Solution of Ordinary Differential Equation System with Unknown Coefficients by Auge 21

COMITÉ EDITORIAL Manuel Arellano Castañeda Lic. en Informática Gerente Tecnologías de Información y Comunicación 3r's de México Erika Uscanga Noguerola Mtra. en Educación Coordinadora de Gestión Ambiental Centro Universitario Hispano Mexicano Maria Fernanda Corona Salazar Maestra Psicóloga en Constelaciones Familiares Dirección de Orientación Educativa Manuel Herrerías Rul Dr. en Derecho Herrerías y Asociados Raúl Vargas Ph.D. Mechanical Engineering College Of Engineering And Computer Science Florida Atlantic University Mtra. Lorena Casanova Pérez Manejo Sustentable de Recursos Naturales Universidad Tecnológica de la Huasteca Hidalguense. Hidalgo, México Mtro. Sérvulo Anzola Rojas Director de Liderazgo Emprendedor División de Administración y Finanzas Tecnológico de Monterrey, Campus Monterrey. Monterrey, México María Leticia Meseguer Santamaría Doctora Europea en Gestión Socio‐Sanitaria Especialista en Análisis socio‐económico de la situación de las personas con discapacidad. Universidad de Castilla‐La Mancha, España. Red RIDES Red INERTE

COMITÉ DE EDITORIAL Raúl Sánchez Padilla Dr. Ingeniería Civil y Arquitectura Gerente General Desarrollos en Ingeniería Aplicada Presidente Comité Editorial Judith Ceja Hernández Ing. Industrial. Gerente de Gestión 3R's de México Vicepresidenta Comité Editorial Juan Manuel Negrete Naranjo Dr. en Filosofía Universidad de Freiburg i Br. Francisco J. Hidalgo Trujillo Dr. en Ingeniería Industrial Universitat Politécnica de Catalunya – FUNIBER Director Sede México Fundación Universitaria Iberoamericana David Vivas Agrafojo Mtro. en Educación Ambiental Universitat de Valencia ‐ Responsable IMEDES Andalucía Antonio Olguín Reza Mtro. Desarrollo de Negocios Jabil Circuit Oscar Alberto Galindo Ríos Mtro. en Ingeniería Mecánica Eléctrica Secretario de la Asociación Mexicana de Energía Eólica Amalia Vahí Serrano Dra. en Geografía e Historia Universidad Internacional de Andalucía Universidad "Pablo Olavide" Ricardo Bérriz Valle Dr. en Sociología Coordinador de Proyecto Regional de Ciudadanía Ambiental Global

Manuel Vargas Vargas Doctor en Economía Especialista en Economía Cuantitativa. Universidad de Castilla‐La Mancha, España Red RIDES Red INERTE

COMITÉ DE ARBITRAJE INTERNACIONAL David Vivas Agrafojo Mtro. en Educación Ambiental Universitat de Valencia ‐ Responsable IMEDES Andalucía Juan Manuel Negrete Naranjo Dr. en Filosofía Universidad de Freiburg i Br., Alemania Delia Martínez Vázquez Maestra Psicologa en Desarrollo Humano y Acompañamiento de Grupos. Universidad de Valencia Erika Uscanga Noguerola Mtra. en Educación Coordinadora de Gestión Ambiental. Centro Universitario Hispano Mexicano Bill Hanson Dr. Ingeniería en Ciencias National Center for Enviromental Innovation. US Enviromental Protection Agency Ph.D. María M. Larrondo‐Petrie Directora Ejecutiva del Latin American And Caribbean Consortium Of Engineering Institutions "Laccei" María Leticia Meseguer Santamaría Doctora Europea en Gestión Socio‐Sanitaria Especialista en Análisis socio‐económico de la situación de las personas con discapacidad. Universidad de Castilla‐La Mancha, España. Red RIDES Red INERTE Manuel Vargas Vargas Doctor en Economía Especialista en Economía Cuantitativa. Universidad de Castilla‐La Mancha, España Red RIDES Red INERTE

Auge21: Revista Científica Multidisciplinaria

ISSN: 1870-8773

Año 7 / No. I / Enero - Junio / 2012

SOLUTION OF ORDINARY DIFFERENTIAL EQUATION SYSTEM WITH UNKNOWN COEFFICIENTS 1Verónica

Adriana Galván-Sánchez (e-mail: vgalvan@gdl.cinvestav.mx) Alberto Gutiérrez-Robles (e-mail: alberto.gutierrez@cucei.udg.mx) 2Miguel Ángel Olmos Gómez (e-mail: miguel.olmos@cucei.udg.mx) 1 Cinvestav-Unidad Guadalajara, México. 2 Universidad de Guadalajara, Departamento de Matemáticas, México.

2José

ABSTRACT

There is more than one technique to fit a set of points or a function over a range. In the first case, the use of the sum properties yields to the minimized error and in the second case the use of the integral formulation is necessary to develop the final normal form of the approximation. The difference between both techniques is the way or criterion chosen in order to minimize a function. One of the useful is the least-square criterion. If one have an over determined problem, the left multiplication by the transposed system is equivalent to the leastsquare approximation. If one has a set of points of real numbers over a range, the use of orthogonal polynomials gives a very stable function that fit the set of points. In the case of complex numbers, a more elaborated methodology is needed. In this paper, the vector fitting technique is used to fit a set of complex over determined set of numbers to solve an ordinary differential equation system (ODES). Key words: ODES, Least-Square Method, complex plane fitting.

II.

BACKGROUND

There are two types of problem considered under this heading; the first one is the problem of interpolation, which involves finding intermediate values when values are given at a finite set of points, and the second problem is the problem of approximating a function over a complete range by a simple function which is more suitable for computations. Clearly, the main goal is that the approximation should make the error as small as possible; different methods arise depending on the way of defining the error, for example [1] n

En   n xi   f i i 0

(1)

where

is the error of the approximation

n x i  is the proposed function fi

is the function to be fitted Fitting the function at exactly the n  1 points will reduce the error En to zero. The error defined in equation (1) can certainly been minimized, but the question remains whether the values at points where x  xi give good approximations. The approximation problem is concerned with the error at all points in the range. For example, a set of equidistant points, xi i  0,1,..., n  are chosen to approximate the function 1 1  25 x 2  over the range  1,1 . It is found that for any point x  xi , where x  0 .726 the error  n x   f x  of the approximation increases without bound as n increases, and this is true even though n xi   f xi  i  0,1,..., n  which means that E n  0 . When considering the error over a whole range, a more satisfactory objective is to make the maximum error as small as possible. This is the minimax type of approximation where the error is defined by [1]: (2) Emax  max  x   f x  a xb

and the function  x  is chosen so that Emax is minimized. It is in this context that the Tchebyshev polynomials have found wide application. The third case, which is our interest, is when the number of points at which values are given is considerably greater than the degree of an approximating polynomial. For example, it may be desirable to use a low-order

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polynomial, a cubic say, as an approximation over a range in which perhaps twelve function values are known. Four points would be sufficient to determine a cubic uniquely and errors would, therefore, arise at the remaining points. In this situation, rather than having a zero error at any particular point, we require that the overall error to be as small as possible. An appropriate choice of error definition is given by [1]: m (3) 2 S m   n xi   f i  , m  n i 0

The least-square fit is obtained by finding a function n x  on a given set of functions which minimizes the quantity Sm . The subscript n implies that the function n x  is dependent on a number of parameters n which can be chosen in an appropriate way in order to obtain the least-square property. In the case of a polynomial these parameters are the coefficients a0 , a1 ,..., an  and the function n1 x  would have n  1 variable parameters. Polynomials are widely used for approximation, so it is worthwhile considering whether they are the most appropriate functions for this purpose. One big advantage is that by using the arithmetic operations available on a digital computer, it is possible to evaluate directly a polynomial or the quotient of two polynomials. However, the calculation of other functions, such as exponential or trigonometric functions is by means of approximation methods. Also, it is easy to evaluate both integrals and derivatives of polynomials by direct calculation. It is also important to know how close an approximation can be obtained using polynomial. Fortunately, there is a theorem due to a Weierstrass which shows that, for any continuous function on a finite interval, the minimax error can be made as small as we please by choosing a polynomial of sufficiently high degree. Another type of approximation which is also significant is the approximation by Fourier series. In this case it can be shown that arbitrarily close approximation can be obtained for a much wider class of functions, such as those satisfying the Dirichlet conditions.

III.

FITTING BY THE LEAST SQUARE METHOD

The fundamental property of this method is that the sum of squares is made as small as possible. Two situations arise, according to whether we approximate a finite set of values or a function defined over a range. In the first case the error will be defined as the sum of the squares of the individual errors at each point and, in the second case an integral formulation is necessary. The latter formulation will be used to illustrate the theoretical basis of the method since most readers are more familiar with integral calculus than summation properties. In its general form the least-square method is based on an approximating functions which depends linearly on a set of parameters a0 , a1 ,..., an  . The integral summation of the squares of the errors is given by [1, 2 ,3]: b

S    f x    a0 , a1 ,..., an , x  dx a

(4)

Since we requires S to be minimum, the first derivatives with respect to the various coefficients will be zero, (5) S ai  0 If the appropriate conditions hold for differentiation under the integral sign, then this gives n  1 equations for the coefficients ai ,

  f x    a0 , a1 ,..., an , xdx  0, i  0,1,..., n (6) a ai Given that  is a linear function of the coefficients the first term of these equations is constant so that the equations can be written as, b b     a , a ,..., a , x  dx  a a 0 1 n a a f x dx, i  0,1,..., n (7) i i These equations are known as the normal equations. As a simple illustration of the method, consider the case where a polynomial approximation is chosen, (8)  a0 , a1 ,..., an , x dx  a0  a1 x    an x n b

 2

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The normal equation then becomes,

 x a b

 a1 x    an x n dx   x i f x dx, i  0,1,..., n b

If these equations are written explicitly u00 a0  u01a1    u0 n an  b0 ...................................

(9)

(10)

u n 0 a0  u n1a1    u nn an  bn then the coefficients u i j of the left hand side matrix U are given by b

ui j   xi  x j dx,

i, j  0,1,..., n

(11)

Ideally, we would like the normal equations to take a simple form which would give an efficient solution to the problem. The simplest form possible is the diagonal form, which would enable the coefficients a0 , a1 ,..., an  to be found directly by dividing by ui i i  0,1,..., n  . This form can be produced by taking advantage of the special properties of orthogonal functions.

III.A.- Discrete least-square approximation From the fundamental property of this method expressed is discrete form, yields to the following relation: m

Em    yi  Pn  xi 

(12)

i 1

where: n



Pn  xi    a j xij



(13)

j 0

By substituting equation (13) into equation (12), one obtains: n  Em    yi   a j xij i 1  j 0 m



   



(14)

Expanding the quadratic term, n m   m  n n  m Em   yi2  2 a j   yi xij    a j ak   xij k  j 0 i 1   i1  j  0 k 0  i1

(15)

The minimum least square error is defined as the derivative of the error respect to the coefficients equal to zero as, Em a j  0 , then n  m  m  n n  m    yi2  2 a j   yi xij     a j ak   xij  k   j 0  i 1  j 0 k  0  i 1   i 1 0 a j

Expanding the solution, in matrix form we have: m m m m 0  m  xi xi1 xi2   xin  yi xi0      i 1 i 1 i 1 i 1  1 i m m m m 1   a0   m  xi3   xin1   a1    yi xi1    xi  xi2    i 1 i 1 i 1  im1   im1  m m m 3 4 n 2   a2     x2 2 xi xi   xi   yx   i     i i  i 1 i 1 i 1 i 1    i 1         m  an   m m m m  n n n 1 n 2 n n   xi  xi xi   xi    y i xi    i 1  i 1 i 1 i 1  i 1  where m is the number of discrete samples.

(16)

(17)

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ISSN: 1870-8773

Año 7 / No. I / Enero - Junio / 2012

ORDINARY DIFFERENTIAL EQUATIONS WITH UNKNOWN COEFFICIENTS

Most of the common physical problems are modelled with an ordinary differential equation system as:

x  Ax  Bu

(18) (19)

y  Cx  Du

where A , B , C and D are, in this case, unknown matrices of coefficients. The solution of equation (18) in Laplace domain is, sXs   x0  A  Xs   B  Fs  Solving for Xs  one obtains, X s   s I  A  x 0   sI  A  B  F s  1

(20b)

1

The solution of equation (19) in the frequency domain is, Ys   C  Xs   D  Fs  By substituting equation (20b) into equation (21a) one has,





Y s   C  s I  A  x 0   s I  A  B  F s   D  F s  1

1

The zero state response of (21b), x0  0 , yields to,





Y s   C  s I  A   B  D  F s  1

(20a)

(21a) (21b) (22)

Given that the transfer function of a system is the relation between the input and the output, one has H s   C  s I  A   B  D

(23)

1

The purpose is then the rational approximation of Hs  by supposing that A is a diagonal matrix and B is a column vector of ones. This is performed by fitting the transfer function in the complex plane as it is explained in section V.

COMPLEX PLANE FITTING

Vector fitting has become a popular tool for the identification of linear systems, it is based on the approximation of a function in the frequency domain as follows [4, 5, 6]: (24)  s  f fit s    f fit s  where: N 1

c f fit s    n  d  sh  h n 1 s  a n N

 s  z  n 1 N

 s  a  n 1

(25b)

c~  s    n~  1  n 1 s  an N

(25a)

 s  ~z  n

n 1 N

 s  a~  n

n 1

N 1

 s  zˆn  cˆ  f fit s    n~  d  sh  h nN1 n 1 s  an s  a~ 

(25c)

 n 1

Solving for f fit s  in equation (24):

N 1

 f fit s  f fit s     s 

 s  zˆ  n 1 N

 s  a~  n

n 1 N

 s  ~z  n 1 N

 s  a~  s  and  s  are the same, then the resulting equation is: n 1

If the poles of  f fit

(26a)

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 f fit s  f fit s   h  s 

ISSN: 1870-8773

(26b)

N 1

 s  zˆ  n 1 N

Año 7 / No. I / Enero - Junio / 2012

 s  ~z  n 1

Equation (26b) indicates that the poles of f fit s  are the zeros of  s  .

VI.A.- Numerical procedure Substituting equation (25a,b,c) into equation (24) yields to: N  N c~n  cˆn       1 f s  ~  d  sh  n1 s  a~  fit n1 s  an n   That is: N N cˆn c~n *       f s f s   fit fit ~ ~  d  sh n 1 s  a n 1 s  a Solving for f

(27b)

* fit

(27a)

s  :

N c~ cˆn  d  sh   n~ f fit s  ~ n 1 s  a n n 1 s  a n Re-writing in matrix form: cˆn  N f   N 1 fit s   d  1 f fit* s    s   ~ ~  n 1 s  a n   h   n 1 s  an ~  c n  Let it denote, N f s   N 1 fit f s    1 s   ~ s a~n    s a n  1 n  1 n  X T  cˆn d h c~n  The notation in matrix form is: N

f fit* s   

f fit* s   f s   X

(27c)

(27d)

(28a) (28b) (28c)

where: 1. f *fit s  is the discretized function to be fitted. 2. s is the Laplace variable (discretized). 3. f s  is the function to perform the fitting. 4. a~n contains the initial poles for the fitting. 5. cˆn , d , h and c~n are the unknown values. Equation (27d) denotes the fitting as a nonlinear problem in which f fit* s  depends on f fit s  . So, it could be solved iteratively [6]. Equation (27d) is the realization of one set of data with an improper function. If one has a proper realization, one need to cancel the term “sh” in equation (27a) and follows the same procedure. By the other hand, if one wants a strictly proper realization, one needs to cancel “d+sh” in equation (27a) and follows the same procedure.

VI.B.- Calculation of the residues c~n Having the equation f fit* s   f s   X and knowing that f fit* s  and f s  are complex, they can be separated in the real and imaginary parts as follows: A1  real  f s 

A 2  imag  f s 

b 1  real  f fit* s 

b 2  imag  f

* fit

s 

(29a) (29b) (29c) (29d)

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Re-arranging these matrices into a new matrix system yields to the following ones: A  A   1 (30a) A 2  and b  b   1 (30b) b 2  That is: b  AX . (31) In order to solve system (31) the matrix A is normalized by its quadratic Euclidean norm which is given by 1 (32a) p p  n normV, p     vi   i 1  where p  2 and n is the number of elements in the vector V . So, applying the Euclidian norm by columns yields to the following equation: A:, m  m  1 : Nc  A:, m   (32b) normA m ,2 Here A m is a column vector. After that, it is used the means square approximation which is equivalent to pre-

 

multiplying by A T , so solving for X one obtains: X  A T  A   A T  b

(33)

1

Finally the quadratic Euclidian norm is applied to the vector X as follows X  X. norm(A).' ; knowing that X T  cˆn d h c~n  , one obtains: (34a) cˆn  X1 : N p 

  h  XN  2 c~  XN  3 : 2 N

(34b)

d  X Np 1

(34c)

(34d)  2 ~ where N p is the length of an . It is important to note that when one has a complex conjugate pair in the initial poles, i.e. a~k , then c~k will contain the real part and c~k 1 will contain the imaginary part of the residues of  s  . n

VI.C.- Calculation of the zeros ~z n of  s  From equation (25b), and given s , c~n and a~n , it is possible to obtain ~z n  from the following equation: N

c~n  ~ 1  n 1 s  an N

 s  ~z  n

n 1 N

(35)

 s  a~  n

n 1



~ There are two possibilities, if one has real initial poles a~1 , a~2  or if one has complex conjugates pair a~1  a~  jb





~ and a~2  a~  jb .

1st CASE.- With to real poles, the procedure to obtain the zeros ~z n  is as follows: c~ c~  s   1 ~  2~  1 sa sa 1

By building a common factor and grouping around the variable s yields to: s 2  c~1  c~2  a~1  a~2 s  a~1a~2  c~1a~2  c~2 a~1   s   s  a~1 s  a~2  z  , so finally one has: From the solution of the quadratic equation  s   0 one obtains ~



(36a)

(36b)

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s   s    s  a~1 s  a~2 

(36c)

z1   and ~z 2   . Alternatively, these zeros could be calculated by the following procedure: where ~ ~ ~z  eigenvalue s  a1 0   1c~ c~   n    1 2  0 a~2  1  So one has:  c~  a~  c~ ~ zn  eigenvalues  1 ~ 1 ~ 2~    c1 a2  c2  The calculus of the eigenvalues could be made by using the equation A  I   0 , ~ ~  c~  1 0  a  c   1 ~ 1 ~ 2~       0 a 2  c2    c1 0 1   Performing the indicated operations and grouping the equation yields to:

2  c~1  c~2  a~1  a~2   a~1a~2  c~1a~2  c~2 a~1   0

(37a)

(37b)

(38a) (38b)

The solution of this equation gives 1 and 2 which are the zeros ~z n  . Obviously 1   and 2   . 2nd CASE.- With two complex conjugate poles, the procedure to obtain the zeros ~z n  is as follows: c~1  jc~2 c~1  jc~2  s    ~ ~ 1 s  a~  jb s  a~  jb Using a common factor procedure and grouping yields to: ~ ~ s 2  2c~1  2a~ s  a~ 2  b 2  2a~c~1  2b c~2  s   ~ ~ s  a~  jb s  a~  jb z  , so finally: From the solution of the quadratic equation  s   0 one obtains ~













(39b)

s   s     s   s  a~ s  a~  1

(39a)

(39c)

where ~z1   and ~z 2   . Alternatively, these zeros could be calculated by the following procedure:  a~ b~  2  ~ z n  eigenvalues  ~ ~    c~1 c~2   b a  0  Then, ~ ~ ~ ~ ~z  eigenvalue s a  2c1 b  2c2    ~ n a~    b

(40a)

(40b)

The calculus of the eigenvalues could be made by using the equation A  I   0 , a~  2c~ b~  2c~  1 0  2   ~ 1 (41a)      0 ~ a    b 0 1  Making the indicate operations and grouping the equation is ~ ~ (41b)  2  2 c~1  2 a~   a~ 2  b 2  2 a~ c~1  2b c~2  0 ~ The solution of this equation gives 1 and 2 which are the zeros z n  . Obviously 1   and 2   .





VI.D.- Calculation of the residues cn The original function to be fitted is: N c f fit s    n  d  sh n 1 s  an

(42a)

From equation (26b), one has that the poles of f fit s  are the zeros of  s  , that means an  ~z n ; this yields to:

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 cn   1 s   d   h   

(42b)

In matrix form:

f fit s   f s   X T

with X T  cn

(42c)



h  and f s   

1 s  an

 n1 real and imaginary parts as follows: Z1  real f s  Z 2  imagf s  y 1  real f fit s 

 1 s . Because f fit s  and f s  are complex, both are separated in their  (43a) (43b) (43c)

y 2  imag f fit s 

(43d)

Re-arranging these matrices into a new matrix system yields to the following ones: Z  Z   1 (44a) Z 2  and y  y   1 (44b) y 2  Hence (45) y  Z  XT In order to solve system (45), the matrix Z is divided by its quadratic Euclidean norm, then it is used the means square approximation which is equivalent to pre-multiplying by Z T , one will obtains: 1 (46) X  Z T  Z   Z T  y Finally the Euclidian norm is applied to the vector X . Knowing that X T  cn cn  X1 : N p  d  XN p  1 h  XN p  2 where N p is the length of an .

h  , one obtains:

(47a) (47b) (47c)

It is important to note that when one has a complex conjugate pair in the initial poles, i.e. a k , then ck will contain the real part and ck 1 will contain the imaginary part of the residue. The final approximation is the analytical formula given by the an poles and the cn residues complemented by the constant term d and the proportional constant h . That means, from the discretized values in the complex plane, this procedure calculates one formula which is the analytical approximation of the set of data. For a practical application one will obtain an analytical model from a set of values and this is main goal of fitting the data in all cases.

VI.

NUMERICAL PROCEDURE

It could be possible to fit a group of data with a strictly proper function, a proper function or an improper function. If one has the set of data is clearly impossible to know a priori the adequate kind and the order of the function to fit this set of data. So, it is necessary to explore around these two variables to the equation to fit, as good as possible, the set of data. For example if one has a transfer function Hs  described by: 0.1  6.2832i  0.1  29.164i   S  0.1  135.37i    0.1  628.32i 

6.4095  0.78136i  6.0217  0.20376i   HS    6.001  0.044304i     6  0.0095491i 

1 1 B  1  1

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where S is the Laplace variable in discrete form, H S is the discretized transfer function and B is the weight factor. By using this technique, the synthesised analytical function is as equation (42a), means we have: N c f s    n  d  sh n 1 s  an where N  gives the order of the synthesis and d  sh  determines the kind of the approximation. Table 1 gives the initial poles according with the order of the approximation and the factor for each initial pole, means an associate number to make a difference between a real pole and the real and imaginary part of a complex pole. The fitting procedure is sensitive to the initial poles, but after a lot of tests the linspace function of the MatLab distributes the set linearly between a minimum and maximum value for the real and imaginary parts. Table 1 is generated for the real part with the limits [-0.01 -1] and the imaginary with [1 100] TABLE 1.- Initial poles for each order of approximation. Order Initial poles Factor for each initial pole. 0 for real, 1 for the real part and 2 for imaginary part of complex pole. st 1 Pinit   49.5 I index  0 2nd

3rd 4th

Pinit   1  100i

 1  100i 

Pinit   0.01  i

 0.01  i

Pinit   49 .5  1  100 i

I index  1 2 

 1  100 i   1  100 i

 1  100i 

I index  0 1 2 

I index  1 2 1 2 

Applying the methodology previously enounced, it is possible to obtains for each order a strictly proper, a proper o an improper function. Table 2 resumes the obtained function according with the order and the kind of approximation. TABLE 2.- Functions for each order and kind of approximation. Kind of function Strictly proper Proper Improper 6.212 6  6.0016  1.5125  10 s f s   s 55588..53396 f s   s 6.32112  6.0016 f  s   .1221 s  3.1232

Order 1st 2nd 3rd

f s   f s   

4th

1.585106



s  2.6416105

2.6696108 s  4.4493107



6.2366 s  3.1845

2.9987  2.3423 i s  2.18411.0374 i

f s  

3.0008 2.5104 i 3.0008 2.5104 i   6 s  2.1635 0.99951i s  2.1635 0.99951i

f s  

3.0008 2.5085 i 3.0008 2.5085 i 8   6  1.5929  10 s s  2.1637 0.99996 i s  2.1637  0.99996 i

f s  

2 2 i 2i    6 s 5 s  3 2 i s  3 2 i

18 2 2 i 2i    6  4.101  10 s s 5 s  3 2 i s  3 2 i

2.9987  2.3423 i s  2.18411.0374 i

f s   

7.369910 20 s 1.231910 20



9.261 11 .23  s 18 .054 s  6.0666

11 .331 s  241 .91

10 f s    3.3113 s  9.0123

10 

2 2 i 2 i  6   s 5 s  3 2 i s  3 2 i

f s   

2.9988  2.3564 i 2.9988  2.3564 i 0.00050299  6.2064 10 5 i   s  2.1819 1.0345 i s  2.1819 1.0345 i s 1100 i

9 0.00050299  6.2064 10  5 i s  6  3.4726  10 s 1100 i

It is possible to have high order or low order fittings, but we have some considerations; for example: 1. The set of data have an origin of certain order and certain kind of function (original problem). 2. Although we don't know the original problem, when we overpass the order in the fitting, one has residues close to cero, complex pure imaginary poles or very large poles. 3. By the other hand, when we have a low order fitting, we will have some terms with very large zeros and very large poles. 4. For these reasons, the measure of the fitting is given in terms of the least square error [4, 5, 6]. The evaluation of the error by the least square error has the problem that by using a fix error more than one function could be right. Like this, the proper functions order 3 and 4, and the improper function order 3 have almost the same error but the functions differ in the order or in the kind. Analyzing the results in a visual way, the Figure 1 shows the absolute value of the data and all the approximations, Figure 2 shows the real parts and Figure 3 shows the imaginary ones. Figures 4, 5 and 6 show the zoom of the functions in order to note which ones are better approximations. It seems that the three mention functions (proper order 3 and 4, and improper order 3) have unnoticeable differences but table 2 shows the differences between them.

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7 6

Absolute value

5 4 3 2 1 0 0 10

Original set of data First order strictly proper function First order proper function First order improper function Second order strictly proper function Second order proper function Second order improper function Third order strictly proper function Third order proper function Third order improper function Fourth order strictly proper function Fourth order proper function Fourth order improper function 1

10 Frequency

FIGURE 1.- Absolute value of the original set of data and all the functions. 7

Real part of the function

6 5 4 3 2 1 0 0 10

10 Frequency

FIGURE 2.- Real part of the functions.

Imaginary part of the function

0 -0.5 Original set of data First order strictly proper function First order proper function First order improper function Second order strictly proper function Second order proper function Second order improper function Third order strictly proper function Third order proper function Third order improper function Fourth order strictly proper function Fourth order proper function Fourth order improper function

-1 -1.5 -2 -2.5 -3 -3.5 0 10

10 Frequency

FIGURE 3.- Imaginary part of the functions.

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Absolute value

6.001 6.001 Original set of data First order strictly proper function First order proper function First order improper function Second order strictly proper function Second order proper function Second order improper function Third order strictly proper function Third order proper function Third order improper function Fourth order strictly proper function Fourth order proper function Fourth order improper function

6.001 6.001 6.001

1.3331

1.3332

1.3333

10 Frequency

1.3334

1.3335

1.3336

FIGURE 4.- Zoom of the original set of data and all the functions (Absolute value).

Real part of the function

6.001 6.001 6.001

6.001 6.001 6.001 10

1.3331

1.3332

1.3333

1.3334

10 Frequency

1.3335

1.3336

Imaginary part of the function

FIGURE 5.- Zoom of the original set of data and all the functions (Real part of the function).

-0.0443 -0.0443 -0.0443

-0.0443 -0.0443 -0.0443 -0.0443 10

1.3331

1.3332

1.3333

10 Frequency

1.3334

1.3335

1.3336

FIGURE 6.- Zoom of the original set of data and all the functions (Imaginary part of the function).

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According to the numerical results, the best approximations are the third order proper function, the third order improper function and the fourth order proper function. By the analysis of these functions, it is noticeable that the third order improper function has the proportional term almost equal to cero and that the fourth order proper function has a residue almost equal to zero, therefore they are practically the same function. Of course if one has a set of measures, one has noise from the measure devices and the fit procedure could be affected deeply, but one could has or expect the same behavior. If one applies the fit procedure iteratively, there is a relocation of poles; it is possible to have a better approximation [6]. By using a high order fit function, if one notes that some of the poles are very high, complex pure imaginary or some of the residues are almost zero, one can use the rest of the poles like a new set of beginning ones. The previous procedure means that one could begins with a high order function and reduce this order until arrives to a good set of poles, residues, constant and proportional terms. The previous example is performed in one loop, the procedure step by step, to obtain a third order proper function is described as follows: 1. One needs a complete set of data, this set is composed by, 0.1  6.2832i  6.4095  0.78136i  1 0    50  0.1  29.164i  6.0217  0.20376i  1              S HS B I index   1  Pinit    1  10i   0.1  135.37i   6.001  0.044304i  1  2   1  10i       0.1  628.32i   6  0.0095491i  1 2. Using this information one obtains matrix A and b for a proper function, as: Bf fit   B A B b  f fit* s   S - Pinit  S P init  Numerically one has, 0.12403+0.031151i 0.641251.1333i 1.9234 0.68385i 0.30851 0.069083i 1  0.019651 0.0024645i 0.077343+0.18625i 6.40950.78136i   0.0264960.0022687i 0.088005+0.055296i 0.0064959+0.46758i 0.20376 i 0.0149080.0086783i 0.0037019 0.077523i 1 0.16002+0.0082628i A  0.0024047 b  66..0217   0.000122030.014855i 0.0064973i 0.00109721.793110 i 1 0.014143+0.039097i 7.421110 +0.089149i 0.0065852+5.899410 i 0010.044304 i  0.00012610.0015815i 5.576910 0.0031839i 5.067310 1.774710 i 1 0.00074152+0.0094902i 3.058210 +0.019104i 0.00030404+5.809610 i   60.0095491i  3. Separate the matrices A and b in real and imaginary parts:  0.019651 0.077343 0.30851 1 0.12403 0.64125 1.9234   6.4095   0.026496 1  0.088005  0.0064959 0.014908 0.0037019 0.16002  0.0024047   66.0217   0.0010972 1  0.014143  7.421110 12.20310 0.0065852 .001  12.6110 5.576910 5.067310 1  74.15210 3.058210 30.40410    6 A   0.0024645 0.18625  0.069083 0 0.031151 b    0.78136  1.1333 0.68385     0.0086783 0.077523 0.0022687 0 0.055296 0.46758 0.0082628     0.20376   0.014855 1.793110 0 0.039097 0.089149 5.899410 044304 0.0064973   00..0095491  0.0015815  0.0031839 1.774710 0 0.0094902 0.019104 5.809610  4. Compute the Euclidian norm of matrix A , which is (by column) norm( A)  0.0272 0.2166 0.3173 2.0000 0.1702 1.3866 2.0477 -5

-6

-5

-7

-5

-6

-7

-5

-6

-5

-6

-5

-7

Solve the system making (by column), A  A norm( A) and the applying X  A T  A   A T  b , and then X  X. norm(A).' to obtain: 1

 45.8182  95 .5909  252  .7384 X   6.0000   7.8041   16.3480    41.9512  One has a third order fit, so the last three terms corresponds to the residues of  s  :

 ~   167..8041 C  41.3480 9512  Knowing the initial poles and the weighted matrix which are, 49.5 0 0  1  W  2  init   0 1 100 and  0 100 1  0  With the residues of  s  , the initial poles and the weighted matrix, one obtains the zeros of  s  as,   49.5 Z ZEROS  eig   0  0

1 100 100 1

 1     20 7.8041

16.3480  41.9512



 

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The zeros of the function  s  are the poles of the transfer function Hs  , so one has:

 5  R ROOTS  3 2i   3  2 i  10. Use these roots to obtain the residues, the constant and the proportional terms by using,  B  A B b  f fit* s  S - RS  Numerically one has,  0.077876 0.15052 0.047319 1   6.4095  0.0073079 0.0045663 1   00.0058183  6.0217  .00027793 0.00033839 0.00021799 1 6.001 1.291810 1.570510 1.013110 1    6 A   0.095943 0.25911 0.071259 0  b   -0.78136     0 . 033271 0 . 068114 98 . 653 10 0    -0.20376  0.0073768 -0.01477 9.991910 0 0.044304  -0.0015914  --0.0095491  -0.0031831 9.997710 0 Solving the system by following steps 4 and 5, one obtains: 2 X  21  6  The first three terms of X are the residues of Hs  and the fourth is the constant term. According with I index the first term of the residues is real, the second term is a real part of a complex pole and the third term is the imaginary part of this pole, so one has, 2 C  2i  d 6 and  2 i  Following this procedure, we have the following EDOS, 0 0  x1  1  5  x1         0  x2   1ut  y  2 2  i 2  i x2   6ut  x   0  3  2i  0 x3  0  3  2i  x3  1 By using equation (23) the analytical function to fit the discretized Hs  function is: 2 2i 2i  f s   Hs    6 s  5 s  3  2i s  3  2i -5

-5

-6 -8

11.

12.

13.

14.

VII.

APPLICATION EXAMPLES

EXAMPLE 1.- The behaviour of a physical phenomenon can be represented by an ordinary differential equation system (ODES); the representation is not unique, that means, it could be possible to have two different ODES with the same solution. In this case an ODES is defined as a benchmark, then it is developed an ODES from the discrete points of Hs  and compare its solution with the first one. If one has the system,

x  Ax  But  so, we have,

and

y  Cx  Dut 

 x1   25  5  x1  10  u t    x     2  x2   0   2   1

and

 y1   2 0   2 y   0  0  2  3    y3    5  1   2       1   x1  0  y4    0  u t   y 5    1 0   x 2  1         0  y6   1 0  y   1  2 0  7      y8   0 2  0

The analytical solution of this ODES is as follows: x1 t   0.4075 1  e 24.7805141726 785 t   0.0437 1  e 2.2194858273 2152 t 

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x2 t   0.0179 1  e 24.7805141726 785 t   0.1996 1  e 2.2194858273 2152 t  Then the analytical solution for the output is: y1  0.81493752 0035717 1  e 24.7805141726 785 t   0.03577344 71600778 1  e 2.2194858273 2152 t   2u t  y 2  1.22240628 005358 1  e 24.7805141726 785 t   0.05366017 07401167 1  e 2.2194858273 2152 t 

y3  1.99351140 37078 1  e 24.7805141726 785 t   0.11027128 7498026 1  e 2.2194858273 2152 t   2u t 

y 4  0.04383239 63814949 1  e 24.7805141726 785 t   0.19970490 53982211  e 2.2194858273 2152 t 

y5  0.40746876 0017858 1  e 24.7805141726 785 t   0.01788672 35800389 1  e 2.2194858273 2152 t   u t 

y 6  0.40746876 0017858 1  e 24.7805141726 785 t   0.01788672 35800389 1  e 2.2194858273 2152 t  y 7  0.49513355 2780848 1  e 24.7805141726 785 t   0.41729653 437648 1  e 2.2194858273 2152 t 

y8  0.08766479 27629897 1  e 24.7805141726 785 t   0.39940981 0796442 1  e 2.2194858273 2152 t  These results are used like a benchmark, but the intention is to construct an ODES by supposing that we know only the transfer function Hs  , which is: T

2 2  2 s 2  34s  70  30s  60 2 s  4s 10 s  17 s  35 10s  20 10s 20  Hs     s 2  27 s  55 s 2  27 s  55 s 2  27 s  55 s 2  27 s  55 s 2  27 s  55 s 2  27 s  55 s 2  27 s  55 s 2  27 s  55  Discretizing the transfer function Hs  with 500 samples, logarithmically distributed between 10 3 10 8  Hz, then making s  j , figure 7 shows the behavior of each term.

|Hs(1)| |Hs(2)| |Hs(3)| |Hs(4)| |Hs(5)| |Hs(6)| |Hs(7)| |Hs(8)|

1.8

Absolute value

1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0

-2

10 Frequency in Hz

FIGURE 7a.- Absolute value of the function H s  .

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Real part of the function

real(Hs(1)) real(Hs(2)) real(Hs(3)) real(Hs(4)) real(Hs(5)) real(Hs(6)) real(Hs(7)) real(Hs(8))

1.5 1 0.5 0 -0.5

-2

10 Frequency in Hz

FIGURE 7b.- Real part of Hs  .

Imaginary part of the function

1.2

imag(Hs(1)) imag(Hs(2)) imag(Hs(3)) imag(Hs(4)) imag(Hs(5)) imag(Hs(6)) imag(Hs(7)) imag(Hs(8))

1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6

-2

10 Frequency in Hz

FIGURE 7c.- Imaginary part of H s  . From a priori knowledge, it is a fact that the system has common poles, that means, all data of H s  comes from the same physical system. In this case using every set of points, one arrives to the same poles. By using the described technique to fit each curve with a second order proper function, one arrives to the following system: 0   x 1   1  x 1   24.7805141726 785  ut  x     2.2194858273 2152   x 2  1 0  2  

 y1   20.1945 0.1945  2 y   0    30 . 2918 0 . 2918  2      y3   50.0432 0.0432  2        y4     0.4432 0.4432   x1   0 u t   y5    10.0972 0.0972   x2  1         y6   10.0972  0.0972 0   y   10.9837  0.9837 0   7      y8    0.8864 0.8864  0

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Analyzing this results, we arrive to a different ordinary differential equation system but, the main goal is the behavior of this ODES. The analytical solution of this ODES is: x1 t   0.0403542877 69482 1  e 24.7805141726 785 t  x2 t   0.4505548031 39609 1  e 2.2194858273 2152 t  Then the analytical solution for the output is: y1  0.81493466 4360809 1  e 24.7805141726 785 t   0.08763290 92106539 1  e 2.2194858273 2152 t   2u t  y 2  1.22240401 42556 1  e 24.7805141726 785 t   0.13147189 1556138 1  e 2.2194858273 2152 t 

y3  2.01945769 370575 1  e 24.7805141726 785 t   0.01946396 749563111  e 2.2194858273 2152 t   2u t 

y 4  0.01788502 03394345 1  e 24.7805141726 785 t   0.19968588 8751475 1  e 2.2194858273 2152 t 

y5  0.40746531 4466016 1  e 24.7805141726 785 t   0.04379392 686517 1  e 2.2194858273 2152 t   u t  y 6  0.40746531 4466016 1  e 24.7805141726 785 t   0.04379392 686517 1  e 2.2194858273 2152 t 

y 7  0.44323939 0573662 1  e 24.7805141726 785 t   0.44321075 9848433 1  e 2.2194858273 2152 t 

y8  0.03577004 0678869 1  e 24.7805141726 785 t   0.39937177 7502949 1  e 2.2194858273 2152 t  By comparing the solution of the original system with the fitted one, the outputs seem to be different but the time solution shows that the real differences are almost zero. Figure 8 shows the numerical solution of both systems and figure 9 shows the percentage error between these two solutions. The maximum error is around 0.01%, this is acceptable because it is lower than the uncertainties in the parameters or the error incorporated by the numerical procedure. 2

Original functions Fitted functions

Amplitude in p.u.

1.5

0.5

0 0

0.2

0.4

0.6

0.8 1 1.2 Time in seconds

1.4

1.6

1.8

FIGURE 8.- Original and fitted solution, numerically solved.

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0.01

Error(y1) Error(y2) Error(y3) Error(y4) Error(y5) Error(y6) Error(y7) Error(y8)

Error in percent

0.005

-0.005

-0.01 0

0.2

0.4

0.6

0.8 1 1.2 Time in seconds

1.4

1.6

1.8

FIGURE 9.- Percent error between the original and the fitted solution. EXAMPLE 2.- The previous example is shown in order to explore the technique, but the proposed application is in the case in which one has only the transfer function in an analytical or numerical way (the technique is the same), for example if one has the ODES as:  x1   a11  x  a  2    21        xn  an1

a12 a22  an 2

 a1n   x1   e1   a2 n   x2  e2  u t                ann   xn  en 

 y1   c11 c11  c1n   x1   d1   y  c      2    21 c22  c2 n   x2    d 2  u t                      y m  cm1 cm 2  cmn   xn  d m  If one knows that the transfer function H s  of this ODES is,  7 s 3  36.5 s 2  196.57 s  188.5   s 3  9.5 s 2  29.51 s  29.785  H s   2 s 3  39 s 2  149.02 s 99.57   1   s 3  9.5 s 2  29.51 s  29.785  H 2 s    4 s 3  2 s 2  98.04 s  89.14       H s 3 2 3 H s    s3  9.5 s 2 29.51 s  29.785     5 s  27.5 s  117.55 s  138.93  H 4 s    s 3  9.5 s 2  29.51 s  29.785       H s   5  10 s 2  40 s  30  s 3  9.5 s 2  29.51 s  29.785   H s   3 s 3  28.5 s 2 108.53 s 109.36   6   s 3  9.5 s 2  29.51 s  29.785 

(48a)

(48b)

(49)

The first step is to discretize the transfer function H s  , in order to compare the effect of the sampling, we use

4 and 10000 samples logarithmically distributed between 10 3 10 5  Hz, making s  j . The second step is to explore the order of the function to fit all the terms, in this case one could arrive to a third order proper transfer function, so one obtains the following roots enounced in table 3.

H 1 s 

H 2 s 

TABLE 3.- Roots using 4 and 10000 samples Roots using 4 samples Roots using 10000 samples r1  3.700000 r2  3.499999 r3  2.300000 r1  3.699999 r2  3.500000 r3  2.299999

r1  3.700000 r2  3.499999 r3  2.299999

r1  3.699999 r2  3.500000 r3  2.300000

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H 3 s 

H 4 s  H 5 s 

H 6 s 

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r1  3.699999 r2  3.500000 r3  2.299999

r1  3.700000 r2  3.499999 r3  2.300000

r1  3.699999 r2  3.500000 r3  2.299999

r1  3.700000 r2  3.499999 r3  2.300000

r1  3.700000 r2  3.499999 r3  2.299999

r1  3.699999 r2  3.500000 r3  2.299999

r1  3.700000 r2  3.499999 r3  2.299999

By the numerical analysis of the groups of roots, it is necessary to choose three roots to fit all the functions with these group. In this case the three chosen roots are r1  3.7 r2  3.5 r3  2.3 . The final results using 4 samples or 10000 are almost the same, there are only slightly differences. By using the chosen roots, we obtain the results showed in figure 10, 11 and 12. In this figure we compare every term of the original transfer function with the fitted one. Figure 10 shows the absolute value of the original and fitted transfer function, figure 11 shows the real part and figure 12 the imaginary.

Absolute value of the function

Original function Resulting of the numerical fitting

7 6 5 4 3 2 1 0

-2

Frequency in Hz

FIGURE 10.- Absolute value of the function Hs  , original and fitted. 8

Original function Resulting of the numerical fitting

Real part of the function

7 6 5 4 3 2 1 0 -1 -2 -3 10

-2

-1

10 10 Frequency in Hz

FIGURE 11.- Real part of the function Hs  , original and fitted.

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Imaginary part of the function

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Original function Resulting of the numerical fitting

2 1 0 -1 -2 -3 -3 10

-2

-1

10 10 Frequency in Hz

FIGURE 12.- Imaginary part of the function H s  , original and fitted.

Finally, by the use of this methodology we arrive to the following ODES: 0 0   x1  1  x1   3.7 x    0  3 . 5 0   x2   1 u t   2    x 2   0  2.3  x3  1 0

 y1    1370.4 1427.1  86.726  7  y    2   83.333  6.6667   2   110  x1     y3    1762.9 1833.3  110.48    4     x2     u t   y4   688.57 708.33  39.762   x  5 3  y5    874.64 927.08  62.44    0       3  y6    192.86 208.33  15.476  This ODES has analytical solution as: x 1 t   0.2702702702 70 1  e 3.7 t 

(50a)

(50b)

x 2 t   0.2857142857 14 1  e 3.5 t 

x 3 t   0.4347826086 951  e 2.3 t  Then the analytical solution for the output is: y 1  370.3667953667 951  e 3.7 t   407.7380952380 94 1  e 3.5 t   37.7070393374 711  e 2.3 t   7 ut  y 2  29.7297297297 29 1  e 3.7 t   23.8095238095 231  e 3.5 t   2.8985507246 37 1  e 2.3 t   2 ut 

y 3  476 .4478764478 76 1  e 3.7 t   523 .8095238095 231  e 3.5 t   48.0331262939 931  e 2.3 t   4ut 

y 4  186 .1003861003 86 1  e 3.7 t   202 .3809523809 52 1  e 3.5 t   17.2877846790 881  e 2.3 t   5ut  y 5  236 .3899613899 611  e 3.7 t   264 .8809523809 511  e 3.5 t   27.1480331262 92 1  e 2.3 t   0ut 

y 6  52.1235521235 52 1  e 3.7 t   59.5238095238 09 1  e 3.5 t   6.7287784679 081  e 2.3 t   3ut  Figure 13 shows the solution of the system described by equation (50a,b).

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Amplitude in p.u.

6 4 y1 y2 y3 y4 y5 y6

2 0 -2 0

0.2

0.4

0.6

0.8 1 1.2 Time in seconds

1.4

1.6

1.8

FIGURE 13.- Solution of the ODES described by equations (50a) and (50b).

This example shows the procedure to obtain an analytical system from a transfer or discrete function, so with this methodology we can solve an ordinary differential equation system with unknown coefficients by construction a similar one, means, an ODES with the same behavior that the transfer function H s  .

VIII. CONCLUSIONS If one has a numerically described transfer function of a physical system, it is possible to build the ODES which has the same behavior via the fitting in complex plane. The solution of this ODES is the solution of the original system. The main purpose of fitting is to obtain an analytical model from a set of data. This paper shows the procedure, step by step, to fit a group of complex data with a transfer function in s domain via the vector fitting procedure.

IX.

BIOGRAPHIES

Verónica Adriana Galván Sánchez. She received her BSEE and M. Sc. degrees from Universidad de Guadalajara (UdG), Mexico, in 2008 and 2011 respectively. Currently, she is a PhD student in Cinvestav-Guadalajara. Her research interests are in power system electromagnetic transients and transient stability analysis. José Alberto Gutiérrez Robles (IEEE, Member 2004). He received his B. Eng. in Mechanical and Electrical Engineering and his M. Sc. degrees from CUCEI-Universidad de Guadalajara in 1993 and 1998, respectively. He received his PhD degree from Cinvestav, Guadalajara Campus, Mexico, in 2002. He currently is a full professor at the Department of Mathematics, CUCEI, University of Guadalajara, México. His research interests are in Applied mathematics, Power System Electromagnetic Transients and Lightning Performance. Miguel Angel Olmos Gómez. He received his B. In Mathematics Universidad de Guadalajara in 1986. He received his PhD degree from Washington State University, Pullman, WA, in 1995. He is a full professor at the Department of Mathematics , CUCEI, University of Guadalajara, Mexico. His research interest is in Numerical Solution of Nonlinear Differential Equations.

REFERENCES

[1] P. W. Williams, "Numerical computation", Nelson Edition, boards: 0-17-761018-2, paper: 0-17-771018-7, 1972, 191 pages. [2] L. R. Burden, J. D. Faires, "Numerical analysis", Brooks/Cole 20 Channel Center Street, Boston MA 002210 USA. Night edition, ISBN-13: 978-0-538-73351-9, CENGAGE Learning. [3] J. D. Hoffman, "Numerical methods for engineers and scientists", McGraw-Hill International Editions ISBN: O-07029213-2.

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[4] B. Gustavsen, A. Semlyen, "A robust approach for system identification the frequency domain", IEEE Transactions on power delivery, vol. 19, no. 3, pp.1167-1173, July 2004. [5] B. Gustavsen, A. Semlyen, "Rational approximation of frequency domain response by vector fitting", IEEE Transactions on power delivery, vol. 14, no. 3, pp.1052-1061, July 1999. [6] A. Semlyen, B. Gustavsen, "Vector fitting by pole relocation for the state equation approximation of nonrational transfer matrices", Circuits Systems Signal Process, vol. 19, no. 6, pp. 549-566,2000.

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