acknowledgrnent - the university of new mexicoece-research.unm.edu/summa/notes/ssn/note93.pdf ·...
TRANSCRIPT
,: -
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
.