remarks: in this report there are present errors and

Remarks: In this Report there are present errors and omissions. This document has been correct and now it constitutes a chapter of the book : “The Wiener-Hopf Method in Electromagnetics” by V.Daniele and R.Zich published by SciTech Publishing (www.scitechpub.com) in 2014

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6/16/2008 D:\Users\vdaniele\libri\WHnew\definitivo\chapter7_1.doc Chapter 7 Wiener-Hopf solution of discontinuity problems in plane- stratified regions 7.1 Introduction The study the electromagnetic propagation in stratified planar regions (fig.1) is very important since it provides all the fundamental concepts in the wave propagation. There are very excellent books devoted to this topics (Felsen and Marcuvitz,ch.5 and Brekhovskikh L.M. ,Waves in layered media, Academic Press, New York,1960). In this report we consider many aspects not considered in the cited books. For instance we introduce planar discontinuities in some sections of the stratified regions . An unified theory will be presented. It is based on the spatial Fourier transforms of the fields and the Wiener-Hopf formulation of all the considered problems

In this chapter, for illustrative purposes, we will consider only isotropic stratified isotropic media . The general formulation that involves arbitrary linear stratified media will be presented in the chapter 8. In presenza di discontinuita’ PEC o PCM o impedenze superficiali o di discontinuita’ scalari. e mezzi isotropi , tutti I casi considerati di incidenza obliqua si riducono alle due polarizzazioni E a H . (sezione5.1) 7.1.2 Spectral transmission line in homogeneous isotropic regions. Let’s consider time harmonic electromagnetic fields with a time dependence specified by the factor tje ω which is omitted. These fields are present in an isotropic homogeneous with permettivity oε and permeability oµ .

2

It is convenient to consider the layer indicated in the fig.1 as a piece of wave guide [Felsen-Marcuvitz, p. 183-207] having the axis y a with unbounded section in the transversal plane x z− For sake of simplicity we suppose that only the components zE , xH and yH are not vanishing and there are no variations of these fields in the z-direction:

( , )z zE E x y= , ( , )x xH H x y= and ( , )y yH H x y= . Since the guide presents an unbounded section, we are involved with a modal continuous spectrum [1] that can be represented through the Fourier Transform of the transversal field:

(7.1.1) ( , ) ( , ) j xzV y E x y e dxηη

∞

−∞= ∫

(7.1.2) ( , ) ( , ) j xxI y H x y e dxηη

∞

−∞= ∫

Applying Maxell equations yields the following transmission equations in the voltage

( , )V yη and the current ( , )I yη :

(7.1.3) ( , ) ( , )cd V y j Z I ydy

η τ η= −

(7.1.4) ( , ) ( , )cd I y j Y V ydy

η τ η= −

where being oo

o

Z µε

= and o ok ω µ ε= respectively the characteristic impedance and

the propagation constant of the medium, the propagation constant τ and the characteristic impedance cZ of the line are defined by: (7.1.5) 2 2( ) kτ τ η η= = −

3

(7.1.6) 2 2

1( )o oc c c

c o

k ZZ Z YZk

ωµ τητ ωµη

= = = = =−

7.1.3 Circuital considerations Solving the transmission line equations (3.3) and (3.4) in the layer bounded by the section 1y y= and 2y y= (fig.1) yields the equations: (7.1.7) 1 2 2cos( ) sin( )( )cV d V jZ d Iτ τ= + − (7.1.8) 1 2 2sin( ) cos( )( )cI jY d V d Iτ τ= + − where 2 1d y y= − , 1,2 1,2 1,2 1,2 1,2 1,2( ) ( , ), ( ) ( , ),V V V y I I I yα α α α= = = = ± It yields the two-port representation of the layer indicated in fig.2 where T is the transmission matrix of the two port:

(7.1.9) cos( ) sin( )

sin( ) cos( )c

c

d jZ dT

jY d dτ ττ τ

=

Knowing the transmission matrix T , circuit theory provides the following matrices that express the impedance matrix and the admittance matrix:

(7.1.10)

1cot( )sin( )

1 cot( )sin( )

c c

c c

j d Z jZd

ZjZ j d Z

d

ττ

ττ

− − = − −

(7.1.11) 1

1cot( )sin( )

1 cot( )sin( )

c c

c c

j d Y jYd

Y ZjY j d Y

d

ττ

ττ

−

− + = = + −

4

The reciprocity property of the two port provide simple circuit representation . For instance fig.3 illustrates the Π representation

The parameters of the T representation are given by:

1 2

3

tan ,2

1sin

c

c

dZ Z jZ

Z jZd

τ

τ

= = −

= −

To conclude this section we observe that if the length d of the layer is infinitely large the impedance seen by the initial section of the layer is the characteristic impedance of the two-port. Of course this impedance is coincident with the impedance cZ of the transmission line 7.1.3 Jump of voltage or current in a section where it is present a discontinuity. Let’s consider that in the section oy y= there is present a planar discontinuity (fig.4).

5

We can model this discontinuity with a current and a voltage generators given by:

(7.1.12) [ ]( ) ( , ) ( , ) ( , ) ( , ) j xo o x o x oA A I y I y H x y H x y e dxηη η η

∞

+ − + −−∞= = − = −∫

(7.1.13) [ ]( ) ( , ) ( , ) ( , ) ( , ) j xo o o o z o z oE E V y V y E x y E x y e dxηη η η

∞

+ − + −−∞= = − = −∫

7.1.5 Jump of voltage or current in a section where it is present a concentrated

source. The circuit description of concentrated source like electric or magnetic dipoles, electrical or magnetic line sources is accomplished in (Felsen-Marcuvitz, p. 205-207). In essence likely the planar discontinuities , the concentrated sources in the point

oy y= introduce in the circuit representation voltage and/or current generators located in the section oy y= . A general expression of the intensities of these generators will be provided in the section…… 7.1.6 Wiener-Hopf Equations in the Laplace domain For problems involving discontinuities such that half-planes that are located in the sections y const= it turn convenient to introduce Laplace transforms instead of Fourier transforms. . The Laplace transforms may present many advantages with respect the Fourier transforms. For instance they are analytical functions and not distributions. Because the presence of discontinuities in 0x = . We will consider the intervals 0x −−∞ < ≤ , 0 0x− +≤ ≤ and 0 x+ ≤ < +∞ . Let’s define:

(7.1.14) 0

( ) ( ) j xx e dxηη ψ−− −∞Ψ = ∫ ,

6

(7.1.15) 0

0 0( ) ( ) j xx e dxηη ψ+

−

Ψ = ∫

(7.1.16) 0( ) ( ) j xx e dxηη ψ

+

+∞

+Ψ = ∫

The integrals in eqs. (3.25)-(3.27) define analytical elements of the Laplace transforms. The analytical continuation provide the analytical functions that will be called respectively minus, finite and plus Laplace transforms. We will define the total Laplace transform as: (7.1.17) 0( ) ( ) ( ) ( )η η η η− +Ψ = Ψ +Ψ +Ψ If no concentrated physical sources are present in 0x = , delta functions are not involved and the entire function 0 ( )ηΨ is vanishing. Since possible discontinuities in the x-direction do not originate delta functions in 0x = , in the following we will suppose

0 ( ) 0ηΨ = : (7.1.18) [ ] [ ]( ) ( ) ( ) ( ) ( ) ( ) ( )L x u x L x u xη η η ψ ψ− +Ψ = Ψ +Ψ = − + where [ ]L stands Laplace transform and ( )u x is the step function:

(7.1.19) [ ]( ) ( ) s x

s jL x x e dx

ηψ ψ

∞ −

−∞ == ∫

Without some exceptional cases that will be discussed in….(stratified media) , the circuit considerations described in the previous section hold also for Laplace transform. However it is important to remember that the inverse Laplace transforms have the form:

(7.1.20) 1( ) ( )2

j x

Bx e dηψ η η

π −

−− −= Ψ∫

(7.1.21) 1( ) ( )2

j x

Bx e dηψ η η

π +

−+ += Ψ∫

where B− is an horizontal line located under the singularities of ( )η−Ψ and B− is an horizontal line located above the singularities of ( )η+Ψ . To put in evidence the difference between the Laplace and the Fourier transform let’s consider a field having a x dependence described by : (7.1.22) ( ) oj xf x e η−= Fourier transform requires real values of oη and it yields the following expression:

7

(7.1.23) 2 ( )oj x j xoe e dxη η π δ η η

∞ −

−∞= −∫

Conversely applying the Laplace transform we have that ( )F η+ is the analytical continuation of:

(7.1.24) 0

1oj x j x

o

e e dx jη η

η η∞ − =

−∫

whereas ( )F η− is the analytical continuation of:

(7.1.25) 0 1

oj x j x

o

e e dx jη η

η η−

−∞= −

−∫

It follows that the complete Laplace transform of a plane wave is vanishing:

(7.1.26) 1 1( ) ( ) 0o o oj x j x j x

o o

L e L e u x L e u x j jη η η

η η η η− − − = − + = − + = − −

It is important to observe that the Laplace transform does exist also for complex values of

oη . This allows us to introduce small losses in the media in order to avoid singularities on the real axis Im[ ] 0η = . Even though we have vanishing Laplace transform for the fields due to remote sources,

notice however that in the previous equation the term 1

o

jη η

−−

must be interpreted as a

minus functions so that, if Im[ ] 0oη < , the inverse transformation introduce a Bromwich

line B− such that 1 1 ( )2

oj xj x

Bo

j e d e u xηη ηπ η η−

−−− = −−∫ .

Conversely the term 1

o

jη η−

must be interpreted as a plus functions so that, if

Im[ ] 0oη < , the inverse transformation introduce a Bromwich line B+ such that 1 1 ( )

2oj xj x

Bo

j e d e u xηη ηπ η η+

−− =−∫ .

Consequently despite its vanishing values the inverse transformation reproduces exactly the field due to the remote source. The dependence oj xe η− is typical of remote source (incident plane wave, incident mode). Since the Laplace transform for remote source are vanishing , we can deduce the Wiener Hopf equations in the spectral domain by ignoring the presence of the remote sources. It will yield the homogeneous Wiener Hopf equations discussed in 1.2.5.

8

In the following we will use this technique that is alternative to the traditional ones described in [Noble, Mittra Lee] Even though the solution of the W-H equation is independent of the location of oη , the inverse Laplace transforms that provides the evaluation of the electromagnetic field in all the points the geometrical domain can pose some problem related to the choice of the integration line on the spectral plane η . This problem does not exists if we work with Fourier transform since in this case the integration line always is the real axis. We overcome this problem by observing that when the expression to invert contain plus or minus conventional W-H functions the integration line is the real axis. Conversely if for the presence of a pole oη a not conventional plus or minus functions is involved, in the inversion problem we must assume a Bromwich line, i.e. a line above the pole oη if is present a not conventional plus function or below the pole oη if is present a not conventional minus function. To validate this observation it is sufficient consider the inversion in the Fourier domain. For instance let’s consider the inversion of the expression: (7.1.27) ( ) ( ) ( )X H Fη η η+=

ˆ ( )( )o

FF ηηη η

++ =

− is plus conventional or not plus conventional if the pole

oη presents

negative or positive imaginary part. To introduce the Fourier domain where the integration line is the real axis we start from conventional plus function where

o ou jη ε= − ( 0ε > ) and introduce the following limiting process:

(7.1.28)0

1 1lim . . ( )( ) o

o o

PV j uu j uε

π δ ηη ε η→ +

= − −− − −

This equation allows to consider that the quantity:

(7.1.29) 1ˆ( ) ( ) ( ) . . ( )oo

N H F PV j uu

η η η π δ ηη+

= − − −

is defined in the Fourier domain. It yields the following inversion equation: (7.1.30)

1 1ˆ( ) ( ) ( ) ( )2

1 1ˆ ˆ. . ( ) ( ) ( ) ( )2 2

o

j xo

o

ju xj xo o

o

n x H F j u e du

jPV H F e d H u F u eu

η

η

η η π δ η ηπ η

η η ηπ η

∞ −+−∞

∞ −−+ +−∞

= − − = −

= −−

∫

∫ =

Now we have: (7.1.31)

9

2

1 1 1 1ˆ ˆ ˆ. . ( ) ( ) ( ) ( ) ( ) ( )2 2 2

oju xj x j xo o

o o

jPV H F e d H F e d H u F u eu u

η η

γη η η η η η

π η π η∞ −− −

+ + +−∞= +

− −∫ ∫ where 2γ is the “frown” real axis Consequently we get: (7.1.32)

2

( )1 1ˆ( ) ( )

2j x

o

n x

H F e du

η

γη η η

π η−

+

=

=−∫

Since the “frown” real axis 2γ permits to be substituted by the real axis when the the pole o ou jη ε= − ( 0ε > ) is located in the upper half-plane , the previous equation can be rewritten: (7.1.33)

( )1 1ˆ( ) ( )

2j x

o

n x

H F e dηη η ηπ η η

∞ −+−∞

=

=−∫ ( 0ε > )

This equation shows that in presence of conventional plus function the integration line remain the real axis also working with Laplace transform. Conversely if we are dealing with not conventional plus functions, since changing with continuity from positive to negative the values of ε , the integral must be continue, the “frown” real axis 2γ must maintains always above the pole oη so that when this pole is located in the upper half-plane is not possible to warp 2γ to the real axis. the “frown” real axis 2γ can be substituted by the Bromwich line B+ , i.e. a straight line above the pole oη We get: (7.1.34)

( )1 1ˆ( ) ( )

2j x

Bo

n x

H F e dηη η ηπ η η+

−+

=

=−∫

where the line B+ is assumed straight line above the pole oη . Similar considerations apply when are present conventional or not conventional minus function If we always wish work with integration lines constituted by the real axis, we first can study the problem with poles oη located in the conventional half-planes . To solve the problem with this limitation

10

is not restrictive since the presence of non conventional poles can be obtained successively by analytical continuation of the results obtained with conventional poles. Alternatively the analytical continuation process for not conventional poles can be avoid by rewriting the W-H equation in the form:

1( ) ( ) ( )G F Fη η η−− +=

By this way we privilege the minus function and the not conventional pole present in the plus function, becomes conventional for the minus function . Many times physical considerations suggest to introduce the techniques indicated before to force the real axis as the integration line in the inversion problem (see half-plane with impedance surfaces), however as it was showed before, it is not mandatory and with a careful choice of the integration path, we safely can deal with not conventional plus of minus functions.

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6/16/2008 D:\Users\vdaniele\libri\WHnew\definitivo\chapter7_2.doc

7.2 The PEC half-plane problem 7.2.1 E-polarization case

The planar discontinuity is constituted by a PEC half plane located in 0y = . The voltage

(7.2.1)

0

( ,0 ) ( ,0 ) ( ,0 ) ( ,0 )

( ,0 ) ( )

j x j xz z

j xz

V V V E x e dx E x e dx

E x e dx V

η η

η

η η

η

∞ ∞

+ − + −−∞ −∞

∞

+ +

= = = = =

=

∫ ∫∫

is continuous since ( , )zE x y is vanishing on the half plane 0, 0y x= ≤ . It means that the voltage generator indicated in fig.4 is vanishing. Moreover the vanishing of ( , )zE x y on the half plane implies also that ( )V V η+= is a plus function. Taking into account that the continuity of ( , )xE x y on the aperture 0, 0y x= ≥ implies:

(7.2.2) 0

( ,0 ) ( ,0 ) [ ( ,0 ) ( ,0 )]

[ ( ,0 ) ( ,0 )] ( )

j xx x

j xx x

A I I H x H x e dx

H x H x e dx A

η

η

η η

η

∞

+ − + −−∞

+ − −−∞

= − = − =

= − =

∫∫

the function ( )A A η−= is a minus function. According to the boundary condition on a PEC surface, ( )A A η−− = − represents the Fourier transform of the total physical current induced on both the faces of the PEC half plane.

12

Since the layers 0y ≥ and 0y ≤ are unbounded we have to consider the impedance

cZ in both the directions. It yields the circuit representation of fig. 5. Analysis of this circuit leads to equations:

(7.2.3) ||2c

c cZV Z Z A A= =

or taking into account that

2 2( ) kτ τ η η= = − 2 2

1( )o oc c c

c o

k ZZ Z YZk

ωµ τητ ωµη

= = = = =−

and ( ), ( )V V A Aη η+ −= = :

(7.2.4) 2 22

( ) ( )k

V Aη

η ηωµ + −

−=

The homogeneous W-H equation can be rewritten in the normal form of section 1.2.6 by taking into account the source constituted by the incident plane wave: (7.2.5) cos( )( , ) ojki

z oE x y E e ρ ϕ ϕ−= On the aperture 0, 0y x= ≥ or 0, xϕ ρ= = it provides the voltage:

(7.2.6) cos

0( ) ojki j x o

oo

EV E e e dx jρ ϕ ηηη η

∞

+ = =−∫

where coso okη ϕ= − Of course the voltage ( ) ( ) ( )s iV V Vη η η+ + += − is due to the current induced on the half plane so that it do not present the pole coso okη ϕ= − . Taking into account eq. (3.17) It follows that the Liouville characteristic part of ( )A A η−= in the pole coso okη ϕ= − is

2 22 o o

o

k Ejη

ωµ η η−

−

Notice that put cosk wη = − (sect.1.2.12.2), coso okη ϕ= − corresponds to o ow ϕ= − in

the w-plane. It yields the determination 2 2 sin sino o ok k w kη ϕ− = − = . Consequently we have:

(7.2.7)2 22 sin2

( )o o o o

o o o

k E Ej jZ

η ϕωµ η η η η−

=− −

13

and the normal form of the W-H equation (3.17) is :

(7.2.8)2 22 sin( ) ( ) 2

( )d o o

o o

k EV A jZ

η ϕη ηωµ η η+ −

−= +

−

where ( )dA η− is regular in the pole coso okη ϕ= − .

The source term sin2( )

p o o

o o

EA jZ

ϕη η− = −

in (3.21) can be obtained directly by considering that

it represents the primary contribution for the total current induced on the half-plane. Physical considerations yield to interpret this term as the geometrical optic current induced on the illuminated face ϕ π= (see fig.6): In fact the total current induced on the PEC half plane is the sum of the contribute of the remote source in presence of a full PEC plane in y=0 (i.e the primary field) plus the contribution of the equivalent magnetic current. Since the last contribute does not present any pole, the characteristic part of ( )A η− is the minus Fourier transform of the current induced from the primary field:

(7.2.9)

( ,0 ) ( ,0 ) ( ,0 )

M Primary field

d gx x xH x H x H x+ + += +

⇑ ⇑

Taking into account the reflected field on the plane 0y = :(see fig.6):

(7.2.10) cos( ) cos( ,0 ) 2 sin 2 sino ojk jk xg o ox o o

o o

E EH x e eZ Z

ρ π ϕ ϕϕ ϕ−+ = − = −

We get :

(7.2.11)0 sin( ,0 ) 2

( )p g j x o o

xo o

EA H x e dx jZ

η ϕη η− +−∞

= =−∫

with coso okη ϕ= − The solution of the W-H equation (3.21) is:

(7.2.12) sin1( ) o o

oo

kEV jk k

ϕηη ηη η+ =−− +

, 2 sin( ) o o

oo o

k EA jZ k

η ϕηη ηη−

+=

−+

14

Since the factorized functions can be chosen within a constant factor, we can assume

arbitrary the branch of ( ) kτ η η+ = − and put successively: 2 2k

kk

ηη

η−

+ =−

. In the

following, if not indicated otherwise, we will adopt as branch the one that satisfies the condition (0) kτ+ = for real values of k. In the w-plane (sect.1.2.12.2) we have:

(7.2.13) ( ) 2 cos2wk kτ η η+ = − = , ( ) 2 sin

2wk kτ η η− = + = −

7.2.1.1 Far field contribution. The solution of the W-H equations allows to evaluate the total field by using a suitable equivalence theorem.

For instance if we close the aperture with a PEC half plane located in y=0,x>0 , taking into account that the effect of the equivalent magnetic currents M are given by the solution of the transmission line equations (3.3) and (3.4) for an indefinite medium:

(7.2.14)

M ( )( , ) ( ) , I ( , ) y>0

( )( , ) ( ) , I ( , ) y<0

j y j yM

c

j y j yM M

c

VV y V e y eZVV y V e y e

Z

τ τ

τ τ

ηη η η

ηη η η

− −++

+ ++

= =

= = −

where the plus W-H unknown ( )V η+ is expressed by (3.25):

(7.2.15) sin1( ) o o

oo

kEV jk k

ϕηη ηη η+ =−− +

Consequently we obtain:

(1.3) (7.2.16) P M

( , ) ( , ) ( , ), I( , ) I ( , ) +I ( , )

P MV y V y V yy y y

η η η

η η η

= +

=

where according to the equivalence theorem ( , )PV yη and ( , )PI yη represent the primary field i.e the Laplace transform of the field due to the source in presence of PEC plane located in y=0. For sources constituted by plane waves for y>0 in the spectral domain

15

( , )PV yη and ( , )PI yη are vanishing. So it is more convenient to rephrase (1.3) in the natural domain. For instance for the first of (1.3): (7.2.17) ( , ) ( , ) ( , ) ( , ) i r M

z z z zE x y E x y E x y E x y= + + where ( , ) ( , )i r

z zE x y E x y+ is the primary field ( the incident plus the reflected wave). Notice that for y<0 we have ( , ) 0, ( , ) 0i r

z zE x y E x y= = The inverse Fourier Transform yields the following expression for the longitudinal field

( , )MzE x y :

(7.2.18) 2 2

sin( , )2 ( )

j k yj xM o oz B

o o

jkE e eE x y dk k

ηηϕ ηπ η η η η+

− −−

=− + −∫

where the line B+ a is straight line above the pole oη Algebraic manipulations allow to reduce this integral to special functions involving Fresnel integrals [Mittra-Lee}. However in different problems as such those considered in the next sections, it is not possible to obtain the inverse Fourier transforms in closed form. Approximate evaluation techniques must be indicated. In particular for the evaluation of far fields, the saddle point method (Felsen-Marcuvitz) is particularly indicated (see 1.2.12.2: Using the mapping indicated in 1.2.12.2:

(7.2.19) cossin

k wk w

ητ= −= −

and considering the cylindrical coordinates:

(7.2.20) cossin

xy

ρ ϕρ ϕ

==

yields the integral:

(7.2.21) cos( )ˆ ( )w

jk wc sr

I k f w e dwρ ϕ−= ∫

where :

16

(7.2.22)

sin sinˆ ( ) ( cos )sin2 2cos sin ( cos cos )

2 2

cos sin2 2

( cos cos )

o os

oo

oo

o

jE wf w f k w wwk w

wjE

k w

ϕϕπ ϕ

ϕ

π ϕ

= − = = − − − +

=− +

and wr is the image of the real axis in the w plane− (fig. 2.4). We can warp the contour path wr in (1.6) into the SDP. Since the eventual pole contribution provides the geometric optics contribution (see later) it means that the SDP contribution has the fundamental physical meaning of representing the diffracted far field. Taking into account the application in the sect. 1.2.12.2 ,we obtain the following contribution of the SDP (the contribution is obtained according to the versus of SDP indicated in 1.2.12. Observe that this versus is opposite with respect that of wr ). :

(7.2.23) ( )/ 4cos cos2 2 2 1

cos cos

oo

j kSDPz

o

EE j e k

kρ π

ϕ ϕ

ρπ ρ ϕ ϕ

− += − >>+

Eq. (1.25) cannot be used when the observation point approaches boundaries of the incident and the reflected waves. In this case the pole is near the saddle point and the approximation (1.19) does not hold longer. In this case we must use a more general technique indicated in (F-M) or (V-S) (see chapter 1.2.12.2) The previous equations provide the evaluation of the integral on the SDP. We can relate this evaluation to the original one on the path wr by warping the contour path wr in (1.6) into the SDP and take into account the eventual singularity contributions located in the region between the SDP and wr . Sometimes is preferable to study the warping directly in the planeη − . In this plane the SDP do not change if we change ϕ with ϕ− . The fig. illustrates the location of the SDP in the planeη − .

17

The versus of the SDP in the planeη − is in according to the versus of the SDP in the w-plane. With this versus we obtained for every value of ϕ , π ϕ π− < ≤ :

( )/ 4cos cos2 2 2 1

cos cos

oo

j kSDPz

o

EE j e k

kρ π

ϕ ϕ

ρπ ρ ϕ ϕ

− += − >>+

If cos oϕ is positive the pole oη lies in the third quadrant. However the Browmich line B+ always is located above oη . The fig. shows that by warping the line B+ on the SDP, the pole oη is always captured

clockwise for o so sk k

η ηη η= < = .

The warping yields the following result: (7.2.24)

2 2

sin( , ) ( , ) 2 Res ( )2 ( )

( , ) ( ) 0( )

( , ) ( ) 0

o

oo

j k yj xM SDP o oz z s o

o o

rj yj x z s o

o s o iz s o

jkE e eE x y E x y j uk k

E x y u yE e e u

E x y u y

ηη

η

τη

ϕπ η ηπ η η η η

η ηη η

η η

− −−

−−

+ = − − =

− + −

− − >= − =

− <

Where according to the correct sign indicated in (3.37) and the boundary conditions we assumed:

sin sino o ok w kτ ϕ= − = ,

( ,0) ( ,0) 0i rz zE x E x+ =

18

It yields:

( , ) ( , ) ( , ) ( , ) ( , ) ( , ) ( ) ( , ), y>0 i r M i r difz z z z z z o s zE x y E x y E x y E x y E x y E x y u E x yη η= + + = + − +

(7.2.25) ( , ) ( , ( , ) ( ) ( , ), y<0 M i dif

z z z s o zE x y E x y E x y u E x yη η= + = − + where:

( ) ( )/ 4cos cos2 2 2 1,

cos cos

oo j kdif SDP

z z oo

EE E j e k

kρ π

ϕ ϕ

ρ ϕ π ϕπ ρ ϕ ϕ

− += − = >> ≠ ± −+

The previous equations show that, according to the geometrical optics ,for 0y > , there is not the reflected wave when 0 oϕ π ϕ< < − . Conversely in the region 0y < there is not the incident wave when oπ ϕ ϕ π− < < − (deep shadow region) 7.2.2 The skew incidence Let’s consider an incident plane wave impinging at skew incidence angle β :

(7.2.26)cos( )o o oj j zi

z oE E e eτ ρ ϕ ϕ α− −= zjj

oiz

ooo eeHH αϕϕρτ −−= )cos(

where k is the propagation constant and coso kα β= , 2 2sino ok kτ β α= = − cos( ) cos sino o ox yρ ϕ ϕ ϕ ϕ− = + Using the Maxwell equations yields the following other components of incident plane wave:

cos( ) cos sino o oj j zi o o o o o o

xo

E Z H kE e eτ ρ ϕ ϕ α α ϕ ϕτ

− − +=

cos( ) sin coso o oj j zi o o o o o o

yo


− − −=

(7.2.27) cos( ) sin coso o oj j zi o o o o o o

xo

E kY HH e eτ ρ ϕ ϕ α ϕ α ϕτ

− − − +=


yo


− − +=

where oZ and 1

o oY Z −= are the impedance and the admittance of the medium

19

To deduce the Wiener-Hopf equations let’s consider the transverse fields (7.2.28) xzt ExEz ˆˆ +=E , xzt HxHz ˆˆ +=H and their Fourier tansforms:

(7.2.29)ˆ( , ) ( , , )

( , ) ( , , )

o

o

j z j xt

j z j xt

y e y x y z e dx

y e x y z e dx

α η

α η

η

η

∞

−∞

∞

−∞

= ×

=

∫∫

V E

I H

The function ( ) ( ,0)η η+ =V V is a plus function since ( ,0, ) 0t x z =E on the PEC half-plane 0y = . Besides we have (Daniele, Internal Report ELT-2006):

(7.2.30)( ) ( ,0 )( ) ( ,0 )

c

c

η ηη η

+ +

+ −

⋅ =⋅ = −

Y V IY V I

where, with the used order of the components of the field, the matrix admittance cY is given by: (Daniele Internal Report ELT-2006):

(7.2.31)2

12 22 2 2

O OoC

Oo

Ykk k

τ η αη α ηη α

− −= =

− −− −cY Z

and the matrix impedance cZ assumes the value:

(7.2.32)2 2

22 2 2oo

o Oo

kZk k

η η αη α τα η

−=

− −cZ

Summing the (7.2.30) yields the homogeneous W-H equation: (7.2.33) 2 ( ) ( )c η η+ −⋅ =Y V A where ( )η−A is the Fourier transform of the total current induces on the PEC half-plane: (7.2.34) ( ) ( ,0 ) ( ,0 )η η η− + −= −A I I The factor cos( )o oje τ ρ ϕ ϕ− evaluated in 0ϕ = induces in the plus function ( )η+V a pole arising from the integral:

20

(7.2.35) cos

0

1o oj x j x

o

e e dx jτ ϕ η

η η∞

=−∫

where:

coso o oη τ ϕ= −

Taking into account that: cos( )o o oj j zi

z oE E e eτ ρ ϕ ϕ α− −=

(7.2.36) cos( ) cos sino o oj j zi o o o o o o

xo


− − +=

we obtain that the residue oT of ( )η+V in this pole is known and it is given by:

(7.2.37) cos sinˆ ˆo o o o o oo o

o

E Z H kj E x zα ϕ ϕτ+

= −T

It induces a residue in ( )η−A given by (Et_Ht.nb): (7.2.38) 1 2Res[ ( )] 2 ( ) ,

o

tc o o o o oY T R R Rη ηη η− = = ⋅ = =A

where:

(7.2.39) 1 2o oR j H= − , 22 cos 2 sino o o

o o oo o o

j jkR H EZ

α ϕ ϕτ τ

= − +

Thus we get the following W-H equation in normal form:

(7.2.40) 2 ( ) ( )s oc

o

Rη ηη η+ −⋅ = +−

Y V A

with ( )s η−A conventional minus function regular in oη η− . The factorization of the kernel:

(7.2.41)2

2 22 2 2( ) 2 2 O Oo

Oo

YGkk k

τ η αη

η α ηη α

−= =

− −− −cY

can be accomplished in three several ways. For example we can rewrite:

21

(7.2.42)( )

( )

10 0

( )0 01

O

OO Oo

OO

O

kYGk kk

k

η ατ ητ τ

ηη αη ητ η

τ η

−−

= + −+ −

+

The factorization problem has been reduced to the factorization of the central matrix

( )

( )

1

1

O

O

O

O

k

k

η ατ η

η ατ η

−+

−−

. This matrix is rational and has a Daniele form so both the

techniques indicated in the sect. and .. , appl fattoriz_Yc.nb and we get straightforward factorization of ( )G η . Usually in the literature for physical interpretation reasons is preferred an alternative “weak” factorization [Seberst and Luneburg] that will be considered next. In fact the matrix

2

2 22 2 2

O Oo

Oo

Ykk k


−=

− −− −cY can be rewritten:

2

12

0( ) ( ) ( ) ( )

0o

w wkY t t

kη η η η

τ τ−

− += ⋅ ⋅ = ⋅c c cY Y Y

where 2 2Oτ τ η= − and ( ) o

o

tη α

ηα η

− =

We obtain the following weak factorized matrices:

(7.2.43)1 0( )( )

0o

woo

kY tk

ηητ ητ η

−

− = ⋅++cY

0 ( )( )0

ow

o o

kY tk

ηητ η τ η+ = ⋅− −cY

The offending poles arise from:

(7.2.44) 12 2

( )( ) a

o

tt ηηη α

− =+

22

where: ( ) oa

o

tη α

ηα η

= −

and they are: 1,2 ojη α= ± . The W-H equation can be rewritten: 12 ( ) ( ) ( ) ( )cw cwη η η η−

+ + − −⋅ =Y V Y A or :

(7.2.45) 1 1 12 ( ) ( ) ( ) ( ) ( ) ( )o ocw cw o cw cw o

o o

R R wη η η η η ηη η η η

− − −+ + − − − −⋅ − ⋅ = − ⋅ =

− −Y V Y Y A Y

Since both the member of the last equation are regular respectively in Im[ ] 0η ≤ and Im[ ] 0η ≥ , apparently it seems that the offensive poles 1,2 ojη α= ± does not produce any effect. However the asymptotic behavior of both the members of the equation shows that the entire vector is constant. To evaluate this constant we must eliminate offending poles present both in ( )η+V and ( )η−A . We have: (7.2.46)

12 2

0( )1( ) . ( )0

o a ocw o

oo oo

kY t Rwk

ηη ητ ηη α η ητ η

−− −

= ⋅ ⋅ + ⋅ ++ −+

A Y

To eliminate the apparent offending pole 1 ojη α= in ( )η+V , we must have :

(7.2.47) 11 1

0( )

0o

ocw o

o o j

k Rw c uη α

ητ η η η

−−

=

⋅ + ⋅ = − −

Y

where 1 1j

u = is the null space of ( )a ot jα

Similarly , to eliminate the apparent offending pole 2 ojη α= − in ( )η−A , we must have :

1

1

12 2

0( ) 2 ( )

0

0( ) . ( )0

2

o ocw o

o o

o a ocw o

oo oo

kY Rwk

kt RwYk

η ητ η η η

τ η η ητ ηη α η η

−

−+ −

−−

= ⋅ + ⋅ = − −

− = ⋅ + ⋅ −+ −

V Y

Y

23

(7.2.48) 12 2

0( )

0o

ocw o

o o j

k Rw c uη α

ητ η η η

−−

=−

⋅ + ⋅ = + −

Y

where 2 1j

u−

= is the null space of ( )a ot jα−

It follows:

11 1 1

1 10( ) 1 10

ocw o

o o

o o o o

jR k kw c u c

jj j

ηα η

τ α τ α

−−+ ⋅ = =

−− −

Y

(7.2.49)

12

1

( ) 1o

cw oo o

o o

jR kw c

jj

ηα η

τ α

−−

−− ⋅ =

+−

Y

The previous equations provide four equation in the unknowns 1

2

ww

w= , 1,2c . First we

obtain 1,2c by eliminating w:

11 2 2 2

1 12( )1 1

o ocw o

o o

o o o o

j jk k j Rc c

j j

αηα η

τ α τ α

−−

++ = − ⋅

+− − +

Y

We get:

(7.2.50)11

2 22

2( ) o ocw o

o o

c j RMc

αηα η

−

− = − ⋅ ⋅ +Y

where:

1 1

1 1

o o o o

j jk kM

j jτ α τ α

=−

− −

Known 1,2c we get w by one of the equations (7.2.49).

24

It is easy to show that by imposing the absence of the offending pole present in ( )η+V and ( )η−A it is equivalent to impose the absence of the poles 1,2 ojη α= ± both in

( )η+V and ( )η−A

25

6/16/2008 chapter7_3.doc

7.3 The planar waveguide problem 7.3.1 The E-polarization case By supposing the planar waveguide indicated in fig.6 constituted by PEC material we obtain the circuit representation indicated in fig.6 Writing the node equations yields directly the W-H equations of the problem:

(7.3.1) 1 3 1 3 2 1

3 1 2 3 2 2

node 1 ( )node 2 +( )

c

c

Y Y Y V Y V AY V Y Y Y V A

+ + −

+ + −

+ + − =− + + =

where: 2 2( ) kτ τ η η= = − 1c

c o

YZ

τωµ

= = , 1 2 31tan ,

2 sinc cdY Y jY Y jY

dτ

τ= = = − and

1 2 and -A A− −− are the Fourier transforms of the total currents induced on the two half planes.

26

Since 2 1Y Y= , by summing and subtracting system (3.26) decouples in the two scalar equations:

(7.3.2) 1 1 2 1 2

1 3 1 2 1 2

( ) ( ) ( 2 )( )

c

c

Y Y V V A AY Y Y V V A A

+ + − −

+ + − −

+ + = ++ + − = −

or considering the matrix kernel of the W-H equation (3.26) 1 1

2 2

( )V A

GV A

η + −

+ −

=i :

(7.3.3) 1 3 3 1

3 1 3 1 3

01 1 1 11( )0 21 1 1 12

c c

c c

Y Y Y Y Y YG

Y Y Y Y Y Y Yη

+ + − + = = − + + + +− −

i i

the factorization problem is reduced to the factorization of the diagonal matrix:

(7.3.4) 1 1

1 3 2

00 0cos( )

( )0 2 0

0sin( )

j a

cj a

c

eY Y ma

MY Y Y mej

a

τ

τ

τωµ τ

ητ

ωµ τ

+ = = = + + −

where:

(7.3.5) 1 1 1( ) ( )cos( )

j aem m ma

ττ η ηωµ τ − += = , 2 2 2( ) ( ) ( )

sin( )

j aem j m ma

ττη η ηωµ τ − += − =

According to the results of sect. 1.3.2.2 and 1.3.2.3 we have: A) factorization of 1( )m η

(7.3.6) 11( ) ( )cm mη ηωµ+ += , 1 ( ) ( )cm mη η− −=

where:

1

11 exp[ log ]( )

cos( ) 1 1c

nA

n

B a jk q A n BA A kmBka eA

ηγ

ηη τ τ ηη ηπα

ηα

∞

+=

− −− Γ − + + − + = Γ + −

∏

27

( ) ( )c cm mη η− += −

22 ( 1/ 2)

nnk

aπα − = −

, 1, , 1, 2,...

2A j B A n

aπ

= − = − =

2log 1a jq jkaπ γ

π = − + −

B) factorization of 2 ( )m η

(7.3.7) 2 21( ) ( ), ( ) ( )s sm j m m m

aη η η η

ωµ+ + − −= − =

where:

log

1

1

1( ) (1 )sin( ) sin( )

(1)

a jk

nj a

ans a

qj

aj ne

e ka am a ka je ea

τ τ ηπ

τ

ηγηπ

ηπηη ηη τ π

τ

− ∞

=

+−

+

−−

− = = Γ −

− Γ

∏

( ) ( )sin( )

j a

s sem ma

a

τ

η ηττ

− +

−

= = −

22

annkaπη = −

1, 2...., Im[ ] 0ann η= <

2log 1a jq jkaπ γ

π = − + −

To obtain the solution of 1 1

2 2

( )V A

GV A

η + −

+ −

=i , we must take into account the source

7.3.1.1 Source constituted by a wave plane

28

Similarly to the half-plane problem, the primary contribution is present only in the terms

1A − and it has the value:

0

1sin( ,0 ) 2

( )p g j x o o

xo o

EA H x e dx jZ

η ϕη η− +−∞

= =−∫

where coso okη ϕ= − It yields the non homogeneous W-H equation:

(7.3.8) 1 1

2 2

sin2 1( )0

o od

oo

EjV A ZGV A

ϕη

η η+ −

+ −

= +−

i

Alternatively putting 1 1 1 2 2 2, s i s iV V V V V V+ + + + + += + = + for the contribution

1 2 and i iV V+ + of the incident field ( , )izE x y on the two apertures 0, 0y x= ≥ and

, 0y d x= − ≥ , we have:

(7.3.9) 1 -i o

o

EV jη η+ = ,

sin

2 -

ojk di o

o

E eV jϕ

η η

−

+ =

Again the characteristic part of 1

2

AA

−

−

in the pole coso okη ϕ= − is identical to its primary

contribution :sin2 1

0

o o

oo

EjZϕ

η η−. In fact we have (campo inc_e prim.nb):

(7.3.10) 1

2

( )i

o i

VG

Vη +

+

=isin2 1

0

o o

oo

EjZϕ

η η−

A third way to evaluate the source term is to put:

1 1 1 2 2 2, d g d gV V V V V V+ + + + + += + = + where 1 2 and g gV V+ + represent the geometrical optic

contribution of the field on the two apertures. Supposing 2oπϕ > , since the aperture

, 0y d x= − ≥ is illuminated only in the region cot od xϕ− < < ∞ , it follows that 2 gV + differs from 2 iV + for the contribution

29

(7.3.11)cot cos

0( ) ( ) o

od jki g j x

oV V E e e dxϕ ρ ϕ ηη η

−

+ +− = ∫

However this contribution is regular in coso okη ϕ= − and consequently we have again

that the characteristic part of 1

2

AA

−

−

is : sin2 1 1

0

o o

o oo o

EjZ Rϕ

η η η η=

− −

Using the solution of the W-H equations obtained in sect.1 .5:

1 1( ) ( ) ( ) oo

o

RF G Gη η ηα η

− −+ + −= ⋅ ⋅

− , 1( ) ( ) ( ) o

oo

RF G Gα α αα α

−− − −=

−

yields the explicit solution: (7.3.12)

1 1 2 21

2

1 1 2 2

1 1( ) ( ) ( ) ( ) sin

1 1 ( )( ) ( ) ( ) ( )

o o o o

o o

o o

m m m mV jEZV

m m m m

η η η η ϕη η

η η η η

− + − ++

+

− + − +

+

=−

−

1 2

1 21

1 22

1 2

( ) ( )( ) ( ) sin( ) ( ) ( )( ) ( )

o o o o

o o

o o

m mm mA jEm m ZAm m

η ηη η ϕη η η ηη η

− −

− −−

− −−

− −

+

=−

−

7.3.1.2 Source constituted by an incident mode The modes in the planar waveguide are defined by the vanishing of cos aτ (present in

the denominator of 1( )m η ) and by the vanishing of sin aττ

(present in the denominator

of 2 ( ))m η . It is consistent with the equation:

(7.3.13) sin sin2 cos 0d a aτ τ ττ τ

= =

that yields:

(7.3.14) 2

2dn

nkdπη = −

1, 2...., Im[ ] 0ann η= <

30

The first mode occurs when n=1:

(7.3.15)2

21 1d k

dπη α = = −

For this mode we have:

(7.3.16) 1( , ) cos j xix oH x y H y e

dαπ −=

Using eq.(3.12), it yields:

(7.3.17) 1 11

s oHA A jη α− −= +−

, 2 21

s oHA A jη α− −= +−

So the W-H equation becomes:

(7.3.18) 1 1

2 12

1( )s

os

o

jHV AG

jHV Aη

η α+ −

+ −

= +−

i

It yields the solution:

(7.3.19) 1 1 11

2 1

1 1 1

1( ) ( )

1( ) ( )

om mV jHV

m m

α ηη α

α η

− ++

+

− +

=−

(7.3.20)

1

2 1 21

2 1 1

2 1 2

( )( ) ( )

( )( ) ( )

o

mm mA jH

A mm m

ηα η

η η αα η

− +−

−

− +

=−

Notice that in this case 1

2

VV

+

+

does not present the pole contribution in 1η α=

7.3. 2 The skew plane wave case Let’s consider an incident plane wave impinging at skew incidence angle β :

31



oiz


where coso kα β= , 2 2sino ok kτ β α= = − cos( ) cos sino o ox yρ ϕ ϕ ϕ ϕ− = + Using the Maxwell equations yields the following other components of incident plane wave:


xo


− − +=


yo


− − −=

(7.3.22)


xo


− − − +=


yo


− − +=

where oZ and 1

o oY Z −= are the impedance and the admittance of the free medium In the spectral domain we introduce the Fourier transforms:

(7.3.23)ˆ( , ) ( , , )

( , ) ( , , )

o

o

j z j xt

j z j xt

y e y x y z e dx

y e x y z e dx

α η

α η

η

η

∞

−∞

∞

−∞

= ×

=

∫∫

V E

I H

where: xzt ExEz ˆˆ +=E , xzt HxHz ˆˆ +=H The previous formulation yield again:

(7.3.24) ( )η ⋅ 1+ 1-

2+ 2-

V AG =

V A

where:

1 ( ) ( ,0)η η+ =V V , 2 ( ) ( , )dη η+ = −V V , ( ) ( ,0 ) ( ,0 )η η η+ −= −1-A I I ,

2 ( ) ( , ) ( , )d dη η η+ −= − − −-A I I

32

(7.3.25) 3( ) c

c

η−

= − 1 3

3 1 3

Y + Y + Y YG

Y Y + Y + Y

2 2( ) oξ ξ η τ η= = − , 1tan ,

2 sincdj j

dξ

ξ−1 2 c 3Y = Y = Y Y = Y ,

2

2 22 2 2

O Oo

Oo

Ykk k


−=

− −− −cY

The factorization of ( )ηG can be accomplished by putting the 4 4× matrix in the form:

(7.3.26) 1 3( ) ( ) ( )c

c

G Rη η η−

= = ⊗ − 3

3 1 3

Y + Y + Y YG

Y Y + Y + Y

2

3 1 3 32 2 2

3 1 3

1c c O O

c c O

Y Y Y YY Y Y Y k

τ η αξ η α η

− + + − − = ⊗ − − + + − −

1 3

3 1 3

Y + Y + Y YY Y + Y + Y

where the first 2 2× matrix ( )G η is the matrix considered in the E-polarization case, ⊗ is the Kronecker product symbol and the rational matrix ( )R η is defined by

(7.3.27)2

2 2 2

1( ) ( )O Oc

oO

kRYk

τ η αη η

ξ ξη α η−

= =− −

Y

Taking into account that (see also sect.5.1.1):

(7.3.28)[ ] [ ]

[ ] [ ]( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

G R G G R R

G R G R

η η η η η η η

η η η η η η− + − +

− − + + − +

= ⊗ = ⋅ ⊗ ⋅ =

= ⊗ ⋅ ⊗ = ⋅

G

G G

we get: (7.3.29) [ ]( ) ( ) ( )G Rη η η− − −= ⊗G , [ ]( ) ( ) ( )G Rη η η+ + += ⊗G Consequently the factorization of ( )ηG is reduced to the factorization of ( )G η and

( )R η that have been obtained in the previous sections. To take the source we put:

(7.3.30) 1

2

( )s

os

o

ηη η

−

−

⋅ +−

1+

2+

V A RG =V A

33

with: coso o oη τ ϕ= − and where, according to the previous considerations , oR can be

evaluated either as the characteristic part of the minus function 1-

2-

AA

or the

characteristic part of ( )η ⋅ 1+

2+

VG

V

We observe that the geometrical optics arises from the following contributes: Half-plane : 0, 0y x= < : incident wave plus reflected wave due to the PEC plane

0,y = Half-plane : , 0y d x= − < : no geometrical optics contribution Half-plane : 0, 0y x= > : incident wave Half-plane : , 0y d x= − > : incident wave The first evaluation is very simple since in this case for 1-A we have the same characteristic part considered in the skew incidence on a single PEC half-plane whereas the characteristic part of 2-A is vanishing. It follows:

(7.3.31)

22 cos 2 sin

00

o

o o oo o

o o oo

j Hj jkH E

Zα ϕ ϕτ τ

−

− +=R

Conversely in the alternative evaluation the characteristic part of 1+

2+

VV

is given by:

(7.3.32) t

oη η−T where taking into account (2) (Et_Ht.nb):

sin

sin

cos sin

cos sino o

o o

o o o o o o

o

ot

j d o o o o o o

o

j do

E Z H kj

jEE Z H kj e

j e E

τ ϕ

τ ϕ

α ϕ ϕτ

α ϕ ϕτ

−

−

+−

= +−

T

34

Of course it is has been proved that (Et_Ht.nb): (7.3.33) ( )o o tη= ⋅R G T

35

6/16/2008 chapter7_4.doc 7.4 The reversed half-planes problem 7.4.1 The E-polarization case semipian_iinver_skew_form.nb

By supposing the two reversed half planes indicated in fig.7 constituted by PEC material we obtain the circuit representation indicated in fig.7 Writing the node equations yields directly the W-H equations of the problem:

(7.4.1) 1 3 1 3 2 1

3 1 2 3 2 2

node 1 ( )node 2 +( )

c

c

Y Y Y V Y V AY V Y Y Y V A

+ − −

+ − +

+ + − =− + + =

where: 1 2 31tan ,


dτ

τ= = = − and 1 2and -A A− +− are the Fourier

transforms of the total currents induced on the two half planes.

36

With respect the problem considered in fig. 6 we deal with different plus and minus function. In fact now the plus functions are 1 2and AV + + and the minus functions are

2 1 and AV − − . It yields the W-H equations:

(7.4.2)

( ) ( )1 1 3 3

1 11 3 1 3

2 23

1 3 1 3

2

1

c

c c

c c

Y Y Y Y Y YV AY Y Y Y Y YA VY

Y Y Y Y Y Y

+ −

+ −

+ ⋅ + +− + + + + ⋅ =

+ + + +

Taking into account of the primary field contribution, for the E-polarization case (normal incidence) we get:

(7.4.3) 1 1

2 2

sin2 1( )0

o odo

oo o

EjV Z A ZGZ A V

ϕη

η η+ −

+ −

= +−

i

where the matrix kernel is given by:

(7.4.4) 2

2

( )1

2

j d

j dj d

ekG

ee k

τ

ττ

τ

η

τ

−

−−

− =

−

and coso okη ϕ= − . Despite the many efforts, up to now this matrix has not been factorized in closed form. 7.4.1.1 Qualitative characteristics of the solution . Taking into account that det[ ( )] 1G η = , we observe that the structural singularities of the W-H unknowns are only the branch points kη = ± . Once we obtained the W-H unknowns 1 2and AV + + and 2 1and AV − − , we can use the following expressions to evaluate ( , )V yη and ( , )I yη and the corresponding transverse components of the electromagnetic field everywhere:

0y >

(7.4.5)-

1( , ) ( , ) ( , ) ( , )2

i r j x j yz z zE x y E x y E x y V y e e dη τη η

π∞ − −

∞= + + ∫

37

(7.4.6)-

1( , ) ( , ) ( , ) ( , )2

i r j x j yx x xH x y H x y H x y I y e e dη τη η

π∞ − −

∞= + + ∫

1

1( , ) , ( , ) ( 0)j y j y

c

VV y V e I y e yZ

τ τη η− −++= = >

(7.4.7) -d< 0y <

1 2sin( ( )) sin( )( , )sin( )

V d y V yV yd

τ τητ

+ −+ −=

1 2cos( ( )) cos( )( , )

sin( )c

V d y V yI y jZ d

τ τητ

+ −+ −=

y d< −

( ) ( )22( , ) , ( , ) ( )j y d j y d

c

VV y V e I y e y dZ

τ τη η− + − +−−= = − < −

These expressions show that ( , )V yη and ( , )I yη involve as singularities only the branch points kη = ± . It seems in the expressions (3.5) that ( , )V yη and ( , )I yη are exponentially unbounded as η → ±∞ . The following alternative expression (Vy_Iy.nb) obtained by circuit considerations do not present this apparent ; they are more suitable for the evaluation in the region (-d< 0)y < through an inverse Fourier transform:

(7.4.8)

j -j ( )1 2

-j j (2 ) j ( ) j ( )1 2

2j 2j

e e( , )2

(e e ) (e e )( , ) (-d< 0)(1-e ) (1-e )

y d y

y d y d y d y

d d

A AV y

V VI y y

τ τ

τ τ τ τ

τ τ

η ωµτ

τ τηωµ ωµ

+− +

+ + −+ −

+=

+ += − <

7.4.1.2 Numerical evaluation of the electromagnetic field The electromagnetic field can be evaluated through the inverse Fourier transforms:

38

(7.4.9) 1( , ) ( , )2

j xzE x y V y e dηη η

π∞ −

−∞= ∫

(7.4.10) 1( , ) ( , )2

j xxH x y I y e dηη η

π∞ −

−∞= ∫

In the region (-d< 0)y < it is sufficient to use the inverse discrete Fourier transform . In the region usare il metodo del punto di sella. Considerare abnche per I problemi precednti In ogni caso bisogna calcolare bene le incognite W_H che sono le piu’ critiche. 3.4.1.3 Numerical Solution of the W-H equations An efficient approximate factorization can be obtained by using the Fredholm integral equation technique (1.6.2) . Firstly we modify ( )G η in the form:

(7.4.11)2

2 0 0( ) ( )

10 0

2

j d

j dj d

k ke

k kkG Me k ke k

k k

τ

ττ

η ητ

η η

τ η η

−

−−

+ − − = = ⋅ ⋅ − + −

where:

(7.4.12) 2

2

( )1

2

j d

j dj d

ke

kM

k eek

τ

ττ

ηη

ηηη

−

−−

−−

+ = + − −

Let’s observe that:

(7.4.13)

2

1

12

( )

2

j dj d

j d

ke ek

Mk

ek

ττ

τ

ηη

ηηη

−−

−

−

−−

+ = + − −

39

is bounded . Consequently the Fredholm equation:

(7.4.14) [ ]( ) ( ) ( )1( ) ( ) .2 o

M t M F t RM F dtj t

ηη η

π η η η+∞ +

+ −∞

−+ =

− −∫

involves a Fredholm kernel : [ ]1 ( ) ( )( )

M t MM

tη

ηη

− −−

that is compact (1.6.4.2).

Consequently we will use the approximate techniques indicated in sect.1.6.2. Since the original integration line (real axis) is very near the branch points k± and the pole oη it is convenient to deform the integration path . According to the indications of the sect. 1.6.2.2.3, in the following we used three paths according to the equations:

A) (7.4.15) 1( ) cos[ ( ) ]2 2

y k gd y j yπη = − − + − − (gudermann line)

with 1( ) cos sgn( )cosh

gd x arc xx

=

B) (7.4.16) 4( ) ,j

y e y yπ

η = −∞ < < ∞ (bisector straight line)

C) (7.4.17) 2

arctan( ) [ ]10(1 )

xx k x jx

η = ++

(slight deformed real axis)

The cases B) and C) involves a physical meaning in the original planeη − . For the case A) the gudermann line is evident working in the w plane− 7.4.1.4 Numerical Simulations In performing numerical solutions we face with the problem to choice the discretization step h and the truncation range A. Due to the compactness of the Fredholm kernel

[ ]1 ( ) ( )( )

M t MM

tη

ηη

− −−

we are sure that by increasing A and by decreasing h the

approximate solution converges to the exact solution (Kantorovich -Krylov). Consequently the problem is to find the best technique the minimize the computer time. Previously we have indicated three techniques A), B) and C) based on the deformation of the original real axis. Working in the w plane− (case A)) implies the possibility to assume small values of A

since the kernel ( , )m tη behaves as 1cosh y

instead as 1y

as y →∞ . Conversely

working in the planeη − (case B) and C)), we observe that the discretization is uniform in the planeη − and it assures that the contribution of integrand is well considered everywhere in the integral.

40

Since it is not possible to compare with an exact solution, in order to have a criterion on the accuracy of the approximate obtained solution, we changed the values of A and h and compared the different solutions obtained with the three techniques A), B) and C) . Case A We considered simulations

42

(per errore vedere Fredholm_semipiani_polo_sopra_2.nb Per valori di d differenti vedere Fredholm_semipiani_d_05_1.nb eccetera Sostituire d con kd e α con / kα 7.4.1.5 A method for increasing the accuracy or the function 2 ( )V η− From the W-H equations we have:

43

(7.4.18) ( )2 1 2

sin( ) ( ) ( )

j dj d

o

e dV e V jkd Z A

d

ττ τ

η η ητ

−−

− + += +

We observed that whereas the accuracy of the approximate evaluations of

1 2( ) and ( )V Aη η+ + are good, it does not happen for the function 2 ( ) V η− . For instance its inverse Fourier transform ( , ) zE x d− presents significant values for 0x > in contrast with the fact that ( , ) zE x d− must be vanishing on the PEC half plane located in

0,x y d> = − The reason of it is due to the fact that, specially if k d is small, the values of

( , ) zE x y on the aperture 0,x y d< = − are very small with respect those on the aperture 0, 0x y> = . Consequently we need an high accuracy on the approximate expressions

of 1 2( ) and ( )V Aη η+ + for having a sufficient accuracy of 2 ( ) V η− obtained by the equation:

(7.4.19) ( )2 1 2

sin( ) ( ) ( )

j dj d

o

e dV e V jk d Z A

d

ττ τ

η η ητ

−−

− + += +

For instance this accuracy must be elevate near kη = , since 1 2( ) and ( )V Aη η+ + present a branch points but 2 ( ) V η− is regular. To overcome this numerical problems we consider the previous equation as a scalar W-H equation having 2 2( ) and ( )V Aη η− + as unknowns. The factorization of the function

(7.4.20)2 2

2 2

sin[ ]( ) j d k dM ek d

τ ααα

− −=

−

has been accomplished in Example 6 of 1.3.2.3 and yields (miste7.nb)

(7.4.21)1

sin( ) 1exp[ log ] 1( )

1 1

A

n

n

kd d j Bq ekd k AM

BA A A n B

αγ ατ τ α ααπα

α α

∞

+=

− −− + Γ + =

Γ − + + − +

∏

(7.4.22) ( ) ( )M Mα α− += − where:

2 2kτ α= −

44

(7.4.23) , 0A j Bdπ

= − =

−+

−= γπ

π12log

kdjdjq

22

nnkdπα = −

1, 2...., Im[ ] 0nn α= < ,

By assuming 621, (1 10 ), 1.12

k j dπ λλλ

−= = − = and a truncated product with

200=bN , we obtain an error ( ) ( ) ( )M M Mα α α− + − that is vanishing in the range 100 100α− ≤ ≤ .

We ascertain also that the function ( )k Mα α+− and its inverse 1( ( ))k Mα α −+−

must behave as a bounded not vanishing constant as α →∞ . The W-H technique yields the following expression of 2 ( )V η− : (7.4.24) 2 ( ) ( ) ( )V M Yη η η− − −=

where { }2 211( ) ( ) ( )j k dY M e Vηη η η− −−

− − +−

= is the minus decomposed function of

2 211( ) ( )j k dM e Vηη η− −−

− + that is expressed by (eq.2 of sect. 1.3.1):

(7.4.25)2

1 ( )1( ) ( )1( )

2

j u dM u e V uY duj u

τ

γη

π η

− −− +

− = −−∫

where:

2 2( ) kτ η η= − By decomposing τ τ τ− += + (eqs. (13) of 1.3.1):

( ) ( )( ) log jj kτ η τ η ητ ηπ+

−=

(7.4.26) ( ) ( )( ) log jj kτ η τ η ητ ηπ−

+=

we get:

45

(7.4.27)

2

2

2

( ) ( )11

( ) ( ) ( )1 11

( ) ( )11

( ) ( )1( )2

( ) ( ) ( )12

( ) ( )2

j u d j u d

j u d j d j u d

j d j u d

M u e e V uY duj u

M u e M e e V udu

j uM e e V u du

j u

τ τ

γ

τ τ η τ

γ

τ η τ

γ

ηπ η

η

π η

ηπ η

− +

− − +

− +

− −−− +

−

− − −− −− − +

− −−− +

= − =−

− − +−

+−

∫

∫

∫

Now the last integral is vanishing since it represents the minus part of a plus function. It yields:

(7.4.28)

( ) ( ) ( )1 111

2

( ) ( ) ( )1( ) ( )2

j u d j d j u dM u e M e e V uM V du

j u

τ τ η τηη η

π η

− − +− − −− −∞ − − +−

− − −∞

− = −−∫

A numerical quadrature yields:

(7.4.29)

/( )1

2 1/

( ) ( ) ( , ) ( )2

A hj h i d

i A h

hM V h i e V h ij

τη η θ ηπ

+−−− − +

=−

= − ∑

where:

(7.4.30) [ ] ( ) ( )( , ) [ , ( ) , ]udu If ud u

λ λ ηθ η η λ ηη η

− −−

−= =

−

with:

(7.4.31) ( )1

1

1exp[ ] 1( ) ( )

sin( ) 1 1

A

nj d

n

Bq eA AM e

kd Bkd A A n B

ηγ

τ η

ηηη αλ η η

η−

−

∞−−

− −=

−− Γ − + + = =

Γ + − +

∏

46

7.4.2 Source constituted by a skew plane wave semipian_iinver_skew_form.nb Let’s consider an incident plane wave impinging at skew incidence angle β :



oiz


where coso kα β= , 2 2sino ok kτ β α= = − cos( ) cos sino o ox yρ ϕ ϕ ϕ ϕ− = + Using the Maxwell equations yields the following other components of incident plane wave:


xo


− − +=


yo


− − −=

(7.4.33) cos( ) sin coso o oj j zi o o o o o o

xo


− − − +=


yo


− − +=

where oZ and 1

o oY Z −= are the impedance and the admittance of the free medium In the spectral domain we introduce the Fourier transforms:

(7.4.34)ˆ( , ) ( , , )

( , ) ( , , )

o

o

j z j xt

j z j xt

y e y x y z e dx

y e x y z e dx

α η

α η

η

η

∞

−∞

∞

−∞

= ×

=

∫∫

V E

I H

where: xzt ExEz ˆˆ +=E , xzt HxHz ˆˆ +=H The previous formulation yield again:

47

(7.4.35) 1

2 2

( )s

os

o

ηη η

−

+ − −

⋅ = +−

1+ 1-

2

V A A RG =A V V

where:

coso o oη τ ϕ= −

1 ( ) ( ,0)η η+ =V V , 2 ( ) ( , )dη η− = −V V , ( ) ( ,0 ) ( ,0 )η η η+ −= −1-A I I ,

2 ( ) ( , ) ( , )d dη η η+ + −= − − −A I I (7.4.36)

( ) ( )

( )2

2

2( )

sin

j dc

j d j dc

ee j e d

ξ

ξ ξηξ

−

− −

⋅ − = =

1 c 1 3 3

1 3 c 1 3 c

3

1 3 c 1 3 c

Y + Y Y + 2Y + Y Y-Y 1Y + Y + Y Y + Y + Y

G1 ZY 1

Y + Y + Y Y + Y + Y

where:

2 2( ) oξ ξ η τ η= = − , 1tan ,2 sincdj j

dξ

ξ−1 2 c 3Y = Y = Y Y = Y ,

2

2 22 2 2

O Oo

Oo

Ykk k


−=

− −− −cY ,

2 2

22 2 2oo

o Oo

kZk k

η η αη α τα η

−=

− −cZ

and 21 is the 2 2× identity matrix. According to the previous considerations , oR can be evaluated either as the

characteristic part of the minus function −

1-

2

AV

or the characteristic part of 2

( )η+

⋅ 1+VG

A

We observe that the geometrical optics arises from the following contributes: Half-plane : 0, 0y x= < : incident wave plus reflected wave due to the PEC plane Half-plane : , 0y d x= − < : no geometrical optics contribution Half-plane : 0, 0y x= > : incident wave plus reflected wave due to the PEC plane

,y d= − Half-plane : , 0y d x= − > : incident wave plus reflected wave due to a PEC plane

,y d= − The first evaluation is very simple since in this case for 1-A we have the same characteristic part considered in the skew incidence on an single PEC half-plane whereas the characteristic part of 2-V is vanishing. It follows:

48

(7.4.37)

22 cos 2 sin

00

o

o o oo o

o o oo

j Hj jkH E

Zα ϕ ϕτ τ

−

− +=R

Conversely in the alternative evaluation the characteristic part of 1+

2+

VV

is given by:

(7.4.38) t

oη η−T where taking into account (2) (Et_Hprova.nb):

Of course it is has been proved that (Et_Hprova.nb): (7.4.39) ( )o o tη= ⋅R G T Taking into account that 1

c−=cZ Y commutes with cY .The factorization of ( )ηG can be

accomplished by putting the 4 4× matrix in the form:

2( ) ( ) ( ) ( )cη η η η= ⊗ + ⊗G A 1 B Y where the matrices ( ) and ( )η ηA B are 2 2× matrices. The key point to reduce the order of matrices to be factorized is the observation that the

polynomial matrix ηααη

ηo

ot−

=)( makes diagonal cY and also cZ . In fact

49

(7.4.40) )(0

0)( 2

1 η

ξεω

εωξ

η tk

tc ⋅⋅= −Z )(0

0)(1 η

µωξ

ξεω

η ttc ⋅⋅= −Z

or :

(7.4.41) )()()(12

22

ηηηταηαηη

tdtk

OO

O ⋅⋅=− −

where:

2

2

00

)(k

dξ

η =

It follows that putting 2

( ) 0( ) 1 ( )

0 ( )t

T tt

ηη η

η= = ⊗ yields the matrix

1( ) ( ) ( )T G Tη η η−⋅ ⋅ to have the form:

(7.4.42) 1

0 00 0

( ) ( ) ( )0 0

0 0

x xx x

T G Tx x

x x

η η η−⋅ ⋅ =

where x means an element not vanishing. Introducing the permutation matrix 1P P−= (see 5.1.1) defined by:

(7.4.43)

1000001001000001

=P

we can reorder in order to have the following quasi diagonal matrix

(7.4.44) 11

2

( ) 0( ) ( ) ( )

0 ( )d

d

GP T G T P

Gη

η η ηη

−⋅ ⋅ ⋅ ⋅ =

where the matrices 1( )dG η and 2 ( )dG η of order two (semipian_iinver_skew_form.nb)

50

In order to have matrices that ,with their inverses, exist and are bounded as η → ±∞ we normalize by introducing the matrix ( )t ηG defined by : (7.4.45)

1 ( )( ) [1, , ,1] ( ) [ ,1,1, ]

( )th

t o o o ote

n diag Y Z P T T P diag Z Y nη

η ηη

−− +

= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =

G 0G G

0 G

where the normalization matrices n± are defined by (semipian_iinver_skew_form.nb) :

(7.4.46) [ , , , ]o o o o

o o o o

n diagτ η τ τ τ ητ τ η τ η τ± ±

=± ±∓

The 2 2× matrices ( )thG η and ( )teG η (that are suitable for the Fredholm factorization ) are expressed by (semipian_iinver_skew_form.nb) (7.4.47)

( )

2

( )

sin

oj d

oth

oj d j d

o

e

e j e d

ξ

ξ ξ

τ ητ η

ητ η

ξτ η

−

− −

+−

− =

− +

G ,

( )

2

( )

sin

oj d

ote

oj d j d

o

e

e j e d

ξ

ξ ξ

τ ητ η

ητ η

ξτ η

−

− −

−−

+ =

+ −

G

In the previous section we have factorized ( )teG η by the Fredholm method. Now we will show that the factorization of ( )thG η reduces to that of ( ) ( ) ( )te te teG G Gη η η− += ⋅ . In fact the following equation holds:

(7.4.48)1 0 1 0

( ) ( )0 0th teo o

o o

G Gη ητ η τ ητ η τ η

= ⋅ ⋅− + + −

It provides the “weak” factorization (sect.2.8 and 5.5.1) of ( ) ( ) ( )th thw thwG G Gη η η− += ⋅ where:

(7.4.49)1 0

( ) ( )0thw teo

o

G Gη ητ ητ η

− −

= ⋅− +

, 1 0

( ) ( )0thw te o

o

G Gη η τ ητ η

+ +

= ⋅ + −

51

The factorizations of ( ) ( ) ( )th th thG G Gη η η− += ⋅ and ( ) ( ) ( )te te teG G Gη η η− += ⋅ yield yields the following “weak” factorization of ( ) ( ) ( )w wη η η− += ⋅G G G :

(7.4.50) [ ] 1 ( )( ) [1, , ,1]

( )th

w o ote

n diag Y Z P Tη

ηη

− −− −

−

= ⋅ ⋅ ⋅ ⋅

G 0G

0 G

(7.4.51)11( )

( ) [ ,1,1, ]( )

thw o o

te

T P diag Z Y nη

ηη

−+ −+ +

+

= ⋅ ⋅ ⋅ ⋅

G 0G

0 G

This factorization is “weak” for the presence of the poles ojα± in

12 2

1( ) o

oo

tη α

ηα ηη α

− =−+

or in 12 2

( )a

o

TT ηη α

− =+

where

(7.4.52)

0 00 0

( )0 00 0

o

oa

o

o

T

η αα η

ηη αα η

−=

−

7.4.2.1 Formal solution Rewritten the W-H equation in the form:

(7.4.53) ( ) ( ) ( ) ( ) o

o

η η η ηη η+ − −⋅ = +−RG F = F X

where:

2

( )η++

= 1+VF

A, 1

2

( )s

sη −−

−

=A

XV

, 1

2

( )η −−

−

=A

FV

the W-H technique yields:

(7.4.54) 1 1( ) ( ) ( ) ( ) ( ) ow w w

o

η η η η ηη η

− −+ + − − −⋅ ⋅ + ⋅

−RG F = G X G

or:

52

(7.4.55) 1 1 1( ) ( ) ( ) ( ) ( ) ( )o ow w o w w o

o o

η η η η η ηη η η η

− − −+ + − − − −⋅ − ⋅ ⋅ − ⋅ =

− −R RG F G = G F G w

Taking into account the form of the factorized matrices ( ) and ( )w wη η− +G G it follows that the vector w is constant. To obtain it we resort to the same procedure used to factorize ( )c ηY in the previous section. For instance w must be chosen in order to assure that the solutions :

(7.4.56) 1 1( ) ( ) ( ) ow w o

o

η η ηη η

− −+ + −

⋅ + ⋅ −

RF = G w G

(7.4.57) 1( ) ( ) ( ) ow w o

o

η η ηη η

−− − −

= ⋅ + ⋅ −

RF G w G

do not present the offending poles ojη α= and ojη α= − respectively. In the following will be important the Null Spaces of ( )a oT jα and ( )a oT jα− . By using MATHEMATICA we get:

1 1 2 2[ ( )]a oNullSpace T j c cα− = +u u

3 3 4 4[ ( )]a oNullSpace T j c cα = − −u u where the constant ic , 1, 2,3,4i = , are arbitrary and the vectors iu , 1, 2,3,4i = , are defined by:

1

00

1j

=−

u , 2

100

j−

=u , 3

00

1j

=u , 4

100

j

=u ,

Looking to the equation(7.4.56), the absence of the offending pole ojη α= in ( )η+F implies that the vector

11( )

[ ,1,1, ] ( ) ( )( )

th o oo o o w o

te o o o

jP diag Z Y n j

j jα

α ηα α η

−+ −

+ −+

⋅ ⋅ ⋅ ⋅ + ⋅ −

G 0 Rw G0 G

must be in the Null Space of ( )a oT jα :

53

(7.4.58)1

13 3 4 4

( )[ ,1,1, ] ( ) ( )

( )th o o

o o o w ote o o o

jP diag Z Y n j c c

j jα

α ηα α η

−+ −

+ −+

⋅ ⋅ ⋅ ⋅ + ⋅ = − − −

G 0 Rw G u u0 G

At the same time, the absence of the offending pole ojη α= − in ( )η−F (7.4.57) implies: (7.4.59)

1 11 1 2 2

( )[1, , ,1] ( ) ( )

( )th o o

o o o w ote o o o

jP diag Z Y n j c c

j jα

α ηα α η

−− −− −

−

− ⋅ ⋅ − ⋅ ⋅ + ⋅ = + − − −

G 0 Rw G u u0 G

The previous eight scalar equations allows to determine the eight unknowns ic , 1, 2,3, 4i = e the four components of w .

We get :

(7.4.60)

1

2 1 12 2

3

4

2 ( ) o ow o

o o

cc

jcc

αηα η

− −−= ⋅ ⋅

+RM G

(7.4.61)

( )1

1 11 1 2 2

( )[1, , ,1] ( ) ( )

( )th o o

o o o w ote o o o

jP diag Z Y n j c c

j jα

α ηα α η

−

−− −− −

−

− = ⋅ ⋅ − ⋅ ⋅ + − ⋅ − − −

G 0 Rw u u G0 G

where the 4 4× matrix M is defined by: (7.4.62) M = 1 2 3 4U , U , U , U where the vector iU , 1, 2,3, 4i = are defined by: (7.4.63)

1

11

( )[1, , ,1] ( )

( )th o

o o ote o

jP diag Z Y n j

jα

αα

−

−−−

−

− = ⋅ ⋅ − ⋅ ⋅ −

1

G 0U u

0 G

1

12 2

( )[1, , ,1] ( )

( )th o

o o ote o

jP diag Z Y n j

jα

αα

−

−−−

−

− = ⋅ ⋅ − ⋅ ⋅ −

G 0U u

0 G

11

3 3

( )[ ,1,1, ] ( )

( )th o

o o ote o

jP diag Z Y n j

jα

αα

−−+

++

= ⋅ ⋅ ⋅ ⋅

G 0U u

0 G

54

11

4 4

( )[ ,1,1, ] ( )

( )th o

o o ote o

jP diag Z Y n j

jα

αα

−−+

++

= ⋅ ⋅ ⋅ ⋅

G 0U u

0 G

7.4.2.2 The relation between the factorized 2 2× matrices of ( ) and ( )th teG Gη η . In the previous section we provided an equation that relates ( ) and ( )th teG Gη η and reduced the “weak”

factorization of ( ) thG η to that of ( )teG η (7.4.49).

In this section we will consider another possibility to the factorize ( ) thG η by taking into account that the following equation hold:

(7.4.64) [ ]1 0 1 0( ) ( ) ( )

0 1 0 1t

th te teG G Gη η η = − = ⋅ ⋅ − −

In effect this equation provides the “right” factorization of ( ) ( ) ( ) th th thG G Gη η η+ −= ⋅# #

and

unfortunately ,known ( ) and G ( ) th thG η η+ −

# #, we are not able to obtain the required

factorization ( ) ( ) ( ) th th thG G Gη η η− += ⋅ . To obtain the “right” factorization of ( ) ( ) ( ) th th thG G Gη η η+ −= ⋅

# # ,we observe that:

(7.4.65)

[ ]

[ ] [ ]

1 0 1 0( ) ( )

0 1 0 1

1 0 1 0( ) ( ) ( ) ( )

0 1 0 1

tth te

t tte te th th

G G

G G G G

η η

η η η η+ − + −

= ⋅ ⋅ = − −

= ⋅ ⋅ = ⋅ − −

# #

where

[ ]1 0( ) ( )

0 1t

th teG Gη η+ +

⋅ = ⋅ −

#, [ ] 1 0

( ) ( )0 1

tth teG Gη η− −

= ⋅ −

#

The previous equation provides a factorization of ( ) ( ) ( )th th thG G Gη η η+ −= ⋅

# #that presents factors that are

inverted with respect ( ) ( ) ( )th th thG G Gη η η− += ⋅ so we must face the problem to obtain

( ) and ( )th thG Gη η− + known ( ) and ( )th thG Gη η+ −

# #. Even though this problem has been solved by

Jones (Report 1-2004) in the case of 2 2× matrices having a Daniele form , we were not able to solve it in general.

55

6/16/2008 C:\libri\WHnew\definitivo\chapter7_5.doc 7.5 The three half planes problem 7.5.1 Introduction tre_semipiani.nb

56

The fig. illustrates the geometry of the problem and the circuit representation in the spettral domain. The circuit nodal analysis provide the three following equations (7.5.1)

( )( )( )

1 3 1 3 2 1

1 3 3 3 2 3

1 3 2 3 1 3 3 2

node 1

node 3

node 2 2

c

c

Y Y Y V Y V A

Y Y Y V Y V A

Y Y V Y V Y V A

+ + −

+ + −

+ + + −

⇒ + + − =

⇒ + + − =

⇒ + − − =

where:

1 2 31tan ,


dτ

τ= = = − , cY τ

ω µ= , 2 2kτ η= − .

1 2 3, and V V V+ + + and 1 2 3, and -A A A− − +− − are respectively the Fourier transforms of the electric field zE on the three apertures 0, , 2y y d y d= = − = − and the total currents induced on the three half- planes. The previous equations constitute a W-H system of third order. The symmetry of the geometry allows to reduce the order. In fact by summing the first two equations we get the system:

(7.5.2)( )

( )1 31 3

22

( )2

ooo

oo

o

kZ A Akk V V

Gkk V Z A

k

ηηη

ηη

− −

+ +

+−

++− +

⋅ =−

+

where:

22

12( ) 1(1 )

2

j d

j dj d j d

eG ee e

τ

ττ τ

η

−

−− −

− = + − −

We have:

(7.5.3)2

1 11( )2 1

j d j d

j d

e eG

e

τ τ

τη

− −−

−

+=

57

Consequently ( )G η and its inverse 1( )G η− are bounded on the real axis as η → ±∞ . It allows to apply the Fredholm factorization to the matrix ( )G η . Taking into account that:

(7.5.4) 1 1sin 12s o o

o o

EA A jZϕ

η η− −= +−

we can rewrite the W-H system in the normal form:

(7.5.5) ( ) ( ) ( ) o

o

RG F Xη η ηη η+ −⋅ = +−

where:

(7.5.6)( )1 3

2

( )2o

o

k V VF

k V

ηη

η+ +

+

+

− +=

−,

2 sin

0

o oo oo

k jEkR

ϕη+=

7.5.2 Factorization of the matrix ( )G η (Fredholm_base 2….nb)

8 131 10 , 2.0, , cos ,04

16, 0.1

o o ok j d k R

A h

πϕ η ϕ−= − = = = − =

= =

58

8 131 10 , 2.0, , cos ,04

16, 0.1

o o ok j d k R

A h

πϕ η ϕ−= − = = = − =

= =

remarks: in this report there are present errors and

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