Electromagnetic Theory
for Electromagnetic
Compatibility Engineers
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Electromagnetic Theory
for Electromagnetic
Compatibility Engineers
Tze-Chuen Toh
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CRC Press
Taylor & Francis Group
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Boca Raton, FL 33487-2742
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Version Date: 20131009
International Standard Book Number-13: 978-1-4665-1816-2 (eBook - PDF)
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To my wise and inspirational parents Nancy and Swee-Hock, my endearing
wife and muse Linnie, my cherished children by marriage, Courtney,
Allison, and Paul . . . and in memory of our cheeky dog Lucky.
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Contents
Preface.......................................................................................................................ix
About the Author....................................................................................................xi
Notations............................................................................................................... xiii
1 Brief Review of Maxwell’s Theory..............................................................1
1.1 Electrostatics...........................................................................................1
1.2 Magnetostatics...................................................................................... 12
1.3 Maxwell’s Equations............................................................................ 16
1.4 Electromagnetic Waves....................................................................... 21
1.5 Worked Problems................................................................................. 32
References........................................................................................................34
2 Fourier Transform and Roll-Off Frequency............................................. 35
2.1 Fourier Series........................................................................................ 35
2.2 Fourier Transform................................................................................ 40
2.3 Roll-Off Frequency.............................................................................. 46
2.4Frequency Response and Filter Theory: A Primer.......................... 50
2.5 Worked Problems................................................................................. 60
References........................................................................................................65
3 Boundary Value Problems in Electrostatics............................................. 67
3.1 Electromagnetic Boundary Conditions............................................ 67
3.2 Image Theory Revisited...................................................................... 71
3.3 Multipole Expansion............................................................................80
3.4 Steady-State Currents..........................................................................84
3.5 Duality................................................................................................... 88
3.6 Worked Problems................................................................................. 91
References........................................................................................................ 98
4 Transmission Line Theory......................................................................... 101
4.1 Introduction........................................................................................ 101
4.2 Transmission Line Equations........................................................... 102
4.3 Characteristic Impedance and the Smith Chart............................ 108
4.4 Impedance Matching and Standing Waves................................... 122
4.5 Worked Problems............................................................................... 135
References...................................................................................................... 141
5 Differential Transmission Lines.............................................................. 143
5.1 Differential Pair: Odd and Even Modes......................................... 143
5.2 Impedance Matching Along a Differential Pair............................ 157
vii
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viii
Contents
5.3 Field Propagation Along a Differential Pair.................................. 160
5.4 Worked Problems............................................................................... 167
References...................................................................................................... 182
6 Cross-Talk in Multiconductor Transmission Lines.............................. 183
6.1 Reciprocity Theorem and Mutual Capacitance............................. 183
6.2 Mutual Inductance and Mutual Impedance.................................. 192
6.3 Multiconductor Transmission Lines and Cross-Talk.................... 199
6.4 S-Parameters: Scattering Parameters.............................................. 210
6.5 Worked Problems............................................................................... 218
References...................................................................................................... 221
7 Waveguides and Cavity Resonance.........................................................223
7.1 Parallel Plate Guides..........................................................................223
7.2 Rectangular Waveguides.................................................................. 236
7.3 Cavity Resonance............................................................................... 243
7.4 Worked Problems............................................................................... 251
References...................................................................................................... 259
8 Basic Antenna Theory................................................................................ 261
8.1 Radiation from a Charged Particle.................................................. 261
8.2 Hertzian Dipole Antenna................................................................. 263
8.3 Magnetic Dipole Antenna................................................................. 270
8.4 Microstrip Antenna: A Qualitative Overview............................... 276
8.5 Array Antenna and Aperture Antenna.......................................... 280
8.6 Worked Problems............................................................................... 295
References...................................................................................................... 302
9 Elements of Electrostatic Discharge........................................................ 303
9.1 Electrostatic Shielding....................................................................... 303
9.2 Dielectric Properties: the Kramers–Kronig Relations..................305
9.3 Beyond Classical Theory................................................................... 311
9.4 Dielectric Breakdown........................................................................ 319
9.5 Worked Problems............................................................................... 324
References...................................................................................................... 328
Appendix A.......................................................................................................... 331
Index...................................................................................................................... 365
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Preface
The irony of a preface is that it is often ignored by the reader; notwithstanding, it is occasionally employed by authors to justify formally the existence of
their work. This primer is an outgrowth, and indeed a much expanded version, of a course I gave at Gateway Inc. to help electromagnetic compatibility
(EMC) engineers be less dependent upon empirical data when solving puzzling electromagnetic interference (EMI) problems. In short, it is a vehicle to
transform practical engineers into theoretical engineers, and in particular,
to cultivate academic skills left by the wayside after they have completed
their tertiary education. Furthermore, as this course is designed for practical
applications, historical developments are bypassed altogether.
This set of lecture notes is written at a level equivalent to that of thirdor fourth-year Honors undergraduate study. It is, in essence, a refresher
for professional engineers who already have their Bachelor’s degree and
to reacquaint them with the power of mathematical rigor in solving realworld problems. Conversely, it also serves to introduce undergraduates to
the basics of EMC encountered in the technology industry. It is completely
self-contained and designed with self-study in mind.
For a condensed course, the recommended topics to cover are Chapters
1, 4, 5, and 6. Having pointed this out, I must nevertheless emphasize that
each topic presented herein is essential for EMC engineers to be competent
EMC theoreticians. Clearly, in a brief course such as this, it is impossible to
cover every aspect of electromagnetic theory. A list of references upon which
parts of this course are based is provided for the reader who wishes to delve
deeper into topics glossed over for want of space. More importantly, it was
written specifically for theoretical physicists and mathematicians new to the
field of EMC, signal integrity and RF design.
Electromagnetic theory is the simplest theory among the four known forces
of the universe. Indeed, it is the first step toward finding the holy grail of theoretical physics: the grand unification of the four known forces. Maxwell’s
theory achieved the unification of the electric field and the magnetic field
into a single entity called the electromagnetic field. Notwithstanding, its
underlying mathematical structure is similar to the other three forces—the
weak nuclear force, strong nuclear force, and gravitational force—to wit,
gauge theory. Gauge theory provides a common arena for the foundational
description of these forces. However, studying gauge theory, fascinating as
it is, will take the reader too far afield from the original intent of these notes.
Engineers, after all, are practical and down to earth, and hold little interest
in the more abstruse mathematical guise of nature.
The primary purpose of this monograph is to integrate theory with practicable engineering applications. As a case in point, it is often difficult to
ix
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x
Preface
find a monograph on electromagnetic theory that expounds elements of
differential transmission line theory, roll-off frequencies, and electrostatic
discharge frequently encountered in the industry. That is, the material that
professional engineers seek is often not found in standard courses on electrodynamics. On the other hand, although references for differential transmission line theory and electrostatic discharge for professional engineers
abound, they often lack mathematical rigor, preferring to cite engineering
rules of thumb and observations. It is hoped that this monograph will bridge
the gap by providing the needed rigor to engineering applications, and more
important, to instill the need for rigorous mathematical science in good engineering practice.
In short, the intent is to appeal to both sets of audiences: to entice the
practical engineer to explore some worthwhile mathematical methods, and
to reorient the theoretical scientist to apply the theory in the technology
industry. For there is much indeed that a professional engineer can profit
from in pure academic pursuit to further the cause of innovation and technology advancement.
Finally, SI units are employed throughout this exposition. Although CGS
units can still be found in some textbooks on electrodynamics, particularly
references prior to the 1970s, I am quite persuaded that the process of converting CGS units back to SI units causes more confusion and frustration
than the convenience it purports to impart to the formalism of electromagnetic theory.
Chuen Toh
Lexington, KY
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About the Author
Tze-Chuen Toh is a theoretical physicist and a consultant. He received his
BSc (Hons I) in mathematics from the University of Queensland in Australia,
and his PhD in theoretical physics from the Australian National University.
In the past, he worked at Lexmark International Inc. as an electrical engineer, and at Dell Inc. and Gateway Inc. as an electromagnetic compatibility
engineer. His research portfolio comprises publications in peer-reviewed
scientific journals and patents. His current research interests lie primarily in
quantum gravity, gauge theory, quantum computing and information, electrodynamics, and mathematical modeling.
xi
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Notations
The following notations are used throughout this monograph.
• The empty set is denoted by ∅.
• Given a set A ≠ ∅, a ∈ A reads as a is an element of (or a member of) A.
• The set A ∪ B denotes the union of sets A and B; A ∩ B denotes the
intersection of sets A and B.
• If A is a set, then 2 A = {S : S ⊆ A} denotes the set of all subsets of A,
that is, the power set of A.
• If A and B are nonempty sets, then A × B is the set of all points*
(a, b) such that a belongs to A and b belongs to B. This is written as
A × B = {( a, b) : a ∈ A, b ∈ B} .
• Given two nonempty sets A and B, a mapping f.
• A → B defines a relation between sets A and B whereby f assigns
each element a ∈ A a unique element b ∈ B called the value of f at a.
This is written as f (a) = b or f: a ↦ b.
• The sets R , R 2 denote, respectively, the real line and Euclidean
2-dimensional space. More generally, the Euclidean n-dimensional
space R n = {( x 1 , , x n ) : x i ∈R , i = 1, , n} .
• A closed interval [a, b] ⊂ R on the real line is the subset [a, b] = {r ∈ R: a
≤ r ≤ b}, and an open interval (a, b) ⊂ R is defined by (a, b) = {r ∈ R: a <
r < b}.
• ∀ symbolizes for all or for every.
• ∃ symbolizes there exists.
• ⇒ denotes implication; that is, P ⇒ Q reads as if P then Q, or equivalently, P implies Q.
• ⇔ denotes equivalence; that is, P ⇔ Q reads as P if and only if Q, or
P is equivalent to Q.
• Given a nonempty set A, B ⊆ A denotes that B is a subset of A. Recall
that B ⊆ A if ∀x ∈ B ⇒ x ∈ A, whereas B ⊂ A means that B is a proper
subset of A; that is, B ⊆ A and B ≠ A.
• a ∉ A: a does not belong to A.
• B ⊄ A: B is not a subset of A.
• Given a nonempty set A, the boundary of A is denoted by ∂A.
*
Technically known as an ordered pair.
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xiv
Notations
• x → x0+ denotes x → x0 such that x > x0 ; that is, x tends toward x0 in
the limit from above.
−
• x → x0 denotes x → x0 such that x < x0 ; that is, x tends toward x0 in
the limit from below.
• a+ ( a− ) or a± reads a+ and a− , respectively.
• In particular, a±  b is short for a+ − b− and a− + b+ , respectively.
• A = {x: Q} reads as A is the set of all x such that proposition Q is
satisfied.
• R = {x:−∞ < x < ∞} = (−∞,∞) is the set of real numbers.
• C = {x + iy: −∞ < x, y < ∞} is the set of complex numbers, where i = −1.
• N = {1,2,3,…} is the set of natural numbers.
2
• Z = {0,±1,±2,…} is the set of integers ∂ x = ∂∂x and ∂2x = ∂∂x2 , and so on.
• ∆ ≡ ∇ 2 denotes the Laplacian operator. For instance, in 3-dimensional
Euclidean space, ∆ = ∂2x + ∂2y + ∂2z in rectangular coordinates.
• R n = ( x 1 , , x n ) : −∞ < x i < ∞ ∀i = 1, , n denotes the n-dimensional
Euclidean space.
• (a, b] ≡ {x: a < x ≤ b}.
• If f: A → B is a function that maps A into B, and C ⊂ A, then the restriction of f to the subset C is written as f|C or f |C .
• Given two sets A, B, the Cartesian product A × B = {(a, b): a ∈ A, b ∈ B}.
• Given a set A and subset B ⊆ A, the complement Bc ≡ A − B =
{ a ∈ A : a ∉ B} .
• Given a nonempty set A, a partition P of A is the collection of subsets
B ⊂ A such that ∀B, B′ ∈P , B ≠ B′ ⇒ B ∩ B′ = ∅ and A = ∪B∈P B .
• The space C(R k ) defines the set of all continuous functions
f : Rk → R .
• The space C1 (R k ) defines the set of all continuous functions f : R k → R
that is continuously one-time differentiable; i.e., f ′( x) = dxd f ( x) is
continuous on R.
• The space C 2 (R k ) defines the set of all continuous functions
f : R k → R that is continuously twice differentiable. That is,
{
}
∂2 f
∂ xi ∂ x j
=
∂2 f
∂ x j ∂ xi
∀i, j = 1, , k
• where ( x 1 , , x k ) ∈R k .
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1
Brief Review of Maxwell’s Theory
In this chapter, Maxwell’s theory is briefly reviewed. In particular, it is shown
that electromagnetic fields exist and they propagate in space as waves.
Historically, the set of Maxwell’s equations were not all derived by Maxwell;
they were attributed to him because, among other things, he correctly added
a displacement current term to Faraday’s equation, and predicted the existence of electromagnetic waves from the set of equations that is now known
as Maxwell’s equations [2,10,11].
As electronic components get smaller, and digital circuits operate in the
microwave frequencies, it is imperative that electromagnetic compatibility
(EMC) engineers possess a sound understanding of electromagnetic theory
in order to root-cause electromagnetic interference issues and to design
printed circuit boards effectively with low electromagnetic emissions in
mind.
An outline of this chapter runs as follows. Electrostatics is introduced in
Section 1.1, followed by a brief overview of steady-state currents in Section
1.2. Maxwell’s equations are derived in Section 1.3, and the significance of
the equations highlighted. Finally, the existence of electromagnetic waves is
illustrated in the last section.
1.1 Electrostatics
It is known empirically that two point charges qi, i = 1,2, exert a force F on
each other. This force is the celebrated Coulomb’s law:
F=
1 q1q2
4 πε 0 r 3
r
(1.1)
where ri = ( xi , y i , zi ) for i = 1,2, is the location of charge qi in the rectangular
coordinate system, r = ( x2 − x1 )2 + ( y 2 − y1 )2 + ( z2 − z1 )2 and r = r2 − r1 . The
constant ε 0 = 8.854 × 10−12 defines the electric permittivity of free-space, in farads per meter (F/m). This fact suggests that a field surrounds a point charge
mediating the force between the two charges. Informally then, the electric
field E generated by a point charge q is defined as follows. Consider a test
1
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2
Electromagnetic Theory for Electromagnetic Compatibility Engineers
charge δq in the vicinity of q and let F denote the force between the pair
(q, δq). Then, in the limit as the test charge δq → 0,
E ≡ lim δ1q F =
δq→ 0
q
1
4 πε 0 r 3
r
The unit of the electric field is in volts per meter (V/m). A more formal derivation is provided later.
In a more realistic scenario, consider the electric field generated by some
charged volume V. First, consider the simpler case wherein finite point
charges q1 ,  , qn are embedded in V. Let qi be located at ri with respect to
some fixed origin where a test charge δq is located. For an arbitrary point r
away from the origin, let the electric field generated by the ith charge qi be Ei.
Because force is a vector, the superposition principle holds. Thus, it follows
from the heuristic definition of the electric field that the electric field at the
origin, in the limit as δq → 0, is:
E ≡ lim δ1q ( F1 +  + Fn ) =
δq→ 0
1
4 πε 0
∑
n
qi
3
i = 1 ri
ri
(1.2)
∑
n
Ei .
where ri = ( x − xi )2 + ( y − y i )2 + ( z − zi )2 ∀i = 1, , n . That is, E =
i=1
Indeed, it is seen more formally later that the electric field obeys the superposition principle because the operators defining Maxwell’s equations are
linear. From (1.2), the extension to a continuous charge distribution in V is
obvious. To see this, suppose that V has a volume charge density ρ. Then,
heuristically, on setting δqi = ρδVi , where δV is a differential volume element
of V, taking the limit carefully as δVi → 0 and n → ∞, such that
n
Q = lim
∑ δq = ∫ ρ(r )d r
3
i
δVi → 0
n→∞ i = 1
V
is the well-defined total charge on V, it follows that the summation in (1.2)
becomes an integral:
E( r ) =
1
4 πε 0
∫
V
ρ( r ′ )
R′ 3
R′d 3 r ′
(1.3)
where R′ = ( x − x′)2 + ( y − y ′)2 + ( z − z′)2 , for all r ′ = ( x ′ , y ′ , z ′) ∈V and r = (x,y,z).
In the above intuitive approach, the limit δq → 0 was invoked in a somewhat
cavalier fashion. In reality, it is known that the electric charge cannot be arbitrarily small. To date, the smallest nonzero electric charge is 31 |e|,* where e is the
*
Subatomic particles called quarks possess a fractional electronic charge composed of integer
multiples of one-third that of an electronic charge.
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3
Brief Review of Maxwell’s Theory
fundamental electronic charge of the electron in Coulomb: e = −1.602 × 10−19 C.
However, as this is an exposition of classical physics rather than quantum
physics, this subtlety may be ignored for all intents and purposes.
Before formally defining an electric field, consider the example below
which is critical in the development of antenna theory; to wit, a static electric
dipole. Further details are explored in Chapter 8. An electric dipole is essentially two oppositely charged (point) particles separated by a fixed small
distance.
1.1.1 Example
Consider two point charges ( ±q , d0 ) separated by a distance d0 > 0. Without
loss of generality, suppose that ± q is located, respectively, at r± = (0, 0, ± 21 d0 ).
Then, the electric field at any point r = ( x , y , z) is given by the superposition of the two charges:
E( r ) =
q
4 π ε0
{
r− r+
r − r+ 3
−
r − r−
r − r− 3
}
q
=
4π ε 0


x


1
y
  2 2 1 2  −3/2 
  x + y +( z− 2 d0 )   z − 1 d0
2



x

1
y
−
−3/2 
 x 2 + y 2 + z+ 1 d 2 

( 2 0 )   z + 1 d
2 0






 

 

where the first term is the contribution from +q and the second term is from
–q. See Figure 1.1.
Now, if d0 << r , then the binomial expansion may be invoked:
1
r − r± 3
=
1
r3
(1  )
d0 z
r2
Electric dipole
z
1d
2 0
− 23
≈
1
r3
(1 ±
3 d0 z
2 r2
)
Cross section of an electric dipole field profile
P
q
0
– 1 d0
2
–q
r
y
x
Figure 1.1
Electric field generated by an electric dipole.
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4
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Whence, substituting the binomial approximation into the above expression yields
E( r ) =
q 3 d0 z
4 π ε0 r 5
 x 
 y 


 z 
for the electric field at any point very far away from the pair of point charges.
Observe in particular that the electric field is identically zero on the plane
z = 0 due to the symmetry of the problem, and for z ≠ 0, it falls of as ~ rz4 from
the pair of charges.
□
A closely related quantity to the electric field is the potential field.
Intuitively, the electric potential is related to the work done against an electric field in moving a charge, located out in infinity, to some fixed arbitrary
position. A more rigorous development is given below.
First, recall that a force F is conservative on some domain Ω ⊆ R 3 if ∃ψ ∈C 2 (Ω)
such that F = –∇ψ. Here, the continuously twice differentiable function ψ on Ω is
called a scalar potential. Geometrically, the gradient ∇ψ is normal to the surface
defined by ψ. Thus, a conservative force is always normal to equipotential surfaces defining the force. By convention, the negative sign is present to indicate
that the direction of the force points from a surface of higher potential toward
a surface of lower potential. That is, a particle in the presence of a conservative
force always traverses from a higher potential to a lower potential surface.
Now, in view of (1.1), it is clear that a conservative electrostatic force acting
on a charge q can be formally defined as follows: F = −q∇φ ⇒ E ≡ −∇φ, for
some scalar potential φ, where in rectangular coordinates, ∇ = (∂ x , ∂ y , ∂ z ).
That is, the electrostatic field is formally defined by the gradient of some
scalar potential generating a conservative electrostatic force: E = −∇ϕ. Thus,
the electric field is, by definition, normal to equipotential surfaces, and
electric charges move from a higher electric potential surface to a lower
electric potential surface.
In this section, only the static case is considered; that is, ∂ t E = 0. In
hindsight, define the static condition for an electric field by ∇ × E = 0; see
Proposition 1.1.2 below, given without proof. Indeed, this stipulation is manifestly apparent in Section 1.3.
1.1.2 Proposition
Suppose an electric field E is defined on R3 such that ∇ × E = 0. Then, ∃ϕ ∈C 2 (R 3 )
such that E = −∇φ on R 3, where φ is some scalar potential field. That is, the electric field E is conservative.
□
The existence of φ in Proposition 1.1.2 is due to a lemma by Poincaré [1,3];
the exact detail is beyond the scope of this exposition. The potential φ can be
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5
Brief Review of Maxwell’s Theory
recast into a more convenient form. 3First, recall that a path in R 3 is defined by
a continuous function γ : [0, 1] → R . In particular, γ defines a loop (or closed
path) if γ(0) = γ(1) (cf. Appendix A.2 for further details). Second, an immediate consequence of Poincaré’s lemma mentioned above (loc. cit.) is needed in
order to complete the equivalent definition for φ. The corollary is stated below
without proof.
1.1.3 Corollary
Let M ⊆ R 3 be a simply connected subspace and let f: M → R be any integrable function; that is, ∫ M | f |< ∞. Then, for any pair of paths γ 1 , γ 2 : [0, 1] → M
such that γ 1 (0) = γ 2 (0) and γ 1 (1) = γ 2 (1), ∫ γ 1 f = ∫ γ 2 f holds.
□
That is, when a space is simply connected, the path integral along any path
connecting some fixed pair of endpoints depends solely upon the endpoints
and is completely independent of the paths connecting the two endpoints.
This leads to the following equivalent definition for an electric potential,
ϕ(r ) = −
∫
γ∞
E ⋅ dr ′
(1.4)
where r is the position vector of some arbitrary point P ∈R 3 and γ ∞ ⊂ R 3
is a path from ∞ to the point P. Then, the potential is said to be conservative. Recall that a path γ ∞ connecting a point at ∞ to a point P is defined by
γ ∞ : (0, 1] → R 3 such that lim γ ∞ (t) → ∞ . Finally note that R 3 is simply cont→ 0
nected; in contrast, a two-dimensional torus is not simply connected; see
Appendix A.2 for details.
The definition for a potential field is well-defined by Corollary 1.1.3. In particular, the potential difference δφ between any two points P1 , P2 ∈R 3 connected
by any path γ such that γ (0) = P1 and γ (1) = P2 , is given by δϕ(r ) = − ∫ γ E ⋅ dr ′.
From Corollary 1.1.3, as R 3 is simply connected, this integral is only determined by the two endpoints and not the path connecting them.
1.1.4 Proposition
Suppose ∇ × E = 0 on R 3. Then, δφ( r ) = − ∫ γ E ⋅ dr ′ = 0 ∀γ ⊂ R 3 , where γ is any
loop in Euclidean 3-space.
Proof
Let γ + = γ |[0, 21 ] be the restriction of γ to the interval [0, 21 ] and set γ − = γ |[ 21 , 1].
Then, by definition, γ = γ − ∗ γ + , where
 γ + (2t) for 0 ≤ t ≤ 1 ,
2

γ− ∗γ+ ≡ 
 γ − (2t − 1) for 21 ≤ t ≤ 1.

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6
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Now, note that if Γ : [0, 1] → R 3 is a path, then Γ (t) ≡ Γ(1 − t) defines the
reverse orientation of Γ: Γ (0) ≡ Γ(1) and Γ (1) ≡ Γ(0). Thus, by Corollary 1.1.3,
Γ (0)
*
for any integrable function f on R 3 , ∫ Γ f = ∫ ΓΓ (1)
(0) f = − ∫ Γ (1) f = − ∫ Γ f . Thus,
from ∇ × E = 0 ⇒ ϕ = − ∫ γ E ⋅ dr ′ , it follows immediately that
−
∫ E ⋅ dr ′ = − ∫
γ
γ−
E ⋅ dr ′ −
∫
γ+
E ⋅ dr ′ = −
∫
γ−
E ⋅ dr ′ +
∫
γ−
E ⋅ dr ′ = 0
as required.
In other words, for a simply connected space (i.e., any two paths with the same
endpoints can be continuously deformed into each other), the potential difference around a loop is zero. This is a rather critical point to note: indeed, in a
space that is not simply connected, for example, a 2-torus, the path integral generally depends upon both the endpoints and the path connecting the endpoints.
More specifically, for a multiply connected space M (viz. a nonsimply connected space) E = −∇ϕ holds locally in the following sense: ∀x ∈ M, there
exists a neighborhood N x ⊂ M of x and a function ϕ x on Nx such that
E | N x = −∇ϕ x holds on Nx for each x ∈ M. The collection {ϕ x } thus defines
the potential on M. Moreover, it is also manifestly clear in this more general formalism for electrostatics, φ is unique up to an arbitrary constant
c: φ → φ + c defines the same electric field as ∇c = 0.
At a more practical level, imagine measuring the potential difference
between two fixed points in a circuit using a voltmeter. If the space within
the vicinity of the circuit were multiply connected, then it is conceivable that
by merely moving the leads without changing the endpoints of the leads,
the voltage reading registered by the meter could change. This clearly renders the voltage measured to be ill-defined. Fortunately, the space within the
vicinity of our solar system appears to be simply connected!
1.1.5 Example
Suppose that
E=
q
1
4 πε 0 r 3
r
for some charge q, where q is taken to be at the origin in R3. Then, the potential generated by q is, from Equation (1.4),
ϕ = − 4 πε1 0
*
∫
P
∞
q
r′3
r ′ ⋅ dr ′ = −
1
4 πε 0
∫
P
∞
q
r′2
dr ′ = 4 πε1 0
q
r
Prove this result without resorting to Lemma 1.1.3. See Exercise 1.5.1(a) for details.
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7
Brief Review of Maxwell’s Theory
Indeed, it is quite clear that if
E=
1
4 πε 0
∑
n
qi
3
i = 1 ri
ri
then by the superposition principle,
ϕ=
1
4 πε 0
∑
n
i=1
qi
ri
whence, if an extended charged body V ⊂ R 3 has a charge density ρ, then
each differential volume element δVi has a charge δqi = ρδVi for i = 1, , m.
In the limit as m→∞ and δVi → 0 such that
Q = lim
m→∞
∑
m
i=1
δqi →
∫ dq = ∫ ρd r
3
V
V
it follows by construction that
ϕ=
1
4 πε 0
∫
V
ρ( r ′ )
R′
d3 r ′
(1.5)
where R ′ = ( x − x ′)2 + ( y − y ′)2 + ( z − z ′)2 , for all r ′ = ( x ′ , y ′ , z ′) ∈V and r =
(x,y,z). See Equation (1.3) above.
□
Equation (1.5) thus furnishes the general definition for a potential at an arbitrary point generated by a charged volume (V,ρ).
This section closes with an extremely powerful method for solving electrostatic problems: the method of images. This technique is essentially a means
to construct a Green’s function to solve Poisson’s and Laplace’s equations in
electrostatics (see Appendix A.3). The principle is rather intuitive. Further
details are elaborated in Chapter 3.
First, consider a point charge q located a distance z0 above an infinite,
perfect electrical conducting (PEC) plane that is grounded (see Figure 1.2).
What is the potential φ at the point P = (x,y,z) above the ground plane? From
q
P
z0
–z0
Ground plane
–q
Figure 1.2
Point charge and its image below a ground plane.
K15149_Book.indb 7
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8
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Appendix A.3, the potential function ϕ ∈C 2 (Ω) ∩ C(Ω) satisfies the following
Laplace equation,
Δφ = 0 on Ω
(1.6a)
subject to the boundary condition
φ = 0 on ∂Ω
(1.6b)
where Ω = R 3+ − {(0, 0, z0 )} and R3+ = {( x , y , z) ∈R 3 : z ≥ 0} is the upper half-space.
Heuristically, the presence of the charge q induces the opposite charge distribution on the ground plane, via the electric field, such that
a) The resultant induced charge on the plane is equal to −q.
b) The potential of the grounded plane is 0.
Then, by construction, ϕ(r ) = ϕ + (r − r0 ) + ϕ − (r − r0′) is the potential at r = (x,y,z),
where ϕ ± is the potential resulting from ±q. Explicitly, from Example 1.1.5,
ϕ(r ) =
1
4 πε 0
{
q
x 2 + y 2 + ( z − z0 )2
−
q
x 2 + y 2 + ( z + z0 )2
}
(1.7)
Now, observe that by definition, (b) is satisfied: φ|∂Ω = 0. Furthermore, it
is an easy matter to verify that Δφ = 0, where ∆ = ∂2x + ∂2y + ∂2z . Indeed, this follows at once from ∆ϕ ± = 0. This is left as a warm-up exercise for the reader.
Because the boundary condition is also satisfied, it follows immediately from
the uniqueness theorem of the Laplace equation that (1.7) is the sought-for solution. That is, the potential φ on Ω is defined by (1.7). Finally, note that (1.7) does
not apply to z < 0, as the image charge −q does not really exist below the ground
plane. The charge is in fact induced on the surface of the ground plane.
1.1.6 Example
Given two infinitely long, perfect electrical conducting cylinders, consider
their cross-section C± = ( x , y , 0) : x 2 + ( y − d± )2 ≤ a±2 , where the axes of the
cylinders are parallel to the z-axis, and a± are the radii of the respective cylinders. Suppose the cylinders are held at a potential of φ ± . Find the potential
at any point external to the cylinders.
The easiest approach to solving this problem is by reducing the charged
cylinders to equivalent line charges satisfying the equipotential boundary
conditions. Specifically, consider each charged cylinder together with its mirror image across an imaginary boundary initially, and then sum the potentials via superposition.
{
K15149_Book.indb 8
}
10/18/13 10:43 AM
9
Brief Review of Maxwell’s Theory
So, provisionally, consider an equivalent line charge λ + located at some distance y + (to be determined) from the origin. The distance y + is then determined as a function of the cylinder radius and charge via the method of
images. For an arbitrary point r = (x, y) away from the line charges, the potential is determined via Equation (1.3) as follows.
E+ =
λ+
4 π ε0
∫
∞
1
{
}
3
−∞ x 2 + ( y − y+ )2 + ( z − z ′ )2 2

x
 y−y
+

 z − z ′

 dz ′


where λ + is an infinite line charge density parallel to the z-axis.
Now, observe that
∫
∞
∫
∞
1
{
−∞ x 2 +( y − y +
{
3
)2 +( z− z′ )2 2
}
z′
−∞ x 2 +( y − y +
3
)2 +( z− z′ )2 2
}
dz′ = −
dz′ =
z − z′
( x 2 +( y − y + )2 ) x 2 +( y − y + )2 +( z− z′ )2
2
2
2
2
( x + ( y − y + ) ) x + ( y − y + ) + ( z − z′ )
=
−∞
∞
− ( x 2 +( y − y + )2 )+ zz′− z 2
2
∞
−∞
=
2
x 2 +( y − y+ )2
2z
x 2 +( y − y+ )2
whence
E+ =
λ+
r+
2 π ε 0 x 2 + ( y − y+ )2
where r+ = ( x , y − y + , 0) defines the electric field in R 3 − {(0, y + , z) : z ∈R}. The
electric field E − follows mutatis mutandis via the replacement
( λ + , y + ) → ( λ − , y − ) : E− =
λ−
r−
2 π ε 0 x 2 + ( y − y− )2
In particular, inasmuch as the fields are independent of the z-component, in
all that follows, we may consider the 2D plane defined by z = 0.
Next, appealing to Definition 1.1.2, via polar coordinates
E± =
λ±
2 π ε 0 r±2
er
and the contribution from λ + thus yields
ϕ=−
K15149_Book.indb 9
∫E
γ
+
⋅ dr ′ = − 2λπ+ε0
∫
dr
r
= − 2λπ+ε0 ln x 2 + ( y − y + )2 + C
10/18/13 10:43 AM
10
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Line charge density λ+
y
r+
y+
ε0
ε0
y–
(x, y)
r
0
x
r–
Line charge density λ–
Figure 1.3
Potential arising from two infinite line charges.
where C is some arbitrary constant of integration. By symmetry, the resultant
potential from the two line charges is:
ϕ(r ) = − 2λπε+ 0 ln rrˆ+ −
λ−
2 πε 0
ln rrˆ−
(1.8)
where r± = x 2 + ( y − y ± )2 , and some constant reference point r̂ . See
Figure 1.3.
For simplicity, assume provisionally that λ − = −λ + : this condition is relaxed
later. Then, φ(x,0) = 0 ∀x ⇒ y − = − y + . Furthermore, from (1.8), at any point
( x , y ) ∈C+ , ϕ + =
λ+
2 πε 0
ln rr−+
by definition, whence, it follows immediately that
r+
r−
= constant ⇒
x 2 + ( y − y + )2
x 2 + ( y + y + )2
=c
for some constant c. However, notice that this expression can be rearranged
into the form
(y −
1+ c 2
1− c 2
y+
)
2
+ x2 =
(
2c
1− c 2
y+
)
2
(1.9)
Derive (1.9) as a warm-up exercise; see Exercise 1.5.1(b). Equation (1.9) is
nothing but the equation of a circle with
radius
center
K15149_Book.indb 10
(
1+ c 2
1− c 2
2 cy +
1− c 2
)
y + , 0 ∈R 2
10/18/13 10:43 AM
11
Brief Review of Maxwell’s Theory
The circles defined by the above radii and centers describe circles of equipotential. In particular, ∃c+ such that
2 c+ y +
= a+
1− c+2
the equipotential circle coincides with C+ and
d+ =
1+ c+2
1− c+2
y+
Thus,
d+2 =
( )(
c+2 + 1
2 c+
2
2 c+
c+2 − 1
y+
) =( ) a
2
c+2 + 1
2 c+
2
2
+
Moreover, observe that
( )y
1+ c+2
1− c+2
2
2
+
− y +2 = y +2
(
1+ c+2
1− c+2
)(
+1
1+ c+2
1− c+2
) ( )
−1 =
2 c+ y +
1− c+2
2
= a+2
and hence, yielding
d+2 = y +2 + a+2
(1.10)
That is, y + = d+2 − a+2 .
Now, removing the assumption that λ − = −λ + , it is clear by symmetry that
y − = d−2 − a−2 . To proceed with the analysis, observe that on the boundary
∂C+ , ϕ = ϕ + . Hence, choosing the point (0, d+ − a+ ) ∈∂C+ ,
ϕ + = − 2λπε+ 0 ln d+ − a+ − y + −
λ−
2 πε 0
ln d+ − a+ − y−
(1.11a)
and likewise, ϕ|C− = ϕ − implies choosing the point (0, d− − a− ) ∈∂C− yields
ϕ − = − 2λπε+ 0 ln d− − a− − y + −
λ−
2 πε 0
ln d− − a− − y−
(1.11b)
Hence, solving for λ ± simultaneously via (1.11) gives
(
λ + = ϕ− − ϕ+
λ− =
K15149_Book.indb 11
ϕ+
β+
−
α+
β+
β−
β+
(ϕ
−
)(α
−
− ϕ+
+ α+
β−
β+
β−
β+
)(α
−
)
−1
+ α+
β−
β+
)
−1
10/18/13 10:43 AM
12
Electromagnetic Theory for Electromagnetic Compatibility Engineers
where
α ± = − 2 πε1 0 ln d± − a± − y + and β ± = − 2 πε1 0 ln d± − a± − y −
□
1.1.7 Remark
The case wherein the electric permittivity ε of a dielectric such that ε > ε 0
can be found in Reference [14] by applying the method of transformation in
a circle [7].
1.2 Magnetostatics
Along a vein similar to the definition for an electric field, the magnetic field
can be defined as the field generated by an element of current. More formally, given a conductor (M, σ), where σ (in units of Ω−1 m −1 ) is the electrical
conductivity, if a constant electric field E is applied across ∂M, it will generate a flow of electrons within M defined by J = ρe ve . Here, J defines the
current density, ρe is the electron charge density, and ve is the average drift
velocity of the electrons. The current density is related to the electric field
via Ohm’s law
J = σE
(1.12)
This defines the conduction current density. Furthermore, if there exist
free mobile charges of density ρ in the medium moving at an average
velocity v, then the conduction current density also contains a convection
current density term: ρv. That is, J = σE + ρv. In most media of interest, this convection term is zero; on the other hand, this term is nonzero
in cases such as the vacuum tube, wherein the primary conduction current density comprises that of convection current density, simply because
σ = 0 in vacuum.
1.2.1 Remark
If M is suspended in a constant electric field, then electrons will migrate
toward ∂ M− ⊂ ∂ M , causing it to be negatively charged, whereas its complement ∂ M+ = ∂ M − ∂ M− becomes positively charged. Because M is initially
uncharged, its total charge must remain zero. Hence, the charges will redistribute on ∂M such that the induced electric field within M generated by the
pair (∂ M− , ∂ M+ ) cancels out the applied electric field. At which point, the flow
of electrons will cease as there is a zero electric field within M. As an immediate corollary static electric field cannot sustain a current in a conductor.
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13
Brief Review of Maxwell’s Theory
Throughout the book, references made to a cross section of a conductor and an axis of a conductor are made. As such, their formal definitions
are provided for future reference. Let M ⊂ R 3 such that ∃γ ⊂ M a simple
path and a collection C = { Σ x ⊂ M : x ∈γ } of compact subsets satisfying
Σ x ∩ Σ x′ = ∅ ∀x , x ′ ∈γ , x ≠ x ′ , and M = ∪ x∈γ Σ x and the tangent vector along
γ at X is normal to ∑x. Then Σ ∈C is called a cross-section of M and M is called
a closed tubular neighborhood of γ. If each Σ ∈C is open with compact closure,
then M is called an open tubular neighborhood of γ. Finally, γ ⊂ M is said to
be the axis of the tubular neighborhood M if for each cross-section Σ ⊂ M
with { x } = γ ∩ Σ , B2 (rˆ ′ , x ′) ⊆ B2 (rˆ , x ) ∀x ′ ∈Σ , where the maximal radius
Rˆ = max {R > 0 : B2 (R , u) ⊆ Σ , u ∈Σ} , is the largest radius such that the twodimensional disc B2 (R , u) is contained in Σ, with B2 (r , u) = B(r , u) ∩ {R 2 + u} ,
{
B(r , u) = y ∈R 3 : y − u < r
}
is a three-dimensional disc in R 3 , and R 2 + u is the hyperplane in R 3 translated by u: {y ′ + u ∈R 3 : y ∈R 2 } . It is easy to see that the axis of a cylinder
coincides with the above more general definition.
1.2.2 Definition
Given a conductor (M, σ), suppose a current density J flows across a crosssection Σ ⊂ M of M. Then, the current I flowing through M is defined by
I=
∫ J ⋅ nd x
2
Σ
where n is a normal vector field on Σ.
1.2.3 Definition
Suppose J is defined on (M, σ). Then, a vector potential A generated by J is
defined on R 3 by
A(r ) =
µ0
4π
∫
M
J (r ′)
R
d3 x
where R = r − r ′ = ( x − x ′)2 + ( y − y ′)2 + ( z − z ′)2 ∀( x ′ , y ′ , z ′) ∈ M and ( x , y , z) ∈R 3.
Here, µ 0 = 4π × 10−7 H/m (in Henry/meter) denotes the magnetic permeability
of free-space. Moreover, the magnetic field density (or to be more technically
precise, the magnetic flux density) B on R 3 is given by B = ∇ × A.
Note in passing that the definition given above for a magnetic field is motivated by the Lorentz force law. Explicitly, in the absence of an electric field,
a charge particle moving at some fixed velocity v in a static magnetic field B
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14
Electromagnetic Theory for Electromagnetic Compatibility Engineers
obeys the Lorentz force law: F = −qv × B. In the presence of an electric field
Lorentz’s law becomes, via Coulomb’s law, F = qE − qv × B. Thus, considering for simplicity, the absence of an electric field, the magnetic field density B
may be defined to be the field acting on a test charge such that it satisfies the
Lorentz force law. More specifically, if the Lorentz force is generated by some
potential field, then, given that ∇ ⋅ B = 0 empirically (see Section 1.3 for further
details), it follows that there exists some vector potential A such that B = ∇ × A;
refer to Appendix A.1. This completes the justification for Definition 1.2.3.
In many instances, evaluating the magnetic field via Definition 1.2.3 is
easier than using the conventional definition often encountered in first-year
electromagnetics courses. This is given below for completeness. Suppose γ is
a loop and I a current circulating around γ. Then, the magnetic flux density
B on R 3 is defined by
B(r ) =
µ0
4π
∫
γ
I (r ′)
R2
dl × eR
(1.13a)
where
R = r − r ′ = ( x − x ′ ) 2 + ( y − y ′ ) 2 + ( z − z ′ )2 , e R =
R
R
, r ′ ∈γ and r ∈R 3
More generally, Equation (1.13a) can be extended to a volume in the following
fashion. Suppose M ⊂ R 3 has a current density J flowing across its cross-section Σ
⊂ M. Then, the magnetic field at r ∈R 3 resulting from the triple (M, J, Σ) is given by
B( r ) =
µ0
4π
∫
Σ
J (r′)
R2
× eR d2 x
(1.13b)
The magnetic field developed above did not depend on time: ∂t B = 0 . That
is, only the static scenario was considered. In view of Definition 1.2.3, this is
equivalent to ∂t A = 0 and hence ∂t J = 0 . Thus a static magnetic field is generated by a constant current.
Reflecting on the treatment of electrostatics in Section 1.1, wherein the equi­
valent case of ∂t E = 0 was defined by ∇ × E = 0 , is there a dual criterion wherein
∇ × B defines a magnetostatic scenario? This question is addressed in Section
1.3. Finally, this section concludes with an application of Equation (1.13a).
1.2.4 Example
Let C = {( x , y , 0) ∈R 3 : x 2 + y 2 = a 2 } define a circle on the (x,y)-plane and suppose that a constant current I is circulating around C; see Figure 1.4. Then,
given any point r ∈R 3 − C, the magnetic field density at r can be determined
via Equation (1.13a) as follows.
K15149_Book.indb 14
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15
Brief Review of Maxwell’s Theory
Cross section of magnetic dipole field
z
P
y
r
C
0
I
a
x
Figure 1.4
Magnetic dipole moment generated by a loop current I.
First, observe that
eR
R2
= ( x + a sin θ, y − a cos θ, z){( x + a sin θ)2 + ( y − a cos θ)2 + z 2 }
− 23
Next, from dl = a(− sin θ, cos θ, 0)dθ ,
dl ×
eR
R2
=
=
1
3
{( x − a cos θ )2 + ( y − a sin θ )2 + z 2 } 2
a
ex
ey
ez
− sin θ
x − a cos θ
cos θ
y − a sin θ
0
z
adθ

z cos θ

z sin θ

 −( y − a sin θ)sin θ − ( x − a cos θ)cos θ
3
{( x − a cos θ )2 + ( y − a sin θ )2 + z 2 } 2

 dθ.


Thus,
µ0 I z
Bx =
4 π a3
By =
4 π a3
µ0 I z
∫
2π
∫
2π
Bz = − 4µπ0aI3
cos θ dθ
(
)
()
)
()
3
2
2
2
 x
y
z 2
 a − cos θ + a − sin θ + a 


(
0
)
sin θ dθ
(
3
2
2
2
 x
y
z 2
 a − cos θ + a − sin θ + a 


(
0
∫
2π
)
( y − a sin θ)sin θ+ ( x − a cos θ)cos θ
(
)
3
2
2
2
 x
y
z 2
 a − cos θ + a − sin θ + a 


(
0
)
()
dθ
Now, suppose that r >> a. Then, on setting
ξ=
K15149_Book.indb 15
( xa − cos θ)2 + ( ya − sin θ) + ( za )2
2
and ξ 0 =
( xa )2 + ( ya ) + ( za )2 = ( ra )
2
10/18/13 10:43 AM
16
Electromagnetic Theory for Electromagnetic Compatibility Engineers
it follows that up to first order in sin θ and cos θ,
ξ ≈ ξ0 1 +
2
ξ 20
(
x
a
y
cos θ + a sin θ
)
whence, by appealing to the binomial expansion,
1
ξ3
≈
1
ξ 30
{1 +
3
ξ 20
(
x
a
y
cos θ + a sin θ
)}
Substituting this approximation into the integrals for Bx , By and Bz yields
Bz = − 34
µ0 I
a 3 ξ 30
{
3 x2
r2
Bx =
3 µ 0 I xz
4 a3 ξ 5
0
=
2
3 µ 0 I a xz
4 r3 r2
(1.14a)
By =
3 µ 0 I yz
4 a3 ξ 5
0
=
2
3 µ 0 I a yz
4 r3 r2
(1.14b)
+
3y2
r2
}
− 2 a 2 = − 34
µ 0 Ia2
r3
{
3 x2
r2
+
3 y2
r2
−2
}
(1.14c)
Indeed, observe from Equation (1.14) that for r >> a, the magnetic field is
directly proportional to the area of the current loop.
□
1.2.5 Remark
The above example defines a magnetic dipole: to wit, a simple loop with a nonzero current circulating around the loop (cf. an electric dipole of Example 1.1.1).
Further details can be found in Chapter 8.
1.3 Maxwell’s Equations
Maxwell’s electromagnetic theory is in fact the first step toward field unification. It unifies the electric field and the magnetic field into a single entity,
giving rise to the electromagnetic field.* In all that follows, unless stated
explicitly, rectangular coordinates r = (x, y, z) are chosen.
*
The question regarding the existence of the electric and the magnetic fields is an interesting
one. Although an electromagnetic field exists globally, it cannot be decomposed into electric and magnetic fields if space–time cannot be split into space and time. The various ways
of splitting up flat space–time (a property known as foliation) yields the Lorentz symmetry
group—this forms the basis for Einstein’s theory of special relativity. In particular, the constancy of the speed of light in any inertial frame is indeed a consequence of the Lorentz symmetry of flat space–time; see, for example, Reference [12].
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Brief Review of Maxwell’s Theory
Let E be an electric field and B some magnetic field defined on (R 3 , ε , µ , σ ),
where ε denotes the electric permittivity, μ the magnetic permeability, and σ
the electric conductivity of the medium. Free charges present in the medium
are defined by the charge density ρ. Then, Maxwell’s equations comprise the
following,
∇ × E(r , t) = − ∂t B(r , t)
(1.15)
∇ ⋅ B(r , t) = 0
(1.16)
∇ × B(r , t) = µJ (r , t) + µε ∂t E(r , t)
(1.17)
∇ ⋅ E( r , t ) =
ρ
ε
(1.18)
where J ≡ σE . That is, the entire theory of classical electromagnetism
involves finding the solution ( E , B) to Maxwell’s equations while satisfying
the appropriate boundary conditions.
As a side comment, define the magnetic field intensity H by B = µH and the
electric displacement by D = εE . Then, the set of Maxwell’s equations is often
presented as
∇ × E(r , t) = − ∂t B(r , t)
(1.15′)
∇ × H (r , t) = J (r , t) + ∂t D(r , t)
(1.16′)
∇ ⋅ D(r , t) = ρ
(1.17′)
∇ ⋅ B(r , t) = 0
(1.18′)
The reason for presenting Maxwell’s equations as (1.15)–(1.18) is elaborated
in Remark 1.3.2.
Here, the medium is assumed to be homogeneous and isotropic; that is,
μ,σ,ε are independent of the x-, y-, z-variables and direction. For simplicity,
the constants μ,σ,ε herein are also assumed to be independent of frequency
and temperature. The frequency dependency of ε and σ is fully explored via
the Kramers-Kronig relations in Chapter 9.
Equation (1.15) is known as Faraday’s law; it shows how a time-varying
magnetic field couples with an electric field. Intuitively, the time variation of
the magnetic field generates the spatial variation in the electric field. Strictly
speaking, this is not a technically precise statement (cf. Section 8.1 for the correct interpretation). Equation (1.16) shows how a time-varying electric field
couples with a magnetic field. Intuitively, the time variation of the electric
field generates the spatial variation of the magnetic field.
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18
Electromagnetic Theory for Electromagnetic Compatibility Engineers
1.3.1 Theorem
Suppose the pair (E, B) is a solution to Maxwell’s equations on a simply connected, open subset Ω ⊆ R 3 satisfying the appropriate boundary conditions
on ∂Ω. Then, the most general expression for the electric field is given by
E = −∇ϕ − ∂t A, where A satisfies B = ∇ × A and φ is some conservative scalar
potential field on Ω.
□
Proof
From (1.15), ∇ × E = − ∂t ∇ × A ⇒ 0 = ∇ × ( E + ∂t A) . Set E = E + ∂t A . Then, Ω
2
is simply connected implies ∃ϕ ∈C (Ω) such that E = −∇ϕ on Ω. Whence,
□
E = −∇ϕ − ∂t A , as required.
It is obvious from Theorem 1.3.1 that the general expression for an electric
field comprises a static part and a time-varying part. In particular, the vector
potential contributes to the time variation of the electric field and the scalar
potential contributes to the conservative aspect of the electric field.
1.3.2 Corollary
Let (E, B) be a solution of Maxwell’s equations. Then, ∂t E(r , t) = 0 = ∂t B(r , t) if
and only if ∇ × E(r , t) = 0 and ∇ × B(r , t) = µJ (r , t).
Proof
Suppose ∂t E = 0 = ∂t B . Then, ∂t B = 0 ⇒ ∂t A = 0 as B = ∇ × A by definition,
whence, Equation (1.15) reduces to ∇ × E(r , t) = 0 , and from ∂t E = 0 ⇒ ∂t D = 0,
it follows at once that Equation (1.17) reduces to ∇ × B(r , t) = µJ (r , t) .
Conversely, ∇ × E = 0 ⇒ ∂t A = 0 from Theorem 1.3.1, and hence, ∂t B(r , t) = 0 .
Likewise, ∇ × B( r , t) = µJ (r , t) ⇒ ∂t E = 0 , as asserted.
□
Thus, the static condition is precisely equivalent to the case wherein the
electric field and magnetic field are decoupled. In particular, the definition of
electrostatics defined in Section 1.1 by the irrotational criterion ∇ × E(r , t) = 0
is completely justified in view of Corollary 1.3.2.
Furthermore, it ought to be pointed out that the pair (φ, A) defining (E, B) is
not unique. Indeed, it is unique up to some arbitrary, differentiable, function
f = f(r, t) as follows: ϕ → ϕ ′ = ϕ + ∂t f and A → A′ = A − ∇f . This can be easily
seen via direct substitution:
E ′ = −∇ϕ ′ − ∂t A′ = −∇ϕ − ∂t ∇f − ∂t A + ∂t ∇f = −∇ϕ − ∂t A = E
B′ = ∇ × A′ = ∇ × A − ∇ × ∇f = ∇ × A = B
as ∇ × ∇f = 0 on R 3 for any twice differentiable function f on R 3 . This property is known as gauge transformation and Maxwell’s theory is said to be
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19
Brief Review of Maxwell’s Theory
invariant under gauge transformation. In particular, fixing the choice of
(ϕ , A) is called gauge fixing. Fixing the choice of gauge will in no way affect
the physical significance of the problem. Thus, there is freedom to choose a
gauge to simplify solving Maxwell’s equations.
1.3.3 Remark
Equations (1.15) and (1.16) are called the first pair of Maxwell’s equations,
and (1.17) and (1.18) are called the second pair of Maxwell’s equations. This is
because in the language of differential geometry, the pair (1.15) and (1.16) can
be expressed as a single equation, and the second pair (1.17) and (1.18) can
be expressed as the dual of the former equation. For those readers interested
in understanding the geometric formalism of Maxwell’s theory, consult
References [1,3,8,9].
The first term of Equation (1.16) is the displacement term governed by bound
charges in the medium; it gives rise to the phenomenon of the displacement
current present in capacitors. The second term, as mentioned in Section 1.2, is
the conduction term: J (r , t) = σE(r , t) is the current density flowing across the
medium surrounding the conductors, and it may also include a convection
term; see Equation (1.12). If the medium has zero conductivity, that is, σ = 0,
then Equation (1.16) is dominated purely by the displacement current term.
Equation (1.17) is Gauss’ law. It states that the electric field is generated by
the presence of a charge density ρ. In contrast, Equation (1.18) states that the
magnetic field is not generated by magnetic charges: there are no magnetic
monopoles.* This equation is based purely upon the observation that magnetic monopoles have never been found in nature.
1.3.4 Theorem (Charge Conservation)
The electric charge is conserved in R 3 . That is,
∇ ⋅ J + ∂t ρ = 0
(1.19)
Proof
From (1.16), let Σ be the boundary of some compact volume M in R 3. Then,
by Corollary A.1.2,
0=
*

∫∫ ∇ × H ⋅ ndS = ε ∂ 
∫∫ E ⋅ ndS + σ 
∫∫ E ⋅ ndS
Σ
t
Σ
Σ
This is a rather lively topic that is open to much theoretical debate: once again, its existence
depends largely upon the topological structure of space–time (cf., e.g., References [1,3,8,9]).
From a mathematical standpoint, magnetic monopoles exist in space–times that have holes.
An even deeper consequence for the existence of magnetic monopoles is that electric charges
must be quantized (i.e., discrete)!
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20
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Because, via the divergence theorem,
∫∫ E ⋅ n dS = ∫∫∫
Σ
M
∇ ⋅ E dV =
1
ε
∫∫∫
M
ρ dV
it follows immediately that
0 = ∂t

∫∫∫
M
ρ dV + σ

∫∫∫
M
∇ ⋅ E dV ≡ ∂t

∫∫∫
M
ρ dV +

∫∫∫
M
∇ ⋅ J dV ,
whence ∇ ⋅ J + ∂t ρ = 0 as claimed.
□
Equation (1.19) is known as the continuity equation for electric charges.
It shows that the electric charge is conserved in the following manner.
Because the divergence ∇ ⋅ J of the current density J denotes the net loss
of charges, if there is a zero net loss of charge within a compact volume
M, that is, ∇ ⋅ J = 0 , then by (1.19), the time variation of the charge density
∂t ρ = 0. That is, the electric charge is conserved because ∂t ρ = 0 implies that
ρ is constant in time.
Conversely, observe that Theorem 1.3.1 implies that ∇ ⋅ J = 0 is precisely the
condition for steady-state current. To see this, it suffices to note that ∂t ρ = 0
implies that the charge remains constant in time, and hence, the current density cannot vary with time. Indeed, a consequence of the continuity equation
is the following. From Equation (1.18), ∇ ⋅ J can be expressed as
∇ ⋅ J = ∇ ⋅ σE = σε ρ
because the electric conductivity σ is assumed constant. Hence, from
Equation (1.19),
σ
ε
ρ + ∂t ρ = 0 ⇒ ρ = ρ0 e− ( σ/ε )t
where ρ0 is the initial charge density within a compact volume M.
Notice the exponential decay of charges with respect to time within the
volume M. This means in particular that any charges placed within a conductor will quickly diffuse onto the surface (i.e., the boundary) of the conductor in accordance with the exponential decay rate given by
ρ(r , t) = ρ0 (r , t)e− ( σ/ε )t
(1.20)
Physically, the charges are distributed on the surface such that the resultant
electric field within the conductor is zero.
Two consequences are discernable from the continuity equation.
(a) For a perfect dielectric where the conductivity σ = 0, ρ = ρ0 .
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21
Brief Review of Maxwell’s Theory
That is, the charges will remain where they are; they are bound charges. On
the other hand, if no charges are added to the dielectric, that is, the dielectric
remains uncharged, then an electric field will polarize the molecules into
electric dipoles.
(b) For a perfect conductor, that is, σ → ∞, the charges diffuse very rapidly
onto the surface of the conductor such that the resultant field within
the conductor is zero.
That is, the charges are mobile. Another way of seeing this is as follows.
By definition, the current density within the conductor (or any medium)
is given by J = σE . Inasmuch as physically, J < ∞, it follows that E → 0 in
the limit as σ → ∞ such that the current density will continue to remain
finite. This fact gives rise to the statement that a perfect conductor has a zero
electric field inside it. There is a zero electric field inside a perfect conductor, therefore it follows that all the charges must reside on the surface of a
perfect conductor and be distributed in such a manner that the fields cancel
within the conductor (viz., E ≡ 0 ).
By appealing to Faraday’s law, it can also be shown that the magnetic flux
Ψ = ∫ Σ B ⋅ dx 2 is conserved—that is the reason why B is called the magnetic
flux density. That is, the sum of the magnetic flux entering and exiting a
compact surface is zero. This in turn implies that magnetic flux forms closed
loops: i.e., no magnetic monopoles.
1.4 Electromagnetic Waves
First, consider a stationary electric charge in space. It was shown above that
∇ × E = 0 defines an electrostatic field because of the absence of a time-varying
field. Consequently, Maxwell’s equations reduce to:
∇× E =0
(1.21)
∇⋅B = 0
(1.22)
∇ × B = µJ
(1.23)
∇⋅E =
ρ
ε
(1.24)
where, for convenience, the boundary conditions are left unspecified for
the moment.
To recapitulate Section 1.3, Equation (1.21) implies that ∂t H = 0; that is, the
electric field does not generate a time-varying magnetic field. The electric
field is decoupled from the magnetic field. Similarly, from Equation (1.22),
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22
Electromagnetic Theory for Electromagnetic Compatibility Engineers
the magnetic field is decoupled from a time-varying electric field: the magnetic field does not generate a time-varying electric field.
From Equation (1.23), if the electric charges are stationary, then J = 0 and
hence, ∇ × B = 0. However, this does not imply that B = 0. What makes B = 0
is Equation (1.22): the absence of a magnetic charge. Hence, in this universe,
∇ × B = 0 implies that B = 0 due to the absence of magnetic monopoles.
Recall that the full Maxwell’s equations state that a time-varying electric
(magnetic) field generates a time-varying magnetic (electric) field, which
in turn generates an electric (magnetic) field and so forth. From this standpoint, it is intuitively clear how an electromagnetic wave can propagate in
free space: it is self-sustaining, wherein each field generates the other field in
space–time, provided both the fields are time-varying. Hence, it is once again
intuitively clear that in the case of static fields, there can be no propagation
of electromagnetic waves.
To study the existence of electromagnetic waves, consider the full
Maxwell’s equations, where, for simplicity, the homogeneous and isotropic
medium is assumed to be charge-free. As always, in the absence of any particular symmetry, rectangular coordinates are used for simplicity. Finally,
for convenience, it is tacitly assumed that ε and μ are scalars.
From Equation (1.15), where {e x , e y , e z } denotes the standard basis in rectangular coordinates,
∇×E=
ex
ey
ez
∂x
∂y
∂z
Ex
Ey
Ez
 ∂ y Ez − ∂ z Ey

=  − ∂ x Ez + ∂ z Ex
 ∂ E −∂ E
x y
y x


 ∂t Bx


 = −  ∂t By

 ∂B
 t z






(1.25)
  ε ∂t Ex + σEx
 
 =  ε ∂t Ey + σEy
  ε ∂ E + σE
t z
z
 





(1.26)
From Equation (1.17),
∇×B=
ex
ey
ez
∂x
∂y
∂z
Bx
By
Bz
 ∂ y Bz − ∂ z By

=  − ∂ x Bz + ∂ z Bx
 ∂ B −∂ B
x y
y x

Because the wave is propagating in a charge-free space, Equation (1.18)
becomes ∇ ⋅ E = 0. This leads to
∂ x Ex + ∂ y Ey + ∂ z Ez = 0
(1.27)
Likewise, Equation (1.18) leads to ∂ x Bx + ∂ y By + ∂ z Bz = 0.
Recall that a wave ψ defined in R 4 satisfies the wave equation:
∆ψ = α ∂t2 ψ + β ∂t ψ
K15149_Book.indb 22
(1.28)
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23
Brief Review of Maxwell’s Theory
1
where α is the speed of the propagating wave and β is the loss present in
the medium. Some comments regarding Equation (1.28) are now due.
• If the medium is lossless, β = 0, and the wave equation reduces to
∆ψ = α ∂t2 ψ
• If α ∂t2 ψ << β ∂t ψ in R 4 , then Equation (1.28) reduces to
∆ψ ≈ β ∂t ψ
This is precisely the diffusion equation, also known as the heat equation.
Under this condition, there are no waves propagating in the medium. This
equation can also be solved using separation of variables if the configuration
possesses nice symmetries. If the medium is very lossy, that is, β >> α such
that α ∂t2 ψ << β ∂t ψ is satisfied, then no waves will propagate through the
medium.
In fact, the astute reader will quickly observe that because ∂t2 ψ < ∞ holds
physically for all times, it follows that α ∂t2 ψ << β ∂t ψ is satisfied for any
|β|< ∞ if α → 0. This means that the effect of any changes made to the wave
will spread almost instantaneously throughout the entire space. This superficially appears to be in violation of causality; food for thought here!
The wave equation for the z-component is worked out explicitly below.
The derivation for the other two components as well as for the magnetic field
follows mutatis mutandis. To begin, differentiating the first line of Equation
(1.25) with respect to y and using the last row of (1.26) yield:
∂2y Ez − ∂ y ∂ z Ey = −µ ∂t ∂ y H x = µ ∂t (ε ∂t Ez + σEz − ∂ x H y )
Next, differentiating the second line of Equation (1.26) with respect to x
and using the last row of (1.26) yield:
− ∂2x Ez + ∂ x ∂ z Ex = −µ ∂t ∂ x H y = −µ ∂t (−ε ∂t Ez − σEz + ∂ y H x )
From Equation (1.27),
∂ z Ez = − ∂ x Ex − ∂ y Ey
Taking the difference of the first two equations yields:
∂2x Ez + ∂2y Ez − ∂ z (∂ x Ex + ∂ y Ey ) = − ∂ x H y + ∂ y H x
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24
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Using (1.27) and the last row of (1.26) yields:
∂2x Ez + ∂2y Ez + ∂2z Ez = µε ∂t2 Ez + µσ ∂t Ez
This expression is precisely the wave equation given by (1.28):
∆Ez = µε ∂t2 Ez + µσ ∂t Ez
Going through the same procedure yields the identical form for the other
components of the electric field and the magnetic field. Explicitly, the fields
are defined by the Helmholtz equations:
−∆E + µε ∂t2 E + µσ ∂t E = 0
(1.29)
−∆B + µε ∂t2 B + µσ ∂t B = 0
(1.30)
From Equation (1.28), it is evident that electromagnetic waves propagate
in any medium at the speed v = 1µε , where ε, μ depend upon the medium.
Diffusion through the medium is related directly to its conductivity and permeability: μσ. In particular, for a perfect—that is, lossless—dielectric, σ ≡ 0,
and Equation (1.29) reduces in a lossless medium to the wave equation:
−∆E + µε ∂t2 E = 0.
(1.31)
For a medium with a very high conductivity, (1.29) reduces to the diffusion
equation. Consequently, electromagnetic waves cannot propagate inside a
good conductor. They will diffuse in a good conductor the way heat diffuses
through a medium. Technically, the electric field will still propagate inside
a good conductor as waves; however, the waves will be severely attenuated.
This, in turn, may be approximated by the diffusion equation (cf. the discussion carried out above). All the comments just made about the electric field
also apply to the time-varying magnetic field. In summary, Maxwell’s equations predict the existence of electromagnetic waves.
Transverse electric and magnetic (TEM) waves possess a number of interesting properties that are outlined below. TEM waves are electromagnetic
waves with no longitudinal (also called axial) field components. That is, if the
direction of propagation is along the z-axis, then the z-components of both
the electric and magnetic fields are identically zero. In all that follows, it is
assumed that the waves are propagating along the z-axis.
For TEM waves, Equation (1.25) reduces to (1.30)
∇×E=
K15149_Book.indb 24
ex
ey
ez
∂x
∂y
∂z
Ex
Ey
0

− ∂ z Ey

=
∂ z Ex
 ∂ E −∂ E
y x
 x y
  −∂ B
t x
 
=
−
∂
 
t By
 
0
 




10/18/13 10:44 AM
25
Brief Review of Maxwell’s Theory
and (1.26) reduces to (1.31)
∇×B=
ex
ey
ez
∂x
∂y
∂z
Bx
By
0

− ∂ z By

=
∂ z Bx
 ∂ B −∂ B
y x
 x y

 ε ∂t Ex + σEx


 = µ  ε ∂t Ey + σEy


0





First, consider the last row in Equation (1.31): ∂ x Ey − ∂ y Ex = 0. Observe that
this can be expressed in the form:
∇ ⊥ × E⊥ ≡
ex
ey
ez
∂x
∂y
0
Ex
Ey
0

0

0
=
 ∂ x Ey − ∂ y Ex

 =0


(1.32)
The notation ∇ ⊥ ≡ (∂ x , ∂ y , 0) defines the transverse (i.e., normal to the
direction of propagation) component of ∇; likewise, E⊥ ≡ (Ex , Ey , 0) defines
the transverse electric field components. In general, ∇ = ∇ ⊥ + ez ∂ z and
E = E⊥ + ez Ez (in rectangular or cylindrical coordinates; cf. Appendix A.1).
There are a number of implications to be drawn from Equation (1.32). The
first is the following. Recall that ∇ ⊥ × E⊥ = 0 implies the existence of some
potential function φ such that E⊥ = −∇ ⊥ ϕ . That is, from (1.32), the transverse
components of a TEM wave are static! Indeed, from Equations (1.30) and (1.32),
it is clear that ∂t Bx = 0 = ∂t By : this is precisely the static condition. Secondly,
can a single conductor (such as a waveguide cavity) sustain TEM wave propagation? This clearly has implications in the design of high-speed digital
circuits.
1.4.1 Theorem
Let Ω ⊂ R 3 be a perfect conductor that is compact and connected. Then Ω
cannot support a TEM wave propagation.
Proof
By (1.32), ∃φ on Ω such that ∆ ⊥ ϕ = 0 is subject to ϕ|∂Ω = ϕ 0 , for some constant
ϕ 0 . Because φ is analytic on Ω, by the maximum modulus principle [5], its
minimum and maximum must lie on ∂Ω. However, ϕ = ϕ 0 is constant on ∂Ω;
hence, φ must be constant on Ω. Thus, as it is a perfect conductor Ω cannot
sustain a TEM wave, as claimed.
□
1.4.2 Corollary
Let Ω = S × R be a perfect conductor such that S ⊂ R 2 is compact and connected. Then, a TEM wave cannot propagate along Ω.
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26
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Proof
Suppose Ω can support a TEM wave. Then, the restriction to any connected
compact subset Ω′ ⊂ Ω must also support a TEM wave, yielding a contradiction by Theorem 1.4.1.
□
Thus, at least two conductors are required to support a TEM wave. In
particular, a TEM wave incident on a waveguide will be transformed
into a transverse electric (TE) or a transverse magnetic (TM) wave as it
propagates along the waveguide. Note that a TEM wave can only exist
along a perfect electrical conductor: σ → ∞. When σ < ∞, there exists a
small longitudinal component of electric field e z Ez near the surface of the
conductor in order to overcome the ohmic loss resulting from the finite
conductivity. Hence, good conductors can at best sustain an approximate
TEM wave known as a quasi-TEM wave. Only perfect conductors can sustain TEM waves.
Returning to Equation (1.30), differentiating the first two rows with respect
to z and using the first two rows of (1.31), the following wave equation ensues,
∂2z E⊥ = µε ∂t2 E⊥ + µσ ∂t E⊥
(1.33)
− γz iωt
To solve this equation, try the following solution: ψ = A( x , y )e e , where
ω = 2 π f is the angular frequency of the propagating wave, with f (in Hertz)
being the frequency of propagation. Substituting into (1.33) yields
γ 2 A( x , y )e− γz eiωt = −ω 2 µεA( x , y )e− γz eiωt + iωµσA( x , y )e− γz eiωt
Verify this as a simple exercise (see Exercise 1.5.2). Hence, this immediately
leads to the following expression:
γ 2 = −ω 2 µε + iωµσ
(1.34)
Set γ = α + iβ . Then, the evaluation of α, β is given as follows. First, observe
that
α 2 − β 2 + i2αβ = γ 2 = −ω 2 µε + iωµσ
implies:
α 2 − β 2 = −ω 2 µε
2αβ = ωµσ
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27
Brief Review of Maxwell’s Theory
Next, noting that (α 2 + β 2 )2 = (α 2 − β 2 )2 + 4α 2β 2 = (ω 2 µε)2 + (ωµσ )2 , it is
thus clear that α 2 + β 2 = (ω 2 µε)2 + (ωµσ )2 . Finally, adding and subtracting
α 2 + β 2 and α 2 − β 2 from each other yield
α=
1
2
β=
1
2
(
(ω 2 µε)2 + (ωµσ )2 − ω 2 µε
)
1
1
2
=ω
µε
2
 1 + σ 2 − 1 2
( ωε ) 

=ω
µε
2
 1 + σ 2 + 1 2
( ωε ) 

and
(
2
2
2
2
(ω µε) + (ωµσ ) + ω µε
)
1
2
1
whence,
1
γ=
1
2
2
σ 2
ω µε  1 + ( ωε
− 1 + i
)


1
1
2
2
σ 2
ω µε  1 + ( ωε
+ 1
)


(1.35)
The solution of (1.33) is thus
ψ ( x , y , z, t) = A( x , y )e
1 

1

2 

σ 2

2
+ 1 z 
i ωt − 1 ω µε  1+ ωε
σ 2
2
− 1 ω µε  1+ ωε
− 1 z 



2




( )
( )
e
(1.36)
for a wave propagating in the +z-direction.
From γ = α + iβ and Equation (1.36), it is clear from (1.35) that α corresponds to the loss associated with the conductivity of the medium. This is
because e−αz is real and α ≥ 0 implies that it decays exponentially along the
direction of propagation. On the other hand, e− iβz is the oscillating (sinusoidal) portion of the wave. This is due to the presence of the imaginary factor
i = −1 . To see this, it will suffice to recall Euler’s rule: eiθ = cos θ + i sin θ .
Thus, Equation (1.36) corresponds to a sinusoidal wave propagating along
the z-direction whose amplitude decays exponentially. In particular, the
negative sign in front of iβz indicates that the wave is propagating forward
along the +z-direction.
As an aside, recall that the one-dimensional wave equation—that is, one
that involves only one space variable—differs from all other partial differential equations in the sense that it can be solved without imposing any
boundary conditions. However, there is a price to pay: the partial differential
equation has infinitely many solutions.
Explicitly, given a one-dimensional wave equation ∂2z ψ = µε ∂t2 ψ in vacuum,
any solution of the form
(
ψ = f z−
K15149_Book.indb 27
1
µε
) (
t + g z+
1
µε
t
)
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28
Electromagnetic Theory for Electromagnetic Compatibility Engineers
where f, g are any functions that are twice differentiable with respect to z
and t, is a solution of the wave equation, as can be easily verified by direct
substitution (cf. Exercise 1.5.3). This is known as the D’Alembert wave solution [4,6].
Returning to Equation (1.36), can the form of A be determined further? First,
recall from (1.33) that A = A( x , y ) must be a solution satisfying E⊥ = −∇ ⊥ ϕ .
Because the wave is propagating in a charge-free region, ∇ ⊥ ⋅ E⊥ = 0 ⇒ ∇ ⊥2 ϕ = 0,
which is Laplace’s equation. Hence, the final form for A = A( x , y ) necessarily
depends on the boundary conditions imposed on ∇ 2⊥ ϕ = 0 . Show that this is
indeed the case; see Exercise 1.5.4.
In summary, a TEM wave in some domain M × R ⊂ R 4 has the solution:
E( x , y , z, t) = E⊥ ( x , y )e
−
1
ω
2
1 


2 
−1

σ 2

 2
+ 1 z 
i ωt − 1 ω µε  1+ ωε
σ 2
2
µε  1+ ωε − 1 z 

 




( )
( )
e
(1.37)
where E⊥ ( x , y ) = −∇ϕ( x , y ) for some twice-differentiable function
ϕ ∈C 2 ( M) ∩ C( M) that depends upon the boundary conditions on ∂M
wherein the wave is propagating. Indeed, the method of images outlined
in Section 1.1 can be applied to solve for E⊥ ( x , y ) = −∇ϕ( x , y ). Inasmuch as
Equation (1.37) is just the solution for a propagating plane wave, it follows
that a TEM wave is an electromagnetic plane wave.
From (1.35), Equation (1.37) can be written more simply as
E( x , y , z, t) = E⊥ ( x , y )e−αz ei(ωt −βz)
(1.38)
This is the solution for a TEM wave propagating in an unbounded medium in
the +z-direction, where E⊥ ( x , y ) → 0 in the limit as x → ∞ or y → ∞ . Observe
that when the conductivity is zero (i.e., in a perfect dielectric) there is zero
attenuation of the plane waves: α = 0 when σ = 0. In particular, when
a)
σ
ωε
<< 1, α ≈
1
2
ω µε 1 +
(
µε ( 1 +
1
2
( ωεσ )2 − 1)
b)
σ
ωε
<< 1, β ≈
1
2
ω
1
2
( ωεσ )2 + 1)
1
2
1
2
≈ 21 σ
≈
1
2
µ
ε
1
ω µε ( 1 + 1) 2 = ω µε
Thus, when the conduction term is small, the displacement term dominates. Hence, for a good dielectric,
E( x , y , z, t) ≈ E⊥ ( x , y )e
K15149_Book.indb 28
µ
− 21 σ ε
ei(ωt −βz ) ≈ E⊥ ( x , y )ei(ωt −βz)
10/18/13 10:44 AM
29
Brief Review of Maxwell’s Theory
σ
In summary, a good dielectric is defined by the condition that ωε
<< 1 , and the
waves are thus approximately plane waves. Conversely, in a good conductor,
TEM waves are attenuated exponentially. In particular, no waves can propaσ
gate in a perfect conductor as σ → ∞ ⇒ α → ∞ So, for ωε
>> 1,
α≈
1
2
ω µε
σ
ωε
=
µωε
2
≡ δ1 and β ≈
1
2
ω µε
σ
ωε
=
1
δ
2
The quantity δ = µωσ
is called the skin depth of a conductor. In other words,
at a depth z = δ from the surface of a conductor, the field falls off to e−1 of its
original magnitude.
Notice that for very good conductors, α ≈ δ1 ≈ β . Hence, for a good conductor,
(
z
− z i ωt − δ
E ( x , y , z, t) ≈ E⊥ ( x , y )e δ e
)
σ
>> 1 and the elecThus, a good conductor is defined by the condition that ωε
tromagnetic field propagates within the conductor via diffusion. It is also
interesting to note that TEM waves are attenuated exponentially when
the magnetic permeability of a medium is high. This follows directly from
Equation (1.35):
1
α=ω
µε
2
 1 + σ 2 − 1 2 → ∞ as µ → ∞
( ωε ) 

That is, electromagnetic fields do not propagate in a high-magnetic permeability medium. Finally, verify directly from (1.35) that for ε >> ωσ , α ≈ 21 σ µε . In
particular, ε → ∞ ⇒ α → 0. However, observe in this instance that the field will
undergo very rapid oscillations as β ≈ ω µε → ∞. Physically, the field becomes
ill-defined and hence will cease to propagate within the medium (cf. the phenomenon within a good conductor wherein the field also ceases to propagate).
To complete the analysis, the magnetic field propagation is determined.
For simplicity, set E = E⊥ e− γz eiωt , where γ = α + iβ. Second, observe from
Equation (1.31) that
e z × (∇ × E ) =
ex
ey
ez
0
− ∂ z Ey
0
∂ z Ex
1
0
 − ∂ z Ex

=  − ∂ z Ey

0
  − ∂t Bx
 
 =  − ∂t By
 
0




Hence, it is clear for TEM waves that
∂t B = − e z × (∇ × E)
K15149_Book.indb 29
(1.39)
10/18/13 10:44 AM
30
Electromagnetic Theory for Electromagnetic Compatibility Engineers
That is, B = −e z × ∫(∇ × E)dt . This yields
∫
∫
B = −e z × (∇ × E)dt = −e z × ∇ × E⊥ e− γz eiωt dt = ωi e z × ∇ × E
(1.40)
Explicitly,
e z × (∇ × E) =
ex
ey
ez
0
− ∂ z Ey
0
∂ z Ex
1
0
 − ∂ z Ex

=  − ∂ z Ey

0


i ( ωt + iγz )
= −γe z × E⊥ ei(ωt + iγz )
 = e z × E⊥ ∂ z e

Hence,
B = − iωγ e z × E⊥ ei(ωt − γz) ≡ − iωγ e z × E
(1.41)
The magnetic field is thus completely defined in terms of the electric field.
In particular, all the comments made for the electric field above also apply
to the magnetic field with one important proviso: that the magnetic field be
nonstatic, as must be the case for the TEM mode.
The above results are briefly summarized below.
• In a perfect dielectric, there is no attenuation for a time-varying
magnetic field.
• In a perfect conductor, there is no time-varying magnetic field
propagation as the electric field is zero. In particular, a time-varying
magnetic field is attenuated exponentially. It is critical to note that
the derivations were made with time-varying magnetic fields. This
condition need not hold for static magnetic fields.
• Increasing the electric permittivity of the medium wherein the
waves propagate decreases the attenuation; however, this leads to
“infinite” oscillation of the field, rendering the field ill-defined.
• Last but not least, the magnetic permeability also introduces an
exponential attenuation.
Furthermore, observe trivially that for a TEM wave, the electric and magnetic fields possess similar attenuation properties; although, perhaps, this
fact is not surprising inasmuch as the electric field and magnetic field are
part of the same physical field called the electromagnetic field.
In short, a good conductor is as effective a shield for a propagating electric
field as it is for a propagating magnetic field of a TEM wave normal to the surface of a conductor. This follows very clearly from the fact that the transverse
K15149_Book.indb 30
10/18/13 10:44 AM
31
Brief Review of Maxwell’s Theory
fields incident normally on a conductor are attenuated exponentially by the
2
skin depth δ = µωσ
.
σ
Finally, for a TEM wave propagating in a medium (ε,μ) such that ωε
<< 1,
β ≈ ω µε and in particular, lim γ = iβ; whence, (1.41) becomes
σ→ 0
B = ωβ e z × E = µη e z × E
(1.42)
where
η=
ωµ
β
=
µ
ε
This is called the TEM wave impedance. In a vacuum, the TEM impedance
η0 = µε00 . Observe that the TEM impedance η is independent of frequency in
the lossless case where σ = 0; in particular, it is real.
1.4.3 Remark
For a general dielectric medium with loss (i.e., σ > 0) the TEM impedance is
given by
η = − ωµ
iγ =
σ
ωε
In particular, for
ωµ
α 2 +β 2
(β + iα)
<< 1, it can be shown (cf. Exercise 1.5.5) that
η≈
µ 1+ i( σ /2 ωε )
ε 1+ ( σ /2 ωε )2
Finally, this chapter closes with a brief description of the power transferred by an electromagnetic wave. First, recall that given an electromagnetic field (E, B) propagating in a dielectric medium, the Poynting vector
S = E × H defines the power flow per unit area, with units of W/m2 . In
short, it represents the power density of an electromagnetic field. Next,
2
recall that the energy density of the electric field is given by 21 D ⋅ E ∗ = 21 ε E ,
2
the energy density of the magnetic field is given by 21 B ⋅ H ∗ = 21 µ H , and
lastly, the ohmic loss in a dielectric medium is given by the ohmic power
2
density J ⋅ E ∗ = σ E . Then, the following result holds.
1.4.4 Theorem (Poynting)
Suppose an electromagnetic field ( E , B) is propagating in (R 3 , µ , ε , σ ) . If
Ω ⊂ R 3 is a compact domain, then
0=
K15149_Book.indb 31
∫
∂Ω
S ⋅ nd 2 r + 21 ∂t
∫ {ε E
Ω
2
+µ H
2
}d r + ∫
3
Ω
2
σ E d3 r
□
10/18/13 10:44 AM
32
Electromagnetic Theory for Electromagnetic Compatibility Engineers
The result states that the power leaving a compact surface is equal to the
integral of the Poynting vector over the compact surface. This is essentially
the principle of energy conservation.
1.5 Worked Problems
1.5.1 Exercise
(a) Given a path γ = γ (t), t ∈[0, 1], show that
(b) Establish Equation (1.9).
∫
γ
f ( z(t))dt = −
∫
γ
f ( z(t))dt
Solution
(a) Set τ = 1 − t . Then, dτ = −dt and hence,
∫
∫
However, this is precisely −
∫
γ
∫
1
f ( γ (t)) ddt γ (t)dt =
f ( z)d z =
0
0
f ( γ (τ))(−γ (τ))(−dτ) = −
1
∫
1
0
f ( γ ( s))γ ( s)d s
f ( z)d z as s is just a dummy variable.
γ
(b) From
x 2 + ( y − y+ )2
x 2 + ( y + y+ )2
= c2
expanding ( y ± y + )2 = y 2 ± 2 yy + + y +2 yields
0 = x 2 (1 − c 2 ) + y 2 (1 − c 2 ) − 2 yy + (1 + c 2 ) + y +2 (1 − c 2 ) ⇔
(
0 = x 2 + y − y+
(
x 2 + y − y+
1+ c 2
1− c 2
1+ c 2
1− c 2
) − (y
2
) = (y
2
1+ c 2
+ 1− c 2
1+ c 2
+ 1− c 2
)
2
)
2
+ y+2
⇔
− y+2 = y+2
{(
1+ c 2
1− c 2
)
2
}
(
− 1 = y+
2c
1− c 2
)
2
via
( )
1+ c 2
1− c 2
as required.
K15149_Book.indb 32
2
−1=
(
1+ c 2
1− c 2
)(
−1
1+ c 2
1− c 2
)
+1
□
10/18/13 10:44 AM
33
Brief Review of Maxwell’s Theory
1.5.2 Exercise
Show that
γ 2 A( x , y )e− γz eiωt = −ω 2 µεA( x , y )e− γz eiωt + iωµσA( x , y )e− γz eiωt
Solution
From Equation (1.33), set E⊥ = Ae− γz eiωt. Then, ∂ z E⊥ = −γAe− γz eiωt and hence,
∂2z E⊥ = γ 2 Ae− γz eiωt. Likewise, ∂t E⊥ = iωAe− γz eiωt ⇒ ∂t2 E⊥ = −ω 2 Ae− γz eiωt . Sub­
stituting these into (1.33) yields the claim.
□
1.5.3 Exercise
Find a general solution to the D’Alembert wave equation ∂2z ψ = µε ∂t2 ψ on R 3.
Solution
Set ξ = z + ct and ζ = z − ct, where c = 1εµ . Let ψ ( z, t) = f (ξ) + g(ζ), for arbitrary twice differentiable functions f , g on R 3. Then, ∂ z ψ = ∂ξ f ∂ z ξ + ∂ζ g ∂ z ζ.
That is, ∂ z ψ = ∂ξ f + ∂ζ g and hence, ∂2z ψ = ∂ξ2 f + ∂ζ2 g . Similarly,
∂t ψ = ∂ξ f ∂t ξ + ∂ζ g ∂t ζ ⇒ ∂t ψ = − c ∂ξ f + c ∂ζ g ⇒ ∂2z ψ = c 2 (∂ξ2 f + ∂ζ2 g ); whence,
µε ∂t2 ψ = ∂2ξ f + ∂ζ2 g = ∂2z ψ , as required.
□
1.5.4 Exercise
For a TEM wave, show that A = A( x , y ) necessarily depends on the boundary
conditions imposed on ∇ 2⊥ ϕ = 0 .
Solution
From Equation (1.33), set E = A( x , y )e− γz eiωt . Then, E⊥ = −∇ ⊥ ϕ ⇒ ∃ϕ twice
differentiable such that ϕ = ϕ ( x , y )e− γz eiωt ; whence, A = −∇ ⊥ ϕ by construction and in particular, ∆ ⊥ ϕ = 0 ⇒ ∆ ⊥ ϕ = 0. Because the unique solution ϕ to
Laplace’s equation, if it should exist, depends upon the boundary conditions
imposed, the assertion follows.
□
1.5.5 Exercise
Establish the approximation
η≈
µ 1+ i( σ /2 ωε )
ε 1+ ( σ /2 ωε )2
given in Remark 1.43.
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10/18/13 10:44 AM
34
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Solution
σ
<< 1, via (1.35), it was established above that α ≈ σ2 µε
For ωε
2
β ≈ ω µε . Thus, from (1.43), α 2 + β 2 ≈ 14 σ 2 µε + ω 2 µε = ω 2 µε 1 + ( 2σωε )
β + iα = ω µε {1 + i 2σωε } . Substituting these quantities into
{
η=
ωµ
α 2 +β 2
}
and
and
(β + iα)
leads to the approximation.
References
1. Baez, J. and Munian, J. 1994. Gauge Fields, Knots and Gravity. Singapore:
World Scientific.
2. Cheng, D. 1992. Field and Electromagnetics. Reading, MA: Addison-Wesley.
3. Choquet-Bruhat, Y., DeWitt-Morette, C., and Dillard-Bleick, M. 1982. Analysis,
Manifolds and Physics, Part I: Basics. Amsterdam: North-Holland.
4. Chester, C. 1971. Techniques in Partial Differential Equations. New York: McGraw-Hill.
5. Churchill, R. and Brown, J. 1990. Complex Variables and Applications. New
York: McGraw-Hill.
6. Farlow, S. 1993. Partial Differential Equations for Scientists and Engineers. New
York: Dover.
7. Lin, W. and Jin, H. 1990. Analytic solutions to the electrostatic problems of two
dielectric spheres. J. Appl. Phys. 67(3): 1160–1166.
8. Nakahara, M. 2003. Geometry, Topology and Physics. Bristol, UK: IOP.
9. Nash, C. and Sen, S. 1983. Topology and Geometry for Physicists. New York:
Academic Press.
10. Neff, Jr., H. 1981. Basic Electromagnetic Fields. New York: Harper & Row.
11. Plonsey, R. and Collin, R. 1961. Principles and Applications of Electromagnetic
Fields. New York: McGraw-Hill.
12. Schröder, U. 1990. Special Relativity. LNP 33. Singapore: World Scientific.
13. Stratton, J. 1941. Electromagnetic Theory. New York: McGraw-Hill.
14. Toh, T.-C. 2011. Potential generated by rotating charged cylinders. Prog.
Electromagn. Res. B. 33: 239–256.
K15149_Book.indb 34
10/18/13 10:44 AM
2
Fourier Transform and Roll-Off Frequency
An informal survey of Fourier analysis is presented here to motivate practical applications in electromagnetic interference (EMI), even though this topic
is too vast for this chapter to do it proper justice. In particular, the subject
matter is presented to provide insight into the origin of harmonics, transients, and methods to suppress harmonics in high-speed digital circuits.
The chapter concludes with a pithy overview of some elementary properties of filter networks. An understanding of filter circuits clearly plays an
important role in EMC mitigation and signal integrity in general. Readers
who are interested in an in-depth exposition on Fourier analysis and its
applications will benefit greatly from References [1,3,7,14,15]. The application
of complex theory to filter theory can be found in References [12,13].
2.1 Fourier Series
Periodic functions can be expressed as an infinite sum of sine and cosine
functions. Recall that a function f : R → R is periodic with period T > 0
if f (t + T ) = f (t) for all t ∈R . The function is said to be even if f (t) = f (−t)
∀t ∈R , and odd if f (t) = − f (t) for all t ∈R .
Note that the concept of even and odd symmetry of a function depends
up­­on the origin chosen. To elaborate further: consider the function f (t) =
sin t. Clearly, choosing the origin to be at t = 0, f (−t) = sin(−t) = − sin t =
− f (t) ∀t ∈R . Hence, f is odd. However, suppose the origin along the t-axis
were translated to t = − π2 ; that is, f (t) → F(t) ≡ f (t + π2 ) . Then, by definition,
F(t) = sin(t + π2 ) = cos t is an even function as cos t = cos(−t) ∀t ∈R . Thus, the
odd/even symmetry of a function is a relative concept and not an absolute
one. By fixing the origin, the concept then becomes an absolute one.
2.1.1 Definition
Given a function f : R → R , it is said to be of bounded variation on [ a, b] ⊂ R ,
if ∃ M > 0 such that sup Σ P f ( xi ) − f ( xi+1 ) ≤ M , where P is the set of all finite
P∈P
partition P = { xi : a ≤ xi < xi + 1 ≤ b , i = 1,  ,|P|} of [ a, b], and the cardinality |P|
is the number of elements in P.
35
K15149_Book.indb 35
10/18/13 10:44 AM
36
Electromagnetic Theory for Electromagnetic Compatibility Engineers
2.1.2 Theorem (Fourier)
Let f : R → R be a periodic function with period T > 0 satisfying (i)
∫ T0 f (t) dt < ∞ , (ii) f is piecewise continuous on [0, T ], and (iii) f is of bounded
variation on [0, T ]. Then, f (t) = 21 ( f (t + ) + f (t − )) on R, where
f (t) =
∑
∞
n= 0
{an cos nωt + bn sin nωt}
(2.1)
ω ≡ 2Tπ and an , bn ∈R ∀n = 0,1,2, … , are constants called the Fourier coefficients
of f defined by
∫
T
2
T
∫
T
1
T
∫
T
an =
2
T
bn =
a0 =
0
0
0
f (t)cos nωtdt for n > 0,
(2.2a)
f (t)sin nωtdt for n > 0,
(2.2b)
f (t)dt
(2.2c)
□
2.1.3 Corollary
Suppose that f : R → R is an aperiodic function—that is, nonperiodic—such
that (i) f (t) = 0 ∀t ∉ (a,b), for some −∞ < a < b < ∞, (ii) ∫ ba f < ∞ , and (iii) f
is of bounded variation on [a, b]. Then, the Fourier expansion defined by
Equation (2.1) exists for f.
Proof
Define a periodic function f : R → R as follows. First, set f = f on [a, b]. Now,
observe that given [α, β] ⊂ R, there exists a homeomorphism τ :[α, β] → [a, b]
defined by
τ(t) =
b− a
β−α
(t − α) + a
called a reparametrization. Moreover, given any t > b, ∃ i > 0 such that
b + (i − 1)a ≤ t ≤ b + ia, set α + = b + (i − 1)a and β + = b + ia. Likewise, for any
t < a, ∃ i > 0 such that (i + 1)a − ib ≤ t ≤ ia − (i − 1)b. So, set α − = (i + 1)a − ib and
β − = ia − (i − 1)b , and define
τ ± (t) =
K15149_Book.indb 36
b− a
β ± −α ±
(t − α ± ) + a
10/18/13 10:44 AM
37
Fourier Transform and Roll-Off Frequency
Thus, define
 f (τ (t)) for t < a,
−


f (t) =  f (t) for a ≤ t ≤ b ,

 f (τ + (t)) for t > b.

Then, this is the desired extension of f to the real line. In particular, observe
that the extension need not be continuous at t = α ± or t = β ± . Clearly if
f (a) = f(b), then f is continuous on the real line. See Figure 2.1.
By construction, f is periodic with period T = b − a and it satisfies the
conditions (i) and (ii) of Theorem 2.1.2. Thus, the conclusion of Theorem 2.1.2
□
applies.
2.1.4 Remark
By appealing to Euler’s formula, to wit, eiϕ = cos φ + i sin φ , (2.1) can be
expressed in complex Fourier coefficients by
f (t) =
∑
∞
n =−∞
cneinωt
(2.3)
where
cn =
1
T
∫
T /2
− T /2
f (t)e− inωt dt
for all integer n. In particular, it will be left as an easy exercise to verify that
cn = 21 ( an − ibn ), n > 0
(2.4a)
f
(i + 1)a – ib
a–b
a
b
(i + 1)b – ia
Figure 2.1
Extending a continuous function defined on the closed interval [a,b].
K15149_Book.indb 37
10/18/13 10:44 AM
38
Electromagnetic Theory for Electromagnetic Compatibility Engineers
c− n = 21 ( an + ibn ), n > 0
(2.4b)
c 0 = a0
(2.4c)
See Exercise 2.5.1 for more details.
The pair ( an , bn ) is called the nth harmonic of f. Specifically, the discrete
spectrum of f is defined by the quantity an2 + bn2 . This then determines the
magnitude of the n-harmonic of f. In the EMC world, the n-harmonic refers
to the frequency n2ωπ , and Equation (2.1) then yields the magnitude of the
n-harmonic. That is, (2.1) determines the strength of the emissions, and if
they exceed regulatory limits, knowing the harmonic provides a means to
target the source of the emission.
2.1.5 Example
Consider the following square wave defined by

1
 1 for 0 ≤ t < 2 τ ,
f =
 0 for 21 τ ≤ t < τ ,

where τ > 0 is the period of the square wave. Then, from (2.1), and setting ω =
a0 =
1
τ
an =
2
τ
bn =
2
τ
1
τ
2
∫
0
∫
1
τ
2
0
∫
1
τ
2
0
dt =
2π
τ
,
1
2
cos nωt dt = 0, for n = 1, 2, ,
sin nωt dt = − n1π (cos nπ − 1) =
2
(2 n − 1) π
, for n = 1, 2,.
That is,
f (t) = 21 +
∑
∞
n= 1
2
(2 n − 1) π
sin(2 n − 1)ωt
is the Fourier decomposition for a square wave. A plot of f, where n = 200
and period τ = 1 s, is given below. The phenomenon of ringing indicated
in the plot is related to discontinuity of the Fourier series; more details are
provided later.
K15149_Book.indb 38
10/18/13 10:44 AM
39
Fourier Transform and Roll-Off Frequency
Approximate Square Wave (τ = 1 s)
1.2
Amplitude
1
0.8
Ringing
0.6
0.4
1.98
1.8
1.89
1.71
1.62
1.53
1.44
1.35
1.26
1.17
1.08
0.9
0.99
0.81
0.72
0.63
0.54
0.45
0.36
0.27
0.18
–0.2
0.09
0
0
0.2
Time (s)
2.1.6 Example
Consider a triangular wave defined as

2
for 0 ≤ t ≤ 21 τ ,
τt

f =
 − 2τ t + 2 for 21 τ ≤ t ≤ τ ,

where τ > 0 is the period of the triangular wave. Then, from Equation (2.1),
a0 =
1
τ
an = (
1
τ
2 2
τ
0
∫
)
2 2
τ
∫
t dt −
1τ
2
0
1
τ
∫
τ
2
1 τ
τ
2
t dt +
2
τ
∫

t cos nωt dt + 2τ − 2τ

τ
1
τ
2
∫
dt = 1
τ
1
τ
2
t cos nωt dt + 2
∫

cos nωt dt  ⇒
1τ
2

τ
 − 24 2 for n odd,

π n
an = 
0
for n even.

Likewise,
bn = ( 2τ )
2
∫
1τ
2
0

t sin nωt dt + 2τ − 2τ

∫
τ
1τ
2
t sin nωt dt + 2
∫

sin nωt dt  = 0 ∀n
1τ
2

τ
Hence,
f (t) = 1 −
4
π2
∑
∞
1
2
n = 1 ( 2 n − 1)
cos(2 n − 1)ωt
The plot of f is obtained by setting n = 200 and τ = 1 s.
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40
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Approximate Triangular Wave (τ = 1 s)
1.2
Amplitude
1
0.8
0.6
0.4
0.2
1.98
1.8
1.89
1.71
1.62
1.53
1.44
1.35
1.26
Time (s)
1.17
1.08
0.9
0.99
0.81
0.72
0.63
0.54
0.45
0.36
0.27
0.18
0
0.09
0
Observe from Example 2.1.5 that there is ringing occurring at t = 0.5 s of
the square wave, where the function transitions discontinuously. This phenomenon is called Gibb’s phenomenon. This can be intuitively seen from
Theorem 2.1.2, where f (ti ) = 21 ( f (ti+ ) + f (ti− )) at the points ti of discontinuities:
f (ti ) ≠ f (ti ). This is because the rate of change as the function approaches the
discontinuity changes rapidly.
Indeed, this has physical implications for EMC engineers: ringing happens
when there is a sharp transition in signal level. For instance, ringing is apparent
if a signal trace is inductive. This can be seen heuristically as follows. When a
line is inductive, and the signal is decomposed into its Fourier series, the higher
harmonics are truncated, and the amplitudes for the lower harmonics are much
larger than those of higher harmonics. On the other hand, for a capacitive line,
ringing is absent as the transitions at the edges are smoothed out: a capacitive
line truncates low harmonics. Refer to Section 2.4 for more details.
2.2 Fourier Transform
Intuitively—or perhaps not!—an arbitrary function that vanishes sufficiently
rapidly away from the origin so that its integral over the real line is finite also
has a Fourier decomposition. In some instances, it might not be possible to
Fourier-decompose the function into a countably infinite sum of sines and
cosines. Instead, the summation is uncountably infinite and hence gives rise
to the Fourier integral.
2.2.1 Theorem
Suppose that f: R → R is a function such that
∫
∞
| f (t)|2 dt < ∞
−∞
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41
Fourier Transform and Roll-Off Frequency
Then, the integral
F(ω ) =
∫
1
2π
∞
−∞
e− iωt f (t)dt
(2.5)
exists and its inverse is given by
f (t) =
∫
∞
−∞
eiωt F(ω )dω
(2.6)
□
2.2.2 Remark
It is crucial to note that if the condition ∫ ∞−∞ | f (t)|2 dt < ∞ were relaxed to
∫ ∞−∞ | f (t)|dt < ∞ , the inverse Fourier transform might not exist.
Equations (2.5) and (2.6) constitute a Fourier transform pair and (2.5) is called
the Fourier transform of f, and (2.6) is called the inverse Fourier transform of F.
Therefore what is the relationship between a Fourier series and a Fourier
T/2
transform? To see this, consider the coefficients of (2.6): cn = T1 ∫ − T/2 f (t)e− inωt dt ,
2π
and recall that ω = T . Clearly, in the limit as T → ∞, ω → 0; and for an aperiodic function, T → ∞. Hence, for very large T, ω may be expressed as an
infinitesimal quantity: ω → δω.
Secondly, as ω → δω, the discrete spectrum nω becomes a continuous nonzero spectrum ω : that is, nω → ω in the limit as ω → 0 and n → ∞ such
that the quantity 0 < ω < ∞ . Then, taking the limit as the pair T,n → ∞, the
quantity cnT → ∫ ∞−∞ f (t)e− iωt dt < ∞ . This limiting process yields, informally,
the Fourier transform.
Conversely, rewriting Equation (2.3) as f (t) = ∑ ∞n=−∞ cnTeinωt T1 , in the limiting
process as T → ∞, and noting from above that lim cnT → F(ω ) converges to
T , n→∞
the Fourier transform of f, 2Tπ → dω and the summation becomes an integral:
1
2π
∑
∞
n =−∞
cnTeinωt
2π
T
→
1
2π
∫
∞
−∞
eiωt F(ω )dω
yielding the inverse Fourier transform. Thus, informally, a Fourier transform takes a function and transforms it into another function.* See Exercise
2.5.1 for details. It is a linear operator that maps a certain set of functions
into another set of functions. For more details on linear operators, refer
to Appendix A.1. For the present, it suffices to recall that linearity means
that both scaling T(af) = aT( f), a ∈ C, and superposition T ( ∑ i fi ) = ∑ i T ( fi ),
are preserved under the mapping.
*
Mathematically, a Fourier transform is a linear isomorphism on a Hilbert space of square
integrable functions (cf. Remark 2.2.2). This factoid is for readers interested in exploring the
wonderful world of Hilbert spaces and their various applications in the field of engineering
and the theory of partial differential equations; see Appendix A.4 for details.
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42
Electromagnetic Theory for Electromagnetic Compatibility Engineers
These two properties—scaling and the superposition principle—make linear systems a lot easier to study than nonlinear systems. In other words, a
scaled solution of a linear system is again a solution of the system, and the
sum of solutions of a linear system is again a solution of the system. These
conclusions are generally false for nonlinear systems.
From an engineering perspective, a system is linear if its response to an
input function is linear. As a concrete example, let an operator T describe
the response of a system under an input voltage (function) v. Explicitly, let
v  v ′ = T ( v). If v1 , v2 are two voltages simultaneously applied into the input
terminal of the system in question, then the resultant output voltage (i.e.,
response of the system to the input voltages) satisfies
v1 + v2 T
→ v1′ + v2′ = T ( v1 ) + T ( v2 )
if and only if T is linear. Thus, the Fourier transform can be employed to
describe linear systems and it cannot, by definition, be used to describe nonlinear systems. The following result from Fourier theory has great applications in the physical sciences.
2.2.3 Theorem (Parseval)
Let f : R → R be a function such that ∫ ∞−∞ | f (t)|2 dt < ∞ . Then,
dω = ∫ ∞−∞ f (t)2 dt.
1
2π
2
∫ ∞−∞ F(ω )
□
This is known as the energy integral, and its proof can be found in any
2
standard reference on Fourier analysis. The integrand F(ω ) has units of
2
joules per Hz. Intuitively then, 21π ∫ aa +δa F(ω ) dω represents the energy content of the wave f in the frequency range [a, a + δa]. The physical significance
of this is considered in the next section.
This section closes with an example regarding an application of the Fourier
transform to analyze Maxwell’s equations and a brief remark on the Laplace
transform, as it is essentially a natural extension of the Fourier transform. It
is clear from the definition of the Fourier transform that even some simple
functions do not possess “proper” Fourier transforms (without resorting to
distribution theory). For instance, the unit function 1, eiωt and cost, do not
possess Fourier transforms that are functions in the strict sense:
1 ↔ 2 πδ(ω ) : F(ω ) =
1
2π
∫
∞
−∞
eit ↔ 2 πδ(ω − 1) : F(ω ) =
1e− iωt dt ≡ 2 πδ(ω )
1
2π
∫
∞
−∞
eit e− iωt dt =
1
2π
∫
∞
−∞
e− i(ω− 1)t dt = 2 πδ(ω − 1)
cos t ↔ π(δ(ω − 1) + δ(ω + 1))
Verify this in Exercise 2.5.2.
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43
Fourier Transform and Roll-Off Frequency
These situations are clearly not very satisfactory at all. Indeed, the Laplace
transform will rectify these unpleasant scenarios. Heuristically, without
going into the development of Laplace transform theory, this can be easily
accomplished by the following replacement: iω → s = σ + iω, where σ is real.
The Laplace transform pair is defined by


F( s) =



1
 f (t) = 2 πi

∫
∞
0
∫
f (t)e− st dt
σ+ i∞
σ− i∞
F( s)e st d s
(2.7)
2.2.4 Theorem
Suppose f : R → R is piecewise continuous and satisfies (a) f is of bounded
variation for all compact interval [a, b] ⊂ R, and (b) ∃α,M,τ > 0 constants such
that e−αt | f (t)|< M ∀t > τ . Then, the Laplace transform F = F(s) of f = f (t) exists
whenever ℜe( s) > inf{α : e−αt | f (t)|< M ∀t > τ}.
□
As a final comment, Fourier and Laplace transforms are often employed
to solve ordinary differential equations and partial differential equations.
By way of example, consider the one-dimensional diffusion equation
∂t ψ = α 2 ∂2x ψ ∀( x , t) ∈R × [0, ∞) subject to the initial condition ψ(x,0) = g(x) on
R, where g is square-integrable.
By definition, f (t) ↔ F(ω ) ⇒ ∂t F(ω ) ≡ 0. In particular,
∂t ( f (t)e− iωt ) = e− iωt ∂t f (t) − iωf (t)e− iωt ⇒ F [∂t f ] = iωF [ f ]
where F [ f ] denotes the Fourier transform of f for convenience. Indeed, on setting h = ∂t f , it follows immediately that F [∂t2 f ] = F [∂t h] = iωF [ h] = −ω 2F [ f ].
Hence, the Fourier transform with respect to the x variable yields
∂t Ψ(ξ , t) = −α 2 ξ 2 Ψ(ξ , t), where Ψ(ξ , t) = 21π ∫ ∞−∞ ψ ( x , t)e− iξ x dx and ξ is an arbitrary parameter. Because ξ is a parameter, eliminating it as a variable makes
the result more transparent: ddt Ψ(t) = −α 2 ξ 2 Ψ(t). Thus, the Fourier transform
reduces the partial differential equation to an ordinary differential equation
2 2
d
dt Ψ(t) = −α ξ Ψ(t) satisfying Ψ(0) = F [ g ](ξ ). The solution Ψ can be easily
solved and finding the inverse Fourier transform yields the desired result,
ψ = F −1[Ψ ].
The astute reader will notice immediately that the Fourier transform operated on the x variable instead of the t variable. Indeed, the Laplace transform
is generally used to transform the time variable instead of the space variable.
This is because (i) the time variable typically lies in the open interval [0, ∞)
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44
Electromagnetic Theory for Electromagnetic Compatibility Engineers
[cf. Equation (2.7)] and (ii) it takes care of the initial conditions. See, for exa­
mple, References [12,13,15] for further details. This section concludes with a
final example.
2.2.5 Example
The Fourier transform can be employed to gain further insight into Maxwell’s
equations [4,10]. Assuming the fields are time harmonic, the Fourier transform of the electric field E(r, t) is defined by
E (k, ω ) =
∫∫ E(r , t)e
− i( k⋅r −ω t )
d 3 rdt
i k⋅r −ω t ) 3
d kdω. Here, k = 2λπ defines the
The inverse is E(r , t) = ( 2 π1 )4 ∫ ∫ E (k, ω )e (
wave number and λ is the wavelength. Then, by definition, noting that
∂ξ ei( k⋅r −ωt ) = ikξ ei( k⋅r −ωt ) for ξ = x , y , z , that is, ∂ξ ↔ ikξ , and ∂t ei( k⋅r −ωt )
= − iωei( k⋅r −ωt ) ⇒ ∂t ↔ − iω , it is easy to see that
 ∇ × E(r , t) = ik × E(r , t)


 ∇ ⋅ E(r , t) = ik ⋅ E(r , t)

 ∇ × B(r , t) = ik × B(r , t)

 ∇ ⋅ B(r , t) = ik ⋅ B(r , t)

and ∂t ( E(r , t), B(r , t)) = − iω( E(r , t), B(r , t)) .
In particular, taking the Fourier transform of the scalar and vector potential,
 (k , ω )
E (k, ω ) = − ikϕ (k, ω ) + iωA
 (k , ω )
B (k, ω ) = ik × A
Whence, Maxwell’s equations can be rewritten as

k × E (k, ω ) = ωB (k, ω )



k ⋅ B (k, ω ) = 0

 ik × B (k, ω ) = µJ − iωµεE (k, ω )


ik ⋅ εE (k, ω ) = ρ (k, ω )

Observe that k is the direction of the wave propagation. Hence, defining e k = |1k| k , it follows at once that E = E ⊥ + E || , where E ⊥ = (e k × E ) × e k
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Fourier Transform and Roll-Off Frequency
45
and E|| = (e k ⋅ E )e k . Then, recalling that ∇ ↔ ik, it is quite clear that
k ⋅ E = k ⋅ E|| ⇒ ∇ ⋅ E ⊥ = 0 , and k × E = k × E ⊥ ⇒ ∇ × E|| = 0 . Physically, this
means that the longitudinal component does not affect the rotation of the
electric field and the transverse component of the electric field does not
affect the divergence of the electric field.
An immediate consequence is the following:
 = 0 ⇒ k⋅ A
|| = 0 ⇒ A
|| = 0 . That is, the longitudinal
• Coulomb gauge: k ⋅ A
component is identically zero in the Coulomb gauge.
• Charge conservation: iωρ = k ⋅ J = k ⋅ J||. That is, charge conservation
is solely influenced by the longitudinal component of the current
density.
Finally, to conclude this example, via Maxwell’s equation (1.17),
∇ × ∇ × A = µJ − µε ∂t (∇ϕ + ∂t A)
the vector identity ∇ × ∇ × A = ∇(∇ ⋅ A) − ∆A leads to the wave equation:
−∆A + ∇(∇ ⋅ A) + µε ∂t2 A = µJ − µε∇ ∂t ϕ
However, instead of invoking the Lorentz gauge ∇ ⋅ A = −µε ∂t ϕ, continue to use the Coulomb gauge. From Poisson’s equation, ε∇ ⋅ ∇ϕ =
ρ ⇒ ε∇ ⋅ ∂t ∇ϕ = ∂t ρ = ∇ ⋅ J and hence,
∇ ⋅ (ε ∂t ∇ϕ) = ∇ ⋅ J ⇒ iωεϕ k = J||
whence substituting this into the wave equation for the vector potential yields
 − ω 2µεA
 = µJ − µJ|| = µJ⊥
−k2 A
In particular, taking the Fourier transform gives −∆A + µε ∂t A = −µJ ⊥ .
Thus, the transverse component of the electric current density contributes to
the vector potential whereas the longitudinal component contributes to the
scalar potential. Finally, from E = −∇ϕ − ∂t A ,
 ⇒ E|| = − iϕ k and E ⊥ = iωA
⊥
E = − iϕ k + iωA
|| = 0 . In other words, the longitudinal component of the electric field is
as A
the result of the scalar potential and the transverse component of the electric
field is the result of the vector potential.
□
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46
Electromagnetic Theory for Electromagnetic Compatibility Engineers
2.3 Roll-Off Frequency
The Fourier transform of a function f is related to the energy content of
2
the function as follows. F(iω ) dω defines the energy content in the infinitesimal frequency bandwidth dω. This fact becomes important to EMC
engineers in the qualitative assessment of the energy contents of radio frequency harmonics [11]. Indeed, the source of electromagnetic interference
is often mitigated by suppressing the harmonic with the highest energy
content.
In summary, the total energy content of a function is given by
2
2
1 ∞
2 π ∫ −∞ F (ω ) dω . By Theorem 2.2.3, f (t) dt defines the energy content of 2the
function within the infinitesimal time interval dt. In particular, ∫ ∞−∞ f (t) dt
defines the total energy (i.e., spectral energy) content of the function by
Parseval’s theorem.
In fact, by plotting the graph of F(ω ) as a function of ω, a piecewise linear
upper bound that envelops the graph F(ω ) defines the roll-off frequency
[5,11]. In this section, the foundational concepts are developed instead of
approaching the subject via the upper bound envelope approximation typically cited in the EMC literature.
2.3.1 Example
2
Consider the function f (t) = e− t on R. Its Fourier transform is
F(ω ) =
∫
∞
−∞
2
e− t e− iωt dt = π e
( )
2
− ω2
− ω
Thus, the spectrum of f is F(ω ) = π e ( 2 ) and the graph of the spectrum is
given below.
2
Fourier Spectrum
3.5
Amplitude
3
2.5
2
1.5
1
9.68
8.8
9.24
8.36
7.92
7.48
7.04
6.6
6.16
5.72
5.28
4.4
4.84
3.96
3.52
3.08
2.2
2.64
1.76
1.32
0.88
0
0
0.44
0.5
Angular Frequency (rad)
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47
Fourier Transform and Roll-Off Frequency
The following example is particularly relevant to EMC engineers: the spectrum of a square wave. A square waveform is an idealized digital wave generated for data transmission. A more cogent example to consider is that of a
trapezoidal wave.
2.3.2 Example
Consider a square pulse defined by


f (t) = 


1 for 0 ≤ t ≤ τ
0 for t ≥ τ
Then, F(ω ) = ∫ τ0 e− iωt dt = 2i (e− iωτ − 1) = e− iωτ ω2 sin ωτ2 = τ e− iωτ sinc ωτ2 , where sincφ =
sin ϕ
ωτ
ϕ . Hence, |F (ω )|=|τ sinc 2 |. A plot for τ = 1 is given below.
Spectrum of a Squre Pulse
1.2
1
Amplitude
0.8
0.6
0.4
19.8
18.9
18.1
17.2
16.3
15.5
14.6
13.8
12
12.9
11.2
10.3
8.6
9.46
7.74
6.88
6.02
4.3
5.16
3.44
2.58
1.72
0
0
0.86
0.2
Angular Frequency (rad)
It is clear from the plot that the majority of the spectral energy lies in the
lower harmonics.
□
2.3.3 Proposition
Let f : R → R be a continuous trapezoidal pulse of period τ, rise time τ + and
fall time τ − defined by




f (t) = 




A
τ+
A
τ−
t
for 0 ≤ t ≤ τ +
A
for τ + ≤ t ≤ τ -τ −
(τ − t)
for τ -τ − ≤ t ≤ τ
0
(2.8)
for t ≥ τ
Then, 0 < τ ± << τ ⇒ F(ω ) ≈ 21 A(τ + − τ − ) .
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48
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Amplitude
Trapezoidal Pulse
Period τ = τ+ + τ– + δτ
A
0
τ+
δτ
τ–
Time
τ
Figure 2.2
A trapezoidal pulse of period τ.
Proof
Consider the waveform defined by Equation (2.8) depicted in Figure 2.2.
The Fourier transform of f is given by
F(ω ) =
∫
∞
−∞
f (t)e− iωt dt =
A
τ+
∫
τ+
0
te− iωt dt + A
∫
τ−τ −
τ+
e− iωt dt +
A
τ−
τ
∫ (
τ−τ −
τ+τ + +τ −
2
)
− t e− iωt dt
Now, noting that ωτ = 2 π ⇒ e ± i2 π = 1 , evaluating the integral term by term
yields:
A
τ+
A
A
τ−
∫
τ+
0
∫
te− iωt dt =
τ−τ −
τ+
∫
τ
τ−τ −
A
τ+ ω 2
{
e− iωτ+ 1 + iωτ + − eiωτ+
}
e− iωt dt = i ωA {eiωτ− − e− iωτ+ }
( τ − t ) e− iωt dt = ω4πτA ei
2
1
2 ωτ −
−
sin ωτ2− + i ω 2Aτ
−
{2e
i 21 ωτ −
sin ωτ2− (1 + 2 π i) − ωτ −eiωτ−
i
ϕ
whence appealing
to (a) sin ϕ = 2i1 (eiϕ − e− iϕ ), and (b) 1 − eiϕ = e 2 (e
ϕ
i2
ϕ
iϕ
1 − e = −2ie sin 2 , it follows at once that
F(ω ) = i 2ωA2
{
1
τ−
e
i 21 ωτ −
sin ωτ2− −
1
τ+
e
− i 21 ωτ +
sin ωτ2+
}
ϕ
−i 2
i
}
ϕ
−e 2) ⇒
(2.9)
Equation (2.9) can be expressed equivalently as
{
F(ω ) = i ωA e
K15149_Book.indb 48
i 21 ωτ −
sinc ωτ2− − e
− i 21 ωτ +
sinc ωτ2+
}
(2.10)
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49
Fourier Transform and Roll-Off Frequency
So, suppose that τ ± << τ . Then, τ + + τ − << τ and hence, via the binomial
approximation,
e
− i 21 ωτ ±
≈ 1 − i 21 ωτ ± + o
(( ) )
ωτ ± 2
2
and sinc
( ) ≈ 1− ( )
ωτ ±
2
1
3!
ωτ ± 2
2
+o
(( ) )
ωτ ± 4
2
and hence,
F(ω ) ≈ A2 (τ + − τ − ) {1 −
i ω
3! 2
(τ + + τ − )}
to first order in 21 ω(τ + + τ − ). That is, F(ω ) ≈ 21 A(τ − + τ + ), as required. Fill in
the details in Exercise 2.5.3.
□
2.3.4 Corollary
Suppose 0 < τ + << τ − ~ τ , then
F(ω ) ≈
A
ω
{1 +
1
2
( )}
ωτ + 2
2
Proof
From Equation (2.10), and τ − ~ τ ⇒ sin ωτ − ≈ 0. Hence,
F(ω ) ≈ − i ωA e
≈ − i ωA
1
− i ωτ +
2
{(
sinc
( )
ωτ +
2
)(
1 − i ωτ2+ 1 −
{
≈ − i ωA 1 −
1
3!
( )
ωτ + 2
2
1
3!
( ) )}
ωτ + 2
2
− i ωτ2+
}
implies immediately that
F(ω ) ≈
as claimed.
A
ω
{1 +
1
2
( )}
ωτ + 2
2
□
In summary, it is clear from the above results that the spectral energy of
radiation emissions is essentially dictated by the fast rise and fall times: that
is, whether the rise/fall time is much less than the period of the digital signal. Hence, from an electromagnetic interference perspective, fast rise/fall
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50
Electromagnetic Theory for Electromagnetic Compatibility Engineers
times have the potential to radiate strongly, indeed, to the extent of violating the upper bounds for radiation emissions set by various international
regulatory agencies. Reducing the rise time is a common practice in reducing
radio emissions in the EMC industry.*
2.4 Frequency Response and Filter Theory: A Primer
The study of system response in terms of the time variable t (known also
as the time domain) can be equivalently studied in terms of the frequency
variable ω (also known as the frequency domain) via Fourier transform.
Oftentimes, transforming a problem into the frequency domain simplifies
the problem tremendously. The time domain solution is then obtained from
the frequency domain solution via the inverse Fourier transform.
By way of motivation, consider the generalization of the frequency variable ω by expressing it as a complex frequency s = σ + iω , where σ , ω ∈R.
Next, consider an exponentially damped sinusoidal voltage given by
v(t) = V0 eσt cos(ωt + θ), where V0 ∈R is the magnitude of the voltage, ω = 2πf
the angular frequency and θ is an arbitrary phase.
• For σ < 0, eσt → 0 in the limit as t → ∞; that is, v(t) decays exponentially in time (exponentially damped).
• For σ > 0, eσt → ∞ in the limit as t → ∞; this is not physically realizable, and hence if σ ≠ 0, it is always chosen to be negative. It represents
the loss as the voltage wave propagates through a lossy medium.
2.4.1 Example
Given the damped sinusoidal voltage v(t) = V0 eσt cos(ωt + θ) , show that this
can be expressed as v(t) = ℜe(V0 e st eiθ ) . To see that this is indeed the case, it
suffices to appeal to Euler’s formula: eiθ = cos θ + i sin θ . Then, it is evident
that
e st eiθ = e st + iθ = eσt + i(ωt +θ) = eσt ei(ωt +θ) = eσt (cos(ωt + θ) + i sin(ωt + θ))
whence ℜe(e st eiθ ) = eσt cos(ωt + θ) ⇒ v(t) = ℜe(Vm e st eiθ ) , as claimed.
*
□
However, EMC engineers are also aware that reducing the rise/fall times too much will result
in compromising the signal integrity of the digital device in question. Indeed, an ingenious
workaround to this problem (typically found in clock generators in integrated circuits) is the
implementation of spread spectrum [8,9].
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Fourier Transform and Roll-Off Frequency
51
Example 2.4.1 leads to the following generalization. Physically measurable
quantities such as current and voltage can be generalized to complex quantities. There are two reasons for making the generalization. First, only physically
measurable quantities are real; however, the imaginary component encodes
information regarding phase and loss. More on this is shown in Chapter 4.
Before proceeding, a small mathematical detour is made below for sinusoidal waves. Let f = f (x, t) be a complex function. Then, f is said to be time
harmonic (or a steady state sinusoidal function of t) if there exists a complex
function F = F (x, θ) such that f ( x , t) = F( x , θ)eiωt , for some phase θ ∈ R. Thus,
there is a clear bijection between f and F; in particular, the time variable
t can be factored from the original function f. This leads to the following
definition.
2.4.2 Definition
The phasor transform of a time harmonic function f ( x , t) = F( x , θ)eiωt is a mapping P : f  F defined by P[ f ]( x , θ) = f ( x , t)e− iωt . The inverse P −1 exists for
time-harmonic functions and is trivially defined by P −1 [ F ]( x , t) = F( x , θ)eiωt .
Notice that the positive sign in eiωt is merely an arbitrary choice. It is equally
valid to use e− iωt in the definition of time harmonicity as long as consistency
is maintained. The positive sign was chosen to be consistent with the definition of the Fourier transform.
2.4.3 Lemma
The phasor transform P satisfies the following properties:
(a) linearity: P[af + bg] = aP[ f ] + bP[g] for all time-harmonic f, g and
scalars a,b ∈ C:
(b) P[∂t f ]( x , θ) = iωP[ f ]( x , θ)
(c) P[∂ x f ]( x , θ) = ∂ x P[ f ]( x , θ)
for some fixed phase θ ∈ R.
Proof
Linearity is clear: P[ af + bg ] = af ( x , t)e− iωt + bg( x , t)e− iωt = aP[ f ] + bP[ g ], where
f, g are any two steady-state sinusoidal functions and a, b are constants.
Property (b) is also easily established:
P[∂t f ]( x , θ) = P[∂t ( F( x , θ)eiωt )] = P[ F( x , θ) ddt eiωt ] = P[iωF( x , θ)eiωt ] ≡ iωP[ f ]( x , θ)
and finally, to prove (c), it is enough to observe that
P[∂ x f ]( x , θ) = ∂ x ( f ( x , t)e− iωt ) = ∂ x F( x , θ) ≡ ∂ x P[ f ]( x , θ) K15149_Book.indb 51
□
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52
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Throughout the analysis, a steady-state sinusoidal propagating wave is
assumed for simplicity. The justification for a sinusoidal assumption is somewhat obvious: (a) most physical waveforms have Fourier decompositions; (b)
the superposition principal can be applied to the propagating waves; (c) the
phasor transform can be applied to convert the time dependency in partial
differential equations into steady-state differential equations. More generally, Fourier and Laplace transforms are often used to solve differential
equations, as pointed out in Section 2.2.
Suppose that a time harmonic function f = f (x, t) is twice differentiable with
respect to x and t. Then, it is very easy to see the following.
P[∂t2 f ] = P[∂t (∂t f )] = iωP[∂t f ] = −ω 2 P[ f ]
and
P[∂2x f ]( x , ω ) = ∂2x ( F( x , ω )e− iωt ) ≡ ∂2x P[ f ]( x , ω )
Thus, consider a simple wave equation ∂2x f ( x , t) = α 2 ∂t2 f ( x , t). Phasor transforming the wave equation yields P[∂2x f ] = P[α 2 ∂t2 f ] = −ω 2 α 2 P[ f ]. Thus, the
partial differential equation is converted into a second-order ordinary differential equation (with respect to x): ∂2x F( x , ω ) = −ω 2 α 2 F( x , ω ).
Hence, for steady-state sinusoidal waves, it is convenient to apply phasor
transform to convert a time domain solution into a frequency domain solution via the following rules.
•
•
∂
∂t
∂2
∂t 2
→ iω under the phasor transform.
→ −ω 2 under the phasor transform.
Furthermore, ∂ x and P commute; that is, P[∂ x f ] = ∂ x P[ f ] for any steady-state
sinusoidal function f.
Finally, define a complex voltage by v(t) = V e st , where V = V0 eiθ for some
real constants V0 , θ. This is known as the phasor representation for the timeharmonic voltage. In the above notation, the impedance of an inductor is represented by X L = sL , and that of a capacitor is represented by X C = sC1 . The
motivation for using phasor representation is to eliminate time dependency,
as pointed out above. Another example is given to illustrate this point.
Consider a voltage induced across an inductor: v(t) = L ddi(tt ) . Substituting
i(t) = Ie st yields ddi(tt ) = sIe st = si(t) . That is, v(t) = sLi(t) ⇒ V = sLI . Because the
time variable has been eliminated, the expression is solely a function of
frequency; the time domain expression has thus been transformed into
the frequency domain expression. Likewise, the voltage induced across a
capacitor is v(t) = C1 ∫ i(t)dt . Substituting i(t) = Ie st yields ∫ i(t)dt = 1s Ie st = 1s i(t).
That is, v(t) = sC1 i(t) ⇒ V = sC1 I.
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53
Fourier Transform and Roll-Off Frequency
2.4.4 Example
Consider a simple series RL-circuit driven by some voltage v (t) = V (θ)e st ,
where V (θ) = V0 eiθ and V0 , θ ∈R are fixed. Here, the system is defined by
the following ordinary differential equation Ri(t) + L ddt i(t) = v(t) . By Lemma
2.4.3, via phasor representation,
Ri(t) + L ddt i(t) = v(t) ⇔ RI(φ) + iωLI(φ) = V (θ)

for some fixed phase ϕ, whence, I(φ) = RV+(iθω)L .
In order to evaluate ϕ, it suffices to note that I(φ) =
real I 0 . Therefore
V ( θ )
R + iωL
= V0eiθ
R − iωL
R 2 +ω 2 L2
= V0eiθ
1
R 2 +ω 2 L2
i(t) = I(φ)eiωt =
V0
2
2 2
R +ω L
R 2 +ω 2 L2
e
(
= I 0eiφ , for some
eiθ′ ≡ I 0eiφ
V0
where θ′ = arctan −ωL
R , φ = θ + θ′ and I 0 =
representation,
V ( θ )
R + iωL
. Thus, in time domain
i θ−arctan ωRL +ωt
)
yielding the solution to the differential equation governing the circuit.
In particular, the current measured by an ammeter is ℜe(i(t)) = 2 V0 2 2
R +ω L
cos(θ − arctan ωRL + ωt).
□
In general, particularly for forcing functions that are not necessarily time
harmonic, the differential equation governing the linear system is solved via
Laplace transform by converting the time domain into a complex frequency
domain. This leads to the concept of a transfer function. Informally, a transfer
function represents the response of a linear system to an input function. Let
H = H(s) characterize a linear system response in the frequency domain.
Then, some examples of transfer functions are:
V1 ( s )
V0 ( s )
= H ( s), where V0 represents the input voltage and V1 represents
the output voltage.

• VI ((ss)) = H ( s); here, H(s) represents the impedance of the system.

• VI((ss)) = H ( s); here, H(s) represents the admittance of the system.
•
Recall that the admittance Y(s) is defined as the inverse of the impedance
Z(s): Y(s) = 1/Z(s).
Before proceeding further, some formal definitions are given for completeness. A circuit (network) is often said to be linear if it comprises
resistors, capacitors, and inductors, that is, if I  V is a linear invertible
mapping. If the properties of the network do not vary with time, it is time
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54
Electromagnetic Theory for Electromagnetic Compatibility Engineers
invariant. And finally, if the network comprises passive linear elements,
then it satisfies the reciprocity theorems (cf. Section 6.1) and hence the network is said to be a reciprocal network.
2.4.5 Definition
A network N = {(E i , Cˆ ij )} comprises circuit elements Ei and conductors Cˆ ij
connecting the pair (E i , E j ).
a) It is linear if each Ei can be represented by an invertible linear operator Li such that Li [ I ( z, t)] = V ( z, t), where the pair (V, I ) is the voltage
and current associated with Ei.
b) The invertible linear operator Li is time invariant if
Li [ I ( z, t + τ)] = V ( x , t + τ) ∀τ . More precisely, if Tτ is a time translation
operator defined by Tτ [ξ( z, t)] = ξ( z, t + τ), then Li is time invariant if
LiTτ = Tτ Li ∀τ .
Some examples are given below for clarification. Consider a simple resistor R. By Ohm’s law, V = RI ; that is, the linear operator is just multiplication
by R. Next, consider an ideal capacitor C. Then, this is represented by the
following linear operator: V = C1 ∫ I (t ′)dt ′ (recall that the integral operator is a
linear operator). Finally, consider an ideal inductor L. This is represented by
the operator: V = − L ddIt ; clearly, the operator ddt is linear and its inverse is the
integral operator.
2.4.6 Definition
Let H : C → C be a complex function. Then, it is called a rational function if it
can be expressed as
( s − sn )
H ( s) = A ( s − s(ns+−1s)(1 )(s −s −sns+22)
)( s − sn + m )
where A, s1 , , sn+ m ∈C are constants, for some n, m ∈N . The complex numbers s1 ,  , sn are called the zeros of H and sn+ 1 ,  , sn+ m are called the poles
of H.
2.4.7 Remark
Let h = h(t) be time harmonic. Then, s = iω is purely imaginary and
2
H ( s)H (− s) = H ( s) . Studying the poles and zeros of the transfer function H
is important in the design of filters for signal integrity and suppressing electromagnetic interference without affecting signal quality.
This chapter concludes with a qualitative sketch of the natural response
of a linear system response in terms of the poles of its transfer function.
Recall that in general, s = σ + iω, for some real σ, ω. The plot of H ( s) as
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55
Fourier Transform and Roll-Off Frequency
a function of s can be analyzed as follows. First, set σ = 0 and plot H(iω )
as a function of ω. Next, set ω = 0 and plot H(σ ) as a function of σ. The
phase of H is determined as follows: set H(iω) = A(ω) + iB(ω), where A(ω),
B(ω) ∈ R. Then, the phase of H(iω) on the complex plane is, by definition,
arg H (iω ) = arctan AB((ωω)) .
A physical interpretation of the poles of a transfer function H = H(s) is
given below:
• An input function (also known as a forcing function) operating at any
one of the poles will result in an infinite response by the system. An
example of a forcing function is an input voltage.
• A natural response is defined by the response of a system when the
input terminals are shorted, that is, when the forcing function is set
to zero. An example is the natural resonance of a series RC-circuit.
On the other hand, the zeros of a transfer function determine the magnitude
and phase of the response in time domain.
As a concrete example, consider a transfer function
H ( s) =
K ( s)
( s − p1 )( s − pm )
where K = K ( s) encodes the zeros z1 ,  , zn of the system. Now, by taking the
partial fraction expansion, this can be expressed as
H ( s) =
K ( s)
( s − p1 )( s − pm )
=
K1
s − p1
++
Km
s − pm
(2.11)
for some constants K1 , , K m ∈C associated with K ( s). It is thus obvious
from Equation (2.11) that the zeros, via { K1 ,  , K m }, determine the magnitude and phase of H. More specifically, converting (2.11) back to the time
domain via inverse Laplace transform yields
h(t) = K1e p1t +  + K m e pmt
(2.12)
The above comments are now apparent: as |eiα|= 1 ∀α ∈R , it follows at once
that K i ∈C ⇒ K i = K i eiφ , for some phase ϕ. Thus, {Ki} encodes the phase and
magnitude information, as asserted.
Next, consider the output voltage resulting from a natural response of the
system. Given the poles sn+ 1 ,  , sn+ m of the transfer function H ( s) , the output
voltage has the form v(t) = A1e sn + 1t + A2 e sn + 2 t +  + Am e sn + mt , where A1 ,  , Am
are constants that depend on the initial conditions of the system. Thus, by
studying the frequency response of a linear system in the frequency domain,
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56
Electromagnetic Theory for Electromagnetic Compatibility Engineers
a qualitative solution in the time domain can be obtained via the poles of the
transfer function. See Exercise 2.5.4 for a concrete example.
A transfer function H = H ( s) is also called a response function. Suppose
that
H ( s) =
P1
s − p1
++
Pn
s − pn
+
Q1
s − q1
++
Qm
s − qm
where Pi , Q j for i = 1,  , n and j = 1,  , m , are constants, { pi } are natural
frequencies and {q j } are the forcing frequencies. That is, the response function can be decomposed as
H ( s) = H 1 ( s) + H 2 ( s)
where H 1 ( s) = s −P1p1 +  + s −Pnpn is the natural response and H 2 ( s) = sQ− q11 +  +
is the forcing response. Taking the inverse Laplace transform yields
Qm
s − qm
h(t) = h1 (t) + h2 (t)
where h1 (t) = P1e p1t +  + Pne pnt and h2 (t) = Q1eq1t +  + Qm eqmt are, respectively,
the natural part and the forcing part of the system in the time domain.
The poles of the system determine the natural response. The poles of
the excitation applied to the system determine the forced response. Now,
observe that the steady-state response of the system is defined by h(t)|t >T ,
where T >> 0 is some arbitrarily large number. In particular, if ℜe( pi ) > 0,
then lim h1 (t) → ∞ and hence the system is unstable; there is no steady-state
t→∞
response. Thus, in order for steady state to exist, it is necessary that the natural
poles { pi } lie on the left-hand side of the complex plane; that is, ℜe( pi ) < 0 ∀i.
Because ℜe( pi ) < 0 ⇒ lim h1 (t) → 0, it follows that h(t) ≈ h2 (t) for large t > 0.
t→∞
Hence, in the study of the steady-state response of a system, only the forced
response need be considered.
2.4.8 Example
Consider the LC-network defined [5] as
Z( s) = H
( s2 +ω 12 )( s2 +ω 23 )( s2 +ω 22 n + 1 )
s( s2 +ω 22 )( s2 +ω 24 )( s2 +ω 22 n )
(2.13)
where 0 < ω i < ω i + 1 ∀i = 1,  , 2 n . Then, via partial fraction expansion, this
can be reduced to
Z( s) = sH +
K15149_Book.indb 56
K0
s
+
sK1
s2 +ω 22
+
sK 2
s2 +ω 24
+
sKn
s2 +ω 22 n
(2.14)
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57
Fourier Transform and Roll-Off Frequency
where K i > 0 ∀i = 0, 1,  , n . Clearly,
{
lim sZ( s) = lim s2 H + K 0 +  +
s→ 0
s→ 0
s2 Kn
s2 +ω 22 n
}= K
0
and likewise,
lim
s→ iω i
s2 +ω 22 i
s
{
Z( s) = lim H ( s2 + ω 22 i ) + + K i + + K n
s→ 0
s2 +ω 22 i
s2 +ω 22 n
} = K , for i = 1,…, n.
i
Now, observe that
sKi
s +ω 22 i
2
where sH ↔ inductance H,
{
s
Ki
+
1
sKi / ω 22 i
}
−1
=
s K1 +
i
1
=
{ } {
K0
s
s2 +ω 22 i
sKi
−1
=
s
Ki
+
↔ capacitance
1
s ( Ki /ω 22 i )
⇒
1
Ki
1
sKi / ω 22 i
1
K0
}
−1
. Hence,
↔ capacitor and
connected in parallel, as it has the form
{ }
1
1
sC + sL
Ki
ω 22 i
↔ inductor
−1
corresponding to the
impedance of a parallel LC circuit. The first term corresponds to an inductor, the second term to a capacitor, and the subsequent terms are parallel LC circuits, all of which are connected in series, as indicated by the
summation in Equation (2.13). Hence, (2.13) has the circuit representation
depicted in Figure 2.3 and it is called the LC Foster I network.
By inspecting (2.13), the poles are {0, ± iω 2 k , ∞ : i = 1,  n} and the zeros are
{ ± iω 2 k − 1 : k = 1,  , n} . Thus, all the finite poles and zeros lie on the imaginary
axis, as expected, for by construction, the resistance of the network is zero,
and also by definition, the poles and zeros alternate with a singularity at the
H
Z(s)
1
K0
K1
ε22
K2
ε24
K2
ε22n
1
K1
1
K2
1
Kn
Figure 2.3
A realization of a general LC Foster I network.
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58
Electromagnetic Theory for Electromagnetic Compatibility Engineers
origin. Moreover, it is also clear that (i) lim Z( s) = ∞ if and only if a series
s→∞
inductor is present, and (ii) lim Z( s) = ∞ if and only if a series capacitor is
s→ 0
present.
To conclude this example, observe that if Equation (2.13) has the following
form instead:
s( s2 +ω 2 )( s2 +ω 2 )( s2 +ω 2 )
(2.15)
Z( s) = H ( s2 +ω 2 )(2 s2 +ω 2 )4( s2 +ω 2 2 n )
1
2 n− 1
3
where 0 < ω i < ω i + 1 ∀i = 1,  , 2 n − 1, then a similar analysis leads to:
(a) A zero instead of a singularity at the origin.
(b) The alternating poles and zeros pattern is the reverse of Equation
(2.13); that is, the poles of (2.13) correspond to the zeros of (2.15) and
vice versa along the imaginary axis.
□
2.4.9 Example
Consider the LC Foster I network defined by (2.14) above. Its admittance is
Y ( s) = Z1(s) . Once again, proceeding as in Example 2.4.8 via partial fraction
expansion:
Y ( s) = sH +
K0
s
+
sK1
s2 +ω 12
+
sK 2
s2 +ω 22
1
Zi
≡
sKi
s2 +ω i2
Zi =
s
Ki
+
Noting by construction that
Example 2.4.8 that
++
(2.16)
sKn
s2 +ω n2
, it follows immediately from
1
K
s 2i
ω
i
1
for each i = 1,  , n . That is, each Zi is a series LC circuit of inductance Ki and
Ki
capacitance ω 2i . Thus, Equation (2.16) has the physical representation illustrated in Figure 2.4.
Y(s)
H
1
K0
1
K1
K1
ω21
1
K2
K2
ω22
1
Kn
Kn
ω2n
Figure 2.4
A realization of a general LC Foster II network.
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59
Fourier Transform and Roll-Off Frequency
This network is called the LC Foster II network. The analysis follows that
of Example 2.4.8 mutatis mutandis. In particular, by inspecting Figure 2.3, it is
evident that lim Y ( s) = ∞ ⇒ K 0 > 0 and the capacitor H is absent. On the other
s→ 0
hand, lim Y ( s) = 0 ⇒ H > 0 and the inductor K10 is absent.
□
s→∞
2.4.10 Example
Consider the LC Cauer network defined [2] in Figure 2.5.
Set ∆ n− 1 = Zn− 1 + Zn = Zn− 1 + Y1n and ∆ n− 2 = Yn− 2 + ∆ n1− 1 . Then, by induction,
it is clear from Figure 2.4, that
 Zk + 1 for k odd,
∆k +1

∆k = 
 Yk + ∆ k1+ 1 for k even,

where by construction, n is even. Observe that by definition, ∆ 2 k defines
admittance whereas ∆ 2 k + 1 defines impedance for k = 1,  n − 1.
The impedance Z is thus given by the following continued fraction
expansion,
Z = Z1 +
Y2 +
1
(2.17)
1
Z3 + 1
Y4 +

1
.
∆n − 1
Then, the LC Cauer I network is defined by
 sL for k odd,

Zk = 
 sC1 for k even.

Two basic properties of the LC Cauer I network can be easily deduced from
Equation (2.17):
(a) lim Z( s) → ∞ ⇔ Z1 = sL ≠ 0
s→∞
(b) lim Z( s) → 0 ⇔ Z1 = sL ≡ 0 and Z2 =
s→∞
Z1
Z
Z3
Z2
1
sC
≠0
Zn–3
Z4
Zn–1
Zn–2
Zn
Figure 2.5
A general LC Cauer network (also known as a ladder network).
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60
Electromagnetic Theory for Electromagnetic Compatibility Engineers
The LC Cauer II network is defined by
 1 for k odd,
 sC
Zk = 
 sL for k even.

a) lim Z( s) → ∞ ⇔ Z1 =
1
sC
≠0
b) lim Z( s) → 0 ⇔ Z1 =
1
sC
≡ 0 and Z2 = sL ≠ 0
s→ 0
s→ 0
The RC Foster I and II networks and Cauer I and II networks can likewise
be constructed and studied. In particular, a general filter is often the synthesis of Foster and Cauer networks. These considerations are not pursued here.
See Exercises 2.5.5–2.5.7 for some concrete examples.
2.5 Worked Problems
2.5.1 Exercise
Given the Fourier expansion f (t) = Σ ∞n= 0 {an cos nωt + bn sin nωt} , (i) show that
it can be expressed as f (t) = Σ ∞n=−∞ cneinωt , where the coefficients are given
by Equation (2.4); (ii) formally deduce Fourier transform from the Fourier
expansion.
Solution
(i) From Euler’s formula, eit = cos t + i sin t , it follows that
K15149_Book.indb 60
f (t) =
∑
=
∑
=
∑
≡
∑
∞
n= 0
∞
n= 0
∞
n= 0
{an cos nωt + bn sin nωt}
1
2
an (einωt + e− inωt ) −
an − ibn
2
∞
n =−∞
einωt +
∑
∞
n= 0
∑
an + ibn
2
∞
n= 0
1
2
ibn (einωt − e− inωt )
e− inωt
cneinωt
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61
Fourier Transform and Roll-Off Frequency
where c0 = a0 , cn = 21 ( an − ibn ) and c− n = 21 ( an + ibn ) ∀n > 0 .
(ii) From (i) and the fact that
cn =
1
T
∫
T /2
− T /2
f (t)e− inωt dt
it follows by direct substitution that
∑
f (t) =
∞
−∞
1
T



Now, on setting ω n = nω and δω =
lim
T , n→∞
whence f (t) =
∑
1
2π
∞
−∞
1
T



∫
T /2
− T /2
∫
T/2
− T/2
2π
T

f (t)e− inωt dt  einωt

, it clearly follows that

f (t)e− iω nt dt  eiω nt →

1
2π
∫
∞
−∞
F(ω )eiωt δω
∫ ∞−∞ F(ω )eiω nt dω , as required.
□
2.5.2 Exercise
Verify the Fourier transform pair: cos t ↔ π(δ(ω − 1) + δ(ω + 1)).
Solution
First, set f (t) = cos t and recall that
− it
it
1
2 (e + e ), it follows that

F(iω ) = 21 

∫
∞
−∞
eit e− iωt dt +
1
2π
∫ ∞−∞ e− iωt dt = 2 πδ(ω ). Then, from cos t =


e− it e− iωt dt  = 21 
−∞


∫
∞
∫
∞
−∞
e− i(ω− 1)t dt +
∫
∞
−∞

e− i(ω+ 1)t dt 

= π ( δ(ω − 1) + δ(ω + 1)) .
2.5.3 Exercise
Derive Equation (2.12) and hence establish (2.14).
Solution
A
τ+
K15149_Book.indb 61
∫
τ+
0
te− iωt dt =
A
τ+ ω 2
{e
− iωτ +
}
(1 + iωτ + ) − 1 =
A
τ+ ω 2
e
− i 21 ωτ +
{e
− i 21 ωτ +
(1 + iωτ + ) − e
i 21 ωτ +
}
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
From sin φ =
i(eiφ − e− iφ ), it follows that
1
2i
e
A
τ+ ω 2
=
A
τ+ ω 2
e
− i 21 ωτ +
− i 21 ωτ +
= − i ω22Aτ e
{e (1 + iωτ ) − e }
{−2i sin + iωτ e }
− i 21 ωτ +
ωτ +
2
− i 21 ωτ +
i 21 ωτ +
+
i 21 ωτ +
+
sin ωτ2+ + i ωA eiωτ+
+
Likewise,
Aτ
τ−
∫
τ
τ−τ −
e− iωt dt = i ωA
=
τ
τ−
4π A
ω 2 τ−
2π A
(1 − eiωτ− ) = i ω 2 τ e
i 21 ωτ −
−
e
i 21 ωτ −
(e
− i 21 ωτ −
−e
i 21 ωτ −
)
sin ωτ2−
Finally,
− τA−
∫
τ
τ−τ −
te− iωt dt = − ω 2Aτ
−
= − ω 2Aτ
−
= i ω 2Aτ
{1 + 2π i − e
iωτ −
}
{− i2e sin + 4π e sin + iωτ e }
{2e sin (1 + 2π i) − ωτ e }
i 21 ωτ −
= i ω22Aτ e
ωτ −
2
i 21 ωτ −
sin ωτ2− −
4π A
i 21 ωτ −
−
i 21 ωτ −
ωτ −
2
i 21 ωτ −
−
− 2 π ieiωτ− + iωτ − eiωτ−
ωτ −
2
−
4 πA
ω 2 τ−
e
i 21 ωτ −
−
iωτ −
iωτ −
sin ωτ2− − i ωA eiωτ−
From this,
F(ω ) = − i ω22Aτ e
− i 21 ωτ +
= i 2ωA2
1
τ−
i 21 ωτ −
= i ωA
i 21 ωτ −
{
{e
+
e
sin ωτ2+ +
ω 2 τ−
sin ωτ2− −
1
τ+
sinc ωτ2− − e
e
− i 21 ωτ +
e
− i 21 ωτ +
sin ωτ2− + i ω22Aτ e
sin ωτ2+
sinc ωτ2+
}
}
− i 21 ωτ −
−
The conclusion of the proof thus follows accordingly.
sin ωτ2− (1 + 2 π i)
□
2.5.4 Exercise
Find the transfer function H ( s) = VV0i ((ss)) for the circuit illustrated in Figure 2.6,
where Vi is the input voltage and V0 is the output voltage.
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63
Fourier Transform and Roll-Off Frequency
R
Vi(s)
C
R0
L
V0(s)
Figure 2.6
A simple RLC filter network.
Solution
The admittance of the RLC network is
Y ′ = sC +
1
sL
+
1
R0
⇒ Z′ =
=
1
Y′
sL
s2 LC +α
where α = 1 + R10 . Hence, the total impedance Z = R + s2 LC +sLs L + 1 . From the total
R0
current I = VZi , it follows that V0 = Vi − IR = Vi ( 1 − RZ ) and hence,
H ( s) = 1 −
(
where α = RLC and β = L 1 +
R
R0
R
Z
=
sβ
sβ+ s2 α+ R
). Performing partial fraction expansion,
sβ
sβ+ s2 α+ R
H ( s) =
=
A
s +ω +
+
B
s +ω −
where
ω± = −
1+ RR
0
2 RC
±
1
2 LC
(1 + )
R
R0
2
− 4LC
are the roots of the denominator s2 α + sβ + R = 0 ,
sβ = A( s + ω − ) + B( s + ω + )
+L
To evaluate for the constants A,B, set s = −ω + . Then, A = ω +ω−ω
. Likewise, set−
ω− L
ting s = −ω − yields B = − ω + −ω − .
Hence, the poles of the transfer function H are {ω + , ω − } and the zero
of H is clearly at the origin s = 0 whenever R ≠ 0. Finally, observe that if
2 LC > 1 + RR0 , R > 0 ⇒ ω ± ∈C , which are not purely imaginary numbers.
On the other hand, for 2 LC ≤ 1 + RR0 ⇒ ω ± ∈R and hence, the poles and
zeros all lie on the real line.
□
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64
Electromagnetic Theory for Electromagnetic Compatibility Engineers
2.5.5 Exercise
Construct a general low-pass filter to allow all frequencies 0 ≤ ω ≤ ω 0 to pass
through with very little attenuation.
Solution
Consider a general transfer function H for a filter:|H (ω )|2 =
2
P(ω 2 )
Q(ω 2 )
2
, where P, Q are
polynomial functions of ω , and 0 ≤|H (ω )| ≤ 1 . Ideally,|H (ω )| = 1 − u(ω 2 − ω 20 ),
where
2
 1 for x > 0,

u( x) = 
 0 for x ≤ 0,

is the Heavyside function. Hence, Q(ω 2 ) ≈ P(ω 2 ) for ω ≤ ω 0 and Q(ω 2 ) >> P(ω 2 )
for ω > ω 0 . Thus, set Q(ω 2 ) = P(ω 2 ) + P ′(ω 2 ) , where P ′(ω 2 ) ≈ 0 for ω ≤ ω 0 and
2
2
P ′(ω 2 ) >> 0 for ω > ω 0 . Then, H (ω ) = 1+ h1(ω 2 ) , where h(ω 2 ) = PP′((ωω2 )) . It thus
remains to determine the functional form of h.
2n
It is obvious that by setting h = ωω0 , then h << 1 for n >> 1 whenever
2
ω < ω 0 . In particular, that filter is maximally flat at ω = 0: H(0) = 1 . This
choice of h is called the Butterworth approximation. To determine the property
2
of H, make the substitution: ω 2 = −s2 . Then, by definition, H ( s)H (− s) = H ( s) ,
2
and rewriting ε 2 = ω1n , it follows that H ( s) = 1+ε2 (1− s2 )n .
( )
0
Now, the poles of H(s) lie precisely on the curve 1 + ε 2 (− s2 )n = 0 . However,
this curve is precisely the equation of a circle in C of radius ε2/1 n . To see this,
it suffices to note that
1 + ε 2 ( − s 2 ) n = 0 ⇔ ( − s 2 )n =
1
ε 2/n
(−1)1/n =
1
ε 2/n
e
iπ/n
⇒s=
1
ε 2/n
e
i π2 nn+ 1
That is,
se
− i π2 nn+ 1
=
1
ε 2/n
1
Thus, the poles lie on the circle of radius ε2/n . And in particular, from
s = σ + iω, it is clear that for physically realizable filters, the poles must lie on
the circle intersecting the left-hand complex plane; that is, σ < 0.
□
2.5.6 Exercise
Construct a general high-pass filter to allow all frequencies ω ≥ ω 0 >> 0 to
pass through with very little attenuation.
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65
Fourier Transform and Roll-Off Frequency
Solution
Now, observe trivially that 1s → 0 as s → ∞ . That is, if H = H(s) is a low-pass
filter, then G(s)=1/H(s) represents a high-pass filter. Hence, from Exercise
2.5.5, make the following transformation s → 1/s. Then, it is clear that
Hˆ ( s) ≡ H ( 1/ s ) defines a high-pass filter. That is, using Exercise 2.5.5,
Hˆ ( s) =
( − s2 )n
( − s2 )n +ε 2
More generally, for a low-pass filter G( s) = QP((ss)) , where P( s) = Σ ni = 0 ai si and
Q( s) = Σ mi = 0 bi s i , with m > n. Then, s → 1s ⇒ P ( 1s ) = s1n Σ ni = 0 ai s n− i and Q ( 1s ) =
1
Σ mi = 0 bi s m− i. This leads to the following high-pass filter:
sm
Gˆ ( s) = G ( 1s ) = sm− n
a0 sn ++ an
b0 sm ++ bn
The physical realization is not difficult to construct. Indeed, it is enough to
observe that s → 1/s transforms an inductive reactance to a capacitive reactance and vice versa. Hence, replacing inductors with capacitors and vice
versa in a low-pass filter immediately leads to a high-pass filter.
□
References
1. Brown, J. and Churchill, R. 2011. Fourier Series and Boundary Value Problems. New
York: McGraw Hill.
2. Campbell, G. 1922. Physical theory of the electric wave-filter. Bell Syst. Tech. J.,
1(2): 1–32.
3. Carslaw, H. 1921. Introduction to the Theory of Fourier’s Series and Integrals.
London: Macmillian.
4. Dressel, M. and Grüner, G. 2002. Electrodynamics of Solids: Optical Properties of
Electrons in Matter. Cambridge: Cambridge University Press, UK.
5. Duff, W. 1988. Fundamentals of Electromagnetic Compatibility (Handbook Series on
Electromagnetic Interference and Compatibility) Vol. 1. Gainesville, VA: Interference
Control Technologies Inc.
6. Foster, R. 1924. A reactance theorem. Bell Syst. Tech. J., 3(2): 259– 267.
7. Hanna, R. and Rowland, J. 1990. Fourier Series, Transforms and Boundary Value
Problems. New York: Wiley-Interscience.
8. Hardin, K., Fessler, J., and Bush, D. 1994. Spread spectrum clock generation
for the reduction of radiated emissions. In Proceedings of the IEEE Int. Symp. on
Electromagn. Compat., 227–231.
K15149_Book.indb 65
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66
Electromagnetic Theory for Electromagnetic Compatibility Engineers
9. Hardin, K., Fessler, J., and Bush, D. 1995. A study of the interference potential
of spread spectrum clock generation techniques. In Proceedings of the IEEE Int.
Symp. on Electromagn. Compat., 624–629.
10. Jackson, J. 1962. Classical Electrodynamics. New York: John Wiley & Sons.
11. Johnson, H. and Graham, M. 1993. High-Speed Digital Design. Upper Saddle
River, NJ: Prentice Hall.
12. Hayt, W. and Kemmerly, Jr., J. 1978. Engineering Circuit Analysis. Sydney:
McGraw Hill.
13. LePage, W. 1961. Complex Variables and the Laplace Transform for Engineers. New
York: Dover.
14. Reed, M. and Simon, B. 1980. Methods of Modern Mathematical Physics. Vol. I:
Functional Analysis. New York: Academic Press.
15. Wylie, C., Jr. 1960. Advanced Engineering Mathematics. New York: McGraw-Hill.
K15149_Book.indb 66
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3
Boundary Value Problems in Electrostatics
No exposition on electrodynamics is complete without delving into some
basic boundary value problems encountered in electrostatics. Indeed, neither
would the exposition be complete if a cursory glimpse of multipole theory
were absent [1,5–8]. The former is crucial to EMC engineers in developing
an intuitive feel for real-world problems, and how simplifying a complicated
scenario via a toy model can greatly help resolve electromagnetic interference problems. The latter is useful in understanding the basis for various
rules of thumb employed by EMC engineers. Unfortunately, as is often the
case, sometimes EMC engineers apply these rules with reckless abandon
without being cognizant of the origins of the rules.
Finally, the power of the method of images is developed further in this
chapter. The technique is particularly useful for solving many problems
encountered by EMC engineers. In particular, for 2-dimensional problems,
utilizing techniques in complex variables [2,10] also come in very handy,
and EMC engineers are encouraged to review the theory of analytic functions to solve two-dimensional Laplace equations encountered in electrostatics and magnetostatics.
3.1 Electromagnetic Boundary Conditions
A brief summary of electromagnetic boundary conditions is collected here
for ease of reference. The derivations can be found in Section A.3 of the
Appendix. These conditions are utilized in subsequent sections to solve
boundary value problems. By way of establishing some notations, let Ω ± ⊂ R 3
be two connected open sets such that ∂Ω0 = Ω+ ∩ Ω− is a 2-dimensional surface. Given the pair (Ω ± , ε ± , µ ± , σ ± ), where ε ± is the electric permittivity, µ ±
is the magnetic permeability, and σ ± is the conductivity on Ω ± consider an
electric field E− in Ω− incident on ∂Ω0 , and the resultant transmitted field
E+ in Ω+ . Finally, the unit normal vector field n± on ∂Ω0 is defined to be
directed into Ω  and let ρ0 denote the free surface charge density and J 0 the
surface current density on ∂Ω0 .
67
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68
Electromagnetic Theory for Electromagnetic Compatibility Engineers
3.1.1 Theorem
Suppose σ ± = 0 and ρ0 = 0 . Then, on setting D± = ε ± E± , the following conditions hold on ∂Ω0 :
(a) n+ ⋅ ( D− − D+ ) = 0
(b) n+ × ( E− − E+ ) = 0
□
Condition 3.1.1(a) states that for lossless dielectrics, the normal component of the electric displacement field is continuous across the boundary; by definition then, the normal component of the electric field across
the interface must be discontinuous. In contrast, 3.1.1(b) asserts that the
tangential component of the electric field across the boundary interface
is continuous. A similar interpretation holds for the subsequent results
stated below.
3.1.2 Corollary
Suppose σ ± ≠ 0, and let J ± = σ ± E be a non-time–varying current density in
Ω ± . Then, on ∂Ω0 , ρ0 ≠ 0 and
(a) n+ ⋅ ( D− − D+ ) = ρ0
(b) n+ ⋅ ( J − − J + ) = 0
(c) n+ ×
(
1
σ−
)
1
J − − σ+
J+ = 0
(d) σ + → ∞ ⇒ n+ × E− = 0 and n+ ⋅ D− = ρ0
□
3.1.3 Theorem
Let σ ± = 0 and J 0 = 0 . Then, the following conditions hold on ∂Ω0 .
(a) n+ ⋅ (B− − B+ ) = 0
(b) n+ ×
(
1
µ−
)
B− − µ1+ B+ = 0
□
3.1.4 Corollary
Suppose σ ± ≠ 0. Then, on ∂Ω0 , J 0 ≠ 0 and
(a) n+ × ( µ1− B− − µ1+ B+ ) = J 0
(b) σ + → ∞ ⇒ n+ × µ1− B− = J 0
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and n+ ⋅ D− = 0
□
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Boundary Value Problems in Electrostatics
69
3.1.5 Remark
A surface charge density ρ0 and a surface current density J 0 are idealizations that do not truly exist across a boundary interface. More precisely,
∃δ > 0 sufficiently small, and some small open neighborhood N δ ⊂ Ω+ ∪ Ω−
such that ∂Ω0 ⊂ N δ and N δ ⊆  x∈∂Ω0 Bδ ( x), wherein a differential volume
charge density and volume current density exist in N δ , instead of an idealized surface charge and current densities on ∂Ω0 . Notwithstanding, for
convenience, surface charge and current densities are used without further
ado. Technically, they only exist when Ω+ is a perfect conductor.
As a final comment, observe from Theorem 3.1.1 and Corollary 3.1.2 that for
lossy dielectrics, at the interface between two media, the continuity requirement is dependent upon the respective conductivities of the media and not
on the electric permittivities under steady-state conditions. Indeed, for lossy
media, the electric permittivities determine the charge density accumulated
at the interface. Note that for general lossy dielectric media, there exists a
transient response, depending upon the charge relaxation times before the
steady-state condition is attained.
Recall from Equation (1.20) that charges placed in a dielectric
will
−t
quickly diffuse to the surface according to ρ(r , t) = ρ0 (r , t)e τ , where τ ≡ σε
is called the charge relaxation time of the dielectric. More specifically, for
time t << τ (charge relaxation time), the dielectric constants determine the
electric field profile. Once charge density accumulates on the interface, the
electrical conductivities of the media dictate how the electric field behaves
in the media; more details follow shortly.
Indeed, it is evident that for an ideal (lossless) dielectric, free charges
injected into the dielectric will remain where they are: they will not diffuse
onto the boundary as τ → ∞. On the contrary, for good conductors, 0 < τ << 1
and hence, charges injected into a conductor will diffuse very rapidly onto
the boundary of the conductor. The response of the fields within a lossy
dielectric for τ > 0 clearly fall under two categories: (i) transient response
when t ≤ τ, and (ii) steady-state response when t >> τ. Transient response is
usually complicated to solve whereas the steady state is somewhat easier.
Examples of steady-state response are given below.
3.1.6 Corollary
Consider a domain Ω ⊂ R 3 such that Ω = (Ω+ , ε + , σ + ) ∪ (Ω− , ε − , σ − ),
Ω+ ∩ Ω− = ∅ , and Γ = Ω+ ∩ Ω− ≠ ∅ is the boundary interface between the
two media. Finally, set ∂Ω = Γ + ∪ Γ − , where ∂Ω ± = Γ ± ∪ Γ . Suppose some
fixed potential φ is applied at the boundary of Ω:
 V+ on ∂Ω+

ϕ=
 V− on ∂Ω−

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70
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Then, the charge ρ on Γ satisfies ρ = (τ − − τ + ) J + , ⊥ , where τ ± = σε ±± is the respective charge relaxation time in Ω ± and J + , ⊥ = J + ⋅ n+ is the normal component
of the current density.
Proof
By definition, ρ = n+ ⋅ ( D− − D+ ) = n+ ⋅ (ε − E− − ε + E+ ) = n+ ⋅ (−ε − ∇ϕ − + ε + ∇ϕ + ) on
Γ. However, by Corollary 3.1.2, 0 = n+ ⋅ ( J − − J + ) = n+ ⋅ (−σ − ∇ϕ − + σ + ∇ϕ + ) on Γ
implies that n+ ⋅ J − = n+ ⋅ J + ⇒ ∂ n+ ϕ − = σσ +− ∂ n+ ϕ + and hence,
(
ρ = n+ ⋅ ∇ϕ + −ε −
σ+
σ−
) (
+ ε + = − σε−− +
ε+
σ+
)σ
+
n+ ⋅ ∇ϕ +
Now, by construction, n+ ⋅ J + = − J + , ⊥ ⇒ ρ = (τ + − τ − ) J + , ⊥ , as claimed.
□
It is thus clear from the above result that ρ = 0 ⇔ τ + = τ − in lossy media.
That is, lossy media appear to behave like lossless dielectric media whenever
their respective charge relaxation times should coincide.
3.1.7 Remark
As a concrete example, consider the domain Ω = (Ω+ , ε + , σ + ) ∪ (Ω− , ε − , σ − ),
where Ω− = (0, c) × (0, a] and Ω+ = (0, c) × [ a, b), for some 0 < a < b . Suppose
the boundary conditions are:
 V0 for y = b ,
 0 for x = a


ϕ=
and ∂ x ϕ = 
 0 for y = 0,
 0 for x = 0


Consider the Laplace solution ∆ϕ 0 = 0 on Ω for the case wherein σ ± = 0,
that is, the lossless case. Finally, let ∆ϕ ∞ = 0 on Ω be the solution for the
steady-state case. Then, for
0 ≤ t << min
{
ε+
σ+
, σε−−
}
the general solution ϕ ≈ ϕ 0 , whereas for
t >> max
{
ε+
σ+
, σε−−
}
ϕ ≈ ϕ ∞ . The solution that lies between the two extremes is technically a
transient solution, and it defines a smooth homotopy between ϕ 0 and ϕ ∞ as
the charge density build-up reaches a maximum on the interface at y = a if
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71
Boundary Value Problems in Electrostatics
ε+
σ+
≠ σε−− . The transient solution can be solved numerically by following coupled equations subject to the above boundary conditions:

−∆ϕ + µ 0 ε ∂t2 ϕ = ρε


 ∂t ρ = σ∆ϕ − σµ 0 ε ∂t2 ϕ

This is easily derived via Gauss’ law and the charge conservation equation,
and exploiting the Lorentz gauge: ∇ ⋅ A + µε ∂t ϕ = 0 . Indeed, the above equation yields the complete solution.
Explicitly, invoking Gauss: ε∇ ⋅ E = ρ ⇒ −ε∆ϕ − ε ∂t ∇ ⋅ A = ρ. Moreover, via
charge continuity: ∂t ρ = −∇ ⋅ J = σ∆ϕ + σ ∂t ∇ ⋅ A . Finally, applying the Lorentz
gauge yields the pair of partial differential equations.
Indeed, it is quite clear from the above discussion that even if the relaxation times are the same, σε++ = σε−− , the field will nevertheless deform from a
pure dielectric solution to that of a steady-state conductive solution. The difference is the absence of charge density on the media interface.
3.2 Image Theory Revisited
In Chapter 1, image theory was introduced to solve some electrostatic problems. In this section, this method is developed in some depth to help EMC
engineers apply the technique to product development and research. The
material herein is organized in a series of examples with various methods
demonstrated for ease of reference.
3.2.1 Example
As a first example, the method of inversion is utilized to find the potential
induced by a point charge outside a grounded conducting solid sphere. Let
B( a) = {( x , y , z) ∈R 3 : x 2 + y 2 + z 2 < a 2 } be a perfect, electrical, conducting solid
sphere located at the origin, and a point charge Q ≠ 0 located at rQ = ( xQ , 0, 0)
without any loss of generality, where xQ > a . See Figure 3.1.
To begin, consider some image charge Q ′ in B(a) located at rQ′ = ( xQ′ , 0, 0)
such that the potential ϕ|B( a) = 0 . Set
ϕ(r ) =
K15149_Book.indb 71
1
4 πε 0
{
Q
r
+ Qr′′
}
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
r'
a
2
r' = ar
r
0
Figure 3.1
Transformation of inversion in a sphere.
where
r = r − rQ
and r ′ = r ′ − rQ
for arbitrary point r = (x, y, z). Then,
ϕ|∂B( a) = 0 ⇒ Qr + Qr′′ = 0 ∀r ∈∂B( a)
Does the pair (Q′ , r ′) exist?
2
Define a mapping ζ : B( a) → R 3 − B( a) by r  r ′ ≡ ar2 r , called the inversion
in a sphere. In short, ζ maps the origin into a point at infinity. Very briefly,
the properties [2,4,9] may be summarized as follows: (a) planes are mapped
into spheres tangent at the origin of the inversion, (b) spheres are mapped
into spheres, (c) points within the circle are mapped into points outside the
circle, and (d) points on the circle are mapped onto themselves. Some insight
into this mapping can be found in Exercise 3.6.1.
Q
Q′
Under the inversion transformation, ∃ (Q ′ , r ′) such that r + r ′ = 0 ∀r ∈∂B( a).
Thus,
Q ′ = − Qa rQ′ = −Q rQa
is the image charge within the sphere such that φ|∂B(a) = 0. Thus, for an arbitrary point r ∈R 3 − B( a), the potential is given by
ϕ( r ) =
1
4 πε 0
{
Q
R
+ QR′′
}
where
R = ( x − xQ )2 + y 2 + z 2
and R ′ = ( x − xQ′ )2 + y 2 + z 2
In spherical coordinates, given an arbitrary point (r , φ, θ) ∈R 3 − B( a) , under
2
the inversion mapping, (r , θ, φ)  ar , θ, φ . Let ϑ denote the angle between r
(
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)
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73
Boundary Value Problems in Electrostatics
and rQ = (rQ , 0, π2 ). Then, R = r 2 + rQ2 − 2 rrQ cos ϑ and R′ = r 2 +
a4
rQ2
2
− 2 r raQ cos ϑ .
In particular, executing a minor algebraic manipulation leads to
R′ =
a
rQ
Q
4 πε 0




( )
rQ r 2
a
+ a 2 − 2 rQ r cos ϑ
and hence,
ϕ(r , ϑ) =
1
r 2 + rQ2 − 2 rQr cos ϑ
−



2
Qr
2
+
a
−
2
r
r
cos
ϑ
( a ) Q 
1
(3.1)
r ⋅r
on R 3 − B( a), where cos ϑ = rQQr = sin φ cos θ . Show this in Exercise 3.6.3.
Finally, from Equation (3.1), it is clear that the Green’s function for a unit
charge external to a grounded conducting sphere is
G(r , rQ ) =
−
1
r 2 + rQ2 − 2 rQ r cos ϑ
In particular, if rQ = (rQ , θQ , φQ ) ,
cos φ cos φQ . See Exercise 3.6.2.
then
2
1
(3.2)
 rQ r  + a 2 − 2 r r cos ϑ
Q
 a 
cos ϑ = sin φ sin φQ cos(θ − θQ ) +
□
3.2.2 Example
Suppose the above conducting solid sphere B( a) is charged at some constant
potential ϕ 0 . What is the resultant potential in R 3 − B( a)?
First, it is enough to observe that as B( a) is a perfect conductor; it follows
that ϕ|B( a) = ϕ 0 , and hence, may represent ϕ 0 by some charge q located at the
center of B( a). Now, observe further that as the external charge Q induces an
image charge Q ′ in B(a), it follows that the replacement charge q → q − Q ′
must be made so that the resultant charge in B(a) in the presence of Q is q.
Then, by the superposition principle, it follows immediately that on
R 3 − B( a),
ϕ(r , ϑ) =
Q
4 πε 0




1
r 2 + rQ2 − 2 rQr cos ϑ
−


+
2
 rQ r  + a 2 − 2 r r cos ϑ
 a 
Q

1
1 q − Q′
4 πε 0 r
At r = a,
ϕ(r , ϑ) =
K15149_Book.indb 73
Q
4 πε 0
×0+
1 q −Q′
r
4 πε 0
= ϕ0
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74
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Hence, q − Q ′ = 4πε 0 ϕ 0 a; and substituting this into the above equation yields
ϕ(r , ϑ) =
Q
4 πε 0




1
r 2 + rQ2 − 2 rQr cos ϑ
−


 + ϕ0
2
 rQ r  + a2 − 2 r r cos ϑ
Q

 a 
1
a
r
(3.3)
As a quick verification, it is seen that by construction, {}|r = a = 0 and
hence, ϕ( a, ϑ) = ϕ 0 ∀ϑ , as required.
□
3.2.3 Example
Consider the space Ω = {( x , y , z) ∈R 3 : 0 < y < a} , for some constant a > 0, and
suppose a point charge Q is located at r0 = ( x0 , y 0 , c) ∈Ω. Suppose ∂Ω =
∂R 3+ ∪ ∂Ω a are perfect electrical conductors, where ∂Ω a = {( x , y , a) : x , y ∈R}.
Solve the Dirichlet boundary value problem:
 −∆ϕ = 1 Qδ 3 (r − r ) on Ω
0
ε0



ϕ = 0 on ∂Ω

Now, as lim ϕ → 0 , it follows that [10, p. 226] if a Green’s function G for the
x , y →±∞
boundary value problem can be determined, then the solution is given by
ϕ( r ) =
∫
Ω
Q
ε0
δ 3 (r − r ′) G (r , r ′)d 3 r ′ =
Q
ε0
G(r , r0 )
To find the Green’s function for Ω, it suffices to consider the space
Ω′ = R 2 × (0, 1) and construct a Green’s function G′ on Ω′. Toward this end,
consider Figure 3.2.
z=3
z=2
z=1
z=0
z = –1
z = –2
–1
2+c
+1
1+c
–1
c
+1
–c
–1
–(1 + c)
+1
–(2 + c)
–1
Figure 3.2
Infinite sequence of image charges induced by a unit charge.
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75
Boundary Value Problems in Electrostatics
Referring to Figure 3.2, set a unit charge at r = ( x , y , c), and consider an
image charge placed at ( x , y , − c), that is, the reflection across the x–y plane
E 0 = R 2. The presence of the image charge preserves the boundary condition
on E 0. Next, consider the reflection across the affine plane E1 = R 2 + {(0, 0, 1)}
independent of E 0 at ( x , y , 2 − c). Placing an image charge at ( x , y , 2 − c) preserves the boundary condition at E1. However, the presence of an image
charge at ( x , y , 2 − c) breaks the boundary condition on E 0 ; hence, a third
image charge must be placed at ( x , y , c − 2) . Likewise, to preserve the boundary condition of E1 as the result of the second image charge across E 0 , a
fourth image charge must be placed at ( x , y , 2 + c). By induction, the process
continues indefinitely, yielding a sequence of image charges at the following
locations:
rk = ( x , y , 2 k + c) and rk = ( x , y , 2 k − c) ∀k ∈ Z
See Figure 3.2.
Thus, on setting
δrk = r ′ − rk = ( x ′ − x)2 + ( y ′ − y )2 + ( z ′ − c − 2 k )2
and
δrk = r ′ − rk = ( x ′ − x)2 + ( y ′ − y )2 + ( z ′ + c − 2 k )2
and observing that for any fixed z′ ∈Z and c ∈ (0,1),
{ z′ + c + 2 k : k ∈ Z} = { z′ − c + 2 k : k ∈ Z} ⇒ ∑ k∈Z { δ1r
k
}
− δ1rk = 0
This suggests at once that
G′(r , r ′) =
1
4π
∑ {
k ∈Z
1
δrk
−
1
δrk
}
is the sought-for Green’s function on Ω′. Moreover, as
1 1
4π r
is the Green’s function for a point charge, that is, −∆ 1r = 4πδ(r ) , and G′(r , r ′)
satisfies the boundary conditions at z ′ = 0, 1, the uniqueness of Poisson’s
equation implies immediately that G′(r , r ′) is the required Green’s function
on Ω′.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
The generalization to Ω is trivial. Transform G′(r , r ′) → G(r , r ′) under the
transformation 2 k → 2 ka . Then, clearly,
{ c + 2 ka : k ∈Z} = { − c + 2 ka : k ∈Z} ⇒ ∑ k ∈Z { δR1
k
−
1
δRk
}= 0
(3.4)
where
δRk = r ′ − Rk = ( x ′ − x)2 + ( y ′ − y )2 + ( z ′ − c − 2 ka)2
δRk = r ′ − Rk = ( x ′ − x)2 + ( y ′ − y )2 + ( z ′ + c − 2 ka)2
with z ′ = 0 . Finally, note that when z ′ = a , Equation (3.4) is once again
satisfied: the construction would fail if k ∈N . The required Green’s function on Ω is thus
∑ {
1
4π
G′(r , r ′) =
k ∈Z
1
δRk
−
1
δRk
}
(3.5)
□
This section ends with two more examples regarding the application of the
method of images: determine the potential resulting from a charged cylinder
over a ground plane and a charged conductive sphere over a lossless dielectric half-space.
3.2.4 Example
Consider a cross-section of an infinitely long conducting cylinder C of
radius a > 0 over a conducting ground plane ∂R 2+ = {( x , y ) ∈R 2 : y = 0} : C =
{( x , y ) ∈R 2 : x 2 + ( y − y 0 )2 ≤ a}, where y 0 > a > 0, and set Ω = R 2+ − C. By construction, the center of the cylinder is (0, y 0 ) . Suppose that the cylinder is set
at a potential ϕ = ϕ 0 , determine the potential in Ω.
Consider a line charge λ located at (0, y + ) ∈C , where the pair (λ , y + ) are to
be determined. By the method of images, consider a line charge density −λ
located at y = (0, − y + ). Recall that the potential in R 2+ − {(0, y + )} is given by
ϕ( x , y ) = −
λ
ln
2 πε 0
x 2 + ( y − y+ )2
(3.6)
x 2 + ( y + y+ )2
To see Equation (3.6), consider, for simplicity, a unit charge at the center of a
ρ
disk Br (0) = {( x , y ) : x 2 + y 2 ≤ r 2 } of radius r. Invoking Gauss’ law, ∇ ⋅ E = ε0 ⇒
δ( x )
−∆ψ = ε0 . By Stokes’ theorem,
−
K15149_Book.indb 76
∫
Br (0)
∆ψd 2 x = −
∫
∂ Br (0)
∇ψ ⋅ er d 2 x = −
∫
2π
0
∂r ψr dθ =
∫
δ( x )
ε0
dx =
1
ε0
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77
Boundary Value Problems in Electrostatics
M = M0
C
Ω
λ
y0
y+
–y+
Mirror image
–y0
–λ
Figure 3.3
Potential generated by a charged infinitely long conducting cylinder.
whence
− ∂r ψ =
1
2 πε 0 r
⇒ ψ = − 2 πε1 0 ln r
and the potential resulting from two line charges yields
ϕ( x , y ) = − 2 πελ 0 ln x 2 + ( y − y + )2 +
λ
2 πε 0
ln x 2 + ( y + y + )2
establishing Equation (3.6). See Figure 3.3 for an intuitive explanation.
Observe that as C is a perfect conductor, it forms a surface of equipotential.
Hence, it suffices to determine surfaces of equipotential defined by the line
charge density. From (3.6), let S(V ) ⊂ R 2+ define an equipotential surface such
that
V = − 2 πελ 0 ln
x 2 + ( y − y+ )2
x 2 + ( y + y+ )2
for all (x,y) ∈ S(V), where V is a constant potential. Set K = e−2 πε0V /λ . Then,
by definition,
K2 =
x 2 + ( y − y+ )2
x 2 + ( y + y+ )2
⇔ x 2 + y 2 + 2 yy +
K2 +1
K2 −1
+ y +2 = 0
Noting that
(y + y
K15149_Book.indb 77
K2 +1
+ K2 −1
)
2
= y 2 + 2 yy +
K 2 +1
K 2 −1
(
+ y+
K 2 +1
K 2 −1
)
2
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78
Electromagnetic Theory for Electromagnetic Compatibility Engineers
it follows clearly that the above equation reduces, after some algebraic
manipulation, to
(
x 2 + y + y+
K 2 +1
K 2 −1
) = (y
2
K 2 +1
+ K 2 −1
)
2
− y +2 =
(
2K
K 2 −1
y+
)
2
(3.7)
However, this is nothing but the equation of a circle centered at
( 0, −
K2 +1
K2 −1
y+
)
below the ground plane and via reflection, at
( 0,
K2 +1
K2 −1
y+
)
above the ground plane. Set
K2 +1
K2 −1
y0 =
y+
Then (0, y 0 ) is the location of the line charge above the ground plane such that
C is at equipotential.
It thus remains to determine λ = λ(ϕ 0 ). From Equation (3.7), it follows
2
immediately that a 2 = K22K− 1 y + . Now, observe that
(
)
a 2 + y +2 = y +2
{
4 K 2 + ( K 2 − 1)2
( K 2 − 1)2
Hence, y + = y 02 − a 2 . Next, set α =
α=
K2 +1
K2 −1
y0
y+
}= y ( ) = y
2
+
K2 +1
K2 −1
2
2
0
. Then,
⇒ K2 =
α+ 1
α− 1
⇒K=
α+ 1
α− 1
whence
K = e−2 πε0V /λ ⇒ λ = −
as required.
2 πε 0 ϕ 0
+1
ln α
α−1
⇒ϕ= −
ϕ0
ln
α+ 1
α− 1
ln
x 2 + ( y − y+ )2
x 2 + ( y + y+ )2
□
3.2.5 Example
Let R 3− = {( x , y , z) ∈R 3 : z ≤ 0} denote an infinite dielectric half-space with
electric permittivity ε − and R 3+ = R − R 3− denote a pure dielectric medium of
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79
Boundary Value Problems in Electrostatics
electric permittivity ε + . Suppose Da ( r0 ) = {( x , y , z) ∈Ω : x 2 + y 2 + ( z − z0 )2 ≤ a 2 }
is a charged PEC sphere of charge Q, the center of which is r0 = (0, 0, z0 )
above the dielectric plane, where z0 > a > 0 . Determine the potential field in
Ω = R 3+ − Da (r0 ) .
Because Da (r0 ) is a perfect conductor, its surface is at equipotential. Hence,
suppose without loss of generality that ∃Q0 is such that V0 is the potential
on the surface of Da (r0 ). Then, by Exercise 3.5.8,
+
∃Q1 = − εε−− −ε
+ε + Q0
3
located at r1 = (0, 0, z1 ) , with z1 = − z0 , such that the potential on ∂R − is zero.
However, the presence of Q1 violates the equipotential condition on ∂Da (r0 ).
Hence, appealing to Example 3.2.1, ∃Q2 = − 2 az0 Q1 at r2 = (0, 0, z2 ) , where
z2 = z0 − 2 az0 , such that ∂Da (r0 ) is once again an equipotential surface.
It is clear by now that the presence of Q2 breaks the equipotential condition on ∂R 3− . Thus, a charge
+
Q3 = − εε−− −ε
+ε + Q2
must be placed at r3 = (0, 0, z3 ) , where z3 = − z2 , such that the equipotential
condition on ∂R 3− is restored. As this in turn violates the equipotential condition on ∂Da (r0 ) , another fictitious charge
Q4 =
a
{
}
2 z4 + a 2 2 z0 − a 2/(2 x0 )
−1
Q3
at
r4 = (0, 0, z4 )
where
z 4 = 2 z0 −
a2
2 z0 − a2 /(2 z0 )
For notational convenience, set
α=
ε − −ε +
ε − +ε +
, d0 = 2 z0
and d1 = d0 −
a2
d0
Then, Q1 = −αQ0 , Q2 = − da0 Q1 , Q3 = −αQ2 , and Q4 = − da1 Q3 , where d2 = d0 −
with z2 = d1 , z3 = − z2 , and z4 = d2 . By induction, it is clear that
Q2 k − 1 = −αQ2 k − 2
and
z2 k − 1 = − z2 k − 2
and
a2
d1
,
Q2 k = − d2 ka− 1 Q2 k − 1 for all k = 1,2,…
z2 k = d2 k − 1 for all k = 1,2,…
and Q = Σ k ≥0Q2 k by construction.
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80
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Sphere of charge Q
Q0
z0
Q2 z2
ε+
0
ε–
Ω
z=0
Q 3 z3
Q1 z1 = –z0
Figure 3.4
Method of images applied to a charged conducting sphere over a dielectric plane.
Indeed, further simplification can be achieved as
(
)
Q2 k + 1 = −αQ2 k = −α − d2 ka− 1 Q2 k − 1 =  = (−1)2 k + 1 α k + 1
ak
d2 k − 1d2 k − 2 d0
Q0
k +1 k
= − d2 k −α1d2 k −a2 d0 Q0
(
)
Q2 k = − d2 ka− 1 Q2 k − 1 = −α − d2 ka− 1 Q2 k − 1 =  = (−1)2 k α k
=
k k
α a
d2 k − 1d2 k − 2 d0
ak
d2 k − 1d2 k − 2 d1
Q0
Q0
Thus, given an arbitrary point r = (x,y,z) ∈ Ω, by the superposition principle,
the electric potential field φ on Ω is given by the summation of all the charges
ϕ=
1
4 πε +
∑
k ≥0
=
Q0
4 πε +
∑
k ≥0
{
Q2 k
x 2 + y 2 + ( z − z2 k )2
α k ak
d2 k − 1d2 k − 2 d0
{
+
Q2 k + 1
x 2 + y 2 + ( z − z2 k + 1 )2
1
x 2 + y 2 + ( z − z2 k )2
−
}
α
x 2 + y 2 + ( z − z2 k + 1 )2
}
See Figure 3.4 for the mathematical representation of fictitious charges.
□
3.3 Multipole Expansion
It is clear by now that the scalar potential resulting from a localized static
charge density ρ in R 3 satisfies the Poisson equation −∆ϕ(r ) = ρ(εr ) ; see Exercise
3.6.5. From Section A.4, it is seen that the Green’s function for a unit point
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81
Boundary Value Problems in Electrostatics
charge is G(r − r ′) = 41π
Poisson’s equation is
1
r −r′
. Now, in R 3 , lim ρ(r ) = 0, whence the solution to
r →∞
ϕ(r ) =
1
4 πε 0
∫
ρ( r ′ )
r −r′
(3.8)
d3 r ′
in free space, where a charge density ρ : R 3 → R with nonempty compact
support, Ωρ = supp(ρ) is assumed in all that follows.
1
Now, observe that for r >> r ′ , the denominator r − r ′ can be Taylor
expanded as follows. Set f (r ) = r −1r ′ . Then, r − r ′ = r 2 + r ′ 2 − 2 r ′r cos θ ,
where cos θ = rrr⋅r ′ . This can be rewritten as
r − r ′ = r 1 + ( rr′ ) − 2 rr′ cos θ ≡ r 1 + ξ
2
where ξ = ( rr′ ) − 2 rr′ cos θ and ξ << 1, whence, Taylor expanding about
1 yields
2
1
{1 + ξ}− 2 = 1 − 21 ξ + 38 ξ 2 − 165 ξ 3 + o(ξ 4 )
That is,
f (r ) =
1
r
{1 +
r′
r
cos θ + ( rr′ )
2 3 cos 2 θ− 1
2
+o
(( ) )}
r′ 3
r
However, this is just the spherical harmonic expansion [5,8,9]:
f (r ) =
1
r
∑
∞
n= 0
Pn (cos θ) ( rr′ )
n
(3.9)
for r > r ′ , where
Pn ( x) =
∑
[ n]
( −1)k (2 n − k )!
n
k = 0 2 k !( n − k )!( n − 2 k )!
x n− 2 k
defines the Legendre polynomials of order n, with


[n] = 


K15149_Book.indb 81
n
2
, for n even
n− 1
2
, for n odd
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82
Electromagnetic Theory for Electromagnetic Compatibility Engineers
It can be expressed more compactly via the Rodrigues’ formula
Pn ( x) =
dn
1
2 n n! dx n
( x 2 − 1)n
The first few terms are evaluated as
P0 ( x) = 1
P1 ( x) = x
P2 ( x) = 21 (3 x 2 − 1)
P3 ( x) = 21 (5 x 3 − 3 x),
For completeness, consider the scenario wherein r < r ′ ; physically, this corresponds precisely to the case where the point is located within the charge
body Ωρ. It can be shown that
f (r) =
1
r′
∑
∞
n= 0
Pn (cos θ) ( rr′ )
n
(3.10)
Together, Equations (3.9) and (3.10) are often expressed in a more compact
fashion as
f (r ) =
1
r>
∑
∞
n= 0
Pn (cos θ)
( )
r< n
r>
(3.11)
where r> = max{r , r ′} and r< = min{r , r ′}.
3.3.1 Definition
The multipole expansion of ϕ(r ) = ∫ Ωρ
ϕ(r ) =
1
4 πε r
∑ ∫
∞
n= 0
Ωρ
ρ( r ′ )
r −r′
d 3 r ′ , for r > r ′ , is defined by
ρ(r ′)Pn (cos θ) ( rr′ ) d 3 r ′
n
(3.12)
The first term corresponds to an electric monopole, the second term a dipole,
the third term a quadrupole, and so forth.
3.3.2 Remark
Notice that whereas the monopole falls off as 1r , the dipole falls off as
1
, the quadrupole falls off as r13 , and so on. It is thus obvious that for
r2
an arbitrary compact source Ωρ , if the distance r away from the source
satisfies r >> diam(Ωρ ), where diam(Ωρ ) denote the diameter of Ωρ (i.e.,
the largest side of Ωρ or diagonal, whichever is larger), and is defined by
diam(Ωρ ) = max { x − x ′ : x , x ′ ∈Ωρ } , then Ωρ may be approximated by a
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83
Boundary Value Problems in Electrostatics
point source plus a small correction term from the dipole contribution that
rapidly becomes negligible for very large r. On the other hand, close to the
charge distribution Ωρ , the higher-order poles dominate, and in particular,
it is incorrect to approximate the extended source by a point source. This is
evident from Examples 3.2.1 and 3.2.5.
From the definition of the magnetic potential
A(r ) =
µ0
4π
∫
J (r ′)
r −r′
d3 r ′
it follows from the above discussion mutatis mutandis that multipole expansion for the magnetic field B = ∇ × A for r > r ′:
A(r ) =
µ0
4π r
∑ ∫ J(r ′)P (cos θ)(
∞
n
n= 0
)
r′ n
r
d3 r ′
(3.13)
Finally, to complete the discussion, consider the special case wherein the
charge (or current) is distributed on the surface of some compact set S. This
leads to the introduction of spherical harmonics Ynm (θ, φ) and they are defined
as follows. First, the associated Legendre polynomials are given by
m
Pnm ( x) = (−1)m (1 − x 2 ) 2
dn + m
dx n + m
( x 2 − 1)n
An important property satisfied by the polynomials is the orthogonality relation:
∫
1
−1
Pkm ( x)Pnm ( x)d x =
2 ( n + m)!
2 n + 1 ( n − m)!
δ kn
and
 1 for i = j

δ ij = 
 0 for i ≠ j

is the Kronecker-delta function. Then, the normalized spherical harmonics are
defined by
Ynm (θ, φ) =
2 n + 1 ( n − m )!
4 π ( n + m )!
Pnm (cos θ)eimϕ
∗
and Yn, − m (θ, φ) = (−1)m Ynm
(θ, φ)
satisfying the orthogonality relation ∫ 20 π dφ ∫ 0π sin θdθYn∗′m′ (θ, φ)Ynm (θ, φ) =
δ n′nδ m′m . In particular, one has the completeness relation:
∑ ∑
K15149_Book.indb 83
∞
n
n= 0
m =− n
∗
Ynm
(θ′ , φ′)Ynm (θ, φ) = δ(φ − φ′)δ(cos θ − cos θ′)
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84
Electromagnetic Theory for Electromagnetic Compatibility Engineers
As mentioned before, an immediate application of spherical harmonics is
to expand a function f : ∂S → R defined on the boundary of some solid sphere
S ⊂ R 3:
f (θ, φ) =
∑ ∑
∞
n
n= 0
m =− n
AnmYnm (θ, φ)
where the coefficients Anm are defined via the orthogonal relation as
Anm =
∫
2π
0
dφ
∫
π
0
sin θ dθYn∗′m′ (θ, φ) f (θ, φ)
Here, f could represent a boundary value or surface charge. This concludes a
brief sketch on the topic of spherical harmonics.
3.4 Steady-State Currents
Recall from the conservation of charge that ∂t ρ + ∇ ⋅ J = 0 in some domain
Ω. Hence, when the charge density is constant, that is, ∂t ρ = 0 ⇒ ∇ ⋅ J = 0 on
Ω, this defines the condition for steady-state current. The intuitive notion
behind this definition is clear: when a constant current flows through a wire,
d
dt I = 0 and hence, the current density is independent of time.
3.4.1 Example
Consider two rectangles (Ω ± , ε ± , σ ± ), where Ω+ = (0, a) × (b− , b+ ), Ω− =
(0, a) × (0, b− ), and ε ± are the respective electric permittivities, and σ ± the
respective conductivities. Set Ω = Ω− ∪ Ω+ ∪ Γ , where Γ = {( x , b− ) : 0 < x < a}.
Find the potential ϕ ∈C 2 (Ω) ∩ C 0 (Ω) satisfying the following Dirichlet boundary value problem:
∆ϕ = 0 on Ω
 V for y = b
+
 +

 V− for y = b−
ϕ=
 0 for x = 0

 0 for x = a

K15149_Book.indb 84
(3.14)
(3.15)
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85
Boundary Value Problems in Electrostatics
First, observe that under steady-state conditions, J + ⋅ e y = J − ⋅ e y on Γ,
where J ± = σ ± E± is the current density defined on Ω ± and E± is the electric
field on Ω ± . Set ϕ ± to be the potential on Ω ± . Then, on Γ, σ − ∂ y ϕ − = σ + ∂ y ϕ + .
From Exercise 3.6.4, the general solution on Ω ± is given by
ϕ ± (x, y) =
∑ {α
n> 0
±
n
}
cosh λ n y + β n± sinh λ n y sin λ n x
(3.16)
where λ n = πan , n = 1, 2, 
Now, the coefficients α n± , β n± are evaluated by imposing Equation (3.15) on
(3.16). Proceeding systematically, consider y = b on Ω+ :
V+ = ϕ + ( x , b) =
∑ {α
n> 0
+
n
}
cosh λ nb + β n+ sinh λ nb sin λ n x
Thus, multiplying both sides by sin λ m x and integrating along (0, a) yields
α +n cosh λ nb + β n+ sinh λ nb = V+
∫
a
0


sin λ n xdx = 

4V+
nπ
for n odd
0 for n even
(3.17a)
Likewise, for y = 0 on Ω–:
α −n cosh λ nb = V−
∫
a
0


sin λ n xdx = 

4V−
nπ
for n odd
(3.17b)
0 for n even
Furthermore, from the continuity of the potential across a boundary,
{α
+
n
}
{
}
cosh λ n a + β +n sinh λ n a sin λ n x = α n− cosh λ n a + β n− sinh λ n a sin λ n x ∀x
That is,
α +n cosh λ n a + β +n sinh λ n a = α −n cosh λ n a + β −n sinh λ n a
(3.17c)
and from the continuity of the normal component of the current density,
σ+
σ−
K15149_Book.indb 85
{α
+
n
}
sinh λ n a + β +n cosh λ n a = α n− sinh λ n a + β n− cosh λ n a
(3.17d)
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Thus, via Equation (3.17), the coefficients can be determined. After some
tedious but trivial routine manipulations, the coefficients to (3.16) are:
α +2 n − 1 =
4
(2 n − 1) π
{
σ+
σ−
{V (1 − ) coth λ
+
σ+
σ−
2 n− 1
a−
2 V−
sinh 2 λ 2 n−1b
}×
( sinh λ 2 n− 1a − cosh λ 2 n− 1a coth λ 2 n− 1b ) − cosh λ 2 n− 1a ( coth λ 2 n− 1a − coth λ 2 n− 1b )}
β 2+n − 1 =
− α +2 n − 1 coth λ 2 n − 1b
4V+
1
(2 n − 1) π sinh λ 2 n−1b
β −2 n − 1 = α +2 n − 1 ( coth λ 2 n − 1a − coth λ 2 n − 1b ) +
α −2 n − 1 =
−1
4
(2 n − 1) π
(
V+
sinh λ 2 n−1b
−
V− coth λ 2 n−1 a
cosh λ 2 n−1b
)
4V−
1
(2 n − 1) π cosh λ 2 n−1b
In particular, the solution on Ω is given by



ϕ( x , y ) = 


∑ (α
+
2 n− 1
cosh λ 2 n− 1 y + β 2+n− 1 sinh λ 2 n− 1 y sin λ 2 n− 1 x on Ω+
∑ (α
−
2 n− 1
cosh λ 2 n− 1 y + β 2−n− 1 sinh λ 2 n− 1 y sin λ 2 n− 1 x on Ω−
n> 0
n> 0
)
)
□
Observe from Example 3.4.1 that when dielectric media are lossy, at the
boundary interface between the two media, in the steady-state limit, the
potential is dictated by the conductivities of the media instead of the electric
permittivities. It is only when the dielectric media are lossless that the electric permittivities dictate how the field behaves (cf. Remark 3.1.7 for details).
This section concludes with a final example.
3.4.2 Example
Consider a composite annulus Ω = (Ω+ , ε + , σ + ) ∪ (Ω− , ε − , σ − ), where ε ± are
the respective electric permittivities and σ ± the respective conductivities
in Ω ± ,
Ω+ = {( x , y ) ∈R 2 : a 2 < x 2 + y 2 < b 2 }
Ω− = {( x , y ) ∈R 2 : b 2 < x 2 + y 2 < c 2 }
Moreover, set Γ = ∂Ω+ ∩ ∂Ω− and Γ ± = ∂Ω ± − Γ . Determine the electric
potential φ on Ω if ϕ = ϕ ± on Γ ± , where ϕ ± are constants.
Inasmuch as the field is static, it must satisfy Laplace’s equation Δφ = 0
subject to the boundary condition at the interface Γ from Corollary 3.1.2.
Employing cylindrical coordinates, where
∆ϕ = 1r ∂r (r ∂r ϕ) +
K15149_Book.indb 86
1
r2
∂θ2 ϕ + ∂2z ϕ
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Boundary Value Problems in Electrostatics
87
∂ z ≡ 0 and the angular symmetry of the problem, that is, ϕ(r , θ) = ϕ(r , θ′)∀θ, θ′ ,
reduces Laplace’s equation to
∆ϕ = 1r ∂r (r ∂r ϕ) = 0 ⇒ r ∂r ϕ = k
for some constant k. Whence, ϕ = k ∫ drr ⇒ ϕ = k ln r + k ′ , where k′ is a constant
of integration. Thus, the general solution is ϕ = a + b ln r .
Now, let ϕ ± = a± + b± ln r on Ω ± , subject to the following boundary condition at the interface: (i) 0 = n− ⋅ ( J + − J − ) ⇒ σ + ∂r ϕ + = σ − ∂r ϕ − on Γ, and
(ii) lim− ϕ + (r ) = lim+ ϕ − (r ) on Γ. Imposing the Dirichlet boundary conditions
r →b
leads to:
r →b
ϕ 0 = ϕ + ( a) = a+ + b+ ln a and 0 = ϕ − (c) = a− + b− ln c
Thus, a− = −b− ln c and a+ = ϕ 0 − b+ ln a . Because there are four unknown
variables to evaluate, two more equations are required. Those are supplied
by conditions (i) and (ii). From (i), ∂r ϕ ± = br± ⇒ σ + b+ = σ − b− , and the continuity relation (ii) yields a+ + b+ ln b = a− + b− ln b .
Direct substitution into (ii) yields ϕ 0 + b+ ln ba = b− ln bc and hence, on setting
γ = σ + ln bc + σ − ln ba , it follows that b+ = ϕ 0 σγ− , b− = ϕ 0 σγ+ , a− = −ϕ 0 σγ+ ln c ,
and a+ = ϕγ0 ( σ + ln bc − σ − ln b ) . Thus, ϕ + = a+ + b+ ln r = ϕγ0 ( σ + ln bc + σ − ln br ) .
It is easy to see that ϕ + ( a) = ϕ 0 , as expected by construction. Similarly,
ϕ − = a− + b− ln r = ϕ 0 σγ+ ln rc . As a sanity check, ϕ − (c) ≡ 0 and ϕ + (b) = ϕ − (b), as
expected. Thus,
 ϕ + on Ω+

ϕ=
 ϕ − on Ω−

is the required solution for Ω.
□
Once again, from the above example, the field is determined by the conductivities of the media, as opposed to pure dielectric media, wherein the
electric permittivities determine the field profile. On the other hand, the
electric permittivities determine the charge density on Γ. Explicitly, the
line charge density is ρ = ε + ∂r ϕ + − ε − ∂r ϕ − on Γ. The current density is
trivially given by
 −σ + ∇ϕ + on Ω+

J=
 −σ − ∇ϕ − on Ω
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Finally, recall that if Ω were a lossless composite dielectric, then ρ|Γ ≡ 0 and
φ is obtained by replacing σ ± with ε ± . Clearly in this instance, the current
density is zero.
3.5 Duality
It is clear by inspecting Maxwell’s equations that they can be rendered symmetric by the introduction of a fictitious magnetic charge density ς and its
associated magnetic current density j; to wit,
∇⋅B = ς
(3.18a)
∇ × E = − ∂t B − j
(3.18b)
whence following the proof of Theorem 1.3.1 mutatis mutandis yields the
equivalent magnetic charge conservation:
∇ ⋅ j + ∂t ς = 0
(3.19)
The symmetrized extensions of Maxwell’s equations are given below for
ease of reference:
∇ × E(r , t) = − j(r , t) − ∂t B(r , t)
(3.20a)
∇ ⋅ B(r , t) = ς
(3.20b)
∇ × H (r , t) = J (r , t) + ∂t D(r , t)
(3.20c)
∇ ⋅ D(r , t) = ρ
(3.20d)
where D = εE and B = μH. Maxwell’s equations are rewritten as (3.20) to display
the symmetry between them.
Now, observe that transforming Equation (3.20a) to (3.20c) requires the
rep­lacement:
E → H, j → −J, B → −D
In particular, this implies that (3.20b) becomes ς = ∇ ⋅ B(r , t) →∇⋅ (− D(r , t)) = −ρ,
and hence, transforming (3.20b) to (3.20d) requires the replacement:
ς → −ρ
Finally, transforming from (3.20c) to (3.20a) yields:
H → −E, J → j, D → B
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Boundary Value Problems in Electrostatics
Likewise, transforming (3.20d) to (3.20b) leads to the replacement:
ρ→ς
In summary, the former four pairs constitute the required transformation
transforming the first pair of Maxwell’s equations into the second pair of
Maxwell’s equations, and vice versa for the latter pairs.
It is clear that the analogy can be carried out further by defining magnetic
conductivity σm as j = σµm B . The four fictitious quantities (ς , σ m , j ) lead in a
natural way to magnetic boundary conditions presented in Section 3.1 mutatis mutandis:
n+ ⋅ (B− − B+ ) = ς
(3.21a)
n+ × ( E− − E+ ) = − j
(3.21b)
For perfect electric conductors (PEC) and perfect magnetic conductors (PMC),
the boundary conditions are summarized as
 n+ × E+ = 0


 n+ × B+ = µJ

(PEC)
 n+ × B+ = 0


 n+ × E+ = − j

(PMC)
Note in passing that PMC is an idealized condition and it does not exist;
however, it is a useful condition to impose when solving radiation problems.
3.5.1 Theorem (Lorentz reciprocity)
Given some open subset (Ω, ε , µ) ⊆ R 3 , suppose (S± , J ± , j± ) are two timevarying current sources on some compact S± ⊂ Ω . Set Ω0 = Ω − (S+ ∪ S− ) . If
the current sources are time harmonic, then

∫∫
∂Ω0
E+ × B− ⋅ nd 2 r =

∫∫
∂Ω0
E− × B+ ⋅ nd 2 r
Proof
Now, from Maxwell’s equations, the source generates ( E± , B± ) via
∇ × E± = − ∂t B± − j±
K15149_Book.indb 89
and ∇ × B± = µ ∂t E± + µJ ±
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Next, consider the pair E+ × B− and E− × B+ and motivated by the vector
identity [3]
∇ ⋅ ( E+ × B− ) = (∇ × E+ ) ⋅ B− − (∇ × B− ) ⋅ E+
it follows that
∇ ⋅ ( E+ × B− − E− × B+ )
= (∇ × E+ ) ⋅ B− − (∇ × B− ) ⋅ E+ − (∇ × E− ) ⋅ B+ + (∇ × B+ ) ⋅ E−
= − ∂t B+ ⋅ B− − j+ ⋅ B− − µ ∂t E− ⋅ E+ − µJ − ⋅ E+ + ∂t B− ⋅ B+ + j− ⋅ B+ + µ ∂t E+ ⋅ E− + µJ + ⋅ E−
However, by assumption, the sources are time harmonic; hence, the fields
generated are time harmonic.* In particular, ∂t ( E± , B± ) = iω( E± , B± ) implies
that the ∂t -terms cancel, yielding
∇ ⋅ ( E+ × B− − E− × B+ ) = j− ⋅ B+ − j+ ⋅ B− + µJ + ⋅ E− − µJ − ⋅ E+
Finally, appealing to the divergence theorem,
∫∫∫
Ω0
∇ ⋅ ( E+ × B− − E− × B+ )d 3 r =
∫∫
∂Ω0
( E+ × B− − E− × B+ ) ⋅ nd 2 r
where n is the unit, outward, normal vector field on ∂Ω0 ,

∫∫∫
Ω0
( j− ⋅ B+ − j+ ⋅ B− + µJ + ⋅ E− − µJ − ⋅ E+ )d 3 r = 0
by construction implies immediately that

∫∫
∂Ω0
( E+ × B− − E− × B+ ) ⋅ nd 2 r = 0
□
3.5.2 Corollary
Given some open, bounded, subset (Ω, ε , µ) ⊆ R 3 , suppose (S± , J ± , j± ) are two
time-varying current sources on some compact S± ⊂ Ω . If ∂Ω satisfies either
the PEC or PMC boundary condition, then
∫∫∫ ( J
Ω
*
+
⋅ E− − µ1 j+ ⋅ B− )d 3 r =
∫∫∫ ( J
Ω
−
⋅ E+ − µ1 j− ⋅ B+ )d 3 r
See Exercise 3.6.7.
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Boundary Value Problems in Electrostatics
Proof
From the proof of Theorem 3.5.1, it suffices to show that
∫∫
∂Ω
( E+ × B− − E− × B+ ) ⋅ nd 2 r = 0
Now, noting [3] the vector identity A × B ⋅ C = A ⋅ B × C, it follows that
( E+ × B− − E− × B+ ) ⋅ n = n × E+ ⋅ B− − n × E− ⋅ B+ = 0
for PEC: n × E± = 0. Conversely, applying the vector identity again,
n × E+ ⋅ B− − n × E− ⋅ B+ = − n × B− ⋅ E+ + n × B+ ⋅ E− = 0
□
for PMC: n × B± = 0 , and the result thus follows.
As an interesting application of the reciprocity theorem, consider an
open subspace Ω ⊂ R 3 bounded by a metal chassis ∂Ω. Suppose also
that there exists a current density J + induced on some compact subset K + ⊂ Ω , where K + is a PEC, and some current density J − on a compact source K − ⊂ Ω . Then, from Corollary 3.5.2, setting j± = 0 yields
∫ ∫ ∫ Ω J + ⋅ E− d 3 r = ∫ ∫ ∫ Ω J − ⋅ E+ d 3 r .
However, ∫ ∫ ∫ Ω J + ⋅ E− d 3 r = ∫ ∫ ∫ K+ J + ⋅ E− d 3 r = 0 as E− |K + = 0 and hence,
∫ ∫ ∫ Ω J − ⋅ E+ d 3 r = 0. Furthermore, noting that as ( J − , K − ) is arbitrary, it follows
at once that ∫ ∫ ∫ Ω J − ⋅ E+ d 3 r = 0 ⇒ E+ ≡ 0 on Ω. That is, induced current on a
PEC does not radiate; this is because the electric field generated on the PEC
precisely cancels out the incident electric field. From an EMC perspective, this
is an interesting example. It demonstrates that conductors within a chassis
do not radiate from currents induced upon them. In particular, the walls of
a chassis do not reradiate from surface current densities induced on them by
Corollary 3.5.2.
3.6 Worked Problems
3.6.1 Exercise
2
Prove that the inversion in a circle ς : B( a) → R 2 − B( a) by r  r ′ ≡ ar such
that ς|∂B( a) = 1∂ B( a ) , is conformal about any deleted neighborhood of
0, and hence, deduce that the inversion in a sphere mapping is also
conformal.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Solution
Recall that a mapping is conformal in some neighborhood N ⊂ C if it is analytic and its derivative is nowhere zero in N. As R 2 ≅ C under the canonical homeomorphism h : ( x , y )  x + iy , it suffices to consider the mapping
ζ : C → C defined by ζ = 1z . However, this mapping is clearly conformal away
from the origin: ζ′ = − z12 ≠ 0 on C − {0}. Hence, the circle inversion mapping
is conformal.
Regarding the inversion in a sphere, it suffices to note that each crosssection of the fixed sphere is a circle. As the inversion in a sphere is merely
the mapping restricted to the respective (circular) cross-section, it follows
immediately that the mapping must also be conformal, as claimed.
□
3.6.2 Exercise
Given r = (r , θ, φ) and rQ = (rQ , 0, π2 ), let ϑ denote the angle between r and rQ .
Show that cos ϑ = sin φ cos θ . Hence, deduce the result for rQ = (rQ , θQ , φQ ) .
Solution
In rectangular coordinates, x = r sin ϕ cos θ, y = r sin ϕ sin θ, z = r cos ϕ. By defir ⋅r
nition, cos ϑ = rQQr . Hence, without loss of generality, we may set r = 1 = rQ .
Then,
r ⋅ rQ = xxQ + y ⋅ 0 + z ⋅ 0 = sin φ cos θ
and the result thus follows. To complete the proof, it is enough to note that
r ⋅ rQ = sin φ cos θ sin φQ cos θQ + sin φ sin θ sin φQ sin θQ + cos φ cos φQ
and invoking the identity cos(a + b) = cos a cos b − sin a sin b, the conclusion
thus follows.
□
3.6.3 Exercise
Consider two infinite strips (Ω ± , ε ± ), where Ω− = R × (0, a) and Ω+ = R × ( a, b),
with ε ± being the respective electric permittivities and ε + ≠ ε − . Find the
potential φ on Ω = Ω− ∪ Ω+ , if
 ϕ 0 if y = b

ϕ=
 0 if y = 0

(3.22)
Deduce the explicit expression for the electric field on Ω.
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Boundary Value Problems in Electrostatics
Solution
First, define ∂Ω0 = {( x , a) : −∞ < x < ∞} , ∂Ω− = {( x , 0) : −∞ < x < ∞}, and ∂Ω+ =
{( x , b) : −∞ < x < ∞} . Then, φ satisfies the Laplace equation Δφ = 0 subject to the
boundary conditions (3.22),
lim ϕ( x , y ) = lim+ ϕ( x , y ) and
y → a−
lim ε − ∂ y ϕ( x , y ) = lim+ ε + ∂ y ϕ( x , y )
y → a−
y→ a
y→ a
The uniformity of the potential along the x-axis suggests that there is no
variation along the x-axis: ∂ x ϕ = 0 on Ω, whence,
∆ϕ = ∂2x ϕ + ∂2y ϕ = ∂2y ϕ = 0 ⇒ ϕ = Cy + D
for some constants C, D.
Furthermore, as ε + ≠ ε − , set ϕ ± = C± y + D± on Ω ± . Then, appealing to the
boundary conditions,
ϕ 0 = C+ b + D+ ⇒ D+ = ϕ 0 − C+ b and 0 = C− 0 + D− ⇒ D− = 0
Also, the two continuity conditions yield
C+ a + D+ = C− a ⇒ ϕ 0 + C+ ( a − b) = C− a
ε − C− = ε + C+ ⇒ C− =
whence C+
{(
ε+
ε−
)
}
− 1 a + b = ϕ 0 ⇒ C+ = ϕ 0
ε+
ε−
C+
{(
ε+
ε−
)
−1 a+b
}
−1
and hence,



 ϕ 0 1 − ε+ b − y  on Ω+
−
a
+
b
1

 ( ε− ) 
ϕ=

y
ϕ 0 εε+− ε+
on Ω−

−
1) a + b
(
ε−

Finally, from E − ∇ϕ, it follows clearly that



E=



K15149_Book.indb 93
(
ε+
ε−
ε+
ε−
(
ϕ0
)
−1 a+b
ε+
ε−
ey
ϕ0
)
−1 a+b
ey
on Ω+
on Ω−
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
As a quick sanity check, it can be easily seen that when ε + = ε − , ϕ = ϕ 0
E = ϕb0 , as expected.
y
b
and
□
3.6.4 Exercise
Given (Ω, ε), where Ω = (0, a) × (0, b), suppose
 V+ for y = b

ϕ =  V− for y = 0
 0 for x = 0, a

(3.23)
Find the potential φ on Ω. How does ε affect φ?
Solution
The solution ϕ ∈C 2 (Ω) ∩ C(Ω) is given by the Dirichlet boundary value
problem Δφ = 0 on Ω satisfying boundary conditions (3.23). The simple geometry of the domain and the simple boundary conditions lead one to attempt
to solve Laplace’s equation via the separation of variables. So, set φ(x,y) =
Φ(x)Θ(y). Then,
0 = ∆ϕ = Θ( y ) ∂2x Φ( x) + Φ( x) ∂2y Θ( y ) ⇒
∂2x Φ
Φ
=−
∂2y Θ
Θ
≡ −λ 2
for some constant λ ∈ R. The negative sign in front of λ 2 was chosen because
of the periodic boundary condition φ(0,y) = 0 = φ(0,a) ∀y. The general solution
is thus Φ(x) = a cos λx + b sin λx.
A solution satisfying the boundary condition is a = 1, b = 0 and Φ(x) =
sin λx, where λ = πna , n = 0, 1, 2,…. Likewise, the general solution satisfying
d2
Θ − λ 2 Θ = 0 is given by
dy 2
Θ(y) = c cosh λy + d sinh λy
Hence, ϕ = ( α cosh λy + β sinh λy ) sin λx satisfies ∆ϕ = 0 . As this holds for
arbitrary integer n, the linearity of the Laplacian operator Δ implies that the
general solution is given by ϕ( x , y ) = Σ n∈Z (α n cosh λ n y + β n sinh λ n y )sin λ n x ,
where λ n ≡ πan to display the explicit dependence of n, and the pair of coefficients (α n , β n ) are determined via the remaining boundary conditions.
Furthermore, as φ ≡ 0 for n = 0, it follows that n ∈ N.
Explicitly, V+ = ϕ( x , b) = Σ n∈N (α n cosh λ nb + β n sinh λ nb)sin λ n x . Whence, multiplying both sides by sin λ m x and integrating yields
α n cosh λ nb + β n sinh λ nb =
K15149_Book.indb 94
2 V+
a
∫
a
0


sin λ n xdx = 

4V+
nπ
for n odd
0 for n even
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Boundary Value Problems in Electrostatics
Likewise,
V− = ϕ( x , 0) =
∑α
n
n


sin λ n x ⇒ α n = 

4V−
nπ
for n odd
0 for n even
Thus,
β 2 n− 1 =
V+ − V− cosh λ 2 n − 1b
4
(2 n − 1) π
sinh λ 2 n − 1b
∀n = 1, 2, 
and the general solution is
ϕ( x , y ) =
4
π
∑
n
1
2 n− 1
( cosh λ
2 n− 1
)
V− cosh λ 2 n − 1b
y + V+ −sinh
sinh λ 2 n− 1 y sin λ 2 n− 1 x
λ 2 n − 1b
Lastly, note that φ is independent of ε in this example. In general, this is not
the case if ε varies in Ω as the boundary condition at the interface wherein ε
changes in Ω must be satisfied by the solution.
□
3.6.5 Exercise
Show that a localized static charge density ρ in R 3 satisfies the Poisson equation −∆ϕ(r ) = ρ(εr ) . And hence, establish that the Poisson equation −∆ϕ = f
describes electrostatics in general. In particular, deduce that steady-state
conditions are also satisfied by Laplace’s equation.
Solution
From Gauss’ law, ∇ ⋅ E = ρε ; substituting E = −∇φ yields −∆ϕ(r ) = ρ(εr ) , as
required. Finally, set f = ρε , and the assertion is established. Next, to establish the last assertion, consider the charge conservation relation ∂t ρ = −∇ ⋅ J .
Under steady-state conditions, ∂t ρ = 0 = −∇ ⋅ J ⇒ 0 = −∇ ⋅ (−σ∇ϕ) ⇒ −∆ϕ = 0 ,
as required.
□
3.6.6 Exercise
Suppose J (r , t) is some current density defined on a compact conductor
Ω ⊂ R 3 above a ground plane ∂R 3+ . Show that its mirror image is − J ( r , t),
where r = ( x , y , − z) with r = (x, y, z).
Solution
By definition, J(r, t) = ρ(r, t)v(r), where ρ(r, t) is the charge density defined on
Ω and v is the average velocity of the charge density circulating on Ω. By
definition, under the mirror image transformation (i.e., reflection on ∂R 3+ ),
r = ( x , y , z) → r ≡ ( x , y , − z) , the image of ρ(r , t) → −ρ( r , t) and v(r , t) → v( r , t).
Hence, the image current is −ρ( r , t)v( r , t) = − J ( r , t) , as
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
3.6.7 Exercise
Consider a segment of an annulus defined by
{
}
Ω = ( x , y ) ∈R 2 : a 2 < x 2 + y 2 < b 2 ∩ ∆
where ∆ = {( x , y ) ∈R 2 : x = b cos θ, y = b sin θ, 0 < θ0 < θ < θ1 } is a wedge of a
disk of radius b. Let ∂Ω a (∂Ωb ) denote the inner (outer) radial boundary of Ω,
and Γ 0 (Γ 1 ) denote the boundary of Ω that subtends the x-axis by an angle
θ0 (θ1 ) . If
 V− on Γ 0

ϕ=
 V+ on Γ 1

such that ∂ n ϕ = 0 on ∂Ω a ∪ ∂Ωb , where n is the outward pointing unit normal vector field on ∂Ω a ∪ ∂Ωb , what is the potential in Ω? Hint: consider the
conformal transformation w = e z in the complex plane, with Ω embedded in
C and utilize the result from Exercise 3.6.4.
Solution
Set z = x + iy and w = e z . Then, w = reiθ ⇒ r = e x and θ = y (modulo 2π). From
this, it is clear that the point (c , y )  (ec , θ). Specifically, if θ0 < y < θ1 and
c = a, then the mapping
{( a, y ) : θ0 < y < θ1 } → ∂Ω a
is a bijection. Likewise, let a < x < b , and set y = θ0 . Then, the mapping
{( x , θ0 ) : a < x < b} → Γ 0
is also a bijection under the conformal mapping.
Thus, under the conformal map w = e z , vertical lines are mapped into
angular arcs of the complex plane, and horizontal lines are mapped into the
radial lines of the complex planes; see Figure 3.5 for details.
Because the Dirichlet problem for the Laplace equation is invariant
under a conformal transformation, it suffices to transform Ω onto the
rectangular domain (via the inverse conformal transformation), solve the
Laplace equation, and then transform the solution back into the original
domain Ω.
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Boundary Value Problems in Electrostatics
Under the conformal mapping, the sides {1, 2, 3, 4}
are mapped respectively onto the sides {1', 2', 3', 4'}
w = ez
3
4
3' Ω 2'
4' 1'
θ0
θ1
2
1
Figure 3.5
Mapping under a conformal transformation.
The solution to the rectangular domain R = (0,a) × (0,b) is precisely that of
Exercise 3.6.4. Because boundary conditions transform invariantly under a
conformal transformation, it follows that the potential ψ defined on R subject
to the following boundary conditions,
 V− on x = 0

ψ=

 V+ on x = a
and ∂ y ψ = 0 on {y = 0} ∪ {y = b}, is given by
ϕ( x , y ) =
4
π
∑
n
1
2 n− 1
( cosh λ
2 n− 1
)
V− cosh λ 2 n − 1b
y + V+ −sinh
sinh λ 2 n− 1 y sin λ 2 n− 1 x
λ 2 n − 1b
Now, the inverse conformal transformation yields x = ln r and y = θ. Hence,
via the composition of maps, ϕ(r , θ) ≡ ψ  w −1 , the desired potential on Ω is
ψ (r , θ) =
4
π
∑
n
1
2 n− 1
( cosh λ
2 n− 1
)
V− cosh λ 2 n − 1b
θ + V+ −sinh
sinh λ 2 n− 1θ sin ( λ 2 n− 1 ln r )
λ 2 n − 1b
□
3.6.8 Exercise
Suppose a point charge Q is located at (0, 0, z0 ) ∈R 3+ above a pure dielectric half-space (R 3− , ε) , where ε is the electric permittivity of the dielectric
medium. Determine the electric potential φ in Ω = R 3+ − {(0, 0, z0 )}, ε 0 , where
ε 0 is the electric permittivity of air. What is the potential in R 3− ? Hint: Apply
the method of images.
(
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Solution
3
The potential φ satisfies Poisson’s equation −ε 0 ∆ϕ = Qδ (r ) satisfying the
boundary condition at the interface:
lim ϕ( x , y , z) = lim+ ϕ( x , y , z)
(i)
z → 0−
z→ 0
lim ε ∂ z ϕ( x , y , z) = lim+ ε 0 ∂ z ϕ( x , y , z)
(ii)
z → 0−
z→ 0
Applying the method of images, let q denote the image charge at (0, 0, −z0 )
∈R 3− . Then, the potential ϕ + in Ω is trivially given by
ϕ + ( x , y , z) =
1
4 πε 0
{
Q
+
x 2 + y 2 + ( z − z0 )2
q
x 2 + y 2 + ( z + z0 )2
}
On the other hand, the solution ϕ − on R3− is obtained by replacing Q with
some charge q′ at (0, 0, z0 ) ∈(R 3 , ε) and replacing the entire space with electric permittivity ε. Then, the potential field defined on R 3− is
ϕ − ( x , y , z) =
q′
1
4 πε
x 2 + y 2 + ( z − z0 )2
Invoking the continuity condition (i) yields 4 πε1 0 (Q + q) =
condition (ii) yields 4επε0 0 (Q − q) = 4επε q ′ , whence
Q+q=
ε0
ε
q′ =
ε0
ε
1
4 πε
q ′ , and applying
0
(Q − q) ⇒ q = − ε−ε
and q ′ =
ε+ε 0 Q
2ε
ε+ε 0
Thus, the electric potential in R 3 is given by




ϕ=



Q
4 πε 0
{
1
x 2 + y 2 + ( z − z0 )2
2ε
1
4 πε 0 ε+ε 0
−
ε−ε 0
ε+ε 0
1
x 2 + y 2 + ( z + z0 )2
Q
x 2 + y 2 + ( z − z0 )2
}
on Ω
on R −3
References
1. Chang, D. 1992. Fields and Wave Electromagnetics. Reading, MA: Addison-Wesley.
2. Churchill, R. and Brown, J. 1990. Complex Variables and Applications. New
York: McGraw-Hill.
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Boundary Value Problems in Electrostatics
99
3. Hsu, H. 1984. Applied Vector Analysis. New York: Harcourt Brace Jovanovich.
4. LePage, W. 1961. Complex Variables and the Laplace Transform for Engineers. New
York: Dover.
5. Jackson, J. 1962. Classical Electrodynamics. New York: John Wiley & Sons Inc.
6. Rothwell, E. and Cloud, M.J. 2001. Electromagnetics. New York: CRC Press.
7. Smythe, W. 1950. Static and Dynamic Electricity. New York: McGraw-Hill.
8. Stratton, J. 1941. Electromagnetic Theory. New York: McGraw-Hill.
9. Wylie, C. Jr. 1960. Advanced Engineering Mathematics. New York: McGraw-Hill.
10. Zachmanoglou, E. and Thoe, D. 1976. Introduction to Partial Differential Equations
with Applications. New York: Dover.
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4
Transmission Line Theory
Transmission line theory is the study of electromagnetic waves propagating
between two or more distinct conductors. The model is obtained by the judicious application of Maxwell’s equations. It is typically derived via discrete
circuit elements. The purpose here is to demonstrate how circuit theory is
derived from field theory.
Indeed, the emphasis placed on the field-theoretic derivation is twofold: it
has general applicability, and more important, it provides the basis for many
engineering approximations and rules of thumb. This is particularly true in
the chapter on antennae. Some useful references regarding the derivation
of transmission lines from Maxwell’s equations can be found in References
[1,6,7], and for an informal and practical approach to the subject, refer to
References [3– 5].
4.1 Introduction
Recall from Theorem 1.4.1 and Corollary 1.4.2 that TEM cannot exist on a single conductor. Because, as shown below, waves defined by the transmission
line equation are in TEM mode, it follows that transmission line structures
comprise at least two conductors.
In practice, for very good conductors, an approximate TEM wave is sustained because conductivity σ < ∞ ⇒ E|| ≠ 0 along the conductors, where E||
is the longitudinal (or axial) component of the electric field, (i.e., E|| is the
field parallel to the direction of the TEM propagation). Physically, E|| ≠ 0 follows from the fact that σ < ∞ implies that the conductors have finite resistance (instead of zero resistance) and hence a driving voltage is needed to
sustain the flow of charges, as energy is lost through the ohmic heating
effect. Thus, the propagating wave is a quasi-TEM wave. It is not a perfect
TEM wave because E|| ≠ 0. For good conductors, a TEM wave solution may
be assumed for simplicity. Finally, a corollary of Section 1.4 is summarized
below for future reference.
101
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
4.1.1. Proposition.
Let M± ⊂ R 3 be two disjoint conductors whose conductivity σ ± >> 1. Then,
a TEM wave ( E , B) propagating between M± induces a potential difference
between M− and M+ . In particular, the TEM wave induces a current density
along M± .
Proof
Without loss of generality, suppose M± is oriented such that the TEM wave is
propagating along the z-axis. From Section 1.4, E = E⊥ by definition, in particular, by Theorem 1.4.1, ∃ϕ such that E⊥ = −∇ϕ . Because M± may be approximated as PECs, the boundary conditions on M± imply that ϕ|M± ≡ ϕ ±
are constants, whence, δϕ = ϕ + − ϕ − is the required potential difference, as
claimed. To complete the proof, it suffices to observe that Bz = 0 implies that
n± × B⊥ = µJ ± . The surface current is thus parallel to ∂ M± , as required.
□
Thus, Proposition 4.1.1 establishes an equivalence between voltage–current
and TEM waves. That is, a propagating TEM wave between two separate
conductors will generate a potential difference between the two conductors
and, equivalently, applying a time-harmonic voltage across two separate
conductors will generate a TEM wave. The above theorem forms the basis
for transmission line theory.
4.2 Transmission Line Equations
In this section, transmission line equations are derived from first principles
via a field-theoretic method. It is then demonstrated that a pair of transmission lines can be approximated by a distributed line model commonly found
in the literature. This thus justifies the use of a distributed line model in
deriving the transmission line equation.
Transmission line equations are also known as the telephone equations. The
following assumptions are made in the derivation.
• The conductors are imperfect; that is, 1 << σ < ∞, where σ is the conductivity of the conductors, and without loss of generality, the conductors are assumed to have the same conductivity.
• The dielectric medium is imperfect; that is, 0 < σ 0 << 1 , where σ 0 is
the conductivity of the surrounding homogeneous dielectric medium.
• The TEM solution holds between the pair of conductors, as
0 < σ 0 << 1 << σ .
• The conductors are arbitrarily long with uniform cross-sections.
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Transmission Line Theory
Circles centred about C(0) and C(1) respectively
with π(C(0)) = z and π(C(1)) = z + δz, where
π(x, y, z) = z is a projection map
Γ'
Plane waves
E, B
γ
The surface area S spanned by the
oriented loop γ + C + γ' + C'
Γ"
S
C
C'
z
Unit normal
γ' Good conductors vector field
on S is directed
into the page
z + δz
Figure 4.1
TEM waves propagating between two infinite transmission lines.
Figure 4.1 shows a TEM wave (or equivalently, some oscillating voltage
source not shown) incident on an infinitely long pair of conductors. The
closed path Γ = γ ∪ C ∪ γ ′ ∪ C ′ is oriented as shown in the figure and S is
the rectangular surface that spans the oriented loop Γ. That is, Γ(0) = γ(0) and
Γ(1) = γ(1). The unit vector n normal to S is directed into the page, consistent
with the orientation of Γ (right-hand rule).
Assume for simplicity that γ is oriented in the e y direction along the y-axis,
C is oriented in the e z direction along the z-axis and n is directed in the
+x-axis direction, and suppose that the angular frequency of the incident
TEM wave is ω. Suppose also that the conductors are very good conductors
(viz., σ >> ωε for the ω in question). Finally, recall that the following approximations are employed: (a) the general solution is a TEM solution and (b) a
very small Ez -field on the surface of the conductors is assumed in order to
overcome the ohmic loss due to finite conductivity of the conductors.
From Equation (1.15), invoking Stokes’ theorem yields
∇ × E ⋅ nd 2 r =
S
E ⋅ l dr and hence,

∫∫
∫
Γ
− ddt
∫∫ B ⋅ n dS = ∫ E ⋅ l d = ∫ E ⋅ e d + ∫ E ⋅ e d + ∫
S
Γ
γ
y
C
z
γ′
E ⋅ e y d +
∫
C′
E ⋅ e z d
(4.1)
where n is a unit normal vector field on S and l is the unit tangent vector
field on Γ. Before proceeding further, recall that
B ⋅ ndS = Ψ is just the
S
magnetic flux that crosses the surface S, and inductance L = L(S) is defined by
Ψ ≡ Li, where i is the current flow around ∂S = γ.
Some comments are due. First, observe that the inductance L = L(S)
depends implicitly on the surface area S via the surface integral. Second,
by definition, the potential difference along the path γ is given by
v( γ ; t) = − ∫ γ E ⋅ e y d, and likewise, the potential difference along the path γ′ is
given by v( γ ′ ; t) = − ∫ γ ′ E ⋅ e y d = − v(− γ ′ ; t), where (− γ ′)(t) ≡ γ ′(1 − t) ∀t ∈[0, 1]
is the reverse orientation of γ′.

∫∫
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
The potential difference v( γ ; t) and v( γ ′ ; t) are well-defined because, by
definition, TEM waves satisfy the Laplace equation and hence, the field is
conservative. That is, the path integral is path-independent in a simply connected space as long as the endpoints are specified. By construction (see
Figure 4.1) we may set v(z, t) = v(γ; t) and v( z + δz, t) = v(− γ ′ ; t).
The potential drop across C resulting from an imperfect conductor is a little bit more involved. For concreteness, let the loop Γ and the paths γ , γ ′ , C , C ′
be parameterized by 0 ≤ s ≤ 1. Fix some s ∈(0, 1) such that Γ(s) = C(0) and a
circle Γ′ centered at C(0) about the upper conductor along C, with π(C(0)) = z,
where π : R 3 → R defined by (x, y, z) ↦ z is a projection map that projects
a 3-vector onto its z-component; that is, π(v ) = v ⋅ e z (see Figure 4.1). Then,
iC ( z, t) = µ1
B ⋅ eφ d defines the flow of current along C, where eφ is a unit
∫
Γ′
vector field tangent to Γ′. By Ohm’s law,
v(C ; t) = −
∫ E dz =
C
z
1
2
RδziC ( z, t)
where 21 R is the resistance per unit length of the respective conductors.
Likewise, along C′, the voltage drop is given by
v(C ′ ; t) = −
∫
C′
Ez d z = − 21 RδziC′ ( z, t)
where, via Kirchhoff’s current law, iC′ ( z, t) + iC ( z, t) = 0 ∀z, t .
Indeed, this suggests that the current flowing from C to C′ along γ′ comprises essentially a displacement current; more on this point later. Thus,
from Equation (4.1), it follows that
− ddt

∫∫ B ⋅ nd x = − v(z, t) + v(z + δz, t) +
2
S
1
2
RδziC ( z, t) − 21 RδziC ' ( z, t)
(4.2)
and observe that iC′ ( z, t) = − i− C′ ( z, t).
Now, recall from the preceding paragraph that the magnetic flux is defined
by Ψ = ∫ S B ⋅ ndS = Li, where L is the inductance and i the current around Γ.
Furthermore, observe trivially that lim S → 0 , where S denotes the area of
δz→ 0
S and δz is the length of S. Hence, this motivates the following definition:
 ≡ lim 1 ∫ S B ⋅ ndS, the magnetic flux per unit length. In particular, it folΨ
δz
δz→ 0

lows that the inductance per unit length L = Ψi is also well-defined. Finally,
from − ddt ∫ S B ⋅ nd 2 x = − L ∂t i and hence, Equation (4.2) becomes
∂ z v( z, t) = − L ∂t i( z , t) − Ri( z , t)
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(4.3)
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105
Transmission Line Theory
Volume M
Surface
S–
S+
Surface
E, B
Plane wave
Good conductors
Surface S0
Figure 4.2
TEM waves propagating between two infinite transmission lines.
where
∂ z v( z, t) = lim v( z +δz ,δtz)− v( z ,t )
δz→ 0
and i( z, t) = 21 iC ' ( z, t) + 21 i− C ' ( z, t)
This is the first equation for the pair of transmission lines. To obtain the
second transmission line equation, consider Figure 4.2.
The boundary of the small volume M is clearly seen to be ∂ M = S+ ∪ S− ∪ S0 .
Moreover, assume the same coordinate axes as in Figure 4.1. Namely, the
z-direction is the direction of the wave propagation and the x–y plane
defines the transverse (i.e., perpendicular) coordinates. Finally, suppose that
the z-component of the surface S+ is z and the z-component of the surface
S− is z + δz, where δz is the length of the cylindrical surface S0.

∫∫∫ ρd x = 
∫∫∫ ∇ ⋅ Jd x .
Invoking the divergence theorem,
∫∫∫ ∇ ⋅ Jd x = ∫∫ J ⋅ d x , and Ohm’s
law, J = σE, it follows that
From the equation of continuity (1.19), − ddt
3
3
M
M
3
2
∂M
M

∫∫
∂M
J ⋅ nd 2 x =
∫∫
S+
J ⋅ n+ d 2 x +
∫∫
S−
J ⋅ n− d 2 x + σ
∫∫
S0
E ⋅ nd 2 x
where σ is the conductivity of the homogeneous medium wherein the
two conductors are embedded. Set i( z, t) = ∫ ∫ S+ J ⋅ n+ d 2 x and i( z + δz, t) =
∫ ∫ S− J ⋅ n− d 2 x , where n+ = − n− is the unit vector normal to S± , respectively,
to be the conduction current across the surfaces S± . Hence, ∫ ∫ S+ J ⋅ (− n− )d 2 x +
∫ ∫ S− J ⋅ n− d 2 x = i( z + δz, t) − i( z, t) .
The third term, σ ∫ ∫ S0 E ⋅ nd 2 x , denotes the transverse conduction contribution (it corresponds to the conductivity of a lossy dielectric medium) where n
is the unit vector field on S0. Moreover, recalling that G(S0 ) = v(σz ,t ) ∫ ∫ S0 E ⋅ nd 2 x
defines the conductance, and noting trivially that lim S0 = 0 it follows that
G(S0 ) ≡ lim δ1z
δz→ 0
σ
v( z , t )
per unit length.
Finally, applying
−ε ddt
∫∫
K15149_Book.indb 105
∂M
δz→ 0
∫ ∫ S0 E ⋅ nd 2 x , is well-defined; this defines the conductance
Gauss’
law,
− ddt

∫∫∫
M
ρd 3 x = −ε ddt

∫∫∫
M
∇ ⋅ Ed 3 x =
E ⋅ d 2 x. Whence, from Q = CV, it follows that the capacitance
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
C(∂ M) = v( zε, t ) ddt ∫ ∫ ∂ M E ⋅ d 2 x and in particular, the capacitance per unit length
C(S0 ) = lim δ1z C(S0 ) is well-defined, as the displacement current is zero across
δz→ 0
S+ ∪ S− . Thus,
− C(S0 )δz ∂t v( z, t) = i( z + δz, t) − i( z, t) + G(S0 )δzv( z, t)
(4.4)
and taking the limit as δz → 0 yields the second transmission line equation:
∂ z i( z, t) = − C(S0 ) ∂t v( z , t) − G(S0 )v(z , t)
(4.5)
As an aside, an alternative derivation of Equation (4.5) via (1.17) is given in
Chapter 5. The above results can be formally epitomized as follows.
4.2.1 Theorem
Given a pair of semi-infinitely long conductors of conductivity σ 0 >> 1
that are parallel to the z-axis, suppose that the conductors have uniform
cross-sections and are embedded in a homogeneous dielectric medium
(ε, σ), where 0 < σ << 1 << σ 0 . Suppose a time-harmonic electromagnetic
plane wave ( E , B, ω ) is incident on the pair of conductors at z = 0, where
σ 0 >> εω . Then, the induced voltage and current waves propagating along
the conductor pair may be approximated by the TEM solution given by
Equations (4.3) and (4.5).
□
What is the explicit expression for the voltage v and current i from the
coupled equations? To briefly see it, differentiate (4.3) with respect to z and
replace ∂ z i in the equation using (4.5):
∂2z v( z, t) = RGv( z, t) + (RC + LG) ∂t v( z, t) + LC ∂t2 v( z, t)
(4.6)
where R,G, L, C are, respectively, resistance, conductance, inductance, and
capacitance per unit length. Likewise, differentiating (4.5) with respect to z
and replacing ∂ z v in the equation with (4.3) yield:
∂2z i( z, t) = RGi( z, t) + (RC + LG) ∂t i( z, t) + LC ∂t2 i( z, t)
(4.7)
Now, observe that Equations (4.6) and (4.7) have identical forms. Indeed, the
equations are precisely the one-dimensional D’Alembert wave equation.
Hence, the voltage and current that propagate along the conductors are pre1
cisely plane waves. By inspection, they propagate at speed LC . In hindsight,
it is almost clear that L, C, G are related to σ, μ, ε in some way.
4.2.2 Corollary
Given the conditions stated in Theorem 4.2.1, LG = μσ and LC = με.
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107
Transmission Line Theory
Proof
Suppose that the conductors are perfect conductors; that is, set R = 0. Then,
Equations (4.6) and (4.7) reduce to
∂2z v( z, t) = LG ∂t v( z, t) + LC ∂2t v( z, t)
(4.8)
∂2z i( z, t) = LG ∂t i( z, t) + LC ∂t2 i( z, t)
(4.9)
Now, recalling the wave Equation (1.28) and noting that v( z, t) = − ∫ γ E ⋅ ld,
(4.8) reduces to
−
∫ ∂ E ⋅ l d = −LG ∫ ∂ E ⋅ l d − LC ∫ ∂ E ⋅ l d
γ
2
z
γ
t
γ
2
t
whence comparing the coefficients with Equation (1.28) yields
∂2z v( z, t) = µε ∂t2 v( z, t) + µσ ∂t v( z , t)
Likewise, recalling that i( z, t) =
(4.10)
∫ H ⋅ l d , (4.9) has the same form as the
C
wave Equation (1.29) for the magnetic flux density, yielding
∂2z i( z, t) = µε ∂t2 i( z, t) + µσ ∂t i( z, t)
whence, by inspection, it is clear that LG = μσ and LC = με, as required.
(4.11)
□
Observe trivially from the above derivation that a transmission line can
be represented by the distributed parameter model illustrated in Figure 4.3.
Indeed, from the equivalence between (v,i) and (E,B), it follows that voltage and current are subject to reflection at a boundary interface, just as
electric and magnetic fields are. This topic is investigated in subsequent
sections.
R
L
G
R
C
L
Figure 4.3
A distributed lumped parameter model representing a pair of transmission lines.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
4.3 Characteristic Impedance and the Smith Chart
In this section, the concept of characteristic impedance is established. It has
special significance in transmission line theory: to wit, the characteristic
impedance is independent of the length of the conductors. This is established
below. In Section 4.2, infinitely long conductors were studied. In contrast,
finite conductors are investigated here. The motivation for this section runs
as follows.
• Under what circumstances do finite conductors appear to be infinitely long conductors to a propagating TEM wave?
• What happens to an incident voltage and current at a boundary
interface?
• What is the impedance along a transmission line from the perspective of an incident propagating wave?
For simplicity, consider a propagating time-harmonic TEM wave. Then,
phasor transforming (v,i) → (V,I), Equations (4.3) and (4.4) can be rewritten as
dV ( z )
dz
dI ( z )
dz
= − RI ( z) − iωLI ( z) = −(R + iωL)I ( z)
(4.12)
= − GV ( z) − iωCV ( z) = −(G + iωC)V ( z)
(4.13)
The equations are now a pair of coupled first-order ordinary differential
equations. The variable ω has been suppressed for convenience. So, differentiating Equation (4.12) with respect to z and substituting the value for
dI ( z )
dz via (4.13):
d2V ( z )
dz 2
= (R + iωL)(G + iωC)V ( z) ≡ γ 2V ( z)
(4.14)
Likewise, differentiating (4.13) with respect to z yields:
d2 I ( z )
dz 2
= (R + iωL)(G + iωC)V ( z) ≡ γ 2 I ( z)
(4.15)
Observe the symmetry between the two equations. Before proceeding further, the wave propagation constant γ ≡ α + iβ is evaluated explicitly.
First note that (α + iβ)2 = γ 2 = (R + iωL)(G + iωC). Hence, using the same
technique as in Chapter 1, α 2 − β 2 + i2αβ = RG − ω 2 LC + iω(RC + LG), leads to
α 2 − β 2 = RG − ω 2 LC and 2αβ = ω(RC + LG)
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109
Transmission Line Theory
Next, noting that (α 2 + β 2 )2 = (α 2 )2 + (β 2 )2 + 2α 2β 2 1= (α 2 − β 2 )2 + 4α 2β 2 , it
follows that α 2 + β 2 = ((RG − ω 2 LC)2 + ω 2 (RC + LG)2 ) 2 . Hence, adding and
subtracting the two equations, respectively, for α 2 ± β 2 give
1
2α 2 = ((RG − ω 2 LC)2 + ω 2 (RC + LG)2 ) 2 + RG − ω 2 LC
1
2β 2 = ((RG − ω 2 LC)2 + ω 2 (RC + LG)2 ) 2 − (RG − ω 2 LC)
In summary,
α=
β=
1
2
1
2
{(RG − ω LC)
2
2
+ ω 2 (RC + LG)2
}
1
2
+ RG − ω 2 LC
(4.16a)
((RG − ω 2 LC)2 + ω 2 (RC + LG)2 ) 2 − (RG − ω 2 LC)
(4.16b)
1
Now, observe trivially that
that whenever
a 2 + b 2 ≥ max{ a, b} ∀a, b ∈R . Thus, it follows
ω << min
{
RG
LC
, RC+1 LG
}
by invoking the binomial expansion, Equation (4.16) reduces to
{
α ≈ RG 1 +
ω2
8
( GC + RL )2 }
and β ≈ RG
ω
2
( GC + RL )
To derive the expressions, it suffices to observe that RG − ω 2 LC ≈ RG and
+ LG
1 + ω 2 ( RCRG
) ≈ 1+
2
ω2
2
( RCRG+ LG )2 = 1 + ω2 ( GC + RL )2
2
In particular,
lim α = RG
and lim β = 0
ω→ 0
ω→ 0
and there is no wave propagation for
ω << min
{
RG
LC
}
, RC+1 LG , as γ ≈ α
Next, consider the other extreme scenario wherein
ω >> max
K15149_Book.indb 109
{
R
L
+
G
C
,
RG
LC
}
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Then, by applying the binomial approximation once again, it follows that
α≈
1
2
{R
C
L
+G
L
C
}
{
and β ≈ ω LC 1 +
1
8ω 2
( RL + GC )2 }
Once again, the derivation is similar to the former expressions: to wit,
RG − ω 2 LC ≈ −ω 2 LC and
+ LG
1 + ( RCωLC
) ≈ 1+
2
( RCLC+ LG )2 = 1 + 2ω1 ( RL + GC )2
1
2ω2
2
In particular,
lim α ≈
ω→∞
1
2
{R
C
L
+G
L
C
}≡ {
1
2
R
R0
+ GR0
}
and lim β ≈ ω LC
ω→∞
where R0 = L/C defines the characteristic impedance of a lossless transmission line.
Thus, it is clear from the above analysis that for very low frequencies,
γ ≈ α ≈ RG , and the line appears to be purely resistive. On the other hand,
for very high frequencies, ℜe( γ ) is independent of frequency. Finally,
for a perfect conductor in a perfect dielectric, α = 0 and β = ω LC ⇒
γ = iω LC ∀ω. These results are summarized formally below for future
reference.
4.3.1 Proposition
Given a pair of infinitely long transmission lines (C± ,L,C, R, G), for any positive number ε > 0, ∃ω ε > 0 such that ∀ω > ω ε ,
( RG0 + GR0 ) < ε
a)
α−
b)
β − ω LC < ε
1
2
where R0 = CL is the characteristic impedance and G0 =
up to first order in ω1 ,
γ≈
1
2
{
( RG0 + GR0 ) + iω
LC 1 +
1
8ω 2
C
L
. In particular,
( RL + GC )2 }
Proof
From the discussion above, given any ε > 0, provisionally choose
ω >> max
K15149_Book.indb 110
{
R
L
+
G
C
,
RG
LC
}
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Transmission Line Theory
For concreteness, set
ω > ω 0 ≡ a max
{
R
L
+
G
C
,
RG
LC
}
for any fixed constant a >> 1. Then, appealing to the binomial approximation, (1 + δ)r ≈ 1 + rδ + r ( r2!− 1) δ 2 + o(δ 3 ), where δ << 1 and r ∈ R, on setting
κ = 21 (RG0 + GR0 ), it follows that up to ω12 ,
α≈
ω LC
{
1
2ω2
{
(
R
L
1
2
≈ κ 1−
≈κ−
1
4ω 2
κ
4ω 2
( RL + GC )2 − 2ω1 ( RL + GC )2 4ω1 ( RL + GC )2 + o ( ω1 )}
2
+
)}
G 2
C
2
6
1
2
( RL + GC )2
whence, given ε > 0, choose ω such that
κ
4ω 2
( RL + GC )2 < ε
In particular, set
ω ′ε = max
{
κ
ε
( RL + GC ) , ω 0 }
Then, ∀ω > ω ′ε ⇒ α − 21 ( RG0 + GR0 ) < ε .
Lastly, for case (b), via the binomial approximation, choose ω > ω 0 , then
β≈
1
2
ω LC 2 +
1
2ω2
( RL + GC )2 ⇒ β − ω
LC ≈ 1 +
LC
8ω
( RL + GC )2 + o ( ω1 )
3
whence choosing
ω > ω ′′ε ≡ max
{
LC
8ε
( RL + GC )2 , ω 0 } ⇒ β − ω
LC < ε
So, it is evident that upon choosing ω ε = max{ω ′ε , ω ′′ε }, both (a) and (b) are
satisfied whenever ω > ω ε , as required.
□
4.3.2 Definition
The wavelength λ of a wave propagating between a pair of transmission
lines is defined by λ = 2βπ , where β = ℑm(γ).
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
For perfect conductors in a perfect dielectric, γ = iω LC and hence,
1
λ = ω 2 πLC . From this, it is clear that the phase velocity is υ = LC
. For non­
ideal conductors and nonideal dielectrics, the phase velocity clearly becomes
much more complicated.
The general solution to Equation (4.14) is
V ( z) = V+ e− γ z + V− e γ z
(4.17)
The coefficient V+ is the coefficient for the forward propagating (i.e., incident) wave. The coefficient V− is the coefficient for the backward propagating
(i.e., reflected) wave. In particular, V− ≡ 0 if the transmission line is infinitely
long.
As a corollary to Proposition 4.3.1, when the frequency of a propagating
wave is sufficiently high, the attenuation of the wave falls off essentially
− 1 ( RG + GR ) z
as e 2 0 0 , irrespective of the frequency. So, what happens should the
line be finite? To answer this question, recall from Section 4.2 regarding the
equivalence between L, G, C and (µ , σ , ε):
LG = μσ
and
LC = με
(4.18)
Hence, changing the values of (μ, ε) will have an impact on the boundary between two media when a TEM wave is incident on the boundary. The
equivalence established by Equation (4.16) implies that changing the values
of the inductance and the capacitance of the line will affect how the voltage
and current waves propagate. In particular, reflection will generally occur
when the voltage wave is incident on the boundary between the conductors
with different values of inductance and capacitance.
Reflection occurs in order to satisfy the boundary condition imposed by a
change in L, C, G, or equivalently, in μ, σ, ε. For example, if a finite conductor is open with respect to ground, a physical boundary condition would be
that the current at the end of the conductor be zero. Conversely, if a finite
conductor were shorted to ground, a physical boundary condition would be
that the voltage at the endpoint of the conductor be zero. A physical solution
for the respective differential equations must then satisfy the given boundary conditions.
4.3.3 Example
Consider a pair of infinitely long perfect conductors such that
 L1
L=
 L 2
K15149_Book.indb 112
for
z < 0,
for
z > 0,
 C1
and C = 
 C2
for
z < 0,
for
z > 0.
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Transmission Line Theory
That is, the conductor consists of a pair of semi-infinite conductors of different L, C. Recall that the propagating voltage wave must satisfy the wave
equation
∂2z Vα ( z) = γ α2 Vα ( z)
where γ α2 = −ω 2 Lα Cα , in conductor Cα , α = 1, 2 .
Let the solution in z < 0 be V ( z) = V1+ e γ 1z + V1− e− γ 1z and the solution in z > 0
be V ( z) = V2+ e− γ 2 z . Note that by assumption, V2− = 0 as there is no reflection
from an infinite line (the absence of a boundary for all finite z > 0). Inasmuch
as the potential must be continuous
at the boundary z = 0, it follows that
2
V1+ + V1− = V2+ . Moreover, as d dVz2( z ) is defined on the transmission line, it follows
that dVdz( z ) must also be continuous at z = 0. Hence,
−γ 1V1+ + γ 1V1− = −γ 2V2+
The voltage V1+ e γ 1z represents the incident wave, V1− e− γ 1z corresponds to the
reflected wave, and V2+ e γ 2 z the transmitted wave.
Now, observe that solving for V1± as functions of V2+ at z = 0 yields
(
V1± = 21V2+ 1 ±
γ2
γ1
)⇒
V1−
V1+
=
γ1−γ2
γ1+γ2
and
V2+
V1+
=
2γ1
γ1+γ2
which are constants. The former constant relates to the reflection of the
wave at the boundary z = 0, whereas the latter relates to the transmission
of the wave at the interface z = 0. The analysis for ( I1± , I 2+ ) follows that of
(V1± , V2+ ) mutatis mutandis, leading to
I1−
I1+
=
γ1−γ2
γ1+γ2
and
V2+
V1+
=
2γ1
γ1+γ2
□
This analysis motivates the details sketched below.
Having had a sneak preview of what’s to follow from Example 4.3.3, consider
for simplicity an infinitely long pair of transmission lines. Then, the general
solutions to Equations (4.15) and (4.16) are, respectively, V ( z) = V0+ e− γ z + V0− e γ z
and I ( z) = I 0+ e− γ z + I 0− e γ z , where (V0+ , I 0+ ) are forward propagating waves and
(V0− , I 0− ) are backward propagating (i.e., reflected) waves.
Next, noting that ddz V = −γ {V0+ e− γ z − V0− e γ z }, it follows from Equation (4.12)
that ddz V = −γ {V0+ e− γ z − V0− e γ z } = −(R + iωL)I . That is, V0+ e− γ z − V0− e γ z = R +γiωL
{ I 0+ e− γ z + I 0− e γ z } implies that
{
0 = V0+ −
K15149_Book.indb 113
R + iωL
γ
}
{
I 0+ e− γ z − V0− +
R + iωL
γ
}
I 0− e γ z
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Because this holds for all z, it follows immediately that V0+ −
V0− + R +γiωL I 0− = 0 ; to wit,
V0+
I 0+
=
R + iωL
γ
R + iωL
γ
I 0+ = 0 and
−
= − VI0−
0
As a side remark, the same result can be obtained via Equation (4.13) and
I ( z) = I 0+ e− γ z + I 0− e γ z :
I ( z) =
γ
R + iωL
{V e
+
− γz
}
− V− e γz =
G + iωC
R + iωL
{V e
+
− γz
− V− e γz
}
as expected. These results motivate the following definition.
4.3.4 Definition
Given a pair of parallel transmission lines (C± , , R, L, C, G) of length ℓ ≤ ∞, its
characteristic impedance is given by Z0 = GR ++ iiωωCL .
The characteristic impedance is thus the impedance looking down a pair
of infinite transmission lines. By definition, the characteristic impedance is,
in general, frequency dependent. Clearly, for ω >> 1 to be sufficiently large,
that is, R << ωL and G << ωC, then Z0 ≈ R0 ≡ CL . That is, the characteristic
impedance is approximately independent of the frequency. Explicitly, via the
binomial expansion, up to first order in 1/ω,
Z0 =
R + iωL
G + iωC
≈
L
C
{1 − 2iω RL } {1 + 2iω GC } ≈ R0 {1 + 2iω ( GC − RL )}
4.3.5 Theorem
The characteristic impedance of a lossless pair of parallel transmission lines
(C± , , L,C) is independent of the angular frequency ω of a propagating timeharmonic wave.
Proof
Because the lines are lossless, R = 0 = G, and hence, from Definition 4.3.4,
Z0 = L/C .
□
From the previous discussion, a pair of transmission lines transmitting a
very high frequency, monochromatic, time-harmonic TEM wave behave as if
the lines were lossless. In particular, this holds for a square wave or trapezoidal wave provided the fundamental frequency is sufficiently large.
In what follows, only finite transmission lines are considered. For simplicity, let ( γ + , γ − ) denote the axes of a pair of parallel transmission lines
(C+ , C− ) such that the lengths γ + = γ − . The pair of transmission lines (C+ , C− )
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115
Transmission Line Theory
C+
z=0
Source
impedance Zs
z=l
Load
ZL impedance
R, L, C, G,
V = V0(ω)
C–
Figure 4.4
A pair of finite
is identified with ( γ + , γ − ) , and where convenient, they are parameterized
by γ ± : [0, 1] → R 3 . See Figure 4.4 for details. Finally, for notational convenience, let (C± , , R, L, C, G) denote the above transmission line pair such
that γ + =  = γ − . Note that although it is unfortunate that the axes of a pair
of transmission lines ( γ + , γ − ) share the same symbol as the wave propagation constant γ, no confusion should arise on this account.
4.3.6 Proposition
Suppose a transmission line pair (C± , , R, L, C, G) is connected to a source
(V0 (ω ), ZS ) and terminated by a load ZL . Then, the input impedance Z(z)
toward the load at any point z ∈ [0, ℓ] on the transmission line is given by
Z( z) = Z0
ZL + Z0 tanh γ (  − z )
Z0 + ZL tanh γ (  − z )
(4.19)
Proof
The general solutions of Equations (4.15) and (4.16) at z = ℓ yield
V () = V0+ e− γ  + V0− e γ 
and
I () = I 0+ e− γ  + I 0− e γ 
−
−
Now, recalling that Z0 = − VI0− , it follows at once that I () = I 0+ e− γ  − VZ00 e γ  .
0
Hence, solving for V0± yield V0± = 21 (V () ± I ()Z0 )e ± γ . Furthermore, by definition, ZL = VI(()) implies that
V0± = 21 I ()(ZL ± Z0 )e ± γ
and hence, substituting into the general solution for V(z),I(z) leads directly to
{
}
(4.20)
}
(4.21)
V ( z) = 21 I () (ZL + Z0 )e γ (  − z ) + (ZL − Z0 )e− γ (  − z )
I ( z) =
K15149_Book.indb 115
1 I ()
2 Z0
{(Z
L
+ Z0 )e γ (  − z ) − (ZL − Z0 )e− γ (  − z )
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
C+
z=0
Source
impedance
V = V0(ω)
Zs
R, L, C, G,
V(0)
Input
impedance
z=l
Source
impedance Zs
Load
ZL
impedance
V = V0(ω)
C–
Z(0)
Z(0)
Figure 4.5
The equivalence between a transmission line and a
Lastly, substituting cosh x = 21 {e x + e− x } and sinh x = 21 {e x − e− x } into the above
two equations, it follows clearly that
Z( z) =
V ( z)
I ( z)
= Z0
ZL cosh γ (  − z )+ Z0 sinh γ (  − z )
Z0 cosh γ (  − z )+ ZL sinh γ (  − z )
= Z0
ZL + Z0 tanh γ (  − z )
Z0 + ZL tanh γ (  − z )
□
In particular, it is obvious that the input impedance as seen by the source at
z = 0 is
Z(0) = Z0
ZL + Z0 tanh γ
Z0 + ZL tanh γ
Its physical significance is best illustrated in Figure 4.5.
4.3.7 Corollary
Suppose (C±( ∞ ) , R, L, C, G) is a pair of infinitely long transmission lines and
let (C± , , R, L, C, G) ⊂ (C±( ∞ ) , R, L, C, G) be an arbitrary compact segment that is
terminated by some load Z. That is, the pair ( γ + (1), γ − (1)) is connected across
Z. Then, (C± , , R, L, C, G) is equivalent to (C±( ∞ ) , R, L, C, G) if and only if Z = Z0
for all  > 0.
Proof
From Proposition 4.3.6, ZL = Z0 if and only if Z( z) = Z0 ∀z ∈[0, ] ,  > 0
arbitrary.
□
The above result thus illustrates an important feature of infinite transmission lines: they can be simulated by terminating a finite pair of transmission
lines with their characteristic impedance. In this case, the finite transmission
line is said to be matched.
Observe from Equation (4.20) and V0± = 21 I ()(ZL ± Z0 ) that
{
V ( z) = V0+ e γ (  − z ) +
K15149_Book.indb 116
ZL − Z0
ZL + Z0
e− γ ( − z)
}
(4.22)
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Transmission Line Theory
for 0 ≤ z ≤ ℓ. In particular,
V0−
V0+
=
ZL − Z0
ZL + Z0
by definition, and
V0+ + V0−
V0+
=
2 ZL
ZL + Z0
Finally, recalling that V0− denotes the reflected wave whereas V0+ denotes the
incident wave, the following definition ensues.
4.3.8 Definition
Given a pair of finite transmission lines (C± , , R,L,C,G) , the reflection coefficient at the load is Γ() = ZZLL −+ ZZ00 and the transmission coefficient at the load
is Τ() = Z2L Z+LZ0 .
Specifically, the reflected wave amplitude is Γ()V0+ and the transmitted wave
amplitude is Τ()V0+ . Furthermore, it is clear from Definition 4.3.8 that 1 + Γ(ℓ) =
Τ(ℓ). That is, the sum of the incident wave and the reflected wave at the load is
equal to the wave transmitted across the load. Lastly, observe that as impedance
is, in general, complex, it is clear that Γ() = Γ() eiθ , where θ = arg Γ().
4.3.9 Lemma

Given (C± , , R,L,C,G) , the local maxima of V occur at z =  − 2 nπ+θ
∀n =
2β
(2 n − 1) π+θ
0, 1, 2,, and local minima of V occur at z = −
∀n = 0, 1, 2, … ,
2β
where γ = α + iβ is the wave propagation constant.
Proof
{
}
From Equation (4.22), V ( z) = V0+ eα (  − z )eiβ (  − z ) 1 + Γ() ei(θ − 2β(  − z ))e−2 α (  − z ) . That is,
V ( z) = V0+ eα (  − z ) 1 + Γ() ei(θ − 2β(  − z ))e−2β (  − z )
Thus, it is quite evident that local maxima occur whenever ei(θ − 2β(  − z )) = 1.
Specifically,
ei(θ − 2β( − z )) = 1 ⇒ θ − 2β( − z) = −2 nπ ⇒ z = − 2 nπ+θ
+ ∀n = 0, 1, …
2β
as z ≥ 0. Likewise, local minima occur whenever ei(θ − 2β(  − z )) = −1. That is,
ei(θ − 2β( − z )) = −1 ⇒ θ − 2β( − z) = −(2 n − 1)π ⇒ z = − (2 n−21)βπ+θ + ∀n = 0, 1, …
□
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
In view of Lemma 4.3.9, given (C± , , R, L, C, G) , suppose  > 0 is such that
α << 1. Then, e−α ≈ 1 and hence,
max V
min V
≈
1+ Γ (  )
1− Γ (  )
which is a constant. This suggests the following definition. Suppose a lossless, finite transmission line, terminated at some load, is embedded in a lossless medium. Then, the voltage standing wave ratio is defined by
V =
max V
min V
=
1+ Γ (  )
1− Γ (  )
Observe that for a low loss transmission line system (C± , , R, L, C, G) ,
VSWR provides a convenient and well-defined way of quantifying standing waves that exist on the transmission line system. Indeed, when the
load is matched with the characteristic impedance, Γ(ℓ) = 0 ⇒ V = 1. On
the other hand, when Γ(ℓ) = 1 (i.e., maximal reflection), then V = ∞. Show,
in Exercise 4.5.1, that if ω ≥ 0 is fixed, then α << 1 ⇔ λ << 1, where λ is the
wavelength of the propagating field.
4.3.10 Proposition
Given (C± , , R, L, C, G) , let γ = α + iβ be the wave propagation constant.
Suppose that λ << 1, where λ is the wavelength of the propagating timeharmonic voltage wave. Then, the following two conditions hold.
a) Let z = z n correspond to the minima of V = V(z), and z = zn correspond
to the maxima of V = V(z), where  < z n < zn < z n+ 1 < zn+ 1 <  ∀n .
Then, zn − z n ≈ λ4 ∀n .
b) Given VSWR V , Γ() = VV −+11 and hence, for a known zk for some k,
θ = 2β( − zk ) − 2 kπ ⇒ Γ() =
V −1
V +1
ei2β( − zk )−2 kπ =
V −1
V +1
ei2β( − zk )
Similarly, Γ(ℓ) can be determined if V and any fixed z k are known.
Proof
(a) From Lemma 4.3.9, zn − z n = 2πβ . For αℓ << 1, by Exercise 4.5.1,

2 π , and the result thus follows.
λ << 1 ⇒ β ≈ ω µε = λ
(b) It is easy to see that
V =
1+ Γ (  )
1− Γ (  )
⇒ V ( 1 − Γ() ) = 1 + Γ() ⇒ Γ =
Finally, Γ = Γ eiθ and zn = −
yields the desired result.
K15149_Book.indb 118
2 nπ+θ
2β
V −1
V +1
∀n = 0, 1, 2, …, from Lemma 4.3.9,
□
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Transmission Line Theory
Because VSWR V = VVmax
, it follows that should VSWR become very
min
large, then Vmax can become very large on parts of the transmission line.
This voltage amplification phenomenon is known as the Ferranti effect.
It is possible for breakdown voltage of the line to be attained via the
Ferranti effect. This would clearly be detrimental in any digital circuit.
However, this is generally more of a concern for power lines than for
digital circuits.
4.3.11 Theorem
Given (C± , , R, L, C, G) , the condition RL = GC constitutes a necessary and sufficient condition for the characteristic impedance of a transmission line to be
independent of the angular frequency of propagation. That is, Z0 ≡ L/C if
and only if R/L=G/C.
Proof
Now,
Z0 ≡
R + iωL
G + iωC
can be rewritten as
R + iωL
G + iωC
=
L
C
R
+ iω
L
G
+ iω
C
Hence,
R
L
=
G
C
if and only if
R+ ω
i
L
G
+ iω
C
=1
and thus yielding Z0 = L/C for any frequency ω ≥ 0.
□
A transmission line satisfying the condition stated in Theorem 4.3.11 is
said to be distortionless. As a trivial corollary, note that for a perfect conductor (R = 0) in a lossless dielectric (G = 0), Z0 = L/C automatically holds.
However, this does not mean that if the conditions of Theorem 4.3.11 should
be satisfied, there is no attenuation to the propagating TEM waves along
(C+ , C− ). Indeed, attenuation will occur whenever R ≠ 0: the energy needed to
overcome the finite electrical conductivity manifests as heat, that is, as Joule
heating. Theorem 4.3.11 merely states that the characteristic impedance will
appear as if the transmission line system were lossless and hence no signal
distortion for all angular frequencies of wave propagation.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
4.3.12 Proposition
Given a pair of finite transmission lines (C± , , R,L,C,G) , suppose ℓ > 0. If
ω > 0 satisfies β = 21 nπ for any n ∈N , then ZL ∈R ⇒ Z(0) ∈R and ZL ∈C ⇒
Z(0) ∈C .
Proof
The above assertions can be easily established by noting that
Z(0) = Z0
ZL + Z0 tanh γ
Z0 + ZL tanh γ
and tanh(α + iβ) =
tanh α + itanβ
1+ i tanh α tan β
where γ = α + iβ. Then, for β = nπ , n ∈N, tan β = 0, and hence,
tanh γ = tanh α ∈R. Thus, ZL ∈R ⇒ Z(0) ∈R and ZL ∈C ⇒ Z(0) ∈C. Next,
if β = 2 n2+ 1 π , n ∈N, then tanh γ = tanh1 α ∈R and hence, ZL ∈R ⇒ Z(0) ∈R
and ZL ∈C ⇒ Z(0) ∈C, once again, as claimed.
□
The above proposition yields a simple condition under which the input
impedance of a transmission line is resistive if the load is resistive. In general,
it is clear from the proof that even if the load is resistive, the input impedance
is complex. This section closes by demonstrating that there exists a conformal transformation mapping the input impedance (or the load impedance)
onto the reflection coefficient. The graph of the mapping is called the Smith
Chart, and it is often used to determine graphically the input impedance or
load impedance, given that the reflection coefficient at the load is known.
4.3.13 Theorem
Given a pair of lossless transmission lines ( γ ± ,  , L,C), set Zˆ ( z) = ZZ(0z ) to be
the normalized impedance along the transmission line. Then, there exists a
conformal transformation ς mapping the ρ-space onto the Ẑ -space given by
ς ( Γ ; z) =
1+ρ( z )
1−ρ( z )
for any fixed z ∈[0, ], where Γ = Γ() and ρ = Γe− iβ(  − z ) . In particular, on setting Zˆ = Rˆ + iXˆ and ρ = σ + iχ , where Rˆ , Xˆ , σ , χ ∈R , and the z-variable has
been suppressed for simplicity,
(σ − )
Rˆ
1+ Rˆ
2
+ χ2 =
1
(1+ Rˆ )2
(
and (σ − 1)2 + χ −
1
Xˆ
)
2
=
1
χ2
Proof
From Equations (4.20) and (4.21), it is clear that
Z( z) =
K15149_Book.indb 120
V ( z)
I ( z)
= Z0
1 + Γe− i2β(  − z )
eiβ(  − z ) + Γe− iβ(  − z )
= Z0
iβ (  − z )
− iβ (  − z )
e
− Γe
1 − Γe− i2β(  − z )
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Transmission Line Theory
Hence,
ς(Γ ; z) = Zˆ ( z) =
1+ρ( z )
1−ρ( z )
az + b
is well-defined. Moreover as the mapping cz
+ d ∀z ∈ C is a conformal transformation [2], it follows by inspection that ς is also conformal.
iχ
To complete the proof, observe by definition that Rˆ + iXˆ = 11+σ+
−σ− iχ . Hence,
equating real and imaginary parts,
1+σ+ iχ
1−σ− iχ
=
1−σ 2 −χ 2
+ (1−σi2)2χ+χ2
(1−σ )2 +χ 2
yields
Rˆ =
and Xˆ =
1−σ 2 −χ 2
(1−σ )2 +χ 2
2χ
(1−σ )2 +χ 2
Rearranging, and noting that
(σ − ) − ( )
Rˆ
1+ Rˆ
2
2 σRˆ
1+ Rˆ
+ σ 2 + χ2 −
Rˆ
1+ Rˆ
2
= σ2 −
2 σRˆ
1+ Rˆ
yields
Rˆ =
1−σ 2 −χ 2
(1−σ )2 +χ 2
⇔0=
Rˆ
1+ Rˆ
(
Likewise, noting that χ −
mutatis mutandis, that
Xˆ =
−
1
Xˆ
)
2
−
2χ
(1−σ )2 +χ 2
1
Xˆ 2
= χ2 −
2χ
Xˆ
1
1+ Rˆ
(
⇔ σ−
Rˆ
1+ Rˆ
)
2
+ χ2 =
1
(1+ Rˆ )2
, it follows, from the above proof
(
⇔ (σ − 1)2 + χ −
1
Xˆ
)
2
=
1
Xˆ 2
□
It is clear that
(σ − )
Rˆ
1+ Rˆ
defines a circle of radius
defines a circle of radius
1
1+Rˆ
1
χ
2
+ χ2 =
centered at
(
1
(1+ Rˆ )2
Rˆ
1+ Rˆ
1
Xˆ
)
(
, 0 , and (σ − 1)2 + χ −
1
Xˆ
)
2
=
1
χ2
( ) . These circles are orthogonal
centered at 1,
to one another at the point wherein they intersect with one another. This
forms the basis for the Smith Chart, and they define a coordinate system on
the complex plane.
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122
Electromagnetic Theory for Electromagnetic Compatibility Engineers
4.3.14 Remark
From the equivalence
Rˆ =
1−σ 2 −χ 2
(1−σ )2 +χ 2
(
⇔ σ−
Rˆ
1+ Rˆ
)
2
+ χ2 =
1
(1+ Rˆ )2
the centers of the circles lie on the χ-axis. Moreover, R̂ = 0 ⇔ σ 2 + χ 2 = 1 ; that
is, R̂ = 0 ⇔ ρ = 1. Next, the equivalence
Xˆ =
2χ
(1−σ )2 +χ 2
(
⇔ (σ − 1)2 + χ −
1
Xˆ
)
2
=
1
Xˆ 2
evinces that the center of the circles lies on the line σ = 1. In particular,
X̂ > 0 ⇔ χ > 0 yields a condition for inductive reactance, and X̂ < 0 ⇔ χ < 0
leads to a condition for capacitive reactance.
4.4 Impedance Matching and Standing Waves
A boundary point along a transmission line is typically the result of a change
in impedance at a point by virtue of a load. The terms load and boundary are
used interchangeably in this section when referring to a transmission line.
4.4.1 Example
Consider a transmission line pair (C± , , R, L, C, G) connected to a source
(V0 (ω ), ZS ) and terminated by a load ZL (see Figure 4.3). What is the input
impedance if (a) the line is shorted and (b) the line is open. Finally, consider
as a special case wherein R = 0 = G, that is, a perfect conductor embedded in
a perfect dielectric medium.
(a) When the line is shorted to ground, the load is zero: ZL = 0 . Hence,
by Equation (4.19),
Z(0) = Z0
Z0 tanh γ
Z0
= Z0 tanh γ
(b) When the line is open, the load is infinite: ZL = ∞ , hence, by (4.19),
Z(0) = Z0
ZL
ZL tanh γ
= Z0 coth γ
Now, set Z(0) = Z0 tanh γ and Z( ∞ ) = Z0 coth γ . Then, Z(0) Z( ∞ ) = Z02 ⇒ Z0 =
Z(0) Z( ∞ ) , a rather startling result! That is, the characteristic impedance can
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123
Transmission Line Theory
be deduced immediately simply by knowing the input impedance when the
line is shorted and when it is open.
Finally, to complete the example, consider the special case wherein R = 0 = G.
Recalling that tanh iϕ = i tan ϕ, it follows at once that Equation (4.19) reduces to
Z( z) = Z0
ZL + iZ0 tan β (  − z )
Z0 + iZL tan β (  − z )
(4.23)
where β = ℑmγ and in this instance, α = ℜeγ = 0. Hence, for case (a),
Z(0) = iZ0 tan β and for (b), Z( ∞ ) = − iZ0 cot β . Thus, depending upon the
length of the transmission line, the input impedance of a short or open circuit oscillates between being inductive and capacitive reactance; see the plot
for illustration.
inductive
reactance
Graphs of Tan x and Cot x
60
40
4.1
Cot x
4.62
3.58
3.06
2.54
2.03
1.51
0.99
–0
0.47
–0.6
–1.1
–1.6
–2.1
–2.6
–3.2
–3.7
Tan x
–4.2
0
–4.7
Amplitude
20
–20
–40
–60
capacitive
reactance
x (rads)
Explicitly, in view of the plot, for the case wherein R = 0 = G,
• Z(0) is inductive whenever  ∈ (n − 1) βπ ,(n − 21 ) βπ  ∀n ∈N .
• Z(0) is capacitive whenever  ∈ (n − 21 ) βπ , n βπ  ∀n ∈N .
• Z( ∞ ) is inductive whenever  ∈ (n − 21 ) βπ , n βπ  ∀n ∈N .
• Z( ∞ ) is capacitive whenever  ∈ (n − 1) βπ ,(n − 21 ) βπ  ∀n ∈N.
Unfortunately, for the general case wherein γ ∈C, no simple relationships
similar to those given above exist.
□
4.4.2 Proposition
Given a pair of finite transmission lines (C± , , R, L, C, G) as shown in
Figure 4.3, suppose (V0 (ω ), ZS ) is the source at z = 0 and the transmission line
terminates at some load ZL . Then,
∀z ∈[0, ] V ( z) =
K15149_Book.indb 123
V0 ( ω ) Z0
e− γ z
Z0 + ZS 1−Γ (0) Γ (  )e−2 γ 
{1 + Γ()e
−2 γ (  − z )
}
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Proof
Recall that the general solution of the voltage and current waves propagating
along the transmission pair (4.20) and (4.21) can be rewritten as
V ( z) = 21 I ()(ZL + Z0 )e γ {e− γ z + Γ()e− γ (2  − z ) }
(4.24a)
(ZL + Z0 )e γ {e− γ z − Γ()e− γ (2  − z ) }
(4.24b)
I ( z) =
1 I ( )
2 Z0
From Figure 4.4, it is clear from Ohm’s law that V0 (ω ) = I (0)ZS + V (0), whence,
appealing to Equation (4.24), V0 (ω ) = ZS 2I (Z0) (ZL + Z0 )e γ {1 − Γ()e−2 γ  } + I (2) (ZL +
Z0 )e γ {1 + Γ()e −2 γ }. Thus,
V0 ( ω ) 2
ZL + Z0 I (  )
where Γ(0) =
ZS − Z0
ZS + Z0
e− γ  =
ZS
Z0
{1 − Γ()e−2 γ  } + 1 + Γ()e−2 γ 
{
+ 1 + Γ()e−2 γ  1 −
=
ZS
Z0
=
ZS + Z0
Z0
ZS
Z0
}
{1 + Γ(0)Γ()e−2 γ  }
. In turn, this implies that
1
2
I ()(ZL + Z0 )e γ =
V0 ( ω )
Z0
ZS + Z0 1+Γ (0) Γ (  )e−2 γ 
Substituting this back into Equation (4.24a) and noting that e− γ z + Γ()e− γ (2 − z ) =
e− γ z {1 + Γ()e−2 γ (  − z ) } , the assertion follows immediately.
□
Provide an alternative proof for Proposition 4.4.2 (cf. Exercise 4.5.4) via the
explicit reflection of waves occurring at each boundary interface, that is, at
the source and at the load, respectively.
4.4.3 Corollary
Given the conditions of Proposition 4.4.2,
I ( z) =
V0 ( ω )
e− γ z
Z0 + ZS 1−Γ (0) Γ (  )e−2 γ 
{1 − Γ()e
−2 γ (  − z )
} , ∀z ∈[0, ].
Proof
The proof follows that of Proposition 4.4.2 mutatis mutandis.
□
In general, wave reflections are to be avoided as they have a tendency to
form (partial) standing waves between the source and the load, leading to
unwanted emissions and thereby rendering digital devices to be potentially
noncompliant with regulatory agency requirements. Moreover, reflected
waves may potentially cause electromagnetic interference by coupling onto
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125
Transmission Line Theory
adjacent lines, switching transistors on and off randomly. The following theorem also demonstrates why impedance matching is important; note, however, that for digital electronics, this aspect is usually immaterial.
4.4.4 Theorem
Given a finite transmission line system (C± , , R,L,C,G), maximal power
transfer occurs at the load if and only if Γ(ℓ) = 0.
Proof
Let p = p(z,t) denote the instantaneous power along γ ± . From Equation (4.22),
p( z) = V ( z)I ∗ ( z) =
V0+
Z0
2
2
{e2 α (  − z ) − Γ() e− iθ ei2β(  − z ) + Γ() eiθ e− i2β(  − z ) − Γ() e−2 α (  − z ) }
where γ = α + iβ. Thus, on setting ⟨ p( z)⟩ = 21 ℜe p( z) , it can be shown (see
V+
2
Exercise 4.5.2) that P() = 20Z0 {1−|Γ()|2 }. Whence, maximal power transfer
□
p() = max ⟨ p()⟩ ⇔ Γ() = 0 , as claimed.
Γ
4.4.5 Corollary
Under the conditions of maximal power transfer along a pair of transmission lines toward a fixed load, the transmitted voltage across the load is
V () = Z0Z+0ZS V0 (ω )e− γ  .
Proof
By Theorem 4.4.4, maximal power transfer implies that Γ(ℓ) = 0. The result
thus follows at once from Proposition 4.4.2.
□
Note that ⟨ p( z)⟩ = 21 ℜe p( z) in the proof of Theorem 4.4.4 defines
the time-average power (see the time-average power density) defined by
⟨S⟩ = 21 ℜe( E × H ∗ ) , where S is the Poynting vector. From Theorem 4.4.4, it
is clear that the absence of reflection results in maximal power transfer: all
the transmitted power is absorbed by the load. Physically, this is obvious as
reflection implies part of the incident energy is redirected away from the
transmitted energy via 1 + Γ = Τ (transmission coefficient).
In spite of the less appealing nature of reflected waves, reflection has useful applications in the high-technology industry, for instance, in the memory
architectures of personal computers. As a concrete example, suppose the
input into an integrated circuit (IC) requires some fixed voltage V0 . Now, if
the input impedance (into the IC) is extremely high relative to the characteristic line impedance, it is possible to utilize reflection at the input impedance
to drive the line at 21 V0 and thereby reduce power consumption. Explicitly,
by designing the circuit such that Γ ≈ 1 at the load, the resultant transmission
coefficient at the input is T = 1 + Γ ≈ 2: thus, 21 V0 → V0 , which is the required
input voltage V0, as claimed. For more details, see Exercise 4.5.3.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
A transmission line is said to be electrically long if |γ| >> 1, where γ is the
wave propagation constant. Finally, define a transmission line to be electrically
short if |γ| << 1. It is clear from
Z( z) = Z0
ZL + Z0 tanh γ (  − z )
Z0 + ZL tanh γ (  − z )
that Z(0) ≈ ZL , where ZL is the load impedance at the end of the transmission
line as |γ| << 1 ⇒ tanh γ ≈ 0 . Thus, if a transmission line is electrically short,
the voltage transmitted across the load is V () = Z0Z+LZL V (0) , where V(0) is the
output voltage from the source appearing at z = 0 of the transmission line. Thus,
for an electrically short line, the load-line system forms a simple voltage divider.
4.4.6 Lemma
Given a pair of lossless transmission lines (C± , , L,C) with some source
(VS , ω , ZS ) and load ZL , there exists an infinite discrete set of angular frequencies {ω n } such that the lines appear to be electrically short with respect to {ω n } .
Proof
For a lossless line, β = ω LC . From (4.23), set ω n =
tan β n  = 0 ∀n ⇒ Zin = Z(0) = ZL , as required.
nπ
 LC
for n = 1, 2,  . Then,
□
Observe from Proposition 4.4.2 that if Γ(0), Γ(ℓ) ≠ 0, waves will be reflected
back and forth along a finite transmission line (see Exercise 4.5.4). Thus,
impedance mismatch at the source and load leads to standing waves on the
transmission line. An example below illustrates the technique of voltage
reflection diagrams.
4.4.7 Example
Consider a finite transmission line system (C± , , R, L, C, G) with reference to
Figure 4.3 and the general equation V ( z) = V0+ e− γ z + V0− e γ z . Suppose without
loss of generality that Γ(0) ≠ 0 ≠ Γ(ℓ), and the lines are lossless for simplic1
ity: R = 0 = G. Set υ = LC
to be the speed of wave propagation and τ = υ .
+
Then, at t = 0, V = V0 . At t = τ, the reflected wave is V0− = Γ()V0+ . This wave
travels back toward the source, whereupon it is reflected by the source
impedance at t = 2τ, and the reflected voltage is V1+ = Γ(0)V0− = Γ(0)Γ()V0+ .
Now, this reflected wave in turn propagates toward the load at t = 3τ, yielding V1− = Γ()V1+ = Γ(0)Γ 2 ()V0+ , and at t = 4τ, the reflected wave is given by
V2+ = Γ(0)V1− = Γ 2 (0)Γ 2 ()V0+ .
In general, it is easy to see inductively that ∀n ≥ 0,
 Vn+ = Γ n (0)Γ n ()V0+ for t = nτ


 Vn− = Γ n+1 (0)Γ n ()V0+ for t = (n + 1)τ
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Transmission Line Theory
This leads to the voltage reflection diagram
t = 3τ
Γ(0)Γ(ℓ)V +0
t = 2τ
Γ(ℓ)V +0
t=τ
V +0
t=0
z=0
z=ℓ
The gradient of the forward propagating wave is 1/v
The gradient of the backward propagating wave is –1/v
To complete the example, consider the plot of the transmitted voltage as
a function of time for some fixed distance 0 < zˆ <  along the transmission
line. For notational simplicity, set τˆ = υzˆ . For t ∈[0, τˆ ), V = 0 at z = ẑ as it takes
finite time for the wave to propagate to z = ẑ. Namely, it takes t = τˆ before
the wave reaches z = ẑ .
For notational convenience, set τ1− = 0, τ1+ = τˆ , τ 2± = 2 τ ± τˆ , , τ n± = nτ ± τˆ ∀n.
This yields the sequence
0 = τ1− < τ 1+ < τ < τ −2 < 2 τ < τ +2 <  < τ −n < nτ < τ +n < 
Then, for t ∈[τ1+ , τ 2− ) , the transmitted wave is V = V0+ . In particular, observe
that at t = τ 1+ , δV ( zˆ ) = lim− V ( z) − lim+ V ( zˆ ) = V0+ as on z ∈[0, zˆ ], V ( z) = V0+
z→( zˆ )
z→( zˆ )
whereas for z ∈( zˆ , ], V(z) = 0 as finite time is required for signal propagation
(see the voltage reflection diagram).
Likewise, at t = τ −2 , it is clear from the voltage reflection diagram
that on z ∈[0, zˆ ], V = V0+ (1 + Γ()) , whereas for z ∈( zˆ , ], V ( z) = V0+ ,
whence, at t = τ −2 , the voltage at z = ẑ is discontinuous by the amount
δV ( zˆ ) = lim− V ( zˆ ) − lim+ V ( zˆ ) = Γ()V0+ .
t→( zˆ )
t→( zˆ )
Thus, by induction, it is clear that at each t = nτˆ , there exists a voltage
discontinuity at z = ẑ given by
 δV2 n ( zˆ ) = Γ n− 1 (0)Γ n− 1 ()V0+


 δV2 n+ 1 ( zˆ ) = Γ n− 1 (0)Γ n ()V0+

for all n ≥ 1. This can be sketched as shown in Figure
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128
Electromagnetic Theory for Electromagnetic Compatibility Engineers
+
+
+
V 0 + Γ(l)V 0 + Γ(l)Γ(0)V 0
V +0 + Γ(l)V +0
+
V0
τ̂
0
τ
τ–2 2τ τ+2
Figure 4.6
Successive reflection of voltage waves between source and load.
In general, the voltage at ẑ for t ∈[τ −n , τ n+ ) is given by
 V + (1 + Γ() + + Γ nˆ − 1 (0)Γ nˆ − 1 ()) if n = 2 nˆ
0

V ( zˆ ) = 
 V0+ (1 + Γ() + + Γ nˆ − 1 (0)Γ nˆ ())
if n = 2 nˆ + 1

□
4.4.8 Lemma
Given (C± , , R,L,C,G), set γ = α + iβ and suppose that β = nπ for any fixed
n = 1, 2, . Then, for arbitrary load ZL ,
(a) α >> 1 ⇒ Z(0) ≈ Z0
(b) α << 1 ⇒ Z(0) ≈ ZL
In particular, Z(0) is purely resistive if ZL is purely resistive.
Proof
From
tanh(α + iβ) =
tanh α + itanβ
1+ i tanh α tan β
,
it is clear that tanh(α + iβ)ℓ = tanh αℓ and the conclusion to (a) and (b) thus
follows from
Z( z) = Z0
ZL + Z0 tanh γ (  − z )
Z0 + ZL tanh γ (  − z )
.
Explicitly, it suffices to observe that αℓ >> 1 ⇒ tanh αℓ ≈ 1 whereas αℓ << 1 ⇒
tanh αℓ ≈ 0.
□
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Transmission Line Theory
4.4.9 Lemma
Given (C± , , R,L,C,G), set γ = α + iβ and suppose that β =
n = 1, 2,  . Then, for arbitrary load ZL ,
2 n− 1
2
π for any fixed
a) α >> 1 ⇒ Z(0) ≈ Z0
b) α << 1 ⇒ Z(0) ≈
Z02
ZL
= ZL
( )
Z0 2
ZL
In particular, Z(0) is purely resistive if ZL is purely resistive.
Proof
The proof is similar to that of Lemma 4.4.8.
□
Observe that the conclusions of Lemma 4.4.8 and Lemma 4.4.9 for the case
wherein αℓ >> 1 are identical. In particular, the input impedance is identical to the characteristic impedance of the transmission lines; that is, the
input impedance is independent of the load. Hence, maximal power transfer
occurs if the source ZS = Z0 . More important, electromagnetic emissions are
minimized under this condition. This is summarized below.
4.4.10 Corollary
Given (C± , , R,L,C,G) with a source (VS , ω , ZS ) and a fixed γ = α + iβ , if ℓ > 0
satisfies β = n2 π for some n ∈N , and αℓ >> 0, then for arbitrary load ZL ,
maximal power transfer is achieved if ZS = Z0 , and in particular, electromagnetic emissions are minimized.
Proof
By Lemma 4.4.8(a) and Lemma 4.4.9(a), the input impedance Z(0) = Z0 . Hence,
setting ZS = Z0 , yields (i) maximal power transfer and (ii) reflection suppression at the source and hence mitigating emissions for arbitrary loads.
□
As a side remark, for a long trace on a printed circuit board (PCB) with a
source (e.g., clocks) in the microwave regime, this corollary is still applicable
provided that at the load the current is enhanced by a current source. For
instance, the current source can be supplied via a parallel termination configuration, with a pull-up resistor to Vcc to supply the required current and a
pull-down to complete the impedance match (see Figure 4.7).
On a side note regarding digital circuit design, is there an optimal placement along a trace for a series surface mount resistor to be placed in order to
suppress emissions, assuming that its value is different from the characteristic impedance of the line for emission suppression? Clearly, placing the resistor R arbitrarily along the trace will result in unwanted reflections along the
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
VDD
R1
Z0
Source
Load
(R, L, C, G)
R2
R1 || R2 = R0
Figure 4.7
Active parallel termination scheme.
line. It is clear from transmission line theory that placing the resistor as close
to the load as possible will optimize emissions suppression. This practice is
commonly employed in the electronics industry.
4.4.11 Lemma
Given a pair of transmission lines (C± , , R,L,C,G), the characteristic impedance Z0 of the line is always inductive. That is, ℑm(Z0 ) ≥ 0.
Proof
Now,
Z0 ≡
R + iωL
G + iωC
= R0
( )
R i
+ω
L
G
+ iω
C
1
2

= R0 

(
RG
+ω 2 + iω G
−R
LC
C L
2
G 2
+ω
C
( )
1
) 2

where R0 = L/C . As before, setting


(
1
RG
+ω 2 + iω G
−R
LC
C L
2
G
+ω 2
C
( )
)  2 = a + ib


gives
a2 − b2 =
RG
+ω 2
LC
G 2
+ω 2
C
( )
and 2 ab =
( )
2
( GC ) +ω2
ω G
−R
C L
whence, noting that
1
 RG +ω 2  2  ω ( G − R )  2  2
2
2 2
2
2 2
2
2
2
( a + b ) = ( a − b ) + (2 ab) ⇒ a + b =  GLC2 2  +  G C2 L 2  
( ) +ω   ( C ) +ω  
 C
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131
Transmission Line Theory
yields
1
a=
1
2
  RG 2  2  G R  2
2
RG
ω( C − L )
+ω
+ω 2 

LC
LC
  G 2 +ω 2  +  G 2 +ω 2  + G 2 +ω 2 
(C) 
  (C)

  ( C )

1
2
  RG 2  2  G R  2
2
RG
ω( C − L )
+ω 2 
+ω

LC
LC
  G 2 +ω 2  +  G 2 +ω 2  − G 2 +ω 2 
(C) 
  (C)

  ( C )

1
b=
Because trivially, α 2 + β 2 − α ≥ α − α = 0 , it follows immediately that b ≥ 0.
That is, ℑm(Z0 ) = R0 b ≥ 0, as required.
□
Thus, it is evident that the characteristic impedance of traces on a PCB
cannot be capacitive regardless of its length. However, depending upon
the load at the end of the transmission line and the length of the transmission line, the input impedance can clearly be made capacitive, inductive, or
purely resistive.
4.4.12 Proposition
There exists at most a countably infinite set of angular frequencies {ω n } such
that the input impedance Z(0) of a pair of lossless transmission lines (C± , , L,C)
terminated by a purely resistive load RL is real.
Proof
It suffices to note that
Z(0) = R0
RL + iR0 tan ω LC 
R0 + iRL tan ω LC 
= RL
if, tan ω LC  = 0, where R0 = L/C . This condition is satisfied if ω =
for each n = 1,2,…. Likewise, if
ω LC =
2 n− 1
2
nπ
 LC
,
π , n ∈N
R2
then Z(0) = RL0 ∈R . Thus, the desired countably infinite set of angular frequencies is precisely
{
K15149_Book.indb 131
nπ
 LC
} {
: n ∈N ∪
2 n− 1 π
2  LC
}
: n ∈N
□
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Now, consider the case wherein a lossless transmission line is terminated to ground: ZL = 0 . That is,
Z( z) = R0
RL + iR0 tan β (  − z )
R0 + iRL tan β (  − z )
= iR0 tan β( − z)
Then, it is evident that along z ∈[0, ] such that β( − z ) = nπ , Z( z ) = 0 and
hence, those points may be shorted to ground without affecting the wave
propagation. On the other hand, at points z ∈[0, ] such that β( − z ) = 2 n2− 1 π ,
Z( z ) = ± i∞ and hence, those points may be cut without affecting the wave
propagation along the line.
A similar analysis can be made for the case wherein a lossless transmission line is open. Indeed, in this instance,
RL → ∞ ⇒ Z( z) = R0
RL + iR0 tan β (  − z )
R0 + iRL tan β (  − z )
= − iR0 cot β( − z)
whence, along z ∈[0, ] such that β( − z ) = nπ , Z( z ) → ± i∞ ; thus, at these
points, the line may be cut without affecting the wave propagation along the
line. Likewise, along points z ∈[0, ] such that β( − z ) = 2 n2− 1 π , Z( z ) = 0 and
hence at these points may be grounded without affecting the wave propagation along the line.
4.4.13 Lemma
Given a pair of transmission lines (C± ,,R, G,L, C), given any ε > 0,
∀ω >
G R
−
1 C L
2 (1+ε )2 − 1
2
+
 GC − RL 
 (1+ε )2 − 1  −
1
2
( )
2 RG 

− LC 
4  (1+ε )2 G
C


(1+ε )2 − 1
⇒ Z0 (ω ) − R0 < R0 1 + iε ,
where R0 = L/C .
Proof
First, observe in general that Z0 (ω ) = GR ++ iiωωCL is a continuous function of ω.
Hence, by continuity, for any given ε > 0, ∃ω ε > 0 satisfying the lemma. Thus,
the proof is complete if ω ε can be determined. Rearranging the expression
for Z0 yields:
R + iωL
G + iωC
=
L
C
(
)
1
1− iR/ωL 2
1− iG/ωC
=
L
C
(
)
1
(1− iR/ωL)(1+ iG/ωC) 2
1+ (G/ωC)2
=
L
1
C (1+ (G/ωC)2 )1/2
(1 +
i
ω
( GC − RL ) + ωRGLC )
1
2
2
Hence, for any given ε > 0, it suffices to seek ω such that
1−
K15149_Book.indb 132
1
1+ (G/ωC)2
{1 +
i
ω
( GC − RL ) + ωRGLC }
2
1
2
<ε
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133
Transmission Line Theory
2
a −||
b ≤|a − b|, it suffices for ω to satisfy
because |1 + iε|= 1 + ε > ε . So, from ||
the inequality:
1
1+ (G/ωC)2
(1 +
i
ω
( GC − RL ) + ωRGLC )
1
2
< 1+ ε
2
This is equivalent to
1+
i
ω
( GC − RL ) + ωRGLC
2
{
< (1 + ε)2 1 + ( ωGC )
2
}
However,
1+
i
ω
( GC − RL ) + ωRGLC
2
≤ 1+
RG
ω 2 LC
+
i
ω
( GC − RL )
Thus, it is enough to have
1+
RG
ω 2 LC
+
−
1 G
ω C
R
L
{
< (1 + ε 2 ) 1 + ( ωGC )
2
}
So, multiplying the inequality by ω 2 yields:
ω 2 ((1 + ε)2 − 1) − ω
G
C
−
R
L
+ (1 + ε)2 ( GC ) −
2
RG
LC
>0
The roots of the equation are:
ω ± ( ε) =
G R
−
1 C L
2 (1+ε )2 − 1
2
±
1
2
 GC − RL 
 (1+ε )2 − 1  −
( )
2 RG 

− LC 
4 (1+ε )2 G
C


(1+ε )2 − 1
Therefore, set ω ε = ω + (ε). Then, for any ε > 0, ω > ω ε implies that
□
Z0 − L/C < L/C 1 + iε as required.
The above lemma asserts trivially that for high enough frequencies, the
characteristic impedance of the transmission line may be approximated with
that of the lossless line up to first order in ε, where ε is some positive number.
Some comments regarding standing waves are in order. Consider
Equation (4.22):
{
V ( z) = V0+ e γ (  − z ) + Γ()e− γ (  − z )
}
Now, consider the case wherein Γ(ℓ) = 1. Then, by definition,
{
}
V ( z) = V0+ e γ (  − z ) + e− γ (  − z ) = 2V0+ cosh γ ( − z)
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134
Electromagnetic Theory for Electromagnetic Compatibility Engineers
and hence, at the load, V () = 2V0+ . That is, twice the incident voltage is transmitted to the load. On the other hand, for Γ(ℓ) = −1,
{
}
V ( z) = V0+ e γ (  − z ) − e− γ (  − z ) = 2V0+ sinh γ ( − z)
and hence, at the load, V(ℓ) = 0 no voltage is transmitted to the load. Finally,
define a standing wave along a transmission line to be a partial standing wave
whenever 0 < Γ() < 1. Here, the transmission line contains both traveling
waves and (partial) standing waves.
Physically, standing waves are the result of energy stored as fields in a
reactive line. Because reactive energy is not transmitted to the load, it is, in
effect, reducing the total energy that the load can absorb from the incident
wave. Indeed, this fact is expressed by the definition of the time-average
power at any point along the transmission line:
⟨ p( z)⟩ ≡
1
T
∫
T
0
p(t , z)dt = 21 Re( P( z , ω ))
where T is a single period of the wave.
Mathematically, the time-average reactive power is always identical to
zero. As mentioned above, this corresponds to the power that is unavailable
to the load. To see this, consider the definition of time-average power, and
noting that c ∈C ⇒ ℜe c = 21 (c + c∗ ),
p( z, t) = ℜe v( z, t)ℜe i( z , t)
=
1
2
=
1
4
=
1
2
=
1
2
{v(z, t) + v (z, t)} {i(z, t) + i (z, t)}
{v(z, t)i(z, t) + v (z, t)i (z, t) + v (z, t)i(z, t) + v(z, t)i (z, t)}
{ℜe(v(z, t)i(z, t)) + ℜe(v(z, t)i (z, t))}
{ℜe(V(z)I(z)e ) + ℜe(V(z)I (z))}
∗
∗
1
2
∗
∗
∗
∗
∗
i2 ωt
∗
The first term corresponds to the reactive term, and it is evident that the time
average of that term vanishes. Hence, it does not contribute towards the load.
This chapter concludes with a brief analysis on time delay along a transmission line. This has strong implications in the design of high-speed
1
digital circuits where timing is critical. Given (C± , , R,L,C,G), set υ = LC
.
Then, the time it takes for a signal to propagate from the source to the load
is τ  =  LC .
There are a number of methods to delay a signal. The easiest is to increase
the length of the trace. However, this will affect the input impedance,
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135
Transmission Line Theory
Time delay: τ = RC
(R, L, C, G)
R
C
Figure 4.8
Implementing a simple RC-delay circuit.
assuming real estate on the PCB is available. A simple alternative is to implement an RC circuit as shown in Figure 4.8.
It is immediately clear that using an RC-delay circuit will affect the
input resistance. Indeed, there will be a nonzero coefficient of reflection
occurring on both sides of the transmission line where the RC-circuit is
implemented. As this is undesirable, it is expedient to implement the circuit close to the load or source. Furthermore, to prevent reflection back
from the source, an option is to add an impedance at the source such that
the resultant time delay from the impedance and the RC-circuit meets the
required specification.
4.4.14 Remark
Consider (C± , ,L,C) and suppose that the shortest data pulse propagating
along the line is τd . If the load and source are not matched with the line,
then clearly multiple reflections along the line will occur. Reflection occurring at the load will return to the load at t = 2τ, where τ =  LC . Finally,
suppose that the minimal pulse width δτ to which the load can respond
is δτ ≥ τd .
If ℓ > 0 is such that 2τ > δτ, then the reflected wave can potentially trigger
the load, leading to multiple false data pulses. Hence, to avoid this problem,
the easiest implementation is to require that 2τ < δτ; then, multiple reflections
will not be seen by the load as multiple data pulses. Suppose a design specification requires that 2 nτ < δτ , for some n ∈N. Then, it will suffice to choose
ℓ > 0 such that 2 n LC < τd ⇒  = 2 nτdLC .
4.5 Worked Problems
4.5.1 Exercise
Given a finite transmission line system (C± , , R,L,C,G), where the angular
frequency ω of a time-harmonic wave is fixed, set α = ℜeγ and β = ℑmγ. Then,
α << 1 ⇔ λ << 1.
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136
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Solution
Recalling that
1
2

σ 2
α = ω µε  21  1 + ( ωε
− 1 
)

 
after some tedious manipulation, it is easy to show that
α << 1 ⇔
2
( ω µε )2
σ
>> 1 + ( ωε
) −1⇔
2
{
2
( ω µε )2
} − 1 >> (
+1
2
)
σ 2
ωε
Exploiting the identity a 2 − b 2 = ( a + b)( a − b) , the above inequality is equivalent to
σ
ωε
<<
1 + ( ω 1µε )2 ⇔ σ <<
2
ω µε
However, ω 2 µε =
4π2
λ2
σ <<
2

ε
µ
1 + ( ω 1µε )2 =
2
2 ωµ
1 + (ω )2 µε
, whence
2
 2 ωµ
1 + 4π 2 ( λ ) ⇒ σ <<
2
1 λ
πυµ 
⇔

λ
<< 1
as claimed, where υ = fλ.
□
4.5.2 Exercise
(V0+ ) {1 − Γ() 2 } in the proof of Theorem 4.4.2.
2 Z0
2
Establish P() =
Solution
First, recall that eiθ = cos θ + i sin θ. Then, from
(V0+ ) {e2α( − z) − Γ() e− iθ ei2β( − z) + Γ() eiθ e− i2β( − z) − Γ() 2 e−2α( − z) }
Z0
2
p(z) = V (z)I ∗(z) =
expanding the expression below and appealing to the identity sin(a + b) = sin a
cos b + cos a sin b,
Γ() {−e− iθ ei2β(  − z ) + eiθ e− i2β(  − z ) }
= Γ() {sin θ cos 2β( − z) − cos θ sin 2β( − z)}2i
= i2 Γ() sin(θ − 2β( − z))
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137
Transmission Line Theory
and hence,
p( z) = V ( z)I ∗ ( z) =
(V0+ ) e2α( − z) − Γ() 2 e−2α( − z) + i2 Γ() sin(θ − 2β( − z))

Z0
2
{
}
(V0+ ) e2α(− z) {1 − Γ() 2 e−4α(− z) }
2 Z0
2
Thus, ⟨ p( z)⟩ = 21 ℜe p( z) =
yielding
(V0+ ) {1 − Γ() 2 } ⇒ ⟨ p()⟩ = max ⟨ p()⟩ ⇔ Γ() = 0
2 Z0
2
⟨ p()⟩ =
Γ
□
by inspection.
4.5.3 Exercise
Consider the schematic diagram, where R = 0 = G.
Source
impedance
Clock
V = λV0(ω)
Zs
Require V0(ω) to latch
the load HIGH
z=0
C+
z=l
(R, L, C, G)
Load
impedance
Vcc
R
Output
C–
Suppose a clock can only supply V = λV0 , for some 0 < λ < 21 , and the required
voltage by the load to signal High (or 1) is V = V0 . In the schematics shown,
the input to the IC only depends upon the voltage level and not the current,
the output is an open-drain, and the required voltage is supplied by Vcc via
some pull-up resistor R. Furthermore, suppose ZS << Z0 << ZL .
(a) Determine the length of the trace C+ and series resistor r such that
V () = V0 (ω ) at the load impedance.
(b) Find the correct series resistor r such that no reflection will occur at
the source.
(c) Are the stipulations of (a) and (b) compatible?
Solution
(a) By Proposition 4.4.2,
V () =
K15149_Book.indb 137
λV0 ( ω ) Z0
e− γ 
Z0 + ZS′ 1−Γ (0) Γ (  )e−2 γ 
{1 + Γ()}
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138
Electromagnetic Theory for Electromagnetic Compatibility Engineers
where ZS′ = ZS + r . By assumption, ZL >> Z0 ⇒ Γ() ≈ 1 . Hence,
V () =
2 λV0 ( ω ) R0
e− iβ 
R0 + ZS′
1−Γ (0)e− i 2 β 
Now, in order for V () = V0 (ω ) , it suffices that
1=
be satisfied. Thus, Γ(0) =
2 λR0
e− iβ 
R0 + ZS′ 1−Γ (0)e− i 2 β 
ZS′ − R0
ZS′ + R0
=
2 λR0
e− iβ 
R0 + ZS′ 1−Γ (0)e− i 2 β 
yields
2 λR0 e− iβ 
ZS′ e− iβ  (eiβ  − e− iβ  )+ R0 e− iβ  (eiβ  + e− iβ  )
=
λR0
ZS′ sin β + R0 cos β
Next, invoking the trigonometric identity a sin x + b cos x = a 2 + b 2 sin( x + y ),
where y = sgn(b)arccos 2a 2 and sgn(x) is the signum function defined by
a +b
 1 if x > 0

sgn( x) = 

 −1 if x < 0
it follows at once that sin(β + φλ ) =
λR0
R02 + ( ZS′ )2

 = β1 arcsin

where ϕ λ = arccos
ZS′
R02 + ( ZS′ )2
and hence,
λR0
R02 + ( ZS′ )2

− ϕλ 

, as desired.
(b) To prevent reflection from the source, 0 = Γ(0) = ZZSS′′ −+ RR00 ⇒ ZS′ = R0 ⇒
r = R0 − ZS .
(c) It is clear that (a) and (b) are compatible ∀r ≥ 0. That is, it is possible
to suppress reflection from the source and obtain the desired voltage
transmission by appropriately selecting the length of the trace.
□
4.5.4 Exercise
Prove Proposition 4.4.2 explicitly via reflection of waves.
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139
Transmission Line Theory
Solution
Set V0+ = ZV0S+ZZ0S e− γ  from the proof of Proposition 4.4.2. Then, via Example
4.4.7, V ( z) = V0+ {e γ (  − z ) + Γ()e− γ (  − z ) } becomes
V ( z) = V0+ {e γ ( − z ) + Γ()e− γ ( − z ) + Γ(0)Γ()e γ (3 − z ) + Γ(0)Γ 2 ()e− γ (3 − z ) + Γ n (0)Γ n ()e γ ((2 n+ 1) − z ) + Γ n (0)Γ n+ 1 ()e γ ((2 n+ 1) − z ) + }
= V0+ e γ ( − z ) (1 + Γ()e−2 γ ( − z ) )(1 + Γ(0)Γ()e−2 γ + Γ 2 (0)Γ 2 ()e−4 γ + )
= V0+ e γ ( − z ) (1 + Γ()e−2 γ ( − z ) ){1 + Γ(0)Γ()e−2 γ + (Γ(0)Γ()e−2 γ )2 + }
= V0+ e γ ( − z ) (1 + Γ()e−2 γ ( − z ) ) 1−Γ (0)Γ1( )e−2 γ yielding
V ( z) =
VS Z0
e− γ z
Z0 + ZS 1−Γ (0) Γ (  )e−2 γ 
{1 + Γ()e−2 γ (  − z ) }
□
as required.
4.5.5 Exercise
Suppose a load on a lossless transmission line (C± , , L,C) is capacitive:
ZL = RL − iX L , for some X C > 0 . Suppose further that the load is a digital logic
input that responds to a minimal pulse width of τd . Find (Z , z ) such that at
z = z , Γ( z ) =
Z ′ ( z )− R0
Z ′ ( z )+ R0
=0
where Z ′( z ) = Z ||Z( z ). Unwanted reflections are thus mitigated by the presence of the impedance Z placed at z = z .
Solution
Consider the schematic diagram, and determine δZ such that the input
impedanceat z = l′ is matched.
z=0
z = l'
δl
z=l
R0
δZ
K15149_Book.indb 139
ZL
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140
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Now, set Y ′ = δY + YL to be the input admittance at z = ′, and let δy = R0 δY ,
y L = R0YL be the normalized admittance, where
YL =
1 R0 + iZL tan βδ
R0 ZL + iR0 tan βδ
Hence, the requirement that
Γ ′( ′) =
Z ′ (  ′ )− R0
Z ′ (  ′ )+ R0
=0
that is, that there be zero reflection at z =  ′ ⇒ y ′ = 1. In other words
1 = δy + y L ⇒ ℜe(δy ) = 1 − ℜe( y L ) and ℑm(δy ) = −ℑm( y L )
It is a somewhat tedious matter to show that
y L =
R0 ZL sec 2 βδ
ZL2 + R02 tan 2 βδ
+i
( ZL2 − R02 )tan 2 βδ
ZL2 + R02 tan 2 βδ
To complete the solution, observe from Remark 4.4.14 that 2 nδ LC < τ d ⇒
δ < 2 nτdLC . Hence, set δ = 4 τdLC ; then, the required matching impedance δZ
is obtained via
ℜe(δy ) = 1 −
R0 ZL sec 2 βδ
ZL2 + R02 tan 2 βδ
and ℑm(δy ) = −
( ZL2 − R02 )tan 2 βδ
ZL2 + R02 tan 2 βδ
Note: It is clear from the above analysis why the Smith Chart was developed
to graphically determine the desired matching load.
□
4.5.6 Exercise
Consider the schematic diagram for series termination versus parallel termination. Describe qualitatively the impact of the respective terminations
on the rise time at the load. Suppose that ℓ >> δℓ, where δℓ is the distance
between the resistor and the load.
R
Z0
(a) Series termination
K15149_Book.indb 140
ZL
Z0
R
ZL
(b) Parallel termination
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141
Transmission Line Theory
Solution

(a) Series termination. Here, set CL = R + CL . Then, ∃ a > 1 such that
L
a CL , whence, τ ≈ αLC > LC = τ , the propagation delay
 =
C
time. That is, the propagation delay of the input voltage wave is τ
whereas the propagation delay of V(0), just after the resistor R, is τ.
From this, it is clear that the voltage reaches a peak at the load before
it reaches a peak at the input side into the series resistor, that is, faster
rise time at the load than at the input terminal. Thus, the far end
switches faster with respect to the input terminal, as the series termination introduces a longer delay at the input voltage. Intuitively,
recalling that a transmission line is inductive, introducing a series
resistor leads to a delay induced by a series RL-time constant at the
input terminal. Beyond the series termination, the propagating wave
sees the characteristic impedance.
(b) Parallel termination. The analysis follows that of (a) mutatis mutandis.
Here, the wave propagation does not see the load until it arrives at
z = ℓ. Hence, a propagation delay is introduced at the load and not
at the input terminal. There is thus a delay in the rise time at the
load with respect to the terminal. The scenario is thus the reverse
of (a). In short, parallel termination is good for slowing down rise
time for emissions reduction; however, series termination is used
instead of parallel termination if rise time is critical. Intuitively, the
wave propagating along the transmission line only sees the characteristic impedance until it arrives at the load; then, it sees a parallel
RZ-impedance which, in the case of a capacitive load, will introduce
an RC-time constant, delaying the rise time.
References
1. Cheng, D. 1989. Field and Wave Electromagnetics. Reading, MA: Addison-Wesley.
2. Churchill, R. and Brown, J. 1990. Complex Variables and Applications. New
York: McGraw-Hill.
3. Johnson, H. and Graham, M. 1993. High-Speed Digital Design. Upper Saddle
River, NJ: Prentice-Hall.
4. Kohonen, T. 1972. Digital Circuits and Devices. Englewood Cliffs, NJ: Prentice-Hall.
5. Krutz, R. 1988. Interfacing Techniques in Digital Design. New York: John Wiley & Sons.
6. Neff, Jr., H. 1981. Basic Electromagnetic Fields. New York: Harper & Row.
7. Plonsey, R. and Collin, R. 1961. Principles and Applications of Electromagnetic
Fields. New York: McGraw-Hill.
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5
Differential Transmission Lines
At the simplest level, a pair of differential lines comprises two transmission
lines such that the current flowing along one line is equal to the current
flowing in the opposite direction along the second line. Informally then, the
resultant current flowing along the pair is zero. Equivalently, the transmitted
voltage on one of the pair is 180 degrees out of phase with the voltage of the
other line. Thus, each line acts as the ground for its counterpart, and in this
sense, they have a common self-referencing ground.
From a theoretical perspective, differential lines are less sensitive to noise
and have fewer emissions than nondifferential lines. Thus, they are often
used in radio frequency (RF) design, especially for clock rates in the microwave regime. Notwithstanding, differential lines must still be matched or
they can cause undesired emissions.
A knowledge of differential pairs is extremely important to EMC engineers,
and an informal treatment can be found in References [8,10]. Differential
transmission line theory is essentially a corollary of transmission line theory,
and can thus be derived there from References [1,3–7]. This is the approach
taken in this chapter, wherein the bulk of the derivation leverages the results
from Chapter 4.
5.1 Differential Pair: Odd and Even Modes
The definition of an ideal differential pair is given below. The pair is assumed
to be surrounded in a homogeneous dielectric medium. In practice, differential pairs are typically routed over a ground plane; more is said later. First,
some notations are established below. In all that follows, where convenient
and unless explicitly stated otherwise, let Ω ⊂ R 3 denote an open subset upon
which circuits are modeled, and in addition, given a pair (C+ , C− ) of transmission lines of uniform cross-section s± , let γ ± : [0, 1] → Ω be a path representing
the axis of C± . Finally, the Euclidean metric ⋅ in R 3 is used whenever a distance function is invoked and it is also assumed that the cross-sectional area is
much less than the length of the transmission line: s± << γ ± .
143
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
5.1.1 Definition
Consider a pair of parallel transmission lines (C+ , C− ) , and denote the
lengths of the transmission lines by γ + , γ − , respectively. Suppose that the
cross-section s+ ( γ + ( s)) = s− ( γ − ( s)) ∀s ∈[0, 1], and (C+ , C− ) satisfies the following criteria:
(a)
γ + =  = γ − , for some  > 0
(b) d( γ + ( s), γ − ( s)) = constant ∀s ∈[0, 1]
where d( x , y ) = x − y denotes the distance separating the points x,y ∈ Ω,
(c) I + ( s) + I − ( s) = 0 ∀s ∈[0, 1]
where I ± ( s) ≡ I ( γ ± ( s)) denotes the current at the point γ ± ( s) along the lines
γ ± , respectively,
(d) R0 (C+ ) = R0 (C− )
where R0 (C± ) denotes the characteristic impedance of C± with respect to a
common ground.
Then, the pair (C+ , C− ) is said to be a symmetric (or balanced) differential pair.
If condition (d) is violated, then the pair forms an asymmetric (or unbalanced)
differential pair.
5.1.2 Remark
In the technology industry, a differential pair is often defined to be a pair
(C+ , C− ) where criterion (b)—which essentially requires that the lines be parallel to each other—is relaxed due to physical constraints and design considerations.* Furthermore, it is trivially true that (d) holds if the pair is isolated
in space, because then the characteristic impedance is defined by the pair.
Where condition (d) comes into play is the following scenario: in a typical circuit layout, a differential pair often has a common ground plane. The characteristic impedance R0 (C± ) is thus measured with respect to the ground plane
in question. This is not to be confused with the differential impedance of the
pair defined shortly.
*
A serpentine differential pair is a case in point.
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Differential Transmission Lines
An idealized differential circuit is depicted in Figure 5.1.
• Zdiff denotes the characteristic differential impedance defined later.
• R = Zdiff for a perfectly matched (symmetric) differential pair.
• The differential receiver is essentially a differential amplifier with
the resultant output being the difference between the impressed
voltages across R.
Let V± (R) denote the voltage developed across R by V± propagating along
C± . Then, the resultant output Vout = V+ (R) − V− (R) (modulo a scaling factor
if the output is amplified by the differential amplifier). It is shown below that
for a symmetrical differential line, Vout = 2V+ (R).
5.1.3 Lemma
Let (C+ , C− ) be a symmetrical differential pair as illustrated in Figure 5.1.
Suppose that the characteristic impedance of C± relative to a fixed ground
plane is Z0 . Then, V+ ( s) = −V− ( s) for 0 ≤ s ≤ 1, where V± ( s) = V ( γ ± ( s)) is the
voltage at γ ± ( s) along γ ± .
Proof
First, recall from Chapter 4 that, heuristically, the voltage on one line can be
induced along the other line via the distributed line model for a transmission
line. This can be represented as
V+ = I + Z++ + I − Z+−
(5.1)
where Z++ = Z0 is the characteristic impedance of C+ relative to some fixed
ground, and Z+− is the transfer impedance that induces a voltage on C+ as a
result of the current I − flowing through C− . By symmetry, the voltage propagating along C− is given by
V− = I + Z−+ + I − Z−−
I+
V+
Differential transmitter
(constant current source)
C+
Zdiff
I–
V–
(5.2)
Differential termination
R
Differential
receiver
C–
Figure 5.1
Schematic representation of an idealized symmetric differential pair.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
where Z−− = Z0 and Z−+ is the transfer impedance that induces a voltage on
C− as a result of the current I + flowing through C+ . Because the dielectric
medium is linear and homogeneous with the magnetic permeability µ = µ 0 ,
the transfer impedance must be symmetric: Z+− = Zˆ = Z−+ , for some Zˆ , where
Z+− ≡ VI−+ |I+ = 0 and Z−+ ≡ VI+− |I− = 0 . Adding Equations (5.1) and (5.2) and using
Definition 5.1.1(c) yield the result: V+ + V− = I − Zˆ + I + Zˆ = 0 ⇒ V+ = −V− .
□
5.1.4 Remark
In the proof of Lemma 5.1.3, let Zˆ = κZ0 , where κ > 0 is some constant called
the differential coupling coefficient. Then, Equation (5.1) becomes
V+ = I + Z11 + I − Z12 = I + (1 − κ )Z0
+
The ratio Zodd
= VI++ = (1 − κ )Z0 is known as the odd mode impedance along
+
−
C+. By symmetry, Zodd
= Zodd = Zodd
. For this reason, it is known as the odd
mode impedance of a differential pair, as I + = − I − . Further details are outlined shortly.
5.1.5. Lemma.
Given a symmetric differential pair, the odd mode impedance is given by
Zodd = Z0 − Ẑ .
Proof
By (5.1), V+ = I + Z11 + I − Z12 = I + (Z0 − Zˆ ) ⇒ Z0 − Zˆ =
V+
I+
= Zodd , as required.
□
The intent of the above brief account was to motivate an intuitive comprehension of a differential pair. A more rigorous approach via Maxwell’s
theory is sketched below. In particular, the odd and even mode impedances
are derived. The analysis is a special case of multitransmission line theory
covered in the following chapter. Indeed, much of the theoretical infrastructure has already been developed in Chapter 4.
Referring to Figure 5.2, consider initially a system of three conductors
embedded in a homogeneous dielectric medium (Ω, ε , µ , σ ), where the middle conductor C0 is grounded. Suppose I ± = I ± ( z, t) is propagating along conductor C± . Now, observe that the results of transmission line theory outlined
in Chapter 4 apply to each pair (C+ , C0 ) and (C− , C0 ) by evaluating the field
around loops γ + and γ − , respectively, defined in Figure 5.2. However, crosscoupling also occurs between the pair (C+ , C− ) , and the analysis carried out
for (C± , C0 ) also applies to (C+ , C− ).
Without loss of generality, consider the electric field E+ of an incident
TEM wave on (C+ , γ + ). Appealing to Stokes’ theorem, the integral version of
Equation (1.15) is obtained:
∫
γ+
K15149_Book.indb 146
E+ ⋅ dl = − ∂t
∫∫
S+
B+ ⋅ dS
(5.3)
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147
Differential Transmission Lines
E+z(z,t)
ey
ex
E+(z,t) 1
ez
I+(z,t) E+(z + δz,t)
z
C+
2
γ+
3 E+(z + δz,t) Ω = (Ω+,ε,µ) U(Ω–,ε,µ)
4
4
E–(z,t) 3
γ–
E–z(z,t)
–
1 E (z + δz,t) γ = γ–(1) + γ–(2) + γ–(3) + γ–(4)
2
C–
γ = γ+(1) + γ+(2) + γ+(3) + γ+(4)
I–(z,t)
E–z(z + δz,t)
Figure 5.2
Differential transmission line pair supporting a TEM wave.
where S+ = S( γ + ) ⊂ R 3 is an arbitrary compact surface spanned by γ + :
∂S+ = γ + , whence, following the argument in Chapter 4 mutatis mutandis, suppressing the (x,y) coordinates in the scalar and vector fields, yields
− ∂t

∫∫
S+
(3)
(2)
B+ ⋅ dS = − v+ ( z, t ; γ (1)
+ ) + v+ ( z + δz , t ; γ + ) + R + δzi+ ( z , t ; γ + )
(5.4)
+ R 0 δzi0 ( z, t ; γ (4)
+ )
where I 0 + I + + I − = 0 by Kirchhoff’s current law, and R 0 is the resistance per
unit length of the ground conductor. Similarly,
E− ⋅ dl = − ∂t
B− ⋅ dS
γ−
S−
yields
∫
− ∂t

∫∫
S−
∫∫
(1)
(2)
B− ⋅ dS = − v− ( z, t ; γ (3)
− ) + v− ( z + δz , t ; γ − ) + R − δzi− ( z , t ; γ − )
+ R 0 δzi0 ( z, t ; γ (4)
− )
(5.5)
where S− = S( γ − ) ⊂ R 3 is a compact surface spanned by γ − .
Now, recall from Chapter 4 that the magnetic flux per unit length
ψ ± = lim δ1z
B± ⋅ dS , where δz = γ (4)
± , is well-defined, and moreover, from
δz→ 0

∫∫
S±
the definition of magnetic flux Ψ = Li, it follows that
ψ + = L ++ δzI + + L +− δzI −
(5.6)
where L ++ is the self-inductance per unit length of conductor Γ + , coupling
the magnetic flux generated by current I + to S+ , and L +− is the mutual
inductance per unit length coupling the magnetic flux generated by current
i− to S+.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Inasmuch as the limit lim δ1z
δz→ 0
follows at once that lim
δz→ 0
with Equation (4.4) yield
1
δz

∫∫
S±

∫∫
S±
B± ⋅ dS is well-defined as lim δ1z S± < ∞ , it
δz→ 0
∂t B± ⋅ dS = L ++ ∂t I + ( z, t) + L +− ∂t I − ( z, t) together
∂ z v+ ( z, t) = −(R + + R 0 )I + ( z, t) − R 0 I − ( z, t) − L ++ ∂t I + ( z, t) − L +− ∂t I − ( z, t). (5.7)
By symmetry,
∂ z v− ( z, t) = −(R − + R 0 )I − ( z, t) − R 0 I + ( z, t) − L −− ∂t I − ( z, t) − L −+ ∂t I + ( z, t). (5.8)
5.1.6 Lemma
Given a system of three parallel conductors (C+ , C− , C0 ) as depicted in
Figure 5.2, if i− ( z, t) + i+ ( z, t) = 0 ∀z, t ≥ 0 , then the triple (C+ , C− , C0 ) is equivalent to the pair (C+ , C− ) where C0 is removed from the system of conductors.
Proof
From Kirchhoff’s current law, 0 = I 0 + I + + I − , I − + I + = 0 ⇒ I 0 ≡ 0 ∀z, t ≥ 0 and
similarly, the potential is also identically zero, whence, C0 may be removed from
the system without affecting the field distribution from C± , as required.
□
The above result is the reason why a differential pair is said to have a selfreferencing ground; that is, it possesses a “virtual” ground whereby the physical
ground conductor may be removed without affecting the electromagnetic field
distribution of the system. Thus, in view of Remark 5.1.7, the ground conductor
may be removed by setting R 0 = 0 when considering a differential pair; that is,
I − ( z, t) = − I + ( z, t) ∀z, t ≥ 0 . This simplifies Equations (5.7) and (5.8) to
∂ z v+ ( z, t) = − R + I + ( z, t) − (L ++ − L +− ) ∂t I + ( z, t)
(5.9)
∂ z v− ( z, t) = R − I + ( z, t) + (L −− − L −+ ) ∂t I + ( z, t)
(5.10)
5.1.7 Remark
Note in passing from Equation (5.7) that for a balanced differential pair
where I − ( z, t) = − I + ( z, t) ∀z, t ≥ 0 , (5.7) reduces automatically to (5.9) without
the need to set R 0 = 0. This implies that R 0 is arbitrary and hence, we may
choose R 0 = 0 without any loss of generality.
Indeed, Equations (5.9) and (5.10) can be rewritten more compactly as
 v+ 
 R+
∂z 
 = −
 v− 
 −R−
L ++ − L +−   I + 


L −+ − L −−   ∂t I + 
This furnishes a field-theoretic proof for Lemma 5.1.3 given above.
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Differential Transmission Lines
5.1.8 Lemma
Suppose (C+ , C− ) is a balanced differential pair (viz., R + = R = R − ), for some
fixed impedance per unit length R. Then, v+ ( z, t) = − v− ( z, t) ∀z, t ≥ 0.
Proof
Adding Equations (5.9) and (5.10) yields
∂ z ( v+ ( z, t) + v− ( z, t)) = ( − R + + R − )I + ( z, t) − (L ++ − L −− ) ∂t I + ( z, t).
Because the differential lines are balanced, RL = σµ = constant ⇒ L ++ = L −−
and hence, ∂t ( v+ + v− ) = 0 implies that v+ = − v− + k for some constant k,
where σ, μ are, respectively, the conductivity and magnetic permeability
of the medium surrounding (C+ , C− ) . Furthermore, “balanced” implies the
pair (C+ , C− ) has identical characteristic impedance with respect to a common ground and hence, k ≡ 0, as required.
□
To complete the analysis, the current variation along the conductors is
developed below. Consider the cylinder C+ = S+ ( z) × [ z, z + δz] shown in
Figure 5.3, where
{
S+ ( z) = ( x , y ) ∈R 2 : ( x − x+ )2 + ( y − y + )2 ≤ r+2
}
( x+ , y + , z) ∈C+ is a point along the axis of C+ . Let γ +z = ∂S+ ( z) and γ +z +δz =
∂S+ ( z + δz) . Finally, set C+ = S+ ( z) × ( z, z + δz).
By appealing to Maxwell’s second equation ∇ × B = µε ∂t E + µσE and
invoking Stokes’ theorem, it follows from Figure 5.3 that
∫∫
∂C +
I+(z,t)
C+
Ω+
I0(z,t)
C0
Ω–
E
S+(z,t)
E+z (z,t)
+(z,t)
∇ × B ⋅ dS =
C+
∫
∂2 C+
B ⋅ dl ≡ 0
S+(z + δz,t)
Ez+(z + δz,t)
E+(z + δz,t)
S++
s+– = s––*s+–+1
where
s++ : [0, 1] Ω+
s–– : [0, 1]
E–(z,t)
C–
I–(z,t)
E–z (z,t)
E–(z
S––
S–(z,t)
C–
+ δz,t)
Ω+
s+–+1(t) = s++(1–t)
Ez–(z + δz,t)
S–(z + δz,t)
Figure 5.3
Current variation of TEM wave along multiple conductors.
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150
Electromagnetic Theory for Electromagnetic Compatibility Engineers
where ∂2 C+ = ∂(∂C+ ) is the boundary of the boundary, which is precisely the
empty set.* Hence the integral vanishes.
−1
−1
(t) = s++ (1 − t) and
In Figure 5.3, s+− = s−− ∗ s++
, where s++

s−− (2t) for 0 ≤ t ≤ 21

s−− ∗ s (t) ≡ 
−1
 s++
(2t − 1) for 21 ≤ t ≤ 1

−1
++
Moreover, observe that as the field is assumed to be quasi-static, the potential
difference between two points is essentially independent of the path chosen,
and hence the choice of paths depicted in Figure 5.3. The paths sαβ , for α,β ∈
{+,−}, are chosen such that they are normal to both ∂C± for simplicity.
For convenience, denote E = E+ + E− and B = B+ + B− , where E± (B± )
defines the electric (magnetic) field generated, respectively, by conductors C± . Furthermore, let i+ ( z, t) denote the current propagating along C+
entering S+ ( z) and i+ ( z + δz, t) the current exiting S+ ( z + δz). Physically, it
can be seen that for 0 < σ < ∞ , some of the current propagating along C+
will also be conducted via the medium to C0 and C− . Hence, it is intuitively clear that i+ ( z + δz, t) ≠ i+ ( z, t) when the medium is lossy. Indeed, even
if the medium were lossless, it is clear via the distributed line model that
the existence of a distributed capacitance renders the inequality to hold in
general.
Now, by Figure 5.3, ∂C+ = ∂C+ ∪ S+ ( z) ∪ S+ ( z + δz) is the boundary of the
differential cylinder of length δz along C+ , whence Maxwell’s equation yields
0=µ
∫
∂ C+
J ⋅ nd 2 x + µε ddt
∫
∂ C+
E ⋅ nd 2 x = µ
∫
∂ C+
J ⊥ ⋅ nd 2 x + µε ddt
∫
∂ C+
E⊥ ⋅ nd 2 x
(5.11)
under the assumption of quasi-TEM propagation: E|| ⋅ n = 0 , where E|| << E⊥
and E = E⊥ + E||, the sum of the transverse and longitudinal components,
respectively.
Next, because the medium surrounding the conductors is an imperfect dielectric of conductivity 0 < σ 0 < ∞ ,
J + ⋅ dS ≠ 0 . Hence,
∫∫
∂ C+
J ⊥ ⋅ nd 2 x = σ 0
∫∫
∂ C+
∫∫
∂ C+
E⊥ ⋅ nd 2 x defines the conduction current through
Ω+ from C+ to C0 and C− across ∂C+ . Thus, by construction, invoking the
*
Intuitively, this can be seen by considering a 2-sphere: the boundary of a three-dimensional
ball. It is obvious that a 2-sphere has no boundary as every two-dimensional neighborhood
about any point on the sphere is completely contained in the sphere. Thus, the boundary of a
2-sphere is empty.
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151
Differential Transmission Lines
{ ∫∫


definition of resistance R =  E⊥ ⋅ dl  σ
 γ

that via Kirchhoff’s current law,
∫
σ0
∫∫
∂ C+
E ⊥ ⋅ dS =
δz
R+ +
∫
s+
E⊥ ⋅ dl +
+ G +− δz
∫
s+ −
δz
R+ −
S
∫
s+ −
E⊥ ⋅ nd 2 x
}
−1
, it follows at once
E⊥ ⋅ dl ≡ G ++ δz
∫
s+
E⊥ ⋅ dl
E⊥ ⋅ dl
(5.12)
where s++ ( s+− ) is a path from C+ ∩ C+ to C0 (C− ) , and R ++ (R +− ) is the resistance per unit length along the segment [z, z + δz].
From Equation (5.11), evaluating term by term yields:
∫
S+ ( z )
∫
J ⊥ ⋅ n+ dS = − I + ( z)
S+ ( z +δ z )
∫
C+
J ⊥ ⋅ n+ dS = I + ( z + δz)
J ⊥ ⋅ n+ d 2 x = σ
∫
C+
E⊥ ⋅ n+ d 2 x
However, noting that ∫ C+ J ⊥ ⋅ n+ d 2 x is the conduction current flowing
across C+ , from Ohm’s law, I = GV , where G is the conductance, it follows
immediately that σ ∫ C+ E⊥+ ⋅ n+ d 2 x = G ++ δzv+ with G ++ denoting the conductance per unit length and δz is the length of the cylinder C+ . Likewise,
σ ∫ C+ E⊥− ⋅ n+ d 2 x = G +− δz( v+ − v− ) . Hence,
σ
∫
C+
E⊥ ⋅ n+ dS = G +− δz( v+ − v− ) + G ++ δzv+
(5.13a)
Next, via Figure 5.3, the displacement current across the segment C+ is
determined by ε ∂t ∫ C+ E⊥+ ⋅ n+ dS . Indeed, Q = CV ⇒ i = ∂t Q = C ∂t V . Hence, on
setting C+− to be the capacitance per unit length between the pair (C+ , C− )
and C++ the pair (C+ , C0 ) , it follows at once that
ε ∂t
∫
C+
E⊥+ ⋅ n+ dS = C+− δz ∂t ( v+ − v− ) + C++ δz ∂t v+
whence, applying Equations (5.13a) and (5.13b) to 0 = µ
E⊥ ⋅ nd 2 x and taking the limit δz → 0 yields
∫
∫
C+
(5.13b)
J ⊥ ⋅ nd 2 x + µε ddt
C+
∂ z I + ( z, t) = −(G +− + G ++ )v+ + G +− v− − (C+− + C++ ) ∂t v+ + C+− ∂t v−
K15149_Book.indb 151
(5.14)
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
By symmetry,
∂ z I − ( z, t) = −(G −+ + G −− )v− + G −+ v+ − (C−+ + C−− ) ∂t v− + C−+ ∂t v+
(5.15)
Indeed, in the case of a balanced differential pair, where v+ = − v− , Equations
(5.14) and (5.15) reduce to
 I +   −(G++ + 2G+− )
∂t 
 =
 I −   G−− + 2G+−
−(C++ + 2C+− )   v+ 


C−− + 2C+−
  ∂ t v+ 
Now, consider a balanced differential pair (C+ , C− ) satisfying the respective conditions:
(a) v+ = − v−
(b) v+ = v−
where the incident field is assumed to be time harmonic; that is,
E± ( z, t) = E± ( z)eiωt and the same symbol is used for convenience should no
confusion arise. Recall that this means the replacement ∂t → iω may be
freely made.
5.1.9 Lemma
Given a symmetric differential pair (C± , C0 ) , where C0 is the ground conductor*, v+ ( z, t) = v− ( z, t) ⇒ I + ( z, t) = I − ( z, t) ∀z, t ≥ 0 .
Proof
For a symmetric pair, L ++ = L −− and R + = R − . The result thus follows trivially from Equations (5.7) and (5.8).
□
5.1.10 Remark
It is clear from the proof of Lemma 5.1.9 that as R 0 does not cancel out, or
equivalently, as I 0 = −2 I + ≠ 0 , R 0 cannot be set arbitrarily, as in the case for
v+ = − v− .
For case (a): This case was examined above. Whence, taking the ratio ∂∂zz vI++
yields
Z′2 ≡
v+ ∂ z v+
I+ ∂z I+
=
R + + iω (L + + − L + − )
G+ + + 2G+ − + iω (C+ + + 2C+ − )
For case (b): From Remark 5.1.10, taking the ratio
Z ′′ 2 ≡
v+ ∂ z v+
I+ ∂z I+
=
∂ z v+
∂z I+
yields
R + + 2R 0 + iω (L + + + L + − )
G+ + + 2G+ − + iωC+ +
These two cases lead to the following respective definitions.
*
In the case of a printed circuit board, this represents the ground plane.
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Differential Transmission Lines
5.1.11 Definition
Given a symmetric differential pair (C± , C0 ) , the odd mode is defined by
the condition v+ ( z, t) = − v− ( z, t) ∀z, t ≥ 0, and the odd mode impedance Zodd is
defined by
Zodd =
R + + iω (L + + − L + − )
G+ + + 2G+ − + iω (C+ + + 2C+ − )
In particular, for a lossless system,
Zodd =
L+ + − L+ −
C+ + + 2C+ −
5.1.12 Definition
Given a symmetric differential pair (C± , C0 ), the even mode is defined by
the condition v+ ( z, t) = v− ( z, t) ∀z, t ≥ 0, and the even mode impedance Zeven is
defined by
R + + 2R 0 + iω (L + + + L + − )
G+ + + 2G+ − + iωC+ +
Zeven =
In particular, for a lossless system,
Zeven =
L+ + + L+ −
C+ +
5.1.13 Proposition
Given a symmetric differential pair (C± , C0 ), under the condition of even
mode propagation, there is no capacitive coupling between (C+ , C− ).
Proof
From Equation (5.14), on setting v+ = v− , the two terms involving C+− cancel
out.
□
Lastly, given a symmetric differential pair (C± , C0 ), the impedance seen by
the wave propagating between the pair (C+ , C− ) for v+ = − v− is called the differential impedance, where C− is considered as a reference. As the potential difference vdiff = v+ − v− = 2 v+ ; whereas the current along C+ is I + = − I − , it follows
at once that the differential impedance is defined by Zdiff = vIdiff
= 2Iv++ = 2 Zodd .
+
Explicitly,
Zdiff = 2
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R + + iω (L + + − L + − )
G+ + + 2G+ − + iω (C+ + + 2C+ − )
(5.16)
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
and for a lossless system,
Zdiff = 2
L+ + − L+ −
C+ + + 2C+ −
By symmetry, the impedance seen by a wave propagating between the
pair (C+ , C− ) for v+ = v− , along a common ground Γ 0 is called the common
mode impedance. Here, the pair (C+ , C− ) is treated as a single conductor and
the analysis reduces to the pair (C , C0 ), where C = (C+ , C− ). Although it is
possible to go through the entire field argument carried out above, it is
easier to observe that (a) the common mode potential is vcm = v+ , and (b)
the common mode current I cm = I + + I − = 2 I + , whence, the common mode
impedance is
Zcm =
vcm
I cm
=
v+
2 I+
= 21 Zeven
That is,
Zcm =
1
2
R + + 2R 0 + iω (L + + + L + − )
G+ + + 2G+ − + iωC+ +
(5.17)
and for a lossless system,
Zcm =
1
2
L+ + + L+ −
C+ +
The derivation of differential impedance and common mode impedance via
distributed lumped model is left as an exercise for the attentive reader; see
Exercises 5.4.1 and 5.4.2.
To complete the analysis of odd and even modes along a symmetric differential pair, consider qualitatively, the fields generated by the respective
modes. First, consider the odd mode wave propagation. Here, recall that the
propagating current and hence voltage field along C+ are 180 degrees out
of phase with respect to the current and voltage fields propagating along
C− . By the right-hand rule, the magnetic field density is as illustrated,
where the odd mode current along C+ is directed into the page, and the
current along C− is directed off the page. Suppose γ ± = ( x± , 0), for some
x− < 0 < x+ with the z-component suppressed for simplicity. Then, along
J 0 = {( x , 0) : x ∈( x− + a, x+ − a)} , where a > 0 is the radius of the cable, the
magnetic field density crowds along J 0 , whereas the magnetic field density
is much weaker on R 2 − J 0 , that is, away from J 0 . Physically, because the
currents are oppositely directed—that is, I + = − I − —by Definition 5.1.11, the
odd mode inductance is reduced by the mutual inductance and hence the
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Differential Transmission Lines
magnetic flux, for a fixed current, must trivially decrease correspondingly
via Ψ B = LI.
C–
C+
Odd mode current propagating along a differential pair.
By contrast, observe that the electric field has the profile shown in the
figure, where odd mode voltage, V− = −V+ , via Definition 5.1.11, leads to
the enhancement of the resultant capacitance via the mutual capacitance.
Physically, the surface charge density crowds around the boundary ∂C± of
the lines about a small neighborhood of J 0 , and away from J 0 , the charge
density on ∂C ± decreases (see the electric field lines in the figure), whence,
for a fixed potential difference between C± , Q = CV implies that the resultant surface charge density on ∂C± is enhanced by the increase by the
mutual capacitance.
C–
C+
Odd mode voltage propagating along a differential pair.
Finally, to complete the picture, it is clear from Definition 5.1.12, that the
scenario for even mode propagation is different from odd mode propagation. First of all, in the case of the even mode current propagation, as
I − = I + , it is clear via the right-hand rule that the magnetic field density
on J 0 cancels, leading to the diagram. In a sense, the magnetic field is
enhanced for even mode propagation. Physically, via Definition 5.1.12, the
even mode inductance is increased by the mutual inductance. Hence, the
magnetic flux, for a fixed current, must clearly increase from Ψ B = LI .
C–
C+
Even mode current propagating along a differential pair.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
The scenario with the voltage profile for even mode propagation can
similarly be sketched. Thus, via Definition 5.1.12 and Q = CV, it is clear
that the even mode capacitance is decreased by the mutual capacitance
and hence, the resultant electric field decreases correspondingly (see the
odd mode propagation electric field). The electric flux is low on J 0 due to
mutual repulsion of the electric field, as illustrated in the diagram (see the
mutual repulsion of the magnetic flux density for the odd mode propagation depicted above).
C+
C–
Even mode voltage propagating along a differential.
5.1.14 Remark
Given a lossless differential pair (C± , C0 , µ , ε) defined by the following
distributed parameter model, where the propagation waves are assumed
without loss of generality to be time harmonic:
 L ++
 v+ 
∂z 
 = − iω 
 v− 
 L −+
L +−   i+ 


L −−   i− 
 i+ 
 C++
∂z 
 = − iω 
 i− 
 − C−+
− C+−   v+ 


C−−   v− 
Set L +− = κ ′ L ++ L −− and C+− = κ ′′ C++ C−− , for some coupling constants
κ ′ , κ ′′ ∈[0, 1] . Then, CL++++ = Z+ Z− and κ ′ = κ ′′ , where Z± is the characteristic impedance of C± with respect to C0 . To see this, it suffices to observe that
L+ +
C+ +
=
µ
ε
=
L− −
C− −
Next, by definition, LC = µεI ⇒ L ++ C+− − L +− C−− = 0 and L +− C++ − L −− C+− = 0,
where
 1
I=
 0
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0 
1 
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Differential Transmission Lines
whence
( )
L+ − 2
C+ −
=
L+ + L− −
C+ + C− −
⇒
L+ −
C+ −
= Z+ Z− =
µ
ε
Moreover,
L+ −
C+ −
=
κ′ µ
κ ′′ ε
⇒ κ ′ = κ ′′
□
as claimed.
5.2 Impedance Matching Along a Differential Pair
In the previous section, the odd and even mode impedance along a differential pair were defined. In this section, matching the impedance of a differential pair is crucial when designing a matched differential line to prevent
unwanted reflections. In particular, as Zodd ≠ Zeven , it follows that informally,
different prescriptions may be required to prevent reflection from odd/
even mode propagation. This is clearly important when designing for signal
integrity and electromagnetic interference.
Indeed, in view of Remark 5.1.4, the coupling constant κ allows the pair
(C+ , C− ) to be analyzed as a single transmission pair wherein C− is taken
to be the reference ground, and the resultant current flowing along C+ is
I + + κI − = (1 − κ )I + . Intuitively, the cross-coupling is the result of mutual
inductance and mutual capacitance between the differential pair; this should
be clear from Chapter 4. This is indeed obvious by expressing Ẑ in terms of
Z0 via the coupling constant: Zˆ = κZ0 .
The transfer impedance Ẑ is a quantitative characterization of the induced
voltage appearing on C+ as a result of current flowing in C− . For instance,
switching occurring on an adjacent trace, called the aggressor, can induce a
voltage (noise) on the trace lying next to it; this trace is called a victim trace.
5.2.1 Lemma
Given a symmetric differential pair (C+ , C− ) , the differential coupling coefficient κ satisfies 0 ≤ κ ≤ 1.
Proof
The pair of Equations (5.1) and (5.2) can be written in matrix form as V = ZI .
Explicitly,
 V+   Z0

 =
 V−   Zˆ
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
Zˆ
I 
 + 
Z0   I − 
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
∗
The time-average power delivered to the load is P = 21 Re(V ⋅ I ) , whence
from
 V+   Z0

 =
 V−   Zˆ


ˆ −
Z0 I + + ZI
I 
 +  =
ˆ + + Z0 I −
  I −   ZI
Zˆ
Z0

ˆ −*
Z0 I +* + ZI
V ⋅I = 
ˆ +* + Z0 I −*
 ZI
∗



t
 
I 
ˆ −* )I + + (ZI
ˆ +* + Z0 I −* )I − = 2 Z0 I + 2 (1 − κ )
  +  = (Z0 I +* + ZI
  I − 
2
Thus, P = Z0 I + (1 − κ ). However, P ≥ 0 ⇒ κ ≤ 1. Clearly, κ ≥ 0 because the
power delivered to the load cannot be greater than that of the power generated by the source. Hence, 0 ≤ κ ≤ 1, as claimed.
□
From the proof of Lemma 5.2.1, define an alternative expression for the
time-average power in terms of forward and backward propagating waves
(cf. Exercise 5.4.4).
The odd mode impedance is the effective characteristic impedance of
C+ (respectively, C− ) of a differential pair (C+ , C− ) if C− (respectively, C+ )
were treated as the reference ground. In particular, because the solution to
Maxwell’s equations is linear, a differential pair (C+ , C− ) can be separately
analyzed by investigating C+ and C− separately and then summing the two
solutions. The summation of the two solutions is again a solution because of
linearity (i.e., the principle of superposition).
Now, recalling that given a symmetric differential pair (C+ , C− ) , the characteristic differential impedance Zdiff = Zodd + Zodd = 2(1 − κ )Z0 , it follows that
the differential lines of Figure 5.1 are matched if R = 2(1 − κ )Z0 . Suppose a
common mode noise of voltage δV is (simultaneously) propagating along the
pair (C+ , C− ) depicted in Figure 5.1. That is, V± → V±′ = V± + δV . Then, the resultant output signal at the receiver is Vout = V+′ − V−′ = V+ − V− . Consequently, a
symmetrical differential pair is immune to common mode noise.
5.2.2 Remark
For completeness, consider Equations (5.1) and (5.2) again. Given a symmetric differential pair (C+ , C− ) , suppose that the currents ( I + , I − ) along the
lines satisfy I + (t) = I − (t) ∀t ∈[0, 1]. Then, the signal ( I + , I − ) is called common
mode. It is trivial to establish that V+ ( z) = V− ( z) ∀z ∈[0, ] for common mode
via (5.1) and (5.2).
5.2.3 Lemma
Given a symmetric differential pair, the even mode impedance is given by
Zeven = Z0 + Ẑ .
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159
Differential Transmission Lines
Differential and common
mode termination
I+ δV
Differential transmitter
(constant current source)
V+
C+
r
R
Zdiff
I– δV
V–
Differential
receiver
r
C–
Figure 5.4
Differential termination scheme with common mode noise.
Proof
By (5.1), V+ = I + Z++ + I − Z+− = I + (Z0 + Zˆ ) ⇒ Z0 + Zˆ =
V+
I+
= Zeven , as required.
□
5.2.4 Proposition
Suppose a common mode noise δV couples onto a symmetric differential
pair (C+ , C− ). In order to prevent δV from reflecting back to the source along
(C+ , C− ) , a proper termination scheme for Figure 5.1 is shown in Figure 5.4.
The π-termination scheme comprises a series resistor placed very close to the
parallel differential termination resistor along each line. The proper values
for (R,r) are:
r = Zeven
R=
Zdiff Zeven
Zeven − Zodd
(5.18)
(5.19)
Proof
First, observe by definition that (C+ , C− ) is at equipotential with respect to
δV. In particular, there is zero (noise) current flowing through R as it is at
the same potential as the pair (C+ , C− ). Thus, the proper termination for the
noise is r = Zeven by Remark 4.2.3.
To complete the proof, recall that the impedance seen by odd mode propa−1
gation is Zodd . Hence, in order to match the impedance, Zodd = { R2 + 1r } ⇒
diff Zeven
2 Zodd r = R(r − Zodd ) . Rearranging yields R = ZZeven
− Zodd , where Zdiff = 2 Zodd via
(5.16) and r = Zeven from (a) evaluated above.
□
5.2.5 Remark
Observe that the termination scheme shown in Figure 5.4 for Proposition
5.2.4 depends on the placement of R. Explicitly, the results of Proposition
5.2.4 do not hold for the layout shown in Figure 5.5.
To see this, note the following. First, by Remark 4.2.3, r = Zeven . Second, by
Definition 4.2.2, r + 21 R = Zodd . And because Zeven = Z0 + Zˆ > Z0 − Zˆ = Zodd , it
thus follows that r + 21 R = Zeven + 21 R > Zodd ∀R ≥ 0. Hence, the placement of
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Differential and common
mode termination
I+ δV
Differential transmitter
(constant current source)
V+
C+
r
Zdiff
I– δV
V–
R
Differential
receiver
r
C–
Figure 5.5
Improper termination scheme for even and odd mode signals.
(R, r) is critical in the proper termination of a differential pair (C+ , C− ) . In
particular, the termination scheme of Proposition 5.2.4 will correctly terminate both odd and even mode propagations, preventing unwanted reflections back along the differential pair. Clearly (R, r) must be placed as close
to the differential receiver as possible to mitigate false triggers arising from
reflections between (R, r) and the load (cf. Remark 4.4.14).
5.2.6 Remark
In the technology industry, the scheme shown in Figure 5.6 is often used
for differential matching: termination via capacitor to ground as depicted
in the schematic diagram, where the presence of the capacitor is to filter out
high frequency noise. In Exercise 5.4.3, the reader is ask to comment on the
validity of the scheme. In particular, is the capacitor necessary to filter highfrequency noise for emissions suppression? If so, under what conditions?
5.3 Field Propagation Along a Differential Pair
For simplicity, consider an infinitely long differential pair over an infinite
ground plane, as portrayed in Figure 5.7(a), where the transmission lines are of
radius a > 0 and the centers of the transmission lines are a distance h > a above
the ground plane. Remember from Chapter 4 that a TEM wave propagating
Differential Zdiff = 2R
termination
I+
Differential transmitter
(constant current source)
V+
C+
R
Differential
receiver
Zdiff
I–
V–
C–
R
Figure 5.6
Differential termination with a bypassing capacitor.
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161
Differential Transmission Lines
–
0
b
Radius a
h
0
(–xλ, yλ)
Radius a
–λ
λ
(xλ, yλ)
Image charge
Ground plane
Ground plane
Image pair
Image charge
(–xλ, –yλ)
(a) Differential pair of fixed radii
Image charge
–λ (xλ, –yλ)
λ
(b) Equivalent line charge representation
Figure 5.7
Potential of a differential pair via the method of images.
along a transmission line satisfies the Laplace equation on the plane normal to
the direction of propagation. In addition, recall from Chapter 3 that a charged,
infinitely long cylinder over an infinite ground plane is equivalent to a suitably placed infinite line charge. This suggests that the scalar potential can be
obtained via Figure 5.7(b), with the boundary of the differential lines suitably placed. That is, via the method of images, Figure 5.7(a) is equivalent to
Figure 5.7(b).
For simplicity, let λ denote the line charge as depicted in Figure 5.7(b). For
notational convenience, set rλ(1) = ( xλ , y λ ), rλ(2) = (− xλ , y λ ), rλ(3) = (− xλ , − y λ ), and
rλ(4) = ( xλ , − y λ ), and also set r ( i ) = r − rλ( i ) ∀i, where r = (x, y). Then, the potential
at an arbitrary point r ≠ rλ( i ) ∀i is given by
{
}
1
1
1
ϕ( x , y ) = − 2λπε ln r(1)
− ln r(2)
+ ln r(3)
− ln r(14) =
λ
2 πε
{ln
r(1)
r (2)
(3)
+ ln rr( 4)
}
(5.20)
Referring to Figure 5.7(b), an equipotential loop about rλ(1) satisfies
K=
r (1) r (3)
r (2) r( 4)
for some constant K > 0. Without loss of generality, set K = K1 K 2 , where
(1)
(3)
(1)
(3)
K1 = rr(2) , K 2 = rr( 4) . Then, evaluating K1 = rr(2) and K 2 = rr( 4) lead to
K15149_Book.indb 161
(x −
K1 + 1
K1 − 1
xλ
)
(x +
K2 + 1
K2 − 1
xλ
)
2
2
+ ( y − y λ )2 =
(
2 K1
K1 − 1
xλ
)
+ ( y + y λ )2 =
(
2 K2
K2 − 1
xλ
)
2
2
(5.21)
(5.22)
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Equations (5.21) and (5.22) describe the equipotential loops about rλ(1) and
rλ(3) , respectively. By inspection, the loops are circles with their respective
centers and radii given by
{(
K1 + 1
K1 − 1
)
2 K1
K1 − 1
xλ , y λ ,
xλ
}
{(−
and
K2 + 1
K2 − 1
)
2 K2
K2 − 1
xλ , − y λ ,
xλ
}
By the method of images, the images must have the same radii as the original
loop about rλ(1) . Hence, K1 = K = K 2 , and in particular,
x0 =
Also, x02 =
K1 + 1
K1 − 1
(
xλ
)
K1 + 1 2
K1 − 1
and
a=
2 K
K −1
xλ2 and a 2 =
However, it is clear that
xλ ⇒ 0 = K − 1 −
4K
( K − 1)2
2 x0
a
xλ2 ⇒ x02 − a 2 = xλ2
( KK +−11 )2 − ( K4−K1)
2
=
K 2 + 1− 2 K
( K − 1)2
K⇒ K=
{(
)
K +1 2
K −1
x0 + x02 − a 2
a
− ( K4−K1)2
}
=1
Hence, xλ = x02 − a 2 . Thus, given the center of the cable ( x0 , y 0 ) ≡ ( x0 , y λ ) , the
location of the equivalent line charge ( xλ , y λ ) can be determined. Indeed, suppose the transmission line is at some fixed potential ϕ 0 . Then,
ϕ0 =
λ
2 πε
 x0 +
ln K ⇒ λ = 2 πεϕ 0 ln

x02 − a 2
a



−1
yielding the line charge as a function of potential to which the transmission line is charged. The above analysis is summarized in the following
lemma.
5.3.1 Lemma
Given a lossless, infinitely long differential pair (C± , a, b, h) , where a > 0 is
the radii of the transmission lines, b > 0 the distance of separation between
the centers of (C+ , C− ) , and h > 0 the distance between the respective centers
of C± and the ground plane [see Figure 5.7(a)] the differential pair and their
images are given, respectively, by C± = B(r0± , a) × R and C±′ = B(r0′ ± , a) × R ,
where
B(r0± , a) = {( x , y ) ∈R 2 : ( x  x0 )2 + ( y − y 0 )2 ≤ a 2 }
B′(r0′ ± , a) = {( x , y ) ∈R 2 : ( x  x0 )2 + ( y + y 0 )2 ≤ a 2 }
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163
Differential Transmission Lines
are represented by infinite line charges ( γ ± , ±λ) and their images ( γ ′± , ±λ)
defined by
γ ± = {( ± xλ , y λ , z) : −∞ < z < ∞} and
γ ′± = {( ± xλ , − y λ , z) : −∞ < z < ∞}
□
with xλ = x02 − a 2 and y λ = y 0
5.3.2 Proposition
Given a pair of infinitely long, lossless, differential transmission lines C±
over an infinite ground plane, suppose that C± are charged, respectively, to
V (ω ) = ±ϕ 0 (ω ) . Set Ω = R 3 − (C+ ∪ C− ) , where C± = B(r0± , a) × R are the differential lines. Then, the potential at r = (x, y, z) ∈ Ω is given by
ϕ( x , y ) = −
{
ϕ0
}
ln x0 + x02 − a 2 − ln a
ln
{( x + xλ )2 + ( y − y λ )2 }{( x − xλ )2 + ( y + y λ )2 }
{( x − xλ )2 + ( y − y λ )2 }{( x + xλ )2 + ( y + y λ )2 }
In particular, for
b+ h
r
where δ = 2 2
rλ
r
<<
1
2
, ϕ( x , y ) <
{
ϕ0
}
ln x0 + x02 − a2 − ln a
δ2
.
Proof
The first assertion follows trivially from Equation (5.20). Thus, it remains to
establish the last assertion. Now,
{( x + xλ )2 +( y − yλ )2 }{( x − xλ )2 +( y + yλ )2 } ≈ 1+ 2( xλ x − yλ y )/r 2
{( x − xλ )2 +( y − yλ )2 }{( x + xλ )2 +( y + yλ )2 } 1− 2( xλ x + yλ y )/r 2
1− 2( xλ x − yλ y )/r 2
1+ 2( xλ x + yλ y )/r 2
Moreover, noting that rλ = 21 b 2 + h2 ≤ 21 (b + h) <<
binomial approximation that
1
2 2
=
1− 4( xλ x − yλ y )2 /r 4
1− 4( r ⋅rλ )2 /r 4
r < r , it follows via the
{
}{
{( x + xλ )2 +( y − yλ )2 }{( x − xλ )2 +( y + yλ )2 } ≈ 1− 4( xλ x − yλ y )2 /r 4 ≈ 1 − 4( xλ x − yλ y )2 1 + 4( r⋅r λ )2
1− 4( r ⋅rλ )2 /r 4
r4
r4
{( x − xλ )2 +( y − yλ )2 }{( x + xλ )2 +( y + yλ )2 }
= 1+ 4
}
( r ⋅r λ )2 − ( xλ x − yλ y )2
r4
Now, noting that ln(1 + δ) ≈ δ for δ << 1, it follows at once that
ln
K15149_Book.indb 163
{( x + xλ )2 +( y − yλ )2 }{( x − xλ )2 +( y + yλ )2 } ≈ 2 (r ⋅rλ )2 −( xλ x − yλ y )2 = 8 yλ
r
r4
{( x − xλ )2 +( y − yλ )2 }{( x + xλ )2 +( y + yλ )2 }
y xλ x
r r r
<8
y λ xλ
r r
<8
( )
rλ 2
r
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
rλ
r
Finally, set δ = 2 2
ln
. Then,
rλ
r
<
1
2 2
⇒ δ << 1 , and
{( x + xλ )2 +( y − yλ )2 }{( x − xλ )2 +( y + yλ )2 } < 8 yλ
r
{( x − xλ )2 +( y − yλ )2 }{( x + xλ )2 +( y + yλ )2 }
xλ
r
≤ δ2
whence
ϕ( x , y ) <
{
ϕ0
}
ln x0 + x02 − a 2 − ln a
δ2
□
as asserted.
5.3.3 Remark
The above proposition is the reason why EMC engineers often state that
fields cancel out for differential lines; that is, away from a differential pair,
the electromagnetic field cancels. Clearly, that is an oversimplification:
the fields do not cancel out identically. However, at a sufficiently large
distance away from the pair, the potential field falls off approximately
2
as rrλ . Thus, a differential pair does mitigate emissions, although it is
clear from the proof that emissions do not vanish, a common incorrect
assumption made by EMC engineers. Indeed, even at an intuitive level, it
is clear that fields do not identically cancel out: a differential pair can be
looked upon as a dipole string (as opposed to point charge dipole). Thus,
the fields do not cancel out, albeit they fall off very much faster than the
field of a single line charge.
( )
5.3.4 Corollary
Given a differential pair stated in Proposition 5.3.2, suppose r − rλ = δ > a.
Then,
{
ln x0 +
2 ϕ0
x02 − a 2
}− ln a
δ 2 + r ⋅r
ln δ2 + 4 rλ ⋅r < ϕ( x , y ) <
λ
{
2 ϕ0
}
ln x0 + x02 − a 2 − ln a
{
ln 1 +
4 r λ ⋅r
δ2
}
Proof
Now,
{( x + xλ )2 +( y − yλ )2 }{( x − xλ )2 +( y + yλ )2 } = δ2 + 4 xλ x δ2 + 4 yλ y
δ2
δ 2 + 4 r λ ⋅r
{( x − xλ )2 +( y − yλ )2 }{( x + xλ )2 +( y + yλ )2 }
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Differential Transmission Lines
Thus, noting the following inequality
δ 2 + r λ ⋅r
δ2
δ2
δ 2 + 4 r λ ⋅r
<
2
δ 2 + 4 xλ x δ + 4 y λ y
δ2
δ 2 + 4 r λ ⋅r
<
δ 2 + 4 r λ ⋅r
δ2
□
yields the desired result at once.
In particular, the above corollary provides a bound for the scalar potential
near one of the differential lines.
5.3.5 Remark
It is clear from the above discussion that the potential generated by the differential pair at an arbitrary point r ∈R 3+ − (C+ ∪ C− ) satisfies V (r ) = ϕ( x , y )e− iβz ,
and hence, the electric field is given by E(r) = − ∇V(r).
Now, observe that the capacitance per unit length C++ of C+ with respect to
the ground plane is found as follows.
By definition, the capacitance per unit length C of two infinitely long cables
of uniform radius a, their axes separated by a distance d, is given by
{
C = πε ln d +
d2 + 4 a2
2a
}
−1
whence it follows at once that
C++ =
λ
ϕ 0 +ϕ 0
{
h2 − a 2
a
= πε ln h+
}
−1
Likewise, the mutual capacitance C+− is, by definition,
{
C+− = πε ln b +
b2 − 4 a2
2a
}
−1
In particular, the total capacitance per unit length
(

C+ = C++ + C+− = πε  ln b +

b2 − 4 a2
2a
) + (ln
−1
h + h2 − a 2
a
)
−1



For odd mode propagation, the mutual inductance can be found via
Equation (5.16):
(

2
2
L +− = L ++ − 14 Zdiff
(C+ + C+− ) = L + − 14 Zdiff
πε  ln b +

b2 − 4 a2
2a
) + 2 (ln
−1
h + h2 − a 2
a
)
−1



Hence, any current noise pulse δ i propagating along one of the differential
pair will induce a voltage spike on the corresponding line via δv = − L+− ddt δ i.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Differential Pair of Fixed Radii
–
C–
0
b
Radius a
h
C+
0
Radius a
Ground plane
Image pair
Figure 5.8
Potential of a differential pair via the method of images.
Finally, observe that for the (lossless, symmetric) differential pair illustrated
in Figure 5.8, the “return” current via the ground plane is nonzero; this fact
is occasionally falsely assumed to be zero by less careful EMC engineers.
5.3.6 Proposition
Given the lossless, symmetric differential pair illustrated in Figure 5.8, the
ratio of the image current on the ground plane of C± with respect to the
current propagating along C± satisfies
I+ −
I+ +
= 2 ln h+
h2 − a 2
a
{ln
b + b2 − 4 a2
2a
}
−1
(5.23)
where I +− is the return current between (C+ , C− ) and I ++ is the return current
between C+ and the ground plane.
Proof
From Q = CV, it is clear that
I+ −
I+ +
=
dQ+ −
dt
(
)
dQ+ + −1
dt
=
Q+ −
Q+ +
From the preceding discussion,
C++ =
λ
ϕ 0 +ϕ 0
{
= πε ln h+
h2 − a 2
a
}
−1
{
and C+− = πε ln b +
b2 − 4 a2
2a
}
−1
whence, noting that δV+− = ϕ + + ϕ + = 2ϕ + is the potential difference between
(C+ , C− ) , and δV++ = ϕ + is the potential difference between C+ and the
ground plane, yields the result at once.
□
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Differential Transmission Lines
It is clear from Proposition 5.3.5 that in a typical scenario, such as for
traces buried within a PCB—known as a stripline as opposed to a microstrip
wherein the traces are on top of a PCB—the following usually holds: 21 b >> h .
Hence, in this instance, I +− < I ++ and most of the return current occurs along
the ground plane.
5.4 Worked Problems
5.4.1 Exercise
Derive the odd and even mode impedance via the distributed parameter
model for a lossless differential pair. Comment on the time delay for the odd
mode versus even mode propagation along a differential pair.
Solution
Consider the following pair of transmission lines depicted in Figure 5.9. The
capacitor coupling the two lines (i.e., mutual capacitance C12 = C21 ) and the
inductor coupling the two lines (i.e., mutual inductance L12 = L21 ) are also
illustrated once for simplicity.
Now, observe that the current ( I1 , γ 1 ) induces, via L12 , a voltage on γ 2 :
V21 = − L12 ddIt1 . By symmetry, ( I 2 , γ 2 ) induces a voltage on γ 1 : V12 = − L12 ddIt2 ,
where for simplicity, ( γ 1 , γ 2 ) represents the respective axes of a pair
(C1 , C2 ) of thin transmission lines. Similarly, the voltage (V2 , γ 2 ) induces
a current, via the mutual capacitance C12 , on γ 1 as follows. The instantaneous charge per unit length induced on γ 1 is δ Q 1 = C12 (V1 − V2 ), whence,
Q 1 = C10V1 + δ Q 1 = C11V1 − C12V2 . Likewise, Q 2 = C20V2 + δ Q 2 = C22V2 − C12V1 .
C+
RS1
VS1
L11
C10
L12
C12
Distributed
RL1 Parameter
model
C11 = C10 + C12
C–
RS2
VS2
C20
L22
C22 = C20 + C12
RL2
Figure 5.9
Mutual inductance and mutual capacitance coupling the two lines.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Thus, appealing to transmission line theory,
∂ z V1 = − L11 ∂t I1 − L12 ∂t I 2 , ∂ z I1 = − C11 ∂t V1 − C12 ∂t (V1 − V2 )
∂ z V2 = − L12 ∂t I1 − L 22 ∂t I 2 , ∂ z I 2 = − C12 ∂t (V2 − V1 ) − C22 ∂t V2
That is, ∂ z V = − L ∂t I and ∂ z I = −C ∂t V , where
 C11 + C12
 I1 
 V1 
V =
,I=
,C=
V
I
− C12

 2 
 2 

 L11
− C12
 and L = 
C22 + C12 
 L12
L12 

L 22 
Now, for an odd mode (i.e., differential) current I1 = − I 2 ⇒∂ z Vk = −(L kk − L12 ) ∂t I k
for k = 1, 2 . For a symmetric differential pair, L1 = L 2 ⇒ L ′ = L kk − L12 ∀k.
Hence, ∂ z Vk = − L ′ ∂t I k for a differential signal. Thus, define Lodd ≡ L ′ to be
the odd mode inductance for a differential pair. Likewise, for odd mode current, V1 = −V2 ⇒ ∂ z I k = −(C kk + 2C12 ) ∂t Vk , and for a symmetric differential
pair, C1 = C2 ⇒ C′ = C kk + 2C12 ∀k . Thus, define Codd ≡ C′ to be the odd mode
capacitance for a differential pair.
In summary, for a differential pair, the transmission line equations for
odd mode become
 ∂ z Vk = − Lodd ∂t I k


 ∂ z I k = − Codd ∂t Vk

for k = 1,2. In particular, by definition, the odd mode impedance is
Zodd =
Lodd
Codd
=
L kk − L12
C kk + 2C12
for the transmission line γ k , as required. As an aside, observe that
Vdiff = V1 − V2 = 2V1 ⇒ Zodd = 2 Zodd , consistent with the derivation of Section
5.1. For symmetric lines, L kk = L 0 and C kk = C0 for all k = 1,2, where L 0 (C0 )
is the inductance (capacitance) per unit length of a single line defining its
characteristic impedance R0 = CL00 .
To complete this problem, consider even mode propagation. Here, I1 = I 2 ⇒
∂ z Vk = −(L kk + L12 ) ∂t I k , and V1 = V2 ⇒ ∂ z I k = − C kk ∂t Vk ∀k = 1, 2 . That is, for
even mode,
 ∂ z Vk = − Leven ∂t I k ≡ −(L kk + L12 ) ∂t I k



∂ z I k = − Ceven ∂t Vk ≡ − C kk ∂t Vk

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Differential Transmission Lines
Hence, the even mode impedance is
Zeven =
Leven
Ceven
=
L kk + L12
C kk
≡
L0 + L12
C0
As a passing remark, it is clear from the above analysis that the time delay in
odd mode versus even mode propagation is, in general, different:
τodd =  Lodd Codd
and τeven =  Leven Ceven
where ℓ is the length of the differential pair. Finally, observe from the above
derivation that
Lodd < L 0 < Leven
and Ceven < C0 < Codd
5.4.2 Exercise
Provide an alternative derivation for the relationship between common
mode and even mode impedance.
Solution
For notational simplicity, the transmission line pair (C+ , C− ) is identified with its respective axes ( γ + , γ − ). This convention is adopted in all
subsequent exercises. First, recall from (4.1) that along γ + , V+ = I + Z0 +
I − Zˆ = I + Z0 + I + κZ0 = I + (1 + κ )Z0 . Thus, the effective characteristic impedance
for common mode waves propagating along γ + is Z+ = VI−+ = (1 + κ )Z0 . That
is, the current I + flowing along γ + sees an effective impedance of Zeven = Z+ .
Because I − = I + along γ − , it follows that the total current conducting along
( γ + , γ − ) is I = I + + I − = 2 I + . Hence, the pair ( γ + , γ − ) appears to be a common
mode signal as parallel lines connected to some virtual ground. That is, the
1
1
1
resultant characteristic impedance of ( γ + , γ − ) is Zcm = Zeve + Zeven ⇒ Zcm =
1
1
2 Zeven . The resultant characteristic impedance Zcm = 2 (1 + κ )Z0 for a common mode signal is called the common mode impedance of ( γ + , γ − ).
□
Exercise 5.4.3
Refer to the question posed in Remark 5.2.6, where the differential pair is
assumed to be symmetric.
Solution
If a high-frequency noise is a common mode noise, then, by assumption,
it will perfectly cancel out at the load, and hence, it is not necessary to add
the capacitor to filter out the noise. On the other hand, if the noise only
appears on either γ + or γ − , then the only way the high-frequency noise
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
can be by-passed without affecting the input impedance is to implement
it as illustrated in Remark 5.2.6 for impedance-matching purposes. To see
this, it suffices to note that by construction, the capacitor is at zero potential
with respect to γ ± , and hence, its presence will not disturb the differential
impedance match afforded by 2R.
In practice, this implementation depends upon the real estate of the printed
circuit board, and for very high frequency noise, the added surface mount
and vias will potentially increase the inductance in the matching circuit, and
may perturb the effectiveness of the impedance matching scheme.
□
Exercise 5.4.4
Given a symmetric differential pair* ( γ + , γ − ) denote γ + ∪ γ − to be the
pair that incorporates mutual interaction between the pair of lines:
γ + ∪ γ − ≡ γ + ⊕ γ − + γ + ⊗ γ − , where γ + ⊕ γ − defines the pair ( γ + , γ − ) in the
absence of mutual interaction; that is, γ + , γ − are over a common ground plane
such that they are infinitely far away from each other, and γ + ⊗ γ − denotes
the mutual interaction between the pair defined by the mutual inductance
and mutual capacitance. For simplicity, assume that the differential pair is
lossless in all that follows, and in particular, for (b) and (c) below, assume
that ZS ± = Z0 = ZL ± .
(a) Show that the time-average power can be expressed as P =
†
†
†
†
∗
∗
1
2 (a a − b b), where a = ( a+ , a− ), a = ( a+ , a− ), b = (b+ , b− ) and b =
∗
∗
(b+ , b− ),
a± =
V± + Z0 I ±
2 Z0
and b± =
V± − Z0 I ±
2 Z0
(b) Define the far-end cross-talk induced on γ − by γ + by δV− , + = k+ ddt
V+ (t − τ) , where k+ is the far end cross-talk coupling coefficient [5,6].
By definition, δV− , + is the induced voltage near the source of γ −
(adjacent to the source of γ + ). Show that
k+ = − 21 τ
(
L+ −
L0
−
C+ −
C0
)
where τ =  µε .
(c) Define the near-end cross-talk induced on γ − by γ + by δV− , + =
k− (V+ (t) − V+ (t − 2 τ)), where k− is the near-end cross-talk coupling
coefficient. By definition, δV− , + is the induced voltage near the source
of γ − (adjacent to the source of γ + ). Show that
k− =
*
1
4
(
L+ −
L0
+
C+ −
C0
)
Recall that transmission lines and their respective axes are identified for simplicity.
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Differential Transmission Lines
Solution
V1+
Z0
(a) By definition, V+ = V1+ e− iβz + V1− eiβz and I + =
a+ ( z ) =
V1+
Z0
e− iβz
a+ ( z ) =
and
−
e− iβz − VZ10 eiβz . So, set
V1+
Z0
e− iβz
Then, substituting back into the original equations yields
 V+  

 =
 I +  
Z0
Z0
−
1
Z0
1
Z0

a 
 + 
  b+ 

and hence, solving for ( a+ , b+ ) via inversion yields

 a+ 
1
=

 2
 b+ 

1
Z0

  V+ 
  I + 

Z0
− Z0
1
Z0
V+
−
By symmetry, V− = V2+ e− iβz + V2− eiβz and I − = Z20 e− iβz − VZ20 eiβz , where V2± = −V1±
from the assumption of a symmetric differential pair, yield

 a− 
1

 = 2
 b− 

1
Z0


  V− 
  I− 

Z0
− Z0
Z0
That is,
a± =
V± + Z0 I ±
2 Z0
and b± =
V± − Z0 I ±
2 Z0
Note that by construction, a, b represents forward propagating and backward propagating waves, respectively.
Next, appealing to the definition of time-average power P = 21 ℜe(V † I )
for the differential pair, it follows at once that P = 21 ℜe(V+∗ I + ) + 21 ℜe(V−∗ I − ),
where V = (V+ , V− ) and I = ( I + , I − ) , whence substituting V± = Z0 ( a± + b± )
and I ± = Z1 0 ( a± − b± ) into V † I yields
V † I = V+∗ I + + V−∗ I − = a+∗ a+ + a−∗ a− − b+∗ b+ − b−∗ b− +
− a+∗ b+ + b+∗ a+ − a−∗ b− + b−∗ a−
2
2
2
2
{
}
= a+ + a− − b+ − b− + i2 ℑm(b+† a+ ) + ℑm(b−† a− )
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
{
2
2
2
2
}
and hence, P = 21 ℜe(V † I ) = 21 a+ + a− − b+ − b− = 21 (a† a − b† b), where
for any complex z = x + iy ⇒ z − z∗ = i2 y was invoked. The time-average
power can thus be expressed as the difference between forward and backward travelling waves.
(b) Referring to Example 4.4.7, the forward propagating wave in timedomain can be expressed as
V + (t −
z
v
) = V0 (t − vz ) + Γ(0)Γ()V0 (t − 2τ − vz ) + with
V0 ( t −
z
v
) = Z Z+ Z
0
0
S
VS (t)e− γz
and the backward propagating wave in time-domain can be expressed as
V − (t +
z
v
) = Γ()V0 (t − 2τ + vz ) + Γ(0)Γ()2 V0 (t − 4τ + vz ) + where v = 1µε is the wave propagation speed and τ = v is the propagation
delay time, thus defining V ( t − vz ) = V + ( t − vz ) + V − ( t − vz ) ⇒ V (t) = V + (t) +
Γ()V + (t − 2 τ), where τ = v is the propagation delay time. In particular, at the
source z = 0,
V (t) = V + (t) + Γ()V + (t − 2 τ)
= V0 ( t ) + Γ(0)Γ()V0 (t − 2 τ) + + Γ(){V0 (t − 2 τ) + Γ(0)Γ()V0 (t − 4τ) + }
{
= V0 (t) + 1 +
1
Γ (0)
} {Γ(0)Γ()V (t − 2τ) + Γ (0)Γ ()V (t − 4τ) + }
2
0
and at the load z = ℓ, noting that V − ( t −
V (t −
v

v
2
0
) = Γ()V + (t + v − 2τ ) = Γ()V + (t − τ)
) = V + (t − v ) + Γ()V + (t − v )
= (1 + Γ()){V0 (t − τ) + Γ(0)Γ()V0 (t − 3τ) + Γ 2 (0)Γ 2 ()V0 (t − 5τ)}
From (a),
a+ =
V+ + Z0 I +
2 Z0
⇒ ∂ z a+ =
1
2 Z0
{∂z V+ + Z0 ∂z I+ }
For simplicity, suppose the waves are time harmonic. Then, from Remark
5.1.14,
∂ z V+ = − iω {L ++ I + + L +− I − } and ∂ z I + = − iω {C++ V+ − C+− V− }
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Differential Transmission Lines
Thus, substituting the above relations yields
∂ z a+ = − 2 iωZ0 {L ++ I + + Z0 C++ V+ + L +− I − − Z0 C+− V2 }
= − iωZ0 C++ a+ + iω Z0 C+ −2V−Z−0L+ − I−
Now, noting that
that
L+ −
C+ −
= Z++ Z−− = Z02 by assumption, it follows immediately
L+ −
V− − Z C
I−
iω Z0 C+ −2V−Z−0L+ − I− = iωZ0 C+−
0 +−
2 Z0
= iωZ0 C+−
V− − Z0 I −
2 Z0
= iωZ0 C+− b−
Hence, ∂ z a+ = − iωZ0 C++ a+ + iωZ0 C+− b− . The evaluation of ∂ z a− follows that of
∂ z a+ mutatis mutandis. In particular, ∂ z a− = iωZ0 C+− b+ − iωZ0 C−− a− . Similarly,
evaluating ∂ z b± yields ∂ z b+ = iωZ0 C++ b+ − iωZ0 C+− a− and ∂ z b− = − iωZ0 C+− a+ +
iωZ0 C−− b− , whence, observing trivially that ωZ0 C ± ± = ω L ± ± C ± ± ≡ ωυ ≡ β,
with υ = L± ±1C± ± being the speed of a wave propagating in a homogeneous
medium—that is, for the case wherein L ± = L 0 and C ± = C0 —it follows
clearly that
 β
 a+ 
∂z 
 = − i 
 a− 
 0
0   a+  
0
+ i


β   a−   ωZ0 C+−
ωZ0 C+−   b+ 


0
  b− 
ωZ0 C+−   a+   β

 + i
0
  a−   0
 b+ 

0
∂z 
 = −i 
 ωZ0 C+−
 b− 
0   b+ 


β   b− 
Next, note that ωZ0 C+− b− = 2 1Z0 ωZ0 C+− (V− − Z0 I − ) . Specifically, motivated by
the fact that the mutual capacitive and mutual inductive coupling induce the
following waves propagating towards the load:
V+− = − iωL +− I + + iωZ0 C+− V+ = − iω
{
L+ −
Z0
}
− Z0 C+− V+ = iωZ0 C+− (V+ − Z0 I − )
this suggests the decomposition:
ωZ0 C+− (V− − Z0 I − ) = − 21 ω
{
L+−
Z0
}
− Z0 C+− (V− + Z0 I − ) + 21 ω
{
L+−
Z0
}
+ Z0 C+− (V− − Z0 I − )
That is,
−ωZ0 C+− b− = 21 ω
K15149_Book.indb 173
{
L+ −
Z0
}
− Z0 C+− a− − 21 ω
{
L+ −
Z0
}
+ Z0 C+− b−
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Likewise, it can be seen by symmetry that
−ωZ0 C+− b+ = 21 ω
{
L+ −
Z0
}
− Z0 C+− a+ − 21 ω
{
L+ −
Z0
}
+ Z0 C+− b+
Thus,
 β
 a+ 
∂z 
 = − i 
 a− 
 0
0   a+   0

 − i
β   a−   κ −
 β
= −i 
 κ −
where κ ± = 21 ω
On setting
{
κ −   a+   0

 + i
β   a−   κ +
L+ −
Z0
κ −   a+   0

 + i
0   a−   κ +
κ +   b+ 


0   b− 
κ +   b+ 


0   b− 
}
± Z0 C+− .
 β
U=
 κ −
κ− 

β 
 0
and V = 
 κ+
κ+ 

0 
the above partial differential equation leads to the pair of transmission line
equations:
∂ z V = − i(U + V)Z0 I
and ∂ z I = − i(U − V) Z10 V
Thus, from ∂ z V = − iωLI and ∂ z I = − iωCV , where
 L ++
L=
 L +−
L +− 

L −− 
 C++
and C = 
 − C+−
− C+− 

C−− 
it is clear by definition that the impedance matrix is given by
Z2 ≡ LC −1 = Z02 (U + V)(U − V) ⇒ Z = Z0 H,
where H = (U + V)(U − V) , and the reflection coefficient matrix by Γ =
(Z − Z0 I)(Z + Z0 I)−1 , with I being the identity 2 × 2 matrix. Finally, substituting
the results from (a) into ∂2z a and ∂2z b yields
∂2z a = −(U + V)(U − V)a and ∂2z b = −(U + V)(U − V)b
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175
Differential Transmission Lines
Now, note first that U, V are diagonalizable and hence, so is H. Furthermore,
observe that for a matrix
 a
M=
 b
b 
a 
its eigenvalues are m± = a ± b, and the diagonalization of M is diag(m+ ,m− ).
Next, in all that follows, by an abuse of notations, denote H by its diagonalization for convenience, because the original solution can easily be obtained
from the inverse similarity transformation diagonalizing M; see Chapter 6
for more details. Thus,
 H+
H→H= 
 0
0 

H− 
where H ± are the two eigenvalues of H, with H ±2 = β 2 + κ 2− − κ 2+ ± 2βκ − =
(β ± κ − )2 − κ +2 .
Specifically,
H +2 = (β + κ − + κ + )(β + κ − − κ + ) ⇒ H + = ω (L 0 + L +− )(C0 − C+− )
and
H −2 = (β − κ − + κ + )(β − κ − − κ + ) ⇒ H − = ω (L 0 − L +− )(C0 + C+− )
are the explicit eigenvalues. Moreover, as a(b) is the forward (backward)
propagating wave, the results of Chapter 4 (cf. Proposition 4.4.2) generalize
immediately to the coupled system of transmission line equations:
V ( z) = VS Z(Z + ZS )−1 e− iHz {1 − Γ (0)Γ ()e− i2 H }−1 {I + Γ ()e− i2 H(  − z ) }
(5.24)
where the source impedance
 ZS +
ZS = 
 0
0 

ZS − 
 VS + 
and VS = 

 VS − 
is the source of γ ± .
Following [5], let e ± denote the normalized eigenvectors of H associated with the eigenvalues H ± , where e ± = 12 (1, ±1) . It is easy to see that
diag(m+ , m− ) = PHP † , where † denotes the conjugate transpose, and P = (e+ , e− ) ,
for all a,b ≠ 0. Intuitively, e+ corresponds to the waves propagating along
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
γ ± in the parallel direction (→, →), e− corresponds to waves along γ ± in the
antiparallel direction (→, ←). That is,

e+ ↔ 

γ + propagation

→ 

γ − propagation

→ 
 γ
+ propagation
→
and e− ↔ 
γ
propagation
−
 ←




Hence, e+ defines the even mode and e− defines the odd mode.
Now B = {e+ , e− } defines a basis as e+ ⋅ e− = 0 , and it decouples γ + ∪ γ − →
γ + ⊕ γ − in the following fashion. By definition, set V = V ( + )e+ + V ( − )e− , where
V ( ± ) = 12 (V+ ± V− ) . Then, noting that
 a
exp 
 0
0   1
=
b   0
0   a
+
1   0
 1 + a + 1 a2 + 
2!
+ = 

0
0 
+
b 
1
2!
 a
 0

0 
b 
2


1 + a + 2!1 a 2 +  
0
that is,
 a
exp 
 0
0   ea
=
b   0
0 

eb 
it follows upon expanding Equation (5.24), and defining
Z± =
L0 ± L+ −
C0  C+ −
and Γ ± =
Z − Z±
Z + Z±
where Z± , Γ ± correspond, respectively, to the characteristic impedance and
coefficient of reflection of the even(odd) modes, that
 H+
Z(Z + ZS ) = 
 0
−1
(0)
where ZS ± =
ZS ±
Z0
(0)
0   H + + ZS +


H−  
0

H − + ZS(0)−
, and likewise, on setting ZL(0)± =




1
0
I − Γ (0)Γ ()e− i2 H = 
−

 0 1 

K15149_Book.indb 176
0
H + − ZS(0)
+
H + + ZS(0)
+
0
0
H − − ZS(0)
−
H − + ZS(0)
−




 
ZL ±
Z0
H + − ZL(0)
+
H + + ZL(0)
+
0



−1


=

H+
H + + ZS(0)
+
0
0
H−
H − + ZS(0)
−




,
0
H − − ZL(0)
−
H − + ZL(0)
−

  e− i2 H+ 
0 


0
e− i2 H+  


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Differential Transmission Lines
and hence, the system is decoupled via the diagonalization scheme, and further simplification thus yields
V ( ± ) = VS( ± )
Z±
Z± + ZS ±
{
e− iH ± z 1 − Γ ± (0)Γ ± ()e− i2 H ± 
} {1 + Γ
−1
±
()e− i2 H ± (  − z )
}
(5.25)
Note, as a side comment, that the speed of wave propagation for the even/
odd mode is given, respectively, by
v± =
1
(L 0 ± L + − )(C0  C+ − )
and hence β ± = ωv± is the respective even/odd mode wave number.
By construction, V = V+ e1 + V− e2 , where e1 = (1, 0), e2 = (0, 1) . The linear
operator mapping the basis B onto the standard basis B0 = {e1 , e2 } can be
shown, after some algebraic manipulation, to be
T=
1
2
 1
 1
1 
−1 
Thus, T −1 = T ⇒ (V+ , V− ) = V ( + ) T(e+ ) + V ( − ) T(e− ). Explicitly, V± = 12 {V ( + ) ± V ( − ) }
and similarly, VS( ± ) = 12 {VS + ± VS − } .
To determine the cross-talk coefficient on γ − , it suffices to assume without
loss of generality that VS − = 0 . Then, the far end cross-talk is determined by
setting z = ℓ. That is,
V− () =
1
2
{V ( + ) () − V ( − ) ()}
Furthermore, observe that
Z±
Z± + ZS ±
=
1 Z± + ZS ± − ZS ± + Z±
Z± + ZS ±
2
= 21 (1 − Γ ± (0))
Hence,
V− () =
1
2
1
2
{
{
}
())}
1
2
(1 − Γ + (0))(VS + + VS − )(1 − Γ + (0)Γ + ()e− i2 H ±  )−1 e− iH+  (1 + Γ + ()) −
1
2
(1 − Γ − (0))(VS + − VS − )(1 − Γ − (0)Γ − ()e− i2 H ±  )−1 e− iH+  (1 + Γ −
Finally, observe that under conditions of weak coupling, L +− << L ± ± and
C+− << C ± ± . Remember moreover, it was assumed herein that L ± = L 0 and
C ± = C0 . Thus,
H ± = ω (L 0 ± L +− )(C0  C+− ) ≈ ω L 0 C0 1 ±
K15149_Book.indb 177
L+ −
L0

C+ −
C0
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
In particular, we may define, up to first order in
L+ −
L0
, CC+0−
(i.e., the conditions of weak coupling) H ± = β ± δH , where
β = ω L 0 C0
and δH = 21 β
(
L+ −
L0
−
C+ −
C0
)
Likewise,
Z± =
L0 ± L+ −
C0  C+ −
= Z0
(1 ± ) (1 ± )
C+ − −1
C0
L+ −
L0
{
≈ Z0 1 ±
1
2
(
L+ −
L0
+
C+ −
C0
)} = Z
0
± 21 Z0
(
L+ −
L0
+
C+ −
C0
)
under conditions of weak coupling.
Moreover, on defining the differential pair reflection coefficients
Γ ± ( z) =
Z± ( z )− Z±
Z± ( z )+ Z±
if the source and load impedances are matched to Z0 , then the individual
uncoupled reflection coefficients under the conditions of weak coupling
along γ ± are:
Γ (0± ) =
Z± − Z0
Z± + Z0
= ± 21
(
L+ −
L0
+
C+ −
C0
) {2 ± (
1
2
L+ −
L0
C+ −
C0
+
)}
−1
≈ ± 14
(
L+ −
L0
+
C+ −
C0
)
Lastly, set τ ± =  Hω± to be the propagation delay time. Then, under weak coupling, τ ± ≈ τ ± δτ , where δτ = ω δH and τ = ω β .
In terms of κ ± , noting that
κ ± = 21 β
(
L+ −
L0
±
C+ −
C0
)
it is clear that
δH = 21 β
(
L+ −
L0
−
C+ −
C0
)=κ
−
, Γ (0± ) ≈ ± 21
κ+
β
In time-domain, from the expansion of V ( t −
this problem outlined above leads trivially to
, δτ = ω κ − = βτ κ −

v
) given at the beginning of
V− () = 21 (1 − Γ + (0))(1 + Γ + ()){V0 (t − τ + ) + Γ + (0)Γ + ()V0 (t − 3τ + ) + }
− 21 (1 − Γ − (0))(1 + Γ − ()){V0 (t − τ − ) + Γ − (0)Γ − ()V0 (t − 3τ − ) + }
whence, under the assumptions of weak coupling, set δΓ = 21 κβ+ , which
by definition, is first order in LL+0− , CC+0− . Hence, define δΓ(0) = δΓ + ε ′ and
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179
Differential Transmission Lines
δΓ() = δΓ + ε ′′ , for some ε ′ , ε ′′ = o(δΓ 2 ) . Then, ε ′ − ε ′′ ~ o(δΓ 2 ) , and via the
assumption ZS ± = Z0 = ZL ± ,
(1 − Γ ± (0))(1 + Γ ± ()) = 1 − Γ ± (0)Γ ± () ≈ 1
Thus, noting via Taylor’s expansion (to first order) that V0 (t − τ ± ) ≈ V0 (t − τ) 
∂t V0 (t − τ)δτ , it is clear that
V− () ≈ 21 (V0 (t − τ + ) − V0 (t − τ − )) =
1
2
{V0 (t − τ) − ∂t V0 (t − τ)δτ − V0 (t − τ) − ∂t V0 (t − τ)δτ} ,
that is, V− () ≈ − ∂t V0 (t − τ)δτ . Thus,
k+ = −δτ = − ω κ − = − 21 ω β
(
L+ −
L0
−
C+ −
C0
) = − τ(
1
2
L+ −
L0
−
C+ −
C0
)
as required.
(c) The near-end cross-talk is determined as (b) above by setting z = 0.
Under the conditions of weak coupling, via V− (0) = 12 {V ( + ) (0)
−V ( − ) (0)}, following the derivation above for V− (),
V− () =
1
2
{
−
1
2
1
2
}
(1 − Γ + (0))(VS + + VS − )(1 − Γ + (0)Γ + ()e− i2 H ±  )−1 (1 + Γ + ()eiH+  )
{
1
2
}
(1 − Γ − (0))(VS + − VS − )(1 − Γ − (0)Γ − ()e− i2 H ±  )−1 (1 + Γ − ()eiH+  )
Noting that V0 (t) = 21 VS + (t) when ZS ± = Z0 = ZL ± , and recalling that
∑
V0 (t) + Γ + ()
m> 0
Γ +m − 1 (0)Γ +m − 1 ()V0 (t − 2 mτ) = V0 (t){1 + Γ + ()eiH+  }{1 − Γ + (0)Γ + ()e− iH+  }−1
weak coupling yields Γ + (0) − Γ − (0), Γ + () − Γ − () ~ δΓ + − δΓ − =
{
(
V− (0) ≈ 21 (1 − Γ + (0)) V0 (t) + 1 +
{
1
Γ + (0)
(
− 21 (1 − Γ − (0)) V0 (t) + 1 +
{
(
= 21 (1 + Γ (0+ ) ) V0 (t) + 1 −
{
(
1
Γ(0+ )
− 21 (1 + Γ (0− ) ) V0 (t) + 1 −
{
(
)
κ+
β
and hence
) Γ (0)Γ ()V (t − 2τ)}
) Γ (0)Γ ()V (t − 2τ)}
+
1
Γ − (0)
+
−
0
−
0
}
Γ (0+ ) Γ (0+ )V0 (t − 2 τ)
1
Γ(0− )
)
)
}
Γ (0− ) Γ (0− )V0 (t − 2 τ)
}
= 21 (1 + Γ (0+ ) ) V0 (t) − 1 − Γ (0+ ) Γ (0+ )V0 (t − 2 τ)
{
(
)
}
− 21 (1 + Γ (0− ) ) V0 (t) − 1 − Γ (0− ) Γ (0− )V0 (t − 2 τ)
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Finally, noting that Γ (0+ ) − Γ (0− ) =
κ+
β
, it is clear that
V− (0) ≈ 21 (Γ (0+ ) − Γ (0− ) )V0 (t) − 21 (Γ (0+ ) − Γ (0− ) )V0 (t − 2 τ)
≈
1 κ+
2 β
(V0 (t) − V0 (t − 2 τ))
and hence, by definition,
k− =
1 κ+
2 β
=
1
4
(
L+ −
L0
C+ −
C0
+
)
□
as desired.
Exercise 5.4.5
The electrical implementation of a transition minimized differential sig­
naling (TMDS) signaling protocol is given below [2]: this is a low-cost imp­
lementation for transmitting high-frequency digital data packages that
minimizes electromagnetic interference. It is typically employed in the
design of graphics interfaces for computers, high-definition televisions,
and graphic cards.
Suppose that the transmission lines are lossless, and the differential
impedance Z2 of the cable differs from that of the differential pair: Z1 ≠ Z2 .
Suppose further that Rs ≠ 21 Z1 is the source impedance. What is the best way
to match the impedance between the transmitter and the TMDS cable in
order to minimize undue reflections? Refer to method (b) versus method
(c) of Figure 5.10.
Vcc
A
Rterm
Vcc
Differential signal pair Termination
resistor
Current source
(V+, I+)
RS
A
+
RS
–I
Z1
–
B
(V–, I–)
Transmitter
Termination
Z2 resistor
Rterm
I
Rdiff
+
–
TMDS cable
B
Receiver
(b)
Vcc
B
R0
R0
Vcc
A
(a)
(c)
Figure 5.10
A simple model for a TMDS data pair.
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181
Differential Transmission Lines
Solution
The only place on the schematics to implement an impedance match is
between A, B. First, observe that if scheme (b) were implemented, where
Rdiff = Z1 , the resultant potential difference along the TMDS cable will be
halved. This can be easily seen (via the superposition principle) schematically as follows.
The discrete lumped equivalent circuit representation for γ + is Z1 in series
with R = 21 Rdiff , whence, the voltage output is that of the potential divider:
VA = V+ R +RZ1 = 21 V+ , where V+ is the incident voltage.
By symmetry, on γ − , VB = V− R +RZ1 = 21 V− , whence Vdiff = VA − VB = V+ ,
which is half the amplitude at the differential load.
Finally, consider (c). Here, on γ + , set R0 = Z1 in order to match
the impedance along γ + . Set Vcc = V+ . Then, VA = Vcc − V+ R +RZ1 = 21 V+
and VB = V− R +RZ1 − Vcc . That is, VB = −V+ R +RZ1 − Vcc = 23 V+ and hence,
Vdiff = VA − VB = 2 V+ , as required.
□
Exercise 5.4.6
Describe an alternative method to terminate a differential line such that both
odd and even modes are properly terminated.
Solution
T-termination of differential lines is an alternative termination scheme; see
the schematics shown in Figure 5.11.
Now, for odd mode, by definition, the potential drop across R is zero.
Hence, 2 R1 = Zdiff ; that is, R1 = Zodd . Next, consider an even mode noise propagating along the differential pair toward the receiver. First, observe that via
the principle of superposition, in the absence of any coupling, the differential
pair γ + ∪ γ − is equivalent to γ + ⊕ γ − , where the direct sum ⊕ denotes the
Differential signal pair
Current source
RS
+
I
RS
–I
(V+, I+)
R1
Z1
R1
–
R
+
Receiver
–
(V–, I–)
Transmitter
Figure 5.11
T-termination scheme for a differential pair.
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182
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Differential signal pair
RS
RS
+
RS +
(V+, I+)
R1
Z1
–
R1
(V–, I–)
R
(V+, I+)
Equivalence
(no cross coupling)
R1
~
R
+
RS
R1
–
(V–, I–)
~
R
Figure 5.12
Equivalent T-termination scheme in the absence of cross-coupling.
absence of cross-coupling (the disjoint union). Here, by definition, the load of
γ ± is Z = R1 + R , where R = R ||R ⇒ R = 21 R (cf. Figure 5.12).
Hence, by construction, Zeven = R1 + R ⇒ R = 2(Zeven − Zodd ), as required. □
References
1. Chang, D. 1992. Field and Wave Electromagnetics. Reading, MA: Addison-Wesley.
2. Digital Visual Interface. 1999. Digital Display Working Group, Rev. 1.0.
3. Knockaert, J., Peuteman, J., Catrysse, J., and Belmans, R. 2009. General equations for the characteristic impedance matrix and termination network of multiconductor transmission lines. In Proceedings of the IEEE Int. Conf. Industrial
Technology, pp. 595–600.
4. Neff, Jr., H. 1981. Basic Electromagnetic Fields. New York: Harper & Row.
5. Orfanidis, S. 2002. Electromagnetic Waves and Antenna. Rutgers University, ECE
Dept., http://www.ece.rutgers.edu/~orfanidi/ewa/.
6. Paul, C. 2002: Solution of the transmission-line equations under the weakcoupling assumption. IEEE Trans. Electromagn. Compat. 44(3), 413–423.
7. Paul, C. 1994. Analysis of Multiconductor Transmission Lines. New York, John
Wiley & Sons.
8. Paul, C. 2006. Introduction to Electromagnetic Compatibility. Hoboken, NJ: John
Wiley & Sons.
9. Plonsey, R. and Collin, R. 1961. Principles and Applications of Electromagnetic
Fields. New York: McGraw-Hill.
10. Sengupta, D. and Liepa, V. 2006. Applied Electromagnetics and Electromagnetic
Compatibility. Hoboken, NJ: John Wiley & Sons.
K15149_Book.indb 182
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6
Cross-Talk in Multiconductor
Transmission Lines
In this section, the phenomenon of cross-talk from transmission lines coupling electromagnetically is derived from Maxwell’s theory. To facilitate the
discussion of cross-talk, the concepts of mutual capacitance, mutual impedance, and mutual inductance are introduced; see, for example, References
[4,5,7,9]. The concept of mutual coupling leads naturally to the study of multiconductor transmission lines. Indeed, a glimpse into the concept of crosstalk was introduced in Chapter 5; in particular, see Exercise 5.4.4.
The last two sections comprise elements of multiconductor transmission line
theory, followed by the concept of scattering parameters. Scattering parameters
have great applications in microwave engineering, the details of which can be
found in References [5,6,8]. In particular, for readers wishing to pursue these
topics in greater depth, Reference [6] is a classic exposition on multiconductor
transmission line theory and that of microwave engineering is given in [8].
6.1 Reciprocity Theorem and Mutual Capacitance
An important result employed in this section is called Green’s reciprocity theorem;
see, for example, References [2,3,9,10] for various equivalent reciprocity theorems.
6.1.1 Theorem (Green’s Reciprocity)
Given a system {(Ci , ρi )}in= 1 of charged conductors, where ρi is the charge density on conductor Ci, and Ci ∩ C j = ∅ ∀i ≠ j, let Vi be a fixed potential on
Ci ∀i = 1, , n. Then, given another system {(Ci′, ρ′i )}ni = 1 of charged conductors,
with Ci′ ∩ C ′j = ∅ and Ci ∩ C ′j = ∅ ∀i ≠ j,
∑ ∫ ρ ϕ′d x =∑ ∫ ρ′ϕ d x
i
Ω
i
i
3
i
Ω
i
i
3
where Ω = R 3 −  i (Ci ∪ Ci′), and the potential φi satisfies Poisson’s equation
−∆ϕ i = 1ε ρi subject to the Dirichlet boundary condition ϕ i |∂Ci = Vi , and the
same also applies to the triple (ϕ ′i , Vi′, Ci′) ∀i .
183
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184
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Proof
The proof is an easy matter via mathematical induction. First, consider the
case where n = 1. Then, via the second Green’s identity, ∫ Ω ϕ 1 ∆ϕ 1′ − ϕ 1′ ∆ ϕ 1 =
∫ ∂Ω ϕ 1 ∂ n ϕ 1′ − ϕ 1′ ∂ n ϕ 1, substituting Poisson’s equation yields
1
ε
∫ (−ϕ ρ′ + ϕ′ ρ ) = ∫
1
Ω
1
i
1
∂Ω
ϕ 1 ∂ n ϕ 1′ − ϕ 1′ ∂ n ϕ 1 = 0
as ϕ 1 |∂Ω, ϕ 1′ |∂Ω are constants implies that ∂ n ϕ 1 |∂Ω = 0 = ∂ n ϕ 1′ |∂Ω , and
hence, ∫ Ω ϕ 1ρ1′ = ∫ Ω ϕ 1′ ρ1. So, suppose that for some fixed k > 1, the following
holds:
∑ ∫ ρ ϕ′d x = ∑ ∫ ρ ϕ′d x
k
i=1
i
Ω
k
3
i
i=1
Ω
i
i
3
Consider n = k + 1.
Then
∑ ∫ ϕ ∆ϕ′ − ϕ′∆ϕ = ∑ ∫
n
n
i=1
Ω
i
i
i
i
i=1
∂Ω
ϕ i ∂ n ϕ′i − ϕ′i ∂ n ϕ i
from Green’s identity. However, noting that
∑ ∫ ϕ ∆ϕ′ − ϕ′∆ϕ = ∑ ∫ ϕ ∆ϕ′ − ϕ′∆ϕ + ∫ ϕ
n
i=1
k
Ω
i
i
i
i
i=1
Ω
i
i
i
i
Ω
k +1
∆ϕ′k +1 − ϕ′k +1∆ϕ k +1 ,
the inductive assumption implies immediately that the first k = n − 1 summation vanishes, leaving
∫ϕ
Ω
k +1
∆ϕ′k +1 − ϕ′k +1∆ϕ k +1 =
∫
∂Ω
ϕ k +1 ∂ n ϕ′k +1 − ϕ′k +1 ∂ n ϕ k +1 .
From which, Gauss’ law and ∂ n ϕ k +1 |∂Ω = 0 = ∂ n ϕ′k +1 |∂Ω together yield
∫ϕ
Ω
ρ′
k +1 k +1
=
∫ ϕ′
Ω
ρ
k +1 k +1
, hence,
∑ ∫ ρ ϕ′d x = ∑ ∫ ρ′ϕ d x ,
n
i=1
Ω
i
i
3
n
i=1
and induction thus follows, as desired.
Ω
i
i
3
□
Indeed, observe trivially as an immediate corollary that for the pair of
systems {(ϕ i , Ci )}ni = 1 and {(ϕ ′i , Ci′)}in= 1, the restriction of ϕ i (ϕ ′i ) to Ci (Ci′) leads
immediately to the relation
∑
n
i=1
QiVi′=
∑
n
i=1
Qi′Vi
(6.1)
where Qi = ∫ Ci ρi d 3 x and Qi′ = ∫ Ci′ ρ′i d 3 x for all i = 1,…, n. Establish this in
Exercise 6.5.1.
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Cross-Talk in Multiconductor Transmission Lines
6.1.2 Example
As an application of Theorem 6.1.1, consider a system {Ci} of disjoint grounded
n conductors and suppose that conductor C0 has a total charge Q 0. Determine
the induced charge Qk on Ck, for some k, as a function of Q 0.
Let V0 be the equipotential on C0 as a result of some free charge Q 0 placed
on the conductor. Let Qi′ ∀i = 1, , n, denote the induced charges on {Ci }ni = 1 ,
if {Ci }ni = 1 were raised to potential {Vi′} , and likewise, let V0′ denote the potential on C0 if Q 0 = 0 and Qi are free charges placed on Ci ∀i = 1, , n. Then,
from the above corollary,
∑
n
i= 0
QiVi′=
∑
n
i= 0
Qi′ Vi ⇒ Q0V0′ + Q1V1′ +  + QnVn′ = Q0V0 + Q1V1 +  + QnVn = 0
as Vi = 0 ∀i = 1,…, n by assumption (the conductors are grounded), and Q 0 =
0. Whence, noting from the conservation of charge that Q0 + Q1 +  + Qn = 0,
it follows that
{ ∑ V ′}
Qi = − Vi′−
k≠i
k
−1

Q0

∑ V ′− ∑
i
i
j≠ i
Qj
∑
( j)
k≠ j

Vk′ 

where ∑(ki ) α k = α 1 +  + α i − 1 + α i + 1 + . To see this, it suffices to expand and
regroup the expression:
0 = Q0V0′ + Q1V1′ +  + QnVn′
= Q0V0′ + Q1V1′ + {Q0 − Q1 − Q3 −  − Qn } V2′ +  + {Q0 − Q1 − Q3 −  − Qn− 1 } Vn′
□
The Lorentz reciprocity theorem stated in Chapter 3 is given in a slightly
different form below with a shorter proof for instructive purposes, and to
preserve the continuity of the exposition.
6.1.3 Theorem (Lorentz Reciprocity)
Given (Ω, ε, μ), where Ω = R 3 − (C1 ∪ C2 ) and (Ci , J i ) , for i = 1,2, are compact
sources with time harmonic current densities Ji defined on Ci. Then, on Ω,
∇ ⋅ ( E × B′ − E ′ × B) = 0. In particular, ∫ C1 ∪ C2 E ⋅ J ′ d 3 x = ∫ C1 ∪ C2 E ′ ⋅ Jd 3 x .
Proof
First, recall the vector identity ∇ ⋅ A × B = B ⋅ ∇ × A − A ⋅ ∇ × B. Then, clearly,
∇ ⋅ ( E × B′ − E′ × B) = B′ ⋅∇ × E − E ⋅∇ × B′ − E′ ⋅∇ × B + B ⋅∇ × E′
= − iωB′ ⋅ B − iωµεE ⋅ E′ − iωµJ ′ −
iωE′ ⋅ E + iωµεB ⋅ B′ + iωµJ
= iωµJ − iωµJ ′
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
whence, on Ω, J , J ′ = 0 ⇒ ∇ ⋅ ( E × B′ − E ′ × B) = 0 . To establish the second
result, it suffices to apply the divergence theorem:
∫ ∇ ⋅ (E × B′ − E′ × B)d x = ∫
3
Ω
∂Ω
( E × B′ − E ′ × B) ⋅ nd 2 x
= iωµ
∫ (E′ ⋅ J − E ⋅ J ′)d x = 0
3
Ω
□
as J|Ω , J ′|Ω ≡ 0 completing the proof.
6.1.4 Remark
From Theorem 6.1.3, and noting that V ∝ E, I ∝ J, it follows that given two
thin conductors (Ci , Vi , I i ), at potential Vi and current Ii, for i = 1,2, with
respect to some fixed ground, then V1 I 2 = V2 I1; see Exercise 6.5.2 for a rigorous proof. More generally, for a system of thin conductors {(Ci , Vi , I i )}
and {(Ci′, Vi′, I i′)} , for i = 1,…, n, ∑ ni = 1 Vi I i′ = ∑ ni = 1 Vi′I i . Establish this in Exercise
6.5.2. Finally, it ought to be pointed out that the full statement of Lorentz
reciprocity involves introducing a fictitious magnetic charge density J ;
see, for example Reference [1]. This was deliberately left out in the proof of
Theorem 6.1.3. To wit, the statement of the theorem in generality is:
∫
∂Ω
( E1 × H 2 − E2 × H 1 ) ⋅ d 2 x =
∫ (E ⋅ J + H ⋅ J
Ω
2
1
1
2
− E1 ⋅ J 2 − H 2 ⋅ J1 )d 3 x
where Ji is the (fictitious) magnetic current density on Ci, and Bi = µH i.
Consider a system {Ci} of N + 1 conductors as illustrated in Figure 6.1. Suppose
that the conductors are of arbitrary cross-sections, and their lengths run parallel to one another. For simplicity, assume that the conductors are of identical
lengths ℓ. Fix one conductor as the ground conductor and denote this by the 0th
conductor C0. All potential references are made with respect to the conductor C0.
V2
ρ2
V2
ρ1
r
2
VN
1
0
V0 = 0
N
ρN
ρ0
Figure 6.1
Cross-section of conductors in a homogeneous dielectric medium.
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Cross-Talk in Multiconductor Transmission Lines
Suppose that the conductor Ci is charged to a uniform charge density of ρi
for each i = 1,…, N. Under electrostatic conditions, the resultant charge on the
system comprising the N + 1 conductors must sum to zero. Whence, the charge
density on the ground conductor C0 (here, denoted by conductor 0) must satisfy
ρ0 = −(ρ1 +  + ρN )
(6.2)
Set Ω = R 3 −  i Ci for simplicity. Then the electric field at an arbitrary
point r ∈ Ω is the sum of the electric field contribution from each conductor:
E(r ) = ∑ iN= 0 Ei (r ). From Chapter 3, it is clear that the electric field E(r) = −∇φ is
defined by the Laplace equation:

∆ϕ = 0 on Ω,

−ε∇ϕ ⋅ ni = ρi ∀i = 0, , N ,

(6.3)
where n i is the unit normal vector field on ∂Ci .
By definition, Vi = − ∫ γ i E ⋅ dl , where γ i ⊂ Ω is a path satisfying γ i (0) ∈C0 and
γ i (1) ∈Ci . Furthermore, from Q = CV, it is clear by invoking Gauss’ law and the
N
divergence theorem, that Vi = ∑ j =0 pijQ j, where Qi = ε ∫ Ci E ⋅ d 3 x = ε ∫ ∂Ci E ⋅ ni d 2 x,
whence,
Vi = −
∫
γi
E ⋅ dl = −
∑ ∫
N
j= 0
γi
E j ⋅ dl =
∑
N
j= 0
pijQ j ⇒ pij = − Q1j
∫
γi
E j ⋅ dl
This leads to the following result.
6.1.5 Lemma
Given a system of charged, thinly insulated conductors illustrated in Figure 6.1,
the charge Qi on the conductor Ci is related to the collection {(Vk , Ck )} via
 V1

  VN
  P11
 =
    PN 1
P1N
PNN
  Q1

   QN




(6.4)
where Pij = pij − pi 0 ∀i, j .
Proof
From (6.1), Q0 + Q1 +  + QN = 0 (charge conservation), whence,
Vi =
∑
as required.
K15149_Book.indb 187
N
j= 0
pijQ j =
∑
N
j=1
pijQ j − pi 0 (Q1 +  + QN ) =
∑
N
j= 0
( pij − pi 0 )Q j ,
□
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
6.1.6 Lemma
The matrix P = ( Pij ) of (6.4) is symmetric: Pij = Pji ∀i, j.
Proof
Now, by (6.1), ∑ ni = 1 QiVi′= ∑ ni = 1 Qi′Vi ; that is, Q′ ⋅ V = Q ⋅ V ′, where
V = (V1 , , VN ), V ′ = (V1′, , VN′ ), Q = (Q1 , , QN ), Q′ = (Q1′ , , QN′ )
Thus, by definition,
Q′ ⋅ V =
∑ ∑ Q′P Q = ∑ Q′P Q + ∑ ∑ Q′P Q + ∑ ∑ Q′P Q
i
j
i ij
j
i
i ii
j
i< j
j
i ij
j
i> j
j
i ij
j
and
Q ⋅V ′ =
∑ ∑ Q P Q′ = ∑ Q P Q′ + ∑ ∑ Q P Q′ + ∑ ∑ Q P Q′
i
j
i ij
j
i
i ii
i
i< j
j
i ij
j
i> j
j
i ij
j
Hence, from Q′ ⋅ V = Q ⋅ V ′, the above expansions imply immediately that
∑ ∑ Q P Q′ = ∑ ∑ Q P Q′ = ∑ ∑ Q P Q′
i< j
j
i ij
j
i> j
j
j
ji
i
i< j
j
i
ji
j
∑ ∑ Q P Q′ = ∑ ∑ Q′P Q = ∑ ∑ Q P Q′
i> j
j
i ij
j
i< j
j
i ij
j
i> j
j
i
ji
j
via relabeling the dummy indices, and hence, Pij = Pji ∀i, j, as claimed.
□
Indeed, it is easy to see from Equation (6.4) that as P is a constant, it clearly
follows from
Vi = Pi 1Q1 + Pi 2Q2 +  + PiN QN
that on setting Qk = 0 ∀k ≠ i , Vi = Pik Qk ⇒ Pik = QVki ; that is, when all conductors
Ck for k ≠ i are discharged except for Ci. Furthermore, recalling that Q = CV,
it is evident that this can be generalized to a multiconductor system {Ci} via
Q = CV. Indeed, based on physical considerations, for passive devices—that
is, devices that are not anisotropic (such as ferromagnetic materials, plasma)
or active generators—it is clear that P is invertible and in particular, P −1 = C.
Moreover, by Lemma 6.1.6, C is symmetric as it is the inverse of a symmetric
matrix. These observations justify the following definition.
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Cross-Talk in Multiconductor Transmission Lines
189
6.1.7 Definition
Given the pair V,Q such that V = PQ, the coefficients Pij are called the coefficients of inductance. Moreover,
Pij =
Vi
Qj
Qk = 0 ∀k ≠ j
Likewise, the coefficients Cij of Q = CV, are called the coefficients of capacitance, and
Cij =
Qi
Vj
Vk = 0 ∀k ≠ j
when all conductors Ck for k ≠ i are grounded.
6.1.8 Remark
Physically, the coefficient of capacitance Cij, for i ≠ j, expresses the mutual
capacitance between Ci, Cj. That is, it represents the capacitive coupling
between Ci, Cj. On the other hand, Cii represents the self-capacitance: it is the
capacitance between Ci ,  j ≠ i C j ∪ C0 , when  j ≠ i C j is grounded. Explicitly,
consider Q = CV :
Qi = Ci 1V1 + Ci 2V2 +  + CinVn ≡ ci 1 (Vi − V1 ) +  + ciiVi +  + cin (Vi − Vn )
where Cij denotes the mutual capacitance between Ci, Cj for i ≠ j, and Cii
denotes the capacitance between Ci,C0. Then, by definition, equating the
coefficients of Vi,


Cij = 


∑
k ≥1
cik
for i = j,
for i ≠ j.
cij
That is, Cii is the sum of the mutual capacitance between Ci,Cj for each j ≠ i.
6.1.9 Lemma
Given the pair Q = CV, Cii > 0 and Cij < 0 ∀i, j.
Proof
By definition,
Cij =
Qi
Vj
Vk = 0 ∀k ≠ j
Hence, suppose without loss of generality that Vj > 0 as Vj < 0 follows mutatis
mutandis. Then, Q j > 0 ⇒ Qi < 0, and Cii = QVii implies that Vi > 0 ⇒ Qi > 0. □
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Radius a1
d
C11 = c11 + c12
C12 = –c12
Radius a2
C22 = c22 + c12
Equivalent
h
h
Schematic
c12
c11
c22
Figure 6.2
Capacitive coupling among conductors and the ground plane.
6.1.10 Example
The above theory is illustrated by applying it to a system of two conductors
illustrated in Figure 6.2, where the conductors
{
Ci = ( x , y ) : ( x +
χi
2
}
d)2 + ( y − hi )2 ≤ ai2 × R
for i = 1,2, and
 1 if i = 1,
χi = 
 −1 if i = 2.
Finally, set Ci (0) = {( x , y ) : ( x + χ2i d)2 + ( y − hi )2 ≤ ai2 }.
By the method of images, the field resulting from Ci(0) (see Chapter 3) can
be determined as follows. Via the principle of superposition, there are two
cases to be considered. First, consider the case wherein C2 is absent. Let ρ1
denote the charge per unit length. Then, the equivalent line charge in C1 is
a2
displaced from the center O1 = (− 21 d , h) to O1′ = (− 21 d , h − d1 ), where d1 = 21h .
By symmetry, the equivalent line charge ρ2 in C2, in the absence of 2C1, is
a
displaced from the center O2 = ( 21 d , h) to O2′ = ( 21 d , h − d2 ) , where d2 = 22h .
Next, consider the pair C1, C2 in the absence of the ground plane. The locations of the equivalent line charges are determined as follows. First, without
loss of generality, relative to a new origin Oδˆ = (δˆ , 0), for some δ̂ such that the
equivalent line charges are located, respectively, at Oˆ 1′ = (− xˆ , h) and Oˆ 2′ = ( xˆ , h).
Recalling from Example 3.2.4, it is clear that the respective x-coordinates of
C1, C2 are
xˆ i2 = xˆ 2 − ai2
(6.5)
for i = 1,2. Hence, xˆ 12 + a12 = xˆ 22 + a22 ⇒ xˆ 12 − xˆ 22 = a22 − a12 . That is,
( xˆ 1 − xˆ 2 )( xˆ 1 + xˆ 2 ) = a22 − a12
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Cross-Talk in Multiconductor Transmission Lines
Furthermore, observe by definition that d = xˆ 1 + xˆ 2 . Whence, xˆ 1 − xˆ 2 =
and thus
a22 − a12
d
xˆ 1 =
d 2 + a12 − a22
2d
(6.6)
xˆ 2 =
d 2 + a22 − a12
2d
(6.7)
Without loss of generality, suppose that a1 < a2. Then, by definition, δˆ < 0
a2 − a2
with respect to x = 0, and in particular, xˆ 1 +|δˆ|= 21 d ⇒|δˆ|= 22 d 1 . That is,
a2 − a2
δˆ = 12 d 2 . Thus, in the absence of the ground plane, the equivalent charges
are located, with respect to the origin O = (0, 0), at O1′′ = (− xˆ + δˆ , h) and
O2′′ = ( xˆ + δˆ , h) , respectively.
From the above analysis, it is clear that the new locations of the equivalent line charges are, respectively, O1 = (− xˆ + δˆ , h − d1 ) and O2 = ( xˆ + δˆ , h − d2 ). To
see this, it suffices to note that in the former case (see Figure 6.3) the presence of the ground plane exerts a force f1′ on the equivalent line charge
located at the axis of C1 such that it is translated to O1′ = (− 21 d , h − d1 ), and
in the latter scenario, the presence of the grounded conductor C2 exerts a
force f1′′ on the equivalent line charge located at the axis of C1 such that it is
translated to O1′′ = (− xˆ + δˆ , h). Hence, the resultant location in the presence of
the ground plane and C1 leads to the translation of the line charge initially
at O1 = (− 21 d , h) to O1 = (− xˆ + δˆ , h − d1 ) under the force f1 = f1′ + f1′′ . The new
location O2 of the line charge in C2 follows that of O1 mutatis mutandis.
Consequently, the resultant equivalent line charges and their respective
image charges are:
O1 = (− xˆ + δˆ , h − d1 ), O2 = ( xˆ + δˆ , h − d2 ), O1∗ = (− xˆ + δˆ , − h + d1 ),
O2∗ = ( xˆ + δˆ , − h + d2 )
C1
C2
1 h)
(– —d,
2
f1'
f1"
f2'
1 h)
(—d,
2
f2"
Equivalent line charge
Figure 6.3
Equivalent line charges for the two respective charged conductors.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Next, recall from Example 3.2.4 that the potential of an infinite line charge
λ is ϕ = − 2 πελ 0 ln rr∞ , for some reference point r∞; it follows at once that the
potential above the ground plane is
Φ = − 2 πελ 0 ln
( x + xˆ −δˆ )2 +( y − h+ d1 )2
( x + xˆ −δˆ )2 +( y + h− d1 )2
+ 2 πελ 0 ln
( x− xˆ −δˆ )2 +( y − h+ d2 )2
( x− xˆ −δˆ )2 +( y + h− d2 )2
, y >0
(6.8)
where, without loss of generality, λ is any fixed line charge on C1 and −λ the
charge on C2. Then, by definition, the capacitance per unit length is between
C1, C2 is C12 = Φλ . That is,

C12 = 2 πε 0 ln

( x + xˆ −δˆ )2 +( y − h+ d1 )2 ( x− xˆ −δˆ )2 +( y + h− d2 )2
( x + xˆ −δˆ )2 +( y + h− d1 )2 ( x− xˆ −δˆ )2 +( y − h+ d2 )2



−1
(6.9)
Similarly, the coefficient of capacitance per unit length between C1 and
ground is

c11 = 2 πε 0 ln

( x + xˆ −δˆ )2 + ( y − h + d1 )2
( x + xˆ −δˆ )2 + ( y + h − d1 )2



−1
(6.10)
and that between C2 and ground is

c22 = 2 πε 0 ln

( x − xˆ −δˆ )2 +( y − h+ d2 )2
( x − xˆ −δˆ )2 +( y + h− d2 )2
−1

 ,

(6.11)
□
as required.
6.2 Mutual Inductance and Mutual Impedance
From Chapter 5, it is clear that cross-talk is also contributed by mutual inductance. Thus, along the vein of Section 6.1, consider a system of N + 1 conductors running parallel to one another, each of length ℓ. Let the 0th conductor be
a grounded conductor and the remaining N conductors be measured relative
to the grounded conductor.
Let the ith conductor Ci carry a current Ii and γ i = ∂Ci (0) be a circular path
encircling the boundary of a cross-section of the conductor. Then, invoking Kirchhoff’s current law, I1 +  + I N = − I 0 . That is, the ground conductor
provides the return current pathway for the circuit.
At any point (x,y,z), the total magnetic field intensity is given by the sum of
magnetic field contributions from the N + 1 conductors:
B( x , y , z) =
K15149_Book.indb 192
∑
N
i= 0
Bi ( x , y , z)
(6.12)
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Cross-Talk in Multiconductor Transmission Lines
where, without loss of generality, set z = 0 in all that follows. Furthermore, for
simplicity, assume that the conductor cross-sectional areas Si′ are constants. Then,
Bi ( x , y , z) =
µ0
4π
∫∫
Si′
{
2
2
2
d S′J i ( xi′ , y i′ , z ′) ( x − xi′) + ( y − y i′) + ( z − z ′)
2
}
− 23
 y i′ − y

 x − xi′



(6.13)
with Ji being the surface current density flowing along conductor Ci,
xi′ = x ′ + xi, y i′ = y ′ + y i and ( xi , y i , z) is the axis of Ci, for some fixed pair
( xi , yi ).
Next, recall that the magnetic flux Ψi per unit length intersecting the area S( γ i )
spanned by a loop γi around Ci(0) is given by Ψ i = lim δ1z ∫ ∫ ∂S( γ i )×[0,δz ] B ⋅ ni d 2 x ,
δz→ 0
where ni is the unit vector normal to ∂S( γ i ) × [0, δz] ; see Chapter 4 on transmission line theory. Hence, by Equation (6.12),
Ψ i = lim
δz→0
∑ ∫∫
k
1
δz
∂ S( γ k )×[0,δ z ]
Bk ⋅ nk d 2 x ≡
∑LI
k
(6.14)
ik k
where Lik is the mutual inductance per unit length between Ci, Ck. See
Figure 6.4. In particular, by Equation (6.13),
L ij =
Ψi
Ij
(6.15)
I k = 0, k ≠ j
That is, the mutual inductance between conductors i and j is defined by making
conductors k ≠ j open circuits. Indeed, Equation (6.15) can be defined by the matrix:
(6.16)
Ψ = LI
where
 Ψ1
Ψ =   Ψ N
 L11

,L=
 

 LN 1
L1N
LNN

 I1
,I=

  I N

Lii
Ci
Schematic
Cj
Ck




Representation
Lij
Mutual
inductance
Lik
Ljj
Lkk
Figure 6.4
Voltage coupling via mutual inductance between two conductors.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
6.2.1 Lemma
Given Equation (6.16), the matrix L is symmetric: L ij = L ji ∀i, j .
Proof
Just as with Cij, Lij depends only on geometry, and in particular, the coefficients are constants. Now, consider a loop γi bounded by (Ci, C0) where C0 is
the ground conductor: γ i ∩ ∂Ci ≠ ∅ and γ i ∩ ∂C0 ≠ ∅ . Let S( γ i ) denote the
surface spanned by γi and ni be the unit normal on S( γ i ) . Then, the magnetic
flux Ψij linking Ci with respect to γi due to current flowing in Cj is, by definition,
Ψ ij =
∫
S( γ i )
B j ⋅ ni d 2 x =
µ0 I j
4π
∫ ∫
S( γ i )
γj
∇×
dl j
R
⋅ ni d 2 x
By Stokes’ theorem,
∫
∇×
Ψ ij =
µ0 I j
4π
∫ ∫
L ij ≡
Ψ ij
Ij
S( γ j )
dl j
R
⋅ ni d 2 x =
∫
dl j
R
γj
⋅ dRli
(6.17)
whence,
∇×
S( γ i )
γj
µ0
4π
∫∫
dl j
R
µ0 I j
4π
∫∫
∫∫
dli ⋅dl j
R
⋅ ni d 2 x =
γi
γj
dl j ⋅dli
R
By definition,
=
γi
γj
dl j
R
⋅ dRli =
µ0
4π
γj
γi
≡ L ji
and hence, the arbitrariness of γi, γj concludes the proof.
□
6.2.2 Example
The above theory is illustrated by applying it to a system of two parallel
conductors of radii a > 0 (cf. Figure 6.5), where the lines are assumed to be
infinitely long, separated by a distance d > a, and their respective axes are a
distance h > 0 above the ground plane.
Determine the resultant voltage at the loads of C1, C2 respectively.
V1
V2
L11 self-inductance
C1
d
r s
δz
C2
Schematic
Representation
L12
Mutual
inductance
L22
Figure 6.5
Voltage coupling via mutual inductance.
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Cross-Talk in Multiconductor Transmission Lines
Now, via superposition principle, it suffices to consider V2 = 0 and determine the induced voltage δV21 along C2. Then, the resultant voltage on C2 is
thus V2 + δV21 to first order, because δV21 will in turn induce a voltage δv21
on C1, and this voltage will in turn induce a voltage δ 2V21 ≡ δ(δV21 ) on C2, ad
infinitum: V2 → V2 + δV21 + δ 2V21 + δ 3V21 +  . By symmetry, on setting V1 = 0,
the resultant voltage is, to first order, V1 + δV21.
Therefore, suppose that V2 = 0 and V1 = V1 (t) is some time harmonic voltage
propagating along C1. Then, δV21 = −L 21 ddIt1 . From Chapter 5, recall that
L 21 = ΨI12
, where Ψ 2 |I2 = 0 δz = ∫ S B|I2 = 0 ⋅ nd 2 x with n being the unit norI2 = 0
mal vector field on S and B|I = 0 is determined by the field along C1 and the
2
return current I 0′′ along C2, where I 0 + I1 = 0 and I 0 = I 0′ + I 0′′ , with I 0′ being
the return current on the ground plane.
By inspecting Figure 6.5 (cf. Example 6.1.8) it is clear that the boundary
charge density induced on C2(0) is given by ρs,2 = −ε 0 ∇Φ ⋅ n2 , where Φ is
defined by Equation (6.8) and n 2 is the unit normal vector field on ∂C2(0) .
Thus, the equivalent return current density along C2 is
J 0′′ =
d
dt
∫
∂ C2 (0)
ρs,2 (t)dl =
dλ ( t ) 1
dt 2 π
∫
∂ C2 (0)
∇ ln
( x + xˆ −δˆ )2 +( y + h− d1 )2 ( x − xˆ −δˆ )2 +( y − h+ d2 )2
( x + xˆ −δˆ )2 +( y − h+ d1 )2 ( x − xˆ −δˆ )2 +( y + h− d2 )2
⋅ n2 dl
(6.18)
Likewise, the current density along C1 is given by
J1 =
d
dt
∫
∂ C1 (0)
ρs ,1 (t) dl =
dλ ( t ) 1
dt 2 π
∫
∂ C1 (0)
∇ ln
( x + xˆ −δˆ )2 +( y + h− d1 )2 ( x− xˆ −δˆ )2 +( y − h+ d2 )2
( x + xˆ −δˆ )2 +( y − h+ d1 )2 ( x− xˆ −δˆ )2 +( y + h− d2 )2
⋅ n1 dl
(6.19)
From this, the magnetic density at an arbitrary point ( x , y ) ∈ R 2+ − (C1 (0) ∪
C2 (0)) resulting from (C1, C2) is
B12 |I2 = 0 =
µ 0 dλ ( t )
2 π dt
πa 2
{
κ1
( x + xˆ −δˆ )2 + ( y − h + d1 )2
+
κ ′′2
( x − xˆ −δˆ )2 + ( y − h + d2 )2
}e
φ
where κ 1 , κ ′′2 are defined by Equations (6.18) and (6.19), respectively:
K15149_Book.indb 195
κ1 =
1
2π
∫
∇ ln
( x + xˆ −δˆ )2 +( y + h− d1 )2 ( x− xˆ −δˆ )2 +( y − h+ d2 )2
( x + xˆ −δˆ )2 +( y − h+ d1 )2 ( x− xˆ −δˆ )2 +( y + h− d2 )2
⋅ n1 dl
κ ′′2 =
1
2π
∫
∇ ln
( x + xˆ −δˆ )2 +( y + h− d1 )2 ( x− xˆ −δˆ )2 +( y − h+ d2 )2
( x + xˆ −δˆ )2 +( y − h+ d1 )2 ( x− xˆ −δˆ )2 +( y + h− d2 )2
⋅ n2 dl
∂ C1 (0)
∂ C2 (0)
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
By definition, the magnetic flux density per unit length across the rectangular loop S = [a,d − a] × [0,δz] at y = h is:
lim Ψ 2 |I2 =0 = lim δ1z
δz→0
δz→0
=
µ 0 dλ
2 π dt
µ 0 dλ
2 π dt
∫ ∫ {

κ 1 ln

δz
d− a
0
a
κ1
( x + xˆ −δˆ )2 + d12
+
κ ′′2
( x − xˆ −δˆ )2 + d22
} dx dz
+ κ ′′2 ln
( d − a− xˆ − dˆ )2 + d22 + d − a− xˆ −δˆ
( a− xˆ − dˆ )2 + d 2 + a− xˆ −δˆ



+ κ ′′2 ln
( d − a− xˆ − dˆ )2 + d22 + d − a− xˆ −δˆ
( a− xˆ − dˆ )2 + d 2 + a− xˆ −δˆ



( d − a+ xˆ − dˆ )2 + d12 + d − a+ xˆ −δˆ
( a+ xˆ − dˆ )2 + d 2 + a+ xˆ −δˆ
1
2
and hence,
L21 =
Ψ2
I1 I = 0
2
=
µ0
2π

κ 1 ln

( d − a+ xˆ − dˆ )2 + d12 + d − a+ xˆ −δˆ
( a+ xˆ − dˆ )2 + d 2 + a+ xˆ −δˆ
1
2
That is, the resultant voltage along C2 is thus

V2 − 2µπ0 κ 1 ln

( d − a+ xˆ − dˆ )2 + d12 + d − a+ xˆ −δˆ
( a+ xˆ − dˆ )2 + d 2 + a+ xˆ −δˆ
1
+κ ′′2 ln
( d− a− xˆ − dˆ )2 + d22 + d− a− xˆ −δˆ
( a− xˆ − dˆ )2 + d 2 + a− xˆ −δˆ
2
 dI1
 dt

By symmetry, the resultant voltage along C1 is

V1 − 2µπ0 κ 1 ln

( d − a+ xˆ − dˆ )2 + d12 + d − a+ xˆ −δˆ
( a+ xˆ − dˆ )2 + d 2 + a+ xˆ −δˆ
1
+κ ′′2 ln
( d− a− xˆ − dˆ )2 + d22 + d− a− xˆ −δˆ
( a− xˆ − dˆ )2 + d 2 + a− xˆ −δˆ
2
 dI2
 dt

□
Having established the mutual inductance and capacitance for a system of conductors, the mutual impedance can now be defined. In view of
Equations (6.4) and (6.16), define
V1 = Z11 I1 + + Z1n I n
(6.20)
Vn = Zn1 I1 + + Zn1 I n
for n-conductors; that is, V = ZI. Then, it is clear by definition that for each
V
fixed i, Vi = Zij I j if I k = 0 ∀k ≠ j . That is, Zij ≡ I ji|Ik = 0 ∀k ≠ j is well-defined as Zij
are constants for all i,j = 1,…, n. Thus, the mutual impedance Zij is obtained
by keeping each conductor Ck ∀k ≠ j an open circuit. The matrix Z is also
called the transfer impedance matrix or simply the Z-matrix.
6.2.3 Example
Consider the schematic diagram shown in Figure 6.6. It illustrates qualitatively the impact of a nonzero, finite, mutual impedance between two conductors C1, C2. In the illustrated scenario, C1 is assumed to be an open circuit
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Cross-Talk in Multiconductor Transmission Lines
I1 = 0
Victim
C1
V1
Z12
C2
V2
I2
False triggering of C1 by the
switching of C2 resulting from
the transfer impedance
Aggressor
Figure 6.6
Induced voltage along victim line via transfer impedance.
at the load (not illustrated): I1 = 0. When C2 is switching and hence I2 ≠ 0,
the transfer impedance induces a potential difference along C1 according
to: V1 = Z12 I2. From an application perspective, if V1 is sufficiently large—for
instance, large enough to switch on a transistor—then this will cause a false
bit to be transmitted, causing signal integrity issues. In this particular scenario, the effect is called a ground bounce wherein the sudden current draw
by C2 induces a voltage spike along C1.
□
Returning to Equation (6.20), it is clear that the transfer matrix Z for passive
physical systems (i.e., networks) encountered is invertible: if this were false,
then a fixed set of currents {Ii} would induce multivalued voltages {Vi( k ) } ,
which is clearly not observed in passive physical networks: that is, the voltages are not uniquely defined. Hence, for most practicable applications,
I = YV = YZI ≡ I ⇔ YZ = I
where I is the identity matrix. That is, Z−1 = Y exists, assuming a linear passive network, and there is thus a one-to-one correspondence between voltage
and current for ohmic systems, as required. For obvious reasons, Y is called
the mutual admittance matrix.
6.2.4 Lemma
Given a linear passive network defined by Equation (6.20), the mutual impedance matrix Z is symmetric.
Proof
Now, for any fixed i, I k = 0 ∀k ≠ j ⇒ Vi = Zij I j ⇔ Vi I j = Zij I 2j . Likewise, for
any fixed j, I k = 0 ∀k ≠ i ⇒ Vj = Z ji I i ⇔ Vj I i = Z ji I i2 . Hence, by the reciprocity
theorem via Remark 6.1.4, Vi I j = Vj I i ⇒ Zij = Z ji on setting I i = I j (as I i , I j are
arbitrary and Zij are constants).
□
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Indeed, it is clear by definition that Zmn ≡ − I1n ∫ γ mn ( Em − En ) ⋅ dl, where
I k = 0 ∀k ≠ n and E k is the electric field resulting from conductor Ck. That is,
the mutual impedance measures the potential difference between Cm and Cn
when current In is conducting along Cn, with the remaining conductors kept
open.
6.2.5 Definition
Given a linear system defined by V = ZI, the system is said to be lossless if
ˆ , where Zˆ ij = ℑm(Zij ) for all i, j. Finally,
ℜe(Zij ) = 0 ∀i, j . In particular, Z = iZ
I
on setting Y = Z−1, I = YV , where Yij = Vij|Vk = 0 ∀k ≠ j ; that is, Ck is grounded ∀k ≠ j.
6.2.6 Remark
For a lossless linear network, the time-average power ⟨ P⟩ = 21 ℜe(V ⋅ I ∗ )
implies that ⟨ P⟩ = 21 ℜe⟨ ZI , I ∗ ⟩ = 21 ℜe ∑ i , j i Zˆ ij I i I ∗j = 0. That is, the net power
delivered to the network is zero. Thus, the energy is stored reactively.
6.2.7 Example
Consider a 2-port network illustrated in Figure 6.7. Is it possible to express
the general 2-port network as Figure 6.7(a) or (b)? Because Z is symmetric, it depends only on three parameters instead of four: {Z11 , Z12 , Z22 }.
However, Figure 6.7(a) or (b) depends only on three parameters {R1 , R2 , R3 }
or {r1 , r2 , r3 }, respectively. Hence, the 2-port network can be represented by
(a) or (b).
Now, consider the T-network depicted in Figure 6.7(a). Express Zij in
terms of Rk for k = 1,2,3. By definition, Z11 = VI11 with I 2 ≡ 0 . Then, inspecting
Figure 6.7(a), Z11 = R1 + R3, as I 2 = 0 ⇒ R2-branch is an open circuit. Likewise,
V1 = V1+ + V1– V2 = V2+ + V2–
I1 = I1+
+
I1–
I2 = I2+
Incident wave
I1+
V1+
V1–
I1–
Reflected wave
+
I1 R1
(a)
Incident wave
I2+
2-port
network
R2 I2
I2–
V2+
V2–
V1
Equivalent
circuit
V2
R3
r3
I1
I2
I2–
Reflected wave
(b)
V1
r1
r2
V2
Figure 6.7
A general 2-port network and its equivalent circuit representation.
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Cross-Talk in Multiconductor Transmission Lines
199
Z22 = VI22 with I1 = 0 leads to the R1-branch of the circuit being open. Hence,
Z22 = R2 + R3 . Finally, for Z12 = VI21|I1 = 0 , I1 = 0 implies that the potential drop
across R3 is V1. Thus, V1 = R3 I 2 ⇒ Z12 = R3 . Indeed, it is easy to see that Z21 =
Z12: I 2 = 0 ⇒ V2 = R3 I1 ⇒ Z21 = R3 = Z12. Equivalently, expressing Ri in terms of
Zij yields: R1 = Z11 − Z12, R2 = Z22 − Z12 , R3 = Z12.
Next, consider Figure 6.7(b): a Π-network. Set Yi = ri ∀i = 1, 2, 3 . Then, in
terms of admittance, via Definition 6.2.5, on setting V2 = 0 and V1 , I1 ≠ 0,
Y11 = VI11 = Y1 + Y3 . Thus,
Z11 =
r1r3
1
Y1 + Y3 r1 + r3
Likewise, on setting V1 = 0 and V2 , I 2 ≠ 0 , Y22 = VI22 = Y2 + Y3, and
hence, Z22 = Y2 +1Y3 = r2r2+r3r3 . Finally, setting V1 = 0 and V2 , I1 ≠ 0 , it follows from Kirchhoff’s current law that I1 + Y3V2 = 0 ⇒ Y3 = − VI12 ≡ Y12
by definition. Moreover, it is easy to see that for V2 = 0 and V1 , I 2 ≠ 0,
I 2 + Y3V1 = 0 ⇒ Y3 = − VI21 ≡ Y21 = Y12, as expected. Finally, for completeness,
expressing Yi in terms of Yij yields:
Y1 = Y11 + Y12 , Y2 = Y22 + Y12
and Y3 = −Y12
In particular,
r1 =
1
Y11 + Y12
, r2 =
1
Y22 + Y12
and r3 = − Y112
□
6.3 Multiconductor Transmission Lines and Cross-Talk
In this section, a pair of transmission lines is generalized to n > 2 transmission lines. For simplicity, assume that the conductors {Ci} are parallel to one
another, embedded in a homogeneous medium (R 3 , µ , ε), where the conductor C0 is taken to be the ground reference. Indeed, the analysis for multitransmission lines parallels that of the differential pair expounded in Section 5.1.
6.3.1 Theorem
Given a system of multitransmission lines {Ci} in Ω = (R 3 , µ , ε) −  ni = 0 Ci ,
where the cross-sections of Ci are constants, and the conductor losses are
sufficiently small so that the TEM approximation may be employed, the system is then defined by
K15149_Book.indb 199
∂ z V = −RI − L ∂t I
(6.21)
∂ z I = −GV − C ∂t V
(6.22)
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200
Electromagnetic Theory for Electromagnetic Compatibility Engineers
where V = (V1 , , Vn ) , I = ( I1 , , I n )
 R1 + R0

R0
R = 

R

0
 L11

L 21
L=
 
 L n1




G= 





∑




C=





∑
L12
L 22
Ln2
n
i=1
G 1i
− G 21
R0
R2 + R0
R0
L 1n
L2 n
L nn
− G12
∑
n
i=1
− G n1
n
i=1
G2 i
C1i
− C21
− C n1


,



R0
R0
Rn + R0






− G 1n
− G2 n
∑
n
− Gn2
− C12
− C1n
− C2 n
∑
n
i=1
C2 i
− Cn 2
∑
i=1
G ni
n
i=1
Cni




,















Ri is the resistance per unit length along Ci, and Lij, Gij, Cij are, respectively,
the mutual inductance, mutual admittance, and mutual capacitance per unit
length between Ci and Cj.
Proof
Essentially, the argument follows that of Section 5.1 mutatis mutandis. First,
recall that I 0 + ∑ ni = 1 I i = 0 , where I0 is the return current along the ground
conductor C0. Then, the voltage variation along Ci is determined as follows.
Consider a differential rectangular loop between Ci and C0; see Figure 5.2,
where C+ corresponds to Ci and C– corresponds to C0. Let E⊥ denote the
electric field from {Ci} normal to {Ci}, and Ei ,|| the electric field on ∂Ci and
parallel to Ci, where, by assumption, ||Ei ,||||<<||E⊥||. Likewise, let B⊥ denote
the magnetic field density from {Ci} normal to {Ci}.
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Cross-Talk in Multiconductor Transmission Lines
Next, invoke Maxwell’s integral equation: ∫ ∂Si Ei ⋅ dl = − ddt ∫ Si B⊥ ⋅ nd 2 x ,
where Ei = E⊥ + Ei ,|| . Then, along the loop ∂Si = γ ↑ ∪ γ → ∪ γ ↓ ∪ γ ←, where
γ↑
→
←
↓
a 
→ b, b γ
→ c, c γ
→ d, and lastly, d γ
→ a; then by definition,
∫ ∂Si E⊥ ⋅ dl yields:
Vi ( z) = −
δVi = −
∫
c
Vi ( z + δz) = −
δVi ,0 = −
∫
b
c
b
∫
b
a
E⊥ ⋅ dl
Ei ,|| ⋅ dl ≡ R iδzI i
b
∫
E⊥ ⋅ dl =
c
∫
c
b
E⊥ ⋅ dl
Ei ,|| ⋅ dl ≡ R 0δzI 0 = − R 0δz
∑
n
k =1
Ik
whence ∫ ∂Si E⊥ ⋅ dl = −Vi ( z) + R i δzI i + Vi ( z + δz) + R 0 δz ∑ nk = 1 I k , where the time
variable is suppressed for notational convenience, and |γ →|= δz =|γ ←|.
Similarly, considering the surface area Si spanned by the loop ∂Si, the magnetic flux across Si is Ψ i = ∫ Si B⊥ ⋅ nd 2 x . Because B⊥ is the contribution from
each Ck, it thus follows by definition Ψ = LI that ψ i ≡ lim δ1z Ψ i = L i 1 I1 +  + L in I n
δz→ 0
defines the magnetic flux per unit length cutting the surface area Si. Thus,
∂t ψ i ≡ lim
δz→ 0
1
∂t Ψ i = lim δ1z
δz→ 0
δz
d
dt
∫
Si
B⊥ ⋅ nd 2 x = L i 1 ∂t I1 +  + L in ∂t I n
where we recall by definition that
L ij =
ψi
Ij
I k = 0 ∀k ≠ j
yielding
∂ z Vi = − R i I i + R 0 I 0 − L i 1 ∂t I1 −  − L in ∂t I n
= − R 0 I1 −  − (R 0 + R i )I i −  − R 0 I n − L i 1 ∂t I1 −  − L in ∂t I n
Likewise, the current variation along Ci is determined by inspecting Figure 5.3, and invoking Maxwell’s equation in integral form
∫ S ∇ × B ⋅ nd 2 x = µ ∫ S J ⋅ nd 2 x + µ ddt ∫ S E ⋅ nd 2 x, for any compact surface S.
Indeed, it is instructive to summarize the argument carried out in Section 5.1
again.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
First, consider some differential cylinder ci of length δz around ci,
let Ci− ( z), Ci+ ( z + δz) ⊂ Ci denote the end caps of the cylinder, and set
∂Ci = ∂Ci ∪ Ci− ( z) ∪ Ci+ ( z + δz). By construction, if ∂Ci denotes the boundary of
the cylinder Ci, then ∂(∂Ci ) ≡ ∂2 Ci ≡ ∅ (as the boundary of a boundary is the
empty set). Hence, by Stokes’ theorem, ∫ ∂Ci ∇ × B ⋅ nd 2 x = ∫ ∂2 Ci B ⋅ dl = 0. Thus,
0=µ
∫
∂ Ci
J ⋅ nd 2 x + µε dtd
∫
∂ Ci
E ⋅ nd 2 x
Now,
∫
∂ Ci
J ⋅ nd 2 x =
∫
∂ Ci
J ⋅ n⊥ d 2 x −
∫
Ci− ( z )
J ⋅ n||d 2 x +
∫
Ci+ ( z+δ z )
J ⋅ n|| d 2 x
where n⊥ ( n||) is a unit normal vector field on ∂Ci (Ci± ) . By definition,
∫ Ci− ( z ) J ⋅ n|| d 2 x = − I i ( z) and ∫ C+ ( z +δz ) J ⋅ n|| d 2 x = I i ( z + δz) . Furthermore, ∫ ∂Ci J ⋅
i
n⊥ d 2 x ≈ σ ∫ ∂Ci E⊥ ⋅ n⊥ d 2 x , where the quasi-TEM approximation is invoked:
E|| ≈ 0. And from V = IR, and noting that ∫ ∂Ci J ⋅ n⊥ d 2 x defines the total conduction current across ∂Ci between Ci, Cj along the segment [z, z + δz] of Ci,
for each j = 1, . . . , n,
σ
∫
∂ Ci
E⊥ ⋅ n⊥ d 2 x ≡
∑
j≠ i
Yij δz(Vi − Vj ) + Yii δz(Vi − 0)
where Vk is the potential of Ck with respect to C0, and Yij is the admittance per
unit length between Ck, Cj, and finally, Yii is the admittance per unit length
between Ci, C0.
Next, µε ddt ∫ ∂Ci E ⋅ nd 2 x = µε ddt ∫ ∂Ci E⊥ ⋅ n⊥ d 2 x. Thus, via Gauss’ theorem and
Q = CV,
ε
∫
∂ Ci
E⊥ ⋅ n⊥ d 2 x = Cii δzVi +
∑
j≠ i
Cij δz(Vi − Vj )
where Cij is the mutual capacitance per unit length between Ci, Cj, and Cij is
the capacitance between Ci, C0. whence, in the limit as δz → 0,
∂ z I i = −  − G i 1Vi −  +

∑
− (− Ci 1 ∂t V1 −  +
j=1
∑
and the theorem is thus established.
K15149_Book.indb 202
n
G ijVj −  − G inVn 

n
j=1
Cij ∂t Vj −  − Cin ∂t Vn )
□
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Cross-Talk in Multiconductor Transmission Lines
203
6.3.2 Corollary
Given the conditions of Theorem 6.3.1, the following wave equations are
satisfied:
∂2z V = RGV + (RC + LG) ∂t V + LC ∂t2 V
(6.23)
∂2z I = GRI + (CR + GL) ∂t I + CL ∂t2 I
(6.24)
In particular, LC = µεI = CL and LG = µσI = GL , where I is the identity
matrix, and hence, the waves propagate along Ci at v = 1µε .
Proof
The proof is similar to that in Chapter 4; the proof is left as a warm-up exercise (see Exercise 6.5.3).
□
6.3.3 Remark
From Corollary 6.3.2, it follows immediately that in a homogeneous medium,
L = µεC −1 and G = (σ/ε)C −1 ; see Chapter 4 for a transmission line pair.
Now, the phenomenon of cross-talk outlined in Chapter 5 can be extended
once again to multiconductor transmission lines. In particular, the analysis
follows that of Chapter 5 along a similar vein. To facilitate the discussion, the
following definition is made for notational convenience.
6.3.4 Definition
Given a system of n transmission lines {Ci}, define  ni = 1 Ci = ⊕ni = 1 Ci + ⊗ni = 1 Ci ,
where the first term represents the isolated coupling of the pair {Ci C0}—that
is, where the cross-talk between {Ci , C j } ∀j ≠ 0, i is absent—and the second
term represents the cross-talk between {Ci , C j } ∀j ≠ 0, i . Call ⊕ the direct sum
and ⊗ the direct product.
The generalization to n transmission lines is carried out briefly below.
First, from Corollary 6.3.2, the generalized wave equation can be easily
obtained. Then, referring to Exercise 5.4.4, the technique to diagonalize
the coefficient matrices can be employed to decouple the multiconductor
lines: γ 1 ∪  ∪ γ n → γ 1 ⊕  ⊕ γ n . First, observe that Corollary 6.3.2 can be
expressed as follows.
6.3.5 Corollary
Under the conditions of Theorem 6.3.1, where the propagating waves are
assumed to be time harmonic, there exists a coordinate transformation P
such that Equations (6.23) and (6.24) can be diagonalized. Explicitly, rewriting (6.21) and (6.22) as
∂2z V = UWV
K15149_Book.indb 203
and ∂2z I = WUI
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204
Electromagnetic Theory for Electromagnetic Compatibility Engineers
where U = R + iωL and W = G + iωC , defining V = PV and I = P t I lead to
the decoupled system of wave equations:
Vk = Vk+ e−
λ k (− z)
+ Vk− e
λ k (− z)
and
Ik = Ik+ e−
λ k (− z)
− Ik− e
λ k (− z)
for i = 1, . . . , n, where {λ 1 , , λ n } is the set of eigenvalues of UW.
Proof
From Proposition 6.3.1, inasmuch as it is not clear in general that R and G
commute, it follows that U and W might not commute. However, because L,
C, R, G are diagonalizable, it follows immediately that U and W are diagonalizable. So, let {P1 , P2 } be the respective coordinate transformations (i.e.,
similarity transformations) that diagonalize U and W:
P1−1UP1 = diag(ξ 1 , , ξ n ) and P2−1WP2 = diag(ζ1 , , ζn )
where ξ i , ζi ∀i = 1, , n are defined later.
Then clearly, (P1−1UP2 )(P2−1WP1 ) = (P2−1WP1 )(P1−1UP2 ) as diagonal matrices
commute. Furthermore, noting that
(P1−1 UP2 P2−1 WP1 )t = P1t W t (P2−1 )t P2t U t (P1−1 )t = (P2−1 WP1 )t (P1−1 UP2 )t ,
as L, C, R , G are symmetric imply that U, V are symmetric. Hence, the fact
that diagonal matrices commute implies at once that (P2−1 WP1 )t (P1−1 UP2 )t =
(P1−1 UP2 )t (P2−1 WP1 )t . That is,
P1t W(P2−1 )t P2t U(P1−1 )t = (P2−1WP1 )(P1−1UP2 ) ⇒ P1t = P2−1 .
.
Equivalently, P2 = (P1−1 )t ⇒ P2t = P1−1 . Hence, setting P = P1−1 ⇒ P2 = P t .
In the light of the above analysis, and noting that U, W are symmetric, it
follows at once that
P t UP = diag(ξ 1 , , ξ n ) and (P −1 )t WP −1 = diag(ζ1 , , ζn ),
where {( w1,i , ξ i )} denotes the eigenvectors and eigenvalues of U, and {( w2,i , ζi )}
the eigenvectors and eigenvalues of W . That is,
Uw1, i = ξ i w1, i
and Ww2, i = ζi w2, i ∀i = 1, , n.
Then, P = [ w1,1 , , w1,n ] and is the required matrix to diagonalise U, W
respectively, with w2,i related to w1,i via P −1 = [ w2,1 , , w2,n ] .
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205
Cross-Talk in Multiconductor Transmission Lines
For notational simplicity, given a diagonalisable matrix D, set D ∆ =
diag(λ 1 , , λ n ) to be the diagonalisation of D, where {λ i } are eigenvalues of
D. Then, on setting D = UW ,
∂2z V = UWV ⇔ ∂2z V = D ∆V ⇒ V = e−
V+ + e
Λ ( − z )
V−
Λ ( − z )
is the general solution of the decoupled system, for some V± . Explicitly,
 V
1

∂2z  V2
 
 Vn
  λ
1
 
0
 =
   
  0
 λ V
1 1

=  λ 2V2


 λ nVn
0
λ2
0
0
0
0
λn


 ⇒ V = V + e−
i
i



  V1
   V2
    Vn
λi (− z)






+ Vi− e
Similarly, the current propagation is: ∂2z I = WUI ⇒ I = e−
some I± . That is,
 I
1

∂2z  I 2
 
 In
  λ
1
 
0
 =
   
  0
 λ I
1 1

=  λ 2 I2


 λ n In
0
λ2
0
0
0
0
λn


 ⇒ I = I + e−
i
i



  I1
   I2
    In
λi (− z)
λi (− z)
Λ ( − z )
∀i.
I+ −e
Λ (− z)
I−, for






+ Ii− e
λi (− z)
∀i.
Whence, the desired solutions for the voltage and currentt propagations
are obtained via the linear transformation γ 1 ∪  ∪ γ n P→ γ 1 ⊕  ⊕ γ n :
V = P −1V and I = P t I , as required.
□
6.3.6 Proposition
The characteristic impedance matrix for the multiconductor transmission line
system is given by Z0 = W −1P t e Λ P .
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206
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Proof
From ∂ z I = − WV , via Corollary 6.3.5,
{
V = − W −1 P t Λ −e−
I − e
Λ (− z) +
I
Λ (− z) −
}=W
−1
{
P t Λ PP t e−
I + e
Λ (− z) +
I
Λ (− z) −
}
whence, motivated by the fact that if M were any symmetric matrix, then
Me ± Λ (  − z ) = e ± Λ (  − z )M , it follows clearly that
{
V = Z0 P t e−
I + e
I
Λ (− z) +
Λ (− z) −
}
where Z0 ≡ W −1 P t Λ P , as asserted. □
Note in passing that at the source and load, respectively,
V (0) = VS − ZS I (0) and V () = VL + Z()I () , via Kirchhoff’s voltage law,
where ZS = diag(ZS,1 , ,ZS,n ) is the source impedance matrix, and
Z() = diag(Z1 (),… ,Zn ()) is the load impedance matrix, Vs is the voltage
source at z = 0 and lastly, VL is the voltage source at z = ℓ. Then, the near-end
and far-end voltages can be determined by the following standard technique
(e.g., see Reference [6]).
First, consider the coupled system expressed in terms of V(0),I(0):
 V ( z)

 I ( z)
  M 11 ( z)
 =
  M 21 ( z)
M 12 ( z)   V (0)

M 22 ( z)   I (0)



(6.25)
for some yet to be determined matrix coefficients M ij ∀i, j = 1, 2 . Then, for z = ℓ,
Equation (6.25) implies at once that
VL + Z()I () = M 11 ()(VS − Z S I (0)) + M 12 ()I (0)
(6.26)
I () = M 21 ()(VS − ZS I (0)) + M 22 ()I (0)
(6.27)
6.3.7 Lemma
Given (6.25), at z = ℓ, the matrix coefficients are defined by
M 11 () = Z0 P t cosh
(
)
Λ  (P t )−1 Z0−1
M 12 () = Z0 P t sinh
M 21 () = P t sinh
(
M 22 () = P t cosh
K15149_Book.indb 206
(
)
Λ  (P t )−1
)
Λ  (P t )−1 Z−01
(
)
Λ  (P t )−1
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207
Cross-Talk in Multiconductor Transmission Lines
Proof
From the proof of Proposition 6.3.6,
{
I + e
V (0) = Z0 P t e−
Λ +
I
Λ −
}
and
{
I (0) = P t e−
I − e
Λ +
I
Λ −
}
Hence, rearranging the two equations to express I ± in terms of V(0), I(0) yields
I ± = 21 e ±
Λ
{
}
(P t )−1 Z−01V (0) ± I (0)
Then, substituting into Equation (6.25) leads to
V () = Z0 P t
{ (e
Λ
1
2
t
= Z0 P cosh
(
+ e−
Λ
)
)(P )
t −1
t −1
Z0−1V (0) +
−1
0
1
2
(e
t
Λ  (P ) Z V (0) + Z0 P sinh
Λ
(
− e−
Λ
)
)(P )
t −1
}
I (0)
t −1
Λ  (P ) I (0),
and
I () = P t
{ (e
Λ
1
2
= P t sinh
(
− e−
)
Λ
)(P )
t −1
Z−01V (0) +
1
2
(e
Λ  (P t )−1 Z−01V (0) + P t cosh
(
Λ
+ e−
Λ
)
)(P )
t −1
}
I (0)
Λ  (P t )−1 I (0)
So, comparing the two expressions with (6.25) and equating coefficients,
M 11 () = Z0 P t cosh
(
)
Λ  (P t )−1 Z0−1
M 12 () = Z0 P t sinh
M 21 () = P t sinh
(
M 22 () = P t cosh
(
)
Λ  (P t )−1
)
Λ  (P t )−1 Z−01
(
)
Λ  (P t )−1
□
6.3.8 Proposition
Given (VS , VL ) , the near-end and far-end currents and voltages of {Ci} satisfy
I (0) = N()−1 {VL + ( Z()M 21 () − M 11 ()) VS }
(
)
I () = M 21 () I + ZS N()−1 M 11 () VS + M 22 ()N()−1 VL
{
}
V (0) = I − ZS N()−1 ( Z()M 21 () − M 11 ()) VS − ZS N()−1 VL
{
}
V () = M 11 () + ( M 12 () − M 11 ()ZS ) N()−1 ( Z()M 21 () − M 11 ()) VS
+ ( M 12 () − M 11 ()ZS ) N() VL ,
−1
where N() = M 12 () − M 11 ()ZS − Z() ( M 22 () − M 21 ()ZS ) .
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208
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Proof
Substituting Equation (6.25) into (6.26),
VL + Z()I () = M 11 ()VS − M 11 ()ZS I (0) + M 12 ()I (0)
= M 11 ()VS + {M12 () − M11 ()ZS } I (0)
and from (6.25),
I () = M 21 ()VS + {M 22 () − M 21 ()ZS } I (0)
whence,
VL + Z()M 21 ()VS + {ZS M 22 () − M 21 ()ZS } I (0) + M 12 ()I (0)
= M 11 ()VS + {M 12 () − M 11 ()ZS } I (0)
⇒
I (0) = N()−1 {VL + ( Z()M 21 () − M 11 ()) VS }
where N() = M 12 () − M 11 ()ZS − Z() {M 22 () − M 21 ()ZS } . Substituting I(0)
into the above expression for I(ℓ) yields the result stated in the proposition.
Finally, substituting I(0) into V (0) = VS − ZS I (0) and V () = M 11VS + {M 12 () −
M 11 ()ZS } I (0) concludes the proof.
□
In many cases, VL ≡ 0, and Proposition 6.3.6 thus yields the near-end and
far-end cross-talk for the multiconductor transmission line {Ci}.
6.3.9 Corollary
Given a system of multiconductor transmission lines {Ci} depicted in Figure 6.8,
the induced near-end cross-talk at z = 0 and far-end cross-talk at z = ℓ on the
system are given by:
I (0) = N()−1 {Z()M 21 () − M 11 ()} VS
{
}
I () = M 21 () I + ZS N()−1 M 11 () VS
{
}
V (0) = I − ZS N()−1 ( Z()M 21 () − M 11 ()) VS
{
}
V () = M 11 () + ( M 12 () − M 11 ()ZS ) N()−1 ( Z()M 21 () − M 11 ()) VS
□
6.3.10 Lemma
Given a matrix A = ( Aij ) , define ||A||= max|Aij|. Suppose that {Ci} is embedded
i, j
in a homogeneous medium such that the transmission lines satisfy ℓ <<
λ, for some time harmonic voltage sources, where λ = min λ k , and λk is
k
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209
Cross-Talk in Multiconductor Transmission Lines
(C1, l)
ZS,1
VS,1
R1
RL,1
RL,2
(C2, l)
ZS,2
VS,2
R2
RL,n
(Cn, l)
ZS,n
VS,n
Rn
(C0, l)
R0
Figure 6.8
Cross-talk induced on a system of multiconductor transmission lines.
the wavelength of the wave propagation along Ck, with  =|Ck|∀k. If
||Z−01 ZS − Z−1 ()(ZS − Z0 )||≤ 1 and {Ci} is lossless, then to first order in λ ,
N−1 () ≈ −(ZS + Z())−1 {I + iβ(ZS + Z())−1 (Z0 + Z()Z0−1 ZS )}.
Proof
□
The proof is left as an exercise; see Exercise 6.5.4.
6.3.11 Proposition
Suppose {Ci} is embedded in a homogeneous medium such that the transmission lines satisfy ℓ << λ, for some time harmonic voltage sources, where
λ = min λ k , and λk is the wavelength of the wave propagation along Ck, with
k
 = Ck ∀k . Suppose that ||Z−01 ZS − Z−1 ()(ZS − Z0 )||≤ 1 and {Ci} is lossless.
Then, for VS ≠ 0, VL = 0 , the cross-talk noise can be approximated to first
order in λ by
{
(
(
)}
)
I (0) ≈ (ZS + Z())−1 I + iβ (ZS + Z())−1 Z0 + Z()Z0−1 ZS − Z()Z−01 V S
{
}
I () ≈ iβZ−01 I − ZS (ZS + Z())−1 VS
{
(
(
V (0) ≈ I − ZS (ZS + Z())−1 I + iβ Ξ() − Z()Z0−1
{
(
))}V
S
)}
V () ≈ I − ZS (ZS + Z())−1 + iβ Z0 (ZS + Z())−1 − ZS (ZS + Z())−1 (Ξ − Z()Z0−1 ) VS
where Ξ() = iβ(ZS + Z())−1 (Z0 + Z()Z0−1 ZS ) .
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210
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Proof
This is an immediate consequence of Corollary 6.3.7 and Lemma 6.3.8. See
Exercise 6.5.5.
□
6.4 S-Parameters: Scattering Parameters
No exposition on electromagnetic theory for EMC engineers is complete
without incorporating a cursory review on S-parameter theory. By way of
introduction, “S” is short for scattering, and the origin of the term is best
illustrated in Figure 6.9. The motivation is to determine the i-load impedance
without knowing the makeup of the black box by observing the scattering
(i.e., reflected) waves traveling back toward the source. Note that this applies
to a linear network system under various steady-state conditions. The special
case is first formulated, and then followed by the general theory. An example
illustrating a 2-port network is also presented in some depth.
In formulating the special case, the characteristic impedance across each
port is assumed to be the same; that is, the characteristic impedance of the
i-port is identical with that of the j-port ∀i,j. In the general case, this stipulation is dropped. So, referring to Figure 6.9, let Tk denote the k-terminal at
along the k-transmission pair Ck± such that the voltage and current phases
are 0, for simplicity. That is, along the pair {Ck+ , Ck− } terminating at the kth
port of the black box, let Tk denote the terminal along {Ck+ , Ck− }, near the port,
such that the phase is zero. This can be easily accomplished via a coordinate
transformation.
(V2–, I2–)
(V2+, I2+)
T2
Ti
(V1+, I1+)
(V1–, I1–)
T1
Black box
(Vi+, Ii+)
(Vi–, Ii–)
Tk = terminal at 0 phase
Vk+ = forward propagating wave
Ik+ = backward propagating wave
Vk– = forward propagating wave
n-port network Nn
Ik– = backward propagating wave
Tn
(Vn+, In+) (Vn–, In–)
Figure 6.9
Propagating waves along n-pairs of transmission lines entering some network.
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211
Cross-Talk in Multiconductor Transmission Lines
6.4.1 Lemma
Given X = QY , where X = ( a1 X 1 , , an X n ) , Y = (b1Y1 , , bnYn ) , and Q = (Qij ),
for some constants ai , bi , i = 1, , n , there exists a transformation ( X , Y ) ( X , Y )
such that X = (X 1 ,… , X n ) and Y = (Y1 ,… , Yn ) .
Proof
The proof is trivial. First, observe that
 a1 X 1


 an X n

 X1
 = diag( a ,… , a ) 
1
n

 
 X n
 b1Y1


 bnYn

 Y1
 = diag(b ,… , b ) 
n
1

  Yn





and




Hence, on setting A = diag( a1 , , an ) and B = diag(b1 , , bn ) , it follows at once
 = A −1QB.
 Y , where Q
that X = Q
□
To continue with the above exposition, suppose in general that the n-port
network is connected to (Ck± , Z0, k ,  i , VS, k , ZS, k ), where  k is the length of
the transmission lines Ck± , Z0, k ↔ γ k is the characteristic impedance of
(Ck+ , Ck− ), and (VS, k , ZS, k ) is the source at zk = 0, where γk is the wave propagation constant. By assumption, Z0, i ≡ Z0 , for some Z0 ≠ 0 . Note further
that by construction, T k is located at zk =  k and that Lemma 6.4.1 is indeed
equivalent to translating the origin along Ck± to T k: zk zk − k . Thus,
recalling that
Vk ( z) = Vk+ e− γ (  − z ) + Vk− e γ (  − z ) ⇒ Vk (0) = Vk+ e− γ k  k + Vk− e γ k  k
I k ( z) = I k+ e− γ (  − z ) − I k− e γ (  − z ) ⇒ I k (0) = I k+ e− γ k  k − I k− e γ k  k
by Lemma 6.4.1, (Vk (0), I k (0)) can be transformed into
(Vk (0), I k (0)) (Vk (0), Ik (0)) = (Vk+ + Vk− , I k+ + I k− )
That is, via Lemma 6.4.1, Tk can be transformed into a zero-phase terminal
for the n-port network. In particular, without loss of generality, we define Tk
to be the zero-phase terminal in all that follows.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Then, for the n-port network illustrated in Figure 6.9, the S-parameter formalism of the network is defined by expressing the reflected waves as a
function of the incident waves:
 V−
1


 Vn−

  S
11
 
=
 
  Sn1

S1n
Snn
  V1+

   Vn+






(6.28)
where S = (Sij ) is the S-matrix (or scattering matrix) defining the n-port network.
Because the S-matrix is a constant, it follows at once by construction that
Sij =
Vi−
Vj+
Vk+ = 0 ∀k ≠ j
(6.29)
Verify this assertion in Exercise 6.5.6. Some comments are due. First, Vk+ = 0
is obtained by (i) setting the source at zk = −  k along Ck± to be zero, and (ii)
matching the source impedance to the characteristic impedance of Ck± . Note
that (i) is clearly necessary, and (ii) is also necessary to ensure that any waves
traveling from the k-port toward the source of Ck± are completely absorbed
by the source (technically, “load”, in this instance). These two conditions are
+
clearly sufficient to ensure that Vk = 0.
In short, each Ck± , for k ≠ j, is connected to a matched load (with the source
removed). Second, Sij for i ≠ j, can be intuitively viewed as the transmission
coefficients, and Sij can be viewed as the reflection coefficients; these two
observations are true when all other ports are matched with their characteristic impedance. For instance, when some of the ports are not matched,
then Sij is not necessarily equal to the reflection coefficient Γi at Ti, that is,
waves traveling out of the i-port will be the superposition of reflected waves
from the nonmatched loads terminating the k-port Similar comments hold
for Sij, i ≠ j, by interchanging reflection with transmission.
6.4.2 Definition
Given a linear time-invariant network defined by V = ZI , the network is said
to be reciprocal if Z is symmetric: Zij = Z ji ∀i, j . And the network is said to be
lossless if ℜe(Zij ) = 0 ∀i, j .
Clearly, as the inverse of a symmetric matrix is symmetric, it follows at
once that Z = Zt ⇒ Y = Y t, where Y = Z−1.
6.4.3 Proposition
The S-matrix S is symmetric; that is, Sij = Sji ∀i, j , whenever the n-port
network is reciprocal.
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213
Cross-Talk in Multiconductor Transmission Lines
Proof
By definition, V ± = 21 (V ± Z0 I ). This follows from the fact that V = V + + V −
and I = I + − I − , where I ± = Z−01V ±, whence, from V = ZI ,
V ± = 21 (Z ± Z0 )I ⇒ (Z − Z0 )I = 21 S(Z + Z0 )I
That is, S = (Z − Z0 )(Z + Z0 )−1, and thus, Z, Z0 are symmetric imply that S is
also symmetric, as asserted.
□
In light of the above proof, the following lemma is evident. For notational
convenience, let N n denote the n-port network illustrated in Figure 6.9.
6.4.4 Lemma
Let S be the S-matrix associated with N n. Then, Skk = 0 ∀k if each k-port of
N n is matched.
Proof
By definition, V − = SV +. Hence, by assumption, Vk− = Skk Vk+ = 0 ⇒ Skk ≡ 0 ∀k. □
6.4.5 Theorem
The network N n is lossless if and only if S† S = I; that is, S is unitary.
Proof
First, recall that the time-average power is defined by
{
}
Pk± = 21 ℜe (Vk± )∗ I k± =
V±
1 k
2 Z0
2
where Z0 is the characteristic impedance of (Ck+ , Ck− ) connected to the k-port of
N n . Here, ⟨ Pk+ ⟩(⟨ Pk− ⟩) denote the incident (reflected) power at the k-port. That
is, ⟨ Pk+ ⟩ represents the time-average power entering N n whereas ⟨ Pk− ⟩ represents the power exiting N n . Thus, N n is lossless if and only if ⟨ P + ⟩ = ⟨ P − ⟩ ,
where ⟨ P ± ⟩ = ∑ k ⟨ Pk± ⟩ . Physically, this means that the power absorbed by N n
must be zero (i.e., no loss!).
Now, the resultant power absorbed by N n is defined by ⟨δP⟩ = ⟨ P + ⟩ − ⟨ P − ⟩.
On setting Z0 = diag(Z0 , , Z0 ) and noting that
P+ =
P− =
K15149_Book.indb 213
∑
k
∑
k
Pk+ = 21 Z−01 (V + )† V +
Pk− = 21 Z−01 (SV + )† SV + = 21 Z0−1 (V + )† S† SV +
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214
Electromagnetic Theory for Electromagnetic Compatibility Engineers
2-port network
Z1
VS,1
ZS,1
Z2
T1
Z3
T2
VS,2
ZS,2
Figure 6.10
Determination of the S-matrix for a T-network.
it follows that lossless condition is equivalent to
{
0 = δP = 21 diag −1 (Z0 , , Z0 ) (V + )† V + − (V + )† S† SV +
{
}
}
= 21 diag −1 (Z0 , , Z0 )(V + )† I − S† S V +
and hence, V + is arbitrary implies at once that I − S† S ≡ 0 , as required.
□
6.4.6 Example
Consider the 2-port network depicted in Figure 6.10. Determine the
S-parameters for the T-filter given passive elements {Z1 , Z2 , Z3 } given that
T1, T2 are matched. Note that in Figure 6.10, only the source impedance is
shown at T1; the source is left out for simplicity. At T2, only a load is present.
Matching ZL at T2 to the T-filter can be accomplished as follows. Set
(1)
Zin
= Z1 +
{
1
Z3
+
1
Z2 + ZL
}
−1
This is the impedance looking into T1. Then, impedance matching port 2 at
T2 yields:
(1)
Zin
= Z1 +
ZL =
1
2
{
1
Z3
+
1
Z2 + ZL
{Z − Z +
1
2
}
−1
= ZL ⇒
(Z1 − Z2 )2 + 4(Z1 Z2 + Z1 Z3 + Z2 Z3 )
}
where the positive solution was chosen for physical reasons. Thus, terminating T2 with ZL leads to
S11 =
K15149_Book.indb 214
V1−
V1+
V2+ = 0
= Γ1 =
(1)
Zin
− ZL
(1)
Zin
+ ZL
=0
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215
Cross-Talk in Multiconductor Transmission Lines
{
}
For S22, interchanging Z1 ↔ Z2 ⇒ ZS = 21 Z2 − Z1 + (Z2 − Z1 )2 + 4(Z1Z2 + Z2 Z3 + Z1Z3 ) ,
(2)
and hence, by assumption, Zin
= ZS,1 yields
S22 =
V2−
V2+
V1+ =
0
= Γ2 =
(2)
Zin
− ZS
=0
(2)
Zin
+ ZS
Finally, regarding S21, observing that T2 is matched implies that V1− ≡ 0 ,
that is, no reflection, and similarly, V2+ ≡ 0 as there can be no reflection from
a matched load. Hence, V1 = V1+ + V1− ⇒ V1+ = V1 and VL = V2+ + V2− ⇒ V2− = VL ,
where V1 = VS Z +ZZS and Z is the resultant impedance defined by
Z = ZS + Z1 +
{
1
Z3
+
1
Z2 + ZL
}
−1
.
Let I denote the current conducting through the source Zs. Then, by
construction,
I = VZS = ZV− 1ZS
So, on setting I = I ′ + I ′′ , where I′ is the current through ZL and I″ the current through Z3, it follows that
I ′ = I Z3 + ZZ23 + ZL
and hence,
VL = ZL I ′ =
Z3
ZL
Z − ZS Z2 + Z3 + ZL
V1
Thus,
S21 =
V2−
V1+
V2+ = 0
=
Z3
ZL
Z − ZS Z2 + Z3 + ZL
= S12
□
by appealing to the symmetry of S.
The general S-parameter theory is considered below. First, recall the
essence of the S-matrix formulation. It is based on measuring the incident
power and reflected power occurring at the ports of a linear network. In the
+
−
present formulation, each transmission line pair (C k , C k ) has characteristic
impedance Z0, k , not necessarily identical for each k. Second, recall that the
time-average power transmitted is defined by P = 21 ℜe(V ⋅ I ∗ ). Hence, along
(C k+ , C k− ), the time-average power is ⟨ Pk± ⟩ =
ak± =
Vk±
Z0, k
∝
2
V±
1 k
2 Z0, k
, motivating the definition:
Pk±
+
−
In what follows, assume for simplicity that (C k , C k ) is lossless and hence,
±
± ± iβ k  k
Z0, k ∈R and Vk ( z) = V0, k e
∀k .
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216
Electromagnetic Theory for Electromagnetic Compatibility Engineers
6.4.7 Remark
Observe that by definition, ak± = 2 Z10, k (Vk ± Z0, k I k ) . To see this, it suffices to recall
that Vk ( z) = Vk+ ( z) + Vk− ( z) and I k ( z) = I k+ ( z) − I k− ( z) , where Vk± = Z0, k I k± . From this,
it follows that Z0, k I k ( z) = Vk+ ( z) − Vk− ( z) ⇒ ak+ ( z) + ak− ( z) = Z10, k (Vk+ + Vk− ) = Z10, k Vk
and ak+ ( z) − ak− ( z) = Z10, k (Vk+ − Vk− ) = Z0, k I k ( z) whence, adding and subtracting the expressions for ak± yields ak± = 2 Z10, k (Vk ± Z0, k I k ), as claimed.
6.4.8 Lemma
+
−
The pair ( ak+ , ak− ) defines the transmitted and reflected power along (C k , C k )
±
± 2
1
by ⟨ Pk ⟩ = 2 |ak| .
Proof
By definition,
Pk± =
2
V±
1 k
2 Z0, k
=
Vk±
1
2
Z0, k
( )=
∗
Vk±
Z0, k
1
2
( )
ak± ak±
∗
=
1
2
ak±
2
□
as required.
6.4.9 Corollary
The time-average power across Tk satisfies ⟨ Pk ⟩ = ⟨ Pk+ ⟩ − ⟨ Pk− ⟩ .
Proof
By definition, ⟨ Pk ⟩ = 21 ℜe(Vk I k∗ ) = 21 ℜe
{(
{
Vk
Z0, k
}
Z0, k I k∗ . From Remark 6.4.6,
)(
Pk = 21 ℜe ak+ + ak− ak+∗ − ak−∗
{
2
)}
}
2
= 21 ℜe ak+ − ak− + ak− ak+∗ − ak+ ak−∗ .
+
−
By assumption, (C k , C k ) is lossless implies that ak± = a0,± k e ± iβk zk , where a0,± k ∈R .
Hence,
(
)
ak− ak+∗ − ak+ ak−∗ = a0,+ k a0,− k e− i2βk zz − ei2βk zz = − i2a0,+ k a0,− k sin 2β k zz
which is purely imaginary. Thus,
Pk =
by Lemma 6.4.8.
K15149_Book.indb 216
1
2
{a
+ 2
k
− ak−
2
}= P
+
k
− Pk−
□
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217
Cross-Talk in Multiconductor Transmission Lines
6.4.10 Definition
Given a linear network {N n , Ck± , Z0, k }, where the transmission lines are lossless, the S-parameters are defined by
 a−
1


 an−

  S
11
 
=
 
  Sn1

S1n
Snn
  a1+

   an+






(6.30)
where
Sij =
ai−
a +j
ak+ = 0 ∀k ≠ j
with each k-port matched for all k ≠ j.
6.4.11 Remark
Provisionally, set a− = Ŝa+ and V − = SV + . Then, by definition,
Sˆ ij =
ai−
a +j
=
Vi−
Z0, i
Z0, j
Vj+
=
Vi−
Vj+
Z0, j
Z0, i
= Sij
Z0, j
Z0, i
Hence, Z0, k = Z0 ∀k ⇒ Sˆ ij = Sij ∀i, j. Thus, this justifies denoting the S-matrix
in Equation (6.30) by S instead of Ŝ. In particular, it is clear that S is symmetric does not imply that Ŝ is also symmetric. This is because unless Z0, i = Z0, j ,
Sij = Sji ⇒
/ Sˆ ij = Sˆ ji , in general. This immediately yields the following theorem.
6.4.12 Theorem
Suppose the network {N n , C k± , Z0, k } is reciprocal. Then, S is symmetric if and
only if Z0, k = Z0 ∀k, for some fixed Z0.
□
On the other hand, by inspecting the proof of Theorem 6.4.4, replacing
diag(Z0 , , Z0 ) with diag(Z0,1 , , Z0, n ) yields the same result. This is summarized below for convenience.
6.4.13 Corollary
The network {N n , C k± , Z0, k } is lossless if and only if S is unitary.
□
This section concludes by showing how Z and S are related. Note in passing that the Z-matrix is determined by either opening or shorting out the
circuit in the case of the Y-matrix. In contrast, the S-matrix is determined by
impedance matching at the ports. From this, it is clear that the S-matrix has
a greater utility over the Y-matrix as shorting out terminals has the potential
to cause irreversible damage to circuit components.
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10/18/13 10:55 AM
218
Electromagnetic Theory for Electromagnetic Compatibility Engineers
6.5.14. Theorem.
Given a network {N n , C k± , Z0, k }, Z and S are related by
{
}
Z = I − Z0 SZ−01 Z0 {I − S}
where Z0 = diag(Z0 ,1 , , Z0 ,n ).
Proof
By definition, a ± = 21 Z−01 {V ± Z0 I }. Thus, from V = ZI and a− = Sa+ ,
{
}
Z−01 {V − Z0 I } = SZ−01 {V + Z0 I } ⇒ I − Z0 SZ0−1 V = Z0 {I + S} I
and hence, V = {I − Z0 SZ−01 }−1 Z0 {I + S} I , as desired.
□
6.5 Worked Problems
6.5.1 Exercise
Establish Equation (6.1) from Theorem 6.1.1.
Solution
Let C =  k ( Ck ∪ Ck′ ). Then, setting Ω = R 3 – C, and noting trivially that
ϕ i |Ci = Vi is a constant for each i, it follows at once that ∑ i ∫ ∂Ci ρi ϕ ′i d 2 x =
∑ i Vi′ ∫ ∂Ci ρi d 2 x = Vi′ Qi , where Qi = ∫ ∂Ci ρi (as charges within a conductor will quickly diffuse to the boundary of the conductor). Likewise,
∑ i ∫ ∂Ci ρ′i ϕ i d 2 x = ∑ i ViQi′ , and Theorem 6.1.1 thus yields the desired result. □
6.5.2 Exercise
For a system of thin conductors {(Ci , Vi , I i )} and {(Ci′, Vi′, I i′)} for i = 1,…, n,
establish that ∑ ni = 1 Vi I i′ = ∑ ni = 1 Vi′I i.
Solution
3
From Theorem 6.1.3, ∫ Ω E ⋅ J ′d 2 x = ∫ Ω E ′ ⋅ Jd 2 x, where Ω = R −  k (C k ∪ C k′ ) .
Under the assumption of thin conductor, restricting to the boundary of the
conductors leads to
∫
Ω
K15149_Book.indb 218
E ⋅ J ′d 3 x →
∑∫
k
∂( Ck ∪ Ck′ )
Ek ⋅ J k′ d 2 x = −
∑∫
k
∂( Ck ∪ Ck′ )
∇ϕ k ⋅ J k′ d 2 x
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219
Cross-Talk in Multiconductor Transmission Lines
where the current density becomes surface current density. Furthermore,
motivated by approximating a thin conductor as a line conductor, by appealing
to integration by parts, ∫ ∂ξ ϕ k J k , ξ dx = ϕ k J k , ξ − ∫ ϕ k ∂ξ J k , ξ dx , it follows clearly that
∑∫
k
∂( Ck ∪ Ck′ )
∇ϕ k ⋅ J k′ d 2 x =

∑  ∫ ϕ J ′ ⋅ n′ d x − ∫
k
k k
k
2
∂( Ck ∪ Ck′ )

ϕ k ∇ ⋅ J ′d 2 x ,

where nk′ is the unit normal to the cross-section of the conductor C k′ .
However, under electrostatic conditions, ∇ ⋅ J ′ = 0, hence, ϕ k |∂C k = Vk is a
constant, implies that ∫ ∂(C k ∪C k′ ) ∇ϕ k ⋅ J k′ dx = ∑ k ϕ k I k′ , where ∫ ∂C k J k′ ⋅ nk′ d 2 x = I k′ .
Hence, by symmetry, the equality ∑ k Vk I k′ = ∑ k Vk′I k follows.
□
6.5.3 Exercise
Establish Corollary 6.3.2.
Solution
From Equation (6.21), substituting (6.22) into ∂t (6.21) :
∂2z V = −R ∂ z I − L ∂t ∂ z I
= −R{− YV − C ∂t V } − L ∂t {− YV − C ∂t V }
= RYV + RC ∂t V + LY ∂t V + LC ∂t2 V
= RYV + (RC + LY ) ∂t V + LC ∂t2 V
Similarly, substituting Equation (6.21) into ∂t (6.22) yields (6.24). To establish the second part of the corollary, compare Equation (4.10) with (6.23),
and note that under perfect TEM conditions, (6.23) reduces to ∂2z V = RYV +
LY ∂t V + LC ∂t2 V , and hence, it is clear that LY = µσI and LC = µεI.
Likewise, under perfect TEM conditions, Equation (6.24) reduces to
∂2z I = YRI + YL ∂t I + CL ∂t2 I and hence, comparing with (4.11) yields:
YL = µσI and CL = µεI . Whence, both Y and C commute with L by inspection. Hence, from the definition of the wave equation, the waves propagate
1
along the conductors at speed µε .
□
6.5.4 Exercise
Prove Lemma 6.3.8.
Solution
From N() = M 12 () − M 11 ()ZS − Z(){M 22 () − M 21 ()ZS }, under the assumption that ℓ << λ, it follows from Lemma 6.3.7, and noting via Taylor expansion that
cosh(ia) = cos a ≈ 1 −
K15149_Book.indb 219
a2
2
{
+ o( a 4 ) and sinh(ia) = i sin a ≈ i a −
a3
3!
}
+ o( a 5 ) ,
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220
Electromagnetic Theory for Electromagnetic Compatibility Engineers
it follows that
{
}
M () = iZ P {βI −
I + } (P )
M () = iP {βI −
I + } (P ) Z
M () = P {I +
I + } (P )
2
M 11 () = Z0 P t I + (β2) I + (P t )−1 Z−01 ≈ I
12
0
(β )3
3!
t
(β )3
3!
t
21
t −1
(β )2
2
t
22
t −1
−1
0
t −1
≈ iβZ0
≈ iβZ0−1
≈I
Substituting these values into N() yields
(
N() ≈ −(ZS + Z()) + iβ Z0 + Z()Z−01 ZS
{
(
)
}
)
= − I − iβ Z0 + Z()Z−01 ZS (ZS + Z())−1 (ZS + Z()).
Furthermore, noting that ||Z−01 ZS − Z−1 ()(ZS − Z0 )||≤ 1 implies, via the triangle inequality, that the following holds: 1+||Z−1 ()(ZS − Z0 )||≥||Z0−1 ZS ||⇒
1+||Z−1 ()(ZS − Z0 )||−||Z0−1 ZS ||≥ 0, hence, βℓ << 1 implies that
{ (
{I + iβ ( Z
)
N−1 () ≈ −(ZS + Z())−1 I − iβ Z0 + Z()Z0−1 ZS (ZS + Z())−1
≈ −(ZS + Z())−1
0
)
+ Z()Z−01ZS (ZS + Z())−1
}
},
−1
□
as asserted.
6.5.5 Exercise
Establish Proposition 6.3.9.
Solution
By the assumption, Lemma 6.3.8 applies. Whence, from Corollary 6.3.7,
{
}
I (0) ≈ N−1 () iβZ()Z−01 − I VS
(
)
≈ (ZS + Z(0))−1 ( I + iβΦ()) I − iβZ()Z−01 VS
{
(
)}
≈ (ZS + Z(0))−1 I + iβ Φ() − Z()Z0−1 VS ,
K15149_Book.indb 220
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221
Cross-Talk in Multiconductor Transmission Lines
where Φ() = (Z0 + Z()Z0−1 ZS )(ZS + Z())−1. Next,
{
{I − Z ( Z
I () ≈ iβZ−01 I − ZS ( ZS + Z())
≈ iβZ−01
S
S
−1
+ Z())
−1
(I + iβΦ())} VS
}V .
S
Similarly,
{
−1
(I + iβΦ()) (I − iβZ()Z−01 )} VS
{
−1
(I + iβ ( Φ() − Z()Z ))}V ,
V (0) ≈ I − ZS ( ZS + Z )
≈ I − ZS ( ZS + Z )
−1
0
S
and
{
)}
(
V () ≈ I + (iβZ0 − ZS ) ( ZS + Z()) (I + iβΦ()) I − iβZ()Z0−1 VS
−1
{
(
{
( Φ() − Z()Z )V
}
(
≈ I + (iβZ0 − ZS )(ZS + Z)−1 I + iβ Φ() − Z()Z0−1
))}V
S
≈ I + (iβZ0 − ZS )(ZS + Z)−1 VS + iβ(iβZ0 − ZS )(ZS + Z)−1 ×
−1
0
{
S
(
≈ I − ZS (ZS + Z)−1 + iβ Z0 ( ZS + Z ) − ZS ( ZS + Z )
as required.
−1
−1
( Φ() − Z()Z ))}V ,
−1
0
S
□
6.5.6 Exercise
Verify Equation (6.29).
Solution
By (6.28), Vk− = Sk 1V1+ +  + SkjVj+ +  + SknVn+ . Inasmuch as Sij are constants,
−
it follows clearly that on setting Vi+ = 0 ∀i ≠ j , Vk− = SkjVj+ ⇒ Skj = VVk+ , as
j
asserted.
□
References
1. Balanis, C. 1982. Antenna Theory: Analysis and Design. New York: John Wiley & Sons.
2. Carson, J. 1924. A generalization of the reciprocal theorem. Bell Sys. Tech. J. 3(3)
July: 393–399.
K15149_Book.indb 221
10/18/13 10:55 AM
222
Electromagnetic Theory for Electromagnetic Compatibility Engineers
3. Carson, J. 1930. The reciprocal energy theorem. Bell Sys. Tech. J. 9(2) April:
325–331.
4. Chang, D. 1992. Field and Wave Electromagnetics. Reading, MA: Addison-Wesley.
5. Orfanidis, O. 2002. Electromagnetic Waves and Antenna. Rutgers University, ECE
Dept., http://www.ece.rutgers.edu/~orfanidi/ewa/.
6. Paul, C. 1994. Analysis of Multiconductor Transmission Lines. New York, John
Wiley & Sons.
7. Plonsey, R. and Collin, R. 1961. Principles and Applications of Electromagnetic
Fields. New York: McGraw-Hill.
8. Pozar, D. 2005. Microwave Engineering. New York: John Wiley & Sons.
9. Rothwell, E. and Cloud, M. 2001. Electromagnetics. Boca Raton, FL: CRC Press.
10. Smythe, W. 1950. Static and Dynamic Electricity. New York: McGraw-Hill.
K15149_Book.indb 222
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7
Waveguides and Cavity Resonance
It was established in Chapter 1 via Theorem 1.4.1 that TEM waves cannot be sustained on a single conductor. In contrast, it is established below
that TE and TM waves can propagate within a single hollow conductor.
Applications of these principles can be found in radars and microwave
ovens, to name a few.
By way of introduction, waves propagating between two disjoint conductors are investigated, followed by attaching the conductors together to form
a single hollow conducting structure. Finally, the hollow conductor is closed
at both ends to form an enclosed cavity. Resonance within a cavity forms as a
result of standing waves sustained within the cavity. Readers who are interested in pursuing the topic of dielectric waveguides (which unfortunately
is not covered here) or more advanced theory and applications may consult
References [1,4,6,8]. The elementary theory can be found in most references
on electromagnetic theory; for example, see References [2,5,7].
7.1 Parallel Plate Guides
By way of motivation, consider waves propagating between two large parallel plates. A pair of transmission lines is a special case of two distinct conductors. It is thus clear that TEM can be sustained between two parallel plates.
However, can two parallel plates sustain TE and TM modes? More precisely,
consider two infinite parallel planes separated by a distance x = a, and for
simplicity, suppose that an incident electromagnetic wave is propagating in
the z-direction. Can the parallel planes support TE and TM modes?
Initially, the transverse electric field propagation is considered. The transverse magnetic field propagation will conclude this introduction. Without
loss of generality, in rectangular coordinates, suppose the direction of pro­
pagation is ez. Then, by definition, Ez = 0 for TE propagation.
Explicitly, consider the space (Ω, μ, ε), where
Ω = {( x , y , z) ∈R 3 : −∞ < x , z < ∞, 0 < y < a}
223
K15149_Book.indb 223
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224
Electromagnetic Theory for Electromagnetic Compatibility Engineers
and suppose that the current density J = 0 on Ω, the charge density ρ = 0 on
Ω, and ∂Ω is a perfect electrical conductor. So, consider Maxwell’s first equation (1.15). Solving for it yields:
∇×E=
ex
ey
ez
∂x
∂y
∂z
Ex
Ey
0

− ∂ z Ey

=
∂ z Ex
 ∂ E −∂ E
y x
 x y

 ∂t Bx


 = −  ∂t By

 ∂B
 t z






Clearly, for TE mode, Bz is not constrained to be zero on Ω. Thus,
∂ z Ey = ∂t Bx
(7.1)
∂ z Ex = − ∂t By
(7.2)
∂ x Ey − ∂ y Ex = − ∂t Bz
(7.3)
Next, solving for Equation (1.17),
∇×B=
ex
ey
ez
∂x
∂y
∂z
Bx
By
Bz
 ∂ y Bz − ∂ z By

=  − ∂ x Bz + ∂ z Bx
 ∂ B −∂ B
x y
y x


 Ex


 = µσ  Ey

 0

 ∂t Ex



 + µε  ∂t Ey


0




That is,
∂ y Bz − ∂ z By = µσEx + µε ∂t Ex
(7.4)
− ∂ x Bz + ∂ z Bx = µσEy + µε ∂t Ey
(7.5)
∂ x By − ∂ y Bx = 0
(7.6)
Equations (7.1)–(7.6) can now be used to establish the properties of TE
wave propagation.
7.1.1 Theorem
Suppose ( E , B)|Ez = 0 propagates in the z-direction in some homogeneous
domain (Ω,μ, ε, σ), where Ω ⊂ R 3. Suppose that the charge density ρ|Ω = 0.
Then, the TE to z-mode is completely characterized by
K15149_Book.indb 224
−∆Bz + µσ ∂t Bz + µε ∂t2 Bz = 0
(7.7)
−∆E⊥ + µσ ∂t E⊥ + µε ∂t2 E⊥ = 0
(7.8)
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Waveguides and Cavity Resonance
225
Proof
Now, ∂ y (7.4) − ∂ x (7.5) yields
∂2y Bz + ∂2x Bz + 2 ∂2z Bz + ∂ z (∂ x Bx + ∂ y By )
= µσ ∂t (∂ y Ex − ∂ x Ey ) + µε ∂t2 (∂ y Ex − ∂ x Ey )
Thus, invoking Gauss’ law ∇ ⋅ E = 0 together with Equation (7.3) give
−∆Bz + µσ ∂t Bz + µε ∂t2 Bz = 0
Next, ∂ x (7.3) + ∂ z (7.1) together with ∇ ⋅ E = 0 yields
∂2y Ey + ∂2x Ey + ∂2z Ey = − ∂t (∂ x Bz − ∂ z Bx ) = µσ ∂t Ey + µε ∂t2 Ey
and likewise, ∂ y (7.3) + ∂ z (7.2) together with Gauss’ law yields
∂2y Ex + ∂2x Ex + ∂2z Ex = − ∂t (∂ y Bz − ∂ z By ) = µσ ∂t Ex + µε ∂t2 Ex
Hence, along a similar vein, together with ∂ y (7.3) + ∂ z (7.1) , −∆E⊥ + µσ ∂t E⊥ +
µε ∂t2 E⊥ = 0 , as required.
□
7.1.2 Proposition
A time-harmonic TE to z-mode wave propagation satisfies E⊥ = ( η/µ)e z × B⊥,
where η is some frequency-dependent parameter.
Proof
A general technique is employed to establish the assertion (see Section 1.4).
First, decompose the fields and operators into transverse and longitudinal
components, and then equate the transverse (longitudinal) components
with the transverse (longitudinal) components. This is done as follows.
Set ∇ = ∇ ⊥ + e z ∂ z, B = B⊥ + e z Bz , and by assumption, E = E⊥. Then, from
Maxwell’s equations and using the convention e− iωt instead of eiωt for time
harmonicity so that ∂t ↔ − iω ,
∇ × E = iωB ⇒ ∇ ⊥ × E⊥ + e z × ∂ z E⊥ = iωB⊥ + iωe z Bz
That is,
K15149_Book.indb 225
∇ ⊥ × E⊥ = iωe z Bz
(7.9)
e z × ∂ z E⊥ = iωB⊥
(7.10)
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
From Equation (7.10), e z × (e z ∂ z × E⊥ ) = − ∂ z E⊥ ⇒ ∂ z E⊥ = − iωe z × B⊥. As a
warm-up exercise, verify this in Exercise 7.4.2. Next, noting that ∂ z e− γz =
−γe− γz ⇒ ∂ z ↔ −γ , it follows that
γE⊥ = iωe z × B⊥ ⇒ E⊥ =
1 iωµ
µ γ
e z × B⊥
(7.11)
□
as required.
7.1.3 Definition
The TE to z-mode wave impedance ηTE for a time-harmonic wave ( E , B)|Ez = 0 is
defined by ηTE = iωµ
γ , where γ depends on the boundary conditions.
Now, consider an electromagnetic field propagating between two infinite
parallel plates illustrated in Figure 7.1.
7.1.4 Proposition
Referring to Figure 7.1, suppose that ( E , B)|Ez = 0 is a time-harmonic wave
propagating in the ez-direction on (Ω,μ,ε), and suppose further that ρ = 0 on
Ω and there is no variation along the ex -direction. Then, the n-mode wave
propagation ( E , B)|Ez = 0 satisfies
Ex , n = E0, n sin ( naπ y ) e− γ n z
By , n =
Bz , n =
γn
ω
(7.12)
E0, n sin ( naπ y ) e− γ n z
i nπ
ω a
(7.13)
E0, n cos ( naπ y ) e− γ n z
(7.14)
for some constant E0,n, and γ 2n = ( naπ ) − ω 2 µε − iωµσ. Furthermore, if σ = 0, there
exists a real sequence (ω n )n of cut-off angular frequencies such that
2
ω > ω n ⇒ e− γ n z = e− iβn z , β n = ℑmγ n
(a)
y=a
ey
ex
ez
x, z = 0
PEC
Direction of wave propagation
Ω
PEC
Figure 7.1
Wave propagation between two infinite parallel planes.
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227
Waveguides and Cavity Resonance
that is, the waves propagate without attenuation,
ω < ω n ⇒ e− γ n z = e−α n z , α n = ℜeγ n
(b)
that is, the waves propagate with exponential attenuation.
Finally, for ω = ω n , standing waves exist between the two parallel planes,
and no wave propagates along the ez-direction.
Proof
Now, applying the assumption ∂x = 0 to Equations (7.3) and (7.6), respectively, yields
∂ y Ex = ∂t Bz ⇒ ∂ y Ex = − iωBz
(7.15)
∂ y Bx = 0
(7.16)
Next, noting that ∂x ∂y = ∂y ∂x as E, B are of class C2, it follows from Equation
(7.16) that ∂ y (7.1) ⇒ 0 = ∂ y ∂ z Ey = ∂ z ∂ y Ey ⇒ Ey is independent of y. Because
Ey |∂Ω = 0, it follows that Ey = 0 on Ω in order to satisfy the boundary conditions. In particular, (7.1) implies at once that Bx = 0 on Ω. Hence, (7.8) reduces
to ∆Ex + (iωµσ + ω 2 µε)Ex = 0. For notational simplicity, set k 2 = ω 2 µε + iωµσ.
The solution of Equation (7.8) can be easily determined via the separation
of variables. Now, by assumption, ∂ x = 0 ⇒ ∆ = ∂2y + ∂2z . Hence, set Ex ( y , z) =
Φ( y )Ψ( z). Then, from Proposition 7.1.2,
∂2y Φ
Φ
∂2 Φ
+
∂2z Ψ
Ψ
+ k 2 = 0 ⇒ − k y2 + γ 2 + k 2 = 0
∂2 Φ
2
y
y
where Φ + k y2 = 0 and ∂zΨΨ − γ 2 = 0 . The
general solution of Φ + k y2 = 0 is
ˆ
∂2z Ψ
2
Φ( y ) = A cos k y y + B sin k y y, and that of Ψ + γ = 0 is given by Ψ = A′e− γz +
γz
− γz
B′e = A′e due to the absence of boundary as z → ∞; that is, B′ ≡ 0 as there
can be no reflected waves. Thus, Ψ = e− γz corresponds to the fundamental
solution.
The boundary condition Ex|∂Ω = 0 implies that Φ(0) = 0 = Φ(a) and hence,
set A = 0 and k y a = πn, for = 0, 1, 2,… . Thus, the fundamental solution is
Φ n ( y ) = sin kn y ≡ sin naπ y . In particular, denoting γ n = γ ,
γ 2n = kn2 − k 2 = ( naπ ) − ω 2 µε − iωµσ
2
(7.17)
So, setting γ n = α n + iβ n, it follows that
2
 2
2
2
 α n − β n = ( naπ ) − ω µε

 2α nβ n = −ωµσ
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228
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Next, noting that (α 2n + β 2n )2 = (β 2n − α 2n )2 + 4α 2nβ 2n , it follows that
((
α n2 + β n2 =
)
nπ 2
a
− ω 2 µε
) + (ωµσ)
2
2
and hence, adding and subtracting from α 2n − β 2n yields


 αn =



 βn =

1
2



((
nπ 2
a
1
2



((
nπ 2
a
)
)
2
− ω µε
− ω 2 µε
) + (ωµσ) − (ω µε − ( ) )
2
) + (ωµσ)
2
2
2
2
+ ω 2 µε − (
nπ 2
a
1
)
nπ 2
a
2


1
2


(7.18)
From Equation (7.18), it is clear that α n , β n ≥ 0 ∀ω ≥ 0 as |a + b|+ a ≥ 0 ∀a ∈R , b ≥ 0
In particular, σ > 0 ⇒ α n , β n > 0. If σ = 0, then (7.18) reduces to the form
f ± (ξ) = ξ  ξ , where f+ ↔ α n and f− ↔ β n . Indeed, it is evident that
(a)
ω 2 µε − ( naπ ) > 0 ⇒ α n = 0 ⇒ e− γ n z = e− iβn z
(b)
ω 2 µε − ( naπ ) < 0 ⇒ β n = 0 ⇒ e− γ n z = e−α n z
2
2
Thus, scenario (a) corresponds to forward wave propagation without attenuation, and scenario (b) corresponds to wave attenuation; these waves are
called evanescent waves, as they die out exponentially for z > β1n .
In summary, Ex , n = E0, n sin( naπ y )eiγ n z, where E0, n is some constant depending on initial conditions. And from Equation (7.2), via ∂ z ↔ −γ n , By , n = γωn E0, n
sin( naπ y )e− γ n z . Moreover, the solution for (7.14) can be easily solved via (7.14):
∂ y Ex = − iωBz ⇒ Bz , n =
inπ
ωa
E0,+ n cos ( naπ y ) eiγ n z
Last, observe from (a) and (b) that ω 2 µε − ( naπ ) = 0 ⇔ ω n = 1µε naπ implies the
existence of a sequence (ωn)n of critical frequencies determining whether
waves are attenuated. Indeed, as ω = ω n ⇒ α n , β n = 0 ⇒ e− γ n z = 1 ∀z , it is
obvious that there is no wave propagation along the z-direction as the field
is independent of z. Thus, the solution (Ex , n , By , n , Bz , n ) shows that standing
waves exist between the two parallel planes when ω = ωn, as required.
□
2
7.1.5 Corollary
The wave impedance for the TE to z-mode of Proposition 7.1.4 is given by
{
ηTE, n = η 1 −
K15149_Book.indb 228
( )
ωn 2
ω
σ
+ i ωε
}
− 21
(7.19)
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229
Waveguides and Cavity Resonance
where η = µ/ε . In particular, when the dielectric is lossless, that is, σ = 0,
then, for each fixed n ∈ N,
{( )
(a)
ω < ω n ⇒ ηTE, n = − iη
(b)
{
ωn 2
ω
ω > ω n ⇒ ηTE, n = η 1 −
(
}
)}
−1
− 21
− 21
ωn 2
ω
Finally, ω >> ω n ⇒ ηTE, n → η. In particular, ηTE, n > η whenever ω > ω n.
Proof
From Proposition 7.1.2, ηTE, n =
γn =
where ω n =
1 nπ
µε a
iωµ
γn
. Furthermore, from (7.17),
( naπ )2 − ω 2 µε − iωµσ = iω
µε 1 −
( )
ωn 2
ω
σ
+ i ωε
. Hence,
ηTE, n =
µ
ε
{
1−
( )
ωn 2
ω
( )
σ
+ i ωε
}
− 21
2
Next, on setting σ = 0 and γˆ n = 1 − ωωn , it is obvious that ω < ω n ⇒ γ̂ n
is purely imaginary, whereas ω > ω n ⇒ γ̂ n is real. To complete the proof, it
2
suffices to note that ω n << ω ⇒ γˆ n → 1 ⇒ ηTE, n → η , and ω > ω n ⇒ 1 − ( ωωn )
< 1 ⇒ ηTE,n > η .
□
It is clear from Corollary 7.1.5, even when the dielectric is lossless,
γˆ n ∈R ⇒ γ n is purely imaginary and hence e− γ n z →
/ 0 as z → ∞; that is, when
the TE mode impedance is real, the fields are traveling waves and will thus
propagate without attenuation. That is, for real TE mode impedance, power
can be transferred.
On the other hand, still considering the lossless dielectric case for simplicity, γ̂ n is imaginary implying that γ n > 0 and hence, e− γ n z → 0 as z → ∞. From
a circuit analogy, the impedance is imaginary and hence reactive; thus, no
power is transferred. Physically, the fields are evanescent waves and are thus
attenuated as they propagate between the parallel planes. Wave propagation cannot be sustained. In particular, when the waves are at the cut-off
frequency ω = ω n, γˆ n = 0 ⇒ γ n = 0 ⇒ e− γ n z ≡ 1 ∀z > 0 and hence, waves cannot
be sustained at the cut-off frequency. Physically, via Corollary 7.1.5, the wave
impedance approaches infinity. See Figure 7.2 for the variation of the TE
mode impedance as the angular frequency varies. The plot is based on the
η
normalized n-mode TE wave impedance TEη ,n for simplicity, where η is the
wave impedance of the dielectric, against the normalized angular frequency
ωn
ωn
ω for some fixed ωn. Lastly, note that ω = 0 ⇔ ω → ∞, for any fixed ω n > 0.
K15149_Book.indb 229
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230
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Normalised n-mode TE Impedance (σ = 0)
Normalised Wave Impedance
3.5
3
2.5
Im(ηTE/η)
2
Re(ηTE/η)
1.5
1
0.5
0.
75
0.
95
1.
1
1.
25
1.
4
1.
55
1.
7
1.
85
2.
05
2.
25
2.
4
2.
55
2.
7
2.
85
0.
6
0.
3
0.
45
0
0.
15
0
Normalised Angular Frequency
Figure 7.2
Normalised n-mode TE wave impedance in a lossless dielectric.
7.1.6 Theorem
Suppose ( E , B)|Bz = 0 propagates in the z-direction in some homogeneous
domain (Ω, μ, ε, σ), where Ω ⊂ R 3 . Suppose that the charge density ρ|Ω = 0.
Then, the TM to z-mode is completely characterized by
−∆Ez + µσ ∂t Ez + µε ∂t2 Ez = 0
(7.20)
−∆B⊥ + µσ ∂t B⊥ + µε ∂t2 B⊥ = 0
(7.21)
Proof
There is a quick way to establish Equations (7.20) and (7.21) via duality
sketched in Section 3.6. Expressing (7.7) as −∆H z + µσ ∂t H z + µε ∂t2 H z = 0,
where Bz = µH z , and invoking the duality transformation H → −E yields
∆Ez − µσ ∂t Ez − µε ∂t2 Ez = 0. Likewise, via E → H, (7.8) transforms into
−∆H ⊥ + µσ ∂t H ⊥ + µε ∂t2 H ⊥ = 0, yielding (7.21) after utilizing B = μH.
□
7.1.7 Proposition
A time-harmonic TM to z-mode wave propagation satisfies B⊥ = (µ/η)e z × E⊥ ,
where η is some frequency-dependent parameter.
Proof
The proof follows Proposition 7.1.2 mutatis mutandis. From
∇ × B = (µσ − iωµε)E ⇒ ∇ ⊥ × B⊥ + e z × ∂ z B⊥ = κE⊥ + κe z Ez
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231
Waveguides and Cavity Resonance
where κ = μσ − iωμε. That is,
∇ ⊥ × B⊥ = κe z Ez
(7.22)
e z × ∂ z B⊥ = κE⊥
(7.23)
From Equation (7.23), e z × (e z ∂ z × B⊥ ) = − ∂ z B⊥ ⇒ − ∂ z B⊥ = κe z × E⊥, whence
∂ z ↔ −γ gives
γB⊥ = κe z × E⊥ ⇒ B⊥ = µ µγκ e z × E⊥
□
as required.
7.1.8 Remark
Clearly, from Proposition 7.1.2, E⊥ = ( η/µ)e z × B⊥ = ηe z × H ⊥ , invoking dual e z × E⊥ ⇒ B⊥ = −(µ/η)e z × E⊥, for some
ity, E → H and H → −E, yields H ⊥ = −η
frequency dependent parameter η̂. However, this would not have yielded
the explicit expression for η̂.
7.1.9 Definition
The TM to z-mode wave impedance ηTM for a time-harmonic wave ( E , B)|Bz = 0 is
defined by ηTM = −(µγ/κ ), where γ depends on the boundary conditions and
κ = μσ − iωμε.
7.1.10 Proposition
Referring to Figure 7.1, suppose that ( E , B)|Bz = 0 is a time-harmonic wave
propagating in the e z -direction on (Ω,μ,ε), and suppose further that ρ = 0 on
Ω and there is no variation along the ex -direction. Then, the n-mode wave
propagation ( E , B)|Bz = 0 satisfies
Bx , n = B0,+ n cos ( naπ y ) e− γ n z
(7.24)
Ey = − γκn B0,+ n cos ( naπ y ) e− γ n z
(7.25)
sin ( naπ y ) e− γ n z
(7.26)
Ez = κ1 B0,+ n
nπ
a
for some constant B0,+ n with κ = μσ − iωμε. Furthermore, if σ = 0, there exists
a real sequence (ω n )n of cut-off angular frequencies such that
(a) ω > ω n ⇒ e− γ n z = e− iβn z , β n = ℑmγ n ; that is, the waves propagate without attenuation
(b) ω < ω n ⇒ e− γ n z = e−α n z , α n = ℜeγ n ; that is, the waves propagate with
attenuation
Finally, for ω = ωn, standing waves exist between the two parallel planes, and
no waves propagate along the ez-direction.
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232
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Proof
The proof mimics that of Proposition 7.1.4; see Exercise 7.4.3.
□
7.1.11 Corollary
The wave impedance for the TM to the z-mode of Proposition 7.1.10 is given by
σ
ηTM, n = η{1 + i ωε
}
−1
1−
( )
ωn 2
ω
σ
+ i ωε
(7.27)
µ
where η = ε . In particular, when the dielectric is lossless (i.e., σ = 0) then,
for each fixed n ∈ N,
( )
ωn 2
ω
(a)
ω < ω n ⇒ ηTM, n = iη
(b)
ω > ω n ⇒ ηTM, n = η 1 −
−1
( )
ωn 2
ω
Finally, ω >> ω n ⇒ ηTM, n → η. In particular, ηTM, n < η whenever ω > ω n , and
hence, ηTM, n < ηTE, n ∀ω > ω n .
Proof
µγ
From Definition 7.1.9, ηTM, n = − κn , where κ = μσ − iωμε. Hence, following the proof of Corollary 7.1.5 mutatis mutandis, and on setting
2
σ
γ n = iω µε 1 − ωωn + i ωε
≡ iω µεγˆ n , it can be shown (see Exercise 7.4.4) that
( )
σ
ηTM, n = η{1 + i ωε
}
−1
1−
( )
ωn 2
ω
σ
+ i ωε
The two cases (a) and (b) are self-evident and (b) implies ηTM, n < η ∀ω > ω n .
Lastly, ηTM, n < ηTE, n ∀ω > ω n follows directly from Corollary 7.1.5.
□
n
The plot of the normalized TM wave impedance ηTM,
versus the normalη
ωn
ized angular frequency ω is shown in Figure 7.3, where ω n is assumed fixed.
7.1.12 Theorem
Given the parallel wave guide illustrated in Figure 7.1, suppose a timeharmonic n-mode TE or TM wave is propagating at some fixed frequency
ω > ω n . Then, the wavelength of the waves within Ω is given by
λ=
2 2π
ω

v

((
)
ωn 2
ω
)
2
−1 +(
)
σ 2
ωε
+ 1−
( )
ωn 2
ω



− 21
(7.28)
In particular, for a lossless medium,
λ=
K15149_Book.indb 232
2π v
ω
{
1−
( )}
− 21
ωn 2
ω
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233
Waveguides and Cavity Resonance
Normalised n-mode TM Impedance (σ = 0)
Normalised Wave Impedance
3
2.5
Im(ηTM/η)
2
1.5
Re(ηTM/η)
1
0.5
7
2.
85
55
2.
2.
4
25
2.
05
2.
85
2.
7
1.
1.
4
1.
55
25
1.
1
1.
1.
75
0.
95
6
0.
0.
3
0.
45
15
0.
0.
0
0
Normalised Angular Frequency
Figure 7.3
Normalised n-mode TM wave impedance in a lossless dielectric.
Proof
Now, an admissible TE or TM wave propagating between two parallel plates
Ω has the form f ( y )e− γ n z e− iωt . Setting γ n = α n + iβ n defined by Equation (7.18),
it is clear that the phase of the wave is defined by e− i(ωt +βn z ) . Thus, the point of
constant phase is Θ = ωt + β n z and the phase velocity is thus
d
dt
Θ = ω + βn
= 0 ⇒ ddzt = − βωn
dz
dt
whence, the n-mode phase speed uˆ n of the wave in Ω is
uˆ n =
= fλ n ⇒ λ n =
ω
βn
2π
βn
(7.29)
Rearranging β n slightly as
1
2



((
)
nπ 2
a

1
ω µε 
2

2
− ω µε
((
ωn
ω
)
2
) + (ωµσ )
2
)
1
2
+ ω µε − (
2
)
nπ 2
a
2
 =

1
2
σ
− 1 + ( ωε
) + 1−
2
( )
ωn
ω
2
2



it is clear that
λn =
K15149_Book.indb 233
2
f

v

((
)
ωn 2
ω
)
2
−1 +(
)
σ 2
ωε
+ 1−
( )
ωn 2
ω



− 21
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234
Electromagnetic Theory for Electromagnetic Compatibility Engineers
where v = 1µε is the phase velocity in (R 3 , µ , ε) . For the lossless case, σ = 0
and the result thus follows.
□
7.1.13 Example
Determine the time-average power transmitted by an admissible n-mode
TE wave propagating in Ω|σ= 0 defined in Figure 7.1, and hence, deduce the
velocity of the energy flow. To determine the power flow, it suffices to consider the time-average power transmitted across a finite cross-section R =
[0,1] × [0,a]. By definition,
⟨ P⟩R =
1
2
∫
R
ℜe( E × H ∗ ) ⋅ nd 2 x =
1
2
1
a
∫ ∫ ℜe(E × H ) ⋅ e dy dx
0
∗
0
From Proposition 7.1.4, ℜe( E × H ∗ ) ⋅ e z = µ1 Ex , n By∗ , n =
γ n = iβ n for lossless Ω. Hence,
⟨ P⟩R =
a 1
4 µ
E0, n
1
µ
E0, n
z
2 βn
ω
sin 2 ( naπ y ) , as
2 βn
ω
is the time-average power transmitted across the cross-section R.
Next, to determine the velocity of energy flow, recall first by definition that
power is the time-rate of the flow of energy. Second, the time-average energy
density per unit length along the direction of wave propagation is given by
⟨w ⟩C = 21 ∫ C ℜe 21 D ⋅ E ∗ + 21 B ⋅ H ∗ d 3 x, where C = R × [0,1]. Observe also that
the assumption of ∂ x = 0 renders the choice of R,C meaningful and hence,
⟨ P⟩R , ⟨w ⟩C meaningful.
By definition, ⟨w ⟩C = 14 ∫ 10 ∫ 10 ∫ 0a (ε|E|2 + µ|H|2 )dydxdz. From Proposition 7.1.4,
(
)
ε E + µ H = ε ( E0, n ) sin 2 ( naπ y ) +
2
2
2
1
µ
and hence,
⟨w ⟩C =
a
8
(E0,n )2
(E ) {( )
2
+
0, n
βn
ω
{ ((
ε+
1
µ
βn
ω
2
}
sin 2 ( naπ y ) + ( ωnπa ) cos 2 ( naπ y )
2
) + ( ) )}
2
nπ 2
ωa
Thus, note that if u n is the n-mode velocity of energy propagation, then
u n ⟨w ⟩C has precisely the dimensions of power. It is intuitively clear that
u n ⟨w ⟩C corresponds to the time-average power: ⟨ P⟩R = u n ⟨w ⟩C . Hence, the
velocity of energy propagation along Ω|σ= 0 is given by
u n =
K15149_Book.indb 234
PR
wC
{
= 2β nω (ω 2 + ω n2 )µε + β n2
}
−1
=
2 βnω
( ω 2 +ω 2n +ω 2 −ω 2n )µε
=
βn
ωµε
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235
Waveguides and Cavity Resonance
However, for σ = 0, ω > ω n ⇒ β n = ω µε 1 −
Hence,
u n =
1
µε
1−
( )
ωn 2
ω
( )
ωn 2
ω
≡ v 1−
, as can be easily verified.
( )
ωn 2
ω
(7.30)
Normalised Phase Velocity
It is clear from the equation that u < v ∀ω > ω n .
Finally, from the proof of Proposition 7.1.12, uˆ n = βωn ⇒ uˆ n u n = µε1 . That is,
the product of the phase velocity and the energy propagation velocity (i.e.,
group velocity) yields precisely the velocity of propagation of a plane wave
in (R 3 , µ , ε). A plot of the normalized phase uˆvn velocity and the normalized
velocity of energy propagation uvn as a function of the normalized angular
frequency ωωn is shown in Figure 7.4.
Normalised n-mode TE Phase Velocity
3
2.5
2
1.5
1
0.5
0
1
1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.05 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9
Normalised Angular Frequency
Normalised Group Velocity
Figure 7.4 a
Normalised phase velocity vs. normalised angular frequency.
3
Normalised n-mode TE Group Velocity
2.5
2
1.5
1
0.5
0
1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9
Normalised Angular Frequency
Figure 7.4b
Normalised group velocity vs. normalised angular frequency.
K15149_Book.indb 235
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236
Electromagnetic Theory for Electromagnetic Compatibility Engineers
7.2 Rectangular Waveguides
In Section 7.1, the concept of TE/TM mode propagation was introduced
between two parallel planes. The parallel planes constitute a waveguide. In
particular, a pair of transmission lines constitutes a waveguide for TEM
propagation. In short, a waveguide is any structure* that supports TE, TM, or
TEM modes. In what follows, let Ω = (0,b) × (0,a) × [0,∞) denote a semi-infinite
rectangular waveguide, and ∂Ω denote its boundary.
7.2.1 Proposition
Suppose that ( E , B)|Ez = 0 is a time-harmonic wave propagating in the ezdirection on (Ω, μ, ε, σ), and suppose further that ρ = 0 on Ω and ∂Ω is a perfect electrical conductor. Then, ( E , B)|Ez = 0 satisfies
Ex , mn = E0, mn cos ( mbπ x ) sin ( naπ y ) e− γ mn z
(7.31)
− γ mn z
mπ
nπ
Ey , mn = − ma
nb E0, mn sin ( b x ) cos ( a y ) e
(7.32)
Bx , mn =
iγ mn ma
ω nb
By , mn =
Bz , mn =
iπ
ω
iγ mn
ω
{
E0, mn sin ( mbπ x ) cos ( naπ y ) e− γ mn z
E0, mn cos ( mbπ x ) sin ( naπ y ) e− γ mn z
E0, mn 1 + ( ma
nb )
2
} cos (
mπ
b
x ) cos ( maπ y ) e− γ mn z
(7.33)
(7.34)
(7.35)
for some constant E0, mn and γ 2mn = ( mbπ ) + ( naπ ) − ω 2 µε − iωµσ . Furthermore,
if σ = 0, there exists a real sequence (ω mn )m , n of cut-off angular frequencies
such that
2
2
(a) ω > ω mn ⇒ e− γ mn z = e− iβmn z , β mn = ℑmγ mn , and hence waves propagate
without attenuation
(b) ω < ω mn ⇒ e− γ mn z = e−α mn z , α mn = ℜeγ mn and hence waves propagate
with exponential attenuation
Proof
From Theorem 7.1.1, ∆Bz + k 2 Bz = 0 subject to the boundary condition
 ∂ x Bz = 0 if x = 0, b

∂ y Bz = 0 if y = 0, a

* More precisely, a single conductor or a set of disjoint conductors.
K15149_Book.indb 236
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237
Waveguides and Cavity Resonance
as ∂Ω is a perfect electrical conductor. So, to begin, consider ∆Ex + k 2 Ex = 0.
Then, via the separation of variables
Bx ( x , y , z) = Φ( x)Ψ( y )Θ( z) ⇒
Set
∂2x Φ
Φ
= − k x2 ,
∂2y Ψ
Ψ
= − k y2 , and
∂2z Θ
Θ
∂2x Φ
Φ
+
∂2y Ψ
Ψ
+
∂2z Θ
Θ
+ k2 = 0
= γ 2. Thus,
Φ = A cos k x x + B sin k x x ⇒ ∂ x Φ = − Ak x sin k x x + Bk x cos k x x
yielding, via the boundary condition, B = 0 ⇒ Φ = cos ( mbπ x ) . Similarly,
Ψ = cos ( naπ y ) and Θ = e− γz . That is, the fundamental solution is Bx ( x , y , z) =
cos ( mbπ x ) cos ( maπ y ) e− γz .
Next, invoking Theorem 7.1.1 again, ∆E⊥ + k 2 E⊥ = 0 . Thus, appealing to the
boundary condition for E⊥ ,
 Ey = 0 if x = 0, b

 Ex = 0 if y = 0, a
the separation of variables applied to the pair Ex , Ey yields:
Ex ~ sin ( naπ y ) { A′ cos k x′ x + B′ sin k x′ x} e− γ ′z
{
}
Ey ~ sin ( mbπ x ) A′′ cos k y′′y + B′′ sin k y′′y e− γ ′′z
Now, appealing to Equation (7.3), ∂ x Ey − ∂ y Ex = iωBz , it follows at once by
invoking the boundary condition for Bz , to wit,
{
∂ x ∂ x Ey − ∂ y Ex
}
x = 0, b
{
= 0 = ∂ y ∂ x Ey − ∂ y Ex
}
y = 0, a
that the following must hold,
γ ′ = γ = γ ′′
A′ cos k x′ x + B′ sin k x′ x ~ cos ( mbπ x ) ⇒ B′ = 0, k x′ =
A′′ cos k y′′y + B′′ sin k y′′y ~ cos ( naπ y ) ⇒ B′′ = 0, k y′′ =
K15149_Book.indb 237
mπ
b
nπ
a
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238
Electromagnetic Theory for Electromagnetic Compatibility Engineers
The results give the fundamental solutions:
Ex = A′ cos ( mbπ x ) sin ( naπ y ) e− γ ′z
Ey = A′′ sin ( mbπ x ) cos ( naπ y ) e− γ ′z
In order to determine the pair A′ , A′′ , it suffices to appeal to Gauss’
law: ∇ ⋅ E⊥ = 0 (by assumption). Hence, ∂ x Ex = − ∂ y Ey ⇒ A′′ naπ = − A′ mbπ ⇒
+
mπ
nπ
A′′ = − ma
nb A ′. Finally, from Equation (7.3), it is clear that iωB0, nm = A ′′ b − A ′ a =
{
− A′π 1 + ( ma
nb )
2
}⇒B
+
0, nm
=
iπ
ω
{
A′ 1 + ( ma
nb )
A′ = E0, mn, some constant,
2
} . From this, it follows that on setting
Ex = E0, mn cos ( mbπ x ) sin ( naπ y ) e− γ mn z
− γ mn z
mπ
nπ
Ey , mn = − ma
nb E0, mn sin ( b x ) cos ( a y ) e
Bz , mn =
iπ
ω
{
E0, mn 1 + ( ma
nb )
2
} cos (
mπ
b
x ) cos ( maπ y ) e− γ mn z
where γ 2mn = km2 + kn2 − k 2 = ( mbπ ) + ( naπ ) − ω 2 µε − iωµσ. From Equation (7.18), it
is obvious that
2


 α mn =



 β mn =

1
2



((
mπ 2
b
1
2



((
mπ 2
b
2
) +( )
2
− ω µε
) + (ωµσ) − (ω µε − (
) +( )
2
) + (ωµσ)
nπ 2
a
nπ 2
a
− ω µε
2
2
2
2
2
+ ω µε − (
) −( )
mπ 2
b
nπ 2
a
)
1
2



1
2
) −( )
mπ 2
b
nπ 2
a
 2


This is easily obtained via the replacement: ( naπ ) → ( mbπ ) + ( naπ ) . Hence, for
2
2
σ = 0, and setting ω 2mnµε = ( mbπ ) + ( naπ ) , it is clear that α mn = 0 if ω > ω mn ,
whereas β mn = 0 if ω < ω mn .
The explicit expressions for the remaining magnetic field density components follow directly from Equations (7.1) and (7.2), respectively:
2
Bx , mn = − iγωmn Ey , mn =
By , mn =
iγ mn
ω
Ex , mn =
iγ mn ma
ω nb
iγ mn
ω
2
2
E0,+ mn sin ( mbπ x ) cos ( naπ y ) e− γ mn z
E0,+ mn cos ( mbπ x ) sin ( naπ y ) e− γ mn z
The proof that resonance occurs when ω = ω mn , the cut-off angular frequency,
follows that of Proposition 7.1.4 mutatis mutandis. In particular, no waves
propagate along the waveguide.
□
K15149_Book.indb 238
10/18/13 10:57 AM
239
Waveguides and Cavity Resonance
7.2.2 Corollary
The TE mn to z-mode wave impedance ηTE, mn for a time-harmonic ( E , B)|Ez = 0
{
is given by ηTE, mn = γiωµ
= η 1−
mn
each fixed (m,n) ∈ N × N,
} . In particular, when σ = 0, for
= − iη{(
) − 1}
= η{1 − (
)}
( )
ω mn 2
ω
σ
+ i ωε
−1
ω mn 2
ω
ω > ω mn ⇒ ηTE, mn
(b)
−1
ω mn 2
ω
ω < ω mn ⇒ ηTE, mn
(a)
− 21
Finally, ω >> ω mn ⇒ ηTE, mn → η. In particular, ηTE, mn > η whenever ω > ω mn .
Proof
Noting that γ mn = iω µε 1 −
ηTE, mn =
( )
ω mn 2
ω
iωµ
γ mn
σ
+ i ωε
, Definition 7.1.3 leads at once to
{
= η 1−
( )
ω mn 2
ω
+i
σ
ωε
}
− 21
The remaining assertions follow the proof of Corollary 7.1.5 mutatis mutandis. □
It is evident from the previous section that traveling waves are sustained
if the frequency is greater than the cut-off frequency. From Corollary 7.2.2,
waves will propagate when the wave impedance is real; when the wave
impedance is imaginary, waves are not sustained—they are attenuated—as
no power is transferred.
7.2.3 Proposition
Suppose that ( E , B)|Bz = 0 is a time-harmonic wave propagating in the ez
-direction on (Ω, μ, ε, σ), and suppose further that ρ = 0 on Ω and ∂Ω is a perfect electrical conductor. Then, ( E , B)|Bz = 0 satisfies
Ez , mn = E0 sin ( mbπ x ) sin ( naπ y ) e− γ mn z
Ex , mn = −γ mn
Ey , mn = −γ mn
Bx , mn = κ
{(
By , mn = −κ
K15149_Book.indb 239
{(
{(
) + ( naπ )2 }
mπ 2
b
) + ( naπ )2 }
mπ 2
b
) + ( naπ )2 }
mπ 2
b
{(
−1
−1
) + ( naπ )2 }
mπ 2
b
−1
nπ
a
−1
mπ
b
E0 cos ( mbπ x ) sin ( naπ y ) e− γ mn z
(7.37)
nπ
a
E0 sin ( mbπ x ) cos ( naπ y ) e− γ mn z
(7.38)
E0+ sin ( mbπ x ) cos ( naπ y ) e− γ mn z
mπ
b
(7.36)
E0+ cos ( mbπ x ) sin ( naπ y ) e− γ mn z
(7.39)
(7.40)
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240
Electromagnetic Theory for Electromagnetic Compatibility Engineers
for some constant E0, κ = μσ − iωμε, and γ 2mn = ( mbπ ) + ( naπ ) − ω 2 µε − iωµσ .
Furthermore, if σ = 0, there exists a real sequence (ω mn )m , n of cut-off angular
frequencies such that
2
2
a) ω > ω mn ⇒ e− γ mn z = e− iβmn z , β mn = ℑmγ mn , and hence waves propagate
without attenuation
b) ω < ω mn ⇒ e− γ mn z = e−α mn z , α mn = ℜeγ mn and hence waves propagate
with exponential attenuation
Proof
From Exercise 7.4.2, the TM mode reduces Maxwell’s equations to
∂ y Ez + γ mnEy = iωBx , − ∂ x Ez − γ mnEx = iωBy , ∂ y Ex = ∂ x Ey
γ mn By = κEx , −γ mn Bx = κEy , ∂ x By − ∂ y Bx = κEz
By Theorem 7.1.6, −∆Ez + µσ ∂t Ez + µε ∂t2 Ez = 0 . So, once again, appealing to
the separation of variables, set Ez = Φ( x)Ψ( y )Θ( z), subject to the boundary
condition:
 x = 0, b
Ez = 0 for 
y = 0, a

By now, it should be obvious that the fundamental solution is
Ez = sin ( mbπ x ) sin ( naπ y ) e− γz . To see this, it suffices to set Φ = A cos k x x + B sin k x x ,
Ψ = A′ cos k y y + B′ sin k y y, and Θ = A′′e− γz + B′′e γz . Then, the boundary conditions and the requirement that the solution be finite yield the desired fundamental solution. Thus, set Ez , mn = E0 sin ( mbπ x ) sin ( naπ y ) e− γ mn z .
Next, substituting Ex = γ κmn By into − ∂ x Ez − γ mnEx = iωBy yields:
By = −
{
γ 2mn
κ
+ iω
}
−1
∂ x Ez = −
{
2
γ mn
κ
+ iω
}
−1
mπ
b
E0+ cos ( mbπ x ) sin ( naπ y ) e− γ mn z
Substituting Ey = − γ κmn Bx into ∂ y Ez + γ mnEy = iωBx leads to:
Bx =
{
γ 2mn
κ
+ iω
}
−1
∂ y Ez =
{
γ 2mn
κ
+ iω
}
−1
nπ
a
E0+ sin ( mbπ x ) cos ( naπ y ) e− γ mn z
and finally, back-substituting yields
K15149_Book.indb 240
Ex = −γ mn
{(
mπ 2
b
) + ( naπ )2 }
Ey = −γ mn
{(
mπ 2
b
−1
) + ( naπ )2 }
−1
mπ
b
E0+ cos ( mbπ x ) sin ( naπ y ) e− γ mn z
nπ
a
E0+ sin ( mbπ x ) cos ( naπ y ) e− γ mn z
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241
Waveguides and Cavity Resonance
And the results thus follow from the simplification: − γ κmn
{
γ 2mn
κ
+ iω
}
−1
= − γ 2γ mn− k 2
mn
and by substituting iωκ = k 2 . So, by definition, γ 2mn − k 2 = ( ) + ( naπ ) yields
the desired results. Lastly, the remaining assertions follow from the proof of
2
2
2
Proposition 7.2.1 mutatis mutandis via the replacement: ( naπ ) → ( mbπ ) + ( naπ ) . □
mπ 2
b
2
7.2.4 Corollary
The TM mn to z-mode wave impedance ηTM, mn for a time-harmonic ( E , B)|Ez = 0
σ
is given by ηTM, mn = η{1 + i ωε
}
for each fixed (m,n) ∈ N × N,
−1
1−
( )
ω mn 2
ω
σ
+ i ωε
. In particular, when σ = 0,
( )
ω mn 2
ω
a)
ω < ω mn ⇒ ηTM, mn = iη
b)
ω > ω mn ⇒ ηTM, mn = η 1 −
−1
( )
ω mn 2
ω
Finally, ω >> ω mn ⇒ ηTM, mn → η. In particular, ηTM, mn < η whenever ω > ω mn.
Proof
Noting that γ mn = iω µε 1 −
( )
ω mn 2
ω
recall that κ = μσ − iωμε and ω mn =
once to
µ
κ
σ
+ i ωε
and
=
( ) +( )
mπ 2
b
1
µε
σ
ηTM, mn = − µγκmn = η{1 + i ωε
}
−1
i
ωε
nπ 2
a
1−
{1 + i ωεσ }−1 ,
where we
, Definition 7.1.9 leads at
( )
ω mn 2
ω
σ
+ i ωε
The remaining assertions are obvious; see the proof of Corollary 7.1.11.
□
Thus, for a lossless rectangular waveguide, ηTM, mn < η < ηTE, mn whenever
ω > ω mn (see Section 7.1 for a pair of parallel plane waveguides). Intuitively,
this is expected, as a pair of parallel plane waveguides is merely a special
case of the rectangular waveguide wherein one dimension (here, the x-direction) extends out to infinity instead of being bounded. The following result,
in view of Theorem 7.1.12 is obvious via the replacement ω n → ω mn.
7.2.5 Theorem
Given an admissible time-harmonic TE mn or TM mn wave is propagating at
some fixed frequency ω > ω mn, the wavelength of the waves within (Ω, μ, ε, σ)
is given by
λ=
2 2π
ω

v

((
)
ω mn 2
ω
)
2
−1 +(
In particular, for a lossless medium, λ =
K15149_Book.indb 241
)
σ 2
ωε
2π v
f
{
+ 1−
1−
( )
ω mn 2
ω
( )}
− 21
ω mn 2
ω



− 21
(7.41)
□
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242
Electromagnetic Theory for Electromagnetic Compatibility Engineers
7.2.6 Remark
It follows directly from the proof of Theorem 7.1.12 and Example 7.1.13 that
the phase velocity ûmn of an admissible TE mn or TM mn mode is given by
uˆ mn

= 2v 

((
)
ω mn 2
ω
)
2
−1 +(
)
σ 2
ωε
+ 1−
( )
ω mn 2
ω



− 21
and the velocity of energy propagation u mn satisfies uˆ mn u mn = v 2 and hence,
u mn =
1
2

v

((
)
ω mn 2
ω
)
1
2
−1 +(
)
σ 2
ωε
+ 1−
( )
ω mn 2
ω
 2


In particular, because the sequence (ω mn )m , n of cut-off angular frequencies is discrete and hence monotonic, the sequence can be ordered
such that {ω m0 n0 < ω m1n1 <  < ω mk nk < } . Hence, given ω > 0 such that
ω mk nk > ω > ω mk − 1nk − 1, for some (mk − 1 , nk − 1 ),(mk , nk ) ∈N × N , the TE or TM wave
propagation is of (mk − 1 , nk − 1 ) -mode. To see this, it suffices to assume that
the wave is of mode (mk − 1 , nk − 1 ) and (m′ , n′) , where ω m′n′ < ω mk − 1 , nk − 1 . Then,
u mk − 1nk − 1 < u m′n′ and uˆ mk − 1nk − 1 > uˆ m′n′ . However, the respective velocities of wave
propagation for a fixed mode are unique. Hence, the wave can only be of
mode (mk − 1 , nk − 1 ) and higher mode waves* and thus transfers energy (or
information) at a slower speed with respect to lower mode waves in a fixed
waveguide.
7.2.7 Example
Consider the semi-infinite rectangular waveguide (Ω, μ, ε, σ) such that ∂Ω is a
perfect electrical conductor. Observe that for (m,0)-mode, where m ≠ 0, the TE
propagation leads, via Equations (7.23)–(7.27) to:
Ex , m 0 = 0 = By , m 0
and (Ey , m 0 , Bx , m 0 , Bz , m 0 ) ≠ 0
and hence, TE m 0 exists. On the other hand, via Equations (7.28)–(7.32) for TM mode,
Ez , m 0 = 0 = Ex , m 0 = By , m 0
and (Ey , m 0 , Bx , m 0 ) ≠ 0
Inasmuch as TEM cannot exist on Ω, it follows at once that TMm0 does not
exist. By symmetry, it can be easily seen that for (0,n)-mode, where n ≠ 0, TE0n
exists but TM0n does not exist.
□
*
In the sense of a higher admissible angular frequency.
K15149_Book.indb 242
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243
Waveguides and Cavity Resonance
7.2.8 Example
Suppose Ω = (Ω′ , µ ′ , ε ′) ∪ (Ω′′ , µ ′′ , ε ′′) , where Ω′ = (0, b) × (0, a) × (0, z0 ] and
Ω′′ = (0, b) × (0, a) × [ z0 , ∞) . Is there a frequency ω > 0 such that there is
no reflection of a TE mode at the boundary interface at z = z0 , where it is
assumed that {µ ′ , ε ′} ≠ {µ ′′ , ε ′′} (as sets)? If so, derive the conditions under
which this can occur by modifying (µ ′′ , ε ′′) in Ω″.
Now, in order for the coefficient of reflection to be zero, it is necessary and
sufficient that η′TE, mn = η′′TE, mn , the wave impedances in Ω′ , Ω′′, respectively.
Hence, by Corollary 7.2.2,
η′
η′′
η′TE, mn = η′′TE, mn ⇔
where η′ =
µ′
ε′
ω2 =
µ ′′
ε ′′
and η′′ =
{
η′
η′′
} {
−1
−1
{
= 1−
( ) }{1 − ( ) }
2 −1
ω mn
′′
ω mn
2
ω mn
′
ω mn
=
ω 2 − ω mn
′2
ω 2 − ω mn
′′ 2
, whence, rearranging yields
1 η′
µ ′ε ′ η′′
−
1
µ ′′ε ′′
}⇒ω ={
η′
η′′
}
−1
− 21
1 η′
µ ′ε ′ η′′
−
1
µ ′′ε ′′
Thus, a unique solution ω > 0 exists if and only if the following two criteria
are satisfied,
η′ > η′′ ⇔ µ ′ε ′′ > µ ′′ε ′
(a)
(b)
1 η′
µ ′ε ′ η′′
−
1
µ ′′ε ′′
>1⇔
µ ′′ε ′′ η′
µ ′ε ′ η′′
=
ε ′′
ε′
µ ′′ε ′′
µ ′ε ′
> 1 ⇔ ε ′′ 3 µ ′′ > ε ′ 3 µ ′
Now, it is clear that conditions (a) and (b) can be satisfied in two ways.
First, suppose that µ ′ , µ ′′ > 0 are fixed. Then, (a) implies that ε ′′ > µµ′′′ ε ′ and
(b) implies that ε ′′ >
( )
1
µ′ 3
µ ′′
ε ′ , whence, set ε ′′ > max
{
µ ′′
µ′
ε′,
( )
1
µ′ 3
µ ′′
}
ε ′ . Then,
criteria (a) and (b) are automatically satisfied by construction. Similarly,
if ε ′ , ε ′′ are fixed, then (a) implies that µ ′′ < εε′′′ µ ′ , and (b) implies that
3
3
µ ′′ > ( εε′′′ ) µ ′ . Therefore, μ″ must satisfy ( εε′′′ ) µ ′ < µ ′′ < εε′′′ µ ′ . In particular,
3
this can only hold if ε ′ < ε ′′. Thus, ( εε′′′ ) µ ′ < µ ′′ < εε′′′ µ ′ and ε ′ < ε ′′ together
will ensure that criteria (a) and (b) are preserved.
Notice the major difference between the two solutions, to wit, either choosing ε″ or μ″ for impedance matching. In the former, µ ′ , µ ′′ > 0 are completely
arbitrary, whereas the latter requires that the pair (ε ′ , ε ′′) satisfies the constraint ε ′ < ε ′′ in order for a solution to exist.
7.3 Cavity Resonance
This chapter closes with a quantitative description of cavity resonance and
its properties. Again, for simplicity, consider a closed rectangular cavity
surrounded by perfect electrical conducting boundaries. More precisely, let
K15149_Book.indb 243
10/18/13 10:58 AM
244
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Ω = (0, b) × (0, a) × (0, z0 ), where ∂Ω is a perfect electrical conductor. Notice that
in this final scenario, the waves are confined in a compact space. Admissible
frequencies are investigated, together with the wave impedance and cavity
wavelength.
7.3.1 Proposition
Given a homogeneous bounded space (Ω, μ, ε, σ), where ∂Ω is a perfect electrical conductor, an admissible time-harmonic TE to z-mode wave propagation
( E , B)|Ez = 0 satisfies:
Bz , mnp = B0 cos ( mbπ x ) cos ( naπ y ) sin
( )}
{
= − {k − ( ) }
= − iω { k − ( ) }
= iω { k − ( ) }
Bx , mnp = − k 2 −
By , mnp
Ex , mnp
Ey , mnp
2
( z)
pπ
z0
(7.42)
−1
pπ 2
mπ pπ
z0
b z0
B0 sin ( mbπ x ) cos ( naπ y ) sin
−1
pπ 2
nπ pπ
z0
a z0
−1
pπ 2
nπ
z0
a
2
−1
pπ 2
mπ
z0
b
2
( z)
pπ
z0
(7.43)
B0 cos ( mbπ x ) sin ( naπ y ) sin
( z)
(7.44)
B0 cos ( mbπ x ) sin ( naπ y ) sin
( z)
pπ
z0
(7.45)
( z)
(7.46)
B0 sin ( mbπ x ) cos ( naπ y ) sin
pπ
z0
pπ
z0
where k 2 = ω 2 µε + iµωσ . Moreover, if σ = 0, there exists a real sequence
(ω mnp )m , n, p of angular frequencies such that the TE mnp mode resonates in Ω,
where ω mnp =
1
µε
{( mbπ )2 + ( naπ )2 + ( naπ )2 }
1
2
.
Proof
The proof should be routine by now via the separation of variables. From
Equation (7.7), let Bz = Φ( x)Ψ( y )Θ( z). Then, imposing the boundary conditions on (7.7) yields
Bz = 0 for
z = 0, z0 , ∂ x Bz = 0 for
x = 0, b, ∂ y Bz = 0
for
y = 0, a
Thus, the fundamental solutions are: Φ( x) = cos ( mbπ x ), Ψ( x) = cos ( naπ y ) and
pπ
pπ
Θ( z) = sin z0 z . That is, Bz , mnp = B0 cos ( mbπ x ) cos ( naπ y ) sin z0 z , for some
constant B0 . Therefore Equation (7.1) yields Bx , mnp = − ωi ∂ z Ey , mnp , and substituting this into (7.5) gives − ∂ x Bz , mnp = κEy , mnp − ωi ∂2z Ey , mnp. However, noting
2
2
that ∂2z ↔ −γ 2 ≡ ( mbπ ) + ( naπ ) − k 2, it follows that
( )
( )
(
− ∂ x Bz , mnp = κ +
K15149_Book.indb 244
iγ 2
ω
)E
y , mnp
=
i
ω
(k
2
)
− γ 2 Ey , mnp =
i
ω
{k − ( ) } E
2
pπ 2
z0
y , mnp
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245
Waveguides and Cavity Resonance
and hence,
{
Ey , mnp = iω k 2 −
( )}
−1
pπ 2
z0
{
= − iω k 2 −
∂ x Bz , mnp
( )}
−1
pπ 2
mπ
z0
b
B0 sin ( mbπ x ) cos ( naπ y ) sin
( z)
pπ
z0
Similarly, Equation (7.2) leads to By , mnp = − ωi ∂ z Ex , mnp , and hence, via (7.4),
{
∂ y Bz , mnp = κ +
iγ 2
ω
}E
x , mnp
=
i
ω
{k − ( ) } E
2
pπ 2
z0
x , mnp
Thus,
{
Ex , mnp = iω k 2 −
( )}
−1
pπ 2
nπ
z0
a
B0 cos ( mbπ x ) sin ( naπ y ) sin
( z)
pπ
z0
and finally,
{
= − {k − ( ) }
Bx , mnp = − k 2 −
By , mnp
2
( )}
−1
pπ 2
mπ pπ
z0
b z0
B0 sin ( mbπ x ) cos ( naπ y ) sin
( z)
−1
pπ 2
nπ pπ
z0
a z0
B0 cos ( mbπ x ) sin ( naπ y ) sin
( z)
{
( )
pπ
z0
pπ
z0
( )}
Lastly, from k 2 = ( mbπ ) + ( naπ ) + z0 , σ = 0 ⇒ ω 2mnp = µε1 ( mbπ ) + ( naπ ) + z0
.
Observe that there are no cut-off frequencies, as the finite boundary allows
the waves to reflect back and forth. The fields attain a state of resonance
when ω = ω mnp . At resonance, there is no power transmitted as they are
stored reactively for the case wherein σ = 0.
□
2
2
pπ 2
2
2
pπ 2
7.3.2 Proposition
Given a homogeneous bounded space (Ω, μ, ε, σ), where ∂Ω is a perfect electrical conductor, an admissible time-harmonic TM to z-mode wave propagation
( E , B)|Bz = 0 satisfies:
Ez , mnp = E0 sin ( mbπ x ) sin ( naπ y ) cos
{
= − {k − ( ) }
Ex , mnp = − k 2 −
Ey , mnp
K15149_Book.indb 245
2
( )}
( z)
pπ
z0
(7.47)
−1
pπ 2
mπ pπ
z0
b z0
E0 cos ( mbπ x ) sin ( naπ y ) sin
−1
pπ 2
nπ pπ
z0
a z0
E0 sin ( mbπ x ) cos ( naπ y ) sin
( z)
pπ
z0
(7.48)
( z)
(7.49)
pπ
z0
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246
Electromagnetic Theory for Electromagnetic Compatibility Engineers
{
= −κ { k − ( ) }
Bx , mnp = κ k 2 −
By , mnp
2
( )}
−1
pπ 2
nπ
z0
a
B0 sin ( mbπ x ) cos ( naπ y ) cos
−1
pπ 2
mπ
z0
b
( z)
B0 cos ( mbπ x ) sin ( naπ y ) cos
pπ
z0
(7.50)
( z)
(7.51)
pπ
z0
where k 2 = ω 2 µε + iµωσ and κ = μσ − iωμε. Moreover, if σ = 0, there exists a
real sequence (ω mnp )m , n, p of angular frequencies such1 that the TM mnp mode
2
2
2 2
resonates in Ω, where ω mnp = 1µε ( mbπ ) + ( naπ ) + ( naπ ) .
{
}
Proof
□
The proof is left for the reader to establish; see Exercise 7.4.6.
It ought to be pointed out that within the rectangular cavity, only standing waves exist, as the waves are reflected from the walls of the cavity, and
hence, a cavity stores electromagnetic energy. However, if there exists an
aperture in the wall, current is induced around the aperture, and, as shown
in the following chapter, the aperture will radiate. Thus, from an EMC perspective, cavity resonance can be a source of annoyance. Clearly, if the walls
have finite conductivity, then the energy is lost over time as ohmic heat. This
leads to the concept of quantifying resonance within a cavity.
Before proceeding to quantify cavity resonance, the concept of surface
impedance is required. First, recall from (1.40), that B⊥ = − iωγ e z × E⊥ ⇒ H ⊥ =
iγ
2
− µω
e z × E⊥ . Hence, set η = iωµ
γ , where γ = i ω µε + iωµσ from (1.32). Then, η
defines the wave impedance for a TEM wave propagating in the z-direction,
ωµ
and η = 2
.
ω µε+ iωµσ
That η defines the wave impedance is not difficult to see via the following equivalent expression. Given a TEM to z-direction wave (Ex , H y ) , from
Chapter 1,
+
−
Ex = E + e− γz + E − e γz ⇒ H y = Eη e− γz − Eη e γz
∫
Intuitively, Ex ↔ V , H y ↔ I ⇒ η ↔ R, where V = − ∫ E ⋅ dl , I = H ⋅ dl and
R = VI .
Denote η = RS + iX S. As a particular instance, this applies to fields penetrating a conductor, and it is called the surface impedance of the conductor.
Indeed, noting trivially that
ω 2 µε + iωµσ = ω µε 1 +
setting
η0 =
µ
ε
,
1+
η = η0
K15149_Book.indb 246
iσ
ωε
= ξ + + iξ − yields ξ ± =
ξ + − iξ −
ξ 2+ +ξ 2−
{
σ
⇒ RS = η0 1 + ( ωε
)
}
1
2 −2
1
2
{
1+(
iσ
ωε
)
σ 2
ωε
}
±1
{
1
2
. Thus, putting
σ
ξ + , X S = −η0 1 + ( ωε
)
}
1
2 −2
ξ−
10/18/13 10:58 AM
247
Waveguides and Cavity Resonance
7.3.3 Lemma
The time-average power density loss as a TEM wave transmits into a conductive
 ⟩=−
boundary at z = z0 is given by ⟨W
2
1
2 η0
{
2
e z Ex + Ey
2
} {
1+(
z = z0
)
σ 2
ωε
}
+1
1
2
.
Proof
(
)
The time-average Poynting vector is ⟨S⟩ z = z0 = 21 ℜe E⊥ × H ⊥∗
. Thus, the
z = z0
 ⟩ z = z = −⟨S⟩ z = z via the conservation
time-average power density loss ⟨W
of
0
0
energy. Evaluating this expression yields:
{
} ℜe ( )
e {E + E }
e {E + E }
{ 1+(
 ⟩ = − 1 e z Ex 2 + Ey
⟨W
2
= − 21
z
= − 2 12
x
z
y
2
1
η∗
z = z0
2
2
x
2
z = z0
RS
RS2 + X S2
2
y
z = z0
1
η0
)
σ 2
ωε
}
+1
1
2
.
□
From the above lemma, it is clear that for a rectangular cavity Ω, the total
 ⟩ ⋅ nd 2 x.
time-average power loss is over the sum of the boundaries: ⟨W ⟩ = ∫ ∂Ω ⟨W
That is,
⟨W ⟩ =
∑∫
i
∂Ωi
 ⟩ ⋅ ni d 2 x = −
⟨W
2
1
2 η0
{E
x
2
+ Ey
2
} {
z = z0
}
σ
1 + ( ωε
) +1
2
where ∂Ωi is the i th -face of ∂Ω.
7.3.4 Definition
Let ωmnp be some fixed resonant angular frequency in a rectangular cavity
(Ω,μ, ε, σ) with (∂Ω, σ ∂Ω ) , and suppose that σ = 0 with 0 < σ ∂Ω < ∞. Furthermore,
let ⟨U ⟩ = ⟨U E ⟩ + ⟨U B ⟩ denote the time-average energy stored in Ω, where
U E = 21 D ⋅ E ∗ (U B = 21 B ⋅ H ∗ ) is the electric (magnetic) energy, and ⟨W ⟩ denotes
2
2
the time-average power dissipated in Ω, with W = 21 J S RS = 21 H ⊥ RS. Then,
U
the quality factor of (m, n, p)-mode is defined by Qmnp = ω mnp W .
7.3.5 Lemma
Given a rectangular cavity (Ω,μ,ε) with (∂Ω, σ ∂Ω ), the time-average power
ω
− 0t
loss satisfies ⟨W ⟩ = ⟨U 0 ⟩ e Q , where ω 0 is some fixed resonant angular
K15149_Book.indb 247
10/18/13 10:58 AM
248
Electromagnetic Theory for Electromagnetic Compatibility Engineers
frequency. In particular, the frequency distribution of the magnitude of the
electric field within the cavity in frequency domain is
E (ω ) =
1
2π
E0
{(
)
ω0 2
2Q
+ (ω − ω 0 )2
}
− 21
for some initial value E0.
Proof
By the conservation of energy, the negative time rate of change of stored electromagnetic energy U is precisely the energy loss on the cavity walls W. Whence,
ω
− 0t
U
from Definition 7.3.4, Q = ω 0 W ⇒ − ddt ⟨U ⟩ = ⟨W ⟩ = ωQ0 ⟨U ⟩ ⇒ ⟨U ⟩ = ⟨U 0 ⟩ e Q .
Thus, the energy decays exponentially in time. In particular, if the electric field E(t) = E0 eiω 0t initially, it will decay in the cavity according to
E(t) → E0 e
ω
0
− 2Q
t iω t
0
e
, as E ∝ U , whence, taking the Fourier transform of E(t),
E(t) =
∫
Substituting E(t) = E0 e
∞
0
E (ω )eiωt dt ⇒ E (ω ) =
ω
∫
1
2π
∞
0
E(t)e− iωt dt
0
− 2Q
t iω t
0
e
into the integral to obtain the electric field
−1
2
2
ω0 2
2
+
(
ω
−
ω
)
,
magnitude in the frequency domain yields: E (ω ) = 2Eπ0
0
2Q
as required. Show this in Exercise 7.4.7.
□
{(
}
)
7.3.6 Example
Given a lossless rectangular cavity (Ω,μ,ε) and (∂Ω,σ), determine the quality
factor Q for an arbitrary TEmnp mode at resonance.
Now, the time-average electric energy stored in Ω is
⟨U E ⟩ = 21 ε
∫
∫
2
Ω
z0
0
E d 3 x = 21 ε
sin 2
( ) B ( ) ∫ cos (
ω mnp
2
( z) dz + ( ) ∫
mπ 2
b
pπ
z0
nπ 2
a
2
0
2
ω mn
b
0
b
2
0
sin 2 ( mbπ x ) d x
∫
mπ
b
a
0
x ) dx
∫
a
0
sin 2 ( naπ y ) d y
cos 2 ( naπ y ) d y
∫
z0
0
sin 2
( z) dz 
pπ
z0
and hence,
}= ( ) B
at resonance, where for notational convenience, ω = {( ) + ( ) }.
⟨U E ⟩ = 21 ε
( ) B {(
ω mnp
2
ω mn
2
2
0
)
nπ 2 abz0
8
a
+ ( mbπ )
2 abz0
8
1
µ
2
mn
K15149_Book.indb 248
ω mnp 2
ω mn
1
µε
2 abz0
0 16
mπ 2
b
nπ 2
a
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249
Waveguides and Cavity Resonance
The time-average magnetic energy stored in Ω is
⟨U B ⟩ =
1
2µ
∫
Ω
2
B d3 x =
1
4
2 µω mn
( µε )2
∫
a
0
(
(

B02 

)∫
2
mπ pπ
a z0
cos 2 ( naπ y ) d y
)∫
2
nπ pπ
a z0
b
0
0
sin 2 ( mbπ x ) d x
( z) dz +
pπ
z0
sin 2
cos 2 ( mbπ x ) d x
0
4
ω mn
(µε)2
∫
z0
b
∫
b
0
∫
a
0
cos 2 ( mbπ x ) d x
∫
sin 2 ( naπ y ) d y
∫
a
0
z0
0
sin 2 ( naπ y ) d y
∫
sin 2
z0
0
2
µε +
and hence, noting trivially by definition that ω 2mnp µε = ω mn
⟨U B ⟩ =
1
4
2 µω mn
µε
B02
{ ((
1
µε
( )
ω
) + ( mbπ )
nπ 2
a
2
)( )
pπ 2 abz0
z0
8
}
4
+ ω mn
µε abz8 0 =
1
µ
( z) dz +
sin 2
pπ
z0
( z) dz 
pπ
z0
( ),
pπ 2
z0
( )B
ω mnp 2
ω mn
2 abz0
0 16
2
B02 abz8 0 is the total stored electromagnetic energy in the
Thus, ⟨U ⟩ = µ1 ωmnp
mn
cavity.
2
To complete the problem, the loss must be computed. From W = 21 H ⊥ RS ,
where RS = ℜeηS is the resistance of the surface ∂Ω, it follows that the time2
average energy loss by the conductive walls is ⟨W ⟩ = 21 ∫ ∂Ω H ⊥ RS d 2 x. Next,
noting that the energy loss on ∂Ω|x = 0 is identical to that of ∂Ω|x = b , and likewise for {y = 0, y = a} and for { z = 0, z = z0 }, it follows that
⟨W ⟩ =
∫
2
∂Ω|x = 0
H ⊥ RS d 2 x +
∫
2
∂Ω|y = 0
H ⊥ RS d 2 x +
∫
∂Ω|z = 0
2
H ⊥ RS d 2 x
Now, ∂Ω|x = 0 = [0, a] × [0, z0 ] ⇒ H ⊥ = ( H y , H z )x = 0 . Hence,
⟨W ⟩|∂Ω x = 0 = RS
1
µ2
RS
B02
1
µ2
(
B02
= RS
K15149_Book.indb 249
1
(
)
2
ω 2mnµε
a
∫∫
z0
0
0
1
µ2
B02
)∫∫
2
nπ pπ
a z0
a
0
z0
0
sin 2 ( naπ y ) sin 2
cos 2 ( naπ y ) sin 2
az0
4
 1
 ω 2 µε 2
 ( mn )
(
( z) dy dz +
pπ
z0
( z) dy dz
pπ
z0
)
2
nπ pπ
a z0

+ 1 .

10/18/13 10:58 AM
250
Electromagnetic Theory for Electromagnetic Compatibility Engineers
On the wall ∂Ω|y = 0 = [0, b] × [0, z0 ] ⇒ H ⊥ = ( H x , H z )y = 0 = (0, H z )x = 0. Hence,
⟨W ⟩|∂Ω y = 0 = RS
1
µ2
B02
b
∫∫
0
z0
0
cos 2 ( naπ y ) sin 2
( z) dy dz = R
pπ
z0
1
S µ2
B02
bz0
4
.
Lastly, ∂Ω|z = 0 = [0, b] × [0, a] ⇒ H ⊥ = ( H x , H y )z = 0 = (0, 0). Hence, ⟨W ⟩|∂Ω z = 0 = 0.
Thus,
⟨W ⟩ =
RS
µ2
B02
z0
4
 
 a 

(
nπ pπ
1
ω 2mnµε a z0
) + 1 + b 
2
yielding the desired quality factor at resonance:
Qmnp = ω mnp
U
W
= ω mnp
µ
RS
( ) (
ω mnp 2 b
ω mn
4



1
2
ω mn
µε
nπ pπ
a z0
)
2

+ 1 + ba 

−1
□
Now, for very good conductors, recall from Chapter 1 for a TEM wave that
γ = α + iβ ≈ δ1 (1 + i)
where δ =
7.1.3,
2
ωµσ
is the skin depth of the conductor, whence, from Definition
η=
iωµ
γ
= iωµδ 12− i = 21 ωµδ(1 + i) ⇒ RS = 21 ωµδ
and so,
Qmnp =
ω mnp µσ
2
( ) (
ω mnp 2 b
ω mn
2



nπ pπ
2
ω mn
µε a z0
1
)
2

+ 1 + ba 

−1
7.3.7 Remark
This chapter closes with a brief word on dielectric waveguides. A brief acc­
ount can be found in References [2,4]. For simplicity, consider a TM wave pro­
pagating within a lossless infinite rectangular dielectric slab (Ω, µ , ε) ⊂ (R 3 ,
µ 0 , ε 0 ) of finite thickness a: Ω = {( x , y , z) : x , z ∈R , 0 < y < a} with ∂ x = 0 assu­
med. Recall that a TM to z mode is characterized by Ez,n. Then, it is clear that
it must satisfy these equations simultaneously:
(a)
∆ ⊥ Ez + (ω 2 µε − β 2 )Ez = 0 on Ω
(b)
∆ ⊥ Ez + (ω 2 µ 0 ε 0 − β 2 )Ez = 0 on R 3 − Ω
where Ω = Ω ∪ ∂Ω.
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Waveguides and Cavity Resonance
For propagation to occur on Ω, it follows that the waves must be evanescent on R 3 − Ω, and hence, via the separation of variables, Ez = Φ( y )e− iβz with
Φ( y ) = E+ cos k y y + E− sin k y y on Ω, where k y2 = ω 2 µε − β 2, whereas on R 3 − Ω ,
 e−αy for y ≥ a

Φ( y ) ~  αy
for y ≤ 0
 e
where α 2 = β 2 − ω 2 µ 0 ε 0 > 0 . The following pair defines the dispersion relations
for the dielectric slab:
 k 2 = ω 2 µε − β 2
 y
 2
2
2
 α = β − ω µ 0 ε 0
□
7.4 Worked Problems
7.4.1 Exercise
(a) Establish that e z × (e z ∂ z × E⊥ ) = − ∂ z E⊥ ⇒ ∂ z E⊥ = − iωe z × B⊥.
(b) Show that a time-harmonic TE to z-mode wave propagation satisω
fies E⊥ = γ 2 +ω 2iµε+
∇ ⊥ × e z Bz .
iωµσ
Solution
(a) This is just a trivial exercise of directly evaluating the expression by
brute force:
e z ∂ z × E⊥ =
e z × (e z ∂ z × E⊥ ) =
ex
ey
ez
0
Ex
0
Ey
∂z
0
ex
ey
ez
0
− ∂ z Ey
0
∂ z Ex
1
0
 − ∂ z Ey

=  ∂ z Ex

0
 − ∂ z Ex

=  − ∂ z Ey

0






 = − ∂ z E⊥

Therefore, from Equation (7.10), e z × (7.10) ⇒ − ∂ z E⊥ = iωe z × B⊥ , as desired.
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252
Electromagnetic Theory for Electromagnetic Compatibility Engineers
(b) From Maxwell’s equations: ∇ × B = κE ⇒ ∇ ⊥ × B⊥ + e z × ∂ z B⊥ + ∇ ⊥ ×
e z Bz = κE⊥, where κ = μσ − iωμε. Because ∇ ⊥ × B⊥ = κEz e z = 0 , it follows that
e z × ∂ z B⊥ + ∇ ⊥ × e z Bz = κE⊥
Upon substituting the result from (a) into the above equation and noting
trivially that ∂ z e z × B⊥ = e z × ∂ z B⊥ as e z is a constant, it is evident that
− ∂2z E⊥ = iωe z × ∂ z B⊥ = iω {κE⊥ − ∇ ⊥ × e z Bz } ⇒ ∂2z E⊥ + iωκE⊥ = iω∇ ⊥ × e z Bz
Finally, observing that ∂ z e− γz = −γe− γz ⇒ ∂ z ↔ −γ , it follows that
E⊥ =
iω
iωκ + γ 2
∇ ⊥ × e z Bz =
1 iωµ
µ iωκ + γ 2
∇ ⊥ × e z Bz
□
7.4.2 Exercise
Fill in the details for the proof of Proposition 7.1.2.
Solution
Now, ∂ z E⊥ = − iωe z × B⊥ was established in Exercise 7.4.1(a). To complete the
details of the remaining proof, observe that ∂2z e− γz = γ 2 e− γz; that is, ∂2z ↔ γ 2,
whence,
∂2z E⊥ − iωκE⊥ = − iω∇ ⊥ × e z Bz ⇔ γ 2 E⊥ − iωκE⊥ = − iω∇ ⊥ × e z Bz
⇔ ( γ 2 − iωκ )E⊥ = − iω∇ ⊥ × e z Bz
So, recalling that κ = μσ − iωμε, it is clear that − iωκ = ω 2 µε − iωµσ , yielding
E⊥ = − − iωκiω+ γ 2 ∇ ⊥ × e z Bz = − µ1
as required.
iωµ
− iωκ + γ 2
∇ ⊥ × e z Bz = − µ1
iωµ
γ 2 +ω 2 µε− iωµσ
∇ ⊥ × e z Bz
□
7.4.3 Exercise
Prove Proposition 7.1.10.
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Waveguides and Cavity Resonance
Proof
For TM mode, Bz = 0. Thus, invoking the assumption that ∂x = 0,
∇×E=
∇×B=
ex
ey
ez
0
∂y
∂z
Ex
Ey
Ez
ex
ey
ez
0
∂y
∂z
Bx
By
0
 ∂ y Ez − ∂ z Ey

=
∂ z Ex

− ∂ y Ex

 − ∂ z By

=  ∂ z Bx
 −∂ B
y x


 Bx


 = − ∂t  By

 0


 Ex


 = µσ  Ey

 Ez







 Ex


 − µε ∂t  Ey

 Ez







Noting that ∇ ⋅ B = 0 ⇒ ∂ y By = 0 as ∂ x = 0 and Bz ≡ 0 , whence, ∂ y (fourth
equation) leads to 0 = µσ ∂ y Ex + µε ∂t ∂ y Ex ⇒ Ex is independent of y. Hence, in
order to satisfying the boundary condition Ex |∂Ω = 0 ⇒ Ex ≡ 0 . In particular,
the fourth equation leads immediately to ∂ z By = 0 ⇒ By is independent of z.
Hence, the only solution to satisfy this is By ≡ 0 .
To summarize for ease of reference, on setting κ = μσ − iωμε, ∂ z → −γ and
∂t → − iω , Maxwell’s equations reduce to
∂ y Ez + γEy = iωBx
∂ z Bx = κEy
− ∂ y Bx = κEz
Therefore, ∂z (second equation) –∂z (third equation) yields
∂2z Bx + ∂2y Bx = κ(∂ z Ey − ∂ y Ez ) = − iωκBx ⇒ −∆Bx + k 2 Bx = 0
where k 2 = ω 2 µε + iωµσ . The boundary condition is
∂ y Bx |∂Ω = 0 ⇔ ∂ y Bx |y = 0 = 0 = ∂ y Bx |y = a
Thus, via the separation of variables with Bx ( y , z) = Φ( y )Ψ( z) and imposing
the boundary conditions, the fundamental solutions are: Φ n ( y ) = cos naπ y and
2
Ψ( z) = e− γ n z, where γ 2n = ( naπ ) − ω 2 µε − iωµσ satisfies Equation (7.18). So, the
+
solution is thus Bx , n = B0, n cos ( naπ y ) e− γ n z .
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254
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Finally, the latter two equations yield, respectively,
Ey = − γκn Bx = − γκn B0,+ n cos ( naπ y ) e− γ n z
and
Ez = − κ1 ∂ y Bx = κ1 B0,+ n
nπ
a
sin ( naπ y ) e− γ n z
To complete the proof, it suffices to observe that as γ n for TM is identical
with that of TE, the properties derived for TE apply equally to TM.
□
7.4.4 Exercise
σ
Establish ηTM, n = η{1 + i ωε
}
−1
1−
( )
ωn 2
ω
σ
+ i ωε
of Corollary 7.1.11.
Solution
From Definition 7.1.9, ηTM,n = − µγκn , where κ = μσ − iωμε. Next, set γn =
iω µε 1 −
( )
ωn 2
ω
σ
+ i ωε
≡ iω µεγˆ n and noting that σ − iωε = − iωε ( 1 +
ηTM, n = − µγκn = iω µεγˆ n
1
iωε
(1 + ωεiσ )−1 =
µ
ε
(1 + ωεiσ )−1
1−
( )
ωn 2
ω
iσ
ωε
),
σ
+ i ωε
□
7.4.5 Exercise
Derive the TE modes for a semi-infinite hollow cylinder Ω = Ba (0) × [0, ∞),
where Ba (0) = {( x , y ) ∈R 2 : x 2 + y 2 < a 2 }, and assume that ∂Ω is a perfect electrical conductor. Furthermore, determine the wave impedance and waveguide wavelength for each fixed mode.
Solution
The appropriate coordinate system to be invoked here is the cylindrical coordinate system. As an aside, Equations (7.7) and (7.8) apply in all coordinate
systems. Hence, it suffices to express the Laplacian Δ in cylindrical coor­
dinates. Recalling that
∆Bz =
(
1
r
∂r + ∂r2 +
1
r2
)
∂φ2 Bz + ∂2z Bz ≡ ∆ rφ Bz + ∂2z Bz
it follows that (7.7) becomes 0 = ∆ rφ Bz + ( γ 2 + k 2 )Bz . So, attempt the separation
of variables method once again: set Bz = Ψ(r )Φ(φ) . Then, upon expanding
out the expression, dividing by Ψ(r)Φ(ϕ) and multiplying by r2,
1 ∂r Ψ
r Ψ
+
∂r2 Ψ
Ψ
+
2
1 ∂φ Φ
2
Φ
r
+ ( γ 2 + k 2 ) = 0 ⇒ r ∂rΨΨ + r 2
∂2r Ψ
Ψ
+ r 2 (γ 2 + k 2 ) = −
∂φ2 Φ
Φ
= n2
for some constant n2, as the left side of the equation is solely a function of r
and the center equation is solely a function of ϕ.
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255
Waveguides and Cavity Resonance
The general solution for Φ is obvious by now: Φ = A cos nϕ + B sin nϕ. The
differential equation
r ∂rΨΨ + r 2
∂2r Ψ
Ψ
+ r 2 ( γ 2 + k 2 ) − n2 = 0 ⇔
1 ∂r Ψ
r Ψ
+
∂r2 Ψ
Ψ
+ (γ 2 + k 2 ) −
n2
r2
=0
is known as Bessel’s equation. The general solution for Bessel’s equation is
well known:
Ψ = CJ n (( γ 2 + k 2 )r ) + DN n (( γ 2 + k 2 )r )
where
J n (qr ) =
∑
( −1)m ( qr )n + 2 m
n+ 2 m
m ≥ 0 m !( n + m )!2
is Bessel’s function of the first kind (there are other equivalent representations), and
N n (qr ) =
cos( nπ ) J n ( qr )− J − n ( qr )
sin nπ
For more details, see References [2,3,7,9].
Now, Φ is periodic in ϕ and hence, n ∈ Z for each n. In particular, sin nπ =
0 implies at once that D ≡ 0 ⇒ Ψ n = J n (( γ 2 + k 2 )r ) is a fundamental solution.
Because the coefficients A,B of Φ depend upon the choice of an arbitrary reference angle ϕ, it follows for simplicity that either A = 0 or B = 0 may be chosen without any loss of generality. So, let Φn = cos nϕ denote the fundamental
solution. Then, the general solution is: Bz , n = B0 J n (( γ 2 + k 2 )r )cos nφ .
Imposing the boundary condition on Bz , n ⇒ ∂r Bz , n |r = a = 0 . This is equivalent to:
2
2
∂r J n (( γ mn
+ k 2 )a) = 0 ⇔ γ mn
= pmn − k 2
where pmn a are the roots of Bessel’s function ∂r J n ( pmn a) = 0 ∀m, n . Thus,
γ 2mn = pmn − ω 2 µε − iωµσ
and hence, setting γ mn = α mn + iβ mn , following the proof of Proposition 7.1.4
mutatis mutandis yields

 α mn =



 β mn =

K15149_Book.indb 255
1
2
1
2
{
{(
(p
2
mn − ω µε
pmn − ω 2 µε
}
) + (ωµσ ) − (ω µε − p )
)
2
2
2
2
mn
+ ( ωµσ ) + ω 2 µε − pmn
2
}
1
2
1
2
10/18/13 10:59 AM
256
Electromagnetic Theory for Electromagnetic Compatibility Engineers
In particular, for σ = 0,
(a) ω 2 µε − pmn > 0 ⇒ α mn = 0 ⇒ e− γ mn z = e− iβmn z (traveling wave),
(b) ω 2 µε − pmn < 0 ⇒ β mn = 0 ⇒ e− γ mn z = e−α mn z (evanescent wave).
Finally, to determine the remaining fields, consider Maxwell’s equations for
TE mode:
∇×E=
∇×B=
1
r
1
r
er
reφ
ez
∂r
∂φ
∂z
Er
rEφ
0
er
reφ
ez
∂r
∂φ
∂z
Br
rBφ
Bz

− r ∂ z Eφ

= 
r ∂ z Er
 ∂r (rEφ ) − ∂φ Er

1
r

 Br


 = iω  Bφ

 Bz






 Er


 = κ  Eφ

 0





 ∂φ Bz − r ∂ z Bφ

=  − r ∂r Bz + r ∂ z Br
 ∂r (rBφ ) − ∂φ Br

1
r
whence, via the correspondence ∂ z ↔ −γ mn and observing that pmn = iωκ + γ 2mn ,
γ mnEφ , mn = iωBr , mn ⇒ Br , mn = − iγωmn Eφ , mn
−γ mnEr , mn = iωBφ , mn ⇒ Bφ , mn =
iγ mn
ω
Er , mn
and substituting the first equation into − ∂r Bz − γ mn Br , mn = κEφ , mn leads to:
{
− ∂r Bz = κEφ , mn + γ mn Br , mn = κ −
iγ 2mn
ω
{
2
Eφ , mn = − iω iωκ + γ mn
}E
φ , mn
}
−1
{
ω
∂r Bz = − pimn
∂r Bz
Similarly, substituting the second equation into
leads to:
1
r
{
∂φ Bz , mn = κEr , mn − γ mn Bφ , mn = κ −
{
iγ 2mn
ω
Er , mn = iω iωκ + γ 2mn
K15149_Book.indb 256
}
}E
r , mn
−1
}
= − ωi iωκ + γ 2mn Eφ , mn ⇒
1
r
∂φ Bz , mn + γ mn Bφ , mn = κEr , mn
{
}
= − ωi iωκ + γ 2mn Er , mn ⇒
∂r Bz =
iω
pmn
∂r Bz
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257
Waveguides and Cavity Resonance
In summary,
Bz , mn = B0 J n ( pmn r )cos(nφ)
mn
Br , mn = − γpmn
B0 (∂r J n ( pmn r ))cos(nφ)e− γ mn z
Bϕ , mn =
γ mn
pmn
B0 (∂r J n ( pmn r ))cos(nφ)e− γ mn z
ω
Eφ , mn = − pimn
B0 (∂r J n ( pmn r ))cos(nφ)e− γ mn z
Er , mn =
iω
pmn
B0 (∂r J n ( pmn r ))cos(nφ)e− γ mn z
Regarding the wave impedance, by appealing to Definition 7.1.3, and defining ω 2mnµε = pmn ,
ηTE, mn =
iµω
γ mn
µ
ε
=
{
1−
( )
ω mn 2
ω
+
( )
iσ
ωε
}
− 21
2
iσ
where γ mn = pmn − ω 2 µε − iωµσ = iω µε 1 − ωωmn + ωε
; see the wave impedance for a rectangular waveguide in Corollary 7.2.2. Specifically, Corollary
7.2.2 applies equally to a cylindrical waveguide, and it is clear that ω mn
defines the cut-off angular frequencies for a lossless cylindrical waveguide.
Lastly, the determination of the wavelength within the guide follows that of Theorem 7.1.12 identically. Set Θ = ωt + β mn z . Then,
uˆ mn ≡ ddzt = βωmn = fλ mn ⇒ λ mn = β2mnπ . Thus,
λ mn = 2 2 π
{(
{
(p
=
2
f
v
2
pmn − ω µε
2
mn − ω µε
whenever ω > ω mn , where v =
1
µε
) + (ωµσ )
2
) + (ωµσ )
2
2
and ω = 2πf.
2
2
+ ω µε − pmn
+ ω 2 µε − pmn
}
}
− 21
− 21
□
7.4.6 Exercise
Establish Proposition 7.3.2.
Solution
Invoking the separation of variables, Equation (7.20) leads to Ez = Φ( x)Ψ( y )Θ( z),
subject to the boundary condition Φ|x = 0,b = 0 = Ψ|y = 0, a, whence, Φ = sin ( mbπ x )
and Ψ = sin ( naπ y ) are the desired fundamental solutions, as should be
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258
Electromagnetic Theory for Electromagnetic Compatibility Engineers
apparent from the proofs given variously in Sections 7.2 and 7.3. In particular,
this leads to: − k x2 − k y2 + γ 2 + k 2 = 0 , where k x = mbπ , k y = naπ . Furthermore, as Ez
is reflected between z = 0 and z = z0 , and Θ|z = 0, z0 ≠ 0 (as there exists a surface
current induced on the boundary), it follows that Θ = Ae− γz + Be γz ⇒ B ≠ − A .
pπ
That is, Θ ≠ sin z0 . Indeed, as the reflection coefficient of a perfect conductor
is −1, it follows that in a lossless medium, B ≡ A and hence, yielding the funpπ
pπ
damental solution Θ = cos z0 z . Thus, Ez , mnp = E0 sin ( mbπ x ) sin ( naπ y ) cos z0 z .
For ease of reference, Maxwell’s six equations are rewritten below for TM
mode (see Exercise 7.4.3) and labeled, respectively, from (1)–(6):
( )
( )
( )
∂ y Ez − ∂ z Ey = − ∂t Bx , − ∂ x Ez + ∂ z Ex = − ∂t By , ∂ x Ey − ∂ y Ex = 0
− ∂ z By = κEx , ∂ z Bx = κEy , ∂ x By − ∂ y Bx = κEz
( ),
(iωκ − ( ) )
Then, following the proof of Proposition 7.3.1, and noting that ∂2z ↔ −γ 2 =
substituting (5) into (1) leads to: ∂ y Ez = iωBx + ∂ z ( κ1 ∂ z Bx ) =
Bx =
1
κ
(κ − ( ) ) B . Thus,
B
= κ {k − ( ) }
2
pπ 2
z0
1
κ
pπ 2
z0
pπ 2
z0
x
−1
pπ 2
nπ
z0
a
2
x , mnp
E0 sin ( mbπ x ) cos ( naπ y ) cos
( z)
pπ
z0
Likewise, substituting (4) into (2) yields:
− ∂ x Ez = iωBy + ∂ z
That is,
{
By , mnp = −κ k 2 −
(
1
κ
)
∂ z By =
( )}
−1
pπ 2
mπ
z0
b
1
κ
(k − ( ) )B
pπ 2
z0
2
y
E0 cos ( mbπ x ) sin ( naπ y ) cos
( z)
pπ
z0
It is now trivial, via (4) and (5), respectively, to evaluate E⊥ :
{
= − {k − ( ) }
Ey , mnp = − k 2 −
Ex , mnp
2
( )}
−1
pπ 2
mπ pπ
z0
b z0
E0 cos ( mbπ x ) sin ( naπ y ) sin
( z)
−1
pπ 2
nπ pπ
z0
a z0
E0 sin ( mbπ x ) cos ( naπ y ) sin
( z)
pπ
z0
pπ
z0
( )
pπ 2
Finally, observe that for a lossless medium, σ = 0, and k 2 − z0 reduces to
( mbπ )2 + ( naπ )2, and in particular, the resonant angular frequencies are
2
ω mnp
=
by construction.
K15149_Book.indb 258
1
µε
{(
) + ( naπ )2 + ( pzπ )
mπ 2
b
0
2
}
□
10/18/13 10:59 AM
259
Waveguides and Cavity Resonance
7.4.7 Exercise
2
Establish that E (ω ) =
E0 2
2π
{( )
ω0 2
2Q
+ (ω − ω 0 )2
frequency ω 0 , given that E(t) = E0 e
ω
0
− 2Q
t iω t
0
e
.
} , for some fixed resonance
−1
Proof
From E (ω ) =
1
2π
E (ω ) =
∫ ∞0 E(t)e− iωt dt , it is clear that
1
2π
E0
∫
0
= − 21π E0
where K =
ω0
2Q
2
E (ω ) =
=
as desired.
∞
{
e
ω
0
− 2Q
t iω t − iωt
0
ω0
2Q
e
e
dt
}
+ i(ω − ω 0 )
−1
∞
e− Kt t
=0
=
1
2π
E0
{
ω0
2Q
}
+ i(ω − ω 0 )
−1
+ i(ω − ω 0 ). Hence,
1
2π
1
2π
E0
E0
2
{
2
ω0
2Q
}{
− i(ω − ω 0 )
{( )
ω0
2Q
2
ω0
2Q
+ (ω − ω 0 )2
}
}{( )
+ i(ω − ω 0 )
ω0
2Q
2
+ (ω − ω 0 )2
}
−2
−1
□
References
1. Balanis, C. 1989. Advanced Engineering Electromagnetics. New York: John Wiley
& Sons.
2. Chang, D. 1992. Field and Wave Electromagnetics. Reading, MA: Addison-Wesley.
3. Farlow, S. 1993. Partial Differential Equations for Scientists and Engineers. New
York: Dover.
4. Jackson, J. 1962. Classical Electrodynamics. New York: John Wiley & Sons.
5. Neff, H. 1981. Basic Electromagnetic Fields. New York: Harper & Row.
6. Orfanidis, S. 2002. Electromagnetic Waves and Antenna. Rutgers University, ECE
Dept., http://www.ece.rutgers.edu/~orfanidi/ewa/.
7. Plonsey, R. and Collin, R. 1961. Principles and Applications of Electromagnetic
Fields. New York: McGraw-Hill.
8. Silver, S. 1949. Microwave Antenna Theory and Design. New York: McGraw-Hill.
9. Wylie, C., Jr., 1960. Advanced Engineering Mathematics. New York: McGraw-Hill.
K15149_Book.indb 259
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8
Basic Antenna Theory
An antenna is essentially a mechanism by which electromagnetic waves
are transmitted or received. It operates on the principle that an accelerating charge particle radiates. A broadcasting antenna comprises a waveguide that carries electromagnetic energy from the source, generating
fields that radiate out into space. The notion of electric dipoles and the
magnetic dipoles introduced in Chapter 1 comprise the building blocks in
antenna analysis [3, 6–8]. More advanced theory on antennae can be found
in References [2, 9].
An important aspect of antenna theory is the radiation field from apertures. This has direct relevance to EMC engineers from a compliance perspective, to wit, radiation escaping from apertures in chassis enclosing
printed circuit boards. The analysis essentially follows from the application
of diffraction theory. Excellent accounts of diffraction theory can be found in
References [4, 5, 10].
8.1 Radiation from a Charged Particle
In Chapter 1, the existence of electromagnetic waves was established.
However, the question regarding their generation was left unanswered.
How is electromagnetic radiation generated? To understand the concept at
an intuitive level, consider a point charge in vacuum. An electric field is genq
erated by the presence of the point charge via Coulomb’s law: E(r ) = 4 πε1 0 r 3 r,
where q, r are the electric charge and distance away from the point charge,
respectively (taken to be located at the origin).
On the other hand, via Biot–Savart’s law, the magnetic field generated by a
moving point charge is B = μεv × E, where v is the velocity of the point charge
in (Ω, µ , ε), Ω ⊆ R 3 . Because a magnetic field forms loops (cf. Chapter 1) and
the magnetic field vector is normal to both the direction of propagation and
the electric field according to v × E, magnetic loops can be envisaged as circulating around the velocity vector.
The above discussion suggests that in order for radiation to be generated,
a charge particle must undergo an acceleration, as (i) a static magnetic field
261
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262
Electromagnetic Theory for Electromagnetic Compatibility Engineers
does not propagate in space, (ii) a static particle cannot generate a magnetic
field (relative to some fixed reference frame), and (iii) a charged particle traveling at some constant velocity only generates a static magnetic field. Thus, by
elimination, acceleration appears to be the only candidate. Before proceeding
further with the discussion, a fundamental result from Chapter 1 is recalled
below.
8.1.1 Lemma
The electromagnetic field of a charged particle moving at some fixed velocity
v can be completely characterized by the pair (A,φ), where A defines a vector
potential and φ a scalar potential associated with the charged particle.
Proof
The assertion follows trivially from Definition 1.2.3 and Theorem 1.3.1. To wit,
given (A, φ), (E, B) can be uniquely derived via E = −∇ϕ − ∂t A and B = ∇ × A. □
8.1.2 Proposition
An accelerating charged particle generates electromagnetic radiation.
Proof
From B = μεv × E, it follows that 0 ≠ µε ( dtd v ) × E = ∂t B – µεv × ∂t E ⇒ ∂t A ≠ 0.
The result is now evident from Equations (1.15) and (1.17); that is, a timevarying electric field resulting from an accelerating charge is related to a
time-varying magnetic field and vice versa, yielding electromagnetic waves;
see Equations (1.28) and (1.29). Indeed, it is obvious that ∂t A ≠ 0 ⇔ ddt v ≠ 0
and hence, dtd v = 0 ⇔ B is a static field.
□
Upon examining the proof of Proposition 8.1.2 more carefully, it is clear
that ∂t A ≠ 0 generates both a time-varying electric field and a time-varying
magnetic field. When a charged particle is not accelerating, the fields are
attached to the particle. On the other hand, upon accelerating, the fields
are detached from the charged particle; the detached field manifests as
radiation! In this sense, it is not precisely correct to think of a time-varying
electric field generating a time-varying magnetic field and vice versa. The
correct perspective is the following: the fields are generated by in accelerating charge* via ∂t A ≠ 0.
In order to consider radiation, relativistic effects must be taken into
account. Without going into detail, with respect to a fixed laboratory frame,
* Strictly speaking, the transformation between an electric field and a magnetic field is a relativistic effect; recall from a footnote in Chapter 1 that the electric and magnetic fields comprise an entity called the electromagnetic field.
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Basic Antenna Theory
the scalar potential and the vector potential of a moving charged particle of
charge q and velocity v are, respectively [11],
ϕ=
q
1
1
4 πε 0 1− v r − r 0
c
and
A=
q
1
1
4 πε 0 1− v r − r0
c
v
These potentials are called the Liénard–Wiechert potentials.
Now, recalling that E = −∇ϕ − ∂t A, it is can be shown (cf. Exercise 8.6.1) that
the electric field can be decomposed into a term involving the acceleration
and a term involving the static term:
E = − 4 πε0 ( 1 −
q
)
v −2 1 d
c
cr dt
v+
q
4 πε 0
(1 − vc )−1 rv v − ∇ϕ ≡ Eaccel + Estatic
2
where
Eaccel = − 4 πε0 ( 1 −
q
)
v −2 1 d
c
cr dt
v and
Estatic =
q
4 πε 0
(1 − vc )−1 rv v − ∇ϕ
2
The key point to note here is that Eaccel ∝ 1r whereas Estatic ∝ r12 . Hence, for
large r > 0, Eaccel >> Estatic. Once again, in the absence of acceleration, only
the static term remains. In the presence of nonzero acceleration, in the farfield regime, the field detaches, as it were, from the particle and propagates
outward. Indeed, it can also be shown [8, 11] for a radiation field (E, B), E⊥B
and the field is thus TEM, and hence, the saying that an accelerating charged
particle radiates.
8.2 Hertzian Dipole Antenna
It was demonstrated above that an accelerating charge generates electromagnetic waves. Using this principle, an antenna can be constructed by considering an antenna to be made up of differential antenna elements; each element
approximates a charged dipole called a Hertzian dipole.
8.2.1 Definition
Consider two charged point particles of charge +q and −q separated by a constant distance d. The pair of charged particles defined by (±q, d) constitutes
an electrostatic dipole (or more simply an electric dipole), where d is the vector
pointing from −q to +q and ||d||= d .
The electrostatic field profile for an electric dipole was worked out in
Example 1.1.1. Another useful concept to know is the following. Given an electric dipole (±q, d), define an electric dipole moment by p = qd. As an example, suppose a dipole oscillates at an angular frequency ω; then q(t) =||e
q iωt, whence,
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
i(t) = ddt q(t) = iω q eiωt and p = − i ωI d , where i(t) = Ieiωt . Indeed, the potential of
an electric dipole can now be expressed in terms of the dipole moment as
V=
1 1
4 πε r 2
p⋅r
where the origin is taken to be the center of the dipole axis and r is the displacement of an arbitrary point from the origin.
8.2.2 Definition
A Hertzian dipole is an electric dipole (±q, d, ω), where d << 1, such that the
charge oscillates at an angular frequency ω along the distance d between
±q. If δzn denotes the differential length along a thin straight conductor* carrying a time-harmonic current per unit length J = J(z; ω), where n = JJ is a
unit vector, then, a Hertzian antenna is defined to be the triple (±q, δzn, J(z; ω)),
where the oscillating current element δI(z; ω) = J(z; ω)δz.
In what follows, suppose without loss of generality that the current I = I(z;ω)
is time harmonic because in most physical applications, the current has a
Fourier series which is, in essence, the infinite sum of time-harmonic currents.
A physical antenna may be approximated by the sum of Hertzian antennae,
or more precisely, may be expressed as the integral over the Hertzian antennae. Indeed, Definition 8.2.2 may be extended to a conductor with a finite
(nonvanishing) cross-section. Then, the current per unit length is replaced
more generally by a current density over a surface area or volume.
8.2.3 Lemma
Suppose without loss of generality that the center of a Hertzian dipole
( ± q , δze z , J ( z ; ω )) is the origin and the current is time harmonic. Then, the
vector potential A = A(r) of the Hertzian dipole is given by
A(r ) =
µ Iδz
4π r
e− iβr e z
(8.1)
where r is the distance of an arbitrary point from the center of the dipole.
Proof
From A(r ) = 4µπ ∫ V J (r ′)e− iβ r − r ′ r −1r ′ d 3 r ′ , set J (r ′) = I ( z ′)δ( x ′)δ( y ′)e z with r′ =
(0, 0, z) along the z-axis connecting the two charges ±q; δ(u) is the Diracdelta distribution satisfying:
 0 if t ≠ 0,
δ(t) = 
 ∞ if t = 0,
* Here, thin approximates a one-dimensional conductor, that is, a conductor with vanishingly
small cross-sectional area.
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Basic Antenna Theory
and ∫ ∞−∞ δ(t)dt = 1 . Furthermore, note that the integral ∫ V d 3 r ′ = ∫ ε−ε dx ′ ∫ ε−ε dy ′
1
δz
∫ −2 1 δz dz ′ , where V is the 1-dimensional current element of length δz and ε > 0
2
is arbitrary. Indeed, because of the definition of the Dirac-delta distribution,
1
δz
the integral can also be defined by ∫ V d 3 r ′ = ∫ ∞−∞ dx ′ ∫ ∞−∞ dy ′ ∫ −2 1 δz dz ′. Finally
2
observe that on setting |||
r|= r = x 2 + y 2 + z 2 (in rectangular coordinates)
and noting that
||r − r ′||= ( x − x ′)2 + ( y − y ′)2 + ( z − z ′)2
(
= x 2 + y 2 + z 2 + x ′ 2 + y ′ 2 + z ′ 2 − 2( xx ′ + yy ′ + zz ′)
= r 1−
2( xx ′+ yy ′+ zz ′ )
r2
+ ( rr′ )
)
2
it follows that
r ′ << r ⇒||r − r ′||≈ r 1 −
2( xx ′+ yy ′+ zz ′ )
r2
{
≈ r 1−
xx ′+ yy ′+ zz ′
r2
}
where the binomial expansion 1 + ε ≈ 1 + 21 ε for |ε|<< 1 was applied.
Thus, substituting the above approximation into the vector potential integral yields
A(r ) = e
µ
z 4π
Iδ( x ′)δ( y ′)e
∞
∞
1
2 δz
−∞
−∞
− 21 δ z
∫ ∫ ∫
∫
1
2 δz
≈ ez
µ
4π
I
≈ ez
µ
4π
I
≈ ez
µ I
4π r
= ez
µ I 2r
4 π r βz
− 21 δ z
dz′
1
2 δz
∫
− 21 δ z
∫
1
2 δz
− 21 δ z
1
r
∫
∞
−∞
δ( x ′)d x ′
∞
−∞
δ( y ′)e
{x
′
− iβr 1− 2 zz
2
r
2
+ y 2 + ( z − z ′ )2
1
r
{1 − }
2 zz ′
r2
−1
}
− 21
dx ′dy ′dz ′
dy ′
{1 + } e { } dz′
2 zz ′
r2
e
− iβr 1− zz2′
r
{ } dz′
− iβr 1− zz2′
r
e− iβr sin β2zδrz ≈
up to first order in
∫
− iβ x 2 + y 2 + ( z − z ′ )2
1
r
µ Iδz
4π r
e− iβr e z
, where the approximation sin θ ≈ θ for ||
θ << 1 was used. □
8.2.4 Theorem
Suppose without loss of generality that the center of a Hertzian dipole
( ± q , δze z , J ( z ; ω )) is the origin and the current is time
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
J (r ′) = I ( z ′)δ( x ′)δ( y ′)e z . If the dipole is placed in a homogeneous dielectric
medium (R 3 , µ , ε), then the electric field and magnetic field of the dipole are
given in spherical coordinates by
i
Er (r , θ) = − 2 πωε
i
Eθ (r , θ) = − 4 πωε
Bφ (r , θ) =
Iδz
r
Iδz
r2
e− iβr cos θ { 1r + iβ}
{
e− iβr sin θ −β 2 +
µ Iδz
4π r
1
r2
+ i βr
(8.2)
}
(8.3)
e− iβr { 1r + iβ} sin θ
(8.4)
Proof
Figure 8.1 shows a vector r represented in two different coordinate systems:
the rectangular coordinates and spherical polar coordinates. The transformations between rectangular and spherical coordinates are given by
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
Hence, dz = ∂r zdr + ∂θ zdθ = cos θdr − sin θrdθ. That is, e z = cos θer − sin θeθ .
From Proposition 8.2.3, expressing the vector potential in terms of spherical coordinates yields:
Ar = Az cos θ =
µ Iδz
4π r
Aθ = − Az sin θ = − 4µπ
e− iβr cos θ
Iδz
r
e− iβr sin θ
z
r
θ
φ
y
(x, y, z)
(r, θ, φ)
Rectangular
coordinate
system
Spherical
coordinate
system
x
Figure 8.1.
Transformation between rectangular and spherical coordinates.
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267
Basic Antenna Theory
Observe that Aϕ = 0 as there is no ϕ-dependency. Moreover, note that in
1
spherical polar coordinates, ∇ = (∂r , 1r ∂θ , r sin
θ ∂φ ) . Hence,
eθ
er
∇× A=
=
∂r
1
r
Ar
Aθ
eφ
∂θ
1
r sin θ
∂φ
=
1
r
0
er
reθ
r sin θ eφ
∂r
∂θ
∂φ
Ar
rAθ
0
1
r 2 sin θ
er
reθ
eφ
∂r
∂θ
1
r sin θ
Ar
rAθ
∂φ
0
= 1r {∂r (rAθ ) − ∂θ Ar }eφ
Now,
∂θ Ar =
µ Iδz
4π r
e− iβr cos θ = − 4µπ
Iδz
r
e− iβr sin θ and ∂r (rAθ ) = iβ 4µπ Iδze− iβr sin θ
So, this gives
∇× A=
µ Iδz
4π r
e− iβr { 1r + iβ} sin θ eφ
.
Hence, from B = ∇ × A,
Bφ =
µ Iδz
4π r
e− iβr { 1r + iβ} sin θ eφ
Next, via ∇ × B = iωμεE as the medium is lossless, it follows that
i
E = − µωε
∇ × B. Hence,
er
∂r
∇×B=
0
=
1
r 2 sin θ
eθ
1
r
∂θ
0
eφ
1
r sin θ
∂ϕ
=
1
r 2 sin θ
Bφ
er
reθ
r sin θ eϕ
∂r
∂θ
∂ϕ
0
0
(r sin θ)Bφ
{∂θ ((r sin θ)Bφ )e r − r ∂ r ((r sin θ)Bφ )eθ }
So, substituting
∂θ (r sin θH φ ) =
∂r (r sin θH φ ) =
K15149_Book.indb 267
I δze− iβr { 1r + iβ}
2 sin θ cos θ
0
4π
sin 2 θ
0
4π
{
I δze− iβr −β 2 +
1
r2
+ i βr
}
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
results in
i
i
E = − ωµε
∇ × B = − 4 πωµε

2 cos θ 1
r { r + iβ}

Iδz − iβr 
sin θ −β 2 + r12 + i βr
r e


0
{





}
□
as required.
β
r
Now, observe trivially that βr << 1 ⇔ <<
o(r2). Hence,
i
Er (r , θ) ≈ − 2 πωε
i
Eθ (r , θ) = − 4 πωε
Iδz
r
Bφ (r , θ) =
i
{1 − iβr } cos θ { 1r + iβ} ≈ − 2 πωε
Iδz
r2
{
{1 − iβr } sin θ −β 2 +
µ Iδz
4π r
Conversely, βr << 1 ⇔ βr <<
i
Er (r , θ) = − 2 πωε
i
Eθ (r , θ) = − 4 πωε
Bφ (r , θ) =
Iδz
r
1
r2
Iδz
r3
cos θ
}
i
+ i βr ≈ − 4 πωε
{1 − iβr } { 1r + iβ} sin θ≈
1
r2
Iδz
r2
µ Iδz
4π r2
Iδz
r3
sin θ
sin θ
, and so,
e− iβr cos θ { 1r + iβ} ≈
{
e− iβr sin θ −β 2 +
µ Iδz
4π r
and βr << 1 ⇒ e− iβr ≈ 1 − iβr +
1
r2
1
r2
β Iδz
2 πωε r 2
}
+ i βr ≈
e− iβr { 1r + iβ} sin θ≈ iβ 4µπ
e− iβr cos θ
iβ 2 Iδz
4 πωε r
Iδz
r
e− iβr sin θ
e− iβr sin θ
8.2.5 Definition
Given a Hertzian dipole ( ± q , δze z , J ( z ; ω )) where J( r′) = I ( z ′)δ( x ′)δ( y ′)e z , the
near field (or near zone) is defined by the criterion βr << 1. In particular, the
electromagnetic field in the near-field regime to first order in 1r satisfies:
i
Er (r , θ) ≈ − 2 πωε
Iδz
r3
cos θ
(8.5)
i
Eθ (r , θ) ≈ − 4 πωε
Iδz
r3
sin θ
(8.6)
Bφ (r , θ) ≈
µ Iδz
4π r2
sin θ
(8.7)
The far field (or far zone) is defined by the criterion βr >> 1. In particular, the
electromagnetic field in the far-field regime, up to first order in 1r , satisfies:
Eθ (r , θ) ≈
Bφ (r , θ)≈
K15149_Book.indb 268
iβ 2 Iδz
4 πωε r
iβµ Iδz
4π r
e− iβr sin θ
e− iβr sin θ
(8.8)
(8.9)
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Basic Antenna Theory
Finally, a zone that is neither a near zone nor far zone is called an intermediate zone.
Notice from Equations (8.5)–(8.7) in the near zone that E ∝ r13 and B ∝ r12 .
Thus, the electric field is much stronger than the magnetic field in the near
1
zone. Specifically, ωε1r 3 >> rµ2 ⇒ ωµε
>> r ; the electric field is much stronger
than the magnetic field, where ω,μ,ε > 0 are fixed. On the other hand, in the
far zone, (8.8) and (8.9) reveal that the fields resemble that of a plane wave: the
electric field and magnetic field for a plane wave are related by Bφ (r , θ) = µη Eθ ,
where β = ω µε was utilized.
Now, recall that the time-average power absorbed by a resistor is given
by ⟨ P⟩ = 21 ||
I 2 R . This is the energy dissipated by the load. In an analogous
manner, the power absorbed by the intervening medium (dielectric) as
radiation propagates across the medium is the energy dissipated from the
source. Call this equivalent load the radiation resistance. In particular, this
means that more power is dissipated by the medium if its radiation resistance is high.
Intuitively, an antenna may be thought of as an energy storage device inasmuch as it supports standing waves; the radiation resistance then affords
a means whereby the energy can be extracted from the standing waves by
transforming them into traveling waves. In the case of a Hertzian dipole, let
⟨ P⟩ denote the time-average outward power flow from a Hertzian dipole. If
I is the time-harmonic current flowing along the dipole, then the radiation
P
resistance of the Herzian dipole is given by R = 2 I 2 .
8.2.6 Proposition
The radiation resistance of a Hertzian dipole ( ± q , δze z , J ( z ; ω )) , where
J (r ′) = I ( z ′)δ( x ′)δ( y ′)e z , in a homogeneous medium (μ,ε) is given by R = 6ηπ (βδz)2 ,
where δz is the length of the Hertzian dipole and η = µε .
Proof
First, recall that the time-average power is defined via the Poynting vector as
⟨S⟩ = 21 Re( E × H * ). Explicitly,
er
E × H * = − ωεi ( 41π
)
δz 2
r
I
2
{
2 cos θ { 1r + iβ}
sin θ −β 2 +
0
= − ωiε (
K15149_Book.indb 269
)
2
1 δz
4π r
eθ
1
r2
+ i βr
0
{

sin 2 θ { 1r − iβ} −β 2 + r12 + i βr

2
I 
sin 2θ { 1r − iβ}{ 1r + iβ}


0
}
}
eϕ
0
{ 1r − iβ} sin θ





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Electromagnetic Theory for Electromagnetic Compatibility Engineers
After some tedious simplification, the components of the Poynting vector are:
{
I {
( ) ||
( E × H * )r = − sinωε θ ( 4δzπ )
2
( E × H * )θ = − i sinωε2 θ
2
}
+β }
(8.10)




(8.11)
||
I 2 i r13 − β 3
1
r2
δz 2 1
4π
r2
2
1
r2
2
From this,
⟨S⟩ =
2
sin θ
2 ωε
( )
δz 2 1
4π
r2
 β3
|I|  0
 0
2
The time-average power ⟨ P⟩ of the dipole is the power radiated radially outwards. That is,
⟨ P⟩ =
Therefore, R = 2
P
I2
=
2π
∫ ∫
0
π
0
⟨S⟩ ⋅ r̂r 2 sin θ dθ dφ
=
1
2 ωε
( 4δzπ )
=
1
2 ωε
( 4δπz )
=
1
12 πωε
1
6 πωε
2
2
2
I β3
2
I β3
2π
∫ ∫
0
2π
12
π
0
sin 3 θ dθ dφ
[ cos 3θ − 9 cos θ]0π
δz 2 ||
I 2 β3
β 3 δz 2 =
η
6π
(βδz)2, where η = µ/ε , as required.
□
It is clear that the radiation resistance of a Hertzian dipole is very small
as δz << 1. In particular, a Hertzian dipole is a poor radiator unless βδz >> 1;
that is, ω >> δz 1µε . Hence, in the microwave range, a Hertzian dipole makes
a poor antenna.
8.3 Magnetic Dipole Antenna
In Section 8.2, an elementary open-ended antenna in terms of an electric
dipole was considered. A closed-ended antenna forms a loop: indeed, this
forms the basis for a loop antenna. An elementary loop antenna is defined
via a magnetic dipole. Informally, this can be viewed as a charged particle of
charge q traversing around a small loop γ with a constant velocity v, where γ
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271
Basic Antenna Theory
spans a differential surface area δS( γ ) ⊂ R 2 and nγ is the unit vector normal
to the area δS(γ); see Example 1.2.4.
8.3.1 Definition
A magnetic dipole is the triple (q , γ , ωnγ ), where nγ = ω1 r × v is the unit normal
on the surface δS( γ ) ⊂ R 2 spanned by γ, and ω is the angular velocity around
γ. The magnetic moment of a magnetic dipole is defined by m = qω|δS( γ )|nγ ,
where |δS( γ )| denotes the surface area of δS(γ).
By way of an example, consider an ideal current loop γ of radius δr with
a time-harmonic current I = I(ω) flowing around the loop, where δS( γ ) ⊂ R 2 .
Then, the magnetic dipole moment of the current loop is m = Iπδr 2 e z . This follows from the fact that I = qω, |δS( γ )|= πδr 2 and nγ = e z . In view of this result,
a magnetic dipole for a current I flowing around an arbitrary loop γ is defined
by the triple (I,γ,δS(γ)), where δS( γ ) =
dipole antenna.
1
2
∫ r × dr . This is also called a magnetic
γ
8.3.2 Proposition
Given a differential magnetic dipole (I,γ,δS(γ)) in an isotropic homogeneous
medium (μ,ε), that is, |δS( γ )|<< 1, the vector potential in spherical coordinates may be approximated by A(r , θ) = µ4mπ e− iβr sin θ r12 {1 + iβr } eφ , where
m = I δS( γ ) .
Proof
Recall that A(r , θ) =
µ
4π
I0
∫
γ
1
R
e− iβR dl for a one-dimensional current-carrying
loop. Next, consider γ to be a planar loop of radius δr, that is, a loop embedded in R 2. Without loss of generality, one may assume that the center of γ
coincides with the origin. In Cartesian coordinates,
dl = δrdφeφ = (− e x sin φ + e y cos φ)δrdφ
R = r − δre r = ( x − δr cos φ)e x + ( y − δr sin φ)e y + ze z
Now, noting that r 2 = x 2 + y 2 + z 2 , and using the fact that r >> δr,
R = ( x − δr cos φ)2 + ( y − δr sin φ)2 + z 2
= r 2 − 2 xδr cos φ − 2 yδr sin φ + o(δr 2 )
≈ r 1−
2 xδr
r
Therefore,
1
R
K15149_Book.indb 271
≈
1
r
cos φ −
{1 +
xδr
r2
2 yδr
r
sin φ
cos φ +
yδr
r2
}
sin φ
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
and hence
dl
R
≈
1
r
{1 +
xδr
r2
cos φ +
yδr
r2
} {− e
sin φ
x
}
sin φ + e y cos φ δrdφ
Hence, evaluating the following integrals:
∫ {sin φ +
xδr
r2
sin φ cos φ +
∫ {cos φ +
xδr
r2
cos 2 φ +
2π
0
2π
0
yδr
r2
yδr
r2
}
yδr
}
xδr
r2
sin 2 φ dφ =
sin φ cos φ dφ =
r2
π
π
In terms of spherical coordinates (cf. Example A.1.6 of the Appendix),
x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ
it follows that
∫
γ
dl
R
2
≈ − πδr 2r sin θ sin φe x +
πδr 2
r2
sin θ cos φe y
It thus remains to transform (e x , e y ) into spherical coordinates. So, referring to the Appendix, given any vector v in some coordinate basis (e1 , e 2 , e 3 ),
that is, v = v 1 e1 + v 2 e 2 + v 3 e 3 , expressing v in terms of another coordinate basis
(e1′ , e ′2 , e ′3 ) is given simply by
v = (v ⋅ e1′ )e1′ + (v ⋅ e ′2 )e ′2 + (v ⋅ e ′3 )e ′3
2
For simplicity, set v = − πδr 2r sin θ sin φe x +
πδr 2
r2
sin θ cos φe y . Then,
2
v ⋅ e r = − πδr 2r {sin 2 θ sin φ cos φ − sin 2 θ sin φ cos φ} = 0
2
v ⋅ eθ = − πδr 2r {sin θ cos θ sin φ cos φ − sin θ cos θ sin φ cos φ} = 0
v ⋅ eφ =
πδr 2
r2
{sin θ sin 2 φ + sin θ cos 2 φ} =
πδr 2
r2
sin θ
Furthermore, observe that
e− iβR = e− iβ( r − r + R ) ≈ e− iβr (1 − iβ(R − r )) = e− iβr ((1 + iβr ) − iβR)
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273
Basic Antenna Theory
whence,
∫ e
γ
− iβR dl
R
∫ (1 + βr)
≈ e− iβr
γ
dl
R
− e− iβr
∫ iβ dl = e ∫ (1 + βr)
− iβr
γ
γ
dl
R
as
∫
 sin φ 

 dφ = 0
cos φ 


2π
0
That is,
∫ iβ dl = 0. Thus, it follows at once that
γ
A(r , θ) =
µI πδr 2
4π r2
(1 + iβr )e− iβr sin θ eφ =
µ
4π
I|δS( γ )| 1+ri2βr e− iβr sin θ eφ
□
as required.
8.3.3 Theorem
Given a differential magnetic dipole (I,γ,δS(γ)), that is, |δS( γ )|<< 1, the electric
field and magnetic field arising from the magnetic dipole are given by
Eφ = − ωεi
µ
4π
2
e− iβr sin θ βr 2 {1 + iβr }
(8.12)
me− iβr cos θ r13 {1 + iβr }
(8.13)
me− iβr sin θ r13 {−(βr )2 + 1 + iβr }
(8.14)
Br =
Bθ =
µ
2π
m
4π
Proof
From B = ∇ × A, where
∇× A=
1
r 2 sin θ
er
reθ
r sin θeφ
∂r
∂θ
∂φ
0
0
r sin θAφ
 ∂θ (r sin θAφ )

1
= r 2 sin
− r ∂r (r sin θAφ )
θ

0






with
∂θ (r sin θAφ ) =
K15149_Book.indb 273
µ
4π
m 1+riβr e− iβr sin 2θ
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
and
r ∂r (r sin θAφ ) =
{
}
µ
4π
me− iβr rβ 2 − 1r − iβ sin 2 θ .
µ
2π
me− iβr
Thus, it follows that
Br =
µ
4π
Bθ =
{
1+ iβr
r3
2
me− iβr − βr +
1
r3
cos θ
}
+ i rβ2 sin θ
Similarly, the electric field follows from ∇ × B = iωμεE, where the medium is
lossless, and
∇×B=
1
r 2 sin θ
er
reθ
r sin θeφ
∂r
∂θ
∂φ
Br
rBθ
0

− ∂φ (rBθ )

1
= r 2 sin θ 
r ∂φ Br
 r sin θ(∂ (rB ) − ∂ B )
r
θ
θ r






Then, noting that ∂φ Bθ = 0 = ∂φ Br as there are no ϕ-dependencies,
{
2
∂r Bθ = − 4µπ m sin θe− iβr − 2rβ2 +
∂θ Br = − 4µπ me− iβr sin θ
whence, ∂r (rBθ ) − ∂θ Br =
µ
4π
me− iβr sin θ
the electric field is thus Eφ = − ωεi
m
4π
{
{
3
− i βr + i 3r 3β
3
r4
2
r3
+ i 2r 2β
}
β2
r
+ iβ 3 =
m
4π
}
}
2
e− iβr sin θ βr {1 + iβr }, and
2
e− iβr sin θ βr 2 {1 + iβr } , completing the proof. □
Once again, observe that in the far zone, (Eφ , Bθ ) ~ o ( 1r ) and Bφ ~ o
hence, only (Eφ , Bθ ) are dominant. Explicitly, for βr >> 1,
1+ iβr
r3
2
≈ i rβ2 and finally, − βr +
result given below.
1
r3
2
{
}
β2
r2
( ),
1
r2
3
{1 + iβr } ≈ i βr ,
2
+ i rβ2 = − βr 1 − (βr1)2 − i β1r ≈ − βr , yielding the
8.3.4 Corollary
Given a differential magnetic dipole (I, γ, δS(γ)), that is, |δS( γ )|<< 1, the electric
and magnetic far fields, to first order in 1r , are given by
Eφ =
3
1 m β
4 π ωε r
e− iβr sin θ
2
K15149_Book.indb 274
Bθ = − 4µπ m βr e− iβr sin θ
(8.15)
(8.16)
□
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275
Basic Antenna Theory
E
In particular, the wave impedance in the far zone is Zfar ≈ µ Bφθ =
expected, where β = ω µε was utilized.
µ
ε
= η, as
8.3.5 Corollary
Given a differential magnetic dipole (I, γ, δS(γ)), that is, |δS( γ )|<< 1 , the nearfield electric and magnetic fields are given by
Eφ = − ωεi
1
4π
2
m βr 2 e− iβr sin θ
(8.17)
Br =
µ
2π
m r13 e− iβr cos θ
(8.18)
Bθ =
µ
4π
m r13 e− iβr sin θ
(8.19)
provided that |δS( γ )|<< r 2 < 1 continues to hold in the limit as r approaches
the loop γ.
Proof
In the near zone, βr << 1. Hence, as long as the loop γ is sufficiently small so
that |δS( γ )|<< r 2 < 1, then it is clear that 1 + iβr ≈ 1 and −(βr )2 + 1 + iβr ≈ 1 . The
result thus follows.
□
Some remarks regarding Corollary 8.3.5 are due. For a sufficiently small
loop γ such that |δS( γ )|<< r 2 << 1 holds, the magnetic field dominates the
electric field: |H|>>|E| as E ∝ r12 whereas B ∝ r13 . Hence the saying that the
magnetic field dominates in the near field for magnetic loops. More important, observe that in the near field, the magnitude of the magnetic field is
independent of the wavelength of the electromagnetic field whereas the
magnitude of the electric field is directly proportional to its frequency (modulo the phase e− iβr ):
Eφ = − ωεi
m
4π
2
e− iβr sin θ βr 2 = − i 4µπ m rω2 e− iβr sin θ
Finally, recall that for a fixed distance r, near-zone approximation implies
that r << λ, where λ is the wavelength of the electromagnetic field; this is
independent of |δS( γ )|. The requirement that |δS( γ )|<< r 2 follows from the
derivation of Proposition 8.3.2.
8.3.6 Definition
Given a fixed Hertzian or magnetic loop antenna, let Ps (θ, φ) ≡ r 2 ⟨S⟩ ⋅ er define
the power per unit solid angle, and set Pr = ∫ 20 π ∫ 20 π Ps (θ, φ)sin θ dθdφ to be the
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
spatial average power radiated across a sphere. Then, the antenna gain function (or the directive gain) is given by
G(θ, φ) =
4 πPs ( θ , φ )
Pr
(8.20)
Finally, the antenna directivity is defined by D = max G(θ, φ) .
θ,φ
8.3.7 Example
Compare the gain and directivity between a Hertzian antenna and a loop
antenna in the far zone. First, consider a Hertzian antenna ( J , δze z ) . From
Equations (8.8) and (8.9),
r 2 ⟨S⟩ =
r2
2µ
ℜe( E × B∗ ) = e z
β3
( 4 π )2 ωε
( Iδz)2 e− i2βr sin 2 θ
3
β
From this, Pr = ∫ 20 π ∫ π0 Ps (θ, φ)sin θ dθdφ = 2 π ( 4 π )2 ωε ( Iδz)2 e− i2βr ∫ π0 sin 3 θdθ . Noting
that ∫ π0 sin 3 θdθ = 34 , it follows that the Hertzian gain is GH (θ, φ) = 23 sin 2 θ . The
Hertzian directivity is clearly DH = 23 , as max{sin 2 θ} = 1 .
θ
To complete the example, the gain and directivity of a magnetic loop antenna
(I, γ, δS(γ)) are evaluated below. However, by comparing Equations (8.15) and
(8.16) with (8.8) and (8.9), it is obvious that their gains and directivities are
identical. Explicitly, from (8.15) and (8.16),
r 2 ⟨S⟩ =
r2
2µ
ℜe( E × B∗ ) = e z
m2 β 5
( 4 π )2 2 ωε
e− i2βr sin 2 θ
2 5
and hence, Pr = ∫ 20 π ∫ π0 Ps (θ, φ)sin θ dθdφ = 2 π ( 4mπ )2β2 ωε e− i2βr ∫ π0 sin 3 θdθ. Thus, the
3
magnetic loop gain is G (θ, φ) = 3 sin 2 θ and the directivity DM = 2 , as
claimed.
M
2
□
8.4 Microstrip Antenna: A Qualitative Overview
A thin electrical conductor on a printed circuit board is called a trace.
Recall from Chapter 5 that a microstrip is defined to be a trace lying on
the top layer of a PCB and bounded from below by a grounded plane
(assumed here to be a perfect electrical conductor), whereas a stripline is
defined to be a trace that is bounded from above and below by ground
planes (or power planes). The dielectric medium above a microstrip is
typically air. Figure 8.2 illustrates the difference between a microstrip
and a stripline.
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277
Basic Antenna Theory
In parctice, dielectric 1 is
different from dielectric 2
Power/ground plane
Dielectric
Dielectric 2
Stripline
Microstrip
Dielectric 1
Ground plane
Ground plane
(a)
(b)
Figure 8.2.
Two kinds of traces on printed circuit boards: (a) microstrip, (b) stripline.
The far-field and near-field effects of a microstrip are analyzed below. The
microstrip is assumed to be perfectly terminated for simplicity. This criterion
can clearly be generalized to an arbitrary load. The length of the microstrip
is ℓ and its height above its ground plane is h. In the interest of simplicity, it
is further assumed that the dielectrics above and below the microstrip are
identical and the trace is a thin solid cylinder of length ℓ.
The method of images is used to solve the far-field effects of a microstrip
(cf. Figure 8.3). Let θ denote the angle between r and the microstrip at z = 0.
Likewise, let r′ denote the vector of the image microstrip at z = 0, and θ′ the
angle between r′ and the image microstrip. Finally, the current I = I 0 e− iβz is
assumed to travel in the +z-direction toward the termination (load). By definition, the image current travels in the opposite direction.
8.4.1 Proposition
Consider a matched microstrip of length ℓ shown in Figure 8.3 and suppose
some current I ( z) = I 0 e− iβz is flowing along the conductor, where the conductor is surrounded by a homogeneous dielectric medium (μ,ε), with I 0 being a
constant. Then, in the far field, the electric field and magnetic field generated
P
R
Source
θ
R'
θ'
Mirror image
r
r'
l
Current flow
Termination
h
h
Image current flow
Figure 8.3.
The electric field of a microstrip in a homogeneous dielectric medium.
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278
Electromagnetic Theory for Electromagnetic Compatibility Engineers
by the trace at a distance r >> ℓ such that
defined in Figure 8.3 is fixed, are
δEθ (r , θ) ≈
I 0β
sin θ
πε r (cos θ− 1)
e− iβr e
hβ
<< sin 2 θ , where the pair (r, θ) as

r
− i r sin θ i 21 β (cos θ− 1) 
e
cos ( 2 β(cos θ − 1)) sin
(
hβ
r
δBφ (r , θ) ≈
iI 0β
4π
sin θ
)
(8.21)
(
)
− iβr
sin θ
η r (cos
1 − eiβ(cos θ− 1)  1 − e
θ− 1) e

2 hβ
− i r sin θ


(8.22)
where η = µ/ε .
Proof
Without loss of generality, consider the top trace depicted in Figure 8.3
(modulo the ground plane). That is, consider the electric field from the
microstrip, where the pair (r,θ) is fixed. From Equation (8.8), dEθ (r , θ) ≈
2
iβ 2 Idz − iβr
I 0  1 − iβr − iβz
sin θ, whence, Eθ (r , θ) ≈ i4βπωε
∫ 0 r e e sin θdz . To evaluate this
4 πωε r e
2
2
2
integral, let R = r + z − 2 rz cos θ , where 0 ≤ z ≤ ℓ, and set L = r cos θ >> ℓ.
Then, by construction,
cos Θ ≡
⇒ Θ = arccos
L− z
R
{
L− z
r 2 + z 2 − 2 rz cos θ
}
Here, Θ is the angle between R and the microstrip. Then, recalling the identity sin(arccos φ) = 1 − φ2 , it follows that
sin Θ = 1 −
( L − z )2
r 2 + z 2 − 2 rz cos θ
= 1 − cos 2 θ {1 −
to first order in
z
r
z
r
≈ 1 − ( Lr ) {1 −
2
2z
r
2z
r
( 1 − cos θ )}
(1 − cos θ )}
. In particular, by assumption,
{
<< sin 2 θ ⇒ sin Θ ≈ sin θ 1 +
z
r
( 1 − cos θ ) cot 2 θ}
where the binomial approximation and the identity 1 = cos 2 θ + sin 2 θ were
invoked.
Hence,
∫

0
1
R
e− iβ( R + z ) sin Θdz ≈ sin θ
up to first order in
again,
(
z
r
0
1
R
e− iβ( R + z ) d z +
cos2 θ
r sin θ
( 1 − cos θ )
∫

0
z
R
e− iβ( R + z ) d z
. Moreover, appealing to the binomial approximation
R ≈ r 1 − zr cos θ + o
K15149_Book.indb 278
∫

(( ) ))
z 2
r
and
1
R
≈
1
r
(1 + zr cos θ) + o (( zr )2 )
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279
Basic Antenna Theory
Thus, evaluating the integrals, and noting that ∫ ze az dz = e az
∫

0
1
R
e− iβ( R + z) d z ≈

∫
1
r
∫

0
e− iβ( R + z) {1 + zr cos θ} d z ≈ r e− iβr
1
e− iβ( R + z ) d z ≈ r e− iβr
∫
Taking the approximation up to
1
r2
z
R
0
∫

0
1
1
R
0
e− iβ( R + z ) sin Θdz ≈
{sin θ cos θ + (cos θ − 1) }
cos 2 θ
sin θ

∫e
− iβ(cos θ− 1) z
0
,
{1 + zr cos θ} dz
ze− iβ(cos θ− 1) z {1 + zr cos θ} d z
,
sin θ
βr (cos θ− 1)
1
β 2 (cos θ− 1)2 r 2

1− az
a2
{
}
e− iβr 1 − eiβ(cos θ− 1) +
{
}
e− iβr eiβ(cos θ− 1) (1 − iβ(cos θ − 1)) − 1
That is, the electric field in the far zone for the microstrip is approximated by
Eθ (r , θ) ≈
iI 0
4 πε
iI 0β
sin θ
4 πε r (cos θ− 1)
{sin θ cos θ + (cos θ − 1) }
2
cos θ
sin θ
{
}
e− iβr 1 − eiβ(cos θ− 1) +
1
β 2 (cos θ− 1)2 r 2
{
}
e− iβr eiβ(cos θ− 1) (1 − iβ(cos θ − 1)) − 1
Similarly, to determine the field from the image trace, it suffices to note that
r ′ = r 2 + 4 h2 + 4rh sin θ
and θ′ = arccos rL′
the image field can be obtained via: r → r ′ , θ → θ′ , I 0 → − I 0 . Hence,
0β
Eθ (r ′ , θ′) ≈ − i4Iπε
{
2
θ′
− 4iIπε0 sin θ′ cos θ′ + (cos θ′ − 1) cos
sin θ ′
sin θ′
r ′ (cos θ′ − 1)
}
{
}
e− iβr ′ 1 − eiβ(cos θ′ − 1) +
1
β2 (cos θ ′ − 1)2 r ′ 2
{
}
e− iβr ′ eiβ(cos θ′ − 1) (1 − iβ(cos θ′ − 1)) − 1
In particular, the resultant electric field at the point P is, to first order in 1r ,
which is equivalent to the condition βr ∗ >> 1 ⇔ rβ∗ >> ( r∗1)2 , where r ∗ ∈{r , r ′} ,
≈
iI 0β
4 πε
{
δEθ (r , θ) = Eθ (r , θ) + Eθ (r ′ , θ′)
sin θ
r (cos θ− 1)
(
)
e− iβr 1 − eiβ(cos θ− 1) −
sin θ′
r ′ (cos θ′ − 1)
(
e− iβr ′ 1 − eiβ(cos θ′ − 1)
)}
In turn, this can be simplified (see Exercise 8.6.4) to first order in 1r , to
δEθ (r , θ) ≈
K15149_Book.indb 279
iI 0 β sin θ
πε r cos θ− 1
e− iβr e
hβ
− i r sin θ i 21 β (cos θ− 1) 
e
sin ( 2 β(cos θ − 1)) sin
(
hβ
r
sin θ
)
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Finally, to complete the proof, it suffices to recall that H φ = η1 Eθ , and hence,
= η yields
1 µ
η ε
δBθ (r , θ) ≈
iI 0
π
η βr
sin θ
cos θ− 1
e− iβr e
hβ
− i r sin θ i 21 β (cos θ− 1) 
e
sin ( 2 β(cos θ − 1)) sin
(
hβ
r
sin θ
)
□
8.4.2 Remark
Indeed, for 0 < h << 1, the pair (δEθ , δBφ ) can be further approximated by
δEθ (r , θ) ≈
via sin
(
hβ
r
)
sin θ ≈
hβ
r
δBθ (r , θ) ≈
iI 0 h β 2 sin 2 θ
πε r 2 cos θ− 1
sin θ + o
iI 0 h
π
2
η βr 2
e− iβr e
hβ
− i r sin θ i 21 β (cos θ− 1) 
e
sin ( 21 β(cos θ − 1) )
(( ) ). Likewise,
β
r
sin 2 θ
cos θ− 1
3
e− iβr e
hβ
− i r sin θ i 21 β (cos θ− 1) 
e
sin ( 21 β(cos θ − 1) )
From this, it is clear that a microstrip is a poor radiator, as the field falls off
as r12 ; and by inference, a differential pair is also a poor radiator. Furthermore,
it is clear that along the axis of the microstrip, the fields are zero, that is,
when θ = 0, whereas the fields are maximal when θ = π2 . Finally, observe that
for θ ≠ 0, sin ( 21 β(cos θ − 1) ) = 0 ⇔ 21 β(cos θ − 1) = nπ for n = 1,2, … . That is,
2 πn
from β = ω µε , δEθ = 0 = δBφ ⇔ ω = µε ( cos
∀n = 0, 1, 2,… .
θ− 1)
8.5 Array Antenna and Aperture Antenna
This chapter ends with a brief account of the array antenna and aperture
antenna. From Section 8.4, it is clear that a linear antenna has the form:
− iβr
E = E0 F(θ, φ) e r
(8.23)
where F(θ, ϕ) is the antenna factor, and E 0 is the initial field strength. The
antenna factor determines the antenna characteristics that depend solely
upon the antenna geometry. This forms the basis for analyzing an antenna
array structure.
Now, Figure 8.4 illustrates a linear (n + 1)-array antenna, where the
k-antenna has current I k = Ck I 0 eikξ ∀k = 0, 1, 2, , with some fixed phase ξ
> 0, constant Ck > 0 , with C0 = 1 , and the antennae are separated by a con− iβr
stant distance d. In particular, Equation (8.23) becomes Ek = E0, k F(θ, φ) e r ,
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Basic Antenna Theory
where E0, k = Ck E0 eikξ . Denote a linear (n + 1)-array antenna of length ℓ by
{( I k , θ k , d , ξ , )}n .
8.5.1 Theorem
Given a linear (n + 1)-antenna array {( I k , θ k , d , ξ , )}n , suppose that
L = r0 cos θ0 >> nd . Then, the cross-sectional far-field profile of the electric
field, to second order in r10 , satisfies
− iβr0
E ≈ E0 F(θ, φ) e r0
∑
n
k=0
{
C k eik (βd cos θ0 +ξ ) 1 +
kd
r0
cos θ0
}
(8.24)
Proof
From Figure 8.4, it is clear that rk2 = r02 + ( kd)2 − 2 kr0 d cos θ0 ∀k and hence,
rk ≈ r0 1 −
and
1
rk
≈
1
r0
(1 −
2 kd
r0
2 kd
r0
(
cos θ0 ≈ r0 1 −
cos θ0
)
− 21
≈
kd
r0
(1 +
1
r0
cos θ0
kd
r0
)
cos θ0
)
Likewise, by assumption,
L >> nd ⇒ cos θ k =
rk
L − kd
≈
rk
L
(1 + kdL ) ≈ cos θ0 {1 + kd ( L1 − cosr θ )}
0
0
P
d = uniform separation
distance
ξ = plase shift
r1
r0
y
z
x
θ0
0
d
rn
rk
θk
θ1
1
k
[
k[
θn
d
n
Figure 8.4.
Linear array of n-antenna.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Thus, L = r0 cos θ0 ⇒ cos θ k ≈ cos θ0 1 +
{
kd
r0
sin θ0 tan θ0
1− cos2 θ0
cos θ0
=
sin 2 θ0
cos θ0
1
cos θ0
− cos θ0 =
} as
= sin θ0 tan θ0
Moreover, noting that
βrk − kξ ≈ βr0 − βkd cos θ0 − kξ = βr0 − k(βd cos θ0 + ξ)
it follows at once that on setting Ek = Ck E0 nk ≡ E0C k with E0 = E0 n0 ≡ E0C 0 ,
where nk = r1k rk are unit vectors,
E=
∑
n
k=0
∑
Ek ≈ E0 F(θ, φ)
n
k=0
− iβr0
≈ E0 F(θ, φ) e r0
Ck
∑
1
rk
n
k=0
e− iβr0 eik (βd cos θ0 +ξ )
{
C k eik (βd cos θ0 +ξ ) 1 +
kd
r0
cos θ0
}
□
8.5.2 Remark
Observe that because min rk >> nd , it follows that nk ≈ n0 and hence,
− iβr0
Equation (8.24) reduces to E ≈ E0 F(θ, φ) e r0 ∑ nk = 0 Ck eik (βd cos θ0 +ξ ) 1 + kd
r0 cos θ 0 .
The quantity
{
|A|=
∑
n
k=0
{
Ck eik (βd cos θ0 +ξ ) 1 +
kd
r0
cos θ0
}
}
is called the array factor up to first order in r10 . More commonly, the array fac1
tor is often defined as the zeroth order of r0 :
|A|=
∑
n
k=0
Ck eik (βd cos θ0 +ξ )
8.5.3 Corollary
Under the conditions of Theorem 8.5.1, if Ck = C0 ∀k and eik (βd cos θ0 +ξ ) ≠ 1 ∀k >
0, then to first order in r10 ,
− iβr0
E ≈ C 0 E0 F(θ, ϕ) e r0 e
K15149_Book.indb 282
{
}
1
i n2 (βd cos θ0 +ξ ) sin 2 ( n + 1)(βd cos θ0 +ξ )
sin 21 (βd cos θ0 +ξ )
{
}
(8.25)
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Basic Antenna Theory
Proof
− iβr0
From Theorem 8.5.1 and Remark 8.5.2, E ≈ C 0 E0 F(θ, φ) e r0 ∑ nk = 0 eik (βd cos θ0 +ξ ) .
Next, noting that ∑ nk = 0 eik (βd cos θ0 +ξ ) = 1 + ei(βd cos θ0 +ξ ) +  + ein(βd cos θ0 +ξ ) defines a
n
geometric series: 1 + z + z n− 1 = zz −−11 ∀z ∈C, z ≠ 1, n < ∞ , it follows that
∑
n
k=0
ei( n+ 1)(βd cos θ0 +ξ ) − 1
ei(βd cos θ0 +ξ ) − 1
eik (βd cos θ0 +ξ ) =
=
ei( n+ 1)(βd cos θ0 +ξ )/2 ei( n+ 1)(βd cos θ0 +ξ )/2 − e− i( n+ 1)(βd cos θ0 +ξ )/2
ei(βd cos θ0 +ξ )/2
ei(βd cos θ0 +ξ )/2 − e− i(βd cos θ0 +ξ )/2
=e
i n2 (βd cos θ0 +ξ )
sin { 21 (n + 1)(βd cos θ0 + ξ)}
sin { 21 (βd cos θ0 + ξ)}
and hence,
− iβr0
E ≈ C 0 E0 F(θ, ϕ) e r0 e
{
}
1
i n2 (βd cos θ0 +ξ ) sin 2 ( n + 1)(βd cos θ0 +ξ )
sin 21 (βd cos θ0 +ξ )
{
}
□
Note finally from Corollary 8.5.3 that by measuring the distance from
the center of the linear array to the point of interest, that is, on setting
r = r0 − 21 nd cos θ0 (cf. Figure 8.5), Equation (8.25) simplifies to
− iβr
E ≈ C 0 E0 F(θ, φ) e r e
{
}
1
i n2 ξ sin 2 ( n + 1)(βd cos θ0 +ξ )
sin 21 (βd cos θ0 +ξ )
{
}
(8.26)
Lastly, the magnetic field follows immediately from Maxwell’s equation:
∇ × E = iωB, where the convention e–iωt is adopted here for time harmonicity
instead of eiωt.
P
if n is odd, take
n to be n – 1
1 nd cos θ
0
2
0
r0
r–
θ0
1 nd
2
1n
2
Figure 8.5.
Approximating distance in a linear array antenna.
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284
Electromagnetic Theory for Electromagnetic Compatibility Engineers
8.5.4 Example
Determine the array factor of a linear 2-antenna array, where we assume for
simplicity that C0 = C1 . By Remark 8.5.2,
|A|= C0
∑
= C0 2e
1
k=0
eik (βd cos θ0 +ξ ) = C0 1 + ei(βd cos θ0 +ξ )
i 21 (βd cos θ0 +ξ ) 1
2
{e
− i 21 (βd cos θ0 +ξ )
+e
i 21 (βd cos θ0 +ξ )
}
= 2C0 cos ( 21 (βd cos θ0 + ξ))
It is clear from the above that A = 0 ⇔ βd cos θ0 + ξ = (2 n − 1)π ∀n = 1, 2, .
Thus, if ω is fixed, then ξ = (2 n − 1)π − βd cos θ0 leads to far-field cancellation.
Likewise, if the phase ξ is fixed, then transmitting at
ω=
(2 n − 1) π−ξ
µε d cos θ0
will also lead to far-field cancellation. Conversely, A is maximal if and only
□
if βd cos θ0 + ξ = 2 nπ ∀n = 1, 2, .
8.5.5 Definition
The irradiance (or intensity) of an electromagnetic field (E,B) is defined by
I ≡ ⟨S⟩ =
1
2µ0
|E × B∗|=
1
2µ0
|E|2
(8.27)
Thus, in view of Example 8.5.4, the intensity is inversely proportional to r2
and directly proportional to |A|2, the array factor. In contrast, the electric
field and the magnetic field fall off—in the far-field regime—as 1r and their
magnitudes are directly proportional to |A|.
8.5.6 Proposition
Given a linear (n + 1)-antenna array {( I k , θ k , d , ξ , )}n with L = r0 cos θ0 >> nd ,
up to first order in r1 , where r = r0 − 21 nd cos θ0 , set ψ = βd cos θ0 + ξ . If
I k = I 0 ∀k , then
(a) The maximal field strength occurs when ψ = 0.
(b) The minimal field strength occurs when ψ = n2 +kπ1 , for all k > 0.
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Basic Antenna Theory
Proof
To establish the assertions, it suffices to consider the array factor |A|. To simplify the analysis, note first of all that I k = I 0 ∀k ⇒ Ck = C0 ∀k . Hence, the
 can be defined from Equation (8.26):
normalized array factor A
|=
|A
{
sin 21 ( n + 1) ψ
1
n+ 1
{ }
sin 21 ψ
}
Then, via Taylor expansion,
sin x = x −
1
3!
x 3 + o( x 5 )
it follows at once that
|= lim 1
lim|A
n+ 1
ψ→ 0
{
sin 21 ( n + 1) ψ
ψ→ 0
sin
{ }
1
ψ
2
} = lim 1
n+ 1
{
} =1
( n + 1) 21 ψ + o ( ψ 3 )
1 ψ + o( ψ 3 )
2
ψ→ 0
and hence, yielding the maximal field strength as claimed in (a). Regarding
(b), it suffices to note that
|= 0 ⇔ 1 (n + 1)ψ = kπ ⇒ ψ =
|A
2
2 kπ
n+ 1
for all k > 0
□
as required.
A linear antenna array factor, for n = 10, is plotted in Figure 8.6 to illustrate
the maxima and minima.
Array Factor (n = 10)
Amplitude |A|
1.2
1
0.8
0.6
0.4
6.03
5.72
5.4
5.09
4.78
4.46
4.15
3.83
3.2
3.52
2.89
2.58
2.26
1.95
1.63
1.32
1.01
0.69
0.38
0
0.06
0.2
Angle (radians)
Figure 8.6.
The antenna array factor for 10 linear radiators.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
The remainder of this section is devoted to a short survey on aperture
antennae. As chassis for electronic devices often have apertures, it is imp­
ortant for EMC engineers to understand the basics of aperture radiators.
Indeed, the essence of aperture radiators originates from the application of
scalar diffraction theory; specifically, the Kirchhoff integral theorem, which
is a particular case of Green’s theorem, stated below without proof [4].
8.5.7 Theorem (Kirchhoff)
Given some compact neighborhood V ⊂ R 3 of an arbitrary point r0 , suppose a scalar field U: V → R satisfies the homogeneous Helmholtz equation
∆U + k 2U = 0 on V. Set G(r ) = 1r eik ⋅r and let n be a unit normal vector field
on ∂V. Then,
U (r ) =
1
4π
∫
∂V
{G(r ′) ∂nU (r ′) − U (r ′) ∂ n G(r ′)} d2 r ′
(8.28)
Moreover, if the requirement that V ⊂ R 3 be compact is relaxed, then U must
satisfy the boundary condition known as the Sommerfeld radiation condition
lim r {∂n U − ikU } = 0 , in order for the integral to converge.
□
r →∞
An immediate application of Theorem 8.5.7 is to consider an infinite conductive plane R 2 at z = 0, with a single rectangular aperture
Ω = [ − 21 δ , 21 δ ] × [ − 21 , 21  ], where δ << ℓ, that is, a slit. For simplicity, set
R 3+ = {( x , y , z) ∈R 3 : z ≥ 0} and R 3− = R 3 − R 3+ . And as a simple example, consider
a single point source (and the field is thus necessarily spherical) at R ∈R 3− and
an arbitrary point r ′ ∈Ω , where by an abuse of notation, Ω is identified with
Ω × {0}. Finally, for notational simplicity, let U be a scalar field satisfying the
Helmholtz equation, ∆U + k 2U = 0, where k = ω µε is the wave number, and
let n denote a unit normal vector field on Ω directed into R 3− .
8.5.8 Lemma (Fresnel–Kirchhoff Diffraction Formula)
Given a homogeneous medium (R 3 , µ , ε), a unit point source at R ∈R 3− and
a slit Ω ⊂ ∂R 3+, suppose that ∂R 3+ − Ω is a perfect electrical conductor. Then,
the field U at an arbitrary point r ∈R 3+ satisfying kr, kR >> 1* may be approximated by
U (r ) = − λi
∫
Ω
1
rR
eik ( r + R ) 21 (cos θ − cos Θ)d x ′ d y ′
(8.29)
where λ = ω2 πµε , r = r − r ′ = ( x − x ′ , y − y ′ , z) with r ′ = ( x ′ , y ′ , 0) ∈Ω being an
arbitrary point, with cos θ = r1 ( r ⋅ n) and cos Θ = R1 ( R ⋅ n) .
* That is, in the far-field regime.
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Basic Antenna Theory
Unit normal
vector field
n
–
–
P' r
R
r'
r
R
Origin Ω
Point source
P1
P0 Observation point
D
Perfect electrical conductor
Figure 8.7.
Derivation of the Fresnel–Kirchhoff diffraction formula.
Proof
Theorem 8.5.7 is applied twice to obtain Equation (8.29); see Figure 8.7 for
details. Here, let the point source be at P1 = (X , Y , Z) and an arbitrary observation point P0 = ( x , y , z). Finally, let P ′ = ( x ′ , y ′ , 0) be an arbitrary point on Ω,
D ⊂ R 3+ an arbitrary bounded neighborhood of P0 such that Ω ⊂ ∂D ∩ ∂R 3+ ,
and n a unit normal vector field on ∂D ∩ Ω.
Now, appealing to Equation (8.28), the field at P0 from a point source at P′ is:
U (r ) =
1
4π
∫ {
∂D
eikr
r
ikr
∂ n U (r ′) − U (r ′) ∂ n e r
}d r ′
2
Now, by definition, ∂n G ≡ ∇G ⋅ n, and
∇ r1 eikr = eikr ∇ r1 + r1 ∇eikr = −eikr
whence kr >> 1 ⇔ k >>
1
r
1 1
r2 r
r + irk eikr
1
r
r = r1 eikr ( − r1 + ik ) r1 r
implies that ∇ r1 eikr ≈ ik r1 eikr
1
r
r , yielding
U (r ) ≈
1
4π
∫ {
=
1
4π
∫
eikr
r
=
1
4π
∫
eikr {∂ n U ( r ) − U ( r )ik r1 r ⋅ n} rsinθ d θ d φ
∂ D −Ω
∂ D −Ω
∂ D −Ω
eikr
r
∂ n U ( r ) − U ( r )ik r1 eikr
1
r
}
r ⋅ n d2 r
{∂n U (r ) − U (r )ik r1 r ⋅ n} r 2 sinθ dθ dφ
Now, invoking the Sommerfield radiation condition,
lim U (r ) = lim 41π
r →∞
K15149_Book.indb 287
r →∞
∫
∂ D −Ω
eikr r {∂ n U ( r ) − U ( r )ik r1 r ⋅ n} sinθ d θ d φ = 0
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
That is, the integral on the boundary at infinity vanishes, and hence, the only
field contribution arises on Ω. Thus, the surface integral reduces to integration on the aperture Ω:
U (r ) ≈
∫
1
4π
Ω
1
r′
eikr ′ {∂ n U (r ′) − U ( r ′)ik r1′ r ′ ⋅ n} d 2 r
Set r1′ r ′ ⋅ n = cos θ . Then, U (r ) ≈ 41π ∫ Ω r1′ eikr ′ {∂ n U (r ′) − ikU (r ′)cos θ} d 2 r .
Next, given that a point source originates at P1 , it follows that U ( R) = R1 eikR
as a point source generates spherical waves. Therefore substituting this
into the integral yields:
U (r ) ≈
1
4π
∫
Ω
1
r′
{ (
eikr ′ ∂ n
1
R
}
)
eikR − ik R1 eikR cos θ d 2 r
Moreover, applying the assumption that kR >> 1 , ∇ R1 eikR ≈ ik R1 eikR R1 R
and on setting R1 R ⋅ n = cos Θ , it follows at once that U (r ) ≈ 4ikπ ∫ Ω r1 eikr R1 eikR
{cos Θ − cos θ}d 2 r, as desired, where k = 2λπ .
□
It is clear that the above result yields good approximation to radiation
wavelengths that are at least as short as the optical wavelengths. That
is, it will not yield good results for radiation in the microwave regime.
Unfortunately, the microwave range is often the range of interest for EMC
engineers; in particular, the source is often close to apertures and hence, the
criterion kR >> 1 is violated. However, this can be partially rectified from the
proof of Lemma 8.5.8.
8.5.9 Lemma
Given a homogeneous medium (R 3 , µ , ε) , a point source of magnitude U 0 at
R ∈R 3−, and a slit Ω ⊂ ∂R 3+ , suppose that ∂R 3+ − Ω is a perfect electrical conductor. Then, the field U at an arbitrary point r ∈R 3+ satisfying kr >> 1 may
be approximated by
U (r ) = − iUλ0
Proof
∂n
and replace
(
eikξ
ξ
1
R
∫
Ω
)
1
rR
eik ( r + R )
(
1
2
{cos θ − (
eikR = R1 eikR − R1 + ik
→ U0
eikξ
ξ
.
)
1
R
i
kR
R⋅n =
}
)
+ 1 cos Θ d x ′ d y ′
ik
R
eikR
(
i
kR
(8.29)
)
+ 1 cos Θ
□
Clearly, if the source is not a spherical radiator, then the integral
U (r ) ≈ 41π ∫ Ω r1′ eikr ′ {∂n U − ikU cos θ}d 2 r must be evaluated for arbitrary source
U. An example is if a chassis possesses a slit aperture, and the source is
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289
Basic Antenna Theory
nonspherical. Then, this integral must be evaluated instead of appealing to
Equation (8.29); often the integral can only be solved numerically.
However, there is another formalism that lends itself more accessiblly to
computation. The proof can be found, for example, in [4], or more informally
in [5]. This is the Huygens–Fresnel principle and it essentially states every point
on an unobstructed wavefront is a secondary spherical source with the same
frequency.* The precise statement is given without proof below.
8.5.10 Theorem (Huygens–Fresnel)
Given a homogeneous medium (R 3 , µ , ε) and a slit Ω ⊂ ∂R 3+ , suppose that
∂R 3+ − Ω is a perfect electrical conductor. Then, the field U at an arbitrary
point r ∈R 3+ satisfying kr >> 1 may be approximated by
U (r ) = − λi
∫
Ω
eikr U ( x ′ , y ′)cos θ d x ′ d y ′
1
r
(8.30)
where U|Ω is the secondary source at each point on Ω, and cos θ = r1 r ⋅ n.
□
8.5.11 Example
Suppose a constant plane wave U = U 0 e− iβz in R 3− is incident on Ω.†
Determine the field strength at P0 = (r0 , θ0 , z0 ) , if r0 >> r ′ = x ′ 2 + y ′ 2 with
( x ′ , y ′ , 0) ∈Ω , where δ,ℓ > 0 are arbitrary, that is, not necessarily a slit. From
Figure 8.7,
r = ( x0 − x ′)2 + ( y 0 − y ′)2 + z02 = r0 1 +
whence
(
r ≈ r0 1 +
)
x0 x ′+ y 0 y ′
r02
2( x0 x ′+ y 0 y ′ )
and
r02
1
r
≈
1
r0
+
x′2 + y ′2
(1 −
r02
≈ r0 1 +
x0 x ′+ y0 y ′
r02
2( x0 x ′+ y0 y ′ )
r02
)
Lastly, noting that cos θ = zr0 and cos θ0 = zr00 , it follows immediately that
x x ′+ y y ′
cos θ = cos θ0 rr0 ≈ cos θ0 1 − 0 r 2 0 . So, substituting these approximations
0
into Equation (8.30),
(
)
U (r0 ) = − λi U 0 e− iβz
∫
≈ − λi U 0 e− iβz
1
r0
Ω
1
r
eikr cos θ d x ′ d y ′
eikr0 cos θ0
∫
Ω
(
eikx0 x′/r0 eiky0 y ′/r0 1 −
x0 x ′+ y 0 y ′
r02
) dx′ dy ′
2
* Physically, this assertion is clearly false as waves cannot generate more waves on their own;
however, the principle yields a decent approximation for most engineering purposes.
† This is equivalent to a point source at infinity.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
ikx0 x ′/r0 iky 0 y ′/r0
(1 −
ikx0 x ′/r0 iky 0 y ′/r0
dx′ dy ′ +
≈ − λi U 0 e− iβz
1
r0
eikr0 cos θ0
∫e
≈ − λi U 0 e− iβz
1
r0
eikr0 cos θ0
∫e
i
λ
U 0 e− iβz
1
r0
= − λi U 0 e− iβz
eikr0 cos θ0
1
r0
e
Ω
e
Ω
∫e
Ω
e
ikx0 x ′/r0 iky 0 y ′/r0
U 0 e− iβz
1
r0
eikr0 cos θ0
2 x0
r02
i
λ
U 0 e− iβz
1
r0
eikr0 cos θ0
2 y0
∫e
∫e
dx′ dy ′
ikx0 x ′/r0 iky 0 y ′/r0
x′ dx′ dy ′ +
ikx0 x ′/r0 iky 0 y ′/r0
y ′ dx′ dy ′
e
Ω
r02
) dx′ dy ′
dx′ dy ′ +
e
Ω
i
λ
r02
ikx0 x ′/r0 iky 0 y ′/r0 2( x0 x ′+ y 0 y ′ )
r02
∫e
eikr0 cos θ0
2( x0 x ′+ y 0 y ′ )
e
Ω
Now, for simplicity, set
U (r0 ) = − λi U 0 e− iβz
1
r0
{
eikr0 cos θ0 I1 (r0 ) −
Then, employing the identity sin φ =
I1 (r0 ) =
=
∫
Ω
r0 2 4
x0 y 0
k
sin
(
1
2
) (
k xr00δ sin
1
2
∫
k
1δ
2
− 21 δ
y0 
r0
∫ e e x′ dx′ dy ′
= {sinc ( k ) − cos ( k )}
K15149_Book.indb 290
r02
}
I 3 (r0 ) .
eikx0 x′/r0 d x ′
∫
(
)
1
2
− 21 
sin φ
φ
,
eiky0 y ′/r0 d y ′
)
Ω
iδ
k
I 3 (r0 ) =
2 y0
ikx0 x ′/r0 iky 0 y ′/r0
I 2 (r0 ) =
=
I 2 (r0 ) −
(eiφ − e− iφ ) and defining sincφ =
1
2i
eikx0 x′/r0 eiky0 y ′/r0 d x ′ d y ′ =
( )
2 x0
r02
x0 δ
r0
1
2
i2 δ
y 0 x0
( )
∫e
r0 2
k
x0 δ
r0
1
2
2 r0
ky 0
sin
1
2
k
y0 
r0
{ ( ) ( )} ( )
sinc
ikx0 x ′/r02
Ω
e
1
2
k xr02δ − cos
iky 0 y ′/r02
0
1
2
k xr02δ
0
sin
1
2
k
y0 
r02
y ′ dx′ dy ′
{ ( ) ( )} ( )
( ) { ( ) ( )} ( )
=
i
k
=
i2 
x0 y 0
sinc
r0 2
k
1
2
k
y0 
r02
sinc
− cos
1
2
k
y0 
r02
1
2
k
2 r02
kx0
y0 
r02
− cos
1
2
k
sin
y0 
r02
1
2
k xr02δ
sin
0
1
2
k xr02δ
0
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Basic Antenna Theory
whence substituting k =
0
yields
{sin (π ) sin (π ) +
(sinc (π ) − cos (π )) sin (π ) − i (sinc (π ) − cos (π )) sin (π )} .
U (r0 ) = − iU 0e− iβzeikr0
− i xr02δ
2π
λ
cos θ0 λr0
π 2 x0 y 0
x0 δ
λr0
y0 
λx0
x0 δ
λr0
y0 
λx0
x0 δ
λr0
y0 
λx0
x0δ
r02
y0 
λx0
x0 δ
λr0
Now, noting that
lim { x sin x1 sin x1 } = x
x→∞
{
1
x
−
1 1
3! x 3
+o
( )} sin
1
x5
1
x
→0
and lim sinx x → 0 , it follows clearly that
x→ 0
(
) (
y 
)
lim λ sin π xλ0r0δ sin π λx00 = 0
λ→∞
( ( ) ( )) ( )
lim λ ( sinc ( π ) − cos ( π )) sin ( π ) = 0
y 
lim λ sinc π xλ0r0δ − cos π xλ0r0δ sin π λx00 = 0
λ→∞
λ→∞
y0 
λx0
y0 
λx0
x0 δ
λr0
That is, for any fixed δ, , r0 , lim U (r0 ) = 0 . The implication is the following:
λ→∞
for any fixed rectangular aperture, longer wavelength radiation implies less
radiation escaping from the aperture.
Next, for r0 >> 1, it is clear that
U (r0 ) ≈ − iU 0 e− iβz eikr0
to first order in
δ, , r0 when
1
r0
cos θ0 λr0
π 2 x0 y 0
) (
(
y 
sin π xλ0r0δ sin π λx00
)
. In particular, the field is zero precisely for each fixed
x0 δ
λr0
= 2n ⇒ λ =
x0 δ
2 nr0
y0 
λr0
or
= 2n ⇒ λ =
x0 δ
2 nr0
for all n = 0, 1, 2, … . This yields the diffraction fringes observed on a screen placed
at z0 away from the aperture. Indeed, U (r0 ) can be expressed more intuitively:
U (r0 ) ≈ − iU 0 e− iβz eikr0
= − iU 0 e− iβz eikr0
cos θ0 π 2 δ
λr0
π2
(
(
)
y 
sinc π xλ0r0δ sinc π λx00
cos θ0 k πδ
π 2 r0 2
(
)
(
)
y 
sinc π xλ0r0δ sinc π λx00
)
It is thus obvious when expressed in this form that long wavelength radiation yields a lower intensity for a given point away from a fixed aperture, a
result obtained less elegantly from the previous paragraph.
□
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292
Electromagnetic Theory for Electromagnetic Compatibility Engineers
8.5.12 Proposition
Consider n identical slits adjacent to one another, each slit separated from the
other by some distance h > 0:
Ω j =  − L2 + ( j − 1)( h + δ), − L2 + ( j − 1)( h + δ) + δ  × [ − 2 , 2 ] × {0}
where L = N(h + δ) + δ, with Ω j lying on the plane z = 0: Ω j ⊂ R 2 × {0} ∀j ∈N.
Suppose that min rj >> max{ h, δ}, and set r0, j = ( δ−2 L + ( j − 1)( h + δ), 0, 0) to be the
j
center of Ω j , and let r0 = ( x0 , y 0 , z0 ) ∈R +3 be an arbitrary point. Then, the field
at r 0 is
U (r0 ) ≈
U0
k2λ
∑{
j
2
1
Λ j ( x , X ) Λ j ( y ,Y ) rj R j
e
{(
ik rj + R j
(
cos θ j − 1 +
) cos Θ )} ×
i
kR j
j
e
(
ikΛ j ( x , X )
∆ −j + δ2
(
iδ cos 21 kΛ j ( x , X )δ
Λ j ( x , X ) Λ j ( y ,Y )
e
ikΛ j ( x , X ) ∆ +j
)
) sin 1 kΛ (x, X )δ cos 1 kΛ (y , Y ) −
(2 j
) (2 j
)
1
rj R j
e
ik rj + R j
{Λ ′ ( x , X ) +
1
− − i 2 kΛ j ( x , X )δ
j
j
(
)
1
Λ j ( x , X ) Λ j ( y ,Y ) rj R j
(
ikΛ j ( x , X ) ∆ −j + δ2
i
kR j
j
{(1 − ikΛ (x, X)) ∆ e
 sin 21 kΛ j ( x , X )δ
e
(8.31)
e
ik rj + R j
{Λ ′ ( y , Y ) +
j
}
Λ ′′j ( x , X )cos Θ j ×
sinc
i
kR j
(
1
2
)
}
kΛ j ( x , X )δ − 1 +
}
Λ ′′j ( y , Y )cos Θ j ×
}
) sinc 1 kΛ (y , Y ) − cos 1 kΛ (y , Y )
)
(2 j
)}
{ (2 j
where
Λ j (u, U ) ≡
uj
rj
−
Uj
Rj
, Λ ′j (u, U ) =
uj
rj2
cos θ j +
and
Λ ′′j (u, U ) =
Uj
R 2j
uj
rj2
u U 
cos Θ j + (cos θ j − cos Θ j )  r 2j − R2j 
 j
j 
−
3U j
R 2j
,
with
u j ∈{ x j , y j , z j }, U j ∈{X j , Yj , Z j }
K15149_Book.indb 292
.
10/18/13 11:04 AM
293
Basic Antenna Theory
Proof
For each fixed Ω j , set rj = r0 − r0, j and rj′ = r ′ + r0, j . Then, rj = rj − r ′ , for each
j = 1,2, … Likewise, set R j = r0, j − R0 and R j = r0,′ j − R0 . Then, R j = R j + r ′.
Recall that r ′ ∈Ω 1 L , where for simplicity, it is assumed without loss of gener2
ality that L2 ∈N , whence
{
rj = ( x j − x ′)2 + ( y j − y ′)2 + z 2j ≈ rj 1 −
{
}
− 21
{
{
{
x j x ′+ y j y ′
rj2
}
1+
x j x ′+ y j y ′
R j = (X j + x ′)2 + (Yj + y ′)2 + Z 2j ≈ R j 1 +
X j x ′+ Yj y ′
1
rj
1
Rj
= ( x j − x ′)2 + ( y j − y ′)2 + z 2j
{
= (X j + x ′)2 + (Yj + y ′)2 + Z 2j
}
≈
− 21
1
rj
≈
rj2
1−
1
Rj
R 2j
}
}
}
X j x ′+ Yj y ′
R 2j
z
Moreover, note that cos θ j = zr0j and cos θ j = r0j ⇒ cos θ j = cos θ j
Rj
Z0
Z0
cos Θ j = R j and cos Θ j = R j ⇒ cos Θ j = cos Θ j R j , whence,
{
cos θ j ≈ cos θ j 1 +
x j x ′+ y j y ′
rj2
}
{
and cos Θ j ≈ cos Θ j 1 −
rj
rj
. Likewise,
X j x ′+ Yj y ′
R 2j
}
Thus after many tedious algebraic manipulations, which the dedicated
reader is encouraged to perform, where the binomial approximation is applied,
≈
1
rj R j
e
ik rj + R j
1
rj R j
e
ik rj + R j
(cos θ j − cos Θ j )
  xj
1 +  r 2 −
j

 xj
 rj2 cos θ j +
Xj
R 2j
Xj
R 2j
 yj

 x ′ +  rj2 −
Yj
R 2j
 
 y ′  cos θ j − cos Θ j +

{
y

cos Θ j  x ′ +  r 2j cos θ j +
 j

Yj
R 2j
 
cos Θ j  y ′ 
 
and
≈
K15149_Book.indb 293
1
rj R j
e
ik rj + R j
i
kR j
cos Θ j
1
rj R j
e
ik rj + R j
i
kR j
 x
cos Θ j 1 +  r 2j −
  j
3X j
R 2j
 yj

 x ′ +  rj2 −
3Yj
R 2j
 
 y ′ 

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294
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Upon expanding and collecting factors for x ′ , y ′ , set
Λ j (u, U ) ≡
uj
rj
Uj
Rj
−
,
where u j ∈{ x j , y j , z j }, U j ∈{X j , Yj , Z j }
uj
Λ ′j (u, U ) =
rj2
uj
Λ ′′j (u, U ) =
Ξ(1)
j =
1
2 rj R j
rj2
e
Ξ(1)
j ( u, U ) =
cos θ j +
−
u U 
cos Θ j + (cos θ j − cos Θ j )  r 2j − R2j 
 j
j 
Uj
R 2j
3U j
R 2j
ik|rj + R j|
(cos θ j − cos Θ j ), Ξ(2)
j =
1
2 rj R j
e
1
2 rj R j
ik|rj + R j|
Λ ′j (u, U ), Ξ(2)
j ( u, U ) =
e
ik ( rj + R j )
1
2 rj R j
e
cos Θ j
i
kR j
ik ( rj + R j )
i
kR j
cos Θ j Λ ′′j (u, U )
Then, to first order in x ′ , y ′ , from Equation (8.30),
1
2 rj R j
(
e
ik rj + R j
)
(cos θ − cos Θ −
j
(2)
≈ Ξ(1)
e
j + Ξj
j
ikΛ j ( x , X ) x ′ ikΛ j ( y ,Y ) y ′
e
(
)
(2)
+ Ξ(1)
j (y , Y ) + Ξ j (y , Y ) e
)
(
)
(2)
+ Ξ(1)
j (x, X ) + Ξ j (x, X ) e
e
∆+
∫
cos Θ j
ikΛ j ( x , X ) x ′ ikΛ j ( y ,Y ) y ′
Furthermore, set I1, j = ∫ ∆ −jj e
and ∆ +j = ∆ −j + δ . Also, set
I 2, j =
i
kR j
∆ +j
∆ −j
e
ikΛ j ( x , X ) x ′
e
ikΛ j ( x , X ) x ′
x′ dx′
e
x′
y′
ikΛ j ( x , X ) x ′ ikΛ j ( y ,Y ) y ′
e
ikΛ j ( x , X ) x ′ ikΛ j ( y ,Y ) y ′
∫
dx ′ , where ∆ −j = − L2 + ( j − 1)( h + δ)

2
− 2
e
ikΛ j ( y ,Y ) y ′
dy ′
and
I 3, j =
∫
∆ +j
∆ −j
dx′
∫

2
− 2
e
ikΛ j ( y ,Y ) y ′
y ′ dy ′
Then, using the identities
sin x =
K15149_Book.indb 294
eix − e− ix
2i
, cos x =
eix + e− ix
2
, sinc x =
sin x
x
, sin 2x = 2 sin x cos x ,
10/18/13 11:04 AM
295
Basic Antenna Theory
and last, ∫ xeikx dx =
that
I1, j =
e
4i
k 2 Λ j ( x , X ) Λ j ( y ,Y )
(
) sin 1 kΛ (x, X )δ cos 1 kΛ (y , Y )
(2 j
) (2 j
)
ikΛ j ( x , X ) ∆ −j + δ2
{(1 − ikΛ (x, X))∆ e
)
(
) eikΛ j ( x , X )( ∆−j + δ2 ) sinc 1 kΛ (y , Y ) − cos 1 kΛ (y , Y )
)
(2 j
)}
{ (2 j
k 2 Λ j ( x , X ) Λ j ( y ,Y )
2i sin 21 kΛ j ( x , X )δ
I 3, j =
{eikx (1 − ikx)}, it can be shown after some routine effort
(
2 δ cos 21 kΛ j ( x , X )δ
I 2, j =
1
k2
e
ikΛ j ( x , X ) ∆ +j
− − i 21 kΛ j ( x , X )δ
j
j
}
sinc ( 21 kΛ j (x , X )δ ) − 1
k 2 Λ j ( x , X ) Λ j ( y ,Y )
whence
U (r0 ) ≈ − iUλ0
≈ − iUλ0
+
(
∑∫
∑
j
Ωj
1
2 rj R j
(
e
ik rj + R j
(2)
4i Ξ(1)
j +Ξ j
)
) (
(2)
1
2 δ Ξ(1)
j ( x , X ) +Ξ j ( x , X ) cos 2 kΛ j ( x , X )δ
2
)
k Λ j ( x , X ) Λ j ( y ,Y )
{(1 − ikΛ (x, X))∆ e
1
− − i 2 kΛ j ( x , X )δ
j
j
+
e
2
j k Λ j ( x , X ) Λ j ( y ,Y )
(
)
(
(2)
1
2i Ξ(1)
j ( y , Y ) +Ξ j ( y , Y ) sin 2 kΛ j ( x , X )δ
2
k Λ j ( x , X ) Λ j ( y ,Y )
)
{
cos θ j −
(
ikΛ j ( x , X ) ∆ −j + δ2
e
ikΛ j ( x , X ) ∆ +j
sinc
e
(
(
1
2
i
kR j
}
)
+ 1 cos Θ j dx ′dy ′
) sin 1 kΛ (x, X )δ cos 1 kΛ (y , Y )
(2 j
) (2 j
)
×
)
}
kΛ j ( x , X )δ − 1
(
ikΛ j ( x , X ) ∆ −j + δ2
) sinc( 1 kΛ (y , Y )) − cos 1 kΛ (y , Y )
(2 j
)}
{ 2 j
□
as required.
8.6 Worked Problems
8.6.1 Exercise
Show that
E = − 4 πε0 ( 1 −
q
)
v −2 1 d
c
r dt
v+
q
4 πε 0
(1 − vc )−1 rv v − ∇ϕ
2
given that
A=
K15149_Book.indb 295
q
1
1
4 πε 0 1− v r − r0
c
v
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296
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Solution
By definition, E = −∇ϕ − ∂t A . Hence, evaluating ∂t A yields:
4 πε 0
q
where v =
d
dt
∂t A = − r12 v ( 1 −
||
r =
)
v+
v −1
c
1
r
(1 − vc )−2 vc ddt v + 1r (1 − vc )−1 ddt v
= − r12 v ( 1 −
v −1
c
)
v+
1
r
(1 − vc )−1 {(1 − vc )−1 vc + 1} ddt v
= − r12 v ( 1 −
v −1
c
)
v+
1
r
(1 − vc )−2 ddt v
d
dt
r is the speed of the particle. Thus,
Eaccel = − 4 πε0 ( 1 −
q
)
v −2 1 d
c
r dt
v and
Estatic =
q
4 πε 0
(1 − vc )−1 rv v − ∇ϕ ,
2
□
as required.
8.6.2 Exercise
Show that in the near zone, the electric field and magnetic field can be
expressed in terms of the dipole moment as
Er (r , θ) ≈
1 p
2 πε r 3
cos θ
Eθ (r , θ) ≈
1 p
4 πε r 3
sin θ
Solution
Recall from Section 8.2 that p = − i ωI δz . Hence, from Definition 8.2.5, substituting this expression into Equations (8.5) and (8.6) yields the desired results. □
8.6.3 Exercise
Suppose a transmitting antenna and a receiving antenna are separated by
a distance much larger than the physical dimensions of either of the two
antennae. Model the two antennae in terms of equivalent circuits, and hence
derive the power absorbed by the receiver.
Solution
The two antennae can be modeled via the following coupled equations:
V1 = I1 Z11 + I 2 Z12
V2 = I1 Z21 + I 2 Z22
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297
Basic Antenna Theory
Transmitter
Z11
I2Z12
I1
Receiver
Z22
+
–
+
–
Equivalent T-network circuit
~
~
Z22
Z11
I2Z12
~
Z12
V1
I2
V2
Figure 8.8.
Equivalent 2-port network circuit representation.
where the transmitting antenna is modeled by (V1 , Z11 ) , Z11 is the load of
the transmitter, and V1 the voltage across the load. The receiving antenna is
modeled by (V2 , Z22 ), where Z22 represents the load of the receiver.
By reciprocity, Z12 = Z21, and hence, the coupled equation, a 2-port network, can be modeled with a 2-port 3-impedance network. That is, the 2-port
network can be represented by a T-network or a Π-network. Representing
the circuit by a T-network is illustrated in Figure 8.8.
Referring to Figure 8.8, Zii corresponds to the input impedance under
open circuit condition. By Kirchhoff’s voltage law,
V1 = Z 11 I1 + Z 12 ( I1 + I 2 ) and V2 = Z 22 I 2 + Z 12 ( I1 + I 2 )
Thus, I 2 = 0 ⇒ V1 = I1 (Z 11 + Z 12 ) and by definition, Z11 = VI11
= Z 11 + Z 12 .
I2 = 0
V2


Likewise, I1 = 0 implies that V2 = I 2 (Z22 + Z12 ) and Z22 = I2 I = 0 = Z 22 + Z 12 .
1
Lastly, V2 = Z 12 I1 when I 2 = 0 implies at once that Z12 = VI12
= Z 12 , whence,
I2 = 0
Z ii = Zii − Z12 ∀i = 1, 2 .
Therefore the antenna transmitter–receiver system may be represented by
an equivalent T-network, where the terminal V2 is attached to a load impedance ZL of the receiver. Under the assumption of weak coupling, that is, the
two antennae are sufficiently far away, I 2 Z12 ≈ 0 and hence, V1 = I1 Z11 represents the transmitting antenna.
To determine the receiving antenna equivalent representation, Thévenin’s
theorem is applied to ZL . Let VT denote the open circuit Thévenin voltage
and Zs the Thévenin source impedance. Then, the Thévenin voltage across
Z12 is obtained by setting ZL = ∞ ⇒ VT = Z12 I1 = ZZ1211 V1 . Likewise, under
open circuit condition, the Thévenin impedance looking from the receiver
Z2
end is: Zs = Z22 − Z12 + Z12 ||(Z11 − Z12 ) = Z22 − Z1211 . Thus, by Kirchhoff’s law,
V2 = ZL (− I 2 ) ⇒ − ZL I 2 = Z12 I1 + Z22 I 2 ⇒ I 2 = − Z22Z12+ ZL I1 . Hence, the time-average power absorbed by the receiving antenna ZL is
⟨ P⟩ = − 21 ℜe(V2 I 2∗ ) =
K15149_Book.indb 297
1
2
2
I 2 ℜe(ZL ) =
1
2
I2
2
2
Z12
Z22 + ZL
ℜe(ZL ).
□
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
8.6.4 Exercise
Show that
δEθ (r , θ) ≈
I 0β
sin θ
πε r (cos θ− 1)
e− iβr e
hβ
− i r sin θ i 21 β (cos θ− 1) 
e
cos ( 2 β(cos θ − 1)) sin
(
hβ
r
)
sin θ .
Solution
From
iI 0β
4 πε
δEθ ≈
{
sin θ
r (cos θ− 1)
(
)
e− iβr 1 − eiβ(cos θ− 1) −
sin θ′
r ′ (cos θ′ − 1)
(
e− iβr ′ 1 − eiβ(cos θ′ − 1)
)} ,
observe from r ′ = r 2 + 4 h2 + 4rh sin θ and θ′ = arccos rL′ , that up to first
order in 1r ,
{
r ′ = r 2 + 4 h2 + 4rh sin θ = r 1 + ( 2rh ) +
1
r′
≈
1
r
{1 + 4rh sin θ}−
1
2
≈
θ′ = arccos rL′ ≈ arccos
1
r
2
4h
r
}
sin θ
1
2
≈ r {1 +
2h
r
sin θ}
{1 − 2rh sin θ}
{ (1 −
L
r
2h
r
}
sin θ ) ≈ arccos Lr = θ
Substituting these approximations into the original first-order expression of
δEθ yields
δEθ (r , θ) ≈
iI 0β
sin θ
4 πε r (cos θ− 1)
(
)
2 hβ
sin θ 
−i
e− iβr 1 − eiβ(cos θ− 1)  1 − e r
.


Finally, to complete the derivation, it suffices to observe that
1 − e− i2 Λ = 2ie− iΛ
eiΛ − e− iΛ
2i
= 2ie− iΛ sin Λ.
Explicitly,
1 − eiβ(cos θ− 1) = e
i 21 β (cos θ− 1) 
= −2ie
1− e
2 hβ
− i r sin θ
=e
− i 21 β (cos θ− 1) 
i 21 β (cos θ− 1) 
hβ
− i r sin θ
giving the desired result.
K15149_Book.indb 298
(e
−e
i 21 β (cos θ− 1) 
)
sin ( 21 β(cos θ − 1) ) ,
 ei hrβ sin θ − e− i hrβ sin θ  = 2ie− i hrβ sin θ sin


(
hβ
r
)
sin θ ,
□
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299
Basic Antenna Theory
8.6.5 Exercise
Work out Example 8.5.11 for a circular aperture of radius δ > 0 using (8.30)
at some observation point P0 = ( x0 , y 0 , z0 ) ∈R 3, where δ << r0 = x02 + y 02 + z02 .
Solution
Fix and suppose that a uniform field U = U 0 e− iβz impinges upon an aperture
Ω = {( x , y , 0) ∈R 3 : x 2 + y 2 ≤ δ 2 }, where ∂R 3+ − Ω is a perfect electrical conducy
tor. Set R = x02 + y 02 . Then, cos Φ = xR0 and sin Φ = R0 . The scalar field at P0
via (8.30) as follows. Let cos θ = − r10 r0 ⋅ n, where n is the normal unit vector
field on Ω pointing towards R 3− . For each r′ ∈Ω , let cos θ = − r1 r ⋅ n , where
y′
r = r0 − r ′. Finally, set cos φ = xr′′ and sin φ = r ′ .
kr
i
From (8.30), U (r ) = − λi ∫ Ω r1 e U ( x ′ , y ′)cos θdx ′dy ′ . To first order in r10 ,
{
r = ( x0 − x ′)2 + ( y 0 − y ′)2 + z02 ≈ r0 1 −
x0 x ′+ y 0 y ′
r02
Furthermore,
{
cos θ = cos θ rr0 ≈ cos θ 1 +
}
and
1
r
≈
1
r0
{1 +
x0 x ′+ y0 y ′
r02
}.
}
x0 x ′+ y0 y ′
r02
and
r ≈ r0 −
x0
r0
r ′ cos φ −
y0
r0
r ′ sin φ,
whence,
1
r
eikr cos θ ≈
cos θ
r0
{
2( x0 x ′+ y 0 y ′ )
eikr 1 +
r02
}
{1 +
− i( k /r0 ) r ′ {x0 cos φ+ y 0 sin φ}
2 x0
r02
=
cos θ
r0
eikr0 e
=
cos θ
r0
eikr0 e− i( k/r0 )R{cos Φ cos φ+ sin Φ sin φ}r ′ 1 +
=
cos θ
r0
eikr0 e− i( k/r0 )R cos(φ−Φ)r ′ 1 +
{
2 x0
r02
{
r ′ cos φ +
2 x0
r02
r ′ cos φ +
2 y0
r02
r ′ cos φ +
2 y0
r02
}
r ′ sin φ
2 y0
r02
}
}
r ′ sin φ
r ′ sin φ .
kR
Next, set φ = ϕ−Φ and ξ = kR
r0 r ′ . Then, dφ = dϕ and dξ = r0 dr ′ . Also, note
from the axisymmetry of the problem that the following integral is invariant
under angular rotation:
∫
2π
0
K15149_Book.indb 299
e− i( kR/r0 )r ′ cos(φ−Φ) dφ =
∫
2 π−Φ
−Φ
e− i( kR/r0 )r ′ cos ϕ dϕ =
∫
2π
0
e− i( kR/r0 )r ′ cos ϕ dϕ.
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300
Electromagnetic Theory for Electromagnetic Compatibility Engineers
Lastly, noting that the zero-order Bessel’s function of the first kind may also be
defined by
J 0 (ζ) =
1
2π
2 x0
r02
2 y0
∫
2π
0
e− iζ cos ϕ dϕ ,
it follows that on setting
∫
Ω
1
r
I0 =
eikr cos θ ≈
δ
2π
∫∫
0
e
0
cos θ
r0
{
eikr0 I 0 +
− i( k / r0 ) Rr ′ cos ϕ
I1 +
r ′ dϕ d r ′ =
r02
δ
I2
}
δ
∫ J ( r ′ )r ′dr ′ = ( ) ∫
0
0
r0 2
kR
kR
r0
0
ξJ 0 (ξ)dξ =
( )
r0 2
kR
δJ1 (δ ),
i 2 π i( ϕ+ζ cos ϕ )
dϕ is the first-order Bessel’s funcwhere δ = kR
r0 δ and J 1 (ξ ) = − 2 π ∫ 0 e
0δ
tion of the first kind in integral representation. Thus, I 0 = rkR
J1 kR
r0 δ .
2π
δ
− i( kR/r0 ) r ′ cos ϕ
The second term I 2 = ∫ 0 dr ′ ∫ 0 r ′dϕ e
r ′ cos(ϕ + Φ) . So, from the
expansion
{
}
( )
cos(φ+Φ) = cos φ cos Φ − sin φ sin Φ,
∫
2π
∫
2π
{
= ( r ′ ) cos Φ
2
0
}
r ′ dϕ e− i( kR/r0 )r ′ cos ϕ r ′ cos(ϕ + Φ)
0
e− i( kR/r0 )r ′ cos ϕ cos ϕ dϕ − ( r ′ ) sin Φ
2
∫
2π
0
e− i( kR/r0 )r ′ cos ϕ sin ϕ dϕ
Set ψ = cos φ ⇒ dψ = −sin φdφ and ψ(0) = 1,ψ(2π) = 1. Thus,
∫
2π
0
e− i( kR/r0 )r ′ cos ϕ sin ϕ dϕ = −
1
∫e
− i( kR/r0 ) r ′ψ
1
dψ = 0
yielding
∫
2π
0
{
}
r ′ dϕ e− i( kR/r0 )r ′ cos φ r ′ cos(ϕ + Φ) = ( r ′ ) cos Φ
2
∫
2π
0
e− i( kR/r0 )r ′ cos φ cos ϕ dϕ
However, by definition,
J1 (ξ) = − 2iπ
∫
2π
0
ei(ϕ+ζ cos ϕ ) dϕ = − 2iπ
∫
2π
0
eiζ cos ϕ cos ϕ dϕ
implies at once that
∫
2π
0
K15149_Book.indb 300
{
}
(
r ′ dϕ e− i( kR/r0 )r ′ cos ϕ r ′ cos(ϕ + Φ) = i2 π ( r ′ ) cos ΦJ1 − kR
r0 r ′
2
)
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301
Basic Antenna Theory
Thus,
I 2 = i2 π xR0
∫
δ
(
)
dr ′ ( r ′ ) J1 − kR
r0 r ′ .
2
0
{
}
The final term I 3 = ∫ δ0 dr ′ ∫ 20 π r ′dϕ e− i( kR/r0 )r ′ cos ϕ r ′ sin(ϕ + Φ) . Once again, via
the expansion
sin(φ + Φ) = sin φ cos Φ + sin Φ cos φ
∫
2π
= ( r ′ ) cos Φ
∫
2π
= i2 π ( r ′ )
J1 − kR
r0 r ′
0
2
0
{
e− i( kR/r0 )r ′ cos ϕ sin ϕ dϕ + ( r ′ ) sin Φ
2
(
2 y0
R
}
r ′ dϕ e− i( kR/r0 )r ′ cos ϕ r ′ sin(ϕ + Φ)
∫
2π
0
e− i( kR/r0 )r ′ cos ϕ cos ϕ dϕ
)
Thus, from Reference [1, Section 11.1.1],
∫
z
0
zpΓ
p
t J q (t)dt =
Γ
(
(
p + 1+ 1
2
q− p+1
2
)
∞
)∑
( q + 2 k + 1) Γ
Γ
k=0
(
(
q− p+1
+k
2
q+ p+3
+k
2
)
)J
q+ 2 k +1
( z)
where Γ( z) = ∫ ∞0 t z − 1e− t dt is the Gamma function, with Γ(n) = (n − 1)! ∀n ∈ N, it
is clear that
δ
∫ (r′)
2
0
J1 ( κr ′ ) dr ′ =
1
κ3
∫
κδ
0
∞
2
t J1 (t)dt =
2 δ2
κ
∑
2( k + 1) Γ ( 1+ k )
2( k + 1)
( k + 1)!
J
(κδ)
k=0
where κ = − kR
r0 .
The complete solution is thus given by
{
U (r0 ) ≈ − iUλ0 e− iβz cos θ r10 eikr0 I 0 +
=−
iU 0
λ
e
− iβz
cos θ e
1
r0
ikr0
2 x0
r02

r0 δ
 kR J1

In particular, to first order in
1
r0
I1 +
r02
( δ) −
kR
r0
I3
}
2
i8 πR r0 δ
kR
r02
∞
∑
2( k + 1) Γ ( 1+ k )
2( k + 1)
( k + 1)!
k=0
J
(−
kR
r0

δ 

)
,
U (r0 ) ≈ − iUλ0 e− iβz eikr0
K15149_Book.indb 301
2 y0
δ
kR
J1
( δ ) cos θ
kR
r0
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302
Electromagnetic Theory for Electromagnetic Compatibility Engineers
( )
( )
kR
and the field vanishes at the zeros of J1 kR
r0 δ ; that is, the roots of J 1 r0 δ = 0 .
As a side remark, Bessel’s functions have a countably infinite number of zeros.
Moreover, for 0 < ε << 1, J1 (ε) ~ 2ε asymptotically, and hence, for λ very
kR
large, kR
r0 δ << 1 ⇒ J 1 r0 δ ≈ 0 . In other words, for very large wavelengths,
very little radiation is emitted from a small circular aperture. Indeed, a similar result was observed for a rectangular aperture.
□
( )
References
1. Abramowitz, M. and Stegun, I. (Eds.) 1972. Handbook of Mathematical Functions
with Formulas, Graphs and Mathematical Tables. Dept. of Commerce, National
Bureau of Standards, AMS 55.
2. Balanis, C. 1982. Antenna Theory: Analysis and Design. New York: John Wiley & Sons.
3. Chang, D. 1992. Field and Wave Electromagnetics. Reading, MA: Addison-Wesley.
4. Goodman, J. 1996. Introduction to Fourier Optics. New York: McGraw-Hill.
5. Hecht, E. 1987. Optics. Reading, MA: Addison-Wesley.
6. Neff, H. 1981. Basic Electromagnetic Fields. New York: Harper & Row.
7. Plonsey, R. and Collin, R. 1961. Principles and Applications of Electromagnetic
Fields. New York: McGraw-Hill.
8. Schwartz, M. 1972. Principles of Electrodynamics. New York: Dover.
9. Silver, S. 1949. Microwave Antenna Theory and Design. New York: McGraw-Hill.
10. Koshlyakov, N., Smirnov, M., and Gliner, E. 1964. Differential Equations of
Mathematical Physics. Amsterdam: North-Holland.
11. Stratton, J. 1941. Electromagnetic Theory. New York: McGraw-Hill.
K15149_Book.indb 302
10/18/13 11:05 AM
9
Elements of Electrostatic Discharge
Electrostatic discharge can often be a source of immense annoyance to
EMC engineers, and more important, it can also be extremely costly for an
electronics manufacturer if countermeasures to mitigate discharge are not
implemented. As a simple example, consider a person walking on a carpet
during winter when the humidity is low who touches a port on the back of
an electronic device. The charge buildup and its subsequent discharge into
the port may potentially destroy sensitive electronic components. Hence,
electrostatic discharge can be potentially damaging to electronic equipment.
This chapter is intended to acquaint the reader with the physics of electrostatic discharge at an informal level.
9.1 Electrostatic Shielding
In order to gain insight into the breakdown of dielectrics under high voltage,
it is necessary to understand the microscopic behavior of the dielectrics under
the influence of an electric field [5,6,12]. In particular, the application of quantum mechanics in the theory of dielectrics is unavoidable: this leads to the
topic of solid-state theory, and the reader unfamiliar with quantum mechanics, but who is nevertheless interested in pursuing more advanced aspects,
may consult some introductory references such as [1,13] geared toward solidstate theory. On the other hand, for the more empirically minded reader,
Paschen breakdown curves have been studied extensively, and describe the
maximal field between two electrodes in a gaseous medium needed to initiate breakdown [7].*
A somewhat simplified view of a dielectric is sketched in Figure A.5
(Appendix). The first question that springs to mind is the following: does
breakdown occur within a dielectric when its surface cannot hold any more
free charges? This in turn begs another question. How much free charges
can a dielectric hold before breakdown occurs? This question implies tacitly
that the dielectric material is embedded in some homogeneous (dielectric)
medium. The simplest approach to the latter is to consider an isolated conducting sphere embedded in a dielectric medium.
* The original work was investigated by Paschen in the nineteenth century [12].
303
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
9.1.1 Proposition
Given a neutral spherical conductor S 2 ⊂ (R 3 , µ , ε) , where S 2 = {r ∈R 3 :||
r ≤ δ},
suppose the electric field strength required to cause dielectric breakdown of
(R 3 , µ , ε) is E0 . Then, the maximal free charge Q that S 2 can hold prior to the
ionization of the dielectric medium is
Q = 4πεE0 δ 2
(9.1)
Proof
Recall from Example 3.2.1, it was observed that when a point charge q is sufficiently close to S 2, the mirror image will attract q to S 2 . In particular, whenever
q is sufficiently close to S 2, it will be attracted to the surface of S 2. In principle,
S 2 can absorb an arbitrarily large charge, as long as there is a sufficiently large
force to drive q sufficiently close to the surface to overcome Coulomb repulsion.
Now, given a charge Q on S 2 , the electric field on the surface is E = 41πε δQ2 .
Because E0 is the dielectric breakdown of (R 3 , µ , ε) , it follows that ionization
of the medium will begin when
E0 =
1 Q
4 πε δ 2
⇒ Q = 4πεE0 δ 2
□
as required.
Indeed, it is clear from Proposition 9.1.1 that even from a classical analysis
perspective, a spherical conductor can hold more charges prior to dielectric
breakdown if the electric permittivity of the medium is large or if the radius
of the spherical conductor is large.
This section ends with a related topic that is frequently employed in EMC
design: electrostatic shielding. This is clearly one way of avoiding dielectric breakdown! Consider Figure 9.1, where Ci ⊂ R 3 denotes a compact conductive boundary of the i th conductor, and for notational convenience, let
D(Ci ) ⊂ R 3 define the compact space bounded by Ci , that is, Ci = ∂D(Ci ) ,
and set D(Ci ) to be the interior of D(Ci ). Suppose conductor C1 ⊂ D(C2 ) and
Ci ⊂ R 3 − D(C2 ) ∀i ≥ 3. Finally suppose that C2 is grounded: V2 = 0.
From Lemma 6.1.3,
Q1 = C11V1 + + C1nVn
Qn = Cn1V1 + + CnnVn
Thus, V2 = 0 ⇒ V1 is completely determined with respect to C2 inasmuch as
all of the electric field from C1 terminates on C2, and not on Ci ∀i ≥ 3 . Thus,
C1i = 0 ∀i ≥ 3 ; that is, the electric field on C1 does not couple with Ci ∀i ≥ 3.
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305
Elements of Electrostatic Discharge
C3
V3
C2
C1
Cn
V1
Vn
V1 = 0
Figure 9.1
Electrostatic screening of a conductor.
Hence, the potential on C1 is independent of the potential on Ci ∀i ≥ 3 as
C1i = Ci 1 ∀i. By reciprocity, Ci ∀i ≥ 3 are shielded from the potential of C1.
What is more interesting to note—albeit it is intuitively obvious—is that if
Vi = 0 ∀i ≥ 2 , then C1i = 0 ∀i ≥ 3 implies at once that the system of equations
reduces to:
Q1 = C11V1
and Q2 = C21V1
In particular, Qi = 0 ∀i ≥ 3, and hence, the charge on C1 will not induce any
charge on Ci ∀i ≥ 3. Moreover, because Q2 = −Q1 as free charges on C1 will
induce the opposite charge on C2, it follows that C11 = −C21 = C12 .
9.2 Dielectric Properties: the Kramers–Kronig Relations
In the first half of this section, the electromagnetic properties of dielectrics
are investigated from a classical perspective. Without loss of generality, time
harmonicity is assumed: e− iωt . Recalling that dielectric polarization is the
result of bound charges, the rearrangement of bound charges within the
dielectric results in “bound” currents, in complete analogy with the movement of free charges resulting in conduction currents. In what follows, define
the conduction current by J = σ 1 E and the displacement current by D = ε1 E .
So, from Maxwell’s ∇ × B = µJ + µε 1 ∂t E, define ∇ × B = −iωμεE, where
ε = ε1 +
iσ 1
ω
≡ ε1 + iε 2
(9.2)
Furthermore, it can be shown [8] classically that conductivity is complex:
g
σ = σ 1 + iσ 2 , σ i ∈R , i = 1, 2 . More specifically, σ 1 = K g 2 +ω 2 and σ 2 = K g 2 ω+ω 2 ,
where K is some real constant and g >> 1 is a damping constant that is
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306
Electromagnetic Theory for Electromagnetic Compatibility Engineers
dependent upon the average collision rate between an electron and the crystal lattice of the conductor. Thus, in view of Equation (9.2), define
ε = ε0 +
iσ
ω
(9.3)
Then, ℜe(ε) = ε 0 − g 2 K+ω 2 and ℑm(ε) = ω ( g 2K+ω 2 ) . For low ω > 0, that is, g >> ω,
σ 1 >> σ 2 ⇒ σ ≈ σ 1. In the microwave regime, g >> ω typically holds and hence,
the electrical conductivity may be assumed real. In the optical regime, this is
false in general.
A more rigorous approach is given below, culminating in the Kramers–
Kronig relations, also known as the dispersion relations, for the electric permeability and electric conductivity of a dielectric. In what follows, the dielectric
is assumed to be isotropic and homogeneous for ease of analysis, and the
response of the dielectric under a forcing function to be linear; this guarantees that the Fourier transform may be invoked.
By way of motivation, consider D = ε1 E = P + ε 0 E ⇒ P = (ε1 − ε 0 )E . Recall
that the polarization field is the result of displaced electron clouds about
the nuclei; see Figure 9.2. The displaced electron clouds generate (atomic or
molecular) dipoles within the dielectric subject to an external electric field.
That is, the polarization field is the response to a driving electric field, and
hence, ε1 − ε 0 is the response function.
Note that the assumption of response linearity implies that the Fourier
transform may be employed:
P(ω ) = ε (ω )E(ω ) ⇒ P(t) =
∫ ε(ω)E(ω)e
− iωt
dt
where ε is the response function. Taking the Fourier transform (in frequency
domain) leads to
F [P ] =
∫ P(t)e
iωt
dt =
∫e
Neutral atom
(zero external field)
iωt
∫
dt ε (t − τ)E(τ)dτ =
∫e
iωτ
∫
dτ ε (t − τ)E(τ)eiω (t −τ ) dt.
Applied electric field
Electron cloud
Dipole
+
–
Nucleus
Figure 9.2
Electron cloud distortion resulting from an applied electric field.
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Elements of Electrostatic Discharge
That is,
F [P ](ω ) =
∫ E(τ)e
iωτ
∫
dτ ε (t − τ)eiω (t −τ ) dt ≡ F [ E ](ω )F [ε ](ω ).
Last, it is assumed tacitly here that response functions are analytic in ω and in
particular, purely on physical grounds, response functions must tend to zero
in the limit as ω → ∞ as a physical system cannot respond quickly enough to
an infinitely large frequency. That is, response functions are assumed to be
bounded and analytic.
9.2.1 Theorem (Kramers–Kronig Relations)
Suppose a dielectric medium is isotropic, homogeneous, and causal,* then
the electric conductivity and permittivity of the medium satisfy
σ 1 (ω ) =
2
π
∫
ω ′f2 ( ω )
ω ′ 2 −ω 2
0
σ 2 (ω ) = − 2πω
ε1 (ω ) = ε 0 +
2
ε 2 (ω ) = − πω
∞
∫
∞
∫
∞
∫
f2 ( ω )
ω ′ 2 −ω 2
0
2
π
∞
0
dω ′
0
dω ′
ε 2 ( ω ′ )ω ′
ω ′ 2 −ω 2
dω ′
( ε1 ( ω ′ )−ε 0 )ω ′ 2
ω ′ 2 −ω 2
dω ′
(9.4)
(9.5)
(9.6)
(9.7)
Proof
Consider the Fourier transform F [ε ]( z) = ∫ ∞−∞ ε (t − τ)eiz(t −τ )dt , where z ∈ C. So,
setting z = x + iy, x, y ∈ R (no bearing on the components of the rectangular
coordinate system), and invoking causality, that is, ε (t − τ) = 0 ∀t < τ, it follows that
F [ε ]( z) =
∫
t
−∞
ε (t − τ)eiz(t −τ ) dt =
∫
t
−∞
ε (t − τ)e− y (t −τ )eix(t −τ ) dt
Now, observe that y > 0 ⇒ e− y (t −τ ) < ∞ ∀t > τ , and the causality assumption implies that F [ε ]( z) = 0 ∀t − τ < 0 ⇒ F [ε ]( z) ≠ 0 ∀y > 0 . That is, F [ε ]( z) is
analytic on the upper half-plane C + = { z ∈C : ℑm( y ) ≥ 0} of C. So, consider
a closed contour Γ ⊂ C + defined by Figure 9.3, where a semi-circular loop
Γ = Γ ∗ ρ− ∗ γ ∗ ρ+ is defined as follows: γ is a semi-circle of radius δ > 0 about
the point z0 on the real line, and Γ is a semi-circle of radius R >> δ, ρ− is the
path from [− R , z0 − δ], and ρ+ is the path from [ z0 + δ , R] . Then, by definition,
* Informally, causality implies that there is zero response for all times prior to the application
of a forcing function; that is, cause and effect.
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308
Electromagnetic Theory for Electromagnetic Compatibility Engineers
iy
Г
z = x + iy
γ
ρ_
z0
0
x
ρ+
Figure 9.3
Contour integral on the upper-half complex plane.
 F [ε ]( z) is analytic
for all points within the space bounded by Γ and on Γ,
and hence, for z0 ∉ D(Γ ) = { z ∈C + :|z|< R} − { z ∈C + :|z − z0|< δ}, the domain
 implies that Fz[−ε ](z z ) is also analytic on D(Γ ) . Hence, by the
bounded by Γ,
0
Cauchy–Goursat theorem,*
∫
0=
Γ
F [ ε ]( z )
z − z0
dz +
∫
F [ ε ]( z )
z − z0
d z = 0 . Thus,
∫
F [ ε ]( z )
z − z0
dz +
Γ
γ
∫
ρ−
F [ ε ]( z )
z − z0
dz +
∫
ρ+
F [ ε ]( z )
z − z0
dz
From Equation (9.3), lim ∫ Γ Fz[−ε ](z0z ) dz → 0 . To see this, it suffices to observe
R→∞
that F [ε ] is analytic and bounded (by assumption) on D(Γ ) and in particular,
′
ℜeF [ε ] ≤ M
and ℜeF [ε ] ≤ MR′′ , for some constants M ′ , M ′′ > 0 . Furthermore,
R2

lim
R→∞
∫
Γ
1
z − z0
d z = lim
π
R→∞
∫ {1 −
0
z0
R
e− iθ
}
−1
dθ = π < ∞
where ∫ {1 − ae− iθ }−1 dθ = − i ln( a − eiθ ) was employed. Thus, lim ∫ Γ Fz[−ε ](z0z ) dz → 0,
R→∞
as claimed. Thus, it remains to evaluate

I 0 ≡ lim
δ→0
R→∞
∫
R
z0 +δ
F [ ε ]( z )
z − z0
d z + lim
δ→0
R→∞
∫
z0 −δ
−R
F [ ε ]( z )
z− z0
d z = − lim
δ→0
∫
γ
F [ ε ]( z )
z− z0
dz
First, observe that as F [ε ] is analytic on C + , it follows that the residue of
is F [ε ]( z0 ) . For notational simplicity, set f = F [ε ]. Then,
F [ ε ]( z )
z − z0
I 0 = − lim
δ→ 0
∫
γ
f ( z )− f ( z0 )+ f ( z0 )
z − z0
= − lim
δ→ 0
∫
γ
f ( z0 )
z − z0
− lim
δ→ 0
∫
γ
f ( z )− f ( z0 )
z − z0
≡ I 0(1) + I 0(2)
* See Reference [3]: if a function is analytic on a complex domain bounded by a simple loop,
then the loop integral is zero.
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Elements of Electrostatic Discharge
Now, the analyticity of f implies that f ( z) = f ( z0 ) + f ′( z0 )( z − z0 ) +  and
hence, I 0(2) = 0. Thus, evaluating I 0(1) completes the initial part of the proof.
The method is routine: set z = δeiθ . Then, dz = iδeiθ dθ and evaluating the
integral yields
∫
γ
=i
dz
z − z0
∫
0
π
{
δeiθ δeiθ − z0
}
−1
dθ = − i
π
∫ {1 − z δe }
− iθ −1
0
0
1
dθ = ln zz00δ−
δ+ 1
whence,
I 0 = − f ( z0 )ln(−1) = − f ( z0 )ln eiπ = − iπf ( z0 ) = − iπ F [ε ]( z0 )
That is, upon replacing z0 → ω, it follows at once that
F [ε ](ω ) = − πi P
∫
∞
−∞
F [ ε ]( ω ′ )
ω ′ −ω
(9.8)
dω ′
where P denotes the Cauchy principal value and it is defined by
P
∫
∞
F [ ε ]( ω ′ )
ω ′ −ω
−∞
dω ′ = lim
r→∞
∫
r
−r
F [ ε ]( ω ′ )
ω ′ −ω
dω ′
Next, setting F [ε ](ω ) ≡ f = f1 (ω ) + if2 (ω ) yields
f1 (ω ) = π1 P
∫
∞
−∞
f2 ( ω ′ )
ω ′ −ω
dω ′ and
f2 (ω ) = − π1 P
∫
∞
−∞
f1 ( ω ′ )
ω ′ −ω
dω ′
Also, observe that a response function, for f ∈ {ε,σ}, satisfies f (−ω ) = f ∗ (ω ),
hence f1 (−ω ) = f1 (ω ) and f2 (−ω ) = − f2 (ω ). This is because
F [ε ](ω ) =
∫
t
−∞
ε (t − τ)eiω (t −τ ) dt ⇒ F [ε ]∗ (ω ) =
∫
t
ε (t − τ)e− iω (t −τ ) dt
−∞
Thus, the real part is an even function whereas the imaginary part is an odd
function of ω, whence,
f1 (ω ) = π1 P
∞
∫

= π1 P 

∫
K15149_Book.indb 309
∫
0
∞
0
∫
dω ′
f2 (ω ′ )
ω ′ −ω
−∞

= π1 P −

= π2 P
f2 (ω ′ )
ω ′ −ω
−∞
∞
0
dω ′ +
− f2 (ω ′ )
− (ω ′ +ω )
ω ′f2 (ω )
ω ′ 2 −ω 2
∫
∞
0
f2 (ω ′ )
ω ′ −ω
( − dω ′ ) +
∫


dω ′  = π1 P −


∞
0
f2 (ω ′ )
ω ′ −ω
∫
−∞
0
f2 (ω ′ )
ω ′ −ω


dω ′  = π1 P 


∫
∞
0
dω ′ +
∫
∞
0
f2 (ω ′ )
ω ′ −ω
f2 (ω ′) ( ω ′1−ω +

dω ′ 

1
ω ′ +ω
) dω ′ 

dω ′
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
and likewise,
f1 (ω ) = π1 P
∫
∞
f2 ( ω ′ )
ω ′ −ω
−∞

= − π1 P −

∫
∫
∞
= − 2πω P
∞
0
0
dω ′
f2 ( ω ′ )
− ( ω ′ +ω )
f2 ( ω )
ω ′ 2 −ω 2
( − dω ′ ) +
∫
∞
f2 ( ω ′ )
ω ′ −ω
0


dω ′  = π1 P 


∫
∞
f2 (ω ′) ( ω ′1−ω −
0
1
ω ′ +ω
) dω ′ 

dω ′
Furthermore, replacing f by σ based on the discussion preceding the theorem, it follows via σ(ω ) = σ 1 (ω ) + iσ 2 (ω ) that
σ 1 (ω ) = π2 P
∞
∫
ω ′f2 ( ω )
2
ω ′ −ω
0
2
dω ′ and σ 2 (ω ) = − 2πω P
∞
∫
f2 ( ω )
ω ′ 2 −ω 2
0
dω ′
Also, noting from definition ε = ε 0 + i ωσ that σ = ωε 2 − iω(ε − ε 0 ), and hence,
substituting ωε1 = σ 1 and −ω(ε − ε 0 ) = σ 2 yields
∫
ε1 (ω ) = ε 0 + π2 P
∞
0
ε 2 ( ω ′ )ω ′
ω ′ 2 −ω 2
dω ′ and ε 2 (ω ) = − 2πω P
∫
∞
0
ε1 ( ω ′ )−ε 0
ω ′ 2 −ω 2
dω ′
Last, noting that as the analytic functions εi are bounded in the limit as ω → ∞,
it follows that the integrals are absolutely convergent and hence,
ε1 (ω ) = ε 0 +
2
π
∫
∞
0
ε 2 ( ω ′ )ω ′
ω ′ 2 −ω 2
2
dω ′ and ε 2 (ω ) = − πω
∫
∞
0
( ε1 ( ω ′ )−ε 0 )ω ′ 2
ω ′ 2 −ω 2
dω ′
□
as required.
The above dispersion relations show how the real and imaginary components are related to one another in the frequency domain.
9.2.2 Corollary
The DC conductivity σ DC of a conductor is given by
σ DC =
2
π
∫
∞
0
(ε 0 − ε1 (ω ))dω
(9.9)
Proof
Because the DC electric conductivity must be real, Equation (9.4) can be
invoked. So, setting ω = 0 for the DC case, and substituting σ 2 = ω(ε 0 − ε1 ),
(9.4) leads immediately to
σ 1 (0) =
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2
π
∫
∞
0
σ 2 (ω ′ )
ω′
dω ′ =
2
π
∫
∞
0
1
ω′
(ε 0 − ε1 (ω ′))ω ′ dω ′ =
2
π
∫
∞
0
(ε 0 − ε1 (ω ′))dω ′ □
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By inspecting Equation (9.6), it is clear that in the limit as ω → 0, the real
∞
part of the electric permittivity becomes ε1 (0) = ε 0 + π2 ∫ 0 ε2ω(2ω ) dω . Indeed,
these expressions, including those from Theorem 9.2.1 and Corollary 9.2.2,
enable the real part of electric permittivity to be determined when the imaginary part of the electric permittivity is known. What is more interesting to
note is the following: the Kramers–Kronig relations display how the real and
imaginary parts are related; they are not independent of one another!
9.2.3 Remark
It can be shown (see Exercise 9.5.1) that the dielectric constant of a material can be determined classically by considering a collection of oscillating
dipoles subject to a time-harmonic monochromatic plane wave:
ε(ω )
ε0
= 1+
Ne 2
ε 0 m0
∑
fj
2
2
j ω j −ω − iγ j ω
(9.10)
where fi denotes the probability that an electronic transition will take place
at the resonant angular frequency ωj of the jth electronic dipole oscillator. It
is a quantum mechanical phenomenon and must clearly satisfy ∑ j f j = 1.
e denotes the electronic charge, m0 the mass of an electron, and γj
Moreover, ||
corresponds to the damping for the jth electronic dipole oscillator.
9.3 Beyond Classical Theory
The dielectric properties of materials cannot be understood completely
within the context of classical physics. As such, a brief venture into the
world of quantum mechanics is unavoidable. Indeed, quantum mechanics
is a necessary evil to understand dielectric breakdown phenomena. Thus, a
short summary is provided in order to establish some notations; the interested reader may satisfy his or her intellectual craving with a plethora of
books on quantum mechanics and solid-state theory. For example, see also
References [2,4] in addition to those given in Section 9.1. Clearly, no pretense is made for completeness; only a brief informal outline is sketched.
9.3.1 Axioms (Quantum Mechanics)
The following axioms describe the basic tenet of quantum mechanics.
(a) The state ψ of a (physical) system is completely described by a unit
vector |ψ⟩ in a Hilbert space.* That is, the inner product ⟨ψ|ψ⟩ = 1 for
the unit vector.
* See Appendix A.4 for the definition of a Hilbert space. For the present, the reader may view
this space as a vector space endowed with an inner product (the “dot” product) structure.
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(b) A physical observable A is a Hermitian operator on the Hilbert space;
that is, A = A † , where, given any two states ψ, φ, A † is defined by
⟨ψ|Aϕ⟩ = ⟨A † ψ|ϕ⟩.
(c) Given an observable A, let {|ψ n ⟩} ⊂ H be a set of eigenvectors of A, that
is, A|ψ n ⟩ ≡|Aψ n ⟩ = an |ψ n ⟩ ∀n = 1, 2, , where H denotes the Hilbert
space of states and an ∈R ∀n. Then, the probability of measuring
the value an when the system is in state ψ is precisely |⟨ψ n|ψ⟩|2 . In
particular, the measurement of an observable corresponds precisely
to one of its eigenvalues. Moreover, the state ψ immediately after the
measurement collapses down to the state ψn : |ψ⟩ →|ψ n ⟩ , so that the
probability of finding the state in |ψ n ⟩ immediately after the measurement is 1.
(d) States are defined by the Schrödinger equation: ddt |ψ⟩ = − i 2hπ H|ψ⟩,
where h = 6.63 × 10−34 J ⋅ s is called the Planck constant and H is the
Hamiltonian operator, and it is associated with the energy of the system; that is, the eigenvalues correspond precisely to the energy states
of the system. The Hamiltonian is defined by H = − 21m  2 ∆ + U , where
 = 2hπ , m is the mass of the particle described by |ψ⟩, U is the potential
energy operator, and when it is purely a function of position, it acts
upon the state function via multiplication; that is, Uψ = Uψ , where
xψ ≡ xψ.
9.3.2 Remark
(a) Classically, the state of a system can be described by the positions and
momenta of all the particles comprising the system. In quantum mechanics,
the state of a closed system can be completely described by a state vector in
Hilbert space. (b) A classical observable is any measureable quantity that can
be attributed to a system, for example, energy, momentum, position, and the
like. (c) The classical Hamiltonian of a system is defined by H = T + U, where T
is the kinetic energy of the system and U is the potential energy of the system.
A critical point to note is the following: the transition from classical to quantum
mechanics transforms classical observables into Hermitian operators acting on
a Hilbert space. Finally, a wave function is described at an abstract level by a
unit vector in Hilbert space, which completely describes the state of a particle.
An important point in introducing the elements of quantum mechanics here
is to illustrate the concept of quantum tunneling. This concept has wide applications in engineering, an example of which is tunneling diodes. Tunneling
is a purely quantum phenomenon that does not exist in the classical setting.
Explicitly, suppose a particle is trapped in a square potential well U with a
finite width potential barrier (see Figure 9.4). If the kinetic energy T of the particle satisfies T < U, then classically, it is impossible for the particle to escape the
potential barrier. However, in quantum mechanics, there is a nontrivial probability that the particle can travel beyond the potential barrier.
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Square potential well with finite width barrier
U = U0
U = U0
T < U0
U=0
x=0
U=0
x=a
x=b
Figure 9.4
Finite potential barrier illustrating quantum tunneling phenomenon.
9.3.3 Definition
A wave function ψ representing the state vector |ψ⟩ of a particle is called the
probability amplitude of the particle, and the probability that the particle exists
in some compact space K ⊂ R 3 is given by Pψ ( K ) = ∫ K ψ ∗ (r )ψ (r )d 3 r , where
∫ R3 ψ ∗ (r )ψ (r )d 3 r = 1.* Moreover, the probability current density of the state is
defined by j = − 2im (ψ ∗∇ψ − ψ∇ψ ∗ ).
Indeed, it can be shown [4] that ∂t ρ + ∇ ⋅ j = 0 , where ρ ≡ ψ ∗ ψ defines the
probability density function. This leads to the conservation of probability. See
the charge conservation equation in Maxwell’s theory, and hence the respective terminologies defined above. Last, the following definitions are required
to describe the phenomenon of tunneling. Suppose ψ 0 is the incident wave,
ψ r the reflected wave, and ψ t the transmitted wave across a barrier. Then,
R = jj0r defines the reflection coefficient whereas S = jj0t defines the transmission coefficient of the wave function.
9.3.4 Example
Consider a one-dimensional scenario, wherein a particle is trapped in a
square potential well of width a as illustrated in Figure 9.4. Let the potential
energy of the barrier be defined by
 U 0 for x ∈ D1 ∪ D3
U=
 0 for x ∈ D2 ∪ D 4
where D1 = (−∞ , 0], D2 = (0, a) , D3 = [ a, b] and D4 = ( a, ∞) .
Suppose the particle has a kinetic energy T = E and mass m. Inasmuch as
the potential energy is independent of time, the particle is in a stationary
state and hence, via Exercise 9.5.3, the solution satisfies the time-independent
* The probability that the particle is somewhere in space is unity. Here, it is implicitly assumed
that the wave function is normalized.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Schrödinger equation H|ψ⟩ = E|ψ⟩ . That is, the one-dimensional Schrödinger
equation for the particle is described by
{−
where E =
2
1 p
2 m
1
2m
}
 2 ∆ + U ψ = Eψ ⇒ − 21m  2
d2
dt 2
ψ = (E − U )ψ
, p = mv is the momentum of the particle.
On D1 ∪ D3 , Schrödinger equation becomes
− 21m  2
d2
dt 2
ψ + U 0 ψ = Eψ
So, on setting β 2 = 2 m ( 1 ) (U 0 − E) , the equation becomes
hence, the general respective solutions are:
2
d2
dt 2
ψ − β 2 ψ = 0 and
ψ 1 = A1eβx on D1
(9.11)
ψ 3 = A3+ eβx + A3− e−βx on D3
(9.12)
where the solution e−βx in Equation (9.11) was discarded as it diverges in the
limit as x → −∞ and hence is nonphysical.
However, on D2 ∪ D4 , the Schrödinger equation becomes
− 21m  2
d2
dt 2
ψ = Eψ
Thus, set k 2 = 2 m ( 1 ) E . Then, the general solutions for the respective
domains are:
2
ψ 2 = A2+ eikx + A2− e− ikx on D2
(9.13)
ψ 4 = A4 eikx on D4
(9.14)
as there is no boundary for x > b to reflect the waves back toward x = b.
d
To solve the coefficients, it suffices to note that mathematically ψ , dx
ψ
must be continuous at the boundaries: x = 0,a,b. So, imposing the boundary
conditions at x = 0,
A1 = A2+ + A2−
(9.15)
βA1 = ikA2+ − ikA2−
(9.16)
whence, ik(9.15) + (9.16) yields (β + ik )A1 = ikA2+ ⇒
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A1
A2+
=
2ik
β+ ik
.
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Elements of Electrostatic Discharge
At x = a, the boundary conditions yield
A2+ eika + A2− e− ika = A3+ eβa + A3− e−βa
(9.17)
ikA2+ eika − ikA2− e− ika = βA3+ eβa − βA3− e−βa
(9.18)
Thus, ik(9.17) + (9.18) yields
2ikA2+ eika = (ik + β)A3+ eβa + (ik − β) A3− e−βa
(9.19)
At x = b, the boundary conditions yield
A4 eikb = A3+ eβb + A3− e−βb
(9.20)
ikA4 eikb = βA3+ eβb − βA3− e−βb
(9.21)
whence, β(9.20) ± (9.21) lead immediately to
A3+ =
β+ ik
2β
A4 eikb −βb
and
A3− =
β− ik
2β
A4 eikb +βb
Substituting these expressions into Equation (9.19) gives, after some algebraic manipulation, and using the identities sinh x = 21 (e x − e− x ) and cosh x =
−x
x
1
2 (e + e ),
2ikβA2+ e− ik (b − a ) = A4 {−(β 2 − k 2 )sinh(β(b − a)) + i2 kβcosh(β(b − a))}
That is,
A2+
A4
{
2
}
2
= eik (b − a ) cosh(β(b − a)) + i β 2−kβk sinh(β(b − a)) . In particular,
A2+
A4
2
= cosh 2 (β(b − a)) +
( ) sinh (β(b − a))
β2 − k 2
2 kβ
2
2
Finally, observe from Definition 9.3.3 that the current density of the incident wave in D2 is
{
2
2
2
2
}
2
j2+ = − 2im  A2+ e− ikx ikeikx + A2+ eikx ike− ikx = − 2im  2ik A2+
and likewise, on D4,
{
}
j4 = − 2im  A4 e− ikx ikeikx + A4 eikx ike− ikx = − 2im  2ik A4
2
Hence, the transmission coefficient across the barrier is
S4 =
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j4
j2+
=
A4
A2+
2

= cosh 2 (β(b − a)) +

( )
β2 − k 2
2 kβ
2

sinh 2 (β(b − a)) 

−1
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It is evident here that as the width of the potential barrier approaches infinity,
the transmission coefficient falls to zero as lim{cosh x , sinh x} → ∞. Likewise,
x →∞
if the height of the potential barrier approaches infinity, the transmission
coefficient falls to zero.
Furthermore, notice also that on D1 ∪ D2, the probability amplitudes are
real and hence, the current density is identically zero by definition. Physically,
even though there is a nontrivial probability that the waves will penetrate
the potential barrier, the waves are evanescent: they decay exponentially. For
A
instance, it was derived above that A+1 = β+2ikik which is nonzero. This is in
2
sharp distinction with classical physics which prohibits a particle from penetrating a potential barrier that is larger than its inherent kinetic energy. It
is easy to see that the penetration is identically zero in quantum mechanics
□
when the potential barrier is infinitely large: lim AA+1 ~ lim β1 → 0 .
U 0 →∞
2
U 0 →∞
In passing, it is known that transistors, FETs, and MOSFETs have leakage
currents. Indeed, from Example 9.3.3, it is somewhat obvious that the source
of the leakage currents is due to quantum tunneling effects. In particular,
quantum tunneling places a lower bound on the gate size in integrated circuits, and hence, Moore’s law* will eventually reach an upper bound.
As a prelude to the topic on dielectric breakdown discussed in the following section, this section concludes with a quick review of the band theory of
semiconductors. By way of introduction, a comparison between a conductor
(i.e., metal) versus semiconductor is drawn. The theory for metals is essentially based on the Drude–Sommerfeld model, and that of semiconductors is
based on the Lorentz model.
Informally, the difference between a conductor and an insulator is
depicted in Figure 9.5. Basically, electrons in the ground state occupy the
valance band. If there is sufficient energy to excite the electrons into the conduction band, and have the electrons remain there, then conduction can take
place. Specifically, when the conduction band overlaps the valance band in
a solid, and hence, electrons are able to diffuse from the valance band into
the conduction band even under low ambient thermal conditions, the material defines a conductor. For a semiconductor, there is a small nonoverlap
between the conduction band and the valance band; the separation is usually
small enough that thermal energy beyond a small value may be sufficient
to excite the valance electrons beyond the forbidden gap, a region wherein
electrons states do not exist, into the conduction band.
On the other hand, for insulators, the forbidden zone is large enough
that thermal energy alone is insufficient to excite the electrons into the
conduction zone. Hence, there are no electrons in the conduction zone—or
at least, very few electrons (perhaps due to tunneling) in comparison to a
*
Moore’s law essentially states that the number of transistors on integrated circuits doubles
every two years.
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Elements of Electrostatic Discharge
Semiconductor: small separation
between conduction and
valance bands
Conductor: overlap between
conduction and valance bands
Conduction band
Band
overlap
Conduction band
Conduction band
Forbidden
zone
Forbidden
zone
Valance band
Insulator: large separation
between conduction and
valance bands
Valance band
Valance band
Figure 9.5
A simplistic description for (a) conductors, (b) semiconductors, (c) insulators.
semiconductor—rendering the material an insulator. Readers interested in
solid-state theory may consult References [1,9,13].
9.3.5 Theorem
Suppose N > 0 energy states exist for a system in equilibrium. Then, the
probability that the system is in a state with energy Ek is P(Ek ) = Z1 e−βEk,
where β = kB1T , with kB = 1.38 × 10−23 J/K being the Boltzmann constant, T
the temperature, and Z = ∑ nN= 1 e−βEn the partition function. In particular, if |k ⟩
denotes the quantum state of a system with energy Ek, then the expected
□
value of an observable A is given by ⟨A ⟩ = Z1 ∑ k ⟨ k|Ak ⟩e−βEk .
Consider a quantum system comprising 0 < N ≤ ∞ particles. If an arbitrary
number of noninteracting particles within the system can occupy any given
state, the particles are called Bose particles (or bosons); if only one particle may
occupy a given state,* then the particles are called Fermi particles (or fermions).
Let ⟨nk⟩ denote the expected number of particles occupying a given state with
energy Ek. At temperature T = 0 K, electrons in a solid occupy the lowest
energy level; the highest energy level occupied by the valance electrons at 0
K is called the Fermi energy level.
9.3.6 Theorem
For any fixed temperature T, the probability that fermions exists in a state
−1
with energy Ek obeys the Fermi–Dirac distribution: P(Ek , T ) = eβ(Ek −µ ) + 1 ,
where μ is the chemical potential defined by P(μ,T) = 0.5.† In particular, the
probability distribution of holes left below the Fermi level when electrons are
−1
thermally excited into the conduction band is 1 − P(Ek , T ) = eβ(µ− Ek ) + 1 . □
{
{
}
}
* That is, obeying the Pauli exclusion principle, such as electrons (on the other hand, photons are
Bose particles).
† For temperatures of practical interest, the chemical potential is approximately equal to the
Fermi energy [13, p.166].
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
A solid is often modeled as a crystal lattice with periodic potential wherein
the valance electrons are nearly free (i.e., plane waves modulated by the periodicity of the potential energy). In particular, for metal, the electrons freed
from the valance band are free to diffuse throughout the metal (via the conduction band). For semiconductors, this holds when the ambient temperature
is not too low. A precise statement can be captured below wherein electrons
are modeled as an ideal gas.
By way of introduction, given a crystal lattice with periodicity r̂ , where
the lattice is defined by fixed nuclei, define the reciprocal lattice vector k̂ by
rˆ ⋅ kˆ = 2 πl , for some l ∈ N. Finally, if p is the momentum of an electron in the
crystal lattice, define the electron wave vector k by p = k.
9.3.7 Theorem (Bloch)
Given a crystal lattice with time-independent periodic potential U (r + rˆ ) = U (r ),
where r̂ is the periodicity of the lattice and the nuclei are fixed at the lattice sites,
the electron energy eigenfunctions |Ψ k ⟩ of the time-independent Schrödinger
equation |HΨ k ⟩ = E(k)|Ψ k ⟩ satisfy Ψ k (r ) = eik ⋅r ψ k (r ) , where k is the wave vec□
tor, ψ k (r + r ) = ψ k (r ) ∀k and E(k + kˆ ) = E(k) ∀k .
9.3.8 Remark
The eigenvalues E(k) define the energy band structure: a surface in k-space
such that the energy on the surface is the Fermi energy level, called a Fermi
surface ∂F ( kF ), where F ( kF ) = {k ∈ R 3 : k ⋅∇ k ⋅ E(k) = 0, k ≤ kF }. If D(k− , k+ ) ⊂ R 3
defines a conduction band in k-space, then for a conductor, D(k− , k+ ) ∩ F ( kF ) ≠ ∅,
whereas for an insulator,
d(∂D(k− , k+ ), ∂F ( kF )) ≡ sup{d(k′ , k′′) : k′ ∈∂D(k− , k+ ), k′′ ∈∂F ( kF )} >> ε(T )
where ε(T) > 0 is some monotonically increasing function of the (ambient)
temperature T and d : R 3 × R 3 → R + is the Euclidean metric (defined in
Appendix A.2). Thus the minimal energy surface for a conduction band lies
well above the Fermi surface for an insulator; see Figure 9.5. Last but not
least, the forbidden zone is the region in k-space wherein electron states cannot
exist. It is sometimes called the band gap.
9.3.9 Definition
The Drude model for a metal assumes that the valance electrons form an ideal
classical gas that diffuses throughout the metal in the conduction zone. The
Drude–Sommerfeld model for a metal is the Drude model wherein the electrons form an ideal Fermi gas, that is, an ideal gas that obeys the Fermi–Dirac
statistics. Finally, the Lorentz model for semiconductors is the Drude model
wherein the valance electrons are bound by some simple harmonic potential.
This section concludes with a brief comment on temperature and material properties. It is intuitively clear from band theory that increasing the
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thermal temperature of the lattice may potentially impart to the valance electrons sufficient energy to tunnel across the band gap. Hence, the electrical
conductivity of a dielectric or semiconductor depends upon the ambient (or
operating) temperature. For further studies, the reader is invited to consult
any standard references on solid state physics.
9.4 Dielectric Breakdown
A brief exposition is sketched here, based on the previous section, to provide
a heuristic understanding of dielectric breakdown, often resulting in electrostatic discharge. Much empirical work has been done on this account, resulting
in the Paschen breakdown curves and characteristics thereof for dielectrics;
this topic is not pursued here. In what follows, a dielectric is assumed to be an
insulator, unless stated otherwise. Furthermore, observe from quantum tunneling effects that an insulator is not a perfect insulator, as electrons from the
valance band can always tunnel across the forbidden zone to the conduction
band, as the temperature increases. Notwithstanding that the probability is
very small. A simple example is worked out below by way of illustration.
9.4.1 Example
Consider a free electron impinging upon a potential barrier of infinite width.
Suppose further that upon the application of a constant external electric field
E0, the barrier falls off according to
U 0 on R − ∪ ( a, b)



U ( x) = U 0 − eE0 ( x − b) on [b , c]


0 on [0, a] ∪ (c, ∞)

where e is the electronic charge. See Figure 9.6.
U = U0
U = U0
E < U0
U=0
x=0
U=0
x = a x = b x = x0 x = c
Figure 9.6
Potential barrier reduction via an applied electric field.
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U = U0
U1 < E < U0
U = U1
U=0
U=0
Figure 9.7
Potential barrier lower than a particle’s kinetic energy.
Now, let the particle kinetic energy T = E. Then, by the continuity of U on
[a,b], the intermediate value theorem implies ∃x0 ∈( a, b) such that U ( x0 ) = E.
In particular, x ∈[ a, x0 ) ⇒ U ( x) − E > 0 and x ∈( x0 , b] ⇒ U ( x) − E < 0. From
Example 9.3.3, for each fixed x ∈[ a, x0 ), the transmission coefficient

S+ = cosh 2 (β + ( x)( x − a)) +

(
β 2+ ( x )− k 2
2 kβ + ( x )
)
2

sinh 2 (β + ( x)( x − a)) 

−1
where β + ( x) = 2m2 (U ( x) − E) on [b , x0 ) .
To determine the transmission coefficient on [ x0 , c] , consider the opposite
scenario to Example 9.3.3; see Figure 9.7. In this instance, the particle kinetic
energy E > U 1, where U 1 is the potential energy of the finite barrier. Then,
following Example 9.3.3, mutatis mutandis, on R − ,
0=
where β1 =
1

ψ 2 + 2m2 Eψ 2 = 0 ⇒ ψ 2 = A2+ eikx + A2− e− ikx
2 mE . On the interval (a,b),
1

0=
where β 2 =
ψ 1 − 2m2 (U 0 − E)ψ 1 = 0 ⇒ ψ 1 = A1eβ1x
2 m(U 0 − E) . On the interval [0,a],
d2
dt 2
where k =
d2
dt 2
1

d2
dt 2
ψ 3 + 2m2 (E − U 1 )ψ 3 = 0 ⇒ ψ 3 = A3+ eiβ2 x + A3− e− iβ2 x
2 m(E − U 1 ) . Finally, on [b,∞),
0=
d2
dt 2
ψ 4 + 2m2 Eψ 4 = 0 ⇒ ψ 4 = A4eikx
The pair of equations at x = a yields
(
)
(
)
2 A2+ = A3+ 1 + βk2 eiβ2 a + A3− 1 − βk2 e− iβ2 a
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Likewise, at x = b,
(
)
2 A3+ eiβ2 b = A4 eikb 1 + βk2
(
and 2 A3− e− iβ2 b = A4 eikb 1 − βk2
)
whence
(
(
)
)
(
(
)
A2+ = 14 eiβ2 a 1 + βk2 A4 ei( k −β2 )b 1 + βk2 + 14 e− iβ2 a 1 − βk2 A4 ei( k +β2 )b 1 − βk2
)
and upon some tedious algebraic manipulation,
A2+
A4
{
= eikb cos(β 2 (b − a)) +
( ) = e {cos(β (b − a)) −
A2+
A4
∗
− ikb
2
}
2
2
i β2 + k
2 2 kβ 2
sin(β 2 (b − a))
2
2
i β2 + k
2 2 kβ 2
}
sin(β 2 (b − a))
and hence,
A4
A2+
2

= cos 2 (β 2 (b − a)) −

1
4
( )
β 22 + k 2
2 kβ 2

sin 2 (β 2 (b − a)) 

2
−1
To complete the solution, it suffices to note that on the closed interval [x0, c],
E − U(x) > 0, and hence, the transmission coefficient is

S− ( x) = cos 2 (β − ( x)( x − a)) −

1
4
( ) sin (β (x)(x − a))
2
β 2− + k 2
2 kβ −
2
−1
−
where β − ( x) = 1 2 m(E − U ( x)) . Thus, the average transmission coefficient
across the finite width potential barrier is
S=
1
c− a



∫
x0
a
S+ ( x)d x +
∫
c
x0

S− ( x)d x 

Observe by construction that S− (c) ≤ S− ( x) < S < S+ ( x) ≤ S+ ( a) on [a,c].
□
It is quite clear from the above example that in the presence of an external
field, the potential barrier can be reduced and hence, increase the corresponding transmission coefficient: S− (c) < S. This means that electrostatic discharge
has a greater probability of occurring when a dielectric is subject to a strong
field. In particular, recalling the band structure from the previous section, it is
clear that by applying a large enough electric field, the potential barrier may
be lowered to a sufficiently small value such that electrons can easily tunnel
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
across the forbidden zone and into the conduction zone. The applied field
accelerates the electrons in the conduction zone, and collisions with atomic
lattices may potentially excite electrons in the valance band and thereby
impart sufficient energy for the valance electrons to jump the forbidden zone
and into the conduction band or, if the electron acquires sufficient energy, to
leave the dielectric surface (ionization). Thus, at an intuitive level, when a sufficient number of electrons are able to jump to the conduction band, the large
current flow in a dielectric results in the phenomenon of breakdown.
As a particular scenario, consider two oppositely charged parallel plate
conductors in a homogeneous dielectric medium connected to a common
source, that is, a capacitor subject to a fixed potential difference V across the
plates. Here, the electric field is just E = Vd , where d is the plate separation.
For V >> 0 such that the potential barrier between the valance band and the
conduction band is sufficiently low, the valance electrons can tunnel easily
into the conduction band under ambient temperature conditions.
Increasing the electric field further by increasing the potential difference
will cause the electrons to accelerate toward the anode; indeed, a sufficiently
high field will impart a large enough kinetic energy to the electrons such
that their collisions will cause ionization of the dielectric medium. Explicitly,
• The applied electric field will reduce the potential barrier of the
dielectric such that the electrons can tunnel across the forbidden
zone into the conduction band.
• The electrons in the conduction band will diffuse toward the positively charged conductor (anode).
• The movement of the electrons in the conduction band will generate
holes in the valance band (positively charged due to the absence of
electrons) which will diffuse toward the negatively charged conductor (cathode).
• The respective charges will accumulate around the boundary of the
conductors, forming a space charge that partially shields the conductors and hence, result in a drop in the field.
Heuristically, as more electrons accelerate toward the respective plates, a
brief electric arc between the plates may occur. However, note that as the
opposite charges form a charge cloud (i.e., space charge) within the vicinity of the plates, the opposing potential—called screening—will effectively
reduce the applied electric field across the dielectric. The potential is thus
modified by the charge cloud. When the electric field falls below some critical level such that the tunneling coefficient becomes negligible, the dielectric ceases to conduct. Thus, electrostatic discharge cannot be a continuous
event unless the potential difference between the two fixed conductors
increases continuously to supply a strong enough electric field to overcome
the screening effects of the ions. The informal picture sketched above can be
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Elements of Electrostatic Discharge
understood at a more rigorous level by considering a simple example given
below: thermionic emission.
9.4.2 Example
Consider a metal conductor surface subject to a high temperature exposure
on its surface. Suppose EF denotes the Fermi energy level of the metal. If
W∞ denotes the work expanded to move an electron from the Fermi level
to infinity, then the work function ϕ of the metal is defined by φ = W∞ − EF .
By appealing to the Drude–Sommerfeld model, it can be shown [10 p. 237]
that the emission current density from the metal surface is given by the
Richardson–Dushman equation
J=
4 πme
h3
( kBT )2 e−φ/( kBT )
where m,e,h,T are, respectively, the mass and absolute charge of an electron,
□
Planck’s constant, and temperature.
9.4.3 Remark
Cold emissions, that is, emissions of electrons under the influence of a strong
electric field, from a metal surface are again the result of a quantum tunneling
effect. It is called Fowler–Nordheim tunneling, and it has applications in metaloxide semiconductors; the current across the junction is controlled via quantum tunneling through the application of an electric field across the junction.
In closing, the electrostatic discharge can be explained by lowering the
potential barrier of a dielectric such that
• The valance electrons can tunnel across the forbidden zone into the
conduction band.
• The applied electric field imparts sufficient kinetic energy to the
electrons in the conduction band to overcome the work function of
the surface.
It is important to note that under the application of a strong electric field,
the migrating electrons often acquire large enough kinetic energy such that
their collisions with the lattice will impart energy to the lattice; the energy
loss manifests as heat. For large enough energy, this leads to the actual breakdown of the dielectric structure within the vicinity of the current flow. At a
practical level, an intense discharge can cause vaporization of the material,
leading to punctures in the dielectric.
9.4.4 Remark
A rough insight into the above example of a dielectric between two parallel plates can be gained by considering a lossless homogeneous dielectric
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
medium (Ω,μ,ε). Let Wε,T denote the minimal energy required to allow a
valance electron to transition into the conduction band for some fixed temperature T. By modeling the dielectric as a lattice, it may be supposed that
an electron has a mean free path ⟨rε ⟩ between lattice collisions, if sufficient
energy is provided to raise the valance electrons into the conduction band.
At temperature T, there may only be a small quantity of electrons in the
conduction band due to tunneling. A minimal applied uniform electric field
Eε needed to give the electrons in the conduction band an average kinetic
energy of Wε,T can clearly be found from Wε ,T = Fε ⟨rε ⟩ , where the required
Wε , T
minimal force Fε =||
e Eε . Hence, by applying an electric field Eε = ||
e ⟨ rε ⟩ across
the plates, the conduction electrons will possess enough energy to free the
valance electrons into the conduction band. A chain reaction takes place
under the applied electric field, causing dielectric ionization to take place
and hence a breakdown of the dielectric. Trivially, the required potential difference across the plates is Vε = Eε d . Last, if the electric field along the plates
varies, such that ∃p ∈ ∂Ω satisfying E( p) ≥ Eε , then breakdown will occur
along some sufficiently small tubular neighborhood of p.
9.5 Worked Problems
9.5.1 Exercise
Establish Remark 9.2.3.
Solution
For simplicity, consider a bound electron about some fixed origin subject to a
restoring force m0 ω 20 and a damping force m0 γ 0 v, where v is the velocity of
the electron. Then, given a plane monochromatic time harmonic electric field
E = E0 e− i( kz +ωt ), the resultant classical expression of the force on the electron is:
m0 
r + m0 γ 0 r + m0 ω 20 r = eE0 e− iωt
where r = v. Then, upon setting r = r0 e− iωt , it follows at once that
−ω 2 r0 − iωγ 0 r0 + ω 02 r0 =
e
m0
E0 ⇒ r =
e
m0
E ω 2 −ω 21 − iγ
0
0ω
Now, by definition, the dipole moment is p = er, and for a single atomic
2
dipole, p = (ε − ε 0 )E , whence, p = (ε − ε 0 )E = er ⇒ ε = ε 0 + me 0 ω 2 −ω 21 − iγ ω . This
0
0
corresponds precisely to a single mode for an electronic oscillator. If there
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Elements of Electrostatic Discharge
exist j = 1,2, … modes per electronic oscillator, let fj denote the probability that an electronic transition will take place at the resonant angular frequency ωj. Then, by definition, ∑ j f j = 1 (as the total probability must sum to
unity) and hence, for a single electronic oscillator,
ε = ε0 +
e2
m0
∑
fj
2
2
j ω 0 −ω − iγ 0 ω
Finally, to complete the discussion, if there exist N dipoles per unit volume,
then trivially, the electric permittivity of the simple homogeneous dielectric
material is given by
ε = ε0 +
Ne 2
m0
∑
fj
2
2
j ω 0 −ω − iγ 0 ω
□
as required.
9.5.2 Exercise
Prove that the eigenvalues of a Hermitian operator are always real.
Solution
Given a Hermitian operator A, suppose |ψn ⟩ is an eigenvector of A; that is,
A|ψ n ⟩ = an |ψ n ⟩ . Then, by definition, ⟨ψ n|Aψ n ⟩ = an ⟨ψ n|ψ n ⟩ = an , and denoting
∗ to be the complex conjugate, ⟨A † ψ n|ψ n ⟩ = ⟨ψ n|A † ψ n ⟩∗ = ⟨ψ n|Aψ n ⟩∗ = an∗ by
the Hermiticity of A. Hence, ⟨A † ψ n|ψ n ⟩ = ⟨ψ n|Aψ n ⟩ ⇒ an∗ = an. Thus, the arbitrariness of |ψn ⟩ implies that all eigenvalues of A must be real.
9.5.3 Exercise
The reason for the Hilbert space formalism of quantum mechanics arises in
the following fashion. Consider Axiom 9.3.1(d), ddt ψ = − i 1 H ψ . Here,
H = − 21m  2 ∆ + U (r , t)
where Δ is the Laplacian, and in general, the potential energy U depends upon
time. In the Schrödinger (coordinate) representation, the state vector |ψ⟩ is represented by a complex function ψ (r , t) ≡ Sˆ (t)ψ (r ) such that ∫ R3 ψ (r )ψ ∗ (r )d 3 r < ∞
and Ŝ is a linear operator satisfying |Sˆ (t)|= 1 . In particular, the function can
be normalized: ψ → ψ = 1N ψ , where ∫ R3 ψ (r )ψ ∗ (r )d 3 r = N . By definition,
ψ ∈L2 (R 3 ) , which is the Hilbert space of all square integrable functions:
⟨ψ|ψ⟩ ≡ ∫ R3 ψ (r )ψ ∗ (r )d 3 r . The wave function ψ(r) is called the Heisenberg (coordinate) representation (and is, by definition, independent of time). Finally, recalling that the Fourier transform F : L2 (R 3 ) ≈ L2 (R 3 ) is a linear isomorphism on
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
a Hilbert space, the momentum representation ψ(p) is the Fourier transform
F [ψ ]( p) = ∫ R3 ψ (r )e− ip⋅ x d 3 r of the coordinate representation ψ(r). Lastly, the
Heisenberg uncertainty principle δpδx ≥ 21 2hπ , where δp(δx) is the uncertainty in
momentum (position), is a direct consequence of the Hilbert space structure
(specifically, the Cauchy-Schwarz inequality): |ψ||ϕ|≥|⟨ψ|ϕ⟩|. This concludes the
motivation for formulating quantum mechanics on (an infinite-dimensional)
Hilbert space.
Show that the time-independent solution—also called stationary states—
of the Schrödinger wave equation is given by H|ψ⟩ = E|ψ⟩ , and hence,
show explicitly that the time-independent Schrödinger wave equation is
{
− 2
2m
}
∆ + U (r ) ψ (r ) = Eψ (r ). In particular, deduce that a solution of the time− i 1 Et
independent Schrödinger equation has the form: Ψ(r , t) = ψ (r )e  .
Solution
From Axiom 9.3.1(d), ddt |ψ⟩ = − i h1 H|ψ⟩ , it follows that in the coordinate representation, ddt ψ = − i h1 Hψ . Because the Hamiltonian is independent of time,
d
dt H = 0 and in particular, U(r, t) = U(r). So, attempt the separation of variables, ψ(r, t) = φ(r)τ(t). Then,
d
dt
ϕ(r )τ(t) = − i 1 Hϕ(r )τ(t) ⇔ − i 1
1 d
τ ( t ) dt
τ(t) =
1
ϕ( r )
Hϕ(r )
Because the left-hand side of the equality is independent of coordinates by
definition, and the right-hand side of the equation is independent of time,
they must both be equal to some constant E. That is,
− i 1
1 d
τ ( t ) dt
τ(t) = E =
1
ϕ( r )
Hϕ(r )
whence
1 d
τ ( t ) dt
τ(t) = iE ⇒ τ(t) = eiEt
is the fundamental solution, and Hϕ(r ) = Eϕ(r ) implies immediately that E is
an eigenvalue of the Hamiltonian operator H and φ(r) is an eigenfunction of
H, whence the stationary Schrödinger wave equation satisfies
{
− 2
2m
}
∆ + U (r , t) ψ = Eψ
i Et
and the full stationary solution thus has the form ψ (r , t) = ϕ(r )e  .
□
9.5.4. Exercise
Given that the momentum operator is defined by p = − ih∇ , show that for a
free particle, that is, potential energy U = 0, traveling along the x-axis,
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Elements of Electrostatic Discharge
the eigenfunction of p is ψ = e ± ikx with eigenvalue ± k . In fact, this is just the
equation for a plane wave.
Solution
Because the particle is traveling along the x-axis, the variation along the y, z
directions are zero and hence, the problem reduces to a one-dimensional
scenario. Thus, p = −ih ddx . Moreover, as the potential energy is zero, the timeindependent Schrödinger equation defines the state of the particle:
− 2 d2
2 m dx 2
whence, on setting k 2 =
0=
2m
2
d2
dx 2
ψ = Eψ
E , it follows that
ψ + k 2 ψ ⇒ ψ + = eikx , ψ − = Be− ikx
are two possible solutions.
Now, observe that E = 21m p 2 ⇒ p = ± k , where p = k denotes the particle
traveling along the x-axis in the positive direction whilst p = − 2π k indicates
the negative direction. Thus,
pψ ± = − i 2π
d
dx
ψ ± = ± kψ ±
implies at once that the eigenvalue of p is p = k for the eigenfunction
ψ + = eikx , and the eigenvalue is p = − k for the eigenfunction ψ − = Be− ikx .
That is, ψ + = eikx denotes the particle traveling along the positive direction (from left to right) along the x-axis, and ψ − = Be− ikx denotes the particle
traveling along the negative direction (from right to left) along the x-axis, as
□
claimed.
9.5.5 Exercise
Consider the domains (Ω0 , µ 0 , ε 0 , σ 0 ) and (Ω1 , µ 1 , ε1 , σ 1 ), where Ωi ⊂ R 2 are
defined by Ω0 = [ − 21 a, 21 a ] × [0, δ] and Ω1 = [− 21 a, 21 a] × [δ , δ + d], and suppose
also that
C− = {( x , 0) ∈R 2 : x ∈R} and C+ = {( x , δ + d) ∈R 2 : x ∈R}
are grounded planes that bound Ωi ∀i = 0, 1. Given that 0 < δ << a, suppose
that a constant surface charge density ρs exists on ∂Ω0 |y =δ ≡ [ − 21 a, 21 a ] × {δ} at
t = 0. Determine the charge relaxation time of Ω = Ω0 ∪ Ω1 . This question is
of particular interest to EMC engineers and material engineers, as occasionally there is a need to know how long it takes to fully discharge a dielectric
material.
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Electromagnetic Theory for Electromagnetic Compatibility Engineers
Solution
Because ρs is a constant and C± are grounded conductors, it is clear by conp
struction that ∀p ∈∂Ω0 |y =δ and ∀p± ∈C± , − ∫ p± E± ⋅ dl = V0 , for some constant
V0. In particular, the assumption 0 < δ << a implies that E± ≈ ±E± e y , where E±
are some constants, away from x = ± 21 a. Hence, Ed+ = V0 = Eδ− ⇒ E+ = δd E− and
from Gauss’ law, ρ1 = δd εε01 ρ0
Now, for simplicity, let σ , ε denote the electric conductivity and electric
permittivity, respectively, for the composite system Ω, whence, from the continuity of charge,
∂t ρs = −∇ ⋅ J s = − σε ρs ⇒ τ =
ε
σ
is the composite relaxation time.
Furthermore, observe also that for some small time t > 0, σ i > 0 ∀i ⇒
ρs = ρ1 + ρ2 , where ρi is in Ωi. That is, the charge on the boundary will diffuse
toward the grounded conductors under the action of E± . Hence,
∂t ρs = ∂t ρ0 + ∂t ρ1 = − σε00 ρ0 −
σ1
ε1
ρ1 = −
{
σ0
ε0
+
σ1 d
ε0 δ
}ρ
0
In addition,
{
σ0
ε0
+
σ1 d
ε0 δ
}ρ
0
= σε ρs =
σ
ε
{1 +
d ε1
δ ε0
}ρ
0
⇒
σ
ε
= {σ 0 + δd σ 1 }{ε 0 + δd ε1 }
−1
Hence, the charge relaxation time of Ω is τ = {ε 0 + δd ε 1 }{σ 0 + δd σ 1 } .
Indeed, this can be rewritten in terms of the charge relaxation constant of Ω0:
−1
{
τ = τ0 1 +
where τ 0 =
ε0
σ0
as required.
d ε1
δ ε0
}{1 +
d σ1
δ σ0
}
−1
□
References
1. Ahn, D. and Park, S.-H. 2011. Engineering Quantum Mechanics. Hoboken, NJ:
IEEE Press, John Wiley & Sons.
2. Bohm, D. 1979. Quantum Theory. New York: Dover.
3. Churchill, R. and Brown, J. 1990. Complex Variables and Applications. New York:
McGraw-Hill.
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Elements of Electrostatic Discharge
329
4. Davydov, A.1976. Quantum Mechanics. Oxford, UK: Pergamon Press.
5. Dressel, M. and Grüner, G. 2002. Electrodynamics of Solids: Optical Properties of
Electrons in Matter. Cambridge: University of Cambridge Press, UK.
6. Fox, M. 2001. Optical Properties of Solids. Oxford, UK: Oxford University Press.
7. Go, D. and Pohlman, D. (2010). A mathematical model of the modified Paschen’s
curve for breakdown in microscale gaps. J. Appl. Phys. 107(10): 103303.
8. Jackson, D. 1962. Classical Electrodynamics. New York: John Wiley & Sons.
9. Jones, W. and March, N. 1973. Theoretical Solid State Physics. Vol. 1: Perfect Lattices
in Equilibrium. New York: Dover.
10. Kittel, C. 1953. Introduction to Solid State Physics. New York: John Wiley & Sons.
11. Paschen, F. (1889). Ueber die zum Funkenübergang in Luft, Wasserstoff und
Kohlensäure bei verschiedenen Drucken erforderliche Potentialdifferenz. Ann.
Phys. 273(5): 69–75.
12. Raju, G. 2003. Dielectrics in Electric Fields. New York: Marcel Dekker.
13. Tang, C. 2005. Fundamentals of Quantum Mechanics for Solid State Electronics and
Optics. Cambridge: Cambridge University Press, UK.
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Appendix A
A.1 Coordinate Transformations
Recall that a vector space V over R is the quadruple (V,+,•, R) such that the following are preserved
a) Vector addition: ∀u, v ∈ V, u + v ∈ V
b) Scalar multiplication: ∀v ∈ V and ∀a ∈ R, a • v ∈ V (this is often
denoted by av ∈ V)
A.1.1 Definition
Given a (real) vector space V, the set of vectors {v1 , , vk } ⊂ V is said to
be linearly independent if ∑ ik= 1 a k vk = 0 ⇔ a k = 0 ∀k . Furthermore, if ∃n > 0
such that {v1 , , vn } ⊂ V is linearly independent and ∀m > n, {u1 , , um } ⊂ V
is not linearly independent, then n is said to be the (real) dimension of V,
and {v1 , , vn } ⊂ V called a basis of V.
Given a basis {v1 , , vn } ⊂ V , ∀v ∈ V, ∃{ a1 , , a n } ⊂ R such that
v = a1 v1 +  + a n vn . Observe that a basis is not unique, and with respect to a
fixed basis v = {v1 , , vn } , the set { a1 , , a n } is the components of v. In particular, the vector v may be represented by v = ( a1 , , a n ) with respect to v .
The elements of R are called scalars. Denote the dimensions of a vector space
V by dimV.* Then, a basis {v1 , , vn } of V is said to span V if n = dimV. This is
also written as V = span{v1 , , vn }.
A.1.2 Definition
Let V, W be a pair of finite-dimensional vector spaces. A linear transformation
(or operator) T: V → W on V is a mapping that satisfies the criteria: T(av + u) =
aT(u) + T(v) ∀u,v ∈ V and a ∈ R. Furthermore, a linear transformation is (a)
injective (or one–one) if ∀u,v ∈ V, T(u) = T(v) ⇒ u = v, and (b) surjective (or onto)
if ∀w ∈ W, ∃u ∈ V such that T(u) = w; that is, T(V) = W. Last, T is said to be a
linear isomorphism if T is both injective and surjective, and this is denoted by
T: V ≈ W.
By definition, a linear transformation on V is completely defined by
its values on a basis {v1 , , vn } of V. To see this, let T: V → W be a linear
* Only real vector spaces are considered here; these are vector spaces over R.
331
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332
Appendix A
transformation and suppose that T (vi ) = vi′ ∀i = 1, , n . Given any v ∈ V, let
v = ∑ i a i vi for some scalars a i ∈R. Then, by linearity,
T (v) = T ( a1 v1 +  + a n vn ) =
∑ a T(v ) = ∑ a v′.
i
i
i
i
i
i
The above linear transformation is invertible if vi′ ≠ 0 ∀i = 1, , n , and its
inverse T −1 is defined by T −1 (vi′) = T −1 (T (vi )) = vi . In other words, T is invertible if Tu = 0 ⇔ u = 0. Equivalently, a linear transformation T is invertible if
it is injective.
A basis {v1 , , vn } of V defines a coordinate system for V. A linear isomorphism T: V → V defines a change in the coordinate system on V; that is, it
transforms one coordinate system into another coordinate system. Finally, if
W is another vector space such that W ≈ V, then W may be identified with V.
A.1.3 Theorem
Let V be an n-dimensional (real) vector space, where n < ∞. Then, V ≈ R n .
Proof
Let {ui : i = 1, , n} be a basis that spans V and let {ei : i = 1, , n} denote
the standard basis spanning R n . Define a linear transformation T on V
by T (ui ) = ei ∀ i. Then, by construction, T is injective. To see this, suppose
u′ , u′′ ∈V. Then, u′ = ∑ i ai′ui and u′′ = ∑ i ai′′ ui, for some unique ai′ , ai′′∈R ∀ i,
whence,
T (u′) = T (u′′) ⇒
∑ a′e = ∑ a′′e ⇔ ∑ (a′ − a′′ )e = 0 ⇒ a′ = a′′∀ i
i
i i
i
i i
i
i
i
i
i
i
and T is thus injective. Finally, to establish that T is surjective, it suffices to
choose any element w ∈R n . Let w = ∑ i wi ei and define, u = ∑ i wi ui . Then, by
definition, u ∈ V and T (u) = ∑ i wiT (ui ) = ∑ i wi ei = w, whence it follows from
the arbitrariness of u that T is surjective, as required.
□
A.1.4 Example
The space R 3 with respect to the standard basis vectors {e1 , e 2 , e 3 }, where
e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1), is a 3-dimensional vector space. A 3-vector
u ∈R 3 is expressed interchangeably by its components:
u = ( a, b , c) = ae1 + be 2 + ce 3
Moreover, the cross-product of two 3-vectors u = (u1, u2, u3) and v = (v1, v2, v3)
is defined by:
u × v = (u2 v 3 − u3 v 2 , − (u1 v 3 − u3 v 1 ), u1 v 2 − u2 v1 )
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333
Appendix A
The inner (or scalar) product of two 3-vectors u = (u1 , u2 , u3 ) and v = ( v 1 , v 2 , v 3 )
is given by
u ⋅ v = u1 v 1 + u2 v 2 + u3 v 3
The inner product is also denoted by ⟨u|v⟩ or ⟨u, v⟩ .* Observe that the crossproduct of two 3-vectors remains a 3-vector whereas the scalar product of
two vectors is a scalar. In particular, ei ⋅ e j = 0 for ∀ i ≠ j.
Note as a side remark that more generally, an orthonormal basis {e x , e y , e z }
of R 3 satisfies (i) e π ( x ) × e π ( y ) = (−1)|π|e π ( z ) , where π:{x, y, z} → {x,y, z} is a permutation and |π| = 0 if the permutation is cyclic, and |π| = 1 otherwise; (ii)
e i ⋅ e j = δ ij , where i, j ∈ {x, y, z}.
Now, given a vector space V with basis v = {v1 , , vn } , v is orthonormal if
vi ⋅ v j = δ ij , where
 1

δ ij = 

 0
if i = j
if i ≠ j
That is, the vectors are unit vectors such that they are normal (i.e., orthogonal)
to one another. In all that follows, all vector spaces under consideration are
restricted to a maximum of three dimensions.
For notational simplicity, denote the partial derivatives with respect to
x, y, z, t by
∂x ≡
∂
∂x
, ∂y ≡
∂
∂y
, ∂z ≡
∂
∂z
, ∂t ≡
∂
∂t
, ∂2x ≡
∂2
∂ x2
, ∂2y ≡
∂2
∂ y2
, ∂2z ≡
∂2
∂ z2
, ∂t2 ≡
∂2
∂t2
The definition of the del operator ∇ in 3-dimensional space R 3 is:
• In rectangular coordinates, ∇ = (∂ x , ∂ y , ∂ z ) , and ∆ ≡ ∇ ⋅ ∇ = ∂2x + ∂2y + ∂2z .
• In spherical coordinates, x = r sin ϕ cos θ, y = r sin ϕ sin θ, z = r cos ϕ ⇒
2
2
1
1
1
1
∇ = (∂r , 1r ∂θ , r sin
θ ∂φ ) and ∆ = r 2 ∂ r (r ∂ r ) + r 2 sin θ ∂θ (sin θ ∂θ ) + r 2 sin 2 θ ∂φ .
Let S be a 2-dimensional surface in R3 such that ∃{u,v}, a basis field, that
spans S. That is, locally, each vector w defined at a point on S can be expressed
as a linear combination of (u, v): w = au + bv, for some real numbers a, b. Let p
be a unit vector that is normal to the surface S at each point on S. Then, on the
volume S × R defined by S via the local coordinates given by the basis vectors
{u, v, p} at each point in S × R,
∇ ≡ ∇⊥ + p ∂ p
where ∇ ⊥ ≡ (∂u , ∂ v ). See Figure A.1.
* A more general definition is given in a later section.
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334
Appendix A
S×R
ez
R3
ey
S
ex
p vu
An infinite volume S × R
imbedded in the Euclidean
3-space R3 defined by the
surface S
Figure A.1
Arbitrary coordinate system defined on S × R.
A.1.5 Proposition
Let f : R 3 → R 3 be a coordinate transformation from {e1 , e 2 , e 3 } -coordinates
onto {e1′ , e ′2 , e ′3 } -coordinates defined by ( x 1 , x 2 , x 3 )  ( y 1 , y 2 , y 3 ) . Then, f has
the following matrix representation u = f(v) ≡ D(f)v, where D( f ) = (∂ j y i ) is a
3 × 3 matrix called the Jacobian of f defined by
 ∂1 y 1

D( f ) = (∂ j y i ) =  ∂1 y 2

 ∂1 y 3

∂2 y 1
∂2 y 2
∂2 y 3
∂3 y 1 

∂3 y 2 

∂3 y 3 

with ∂ j y i ≡ ∂ x j y i for each i, j = 1, 2, 3, u = ( y 1 , y 2 , y 3 ) and v = ( x 1 , x 2 , x 3 ). Explicitly,
 y1

 y2

 y 3
  ∂ y1
1
 

 = ∂1 y 2
 
  ∂1 y 3

∂2 y 1
∂2 y
2
∂2 y 3
∂3 y 1   x1


∂3 y 2   x 2

3
∂ 3 y 3   x





□
A.1.6 Example
Derive the coordinate transformation between rectangular coordinates
and spherical coordinates in R 3, where in the spherical coordinate system, u =
(r, θ, ϕ) is defined with respect to rectangular coordinates by r = x 2 + y 2 + z 2 ,
y
x2 + y 2
θ = arctan x , φ = arctan z , and the coordinate transformation from rectangular coordinates to spherical coordinates are related by: x = r sin θ cos ϕ, y =
r sin θ sin ϕ, z = r cos θ.
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335
Appendix A
Let f:(r, θ, ϕ) ↦ (x, y, z), where x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ. On
setting y 1 = x , y 2 = y , y 3 = z and applying Proposition A.1.5:
 ux

 uy
 uz

  ∂r x
 
 =  ∂r y
  ∂r z
 
∂θ x
∂θ y
∂θ z
∂ φ x   vr

∂φ y   vθ

∂φ z   vφ





(A.1)
where ∂r y 1 = ∂r x = ∂r (r sin θ cos φ) = sin θ cos φ, ∂θ y 1 = ∂θ (r sin θ cos φ) = r cos θ
cos ϕ, and so on. That is,
 sin θ cos φ

D( f ) =  sin θ sin φ

cos θ

cos θ cos φ
cos θ sin φ
− sin θ
− sin φ
cos φ
0





upon setting e r = rˆ , eθ = rθˆ and eφ = r sin θφˆ (in order for the unit vectors to
have the same dimensions of length). This (invertible) linear transformation
D(f) leads to the transformation of vector components in spherical coordinates into rectangular coordinates:
v = vr e r + vθ eθ + vφ eφ
and u = ux e x + uy e y + uz e z
Now, the above transformation, which transforms spherical coordinates into
rectangular coordinates, is in fact the inverse operation transforming rectangular coordinates into spherical coordinates. Hence,
 sin θ cos φ

D( f ) =  cos θ cos φ

− sin φ

−1
sin θ sin φ
cos θ sin φ
cos φ
cos θ
− sin θ
0





is the required linear transformation taking vector components in rectangular coordinates into components in spherical coordinates. Explicitly,
 vr

 vθ
 vφ
K15149_Book.indb 335
  sin θ cos φ
 
 =  cos θ cos φ
 
− sin φ

sin θ sin φ
cos θ sin φ
cos φ
cos θ
− sin θ
0
  ux

  uy
 u
 z




 □
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336
Appendix A
z
x = ρ cos φ
y = ρ sin φ
z = r cos θ
ρ2 = x2 + y2
ρ
y
r
θ
φ
x
Figure A.2
Cylindrical coordinate system.
A.1.7 Example
In this final example, the coordinate transformation between rectangular
and cylindrical coordinates is derived. First, recall that in terms of cylindrical coordinates (see Figure A.2):
x = ρ cosφ, y = ρ sinφ, z = r cos θ, ρ = x 2 + y 2
On setting y 1 = x , y 2 = y , y 3 = z and applying Proposition A.1.5, f(v) = u ⇒
  ∂ρ x
 
 =  ∂ρ y
  ∂ρ z
 
 ux

 uy
 uz

∂θ x   vρ

∂θ y   vφ

∂θ z   vθ

∂φ x
∂φ y
∂φ z





(A.2)
where ∂ρ y 1 = ∂ρ x = ∂r (ρ cos φ) = cos φ, ∂φ y 1 = ∂φ (ρ cos φ) = −ρ sin φ , and so on.
Substituting the values yields the transformation from cylindrical coordinates to rectangular coordinates:
 cos φ

D( f ) =  sin φ

0

−ρ sin φ
ρ cos φ
0


0

− r sin θ 
0
Inverting D(f) yields the transformation from rectangular coordinates to
cylindrical coordinates:
 cos φ

D( f ) =  − ρ1 sin φ

0

−1
K15149_Book.indb 336
sin φ
0
cos φ
0
1
ρ
0
1
− r sin
θ





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337
Appendix A
This section concludes with a brief list of some common identities employed
in vector analysis [3,12,14,18,19,21]. In what follows, scalars are denoted by
lower case Greek letters, and vectors in upper case bold letters.
•
•
•
•
•
•
∇(αA + B) = A ⋅ ∇α + α∇ ⋅ A + ∇ ⋅ B
∇ × (αA + B) = −A × ∇α + α∇ × A + ∇ × B
∇ × (∇ × A) = ∇(∇ ⋅ A) − ΔA
∇ × ∇α = 0
∇ ⋅ (∇ × A) = 0
∇ × (A × B) = A(∇ ⋅ B) + (B ⋅ ∇)A − B(∇ ⋅ A) − (A ⋅ ∇)B
A.2 Basic Point-Set Topology: A Synopsis
In this section, a sketch of the elements of general topology used in analysis, theoretical physics, and engineering is presented. Some concepts and terminologies
introduced in this section are used throughout the text for mathematical convenience and to make description precise. First, let f: U → V be a mapping. Then, (a)
f is injective if ∀u, v ∈ U, f(u) = f(v) ⇒ u = v; (b) f is surjective if ∀v ∈ V, ∃ u ∈ U such
that f(u) = v; (c) f is bijective if f is both injective and bijective. Second, by way of
motivation, recall the definition of the continuity of functions from real analysis.
A.2.1 Definition
Let f: R → R be a function and fix x0 ∈R . Then, f is said to be continuous
at x0 if ∀ε > 0, ∃δ > 0 such that |x − x0|< δ ⇒ |f ( x) − f ( x0 )|< ε . Moreover, if
f is continuous ∀x ∈ Σ, where Σ ⊆ R, then f is said to be continuous on Σ.
Now, on setting N δ ( x0 ) = ( x0 − δ , x0 + δ) and N ε ( f ( x0 )) = ( f ( x0 ) − δ , f ( x0 ) + δ),
it is clear from Definition A.2.1 that the continuity of f at x0 is equivalent to the following condition: x ∈ N δ ( x0 ) ⇒ f ( x) ∈ N ε ( f ( x0 )); that is,
f ( N δ ( x0 )) ⊂ N ε ( f ( x0 )). See Figure A.3(a). Note that by construction, N δ ( x0 ) is
a δ-interval of x0 and N ε ( f ( x0 )) is an ε-interval of f ( x0 ) .
y+
f(t0)
y–
g
f
N(g(u0))
g(u0)
Generalisation
N(u0)
t–t0 t+
(a)
u0
(b)
Figure A.3
Generalizing the concept of continuity in real analysis.
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Appendix A
This motivates the concept of continuity on spaces that are more general
than the real line. That is, the generalization of intervals, called neighborhoods,
can be modeled on more general spaces to define continuity. Refer to Figure
A.3(b), where (U,V) is a pair of topological spaces, g:U → V is a continuous
map, N ( x0 ) is a neighborhood of x0 in U, and N ( g( x0 )) is a neighborhood
of g( x0 ) in V. The continuity of g guarantees that there exists a neighborhood
N ( x0 ) of x0 such that g( N ( x0 )) ⊂ N ( g( x0 )) . Thus, this leads to the notion of
constructing neighborhoods in abstract spaces.
A.2.2 Definition
Given a nonempty set Ω, suppose τ ⊂ 2 Ω satisfies the following criteria.
(a) ∅, Ω ∈ τ.
(b) ∀U k ∈τ ,  k U k ∈τ (arbitrary union of subsets).
(c) ∀U k ∈τ , k = 1,… , n < ∞ , ∩ nk = 1 U k ∈τ (finite intersection of subsets).
Then, the pair (Ω, τ) is called a topological space, and τ defines the topology of Ω.
Furthermore, the elements of τ are called open subsets (or open neighborhoods)
of Ω. Finally, N ⊂ Ω is a neighborhood of x ∈ Ω if x ∈ N and N ∈ τ.
An example of an open subset in the real line is the open interval (a,b),
where a < b. Observe that given an open interval (a,b) and any point x ∈ (a,b),
there exists an open interval J such that x ∈ J and J ⊂ (a,b). To see this, choose
c = min { x − a +2 x , x 2+ b − x}. Then, by construction, J ≡ (x − c, x + c) ⊂ (a, b) is the
sought-for neighborhood about x that is contained in (a, b). More generally,
the following result can be established.
A.2.3 Lemma
Given a topological space (Ω, τ), a subset N ⊂ Ω is open if and only if ∀x ∈
N, ∃M ∈ τ such that x ∈ M and M ⊂ N.
□
The notion of a closed interval [a, b] in the real line can also be generalized to an arbitrary topological space. A subset N ⊂ Ω is closed if its complement N c = Ω − N is open. Informally, a closed subset contains its boundary
points. Thus, if N ⊂ Ω is open, then N ∪ ∂N is closed. Conversely, given a
subset N ⊂ Ω, the interior N  of N is the subset of all x ∈ N such that ∃U ∈ τ
with x ∈ U ⊂ N. Hence, by definition, N  ∈τ and in particular, N  ∩ ∂ N = ∅ .
A closely related concept defining a closed set is a sequence. Intuitively, a
sequence is a set of ordered numbers: a1 , a2 , , an . A more precise definition
is given below. First, an ordered set (Λ, ≤) is a set such that ∀i, j ∈ Λ, either i ≤ j
or j ≤ i. In addition, if Λ ⊆ N, then the pair (Λ, ≤) is called a countable indexing
set. Finally, a function f: N → R that assigns k ↦ f(k) defines a real sequence,
where it is clear that N inherits a natural order defined by “greater than or
equal to.” A sequence is often represented by (an), (an)n, or {an}n, where an = f(n).
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Appendix A
339
In the case of a topological space Ω, a sequence is defined similarly by a mapping f: Λ → Ω, where Λ is a countable indexing set.
In what follows, only topological spaces whose topology can be characterized by sequences, such as the finite-dimensional Euclidean spaces, are considered.* For this class of topological spaces, continuity can be meaningfully
defined via convergent sequences. In particular, all indexing sets considered
herein are countable.† Again, recall from real analysis that a real sequence (an)n is
said to converge to a0 ∈R . if ∀ε > 0, ∃N > 0 such that all n > N ⇒|an − a0|< ε.
This has the following obvious generalization to topological spaces.
A.2.4 Definition
Let (Ω,τ) be a nonempty topological space and ( an )n∈Λ a sequence in Ω, where
Λ is some indexing set. Then, ( an )n∈Λ converges to the limit a0 ∈Ω if ∀U ∈ τ
with a0 ∈U , ∃nU ∈Λ such that all n ≥ nU ⇒ an ∈U . The limit a0 is said to be a
limit point of the sequence (an)n.
From this definition, it can be shown that a subset S ⊂ Ω is closed if and
only if S contains all of its limit points. And if U ⊂ Ω is open, then its closure
U is defined to be the union of U and all of its limit points. Thus, an equivalent characterization of a closed set is the following: for any sequence (an)n
in S ⊂ Ω such that an → a0 ∈Ω ⇒ a0 ∈S. That is, S contains the limit point of
every sequence in S that is convergent in Ω.
A.2.5 Definition
Let (Ωi , τ i ), for i = 1,2, be nonempty topological spaces and f : Ω1 → Ω2
be a map. Then, f is continuous at x0 ∈Ω1 if for any neighborhood
N ( f ( x0 )) ⊂ Ω2 of f ( x0 ) , there exists a neighborhood U ( x0 ) ⊂ Ω1 of x0 such
that f (U ( x0 )) ⊂ N ( f ( x0 )). In particular, f is continuous on Ω1 if ∀N ∈τ 2 ,
f −1 ( N ∩ f (Ω1 )) ∈τ1.
It is easy to see from Definitions A.2.4 and A.2.5 that f is continuous
at x0 ∈Ω1 if and only if for any convergent sequence xn → x0 in Ω1 , the
sequence ( f ( xn ))n in Ω2 converges to f ( x0 ) ∈Ω2 . As mentioned earlier, this
technically holds for first countable spaces [4,7,11]. Finally, suppose U,V are
two topological spaces and f : U → V is a continuous bijection such that its
inverse f −1 : V → U is also continuous. Then, f is called a homeomorphism and
the two spaces U, V may be identified (topologically) with each other. This is
denoted by U ≅ V.
Given a topological space (Ω, τ) and any subset U ⊆ Ω, an open cover C ⊂ τ
of U is defined by U ⊆  N ∈C N. The open cover is finite if |C| < ∞, where the
*
†
These spaces are called first countable spaces. See, for example, References [7,11] for more
details.
For an uncountable indexing set, the associated “sequence” is called a net. This is the generalization of a sequence (ibid. for further details) required for more general topological spaces;
that is, spaces that do not satisfy the axiom of first countability.
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Appendix A
cardinality |C| denotes the number of elements in C. If C is an open cover of
U ⊆ Ω, and C ′ ⊂ C such that U ⊆  N ∈C′ N, then C′ defines a subcover of U.
A.2.6 Definition
Given a topological space (Ω,τ), a subset U ⊆ Ω is said to be compact if every
open cover of U has a finite subcover.
A.2.7 Proposition
Let Ω be a compact space and K ⊆ Ω Then, K is compact if and only if K is
closed.
□
Intuitively, a compact set is a closed set with “finite volume.” To make this
concept concrete, consider a finite-dimensional Euclidean space R n . Next,
define a metric (or distance function) ρ : R n × R n → R to be a nonnegative function satisfying
(a) ρ( x , y ) ≥ 0 ∀x , y ∈ R n with ρ( x , x) = 0 ∀x ∈R n
(b) ρ( x , y ) = ρ( y , x) ∀x , y ∈R n (symmetry)
(c) ρ( x , y ) ≤ ρ( x , z) + ρ( z, y ) ∀x , y , z ∈ R n (triangle inequality)
An ε-disk Bε ( x0 ) ⊂ R n of x0 is defined by Bε ( x0 ) = { x ∈R n : ρ( x , x0 ) < ε} . This
defines an open subset in R n . An example of a metric on R n is the Euclidean
metric ρ( x , y ) =||x − y||, where ||x − y||= ( x 1 − y 1 )2 +  + ( x n − y n )2 .
Given a subset U ⊂ R n , if ∃ε > 0 such that ∀x ∈U , U ⊂ Bε ( x), then U is
bounded in R n . In R n , a subset U ⊂ R n is compact if and only if it is both
closed and bounded. Thus, every closed ε-disk Bε ( x0 ) is compact in R n .
Compact sets possess some pleasant properties. An instance is given below.
A.2.8 Theorem
Let (Ω,τ) be a topological space and f:K → R be a continuous function, where
K ⊂ Ω is compact. Then, f attains its maximum and minimum on K. That is,
□
∃u , u ∈ K such that f (u) ≤ f ( x) ≤ f (u) on K.
This section closes with the concept of connectedness. Let Ω be a topological space. Then, Ω is said to be connected if ∀U,V ⊂ Ω open with U ∩ V = ∅, Ω
≠ U ∪ V. Furthermore, if ∀u,v ∈ Ω, ∃γ: [0,1] → Ω continuous, such that γ(0) =
u, γ (1) = v, then Ω is path connected. That is, a path-connected space is a space
such that any two points within the space can be connected by a (continuous) path. It can be shown that path connectedness implies connectedness;
however, the converse is false. A path γ such that γ(0) = γ(1) is called a loop or
a closed path.
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Appendix A
A.2.9 Definition
Let Ω be a path-connected topological space and let x,y ∈ Ω be a pair of arbitrary
points. Suppose γ , γ ′ : [0, 1] → Ω are two arbitrary paths connecting x,y. That
is, γ (0) = x = γ ′(0) and γ (1) = y = γ ′(1) . If there exists a continuous mapping
H: [0,1] × [0,1] → Ω such that H(0,t) = γ(t) ∀t ∈ [0,1] and H (1, t) = γ ′(t) ∀t ∈[0, 1],
then Ω is said to be simply connected. The mapping H is called a homotopy.
A multiply connected space is a space that is not simply connected. Lastly, a
loop is null-homotopic if there exists a homotopy mapping the loop into a
point, that is, if the loop can be continuously shrunk to a point.
In short, a homotopy defines a deformation that deforms one path into
another in a continuous fashion. Examples of simply connected spaces are
Rn and the 2-sphere S2. On the other hand, the 2-torus T2 (i.e., a donut) is not
simply connected: the presence of a hole in the torus prevents the loops λ,η
encircling the torus from deforming into one another (see Figure A.4). A hole
thus presents an obstruction toward the existence of a homotopy deforming
λ into η, where λ(0) = η(0) = P = λ(1) = η(1). On the other hand, the two paths
γ , γ ′ in R 2, where γ (0) = γ ′(0) = P = γ (1) = γ ′(1) , can be deformed into each
other. However, observe that if the point Q should be deleted from R 2, then
γ cannot be deformed into γ′ (and vice versa).
Regarding the 2-sphere S2 = {( x , y , z) ∈R 3 : x 2 + y 2 + z 2 = 1} in Figure A.4, it is
also clear that γ , γ ′ can be deformed into each other in S2. However, by deleting
the point Q ∈S2, it can be intuitively seen that γ ′ can be deformed into γ via a
homotopy that traverses around the sphere about the point P ∈S2. Topologically,
by deleting the point Q ∈S2 from S2 and stretching out the deleted point in all
directions to infinity, S2 − {Q} can be transformed into the plane R2.
A.2.10 Remark
Suppose K ⊂ R 3 is a compact (2-dimensional) surface such that ∂K = ∅. Then,
K is called a closed 2-surface. Intuitively, the boundary of a boundary is empty.
In what follows, a surface in R 3 always refers to a 2-surface. Finally, a compact
surface K ⊂ R 3 is said to be spanned by a simple loop γ ⊂ R 3 if γ = ∂K. Recall
P
P
γ
P
Q
γ'
R2
γ
Q γ'
η
λ
Q
T2
S2
Euclidean 2-space
2-sphere
2-torus
(a)
(b)
(c)
Figure A.4
Simple connectedness: (a) and (b); multiple connectedness: (c).
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Appendix A
that a loop is simple if γ(t) ≠ γ(s) ∀ t,s ∈ (0,1); that is, if the loop does not have any
self-intersection.
An example of a closed surface is a 2-sphere, whereas a hemisphere is
clearly a surface spanned by a loop. Suppose M is a path-connected topological space and γ i : [0, 1] → M are any two paths, for i = 1,2, such that
γ 1 (1) = γ 2 (0). Then, the operation ∗ can be defined on the pair ( γ 1 , γ 2 ) to construct a new path γ 1 ∗ γ 2 in the following manner.
 γ 1 (2t) for 0 ≤ t ≤ 1
2


γ 1 ∗ γ 2 (t) = 

1
 γ 2 (2t − 1) for 2 ≤ t ≤ 1
(A.3)
That is, Equation (A.3) defines a new path by joining γ 1 to γ 2 .
Indeed, this process can be continued for the sequence of paths ( γ i )i in
M if they satisfy γ i (1) = γ i + 1 (0) ∀i : γ 1 ∗ γ 2 ∗  ∗ γ i ∗  . Finally, the inverse γ −1
of the path γ is given by γ −1 (t) = γ (1 − t) . That is, the inverse path is just the
reverse orientation of the original path. In particular, γ ∗ γ −1 defines a degenerate loop.
A.3 Boundary Conditions for Electromagnetic Fields
The following two important theorems play a central role in proving certain
elementary properties of electromagnetism. They are stated without proof;
the proofs can be found in the numerous literature on electrodynamics
[3,13,18,19] or [12]. The first theorem is called Stokes’ theorem, and the second
theorem is called the divergence theorem.
Recall that a continuous mapping f : R n → R m is a called an m-vector field
(or simply a vector field) on R n . If m = 1, f is called a scalar field. Finally, for
notational convenience, let C k (R n , R m ) denote the space of k-times continuously differentiable vector fields on R n . Intuitively, think of this space as the
space of vector fields f = ( f 1 ( x ), , f m ( x )) satisfying
∂k f j
∂ x i1 ∂ x ik
=
∂k f j
∂ x σ ( i1 ) ∂ x σ ( ik )
∀ i1 ,… , ik ∈{1,… , n}, ∀j = 1,… , m
and all permutations σ: {1,…,n} → {1,…,n}. Note that a permutation is just a bijection from a countable set onto itself. The space C k (R n , R m ) is often abbreviated by Ck should no confusion arise, and the mappings in C k (R n , R m ) are
said to be of class Ck.
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Appendix A
Some examples are considered below. Let f : R 3 → R be continuous
such that ∂ xi f exists and is continuous on R 3 for i = 1,2,3. Then, f is of
class C1. In addition, if ∂ xi x j f exists and is continuous ∀ i,j = 1,2,3—equivalently, ∂ xi x j f = ∂ x j xi f ∀ i, j —on R 3 , then f is of class C2 . Finally, f ∈C0 (R 3 )
denotes f is a continuous function on R 3: this space is commonly denoted
by C(R 3 ).
A.3.1 Theorem (Stokes’ Theorem)
Let f be a vector field of class C1 on a compact surface Σ ⊂ R 3 in R 3 spanned
by ∂Σ ≠ ∅, l a unit vector field tangent to ∂Σ in R 3 , and N a unit vector field
normal to the surface Σ in R 3 pointing away from the interior of Σ.
Then,
∫
∂Σ
f ⋅ l d =
∫∫ ∇ × f ⋅ n dS.
□
Σ
A.3.2 Corollary
If Σ is a closed compact surface in R 3, then

∫∫ ∇ × f ⋅ ndS = 0 .
Σ
Proof
By assumption, ∂Σ = 0. Hence, invoking Stokes’ theorem,
∫∫ ∇ × f ⋅ n dS = ∫
Σ
∂Σ
f ⋅ l d ≡ 0
□
as asserted.
Heuristically, Corollary A.3.2 is evident: it suffices to consider S2 ⊂ R 3 and
note that ∂S2 = ∅ . That is, a sphere has no boundary. Thus, by Theorem
A.3.1, the surface integral must vanish, as the integral is over an empty set.
A.3.3 Theorem (Divergence Theorem)
If a vector field f on a compact set Ω ⊂ R 3is of class C1 , where Ω is simply
connected and ∂Ω is closed, then
∫∫
∂Ω
f ⋅ n dS =
outward, normal, unit vector field on ∂Ω.
∫∫∫ ∇ ⋅ f dV, where n is the
Ω
□
As a side remark, Stokes’ theorem and the divergence theorem are actually
special cases of a more general theorem attributed to Stokes, which relates
the volume integral of an n-dimensional volume to the surface integral of the
associated (n − 1)-dimensional boundary of the n-dimensional volume. For
more details, refer to any standard reference on differential geometry [4,17].
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Appendix A
Faraday’s law follows as a direct consequence of (1.1) via Stokes’ theorem:
∫
∂Σ
E ⋅ l d =
∫∫ ∇ × E ⋅ n dS = − ∂ ∫∫ B ⋅ n dS ≡ − ∂ Ψ
t
Σ
t
Σ
where Ψ ≡ ∫ ∫ Σ B ⋅ dS is the magnetic flux flowing across the surface Σ.
Notice that when the compact surface Σ is closed, Faraday’s law implies that
− ∂t Ψ = 0 . That is, the magnetic flux is conserved because it does not change
with time.
Furthermore, observe that the sum of magnetic flux entering and exiting a
closed surface is zero. In other words, magnetic flux forms loops. To see this,
it suffices to appeal to the divergence theorem: Ψ = ∫∫ Σ B ⋅ dS = ∫∫∫ M ∇ ⋅ BdV ,
where ∂M = Σ. From Maxwell’s equation, ∇ ⋅ B = 0, it follows at once that Ψ =
0. Thus, any magnetic flux lines that exit a compact surface must return to the
same surface.
Finally, recall two important identities from vector calculus. They are stated
without proof below. Suppose φ is a scalar field on R 3 that is of class C2, and f
a vector field on R 3 that is also of class C2. Then, the following identities hold.
∇ × ∇φ = 0
(A.4)
∇⋅∇×f=0
(A.5)
The contents of the remainder of this section comprise (i) the boundary
conditions for the electrostatic field, and (ii) the boundary conditions for the
magnetostatic field. So, first of all, consider a perfect dielectric medium; this
is an idealized medium whose conductivity is identically zero. In the presence of an applied electric field, the molecules comprising the dielectric are
polarized to form dipoles.
In complete analogy with the electric field, define the electric polarization
vector P on a dielectric medium Ω ⊂ R 3 to be the electric field induced by
the dipoles via an applied external electric field; see Figure A.5. Informally,
–
Zero
applied
electric
field
–
+
–
+ –
–
+
+
–
Electric
dipole
–
–
+
+
+
Random
orientation
–
+ –
–
–
Polarisation
+
–
field
–
+
+
+
+
Applied
external
electric
field
+
Figure A.5
Alignment of electric dipoles under an external electric field.
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Appendix A
under the influence of an applied electric field, the electric dipoles of a dielectric medium align themselves according to the applied field. The aligned
dipoles thus induce a polarization field of their own.
A.3.4 Definition
Let Ω ⊂ R 3 be a dielectric medium that is uncharged. Then, bound charges
are charges within Ω such that they lie within the valence band.* An electric
polarization field P generated by Ω is a C1 vector field satisfying
(A.6)
∇ ⋅ P + ρ = 0
where ρ is the bound charge density in Ω induced by the dipoles. In particular,
a bound surface charge density σ on Ω is defined by
(A.7)
P ⋅ n = σ
where n is the normal vector field on ∂Ω.
A.3.5 Lemma
Equations (A.6) and (A.7) are well-defined.
Proof
The total charge QΩ in Ω is clearly given by QΩ =
Theorem A.3.3,
0 = QΩ =
∫∫
∂Ω
∫∫
∂Ω
P ⋅ nd 2 x =
P ⋅ n d2 x −
∫∫
∂Ω
σ d 2 x +
∫∫∫ ρ d x . By
3
Ω
∫∫∫ ∇ ⋅ Pd x implies trivially that
3
Ω
∫∫∫ ∇ ⋅ P d x = ∫∫
3
Ω
∂Ω
σ d 2 x +
∫∫∫ ρ d x
as Ω is assumed to be uncharged, establishing the assertion.
3
Ω
□
A more general proof can be obtained (see, e.g., Reference [3, p. 107]) via the
vector identity ∇(αA + B) = A ⋅ ∇α + α∇ ⋅ A + ∇ ⋅ B given in Section A.1.
Indeed, it is evident from Gauss’ law ∇ ⋅ ε 0 E = ρ and the proof of Lemma
A.3.5, that the negative sign in Equation (A.6) is required in order for the
pola­rized charge in an electrically neutral dielectric to remain zero; that is,
∇ ⋅ P = −ρ . In a sense, Lemma A.3.5 furnishes a motivation for Definition A.3.4.
Some comments are due. First, it is clear from Definition A.3.4 that the presence of a polarization field will perturb the existing applied electric field; see
* Charges that lie in the conduction band, such as in a conductor, can flow freely within the
medium. However, charges that are in the valence band cannot flow freely throughout the
medium; they are thus bound.
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Appendix A
Figure A.5. To wit, the resultant field is the superposition of the applied field and
the field induced by the aligned dipoles. Second, by Gauss’ law, the total volume
charge density must satisfy ∇ ⋅ ε 0 E = ρ + ρ , whence, by (A.6), ρ = ∇ ⋅ (ε 0 E + P ).
This leads naturally to the following definition.
A.3.6 Definition
Given a dielectric medium Σ ⊂ R 3 subject to an external electric field E, the
electric displacement D is a vector field in R 3 given by D = ε 0 E + P , and the
electric permittivity ε of Ω is defined by P = (ε − ε 0 )E, where ε0 is the electric
permittivity of free space. The ratio εε0 is called the dielectric constant (or
relative permittivity) of Ω.
A.3.7 Proposition
Given a dielectric Ω ⊂ R 3 and an applied external electric field E, the electric
displacement satisfies D = εE.
Proof
By definition, P = (ε − ε 0 )E ⇒ D = ε 0 E + (ε − ε 0 )E = εE , as claimed.
□
A.3.8 Theorem
Let (Ω ± , ε ± ) ⊂ R 3 be two simply connected dielectric media such that
∂Ω+ = S = ∂Ω−. Suppose an electric field E is directed from Ω+ across the
boundary S into Ω− . Then, n × ( E+ − E− ) = 0 , where n is a unit, normal, vector
field on S directed into Ω− and E± = E|Ω ± .
Proof
Recall that for a simply connected space,
∫ E ⋅ dl = 0, where l is the vector
γ
field tangent to a loop γ, thus, choose a sufficiently small rectangular loop
γ = γ ↑ ∪ γ → ∪ γ ↓ ∪ γ ← , where γ → ⊂ Ω− , γ ← ⊂ Ω+ are paths parallel to a vector
v tangent at some fixed point x ∈ S, and γ ↑ , γ ↓ are paths normal to v(x), for
a sufficiently small loop γ. That is, the path γ → ( γ ↑ ) is of the same length and
is oppositely oriented to γ ← ( γ ↓ ) . Then, by construction,
0=
∫ E ⋅ dl = ∫
γ
γ↑
E ⋅ dl +
∫
γ→
E ⋅ dl +
∫
γ↓
E ⋅ dl +
∫
γ←
E ⋅ dl
In particular, in the limit as the lengths of γ ↑ , γ ↓ go to zero, where γ → , γ ←
are held constant,
0=
K15149_Book.indb 346
∫
γ→
E ⋅ dl +
∫
γ←
E ⋅ dl =
∫
γ→
( E− − E+ ) ⋅ dl
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347
Appendix A
where, recall that in the limit as the lengths of γ ↑ , γ ↓ tend to zero,
∫
γ→
E ⋅ dl = −
∫
γ←
E ⋅ dl
(as they will then have the same endpoints in passing to the limit). As Ω is
simply connected, this is possible; a more rigorous construction via homotopy is not pursued here.
Thus, 0 = ∫ γ → ( E− − E+ ) ⋅ dl ⇒ ( E− − E+ ) ⋅ l = 0, where l is the unit tangent vector field on γ →. Because the scalar product of the normal component of E±
with l is zero by construction, it follows from the arbitrariness of E that the
E+ ≡ E−. Finally, as n is normal to l, the result is equivalent to n × ( E− − E+ ) = 0.
That is, E is continuous across the boundary interface S, as required.
□
A.3.9 Corollary
Suppose Ω− is a perfect electrical conductor (PEC). Then, n × E+ = 0 on S.
Proof
In a smuch as Ω− is PEC, it follows that the static electric field in Ω− is zero:
E− = 0 . Thus, the conclusion follows.
□
Now, from Proposition A.3.8, it is clear that the tangential component of
D ± on S is thus discontinuous across S unless ε + = ε − . Indeed this is quite
evident: n × E+ = n × E− ⇒ n × D+ ε1+ = n × D− ε1− ⇒ D+ = εε+− D− , as claimed.
Physically, this relates to the dipoles induced on the boundary of Ω ± . The
tangential component of the electric displacement is thus reduced by the
dielectric constant of the medium. Finally, from the definition of potential,
ϕ = − ∫ E ⋅ dl , invoking Theorem A.2.8 leads to the continuity of the potential
across S. To see this, it suffices to observe from the proof of Theorem A.2.8
that 0 = ∫ γ → E ⋅ dl + ∫ γ ← E ⋅ dl = ∫ γ → ( E− − E+ ) ⋅ dl = ϕ − − ϕ + , whence ϕ − |S = ϕ +|S .
Thus, Theorem A.3.8 is equivalent to the continuity of the potential across a
dielectric interface. This is formally stated below.
A.3.10 Corollary
Given any x ∈ S, lim ϕ − (y) = lim ϕ + (y ′), where y ∈Ω− and y ′ ∈Ω+ .
y→ x
y ′→ x
□
The next result establishes the normal component of the electric field
across a dielectric boundary.
A.3.11 Theorem
Let (Ω ± , ε ± ) ⊂ R 3 be two simply connected dielectric media such that
∂Ω+ = S = ∂Ω− . Suppose an electric field E is directed from Ω+ across the
boundary S into Ω− . Then, n ⋅ ( D+ − D− ) = 0 , where n is a unit vector field
normal to S pointing into Ω− , and D± = D|Ω ± .
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348
Appendix A
Proof
Across S, fix some point x 0 = ( x01 , x02 , x03 ) ∈S and consider a small cylinder
V (r , δ) = B2 ( x 0 , r ) × [ x03 − δ , x03 + δ], where
{
B2 ( x 0 , r ) = x ∈R 3 : ( x 1 − x01 )2 + ( x 2 − x02 )2 ≤ r , x 3 = x03
}
is a circular surface of radius r centered about x 0 . Set V± (r , δ) = Ω ± ∩ V (r , δ).
Then, by Gauss’ law,
∫
V (r ,δ)
∇ ⋅ Dd 3 x =
∫
V (r ,δ)
ρd 3 x
where ρ is the charge density in V(r, δ). Invoking Theorem A.3.3,
∫
V (r ,δ)
∫
∇ ⋅ Dd 3 x =
∂V↑ ( r , δ )
∫
∂V0+ ( r , δ )
D− ⋅ n↑ d 2 x +
D− ⋅ n0 d 2 x +
∫
∫
∂V↓ ( r , δ )
∂V0− ( r , δ )
D− ⋅ n↓ d 2 x +
D ⋅ n0 d 2 x
where ∂V↑ (r , δ) = B2 ( x 0 , δ) × { x03 + δ}, ∂V↓ (r , δ) = B2 ( x 0 , δ) × { x03 − δ} are the end
caps of the cylinder V(r, δ),
∂V0+ (r , δ) = (∂V (r , δ) ∩ Ω+ ) − ∂V↓ (r , δ) and ∂V0− (r , δ) = (∂V (r , δ) ∩ Ω− ) − ∂V↑ (r , δ)
are the cylindrical boundary modulo the end caps of the cylinder. The unit
vector field n↑ ( n↓ ) is the outward normal on ∂V↑ (r , δ) ( ∂V↓ (r , δ) ) and n0 is
the outward normal unit vector field on ∂V0− (r , δ) ∪ ∂V0+ (r , δ) .
Then, by construction, lim ∫ ∂V ± ( r ,δ ) D± ⋅ n0 d 2 x = 0 , and hence, for r > 0
0
δ→ 0
sufficiently small, and noting that on S, n↑ = − n↓ ,
ρSπr 2 ≈ lim
δ→ 0
∫
V (r ,δ)
∇ ⋅ Dd 3 x =
∫
∂V↑ ( r , δ )
D− ⋅ n↑d 2 x +
∫
∂V↓ ( r , δ )
D+ ⋅ n↓d 2 x ≈ n↑ ⋅ (D− − D+ )πr 2
where ρS is the surface charge density on ∂V(r, δ). Hence, n↑ ⋅ ( D− − D+ ) = ρS .
In particular, in the absence of free charges, ρS = 0 ⇒ n↑ ⋅ ( D− − D+ ) = 0, as
desired.
□
The electric field normal to a dielectric interface is thus discontinuous, and
the discontinuity is the result of the formation of the induced dipoles at the
interface caused by the applied electric field. Moreover, note that in a conductor, the valence electrons are not bound to the crystal lattice: they are free
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349
Appendix A
charges that may be approximated by a cloud of plasma in the conduction
band. This observation leads to the following corollary.
A.3.12 Corollary
If Ω− is a PEC, then n↓ ⋅ D+ = ρS , where n↓ is the unit normal vector field
on S pointing into Ω+ and ρS is the surface charge induced by the applied
electric field.
Proof
Because Ω− is a PEC, it follows that the static electric field within Ω− is zero
and hence, − n↑ ⋅ D+ = n↓ ⋅ D+ = ρS , as required.
□
Some comments regarding the proof of Theorem A.3.11 are due. An
uncharged pure dielectric can only have bound charges. These charges are
induced by the presence of an applied electric field that polarizes the molecules to form dipoles. However, if free charges were placed on the boundary
S, then n↑ ⋅ ( D− − D+ ) = ρS would hold.
Conversely, a conductor has, by definition, free charges present in the
conduction band in order to conduct electrons.* Hence, even if the conductor were initially neutral, in the presence of an electric field, the electric field would cause the free charges in Ω− to migrate onto S such that
the resultant static electric field in Ω− is zero. This is encapsulated in
Corollary A.3.12.
A.3.13 Example
Consider a point charge above a dielectric interface defined as follows. Given
(Ω ± , ε ± ) ⊂ R 3 , where Ω+ = R 3+ and Ω− = R 3 − Ω+ , set S = ∂Ω+ and suppose
that a charge Q ≠ 0 is located at xQ = (0, 0, d) ∈Ω+. What is the potential in R3?
Using the method of images, consider an image charge Q ′ ∈Ω− located at
xQ′ = (0, 0, − d) . Then, the potential on Ω+ is given by
ϕ+ =
1
4 πε +
{
Q
x 2 + y 2 + ( z − d )2
+
Q′
x 2 + y 2 + ( z + d )2
}
(A.8)
Next, given that Ω− is a dielectric, the field within Ω− is nonzero. To determine the field in Ω− , suppose ∃Q ′′ ∈Ω+ located at xQ = (0, 0, d) , where the electric permittivity of the whole space is ε − . Then, the potential in Ω− is given by
ϕ− =
1
4 πε −
Q ′′
x 2 + y 2 + ( z − d )2
(A.9)
* More precisely, the intersection of the conduction band and the valence band is nonempty.
Hence, the valence electrons become available for conduction as they also exist in the conduction band.
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350
Appendix A
It thus remains to determine (Q ′ , Q ′′) . To do so, the boundary conditions
must be invoked. By Theorem A.3.8, lim+ ε - n ⋅ E− ( x , y , z) = lim− ε + n ⋅ E+ ( x , y , z),
z→ 0
z→ 0
where n is the unit normal vector field on S directed toward Ω− . From E =
−∇φ, it follows that n ⋅ E± = − ∂ z ϕ ± . Thus,

Q( z − d )
Q′( z + d )
∂ z ϕ + = − 4 πε1 + 
3 +
3
2
2
2 2
2
2
(
)
x
+
y
+
z
−
d
x
+
y
+ ( z + d )2 ) 2
(
)
(

∂ z ϕ − = − 4 πε1 +
Q ′′ ( z − d )
(x
2
)



3
+ y 2 + ( z − d )2 2
and on S, the boundary condition yields −Q + Q ′ = −Q ′′ ⇒ Q ′′ = Q − Q ′ .
Next, invoking Corollary A.3.10, lim+ ϕ + = lim− ϕ −. Thus, from (A.8) and (A.9),
z→ 0
1
ε+
(Q + Q ′ ) =
1
ε−
z→ 0
Q ′′ ⇒ Q ′′ =
ε−
ε+
(Q + Q ′ )
whence solving the pair of equations yields
Q′ =
ε + −ε −
ε + +ε −
Q
(A.10)
Q ′′ =
2 ε−
ε + +ε −
Q
(A.11)
□
It is interesting to note from Equation (A.10) that when ε + > ε − , Q′ has the
same sign as Q, whereas for ε + < ε − , Q′ has the opposite sign as Q. This also
explains why the mirror image induced by a conducting ground plane is oppositely charged, as the electric permittivity of a PEC is infinity. On the other
hand, when a charge is embedded in a dielectric medium, its image charge
induced across the boundary in a medium with a lesser electric permittivity is
of the same sign as that of the original charge. This is clearly counterintuitive.
A.4 Elements of Partial Differential Equations
In this brief primer, two topics in partial differential equations (PDE) that
are of primary importance to electromagnetic theory are reviewed. They
are the Poisson equation and the wave equation. Readers interested in pursuing more advanced topics in PDE may refer to some excellent references
[1,8,10,16,23] for a more general and abstract approach, or [9,26] for a more
applied approach. An elementary knowledge of complex analysis is assumed
(see Reference [4] for an excellent exposition).
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Appendix A
Recall that the concept of distance on the real line R is defined by the
following metric, d : R × R → R + , d( x , y ) =|x − y|. More generally, on R 3 ,
d3 : R 3 × R 3 → R + may be defined by the Euclidean norm
d3 ( x , y) =||x − y||3 = ( x 1 − y 1 )2 + ( x 2 − y 2 )2 + ( x 3 − y 3 )2
There are other equivalent definitions, however, the above suffices for the
present discussion.
A.4.1 Definition
Let f : R 3 → R be a function. If K ⊂ R 3 is compact and f = 0 on R 3 − K , then
f is said to have a compact support supp(f) = K.
A.4.2 Definition
Suppose ψ: M → R is of class C2, where M ⊂ R 3 is compact. Furthermore, let
f: M → R be continuous on M. Then, −Δψ = f on M is called the Poisson equation on M, and if f = 0, it is known as the Laplace equation. The operator Δ is
called the (3-dimensional) Laplacian. Moreover, if −Δψ = f satisfies
(a) ψ = g on ∂M, where g ∈ C(∂ M), then it is said to satisfy the Dirichlet
boundary condition
(b) ∂ n ψ = g on ∂M, where n is a normal vector field on ∂M (pointing
outward from M) and some g ∈ C(∂ M), then it is said to satisfy the
Neumann boundary condition
It is crucial to note that −Δψ = f has no solution if both boundary conditions
(a) and (b) are simultaneously satisfied on ∂M. In order for a solution to exist,
it is necessary that either (a) or (b) hold. A solution ψ of the Laplace equation is called a harmonic function. Furthermore, Definition A.4.2 also holds for
M = R 3 if ψ has compact support in M.
A.4.3 Definition
The Poisson equation −Δψ = f on M subject to the Dirichlet boundary condition ψ = g on ∂M is said to be well-posed if
(a) A unique solution ψ exists, and
(b) Given g i ∈ C(∂ M), i = 1,2, and unique solutions ψ i to the Dirichlet
problem −Δψ = f on M subject to ψ = g i on ∂M, ∀ε > 0, |g1 − g 2|< ε
on ∂M implies that |ψ 1 − ψ 2|< ε .
Condition A.4.3(b) is known as the continuous dependence on data for the Dirichlet
problem. Essentially, it states that if the data undergo a small perturbation, then
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352
Appendix A
the solution will also change by a small perturbation. That is, deforming the
data continuously from g1 into g2 will deform the solution ψ1 continuously into
ψ2. Finally, the solutions considered in this monograph are classical solutions.
More general solutions, called weak solutions [8,23,25], are not considered here
although they are briefly explored in the last section for the reader interested in
applying finite element analysis [14,22,25] to electromagnetics.
A.4.4 Example
Consider r : R 2 → R defined by r( x ) = ( x 1 )2 + ( x 2 )2 . Then, for x ≠ 0,
∂ xi r( x ) =
xi
( x1 )2 + ( x 2 )2
=
xi
r( x)
for i = 1, 2.
And
∂2xi r( x ) =
xi
( x1 )2 + ( x 2 )2
=
1
r( x)
−
( x i )2
r3 ( x)
Thus, for any twice differentiable function f on R, the composition
f  r : R 2 → R is of class C2 and hence, on setting ψ = f ∘ r, ∆ψ = ψ ′′ + 1r ψ ′ ,
where ψ ′ = drd ψ . So, when Δψ = 0 on R 2, setting v = ψ ′ ⇒ vv′ = − 1r ⇒ v = 1r , as
− ln r = ln 1r , for v, r ≠ 0. Thus, ddr ψ = 1r ⇒ ψ = ln r .
Now, as Δ is a linear operator, it follows that the most general solution is
of the form ψ = a ln r + b, where a,b are arbitrary constants. In particular, the
solution
ψ = − 21π ln r
(A.12)
is called the fundamental solution of the 2-dimensional Laplace equation
Δψ = 0.
□
A.4.5 Example
Emulating Example A.3.3, consider
r : R3 → R
defined by r (x) =
( x ) + ( x ) + ( x ) and for any twice differentiable function f on R, form
the composition f  r : R 3 → R . Then, f is of class C2 and in view of Example
A.4.4,
1 2
2 2
3 2
∆ψ = ψ ′′ + 2r ψ ′
where ψ = f ∘ r. Thus, as before, on setting v = ψ ′ ⇒ vv′ = − 2r ⇒ v = r12 That is,
the general solution to the 3-dimensional Laplace equation Δψ = 0 is given by
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353
Appendix A
ψ = ar + b , for r ≠ 0 and some arbitrary constants a,b. In particular, the fundamental solution of the 3-dimensional Laplace equation is defined by
ψ = 41π
(A.13)
□
1
r
It is clear that the solutions defined in Examples A.4.4 and A.4.5 are radial
solutions. An important property satisfied by the solution of Laplace’s equation is the mean value property.
A.4.6 Definition
Let M ⊂ R 3 be open and ψ: M → R be continuous. Suppose ∀x ∈ M, ∃r > 0
such that B( x , r ) ⊂ M and ψ ( x ) = 4 π1r 2 ∫ ∂ B( x , r ) ψ (y)d 2 y . Then, ψ is said to satisfy the mean value property. Finally, for M ⊂ R 2 , the mean value property is
defined by ψ ( x ) = 2 π1 r ∫ ∂ B( x , r ) ψ (y)dy .
A.4.7 Remark
It can be established that if ψ is analytic on M ⊂ R 3 , then it possesses the
mean value property. In particular,
ψ( x) =
1
4π r2
∫
∂ B( x , r )
ψ (y)d 2 y =
3
4π r3
∫
B( x , r )
ψ (y)d 3 y
Moreover, the Dirichlet problem of Definition A.4.2(a) has at the most one
solution; that is, if a solution exists, it is unique. The existence of a solution
is often a very difficult task to establish, and the existence depends upon the
smoothness of the boundary [8,10,15]. Clearly, in this appendix, the boundary is always assumed to be sufficiently smooth (the term regular is often
used interchangeably in the literature) to avoid pathological problems with
the question of existence.
A subset M ⊂ R 3 is called a domain if it is open, bounded, and path connected. Now, let M ⊂ R 3 be a domain and consider the Dirichlet problem
−Δψ = f on M
(A.14a)
ψ = g on ∂M
(A.14b)
where g ∈ C(∂ M) and suppose that a solution ψ ∈ C2 (M) ∩ C(∂ M) for (A.14)
exists. Is there a formal representation for the solution? This leads to the
concept of Green’s function.
In all that follows, the analysis is restricted to R n , for n = 2,3. Given a domain
M ⊂ R n, let Ψ n = Ψ n (|x ′ − x|) denote the fundamental solution on R n , and
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354
Appendix A
x ′ , x ∈ M , where x ′ ≠ x . This is often written as Ψ n = Ψ n ( x ′ , x ) . By definition, the function is symmetric: Ψ n ( x , x ′) = Ψ n ( x ′ , x ) ; see Equation (A.12) and
(A.13), where r =||x ′ − x||=||x − x ′||. Furthermore, suppose ∃Φ n = Φ n ( x ′ , x ) ,
symmetric, such that it is the solution to the following Laplace equation, for
each fixed x ∈ M,
∆ ′Φ n ( x ′ , x ) = 0 ∀x ′ ∈ M
(A.15a)
Φ n ( x ′ , x ) = −Ψ n ( x ′ , x ) ∀x ′ ∈∂ M
(A.15b)
where Δ′ denotes the Laplacian of the primed coordinates x′. For instance, if
∆ = ∑ i ∂2xi in rectangular coordinates, then ∆ ′ = ∑ i ∂2x′i . Then,
G( x ′ , x ) = Ψ n ( x ′ , x ) + Φ n ( x ′ , x )
(A.16)
defines the Green function for the Dirichlet problem (A.14). By construction,
Green’s function is symmetric in its variables.
It can be shown that the solution to Equation (A.14) has the representation:
ψ( x) =
∫
M
f ( x ′)G( x ′ , x )d n x ′ −
∫
∂M
g( x ′) ∂ n′ G( x ′ , x )d n− 1 x ′ ,
(A.17)
where n = 2,3, and n′ is the unit normal vector field on ∂M pointing outward
from M; that is, exterior to M. From an electrostatic perspective, the Green
function denotes an impulse function acting on a unit point charge located
at x′.
A.4.8 Example
Consider the example of a charge above a grounded plane immediately following Example 1.1.5. Recasting the problem in the form given by (A.14): M = R 3+ ,
f = δ 3 ( x ′ − x ) ≡ δ( x ′ 1 − x 1 )δ( x ′ 2 − x 2 )δ( x ′ 3 − x 3 ) is the Dirac-delta distribution
defined by
 0 for x ′ ≠ x

δ( x ′ − x)d x ′ = 1 and δ( x ′ − x) = 
−∞

 ∞ for x ′ = x
∫
∞
and lastly, g = 0. Thus, by Equation (A.15),
ψ( x) =
K15149_Book.indb 354
∫
M
f ( x ′)G( x ′ , x )d n x ′ = G(0, x )
□
10/18/13 11:11 AM
355
Appendix A
The fundamental solution in R 3 is given by Equation (A.13):
Ψ3 =
1
1
4 π ||x ′− x||
where x ′ = (0, 0, d) and x ∈ M is any fixed arbitrary point. Returning to
Example A.4.8, choose Φ 3 = − 41π ||x ′′−1 x|| , where x ′′ = (0, 0, − d) . Then, on the
plane ∂R 3+ , G( x ′ , x ) = Ψ 3 ( x ′ , x ) + Φ 3 ( x ′ , x ) = 0 ∀x ∈∂R 3. This then is the essence
of the method of images.
Before proceeding to the study of wave equations, some definitions must
be recalled to facilitate the discussion. First, a general linear second-order PDE
has the form:
∑
3
i, j= 0
aij ( x) ∂ij ψ ( x) +
∑
3
j= 0
b j ( x) ∂ j ψ ( x) + c( y )ψ ( x) = f ( y ) ∀x ∈ M ⊂ R 3+ 1
where x0 of the coordinate x = ( x 0 , , x 3 ) denotes time, and aij , b j , c ∈ C( M)
such that aij = a ji ∀i, j. For notational simplicity, ∂ k = ∂ xk . Then, a nonconstant
function ξ ∈C2 ( M) defines a characteristic surface S = { x ∈ M : ξ( x) = 0} if the
following criterion is satisfied: ∑ i3, j = 0 aij ( x) ∂i ξ( x) ∂ j ξ( x) = 0 on S. Otherwise,
the surface is said to be noncharacteristic with respect to the linear secondorder PDE. The quantity Qξ ( x) ≡ ∑ i3, j = 0 aij ( x) ∂i ξ( x) ∂ j ξ( x) is called the associated ξ-quadratic form.
A.4.9 Remark
Consider the matrix A = ( aij ) of the general second-order linear PDE. For a
fixed ( x0 , t0 ) ∈ M × [0, ∞), if the set σ(A) of eigenvalues of A satisfies
(a) sgn(λ i ) = sgn(λ j ) ∀λ i , λ j ∈σ(A) , then the PDE is elliptic at ( x0 , t0 )
(b) sgn(λ i ) = − sgn(λ iˆ ) ∀λ i ∈σ(A) − {λ iˆ } for some fixed λ î , then the PDE
is hyperbolic at ( x0 , t0 )
(c) ∃λ = 0 for some λ ∈σ(A), then the PDE is parabolic at ( x0 , t0 )
Furthermore, if the general second-order linear PDE is respectively, elliptic, hyperbolic, or parabolic on M × [0,∞), then the PDE is said to be, respectively, elliptic, hyperbolic, or parabolic.
A.4.10 Definition
Suppose ψ: M × (0,∞)→R is of class C2, where M ⊂ R 3 is compact. Furthermore,
let f: M × [0,∞) → R be continuous on M. Then, the hyperbolic equation
∂t2 ψ − c 2 ∆ψ = f
K15149_Book.indb 355
on
M
(A.18a)
10/18/13 11:11 AM
356
Appendix A
satisfying the following initial value conditions
 ψ = g

 ∂t ψ = h
on
M × {0}
(A.18b)
for some continuous functions g, h on M, is called the inhomogeneous wave
equation on M, and if f ≡ 0, it is known as the wave equation. The coefficient c is
the speed of the wave propagation.
Now, given the wave equation ∂t2 ψ − c 2 ∆ψ = 0 on M × (0,∞), consider
the associated quadratic form Qξ = (∂t ξ)2 − c 2 |∇ξ|2 on M × [0,∞), where
∇ = (∂1 , ∂2 , ∂3 ) and ∂t = ∂0 . Next, consider the function ξ: M × [0,∞) → R
2
defined by ξ( x , t) = c2 (t − τ)2 − 21 |x − y|, for some fixed (y, τ) ∈ M × [0,∞). Then,
by definition,
Qξ = (∂t ξ)2 − c 2 |∇ξ|2 = c 4 (t − τ)2 − c 2 |x − y|2 = 2 c 2 ξ
whence the characteristic surface Sξ = {( x , t) ∈Ω × [0, ∞) : c|t − τ|=|x − y|}, as
Qξ = 0 implies at once that c 4 (t − τ)2 − c 2 |x − y|2 = 0 ⇒ c 2 (t − τ)2 −|x − y|2 = 0 .
See Figure A.6 for an illustration of the characteristic surface based at some
(y,τ) ∈ M × (0,∞).
A.4.11 Remark
Given the wave equation ∂t2 ψ − c 2 ∆ψ = 0 on M × (0,∞) satisfying the
initial value condition ψ|t = 0 = g and ∂t ψ|t = 0 = h on M, let Sξ ( x0 , t0 )
denote the closure of the characteristic cone of the wave equation:
Sξ = {( x , t) ∈ M × [0, ∞) : c(t0 − t) ≥|x − x0|} at ( x0 , t0 ) ∈ M × (0, ∞) . Then, the
domain of dependence at the point ( x0 , t0 ) is defined by the closed disk
B( x0 ) = Sξ ( x0 , t0 ) ∩ M × {0} . The reason for this is the following: the wave
solution at ( x0 , t0 ) only depends on its initial value on B( x0 ) . This is clarified
by the following theorem.
Forward
characteristic
cone
t
Cone apex
(y, τ)
Backward
characteristic
cone
x
Figure A.6
Characteristic surface of a wave equation.
K15149_Book.indb 356
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357
Appendix A
A.4.12 Theorem (Domain of Dependence Inequality)
Given the wave equation ∂t2 ψ − c 2 ∆ψ = 0 on M × (0,∞) satisfying the initial
value condition ψ|t = 0 = g and ∂t ψ|t = 0 = h on M, let Sξ ( x0 , t0 ) denote its closed
characteristic cone at ( x0 , t0 ) ∈ M × (0, ∞), and set
B( x0 ; τ) = Sξ ( x0 , t0 ) ∩ Ω × {τ}
Then,
∫
B ( x0 ; τ )
{|∇ψ| + ∂ ψ }
2
2
t
t =τ
d3 x ≤
∫ {|∇ψ| + ∂ ψ }
2
B ( x0 )
{
2
⟨ψ ( x ; r , t)⟩∂ B( x , r ) ≡ |∂ B(1x , r )|
∫
2
t
t=0
d 3 x ∀τ ∈[0, t0 ]
□
}
The quantity e( x0 ; τ) = 21 ∫ B( x0 ;τ ) |∇ψ|2 + ∂t ψ
d 3 x is called the energy of the
t =τ
wave in B( x0 ; τ) .
Now, consider (A.18) and define the average
∂ B( x , r )
ψ ( y , t)d 2 y
where ∂B(x,r) is a sphere of radius r > 0 centered about x, and |∂B( x , r )| denotes
the surface area of ∂B(x,r). To express the solution of Equation (A.18) as a
function of g,h, recall first, the following lemma [8, p.70].
A.4.13 Lemma (Euler-Poisson-Darboux)
Fix x ∈R 3 and some r > 0, and suppose that ψ is a solution of (A.18). Then,
relative to polar coordinates, ∂t2 Ψ − ∂2r Ψ − 2r ∂r Ψ = 0 on R + × (0, ∞) with
Ψ = ⟨ g( x ; r , t)⟩∂ B( x , r ) and ∂t Ψ = ⟨ h( x ; r , t)⟩∂ B( x , r ) defined on R + × {0} , where
Ψ ≡ ⟨ψ ( x ; r , t)⟩∂ B( x , r ) .
□
Indeed, it can be shown via Lemma A.4.13 that the solution of Equation
(A.18) for Ω ⊆ R 3 is given by the following Kirchhoff formula,
ψ ( x , t) = ⟨th( y ) + g( y ) + ∇g ⋅ ( y − x)⟩∂ B( x ,t )
(A.19)
for t > 0, and for Ω ⊆ R 2 , the solution is given by the Poisson formula:
ψ ( x , t) =
tg ( y )+ t 2 h( y )+ t∇g ⋅( y − x )
t 2 −|y − x|2
(A.20)
B( x , t )
The interested reader may pursue the excellent references [8,16,23,26] for the
details of the derivation.
This section closes with a cursory generalization of PDE solutions. From
an application perspective (e.g., by numerical scientists and engineers) its
utility arises in the formulation of the finite element method [14,22,25] for
K15149_Book.indb 357
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358
Appendix A
Ω
φ=0
∆φ = 0
0
φ=1
Figure A.7
No classical solution exists for the Dirichlet problem.
numerical computation. In essence, this final section paves the foundation
for EMC engineers to apply the method of finite element analysis to solve
Maxwell’s equations numerically.
By way of motivation, consider Poisson’s equation on some domain M:
 −∆ψ = f on M



 ψ = g on ∂ M
(A.21)
If a solution ψ ∈ C2 ( M) ∩ C( M) of Equation (A.21) exists, then it is called a
classical (or strong) solution. Before proceeding to weaken the definition of the
solution of (A.21), some background information is required.
Given a subset Ω ⊆ R 3 , define its volume by μ(Ω), where μ is the Lebesgue
measure; see, for example, References [4,6,20] for details. Note that if μ(N) = 0
for some N ⊂ R 3 , then N is set to be a null set (or a set of measure zero), and μ(Ω
∪ N) = μ(Ω) by the definition of Lebesgue measure. In particular, if N ⊂ R 3 is
countable, then it is a set of measure zero. Informally, the Lebesgue measure
is roughly the generalisation of the Riemann integral (encountered in undergraduate engineering calculus) extended to a collection of null sets such that
if N is a null set, then so is M ⊂ R 3 ∀M ⊂ N. For a precise definition, consult
the previously cited references.
Now, given a function f: M → R, the function is Lebesgue integrable if
∫ Ω |f ( x)|dµ( x) < ∞ . Moreover, if f, g: Ω → R such that f = g on Ω − N for some
N ⊂ Ω such that μ(N) = 0, then, f = g almost everywhere (a.e.). Next, define the
space of square integrable functions by L2 (Ω) = f : ∫ Ω |f ( x)|2 dµ( x) < ∞ (modulo functions that equal one another μ-a.e.). The space L2 (Ω) is a Hilbert
space.
{
}
A.4.14 Definition
Given a real vector space V, an inner product (⋅, ⋅) : V × V → R is a continuous
function satisfying
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359
Appendix A
(a) Positive-definiteness: (u, u) ≥ 0 ∀u ∈V with (u, u) = 0 if u = 0.
(b) Symmetry: (u, v) = ( v , u) ∀u, v ∈V.
(c) Linearity: (αu + v , w) = α( v , w) + ( v , w) ∀u, v , w ∈V and α ∈ R.
Then, V is called a pre-Hilbert space if it admits an inner product structure. An
inner product defines a metric or norm on V in an obvious manner: |u|= ( u, u) .
Furthermore, recall that a sequence (un ) ⊂ V is Cauchy if ∀ε > 0, ∃N > 0 such
that n, m > N ⇒|un − um|< ε ; and a sequence (un ) ⊂ V converges (un ) → u0 ∈V ,
if ∀ε > 0, ∃N > 0 such that n > N ⇒|un − u0|< ε . The point u0 is called a limit
point. Then, V is complete with respect to the norm if every Cauchy sequence in
V converges in V.
In view of Definition A.4.11, a Hilbert space is a pre-Hilbert space that is
complete with respect to the inner product. It can be shown that the Lebesgue
measure μ on M ⊆ R 3 defines an inner product on L2 ( M) as follows.
( u, v ) =
∫
M
u( x)v( x)dµ( x)
(A.22)
It particular, L2 ( M) endowed with (A.22) defines a Hilbert space.
A.4.15 Example
Define H 1 ( M) = {u ∈ L2 ( M) : ∂i u ∈ L2 ( M), i = 1, 2, 3} . Then, an inner product
can be defined on H 1 ( M) as follows,
( u, v )H 1 ( M ) = ( u , v ) +
∑ ( ∂ u, ∂ v )
i
i
i
(A.23)
with norm given by |u|H1 ( M ) = (u, v) + ∑ i (∂i u, ∂i v) . It can be shown that the
space is complete with respect to the inner product and it is thus a Hilbert
space. This is a space wherein the function together with its partial derivatives are square-integrable.
□
Now, in hindsight, define H 10 ( M) = {u ∈ H 1 ( M) : u|∂ M = 0} . Then, in view of
the homogeneous Dirichlet boundary problem, H 10 ( M) is constructed as a
possible solution space for the homogeneous Dirichlet problem. The space
H 10 ( M) admits an inner product inherited from H 1 ( M) via Equation (A.23).
Finally, define H −1 ( M) ≡ H 10 ( M)′ to be the topological dual of H 10 ( M) ; this
is the space of bounded linear functionals ξ : H 10 ( M) → R endowed with
appropriate topology that is outside the scope of this exposition [2,4].*
Returning to Equation (A.21), by way of motivation, consider two commonly cited examples due to Zaremba and Lebesgue, respectively: the former can be found in Reference [15, p. 285] and the latter in [26, p. 198]. Given a
* These spaces are more generally known as Sobolev spaces, the properties of which can be
found in References [1,8,10].
K15149_Book.indb 359
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360
Appendix A
unit 2-disk B(0, 1) ⊂ R 2, set Ω = B(0,1) − {0}, illustrated in Figure A.7 no classical solution exists for the Dirichlet problem Δφ = 0 on Ω subject to the boundary conditions:
 1 for x = 0,

ϕ=
 0 on ∂B(0, 1)

Indeed, to see the nonexistence of a classical solution, suppose
∃ϕ ∈ C2 (Ω) ∩ C(Ω) satisfying the above boundary conditions. By definition, φ
is analytic, hence, if φ were analytic at 0, then φ ≡ 0 on Ω by the maximum
modulus principle, yielding a contradiction. Thus, the origin 0 must at most be
a removable singularity by Riemann. Recall that Riemann’s theorem on removable singularity states that if a function is analytic on Ω, then either the function
is analytic at 0 or 0 is a removable singularity of the function [5]. Hence, a suitable value can be assigned to φ(0), rendering φ analytic at 0. However, in order
for φ to satisfy the boundary condition at 0, φ must be discontinuous at 0 as analyticity implies that φ ≡ 0 on Ω. Hence, no classical solution exists, as asserted.
The second example due to Lebesgue is more complicated, and hence, only
a heuristic argument is sketched. See Figure A.8, where the exponential cusp
is generated by rotating the following exponential curve about the z-axis,
 − z −1z0
for z > z0
 e

x=

 0 for z = z0
for some z0 > 0.
Rotation about the z-axis
to generate a sphere in R3
z0
f=0
on the boundary
around the cusp
f is continuous
on the boundary
∆φ = 0
z
Exponential
conical cusp
x
Ω
f assumes some large
constant away from
the boundary of the cone
Figure A.8
Cross-section of a sphere with an exponential cusp.
K15149_Book.indb 360
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361
Appendix A
Suppose that f = 0 on ∂Ω ∩ B( z0 , ε) , for some ε > 0 small, and f = ϕ 0 >> 0
away from the cusp, for some large constant ϕ 0 . To see why a solution cannot
exist, suppose for concreteness that φ represents steady-state temperature.
Then, about ∂Ω ∩ B( z0 , ε) , for a sufficiently small ε > 0, the cusp at z0 cannot
absorb heat sufficiently fast to keep the temperature close to 0 as required by
the boundary condition at the cusp due to the lack of surface area about the
cusp to conduct heat away. Hence, if a solution exists, it cannot be continuous
at z0, yielding a contradiction. In short, no solution exists, as required.
Thus, the above two examples amply demonstrate that the requirement
ψ ∈ C2 ( M) ∩ C( M) for the existence of a solution to Equation (A.21) is too
stringent. The question thus leads to the conditions under which a meaningful solution can exist when the smoothness requirement is relaxed.
First, define L2loc ( M) to be the set of all functions f: M → R such that
∫ K |f|2 dµ < ∞ for all compact K ⊂ M. This is the set of all locally square integrable functions, and let C∞0 ( M) denote the set of all infinitely differentiable funcδ (t)
tions on M with compact support. For M = R, define the weak derivative ddt
by
∫
∞
−∞
dδ ( t )
dt
v(t)dµ(t) ≡ −
∫
∞
−∞
δ(t) dvd(tt ) dµ(t)
∀v ∈ C∞0 (R). The space C∞0 (R) is called the test function space.
Motivated by this endeavor, applying Green’s theorem to − ∫ M ∆ϕvdµ,
assuming the integral exists, leads to:
−
∫
M
∆ϕv dµ =
∫
M
∇ϕ ⋅ ∇v dµ −
∫
∂M
ϕ∇v ⋅ n dλ
(A.24)
for all v ∈ C∞0 ( M), where λ is the Lebesgue measure on ∂M. Then, applying
this to (A.14) with g = 0, that is, the homogeneous Dirichlet problem, yields:
∫
M
∇ϕ ⋅ ∇v dµ =
∫
M
fv dµ
(A.25)
If the equality (A.25) holds ∀v ∈ C∞0 ( M), then φ defines a solution to Equation
(A.14). In particular, φ,v can be relaxed to ϕ , v ∈ H 10 ( M) and f ∈ H −1 ( M) in
order for (A.25) to hold. Then, ϕ ∈H 10 ( M) is said to be the weak solution of (A.21).
A.4.16 Remark
Suppose that Equation (A.21) is inhomogeneous; that is, g ≠ 0. This can be
converted easily into a homogeneous Dirichlet problem as follows. Construct
any function ψ ∈ C2 ( M) ∩ C( M) such that ψ|∂M = g. Set ϕ = φ − ψ. Then, by
construction,
−Δϕ = −Δφ − Δψ = f − Δψ on M
K15149_Book.indb 361
and
ϕ = φ − ψ = g − g = 0 on ∂M.
10/18/13 11:12 AM
362
Appendix A
That is, the inhomogeneous Dirichlet problem is transformed into the homogeneous Dirichlet problem:
 −∆φ = f − ∆ψ on M



φ = 0 on ∂ M

References
1. Adams, R. 1975. Sobolev Spaces. New York: Academic Press.
2. Berberian, S. 1974. Lectures in Functional Analysis and Operator Theory. New York:
Springer-Verlag.
3. Cheng, D. 1992. Field and Wave Electromagnetics. Reading, MA: Addison-Wesley.
4. Choquet-Bruhat, Y., DeWitt-Morette, C., and Dillard-Bleick, M. 1982. Analysis,
Manifolds and Physics, Part I: Basics. Amsterdam: North-Holland.
5. Churchill, R. and Brown, J. 1990. Complex Variables and Applications. New York:
McGraw-Hill.
6. Cohn, D. 1980. Measure Theory. Boston: Birkhäuser.
7. Engelking, R. 1989. General Topology. Berlin: Heldermann Verlag (Sigma Series in
Pure Mathematics, Vol. 6).
8. Evans, L. 1998. Partial Differential Equations. Providence, RI: American
Mathematical Society (GSM Vol. 19).
9. Farlow, S. 1993. Partial Differential Equations for Scientists and Engineers. New York:
Dover.
10. Grisvard, P. 1985. Elliptic Problems in Nonsmooth Domains. Boston: Pitman.
11. Hocking, J. and Young, G. 1961. Topology. New York: Dover.
12. Hsu, H. 1984. Applied Vector Analysis. San Diego: Harcourt Brace Jovanovich.
13. Jackson, J. 1962. Classical Electrodynamics. New York: John Wiley & Sons.
14. Johnson, C. 2009. Numerical Solution of Partial Differential Equations by the Finite
Element Method. New York: Dover.
15. Kellogg, O. 1953. Foundations of Potential Theory. New York: Dover.
16. Koshlyakov, N., Smirnov, M., and Gliner, E. 1964. Differential Equations of
Mathematical Physics. Amsterdam: North-Holland, UK.
17. Nakahara, M. 2003. Geometry, Topology and Physics. Bristol: IOP.
18. Neff, H. Jr. 1981. Basic Electromagnetic Fields. New York: Harper & Row.
19. Plonsey, R. and Collin, R. 1961. Principles and Applications of Electromagnetic
Fields. New York: McGraw-Hill.
20. Rao, M. 1987. Measure Theory and Integration. New York: John Wiley & Sons.
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Appendix A
363
21. Rothwell, E. and Cloud, M. 2001. Electromagnetics. Boca Raton, FL: CRC Press.
22. Sadiku, M. 2001. Numerical Techniques in Electromagnetics. Boca Raton, FL:
CRC Press.
23. Sauvigny, F. 2006. Partial Differential Equations 1: Foundations and Integral
Representations. Heidelberg: Springer-Verlag.
24. Smythe, W. 1950. Static and Dynamic Electricity. New York: McGraw-Hill.
25. Solin, P. 2006. Partial Differential Equations and the Finite Element Method. Mineola,
NY: John Wiley & Sons.
26. Zachmanoglou, E. and Thoe, D. 1986. Introduction to Partial Differential Equations
with Applications. New York: Dover.
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10/18/13 11:12 AM
Index
A
Antenna factor, 280
Antennas, aperture, 280
Antennas, array, 280–281
Antennas, Hertzian. See Hertzian
antennas
Antennas, magnetic dipole. See
Magnetic dipole antennas
Aperiodic functions, 36
Aperture antennas, 280
Array antennas, 280–281
Associated Legendre polynomials, 83
B
Band gaps, 318
Biot-Savart’s law, 261
Bose particles, 317
Bosons, 317
Bound charge density, 345
Bound surface charge density, 345
Bounded variation, 35
C
Cardinality, 35
Cauchy-Goursat theorem, 308
Cauer networks, 59, 60
Cavity resonance, 243–244, 245
Characteristic impedance, 108
Characteristic impedance matrix, 205,
206, 207
Charge density, 7, 81
Charge relaxation time, 69
Charged particles
accelerating, 262
electromagnetic radiation from,
262–263
electrostatic dipoles, 263
Coefficients of inductance, 189
Coefficient matrices, 203
Cold emissions, 323
Common mode impedance, 154, 169
Common mode noise, 159
Conduction current density, 12
Conductivity, electric, 20
Conductivity, medium, 105
Conductor, axis of, 13
Conductors, 316
shield, use as, 30–31
skin depth of, 29
surface of, 30
Continuity equation, 20
Convection current density, 12
Coulomb gauge, 45
Coulomb’s law, 1, 14
Coupling constants, 156
Cross-talk, 192, 199, 208, 209
D
D’Alembert wave solution, 28
Decay rate, 20
Dielectric constant, 346
Dielectric media, 86, 105
Dielectric waveguides, 250
Dielectrics
behavior, under high voltage, 303
breakdown of, 303
electromagnetic properties, 305
Differential coupling coefficient, 146
Differential impedance, 153, 154
Differential transmission lines
asymmetric, 144
defining, 143, 144
even modes, 143, 157, 158, 160, 169
field propagation, 160–167
matching impedances, 157–158
odd modes, 143, 146, 157, 158, 159
parallel, 144
symmetric, 144, 158, 169, 170
terminations, 159–160, 181–182
Diffraction fringes, 291
Diffraction theory, 261
Diffusion equation, 23
365
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366
Dirac-delta distribution, 264, 265
Direct product, 203
Direct sum, 203
Dirichlet boundary condition, 74, 351,
359, 362
Discrete spectrums, 38
Displacement current, 106
Distributed parameter model, 107, 156
Divergence theorem, 20, 90, 105, 187,
343–345
Domain of dependence inequality, 357
Drude–Sommerfeld model, 316
E
Electric dipole moment, 263
Electric dipoles, electric fields
generated by, 3–4
Electric fields
electric dipoles, generated by, 3–4
longitudinal component, 45
time-varying, 17
units of measure, 2
Electric permittivity, 84
dielectric, of, 12
free-space, of, 1
Electromagnetic boundary conditions
non-time-varying current density, 68
overview, 67, 68
Electromagnetic compatibility (EMC),
1, 50
Electromagnetic fields, boundary
conditions, 342–343
Electromagnetic fields, irradiance of, 284
Electromagnetic waves
lossless medium, 23
lossy medium, 23
rectangular coordinates, 22
study of, using full Maxwell’s
equations, 22
transverse electric and magnetic
(TEM) waves; see Transverse
electric and magnetic (TEM)
waves
Electrostatic dipoles, 263
Electrostatic fields, 21
Electrostatic shielding, 304
Electrostatics, 1–2
Equipotential condition, 79
K15149_Book.indb 366
Index
Euler’s formula, 27, 37
Euler-Poisson-Darboux lemma, 357–358
F
Far field, 268
Far zone, 268
Faraday’s law, 17, 21
Fermi energy level, 317
Fermi particles, 317
Fermi-Dirac distribution, 317
Fermions, 317
Ferranti effect, 119
First-order ordinary differential
equations, 108
Forbidden zones, 318
Forcing function, 55
Foster networks, 58, 59, 60
Fourier integral, 40
Fourier series. See also
Fourier transform
Fourier coefficients, 36, 37
harmonics, 40
overview, 35
square wave, 38
symmetry of a function, 35
triangular wave, 39, 40
Fourier transform, 307, 325. See also
Fourier series
“proper” transforms, 42
decomposition, 52
defining, 51
inverse, 41
Maxwell’s equations, analysis of,
42–43, 43–44
overview, 40, 41
pair, 41
roll-off frequency, 46–47, 48–50
Fowler-Nordheim tunneling, 323
Frequency domain, 53
Frequency response, 50
Fresnel-Kirchhoff diffraction formula,
286, 287
G
Gain function, 276
Gauge transformation, 18–19
Gauss’ law, 19, 76, 105, 187, 225, 328
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367
Index
Gibb’s phenomenon, 40
Green’s function, 7, 74, 75, 76, 80–81
Green’s identity, 184
Green’s reciprocity, 183
Ground bounce, 197
H
Hamiltonian operators, 312, 326
Harmonic function, 351
Heat equation, 23
Helmholtz equations, 24
Hermitian operators, 325
Hertzian antennas, 276
Hertzian dipoles, 263, 264, 268, 269,
270, 275
Hilbert space, 312, 325
Huygens-Fresnel formula, 289
Hyperplanes, 13
L
Laplace transform, 43, 70
Laplace’s equations, 7, 28, 86–87
LC Cauer I network, 59
LC Cauer II network, 59
LC Cauer network, 59
LC Foster I network, 58
LC Foster II network, 59
LC-networks, 56–58
Lebesgue integrable, 358, 360
Legendre polynomials, 81
Linear-array antennas, 280–281
Linearity, 41, 51
Lorentz force law, 13–14
Lorentz gauge, 45
Lorentz reciprocity, 89, 185, 186
Lossless linear network, 198
Lossless transmission lines, 110
I
M
Image charge, 8, 71
sphere, within, 72
Image theory, 71
Images, method of, 7–8
Impedance matching, 122–123
Input function, 55
Insulators, 316
Inverse Laplace transform, 56
Inversion in a sphere, 72
Inversion mapping, 72
Inversion transformation, 72
Irrational criterion, 18
Isomorphism, 325
Magnetic current density, 88
Magnetic density, 195
Magnetic dipole, 16
Magnetic dipole antennas, 270–271,
274, 275
directive gain, 276
gain function, 276
Magnetic field
density; see Magnetic field density
electric field, defining in terms of, 30
intensity; see Magnetic field
intensity
Magnetic field density, 13, 14, 200
Magnetic field intensity, 17
Magnetic flux, 147
Magnetic flux density, 13, 14, 156, 196
Magnetic flux, defining, 104
Magnetic monopoles, 19
Magnetic potential, 83
Magnetostatics, 12
Maxwell’s electromagnetic theory
differential pairs and, 146
equations, 17, 18, 21, 22, 42, 44, 88, 89
field unification, role in, 16
first equation, 224
first pair, 19
integral equation, 201
J
Jacobians, 334
Joule heating, 119
K
k-port, 212
Kinetic energy, 313
Kirchhoff’s current law, 104, 148, 151,
192, 199
Kramers-Kronig relations, 17, 305,
307–308, 311
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368
invariance, under gauge
transformation, 18–19
overview, 1
second equation, 149, 150
second pair, 19
Microstrips, 167, 276, 277, 278
Momentum operator, 326
Moore’s law, 316
Multiconductor transmission lines
characteristic impedance matrix,
205, 206, 207
far-end cross-talk, 208
near-end cross-talk, 208
overview, 199–200
proof, 200–201
Multipole expansion, 82
Mutatis mutandis, 23, 59, 83, 89, 113, 121,
173, 191, 200, 320
Mutual admittance matrix, 197
Mutual capacitance, 189, 196
Mutual inductance, 147, 192–193, 196
N
n-port network, 211, 212
Network, definition of, 54
Neumann boundary condition, 351
Nonzero electric charge, 2–3
Nth harmonic, 38
O
Odd mode impedance, 146
Ohm’s law, 12, 54, 104
Orthogonality relation, 83, 84
P
Parallel plate guides, 223–224
Parallel transmission lines. See
Transmission lines
Partial fraction expansion, 58
Partial standing wave, 134
Paschen breakdown, 319
Perfect electric conductors (PECs), 89,
90, 347
Perfect magnetic conductors (PMCs),
89, 90
Periodic functions, 36
K15149_Book.indb 368
Index
Phasor transform, 51, 52
Poisson’s equations, 7, 45, 75, 80, 81, 351
Potential difference between two
points, 5
Potential field
case examples, 6–7, 8–12
defining, 4, 5
Poynting vector, 31–32, 125, 247, 269
Pre-Hilbert space, 359
Probability amplitude, 313
Propagating waves, backward, 158,
172, 203
Propagating waves, forward, 158
Q
Quadrupole, 82
Quantum mechanics, axioms of, 311,
312, 325
R
Radiation resistance, 269
Radio emissions, 50
Rational functions, 54
Rectangular waveguides, 242
Reflected waves, 125
Reflection coefficients, 313
Reflection suppression, 129
Reparametization, 36
Riemann’s theorem, 360
Rodrigues’ formula, 81
S
S-matrix, 215
S-parameters
defining, 217
overview, 210
proofs, 211, 212, 213
Scalar field, 342
Scalar potential, 4, 80, 263
Scaling, 41, 42
Scattering parameters. See S-parameters
Schrodinger equation, 318, 325, 326
Self-inductance, 147
Self-referencing ground, 148
Shielding, electrostatic, 304
Skin depth of conductors, 29
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Index
Smith chart, 108, 120, 121
Sommerfield radiation condition, 287
Spherical harmonic expansion, 81, 84
Square integrable functions, 358
Standing waves, 134
Steady-state currents
charge density, 84, 85
continuity of potential, 85
current density, 87
Steady-state sinusoidal function, 51, 52
Stokes’ theorem, 76, 103, 146, 149, 343
Striplines, 167, 276
Superposition principle, 72, 195
Surface charge density, 69
Surface impedance of a conductor, 246
T
Taylor expansion, 285
Taylor’s expansion, 179
TE to z-mode wave propagation, 244
Telephone equations. See Transmission
line equations
TEM waves. See Transverse electric and
magnetic (TEM) waves
Time domain, 50
Time harmonics, 51, 55, 89, 90, 156, 195,
203, 209, 239, 269
Time-average power density, 125
Time-invariant networks, 212
Time-varying fields, 21, 22
TM to z-mode impedance, 231, 241
Transfer functions
input function, 55
natural response, 55
poles of, 55
Transfer impedance, 197
Transition minimized differential
signaling (TMDS), 180, 181
Transmission coefficients, 313
Transmission line equations, 102–106
Transmission line theory
defining, 101
resistors, closeness to source, 130
Transmission lines
differential; see Differential
transmission lines
K15149_Book.indb 369
distortionless, 119
finite, 117–118, 120, 123, 126
inductive impedance, 130
infinitely long, 116–117
load impedance, 126
lossless, 119, 120, 131, 132
multi-; see Multiconductor
transmission lines
parallel, 114–115
time-average power along, 134
Transmitted wave amplitude, 117
Transverse conduction contribution, 105
Transverse electric and magnetic
(TEM) waves
attenuation, 29
conductors, 101, 102
dielectric, 29, 30, 31
existence, proving, 24–25
medium conductivity, 26, 28, 31
propagation, 101, 108
quasi waves, 26, 150
transmission line equations, use for
solving, 103
wave impedance, 31
wave propagation, 25, 26–27, 28,
31, 246
V
Vector identity, 90
Vector potential, 13, 45, 263, 265
Vector space, 331, 332, 333
Voltage profiles, 156
W
Wave amplitude, 117
Wave equations, 22–23
one-dimensional, 27
Wave propagation, 224, 246
Wave propagation constant, 108, 117,
118, 154, 209
Wave propagations, 109, 111–112
Z
Zero-phase terminal, 211
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