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Journal of Applied Mathematics

Volume 2012 (2012), Article ID 840603, 32 pages

http://dx.doi.org/10.1155/2012/840603

## Interval Continuous Plant Identification from Value Sets

^{1}Departamento de Sistemas de Comunicación y Control, UNED, c/Juan del Rosal 16, 28040 Madrid, Spain^{2}Departamento de Tecnología de Computadores y Comunicaciones, Universidad de Extremadura, 06800 Madrid, Spain

Received 13 April 2012; Revised 1 September 2012; Accepted 1 September 2012

Academic Editor: Zhiwei Gao

Copyright © 2012 R. Hernández et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

This paper shows how to obtain the values of the numerator and denominator Kharitonov polynomials of an interval plant from its value set at a given frequency. Moreover, it is proven that given a value set, all the assigned polynomials of the vertices can be determined if and only if there is a complete edge or a complete arc lying on a quadrant. This algorithm is nonconservative in the sense that if the value-set boundary of an interval plant is exactly known, and particularly its vertices, then the Kharitonov rectangles are exactly those used to obtain these value sets.

#### 1. Introduction

In reference to the identification problem, these have been widely motivated and analysed over recent years [1]. Van Overschee and De Moor in [2] explains a subspace identification algorithm. In [3] the authors present a robust identification procedure for a priori classes of models in ; the authors consider casual, linear time invariant, stable, both continuous or discrete time models, and only SISO systems.

Interval plants have been widely motivated and analysed over recent years. For further engineering motivation, among the numerous papers and books, [4–9] must be pointed out and the references thereof.

The identification problem using the interval plant framework, that is, to compute an interval plant from the frequency response, has not been completely solved. Interval plant identification was investigated by Bhattacharyya et al. [5], who developed a method in which identification is carried out for interval plants so that the numerator and denominator have the same degree, starting from the variation of the coefficient values of a nominal transfer function at certain intervals. So, the identification of a nominal transfer function is carried out first, and then the intervals of variation of the coefficients are determined.

A different approach was developed by Hernández et al. [10] studying the problem from the extreme point results point of view. This was a first step for the identification of an interval plant, showing three main properties to characterize the value set lying on a quadrant. Then an algorithm for the identification of interval plants from the vertices of the value sets is obtained. However, this algorithm solves the identification problem when the value set contains at least five vertices in a quadrant.

This paper improves the results obtained in [10] and shows how to obtain the values of the numerator and denominator Kharitonov polynomials when the value sets have less than five vertices in the same quadrant. Identification with such an interval plant allows engineers predict the worst case performance and stability margins using the results on interval systems, particularly extreme point results.

#### 2. Problem Statement

Let us consider a linear interval plant of real coefficients, of the form where and are interval polynomials given as with ,,, and where vectors , and are the uncertainty parameters that lie in the hyperrectangles and , respectively.

Numerator and denominator polynomial families are characterized by their respective Kharitonov polynomials, and they can be expressed in terms of their even and odd parts, at , as follows:

Family : where

Family : where

As is well known, the values of the complex plane obtained for the transfer function at a given frequency are denominated as a *value set.* The identification of the system consists in determining the transfer function coefficients from the value set.

As can be observed in [10], when the values and are known, then the system of equations given in [10, equation 14] can be solved and therefore the interval plant is identified (see [10] for details).

As is shown [10] the vertices of the value-set boundary of an interval plant can be assigned as
where and are the assigned polynomials numerator and denominator, respectively. When they are in the same quadrant they are a *Sorted Set of Vertices (SSV)*.

As is well known, the Kharitonov polynomials values can be obtained from It must be pointed out that the results presented in [10] must be considered as the background necessary for this work. Thus, the geometry of the value set is described in [10] and the concepts necessary for its description are defined, (such as the successor, predecessor element, etc.) and the fundamental properties on which this work is based are proven.

This paper is organized as follows. Section 3 shows how to determine the assigned polynomial with the only condition that there is a complete segment in a quadrant. Similarly Section 4 shows it when there is an arc in a quadrant. Section 5 illustrates the algorithm and examples. Finally, the conclusions are shown in Section 6.

#### 3. Assigned Polynomial Determination When There Is a Complete Segment in a Quadrant

In order to determine the polynomials numerator and denominator associated to a vertex of the value set boundary with the minimum number of elements, the situation of a segment in a quadrant will be considered. So, let be a segment of the value-set boundary with vertices and . Continuity segment-arc in a quadrant (see [10, Theorem 2]) implies that there will be a successor arc with vertices , counter-clockwise and a predecessor arc with vertices counter-clockwise. When these arcs are completed the denominators are vertices of the Kharitonov rectangle. Figures 1 and 2 show this situation.

As was shown, the values of ,, and can be calculated from the complete segment based on a normalization (see [10, Theorem 4]). The following normalization simplifies the nomenclature.

Lemma 3.1 (segment normalization). * Let be a complete segment of the value-set boundary with vertices and and the normalization , where being the argument of the segment . Then ,,, and , where () is any point of the next (previous) arc of the segment .*

*Proof. *It is trivial. This normalization is one of the infinite possible solutions [10] for a value set. This normalization implies fitting with modulus and angle so that the segment of the Kharitonov polynomial numerator with vertices and will be parallel to the real axis counter-clockwise. Thus, from the information with a complete segment in a quadrant the values of ,,,, and can be calculated.

This paper deals with the general case where ,,, and .

Given a vertex in a quadrant, the target is to determine the polynomials and . The vertex belongs to a part of a segment and a part of an arc, due to the continuity segment-arc in a quadrant. So, will be the vertex of two elements, arc-segment (Figure 3) or segment-arc (Figure 4).

The following Lemma shows the necessary conditions on the denominator to be a solution of .

Lemma 3.2 (denominator condition). * Let be a complete segment in a quadrant and let be the denominator of a vertex in a quadrant. Then it is a necessary condition that satisfies one of the following conditions:*(1)* and or or or ,*(2)* and or or or ,*(3)* and or or or ,*(4)* and or or or ,**
where is the real part of and is the imaginary part of , and the corresponding assigned denominator is shown between brackets.*

*Proof. * The proof is obtained directly from the information of a complete segment in a quadrant and the properties of the Kharitonov rectangle. So, from the complete segment and the normalization (Lemma 3.1), the values of ,, and are known. Then, can be established as ,,**,** or . If then is . Given a value , it will be a vertex of the Kharitonov rectangle denominator only if ( is ) or ( is ) or ( is ) or ( is ). (Figures 5(a), 5(b), 5(c), and 5(d)). Note that if any of these conditions is not satisfied, then cannot be a solution. For example, if , does not belong to the rectangle with vertex , , and are elements of the successor and predecessor edges. Figure 6 shows these considerations. Similarly, if then is . Given a value , it will be a vertex of the Kharitonov rectangle denominator only if ( is ) or ( is ) or ( is ) or ( is ). If then is . Given a value , it will be a vertex of the Kharitonov rectangle denominator only if ( is ) or ( is ) or ( is ) or ( is ). Finally, if then is . Given a value , it will be a vertex of the Kharitonov rectangle denominator only if ( is ) or ( is ) or ( is ) or ( is ).

On the other hand, the behaviour of a segment on the complex plane when divided by a complex number is well known. The following property shows this behaviour.

*Property 1. *Let be a segment on the complex plane with vertices and counter-clockwise where is a segment with vertices and counter-clockwise. Let be a complex number with argument . Let be . Then the relation between the argument of and , is given by(1) if and only if ,(2) if and only if ,(3) if and only if ,(4) if and only if .

The following Theorem shows how to characterize and calculate the polynomials and associated with a vertex from the information of the boundary with a segment in a quadrant, belonging to a segment-arc.

Theorem 3.3 (predecessor). *Let be a complete segment of the value-set boundary with vertices and , the successor arc with vertices , counter-clockwise, and the predecessor arc with vertices , counter-clockwise. Let be a segment with vertices and counter-clockwise, where belongs to the intersection of and an arc of the boundary (Figure 7). Then*(1)* (condition C1) and the denominator of defined by satisfies the denominator condition (Lemma 3.2), if and only if and cannot be any other assigned polynomial,*(2)*when , (condition C2) and the denominator of defined by satisfies the denominator condition (Lemma 3.2) if and only if and cannot be any other assigned polynomial,*(3)*when and , (condition C3), and the denominator of defined by satisfies the denominator condition (Lemma 3.2) if and only if and cannot be any other assigned polynomial,*(4)*when , , and , (condition C4), and the denominator of defined by satisfies the denominator condition (Lemma 3.2) if and only if .*

*Proof. *From the complete segment using the normalization (Lemma 3.1) the values of ,,,, and are known. Obviously the value is known.

If the value of can be calculated and the denominator condition (Lemma 3.2) is satisfied. On the other hand, the quotient of the vertices and is , and . , where is part of the segment with vertices and , then (normalization). Thus (Property 1) and ; Theorem 3.3(C1) is satisfied.

In order to demonstrate the “only if” part, it must be proven that if Theorem 3.3(C1) and the denominator condition are satisfied then the solution , is unique. It must be noted that Theorem 3.3(C1) can be satisfied when (a) , (b) or (c) and in all the cases, the value of determined, verify the denominator condition.

Let be the denominator of determined by , verifying Theorem 3.3(C1), and denominator condition, and let where is part of the segment with vertices and , .

Let be the denominator of determined by . Then where is part of the segment with vertices and , (normalization) and using Property 1 . As is the same vertex, then , and . verify Theorem 3.3(C1), because
Let with . Then and (by normalization ). Thus . Moreover , and if then . As and , then and they have equal real and imaginary parts.

then
Thus

then
Thus .

Taking into account both conditions, . This relation is impossible. Therefore, if is a solution then is not, and is not a solution.

Let be the denominator of determined by . Then where is part of the segment with vertices and , (normalization) and using Property 1 . As is the same vertex, then and . verify Theorem 3.3(C1), because
In this case the demonstration is trivial noting that . This is not possible because the Kharitonov polynomial denominator cannot contain the zero.

Let be the denominator of determined by . Then where is part of the segment with vertices and , (normalization) and using Property 1 . As is the same vertex, then , and . verify Theorem 3.3(C1), because
Let with . Then and (by normalization ). Thus . Moreover , and if then . As and , then and they have equals real and imaginary parts.

then
Thus .

then
Thus .

Taking into account both conditions, . This relation is impossible. Therefore, if is a solution, is not and cannot be a solution.

If the value of can be calculated and the denominator condition (Lemma 3.2) is satisfied. On the other hand, the quotient of the vertices and is , and . where is part of the segment with vertices and , then (normalization). Thus (Property 1) and ; Theorem 3.3(C2)is satisfied.

In order to demonstrate the “only if” part, it must be proven that if Theorem 3.3(C2) and the denominator condition are satisfied then the solution , is unique. It must be noted that Theorem 3.3(C2) can be satisfied when (a) or (b) and in all the cases, the value of determined, verify the denominator condition.

Let be the denominator of determined by , verifying Theorem 3.3(C2), and denominator condition, and let where is part of the segment with vertices and , .

Let be the denominator of determined by . Then where is part of the segment with vertices and , (normalization) and using Property 1 . As is the same vertex, then and . verify Theorem 3.3(C2), because

In this case the demonstration is trivial noting that . This is not possible because the Kharitonov polynomial denominator cannot contain the zero.

Let be the denominator determined by . Then where is part of the segment with vertices and , (normalization) and using Property 1. As is the same vertex, then , and . verify Theorem 3.3(C2), because

Let with . Then and (by normalization ). Thus . Moreover , and if then . How and , then and they have equals real and imaginary parts.

Thus .

and finally .

Taking into account both conditions, . This relation is impossible. Therefore, if is a solution, is not and is not a solution.

If then cannot be directly calculated because is not known. First, Theorem 3.3(C3) is developed. If then where is part of the segment with vertices and and . Thus (Property 1) and .

As , then . On the other hand, is greater than because it is counter-clockwise. Therefore (Theorem 3.3(C3)) is satisfied and can be calculated by the expression .

In order to demonstrate the “only if” part, it must be proven that if Theorem 3.3(C3) and the denominator condition are satisfied then the solution , is unique. If and , it must be noted that Theorem 3.3(C3) can be satisfied when .

Let be the denominator of determined by verifying Theorem 3.3(C3) and denominator condition. where is part of the segment with vertices and , .

Let be the denominator of determined by . Then where is part of the segment with vertices and , (normalization) and using Property 1 . Thus .

As is the same vertex, , and then . Let , then , and because verifies (by normalization), Theorem 3.3(C3) is satisfied.

. If then . As and , then and they have equal real and imaginary parts.

and finally .

and finally .

Taking into account both conditions, . This relation is impossible. Therefore, if is a solution, is not, and is not a solution.

If then cannot be directly calculated because is not known. First, Theorem 3.3(C4) is developed.

If then where is part of the segment with vertices and verifying that . Thus (Property 1) and . Moreover, . Then . On the other hand, is greater than because it is counter-clockwise.

Therefore the condition Theorem 3.3(C4) is satisfied and can be calculated using the expression .

If , and it is .

*Remark 3.4. *This theorem is used in the example of Section 5, for the value set III (frequency ) in order to assign the second and fifth vertices.

The following Theorem is analogous to Theorem 3.3 when is a segment with vertices and counter-clockwise, and belonging to an arc-segment.

Theorem 3.5 (successor). *Let be a complete segment of the value-set boundary with vertices and , the successor arc to , with vertices , counter-clockwise, and the predecessor arc to with vertices , counter-clockwise. Let be a boundary segment with vertices and counter-clockwise, where belongs to the intersection of an arc of the boundary and . Then *(1)* (condition C1) and the denominator of defined by satisfies the denominator condition (Lemma 3.2), if and only if and cannot be any other assigned polynomial,*(2)*when , (condition C2) and the denominator of of defined by satisfies the denominator condition (Lemma 3.2) if and only if and cannot be any other assigned polynomial,*(3)*when and , (condition C3), and the denominator of defined by satisfies the denominator condition (Lemma 3.2) if and only if and cannot be any other assigned polynomial,*(4)*when , , and , (condition C4), and the denominator of defined by satisfies the denominator condition (Lemma 3.2) if and only if .*

* Proof. * Analogous to Theorem 3.3.

*Remark 3.6. * This theorem is used in the example of Section 5, for the value set III (frequency ) in order to assign the third, fifth, and sixth vertices.

#### 4. Assigned Polynomial Determination When There Is a Complete Arc in a Quadrant

In order to determine the polynomials numerator and denominator associated to a vertex of the value set boundary with the minimum number of elements, the situation of an arc in a quadrant will be considered. So, let be an arc of the value-set boundary with vertices and . A continuity arc-segment in a quadrant (see [10, Theorem 2]) implies that there will be a successor segment with vertices , counter-clockwise and a predecessor segment with vertices and counter-clockwise.

When these segments are completed the denominators are vertices of the Kharitonov rectangle. Figure 8 shows this situation.

As was shown, the values of , , and can be calculated from the complete arc based on a normalization (see [10, Theorem 5]). The following normalization simplifies the nomenclature.

Lemma 4.1 (arc normalization). *Let be a complete arc of the value-set boundary with vertices and , the normalization , where , being the argument of the segment . Then , , , and , where () is any point of the next (previous) segment of the arc .*

*Proof. * It is trivial. This normalization is one of the infinite possible solutions for a value set. This normalization implies fitting with modulus and angle so that the segment of the Kharitonov polynomial denominator with vertices and will be parallel to the real axis counter-clockwise. Thus, from the information with a complete arc in a quadrant the values of , , , , and can be calculated.

This paper deals with the general case where , , , and .

Given a vertex in a quadrant, the target is to determine the polynomials and . The vertex belongs to a part of an arc and a part of a segment, due to the continuity arc-segment in a quadrant. So, will be the vertex of two elements, segment-arc (Figure 9(a)) or arc-segment (Figure 9(b)).

The following Lemma shows the necessary conditions on the denominator to be a solution of .

Lemma 4.2 (numerator condition). *Let be a complete arc in a quadrant and let be the numerator of a vertex in a quadrant. Then it is a necessary condition that satisfies one of the following conditions:*(1)* and or or or ,*(2)* and or or or ,*(3)* and or or or ,*(4)* and or or or ,**
where is the real part of and is the imaginary part of , and the corresponding assigned numerator is shown between brackets.*

*Proof. *The proof is obtained directly from the information of a complete arc in a quadrant and the properties of the Kharitonov rectangle. So, from the complete arc and the normalization (Lemma 3.2), the values of , , and are known. Then, can be established as , , , or . (1)If then is . Given a value , it will be a vertex of the Kharitonov rectangle numerator only if ( is ) or ( is ) or ( is ) or ( is ). Note that if any of these conditions is not satisfied, then cannot be a solution. For example, if , does not belong to the rectangle with vertex , , and are elements of the successor and predecessor edge.(2)Similarly, if then is . Given a value , it will be a vertex of the Kharitonov rectangle numerator only if ( is ) or ( is ) or ( is ) or ( is ).(3)If then is . Given a value , it will be a vertex of the Kharitonov rectangle numerator only if ( is ) or ( is ) or ( is ) or ( is ). (4)Finally, if then is . Given a value , it will be a vertex of the Kharitonov rectangle numerator only if ( is ) or ( is ) or ( is ) or ( is ).

On the other hand, the behaviour of an arc on the complex plane when it is divided by a complex number is well known. The following property shows this behaviour.

*Property 2. *Let be an arc on the complex plane with vertices and counter-clockwise where is a segment with vertices and counter-clockwise. Let be a complex number with argument . Let be . Then the relation between the argument of and , is given by(1) if and only if ,(2) if and only if ,(3) if and only if ,(4)