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NATURAL RESONANCE REPRESENTATION OF THE TRANSIENT
FIELD REFLECTED BY A PLANAR LAYERED LOSSY DIELECTRIC

presented by

JONG CHAN OH

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NATURAL RESONANCE REPRESENTATION OF THE TRANSIENT FIELD
REFLECTED BY A PLANAR LAYERED LOSSY DIELECTRIC

By
JONG CHAN OH

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY
Department of Electrical and Computer Engineering

2002

ABSTRACT
NATURAL RESONANCE REPRESENTATION OF THE TRANSIENT FIELD
REFLECTED BY A PLANAR LAYERED LOSSY DIELECTRIC
By
JONG CHAN OH

The impulse response of a conductor-backed or air-backed lossy dielectric slab with
frequency independent or dependent (Debye type) material parameters is evaluated
analytically. It is shown that the impulse response consists of a specular reflection
from the interface between free-space and the dielectric slab during the early—time
period, and a natural mode series, which is a pure sum of damped sinusoids whose
frequencies are determined by the poles of the complex s—plane reflection coefficient,
during the late-time period. Time—domain responses using a truncated Gaussian
pulse as an input with an arbitrary incident angle and with parallel or perpendicular
polarization are compared to responses found by the inverse fast Fourier transform.
The results may be applied to material characterization using the E—pulse method,
and also give physical insight into the nature of transient scattering by a layered

medium.

For my wife Sun-Kyung and our children Sang-Jun and Sang-Gun

iii

Acknowledgements

I would like to express my sincere gratitude to the Republic of Korea Air Force
for providing me the opportunity to study abroad. I would also like to thank those
who taught, guided, and encouraged me throughout my academic pursuit. Foremost,
I am deeply indebted to my academic advisor Professor Edward Rothwell for his in-
valuable guidance and encouragement. I would like to thank him for enabling me to
successfully complete my research endeavors. I am very grateful to Professor Den-
nis Nyquist for his generosity, and in providing motivation through his inspirational
lectures. Special thanks must be given to Professor Leo Kempel and Mrs. Kempel
for their kindness and encouragement in helping me to overcome difficulties in times
of need. I would like to express my thanks to Professor Byron Drachman for his
guidance and generosity. I also would like to extend my thanks to Professor Shanker
Balasubramaniam who joined my guidance committee in place of Professor Nyquist.
I am very much thankful to Hyun Park and the colleagues in EM group, especially
Christopher Coleman, Jeffery Meese, and Brad Perry for their help and invaluable
discussions.

Finally, I would like to reserve my deepest gratitude to my wife, my best friend
Sun-Kyung, and my sons Sang-Jun and Sang-Gun for their encouragement and pa-

tience. Without their support, this research would not have been possible.

iv

Table of Contents

LIST OF TABLES vii
LIST OF FIGURES ix
1 Introduction 1

2 Frequency Domain Analysis of Plane-Wave Propagation in a General

Planar Layered Medium 4
2.1 The frequency domain wave equation .................. 4
2.2 Reflection from a single interface .................... 7
2.2.1 Perpendicular polarization .................... 7
2.2.2 Parallel polarization ....................... 10

2.3 Reflection from multiple layers ...................... 12
3 Natural Mode Analysis of a Conductor-Backed Slab 23
3.1 Formulation of the frequency-domain reflection ............. 23
3.1.1 The frequency—domain reflection coefficient ........... 23
3.1.2 The impulse response ....................... 25

3.2 Singularities and the branch cut ..................... 26
3.3 Evaluation of r(t) ............................ 32
3.3.1 Case I: t > To ........................... 32
3.3.2 Case II: t < To .......................... 37

v

3.4 Results ................................... 39

4 Natural Mode Analysis of an Air-Backed Slab 75
4.1 Formulation of the frequency-domain reflection ............. 75
4.2 Singularities and the branch cut ..................... 77
4.3 Evaluation of 'r(t) ............................ 82

4.3.1 Case I: t > To ........................... 83
4.3.2 Case II: t < To .......................... 88
4.4 Results ................................... 89

5 Natural Mode Analysis of a Single Layer with Debye Material 119

5.1 Formulation of the frequency—domain reflection ............. 119
5.2 Singularities and the branch cut ..................... 124
5.3 Evaluation of r(t) ............................ 128
5.3.1 Case I: t > To ........................... 128

5.3.2 Case II: t < To .......................... 132

5.4 Results ................................... 134

6 Conclusions 158
BIBLIOGRAPHY 160

vi

3.1

3.2

3.3

3.4

3.5

3.6

3.7

3.8

3.9

4.1

4.2

4.3

4.4

List of Tables

Poles and correslgxmding complex natural mode amplitudes (Ak), lose-
less case (6,. = 9, A = 2 cm, 6,- : 0°) ...................

Poles and corresponding complex natural mode amplitudes (Ak), ||
polarization (6,. = 9, A = 2 cm, 6,; = 20°, or = 0.5 S/m) ........

Poles and corresponding complex natural mode amplitudes (Ak), _L
polarization (6,. = 9, A = 2 cm, 6,- = 20°, 0 = 0.1 S/m) ........

Poles and corresponding complex natural mode am]:)litudes (Ak), H
polarization (6, : 9, A = 2 cm. 6, = 30°, 0 = 1.0 S/m) ........

Poles and corresponding complex natural mode amplitudes (Ak), J.
polarization (6,. = 9, A = 2 cm, 6,- : 30°, 0 = 1.5 S/m). .......

Poles and corresponding complex natural mode amplitudes (Ak), ||
polarization (6,- = 9, A = 2 cm, 9,- = 70°, 0 = 0.5 S/m) ........

Poles and corresponding complex natural mode amplitudes (At), _L
polarization (6,. = 9, A = 2 cm, 6,- = 70°, 0 = 0.5 S/m) ........

Poles and corresponding complex natural mode amplitudes (At), II
polarization (6., = 9, A = 2 cm, 6,- : 85°, 0 = 0.5 S/m) ........

Poles and corresponding complex natural mode amplitudes (Ak), J.
polarization (6,- = 9, A = 2 cm, 6?,- = 85°, 0 = 0.5 S/m) ........

Poles and corresponding complex natural mode amplitudes (Ak), lose-
less case (6,- = 9, A = 2 cm, 6,- : 0°) ...................

Poles and corresponding complex natural mode amplitudes (Ak), ||
polarization (6,- : 9, A = 2 cm, 6,- : 20°, 0 = 0.5 S/m) ........

Poles and corresponding complex natural mode amplitudes (A1,), _L
polarization (6, = 9, A = 2 cm, (9,- = 20°, 0 = 0.1 S/m) ........

Poles and corresponding complex natural mode amplitudes (Ah), [I
polarization (6,. = 9, A = 2 cm, 6, : 30°, 0 = 1.0 S/m) ........

vii

41

42

43

44

46

47

48

49

91

92

93

4.5

4.6

4.7

5.1

5.2

5.3

5.4

5.5

5.6

Poles and corresponding complex natural mode amplitudes (Ak), J.
polarization (6, = 9, A = 2 cm, 0, = 30°, 0 = 1.5 S/m). ....... 95

Poles and corresponding complex natural mode amplitudes (Ak), ||
polarization (6,. = 9, A = 2 cm, 6,- : 70°, 0 = 0.5 S/m) ........ 96

Poles and corresponding complex natural mode amplitudes (Ak), _L
polarization (6,- : 9, A = 2 cm, 6,- : 70°, 0 = 0.5 S/m) ........ 97

Poles and corresponding complex natural mode amplitudes (A1,), conductor-
backed case, normal incidence (63 = 78.360, 600 = 5.0(0, E = 9.6 x 10—12
sec, A = 2 cm, 0,; = 0°, 0 = 0.0 S/m) .................. 136

Poles and corresponding complex natural mode amplitudes (Ak), conductor-
backed case, J. polarization (6,, = 78.360, 600 = 5.060, 5 = 9.6 x 10‘12
see, A = 2 cm, 6,- : 30°, 0 = 0.1 S/m) ................. 137

Poles and corresponding complex natural mode amplitudes (A1,), conductor-
backed case, M polarization (6, = 78.360, 600 = 5.060, g = 9.6 x 10'12
see, A = 2 cm, 6,- = 30°, 0 = 0.5 S/m) ................. 138

Poles and corresponding complex natural mode amplitudes (Ak), air—
backed case, normal incidence (es = 78.360, 600 = 5.060, 5 = 9.6 x 10‘12
sec, A=2cm, I9,- =0°, 0:0.0 S/m) .................. 139

Poles and corresponding complex natural mode amplitudes (At), air-
backed case, I polarization (6,, = 78.360, 600 = 5.060, E = 9.6 x 10‘12
see, A = 2 cm, 6,- : 30°, 0 = 0.1 S/m) ................. 140

Poles and corresponding complex natural mode amplitudes (Ak), air-
backed case, M polarization (es 2 78.360, 600 = 5.060, 6 = 9.6 x 1042
see, A = 2 cm, 6,: = 30°, 0 = 0.5 S/m) ................. 141

viii

2.1

2.2

2.3

2.4

3.1

3.2

3.3

3.4

3.5

3.6

3.7

3.8

3.9

List of Figures

Uniform plane wave incident on a planar interface between two lossy
regions of space. Perpendicular polarization. .............

Uniform plane wave incident on a planar interface between two lossy
regions of space. Parallel polarization. .................

Interaction of a uniform plane wave with a multi-layered material. . .

Wave flow diagram showing interaction of incident and reflected waves
for region 72,. ...............................

A uniform plane wave incident from free space upon a conductor-
backed lossy slab. ............................

Pole trajectories of the reflection coefficient, I polarization (6,- = 9,
A = 2 cm, 6,- = 30°). ...........................

Pole trajectories of the reflection coefficient, H polarization (6, = 9,
A = 2 cm, 0,- : 30°). ...........................

The amplitude of the imaginary parts of poles vs. conductivity, _L
polarization (6,. = 9, A = 2 cm, 9,- : 30°, 0 = 0 —> 10.0 S/m).

The amplitude of the imaginary parts of poles vs. conductivity, ||
polarization (6,. = 9, A = 2 cm, 6,- = 30°, 0 = O —+ 10.0 S/m).

Pole trajectories of the reflection coefficient vs. angle of incidence, J.
polarization (6,. = 9, A = 2 cm, 0 = 0.1 S/m). .............

Pole trajectories of the reflection coefficient vs. angle of incidence, ||
polarization (6,. = 9, A = 2 cm, a = 0.1 S/m). .............

Characteristic function vs. real 3, _L polarization (6,. = 9, A = 2 cm,
0 = 5.0 S/m, 9, = 30°). .........................

Characteristic function vs. real 3, _L polarization (6,. = 9, A = 2 cm,

0 = 0.5 S/m, 6,- = 30°). .........................

ix

19

20

21

22

50

51

52

53

54

55

56

57

3.10

3.11

3.12

3.13

3.15

3.16

3.17

3.18

3.20

3.21

3.22

3.23

3.24

Amplitudes of the reflection coefficients (6,. = 5, A = 2 cm, a = 1.5
S/m, 61': 30°). ..............................

Real pole (s < 30), (6r 2 5, A = 2 cm, 0 = 1.5 S/m, 6,- : 30°).
Real pole (s < 50), (6, = 5, A = 2 cm, 0 21.5 S/m, 6, = 30°). . . . .

Complete sketch of the pole trajectories showing both complex and
real poles, J. polarization (6,. = 9, A = 2 cm, 6,- : 20°). ........

Complete sketch of the pole trajectories showing both complex and
real poles, H polarization (6, = 9, A = 2 cm, 6, = 20°). ........

The integration contour when t > T0. lal, Ibl —-> oo, 0 < c < 00 .....
The integration contour when t < 7’0. R —-> oo, 0 < a < oo. ......

Time domain response, lossless case (6,. = 9, A = 2 cm, 6, 2 0°). Inset
shows equivalent transmitted waveform. ................

Time domain response, || polarization (6,. = 9, A = 2 cm, 6,- : 20°,
0 = 0.5 S/m). ...............................

Time domain response, J. polarization (6,. = 9, A = 2 cm 6,- = 20°,
0 = 0.1 S/m). ...............................

Time domain response, I] polarization (er 2 9, A = 2 cm, 6, = 30°,
0 = 1.0 S/m). ...............................

Time domain response, I polarization (6,. = 9, A = 2 cm, 6, = 30°,
0 = 1.5 S/ m). ...............................
Time domain response, || polarization (6,. = 9, A = 2 cm, 6,- = 70°,
0 = 0.5 S / m). ...............................
Time domain response, J. polarization (6,- : 9, A = 2 cm, 6,- = 70°,

0 = 0.5 S/m). ...............................

Time domain response, || polarization (6,. = 9, A = 2 cm, 6.,- = 85°,
0 = 0.5 S/m). ...............................

59

60

61

62

64

65

66

67

68

69

70

71

72

73

3.25

4.1

4.2

4.3

4.4

4.5

4.6

4.7

4.8

4.9

4.10

4.11

4.12

4.13

4.14

Time domain response, J. polarization (6,. = 9, A = 2 cm, 6,- = 85°,
0 = 0.5 S/m). ...............................

A uniform plane wave incident from free space upon a Air-backed lossy

slab. ....................................

Pole trajectories of the reflection coefficient, J. polarization (r,r = 9,

A = 2 cm, 6,- : 30°). ...........................

Pole trajectories of the reflection coefficient, I] polarization (6, = 9,
A = 2 cm, 6,- : 30°). ...........................

The amplitude of the imaginary parts of poles vs. conductivity, .L
polarization (6,. = 9, A = 2 cm, 6,- : 30°, 0 = 0 ——> 10.0 S/m).

The amplitude of the imaginary parts of poles vs. conductivity, ll
polarization (6,. = 9, A = 2 cm, 6,- = 30°, 0 = 0 ——> 10.0 S/m).

Pole trajectories of the reflection coefficient vs. angle of incidence, I
polarization (6,. = 9, A = 2 cm, a = 1.0 S/m). .............

Pole trajectories of the reflection coefficient vs. angle of incidence, II
polarization (6,- = 9, A = 2 cm, a = 1.0 S/m). .............

Amplitudes of the reflection coefficients (6, = 5, A = 2 cm, 0 = 3.0

S/m, 6,- = 30°). ..............................
Real pole (s < .30), (6,. = 5, A = 2 cm, 0 = 3.0 S/m, 6,- = 30°).
Real pole (s < 50), (6,. = 5, A = 2 cm, a = 3.0 S/m, 6,- = 30°).

Complete Sketch of the pole trajectories showing both complex and
real poles, J. polarization (6,. = 9, A =2 2 cm, 6,- : 30°). ........

Complete sketch of the pole trajectories showing both complex and
real poles, || polarization (6,. = 9, A = 2 cm, 6,; = 30°). ........

The integration contour when t > 7'0. |a|, Ibl —> oo, 0 < c < oo .....

The integration contour when t < 7'0. R ——> 00, 0 < a < oo. ......

xi

74

99

100

101

102

103

104

105

106

107

108

110

111

4.15

4.16

4.17

4.18

4.19

4.20

4.21

5.1

5.2

5.3

5.4

5.5

Time domain response, lossless case (6,. = 9, A = 2 cm, 6,- : 0°). Inset
shows equivalent transmitted waveform. ................ 112

Time domain response, || polarization (6,. = 9, A = 2 cm, 6,- : 20°,
0205 S/m). ............................... 113

Time domain response, J. polarization (6, = 9, A = 2 cm, 6, = 20°,
0:0.1S/m). ............................... 114

Time domain response, H polarization (6r 2 9, A = 2 cm, 6,- = 30°,
0 =1.0 S/m). ............................... 115

Time domain response, J. polarization (6,- = 9, A = 2 cm 6,- : 30°,
0 =1.5 S/m). ............................... 116

Time domain response, H polarization (6,. = 9, A = 2 cm 6,- : 70°,

=0.5 S/m). ............................... 117

Time domain response, I polarization (6r :2 9, A = 2 cm 6,- = 70°,
0 = 0.5 S/m). ............................... 118

The amplitude of the real parts of poles vs. the ratio of c, to 600, H
polarization (6, = 5.060 —> 78.360, 600 = 5.060, 5 = 9.6 x 10"12 see,
A = 2 cm, 6,; = 30°, 0 = 0.5 S/m). ................... 142

The amplitude of the imaginary parts of poles vs. the ratio of 63 to
600, || polarization (6,, = 5.060 —+ 78.360, coo = 5.060, 5 = 9.6 x 10’12
see, A = 2 cm, 6, = 30°, 0 = 0.5 S/m). ................. 143

Pole trajectories of the reflection coefficient, H polarization (63 = 5.060 —+
78.360, 600 = 5.060, 5 = 9.6 x 10’12 see, A = 2 cm, 6,: = 30°, 0 = 0.5
S/m). ................................... 144

The amplitude of the real parts of poles vs. the ratio of 5, to 600, J.
polarization (6, = 5.060 —> 78.360, 600 = 5.060, 5 = 9.6 x 10“12 see,
A = 2 cm, 6,- : 30°, 0 = 0.5 S/m). ................... 145

The amplitude of the imaginary parts of poles vs. the ratio of e, to
600, .L polarization (6,, = 5.060 —> 78.360, 600 = 5.060, 5 = 9.6 x 10’12
see, A = 2 cm, 6,- : 30°, 0 = 0.5 S/m). ................. 146

xii

5.6 Pole trajectories of the reflection coefficient, .L polarization (68 =
5.060 —> 78.360, 600 = 5.060, 5 = 9.6 x 10‘12 see, A = 2 cm, 6, = 30°,
0 = 0.5 S/m). ..............................

5.7 Amplitude of the reflection coefficient vs. frequency for J. polarization
(63 = 5.060, 600 = 3.060, 5 = 9.6 x 10’12 see, A = 2 cm, 6, = 30°,
0 = 1.0 S/m). ..............................

5.8 Amplitude of the reflection coefficient vs. frequency for I] polarization
(6,, = 5.060, 600 = 3.060, 5 = 9.6 x 10‘12 see, A = 2 cm, 6, = 30°,
or 21.0 S/m). ..............................

5.9 The integration contour when t > T0. |a|, Ibl —> 00, 0 < c < oc .....
5.10 The integration contour when t < T0. R ——> oo, 0 < a < oo. ......

5.11 Time domain response, conductor—backed case, normal incidence, (68 =

150

151

78.360, 6.30 = 5106025: 9.6 x 10’12 sec, A = 2 cm, 6, = 0°, 0 = 0.0 S/m).152

5.12 Time domain response, conductor-backed case, I polarization, ((3 =
78.360, 600 = 5.060, 5 = 9.6 x 10‘12 see, A = 2 cm, 6, = 30°, 0 = 0.1
S/m). ...................................

5.13 Time domain response, conductor-backed case, M polarization, (6, =
78.360, 600 = 5.060, 5 2 9.6 x 10'1'2 see, A = 2 cm, 6, = 30°, 0 = 0.5
S/m). ...................................

5.14 Time domain response, air-backed case, normal incidence, (6, = 78.360,
600 = 5.060, 5 = 9.6 x 10”12 see, A = 2 cm, 6, 2 0°, 0 = 0.0 S/m).

5.15 Time domain response, air-backed case, .L polarization, (6, = 78.360,
600 = 5.060, 5 = 9.6 x 10’“2 see, A z 2 cm, 6, = 30°, 0 = 0.1 S/m).. .

5.16 Time domain response, air-backed case, || polarization, (6, = 78.360,
600 = 5.060, 5 = 9.6 X 10"12 see, A = 2 cm, 6, = 30°, 0 = 0.5 S/m).. .

xiii

155

156

157

 

Chapter 1: Introduction

Time domain electromagnetics has been utilized for various applications such as
target-signature analysis and the determination of the intrinsic properties of
materials [1], mainly due to its adaptability to the broadband signals. The
singularity expansion method (SEM) [2] which views the late-time period transient
response of a scatterer as a series of the residues at the poles of the Green function
[3], has been studied by many researchers for the target—signature analysis purpose.
Here, the late-time period refers to the time period after the forcing function
completely excites the entire body of a scatterer, whereas the early-time period
refers to the time period before the late-time begins. The applications of the
singularity expansion method and its limitations are found in [4H6].

On the other hand, the E-pulse method [7H9] which is one of the most
successful target identification schemes using the singularity expansion method, can
be used for non-destructive material testing [10], [11]. Here, the E—pulse refers to
the signal that is synthesized to annihilate the output response when it is convolved
with the late-time period target response, such that it can provide a measure to
determine the closeness of the unknown target signature to the library of the known
target signatures.

Layered materials are often applied to conducting surfaces for the purpose of
reducing the scattered field strength within specific frequency bands. Because of the
band-limited nature of the reflected field, a wideband pulse can be used to
interrogate the layered structure so as to characterize the materials or determine
whether the materials have degraded. The E—pulse method can be used to
determine whether the material properties have changed compared to baseline
values determined by previous measurements or constructed by analytically

obtained natural resonance series, provided that the late-time response can be

 

expressed as a natural resonance series. Therefore, it is important to identify the
natural frequencies, i.e., the poles of the reflection coefficient in the complex s-plane,
and to show that the late-time response is indeed a sum of damped sinusoids.

So far, there has been no detailed analysis of the transient response of a layered
medium as a resonance series, except the work done by Tihuis and Block in 1984
[12], [13]. They calculated the transient scattering of a normally incident plane wave
by an air-backed lossy dielectric slab using the singularity expansion method.
However, their approach was semi-analytic and did not clearly show that the
late—time response can be expressed as a pure natural resonance series, due to the
difficulty in evaluating the contributions from the branch cut and the closing
contours at infinity when evaluating the Laplace inversion integrals.

In this thesis, we will consider a conductor—backed or air-backed lossy dielectric
slab with frequency independent or dependent (Debye type) material parameters,
and a plane wave excitation of both TE and TM polarization at arbitrary incidence
angle. It will be shown that the integral contribution from the closing contours at
infinity vanishes as long as the input waveform is at least APC (almost piecewise
continuous) and time limited [14]. When viewed as a functional, the delta function
can be considered APC, and thus the impulse response can be computed directly.
Also, it will be shown analytically that the branch-cut contribution for the current
problem vanishes.

By evaluating the impulse response, it will be shown that the transient field
reflected by a conductor-backed or air-backed lossy dielectric slab can be
represented as the sum of early-time and late-time components. The early-time
component consists of the specular reflection from the air-slab interface, which
persists until the arrival of the field reflected by the backing material. The ensuing
multiple reflections can be viewed as the late-time component, which is written as a

pure sum of natural resonance modes. The scattered field impulse response will be

determined as an inverse Laplace transform of the frequencydomain reflection
coefficient, and will be shown that the branch-cut contribution vanishes during the
late—time period, except for the residual contributions at the poles on the
branch-cut, allowing a pure natural resonance representation.

Results computed using the resonance formulation are verified by comparing to
the direct inverse fast Fourier transform (IFF T). The distribution of resonance
frequencies in the complex plane, and their dependence on conductivity and
incidence angle is examined.

This thesis is organized in 6 chapters. In Chapter 2, the frequency domain
formulation of plane—wave propagation in a. general planar layered medium is
presented. In particular, the interfacial reflection coefficient F (w) and the reflection
coefficient R(w) are distinctively defined, and the definitions are applied throughout
the thesis.

In Chapters 3 and 4, the transient. responses from the conductor-backed and
the air-backed cases with frequency independent material parameters are examined.
The complex s-plane poles of the reflection coefficient are found numerically and the
residues at the poles are calculated. The transient responses are constructed from
the residue series and they are compared to the responses obtained by the IFFT.

In Chapter 5, the formulation of the frequency domain reflection with a
frequency dependent Debye type permittivity [28]—[32] for both conductor—backed
and air-backed cases is presented. The transient responses from water with varying
conductivity are constructed and compared to the responses from the IFFT.

Finally, the conclusions and the proposed future research areas are presented in

Chapter 6.

Chapter 2: Frequency Domain
Analysis of Plane-Wave
Propagation in a General Planar

Layered Medium

The frequency domain rejn'esentation of plane waves in a planar layered
medium can be found in [15]. In this chapter, these representations will be reviewed
and specialized for the current problem. For the subsequent discussion, only

source—free, linear, isotropic, homogeneous materials are considered.

2.1 The frequency domain wave equation

In a source-free and simple material, Maxwell’s equations in terms of E and H

are

V x E = —jw,uH, (2.1)
V X H = jweCE, (2.2)
V - E = 0, (2.3)

v - H = 0, (2.4)

where EC is the frequency-dependent complex permittivity which is a combination of

the conductivity 0(w) and the permittivity 6(a)). The curl of (2.1) is then,

V x (V x E) = —jwn(V x H) = -jw,u(jwc°E).

Since, V X (V X E) = V(V - E) — VZE, using (2.3) gives
VQE + k2E = 0 (2.5)

Here, k = tin/p.66 is the wave number of the medium. Similarly, taking the curl of
(2-2),
V X (V X H) = jwc°(V X E) = jw6°(—jwnH),

and thus

V2H + sz = 0. (2.6)

Equations (2.5) and (2.6) are the homogeneous vector Helmholtz equations. Thus,

the rectangular components of E and H satisfy the scalar wave equation
V21!) + [1721/] Z 0)

whose solution is a linear combination of the harmonic functions,

—jkr.

sinkr, coskgr, 8"”, 6

Consider a propagating-wave solution to the homogeneous vector Helmholtz
equation,

E _—_ EOE—jere—jkyye—jkzz
where E0 is the vector amplitude spectrum. If we define the wave vector
k = xk, + yr, + 2k“

then,

E = EOe-J‘k'r (2.7)

where

r = xx + yy + 22
If k is real, the vector phase constant 8', which is defined by B = —V<I> [16], where (1)
represents the phase of (2.7), becomes

8': —V(—k-r) = k.

Hence, the equiphase surfaces are planes perpendicular to k, and (2.7) represents a

uniform plane wave. If k is complex, it can be represented as

k = ('3 — ja,

where both [If and 62 are real vectors. Then, the wave propagation constant 7' is [16]
a: 479 = —V(—jk-r) =jk= awe

Therefore, the equiphase surface is perpendicular to ,6 and the eqiamplitude surface
is perpendicular to CY. In general, it is not a uniform plane wave unless (Y and 6 are
in the same direction.

Now, from (2.1) and (2.7),

v x E = v x (Eoe‘jk'r)
z e-J‘” (v x E0) — E0 x (ye-3"")
2 "E0 X (“-jkC—jkir)

= —jk x E = —jw,u,H.

Therefore,

_kXE

 

 

 

 

 

H (2.8)
can
Taking the cross product. to (2.8) with k,
kXH= kx (kXE) = k(k-E)-E(k-k).
can. top.
Here, k - E = 0, since
v - E = v - (EUe—Jk'r)
—_— e-jk'rv . E0 — jk . Egg-1"" = 0
As a result,
k H
E = — X p . (2.9)
we
If we assume a uniform plane wave, i.e., k = kk, from (2.8) and (2.9)
k X E k E .
= = —x °e-J'”, (2.10)
77 '7]
E = —nkXH, (2.11)

where 77 = CUM/k = ,/ n/ec. Clearly, these equations represent a TEM wave.

2.2 Reflection from a single interface

2.2.1 Perpendicular polarization

Consider a uniform plane wave incident on a planar interface between two lossy
regions of space as shown in Figure 2.1. Here, the incident wave is perpendicularly
polarized to the plane of incidence, which is defined as the plane containing the

wave propagation vector k and normal to the interface. At this point it should be

noted that, for a uniform plane wave, the field can be decomposed into two
orthogonal components, one parallel and the other perpendicular to the plane of
incidence. Therefore, the solutions from each case are sufficient to characterize any
arbitrary incident uniform plane wave. For the perpendicularly incident uniform

plane wave, the incident. fields are

Ei __: yEée—jkl(.rsin6,+zcos6,),
Hi _ _Q (_i COS 9, + E 8111 9i)e—jfi71(.’r 81116, + Z COS 6,)
"771

Although the reflected field and the transmitted field are not known at. this point,
they cannot have vector components not present in the incident field, in order to

satisfy the boundary conditions at the planar surface. Therefore, the reflected and

transmitted fields are

Er _ yESe—jk,(zsin6r—zcos6,.)

7

ET _ - ., . _ .
Hr = —0- (x cos 6,. + zs1n6r) e 31‘1” 51119,. 300° 6’”),
771

and

Et __ yEée—jk2(rsin6,+zcos6,),
E’ . . .

Ht = J(—xcos6,+zsm6,
772

)e_jl,~2(:r sin 6, + 2 cos 9:)

9

respectively. In order to satisfy the continuity of tangential E and H over the entire

interface, the Lit-variation of all three partial fields must be the same, i.e.,

k1 sin 6, = k, sin 6,. 2 k2 sin 6,.

As a result,

 

 

 

 

 

 

 

 

 

 

9,, = 9,, (2.12)
Sing’ = 5‘— : Eff“ . (2.13)
8111 6, k2 63,112
. . Ey . .
Also, z-du‘ected wave impedance Z = ——H— should be contnmous at the 1nterface,
that is
21 '320 = Z2lz=0 - (214)
erre,
(.1) i r
‘ Hf.” (E5 — 55) s a/m
Ei Er
T]l( Oil— 0) 3 (215)
cos 6,-(E6 —- E6)
E(2) l. 77‘)
Zr ,: = ——-—-y—,—- =-——————==————“——-. 2.16
2L 0 H42) _0 — cos 6,/772 cos 6! ( )
From (2.14) - (2.16),
771(E6 + E6) = 712
cos 6,-(E’5 — E6) cos 6,3
or
E3072 cos 6, — 7), cos 6,) 2 E5072 cos 6, + 77, cos 6,).
Therefore,
E6 7);, cos 6, — 7}, cos 6,
1“ z __, =
E5 77;, cos 6, + 77, cos 6,
772 _ 771 (2) (1)
= cos 6, cos 6, : ii (2.17)

722 + 771 213+ zill'
cos 6, cos 6,

 

Here, F is the interfacial reflection coefficient, and

 

771 772
and are the
cos 6, cos ,
z-directed wave impedances in region 1 and region 2 respectively. In order to find

the transmission coefficient T, use

 

2 X (E + ET)|,=0 = 2 X E’

 

 

2:03
and thus
Ef,(1 + F) 2 E5.
Therefore,
E"
T = —9. = 1 r. 2.18
E5 + < )
Finally, collecting the results from (2.17) and (2.18),
2(2) __ Z“)
FJ. Z —‘L+———i—-, Ti 2 1 + F
2(2) + 2(1)
1 J. (2.19)
. . . k,
2(1) _ 771 2(2) _ 772 _ 772 2
i cos 6,’ f

 

cos 6, k2,, ’

where 132,, 2 [62 cos 6, is the z-component of kg.

2.2.2 Parallel polarization

Analogous to the perpendicular incidence case, Figure 2.2 shows a uniform

plane wave incident on a planar interface between two lossy regions of space with

10

parallel polarization. In this case the fields are

D - A A . —,. I . V. . z ‘
E' 2 E6 (x cos 6, —- zsm 6,) e 1‘1““ ”“61 + (0°61),

Hi — y.Eie—jf€1(rsin6, + zcos 6,)

"71
Er = E6 (x cos 6,. — 251119,.) e-JA'1(4L 51116,. — 4 cos 9r),
r , .
H' : _yflP—jkfirsmfi—zcosm)
7h 1
A A' _3 z:.‘
Et = E5(xcos6,—zsm6,)e JAzil" 1116t+ cos6,),

Ht = ygdp—jkzfir sin 6, + 3 cos 6,)
772
As with the perpendicular polarization case, 6,. = 6,, and the z-directed wave

. E . . . .
Impedance Z = -H—‘T should be continuous at the Interface. In parallel polarization,
y

 

 

 

 

Z, ~=0 ___ Ejl) _ cosf),(E€, + E6)
f H51) 2: 0 (E5 — E5l721
E' Er
: 711 (:0ng E'— + E5 (2.20)
E?)
Z2lz=0 (2) = 772 cos 6,. (2.21)
Hy 2:0

 

Then, in order to satisfy the continuity of the tangential wave impedance at the

interface,
E1 + E6
7 cos6,—-— —' - c086,
ll IEO_ E6 712 t
or
17, cos 6,(E(', + E6) = 772 cos 6,(E(’, — E6).
Hence,

E6 772 COS 6t _ 771 COS 6,; 262) — Z6:
E5 772 COS 9: + ’01 cos 6,- Zlig) + Zjl

 

11

In (2.22), 7], cos 6, and 772 cos 6, are the z-directed wave impedances in region 1 and

region 2 respectively. In order to find the transmission coefficient T, use
A 7 A t
Z X (El + Er)lz:0 : Z X E IZZOa

and thus

(1+ T) cos 6,-E5 = cos 6,E6.

Therefore,
t

T 1% (1+P)

cos 6, 2772 cos 6,
E0

 

_ _ 2.23
cos 6, n2 cos 6, + 77, cos 6, ( >

Note that (2.23) is the ratio of the total electric fields at the interface. Hence, later

when we apply the tangential boundary conditions at the interfaces in order to

obtain the global reflection coefficients of the layered medium, we use

 

’”:1+F. 92%

This is the tangential ratio of the transmitted to incident electrical field.

Finally, collecting the results from (2.22) and (2.24),

 

F” = M T” 2 H p
(2) (1)’ ’
Zn + Zn (2.25)
. k: 2
2(1) 2 771 cos6,, 2(2) 2 772 cos6, = 772 2’ .
n n k,

2.3 Reflection from multiple layers

In this section, the approach used in [15] will be reviewed, and the results will
be specialized for the current problem.
Consider N + 1 regions of space separated by N plannar interfaces as shown in

Figure 2.3, and assume that a uniform plane wave is incident on the first interface

12

at angle 6,. Each region is assumed isotropic and homogeneous with a
frequency-dependent complex permittivity and permeability. As shown in the single
interface case, in order to satisfy the boundary conditions, each region, except

region N, contains an incident-type wave of the form
1' _ 2' —jk"-r
E — E06
and a reflected-type eve of the form

r _ r ,—jkr-r
E — 06 .

In region 71, the wave vectors are described as

Here, in order to satisfy the boundary conditions, Snell’s law of reflection should
hold. Therefore,
km 2 km 2 k0 sin 6,. (2.26)

For the case of perpendicular polarization, the electrical field in region n,
O_<_n_<_ N— 1, isE=Efl+Efl where

E; _ yan+le—Jk:r,nxe—Jkz,n(z—zn+1)

\
0

ET : ybn+1(“g—jkxmxe'l‘jkznxz _ Zn+l),

n

and the magnetic field is H = H; + H1”, where

 

 

H2, = —szl: + 2km (111+le—jkm,nate—jkz,n,(z -— Zn+1),
nnn

Hf, : +xk‘z; + zkm bnl+le—jkr,nate+jkrz,n(z — 2H,).
~n7ln

13

When n = N there is no reflected wave, therefore

EN = S’GNHB—JkI’Nle‘Jk‘zv-N(“ — “'N)

9

"" A f~ 7 AA‘ ’ r. A I '_ ,,
HN : xk.,.~ + Z “Vaw“e—ka,l~'-Fe—Jfrz,;v(Z _ ZN).

knnn

 

Since (1.1 is the known amplitude of the incident wave, there are 2N unknown wave
amplitudes, and 2N simultaneous equations by applying the boundary conditions at
each of the interfaces. At interface n located at z = 2”, l S n S N — 1, from the

continuity of tangential electric field

an + b, = an+1e—jkzv"(z"— Zn“) + bn+le+jkzfi(z" — Zn“) (2.27)

while from the continuity of magnetic field

 

 

k n l A n 1 k n k 7
Z, — ‘z! — Z, -l » 4w - Z
_'an_:_—__ biz-if— : -(l,,+1 .. 6 J z,n( n n+1) +
kn— 1 Tin—1 An — 1 7711—1 A717)"
k. n ' ,. 7 _
er-l ‘ 6+]f'»z,n(~n Zn+1) . (228)
'n 11

Notng that the wave impedance of region 71 for the perpendicular polarization is

knnn
kz n

7

 

Zln :
and defining the region 71 propagation factor as

Pn = e_jkzv"A" (2.29)

14

where An = zn+1 — Zn, (2.27) and (2.28) become

anPn + ann 2 an.“ + bn+1f)3, (2.30)
Z n— Z n— .
—a,,P,, + w, = —a,,+1—i—1'- + (twp—#123. (2.31)
Zin 2111

When 2 = ZN, (2.30) and (2.31) hold if we set bN+1 = 0 and PN = 1. The 2N
simultaneous equations (2.30) and (2.31) may be solved using standard matrix
methods. However, through a little manipulation, these equations can be solved by

recursion. By subtracting (2.31) from (2.30),

 

 

 

 

Z n_ Z n-
2a,,Pn = a,+1 [1+ i 1] +bn+1P3 [1— i I] . (2.32)
in Z_Ln
Defining
Z n _ Z n—
,, i L 1 (2.33)
Zin + Z1n—1
as the interfacial reflection coefficient for interface 72., and
2Z1n
Z1n ‘1' 2171-1 ( )

as the interfacial transmission coefi‘icient for interface n, (2.32) can be written as

an.“ 2 anTnPn + bn+1Pn(—Fn)Pn. (2.35)

Finally, if we define the global reflection coefficient R, for region n as the ratio of

the amplitudes of the reflected and incident waves,

15

(2.35) can be written as

an+1 = a'nTnPn + an+an+1B1(—Fn)Pn- (236)

If we choose to eliminate an“ from (2.30) and (2.31) we find that

bn : (InFR + R‘Il+1Pn(1 _‘ Fn)(1,,+1. (2.37)

Equations (2.36) and (2.37) have nice physical interpretations. Consider Figure
2.4, which shows the wave amplitudes for region 77. We may think of the wave
incident on interface n + 1 with amplitude an“ as consisting of two terms. The first
term is the wave transmitted through interface n. This wave must propagate
through a distance An to reach interface n + 1 and thus has an amplitude anTnPn.
The second term is the reflection at interface n of the wave traveling in the —z
direction within region n. The amplitude of the wave before reflection is merely
b.,,+1P,,, where the term P" results from the propagation of the negatively-traveling
wave from interface n + l to interface 77.. Now, since the interfacial reflection
coefficient at interface n for a wave incident. from region n is the negative of that for
a wave incident from region n. - 1, and since the reflected wave must travel through
a distance An from interface n back to interface n + 1, the amplitude of the second

term is b.n+1Pn(—Fn)Pn. Finally, remembering that b,,+1 = R,,+1a,,+1, we can write

an+1 = anTnPn + an+1Rn+1Pn(_Pn)Pn-

This equation is exactly the same as (2.36) which was found using the boundary
conditions. By similar reasoning, we may say that the wave traveling in the ——z
direction in region n —- 1 consists of a term reflected from the interface and a term

transmitted through the interface. The amplitude of the reflected term is merely

16

anFn. The amplitude of the transmitted term is found by considering

b,,+1 = R,,+1a,,+1 propagated through a distance An and then transmitted
backwards through interface 72. Since the transmission coefficient. for a wave going
region n to region n — 1 is 1 + (—F,,), the amplitude of the transmitted term is

Rn+1P,,,(1 —— Fn)a,,+1. Thus

bn : Friar: + Rn+an(1 — P71)(l‘n+la

which is identical to (2.37).

Now, from (2.34) and (2.36)

 

 

(1 + I‘n)P,,
..,, = , n. 2.38
a +1 1+ rrlR'n+1P,fa ( )
Substituting this into (2.37)
b + +1 " an. (2.39)

n = 1+ I‘11F2n+11jy?

Using this expression we find a recursive relatirmship for the gloval reflection

coefficient:

bn r71. + R71+1P2
R, = -— = n, . 2.40
an 1+ Pan-prf ( )

 

The procedure is now as follows. The global reflection coefficient for interface N is

obtained from (2.40) with RN+1 = 0. We next find RN_1:

2
1“IV—1+ RNP-v_1

I

R _ = ’ '
N 1 1+I‘N_1RNID,<'-1

 

This process is repeated until reaching R1, whereupon all of the global reflection
coefficients are known. We then find the amplitudes beginning with al, which is the

known incident field amplitude. From (2.40) we find I), = 0.1121, and from (2.38) we

17

find
(1 + F1)P1
0.2 = _———-—2 a].
1 + FleP,
This process is repeated until all field amplitudes are known.

Note that the process outlined above holds equally well for parallel polarization

as long as we use the parallel wave impedances

 

when computing the interfacial reflection coefficients.
Now, specializing (2.40) for the problems of interest, first consider a
PEG-backed lossy slab. In this case N = 2 and R2 = ——1, thus from (2.40) the

global reflection coefficient in region 0, which is free-space, becomes

1‘, — P,2
R = ————.. 2.41
1 1 — 1“le ( )
Next, for an air—backed lossy slab where R2 = —F 1, the global reflection coefficient
in region 0 becomes
1‘1 (1 — P?)
R = ———.——. 2.42
1 1— rip; ( l

18

 

. Z: 0
region 1: cf, n,

V

‘ C
reg1on 2: a, , u,

Figure 2.1: Uniform plane wave incident on a planar interface between two lossy

regions of space. Perpendicular polarization.

19

 

z = 0
region 1: 810, u, region 2: 8;, u,

Figure 2.2: Uniform plane wave incident on a planar interface between two lossy
regions of space. Parallel polarization.

20

 

 

 

 

2:2, 2:22

 

an+l

——>

<——

bn+l

 

Z:le+l

a
__N,

{-—

bN

z=zN

 

Figure 2.3: Interaction of a uniform plane wave with a multi-layered material.

21

 

 

('1)

2:2,

(n+1)

ZZZIHI

 

n+1

n+1

Figure 2.4: Wave flow diagram showing interaction of incident and reflected waves

for region n.

22

Chapter 3: Natural Mode Analysis
of a Conductor-Backed Slab

3.1 Formulation of the frequency-domain

reflection

3.1.1 The frequency-domain reflection coefficient

Consider a plane wave of frequency w incident from free space onto an interface
between free space and a conductor backed slab of material with frequency
independent material parameters 6 = 6,60, 11,0, and a, and with thickness A, as
shown in Figure 3.1. In the figure, region 0 and region 1 correspond to free space
and the dielectric slab respectively. Likewise, for the subsequent discussion, the
subscripts 0 and 1 correspond to free space and dielectric, respectively. As seen in

(2.29), (2.33) and (2.41), the reflection coefficients have the general form

2 PM — PM
1 — I‘(w)P2(w)

 

R(w)

where F(w) is the interfacial reflection coefficient, given in general form by

Z,(w) — Z0

”“2 = m

and P(w) is the propagation factor, defined by

P(w) = 6—].sz A.

23

 

Since

.52 __ .2 .‘2
A'1 — kaml + Az,1
where

.. (7
[cf 2 wgnoec, 6C 2 e + —_-—,
3w

A7,, 2 k0 sin 6, = w 1106081116,,

the z—component of the wave number k, becomes

k3,, = Mk? — k3, = (AP/£06 — rug/.1060 sin2 6,

0'

= \flletok + —) — w2uueo sin2 6,
ya)

 

 

 

 

. 0 . .
= LUZ/1,060“, + _ — sm‘2 6,)
Jw’fo

a . w a
= k0\/6, + _ — sin2 6, = —\/e,. + , — sin2 6,. (3.2)
33160 c jwro

 

 

 

 

Defining E = 6,. — sin'2 6,, (3.2) can be written as

 

 

 

 

k3,, _ gFa), Fe») = 5+ .0 .
c wao
Hence, for perpendicular polarization
kim 770 720
Z ‘ E Z : = = . Z E Z = . 3.3
‘M M”) 1,, F(w)' 0 0t c056, ( l
and for parallel polarization
k, 7 F u)
Z1(w) E Z|](w) = _k1_771 = M, Z0 E Zn” 21706080,. (3.4)
1 6r + jwco

24

As a result, (3.1) is written as

I‘(w) — 6’31”“)

 

R(w) = 1 _ I‘(w)6—jWT(“’), (3.0)
where
7(3)) 2 ZéF(w).
0

3.1.2 The impulse response

Since R(w) exists, its Laplace domain representation also exists [21H23] and is

given by
F(S) _ e—sr(s)

 

 

 

R(8) : R(wl)lg,v:s/j : 1 __ F(S)e—3T(s)l (3-6)
where
37(8) = jw7'(w)|w:S/j
2A 2A
= J'w—F(w)l..v=s/j = s— a + 3—
C C 860
2A
= _fifi S + ‘—_
C 606
2A
= _fim, (3 7)
21(8) — Z0
F . = —. 3.8
(9) 21(3) + Z0 ( )
Here 1/ = c/fi and so = —o/(€0E). For perpendicular polarization the z-directed
impedance 21(3) is
770 770
21(8) = F( > = _
w w=s/j e + f;
— "”3 (3.9)

 

25

and for parallel polarization

770, [6+ 33—0

_ 0
fr + 560

770117191)

0
6,. + jwéo

2.,(3) =

 

w=s/j

‘Uofififl

ser+i

 

(3.10)

Therefore, the interfacial reflection coefficient F(s) in (3.8) can be written as

scos 6, — 63/563 — so

P _ scos 6, + fifim' (_L p01.) 31
(s)— fififl—cos6,(ser+%) (ll 1) ( . 1)
p0.

fifivs — so + cos 6.,(56, + 203).

 

 

The reflection coefficient R(s) is the transfer function of the system. Thus, the

impulse response 'r(t) is obtained by

r(t)=£‘1{R(s)}— 1 R(s)eStds (3.12)

— .72” 87'

where Br indicates the Bromwich path. This integral can be evaluated by contour
integration. We must thus examine the singularities and define an appropriate

branch for the integrand.

3.2 Singularities and the branch cut

From the complex square roots in 8T(8) and Z1(s), the branch points are
located at s = 0 and s = so. In order to ensure the continuity of R(s), the branch
cut is taken along the negative real axis between the two branch points.

From (3.6), the poles are the roots of
1 — F(s)e—ST(3) = 0. (3.13)

26

Taking the logarithm of this equation and rearranging gives

1n (“fie—37(3)] = 1n |F(s)e'ST(S)

 

+jarg [F(s)e—ST(S)] :l: j2mr = 0, (n. = 0, l, 2, - - - ).

Since —7r < arg [F(s)e‘”(5)] < 7r and the goal is to find roots, n can be set to 0.

Therefore, the poles should satisfy

ln |F(s)e_ST(5)

 

+ j arg [F(s)e‘ST(S)] = 0. (3.14)

Except for the lossless case (a = 0), wherein the poles can be calculated
analytically, the above equation needs to be solved numerically. For the lossless
case, both F(s) and T(s) are independent of s, and *1 < F < 0, i.e. arg(F) = 7r. As

a result, from (3.14), the lossless case poles are
1 .
s = —T- [111 IF] i 3(2n —1)7r], n = 1,2,3, - .. (3.15)

From (3.14) and (3.15), it is clear that the poles occur in complex conjugate pairs as
expected for real signals. Table 3.1 shows the lossless case poles, where 6, = 0°,

6,. = 9 and A = 2 cm. In the table, only poles in the upper-half complex plane are
shown.

When the conductivity is nonzero, equation (3.14) can be solved by 2-D root
search algorithms such as the N ewton-Rahpson method [17], which was used in this
work using the lossless case poles as the initial guesses. At this point, it should be
emphasized that the real part of a pole should be less than or equal to zero for a
passive system. This can be shown to be true by noting that the magnitude of
I‘(s)e“"(s) in (3.14) is always less than 1 if Re{s} is greater than zero, since both
the magnitudes of F(s) and 637(3) are less than 1 as can be shown using (3.7) and

(3.11). Here, it should be remembered that so = ——0/(eoE) is a non-positive

27

constant. As a result, no solution of (3.14) can exist if Re{s} is positive.

Figure 3.2 and Figure 3.3 show several pole trajectories obtained by the
Newton-Rahpson method when the conductivity 0 varies from 0 to 10 S / m for
perpendicular and parallel polarization respectively. When the conductivity reaches
a certain value, each pole becomes purely real. In the figures, the conductivity
values at which the poles become real are shown, and the corresponding pole
locations in the complex 3 plane are shown as the cross—marks. As shown in the
figures, when the conductivity increases, the magnitudes of the imaginary parts of
the poles decrease whereas those of the real parts increase until the poles approach
the real axis and become purely real. This phenomenon is clearly seen in Figure 3.4
and Figure 3.5. Note the difference between the mode 0 pole trajectories of the
perpendicular and parallel polarization cases.

These pole trajectories of both polarizations suggest that when the
conductivity becomes higher, the amplitude decay factors become larger, resulting
in a smaller contribution to the late-time response.

In Figure 3.6 and Figure 3.7, the pole trajectories for varying angle of incidence
are shown, where e, = 9 and 0 = 0.1 S / m. As seen in Figure 3.6, which depicts the
pole amplitudes of the perpendicular polarization case, as the angle of incidence
increases, the magnitudes of the real parts of the poles monotonically decrease
whereas those of the imaginary parts are relatively unchanged. At this point, it
should be emphasized that the pole locations do not solely determine the late-time
transient responses. In order to obtain the correct transient responses, the residues
corresponding to the poles should be considered. For the parallel polarization case,
as the angle of incidence increases, the magnitudes of the real parts of the poles

show a dip around 70° of incidence angle which approximately corresponds to the

28

Brewster angle calculated as [15]

. _ E2 . _ 9
6 = 8111 1 2 sm 1 —— = 716°,
8“ 514,52 V 1+9

where we assume the dielectric material is lossless and 6,- : 9. Also, it should be

 

noticed that the mode 0 pole becomes real approximately at this angle. This
behavior evidently suggests that the late-time response may be affected by the
Brewster angle. Further discussions 011 this behavior will be made in the result
section of this chapter. The existence condition for the real poles and the algorithm
for finding such poles are examined next.

Let us assume a complex frequency 8 to be negative real and denote its

magnitude as :r, where :1: > 0. Then, ifs > so, i.e., a > —seo'€, from (3.7) and (3.11)

— —j—\/_.\/—( 1+ so), (3.16)

and the reflection coefficient becomes

—:rcos6, — jfiflV—(IE-l- so)

_srcos6, +jfifim (i pol.)
m8) 2 (3.17)

jfifim—cw‘M-mfi) (n 01>
jfifime cos 611‘“? + €05). p '

 

It is now clear that the magnitudes of both 6""(3) and F(s ) are 1. As a result,
ln |I‘(s)e"“(3)| = 0, and the poles need to satisfy only the imaginary part of (3.14),
i.e., arg [F(s)e‘"(3)] = 0. Therefore, if we denote —(2A/z/)\/E\/—(.r + so), which is

—sr(s), as d), and arg [F(s)] as Q52, the poles need to satisfy
C61+¢2+2TL7T=0 n=0,1,2,--- (3.18)

Here, only the positive sign in front of n is necessary, since (1), is always negative,

29

when 3 > so and —7r < gbg < 7r. The above equation, which we call the characteristic
function, can be solved by any l-D root search algorithm, such as the secant
method, which was used in this work. Here, it is very important to note that (3.18)
does not always have solutions, and that the number of solutions is dependent upon
the conductivity, permittivity and incident angle. In addition, great care should be
taken when finding the real roots, especially for the n = 0 case where the
characteristic function can have more than one root, which may cause the usual
secant method to fail. Figure 3.8 and Figure 3.9 show the amplitude of the
characteristic function versus the amplitude of s, which is purely real. In Figure 3.8,
the 0 amplitude line was crossed 3 times by the characteristic function, and the
corresponding values of s are the poles. Here, the left end point of the amplitude
curve of the n = 0 case corresponds to the removable pole located at s = so as
shown in the following discussion. On the other hand, there is no such crossing in
Figure 3.9 implying no pole exists for that case.

If s < so, ST(S) becomes negative real causing €_ST(S) to be greater than 1, and
F(s) becomes real. Therefore, in (3.14), arg [F(s)e”"(3)] = 0 can only be satisfied
when F(s) > 0. If s < so and F(s) > 0, the poles only need to satisfy
ln |F(s)e‘"(s)| = 0. Figure 3.10 shows the amplitudes of the reflection coefficients
for both polarizations, where s < so. As shown in the figure, for perpendicular
polarization, the reflection coefficient F(s) = 1 when 5 = so and monotonically
decreases as 3 decreases. In particular, it becomes negative when 8 < 60;, since

(1‘57')

the reflection coefficient has roots

0

= 0 —.
S 1 60(1— 61‘)

As a result, one additional real pole can exist when < s < so for

__£’_
60(1—Cr)

perpendicular polarization. Figure 3.11 shows the amplitudes of —sr(s), ln|l‘(s)|

30

and their sum for perpendicular polarization. In the figure, the point where

ln|F(s)| — sr(s) crosses the 0 amplitude line represents the pole location. It should
be noted that the apparent pole s = so where both €_ST(S) and F(s) are 1 is a
removable pole, since the numerator of the reflection coefficient becomes identical to
the denominator.

For parallel polarization, F(s) = —1 when s = so, and it has two real roots at

0' 0'

s = —, .
eo(1 — 6,) eo(tan2 6, — 6,.)

 

Therefore, depending on 6,. and 6,, the amplitude of the reflection coefficient
increases initially, then stays positive or monotonically decreases as 3 decreases. As
a result, at most 2 real poles can exist when 3 < so for parallel polarization. Figure
3.12 shows the case where 2 real poles exist. However, it should be noted that the
contribution to the late time response from the real poles located where s < so are
dominated by the contributions from the other poles, since their real parts are much
smaller than those of the dominant poles.

A complete sketch of the poles is shown in Figure 3.13 and Figure 3.14. In the
figures, the solid lines represent the trajectories of the complex poles, and the
cross~marks represent the real poles. The real poles consist of the poles with
magnitudes less than so and those with magnitudes greater than so. Since the
complex pole trajectories from different modes follow approximately the same curve
in the complex 3 plane (before each mode becomes real and diverges), it is difficult
to resolve each complex pole trajectory from the figures. As previously discussed,
the complex poles gradually become real poles as the conductivity increases. Note

that, except for mode 0, each single complex pole splits into a pair of real poles.

31

3.3 Evaluation of 'r(t)

Since all poles and branch points lie in the left half plane including the
imaginary axis, the region of convergence is the right half plane. Therefore, the
Bromwich path is put in the right half plane and the Laplace inversion integral is
evaluated by contour integration. The evaluation is accomplished in two different

time intervals, corresponding to the early-time period and the late—time period. The

2A

beginning of the late-time period is To = 7, which represents the two-way transit

time of the wave inside the dielectric slab.

3.3.1 Case I: t > T0

W'hen t > To, the integration contour is closed in the left half plane as shown in
Figure 3.15. Inside the contour, the branch cut lies from 0 to so and there is a
closed path enclosing the branch cut and several possible real poles on the branch
cut (two of which are shown). We denote the outer integration contour which
includes the Bromwich path as Co and the closed path which encloses the branch

cut as C,, such that

Co _—; Br+Foo+L1+L2.

01 ’7'1+’)"2+"°+"f6+11+12+"‘+15.

Then, by Cauchy’s residue theorem,

/ R(s)eStds = fin: Res[R(s)eS‘, poles] (3.19)
Co+C1

Therefore, if the integral contribution from each path is known, the impulse

response r(t) can be determined.

32

The contribution from [‘00

On Foo, from (3.7),

R€{T(S)} = Re{ lim 3A— 1+_0_}

Re{s}—»—oo z/ seoE

2A

— =71)
1/

I 2

If we let t1 = t — To, which is positive since we are evaluating the integral when

t > To, then

P , (8T0 _ —sr(s) ’37-0 .
/mR@M“mwi[x(?:p@;flwj €“mwj/mneeMs

Since f(s) —+ 0 on Foo, by Jordan’s Lemma [20], [21],

 

/ R(s)eS’~ d3 = 0 (3.20)
Foo

The contribution from L1 and L2

On L1 and L2, it is not possible to apply Jordan’s Lemma without imposing
some restrictions that will be introduced in subsequent discussion, since R(s) does
not approach zero over these paths. Fbrthermore, it would be very difficult to
evaluate the inversion integral directly. However, it is possible to apply Jordan’s
Lemma if we use the following theorem [14]:

Theorem Let f (t) be a function which is APC { almost piecewise continuous) and
which is identically zero fort greater than some number T. Then the Laplace
transform of f (t) approaches zero uniformly as 3 becomes infinite in a right half
plane,

wee—ans;

where co = oo + jwo is any complex constant in the 5 plane.

33

In this theorem, the AFC function is piecewise continuous except at a finite
number of isolated points. Let the time response of the incident wave be given by
g(t), and let this function be AFC. Then the response of the system is given by
G(s)R(s), where 0(8) 2 £{g(t)}. Since C(s) approaches zero 011 L, and L2, we can
use Jordan’s Lemma to show that the contribution from L1 and L2 is zero for an
APC input waveform. W’hen considering the impulse. response, we view the delta

function as a. functional that is AFC [21], [24], and thus

/ R(s)cS' ds = 0 (3.21)
L1.L2

The contribution from C1

The segments of the contour C, enclosing the branch cut can be divided into
three groups. The first group consists of “y, and “/4 that enclose the branch points so
and 0 respectively. The second group consists of the straight lines immediately
below and above the branch cut, and those are designated 1,, [2, . - - , l6. The last
group consists of 72, 73, “/5 and ”)6 that enclose the real poles on the branch cut.
These groups will be examined separately.

Denote the radius of ”)1 as r, and let 6 to be the angle measured
counterclockwise from the real axis to the point on ”)1. Then, any point 011 ”)1 can be
represented as

s = so + r619.
The reflection coefficient on 7, with r -—> 0, i.e. s ——> so, becomes

3770 ’ Zo1\/_E\/§v3 '— 80

r( ) 8770 + Zufifi s — 80 (3 22)
s = .

Thfl/fiJS—VS — 80 — 20]](86r + %)

nofifivs — so + Zo],(s€,. + %)

 

——> 1, _L- pol

 

——-> —-1. || -1)01

34

Then,

 

I‘ _ ’—sr(s) est ’ _L _ p01
R<8>€St : 1 (51‘ ‘e —sr(s)68t =
_ (5)8 —es‘ . ll - p01
and
]R(S)€St] S (,(so+'r)t'
Therefore,

_<_ 27rre<s°+rlt —> 0, (r ——> 0).

 

 

/ R(s)e8’ ds
71

As a result,

/ R(s)e5" ds = 0. (3.23)

’71
Similarly, it can be shown

/ R(s)e3‘ ds = 0. (3.24)

A!

I4

For the second group, it. is necessary to determine F(s) and sr(s) along a path
immediately above the branch cut, which we denote as B+, and immediately below
the branch cut, which we denote as B“. If we denote the magnitude of s as :1:

(so +r < —:r < —r < 0), then on B+,

\/§vs— 30 =jfiV—1" “30,
and on B“,
\/-8-\/8 — 80 = —j\/.I'_-\/ —'$ — 80.

Now, let Z(s) on B+ be denoted as Z+, and Z(s) on B“ as Z“. Then, using (3.11),
it can be shown that

2+ = —Z“. (3.25)

35

Since,

 

 

R l.
(S) 1 — I‘(s)c—ST(5)
_ [2(3) - Zak-28"” — [2(8) + Zola-lets)
[2(5) ‘1‘ 21)]833T(3) - [2(3) — ZO]€—%sr(s) ’
we have,
2+ - Z J" — Z+ Z _i—m
R(3)]B+ :: [ 01( 1 + 0]?

 

[Z+ + Zo]eJ‘-" — [Z+ — Zo]e“j9”
jZ+ sin (b — Zo cos a)
jZ+ sin (b + Zo cos cb

 

where jrb 2 597(3), and

[Z“ — Zo]e“j‘l’ — [Z“ + Zo]cj‘l’
[Z“ + Zo]€“—7¢’ — [Z“ — Zo]eJ¢
—jZ“ singb — Zocoscb
—jZ‘ sin¢b+ Zocosqb

 

R(8)]B- =

 

Therefore, by (3.25)
3(8)|B+ = “fills--

As a result,

/ R(s)|B+ 6“ ds +/ R(s)|3— e” ds = 0 (3.26)
l1+12+13

l4+fs+16
The integral contributions from the third group and the other complex poles
can be determined by calculating the residues of R(s)e5‘ at the poles. It is found

that all of the poles of R(s) are of first order and thus

es' =Akesk’, (3.27)

 

P _ —sr(s)
R68 [R(S)eStl [90188:] [328k = 3121511,.(8 — 8k) 1 E-S)F(5686—6266

36

where

F(.Sk) _ e-SkT(-9k)
A, = — (3.23)
2143' [r(sle—ST(S)1 13:3,,

is the complex natural mode amplitude. Carrying out the details gives

 

. . . A 2. — -
Ak:e_%‘/;‘/5_‘SO G(s)——F(s)——S 5° , (3.29)

where

2cos6\/:\/_\/S— COSQV—fim
32cos26,+2s\/§\;§\/8—80+68( (8’80) 1

for perpendicular polarization and

C(s) =

0 2s — so
cos 6,(se,.+ +—)\/é'————— —26, cos 6,\/%\/s\/s — so
0(8) : V563 — so

Es(s——so)+2cos6,( ser+ 6(:))\/:\/s\/s—so+cos2 6,( (se,+ +3)

60

2

for parallel polarization.
Collecting the results from (3.20)-(3.29), the impulse response r(t ) becomes

1
r(t): Jt—QW/BXS) e—“ds — Z Akes” t > TO, (3.30)

Br

and thus the late-time period is a pure natural resonance series, and contains no

branch-cut contribution.

3.3.2 Case II: t < T0

For t < T0, r(t) is found by computing the inverse Laplace transform of

R(s) = 1(9) + B(s), where

 

Then
1 -—sr(s)

' 8
Z —_ F2 1" — 1 {st 1 ‘.
j27l- Ar[ (8) 1 1_ F(.S)€_8T(3)P ( .5

To compute this integral, the integration contour is closed in the right half plane, as

so) = c’1{fi(s)}

 

shown in Figure 3.16. On Foo, from (3.7),

213
Re {7(5)} = R0 {Elegy—Loo 7 1+ Ego—F}
2A
1’ -— = 7'0
1/

If we let t, = t — To, which is negative since we are evaluating the integral when

t < 7'0, then

—ST(.S)CST0

— S e 81 S] ,
/ R(s)e 'ds 2/ [F2(s) — 1] 1_ F(s)e—ST(S)6 ' ds = (s)e ’ ds

Foo

 

Here, it is not possible to apply Jordan’s Lemma directly, since f (s) does not
approach zero over the entire contour Foo. However, if we use the theorem

introduced earlier, it can be argued that

/ B(s)e3t ds = 0

00

The inverse transform of F(s), denoted as F(t), can be found in [26] and [27]. As a

result, when t < T0, r(t) becomes
r(t) = F(t), t < TO. (3.31)

Combining the results (3.30) and (3.31), the impulse response r(t) is

Z Akeskt, t > To
r(t) = (3.32)
F(t), t < 7'0.

38

3.4 Results

It has been shown analytically that the early-time response is a specular
reflection from the interface between free-space and the dielectric slab, and the
late-time response is a pure sum of damped sinusoids. In order to verify these
results, the natural mode series is compared to the direct IFFT using a truncated
Gaussian pulse as the input waveform. This waveform is given by f (t) = e“"(‘“")2/72,
where r and u are 0.1 and 0.15 ns respectively, and is shown in the inset of Figure
3.17. Figures 3.17-3.25 show the transient responses calculated by the natural mode
series and the IFFT. In addition, the IFFT of the interfacial reflection coefficients
F(s) are also shown in order to explain the early-time behavior. The natural
resonant frequencies of the corresponding cases are shown in Table 3.1-3.9.

As an example of the results, in Figure 3.19 the incident wave is in
perpendicular polarization with an incidence angle of 20°, and the material
parameters are A = 2 cm, 6,. = 9 and o = 0.1 S / m. Therefore, the beginning of the
impulse response late-time, i.e., the two-way transit time of the wave inside the slab,
is To = 0.397 ns. As shown in Figure 3.19, during the early-time period, before To,
the IF F T of R(w)F(w) (where F (a) is the spectrum of the input waveform)
matches well with the IF FT of F(w)F(w), which is the specular reflection from the
interface. During the late-time, the IFF T of R(w)F(w) matches well with the
natural mode series. It should be noted that when t < To the response from the
natural mode series does not have any meaning; i.e., the natural mode series is valid
only when t > T0, the late—time period. Since the frequency band of the input
waveform is roughly 0 — 20 GHz, it was found that the first 7 natural modes, which
are shown in Table 3.3, are sufficient to represent the late-time reflected field.

Figure 3.17 shows the response from the lossless, normal incidence case. As
seen in the figure, there are equally spaced replicas of the input waveform with

gradually diminishing amplitudes during the late-time period. Note that the

39

amplitude of the first peak is greater than that of the specular reflection in this case.

Comparing Figures 3.20 and 3.21 to Figures 3.18 and 3.19, due. to the relatively
higher conductivity in the Figure 3.20 and 3.21 cases, relatively smaller late-time
responses are observed. The corresponding natural resonance frequencies for these
cases are shown in Tables 3.4 and 3.5, and it is seen that the first mode poles of
these cases are purely real.

The responses from both polarizations with the incidence angle 6., = 70°, which
is close to the lossless case Brewster angle, are shown in Figures 3.22 and 3.23. As
seen in Figure 3.22, the interfacial reflection is very small compared to the first peak
of the late—time response for the parallel polarization case. Also, note that no
subsequent late-time response is observed for this case. This property is not
observed for perpendicular polarization case, as shown in Figure 3.23.

Figures 3.24 and 3.25 show the responses when the incidence angle is large
(85°) for parallel and perpendicular polarization respectively. As seen in Table 3.8,
the parallel polarization case has a real pole with a relatively smaller amplitude. As
a result, the late-time response from the parallel case shows a slower decay rate
than the perpendicular case. On the other hand, it is observed that the late—time
response from the perpendicular case is smaller than the parallel case due to its high

interfacial reflection.

40

Table 3.1: Poles and corresponding complex natural mode amplitudes (Ak), loseless
case (6,. = 9, A = 2 cm, 6, 2 0°)

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

mode 0 —.17329 x 1010 .78540 x 1010 .37500 x 1010 .70819 x 10“6
mode 1 —.17329 x 1010 .23562 x 1011 .37500 x 1010 .26797 x 10“5
mode 2 —.17329 x 1010 .39270 x 1011 .37500 x 1010 .29858 x 10“5
mode 3 —.17329 x 1010 .54978 X 1011 .37500 x 1010 .10715 x 10“5
mode 4 —.17329 x 1010 .70686 x 1011 .37500 x 1010 .58186 x 10“5
mode 5 -.17329 x 1010 .86394 x 1011 .37500 x 1010 .15007 x 10“4
mode 6 -.17329 x 1010 .10210 x 1012 .37500 x 1010 .64309 x 10-5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

41

Table 3.2: Poles and corresponding complex natural mode amplitudes (A1,), [I polar-
ization (6,. = 9, A = 2 cm, 6, = 20°, 0 = 0.5 S/In)

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

mode 0 —.52539 x 1010 .62604 x 1010 .50445 x 1010 .26990 x 1010

 

mode 1 -.50527 X 1010 .23232 X 1011 .41113 x 1010 .69726 x 109

 

mode 2 —.50419 X 1010 .39239 X 1011 .40715 X 1010 .41149 X 109

 

mode3 —.50390x1010 .55133x1011 .40610X1010 .29261x109

 

mode 4 —-.50378 X 1010 .70990 X 1011 .40567 X 1010 .22717 X 109

 

mode 5 —.50372 X 1010 .86830 X 1011 .40546 X 1010 .18569 X 109

 

 

 

 

 

 

 

 

mode 6 ——.50369 X 1010 .10266 X 1012 .40534 X 1010 .15704 X 109

 

42

Table 3.3: Poles and corresponding complex natural mode amplitudes (Ak), J. polar-
ization (e, = 9, A = 2 cm, 6, = 20°, 0 = 0.1S/m)

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

mode 0 —.23079 X 1010 .77397 X 1010 .36509 X 1010 .32595 X 109
mode 1 -—.22824 X 1010 .23661 X 1011 .35377 X 1010 .11479 X 109
mode 2 —.22804 X 1010 .39494 X 1011 .35288 X 1010 .69136 X 108
mode 3 —-.22799 X 1010 .55315 X 1011 .35264 X 1010 .49433 X 108
mode 4 —.22797 X 1010 .71131 X 1011 .35254 X 1010 .38464 X 108
mode 5 —.22796 X 1010 .86946 X 1011 .35249 X 1010 .31477 X 108
mode 6 —.22795 X 1010 .10276 X 1012 .35246 X 1010 .26638 X 108

 

 

 

 

 

 

 

 

 

 

 

 

 

 

43

Table 3.4: Poles and corresponding complex natural mode amplitudes (4),), I] polar-

ization (6,. = 9, A = 2 cm, 6, = 30°, 0 = 1.0 S/m)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Pole Amplitude Natural Mode Amplitude
Real Part Imaginary Part Real Part Imaginary Part
—.52600 X 1010 0.0 —.68433 X 1010 —.51876 X 103
mode 0
—-.12581 x 1011 .19393 x 1010 .10205 x 1011 —-.16967 x 1010
mode 1 -—.85250 x 1010 .22393 x 1011 .45539 x 1010 .16527 x 1010
mode 2 —.85010 X 1010 .38947 X 1011 .45153 X 1010 .94467 X 109
mode 3 —.84948 X 1010 .55133 X 1011 .45059 X 1010 .66634 X 109
mode 4 —.84923 X 1010 .71204 X 1011 .45022 X 1010 .51564 X 109
mode 5 —.84910 X 1010 .87223 X 1011 .45004 X 1010 .42081 X 109
mode 6 —.84903 X 1010 .10322 x 101'2 .44993 X 1010 .35555 X 109

 

44

 

Table 3.5: Poles and corresponding complex natural mode amplitudes (Ak), _L polar-
ization (6,- = 9, A = 2 cm, 6, = 30°, 0 21.5 S/m).

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

mode 0 —.33106 x 1010 0.0 —.13299 x 1010 .20254 x 102
mode 1 -—.11233 x 1011 .21104 x 1011 .32151 x 1010 .17600 x 1010
mode 2 —.11227 X 1011 .38219 X 1011 .32366 X 1010 .97606 X 109
mode 3 -.11225 x 1011 .54621 x 1011 .32421 x 1010 .68370 x 109
mode 4 —.11224 x 1011 .70808 x 1011 .32443 x 1010 .52763 x 109
mode 5 —.11224 x 1011 .86900 x 1011 .32454 x 1010 .43002 x 109
mode 6 —.11224 x 1011 .10294 x 1012 .32460 x 1010 .36305 x 109

 

 

 

 

 

 

 

 

 

 

 

 

 

 

45

Table 3.6: Poles and corresponding complex natural mode amplitudes (Ak), [I polar-
ization (6,. = 9, A = 2 cm, 6., = 70°, 0 = 0.5 S/m)

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

 

 

mode 0 —.37966 x 1010 0.0 —.58702 x 1010 —.55113 x 103
mode 1 —.10993 X 1011 .21160 X 1011 .10655 X 1011 .48499 X 1011
mode 2 —.11619 x 1011 .39009 x 1011 .39566 x 1011 .45363 x 1011
mode 3 —.11821 x 1011 .56182 x 1011 .52026 x 1011 .36984 x 1011
mode 4 —.11908 x 1011 .73090 x 1011 .57936 x 1011 .30405 x 1011
mode 5 —.11953 x 1011 .89869 x 1011 .61121 x 1011 .25589 x 1011
mode 6 —.11979 x 1011 .10658 x 1012 .63104 x 1011 .22008 x 1011

 

 

 

 

 

 

 

 

 

 

 

 

 

 

46

Table 3.7: Poles and corresponding complex natural mode amplitudes (Ak), _L polar-
ization (cr = 9, A = 2 cm, 6i = 70°, 0 = 0.5 S/m)

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

mode 0 —.41396 x 1010 .71833 x 1010 .13691 x 1010 .67111 x 109
mode 1 —.41206 x 1010 .24473 x 1011 .12906 x 1010 .18876 x 109
mode 2 —.41192 x 1010 .41150 x 1011 .12854 x 1010 .11193 x 109
mode 3 —.41188 x 1010 .57748 x 1011 .12840 x 1010 .79695 x 108
mode 4 —.41187 x 1010 .74320 x 1011 .12835 x 1010 .61904 x 108
mode 5 —.41186 x 1010 .90880 x 1011 .12832 x 1010 .50615 x 108
mode 6 —.41186 x 1010 .10743 x 1012 .12830 x 1010 .42811 x 108

 

 

 

 

 

 

 

 

 

 

 

 

 

 

47

Table 3.8: Poles and corresponding complex natural mode amplitudes (Ak), || polar-
ization (e, = 9, A = 2 cm, 9,- : 85°, 0 = 0.5 S/m)

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

mode 0 —.11692 X 1010 0.0 -—.21180 X 1010 .20153 X 103
mode 1 —.50128 X 1010 .16532 X 1011 —.30588 X 1010 .57062 X 109
mode 2 ——.50326 X 1010 .33249 X 1011 —.31490 X 1010 .30084 X 109
mode 3 —.50365 X 1010 .49921 X 1011 -—.31678 X 1010 .20275 X 109
mode 4 —.50379 X 1010 .66584 X 1011 —.31745 X 1010 .15265 X 109
mode 5 —.50386 X 1010 .83242 X 1011 —.31777 X 1010 .12234 X 109
mode 6 —.50390 X 1010 .99899 X 1011 —-.31794 X 1010 .10205 X 109

 

 

 

 

 

 

 

 

 

 

 

 

 

 

48

Table 3.9: Poles and corresponding complex natural mode amplitudes (Ak), _L polar-
ization (Gr 2 9, A = 2 cm, 9,- = 85°, 0 = 0.5 S/m)

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

mode 0 —.36956 X 1010 .74632 X 1010 .33232 X 109 .15734 X 109
mode I —.36944 X 1010 .24705 X 1011 .32742 X 109 .46879 X 108
mode 2 —.36943 X 1010 .41468 X 1011 .32704 X 109 .27899 X 108

 

 

 

mode 3 —.36943 X 1010 .58168 X 1011 .32694 X 109 .19883 X 108

 

mode 4 —-.36943 X 1010 .7484? X 1011 .32690 X 109 .15451 X 108

 

mode 5 —.36943 X 1010 .91517 X 1011 .32688 X 109 .12636 X 108
mode 6 ——.36943 X 1010 .10818 X 1012 .32686 X 109 .10689 X 108

 

 

 

 

 

 

 

 

 

49

x
Free Space ADielectricg
\(region 0) (region 1)

 

 

A
7

g
Z

0, /
A10 8 , 90, 6

z;0 z=A

 

Figure 3.1: A uniform plane wave. incident from free space upon a conductor-backed
lossy slab.

50

 

Imaginary Part

N

 

 

 

 

Real Part X 1°

Figure 3.2: Pole trajectories of the reflection coefficient, J. polarization (6,. = 9, A = 2
cm, 91‘ = 30°).

51

x101o

 

A
j

Imaginary Part

 

 

 

 

2 r a
1 .- .1
0
—6 0
10
Real Part x ‘0

Figure 3.3: Pole trajectories of the reflection coefficient, || polarization (6,. == 9, A = 2
CHI, 61' = 30°).

52

x101o

 

mode 3

Imaginary Part

 

 

 

 

Conductivity (S/m)

Figure 3.4: The amplitude of the imaginary parts of poles vs. conductivity, _L polar-
ization (e, = 9, A = 2 cm, 0, = 30°, 0 = 0 ——> 10.0 S/m).

53

10

x10

 

mode 3

Imaginary Part

 

 

 

 

5 9

Conductivity (S/m)

Figure 3.5: The amplitude of the imaginary parts of poles vs. conductivity, H polar-
ization (6r : 9, A = 2 cm, 9,- = 30°, 0 = 0 ——> 10.0 S/m).

54

 

 

 

 

 

 

 

 

 

 

 

 

 

10
6 x . . .
t o o 0 o
:3.“ 0° 30° 45 60 75 3?
4 _ A; i L i f. _;
a 15° mode 2
.g a; J; 1‘. i L 1:
$2 - mode 1 J
g A A a 2. _A. a a
__ . . mode 0
3.5 -2 —1.5 -1 -0.5
X 109 Real Part x 109
—0.5 . . 1
mode 0
1: -1 -
CU
0' 1 5
To _
(D
CC .

 

 

     

 

—2.5 ‘ L
o 20 40 60 80 90
Incident Angle (Degree)

Figure 3.6: Pole trajectories of the reflection coefficient vs. angle of incidence, _L
polarization (6,. = 9, A = 2 cm, 0 = 0.1 S/m).

55

30°

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

10
1
5 X 0 I l I I T 1 I
t mode 2 62 4? 0°
6.“ 4 C . “ .
as - ‘° ‘° .
«3 mode 1 75 L A I. I 89
.E 2 l- C -l
o) a a
m
g 1 '71. mode 0 a a m ‘
O (71 1 4 I 1 1 1
-16 —14 —12 —1 O —8 —6 —4 —2 0
9 Real Part x 109
1
O x O I I T I I I I I I?
II"
I"
g -5 " ,I’ ‘
D. ,1
_ .I
8 -- mode 0 i1
a; -10 ---- mode 1 3 ‘
- - mode 2
a real poles
__1 5 I I l l l l

0 10 20 30 40 50 60 70 80 9o
Incident Angle (Degree)

Figure 3.7: Pole trajectories of the reflection coefficient vs. angle of incidence, H
polarization (6,- = 9, A = 2 cm, a = 0.1 S/m).

56

 

20 I I I I I I

 

 

O

   

 

Amplitude

 

l.
o

 

 

 

 

 

—1 5
-20
_25 1 1 1 1 1
—7 —6 —5 -4 -3 —2 —1 o
3 (real) me”

Figure 3.8: Characteristic function vs. real 3, _L polarization (6,. = 9, A = 2 cm,

a = 5.0 S/m, 6.,- = 30°).

57

 

20 I I
15

10-.

Amplitude

.................................
n

.................................

-10-

 

 

 

-15
-6 -5
9

—7
x10

Figure 3.9: Characteristic function vs. real 3, _L polarization (6,. = 9, A = 2 cm,

a = 0.5 S/m, 0, = 30°).

58

 

 

 

 

 

 

Amplitude

 

 

 

—0.4 ~ -
s = o/[eo(1 — er)]
-0.8 - a
:46 —4.4 —4.2 —4 -3.8 —3.6 —3.4

8 (Real) x 1010

Figure 3.10: Amplitudes of the reflection coefficients (6,. = 5, A 2: 2 cm, 0‘ = 1.5
S/m, 61' = 300).

59

 

   

 

 

 

 

 

 

 

 

 

6 E I I T -
4 — ‘ ‘ ~ ~ - - § .
E ‘ ~ ~ - s = s() ":
:V~ _ _ ‘ ~ ~ ‘
2 _ 3 s — o/[eo(1 er)] \ ~ g
a) 0
'D
2
'5. —2
E
_4 .
\, - I
\. 2 .,°
—6 )— El 4
2 - - —sr(s)
"8‘ ‘ ._.- lnII‘J-(SH H
j _ ln|I‘i(s)| — 813(3)
_10 l ' 1 ¥ 1 1 l 1 I
-4.4 —4.3 —4.2 -4.1 —4 —3.9 -3.8 —3.7 —3.6 —3.5
8 (Real) x 10‘°

Figure 3.11: Real pole (s < 80), (6,. = 5, A = 2 cm, a 21.5 S/m, 6, = 30°).

60

 

I T I

6

.-
~ '
“--
--
" ~
- ~ -
.-
’

 

Amplitude

 

- - -sr(s)
Inlr, ,(sII L

_ ln|I‘| l(s)| — 51(3)

 

 

 

 

 

L l

—12 Z 1
-4.4 —4.3 —4.2 —4.1 —4 —3.9
8 (Real) x 10

 

 

-3.8 —3.7 -3.6 —3.5
10

Figure 3.12: Real pole (s < 80), (6r = 5, A = 2 cm, 0 = 1.5 S/m, 9,- = 30°).

61

 

Pole Amplitude (Real Part)

 

 

 

 

_14 4 4 4 . a.
o 2 4 6 8 10
Conductivity (S/m)

Figure 3.13: Complete sketch of the pole trajectories showing both complex and real
poles, _L polarization (6,. = 9, A = 2 cm, 6,- : 20°).

62

 

 
   

 

 

 

x10
0 ',gaxba,—,§WWHW
u modeo ”',,gusuuannsuaaasssuss
_2-
’t‘
(13
Q —4~
(_U
(D as
E -6l’ 8
(D a a
U ”a: 3;: ”' '
a -8_ S—SO ".." fl” . _
E " .03.: '2. a
< “"‘.{{ " '.
2-105 33‘ {g 8' a-
Do. ‘u 5‘” ',
a“ .' g
a: ." "
.3 a
-12” '3 “N d
a” N
at.' '“
a
_14 1 1 1 l 1 1 l 1 1 L
0 1 2 3 4 5 6 7 8 9 10

Conductivity (S/m)

Figure 3.14: Complete sketch of the pole trajectories showing both complex and real
poles, I] polarization (6,» = 9, A = 2 cm, 6,- = 20°).

63

 

 

 

 

 

 

 

 

 

 

L1
( s=jb
Foo BI‘
A
Refi}
s=-I)
L2)r J
s=a s=c

 

Figure 3.15: The integration contour when t > T0. |a|, |b| -—+ 00, 0 < c < oo.

64

Im{s}4

Re
0 Brf‘ Re{s}

 

 

 

sia

Figure 3.16: The integration contour when t < T0. R —> oo, 0 < a. < oo.

65

 

 

 

 

 

 

 

 

 

 

1 I I I I I
- - IFFT I‘(o))F(00)
0,3 . ii a IFFT R(u))F(0)) _
— Natural Mode Series
0.6 r .
0.4 P 1

a) ' I:
_O 0.2
3

a 0 - " " "

E d

< _0 2 . I ‘ ..

' 2 2
_0.4 , I: f(t) = 97“”) h .
b I = 0.10 nsec
_0'6 h 11 = 0.15 nsec .
-0'8 I 0.3 nsec '
_1 l l l l
0.5 1 1.5 2 2.5 3
Time (nsec)

Figure 3.17: Time domain response, lossless case (6, = 9, A = 2 cm, 6, 2 0°). Inset

shows equivalent transmitted waveform.

66

I

 

-- lFF'l: F(0))F((0)
_ A . IFFT R((0)F(00)

 

 

 

— Natural Mode Series '

 

0.8

0.6 '-

0.4 '-

 

 

Amplitude
O
'?

  

 

 

 

0
—0.2- I,
, I‘
‘ I
—0.4_ l, f .
I
‘as
-0.6 I I I I I I I
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

Time (nsec)

Figure 3.18: Time domain response, || polarization (6,. : 9, A = 2 cm, 6,- = 20°,

0 = 0.5 S/m).

67

 

 

1 I I I
- - IFFl’ I"(m)F(w)
* IFFT R(co)F(m)
— Natural Mode Series

0.8- n
.l

 

 

 

.9
m

.0
.p.

 

 

Amplitude
O
"3

 

 

 

0 --..
a
02 i I
—. . ' I
. f
t ‘f
-0.4 5' -
V .
_0.6 I I I
0 0.5 1 1.5 2

Time (nsec)
Figure 3.19: Time domain response, I polarization (er 2 9, A = 2 cm, 6,- = 20°,

0 = 0.1 S/m).

68

 

 

1 . - .

-- IFFT I‘(0))F((0)
'3 IFFT R((0)F((0)
Natural Mode Series '

 

 

 

 

I

0.8

0.6 -

Amplitude
O
't’

 

 

 

o .
I
‘t .

-o.2 a , , .

\ l

‘ I

,
_04 - “ ‘ at
*J
-O.6 I l I L
0 0.2 0.4 0.6 0.8 1

Time (nsec)

Figure 3.20: Time domain response, || polarization (6.,- = 9, A = 2 cm, 6,- = 30°.

0 =1.0 S/m).

69

 

 

 

 

 

 

 

 

 

0.3‘ I I I i 1 r 1
- - IFFT F(m)F(m)
0,2 - '3 IFFT R((0)F(a)) ~
Natural Mode Series
0.1 - .
I ~ _ .. _
‘ .. -
'1
<1) —0.1 - k .
'o I '
a I, 1
3—0 2 .. 1 ' i
E I; ’
< —0.3 L , ~
‘, .
—0.4 _ . ,' 1
‘. .'
-O.5 - g. ‘ -
| l
—0 6 - 1‘ ,5 .
J
-0.7 l l I l l l l l l
O O 1 0.2 0 3 0.4 0.5 0.6 0.7 0.8 0.9 1
Time (nsec)

Figure 3.21: Time domain response, I polarization (6, = 9, A = 2 cm, 0,- = 30°,

0 =1.5 S/m).

7O

 

Amplitude
<2»

 

 

 

 

 

 

 

 

 

 

 

1.4

_4 .
_5 . .
-5 - .
- - IFFT l"(0))F(m)
-7 ' . IFFT R(m)F(co)
Natural Mode Series
_8 l I l l l
0 0.2 0.4 0.6 0.8 1 1.2
TIme (nsec)

Figure 3.22: Time domain response, || polarization (6,. = 9, A = 2 cm, 0,- = 70°,

0 = 0.5 S/m).

71

 

 

 

 

 

 

 

 

 

1.4

 

0.4 .
d
m I
'D—O.2- I
3 I I
a I 7
* l
E_ .
<1: 0'4 l f 1
§ l
. t
—0.6- | , -
'l II
II
-0.8- '2‘ -- IFFT I‘(00)F(co) r
' IFFT R(c0)F(c0)
Natural Mode Series
_1 l l l j l l
0 0.2 0.4 0.6 0.8 1 1.2
TIme(nsec)

Figure 3.23: Time domain response, I polarization (6,. = 9, A = 2 cm, 6,- = 70°,

0 = 0.5 S/m).

72

 

L L i” (
O 0 0 0 4 0 (U
. . ~ 9
1rd
‘l
.1010
Er.

0.6 '

®U3u:QE<

 

 

 

 

 

 

 

 

 

 

 

 

0.6 I I I l I j l l
‘1. - - IFFT 1“(w)F(co)
,1 I '3 IFFT R(c0)F(c0)
0.4 - I 1 Natural Mode Series ‘-
' b
‘ I
_ l
0.2 ‘ l, 4
l I
(D I
0 t
3 II
E‘
—O.2 -
<
—0.4 - -
_0.6 I- U u
_0.8 l I l l l l l l l
0 0.2 0.4 0.6 0.8. 1 1.2 1.4 1.6 1.8 2
TIme (nsec)

Figure 3.24: Time domain response, || polarization (6, = 9, A = 2 cm, 6,- = 85°,

0 = 0.5 S/m).

73

0.4 r-

0.2 '

Amplitude
g.)
l\) O
- I

l
_O
J)

I

I
_O
O)

I

 

0 r. 05 S

 

 

 

 

 

 

 

 

0.4 . . . . . . . .
- - IFFT I“(oo)F((0)
'3 IFFT R(co)F(m)
0.2 - — Natural Mode Series
0 - .. _
‘
S
a)
'0—02- I, f -
r3 8 ,fi
3" 04 “ '
<" ' l: J
‘ I
‘ l
—0.6- 3‘ ,' 4
‘ I
‘ l
-03. 'I‘ f .
* I
‘f
0.2 0.3 0.4. 0.5 0.6 0.7 0.8 0.9 1
TIme (nsec)

Figure 3.25: Time domain response, I polarization (6,. =2 9, A = 2 cm, 6,- = 85°,

0 = 0.5 S/m).

74

Chapter 4: Natural Mode Analysis
of an Air-Backed Slab

In the previous chapter, the transient response from the conductor backed slab
has been discussed. For the air backed slab case, most of the previous developments
can be used with a few modifications. The main difference from the previous case is
that the reflection coefficient has the different form due to its different backing
material. For the detail of the following discussion, refer to the conductor-backed
case in the previous chapter. However, in some parts, the discussion will be

repeated in order to clarify the developments.

4.1 Formulation of the frequency-domain
reflection

Consider a plane wave of frequency w incident from free space onto an interface
between free space and an air backed slab of material with frequency independent
material parameters 6 = 6,60, 11,0, and a, and with thickness A, as shown in Figure
4.1. In the figure, region 0 and region 1 correspond to free space and the dielectric
slab respectively. Likewise, for the subsequent discussion, the subscripts O and 1
correspond to free space and dielectric, respectively. As shown in (2.41), (3.3) and

(3.4), the reflection coefficients have the general form

 

w _e—jmw
R(w) _ 1“( )l1 l

— 1 — F2(w)e‘j“”(“) (4'1)

where F (w) is the interfacial reflection coefficient, given in general form by

 

 

Z , -— Z
f(w) __. _1(_°‘J_)__0,
210.11) + 20
and
2A
7(6)) 2 ———F(w), F(w)= 6+ .0 .
c 30160
Since 1?(<.u) exists, its Laplace domain representation also exists [21], and is
given by
F(s)[1 — e‘STlsll
R(3) "‘ R(w)lw=s/j _ 1_ F2(S)€—ST(5), (42)
where

2A
ST(S) = TVSVS '— 80, (4.3)
21(3) — 20
21(8) ‘l' 20 l
as shown in (3.7) and (3.8). Here l/ = C/\/:€: and so = —0/(60€).

For perpendicular polarization the z-directed impedance Z1(s) is

 

 

 

 

 

77 77
Z.L('S) : F 0 = O
(W) w=s/j E + it
7708
— I 4.5
\/-€-\/§\/S — 80 ( )
and for parallel polarization
77013101) 710 E + fo-
Zu(8) = ——_0 = ___a
61‘ + T— 67“ + —
JWCO w=s/j SEQ
_ "ofifivs " 80 4 6
— 86 + 1 l ( ' )
r 60

76

Therefore, the interfacial reflection coefficient Us) in (4.4) can be written as

scos 9, — fifivs — so

 

P( ) — SCOS 91' + fifim . (J- pOI.) (4 7)
'~ ‘ x/Efix/s—Ts—o — “Bails“ + i) (ll P01.) .

 

«Rwy—TO + cosfizlser + 2%) l

The reflection coefficient R(s) is the transfer function of the system. Thus, the

impulse response r(t) is obtained by

r(t) = [1'1 {R(s)} — 1 [B R(s)es‘d3 (4.8)

_E

where Br indicates the Bromwich path. This integral can be evaluated by contour
integration. We must thus examine the singularities and define an appropriate

branch for the integrand.

4.2 Singularities and the branch out

From the complex square roots in ST(S) and 21(8), the branch points are
located at s = O and s = so. In order to ensure the continuity of R(s), the branch
cut is taken along the negative real axis between the two branch points.

From (4.2), the poles are the roots of
1 — F2(s)e—ST(S) = 0. (4.9)
Taking the logarithm of this equation and rearranging gives
1n [F2(s)e"37(3)] = 1n |I‘2(s)e—ST(3)|+j arg [F2(s)e—ST(S)]:l:j2mr = 0, (n = O, 1, 2, ~ ~ - ).

Since —7r < arg [1‘2(s)e’37(8)] < 7r and the goal is to find roots, n can be set to 0.

77

Therefore, the poles should satisfy
111 [r2(s)e-ST<3>| + j arg [1‘2(s)e‘“<8>] z 0. (4.10)

Except for the lossless case (a = 0), wherein the poles can be calculated
analytically, the above equation needs to be solved numerically. For the lossless
case, both F(s) and 7(3) are independent of s, and —1 < F < O, i.e. arg(F) = 7r. As

a result, from (4.10), the lossless case poles are
2 .
s: —[ln|F| :l: jmr], n=0,1,2,3,-~- (4.11)
T

Note that, unlike the conductor-backed case, (4.11) has a real pole even for the
lossless case. From (4.10) and (4.11), it is clear that. the poles occur in complex
conjugate pairs as expected for real signals. Table 4.1 shows the lossless case poles,
where 6,- : 0°, 6,. = 9 and A = 2 cm. In the table, only poles in the upper-half
complex plane including the real axis are shown.

When the conductivity is nonzero, equation (4.10) can be solved by 2-D root
search algorithms such as the Newton-Rahpson method [17] using the lossless case
poles as the initial guesses. Again, it should be emphasized that the real part of a
pole should be less than or equal to zero for a passive system. This can be shown to
be true by noting that the magnitude of F2(s)e'37(3) in (4.10) is always less than 1 if
Re{s} is greater than zero, since both the magnitudes of F(s) and 6’37“) are less
than 1 as can be shown using (4.3) and (4.7). Here, it should be remembered that
so = —0/(eoE) is a non-positive constant. As a result, no solution of (4.10) can exist
if Re{s} is positive.

Figure 4.2 and Figure 4.3 show several pole trajectories obtained by the
N ewton-Rahpson method when the conductivity 0 varies from 0 to 10 S/m for

perpendicular and parallel polarization respectively. Here, only the complex poles

78

are shown and the mode number 0 is assigned to the pole with the smallest
imaginary part and the mode number 1 for the next smallest, and so on. Similar to
the conductor—backed case, when the conductivity reaches a certain value, each pole
becomes purely real. In the figures, the conductivity values at which the poles
become real are shown, and the corresponding pole locations in the complex 8 plane
are shown as the cross-marks. As shown in the figures, when the conductivity
increases, the magnitudes of the imaginary parts of the poles decrease whereas those
of the real parts increase until the poles approach the real axis and become purely
real. This phenomenon is clearly seen in Figure 4.4 and Figure 4.5. Note the
difference between the mode 0 pole trajectories of the perpendicular and parallel
polarization cases.

These pole trajectories of both polarizations suggest that when the
conductivity becomes higher, the amplitude decay factors become larger, resulting
in a smaller contribution to the late—time response.

In Figure 4.6 and Figure 4.7, the pole trajectories for varying angle of incidence
are shown, where fr 2 9 and a = 1.0 S / In. As seen in Figure 4.6, which depicts the
pole amplitudes of the perpendicular polarization case, as the angle of incidence
increases, the magnitudes of the real parts of the poles monotonically decrease
whereas those of the imaginary parts are relatively unchanged. Again, it should be
emphasized that the pole locations do not solely determine the late-time transient
responses. In order to obtain the correct transient responses, the residues
corresponding to the poles should be considered. For the parallel polarization case,
as the angle of incidence increases, the magnitudes of the real parts of the poles
show a dip around 70° of incidence angle which approximately corresponds to the
Brewster angle calculated as [15]

1 9

= sin’ —— = 71.6°,
1

6 = sin—1
3” + 9

 

79

where we assume the dielectric material is lossless and 6,- : 9. Similar to the
conductor-backed case, for the air—backed case, the mode 0 pole also becomes real
approximately at this angle. However, unlike the conductor-backed case where the
mode 0 pole remained real beyond this incidence angle, it becomes complex pole
again (in this case, it becomes real at 68° and becomes complex again at 72° of
incidence angle), and it follows the trajectory of the bottom arc of the higher mode
(mode 1) as shown in the top figure of Figure 4.7. The other modes also show the
similar behavior, i.e. at a certain angle of incidence, they follow the bottom are of
the trajectories of the next higher modes. It seems that this particular behavior is
due to the periodic nature of the arguments of the complex numbers. The behavior
of the parallel polarization case evidently suggests that the late-time response may
be affected by the Brewster angle. The existence condition for the real poles and the
algorithm for finding such poles are examined next.

Let us assume a complex frequency 5 to be negative real and denote its

magnitude as :r, where :r > 0. Then, ifs > so, i.e., a > —seoE, from (3.7) and (3.11)

8(8) )=j—\/:?\/—( ()a:+80 (4.12)

and the reflection coefficient becomes

( —:ccos€,- -- jfifim

—mcos€,+j\/Efim' (~L p01)
r(s) : (4.13)

jff\/—( ()a:+so )—cos€( —:re,.+ 695) (H 1)
. 0.
K j\/‘_€\/_ —(;1:+so)(+c080,-(—-:rer+ :—O) p

 

 

It is now clear that the magnitudes of both 63—87(3) and F(s) are 1. As a result,

In |F(s) )2 e‘msl )l— -- 0, and the poles need to satisfy only the imaginary part of (4.10),
i.e., arg [Us )2e ‘ST(S)]— — 0. Therefore, if we denote —(2A/1/)\/:1—:\/—-(:r + so), )which

80

is —ST(S), as (151 and arg [F2(s)] as do, the poles need to satisfy
gbl+g§2+2mr=0. n=0,1,2,--- (4.14)

Here, only the positive sign in front of 71 is necessary, since 6.51 is always negative,
when 8 > so and —7r < 9'92 < 7r. The above equation, which we call the characteristic
function, can be solved by any 1-D root search algorithm, such as the secant
method. Here, it is very important to note that (4.14) does not always have
solutions, and that the number of solutions is dependent upon the conductivity,
permittivity and incident angle. In addition, great care should be taken when
finding the real roots, especially for the n = 0 case where the characteristic function
can have more than one root, which may cause the usual secant method to fail.

If s < so, ST(8) becomes negative real causing 6‘37"“) to be greater than 1, and
F(s) becomes real. Therefore, in (4.10), arg [F(s)ze'”(3)] = 0. As a result, the poles
only need to satisfy 2 ln |F(s)| — 37(5) = 0. Figure 4.8 shows the amplitudes of the
reflection coefficients for both polarizations, where s < so. As shown in the figure,
for perpendicular polarization, the reflection coefficient F(s) = 1 when 3 = so and
monotonically decreases as 8 decreases. In particular, it becomes negative when

s < since the reflection coefficient has roots

___9__
60(1—er)’

0'

. = 0 —.
9 , 60(1— 6,.)

As a result, two additional real pole can exist when 3 < so for perpendicular
polarization. Figure 4.9 shows the amplitudes of -sr(s), ln|F(s)| and

2ln |F(s)| —— 37(3) for perpendicular polarization. In the figure, the points where
2ln IF (s)| -— 8T(S) cross the O amplitude line represent the pole locations (there are
2). It should be noted that the apparent pole s = so where both 6‘37“) and F2(s)

are 1 is a removable pole, since the numerator of the reflection coefficient becomes

81

identical to the denominator.

For parallel polarization, F(s) = -1 when 3 = so, and it has two real roots at

0 0
60(1— 6,) eo(tan'2 6,- — 5,)

 

8:

Therefore, depending on er and 9,, the amplitude of the reflection coefficient
increases initially, then stays positive or monotonically decreases as 3 decreases. As
a result, at most 4 real poles can exist when 8 < so for parallel polarization. Figure
4.10 shows the case where 3 real poles exist.

A complete sketch of the poles is shown in Figure 4.11 and Figure 4.12. In the
figures, the solid lines represent the trajectories of the complex poles, and the
cross-marks represent the real poles. The real poles consist of the poles with
magnitudes less than so and those with magnitudes greater than so. Since the
complex pole trajectories from different modes follow approximately the same curve
in the complex 8 plane (before each mode becomes real and diverges), it is difficult
to resolve each complex pole trajectory from the figures. As previously discussed,
the complex poles gradually become real poles as the conductivity increases. Note

that, except for mode 0, each single complex pole splits into a pair of real poles.

4.3 Evaluation of r(t)

Since all poles and branch points lie in the left half plane including the
imaginary axis, the region of convergence is the right half plane. Therefore, the
Bromwich path is put in the right half plane and the Laplace inversion integral is
evaluated by contour integration. The evaluation is accomplished in two different

time intervals, corresponding to the early-time period and the late—time period. The

2A

beginning of the late-time period is T0 = 7, which represents the two-way transit

time of the wave inside the dielectric slab.

82

4.3.1 Case I: t > To

When t > 7o, the integration contour is closed in the left half plane as shown in
Figure 4.13. Inside the contour, the branch cut lies from 0 to so and there is a
closed path enclosing the branch cut and several possible real poles on the branch
cut (two of which are shown). We denote the outer integration contour which
includes the Brmnwich path as Co and the closed path which encloses the branch

cut as C1, such that

C0 2 Br+roo+L1+L21

C1 = "1’1+)2+"'+’)’6+11+12+"'+16-

Then, by Cauchy’s residue theorem,

/ R(s)eStds = j27rZ Res[R(s)es‘, poles] (4.15)
Co-l-Cl

Therefore, if the integral contribution from each path is known, the impulse

response r(t) can be determined.

The contribution from [‘00

011 Foo, from (4.3),

2A 0
R. . = R l" — 1 —
e {T(S)} C {Re{sl}IB—oc I/ + 860?}
2A
2 — 2 7o
1/

83

If we let t1 = t — 7o, which is positive since we are evaluating the integral when

t > 7o, then

. I“(s)es'rO — F(s)e“°e’”(s)
, _Jst. _ __st1 ,_ stl
foo R(s)c ds _ [0° 1— 1‘2(s)e‘37(5) e d5 —./1‘oo f(s)c ds

 

Since f(s) -—> 0 on Foo, by Jordan’s Lemma [20], [21],

/ R(s)e“ ds = 0 (4.16)

Foo
The contribution from L1 and L2

As seen in the conductor backed case, for L1 and L2, the same theorem [14] can

be used to apply Jordan’s Lemma. As a result, it can be shown that

/ R(s)eSt ds = O (4.17)
14,14

The contribution from Cl

The segments of the contour C] enclosing the branch out can be divided into
three groups as did in the conductor backed case. The first group consists of 71 and
74 that enclose the branch points so and 0 respectively. The second group consists
of the straight lines immediately below and above the branch out, and those are
designated ll, [2, - - . , lo. The last group consists of 72, ’73, 75 and yo that enclose the
real poles on the branch out. These groups will be examined separately.

Denote the radius of 71 as r, and let 0 to be the angle measured
counterclockwise from the real axis to the point on 71. Then, any point on '71 can be
represented as

s = so + r630.

84

The reflection coefficient on 71 with r —> O, i.e. s —> so, becomes

5770 — ZOLfifiVS — 50
3770 + Zo1\/f\/§V3 — 30

 

—> 1, _L-pol

 

 

 

I‘ s = - a
( ) nofififl - ZOMSEr + a)
_- 0 ‘—_’ _11 ll - p0]
,]0\/;\/§\/3——5_o + Zoll(5‘r + 5)
Then,
F 1— 787(8) eat ’ "L I p01
R(5)€St : 1(S)i“2 6’ —sr(8)]88t :
— (8)6 —eSt ) ll ' p01
and
lR(s)eS‘ S €l$0+r)t-
Therefore,

_<_ 27rre(30+r)t —> O, (r ——> 0).

./ R(3)c3t ds

 

As a result,

/ R(s)eS’ d3 = 0.

ll

Similarly, it can be shown

/ R(s)eSt ds = 0.

7’4

(4.18)

(4.19)

(4.20)

For the second group, it is necessary to determine F(s) and 37(3) along a path

immediatelv above the branch cut which we denote as B+ and immediately below
U 9 3 .

the branch cut, which we denote as B‘. If we denote the magnitude of s as I

(so + r < —:r < —r < 0), then on B+,

x/EVS—SozjfiV—x—So,

and on B‘,

WEE: —j\/E\/-—CE_‘?0-

85

Now, let Z(s) on B”r be denoted as Z+, and Z(s) on B‘ as 2'. Then, using (4.7), it

can be shown that

 

 

2+ = —Z_' (4.21)

Since,

1‘2(5)[1 — 62—87(8)]

12(5) — 1__ F2(3)€—37(.s)
: [22(8) " Zg]e%87(s) __ [Z2(S) + Z§]e’%37(8)

[2(8) + ZO]2€%ST(S) — [Z(s) — Zol28’%87(s) ’

we have,
R(5)ls+ — [22+ _ Ziilc'm — [Z2+ + gag—N,

 

[2+ + Zo]2ej¢ — [Z+ — Zo]Qe“J¢’
2j[Z2+ — 23] 81nd)
2j[Z2+ + Z3] sino + 2Z+Zo cosqb

 

where jgb = 557(3), and

[Z2‘ — 23k?” — [ZZ— + Z8]e’¢

[Z‘ + Zo]2e‘39’ — [Z’ —- Zo]‘~’ej¢
—2j[22- — 231mm

—2j[Z'2‘ + 231mm + 2Z“Zo cosqb

 

R(8)l3-

 

Therefore, by (4.21)
R(5)l8+ 2" R(S)l[3--

As a result,

/ R(s)|3+e“ ds +/ R(s)|B—e“ d3 = 0 (4.22)
11+12+13

14+15+16
The integral contributions from the third group and the other complex poles

can be determined by calculating the residues of R(s)263‘ at the poles. It is found

86

that all of the poles of 1?.(3) are of first order and thus

2 3t _ - ,_ 'l l 3! _ 3),! r
Res [R(s )e ,pole s] lszsk — 312141,,“ 3),) 1_ F2(3)e‘37(3)e — Ake , (4.23)

 

where

F(.9k)[1 — €_SkT(Sk)]
Ak = — . _ . (4.24)
% [F2(8)€—.57(s)] |S=Sk

 

is the complex natural mode amplitude. Carrying out the details gives

A 28 — 90

A, = ran-H8“ 2G( )——-F(3) )T—fm (4.25)

where

2cos6 \/f s,/s —cos9 \/:———
f \/-\/898 — 8()
3'2 cos2 6,- + 23\/E_\/(js\/s — 80 + 68(8 — 80) l

for perpendicular polarization and

0(5) =

0 2s — so
cos 6,(ser+ O)\/:——— —2er cos 6,\/F\/s\/s — so
C(s) _ \/s\/s -— so

_ 0 2
’6s(3 — so) + 2cos 06,-(36,.+ :0)\/:,\/—\/s — so + cos2 (86,. + —)

60

for parallel polarization.

Collecting the results from (4.16)-(4.25), the impulse response r(t ) becomes

1
r(t): J2” s)eS‘ ds— - Z Akew t > TO, (4.26)
Br

and thus the late-time period is a pure natural resonance series, and contains no

branch-cut contribution.

87

4.3.2 Case II: t < T0

For t < 7o, r(t) is found by computing the inverse Laplace transform of

[f(s) = F(s) + Eb), where

 

Then

__ _ _ 1 . F(s)e“3T(3) .
R t = 1 R = —— F2 - 1 “Std“
() L i (5)} j27T Ari (5) l 1_ P2(5)€—s7(s)e 9

 

To compute this integral, the integration contour is closed in the right half plane, as

shown in Figure 4.14. 011 Foo, from (4.3),

2A
Re{7(3)} = Re {R {lii}n — 1+;}
e 3 —+00 V seoe
2A
2 — = 7'0
1/

If we let t1 = t — 70, which is negative since we are evaluating the integral when

t < 7o, then

 

— . . F(s)e‘”(sle“°
R; 8&1 = 1‘21—1 st1d. = “Id
foo (s): s [00 [ (s) ] 1_ F2(3)e—ST(5)€ 3 Foo (s)e 3

Here, it is not possible to apply Jordan’s Lemma directly, since f (s) does not
approach zero over the entire contour Foo. However, if we use the theorem

introduced earlier, it can be argued that

/ Task“ d3 = 0

Foo

88

The inverse transform of F(s), denoted as F(t), can be found in [26] and [27]. As a

result, when t < 7o, r(t) becomes

r(t) = F(t), t < 7o. (4.27)

Combining the results (4.26) and (4.27), the impulse response r(t) is

Z Akeskt, t > T0
r(t) = (4.28)
F(t), t < T0.

4.4 Results

As with the conductor—backed case, it has been shown analytically that the
early—time response is a specular reflection from the interface between free-space and
the dielectric slab, and the late-time response is a pure sum of damped sinusoids. In
order to verify these results, the natural mode series is compared to the direct IFFT
using a truncated Gaussian pulse as the input waveform. This waveform, which is
identical to the conductor—backed case, is shown in the inset of Figure 4.15. Figures
4.15-4.21 show the transient responses calculated by the natural mode series and the
IFF T. In addition, the IF FT of the interfacial reflection coefficients F(s) are also
shown in order to explain the early-time behavior. The natural resonant frequencies
of the corresponding cases are shown in Table 4.1-4.7.

As an example of the results, in Figure 4.17 the incident wave is in
perpendicular polarization with an incidence angle of 20°, and the material
parameters are A = 2 cm, 6, = 9 and a = 0.1 S / 111. Therefore, the beginning of the
impulse response late—time, i.e., the two-way transit time of the wave inside the slab,
is To = 0.397 us. As shown in Figure 4.17, during the early-time period, before 7o,

the IF FT of R(w)F(w) (where F (to) is the spectrum of the input waveform)

89

matches well with the IF F T of F(w)F(w), which is the specular reflection from the
interface. During the late-time, the IFFT of R(w)F(w) matches well with the
natural mode series. Again, it should be noted that when t < 7o the response from
the natural mode series does not have any meaning; i.e., the natural mode series is
valid only when t > 7o, the late-time period. Since the frequency band of the input
waveform is roughly 0 —— 20 GHz, it was found that the first 7 natural modes, which
are shown in Table 4.17, are sufficient to represent the late-time reflected field.

Figure 4.15 shows the response from the lossless, normal incidence case. As
seen in the figure, there are equally spaced replicas of the input waveform with
gradually diminishing amplitudes during the late—time period.

Comparing Figures 4.18 and 4.19 to Figures 4.16 and 4.17, due to the relatively
higher conductivity in the Figure 4.18 and 4.19 cases, relatively smaller late-time
responses are observed. The corresponding natural resonance frequencies for these
cases are shown in Tables 4.4 and 4.5.

The responses from both polarizations with the incidence angle 0,- = 70°, which
is close to the lossless case Brewster angle, are shown in Figures 4.20 and 4.21. As
seen in Figure 4.20, there is neither interfacial reflection nor the late-time response
for the parallel polarization case whereas both the interfacial reflection and the

late-time response are observed for the perpendicular polarization case, as shown in

Figure 4.21.

90

Table 4.1: Poles and corresponding complex natural mode amplitudes (Ak), loseless
case (6,— = 9, A = 2 cm, 6,- : 0°)

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

mode 0 —-.34657 X 1010 .15708 X 1011 .37500 X 1010 .14164 X 1075
mode 1 —.34657 X 1010 .31416 X 1011 .37500 X 1010 .28327 X 10—5
mOde 2 —.34657 X 1010 .47124 X 1011 .37500 X 1010 .53593 X 10—5
mode 3 —.34657 X 1010 .6283? X 1011 .37500 X 1010 .56655 X 10"5
mode 4 —.34657 x 1010 .78540 x 1011 .37500 x 1010 .59717 x 10‘5
mode 5 —.34657 X 1010 .94248 X 1011 .37500 X 1010 .10719 X 10—4
mode 6 —.34657 X 1010 .10996 X 101‘2 .37500 X 1010 .21431 X 10—5
real -.34657 X 1010 0.0 .37500 X 1010 0.0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

91

Table 4.2: Poles and corresponding complex natural mode amplitudes (Ah), [I polar-
ization (6,. = 9, A = 2 cm, 6.,- = 20°, 0 = 0.5 S/m)

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

mode 0 —.70815 X 1010 .14659 X 1011 .44840 X 1010 .10437 X 1010
mode 1 —.69320 x 1010 .31060 X 1011 .41426 X 1010 .51448 X 109
mode 2 —.69078 X 1010 .47060 X 1011 .40902 X 1010 .34128 x 109
mode 3 —.68995 X 1010 .62965 X 1011 .40725 X 1010 .25549 X 109
mode 4 —.68957 X 1010 .78832 X 1011 .40644 X 1010 .20421 X 109
mode 5 —.68937 X 1010 .94680 X 1011 .40601 X 1010 .17009 x 109
mode 6 —.68925 x 1010 .11052 x 1012 .40575 X 1010 .14575 x 109
real -.10290 X 1011 0.0 .47555 X 1010 0.0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

92

Table 4.3: Poles and corresponding complex natural mode amplitudes (A1,), _L polar—
ization (6,. = 9, A = 2 cm, (9,- : 20°, 0 =2 0.1 S/m)

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

mode 0 —.39505 X 1010 .15660 x 1011 .35867 X 1010 .15824 x 109
mode 1 —.39292 X 1010 .31545 X 1011 .35399 X 1010 .84744 X 108
mode 2 —.39252 X 1010 .47382 X 1011 .35310 X 1010 .57184 X 108
mode 3 -—.39239 X 1010 .63206 X 1011 .35279 X 1010 .43069 X 108
mode 4 —.39232 X 1010 .79024 X 1011 .35264 X 1010 .34521 X 108
mode 5 —.39229 x 1010 .94841 X 1011 .35256 X 1010 .28799 X 108
mode 6 -.39226 X 1010 .11066 X 1012 .35252 X 1010 .24700 X 108
real —.46054 x 1010 0.0 .36268 X 1010 0.0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

93

Table 4.4: Poles and corresponding complex natural mode amplitudes (Ak), H polar-
ization (6,. = 9, A = 2 cm, 6, = 30°, 0 = 1.0 S/m)

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

mode 0 —.11099 x 1011 .30179 X 1011 .58349 X 1010 .16527 X 1010
mode 1 —.10616 X 1011 .30297 X 1011 .46827 X 1010 .12227 X 1010
mode 2 ——.10557 X 1011 .40767 X 1011 .45729 X 1010 .78779 X 109

 

 

 

mode 3 —.10538 X 1011 .62959 x 1011 .45384 X 1010 .58407 X 109
mode 4 —.10529 X 1011 .79044 X 1011 .45231 X 1010 .46481 X 109
mode 5 ——.10525 X 1011 .95077 X 1011 .45149 X 1010 .38624 X 109
mode 6 —.10522 X 1011 .11108 X 1012 .45100 X 1010 .33050 X 109
real ——.17321 X 1011 0.0 .63125 X 1010 0.0

 

 

 

 

 

 

 

 

 

 

 

 

94

Table 4.5: Poles and corresponding complex natural mode amplitudes (Ak), _L polar-
ization (6,. = 9, A = 2 cm, 6,: = 30°, 0 21.5 S/m).

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

mode 0 —.13152 X 1011 95409 X 1010 .41542 X 1010 .45363 X 1010
mode 1 —.12817 X 1011 .29278 X 1011 .33528 X 1010 .13039 X 1010

 

 

mode 2 —.12780 X 1011 .46117 X 1011 .32903 X 1010 .81773 X 109
mode 3 —.12768 X 1011 .62478 X 1011 .32709 X 1010 .60124 X 109

 

 

mode 4 —.12762 X 1011 .78662 X 1011 .32623 X 1010 .47670 X 1()9
mode 5 —.12759 X 1011 .94760 X 1011 .32578 X 1010 .39535 x 109
mode 6 —.12758 X 1011 .11081 X 1012 .32550 X 1010 .33790 X 109

 

 

 

real —.23152 X 1011 0.0 .49068 X 1010 0.0

 

 

 

 

 

 

 

 

95

Table 4.6: Poles and corresponding complex natural mode amplitudes (Ak), || polar-
ization (e.,. = 9, A = 2 cm, 6.,- = 70°, 0 = 0.5 S/m)

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

mode 0 -.55910 X 1010 0.0 —.11317 X 1010 —.65568 X 101
mode 1 —.20677 X 1011 .26650 X 1011 .31069 X 1011 .77117 X 1011
mode 2 —.20636 X 1011 .45526 X 1011 .53730 X 1011 .50664 X 1011
mode 3 —.20624 X 1011 .63136 X 1011 .60330 X 1011 .37697 X 1011
mode 4 —.20619 X 1011 .80298 X 1011 .63188 X 1011 .30038 X 1011
mode 5 —.20616 x 1011 .97246 X 1011 .64693 X 1011 .24975 x 1011
mode 6 —.20614 X 1011 .11408 X 1012 .65584 X 1011 21378 X 1011

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

—.37516 X 1011 0.0 .58574 X 1012 0.0
real —.21023 X 1011 0.0 —.15778 X 1012 0.0
—.39937 X 1011 0.0 .41373 X 1012 0.0

 

96

Table 4.7: Poles and corresponding complex natural mode amplitudes (Ak), J. polar-
ization (6,. = 9, A = 2 cm, 6, = 70°, 0 = 0.5 S/m)

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

mode 0 —.47744 x 1010 .15888 x 1011 .13251 x 1010 .29387 x 109
mode 1 —.47586 x 1010 .32760 x 1011 .12925 x 1010 .14088 x 109
mode 2 -.47558 x 1010 .49408 x 1011 .12869 x 1010 .93218 x 108
‘mode 3 —.47548 x 1010 .66002 x 1011 .12850 x 1010 .69733 x 108
mode 4 —.47544 x 1010 .82574 x 1011 .12841 x 1010 .55719 x 108
mode 5 —.47541 x 1010 .99136 x 1011 .12837 x 1010 .46403 x 108
mode 6 —.47540 x 1010 .11569 x 1012 .12834 x 1010 .39758 x 108
real —.82716 x 1010 0.0 .13535 x 1010 0.0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

97

x
Free Space jDielectricl Free Space

V

\\\\\

‘\
.Q
‘Q
‘Q
‘\
Ԥ
Q
‘\
\
r ~,
Q
0"
'0
O
1 o'
'0
'0
/.
'l
'0
I
l
I

80:“0 ligalvloao-
z=0 z=A

Figure 4.1: A uniform plane wave incident from free space upon a Air-backed lossy
slab.

98

x1010

 

Imaginary Part
0) hs 01

N

 

 

 

 

97 -4 -3
Real Part

 

Figure 4.2: Pole trajectories of the reflection coefficient, _L polarization (6,. = 9, A = 2
cm, 6, = 30°).

99

x1010

 

 

 

 

 

 

 

7 I r I r I 1
mode3
6' ; 1 6:0 4
.. 3
4
5~ ’ 5 mode2 ‘ .
E 6 ‘
n. I. .
a4 7
cu mode1
E
Ef3' 8 ‘
_E
21-
-9
123:4. 0:9.24 0:6.77 '
—7 -6 —5 —4 o
10
RealPan x10

Figure 4.3: Pole trajectories of the reflection coefficient, || polarization (6,. = 9, A = 2
cm, 6i = 30°).

100

x101O

 

mode 3

 

Imaginary Part

 

 

 

 

5 10
Conductivity (S/m)

Figure 4.4: The amplitude of the imaginary parts of poles vs. conductivity, _L polar-
ization (6,. = 9, A = 2 cm, 6, = 30°, 0 = O -——> 10.0 S/m).

101

x1010

 

mode 3

Imaginary Part

 

 

‘

 

 

 

 

O 1 1 l 1 l 1 l l l
o 1 2 3 4 5 6 7 8 9 1o
Conductivity (S/m)

Figure 4.5: The amplitude of the imaginary parts of poles vs. conductivity, || polar~
ization (6,. = 9, A = 2 cm, 6,- = 30°, 0 = 0 —->10.0 S/m).

102

RealPan

Figure 4.6: Pole trajectories of the reflection coefficient vs.

x10

10

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6 I 1 o I I I 0
0° 69 75 8?
t 3.2— - T
a? 4 _ 15.. mode 2 _
E W i 2‘
.g Inodei
0')
co 2- -
E a; a i i d
" mode 0
0 l l l 1 L
-—10.5 —10 -9 —8.5 —8 —7.5 -7
9 RealPan x109
x10
“'7 I l l I I
mode 0
_8, 8 rnode1 _
- - mode 2
-9- 4
—1 0 -
_1 1 1 l l l l l
0 10 2O 40 50 60 7O 80 90

polarization (e.,. = 9, A = 2 cm, a = 1.0 S/m).

103

angle of incidence, i

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

10
6 X I I I I I I 1 I I
t - ................... £- ‘
a 4 _ < mode 2 _
2‘ m”W A
.5 - 7 O o d
22 ‘37 33
’i—w
!- o _+— o o "
—1 68......... 60° 45 0 ”me 0
0 ' I I I I A J I I
—2 6 —2.4 -2 2 -2 —1.8 —1.6 —1.4 —1.2 -1 -0.8 —0.6
Real Part 10
x 10
10
1
-0.5 x 0 I I I I I I I I
_1 Isn't-I "HRH-.1- HITII'. 9.9.33- u. .. I! d
t o......'.‘,',‘:.'.-.-:'~.m“a~ J
g _1.5 - ....... '3‘,“ 3;). -
m o w. ‘ -I
a) -2 " "‘
m mode 0 ~
—2.5 ~ iiiiii mode 1 -
- - mode 2
_3 L l l L l l l
0 10 20 30 40 50 60 70 80 90
lncrdent Angle (Degree)

Figure 4.7: Pole trajectories of the reflection coefficient vs. angle of incidence, ||
polarization (6, = 9, A = 2 cm, 0 = 1.0 S/m).

104

 

 

0.8

J

 

 

 

0.6 ~

0.4-

 

Amplitude

 

 

\
\

\

\

I

l

I
I-
r
- l
: I

—0.8 - \ s = o/[eo(1-er)] s = s /': .

: I
° I

13.8 -8.6 —8.4 —8.2 —8 -7.8 —7.6 —7.4 -72 —7
S(Fiea|) x1010

 

Figure 4.8: Amplitudes of the reflection coefficients (e,r = 5, A = 2 cm, 0 = 3.0 S/m,
61' = 300).

105

 

 

15 I I I I I I I I I f
: - - —sr(s)
....- 2|nll“,(8)l

_ 2|n|I‘i(s)| - 51(3)

I

 

 

 

 

 

Amplitude

-15.

s = o/[eo(1-er)] s = 30 ‘

 

 

 

_2 l l 1 J l 1 .
3.3 -8.6 —8.4 —3.2 —8 —7.8 -7.6 —7.4 —7.2 _7
5 (Real) x1010

Figure 4.9: Real pole (3 < so), (6,,. = 5, A = 2 cm, 0 = 3.0 S/m, 6, = 30°).

106

I f I '

 

 

15 I

 

 

 

 

- - -s'r(s)
2Inlr,(s)I

_ 2lnlf‘i(s)| —s'c(s) '

 

 

 

 

 

 

(D
'C
3
E.
E
<
'8 2
i E
—20r i s=c/[80(1—er)] fl
- s = c/[eo(tan9i—er)] § 3 =.s0
_2 L ' l L l I l l ' J '
3.8 -8.6 —8.4 -8.2 —8 —7.8 —7.6 —7.4 -7.2 —7
S(Real) x10‘°

Figure 4.10: Real pole (s < so), (6,. = 5, A = 2 cm, a = 3.0 S/m, 6, = 30°).

107

 

 
   

 

 

x10
0 I ' ;'g3';8flflfl;8833§""£""*'."f”u"
" gauaaausuauauauuuun
‘
_2_
E"
(U
o. -4-
70
6‘6
v _6..
g B 2“ 8' "
.H I ‘ 8 '
:.'= —8r . ‘~ ' .t d
o- . \ " ' .
E .. ‘ a a“
. ‘3' '
< a ‘ ' '
m-1OP _ ' ‘fta: " _
B S—So ' “a 7..
.
CL '5. 4‘“ '2.
‘ II
12’ at ‘J A
— '. ‘.fl
. s
a w
3
'
-14 l l I l L I I I I L

 

1 2 3 4 5 6 .7 8 9 10
Conductivity (S/m)

Figure 4.11: Complete sketch of the pole trajectories showing both complex and real
poles, J. polarization (6, = 9, A = 2 cm, 6, = 30°).

108

 

    
   

 

 

0 r r g.;"";""§.gguhuslfliaxaakxaaxhxua:
a as:
. " ssuusuusuxufi*"""'
_2_
7c?
(U
D. —4“
'53
GJ
EB .5L
0)
g ':;:\' 8
-o—I BF} * '
2: _3- 8 3 ~
g- F"* ”a ff
= a 1" "
< 5 SO ..g~'\.‘ 8'
52—10“ ”. '} ”a 7
O .' }§ !
fl. ' i “a
II Oi, .
" e
_12_ II.' ,p _
a *5,
u' ,~
a
a
_14 I 1 I L I I I I I L

 

1 2 3 4 5 6 7 8 9 10
Conductivity (S/m)

Figure 4.12: Complete sketch of the pole trajectories showing both complex and real
poles, || polarization (6, = 9, A = 2 cm, 6, = 30°).

109

 

 

 

 

 

 

 

 

 

 

 

L1
( s=jb
Foo BI‘
A A
Re{s}
s=jb
s=a s=c

 

Figure 4.13: The integration contour when t > To. |a|, |b| —+ 00, 0 < c < 00.

110

Im{s}4

V 130

0 131A Re{s}

 

 

 

s¥=a

Figure 4.14: The integration contour when t < 7o. R —> oo, 0 < a < 00.

111

 

- - IFFT- I‘(m)F(m)
IFFT R(co)F(m)

l

 

 

 

 

 

 

8
-- Natural Mode Series

 

 

 

1
0.8 - A
0.6 -
0.4 -
a) .
‘O 0.2
3
E .. v
< —O 2 l' f. I -
2 2
I: II f(t) = e—n (t—u) /t
—0.4 - inf ,
1: = 0.10 nsec
‘05 ' p. = 0.15 nsec '1
—0.8 -
0.3 nsec
0.5 1 1.5 2
time(nsec)

 

Figure 4.15: Time domain response, lossless case (6,. = 9, A = 2 cm, 6, 2 0°). Inset

shows equivalent transmitted waveform.

112

- - IFFT I‘(w)F(m)
A . IFFT R(m)F(m) ,
— Natural Mode Series

 

 

 

 

 

0.8 -

0.6 '

0.4

 

 

Amplitude
O
"3

 

 

 

 

o
—o.2 -
—o.4 -
‘0.6 l I l L l l
o 0.2 0.4 0.6 . 0.8 1 1.2 1.4 1.6
Time (nsec)

Figure 4.16: Time domain response, || polarization (6,. = 9, A = 2 cm, 6, = 20°,

0 = 0.5 S/m).

113

 

 

 

 

 

 

 

 

 

 

 

1 i I I i i 1 J_ I
-- IFFT I"((I))F((II)
.. IFFT R(o))F(co) _
08' fl — Natural Mode Series
0.6- '
m 0.4- 1
“C
3
'5- 0.2”
E
0
t
02 i 3
_ . F ' d
, i
V i
-O.4- I): .
I!
0.8 1 1.2 1.4 1.6 1.8 2

‘0'60 0.2 0.4 0.6 .
Time(nsec)

Figure 4.17: Time domain response, J. polarization (6,. = 9, A = 2 cm, 6, = 20°,

0 = 0.1S/m).

114

 

 

 

 

 

 

1 t t I I I I I J
-- IFFT r(m)F(m)
“ IFFT R((I))F((I))
03' — Natural Mode Series '
0.6- ‘
m 0.4- '
“o
3
'3 0.2- 1
E
0 - ......
u
‘i x
—0.2' . 3 .
\ i
‘ I
f
—o.4- i, ,, '
"J
0.9 1

 

 

_0.6 l I l 1 l
o 0.2 0.3 0.4. 0.5 0.6 0.7 0.8
Time (nsec)

Figure 4.18: Time domain response, || polarization (E,- = 9, A = 2 cm, 6, = 30°,

0 =1.0 S/m).

115

I

 

 

0.8 I I T fi I l I
- - IFFT I‘((o)F(w)
8' IFFT R((o)F(co)
— Natural Mode Series

 

 

 

T

0.6

.o .o
N J5

Amplitude
O

—0.2

 

 

_o.8 I I I I I I I
0.1 0.2 0.3 0.4. 0.5 0.6 0.7 0.8 0.9 1
Time (nsec)

 

 

Figure 4.19: Time domain response, i polarization (6,. = 9, A = 2 cm, 6, = 30°,

0 21.5 S/m).

116

 

 

18 . . . fi 1 fl . .
A - - IFFT F(m)F(w)

 

 

 

15 - '3 IFFT R(co)F(co) -
— Natural Mode Series
14 - ..
12 ~ -
Q) 10 ' 1
.0
2
'5. 8' ‘
E
< 6

 

 

 

 

 

A

_2 l l I I l
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Time (nsec)

Figure 4.20: Time domain response, || polarization (6,,. = 9, A = 2 cm, 6, = 70°,
0 = 0.5 S/m).

117

 

0.4 . . . . .
- - IFFT I‘(w)F((o)
' IFFT R(m)F(m)

 

 

 

 

 

 

 

 

 

0.2 - Natural Mode Series I
I.
s 02. l i
3_ ' I I l
:6- I f
I I
E- . I .
< 0'4 I f
i i
—0.6b ‘| , -
I I
I I
-0.8' ‘s‘ "
_1 I I I I I I
0 0.2 0.4 (1.6 0.8 1 1.2 1.4
Time (nsec)

Figure 4.21: Time domain response, .1. polarization (6, = 9, A = 2 cm, 6, = 70°,

0 = 0.5 S/m).

118

Chapter 5: Natural Mode Analysis
of a Single Layer with Debye

Material

In Chapter 3 and Chapter 4, a lossy slab of material with frequency
independent material parameters 6 = ereo, no, and 0 was considered. However, the
time and frequency dependence of the dielectric response necessitates the use of the
time and frequency dependent permittivity models such as the Debye and the
Cole-Cole models [28]-[31]. In this chapter, the Debye model, which is especially
suitable for representing liquid polar materials, will be considered. Note that the
primary purpose of this chapter is to show the validity and the applicability of the
natural resonance representation of the transient dielectric response, rather than to
discuss the theory of the dielectric properties. In the following discussion, both the
conductor-backed and the air-backed cases will be considered, and the developments
made in the previous chapters will be applied without discussing the details.
However, in some parts the discussions will be repeated in order to clarify the

developments.

5.1 Formulation of the frequency-domain
reflection

The geometry of the Debye material case is the same as the

frequency-independent material case except for the frequency-dependent

119

permittivity represented by the Debye equation [15], which is written as

€S_€x
.:= w ————a r1
(a) e-+1+5€, (a)

where s = jw, 5 is the relaxation time of the dielectric, 63 is the real static
permittivity obtained when w ——+ 0, and 600 is the real optical permittivity describing
the high frequency behavior. Note that 6, > 600 in order to describe the dielectric
loss. Also, note that both w and s will appear in the following discussions for
clarifying the develI’Ipn'Ients.

The Debye model often includes a conductivity term. Then the complex

permittivity becomes [29], [32]

1+3€ (56

66(8) 2 6x. +

Here, the last term represents the loss due to the conduction current. It cannot be
absorbed into (5.1), since they represent different processes. Hence, the 2 directed
wave number of the dielectric slab becomes

.. _ -2 .2
A” 1 _ A'1 _ A131

“I

 

= \/w2,uoec(w) — w'zitoco sin2 6,

 

— sin2 6,

 

 

#060-

_ __Sses(—630)+0(1+35)
— gal/00 in 2,9 + 608(1+S€)

— 6081112,6)(S + 325) + 8(6S — 60¢) + 0(1+s£)
€o3(1+s§)

  

 

 

 

 

Z—J:

 

_ .3 82€<£oo — 50 SlIl“2 6,) + 8(63 + 0f __ 60 81112 61'.) _+_ 0
_ JC €os(1 + sf)

 

 

= _fl— f \/3 +Bs+C (53)

(«3+ 1/€

120

where

 

€00 , .
X = — — sm2 6,,
60

68 + £0 — 60 SIII2 6,
{(600 — 6081112 6,) l
a

5(600 — co si112 6,)

 

 

The roots of the quadratic equation in (5.3) are

_ —B:l: «132—40
_ 2 ,

 

5

Here, it can be shown that these roots are non-positive real, since

(6, + £0 -— so sin2 6,)2 — 460(60O —— 60 sin2 6,)

62(600 -— (0 sin2 6,)2 ’

132—40:

 

and the numerator of this equation becomes

(63 + £0 — co sin? 6,)2 —— 450(600 — 60 sin2 6,)
= 6: + 268€U + £202 — 26360 sin2 6, — 2Eaeo sin2 6, + 6.381114 6, — 450600 + 4Eaeo sin2 6,
—22 ”2 ”02 '29- 2'464 4
— (s — 26,60 +€ a —— €560 sm , + éaeo 3111 ,+ so 8111 ,— {0600 + 6,50

2: (Es — £0 — 60 81112 6,)2 ‘l‘ 4€0(Es — 600)»

which is always positive (63 > 600).

As a result,

 

 

I

k ___ _jngEVS—IfiV-S—Ss
21 C r_S—82 I

121

where

 

—B+\/B2—4C

2
1
82 : _Ea
—B — I/B‘Z — 4C
83 = .

 

2

Here, it is found that 33 < 32 < 31 _<_ O by substituting 32 for 3 into the quadratic
equation in (5.3) and examining the sign of the result.

Therefore, by (3.3), (3.4) and (3.2) from Chapter 3,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

, 2A 3 3 — 3 s —— s
37(3) = ykzo2A = —X \/—\/ ll/ 3, (5.6)
c m
k 7
Z1(3) E Zj(s) : k1 71
2,1
: floSVS — 82 (5 7)
X\/§\/5 - 31\/3 - 83’
(C 1771
21(8) E ZH(8) = Z];
1
meow/3J3 — 31\/3 — 32\/3 — 33
= . (5.8)
3(363c -— 8263) + 0(3 — 32)
Hence for pI—érpendicular polarization, the interfacial reflection coefficient is
written as
F ( ) cos6,\/3\/3—32—X\/8—81\/8—83 (.- 9)
3 = , o.
L cos6,\/3\/3—32+X\/3—31\/s-—33
and for parallel j.)olarization
eox\/3\/s -— 31\/s — 32\/s — 33 — cos 6,[s(scx. — 326,.) + 0(3 — 32)] r
FH(3) = . (5.10)
€0X\/S\/8 — 31\/3 — 32\/3 — 33 + cos 6,[3(360c — 326,) + 0(3 — 32)]

The reflection coefficients 17(3) remain the same as in previous chapters, such that

__ r(s) —— €_ST(S)
_ 1 —- I“(3)e—°‘T(S)’

 

[f(s) (5.11)

122

for the conductor—backed case and

 

R = _ .1
(8) 1 _ F2(S)e—37(s) (5 2)
for the air-backed case.
Thus, the impulse response r(t) is obtained by
1
r(t) .—_ £‘1{R(3)} = —/ R(s)eStds (5.13)
.7271. Br

where Br indicates the Bromwich path. This integral can be evaluated by contour
integration. We must thus examine the singularities and define an appropriate

branch for the integrand.

Comparison with the frequency—independent material parameters case

If we let the static real permittivity 6, be the same as the optical permittivity

600, then from (5.4),

 

 

_ _ '2 ,2
32—40 : \/(€OO £0 eosm 6,)

€2(€oo — GO 81112 6,)2
(1500 —— £0 — so sin2 6,)

£(e00 — 60 31112 6,)

 

Therefore, 33 becomes
—B — \/ B2 - 4C 1
83 = 2 = —E, (5.14)

 

which is equal to 32, and

 

_—B+\/B2-4C_ 0 0

31 —- , . = ——. ( .
2 600 — 60 $1112 6, 60X

 

Cf!
)__I
01
v

123

 

Here, X = \/Eoo/€0 —- sin2 6,- is same as fl, where E = 6,. — sin? 6., as defined in
Chapter 3, by letting 600/60 2 6,. Note that m and m in (5.5) cancel
each other, since .93 = 82. As a result, (5.5)-(5.8) become the same as (3.2), (3.7),
(3.9) and (3.10) respectively; i.e., the current case becomes the frequency-
independent material parameters case if 68 = 600.

On the other hand, if the DC conductivity 0 is 0, then 31 becomes 0 resulting in
(5.16)

It should be noted that there are only two branch points left for the above

special cases.

5.2 Singularities and the branch out

From the complex square roots in ST(S) and 21(3), the branch points are
located at s = 0, s = 81, s = 32 and s = 33. These branch points are non-positive
real, and are related by 33 < 32 < 31 S 0. It should be noted that R(s) is continuous
about the negative real axis between 31 and 32, i.e., 57(5) and I‘(s) are continuous.
Therefore, unlike the previous cases, in order to ensure the continuity of R(8), two
separate branch cuts are made, such that one is taken along the negative real axis
between 0 and 81 and the other is taken along the negative real axis between 32 and
83.

r

From (0.11) and (5.12), the poles should satisfy

ln |I“(s)e’”(8)

 

+ j arg [F(s)e—ST(S)] = 0, (5.17)

124

for the conductor backed case, and
111 |F2(s)e_87(3)| + j arg [F2(s)e—3T(S)] : 0, (5.18)

for the air-backed case.

Except for the lossless case (es = 600 and a = 0), wherein the poles can be
calculated analytically, the above equations need to be solved numerically. “’hen
calculating the poles numerically, it is very important to have good initial guesses
and to know the approximate pole behavior. In this work, the lossless case poles
were used as the initial guesses. In order to trace the poles correctly, first the poles
are traced by gradually increasing the conductivity while keeping es the same as 600.
\Vhen the conductivity reaches the target value, 63 is then increased to its desired
value.

Figures 5.1-5.3 show the pole trajectories obtained by the Newton—Rahpson
method for the conductor-backed, parallel polarization case, when the 63 varies from
5.060 to 78.360, where 6.30 = 5.060, f = 9.6 x 10"12 sec and 0 = 0.5 S/m. In the
figures the amplitudes vs. the ratios of 63 to 600 are shown. As seen in the figures,
when the real static permittivities 65 become larger, the magnitudes of the real and
imaginary parts of the poles become smaller suggesting that the amplitude decay
factors become smaller.

The pole trajectories for the air-backed, perpendicular polarization case with
the same conditions are shown in Figures 5.4-5.6. The air-backed case also shows
the same behavior as the conductor-backed case. The existence condition for the
real poles and the algorithm for finding such poles are examined next.

Let us assume a complex frequtmcy s to be negative real and denote its

125

magnitude as :13, where 1: > 0. Then, ifs > s], from (5.6), (5.9) and (5.10)

 

_2A :r.(;1:+81)(;r+s3)
8T(8) = J—, X\/ ,
(r —(.T + 82)

 

and the reflection coefficient becomes

F()_jcos9,\/—:L(LI?+82)—X\/(I+81(I+93)
_LS
jcos0n/—.r( r+52) )+X\/(SC+81) )(113‘1'33)

for perpendicular polarization, and

 

 

 

j€0H\/‘( (.r + 51)(;r. + .92)(17 + .93) — cos 62[J*(;l*€oo + 8263) — 0C? + 32ll

I21(5):
I j€0X\/—.r (.1: + .91) (J: + 52)(3: + 33) + cos 041135600 + 8263) — 0(1‘ + 82)]

 

 

for parallel polarization.

As seen from the above equations, the magnitudes of both 6’3““) and 1(9) are
1. As a result, the real parts of (5.17) and (5.18) become 0, i..,e 111 [1(9) e‘”(3)| = 0
and In |F(s)28‘ST(S)| = 0. Therefore, the poles need to satisfy only the imaginary
part of (5.17) or (5.18), i.e., arg [F(s)e‘57(s)] = 0 or arg [F(8)28—ST(S ]— - 0 for the
conductor-backed and the air-backed case respectively. Here, if we denote —ST(S) as
(3)1 and arg [1(9)] or arg [F2(s)] as (.62 for the conductor—backed or the air-backed case

respectively, the poles need to satisfy
951+c52+2n7r=0. n =0,1,2,--- (5.19)

Here, only the positive sign in front of n is necessary, since (251 is always negative,
when 5 > 81 and —7r < 6.52 < 7r. The above equation, which we call the characteristic
function, can be solved by any l-D root search algorithm, such as the secant
method. Again, it is very important to note that (5.19) does not. always have

solutions, and that the number of solutions is dependent upon the conductivity,

126

permittivities and incident angle.

If 32 < s < 51, i.e., —81 < I < —s2,

 

 

37(3) 2 —

   

C $+82

2A /\/l‘(1‘ + 51)(I + 83)

and the reflection coefficient becomes

F (9) cos61/— I(I+s2) ——(X\/ (I+s1)(I+83)
1‘ =
cos611/—I(I+s2) )()+X\/—;L‘+81( )(I+s3)

for perpendicular polarization, and

 

 

 

—€0X\/17( (I + s1)(I + s2)(I + 83) — cos 92[I(Icoo + 8265) — 0(33 + 82)]
—€0X\/l‘(.T + 31) (I + s2)(I + 53) + cos 91[IE(CIT€OO + 826.3) — (I(I + 32)]

 

 

FMS):

for parallel polarization.
As seen from the equations, when .92 < s < 81, sr(s ) becomes negative real

causing 6737(8)

to be greater than 1, and F(s) becomes pure real. Therefore, the
imaginary parts of (5.17) is zero as long as F(s) is positive, whereas that of (5.18) is
always zero regardless of the sign of F (s) W hen the imaginary parts of (5.17) and
(5.18) are zero, the poles only need to satisfy the real parts of the equations, i.e.
ln IF(s)e ’3” 8)-——| — 0 and In |F() )26 ‘3“ ”I = 0 for the conductor-backed and the
air-backed case respectively.

The cases where 53 < s < 32 and 3 < 83 are analogous to the cases where
.91 < s < O and 52 < s < .31 respectively. Figure 5.7 and Figure 5.8 show the
amplitudes of the interfacial reflection coefficients for the perpendicular and the
parallel polarization case, where es 2 5.060, 6.00 = 3.060 and 0 = 1.0 S / m. From the
figures, it can be observed that F(s) follows the above discussion.

Finally, it should be emphasized that the real poles in the region where s > 81

dominate the other real poles due to their relatively smaller magnitudes. The

127

greatest care should be given to calculating these poles.

5.3 Evaluation of r(t)

Since all poles and branch points lie in the left half plane including the
imaginary axis, the region of convergence is the right half plane. Therefore, the
Bromwich path is put in the right half plane and the Laplace inversion integral is
evaluated by contour integration. The evaluation is accomplished in two different
time intervals, corresponding to the early—time period and the late—time period. The
beginning of the late-time period is To = 2? X. It should be emphasized that, unlike
with the previous cases, 7'0 does not represents the two-way transit time of the wave
inside the dielectric.

In this section, the evaluation will be done for the conductor-backed case. The
air-backed case follows the same procedure giving similar results; for the details

refer to Chapter 4.

5.3.1 Case I: t > To

When t > T0, the integration contour is closed in the left half plane as shown in
Figure 5.9. Inside the contour, the branch cuts lie from 0 to s1 and from 32 to 33,
and there are corresponding closed paths enclosing the branch cuts. Note that the
possible real poles are not shown in the figure. We denote the outer integration
contour which includes the Bromwich path as CO and the closed paths which enclose

the branch cuts as C1, such that

Cl 1 71+72+73+74+11+12+13+l+

128

Then, by Cauchy’s residue theorem,

/ R(s)eStds = j27rZ Res[R(s)eSt, poles] (5.20)
C0+C1

Therefore, if the integral contribution from each path is known, the impulse

response r(t) can be determined.

The contribution from Foo

On Foo, from (5.6),

2A . _ _(
Refsflsli = Re{ lim —X‘/3\/S 51\/3 93}
2A

2 S—X = 870
C

 

 

If we let t1 = t — 7'0, which is positive since we are evaluating the integral when

t > T0, then

I‘ 2 (570 _ ——3'r(s) Isro ‘
/oc R(s)es’ ds = [so (51): F(sfe—“(Sf 6"“ ds = /00 f(s)eSt1ds

Since f(s) ——> 0 on Foo, by Jordan’s Lem-ma [20], [21],

 

/ R(s)es" ds = O (5.21)
F00

The contribution from L1 and L2

As seen in the previous chapters, for L1 and L2, the same theorem [14] can be

used to apply Jordan’s Lemma. As a result, it can be shown that

/ R(s)e“" ds = 0 (5.22)
1.1.1.2

129

The contribution from C1

The segments of the contour C1 enclosing the branch cut. can be divided into

three groups. The first group consists of 71, 7'2, 73 and 74 that enclose the branch

points 33, s2, s1 and 0 respectively. The second group consists of the straight lines

immediately below and above the branch cut, and those are designated l1, l2, [3 and

1.1. The last. group consists of the paths that enclose the real poles on the branch cut

which are not shown in Figure 5.9. For the third group, refer to the previous

chapters. The first two groups will be examined separately. F-
Denote the radius of 71 as r, and let 6 to be the angle measured

counterclockwise from the real axis to the point on 71. Then, any point on 71 can be

represented as

s = so + re”.

The reflection coefficient on 71 with r -—> 0, i.e. s —> 80, becomes —1 for parallel
polarization and 1 for perpendicular polarization. Then,
F(s) — e-sfls) 6“ , i — pol

R c 68' = e“ =
(s) 1 -— F(s)e—ST(S) st

 

and

|R(8)€Sti S e(so+r)t.

Therefore,

3 27n~e(~‘>‘0+">t —> 0, (r —+ 0).

 

 

/ R(s)es‘ d8
71

As a result,

/ R(s)es‘ ds = 0. (5.23)
n

130

Similarly, it can be shown

/ lamest as = o. (5.24)
72173174

For the second group. it is necessary to determine F(s) and 37(5) along a path
immediately above the branch cut, which we denote as 8+, and immediately below
the branch cut, which we denote as 3‘. Also, if we let Z (s) on B+ be denoted as

2+, and Z(s) on B” as Z‘, using (5.7) and (5.8), it can be shown that
2+ = ——Z". (5.25)

Since,

 

 

R ‘_ _
(.1 1_ mac-w
_ [Z(b) - Z{)[(’%ST(S) _ [2(6) + Zo]€—%ST(S)
[2(9) + Z0]( 537(3) _ [Z(s) _ Zo]e—%sr(s)’
we have,
Z+—Z VJI¢__ Z+ Z (_jq')
R(s)|3+ _ 1 016 l + 0k»

 

[Z+ + Z0]ej¢ — [Z+ — Z0]e‘j¢
jZ+ sin 96 — Z0 cos g5
jZ+ sin (15+ 20 cos (,9

 

57(5), and

MI»—

where jgb =

[Z‘ — Zak—34’ -— [Z— + Z0]ej¢
[Z‘ + Zo]e‘1¢ — [Z‘ — Zo[cJ¢’
—jZ‘ sin (,5 — Z0 cos (5)
—jZ‘ sin¢5+ Zocoscj)

 

Rfslls- =

 

Therefore, by (5.25)
Rf8)|3+ = RfS)|3—-

131

As a result,

/ R(8)[]3+ 88" C18 +/ R(8)[B— 88’ (18 = 0 (5.26)
11+12

13+l4
The integral contributions from the third group and the other complex poles
can be determined by calculating the residues of R(s)eS’ at the poles. It. is found

that all of the poles of R(s) are of first order and thus

 

I“ _ —sr(s)
Res [R(s)e‘9", poles] [5:3, = lim (3 — 3k) (8) 6

”st =A skt, 527
s—.s1. 1 _ F(s)e—ST(S)€ 1e ( l

where
F(sk) — e‘skaS’“)

dis [F(3)e_ST(S)] [8:81.-

A1:—

 

is the complex natural mode amplitude. In this chapter, Ak is calculated by using
the numerically obtained Jacobian at the pole locations.

Collecting the results from (5.21)-(5.27), the impulse response r(t) becomes

1

r(t) = fi/R(s)est ds = Z Akesk‘, t > 7'0, (5.28)
B.

and thus the late—time period is a pure natural resonance series, and contains no

branch-cut contribution.

5.3.2 Case II: t < To

For t < r11, r(t) is found by computing the inverse Laplace transform of

R(5) = F(s) + F(s), where

 

Then

 

To compute this integral, the integration contour is closed in the right half plane, as

shown in Figure 5.10. On Foo, from (5.6),

. _ .1 2A \/.§\/s-81\/§_53
Re{37(b)} _ RC {Refiir-Loo C X \/S_—52 }

 

 

If we let t1 = t — To, which is negative since we are evaluating the integral when

t < r0, then

 

__ q. e—sr(s) 83m 3 s
/ R(s)e‘tds 2/ [F2(s) — 1] 1_ F(s)e"3T(S)€ ‘1 d3 = r f(s)e ‘1 d3

Here, it is not possible to apply Jordan’s Lemma directly, since f (s) does not
approach zero over the entire contour FOO. However, if we use the theorem

introduced in the previous chapters, it can be argued that

/ Tasha“ d8 = 0
F00

The inverse transform of F (s), denoted as F (t), is the interfacial reflection coefficient
and can be obtained numerically. Note that we are interested in the late-time
responses, rather than the early-time specular reflections. As a result, when t < T0,

r(t) becomes

r(t) = F(t), t < 10. (5.29)

Combining the results (5.28) and (5.29), the impulse response r(t) is

Z Akesk‘, t > T0

F(t), t < To.

133

5.4 Results

As with the. frequency-independent parameter cases, it has been sl1(,)wn
analytically with the Debye material that the early-time response is a specular
reflection from the interface between free-space and the dielectric slab, and the
late-time response is a. pure sum of damped sinusoids. In order to verify these
results, the natural mode series is compared to the direct IFFT using the same
input waveform used for the previous cases. Figures 5.11-5.16 show the transient
responses calculated by the natural mode series and the IFFT. In addition, the
IFFT of the interfacial reflection coefficients F(s) are also shown in order to explain
the early-time behavior. The natural resonant. frequencies of the corresponding
cases are shown in Table 5.1—5.6.

Figure 5.11 and Figure 5.14 show the responses from the pure water at room
temperature from the conductor-backed and the air-backed case respectively, where
63 = 78.360, 600 = 5.060, 6 = 9.6 x 10‘12 sec, 0 = 0 S/m, 6.,- = 0° and A = 2 cm. For
these cases, the beginning of the impulse response late-time is To = 0.291 ns. As
shown in figures, during the early-time period, before To, the IF F T of R(w)F(w)
(where F (I) is the spectrum of the input waveform) matches well with the IFFT of
F(w)F(w), which is the specular reflection from the interface. During the late-time,
the IFFT of R(w)F(w) matches well with the natural mode series. Note that, even
though T0 = 0.291 us, the first amplitude peak during the late-time occurs at about
1.3 ns due to the high real static pern‘iittivity 63. Again, it should be noted that
when t < T0 the response from the natural mode series does not have any meaning;
i.e., the natural mode series is valid only when t > T0, the late-time period. Since
the frequency band of the input. waveform is roughly 0 — 20 GHz, the first 7 natural
modes, which are shown in Table 5.1 and Table 5.4, are sufficient to represent the
late—time reflected field.

Figure 5.12 and Figure 5.15 show the responses from the same material except

134

for nonzero conductivity 0 = 0.1, and with the incidence angle 6.,- = 300 for
perpendicular polarization. Figure 5.13 and Figure 5.16 represent the responses
obtained with the same condition except for a higher conductivity 0 = 0.5 and for
parallel polarization.

Comparing Figures 5.12 and 5.14 to Figures 5.13 and 5.16, due to the relatively
higher conductivity in the Figure 5.13 and 5.16 cases, relatively smaller late-time
responses are observed. The corresponding natural resonance frequencies for these
cases are shown in Tables 5.3 and 5.6. Note that there is one real pole for each

air-backed case.

135

Table 5.1: Poles and corresponding complex natural mode amplitudes (Ak),
conductor-backed case, normal incidence (63 = 78.360, 600 = 5.060, E = 9.6 x 10~12
sec, A = 2 cm, 6,- : 0°, 0 = 0.0 S/ni)

 

Pole Amplitude Natural Mode Amplitude

 

Real Part Imaginary Part Real Part Imaginary Part

 

 

 

mode 0 —-.22382 x 109 .26557 x 1010 .38655 x 109 .13994 x 108
mode 1 ——.47802 x 109 .79638 x 1010 .38497 x 109 .41972 x 108
mode 2 —.98686 x 109 .13262 x 1011 .38180 x 109 .69914 x 108
mode 3 —.17512 x 1010 .18543 x 1011 .37703 x 109 .97797 x 108
mode 4 —.27725 x 1010 .23801 x 1011 .37063 x 109 .12560 x 109
mode 5 —.40525 x 1010 .29028 x 1011 .36264 x 109 .15329 X 109
mode 6 —.55934 x 1010 .34217 x 1012 .35291 x 109 .18084 x 109

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

136

 

Table 5.2: Poles and corresponding complex natural mode amplitudes (A1,),
conductor-backed case, I polarization (6, =- 78.360, 600 =2 5.060, 6 = 9.6 x 10‘12
sec, A = 2 cm, 61: 30°, 0 = 0.1S/m)

 

Pole Amplitude [[ Natural Mode Amplitude

 

Real Part Imaginary Part ]] Real Part Imaginary Part

 

 

 

mode 0 —.27087 x 109 .26527 x 1010 .33517 x 109 .21358 x 108
mode 1 —.52625 x 109 .79695 x 1010 .33277 x 109 .39472 x 108
mode 2 —.10379 x 1010 .13273 x 1011 .32992 x 109 .62452 x 108
mode 3 —.18065 x 1010 .18559 x 1011 .32573 x 109 .86082 x 108
mode 4 —.28333 x 1010 .23820 x 1011 .32013 x 109 .10987 x 109
mode 5 -.41202 x 1010 .29050 x 1011 .31316 x 109 .13366 x 109
mode 6 —.56692 x 1010 .34240 x 1011 .30466 x 109 .15738 x 109

 

 

 

 

 

 

 

 

 

 

 

 

 

137

Table 5.3: Poles and corresponding complex natural mode amplitudes (A1,),
conductor-backed case, I] polarization (6, = 78.360, 600 = 5.060, 5 = 9.6 x 10~12
sec, A = 2 cm, 6,- = 30°, 0 = 0.5 S/m)

 

Pole Amplitude [[ Natural Mode Amplitude

 

Real Part Imaginary Part [I Real Part Imaginary Part

 

 

mode 0 -.61336 x 109 .25908 x 1010 .45231 x 109 .79609 x 108
mode 1 —.86535 x 109 .79163 x 1010 .44129 x 109 .69074 x 108
mode 2 —.13741 x 1010 .13202 x 1011 .43689 x 109 .92938 x 108
mode 3 —.21385 x 1010 .18466 x 1011 .43112 x 109 .12148 x 109
mode 4 —.31597 x 1010 .23704 x 1011 .42357 x 109 .15149 x 109
mode 5 -—.44396 x 1010 .28911 x 1011 .41418 x 109 .18208 x 109
mode 6 —.59800 x 1010 .34078 x 1011 .40280 x 109 .21285 x 109

 

 

 

 

 

 

 

 

 

 

 

 

 

138

Table 5.4: Poles and corresponding complex natural mode amplitudes (A1,), air-
backed case, normal incidence (63 = 78.360, 600 = 5.060, 5 = 9.6 X 10~12 see, A = 2

cm, 6,- : 0°, 0 = 0.0 S/m)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Pole Amplitude Natural Mode Amplitude
Real Part Imaginary Part Real Part Imaginary Part
mode 0 —.51014 X 109 .52968 X 1010 .38462 x 109 .27874 X 108
mode 1 —.89043 X 109 .10587 X 1011 .38225 X 109 .55725 X 108
mode 2 —.15250 X 1010 .15864 X 1011 .37830 X 109 .83529 X 108
mode 3 —.24149 x 1010 .21120 X 1011 .37274 X 109 .11126 X 109
mode 4 —.35617 X 1010 .26350 X 1011 .36554 X 109 .13889 x 109
mode 5 —.49676 x 1010 .31545 x 1011 .35676 x 109 .16641 x 109
mode 6 —.66349 X 1010 .36699 X 1012 .34621 X 109 .19376 X 109
real —-.38345 X 109 0.0 .38540 X 109 0.0

 

139

 

Table 5.5: Poles and corresponding complex natural mode amplitudes (A1,), air—
backed case, _J_ polarization (6,, = 78.360, 600 = 5.060, g = 9.6 x 10“12 see, A = 2

cm, 6,- : 30°, 0 : 0.1S/m)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Pole Amplitude [[ Natural Mode Amplitude
Real Part Imaginary Part [[ Real Part Imaginary Part
mode 0 -.53216 x 109 .15660 X 1010 .33307 X 109 .28773 x 108
mode 1 —.91452 x 109 .10598 x 1011 .33058 x 109 .50640 x 108
mode 2 —.15528 x 1010 .15882 X 1011 .32704 X 109 .73977 X 108
mode 3 —.24479 x 1010 .21145 x 1011 .32215 x 109 .97629 x 108
mode 4 —.36014 X 1010 .26379 x 1011 .31585 X 109 .12134 x 109
mode 5 —.50153 X 1010 .31579 X 1011 .30818 X 109 .14503 x 109
mode 6 —.66918 x 1010 .36735 X 1011 .29896 x 109 .16859 x 109
real —.47724 X 109 0.0 .33412 x 109 0.0

 

 

140

Table 5.6: Poles and corresponding complex natural mode amplitudes (A1,), air-
backed case, [| polarization (6, = 78.360, 600 = 5.060, g = 9.6 X 10’12 sec, A = 2

cm, 0,: = 30°, 0 = 0.5 S/m)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Pole Amplitude Natural Mode Amplitude
Real Part Imaginary Part Real Part Imaginary Part
mode 0 —.92647 X 109 .52329 x 1010 .44466 X 109 .63070 X 108
mode 1 —.13048 x 1010 .10523 x 1011 .43808 x 109 .79647 x 108
mode 2 —.19388 X 1010 .15784 X 1011 .43277 x 109 .10647 X 109
mode 3 —.28282 X 1010 .21022 X 1011 .42603 x 109 .13578 x 109
mode 4 —.39743 X 1010 .26230 X 1011 .41752 X 109 .16599 X 109
mode 5 -—.53792 X 1010 .31403 X 1011 .40719 X 109 .19656 X 109
mode 6 —.70450 X 1010 .36532 X 1011 .39484 X 109 .22720 X 109
real —.11590 X 1010 0.0 .44881 X 109 0.0

 

141

 

 

 

x10

 

ll]

mode 0

mode 1 mode 2

mode 3

Real Part

—10

l
l

_12— _

 

 

1..
1—
p-
r—
_

 

-14
0

Figure 5.1: The amplitude of the real parts of poles vs. the ratio of e, to 600, [I
polarization (6, = 5.060 ——> 78.360, 600 = 5.060, (E = 9.6 X 10‘12 sec, A = 2 cm,
6, = 30°, 0 = 0.5 S/m).

142

 

x 1010

 

Imaginary Part
1‘ 9‘ °.’

00
1

 

 

 

 

 

l l l

0 2 4 6 8 1O 12 14 16

8/8
500

 

Figure 5.2: The amplitude of the imaginary parts of poles vs. the ratio of e, to 600,
I] polarization (6, = 5.060 —~> 78.360, 600 = 5.060, 5 = 9.6 X 10‘12 sec, A = 2 cm,
9,- : 30°, 0 = 0.5 S/m).

143

x101O

 

CD
I

01
r

Imaginary Part
“3 ‘?

 

 

 

 

 

O 4 4L
—1 4 —1 2 —1 o —8 —6 —4 —2 0
Real Part x109

Figure 5.3: Pole trajectories of the reflection coefficient, || polarization (63 = 5.060 —->
78.360, 600 = 5.060, E = 9.6 x 10‘12 see, A = 2 (:In, 61- : 30°, 0 = 0.5 S/ni).

144

 

x10

 

lll

 

mode 3

_4 _ mode 2 mode 1 mode 0 —

Real Part
('30

_1o~ -

_14— a

 

 

 

Figure 5.4: The amplitude of the real parts of poles vs. the ratio of es to 600, _L
polarization (53 = 5.060 -—+ 78.360, 600 = 5.060, E = 9.6 x 10“12 see, A = 2 cm,
62- :: 30°, 0 = 0.5 S/m).

145

 

10

x10

 

Imaginary Part
‘? ‘3‘ ‘3’

(A)
I

 

 

 

 

 

 

l l l

0 2 4 6 8 10 12 14 16

8/8
300

 

Figure 5.5: The amplitude of the imaginary parts of poles vs. the ratio of es to 600,
_L polarization (63 = 5.060 —> 78.360, 600 = 5.060, E = 9.6 x 10‘12 sec, A = 2 cm,
61- = 30°, 0 = 0.5 S/In).

146

10

 

x10
9 I I I I I I I
mode3
E/E =1
8~ 3 °° a
7- a
mode2
6i _

01
I
1

Imaginary Part
A
3
O
O.
(D
8???
II
.a-
on

(D
1

 

 

 

 

2 ~ a “‘ -
mode 0 '

1 — \NM _
O l l l l l l l

—16 -14 —12 —10 —8 —6 —4 -2 0

Real Part x 109

Figure 5.6: Pole trajectories of the reflection coefficient, _L polarization (63 = 5.060 —>
78.360, 600 = 5.060, 5 = 9.6 x 10‘1'2 sec, A = 2 cm, 6i 2 30°, 0 = 0.5 S/m).

147

 

 

 

 

 

 

 

 

 

 

 

1 ~ ' ‘
0.8~
0.6-
0.4»
(D _ ‘1
.0 0.2
3
a o~
E
< —o.2~
—0.4~
o 6 3 ‘x
— - - - Re[l‘i(s)] \
_0.8 H """ |m[ FJ:(S)] \\ j
_ Magnitude[1“i(s)] \ '
_1 I 1 \\1 1 1 - \
—2.5 —2 —1.5 -1 -O.5 O
8 (Real) x 10‘1

Figure 5.7: Amplitude of the reflection coefficient vs. frequency for _L polarization
(6,,. = 5.060, £00 2 3.060, 5 = 9.6 ><1O’12 sec, A = 2 cm, 61- : 30°, 0 21.0 S/m).

148

 

 

 

 

 

 

 

 

 

 

 

 

1 P {7‘ of ““3
' \ 3' 's
0.8- : \V:
' :\ ~.
0.6" : ;’ \ a.
I " \ ‘
O.4~ l ‘ \ .. :
l \ 1
8 0.2” t: \\ ." 1
3 i' \ :1 |
E 0_ .. .1 3 \ 3 I
Q. ‘3 ' \ ' If I
E as x :
_0 2— E 5 \\ i a” E ‘1 d
_ Z / 1
-O 42 8"S3 \\ g ,” I 1 .1
. z \‘ 3 ’1’ I “
-0.6 4 \ 5:52. ,I’ S=S1i 1 _
Re[FII(S)] \ I : ‘
——0.8 ~ """ lml F”(S)] \\ El, I ‘\ 4
__ Magmtude[I‘ (3)] \ I ' \
|| \1 ' \
_1 E 1 1 1 _
-2.5 -2 —1.5 —1 —O.5 0
8 (Real) x 1011

Figure 5.8: Amplitude of the reflection coefficient vs. frequency for H polarization
(63 = 5.060, 600 = 3.060, f = 9.6 x 10‘12 sec, A = 2 cm, 6i = 30°, 0 21.0 S/m).

149

Foo

W8}

 

 

 

 

 

 

 

 

 

 

L1
< s=jb
x Br
Y] Y2 Y3 [2 Y4
% : G=,=<} A R Av
33 S2x S1 4 O 313}
x i><:poles
s=-1)
L2)I J
s=a s=c

 

Figure 5.9: The integration contour when t > T0. |a|, |b| ——> oo, 0 < c < 00.

150

 

1mm?

/ I30

0 Brl‘ RA’eiSl

 

 

 

sia

Figure 5.10: The integration contour when t < 7'0. R —-> oo, 0 < a < 00.

151

 

 

 

 

 

 

 

 

 

0.6 . f A
- - lFl-‘l' F(m)F((o)
x IFFT R(w)F((n)
0.4- — Natural Mode Series -l
0.2- .1
(D
'0 0 - — -
3 .
= I
D. 1
E_ '- ,
< 0.2 |:
:l
I
-O.4 1| II
l
E
-0.6Fl .1
I
I
l
- .8; J I l I
0 O 2 4 6 8 10

Time (nsec)

Figure 5.11: Time domain response, conductor-backed case, normal incidence, (63 —
78.360, 600 = 5.060, E = 9.6 X 10”12 sec, A = 2 cm, 9,- : 0°, 0 = 0.0 S/m).

152

 

 

0.4 . . 1
- - IFFT F(w)F((o)

x IFFT R(o))F(w)
-— Natural Mode Series ll

 

 

 

0.2 1-

 

 

 

 

 

G)

.0

2

'5.

E

<
I.
'1
I

-0.8 -l .
_1 l I l g

o 2 4 6 8 10

Time (nsec)

Figure 5.12: Time domain response, conductor-backed case, .1. polarization, (es
78.360, 600 = 5.060, E = 9.6 x 10’].2 sec, A = 2 cm, 6,- = 30°, 0 = 0.1 S/m).

153

 

 

 

0.6 . . r
- - IFFT F(m)F((o)
X IFFT R(w)F(m)
— Natural Mode Series

 

 

 

Amplitude

 

 

 

 

 

‘0'80 1 2 3 4 5
Time (nsec)

Figure 5.13: Time domain response, conductor-backed case, M polarization, (6,
78.360, 600 = 5.060, f = 9.6 x 10“12 sec, A = 2 cm, 6.,- = 30°, 0 = 0.5 S/m).

154

 

 

 

0.6 . . L .
- - IFFT F(co)F(m)
x IFFT R(m)F(w)
— Natural Mode Series

I

 

 

 

0.4 -

Amplitude

 

 

_ _1 1 .
0'8 3 4 5 6
Time (nsec)

 

 

Figure 5.14: Time domain response, air-backed case, normal incidence, (63 = 78.360,
600 = 5.060, g = 9.6 x 10‘12 sec, A = 2 cm, 0,- : 0°, 0 = 0.0 S/In).

155

 

 

 

 

 

 

 

 

 

 

0.4 I I I I I
-- IFFT F(m)F(co)
x IFFT R(w)F(w)
0.2 - — Natural Mode Series -
I l
I I
a: I I
'0 —0 2 -| I I
a --
= I |
6 --
_ .I .
< 0.4 ':
H
‘1
I
—0.6-,: -
H
II
—o.6-: -
_1 l l l l j
0 1 2 3 4 5 6

Time (nsec)

Figure 5.15: Time domain response, air-backed case, _L polarization, (63 = 78.360,
600 = 5.060, g z: 9.6 ><10‘l2 sec, A = 2 cm, 6,- : 30°, 0 = 0.18/m).

 

I I

0.6 . . . .
- - IFFT F(m)F(m)
x IFFT R(w)F(w)
— Natural Mode Series

 

 

 

L

 

 

 

 

Amplitude

 

 

_O.8 l l l l
0.5 1 1.5 2 2.5 3 3.5 4
Time (nsec)

 

 

 

Figure 5.16: Time domain response, air-backed case, [I polarization, (68 = 78.360,

600 = 5.060, g = 9.6 x 10‘12 sec, A = 2 cm, 6,- : 30°, 0 = 0.5 S/m).

157

Chapter 6: Conclusions

The natural mode representations of the field reflected by conductor-backed
and air—backed lossy layers with frequency independent or dependent (Debye type)
material parameters have been analyzed and obtained numerically. It has been
shown that the late time response can be represented as a. pure series of damped
sinusoids. These results are not only useful in the practical application to material
characterization using the E—pulse method, but also give us a significant physical
insight into the nature of the transient scattering.

In this thesis, a complete set of complex 3 domain poles was determined for each
case. In particular, the existance conditions of the real poles that are indispensable
for calculating the transient response were analyzed, and the determining equation
was formulated and numerically implemented. It should be emphasized that, when
calculating the poles numerically, it is very important to have good initial guesses
and to know the approximate pole behavior in order to bound the pole locations.

Unlike in the previous work [12], [13], the evaluation of the Laplace inversion
integral was accon'iplished analytically without resorting to numerical evaluation. It
was clearly justified that the contributions from the deformation contour located at
infinity and the branch cuts along the negative real axis vanishes. Here, it should be
noted that the assumption of a practically realizable signal which belongs to the
class of almost piecewise continuous (AFC) signal [14] is fundamentally crucial in
evaluating the Laplace inversion integrals. In the previous work [12], [13], the
assumption of an unrealistic signal (a rectangular pulse with instant rising time) led
to inaccurate and computationally expensive results.

Also in this thesis, the transient. responses from the frequency dependent Debye
type material were evaluated analytically showing the natural resonance

representation works equally well for the frequency dependent material parameters

158

 

case. This result can be used for developing different permittivity models.

In future work, a multiple layer material can be examined with the reflection
coefficient formulated in Chapter 1. Also, it may be possible to develop a different
permittivity model that can better represent a general class of dielectrics. For this
purpose, a series of known permittivity models such as the Debye model can be
considered [32]. As a common structure, a curved structure also deserves to be
examined. Finally, the parameter extraction problem, which is to find material

parameters from a known time-domain response, is worthy of investigation.

159

 

BIBLIOGRAPHY

160

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163

lll[l[l[[[l[l,l[[[[[[Illl