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Robust stability of fractional order system with general interval uncertainties
Systems & Control Letters 99 (2017) 1–8
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Systems & Control Letters journal homepage: www.elsevier.com/locate/sysconle
Robust stability of fractional order system with general interval uncertainties Shiqi Zheng School of Automation, China University of Geosciences, No. 388 Lumo Road, Wuhan, China
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info
Article history: Received 29 February 2016 Received in revised form 18 June 2016 Accepted 3 November 2016
Keywords: Fractional order control system Interval uncertainty Robust stability
a b s t r a c t This paper focuses on the analysis of robust stability of fractional order system with general interval uncertainties. The concept of general interval uncertainties means that the interval uncertainties exist both in the coefficients and orders of the fractional order system. Necessary and sufficient conditions are proposed to check the robust stability of general interval fractional order system. According to the proposed stability criterion, it is interesting to find that the Edge Theorem, which was initially proposed for integer order system, cannot be directly extended to test the stability of fractional order system with general interval uncertainties. Examples are followed to verify the validity of the proposed method. © 2016 Elsevier B.V. All rights reserved.
1. Introduction Recently, fractional order calculus has gained much interest both from academic and industrial point of view. It proves to be a modern powerful tool in analyzing various physical phenomena. A growing number of physical systems, including servo motor system, chaotic system, biological system etc., have been verified that their behaviors can be better described by fractional order systems [1,2]. Meanwhile, control of these fractional order systems has also become a hot topic. Some typical research about fractional order control system, such as CRONE control, fractional order PID etc., can be found in [3–7]. A critical aspect of the fractional control system is the robust stability. Robust stability refers to the preservation of stability in face of different classes of uncertainties. This paper aims to investigate the robust stability of fractional order control system with interval uncertainties [8–13]. The interval uncertainties mean that the parameters of the fractional order control system lie in specified intervals. There are mainly two classes of methods to check the stability of interval fractional order system, i.e. Linear Matrix Inequalities (LMI) method [14–16] and graphical method [8–13]. A nice feature of LMI method is its ability to accommodate uncertainty in a relatively straightforward manner [3]. However, the LMI method may be only limited to fractional order system with commensurate order. Moreover, solving LMI sometimes can cause conservatism in stability test [10,17]. For graphical approach, it firstly depicts the value set of the interval system in the complex plane, and then tests the stability by applying the Zero Exclusion Principle [1,18]. The graphical method turns out to be effective and accurate. E-mail address: [email protected]. http://dx.doi.org/10.1016/j.sysconle.2016.11.001 0167-6911/© 2016 Elsevier B.V. All rights reserved.
It is noted that the methods mentioned above only considered the interval uncertainties in the coefficients of fractional order control system. Yet, the orders of the fractional order system may also suffer from interval uncertainties. For example, in system modeling, the parameters of a fractional model, including coefficients and orders, could all vary as the working conditions of the control system change. The interval uncertainties in the fractional orders have been mentioned in [9,19,20]. Here we refer to interval uncertainties both in the coefficients and orders as general interval uncertainties. It is worth noting that the stability problem of fractional order system with general interval uncertainties is still open to our best knowledge. Recently the Edge Theorem [9,18] is adopted to test the stability of fractional order system with interval uncertainties only in orders. However, it will be seen in the sequel that the Edge Theorem, which was initially proposed for integer order system, cannot be directly extended to test the stability of fractional order system with general interval uncertainties. In fact, the hypothesis in the Edge Theorem may be violated in presence of general interval uncertainties (see Example 3). In this paper, we present some necessary and sufficient stability criteria for fractional order system with general interval uncertainties based on graphical method. According to the proposed stability criteria, it can be concluded that the Edge Theorem is applicable only when some certain conditions of the interval uncertainties are satisfied. The rest of the paper is as follows. Section 2 formulates the problem. Main results are presented in Section 3. Section 4 gives some discussions on the main results. Examples are followed. Section 5 concludes the paper.
2
S. Zheng / Systems & Control Letters 99 (2017) 1–8
2. Problem formulation
Note that let s = jω in Si (s) where ω is a fixed frequency, then it can be seen:
Notation. R is the set of real numbers. N is the set of natural numbers. ∅ stands for empty set. ∂ (P) denotes the boundary of value set P in complex plane.
Si (jω) =
Definition 1. Consider the following uncertain real fractional order system
δ (s, q, α) = q0 sα0 + q1 sα1 + · · · + qn sαn
(1)
where s is the Laplace operator, q = (q0 , q1 , . . . , qn ) and α = (α0 , α1 , . . . , αn ) are the coefficient and order vectors respectively. They vary in the following hypercubes: + q ∈ Q = {q ∈ Rn | qi ∈ [q− i , qi ], i = 0, 1, . . . , n},
α ∈ A = {α ∈ Rn | αi ∈ [αi− , αi+ ], αi− ≥ 0, i = 0, 1, . . . , n} + − + where [q− i , qi ] and [αi , αi ] are specified intervals. Since the coefficients and orders of (1) both vary in independent intervals, (1) is referred to as fractional order system with general interval uncertainties.
We give the following assumption throughout this paper: Assumption 1. Assume (1) satisfies: (1) αn > αm (m = 0, 1, 2, . . . , n − 1) and qn ̸ = 0; (2) |αi+ − αi− | ≤ 2 (i = 0, 1, 2, . . . , n).
−
[αi q− i (jω )
Based on the above facts, we formulate the following design problem:
,αi+ ]
−
[αi , q+ i (jω )
,αi+ ]
©
.
Therefore, the value set Si (jω) in complex plane contains two line segments and two arc segments. Our main results are based on the following lemma. Note that all the proofs in Sections 3 and 4 are put in Appendix. Lemma 1 (Zero Exclusion Principle). δ (s, q, α) is Hurwitz stable if and only if δ (s, q, α) contains at least one Hurwitz stable polynomial and 0 ̸ ∈ δ (jω, q, α) for ∀ω ∈ [0, +∞), or equivalently 0 ̸ ∈ ∂δ (jω, q, α) for ∀ω ∈ [0, +∞). From this lemma, it can be seen that the main problem here is how to construct the boundary of δ (s, q, α) efficiently. First we will consider the stability of δ (s, q, α) when n = 2 in (1). 3.1. Stability criterion for δ (s, q, α) when n = 2 Lemma 2. Assume in (1), for a fixed ω we have:
∂δ (jω, q, α) = ∂[q0 (jω)α0 + q1 (jω)α1 ] ⊆ S0 (jω) + S1 (jω) α α = [I0q (jω)V0 + I1q (jω)V1 ] α α α q q Vα ∪ [I0 (jω) 0 + V1 (jω)I1 ] ∪ [V0q (jω)I0 + I1q (jω)V1 ] q q Iα Iα ∪ [V0 (jω) 0 + V1 (jω) 1 ] where Vi , Ii , Viα , Iiα , Si (i = 0, 1) are from Definition 2. q
Remark 1. Condition (1) in Assumption 1 is standard. It guarantees that αn is always the highest order in δ (s, q, α) and the term qn sαn never vanishes [18]. Condition (2) in Assumption 1 assumes that the variation of the order in δ (s, q, α) belongs to [0, 2], i.e. ∆αi ∈ [0, 2]. This seems to be a little conservative. However, the control system can be very sensitive to the order change, say the dynamic behavior of the zero order system is very different from second order system. Therefore, ∆αi ∈ [0, 2] in fact can represent a lot of different dynamics (see the claim in Example 1 in [19]). Moreover, if in some situations, the range of ∆αi exceeds [0, 2], one can separate the interval into some smaller intervals and consider them respectively.
+
−
αi αi + + , , [q− [q− i , qi ](jω ) i , qi ](jω )
¶
q
From Lemma 2, it can be seen that ∂δ (jω, q, α) is included in S0 (jω) + S1 (jω). However, directly constructing value set S0 (jω) + S1 (jω) is still not an easy task since they are interval elements plus interval elements. Note that S0 (jω)+S1 (jω) contain three parts: line segment plus line segment, line segment plus arc segment and arc segment plus arc segment. Next, we will have a further analysis on these three parts. Lemma 3 (Line Segment Plus Line Segment [8,18]). α
α
∂[I0q (jω)V0 + I1q (jω)V1 ] α α α α ⊆ [V0q (jω)V0 + I1q (jω)V1 ] ∪ [I0q (jω)V0 + V1q (jω)V1 ] α α = [K0 (jω) + I1q (jω)V1 ] ∪ [I0q (jω)V0 + K1 (jω)].
Problem 1. Develop an efficient method to check the Hurwitz stability of the general interval fractional order system δ (s, q, α) by (1).
This lemma shows that the boundary of line segment plus line segment is contained in some line segments. For line segment plus arc segment, we first give the following definition:
3. Main results
Definition 3. For a fixed ω, a vertex set Viα = {αi− , αi+ } and a specified α , define A(ω, Viα , α ) ∈ R: A(ω, Viα , α )
First, we give the following definition: Definition 2. For a term qi s define: q
αi
= qi s
[αi− , αi+ ]
(i = 0, 1, . . . , n) in (1),
Viα = {αi− , αi+ }, Iiα = [αi− , αi+ ], Vi and Viα are both sets of two end points, Ii and Iiα are both intervals. Also define: q
q
α
−
+
−
Si (s) =
∪
q α Vi sIi −
= {[qi , qi ]s −
+
αi
, [qi , qi ]s −
+
αi+
− [αi− ,αi+ ]
, qi s
+ [αi− ,αi+ ]
, qi s
where [·] is rounding function, β = arctan[π/(2 ln ω)]. For a term qi sαi in (1) define the vertex set α ,α )
K¯ i (jω, α ) = Ki (jω) ∪ Vi (jω)A(ω,Vi q
K¯ i is a set of six vertex terms.
+
αi αi αi αi Ki (s) = Vi sVi = {q− , q− , q+ , q+ }, i s i s i s i s q α Ii sVi
if [(αi − α )/2 + β/π] = [(αi − α )/2 + β/π] + 1
⎪ ⎩none exist +
if [(αi − α )/2 + β/π] ̸ = [(αi− − α )/2 + β/π] + 1
q
+ − + Vi = {q− i , qi }, Ii = [qi , qi ],
q
=
⎧ + 2[αi+ /2 + β/π − 1]/π ⎪ ⎨α − 2β/π + −
}
Ki (s) is a set of four vertex terms, Si (s) contains four interval elements.
A(ω, Viα , α ) is in fact a point in interval [αi− , αi+ ]. This point will contribute to the boundary of line segment plus arc segment. Further details about how A(ω, Viα , α ) is computed can be found in the Proof of Lemma 4. Based on Definition 3, we have the following result:
S. Zheng / Systems & Control Letters 99 (2017) 1–8
than two terms in δ (s), the terms 1.9 − 2.9s + s2 are fixed. Thus, Lemma 5 can still be adopted. Then from Lemma 5, we have
Lemma 4 (Line Segment Plus Arc Segment). Vα
Iα
∂[I0q (jω) 0 + V1q (jω) 1 ] + − ⊆ [I0q (jω)α0 + K¯ 1 (jω, α0− )] ∪ [I0q (jω)α0 + K¯ 1 (jω, α0+ )] α ∪ [K0 (jω) + V1q (jω)I1 ]. This lemma shows that the boundary of line segment plus arc segment is contained in some line segments and arc segments. Example 1. Consider δ (s) = q0 s0.1 + sα1 where q0 ∈ [3, 4], α1 ∈ [1, 2]. Then using Lemma 4, we have ∂δ (jω) ⊆ {[3, 4](jω)0.1 + K¯ 1 (jω, 0.1)}
∪ {3(jω)0.1 + (jω)[1,2] } ∪ {4(jω)0.1 + (jω)[1,2] } where K¯ 1 (jω, 0.1) = {(jω)1 , (jω)2 , (jω)A(ω,{1,2},0.1) }, A(ω, {1, 2}, 0.1) can be computed according to Definition 3 and it is dependent on ω. For instance, A(2, {1, 2}, 0.1) = 1.364, thus K¯ 1 (jω, 0.1) = {(jω)1 , (jω)2 , (jω)1.364 }; A(0.5, {1, 2}, 0.1) does not exist, thus K¯ 1 (jω, 0.1) = K1 (jω) = {(jω)1 , (jω)2 }. For arc segment plus arc segment, we first give the following definition: α
Definition 4. For a vertex set Vi ¯ i: ki , k¯ i , Mi , M ¯ ki , ki ∈ N are determined by:
=
{αi , αi }, define −
+
if 2 < αi − 2ki < 4.
¯ i ]. [αi− , αi+ ] = [2ki + Mi ] ∪ [2k¯ i + M Any term qi sαi in (1) can be given as: ¯
= qi {s2ki · sMi ∪ s2ki · sMi }.
¯ i are both intervals that satisfy Mi , M ¯ i ⊆ [0, 2]. Note that Mi , M [αi− ,αi+ ] ¯ can be separated into Using Mi , Mi , the arc segment qi (jω) two segments: one is in the upper half plane, the other is in the ¯ i can be found in the lower half plane. Further details about Mi , M Proof of Lemma 5. Based on Definition 4, we have: Lemma 5 (Arc Segment Plus Arc Segment). α
∂[V0q (jω)I0 + V1q (jω)I1 ] α
α
⊆ [V0q (jω)I0 + K1 (jω)] ∪ [K0 (jω) + V1q (jω)I1 ] ∪ [f (jω; V0q , V1q ; V0α , V1α )] where function f (·) is expressed as: f (s; V0 , V1 ; V0α , V1α ) q
¯
¯
= (V0q s2k0 + V1q s2k1 )sM0 ∩M1 ∪ (V0q s2k0 + V1q s2k1 )sM0 ∩M1 ∪
+
∪
q ¯ (V0 s2k0
+
1
1
1
α
where Vi , Ii , Viα , Iiα , Ki (i = 0, 1) are from Definition 1, K¯ i are from Definition 3, function f (·) is from (2).
¯ ¯ q ¯ V1 s2k1 )sM0 ∩M1
(2)
.
q
Example 3. Consider the SISO control system in [13]. Suppose the controller is a proportional gain Kp = 2, the plant P(s) =
¯
+
α
q
Adopting the above definition, interval [αi− , αi+ ] can be expressed as
¯ q ¯ V1 s2k1 )sM0 ∩M1
where
= [V0q (jω)I0 + K1 (jω)] ∪ [K0 (jω) + V1q (jω)I1 ] ∪ [f (jω; V0q , V1q ; V0α , V1α )]
if 0 < αi − 2ki < 2
q (V0 s2k0
¯ (jω) = ∆ ¯ line (jω) ∪ ∆ ¯ arc (jω) ∆
1
− ¯ i = [0, α + − 2ki − 2] M ⎪ i , 2], i ⎩Mi = [αi − 2k +
q
Theorem 1. Assume n = 2 in (1), then δ (s, q, α) is Hurwitz stable if and only if δ (s, q, α) contains at least one Hurwitz stable polynomial ¯ (jω) for ∀ω ∈ [0, +∞) where ∆ ¯ (jω) is given as: and 0 ̸ ∈ ∆
¯ arc (jω) ∆
⎧ ¯i = ∅ M = [αi− − 2ki , αi+ − 2ki ], M ⎪ ⎨ i +
α
According to Lemmas 1–5, we have the stability criterion for
δ (s, q, α) when n = 2:
−
¯ i ⊆ [0, 2] are intervals and are calculated as: Mi , M
,αi+ ]
where F (s) = 1.9 − 2.9s + s2 . To compute f (jω), based on Definition 4 we have k0 = 0, k¯ 0 = ¯ 0 = ∅, k1 = 0, k¯ 1 = 1, M1 = [0.2, 1.7], 1, M0 = [0.1, 1.5], M ¯ 1 = ∅. Therefore, f (jω) = (2 + 1)(jω)[0.2,1.5] . M
= [I0q (jω)α0 + K¯ 1 (jω, α0− )] ∪ [I0q (jω)α0 + K¯ 1 (jω, α0+ )] , + − ∪ [K¯ 0 (jω, α − ) + I q (jω)α1 ] ∪ [K¯ 0 (jω, α + ) + I q (jω)α1 ]
ki = [αi /2], k¯ i = ki + 1.
−
∂δ (jω) ⊆ {2(jω)0.1 + (jω)[0.2,1.7] + F ((jω))} ∪ {2(jω)1.5 + (jω)[0.2,1.7] + F ((jω))} ∪ {2(jω)[0.1,1.5] + (jω)0.2 + F ((jω))} ∪ {2(jω)[0.1,1.5] + (jω)1.7 + F ((jω))} ∪ [f (jω; 2, 1; {0.1, 1.5}, {0.2, 1.7})]
¯ line (jω) ∆
−
qi sαi = qi s[αi
3
sα0 + 0.5 sα1 + 0.9 − 2.9s + s2
where α0 ∈ [0.1, 1.5], α1 ∈ [0.2, 1.7]. The characteristic polynomial of the control system can be expressed as:
δ (s) = 2sα0 + sα1 + 1.9 − 2.9s + s2 .
(3)
δ (s) is the same as that in Example 2. First, we can easily check that δ (s) contains one stable polynomial, i.e. α0 = α1 = 1. Then, ¯ (jω) in the complex plane. As by using Theorem 1, we can plot ∆ ¯ (jω). illustrated in Fig. 1, the zero is included in the value set ∆ Therefore, it can be concluded that δ (s) is not always stable for any α0 ∈ [0.1, 1.5], α1 ∈ [0.2, 1.7]. Meanwhile, in Fig. 2 we have also plotted the value set δ (jω) (blue points). It is shown that the ¯ (jω), and δ (jω) also includes boundary of δ (jω) is contained in ∆ zero. Consequently, by Lemma 1 (Zero Exclusion Principle), δ (s) is unstable. This verifies the validity of Theorem 1. Note that if one uses the Edge Theorem [9,18] to test the stability of δ (s) the result may not be reliable. According to Edge Theorem, δ (s) is Hurwitz stable if q
α
q
α
0 ̸ ∈ [V0 (jω)I0 + K1 (jω)] ∪ [K0 (jω) + V1 (jω)I1 ] This lemma shows that the boundary of arc segment plus arc segment is contained in some arc segments. α0
α1
Example 2. Consider δ (s) = 2s + s + 1.9 − 2.9s + s where α0 ∈ [0.1, 1.5], α1 ∈ [0.2, 1.7]. Note that though there are more 2
= {2(jω)0.1 + (jω)[0.2,1.7] + F ((jω))} ∪ {2(jω)1.5 + (jω)[0.2,1.7] + F ((jω))} ∪ {2(jω)[0.1,1.5] + (jω)0.2 + F ((jω))} ∪ {2(jω)[0.1,1.5] + (jω)1.7 + F ((jω))}
(4)
4
S. Zheng / Systems & Control Letters 99 (2017) 1–8 +
∪ [I0q (jω)α0 + K¯ 1 (jω, α0+ ) + · · · K¯ n (jω, α0+ )] − ∪ [K¯ 0 (jω, α − ) + I q (jω)α1 + · · · K¯ n (jω, α − )] 1
1
1
+
∪ [K¯ 0 (jω, α1+ ) + I1q (jω)α1 + · · · K¯ n (jω, α1+ )] ∪ ··· − ∪ [K¯ 0 (jω, αn− ) + K¯ 1 (jω, αn− ) + · · · Inq (jω)αn ] + ∪ [K¯ 0 (jω, αn+ ) + K¯ 1 (jω, αn+ ) + · · · Inq (jω)αn ]. Definition 6. For qi sαi (i = 0, 1, . . . , n) in (1), define function f (·) which is a generalization of (2) f (s; V0 , V1 , . . . , Vnq ; V0α , V1α , . . . , Vnα ) q
q
= (V0q s2k0 + V1q s2k1 + · · · + Vnq s2kn )sM0 ∩M1 ∩M2 ···∩Mn ¯
¯
¯ (jω). Fig. 1. Value set of ∆
∪ (V0q s2k0 + V1q s2k1 + · · · + Vnq s2kn )sM0 ∩M1 ∩M2 ···∩Mn ¯ ¯ ∪ (V0q s2k0 + V1q s2k1 + · · · + Vnq s2kn )sM0 ∩M1 ∩M2 ···∩Mn ¯
¯
¯
¯
¯
¯
¯
¯
∪ (V0q s2k0 + V1q s2k1 + · · · + Vnq s2kn )sM0 ∩M1 ∩M2 ···∩Mn ∪ ··· ¯
¯
¯
∪ (V0q s2k0 + V1q s2k1 + · · · + Vnq s2kn )sM0 ∩M1 ∩M2 ···∩Mn ¯i where Vi , Viα (i = 0, 1, . . . , n) are from Definition 1, ki , k¯ i , Mi , M α (i = 0, 1, . . . , n) are computed by Definition 4 according to Vi . q
Then using Definition 6, we have Definition 7. For δ (s, q, α), ∆arc (jω) only contains arc segments and is defined as: ∆arc (jω) = ∆a (jω) ∪ ∆af (jω)
where ∆a (jω)
α
= [V0q (jω)I0 + K1 (jω) + K2 (jω) + · · · + Kn (jω)] α
Fig. 2. Value set of δ (jω) and edges. (For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.)
for ∀ω ∈ [0, +∞) where F (s) = 1.9 − 2.9s + s2 . The red line in Fig. 2 α α q q represents the value set [V0 (jω)I0 + K1 (jω)] ∪ [K0 (jω) + V1 (jω)I1 ]. It can be seen that zero is excluded by this value set, which implies that δ (s) is Hurwitz stable. However, this is contradicted against the Zero Exclusion Principle. In fact, it can be easily checked that when α0 = α1 = 0.5, the given fractional system is unstable, i.e. the roots of the characteristic function δ (s) (3) do not all lie in the left half plane. This means that Edge Theorem cannot be directly applied to test the stability of general interval fractional system. In [8,21], the authors pointed out that the Kharitonov theorem failed to check the stability of fractional order interval system. Here, it has been shown that the Edge Theorem is not applicable for general interval fractional order system. The reason for this may lie on the fact that the Edge Theorem generally assumes the order of the control system is fixed [18]. Thus, the Edge Theorem may not hold true. 3.2. Stability criterion for δ (s, q, α) when n ≥ 2 In this section, we will consider the case when n ≥ 2 for δ (s, q, α). It is a generalization of Theorem 1. To begin with, we give
∪ [K0 (jω) + V1q (jω)I1 + K2 (jω) + · · · + Kn (jω)] ∪ ··· α ∪ [K0 (jω) + K1 (jω) + +K2 (jω) + · · · + Vnq (jω)In ], ∆af (jω) = ∆2af (jω) ∪ ∆3af (jω) ∪ · · · ∪ ∆naf (jω), ∆2af (jω) = ∪0≤i
{
q
q
+ Ki−1 + K{i+1 + · · · + Kj−1 + Kj+1 + · · · + Kn q q q q ∆3af (jω) = ∪0≤i
}
According to Definitions 5–7, we have: Theorem 2. δ (s, q, α) is Hurwitz stable if and only if δ (s, q, α) contains at least one Hurwitz stable polynomial and 0 ̸ ∈ ∆(jω) for ∀ω ∈ [0, +∞) where ∆(jω) = ∆line (jω) ∪ ∆arc (jω). Note that ∆(jω) contains a finite number of interval elements. Meanwhile, each interval element only has one parameter varying in an interval. Therefore, constructing ∆(jω) will be much more efficient than directly constructing δ (jω, q, α). See Example 4 for more information.
the following definitions.
Example 4. Consider the SISO control system in [13]. Suppose the controller is a PD controller which is given by:
Definition 5. For δ (s, q, α), ∆line (jω) only contains line segments and is defined as:
C (s) = 1 + 2s.
∆line (jω) −
= [I0q (jω)α0 + K¯ 1 (jω, α0− ) + · · · K¯ n (jω, α0− )]
The plant is given by: P(s) =
1 q0 + 0.5sα1 + q2 sα2
S. Zheng / Systems & Control Letters 99 (2017) 1–8
5
Theorem 3. The Edge Theorem is applicable if one of the following conditions hold: (1) The order vector α = (α0 , α1 , . . . , αn ) is fixed; (2) The coefficient vector q = (q0 , q1 , . . . , qn ) is fixed and for ∀0 ≤ i < j ≤ n, f (s; Viq , Vjq ; Viα , Vjα ) ≡ ∅. 4.2. Finite test frequency interval and auxiliary function In Theorem 2 the test frequency interval is infinite. Hence, in some situations it may be difficult to verify the condition 0 ̸ ∈ ∆(jω) [10–12]. We present the following result to limit the test frequency interval. Lemma 6. If ω ∈ (0, M − ) or (M + , +∞), then 0 ̸ ∈ δ (jω, q, α). M − and M + are defined as: Fig. 3. Value set of ∆(jω) and δ (jω). (For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.)
where q0 ∈ [0.8, 1.2], α1 ∈ [0.6, 1], q2 ∈ [0.9, 1.4], α2 ∈ [1.5, 2.5]. The characteristic polynomial of the control system can be expressed as:
δ (s) = q0 + 0.5s
α1
+ q2 s
α2
+ 1 + 2s.
(5)
−
M + = max{1, C1 1/(αn −ϕ1 ) }, M − = min{1, C2 1/ϕ2 } where C1 =
n−1 ∑
+ − + max{|q− i |, |qi |}/min{|qn |, |qn |},
i=0
ϕ1 = max{α0+ , α1+ , . . . , αn+−1 }, n ∑
Let F (jω) = 1 + 2(jω), then ∆line (jω) corresponding to δ (s) is constructed as:
+ C2 = min{|q− 0 |, |q0 |}/
∆line (jω)
ϕ2 = min{α1− , α1− , . . . , αn− }.
= {[0.8, 1.2] + K¯ 1 (jω, 0) + K¯ 2 (jω, 0) + F (jω)} ∪ {K¯ 0 + K¯ 1 (jω, 1.5) + [0.9, 1.4](jω)1.5 + F (jω)} ∪ {K¯ 0 + K¯ 1 (jω, 2.5) + [0.9, 1.4](jω)2.5 + F (jω)}. ∆arc (jω) is constructed as: ∆arc (jω) = ∆a (jω) ∪ ∆af (jω)
where ∆a (jω)
= [K0 + 0.5(jω)[0.6,1] + K2 (jω) + F (jω)] ∪ [K0 + K1 (jω) + 0.9(jω)[1.5,2.5] + F (jω)] ∪ [K0 + K1 (jω) + 1.4(jω)[1.5,2.5] + F (jω)], ∆af (jω) = ∅.
The value set ∆(jω) for different ω is depicted in Fig. 3 (red lines). It can be seen that zero is excluded by ∆(jω), thus δ (s) is Hurwitz stable. We also plot δ (jω) when ω = 5 (blue points). It is shown that ∂δ (jω) is contained in ∆(jω). This verifies the validity of Theorem 2. It is noted that for each ω, ∆(jω) contains 46 segment at most. Thus, if N points for one segment, then to construct ∆(jω) one only needs to plot 46N points at most for each ω. On the other hand, if one directly constructs δ (jω) by sweeping over the hypercubes Q and A, then there will be N 4 points which should be plotted for each ω. The computational efficiency has been improved a lot by our proposed method. 4. Discussions on main results 4.1. Conditions under which Edge Theorem is applicable In Example 2, we have shown that the Edge Theorem is not applicable for general interval fractional order system. However, when some certain conditions of the interval uncertainties are satisfied, the Edge Theorem can be adopted.
+ max{|q− i |, |qi |},
i=1
Since the value set ∆(jω) in Theorem 2 is non-convex, there is a need to construct an auxiliary function [10–12] which can directly indicate the position relationship between the origin and ∆(jω). Note that for a fixed ω > 0, any element ϑ (jω) ∈ ∆(jω) is a line or arc segment. Hence, ϑ (jω) can be expressed in the following two forms: If ϑ (jω) is a line segment:
ϑ (jω) = Aline (jω)λline + Bline (jω). If ϑ (jω) is an arc segment:
ϑ (jω) = Aarc (jω) · (jω)λarc + Barc (jω) + − + where λline ∈ [λ− line , λline ], 0 ≤ λarc ∈ [λarc , λarc ], Aline (jω ), Bline (jω ), Aarc (jω), Barc (jω) are only dependent on ω. Based on the above two equations, we define the following auxiliary function:
Definition 8. For any element ϑ (jω) ∈ ∆(jω), ω > 0, define: If ϑ (jω) is a line segment: − F (ϑ ) = |Bline + Aline λ+ line | + |Bline + Aline λline | − −|Aline λ+ line − Aline λline |.
If ϑ (jω) is an arc segment:
F (ϑ ) =
⎧ | ln(−Aarc /Barc ) + ln(jω)λ+ ⎪ arc | + | ln(−Aarc /Barc ) ⎪ ⎪ − + − ⎪ + ln(j ω ) λ | − | ln(j ω ) λ ⎨ arc arc − ln(jω )λarc |, if Aarc ̸ = 0, Barc ̸ = 0
⎪ ⎪ ⎪ 1 ⎪ ⎩ 0
if Aarc = 0, Barc ̸ = 0 or Aarc ̸ = 0, Barc = 0 if Aarc = 0, Barc = 0.
Based on Definition 8 and Lemma 6, we have the following results: Lemma 7. For a fixed ω, F (ϑ (jω)) > 0 is equivalent to 0 ̸ ∈ ϑ (jω). Theorem 4. δ (s, q, α) is Hurwitz stable if and only if the following two conditions hold:
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S. Zheng / Systems & Control Letters 99 (2017) 1–8
Acknowledgments The author thanks the editors and the reviewers for constructive criticism that results in significant improvements to the paper. Appendix
Fig. 4. (a) Stability test. (b) Stability margin test. K = 1 (blue solid lines), K = 2 (red dashed lines), K = 3 (green dotted lines). (For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.)
(1) δ (s, q, α) contains at least one Hurwitz stable polynomial; (2) For any element ϑ (jω) ∈ ∆(jω), F (ϑ (jω)) > 0 for ∀ω ∈ [M − , M + ]. Example 5. Example 4 continued. Let q0 = 0.8, α1 = 0.6, q2 ∈ [0.9, 1.4], α2 ∈ [1.5, 2.5] in (5). Using Lemma 6, we can calculate [M − , M + ] = [0.2756, 22.82] rad/s. The ∆(jω) in this case contains two line segments and two arc segments. Their corresponding F (ϑ (jω)) are shown in Fig. 4(a). It can be seen that F (ϑ (jω)) is always larger than zero implying that 0 ̸ ∈ ∆(jω). Hence, it can be concluded that δ (s) is Hurwitz stable.
Proof of Lemma 1. The proof can be obtained from the Zero Exclusion Condition in [1,12]. According to [1,12], we can directly conclude that δ (s, q, α) is stable if and only if δ (s, q, α) contains at least one Hurwitz stable polynomial and 0 ̸ ∈ δ (jω, q, α) for ∀ω ∈ [0, +∞). In addition, from Assumption 1, it can be concluded that when ω → ∞, δ (s, q, α) ∼ qn sαn . Since qn ̸ = 0 and αn > 0, 0 ̸∈ δ (∞, q, α). Note that δ (jω, q, α) is continuous for ω ∈ [0, +∞). Therefore, 0 ̸∈ δ (jω, q, α) for ∀ω ∈ [0, +∞) if and only if 0 ̸ ∈ ∂δ (jω, q, α) for ∀ω ∈ [0, +∞). It completes the proof. Proof of Lemma 2. To prove this lemma, we need the following two lemmas: Lemma A.1. For any term qi sαi (i = 0, 1, . . . , n) in δ (s, q, α), we have
∂[qi (jω)αi ] = Si (jω) where ω is a fixed frequency, Si (s) is from Definition 2. Proof of Lemma A.1. Note that qi (jω)αi = qi (ω)αi ejπαi /2 + − + where qi ∈ [q− i , qi ], αi ∈ [αi , αi ]. Therefore, it can be checked αi that the value set of ∂[qi (jω) ] is composed of two line segments α α q q and two arc segments, which are Ii (jω)Vi ∪ Vi (jω)Ii = Si (jω). This completes the proof.
4.3. Stability margin The following example will show how our proposed results can be extended to address stability margin of the control system. Example 6. Example 5 continued. To consider the gain margin, a variable gain is inserted in the closed loop. The characteristic function (5) is revised into:
δ (s) = q0 + 0.5sα1 + q2 sα2 + K (1 + 2s) where K represents the gain margin [18]. When K selects 1, 2, 3, we can plot the corresponding F (ϑ (jω)) as shown in Fig. 4(b). It can be concluded from Fig. 4(b) that the closed control system is always stable when K varies in interval [1,3]. In addition, to consider the phase margin, the variable gain K α α can be replaced by the function ( ωs − 1)/( ωs + 1) where ω0 is a 0 0 very low value and α ∈ [α − , α + ]. The above result shows that our proposed method can be used in the robust control theory [6,7], such as evaluating gain/phase margin of the control system. 5. Conclusions and future works This paper proposes efficient graphical methods to check the robust stability of general interval fractional order system. Our proposed method provides necessary and sufficient conditions for this stability problem. It is worthwhile to compare our work with [21]. Sondhi et al. [21] proposed a simple technique for the relative stability analysis of commensurate/incommensurate fractional order interval systems based on the Kharitonov’s Theorem. Compared with [21], we do not consider the relative stability of the fractional order system. This is a very good research line which should be our future works. Another future work may include how to adopt our proposed results in robust control strategies and real practical applications.
Lemma A.2 ([22]). ∂ (D1 + D2 ) ⊆ ∂ D1 + ∂ D2 where D1 and D2 are two bounded and closed value sets in the complex plane. Next, for the proof of Lemma 2, by Lemmas A.1 and A.2 we have
∂δ (jω, q, α ) = ∂[q0 (jω)α0 + q1 (jω)α1 ] ⊆ ∂[q0 (jω)α0 ] + ∂[q1 (jω)α1 ] = S0 (jω) + S1 (jω). Then using Definition 2, we can complete the proof. Proof of Lemma 4. Consider the following value set α
δal (jω) = I0q (jω)α0 + q1 (jω)I1
where ω, α0 , q1 are fixed. δal (jω) is an addition of a line and arc segments. As shown in Fig. A.5, CD represents the value set of the line q Ù represents arc segment q1 (jω)I1α . We have segment I0 (jω)α0 and AB
− →′
− →
− →
′ B′ along the direction of AA′ , the Ù translates to Aˆ AA = CD. When AB ′ ′ ′ area APP B BA A is the value set δal (jω). ′ B′ , AA′ , BB′ , Ù , Aˆ It can be seen that the ∂δal (jω) is contained in AB PP ′ . By simple analytic geometry, we have −
AA′ = I0 (jω)α0 + q1 (jω)α1 , q
+
BB′ = I0 (jω)α0 + q1 (jω)α0 , q
α
α0 Ù = q− AB + q1 (jω)I1 , 0 (jω ) α
′ B′ = q+ (jω)α0 + q (jω)I1 . Aˆ 1 0
− →
− →
Ù and PP ′ = AA′ , To compute PP ′ , first note that P is a point on AB − →′
thus PP can be expressed as: PP ′ = I0 (jω)α0 + q1 (jω)Θ (ω) q
S. Zheng / Systems & Control Letters 99 (2017) 1–8
7
If follows that
α1 = α0 + 2k, k ∈ Z. Therefore, we can conclude that the boundary point P is achieved at the following three situations: α0∗ = V0α , α1∗ = V1α or α1∗ = α0∗ + 2k, k ∈ Z. Thus, we have: α
α
α
α
∂δaa (jω) ⊆ [q0 (jω)I0 + q1 (jω)V1 ] ∪ [q0 (jω)V0 + q1 (jω)I0 ] ∪ [q0 (jω)α + q1 (jω)α+2k ] where α ∈ [α0− , α0+ ], α + 2k ∈ [α1− , α1+ ], k ∈ Z. Recall Definition 4 and (2), we have:
ÙB plus line CD. Fig. A.5. Value set of arc A
q0 (jω)α + q1 (jω)α+2k ≡ f (jω; V0 , V1 , V0α , V1α ) q
where Θ (ω) is dependent on ω and is a point in the interval [α1− , α1+ ]. Ù and the slope of PP ′ is equal to Next, since PP ′ is tangent to AB the slope of CD, we have: kPP ′
(α π ) ∂ (ωΘ sin(Θ π /2))/∂ Θ 0 = = k = tan . CD ∂ (ωΘ cos(Θ π /2))/∂ Θ 2
q
where f (jω; q0 , q1 ; V0α , V1α ) is from (2). According to the above two equations, we have α
α
α
α
∂δaa (jω) ⊆ [q0 (jω)I0 + q1 (jω)V1 ] ∪ [q0 (jω)V0 + q1 (jω)I0 ] ∪ [f (jω; q0 , q1 ; V0α , V1α )] where f (jω; q0 , q1 ; V0α , V1α ) is from (2). Repeatedly using the above equation, we can complete the proof.
By solving this equation, and note that Θ (ω) ∈ [α1− , α1+ ], we obtain:
Proof of Theorem 1. The proof is obtained by using Lemmas 1–5. Note that the value set δ (jω, q, α) is continuous for ∀ω ∈ [0, +∞).
Θ (ω) = A(ω, {α1− , α1− }, α0 )
Proof of Theorem 2. The proof is obtained by repeatedly using Theorem 1.
where A(ω, {α1 , α1 }, α0 ) is from Definition 3. Based on the above analysis, we have: −
−
α ∂δal (jω) ⊆ [I0q (jω)α0 + K¯ 1 (jω, α0 )] ∪ [K0 (jω) + q1 (jω)I1 ].
Repeatedly using the above equation, we can complete the proof. Proof of Lemma 5. Consider the following value set α
α
δaa (jω) = q0 (jω)I0 + q1 (jω)I1
where ω, q0 , q1 are fixed. δaa (jω) is an addition of two arc segments. Suppose O is the origin in the complex plane, P is a point on − → ∂δaa (jω). Then OP can be computed as
− →
OP = q0 (jω)α0 + q1 (jω)α1 ∗
∗
where α0∗ ∈ I0α , α1∗ ∈ I1α . Since P is on the boundary, then the distance |OP | is the max− → imum/minimum value along the direction OP for any point in the ∗ ∗ value set δaa (jω). More precisely, α0 and α1 are the solutions to the following optimization problem: max/min d(α0 , α1 ) = q0 (ω)α0 cos s.t . α0 ∈ I0α , α1 ∈ I1α g(α0 , α1 ) (
= q0 (ω)α0 sin
(α π 0
α0 π 2
2
)
− θ + q1 (ω)α1 cos
)
− θ + q1 (jω)α1 sin
(α π 1
(α π 1 2
2
−θ
)
−θ
)
=0 − →
where d(α0 , α1 ) represents the distance along the direction OP for − → any point in the value set δaa (jω), θ is the angle of OP. Based on Lagrange Multiplier Approach, the solutions to the following equations are possible α0∗ and α1∗ :
∂ g(α0 , α1 ) ∂ d(α0 , α1 ) +λ =0 ∂α0 ∂α0 ∂ d(α0 , α1 ) ∂ g(α0 , α1 ) +λ =0 ∂α1 ∂α1 where λ ∈ R. Solving the above equation, we obtain
[(ln ω)2 + (π /2)2 ] sin[(α1 − α0 )π/2] = 0.
Proof of Theorem 3. Note that if for ∀0 ≤ i < j ≤ n, q q f (s; Vi , Vj ; Viα , Vjα ) ≡ ∅, then ∆af (jω) = ∅. This completes the proof. Proof of Lemma 6. The proof can be obtained from [1,10]. It is noted that here we do not have delay terms. Meanwhile, the uncertain parameters qi and αi (i = 0, 1, . . . , n) vary in independent intervals. Proof of Lemma 7. First assume ϑ (ω) is a line segment. We can see − that |Aline λ+ line − Aline λline | is the length of the line segment. |Bline + | and | B + Aline λ+ Aline λ+ line line | represent the distances between line the origin and the endpoints of the line segment. Therefore, by resorting to [10], we have 0 ̸ ∈ ϑ (ω) ⇔ F (ϑ ) > 0. Next, assume ϑ (jω) is an arc segment, then when Aarc ̸ = 0, Barc ̸ = 0 we have Aarc (jω) · (jω)λarc + Barc (jω) ̸ = 0 ⇔
λarc ln(jω) + ln(−Aarc /Barc ) ̸= 0. Note that the value set of λarc ln(jω) + ln(−Aarc /Barc ) is a line segment. Hence, we have 0 ̸ ∈ ϑ (ω) ⇔ F (ϑ ) > 0. This completes the proof. Proof of Theorem 4. The proof is directly from Theorem 2, Lemmas 6 and 7. References [1] K. Moornani, M. Haeri, On robust stability of LTI fractional-order delay systems of retarded and neutral type, Automatica 46 (2) (2010) 362–368. [2] C. Yeroglu, B. Senol, Investigation of robust stability of fractional order multilinear affine systems: 2q-convex parpolygon approach, Systems Control Lett. 62 (10) (2013) 845–855. [3] J.Y. Kaminski, R. Shorten, E. Zeheb, Exact stability test and stabilization for fractional systems, Systems Control Lett. 85 (2015) 95–99. [4] Y. Luo, Y.Q. Chen, C.Y. Wang, Y.G. Pi, Tuning fractional order proportional integral controllers for fractional order system, J. Process Control 20 (7) (2010) 823–831. [5] J. Sabatier, A. Oustaloup, A. Iturricha, P. Lanusse, CRONE control: Principles and extension to time-variant plants with asymptotically constant coefficients, Nonlinear Dynam. 29 (1) (2002) 363–385.
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[6] A. Oustaloup, P. Lanusse, J. Sabatier, P. Meclchior, CRONE control: Principles extensions applications, J. Appl. Nonlinear Dynam. 2 (3) (2013) 207–223. [7] C. Farges, L. Fadiga, J. Sabatier, H∞ analysis and control of commensurate fractional order system, Mechatronics 23 (7) (2013) 772–780. [8] N. Tan, O. Ozguven, M. Ozyetkin, Robust stability analysis of fractional order interval polynomials, ISA Trans. 48 (2) (2009) 166–172. [9] B. Senol, A. Ates, B. Alagoz, C. Yeroglu, A numerical investigation for robust stability of fractional-order uncertain systems, ISA Trans. 53 (2) (2014) 189–198. [10] Z. Gao, Robust stabilization criterion of fractional-order controllers for interval fractional-order plants, Automatica 61 (2015) 9–17. [11] Z. Gao, Robust stability criterion for fractional-order systems with interval uncertain coefficients and a time-delay, ISA Trans. 58 (2015) 76–84. [12] Z. Gao, Robust stability criterion of fractional-order functions for interval fractional-order systems, IET Control Theory A 7 (1) (2013) 60–67. [13] S. Zheng, X. Tang, B. Song, Graphical tuning method of FOPID controllers for fractional order uncertain system achieving robust D-stability, Int. J. Robust Nonlin. 26 (5) (2015) 1112–1142. [14] H.S. Ahn, Y.Q. Chen, Necessary and sufficient condition of fractional-order interval linear systems, Automatica 44 (11) (2008) 2985–2988.
[15] J.G. Lu, Y.Q. Chen, Robust stability and stabilization of fractional-order interval systems with the fractional order α : The 0 < α < 1 case, IEEE Trans. Automat. Control 55 (1) (2010) 152–158. [16] C. Farges, J. Sabatier, M. Moze, Fractional order polytopic systems: robust stability and stabilization, Adv. Differential Equations 35 (2011) 1–10. [17] B. Aguiar, T. Gonzolez, M. Bernal, Comments on ‘‘Robust stability and stabilization of fractional-order interval systems with the fractional order α : The 0 < α < 1 case’’, IEEE Trans. Automat. Control 60 (2) (2015) 582–583. [18] S.P. Bhattacharyya, H. Chapellat, L.H. Keel, Robust Control: The Parametric Control, Prentice Hall, 1995. [19] F. Padula, R. Vilanova, A. Visioli, H∞ optimization-based fractional-order PID controllers design, Int. J. Robust Nonlin. 24 (17) (2014) 3009–3026. [20] F. Khemane, R. Malti, T. Raissi, X. Moreau, Robust estimation of fractional models in the frequency domain using set membership methods, Signal Process. 92 (7) (2012) 1591–1601. [21] S. Sondhi, Y. Hote, Robust stability test for fractional-order interval systems using Kharitonov’s Theorem, J. Control Autom. Electr. Syst. 27 (1) (2016) 1–9. [22] M. Fu, S. Dasgupta, V. Blondel, Robust stability under a class of nonlinear parametric perturbations, IEEE Trans Automat. Control 40 (2) (1995) 213–223.
Contents lists available at ScienceDirect
Systems & Control Letters journal homepage: www.elsevier.com/locate/sysconle
Robust stability of fractional order system with general interval uncertainties Shiqi Zheng School of Automation, China University of Geosciences, No. 388 Lumo Road, Wuhan, China
article
info
Article history: Received 29 February 2016 Received in revised form 18 June 2016 Accepted 3 November 2016
Keywords: Fractional order control system Interval uncertainty Robust stability
a b s t r a c t This paper focuses on the analysis of robust stability of fractional order system with general interval uncertainties. The concept of general interval uncertainties means that the interval uncertainties exist both in the coefficients and orders of the fractional order system. Necessary and sufficient conditions are proposed to check the robust stability of general interval fractional order system. According to the proposed stability criterion, it is interesting to find that the Edge Theorem, which was initially proposed for integer order system, cannot be directly extended to test the stability of fractional order system with general interval uncertainties. Examples are followed to verify the validity of the proposed method. © 2016 Elsevier B.V. All rights reserved.
1. Introduction Recently, fractional order calculus has gained much interest both from academic and industrial point of view. It proves to be a modern powerful tool in analyzing various physical phenomena. A growing number of physical systems, including servo motor system, chaotic system, biological system etc., have been verified that their behaviors can be better described by fractional order systems [1,2]. Meanwhile, control of these fractional order systems has also become a hot topic. Some typical research about fractional order control system, such as CRONE control, fractional order PID etc., can be found in [3–7]. A critical aspect of the fractional control system is the robust stability. Robust stability refers to the preservation of stability in face of different classes of uncertainties. This paper aims to investigate the robust stability of fractional order control system with interval uncertainties [8–13]. The interval uncertainties mean that the parameters of the fractional order control system lie in specified intervals. There are mainly two classes of methods to check the stability of interval fractional order system, i.e. Linear Matrix Inequalities (LMI) method [14–16] and graphical method [8–13]. A nice feature of LMI method is its ability to accommodate uncertainty in a relatively straightforward manner [3]. However, the LMI method may be only limited to fractional order system with commensurate order. Moreover, solving LMI sometimes can cause conservatism in stability test [10,17]. For graphical approach, it firstly depicts the value set of the interval system in the complex plane, and then tests the stability by applying the Zero Exclusion Principle [1,18]. The graphical method turns out to be effective and accurate. E-mail address: [email protected]. http://dx.doi.org/10.1016/j.sysconle.2016.11.001 0167-6911/© 2016 Elsevier B.V. All rights reserved.
It is noted that the methods mentioned above only considered the interval uncertainties in the coefficients of fractional order control system. Yet, the orders of the fractional order system may also suffer from interval uncertainties. For example, in system modeling, the parameters of a fractional model, including coefficients and orders, could all vary as the working conditions of the control system change. The interval uncertainties in the fractional orders have been mentioned in [9,19,20]. Here we refer to interval uncertainties both in the coefficients and orders as general interval uncertainties. It is worth noting that the stability problem of fractional order system with general interval uncertainties is still open to our best knowledge. Recently the Edge Theorem [9,18] is adopted to test the stability of fractional order system with interval uncertainties only in orders. However, it will be seen in the sequel that the Edge Theorem, which was initially proposed for integer order system, cannot be directly extended to test the stability of fractional order system with general interval uncertainties. In fact, the hypothesis in the Edge Theorem may be violated in presence of general interval uncertainties (see Example 3). In this paper, we present some necessary and sufficient stability criteria for fractional order system with general interval uncertainties based on graphical method. According to the proposed stability criteria, it can be concluded that the Edge Theorem is applicable only when some certain conditions of the interval uncertainties are satisfied. The rest of the paper is as follows. Section 2 formulates the problem. Main results are presented in Section 3. Section 4 gives some discussions on the main results. Examples are followed. Section 5 concludes the paper.
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2. Problem formulation
Note that let s = jω in Si (s) where ω is a fixed frequency, then it can be seen:
Notation. R is the set of real numbers. N is the set of natural numbers. ∅ stands for empty set. ∂ (P) denotes the boundary of value set P in complex plane.
Si (jω) =
Definition 1. Consider the following uncertain real fractional order system
δ (s, q, α) = q0 sα0 + q1 sα1 + · · · + qn sαn
(1)
where s is the Laplace operator, q = (q0 , q1 , . . . , qn ) and α = (α0 , α1 , . . . , αn ) are the coefficient and order vectors respectively. They vary in the following hypercubes: + q ∈ Q = {q ∈ Rn | qi ∈ [q− i , qi ], i = 0, 1, . . . , n},
α ∈ A = {α ∈ Rn | αi ∈ [αi− , αi+ ], αi− ≥ 0, i = 0, 1, . . . , n} + − + where [q− i , qi ] and [αi , αi ] are specified intervals. Since the coefficients and orders of (1) both vary in independent intervals, (1) is referred to as fractional order system with general interval uncertainties.
We give the following assumption throughout this paper: Assumption 1. Assume (1) satisfies: (1) αn > αm (m = 0, 1, 2, . . . , n − 1) and qn ̸ = 0; (2) |αi+ − αi− | ≤ 2 (i = 0, 1, 2, . . . , n).
−
[αi q− i (jω )
Based on the above facts, we formulate the following design problem:
,αi+ ]
−
[αi , q+ i (jω )
,αi+ ]
©
.
Therefore, the value set Si (jω) in complex plane contains two line segments and two arc segments. Our main results are based on the following lemma. Note that all the proofs in Sections 3 and 4 are put in Appendix. Lemma 1 (Zero Exclusion Principle). δ (s, q, α) is Hurwitz stable if and only if δ (s, q, α) contains at least one Hurwitz stable polynomial and 0 ̸ ∈ δ (jω, q, α) for ∀ω ∈ [0, +∞), or equivalently 0 ̸ ∈ ∂δ (jω, q, α) for ∀ω ∈ [0, +∞). From this lemma, it can be seen that the main problem here is how to construct the boundary of δ (s, q, α) efficiently. First we will consider the stability of δ (s, q, α) when n = 2 in (1). 3.1. Stability criterion for δ (s, q, α) when n = 2 Lemma 2. Assume in (1), for a fixed ω we have:
∂δ (jω, q, α) = ∂[q0 (jω)α0 + q1 (jω)α1 ] ⊆ S0 (jω) + S1 (jω) α α = [I0q (jω)V0 + I1q (jω)V1 ] α α α q q Vα ∪ [I0 (jω) 0 + V1 (jω)I1 ] ∪ [V0q (jω)I0 + I1q (jω)V1 ] q q Iα Iα ∪ [V0 (jω) 0 + V1 (jω) 1 ] where Vi , Ii , Viα , Iiα , Si (i = 0, 1) are from Definition 2. q
Remark 1. Condition (1) in Assumption 1 is standard. It guarantees that αn is always the highest order in δ (s, q, α) and the term qn sαn never vanishes [18]. Condition (2) in Assumption 1 assumes that the variation of the order in δ (s, q, α) belongs to [0, 2], i.e. ∆αi ∈ [0, 2]. This seems to be a little conservative. However, the control system can be very sensitive to the order change, say the dynamic behavior of the zero order system is very different from second order system. Therefore, ∆αi ∈ [0, 2] in fact can represent a lot of different dynamics (see the claim in Example 1 in [19]). Moreover, if in some situations, the range of ∆αi exceeds [0, 2], one can separate the interval into some smaller intervals and consider them respectively.
+
−
αi αi + + , , [q− [q− i , qi ](jω ) i , qi ](jω )
¶
q
From Lemma 2, it can be seen that ∂δ (jω, q, α) is included in S0 (jω) + S1 (jω). However, directly constructing value set S0 (jω) + S1 (jω) is still not an easy task since they are interval elements plus interval elements. Note that S0 (jω)+S1 (jω) contain three parts: line segment plus line segment, line segment plus arc segment and arc segment plus arc segment. Next, we will have a further analysis on these three parts. Lemma 3 (Line Segment Plus Line Segment [8,18]). α
α
∂[I0q (jω)V0 + I1q (jω)V1 ] α α α α ⊆ [V0q (jω)V0 + I1q (jω)V1 ] ∪ [I0q (jω)V0 + V1q (jω)V1 ] α α = [K0 (jω) + I1q (jω)V1 ] ∪ [I0q (jω)V0 + K1 (jω)].
Problem 1. Develop an efficient method to check the Hurwitz stability of the general interval fractional order system δ (s, q, α) by (1).
This lemma shows that the boundary of line segment plus line segment is contained in some line segments. For line segment plus arc segment, we first give the following definition:
3. Main results
Definition 3. For a fixed ω, a vertex set Viα = {αi− , αi+ } and a specified α , define A(ω, Viα , α ) ∈ R: A(ω, Viα , α )
First, we give the following definition: Definition 2. For a term qi s define: q
αi
= qi s
[αi− , αi+ ]
(i = 0, 1, . . . , n) in (1),
Viα = {αi− , αi+ }, Iiα = [αi− , αi+ ], Vi and Viα are both sets of two end points, Ii and Iiα are both intervals. Also define: q
q
α
−
+
−
Si (s) =
∪
q α Vi sIi −
= {[qi , qi ]s −
+
αi
, [qi , qi ]s −
+
αi+
− [αi− ,αi+ ]
, qi s
+ [αi− ,αi+ ]
, qi s
where [·] is rounding function, β = arctan[π/(2 ln ω)]. For a term qi sαi in (1) define the vertex set α ,α )
K¯ i (jω, α ) = Ki (jω) ∪ Vi (jω)A(ω,Vi q
K¯ i is a set of six vertex terms.
+
αi αi αi αi Ki (s) = Vi sVi = {q− , q− , q+ , q+ }, i s i s i s i s q α Ii sVi
if [(αi − α )/2 + β/π] = [(αi − α )/2 + β/π] + 1
⎪ ⎩none exist +
if [(αi − α )/2 + β/π] ̸ = [(αi− − α )/2 + β/π] + 1
q
+ − + Vi = {q− i , qi }, Ii = [qi , qi ],
q
=
⎧ + 2[αi+ /2 + β/π − 1]/π ⎪ ⎨α − 2β/π + −
}
Ki (s) is a set of four vertex terms, Si (s) contains four interval elements.
A(ω, Viα , α ) is in fact a point in interval [αi− , αi+ ]. This point will contribute to the boundary of line segment plus arc segment. Further details about how A(ω, Viα , α ) is computed can be found in the Proof of Lemma 4. Based on Definition 3, we have the following result:
S. Zheng / Systems & Control Letters 99 (2017) 1–8
than two terms in δ (s), the terms 1.9 − 2.9s + s2 are fixed. Thus, Lemma 5 can still be adopted. Then from Lemma 5, we have
Lemma 4 (Line Segment Plus Arc Segment). Vα
Iα
∂[I0q (jω) 0 + V1q (jω) 1 ] + − ⊆ [I0q (jω)α0 + K¯ 1 (jω, α0− )] ∪ [I0q (jω)α0 + K¯ 1 (jω, α0+ )] α ∪ [K0 (jω) + V1q (jω)I1 ]. This lemma shows that the boundary of line segment plus arc segment is contained in some line segments and arc segments. Example 1. Consider δ (s) = q0 s0.1 + sα1 where q0 ∈ [3, 4], α1 ∈ [1, 2]. Then using Lemma 4, we have ∂δ (jω) ⊆ {[3, 4](jω)0.1 + K¯ 1 (jω, 0.1)}
∪ {3(jω)0.1 + (jω)[1,2] } ∪ {4(jω)0.1 + (jω)[1,2] } where K¯ 1 (jω, 0.1) = {(jω)1 , (jω)2 , (jω)A(ω,{1,2},0.1) }, A(ω, {1, 2}, 0.1) can be computed according to Definition 3 and it is dependent on ω. For instance, A(2, {1, 2}, 0.1) = 1.364, thus K¯ 1 (jω, 0.1) = {(jω)1 , (jω)2 , (jω)1.364 }; A(0.5, {1, 2}, 0.1) does not exist, thus K¯ 1 (jω, 0.1) = K1 (jω) = {(jω)1 , (jω)2 }. For arc segment plus arc segment, we first give the following definition: α
Definition 4. For a vertex set Vi ¯ i: ki , k¯ i , Mi , M ¯ ki , ki ∈ N are determined by:
=
{αi , αi }, define −
+
if 2 < αi − 2ki < 4.
¯ i ]. [αi− , αi+ ] = [2ki + Mi ] ∪ [2k¯ i + M Any term qi sαi in (1) can be given as: ¯
= qi {s2ki · sMi ∪ s2ki · sMi }.
¯ i are both intervals that satisfy Mi , M ¯ i ⊆ [0, 2]. Note that Mi , M [αi− ,αi+ ] ¯ can be separated into Using Mi , Mi , the arc segment qi (jω) two segments: one is in the upper half plane, the other is in the ¯ i can be found in the lower half plane. Further details about Mi , M Proof of Lemma 5. Based on Definition 4, we have: Lemma 5 (Arc Segment Plus Arc Segment). α
∂[V0q (jω)I0 + V1q (jω)I1 ] α
α
⊆ [V0q (jω)I0 + K1 (jω)] ∪ [K0 (jω) + V1q (jω)I1 ] ∪ [f (jω; V0q , V1q ; V0α , V1α )] where function f (·) is expressed as: f (s; V0 , V1 ; V0α , V1α ) q
¯
¯
= (V0q s2k0 + V1q s2k1 )sM0 ∩M1 ∪ (V0q s2k0 + V1q s2k1 )sM0 ∩M1 ∪
+
∪
q ¯ (V0 s2k0
+
1
1
1
α
where Vi , Ii , Viα , Iiα , Ki (i = 0, 1) are from Definition 1, K¯ i are from Definition 3, function f (·) is from (2).
¯ ¯ q ¯ V1 s2k1 )sM0 ∩M1
(2)
.
q
Example 3. Consider the SISO control system in [13]. Suppose the controller is a proportional gain Kp = 2, the plant P(s) =
¯
+
α
q
Adopting the above definition, interval [αi− , αi+ ] can be expressed as
¯ q ¯ V1 s2k1 )sM0 ∩M1
where
= [V0q (jω)I0 + K1 (jω)] ∪ [K0 (jω) + V1q (jω)I1 ] ∪ [f (jω; V0q , V1q ; V0α , V1α )]
if 0 < αi − 2ki < 2
q (V0 s2k0
¯ (jω) = ∆ ¯ line (jω) ∪ ∆ ¯ arc (jω) ∆
1
− ¯ i = [0, α + − 2ki − 2] M ⎪ i , 2], i ⎩Mi = [αi − 2k +
q
Theorem 1. Assume n = 2 in (1), then δ (s, q, α) is Hurwitz stable if and only if δ (s, q, α) contains at least one Hurwitz stable polynomial ¯ (jω) for ∀ω ∈ [0, +∞) where ∆ ¯ (jω) is given as: and 0 ̸ ∈ ∆
¯ arc (jω) ∆
⎧ ¯i = ∅ M = [αi− − 2ki , αi+ − 2ki ], M ⎪ ⎨ i +
α
According to Lemmas 1–5, we have the stability criterion for
δ (s, q, α) when n = 2:
−
¯ i ⊆ [0, 2] are intervals and are calculated as: Mi , M
,αi+ ]
where F (s) = 1.9 − 2.9s + s2 . To compute f (jω), based on Definition 4 we have k0 = 0, k¯ 0 = ¯ 0 = ∅, k1 = 0, k¯ 1 = 1, M1 = [0.2, 1.7], 1, M0 = [0.1, 1.5], M ¯ 1 = ∅. Therefore, f (jω) = (2 + 1)(jω)[0.2,1.5] . M
= [I0q (jω)α0 + K¯ 1 (jω, α0− )] ∪ [I0q (jω)α0 + K¯ 1 (jω, α0+ )] , + − ∪ [K¯ 0 (jω, α − ) + I q (jω)α1 ] ∪ [K¯ 0 (jω, α + ) + I q (jω)α1 ]
ki = [αi /2], k¯ i = ki + 1.
−
∂δ (jω) ⊆ {2(jω)0.1 + (jω)[0.2,1.7] + F ((jω))} ∪ {2(jω)1.5 + (jω)[0.2,1.7] + F ((jω))} ∪ {2(jω)[0.1,1.5] + (jω)0.2 + F ((jω))} ∪ {2(jω)[0.1,1.5] + (jω)1.7 + F ((jω))} ∪ [f (jω; 2, 1; {0.1, 1.5}, {0.2, 1.7})]
¯ line (jω) ∆
−
qi sαi = qi s[αi
3
sα0 + 0.5 sα1 + 0.9 − 2.9s + s2
where α0 ∈ [0.1, 1.5], α1 ∈ [0.2, 1.7]. The characteristic polynomial of the control system can be expressed as:
δ (s) = 2sα0 + sα1 + 1.9 − 2.9s + s2 .
(3)
δ (s) is the same as that in Example 2. First, we can easily check that δ (s) contains one stable polynomial, i.e. α0 = α1 = 1. Then, ¯ (jω) in the complex plane. As by using Theorem 1, we can plot ∆ ¯ (jω). illustrated in Fig. 1, the zero is included in the value set ∆ Therefore, it can be concluded that δ (s) is not always stable for any α0 ∈ [0.1, 1.5], α1 ∈ [0.2, 1.7]. Meanwhile, in Fig. 2 we have also plotted the value set δ (jω) (blue points). It is shown that the ¯ (jω), and δ (jω) also includes boundary of δ (jω) is contained in ∆ zero. Consequently, by Lemma 1 (Zero Exclusion Principle), δ (s) is unstable. This verifies the validity of Theorem 1. Note that if one uses the Edge Theorem [9,18] to test the stability of δ (s) the result may not be reliable. According to Edge Theorem, δ (s) is Hurwitz stable if q
α
q
α
0 ̸ ∈ [V0 (jω)I0 + K1 (jω)] ∪ [K0 (jω) + V1 (jω)I1 ] This lemma shows that the boundary of arc segment plus arc segment is contained in some arc segments. α0
α1
Example 2. Consider δ (s) = 2s + s + 1.9 − 2.9s + s where α0 ∈ [0.1, 1.5], α1 ∈ [0.2, 1.7]. Note that though there are more 2
= {2(jω)0.1 + (jω)[0.2,1.7] + F ((jω))} ∪ {2(jω)1.5 + (jω)[0.2,1.7] + F ((jω))} ∪ {2(jω)[0.1,1.5] + (jω)0.2 + F ((jω))} ∪ {2(jω)[0.1,1.5] + (jω)1.7 + F ((jω))}
(4)
4
S. Zheng / Systems & Control Letters 99 (2017) 1–8 +
∪ [I0q (jω)α0 + K¯ 1 (jω, α0+ ) + · · · K¯ n (jω, α0+ )] − ∪ [K¯ 0 (jω, α − ) + I q (jω)α1 + · · · K¯ n (jω, α − )] 1
1
1
+
∪ [K¯ 0 (jω, α1+ ) + I1q (jω)α1 + · · · K¯ n (jω, α1+ )] ∪ ··· − ∪ [K¯ 0 (jω, αn− ) + K¯ 1 (jω, αn− ) + · · · Inq (jω)αn ] + ∪ [K¯ 0 (jω, αn+ ) + K¯ 1 (jω, αn+ ) + · · · Inq (jω)αn ]. Definition 6. For qi sαi (i = 0, 1, . . . , n) in (1), define function f (·) which is a generalization of (2) f (s; V0 , V1 , . . . , Vnq ; V0α , V1α , . . . , Vnα ) q
q
= (V0q s2k0 + V1q s2k1 + · · · + Vnq s2kn )sM0 ∩M1 ∩M2 ···∩Mn ¯
¯
¯ (jω). Fig. 1. Value set of ∆
∪ (V0q s2k0 + V1q s2k1 + · · · + Vnq s2kn )sM0 ∩M1 ∩M2 ···∩Mn ¯ ¯ ∪ (V0q s2k0 + V1q s2k1 + · · · + Vnq s2kn )sM0 ∩M1 ∩M2 ···∩Mn ¯
¯
¯
¯
¯
¯
¯
¯
∪ (V0q s2k0 + V1q s2k1 + · · · + Vnq s2kn )sM0 ∩M1 ∩M2 ···∩Mn ∪ ··· ¯
¯
¯
∪ (V0q s2k0 + V1q s2k1 + · · · + Vnq s2kn )sM0 ∩M1 ∩M2 ···∩Mn ¯i where Vi , Viα (i = 0, 1, . . . , n) are from Definition 1, ki , k¯ i , Mi , M α (i = 0, 1, . . . , n) are computed by Definition 4 according to Vi . q
Then using Definition 6, we have Definition 7. For δ (s, q, α), ∆arc (jω) only contains arc segments and is defined as: ∆arc (jω) = ∆a (jω) ∪ ∆af (jω)
where ∆a (jω)
α
= [V0q (jω)I0 + K1 (jω) + K2 (jω) + · · · + Kn (jω)] α
Fig. 2. Value set of δ (jω) and edges. (For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.)
for ∀ω ∈ [0, +∞) where F (s) = 1.9 − 2.9s + s2 . The red line in Fig. 2 α α q q represents the value set [V0 (jω)I0 + K1 (jω)] ∪ [K0 (jω) + V1 (jω)I1 ]. It can be seen that zero is excluded by this value set, which implies that δ (s) is Hurwitz stable. However, this is contradicted against the Zero Exclusion Principle. In fact, it can be easily checked that when α0 = α1 = 0.5, the given fractional system is unstable, i.e. the roots of the characteristic function δ (s) (3) do not all lie in the left half plane. This means that Edge Theorem cannot be directly applied to test the stability of general interval fractional system. In [8,21], the authors pointed out that the Kharitonov theorem failed to check the stability of fractional order interval system. Here, it has been shown that the Edge Theorem is not applicable for general interval fractional order system. The reason for this may lie on the fact that the Edge Theorem generally assumes the order of the control system is fixed [18]. Thus, the Edge Theorem may not hold true. 3.2. Stability criterion for δ (s, q, α) when n ≥ 2 In this section, we will consider the case when n ≥ 2 for δ (s, q, α). It is a generalization of Theorem 1. To begin with, we give
∪ [K0 (jω) + V1q (jω)I1 + K2 (jω) + · · · + Kn (jω)] ∪ ··· α ∪ [K0 (jω) + K1 (jω) + +K2 (jω) + · · · + Vnq (jω)In ], ∆af (jω) = ∆2af (jω) ∪ ∆3af (jω) ∪ · · · ∪ ∆naf (jω), ∆2af (jω) = ∪0≤i
{
q
q
+ Ki−1 + K{i+1 + · · · + Kj−1 + Kj+1 + · · · + Kn q q q q ∆3af (jω) = ∪0≤i
}
According to Definitions 5–7, we have: Theorem 2. δ (s, q, α) is Hurwitz stable if and only if δ (s, q, α) contains at least one Hurwitz stable polynomial and 0 ̸ ∈ ∆(jω) for ∀ω ∈ [0, +∞) where ∆(jω) = ∆line (jω) ∪ ∆arc (jω). Note that ∆(jω) contains a finite number of interval elements. Meanwhile, each interval element only has one parameter varying in an interval. Therefore, constructing ∆(jω) will be much more efficient than directly constructing δ (jω, q, α). See Example 4 for more information.
the following definitions.
Example 4. Consider the SISO control system in [13]. Suppose the controller is a PD controller which is given by:
Definition 5. For δ (s, q, α), ∆line (jω) only contains line segments and is defined as:
C (s) = 1 + 2s.
∆line (jω) −
= [I0q (jω)α0 + K¯ 1 (jω, α0− ) + · · · K¯ n (jω, α0− )]
The plant is given by: P(s) =
1 q0 + 0.5sα1 + q2 sα2
S. Zheng / Systems & Control Letters 99 (2017) 1–8
5
Theorem 3. The Edge Theorem is applicable if one of the following conditions hold: (1) The order vector α = (α0 , α1 , . . . , αn ) is fixed; (2) The coefficient vector q = (q0 , q1 , . . . , qn ) is fixed and for ∀0 ≤ i < j ≤ n, f (s; Viq , Vjq ; Viα , Vjα ) ≡ ∅. 4.2. Finite test frequency interval and auxiliary function In Theorem 2 the test frequency interval is infinite. Hence, in some situations it may be difficult to verify the condition 0 ̸ ∈ ∆(jω) [10–12]. We present the following result to limit the test frequency interval. Lemma 6. If ω ∈ (0, M − ) or (M + , +∞), then 0 ̸ ∈ δ (jω, q, α). M − and M + are defined as: Fig. 3. Value set of ∆(jω) and δ (jω). (For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.)
where q0 ∈ [0.8, 1.2], α1 ∈ [0.6, 1], q2 ∈ [0.9, 1.4], α2 ∈ [1.5, 2.5]. The characteristic polynomial of the control system can be expressed as:
δ (s) = q0 + 0.5s
α1
+ q2 s
α2
+ 1 + 2s.
(5)
−
M + = max{1, C1 1/(αn −ϕ1 ) }, M − = min{1, C2 1/ϕ2 } where C1 =
n−1 ∑
+ − + max{|q− i |, |qi |}/min{|qn |, |qn |},
i=0
ϕ1 = max{α0+ , α1+ , . . . , αn+−1 }, n ∑
Let F (jω) = 1 + 2(jω), then ∆line (jω) corresponding to δ (s) is constructed as:
+ C2 = min{|q− 0 |, |q0 |}/
∆line (jω)
ϕ2 = min{α1− , α1− , . . . , αn− }.
= {[0.8, 1.2] + K¯ 1 (jω, 0) + K¯ 2 (jω, 0) + F (jω)} ∪ {K¯ 0 + K¯ 1 (jω, 1.5) + [0.9, 1.4](jω)1.5 + F (jω)} ∪ {K¯ 0 + K¯ 1 (jω, 2.5) + [0.9, 1.4](jω)2.5 + F (jω)}. ∆arc (jω) is constructed as: ∆arc (jω) = ∆a (jω) ∪ ∆af (jω)
where ∆a (jω)
= [K0 + 0.5(jω)[0.6,1] + K2 (jω) + F (jω)] ∪ [K0 + K1 (jω) + 0.9(jω)[1.5,2.5] + F (jω)] ∪ [K0 + K1 (jω) + 1.4(jω)[1.5,2.5] + F (jω)], ∆af (jω) = ∅.
The value set ∆(jω) for different ω is depicted in Fig. 3 (red lines). It can be seen that zero is excluded by ∆(jω), thus δ (s) is Hurwitz stable. We also plot δ (jω) when ω = 5 (blue points). It is shown that ∂δ (jω) is contained in ∆(jω). This verifies the validity of Theorem 2. It is noted that for each ω, ∆(jω) contains 46 segment at most. Thus, if N points for one segment, then to construct ∆(jω) one only needs to plot 46N points at most for each ω. On the other hand, if one directly constructs δ (jω) by sweeping over the hypercubes Q and A, then there will be N 4 points which should be plotted for each ω. The computational efficiency has been improved a lot by our proposed method. 4. Discussions on main results 4.1. Conditions under which Edge Theorem is applicable In Example 2, we have shown that the Edge Theorem is not applicable for general interval fractional order system. However, when some certain conditions of the interval uncertainties are satisfied, the Edge Theorem can be adopted.
+ max{|q− i |, |qi |},
i=1
Since the value set ∆(jω) in Theorem 2 is non-convex, there is a need to construct an auxiliary function [10–12] which can directly indicate the position relationship between the origin and ∆(jω). Note that for a fixed ω > 0, any element ϑ (jω) ∈ ∆(jω) is a line or arc segment. Hence, ϑ (jω) can be expressed in the following two forms: If ϑ (jω) is a line segment:
ϑ (jω) = Aline (jω)λline + Bline (jω). If ϑ (jω) is an arc segment:
ϑ (jω) = Aarc (jω) · (jω)λarc + Barc (jω) + − + where λline ∈ [λ− line , λline ], 0 ≤ λarc ∈ [λarc , λarc ], Aline (jω ), Bline (jω ), Aarc (jω), Barc (jω) are only dependent on ω. Based on the above two equations, we define the following auxiliary function:
Definition 8. For any element ϑ (jω) ∈ ∆(jω), ω > 0, define: If ϑ (jω) is a line segment: − F (ϑ ) = |Bline + Aline λ+ line | + |Bline + Aline λline | − −|Aline λ+ line − Aline λline |.
If ϑ (jω) is an arc segment:
F (ϑ ) =
⎧ | ln(−Aarc /Barc ) + ln(jω)λ+ ⎪ arc | + | ln(−Aarc /Barc ) ⎪ ⎪ − + − ⎪ + ln(j ω ) λ | − | ln(j ω ) λ ⎨ arc arc − ln(jω )λarc |, if Aarc ̸ = 0, Barc ̸ = 0
⎪ ⎪ ⎪ 1 ⎪ ⎩ 0
if Aarc = 0, Barc ̸ = 0 or Aarc ̸ = 0, Barc = 0 if Aarc = 0, Barc = 0.
Based on Definition 8 and Lemma 6, we have the following results: Lemma 7. For a fixed ω, F (ϑ (jω)) > 0 is equivalent to 0 ̸ ∈ ϑ (jω). Theorem 4. δ (s, q, α) is Hurwitz stable if and only if the following two conditions hold:
6
S. Zheng / Systems & Control Letters 99 (2017) 1–8
Acknowledgments The author thanks the editors and the reviewers for constructive criticism that results in significant improvements to the paper. Appendix
Fig. 4. (a) Stability test. (b) Stability margin test. K = 1 (blue solid lines), K = 2 (red dashed lines), K = 3 (green dotted lines). (For interpretation of the references to color in this figure legend, the reader is referred to the web version of this article.)
(1) δ (s, q, α) contains at least one Hurwitz stable polynomial; (2) For any element ϑ (jω) ∈ ∆(jω), F (ϑ (jω)) > 0 for ∀ω ∈ [M − , M + ]. Example 5. Example 4 continued. Let q0 = 0.8, α1 = 0.6, q2 ∈ [0.9, 1.4], α2 ∈ [1.5, 2.5] in (5). Using Lemma 6, we can calculate [M − , M + ] = [0.2756, 22.82] rad/s. The ∆(jω) in this case contains two line segments and two arc segments. Their corresponding F (ϑ (jω)) are shown in Fig. 4(a). It can be seen that F (ϑ (jω)) is always larger than zero implying that 0 ̸ ∈ ∆(jω). Hence, it can be concluded that δ (s) is Hurwitz stable.
Proof of Lemma 1. The proof can be obtained from the Zero Exclusion Condition in [1,12]. According to [1,12], we can directly conclude that δ (s, q, α) is stable if and only if δ (s, q, α) contains at least one Hurwitz stable polynomial and 0 ̸ ∈ δ (jω, q, α) for ∀ω ∈ [0, +∞). In addition, from Assumption 1, it can be concluded that when ω → ∞, δ (s, q, α) ∼ qn sαn . Since qn ̸ = 0 and αn > 0, 0 ̸∈ δ (∞, q, α). Note that δ (jω, q, α) is continuous for ω ∈ [0, +∞). Therefore, 0 ̸∈ δ (jω, q, α) for ∀ω ∈ [0, +∞) if and only if 0 ̸ ∈ ∂δ (jω, q, α) for ∀ω ∈ [0, +∞). It completes the proof. Proof of Lemma 2. To prove this lemma, we need the following two lemmas: Lemma A.1. For any term qi sαi (i = 0, 1, . . . , n) in δ (s, q, α), we have
∂[qi (jω)αi ] = Si (jω) where ω is a fixed frequency, Si (s) is from Definition 2. Proof of Lemma A.1. Note that qi (jω)αi = qi (ω)αi ejπαi /2 + − + where qi ∈ [q− i , qi ], αi ∈ [αi , αi ]. Therefore, it can be checked αi that the value set of ∂[qi (jω) ] is composed of two line segments α α q q and two arc segments, which are Ii (jω)Vi ∪ Vi (jω)Ii = Si (jω). This completes the proof.
4.3. Stability margin The following example will show how our proposed results can be extended to address stability margin of the control system. Example 6. Example 5 continued. To consider the gain margin, a variable gain is inserted in the closed loop. The characteristic function (5) is revised into:
δ (s) = q0 + 0.5sα1 + q2 sα2 + K (1 + 2s) where K represents the gain margin [18]. When K selects 1, 2, 3, we can plot the corresponding F (ϑ (jω)) as shown in Fig. 4(b). It can be concluded from Fig. 4(b) that the closed control system is always stable when K varies in interval [1,3]. In addition, to consider the phase margin, the variable gain K α α can be replaced by the function ( ωs − 1)/( ωs + 1) where ω0 is a 0 0 very low value and α ∈ [α − , α + ]. The above result shows that our proposed method can be used in the robust control theory [6,7], such as evaluating gain/phase margin of the control system. 5. Conclusions and future works This paper proposes efficient graphical methods to check the robust stability of general interval fractional order system. Our proposed method provides necessary and sufficient conditions for this stability problem. It is worthwhile to compare our work with [21]. Sondhi et al. [21] proposed a simple technique for the relative stability analysis of commensurate/incommensurate fractional order interval systems based on the Kharitonov’s Theorem. Compared with [21], we do not consider the relative stability of the fractional order system. This is a very good research line which should be our future works. Another future work may include how to adopt our proposed results in robust control strategies and real practical applications.
Lemma A.2 ([22]). ∂ (D1 + D2 ) ⊆ ∂ D1 + ∂ D2 where D1 and D2 are two bounded and closed value sets in the complex plane. Next, for the proof of Lemma 2, by Lemmas A.1 and A.2 we have
∂δ (jω, q, α ) = ∂[q0 (jω)α0 + q1 (jω)α1 ] ⊆ ∂[q0 (jω)α0 ] + ∂[q1 (jω)α1 ] = S0 (jω) + S1 (jω). Then using Definition 2, we can complete the proof. Proof of Lemma 4. Consider the following value set α
δal (jω) = I0q (jω)α0 + q1 (jω)I1
where ω, α0 , q1 are fixed. δal (jω) is an addition of a line and arc segments. As shown in Fig. A.5, CD represents the value set of the line q Ù represents arc segment q1 (jω)I1α . We have segment I0 (jω)α0 and AB
− →′
− →
− →
′ B′ along the direction of AA′ , the Ù translates to Aˆ AA = CD. When AB ′ ′ ′ area APP B BA A is the value set δal (jω). ′ B′ , AA′ , BB′ , Ù , Aˆ It can be seen that the ∂δal (jω) is contained in AB PP ′ . By simple analytic geometry, we have −
AA′ = I0 (jω)α0 + q1 (jω)α1 , q
+
BB′ = I0 (jω)α0 + q1 (jω)α0 , q
α
α0 Ù = q− AB + q1 (jω)I1 , 0 (jω ) α
′ B′ = q+ (jω)α0 + q (jω)I1 . Aˆ 1 0
− →
− →
Ù and PP ′ = AA′ , To compute PP ′ , first note that P is a point on AB − →′
thus PP can be expressed as: PP ′ = I0 (jω)α0 + q1 (jω)Θ (ω) q
S. Zheng / Systems & Control Letters 99 (2017) 1–8
7
If follows that
α1 = α0 + 2k, k ∈ Z. Therefore, we can conclude that the boundary point P is achieved at the following three situations: α0∗ = V0α , α1∗ = V1α or α1∗ = α0∗ + 2k, k ∈ Z. Thus, we have: α
α
α
α
∂δaa (jω) ⊆ [q0 (jω)I0 + q1 (jω)V1 ] ∪ [q0 (jω)V0 + q1 (jω)I0 ] ∪ [q0 (jω)α + q1 (jω)α+2k ] where α ∈ [α0− , α0+ ], α + 2k ∈ [α1− , α1+ ], k ∈ Z. Recall Definition 4 and (2), we have:
ÙB plus line CD. Fig. A.5. Value set of arc A
q0 (jω)α + q1 (jω)α+2k ≡ f (jω; V0 , V1 , V0α , V1α ) q
where Θ (ω) is dependent on ω and is a point in the interval [α1− , α1+ ]. Ù and the slope of PP ′ is equal to Next, since PP ′ is tangent to AB the slope of CD, we have: kPP ′
(α π ) ∂ (ωΘ sin(Θ π /2))/∂ Θ 0 = = k = tan . CD ∂ (ωΘ cos(Θ π /2))/∂ Θ 2
q
where f (jω; q0 , q1 ; V0α , V1α ) is from (2). According to the above two equations, we have α
α
α
α
∂δaa (jω) ⊆ [q0 (jω)I0 + q1 (jω)V1 ] ∪ [q0 (jω)V0 + q1 (jω)I0 ] ∪ [f (jω; q0 , q1 ; V0α , V1α )] where f (jω; q0 , q1 ; V0α , V1α ) is from (2). Repeatedly using the above equation, we can complete the proof.
By solving this equation, and note that Θ (ω) ∈ [α1− , α1+ ], we obtain:
Proof of Theorem 1. The proof is obtained by using Lemmas 1–5. Note that the value set δ (jω, q, α) is continuous for ∀ω ∈ [0, +∞).
Θ (ω) = A(ω, {α1− , α1− }, α0 )
Proof of Theorem 2. The proof is obtained by repeatedly using Theorem 1.
where A(ω, {α1 , α1 }, α0 ) is from Definition 3. Based on the above analysis, we have: −
−
α ∂δal (jω) ⊆ [I0q (jω)α0 + K¯ 1 (jω, α0 )] ∪ [K0 (jω) + q1 (jω)I1 ].
Repeatedly using the above equation, we can complete the proof. Proof of Lemma 5. Consider the following value set α
α
δaa (jω) = q0 (jω)I0 + q1 (jω)I1
where ω, q0 , q1 are fixed. δaa (jω) is an addition of two arc segments. Suppose O is the origin in the complex plane, P is a point on − → ∂δaa (jω). Then OP can be computed as
− →
OP = q0 (jω)α0 + q1 (jω)α1 ∗
∗
where α0∗ ∈ I0α , α1∗ ∈ I1α . Since P is on the boundary, then the distance |OP | is the max− → imum/minimum value along the direction OP for any point in the ∗ ∗ value set δaa (jω). More precisely, α0 and α1 are the solutions to the following optimization problem: max/min d(α0 , α1 ) = q0 (ω)α0 cos s.t . α0 ∈ I0α , α1 ∈ I1α g(α0 , α1 ) (
= q0 (ω)α0 sin
(α π 0
α0 π 2
2
)
− θ + q1 (ω)α1 cos
)
− θ + q1 (jω)α1 sin
(α π 1
(α π 1 2
2
−θ
)
−θ
)
=0 − →
where d(α0 , α1 ) represents the distance along the direction OP for − → any point in the value set δaa (jω), θ is the angle of OP. Based on Lagrange Multiplier Approach, the solutions to the following equations are possible α0∗ and α1∗ :
∂ g(α0 , α1 ) ∂ d(α0 , α1 ) +λ =0 ∂α0 ∂α0 ∂ d(α0 , α1 ) ∂ g(α0 , α1 ) +λ =0 ∂α1 ∂α1 where λ ∈ R. Solving the above equation, we obtain
[(ln ω)2 + (π /2)2 ] sin[(α1 − α0 )π/2] = 0.
Proof of Theorem 3. Note that if for ∀0 ≤ i < j ≤ n, q q f (s; Vi , Vj ; Viα , Vjα ) ≡ ∅, then ∆af (jω) = ∅. This completes the proof. Proof of Lemma 6. The proof can be obtained from [1,10]. It is noted that here we do not have delay terms. Meanwhile, the uncertain parameters qi and αi (i = 0, 1, . . . , n) vary in independent intervals. Proof of Lemma 7. First assume ϑ (ω) is a line segment. We can see − that |Aline λ+ line − Aline λline | is the length of the line segment. |Bline + | and | B + Aline λ+ Aline λ+ line line | represent the distances between line the origin and the endpoints of the line segment. Therefore, by resorting to [10], we have 0 ̸ ∈ ϑ (ω) ⇔ F (ϑ ) > 0. Next, assume ϑ (jω) is an arc segment, then when Aarc ̸ = 0, Barc ̸ = 0 we have Aarc (jω) · (jω)λarc + Barc (jω) ̸ = 0 ⇔
λarc ln(jω) + ln(−Aarc /Barc ) ̸= 0. Note that the value set of λarc ln(jω) + ln(−Aarc /Barc ) is a line segment. Hence, we have 0 ̸ ∈ ϑ (ω) ⇔ F (ϑ ) > 0. This completes the proof. Proof of Theorem 4. The proof is directly from Theorem 2, Lemmas 6 and 7. References [1] K. Moornani, M. Haeri, On robust stability of LTI fractional-order delay systems of retarded and neutral type, Automatica 46 (2) (2010) 362–368. [2] C. Yeroglu, B. Senol, Investigation of robust stability of fractional order multilinear affine systems: 2q-convex parpolygon approach, Systems Control Lett. 62 (10) (2013) 845–855. [3] J.Y. Kaminski, R. Shorten, E. Zeheb, Exact stability test and stabilization for fractional systems, Systems Control Lett. 85 (2015) 95–99. [4] Y. Luo, Y.Q. Chen, C.Y. Wang, Y.G. Pi, Tuning fractional order proportional integral controllers for fractional order system, J. Process Control 20 (7) (2010) 823–831. [5] J. Sabatier, A. Oustaloup, A. Iturricha, P. Lanusse, CRONE control: Principles and extension to time-variant plants with asymptotically constant coefficients, Nonlinear Dynam. 29 (1) (2002) 363–385.
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