NASA/CR-2000-210297

Electromagnetic Field Penetration Studies

M. D. Deshpande

NYMA, Inc., Hampton, Virginia

June 2000
The NASA STI Program Office . . . in Profile

Since its founding, NASA has been dedicated

to the advancement of aeronautics and space

science. The NASA Scientific and Technical

Information (STI) Program Office plays a key

part in helping NASA maintain this

important role.

The NASA STI Program Office is operated by

Langley Research Center, the lead center for

NASA’s scientific and technical information.

The NASA STI Program Office provides

access to the NASA STI Database, the

largest collection of aeronautical and space

science STI in the world. The Program Office

is also NASA’s institutional mechanism for

disseminating the results of its research and

development activities. These results are

published by NASA in the NASA STI Report

Series, which includes the following report

types:

• TECHNICAL PUBLICATION. Reports of

completed research or a major significant

phase of research that present the results

of NASA programs and include extensive

data or theoretical analysis. Includes

compilations of significant scientific and

technical data and information deemed

to be of continuing reference value. NASA

counterpart of peer-reviewed formal

professional papers, but having less

stringent limitations on manuscript

length and extent of graphic

presentations.

• TECHNICAL MEMORANDUM.

Scientific and technical findings that are

preliminary or of specialized interest,

e.g., quick release reports, working

papers, and bibliographies that contain

minimal annotation. Does not contain

extensive analysis.

• CONTRACTOR REPORT. Scientific and

technical findings by NASA-sponsored

contractors and grantees.

• CONFERENCE PUBLICATION.

Collected papers from scientific and

technical conferences, symposia,

seminars, or other meetings sponsored or

co-sponsored by NASA.

• SPECIAL PUBLICATION. Scientific,

technical, or historical information from

NASA programs, projects, and missions,

often concerned with subjects having

substantial public interest.

• TECHNICAL TRANSLATION. Englishlanguage translations of foreign scientific

and technical material pertinent to

NASA’s mission.

Specialized services that complement the

STI Program Office’s diverse offerings include

creating custom thesauri, building customized

databases, organizing and publishing

research results . . . even providing videos.

For more information about the NASA STI

Program Office, see the following:

• Access the NASA STI Program Home

Page at http://www.sti.nasa.gov

• Email your question via the Internet to

help@sti.nasa.gov

• Fax your question to the NASA STI

Help Desk at (301) 621-0134

• Telephone the NASA STI Help Desk at

(301) 621-0390

• Write to:

NASA STI Help Desk

NASA Center for AeroSpace Information

7121 Standard Drive

Hanover, MD 21076-1320
National Aeronautics and

Space Administration

Langley Research Center

Hampton, Virginia 23681-2199

NASA/CR-2000-210297

Electromagnetic Field Penetration Studies

M. D. Deshpande

NYMA, Inc., Hampton, Virginia

June 2000

Prepared for Langley Research Center

under Contract NAS1-96013
Available from:

NASA Center for AeroSpace Information (CASI) National Technical Information Service (NTIS)

7121 Standard Drive 5285 Port Royal Road

Hanover, MD 21076-1320 Springfield, VA 22161-2171

(301) 621-0390 (703) 605-6000
1

Contents

List of Figures 2

List of Symbols 5

Abstract 9

1. Introduction 10

2. Formulation of Problem 12

2.1 Electromagnetic Field Outside Enclosure 15

2.1.1 Electromagnetic Field Due to Incident Field 15

2.1.2 Electromagnetic Field Scattered Due to Apertures 16

2.2 Electromagnetic Field Inside Enclosure 19

2.2.1 Electromagnetic Field Due to Aperture on Plane 20

2.2.2 Electromagnetic Field Due to Aperture on Plane 23

2.3 Derivation of Integral Equation 26

2.3.1 Integral Equation for Apertures on Plane 26

2.3.2 Integral Equation for Apertures on Plane 30

3. Numerical Results & Discussions 35

3.1 Single Aperture Case 35

3.2 Two or Multi Apertures Case 41

3.3 EM Field Penetration Inside B-757Aircraft Configuration 47

3.4 Convergence Test for Large Size Cavities 55

4. Conclusion 60

Acknowledgement 61

References 61

Appendix I 62

Appendix II 62

Appendix III 63

Appendix IV 63

z = 0

z c =

z = 0

z c =
2

List of Figures

Figure 1 Geometry of rectangular enclosure with rectangular apertures illuminated by a

plane wave at normal incidence.

Figure 2 Geometry of equivalent problem with apertures replaced by magnetic current

sources.

Figure3.1 Electric field shielding at the center of (30. x 12. x 30.) cm3 enclosure with (10. x

0.5) cm2 aperture located at (15.,6.) cm in z = 0 plane with dominant cavity

mode considered.

Figure 3.2 Electric field shielding at the center of (30. x 12. x 30.) cm3

 enclosure with (10.

x 0.5) cm2

 aperture located at (15., 6.) cm in z = 0 plane with dominant and

higher order cavity mode considered.

Figure 4 Electric field shielding at the center of (30. x 12. x 30.) cm3

 enclosure with (20.

x 3.0) cm2 aperture located at (15. 6.0) cm in z = 0 plane with dominant cavity

mode considered.

Figure 5 Electric field shielding at the center of (30. x 12. x 30.) cm3

 enclosure with (20.

x 3.0) cm2

 aperture located at (15., 6.) cm in z = 0 plane with dominant and

higher order cavity mode considered.

Figure 6 Electric field shielding at the center of (30. x 12. x 30.) cm3

 enclosure with (20.

x 3.0) cm2 aperture located at (15., 6.0)cm in z = 0 plane with two aperture

modes considered.
3

 Figure 7 Electric field shielding at the center of (30. x 12. x 30. )cm3 enclosure with (10.

x 3.0) cm2

aperture located at ( 15., 6.0) cm in z = 0 plane with three aperture

modes considered.

Figure 8 Electric field shielding at the center of (30. x 12. x 30.) cm3

 enclosure with two

identical apertures with (20. x 3.0) cm2 size. Aperture one located at ( 15., 6.,

0.0) cm, and aperture two located at (15., 6., 30.) cm.

Figure 9 Effect of higher order aperture modes on electric field shielding at the center of

(30. x 12. x 30.) cm3

 enclsoure with two identical apertures of size (20. x 3.0)

cm2 . Aperture one located at ( 15., 6., 0.0) cm, and aperture two located at ( 15.,

6., 30.) cm.

Figure 10 Electric field shielding of rectangular enclosure of size (30 x 12 x 30 ) cm3 with

two identical rectangular apertures (one on each side) of various sizes when illuminated by incident plane wave at normal incidence. Comparison with FEM/

MoM calculation.

Figure 11 Rectangular cavity with four identical apertures and illuminated by plane wave

with normal incidence. All dimensions in cm.

Figure 12 Electric field shielding of rectangular cavity shown in Figure 11. Observation

point ( 30, 6, 15) cm. Comparison with FEM/MoM calculation.

Figure 13 Geometry of B-757 aircraft configuration with cross sectional and longitudinal

approximated cross section.

Figure 14(a) Closed rectangular cavity approximating a B-757 fuselage with rectangular apertures on side walls representing passenger windows.

Figure 14(b) Closed rectangular cavity approximating a B-757 fuselage with two rectangular
4

apertures on side walls representing passenger windows.

Figure 15 Electric field shielding (for vertical polarization) of a rectangular cavity for vertical polarization approximating fuselage of a B-757 aircraft. Two windows on

each side of cavity were considered. Cavity size ( 3600. x 400. x 400.) cm3

, window size ( 25. x 35. ) cm2

, window locations (1800., 200., 0.0) cm, and (1800.,

200., 400.)cm, field location point ( 1800., 120., 200.) cm.

Figure 16 Electric field shielding (for vertical polarization) of a rectangular cavity approximating fuselage of a B-757 aircraft when two aperture modes were considered.

Two windows on each side of cavity were considered. Dimensions of cavity,

aperture and field point are as given in Figure 15.

Figure 17 Electric field shielding (for vertical polarization) of a rectangular cavity approximating fuselage of a B-757 aircraft when three aperture modes were considered.

Two windows on each side of cavity were considered. Dimensions of cavity,

aperture and field point are as given in Figure 15.

Figure 18 Electric field shielding (for horizontal polarization) of a rectangular cavity

approximating fuselage of a B-757 aircraft. Two windows on each side of cavity

were considered. Dimensions of cavity, aperture and their locations are as given

in Figure 15.

Figure 19 Electric field amplitude at ( 2700., 300., 300.)cm inside a rectangular cavity of

size ( 5400. x 600. x 600.) cm3 illuminated by plane wave at normal incidence

through rectangular windows (25. x 35. )cm2 as a function of cavity mode index

n. Other parameters: cavity mode index m = 1200, frequency = 2.87510864

GHz.
5

Figure 20 Electric field amplitude at ( 2700., 300., 300.)cm inside a rectangular cavity of

size ( 5400. x 600. x 600.) cm3 illuminated by plane wave at normal incidence

through rectangular windows (25. x 35. )cm2 as a function of cavity mode index

m. Other parameters: cavity mode index n = 200, frequency = 2.87510864

Figure 21 Electric field amplitude at ( 2700., 300., 300.)cm inside a rectangular cavity of

size ( 5400. x 600. x 600.) cm3 illuminated by plane wave at normal incidence

through rectangular windows (25. x 35. )cm2 as a function of number of aperture

modes. Other parameters: cavity mode index n = 200, m = 1200, frequency =

2.87510864 GHz GHz.

List of Symbols

x-dimension of rectangular cavity/approximate length of fuselage

complex modal amplitude of mode for aperture on plane

(for x-directed aperture field)

y-dimension of rectangular cavity/approximate height of fuselage

complex modal amplitude of mode for aperture on plane

(for y-directed aperture field)

z-dimension of cavity/approximate width of fuselage

incident electric field

tangential aperture electric field

electric field due to in region I

electric field due to and in region II

electric field due to in region III

electric vector potential for region I

electric vector potential for region II

x-component of electric vector potential

y-component of electric vector potential

a

Arpq pq

th

z c =

b

Brpq pq

th

z c =

c

Ei

Eapt

E

I

( ) Mr1 Mr1

E

II

( ) Mr Mr1 Mr2

E

III

( ) Mr2 Mr2

F

I

F

II

Fx

II

Fy

II
6

dyadic Green’s function for rectangular enclosure

x-component of dyadic Green’s function

y-component of dyaduc Green’s function

incident magnetic field

 component of incident magnetic field

 component of incident magnetic field

x-, y-, and z- component of incident magnetic field

magnetic field due to in region I

magnetic field due to and in region II

magnetic field due to in region III

rectangular components of vector

coupling of testing function with x- component of incident magnetic field

coupling of testing function with y- component of incident magnetic field

unit dyadic

free-space wave number

propagation constant

wave numbers in x-, y-, and z-directions, respectively

x-dimension of aperture/window

equivalent magnetic current on aperture

equivalent magnetic current for apertures on plane

equivalent magnetic current for apertures on plane

mode index numbers used for cavity modes

integers used for aperture modes

number of apertures on each side

complex modal amplitude of mode for aperture on plane

(for x-directed aperture field)

G ̃ m

Gmxx

Gmyy

Hi

Hθi θ

Hφi φ

Hxi Hyi Hzi ( ) , ,

H

I

( ) Mr1 Mr1

H

II

( ) Mr Mr1 Mr2

H

III

( ) Mr2 Mr2

Hx

II HyII HzII , , H

II

( ) Mr

Ir' p'q'xi

Ir' p'q'yi

I

j –1

k0

kI

kx ky kz ( ) , ,

Lr r

th

Mapt

Mr1 z = 0

Mr2 z c =

( ) m n,

( ) p q,

R

Urpq pq

th

z = 0
7

complex modal amplitude of mode for aperture on plane

y-dimension of aperture

unit vectors along x- and y-directions, respectively

coordinates of center point of aperture

coordinates of a point in the rectangular coordinate system

mutual admittance between x-components of currents on plane

mutual coupling between x- and y-components of currents on plane

mutual coupling between x- component of currents on plane and ycomponent of current on plane.

mutual coupling between x- component of currents on plane and xcomponent of current on plane.

mutual admittance between y- and x-components of currents on plane

mutual coupling between x- component of currents on plane

mutual coupling between y- component of currents on plane and ycomponent of current on plane.

mutual coupling between y- component of currents on plane and xcomponent of current on plane.

mutual admittance between x-component of currents on plane and xcomponent of current on plane

mutual coupling between x- component of currents on plane and ycomponent of current on plane

mutual coupling between x- and y-components of currents on plane

mutual coupling between x- components of currents on plane

mutual admittance between y-component of currents on plane and xcomponent of current on plane

mutual coupling between y- component of currents on plane and ycomponent of current on plane

mutual coupling between y-components of currents on plane

Vrpq pq

th

z = 0

Wr r

th

xˆ, yˆ

xrc yrc zrc ( ) , , r

th

( ) xyz , ,

Yrpqr' p'q'

x1x1

z = 0

Yrpqr' p'q'

x1y1

z = 0

Yrpqr' p'q'

x1y2

z = 0

z c =

Yrpqr' p'q'

x1x2

z = 0

z c =

Yrpqr' p'q'

y1x1

z = 0

Yrpqr' p'q'

y1y1

z = 0

Yrpqr' p'q'

y1y2

z = 0

z c =

Yrpqr' p'q'

y1x2

z = 0

z c =

Yrpqr' p'q'

x2x1

z c =

z = 0

Yrpqr' p'q'

x2y1

z c =

z = 0

Yrpqr' p'q'

x2y2

z c =

Yrpqr' p'q'

x2x2

z c =

Yrpqr' p'q'

y2x1

z c =

z = 0

Yrpqr' p'q'

y2y1

z c =

z = 0

Yrpqr' p'q'

y2y2

z c =
8

mutual coupling between y- and x-components of currents on plane

 modal function for aperture for y-directed magnetic current

 modal function for aperture for x-directed magnetic current

Fourier transform of

Fourier transform of

unit vectors in -and -directions

angle of incident plane wave

Neumann’s number

permittivity of free space

permeability of free-space

angular frequency

Kronecker delta function

gradient operator

Yrpqr' p'q'

y2x2

z c =

Φrpq ( ) p q, th

r

th

Ψrpq ( ) p q, thr

th

φrpqy Φrpq

ψrpqx Ψrpq

θˆi φˆi () θφ ,

θi φi ( ) ,

ε0m

ε0

μ0

ω

δ( ) .

∇
9

Abstract

A numerical method is presented to determine electromagnetic shielding

effectiveness of rectangular enclosure with apertures on its wall used for input and output

connections, control panels, visual-access windows, ventilation panels, etc. Expressing

electromagnetic fields in terms of cavity Green’s function inside the enclosure and the free

space Green’s function outside the enclosure, integral equations with aperture tangential

electric fields as unknown variables are obtained by enforcing the continuity of tangential

electric and magnetic fields across the apertures. Using the Method of Moments, the integral

equations are solved for unknown aperture fields. From these aperture fields, the

electromagnetic fields inside a rectangular enclosure due to external electromagnetic sources

are determined.

Numerical results on electric field shielding of a rectangular cavity with a

thin rectangular slot obtained using the present method are compared with the results obtained

using simple the transmission line technique for code validation. The present technique is

applied to determine field penetration inside a Boeing -757 by approximating its passenger

cabin as a rectangular cavity filled with a homogeneous medium and its passenger windows

by rectangular apertures. Preliminary results for, two windows; one on each side of fuselage

were considered. Numerical results on electromagnetic field penetration through passenger

windows of a Boeing -757 at frequencies 26 MHz, 171-175 MHz, and 428-432 MHz are

presented.
10

1.0 Introduction

Apertures of various sizes and shapes on a metallic enclosure are used for reasons such as

input and output connections, control panels, visual-access windows, ventilation panels, etc.

Since these apertures at appropriate electromagnetic (EM) frequencies behave as very efficient

antennas, they also become sources of electromagnetic interference (EMI) problems for both EM

emission and susceptibility. It is important to know the EM shielding effectiveness of these

enclosures in presence of the apertures. The EM shielding effectiveness study may also help in

locating these apertures at proper places to reduce the EM emission or improving the immunity

of electronic components present inside the metallic enclosure[1]-[5]

Shielding effectiveness can be calculated using numerical or analytical methods. Numerical methods such as the Finite Difference Time Domain (FDTD) method [6], and the hybrid

Finite Element/ Method of Moments [7] can model complex structures inside enclosure but often

requires large computing time and memory. For electrically large size enclosures and apertures

these methods , though more accurate, become difficult for designers to use to investigate effects

of EM shielding on the design parameters.

Pure analytical formulations described in [3]-[5] even though provide a much faster means

of calculating shielding effectiveness are based on various simplifying assumptions whose validity may be questionable at high frequencies.

In the present paper a method which is suitable for large but regular shaped enclosure and

apertures is described. Using the equivalence principle as described in earlier papers [1], [2], the

apertures are replaced by equivalent magnetic current sources. By matching the tangential electromagnetic fields across the apertures, coupled integro-differential equations with the magnetic

currents at the apertures as an unknown variables are obtained. The coupled integro-differential
11

equation in conjunction with the method of moment are then solved for unknown magnetic current amplitudes. One of the advantages of the present approach is that it gives an appropriate

model for EM shielding effectiveness studies and avoids some of approximations or assumptions

which are made in analytical formulations to arrive at closed form formulas for EM shielding.

Furthermore, by representing the magnetic current over the apertures in terms of entire domain

basis functions, three to five number of unknowns per aperture are required for convergence. The

present method, therefore, requires considerable less computer time compared to other numerical

methods. The CPU time taken to determine EM shielding for a rectangular cavity illuminated

through two rectangular apertures with three entire domain modes on each is around 23.00 seconds on a SGI Origin 2000 machine.

The remainder of this report is organized as follows. A general formulation of the problem of a rectangular cavity with a series of rectangular apertures on its broad walls which are

illuminated by plane wave at normal incidence is presented in section 2. Section 2 describes: (1)

use of cavity Green’s function to determine electromagnetic fields scattered due to the apertures

inside the cavity, (2) use of free space Green’s function in plane wave spectrum form to determine

scattered field outside the cavity due to apertures, (3) setting of integral equation and use of

method of moments to determine matrix equation with aperture fields as unknown variables.

Numerical results on electric field shielding due to horizontally and verically polarized incident

waves for various sizes of rectangular cavity and rectangular apertures are given in section 3 along

with earlier published data for comparison and validation. The report concludes in section 4 with

remarks on the advantages and limitations of the present method. Section 4 also gives a recommendation for future work to be persued.
12

2.0 Formulation of Problem

Figure 1 shows a geometry of rectangular enclosure with rectangular apertures on its two

opposite walls. These apertures may be used for input and output connections, control panels,

visual access windows, etc. Since the rectangular apertures are very efficient electromagnetic

radiators, they also act as sources of electromagnetic interference both for emission and susceptibility. The electromagnetic energy radiated by conducting wires and loops inside the enclosure

may escape through these apertures and cause an interference with another electronic

components. The electromagnetic energy from sources outside the enclosure will get through

these apertures into the enclosure and may cause interference with electronic circuits/components

x

y

z

Ei

Hi

Incident Plane

Wave

Rectangular

Apertures

Rectangular

Metal Cavity

a

b c

Figure 1 Geometry of rectangular enclosure with rectangular apertures

 illuminated by a plane wave at normal incidence.

x'

y'

r

th Aperture

Lr

Wr
13

present inside. In this report the latter problem is considered. To determine electromagnetic

shielding effectiveness of the enclosure shown in figure 1 due to external EM radiation, a plane

wave at normal incidence is assumed. This assumption is valid for worst case study. The tangential electric field induced on the apertures due to the incident field may be represented by

(1)

 where are the coordinates of center of aperture, and are the

unknown voltages of mode on aperture, are the width and length of aperture,

respectively, are the unit vectors in directions. In equation (1) it is assumed that there

are apertures on the plane. The unknown modal amplitudes and for

, , and otherwise. Using the

equivalence principle, the apertures on the plane can be replaced by magnetic currents

given by

(2)

Likewise the apertures on the plane can be replaced by equivalent magnetic currents given

by

(3)

where

Eapt yˆUrpq

pπ

Lr

------ Lr

2

----- x xcr + –         sin qπ

Wr

------- Wr

2

------- y ycr + –         cos

q

∑

p. r = 1.∑ R = ∑

xˆVrpq

pπ

Lr

------ Lr

2

----- x xcr + –         cos qπ

Wr

------- Wr

2

------- y ycr + –         sin

q

∑

p. r = 1.∑ R

∑

xcr ycr ( ) , r

th Urpq Vrpq

pq

th

r

th Wr Lr , rth

xˆ, yˆ x y,

R z = 0 Urpq ≠ 0 Vrpq ≠ 0

xcr

Lr

2 – ----- x xcr

Lr

2 ≤ ≤ + ----- ycr

Wr

2 – ------- y ycr

Wr

2 ≤ ≤ + ------- Urpq Vrpq = = 0

z = 0

Mapt Vrpq –yˆΦrpqy ( )

q

∑

p r = 1∑ R

∑ UrpqxˆΨrpqx

q

∑

p r = 1∑ R + ∑ Mr1 r = 1

R = = ∑

z c =

Mapt ArpqyˆΦrpqy

pq r = 1∑ R

∑ Brpq –xˆΨrpqx ( )

pq r = 1∑ R + ∑ Mr2 r = 1

R = = ∑
14

and , are the unknown modal amplitudes corresponding to the apertures on

plane and they satisfiy , and for ,

, and otherwise. The problem of EM coupling to a

rectangular enclosure through apertures therefore can be split into internal and external regions as

shown in Figure 2. Assuming the and planes are infinite, the entire problem can be

split into three regions: region I for , region II for , and region III for . It

should be noted that because of the presence of the infinite ground planes at and ,

the apertures on the plane do not couple to the apertures on the externally.

The internal problem consists of a rectangular volume enclosed by the enclosure and illuminated by various equivalent magnetic current sources. The external problem consists of magnetic current sources backed by infinite ground plane for each wall of the rectangular enclosure.

The unknown amplitudes , , , and appearing in (2) and (3) are determined by

setting up coupled integral equations. The expressions for the EM fields required to form integral

equation are determined in the following sections.

Φrpqy

pπ

Lr

------ Lr

2

----- x xcr + –         qπ

Wr

------- Wr

2

------- y ycr + –         = cos sin

Ψrpqx

pπ

Lr

------ Lr

2

----- x xcr + –         sin qπ

Wr

------- Wr

2

------- y ycr + –         = cos

Arpq Brpq z c =

Arpq ≠ 0 Brpq ≠ 0 xcr

Lr

2 – ----- x xcr

Lr

2 ≤ ≤ + -----

ycr

Wr

2 – ------- y ycr

Wr

2 ≤ ≤ + ------- Arpq Brpq = = 0

z = 0 z c =

z ≤ 0 0 ≤ ≤z c z c ≥

z = 0 z c =

z = 0 z c =

Vrpq Urpq Arpq Brpq
15

2.1 Electromagnetic Field Outside Enclosure

The total electromagnetic field outside the enclosure is obtained by superposition of the

field due to incident wave and the scattered field due to apertures. In the following sections the

EM field due to incident wave and the scattered field due to aperture are determined.

2.1.1 Electromagnetic Field Due to Incident Wave

The incident field with a time variation of may be written as

(4)

x

y

z

Ei

Hi

Incident Plane

M11 M21

MR1

M12 MR2

Wave

Figure 2 Geometry of equivalent problem with apertures replaced by

 magnetic current sources.

e jωt

Hi θˆiHθi φˆiHφi ( ) + e

–ki • r =

θˆi Hi α0 cos( ) φˆi Hi α0 ( ) + sin( ) e

–ki • r =
16

where , is the free-space wave

number, and are the angle of incident plane wave. From equation (4), the x-, -y, and zcomponents of the incident magnetic field may be written, respectively as

(5)

(6)

(7)

For normal incidence, with , , and , the incident field in the is

given by , , and .

2.1.2 Electromagnetic Field Scattered due to Apertures:

Consider the aperture on the plane, the EM field scattered due to can be

obtained from [8]

(8)

(9)

where the electric vector potential is given by

(10)

The factor 2 in equation (9) is considered to take into account the image. In deriving (9) it is

assumed that the plane is an infinite. Expressing in terms of plane waves [8] as

ki • r k0 θi ( ) x φi ( ) y φi ( ) cos + sin( ) zk0 θi = – sin – cos( ) k0

θi φi ( ) ,

Hxi Hθi θi () φi ( ) Hφi φi = cos cos – sin( )

Hyi Hθi θi () φi ( ) Hφi φi = cos sin( ) + cos( )

Hzi –Hθi θi = sin( )

α0 = 0 θi = 0 φi = 0 z = 0

Hxi = Hi Hyi = 0 Hzi = 0

r

th

z = 0 Mr1

E

I

( ) Mr –1

ε0

----- F

I

= ∇×

H

I

( ) Mr – jω

k0

2 --------- k0

2

F

I

F

I

= ( ) + ∇( ) ∇•

F

I

F

I ε0

4π

------ 2Mr

e

jk0 – r r – '

r r – ' ----------------------ds

Apt

∫ ∫ =

z = 0 e

jk0 – r r – '

r r – ' ----------------------
17

the electric vector potential can be written in form

(11)

where

 for and

 for

Substituting (11) in equations (8) and (9), the scattered electromagnetic field dueto the aperture is obtained. Superposition of the scattered electromagnetic field due to all apertures on the

 gives the total scattered field as

(12)

(13)

(14)

e

jk0 – r r – '

r r – ' ---------------------- 1

2πj -------- e

jkz – ( ) z z – '

kz

-------------------------e

jkx( ) x x – ' jky + ( ) y y – ' kx d ky d

–∞

∞

∫ –∞

∞

∫ =

F

I ε0

4π

2 -------- e

jkz – ( ) z z – '

jkz

-------------------------mr

     

e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ =

mr1 Mr1e

jkx( ) –x' jky + ( ) –y'

ds

Apt

∫ ∫ =

kz k0

2 kx2 ky2 = – – k02 kx2 ky2 ≥ +

j kx

2 ky2 k02 = – + – k02 kx2 ky2 ≤ ( ) +

r

th

z = 0

Ex

I

p q,

∑ r = 1

R

∑ Vrpq –

4π

2 ------------- e

jkz – ( ) z z – ' φrpqyejkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ =

Ey

I

pq

∑ r = 1

R

∑ Urpq –

4π

2 -------------- e

jkz – ( ) z z – ' ψrpqxejkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ =

Ez

I

pq

∑ r = 1

R

∑ –1

4π

2 -------- e

jkz – ( ) z z – ' Vrpqφrpqykx Urpqψrpqxky ( ) +

kz

-----------------------------------------------------------------------e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ =

Hx

I

pq

∑ r = 1

R

∑ ωε0Urpq

4π

2

k0

2 --------------------- e

jkz – ( ) z z – ' ψrpqx

k0

2 kx2 ( ) –

kz

-------------------- e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ =
18

(15)

(16)

(17)

 In expressions (12)-(17) is the Fourier transform of and is the Fourier transform of . The expressions for and are given in Appendix III.

Consider the aperture on the plane, the EM field scattered due to can be obtained

following the same procedure

(18)

(19)

(20)

pq

∑ r = 1

R

∑ ωε0Vrpq –

4π

2

k0

2 ----------------------- e

jkz – ( ) z z – ' φrpqy

kxky ( ) –

kz

------------------e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ +

Hy

I

pq

∑ r = 1

R

∑ ωε0Vrpq –

4π

2

k0

2 ----------------------- e

jkz – ( ) z z – ' φrpqy

k0

2 ky2 ( ) –

kz

-------------------- e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ =

pq

∑ r = 1

R

∑ ωε0Urpq

4π

2

k0

2 --------------------- e

jkz – ( ) z z – ' ψrpqx

kxky ( ) –

kz

------------------e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ +

Hz

I

pq

∑ r = 1

R

∑ ωε0 –

4π

2

k0

2 -------------- e

jkz – ( ) z z – ' Urpqψrpqxkx Vrpqφrpqyky ( ) – ejkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ =

φrpqy Φrpqy ψrpqx

Ψrpqx φrpqy ψrpqx

r

th

z c = Mr2

Ex

III

p q,

∑ r = 1

R

∑ Arpq

4π

2 ---------- e

jkz – ( ) z z – ' φrpqyejkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ =

Ey

III

pq

∑ r = 1

R

∑ Brpq

4π

2 ---------- e

jkz – ( ) z z – ' ψrpqxejkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ =

Ez

III

pq

∑ r = 1

R

∑ 1

4π

2 -------- e

jkz – ( ) z z – ' Arpqφrpqykx Brpqψrpqxky ( ) +

kz

----------------------------------------------------------------------e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ =

Hx

III

pq

∑ r = 1

R

∑ –ωε0Brpq

4π

2

k0

2 ----------------------- e

jkz – ( ) z z – ' ψrpqx

k0

2 kx2 ( ) –

kz

-------------------- e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ =
19

(21)

(22)

.

2.2 Electromagnetic Field Inside Enclosure

The equivalent magnetic currents, present on the apertures of an enclosure radiate electromagnetic fields inside the enclosure. The total EM field at any point inside is obtained by a superposition of fields due to each equivalent magnetic current source. Consider the aperture, the

electromagnetic field inside the enclosure due to the aperture is obtained from

(23)

(24)

where the assumed variation has been suppressed. The electric vector potential appearing in

(23) and (24) satifies the inhomogeneous wave equation

(25)

If is the dyadic Green’s function for the rectangular enclosure for a unit

dyad inside the enclosure, then the electric vector potential can

be written in the form

pq

∑ r = 1

R

∑ ωε0Arpq

4π

2

k0

2 -------------------- e

jkz – ( ) z z – ' φrpqy

kxky ( ) –

kz

------------------e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ +

Hy

III

pq

∑ r = 1

R

∑ ωε0Arpq

4π

2

k0

2 -------------------- e

jkz – ( ) z z – ' φrpqy

k0

2 ky2 ( ) –

kz

-------------------- e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ =

pq

∑ r = 1

R

∑ –ωε0Brpq

4π

2

k0

2 ----------------------- e

jkz – ( ) z z – ' ψrpqx

kxky ( ) –

kz

------------------e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ +

Hz

III

pq

∑ r = 1

R

∑ ωε0 –

4π

2

k0

2 -------------- e

jkz – ( ) z z – ' Brpqψrpqxkx Arpqφrpqyky ( ) – ejkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ =

r

th

r

th

E

II 1

ε0

---- F

II

= – ∇×

H

II – jω

k0

2 --------- k0

2

F

II

F

II

= ( ) + ∇∇•

e jωt

F

II

∇ ( ) xyz , , 2 k0

2

F

II

+ ε = – 0Mr( ) xyz , ,

G ̃ m( ) xyz , , ⁄ x', , y' z'

I x( ) ', , y' z' = xˆ xˆ + + yˆ yˆ zˆzˆ F

II
20

(26)

Substitution of (26) in (25) yields

(27)

Since the apertures are in the xy plane and can have both the x- and y-directions,

, hence equation (27) can be written in component forms as

(28)

(29)

2.2.1 Electromagnetic Fields Scattered Due to Apertures on Plane

Considering the x-component of the magentic current and using the proper boundary conditions, the solution of equation (28) can be written as

(30)

for the aperture located in the plane. In expressions (30) ,

 for and for

, for and for . Substituting (30) into (26)

the electric vector potential due to the x-component of on the plane is obtained

as

F

II

( ) xyz , , G ̃ m( ) xyz , , ⁄ x', , y' z' • Mr( ) x', , y' z' d y x'd z 'd '

Source

∫∫ ∫ =

Gm

 ̃ ∇2 k0

2

G ̃ + m ε0 = – Iδ( )δ x x – ' ( )δ y y – ' ( ) z z – '

Mr

I x( ) ', , y' z' = xˆ xˆ + yˆ yˆ

∇ Gmxx 2 k0

2

Gmxx + ε0 = – δ( )δ x x – ' ( )δ y y – ' ( ) z z – '

∇ Gmyy

2 k0

2

Gmyy + ε0 = – δ( )δ x x – ' ( )δ y y – ' ( ) z z – '

z = 0

Gxx

ε0

kI

---- ε0mε0n

ab ---------------- mπx'a -----------     sin nπy'b

----------     kI cos( ) ( ) z c –

kI sin( ) c ---------------------------------- mπx

a

----------     sin nπy

b

---------     cos cos δ( ) z'

m n,

∞

= ∑

z = . 0 ( )

m n,

∞

∑ ( ) .

n = 0

∞

∑ m = 0

∞

= ∑

kI k0

2 mπ

a

-------     2 nπ

b

------     2

= – – k0

2 mπ

a

-------     2 nπ

b

------     2

≥ + kI j mπ

a

-------     2 nπ

b

------     2

k0

2 = – + –

k0

2 mπ

a

-------     2 nπ

b

------     2

≤ ε + 0m = 1 m = 0 ε0m = 2 m ≠ 0

F

IIx0

Mr z = 0
21

(31)

The total electric vector potential due to all the apertures is then obtained by superposition as

(32)

Since , expression (32) can be written in component form as

(33)

The total magnetic field inside the enclosure is then obtained from (24) and (33) as

(34)

(35)

(36)

Fx

IIx0 ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- kI cos( ) ( ) z c – mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

= ∑

( ) Mr( ) x', , y' 0 • xˆ mπx'

a -----------     sin nπy'

b

----------     cos     d y x'd ' ∫

q

∫

Fx

IIx0 Urpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- kI cos( ) ( ) z c – mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

( ) Ψrpq( ) x', y' • xˆ mπx'

a -----------     sin nπy'b

----------     cos d y x'd ' ∫

q

∫

Ψrpq( ) x', y' xˆΨrpqx = ( ) x', y'

Fx

IIx0 Urqp

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- kI cos( ) ( ) z c – mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

Ψrpqx( ) x', y' mπx'

a -----------     sin nπy'b

----------     cos d y x'd ' ∫

q

∫

Hx

IIx0 – jω

k0

2 --------- Urpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 mπ

a

-------     2

–     mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI cos( ) ( ) z c – Irpqmnx

Hy

IIx0 – jω

k0

2 --------- Urpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------mπa

------- nπ

b –------     mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI cos( ) ( ) z c – Irpqmnx

Hz

IIx0 – jω

k0

2 --------- Urpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------mπa

------- kI ( ) – mπx

a

----------     cos

nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI sin( ) ( ) z c – Irpqmnx
22

In (34)-(36), is equal to and can be

expressed in closed form (see Appendix I ).

Likewise, considering the y-component of the magnetic current and using the proper

boundary conditions, the solution of equation (29) can be written as

(37)

for the aperture located in the plane Substituting (37) into (26) the electric vector potential due to the y-component of due to apertures in the plane is obtained as

(38)

The total electric vector potential due to all apertures on is then obtained by superposition

as

(39)

Since , expression (39) can be written as

(40)

The total magnetic field inside the enclosure is then obtained from (24) and (40) as

Irpqmnx Ψrpqx( ) x', y' mπx'

a -----------     sin nπy'b

----------     cos d y x'd ' ∫

q

∫

Gmyy

ε0 –

kI

------- ε0mε0n

ab ---------------- mπx'a -----------     cos

nπy'

b

----------     kI cos( ) ( ) z c –

kI sin( ) c ---------------------------------- mπx

a

----------     cos

nπy

b

---------     sin sin δ( ) z'

m n,

∞

= ∑

z = 0

F

IIy0

Mr z = 0

Fy

IIy0 ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- kI cos( ) ( ) z c – mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

= ∑

( ) Mr( ) x', y' • yˆ mπx'

a -----------     cos

nπy'

b

----------     sin     d y x'd ' ∫

q

∫

z = 0

Fy

IIy0 Vrpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- kI cos( ) ( ) z c – mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

( ) Φrpq( ) x', y' • yˆ mπx'

a -----------     cos

nπy'

b

----------     sin d y x'd ' ∫

q

∫

Φrqp( ) x', y' –yˆΦrqpy = ( ) x', y'

Fy

IIy0 Vrpq – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- kI cos( ) ( ) z c – mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

Φrpqy( ) x', y' mπx'

a -----------     cos

nπy'

b

----------     sin d y x'd ' ∫

q

∫
23

(41)

(42)

(43)

where appearing in (41)-(43) is equal to

and can be expressed in closed form (see Appendix II).

2.2.2 Electromagnetic Fields Scattered Due to Apertures on Plane

Considering the x-component of the magnetic current and using the proper boundary conditions, the solution of equation (28) can be written as

(44)

for the aperture located in the plane. Substituting (44) into (26) the electric vector potential due to the x-component of on the plane is obtained as

(45)

The total electric vector potential due to all apertures is then obtained by superposition as

Hx

IIy0 – jω

k0

2 --------- Vrpq – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- mπ

a –-------     nπ

b

------ mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI cos( ) ( ) z c – Irpqmny

Hy

IIy0 – jω

k0

2 --------- Vrpq – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 nπ

b

------     2

–     mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI cos( ) ( ) z c – Irpqmny

Hz

IIy0 – jω

k0

2 --------- Vrpq – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- nπb

------ kI ( ) – mπx

a

----------     cos

nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI sin( ) ( ) z c – Irpqmny

Irpqmny Φrpqy( ) x', y' mπx'

a -----------     cos

nπy'

b

----------     sin d y x'd ' ∫

q

∫

z c =

Gxx

ε0 –

kI

------- ε0mε0n

ab ---------------- mπx'a -----------     sin nπy'b

----------     kI cos( )z

kI sin( ) c -------------------- mπx

a

----------     sin nπy

b

---------     cos cos δ( ) z c –

m n,

∞

= ∑

z c =

F

IIxc

Mr z c =

Fx

IIxc ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- kI cos( )z

mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

= ∑

( ) Mr( ) x', , y' c • xˆ mπx'

a -----------     sin nπy'b

----------     cos     d y x'd ' ∫

q

∫

Fx

IIxc Brpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- kI cos( )z

mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

= ∑
24

(46)

Since , expression (46) can be written in component form

as

(47)

The total magnetic field inside the enclosure is then obtained from (24) and (47) as

(48)

(49)

(50)

Considering the y-component of the magnetic current and using the proper boundary conditions, the solution of equation (29) can be written as

(51)

for the aperture located in the plane. Substituting (51) into (26) the electric vector potential due to the y-component of due to apertures in the plane is obtained as

( ) Ψrpq( ) x', , y' c • xˆ mπx'

a -----------     sin nπy'b

----------     cos d y x'd ' ∫

q

∫

Ψrpq( ) x', , y' c –xˆΨrpqx = ( ) x', y'

Fx

IIxc Brqp ( ) – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- kI cos( )z

mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

Ψrpqx( ) x', y' mπx'

a -----------     sin nπy'b

----------     cos d y x'd ' ∫

r

∫

Hx

IIxc – jω

k0

2 --------- Brqp ( ) – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 mπ

a

-------     2

–     mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI cos( )z Irpqmnx

Hy

IIxc – jω

k0

2 --------- Brqp ( ) – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------mπa

------- nπ

b –------     mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI cos( )z Irpqmnx

Hz

IIxc – jω

k0

2 --------- Brqp ( ) – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------mπa

------- kI ( ) – mπx

a

----------     cos

nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI sin( )z Irpqmnx

Gmyy

ε0 –

kI

------- ε0mε0n

ab ---------------- mπx'a -----------     cos

nπy'

b

----------     kI cos( )z

kI sin( ) c -------------------- mπx

a

----------     cos

nπy

b

---------     sin sin δ( ) z c –

m n,

∞

= ∑

z c =

F

IIyc

Mr z c =

Fy

IIyc ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- kI cos( )z

mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

= ∑
25

(52)

The total electric vector potential due to all apertures on is then obtained by superposition

as

(53)

Since , expression (53) can be written in component form as

(54)

The total magnetic field inside the enclosure is then obtained from (24) and (54) as

(55)

(56)

(57)

For a unique solution the electromagnetic fields in various regions must satisfy continuity conditions over their common surfaces. The tangential electric fields over the apertures are continuous.

The tangential magnetic fields over the apertures must also be continuous, thus yielding coupled

( ) Mr( ) x', , y' c • yˆ mπx'

a -----------     cos

nπy'

b

----------     sin     d y x'd ' ∫

r

∫

z c =

Fy

IIyc Arpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- kI cos( )z

mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

( ) Φrpq( ) x', , y' c • yˆ mπx'

a -----------     cos

nπy'

b

----------     sin d y x'd ' ∫

r

∫

Φrqp( ) x', , y' c yˆΦrqpy = ( ) x', y'

Fy

IIyc Arpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- kI cos( )z

mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

Φrpqy( ) x', y' mπx'

a -----------     cos

nπy'

b

----------     sin d y x'd ' ∫

r

∫

Hx

IIyc – jω

k0

2 --------- Arpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- mπ

a –-------     nπ

b

------ mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI cos( )z Irpqmny

Hy

IIyc – jω

k0

2 --------- Arpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 nπ

b

------     2

–     mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI cos( )z Irpqmny

Hz

IIyc – jω

k0

2 --------- Arpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- nπb

------ kI ( ) – mπx

a

----------     cos

nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI sin( )z Irpqmny
26

integral equations with the magnetic currents as unknown variables. The coupled integral equation in conjunction with the method of moments can be solved for the amplitudes of magnetic currents.

2.3 Derivation of Integral Equation

The total tangential magnetic fields inside the cavity from apertures on both sides are written as

Using continuity of tangential magnetic fields across the apertures in the plane yields

(58)

(59)

And matching the tangential magnetic fields across the apertures in plane yields

(60)

(61)

In deriving equations (58)-(61) it is assumed that the cavity is excited by a plane wave incident

from .

2.3.1 Integral Equation for Apertures on Plane

Using the field expressions derived in the earlier sections, equation (58) can be written as

Hx

II HxIIx0 HxIIy0 HxIIxc HxIIyc = ( ) +++

Hy

II HyIIx0 HyIIy0 HyIIxc HyIIyc = ( ) +++

z = 0

Hxi

z = 0

Hx

I

z = 0

+ Hx

IIx0 HxIIy0 HxIIxc HxIIyc ( ) +++

z = 0 =

Hyi z = 0Hy

I

z = 0

+ Hy

IIx0 HyIIy0 HyIIxc HyIIyc ( ) +++

z = 0 =

z c =

Hx

IIx0 HxIIy0 HxIIxc HxIIyc ( ) +++

z c = Hx

III

z c = =

Hy

IIx0 HyIIy0 HyIIxc HyIIyc ( ) +++

z c = Hy

III

z c = =

z = –∞

z = 0

Hxi

pq

∑ r = 1

R

∑ ωε0Urpq

4π

2

k0

2 --------------------- ψrpqx

k0

2 kx2 ( ) –

kI

-------------------- e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ +
27

(62)

Now selecting as a testing function and use of Galerkin’s method reduces the above equation to

(63)

Rearranging the terms in (63) we get

pq

∑ r = 1

R

∑ ωε0Vrpq –

4π

2

k0

2 ----------------------- φrpqy

kxky ( ) –

kI

------------------e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ +

– jω

k0

2 --------- Urpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 mπ

a

-------     2

–     mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI cos( ) c Irpqmnx

– jω

k0

2 --------- Vrpq – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- mπ

a –-------     nπ

b

------ mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

∑

kI cos( ) c Irpqmny

– jω

k0

2 --------- Brqp ( ) – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------Irpqmnx     k0

2 mπ

a

-------     2

–     mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

∑

– jω

k0

2 --------- Arpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------Irpqmny     mπ

a –-------     nπ

b

------ mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

∑

Ψr' p'q'x

Ir' p'q'xi

pq

∑ r = 1

R

∑ ωε0Urpq

4π

2

k0

2 --------------------- ψrpqxψr' p'q'x

* k0

2 kx2 ( ) –

kI

-------------------- kx d ky d

–∞

∞

∫ –∞

∞

∫ +

pq

∑ r = 1

R

∑ ωε0Vrpq –

4π

2

k0

2 ----------------------- φrpqyψr' p'q'x

* ( )

kxky ( ) –

kI

------------------ kx d ky d

–∞

∞

∫ –∞

∞

∫ +

– jω

k0

2 --------- Urpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 mπ

a

-------     2

–     kI cos( ) c IrpqmnxIr' p'q'mnx

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

– jω

k0

2 --------- Vrpq – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- mπ

a –-------     nπ

b

------ kI cos( ) c IrpqmnyIr' p'q'mnx

m n,

∞

∑

p q,

∑ r = 1

R

∑

– jω

k0

2 --------- Brqp ( ) – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnxIr' p'q'mnx     k0

2 mπ

a

-------     2

–    

m n,

∞

∑

p q,

∑ r = 1

R

∑

– jω

k0

2 --------- Arpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnyIr' p'q'mnx     mπ

a –-------     nπ

b

------

m n,

∞

∑

p q,

∑ r = 1

R

∑
28

(64)

where

(65)

(66)

(67)

(68)

(69)

Using expression (59) we get

Ir' p'q'xi

pq

∑ r = 1

R

∑ UrpqYrpqr' p'q'

x1x1 VrpqYrpqr' p'q'x1y1 ArpqYrpqr' p'q'x1y2 BrqpYrpqr' p'q'x1x2 = ( +++ )

Yrpqr' p'q'

x1x1 – jω

k0

2 --------- ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 mπ

a

-------     2

–     kI cos( ) c IrpqmnxIr' p'q'mnx

m n,

∞

= ∑

ωε0

4π

2

k0

2 –-------------- ψrpqxψr' p'q'x

* k0

2 kx2 ( ) –

kz

-------------------- kx d ky d

–∞

∞

∫ –∞

∞

∫

Yrpqr' p'q'

x1y1 jω

k0

2 ------ ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- mπ

a –-------     nπ

b

------ kI cos( ) c IrpqmnyIr' p'q'mnx

m n,

∞

= ∑

ωε0

4π

2

k0

2 -------------- φrpqyψr' p'q'x

* ( )

kxky ( ) –

kz

------------------ kx d ky d

–∞

∞

∫ –∞

∞

∫ +

Yrpqr' p'q'

x1x2 jω

k0

2 ------ ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnxIr' p'q'mnx     k0

2 mπ

a

-------     2

–    

m n,

∞

= ∑

Yrpqr' p'q'

x1y2 – jω

k0

2 --------- ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnyIr' p'q'mnx     mπ

a –-------     nπ

b

------

m n,

∞

= ∑

Ir' p'q'xi HxiΨr' p'q'xdxdy

r' p'q'

∫ ∫ =

Hyi

pq

∑ r = 1

R

∑ ωε0Vrpq –

4π

2

k0

2 ----------------------- φrpqy

k0

2 ky2 ( ) –

kz

-------------------- e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ +

pq

∑ r = 1

R

∑ ωε0Urpq

4π

2

k0

2 --------------------- ψrpqx

kxky ( ) –

kz

------------------e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ +

– jω

k0

2 --------- Urpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------mπa

------- nπ

b –------     mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI cos( ) c Irpqmnx
29

(70)

Now selecting as a testing function and use of Galerkin’s method yields

(71)

Rearranging the terms in (71) we get

(72)

where

– jω

k0

2 --------- Vrpq – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 nπ

b

------     2

–     mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

∑

kI cos( ) c Irpqmny

– jω

k0

2 --------- Brqp ( ) – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------Irpqmnx     mπ

a

------- nπ

b –------     mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

∑

– jω

k0

2 --------- Arpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------Irpqmny     k0

2 nπ

b

------     2

–     mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

∑

–Φr' p'q'y

Ir' p'q'yi

pq

∑ r = 1

R

∑ –ωε0Vrpq

4π

2

k0

2 ----------------------- φrpqyφr' p'q'y

* k0

2 ky2 ( ) –

kz

-------------------- kx d ky d

–∞

∞

∫ –∞

∞

∫ =

pq

∑ r = 1

R

+ ∑ ωε0Urpq

4π

2

k0

2 --------------------- ψrpqxφr' p'q'y

* ( )

kxky ( ) –

kz

------------------ kx d ky d

–∞

∞

∫ –∞

∞

∫

jω

k0

2 ------ Urpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------mπa

------- nπ

b –------     kI cos( ) c IrpqmnxIr' pq'mny

m n,

∞

∑

p q,

∑ r = 1

R

+ ∑

jω

k0

2 ------ Vrpq – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 nπ

b

------     2

–     kI cos( ) c IrpqmnyIr' pq'mny

m n,

∞

∑

p q,

∑ r = 1

R

+ ∑

jω

k0

2 ------ Brqp ( ) – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnxIr' pq'mny     mπ

a

------- nπ

b –------    

m n,

∞

∑

p q,

∑ r = 1

R

+ ∑

jω

k0

2 ------ Arpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnyIr' pq'mny     k0

2 nπ

b

------     2

–    

m n,

∞

∑

p q,

∑ r = 1

R

+ ∑

Ir' p'q'yi

pq

∑ r = 1

R

∑ UrpqYrpqr' p'q'

y1x1 VrpqYrpqr' p'q'y1y1 ArpqYrpqr' p'q'y1y2 BrqpYrpqr' p'q'y1x2 = ( +++ )

Yrpqr' p'q'

y1x1 ωε0

4π

2

k0

2 -------------- ψrpqxφr' p'q'y

* ( )

kxky ( ) –

kz

------------------ kx d ky d

–∞

∞

∫ –∞

∞

∫ =
30

(73)

(74)

(75)

(76)

(77)

2.3.2 Integral Equations for Apertures on Plane

Using (60), we get

jω

k0

2 ------ ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------mπa

------- nπ

b –------     kI cos( ) c IrpqmnxIr' pq'mny

m n,

∞

+ ∑

Yrpqr' p'q'

y1y1 –ωε0

4π

2

k0

2 -------------- φrpqyφr' p'q'y

* k0

2 ky2 ( ) –

kz

-------------------- kx d ky d

–∞

∞

∫ –∞

∞

∫ =

– jω

k0

2 --------- ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 nπ

b

------     2

–     kI cos( ) c IrpqmnyIr' pq'mny

m n,

∞

∑

Yrpqr' p'q'

y1x2 – jω

k0

2 --------- ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnxIr' pq'mny     mπ

a

------- nπ

b –------    

m n,

∞

= ∑

Yrpqr' p'q'

y1y2 jω

k0

2 ------ ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnyIr' pq'mny     k0

2 nπ

b

------     2

–    

m n,

∞

= + ∑

Ir' p'q'yi HyiΦr' p'q'xdxdy

r' p'q'

∫ ∫ = –

z c =

– jω

k0

2 --------- Urpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------Irpqmnx     k0

2 mπ

a

-------     2

–     mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

∑

– jω

k0

2 --------- Vrpq – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------Irpqmny     mπ

a –-------     nπ

b

------ mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

∑

– jω

k0

2 --------- Brqp ( ) – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 mπ

a

-------     2

–     mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

∑
31

(78)

Now selecting testing function and use of Galerkin’s method yields

(79)

Rearranging the terms in (79) we get

(80)

kI cos( ) c Irpqmnx

– jω

k0

2 --------- Arpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- mπ

a –-------     nπ

b

------ mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

∑

kI cos( ) c Irpqmny

pq

∑ r = 1

R

∑ –ωε0Brpq

4π

2

k0

2 ----------------------- ψrpqx

k0

2 kx2 ( ) –

kI

-------------------- e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ =

pq

∑ r = 1

R

∑ ωε0Arpq

4π

2

k0

2 -------------------- φrpqy

kxky ( ) –

kI

------------------e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ +

–Ψr' p'q'x

0 – jω

k0

2 --------- Urpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnxIr' p'q'mnx k0

2 mπ

a

-------     2

–    

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

jω

k0

2 ------ Vrpq – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnyIr' p'q'mnx

mπ

a –-------     nπ

b

------

m n,

∞

∑

p q,

∑ r = 1

R

+ ∑

– jω

k0

2 --------- Brqp ( ) – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 mπ

a

-------     2

–     kI cos( ) c IrpqmnxIr' p'q'mnx

m n,

∞

∑

p q,

∑ r = 1

R

∑

– jω

k0

2 --------- Arpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- mπ

a –-------     nπ

b

------ kI cos( ) c IrpqmnyIr' p'q'mnx

m n,

∞

∑

p q,

∑ r = 1

R

∑

pq

∑ r = 1

R

∑ ωε0Brpq

4π

2

k0

2 -------------------- ψrpqxψr' p'q'x

* k0

2 kx2 ( ) –

kz

-------------------- kx d ky d

–∞

∞

∫ –∞

∞

∫ +

pq

∑ r = 1

R

– ∑ ωε0Arpq

4π

2

k0

2 -------------------- φrpqyψr' p'q'x

* kxky ( ) –

kz

------------------ kx d ky d

–∞

∞

∫ –∞

∞

∫

0

pq

∑ r = 1

R

∑ UrpqYrpqr' p'q'

x2x1 VrpqYrpqr' p'q'x2y1 ArpqYrpqr' p'q'x2y2 BrqpYrpqr' p'q'x2x2 = ( +++ )
32

where

(81)

(82)

(83)

(84)

From (61) we get

Yrpqr' p'q'

x2x1 – jω

k0

2 --------- ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnxIr' p'q'mnx k0

2 mπ

a

-------     2

–    

m n,

∞

= ∑

Yrpqr' p'q'

x2y1 jω

k0

2 –------ ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnyIr' p'q'mnx

mπ

a –-------     nπ

b

------

m n,

∞

= ∑

Yrpqr' p'q'

x2x2 jω

k0

2 ------ ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 mπ

a

-------     2

–     kI cos( ) c IrpqmnxIr' p'q'mnx

m n,

∞

= ∑

ωε0

4π

2

k0

2 -------------- ψrpqxψr' p'q'x

* k0

2 kx2 ( ) –

kz

-------------------- kx d ky d

–∞

∞

∫ –∞

∞

∫ +

Yrpqr' p'q'

x2y2 – jω

k0

2 --------- ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- mπ

a –-------     nπ

b

------ kI cos( ) c IrpqmnyIr' p'q'mnx

m n,

∞

= ∑

ωε0

4π

2

k0

2 –-------------- φrpqyψr' p'q'x

* kxky ( ) –

kz

------------------ kx d ky d

–∞

∞

∫ –∞

∞

∫

– jω

k0

2 --------- Urpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------Irpqmnx     mπ

a

------- nπ

b –------     mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

∑

– jω

k0

2 --------- Vrpq – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------Irpqmny     k0

2 nπ

b

------     2

–     mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

∑

– jω

k0

2 --------- Brqp ( ) – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------mπa

------- nπ

b –------     mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

∑
33

(85)

Now selecting as a testing function and use of Galerkin’s method yields

(86)

Rearranging the terms in (84) we get

(87)

kI cos( ) c Irpqmnx

– jω

k0

2 --------- Arpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 nπ

b

------     2

–     mπx

a

----------     cos

nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

∑

kI cos( ) c Irpqmny

pq

∑ r = 1

R

∑ ωε0Arpq

4π

2

k0

2 -------------------- φrpqy

k0

2 ky2 ( ) –

kz

-------------------- e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ =

pq

∑ r = 1

R

∑ –ωε0Brpq

4π

2

k0

2 ----------------------- ψpqx

kxky ( ) –

kz

------------------e

jkx x jky + y kx d ky d

–∞

∞

∫ –∞

∞

∫ +

Φr' p'q'y

0 jω

k0

2 ------ Urpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnxIr' p'q'mny

mπ

a

------- nπ

b –------    

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

jω

k0

2 ------ Vrpq – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnyIr' p'q'mny k0

2 nπ

b

------     2

–    

m n,

∞

∑

p q,

∑ r = 1

R

+ ∑

jω

k0

2 ------ Brqp ( ) – ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------mπa

------- nπ

b –------     kI cos( ) c IrpqmnxIr' p'q'mny

m n,

∞

∑

p q,

∑ r = 1

R

+ ∑

pq

∑ r = 1

R

∑ –ωε0Brpq

4π

2

k0

2 ----------------------- ψpqxφr' p'q'y

kxky ( ) –

kz

------------------ kx d ky d

–∞

∞

∫ –∞

∞

∫ +

jω

k0

2 ------ Arpq

ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 nπ

b

------     2

–     kI cos( ) c IrpqmnyIr' p'q'mny

m n,

∞

∑

p q,

∑ r = 1

R

+ ∑

pq

∑ r = 1

R

∑ ωε0Arpq

4π

2

k0

2 -------------------- φrpqyφr' p'q'y

k0

2 ky2 ( ) –

kz

-------------------- kx d ky d

–∞

∞

∫ –∞

∞

∫ +

0

pq

∑ r = 1

R

∑ UrpqYrpqr' p'q'

y2x1 VrpqYrpqr' p'q'y2y1 ArpqYrpqr' p'q'y2y2 BrqpYrpqr' p'q'y2x2 = ( +++ )
34

where

(88)

(89)

(90)

(91)

Equations (64), (72), (80), and (87) can be written in a matrix form as

(92)

The matrix equation (92) can be numerically solved for the unknown amplitudes of equivalent

magnetic currents induced on the apertures due to given incident field. From the knowledge of

these amplitudes electromagnetic field inside as well as outside the cavity can be determined. The

shielding effectiveness of the enclosure with rectangular apertures is then determined from the

Yrpqr' p'q'

y2x1 jω

k0

2 ------ ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnxIr' p'q'mny

mπ

a

------- nπ

b –------    

m n,

∞

= ∑

Yrpqr' p'q'

y2y1 – jω

k0

2 --------- ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------IrpqmnyIr' p'q'mny k0

2 nπ

b

------     2

–    

m n,

∞

= ∑

Yrpqr' p'q'

y2x2 – jω

k0

2 --------- ε0 –

kI

------- ε0mε0n

ab kI sin( ) c --------------------------mπa

------- nπ

b –------     kI cos( ) c IrpqmnxIr' p'q'mny

m n,

∞

= ∑

–ωε0

4π

2

k0

2 -------------- ψpqxφr' p'q'y

* kxky ( ) –

kz

------------------ kx d ky d

–∞

∞

∫ –∞

∞

∫ +

Yrpqr' p'q'

y2y2 jω

k0

2 ------ ε0 –

kI

------- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 nπ

b

------     2

–     kI cos( ) c IrpqmnyIr' p'q'mny

m n,

∞

= ∑

ωε0

4π

2

k0

2 -------------- φrpqyφr' p'q'y

* k0

2 ky2 ( ) –

kz

-------------------- kx d ky d

–∞

∞

∫ –∞

∞

∫ +

Yrpqr' p'q'

x1x1 Yrpqr' p'q'x1y1 Yrpqr' p'q'x1y2 Yrpqr' p'q'x1x2

Yrpqr' p'q'

y1x1 Yrpqr' p'q'y1y1 Yrpqr' p'q'y1y2 Yrpqr' p'q'y1x2

Yrpqr' p'q'

x2x1 Yrpqr' p'q'x2y1 Yrpqr' p'q'x2y2 Yrpqr' p'q'x2x2

Yrpqr' p'q'

y2x1 Yrpqr' p'q'y2y1 Yrpqr' p'q'y2y2 Yrpqr' p'q'y2x2

Urpq

Vrpq

Arpq

Brqp

Ir' p'q'xi

Ir' p'q'yi

0

0

=
35

EM fields using following expression [3]

(93)

where is the total electric field at a given point inside the enclosure and is the field at the

same point in absence of the enclosure. The expressions for electric field components inside the

cavity due external EM sources are given in Appendix IV.

3.0 Numerical Results & Discussions

3.1 Single Aperture Case

In this section we consider a variety of simple cases and compare the numerical results

obtained using the present method with results published earlier [3] for validation of the computer

code.

Consider a rectangular enclosure with single rectangular aperture on the plane and

illuminated by a plane wave at normal incidence polarized in the x-direction (as shown in Figure

3). The matrix equation (92) for this case reduces to

(94)

For normal incidence

E– Shielding dB ( ) 20. Eint

Eext

------------

      = – log

Eint Eext

z = 0

Yrpqr' p'q'

x1x1 Yrpqr' p'q'x1y1

Yrpqr' p'q'

y1x1 Yrpqr' p'q'y1y1

Urpq

Vrpq

Ir' p'q'xi

0

=

Ir' p'q'xi

HxiWr

1 – cos( ) p'π

( ) p'π Lr' ⁄ ------------------------------    , for q( ) ' 0 =

  0, for q( ) ' 0 ≠

    =
36

 Using the orthogonality of expansion functions it can be shown ,

hence equation (93) simplifies to

(95)

 Furthermore, it has been found from numerical results that the expansion modes ,

and are strongly excited. In equation (95) and are and

For a numerical solution of (94) we consider a rectangular cavity with cms,

cms, and cms with a rectangular aperture of size ( ) cms2

, located at

 cms, and illuminated by a plane wave at normal incidence. Assuming only

 expansion mode on the aperture and considering only dominant

mode inside the cavity, electric field shielding obtained using expression (93) is plotted in Figure

3.1 along with the results from [3]. It is observed that the numerical data obtained using the

present method agrees well with the earlier published results. Experimental data from [3] is also

reproduced in Figure 3.1.

Yrpqr' p'q'

x1y1 Yrpqr' p'q'y1x1 = = 0

Yrpqr' p'q'

x1x1 Urpq Ir' p'q'xi =

p = 1 3 5 ,... , ,

q = 0 p' q' p' 1 3 5 ,... = , , q' 0 =

a = 30

b = 12 c = 10 30 x0.5

( ) 15 6 0 , ,

p = = 1, q 0 m = = 1, n 0
37

0 200 400 600 800 1000 -40

-20

0

20

40

60

80

100

Ey

Hx

30cm

12cm

30cm

Rectangular

Figure 3.1 Electric field shielding at the center of 30 cm x 12 cm x 30 cm enclosure

 10 x 0.5 cm2 aperture located at 15 x 6 cm in z=0 plane with dominant

Frequency (MHz)

Electric Field Shielding (dB)

Present Method

Calculation [3]

Apt.( 10 x0.5) cm2

Measurment [3]

 cavity mode considered.

 with
38

In order to study the effect of higher order modes, the electric field shielding is calculated

using the present method with expansion mode and on the aperture

and considering modes including dominant mode inside the

cavity. The electric field shielding obtained using the present method is plotted in Figure 3.2

along with the results from [3]. The numerical data in Figure 3.2 shows that consideration of

higher order modes inside the cavity gives results closer to the measured values. Through numerical calculations, it was also noted that inclusion of higher order modes on the aperture did not

alter the electric field shielding values shown in Figure 3.2.

To validate the present method for a wider aperture, the electric field shielding of a rectan0 200 400 600 800 100 -40

-20

0

20

40

60

80

100

Ey

Hx

30cm

12cm

30cm

Rectangular

Frequency (MHz)

Electric Field Shielding (dB)

Present Method(Dominant Mode)

Calculation [3]

Apt.( 10 x0.5)cm2

Measurment [3]

 higher order modes inside cavity and single mode on aperture were considered.

Figure 3.2 Electric field shielding at the center of 30 cm x 12 cm x 30 cm enclosure

 10 x 0.5 cm2 with aperture located at 15 x 6 cm in z=0 plane when

..............(Higher Order Modes)

p = = 1, q 0 p' 1 = = , q' 0

m max – 1200 = = , n max – 200
39

0 200 400 600 800 1000 -20

0

20

40

60

80

gular enclosure ( 30 cms x 12 cms x 30 cms) with rectangular aperture (20 cms x 3 cms) as shown

in Figure 4 is determined as a function of frequency and presented in Figure 4 along with earlier

published data [3]. It may be observed from the data presented in Figure 4 that numerical results

obtained by the simple transmission line model used in [3] predicts less electric field shielding

than the present method. However, the location of minimum shielding predicted by the present

methods agrees better with the measured data compared to transmission line model of [3].

In the numerical data presented in Figure 4 only a single mode (i.e. ) over

the aperture and dominant mode inside the cavity were considered. To study the effects of higher

order cavity modes on the electric field shielding, a numerical example with the cavity and aper0 200 400 600 800 1000 -20

0

20

40

60

80

Figure 4 Electric field shielding at the center of (30 x 12 x 30) cm3 enclosure with

 (20 x 3.0) cm2 aperture located at 15 x 6 cm in z=0 plane

Frequency (MHz)

Electric Field Shielding (dB)

Present Method

Calculation [3]

Measured [3]

Ey

Hx

30cm

12cm

30cm

Rectangular

Apt.( 20 x3.0)cm2

p = = 1, q 0
40

ture sizes same as shown in Figure 4 is considered. The electric field shielding calculated at the

center of the rectangular enclosure considering maximum value of equal to 100 and maximum

value of equal to 100 is shown in Figure 5 along with other numerical data. From the numerical

data presented in Figure 5 it may be concluded that inclusion of higher order cavity modes

improves electric field shielding estimation.

Figures 6 and 7 show the electric field shielding when more than a single mode is considered at the aperture. The dimensions of the enclosure and the rectangular aperture for Figures 6

and 7 were the same as shown in Figure 4. From Figures 6 and 7 it may be concluded that for the

m

n

0 200 400 600 800 1000 -20

0

20

40

60

80

Figure 5 Electric field shielding at the center of (30 x 12 x 30) cm3 enclosure with

 (20 x 3.0) cm2

 aperture located at 15 x 6 cm in z=0 plane

Frequency (MHz)

Electric Field Shielding (dB)

Present Method(Dominant Mode)

(Higher Order Mode) Calculation [3]

Measured [3]

Ey

Hx

30cm

12cm

30cm

Rectangular

Apt.( 20 x3.0)cm2

 ”
41

aperture size and the frequency range considered, a single dominant mode at the aperture is adequet for achieving convergent results.

0 200 400 600 800 1000 -20

0

20

40

60

80

Present Method (1,0 Apt. mode) -----(1,0 + 3,0 Apt. modes) Calculation [3]

Measured [3]

Figure 6 Electric field shielding at the center of (30 x 12 x 30) cm3 enclosure with

 (20 x 3.0) cm2

 aperture located at 15 x 6 cm in z=0 plane

Frequency (GHz)

Electric Field Shielding (dB)

Ey

Hx

30cm

12cm

30cm

Rectangular

Apt.( 20 x3.0)cm2
42

3.2 Two Aperture Case

Consider a rectangular enclosure with two rectangular apertures, one on the and

another on the planes. It is assumed that the enclosure is illuminated by a plane wave at

normal incidence on the aperture at the plane. Due to assumed polarization of the incidence wave it can be shown that and . Hence the matrix equation (92)

reduces to

(96)

0 200 400 600 800 1000 -20

0

20

40

60

80

Figure 7 Electric field shielding at the center of (30 x 12 x 30) cm3 enclosure with

 (20 x 3.0) cm2 aperture located at 15 x 6 cm in z=0 plane

Frequency (GHz)

Electric Field Shielding (dB)

Apt. Mode (1,0)

Apt. Modes (1,0 + 3,0)

Apt. Modes (1,0 + 3,0 + 5,0

Calculation [3]

Measured [3]

Ey

Hx

30cm

12cm

30cm

Rectangular

Apt.( 20 x3.0)cm2

z = 0

z c =

z = 0

Vrpq = 0 Arpq = 0

Yrpqr' p'q'

x1x1 Yrpqr' p'q'x1x2

Yrpqr' p'q'

x2x1 Yrpqr' p'q'x2x2

Urpq

Brpq

Ir' p'q'xi

0

=
43

The matrix equation (96) is solved for and using numerical methods. From

 and the electric field shielding can be determined using expression (93). For numerical solution of (96) we consider a rectangular cavity with cms, cms, and

cms with two rectangular apertures. One of the rectangular apertures was of size

( ) cm2, located at cms, and another rectangular aperture of the same size as the

first aperture but located at cms. The rectangular cavity is illuminated by plane wave at

normal incidence to the first aperture. Assuming only expansion mode on the aperture and considering only dominant mode inside the cavity, the electric field

shielding obtained using the present method is plotted in Figure 8 (Solid Line). The numerical

data represented by the solid line in Figure 8 is obtained with the internal coupling between the

apertures not zero (i.e. setting ). The electric field shielding of the

rectangular enclosure with two apertures is also obtained by extending the simple transmission

line model described in [3]. In this model the expression for given in equation (6) of reference

[3] is modified to

(97)

to account for the second aperture. The numerical data obtained using the simple transmission

line are also plotted in Figure 8. General behaviour of the electric field shielding as a function of

frequency obtained by the present method and the transmission line method [3] is in good agreement. It is also observed that when the mutual coupling between the apertures is not considered

the present method gives shielding results identical to the results obtained using transmission line

Urpq Brpq

Urpq Brpq

a = 30 b = 12

c = 30

20x3.0 15 6 0 , ,

15 6 30 , ,

p = = 1, q 0

m = = 1, n 0

Yrpqr' p'q'

x2x1 Yrpqr' p'q'x1x2

≠ ≠ 0

Z3

Z3 Zg

Zap jZg kg ( ) + tan( ) ( ) d p –

Zg jZap kg + tan( ) ( ) d p – = ----------------------------------------------------------------
44

method.

In Figure 9 the electric field shielding for the enclosure and aperture dimensions as given

in Figure 8 is studied when higher order modes inside the cavity and on the apertures are considered.

0 200 400 600 800 1000

-20

-10

0

10

20

30

40

50

60

70

80

Calculation [3]

Present Method (With Internal Coupling)

Present Method (Without Internal Coupling)

Frequency (GHz)

Electric Field Shielding (dB)

Figure 8 Electric field shielding at the center of (30 x 12 x 30) cm3 enclosure with

 two identical apertures with (20 x 3.) cm2

 size. Aperture one located at

 (15 ,6 , 0) cms, aperture two located at (15, 6 , 30) cms.
45

0 200 400 600 800 1000

-20

-10

0

10

20

30

40

50

60

70

80

Frequency (GHz)

Electric Field Shielding (dB)

Figure 9(a) Effect of higher order aperture modes on electric field shielding

 at the center of (30 x 12 x 30) cm3 enclosure with

 with (20 x 3.) cm2 size. Aperture one located at

 aperture two located at (15 , 6 , 30) cms.

 two identical apertures

 (15 , 6 , 0) cms,

Apt. Mode (1,0), Cavity Modes (1,0)

Apt. Mode ( (1,0) + (3,0) ), Cavity Mode(1,0)

Calculation [3]
46

To validate the Modal Method (MM) presented in this report, in the following section, we

compare the electric field shielding data obtained using the MM method with the FEM-MoM [8]

techniques. For numerical simulation we consider a rectangular cavity with cms,

cms, and cms with two rectangular apertures. One of the rectangular apertures

was of size ( ) cm2

, located at cms, and another rectangular aperture of the same

size as the first aperture but located at cms. The rectangular cavity is illuminated by a

plane wave at normal incidence to the first aperture. Assuming two aperture modes and m-max =

0 200 400 600 800 1000

-20

-10

0

10

20

30

40

50

60

70

80

Frequency (GHz)

Electric Field Shielding (dB)

Apt. Mode (1,0), Cavity Modes (1,0)

Apt. Mode ( (1,0) + (3,0) ),Cavity Mode(1,0)

Calculation [3]

Apt. Mode ( (1,0) + (3,0) ) #

Figure 9(b) Effect of higher order aperture modes on electric field shielding

 at the center of (30 x 12 x 30) cm3 enclosure with

 with (20 x 3.) cm2 size. Aperture one located at

 aperture two located at (15 , 6 , 30) cms. # Cavity modes considered

 two identical apertures

 (15 , 6 , 0) cms,

(100 x100).

a = 30

b = 12 c = 30

20x3.0 15 6 0 , ,

15 6 30 , ,
47

100, and n-max = 100, the electric field shielding at the center of the cavity is calculated using the

present method and presented in Figure 10. In Figure 10, the numerical data on electric field

shielding of the cavity using the hybrid Finite Element Method-Method of Moments (FEM/MoM)

[ 8] is also presented for various sizes of the apertures. Eexcellent agreement between the results

obtained from the present and FEM/MoM methods validates the accuracy of the present method.

 Figure 11 shows a rectangular cavity excited through four apertures; two on each side. The cavity and aperture sizes used in Figure 11 are small enough so that shielding effectiveness of the

cavity can be calculated using FEM/MoM for validation purpose. The electric field shielding of

the cavity shown in Figure 11 is calculated using the present and FEM/MoM methods and pre0 200 400 600 800 1000 -20

0

20

40

60

Aperture Size

( ) 20 3 × cm

2

( ) 20 8 × cm

2

( ) 30 12 × cm

2

( ) 30 12 × cm

2

( ) 20 8 × cm

2

( ) 20 3 × cm

2

}

}

Modal Method

FEM/MoM [8]

Frequency (MHz)

E_Field Shielding (dB)

Figure 10 Electric field shielding of rectangular enclosure of size (30 x 12 x 30 ) cm3 with

 two identical rectangular apertures ( one on each side) of various sizes when

 illuminated by incident plane wave at normal incidence.
48

sented in Figure 12. A good agreement between the two results further validate the present

method. The small disagreement at higher frequencies between the two methods may be attributed to the numerical inaccuracy involved in the FEM/MoM methods due to discretization level.

6

20

8

20

6

8

x

y

z

30

60

12

6

Figure 11 Rectangular cavity with four identical apertures and illuminated by

 plane wave with normal incidence. All dimensions in cm

Ey

Hx Incident

Wave

Rectangular

Cavity

Rectangular

Apertures

20
49

3.3 EM Field Penetration Inside B-757 Aircraft Configuration

A typical cross section of a B-757 aircraft with various dimensions is shown in Figure 13.

For simplicity, the inner complexities of passenger seat arrangement and overhead compartments

are assumed not to be present. The cross section shown in Figure 13 can be approximated by a

rectangular crosssection as shown in Figure 13. Therefore the entire fuselage of a B-757 can be

approximated by a rectangular cavity of dimensions shown in Figure 14(a). An external electromagnetic source would cause EM field to penetrate through passenger windows and cockpit windows into the fuselage. The EM field present inside the fuselage due to any external EM source

may coupled to sensitive instrumentation wiring and hence cause electronic upset. Furthermore,

if the EM field strength inside a fuselage is sufficiently large it may cause electric spark on broken

0 200 400 600 800 1000

-10

0

10

20

30

40

50

Apt. Modes ( (1,0) + (3,0))

Apt. Modes ((1,0) +(3,0) + (5,0))

FEM/MoM [8]

Figure 12 Electric field shielding of rectangular cavity shown in Figure 11. Observation

 point ( 30 , 6, 15 ) cm.

Frequency (MHz)

E_Field Shielding (dB)
50

wires. It is therefore important to estimate electromagnetic field shielding of a B-757 fuselage or

for that matter any aircraft passenger cabin. The external EM field may also penetrate a B-757

fuselage through the cockpit windows. However for an aircraft illuminated from sides passenger

windows will be dominant contributors. Hence in this report only side windows are considered.

Approximating the B-757 fuselage by a rectangular cavity with rectangular apertures representing

passenger windows the computer code developed here is used to determine the field strength

inside the passenger cabin due to any external EM source. For the worst case scenario, it is

assumed that the cavity is excited by an electromagnetic plane wave at normal incidence with veritcal as well as horizontal polarizations.

For the numerical estimate of electric field shielding of a B-757 fuselage, we first consider

the effect of two windows, one on each side of a rectangular cavity ( 3600 x 400 x 400 cms) as

shown in Figure 14(b). The window dimensions selected for the numerical simulation were ( 25 x

35) cm2

. The windows were assumed to be located with their centers at (1800 , 200 , 0.0) cms and

(1800, 200, 400) cms. Using the computer code developed here, the electric field shielding calculated at (1800, 120, 200) cms as a function of frequency is presented in Figure 13. In the numerical simulation it is assumed that the incident wave is polarized in the y-direction. The numbers

of cavity modes considered in this example were and

. Any further addition of cavity modes was found to have no appreciable

effect on the shielding calculation. For the numerical results shown in Figure 15 only a single

dominant mode (i.e. was considered on the window apertures. For comparison,

the measured values of electric field shielding for a B-757 aircraft at two separate frequencies are

also shown in Figure 15. However, these measurements were done under different incident wave

conditions and hence are shown here only for illustrating that the numerical data obtained using

m = 0 1 2 .....,1200 ,,,

n = 0 1 2 ....200 ,,,

p = = 1, q 0
51

the present method are in the range of measured values.

Rectangular

Cross Section

Figure 13 Geometry of B-757 aircraft configuration with cross sectional and

 longitudinal view. Dotted line shows approximated cross section.
52

400 cms

400cms

3600cms

X

Y

Z

Passenger

Windows

Figure 14(a) Closed rectangular cavity approximating B-757 fuselage with

 rectangular apertures on side walls representing passenger windows.

Ey

Hx

Incident

Plane

Wave

Rectangular

Cavity

25cms

35cms
53

Figures 15-17 shows the electric field shielding when higher order modes on the apertures

were considered. In Figures 15-17, the incident wave was considered to be parallel to the longer

dimensions of the aperture windows. For the frequency range shown in Figures 15-17 it is clear

that single mode on the aperture is adequate for obtaining convergent results. Figure 18 shows the electric field shielding of a rectangular cavity with two apertures when the incident wave was polarized parallel to the shorter dimension of the window. The dimensions of the

400 cms

400cms

3600cms

X

Y

Z

Passenger

Windows

Figure 14(b) Closed rectangular cavity approximating B-757 fuselage with

 two rectangular apertures on side walls representing passenger windows.

Ey

Hx

Incident

Plane

Wave

Rectangular

Cavity

25cms

35cms

p = = 1, q 0
54

cavity and aperture used in numerical data shown in Figure 18 were the same as given in Figure

15. Only a single aperture mode was considered in the numerical data shown in Figure 18.

0 200 400 600 800 1000 -20

0

20

40

60

80

100

Figure 15 Electric field shielding of rectangular cavity approximating fuselage of

 B-757 aircraft. Two windows on each side of cavity were considered.

 Cavity size ( 3600 x 400 x 400) cm3, window size (25 x 35) cm2, window

 locations ( 1800, 200, 0.0) cms and (1800, 200, 400) cms.

 Field point location (1800, 120, 200) cms.

Frequency (MHz)

Electric Field Shielding (dB)

Measured Value at 176 MHz

Measured Value at 430 MHz
55

0 200 400 600 800 1000 -20

0

20

40

60

80

Figure 16 Electric field shielding of rectangular cavity approximating fuselage of

 B-757 aircraft. Two windows on each side of cavity were considered.

 Dimensions of Cavity, aperture and field point are as given in Figure 15

Frequency (MHz)

Electric Field Shielding (dB)

(1,0) Mode on Aperture

(1,0 + 3,0) Modes on Aperture
56

200 400 600 800 1000

0

20

40

60

80

Frequency (MHz)

Electric Field Shielding (dB)

Figure 17 Electric field shielding of rectangular cavity approximating fuselage of

 B-757 aircraft. Two windows on each side of cavity were considered.

 Dimensions of cavity, aperture and field point are as given in Figure 15

(1,0) Mode on Aperture

(1,0 + 3,0)Modes on Aperture

(1,0 + 3,0 + 5,0)Modes on Aperture
57

3.4 Convergence Test For Large size Cavities

The numerical estimate of electric field shielding in an electrically large size cavities illuminated through large size the apertures depends upon the number of cavity modes as well as

aperture modes considered in the numerical simulation. An adequate number of cavity modes and

aperture modes must be considered to achieve numerical stable results. In this section, dependence of the electric field shielding on number of cavity modes and aperture modes is studied

through numerical examples.

 For numerical estimate of electric field shielding, we first consider a rectangular cavity Electric Field Shielding (dB)

Frequency (MHz)

Figure 18 Electric field shielding (for horizontal polarization) of rectangular cavity

 B-757 aircraft. Two windows on each side of

 Dimensions of cavity, aperture and field point

 approximating fuslelage of

 cavity were considered.

are as given in Figure 15

0 200 400 600 800 1000 -20

0

20

40

60

80

(1,0 + 3,0 + 5,0)Modes on Aperture

Frequency (MHz)
58

 ( 5400 x 600 x 600 ) cm3, which approximates the fuselage of a Boeing-747. For numerical simulation only two windows; one on each side of fuselage were considered. The aperture dimensions selected for the numerical simulation were ( 25 x 35) cm2

. The window was assumed to be

located with their centers at (2700, 300, 0.0) cms and (2700, 300, 600) cms. Using the computer

code developed here, the electric field in volts/meter calculated at (2700, 300, 300) cms as a function of mode index is presented in Figure 19. In the numerical simulation it is assumed that

the incident wave is polarized in the y-direction. In Figure 19, stable results are obtained for the

cavity mode index higher than 200. For Figure 19, only a single mode at the aperture was considered.

For the cavity and window dimensions as shown in Figure 19, the electric field at the center of cavity is calculated and shown in Figure 20 as a function of cavity mode index . From

Figure 20 it is clear that achieves numerically stable results.

After selecting the cavity mode indices for numerically stable results, the electric

field at the center of the cavity with dimensions as shown in Figure 19 is calculated as a function

of aperture modes and shown in Figure 21. From Figure 21 it may be concluded that for normal

incidence the 4 or 5 aperture modes are sufficient to obtain numerically stable results. Consideration of higher order modes on the aperture improves the accuracy of the numerical calculation

but increases the computational time significantly. For fast computation of electric field shielding,

we consider only single mode at the aperture.

n

n

m

m ≥ 1200

m n,
59

Mode Index n

Electric Field Amplitude (V/m)

Figure 19 Electric field amplitude at (2700, 300, 300) cms inside a rectangular cavity

 ( 5400x600x600 ) cm3

 illuminated by plane wave at normal incidence through rectangular windows (25 x 35) cm2

 as a function of cavity mode index n.

 Other parameters: cavity mode index m = 1200, frequency = 2.87510864 GHz.

Real Part

Imaginary Part

50 100 150 200 -5

-4

-3

-2

-1

0

1

2

3

4

5

Real Part of Ey

Imaginary Part of Ey
60

Mode Index m

Electric Field Amplitude (V/m)

Figure 20 Electric field amplitude at (2700, 300, 300) cms inside a rectangular cavity

 ( 5400x600x600) cm3

 illuminated by plane wave at normal incidence through

 rectangular windows (25 x 35) cm2 as a function of cavity mode index m.

 Other parameters: cavity mode index n = 200, frequency = 2.87510864 GHz.

Real Part

Imaginary Part

200 400 600 800 1000 1200 -5

-4

-3

-2

-1

0

1

2

3

4

5

Real Part of Ey

Imaginary Part of Ey
61

Electric Field Amplitude (V/m)

Real Part

Imaginary Part

No. of Modes on Apertures

Figure 21 Electric field amplitude at (2700, 300, 300) cms inside a rectangular cavity

 ( 5400x600x600) cm3

 illuminated by plane wave at normal incidence through

 rectangular windows (25 x 35) cm2

 as a function of number of modes on apertures.

 Other parameters are cavity mode index n = 200, m = 1200,

 frequency = 2.8745GHz.

1 2 3 4 5 6 7 8 9 10 -5

-4

-3

-2

-1

0

1

2

3

4

5

Real Part of Ey

Imaginary Part of Ey
62

4.0 Conclusion

A method suitable for estimation of electric field shielding effectiveness of large but regular shaped rectangular cavity with multiple rectangular apertures is presented. Assuming appropriate electric field distribution on the aperture, fields inside the cavity are determined using

rectangular cavity Green’s function. Electromagnetic fields outside the cavity and scattered due

to the aperture are obtained using the free space Green’s function. Matching the tangential magnetic field across the apertures, the integral equation with aperture fields as unknown variables is

obtained. The integral equation is solved for unknown aperture fields using the Method of

Moments. From the aperture fields the electromagnetic shielding effectiveness of the rectangular

cavity is determined.

Numerical results on electric field shielding of a rectangular enclosure are validated with

data available in the literature. Effects of cavity and aperture modes for large size enclosure are

studied. Approximating a Boeing-757 aircraft cabin by a rectangular cavity and passenger windows by rectangular aperture electromagnetic field penetrations inside the cabin due to external

plane wave at normal incidence are estimated. Electromagnetic field penetrations due to horizontal as well as vertical polarizied incident fields are studied. Though in the present report electromagnetic field penetration through two apertures, one on each side of fuselage is estimated, the

computer code can be used to simulate effects of multiple apertures on each side of fuselage.

One advantage of the present method is that the order of matrix equation to be solved in

the present approach is very small compared to the order of matrix equation one encounters in the

Finite Element Method. Another advantage is that the computer storge requirement for the

present approach is negligibly small compared to the Finite Element Method and Finite Difference Time Domain method. However, because of the use of the cavity Green’s function it is very
63

difficult to take into account the complex geometries and objects that may be present in the passenger cabins. Also, the method does not take into account the losses that will be present in practical aircraft cabin. Inspite of its inadequateness in handling losses in the system and complex

geometries, the present method gives quick and fairly accurate estimates of electromagnetic

shielding effectiveness of rectangular enclosures.

Acknowledgement

The author is grateful to Dr. C. J. Reddy, Hampton University, Hampton, Virginia, for providing FEM /MoM numerical results shown in Figures 10 and 12.

References

[1] H. A. Mendez, “ Shielding theory of enclosures with apertures, ” IEEE Trans. on Electromagnetic Compatibility, Vol. EMC -20, No. 2, pp. 296-305, May 1978.

[2] G. Cerri, R. D. Leo, and V. M. Primiani, “ Theoretical and experimental evaluation of the electromagnetic radiation from apertures in shielded enclosures, ” IEEE Trans. on Electromagnetic Compatibility, Vol. EMC -34, No. 4, pp. 423-432, Nov. 1992.

[3] M. P. Robinson, et al., “Analytical formulation for the shielding effectiveness of enclosures

with apertures, ” IEEE Trans. on Electromagnetic Compatibility, Vol. EMC-40, No. 3, pp.

240-247, August 1998.

[4] M. P. Robinson, et al., “Shielding effectiveness of a rectangular enclosure with a rectangular

aperture, ” Electronics Letter, Vol. 32, No. 17, pp. 1559-1560, 15th August 1996.

[5] D. A. Hill, et al., “Aperture excitation of electrically large, lossy cavities, ” IEEE Trans. on

Electromagnetic Compatibility, Vol. EMC 36, No. 3, pp. 169-178, August 1994.

[6] K. S. Kunz and R. J. Luebbers, “The finite difference time domain methods for electromagnetics, ” Orlando, FL, CRC, 1993.

[7] C. J. Reddy et al., “A combined FEM/MoM/GTD technique to analyze elliptically polarized

cavity-backed antenna with finite ground plane, ” NASA Technical Paper 3618, Nov. 1996.
64

[8] C. J. Reddy and M. D. Deshpande, “User’s Manual for CBS3DS Version 1.0, ” NASA Contractor Report 198236, Oct. 1995.

Appendix I

In this appendix we evaluate the integral in a

closed form. Substituting for the expression for can be written as

In above equation are the coordinates of the center of aperture. Performing the integration expression for can be written as

(98)

Appendix II

In this appendix we evaluate the integral in a

closed form. Substituting for the expression for can be written as

In above equation are the coordinates of the center of aperture. Performing the integration expression for can be written as

Ψrpqx( ) x', y' mπx'

a -----------     sin nπy'b

----------     cos d y x'd ' ∫∫

Ψrpqx( ) x', y' Irpqmnx

pπ

Lr

------ Lr

2

----- + x'         sin qπ

Wr

------- Wr

2

------- + y'         mπ x' xr ( ) +

a ---------------------------     sin

nπ y' yr ( ) +

b --------------------------     cos cos d y x'd '

Lr

2 –----

Lr

2

----

∫

Wr

2 –------

Wr

2

------

∫

xr yr , r

th

Irpqmnx

pπLr

( ) pπ

2 mπLr

a -------------     2

–

----------------------------------------- cos( ) pπ

mπ

a

------- Lr

2

----- xr +         sin mπ

a

------- Lr

2

----- xr –         + sin

      –

–( ) nπ ⁄ b

qπ

Wr

-------     2 nπ

b

------     2

–

------------------------------------ cos( ) qπ

nπ

b

------ Wr

2

------- yr +         sin nπ

b

------ Wr

2

------- yr –         + sin

     

Φrpqy( ) x', y' mπx'

a -----------     cos

nπy'

b

----------     sin d y x'd ' ∫∫

Φrpqy( ) x', y' Irpqmny

pπ

Lr

------ Lr

2

----- + x'         cos qπ

Wr

------- Wr

2

------- + y'         mπ x' xr ( ) +

a ---------------------------     cos

nπ y' yr ( ) +

b --------------------------     sin sin d y x'd '

Lr

2 –----

Lr

2

----

∫

Wr

2 –------

Wr

2

------

∫

xr yr , r

th

Irpqmny

qπWr

( ) qπ

2 nπWr

b --------------     2

–

------------------------------------------ cos( ) pπ

nπ

b

------ Wr

2

------- yr +         sin nπ

b

------ Wr

2

------- yr –         + sin

      –
65

(99)

 Appendix III

The Fourier transforms and of and , respectively are derived in

this appendix. Using the defination , the Fourier transform can be written as

Following the same procedure the Fourier transform of can be written as

The expressions for and are obtained from expressions for and , respectively by replacing by respectively.

Appendix IV

 In this appendix, the expressions for electric field inside the cavity due to magnetic current sources present on its walls are given. Using the electric vector potential given in (31) and

expression (23), the electric field components inside the cavity due to the x component of magnetic current sources at plane are obtained as

–( ) mπ ⁄ a

pπ

Lr

------     2 mπ

a

-------     2

–

------------------------------------ cos( ) pπ

mπ

a

------- Lr

2

----- xr +         sin mπ

a

------- Lr

2

----- xr –         + sin

     

φrpqy ψrpqx Φrpqy Ψrpqx

ψrpqx kx ky () Ψ , rpqye

jkx – x jky – y

d y xd

r

∫∫ =

ψrpqx

ψrpqx

LrWr

4 j ------------ e

jkx x – r jky y – r e

j

pπ

2

------

pπ

2

------ kxLr

2 – ----------     sin

pπ

2

------ kxLr

2 – ----------

------------------------------------- e

– j

pπ

2

------

pπ

2

------ kxLr

2 + ----------     sin

pπ

2

------ kxLr

2 + ----------

– -------------------------------------

         

=

e

j

qπ

2

------

qπ

2

------ kyWr

2 – ------------     sin

qπ

2

------ kyWr

2 – ------------

--------------------------------------- e

– j

qπ

2

------

qπ

2

------ kyWr

2 + ------------     sin

qπ

2

------ kyWr

2 + ------------

+ ---------------------------------------

         

Φrpqy

φrpqy

LrWr

4 j ------------ e

jkx x – r jky y – r

e

j

pπ

2

------

pπ

2

------ kxLr

2 – ----------     sin

pπ

2

------ kxLr

2 – ----------

------------------------------------- e

– j

pπ

2

------

pπ

2

------ kxLr

2 + ----------     sin

pπ

2

------ kxLr

2 + ----------

+ -------------------------------------

         

=

e

j

qπ

2

------

qπ

2

------ kyWr

2 – ------------     sin

qπ

2

------ kyWr

2 – ------------

--------------------------------------- e

– j

qπ

2

------

–

qπ

2

------ kyWr

2 + ------------     sin

qπ

2

------ kyWr

2 + ------------

---------------------------------------

         

ψrpqx

* φrpqy* ψrpqx φrpqy

kx ky , kx ky – , –

z = 0
66

Likewise the electric field components due to the x-component of magnetic current sources at

 plane are obtained as

Ex

IIx0 = 0

Ey

IIx0 Urpq

1

kI

---- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 nπ

b

------     2

–     mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI sin( ) ( ) c z – Irpqmnx

Ez

IIx0 Urpq

1

kI

---- ε0mε0n

ab kI sin( ) c -------------------------- nπb

------ mπx

a

----------     sin nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI cos( ) ( ) c z – Irpqmnx

z c =

Ex

IIxc = 0

Ey

IIxc Brpq

1

kI

---- ε0mε0n

ab kI sin( ) c -------------------------- k0

2 nπ

b

------     2

–     mπx

a

----------     sin nπy

b

---------     cos

m n,

∞

∑

p q,

∑ r = 1

R

= – ∑

kI sin( )z Irpqmnx

Ez

IIxc Brpq

1

kI

---- ε0mε0n

ab kI sin( ) c -------------------------- nπb

------ mπx

a

----------     sin nπy

b

---------     sin

m n,

∞

∑

p q,

∑ r = 1

R

= ∑

kI cos( )z Irpqmnx
REPORT DOCUMENTATION PAGE Form Approved

OMB No. 0704-0188

Public reporting burden for this collection of information is estimated to average 1 hour per response, including the time for reviewing instructions, searching existing data

sources, gathering and maintaining the data needed, and completing and reviewing the collection of information. Send comments regarding this burden estimate or any other

aspect of this collection of information, including suggestions for reducing this burden, to Washington Headquarters Services, Directorate for Information Operations and

Reports, 1215 Jefferson Davis Highway, Suite 1204, Arlington, VA 22202-4302, and to the Office of Management and Budget, Paperwork Reduction Project (0704-0188),

Washington, DC 20503.

1. AGENCY USE ONLY (Leave blank) 2. REPORT DATE

June 2000

3. REPORT TYPE AND DATES COVERED

Contractor Report

4. TITLE AND SUBTITLE

Electromagnetic Field Penetration Studies

5. FUNDING NUMBERS

C NAS1-96013

 

6. AUTHOR(S)

M. D. Deshpande

WU 522-31-21-04

7. PERFORMING ORGANIZATION NAME(S) AND ADDRESS(ES)

NYMA, Inc.

NASA Langley Research Center

Hampton, VA 23681-2199

 

 

8. PERFORMING ORGANIZATION

REPORT NUMBER

 

9. SPONSORING/MONITORING AGENCY NAME(S) AND ADDRESS(ES)

National Aeronautics and Space Administration

Langley Research Center

Hampton, VA 23681-2199

10. SPONSORING/MONITORING

AGENCY REPORT NUMBER

NASA/CR-2000-210297

11. SUPPLEMENTARY NOTES

M. D. Deshpande, FDC/NYMA Inc., Hampton, VA

Langley Technical Monitor: Fred B. Beck

12a. DISTRIBUTION/AVAILABILITY STATEMENT

Unclassified-Unlimited

Subject Category 32 Distribution: Nonstandard

Availability: NASA CASI (301) 621-0390

12b. DISTRIBUTION CODE

13. ABSTRACT (Maximum 200 words)

A numerical method is presented to determine electromagnetic shielding effectiveness of rectangular enclosure

with apertures on its wall used for input and output connections, control panels, visual-access windows,

ventilation panels, etc. Expressing EM fields in terms of cavity Green's function inside the enclosure and the

free space Green's function outside the enclosure, integral equations with aperture tangential electri fields as

unknown variables are obtained by enforcing the continuity of tangential electric and magnetic fields across the

apertures. Using the Method of Moments, the integral equations are solved for unknown aperture fields. From

these aperture fields, the EM field inside a rectangular enclosure due to external electromagnetic sources are

determined. Numerical results on electric field shielding of a rectangular cavity with a thin rectangular slot

obtained using the present method are compared with the results obtained using simple transmission line

technique for code validation. The present technique is applied to determine field penetration inside a Boeing757 by approximating its passenger cabin as a rectangular cavity filled with a homogeneous medium and its

passenger windows by rectangular apertures. Preliminary results for, two windows, one on each side of fuselage

were considered. Numerical results for Boeing -757 at frequencies 26 MHz, 171-175 MHz, and 428-432 MHz

are presented.

14. SUBJECT TERMS

EM Penetration, EM Shielding, Aperture Coupling, EM Susceptibility

15. NUMBER OF PAGES

71

 16. PRICE CODE

A04

17. SEC U RITY CL ASSIF IC AT ION 

O F REPO R T

Unclassified

18. SEC U RITY CL ASSIF IC AT ION 

O F TH IS PA GE

Unclassified

19. SECURITY CLASSIFICATION

 OF ABSTRACT

Unclassified

20. LIMITATION

 OF ABSTRACT

 UL

NSN 7540-01-280-5500 Standard Form 298 (Rev. 2-89)

Prescribed by ANSI Std. Z-39-18

298-102