11764-51 O-M = RL-2224 011764-510-M 8 November 1973 MEMO TO: File FROM: Jovan Zatkalik SUBJECT: Non-Specular Radar Cross Section of the Discontinuity in Surface Impedance. Part B: The Case of Real Impedance. In this memo the far field scattering from the discontinuity in surface impedance is treated for the case of real impedances. The case of reactive impedances was treated in the memo numbered 011764-508-M (Ref. 1). The reason for such a separation lies in the fact that in the case of reactive impedances, the exact evaluation of some integrals can be carried out, while in the other cases the exact solution cannot be obtained except in the convergent infinite series form. Besides this form of solution, it is attempted here to get the approximate solution for the far field in the closed form, along with the estimation of error. The forms of integrals defining the decomposition functions (-4) and F (g) (Ref. 1, page 15) depend on the choice of the integration contours 2 and branch cuts. We have tried a lot of contours seeldking to find the solution in the simplest form and have finally chosen two contours: one the same as in Fig. 5 of Ref. 1, and the second as in Fig. 1 of this memo. The notation and the meaning of various quantities are the same as in Ref. 1 to which we shall often make reference. The First Integration Contour In this case we shall put: ^= R1 2= R2 DISTRIBUTION Hiatt/File Liepa Senior Knott Sengupta

011764-510-M and for the function F1 (), where = Kcos 0, or 5 = Kcos 0, we have the following expression (Ref. 1, Eq. (33)): ia00 JKR 1 jKR1 F(In (1- 77. (1) It is easy to show that F1 () is real, if |j < K. Namely, by taking the same contour as in Fig. 5 of Ref. 1 we obtain from Eq. (1): 2r i 2~, v-~2K -J'KR1 A F1 ().=-.F In XK.j- - c-c. (2) 1jkR In Eq. (2) the real part of the integral is zero, for: a-jb 1 a2+b2 In: 1ni: =0 a+jb 2 n 2 b2 so we have: -KR KR d F1 )()=- arctg ( -arctg( d 4 2 K 2 K 2 2K2 and F (M) is real. ao Kchx R1= sin dc=Kshx = cos0 K 2

011764-510-M the function F1 () becomes: F(Kcss9inoc~ shx Fl (Kcos e) = - arctg hxchx cos dx (4) 1 )T shx chx-cos Consider the infinite series: p2i-1 sin(2i-1)t 1arc 2 (5) 2i- 1 -2 i=l \l-p given in Ref. 2, page 41, which is convergent for 0 < t < 2ir, and p2 < 1. If we put: -X p = e (x >0, and p < 1 accordingly) t'= a we obtain from Eq. (5): -(2i-l)x sin(2i-1)a 1 arctge - sina 1 sctgina 2 e 2i — 2 arctg -2x arctg= h ) 2i-1 2 1 e-22x 2 sh i=l which is precisely (save factor -) the first factor in the integrand for 2 F (Kcos e). So we have from Eq. (4): F1(Kcos) = 2i-1 chxcosdX (a) 2 e-(2i-l)x sin(2i-1)a sh x 1 T 2i- 1 chx - cos e i=1 It is easy to prove that in Eq. (5a) the conditions for changing the order of integration and summation are fulfilled, so we have: 3

011764-510-M CD F-(Kcos 0) = - i=l e-(2i-1) x shx chx - cos 0 (6) The integral in the brackets we shall evaluate on the basis of integral: ID OD e-,x 2 X sinnt chx-cost sin t +u+n n=l (see Ref. 2, page 357; no. 3545.2). After some manipulation we obtain: -(2i-1)x shx dx chx-cosO \ 1 (sinO sinO 2i-1 sin 20 + 2i CD 00 1 i sinnO _ sinn0\ sin0 2i-2+n 2i+n2 ' n=3 n=l (7) The terms in the brackets, which we shall designate by F (0, i), can be rearranged as follows: F(, i): = sin(2+n) 0- sinnS = 2 sine 2i+n cos (n+l)0 = 2i +n n=l co =2sinO sin(2i-1)0 6 s m=2i+l n=l 00 in m + 2 sin cos(2i-1)0 o cos m m i m m-2i+1 where we have put: m = 2i +n. We know that: ~D sinKx 7r- x K 2 K=l 4

011764-510-M cosKx 1 sin x K 2 2 K=1 (see, for example, Ref. 2, page 38) so we have after some manipulation: F(8, i) = 2sin sin(2i-1)8 -2 sin cos(2i-1) ( (8) m=l With Eqs. (6), (7) and (8) the function F' (Kcos 8) becomes: F1 (Kcos 0)= 00 - sin (2i-) 2i-1 i=l + 2 f (, i) sin(2icos 8 + 2 fl (80 i) sin (2i-1) 8 = 2i-1+ i - 2 f (0, i) cos (2i-0) = ao 2 sin(2i-a) i1 (2i-1)2 00 2 cos 0 sin(2i-ca) r - i(2i-a) i=l oD + 2 cos [2i-1) (a+0i + ) 2i-1 i=l CO 2i2 in[2i-1) (a+0j +r; 2i-1 i=l cos i2i-1)(a-0 f (,i)+ 2i-1 f1(0) sin i2i -1)(a -8f (9I i) 2i-1 J2 (9) where: 5

011764-510-M f (e, iTO '1' 2 2i ~sin mo Lim m4l 2i f2 (0, )= In sin 0+XCos-me m=1 Using relations: j=1 cos (2i-1) x 2i-1 = -tInctgx 2 2 Xsin (2i-1)x IT As 2i-1 4 i= 1 we can transform Eq. (9) into: F0.(KcosO)=f (9)+f (9)tn 1,si 1 2 1 R i-s inO + f 3 (a) Cos + f4(a) (1 0) where: f (6) -1 1 IT OD i = 1 i=1 sin mO m1 OD f(6) = i=1 i=1 m1j f (a)= 2Xsin(2i-1)a i=1 (2i-1) 6

011764-510-M 00 f2, 2 sin(2i-1) a (1) f() = r (2i-1)i (11) i=l with a= arc sinR1. In this way, all functions f. in the above expression are the functions 1 of one variable only (0 or a). The corresponding expressions for F1 (Kcos0 ), F2 (Kcos 0) and F (Kcos 0 ) can be obtained from Eq. (10) by simply changing 0e-0 and 2 o o R - R2. With the aid of Eq. (39) in Ref. 1 we can obtain the desired expression for radar cross section a. The Second Integration Contour As it was pointed out in the beginning, the form of the integral to be evaluated is strongly dependent on the contour chosen. Here we are going to take the integration contour and branch cut as shown in Fig. 1. This contour represents the proper deformation of the basic integration contour from Fig. 4 in Ref. 1. I ii I a' = a + jT plane l', - - branch cut i- r integration contour ' l = -K a = (pole) a = K Fig. 1 7

011764-510-M We now have the pole on the contour at c- = I, so we introduced the indentation above it. Along the right and left-hand sides of the jT axis, the function y has the values: + (j2 -K2=+j/ +K 22 22 -(T) -K -j K respectively, while along the upper and lower side of the a axis, y is, respectively: 1 = -/7 y-j K-aSo we have from Eq. (1) and Fig. 1: 1 ^ +K -KR In + K —+ 2 2- 2 n2 R en K (12) 1 2_~2KR K f^,.-KR 2 2+KR 2~ or2n K where we used the well known formula: b (a) daJrf()+Pf fa_) d C a in which C is the part of a axis between a and b, with indentation above pole a = I, and P stands for the Cauchy principle value of the integral. From Eq. (12) we obtain, after simple manipulation: 8

01 1764-510-M F 0-( 1 00 Il dT T2+2 7 +e _27 2Tr TdT 1 + i u. 1 d (cr rd -r i +t 2,Iaa (13) T2+ 2 2_r K-C2+KR1 The first and second integrals in Eq. (13) are real. Namely, T in them goes from 0 to oo, and we take R1 < 1, so the factor under log cannot be negative. The third integral is real from a = 0 to a = K -R2, and from K 1-R2 to K it is complex because the factor under log is negative. However, from Eq. (3) we know that Fo1() must be real, so the imaginary parts in Eq. (13) have to cancel each other and we have: F-(~) 1 dT 1 T2+g 2 dar cra -?e 2 +2-r log 1 2 - 2 T +-T 27r 1 2r 2 (14) 9 where T1 and T2 represent corresponding integrals evaluate integral T first. We have: in Eq. (14). Let us T = 2rP | - 7T lim 2- 0 0 - R 2 iKW1_~ 1 -ja + it- * J - 9

Q 011764-510-M 0 7=Tn K (15) K/1-R2 -Integral T1 is much more difficult to evaluate. Moreover, it turned out, after many attempts, that integral T1 cannot be expressed as a finitel combination of elementary functions. So, we shall give here two solutions: an exact one, but in the form of series, and an approximate one which is in closed form, more suitable for our purposes. We have: 2+ K2 - KR1 d T n 1 d T +K2+KR1 T2+2 1 If we put: To= Kshx we obtain: O Chx-R1 ChR1 Chx dx 1 n Chx+R1 2 sh2x+( ) If we take 0 < R < 1, we can put: R = sina (16) where a is real and 0 < a < 7. On the other hand, < 1,so we can write: 2 KI fe\2 2 1- () cos (17) and T1 becomes: T1= en Chx-sina. Ch dx 18 n Chx+sina Ch 2 (18) 0^ Ch x-cos 13 10

011764-510-M which is the more suitable form for evaluation. From Bateman's Tables of cosine transform (Ref. 3, p. 36, No. 50) we can write for the first factor under integral in Eq. (18): (Chx-sina sh In h xsin 2! 1 cosxydy. (19) yChTry shcy Ch x - The integral T2T = - Ch x C -2 - cosxyd 2x0 1 s Ch xy s s* Ch2x - cos23 cos x y d can be evaluated exactly on the basis of the integral: T cos ax chx dx -r ctgb shaf 4 = Chx - cos f3 sh a 7r given in Ref. 2, p. 506, No. 3. 984. 2. After simple manipulation we obtain: 2sin sh yv+sh(7r-3)y] 7r Ch(1-7r/2)y 3 2 sing sh 7y 2 sing Ch[ sh ry = 2sinI h which, after introducing in Eq. (20), gives: shay. Ch(/3- )y sh2t. Ch(2- - l)t dt 1 sin r sin y Ch2 y t- Ch2t 11

011764-510-M where we put y = t. Using the relation: 2 sh a. chb= - sh(a+b)+ sh(a-b) we can write: T1 2 sinfl T6] (21) where: Tshpt dt (22) ok tch t T sh t dt (23) 6 tch t with: 2 (a+3) 7r (24) q2(a- 3)+ 7T Integrals T and T6 cannot be expressed by finite combination of elementary 5 6 functions, but they may serve as the basis for approximate evaluation, as will be shown later. The exact expression for T. and T6 can be obtained in the 5 6 form of convergent power series in the following way. Consider the integrals: T7 = 7- dt; T= = dt 7 ac ir 2'' 8 aa ir 2 ch t ch t which can be evaluated exactly on the basis of integral No. 3. 514. 3 given in Ref. 2, p. 345. We have, after simple manipulation, 12

011764-510-M 7r 2 2 1 2 a+, T =7 7T. ~r cos(a+3) ~ cos(ma+p) sin. p (25) 7r 2 1 2 aT = 8. 7r cos(a-3) T cos(a-,3) sins q It is known (see Ref. 2, p. 190 and Ref. 4, p. 47) that: 2i 2 4 6 8 10 x 2(2 x x 5x 61x 1385 x T I =- + + + + +... 9 cosx (2i+2)(2i!) 2 4.2! 6.4! 8.6! 10 8! b=0 where E are Euler numbers which are well tabulated in standard Mathematical n Tables (for example, Ref. 5, p. 810). On the other hand we have (Ref. 4, p. 40): _ X 1 1+sinx dx = Intg (7r/4+x) = log sinx So we have: T5+T T u da a+ ^T da8 - 6 7 8 cos (a,+ coss(a + -3 2 a'+1, _1 -da Jcos (a+) cos (a-3) 1 1 + sin - 1 +-sin 2 -sin 2 2 00 I I 2^ 2i L.2i+2 ( o2i+2 7r (2i+2)(2i!) ) i=0 After some manipulation we obtain, with the aid of Eq. (21): 13

011764-510-M sina+cos2 0 IE I i 2 2 1 E2i 1 2 In a sin a n (2i+2) (2i) sin — cos 2 2i r IV(i+2\ 2(i-n)+ 12n+l n+l(26) n=O (26) Taking into account Eqs. (14), (15) and (26), we can write for the function F (c): 1F Fi(~)= In 1 2 1 +4n 4 a 2 sin- cos 2 2 KOD(2i+1l)E2 - = F.(a, 3) i=O (27) where: i 2 (i-n)+ 1 2n+ 1 ) (2n+l)! (i - n)! n=O If we put in Eq. (27): sina= R1 sin = = cos K We obtain for the function F(-Kcos We obtain for the function F (-K cos 0): 1 1 F1(-Kcos 0)= - n 1. 2 /1 - 1-R 2+ +l++sin R 2 2 1 1 + cos 0 J1- - + sin0 - RC+cos 0 1 1 -( (R1, e) (28) 14

011764-510-M where: 1 c(2i+1)1E (R 1' ) cos 2 1' ) i=0 and: 2 (i-n)+l 2n+1 i (arcsin R)2 (if/2- ) 1(R1' 0) (2n+ 1) (i-n)! n=0 The remaining functions we need, i.e., F1 (-cos 0 ), F (-cos 0) and 1 0 2 F2 (-cos 6 ), we can obtain from Eq. (28) by simply putting cos 0 instead of 2 o o cos 0 and R2 instead of R1. For evaluation Of ao we need not directly F1 (-cos 0) 2F1 (-cos 0) but e, according to formula (39) in Ref. 1. So we have: 2F1 (-Kcos0) sin0-R 1-. + 1+sin -2 (R1,0) e1 1 1 +cos 0 1 - - -R- +sin0 - R+cos0 1 1 The obtained formulas are rather cumbersome and are not suitable for practical application, so we are going to find some approximate and simpler ones. We shall start with the integral T5 and T6 in Eqs. (22) and (23). For T we have: 5 OD T shpdt shpt dt T pI 2 dt =.ch2t 1 ' tch t ch2t - 2 2 For approximate evaluation of T we shall try the following method: find an 5 integral, say T10, for which we can prove that T5 < T10, and a second integral, say T 0, for which we can prove that T > Tv. Suppose that we 10, 5 10 can evaluate T10 and T! exactly and find the difference between them. It may happen that this difference is small in which case the difference between10 may happen that this difference is small in which case the difference between 15

011764-510-M T and T, or T and T' is small also and we can take T 5T10, or 5 10' 5 10 50 or T T T1 with a good estimation of error. 5 10 Consider now the integral: = shptdt 10 I Ch2t 2 1 which differs from T in missing the term - in the denominator of T5. It is clear that: T < T1 5 10 On the other hand, from Eq. (25) we know that the exact form of T5 is: 5 r5 sh ptd T dp= dt dp - -0 5 JT5 >p d p f_2 p2 2 (22 1) 1 2K+] dp= -- 2K+ 2 snp f (2K+ 1 B2K 1 sinm p Tp p)3 1 +7 )5.5 1 +31 ( P)7 1. (30) where B. are Bernoulli numbers, tabulated in standard Mathematical Tables 10 (for example, in Ref. 5, p. 810). If we denote by T0 the sum of the first few terms in the above infinite series, we are sure that: T > Tp 5 10 because all the terms are positive. So we have: T' < T < T 10 5 10 16

011764-510-M If we can now prove that the difference A between T1 and T1 is 10 10 small for the whole range of p, then the difference A1 between T10 and T is still smaller and we can have: 5 T10 which has the exact and relatively simple solution: 1+ tg p T10= 21n tg ( _E+ ) = 2 In 8 10 8 4 1 g 1-tg 8 (see, for example, Ref. 2, p. 351, No. 254.24). After some computation we found that taking the first four member of series T, the maximum value of: 5 = 100 percent = (1- - T ) 100 percent 10 10 which represents the maximum relative error in percent, is less than 8 percent, for all values of p between 0 and 1. So we have: 1 +tg 1+ 8 T n 8 T5, 21n - 2 -n + - tg P 1 - p 8 8 with additional error of about 2 percent, and 1+q T 2n -8 T6 ' 2 1 qr 1-8 8 The integral T1 after some manipulation becomes: 2 4(a+2)2-(I- )2 inr In )2 2 4 (a - 2)2-( - )2 T1~ in B 4(<x2 )- (3 -T ) 17

011764-510-M The function F1(-Kcos 0) is now: 2 0 2 sin-R1 1+os (arcsin Ri+2) -( —) 1_____1 1 + cos Q ___1__2_ i 2 sin +R 2 \ F (-co s01 l-R +cos0 (arcsinR1 2) - (-) 12 while the function K1 (-Kcos 0)2 is: 2 2 2F (-cos0) sin0-R1 l+cos (arcsinR +2) -2 - 2 1 1 1+cos 1 2 K (-Kcos0) ne + 2.2 6 2 1 1-R +cos0 (arcsinR-2) -(-) 1 2 which is much simpler than Eq. (28) and accurate enough. The corresponding function K (-cos 0 ), K (-cos 0) and K2 (-cos 0 ) can be obtained from K1(-cos 0) 1 0 2 2 0 1 by simple change of variables. With the aid of formula (39) in Ref. 1 we can obtain the desired expression for radar cross section a. References 1. J. Zatkalik: Non-Specular Radar Cross Section of the Discontinuity in Surface Impedance, Part A: The Case of Reactive Impedances, Radiation Laboratory Memo No. 011764-508-M, 16 October 1973. 2. I. Gradshteyn, I. Ryzhik: Tables of Integrals, Series and Products, Academic Press, New York, 1965. 3. H. Bateman: Tables of Integral Transforms, McGraw-Hill, New York 1954. 4. B. Pierce: A Short Table of Integrals, Ginn and Company, Boston, 1929. 5. M Abramowitz: Handbook of Mathematical Functions, National Bureau of Standards, 1964. 18