. .. 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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