# Stabilization of Linear Sampled-Data Systems by  a Time-Delay Feedback Control

Authors: F. Ricardo García; Baltazar Aguirre; Rodolfo Suárez
Journal: Mathematical Problems in Engineering
 (2008)

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## Abstract

We consider one-dimensional, time-invariant sampled-data linear
systems with constant feedback gain, an arbitrary fixed time delay, which
is a multiple of the sampling period and a zero-order hold for reconstructing
the sampled signal of the system in the feedback control. We obtain
sufficient conditions on the coefficients of the characteristic polynomial associated
with the system. We get these conditions by finding both lower
and upper bounds on the coefficients. These conditions let us give both an
estimation of the maximum value of the sampling period and an interval
on the controller gain that guarantees the stabilization of the system.


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## Body

## 1. Introduction

The sampled-data systems are particular cases of a
general type of systems  called networked control systems, that have an
important value in applications (see Hespanha et al. [1], Hespanha et al. [2], Hikichi et al. [3], Meng et al. [4], Naghshtabrizi and Hespanha [5] Ögren et al. [6], Seiler and Sengupta [7] and Shirmohammadi and Woo [8]). The networked control systems can be studied either
from the approach of control theory or communication theory (see Hespanha et al. [1]). Among the
reported papers in control theory that have researched about networked control
systems it is worth to mention the works by Zhang, Tipsuwan, and Hespanha (see
Zhang et al. [9], Tipsuwan and Chow [10], and Hespanha et al. [1]).When in networked control systems it is satisfied that
the plant outputs and the control inputs are delivered at the same time, then
we obtain a sampled-data system. In this paper, we focus our attention on
sampled-data systems. These systems have widely been studied due to their
importance in engineering applications (see Åström and Wittenmark [11], Chen and Francis [12], Franklin et al. [13], and Kolmanovskii and Myshkis [14]).A sampled-data linear system with fixed time delay in
the feedback is a continuous plant such that the feedback control of the closed
loop system is discrete and has a delayr,
namely, (1.1)x˙=Ax(t)+Buk−r(t),(1.2)uk−r(t)=Kx([th]h−r),h=tk+1−tk, where [α] denotes the integer part of α,A is an n×n matrix, B∈Rn,r∈R, and h is the interval between the successive sample
instants tk and tk+1.
If h is a constant, it is called the sampling
period and tk=kh. Recommendable references about time-delay
systems are the books by Hale and Verduyn Lunel [15], and Kolmanovskii and Myshkis [14]. On the other hand, the
theory about n-dimensional sampled-data control systems can be studied in the
books by Åström and Wittenmark [11] or
Chen and Francis [12].
In relation with the study of sampled-data systems and the problem of proving
the existence of a stabilizing control, it is worth to mention the work by
Fridman et al. [16],
which is based on solving a linear matrix inequality. The application of this
approach has been very successful in subsequent works (see Fridman etal. [17], and Mirkin [18]). Another idea is to propose a control depending on a
parameter ϵ and then prove that the control stabilizes the
system when ϵ is small enough. This idea was developed by
Yong and Arapostathis [19]. Since the existence has been proved for these last
authors, now we focus on estimating an interval for ϵ.
In order to reduce the difficulty of the problem, we will restrict our study to
the one-dimensional sampled-data systems. These systems have attracted the
attention of   several researchers as they
can model interesting phenomena in engineering (see, e.g., Busenberg and Cooke [20] and Cooke and Wiener [21]). We will consider the
one-dimensional case of (1.1), that is, we will study the differential equation (1.3)x˙=ax(t)+buk−r(t), where a and b are given constants. Our problem is to find
the values of the (gain) parameter K and of the period h so that the discrete control (zero-order hold)
with delay r(1.4)uk−r=Kx([th]h−r) makes the system (1.3) an
asymptotically stable one. The time delay is considered an integer multiple of
the sampling period h in the sense that r=Nh, where N is a natural number.Fort∈[kh,(k+1)h),k∈Z+, the function x([t/h]h−Nh) is constant and the solution of the
differential equation (1.3) is (1.5)x(t)=ea(t−kh)x(kh)+∫0t−kheaτdτbKx((k−N)h). Therefore by continuity(1.6)x((k+1)h)=eahx(kh)+∫0heaτbKx((k−N)h)dτ. We now define (1.7)Ad=eah,Bd=b∫0heaτdτ,ε(k)=x(kh). From (1.6) we obtain the following
difference equation: (1.8)ε(k+1)=Adε(k)+BdKε(k−N). By making the change of variable J=k−N,
the difference equation (1.8) becomes a homogeneous difference equation of order N+1,
namely, (1.9)ε(J+N+1)−Adε(J+N)−BdKε(J)=0. This homogeneous difference
equation of order N+1 can be rewritten as the following system of N+1 difference equations of order one. Indeed let (1.10)ε(J)=x1(J)ε(J+1)=x1(J+1)=x2(J)⋮ε(J+N)=xN(J+1)=xN+1(J)ε(J+N+1)=xN+1(J+1). Using (1.9), we obtain the
following system of difference equations: (1.11)x1(J+1)=x2(J)x2(J+1)=x3(J)⋮xN(J+1)=xN+1(J)xN+1(J+1)=AdxN+1(J)+BdKx1(J), which in matrix form
becomes (1.12)X(J+1)=AX(J), where (1.13)A=[01⋯0000⋯0000⋯00⋮⋮⋮00⋯01BdK0⋯0Ad],X(J)=[x1(J)x2(J)x3(J)⋮xN(J)xN+1(J)].To give stability conditions of the system of
difference equations, we first obtain the characteristic polynomial of the
matrixA: (1.14)P(λ)=λN+1−AdλN−BdK. Thus the problem of stabilizing
system (1.3) is equivalent to giving conditions on the coefficients of the
characteristic polynomial (1.14) so that this polynomial is Schur stable. The
problem of characterizing the stability region of (1.12) [or equivalently (1.14)] is
considered an interesting problem [1] although it is   known that
it is very difficult [9]. Our objective in this paper is to find information
about the stability region, which is explained below.System (1.1) has been studied, and necessary and
sufficient conditions on [A,B] for the r-stabilization of the system have been
obtained (see Yong and Arapostathis [19]), but they are not easily verifiable. For the
one-dimensional case (1.3), their result is the following. Suppose −(N+1)/N<−Ad<−1.
Then the polynomial (1.14) is Schur stable if −BdK=−(−Ad)−1+ϵ for a sufficiently small ϵ.
However in a design problem we need to say how to find such an ϵ,
or to obtain an estimation of the maximum sampling interval for which the
stability is guaranteed, that is very important (see Hespanha et al. [1]).In this paper, we find aϵ˜˜0 such that the polynomial (1.15)P(λ)=λN+1−AdλN−(−Ad)−1+ϵ is Schur stable if 0<ϵ<ϵ˜˜0.
That is, we get an estimation of the largest ϵmax with the property that the polynomial (1.15) is
Schur stable for 0<ϵ<ϵmax.Some general results about the stability for retarded
differential equations with piecewise constant delays were obtained by Cooke
and Wiener [21].
Problems (1.1) and (1.3) for continuous-time systems were studied by Yong [22, 23] with an analogous approach.


## 2. Main Result

Consider a
polynomialP(z)=anzn+an−1zn−1+a0 such that −n/(n−1)<an−1/an<−1.
Our objective is to give values of the coefficient a0 such that P(z) is Schur stable. The result is the following.
Choose a0=−an−1+an(ϵ−1),
then P(z) is Schur stable if ϵ satisfies the inequality 0<ϵ<3n/(2n−1)+(3(n−1)/(2n−1))(an−1/an).We begin by establishing the result when the degree ofP(z) is two (in fact, we have here necessary and
sufficient conditions).Theorem 2.1.
LetP(z)=a2z2+a1z+a0 a polynomial such that −2<a1/a2<−1,
where a0=−a1+a2(ϵ−1).
Then P(z) is Schur stable if and only if (2.1)0<ϵ<2+a1a2.Proof.
P(z) is Schur stable if and only if its
coefficients satisfy [24] the following: (2.2)|a2|>|a2(ϵ−1)−a1|,|a1|<|a2+a2(ϵ−1)−a1|; or equivalently (2.3)0>a22ϵ2−(2a22+2a2a1)ϵ+2a2a1+a12,0<ϵ(a22ϵ−2a2a1).
To prove this last part, we define(2.4)g(ϵ)=a22ϵ2−(2a22+2a2a1)ϵ+2a2a1+a12. Then (2.5)g(ϵ)=0⇔(ϵ=2+a1a2orϵ=a1a2). Since the coefficient of ϵ2 is positive, g(ϵ)<0 if and only if (2.6)a1a2<ϵ<2+a1a2.
We have, it holds that.
On the other hand,ϵ(a22ϵ−2a2a1)>0 if and only if (ϵ>0 and ϵ>2(a1/a2)). Now since a1/a2<−1, it holds that
2(a1/a2)<−2.
Therefore, ϵ(a22ϵ−2a2a1)>0 if and only if ϵ>0,
so that [g(ϵ)<0 and ϵ(a22ϵ−2a2a1)>0] if and only if 0<ϵ<2+a1/a2.The arbitrary degree proof depends on the following
lemma and several technical propositions that can be checked in the appendix.Lemma 2.2.
Fix an arbitrary integern>2.
Given P(z)=an+1zn+1+anzn+a0 with −(n+1)/n<an/an+1<−1 and a0=−an+an+1(ϵ−1),
define Q(z)=a0zn+1+anz+an+1 and (2.7)R(z)=1z[P(z)−a0an+1Q(z)]=1an+1[Anzn+An−1zn−1+A0], where An=an+12−a02,An−1=an+1an and A0=−a0an.
If ϵ satisfies 0<ϵ<3(n+1)/(2n+1)+(3n/(2n+1))(an/an+1),
then (|an+1|>|a0| and |An|>|A0|).Proof.
We
have that|an+1|>|a0| if and only if (Proposition A.1) (2.8)0<ϵ<2+anan+1. Hence to prove the lemma, it is
sufficient to show that (2.9)3(n+1)2n+1+3n2n+1anan+1<2+anan+1. A straightforward calculation
shows that inequality (2.9) holds if and only if ((n−1)/(2n+1))(an/an+1)<(n−1)/(2n+1),
which is true because an/an+1<−1.
We now show that|An|>|A0|.
It can be seen that An=an+12−a02 and A0=−a0an,
from where (2.10)|An|>|A0|⇔|an+12−a02|>|a0an|⇔(an+12−a02)2>(a0an)2⇔[(an+12−a02)−a0an][an+12−a02+a0an]>0⇔{[(an+12−a02)−a0an>0,an+12−a02+a0an>0]or[(an+12−a02)−a0an<0,an+12−a02+a0an<0]}.
We will split the analysis into the following two
cases:(2.11)[(an+12−a02)−a0an>0,an+12−a02+a0an>0] or(2.12)[(an+12−a02)−a0an<0,an+12−a02+a0an<0].
We analyze (2.11). By Proposition A.2, the first
inequality  in (2.11) is satisfied if and only if (2.13)1+12anan+1−1+14(anan+1)2<ϵ<1+12anan+1+1+14(anan+1)22. Since 1+(1/2)(an/an+1)>0,
it follows that (2.14)1+12anan+1+1+14(anan+1)2>0. By straightforward
calculations, (2.15)1+12anan+1−1+14(anan+1)2<0, and since ϵ>0, it must satisfy (2.16)0<ϵ<1+12anan+1+1+14(anan+1)2.
For the second inequality in (2.11), we use Proposition
A.3. So an+12−a02+a0an>0 if and only if (2.17)1+3an2an+1−1+14(anan+1)2<ϵ<1+3an2an+1+1+14(anan+1)2. Since −(n+1)/n<an/an+1<−1,
we have the following two inequalities: (2.18)1+3an2an+1−1+14(anan+1)2<0,1+3an2an+1+1+14(anan+1)2>0. Now, since we are interested in ϵ>0,
it must satisfy (2.19)0<ϵ<1+3an2an+1+1+14(anan+1)2.
By straightforward calculations, it follows
that(2.20)1+3an2an+1+1+14(anan+1)2<1+12anan+1+1+14(anan+1)2. Therefore both inequalities in
(2.11) are satisfied if and only if (2.21)0<ϵ<1+3an2an+1+1+14(anan+1)2. Note that not depending on (2.12),
we get that |An|>|A0| if (2.21) is satisfied, so we can omit the
analysis of (2.12).
Now by hypothesisϵ<3(n+1)/(2n+1)+(3n/(2n+1))(an/an+1) and by Proposition A.4, it holds that(2.22)3(n+1)2n+1+3n2n+1anan+1<1+3an2an+1+1+14(anan+1)2. It follows that (2.23)ϵ<1+3an2an+1+1+14(anan+1)2, and consequently |An|>|A0|.We now prove the main result for an arbitrary degree.Theorem (fix an arbitrary integern≥2).
LetP(z)=anzn+an−1zn−1+a0 be a polynomial such that −n/(n−1)<an−1/an<−1,
where a0=−an−1+an(ϵ−1).
If ϵ satisfies 0<ϵ<3n/(2n−1)+(3(n−1)/(2n−1))(an−1/an),
then we have that |an|>|a0| and P is a polynomial Schur stable.Proof.
We
make induction overn.
The case n=2 is part of Theorem 2.1. Now suppose that the
theorem holds for n≥2,
and let P(z)=an+1zn+1+anzn+a0 be a polynomial of degree n+1 such that (2.24)−n+1n<anan+1<−1,a0=−an+an+1(ϵ−1).
If we define the polynomialsQ and R as in lemma, then replacing P and Q in the polynomial R,
we obtain (2.25)R(z)=(an+12−a02)zn+an+1anzn−1−a0anan+1. If |an+1|>|a0|,
then an+1R(z)=(an+12−a02)zn+an+1anzn−1−a0an is a Schur stable polynomial if and only if P is Schur stable [25]. The inequality |an+1|>|a0| was proved in the lemma.
If we defineAn=an+12−a02,An−1=an+1an and A0=−a0an and since the inequality |An|>|A0| is satisfied (which was proved in the lemma),
then by induction hypothesis the polynomial an+1R(z)=Anzn+An−1zn−1+A0 is Schur stable if A0=−An−1+An(ϵ˜−1) and ϵ˜ satisfies 0<ϵ˜<3n/(2n−1)+3(n−1)/(2n−1)(An−1/An). From the equality A0=−An−1+An(ϵ˜−1), it follows that (2.26)A0=−An−1−An+Anϵ˜. By (2.24), −a0an=an2+anan+1−anan+1ϵ or equivalently (2.27)−a0an=−anan+1−(an+12−a02)+[(an+12−a02)+an2+2anan+1−anan+1ϵan+12−a02](an+12−a02). That is (2.28)A0=−An−1−An+[(an+12−a02)+an2+2anan+1−anan+1ϵan+12−a02]An. Comparing this with (2.26), we see
that (2.29)ϵ˜=an+12−a02+an2+2anan+1−anan+1ϵan+12−a02. Moreover by induction hypothesis ϵ˜ must satisfy the condition (2.30)0<ϵ˜<3n2n−1+3(n−1)2n−1(An−1An). Substituting ϵ˜,An−1 and An into (2.30), we obtain (2.31)0<an+12−a02+an2+2anan+1−anan+1ϵan+12−a02<3n2n−1+3(n−1)2n−1an+1anan+12−a02. The first inequality in (2.31) is
equivalent to (2.32)0<an+12−a02+an2+2anan+1−anan+1ϵ. And by Proposition A.5 this holds
if and only if (2.33)0<ϵ<2+anan+1.
Now we will analyze the second inequality in (2.31)
which is equivalent to(2.34)an+12−a02+an2+2anan+1−anan+1ϵ<3n(an+12−a02)2n−1+3(n−1)2n−1an+1an. By Proposition A.6, inequality
(2.34) is obtained if and only if (2.35)1+4n+12(n+1)c−Hn(c)<ϵ<1+4n+12(n+1)c+Hn(c), where c=an/an+1 and Hn(c)=1+((n−2)/(n+1))c+((n−1/2)/(n+1))2c2.
By PropositionA.7, (2.36)1+4n+12(n+1)c−Hn(c)<0. So that (2.34) is satisfied if and
only if (2.37)0<ϵ<1+4n+12(n+1)c+Hn(c).Moreover by Proposition
A.8, (2.38)1+4n+12(n+1)c+Hn(c)≤2+c∀n≥1. Thus (2.32) and (2.34) are satisfied
if and only if (2.39)0<ϵ<1+4n+12(n+1)c+Hn(c). We now analyze the right-hand
side of (2.39). Let (2.40)F(c)=1+4n+12(n+1)c+Hn(c). By Proposition A.9, it holds
that F(c) is increasing and
convex, F(−(n+1)/n)=0 and F′(−(n+1)/n)=3n/(2n+1).
We now get the equation of the tangent line of the
functionF at the point c=−(n+1)/n.
To do this, we use fact hat F(−(n+1)/n)=0 and F′(−(n+1)/n)=3n/(2n+1).
So that the equation of the tangent line passing through the point (−(n+1)/n,0) is y=3(n+1)/(2n+1)+(3n/(2n+1))c.
Therefore if 0<ϵ<3(n+1)/(2n+1)+(3n/(2n+1))c,
then (2.41)0<ϵ<1+4n+12(n+1)c+Hn(c) from which Theorem 2.3 follows.Remark 2.4.
Note
that the inequality−(N+1)/N<−Ad<−1 implies that the number a in (1.3) must be positive since Ad=eah with h>0 and then: −eah<−1 is satisfied if and only if a>0.The next corollary is a consequence of our results.Corollary 2.5.
Suppose that the system (1.3) has a proportional control (1.4) with delay r=Nh and suppose that a,b>0.
If the sampling period and the gain of the controller satisfy (2.42)h<ln[3(N+1)/3N]a,−ab[1−3N2N+1+3(2N+1)(eah−1)]<K<−ab, then the sampled-data system is
stabilizable.


## 3. Example

We consider the
sampled-data system(3.1)x˙=x(t)+uk−r(t),uk−r(t)=Kx([th]h−4h), where the values of the
parameters are a=1,b=1,N=4,
and r=4h.
The difference equation (1.8) is ε(k+1)=ehε(k)+(eh−1)Kε(k−4) and the characteristic polynomial (1.12)
associated with the system is P(λ)=λ5−ehλ4+eh−1+ϵ which is Schur stable for 0<ϵ<(5−4eh)/3 by Theorem 2.1. Furthermore by Corollary 2.5 the
maximum sampling period is h<ln(5/4) and the interval for the gain of the
controller is (3.2)eh−23(eh−1)<K<−1. Now for h=0.22,
the interval of the gain that guaranties the stabilization of the system is (3.3)−1.02<K<−1. For K=−1.01 the sampled-data system is stable as the
characteristic polynomial has roots with modulus less than one: λ1=−0.605301,λ2=−0.0721534−0.637776i, λ3=−0.0721534+0.637776i, λ4=0.997839−0.031305i,
and λ5=0.997839+0.031305i.



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