mnp mode. Assume
n1r (Ok Az(x,y,z) = sin a x sin b yA'e' zZ m1r
+ B'e-'
zZ
)
(4.7) (4.8)
F z (x, y, z) = 0 .
The tangential electric fields at z boundary conditions
ok
o and
d must vanish. Therefore, the.
4.1 Rectangular Cavity Resonator
Exl
=
z=O,d =
E Y
z=O,d
iFAz(x,y,z) WJ-LE 8x 8z i
85
=0
(4.9)
=0
(4.10)
z=O,d
2
8 A z (x,y,z) WJ-LE 8y 8z i
z=O,d
yield
"
m1l"
.
n1l"
rm
A z (X, y, z ) = C sm a x sm b y cos d z
(4.11)
where m, n = 1,2,3, ... and p = 0,1,2,· ... The resonant wavenumber k mnp for the TEmnp and TM mnp modes is given by
=W# where
W
(4.12)
is the resonant angular frequency.
4.1.3 Quality Factor
When conducting walls that surround a cavity are imperfect with finite conductivity (J, ohmic energy loss due to the finite (J occurs. To measure the energy storage and loss for a resonant cavity, the quality factor Q of a resonant cavity is used as (4.13) where U is the energy storage and WI is the time-average power loss. Let us evaluate Q for a rectangular cavity that is surrounded by imperfect but good conducting walls. The field configuration near the surface is shown in Fig. 4.2. Assume that current flow through good conductors exists almost near the conducting surface due to the small skin depth. For a good conductor like 3 copper, the skin depth is 6 = 6.6 X 10- (mm) at 100 (MHz). The surface current density J s on a good conductor is approximated by the surface current on a PEC surface as
•
-J
8
:::::!
nx
-H
(4.14)
where H is the magnetic field on an idealized PEC surface. The average power loss on the imperfect conductor surface S is then 1
WI =-Re
2
(4.15)
86
4 Cavity Resonators and Coupler
n'"
-
II
cavity interior (Il, €) '/j
goo con uctor (Ill' cr)
Fig. 4.2. Fields in good conducting surface. I
According to (2.29), the relation between the electric and magnetic fields inside a good conductor is given in terms of the wave impedance Z as E~ZnxH
(4.16)
where WJ.LJ
Z = (1 - i)
2(J
•
(4.17)
ds
(4.18)
Hence 1
-
R J8 WI =2 s where R =
WJ.Ll
2
is the surface resistivity of a good conductor.
2(J For illustration, let us consider a case when the TE 101 mode is excited within a rectangular resonant cavity. Substituting the electric vector potential for the TE 101 mode
FzC
x, y, z) = C cos ex) sin Gz)
(4.19)
into Table 3.1 of Section 3.1 results in the field components. They are given by Ex = E z = 0
(4.20)
(4.21)
87
4.1 Rectangular Cavity Resonator
(1f) (1f) C ad sin ;;x cos d
i
Hx = -
1f2
WJ-L€
(4.22)
Z
(4.23)
H y =0
(1f)2 cos (1f). (1f a x sm d
i Hz = WJ-L€ C a
Z
)
(4.24)
•
First, consider the current flow on the top surface (z = d) as
Js
= z=d
nx H
z=d
+ zHz )
= -z x (xHx = -fjHx
z=d
.
= -y"
z=d
Z
WJ-L€
2
C1fd sm . (1f-x ) a
a
(4.25)
.
The power loss from the top surface is -
R Js
2
z=d
ds
S
-
1
2
a
b
o
0
1
R
2
WJ-L€
(1f2 ) 2 (1fX ) IC/ ad sin a dx dy 2
2
(4.26)
where w=
1f
Ja 2 +d2
(4.27)
----~-.
#
ad
Similarly, it is possible, to evaluate the power loss associated with the bottom and side surfaces. The total power loss is shown to be
ICI1f
2
WJ-L€
2
2b(a3
+ d3) + ad(a + ~) 2
a 4 d2
•
(4.28)
The energy storage is the sum of the electric and magnetic energy as
u=~4
(4.29)
88
4 Cavity Resonators and Coupler
where d
1 -2 1 4 V fiE I dv = 4
0
b
o o (4.30)
(4.31) Note that the stored time-average electric energy is equal to the stored timeaverage magnetic energy. Substituting U and WI into the quality factor Q finally yields Q = wU WI
1f 2R
/-L
f
b(a 2 + ~)1.5 2b(a 3 + d 3 ) + ad(a 2
+ d2)
.
(4.32)
For a cube cavity (a = b = d),
Q = .j21f 6R
/-L
-
•
(4.33)
4.2 Circular Cavity Resonator A hollow circular cylinder is another simple cavity structure often used for confining electromagnetic wave energy. This section analyzes electromagnetic resonance within a circular cavity resonator (/-L and f) covered with perfect conducting walls, as shown in Fig. 4.3. 4.2.1 TM Mode Consider a TM mn mode that propagates along the ±z-directions in a circular cavity resonator. The vector potentials associated with a TM mode are F(p, 0 is given by -
.
€
F(r) = 411"
~ -
S'
€
411" S'
eik1r-r' 1
2f1. x
It (x', y', 0+) 1_r - _'I ds' r
2f1. x
F
€eikr
(x', y', 0+) ~exp [ik(r - r' . f)] ds'
b/2
a/2
-b/2
-a/2
r
--,-----
411"r
•
2f1. x
It (x', y' ,0+) exp (-ikr' . f) dx' dy'
.
(6.36)
Let us evaluate the electric vector potential using spherical coordinates (r, B, ¢J). Substituting
138
6 Electromagnetic Theorems
z
,.,
n
:....---.... y
a
b x
Fig. 6.9. Perfectly conducting rectangular plate of size (a x b) . •
n x If' (x',y',O+)
= f)Eoexp(iky'sin(}i)
(6.37)
r' . f = (Xx' + f)y') . (x sin (} cos 4> + Ysin (} sin 4> + Z cos (}) = x' sin (} cos 4> + y' sin (} sin 4>
(6.38)
into F(r) and performing integration gives
F(r) =
(f sin (} sin 4> + {; cos(} sin 4> + ¢ cos 4» .
f.e
ikr
21rr
Eoab A
sin Q Q
sin f3
f3
A
+ (}Fe + 4>FcfJ
(6.39)
Q
ka . (} ..I. = 2 sm cos 'I'
(6.40)
f3
=
= fFr where
k; (sin (} sin 4> -
sin (}i) .
(6.41)
6.5 Duality Theorem
-
139
-
The far-zone scattered fields E and H are given by
E =
ik ( . -~ -()F
.) + ¢Fe
H=iw(OFe+¢F
representing an outward-traveling wave that vanishes at infinity. Similarly, the transmitted magnetic vector potential can be written as 00
A;(p,4» =
L
i-nbnJn(klP)ein.
(7.6)
n=-oo
To determine the unknown coefficients an and bn , the boundary conditions for the E z and H field continuities must be enforced at p = a. The field components are obtained by substituting A~(p, 4», A:(p, 4», and A;(p, 4» into Table 3.2 in Section 3.1. The continuity of the tangential electric field at p = a
7.1 Dielectric Circular Cylinder
149
•
reqUIres (7.8) . The continuity of the tangential magnetic field at p = a (7.9) yields
(7.10) where the symbol
I
denotes differentiation with respect to the argument
J' (k ) = dJn(kp) n a d(kp)
•
p=a
Solving (7.8) and (7.10) for an and bn yields
~Jn(k1a)J~(ka) lO1
J.ll
lO1
1
J~(k1a)Jn(ka)
J.l
J.l1
J~ (k 1a)H~l) (ka) -
~Jn (k 1 a)H~l)1 (ka) J.l
(7.11)
--(7.12)
where the Bessel function Wronskian
(7.13) is used to simplify bn . The scattered field in the far zone (kp asymptotic expression for H~l) (kp) as
»
1) is written in terms of an
00
E;
kp:oo
Eo
2: n=-ex:>
2 a
~kp
eikp+in(tj>-1f) -i1f / 4 n
.
(7.14)
150
7 Wave Scattering
When a circular cylinder is a perfect electric conductor (PEC, an incident wave cannot penetrate the cylinder. Thus, bn=O and
£)
-+ 00),
(7.15) The electric current density that is induced on the cylindrical surface at p = a • )s
-J=nxH -
p=a p=a
= i [H~(a, ¢) 2' -
-
" Z
+ H;(a, ¢)] in¢
00
t
""'
L....J
J.L1ra n=-oo
'-n e t --;::-;---
H~))(ka)'
(7.16)
7.1.2 TE Scattering
Assume that an incident TE wave impinges on a cylinder that is infinitely long in the z-direction. The incident TE wave has nonzero Hz, E p , and E¢ components. The analysis for TE scattering is somewhat similar to that of the previous TM scattering case. The incident electric vector potential is written as
F;(p, ¢) = e- ikx = e-ikpcos¢ 00
=
L
in en In(kp)e ¢ .
(7.17)
n=-oo
Similarly, the scattered and transmitted electric vector potentials are, respectively 00
F:(p, ¢) =
L
encnH~l)(kp)ein¢
(7.18)
i-ndnJn(klP)ein¢.
(7.19)
n=-oo 00
F;(p, ¢) =
L
n=-(X)
The continuity of the tangential magnetic field at p = a H~(a, ¢)
+ H:(a, ¢)
= H;(a, ¢)
(7.20)
7.1 Dielectric Circular Cylinder
151
yields (7.21) The continuity of the tangential electric field at p = a E~(a, ¢)
+ E;'(a, ¢) =
E~(a, ¢)
(7.22)
yields (7.23) where the symbol I denotes differentiation with respect to the argument. Solving (7.21) and (7.23) for C n and d n yields j), Cn
to
= j),l tOl
In(kla)J~(ka) -
1
J~(kla)HAl)(ka) _
j),l tOl
J~(kla)Jn(ka) (7.24)
j), In(kla)H~l)'(ka)
to
Note that the modal coefficients en and dn are a dual set of an and bn since en and dn can be obtained from (7.11) and (7.12) (an and bn ) by interchanging j), and to. 7.1.3 Electrostatic Fields Let us consider the field distribution due to a circular cylinder when the applied field is static (k = 0 and E~ = Eo). It is convenient to work with the electrostatic potential ¢e using Laplace's equation
\72¢e = 0
(7.26)
which is rewritten as
+
[P¢e 8¢2 = O.
(7.27)
152
7 Wave Scattering
y • • •• • • •• . .... . . ... • • • • • •
• • ••
....
• •
•
•
•
.
• ••• • ••••••
.
p==a
•
••
•
••
•
••
_+..~.!.. -';;':':':.1-X 1':.:. z ::.:.: r:.:.;-:. •
••
•
static E-field .. ....... . • • • • ••
• ••••••
.
Fig. 7.2. Cross section of circular cylindrical dielectric cylinder.
Note that there is no field variation in the z-direction
O¢e
oz
= 0 . Laplace's
equation is a special form of Poisson's equation in Section 1.6, where the electric charge is assumed to be zero (Pe = 0). A solution is available from the separation of variables technique, as was discussed in Section 3.4. Substituting the product form for ¢e (p, ¢)
¢e(P, ¢) = R(p) p(¢)
(7.28)
into Laplace's equation and dividing by R(p) p(¢) yields
2cPR (P) dR(p) R(p) P dp2 +p dp 1
1 cPp(¢) _ 0 + p(¢) d¢2 - .
(7.29)
'-
Thus
cP R(p) p dp2 2
dR(p) _ 2R() - 0 + P dp m p-
cP;~¢)
2 + m p(¢) = 0 .
(7.30)
(7.31)
Equation (7.30) is a type of Euler-Cauchy equation. Substituting the assumption
R(p) = pP
(7.32)
into (7.30) yields (7.33)
7.1 Dielectric Circular Cylinder
153
Therefore, p = ±m and
R(p)
= Apm + Bp-m .
(7.34)
Since p( ¢) is periodic in ¢ with 27r periodicity, m = 1, 2,3, .. " and the solution to (7.31) is
p(¢) = Ccosm¢+Dsinm¢
(7.35)
for every integer m > 1. The total potential for the region p > a consists of the applied and scattered potentials ¢~ (p, ¢) and ¢: (p, ¢), respectively, ¢~(p, ¢)
= -Boy = -Eopsin¢
(7.36)
00
L
¢:(p,¢) =
BmP-msinm¢.
(7.37)
m=l
The transmitted potential for p
~
a is
00
¢~(p, ¢) =
L
Ampm sin m¢ .
(7.38)
m=l
To determine the unknown modal coefficients Am and B m , the boundary conditions must be enforced. The continuity boundary conditions (7.39) (7.40) p=a
p=a
yield (7.41) (7.42) and all other Am = B m = 0 for m
~
2. Hence
Al = _ 2€Eo €l
BI =
€l €l
+€ €
+€
2
a Eo .
Therefore, the electric field can be obtained from
(7.43) (7.44)
154
7 Wave Scattering S
E = -\hP: = t
E = =
(psin a consists of the applied and scattered potentials ¢~(r,(),¢) and ¢:(r,(),¢), respectively, whereas the potential for the region r < a is given by the transmitted potential ¢~(r,(),¢). The continuity boundary conditions require ¢~ (a, (), ¢)
+ ¢: (a, (), ¢)
= ¢~ (a, (), ¢)
(7.90)
•
(7.91)
r=a
Since the applied potential is in the form of ¢~(r,(),¢)
= -Eox = -Eorsin()cos¢
(7.92)
the scattered potential takes the form of 00
¢:(r, (), ¢) =
L
n [Anr + Bnr-(nH)] P~(cos()) cos ¢ .
(7.93)
n=O
In view of the associated Legendre polynomials P~(cos()) = 0
(7.94)
pl (cos ()) =
sin ()
(7.95)
pi (cos ()) =
3 sin () cos ()
(7.96) (7.97)
the scattered potential can be written as
(7.98) Furthermore, since the scattered potential vanishes at infinity (r -t 00), Al = o and
7.3 Step in Parallel-Plate Waveguide
Similarly, the transmitted potential for r 00
¢;(r, (), ¢) =
L
161
:s a can be written as
n Cnr P~ (cos ()) cos ¢
n=O
(7.100) Substituting ¢~(r,(),¢), ¢:(r,(),¢), and ~(r,(), 0). Considering the contour which consists of r 1 , 2 , ... , 7 in the upper half-plane, the residue theorem gives
r,
r
f«()d(=
r
r
(7.160)
f«()d(=O.
The paths n and r3 denote infinitesimal semicircles and the path r6 denotes an infinitesimal circle. The path r 4 denotes an infinite semicircle. Note
f«() d( = I(k o)
(7.161)
(7.162) •
(7.163)
r.
f«()d( = 0
(7.164)
7.4 Slit in Conducting Plane
171
1m (~) branch cut
14
17
G
12
-k 0
16
-a m 11 am k o
Re
(~)
branch cut
Fig. 7.6. Complex (-plane.
rs
ra
f(() d( = K j
(7.165)
f(() d( = 0
(7.166)
(7.167)
Hence (7.168)
where K 1 and K 2 are integrations along the branch cut r 5 and n associated with the branch point ko. In order to evaluate K j and K 2 , consider KO = Jk~ - (2. Introducing the polar forms
(- leo
(7.169)
ei02 r2
(7.170)
= rje
( + ko = leads to
i1h
172
7 Wave Scattering
(7.171)
The corresponding phase diagram is illustrated in Fig. 7.7. The phase variations along the branch cut path AI, A 2 , A 3 , and A 4 are in Table 7.1, showing that the difference in 1/J at Al and A 4 (or at A 2 and A 3 ) is 1r. Therefore (7.172) (7.173)
1m (~) (infinity)A 4
Al (infinity)
k·-O I
Re k·-O I
Fig. 1.7. Phase diagram.
(~)
7.4 Slit in Conducting Plane
173
Table 7.1. Phase variations along branch cut
Al
A2 A3 A4
(h
(12
1r
1r
-2
1r
-2
31r
-2
0
31r
1r
2
1r
31r
0
-2 --
1/J
4 1r
--4
2
0
7.4.4 Thin Slit Within High-Frequency Limit The previous subsections performed a rigorous analysis of the scattering from a slit in a conducting plane. While a rigorous scattering analysis provides an exact solution, it requires further tedious computations to obtain the scattering characteristics. This subsection presents an alternative approximate solution to the problem based on Love's equivalence principle and image method. Consider an incident wave E~(x, y) that impinges on an infinitesimally thin slit (d = 0), as shown in Fig. 7.8 (a). Assume that the medium wavenumber is k( = wJJii.) throughout. Let the transmitted (x > 0) wave be denoted by -t
E (x,y). Based on Love's equivalence principle, the original problem (a) can be transformed into an equivalent problem (b). In problem (b), the surface im-t -t pressed currents J. = fi x H and M. = -fi x E are placed at x = 0, thereby -t -t radiating E and H for x > 0 and the null field for x < O. When a perfect electric conductor is placed at x = 0 beneath the impressed current in problem (c), the impressed electric current ]. becomes short-circuited on the PEC. The impressed current M. on the infinite PEC plane in problem (c) is equivalent to the impressed current -t
(7.174)
2M. = -2x x E (O,y) t
t
in a medium of infinite extent in problem (d), as far as the field E and H for x > 0 is concer ed. t To determine - 2x x E (0, y)], the tangential field continuities at x = 0 must be used. The continuity of the tangential electric field for E~(O, y)
+ E~(O, y) + E;(O, y)
= E;(O, y) .
When the operating frequency is high enough that ka
•
»
Iyl < a requires (7.175)
1, the condition
174
7 Wave Scattering
e· I
:zz:-- y
2a
e x
original problem (a)
-
null field
~===="-+Y 1+--+--+1 "-
n
2a
•
x
equivalent problem (b)
PEe /
2a x
equivalent problem (c)
?JJs ':------+y
1+--+--+1
2a
x
equivalent problem (d)
Fig. 1.8. Thin slit in conducting plane.
7.4 Slit in Conducting Plane
IE~(O,y)+E;(O,y)l«
applies for -a
E~(O,y)1
175
(7.176)
< y < a. Therefore E;(O,y) ~ E~(O,y)
(7.177)
•
2M s (0,y') = ii'2E(0,y') .
(7.178)
The electric vector potential F(p) due to 2 M s (0, y') in a medium of infinite extent is given in terms of the free-space Green's function in Section 1.5.2. The electric vector potential F(p) is thus written as a
F(p) =
E
.
2M s (0, yl)~H61) (kip - p' I) dy'
-a
(7.179) •
where ~ H6 ) (kip - P'I) is the two-dimensional free-space Green's function. Substituting the asymptotic form 1
'Ir~P exp [i (kp -
1
H6 )(klp - p'l) -+ for the far zone p » p' and kp
F(p)=yiE
»
ky' sinO -
~)]
(7.180)
1 into F(p) gives
2 eXP[i(kp_'lr)]Sin[k(sinO-sinOi)a]. 'lrkp 4 k(sin 0 - sin Oi)
(7.181)
.
Fy
Since y = f; sin () + 0cos (), the electric vector potential F(p) using cylindrical coordinates (p, (), z) is
F(p) = f;Fy sin() + OFy cosO. In the far zone kp -
1
,;p
»
1, the radiation field is of the
(7.182) 1
,;p order. Taking the
order terms from -t
1
-
E =--\7xF
(7.183)
E
gives the far-zone field t
EZ
~
1.
--zkFo E
= cosO
2 exp 'lrkp
[i (kp- ~)] sin[k~sinO -.sinO
i)
4
(sm 0 - sm Oi)
a] . (7.184)
176
7 Wave Scattering
7.5 Circular Aperture: Electrostatic Case The subject of wave penetration into a circular aperture in a thick conducting plane is important in electromagnetic interference problems. When the wave frequency is relatively low, low-frequency fields can approximately be described in terms of the static potentials. This section investigates electrostatic potential penetration into a circular aperture in a thick perfectly conducting plane, as shown in Fig. 7.9. A similar discussion is available in [10]. Regions (I) (z ? 0), (II) (-d ~ z ~ 0 and p ~ a), and (III) (z < -d) represent the upper half-space, circular aperture, and lower half-space, respectively. In region (I) (z > 0) the primary potential ~(p, z) is applied to a circular aperture with radius a and depth d in a thick perfectly conducting plane at a zero potential. The electrostatic potential 0) field for the thin slit (d = 0). 4. Derive (7.206).
8
Green's Functions: Fundamentals
8.1 Delta Function and Sturm-Liouville Equation The delta function is an important concept in dealing with point charges and currents. The Sturm-Liouville equation also plays an important role in the study of wave propagation and radiation. For instance, Bessel's equation is of the Sturm-Liouville equation type. This section introduces the fundamentals of the delta function and Sturm-Liouville equation, which are useful in radiation and scattering formulation. 8.1.1 Delta Function
The delta function c5(r - r') is defined as
J(r - r') = 0
v
J(r - r') dv = 1
when r
=I r'
when V contains r'.
(8.1) (8.2)
The delta function has the sifting property
v
f(r)J(r - r') dv = fer') .
(8.3)
Let us represent the delta function using rectangular, cylindrical, and spherical coordinates, respectively. •
Consider the delta function c5(r - r') where the source is located at the position x = x'. Then
J(r - r') = J(x - x') .
(8.4)
182
8 Green's Functions: Fundamentals Similarly, in two- and three-dimensional rectangular coordinates
•
8(1' - 1") = 8(x - x')8(y - y')
(8.5)
8(r - r') = 8(x - x')8(y - y')8(z - z') .
(8.6)
Consider the delta function 8(1'-r') where the source is located at the position (p', . = k . The Sturm-Liouville equation is given by
::2 + k
2
g(x; x') = -8(x - x')
(8.41)
subject to the radiation condition g(±oo; x') = o. The response g(x; x') at x is due to the delta source at x', as shown in Fig. 8.1. In the following, two different approaches are introduced to derive g(x; x').
s (x-x') --I
g(x;x')
I
•x
x = x' .
Fig. 8.1. Free-space Green's function.
8.2 One-Dimensional Green's Function
187
Approach 1
It is possible to represent g(Xj x') in terms of eigenfunctions. Since the domain is open (-00 < x < 00), g(x; x') is represented in the inverse Fourier transform 00
1 g(XjX ' ) = 21T
g«(j x')ei(z d( .
(8.42)
-00
Substituting (8.42) and the identity 00
J(X _ x') = 1
(8.43)
21T
-00
into (8.41) yields 1 ei«z-z')
00
g(X; x') =
-00
21T «(2 _ k 2 ) d( . , .,
(8.44)
.
f«() The physical condition requires that the delta source response g(Xj x') be an outgoing wave in the form of e±ikz that vanishes as x ---+ ±oo. This condition can be met if the medium is assumed to be slightly lossy, where the medium wavenumber k = k r + iki has an infinitesimally small positive imaginary part (k i > 0). For x-x' > 0, let us use the residue theorem in the complex (-plane, as shown in Fig. 8.2 (a). Performing contour integration along the path r 1 and r 2 yields
n Since
n
f «() d(
f«() d( +
---+ 0 and
r}
n
f«() d( = 21Ti Res f«()
(8.45)
(=k
f( () d( = g(Xj x') as R ---+ 00, •
Z eik(z-z') g( x', x') = 2k .
When x - x' so as to make
n
(8.46)
< 0, the semicircle in the lower half-plane is chosen for
n
f«() d( ---+ 0 as R ---+ 00, as shown in Fig. 8.2 (b). Contour
integration gives
•
I
'k(
Z
g(x; x) = 2k e-'
z-z
')
.
(8.47)
The one-dimensional free-space Green's function is, therefore, given by •
Z g(x; x') = 2k eiklz-z'l
.
(8.48)
188
8 Green's Functions: Fundamentals
k x
x
Re
(~)
-k
contour (a)
1m (~)
k x x
-k
R
contour (b)
Fig. 8.2. Complex (-plane.
Approach 2
The solution g(x; Xl) can also be obtained by another approach as follows: Consider the homogeneous differential equation
cP 2 dx 2 + k
I) 9 X; X = 0 (
(8.49)
when X f:. Xl. The Green's function that satisfies the radiation condition at X = ±oo is
8.2 One-Dimensional Green's Function ikz Ae
for x
189
> x'
g(x; x') =
(8.50) ilcx Be-
for x
< x'
.
The unknown constants A and B can be determined by the boundary conx' ± .11, where .11 denotes an infinitesimally small interval. ditions at x Integrating (8.41) from x = x' - .11 to x = x' + .11 yields
=
dg(x; x') dx
x'+A
x'+A 2
+
x'-A
k g(x; x') dx = -1 .
(8.51 )
x'-A
· dg(x; x') . di t' d ( ') . . I h b d SIDce dx IS scon IDUOUS an 9 x; x IS contlDuous at x ,t e oun ary conditions
dg(x; x') dx
x'+A
= -1
(8.52)
= 0
(8.53)
x'-A x'+A
g(x; x')
x'-A
•
glVe
ik
ikz (Ae '
+
ikz Be')
= -1
(8.54) (8.55)
Solving (8.54) and (8.55) for A and B finally gives •
~ eilc(x-x')
2k
g(x; x') =
for x> x'
•
~ e -ilc(x-x')
for x
< x'
2k •
- 2~k eilclx-x'i .
(8.56)
8.2.2 Half Space
Let us investigate the Sturm-Liouville equation in one-dimensional half space 2 with the parameters p(x) = r(x) = 1, q(x) = 0, and A = k . These types of problems are often encountered in scattering and radiation problems dealing with perfectly conducting boundaries. Consider the half-space Sturm-Liouville equation
:::2 + k
2
g(x; x') = -8(x - x')
(8.57)
for x > 0 subject to the boundary conditions g(O; x') = 0 and g(oo; x') = 0, as shown in Fig. 8.3. Let us introduce four different approaches to obtain g(x; x').
190
8 Green's Functions: Fundamentals
o(X-X')
g (x; X ')
X =X '
X=o
Fig. 8.3. Half-space Green's function.
Approach 1
Due to reflection from the boundary at x = 0, g(X; x') takes a standing wave sin kx for 0 < x < x' and a traveling wave e ikx for x > x'. Assume
Ae
ikx
for x
> x'
g(x; x') =
(8.58)
B sin kx
for 0 -< x
< x'
which satisfies the boundary conditions g( 00; x') = 0 and g(O; x') = O. Integrating (8.57) gives
x'+.a
= -1
(8.59)
x'-.a (8.60) Solving (8.59) and (8.60) for A and B yields
A = sin kx ' k
(8.61 )
eikx ' B = --:-k .
(8.62)
Hence, the half-space Green's function is
. k I sm x
k
ikx
e
for
x> x' (8.63)
e
ikx'
k
sin kx
for 0
< x < x'
.
8.2 One-Dimensional Green's Function
191
Approach 2
It is also possible to obtain g(x; x') using eigenfunction expansions. Since the domain is open for x > 0 and g(O; x') = 0, it is convenient to represent g(x; x') in terms of the Fourier sine transform. Substituting 00
2 g(x; x') = 11'
J(x - x') =
(8.64)
sin (x sin (x' d( .
(8.65)
0 00
~ 11'
g((; x') sin (xd(
0
into (8.57) yields
_(c. ') = sin (x' 9 ,x (2 _ k2 .
(8.66)
Therefore
g(x; x') = 2
o
11'
1
00
-:;
-
sin (x sin (x' d( (2 - k 2
00
-00
sin (x sin (x' (2 _ k2 d(
1
00
411'
-00
ei(x+x') _ ei(x-x') _ ei( -x+x')
+ ei(
(2 _ k2
-x-x')
d(. (8.67)
In view of the free-space Green's function representations (8.44) and (8.48), g(x; x') is written as •
t
g(x; x') = - 4k
(eiklx+x'i _ eiklx-x'i _ eikl-x+x'l
. k x ' ikx sm k e
+ eikl-x-x/l)
for x> x' (8.68)
ikx' e
k sin kx
for 0
< x < x' .
Approach 3
Let us show an approach based on the image method, as shown in Fig. 8.4, where the image source -J(x + x') is added to satisfy the boundary condition g(O; x') = O. Since both J(x - x') and -J(x + x') are assumed to be in free
192
8 Green's Functions: Fundamentals I I
I
-0 (x+x')
0 (x-x')
g (x; x')
-Ir---I;.....---I------I I
.
x
x =x'
I
I I I
X
=0
Fig. 8.4. Half-space Green's function based on image theorem.
space, the response is the sum of two free-space Green's functions •
z g(x; x') = 2k (eiklx-xll - eiklx+xll)
. k I sm x ikx k e
for x> x'
eikx' ----:-- sin kx k
(8.69)
for 0
< x < x' .
Approach 4
A solution to (8.57) consists ofthe incident [gi(X; x')] and scattered [g8 (x; x')] Green's functions. The incident Green's function is a response to a delta source in free space and the scattered one results from reflection from the boundary at x = O. The incident term is the one-dimensional free-space Green's function and the scattered one is given in terms of the reflected waves as •
'kl 'I g'(x; x') = 2k e' x-x .
Z
(8.70) (8.71)
Since
(8.72) the coefficient A is given by (8.73)
8.2 One-Dimensional Green's Function
193
Therefore
g(x; x') = gi(X; x')
+ g8(X; x')
. k I sm x ikz k e
for x> x' (8.74)
eikz ' k sin kx
for 0
< x < x' .
8.2.3 Closed Space
This subsection investigates the Sturm-Liouville equation in one-dimensional closed space. Consider cf2
dx 2
+k
2
g(x; x') = -')
= _ J(p - p') J(¢> _ ¢>') .
(8.95)
p
y
-p - -p'
-
~---J'--
-
p
..
X
Fig. 8.6. Two-dimensional free-space Green's function.
8.3 Two-Dimensional Green's Function
197
The problem geometry is shown in Fig. 8.6. Let us find a response at p due to a delta source at (l using three different approaches. 8.3.1 Approach 1
Based on the separation of variables technique, the Green's function and delta function are given by 00
9(P, ¢; p', ¢') =
L m=-oo 00
o(¢-¢') =
L
m=-oo
im 9m (p; p', ¢')e ¢
(8.96)
1 e im (¢-¢') . 21T
(8.97)
Substituting 9(P, ¢; p', ¢') and o( ¢ - ¢') into (8.95) gives 1 d
d
pdp
dp
-- P
2
_ m 2 p
k2
+
i:(
')
-im¢'
-1,') __ u p - p e () 9m P,P,'I' - 2 - ' 8.98 p 1T (.,
Let us use an approach based on the cylindrical wave propagation characteristics, as shown in Section 3.4. When p:j:. p', (8.98) reduces to Bessel's equation where its solutions 9m(P; p', ¢') are the Bessel and Hankel functions for p
., ,
< p' (8.99)
9m(P; P , ¢ ) = BmHg) (kp)
for p> p' .
Multiplying (8.98) by p and integrating from p = p' - Ll to p = p'
d9m (p; p' , ¢') p dp e
. ..., -,mY"
-----,--
21T
p' +L1
p'+L1
+ p'-L1
p'-L1
2
m -;;"'2 P
+
k2
+ Ll yields
9m(P; p', ¢')p dp
.
(8.100)
Since d9m (Pi p', ¢') and 9m (p; p', ¢') are discontinuous and continuous at p = dp p', respectively, the boundary conditions are written as .
...'
-1m",
p' - AmkJ:n(kp')
+ BmkHgl' (kp')
= - e 21T
(8.101) (8.102)
, ( ') dJm(kp) where Jm kp = d(kp)
. Solving (8.101) and (8.102) for Am yields p=p'
198
8 Green's Functions: Fundamentals
Since
-J:n (kp')H~) (kp')
+ J m (kp')H~)' (kp')
. 2 = ~ 'Trkp'
(8.104)
the coefficients Am and B m as well as the Green's function gm (Pi p', ¢') are given by •
im Am = ~ e- 4>' H~) (kp')
(8.105)
•
im B m = :"e- 4>' Jm(kp')
(8.106)
4
for p
•
gm (p; P" ,¢) = 4"~
< p' (8.107)
for p> p' . Therefore
g(p, ¢; p', ¢') 00
L
=
im gm (p; p', ¢')e 4>
m=-oo 00
L •
-
~
H~)(kp')Jm(kp)eim(4>-4>')
for p
< p'
m=-oo
-
4
(8.108) 00
L
Jm(kp')H~) (kp)e im (4>-4>')
for p
> p' .
m=-oo
By the summation theorem for the Hankel function [11, page 979], the twodimensional free-space Green's function is written as •
g(p,¢;p',¢') = ~H~l) (kip-pI) .
(8.109)
Let us rewrite g(p, ¢; p', ¢') in terms of the superposition of plane waves in rectangular coordinates. Rewriting (8.94) in rectangular coordinates (x, y) • glVes
8.3 Two-Dimensional Green's Function
8
2
8
2
8x 2 + 8 2 + k y
2
l g(x, y; Xl, yl) = -
It is also possible to derive (8.107) based on the residue calculus and Hankel transform approach. Substituting the Hankel transform representations 00
gm (p; p',
The Helmholtz equation (8.144) is explicitly rewritten using spherical coordinates as 1 a
r2
ar
2ag r -=a=-r
a
1
. ()ag
sm
+ r 2 sin () a()
o(r - r')o(() - ()')o( ¢J - ¢J') r 2 sin ()
ag 2
1
2_
a() + r 2SID ' 2 () a¢J2 + k g -
(8.162)
•
It is expedient to represent the Green's function and delta function using spherical harmonics P:;'(cos())e im¢, as shown in Appendix E. The Green's
function g and the delta function
o(() - ()')o(¢J - ¢J')
00
g(r,(),¢J;r',()',¢J') =
. () sm
n
L L
are written as
im gnm(r;r',()',¢J')P:;'(cos())e ¢ (8.163)
n=Om=-n and
o(() - ()')o(¢J - ¢J') _ sin () 00
n
L L
im AnmP::' (cos ()')P::' (cos ())e (¢-¢')
(8.164)
= 2n + 1 (n - m)! . nm 47l' (n + m)!
(8.165)
n=Om=-n where
A .
.
Substltutmg g(r, (), ¢J; r', ()', ¢J') and
o(() - ()')o(¢J - ¢J') . . ()
sm
.
mto (8.162) gives
208
8 Green's Functions: Fundamentals
cPgnm
2 dg nm
+r
-d-::-r-'-;;:2~
+
dr
gnm =
. ..' ~(r - r') m -A nm pm n (cos ()')e-· ", 2 r
Introducing gnm
•
(8.166)
= krg nm gives cPgnm dr 2
+
k2
-
n(n + 1) gnm = 2 r A
.m..' ~(r - r') -AnmP::'(cosB')e-· ", k--'------'- . r
(8.167)
The solution to (8.167) for r f:. r' is the spherical Bessel functions and spherical Hankel functions. Let us assume A
BJn(kr) gnm =
for r
< r' (8.168)
CfI~I)(kr)
for r
> r'
where In(kr) and fI~I)(kr) are the spherical Bessel function of the first kind and the spherical Hankel function of the first kind, respectively. Multiplying (8.167) by r and integrating from r' - L\ to r' + L\ gives
dg nm r dr
r ' +.6
= -AnmP::'(cosB')e-imtfJ' k
(8.169)
= 0
(8.170)
r ' -.6
r ' +.6
gnm
r ' -.6
which are then rewritten as
r' [cfI~I)' (kr') - BJ~(kr')] = -AnmP::'(Cos()')e-imtfJ' CfI~I)(kr') - BJn(kr') = 0 (I) dHn (kr)
(8.171) (8.172)
A
A
(1)'
,
_
where H n (kr) -
d(kr)
. Solving (8.171) and (8.172) for B and C r=r'
•
gIVes
B =
imtfJ' -A nm iI(1)(kr')pm(cos()')en n
r' [In(kr')fI~I)' (kr') - J~(kr')fI~I)(krl)] •
imtfJ fI(1) (kr')pm(cos B')e- '
(8.173)
C = ';AnmJn(kr')P::,(cosB')e-imtfJ' . r
(8.174)
= ~A I nm
r
n
n
•
8.4 Three Dimensional Green's Function
209
Substituting B and C into (8.168) gives •
g(r,B,¢;r',B',¢') = k:r'
00
n
l:: l::
n=Qm=-n AnmP;:' (cos B)P;:' (cos B')e im (t/>-t/>')
.
(8.175)
The Bessel function summation theorem in [11, page 980] shows that (8.175) reduces to (8.151). 8.4.3 Approach 3 The Helmholtz equation (8.144) is rewritten using rectangular coordinates as 2
2
2
8 8 8 2 8x 2 + 8 y2 + 8z 2 + k
9
(
"')
x, y, z; x , y , z =
-o(x - x')o(y - y')o(z - z') .
(8.176)
Substituting the three-dimensional Fourier transform representations
g(x,y,z;x',y',z') = 00
-IX)
00
00
-<Xl
-00
g((, TJ, ~)ei(("'-"")+i1J(Y-Y')+ie(z-z') d( dry ~
(8.177)
o(x - x')o(y - y')o(z - z') = 00
00
-00
-00
00
-(X)
(8.178) into (8.176) results in (8.179)
210
8 Green's Functions: Fundamentals
g(X,y,z;xl,yl,Z') = 00
1
00
00
e i«(x-x')+i1)(y-y')+i{ (%_%') 1"2
(21r )3
-00
-00
-00
..
t2
2
+ TJ + '" -
d( dTJ df.
k2
. (8.180)
Let
R = ((x - x') _
A
+ fJ(y -
y')
+ t(Z -
Zl)
(8.181)
A
K = (( + fJTJ
+ ~~ .
(8.182)
Then
exp(iK.R) 9 ( x,y,z;x ,y ,z = (21r)3 V K2 _ k 2 dv. I
1
')
I
Let us introduce rectangular coordinates ((', TJ ' , ~') whose with the vector R, as shown in Fig. 8.11. Therefore
() 8.183
e-axis coincides
(' = K sin Bcos
(8.184)
TJ' = K sin Bsin
(8.185)
(=KcosB.
(8.186)
Using spherical coordinates (K, B, p'.
Multiplying (9.19) by p and integrating from p = pi - ..1 to P = pi
dA ( . ') p zP,P dp
pi +L1
pi +L1
+
p' -L1
pi -L1
+ ..1 yields
I 2 k A z (p;pl)pdp=-J-L.
(9.21)
21l"
· (p; pi) an d A z (p; P') are discant'muous and ' . I SIDce dAz dp contmuous, respective y, at p = pi,
pdAz(p; pi) dp
p'+L1 pi _ L1
[-AkJ~(kpl) + BkHd1)1 (kpl)]
= pi
p'+L1
Az(p;p')
= _1.L1
(9.22)
21l"
.
=-AJo(kpl)+BHdl)(kp') =0.
(9.23)
p'-L1
Solving (9.22) and (9.23) for A and B, and substituting A and B into Az(p; pi) • gIVes
1 Hd )(kp')Jo(kp)
')
A z (p; P = iJ-LI 4
1 JO(kp')Hd )(kp)
for p
< pi (9.24)
for p> pi .
Approach 3
In this approach, A z (p; pi) is derived starting from a transform approach. It is expedient to express Az(p; pi) based on the Hankel transform representation 00
(9.25)
o 00
Az «(; pi)
Az(p; p')Jn«(p)pdp .
=
(9.26)
o
Since A z (p; pi) has no azimuthal variation in ¢, the parameter n = 0 is chosen in view of Bessel's equation. Hence 00
Az(p; p') = J(p - pi) _ P
-
(9.27)
o 00
o
Jo«(p) Jo «(p')( d( .
(9.28)
218
9 Green's Functions: Applications
Substituting Az(Pi p') and 6(p - p') into (9.19) and noting P 1 d
dJo((p) P dp p dp
(9.29)
= _(2 Jo((p)
yields (9.30)
Let us first evaluate Az(Pi pi) when p < p'. When Az((j p') and
JO((p') =
~ [H~I)((P') + H~2)((pl)]
(9.31)
are substituted into (9.27), Az(PiP') is written as
( z A PiP
')
J-LI = 27r +J-LI 27r
Since H~I)(-u)
= -H~2)(u) A z (Pi p') =
00
Jo((p)H~I)((pl)(d( 2((2 - P)
0 00
Jo((p)H~2)((pl)(d(
0
and Jo(-u)
2((2 - k 2 )
•
(9.32)
= Jo(u),
~~
(9.33)
Contour integration is performed in the complex (-plane, as shown in Fig. 8.7. Note that the integral evaluation along the path r 2 vanishes since P < p'. The result is
') iJ-LI (1)( ') ( ) ( A z Pi P = 4 H o kp J o kp . Similarly, when p
(9.34)
> pi, A z (Pi P') = iJ-LI 4 H o(1)( kp ) J o( kp') .
(9.35)
Note that the solution Az(PiP') is identical with (9.17).
9.2 Line Current in Rectangular Waveguide This section investigates three cases dealing with radiation from line currents in a rectangular waveguide. The first is radiation in a parallel-plate waveguide, the second is radiation in a shorted parallel-plate waveguide, and the third is radiation in a rectangular waveguide.
9.2 Line Current in Rectangular Waveguide
219
9.2.1 Radiation in Parallel-Plate Waveguide
Let us consider electromagnetic radiation from an infinitely long electric line current J(x', y') within a conducting parallel-plate waveguide. The geometry of the radiating problem is shown in Fig. 9.3. Two different approaches are presented to analyze the problem. Approach 1
The Helmholtz equation that governs radiation from J(x', y') - iJJ(x x')J(y - y') is (9.36)
Using rectangular coordinates (x, y), it can be rewritten as 2
8 8x 2
+
2
8 8 2 y
+k
2
Az(x, y; x', y') = -j.JJJ(x - x')J(y - y').
(9.37)
The boundary condition requires
Ez(x,y;x',y')
x=O,a
=iwAz(x,y;x',y')
x=O,a
=0.
(9.38)
Let us choose eigenfunctions
2 . m7r -sm x a a
(9.39)
and expand A z (x, y; x' ,y') in terms of "pm (x) as 00
Az(x,y;x',y') =
L
am(y;x',y')"pm(x) .
m=l
x
PEe
a
®
J (x', y')
z
Fig. 9.3. Line current J in PEC parallel-plate waveguide.
(9.40)
220
9 Green's Functions: Applications
The delta function o(x - x') is also written as 00
o(x - x') =
L
1/Jm(x)1/Jm(X' ) .
(9.41)
m=l
Substituting A z (x, y; x', y') and o(x - x') into (9.37) yields
dcf22 y
2 (m1l")2 k a + am (y;x,y I
-
')
I = -j.LJ1/Jm ( x ') o(y - y) .
(9.42)
The solution to (9.42) is the following one-dimensional free-space Green's function
ij.LJ1/Jm (x') (. I 'I) am y; x , y = 2(m exp ~(m Y - Y I
(
where
em =
k
2
-
')
(9.43)
m1l") 2 (a .
The vector potential and electric field inside the waveguide are (9.44)
E = zEz = ziwAz(x, y; x', y') .
(9.45)
Approach 2
It is also possible to obtain A z (x, y; x', y') using different expressions for the Green's function. The second approach uses the continuous mode representations for Az(x,y;x',y') and o(y - y') based on the Fourier transform along the y-direction. Let 00
(
I
A z x, y; x , y
')
1 = 21l"
Azei(y d(
(9.46)
ei«(y-y') d( .
(9.47)
-00
1 o(y - y') = 21l"
00
-00
Substituting A z (x, y; x', y') and o(y - y') into (9.37) gives
cf2
dx 2
Jk
+K
2
-
"( ,
(9.48)
A z = -j.LJo(x - x')e-' y
2 - (2. Since A vanishes at x = 0 and a, where K = z one-dimensional closed-space Green's function
Az
is given by the
9.2 Line Current in Rectangular Waveguide l sin K,(x - a) .
. K, sm K,a
sm K,X
for 0
221
< x < Xl (9.49)
I
sin K,X . sin K,(x - a) K,sm Ka Let us rewrite A z (x, y; Xl, yl) for x - Xl
>0
for Xl
<X O. In view of Fig. 9.4, contour integration is performed to obtain
«(
Jp -
00
AZ(X,y;xl,yl)
+
r2
f(() d( = 21ri
L
> 0 and x -
Note that the function f«() is bounded when y - yl
f(() d( ---+ 0 as R ---+
(9.51)
Res f«() m=l (-('" Xl
> O. Since
AAx, y; Xl, yl) is given as follows:
00,
r2 00
Az(x,y;xl,yl) = 21ri
L
Res f(()
m=l
(=(",
00
= 21ri
I
L
Res
m=l
(=(",
jJ-J sin K,(x - a) sin K,X i«(y-y') e 21rK, sin K,a I
sin K,(x - a) sin K,X i«(y-y') d(sin Ka) e K, d( 00
= -ijJ-J
L
m=l
sin K,(x - a) sin K,X
I
----''--:-:---e
(=(",
i«(y-y')
-a(cosKa
(=(m
(9.52) Similarly, it is possible to show that Az(x, y; Xl, yl) for X - Xl (9.52) . Equation (9.52) agrees with (9.44) for y - yl > O.
< 0 reduces
to
222
9 Green's Functions: Applications
1m
(~)
~m
x··· -.:"':."':"'.vx-;x~x:""""':::x~I--~~r"':l~:""--+l..--+Re X
-
X X
(~)
-~ m x • • •
Fig. 9.4. Complex (-plane with simple poles.
For y - y' < 0, A z (x, Yi x', y') is also shown to be
9.2.2 Radiation in Shorted Parallel-Plate Waveguide
Let us consider radiation when a perfect conducting plane is placed at y = 0 in a parallel-plate waveguide. The problem geometry is shown in Fig. 9.5. The Helmholtz equation that governs radiation from J(x', y') = zJo(x-x')o(y-y') •
IS
x
®J(x',y')
Fig. 9.5.
Line current J
a
in shorted parallel-plate waveguide.
9.2 Line Current in Rectangular Waveguide
(\7 + k 2
2
/ A(x, Yi x', y') = - p.J(x , y') .
)
223
(9.54)
The boundary conditions require
AZ(X,y;x/,y')
= Az(X,YiX/,y') :z:=O,a
=0.
(9.55)
y=O
Approach 1
Following the previous procedure, the solution can be written as 00
Az(x,y;x/,y') =
L
am(y;x/,Y')'l/Jm(x)
(9.56)
m=l
. m7r x. Th en, am ( . fy am (0i x I ,Y') = 0 -2 sm Yi " x ,) y must sabs
where 'l/Jm(x) = and for y
a
>0
a
~2 - (r: f 7r
+k
2
am(Yjx',y') = -f..LJ'l/Jm(xl)J(y_y').
(9.57)
The solution to (9.57) is the one-dimensional half-space Green's function that satisfies the radiation condition at y = 00 I
')
am ( Yi x, Y =
k2
where (m =
f..LJ'l/Jm(x ' )
sin (mY I ei(m y
for y'
p'
where (9.75) The choice (9.74) automatically satisfies the boundary condition at p = a since ilm(ka) = O. It is necessary to impose another boundary condition near p = p'. Multiplying (9.73) by p and integrating from p = pi - L1 to P = p' + L1 yields
pdA~(p;p') dp Since
dA~(p;
p')
p'+L1
+
p'_L1
and
/+L1
m
2
P
p'_L1
A~(p; p')
2
+
k2
Am(p; p')p dp = z
1 2~
. (9.76)
are discontinuous and continuous, respec-
dp tively, at p = p', the boundary conditions are kp' [-Am J:" (kp')
+ Bmn~(kp')]
= -
2~
(9.77)
AmJm(kp') = Bmilm(kp') .
(9.78)
Solving (9.77) and (9.78) for Am and B m gives
Jm(kp') m B = 2dp' [nm(kp')J:"(kp') - Jm(kp')n:"(kp')] _ _ ---,,-J-..:..:m::;(-,:kP:-'~)----:--:-:---,---,,..,---;77 - 2~kp'Jm(ka) [Jm(kp')N:"(kp') - Nm(kP')J:"(kp')] ,
y
#
2 ~kp'
Jm(kp') 4Jm (ka) Am = nm(kp') . 4Jm (ka) Substituting Am and B m into (9.74) gives
(9.79) (9.80)
228
9 Green's Functions: Applications 00
2:
A z (p, ¢>; p', ¢>') = J1-:
m=-oo
[} (k ') Jm(kp) im(¢-¢') m p Jm(ka) e
for p
< p' (9.81 )
[}m (kp/m (kp')
eim (¢-¢/)
for p > p' .
Jm(ka) The electric field is, therefore
E = zEz = ziwAAp, ¢>; p' ,¢>') .
(9.82)
9.3.2 Approach 2
A solution to (9.69) consists ofthe incident (A~) and scattered (A:) potentials, where the incident term is a response to a delta source in free space and the scattered term is due to the presence of a boundary at p = a. The incident potential is the two-dimensional free-space Green's function and the scattered potential is given in terms of Bessel functions of the first kind. Hence
iJ1-J ~ A z p, ¢>; p , ¢> = 4 LJ i (
")
m=-oo
(9.83)
Jm(kp')Hg) (kp)e im (¢-¢')
A~(p,¢>;p',¢>')
=
i~J f:
for p > p'
im AmJm(kp)e (¢-¢/).
(9.84)
m=-oo
Therefore, the total potential A z (p, ¢>; p', ¢>') is
= A~ (p, ¢>; p' ,¢>')
+ A~ (p, ¢>; p', ¢>')
im [Hg) (kp')Jm(kp) + AmJm(kp)] e (¢-¢')
for p < p' (9.85)
im [Jm(kp')Hg)(kp) + AmJm(kP)] e (¢-¢')
for p> p'.
9.4 Sheet Current in Parallel-Plate Waveguide
229
To determine the unknown coefficient Am, the boundary condition must be utilized. The boundary condition at a PEC waveguide surface requires A z (p, x' are given by
p7r sm. (p7r) ['( ( ')] z expt"x-x a
E z = - Jo (" cos 2WE
(9.106)
a
(p7ra z) exp [i(,,(x -
x')]
(9.107) (9.108) (9.109)
232
9 Green's Functions: Applications
The radiated power delivery along the x-direction for the real (p is 1 (-.) s 2 Re E x H .
Prad =
as
a
1
= - Re 2
o (9.110)
The power supplied by the current source J is 1
Pi =-2 v
-E· -. J dv
(9.111) The radiated power delivery along the ±x-direction is 2Prad , and the power conservation can be checked through the relation 2Prad = Pi.
Problems for Chapter 9 1. A constant magnetic current sheet M = 2M r5 (x) is placed in free space. The geometry of the radiating problem is shown in Fig. 9.10. Evaluate its radiation field for x > 0 and x < O.
z
x
Fig. 9.10. Current sheet in free space.
Problems for Chapter 9
233
2. Prove (9.35) using residue calculus. 3. Consider an electric line current J(x', y') = ic5(x - x')c5(y - y') within a PMC parallel-plate waveguide, as shown in Fig. 9.11. Evaluate its radiation field. x PMC
a
® J(x',y')
Fig. 9.11. Line current J in PMC parallel-plate waveguide
10
Antenna Radiation
10.1 Antenna Fundamentals Antennas have been used as electromagnetic radiating devices in wireless communications, radars, and nondestructive testing. They are regarded as energy transducers that transform an electric current into an electromagnetic wave for the transmitting antenna and vice versa for the receiving antenna. The macroscopic relation between an electromagnetic wave and an electric current can be described in terms of Maxwell's equations. Thus, to understand the characteristics of the radiation from antennas, Maxwell's equations must be solved. Let us first consider the radiation from an antenna that is modeled as the current sources J and M in free space. This is a common approach to model antennas. The problem geometry is shown in Fig. 10.1. The radiation
z
v'
-r - -r'
••
... .
•••
-r' ~----------y
x
Fig. 10.1. Vector potential and current density in free spa{;e.
236
10 Antenna Radiation
field is governed by the Helmholtz equations for the vector potentials A and Fas (10.1) (10.2) where k ( = w.jjii) is the wavenumber of the medium. The free-space solutions in three dimensions are
v' -
€
F(r) = 4'IT
v'
_ eik1r-r' 1 J(r ' ) I- - I I dv'
(10.3)
iklr-r/l ') e , ( M r 1_ -, I dv
(10.4)
r - r
r - r
where r and r' designate the observation and source points, respectively. Let us consider radiation in a lossless medium where J.t and € are real. It is of practical interest to evaluate the radiation field in the far zone kr» 1. The field in the far zone is called the far field and decays as
»
the far-zone approximation using r
Ir -
r'
~. Let us consider r
r' I = v(r - r') . (r - r') -
-
2rr' cos"l/J ~
r - r' cos"l/J .
(10.5)
It is convenient to represent the far field using spherical coordinates (r, (), ¢). Substituting (10.5) into A(r) and F(r) gives
_ A(r)
~
J.te ikr '-:4'lTr
J(r')e- ikr' cos'" dv' ,
v'
•
function of () and ¢ ~
= fAr
+ ()AII + ¢A:.. ::J :::>>:1/ ~"• 1""1' 1 » • .. ...'. . . ~"""""""""'" ················m •
•
•
•
•
•
•
•
•
•
•
•
•
•
..,...,..,. rrr rrrr ,..--.... X
'l'h0
d
PEC
•
~
~//////.
;ffi
.4---- 2a - - - _ " Fig. 10.5. Rectangular groove in perfectly conducting plane.
upward. Regions (I) and (II) denote the groove interior (-a < x < a and -d < y < 0) and half space (y > 0) above a conducting plane, respectively. The magnetic vector potential in region (I) satisfies (10.63)
where 0(') is the delta function and k1 (= W..jJ.l.1Ed is the wavenumber in region (I). The relation (10.64)
results in
(\7 + kn E; (x, y; x', y') 2
= -iwJ.l.do(x - x')o(y - y') .
(10.65)
Note that the radiated electric field only has a z-component and there is no field variation in
z :z = a . The following are two different approaches to
derive the solution to (10.65). 10.4.1 Approach 1
It is convenient to separate the original problem into two parts based on the superposition principle, as shown in Fig. 10.6. The electric field in region (I) consists of the primary and scattered components as 1 ( Ez
, ') = E" . I ' ) + E8z (x,Yi x I ,Y') x,y;x,y z (x,y,x,y
(10.66)
where the primary field Ef (x, y; x', y') is considered as the Green's function resulting from a line current placed in a PEC rectangular box. An analysis
10.4 Groove-Backed Antenna
E;
247
region (II)
-------~
~ region (I)
Wff.,.0;-
E~+E;
II
E;
region (II)
+
Fig. 10.6. Equivalent problem. l for the primary field E~ (x, Yi Xl, y ) is available in Section 9.2.3. The primary
electric field is represented in terms of the rectangular waveguide modes as p ( .' ') E zX'y,x,y
__ . -
ZWj.Ll
J ~ ~
Wm (x) Wm (Xl) (:
0), respectively, and their respective wavenumbers are k j = WJJ-LEj for j = 1 and 2. Let us assume a TEM wave is incident from below a flanged coaxial line as .
eikl Z
H'(p z) - -
'
-
P
(10.183)
(10.184)
where TJ1 =
.!!:... Since the incident wave has no variation in the azimuthal E1
iK.Z
fI(O e 2 H~l) (Cp)C dC .
(10.229)
-00
To evaluate the far field in region (II) at observation point (r, 0) using spherical coordinates, the saddle-point method can be used. Assume that z = r cos 0, p = r sin 0, C= ~ sin 'l/J, and K = k2 cos 'l/J. Since
31r 2 . ,. ,. exp z "p 4 1r"p
(10.230)
the far field in region (II) is 00
-
H(O -00
1
21rCp
exp ik2 r cos( 'l/J - 0) - i 3; CdC (10.231 )
Problems for Chapter 10
275
Problems for Chapter 10 1. Derive the far-zone expressions (10.11) and (10.13). 2. The magnetic Hertzian dipole has an infinitesimally small magnetic current M = zc5(x - x')c5(y - y')c5(z - z') in free space. Evaluate its radiation field in the far zone. 3. Consider the integral representation (10.111). Using residue calculus and contour integration, show that (10.111) is given by (10.112). 4. Consider TM wave radiation from an N number of slits of width 2a and depth d in a parallel-plate waveguide. The problem geometry is shown in Fig. 10.12. Assume that an incident wave H;(x,z) excites region (1). Throughout regions (1), (II), and (III), the magnetic field has a y-component. There is no field variation with respect to y. Derive the radiation field H;(x, z).
x
PEe
H;VV-
region (I)
incident wave
-a-lT
--.J\.fV II':.
x=-b
Y
region (II)
p
es
I-
I-
2a
-I T
a+lT
-I
region (III)
Fig. 10.12. Slit parallel-plate waveguide
11
Radiation Above Half Space
Antenna radiation above the earth's surface is an important subject in radio wave communications. The inhomogeneous earth terrain can be approximately modeled as a homogeneous half space. The boundary-value problem of radiation from a dipole antenna above a dielectric half space was first investigated by Arnold Sommerfeld and it thus became known as the Sommerfeld dipole problem. Electromagnetic wave radiation above a dielectric half space has been studied in [2, pp. 447-485] and [18]. This chapter investigates radiation from line currents and dipole sources in a half space, using a Fourier transform in the spectral domain. Some relevant discussion is available in [18].
11.1 Electric Line Source Consider electromagnetic radiation from a horizontal electric line source J=yJ(x)J(z - Zl) that is located at (0, z') in a half space. The geometry of the radiation problem is shown in Fig. 11.1. The magnetic vector potentials in the half space satisfy the Helmholtz equations for z
>0
(11.1)
for z
0 and z < 0, respectively. No field variation is assumed with respect to y, such as
8 ~ =0 . 8y Since the problem domain in the x-axis is unbounded, it is expedient to express the vector potentials and delta function in terms of the one-dimensional inverse Fourier transform
278
11 Radiation Above Half Space
z
x
o(X) o(Z-z')
y ·................................ x. · .. . ··... ...... ........ ..... .......... ......•. ........ .... . -.. . ... • •
•
• •
•
• • ••• ••• • • • ••• • •• • •
• ••
•
• • • ••
. • •••••••••••••••••••••••••••••••••••••• . . . ... .. . ..
x
Fig. 11.1. Horizontal electric line source in half space.
1 AO,I(X z) = y' 271"
00
A~,l ((,
z)ei(x d(
(11.3)
-00 00
1 J(x) = 271"
ei(x d( .
(11.4)
-00
Substituting A~,l (x, z) and J(x) into (11.1) and (11.2) gives, respectively,
Ag((, z) -
A~((,
Jk3 -
= -J..LoJ(z - z')
(11.5)
z) = 0
(11.6)
Jk; -
where KO = (2 and KI = (2. The solution to (11.5) consists of the incident and scattered potentials. The incident potential is a response to the delta source in free space, while the scattered potential is due to reflection from the boundary at z = O. A graphical representation of the incident [At((,z)] and scattered [A~((,z)] potentials is shown in Fig. 11.2. The total potential Ag((, z) is, therefore, the sum of the two components
-
-
Ag((,z) = At((,z)
-
+A~((,z)
•
= tJ..Lo exp (iKolz - z'l) 2Ko
+ B exp(iKQz)
.
(11. 7)
Note that the choice exp(iKoz) for the scattered potential satisfies the radiation condition, which requires that the field vanish at z = 00. The solution to (11.6) is a wave that propagates in the -z-direction
11.1 Electric Line Source
279
z
Aiy
x 8(x)8(z-z')
y ................................. • • • • • • • • • • • • • • • •• • • • • • • • • • • • • • • •
"' ..."7 .."' ..."' ..."7 .."' ..."' ..."7 .."' ..."' ..."7 .."' ..."' ..."7 .."'.----to • •••••••••••••••••••••••••••••••••••••
·...... . ................ ... .... .... . , . • • •.• • • • • • • • • • • • • • • • • • • • • • •• • • • • • •• •.••••••••••••••••••••••••••••••••••••••• .. .. . . . .. . . •
X
Fig. 11.2. Incident and scattered potentials.
At((,z)
= Cexp(-iKIZ)
(11.8)
where the choice exp( -iKIZ) satisfies the radiation condition at Z = -00. To determine the unknown coefficients B and C, the boundary conditions for the tangential field continuities must be enforced at z = O. Two boundary conditions, Hz and E y continuities, are required. The field components can be evaluated from the magnetic vector potential A. The nonzero field components are seen to be
E y = iwA y
The Hz continuity at
-
Z
(11.9)
y Hz = _~ 8A J.L 8z
(11.10)
Hz = ~ 8A y J.L 8x
(11.11)
•
= 0
1 8A~(x, z)
8z
J.Lo
1 8A~(x, z)
-
8z
J.Ll
z=o
(11.12)
z=o
•
reqmres
-
-
1 8Ag((, z)
J.Lo
8z
1 8At((,z)
z=o
J.Ll
8z
•
z=o
(11.13)
280
11 Radiation Above Half Space
The E y continuity condition at
Z
= 0
iwA~(x, 0) = iwAt(x, 0)
(11.14)
•
gIves
-
-
A~((,O) = A~((,O) .
(11.15)
Solving (11.13) and (11.15) for B and C yields
B=
J.L 1 "'0 - J.Lo '" 1 J.L 1 "'0 + J.Lo'" 1
iJ.Lo exp( i"'oz') 2"'0
C=
2J.Ll "'0
iJ.Lo exp(i"'oz') 2"'0
(11.16) (11.17)
•
Substituting B and (11.7) into (11.3) finally gives A~(x,z) •
00
exp(i(x)
-00
"'0
Z
-411" .
exp(i"'olz - z'l)
J.Lo
+
exp [i"'o(z
+ z')]
d( . (11.18)
Considering the identity •
•
00
'!...H(ll(k ) = Z 4 0 "'UPl 411"
ei(z
exp(i"'olz - z'l) d(
-00
(11.19)
"'0
where Pl = Jx 2 + (z - Z')2, A~(x,z) can be rewritten as A~(x,z)
•
ZJ.Lo + 411"
00
exp [i(x
+ i"'o (z + z')] d( "'0
-00
.
(11.20)
Similarly, substituting C into (11.3) gives A 1 (x, z) = iJ.LIJ.Lo y 211"
00 -00
exp [i ((x - "'lZ + "'oz')] d( . (J.Ll"'o + J.Lo"'l)
(11.21)
Equation (11.20) consists of two contributions, namely, the direct term •
~ Hgl) (kopd and the scattered term containing the integral. The direct term
11.2 Vertical Electric Dipole
281
accounts for the primary wave radiating from the electric line current in the absence of a boundary, while the scattering term is due to the reflected wave from the boundary at z = O. The integrals in (11.20) and (11.21) must be evaluated to further investigate the characteristics of half-space radiation from the line current.
11.2 Vertical Electric Dipole Consider electromagnetic radiation from a vertical electric dipole at (0,0, z') as J = 28(1' - r') = 28(x)8(y)8(z - z') .
(11.22)
The geometry of the radiation problem is shown in Fig. 11.3. The magnetic vector potentials in a half space satisfy the Helmholtz equations
(\7 + k5)
A~(x,y,z) = -J108(x)8(y)8(z - z')
for z> 0
(11.23)
(\7 + kn
A~(x, y, z) = 0
for z
0 and z < 0, respectively. Since the x-y plane is unbounded, it is expedient to express the magnetic vector potentials in terms of the two-dimensional inverse Fourier transform 00
00
(11.25) -00
-00
z
O(x)O(y)O(z-z ,) • • • • • •• •••••••••••••••••• • ••••••••••••••••••••••••••• . . . . .. .. -. ... .... .. ... .. • ••.. • • • • • • • • •• •••••••••••••••• • • • • • • •• • • • • •• •••• • • • • • • •• • • • • • • • • • •• ••••••••••••••• • • • • • • • • • • • • • • • • •• • • • •• • • • • • •• .......... , ,... .. . .. . .. .. .. .. . .. . .. .. .. . .... .. . . .... .. . .. . .. . . . ................ . • • • • • • • • • • • •• •• • •• •••••••• • • • • • • • •• •• • • • • • • • • • • • • • • • • • ••••••• • • • • • • • • • • • • • • • • •• ••••••• • • •• • • • •••• •• ••• • • • • •• • • • • • .. . . .. ..... .... .. ." .. ... ... .. . . ... .. .. ........ . • •••••• . • • • • • • • • • • • • • • • • • • • •• •••• • • • • • • • • • • • • • • •• • • • • •• •• ••• • •••••• ..... .. . .. ... ....... ... .. ••.••..•... . ..... ........ .... .... . . • • •• • • • • • • • • • • • • • • • • • • • • • • • • • • •• • •••• • •• • • • • • • • • • •••••• --, ••••••• ••••••• •••••••••••• • ••• , , ••• , •• ,...... ••• " •••,':7'"'"-'--+ Y ............... ,...., ..... ,.. ...... .. .. ..... ..........,.. . , . .. :"':":"... . .. .. . ... . .. .. . . ,... .... . ..................... .. .. . . .. ........ . .. . . . . . . . . ......... , , , . . .... . .......,..........,....... .. .......... ...... ......... . ... ........ .. .. .......................... " . ........... ........ , ... .. . . .. .. .. .. ... . ,....... . . ...................................... ........... ............. ..,. . ,. .. .. ........ ......... . ... .... .. ... .. .. .. .. . .. .. . .. . . . .. . . . . . . . . ·.·... .. ... .. .. .. ..,..... ..... . .. .... . .... ...... .. .. .,., . . ................ ... ..... ......- .. .. . .. ..· ... . ..... . .. .... . .... .......... •
x
Fig. 11.3. Vertical electric point source in half space.
282
11 Radiation Above Half Space
Substituting A~·1(X,y,Z) into (11.23) and (11.24) gives, respectively,
ff2
2
ff2
2
dz 2 +/\'o
dz 2
+ /\'1
-
A~((, 1],
z) = -j.Lot5(z - z')
-
A~((,1],z)
Jk'J -
(11.26)
= 0
(11.27)
Jk? -
where /\'0 = (2 _1]2 and /\'1 = (2 _1]2. The solution to (11.26) consists of the incident and scattered potentials
A~((,1],z) = AH(,1],z) +A~((,1],z) •
= ;J.LO exp(iKolz - z'l)
+ B exp(iKoz) .
(11.28)
/\'0
The solution to (11.27) is assumed to be
-
A~((,1],z)
= Cexp(-i/\'1Z).
(11.29)
To determine the unknown coefficients B and C, the boundary conditions for the tangential fields E"" E y , H"" and H y continuities must be enforced at z = O. The tangential field components can be evaluated by substituting A~·1 (x, y, z) into Table 3.1 in Section 3.1. The H", and H y continuities 1 8A~(x,y,z)
8y
j.Lo
-
8x
8y
J.L1
%=0
1 8A~(x,y,z) j.Lo
1 8A;(x, y, z) %=0
1 8A;(x, y, z)
-
8x
J.L1
%=0
(11.30)
(11.31) %=0
•
reqmre 1 -
-A~((,1],O)
J.Lo
=
1 -
(11.32)
Ai((,1],O).
J.Ll
Similarly, the E", and E y continuities 1 8 A~(x, y, z) 2
8x8z
J.Lo€o
1
1 8 A~(x, y, z) 2
J.Lo €o
J.L1 €1
%=0
8y8z
%=0
2
8 A;(x,y,z) 8x8z
(11.33) %=0
2
1 8 A;(x, y, z)
J.L1 €1
8y8z
(11.34) %=0
yield 1
j.Lo€o
-
8A~((, 1],
8z
_
z)
%=0
1
J.L1 €1
8Ai((,1],z) 8z
•
%=0
(11.35)
11.2 Vertical Electric Dipole
283
Solving (11.32) and (11.35) for B and C gives •
B =
~/-Lo
(11.36)
2Ko exp(iKoz') .
(11.37)
Substituting B into A~ (x, Y, z) yields
o ) 1 AAx, y, z = (27l")2
00
00
iei«x+'1Y)
-00
-00
211:0
f.Lo
. exp(iKolz - z'l)
The identity for a three-dimensional free-space Green's function shows 1
(27l")2
00
00
-00
-00
.
~ exp(ill:olZ - z'l)e i (x+i'1Y d( d1)
211:0
(11.39) where r1 =
J x 2 + y2 + (z O(
) _
z')2. Hence, A~(x, y, z) is rewritten as
A z x,y,z - /-Lo
exp(ikord 4
7l" r 1 00
00
-00
-00
•
.exp [i(x
lO111:o -
lO1KO
lOOK1
+ lOOK1
+ 1}Y) + iKo(z + z')]
d( d1} .
(11.40)
11:0
Similarly, substituting C into A~(x,y,z) gives 00
-00
/-L1
+ 1}Y - K1 Z + II:oZ')] d; d ( ) '> 1}. E1KO + lOOK1
exp [i (x
(11.41)
284
11 Radiation Above Half Space
Let us introduce the polar coordinates
( =
(~, 'l/J)
and (p, (1)
~ cos 'l/J
(11.42) (11.43)
= ~ sin 'l/J
'TJ
x = pcos41
(11.44) (11.45)
y = psin{JJ
to further simplify A~(x,y,z) and A~(x,y,z). First, consider integration with respect to 'l/J as 211"
211"
e
i
«(:Z:+'1Y)
d'l/J =
o
exp [ip~ cos('l/J - (1)] d'l/J
0
211"
o
00
L
in In(~p)ein(1/J-cf>) d'l/J
n=-oo
= 21fJo(p~) .
Hence, the final polar forms for the potentials for z respectively,
A O( ) z p, z =
(11.46)
>
0 and z
< 0 are,
exp(ikoTl) J.Lo 41fT}
+
•
00
J.Lo...::.
€}KO -
....:..t
41f
0
€}KO
. exp [iKo(Z
€OK}
+ €OKI
+ Zl)] Jo(p~)~ ~
(11.47)
KO
(11.48) where
P=VX2 +y2 = Jk1J - ~2 = Jkr - e .
(11.49)
KO
(11.50)
K}
(11.51)
11.3 Horizontal Electric Dipole The geometry of the radiation problem is shown in Fig. 11.4. Consider electromagnetic radiation from a horizontal electric dipole located at (0,0, z') as
•
11.3 Horizontal Electric Dipole
285
z
8 (x) 8 (y) 8 (z-z') .. , • •• •••••••••••••••• • • • • • • • • • • •• ~ 0' € 0 •• • • • • • •• • • • • • • • •• • •• • ••••••••••••• • •••• • •• • •••• • • • •• • • • •• • • • • • • • •• •••••••••• • • • • • • • •••••• .. .....-....•.•.•.•.. ..... . . . . . . .... .. . . . . . . .. .. . . . • • • • • • • • • .......•.•..•........ . .. . .. . . .. .. .... ... .... . . .. .. . . . .. .. .. . . .. .. . . .. .. ..................... .. . ... ....................... . ...... .. .. .. ............... ... .... ......... ... ..... .................... .................. ,,-, ................. ...'. :';';,'''-'':' .. y •••••• • • • • • • • • • • •• •• • • • •• • • • • •• ••••••••••••••• ••• •• • • •• • • •• • • • • • • • • • •• ••••• • • •• • • •• • •• •• • ••••• ••• •••••••••••• • • •• ••• •••• •••• • •• • • ••••• • • • •• •• • •• • •• • •• • • • ••• • • • • •• • • • • ••••• •••• ••• •• •••••••• ••• • • • • • •• •••••• • •• •••••••
...........................' . ........••••.•.•••........ . ..... .. .... . . .. .. ..
. . . .. .. . . . ...... .... . ....... . .................. ... ... .... .. .. . .. .. . .. .. .. .. . . .......... ..... . ......... . .. .... ... . . . . . .. .. .. . . . ... .. . .. .. . . ...... .. . . . .. .. .. .. . . . . ..... .... . .. .... ...... ............. ... . ..... ..... ..... .................... .. ........... .......................... ·....... .. ....••••••••••• ......... .. .. ..... ..... . ..... . ....... ·.............. .. .. . . .. . . .. . .. . . . ... -. . . . . . . . . . . . ... • •••••••••••••
..
. . . .. .. .. . .. . .. . . . . . .. .. . . .. •••
•••
.. . .. :-::"
••
• •••••••••••••••••••••••••••••••••••
... -
Fig. 11.4. Horizontal electric point source in half space. J = xo(r - r') = xJ(x)J(y)J(z - z') .
(11.52)
The magnetic vector potentials in the half space satisfy the Helmholtz equations
(\7 + k~) lfl(x,y,z) = -XJ.LoJ(x)J(y)J(z - z')
for z
> 0 (11.53)
(\7 + kD A (x, y, z)
for z
0
(11.55)
(\7 + k~) A~(x,y,z) = 0
for z
>0
(11.56)
(\7 + kD A~(x,y,z) = 0
for z
0 (B.8)
Nv(x) ....., _ (v - I)!
-
x
1r
For x -t
2
v
(B.9)
•
00
Jv(x) .....,
2
1rX
cos
2 sin 1rX
(
1r X -
4 -
V1r)
(x _
1r _
V1r)
4
(B.10)
2
2
•
(B.ll)
302
B Bessel Functions
B.lo2 Wronskian
The Wronskian of the Bessel functions gives (B.12) (B.13) B.lo3 Generating Function
The generating function for the Bessel function gives [11, page 973] 00
eixcos