THE UNIVERSITY OF MICHIGAN 1363-1-T AFCRL-68-0349 NETWORK FUNCTION DETERMINATION FROM PARTIAL SPECIFICATIONS by Arthur R. Braun and E. Lawrence McMahon The University of Michigan Radiation Laboratory 201 Catherine Street Ann Arbor, Michigan 48108 June 1968 Scientific Report No. 1 Contract No. F19628-68-C-0071 Project 5635 Task 563505 Work Unit No. 56350501 Contract Monitor: Philipp Blacksmith Microwave Physics Laboratory Prepared for Air Force Cambridge Research Laboratories Office of Aerospace Research L. G. Hanscom Field Bedford, Massachusetts Distribution of this document is unlimited. It may be released to the Clearinghouse, Department of Commerce, for sale to the general public.

,..d.:~c/ THE UNIVERSITY OF MICHIGAN 1363-1-T ABSTRACT In certain problems of Network Theory the real and imaginary parts of a driving-point impedance function may be independently given in a band of interest (wl < w < w2), leaving its continuations onto _ -1 2 (0 < t < 1) and (w 2 <w < 00) completely unspecified. By manipulating the Hilbert Transforms, relating the real and imaginary parts of a function of a complex variable (p = o + jw) having no poles on the jw axis or in the right-half plane, this report shows that if the continuations exist they are unique and readily obtained. Three necessary conditions for the existence of the continuations to the given parts (of the driving-point impedance function) are obtained. Further, if these three conditions are satisfied the continuations may be obtained as a Fourier series. Four known impedances were used as examples and their continuations determined. The results obtained were excellent. The agreement between the Fourier series solution (using the first six terms at most) and the exact expression was good up to the second significant figure. In the appendix a computer program which obtains the Fourier coefficients of the unknown continuations, from the given real and imaginary parts, is furnished. ii

THE UNIVERSITY OF MICHIGAN 1363-1 -T TABLE OF CONTENTS LIST OF PRINCIPAL SYMBOLS iv LIST OF TABLES vi LIST OF ILLUSTRATIONS vii LIST OF APPENDICES viii CHAPTER ONE: INTRODUCTION 1 CHAPTER TWO: THEORETICAL FORMULATION 7 2.1 Hilbert Transforms 7 2.2 Necessary Conditions for Positive RealContinuations 10 2. 3 Uniqueness of Continuations 12 2.4 Solution for Continuations 13 2. 5 Solution from a Graphical Representation 15 2. 6 Extension to the Band Case 21 2. 7 Measure of Truncation Error 25 2. 8 Extension to Transfer Impedances 26 CHAPTER THREE: NUMERICAL COMPUTATIONS 27 3.1 Transformation of Equations 27 3.2 First Example 30 3. 3 Second Example 33 3. 4 Third Example 35 3. 5 Fourth Example 38 CHAPTER FOUR: COMMENTS, CONCLUSIONS AND SUGGESTIONS FOR FURTHER WORK 41 4.1 Minimum Amount of Information Required to Specify a Passive Driving-point Impedance Function 41 4.2 Uses of this Method 41 4. 3 Problems in Computation 42 ACKNOWLEDGMENTS 48 CHAPTER FIVE: APPENDIX 49 REFERENCES 62 DD 1473 63

PRINCIPAL SYMBOLS Symbol Description First Reference Z(p) Driving-point impedance function Eq. (1.1) R (o) Real part of driving point impedance function for p = jw Eq. (1.1) X (W) Imaginary part of driving-point impedance function for p = jw. Eq. (1.2) R *(0) Real part of driving-point impedance function in the 0 domain; i.e., w = sec 0/2. Eq. (2. 13) k Coefficients of power series for f(w) Eq. (2. 19) _ n =E. ( B A Eq. (2. 25) 22n-1 A A! B) (A - B)! B Eq. (2. 26) A. Fourier coefficients for unknown R*(0) Eq. (2. 28) H Coefficients of power series for g (w) Eq. (2. 36) n E *(0) Imaginary part of driving-point impedance function multiplied by cos 0/2 in the 0 domain Eq. (2.37) T Fourier coefficients for unknown function E*(p) Eq. (2.37) 2nn-1 = 22n-1; n = m H ~nm A _)n-m 2n 2m-1 n + m - 1) in - m -m- 1 nvm Eq. (2.43) iv

List of Principal Symbols (Cont'd) Symbol Description First Reference R *(a) Real part of driving-point impedance function for 0 < w < 1 and w = cos a/2 Eq. (2.41) R Coefficients of power series about the origin for R(w) Eq. (2. 45) Bn Fourier coefficients for R*(a) Eq. (2.41) (2j -)2 1* 3. 5.. (2j-1) Eq. (2.48) (2j)!! 4 6 (2j) C Fourier coefficients for cX(w) in n (0< w < ) Eq. (2.54) D ~ Fourier coefficients for R( ) in (W1 < w < W2) when w = (w2 -1 ) cos2 L Eq. (2. 74)

LIST OF TABLES Table No. Caption Page 3.1 Comparison between points from the exact expression and those from the Fourier series solution for Example One. 31 3.2 Comparison between points from the exact expression and those from the Fourier series solution for Example Two. 35 3. 3 Comparison between points from the exact expression and those from the Fourier series solution for Example Three. 36 3. 4 Comparison between points from the exact expression and those from the Fourier series solution for Example Pour. 40 vi

ILLUSTRATIONS Figure No. Caption Page 1-1 Impedance function necessary to reduce the radar cross section of a metallic sphere by 15 db. 2 2-1 Contour of integration for Eq. (2. 2) 8 3-1 Plot of R () for Example One. 32 3-2 Plot of R(M) for Example Two. 34 3-3 Plot of R(L) for Example Three. 37 3-4 Plot of R (w) for Example Four. 39 vii

APPENDICES Title Page Appendix A. Main Program 48 Appendix B. Subroutine F p RIT 52 Appendix C. Subroutine to obtain points from a given 54 function Appendix D. Subroutine to read the given points into subroutine F 0 RIT. 56 Appendix E. 58 Appendix F. 60 viii

Chapter One INTRODUCTION It has been known for some time that the back scattering radar cross section of a metallic object may be controlled by the method of im1 pedance loading. That is, the return from an object may be increased or decreased, depending on the loading impedance. By properly loading an object its return can, in some instances, be reduced considerably. For example, in the case of a dipole antenna, Chen and Liepa have shown that the return can be reduced by 33 db. at broadside incidence. For the case of a sphere, Liepa and Senior have shown that the return can be reduced by 22 db. Using a different approach, Chang and Senior have obtained a 15 db. reduction in the case of a sphere. The loading impedance function necessary to achieve this reduction (for the sphere case) is shown in Fig. -1. As canbe seen, the impedance function is not positive real (the real part is not positive for all frequencies) and icannot be realized with passive elements. However, considering that the real part of this function remains positive in the band 0 < ka < 1.5, we might hope to change its behavior outside the band such that the function obtained is positive real. However, once the behavior of an impedance function is given in a band of interest, how much freedom is there as to the choice of its continuations outside the band? Is the behavior of an impedance function completely determined for all frequencies by specifying its behavior in a band of interest? In other words: Given independently (in the form of a graph) the real and imaginary parts of a passive driving-point impedance function in a band (0, 1)* (a) What can be said about the compatibility of the given parts; i. e. are they the real and imaginary For simplicity, the band of interest has been taken as (0, 1). However, in a later sections the solution to the problem will be extended to the band,.1' 2) For this case a: radius of sphere, k: f (f in MHz). 1

0. 4 ohms R(w) -0.4 20 10 0 - X(S) ohms -10 -20 1. I I w. 0 1.0 ka 2.0 FIG. 1-1: IMPEDANCE NEEDED TO REDUCE THE RETURN FROM A SPHERE BY 15 db. (after Chang and Senior).

3 parts of a positive real function in the band (0, 1)? (b) What can be said about the continuations onto (1,00)? Are they also the real and imaginary parts of the same positive real function in (1, 0)? (c) What conditions must be imposed on the given parts, in the band (0, 1), such that (a) and (b) are satisfied? (d) Are the continuations unique? (e) What are the continuations onto (1,00)? Assuming that the impedance function in question: (i) has no poles on the jw axis or in the right-half plane this report gives an adequate answer to each of the above questions. This assumption is necessary for the sole purpose of uniqueness; i.e. a pole on the jw axis may be partially or completely removed without changing the behavior of the real part (thereby making the impedance function not unique). In answering question (a), (b) and (c) a set of three necessary (though not sufficient) conditions are obtained which test whether the given real and imaginary parts are compatible in the band of interest. Unfortunately, the solution for the continuations of the given parts can not be obtained in a closed form due to the nature of the 13 integral equations r lating the given parts and their continuations. However, a Fourier series solution can be obtained. This series, as the example solved in a later chapter will show, converges very rapidly. Before going into the solution of the problem, let us review some of the work that has been done previously.

4 Bode5 pointed out that having the real or imaginary part of a passive driving-point impedance function as a rational function of frequency would completely specify the impedance function for all frequencies. He noted that in going from the real to the complex frequency domain (( - p/J) one would obtain: R(w -p/j) = Ev Z(p) = 1/2 [Z(p) + Z(-p)] (1.1) and X(w - p/j) = Od Z(p) = 1/2 [Z(p) - Z(-p)]. (1.2) Hence, expanding the even or odd part of the impedance function (whichever is known and is rational) in a partial fraction expansion, one can identify those terms having left-half plane poles with Z (p) and those with right-half plane poles Z (-p). Guillemin pointed out that by making use of the Hilbert Transforms the real (or imaginary) part of a driving-point impedance function may be obtained from the imaginary (or real) part. The Hilbert Transforms assume that the impedance function in question has no poles in the right-hail plane or on the jw axis. These transforms are: R(w) = -R() (1.3) 2 2 X -t and 2w R(k) dX 2T- (1 4) V 2 2 0-( Both of these methods require some information for all frequencies; i.e. the real or imaginary part of the impedance function must be known throughout the whole spectrum. However, for the problem of impedance loading the real and imaginary parts of the impedance function are given in a band of frequencies wl1 < w<

5 leaving its continuations onto 0 < wo < 1 and w2 < w < C( completely unspecified. 7 For this case Calahan has suggested matching the known impedance function behavior with an expansion of positive-real, rational functions in the band wl < < w2; the coefficients of these rational functions being determined by computer optimization techniques. Another method (due to Redheffer ) of solving this problem makes use of the change of variable w = tan 2. Then, the band of interest -1 22 -1 w1 < W <_ W2 becomes 01 =2 tan l < 0 < 02 = 2tan w2. In the domain Fourier series which give the best fit (in the least square sense) to the real and imaginary parts in 01 < 0 < 02 can be found (subject to the Weiner-Lee criterion for physical realizability'). Unfortunately, the equations for the Fourier coefficients are rather difficult to generate and, as pointed out by Redheffer, it may be shown that the Fourier series will not in general be bounded for 02 < 0 r. Still another method is that suggested by Zeheb and Lempel. If the behavior of an impedance function is known for a finite number of frequencies, a passive network can be synthesized such that its impedance takes on the prescribed values at the given frequencies. However, there is no restriction as to the behavior of the impedance other than at the given frequencies, thereby allowing more than one solution. This report is concerned with the problem of determining the behavior of an impedance function outside of the band where its behavior is completely specified. That is, given an impedance function with a prescribed behavior in a band wl o w < w2 it is desired to find the solution for the function's continuation onto 0 < o < 1 and o2 < < OO. 00 00 =4 If R() = A cos an a, then: n n

In contrast to the methods outlined above, this report does not use any approximation techniques (other than the truncation of an infinite series) and if certain necessary conditions are satisfied, it gives bounded and unique solutions for the continuations.

Chapter Two THEORETICAL FORMULATION 2.1 Hilbert Transforms If Z(p) (where p is the complex variable p = a + jw) is a drivingpoint impedance function, then it is positive-real and by definition: a) Z(p) is analytic in the right half plane b) Re [z(w)]> 0 for all c) Z(p) is real for p real. Given a function G(p), having no poles on the jw axis or the right half plane, we define the function F(p) by: F(p) (2.1) p - Jw(1 Then from the Cauchy-Goursat integral theorem we have: - F(p) dp = 0 (2.2) where c is the contour shown in Fig. 2-1. Writing G(p) in the form: G(p) = H1 (a,w) + jH2 (a,w) (2.3) we have from (2.2) H1(a ) + j H2 (a, w).1. H )jH( dp =. (2.4) Carrying out the contour integration, letting r - 0 and R - 00 and separating real and imaginary parts, we obtain: [H1(0,W) + iHH2(0 0) =- (IH(0 ) A) (2.5)

Let: G(p) Z(p) R+JX (2.6) l+p l+p so that: R(w) 0< (< 1 H (O,w) Hl(W) = (2.7) H (0,) = H() = (2.8) 2 R()1 < w< 00 It should be noted that since Z(p) has no poles on the jw axis or in the right-half plane, then: lim G(p) = O for a> O p OD - jw r _- -" R a

Since we have assumed that Z(p) is real for p real, it follows that R(w) and X(S) are even and odd functions of frequency respectively. Using Eqs. (2.7), (2.8), (2.5) and (2.6) we obtain: 2 R(w) + X(X) dX R() dX f < 1 2 / 2 ( /17T f2 2 (X2i 2 - (2.9) X(S) _ R(X) dX X(X) dX A 2) /1YT -Jo (X 2) 1I f (X2 w2) (2.10) These integrals are to be evaluated in the Cauchy principal value sense; i.e. R(X)X _ lim I R(X)2dX lim L (X2 dX2 +2) d 0< a< 1 (2.11) and so on. Eqs. (2.9) and (2.10) are Huibert Transforms relating the real and imaginary parts of a function of the complex variable p = a + jw for a = 0 and 0 < w < 1. Note the following: 1) Since R(w) and X(w) are given in the band (0, 1) then the functions f( ) and g(w) are known. 2) We are seeking a solution for R(X) and X(X) in (1, 00). 3) Since X(M)/w is even, o X(w) is even and w R(w) is odd, Eqs. (2.9) and (2. 10) have the same functional representation. Hence, we need solve only one of these equations. For our purposes, we will solve equation (2.9) which is reproduced here for convenience.

10 (0 f() =R(X dX 0< < 1. (2.9) 2.2 Necessary Conditions for Positive Real Continuations. Making the change of variable X = c 2 (2.12) in Eq. (2.9) we obtain: 2 o0_ < r f(w) ~ - 2 Cos2 0<2 ~ 2 {Oc~ia 1 (2.13) where: iRN0 = R(X=seec). (2.14) If R*(p) is positive for all, then we must have that f(w) is non-negative for all 0 < w < 1. Hence, in order for R(w) and X(w) to be the real and imaginary parts of a passive driving-point impedance for 0 < w < 1, Eq. (2.9) implies that the following inequality must be satisfied. _ 2 A.-wJ hXOL) d R ___1 XX(X) d-; O <w<_ 1. (2.15) Expanding the kernel of Eq. (2.13) in a power series: _ 1 L w2n 2n 1 2 2I i 1 2 2 Cos 2 WCos <1(2. 16) 2 2 C 2' 1 - w cos 2 n = 0 and substituting this expansion in Eq. (2. 13) we obtain

11 rZ~2n 2n f (=) 1 R*() z cos 2 d2. (2.17) nJO However, since the series in Eq. (2.17) converges within the limits of integration, we may integrate term by term and obtain: f() = 2nJ R*() cosn d. (2.18) n 0 0 Let us now expand f(u) in a power series about the origin f(to): _ K W:2 R*(0) cos d n 2 (2.19) Equating coefficients of like powers in Eq. (2.19) we obtain: K =1 R*(0) cosn d. (2.20) n 2 2 From Eq. (2.20) we note that if R':'(O) is positive for all 0, as it must be for a passive impedance, then: K > 0 for all n. (2.21) n 2n >2 2n+2 0 > Also note that since cosn cos 2 it follows that:.....> K >K >K >.... (2.22) n-1 n n+1 Finally then, we conclude that for the real and imaginary parts of an impedance given in a band [0, 1i to be the real and imaginary parts of a positive real function in that band we must require that:

f(w) > 0 or R(w) > J XX(); < < 1 " ~(X2_ 2) -7K > 0 for all n (2.23) n.. > K >K >K >... for all n. n-1 n n+1 Clearly conditions (2.23) are only necessary and not sufficient conditions. At this point we have made use of the fact that: R(w) > 0, O0 < < D Z(p) is real for p real Z(p) is analytic on the jw axis and in the right half plane. 2.3 Uniqueness of Continuations. If there exists another solution R. (O) to Eq. (2.20) then: = 21 [R(0)-Rc*( cos2n I d0 foralln> 0. (2.24) 2 Jo2 However, replacing R* (0) and R*(0) by their respective Fourier series it can be shown that both series have the same coefficients, implying that there is a unique solution to (2.20). Making in (2.24) the transformation n C n 2 7 B cos(n-m) (2.25) where: 2n am /2a\ 1

13 and replacing R'1'() and R:'(p) by their Fourier series one obtains: n n aO 0 nmB (A. - B.) cos j cos(n-m) d0. (2.26) M=O 0 j:O Integrating (2. 26), noting that the integral vanishes for j (n-m) n 1 (2n\ 0 2n (A -B ) ( or all n > 0 2n n-m n-m,oO which implies that A B. Q.E.D. n-m n-m 2.4 Solution for Continuations Making use of (2.25) Eq. (2.20) becomes: n r K - -' E Bn | R*(0) cos (n-m) 0 d. (2.27) m=0 However, since R* () is a continuous even function, we can expand it in a convergent Fourier cosine series as follows: R'(0) = A.cosjp; (2.28) j=0 A. = 2 R'(g) cos j 9 d9 - n0 Substituting (2.28) into (2.27) we obtain: n OB7r D Kn = B nm Aj c Cjcos (n-m) 0 d. (2.29) n=0 nm

14 Making use of the orthogonality relation: if j =n-m= 0 Cos (n-m) p cos j d r/2 if j = n-m 0O (2.30) o0 l if j n-m we obtain for (2.29): r ~n Kn = B A() cos (n-m) p d (2.31) 0 m=0 or, replacing B from (2.25) nmn n I' 2n ~2 K = A ) (2.32) 7r n 2n (n-m) In Equation (2.32) is then the recursion relation for the coefficients of the Fourier series (2.29) for R* (). Solving Eq. (2.32) for An we obtain K- 2n 2n A 22n 2 2( - (2.33) nr n mr=1 22n rn or n IT n m (n-M) In order to solve for the continuation of the imaginary part onto (1,00), we make use of Eq. (2. 10) which is reproduced here for convenience: rO ~g(w) J X2 w2) ~;~ ~0<w<1. 1(2.10) F1- )X

15 Letting X(A) = X E(X); where E(X) is an even function, Eq. (2.10) becomes: 00 XE(X) d. g(w). 0< 0w< 1. (2.35) Note that Eq. (2.35) has the same functional representation as does Eq. (2.9). Hence, we should expect to obtain a similar solution; i. e. 0D g(W) = H w2n (2.36) n=0 D0 E(x) = E*() = T cos m (2.37) m=0 and from (2.34): 2 2nn2> T = 2 Hn - Tm) (2.38) n m=1 where: 00 X*(p) = cos T cos m. (2.39) m=0 So far a functional representation for the real and imaginary parts of the driving point impedance in [0, 1 ] has been used. However, our goal is to be able to find the continuations from a graphical representation of the real and imaginary parts in [ 0, 1. To this we now turn our attention. 2.5 Solution From a Graphical Representation. From a graphical representation of the real and imaginary parts of a driving point impedance in [ 0, 1], we can always find a Fourier series for these parts by making use of the following change of variable: = cos 2 (2.40)

16 obtaining a) R(w) = R(cos) = R*(a) = B cos na n=O0 (2.41) for-r <a<O Similarly: X(a) = E(w) = E*(a) OD (2.42) = C cosa Z O Using the trigonometric identity: cos n a = o b2m co (2.43) where: 2n-1 A b =2 b 1 nln 00 b 1)n-m 2n n+m 2m- 1 b (...; m n am n-mfn \n-m- 1 we obtain for Eq. (2.41) OD n R* (a) = CB o am 2 (2.44) n O nm 2mO If the summation signs in (2.44) are inverted with the proper change of subscripts, we obtain: CD OD OD m' 2 am 2 R*() = a Bn cosos (2 45)

17 where 00 R = B b (2.46) m n nm n = m Using (2.40) in (2.45) we obtain: 00 2m R* (a) = R = R(o). (2.47) m=0 m This is a power series for R(w) about the origin. Expanding in a power series about the origin the term 00! ( 2j- l) CIWZ E~ 2j- ( 1!'(2.48) we obtain for the first term in the right hand side of equation (2.9) (i)) = m _ (2) (2.49) or: r R(w) 2m ZR (2j -1)" 2 ()2.' 2 tj M~ m=O j=o or: OD m TR( 7). (2mI -2 2 S 0 (m-J) (2j) (2.50) (2.51) We are now left with the task of finding a power series for the second term oven the right hand sid e of Eq. (2.9) so that by equat ing coefficients of like powers, the needed K's of equation (2.34) can be founde i.e. n 1 00 (2.51) However, Eq. (2.51) as it stands has a singular integrand. In order to removev 0. thi snglaity w mkeus ofthIfac tha adin or subtractn

integral* Since J T O2 = 0; O<< w 1 (2.52) we can write xX(X) - W X(w) Job I(A2-W2) e O< 2< 1~-. (2.53) Equation (2.53) has no singularities within the limits of integration. We may now use the substitution (2.42) as follows: nO W X(w) = E C cos na (2.54) 00 x(X) = C cos n n0n with the following change of variable: X = cos (2.55) Then (2. 53) becomes: Z7t1 E Cn (cos n - cos n a) n=0 n -CSos2 2a d. (2.56) 0 Cos 2co 2 Again, since the series of (2.56) converges within the limits of integration, we may integrate term by term and obtain * For a detailed proof of this, see appendix E. ** For a detailed proof of this, see appendix F.

19 (00 I 2 e C | Cos n 5 - cos n a da (2. 57) n=O C08os2 - cos082 2 2 2 However, using expression (2.43) we have: n Cos 2m I- COS 2m cos n - cos n a ( 2mc 2 2 cos os 1 (cos2 os 2 2 22 and: 2m r 2m a m.CO 2 2 2(i- 1) a 2(m-i) (2 cos -2 cos (2 59) 2 2a 2 2 COs - COs 2 2 Using (2.59) in (2.58) we obtain: n m cos n a1) CO 2(mn - coi) CO 2 - b nm2 2 2 (260) cos - cos 2 m=1 i = 1 2 2 Substituting (2.60) in (2.57) we obtain: 00 n n 1 2 COS~2(i-1) ca 2 (m-i) n =1 m n -i n 2 n n n n=i- 1) r2 - n= nn z2 i nm 2 [2 (m-i). n m =1n00 D0 n = I; 2i [ C b [2(im-i) -31 t 2 n + / (m - i)- 2n (-i)2 (2.62) 0 = 0 n =ni1 = where we have gone back to the variable o instead of cos - Comparing Eq. (2.62) with (2.51), we see that:

20 D n -- E n [2(m-i) (263) i n=+l n MC bnm 2(m-i) - 2]t (2.63 and we can write: C00 m D 00 2 W2m R(mj) j2i C m=0 jO (2j) 1 i=0 n=i+l n n (2.64) b (2(m-i) - 3n]'. K 2n Equating coefficients of like powers in (2.64) we obtain: = [IRill+ Z b p R(P) [L2jh + 1 C mp1 nm [2(m-p)-2]!1 (2.65) If we wish to obtain a similar expression for the imaginary part, we proceed as follows: From Eq. (2. 10) we have: l X(w) 7r X(w) r E(w) (2.66) n= 0 so that (2. 66) becomes: 7rX(w) _X2m 7' _m_ 2 Z ~" 2 (m-j) ( 20 1- __ 2 - AL mJ I mt n (2j ) (2.68) where: CD P =,> b Cb. -~ (2.69) m n nm2

21 Making use of Eq. (2. 52) we note that: R(X) - R(o) d R(X) dX 0< < 1 (2.70) and, carrying out a derivation similar to that in Eq. (2.57) to (2. 65) we obtain: p 09 n 2H =P - I)'2m-p)-3 -H p (p-j) (2j) — + Bn b 2(m- ) - ir. n = p+1 m p4 1 nm 2(m-p) -2 (2.71) With the aid of Eq. (2.71) we can use Eq. (2.38) to find the Fourier coefficients of the continuation to the imaginary part onto (1, (O). Let us at this point rewrite expressions (2.33) and (2.38) for convenience: A - 2 K - A() ( ) (2.33) n n n (n-m) ~~2 22nn T 2 22n H TonA. (2.38) n r n (n) m= m Note that from the graphical representation of the real and imaginary parts of a passive driving-point impedance function in (0, 1) we can always find R(pj)0 P(pj), B and C for any p,j and n. Hence, knowing (p-i)'. (p-)' n n these constants, we can always find the constants A and T which give n n the desired solution. 2.6 Extension to the Band Case. So far we have been using (0, 1) as the band of interest. If we wish to extend this method to the case where the band of interest is (Wl, w2) this can be done by first finding the continuations onto 0 < w < wl and then knowing the real and imaginary parts for 0 < w < t1 use the method described in section (2.4) to find the continuations onto (2 <_ w < 00 )

22 Let: 2 2 2o -W1 cos 2 (2.72) 2 2 2 W 2 - W1 so that: W= 2 2-2 cos2 + 1 (2.73) and: R(w) = R [(W2 -W) cos + = D cosn3 (2.74) i. e. we can always find a Fourier series which describes the behavior of the given real part in the band w1 w w2 However, we know from (2.43) that: 79 —' 2m cos n b Cos2 (2.75) m =O so that (2. 74) becomes: OCDO R(w) = os 2 b 2 nm n m=O n=m (2.76) D 2 2\ m D. 2 21 n Dn Z b D = m n=m bnm+I Dn

23 Thus, for Eq. (2.76) to converge, we require that: 2 2 W -1.2 2 < 1. (2.78) 2 2 2 - W1 However, we are interested in representing R(W) down to w = 0, so that we can apply the method outlined in the preceding pages. In order for (2. 76) to extend its radius of convergence all the way down to: = 0, we require that: 2W 2 W2 2(2.79) - 2 1 or: w2 > 22 - 2 (2.80) 1 Hence, for W = 0 we must have that: 2 2 > 2 (2.81) 1 If the band of interest is such that (2. 81) holds then we are assured that we can always find the behavior of the function R(W) all the way down to the origin. However, if this should not be the case, then we can find the behavior of R(W) down to some new frequency, say w3, such that W3 < Wl, and then use Eq. (2. 74) with Wl replaced by 3, and so on until we have the representation of the function R(w) down to the point w = 0. In order to find the behavior of the imaginary part for 0 < w < 1 we use a technique similar to that used for finding R(w). However, in this case we let: X(w) =E (W) (2.82) where E(w) is an even function of W.

24 Then using the change of variables (2.72) we obtain: 0O 21'. 2 C2I E(t) = E (2 - ) cos 2+ = C cos. (2.83) Using (2.75) in (2.83) and inverting the summation signs we obtain: (X 2 2 o E(w) = L, 2 b C (2.84) n 2nm m= O W2 - a1 n=m or: OD 2 2 02 X(w) =[ (-2 E b (2.85) 2 2 a nm m=O ~2 - W1 n =m Note that (2.85) is similar to (2.76). Hence, we can conclude that in order for (2.85),to extend its radius of convergence all the way down to W = O. we require that: 2 > 2. (2.81) The statements made in regards to R(M), following Eq. (2.81) also hold for X(w). Hence, in practice, we only need to know the behavior of the real and imaginary parts of the driving point impedance to a passive network in a finite range of frequencies in order to find its behavior outside this band. Up to now, we have been using infinite series without being concerned with the actual solution to a practical problem. However, if we wish to solve a problem using the method outlined in the preceding pages, we must truncate the infinite series. This truncation introduces an error, for which we can obtain a measure as follows:

25 2.7 Measure of Truncation Error. From Eq. (2.70) we can see that truncating the power series for f(i) is equivalent to truncating the infinite series in the right hand side of (2. 70). Hence, we can define a measure of the error as: N f(w) - K wn (2.86) this presents some difficulties since f(w) is singular at w = 1. In order to remove this singularity, we can multiply f(w) by the factor 1 - and obtain: F(w) = 1- f () (2.87) or: N E(w) F(w)- - K n (2.88) n=0 n We have shown, in preceding pages, that: f(o) > 0; 0< < 1 (2.89) and K > 0; for alln. (2.90) n Then we note that: F (w) = 1-2 f(w)> 0 for 0<O< w. (2.91) Since in (2. 88) we are only taking the first N terms of the power series for f(O) (all the Kn's are positive) we can conclude that: E ()> 0 for 0< w < 1 And then a measure of the normalized error becomes:

26 N E(w) dw K (2n-1) (2n-3)... (3) (1) 2n = n (2n2) (2n)....(4) (2) E = - = 1-.... - -... (2.92) F (w) dw F() dw This is not an accurate measure of the error. However, it gives some measure of the error incurred in the truncation of the infinite series. An exact expression for the truncation error can not be obtained since we do not know anything about R*(w); in addition it appears only under an integral sign. Therefore, the usual least square error technique cannot be applied here. 2.8 Extension to Transfer Impedances. The method described in section 2.4 readily extends to the case of a transfer impedance function. We note that if the function in question has no poles in the right half plane or the iw axis of the complex plane, then we may use the Hilbert Transforms of Eqs. (2.9) and (2.10). These equations require no restrictions on the location of the zeros of the transfer impedance function. Thus, it follows that the method derived in section 2.4 is indeed applicable to the case of a transfer impedance function if and only if this function has no poles on the jw axis or the right half plane. Uniqueness will be achieved only up to an additive constant, since the concept of a minimum resistance function is not meaningful for a transfer impedance. Since there are no restrictions on the behavior of the real part of a transfer impedance function, we can not state any conditions, necessary and/or sufficient, for the compatibility of the given parts. Hence, we must obtain the continuations of the given parts in order to find out whether they are compatible or not; i.e. whether the Fourier series obtained converge or diverge.

Chapter Three NUMERICAL COMPUTATIONS 3.1 Transformation of Equations Several examples have been used to verify the method outlined in the preceeding pages. Functions whose behavior was known outside the band of interest were used as the impedance functions and the results obtained were excellent. A computer program was written to solve for the unknown parts using 4N + 5 points as data; i. e. R(w) and X (M) were given at 4N + 5 different frequencies equally distributed along the band of interest. The program gives as results the first N + 1 Fourier coefficients for R*(0) and X*(0). The reason why we only obtain N + 1 coefficients from 4N + 5 points relates'to the nature of the subroutine used to obtain the Fourier coefficients of R (w) and wX (w) in the band (0, 1). From 4N + 5 points we can obtain only the first 2N + 3 Fourier coefficients for R (w) and X () in the band (0, 1). It will become apparent that this leads to N + 1 coefficients for R*(0) and cos X*(). For convenience we reproduce here the expression for the coefficients of the power series for f(w) about the origin. k 0D n 2 2KR(k) (2n I1)'! + C bn 2(j -k)3]:k 0 (k n) (2n) " n 1 nj 2(j-k) - 21 " K=n=0(n! n k+l =k+ (3.1) where 00 R i b (3.2) i j Looking at the second term in the right hand side we note that b varies as 22 whereas the double factorial varies as 1/22j nj Hence, we would like to remove this term from both b. and the double factorial. 27

28 In order to do that we note that: b 22n- 1 j * 0 nj b- jn -; 2 n+j- 1 2j-1i;n ( A1 n-jO b - 1; n = j = O3) also, L,-3! = 1 ((3.k4) 2(j -k)-2i!! 22(j-k)-2 \ -k 1 Thus we have: b 2(j -k)- 3! 22n-1 1 1 (+ j= nj 2(j-k)-2 2 jn-k(3. 5) n-1 note that when j — N j=k+l n-j 1 2 j -k- 1 (-22k-1 1 22k+1 /2(n-k-1) n-l an-k- 1 j =k+l -k- 1 (3.6) Referring to Eq. (2. 1) we note that it contains two infinite series. N terms. In order to do this and maintain a negligible truncation error we must-use the first N terms of each of the infinite series. However, note that when j - N

29 N RN ib iN- BNN N (3. 7) Hence, we are only using one term to compute RN, thereby making the error not negligible. To be able to truncate the infinite series and maintain the truncation error i ngligible we make; in (3.2) the following change of subscripts 00 R ZB b(3.8) 1Rj = Bi + j (i + j)j (3.8) and using (3.6) in (3. 1): k 2 4 (2n-1)" -K + T (3.9) 7r k 0 (k-n) (2n)! k n=0 ~2k00 n+k n+k+j 2j+2k E;:2 2k+ 1 + ( - )n+k+ j 2(n+k+ l) Tk= ZCn+k+l K + +k+l-j n+k-j ( j-k-1j n=0 j=k+l where: (3.10) Tk C 2k+ + (_l)n+k+l-j 2(n+k+l) 2(-k-l) n=0 j =k+1 n+k+lj n+k - - -k - 1 (3. 11) Making another change of subscripts in (3. 11) nT 0 Cntk-l [2 +kn+1 + 2k+l) n=O n j:O n-j-1 j (3. 12) Now we can truncate Eq. (3. 10), using (3. 8) and (3. 12) to obtain:

_K k R + Z 22k+1 Tr Kk= 3 (k-n) (2n! A n+k+ 1 n=0 n=0 + (l)n- 2(n+k+ 1 2(3.13) n n-Ill where as before: R B(i + j) b(i + j) (3 14) i=0 3. 2 First Example For the first example the pair of functions used were: 1 -to R (w)2 X() 2 (3. 15) 2 2 1 + 1 +W For this case N -~ 4 was used. The coefficients for the power series of the function t(w) are then: n K n 0 0.292 893 2 1 0. 207 106 5 2 0. 167 894 5 (3.16) 3 0.144 602 9 4 0. 128 839 3 Note that as predicted earlier, the Kn's are all positive and monotonically decreasing, and the necessary conditions for compatibility are satisfied. For this case the A's Fourier coefficients for the unknown n R *()) are found to be:

31 n A 0 0.292 893 2 1 0.242 639 6 2 -0.041 639 0 (3. 17) 3 0.006 761 5 4 0. 003 385 6 Figure (3-1) shows a plot; of R*(0) as compared with R(w) given in Eq. (3. 15) (where w = sec 2 ). For this case it is not easy to distinguish the points of the exact expression (3. 15) and those from the Fourier series. Therefore a comparison of both points is given in Table 3. 1 below. The points obtained from the exact expression are labeled R(0) and those from the Fourier series R*(0). (only for 0 < 0 _ r). 0 R(0) R*(0) 0 0.5 0. 504 073 9 /r /4 0. 460 49 0. 459 658 9 r /2 0. 333 333 0. 337 884 8 3-r /4 0. 122 0. 122 749 9 X7r 0. 000 0. 005 271 7 Table 3. 1: Comparison Between Points from the Exact Expression and those from the Fourier Series Solution for Example One. From this table we see that there is good agreement between the exact expression for R(0) and its Fourier series, obtained by the method described in the preceeding pages. As a matter of fact the agreement is good up to the third significant figure.

0., -.5040 - 0.8 w=sec 4 2 4 4 2 4 FIG. 3-1' PLOT OF R(w) FOR EXAMPLE NUMBER ONE. POINTS FROM EXACT EXPRESSION, APOINTS FROM FOURIER SERIESSOLUTION. EXPRESSION, A POINTS FROM FOURIER SERIES SOLUTION.

33 3. 3 Second Example As a second example the following functions were used: R() = (w2- 4)2 7w - 68w ~;R (w) (W) (3.18) (1+ w2)(4 + w) 3(1+ 2)(4 + w2) N = 4 was used for this case and the K's obtained were: n n K n 0 0.492 582 8 1 0.296 269 6 2 0.256 525 3 (3. 19) 3 0. 230 256 2 4 0.210 581 7 Once again we see that the K's are all positive and monotonically decreasing. As expected, the necessary conditions are satisfied. For this example the Fourier coefficients (A ) corresponding to n the real part R*(0) are: n A 0 0.492 582 8 1 0. 199 912 6 2 0.349 257 5 (3.20) 3 0.209 492 8 4 0. 129 740 8 Figure (3-2) shows a plot of R *(0) as compared with the exact expression. Once again we observe that, the points obtained using the Fourier series technique are almost indistinguishable from those of the exact expression. Therefore, a comparison of both points is given in Table 3.2 below.

O 0< <1 WCos 2 0 \ 1< _< 00 W =sec 3 1 t \O ~~~~~~~~-20.962 0.981 116 Rt~~~~~~~~~~~ Os Ha 0.960.273 0661 0.652 291 _' 0.073 392 0 I -0.056365 3 1 _r _ r 3-?0 I 4 2 4 4 2 4 FIG. 3-2: PLOT OF R(w) FOR EXAMPLE NUMBER TWO. ) POINTS FROM EXACT EXPRESSION, A POINTS FROM FOURIER SERIES SOLUTION.

35 0 ~R(0) | R*(0) 0 0.900 0.962 001 0 ir/4 0. 716 0. 652 241 6 * /2 0.222 222 2 0.273 066 1 37r /4 0. 094 379 0. 073 347 3 Xr 1.000 000 0.981 116 1 Table 3.2: Comparison between Points from the Exact Expression and those from the Fourier Series Solution for Example tw o. Once more we see that the Fourier series technique does indeed give a solution which agrees with the exact solution. Here however, we note that the discrepancy occurs in the second significant figure. This is due to the fact that for this example N = 4 is not enough; i.e. more terms of the Fourier series are required. 3.4 Third Example: As a third example we take the transfer impedance function whose real and imaginary parts are given by: 6 -w 2- 5w R(w) 6= 4 2 X() = 4 2 (3.21) w + 13w +36 w +13w +36 For this case N = 5 was used. The K's obtained were: n n K n 0 0.048 469 12 1 0.039 559 93 2 0.032 979 22 (3.22) 3 0.028 658 62 4 0.025 627 36 5 0.023 370 66

36 and the Fourier coefficients corresponding to the unknown real part R *() are: n A n 0 0. 048 469 1 1 0.061 301 47 2 -0.008 353 18 (3.23) 3 -0.004 633 85 4 0. 005 843 25 5 -0.003 502 56 Figure (3-3) shows a plot of the real part R(0) for the range 0 <: < r (1 < w < 00). Note that the points from the exact expression and those from the Fourier series (using six terms) are almost indistinguishable. Again a comparison of both sets of points is given below in table 3-3. R()R*( 0 0. 100 0.099 124 25 r /4 0.091 816 0.091 726 10 r /2 0.0606 0.062 665 55 3r /4 -0.001 102 -0.001 514 09 Vr 0.000 -0.007 205 87 Table 3-3: Comparison Between Points from the Exact Expression and those from the Fourier Series Solution for Example Three. From this table we see that the Fourier series technique gives a solution which indeed agrees with the exact expression, for the continuation. Once again we observe that the discrepancy occurs in the third significant figure.

lI I I I I I I I O< < t< 2 COS: -0.15 -02 0.15 1 < w < 00 w sec 2 0.1 O w~s__. 0.099 124 0.062 665 0.05 0 R(0) -0.001 514 -0. -0.05 - 37r 7 X X-r 0 7 3 1 4 2 4 4 2 4 FIG. 3-3: PLOT OF R(w) FOR EXAMPLE NUMBER THREE. 0 POINTS FROM EXACT EXPRESSION, A POINTS FROM FOURIER SERIES SOLUTION.

38 3. 5 Fourth Example As a fourth example the following transfer impedance function was used. 24 - 9 w2 R(w) 2 4 6 576 + 244 + 29w + w and (3.24) 3 X(w) - 26 w 576 + 244 w + 29 w + w6 For this case N = 4 was used. The K's obtained were n n K 0 0.004 955 57 1 0.005 691 39 2 0.005 180 53 (3.25) 3. 0. 004 671 78 4 0.004 261 05 and the Fourier coefficients corresponding to the unknown real part R *() were: n A n 0 0. 004 955 57 1 0.012 854 42 2 0. 001 737 44 (3.26) 3 -0.003 358 14 4 0.002 308 59 Figure (3-4) shows a plot of the real part R(p) for the range 0 < p < r (1 < W < 00). Note that the points from the exact expression and those from the Fourier series (using the first five terms) are almost

I I I I ~~~~o<W<I I Co I 0,~~~~~~~~~~~~~~~~~~~~~~ 1 < w < 00 WCOSe 2 0.03 Nc 0. 018 497 0.02 0.014 111 0. 01 0. 005 526 -6 0 R( ) -0.000 494 -0. 008 817 0.01 -T 2 70 7 37T 4 2 4 4 2 4 FIG. 3-4: PLOT OF R(w) FOR EXAMPLE NUMBER FOUR. 0 POINTS FROM THE EXACT EXPRESSION, A POINTS FROM FOURIER SERIES SOLUTION.

indistinguishable. Again a comparison of both sets of points is given below in Table 3-4: 0 RR(B) RR*() 0 0.017 647 06 0.018 497 89 r /4 0.014 923 36 0.014 111 04 r /2 0.005 050 50 0.005 526 72 37 /4 -0.009 458 00 -0.008 817 08 w 0.000 -0.000 494 68 Table 3-4:. Comparison Between Points from the Exact Expression and those from the Fourier Series Solution for Example Four. From this table we see that the Fourier series technique gives a solution which indeed agrees with the exact expression, for the continuation.

Chapter Four COMMENTS AND SUGGESTIONS FOR FURTHER WORK 4.1 Minimum Amount of Information Required to Specify an Impedance Function. 6 Guillemin has shown that knowing the real (or imaginary) part of a positive real function, for all frequencies, the imaginary (or real) part is completely specified for all frequencies. Further, he showed that knowing the real (or imaginary) part in a band of frequencies, and the imaginary (or real) part in the rest of the spectrum, the unknown parts are completely specified. That is, it is necessary to have some information for all frequencies. However, in this report, it has been shown that the minimum amount of information which is sufficient to completely and uniquely specify the behavior of a passive driving-point impedance function or a transfer impedance function (assuming these functions have no poles in the right-half plane or on the jw axis) is the knowledge of the function's behavior for a finite (no matter how narrow) band of frequencies. Thus, information is only required in a band. In most problems we have a complete description of the impedance function in a band of interest and we are interested in obtaining its behavior outside this band of interest. However, there is no assurance that there exists an impedance function whose real and imaginary parts, in the band of interest, coincide with the given ones. If such a function exists, then it is sufficient to know its behavior in the band of interest. 4.2 Uses of This Method. In problems similar to that of impedance loading, the impedance is known in a band of interest, leaving its continuations completely unspecified. These continuations may be desired to either synthesize the impedance or to calculate something else about the device in question. Thus, one can use the method described in this report to obtain the desired continuations. 41

42 In other problems the behavior of the impedance function may be measured in only a band, and we may want to know its behavior outside of this band. A typical example of this type of problems may be that of trying to find the transfer characteristics of a transistor. As frequency increases it becomes more difficult to measure the transistor characteristics. Hence, this method may be used to compute the behavior of these characteristics, beyond the frequency where it becomes difficult to measure them. In general, then whenever we know the behavior of the driving-point impedance function, or the transfer impedance function, of a stable system (no poles in the right-half plane or on the jw axis) for a range of frequencies, its behavior for all frequencies is completely determined and may be obtained by the method described in this report. 4.3 Problems in Computation. There are three main problems in computing the impedance continuations, using the method described in this report. These arise mainly because in Eqs. (2.76), (2. 33), (2. 82) and (2.38) we multiply large numbers by small numbers to obtain small numbers. These small numbers are added or subtracted to obtain other small numbers. As can be seen from Eqs. (2.33) and (2.38) the first error in computation derives from the fact that two large numbers n 2n (2n 2n Kn and A(nm) are subtracted to obtain a small number (An). Thus, any slight error in 2~~n~~n the given data (Kn being obtained from the given data) will be enhanced by the factor 2. Data with an error of 5 percent to 10 percent was used and the continuations obtained were completely erroneous. Therefore, great care should be taken in using accurate data. The second source of error in computation arises when the radius of convergence of the power series for R(u) about the origin is less than one. If this should be the case, the coefficients of the power series will not necessarily

43 be monotonically decreasing; indeed, they may be monotonically increasing. Therefore, if the coefficients of this power series are Rn, the first term in the right-hand side of Eq. (2. 76) p R (]-1(4.1) (p_-J) (2j) (4.1) will increase as p increases. However, the second term in the righthand side of Eq. (2.76) (an infinite series), which is truncated for computation purposes, will not increase at the same rate as does the first term; thereby introducing an error. Moreover, when the Kn obtained from (2.76) (containing the error) is substituted into Eq. (2.33) the previously incurred 2n error will be enhanced by the factor 2 Accordingly, if the given impedance function has one or more poles within the left-hand unit semicircle, the computation error incurred may make the continuations obtained completely erroneous. The third source of error comes from the fact that to adequately represent a function with sharp variations by its Fourier series requires more than the first six or ten terms. However, if more than say the first ten terms of the Fourier series are required to represent the function R*(0), the computation error existing in the K's obtained will be enhanced by the factor 2 as can be seen from Eq. (2.33). (Note that for n = 10; 2 e- 10 ). For this reason the method described in this thesis is limited to functions having no sharp variations. It has been brought to the author's attention that the problems in computation may have arisen from expanding both sides of Eq. (2.9) in a power series about the origin. To circumvent some or all of the problems Calahan has suggested using, for the expansion of both sides of Eq. (2.9), a series with a different (larger) radius of convergence. Two expansions which show some promise are Chebyshev polynomials and Fourier series. It will be shown, however that both of these expansions lead to the same result.

44 A Chebyshev polynomial is defined* as: T (w) = cosn(cos ); 0<w< 1. (4.2) Using the change of variable: 0 0< W I cos 2 w for (4.3) 0< o< r one obtains for the Chebyshev polynomials: T (=cos ) = cos (4.4) n 2 2 Thus, using the change of variable (4.3) in (2.9): f (w = cos 9) I'(d (4.5) (u:cos 2_s -: Making once again the change of variable X = sec 2 (4.6) in (4.5), f(o) = R,(P2) d0 (4.7) 2 2- 2' 0 - cos cos 2 Note that f(o) is an even function of frequency, and if expanded in a series of Chebyshev polynomials one obtains: K cos n 1 R*() d2 (4.8) 0 - 2 2 8 2 ~ J n 2 2 * Weinberg, L.,"Network Analysis and Synthesis, McGraw-Hill (1962), pp. 448.

45 However, note that the expansion for f(0) in a series of Chebyshev polynomials is equivalent to the Fourier series expansion of this function. Therefore, we may conclude that expanding both sides of Eq. (2.9) in a series of Chebyshev polynomials is equivalent to a Fourier series expansion of these terms. Expanding the kernel of Eq. (4.8) in a Fourier series of two variables: 2 0 2 T i Tncosn0 Cos m (4.9) 1-cos 2cos 2 n=O m=O where: nmT =4 J cos n 0 cos m 0 dO dp (4.10) O 1- cos 2 cos 2 and substituting (4. 9) in (4. 8) one obtains: 0D 00 0O Z Kn cos = cos n e Tnm 2 R*(N) cos m 0 d 0 n= n=O m=0 n n = 0 n 0 M=O (4.11) or: K ZT nm R*(0)cos m d0. (4.12) Replacing R*(p) by its Fourier series, (4.12) becomes: K = T A. cos. cos m j d (4.13) n Tnm

46 Integrating (4.13), OD K - A +T -A (4.14) n m - nm 4 m no 2 o or: K0 T T 2A TO00 01 ON 0 K1 T10 T11'iN A1I 4.. (4.15) KN TN TN1 TNN N Consequently, we have replaced the problem of solving an integral equation by that of solving an infinite system of equations. As far as the computation problems are concerned, we have gained very little or nothing at all since we are left with problems of different nature which may or may not be easier to circumvent. These problems are: (i) We don't know if there exists an inverse matrix to the system of equations. (ii) If the inverse matrix exists, we can only truncate the system and obtain an approximate solution. (iii) The coefficients T must be obtained numerically. nm (iv) The function f(w = cos 0/2) must be expanded in a Fourier series. (v) If more terms in the Fourier series for R*(0) are required, an enlarged version of the truncated system of equations must be solved. However, it should be noted that even though using a Fourier series expansion the gain may have been little or nothing, there may very well be another complete set of orthogonal functions which does indeed simplify, or at least reduce, the problems in computation. Therefore, it is suggested that further work should be in this direction, that is, finding other suitable

47 complete sets of orthogonal functions which would reduce the degree of the computational problems. Another approach which may be of interest for further work deals with the fact that the method derived in this thesis began from the Hilbert Transforms (2.9) and (2. 10). However, as pointed out by BodeS, additional relationships between the real and imaginary parts of a positive real function may be obtained by performing a contour integration (including the whole right-half plane) on different functions of the impedance Z (p). A number of these relations have been obtained and are given by Bode.

THE UNIVERSITY OF MICHIGAN ACKNOWLEDGMENTS The authors wish to express their appreciation to Mr. Peter Wilcox for computer programming and coumptations and to Mr. Gus Antones for technical illustrations. 48

Chapter Five APPENDIX A. Main Computer Program. This program gives the desired coefficients of the Fourier series for R*(O), from the knowledge of the real and imaginary parts for 0 < w < 1. The data is given as the Fourier coefficients (B and C ) for the real and imaginary parts in the range 0 < w < 1; or - r < 0. O The result obtained is the first N + 1 Fourier coefficients for R*(O), 0< r< ff (corresponding to 1< w < (a)). 49

50 G COMPI LE R. AI 1....... 04-23-68 15..5045 IIPLICIT REAL*8 (A-F,C-Z) RFAL*8 K(25),.(25) e(45s c (45 vR(90o,X(90) REAL*8 8B(4 5,25 )Y(25) Z1(451,BC1(25) C TA / 3.4i5925358979/ CALL FCV1HB(l) 10 PRI T S9q SS FCRPAT (' 1 ) C GTE** E R gX ANG ICIAP CALL RXFUN (R,X,CAPNBC, t500 NEt = NBC41 Ne2 = NBl NE3 = NPIC4t.1 ASEP = PI/hE3 NNI'C = N83 4 2 NP1 = NCtP+l NC2 c -NCP -- hRITE (q,9CC3) ( I,R(I ),X(I ), I=lNBl) 1003 FORVA'T ("- T'6X'R(I)'lXX(I/IX"/(15'2G'15'.7 DC 100 l=hE2,833 X(I) = XldDC-I) * (-DCCS(ASEP*( I-1)))!Y C X ( N-N'C-' -I' "X-Cit C *** GET e ANC C C L L FCRITI (R'h'NeC,KNeC,N1,z I:R IF (IFR.EC.0) CC TC 110 105 PR INT lP01 t[ER 1001 FOPMAT ('- *** IER ='15) GC TC IC 110 PRINT 1002, (,B( I )Z( I),I=1,NB1) 1002 FCRIV6T ('- I'6X'I e()111X'( I1)'/lX/( 15,2G15.7)) CALL FCRI (X,r\EC, BCC, Z,I ER) iF (IEP.NhE.C ) CC'TO 105 PPI1T 10049 (I,C(Il),Z(I),1=tNB1I 1C04 FCRM-A- (- - ---'6X'C( I'I1X'Z I I) /1X/( 15,2Gl5*71 C *** CALCULATE EE ANC Y IF NEEDED DC 150 h=lKNP T FaC = 4.CCE*TFAC Re(N,N) I FAC I iF (b;J'Eo- B-ee ( t I,= 1.00 IS = h+1 F[I = -TFAC*(N+N) FN = N-1 DC 150 I=IS thl BE(I,N) = F11 FI = I-1 150 Fl = -FII* ( FI+F)*(FI+1 DO) / (IFI-FN DO ) *F I Y(1) = loCe EC1(1) = 1.CO CC 160 1=1d vP FIi = I F1 = F II4F I1. C1{I+41) = BCI(1 )*(FI+FI-2.DO)/FIl WR I E 69,1009 )

51 G COMPILER PAIN 04-23-68 15250.45 1009 FCRMIT ('- O3 8 DC....170 ri-l= -hPI........... 170 kR IF (Sl OC6e) t,(BB(IN)t. IlN.NB1) 1-i006 FCRPMIAT (Id'-I.4Gt5.T-/(6X,4G,5.4 7... WRITE (9,1020) (lYII)tBCl (1),=ll1lNPl) 1020 FURPAT (I-."I'6Y( I )"tlOX'iBC1( I)'iX/l ( 159ZGL S.71'C *** CALCULATE F(N) 200 CC 250 N=-l,N'PI R(NI = O.CC - -r-c — 2 5'F- l....... 2.0~ R(N) - R(hI) + BeI(+!K-INI*B(IN-I) PRINT']10OO9 I(9R I'R1i=loNPi) I. 1005 FCRMAT ('- I'6X'R(l /XIX/ (IG15tGl7 -C A'L'CL-AT'E -— ~ - TFAC.SODC dC~ 3iC- i-.-,t2 TFAC = TFAC*4.DC K(') = O.OC FN = N-I F TN = FN-+FN CC 310 I1=1N -i0 (K-NA I ( I + R - l * SI = 0.00 TFIJ =.CC cc 325 j-I NPTFJ = JF1 = C.CO IF (J.EC.I! CC TC 320.IS = TIF*(FJ+FN4+.DO)/(FJ+IoDO)*TIJ FJN = FJ4FTN+2.CO DC 315 I=1,JdI1 FI = I-1 F I = FI+F I+1.00 FJI = FJ-FI TIS = IIS*(FJI+.DC)/FJI S2 -= S2 + 1IS*BC1( ) T IS = -TIS*((FJN+ FI)*(FJI-l..bOI/((FIN+FII+2.00) 6 *(FTN4F I 13.DO)! 315 CCNhINIUE TIJ = I7J*FJN/FJ 320 Sl = SI + CIJ+N)*(BC(IJ)+S2) TIF = -IF 325 CCITINUE K(N) = K(1 ) 4 TFAC*S] 330 CCCNTINUE PR INT ICC1, ( I, K {I t I",I=,N'02) 1007 FCRMAT ('- I'6X'K(1)'/lX/(15,G15.7)):C: **' C-/i-LC-UL aTE A (N? TFAC I.CC SL'P = O. JS = W-1

52 G CCOPILElR PAtN 04-23-68 15:50.45 F 2 JS+JS F' 1-1 —'~ —i V / TFFAC nC 41C J=lJS SL - SUM 4F11*({- J I FJ = J 410 ril a FII*(FK-FJ)/(FJ+1.CCO 415 A(h) = TFAC*(K(h)-SUP) 450 TFAC.F aC*4.OC PRINT 100E, ( 1,6 1A I,I=1,N02) 1008 F CRPATA' ( i —' /i'x'i Y-,' i T5.X T(i C *** DC PART 4 IF CESIREC AN ENC CF FILE hILL SKIP "1 PRIKT 101C C —T-O F C'RIA-'- (' —P —- 4?'- - READ (,IClltEND=101 Z(l) ICli FCRMIYT (f8 PR. IT 1012 1012 FCPMAT ('-'5X'P'I'IlOX'R*(PHI I1X'/ ) OC 600 I=1,2C I = l.0O/I PFI = 2.00*DARCCS () SLp = O.CO CC 61C K=1,hC2 EIC SUM = SUM + AI(N *DCCS( (N-1 )*PHI 60C PPThT 1013, PP1I,SUM C 13 FCR6 T (2G15*7) GG TG 10 500 CALL SYSTET EN.

53 B. Subroutine Forit. The following subroutine obtains the Fourier coefficients B and n Cn of Eqs. (2.52) and (2.53). The data is given as the value of the real (R( ))and imaginary (X())parts, for 0< w_< 1, (-r < 0< O) at 4N + 5 equally spaced frequencies in the 0 range. As a result one obtains the first 2N + 3 Fourier coefficients for R(0) and cos I X().

54 MPILER FOR I 04-23-68 15: 51.13 PAGE 0001 SLBROUTINE FCRI{(FIT,,MA,,BtIER) IMPLICIT REAL*8 (A-H,O-Z) C IPENSICh A(ll,e(ll),FNT l FORIT COtPLTES FCURIER COEFICIENTS FROM A TABULATED SET OF DATA POINTS TI-ESE FUNCTICK VALUES ARE IN'FN1 AND FNT(I) MUST HAVE THE VALUE AT 2*PI *( 1-1 )/(21 +1), 11 2.t...,2*N+1'M' IS TFE CESIRED OROER OF THE COEFICIENTSt SO M+I ARE RETURNED IN'A' ANC'P." (B(1) - Ot ALWAYS) FNI'A' AND e' ARE ALL REAL*8'IER' FAS RPETURN COCE: C - ACK 1 -'N' LESS 1FEN'M' 2 - LESS THAN ZERO IER = 2 IF (i.LT. ) RETURN IER=1 IF (N.LT.Fw) RETURN IfR=O NC = N+N+1 CCEF = 2.CC/NC CCNST = 3.1415926535e979*COEF.S1 = CSIN(CCNSI) CI = CCCS(CUNST) C = 1.CO S = c.CO J — I 1 FNTZ = FNT(1) 0(1) = C.OC 70 U2 = C.CC Ut = COCO I = NC U73Z = F1T(I +)2.CO*C*UI-U2 Ij2 = LI Ul = LZ iF (I.CT.I) GC TC 75 A(J) = CCEF*(FNIZ+C*UI-U2) P.(J) = CCEF*S*U. IF (J.GT.P) GC TC 100 0 = C1*C - S1*S S - C1*S 4 SI*C C = 0 GC I-C 70 100 = A(1) A(l)*.500 END L3II-lr~ — ^ _ _ r _ _

55 C. Subroutine to go From The w Domain to the 0 Domain. The following subroutine transforms any function from the w - domain (O < w < 1) to the - domain (- ir < 0 < 0) by the transformation W = 082 It also obtains 4N + 5 points of the function equally spaced in the 0- domain.

56 G COMPILER PXFUN 04-23-68 15152.07 SUBROUTINE RXFUN(R, XNCAP,C NBC IPPLICIT REAL*8 (A-H,a-Z) DIMENSION R(SC),X(90) CATA P I/3.14 1 926.53 e97tq/ 10 PPINT 1000 1COO0 FCR'PAI (I NCAP "' REAC (191CO1,ENC=500) NCAP ICOl FCRM -AT (15) IF (NCAP.EC.C) GC t0 10 PR INT 100C2,CAP 1002 FCRIAT ('-hCAP = 15) NCAP = hCAF+KCAP+2 NIC = _CAF+KCAP NP2 = h C 4 1 ASEP = P1/(NP2+NBC) DC 100 I=1,KP2 CP = CCS ( SFP*(I-1) ) CP2 CP*CP. R(I) = 1.CO - CP/2.1CO X(I) = (DLCG((2.0C-CP)/(2.DO+CP)) + CP*oDLOG((4.00-CP2)/CP2)/2.DO)/PI 100 CTNIINUE RETURN 500- RETTURN 1'

57 D. Subroutine Used to Read Points From Subroutine C Into the Subroutine Forit.

58 1 COMPILER RXFUN 04-23-68 15:5z2.26 SUBROUTINE RXFU( RX,NCAPNC ) IPPLICIT REAL*8 (A-I,C-Z) DIMENSION P(qo90)(90) OP TA PI /3. o 1 rq 19 2 e 5 5 e 9 7 9 10 PRINT 1000 1000 F(CRMAI (I NCAP?') R EAC (19lCC1,ENO=500) NCAP 1001 FCRMA, (15) IF (NCAP.EC.C) GC tC 10 PRINT 1002,NCAP IC02 FCRTAI ('-ICAP ='15) NCAP = NCAP-NCAP+2 ec = NCA.F+NCAP NF2 = NBC + I ASEP = PI/(NP2+NBC) AGO = ASEP*360./PI PPRINT 1003, NP2,AGC 1003 FCRIM.AT ('-'ENIER'I2'"'R AND X POINTS SEPAR~ATECI3' /' BY'G1o.7' DEGREES') REAC (1,lCC4 (R(I),X l)t=l P2 1004 FORMAT (2GIC.0O) RETURN SoC RETIRN A1 ENDO

59 E. To show that: J..2 )Z =_0 for O <w<1 (. 1) 2 2) J (A -w ) make the following change of variable: 2 2 _= x (E. 2) 2 1 +x CdL = dx (E.3) y1J-X- (1 + x2 So that (E. 1) becomes: dx 0_ (E.4) J0 (1 - 2)x 2 Note that (E. 1) must be evaluated in the Cauchy principal value sense; i.e. dx = L dx + J dx (E. 5) 0 o o +E So that (E.4) becomes: 0 imdx + d (E. 6) E-L0

60 Integrating (E. 4) one obtains: 0 2 Ln -- -... (E. 7) 1W2 2w' 1 -w I, x+ 1_ ~'? Making use of (E. 6) one obtains for (E. 7): -- V1 -0 1 ur0 Ln (1)0 _ _ _ (+E W t -+(w+E) 2-w ~( 1) I (E.8) +E W E when the limit is taken one obtains: 0 0. Q. E. D.

61 F. To show that equation (2. 64) has no singularities we make use of the fact that X(X) is the imaginary part of a function analytic in the righthalf plane, and hence has a power series about the origin. Thus: 00 XX(X) = X X k2n (F.1) n n=l D00 n2 X(w) = X W (F.2) n=1 So that: 0D X X(X) -wX(~) = X (k2 - n n) (F.3) n=1 n Equation (2.64) then becomes: Xn( -2n )n) dX Jo ns we (F. 4) (X 2 2 But: 2n 2n n 2 2 w x2 2(i-1) 2(m-i) (F.5) X - i=l then: 1 3" Xn X 2(i-1) 2(m-i)d (F. 6) nh1 hsn g-a i=1 which has no singularities Q. E. D.

RE FERENCES 1 Harrington, R.F. (1964) "Theory of Loaded Scatterers," Proc. IEEE (London) III, pp. 617-623. 2 Chen, K-M, and V.V., Liepa (1964) "Minimization of Backscattering of a Cylinder by Central Loading," IEEE Trans. Ant. and Prop AP-12, pp. 576-582. 3 Liepa, V.V. and T.B.A. Senior (1964) "Modification to the Scattering Behavior of a Sphere by Reactive Loading, " The Univeristy of Michigan Radiation Laboratory Report No. 5548-2-T, A FCRL-64-915. 4 Chang, S., and T.B.A. Senior (1967) "Study of the Scattering Behavior of a Sphere with an Arbitrary Placed Circumferential Slot," The University of Michigan Radiation Laboratory Report No. 5548-6-T, AFCRL-67-0111. 5 Bode, H.W. (1945) Network Analysis and Feedback Amplifier Design, ~ New York, Van Nostrand. 6 Guillemin, E.A. (1949) The Mathematics of Circuit Analysis, Cambridge Technology Press, New York. 7 Calahan, D. A. Private communication. 8 Redheffer, R.M. (1949), "Design of a Circuit to Approximate a Prescribed Amplitude and Phase," J. of Math. Phys; 28, pp. 140-147. 9 Zeheb, E. and A. Lempel (1966) "Interpolation in the Network Sense," IEEE Trans. in Circuit Theory. March, 1966 pp. 118-119. 10 Churchill, R.V. (1960) Complex Variables and Application, McGraw Hill Book Co. Inc. pp. 106. 11 Gradshteyn, I.S. and I.M. Ryzhik (1965) Table of Integrals, Series and Products, Academic Press, pp. 27. 12 Churchill, R. V. (1963), Fourier Series and Boundary Value Problems, McGraw-Hill Book Co. Inc. pp. 90. 13 Tricomi, F.G. (1957) Integral Equations, Interscience Publishers, Inc. New York, pp. 143. 62

Unclassified Security Classification DOCUMENT CONTROL DATA - R & D (Security classification of title, body of abstralct l Itfd i,,ld!xtld.nnltotation nmt.st be eltered when the overall report 1. classallied) I. ORIGINA TING AC'TIVITY (Corporate,author) 12a. REPORT SECURITY CLASSIFICATION The University of Michigan Radiation Laboratory, Dept. of UNCLASSIFIED Electrical Engineering, 201 Catherine Street, 2b. GROUP Ann Arbor, Michigan 48108 3. REPORT TITLE NETWORK FUNCTION DETERMINATION FROM PARTIAL SPECIFICATIONS 4. DESCRIPTIVE NOTES (Type of report and inclusive datos) Scientific Report No. 1 5. AUTHOR(S) (First name, middle iitial, last name) Arthur R. Braun E. Lawrence McMahon 6. REPORT DATE 7i0. TOTAL NO. OF PAGES 7b. NO. OF REFS " June 1968 62 13 8a. CONTRACT OR GRANT NO. 9a. ORIGINATOR'S REPORT NUMBER(S) F19628-68-C-0071 1363-1-T b. PROJEC T NO. Project, Task, Work Unit Nos. C. 5635-05-01'h. iOT HR REPORT N0(5) (Any othelr nrumbers thof m,,y bhe ass.ined 5635-05-01 this report) d. | AFCRL-68-0349 10. DISTRIBUTION STATEMENT 1 - Distribution of this document is unlimited. It may be released to the Clearinghouse, Department of Commerce, for sale to the general public. II. SUPPLEMENTARY NOTES 12. SPONSORING MILITARY ACTIVITY Air Force Cambridge Research Laboratories( CR Tech, Other L.G. Hanscom Field Bedford, Massachusetts 01730 13. ABSTRACT In certain problems of Network Theory the real and imaginary parts of a drivingpoint impedance function may be independently given in a band of interest (wl < t < W)2 leaving its continuations onto (0 < w < wl ) and (w2 < w < 00) completely unspecified. By manipulating the Hilbert Transforms, relating the real and imaginary parts of a function of a complex variable (p = a + jw) having no poles on the jw axis or in the right-half plane, this report shows that if the continuations exist they are unique and readily obtained. Three necessary conditions for the existence of the continuations to the given parts (of the driving-point impedance function) are obtained. Further, if these three conditions are satisfied the continuations may be obtained as a Fourier series. Four known impedances were used as examples and their continuations determined. The results obtained were excellent. The agreement between the Fourier series solution (using the first six terms at most) and the exact expression was good up to the second significant figure. In the appendix a computer program which obtains the Fourier coefficients of the unknown continuations, from the given real and imaginary parts, is furnished. Unclassified StLcuri\' (ls;I I i (111iol

Unclassified Security Classification 14. | LINK A LINK B LINK C KEY WORDS.... ROLE WT ROLE WT R OLE WT NETWORK FUNCTIONS HILBERT TRANSFORMS INTEGRAL EQUATIONS FOURIER METHODS IMPEDANCE LOADING Unclassified

3 9015 02653 5859