# Interval Continuous Plant Identification from  Value Sets

Authors: R. Hernández; J. A. García; C. Mañoso
Journal: Journal of Applied Mathematics
 (2012)

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## 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.


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## Body

## 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 H∞; 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(2.1)P(s,a,b)=Np(s,a)DP(s,b),
where Np(s,a) and DP(s,b) are interval polynomials given as
(2.2)Np(s,a)=amsm+am-1sm-1+⋯+a0,a∈A={a:ai-≤ai≤ai+,i=0,…,m},DP(s,b)=bnsn+bn-1sn-1+⋯+b0,b∈B={b:bi-≤bi≤bi+,i=0,…,n},
with m≥1,n≥1,0∉Dp(s,b), and where vectors a=[a0,a1,…,am],am≠0, and b=[b0,b1,…,bn],bn≠0 are the uncertainty parameters that lie in the hyperrectangles A and B, 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, ats=jω, as follows:FamilyNp(s):
(2.3)kn1=pemin(jω)+jpomin(jω),kn2=pemax(jω)+jpomin(jω),kn3=pemax(jω)+jpomax(jω),kn4=pemin(jω)+jpomax(jω),
where
(2.4)pemin(jω)=a0--a2+ω2+a4-ω4-a6+ω6+⋯,pemax(jω)=a0+-a2-ω2+a4+ω4-a6-ω6+⋯,pomin(jω)=a1-ω-a3+ω3+a5-ω5-a7+ω7+⋯,pomax(jω)=a1+ω-a3-ω3+a5+ω5-a7-ω7+⋯.FamilyDp(s):
(2.5)kd1=qemin(jω)+jqomin(jω),kd2=qemax(jω)+jqomin(jω),kd3=qemax(jω)+jqomax(jω),kd4=qemin(jω)+jqomax(jω),
where
(2.6)qemin(jω)=b0--b2+ω2+b4-ω4-b6+ω6+⋯,qemax(jω)=b0+-b2-ω2+b4+ω4-b6-ω6+⋯,qomin(jω)=b1-ω-b3+ω3+b5-ω5-b7+ω7+⋯,qomax(jω)=b1+ω-b3-ω3+b5+ω5-b7-ω7+⋯.As is well known, the valuesG(jω) of the complex plane obtained for the transfer function G(s) 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 {kn1(jω),kn2(jω),kn3(jω),kn4(jω)} and {kd1(jω),kd2(jω),kd3(jω),kd4(jω)} 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
(2.7)vi=njdk,
where nj,j=1,2,3,4 and dk,k=1,2,3,4 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(2.8)kn1(jω)=min[Re(n1,n3)]+jmin[Im(n1,n3)],kn2(jω)=max[Re(n1,n3)]+jmin[Im(n1,n3)],kn3(jω)=max[Re(n1,n3)]+jmax[Im(n1,n3)],kn4(jω)=min[Re(n1,n3)]+jmax[Im(n1,n3)],kd1(jω)=min[Re(d1,d3)]+jmin[Im(d1,d3)],kd2(jω)=max[Re(d1,d3)]+jmin[Im(d1,d3)],kd3(jω)=max[Re(d1,d3)]+jmax[Im(d1,d3)],kd4(jω)=min[Re(d1,d3)]+jmax[Im(d1,d3)].
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. Section3 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, letS1 be a segment of the value-set boundary with vertices v1=n1/d1 and v2=n2/d1. Continuity segment-arc in a quadrant (see [10, Theorem 2]) implies that there will be a successor arc with vertices v2=n2/d1,v2succ=n2/d2λ counter-clockwise and a predecessor arc with vertices v1pred=n1/d4λ counter-clockwise. When these arcs are completed the denominators are vertices of the Kharitonov rectangle. Figures 1 and 2 show this situation.Figure 1
egment and complete arcs.Figure 2
Segment and no complete arcs.As was shown, the values ofn1,n2, and d1 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).
LetS1 be a complete segment of the value-set boundary with vertices v1=n1/d1 and v2=n2/d1 and the normalization d1=cos(φ(d1))+jsin(φ(d1)), where φ(d1)=360°-arg(v2-v1)arg(v2-v1) being the argument of the segment v2-v1. Then n1=v1d1,n2=v2d1,d2λ=n2/v2succ, and d4λ=n1/v1pred, where v2succ (v1pred) is any point of the next (previous) arc of the segment S1.Proof.
It is trivial. This normalization is one of the infinite possible solutions [10] for a value set. This normalization implies fitting d1 with modulus |d1|=1 and angle so that the segment of the Kharitonov polynomial numerator with vertices n1 and n2 will be parallel to the real axis counter-clockwise. Thus, from the information with a complete segment in a quadrant the values of d1,n1,n2,d2λ, and d4λ can be calculated.This paper deals with the general case wheren2R≠0,n2I≠0,n1R≠0, and n1I≠0.Given a vertexvx=nx/dx in a quadrant, the target is to determine the polynomials nx and dx. The vertex vx belongs to a part of a segment and a part of an arc, due to the continuity segment-arc in a quadrant. So, vx will be the vertex of two elements, arc-segment (Figure 3) or segment-arc (Figure 4).Figure 3
v
x vertex of two elements, arc-segment.Figure 4
v
x vertex of two elements, segment-arc.The following Lemma shows the necessary conditions on the denominatordx to be a solution of vx=nx/dx.Lemma 3.2 (denominator condition).
LetS1 be a complete segment in a quadrant and let dx be the denominator of a vertex vx=nx/dx in a quadrant. Then it is a necessary condition that dx satisfies one of the following conditions:(1)
(d1R<d2λRandd1I<d4λI) and {(dxR=d1RanddxI=d1I)[dx=d1]or (dxR=d1RanddxI≥d1I)[dx=d4] or (dxI=d1IanddxR≥d1R)[dx=d2] or (dxR>d1RanddxI>d1I)[dx=d3]},(2)
(d1R>d4λRandd1I<d2λI) and {(dxR=d1RanddxI=d1I)[dx=d1] or (dxR=d1RanddxI≥d1I)[dx=d2] or  (dxI=d1IanddxR≤d1R)[dx=d4] or (dxR<d1RanddxI>d1I)[dx=d3]},(3)
(d1R>d2λRandd1I>d4λI) and {(dxR=d1RanddxI=d1I)[dx=d1] or (dxR=d1RanddxI≤d1I)[dx=d4] or  (dxI=d1IanddxR≤d1R)[dx=d2] or (dxR<d1RanddxI<d1I)[dx=d3]},(4)
(d1R<d4λRandd1I>d2λI) and {(dxR=d1RanddxI=d1I)[dx=d1] or (dxR=d1RanddxI≤d1I)[dx=d2] or  (dxI=d1IanddxR≥d1R)[dx=d4] or (dxR>d1RanddxI<d1I)[dx=d3]},
wherediR is the real part of di and diI is the imaginary part of di, 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 (Lemma3.1), the values of d1,d2λ, and d4λ are known. Then,d1 can be established  as kd1,kd2,kd3, or kd4. (1)
If(d1R<d2λRandd1I<d4λI) then d1 is kd1. Given a value dx, it will be a vertex of the Kharitonov rectangle denominator only if dxR=d1RanddxI=d1I (dx is d1=kd1) or dxR=d1RanddxI>d1I (dx is d4=kd4) or dxI=d1IanddxR>d1R (dx is d2=kd2) or dxR>d1RanddxI>d1I (dx is d3=kd3).  (Figures 5(a), 5(b), 5(c), and 5(d)).Note that if any of these conditions is not satisfied, thendx cannot be a solution. For example, if dxR=d1RanddxI<d1I, dx does not belong to the rectangle with vertex d1, d2λ, and d4λ are elements of the successor and predecessor edges. Figure 6 shows these considerations.(2)
Similarly, if(d1R>d4λRandd1I<d2λI) then d1 is kd2. Given a value dx, it will be a vertex of the Kharitonov rectangle denominator only if dxR=d1RanddxI=d1I (dx is d1=kd2) or dxR=d1RanddxI>d1I (dx is d2=kd3) or dxI=d1IanddxR<d1R (dx is d4=kd1)  or dxR<d1RanddxI>d1I (dx is d3=kd4).(3)
Ifd1R>d2λRandd1I>d4λI then d1 is kd3. Given a value dx, it will be a vertex of the Kharitonov rectangle denominator only if dxR=d1RanddxI=d1I (dx is d1=kd3) or dxR=d1RanddxI<d1I (dx is d4=kd2) or dxI=d1IanddxR<d1R (dx is d2=kd4)  or dxR<d1RanddxI<d1I (dx is d3=kd1).(4)
Finally, ifd1R<d4λRandd1I>d2λI then d1 is kd4. Given a value dx, it will be a vertex of the Kharitonov rectangle denominator only if dxR=d1RanddxI=d1I (dx is d1=kd4) or dxR=d1RanddxI<d1I (dx is d2=kd1) or dxI=d1IanddxR>d1R (dx is d4=kd3)  or dxR>d1RanddxI<d1I (dx is d3=kd2).Cases wheredx is a vertex of the kharitonov rectangle denominator.
(a)
(b)
(c)
(d)Figure 6
Cases wheredx is not a vertex of the kharitonov rectangle denominator.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.
LetSx=S/dx be a segment on the complex plane with vertices vx1 and vx2 counter-clockwise where S is a segment with vertices na and nb counter-clockwise. Let dx be a complex number with argument arg(dx). Let φ(Sx) be φ(Sx)≡arg(vx2-vx1). Then the relation between the argument of dx and φ(Sx), is given by(1)
arg(dx)=-φ(Sx) if and only if arg(nb-na)=0°,(2)
arg(dx)=90°-φ(Sx) if and only if arg(nb-na)=90°,(3)
arg(dx)=180°-φ(Sx) if and only if arg(nb-na)=180°,(4)
arg(dx)=270°-φ(Sx) if and only if arg(nb-na)=270°.The following Theorem shows how to characterize and calculate the polynomialsnx and dx associated with a vertex  vx=nx/dx from the information of the boundary with a segment Sx in a quadrant, vx=nx/dx belonging to a segment-arc.Theorem 3.3 (predecessor).
LetS1 be a complete segment of the value-set boundary with vertices v1=n1/d1 and v2=n2/d1, the successor arc with vertices v2=n2/d1, v2succ=n2/d2λ counter-clockwise, and the predecessor arc with vertices  v1pred=n1/d4λ,  v1=n1/d1 counter-clockwise. Let Sx be a segment with vertices vxpred=nxpred/dx and vx=nx/dx counter-clockwise, where vx belongs to the intersection of  Sx and an arc of the boundary (Figure 7). Then(1)
arg(vx/v2)=arg(d1)+φ(Sx) (condition C1) and the denominator dx of vx defined by n2/vx satisfies the denominator condition (Lemma 3.2), if and only if nx=n2 and cannot be any other assigned polynomial,(2)
whennx≠n2, arg(vx/v1)=arg(d1)+φ(Sx)+90° (condition C2) and the denominator dx of vx defined by n1/vx satisfies the denominator condition (Lemma 3.2) if and only if nx=n1 and cannot be any other assigned polynomial,(3)
whennx≠n1 and nx≠n2, tan(arg(vx)-φ(Sx)+90°)n2R>n2I (condition C3), and the denominator dx of vx defined by n2R[1+jtan(arg(vx)-φ(Sx)+90°)]/vx satisfies the denominator condition (Lemma 3.2)  if and only if nx=n3=n2R[1+jtan(arg(vx)-φ(Sx)+90°)] and cannot be any other assigned polynomial,(4)
whennx≠n1, nx≠n2, and nx≠n3, tan(arg(vx)-φ(Sx)+180°)n1R>n1I (condition C4), and the denominator dx of vx defined by n1R[1+jtan(arg(vx)-φ(Sx)+180°)]/vx satisfies the denominator condition (Lemma 3.2)  if and only if nx=n4=n1R[1+jtan(arg(vx)-φ(Sx)+180°)].Figure 7
Vertices for the conditions of the Theorem3.3.Proof.
From the complete segmentS1 using the normalization (Lemma 3.1) the values of d1,n1=v1d1,n2=v2d1,d2λ=n2/v2succ, and d4λ=n1/v1pred are known. Obviously the value vx is known.
(
1
)
⇐ If nx=n2 the value of dx=n2/vx can be calculated and the denominator condition (Lemma 3.2) is satisfied.  On the other hand, the quotient of the vertices vx=n2/dx and v2=n2/d1 is vx/v2=d1/dx, and arg(vx/v2)=arg(d1)-arg(dx). Sx=S2/dx, where S2 is part of the segment with vertices n1 and n2, then arg(n2-n1)=0° (normalization). Thus arg(dx)=-φ(Sx) (Property 1) and arg(vx/v2)=arg(d1)+φ(Sx);  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 dx=n2/vx,  nx=n2 is unique. It must be noted that Theorem 3.3(C1) can be satisfied when (a) nx=n3, (b) nx=n4 or (c) nx=n1 and in all the cases, the value of  dx determined, verify the denominator condition.
Letdx be the denominator of vx determined by n2/vx, verifying Theorem 3.3(C1), and denominator condition, and let Sx=S2/dx where S2 is part of the segment with vertices n1 and n2,  arg(n2-n1)=0°.
(
a
) Let dx* be the denominator of vx determined by n3/vx. Then Sx=S3/dx* where S3 is part of the segment with vertices n2 and n3, arg(n3-n2)=90° (normalization) and using Property 1  arg(dx*)=90°-φ(Sx). As vx is the same vertex, then arg(n3/dx*)=arg(n2/dx), and arg(n3)=arg(n2)+90°. nx=n3  verify Theorem 3.3(C1), because
(3.1)arg(vxv2)=arg(n2)+90°-arg(n2)+arg(d1)-90°+φ(Sx)=arg(d1)+φ(Sx).
Let α=arg(n2) with tan(α)=n2I/n2R. Then arg(n3)=α+90° and tan(α+90°)=n3I/n3R=n3I/n2R (by normalization n3R=n2R). Thus n3=n2R+jn3I=n2R+jtan(α+90°)n2R=n2R-j(n2R2/n2I). Moreover arg(dx*)=90°+arg(dx), and if dx=dxR+jdxI  then dx*=ρej(π/2)dx=-ρdxI+jρdxR. As vx=n2/dx and vx=n3/dx*, then n2dx*=n3dx and they have equal real and imaginary parts.
Re
[
n
2
d
x
*
]
=
Re
[
n
3
d
x
]  then
(3.2)-ρdxIn2R-ρdxRn2I=n2RdxR+n2R2n2IdxI,-ρdxIn2Rn2I-ρdxRn2I2=n2Rn2IdxR+n2R2dxI,-(ρn2I+n2R)dxRn2I=(n2R+ρn2I)dxIn2R.
Thus dxI/dxR=-n2I/n2R
Im
[
n
2
d
x
*
]
=
Im
[
n
3
d
x
]  then
(3.3)ρdxRn2R-ρdxIn2I=dxIn2R-dxRn2R2n2I,ρdxRn2Rn2I-ρdxIn2I2=dxIn2Rn2I-dxRn2R2,(n2Iρ+n2R)n2RdxR=dxIn2I(n2R+ρn2I).
Thus  dxI/dxR=n2R/n2I.
Taking into account both conditions,n2R/n2I=-n2I/n2R⇔n2R2<0. This relation is impossible. Therefore, if dx is a solution then dx* is not, and nx=n3 is not a solution.
(
b
) Let   dx* be the denominator of vx determined by n4/vx. Then Sx=S4/dx* where S4 is part of the segment with vertices n3 and n4, arg(n4-n3)=180° (normalization) and using Property 1  arg(dx*)=180°-φ(Sx). As vx is the same vertex, then arg(n4/dx*)=arg(n2/dx) and arg(n4)=arg(n2)+180°. nx=n4  verify Theorem 3.3(C1), because
(3.4)arg(vxv2)=arg(n2)+180°-arg(n2)+arg(d1)-180°+φ(Sx)=arg(d1)+φ(Sx).
In this case the demonstration is trivial noting that arg(dx*)=180°+arg(dx). This is not possible because the Kharitonov polynomial denominator cannot contain the zero.
(
c
) Let dx* be the denominator of vx determined by n1/vx. Then Sx=S1/dx* where S1 is part of the segment with vertices n4 and n1, arg(n1-n4)=270° (normalization) and using Property 1  arg(dx*)=270°-φ(Sx). As vx is the same vertex, then arg(n1/dx*)=arg(n2/dx), and arg(n1)=arg(n2)+270°. nx=n1  verify Theorem 3.3(C1), because
(3.5)arg(vxv2)=arg(n2)+270°-arg(n2)+arg(d1)-270°+φ(Sx)=arg(d1)+φ(Sx).
Let α=arg(n2) with tan(α)=n2I/n2R. Then arg(n1)=α+270° and tan(α+270°)=n1I/n1R=n2I/n1R (by normalization n3R=n2R). Thus n1=n1R+jn2I=(n2I/tan(α+270°))+jn2I=-(n2I2/n2R)+jn2I. Moreover arg(dx*)=270°+arg(dx), and if dx=dxR+jdxI  then dx*=ρej3(π/2)dx=ρdxI-jρdxR. As vx=n2/dx and vx=n1/dx*, then n2dx*=n1dx and they have equals real and imaginary parts.
Re
[
n
2
d
x
*
]
=
Re
[
n
1
d
x
] then
(3.6)+ρdxIn2R+ρdxRn2I=-n2I2n2RdxR-n2IdxI,(n2I+n2Rρ)dxIn2R=-(n2Rρ+n2I)dxRn2I.
Thus dxI/dxR=-n2I/n2R.
Im
[
n
2
d
x
*
]
=
Im
[
n
1
d
x
] then
(3.7)-ρdxRn2R+ρdxIn2I=-dxIn2I2n2R+dxRn2I,-ρdxRn2Rn2R+ρdxIn2In2R=-dxIn2I2+dxRn2In2R,(n2I+ρn2R)dxIn2I=dxRn2R(ρn2R+n2I).
Thus  dxI/dxR=n2R/n2I.
Taking into account both conditions,n2R/n2I=-n2I/n2R. This relation is impossible. Therefore, if dx is a solution, dx* is not and nx=n1 cannot be a solution.
(
2
)
⇐ If nx=n1 the value of dx=n1/vx can be calculated and the denominator condition (Lemma 3.2) is satisfied.  On the other hand, the quotient of the vertices vx=n1/dx and v2=n1/d1 is vx/v1=d1/dx, and arg(vx/v1)=arg(d1)-arg(dx). Sx=S1/dx where S1 is part of the segment with vertices n4 and n1, then arg(n1-n4)=270° (normalization). Thus arg(dx)=270°-φ(Sx) (Property 1) and arg(vx/v1)=arg(d1)+φ(Sx)+90°; 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 dx=n1/vx,  nx=n1 is unique. It must be noted that Theorem 3.3(C2) can be satisfied when (a) nx=n3 or (b) nx=n4 and in all the cases, the value of  dx determined, verify the denominator condition.
Letdx be the denominator of vx determined by n1/vx, verifying Theorem 3.3(C2), and denominator condition, and let Sx=S1/dx where S1 is part of the segment with vertices n4 and n1,  arg(n2-n1)=270°.
(
a
) Let  dx* be  the denominator of vx determined by n3/vx. Then Sx=S3/dx* where S3 is part of the segment with vertices n2 and n3, arg(n3-n2)=90° (normalization) and using Property 1   arg(dx*)=90°-φ(Sx). As vx is the same vertex, then arg(n3/dx*)=arg(n1/dx) and arg(n3)=arg(n1)+180°. nx=n3 verify Theorem 3.3(C2), because
(3.8)arg(vxv1)=arg(n1)+180°-arg(n1)+arg(d1)-90°+φ(Sx)=arg(d1)+φ(Sx)+90°.
In this case the demonstration is trivial noting thatarg(dx*)=-180°+arg(dx). This is not possible because the Kharitonov polynomial denominator cannot contain the zero.
(
b
) Let dx* be the denominator determined by n4/vx. Then Sx=S4/dx* where S4 is part of the segment with vertices n3 and n4, arg(n4-n3)=180° (normalization) and using Property 1arg(dx*)=180°-φ(Sx). As vx is the same vertex, then arg(n4/dx*)=arg(n1/dx), and arg(n4)=arg(n1)+270°. nx=n4 verify Theorem 3.3(C2), because
(3.9)arg(vxv1)=arg(n1)+270°-arg(n1)+arg(d1)-180°+φ(Sx)=arg(d1)+φ(Sx)+90°.
Letα=arg(n1) with tan(α)=n1I/n1R. Then arg(n1)=α+270° and tan(α+270°)=n4I/n4R=-n1R/n1I (by normalization n1R=n4R). Thus n4=n4R+jn4I=n1R+jn2Rtan(α+270°)=n1R-j(n1R2/n1I). Moreover arg(dx*)=-90°+arg(dx), and if dx=dxR+jdxI  then dx*=ρej3(π/2)dx=ρdxI-jρdxR. How vx=n1/dx and vx=n4/dx*, then n1dx*=n4dx and they have equals real and imaginary parts.
Re
[
n
1
d
x
*
]
=
Re
[
n
4
d
x
]
(3.10)
ρ
d
x
I
n
1
R
+
ρ
d
x
R
n
1
I
=
+
n
1
R
2
n
1
I
d
x
I
+
n
1
R
d
x
R
,
ρ
d
x
I
n
1
R
n
1
I
+
ρ
d
x
R
n
1
I
n
1
I
=
+
n
1
R
2
d
x
I
+
n
1
R
d
x
R
n
1
I
,
(
ρ
n
1
I
-
n
1
R
)
d
x
I
n
1
R
=
(
n
1
R
-
ρ
n
1
I
)
d
x
R
n
1
I
.
Thus dxI/dxR=-n1I/n1R.
Im
[
n
1
d
x
*
]
=
Im
[
n
4
d
x
]
(3.11)
-
ρ
d
x
R
n
1
R
+
ρ
d
x
I
n
1
I
=
-
d
x
R
n
1
R
2
n
1
I
+
d
x
I
n
1
R
,
-
ρ
d
x
R
n
1
R
n
1
I
+
ρ
d
x
I
n
1
I
n
1
I
=
-
d
x
R
n
1
R
2
+
d
x
I
n
1
R
n
1
I
,
(
-
ρ
n
1
I
+
n
1
R
)
d
x
R
n
1
R
=
(
n
1
R
-
ρ
n
1
I
)
d
x
I
n
1
I
.
and finally dxI/dxR=n1R/n1I.
Taking into account both conditions,-n1I/n1R=n1R/n1I. This relation is impossible. Therefore,  if dx is a solution, dx* is not and nx=n4 is not a solution.
(
3
)
⇐ If nx=n3 then dx=n3/vx cannot be directly calculated because n3 is not known. First, Theorem 3.3(C3) is developed. If nx=n3 then Sx=S3/dx where S3 is part of the segment with vertices n2 and n3 and arg(n3-n2)=90°. Thus arg(dx)=90°-φ(Sx) (Property 1) and arg(n3)=arg(vx)+arg(dx)=arg(vx)+90°-φ(Sx).
Asn2R=n3R, then n3=n3R+jn3I=n2R+jn2Rtan(arg(vx)+90°-φ(Sx)).  On the other hand, n3I is greater than n2I because it is counter-clockwise. Therefore tan(arg(vx)-φ(Sx)+90°)n2R>n2I (Theorem 3.3(C3)) is satisfied and dx can be calculated by the expression dx=n3/vx=n2R[1+jtan(arg(vx)-φ(Sx)+90°)]/vx.
⇒  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 dx=n3/vx,  nx=n3 is unique. If nx≠n2 and nx≠n1, it must be noted that Theorem 3.3(C3) can be satisfied when nx=n4.
Letdx be the denominator of vx determined by n3/vx verifying Theorem 3.3(C3) and denominator condition. Sx=S3/dx* where S3 is part of the segment with vertices n2 and n3, arg(n3-n2)=90°.
Letdx* be the denominator of vx determined by n4/vx. Then Sx=S4/dx* where S4 is part of the segment with vertices n3 and n4, arg(n4-n3)=180° (normalization) and using Property 1   arg(dx*)=180°-φ(Sx)=arg(dx)+90°. Thus dx*=ρej(π/2)dx=-ρdxI+jρdxR.
Asvx is the same vertex, arg(n4/dx*)=arg(n3/dx), and then arg(n4)=arg(n3)+90°.  Let α=arg(n3), then α+90°=arg(n4)=arg(vx)+arg(dx*)=arg(vx=)+180°-φ(Sx), and because arg(n3) verifies n3=n2Rtan(α)>n2I (by normalization), Theorem 3.3(C3) is satisfied.
n
3
=
n
2
R
+
j
tan
(
α
)
n
2
R
=
n
2
R
+
j
n
2
R
(
n
3
R
/
n
3
I
). If nx=n4 then  n4=n1R+jtan(α+90°)n1R=n1R-jn1R(n3R/n3I).  As vx=n3/dx and vx=n4/dx*, then n2dx*=n3dx and they have equal real and imaginary parts.
Re
[
n
3
d
x
*
]
=
Re
[
n
4
d
x
]
(3.12)
-
n
2
R
ρ
d
x
I
-
n
3
I
ρ
d
x
R
=
n
1
R
d
x
R
+
d
x
I
n
1
R
n
2
R
n
3
I
,
-
n
2
R
n
3
I
ρ
d
x
I
-
n
3
I
n
3
I
ρ
d
x
R
=
n
1
R
n
3
I
d
x
R
+
d
x
I
n
1
R
n
3
R
,
-
(
n
3
I
ρ
+
n
1
R
)
n
2
R
d
x
I
=
(
n
1
R
+
n
3
I
ρ
)
n
3
I
d
x
R
,
and  finally dxI/dxR=-n3I/n3R.
Im
[
n
3
d
x
*
]
=
Im
[
n
4
d
x
]
(3.13)
-
n
3
I
ρ
d
x
I
+
n
2
R
ρ
d
x
R
=
d
x
I
n
1
R
-
d
x
R
n
1
R
n
2
R
n
3
I
,
-
n
3
I
n
3
I
ρ
d
x
I
+
n
3
I
n
2
R
ρ
d
x
R
=
d
x
I
n
1
R
n
3
I
-
d
x
R
n
1
R
n
2
R
,
-
(
n
3
I
ρ
+
n
1
R
)
d
x
I
n
3
I
=
-
(
n
3
I
ρ
+
n
1
R
)
d
x
R
n
2
R
and finally dxI/dxR=n3R/n3I.
Taking into account both conditions,-n3I/n3R=n3R/n3I. This relation is impossible. Therefore, if dx is a solution, dx* is not, and nx=n3 is not a solution.
(
4
)
⇐ If nx=n4 then dx=n4/vx cannot be directly calculated because n4 is not known. First, Theorem 3.3(C4) is developed.
Ifnx=n4 then Sx=S4/dx where S4 is part of the segment with vertices n3 and n4 verifying that  arg(n4-n3)=180°. Thus arg(dx)=180°-φ(Sx) (Property 1) and arg(n4)=arg(vx)+arg(dx)=arg(vx)+180°-φ(Sx). Moreover, n1R=n4R. Then n4=n4R+jn4I=n1R+jn1Rtan(arg(vx)+180°-φ(Sx)). On the other hand, n4I is greater than n1I because it is counter-clockwise.
Therefore the conditiontan(arg(vx)-φ(Sx)+180°)n1R>n1I Theorem 3.3(C4) is satisfied and dx can be calculated using the expression dx=n4/vx=n1R[1+jtan(arg(vx)-φ(Sx)+180°)]/vx.
⇒ If nx≠n2, nx≠n1 and nx≠n3 it is nx=n4.Remark 3.4.
This theorem is used in the example of Section5, for the value set III (frequency w=1.2) in order to assign the second and fifth vertices.The following Theorem is analogous to Theorem3.3 when Sx is a segment with vertices vx=nx/dx and vxsucc=nxsucc/dx counter-clockwise, and belonging to an arc-segment.Theorem 3.5 (successor).
LetS1 be a complete segment of the value-set boundary with vertices v1=n1/d1 and v2=n2/d1, the successor arc to S1, with vertices v2=n2/d1, v2succ=n2/d2λ counter-clockwise,  and the predecessor arc to S1 with vertices v1pred=n1/d4λ,   v1=n1/d1 counter-clockwise.  Let  Sx be a boundary segment with vertices  vx=nx/dx and vxsucc=nxsucc/dx counter-clockwise, where vx  belongs to the intersection of an arc of the boundary and Sx. Then (1)
arg(vx/v2)=arg(d1)+φ(Sx)-90° (condition C1)  and the denominator dx of vx defined by n2/vx  satisfies the denominator condition (Lemma 3.2), if and only if nx=n2 and cannot be any other assigned polynomial,(2)
whennx≠n2,arg(vx/v1)=arg(d1)+φ(Sx) (condition  C2) and the denominator  dx of vx  of  defined by n1/vx satisfies the denominator condition (Lemma 3.2) if and only if nx=n1 and cannot be any other assigned polynomial,(3)
whennx≠n1 and nx≠n2, tan(arg(vx)-φ(Sx)+180°)n2R>n2I (condition  C3), and the denominator dx of vx  defined by n2R[1+jtan(arg(vx)-φ(Sx)+180°)]/vx satisfies the denominator condition (Lemma 3.2)  if and only if nx=n3=n2R[1+jtan(arg(vx)-φ(Sx)+180°)] and cannot be any other assigned polynomial,(4)
whennx≠n1, nx≠n2, and nx≠n3,  tan(arg(vx)-φ(Sx)+270°)n1R>n1I (condition  C4), and the denominator dx of vx defined by n1R[1+jtan(arg(vx)-φ(Sx)+270°]/vx satisfies the denominator condition (Lemma 3.2)  if and only if nx=n4=n1R[1+jtan(arg(vx)-φ(Sx)+270°)].Proof.
Analogous to Theorem3.3.Remark 3.6.
This theorem is used in the example of Section5, for the value set III (frequency w=1.2) 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, letA1 be an arc of the value-set boundary with vertices v1=n1/d1  and v2=n1/d2. A continuity arc-segment in a quadrant (see [10, Theorem 2]) implies that there will be a successor segment with vertices v2=n1/d2,  v2succ=n2λ/d2 counter-clockwise and a predecessor segment with vertices v1=n1/d1 and  v1pred=n4λ/d1 counter-clockwise.When these segments are completed the denominators are vertices of the Kharitonov rectangle. Figure8 shows this situation.Figure 8
Arc and two complete segments.As was shown, the values ofd1,  d2, and n1 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).
LetA1 be a complete arc of the value-set boundary with vertices  v1=n1/d1 and v2=n1/d2, the normalization n1=cos(φ(n1))+jsin(φ(n1)), where  φ(n1)=360°-arg(1/v2-1/v1), arg(1/v2-1/v1) being the argument of the segment 1/v2-1/v1.  Then d1=n1/v1, d2=n1/v2, n4λ=d1v1pred, and n2λ=d2v2succ, where v2succ (v1pred) is any point of the next (previous) segment of the arc A1.Proof.
It is trivial. This normalization is one of the infinite possible solutions for a value set. This normalization implies fittingn1 with modulus |n1|=1 and angle so that the segment of the Kharitonov polynomial denominator with vertices d1 and d2 will be parallel to the real axis counter-clockwise. Thus, from the information with a complete arc in a quadrant the values of d1, d2, n1, n2λ, and n4λ can be calculated.This paper deals with the general case whered2R≠0, d2I≠0, d1R≠0, and d1I≠0.Given a vertexvx=nx/dx in a quadrant, the target is to determine the polynomials nx and dx. The vertex vx belongs to a part of an arc and a part of a segment, due to the continuity arc-segment in a quadrant. So, vx will be the vertex of two elements, segment-arc (Figure 9(a)) or arc-segment (Figure 9(b)).(a)vx vertex of two elements, segment-arc. (b) vx vertex of two elements, arc-segment.
(a)
(b)The following Lemma shows the necessary conditions on the denominatordx to be a solution of vx=nx/dx.Lemma 4.2 (numerator condition).
LetA1 be a complete arc in a quadrant and let nx be the numerator of a vertex vx=nx/dx in a quadrant. Then it is a necessary condition that nx satisfies one of the following conditions:(1)
(n1R<n2λRandn1I<n4λI) and {(nxR=n1RandnxI=n1I)[nx=n1] or (nxR=n1RandnxI≥n1I)[nx=n4] or (nxI=n1IandnxR≥n1R)[nx=n2] or (nxR>n1RandnxI>n1I)[nx=n3]},(2)
(n1R>n4λRandn1I<n2λI) and {(nxR=n1RandnxI=n1I)[nx=n1] or (nxR=n1RandnxI≥n1I)[nx=n2] or  (nxI=n1IandnxR≤n1R)[nx=n4] or (nxR<n1RandnxI>n1I)[nx=n3]},(3)
(n1R>n2λRandn1I>n4λI) and {(nxR=n1RandnxI=n1I)[nx=n1] or (nxR=n1RandnxI≤n1I)[nx=n4] or  (nxI=n1IandnxR≤n1R)[nx=n2] or (nxR<n1RandnxI<n1I)[nx=n3]},(4)
(n1R<n4λRandn1I>n2λI) and {(nxR=n1RandnxI=n1I)[nx=n1] or (nxR=n1RandnxI≤n1I)[nx=n2] or  (nxI=n1IandnxR≥n1R)[nx=n4] or (nxR>n1RandnxI<n1I)[nx=n3]},
whereniR is the real part of ni and niI is the imaginary part of ni, 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 (Lemma3.2), the values of n1, n2λ, and n4λ are known. Then, n1 can be established as kn1, kn2, kn3, or kd4. (1)
If(n1R<n2λRandn1I<n4λI) then n1 is kn1. Given a value nx, it will be a vertex of the Kharitonov rectangle numerator only if nxR=n1RandnxI=n1I (nx is n1=kn1) or nxR=n1RandnxI>n1I (nx is n4=kn4) or nxI=n1IandnxR>n1R (nx is n2=kn2) or nxR>n1RandnxI>n1I (nx is n3=kn3). Note that if any of these conditions is not satisfied, then nx cannot be a solution. For example, if nxR=n1RandnxI<n1I, nx does not belong to the rectangle with vertex n1, n2λ, and n4λ are elements of the successor and predecessor edge.(2)
Similarly, if(n1R>n4λRandn1I<n2λI) then n1 is kn2. Given a value nx, it will be a vertex of the Kharitonov rectangle numerator only if nxR=n1RandnxI=n1I (nx is n1=kn2) or nxR=n1RandnxI>n1I (nx is n2=kn3) or nxI=n1IandnxR<n1R (nx is n4=kn1)  or nxR<n1RandnxI>n1I (nx is n3=kn4).(3)
Ifn1R>n2λRandn1I>n4λI then n1 is kn3. Given a value nx, it will be a vertex of the Kharitonov rectangle numerator only if nxR=n1RandnxI=n1I (nx is n1=kn3) or nxR=n1RandnxI<n1I (nx is n4=kn2) or nxI=n1IandnxR<n1R (nx is n2=kn4)  or nxR<n1RandnxI<n1I (nx is n3=kn1).(4)
Finally, ifn1R<n4λRandn1I>n2λI then n1 is kn4. Given a value nx, it will be a vertex of the Kharitonov rectangle numerator only if nxR=n1RandnxI=n1I (nx is n1=kn4) or nxR=n1RandnxI<n1I (nx is n2=kn1) or nxI=n1IandnxR>n1R (nx is n4=kn3)  or nxR>n1RandnxI<n1I (nx is n3=kn2).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.
LetAx=nx/S be an arc on the complex plane with vertices vx1 and vx2 counter-clockwise where S is a segment with vertices da and db counter-clockwise. Let nx be a complex number with argument arg(nx). Let φ(Ax)  be  φ(Ax)≡arg(1/vx2-1/vx1). Then the relation between the argument of nx and φ(Ax), is given by(1)
arg(nx)=-φ(Ax) if and only if arg(db-da)=0°,(2)
arg(nx)=90°-φ(Ax) if and only if arg(db-da)=90°,(3)
arg(nx)=180°-φ(Ax) if and only if arg(db-da)=180°,(4)
arg(nx)=270°-φ(Ax) if and only if arg(db-da)=270°.The following Theorem shows how to characterize and calculate the polynomialsnx and dx associated with a vertex  vx=nx/dx from the information of the boundary with an arc Ax in a quadrant, belonging to an arc-segment.Theorem 4.3 (predecessor).
LetA1 be an arc of the value-set boundary with vertices v1=n1/d1 and v2=n1/d2, the successor segment with vertices v2=n1/d2, v2succ=n2λ/d2 counter-clockwise, and the predecessor segment with vertices  v1pred=n4λ/d1,  v1=n1/d1 counter-clockwise  Let Ax be an arc with vertices vxpred=nx/dxpred and vx=nx/dx counter-clockwise.  Then(1)
arg(v2/vx)=arg(n1)+φ(Ax) (condition  C1) and nx satisfies the numerator condition, where nx=d2vx,  if and only if dx=d2 and cannot be any other assigned polynomial,(2)
whendx≠d2, arg(v1/vx)=arg(n1)+φ(Ax)+90° (condition  C2) and nx=d1vx satisfies the numerator condition if and only if dx=d1 and cannot be any other assigned polynomial,(3)
whendx≠d1 and dx≠d2, tan(arg(1/vx)-φ(Ax)+90°)d2R>d2I (condition  C3), and nx=d2R[1+jtan(arg(1/vx)-φ(Ax)+90°)]vx satisfies the numerator condition if and only if dx=d3=d2R(1+jtan(arg(1/vx)-φ(Ax)+90°)) and cannot be any other assigned polynomial,(4)
whendx≠d1,dx≠d2, and dx≠d3, tan(arg(1/vx)-φ(Ax)+180°)d1R>d1I (condition  C4), and nx=d1R[1+jtan(arg(1/vx)-φ(Ax)+180°)]vx satisfies the numerator condition if and only if dx=d4=d1R(1+jtan(arg(1/vx)-φ(Ax)+180°)).Proof.
Analogous to Theorem3.3.Remark 4.4.
This theorem is used in the example of Section5, for the value set I (frequency w=1.0) in order to assign the fifth and seventh vertices, and for the value set II (frequency w=1.1) to assign the third, fifth, and seventh vertices.The following theorem is analogous to Theorem4.3 when Ax is an arc with vertices vx=nx/dx and vxsucc=nx/dxsucc counter-clockwise, and belonging to a segment-arc.Theorem 4.5 (successor).
LetA1 be a complete arc of the value-set boundary with vertices v1=n1/d1 and v2=n1/d2, the successor segment with vertices v2=n1/d2, v2succ=n2λ/d2 counter-clockwise and the predecessor segment with vertices  v1pred=n4λ/d1,  v1=n1/d1 counter-clockwise.  Let Ax be an arc with vertices vxsucc=nx/dxsucc and vx=nx/dx counter-clockwise
Then(1)
arg(v2/vx)=φ(Ax)+arg(n1)-90° (condition  C1) and nx satisfies the numerator condition, where nx=d2vx, if and only if dx=d2 and cannot be any other assigned polynomial,(2)
whendx≠d2, arg(v1/vx)=φ(Ax)+arg(n1) (condition  C2) and nx=d1vx satisfies the numerator condition if and only if dx=d1 and cannot be any other assigned polynomial,(3)
whendx≠d1 and dx≠d2, tan(arg(1/vx)-φ(Ax)+180°)d2R>d2I (condition  C3), and nx=d2R[1+jtan(arg(1/vx)-φ(Ax)+180°)]vx satisfies the numerator condition if and only if dx=d3=d2R(1+jtan(arg(1/vx)-φ(Ax)+180°)) and cannot be any other assigned polynomial,(4)
whendx≠d1, dx≠d2, and dx≠d3, tan(arg(1/vx)-φ(Ax)+270°)d1R>d1I (condition  C4), and nx=d1R[1+jtan(arg(1/vx)-φ(Ax)+270°)]vx satisfies the numerator condition if and only if dx=d4=d1R(1+jtan(arg(1/vx)-φ(Ax)+270°)).Proof.
Analogous to Theorem3.3.Remark 4.6.
This theorem is used in the example of Section5, for the value set I (frequency w=1.0) in order to assign the third, fourth, and sixth vertices, and for the value set II (frequency w=1.1) to assign the fourth and sixth vertices.Finally, the following theorem points out the necessary and sufficient condition.Theorem 4.7.
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 when the normalized edge satisfiesn2R≠0,n2I≠0,n1R≠0, and n1I≠0 or the normalized arc satisfies d2R≠0,d2I≠0,d1R≠0, and d1I≠0.Proof.
It is obvious from Theorems3.3–4.5.


## 5. Algorithm and Examples

Algorithm 5.1.
Given a value set with a complete segment or a complete arc in a quadrant, to obtain the Kharitonov polynomials the following.(1)
If there is a complete segment in a quadrant,S1, with vertices v1=n1/d1 and v2=n2/d1, the successor arc with vertices v2=n2/d1,v2succ=n2/d2λ counter-clockwise and the predecessor arc with vertices  v1pred=n1/d4λ,v1=n1/d1 counter-clockwise then for all vertex vx=nx/dx:
(a)
ifvx=nx/dx is a vertex intersection of a segment and an arc counter-clockwise, then the assigned polynomials numerator and denominator, nx and dx, determine applying Theorem 3.3,(b)
ifvx=nx/dx is a vertex intersection of an arc and a segment counter-clockwise, then the assigned polynomials numerator and denominator, nx and dx, determine applying Theorem 3.5.(2)
If there is a complete arc in a quadrant,A1, with vertices v1=n1/d1 and v2=n1/d2, the successor segment with vertices v2=n1/d2,v2succ=n2λ/d2 counter-clockwise and the predecessor segment with vertices  v1pred=n4λ/d1,  v1=n1/d1 counter-clockwise, then given a vertex vx=nx/dx:
(a)
ifvx=nx/dx is a vertex intersection of an arc and a segment counter-clockwise, then the assigned polynomials numerator and denominator, nx and dx, determine applying Theorem 4.3,(b)
ifvx=nx/dx is a vertex intersection of a segment and an arc counter-clockwise, then the assigned polynomials numerator and denominator, nx and dx, determine applying Theorem 4.5.(3)
Calculate the values of the assigned polynomialsnj, dk, solving the equation system (2.7):
(5.1)vi=njdk.(4)
Calculate the numerator and denominator rectangles with Kharitonov polynomial valuesN=(kn1(jω),kn2(jω),kn3(jω),kn4(jω)), D=(kd1(jω),kd2(jω),kd3(jω),kd4(jω)) applying (2.8).Example 5.2.
Figure10  shows three value sets of an interval plant. The necessary information (Table 1)  is(i)
the vertices,(ii)
the intersections with the axis,(iii)
the shape of the boundary's elements: arc or segment.Table 1
Value set boundary information.
ω
=
1.0
ω
=
1.1
ω
=
1.2
(a)
(b)
(c)
(a)
(b)
(c)
(a)
(b)
(c)
v
1
1.5676
+
2.5946
j
0
v
1
-
2.8422
+
2.9830
j
0
v
1
6.1015
+
5.2779
j
1
v
2
2.0000
+
8.0000
j
1
v
2
-
0.9808
+
2.4599
j
1
v
2
6.5135
+
6.8573
j
0
v
3
0.8000
+
10.4000
j
0
0
+
3.0420
j
0
0
+
8.5560
j
0
0
+
10.0000
j
1
v
3
0.4996
+
3.0386
j
1
v
3
-
3.0339
+
6.1294
j
1
v
4
-
4.8000
+
7.6000
j
0
v
4
2.3317
+
3.0261
j
0
v
4
-
2.2110
+
5.1007
j
0
v
5
-
3.5862
+
1.0345
j
1
v
5
5.1859
+
6.6181
j
1
v
5
-
0.4710
+
3.4462
j
1
v
6
2.5517
+
0.6207
j
0
v
6
5.2164
+
8.6623
j
0
0
+
3.6463
j
0
v
7
-
1.3443
+
1.2131
j
1
0
+
8.7404
j
0
v
6
1.4690
+
3.4428
j
1
0
+
2.3336
j
1
v
7
-
3.8291
+
3.7385
j
1
v
7
2.9559
+
3.2369
j
0
(a): Vertex (vi) or cut point (blank) with an axis. (b): Value of the vertex or cut point.
(c): Edge (2.1) or arc (0) between this element and the next element. If the element is the last, the next element is the first.Three value sets of an interval plant.
(a)
(b)
(c)This example illustrates how to obtain the assigned polynomials and the numerator and denominator rectangles for each value set, and remarks the theorem used in each step.


### 5.1. Value Set at Frequencyω=1.0

The complete arc with verticesv1=n1/d1=1.5676+2.5946j and v2=n1/d2=2.0000+8.0000j is taken as initial element. Then Theorems 4.3 and 4.5 will be applied. So
(5.2)v2succ=n2λd2=0.8000+10.4000j,v1pred=n4λd1=2.3336j.
Applying the arc normalization (Lemma 4.1) the following data are obtained
(5.3)φ(n1)=229.40,n1=-0.6508-0.7592j,d1=-0.3254+0.0542j,d2=-0.1085+0.0542j,n4λ=-0.1266-0.7594j,n2λ=-0.6508-1.0846j.
Then, all the other vertices are assigned as follows.(
1
) Vertex v3=vx=nx/dx=0.8000+10.4000j. Then vxpred=nx/dxpred=2.0000+8.0000j. These are the vertices of an edge, and Theorem 4.5 is applied, vxsucc=10.0000j, φ(Ax)=210.97.Case1. Theorem 4.5(C1) is satisfied: arg(v2/vx)=φ(Ax)+arg(n1)-90=350.36 and nx=d2vx=-0.6508-1.0846j satisfies the Numerator Condition (Lemma 4.2(4),nx=n2):
(5.4)(n1R=-0.6508<n4λR=-0.1266,n1I=-0.7592>n2λI=-1.0846),(nxR=n1R=-0.6508,nxI=-1.0846≤n1I=-0.7592).
Then dx=d2=-0.1085+0.0542j. Therefore v3=vx=n2/d2.(
2
) Vertex v4=vx=nx/dx=-4.8000+7.6000j. Then vxpred=10j. These are the vertices of an edge, and Theorem 4.5 is applied: vxsucc=-3.5862+1.0345j, φ(Ax)=174.29.Case1. Theorem 4.5(C1) is satisfied: arg(v2/vx)=φ(Ax)+arg(n1)-90=313.69 and nx=d2vx=0.1084-1.0847j satisfies the Numerator Condition (Lemma 4.2(4), nx=n3). Then dx=d2=-0.1085+0.0542j. Therefore v4=vx=n3/d2.(
3
) Vertex v5=vx=nx/dx=-3.5862+1.0345j. Then vxpred=nx/dxpred=-4.8000+7.6000j. These are the vertices of an arc, and Theorem 4.3 is applied: φ(Ax)=174.29.Case1. Theorem 4.3(C1) is not satisfied: arg(v2/vx)=272.06≠arg(n1)+φ(Ax)=43.69.Case2. Theorem 4.3(C2) is not satisfied:arg(v1/vx)=254.95≠arg(n1)+φ(Ax)+90=133.69.Case3. Theorem 4.3(C3) is satisfied: tan(arg(1/vx)-φ(Ax)+90)d2R=0.2712>d2I=0.0542 and nx=0.1085-1.0846j satisfies the Numerator Condition (Lemma 4.2(4)) nx=n3. Then dx=d3=-0.1085+0.2712j   v5=vx=n3/d3.(
4
) Vertex v6=vx=nx/dx=-2.5517+0.6207j. Then vxpred=nx/dxpred=-3.5862+1.0345j. These are the vertices of an edge, and Theorem 4.5 is applied: vxsucc-1.3443+1.2131j, φ(Ax)=261.87.Case1. Theorem 4.5(C1) is not satisfied: arg(v2/vx)=269.64≠φ(Ax)+arg(n1)-90=41.27.Case2. Theorem 4.5(C2) is not satisfied: arg(v1/vx)=252.53≠φ(Ax)+arg(n1)=131.27.Case3. Theorem 4.5(C3) is satisfied: tan(arg(1/vx)-φ(Ax)+180)d2R=0.2712>d2I=0.0542 and nx=0.1085-0.7592j  satisfies the Numerator Condition (Lemma 4.2(3)) nx=n4: then dx=d3=-0.1085+0.2712j and v6=vx=n4/d3.(
5
) Vertex v7=vx=nx/dx=-1.3443+1.2131j. Then vxpred=nx/dxpred=-2.5517+0.6207j. These are the vertices of an arc, and Theorem 4.3 is applied: φ(Ax)=261.87.Case1. Theorem 4.3(C1) is not satisfied: arg(v2/vx)=298.03≠arg(n1)+φ(Ax)=131.27.Case2. Theorem 4.3(C2) is not satisfied: arg(v1/vx)=280.93≠arg(n1)+φ(Ax)+90=221.27.Case3. Theorem 4.3(C3) is not satisfied: tan(arg(1/vx)-φ(Ax)+90)d2R=-0.1302<d2I=0.0542.Case4. Theorem 4.3(C4) is satisfied: tan(arg(1/vx)-φ(Ax)+180)d1R=0.2712>d1I=0.0542 and nx=0.1085-0.7592j satisfies the Numerator Condition (Lemma 4.2(4)) nx=n4. Then dx=d4=-0.3254+0.2712j; v7=vx=n4/d4.In summary, the assigned polynomials are(5.5)v1=n1d1,v2=n1d2,v3=n2d2,v4=n3d2,v5=n3d3,v6=n4d3,v7=n4d4,
and the values can be calculated: from normalization,
(5.6)n1=-0.6508-0.7592j,d1=-0.3254+0.0542j,d2=-0.1085+0.0542j,
and from the vertices,
(5.7)v3:n2=-0.6508-1.0846j,d2=-0.1085+0.0542j,v4:n3=0.1084-1.0847j,d2=-0.1085+0.0542j,v5:n3=0.1085-1.0846j,d3=-0.8464+2.0152j,v6:n4=0.1085-0.7592j,d3=-0.1085+0.2712j,v7:n4=0.1085-0.7593j,d4=-0.3254+0.2712j.Then(5.8)kn1(jω)=-0.6508-1.0847j,kn2(jω)=0.1085-1.0847j,kn3(jω)=0.1085-0.7592j,kn4(jω)=-0.6508-0.7592j,kd1(jω)=-0.3254+0.0542j,kd2(jω)=-0.1085+0.0542j,kd3(jω)=-0.1085+0.2712j,kd4(jω)=-0.3254+0.2712j.Table2 shows the results of the algorithm for the value set at frequency ω=1.0.Table 2
Results of the algorithm for the value set at frequencyω=1.0.
v
1
-
v
2 arc
v
3
v
4
v
5
v
6
v
7
0.8000 + 10.4000j
−4.8000 + 7.6000j
−3.5862 + 1.0345j
−2.5517 + 0.6207j
−1.3443 + 1.2131j
Kharitonov rectangles calculated
v
1
1.5676 + 2.5946j
Theoremapplied
Theorem4.5
Theorem4.5
Theorem4.3
Theorem4.5
Theorem4.3
k
n
1
(
j
ω
)
=
-
0.6508
-
1.0847
j
v
2
2.0000 + 8.0000j
v
x
pred
2.0000 + 8.0000j
10.0000j
−4.8000 + 7.6000j
−3.5862 + 1.0345j
−2.5517 + 0.6207j
k
n
2
(
j
ω
)
=
0.1085
-
1.0847
j
v
2
succ
0.8000 + 10.4000j
v
x
succ
10.0000j
−3.5862 + 1.0345j
−2.5517 + 0.6207j
−1.3443 + 1.2131j
2.3336j
k
n
3
(
j
ω
)
=
0.1085
-
0.7592
j
v
1
pred
0 + 2.3336j
φ
(
A
x
)
210.97
174.29
174.29
261.87
261.87
k
n
4
(
j
ω
)
=
-
0.6508
-
0.7592
j
φ
(
n
1
)
229.40
Conditionverified
Theorem4.5(C1)
Theorem4.5(C1)
Theorem4.3(C3)
Theorem4.5(C3)
Theorem4.3(C4)
k
d
1
(
j
ω
)
=
-
0.3254
+
0.0542
j
n
1
−0.6508 − 0.7592j
n
x
−0.6508 − 1.0846j
0.1084 − 1.0847j
0.1085 − 1.0846j
0.1085 − 0.7592j
0.1085 − 0.7592j
k
d
2
(
j
ω
)
=
-
0.1085
+
0.0542
j
d
1
−0.3254 + 0.0542j
d
x
−0.1085 + 0.0542j
−0.1085 + 0.0542j
−0.1085 + 0.2712j
−0.1085 + 0.2712j
−0.3254 + 0.2712j
k
d
3
(
j
ω
)
=
-
0.1085
+
0.2712
j
d
2
−0.1085 + 0.0542j
v
x
n
2
/
d
2
n
3
/
d
2
n
3
/
d
3
n
4
/
d
3
n
4
/
d
4
k
d
4
(
j
ω
)
=
-
0.3254
+
0.2712
j
n
4
λ
−0.1266 − 0.7594j
n
2
λ
−0.6508 − 1.0846jFrom these Kharitonov rectangles the value set given in Figure11(a) is directly obtained.(a)
(b)
(c)


### 5.2. Value Set at Frequencyω=1.1

The complete arc with verticesv1=n1/d1=-2.8422+2.9830j and v2=n1/d2=-0.9808+2.4599j is taken as initial element. Then Theorems 4.3 and 4.5 will be applied. So
(5.9)v2succ=n2λd2=3.0420j,v1pred=n4λd1=-3.8291+3.7385j.
Applying the arc normalization (Lemma 4.1) the following data are obtained:
(5.10)φ(n1)=360-arg(1v2-1v1)=81.05,n1=0.1556+0.9878j,d1=n1v1=0.1475-0.1927j;d2=n1v2=0.3247-0.1927j,n4λ=d1v1pred=0.1556+1.2895j,(5.11)n2λ=d2v2succ=0.5862+0.9878j.
Then, all the other vertices are assigned as follows.(
1
) Vertex v3=vx=nx/dx=0.4996+3.0386j. Then vxpred=nx/dxpred=3.0420j. These are the vertices of an arc, and Theorem 4.3 is applied: φ(Ax)=8.95.Cases1and2. Theorem 4.3(C1) and (C2) are not satisfied.Case3. Theorem 4.3(C3) is satisfied: tan(arg(1/vx)-φ(Ax)+90)d2R=0.0022>d2I=-0.1927 and nx=0.1555+0.9878j satisfies the Numerator Condition (Lemma 4.2(1)) nx=n1.Then dx=d3=0.3247+0.0022jv3=vx=n1/d3.(
2
) Vertex v4=vx=nx/dx=2.3317+3.0261j. Then vxpred=nx/dxpred=0.4996+3.0386j. These are the vertices of an edge, and Theorem 4.5 is applied: vxsucc=5.1859+6.6181j and φ(Ax)=127.23.Cases1and2. Theorem 4.5(C1) and  (C2) are not satisfied.Case3. Theorem 4.5(C3) is satisfied: tan(arg(1/vx)-φ(Ax)+180)d2R=0.0022>d2I=-0.1927 and nx=0.7505+0.9878j  satisfies the Numerator Condition (Lemma 4.2(1)) nx=n2. Then dx=d3=0.3247+0.0022j.  v4=vx=n2/d3.(
3
) Vertex v5=vx=nx/dx=5.1859+6.6181j. Then vxpred=nx/dxpred=2.3317+3.0261j. These are the vertices of an arc, and Theorem 4.3 is applied: φ(Ax)=127.23.Cases1,2,and3. Theorem 4.3(C1), (C2), and (C3) are not satisfied.Case4. Theorem 4.3(C4) is satisfied: tan(arg(1/vx)-φ(Ax)+180)d1R=0.0022>d1I=-0.1927 and nx=0.7505+0.9878j satisfies the Numerator Condition (Lemma 4.2(1)) nx=n2. Then dx=d4=0.1475+0.0022j  v5=vx=n2/d4.(
4
) Vertex v6=vx=nx/dx=5.2164+8.6623j. Then vxpred=nx/dxpred=5.1859+6.6181j. These are the vertices of an edge, and Theorem 4.5 is applied: vxsucc=8.7404j, φ(Ax)=210.20.Cases1,2,and3.  Theorem 4.5(C1), (C2), and (C3) are not satisfied.Case4. Theorem 4.5(C4) is satisfied: tan(arg(1/vx)-φ(Ax)+270)d1R=0.0022>d1I=-0.1927and nx=0.7505+1.2895j  satisfies the Numerator Condition (Lemma 4.2(1)) nx=n3: then dx=d4=0.1475+0.0022j, v6=vx=n3/d4.(
5
) Vertex v7=vx=nx/dx=-3.8291+3.7385j. Then vxpred=nx/dxpred=8.7404j. These are the vertices of an arc, and Theorem 4.3 is applied: φ(Ax)=186.88.Case1. Theorem 4.3(C1) is not satisfied.Case2. Theorem 4.3(C2) is satisfied: arg(v1/vx)=arg(n1)+φ(Ax)+90=357.93and nx=d1vx=0.1556+1.2895j satisfies the Numerator Condition (Lemma 4.2(1)) nx=n4. Then dx=d1=0.1475-0.1927j, v7=vx=n4/d1.In summary, the assigned polynomials are(5.12)v1=n1d1,v2=n1d2,v3=n1d3,v4=n2d3,v5=n2d4,v6=n3d4,v7=n4d1
and the values can be calculated: from normalization,
(5.13)n1=0.1556+0.9878j,d1=0.1475-0.1927j,d2=0.3247-0.1927j,
and from the vertices,
(5.14)v3:n1=0.1555+0.9878j,d3=0.3247+0.0022j,v4:n2=0.7505+0.9878j,d3=0.3247+0.0022j,v5:n2=0.7505+0.9878j,d4=0.1475+0.0022j,v6:n3=0.7505+1.2895j,d4=0.1475+0.0022j,v7:n4=0.1556+1.2895j,d1=0.1475-0.1927j.
Then
(5.15)kn1(jω)=0.1555+0.9878j,kn2(jω)=0.7505+0.9878j,kn3(jω)=0.7505+1.2895j,kn4(jω)=0.1556+1.2895j,kd1(jω)=0.1475-0.1927j,kd2(jω)=0.3247-0.1927j,kd3(jω)=0.3247+0.0022j,kd4(jω)=0.1475+0.0022j.Table3 shows the results of the algorithm for the value set at frequency ω=1.1.Table 3
Results of the algorithm for the value set at frequencyω=1.1.
v
1
-
v
2 arc
v
3
v
4
v
5
v
6
v
7
0.4996
+
3.0386
j
2.3317
+
3.0261
j
5.1859
+
6.6181
j
5.2164
+
8.6623
j
-
3.8291
+
3.7385
j
Kharitonov rectangles calculated
v
1
-
2.8422
+
2.9830
j
Theoremapplied
Theorem4.3
Theorem4.5
Theorem4.3
Theorem4.5
Theorem4.3
k
n
1
(
j
ω
)
=
0.1555
+
0.9878
j
v
2
-
0.9808
+
2.4599
j
v
x
pred
3.0420
j
0.4996
+
3.0386
j
2.3317
+
3.0261
j
5.1859
+
6.6181
j
8.7404
j
k
n
2
(
j
ω
)
=
0.7505
+
0.9878
j
v
2
scuss
0
+
3.0420
j
v
x
succ
2.3317
+
3.0261
j
5.1859
+
6.6181
j
5.2164
+
8.6623
j
8.7404
j
-
2.8422
+
2.9830
j
k
n
3
(
j
ω
)
=
0.7505
+
1.2895
j
v
1
pred
-
3.8291
+
3.7385
j
φ
(
A
x
)
8.95
127.23
127.23
210.20
186.88
k
n
4
(
j
ω
)
=
0.1556
+
1.2895
j
φ
(
n
1
)
81.05
Conditionverified
Theorem4.3(C3)
Theorem4.5(C3)
Theorem4.3(C4)
Theorem4.5(C4)
Theorem4.3(C2)
k
d
1
(
j
ω
)
=
0.1475
-
0.1927
j
n
1
0.1556
+
0.9878
j
n
x
-
0.1555
+
0.9878
j
0.7505
+
0.9878
j
0.7505
+
0.9878
j
0.7505
+
1.2895
j
0.1556
+
1.2895
j
k
d
2
(
j
ω
)
=
0.3247
-
0.1927
j
d
1
0.1475
-
0.1927
j
d
x
0.3247
+
0.0022
j
0.3247
+
0.0022
j
0.1475
+
0.0022
j
0.1475
+
0.0022
j
0.1475
-
0.1927
j
k
d
3
(
j
ω
)
=
0.3247
+
0.0022
j
d
2
0.3247
-
0.1927
j
v
x
n
1
/
d
3
n
2
/
d
3
n
2
/
d
4
n
3
/
d
4
n
4
/
d
1
k
d
4
(
j
ω
)
=
0.1475
+
0.0022
j
n
4
λ
0.1556
+
1.2895
j
n
2
λ
0.5862
+
0.9878
jFrom these Kharitonov rectangles the value set given in Figure11(b) is directly obtained.


### 5.3. Value Set at Frequencyω=1.2

The complete edge with verticesv1=n1/d1=6.1015+5.2779j and v2=n2/d1=6.5135+6.8573j is taken as initial element. Then Theorems 3.3 and 3.5 will be applied. So
(5.16)v2succ=n2d2λ=8.5560j,v1pred=n1d4λ=2.9559+3.2369j.
Applying the edge normalization (Lemma 3.1) the following data are obtained:
(5.17)ϕ(d1)=360-arg(v2-v1)=284.62,d1=cos(φ(d1))+jsin(φ(d1))=0.2524-0.9676j,n1=v1d1=6.6471-4.5717j,n2=v2d1=8.2793-4.5717j,d2λ=n2v2succ=-0.5343-0.9677j,d4λ=n1v1pred=0.2524-1.8230j.
Then, all the other vertices are assigned as follows.(
1
) Vertex v3=vx=nx/dx=-3.0339+6.1294j. Then vxpred=nx/dxpred=8.5560j. These are the vertices of an arc, and Theorem 3.5 is applied: vxsucc=-2.2110+5.1007j and  φ(Sx)=arg(vxsucc-vx)=308.66.Cases1and 2. Theorem 3.5(C1) and (C2) are not satisfied.Case3. Theorem 3.5(C3) is satisfied: tan(arg(vx)-φ(Sx)+180)n2R=-1.8087>n2I=-4.571 and dx=n2R[1+jtan(arg(vx)-φ(Sx)+180)]/vx=-0.7740-0.9676j satisfies the Denominator Condition (Lemma 3.2(3)) dx=d2: then nx=n3=n2R[1+jtan(arg(vx)-φ(Sx)+180)]=8.2793-1.8087j;  v3=vx=n3/d2.(
2
) Vertex v4=vx=nx/dx=-2.211+5.1007j. Then vxpred=nx/dxpred=-3.0339+6.1294j. These are the vertices of an edge, and Theorem 3.3 is applied: φ(Sx)=arg(vx-vxpred)=308.66.Cases1and2. Theorem 3.3(C1) and (C2) are not satisfied.Case3. Theorem 3.3(C3) is satisfied: tan(arg(vx)-φ(Sx)+90)n2R=30.4258>n2I=-4.5717 but dx=n2R[1+jtan(arg(vx)-φ(Sx)+90)]/vx=4.4292-3.5431j does not satisfy the Denominator Condition: (d1R=0.2524>d2λR=-0.5343andd1I=-0.9676>d4λI=-1.8230) (Case 3) but (dxR=4.4292≠d1R=0.2524) then dx≠d1 and dx≠d4(dxI=-3.5431≠d1I-0.9676) then dx≠d2(dxR=4.4292>d1R=0.2524anddxI=-3.5431<d1I=-0.9676) then dx≠d3.Case4. Theorem 3.3(C4) is satisfied: tan(arg(vx)-φ(Sx)+180)n1R=-1.8088>n1I=-4.5717 and dx=n1R[1+jtan(arg(vx)-φ(Sx)+180)]/vx=-0.7741-0.9676j satisfies the Denominator Condition (Lemma 3.2(3)) dx=d2:
(5.18)(d1R=0.2524>d2λR=-0.5343,d1I=-0.9676>d4λI=-1.8230),(dxI=d1I=-0.9676,dxR=-0.7740≤d1R=0.2524).
Then nx=n4=n1R[1+jtan(arg(vx)-φ(Sx)+180)]=6.6471-1.8088j, v4=vx=n4/d2.(
3
) Vertex v5=vx=nx/dx=-0.47099+3.4462j. Then vxpred=nx/dxpred=-2.2110+5.1007j. These are the vertices of an arc, and Theorem 3.5 is applied: vxsucc=3.6463j and  φ(Sx)=arg(vxsucc-vx)=23.01.Cases
1
,
2
,  and 3. Theorems 3.5(C1) and (C2) are not satisfied. Theorem 3.5(C3) is satisfied but dx=8.3397-3.5422j does not satisfy the Denominator Condition (Lemma 3.2(3)).Case4. Theorem 3.5(C4) is satisfied: tan(arg(vx)-φ(Sx)+270)n1R=-1.8098>n1I=-4.5717 and dx=n1R[1+jtan(arg(vx)-φ(Sx)+270)]/vx=-0.7743-1.8230j satisfies the Denominator Condition (Lemma 3.2(3)).Thennx=n4=n1R[1+jtan(arg(vx)-φ(Sx)+270)]=6.6471-1.8098j,  v5=vx=n4/d3.(
4
) Vertex v6=vx=nx/dx=1.469+3.4428j. Then vxpred=nx/dxpred=3.6463j. These are the vertices of an arc, and Theorem 3.5 is applied: vxsucc=2.9559+3.2369j and φ(Sx)=arg(vxsucc-vx)=352.12.Cases1,2,and 3. Theorem 3.5(C1) and (C2) are not satisfied. Theorem 3.5(C3) is satisfied but dx=8.3440+1.1554j does not satisfy the Denominator Condition (Lemma 3.2(3)).Case4. Theorem 3.5(C4) is satisfied: tan(arg(vx)-φ(Sx)+270)n1R=-1.8089>n1I=-4.5717 and dx=0.2524-1.8230j satisfies the Denominator Condition (Lemma 3.2(3)) dx=d4.Thennx=n4=n1R[1+jtan(arg(vx)-φ(Sx)+270)]=6.6471-1.8089j,  v6=vx=n4/d4.(
5
) Vertex v7=vx=nx/dx=2.9559+3.2369j. Then vxpred=nx/dxpred=1.4690+3.4428j. These are the vertices of an edge, and Theorem 3.3 is applied: φ(Sx)=arg(vx-vxpred)=352.12.Case1. Theorem 3.3(C1) is not satisfied.Case2. Theorem 3.3(C2) is satisfied: arg(vx/v1)=arg(d1)+φ(Sx)+90=6.74 and dx=n1/vx=0.2524-1.8230j satisfies the Denominator Condition (Lemma 3.2(3)) dx=d4. Then nx=n1=6.6471-4.5717j, v7=vx=n1/d4.In summary, the assigned polynomials are(5.19)v1=n1d1,v2=n2d1,v3=n3d2,v4=n4d2,v5=n4d3,v6=n4d4,v7=n1d4,
and the values can be calculated: from normalization,
(5.20)d1=0.2524-0.9676,n1=6.6471-4.5717j,n2=8.2793-4.5717j,
and from the vertices,
(5.21)v3:n3=8.2793-1.8087j,d2=-0.7741-0.9676j,v4:n4=6.6471-1.8087j,d2=-0.7741-0.9676j,v5:n4=6.6471-1.8087j,d3=-0.7743-1.8230j,v6:n4=6.6471-1.8087j,d4=0.2524-1.8230j,v7:n1=6.6471-4.5717j,d4=0.2524-1.8230j.
Then
(5.22)kn1(jω)=6.6471-4.5717j,kn2(jω)=8.2793-4.5717j,kn3(jω)=8.2793-1.8087j,kn4(jω)=6.6471-1.8087j,kd1(jω)=-0.7743-1.8230j,kd2(jω)=0.2524-1.8230j,kd3(jω)=0.2524-0.9676j,kd4(jω)=-0.7743-0.9676j.Table4 shows the results of the algorithm for the value set at frequeny w=1.2.Table 4
Results of the algorithm for the value set at frequencyω=1.2.
v
1
-
v
2 edge
v
3
v
4
v
5
v
6
v
7
-
3.0339
+
6.1294
j
-
2.211
+
5.1007
j
-
0.4710
+
3.4462
j
-
1.469
+
3.4428
j
2.9559
+
3.2369
j
Kharitonov rectangles calculated
v
1
6.1015
+
5.2779
j
Theoremapplied
Theorem3.5
Theorem3.3
Theorem3.5
Theorem3.5
Theorem3.3
k
n
1
(
j
ω
)
=
-
6.6471
-
4.5717
j
v
2
6.5135
+
6.8573
j
v
x
pred
8.5560
j
-
3.0339
+
6.1294
j
-
2.211
+
5.1007
j
-
3.6463
j
1.4690
+
3.4428
j
k
n
2
(
j
ω
)
=
8.2793
-
4.5717
j
v
2
succ
8.5560
j
v
x
succ
-
2.211
+
5.1007
j
-
0.4710
+
3.4462
j
3.6463
j
2.9559
+
3.2369
j
6.1015
+
5.2779
j
k
n
3
(
j
ω
)
=
8.2793
-
1.8087
j
v
1
pred
2.9559
+
3.2369
j
φ
(
S
x
)
308.66
308.66
23.01
352.12
352.12
k
n
4
(
j
ω
)
=
6.6471
-
1.8087
j
φ
(
d
1
)
284.62
Conditionverified
Theorem3.5(C3)
Theorem3.3(C4)
Theorem3.5(C4)
Theorem3.5(C4)
Theorem3.3(C2)
k
d
1
(
j
ω
)
=
-
0.7743
-
1.8230
j
d
1
0.2524
-
0.9676
j
d
x
-
0.7741
-
0.9676
j
-
0.7741
-
0.9676
j
-
0.7743
-
1.8230
j
0.2524
-
1.8230
j
0.2524
-
1.8230
j
k
d
2
(
j
ω
)
=
0.2524
-
1.8230
j
n
1
6.6471
-
4.5717
j
n
x
8.2793
-
1.8088
j
6.6471
-
1.8088
j
6.6471
-
1.8098
j
6.6471
-
1.8089
j
6.6471
-
4.5717
j
k
d
3
(
j
ω
)
=
0.2524
-
0.9676
j
n
2
8.2793
-
4.5717
j
v
x
n
3
/
d
2
n
4
/
d
2
n
4
/
d
3
n
4
/
d
4
n
1
/
d
4
k
d
4
(
j
ω
)
=
-
0.7743
-
0.9676
j
d
4
λ
0.2524
-
1.8230
j
d
2
λ
-
0.5343
-
0.9677
jFrom these kharitonov rectangles the value set given in Figure11(c) is directly obtained.Finally, solving the equation system [10, equation (16)], the interval plant is obtained:
(5.23)Gp(s)=[1011]s3+[78]s2+[66.5]s+[57.5][0.751.25]s3+[22.5]s2+[1.52]s+[11.5].
Applying Gp(s=jω) at ω=1.0, ω=1.1 and ω=1.2 the value sets given in Figure 12 are obtained.Value sets obtained atw=1.0, w=1.1, and w=1.2.
(a)
w
=
1.0
(b)
w
=
1.1
(c)
w
=
1.2


## 5.1. Value Set at Frequencyω=1.0

The complete arc with verticesv1=n1/d1=1.5676+2.5946j and v2=n1/d2=2.0000+8.0000j is taken as initial element. Then Theorems 4.3 and 4.5 will be applied. So
(5.2)v2succ=n2λd2=0.8000+10.4000j,v1pred=n4λd1=2.3336j.
Applying the arc normalization (Lemma 4.1) the following data are obtained
(5.3)φ(n1)=229.40,n1=-0.6508-0.7592j,d1=-0.3254+0.0542j,d2=-0.1085+0.0542j,n4λ=-0.1266-0.7594j,n2λ=-0.6508-1.0846j.
Then, all the other vertices are assigned as follows.(
1
) Vertex v3=vx=nx/dx=0.8000+10.4000j. Then vxpred=nx/dxpred=2.0000+8.0000j. These are the vertices of an edge, and Theorem 4.5 is applied, vxsucc=10.0000j, φ(Ax)=210.97.Case1. Theorem 4.5(C1) is satisfied: arg(v2/vx)=φ(Ax)+arg(n1)-90=350.36 and nx=d2vx=-0.6508-1.0846j satisfies the Numerator Condition (Lemma 4.2(4),nx=n2):
(5.4)(n1R=-0.6508<n4λR=-0.1266,n1I=-0.7592>n2λI=-1.0846),(nxR=n1R=-0.6508,nxI=-1.0846≤n1I=-0.7592).
Then dx=d2=-0.1085+0.0542j. Therefore v3=vx=n2/d2.(
2
) Vertex v4=vx=nx/dx=-4.8000+7.6000j. Then vxpred=10j. These are the vertices of an edge, and Theorem 4.5 is applied: vxsucc=-3.5862+1.0345j, φ(Ax)=174.29.Case1. Theorem 4.5(C1) is satisfied: arg(v2/vx)=φ(Ax)+arg(n1)-90=313.69 and nx=d2vx=0.1084-1.0847j satisfies the Numerator Condition (Lemma 4.2(4), nx=n3). Then dx=d2=-0.1085+0.0542j. Therefore v4=vx=n3/d2.(
3
) Vertex v5=vx=nx/dx=-3.5862+1.0345j. Then vxpred=nx/dxpred=-4.8000+7.6000j. These are the vertices of an arc, and Theorem 4.3 is applied: φ(Ax)=174.29.Case1. Theorem 4.3(C1) is not satisfied: arg(v2/vx)=272.06≠arg(n1)+φ(Ax)=43.69.Case2. Theorem 4.3(C2) is not satisfied:arg(v1/vx)=254.95≠arg(n1)+φ(Ax)+90=133.69.Case3. Theorem 4.3(C3) is satisfied: tan(arg(1/vx)-φ(Ax)+90)d2R=0.2712>d2I=0.0542 and nx=0.1085-1.0846j satisfies the Numerator Condition (Lemma 4.2(4)) nx=n3. Then dx=d3=-0.1085+0.2712j   v5=vx=n3/d3.(
4
) Vertex v6=vx=nx/dx=-2.5517+0.6207j. Then vxpred=nx/dxpred=-3.5862+1.0345j. These are the vertices of an edge, and Theorem 4.5 is applied: vxsucc-1.3443+1.2131j, φ(Ax)=261.87.Case1. Theorem 4.5(C1) is not satisfied: arg(v2/vx)=269.64≠φ(Ax)+arg(n1)-90=41.27.Case2. Theorem 4.5(C2) is not satisfied: arg(v1/vx)=252.53≠φ(Ax)+arg(n1)=131.27.Case3. Theorem 4.5(C3) is satisfied: tan(arg(1/vx)-φ(Ax)+180)d2R=0.2712>d2I=0.0542 and nx=0.1085-0.7592j  satisfies the Numerator Condition (Lemma 4.2(3)) nx=n4: then dx=d3=-0.1085+0.2712j and v6=vx=n4/d3.(
5
) Vertex v7=vx=nx/dx=-1.3443+1.2131j. Then vxpred=nx/dxpred=-2.5517+0.6207j. These are the vertices of an arc, and Theorem 4.3 is applied: φ(Ax)=261.87.Case1. Theorem 4.3(C1) is not satisfied: arg(v2/vx)=298.03≠arg(n1)+φ(Ax)=131.27.Case2. Theorem 4.3(C2) is not satisfied: arg(v1/vx)=280.93≠arg(n1)+φ(Ax)+90=221.27.Case3. Theorem 4.3(C3) is not satisfied: tan(arg(1/vx)-φ(Ax)+90)d2R=-0.1302<d2I=0.0542.Case4. Theorem 4.3(C4) is satisfied: tan(arg(1/vx)-φ(Ax)+180)d1R=0.2712>d1I=0.0542 and nx=0.1085-0.7592j satisfies the Numerator Condition (Lemma 4.2(4)) nx=n4. Then dx=d4=-0.3254+0.2712j; v7=vx=n4/d4.In summary, the assigned polynomials are(5.5)v1=n1d1,v2=n1d2,v3=n2d2,v4=n3d2,v5=n3d3,v6=n4d3,v7=n4d4,
and the values can be calculated: from normalization,
(5.6)n1=-0.6508-0.7592j,d1=-0.3254+0.0542j,d2=-0.1085+0.0542j,
and from the vertices,
(5.7)v3:n2=-0.6508-1.0846j,d2=-0.1085+0.0542j,v4:n3=0.1084-1.0847j,d2=-0.1085+0.0542j,v5:n3=0.1085-1.0846j,d3=-0.8464+2.0152j,v6:n4=0.1085-0.7592j,d3=-0.1085+0.2712j,v7:n4=0.1085-0.7593j,d4=-0.3254+0.2712j.Then(5.8)kn1(jω)=-0.6508-1.0847j,kn2(jω)=0.1085-1.0847j,kn3(jω)=0.1085-0.7592j,kn4(jω)=-0.6508-0.7592j,kd1(jω)=-0.3254+0.0542j,kd2(jω)=-0.1085+0.0542j,kd3(jω)=-0.1085+0.2712j,kd4(jω)=-0.3254+0.2712j.Table2 shows the results of the algorithm for the value set at frequency ω=1.0.Table 2
Results of the algorithm for the value set at frequencyω=1.0.
v
1
-
v
2 arc
v
3
v
4
v
5
v
6
v
7
0.8000 + 10.4000j
−4.8000 + 7.6000j
−3.5862 + 1.0345j
−2.5517 + 0.6207j
−1.3443 + 1.2131j
Kharitonov rectangles calculated
v
1
1.5676 + 2.5946j
Theoremapplied
Theorem4.5
Theorem4.5
Theorem4.3
Theorem4.5
Theorem4.3
k
n
1
(
j
ω
)
=
-
0.6508
-
1.0847
j
v
2
2.0000 + 8.0000j
v
x
pred
2.0000 + 8.0000j
10.0000j
−4.8000 + 7.6000j
−3.5862 + 1.0345j
−2.5517 + 0.6207j
k
n
2
(
j
ω
)
=
0.1085
-
1.0847
j
v
2
succ
0.8000 + 10.4000j
v
x
succ
10.0000j
−3.5862 + 1.0345j
−2.5517 + 0.6207j
−1.3443 + 1.2131j
2.3336j
k
n
3
(
j
ω
)
=
0.1085
-
0.7592
j
v
1
pred
0 + 2.3336j
φ
(
A
x
)
210.97
174.29
174.29
261.87
261.87
k
n
4
(
j
ω
)
=
-
0.6508
-
0.7592
j
φ
(
n
1
)
229.40
Conditionverified
Theorem4.5(C1)
Theorem4.5(C1)
Theorem4.3(C3)
Theorem4.5(C3)
Theorem4.3(C4)
k
d
1
(
j
ω
)
=
-
0.3254
+
0.0542
j
n
1
−0.6508 − 0.7592j
n
x
−0.6508 − 1.0846j
0.1084 − 1.0847j
0.1085 − 1.0846j
0.1085 − 0.7592j
0.1085 − 0.7592j
k
d
2
(
j
ω
)
=
-
0.1085
+
0.0542
j
d
1
−0.3254 + 0.0542j
d
x
−0.1085 + 0.0542j
−0.1085 + 0.0542j
−0.1085 + 0.2712j
−0.1085 + 0.2712j
−0.3254 + 0.2712j
k
d
3
(
j
ω
)
=
-
0.1085
+
0.2712
j
d
2
−0.1085 + 0.0542j
v
x
n
2
/
d
2
n
3
/
d
2
n
3
/
d
3
n
4
/
d
3
n
4
/
d
4
k
d
4
(
j
ω
)
=
-
0.3254
+
0.2712
j
n
4
λ
−0.1266 − 0.7594j
n
2
λ
−0.6508 − 1.0846jFrom these Kharitonov rectangles the value set given in Figure11(a) is directly obtained.(a)
(b)
(c)


## 5.2. Value Set at Frequencyω=1.1

The complete arc with verticesv1=n1/d1=-2.8422+2.9830j and v2=n1/d2=-0.9808+2.4599j is taken as initial element. Then Theorems 4.3 and 4.5 will be applied. So
(5.9)v2succ=n2λd2=3.0420j,v1pred=n4λd1=-3.8291+3.7385j.
Applying the arc normalization (Lemma 4.1) the following data are obtained:
(5.10)φ(n1)=360-arg(1v2-1v1)=81.05,n1=0.1556+0.9878j,d1=n1v1=0.1475-0.1927j;d2=n1v2=0.3247-0.1927j,n4λ=d1v1pred=0.1556+1.2895j,(5.11)n2λ=d2v2succ=0.5862+0.9878j.
Then, all the other vertices are assigned as follows.(
1
) Vertex v3=vx=nx/dx=0.4996+3.0386j. Then vxpred=nx/dxpred=3.0420j. These are the vertices of an arc, and Theorem 4.3 is applied: φ(Ax)=8.95.Cases1and2. Theorem 4.3(C1) and (C2) are not satisfied.Case3. Theorem 4.3(C3) is satisfied: tan(arg(1/vx)-φ(Ax)+90)d2R=0.0022>d2I=-0.1927 and nx=0.1555+0.9878j satisfies the Numerator Condition (Lemma 4.2(1)) nx=n1.Then dx=d3=0.3247+0.0022jv3=vx=n1/d3.(
2
) Vertex v4=vx=nx/dx=2.3317+3.0261j. Then vxpred=nx/dxpred=0.4996+3.0386j. These are the vertices of an edge, and Theorem 4.5 is applied: vxsucc=5.1859+6.6181j and φ(Ax)=127.23.Cases1and2. Theorem 4.5(C1) and  (C2) are not satisfied.Case3. Theorem 4.5(C3) is satisfied: tan(arg(1/vx)-φ(Ax)+180)d2R=0.0022>d2I=-0.1927 and nx=0.7505+0.9878j  satisfies the Numerator Condition (Lemma 4.2(1)) nx=n2. Then dx=d3=0.3247+0.0022j.  v4=vx=n2/d3.(
3
) Vertex v5=vx=nx/dx=5.1859+6.6181j. Then vxpred=nx/dxpred=2.3317+3.0261j. These are the vertices of an arc, and Theorem 4.3 is applied: φ(Ax)=127.23.Cases1,2,and3. Theorem 4.3(C1), (C2), and (C3) are not satisfied.Case4. Theorem 4.3(C4) is satisfied: tan(arg(1/vx)-φ(Ax)+180)d1R=0.0022>d1I=-0.1927 and nx=0.7505+0.9878j satisfies the Numerator Condition (Lemma 4.2(1)) nx=n2. Then dx=d4=0.1475+0.0022j  v5=vx=n2/d4.(
4
) Vertex v6=vx=nx/dx=5.2164+8.6623j. Then vxpred=nx/dxpred=5.1859+6.6181j. These are the vertices of an edge, and Theorem 4.5 is applied: vxsucc=8.7404j, φ(Ax)=210.20.Cases1,2,and3.  Theorem 4.5(C1), (C2), and (C3) are not satisfied.Case4. Theorem 4.5(C4) is satisfied: tan(arg(1/vx)-φ(Ax)+270)d1R=0.0022>d1I=-0.1927and nx=0.7505+1.2895j  satisfies the Numerator Condition (Lemma 4.2(1)) nx=n3: then dx=d4=0.1475+0.0022j, v6=vx=n3/d4.(
5
) Vertex v7=vx=nx/dx=-3.8291+3.7385j. Then vxpred=nx/dxpred=8.7404j. These are the vertices of an arc, and Theorem 4.3 is applied: φ(Ax)=186.88.Case1. Theorem 4.3(C1) is not satisfied.Case2. Theorem 4.3(C2) is satisfied: arg(v1/vx)=arg(n1)+φ(Ax)+90=357.93and nx=d1vx=0.1556+1.2895j satisfies the Numerator Condition (Lemma 4.2(1)) nx=n4. Then dx=d1=0.1475-0.1927j, v7=vx=n4/d1.In summary, the assigned polynomials are(5.12)v1=n1d1,v2=n1d2,v3=n1d3,v4=n2d3,v5=n2d4,v6=n3d4,v7=n4d1
and the values can be calculated: from normalization,
(5.13)n1=0.1556+0.9878j,d1=0.1475-0.1927j,d2=0.3247-0.1927j,
and from the vertices,
(5.14)v3:n1=0.1555+0.9878j,d3=0.3247+0.0022j,v4:n2=0.7505+0.9878j,d3=0.3247+0.0022j,v5:n2=0.7505+0.9878j,d4=0.1475+0.0022j,v6:n3=0.7505+1.2895j,d4=0.1475+0.0022j,v7:n4=0.1556+1.2895j,d1=0.1475-0.1927j.
Then
(5.15)kn1(jω)=0.1555+0.9878j,kn2(jω)=0.7505+0.9878j,kn3(jω)=0.7505+1.2895j,kn4(jω)=0.1556+1.2895j,kd1(jω)=0.1475-0.1927j,kd2(jω)=0.3247-0.1927j,kd3(jω)=0.3247+0.0022j,kd4(jω)=0.1475+0.0022j.Table3 shows the results of the algorithm for the value set at frequency ω=1.1.Table 3
Results of the algorithm for the value set at frequencyω=1.1.
v
1
-
v
2 arc
v
3
v
4
v
5
v
6
v
7
0.4996
+
3.0386
j
2.3317
+
3.0261
j
5.1859
+
6.6181
j
5.2164
+
8.6623
j
-
3.8291
+
3.7385
j
Kharitonov rectangles calculated
v
1
-
2.8422
+
2.9830
j
Theoremapplied
Theorem4.3
Theorem4.5
Theorem4.3
Theorem4.5
Theorem4.3
k
n
1
(
j
ω
)
=
0.1555
+
0.9878
j
v
2
-
0.9808
+
2.4599
j
v
x
pred
3.0420
j
0.4996
+
3.0386
j
2.3317
+
3.0261
j
5.1859
+
6.6181
j
8.7404
j
k
n
2
(
j
ω
)
=
0.7505
+
0.9878
j
v
2
scuss
0
+
3.0420
j
v
x
succ
2.3317
+
3.0261
j
5.1859
+
6.6181
j
5.2164
+
8.6623
j
8.7404
j
-
2.8422
+
2.9830
j
k
n
3
(
j
ω
)
=
0.7505
+
1.2895
j
v
1
pred
-
3.8291
+
3.7385
j
φ
(
A
x
)
8.95
127.23
127.23
210.20
186.88
k
n
4
(
j
ω
)
=
0.1556
+
1.2895
j
φ
(
n
1
)
81.05
Conditionverified
Theorem4.3(C3)
Theorem4.5(C3)
Theorem4.3(C4)
Theorem4.5(C4)
Theorem4.3(C2)
k
d
1
(
j
ω
)
=
0.1475
-
0.1927
j
n
1
0.1556
+
0.9878
j
n
x
-
0.1555
+
0.9878
j
0.7505
+
0.9878
j
0.7505
+
0.9878
j
0.7505
+
1.2895
j
0.1556
+
1.2895
j
k
d
2
(
j
ω
)
=
0.3247
-
0.1927
j
d
1
0.1475
-
0.1927
j
d
x
0.3247
+
0.0022
j
0.3247
+
0.0022
j
0.1475
+
0.0022
j
0.1475
+
0.0022
j
0.1475
-
0.1927
j
k
d
3
(
j
ω
)
=
0.3247
+
0.0022
j
d
2
0.3247
-
0.1927
j
v
x
n
1
/
d
3
n
2
/
d
3
n
2
/
d
4
n
3
/
d
4
n
4
/
d
1
k
d
4
(
j
ω
)
=
0.1475
+
0.0022
j
n
4
λ
0.1556
+
1.2895
j
n
2
λ
0.5862
+
0.9878
jFrom these Kharitonov rectangles the value set given in Figure11(b) is directly obtained.


## 5.3. Value Set at Frequencyω=1.2

The complete edge with verticesv1=n1/d1=6.1015+5.2779j and v2=n2/d1=6.5135+6.8573j is taken as initial element. Then Theorems 3.3 and 3.5 will be applied. So
(5.16)v2succ=n2d2λ=8.5560j,v1pred=n1d4λ=2.9559+3.2369j.
Applying the edge normalization (Lemma 3.1) the following data are obtained:
(5.17)ϕ(d1)=360-arg(v2-v1)=284.62,d1=cos(φ(d1))+jsin(φ(d1))=0.2524-0.9676j,n1=v1d1=6.6471-4.5717j,n2=v2d1=8.2793-4.5717j,d2λ=n2v2succ=-0.5343-0.9677j,d4λ=n1v1pred=0.2524-1.8230j.
Then, all the other vertices are assigned as follows.(
1
) Vertex v3=vx=nx/dx=-3.0339+6.1294j. Then vxpred=nx/dxpred=8.5560j. These are the vertices of an arc, and Theorem 3.5 is applied: vxsucc=-2.2110+5.1007j and  φ(Sx)=arg(vxsucc-vx)=308.66.Cases1and 2. Theorem 3.5(C1) and (C2) are not satisfied.Case3. Theorem 3.5(C3) is satisfied: tan(arg(vx)-φ(Sx)+180)n2R=-1.8087>n2I=-4.571 and dx=n2R[1+jtan(arg(vx)-φ(Sx)+180)]/vx=-0.7740-0.9676j satisfies the Denominator Condition (Lemma 3.2(3)) dx=d2: then nx=n3=n2R[1+jtan(arg(vx)-φ(Sx)+180)]=8.2793-1.8087j;  v3=vx=n3/d2.(
2
) Vertex v4=vx=nx/dx=-2.211+5.1007j. Then vxpred=nx/dxpred=-3.0339+6.1294j. These are the vertices of an edge, and Theorem 3.3 is applied: φ(Sx)=arg(vx-vxpred)=308.66.Cases1and2. Theorem 3.3(C1) and (C2) are not satisfied.Case3. Theorem 3.3(C3) is satisfied: tan(arg(vx)-φ(Sx)+90)n2R=30.4258>n2I=-4.5717 but dx=n2R[1+jtan(arg(vx)-φ(Sx)+90)]/vx=4.4292-3.5431j does not satisfy the Denominator Condition: (d1R=0.2524>d2λR=-0.5343andd1I=-0.9676>d4λI=-1.8230) (Case 3) but (dxR=4.4292≠d1R=0.2524) then dx≠d1 and dx≠d4(dxI=-3.5431≠d1I-0.9676) then dx≠d2(dxR=4.4292>d1R=0.2524anddxI=-3.5431<d1I=-0.9676) then dx≠d3.Case4. Theorem 3.3(C4) is satisfied: tan(arg(vx)-φ(Sx)+180)n1R=-1.8088>n1I=-4.5717 and dx=n1R[1+jtan(arg(vx)-φ(Sx)+180)]/vx=-0.7741-0.9676j satisfies the Denominator Condition (Lemma 3.2(3)) dx=d2:
(5.18)(d1R=0.2524>d2λR=-0.5343,d1I=-0.9676>d4λI=-1.8230),(dxI=d1I=-0.9676,dxR=-0.7740≤d1R=0.2524).
Then nx=n4=n1R[1+jtan(arg(vx)-φ(Sx)+180)]=6.6471-1.8088j, v4=vx=n4/d2.(
3
) Vertex v5=vx=nx/dx=-0.47099+3.4462j. Then vxpred=nx/dxpred=-2.2110+5.1007j. These are the vertices of an arc, and Theorem 3.5 is applied: vxsucc=3.6463j and  φ(Sx)=arg(vxsucc-vx)=23.01.Cases
1
,
2
,  and 3. Theorems 3.5(C1) and (C2) are not satisfied. Theorem 3.5(C3) is satisfied but dx=8.3397-3.5422j does not satisfy the Denominator Condition (Lemma 3.2(3)).Case4. Theorem 3.5(C4) is satisfied: tan(arg(vx)-φ(Sx)+270)n1R=-1.8098>n1I=-4.5717 and dx=n1R[1+jtan(arg(vx)-φ(Sx)+270)]/vx=-0.7743-1.8230j satisfies the Denominator Condition (Lemma 3.2(3)).Thennx=n4=n1R[1+jtan(arg(vx)-φ(Sx)+270)]=6.6471-1.8098j,  v5=vx=n4/d3.(
4
) Vertex v6=vx=nx/dx=1.469+3.4428j. Then vxpred=nx/dxpred=3.6463j. These are the vertices of an arc, and Theorem 3.5 is applied: vxsucc=2.9559+3.2369j and φ(Sx)=arg(vxsucc-vx)=352.12.Cases1,2,and 3. Theorem 3.5(C1) and (C2) are not satisfied. Theorem 3.5(C3) is satisfied but dx=8.3440+1.1554j does not satisfy the Denominator Condition (Lemma 3.2(3)).Case4. Theorem 3.5(C4) is satisfied: tan(arg(vx)-φ(Sx)+270)n1R=-1.8089>n1I=-4.5717 and dx=0.2524-1.8230j satisfies the Denominator Condition (Lemma 3.2(3)) dx=d4.Thennx=n4=n1R[1+jtan(arg(vx)-φ(Sx)+270)]=6.6471-1.8089j,  v6=vx=n4/d4.(
5
) Vertex v7=vx=nx/dx=2.9559+3.2369j. Then vxpred=nx/dxpred=1.4690+3.4428j. These are the vertices of an edge, and Theorem 3.3 is applied: φ(Sx)=arg(vx-vxpred)=352.12.Case1. Theorem 3.3(C1) is not satisfied.Case2. Theorem 3.3(C2) is satisfied: arg(vx/v1)=arg(d1)+φ(Sx)+90=6.74 and dx=n1/vx=0.2524-1.8230j satisfies the Denominator Condition (Lemma 3.2(3)) dx=d4. Then nx=n1=6.6471-4.5717j, v7=vx=n1/d4.In summary, the assigned polynomials are(5.19)v1=n1d1,v2=n2d1,v3=n3d2,v4=n4d2,v5=n4d3,v6=n4d4,v7=n1d4,
and the values can be calculated: from normalization,
(5.20)d1=0.2524-0.9676,n1=6.6471-4.5717j,n2=8.2793-4.5717j,
and from the vertices,
(5.21)v3:n3=8.2793-1.8087j,d2=-0.7741-0.9676j,v4:n4=6.6471-1.8087j,d2=-0.7741-0.9676j,v5:n4=6.6471-1.8087j,d3=-0.7743-1.8230j,v6:n4=6.6471-1.8087j,d4=0.2524-1.8230j,v7:n1=6.6471-4.5717j,d4=0.2524-1.8230j.
Then
(5.22)kn1(jω)=6.6471-4.5717j,kn2(jω)=8.2793-4.5717j,kn3(jω)=8.2793-1.8087j,kn4(jω)=6.6471-1.8087j,kd1(jω)=-0.7743-1.8230j,kd2(jω)=0.2524-1.8230j,kd3(jω)=0.2524-0.9676j,kd4(jω)=-0.7743-0.9676j.Table4 shows the results of the algorithm for the value set at frequeny w=1.2.Table 4
Results of the algorithm for the value set at frequencyω=1.2.
v
1
-
v
2 edge
v
3
v
4
v
5
v
6
v
7
-
3.0339
+
6.1294
j
-
2.211
+
5.1007
j
-
0.4710
+
3.4462
j
-
1.469
+
3.4428
j
2.9559
+
3.2369
j
Kharitonov rectangles calculated
v
1
6.1015
+
5.2779
j
Theoremapplied
Theorem3.5
Theorem3.3
Theorem3.5
Theorem3.5
Theorem3.3
k
n
1
(
j
ω
)
=
-
6.6471
-
4.5717
j
v
2
6.5135
+
6.8573
j
v
x
pred
8.5560
j
-
3.0339
+
6.1294
j
-
2.211
+
5.1007
j
-
3.6463
j
1.4690
+
3.4428
j
k
n
2
(
j
ω
)
=
8.2793
-
4.5717
j
v
2
succ
8.5560
j
v
x
succ
-
2.211
+
5.1007
j
-
0.4710
+
3.4462
j
3.6463
j
2.9559
+
3.2369
j
6.1015
+
5.2779
j
k
n
3
(
j
ω
)
=
8.2793
-
1.8087
j
v
1
pred
2.9559
+
3.2369
j
φ
(
S
x
)
308.66
308.66
23.01
352.12
352.12
k
n
4
(
j
ω
)
=
6.6471
-
1.8087
j
φ
(
d
1
)
284.62
Conditionverified
Theorem3.5(C3)
Theorem3.3(C4)
Theorem3.5(C4)
Theorem3.5(C4)
Theorem3.3(C2)
k
d
1
(
j
ω
)
=
-
0.7743
-
1.8230
j
d
1
0.2524
-
0.9676
j
d
x
-
0.7741
-
0.9676
j
-
0.7741
-
0.9676
j
-
0.7743
-
1.8230
j
0.2524
-
1.8230
j
0.2524
-
1.8230
j
k
d
2
(
j
ω
)
=
0.2524
-
1.8230
j
n
1
6.6471
-
4.5717
j
n
x
8.2793
-
1.8088
j
6.6471
-
1.8088
j
6.6471
-
1.8098
j
6.6471
-
1.8089
j
6.6471
-
4.5717
j
k
d
3
(
j
ω
)
=
0.2524
-
0.9676
j
n
2
8.2793
-
4.5717
j
v
x
n
3
/
d
2
n
4
/
d
2
n
4
/
d
3
n
4
/
d
4
n
1
/
d
4
k
d
4
(
j
ω
)
=
-
0.7743
-
0.9676
j
d
4
λ
0.2524
-
1.8230
j
d
2
λ
-
0.5343
-
0.9677
jFrom these kharitonov rectangles the value set given in Figure11(c) is directly obtained.Finally, solving the equation system [10, equation (16)], the interval plant is obtained:
(5.23)Gp(s)=[1011]s3+[78]s2+[66.5]s+[57.5][0.751.25]s3+[22.5]s2+[1.52]s+[11.5].
Applying Gp(s=jω) at ω=1.0, ω=1.1 and ω=1.2 the value sets given in Figure 12 are obtained.Value sets obtained atw=1.0, w=1.1, and w=1.2.
(a)
w
=
1.0
(b)
w
=
1.1
(c)
w
=
1.2


## 6. Conclusions

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, that is, if there are two vertices in a quadrant. This necessary and sufficient condition is not restrictive and practically all the value sets satisfy it. Finally, the interval plant can be identified solving the equation system between the Kharitonov rectangles and the parameters of the plant.The algorithm has been formulated using the frequency domain properties of linear interval systems. The identification procedure of multilinear (affine, polynomial) systems will be studied using the results in [11].



---

*Source: 840603-2012-12-31.xml*