SSN 185, D. L. Wright and J. F. Prewitt, Transmission Line Model of Radiating Dipole with Special Form of Impedance Loading, Sep 73, Dikewood Corp.

SSN 185, D. L. Wright and J. F. Prewitt, Transmission Line Model of Radiating Dipole with Special Form of Impedance Loading, Sep 73, Dikewood Corp.
.
Sensor
and Simulation
Note
185
September
Transmission
with Special
Notes
1973
Line Model of Radiating
Dipole
Form
of Impedance
Loading
David L. Wright
James
F. Prewitt
The Dikewood
Corporation
Albuquerque,
New Mexico
87106
\
Abstract
This
pattern
note
from
a long,
approximation
where
considers
thin,
when
driven
u(t) is the unit
edance
which
the far-field
is taken
step
cylindrical
h is the antenna
tance
measured
mate
antenna
biconical
lytical
are
along
antema,
and numerical
discussed.
inductive
and
loading
transmission
line
of waveform
antenna
is loaded
Vou(t)
with an imp-
2ZC06(U)
h-Y
~ is the
from
impedance
6(LJ) is a complex
In particular,
are
=
antenna
solutions
in the
and field
the form
half-length,
characteristic.
waveform
generator
The
function.
to have
the
antenna
by a voltage
z’
where
radiated
the
effects
center,
based
field
value
of the dis -
ZCUis an approxi-
on that
function
for the far
the
absolute
of a long,
thin
of frequency.
pattern
of resistive
Ana-
and waveform
with parallel
discussed.
Q~
—
>,
.
%“
.-l ,
..\_
/3. ,-!/,
Sensor
and Simulation
Note
185
September
Transmission
with Special
Notes
1973
Line Model of Radiating
Dipole
Form of Impedance
Loading
David L. Wright
James
F. ~rewitt
The Di&ewood Corporation
Albuquerque,
New Mexico 87106
Abstract
This
pattern
note
from
a long,
approximation
where
considers
the far-field
thin,
when driven
u(t) is the unit step
pedance
which
is taken
cylindrical
h is the antenna
tance
measured
mate
antenna
biconical
lytical
are
along
discussed.
inductive
loading
=
half-length,
in the transmission
generator
of waveform
is loaded
line
Vou(t)
with an im-
2ZCY36(L9)
h-Y
K is the absolute
the antenna
from
impedance
solutions
In particular,
are
and field
to have the form
and 6(w) is a complex
and numerical
waveform
The antenna
function.
characteristic
antenna,
antenna
by a voltage
z!
where
radiated
discussed.
the center,
based
the effects
of the dis-
Zcu is an approxi-
on that
function
for the far field
value
of a long,
of frequency.
pattern
of resistive
thin
Ana-
and waveform
with parallel
●
Acknowledgment
The
suggestions
authors
would
and interest
like
to thank
in this
work.
2
Dr.
Carl
E.
Baum
for
his
helpful
I.
One of the problems
selecting
loading
waveform
of pulse
on the antenna
in terms
of some
desired
resistor
the far
zone
pairs
approximates
some
Another
continuously
dipoles
example
Baum
a special
This
form
main
broadband
ance
of this
3.
line
first
of resistive
property
measured
of this
The present
study
functional
is no longer
fixed
is permitted
corresponding
values.
Normalized
given
in both frequency
angle
with respect
type
form
length
case
loading
the center
(driving
considered
of frequency
in pulse
which
loaded
of the antenna
previously
an extension
do-
for the reappearstudies
permits
is the
pulse
variation
domain
to the antenna.
3
of the work
is retained,
Variation
distributions
and time
the currents
smoothing
simulation.
of loading
value.
radiating
special
The reason
of loading
to physical
current
field
resistively
the standpoint
loading
is largely
pulse
from
properties.
waveform
at a single
so that
who considers
This
has been
of resistive
in EMP
antenna
for finding
a uniformly
loading
and directional
form
for selecting
symmetric
approximation
considers
desirable
The same
and lengthwise
on the antenna
special
may be
zone magnetic
of C. E. Baurn3
and Shen and WU5 from
by Wu and King4
of a kind
or near
case
and z is the position
broadbanding
pulse
the radiated
of D. E. Mere-
a cylindrical
of non-uniform
loading.
-1
where h is the half
to (h - IzI)
is proportional
point).
is developed
load
is the work
axially
is the work
of
waveshape.
a transmission
on the antenna.
and then
prescribed
loaded
using
procedure
field
is that
This problem
waveshape.
to symmetrically
electromagnetic
antennas
a way as to optimize
One example
in a number
of ways.
1, 2
Wether
in which a synthesis
dipole
radiating
in such
attacked
lumped
Introduction
over
but the proportionality
real
over
and radiated
as a function
of reference
and complex
resistive
electric
of loading
values
and reactive
fields
are
and observer
.
.
-Y
11.
The
our
analysis
Formulation
antenna
and Solution
geometry
we require
is shown
the antenna
of the Current
in figure
current.
to the current
has
we employ
a transmission
3, 4
that in this zero-order
shown
been
imation
the wave
written
as
equation
a 27’
—.
2
ar
for
the
line
1.
Equation
As the first
TO obtain
an approximation
model
the antenna.
for
or transmission
current
on a thin
702 +sciz~
dipole
line
antenna
r=o
r~m
I
2a
I
I
“z’
z!=h
Zr=o
\
Figure
where
s is the Laplace
1.
Geometry
variable,
7.=:
4
in
It
approxmay
be
(1)
)
(
step
of the
Problem
.
●
1
c’ =
z!
Cz
equivalent
Z.
—
?r
z=
m
and the tilde
2h
7
()
line
h>>a
loading
point
of free
we are
that
per
unit length
2Z
where
6 is in general
Ofr.
Substituting
space
in the Laplace
or frequency
(3)
h-;
a complex
function
of frequency,
—
for C‘ and Z‘ in equation
$(,:+
the differential
equation
for ?
——
~’),=o
for the current
T(s,
but is independent
—
—-
1, the wave
—— —.
and equation
domain.
6
becomes
solution
(2)
take
z’=—
If a trial
—.
..
unit length
characteristic
impedance
of
long, thin biconical
antenna
exclusive
of loading
impedance
indicates
At this
per
length
#o
~
o
r
capacitance
radius
2h = antenna
Zo=
transmission
In
antenna
a=
line
m
equivalent
=
transmission
t)
5 substituted
=(h-~)e
into
(4)
is written
in the form
—.
.
-;.!
@(s, K)
equation
equation
5
4 we find
that
(5)
Q(s, E ) satisfies
●
✎
(6)
Making
a change
of variable
y =
-2yo(h
- ~ ),
(7)
we obtain
2
~
Y
8 y“
This
for
+ (2-y)
is the Kummer
example,
where
=+(6ay
or confluent
by equation
This
differential
@(y)
=CIM
M and U are
Thus,
l)@=o.
equation
of Abramowitz
as ‘given,
and Stegun. 6
has
of the form
- 6,
2;y)
(9)
hypergeornetric
confluent
solution
equation
a solution
2;y)+C2U(l
Kummer’s
a general
hypergeometric
13. 1.1
(l-6,
(8)
for the
current
functions.
is
.——
__
-7’.3
~(s,~
) = (h-~)
e
CIM
![
1-6,
2; -2yo(h-~
) + C2U
1-6, 2;-2~o(h-r
1[
—
Requiring
O, we must
the
set
current
to vanish
at the
ends
C2 = O since
U(a, 2;2) --0
()$
as z +0
and
6
.
II
. ..—,
of the antenna,
)
so that
(lo)
? (s, h) =
_
.
,
M(a,2;
O) = 1.
*
Therefore
-Yot
~(s,
CIM
t)=(h-f)e
1-
(11)
6, 2;-2~o(h-0
1
[
Writing
Cl in terms
of the current
at the center
of the antenna
-T(S, o)
c1
hM(l-6,
=
(12)
2;-2TOM
so that
M
-Tot
= (1-f /h)
T(s, t)
r(s,
e
o)
1-6, 2; -2 To(h-t )
1
[
M(l-6, 2; -2~oh)
(13)
—
To evaluate
transmission
equations
line
we may
the driving
point
current
From
approximation.
we continue
to use the
one of the transmission
line
write
(14)
Substituting
the expression
~o(h-?
T(s,
t)
=
[
)+1
for ~ (s, c ) of equation
M
-2*/o(h-f
1[
sC’h
1
) -2~o(h-Y
M(-2yoh)
13 into equation
[
) M’ -2~o(h-?
1
14
)
--(. ?
T(s,
O)e
(15)
●
The
voltage
at the
center
of the antenna
(~oh+l)
7(s,0)=
Here
for
.
.-\
be written
M (-2~oh)-2yohM’(
-&
simplicity
can then
-2~oh)
r(s,
M(-2~oh)
we have
written
M(a, b;z)
0)
(16)
?)M(a, b;z}
~~
as JM(z) and
as M’(z)
Then
the
antenna
Za =
0(s, o)
y(s, 0)
impedance
in the transmission
line
approximation
is
1
‘
(~oh+l)
M(-2yoh)
m
+yoh
M’(-270h)
M(-2~oh)
(17)
= &
(~oh+l)
- 270h
[
M’(-27
h)
M ~-2T0h)
o 1
Defining
M’(-2-yoh)
f6=2
(18)
M(-2~oh)
and remembering
that
-yO = ~
and Ct = --&
, we have
finally
CQ
(19)
where
s
sh
hs~-
8
.
*
Thus
equation
13 may
be written
in terms
of ~ (s, O) as
-?ot’
Y(s, 0)(1-~
/h) e
M
[
Y(s, r ) =
z
Equation
antenna
20
is then
for an arbitrary
approximation
~+l.
w [ Sh
the solution
driving
and with no special
f6
1[
1-6, 2;-2@I-~)
M 1-6, 2;-2~oh
1
(20)
1
for the frequency
domain current
function
V(s) in the transmission
treatment
for the feedpoint
on the
line
geometry.
HT.
Equation
an arbitrary
the
Current
20 gives
driving
antenna
Solution
for
the solution
Step
Input
for the freauency
At this
function.
Function
point
.
domain
we specialize
to the
current
case
for
.
o
where
is driven
by an idealized
capacitive
pulse generator
of capaciV.
1
and voltage
output Y’(s) = ~
.
impedance
Z = —
tance C
Scg ‘
g
#
Schematically,
the situation
is shown in Figure
2, Note that the generator
.
.,
z
g =-J-Scg
z
a
I
Figure
Schematic
impedance
~g(s)
appears
is not identical
we may
Representation
Driving
in series
to V(S,
of Capacitive
Antenna
with
O).
that
Generator
of the antenna
From
the circuit
and that
of Figure
therefore
2, however,
write’
r’(s,
o) =
v
+
-1
Zg+z
[1
Substituting
2
for
Za from
equation
19’ into
10
(21)
a
equation
21 we obtain
.
-1
v
T’(s,
;
o)=
*
+Z
+-
C
= C’h,
a
Define a parameter
co
a
~
[
where
+
c
a ~ 1 + &
.
- Zm f6
1
liquation
(22)
22 becomes
g
T(S,
Substituting
equation
v
+
r(y)=
23 into equation
[sh (l~f6)+
Figures
figure
3 through
Figures
It may be observed
are
true
values
for other
from
fore
is to be made
first
calculating
from
current
NI(I-6,
u ❑ck,
relative
24 with the results
1
along
the loading,
phase,
arg[I(z’
to be unity.
in figures
normalized
that T(O) is a function
(24)
the antenna.
both the magnitude
The curves
)
2;-2yoh)
and treats
if 6 is taken
of loading.
1-6, 2;-2~/o(h-?
distribution
frequency,
functions
however,
no comparison
e
for any frequency
linear
equation
be remembered,
i
3 and 4 include
that
of the current
1-
9 show
is for a discrete
as a parameter.
obtained
+ [
[1
CCZ]
(23)
13, we obtain
M
co
Each
W:+l-’J1
o) =
) /1(0)] .
and phase
This
is not
3 through
9 are
to T(O).
It should
of frequency.
There-
of absolute
values
of currents
at different
frequencies
the figures.
This
information
is available
only by
X 6, O) for some
particular
11
s=@ by using
equation
6,
23.
.
.
2.0
.
1(Z ‘)
m
1.0
0
0
1. 0
0.5
z’fh
3.
Figure
Normalized
kh=rand&=.2,
Current
Distribution
.4,1.
O
,—..
9
for
27r
1.5
%’
cd
1, 0
T
1(2’)”’
m
I
0.5
0
1.0
0
1:’”~hl
Figure
4.
Normalized
kh=2r,6=.2,
Current
Distribution
.4,1.
O
12
for
5
.
1.5
1.0
I(zr)
m
0, 5
I
0
I
I
!
!
t
I
Figure
5.
Magnitude
of Normalized
for kh = 5~and6=,2,1.
I
1
1 .0
.5
0
1
Current
O
Distribution
1.5
1.0
I(zf)
m
o, 5
0
Figure
1.0
.5
z’/h
0
6.
Magnitude
of Normalized
for kh = 10~and6=.2,1.
13
Current
O
Distribution
.
..
1.5
1.0
Hz’)
m0.5
I-1
1. 0
.5
~zffh~
“o
Figure
7.
-.-— .—..
I
1
-
1.5
Magnitude
of Current
kh=15rand6=.2,1.
--
b
I
I
Distribution
O
I
I
for
—.—.
I
I
i. O
Hz’)
1(0)
I
0.5
0
0
Figure
‘
8.
Magnitude
forkh=20r
.l,u
iz~jhl
of Normalized
and6=.2,1.
Current
O
Distribution
.
—
1.5
-,
.-
1.0
I(z)
m
0.5
0
Figure
1.0
.5
\z’/h/
0
9.
Magnitude
of Normalized
for kh = 25~and6=.2,1.
15
Current
O
Distribution
IV.
Using
on the
is then
a thin-wire
z’ axis.
Following
calculated
Radiated
Field
we assume
approximation,
reference
3 the normalized
that
I is concentrated
radiated
waveform
as
h
P
<(e)’fi
+
o
1
-h
~oz’cos(e)
~(z’)
dz I
e
(25)
where
t*
is the
retarded
time
given
by
(26)
The
normalized
electric
field
waveform
(only
Ef
in equation
a o component)
25 is related
to the far
or radiated
by
e
(2’7)
In addition
16
.
z
where
fg~~,
considered
at the center
beginning
ths
z*
u~ing
~.
retarded
Note
wave
a waveform
that
initiated
? is
at t = O
at the observer
[’-w
-h
—— —. -
transform
the time
inversions
have
carried
of frequency
been
Curves
10 through
value,
but in general,
result.
obtained.
for various
loading
25,
.:
-—
of the far field
domain
out.
In figures
of equations
‘:(-’
z’’+coso)dz
’28)
[“2~o(h-’z’”l
..—
28 yields
K = ] z‘ I into the first
27 we obtain
1
The inverse
maximum
~.
a current
will produce
.—
for ~(! ) with
equations
x
resistive
so that
et-r
Th s
at t* = O.
and using
in figures
define
time
of the antenna
Substituting
were
Also
frequency
For
For
other
of magnitude
values
certain
domain
cases
cases
analytic
numerical
and phase
of observer
result
angle
of 7($
in equation
closed
form
inversions
) as a function
o and loading
6 are
given
15.
10 through
15 the patterns
as a parameter,
so that
the peak
All patterns
directionality
value
are
of E
8
for fixed
are
normalized
as a function
of loading
is not the same
of 6.
17
frequencies
for various
with
to the
is indicated,
values
.
.
–=%
—..
1.0
E(6)
E
max
0.5
0
45
30
15
0
60
75
90
e
Figure
10.
1.0
NormaEzed
Radiation
6 =.2,.4,.6,.8,1.0
I
Pattsrn
at kh = 27r with
e
—
E(0)
E
max
0.5
f-l
r
“o
I
I
I
I
I
15
30
45
e
60
73
J
gfJ
o
Figure
11.
Normalized
6=.2,
1.0
Radiation
18
Pattern
for
kh = 57 and
1.0
E(9)
E
max
0.5
0
0
Figure
15
12.
30
Normalized
and 6=.2,1,0
45
e
Radiation
60
Pattern
75
90
for kh = 10~
1,0
E(9)
E
max
0.5
“o
15
30
60
45
75
Q
Figure
13,
Normalized
Radiation
and 6 = .2,1.0
.19
Pattern
for kh = 157
Qo
.
-\
1,0
0.5
0
0
Figure
15
14.
30
Normalized
and 6=.2,
60
45
Radiation
1.0
Pattern
75
90
for kh = 20r
1.0
E(0)
E
max
0.5
0
15
0
Figure
15.
30
Normalized
andtj =.2,
Radiation
1.0
20
60
45
e
Pattern
75
for kh = 25:
90
.
.
.
V.
Analytic
Consider
Studies
the integrandof
function
by its series
6
and Stegun.
M
definition
=
1[
-2~o(h-
‘x
n=O
~
M
I 2’1)
[
co
of Frequency
equation
as given
1-6,2;
4,(s,6)
We re~lace
by equation
Field
the Kummer
13. 1.2 in Abramowitz
1
-270( h-l
z’1 )
= M(a,
b;z)
(29)
(b)nn !
(a)n ~ a (a+l)(a+2)
equation
28.
Radiated
(a)nzn
with a = 1 - 6, b = 2, z = -270( h-l
Using
Domain
29,
000 (a+n-1),
the integral
= ~
z’!),
part
and with
(a).
(30)
z 1
of eauation
28 may
be written
as
[1-~] ‘$ “-~~(-~~o(h-’
“!)n
e:(-~
‘“‘Z’cose)d
n
n=O
(31)
Interchanging
the order
of integration
to [0, h]
gives
(l-d)n
(-270)n
co
4.6(s,0)
=
I
n=O
of integration
and summation
and changing
the range
us
(2)n n! h
h
(h-z f)n+le
I
o
-Szl
c
(
Sz I
- —cOse
e
c
Szt
—Cose
+e c
cc
=
I
n=O
(32)
x (s) ‘=& (S’o)
n
21
) dz ‘
.
..
where
~~ represents
the factor
only
in iron-t
of the
the integral
part
of equation
If we change
integral.
“32 and x represents
variables
in ~’ (s, 6) by
LL
letting
v = ~-zl
P=
: ( l+COSO)
(33)
q = ; (l-coSe)
Then
~n~ (s, 6) becomes
h
Zn’(s,
(3) =
1 J’+l (
e
p(v-h)+eq(v-h))
~V
( 34)
o
#n’ may
then
be expressed
as the
sum
a
J
xnebxdx
=
0
Thus
if we watch
our
(-l)n+ln!
n+ 1
b
p’s and q’s,
+
e
T
of two integrals
ab
equation
22
each
of the form
n
n ! an-j
2
j=O
‘-1)3
32 becomes
(~-j)!bj
(35)
(1-6) n(-2yo)n
(2)n n!
h
n =0
.—
x
— .
H
d’h
—.—=
n+ 1
(-l)n::p:+2 ~
P
P
.. .
(-J
(n+l-j)!
j =0
_-
—
‘n+l)!‘n+l-J
pj
_
n+ 1
+ e-qh
(-l)n:::+l)!
-—.
-.
—+.
,,_
with 46(s,
13)given
and using
the relation
we
. . _ ——..—.
then be written
by equation
=
~M (a+l,
)11
(n+l)~hn+l-J
(n+l-j)!
qj
-.—
28 may
Z)
(.-l)j
j=O
-- -.
Equation
M’(a, b;
~
q
~
(
.—.
+s
(36)
)
-—
as
36.
b+l;
Substituting
for fa from
equation
18
z)
have
F(e) = *
S+*9)
(ca+sh)
M(l-d,
2; -2 yoh) - sh(l-6)
23
M(2-6,
3; -2yoh)
(39)
._.
.
.
—....— .. ..—
for the
and substituting
~(e)
—
M’s gives
.-
—.—
——.
.-
___
... _
us
S4JS,0)
sinfl
. =-j=
03
(
(1-@n(-2yoh)n
I
(ca+sh)
~
n=O
m
- sh(l-ti)
(2)n n !
(2-6) n(-2yoh)n
I
n=O
(3)n
n!
]
1
(40)
Recalling
aside
that
from
denominator
true
for
functional
dation
yo= s/c,
factors
are
arbitrary
form
in the time
p= s/c
(l+cos~),
of the form
made
ex P ~ ‘Sh( 1*C6-S o)]
up of infinite
continuous
of course).
q = s/c(l-cos6),
series
passive
The
exponential
domain.—.
both
in powers
element
we see
‘merator
of s.
This
(with
the
loading
correspond
to time
that
and
remains
~ h-~z’11 -1
retar
-
—... — -
24
0
.=
L
VI.
Although
antenna,
for
equation
bidding,
there
and analyses
Analysis
a general
special
have been
infinite
will all become
positive
upper
be analytically
integer,
limit
summations,
on the range
Three
the 6 = O case,
of purely
finite
cases
are
which
for-
simplifications
may,
in principle
If6
except
is replaced
integer,
the various
domain.
unchanged
given
quite
further
loading,
into the time
of summations
on the
6 to be a positive
resistive
36 and 40 are
special
for which
polynomials
loading
36 appears
If we choose
transformed
equations
1.
For
cases
to some
by equation
of interest
accomplished.
to certain
at least,
6) defined
cases
corresponding
series
Cases
6 corresponding
40 with ~a(s,
are
of Special
that
isa
~ as an
by 6-1 on those
below
with
6 = O, 1, 2.
6=0,6=;
we go back
to the integral
definition
l,2;
)
1]
for 4.(s,6),
namely
h
.&’o(s, (3) = *
yo(-l
/
-h
(h-1z’1
) M
I
-2yo(h-lz’
Z’1 +Z’coso)
e
(iZ’
(41)
Using
the relation
M(1,2;2z)
which
e=
is equation
—
$ case and
13.6.
= :2
sin z
(42)
14 in Abramowitz
with q =’-”:yo(h-z~)
and Stegun6,
we obtain
25
specializing
to the
o
= -
4.(s,6)
+
In the
-yoh
1
hy
2
2
(43)
l-e
[1
of equation
b+l;
39 we may
z) = &
lb (1-b+z)
+b (b-l) M (a-1,
which
is equation
d~
o
denominator
M (a+l,
e-q
-1) e
[
-yoh
Y~ h
=
-yoh
&?
13. 4.7
M (2, 3; -2yoh)
=
a recursion
b-1:
z) I
namely
(44)
and Stegun6
l(l+2y
relation,
M (a, b;z)
in Abramowitz
&h
use
oh) M(1,
to obtain
2; -2vOh)
-1 I
(45)
o
so
that
settings
c
= 1 + &
a
*
1 for C
>>
~
C
a’
[1
-2yoh
(c+sh)
M (1, 2: -2yoh)
-sh M (2, 3; -2yoh)
—.
Thus,
for
6 = O, 0 =
~ , and
CY = 1, we obtain
26
= ~
e
+1
(46)
. ..
.
.
2
[1
-yOh
~’:=~h
.()
which
agrees
expand
44
this
0 +1
e
[1
with equation
13aum we may
(47)
1:2jh
o
of reference
result
3 with (3 = O, CY= 1.
as a geometric
series
Following
giving
C71
~($)=+[1-2e-y$+e-2y~l
~ ‘-’)”
;2”’0’
ll=o
al
=— 1
‘h
z
1
1+2
-(2n-l)yoh
(-l)n
e
(48)
I
n=O
So that
Cu
= u(7h
)+2
u(t) is the unit
step
$’
where
;
()
The result
in equation
since
ignores
result
of the transmission
line
would
expect
be greatest
except
imation
that
the deviation
line
model
results
for very
should
Th- (2n-1)1
(49)
square
for the radiating
radiation
model
damping,
the result
of a more
case.
For
early
times,
become
more
27
for
but it is the
limit.
obtained
accurate
resistively
the transmission
accurate
wave
cylindrical
in the no loading
between
and that
for the no loading
perhaps
[
an undamped
result
the model
transmission
u
function.
49 is simply
Th > 0 and is not a precise
antenna
(-l)n
x
n=l
One
in the
model
would
loaded
cases,
line approx-
as the loading
increases.
2.6=1
For
in each
this
.
case,
using
summation,
Y(e)
equations
40 and
= =
2(sh+Q)
/[
~-sh(l+cOs$)
1
(m)
‘h
inverse
result
only
the n=O term
we obtain
transform
to equation
time
+1
Sh( l-cOse)
[
is identical
+1
Sh(l+cose)
e-sh(l-cOse)
This
36 with
domain
‘CYT
II
=
’77 of reference
result
l-e
l:cos(e)
-
2i7f
g
tvoe
h
3.
r%
(50)
yor
fg
Therefore,
the
is
*T
h
1
h
dTh)
a( i-c0~(e))2
-#[7 L-(1 -cos(e))]
“
l-e
-1-~
CY
‘(rh-[
)
I-coS($)]
(hcos(e)?
(51)
-cl r
‘C1T
h
e
-1l+cOs(e)
l-e
-
h
a(l+c0s(e))2
-a[T=-(l+cOs(e))]
+
1
l-e
“
‘(rh-[
CY
l+cos(e)]
)
(l+cos(@)2
1
= 2Tf
g
rEf9
—
V.
28
.
.
3.6=2
For
this
case
the frequency
.-
domain
result
is
.
‘h
[(sh(sh+z)+dsh+l)]
r’(e) = +
(52)
-Sh(l-coso)
(
x
2
l-cosf?
1-
)
2(1-COS9)2
‘h
[[
+
-1
e
(
-— 2
1 1+COS6
+
1
.
1
+-
Sh(l-coso)
(i-cosf3)
1
)[
“s (1+ COS6)
eh
-1
+
1
Shz(l+cose)z
1
Sh(l+cose)
.—
+
1 1
.—
(1+COS6)
If we write
sh(sh+2)
rewrite
equation
+a
= (sh+Qt)(sh+
;, )
(53)
CY2
+1
K 2
CY’= ;+1+
where
we may
= Shz +(a+z)sh
-t’a (Sh+l)
52 in the form
+:)
?“(e)= %p(Sh+d);sh
(54i
x-
I
-Sh(l-cose)
(l+cosf3)
e
+
(1-cos9)3
“—(1-cOse)
(l+COS@3
‘h
+
‘h
2(1-cos8)2-shsin26
+(l+cos
sh(l-c0se)3
-sh(l+~Ose)
e
‘h
‘h
2(l+cos@)2-shsin20
+(l-cos$)
sh(l+cos6)’
31
O)
Taking
the
inverse
the time
transform,
domain
waveform
becomes
—. .—.
,.
L
l+cOse
●U
2(1-@))
2+CI’Sin2
CY’
~
I~h-(l-cOse)]+
fl(l-cosg
~1 ~1-5
)3
~
[
(1-COS$)2+
~
sin26
+ (l+cos~)
~1
‘~
e-a’Th
6+(l+COSf))
(1-cose)3
Th
e
+
+
/
\
l-cosg
a(l+cose
,
I
sin26
a’2(l+cos@)2+a’
~lal-s
)3
~,
+ (l-cos
(I+cose
O)
‘CYIT
e
h
)3
()
C22 (l+cose)2+
+
(-)
sin2$
~
+(l-cos
O)
‘~
Th
(:)(@)
’’+cose”
‘~
a’
l-cOse
(l+cose)3
1
U(rh)
“
e
[Th-(l+COS@]
-C&
)( al
l-e
fld-~
()
[(
l-e
-a’[Th-(l+cos@)l
)
1
(55)
30
.
.
Note
quadratic
the symmetries
[sk(s~+2)
with
+a(sL+l)],
La 11
respect
-( Th-l)
2
/d’
l-e
I
-~fr
d2-
1 + (Q’2-i-Qf
waveform
of the general
but finding
analytical
less
roots
obtained
ofaf
convenient.
-’T
hh
-e
+
(56)
from
routine
as presented
equations
54,
55 or 56, the numerical
in figures
18 through
21 agree
results.
algebraically
to continue
becomes
to get the time
Since
/CYf
)1 u(7h) 1 +
directly
by the general
for which
(e
it is possible
inversions
attractive.
calculated
2’
U(rh -1)
at 7h ~ .51 and a minimum
of $’ ~ -O.1 at
3+6
forcy=l iscrl =r
= 2.618.
Although no
well with the analytical
In theory,
e
+1)
has a zero
~ .86,
The value
‘h
curves
were obtained
very
a = 1,
\
r
results
of the
to O = :
(:a(Th-l)),
)
..—— .
This
a’ and a/a’
specialize
If we further
~f’(
we obtain
to the roots
these
special
routine,
an analytic
closed
to higher
integer
increasingly
domain
cases
they
result
clumsy,
becomes
do agree
well
lend confidence
form
solution
values
of 6,
and attempting
correspondingly
with the results
to the other
is not available
results
or
.
VII.
Numerical
Results
./
.
Figures
between
16, and
.2 and
is diminished
lightly
tion
‘h
less
1.0 for
from
the undershoot
which
frequency
of $.
O, the first
zero
crossing
domain
field
e ) - 1; however,
dependence
of the
on e is discussed
antema
increases
ends
on the
the loading
the time
the
end effects
is a function
the
far
center
lated
that
appear
1.866.
at 6’ = ~,
at
T
h
For
discontinuities
the
waveform
Other
consider
unit
the
point
end effects
example,
and
see
in the slope
cases
zero
3.
are
are
19,
of E‘ have
note
angle
the
field
at T
21,
and 27.
removed
the
effects
as one
rapidly
of arrival
of
that
we are in
the
antenna
It may
be calcu-
at r
h
For
= . 134 and
6 = 2,
at the cost
3 the
of damping
rapidly.
of interest
equivalent
arise
when
transmission
6 is a complex
line
to have
number,
an inductive
load
If we
per
length
2zm
L’==~h%=2pofg[l
-
32
[! :’
] ‘1%
The
= 1, at 6 = ~ end effects
end effects
25,
radiated
more
from
(1
is sin$.
that
length.
loss-
loading
smoothed
h
occur
been
The
normalized
the time
antenna
23,
point.
Remembering
simultaneous
~te=+
figures
Also
solu-
+1 to -1 at
the dependence
r is distance
and 2h is the
for
line
times
the radiated
where
is that
the observer
progressively
angle.
1 and
resistive
early
6
toward
31 give
We may
crossing.
>>1
$h
=.5andl.5;
more
field
of observer
of the antenna,
times
of damping
of first
to the observation
late
in reference
at the cost
and decreasing
zone
for
radiat~d
from
various
for very
as
transmission
13 through
that
that
is happening
to +1 at that
for
6 ranging
moves
O. 2 and 3 with
be observed
.
back
E field
of 6 between
It may
We see
discontinuously
Figures
radiated
as a parameter.
as (sin
jumps
49.
parameter
the lossless
~h”= 3, and jumps
to values
goes
what
approaching
by equation
and time
corresponding
angle
~h <1,
0<
at -1 until
is given
observer
Physically,
we are
is -I-1 from
case
a fixed
increases.
cases
= 1, stays
E ‘ and $ ‘ for the loading
1 toward
also
loaded
17 give
(57)
.
101
‘
.——
I
k
i
I t f 1!11
I 8 I u111
I
t
I I 1 1{111
I
t I I II I
6=;
6
‘o”z~
0.4
0.6
0.8
10°
*L
1.0
-1
10
10-2
“3
10
I I ! I 1111
t
10-2
1 I I I t 1111 I ! ( I I I 111
10-1
I
Figure
.—
10°
16a.
I ( I I 111[
101
~?’1 fo~t~ = .2,
I
1 f I 1 1Ifl
I
1.5
.4,
,6,
1 I I 1I !1
102
.8,
1
1.0
I 1 [ f 1[i
i
e=;
1.0
6 =
t 1 ! I 1111
0.8
0.6
0.4
1.0
.5
1
()
arg ~
0
-. 5
-1.0
.
-1,5
-2.0
-2.5
I
1 0-2
I
I
I I 11111
10
Figure
I
I 1 111111
-1
16b.
I
10°
Wt
h
Phase
33
of~’for
I I I 1111!
t
I I I 111!1
101
6=.2,
102
.4,
.6,
.8,
1,0
1.0
.75
.5
—
.25
o
~!
-.25
-.
.
5
-.75
I
I
1
I
I
~
1
I
3
‘b
Figure
17.
$1 for
6 = .2,
.4,
.6,
.8,
1.0
I
.
.
101
1
k
I 1 I I I Ill
I
i
I
.
I 1{11
I
I
..-. — ..
.
I 1 1! 1ii
i
1 r I 1i r
I
6;=
10°
-1
10
10-2
L
10
-3
-1
t I I I l.’~l
I I t I I ttll
10°
10-2
10-1
IJJt
I ‘I’f
I 1 I I II
I
’l
101
102
h
Figure
4
lfla.
-1—lml=
1~1 for
1
I
6 =:,
~, ~
with
6 = 0.2
I I 1{1111 I I I 111111 I 1 1 I I II
3
t
1
2
,1
arg(~’)
0
1 I
IO-2
Figure
I I 1 11!11
I
!
I
I 11111
Phase
35
of?’
for
t
I I I I I Ill
~ol
loo
U,t~
~o-1
18b.
I
6 =;,
$
I I t I Ild
I
102
~ with 6 = 0.2
2
1
k.
, p
0.
.
-1
1
I
i
1
I
2
1
I
3
I
4
‘h
Figure
19.
~’ for
O = ~, ~
23’6
I
with
6 = 0.2
,.,
1
I
.
‘.
101
E’
.——
I
1
I
I I 1111
I
I I i I 4I II
I I I I I )4
I
1 I I III
9: =_
0
iO
I?l
i
i
10-1
10
-2
10-3
(d-t
Figure
4
20a.
17’ I for
——. ———.—_.
1
I I I I 111[
_
(? =h~, ~, $
with
6 = O. 4
[
I I I I 111(
I
I i I 1{111
t
t I I III
I
I I I tllll
I
I I I 11111
I
I I I II IL
3
2
1
arg(’1”)
o
-1
-2
“3
-4
I
I
I
I !11 II
1*-2
Figure
10°
k?th
10-1
20b.
Phase
37
of?!
for
101
@ = $, ~, ~
102
with 6 = 0.4
2
1
,$1
0
-1
1
3
4
Th
Figure
21.
t’ for
O = $,
-$, ~
with
6 = 0.4
,
.
.
101
10°
—.——
I
E ‘ “’’’’”
I
I
I
1111[
I
I
I
I I
1111
.
I I I 1I I
I
tI
-iI
1-
-1
10-1
10-2
10-3
I I I 1I 111[ 1 1 I I 1!111 I I 8 I I I !11
10-2
10-1
100
I
I I I I II
101
102
(dth
Figure
17’I
I I I 11111
I
4
22a.
for6=~,$~
I
I I 1 i #11[
with
—..
8
6 =0.6
I 8 1 I 111[
I
I I I 1I t r“
3
2
1
arg(~’
)
0
-1
-2
-3
I
-4 I
~()-2
I
Figure
1
I
1
Illtl
I
10-1
22b.
Phase
I
I
I
111!1
~~o
of ?“ for
!
I
I
I
11111
101
t
@ = ~, ~, ~ with
I I I 1 1!
~02
6 =0.6
2
1
0
-1
0
I
1
I
I
a
I
I
1
.
E
101
I
I
I
I
I 1111
I
I
I
1 I
I 111
.—
-.—_-—
I
i 1 I I 1Ill
I
-—
—
I J I I [1
‘1
t
1
t
10°
J-q
10-1
10 -2
10-3
1-
-i
I
I
I I 1!11
I
10-2
10
Figure
2
I
24a RI
I
I
t
1 i ! I 1111
1
-1
9
I
I
lo”wth
I
/ I 1 1IIIj
I !11
I I I I [1!
t
101
for o = ~, ~, ~ with
I I Iil
I I
I
102
6 = O. 8
I I I I 111]
I
I I I I
rlT
1
arg
(?’)
0
-1
-2
10-2
Figure
10-1
24b
Phase
10°
dh
of ~’ for
101
6 = ~, $, ~, with
102
d = O. 8
2
1
0
-1
1
Figure
25
{’forfl
T
a, ,
4
.
.
101
LE
10
I I i 1f I t
i
i i 1 I 1111
t
1 I ! t 1!1(
I
-.
I 1 1 II f
4
I
100
~1
I
1
-1
10-2
-1
t
10-3 I
I
t I t t IIlt
10-2
8
t I ! t 1111
10-1
Figure
26a
T
I
I I I ! I Ill
#
1 8 ? 8I 1!
101
10°
Wth
102
fore
=x
7 v with 6=1 0
z,~~
.—
——.
1 I I r +111/ I I f I I I 18
1 i I I I 11!
●
[
2
I f f I 1111
1
arg
(T’)
0
-1
-2 ~+d
1
10-2
10-1
10°
w
I
I
I
I
1111
I
I I I Ill
101
102
h
Figure
26b
Phase
43
tffor
O =~~
Z with~=l
2’ 3’ 6
.
O
2
1
0
-1
0
1
2
‘h
Figure
27
f’for6
=;,
~, ~
,
with
0= 1.0
3
4
101 ~
{ I 1 , 1111
1“-71-IIIJIJ
1 { 1 I 1111
--—I 1 I I tI
10°
10-1
10-2
u\
Figure
28a.
~ for 6 =r/2,
2
I
i
I
1 I
..
I t I 111[
r/3,
I
I r I I 1111
I
i
r/6with
I
6=2,0
I 1 I I 111~
I
I I I II
I 1! 111
I
! I 1 IL
1
-1
“2
I
1
I 1.1!
10-1
10-2
I
I !ill!
i !
I
101
10°
102
Wth
Figure
28b,
Phase
of @ for 6 = 7r/2,
t/3,
7r/6 with
6 = 2, 0
45
.
2
L
1
E’
o
0
1
2
3
4
‘h
●✎✿
),
r
*
i
10L
~-
I I i I I :II
1-’~-1 ltlfl{
t
-.. ..— 1..
I 1 I [ 1111
—.
t 1 I t t 19
1.
10°
F
10
10
-1
-2
-1
!-
10-3~J
1 ! 1 I 111,
~ : I *I, II
10-2
10-1
10°
! I I 1 1I I 1
101
102
fdh
.
1 ‘ “’’’’”‘ “’’’’”‘ “’’’’”“’”r
Figure
30a.
—-
2
IF
fore
.-
—
=r/2,
r/3,
~/6with
6=3,0
-.
1
arg(~)
0
-1
-2
I
10-2
I I I I 11!1
IQ-1
I
I I 1 I 1111
~oo
t
I I I 111!1
101
t
I I I I!
102
(dh
Figure
30b.
Phase
of ~ for @ =r/2,
47
r/3,
r/6with
6=3,
O
2
I
I
I
1
I
I
I
1
.&
‘!!’
o
1
o
I
I
I
I
1
I
3
2
I
4
‘h
Figure
31.
~’ for o =~/2,8/3,
~/6with
6=3.
O
e . .j.
.
in parallel
with a resistance
per
unit length
2Z
R’=—
such
that
the total
h-l
Z:!
impedance
p
per
(58)
‘
unit length
is given
by equation
3,
2Z
“=ti~’
then
6 is related
to p and k by
Shkp
&
In order
to separate
‘Sh%+
(59)
p
6 into its real
and imaginary
parts,
we set the
variable
‘h=@th
so that
6 becomes
P(I + j P/wtL%)
(60)
1 + (pk. )th?d’
Notice
that
in the high frequency
and the impedance
Z‘ appears
limit,
purely
49
resistive.
In the low frequency
limit
so that
the antenna
and imaginary
figure
appears
components
to be unloaded.
The
of 6 /P as a function
behavior
of ~/P Wt
of the
h
real
is shown
in
32.
1.0
I
t 1 111111
I
1 I Itlil[
i
I I I
0.5
0.
10-2
100
10-~
101
102
and #Ire(6)
as a Function
103
x/PkJth
Figure
32.
Behavior
of ~ Re(~)
50
of A/ R&
C
—
Figures
parallel
33 through
with
the far field
-.
40 show
resistance.
All these
and at an angle
value
of p = 1.
6(Sh)
is
Figures
now given
some
results
curves
assume
Figures
@=~ .
39 and 40-are
of adding
inductance
the observer
33 through
38 are
in
is in
for a
The loading parameter
for p = .5.
by
Shpk
d(sh)
where
=
p is the normalized
parallel
inductance
modifies
Th = 1.
affected.
parameter
that
the inclusion
the radiated
zero
Secondly,
Also
(60)
p
and k is the corresponding
parameter.
for P = 1, the first
than
+
resistance
It may be observed
significantly
sh*
wave
crossing
in a number
time
the undershoot
the discontinuity
of parallel
may
of respects.
be moved
and second
in slope
inductance
at ~
h
❑
to values
crossover
First,
greater
time
1 is somewhat
are
reduced
and ~ = 1 is in general
no longer the minimum
point on the curve.
h
It may be remarked
that even for relatively
large parallel
inductance,
the value
ifk=6,
the
of :! for late
p=
P = 1
I,g’at~h
case
with
The frequency
the magnitude
This
highly
oscillatory
for
large
loaded
(large
for which
by almost
results
a factor
for the same
domain
cases
be seen
to analytically
of
g‘
behavior
For
example,
of 2 than
approaches
from
domain.
loading
of the purely
both frequency
determine
increased
in the time
that
behavior
is reflected
As the inductive
33a and 34.
the behavior
show filtering
of the peak
behavior
Attempts
of loading
domain
frequency
as may
by the inductance,
for
%= co.
resonant
~,
case,
is affected
= 4 is larger
and the location
figures
example,
times
some
and time
with both
as X is decreased.
by the highly
Compare,
becomes
resistively
domain
kind of optimum
data.
choice
Instead,
a few values of it were selected
unsuccessful.
..
the second zero crossing
(for P = 1) had not yet occurred
at
were
51
.+
F
10
-1
“1
10-2
-3[
ii-)
, !’, ,,,,*I
I
I t 1 I 1Ill
I
! $ t I I 1!1
t
4 1 I fill
.LU
10-1
10-2
Figure
33a.
~
10°
.—
for Various
@
X w~h
101
102
p = 1
3
t
2 -
.18
K=,
lF
60
.30
.24
arg(~)
0
-1 -“’
-2 -3 I
-4
10-1
10-2
Figure
I I I t f 1111
i t ! Illtl
33b.
Phase
52
1 1 I t tl tll
10°
Wth
of ~ for Various
I
1 I 1 11’
102
101
Xwith
P= 1
o
1.0
I
I
I
I
I
I
I
0.5
E’
o
-0.5
-1.0
o
1
2
‘h
Figure
34.
~1 for Various
K with p = I
3
4
101
%=0.
10°
1.
1.
2.
10-1
,
-2
10
-1
t
10
-3
,
1 I I t I Itl
10-2.
1
I 1 t 1 Iltl
t
1
I t I Ittl
1 I 1 $11s
101
10°
10-1
I
102
GJth
Figure
35a.
. 6,
1~1 for%=
1.2,
1.8,
2.4and9=l
3
2
1
arg( ~ )
0
-1
-2
-3
I
f f i
10-1
10=2
Figure
1 I 1flt
35b,
Phase
t
t
I
I I 11$1
I
I 1 ! t 11!1
10°
W
h
of ~ forfi
I
I t } t I It
102
101
= 0.6,
1.2,
1.8,
2,4
with
F = 1
*
—
?.-
.
1
1.0
I
,.,
1
I
I
I
I
I
I
I
I
0.5
k!’
o
%=0. 6
-0.5
—
-1.0
I
I
I
1
I
I
I
I
I
I
I
1
2
3
4
5
6
7
8
9
10
11
‘h
Figure
36.
E’ for Various R with c = 1
12
10
1
I
I
I
I
I
I
9
8
.
~.m
asymptote
7
/
6
.
4
1st zero
crossing
zero
crossing
I
3
I
I
2
1
,..
I
I\
I
L
0
(,69) 1
I
I
2
3
1
4
I
5
1
7
I
6
‘h
Figure
37.
Variation
of First
and Second Zero
Crossing
Times
with X witl] P = 1
.!
I
.
P
x=2.
o
.
c
.
.
2
3
2
4
5
6
‘h
Figure
38.
Relative
Field
Strength
with
X as a Parameter
7
8
k=
.06
1
,
, 12
.24
.30
6
“I
F
10
-1
10-2
10
-3
f
1 &, I II 3
I I I I ! t tll
10-1
10-2
Figure
39a.
I
10°
IF\ for various
t I 1 ! I tl
!
! I f * ! !1
102
101
~th
x with
p = .5
—.
3
2
.60
X=6, O
.
.30
1
arg(~
1
.06
.12
, 18
.24
)
o
-1
-2
-3
1
I
I
I
I lllt
39b.
!
t
t
1111!
1
J
10°
dh
10-1
10-2
Figure
I
Phase
58
of ~ for Various
I
I
II
Irl
t I ! I tl!l
102
101
~ with
p = .5
●
✌✎
1.0
.06
, 12
.18
.24
.30
.60
6.0
0.5
/
/
/
E’
o
/A
-0.5
-1.0
0
3
’11
l’igurc
40.
or Various A wit L p = . 5
4
*
Th
= 4.
~ime
dorna~n
with
the results
that
the
second
the
zero
clarify
zero
out.
shown
second
To further
zero
the numerical
the
second
may
zero
delay
crossing
radiated
electric
field
radiated
cases.
The
times
for
improvement
time
Iate
instead
of 1.
The
(1) The
equation
mation
geometric
are
The
may
due to the
if a second
and
similar
of the numerical
of the confluent
(2) the number
the frequency
function
is calculated
of the
p = 1 for all
zero
crossing
of the
loading
at
as
shown.
some
finite
will
increase
the
second
zero
crossing
time.
34,
but with
33 and
P = .5
in the two cases.
depends
hypergeometric
of frequency
points
domain.
by a power
of field
ratio
as a measure
crossing
results
to the time
with
inductance
similar
one
to the electric
for the times
on the
to figures
is quite
the
and second
zero
of parallel
dependent
but that
cases
loaded
case
k = x provides
is a finite
resistive-inductive
loaded
is a
at the expense
loaded
be viewed
of p = 1,
P = 1, % # zx cases.
38 shows
the first
through-
of p such
p = 1,
2 there
Figure
resistively
resistively
behavior
k.
between
ratios
by an amount
calculation
from
purely
value
the
and
analytically,
p = O. 8 there
value
inductively
the addition
accuracy
28,
various
is that
39 and 40 are
large
first
loading
Thus,
-.
Th = 6.
P = 1 assumed
if P = 1 is assumed]
enough
strength
conclusion
fields
For
A
b~: crmcludi:rl
not proven
for any finite
(at least
The
case.
to the purely
time
Figures
for
shown
is acceptable,
AIthough
12
:
37 shows
?esistive
H 4. 5, for example.
h
against
which to compare
that
Th
exceeding
p = 1 is the minimum
that
a large
in field
The
of ~ with
for pure
to
36 it may
figure
is at infinity.
by an antenna
times
since
to any arbitrarily
field
each
compared
as a function
time,
time
by selecting
question,
suggest
cases
to values
at 7
case
strength
crossing
crossing
~hese
1{’rorn figwrc
is at infinity.
re suits
this
ior
be pushed
suggest
crossing
The numerical
36.
is ck.asen
zero
reference
exknded
may
times
results
zero
a useful
the zero
crossing
second
second
crossing
p = 1 value
second
were
in Figure
crossing
The
that
dda
series
The
upon
function
appearing
used
in the transfor-
confluent
.
two things:
hyper
in
,
-
in the argument
0
60
-2yo (h- z’ ), and for large
-270
{h-~zf ).
The number
nun-her
of frequency
function
~’ agrees
for the case
points
argument
of terms
in either
required
are
to within
of purely
by an asymptotic
resistive
at least
series
chosen
such
expansion
together
that
1% with the analytical
loading.
61
in
with the
the time
results
domain
at ~h= O
*
Application
VIII.
-.
“
a’
In order
the impact
to iHustrate
antenna
systems,
we consider
to a specific
antenna.
We choose
real
Antenna
half-length
Effective
antenna
Peak
From
dance
of this
so that
f
source
cage
=Zwlzo
the following
radius
of these
on
results
specifications:
- a = 2.5
(assuming
2, above
antenna
the application
calculations
- h = 50 meters
voltage
equation
of the above
meters
capacitive
we calculate
the
generator)
characteristic
- V. =5 x 106 volts
impe-
as
=1.17
g
The
is,
relation
from
between
equation
For
this
the far
electric
field
and the normalized
waveform
27,
antenna,
Ef ~
An observer
electric
a distance
=6.78
x105~
500 meters
from
the antenna
field
Efo
= 1.36
x 103 :! volts/meter
62
experiences
an
4
F
@
The retarded
onset)
is related
time
for this
observer
to the normalized
(time
measured
from
pulse
7 h, by
time,
h7h
>:
t=~
_ 5000
‘h nanoseconds
3
Choosing
the impedance
inductive
loading
loading
+’=442~
L=
in parallel
with a resistive
We can approximate
inductor
To obtain
5 meter intervals.
will be required
we take
at this
location
this
oiJ@in
an
gH/meter
for the elements
from
o ‘meter”
continuous
we assume
The largest
is taken
P = 1, K ‘ 6, wc
loading
on the required
elements,
to be 45 meters
11-lz’i
= 442&
a bound
and resistor
that
loading
R=%
elements.
such
values
by discrete
of these
the antenna
values of both inductance
nearest
the antenna
to be
R =486Q
63
loading
discrete
to be loaded
and resistance
the end of the antenna,
center.
The
at
required
which
resistance
1.36x103
Ef (V/m)
e
0.0
m
p=l
X=m
.b
I
1
I
I
I
I
and the required
inductance
is taken
as
7. 5
L =:=10-6Z
i
z
A
‘z
L =0.486
The time domain electric
r = 500 meters
only the resistive
is shown
in figure
loading.
inductive
load included
resistive
loading.
field produced
Notice
by this
loading
41,
together
that
the cross-over
is approximately
65
500ns
at
with that
later
O= n/2,
produced
time
than
by
with the
that
with only
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