SSN 235, K.-M. Lee, An Analytical Investigation of the Method of Using an Extrapolation Function in Finding Criteria Response from Simulation Response, 12 Dec 77, Mission Research Corp.

SSN 235, K.-M. Lee, An Analytical Investigation of the Method of Using an Extrapolation Function in Finding Criteria Response from Simulation Response, 12 Dec 77, Mission Research Corp.
.
..
Sensor and Simulation
iiotes
-
-
Note 235
12”December 1!377
AN ANALYTICAL INVESTIGATION OF THE METHOD OF USINGAN EXTRAPOLATION
FUi4CTIOil
1)!FIi4DINGCRITERIA RESPONSE FRO!4SIMULATION RESPONSE
Kuan Min Lee
?IissionResearch Corporation
-Albuquerque, Ne\~ifexico87108
.
Abstract
The extrapolation techniques outlined in Sensor and Simulation Note
222 are discussed. “The methods are applied to an infinite cylinder problem as well as a finite cylinder problem using theoretical models. The
effects of the angle of incidence “isalso examined using the infinite
cylinder model. Some conclusions about the utility of the methods in
error estimations are drawn based on these analytical investigations.
-L
--
.:
..
,.
.
.
-——.—
——.—
—
—
.—-—
f~?
%
[//q
.
TABLE OF CONTENTS
PAGE
SECTION 1
-
AN INVESTIGATION OF THE METHOD OF USING EXTRAPOLATION FUNCTION IN FINDIN~ CRITERIA RESPONSE FROM
SIMULATION RESPONSE
6
1.1
INTRODUCTION
6
1,2
NOTATIONS AND DEFINITIONS
9
1.3
ATYPICAL
1.4
APPLICATION OF THE EXTRAPOLATION FUNCTION
TO THE ABOVE PROBLEM
20
CONCLUSION
32
1.5
SECTION 2
-
AN APPLICATION OF THE EXTRAPOLATION FUNCTION
TECHNIQUE TO THE FINITE CYLINDER PROBLEM
33
2.1
INTRODUCTION
33
2.2
FORMULATION OF THE FINITE CYLINDER PROBLEM
33
2,3
APPLICATION OF THE EXTRAPOLATION FUNCTION
TO THE FINITE CYLINDER PROBLEM
37
CONCLUSION
45
2.4
SECTION 3
-
12
PROBLEM
THE INFINITE CYLINDER PROBLEM IN THE PRESENCE OF
A PLANE WAVE WITH DIFFERENT INCIDENT ANGLE
3.1
3.2
SCATTERING OF AN INFINITE CONDUCTING CYLINDER
FOR AN INCIDENT WAVE WITH INCID T ANGLE i3
Y
AN INFINITE CONDUCTING CYLINDER WITH A PERFECT
CO!JDUCTINGGROUND PLANE IN THE PRESENCE OF AN
INCIDENT PLANE WAVE WITH INCIDENT ANGLE $
50
52
3.3
CONSIDER A SPECIAL CASE
55
3.4
APPLICATION OF THE EXTRAPOLATION FUNCTION
TECHNIQUE TO THE ABOVE PROBLEM
59
61
REFERENCES
APPENDIX A -
50
SUPPLEMENTAL RESULTS FOR THE PROBLEM IN SECTIONS
7.3 and 1.4
2
63
t“
.
.
0
-=
LIST_OF ILLUSTRATIONS
FIGURE
la
.Ib
PAGE
Airplane in Free Spa_ce
-..
7
8
Airplane Near Ground
An infin te cylindrical conductor in the presence of an
incident plane wave, (a) without a ground plane and (b)
with a ground plane,
13
3
Coordinate for Using Additional Theorem
1.7
4a
J(FF) as a function of frequency at different angle @
fo; a cylinder with radius a = 0.5m.
21
J(HPD) as a function of frequency at different angle o
fo$ a cylinder with a = 0.5m, d = lm, calculated using
first approximation.
22
JjHpD) as a function of frequency at different angle +
for a cylinder with a = 0.5m, d = Im, calculated using
second approximation.
23
The ratio ]Qi(~)[ at different angle @i calculated using
first ap~Foximation and a reflection coefficient Refl =
-1.0 for’the ground, and a = 0.5m, d = lm.
25
The ratio lQi(m)\ calculated using second approximation
and a reflection coefficient Refl = -1.0 for the ground,
a = 0.5m, d = lm.
26
The ratio /Qi(~)l calculated using the first-approximation and a reflection coefficient-Refl = -1,0, a = 0,5m,
-d = 5m.
27
The ratio \Qi(~)l at different angle @i calculated using
first ap~roximation and a reflection coefficient Refl =
-0.75 for the ground, a = 0.5m, d = lm.
28
The ratio lQi(~)l at aifferent angle @i calculated using
second approximation and a reflection coefficient Refl =
-0.75 for the ground, and a = 0.5m, d = lm,
29
The ratio \Qi(~)l calculated using first approximation
and a reflection coefficient Refl = -0.75 normalized
to the incident field at different locations, a = fl.5m,
d = lm.
30
2
4b
4C
5a
5b
6
7a
7b
8
3
—
●
LIST OF ILLUSTRATIONS (continued)
FIGURE
9?i
PAGE
The function Fl(w) for a cylinder in free space with
a = 0.5m.
31
The ratio lQi(~)l for a cylinder with a = 0.5m at
different angle $.
31
A finite cylindrical conductor in the presence of
an incident plane wave without a ground plane.
34
A finite cylindrical conductor in the presence of
an incident plane wave with a ground plane.
36
12
The axial current density Jx‘FF)(@i,t) at x = -30125m.
38
13
The magnitude of the Fourier transform of Jx‘FF)(+j,t)
at x = -3.125m.
39
The ratio lQli(~)\ for a cylinder in free space for
two incident waves with incident angle different by 90°.
41
The ratio \Qli(M)l for a cylinder in free space for
two incident waves with incident angle different by
90°, calculated using 3-D code.
42
15
The axial current densith Jx‘HpD)(@ijt) at x = -3,125m.
43
16
The magnitude of the Fourier transform of Jx(HpD)(oi,t}
at x = -3.125m,
4&
The ratio /Qli(~)] at different angle @i at x = -3.125m
using data given in Figure 13 and Figure 16.
46
The magnitude of the Fourier transform ‘(
Jx FF)(@j ,~) at
x = Om.
47
The magnitude of the Fourier transform –(HpD)(@i,w)
J
at x = Om.
48
The ratio lQlj(ti)lat different angle @i at x = Om,
using data given in Figure 18 and Figure 19.
49
9b
10
11
14a
14b
17
18
19
20
21
An infinite cylindrical conductor in the presence of an
incident plane with incident angle B.
0
—
LIST OF ILLUSTRATIONS (continued)
PAGE
FIGURE
22
23
24
A-1
A-2
A-3
@
A-4
A-5
An infinite cylindrical conductor with a ground plane in
the presence of an incident plane wave with incident
angle B.
The ratio_lQli(o)\ at different angle (jjwith a = 0.5m,
d = lm, R1 = -1.0 and $ = -Tr/4,O, n/4.
.–
The ratio lQli(~)l at different angle @i with a = 0.5m,
d = lm, R1 = -0.75 and 6 = -Tr/4,O, n/4.
53
57
58
The ratio IQ1(u)I at the angle + = 0°, reflection coefficient Refl = -1.0, a = 0.5m, d = lm.
64
The ratio \Ql(u)l at the angle$= 0°, reflection coefficient Refl_=_-1..O,_
a :.Ll,5m,and d = 5m.
65
The ratio /Qi(~)] at different angle +1 calculated
using numerical solution of Eq. (25) with N = 7 and
a reflection..coefficientRefl = -1.0 for the ground,
and a = 1.5m, d = lm.
66
The ratio lQi(ti)lcalculated using numerical solution
of Eq. (25) with N = 4 and a reflection coefficient
..of Re.fl.
= -1.0 for the ground, a = 0.5m, d = 5m.
67
The ratio [Qi(~)l at different angle @j calculated
using numerical solution of Eq. (25) with N = 7 and a
reflection coefficient Refl = -0.75 for the ground,
a = 0.5m, d= lm.
68
—
—
...
-—
.
SECTION I
AN INVESTIGATION OF THE METHOD OF USING EXTRAPOLATION FUNCTION
IN FINDING CRITERIA RESPONSE FROM SIMULATION RESPONSE
1.1
INTRODUCTION
In the analysis of the electromagnetic field problems, it is often
necessary to use the principle of electrodynamicssimilitude in the design
of apparatus.
It has been shown using Maxwell’s equations that, in order
that two electromagnetic boundary-value problems be similar, it is necessary and sufficient that certain coefficients be identical in both (ref. 1).
This has always been the guide line for designing experiments.
However,
suppose that the true or criteria environment is so difficult to construct
so that the simulation environment is not quite similar to the criteria
environment, what is the remedy to extrapolate useful information from
the simulation response?
In particular, consider the problem of finding
the EMP response in an electronic circuit inside an airplane in free space
due to a nuclear burst (Fig. la).
Instead of simulating the awkward true
situation, measurements were made in an airplane near ground with a pulse
generator as shown in Figure lb.
Is it possible to find the criteria re-
sponse from the measured simulation response inside the airplane if the
scattered fields outside the airplane are given but the transfer functions
between these fields outside and the response inside are not known?
It
has been proposed recently that this problem could be solved by making use
of certain ‘Extrapolation functions” if some error bounds in the result
were allowed (ref. 2).
It is the purpose of this study to investigate this proposal byapplyjng
the suggested method to a typical problem,
6
.
-.
b--
Y
x
Region 3, exterior
1, interior
o
,,:.,.,.:.,.,.,.,,,,.::,,....,,,,.:.:,.,..,.,,,:.,.:j
Region 2, surface
o
Figure la.
Airplane in Free Space.
7
-.
.
.
a
‘;--f‘~y
kx
Region 3, exterior
;(sim)(;3j)
,,
.,,,,.,
,,.,
Region 1, interior
.,,.,,,,,,.,,.,.,.,,
...............~
o
Region 2, surface
~Wooden
o
//’//////////////////
Stand
//////////
Ground
Figure lb.
Airplane Near Ground.
8
,-
1.2
NOTATIONS AND DEFINITIONS (ref; 2).
.
Some notations and definitions used in this report will be summarized
briefly in this section, --Consider the example in Figures la and lb again,
Let the space be divided into three regions where region 1 is the interior
region inside the airplane with position vector~l,
region 2 is the aircraft
surface region with the position vector ;9, and region 3 is the space exterior to the aircraft with the position ve;tor ;3.
measured in Me
The voltage response
k[h circuit component in region 1 will be called V(;lk).
The current -rnea::::ed-at
the ith location of the aircraft surface (region 2)
.th
will be called J($21). The electric field measured at J
location in region
3will
be called=E(r3i),
In order to distinguish the criteria situation
from the simulation s;tuation, we associate the superscript “sire”with
the quantities in the simulation environment and “cri” with the quantities
for a measured voltage on the kth circuit
in the criteria environment. .Thus,
.
in region 11 in the.c.riteriaenvironment, the notation is V (cri)(~lk),
The field in region 1 is induced by the electromagnetic fields in
regions 2 and 3 through the port of entry.
fer functions T(~2i,~lk)> etc.
They are related by the trans-
The transfer functions relating fields
between regions 2 and 3 can be determined by solving the exterior problem
(refs. 3 and 4) and will not be discussed in this report.
Therefore, all the transfer functions considered here will be those
between region 1 and region 2 with the first position vector referring
to those in region 2 and the second position vector referring to those
in region 1.
Note that the transfer functions are varied with frequencies
as well as positions.
Thus, for measured quantities V ‘sim)(~,k) and
~(sim)(;2i) in the frequency domain, we have
N
V(sim)(;,k,ti)=
E
T(sim)(;2j, ‘lkY ~,
i=l
9
J(sim)+
(r~i,~)
.
(1)
.
If we restrict the discussion of the transfer functions to those between
region 1 and 2, and note that the left hand member in equation (1) is always
(sire))is
in region 1, whereas the right hand response (in this case, J
always in region 2, we may simplify the notations by dropping the superscript and write equation (1) as
N
(sim)~ti)J~sim) (w)
V~sim)(U) ‘~Tik
(2)
i=l
where i and k are simplified notations for ;i and ;k.
An equation similar to (2) can be written down for the criteria
situation,
N
V~cri )(w) = ~
T\;ri )(u) J$cri )(u)
●
i=l
(3)
We now make an assumption that the transfer functions in equations (2) and
(3) are the same so that
o
.
(4)
This assumption will be reasonable if the interior configuration as well
as the surface configuration remains unchanged in the criteria environUsing equation (4) in equations (2)
ment and the simulation environment.
and {3) we have
(5)
N
V[cri)(u) .~
Tik(u) J[cri)(ti)
i=l
.
(6)
o
\
and Ji‘cri)(uJ)can be calculated or measured from
The quantities Ji‘sim)((JJ)
(sire)(w)can be measured, and V(cri) (LO)
the exterior problem., .The q!antitY ‘k
is the unknown quantity that we are seeking. If the values of Tjk(Ol are
given, the problem can be easily solved.
Difficulties arise in a real
situation where the point of entry and the transfer function is not known.
In_order.to
circumvent this.difficulty, a quantity called “extrap—
elation function” Ri(~) is introduced> where
)(u)
J(cri
Ri(~)
‘~-
●
(7)
i
Using this definition, equation (6) becomes,
N
Ji‘Sire)(u)
Ri(w)
V(cri )(u) =~Tik(~)
k
.
(8)
i=l
In general , Ri (w) is a function of both frequency and the position and is
not a constant.
—s--—
If there is only one port of entry, equation (8) reduces to
(Sire)(u)
Ri(IJ)
v(cri )(~) = Tik(~) Ji
k
(9)
and from equation (5)
v(sim)(~)= Tjk(bJ) ‘j(Sire)(u)
(10)
-k
so that
“~crf)(o) =
“~‘ire)(u)
Ri(u)
and V\cri) (U) can be determined,
—.
11
(11)
.
Another special case is when Ri(ti)= Ca
constant, then equation (8)
becomes,
N
In particular, when C = 1, the two problems are entirely similar.
It is
easy to see from equation (12) that, if Ri(u) is a constant, the criteria
response Vk‘Cri)(u) can be derived from the simulation response in a simple
manner.
In genera?, if some constant (say, a number obtained by taking
the average of Ri(w) over all positions i) is used in replacing each individual Ri(~), it will result in some error since Ri(ti)is not a constant
under most circumstances.
However, due to the simplicity of equation (12),
it is interesting to study the above procedure and the possible error bounds
more colsely.
The error can be obtained roughly by examining the ratio
Ri(~)/F(ti)as a function of frequency, where F(M) is a selected testing
function which is not a function of positions.
In this study, we shall
select the following test function:
($%logeRi(”)
‘
F1(uJ)= EXP1
1.3
ATYPICAL
(13)
PROBLEM
Consider the problem of an infinite cylindrical conductor in the
presence of an incident plane wave with and without a ground plane as shown
in Figures 2a and 2b.
Let the electric field component of the incident
plane wave be parallel to the axis of the cylinder, with the wave propagating
along positive x direction.
The coordinate systems are as shown in Figures 2a and 2b.
Then, the
incident electric field can be written as
●
12
(14)
e
1
I
x
kx
-Y
E;
H
Y
I
,7
~Y
kx
I
d
-Y
xy
I
////~////’/,
----7-
‘
d
P(p,@)
1’
x
(a)
Figure 2.
!/
(:)
An infinite cylindrical conductor in the presence of an incident
..:+
hn, ,+
ntia,,n,l
nlamA
-HA
rh~
,.,:+L
--a,,..,l
fil--A
..1-.”.,.
.,-..,,.
WILII
a 91UU11U
plullc.
plallc
Wave>
(a)
Wlt,lluu(,
a yluullu
plullc
Urlu \u)
P(P>O)
“’’’’/’/
Perfectly
Conducting
Plane
To solve the problem shown in Figure 2a, equation (14) can be expanded in
+j~t
the following form (refs. 1 and 5) with e
E:= E. ~
suppressed,
j-n Jn(kp) ejn$
.
-m
(15)
The scattered field can be assumed to be
m
E; = E.
x
“-n
Jn
~._a.l
~~2)(kp)
a
ejno
(16)
where Jn(kp) is the Bessells function of order n, and Hn‘Z)(kp) is the Hankel’s
function of the second kind with order n.
Et = E: + E; = O, hence
z
Jm(ka)
a
n ‘-*,
At p = a, the total field
.
(17)
o
Using equations (17), (16) and (15), the total electric field can be
determined.
The total magnetic field is given by
+
.
The current density on the cylinder is related to equation (18) by
Jz = [email protected]~ atp=a.
Combining the above results, it is easily found that
(refs. 5 and 6)
2E0
Jz=—
m
.-n
ejn+
u_q..rna
z
n.-a ‘*
To solve the problem shown in Figure 2b, it is necessary to assume
two more scattered fields due to the presence of the perfect conducting
ground plane.
Thus we have (let R, = -1)
14
(18)
m
E:= E.
J“-n Jn(kp) ejn+
x
(20a)
n..02
03
‘: =
..Z
jn e ‘.jzkd
‘1 ‘o
Jn(kp)
e-jnq
(20b)
n=-m
co
El = E.
—
E
J“-n an H!z)(kp) ejn$
(20C)
n.-m
w
sr
Ez
= ‘1 ‘o z
.Jnann
jn$l
H(2)(kp1) e
(20d)
.
E;=
E;+E;+E:+E:r
(20e)
where E1 is the incident field, E; is the reflected field due to the ground
plane, ~~ is the scattered field due to the cylinder, E~ris the scattered
field due to the image of the cylinder and E; is the total field.
The
origin O is taken to be the center of the cylinder and the ground plane
is located at a distance d away from the origin.
The coordinates p, $, PI
and $1, 0 are as shown in Figure 2b.
To show that the boundary conditions
can be satisfied by these fields, note that equations (20a) and (20b) can
be written as (refs. 1 and 5)
E; = EO e-j ‘p [email protected] = E e-jkx
o
El = R,
E. e-jzkd e+jkx
(21a)
.
15
(21b)
For RI = -1 and X = +d, these lead to E; + E; = 0.
Furthermore, for points
on the ground plane, @ = @l and p = P, so that E; i-E~r= O from equatfon
(20c) and (20d).
Hence, the boundary condition on the ground plane is
satisfied.
.
The second boundary condition is E; + E; + E: + E~r = O at p = a.
To show this, it is convenient to express all fields in term
able p and $.
of the vari-
This can be done by expanding equation (20d) in a different
form using the addition theorem of Bessel’s function (ref. 1).
Referring to
Figure 3, the additional theorem gives.
m
-im(@-6a) - ine .(22)
-in6~ =
H(z)(krl) e
H(2)(kro) Jm+n(kr) e
n
In
z
m. -m
In our case, 00 = O, ro”= 2kd, rl = p,, r=
p, 6 [email protected],
01 = m-$l(see Figures
2b and 3), from (20d) we have,
co
sr
Ez
J‘“nan H~2)(kP1) e
= ‘1 ‘o E
=-m
n
o
jn(m-el)
.
(23)
Applying the boundary condition a-tp = a,
z
z
J“-n Jn(ka) ejn$ +
n.-m
RI e-j2kd
jn
Jn(ka)
ejn$
E
n.--co
m
+
n.-~
(2)(ka) [email protected]
J“-nan Hn
‘[email protected]&
n.-~
m. -m
El(2)(2kd)Jn+m(ka) e
m
[email protected] .0
.
16
(24)
0
-.
.+
—
P-(r,
@)
r
I
o
——
——.
—.—
T
Figure 3.
x
90
Coordinate for Using Additional Theorem.
17
,
Equation (24) determines the proper coefficients an for this problem.
If we change the index in the last summation in (24), and co’Iecting terms
associated with ejno, we have
~
J-n(ka)
j-nan Hn(2~(ka) + R, j-na-n H(2)(2kd)
x+
03
w
+
j-(n+m)
a-(n+m)H$) (2kd)J-n(ka) +
m= 1
=
-(j-n -f-R,e-j2kdjn)
Jn(ka)s
z
Rljm-nam-nH(2)
m (2kd)(-l)m-nJn(ka)
m=1
II = ‘mYCO”.*Yo~”””.”’ ‘m
This is a matrix equation with infinite elements.
mination of an from (25) is difficult.
(25)
.
A complete deter-
Various approximations can be used
in solving (25) depending on the accuracy of the result we are seeking.
We
shall consider the following approximations:
(1)
A first approximation is to neglect the effects due to
the scattered field of the image cylinder so that the
terms with Hm‘2)(2kd) Jn(ka) in equation (25) can be
omitted.
This leads to
~ ~1 e-j2kdjn)
(j-n
Jn(ka)
a =
.n
J n H~2)(ka}
(2)
(26)
The next simple case is to use the diagonal element in
the matrix in solving equation (25) (ref. 7)
a
~
. .
n
(j-n+R1
e -j2kdjn) Jn(ka)
j-n[H~2)(ka) + R, H\~)(2kd) Jn(ka)]
.
18
(27)
.
(2)
This reduces to equation (26) when H2n (2kd) Jn(ka) is
neglected.
(3)
If-great_accuracy is desired, it is possible to solve equation
(25) for N finite_rnode_s.
For the present purpose, approximations (1) and (2) are suff-icient.
,
The current density is given as follows:
(1)
Using equations (26), (20) and (18),
.—.
aE;
1
JZ=H
Jti~
-$lp=a = -
2E0
m
—
ap p.a
(j-n
+
R,
e-jzkdjn)
=—
(28)
~v~a
____ x
n..m
(2)
tl~2)(ka)
us ng equations (27), (20) and (18)
-j
dz=———
E.
cPO
-j E.
W.
w
z
n=-m
J“-n Jn”(ka) eJnO +
m
R, e
X
-j2kd jn Jn’(ka) ejn$ +
n.-m
#z
co
-j
E.
Jnann
H(2)’(ka) e
jnb +
n.-m
cc
-j
E.
jn
an
H(2)(2kd) Jn+’(ka)
m
x
m.
-m
e-j(m+n)$
.
IIIfact, Equation (29) is the general solution to this problem if
the correct an’s are used.
In the next section, we shall discuss the
application of extrapolation functions using the results der;ved in this
section.
1.4
APPLICATION OF THE EXTRAPOLATION FUNCTIONTO
THE ABOVE PROBLEM
To apply the extrapolation function concept to the above problem, we
make use of Equations (19) and (29) (or Eq. (28)), and derive the following
function from them
J(FF)(@i,@
Ri(~) = ~~HpD
‘($i,”)
z
(30)
where J(FF)($. ,u) is used to denote the current density obtained for
Figure ;a (Eq~ (19)), and J$HPD)($ ,OJ)is used to denote the current density
i
obtained for Figure 2b (Eqs. (28) or (29)). next, the average function
Fl(u) is formulated according to Equation (13), that is,
.
N
F1(u) = ~R1(u)R2(w)R3(u) ... RN(u)’
.
(31)
Then, the following ratio is calculated and plotted as a function of
frequency,
.
(FF)
In Figure 4, the absolute value of Jz
of frequency with angle o as parameters.
and
~:HpD)is
-(32)
plotted as a function
Figure 4a is obtained using equa-
tion (19), Figure 4b is obtained using equation (28), and Figure 4C is
obtained using equation (29). Note that, at low frequency range, the cur(FF)
(HpD)
is nearly uniform.
is not uniform whereas Jz
rent density Jz
2Q
I
‘Qz
kx
(Y
0=
x
10
co
o
.
x
-—
k
-N
7
0
1-
1-
~~
o
10
20
30
40
50
60
70
80
90
100
Frequency (MHz)
Figure 4a.
J(FF) as a function OF frequency at different angle @
fo; a cylinder with radius a = 0.5 m.
21
.
100
0
1
I
t
I
I
I
I
I
-1
@q-
X
)
Frequency (MHZ)
Figure 4b.
~
J(HPD)
as a function of frequency at different angle @
for a cylinder with a = 0.5 m, d = 1 m, calculated using
first approximation.
22
100
I
I
I
1
I
I
I
i
I
P
4
a
k~
H
F’
($=0
i-————d———+
v
10 —
r
~/x’-
x—
U
x’
x
/
+
<
x—
x
x/x
/
I
I
L
0.11
0
1
10
20
I
I
30
4CI
I
50
I
60
I
70
\
80
1
90
Frequency (MHz)
Figure 4c.
J(HPD) as a function of frequency at different angle @ for a
‘ f’
cy
Inder with a = 0,5 m, d = 1 m, calculated using second
approx_imationt
—
.
The reason for this is that at low frequency, the first term in equation (19) dominates (which is a constant term) whereas the first term in
equation (28) tends to be zero due to the reflected wave so that the second
term dominates (which is proportional to cos o).
In Figure 5, the function Qli(ti)is shown graphically for a cylinder
with radius a =0.5mandd=lm.
The variation is almost a factor of
three in Figure 5a and a factor of five in Figure 5b.
In Figure 6, the function Qli(u) is shown for the same cylinder at
a distance d = 5 m away from the ground plane.
For certain frequencies
(e.g., 30 MHz, 60 MHz, 90 MHz), the incident and the reflected plane wave
cancels so that a null in the graph is observed.
This makes the error
bounds difficult to estimate.
In Figure 7, an assumed relfected coefficient, REFL = -0.75, for an
imperfect ground plane is used.
For the case of a = 0.5 m, d = 1 m.
The
overall shape still has a factor of three variation.
In Figure 8, a different normalization is used before forming the
-jkpcos$ at
ratio Ri(w). The quantities Jz‘FF)($.,LO)are divided by E’oe
1 (HPD)($.,u) are divided by the collocation (a, $), and the quantities JA
l-jkpcos$-e-j2kd ~+jkpcos$,
bination of incident and reflected wave, Eo(e
5
at location (a, @i) and then the ratio Ri(LO)is formed.
The resulting
Qli(~) seems to be worse than previous results if one looks at the variations.
A last example is to consider the ratio
J(FF)(@i,LIJ)
Ri(~) =JZFF
: )(OO,W)
(33)
and calculate the function Fl(w) and Qli(u) according to equation (31)
and (32).
This is shown in Figure 9.
This result indicates that the method
applied to a cylinder in free space with two different angles of incidence
will result in an error nearly a factor of three.
24
1o~
I
I
I
I
I
I
I
.
00
x$ = 45e
•~ = 90°
[email protected]
= 135°
[email protected] = 180°
~$
1
1-
i
‘x\\/
“=’-=-
1,,,,,,,
0.1
1’0
0
2:
30
40
50
60
70
-1
,,7
80
90
100
Frequency (MHz)
Figure 5a.
The ratio ]Qi(ti)lat different angle @i calculated usin9 first
approximation and a reflection coefficient Refl = -1.0 for the
ground, and a = 0.5 m, d = 1 m.
*
I
A.
.
-2
r
t
0.1
0
10
I
20
1
30
1
40
I
50
I
60
I
70
I
80
1
90
o
Frequency (MHz)
#
Figure 5b.
The ratio \Q.(u)l calculated using second approximation and a
reflection coefficient Refl = -1.0 for the ground, a = 0.5 m,
d=lm.
26
.
0
1
.
L
\
-3
o.
——
—.—
E .——
1
I
-!
t-
o
. . . . . . . . . . . . ..
—... —
20
I
40
60
-i
I
80
100
Frequency (MHZ)
Figure 6.
The ratio lQi(U)l calculated using the first approximation and
a reflection coefficient Refl = -1,0, a = 0.5 m, d = 5 m.
27
,
d
x
3
&“
1
‘x.
-i
)
0.1
o
I
10
I
20
I
30
I
40
I
50
60
I
70
80
I
90
100
Frequency (MHz)
Figure 7a.
The ratio ]Qi(~)l at different angle @i calculated using first
approximation and a reflection coefficient Refl = -0.75 for the
ground, a = 0.5 m, d = 1 m.
2“8
—
r
I
I
—
—
0($)=00
x$= 45°
●$= 90°
[email protected]= 135°
0$= 180°
—
—
—
—
—
/-\
—
-3
—&r
—
—
r
—
o.l~
o
10
20
30
40
50
60
70
80
90
1
o
Frequency (MHz)
Fiqure 7b.
The ratio lQi(ti)[.atdifferent angle @i calculated us ng
second approximation and a ref”ection coefficient Ref’ = -0.75
for the ground, and a = 0.5 m, d=lm.
29
o
/--x
/x
\
x
t
~
L
—
-3&“
—
1
x
\
‘\x
./x
,
P’---’-’O’
0
}
o
Frequency (MHz)
Figure 8.
The ratio lQi(w)l calculated using first approximation and a
reflection coefficient Refl = -0.75 normalized to the incident
0.5 m, d = 1 m.
field at different locations, a
❑
3(3
1
1
r
1
1
t
1
10
20
30
40
50
’60
70
80
90
)
100
Frequency (MHz)
Figure 9a,
10
The function F1(u) for a cylinder in free space with a = 0.5 m.
I
I
I
I
I
~~.(y
x $ = 45°
: $ . 900
e 135Q
180°
*$=
-3
&“
I
I
I
i
z
‘Q-
P
kx
$
$.00
e:
.
()
1
—
0.1
o
I
o
10
20
I
30
I
4(I
50
6’()
;0
Jo
x—>
J
g’() 100
Frequency (MHz)
Figure 9b.
The rat-io IQ.(o)] for a cylinder with a = 0.5 m
at differentlangle $.
31
.
1.5
CONCLUSION
According to the discussion in the last section, the method of using
an average extrapolation function Fl(w) to replace the individual extrapolation function Ri(w) will result in an error of at least three to five
times the correct results for that particular example.
If the angle of
incidence is not known and the comparison is made by selecting one arbitrary angle as reference, then the result could be subjected to an error
as large as three times more,
Since the example considered here is a simple
two dimensional case, the result may or may not hold for the real airplane
or three dimensional objects.
However,”it is highly unlikely that the
error will be less if the object becomes more complex.
For example, for
a finite cylinder at resonance frequency, the currents near the center is
very much larger than those near the ends.
This brings in another error
which could be larger than the variation as a function of angle.
To esti-
mate these errors, one would have to solve the exterior problem for each
case and investigate the error bounds as illustrated in Section 1.4.
It
seems that the procedure is not general enough to be able to extrapolate
information from one case to another unless one solves a similar problem.
In other words, the specific results depend on the particular example and
the boundary conditions used.
This fact is not surprising since the prin-
ciple of electrodynamicssimilitude is not fully obeyed in this method.
32
.
SECTION 2
‘AN APPLICATION OF THE-EXTRAPOLATION FUNCTION
TECHNIQUE TO THE FINITE CYLINDER PROBLEM
2.1
INTRODUCTION
In Section 1 the motivations and the definitions of the method using
extrapolat-ionfunction in finding criteria response from simulation response
are given.
The method is then applied to a typical two-dimensional case
of an infinite conducting cylinder in the presence of a plane incident
wave with and without a ground plane.
an error of a factor of nearly 3 to 5.
The result in that study indicates
In order to obtain a better esti-
mation of the error bound concerning three dimensional objects and aircraft
using this method, we extend the study to a cylinder with finite length in
the presence of a plane incident wave in this report.
The analytical solu-
tion of a finite cylinder in the presence of an incident wave is not easily
o
available except when the cylinder is electrically thin (ref. 9).
Numerical
results have been obtained in several reports (refs. 10, 11, 12 and 13).
The problem of a finite length cylinder with a ground plane in the presence
of an incident field has not been studied extensively.
The available re-
sults are those due to finite difference technique (ref. 14).
In
the
next-section, a brief summary of the numerical solutions of both problems
using three dimensional finite difference method is given.
These results
are then used with the extrapolation function technique in the study of the
error estimation.
2.2
FORMULATION OF THE FINITE CYLINDER PROBLEM (ref. 14, 12)
Figure 10 shows the configuration of a finite cylindrical conductor
in the presence of an incident plane wave without a ground plane,
conductor is selected to be rectangular for convenience.
33
The
The origin is
incident
wave
I
I
)
AIR
Y
t
/———
c1.88—4’
n
A
-#
O.88m
1
/
.3.~25
~x
I
I
Test
Point
z
(a)
Side View of the Pipe
@=p
AIR
Y
incident
wave
(b)
Tail View of the Pipe
Figure 10. A finite cylindrical conductor in the presence
of an incident plane wave without a ground plane.
34
.
the pipe center.
An incident wave with an electric field Ei parallel to
A
the x-axis and a k-vector forming a 45° angle with both y-axis Rnd ~-axis
was selected for some practical reason.
In the calculation, the length
of the pipe is 10 m and the width is 0.88 m as shown in Figure ~Q.
The
corresponding configuration of a finite cylinder in the presence of an
incident plane wave with a ground plane is shown in Figure 11.
eters of the ground are assumed to be c
= 10 ando=
0.02,
The param.
The pipe is
located at a height d = 1 m above the g;ound in the sample calculation.
The incident plane wave is taken to be E1nc(t) = 5.94X104x[e-4,08x10%
e-3.50X108t].
To use the finite difference method to solve for the electromgignetic
fields, the Maxwell equations are expressed in a three-dimensional finite
difference form.
The scattered field caused by the pipe is found by solv-
ing these finite-difference equations subject to the boundary conditions.
The boundary conditions are (1) on the surface of the pipe, Etan
‘Catt(t)
‘
=
-E~!~(t), (2) inside the pipe, Escatt (t) = O, (3) appropriate radiation
conditions are applied on the outer boundary of-the space.
Where
The induced surface current on the-pipe is given by ~ = fix?!,
~-= fiscatt + ~inc is the total magnetic field at the desired location,
A detailed discussion of the implementation of this method is gjv?n In
other reports (refs. 14 and 2) and will not be repeated here,
It Is noted
that, using this method, the response of the finite cylinder with a qrourlc!
plane can also be determined approximately if an assumed reflection coefficient is used for the ground.
Since the response is calculated in
domain using this method, a Fourier transformation has to be used in order
to obtain the response as a function of frequency,
In the following calcu-
Iation, the reflection coefficient for the ground is assumed to be
-ti200x10-g with t in sec.
Refl = -1 + 0.25 e
35
incident wave
~
E’
x
45”
PIP
x-
Im
z
#
ground
= 10, (J= 0.02
‘r
Figure 11. A finite cylindrical conductor in the presence of an
incident plane wave with a ground plane. The dimensions
of the pipe are given in Figure 10.
36
?
2,3
APPLICATION OF THE EXTRAPOLATION FUNCTION TO THE
FINITE CYLINDER PROBLEM”
“Making use of the definitions and notations given in Section 1,
the following extrapolation function is obtained,
(34)
where ;(FF) (o.,w) is the axial current density obtained for the case with1
X
- (ATH)($i,@)
out a ground plane (Figure 10) in frequency domain, and Jx
is the axial current density obtained for the case with a ground plane
(Figure 11) in frequency domain.
An average function F1(u) is formulated
as follows,
,.
F1(u) = N~R,(w)
R2(w) R3(u) ... RN(u)
.
(35)
Then, the following ratio is calculated and plotted as a function of
frequency,
.Qli(w) =
Ri (u)
~
(36)
In all the examples given below, N = 4 and i = 1,2,3,4 corresponding to a
test point located at @i = 0°, 90°, 180° and 270° (see Figure 10b) at the
position x respectively.
In Figure 12, the axial current density Jx(FF)(OO,t) is shown as a
function of time.
F-igure13,
The corresponding Fourier transformation is shown in
A “tail” has been added to the time-domain response from
550 nsec to 1000 nsec in order to minimize the resulting error in the
Fourier transform.
Also, in Figure 12, the axial current density
J(FF)(900,t) is shown.
The corresponding Fourier transform is shown in
F;gure 13.
37
600“
I
I
I
I
I
600
800
400 —
J(FF)(t)O,t)
‘/
~
x
200 –
+.
-&
L
L
-x
T
it
~\
I
\ /n\ /q\//-e
-200—
I/bi
II
~FF)(90Q,t)
~ l-J~
\j
-4000
I
200
1
400
1000
Time (ns)
q$i,t)
Figure IZ. The axial current density Jx
at x = -3,125 m.
38
1 .OE-04
I
I I I I I II
I
I
I
I
I I I I
i
I
I
1 1 1I
J(FF)(oO,ti)
x
I
1.OE
I
J(FF)(900,Ld)
x
1. OE-06
1
1 .0E-07
1
1 .OE-08
1
A
I
I
I
I I I I I
I
I
I
I
I I 1 I
10
I
I
100
Frequency (MHz)
figure 13. The magnitude of,the Fourier transform
of J(FF-)($ijt)at x = -3.125 m.
x
39
I
I I I I I
1000
o
Note that in the case of a cylinder in free space, J(~F)(oO,t) =
(FF)(900,t) = Jx
J(FF)(2700.t), and Jx
‘FF)(1800,t) due to th; incident
f!eld selected and the symmetry configuration as shown in Figure 10.
(All the above curves are calculated at x = -3.125 m),
Next, let us consider the ratio,
(37)
and calculate the function Fl(m) and Qli(ti)according to equation (35)
and equation (36).
This equivalent to applying the extrapolation function
technique to a cylinder in free space for two plane incident wave with
incident angles differing by 90°.
Figure 14.
The function lQli(w)l is shown in
Note that Qli(u) = Qli(w) = 1 due to symmetry,
The function
Qli(u) is the inverse of Qli(w), which can be easily deduced from their
definitions.
The combination of all these curves shows a variation of a
factor near 10.
On the same graph, typical values obtained by using
Sancer’s code (ref. 11) are also shown in open and closed dots.
agree with the results of 3-D code.
They
In using Sancer’s code, the incident
wave is taken to be an impulse, and the cylinder in free space is a circular
cylinder with radius a = 0.5 m and length 10 m.
volume as that used in 3-D code.
This cylinder has the same
These two results are comparable since
the normalized quantities (i.e., the ratio of two responses) are used in
the calculation.
In Figure 15, the axial current density Jx~ATH)($.,t) is shown for
1
@i = 0°, 90°, 180°, 270° at x = -3.125 m. The corresponding Fourier
‘ATH)(Oo,t) is not equal
The current Jx
(ATH) (gO”;t) is not equal to Jx(ATH)(1800,t) beto J(ATH)(2700,t), and Jx
x
cause the presence of the ground plane disturbs the symmetry conditions.
transforms are shown in Figure 16.
40
.
t
J
0
I
I
I
I
20
40
60
80
100
Frequency (MHz)
Figure 14a. The ratio lQli(ti)lfor a cylinder in free space for two
incident waves with incident angle different by 90°.
calculated using 3-D code
---o
--*-with @,=OO,
$2=
calculated using Sancer’s code
90°, @3=1800,
41
$4=2700.
6,8—--—6,8-6,8-.,8—
1.0
E?i
/4
6,8
+
~~4—
‘A
4? AH
4
‘“/
“3
/
-1
f
I
v
0.1
1 $=0°
2 @=90”
3 @=180°
4 (j= 270”
t
1
o
20
60
40
54)=45°
6 $=135°
7 $ = 225°
8 $=315°
I
80
i
Frequency (MHz)
Figure 14b. The ratio \Qli(ti)]for a cylinder in free space for two
incident waves with incident angle different by 90°,
calculated using 3-D code.
42
I
100
1)
401
I
I
I
I
I
,—
’20(
‘2
5
Q
.
r
II
.,1
..
-.,..1
.
.
,.’
I
.
..
.
-20(
-40(
3 - 180°
4 - 270°
-60(
I
I
I
I
I
200
400
600
800
1000
T me (ns)
Figure 15. The axial current density Jx‘ATH)($.,t) at–x = -3,125 m.
1
43
.
‘“”’-o’~
-P
.
.F.
%
——
1.OE-08;
1
I I t I I II I
10
I
1-0°
2-90”
3-180°
4-270°
!
I I I 1 f II1
100
Frequency (MH.z)
Figure 16. The magnitude of the Fourier transform
of J(ATH)(@i, t) at x = -3.125m.
x
44
o
In Figure 8, the ratio
(38)
is used to calculate the function lQli(u)l according to equation (35)
and equation_{36).
It_is seen that ~he function lQli(w)l is not-a-con-
stant at low frequency end, which is similar to the results obtained for
the infinite cylinder problem in reference 8.
The total variation over
the frequency range from 1 MHz to 100 MHz is nearly a factor of 5 as
shown in Figure 17.
‘(FF=)(~.,u)l at x = O are shown,
In Figure 18, the Fourier transforms IJX
1
and in Figure 19, the Fourier transforms lJ~TH)($i,w)l at x = O are shown,
The calculated lQli(u)\ using these curves and equations (34), (35) and
(36) is shown in Figure 20.
The variation over the frequency range from
1 MHz to 100 MHz is roughly a factor of 6.
2.4
CONCLUSION
In conclusion, when the extrapolation concept is generalized and
applied to a cylinder in free space for two incident waves with 90° difference, it shows an error of about a-factor of 10.
When the method is
applied to compare the currents on the cylinder in free space and those
on the cylinder near a ground plane, it shows an error of nearly a factor
of 6.
As pointed out in the concluding remarks in reference 8, this kind
of–error is due to the difference in the boundary condjtjon and cannot be
removed effectively unless the boundary conditions are modified to be
similar to each other in the two problems.
45
..
10
I
I
i
I
-1
“..
“e
.
‘\ ~
1.0 —
3
*
v
0.l—
1 -o”
2-90°
3-180°
4-270’
o
I
20
I
40
I
60
I
80
100
Frequency (MHz)
Figure 17. The ratio lQli(W)l at different angle @i at x = -3.125
using data given in Figure 13and Figure 16.
46
m,
‘ \,
‘\
‘\
‘\
.
ID
1.OE-04
.
I
I I I I I I1
I
I I I I II
!
0
\
1 .OE-06~
L
\
I
\
\
1
.OE-07:
1-0°
2-90”
3-180°
4-270°
LOE-08,
I
I I I II
II
I
10
I
I
I
1 Ill
100
Frequency (MHZ)
Figure 18. The magnitude of the Fourier transform
~(FF)(@i,w) at x = O m.
x
47
““’-o’~
0’-05:
—
.
..
1. 0’-06 ~
—.
OE-07 T
.
.
:‘
.
~-p
2-90°
3-180*
4-270°
1. OE-08~
1
10
T00
Frequency (MHz)
Figure 19.
The magnitude of the Fourier transform
3(ATH)($i,o) at x = Om.
48
‘\
\.
L..
\,
1[
.
: “.
. .
. .
1
1.
——
T.
,r
z
—
0.
1$,=0”
i
2 lj2= 90”
3 $3= 180°
4 $4=2700
1
20
1
40
I
60
I
80
1
1
I
100
Frequency (NIHz)
Figure 20.
The ratio lQli(~)l at different angle @i at x = O m,
using data given in Figure18 and Figure 19.
49
SECTION 111
THE INFINITE CYLINDER PROBLEM IN THE PRESENCE OF
A PLANE WAVE NITH DIFFERENT INCIDENT ANGLE
3.1
SCATTERING OF AN INFINITE CONDUCTING CYLINDER FOR
AN INCIDENT WAVE WITH INCIDENT ANGLE (3
Considering Figure 21, an infinite cylindrical conductor is illumi-
nated by a plane incident wave with incident angle B as shown.
Let the
incident wave have its electric field component.Ez parallel to the axis
of the cylinder (z-axis),
Mathematically, the incident field can be
written as
E:= E. e-jkP Cos
(6+$)
e+jut
.
(39)
The scattered field E; due to the cylinder can be determined by expanding
the scattered and incident fields in terms of cylindrical wave functions
total
=E:+E:=
and using the boundary condition E7
O at p = a. The rei.
L
L
suit is (with e‘JWt suppressed)
co
E; = -E.
E
Jn(ka)
‘-n H~2)(kp) e
J
n=m=
-jn([email protected])
(40)
The total magnetic field is then,
~total =- 1
JUP
+
~Etotal
z
ap
.
(41)
.
(42)
The current density on the cylinder is given by
Jz=H~otal
,atp=a
50
\
\
G
a
\\
—
x
,
Figure 21. An infinite cylindrical conductor in the presence of an
incident plane wave with incident angle ~.
51
.
Combining (39), (40), (41), and (42), it can be shown that
(43)
3.2
AN INFINITE CONDUCTING CYLINDER WITH A PERFECT CONDUCTING
GROUND PLANE IN THE PRESENCE OF AN INCIDENT PLANE WAVE
WITH INCIDENT ANGLE 6
Considering Figure 22, an infinite cylindrical conductor with a
perfect conducting plane is illuminated by a plane incident wave with
incident angle (3.
Let R, be the reflection coefficient (R, = -1 for perfect conductor),
and let
02
E: = E.
x
n=-cu
J“-n Jn(kp) e
-jn(&@)
(44)
m
~~
=
I+
~-jzk~ Cos 6
E.
z
-!-jn([email protected])
J‘*n Jn(kp) e
(45)
n=-m
co
E; = E. ~
an j-n H~2)(kP) [email protected]
n.-ca
SkEz
an j+n~
H~2)(2kd) Jn+m(kp)
= ‘1 ‘o ~
=.m
=-m
m
n
(46)
{.jn$-jm$
(47}
.
52
‘z
R
Hi
Y
B
x
k
\
“’\
\
A
d
1I
/ //~////
x
d
\‘
1
Uimage
Figure .22. An infinite cylindrical conductor .with a ground plane in the
presence of an incident plane wave with incident angle @.
53
.
=E:+E; +~; +E;r
~total
z
where E~tal
(48)
.
is the total electric field in the space, E; is the plane
incident wave expanded in cylindrical wave functions, E; “s the reflected
plane wave due to the ground expanded in cylindrical wave functions, E;
and E~r are scattered fields due to the cylinder and its
It can be easily shown that, on the ground plane
unknown constant a..
(x=
mage with the
d), E;Otal = ; using (44), (45), (46), (47) and (48) and the addi-
tional theorem of Bessel’s functions.
The constant an can be determined by the following boundary condition,
Etotal = ~,
z
(49)
atp=a
From (44) through (48) it is found that
w
x
-1-
an j-
[email protected]
nH(2)(ka)
e
n
m
+ RI
z
.+n
J
an
x
~
~(2)(2kd) Jn+m
m
(ka) e-jn$-jm$ = O S(50)
Rearranging the last summation and collecting terms associated with [email protected],
we find the following equat”on for determining an,
54
.Jnn a
J-n(ka)
H~2)(ka) + R, j-n a-n H(2)(2kd)
~
a
x
+–
RI
j-(n+m)a
-(n+m)H(2)(2kd)J-n(ka)
m
m= 1
m
.-x
R, jm-nam-n H(2)
m (2kd)(-l)m-nJn(ka)
+
=—
,-
-jn~ + R, e-j2kd Cos B j+ne+Jn6) Jn(ka)
Jne
.
(5”)
For simplicity, we consider an approximation with H(2)(2kd) Jn(ka) terms
m
neglected in (51). This is equivalent to neglect the effects due
image cylinder.
\
When this is done, one finds
(
j-ne-jn~
+
R
)
-j2kd cos B j+ne+jnflJn(ka)
le
,J n H~2)(ka)
(52)
Using (52), (44 ) through (48), (41) and (42), the current density is given
by
w
(j-n”e-jri(~-$) +
R,
e -j2kd
Jz &--
(53)
n.-w
3.3
cos 6 j+ne+jtl(6+$))
H:* ‘(ka)
CONSIDER A SPECIAL CASE
Assuming that the infinite cylinder is an ideal version of certain
...
“two-dirnenijonala~rplane” so_that, in Figure 21, the airplane is in the
.—
!55
.
free space and in Figure 22 the airplane is near a ground plane.
.
When
the “airplane” is illuminated by an incident EMP wave, the electromagnetic energy can be coupled into the “airplane” body through the possible
This coupling energy is related to the current density Jz
port of entry,
through some transfer functions.
We shall consider a case where the inci-
dent wave can come from any angle Bwith
-m/2 [email protected]~m/2.
Let J~FF)(Oj,Bk) be the current density given in (43) and J~HpD)(@j,Bk)
be that given in (53), where $j is the location of the current density on
Let a
the cylinder (with radius a), and Bk is the angle of incidence.
maximum number be defined from a set of k numbers as follows:
Jz($j)max = max Jz(@j>61)*
{
Jz(@j~B2), .... Jz($j38k)
(54)
.
}
At a particular angle @j, we only select one current density,
Jz(@j)nlax~ due
to certain incident field among all the possible incident
waves, and assuming this quantity will dominate the coupling effect.
j=l,
For
.... m, we obtained a set of numbers which contains the maximum
current density at each particular location when b,,is varied.
K
Thus, from
J(HpD)($j,Bk) we have the following set
z
A=
J$HpD)($l)max,
J(HPD)
$2)max,
z
.... J:HpD)(@mmax
{
}
.
(55)
Note that 13kcan only be varied in -m/2 ~ 13k~m/2,
(@j)max
Similarly, we can construct a set of numbers Jz(FF)
J(FF)(@j,Bk).
from
In order to make a fiar comparison, we restrict Bk in this
c;se also between -Tr/2to +m/2.
From the result of Dart one in this sec-
tion, it is not hard to see that the maximum current density for an infinite
cylinder illuminated by a plane wave occurs at the illuminated side.
For
example, when the wave is coming from @j = 180° with 13k= 0°, the maximum
current density is that located at @j = 180°.
56
Since we allow the incident
o
.
10.0
I
I
I
3.0 –
2.0
– 3.0
2.0
\37
t
~8,2
—
‘3
.r
o,3p
—
1
0.1
0
0.’1
F
1
1$=0”
2+=45°
3$=90”
4 + = 135°
—
o
20
5
6
7
8
$
$
@
$
=
=
=
=
180°
225°
270°
315°
1
I
40
I
I
60
80
I
100
Frequency (MHz)
--
Figure 23. The ratio IQ1.(u)I at different angle @i with a = 0.5 m,
d = 1 m, RI =l-l.O and @ = -m/4, O, m/4.
57
10.0-
I
i
3.0 –
– 3.0
2.0
– 2.0
‘/
8’2~82_
*
1.0
0.9 –-
0.3–
–-0.3
0,2–
– 0.2
—
0.1—
l$=OG
2+=45”
3$=90°
4 @ = 135°
0
20
60
40
0.1
5$=180°–
6 @ = 225°
7$=270”_
8 $ = 315°
80
100
Frequency (MHz)
Figure 24. The ratio IQ1.(u)I at different angle $i with a = 0.5 m,
d = 1 m, R] =T-0.75 and @ = -n/4, O, m/4.
58
.
0
fields to be varied from Bk = -Tr/2to +Tr/2,the maximum current density
will be a constant equal to Jz‘FF)(180”,00) for n/2 ~ @,i~ 3Tr/2. The
current density in the shadow side can also be determined easily,
Thus,
the following set can be obtained from Jz‘FF)(@j>fi~) for.~ = 1, 2, ,.., m
‘-”-”
\
B =- J~FF)($l)max, J:FF)(02)max, .... J$FF)($ )
m max I
q
(56)
Using the numbers in set A and B, one is ready to apply the method
of-extrapolation function to obtain an error estimation.
3.4
APPLICATION OF THE EXTRAPOLATION FUNCTION TECHNIQUE
TO THE ABOVE PROBLEM
From (55) and (56), we construct the following extrapolation function
J(FF)($i)max
Ri(~) = JZATH
‘($i)max
z
(57;
Fror (57) we obtain an average function Fl(u) as follows:
F1(cIJ)=
~R,(d R2(LII) . . .
RN(u)
(58;1
Then the fcllowing ratio Qli(LU) is calculated
Qli(u)
=
R.(w)
~
(59)
In the examples given in Figures 23 and 24 only three angles of incidence are used, i.e., Pk = Tr/4,o, -IT/4,in calculating Jz(ATH)(~i)max.
the other hand, incident waves with Bk which varied from -Tr/2to +7r/2,
are used to calculate Jz‘FF)(Oi)max.
The function lQli(u)l is plotted as
59
On
a function fo frequency in Figure 23 for a cylinder with d = 1 in,a = 0.5 m,
R=-l.
The variation is only a factor near 2.
In Figure 24 the same calculation is applied to the cylinder, however,
with Rl = -0.75 for an assumed imperfect ground reflection.
is a factor of 2.5 for eight positions.
The variation
This is a better error bound as
compared to those obtained for a single incident wave.
However, these
conclusions are obtained under the assumption that the maximum current
density can be used in estimating the error.
60
o
REFERENCES
1,
2.
Stratton, J. A., Electromagnetic Theory, McGraw-Hill:
Chapters VI and IX.
New York, 1941,
Baum, C. E;, “Extrapolation Techniques for Interpreting the_Results of
Tests in EMP Simulators in Terms of EMP Criteria,” EMP Sensor and Simulation Note 222, AFWL, 20 March 1977.
3.
Merewether, D. E., J. F. Prewitt and C. E. Baum, “Characterization
of–Errors in the Extrapolation of Data fron an EMP Simulator to an
EI!PCriterion,” EMP Sensor and Simulation Note 232, AFWL, 25 October
1977.
4.
Lee, K. M., R. Holland and K. Kunz, Calculation of the Transient Currents
Induced on an Aircraft by 3-D Finite Difference Method, Mission Researcl~
Corporation, AMRC-N-45, October 1976, presented at APS-URSI Symposium,
Stanford University, CA, June 1977.
5.
Taylor, C. D., K. T. Chen and l-.T. Crow, An Improvement on Wire
Modeling for Determining the EMP Interaction with Aircraft, Air Force
Weapons Laboratory, AFWL-TR-74-217, May 1976.
6.
Barrington, R. F., Time-Harmonic Electromagnetic Fields.
New York, 1961, pp. 232-238.
7.
King, R. W P. and T, T. Wu, The Scattering and Diffraction of Waves.
Harvard Un versity Press: Cambridge, Mass., 1959, Chapters 2 and 4.
8.
Row, R. V. “Theoretical and Experimental Study of Electromagnetic
Scattering by Two Identical Conducting Cylinders”, ~. Appl. Phys.,
~, 6, pp. 666-675, June 1955.
9.
King, R. W. P ., The Theory of Linear Antenna: Harvard University
Press. Cambridge, Massachusetts, 1956 (Chapter IV).
McGraw-Hill:
10.
Kao, C. C., “Electromagnetic Scattering from a Finite Tubular
Cylinder: Numerical Solutions”, Radio Science, 5, 3, pp. 617-624
March 1970.
11.
Sancer, M. I. and A. D. Varvatsis (Northrop Corporate Labs), ’’Calculation
of the Induced Surface Current Density on a Perfectly Conducting Body
of Revolution”, EMP Interaction Note 101, Air Force Weapons Laboratory,
April 1972.
12.
Merewether, D. E.,“Transient Currents Induced on a Metallic Body of
Revolution by an Electromagnetic Pulse”, IEEE Trans. EMC, EMC-13, 2,
pp. 41-44, May 1971.
61
REFERENCES (Continued)
13.
Perala R. A.,[’Integral Equation Solution for Induced Surface Currents
IEEE Trans. —EMC, EMC-16, 3, pp. 172-177,
on Bodies of Revolution”, .—
August 1974,
14.
Holland, R., TtlREDE: A Free-Field EMP Coupling and Scattering Code,
AklRC-R-85,Mission Research Corporation, September 1976.
62
.
@
APPENDIX A
SUPPLEMENTAL RESU~TS FOB THE_
PROBLEM IN SECTIONS 1.3 AND 1.4
In this appendix, additional data are presented which supplement
those presented in Sections 1.3 and 1.4.
In Figure A-1, the ratios lQ,(u)\ at the angle $ = 0°, refl = -1.0,
a = 0.5 m, d = 5 m are compared using numerical solution of equat-ion (25-)
with N = 7 and two approximations.
(The results using N = 7 differ less
than 0.1% as compared to those using N = 6.)
In Figure A-2, the ratios IQ,(u)] at the angle $ = 0°, relf = -1.0,
a = 0.5 m and d = 5 m are compared using numerical solution of equation
(25) with N = 4 and two approximations.
in previous approximations disappear.
Note that the deep nulls shown
(The results using N = 4 differ
less than 0.1% as compared to those using N = 3.)
In Figure A-3, the ratios \Qi(~)lat different angles ~i are calculated
using numerical solution of equation (25) with N = 7, refl = -1.0,
a = 0.5 m and..d= 1 _rn_._-.Comparing
with figure 5 of section 1.4, this shows
an error bound smaller than previous results.
-—=—
—
In Figure A-4, the ratios lQi(m)l at different angles @i are calculated
using numerical solution of equation (25) with N = 4, refl = -1.0, a = 0,5”m
andd=5m.
This also shows asrnaller error bound than previous results.
The deep nulls and peaks shown in Figure 6 of Section 1.4 disappear here.
In Figure A-5, the ratios lQi(~)\ at different a“ngles$i are calculated
using numer!cal solution of equation (25-)with N = 7 and refl = -0.75
for the ground, a = 0.5 rn,d = l_rn. Comparing with Figure 7 of Section l.~,
this also shows an error bound smaller than previous results.
63
.
Frequency (MHz)
Figure A-1.
The ratio lQ~(w)l at the angle @ = 0°, reflection
coefficient Refl = -l.O, a=0.5m,
d=lm.
—using
—“—using
–––using
numerical solution of Eq. (25) with N = 7
first approximate solution of Eq. (25)
second approximate solution of Eq. (25)
(cf. Figure 5, Section 1.4)
64
1=
=1
L
--4
--
G’
—
-J
‘$’$” “
(
‘1
1
\,
Ii
o
20
60
40
80
100
Frequency (MHZ)
Figure A-2.
The ratio IQ (u)I at the angle @ = 0°, reflection
coefficient Aefl = -1.0, a = 0.5 m, and d = 5 m.
using numerical solut-ionof Eq. (25) with N = 4
—“—using
first approximate solution of Eq. (25) see Eq. (26))
— ––using
second approximate solution of Eq, (25) (see Eq. (27))
(cf. Figure 6, Section 1.4)
65
.
0_
I
I
1
.—
-N
-.—.
\_
—_
_
------
./
--.~
—--—
=--~
%4:>””
P
-.
-.
-
-—.
-—
——
.3
F
r
.1
o
-i
I
20
I
40
60
80
100
Frequency (MHZ)
1$1=00
‘[email protected] ‘ 45°
“3 $13= 90°
4 44 = 135°
5 $5 = 180°
Figure A-3.
The ratio IQ.(w)I at different angle @i calculated
using numeri~al solution of Eq. (25) with N = 7 and
a reflection coefficient Refl = -1,0 for the ground,
and a = 0.5 m, d = 1 m (cf, Figure 5, Section 1.4).
66
0
o
—
-3
&“
—
.11
[
o
20
I
40
I
60
I
80
I
100
Frequency (MHz)
1 +, = 0°
2 $2 = 45°
3 $3 = 90°
404=
135°
5 45 = 180°
Figure A-4.
The ratio Qi(~)l calculated using numerical solution
of Eq. (25! with N = 4 and a reflection coefficient of
Refl = -1,0 for the ground, a = 0.5 m, d = 5m (cf., Fig–ure 6, Section 1.4).
67
10 _
I
i-
--i
—
‘3
//
&“
—
——-.
.—
.,~
o
20
_’~=_–_~_y
40
60
—
80
100
Frequency (MHz)
o
h$l=o~
2
o~ =
45°
3 1$3= 90”
4 $4 = 135°
5 (j5= 180°
Figure A-5.
The ratio lQi(~)~ at different angle $j calculated using
numerical solution of Eq. (25) with N = 7 and a reflection
coefficient Refl = -0.75 for the ground, a = 0.5 m, d = 1 m.
(cf. Figure 7, Section 1.4).
68
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