. 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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