Solution of Ordinary Differential Equation System with Unknown Coefficients

COMITÉ EDITORIAL

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 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

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

113

SOLUTION OF ORDINARY DIFFERENTIAL EQUATION SYSTEM WITH UNKNOWN COEFFICIENTS

1Verónica Adriana Galván-Sánchez (e-mail: [email protected])

2José Alberto Gutiérrez-Robles (e-mail: [email protected]) 2Miguel Ángel Olmos Gómez (e-mail: [email protected])

1 Cinvestav-Unidad Guadalajara, México. 2 Universidad de Guadalajara, Departamento de Matemáticas, México.

I. 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 least-square 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

iiinn fxE

0

(1)

where nE is the error of the approximation

in x is the proposed function

if is the function to be fitted

Fitting the function at exactly the 1n points will reduce the error nE to zero. The error defined in equation (1) can certainly been minimized, but the question remains whether the values at points where ixx give good approximations. The approximation problem is concerned with the error at all points in the range. For example, a set of equidistant points, nixi ,...,1,0 are chosen to approximate the function 22511 x over the range 1,1 . It is found that for any point ixx , where 726.0x the error xfxn of the approximation increases without

bound as n increases, and this is true even though iin xfx ni ,...,1,0 which means that 0nE . 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]:

xfxEbxa

maxmax (2)

and the function x is chosen so that maxE 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


114

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]:

nmfxSm

iiinm

,0

2 (3)

The least-square fit is obtained by finding a function xn on a given set of functions which minimizes the quantity mS . The subscript n implies that the function xn 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 naaa ,...,, 10 and the function xn 1 would have 1n 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 naaa ,...,, 10 . The integral summation of the squares of the errors is given by [1, 2 ,3]:

b

a

n dxxaaaxfS 2

10 ,,...,, (4)

Since we requires S to be minimum, the first derivatives with respect to the various coefficients will be zero, 0 iaS (5)

If the appropriate conditions hold for differentiation under the integral sign, then this gives 1n equations for the coefficients ia ,

nidxxaaaxfa

b

a

n

i

,...,1,0 ,0,,...,,2 10

(6)

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,

nidxxfa

dxxaaaa

b

a

b

a i

n

i

,...,1,0 ,,,...,, 10

(7)

These equations are known as the normal equations. As a simple illustration of the method, consider the case where a polynomial approximation is chosen,

n

nn xaxaadxxaaa 1010 ,,...,, (8)


115

The normal equation then becomes,

nidxxfxdxxaxaaxb

a

b

a

in

n

i ,...,1,0 ,10 (9)

If these equations are written explicitly

nnnnnn

nn

bauauau

bauauau

1100

00101000

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (10)

then the coefficients jiu of the left hand side matrix U are given by

njidxxxub

a

ji

ji ,...,1,0, , (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 naaa ,...,, 10 to be found directly by dividing by niu ii ,...,1,0 . 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

iinim xPyE

1

2

(12)

where:

n

j

jijin xaxP

0

(13)

By substituting equation (13) into equation (12), one obtains:

m

i

n

j

jijim xayE

1

2

0

(14)

Expanding the quadratic term,

n

j

n

k

m

i

kjikj

n

j

m

i

jiij

m

iim xaaxyayE

0 0 10 11

2 2 (15)

The minimum least square error is defined as the derivative of the error respect to the coefficients equal to zero as, 0 jm aE , then

0

20 0 10 11

2

j

n

j

n

k

m

i

kj

ikj

n

j

m

i

j

iij

m

ii

a

xaaxyay

(16)

Expanding the solution, in matrix form we have:

m

i

n

ii

m

iii

m

iii

m

iii

nm

i

nn

i

m

i

n

i

m

i

n

i

m

i

n

i

m

i

n

i

m

ii

m

ii

m

ii

m

i

n

i

m

ii

m

ii

m

ii

m

i

n

i

m

ii

m

ii

m

ii

xy

xy

xy

xy

a

a

a

a

xxxx

xxxx

xxxx

xxxx

1

1

2

1

1

1

0

2

1

0

11

2

1

1

1

1

2

1

4

1

3

1

2

1

1

1

3

1

2

1

1

11

2

1

1

1

0

(17)

where m is the number of discrete samples.


116

IV. ORDINARY DIFFERENTIAL EQUATIONS WITH UNKNOWN COEFFICIENTS Most of the common physical problems are modelled with an ordinary differential equation system as:

BuAxx (18)

DuCxy (19)

where A , B , C and D are, in this case, unknown matrices of coefficients. The solution of equation (18) in Laplace domain is,

ssss FBXAxX 0 (20a)

Solving for sX one obtains,

ssss FBAIxAIX 11 0 (20b)

The solution of equation (19) in the frequency domain is, sss FDXCY (21a)

By substituting equation (20b) into equation (21a) one has, sssss FDFBAIxAICY 11 0 (21b)

The zero state response of (21b), 00 x , yields to,

sss FDBAICY 1 (22)

Given that the transfer function of a system is the relation between the input and the output, one has DBAICH 1 ss (23)

The purpose is then the rational approximation of sH 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.

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]:

sfsfs fitfit (24)

where:

N

nn

N

nnN

n n

nfit

as

zshshd

as

csf

1

1

1

1

(25a)

N

nn

N

nnN

n n

n

as

zs

as

cs

1

1

1 ~

~

1~

~

(25b)

N

nn

N

nnN

n n

nfit

as

zshshd

as

csf

1

1

1

1 ~

ˆ

~ˆ

(25c)

Solving for sf fit in equation (24):

N

nn

N

nn

N

nn

N

nn

fit

fit

as

zs

as

zsh

s

sfsf

1

1

1

1

1

~

~

~

ˆ

(26a)

If the poles of sf fit and s are the same, then the resulting equation is:


117

N

nn

N

nn

fit

fit

zs

zsh

s

sfsf

1

1

1

~

ˆ

(26b)

Equation (26b) indicates that the poles of sf fit are the zeros of s .

VI.A.- Numerical procedure Substituting equation (25a,b,c) into equation (24) yields to:

shdas

csf

as

c N

n n

nfit

N

n n

n

11

~ˆ

1~

~

(27a)

That is:

shdas

csfsf

as

c N

n n

nfitfit

N

n n

n

1

*

1~

ˆ~

~

(27b)

Solving for sf fit* :

sfas

cshd

as

csf fit

N

n n

nN

n n

nfit

11

*

~

~

~ˆ

(27c)

Re-writing in matrix form:

n

n

N

n n

fitN

n n

fit

c

h

d

c

as

sfs

assf

~

ˆ

~1~1

11

*

(27d)

Let it denote,

N

n n

fitN

n n as

sfs

ass

11~1~

1f

(28a)

nn

T chdc ~ˆX (28b)

The notation in matrix form is: Xff ssfit

* (28c)

where: 1. sfit

*f is the discretized function to be fitted. 2. s is the Laplace variable (discretized). 3. sf is the function to perform the fitting. 4. na

~ contains the initial poles for the fitting. 5. nc , d , h and nc

~ are the unknown values. Equation (27d) denotes the fitting as a nonlinear problem in which sfit

*f depends on sfitf . 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 nc~

Having the equation Xff ssfit

* and knowing that sf fit* and sf are complex, they can be separated in the

real and imaginary parts as follows: sfreal1A (29a)

sfimag2A (29b) sfreal fit

*

1 b (29c)

sfimag fit

*

2 b (29d)


118

Re-arranging these matrices into a new matrix system yields to the following ones:

2

1

A

AA

(30a)

and

2

1

b

bb

(30b)

That is: XAb . (31)

In order to solve system (31) the matrix A is normalized by its quadratic Euclidean norm which is given by

pn

i

p

ivpnorm

1

1

,

V (32a)

where 2p and n is the number of elements in the vector V . So, applying the Euclidian norm by columns yields to the following equation:

Ncm

norm

mm

m

:1 2,

:,:,

A

AA

(32b)

Here mA is a column vector. After that, it is used the means square approximation which is equivalent to pre-

multiplying by TA , so solving for X one obtains: bAAAX TT 1 (33)

Finally the quadratic Euclidian norm is applied to the vector X as follows ).'(AX.X norm ; knowing that

nn

T chdc ~ˆX , one obtains:

pn Nc :1ˆ X (34a)

1 pNd X (34b)

2 pNh X (34c)

22:3~ ppn NNc X (34d)

where pN is the length of na~ . It is important to note that when one has a complex conjugate pair in the initial

poles, i.e. ka~ , then kc

~ will contain the real part and 1~

kc will contain the imaginary part of the residues of s .

VI.C.- Calculation of the zeros nz~ of s

From equation (25b), and given s , nc~ and na

~ , it is possible to obtain nz~ from the following equation:

N

nn

N

nnN

n n

n

as

zs

as

c

1

1

1 ~

~

1~

~

(35)

There are two possibilities, if one has real initial poles 21~,~ aa or if one has complex conjugates pair bjaa

~~~1

and bjaa~~~

2 .

1st CASE.- With to real poles, the procedure to obtain the zeros nz~ is as follows:

1~

~

~

~

2

2

1

1

as

c

as

cs

(36a)

By building a common factor and grouping around the variable s yields to:

21

1221212121

2

~~

~~~~~~~~~~

asas

acacaasaaccss

(36b)

From the solution of the quadratic equation 0s one obtains nz~ , so finally one has:


119

21

~~ asas

sss

(36c)

where 1~z and 2

~z . Alternatively, these zeros could be calculated by the following procedure:

21

2

1 ~~1

1~0

0~~ cc

a

aseigenvaluezn

(37a)

So one has:

221

211

~~~

~~~~

cac

ccaseigenvaluezn

(37b)

The calculus of the eigenvalues could be made by using the equation 0IA ,

010

01~~~

~~~

221

211

cac

cca

(38a)

Performing the indicated operations and grouping the equation yields to:

0~~~~~~~~~~1221212121

2 acacaaaacc (38b)

The solution of this equation gives 1 and 2 which are the zeros nz~ . Obviously 1 and 2 .

2nd CASE.- With two complex conjugate poles, the procedure to obtain the zeros nz~ is as follows:

1~~

~~

~~

~~2121

bjas

cjc

bjas

cjcs

(39a)

Using a common factor procedure and grouping yields to:

bjasbjas

cbcabasacss ~~~~

~~2~~2

~~~2~2 21

22

1

2

(39b)

From the solution of the quadratic equation 0s one obtains nz~ , so finally:

21

~~ asas

sss

(39c)

where 1~z and 2

~z . Alternatively, these zeros could be calculated by the following procedure:

21

~~0

2~~

~~~ cc

ab

baseigenvaluezn

(40a)

Then,

ab

cbcaseigenvaluezn ~~

~2~~2~

~ 21 (40b)

The calculus of the eigenvalues could be made by using the equation 0IA ,

010

01~~

~2~~2~

21

ab

cbca

(41a)

Making the indicate operations and grouping the equation is 0~~

2~~2~~~2~2 21

22

1

2 cbcabaac (41b)

The solution of this equation gives 1 and 2 which are the zeros nz~ . Obviously 1 and 2 .

VI.D.- Calculation of the residues nc

The original function to be fitted is:

shdas

csf

N

n n

nfit

1

(42a)

From equation (26b), one has that the poles of sf fit are the zeros of s , that means nn za ~ ; this yields to:


120

h

d

c

sas

sfn

N

n n

fit 11

1

(42b)

In matrix form: T

fit ss Xff (42c)

with hdcnT X and

sas

sN

n n

11

1

f . Because sfitf and sf are complex, both are separated in their

real and imaginary parts as follows: sreal fZ 1 (43a)

simag fZ 2 (43b) sreal fitfy 1 (43c)

simag fitfy 2 (43d)

Re-arranging these matrices into a new matrix system yields to the following ones:

2

1

Z

ZZ

(44a)

and

2

1

y

yy

(44b)

Hence TXZy (45)

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 TZ , one will obtains:

yZZZX TT 1 (46)

Finally the Euclidian norm is applied to the vector X . Knowing that hdcnT X , one obtains:

pn Nc :1X (47a) 1 pNd X (47b) 2 pNh X (47c)

where pN is the length of na .

It is important to note that when one has a complex conjugate pair in the initial poles, i.e. ka , then kc will contain the real part and 1kc will contain the imaginary part of the residue. The final approximation is the analytical formula given by the na poles and the nc 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 sH described by:

i

i

i

i

32.6281.0

37.1351.0

164.291.0

2832.61.0

S

i

i

i

i

S

0095491.06

044304.0001.6

20376.00217.6

78136.04095.6

H

1

1

1

1

B


121

where S is the Laplace variable in discrete form, SH 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:

shdas

cs

N

n n

n

1

f

where N gives the order of the synthesis and shd 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. 1st 5.49initP 0indexI

2nd iiinit 10011001 P 21indexI 3rd iiinit 100110015.49 P 210indexI 4th iiiiinit 1001100101.001.0 P 2121indexI

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

Order Strictly proper Proper Improper 1st 396.88

53.555 ssf 0016.61221.3

2112.6 ssf sss 6

105125.10016.61232.3

212.6 f

2nd 1845.3

2366.65106416.2

610585.1

sssf 699951.01635.2

5104.20008.3

99951.01635.2

5104.20008.3

is

i

is

isf sis

i

is

is 8105929.1699996.01637.2

5085.20008.3

99996.01637.2

5085.20008.3

f

3rd is

i

is

i

ss

0374.11841.2

3423.29987.2

0374.11841.2

3423.29987.27104493.4

8106696.2

f

623

2

23

2

5

2

is

i

is

i

ssf sis

i

is

i

ss 1810101.4623

2

23

2

5

2

f

4th 0666.6

23.11

054.18

261.9

91.241

331.1120102319.1

20103699.7

ss

sssf

623

2

23

2

5

2

0123.9

10103113.3

is

i

is

i

sssf

sis

i

is

i

is

i

is

is9

104726.361001

5102064.600050299.0

1001

5102064.600050299.0

0345.11819.2

3564.29988.2

0345.11819.2

3564.29988.2

f

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.


122

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

FIGURE 2.- Real part of the functions.

FIGURE 3.- Imaginary part of the functions.

100

101

102

0

1

2

3

4

5

6

7

Frequency

Ab

solu

te v

alue

Original set of dataFirst order strictly proper functionFirst order proper functionFirst order improper functionSecond order strictly proper functionSecond order proper functionSecond order improper functionThird order strictly proper functionThird order proper functionThird order improper functionFourth order strictly proper functionFourth order proper functionFourth order improper function

100

101

102

0

1

2

3

4

5

6

7

Frequency

Rea

l par

t of

the

fun

ctio

n


100

101

102

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

Frequency

Imag

inar

y pa

rt o

f th

e fu

nct

ion



123

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

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

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

101.3331

101.3332

101.3333

101.3334

101.3335

101.3336

6.001

6.001

6.001

6.001

6.001

Frequency

Abs

olut

e va

lue


101.3331

101.3332

101.3333

101.3334

101.3335

101.3336

6.001

6.001

6.001

6.001

6.001

6.001

Frequency

Rea

l par

t of t

he fu

ncti

on


101.3331

101.3332

101.3333

101.3334

101.3335

101.3336

-0.0443

-0.0443

-0.0443

-0.0443

-0.0443

-0.0443

-0.0443

Frequency

Imag

inar

y pa

rt o

f the

func

tion



124

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,

i

i

i

i

32.6281.0

37.1351.0

164.291.0

2832.61.0

S

i

i

i

i

S

0095491.06

044304.0001.6

20376.00217.6

78136.04095.6

H

1

1

1

1

B

i

iinit

101

101

50

P

210

indexI

2. Using this information one obtains matrix A and b for a proper function, as:

init

fit

init P-S

Bf-B

P-S

BA sf fit

*b

Numerically one has,

i105.8096+0.000304040.019104i+103.05820.0094902i+0.000741521i101.7747 105.06730.0031839i 105.5769 0.0015815i0.0001261

i105.8994+0.00658520.089149i+107.42110.039097i+0.0141431i101.79310.00109720.014855i0.00012203 0.0064973i0.0024047

0.0082628i+0.160020.46758i+0.00649590.055296i+0.08800510.0022687i0.026496 0.077523i 0.00370190.0086783i0.014908

0.68385i1.92341.1333i0.641250.031151i+0.1240310.069083i 0.30851 0.18625i+0.0773430.0024645i 0.019651

7-6-7-5-6-

5-5-5-A

i

i

i

i

0095491.06

044304.0001.6

20376.00217.6

78136.04095.6

b

3. Separate the matrices A and b in real and imaginary parts:

7-7-

5-5-

5-6-5-5-6-5-

5-5-

105.80960.0191040.00949020101.77470.00318390.0015815

105.89940.0891490.0390970101.79310.0148550.0064973

0.00826280.467580.05529600.00226870.0775230.0086783

0.683851.13330.03115100.0690830.186250.0024645

1030.404103.05821074.1521105.0673105.57691012.61

0.0065852107.42110.01414310.00109721012.2030.0024047

0.160020.00649590.08800510.0264960.00370190.014908

1.92340.641250.1240310.308510.0773430.019651

A

0095491.0

044304.0

20376.0

78136.0

6

001.6

0217.6

4095.6

b

4. Compute the Euclidian norm of matrix A , which is (by column) 0477.23866.11702.00000.23173.02166.00272.0)( Anorm

5. Solve the system making (by column), )(AAA norm and the applying bAAAX TT 1 , and then ).'(AX.X norm to obtain:

9512.41

3480.16

8041.7

0000.6

7384.252

5909.95

.818245

X

6. One has a third order fit, so the last three terms corresponds to the residues of s :

9512.41

3480.16

8041.7~C

7. Knowing the initial poles and the weighted matrix which are,

11000

10010

005.49

init and

0

2

1

W

8. With the residues of s , the initial poles and the weighted matrix, one obtains the zeros of s as,

9512.413480.168041.7

0

2

1

11000

10010

005.49

eigZEROSZ


125

9. The zeros of the function s are the poles of the transfer function sH , so one has:

i

iROOTS23

23

5

R

10. Use these roots to obtain the residues, the constant and the proportional terms by using,

B

R-S

BA

S

sf fit*b

Numerically one has,

0109977.90.0031831-0.0015914-

0109.99190.01477-0073768.0

010653.98068114.0033271.0

0071259.025911.0095943.0

1100131.1101.5705101.2918

100021799.000033839.000027793.0

10045663.00.00730790.0058183

1047319.015052.0077876.0

8-

6-

5-

5-5-5-

A

0.0095491-

0.044304-

0.20376-

0.78136-

6

6.001

6.0217

6.4095

b

11. Solving the system by following steps 4 and 5, one obtains:

6

1

2

2

X

12. The first three terms of X are the residues of sH and the fourth is the constant term. According with

indexI 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,

i

i

2

2

2

C and 6d

13. Following this procedure, we have the following EDOS,

tuxxx

ii

111

23000230005

3

2

1

x tuxxx

ii 6222

3

2

1

y

14. By using equation (23) the analytical function to fit the discretized sH function is:

623

223

25

2

is

iis

is

ss Hf

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 sH and compare its solution with the first one. If one has the system,

tuBAxx and tuDCxy so, we have,

tux

x

x

x

0

10

21

525

2

1

2

1

and tu

x

x

y

y

y

y

y

y

y

y

0

0

0

1

0

2

0

2

20

21

01

01

10

15

03

02

2

1

8

7

6

5

4

3

2

1

The analytical solution of this ODES is as follows: tt eetx 21522194858273.27857805141726.24

1 10437.014075.0


126

tt eetx 21522194858273.27857805141726.24

2 11996.010179.0 Then the analytical solution for the output is: tueey tt 21716007780.03577344100357170.81493752 21522194858273.27857805141726.24

1

tt eey 21522194858273.27857805141726.24

2 1074011670.0536601710053581.22240628

tueey tt 2174980260.110271281370781.99351140 21522194858273.27857805141726.24

3

tt eey 21522194858273.27857805141726.24

4 153982210.199704901638149490.04383239

tueey tt 21522194858273.27857805141726.24

5 1358003890.01788672100178580.40746876

tt eey 21522194858273.27857805141726.24

6 1358003890.01788672100178580.40746876

tt eey 21522194858273.27857805141726.24

7 14376480.41729653127808480.49513355

tt eey 21522194858273.27857805141726.24

8 107964420.399409811276298970.08766479 These results are used like a benchmark, but the intention is to construct an ODES by supposing that we know only the transfer function sH , which is:

T

ssss

s

ss

s

ss

ss

ssss

ss

ss

s

ss

sss

55272

20

55272

10

55272

2010

55272

35172

55272

10

55272

42

2

55272

6030

55272

70342

2H

Discretizing the transfer function sH with 500 samples, logarithmically distributed between 83 1010 Hz, then making js , figure 7 shows the behavior of each term.

FIGURE 7a.- Absolute value of the function sH .

10-2

100

102

104

106

1080

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Frequency in Hz

Abs

olut

e va

lue

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


127

FIGURE 7b.- Real part of sH .

FIGURE 7c.- Imaginary part of sH .

From a priori knowledge, it is a fact that the system has common poles, that means, all data of sH 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:

tuxx

xx

11

21522194858273.2007857805141726.24

2

1

2

1

tux

x

y

y

y

y

y

y

y

y

0

0

0

1

0

2

0

2

8864.08864.0

9837.09837.10

0972.00972.10

0972.00972.10

4432.04432.0

0432.00432.50

2918.02918.30

1945.01945.20

2

1

8

7

6

5

4

3

2

1

10-2

100

102

104

106

108-0.5

0

0.5

1

1.5

2

Frequency in Hz

Rea

l par

t of t

he fu

ncti

on

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

10-2

100

102

104

106

108-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

Frequency in Hz

Imag

inar

y pa

rt o

f the

func

tion

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


128

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: tetx 7857805141726.24

1 1694820403542877.0 tetx 21522194858273.2

2 1396094505548031.0 Then the analytical solution for the output is: tueey tt 21921065390.08763290143608090.81493466 21522194858273.27857805141726.24

1

tt eey 21522194858273.27857805141726.24

2 115561380.131471891425561.22240401

tueey tt 21749563110.0194639613705752.01945769 21522194858273.27857805141726.24

3

tt eey 21522194858273.27857805141726.24

4 187514750.199685881033943450.01788502

tueey tt 21522194858273.27857805141726.24

5 16865170.04379392144660160.40746531

tt eey 21522194858273.27857805141726.24

6 16865170.04379392144660160.40746531

tt eey 21522194858273.27857805141726.24

7 198484330.44321075105736620.44323939

tt eey 21522194858273.27857805141726.24

8 175029490.39937177106788690.03577004 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.

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

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5

2

Time in seconds

Am

plit

ude

in p

.u.

Original functionsFitted functions


129

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:

tu

e

e

e

x

x

x

aaa

aaa

aaa

x

x

x

nnnnnn

n

n

n

2

1

2

1

21

22221

11211

2

1

(48a)

tu

d

d

d

x

x

x

ccc

ccc

ccc

y

y

y

mnmnmm

n

n

m

2

1

2

1

21

22221

11111

2

1

(48b)

If one knows that the transfer function sH of this ODES is,

ssssss

s

sss

sss

sss

ss

sss

sss

sss

sss

sss

sss

sss

sss

6

5

4

3

2

1

785.2951.295.9

36.10953.1085.283

785.2951.295.9

304010

785.2951.295.9

93.13855.1175.275

785.2951.295.9

14.8904.9824

785.2951.295.9

57.9902.149392

785.2951.295.9

5.18857.1965.367

23

23

23

2

23

23

23

23

23

23

23

23

H

H

H

H

H

H

H

(49)

The first step is to discretize the transfer function sH , in order to compare the effect of the sampling, we use

4 and 10000 samples logarithmically distributed between 53 1010 Hz, making js . 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.

TABLE 3.- Roots using 4 and 10000 samples Roots using 4 samples Roots using 10000 samples

s1H 300000.2499999.3700000.3 321 rrr 299999.2500000.3699999.3 321 rrr s2H 299999.2499999.3700000.3 321 rrr 300000.2500000.3699999.3 321 rrr

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-0.01

-0.005

0

0.005

0.01

Time in seconds

Err

or in

per

cent

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


130

s3H 299999.2500000.3699999.3 321 rrr 300000.2499999.3700000.3 321 rrr s4H 299999.2500000.3699999.3 321 rrr 300000.2499999.3700000.3 321 rrr s5H 300000.2499999.3700000.3 321 rrr 299999.2499999.3700000.3 321 rrr s6H 299999.2500000.3699999.3 321 rrr 299999.2499999.3700000.3 321 rrr

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 3.25.37.3 321 rrr . 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.

FIGURE 10.- Absolute value of the function sH , original and fitted.

FIGURE 11.- Real part of the function sH , original and fitted.

10-2

100

102

104

0

1

2

3

4

5

6

7

8

Frequency in Hz

Abs

olut

e va

lue

of t

he f

unct

ion

Original functionResulting of the numerical fitting

10-3

10-2

10-1

100

101

102

103

104

105

-2

-1

0

1

2

3

4

5

6

7

8

Frequency in Hz

Rea

l par

t of

the

fun

ctio

n



131

FIGURE 12.- Imaginary part of the function sH , original and fitted.

Finally, by the use of this methodology we arrive to the following ODES:

tux

x

x

x

x

x

1

1

1

3.200

05.30

007.3

3

2

1

2

2

1

(50a)

tux

x

x

y

y

y

y

y

y

3

0

5

4

2

7

476.1533.20886.192

44.6208.92764.874

762.3933.70857.688

48.1103.18339.1762

6667.6333.83110

726.861.14274.1370

3

2

1

6

5

4

3

2

1

(50b)

This ODES has analytical solution as: tetx 7.3

1 1702702702702.0

tetx 5.32 1142857142857.0

tetx 3.23 1954347826086.0

Then the analytical solution for the output is: tueeey ttt 71717070393374.371947380952380.0741953667953667.703 3.25.37.3

1

tueeey ttt 21378985507246.21238095238095.231297297297297.29 3.25.37.32

tueeey ttt 41930331262939.481238095238095.5231764478764478.476 3.25.37.33

tueeey ttt 51882877846790.171523809523809.2021861003861003.186 3.25.37.34

tueeey ttt 01921480331262.271518809523809.2641613899613899.236 3.25.37.35

tueeey ttt 31087287784679.61095238095238.591521235521235.52 3.25.37.36

Figure 13 shows the solution of the system described by equation (50a,b).

10-3

10-2

10-1

100

101

102

103

104

105-3

-2

-1

0

1

2

3

Frequency in Hz

Imag

inar

y pa

rt o

f the

func

tion



132

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 sH .

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.

X. 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-07-

029213-2.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-2

0

2

4

6

8

Time in seconds

Am

plit

ude

in p

.u.

y1y2y3y4y5y6


133

[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.

Solution of Ordinary Differential Equation System with Unknown Coefficients

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