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"£21, because the parameter A may, for example, appear in an exponent or under a squareroot sign. If the assigned value of the parameter Pn does not coincide with any of the eigenvalues Ai of the homogeneous problem, the solution of the deteministic problem, if it exists, is unique. On the other hand, if the parameter value coincides with one of the eigenvalues, then we may add any multiple of the corresponding eigensolution to a solution of the deteministic problem and the result is again a solution, which thus is nonunique. For example, for a problem of a source inside a metallic enclosure, the field is unique if the operating frequency does not equal one of the resonance frequencies of the enclosure; otherwise we may add to the solution the field of the resonance mode with any amplitude. The idea of uniqueness is easily understood but to check any specific problem for uniqueness is a difficult task in general, because it requires solving the homogeneous (sourceless) problem for eigenvalues. For some cases the uniqueness can, however, be simply established through a suitable integral theorem which ascertains that for some range of parameter values the homogeneous problem only has the null solution. Although these theorems do not give all possible values of parameters for uniqueness, they usually cover a range of practical values. A few such theorems are considered in the following. 3.5.1
Electrostatic problem
Let us consider the sourceless electrostatic problem, the scalar potential obeying the Laplace equation \124>0 = O. For a closed problem with a surface S bounding the volume V of interest, the following integral expression can be written:
J
1'\7¢012dV
v
=
J
J
f
v
v
s
'\7 . (¢o'\7¢o)dV 
¢o'\72¢odV =
¢o(n· '\7¢o)dS.
(3.203) If the surface integral vanishes because of suitable boundary conditions, we have TV 4>0 == 0 in V, which makes the original problem unique since the corresponding sourceless electric field Eo =  \1 ¢o vanishes in V. We can list a number of boundary conditions giving unique solutions.
CHAPTER 3. FIELD EQUATIONS
86
• :  4>:n. V4>o)dS.
(3.206)
s we see that the surface integral vanishes for the Dirichlet and Neumann boundary conditions, 1> = F(r) and n . V1> = F(r). The volume integral vanishes for sure only if k 2 = w2 tu. is nonreal, or if the medium is lossy and the frequency is real. Also, for impedance conditions a1> + Bn V1> = F(r) on S the surface integral can be shown to vanish provided a/ {3 is real. k 2 real: In this case we apply the identity
a
f
l4>ol2dS =
f 4>:
(a4>o+(3n,V4>o)dS(3 j(IV4>oI2_ k214>oI2)dV. (3.207)
s s v Assuming ~(a/{3) 1= 0 (and hence a 1= 0), we see that for the lossy impedance boundary condition a1> + (3n . V1> = F( r) on S, the second surface integral vanishes for the homogeneous solution 1>0. Because the quantities on each side of the equation have different phase angles, the integrals must be zero. Thus, 0 active medium (medium gives energy to the field);
~{V'.
5}
~{V·
S} $ 0 passive medium (medium does not give energy to the
field);
< 0 lossy medium (field gives energy to the medium);
•
~{V'.
•
~{V'. S} = 0 lossless medium (energy is not transferred between the medium and the field).
S}
These energy conditions are transferred to conditions for the medium parameters of a linear medium if we write from (3.211) for a point r without sources J, J m :
* H jw ~{V·S } =~ { (E·n 2
JW (E. 4
jw (E H). ( 4
* ·B) } =
n*  E* . D + H· B*  H* . B) =
!:~T r (T ) . ( 7 ~

"""iT ~
=* =:;T Jl  Jl
E ) *, H
(3.213)
and require that this quantity be positive, negative or zero, for all fields E, H, corresponding to the different cases above.
3.6. CONDITIONS FOR MEDIUM PARAMETERS
91
Lossless media For lossless media, ~{\7 . S} = 0 must be valid for all vectors E, H. For example, when H = 0, this should be valid for all vectors E. Invoking the theorem A : aa* = 0 for all a => A = 0 we see that the dyadic j(f* _ 'fT) must vanish. Thus, € = f.T* is hermitian, because its antihermitian part vanishes. In a similar way, other conditions are obtained and the conditions for lossless media can be written as (3.214)
For biisotropic media, the conditions simplify to
e* = (, J.L*
€* = €,
= J.L.
(3.215)
e
Writing = (X + jK,)JP,o€o and ( = (X  j~)JJ.Lof.o, we see that, for a lossless biisotropic medium all four parameters €, u, X and Ii, must be real.
Lossy media For lossy bianisotropic media, the antihermitian Eart of the 6 x 6 matrix Ji, null in the lossless defined by the medium parameter dyadics f, case, must be positive definite. For example, for a conductive anisotropic medium, the conduction dyadic ~ = jw(i  ~T*) must be positive definite for lossy media. For a lossy biisotropic medium we can write an inequality limiting the magnitudes of X and ~ parameters. In fact, because the matrix
e, (,
j(€  €*) ( j«(  e*)
j(e  (*) ) j(J.L  J.L*)
(3.216)
must be negative definite, its eigenvalues must be negative numbers. Writing the parameters in terms of real and imaginary parts, €=€re+j€im,
e=ere+jeim,
(=(re+j(im,
J.L=J.Lre+jJ.Lim, (3.217)
we actually arrive at the three conditions €im
< 0,
J.Lim
< 0,
(eim
+ (im)2 + (ere 
(re)2
< 4€imJLim,
(3.218)
of which the last one can be written 2
€" + Kim 2 < rlm rm . 11.'
(3.219) J.Lo€o Thus there is a physical limit for the magnitudes of the imaginary parts of the parameters X and ~ due to the passivity of the medium. The medium cannot be lossy through the X and ~ parameters alone, because JLim = 0 and €im = 0 are seen to imply Xim = 0 and ~im = O. Xim
CHAPTER 3. FIELD EQUATIONS
92
Active media For active media, all the conditions given for lossy media are valid with inequality signs turned around. In particular, the conduction dyadic (j = jw(f.  f.T*) must be negative definite for active media, which corresponds to negative resistance in circuit theory.
Positive energy density function Yet another condition for the medium parameters is obtained by requiring that the energy function be positive. For a dispersive and lossless biisotropic medium, the following is an extension of the energy density expression in a dispersive anisotropic medium given by YEH and LIU (1972), and CHEN (1983):
W=!(EH).~( 8
dw
w(€+€*) w«(+e*)) (E)* = w(€+(*) w(JL+JL*) H
!(E H) . ( . E (X + j2K)VJ.Lo€o ) ( E ) * . 4 (XJ2K,)y'JL of o JL H
(3.220)
In the last expression we have assumed that the real parameters f, J.l and X are independent and the imaginary parameter j K, is linearly dependent on w, which appears a fair assumption for low enough frequencies. To have a positive energy function, the matrix should be positive definite, which leads to the following three conditions for the real parameters, (3.221) which in particular gives a limiting condition for the magnitudes of the parameters X and 1'\,. It must be emphasized, however, that these are valid for low frequencies only and the w dependence of the parameters according to any particular model of the medium affects the inequalities at higher frequencies.
Boundary conditions The power absorbed by a boundary surface with the impedance dyadic can be obtained from
~{n· S} = ~n. (E x H* + E* x H) = ~H. (Zs * + "is T) . H*· 4
4
Zs
(3.222)
Here the unit vector n points towards the impedance surface. If (3.222) is positive for all fields H, the power is flowing from the fields into the
3.6. CONDITIONS FOR MEDIUM PARAMETERS
93
absorbing boundary and not the other way_ Thus, the condition for a lossy boundary is positive definiteness of the hermitian part of the surface impedance dyadic. This means a positive real symmetric part and a negative imaginary antisymmetric part. For a lossless boundary the hermitian part of the surface impedance dyadic is zero. In other words, the impedance satisfies
(3.223) or Zs must be an antihermitian dyadic. For a lossless biisotropic = + Zbn x we thus see that Za must be imagiimpedance, nary and Zb real. In the isotropic case, the boundary impedance must be imaginary to be lossless, as is well understood from circuit theory.
z.t,
Zs
3.6.2
I,
Reciprocity conditions
Besides energy quantites, reaction is another quadratic function of electromagnetic fields and sources. It was introduced by RUMSEY in 1954 to account for the interaction of fields and antennas. Let us consider two sources ga and gb and their fields fa, fb. The reaction of the field a on the source b is a complex number defined by
< fa, gb >=
J
(E a · .r,
 n, . Jmb)dV,
(3.224)
Vi,
where the volume Vb contains all of the sources b. (The original notation < a, b > does not distinguish between the source and the field.) The reaction is a measurable quantity. For example, if the source b is a current element J = u1bL6(r  rs ), the reaction of the unknown field Ea(r) is (3.225) which is Ib times the voltage induced by the field E a in the dipole b. In general, setting a test source b in the field due to a source a and measuring the reaction gives a method to determine the field a. Reaction is symmetric under conditions called reciprocal. The difference of two reactions can be expanded from the Maxwell equations
J
(E, · J b
V

H, . J mb  E, . J a + H, . J ma)dV
=
94
CHAPTER 3. FIELD EQUATIONS
J J(s, ·
n· (E a X n,  Eb X Ha)dS+
s
jw
(IT ~) · s,
 s, · ((7' + e) ·n, +
V
n, · (P' + () · s,  n, · (IV  If) · n,) dV,
(3.226)
where S is the surface of the volume V containing all sources and n is the outward normal unit vector of S. Reciprocity requires (3.226) to vanish for all fields. Being a property of the medium, we can see the conditions for the medium to be reciprocal: =
=!T €=€,

(=~,
:;;T
]i='ji,
(3.227)
because the volume integral in (3.226) vanishes. For the biisotropic medium, this reduces to ~ = (, or X = o. Thus, the parameter X can be called the nonreciprocity parameter. Because a medium with X # 0 was introduced by TELLEGEN in 1948, it is also called the Tellegen parameter. Requiring the surface integral of (3.226) to vanish, we can obtain conditions for a reciprocal boundary. Inserting the impedance boundary condition (3.228) E t = Zs · (n x H), the boundary term in (3.226) can be written as
J
n (E a X
s
J
n,  s, X Ha)dS = n. · iz, 
ZS T) · HadS,
(3.229)
s
which is seen to vanish for any fields H a , H, with the condition
= =r Zs = Zs ·
(3.230)
Thus, a symmetric impedance dladic is reciprocal. The biisotropic impedance dyadic Zs = z.t, + Zbn x I is reciprocal only for Zb = 0, or when the impedance is actually isotropic. References BOOKER, H.G. (1982). Energy in electromagnetism. Peter Peregrinus, London. CHEN, H.C. (1983). Theory of electromagnetic waves, pp. 6574. McGrawHill, New York.
3.6. CONDITIONS FOR MEDIUM PARAMETERS
95
HAMMOND, P. (1981). Energy methods in electromagnetism. Clarendon Press, Oxford. HARRINGTON, R.F. (1961). Timeharmonic electromagnetic fields, pp. 11620. McGrawHill, New York. JACKSON, J.D. (1975). Classical electrodynamics, (2nd edn), pp. 23644. Wiley, New York. KONG, J .A. (1986). Electromagnetic wave theory, pp. 396415. Wiley, New York. LINDELL, I.V. (1972). Some properties of lossless bianisotropic media. Proceedings of the IEEE, 60, (4), 4634. MONTEATH, G.D. (1973). Applications of the electromagnetic reciprocity principle. Pergamon, Oxford. RUMSEY, V.H. (1954). Reaction concept in electromagnetic theory. Physical Review, 94, (6), 148391. TELLEGEN, B.D.H. (1948). The Gyrator, a new electric network element. Philips Research Reports, 3, 81101. YEH, K.C. and LID, C.H. (1972). Theory of ionospheric waves, pp.4956. Academic Press, New York.
Chapter 4
Field transformations There exist various transformations which can be applied to electromagnetic field problems to transform them into other electromagnetic problems. Thus, from known solutions of certain problems solutions of certain other problems can be obtained without going through the solving procedure. For example, the duality transformation changes an electromagnetic problem to a magnetoelectric problem and the affine transformation changes the metric of the space so that an isotropic space becomes anisotropic. Of course, only transformations which do not change the Maxwell equations are of interest, because otherwise the transformed fields would not be electromagnetic fields. A few such transformations are considered in this chapter.
4.1
Reversal transformations
The simplest examples of field transformations are changes of sign in polarity, time, space and frequency, which transform sources, fields and medium parameters so that the Maxwell equations remain invariant. Each of them will be discussed briefly below. 4.1.1
Polarity reversal
In this transformation all electromagnetic source and field quantities q are changed as q ~ qc, starting from the electric charge, which reverses its sign: flo
= e·
qc
= q.
(4.1) Assuming that the medium and boundary parameters do not change, from the linearity of the Maxwell equations it is easily seen that they do not change if all source and field quantities q change sign in this transformation:
(4.2)
The sign of the electric charge is actually a convention which could be changed, the present choice being due to Benjamin Franklin from the eighteenth century. If the medium and boundary quantities are the same,
98
CHAPTER 4. FIELD TRANSFORMATIONS
the fields for sources with change in sign are also changed in sign. Products of these quantities do not change because the two minus signs cancel. Thus power, energy and impedance remain unchanged in this transformation. For example, the force between two sources equals that of two similar sources of opposite sign.
4.1.2
Time reversal
In this transformation, the sign of time is reversed: tT = to Let us assume that the electric charge does not change sign in this transformation: eT(t) = e(t). It turns out that all other transformed quantities can be determined from the Maxwell equations. For example, we have from \l . JT = 8eT/8tT the rule JT(t) = J( t) and from V . DT = or the rule DT(t) = D( t). Assuming €T = € and "=fiT = we s~ from ~ediu~ equations that the other parameters must obey eT = and (T = (. However, for dispersive media, the relation is not so simple. Also, if assuming magnetic charges, they must obey the law gmT = Um, JmT = J m . Thus, all quantites fall into two groups: those which are invariant in the T transformat!on, e,}m, E, D, € and~, and those which are antiinvariant, em, J, H, B, ~ and (. Combining time reversal with charge reversal gives us a similar transformation in which the magnetic charge is invariant, (}mCT = Um, and the electric charge antiinvariant, ocr = g. Because reversal of time corresponds to changing the direction of motion of electric charges, it reverses the currents and magnetic fields but does not change the electric fields. The change of sign in magnetic sources can be understood if magnetic monopoles are omitted as nonphysical and magnetic dipoles are replaced by electric current loops, which change sign in time reversal. In the frequency domain, the T transformation corresponds to change of sign of the frequency as is obvious from the Fourier integral representation
F,
J
e
00
qT(W) =
qT(t)ejwtdt = q(w) = q*(w).
(4.3)
00
Because the T transformation also implies conjugation, it is seen that the real parts of complex functions are even and the imaginary parts are odd functions of frequency.
4.1.
REVERSAL TRANSFORMATIONS
4.1.3
99
Space inversion
In this transformation, space points r are reversed in the origin to points r p = r. Assuming again that the sign of the electric charge is not changed in the reversal, (}p(r) = {}(r), we can find from the Maxwell equations, again, that the quantities H2 ~, J m , €, Ii do not change sign, wheras the quantities J, E, D, em, ~, "( transform like Jp(r) = J(r). Space inversion is a special case of the affine transformation discussed in more detail in a subsequent section. 4.1.4
Transformations of power and impedance
To see how power flow changes as a result of these transformations, let us study the Poynting vector S(t) = E x H. It is easy to see that we have
Sc
= S,
(4.4)
or the power flow is hot dependent on the sign of the charges nor the direction of the current. On the other hand we have
ST(t) = S( t),
(4.5)
which means that the sign of the energy flow is reversed in the time reversal, as is most obvious. Finally, we can write
Sp(r)
= S( r).
(4.6)
The frequency sign change has an effect on the complex Poynting vector:
ST(W)
= S?fc(w).
(4.7)
Thus, the real power flow changes sign in this transformation, whereas the reactive imaginary part does not. Similarly, all impedance quantities change like ZT(W) = Z*(w). (4.8) Again, the real part changes sign in time reversal. This can be understood so that power absorbing passive medium becomes active in time reversal, feeding power to the outside world, which is just like watching a movie backwards. References JACKSON, J.D. (1975). Classical electrodynamics, (2nd edn), pp. 24551. Wiley, New York. MITTER, H. (1990). Elektrodynamik, (2nd edn), pp. 20915. BI Wissenschaftsverlag, Mannheim.
CHAPTER 4. FIELD TRANSFORMATIONS
100
4.2
Duality transformations
The duality transformation makes use of the symmetry between the electric and magnetic quantities in the Maxwell equations. In its classical form, the electric quantities of an electromagnetic problem are changed to magnetic quantities and vice versa. In changing the fields and sources of the problem the parameters of media and boundaries are also changed. In many cases it is necessary that the medium remains unchanged, which makes the transformation dependent on the medium in question. However, in the form considered here, duality transformations can only be found for biisotropic media. 4.2.1
Simple duality
In the literature, a simple form of duality is often applied to transform an equation corresponding to a certain electromagnetic problem. The outcome is another equation corresponding to a dual problem. If the solution of the former is known, there is no need to solve the latter. Instead, by performing the duality transformation to the known solution, the unknown dual solution can be readily obtained. In the simple version, one just makes the substitution E + H and conversely H + E, comp~eted ~ the additional substitutions B ~ D, J ~ J m and € ~ ii. ~ ~ (. However, in this form the duality transformation changes the nature of the quantities so that their dimensions are changed. For example, the electric field is transformed into a magnetic field. Thus, it is not possible to add the original and transformed field quantities. Let us consider duality transformations which transform quantities so that their nature does not change. Thus, the electric field is transformed into another electric field. Let us also limit the theory to the frequency domain.
4.2.2
Duality transformations for isotropic media
The original idea behind duality is to interchange electric and magnetic quantities so that one of the Maxwell equations is transformed to the other one and conversely. Because the equations can be multiplied by nonnull constants without changing their validity, let us require that the corresponding quantities in the two Maxwell equations are transformed into each other through multiplying by constant factors a and {3:
\7
x H == jwD
+J
+
\7 X Ed == jwBd  J md.
(4.10)
4.2. DUALITY TRANSFORMATIONS
101
The equation pair is invariant if we define Ed
= oH, B d = aD,
Jmd
= aJ,
(4.11 ) (4.12)
The divergence equations transform to (4.13) (4.14) if we define (4.15) Further, from the condition that the medium equations transform to one another the parameters of the dual problem can be identified:
(~ ~) (~ d
=
;E)'
(4.16)
Finally, from the impedance boundary conditions Et =
Zs . (n
x H)
(4.17)
with the unit normal vector n pointing out from the boundary, we have for the dual dyadics
(4.18) 4.2.3
Lefthand and righthand transformations
The definitions of the dual quantities above are valid for any values of the constants Q, (3. The simple duality transformation, applicable for the derivation of formulas, is obtained by taking a {3 = 1. The coefficients can also be determined by imposing certain conditions for the transforrnation. It appears natural to require the following properties.
=
1. The duality transformation is an involution, i.e., the dual of a dual quantity equals the original quantity itself. This implies a{3 = 1, as is seen by checking through all the equations.
2. A chosen isotropic medium is defined to be selfdual: its parameters €s, ue are not changed in the duality transfo~ation. T~e ~edium chosen is usually the free space with fS = €a!, Ps = /lol, = 0,
es
CHAPTER 4. FIELD TRANSFORMATIONS
102
O. The conditions fSd = Eol, JlSd = JLol, ~Sd = 0, (Sd = 0 imply the condition a//3 = JLo/E o = TJ~. More generally, we could actually take any isotropic chiral medium with the parameters fS, JLs, ~s = (s = j"'SJJLofo as the selfdual medium, because, as can be seen from (4.16), the dual of the chirality parameter coincides with that of the original medium: Kd = KS.
(S =
From the two conditions a(3 = 1 and a/ (3 for the parameters: I Q
.
= 11]s,
at = j"7S,
= 1]~ we have
two solutions
f3' = _1_ . ,
(4.19)
f3T = ~.
(4.20)
J"7s
J1]s
Two possible solutions means that, corresponding to a chosen selfdual medium, there in fact exist two duality transformations, neither of which is a more 'natural' one than the other. These two transformations will be denoted as follows: let the duality tranformation corresponding to the superscript l be called the lefthand duality transformation, and that corresponding to r the righthand duality transformation. These labels refer to polarizations of two selfdual propagating fields, as will be explained su bsequently. From the previous knowledge we can write the following double duality transformation table for the fields and sources f t t' and f t IT:
It j"7sH jTJsD jTJsJ j'TJse
I
IT
E
j1]sH jTJsD jTJsJ j'TJse
B
Jm em
It E/ins B/jTJs J m / j "7s em/j'TJS
I
IT
H D
E/j1]S B/j"7s Jm/jflS em/j'TJS
J
e
and for the medium and impedance parameters, which transform in the same way in both transformations
I
It = t: Jl/TJ.~ (
f
~ Ys
I Jl (
2Zsxx n n Zs
fls
t' = IT fI~! ~
2= x
fls Y s x nn
It is seen clearly from the above table that the transformation equals its inverse whence only one half of the table is actually needed. The two duality transformations possess the following properties.
4.2. DUALITY TRANSFORMATIONS
103
• For a lossless medium (real1Js) the dual of Ex H* is E* x H, whence the real part of the Poynting vector is self dual. Thus, the field pattern representing the energy flow in a lossless medium is unchanged in both duality transformations. • The PEC and PMC boundaries are dual to each other. • The dual of a dielectric object in air with f.s = f.o, tts = tto, with the relative dielectric factor f. r = A, is a magnetic object with the relative permeability ttr = A, since the dual of €/ f. o in this case equals =PI tto and conversely. • The wave number of a plane wave, k = w#' in the dual of an isotropic medium equals the original propagation factor kd = k. • The wave impedance "l of a propagating plane wave in the dual of an isotropic medium equals "ld = "l~/"l. • The dual of an electric dipole J = uIL6(r) is the magnetic dipole J m = juT}sIL6(r). Application of the duality transformations
4.2.4
Either of the two duality transformations can be applied for an electromagnetic problem in two ways because of the involutory property of the transformation: 1. attach the label d (representing either l or r superscript) to each quantity of the problem and then replace these quantities from the above definitions,
2. replace the original quantities in terms of transformed quantities from the definitions, substitute these in the problem and discard the labels
d. Both ways lead to the same result. It is, however, quite easy to get confused with the two possibilities and end up with a wrong result. This means that one has to be careful and pursue only one path. Of course, only one of the two possible duality transformations, either the lefthand or the righthand transformation, must be systematically applied. Dipole fields
Let us consider as an example of the method 1 above the problem of finding the far field expression of a magnetic dipole from that of an electric dipole.
CHAPTER 4. FIELD TRANSFORMATIONS
104
The electric field of an electric dipole J (r) == uIL8( r) in the far field region in air is e jkr E == uojwJtoILsinO. (4.21) 41fT
Since the selfdual medium is air, we have tts = 'fJo. Making the lefthand duality transformation (i.e. adding the superscript I and replacing transformed quantities from their definitions), for every quantity of this equation we obtain
(4.22) or
jkr
H = uojW€o_eImLsinO,
(4.23)
41fT
which is the expression for the magnetic field arising from a magnetic dipole in the far field region. The same result is obtained for the righthand transformation and also for the simple duality transformation.
Scattering problem As a second example, let us consider the duality transformation of a problem of electromagnetic plane wave scattering from a dielectric obstacle in air with e; = A, Jtr = 1. Let the selfdual medium be free space and the incident wave
E t·( r )  E oejk.r ,
H i () r
= H oejk.r .
(4.24)
Both duality transformations change the obstacle to a magnetic scatterer with J.1r = A and e; = 1. Moreover, they also change the incident plane wave to
Ei(r)
= ±j7/oHoejkr,
Hi(r) = ±
~o e jk r.
(4.25)
)"'0
Also, the scattered field Esc, H sc of the first problem can be transformed to give the scattered field of the second problem. It must be noted, that in this way we obtain a solution to the dual problem involving the dual scatterer and the dual incoming field, which in general is different from the solution for the original incoming field, unless the incoming field is self dual. This can be applied as follows. • If the incoming field is lefthand self dual, i.e. it coincides with its own lefthand transformation, the lefthand duality transformation will not change the incoming field and the dual scattering problem has the original incoming field.
4.2. DUALITY TRANSFORMATIONS
105
• The same is valid if the incoming field is righthand self dual, in which case we can apply the righthand duality transformation. • If the incoming field is none of these, we can write it as a sum of lefthand and righthand selfdual fields, treat these separately as above and add the results. Thus, the scattered field for the dual scatterer and original incoming field can be obtained through two duality transformations, one for each selfdual parts of the problem.
4.2.5
Selfdual problems
Quantities that are invariant in either of the two duality transformations are called selfdual quantities. The two selfdual concepts will be distinguished by the subscript + corresponding to the rand  corresponding to the l transformation. Medium parameters and impedances are not affected by the handedness of the transformation. Obviously the sum of any quantity and its proper dual is a selfdual quantity and their difference is an antiselfdual quantity. It is easy to see that a quantity self dual with respect to the l transformation is antiself dual with respect to the r transformation and conversely. Any quantity can be written as a sum of a selfdual part and an antiselfdual part, or, what is the same thing, as a sum of two selfdual parts (4.26) Selfdual fields
The selfdual electromagnetic fields can be written as
H±
1
1
= 2(H =f .. E),
(4.27)
J'TJs
and they are seen to satisfy (4.28) Further, we can write
B± =
~(B ± j"1sD),
D±
1
1
1
2
J'TJs
J'TJs
= (D ± . B) = ±.B±.
(4.29)
As an example of selfdual fields (also called wave fields) let us consider a selfdual plane wave propagating in an isotropic chiral medium in the direction of the unit vector u. Because the fields of a plane wave satisfy (4.30)
CHAPTER 4. FIELD TRANSFORMATIONS
106
with
1
H, ==  u x Eo, 71s
Eo == f/sU x H o,
(4.31 )
~(E 1= i'7s H ) = ~(E 1= iu x E). 2 2
(4.32)
the selfdual fields are of the form
E± =
The two selfdual fields corresponding to the two duality transforms are obviously circularly polarized, because they satisfy E± . E± == O. We can form the polarization vectors for the two fields and obtain
p(E±) == ±u.
(4.33)
The upper sign corresponds to a righthand polarized and the lower sign a lefthand polarized CP wave when looking in the direction of propagation U, thus justifying the labels 'lefthand duality transformation' and 'righthand duality transformation' with respect to which these two respective polarizations are self dual. Selfdual sources and media Selfdual sources can be written similarly as 111 J m± = 2(J m ± i'7sJ) = ±j'7sJ±,
J± = 2(J ± i'7S J m ) ,
(4.34)
111 (4.35) f2± = 2 (f2 ± .em), f2m± = 2 (em ± jT/se). == ±jT/Sf2±. JT/s A single electric or magnetic current cannot be a selfdual source, instead, there must be a suitable combination of both of them. The selfdual source combination is obtained from any electric current source J, by adding the magnetic current source J m± == ±jf/sJ. For example, a dipole current J == uIL8(r), requires the magnetic dipole current J m == ±uj71sIL6(r) for a selfdual source. The magnetic dipole must be parallel to the electric dipole and have ±90° phase shift and amplitude 71s1. For fields outside the sources, the magnetic dipole can be replaced by an equivalent electric current loop perpendicular to the electric dipole. The isotropic chiral medium assumed self dual with respect to the two duality transformations was denoted by the parameters f.s, Jls, ~s == (s == jKJJlo€o. Since only the impedance ne == VJ.ls/€S is needed in defining the two duality transformations, all isotropic chiral media with the same impedance appear self dual with respect to these transformations.
4.2. DUALITY TRANSFORMATIONS
107
Selfdual boundaries
The selfdual boundary impedance dyadic 
Zsd
2x = 7Js(Ys xnn)
Zs
must satisfy the equation
= Z T= Zs. spm Z,
2 s = 7Js=:
(4.36)
Taking the spm operation of this, we have (spmZ s )2 = 7J~, which has two solutions spmZ s = ±1J~' Substituting in (4.36), we have the condition = whence the impedance dyadic must be either symmetric or antisymmetric to be self dual. Let us study these cases separately. T~ antisyrnmetric selfdual twodimensional dyadic is obviously of the form Zs = Zsll X 7, where n is the unit vector normal to the surface. Equating spmZs with 7J~ gives us finally two antisymmetric selfdual surface impedance solutions
Z;
±Zs,
». = ±j'T/sn x I.
(4.37)
Studying the most general symmetric boundary impedance dyadic (4.38) where Ul and U2 are the two orthonormal eigenvectors of Zs perpendicular to n, from spmZ. 11~ we have Z lZ2 11~ for the sel!:.duality_condition. As one limiting case we have the isotropic impedance Zs = 'T/s7t, like the SilverMuller radiation condition at the spherical boundary in infinity. As another limit we have the anisotropic surface with Zl ~ 0, Z2 ~ 00, which approximates a dense tuned corrugated surface, i.e. a conducting surface with grooves parallel to U1 and a quarter wavelength of depth.
=
4.2.6
=
Selfdual field decomposition
The decomposition of electromagnetic fields in selfdual constituents E+ and E_ in (4.27), also called wave fields, leads to a useful way of treating electromagnetic problems in isotropic chiral media, instead of the conventional analysis based on electric and magnetic fields. In fact, substituting fields and sources in terms of selfdual components, for homogeneous isotropic chiral media, the Maxwell equations are seen to split into two noncoupled sets, each selfdual system of fields and sources acting in an isotropic nonchiral medium of its own. The governing equations can be written together with double subscripts as follows: (4.39)
CHAPTER 4. FIELD TRANSFORMATIONS
108
(4.40)
where the effective nonchiral parameters of the two media are denoted by K.
f± == f(1 ± ),
n
K,
It± == 1t(1 ± ), n
n
==
Jfrltr.
(4.41 )
It must be emphasized, however, that this is valid only for homogeneous media, in inhomogeneous media the two selfdual fields couple to each other. It is also seen that in an isotropic chiral medium the wave impedances of the two selfdual fields are the same and independent of the chirality parameter K "l± ==
.: J¥  == f±
(4.42)
 =="l, e
whereas the wave numbers of plane waves in the two media are different: (4.43)
The total electric and magnetic fields can be obtained any time from the wave fields as follows: (4.44)
Power relations Consider the real part of the Poynting vector in terms of the selfdual wave fields for a homogeneous lossless chiral medium: (4.45)
It is seen that the power is propagated independently in the two selfdual fields because there are no cross terms present. Applying the real polarization vector concept from Chapter 1, of a complex vector 3, p(a) == j(a x a*)/(a· a"), we can write (4.46)
Here, the quantities W±
f±E ±. E*± = 2 It±H±. H*±' =2
v±
1
= 
JJL±€±
(4.47)
109
4.2. DUALITY TRANSFORMATIONS
can be identified as energy densities and phase velocities of the two wave fields in their respective isotropic nonchiral media. The two real polarization vectors p(E±) point to the positive (righthand) directions of the two wavefield vectors and their lengths, which give the ratio of energy velocity to the phase velocity, vary from 0 for LP to 1 for CP fields. Thus, we can state that the power in each of the selfdual fields propagates in a direction which is perpendicular to the plane of polarization and in the righthand direction of the '+' field and the lefthand direction of the '' field. This further justifies the basis for labeling the two duality transformations as 'right hand' and 'left hand'. It is seen that selfdual LP fields are standing waves because the power does not propagate at all. The energy velocities of the two wave fields are different from the phase velocities except for CP waves. Circularly polarized selfdual fields
We can make further conclusions by splitting each of the wave fields into two CP components with labels R, L to be defined shortly. Let us write (4.48)
with two pairs of CP unit vectors u~, u~. The vectors are pairwise coplanar, u~, u~ with E+ and u~, u~ with E_ and they satisfy (4.49) 2
R . u± L _ 1u± R 1 u±
I
2
L1 u±
= 1.
(4.50)
The polarization vectors of the CP vectors are unit vectors and parallel or antiparallel to those of the corresponding E± vectors. They can be seen to obey the relations
L R) = p (u±L) = Ju± . R x u±, P (u±
(4.51 ) (4.52)
Thus, the total power flow expression can be written in terms of CP components of the two selfdual fields as 1
*
2!R{E x H }
IE~12
R
IE~12
L
= w,», IE+1 2 p(u+) + w,», IE+1 2 p(u+)
IE~12 R IE~ 12 L  W_vIE_12P(u)  W_vIE_12P(u).
(4.53)
110
CHAPTER 4. FIELD TRANSFORMATIONS
This shows us that the field can be written in terms of four components, two selfdual fields each with two coplanar CP components, which do not couple energy between one another in a homogeneous medium. From the signs of the different terms it is seen that the + waves are right handed and the  waves left handed with respect to their individual directions of propagation. However, since u~ point oppositely to u~, the polarizations of the total + and  fields depend on the amplitude ratios of the partial waves. Defining the labels so that IE~ I ~ IE~ I, the direction of the energy flow of the + wave points along p(u~) wave whereas for IE~I ~ IE~I the energy of the  wave flows in the direction of p(u~). (This makes sense only if the selfdual fields are not LP fields, which do not propagate at all.) Thus, it is seen that the power in the electromagnetic field in a homogeneous isotropic chiral medium propagates in two selfdual fields each with two uncoupled CP components so that the + wave contains two oppositely propagating righthand waves and the  wave two oppositely propagating lefthand waves. The velocities of the two + waves are different from those of the two  waves if the chirality parameter is nonzero. If both selfdual fields carry energy in the same direction, it is a consequence that the components E~ and E~ are two forward waves and the components E~ and E~ two backward waves with different velocities of propagation.
Fig. 4.1 Electromagnetic field in an isotropic chiral medium can be decomposed into its selfdual parts each with two oppositely polarized CP components.
4.2.7
Duality transformations for biisotropic media
A nonreciprocal biisotropic medium cannot serve as the selfdual medium in the previous duality transformation. However, a more general form for the duality transformation can be obtained by writing (LINDELL and
4.2. DUALITY TRANSFORMATIONS VIITANEN,
111
1992)
(:)d=(~ ~)(:),
(4.54)
(~)d=(~ ~)(~),
(4.55)
(
J~ )
d
= .:
i) ( J~ ) ·
(4.56)
Inserting these in the Maxwell equations for dual quantities
we should obtain the original Maxwell equations, which gives rise to equations for the coefficients A, ...,L:
H=L=A, G=K=B, F=J=C, E=I=D.
(4.58)
For the transformation to be an involution: ()dd = (), we must have
A 2 + BC
= D2 + BC =
1,
C(A + D)
= B(A + D) =
O.
(4.59)
One possible solution is G = B = 0, which is not, however, very interesting because it reproduces the original problem. The second solution satisfies A = D
= ±Vl  BG.
(4.60)
The coefficients Band C can be determined if we choose one particular hiisotropic medium with the parameters fS, J.Ls, = (xs  jItS)JJ.Lof o and (s = (Xs + jlts}JJLo€o as the selfdual medium. Solving Band C we arrive, again, at two duality transformations
es
(:)d
=
c~O (~~/~S ~~o) (:) ·
sIne
JJ.Lof o = Xs=
I
(4.61)
XS .
(4.62) tisee ns These expressions generalize the previous duality transformations in the special case of a reciprocal chiral medium, as is seen if XS = 0, implying e = 0, is substituted in (4.61): (
~ ) d = (±J7JS =Fjt7JS ) ( : ) ·
(4.63)
112
CHAPTER 4. FIELD TRANSFORMATIONS
Again, the upper sign can be called the lefthand transformation and the lower sign the righthand transformation in analogy with the previous convention. The selfdual fields are thus 1
'0
E± == (e=FJ E =f j"lsH), 2 cos ()
(4.64)
which are seen to be generalizations of the selfdual fields for reciprocal biisotropic media. References HARRINGTON, R.F. (1961). TimeHarmonic electromagnetic fields, pp. 98100. McGrawHill, New York. JACKSON, J.D. (1975). Classical Electrodynamics, (2nd edn), p. 252. Wiley, New York. JAGGARD, D.L., SUN, X. and ENGHETA, N. (1989). Canonical sources and duality in chiral media. IEEE Transactions on Antennas and Propagation, 36, (7), 100713. KONG, J.A. (1986). Electromagnetic wave theory, pp. 36776. Wiley, New York. LINDELL, I.V. (1981). Asymptotic highfrequency modes of homogeneous waveguide structures with impedance boundaries. IEEE Transactions on Microwave Theory and Techniques, 29, (10), 108793. LINDELL, I.V. and SIHVOLA, A.H. (1991). Generalized WKB approximation for stratified isotropic. chiral structures. Journal of Electromagnetic Waves and Applications, 5, (8), pp. 85772. LINDELL, I.V. and VIITANEN, A.J. (1992). Duality transformations for general biisotropic (nonreciprocal chiral) media. IEEE Transactions on Antennas and Propagation, 40, (1).
4.3
Affine transformations
An affine transformation changes the metric of the space. An example of such a transformation is one that squeezes the space so that all distances in one direction are changed in the same proportion. A sphere is then transformed to an ellipsoid and a cube to a parallelepiped. Also, making the space the mirror image of itself in a plane is an affine transformation. !.he affine transformations considered here are defined by a constant dyadic A, which moves every space point r to another point according to the law r+ra=A·r.
(4.65)
4.3. AFFINE TRANSFORMATIONS
113
The question remains how to transform sources, fields, media and boundary conditions so that the Maxwell equations remain valid in the transformed space. Knowing this, it is possible to obtain the solution for a transformed problem from that of the original problem by just transforming the solution.
Transformation of fields and sources
4.3.1
Let us assume that in conjuction with the affine transformation of the space, the fields are transformed as
E(r) ~ Ea(r) = B . E(r a) = B . E(A . r),
(4.66)
= C . H(A· r).
(4.67)
H(r) ~ Ha(r) = C· H(r a)
The dyadics Band C should be chosen to depend on A so that the transformed fields satisfy the Maxwell equations. In fact, after the affine transformation the equations in the frequency domain should read as \7
x Ea(r) = jwBa(r)  Jma(r), \7 x Ha(r)
= jwDa(r) + Ja(r).
(4.68) (4.69)
To be sure of this, we apply the knowledge that the original fields satisfy the Maxwell equations in the fa space. The expression of the gradient in the transformed space, \7 a, can be obtained from the definition (4.70)
Replacing
fa
by
A· r = r· AT, we have (4.71)
whence
=IT
\7 a = A
. \7,
or
=T
\7 = A
. \7 a.
(4.72)
The curls of a field vector in the original and transformed spaces are related through the dyadic identity 1(K~K) . (a x b) = K(2) . (a x b) = (K . a) x (K . b), 2
which allows us to write
(4.73)
CHAPTER 4. FIELD TRANSFORMATIONS
114
= (detA)A I . [Va
X
(AIT. B· E(r a ) ) ] .
(4.74)
Thus, the equations (4.68), (4.69) can be written as Va
=  IT X [A
=  IT
X [A
=
] = =. jwA Ba(r) 
. B· E(r a )
A =. Jma(r),
detA
(4.75)
detA
= ] . C . H(r a)
jwA A = = .Da(r) + = .Ja(r).
Va
= jwB(r a) 
(4.76) detA When these equations are compared with the Maxwell equations in the transformed space, \7 a
X
Va
E(r a ) X
detA
H(r a )
Jm(r a ) ,
= jwD(r a ) + J(ra ) ,
(4.77) (4.78)
by identifying the corresponding terms, t~ following relation between the dyadics Band C and the original dyadic A are obtained: _
B
=T
_
= aA,
C
=T
= {3A
,
(4.79)
where a and (3 may be any scalar constants. Thus, we arrive at the following affine transformation for the fields and sources: Ea(r) = aAT . E(A. r), (4.80) =T
=
Ha(r) = (3A . H(A . r),
(4.81)
= a(detA)A 1 . B(A . r),
(4.82)
Da(r) = (3(detA)A 1 • D(A. r),
(4.83)
Jma(r) = 0:'(detA)A1 . Jm(A . r),
(4.84)
Ba(r)
Ja(r)
= (3(detA)A 1 . J(A. r),
(4.85)
which depends on the dyadic A and two scalars 0:', {3. The coefficients 0:', {3 control the amplitudes of the transformed fields. Let us restrict their values by requiring that the affine transformation defined by the inverse dyadic AI giv~the inverse transformation, i.e. return the quantities transformed through A back to the original ones. It is readily seen that this property requires the condition for the parameters (4.86)
4.3. AFFINE TRANSFORMATIONS
115
to be valid. Several important transformations fall under the general case of the affine transformation. For example, rotation around an axis by an angle, space reversion at a point and reflection at a plane are special cases of the affine transformation. 4.3.2
Transformation of media
The affine transformation also changes the medium parameters and their formulas are obtained by comparison of terms. If the constitutive equations for the general bianisotropic medium are written after the affine transformation as (4.87) D, = fa . E a + ~a . H a,
B, = (a . E,
+ Ji a . H,;
(4.88)
and the transformed fields are substituted from the above equations, we are able to identify the transformed medium dyadics in terms of the original medium dyadics in the form (4.89)
ea = (detA)A e· AIT, 1
(4.90)
.
(a = (detA)A 1 . ( . AIT, ~a = ')'(detA)A 1 . J7. AIT, where the parameter "y = af3 has either the value of 1 or 1.
(4.91) (4.92)
The above formulas (4.89)  (4.92) show that an isotropic medium with parameters e, J.L is transformed to an anisotropic medium in the general affine transformation: = r€(detA)(AT . A)I,
(4.93)
~a = T'J.L(detA)(AT . A)I.
(4.94)
fa
From these expressions it is seen that the transformed fa and Ila dyadics are multiples of the same symmetric dyadic and they satisfy the relation 
1
1

€
=
€a·Jia =lla ·€a=I, J.L
or
fa
lla
e
J.L
(4.95)
It is alluring to search for an effine transformation that would transform a given anisotropic medium into an isotropic medium. However, from the previous equations we see that this is only possible for a limited class of
116
CHAPTER 4. FIELD TRANSFORMATIONS
anisotro£ic media. In fact, solving for €, Jl from (4.89), (4.92) with fa = €aI,
Jia = J.Lal shows us that the original medium dyadics f and Ji must satisfy an equation of the form
= =1 E·Jl
€a = == 1, Jla
(4.96) T
or € and Jl must be multiples of the same dyadic. Also, because A .A is a symmetric dyadic.?both € and Ii must be multiples of the same symmetr~ complete dyadic S. Only in this case it is possible to find a dyadic if defining the affine transformation needed..!o make the anisotropic medium isotropic. It is easy to see that the dyadic A can then be taken as symmetric and a multiple of the dyadic (1/2. In fact, we can write (4.97) Thus, only a very special type of anisotropic medium can actually be transformed into an isotropic medium and the corresponding class can be called affinely isotropic. This can be generalized to a relation for bianisotropic media which can be transformed to biisotropic media by adding a similar requirement to the skew parameter dyadics, i.e, that they be multiples of the same symmetric dyadic as € and Ii. Such a medium might be called affinely hiisotropic. Finally, we might be interested to study what kind of anisotropic medium can be transformed to a symmetric uniaxial anisotropi~ medium with medium dyadics of the form fa == €lI + €2 U U and Jia == J.L1I + Jl2 U U. From the above expressions it can be shown, after some algebra, that the original medium dyadics must be of the general symmetric form (4.98) where S is a symmetric complete dyadic and ~ is a certain unit vector. In this case, the required transformation dyadic A can be taken as a multiple =1/2
=
of S and the unit vector u will be a multiple of S1/2 . v. A medium of this kind of can be called affinely uniaxial. 4.3.3
Involutory affine transformations
Finally, let us consider affine transformations that are involutions like the duality transformation, which means that the original electromagnetic field is obtained through two consecutive transformations, or, the equivalent,
4.4.
REFLECTION TRANSFORMATIONS
117
the affine transformation equals its inverse. To find the condition, let us consider the double transformation (4.99) from which we have the condition 2

11 =/.
(4.100)
The same condition is obtained from all transformation formulas. Of course, the condition (X2 = {32 = 1 must also be valid. Thus, for an involutory affine transformati~ the coefficients (X and {3 can only have values +1 or =1 and the dyadic A must be ~ squa~e root of the unit dyadic, whence detA = ±1. The trivial solutions A = ±1 are two special cases of the most general square root of the unit dyadic, which is of the uniaxial form
A = ±(/ 4.4
ab)
with
a ·b
= 2.
(4.101)
Reflection transformations
As an important special case of the involutory affine transformation we consider the most general symmetric square root of the unit dyadic, which can be written in the form
c = 1 2uu,
(4.102)
where u is a real unit vector. This actually defines a reflection transformation, which maps any vector r to the vector r  2u(u . r), i.e. its mirror image with respect to the plane with r . u = O. The uniaxial reflection dyadic C satisfies the basic conditions =2
=
C =1,
4.4.1
CT
= C 1 = C ,
detC
= 1.
(4.103)
Invariance of media
The reflection transformation is often applied in conjunction with an image principle, in which original and transformed fields are combined. This makes sense only if the medium does not change in the reflection transformation, a property which is not valid for arbitrary media. Let us~tudy what kind of media are invariant in reflection. Requiring ~a = ~ for A = C in (4.89), we obtain the condition (4.104)
CHAPTER 4. FIELD TRANSFORMATIONS
118
Taking the determinant function of both sides and assuming that det€ "I 0 (otherwise the medium could not be polarized in a certain direction), we arrive at 'Y == 1. Thus, for an involutory transformation we have two possibilities: either a == 1 and (3 = 1 or a = 1 and (3 = 1. Requiring the invariance of medium parameters in reflection transformation gives us four dyadic equations for the bianisotropic parameters: (4.105) (4.106) can be any dyadics which commute, and ~ and "( any dyadics
Thus, f and Ji that anticommute, with the reflection dyadic C. Writing each dyadic in terms of components parallel and transverse to U we can see that the medium parameter dyadics must be of the form (4.107) "( == UCt
+ d.u,
Ii = JLuu u u + lit,
(4.108)
where the dyadics Et, lit and the vectors at, ... .d, are transverse to u. From the above the following can be seen. • A biisotropic medium is invariant in the reflection transformation if and only if it is isotropic, i.e. if == ( == O. In fact, a chiral medium changes handedness in reflection and thus cannot be invariant.
e
• A bianisotropic medium is invariant only if u is one of the eigenvectors of the medium dyadics €, ii from both left and right and the other eigenvectors are transverse to u. This is the case, for example, for a uniaxial symmetric medium with an_ o~tical axis normal to the reflection plane. Moreover, the dyadics ~, "( must be planar and of a special form to be invariant in reflection. 4.4.2
Electric and magnetic reflections
It was seen that there exist two reflection transformations of electromagnetic fields and sources. Let us call the transformation defined by a == 1 and (3 == 1 the electric reflection, because the electric fields and sources are transformed to their mirror images, whereas the magnetic fields and sources are transformed to their negative mirror images:
Ec(r) == C· E(C· r),
Hc(r) == C· H(C· r),
(4.109)
4.4.
REFLECTION TRANSFORMATIONS
Be(r) == C· B(C· r),
(4.110)
Jme(r) == C· Jm(C . r).
(4.111)
De(r) == C· D(C· r), Je(r) == C· J(C· r),
119
The second possible reflection transformation with a == 1, f3 == 1 is called the magnetic reflection and defined by
Ee(r) == C· E(C· r),
He(r) == C· H(C· r),
(4.112)
De(r) == C . D(C . r), Je(r) == C· J(C· r),
Be(r) == C· B(C· r), Jme(r) == C . Jm(C . r).
(4.113) (4.114)
It is important to note that the naive reflection transformation where all fields and sources are transformed either through C or C alone, does not satisfy the Maxwell equations in a medium which is invariant in the reflection transformation. Without different signs in electric and magnetic field vectors, the Poynting vector S == (1/2)E x H* would not transform to its mirror image but to the negative of its mirror image. 4.4.3
The mirror image principle
Any electromagnetic field and the corresponding source quantity q can be written as a sum of two selfreflecting parts, each invariant in one of the two reflection transformations: (4.115)
Here, the superscript e denotes the quantity invariant in the electric reflection (electrically symmetric) and m, in the magnetic reflection (magnetically symmetric). For example, the electric and magnetic fields can be written as (4.116) with 1

1

He(r) == 2[H(r)C.H(C.r)],
p
1

1

Em(r) == 2[E(r)C.E(C.r)], (4.117)
Ee(r) == "2[E(r)+C.E(C.r)],
Hm(r) == 2[H(r)+C.H(C.r)]. (4.118)
Denoting r == u( u . r) + p, on the plane u r == 0 we have C . r == C . p == == r, whence the arguments of the field vectors and their reflection images
are the same. Thus, we can write
Ee(p)
1


= "2[1 + CJ· E(p) = It . E(p),
(4.119)
CHAPTER 4. FIELD TRANSFORMATIONS
120
1

1

Em(p) = 2[1 C] · E(p) = uu E(p), He(p)
= 2[1 C] · E(p) = uu· H(p),
Hm(p)
= 2[1 + C]· H(p) = It·
1


H(p).
(4.120) (4.121) (4.122)
From the conditions u· E e == 0,
u x He == 0,
(4.123)
we see that electrically symmetric fields satisfy the PMC boundary conditions on the plane of symmetry. Correspondingly, from
u x Em == 0,
u· H'" == 0,
(4.124)
we see that magnetically symmetric fields satisfy the PEC boundary conditions. Electromagnetic fields can be decomposed into electrically and magnetically symmetric parts by decomposing their sources into symmetric parts 1

Je(r) == 2[J(r)
+ C· J(C· r)],
1

J~(r)
== 2[J m(r)  C· Jm(C· r)], 1

Jm(r) == 2[J(r)  C· J(C· r)], 1

J:(r) == 2[J m(r) + C· Jm(C· r)].
(4.125) (4.126) (4.127) (4.128)
Because the symmetric fields satisfy either PMC or PEC boundary conditions at the reflection plane u · r == 0, the problem 'original source and plane boundary' can be replaced by the problem 'original and reflectiontransformed sources with no plane boundary'. This is called the mirror image principle because the transformed source is either the mirror image or the negative of the mirror image of the original source. The image sources for the PEC plane are obtained through magnetic reflection: (4.129) and for the PMC plane through electric reflection (4.130)
4.4. REFLECTION TRANSFORMATIONS
121
The advantage of the image principle is in replacing a boundary value problem by a source problem, which in most cases is easier to handle. The present image principle is, however, only associated with PEC or PMC planes and cannot be simply extended to impedance surfaces or interface problems. A method for solving problems of this kind in terms of a more complicated image principle will be discussed in Chapter 7. 4.4.4
Images in parallel planes
The reflection transformation above was defined with respect to a plane passing through the origin. A more general reflection transformation can be written in the form r
+
rc
= C . r + 2uu· r o ,
(4.131)
which defines reflection with respect to a plane with the normal unit vector u and passing through the point roo The image sources corresponding to a PMC plane are now (4.132) Jmi(r)
= C . Jm(C . r + 2uu· r o ) ,
(4.133)
and corresponding to a PEe plane, Ji(r) = C . J(C . r Jmi(r)
+ 2uu . r o ) ,
= C· Jm(C· r + 2uu· r
(4.134) (4.135)
o) .
These expressions make it possible to extend the image theory to a region bounded by two parallel PEC or PMC planes. As a simple example take the case of two PEe planes 1 and 2 going through the points rl = UTl and r2 = ur2 with the distance d = TI + T2. If the source J(r) lies between the planes, we can substitute the plane 1 by an unknown image source J 1 (r) and the plane 2 by another unknown image source J 2 (r ). The original plus the image sources must be magnetically symmetric with respect to each of the two planes: JI(r) = C· J(C· r
+ 2rl)  C· J 2(C· r + 2rl),
(4.136)
J 2(r) = C· J(G· r
+ 2r2)  C· JI(G· r + 2r2)'
(4.137)
From these equations we can obtain, by elimination, equations for a single unknown source, either J 1 or J 2: Jl(r)  J 1(r  2ud) = C· J(C· r  2ud)
+ J(r 
2ud),
(4.138)
CHAPTER 4. FIELD TRANSFORMATIONS
122
J 2(r )  J 2(r
+ 2nd) = C· J(C· r + 2nd) + J(r + 2nd).
(4.139)
The difference equations for the unknown image currents (4.138), (4.139) can be solved to give
J1(r) = C·
J 2(r) = C·
00
00
n=O
n=1
00
00
n=O
n=1
L J(C· r  2nud + 2rl) + L J(r  2nnd), L J(C· r + 2nud + 2r2) + L J(r + 2nud).
(4.140)
(4.141)
Each of these sources consists of two sets of periodic sources extending to infinity. The theory also works backwards: an infinite periodic structure can be replaced by a finite structure bounded bY,PEC or PMC planes. In addition to sources, boundaries and obstacles also transform by reflection with respect to a PEC or PMC plane. To see this, we can temporarily replace the obstacle by its equivalent source, for example, a dielectric obstacle by its polarization current J p = jw( f  f o )E. Since the polarization current transforms like the electric field, the transformed electric field induces exactly the transformed polarization current at the mirror image position of the dielectric sphere. 4.4.5
Babinet's principle
A disadvantage of the duality transformation is that a PEC boundary is always transformed to a PMC boundary, which is unphysical. However, if the boundary is planar, the duality transformation can be applied in conjunction with the image principle, so that PMC boundaries can be avoided. This combination leads to a method called Babinet's principle, which relates two problems involving complementary PEC boundary planes. Two planes containing PEC and open regions (holes) are called complementary if the holes in one of the planes correspond to PEC regions in the other and conversely. It is possible to obtain a solution for the diffraction field due to a metallic planar structure, a disk for example, if the solution is known for the corresponding complementary structure, such as a hole in the metallic plane. However, the sources of the two problems must be dual to each other. If selfdual, they are the same sources in both problems. Let the original source be J (r) = f (r) in the half space u r > 0 bounded by the plane 8 defined by u . r = o. Let 8 = 8 1 + 8 2 consist of a PEC (metal) part 8 1 and open part 8 2 . If the original source is written as a sum of two combined sources JA(r), In(r) defined by
4.4. REFLECTION TRANSFORMATIONS
1

JA(r)
= 2[f(r) + C· f(C· r)],
JB(r)
= "2 [f(r)
1
 C· f(C· r)],
123
(4.142) (4.143)
the problem can be split into two problems. The source J B is magnetically symmetric and the part 8 2 of the boundary plane S can be covered with PEC because the fields satisfy the correct boundary conditions. Thus, B corresponds to a problem of reflection from an intact metallic plane 8. In the A problem, the 8 2 part of the interface can be covered by PMC, while 8 1 is still a PEC surface, or it is a problem with an inhomogeneous plane boundary. The original problem of whole space and an interface plane has thus been transformed into two problems with plane boundaries. The complementary boundary problem with 8 1 empty and PEe on 8 2 can be approached through a similar method backwards after making the duality transform to problem A with the inhomogeneous boundary. Let us start from a problem with the complementary boundary and the dual source: (4.144) Likewise, this source can be split into two parts
Jmc(r)
jrJ = = = "2 [f(r) + C· f(C· r)],
(4.145)
JmD(r)
jrJ = = = "2[f(r)  C· f(C· r)],
(4.146)
of which the first one (C) corresponds to PEe boundary conditions on the whole plane, whence the plane can be either completed or removed without changing the fields. The second D source gives rise to PMC conditions on the 8 1 part of the surface. This problem is thus dual to the A problem with both dual sources and dual boundary conditions. Hence, the fields are also dual: (4.147) On the other hand, the fields Band C are easily calculated because of the planar PEe conditions. Thus, the original problem 1 and the problem 2 with complementary boundary and dual source are related through
(4.148)
124
CHAPTER 4. FIELD TRANSFORMATIONS
This means that there is a simple relation between the diffraction patterns of the two complementary problems, because the problems Band D involve no diffraction. However, the sources must be dual in the complementary problems, which is a limitation. For example, the diffraction from a hole in a metallic plane for a horizontally polarized plane wave has the same pattern as the diffraction from a complementary metallic disk for a vertically polarized plane wave. If the original source is a selfdual or antiselfdual source, this is no problem, because the dual source is then either the original source or the negative of it. The sign of the selfdual source depends of course on the chosen duality transformation. For an incoming plane wave the selfdual field is circularly polarized.
References BOOKER, H.G. (1946). Slot aerials and their relation to complementary wire areals (Babinet's principle). lEE Proceedings, 93, (3A), 6206. JONES, D.S. (1964). The theory of electromagnetism, pp. 56972. Pergamon, Oxford. KONG, J.A. (1986). Electromagnetic wave theory, pp. 36776. Wiley, New York. SENIOR, T .B.A. (1977). Some extensions of Babinet's principle in electromagnetic theory. IEEE Transactions on Antennas and Propagation, 25, (3), 41720.
Chapter 5
Electromagnetic field solutions In this chapter we consider solutions of the timeharmonic Maxwell equations for simple sources like the point source, line source and plane source in various media. Solutions for more complicated sources can be constructed from solutions corresponding to these simple sources.
5.1
The Green function
The field due to a source of unit amplitude plays a basic role in electromagnetics, because fields for arbitrary sources in linear media can be obtained by integrating such a field function. If the source is a unit point source, line source or plane source, the corresponding fields are called, respectively, the threedimensional, twodimensional or onedimensional Green functions. The equation for the threedimensional scalar Green function in homogeneous medium is written as L(\7)G(r  r')
= 
It is seen that the sum of these expressions equals J, whence the NR currents discarded in the course of the analysis have in fact cancelled each other.
Decomposition of a dipole Let us apply the decomposition of a dipole current J(r) = vIL6(r). Because the u directed component produces a TM field, we may only concentrate on a transversal dipole with v . u = O. From the integral expressions (6.112), (6.113) we immediately have the decomposition:
JTM(r)
rv
uIL[v· V'6(p)]u. V'Gt(z),
JTE(r)
rv
(6.118)
vIL6(r)  JTM(r).
(6.119)
These expressions can be interpreted through a transmissionline analogy. In fact, a twoconductor transmission line with line separation L, parallel to u and in the plane of v, can be characterized in space by the current density function
J(r) = uLI(z)v· \76(p).
I(z)
(6.120)
I(z) IL
)U
Fig. 6.2 TE component of dipole decomposition in line sources consists of the original dipole plus a transmissionline source.
196
6.5.3
CHAPTER 6. SOURCE EQUIVALENCE
Plane source decomposition
Another possibility for decomposing the current J(r) is in terms of planar currents. This can be done by again identifying from the identity (6.107) the TM, TE and NR components on the righthand side and writing the following equations for the corresponding current components:
\7;J™
= uu· \7 x (\7 x J) = \7;u(u· J)  uu \7(\7 t . J),
V';J T E == (uu~V'V')· J = \7;J  V'(V't . J)
+ uu \7(V't
. J).
(6.121) (6.122)
The equation for the NR current is not of interest. The solutions for (6.121) and (6.122) can be written in terms of the twodimensional Green function satisfying the stationary plane current equation
V';G 2(p  p') == 8(p  p'), G2(p  p') =
21n (k lp 21r
(6.123)
p'l).
(6.124)
The solutions are JTM (r) '" uu . / (''\7 x '\7 x J)G 2 (p  p')dS'
=
s
uu· J
+ uu· '\7 /('\7t' J)G 2 (p 
p')dS',
(6.125)
S
JTE(r) '" u x / '\7(u x '\7 . J)G 2 (p  p')dS'
=
s Jt
 uu
'\7 /('\7t' J)G 2 (p  p')dS',
(6.126)
s Here, under the integral sign, J depends on the variables p', Z, and \7 also operates on the same variables. Again, the current components are of the right form (6.106) to produce TM and TE fields as seen in the expressions (6.121), (6.122). Also, the sum of the final expressions equals J, whence the NR current terms discarded earlier have actually cancelled each other.
Decomposition of a dipole Taking the dipole source J = vIL8(r) with v . u = 0, without losing the generality, we can write from (6.125), (6.126), neglecting the NR terms v·p
JTM(r) == uIL8' (z )(v . VG 2(p)) = uIL8' (z ) 2 21rp
r..I
6.5. TE/TM DECOMPOSITION OF SOURCES
=
v
IL8(z)[I  2u pup ]
· 2
21rp
197
1
+ vIL8(r),
(6.127)
2
JTE(r) = vIL8(r)  JTM(r).
(6.128)
The last expression of the surface current term in (6.127) can be identified as a d.c. surface current flowing on a resistive sheet as excited by a dipole current source at the origin. The flow lines on the plane can be shown to be circles, each starting and ending at the origin. Thus, the TE component of the dipole consists of onehalf of the original dipole plus the stationary surface current generated by the original dipole. The TM part consists of the other half of the original dipole plus the negative of the previous surface current. This decomposition in terms of planar currents was given in Fourier components in the classic paper by CLEMMOW in 1963. Point source decornposit ion
6.5.4
The third possibility for decomposing a current J(r) can be made by applying (6.107) recursively, substituting J in the last term over and over again. The resulting series expansions can be written in the following forms: J
TM
J + kU'\?t .L •
= uu· J
(X)
(
u· ")2n+l :;;J, v
J
j
TE,
== J t

k"u\7t
.
(6.129)
J
n=O
~
(u.
n=O
J
\7)2n+l
c: :;;
(6.130)
J.
These decompositions are limited in the region of the original source J(r), because only differential operations are involved.
Decomposition of a dipole If these expressions are applied to a transversal dipole J == v I L8( r), the following multipole expansions are obtained:
JT M (r) J
TE
(r)
=
U
IL[ . \1~( )][6'(z) _ 6(3)(z) k v U P k k3
IL
6' (z)
+
6(5)(Z) _ k5
6(3) (z)
] ..• ,
6(5)(z)
= vIL8(r)uT[v.\76(p)][kk3+k5 ...].
(6.131 )
(6.132)
It is not evident that the point decomposition will converge in all cases. In fact, comparing with the delta expansion (6.43), we see that the amplitude of the moments of the multipole increases as n!/k n + 1 . However, the method can be useful, as was demonstrated by LINDELL (1988).
198
CHAPTER 6. SOURCE EQUIVALENCE
References CLEMMOW, P.C. (1963). The resolution of a dipole field into transverse electric and magnetic waves. lEE Proceedings, 110, (1), 10711. HARRINGTON, R.F. (1961). TimeHarmonic electromagnetic fields, pp. 12932. McGrawHill, New York. LINDELL, I.V. (1988). TE/TM decomposition of electromagnetic sources. IEEE Transactions on Antennas and Propagation, 36, (10), 13828. WILTON, D.R., (1980). A TETM decomposition of the electromagnetic field due to arbitrary sources radiating in unbounded regions. IEEE Transactions on Antennas and Propagation, 28, (1), 1114.
Chapter 7
Exact image theory Image sources can be defined as equivalent sources replacing physical structures such as regions of different media or boundaries with impedance conditions. As a classical example we have the mirror image due to a perfectly conducting plane surface. In this case, the image source located at its mirror image position replaces the conducting plane. It was seen in Chapter 4 that by applying the reflection transformation, the correct boundary conditions on the plane are ensured through the introduction of this kind of image source. The image principle has also been applied in electrostatics to problems with charges in front of a dielectric half space. In this case, the image charge is in the mirror image position and its amplitude depends on the dielectric constants of the two media. In this case, the image source is determined through the property that the correct interface conditions for the electric field are satisfied. A similar principle is also valid in magnetostatics. When trying to extend these to time dependent electromagnetic problems, trouble arises because a simple image source at the mirror image location cannot satisfy the correct interface conditions for both the electric and the magnetic field. Thus, the image concept must be generalized in one way or another. In contrast to different existing approximate theories, the present exact image theory (EIT) was constructed in the 1980s.
7.1
General formulation for layered media
In the present section an introduction to the EIT theory will be given in a general form applicable to layered media with plane parallel interfaces. In the following sections, specific geometries are presented as special cases of the general formulation.
7.1.1
Fourier transformations
Let us consider the problem of piecewise homogeneous media, with all interfaces and boundaries parallel to the xy plane, in terms of Fourier transformation in the two dimensions of the plane. The transformation
CHAPTER 7. EXACT IMAGE THEORY
200
does not seem to be necessary, however, and it would be interesting to find a method of image construction without applying the Fourier transformation at all. As indicated in a paper by KELLER in 1981, some problems of this kind can be handled directly without the Fourier transformation, but the shortcut requires a certain theorem. In waiting for such an approach for the general problem, the Fourier method will be applied here to produce the results with some labour. z
z·············?' 0/
 p
Fig. 7.1 Geometry of a multilayered medium with homogeneous layers of isotropic material.
Vector transmissionline equations The fields in the layer n with parameters satisfy the Maxwell equations
En, JLn
and sources J m n , I
n
(7.1) \7 x H; =
jWEnE n
+I n.
(7.2)
The twodimensional Fourier transformation is defined as
fer)
+
F(z,K) =
J
f(r)eiKPdS p ,
(7.3)
SfJ
with the inverse transformation
F(z,K)
+
fer) =
(2~)2
J
F(z,K)eiKPdSK ,
(7.4)
SK
where the integration domains Sp, SK are the p and K planes, respectively. Applying the transformation to (7.1), (7.2), we obtain the ordinary differential equations
(7.5)
7.1. GENERAL FORMULATION FOR LAYERED MEDIA
u, x H~  jK x H n

201
(7.6)
jW€nEn = I n ,
where the prime denotes differentiation with respect to the coordinate z. To help in finding solutions, these equations are written as transmissionline equations for the vector field components transverse to the normal unit vector u;, denoted bye, h, by eliminating the normal components u, . E, U Z • H. The resulting equations are ,
en
+ J.(3n=Zn' ( U z X
h n) = u ,
X
· Jmn
. + 1Jn k K Jen,
(7.7)
n
 u,
X
h in
· + J'(3n Y n' en = Jen
1 uz kTJn
X
K·Jmn,
(7.8)
n
where the electric source has been denoted by I n = jen + uzjen, and the magnetic source by J m n = jmn + uzjmn in components transverse and parallel to U z . Also, we denote 'T/n=
g, V4:
(7.9)
and the dyadics in (7.7), (7.8) have the form
can be identified as the propagation factor !!! the Fourier plane wave component in u, direction. The planar dyadics Zn, Y n can easily be seen to satisfy the relation
fin
=Zn' =Y n
_ 
1 4= 2 2 x k 2 {32 (knIt  knKK  knuzuzxKK)
= =It,
(7.11)
n n
because of the identity (7.12)
Y n is the twodimensional inverse ofZn and conversely. Thus, we can write = 1 = 1 Yn=2Uzuz~Zn=Zn, TJ n
spm Z; =
=
2
=
=1
Zn=""nuzuz~Yn=Yn, 1
_
spmr";
= ".,~,
(7.13) (7.14)
in accord with the expression for the inverse of twodimensional dyadics as discussed in Chapter 2.
CHAPTER 7. EXACT IMAGE THEORY
202
The equations (7.7), (7.8) are obviously vector counterparts of transmissionline equations with the propagation factor f3n in the z direction. A comparison can be made with the scalar transmissionline equations
+ j{3ZcI(z)
= u(z),
(7.15)
I'(z) + j{3YcU(z)
= i(z),
(7.16)
U'(z)
with characteristic impedance Zc = l/Ye and propagation factor {3, as well as distributed generator voltage u and current i quantities. This shows us that in the present case, en can be interpreted as a vector voltage and U z x h., as a vector current. The ch~acteristic impedance of the vector transmission line is a dyadic quantity Zn. Eliminating the vector u, x h n , an equation for en can be derived from (7.7), (7.8): (7.17) with the distributed source function 8n
2 'f/n K ., . , . (UzX K) Jmn· . (7 . 18) = J.n« (k n =] e>: KK)· ·Jen+k Jen+uzxJmn+J
kn
n
Likewise, we may derive (7.19) with the distributed magnetic source function 8m n
=
U z
. x J.,en  J.( u, x K) Len
j (k n2=I t + kn"1n

KK)· 1 K"J · . Jmn + kmn n"1n
(7.20) Alternatively, these complicatedlooking source expressions can be derived directly from the Helmholtz equation expressions, in homogeneous space outside the sources. For the electric field we can write (7.21) In fact, Fourier transforming the transverse component of this equation gives us exactly the vector transmissionline equation (7.17) with the expression (7.18) for the source above. In the same manner, the magnetic source is obtained from the magnetic field expression. The sources 8 and 8 m can be seen to correspond to each other through the duality transformation: Sd
= j'f/Sm.
7.1. GENERAL FORMULATION FOR LAYERED MEDIA
203
Reflection and transmission Solutions of these equations in each region n can be written in terms of two waves propagating in opposite directions: "(7.22) (7.23) For a wave propagating outside the sources in medium n there is a linear relation between the electric and magnetic field amplitudes. In fact, we can write from (7.7), (7.8) for sourceless regions (7.24) (7.25) with Y n, Zn defined in (7.10). The amplitude vectors depend on the sources and the boundary conditions. At the interfaces, there is the continuity condition for the tangential fields. For example, at the boundary between media 1 and 2 at z = d we can write (7.26) (7.27) At the interface, the boundary condition gives rise to a linear relation for the four amplitude vectors at, aI, at, a2" of the form (7.28) The reflection and transmission dyadics are easily obtained for d = 0 from the above relations in the form
(7.29)
= = = = = = 1 T21=It+Rll=2Z2·(Z2+Z1) = = = = = =1 T = It + R = 2Z (Z2 + Zl)12
22
1·
=
=
=1 Y2 ) ,
=
=
=1
= 2Y l · (Y 1 +
= 2Y 2 · (Y 1 + Y 2) .
(7.30) (7.31 )
If, instead of z = 0, the interface is at z = d, there are additional phase factors in the coefficient dyadics: (7.32)
CHAPTER 7. EXACT IMAGE THEORY
204
R22 ( d) == R22(O)ej{J22d,
T12 (d) = T12 (0)e j (,Bl  ,B2)d , T2 1 (d) = T2 1 (O)ej(,Bl,B2)d.
(7.33)
(7.34) (7.35)
The reflection dyadic expression (7.29) can be generalized to the case where, instead of the medium 2, in the half space z < 0 there is a stratified structure of some kind and in the half space z > 0, theJ.1omogeneous medium 1, if only we know the 'loading' admittance dyadic Y L due to the stratified structure, which is defined through the tangential fields at z == 0 by uzxh=YL·e, or e==ZL·(uzxh), (7.36) ~ith
1
Z L == Y L
R· aI'
.
In fact, from reasoning similar to that above and
at ==
we can write
from which
R can be solved in the form (7.38)
This expression is the same as (7.29) above if we identify Y L w~h Y 2, ~ut the last expression is only valid if the impedance dyadics Zl and Z L commute. In the more general case (for example, anisotropic loading impedance) the impedance expression reads (7.39)
By substituting the freespace admittance or impedance dyadics (7.10) for Y n,
Zn we have y
_. ~ KK n 
a K2 Pn"ln
~ uzuz~KK
+ k n"ln
K2'
zn  f3n"ln KK kn"ln uzuz~KK k K2 + f3n K2 .
(7.40)
(7.41)
n
Thus, the eigenvectors of these twodimensional dyadics are K, and u; x K, which correspond to TM and TE fields, respectively. (To check
this, we may note that, outside of sources, from the divergence condition \7 . E = 0, for a TE field in Fourier space we have K . e == 0, implying that
7.1. GENERAL FORMULATION FOR LAYERED MEDIA
205
e must be parallel to the vector u, x K. Analogously, for a TM field we have u; . h = 0 or U z x K . e = 0, whence e must be parallel to K.) The eigenvectors K and u, x K are also shared by the twodimensional reflection and transmission dyadics, provided the admittance and impedance dyadics of all media and impedance boundaries have the same eigenvectors. In this case, these symmetric dyadics can be written in terms of their eigenvalues as (7.42)
T=TTM K K +TTEUzUz~KK K2 K2'
(7.43)
For example, substituting (7.41) in (7.38), the reflection dyadic associated with an interface of two media takes on the form _
R 11 
{32112 _ {3111 1 KK k2 k1 __ {32!l2 lb.!l1. K2 k2 k1
+
+
k 2 Tl2
{32 k 2!l2 {32
ssu. {31 + ssss. 
xKK K2
UzUzx
(7.44)
{31
The problem can actually be handled in scalar form if we make the TE/TM decomposition to the original source, since the two polarizations do not couple to each other except if there are anisotropic media or surface impedances which possess other eigenvectors.
7.1.2
Image functions for reflection fields
To find the fields in physical space one must perform the inverse Fourier transformation. The result, often called the Sommerfeld integral, cannot be written in closed form because the reflection and transmission coefficients are too complicated functions of the Fourier parameter K. There have been attempts to approximate the reflection coefficients by simpler functions for which the inverse transformation can be found, for example by PARHAMlet ale (1980). However, the accuracy of the result depends on the approximations and cannot be easily predicted. The Sommerfeld integrals, despite their long history, are still a challenge to computers, since the integrands in most cases are highly oscillating. The idea behind EIT is to represent the reflection coefficient functions first as certain integral transformations of other functions which allow the inverse Fourier transformation from K space to p space to be performed exactly and what remains is the new transformation integral. This may just seem to be a substitution of one integral by another, but because the result can be interpreted as integration of an image source, it gains quasiphysical insight, which is helpful in setting up equations. Also, there is some numerical
CHAPTER 7. EXACT IMAGE THEORY
206
advantage in this approach. Thus, it becomes an interesting problem to find image sources corresponding to different geometries. This method was possibly first sketched, although not labelled in terms of images, for the classical halfspace 'problem by BOOKER and CLEMMOW in 1950. In fact, their result can be interpreted in terms of an exponentially diverging image. More recently, KUESTER and CHANG (1979) and MOHSEN (1982) applied a similar idea by making approximations for the reflection coefficient functions. The converging exact form of the image principle was introduced in 1983 by LINDELL and ALANEN. The EIT method can be described for any layered structures, although it was first applied to the Sommerfeld problem of two homogeneous halfspaces. Its success depends on whether the reflection coefficient functions can be represented analytically in terms of suitable integral transformations. The basic image functions can then be obtained by comparing the resulting field expression to the freespace field expression. Electric source problem
The EIT image functions are obtained by comparing actual field expressions with those arising from a line current source along the z axis. The freespace transverse field arising from a dipole source J(r) = vIL8(r  uzh), where v is a unit vector not necessarily equal to u ,, in physical space, obeys the law
e(r)
= jkTJ ( =It + k12 \7 t \7 )
. G(r  uzh)vIL.
(7.45)
Making the Fourier transformation leaves us with the expression =
e(K, z)
= jkTJOe "
ejt3lzhl
"/3
2)
vIL,
(7.46)
where the (electric) operator 0 e is defined as
=
Oe
= KK .Ku, d = It  k2 )/il dz.
The corresponding field from a line current I(z) z = Zo to 00, can be written as the integral
= vI(z),
(7.47)
extending from
(7.48)
7.1. GENERAL FORMULATION FOR LAYERED MEDIA
and in Fourier space in the region =
z
>
with no sources:
Zo
J
00.
e(K, z) = jk.,.,Oe .
207
,
e J /3(z+z )
2jf3
I(z')dz'.
(7.49)
Let us now try to compare the freespace field expression (7.49) with the expression of the reflected field in the homogeneous halfspace z > 0, due to a dipole source at height h: J(r) = vIL6(p)6(z  h), when the stratified structure exists for z < O. The reflected field denotes the difference of the total field and the field from the dipole in free space (the incident field). The effect of the stratified half space is characterized, in Fourier space, through the reflection dyadic R at z = O. Another analysis for the transmission field for~he region :...< 0 will be subsequently made with the transmission dyadic if replacing R. The field incident from the dipole to the stratified structure, in the region h > z > 0, is
e·(K z) t,
=
e j /3 h
.
.
J /3 z = aeJ {3z = Jok'YlO e · vILe 2j{3 , "/
(7.50)
giving rise to the reflected field .
=.
==
= e j /3(z+h)
er(K,z) = a+e J ,8z = R·ae J ,8z = jk.,.,R·Oe·C·
2jf3
vIL. (7.51)
The mirror reflection dyadic (7.52) arises because the differentiation d/dz in the operator O; now operates on exp( j{3z) and not on exp(j{3z). To obtain an image source giving the exact reflected field (7.51) as radiating from a line source in free space, in the form (7.49), the operator Oe should appear in front of the expression. Applying the eigenvector expansion of the reflection dyadic R, with eigenvectors K and U z x K corresponding to the respective TM and TE polarizations, we may write
k2 (1 _ KK t
_ oKUz~) J k 2 dz
. (RTEUZUz~KK R™(KK )) K2 + K2 + u, u; , (7.53)
208
CHAPTER 7. EXACT IMAGE THEORY
or
o; = o; . Re ,
R·
(7.54)
with the modified (electric) reflection dyadic defined as
Re = RTEUzUz~KK ) K2 + R™(KK K2 + u, u, .
(7.55)
This in fact implies that the operator Oe commutes with the modified reflection dyadic R e • Now we are able to write (7.51) in a form resembling that of (7.49): =
er(K,z)=jkr,oe'
e j /3( z + h ) = = 2j{3 Re·C·vIL,
(7.56)
allowing a representation for the reflected field as arising from an image current source.
Dyadic image function The EIT method is based on an integral representation of the modified reflection dyadic, defined by a function H( (): 00
It(K) = flAK, ()e j /3 H«() d( .
(7.57)
o Here, ( is an integration variable and H(() is~ function of ( but not of the Fourier parameter K. The dyadic function leeK, () is called the electric dyadic image function and its functional form depends on the stratified structure in question. Note that for the moment we consider K and {3 as ~dependent variables so that K only ~pears in the vector form. Thus, R; is a function of both fJ and K, and depends on K but not on (3. If a representation of the above kind cannot be found, it is sufficient to define a more general form
Ie
00
Re(K) =
L n=l
00
jlen(K, ()e j /3 H" «() d( , o
(7.58)
which gives us the possibility of expressing the structure in terms of a series of image sources. Inserting (7.57) in (7.56), its form becomes similar to that of (7.49): (7.59)
7.1. GENERAL FORMULATION FOR LAYERED MEDIA
209
Defining the following relation between the two integration variables z' and (:
z'
= h+H«(),
(7.60)
we can finally identify the image current function in the Fourier space in the form I 1 = = (7.61) I(z) H,«(/AK,(). C· vIL.
=
The inverse Fourier transformation to the transverse component of the reflection field can now be performed exactly, from which the total reflected electric field can be obtained in the form 00
Er(r)
= jk1]
(1 + k~ vv) ·/ / G(r  r' + v
uzH«())Ji(r',()dV'd(.
0
'\1.'CI1'J Defining the mirror reflection dyadic C by (7.52), the image source corresponding to the original dipole can obviously now be written in physical space as (7.63) Thus, thefunctional form of the image source depends on the dyadic image function fe' which again depends on the geometry of the problem. It is obvious from linearity that for an original volume source J(r) the image source has the general form (7.64) with the mirror image of the original source defined as
Jc(r)
= C · J(C . r).
(7.65)
The dyadic image function is actually an operator operating on the mirror image of the primary source. The field from the image source must be calculated as a fourfold integral over the three space coordinates and the parameter (. For a point source, the image is a line source, which starts from the mirror image point at z = h and proceeds along a line parallel to the z axis: (7.66) z = h H«(). In many cases, H«() may take complex values so that the image line lies in complex space. This does not, however, essentially complicate the field calculation process.
CHAPTER 7. EXACT IMAGE THEORY
210
Scalar image functions Since the reflection dyadic can be written in terms of its eigenvalues, the dyadic image function can also be written in terms of certain scalar functions. In fact, let us write the modified reflection dyadic (7.55) as
R
x n; k12 uzuzxKK,
TM=
I 
(7.67)
with the third reflection coefficient (7.68)
For each scalar reflection coefficient function, corresponding scalar image functions ITE ( (), ITM (() and f 0 ( ( ) can be defined through
Jf (X)
RTE(IJ) =
TE«( )e i (3 H «) d ( ,
(7.69)
fTM «()ei (3 H ((} d (,
(7.70)
o
J (X)
R™ (/3) =
o
J (X)
Ro(lJ) =
fo«()e i (3 H (() d ( .
(7.71 )
o In terms of these scalar functions, the dyadic image operator can be represented as (7.72)
Thus, the image source can be explicitly written as (7.73)
This form is not unique, since the equivalent source is not unique. Other possible forms for the image source will be given subsequently.
211
7.1. GENERAL FORMULATION FOR LAYERED MEDIA
Duality transformations The image functions have certain relations through the simple duality transformation (Section 4.2.1), following from corresponding relations between the reflection coefficients. The duality transformation of the reflection dyadic can be obtained from a consideration of incident and reflected fields, related through e+=R·e. (7.74) The dual of this relation can be written
h+:::.:: Rd· h ", Inserting
U z
x h±
 u, x h+ = (uzuz~Rd)· (u z x h).
or
= ±Y . e±
(7.75)
leaves us with the relation
UzUz~Rd =


1
y. R· y
(7.76)
This takes a simple form provided the eigenvectors of Y and R are the same, which is the case for example in isotropic problems, because the dyadics then commute. Thus, in this case, the relation reads (7.77)
This corresponds to the following duality relations between the scalar reflection coefficients: R™ d


RT E ,
k2
Rod = K2(RI
M

RI
E
)
= Ro. (7.78)
These, finally, lead to duality relations between the image functions:
!od(()
= !o(().
(7.79)
Note that the dual of a dual always gives the original quantity. Thus, in dealing with the general isotropic medium, we need only derive, for example, the functions ITE, 10 since ITM can be obtained through the duality transformation. However, if the analysis is limited to, for example, dielectric media, this property cannot be applied.
Magnetic source problem The duality transformation analysis can be applied for the construction of the image of a magnetic current source Jm(r). In fact, making the duality transformation for the image expression of the electric source (7.64) gives us (7.80)
212
CHAPTER 7. EXACT IMAGE THEORY
with
=
=
f m(j\7 t , () == f ed(jV t , () ==  f
TE
=
1
(()! + fo(() k 2 u, u, ~VV.
(7.81 )
The total reflection field due to the magnetic image source can written similarly to that in (7.62):
JJ 00
Er(r) = V' x
G(r  r' + uzH«())Jmi(r', ()dV'd(.
v
(7.82)
0
Again, the image line corresponding to a magnetic point source lies on a line parallel to the z axis: z == h  H((),
(7.83)
extending from the mirror image point z == h to infinity, probably on the complex z plane.
Further properties of the image functions The three image functions fTE((), fTM(() and fo(() depend on the reflection coefficient functions and, thus, on the geometry of the problem. This being the case, they must be determined for each problem separately. However, they share certain general properties in all problems. Since there exists a relation between the three reflection coefficients R T E , R™ and R o , we might expect a relation between the corresponding image functions, aside from that due to the duality properties. To find this, we write K 2 == k 2  {32, and apply the relation (7.84) which leads to
J
J
o
0
00
00
k2!o«( )e i f3 H« )d( 
J
!o«(){32e i f3 H«) d( =
00
k 2[JTM«)

jTE«)]e j ,8H ({ ) d ( .
(7.85)
o Since all the image functions are zero together with their derivatives at ( ~ 0, we can easily make partial integrations in the second integral term and discard the terms at the endpoint ( == o. Also, we may anticipate
7.1. GENERAL FORMULATION FOR LAYERED MEDIA
213
functions vanishing at infinity, whence the corresponding endpoint terms are also discarded. Thus, equating the functions in the transformation integrands we finally have the resulting relation for the image function fo(() in the form of a differential equation, the prime denoting differentiation with respect to (: (7.86) For the special case H(() = ( this becomes simply (7.87) with the boundary conditions 10(0) = 10(00) = o. Another form of this condition can be simply written after inverse Fourier transformation as the operator equation, which is valid inside a field integral, where ~ t operates on the current function: (7.88) This together with the following identity
u, u, ~ ~~ + ~~ t
+ (~2 + k2 )u z u,
= ~;I + u, U z . (~~
+ k 2 I),
(7.89)
which is easy to verify, giYes us the possibility of transforming the dyadic image operator function le given in (7.72) in the form (7.90) in a different form
= . le(J'7 t , ( )
= 1
TE
=
=
1
(()1 + fo(()uzu z· (I + k 2 '7'7).
(7.91)
To arrive at this expression, one must note that the terms with '7'7 t and '7 2 + k 2 in the image source function result in NR image components and can be omitted, as explained in Chapter 6. Further, we can easily develop other possible forms, for example: (7.92) For the magnetic dyadic image operator we can correspondingly write from duality (7.93)
214
CHAPTER 7. EXACT IMAGE THEORY
=
.
!m(JV t , ( ) =  !
f m(rVt, () =
TM
=
=
1
(()I+!o(()uzuz·(I+ k2VV),
fTM «()It 
U z u,
(7.94)
. (fTE«()I + k12 fo«()V'V't) .
(7.95)
With any of the above expressions for the dyadic image operators, the image sources can be compactly expressed as Ji(r, () = fe(jV t , () . Jc(r), Jmi(r, ()
(7.96)
= 7m(jV t , () . Jmc(r).
(7.97)
The difference of the electric and magnetic dyadic image operators is seen from previous expressions to be (7.98)
Ie
which shows that the antiselfdual part of the operator is just a function of ( times the unit dyadic. Correspondingly, we have
== fe(jV t , (
)
1 += !m(jV t , ( ) = k2fo(()[2uzuz~VV 
:2fo«()UzUz~V'V' + (fTM(() for twice the selfdual part of 7.1.3
fTE(()]I,
2=
VtI]
= (7.99)
7e '
Image functions for transmission fields
The EIT method can also be applied to problems of fields traversing a stratified layer. Let us assume that the transverse electric field (Fourier component) penetrating into medium 2 through a layer of any kind from rnedium 1 is eT(K, z) = a"2 ej 132z = a 1ej 132Z • (7.100)
T.
For simplicity in notation, the thickness of the layer in between the media 1 and 2 is taken as zero. This can be understood as a convention for the coordinates in the two media: medium 1 is defined for z > 0 and medium 2 for z < O. All information on effects du~ to the layer on transmitted waves is included in the transmission dyadic T. Let us follow a similar line of reasoning as in the derivation of the reflection image expressions. The transmitted field can be written in Fourier space as (7.101)
7.1. GENERAL FORMULATION FOR LAYERED MEDIA
215
in analogy with (7.50), (7.51). The indice~1 and 2 denote parameters corresponding to media 1 and 2. The dyadic Del in (7.47), when operating on the incident field a 1ei,Bl z, can be written as (7.102) Since the transmission image source must be in homogeous space filled with material 2, we should be able to write (7.101) in the form resembling (7.56) (7.103)
It is seen that the exponential term containing the 'wrong' f31 coefficient must be treated otherwise than in the reflection problem, unless the media 1 and 2 are the same. The modified transmission dyadic turns out to have quite a complicatedlooking appearance in the general case:
(7.104) with
Tu €2 T ™
(7.105)
,
tl
T[
= p,d32 T T E ,
(7.106)
J.L2/31
To == (€2{31 T ™ _ JLl(32 T T E) €1(32
JL2(31
ki .
K2
(7.107)
It can be argued that, in To, K2 in the denominator is cancelled by the preceding bracketed factor for any layer. In fact, if the term in brackets is written as a Taylor expansion according to powers of K, it is clear that the series only contains even powers, because the terms are independent of the sign of K (actually they are independent of the direction of the vector K). Thus, the series starts by A + BK 2 + .... Because for K = 0 (normal planewave incidence), the TE and TM polarizations are both TEM, the transmission coefficients are actually the same, and from (31 = k1 and {32 =: k 2 the bracketed term can easily be shown to vanish. Thus, A = 0 and K 2 cancels out.
216
CHAPTER 7. EXACT IMAGE THEORY
Transmission into a similar medium We can now define the transmission dyadic image function he analogously with (7.57). For the special case when the media 1 and 2 are the same, whence k I = k2 = k, {3I = {32 = {3 and T; = TTM, T[ = TTE, we have 00
T e = fhe(K, ()e i{3H«)d(,
(7.108)
o
with (7.109)
!
00
h T E(()e i{3H((}d(,
(7.110)
TTM ({J) = f h™ (()e i{3H((}d(,
(7.111)
TTE({J) =
o 00
o
To({J) =
~: (TTM 
!
00
TTE) =
ho(()ei13H(Od(.
(7.112)
o
Transmission into a different medium For the general case, the previous expressions do not work and, instead, we must define a dyadic image function based not on the transmission dyadic Te but on Te e i {31h , which means that the image functions also become functions of the distance parameter h: 00
Teeil31h = !h.e(K, h, ()e i 132H(h,Od(.
(7.113)
o Applying the expression (7.104) in the form (7.114) the dyadic image function can be expressed in terms of three scalar image functions, in the corresponding form
7.1. GENERAL FORMULATION FOR LAYERED MEDIA
217
uZ~ + Uhu z + K)ho(h, () fJ~?'
(7.115)
ho(h, ()
The scalar transmission image functions are defined through
J ex>
Tu,e j {3 1h
= €2 T ™
e j {31h
=
€1
hu,(h, ()e j {32 H (h '( )d (
,
(7.116)
o
J ex>
Tle j (31h = J.1.1fJ2 TTE e j (31h
=
JL2f31
hI(h, ()e j {32 H (h '( )d ( ,
(7.117)
o
J ex>
Toe j (31h = fJ2 To e j (31h =
ho(h, ()e j (32 H (h'() d( .
(7.118)
o
Other forms for the transmission image dyadic are also possible. The last term in (7.115) will be omitted, because a term starting with jK jf32uz, or in physical space with V', corresponds to an NR source. Thus, we can neglect the corresponding term in the transmission dyadic as well, and apply the expression (7.119)
Also, note the difference between the coefficients To and To = f32To. Application of this expansion leads to a transmission image theory in which the form of the image function corresponding to a point source is dependent on the position of the source, which was not the case for the reflection image. In fact, writing the transverse component of the transmitted electric field in the Fourier space
J ex>
 jk2112
•
eJf32z
Oe2 · 2jfJ2 heCK, h, ()e i 132H (h,() · vILd(
(7.120)
o
we have for the total transmitted field in physical space corresponding to the original dipole J(r') = vIL6(r'  uzh), ex>
ET(r) = jk2 112
JJ
C(D«()) . Ji(r', ()dV'd(,
o v
(7.121)
CHAPTER 7. EXACT IMAGE THEORY
218
with
= Vp2 + [z 
D(()
Ji(r', () =
H(z', ()]2,
he (jV'~, Z', () . V I L8(r', (),
(7.122) (7.123)
which is dependent on the position z' = h of the dipole. Note that this Green dyadic depends on the parameters of medium 2. More generally, for a threedimensional source J(r') we obtain the fourdimensional image (7.124)
7.1.4
Green functions
Instead of applying image functions each time when fields are being computed, one can define a dyadic Green function which takes into account the reflection or change in transmission due to the stratified geometry in question. This gives us an opportunity to apply solution routines designed for freespace problems by adding to the freespace dyadic Green function a new dyadic Green function.
Reflection problem A new Green dyadic K(r) corresponding to the reflection from the structure can be obtained by integrating out the parameter ( in the field expressions. It can be seen that, to construct the dyadic image Green function, two scalar Green functions are required instead of only one as in the homogeneousspace problem. The reflection image Green dyadic is obtained from the reflection field expression: oo
Er(r)
= jkT]
JJ
G(r  r'
v jkT] [
(i
+ uzH(()) . Ji(r', ()dV'd( =
0
G(r  r'
+ uzH(()) ·
 jkT]
fe(j'\7~, ()d() . Jc(r')dV' =
J
K(r  r') · Jc(r')dV',
(7.125)
+ uzH(()) . !e(j'\7~, ()d(.
(7.126)
v oo
K(r) =
J
G(r
o
7.1. GENERAL FORMULATION FOR LAYERED MEDIA
219
The expression of the image current Ji contains functions of the ( variable and the V operator as separate factors, which makes it possible to integrate the ( variable out and define the scalar Green functions
! ! ! 00
KTE(r) =
G(r + uzH((n!TE(()d(,
(7.127)
G(r + uzH((n!TM (()d(,
(7.128)
G(r + uzH((n!o(()d(.
(7.129)
o
00
K™ (r) =
o
00
Ko(r) =
o The \7' operators can be moved in front of the Green function through the following manipulations in the field integrals, recalling that the field point r never coincides with the image source point r':
!
Ko(r 
r')(uzuz~V'V'). Jc(r')dV' =
v u, x
! v
u, x V
V' [Ko(u z
X
V')· Jc]dV'  u, x !(V'Ko)(U z' V' x Jc)dV'
v
J v
Ko(u z X V' . Jc)dV' =
(uzuz~VV) ·
=
Jv
Ko(r  r')Jc(r')dV'.
(7.130) Thus, the dyadic image Green function can be defined in terms of these scalar image Green functions as
=K(r)_
(=
1 V V ) K TM (r)+k2(uZuZxVV)Ko(r), 1 x I+ k2
or, taking another representation for the dyadic operator
(7.131 )
7e , (7.132)
The reflected field due to the image source can be expressed simply as
Er(r)
= jk.,.,
J
K(r  r') · Jc(r')dV'.
v
(7.133)
CHAPTER 7. EXACT IMAGE THEORY
220
Note that the source here is the mirror image of the original source and not the original source. We could have included the mirror image operator c in the definition of the image Green dyadic. It is seen that only two image scalar Green functions are needed in the computation of the reflected fields, either of the functions K™ and K TE together with the function K o. From duality, there exist obvious relations between the image Green functions: KIM (r) = _KTE(r), (7.134)
KJE(r) = _K™ (r),
(7.135)
Kod(r) = Ko(r).
(7.136)
Thus, values for the Green functions K T E and K™ can be computed applying the same procedure with dual parameter values.
Asymptotic expressions for the reflection problem Asymptotic expressions of the Green functions are of interest when computing fields far from the source region. Also, extracting the asymptotic term from the Green function leaves a function which is more easily approximated because it decays more quickly than the original Green function. If the argument r becomes large, so that we may assume IH(()I « [r] for those ( values for which the image function is not negligible, we may write the twoterm Taylor expansion for the distance function as
(7.137) which inserted in the Green function gives us the basic approximation (7.138) Thus, we have for the firstterm approximation of any of the Green functions (7.127)(7.129)
J
J ~
00
G(r + uzH(())f(()d(
o
~ G(r)
ejkUr·U.H(()
f(()d(.
(7.139)
0
The last integral expression can be interpreted in terms of the reflection coefficient function, applying the definitions (7.69)(7.71). In fact, we can write 00
J
ejkUT.·UzH«)
o
f(()d(
= R({3)I l3=k cos ()
= R(kcosO),
(7.140)
7.1. GENERAL f'ORMULATION FOR LAYERED MEDIA
221
denoting the angle of the vector r and the z axis by 0, whence we have the asymptotic expressions for the image Green functions
KTE(r) ~ RTE(kcosO)G(r),
(7.141 )
K™ (r) ~ R™ (k cos O)G(r),
(7.142)
Ko(r)
~
Ro(k cos O)G(r).
(7.143)
The corresponding approximation for the reflection dyadic image Green function
K(r)
~ R™(kcosO)
R o ( k cos 0)
(1 + :2 V'V') G(r)+
:2(
u, u, ~ V'V')G( r).
(7.144)
can be written, if we make the far field approximation \7 t jku r and apply (7.68) (note that the Fourier parameter is now K = k sin 0), in the form (7.145) This can be called the reflectioncoefficient method, or ReM approximation, in which the reflection far field is obtained by replacing the exact image by the mirror image of the original source, and multiplying it by the dyadic R T E u., u., + R T M uo.uo. When computing the reflection field, the argument r of the Green function is replaced by the difference r  r', where r' denotes a point of the mirror image source. The angles and distances are then measured from the point r' and not the origin.
Transmission problem The transmission field can also be written in terms of a new transmission Green dyadic taking care of the geometry behind the transmission plane. In fact, writing the transmission field (7.121) in the form
 jk2 1/2
J
KT(r  r') · J(r')dV',
v
(7.146)
222
CHAPTER 7. EXACT IMAGE THEORY
we can define the transmission Green dyadic by
J 00
KT(r  r')
=
G(D«(» . he(j'V~, z', ()d(,
(7.147)
o
D(() = J[r  p'  UZH(Z', ()] . [r  p'  UZH(ZI, ()].
(7.148)
The asymptotic far field expression for the new Green dyadic can be written similarly with the reflection problem: D(() ~ r  u, . [p' + uzH(z', ()],
(7.149)
00
KT(r  r')
~ G(r)eik2U,·P' . Jhe(j'V~, z', ()eik2U.OU.H(z"()d(.
(7.150)
o
The last integral can be interpreted in terms of the definition of the transmission dyadic, (7.113): (7.151 )
with the parameters in the transmission dyadic and exponent expressions given as (7.152) (7.153)
(7.154)
z.. z·
Fig. 7.2 The asymptotic far field approximation for the transmission image expression can be interpreted in terms of refracted rays starting from a source point at an apparent height z".
7.1. GENERAL FORMULATION FOR LAYERED MEDIA
The angles we can write
()1
and
()2
223
are defined above. To obtain an interpretation,
= ,== Kr(r  r) ~ G(D) . T e ,
(7.155)
D ~ J(r  pi  uzz") . (r  pi  uzz"),
(7.156)
"
{31
I
z = z = {32
k1 cos ()1 I z k2 cos ()2
tan ()2 =z . tan ()1 I
(7.157)
This expression can be interpreted geometrically as representing the point on the z axis from which a ray in the direction u, is emanating, when the layers between the planes in the medium 1 and 2 are squeezed to a plane.
References
P. (1986). Applications of complex image theory. Radio Science, 21, (4), 60516. BOOKER, H.G. and CLEMMOW, P.C. (1950). A relation between the Sommerfeld theory of radio propagation over a flat earth and the theory of diffraction at a straight edge, lEE Proceedings, 97, pt.III, 1827. KELLER, J.B. (1981). Oblique derivative boundary conditions and the image method. SIAM Journal of Applied Mathematics, 41, (2), 294300. KUESTER, E.F. and CHANG, D.C. (1979). Evaluation of Sommerfeld integrals associated with dipole sources above the earth. University of Colorado Electromagnetics Laboratory, Scientific Report 43. LINDELL, I.V. and ALANEN, E. (1983). Exact image theory for the Sommerfeld halfspace problem with a vertical magnetic dipole, Proceedings of the 13th European Microwave Conference, Nuremberg, pp. 72732. Microwave exhibitions and publishers, Tunbridge Wells. LINDELL, I.V., NIKOSKINEN, K.I., ALANEN, E. and HUJANEN, A.T. (1989). Scalar Green function rnethod for microstrip antenna analysis based on the exact image theory, Annales des Telecommunications, 44, BANNISTER,
(910), 53342. MOHSEN, A. (1982). On the evaluation of Sommerfeld integrals. lEE Proceedings, 129H, (4), 17782. PARHAMI, P., RAHMATSAMII, Y. and MITTRA, R. (1980). An efficient approach for evaluating Sommerfeld integrals encountered in the problem of a current element radiating over lossy ground. IEEE Transactions on Antennas and Propagation, 28, (1), 1004. WAIT, J.R. (1985). Electromagnetic wave theory. Harper and Row, New York.
CHAPTER 7. EXACT IMAGE THEORY
224
7.2
Surface problems
As an application of the general theory, let us consider problems involving planar impedance surfaces and sheets. If the surface is impenetrable, it can be characterized by an impedance surface condition, as discussed in Chapter 3. For penetrable surfaces, the corresponding impedance sheet characterization is valid. As a last example in this section, the thin metallic grid problem is considered. In all these cases, the image functions can be quite easily written in explicit form involving only simple exponential functions. 7.2.1
Impedance surface
The planar impedance surface serves as the simplest example of a problem for which the image method can be applied. The idea has also been applied in the book by FELSEN and MARCUVITZ (1973, pp. 557559). Reflection coefficients
In the case of an isotropic planar impedance su.:!ace at !. = 0 with surf~e impedance Zs, defining the impedance dyadic Zs = Zslt, the dyadics Zs and Z obviously commute and the reflection dyadic can be written from (7.38)
R
= (Z s + Z)l. (Z s  Z) = RTEUzUz~KK R™KK K2 + K2 '
(7.158)
with the eigenvalues RTE
=Z
_!l!! S
Z,
{3
2!l!! = 1 __Z_tc_ (3 +
+W
r'

RTM _
Z  f1!J. S
Zs
k

+~ 
1
+
2~ 17
(3 +
¥.
(7.159) These coefficients obviously satisfy the duality condition (7.78) because the dual of Zs/TJ is TJ/Zs. As another check, both coefficients can be seen to tend to 1 as Z, ~ 0 and to +1 as Z, ~ 00, corresponding respectively to perfect electric and magnetic conductor surfaces. The third reflection coefficient takes the following forms:
(7.160)
7.2. SURFACE PROBLEMS
225
and it is seen to vanish in both the perfect electric and magnetic conductor limiting cases. Image functions The image functions corresponding to the reflection coefficients can be found in a straightforward manner if the transformation is defined through the integral
J 00
R(f3) =
f«()ej(j(d(.
(7.161)
o
This means that the function H(t) of (7.57) in this case is simply defined as H«() = (. The image functions can then be written as (7.162)
fTM «()
= 6+«() + 2j Zsk ejZ,k(/.,U+«(),
(7.163)
rJ
fo«() =
~k~~;
(ejZ,k(/.,  ej.,k(/Z.) U+«().
(7.164)
The unit step function U+«() and the corresponding delta function 6+«() are defined at the limits U«(  ~) and 6«(  ~) as Ll + O. Thus, all image functions vanish for ( ~ o. They can also be checked to satisfy the condition (7.87). In terms of these functions, the reflected field can be expressed as arising from an image ~ource, by applying one of the many forms for the dyadic image function fe. For example, with (7.92) we have the following explicit expression for the image source corresponding to a primary source J(r): .
Ji(r, () (fTM«()UzUz
= fe(j~t' () . Je(r) =
+ fTE«()I t + u, f~~()Vt) · Jc(r),
(7.165)
in which the image functions must be substituted from (7.162)(7.164).
Convergence of image functions To obtain a meaningful field integral, the integrand should converge. Because the Green function can be made converging through a proper choice of the branch of the distance function, it suffices that the image functions
226
CHAPTER 7. EXACT IMAGE THEORY
are nondivergent. It is seen immediately that for real parameter ( values, for a lossless medium (real and positive k and T/), the exponential functions are all nondiverging when Z; is real. On the other hand, if Z, has an imaginary part, some of the exponential functions turn out to be diverging. In particular, for a lossless medium with real k and "7, the condition for the TE image (7.162) to be nondivergent is obviously ~{(/Zs} == ~{ (ys} ~ 0, or the surface impedance should be inductive or pure resistive if ( is real and positive. Correspondingly, for the TM image, from (7.163) we have the condition ~{(Zs} ~ 0, or the surface impedance should be capacitive or pure resistive. Thus, we see that unless the surface is purely resistive, one of the functions fT E((), fT M(() diverges for real and positive ( and the fo( () function diverges. The image source diverging for real and positive ( can be interpreted as giving rise to a surfacewave field. The existence of a surface wave is associated with the poles of the reflection coefficient. Writing
A R({J) = {J  (Jp'
(7.166)
a surface wave with z dependence as exp( jf3p z ) exists if the condition ~{,Bp} < 0 is valid. Because the pole corresponds to either ,BJE == k".,/Zs or f3;M = kZs/T/, there exists a TE surface wave for ~{Ys} > 0, which defines a capacitive surface impedance. On the other hand, there exists a TM surface wave for ~{Zs} > 0, which defines an inductive surface impedance. These conditions are the same as those for diverging image functions for real and positive t. Exponentially diverging image functions corresponding to surface waves along the structure will also be encountered in other problems. To obtain nondiverging image functions we may choose an integration path in complex ( plane. It is sufficient to choose a path along the imaginary axis ( = j(', although this does not make the image function converge. However, since the distance function D is now complex, its branch can be chosen so that the Green function becomes exponentially decaying, whence the field integrand converges. However, for imaginary (, there is a singularity in the Green function when the source and field points are at the impedance plane, because the distance function D becomes zero for p = ('. This singularity is clearly demonstrated by the following example. Vertical electric dipole above an impedance plane Let us consider a horizontal impedance plane and the field from a vertical electric dipole (VED) J(r) = u zIL6(z  h). According to the EIT method,
227
7.2. SURFACE PROBLEMS
we can replace the impedance surface with the image source Ji(r, () = fe(j\lt' () . [u zIL8(p)8(z
+ h)] =
 jTM«()u zIL8(p)8(z + It).
(7.167)
+It!·~~
0.9 If+++
I+I,+!:
Integrand Ifor the field due! to YED
0.8toft·r~"tlecteif·trom·· an···'.pedancc····sunaeel:.....1 fOf krho :k 1, kh *= 0, Zs/eta = 1 :
0.71HJ++++=+=+++f
0.6tH4of+of+frf+t
0.5 IH+++++++++~ kz=O.pl 0.4 t++++++++++.+..1
I 0.31+++++++++++..1
/;
0.2 0.1
kz=OJl
~ \ kz~0.3 .....
~
kz=l
"~I
4
3
5
6
8
7
9
10
Fig. 7.3 Normalized integrand of the reflected field integral without the delta term, as a function of k( for different values of kz, at h = 0 and p = o. A peak arises when z + h ~ 0, i.e, the source and field points are both close to the surface.
The image can be expressed in terms of the function ITM alone, because the VED source produces a TM field. The reflected field can be written as
JJ 00
Er{r)
= jk'fJ
G{r  r' + u z () · Ji(r',()dV'd( =
v
0
. (= + k1)
JkrJ I
2
\7\7
. uzIL K TM (r
+ uzh),
when the Green function is defined as in (7.128) with H«() = (:
J 00
K™(r+ u h) = z
o
jkD«)
e fTM(()d( 41rD«()
=
(7.168)
228
CHAPTER 7. EXACT IMAGE THEORY
J » 00
Z _ G(D(O)) + 2jk!..
e
o with the distance function defined by D(() =
jkD«)
47rD(()
e j Z t< k
G(D)JTM(_j(')d('
o
~
7.3. THE SOMMERFELD HALFSPACE PROBLEM
4jVl
k
+ 4K 2
Vik
21r P
(seW~erfc(wd + ~eW~erfC(W2)) ,
235
(7.216)
S
with erfc denoting the complementary error function and
Wl
=
V2ipk(z+hiPS),
W2
=
V2ipk(Z+h+i~).
(7.217)
This can be identified as a sum of two Sommerfeld surfacewave solutions with imaginary numerical distances  j k p S 2 /2 and j kp/28 2 , as discussed in FELSEN and MARCUVITZ (1973, pp. 509510). In the region 2~2 «: kp «: 2/~2 with k(z + h) «: 1 we have IWII ~ 1 and IW21 « 1, whence (7.216) can be shown to reduce to
(7.218)
which is obviously of the form of a radial surface wave, decaying in the z direction and propagating in the radial direction along the grid.
References
N. (1973). Radiation and scattering of waves. PrenticeHall, Englewood Cliffs, N J . KONTOROVICH, N.I. (1963). Averaged boundary conditions at the surface of a grating with square mesh. Radio Engineering and Electronic Physics, 8, (9), 144654. LINDELL, I.V., AKIMOV, V.P. and ALANEN, E. (1986). Image theory for a dipole excitation of fields above and below a wire grid with square cells, IEEE Transactions on Electromagnetic Compatibility, 28, (2), 10710. WAIT, J.R. (1978). Theories of scattering from wire grid and mesh structures. In Electromagnetic scattering (ed. P. Uslenghi), pp.25387. Academic Press, New York. FELSEN, L.B. and MARCUVITZ,
7.3
The Sommerfeld halfspace problem
The simplest electromagnetic field problem involving sources and material media, beyond that of the homogeneous space, is the problem with a source in front of a planar interface of two homogeneous half spaces. This problem
CHAPTER 7. EXACT IMAGE THEORY
236
was first studied seriously by SOMMERFELD in 1909, which is why it is often called the Sommerfeld halfspace problem. Let us find the image functions associated with two halfspaces 1 (z > 0) and 2 (z < 0) with respective medium parameters denoted by €I€o, J.LIJ.Lo and €2€, J.L2J.L. The primary source is assumed to lie in the halfspace 1. Let us denote the parameter ratios by € = €2/€1, J.L = J.L2/J.Ll' For static fields, the image principle is well documented in elementary textbooks. For example, if a point charge Q is located at the point r = uzh in the air (€o,J.Lo), the field reflected from the dielectric ground medium (€€o, J.Lo) can be shown to equal that arising from the reflection image Qr located at the mirror image point ri, when the medium is all air: Qr
€1
=  € + 1 Q,
ri
= C . uzh =
uzh.
(7.219)
Also, the field transmitted into the ground in the static case can be found from the transmission image charge QT, 2€ QT =
f
+ 1Q,
ri
=
uzh,
(7.220)
which appears at the location of the original charge when the medium is all ground. It turns out that the correct interface condition, continuity of the transversal total electric field, is exactly satisfied with these two image sources. For the timedependent problem the image solution is not that obvious. If the original static charge is set in harmonic motion, the idea of setting the static image in the corresponding harmonic motion does not work, since, at the interface, only the condition for the electric field and not for the magnetic field is guaranteed. Thus, possible image sources in this case must necessarily be more complicated, which gives a motivation for the EIT method.
7.3.1
Reflection coefficients
As was seen in the previous section, construction of the image sources requires three reflection coefficient functions and their integral transforms to be found. In the case of an interface between two homogeneous and isotropic media with the parameters €l, ftl' €2, ft2, the reflection coefficient functions were already given in (7.44) based on the wave impedances of the TE and TM polarized fields: ZTE
= kTJ , f3
ZTM =
{3TJ I k
(7.221)
7.3.
THE SOMMERFELD HALFSPACE PROBLEM
237
(7.222)
(7.223)
Here we denote
JL2
(7.224)
JL=, JLl
and q
= ~,
B = jk~  k? = k1 Jw  1,
(7.225)
where k, = WVf.iJLi. The reflection function R( a, q) applied above is defined by
H+1
 aqR( a, q) 1'::271' aq
+ Vq2 + 1
(7.226)
The reflection coefficients obviously satisfy the duality condition (7.78). For the third reflection coefficient we can evaluate after some effort
with
€1 E=f.+ i '
JLI i
M JL+
(7.228)
The image functions can be found in a convenient series form if we expand the reflection coefficient functions in terms of a reflection parameter r defined by r  R(I ,q) 
q
H+1 .
q+H+1
(7.229)
In fact, the following expansion can be easily verified: A+1' 1 + Ar'
Rio: q) =  ,
aI
A=. a+l
(7.230)
CHAPTER 7. EXACT IMAGE THEORY
238
Applying the Taylor expansion, we can write 00
R(a,q) = (A
+ r) "(Art = ALJ n=O
+ 4
00
"(Ar)n.
a 1L.t
Obviously, the series expansion converges best for small 7.3.2
(7.231)
n=l
IArl
values.
Reflection image functions
The reflection image sources are obtained in terms of a function of two variables, f( a, p), which can be identified through the following integral identity, found in any collection of Laplace transforms, for example, ABRAMOWITZ and STEGUN (1964).
J J2~(P) 00
2n
eqpdp =
[R+! q]2n = (rt,
(7.232)
o
where J 2n (p) is the Bessel function of order 2n. In fact, if we write
J 00
R(a,q)
=
[f(a,p)
+:
~~ 8+(p)] epqdp,
(7.233)
o
the image function f(a,p) can be identified from the integral identity and the series expansion of the reflection coefficient in the form (7.234) The special case a = 1 gives us (7.235)
Values for the image function f(a,p) can be easily calculated from the above series, because Bessel functions with the same argument but different indices can be quickly computed with the aid of simple recursive formulas, as explained in ABRAMOWITZ and STEGUN (1964, pp.385386). There exists a differential equation for the image function f( a, p), from which it is easy to derive the Taylor expansion for the same function. In
7.3. THE SOMMERFELD HALFSPACE PROBLEM
239
fact, it can be easily checked that the reflection coefficient function satisfies the algebraic equation 01 0+1
R(o,q)=a
(q 
#+1)2 
(2 )2 o 1q 1
~+~
r + ~+~ )2,(7.236) 0 lql
=a(2
which inserted in the integral identity (7.233) gives rise to the following differential equation for the image function:
d2 (o?  l)d2!(0,P)  !(o,p) p
a1
b+(P)]. = a[!(l,p) + 1 a+
(7.237)
To define the solution uniquely, the following boundary conditions are needed: (7.238) !(0,0) = !(a,oo) = O. It is seen from the differential equation that the function f( a, p) is continuous at p = 0+ but its first derivative (with respect to p) experiences a step discontinuity (7.239) Taylor series expansion
Applying the differential equation (7.237), a Taylor series expansion can be written for the image function. The function /(1, p) has the known expansion
f( l p) ,
= _2 J2(P)U P
+
(p)
=_~
L...J n=O
(l)n(p/2)2n+l U ( ) '( + 2)'. +P , n.n
(7.240)
which can be substituted in the righthand side of the differential equation (7.237). Obviously, like !(1,p), !(o,p) is also an odd function ofp and can be written as a power series with unknown coefficients: (7.241) The coefficients can be solved by equating the factors of equal powers of p on each side of the differential equation (7.237), which gives us a recursive formula. The second derivative gives rise to a delta discontinuity: (7.242)
240
CHAPTER 7. EXACT IMAGE THEORY
Equating the coefficients of the delta functions gives us first
(7.243)
Further coefficients are obtained from the recursive formula
A
_ n+l 
a
An
+
a2 1
+ I)! + 2)!n!'
(1)n(2n
a 2 1 22n+l(n
(7.244)
whose solutions start with
A __ a(a2 + 4a + 5)
A _ a(a+3) 1 
4(a+ 1)3'
A _ a(50 3 3 
8(a
2 
+ 1)4
+ 250 2 + 470 + 35) 64(0 + 1)5 .
'
(7.245)
Asymptotic expansion To write an expansion for the image function f( 0, p) valid for large p values, we must first take out the term corresponding to the pole of the reflection coefficient. In fact, writing (7.226) in the form
_ _ 1_(f.q #+1)2 R(a, q)  2 1 f.
1
q2  f21
(7.246)
shows us that there are poles at 1
qp ±Vf2=1' 
(7.247)
which correspond to exponential image functions. With the convention ~{Vf2=1} > 0 the pole qp = 1/ Vf2=11eads to a converging image, while
7.3. THE SOMMERFELD HALFSPACE PROBLEM
241
the other one can be shown to be cancelled by a zero of the numerator.
0.3rrr~r~.,
+++._+~
0.2t    +  f    \   t       t       +  
0.1 tfft~__t_~_t+__f_~
0.05H~+~t·~=::::t+~
10
5
15
20
25
30
Fig. 7.4 Examples of the image function !(a,p) (absolute value) for different complex values of the parameter
Q.
Extracting the pole leaves us with an expression for the reflection coefficient function that has branch points at qb = ±j. The image function can be obtained by applying the Mellin inverse and Watson's lemma as was done by LINDELL and ALANEN (1984, part II) to produce an expression valid for large p values:
p/~ 4 . ( 1r') f( ) ~  (a 2 2a2 a,p _1)3/2 e + /'iiap3 / 2 sm P+"4 
3 3(3a  8) ( 1f)  . . . . . . . ....cos P + 3
2/'iia p5 / 2
105(91a6
4

15{23a4
472a4 + 896a 2 . IiC 256y 21fo? p9/2 

144a2
+ 128) SIn . ( 1f) p+ +
32V2ia5 p7 / 2

512)
( 1f) cos P +  ... 4
4
(7.248)
CHAPTER 7. EXACT IMAGE THEORY
242
Fig. 7.5 Approximate values of the image function f(a:,p) for a: = 10, as calculated from the 3 term Taylor expansion and 3 term asymptotic expansion, compared with the 15 term Bessel expansion, which may be considered as exact for p < 30. In Fig. 7.4, values for the function 1(0'., p) are seen for different values of the parameter a. For real a the function is real, for complex a, only absolute values are given. It is seen that for large 10'.1, the oscillations in the function are reduced and the exponential term in (7.248) is dominant.
TE and TM image functions The relation between the image function f(a,p) and the functions fTE((), fTM (() defined in the Section 7.1 can be found if the integral expression (7.233) is compared with (7.69), (7.70). In this case we have
H(() == (,
(7.249)
and the relation between the integration parameters p and ( is pq ==
{31P B
==
. a /:
JfJl~,
(7.250)
or (7.251)
243
7.3. THE SOMMERFELD HALFSPACE PROBLEM
O.02ryr:=yyr, O...+++~+__+~f"'o...~~'7"_i_~~"7'I
O.02Htf+O.04Ht+++~~++4
0.06 . .   +     +   1     + _ _  +  7 '   ..    + _ _     + _ _   _ _ _ _ 4 O.08~++~'f+++
0.1 t   +     +   #  +    , + _ _    + _    +      +        1
p
15
10
5
20
25
30
Fig. 7.6 Approximate values of the image function f(a,p) as calculated from the Bessel expansion with different numbers of terms. The accuracy of the Bessel series breaks down at approximately p = 2N, where N is the number of terms in the series.
Thus, we can write for the two image functions fTE and expressions fTE() = jBf(JL,.iB() + M6+() =
_~ fn (J.Ll)n + J..l2  1
n=l
fTM ( ) =
~~n t2

1 L..J
n=l
1
J..l
+
(
p.lfy ((), JL + 1 +
jBf(€,jB()  E6+()
(€  l)n J t
J 2n (j B( )
+1
2n (j B ( )
C
_
€ 
t
1
the
(7.252)
=
fy (().
+1 +
fTM
(7.253)
The third image function fo() can be obtained directly through the definition of R o {(31): (7.254)
(7.255)
244
CHAPTER 7. EXACT IMAGE THEORY
It is readily seen that the fo«() function is invariant in the duality transformations, while f( €, p) and f(J.L, p) transform to one another. It can also be verified, with some effort, that the differential equation (7.87) for foe(), valid for any layered structure, is really satisfied in the present Sommerfeld halfspace problem case. To prove this, we must invoke the differential equation (7.237):..The dyadic image operator fe' needed in the construction of the image source, can be written in any of the forms (7.90)(7.92), for example, according to (7.92) as
=fe(J\lt, . ()
= f
TM
«()uzu z + I
TE=
It
+ lo«() k12 u. u, . vv..
(7.256)
Making a partial integration in the field integral, it is possible to transfer z differentiation u, . \7 to ( differentiation, whence it is possible to write instead of (7.256) (7.257) which appears to be a very useful form, because it involves only one differentiation of the current source J. The derivative of the function fo«() can be found analytically.
Convergence Since the Bessel function In(p) is known to converge for Ipl + 00 only for real values of p, to obtain a converging image, the integration variable ( should be chosen so that the parameter p in (=L.= P jB jk1VJ.Lfl
(7.258)
is real. Thus it is wise to keep p as the integration parameter since p = 0... 00 defines the image line. When B is not imaginary, ( becomes complex, which means that the image line is defined on the complex ( plane by
z
= h 
p
(= h  jB'
(7.259)
The image source and the reflection field The dyadic image operator fe(j\lt, () is now known and the expression for the image source in the Sommerfeld halfspace problem can be expressed in terms of the image functions and the original source J (r) in the form (7.260)
7.3. THE SOMMERFELD HALFSPACE PROBLEM
245
whence the reflection field can be written as 00
Er(r)
= jk l 171
(1 + :? vV').J JG(rr' +Uz()Ji(r', v
()dV'd(. (7.261)
0
This is the general solution for the reflected field and includes all special cases of sources, medium parameters and field points. For example, a VED source (vertical electric dipole above a horizontal interface) J (r) = uzIL8(r  uzh) gives rise to a vertical image source: (7.262) Because the field is everywhere transverse magnetic, only the function fT M appears. On the other hand, a HED source (horizontal dipole) J(r) = vIL8(r  uzh) with v· u, = 0 gives rise to both horizontal and vertical image currents:
IL
Ji(r, () = v fTE (()I tsi» )o(z + h)  uzf~(() k2 lv V'o(p)]o(z+ h). (7.263) 1
The vertical component is a double line current, like a transmission line, because of the term v· V'6(p).
Limiting cases of the theory Let us consider some limiting cases to test these results. For e ~ 1 and J.L ~ 1 we obviously have E ~ 0, M.~ 0 and B ~ 0, whence fTM«() ~ 0, fTE«) ~ 0 and fo«) ~ O. Thus, the image source vanishes, which is in accord with the vanishing interface. Letting € ~ 00 and J.L = 1, we have fTE«() = jBf(l,jB(), which for oo B ~ k 2 ~ 00 acts like 6+ «) f(l, p)dp = 8+ «) in the field integral. Also, we have fTM «) ~ b+«) because the continuous term in (7.253) vanishes. Further, we have fo«() + 0 for the same re~on, whence th.:
fo
dyadic operator from (7.256) can be simply written as le(jV't' () ~
1
and the image source becomes Ji(r, () ~ Jc(r) = C . J(C . r). This is obviously the image in a PEe plane. For large but finite €, we can take the first term of the asymptotic expansion (7.248) to approximate fTM = 8+(()  jBj(€,jB()
_ o+(()
~ o+(() + (€22jB€2 e j B(!V,2 1 ~ _ 1)3/2 + 2jB e j B( / , . e
(7.264)
CHAPTER 7. EXACT IMAGE THEORY
246
This is similar to the impedancesurface function (7.163), if we identify Zs = VJl2/€2, which is a small quantity. In the same way, fTE and I; functions can be checked for the same result. This shows us clearly, that at the limit of large e the interface can be approximated by an inpedance surface. Finally, the reflectioncoefficient method (ReM) can be shown to arise as the far field limiting case of the EIT formalism. This was already done in general form for the image Green dyadic in Section 7.1 and is not repeated here.
Green functions The different Green functions defined in Section 7.1.4 can be expressed in terms of just one scalar Green function K (a, o, z) for the Sommerfeld halfspace problem. In fact, defining
J 00
K(a,p,z)
=
G(D)f(a,p)dp,
(7.265)
o with (7.266) we can write
K™(p,Z)
= K(€,p,z) 
KTE(p, z) = K(Jt, p, z)
EG(r),
(7.267)
+ MG(r),
(7.268)
(7.269) The Green function can be computed for different argument p, z values and stored for computer memory instead of working with the image functions. For large arguments we can approximate as in Section 7.1.4, which results in the limiting case of the Green function
K(a,p,z)
t
(
z
aI) .
G(r) R(a, JJU=l) +  r
JL€  1
a
+1
(7.270)
7.3. THE SOMMERFELD HALFSPACE PROBLEM
247
0.1 /" ........ ",
0.05 , I
~
I I
o
~j "'"\?"/4
u; cos(k1x)dx.
(7.272)
248
CHAPTER 7. EXACT IMAGE THEORY
Writing the total field in terms of the incident and reflected fields E = E, + E r , the impedance functional can be written as Z = Zi + Zr, where Zi is the impedance of the dipole in free space and Zr represents the effect of the ground. To compute Zr, E can be replaced by the expression (7.261) for E r in (7.272) in terms of the image source applying (7.257)
= fe(j\l t, () · Jc(r) =
Ji(r, () [fTE«()u:c1o COS(klX)
+ f~«()uz ~: sin(k 1 x)]8(y)8(z),
(7.273)
valid for [z] < >../4. (7.272) can thus be written as A/4
A/4
J JJ
z: = jk1'f/l
00
G(D) [fTE«() cos(k1x')cos(k1x)+
A/4 A/4 0
fi:() J JJ
sin(k1x') COS(k1X)] d(dxdx'
>/4 A/4
jk1'f/l
00
G(D)
A/4
[!€ fTM «() +
=
[f TE«( ) cos(k1x) cos(k1x')+
>/4 0
f €
1 8+«()] sin(k1x)sin(k1X')] d(dxdx'.
(7.274)
For the last expression we have made partial integrations and applied (7.87) for the expansion of f~' (() to obtain
:2 f:«() =  fo«() + fTM «() 
fTE«() =
1
_! fTM «() _ f
f f
1 8+«( ) _ fTE«().
(7.275)
At this point it is wise to change the integration variables from x, x' to and r = x  x', The distance D does not depend on the parameter ~ at all: D = J(x  x')2 + (2h + ()2 = Jr 2 + (2h + ()2. (7.276)
e = x + x'
Writing 2 cos(k1x) cos(k1x') = COS(kl~)+cos(kl r) and 2 sin(k1x) sin(k1x') =  cos(k1e) + COS(k1T), the parameter e can be integrated out in (7.274). Denoting e = k1r we have 001r
Zr=jZ~JJ G(D) [fTE(SinO + (1rO)cosO)+ o
0
249
7.3. THE SOMMERFELD HALFSPACE PROBLEM
1
[ fTM f
e 1 + 6+]( 
sinO + (1r  0) cosO) ] dOd(.
f
(7.277)
Values for this integral are quite easily computed for different values of complex € and height h. The ( integration can also be performed separately and the final result expressed in terms of the image Green function K( f, 0) as defined in (7.265) and in LINDELLet al. (1985):
z. = j2k~1
J 1r
+ (11" 
[K(l,O)(sinO
0) cos 0)+
o 1 [K(f,O) €
€  1 + G(Do(O))](sinO + (1r f
0) cosO) ] dO,
J
(7.278)
00
K(f,O) =
G(D)j(f,p)dp,
(7.279)
o
klD = J02 klD o = k1luz(x  x')
+ (2k l h + k1( )2,
+ u z 2hl =
J02
+ (2k 1 h )2.
(7.280)
For large k1h values, (7.278) can be shown to give the limiting value j'f/l
V€ 
1
Z +e r 21rk1hV€+1
j2k 1 h
(7.281)
'
which can also be directly obtained by applying the reflectioncoefficient method. 7.3.4
Transmission coefficients
Considering the problem of wave transmission through a planar interface between two isotropic media 1 and 2, the transmission coefficients can be written in the form (7.282)
TTE
= 1 + RT E =
2p,{J1 Jt/31
T. = o
kr (TTM _ TTE) = K2 kr (R™
K2
(7.283)
+ /32 '
_ RT E ) = R
o·
(7.284)
CHAPTER 7. EXACT IMAGE THEORY
250
The transmission image s~urce can be written in terms of the dyadic transmission image function he (7.115), which can be found if the transmission dyadic is first written in the form
(7.285) and then in not exactly the same but the equivalent form in the sense that it produces the same transmitted field:
(7.286) where, denoting q
= {31/B
T«
and applying the definition (7.226), we have
= fT TM = €f32€{32 f3 = f[l 1 + 2
Ri«, q)],
(7.287) (7.288)
(7.289) Note, again, the difference between To and To. Only the latter is needed here. The expressions (7.287)(7.289) will be expanded in terms of powers of the reflection parameter r as given above, to find the transmission image functions needed for the construction of the transmission image source.
7.3.5
Transmission image functions
The transmission image source corresponding to a volume source J(r /) was shown in Section 7.1 to be of the form
(7.290) giving the total transmission field in the form
JJ 00
ET(r)
= jk2 172
G(D) · Ji(r', ()dV'd(,
(7.291)
o v
D
= V(p 
p')2 + (z  H(z', ())2.
(7.292)
7.3. THE SOMMERFELD HALFSPACE PROBLEM
251
The dyadic transmission image function he(K, h, () and the function H(z, () are defined through the integral expression
T
j f31h e e
00
Jh (K e
"h
()e j f32 H ( h "
) d ~, l"
(7.293)
o
with the definitions (7.115)(7.118), written again for convenience,
(7.294)
J J J 00
T"e i 13t h
=
hu(h, ()e i 132H{h,Od(,
(7.295)
h[(h, ()e i 132H{h'()d(,
(7.296)
ho(h, ()e i{32H{h,Od(.
(7.297)
o
00
T[e i 13t h =
o
00
Toeil3th =
o Because of the exponential factor exp(j{31h), an integral identity of a more general type than that applied for the reflection problem must be found. As such we adapt an identity which can be found, for example, in the reference GRADSHTEYN and RYZHIK (1980, equation 6.646), and through a change of variables can be written in the form
(7.298) where ( is the integration parameter, r = ({31  (32)/«(31 parameter and H denotes the function H(h, ():
,
d
H = d(H
()
n.;
+ ..82) the reflection (
= H(h,()
(7.299)
Let us now define a transmission image function F( a, h, () through the integral expression
(7.300)
CHAPTER 7. EXACT IMAGE THEORY
252
From the expansion (7.231) applied in (7.298) and (7.300), the transmission image function can be written explicitly in the form 20'
F(O', h, () == Jo(B()0'+1
0
24: 1
~ (  : : ~) (~: ~) n
n
J2n (B() .
(7.301 )
By making partial integration in (7.300), we can write
J(0~16+(()+F'(0,h,(») 00
e
j 2H {3 d(
.
(7.302)
o
In these expressions, the prime denotes differentiation with respect to the parameter (. (7.302) can be applied directly to find the image functions corresponding to the transmission coefficients T u and T] defined by (7.287), (7.288):
hu(h,()
= ~6+(() + F'(f,h,(), €+1
h/(h,() == _2_ 6+(() + ~F'(/L,h,(). J.L + 1 J.L
(7.303) (7.304)
The third coefficient To defined by (7.289) corresponds to the image function ho which can be written with the aid of (7.300) as follows:
ho(h,()
= jH'(h, ()~[F(/L,h, () jtf
F(f, h, ()].
(7.305)
Thus, substituting in (7.294), the transmission image source for the general current source J (r) is finally obtained in the form
2 ( b+() J.L + 1 I
J.l€  1
1 (J.L, z, () + F I
J.L
)=ItJ + t
H (z, ()[F(J.L, z, ()  F(€, z, ()]u z(\7 t . J). J.l€
(7.306)
7.3. THE SOMMERFELD HALFSPACE PROBLEM
253
The transmission field can now be computed from the expression
JJ oo
ET(r)
= jk2 Tf2
G(D) Ji(r',()dV'd(, 0
(7.307)
o v
D= V(pp')2+(zH(z',())2.
(7.308)
It is noted that, as for the reflection case, a VED source gives rise to a vertical (u, directed) image source while a horizontal source brings about both horizontal and vertical image components, unless the horizontal source is solenoidal (V' . J = 0), in which case the vertical component vanishes. Corresponding expressions for a magnetic source can be directly written applying the duality transformation. Asymptotic expression
Similarly to the reflection field case, we can write a simple asymptotic far field expression for the transmission field starting from the approximation
D r
~r
+ u , . [p' + uzH(z', ()] =
+ r' sin fJ sin fJ' cos(
,
tanhn(O  4»
(7.365)
k=O and it obeys the following rules
= nC~(x),
kC;:(x) C~(x)
= (_x)k+l
n, k
 (_x)kl,
> 0,
(7.366)
k > 0,
(7.367)
C[;(x) = (_x)n,
(7.368)
Cf(x) = n[(_x)n+l  (_x)nl].
(7.369)
With this function we can write the reflection coefficient function in the following form 00
R™
=
00
00
L C~(E)rk + L L [C;+l(E)  C~l(E)] r ke {32(2nd). j
k=O
n=lk=O
(7.370) Now applying the integral identity (7.352) we finally have an expression for the TM image function: 00
fTM «()
= C~(E)6+«() + L
[C:+1(E)  C: 1 (E )] 6+«(  2nd)+
n=O
00
00
00
L Cl(E)Fk(O, o + L L k=l
n=lk=l
[Ck+1(E)  Ck1(E)] Fk(2nd, ().
(7.371)
CHAPTER 7. EXACT IMAGE THEORY
266
The image function fo The image function fo«() needed in the EIT method can be constructed much in the same way but with somewhat more complicated expressions. Writing again for the reflection coefficient R; a Taylor series expression
L 00
L 00
(rTEe2iI32d)m
m=O
f:
(rTMe2iI32d)n =
n=O
1
f
[e2im,82d _ e 2i(m+2),82 d]
x
m=O
m
L {rTE)mi [E{r™)i + (r™)i+l] ,
(7.372)
i=O
we may again apply the identity if the reflection coefficient r™ is expressed in terms of the parameter r. The resulting image function is
fo(()
= f+1 fffD~(E)X f.
m=O k=l i=O
[Fm +k i (2m d, ()  Fm +k i(2(m
+ 2)d, ()],
(7.373)
D~(E)
(7.374)
with
Di(E)
= Ect(E) + ct+ 1(E),
k
> 0,
= o.
These formulas are not too difficult to use in practice although the modified Bessel functions In(x) are diverging, because the sum functions can be seen to converge, except for simple exponentially diverging parts, which can be extracted and handled otherwise. In Fig. 13, examples of the
7.4. MICROSTRIP GEOMETRY
267
image functions are given for some parameter values.
£=3.0 ··..········£=2.2
N
o
d
= o.ix,
_·_··£=1.5
......
_._ ....~':::':"': __ 1.5
2.0
2f············3.0
".:::::::::•..•.:.::_
let'.
._""""""'"''''
4.0
3.5
1
. d I
CD
oI
Fig. 7.13 Normalized derivative of the image function 10 for the microstrip for different relative permittivities
£
of the substrate.
There is the connection (7.87) between the three image functions: (7.375)
which will save us from numerical differentiation of the fo function. The fTM(() and fo(() functions do not, however, converge as such. The lack of convergence is due to certain poles of the reflection coefficient, which correspond to propagating waveguide modes of the grounded slab waveguide, a situation similar to that of the impedance surface. Their contribution can, however, be extracted and made to converge by reorientating their location on the complex plane.
7.4.2
Fields at the interface
The image dyadic operator can be written in the many ways discussed in Section 7.1, for example, as =
.
le(JV't, ()
= I TE (()/= + fo(()uzu z · (/= + k12 VV),
(7.376)
1
which is perhaps the most preferable because it contains the simple function ITE. The reflected fields in the halfspace z > 0 due to the image sources can be written as
JJ 00
Er(r)
= jwllo
G(D)· Ji(r', ()dV'd(,
v
0
(7.377)
CHAPTER 7. EXACT IN/AGE THEORY
268
with
D == D(r, r', () == J(p  p')2 + (z 
Z, 
()2,
~{D} ::;
o.
(7.378)
Surface image sources Restricting the original current source on the planar interface z == 0, J (r) == J s (p)8(z), the transversal component of the electric field due to the image source represents the reaction of the grounded slab to the original field and is of importance for microstrip circuit analysis. This field added to the field due to the original source in free space gives the total transversal field, which is required for formulation of the integral equation of the original source problem. The formulation is made simpler by explicitly expressing the surface charge density
es
\7. J s
==   . JW
(7.379)
in the image source and field expressions (7.64), (7.377)
Ji(r, ()
= !TE(()J s(p)t5(z) 
u, {~ !o(()l!s(p)t5'(z),
(7.380)
1
l!i(r, () = !TE(()l!s(p)t5(z) + :2!O(()l!s(p)t5I1(z),
(7.381 )
1
!!
00
Er(r) = jwlLo
v
G(D)Ji(r',()dV'd(+
0
00
f~! !l\7'G(D))l!i(r', ()dV'd(.
(7.382)
v 0
Here, the integration volume V is any volume containing the original source surface S. Note that the double differentiation within the Green dyadic of (7.377) is reduced through partial integration in (7.382) so that only one differentiation of G(D) is present. This is necessary in reducing the singularity for the practical computation of the integrals. When taking the integration volume V large enough so that the original sources are zero on its boundary, we may perform further partial integrations with respect to z and ( variables so that the differentiations can be transferred from the delta functions to the functions of (. In this case, the equivalence 8/8z == a/a( can be applied when operating on the source function of both ( and r inside the field integral (7.382),
7.4. MICROSTRIP GEOMETRY
269
allowing us to replace (7.380) and (7.381) by the following equivalent surface image sources
Jsi(p, () = fTE(()Js(p)  u, {~ f~(()es(p),
(7.383)
1
(7.384)
The transverse electric field When calculating only the transverse component e of the electric field E, the u, directed terms in the expressions of J si can be omitted. The z integration in (7.382) can now be easily performed because of the delta functions and the transverse reflection field at the interface z = 0 reads 00
er(p)
= jwJlo j S
j G(D)Jsi(p',()dS'd(+ 0
00
€~ j s
j['\1'G(D)]ei(p',()dS'd(,
(7.385)
0
with (7.386) For the double differentiation of the fo(() function needed in the above expressions, an analytic form is desirable, to save us from numerical differentiation in the actual calculation
with a new function defined as
Qn(x, ()
= F n +1 (x, () + 2Fn (x, () + F n  1 (x, ().
(7.388)
CHAPTER 7. EXACT IMAGE THEORY
270
Denoting Do = Ip  p'l, the transverse total field in the same plane as the surface currents can also be written by defining twodimensional Green functions:
e(p) = jWJ.Lo j[G(D o) + KTE(Do)]Js(p')dS'+ s
~ f. o
j V'[G(Do) + L(Do)]es(p')dS'.
(7.389)
s
It is seen that the field can be expressed in terms of the freespace Green function G for the direct field and two new Green functions KTE and L for the reflected field. These Green functions can be calculated in terms of the image functions as follows: 00
KTE(D o) = j G(D)fTE«()d(
= G(Dt}+
o
(7.390)
J 00
L(Do) =
G(D) [JTE«() +
:rf~'«()] d( = KTE(Do) 
EG(Do)+
o
00
j G(D)[QmHi(2md, ()  Qm+ki(2(m + 2)d, ()]d(,
(7.391)
o
with (7.392)
Note that the definition of the Green function L can be expressed as K T M K; in earlier notation. In solving integral equations for the surface current problem it is not necessary to deal with the image sources at all if we directly compute the Green functions defined above and store them in the computer memory for further use.
7.4. MICROSTRIP GEOMETRY
7.4.3
271
The guided modes
As shown by LINDELLet ale (1987), the image function fo«() does not converge as ( approaches infinity owing to certain poles of the reflection coefficient. To obtain convergence, it is necessary to extract a finite number of terms, responsible for the loss of convergence, and treat them separately. Each of these terms corresponds to a propagating mode guided by the grounded slab. For a sufficiently small thickness d and/or if e is close to 1, there is only one such mode. This case is of particular interest to microstrip antenna design, because the guided mode increases coupling between the antenna elements, which is not generally what is wanted. If only one guided mode propagates, we can write the image function fo«() in two parts (7.393) with foe converging and fod exponentially diverging
(7.394) Here, a is the basic solution of the modal equation: (7.395) The amplitude coefficient A can be obtained from the residue of the reflection coefficient R o :
A_
j2€k~ad2[(€
 (t  1)[(k1d)2
 1)(k1d)2  (ad?]
+ (ad)2][(t + ad)(k1d)2 + (€ + 1)(ad)3]'
(7.396)
Also, the corresponding Green function L(D o ) can be calculated in two parts: (7.397) with
:r J 00
Ld(D o ) =
J 00
G(D)f:d(()d( = ja::
o
G(D)eOl(d(.
(7.398)
0
Because this integral does not converge on the negative z axis, we must change the integration path in an imaginary direction by writing ( = j(':
a:: J 00
Ld(D o ) =
G(D')e jOl('
o
«.
(7.399)
CHAPTER 7. EXACT IMAGE THEORY
272
(7.400) Although the exponential term now oscillates and does not converge, the integral converges because the argument of the Green function exponent becomes imaginary when (' passes the value Ip  p/l. Thus, the determination of the Green function L(Dd) is defined in two converging integrations.
7.4.4
Properties of the Green functions
Singularities
Finally, the nature of the singularity of the new Green functions in the transversal field integrals can be questioned. Applying the Green dyadic expression, the doublenabla term is seen to be too singular to give the field within the surface current source itself, even with the definition of principal value integral, which is applied in volume current integrals by YAGHJIAN (1980). When one of the nablas is transferred by partial integration in front of the current function, the principal value can be defined. In fact, we may write for the limit of the field point r approaching the surface S from the normal direction n according to VAN BLADEL (1964)
lim j[VIG(D)]!(r/)dS I
r~s
s
= PV j[V"G(D)lf(r')dS' + !nf(r). (7.401) 2 s
Thus, the normal component of the field is discontinuous through the surface current. Considering only the transverse field component, we see that the principal value integral equals the limit when the field point approaches the surface. Thus, we may write for a point p on the surface S in the transversal field expression (7.385)
JJ ex>
e(p) = jWJLoPV
G(D)Jsi(p',()dS'd(+
s :oPV
! 5
0
ex>
!l\7I G(D)]ei(pl,()dS'd(.
(7.402)
0
For the Sommerfeld problem with a planar interface between two homogeneous half spaces and the present microstrip problem, the same singularity properties can be seen to apply. Thus, the freespace Green function G in (7.402) must be replaced by the correct Green functions, taking into account the properties of the half space z < o.
7.4. MICROSTRIP GEOMETRY
273
Computation of Green functions For numerical computation of the Green functions, it is wise first to extract their asymptotic far field limit expressions, leaving simpler difference functions for raw computation. The asymptotic forms of KTE(p), L(p) are obtained from in the limit p + 00 by approximating the distance function (7.403) whence we may write
J 00
KTE(p) =
J 00
G(D)fTE(()d(
~ G(p)
o
fTE(()d(.
(7.404)
0
From the definition of the reflection coefficients (7.348) we have
J 00
fTE(()d(
= R TE({31) I
= 1.
(7.405)
~l=O
o
Since G(p) is singular at p = 0 whereas KTE(p) is not, we can write (7.406) because (7.407) approaches the same limit p as D. The function ~K vanishes quickly and thus takes little storage space. Likewise, we can write for the second Green function L(p) 00
L(p)
~ KTE(p) + :~G(p) J f~«()d(.
(7.408)
o Applying (7.375), implying f~'
= kr(fTM 
fTE  fo),
(7.409)
gives us 00
L(p)
~ G(p) JU™(() 
fo«()]d( = G(p) [R™
 RoL,=o
= G(p).
o
(7.410)
CHAPTER 7. EXACT IMAGE THEORY
274
E; = 1.5 real part. d=O.1A
N
ci I
piA
•
cilt,~ryr,..r,....., 0.0 1.0 2.0 ....0 J.O
Fig. 7.14 Microstrip Green function KTE(p) for two values of the slab thickness d normalized by the freespace wave length.
The second Green function can now be written
L(p) == G(D 1 )
+ ~L(p),
(7.411 )
with rapidly vanishing term ~L(p). The waveguide mode term must be taken care of separately. For large p values we can write the limiting expression
(7.412)
In Fig. 7.14, examples of computed Green functions are given for some parameter e and d values. It is seen that the Green functions are unsensitive to the thickness d of the substrate at large distances p except for the waveguide term.
References ALANEN, E., LINDELL, I.V. and HUJANEN, A.T. (1986). Exact image method for field calculation in horizontally layered medium above a conducting ground plane. lEE Proceedings, 133H, (4),297304. BLADEL, J. VAN (1964). Electromagnetic fields. McGrawHill, New York.
7.5. ANISOTROPIC HALF SPACE
275
LINDELL, I.V., ALANEN, E. and HUJANEN, A.T. (1987). Exact image theory for the analysis of microstrip structures. Journal for Electromagnetic Waves and Applications, 1, (2), 95108. LINDELL, I.V .. NIKOSKINEN, K.I., ALANEN, E. and HUJANEN A.T. (1989). Microstrip antenna analysis through scalar Green functions. Annales des Telecommunications, 44, (910), 53342. YAGHJIAN, A.D. (1980). Electric dyadic Green's functions in the source region. Proceedings of the IEEE, 68, (2), 24863.
7.5
Anisotropic half space
To extend the image theory to problems with anisotropic media, let us consider the simple case of two homogeneous halfspaces of which one (1) is isotropic and the other one (2) uniaxially anisotropic with the axis normal to the planar interface. The dielectric dyadic of the medium 2 can thus be written as (7.413) In this special anisotropic case, the TE and TM fields do not couple to each other, which simplifies the analysis. As a practical problem involving this type of medium the sea ice may be mentioned, since it contains almost vertically elongated pockets of brine, which makes the medium approximately macroscopically anisotropic. The theory given in previous sections must now be modified since the parameter €2 is not a scalar. Let us very briefly consider the transmissionline analogy in the present uniaxially anisotropic case. From the Maxwell equations, the following Fouriertransformed vector equations for the transverse field components in medium 2 can be straightforwardly derived:
/
=
e2(K, z) = A 2 . (u z x h2)
 u, x
h~(K, z) = B 2 · e2 
1 + U z X j2m + Kj2e,
(7.414)
W€2z
j2e 
_l_(u z
X
K)j2m.
(7.415)
WJ.L2
Here, the dyadics A2 and B 2 are defined by (7.416) B 2 = jW€2t
+ _Juzuz~KK = j=g2 . Y2.
(7.417)
WJ.L2 All these dyadics obviously have the same eigenvectors K, U Z x K as before, whence they commute in the dot product. Defining Y 2 to be the
CHAPTER 7. EXACT IMAGE THEORY
276
twodimensional inverse of Z2, we can derive from the transmissionline equations the secondorder equation for the transverse electric field:
with the propagation dyadic ~2 satisfying
({3fM)2 ~~ + ({3fE)2 U U12
KK
Z
.
(7.419)
The eigenvalues correspond to the two different propagation factors of the two eigenpolarizations TM and TE. They have the explicit expressions
{3fE = /k~t
 K2, (7.420)
with the two wave numbers defined as
k2 z
= WJJl2€2z,
(7.421)
The TE case appears simpler, because it corresponds to an isotropic medium with the parameters E2t, {to For the isotropic limiting case €2t + E2z + E2, these expressions obviously reduce to the earlier isotropic results. Finally, the impedance dyadic can be solved from
(
Z T M )2 KK
2
K2
+
~KK K2
(ZTE)2 UzU z 2
(7.422)
in terms of its eigenvalues ZTE _ 2

WJ.l2
f3'[E'
Z2T M
TM {32_ __ 
•
(7.423)
WE2t
The reflection image Let us now consider reflection from the planar interface of an isotropic medium with scalar parameters El, {t ~nd a uniaxial anisotropic medium with the dyadic parameters E2 = EEl, {tI. The relative permittivity dyadic
7.5. ANISOTROPIC HALF SPACE
277
is denoted by € = f.2/ fl = f z u, u, h. ftlt. If the source is in medium 1, the reflection dyadic can be written in terms of its TE and TM eigenvalues as
RT E R™
Z 2T E
a
ZTE

1
aTE
fJ 1 
fJ2
()
= Zr E + Zr E = fJ1 + fJrE '
_ Z 2T M
ZTM

1
_
{3TM 2.
7.424
a
 ftfJl
 Zr M + Zr M  fJrM + ftfJ1 .
(7.425)
The reflection image functions can now be identified by comparison with the isotropic problem of Section 7.3. Defining the integral transforms by
J 00
R(fJd =
f(()e if3t ( d( ,
(7.426)
o we are able to find the image functions corresponding to the reflection coefficients. The TE case is straightforward since the anisotropic medium acts just like an isotropic medium with €2 = €2t, which gives us the possibility of writing the image function in terms of the !(a,p) function defined in (7.234):
ts, = Jk~t
 kr·
The path of integration goes in the complex ( plane from 0 to the parameter Pt defined by Pt
= jB t (
(7.427) 00
so that (7.428)
is real and positive for fTE(() to converge. Also, writing the TM reflection coefficient in the form
R T M = _ y'f;..flifJ1 y'f;..fli{31 
J fJ~ + B'1 , J f3r + B;
(7.429)
the corresponding image function can again be written from analogy with isotropic media as fTM (()
= ~ ~ ~ o+(() €z€t
jBzf( .fi;€i,jBz().
(7.430)
The image corresponds to that of an isotropic medium with effective parameters f e = J€z€tfl' J.le = J€z/€tJ.ll' The path of integration must now follow another line such that the parameter (7.431)
CHAPTER 7. EXACT IMAGE THEORY
278
is real and positive. Thus, there is a difference compared with the isotropic medium case in that the images of TE and TM sources may take different paths in complex space. However, for lossless media both lines coincide. To find the image for the general source, the third reflection coefficient
R; should be expanded in a form from which the image function can be identified:
R
o
=
2k2 {3T E «r M {32 k 12 T M TEl 2 fJ2  €t 1 K2 (R  R ) = K2 (,BrM + ft,Bt}(,Br E + ,Bt).
(7.432)
After a few hours of algebra, the following form for R; can be found:
(7.433)
This expression requires some checks to be credible. For the limiting case f.t + f. z = f. the medium becomes an isotropic medium. In this case the expression (7.433) can be shown to reduce to
(?434)
which coincides with the corresponding expression of the isotropic Sommerfeld problem (7.255) given in Section 7.3. As another check we may take the limit (31 + 0, whence we can show that both (7.432) and (7.433) tend to the same value R; + 2. Finally, for K + 00, both (7.432) and (7.433) can be seen to give zero. The limit zero means that in the corresponding image function 10«() there
7.5. ANISOTROPIC HALF SPACE
279
are no delta singularities. 0.4 ..~,..___.,..._._____r__r___r____. 0.351+~~,~l;J~..:+1++++f
0.25 ~I++++++\+___.;lttPo~
144+1
0.21ll../I..J~+l~+\~\. ~I..lor+__+++l
0.15 Il~I+I'\ +\++I~++___+____f 0.1 ~I/++_\__+~_+_\If___+_~__++_._+___I
0 0
1
2
3
4
5
6
7
8
9
=
/Bzl for f2t 2fl, The argument is p = jk 1 ( . The delta function part is not shown.
Fig. 7.15 Examples of the normalized image function li™ J..L2
10
= J..Ll and different values of
fz
=
f2z/fl.
o
The image function fo{ associated with the reflection coefficient R; can now be written without problem, because the terms in (7.433) are of the same form as (7.429): fo«() = _8_. ft  1
{f: (v'V€i€t + l)n n=l
1
nJ2n (j B t (
)
+
(
References CLEMMOW, P.C. (1963). The theory of electromagnetic waves in a simple anisotropic medium. lEE Proceedings, 110, (1), 1016. LINDELL, I.V., SIHVOLA, A.H. and VIITANEN, A.J. (1990). Exact image theory for uniaxially anisotropic dielectric halfspace. Journal of Electromagnetic Waves and Applications, 4, (2), 12943.
Appendix A Problems The following set of problems covers a selection of topics discussed in Chapters 1  6. The symbols used in the problems are defined in the corresponding Chapters.
Complex vectors 1.1 Show through the corresponding timeharmonic vector that the com
plex vector a = al + ja2 with real vectors aI, a2 satisfying al ·a2 = 0, is in axial form, Le., the vectors al and a2 are on the two symmetry axes of the ellipse. 1.2 Show that the transformation a . eitPa, with real ¢, does not change the ellipse of the complex vector but moves the phase vector A(O) on the ellipse. 1.3 Show that the length of the polarization vector p(a) of a complex
vector a has the following geometrical properties (a) Ip(a)1 = sin '1/;, where 1/J is any angle of the equilateral quadrangle whose diagonals are the axes of the ellipse of a, (b) Ip(a)1
= 2A/1rlaI 2 , where A is the
area of the ellipse.
1.4 Study the polarization of the following complex vectors a in terms of
the polarization vector p(a): (a) a =
Ux
cos a
+ ju y sin o,
(a a real number)
(b) a = b + ju x b, (b a complex vector, u a real unit vector satisfying u · b = 0) (c) a = b x b*, (b a complex vector) 1.5 Show that p(a x p(a)) = p(a) when a is not a linearly polarized vector.
APPENDIX A PROBLEMS
282
1.6 Show that any complex vector a can be written as the projection of a circularly polarized vector b on the plane of a. Find the possible expressions for b.
1.7 Show that any complex vector a can be written as a = b and C are circularly polarized and [b] = [c].
+ c, where b
1.8 Find the most general complex vector b satisfying p(b) = p(a) when a is a given complex vector.
1.9 Show that if a· b = 0 and a b" = 0, one of the vectors a, b must be linearly polarized or zero. 1.10 Show that the reciprocal basis of the reciprocal basis equals the original basis of complex vectors aj , a2, a3.
1.11 Determine the reciprocal basis of at = a, a2 = a", a3 a is circularly polarized.
= a x a"
when
1.12 Show by expanding in the base a, a", ax a" that the solutions vectors b± to the equations a x b± = ±jb± are of the form
when a is not a circularly polarized vector. The coefficients Q:± may be arbitrary. 1.13 Study the relation between the real and imaginary parts of the complex vector k when it satisfies k . k = k~ with real ko .
Dyadics 2.1 Prove the following identity:
axI=Ixa 2.2 Prove the following identity: (a x
I) : (b x I) =
2a . b
2.3 Prove the following identity: (a x A) : (B x b) = a (A~B) . b
APPENDIX A PROBLEMS
283
2.4 Prove the following identity:
(a x
I) ~ (b x I) = ab + ba
2.5 Prove the following identity:
(A · a) x
(:4. b) =
fI(2) .
(a x b)
2.6 Show that det(A~A) = 8(detA)2. This implies that A~A is complete
only when A is complete. 2.7 Expand the inverse of the dyadic
A
Check that A · AI = 2.8 Study the solutions
Q,
I
= aI +a X I.
is really satisfied.
A of the following dyadic equation: A~A = aA,
when A is restricted to be a symmetric dyadic. 2.9 Solve the following dyadic equation for the dyadic X:
(aI + a x I)~X =
a x
I.
2.10 Defining the uniaxial dyadic as
D
= aft + (3uu,
write its CayleyHamilton equation and find the eigenvalues and eigenvectors. 2.11 Defining the gyrotropic dyadic as
G({3, R, 0) = (3Ull + Re Jo,
J = II X I,
where the dyadic exponential function is understood as
= = It + J() + ,11J 2 ()2 + ,J3()3 + ... 2. 3.
eJ 9
= It cos() + J sinO, It = I derive its eigenvalues and eigenvectors.
uu,
284
APPENDIX A PROBLEMS
2.12 Show that the two conditions ]iI
= RT ,
detR = 1
for a real dyadic R are sufficient to guarantee that if b = R· a, we have Ibl2 = lal 2 and [b x b*I 2 = [a x a*1 2 . These mean that in the transformation a + b = R · a the magnitude and polarization of the complex vector a do not change. Thus, the transformation only moves the ellipse to another position and can be interpreted as a rotation operation. 2.13 The gyrotropic dyadic can be defined by
G«(3, R, 0) =
(3Ull
+ ReJo.
Determine its square root dyadic satisfying the condition = = R, 0)]1/2 = =G(f3, R, 0). [0(,6, R, 0)]1/2 . [G«(3, 2.14 Show the following properties of the dyadic
A:
(a)
11: aa =
(b)
A: (ab  ba) = 0 for all vectors a, b implies A symmetric. A: aa" = 0 for all vectors a implies A = O.
(c)
0 for all vectors a implies A antisymmetric.
Field equations 3.1 Derive the Helmholtz dyadic operator H eC'J) for the biisotropic medium and show that it can be factorized in the form
What are the operators Hi (\7) and H2(V)? 3.2 An electromagnetic shield is comprised of three layers of media: two
dielectric layers of permittivity f1 and thickness t 1 and, in the middle, a third magnetic layer with permeability 1£2 and thickness t2. Determine the relation between these parameters so that there would be no reflection of a normally incident plane wave from the shield. The thicknesses are assumed to be very small and flt1, 1£2t2 finite. Hint: consider the input impedance of the equivalent network.
APPENDIX A PROBLEMS
285
3.3 In some frequency regions the bianisotropic medium can be approxi
mated by a lossless and nondispersive medium whose rnedium parameter dyadics are independent of frequency. Consider the expression of the energy density
w = "41 (E
E* ) , H) · M· ( H*
M=
(€(=p,e) .
and require that the energy density be positive (W > 0) for all possible fields E, H. Show that this leads to the condition that the following four dyadics connected to the material sixdyadic M must be positive definite:
3.4 What is the condition for the parameters corresponding to that above if we require a sharper condition W ~ W o , where
is the energy density in vacuum? 3.5 Consider the special case of a biisotropic medium of the previous problems 3.3 and 3.4. Derive the conditions of losslessness for the scalar medium parameters €, j.L, K, X without applying the result of the problem 3.3. 3.6 Find the conditions of losslessness for the impedance parameters Zl
and Z2 of a hiisotropic impedance surface with the impedance dyadic
when the boundary consition is
nxE=
Zs ·H.
Field transformations 4.1 Study the special duality transformation
APPENDIX A PROBLEMS
286
defined by the transformation matrix
T(o:)
1 = VI sin2a
(
2
f!
1 cos a
y'2sina'f]o ) 1 .
(a) Show that detT(a) = 1. (b) Show that T 1(a) = T(o + 1r).
(c) Derive the transformation_rul~sfor the medium parameters and show that the parameter transforms to itself for all a.
e.(
(d) Show that a reciprocal isotropic medium with parameters E and JL is in general transformed to a nonreciprocal biisotropic medium with ~d, (d =1= O. (e) Which a transforms =t1 to itself? What are the other transformed parameters? (f) Study whether a given nonreciprocal biisotropic medium can always be transformed to a reciprocal isotropic medium with Xd = (~d + (d)/2y!J.Lo€o = O. Find the dependence of the angle a on the parameters u, € and X in this case. 4.2 Applying a suitable affine transformation, solve the basic electrostatic
problem in an anisotropic dielectric: point charge Q at the origin Er is assumed symmetric, real and positive definite. r == 0 in a medium with the permittivity dyadic frEo. The dyadic
In particular, solve the scalar potential ¢( r) satisfying the Poisson equation
\7.
~r • \7¢(r)] =

Q 8(r)
Eo
together with the electric field E(r) = \l¢(r) and the flux density vector D(r) = €. E(r). The following identity may be of some help: 8(A · r)
=
8(rl, IdetAI
where A is a real dyadic. 4.3 Find the image charge for the previous problem when the original
charge lies at the point r = uzh, h > 0, and the previous anisotropic medium is bounded by a perfectly conducting plane at z == O.
287
APPENDIX A PROBLEMS
Electromagnetic field solutions 5.1 Find the twodimensional Green dyadic for the hiisotropic medium.
In particular, find the solution for H(\1) . G(p) = 8(p)I,
with the Helmholtz operator defined by H(\1) = (\1 x
I  jweI) · (\7 x I + jw(I) + k 2I.
Apply the symbols
and k± = k(cos'l9 ± ~r),
where
X
= nsin'l9,
~
= nn«,
n = JJ.Lr€r.
Note that the dyadics are not twodimensional, only the Green dyadic does not depend on z. 5.2 Find the delta singularity of the Green dyadic in the biisotropic medium by comparing its expression to that of the isotropic medium. Note that delta singularities arise from the scalar Green functions obeying 1/r law in the double differentiations. 5.3 Find the field in real space from a dipole in complex space. Assume a
dipole with the current density
J(r) = u yI L8(x )8(y)8(z  ja) where a » A is real. Find the field (amplitude and polarization) close to the z axis for large Izi values. 5.4 Show that any plane wave propagating in the general uniaxial anisotropic medium with the parameter dyadics
is either TE or TM to v, Le., either satisfies v . E = 0 or v . H = 0, unless the material satisfies the condition /Lt€v = €tj.,£v. Explain the special behavior occurring at this special material condition.
288
APPENDIX A PROBLEMS
5.5 Study the planewave propagation in a bianisotropic medium with the
parameter dyadics defined as
with This kind of a medium can be fabricated by taking similar righthanded and lefthanded helices and mixing them in a base medium so that N lefthanded helices are parallel to the z axis and 2N right handed helices are isotropically orthogonal to the z axis. In particular, find the wavenumber surfaces of the two plane waves. Study the optical axis directions in which the wave numbers are the same. What happens when the parameter K, approaches the value n = J tLr€r? (n is the refraction factor of a plane wave in isotropic medium with Kr == 0.) Also determine the eigenpolarizations for propagation along the z axis. Because of axial symmetry, write k == ukN(8), with k = wJ/i€, u is a unit vector which makes the angle () with the z direction. N(O) is the refraction factor of the plane wave propagating at the angle () in the present chiral medium.
Source equivalence 6.1 Find the equivalent electric source corresponding to a magnetic dipole,
J m == u z l mL6(r), in an isotropic chiral medium with parameters J.t and K, == K,rk/ko.
€,
6.2 Show that a radial source of the form J(r) == urf(r) does not radiate outside the support of the function f(r).
6.3 Find the equivalent magnetic source of an electric surfacecurrent source
6.4 Find the equivalent magnetic volume current of the coaxial current
Assume that the equivalent magnetic current is in the volume between the surface currents and that it is of the form Jm(r) == utpJm(p).
APPENDIX A PROBLEMS
289
6.5 Show that the approximation of a current sourceJ(r) by a dipole of moment P at a position r = a can also be done by minimizing the error function R(r) = J(r) P6(ra) in the following sense: require JR(r)dV = 0 and minimize the norm of JrR(r)dV. 6.6 Formulate the Huygens principle to electrostatic problems with charges and magnetic currents as sources. The space is assumed homogeneous and isotropic. Start from the equations
"V x E
= Jm ,
\l · D = g,
D
= fE.
Write the expression for the electric field in a volume V surrounded by the surface S with sources truncated in V and Huygens sources on S. Study the possibility to replace the magnetic Huygens current by equivalent electric charge on the surface S.
Appendix B Solutions The following set of solutions covers most of the problems given in Appendix A. The solutions are given in compact form to save space.
Complex vectors 1.1 The extremal values of the square of the corresponding timeharmonic vector are found from
~ [A(t) · A(t)] = ~[a~ cos2 wt + a~ sin 2 wt] = 0, which reduces to (a~  a~) sin wt cos wt = o. For a circularly polarized vector (a~ = a~) this is satisfied for all t, which means that it has discrete axes. Otherwise we have solutions t = nat12w. For n even the axis occurs at ±al, for n odd at ±a2. 1.2 The timeharmonic vector of aejc/J equals ~{aejw(t+c/J/w)}, which shifts
the time origin by /w but otherwise does not change the ellipse of the vector a. 1.3 Assume first a is not circularly polarized. Write b = aei 8 in axial representation, with 8 such that b· b = lal2 • Because p(a) = p(b), the axial vectors b re and b im satisfy
(a) From the geometry we have tan('l/J/2) = Ibimi/ibrel and 2 tan ('l/J/2)
Ip(a)1 = 1 + tan 2 (t/J/ 2)
.
= sin t/J.
APPENDIX B SOLUTIONS
292
(b) The area of the ellipse can be written as
which equals the required relation. These results are also valid in the limit when a is circularly polarized. 1.4 The polarization vector p(a) can be expanded in each of the cases as
(a) p(a) = uz2sino:coso: = u zsin20:. The axial ratio of the ellipse is e = tan 0:. It is seen that all polarizations are obtained because Ip(a)1 goes through all values between 0 and 1 when 0: is changed. Linear polarization (e = 0 or ±oo) corresponds to Q == n1r/2 and circular (e == ±1 to Q == n1r/4), n odd.
(b) Assuming b not linearly polarized, p(b) is parallel to u and we can write b x b*
+ (u x b)
x (u x b*)  jb x (u x b*)  jb* x (u x b) 2j(lbl2 + ju · b x b*)
p(b)  u u. p(b)
= 1
= u = p(a).
Thus, a is circularly polarized, which is easily checked: aa = O. This is valid in the limit also when b is linearly polarized. (c) Because (b x b*) x (b x b*)* = 0 and, hence, p(a) == 0, a is linearly polarized (or zero). 1.5 a· p(a) == a" · p(a) = 0 in all cases. Expanding
[axp(a)] x [axp(a)]* = p(a)[axa* ·p(a)] == jp(a)(a.a*)[p(a)·p(a)], [a x p(a)] · [a x p(a)]* == (a· a*)[p(a) · p(a)]
we have
[a x p(a)] x [a x p(a)]* j[a x p(a)] . [a x p(a)]*
=. JP
(a) (a. a*)[p(a) · p(a)] j(a· a*)[p(a) . p(a)]
= p(a).
1.6 If a is not linearly polarized, p(a) ::I 0 and we can define a unit vector n = p(a)/lp(a)1 normal totheplaneofa. Ifb = a+no:, from bb == 0 we can solve for 0: and have two possibilities: b == a±jnva:a. This
is also valid for the linearly polarized case when n is any unit vector in the plane orthogonal to a.
APPENDIX B SOLUTIONS
293
1.7 For a circularly polarized we can write b = e = a/2. Applying the
result of the previous problem, we can directly write b
= ~(a + jn..;a:i ),
because [b] =
lei, as can
c
= ~(a 
jn..;a:i ),
be readily checked.
1.9 Because a x (b x b") = bfa b")  b*(a. b) = 0, either b x b* = 0, whence b is linearly polarized, or a is a multiple of b x b", which is known to be a linearly polarized vector. 1.11 For a circularly polarized, p(a) = u is a real unit vector. Thus, a3 = jlaI 2p (a ) = jlal 2u and u x a = (a x a*) x a/jlal 2 = ja. The
reciprocal basis is
1.13 Writing k re + jk im, we have by separating the real and imaginary parts of k . k = k~,
Thus, the vectors k re and kim are orthogonal and Ikrel > Ikiml. The equation Ikre l2 lkiml2 = k~ is that of a hyperbola. In parametrized form, the most general solution for k is k
= ko(ucosh 8 + jv sinh 8),
where u and v are any two real orthogonal unit vectors.
Dyadics 2.1 Consider mapping of a vector b:
[a x
I  I x a] . b =
ax
(I. b)  I· (a x b) = a x b 
a xb
=0
Since the result is zero for any vector b, the dyadic in square brackets must be zero, which gives the identity.
294
APPENDIX B SOLUTIONS
2.2 Using the result of the previous problem, we can write
(a x 1) :(1 x b)
= L L[(a x Ui)Ui] : [Uj(Uj i
==
L L[(a x Ui) . Uj][Ui · (u, x b)] j
i
== a LL(Ui x Uj)(Ui X Uj)' b == a (I~I). b == i
x b)]
j
a (21). b == 2a· b.
j
2.3 It follows directly from
(a x cd) : (ef x b) == a (c x e)(d x f) . b
==
a·
(cd~ef) . b
2.4 Start from the expression
(a x uu)~(b x vv)
= [(a x u)
= [ba (u x v)
+ (a
x (b x v)](u x v)
x b) . uv](u x v)
= ba (uu~vv)  vv x uu· (a x b)
Replace uu and vv by (a x == 2ba
I:
I) ~ (b x I) == ba · (I~ I)  I x I . (a x b) 1 x (a x b)
= 2ba  ba + ab == ab
+ ba
2.5 Start from the expression (cd~ef)
· (a x b) = (c x e)[(d x f) . (a x b)]
== (c x e)[(d· a)(f· b)  (d· b)(f· a)]
= (cd. a) x (ef. b) Replace cd and ef by
(cd b) x (ef· a)
A:
(A~A) · (a x b) == 2(A. a) x (A. b). 2.8 One can immediately see that a = 0 and all linear dyadics A make one set of solutions. For a =1= 0, denoting B = 2A/a we have a new problem B(2) = B. Its solutions B =1= 0 must be complete dyadics. In fact, assuming B planar, B(2) must be linear, whence B is also
APPENDIX B SOLUTIONS
295
linear implying B == B(2) == 0, a contradiction. Since detB f.: 0 for a complete dyadic, taking the determinant gives us detB(2) = det 2 B == detB, or detB = 1. Hence, B(2) == B1 T == B. Because B is assumed symIEetric, it must be a square root of the unit dyadic, i.e., of the form I  2(abja· b) with a b # o. 2.10 The invariants are
whence the CayleyHamilton equation can be written as
Thus, the eigenvalues are Al == (3 and A2,3 == a. The eigenvector == u and the other two a2,3 are any vectors orthogonal to u.
a1
2.11 Because the dyadic G consists of a transverse part and a multiple of uu, it is obvious that one of the eigenvectors is u and the rest are orthogonal to u. The first eigenvalue is Al == (3 and a1 == u. Thus, we are left with the twodimensional eigenvalue problem
=
eJ O • a


A
= (1t cos 0 + J sin 0) · a = R a.
Because we have
(eJO)~(eJO) = 2uu,
spm(eJo) = 2,
the CayleyHamilton equation reads
and the solutions A f.: 0 are
A2,3
== Re±j(J. The eigenvectors satisfy
This equals u x a2,3 == ±ja2,3, whence the eigenvalues are circularly polarized (a2,3·a2,3 = 0), with a2 righthand and a3 lefthand rotation with respect to the u direction.
296
APPENDIX B SOLUTIONS
2.12 We have first =T= b b * =a·R ·R·a* =a·a*, v
whence the magnitude of the vector is not changed. Secondly, because of 
1
R
R(2)T
(2)T
==:=R
detR
T
=R,
we have
b x b"
= (R· a) x (R· a") = R(2) . (a x a") = R· (a x a"),
and thus
(bxb*).(bxb*)*
=
(axa*).RT.R.(axa*)*
= (axa*).(axa*)*,
which tells us that the form of the ellipse has not changed. 2.13 Since the dyadic consists of orthogonal axial and transverse parts, both of these can be treated separately. The squareroot of the axial part is obviously ±.Jljuu sin~ its square is {Juu. The transverse part
gives correspondingly ±J[ieJ 9 / 2 , whence the following four dyadics (double signs are not related)
are square roots of the gyrotropic dyadic. 2.14 For each case we apply the fact that if a· A· b
= 0 for all a,
b, then
A=O. (a) A: (a+b)(a+b) = 0 for all vectors a, b implies A : (ablba) = a· (A + AT) . b = 0, whence A = AT and A is antisymmetric. (b)
A: (ab 
(c)
= 0 and A : (a+ jb)(a+ jb)* = jA : (ab"  ba") = 0 imply A : ab" = 0 for all vectors a, b *,whence =A = O.
ba) = asymmetric.
A: (a+ b)(a+ b)*
(A 
AT) · b
= 0, whence A =
= A: (ab" + ba")
AT and A is
APPENDIX B SOLUTIONS
297
Field equations 3.1 The Helmholtz operator can be written as
JlH e = (\7 x I  jweI) · (\7 x I = (\7 x 1)2 + jw(  €)\7 x 1 
+ jw(I) W
2
W
2
Jl€I
(jl€  €()1
This is a polynom of \7 x 1 and can be written in the form
JlH 1(\7) · H 2(\7) = (\7 x
I + af) . (\7 x I + (3I).
= (\7 x 1)2 + (a + {j)\7 x I + a{jI The coefficients a and {3 can be solved by comparing the expressions and the result is
In the reciprocal chiral (Pasteur) medium with ( = the coefficients reduce to
e =
j"'JJ.to€o
3.2 From the 7r network with shunt admittances Y1 and series impedance Z2 with loading impedance 1Jo we have the input admittance
When this is equated with l/'TJo, we can solve for Z2 in the form
2'TJ;Y1 Z2 = 1 _ 2y;2· 'TJo 1 Substituting Z2 = jkot(jlr  l)'TJo, Y1 = jkot(€r  1)/"10, we have k ot2(JLr  1)
2kotl(fr  1)
= 1 + k~tH€r 
1)2'
3.3 Because of the lossless character, the medium sixdyadic M is hermitian, which implies that all its eigenvalues are real. The positive energy function requires that M is positive definite (PD), whence its
298
APPENDIX B SOLUTIONS
eigenvalues are positive. Also, its diagonal elements € and =p, must be PD. From matrix algebra we know that the inverse M1 must be PD, whence its diagonal dyadics are PD. Solving for the inverse dyadics by solving two vector equations, we arrive at the required condition that
€, ~, €
e·~1 .(,
Ii  ( .€1 .
e
must all be positive definite dyadics. 3.4 Same as above when € and Ii are replaced by €  €o! and Ii  /Lo!, respectively. However, because the sign > was replaced by ~, PD (positive definite) must be replaced by PSD (positive semidefinite). In fact, for the vacuum, the equality sign must apply. 3.5 Consider the energy expression of the problem 3.3. Setting H = 0 and E = 0 we conclude that € > 0 and JL > 0, respectively, corresponding to f ~ f o and /L ~ /Lo of the problem 3.4. Denoting a
= v'€E,
b = JjLH,
we can write the inequality W > 0 as
lal 2 + Ibl 2 ± 2IXrl~{a· b"} ± 21"'rl~{a· b*} > 0, which should be valid for all vectors a, b. Here we have denoted (Xr  j"'r)y'ii€ and ( = (Xr + j"'r)y'ii€. Choosing a = ejab, the inequality becomes
e=
IXrl cos a + I"'r\ sin a < 1, which is valid for all real angles a if 1, Xr2 +2 «;
0 and (€ €o) (JL  JLo) ~ 0, to the respective conditions €JL which coincide with the results derived with a change of symbols.
e.
e(
e(
APPENDIX B SOLUTIONS
299
3.6 The condition of losslessness is n ~{E x H*} = 0 for all possible fields on the surface. Evaluating n· E x H*
= H* . Zs' H = ZllHI 2 + Z2n· H x H*
= (Zl
+ jZ2n· p(H))IHI 2 = o.
The real part of the bracketed term must be null for any field H. Writing Zl = R I + jXI and Z2 = R2 + jX2, we have
Because the real vector p(H) can have any direction, we must have separately R 1 = 0 and X 2 = O. The lossless surface impedance dyadic is of the form Zs = jX1I + R2 n x I, where Xl and R 2 have real values.
Field transformations 4.1
(a) detT(a)
=
lsfn2a(l 2sinacoso:)
= 1.
(b) Inverting the T(a) matrix changes the signs of sin a and cos a in T(o}, which is tantamount to replacing a by a + it . (c) Denoting T =
e+ (, the transformation formulas are
~ 2 ~¥ (1  sin2a)€d = € + 22' cos a  2v2 cos o,
~o
(1  sin 2O:)~d =
(1  sin 20:)1'd =(1
~o
Ji + 2€1J~ sin 2 0: 
+ sin 20:)1' 
2J2T~0 sin o,
VZ€1Jo sin 0:
J2 7l
cos 0:
'fJo
(d) From the previous equations it is seen that if T = 0, in general we have Td =1= 0, which means that a reciprocal medium is transformed to a nonreciprocal medium. (e) =tid = =ti when sin 0: las read
= 0, in which case the transformation formu
300
APPENDIX B SOLUTIONS
lid
=
Ji,
ed  (d =
e(
The double sign corresponds to cos a == ±1. (f) Requiring in the biisotropic case Td == TdI = 0 and denoting T = XJP,of o, 11 = Vp,/f, we have the equation
.£ = yi217 1 sin a + 17 cos Q; /Ji€ 1 + sin 2a for the angle a. The righthand side can be seen to obtain continuously all values from 00 at a = 1r / 4 to +00 at a = 31r/ 4. Thus, there is a unique solution a in the interval 1r/ 4 < a < 31r /4 corresponding to any value of X and a transformation from any nonreciprocal biisotropic medium to a reciprocal biisotropic medium exists. 4.2 Writing \I . [fr . \I¢(r)] =
(11 . \I) . (11 . \I)¢(r),
where A == €r 1/2 is a real, symmetric and positive definite dyadic, we denote A · V = V', whence r' = A 1 . r. Writing also
't/J(r') = (r) == (A· r"), the Poisson equation becomes
V,2'l/J{r')
=  Q