,: -

y-l

Sensor and Simulation Notes

Note 93

September 1969

Minimization of Current Distortions on A CylindricalPost Piercing A Parallel-Plate Waveguide

by

R. W. Latham, K. S. H. Lee, and R. W. SassmanNorthrop Corporate @bOZaEqTie5 ...._.... ... ... -

Pasadena, California

Abstract

The time behavior is obtained of the total current induced on a cylindrical

post by a step-function plane wave traveling between a parallel-plate waveguide

and impinging on the post which protrudes through one of the guide’s walls.

This geometry may be considered the idealized shape of some antenna-like

structure in a parallel-plate transmission line or surface transmission line.

The resonance frequency and the decay time constant of the fundamental

mode (i.e., the mode with the longest wavelength) of the post are found from ‘

the response of the post to a step-function incident plane wave and are plotted

against the size of the hole in the wall through which the post protrudes. It

is found that as the size of the hole is increased, the current amplitude of

the fundamental mode decreases away from the corresponding value when the post

of twice the length

decay time constant

is in free space, whereas the resonance frequency and the

increase toward the free-space values.

CLEARE12FORPU3LIC RELEASE

Q&=N’\(Y49,[3DQsYy

Acknowledgrnent—.

We thank Dr. Carl Baum for his helpful suggestions and discussions

throughout the course of this work and Mrs. G. Peralta for typing the

manuscript and drawing the curves.

2

●

1

I. Introduction.—..

The physics of the surface transmission-line simulator is now fairly

well understood, especially when it is used to simulate the low-frequency

content of a nuclear EMP. In some instances, there are site structures that

rise higher than or as high as the top plate of a surface transmission line.

In such cases, schemes are called for to build the line around such structures

for minimum electromagnetic interactions between the line and the structures.

If the structures are not in electrical contact with the line (that is, they

do not physically touch each other), one can be sure that the interactions

will be minimal in the low-frequency regime. Thus , the logical way to start

is to make holes in the top plate of the line through which the structures

protrude, and then to analyze the high-frequency interactions of such structures

with the line.

The structures considered in this note will be idealized to be cylindrical

in shape and the holes will be taken to be circular. The ground will be

assumed to be perfectly conducting and the line will be taken to be two

perfectly conducting parallel plates of infinite extent. Throughout this note

the cylindrical structures are considered taller than the top plate of the line;

the opposite case where the top plate is taller than the structures may be

considered in a future note.

In Section 11 the problem is formulated in terms of a coupled set of

integral equations, the unknowns being the aperture electric field in the hole

and the total axial current on the part of the post sticking out of the wave-

guide (see Fig. la). With a knowledge of this aperture electric field the

current on the part of the post between the plates can be determined by

integration. The Green’s functions used in Section II are derived in Section

III. Numerical results are presented in graphical form in Section IV. An

appendix is devoted to the derivation of a Greenls theorem applicable to

bodies of revolution and axially symmetric fields.

II. Mathematical Formulation—.

Consider the situation depicted in figures la and lb where a time-

harmonic plane wave travels between two perfectly conducting plates of

infinite extent and impinges on the perfectly conducting,

that protrudes through a circular hole in the top plate.

the total axial current induced on the post by this plane

later be generalized to be a step-function pulse. Let us

cylindrical post

We wish to calculate

wave, which will

recall the well-

known fact that, if only the induced total axial current on an axi-symmetric

body is considered, one can restrict his attention only to the axi-symmetric

mode of the fields, i.e., the mode whose only non-vanishing field components

are H Ez and E0’

independent of the azimuthal variation $. H is easilyP’ 4

seen from Maxwell’s equations to satisfy (A*l) (see Appendix).

An application of (A.7) to region (2) (see Fig. lb) gives

b

H(2)(p,z)==fc(~) -+H~ef(Q) + iwe+ \

G2(p,z;p ’,s)Ep(p’)p’dp’ (L)

a

where G2

satisfies (A-6) with boundary conditions

* (pG2) = o , when p = a

(2)

+G2=0, when z = O , s

refand H

@denotes the field reflected from an infinitely long, perfectly

conducting cylinder and will be determined shortly.

The magnetic ’field vector of the incident plane wave is given by

incH =-eHe

ikx— ~o

4

the axi-symmetric mode of which is —.

21T

j‘nCd@ = - iHoJl(kp)H~nc(p) ‘~ ~ “ ~ (3)

o

To find Href~ (P) h (1) we take (3) as the wave incident on an infinite,

perfectly conducting cylinder. One can easily

[uJI(u)]’ref

‘$(p) = iHo

[llHf’)(u)]’

find that

H;l)(kp) (4)

where the prime denotes differentiation with respect to u(=ka).

Substitution of (3) and (4) into (1) gives

[uJI(u)]’b

h(2)~ (P,z) ‘- iJl(kp) + i - ‘l)(kp) + ik

‘1 j ‘2epp ’alp’ (5)[uH~l) (u)]’

a

(2)= H(2)/Ho and ep = Ep/(ZoHo).where u = ka, h

$ +We now apply (A.i’)to region (1) (see Fig. lb) and obtain

bh(l)~ (p,z)=-ik

~Gl(p,z;p ’,s)ep(p ’)p’dp’

a

h

+~

h(l) (a,z’){~~,4

— (P’G’l)} dz’ + (o..) ,forssz<h (6)

spf=a

where (o.”) denotes the integral over the end cap of the post and has been,1

written out explicitly in a previous note. The Green’s function GI satisfies

(A*6) with boundary condition

+G1=O , whenz=s (7)

and the radiation condition at infinity.

In the hole (z = s, a < p < b) we have h(1) = & Subtracting (5)+

from (6) in the hole we obtain the first relationship between h(1)

4and ep,

name ly,

h b

~K(p,s;a,z’)h ~l)(a,z’)dz’ + ik

){G1(p,S;P’,S) -f-G2(p,s;p’,s)}eO(p’) p’dp’

s a

[uJI(u)]’‘~)(k~) + (*-*)= iJ1(ko) - i , fora<psb (8)

[UHIYU)l’ ‘1

where

K(p,z;a,z’) = - E-& (P’G1)] .~l=a

‘1) and e is givenby (6) whenp = a.The second relationship between h$ P

Thus

h b

\ J~ h~l)(a,z) + K(a,z;a,z’)h~l)(a, z’)dz’ + ik G1(a>z;p’s)ep(Q’)p’dp’

s a

= (“’”) , fors~z<h (9)

Equations (8) and (9) constitute a coupled set of integral equations between

h$l)(a,z) and ep(p). Once eP

is obtained by numerically solving (8) and (9),

6

~(2)

$can be found from (5).

-—In the next section we shall find the explicit forms of the Green’s

functions GI and G2.

7

111. Green~s Functions

The Green’s function Gl that satisfies~.he equation (A-6) and the

boundary condition (7) can be written down by inspection. Thus,

(10)

where

22R2= (z-z’) +p-t-p

,2- 2pp’ Cos @

R2 = (2s - z- z’)2+L?2+p’2- 2p9’ Cosl$s

The explicit form of the kernel K in (8) and (9) is then given by

K(p,z;a,z’) = - [~ (P’GI)]p~=a

2TikR

J’

s

(P -(ikR - l)eikR + (i~~ - l)e. Cos +){ } d+ (Ll)

4TrR3 4mR:o p’=a

To find G2 that satisfies

~21a 22

(~+–—- ++— + k2)G2 = - (s(2- z’)6(P -P’)

app ap

P 322 P’ (12)

and

:G2=0 when Z=o, s.—.

$ @22) = O when Q = a

we proceed in the following usual way. Let

-.

G; = ~ AnH(l) (Anp)cos(nnzls) 9n=o

forp>p’ (13a)

G~l = f a A [P(A a)Jl(Anp)-(1)

- Q(~na)Hl (inO)lcos(n~z/s) , for P < p’ (13b)nn n

n=o

‘l)(X)]?, Q(x) = [xJI(x)]’,where P(x) = [xHI

(1)

‘1 (xnP’)CY =n

%#)9

P(Ana).J1(Anp ’) - Q(Ana)H1

2and ~ = k2 - (nm/s)2. Note that (13a) and (13b) satisfy all the boundary

nconditions (12). To determine the constants An , multiply the differential

equation (12) by cos(nmz/s) and perform

s p’+%

LimJJ

(**.)pdpdz .E%

o 0’-s

After some straightforward manipulations we obtain

W@~i ‘1

aAA=— cos(n~z’/s)nnnsc

n H~l)(Ana)

9

where E =lifn>O, andc =2. Substitu~ing this into (L3a) and (13b)n o

we then have

.

G2(P, Z:P’,Z’) =: ; ‘$-[J1(A*P<)HO‘1)(~na)n=o n

- J (1.a)ll~’)(knp<)] ‘~~~(Anp>) cos(nnz/s) cos(nnz’ /s) (14)on

Ho (Ana)

where [j< (p,) denotes the smaller (larger) of p and p’.

10

Iv. Numerical Results—.

Equations (8) and (9) were solved on a CDC 6600 computer for two values

of a/h, f-ourvalues of s/h, five values of b/h, and a wide range of kh.

Equation (5) was then used to compute h(2)

from the numerical results of e .4 P

Figures 3a-4d show the variations of the current amplitude along the

post with the radius of the hole as a parameter when the frequency of the

incident wave is around the resonance frequency of the

the post, i.e., the mode with the longest wavelength.

quantity 111/10 is related to the field h~ by lll/l.=

For a step-function incident wave, ~ ‘nc(x,t) =-

fundamental mode of

The dimensionless

2nalh41/h.

eHU(t- x/c), the-vO

time behavior of the current at z = O is given in figure; 5a-6d with the

radius of the hole as a parameter, For comparison purposes, the response

of the post without the top plate is also given in those figures.

Figures 7a and 7b summarize the results derived from figures 3a-4d.

The resonance frequency as well as the current amplitude at z = O of the

fundamental mode are plotted against the radius of the hole with the plate

spacing as a parameter.-1

The decay time constant,a , of the fundamental mode given in figures

8a and 8b was obtained by fitting the envelopes of the curves in figures

5a-6d to an exponential curve of the form exp(- act/h). They show that,-1

except for the expected result that a increases with the hole’s size for a-1

given plate spacing, a decreases as the plate spacing is increased for

fixed hole size,

a

In figures 3a-4d and in figures 7a and 7b, k. is the wave number

corresponding to the resonance frequency,

11

Appendix.

—.Let u(p,z) satisfy the equation

22

(~+:$-++— + kz)u = O

ap P a.22

in the space exterior to V, the volume of a body of revolution (see Fig. 2),

and let u satisfy the radiation condition at infinity. We wish to express

u in terms of its value and its normal derivative on S. Since u(p,z)cos @

satisfies the three-dimensional Helmholtz equation

an application of the ordinary Green’s theorem gives

(As1)

where G is the free space Green’s function given by

2 ‘2~ikdp + p - 2pp’ Cos(+ - $’) + (z - Z’)2

G(p,z,@;p’,z’,~’) = .

4Ti’p2+p’2 - 2pp’ Cos ($ - $’) -1-(z- Z’)2

(A92)

Multiplying (A”2) by cos $ and integrating the resulting equation with respect

to $ from O to 2m we get

12

.

2TI

U(p,z) =+J 1

{aGCOS $ d4 ucos$’—-

—an’Gcos @’&} U dSv . (A-3)

o s

For a function F(o) having the properties F(+) = F(- $) and F(+) = F(2Tr+ $),

it can be easily shown that

2n 2V

\

27r

COS @ d$J

F($ - $’)COS ~’ d$’ =n1

F(I/J)COS~ d~ .J

o 0 0

Since G and (3/an’)G have the same properties as F, (A-3) becomes

U(p,z) =I

where

2Tr

G(2) (p,z;p~,zf) =~

Cos ljl

o

(A”4)

ik~(z- Z’)’ +p’+p ’z- 2pp’ COSYPd+ (A05)

4n4(z- Z’)’ +p’+p”- ,2pp’Cos$

which can be shown to satisfy the equation

Let us consider the case where the only non-vanishing magnetic

. (A06)

field

component is H4

and has the same properties as u, and where the volume V is

a right circular cylinder. Remembering that iucE = (a/az)H@ andP

iu&E = - (a/aP + I/P)H@, we have from (A-4)z

13

! u

J‘G ‘2) - iwEEp~(2))~’d~’

HJP, Z) = (g’ ● ~e !) {H+ ~zl___

z 1=comstant

+J

{H+*@’G

‘2)) + iusp’EZG ‘Z)}dz’ .

pt=constant

J

14

(A-7)

—.

Figure la. The geometry of the problem.

(1)

‘incL(2,

z

!- 2b

h

■

A

s

Figure lb. Side view of the geometry of the problem.

15

—.

z

Figure 2. Body of revolution.

16

5

4

7

NOVHH

3

2

1

0

.

II/h= .15

. !

.4koh = 1.2

.

0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0

z/h

FIGURE 3a. Current vs. position around resonance frequency.s/h = .3, a/h = .1,

5

blh = .15

“ b/h= .2525 kh= 1.10

b/h = .6 koh= 1.1

.

3

2

1

0 .1 .2 ?.. .4 .5 .6 .7 .8 .9 1 .J

z/h

FIGLP.T.3b. Current vs. -os~:ion around resonance frequency.

s/h = .5, a/h = .1.

J ,

l--w

s

5

3

2

1

0

bfh = .15 kh =.9o

b/h = .3125 koh = 1.1

“ blh = .8 koh = 1.2

o .1 .2 .3 .4 .5 .6 .7 .8 .9zjh

FIGURE 3~. Current vs. position around resonance frequency.s/h = .7, a/h= .1.

5

3

2

1

0

I 1 I I----- , .—... ..- —r , ---

—T

b/h = .5750

.1 .2 .3 .4 .5 .6 .7 .8 .9 1.0

z/h

FIGURE 3d. Current vs. -osition around resonance frequency. \s/h = ,9, a/h = .1. .A

<

-r

~

NO‘-’ I-4H

3

2

1

(1

b/h = .015 koh = 1.4

.

I I 1 I 1 I 1 I 1

.0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.5

z/h

FIGURE 4a. Current vs. position around resonance Ereauency.a/h = .01, s/h = .3.

5

4

7 “’

NO=HH

3

2

1

0

I

b/h = .015, k_h = 1.3u

.

b/h = .51, koh = 1.5

.

I 1 1 I 1 1 1 1 1

0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0

z/h

FIGURE 4b. Current vs. position around resonance frequency.

a/h = .01, s/h = .5.:!

,-.,I

5

2

1

blh = .015 koh = 1.2

b/h = .3625 koh = 1.3

I

.

co .1 .2 .3 .4 .5 .6 .7 .8 .9 1.0

zfh

FIGURE 4c. Current vs. position around resonance frequency.a/h = .01, s/h = .7.

5

2

1

I

o0 .1 .2 .3 .4 “5 .6 .7 .8 .9 1.0

z/h

FIGURE 4d. Current vs. position around resonance freauencv.a/h = .01, s/h = .9. ..

f’

1,

1.2

1.0

.8

.6

.4

.2

+340R o

-. 2

-. 4

-. 6

-.8

-1.0

–1.2

=

,—

.

0 2 4 6 8 10 12 14 16 18 20et/h

FIGURE 5a. CURRENT VS TIME AT z = O FOR a/h = .1, s/h = .3

1.6

1.4

1.2

1.0

.8

.6

I (o, t)10 .4

.2

.0

-. 2

-. 4

-.6

-.8

0 2 4

FIGURE 5b. ~~

6 8 10 12

et/h

CURRENT VS TIME AT z = O FOR a/h= .1,

14 16 18 20

s/h = .5 ‘,

3.0

2.5

2.0

1.5

1.0

.5

L$.&o

N-4 -.5

-1.0

-1.5

-2.0

-2.5

-3.0

I i I I I I I 1 I I I I I I I I I I I

I I I I 1 1 1 I I I 1 1 1 1 1 I 1 I 1 10 2 4 6 8 10 12 14 16 18 20

et/h

FIGURE 5c. POST CURRENT VS TIME AT z = O FOR a/h = .1, s/h = .7

3.0

2.5

2.C

1.5

1.0

.5

Uwo10

-. 5

-1.0

-1.5

-2. (

-2.5

-3. (

-3. !

1 I I 1 I I i i I I I I I I I I I 1 I

FREE SPACE

I I I I 1 1 i 1 I 1 I 1 I 1 1 1 1 I I

2 4 ,6 8 10 12 14 16 18 20

et/h

FIGURE 5d. POST CURRENT VSTIMEATz=OFOR a/h= .1, s/h=.9!

, .

1.5

1.0

.5

M2zI )()10

r.1u)

-. 5

-1.0

-1.5

1 1 I I I I I I I 1 1 1 I 1 I I I I 1 1

[“I I I I I I I 1 I I I I I I I I I I

o 2 4 6 8 10 12 14 16 18 20

et/h

FIGURE 6a. l?OSTCURRENI’ VSTIMEAT z= OFORa/h =.01, s/h= .3

1.5

1.0

.5

-. 5

-1. C

-1.:

I 1 T I I I “~-– ! I ~-l-----

-1 r- i,— - ,

I

1 I I 1 I I I I I I I 1 1 I 1 I I I 1

2 4 6 8 10 12 14 16 18

et/h

FIGURE 6b. POST CURRENT VS TIME AT z = O FOR a/h= .01, s/h = .5

.

I I I I r 1 I I I 1 I J I I I 1 I i I

1.5 -

mt-

-1.0

-1.5

I 1 1 1 1 I 1 1 I 1 1 I I I 1 1 t 1 1 1

. I

0 2 4 6 8 10 12 14 16 18 20

et/h

FIGURE 6c. POST CURRENT VS T~E AT Z = O FOR a/h = ,01, S~h = .7

1.5

1.0

.5

XM()10

wr--l

-. 5

-1.0

-1.5

I I I I I I I I I I i I I I I I I I I

.015

.2387

.91

I I I I 1 1 1 I 1 I 1 1 1 I I I I I I

o 2 4 6 8 10 12 14 16 18

et/h

FIGURE 6d. POST CURRENT VS TIME AT Z = O FOR a/h= .01, s/h= .9

)

.

5

4

3

0H---

I

Slh = .3

.5

.7

Slh = .7

.

,,

koh

1.3

1.2

1.1

1.0

0.9

1 1 1 1 1 1 1 I 1.1 .2 .3 .4 .5 .6 .7 .8 .9 1.0

b/h

FIGURE 7a. Resonance frequency and current at resonance frequency vs. hole’s size for a/h = .1.(Post in free space: 11(0)~/I = 8.4, k h = 1.23.)

o 0

4

:

0H

lg

.0

H

I I I I I I I I I

.

. 1

SI’n= .3—

..5

.7

I 1 1 I I I I 1 1

.1 .2 .3 .4 .5 .6 .7 .8 .9

b/h

FIGURE 7b. Resonance frequency and current at resonance frequency vs. hole~s size for afh = .01.

(Post in free space: 11(0)1/10= 7.2, koh = 1.45.)

cot-l

1.5

1.4

1.3

1.2

1.1

1.0

. .*

.

I I I I I I I 1 1

4.0

-1a

3.0

2.(

.1 .2 .3 .4 .5 .6 .7 .8 .9 1.0

b/h

FIGURE 8a. Decay time constant vs. hole’s size for a/h = .1.(Free space: a‘1 = 4.6.)

.

<

um

6.C

5.(

-1u

4.(

Slh = .3

I 1 1 I I 1 1 I I

.1 .2 .3 .4 ●5 .6 .7 .8 .9

FIGURE 8b. Decay

(Free

1.0b/h

time constant vs. hole’s size for a/h = .01.

space: ~-1 = 7.8.)

.

*

.*..

Zrn00I-t@ID w

CL~

. ‘2

.

-4

‘%

(m.

.)I-J

.

w.

..

?i

.