22 Kernel Evaluation - Electromagnetic diffraction through a square slot into a cavity

- The integral of the kernel over a grid square is approximated by the value of the kernel at the centerpoint of the grid square with the exception of the grid square that is sampled. At the sampled grid square the integral of the kernel would go singular if this method were applied because rQ = 0 and the term in rQ would , consequently, blow up. On this square the integral is approximated as

,£SN fDSN

_ikr .

^,2tt rDSN/2 -ikr e o i = £ ^ 2 d x d y , i

•'-DSN*'-DSN o 0 ro

where DSN is the length of the grid square, rQ2=r2= x z+yz , and the factor of 4/tt arises from the ratio of the areas of the square and the circle. Thus:

-ikr DSN/2

I = 8iSL— °

This term is finite. The integral of the rest of the terms of the kernel for this grid square are evaluated by the centerpoint method.

The kernel series converges rather slowly. Also, when n is large the (x-£) term and (y-n) term become insig

nificant. For these reasons, the value of 00 -ikr

1 ^ 4 — ^ ; rn = 2nW

NGI n

is evaluated separately, NGI,being large, and is then added r dr d0

23 onto:

NGI-1 -ikr

I — -— - ; r 2 = (x-C) 2 + (y-n) 2 + (2nW) 2

n=0 n n

The savings in computational time more than make up for the small loss of accuracy when this technique is used.

Solution for EA and AA

x v

The integral equations can now be inverted, and a numerical solution can be attempted. Let EA , 11 the

xp particular TE^^ solution” , be the solution to:

A -2E

// Exp (g1 + g2) dA = (on A)

As noted before, E A does not meet the boundary conditions xp

at the edge of the slot. Suppose we found E ^ , "homoge

neous TE. solution", the solution to:

to y

// Exh (g1 + g2) dA = cos(ky) (on A)

Since any constant factor times this meets all the boundary conditions and continuity requirements, excepting, of

course, the jump requirement in which is met by Ex p , A we might try to find B such that:

♦ « 4 ,

meets the edge conditions. This approach is the classical diffraction approach when working with a single mode solu

tion, e.£., see Suzuki (1956). For this problem, however, a proper B cannot be found to meet all the boundary

conditions. Let A , "homogeneous solution", be the

solution to:

f f A^h (g1+g 2) dA = x sin (ky) (on A) Note:

+ |yr + k 2) x sin (ky) = 0

Thus any multiple of A ^ meets all the boundary and conA tinuity requirements excepting the jump in . Note also that A ^ is made odd so that (9 2/8x3y) A ^ will be even, and therefore meet symmetry requirements on . We are now tempted to try to find B and C such that:

A A ^ 2 *

E (x ,y, 0) = + B E_, + C:

x xp xh 3x9y yh

meets the E^ edge conditions, and Ey (x,y,0) = ( 1^, + k 2)

meets the E^ edge conditions. This solution will not work in general again, however, through all these blind alleys we are leading up to a method of solution.

Suppose that there are 2nx2n grid squares. Symmetry will reduce the E^ edge conditions to n equations and the Ey edge conditions to n more equations. We thus have 2n equations to meet with only three solutions. If n is not trivially small, then we need more solutions. Solutions with coupled continuity across A provide a source of these needed solutions.

25 Let E ,A be the solution to:

x X

// k (g^ + g 2) dA = cos(k^x) cos(ky) (on A)

and let A ,A be the solution to:

y x

A i

// Ay k Cg1 +g2) dA = f f h k~ sin(kx £) sin(kn)jHq (kr)dA where, r2 = (x-C) 2 + (y-n) 2 , and dA = d£ dri . Together

A A

( E , , A , ) , " a coupled solution" , meet continuity

x y x

and boundary conditions, excepting the jump in The solutions have been chosen so that (x,y,0) is symmetric.

The number of solutions obtained in this manner is unlimited because of the ability to change k^ and obtain a new solu

tion. On a 2nx2n grid we need 2n-2 coupled solutions.

Suppose that 2n=6. Solution to the diffraction goes as follows: Define 4 = 2n-2 values of k ; k , ,k 0 ,k , ,k . •

x xl x2 x3 x4

A A

Find associated A . and E , solutions. Also, find y X j x X j

Ayh ' EXh ' and EXP • Then!

B Exh + ci w A? h +

X

D j ( E x kxj + kx j > = -Exp

C at y = s/2)

B (0) + C Ayh +

U.

^Aj ^{= 0}

^ ^ ( at x = s/2) represent six equations with six unknowns. Note that we

26 have set to zero at x = s/2 by setting the total field to zero there. This procedure eliminates any numerical in

stability in taking O 2/8y2 + k 2) . The unknown coeffi

cients can be solved for by using matrix routines. The result is a set of fields that meet all boundary conditions and continuity requirements, and hence, an approximation of the actual solution of the diffraction problem.

Removal of the Singularity of Hq (kr) The integral:

f f A Ho (kr) sin(kx £) sin (kn) d£ dn

is performed by a Riemann sum. Center point values of sin(kx £) sin(kn) were used as were all but the sampled grid square values of Hq (kr). The excepted value would be singular on the sampled grid if such a technique were used.

The desired value is found in a manner similar to the re

moval of the singularity of exp(-ikr^) / rQ :

;/A 1 Ho (kr) dX dy “ W (

2tt /-DSN/2 .

~ Hq (kr) r dr d

t x H1(x)

DSN/2 ] 0

where r = x +y , and is the first order Hankei func

tion of the second kind. The diffraction problem is now suitable for programming.

CHAPTER 4

RESULTS AND FUTURE WORK

Results have recently been obtained using the method discussed in chapter 3. Most significant was that the mag

nitude of the effect of the total TM, field was three or to y

four orders less than the magnitude of the effect of the total TE. field on the resultant E aperture field.

to y x c

Also, the coefficient of the homogeneous TM^_ field, A^\ ,

to y yn

was essentially zero. Based on these results we might con

clude that a pure TE^_o solution would yield valid results at lower frequencies if the edge conditions could be met.

Results are shown in Figs. 4-11. At low frequencies the magnitude of the electric field goes linearly with k.

Near the value of k where the slot length is a half-wave

length, however, the fields increase in a resonance effect;

see Fig. 12. The effect of changing W seems small away from cavity resonance; see Figs. 9 and 10.

Suggested for future work is a time domain analysis of the EMP problem. This task will require a fairly large computer run as far as computational time is concerned. The program runs approximately 4 0 decimal seconds for one value of k with a 12 by 12 grid over the aperture, and it requires 60K of core. A recommendation: in such an analysis k values

s—1

S = 1

J

Fig. 3. The sampling points of the program.

Figures 4 through 11: Calculated fields' for various values of k and W ( k= .01 - "3.0; ; W = 10 or 20 ).

. . . The numbers..in the arrays, . Figures 4. through 11/ . represent the magnitudes of their respective fields. The numbers are arrayed to give the effect o f .a..two-dimensional plot. They represent values taken over the grid squares of the fourth quadrant. These sampling points are indicated in Fig. 3. Thus, reading, down an array means reading in the minus-y direction, and reading across from left.to right, means reading in the direction of increasing x . The. origin

is in the upper left corner of the array. ■

It is important to note that the values of the y-component of the vector A are magnified by a factor of : 100,000. The magnitude of the incident electric field is taken to be unity, and the initial direction of,this field is in the x- direction. Both electric fields are symmetric in both x and y , while the y-component of A is anti- . ■ symmetric i n .the two variables.

29 Total A Field

o 0000 . 0000 . 0000 . 0000 . 0000 .0000

.0000 .0000 .0000 ,0000 .0000 .0000

. 0000 .0000 .0000 .0000 . .0000 . 0000

. 0000 . 0000 . 0000 . 0000 . 0000 .0000

.0000 . 0000 .0000 . 0000 .0000 .0000

.0000 .0000 .000 0 . 0000 .0000 .0000

i Component of F

y Field

.0032 . 0032 .0032 . 0033 .0036 .0041

.0030 .0030 .0031 .0032 .0034 .0039

. 0027 . 0027 .0027 . 0028 . 0030 . 0034 . 0021 . 0021 .0022 . 0022 .0023 .0027 . 0013 . 0013 . 0013 . 0013 . 0014 . 0016 . 0000 . 0000 .0000 . 0000 . 0000 . 0000

i Total E Field x

. 0032 . 003 2 . 0032 . 0033 .0036 . 0041 .0030 . 0030 . 0031 ,0032 . 0034 . 003 9

. 0027 .0027 ,0027 . 0028 . 0030 . 0034

. 0021 , 0021 ..0022 . 0022 ,0023 . 0027

. 0013 .0013 . 0013 .0013 . 0014 ,0016

. 0000 .0000 ,0000 .0000 . 0000 . 0000

Fig. 4. Calculated fields for wavenumber of .01 and a cavity dimension of 20.0

.0317 . 0320 . 0325 .0335 .0357 .0415

,0301 .0303 . 0308 .0317 .0337 . 0391 j

0317 , 0320 .0325 .0335 .0357 .0415

0301 .0303 . 0308 .0317 . 0337 .0391

0267 . 0269 ,0273 . 0281 .0297 .0342

0213 . 0214 .0217 . 0222 .0234 . 0266

31 Total A Field

y ^.

. 0018 . .0014 .0003 . 0001 .0001 .0000

.0015 .0011 . 0003 .0001 .0001 .0000

.0012 . 0008 . 0002 . 0001 . 0000 .0000 1 . 0009 . 0006 . 0002 . 0001 . 0000 .0000 1 .0005 . . 0004 . 0001 .0000 . 0000 .0000 j . 0002 .0001 .0000 , . 0000 .0000 •oooo j

IE. Field

.2758 .2775 • .2817 . 2902 .3084 .3575

.2615 .2631 .266 9 .2746 .2914 .3367

.2317 .2330 .2362 . 2426 .2565 .2943

.1839 .1849 .1872 .1917 .2015 .2284

.1121 .1126 .1139 .1162 .1212 .1346

.

0000

.

0000 .0000

.

0000

.

0000 .0000.

JTotal E Field

I ^

2758 .2775 ,2817 .2902 .3084 .3575

2615 .2631 .2669 . 2746 ,2914 .3367 !

2317 .2330 .2362 .2426 .2565 .2943 !

1839 .1849 ,1872 .1917 .2015 .22 84 !

1121 .1126 . .1139 .1162 '.1212 .1346 i 0000

.

0000

.

0000

.

0000

.

0000 .0000 1

Fig. 6. Calculated fields for wavenumber of .8 and a cavity dimension of 2 0.0.

See legend of Fig. 4.

32 Total A Field

.0059 .0030 . .0005 . 0001 . 0002 .0000 . 0050 . 0024 .0004 . .0001 , 0002 .0000 . 0040 .0018 . 0004 . ,0001 .0 0 0 1. : . 0000 .0030

.

⁰⁰¹³ ^{.0 003} ^.0001 ^.⁰⁰⁰¹ ,0000 .0018 .0008 . 0002 .0000 . 0001 . 0000

.0006 .0003 . 0001 .0000

.

⁰⁰⁰⁰ ^.0000

j E Component of F Field .3615

.3637 .3688 .3795 .4028 .4658

.3425 .3445 .3493 .3591 .3804 . 4385

.3033 .3049 .3089 ,3170 " .3345 .3830

.2404 .2416 .2444 .2501 .2624 ,2968

.1463 .1469 .1484 .1514 ,1575 ,1746

. 0000 . 0000 . .0000 . 0000 . 0000 . 0000

Total Ex Field

.3615 .3637 .3688 ;3795 . .4028 .4658

,3425 .3445 .3493 .3591 .3804 ,4385

.3033 ,3049 .3089 .3170 .3345 .3830

.2404 .2416 .2444 ,2501 ,2624 .2968

.1463 .1469 ,1484 .1514 .1575 .174 6

. 0000 . 0000 . 0000 . 0000 .0000 , 0000

Fig. 7. Calculated fields for wavenumber of 1.0 and a cavity dimension of 20.0.

See legend of Fig. 4.

I .4099 .4119 .4166 .4266 .4489 . 5118

1 .3242 .3256 .3289 .3359 .3514 .3959

Total A Field

.01.83 . 0178 .0118 . 0036 .0022 .0000

. 0063 . 0056 .0035 .0016 . 0009 .0000

.

001

Fig.

The f

12. Graph of field versus wave number.

ield is evaluated at the point nearest the center of A.

38 that give cavity resonances, should be used as much as pos

sible. Resonances of the cavity will play a dominant role in the response of the system. Also., corresponding fields are found relatively easily.

Also.suggested is a closer look at the theoretical approach. Even the more basic theorems, such as, a proof of the uniqueness of the integral equation inversion, seem to be lacking. Also, through a theoretical approach, the validity of the given numerical solution might be corrob

orated.

’

-In conclusion, we submit that another interesting.

and fruitful area would be generalizations of the problem to, for example, propagation through a square aperture into various types of cavities. Such a problem would require two things: a changing of the kernel of the integral equations by the addition of a different green’s function and a re

examination of the resonance solution. Eventually, through enough such generalizations, the problem of diffraction into an actual structure might become in some sense solvable.

APPENDIX

PROGRAM FOR COUPLED MODE SOLUTION

The numerical scheme set forth in chapter 3 for finding the solution to the diffraction problem was pro

grammed. This program is presented here preceded by a list of the more important symbols:

AK the wavenumber, k

A M (1,3) initially the storage for anti-symmetric-field- mat rix inversion, later storage for finding coefficients of field solutions

AN (Iz J ) storage for symmetric-field-matrix inversion B C (1,J,K) - storage for field solutions; in K dimension

first 2 (NPS-2) for coupled fields; the two homo geneous fields, particular field, and two total 'fields are next in K; J gives y-value; and 1

gives x-vaiue

Cl the square root of minus one DSN the grid square dimension

storage for g^ + g^

FRR storage for zeroth order Hankel function of second kind

NBK parameter in finding TAIL, roughly, the number of half-cycles of the series before a 2-cycle ■ averaging routine starts

NGI starting point for the calculation of TAIL NN the number of grid points along a side of a

quadrant

. 40 NPS the number of grid points along a side of the

aperture which equals 2*NN

NT ' the maximum number of terms per half-cycle allowed in the calculation of TAIL

SS the aperture side length

SUMO the value of the singularity of g, + g_ ; see page of text

TAIL . the result of summation from NGI to infinity of (exp(-ikrn ) )/rn , where r^ = ,2nW ; see pp.

and of text

W the distance between the plates of the cavity

P R O G R A M C H A R L Y ( I N P U T * O U T P U T )

C T H E A P E R T U R E IS D I V I D E D I N T O F O U R Q U A D R A N T S A N D S Y M M E T R Y IS

C A P P L I E D . N N IS T H E N U M B E R O F G R I o S Q U A R E S A L O N G T H E E D G E O F A S I D E C S U C H A Q U A D R A N T . N F S IS T H E N U M 3 E P A L O N G T H E E D G E O F T H E A P E R T U R E C AK IS T H E W A V E N U M B E R .

C O M P L E X T A I L C O M P L E X Z H A T

c o m p l e x S U M O

c o m p l e x a m i

C O M P L E X b c c o m p l e x f r r

C O M P L E X t e m p,b e t a*s u m

c o m p l e x F R , C , A M , B , A E X , Cl C O M P L E X A N * A S U M

D I M E N S I O N S T O (6*4) D I M E N S I O N B C ( 6 * 6 , 2 6 )

D I M E N S I O N F R (78) , F R R (78)

D I M E N S I O N A M ( 3 6 , 3 6 ) , B ( 3 6 ) , C ( 3 & ) D I M E N S I O N A N ( 3 6 * 3 6 )

D I M E N S I O N A M I ( 3 6 , 3 6 ) D I M E N S I O N A A A (10) N N s 4

N N c ³ N N k 5 N N = ⁶ N N 1 = N N - 1 N F S = ² <*NN

N P S l s N P S - 1 N P T S a N N * * 2 N V = N P S M V l = N V - 2 F N V a N V

Cl ■ CMPLX(0.tl.) PI e 3.141592653589 W » 200000.

w * I5 . w=_{1 0};i

w 20.

ss 1.

S2 55/2 AK s .005

AK .05 AK 6.

AK r 3.

AK := 1.

AK .1 A K = 2 . 5 A K = 3 . 5

Z H A T = C M P L X ( 0 . , 2 . ) P H I NT 1 7 0 , A K , N N

1 7 0 h O R M A T («- W A V £ N u M B E R < » , F10 . 5 » o I N D E X * ,110) F N P S = N P S $ D S N = S S / F N P s

A D 5 N = A K * D S N

D S N 2 = 2 , * D S N $ D 2 = n S N * * 2 A A = D S N / 2 , $ D 2 = D 2 / P I

D O 4 0 0 N = 2 , N P S $ F N = N

F N c ( ( F M - 1 . ) " D S N ) * * 2 5 Nl = ( N - l ) * N / 2 D O 4 0 1 M = 1 * N $ F M s M

N M a N1 ♦M

F M = ( ( F M ~ 1 . ) * D S N ) * » 2 $ F P = A K * S Q R T (F N * F M ) F H R ( N M ) = , 2 5 * D 2 * C M P L X ( B E S 6 Y ( F o . 0 . ,0) .0 E S 6 J (F P , 0 . , 0 ) ) 4 0 1 C O N T I N U E

4 0 0 c o n t i n u e

F P = AK * A A

F R R ( l ) » ( D S N O C M P L X ( B E S⁶Y ( F P t 1, . 0 ) »⁸E S⁶J (F P » 1 . » 0 ) ) ) / A K

. A C J = A B S ( C J )

I M S a 0

Nl « ( N - l ) * N / 2

= ( ( F N m l , ) " D S N ) * * 2

D O 1 3 6 M = ¹ ,N $ F M = M N M = Nl + M

F M = ( ( F M - 1 . ) * D S N ) * * 2 $ F P = F N + F M S U M = C M P I „ X ( 0 . » 0 , )

D O 1 5 5 J = I J » N⁶ I $ F J = J F J = ( ( F J - 1 , ) * W 2 ) * * 2

R J = S O R T (F J + F P ) $ A R J = A K * R J S U M = S U M ♦ C E X P ( - A R J * C I ) / R J

1 5 5 C O N T I N U E '

FR(Nvi) = SUM<>D2 ^* T A I L IU ~ ¹

1 3 6 C O N T I N U E 1 3 5 C O N T I N U E

F R (1) = F R ( 1 ) ♦ S U M O

DO 163 N s 1,6 $ Nl c NO(N,l)/2 DO 164 M = 1,N $ NM = Nl +M AAA(M) = C A B S(FR(NM))*100.

164 C O N T I N U E

PRINT 160,(AAA(MA), MA » 1,N) 163 CONTINUE

02 = OSN**2

FNN = NN $ FNN = FNN * ,5

DO 100 N a 1,NN $ Nl = (M-1)#NNJ DO 101 M = 1,NN $ NM = N%+M

DO 102 L = 1 • NPS $ FL a L C SP = 1.

IF( L .LE. NN) CSP » -1.

FLA = ABS(FNN-FL) -.5 $ LA * FLA 00 103 K s 1 ,NPS $ FK = K

C C P = le

00 265 M a I f N N $ 0 » ST0(Vt*3)

D O ? R 7 N s 1 , N N $ S = STO(N',l>

294 CONTINUE

6 1 0 A M ( U J> * T E M P Te m p = c ( K K ) C(K><) B C ( L ) C ( L ) a T E M P

6 20 DO 6 9 3 I = K K P 1 , N P T S

B E T A a A M ( I , K K ) / A M ( K K » K K >

D O 6 5 0 J s K K P l t N P T S

6 5 0 A M ( I , J ) - A M ( I f J) - B E T A * A M ( K K , J ) 6 9 3 C ( I ) = C ( I > - B E T A * C ( K K )

c b a c k s o l u t i o n

B ( N P T S ) = c ( N P T S ) / A M ( N P T S , N P T S ) I = N M l

7 1 0 IP1 = I ♦ 1 S U M = ( 0 , , 0 * )

D O 7 0 0 J = IP1 , N P T S 7 0 0 S U M = S U M + A M ( I »J) * B (J)

B (I ) = ( c m - S U M ) / A M ( I , n 1 = 1 - 1

IF ( I . G E , 1) G O TO 7 1 0 C E N D OF M A T R I X S O L U T I O N

D O 2 1 5 N = I f N N $ N1 = ( N - l ) * N N D O 2 1 6 M r 1 , N N $ N M = m I + M

B C ( M , N , N L O ) = B ( N M ) A A A (M) a C A B S ( B ( N M ) ) 2 1 6 C O N T I N U E

P R I N T 1 6 0 , ( A A A ( M A ) * M A * 1 ,NN) 2 1 5 C O N T I N U E

2 9 3 C O N T I N U E

P = D S N * D S N 2

D O 3 6 2 N L O = 2 , N « K 1 , 2

C A L C U L A T E T H E E X ‘F I E L D C O M P O N E N T O F T H E T M T O Y S O L N . D O 3 6 3 N = 1 , N N

3 6 3 A M ( N L O , N ) » (B C (N »N N l ,N L O ) - B C ( N , N N , N L O ) )

D O 3 6 4 N = 2 , N N 1 55 N L e N * 1

C a m M A T R I X S O L U T I O N .. G A U S S - J O R D A N

IF ( I , G E . I) G O T O 1 7 1 0

P = D 5 N 2 " D S N 2

LIST OF REFERENCES

Harrington, Roger F. Time Harmonic Electromagnetic Fields.

New York: McGraw-Hill Book Co., 1961.

Stakgold, Ivar. Boundary Value Problems of Mathematical Physics, vols. I and II. New York: The MacMillan Company, 1968.

Stratton, Julius Adams. Electromagnetic Theory. New York:

McGraw-Hill Book Co., Inc., 1941.

Suzuki, Michio. "Diffraction of Plane Electromagnetic Waves by a Rectangular Aperture," I.R.E. Transactions on

Antennas and Propagation, volume AP-4, (April 1956), pp. 149-155.

In document Electromagnetic diffraction through a square slot into a cavity (Page 29-64)