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

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.

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 delay 𝑟, namely, 𝑥̇=𝐴𝑥(𝑡)+𝐵𝑢_𝑘−𝑟𝑢(𝑡),_𝑘−𝑟𝑡(𝑡)=𝐾𝑥([ℎ]ℎ−𝑟),ℎ=𝑡_𝑘+1−𝑡_𝑘, (1.1)(1.2) where [𝛼] denotes the integer part of 𝛼,𝐴 is an 𝑛×𝑛 matrix, 𝐵∈𝑅^𝑛,𝑟∈𝑅, and ℎ is the interval between the successive sample instants 𝑡_𝑘 and 𝑡_𝑘+1. If ℎ is a constant, it is called the sampling period and 𝑡_𝑘=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 𝑥̇=𝑎𝑥(𝑡)+𝑏𝑢_𝑘−𝑟(𝑡), (1.3) 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 𝑟𝑢_𝑘−𝑟𝑡=𝐾𝑥([ℎ]ℎ−𝑟) (1.4) 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.

For 𝑡∈[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 𝑥(𝑡)=𝑒^{𝑎(𝑡−𝑘ℎ)}𝑥(𝑘ℎ)+₀^{𝑡−𝑘ℎ}𝑒^𝑎𝜏𝑑𝜏𝑏𝐾𝑥((𝑘−𝑁)ℎ). (1.5) Therefore by continuity𝑥((𝑘+1)ℎ)=𝑒^𝑎ℎ𝑥(𝑘ℎ)+^ℎ₀𝑒^𝑎𝜏𝑏𝐾𝑥((𝑘−𝑁)ℎ)𝑑𝜏. (1.6) We now define 𝐴_𝑑=𝑒^𝑎ℎ,𝐵_𝑑=𝑏^ℎ₀𝑒^𝑎𝜏𝑑𝜏,𝜀(𝑘)=𝑥(𝑘ℎ). (1.7) From (1.6) we obtain the following difference equation: 𝜀(𝑘+1)=𝐴_𝑑𝜀(𝑘)+𝐵_𝑑𝐾𝜀(𝑘−𝑁). (1.8) By making the change of variable 𝐽=𝑘−𝑁, the difference equation (1.8) becomes a homogeneous difference equation of order 𝑁+1, namely, 𝜀(𝐽+𝑁+1)−𝐴_𝑑𝜀(𝐽+𝑁)−𝐵_𝑑𝐾𝜀(𝐽)=0. (1.9) 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 𝜀(𝐽)=𝑥₁(𝐽)𝜀(𝐽+1)=𝑥₁(𝐽+1)=𝑥₂⋮(𝐽)𝜀(𝐽+𝑁)=𝑥_𝑁(𝐽+1)=𝑥_𝑁+1(𝐽)𝜀(𝐽+𝑁+1)=𝑥_𝑁+1(𝐽+1). (1.10) Using (1.9), we obtain the following system of difference equations: 𝑥₁(𝐽+1)=𝑥₂𝑥(𝐽)₂(𝐽+1)=𝑥₃⋮𝑥(𝐽)_𝑁(𝐽+1)=𝑥_𝑁+1𝑥(𝐽)_𝑁+1(𝐽+1)=𝐴_𝑑𝑥_𝑁+1(𝐽)+𝐵_𝑑𝐾𝑥₁(𝐽), (1.11) which in matrix form becomes 𝑋(𝐽+1)=𝐴𝑋(𝐽), (1.12) where 𝐵𝐴=01⋯0000⋯0000⋯00⋮⋮⋮00⋯01_𝑑𝐾0⋯0𝐴_𝑑𝑥,𝑋(𝐽)=¹𝑥(𝐽)₂𝑥(𝐽)₃⋮𝑥(𝐽)_𝑁𝑥(𝐽)_𝑁+1(𝐽). (1.13)

To give stability conditions of the system of difference equations, we first obtain the characteristic polynomial of the matrix 𝐴: 𝑃(𝜆)=𝜆^𝑁+1−𝐴_𝑑𝜆^𝑁−𝐵_𝑑𝐾. (1.14) 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<−𝐴_𝑑<−1. Then the polynomial (1.14) is Schur stable if −𝐵_𝑑𝐾=−(−𝐴_𝑑)−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 a ^0𝑥002𝑑𝜖^{𝑐0𝑥002𝑑𝑐}₀ such that the polynomial 𝑃(𝜆)=𝜆^𝑁+1−𝐴_𝑑𝜆^𝑁−(−𝐴_𝑑)−1+𝜖 (1.15) is Schur stable if 0<𝜖<𝜖^{0𝑥002𝑑𝑐0𝑥002𝑑𝑐}₀. That is, we get an estimation of the largest 𝜖_max with the property that the polynomial (1.15) is Schur stable for 0<𝜖<𝜖_max.

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 polynomial 𝑃(𝑧)=𝑎_𝑛𝑧^𝑛+𝑎_𝑛−1𝑧^𝑛−1+𝑎₀ such that −𝑛/(𝑛−1)<𝑎_𝑛−1/𝑎_𝑛<−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=−𝑎_𝑛−1+𝑎_𝑛(𝜖−1), then 𝑃(𝑧) is Schur stable if 𝜖 satisfies the inequality 0<𝜖<3𝑛/(2𝑛−1)+(3(𝑛−1)/(2𝑛−1))(𝑎_𝑛−1/𝑎_𝑛).

We begin by establishing the result when the degree of 𝑃(𝑧) is two (in fact, we have here necessary and sufficient conditions).

Theorem 2.1. Let 𝑃(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 𝑎0<𝜖<2+¹𝑎₂. (2.1)

Proof. 𝑃(𝑧) is Schur stable if and only if its coefficients satisfy [24] the following: |𝑎₂|>|𝑎₂(𝜖−1)−𝑎₁|,|𝑎₁|<|𝑎₂+𝑎₂(𝜖−1)−𝑎₁|; (2.2) or equivalently 0>𝑎²₂𝜖²−(2𝑎²₂+2𝑎₂𝑎₁)𝜖+2𝑎₂𝑎₁+𝑎²₁,0<𝜖(𝑎²₂𝜖−2𝑎₂𝑎₁). (2.3)
To prove this last part, we define 𝑔(𝜖)=𝑎²₂𝜖²−(2𝑎²₂+2𝑎₂𝑎₁)𝜖+2𝑎₂𝑎₁+𝑎²₁. (2.4) Then 𝑎𝑔(𝜖)=0⇔(𝜖=2+¹𝑎₂𝑎or𝜖=¹𝑎₂). (2.5) Since the coefficient of 𝜖² is positive, 𝑔(𝜖)<0 if and only if 𝑎₁𝑎₂𝑎<𝜖<2+¹𝑎₂. (2.6) 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 integer 𝑛>2. Given 𝑃(𝑧)=𝑎_𝑛+1𝑧^𝑛+1+𝑎_𝑛𝑧^𝑛+𝑎₀ with −(𝑛+1)/𝑛<𝑎_𝑛/𝑎_𝑛+1<−1 and 𝑎₀=−𝑎_𝑛+𝑎_𝑛+1(𝜖−1), define 𝑄(𝑧)=𝑎₀𝑧^𝑛+1+𝑎_𝑛𝑧+𝑎_𝑛+1 and 1𝑅(𝑧)=𝑧𝑎[𝑃(𝑧)−⁰𝑎_𝑛+11𝑄(𝑧)]=𝑎_𝑛+1[𝐴_𝑛𝑧^𝑛+𝐴_𝑛−1𝑧^𝑛−1+𝐴₀], (2.7) where 𝐴_𝑛=𝑎²_𝑛+1−𝑎²₀,𝐴_𝑛−1=𝑎_𝑛+1𝑎_𝑛 and 𝐴₀=−𝑎₀𝑎_𝑛. If 𝜖 satisfies 0<𝜖<3(𝑛+1)/(2𝑛+1)+(3𝑛/(2𝑛+1))(𝑎_𝑛/𝑎_𝑛+1), then (|𝑎_𝑛+1|>|𝑎₀| and |𝐴_𝑛|>|𝐴₀|).

Proof. We have that |𝑎_𝑛+1|>|𝑎₀| if and only if (Proposition A.1) 𝑎0<𝜖<2+^𝑛𝑎_𝑛+1. (2.8) Hence to prove the lemma, it is sufficient to show that 3(𝑛+1)+2𝑛+13𝑛𝑎2𝑛+1𝑎^𝑛_𝑛+1𝑎<2+^𝑛𝑎_𝑛+1. (2.9) A straightforward calculation shows that inequality (2.9) holds if and only if ((𝑛−1)/(2𝑛+1))(𝑎_𝑛/𝑎_𝑛+1)<(𝑛−1)/(2𝑛+1), which is true because 𝑎_𝑛/𝑎_𝑛+1<−1.
We now show that |𝐴_𝑛|>|𝐴₀|. It can be seen that 𝐴_𝑛=𝑎²_𝑛+1−𝑎²₀ and 𝐴₀=−𝑎₀𝑎_𝑛, from where |𝐴_𝑛|>|𝐴₀|⇔|𝑎²_𝑛+1−𝑎²₀|>|𝑎₀𝑎_𝑛|⇔(𝑎²_𝑛+1−𝑎²₀)²>(𝑎₀𝑎_𝑛)²⇔[(𝑎²_𝑛+1−𝑎²₀)−𝑎₀𝑎_𝑛][𝑎²_𝑛+1−𝑎²₀+𝑎₀𝑎_𝑛]>0⇔{[(𝑎²_𝑛+1−𝑎²₀)−𝑎₀𝑎_𝑛>0,𝑎²_𝑛+1−𝑎²₀+𝑎₀𝑎_𝑛>0]or[(𝑎²_𝑛+1−𝑎²₀)−𝑎₀𝑎_𝑛<0,𝑎²_𝑛+1−𝑎²₀+𝑎₀𝑎_𝑛<0]}. (2.10)
We will split the analysis into the following two cases: [(𝑎²_𝑛+1−𝑎²₀)−𝑎₀𝑎_𝑛>0,𝑎²_𝑛+1−𝑎²₀+𝑎₀𝑎_𝑛>0] (2.11) or[(𝑎²_𝑛+1−𝑎²₀)−𝑎₀𝑎_𝑛<0,𝑎²_𝑛+1−𝑎²₀+𝑎₀𝑎_𝑛<0]. (2.12)
We analyze (2.11). By Proposition A.2, the first inequality in (2.11) is satisfied if and only if 11+2𝑎_𝑛𝑎_𝑛+1−11+4(𝑎_𝑛𝑎_𝑛+1)²1<𝜖<1+2𝑎_𝑛𝑎_𝑛+1+²11+4(𝑎_𝑛𝑎_𝑛+1)². (2.13) Since 1+(1/2)(𝑎_𝑛/𝑎_𝑛+1)>0, it follows that 11+2𝑎_𝑛𝑎_𝑛+1+11+4(𝑎_𝑛𝑎_𝑛+1)²>0. (2.14) By straightforward calculations, 11+2𝑎_𝑛𝑎_𝑛+1−11+4(𝑎_𝑛𝑎_𝑛+1)²<0, (2.15) and since 𝜖>0, it must satisfy 10<𝜖<1+2𝑎_𝑛𝑎_𝑛+1+11+4(𝑎_𝑛𝑎_𝑛+1)². (2.16)
For the second inequality in (2.11), we use Proposition A.3. So 𝑎²_𝑛+1−𝑎²₀+𝑎₀𝑎_𝑛>0 if and only if 1+3𝑎_𝑛2𝑎_𝑛+1−11+4(𝑎_𝑛𝑎_𝑛+1)²<𝜖<1+3𝑎_𝑛2𝑎_𝑛+1+11+4(𝑎_𝑛𝑎_𝑛+1)². (2.17) Since −(𝑛+1)/𝑛<𝑎_𝑛/𝑎_𝑛+1<−1, we have the following two inequalities: 1+3𝑎_𝑛2𝑎_𝑛+1−11+4(𝑎_𝑛𝑎_𝑛+1)²<0,1+3𝑎_𝑛2𝑎_𝑛+1+11+4(𝑎_𝑛𝑎_𝑛+1)²>0. (2.18) Now, since we are interested in 𝜖>0, it must satisfy 0<𝜖<1+3𝑎_𝑛2𝑎_𝑛+1+11+4(𝑎_𝑛𝑎_𝑛+1)². (2.19)
By straightforward calculations, it follows that 1+3𝑎_𝑛2𝑎_𝑛+1+11+4(𝑎_𝑛𝑎_𝑛+1)²1<1+2𝑎_𝑛𝑎_𝑛+1+11+4(𝑎_𝑛𝑎_𝑛+1)². (2.20) Therefore both inequalities in (2.11) are satisfied if and only if 0<𝜖<1+3𝑎_𝑛2𝑎_𝑛+1+11+4(𝑎_𝑛𝑎_𝑛+1)². (2.21) Note that not depending on (2.12), we get that |𝐴_𝑛|>|𝐴₀| if (2.21) is satisfied, so we can omit the analysis of (2.12).
Now by hypothesis 𝜖<3(𝑛+1)/(2𝑛+1)+(3𝑛/(2𝑛+1))(𝑎_𝑛/𝑎_𝑛+1) and by Proposition A.4, it holds that3(𝑛+1)+2𝑛+13𝑛𝑎2𝑛+1𝑎^𝑛_𝑛+1<1+3𝑎_𝑛2𝑎_𝑛+1+11+4(𝑎_𝑛𝑎_𝑛+1)². (2.22) It follows that 𝜖<1+3𝑎_𝑛2𝑎_𝑛+1+11+4(𝑎_𝑛𝑎_𝑛+1)², (2.23) and consequently |𝐴_𝑛|>|𝐴₀|.

We now prove the main result for an arbitrary degree.

Theorem (fix an arbitrary integer 𝑛≥2). Let 𝑃(𝑧)=𝑎_𝑛𝑧^𝑛+𝑎_𝑛−1𝑧^𝑛−1+𝑎₀ be a polynomial such that −𝑛/(𝑛−1)<𝑎_𝑛−1/𝑎_𝑛<−1, where 𝑎₀=−𝑎_𝑛−1+𝑎_𝑛(𝜖−1). If 𝜖 satisfies 0<𝜖<3𝑛/(2𝑛−1)+(3(𝑛−1)/(2𝑛−1))(𝑎_𝑛−1/𝑎_𝑛), then we have that |𝑎_𝑛|>|𝑎₀| and 𝑃 is a polynomial Schur stable.

Proof. We make induction over 𝑛. The case 𝑛=2 is part of Theorem 2.1. Now suppose that the theorem holds for 𝑛≥2, and let 𝑃(𝑧)=𝑎_𝑛+1𝑧^𝑛+1+𝑎_𝑛𝑧^𝑛+𝑎₀ be a polynomial of degree 𝑛+1 such that −𝑛+1𝑛<𝑎_𝑛𝑎_𝑛+1<−1,𝑎₀=−𝑎_𝑛+𝑎_𝑛+1(𝜖−1). (2.24)
If we define the polynomials 𝑄 and 𝑅 as in lemma, then replacing 𝑃 and 𝑄 in the polynomial 𝑅, we obtain 𝑅(𝑧)=(𝑎²_𝑛+1−𝑎²₀)𝑧^𝑛+𝑎_𝑛+1𝑎_𝑛𝑧^𝑛−1−𝑎₀𝑎_𝑛𝑎_𝑛+1. (2.25) If |𝑎_𝑛+1|>|𝑎₀|, then 𝑎_𝑛+1𝑅(𝑧)=(𝑎²_𝑛+1−𝑎²₀)𝑧^𝑛+𝑎_𝑛+1𝑎_𝑛𝑧^𝑛−1−𝑎₀𝑎_𝑛 is a Schur stable polynomial if and only if 𝑃 is Schur stable [25]. The inequality |𝑎_𝑛+1|>|𝑎₀| was proved in the lemma.
If we define 𝐴_𝑛=𝑎²_𝑛+1−𝑎²₀,𝐴_𝑛−1=𝑎_𝑛+1𝑎_𝑛 and 𝐴₀=−𝑎₀𝑎_𝑛 and since the inequality |𝐴_𝑛|>|𝐴₀| is satisfied (which was proved in the lemma), then by induction hypothesis the polynomial 𝑎_𝑛+1𝑅(𝑧)=𝐴_𝑛𝑧^𝑛+𝐴_𝑛−1𝑧^𝑛−1+𝐴₀ is Schur stable if 𝐴₀=−𝐴_𝑛−1+𝐴_𝑛(𝜖^{0𝑥002𝑑𝑐}−1) and ^0𝑥0𝜖^02𝑑𝑐 satisfies 0<𝜖^{0𝑥002𝑑𝑐}<3𝑛/(2𝑛−1)+3(𝑛−1)/(2𝑛−1)(𝐴_𝑛−1/𝐴_𝑛). From the equality 𝐴₀=−𝐴_𝑛−1+𝐴_𝑛(𝜖^{0𝑥002𝑑𝑐}−1), it follows that 𝐴₀=−𝐴_𝑛−1−𝐴_𝑛+𝐴𝜖_𝑛^{0𝑥002𝑑𝑐}. (2.26) By (2.24), −𝑎₀𝑎_𝑛=𝑎²_𝑛+𝑎_𝑛𝑎_𝑛+1−𝑎_𝑛𝑎_𝑛+1𝜖 or equivalently −𝑎₀𝑎_𝑛=−𝑎_𝑛𝑎_𝑛+1−(𝑎²_𝑛+1−𝑎²₀)+[(𝑎²_𝑛+1−𝑎²₀)+𝑎²_𝑛+2𝑎_𝑛𝑎_𝑛+1−𝑎_𝑛𝑎_𝑛+1𝜖𝑎²_𝑛+1−𝑎²₀](𝑎²_𝑛+1−𝑎²₀). (2.27) That is 𝐴₀=−𝐴_𝑛−1−𝐴_𝑛+[(𝑎²_𝑛+1−𝑎²₀)+𝑎²_𝑛+2𝑎_𝑛𝑎_𝑛+1−𝑎_𝑛𝑎_𝑛+1𝜖𝑎²_𝑛+1−𝑎²₀]𝐴_𝑛. (2.28) Comparing this with (2.26), we see that _{0𝑥002𝑑𝑐}𝑎𝜖=^2𝑛+1−𝑎²₀+𝑎²_𝑛+2𝑎_𝑛𝑎_𝑛+1−𝑎_𝑛𝑎_𝑛+1𝜖𝑎²_𝑛+1−𝑎²₀. (2.29) Moreover by induction hypothesis ^0𝑥0𝜖^02𝑑𝑐 must satisfy the condition 0<𝜖^{0𝑥002𝑑𝑐}<3𝑛+2𝑛−13(𝑛−1)(𝐴2𝑛−1𝐴^𝑛−1_𝑛). (2.30) Substituting ^0𝑥0𝜖^02𝑑𝑐,𝐴_𝑛−1 and 𝐴_𝑛 into (2.30), we obtain 𝑎0<^2𝑛+1−𝑎²₀+𝑎²_𝑛+2𝑎_𝑛𝑎_𝑛+1−𝑎_𝑛𝑎_𝑛+1𝜖𝑎²_𝑛+1−𝑎²₀<3𝑛+2𝑛−13(𝑛−1)𝑎2𝑛−1^𝑛+1𝑎_𝑛𝑎²_𝑛+1−𝑎²₀. (2.31) The first inequality in (2.31) is equivalent to 0<𝑎²_𝑛+1−𝑎²₀+𝑎²_𝑛+2𝑎_𝑛𝑎_𝑛+1−𝑎_𝑛𝑎_𝑛+1𝜖. (2.32) And by Proposition A.5 this holds if and only if 𝑎0<𝜖<2+^𝑛𝑎_𝑛+1. (2.33)
Now we will analyze the second inequality in (2.31) which is equivalent to𝑎²_𝑛+1−𝑎²₀+𝑎²_𝑛+2𝑎_𝑛𝑎_𝑛+1−𝑎_𝑛𝑎_𝑛+1𝜖<3𝑛(𝑎²_𝑛+1−𝑎²₀)+2𝑛−13(𝑛−1)𝑎2𝑛−1^𝑛+1𝑎_𝑛. (2.34) By Proposition A.6, inequality (2.34) is obtained if and only if 1+4𝑛+12(𝑛+1)𝑐−𝐻_𝑛(𝑐)<𝜖<1+4𝑛+12(𝑛+1)𝑐+𝐻_𝑛(𝑐), (2.35) where 𝑐=𝑎_𝑛/𝑎_𝑛+1 and 𝐻_𝑛(𝑐)=1+((𝑛−2)/(𝑛+1))𝑐+((𝑛−1/2)/(𝑛+1))²𝑐².
By Proposition A.7, 1+4𝑛+12(𝑛+1)𝑐−𝐻_𝑛(𝑐)<0. (2.36) So that (2.34) is satisfied if and only if 0<𝜖<1+4𝑛+12(𝑛+1)𝑐+𝐻_𝑛(𝑐). (2.37)Moreover by Proposition A.8, 1+4𝑛+12(𝑛+1)𝑐+𝐻_𝑛(𝑐)≤2+𝑐∀𝑛≥1. (2.38) Thus (2.32) and (2.34) are satisfied if and only if 0<𝜖<1+4𝑛+12(𝑛+1)𝑐+𝐻_𝑛(𝑐). (2.39) We now analyze the right-hand side of (2.39). Let 𝐹(𝑐)=1+4𝑛+12(𝑛+1)𝑐+𝐻_𝑛(𝑐). (2.40) 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 function 𝐹 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 0<𝜖<1+4𝑛+12(𝑛+1)𝑐+𝐻_𝑛(𝑐) (2.41) from which Theorem 2.3 follows.

Remark 2.4. Note that the inequality −(𝑁+1)/𝑁<−𝐴_𝑑<−1 implies that the number 𝑎 in (1.3) must be positive since 𝐴_𝑑=𝑒^𝑎ℎ with ℎ>0 and then: −𝑒^𝑎ℎ<−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 ℎ<ln[3(𝑁+1)/3𝑁]𝑎,−𝑎𝑏[1−3𝑁+32𝑁+1(2𝑁+1)(𝑒^𝑎ℎ𝑎−1)]<𝐾<−𝑏, (2.42) then the sampled-data system is stabilizable.

3. Example

We consider the sampled-data system 𝑥̇=𝑥(𝑡)+𝑢_𝑘−𝑟𝑢(𝑡),_𝑘−𝑟𝑡(𝑡)=𝐾𝑥([ℎ]ℎ−4ℎ), (3.1) 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)=𝑒^ℎ𝜀(𝑘)+(𝑒^ℎ−1)𝐾𝜀(𝑘−4) and the characteristic polynomial (1.12) associated with the system is 𝑃(𝜆)=𝜆⁵−𝑒^ℎ𝜆⁴+𝑒^ℎ−1+𝜖 which is Schur stable for 0<𝜖<(5−4𝑒^ℎ)/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 𝑒^ℎ−23(𝑒^ℎ−1)<𝐾<−1. (3.2) Now for ℎ=0.22, the interval of the gain that guaranties the stabilization of the system is −1.02<𝐾<−1. (3.3) 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.

Appendix

In what follows we prove several inequalities.

Proposition A.1. If −(𝑛+1)/𝑛<𝑎_𝑛/𝑎_𝑛+1<−1 and 𝑎₀=−𝑎_𝑛+𝑎_𝑛+1(𝜖−1), then |𝑎_𝑛+1|>|𝑎₀𝑎|⇔0<𝜖<2+^𝑛𝑎_𝑛+1. (A.1)

Proof. Replacing the value of 𝑎₀, we see that |𝑎_𝑛+1|>|𝑎₀|⇔𝑎²_𝑛+1>[−𝑎_𝑛+𝑎_𝑛+1(𝜖−1)]²,⇔𝑎²_𝑛+1𝜖²−2𝑎_𝑛+1(𝑎_𝑛+𝑎_𝑛+1)𝜖+𝑎_𝑛(2𝑎_𝑛+1+𝑎_𝑛)<0. (A.2)
Let ℎ(𝜖)=𝑎²_𝑛+1𝜖²−2𝑎_𝑛+1(𝑎_𝑛+𝑎_𝑛+1)𝜖+𝑎_𝑛(2𝑎_𝑛+1+𝑎_𝑛). The roots of the equation ℎ(𝜖)=0 are 𝜖₁=2+𝑎_𝑛/𝑎_𝑛+1 and 𝜖₂=𝑎_𝑛/𝑎_𝑛+1. Since the coefficient of 𝜖² is positive, then ℎ(𝜖)<0 if and only if 𝑎_𝑛/𝑎_𝑛+1<𝜖<2+𝑎_𝑛/𝑎_𝑛+1. But since 𝑎_𝑛/𝑎_𝑛+1<0, we obtain 0<𝜖<2+𝑎_𝑛/𝑎_𝑛+1.

Proposition A.2. If 𝑎₀=−𝑎_𝑛+𝑎_𝑛+1(𝜖−1), then (𝑎²_𝑛+1−𝑎²₀)−𝑎₀𝑎_𝑛1>0⇔1+2𝑎_𝑛𝑎_𝑛+1−11+4(𝑎_𝑛𝑎_𝑛+1)²1<𝜖<1+2𝑎_𝑛𝑎_𝑛+1+11+4(𝑎_𝑛𝑎_𝑛+1)². (A.3)

Proof. Substituting 𝑎₀ into the first inequality, we see that 𝑎²_𝑛+1−[−𝑎_𝑛+𝑎_𝑛+1(𝜖−1)]²−[−𝑎_𝑛+𝑎_𝑛+1(𝜖−1)]𝑎_𝑛>0 (A.4) if and only if −𝑎²_𝑛+1𝜖²+(2𝑎²_𝑛+1+𝑎_𝑛𝑎_𝑛+1)𝜖−𝑎_𝑛𝑎_𝑛+1>0. (A.5) Let 𝑔(𝜖)=−𝑎²_𝑛+1𝜖²+(2𝑎²_𝑛+1+𝑎_𝑛𝑎_𝑛+1)𝜖−𝑎_𝑛𝑎_𝑛+1, then 1𝑔(𝜖)=0⇔𝜖=1+2𝑎_𝑛𝑎_𝑛+1±11+4(𝑎_𝑛𝑎_𝑛+1)². (A.6) Since the coefficient of 𝜖² is negative, 1𝑔(𝜖)>0⇔1+2𝑎_𝑛𝑎_𝑛+1−11+4(𝑎_𝑛𝑎_𝑛+1)²1<𝜖<1+2𝑎_𝑛𝑎_𝑛+1+11+4(𝑎_𝑛𝑎_𝑛+1)². (A.7)

Proposition A.3. If 𝑎₀=−𝑎_𝑛+𝑎_𝑛+1(𝜖−1), then 𝑎²_𝑛+1−𝑎²₀+𝑎₀𝑎_𝑛>0 if and only if 1+3𝑎_𝑛2𝑎_𝑛+1−11+4(𝑎_𝑛𝑎_𝑛+1)²<𝜖<1+3𝑎_𝑛2𝑎_𝑛+1+11+4(𝑎_𝑛𝑎_𝑛+1)². (A.8)

Proof. Replacing 𝑎₀ into the inequality 𝑎²_𝑛+1−𝑎²₀+𝑎₀𝑎_𝑛>0, we get 𝑎²_𝑛+1−[−𝑎_𝑛+𝑎_𝑛+1(𝜖−1)]²+[−𝑎_𝑛+𝑎_𝑛+1(𝜖−1)]𝑎_𝑛>0. (A.9) That is, −𝑎²_𝑛+1𝜖²+(2𝑎²_𝑛+1+3𝑎_𝑛𝑎_𝑛+1)𝜖−(2𝑎²_𝑛+3𝑎_𝑛𝑎_𝑛+1)>0. (A.10) Let ℎ(𝜖)=−𝑎²_𝑛+1𝜖²+(2𝑎²_𝑛+1+3𝑎_𝑛𝑎_𝑛+1)𝜖−(2𝑎²_𝑛+3𝑎_𝑛𝑎_𝑛+1). (A.11) Then ℎ(𝜖)=0⇔𝜖=1+3𝑎_𝑛2𝑎_𝑛+1±11+4(𝑎_𝑛𝑎_𝑛+1)². (A.12) Since the coefficient of 𝜖² is negative, ℎ(𝜖)>0⇔1+3𝑎_𝑛2𝑎_𝑛+1−11+4(𝑎_𝑛𝑎_𝑛+1)²<𝜖<1+3𝑎_𝑛2𝑎_𝑛+1+11+4(𝑎_𝑛𝑎_𝑛+1)². (A.13)

Proposition A.4. If 𝑛≥2 and −(𝑛+1)/𝑛<𝑎_𝑛/𝑎_𝑛+1<−1, then 3(𝑛+1)+2𝑛+13𝑛𝑎2𝑛+1𝑎^𝑛_𝑛+1<1+3𝑎_𝑛2𝑎_𝑛+1+11+4(𝑎_𝑛𝑎_𝑛+1)². (A.14)

Proof. If we let 𝑐=𝑎_𝑛/𝑎_𝑛+1, then the previous inequality becomes 3(𝑛+1)/(2𝑛+1)+(3𝑛/(2𝑛+1))𝑐<1+(3/2)𝑐+1+(1/4)𝑐², which is satisfied if and only if [(2𝑛+1)²−9]𝑐²+12(𝑛+2)𝑐−4[(𝑛+2)²−(2𝑛+1)²]>0. (A.15) But this is true because the discriminant −4𝑛⁴−4𝑛³+15𝑛²+16𝑛+4 of this quadratic function is negative for 𝑛≥3 and the coefficient of 𝑐² is positive.
If 𝑛=2, the assumption for 𝑎₂/𝑎₃ becomes −3/2<𝑎₂/𝑎₃<−1 and since 𝑐=𝑎₂/𝑎₃ we get that −3/2<𝑐, that is, 0<2𝑐+3 then (2𝑐+3)²>0. On the other hand, the left-hand side of inequality (A.15) becomes 16𝑐²+48𝑐+36=4(2𝑐+3)² which is positive and the proposition follows.

Proposition A.5. If −(𝑛+1)/𝑛<𝑎_𝑛/𝑎_𝑛+1<−1 and 𝑎₀=−𝑎_𝑛+𝑎_𝑛+1(𝜖−1), then 0<𝑎²_𝑛+1−𝑎²₀+𝑎²_𝑛+2𝑎_𝑛𝑎_𝑛+1−𝑎_𝑛𝑎_𝑛+1𝜖 in and only if 0<𝜖<2+𝑎_𝑛/𝑎_𝑛+1.

Proof. If 𝑎₀=−𝑎_𝑛+𝑎_𝑛+1(𝜖−1) is replaced in the first inequality, we obtain 0<𝑎²_𝑛+1−[−𝑎_𝑛+𝑎_𝑛+1(𝜖−1)]²+𝑎²_𝑛+2𝑎_𝑛𝑎_𝑛+1−𝑎_𝑛𝑎_𝑛+1𝜖⇔0<𝜖[−𝑎²_𝑛+1𝜖+2𝑎²_𝑛+1+𝑎_𝑛𝑎_𝑛+1]𝑎⇔0<𝜖<2+^𝑛𝑎_𝑛+1. (A.16) By hypothesis −(𝑛+1)/𝑛<𝑎_𝑛/𝑎_𝑛+1<−1 or equivalently (𝑛−1)/𝑛<2+𝑎_𝑛/𝑎_𝑛+1<1. Since (𝑛−1)/𝑛>0 for all 𝑛≥2,2+𝑎_𝑛/𝑎_𝑛+1>0. Therefore the first inequality is satisfied if and only if 0<𝜖<2+𝑎_𝑛/𝑎_𝑛+1.

Proposition A.6. If 𝑎₀=−𝑎_𝑛+𝑎_𝑛+1(𝜖−1), then 𝑎²_𝑛+1−𝑎²₀+𝑎²_𝑛+2𝑎_𝑛𝑎_𝑛+1−𝑎_𝑛𝑎_𝑛+1𝜖<3𝑛(𝑎²_𝑛+1−𝑎²₀)+2𝑛−13(𝑛−1)𝑎2𝑛−1^𝑛+1𝑎_𝑛⇔1+4𝑛+12(𝑛+1)𝑐−𝐻_𝑛(𝑐)<𝜖<1+4𝑛+12(𝑛+1)𝑐+𝐻_𝑛(𝑐), (A.17) where 𝑐=𝑎_𝑛/𝑎_𝑛+1 and 𝐻_𝑛(𝑐)=1+((𝑛−2)/(𝑛+1))𝑐+((𝑛−1/2)/(𝑛+1))²𝑐².

Proof. The inequality 𝑎²_𝑛+1−𝑎²₀+𝑎²_𝑛+2𝑎_𝑛𝑎_𝑛+1−𝑎_𝑛𝑎_𝑛+1𝜖<3𝑛(𝑎²_𝑛+1−𝑎²₀)+2𝑛−13(𝑛−1)𝑎2𝑛−1^𝑛+1𝑎_𝑛 (A.18) is equivalent to −𝑎_𝑛𝑎_𝑛+1𝜖<𝑛+12𝑛−1(𝑎²_𝑛+1−𝑎²₀)+−(𝑛+1)𝑎2𝑛−1^𝑛+1𝑎_𝑛−𝑎²_𝑛. (A.19) Replacing 𝑎₀ into the last inequality, we see that this is equivalent to say that 0<𝑓(𝜖), where 𝑓(𝜖)=−(𝑛+1)𝑎²_𝑛+1𝜖²+[2(𝑛+1)𝑎²_𝑛+1+(4𝑛+1)𝑎_𝑛𝑎_𝑛+1]𝜖−3𝑛𝑎²_𝑛−3(𝑛+1)𝑎_𝑛𝑎_𝑛+1. (A.20) Since 𝑓(𝜖)=0⇔𝜖=1+4𝑛+12(𝑛+1)𝑐±𝐻_𝑛(𝑐), (A.21) and the coefficient of 𝜖² is negative, it holds that 𝑓(𝜖)>0⇔1+4𝑛+12(𝑛+1)𝑐−𝐻_𝑛(𝑐)<𝜖<1+4𝑛+12(𝑛+1)𝑐+𝐻_𝑛(𝑐). (A.22)

Proposition A.7. Fix an arbitrary 𝑛∈𝑁. If −(𝑛+1)/𝑛<𝑐<−1, then 1+4𝑛+12(𝑛+1)𝑐−𝐻_𝑛(𝑐)<0∀𝑛≥1. (A.23)

Proof. We have that −(𝑛+1)/𝑛<𝑐≤−1 if and only if−(4𝑛+1)2𝑛+1<4𝑛+12(𝑛+1)𝑐+1≤−4𝑛+12(𝑛+1)+1=−2𝑛+12(𝑛+1)<0, (A.24)for all 𝑛≥1, and the result follows.

Proposition A.8. If −(𝑛+1)/𝑛<𝑐<−1, it holds that 1+4𝑛+12(𝑛+1)𝑐+𝐻_𝑛(𝑐)≤2+𝑐∀𝑛≥1. (A.25)

Proof. We have that 1+4𝑛+12(𝑛+1)𝑐+𝐻_𝑛(𝑐)≤2+𝑐 (A.26) if and only if 𝐻_𝑛(𝑐)≤1+[1−4𝑛+12(𝑛+1)]𝑐. (A.27)
From the definition of 𝐻_𝑛(𝑐) this is true if and only if 1+(𝑛−2)(𝑛+1)𝑐+(𝑛−1/2)𝑛+1²𝑐²≤1+(−2𝑛+1𝑛+1)𝑐+(−2𝑛+1)²4(𝑛+1)²𝑐²; (A.28) if and only if ((3𝑛−3)/(𝑛+1))𝑐≤0. Since 𝑐<0 and 3𝑛−3≥0∀𝑛≥1. Then the inequality is satisfied.

Proposition A.9. Let 𝐹(𝑐)=1+4𝑛+12(𝑛+1)𝑐+1+(𝑛−2)(𝑛+1)𝑐+(𝑛−1/2)𝑛+1²𝑐², (A.29) for 𝑐>−(𝑛+1)/𝑛. Then
(a) 𝐹^′(𝑐)>0; (b) 𝐹^′′(𝑐)>0 (𝐹 is convex); (c) 𝐹(−(𝑛+1)/𝑛)=0, and 𝐹^′(−(𝑛+1)/𝑛)=3𝑛/(2𝑛+1).

Proof. It is elementary.