1

Stabilization of networked control systems underclock offsets and quantization

Kunihisa Okano, Masashi Wakaiki, and Joao P. Hespanha

Abstract—This paper studies the impact of clock mismatchesand quantization on networked control systems. We consider ascenario where the plant’s state is measured by a sensor thatcommunicates with the controller through a network. Variablecommunication delays and clock jitter do not permit a perfectsynchronization between the clocks of the sensor and controller.We investigate limitations on the clock offset tolerable forstabilization of the feedback system. For the scalar plant case,we show that there exists a tight bound on the offset abovewhich the closed-loop system cannot be stabilized with any causalcontrollers. For higher dimensional plants, if the plant has twodistinct real poles, then the effect of clock mismatches can becanceled with a finite number of measurements, and hence thereis no fundamental limitation. We also consider the case wherethe measurements are subject to quantization in addition to clockmismatches. For scalar plants, we present necessary conditionsand sufficient conditions for stabilizability, which show that alarger clock offset requires a finer quantization.

I. INTRODUCTION

The components of networked control systems are oftenspatially distributed and hence need to communicate throughdigital networks. A significant challenge in such systemsis that the processors in the components may not share acommon notion of time, because their local clocks are notproperly synchronized; see survey papers [1], [2]. Quantizationis another significant challenge to the design of networkedcontrol systems, as surveyed in [3], [4] and motivated the studyof bit-rate constraints in channels [5] and limited capabilitiesof sensors/actuators [6]. In this paper, we consider clockmismatches and quantization simultaneously and study theimpact of such imperfections on the stability of feedbackcontrol systems.

With a growing demand for digital communication, therehas been a vast amount of research providing clock syn-chronization methodologies over networks; see, e.g., [7]–[9] and the survey papers [10], [11]. Moreover, the GlobalPositioning System (GPS) has allowed us to have access to

This material is based upon work supported by the National ScienceFoundation under Grant No. CNS-1329650.

K. Okano is with the Department of Intelligent Mechanical Systems,Graduate School of Natural Science and Technology, Okayama Univer-sity, Okayama, 700-8530, Japan. M. Wakaiki is with the Departmentof Electrical and Electronic Engineering, Chiba University, Chiba, 263-8522, Japan. J. P. Hespanha is with Center for Control, Dynamical-systems, and Computation (CCDC), University of California, Santa Bar-bara, CA, 93106-9560 USA. E-mails: kokano@okayama-u.ac.jp;wakaiki@chiba-u.jp; hespanha@ece.ucsb.edu.

The work of K. Okano was supported by JSPS Postdoctoral Fellowshipsfor Research Abroad and by JSPS KAKENHI Grant Number JP16H07234.M. Wakaiki acknowledges The Kyoto University Foundation for the supportof this work.

a global clock that has been used in many practical systems.However, we cannot attain perfect synchronization in mostpractical situations. In fact, the work [12] has shown thatsynchronization of clocks with unknown skews and offsets isimpossible in the presence of unknown communication delays.Furthermore, recent works [13], [14] have pointed out thatGPS signals can be spoofed and hence are vulnerable againstattacks.

In this paper, we consider feedback systems where the sen-sor measuring the plant state and the controller do not share acommon clock. This asynchronism results in measurement thatcan cause fundamental limitations in our ability to stabilize asystem through feedback. We consider the situation in whichthe clock offset between the sensor and the controller is knownto be bounded, but its precise value is unknown and potentiallytime varying. Our objective is to clarify limitations on thetolerable clock offsets for stability. We first consider the casein which the communication channel has unlimited bandwidthand thus can convey real-valued state measurements. In suchconfiguration, for the case of a scalar plant, we derive anecessary and sufficient condition for stabilizability, whichgives a tight upper bound on the clock offset beyond whichstabilization is impossible. In contrast, for the case of plantswith dimension two or larger, there exists no fundamentallimitation on the offset if the plant has at least two distinctreal modes.

We also consider the scenario where the sensor measure-ments must be quantized to discrete values. We considerquantizers that are static piecewise constant functions, whichmap the observed state to a discrete set of values. We focusour attention on two types of quantizers that have been widelyconsidered in the literature: logarithmic quantizers [15]–[17]and uniform quantizers [18]–[20]. It has been shown that, byitself, quantization imposes fundamental limitations on stabi-lizability [15], [20], [21]. Here, we introduce the additionalcomplexity of clock mismatch and derive necessary conditionsand sufficient conditions on the fineness of the quantizer andthe maximum clock offset for stabilizability. The conditionsprovide explicit relationships between the tolerable clockoffsets and the required quantization levels for stabilizability.

Systems with uncertain sampling intervals have been studiedfrom the perspective of robust control; see, e.g., [22]–[27].These papers restrict their attention to linear time-invariantcontrollers whose inputs are updated at uncertain time instantsbecause of the variable sampling period. For the resultingfeedback systems, sufficient conditions for stability have beenderived via linear matrix inequalities. In recent works, weemployed a time-stamp aware estimator and investigated the

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Plant

Controller

Sensor

u(t)

x(t)

yk=x(kT+�k)

Fig. 1: Feedback system.

stabilization problem and stability analysis of closed-loopsystems under clock offsets [28], [29]. Now we considergeneral causal controllers, and present necessity results ratherthan only sufficiency ones. These results thus characterizefundamental requirements on the accuracy of local clocks andthe fineness of quantization for stabilizability.

This paper is organized as follows. We formally state theproblem in Section II and introduce the notion of tight stateestimation sets in Section III. In Section IV, we consider thecase where there is no quantization and present conditions forstabilizability under clock offsets. The setup is extended to thecase of quantized measurements in Section V, and stabilizationunder both clock offsets and quantization is considered inSection VI. Finally, we provide concluding remarks in Sec-tion VII. A subset of the results in Section IV appeared in theconference paper [30].

Notations: Z+ denotes the set of nonnegative integers; 0(bold-faced zero) stands for the zero vector of appropriatedimension; and ∥ · ∥ represents the Euclidean norm.

II. PROBLEM FORMULATION

Consider the feedback system depicted in Fig. 1. The plantto be controlled is the following continuous-time linear time-invariant process:

x(t) = Ax(t) +Bu(t), (1)

where x(t) ∈ Rn is the state and u(t) ∈ Rm is the input.The unknown initial state x(0) may be anywhere in Rn. Weintroduce the following assumption on the plant.

Assumption 1: The system (1) is unstable and controllable,and A is invertible.

It is common in the literature (e.g., [20], [21]) to considerunstable and controllable systems to make the stabilizationproblem non-trivial, as any component of the state that belongsto a stable invariant subspace of A will converge to zero,provided that the control signal converges to zero. The non-singularity of A is only assumed to simplify the constructionof the state estimation sets in Section III.

The sensor attempts to sample the state periodically withperiod T , but does not necessarily achieve it due to clockerrors. Let yk, k ∈ Z+, be the kth observed state, and letT > 0 be the desired sampling period from the perspectiveof the controller clock. Fig. 2 illustrates a time chart ofsampling: The actual time instants kT + δk at which thesamples yk are generated are represented by the crosses,whereas the gray boxes represent the uncertainty δk inherentto the timing information due to the offset between the clocks

TimekT (k+1)T

�k

yk yk+1

��

�k+1

must remain constant (per Assumption 3)u(t)

Fig. 2: Time chart.

of the sensor and the controller. We consider the case in whichthe magnitude of the clock offset is bounded.

Assumption 2: For all time k ∈ Z+, δk is bounded as

0 ≤ δk ≤ ∆ < T (2)

with a constant ∆.There are various algorithms for the clock offset estimation,

and the resulting estimation errors have been analyzed; see,e.g., [31]. These results can be used to determine the offsetbound ∆. Under Assumption 2, the sequence yk of observedstates is produced by the following model:

yk = x(kT + δk), δk ∈ [0,∆], (3)

where the actual sampling instant kT + δk is unknown be-cause of the unknown jitter δk. We note that the controllercannot determine the precise value of δk based on the timeinstant at which the sample yk arrives because of unknowncommunication delay.

Using the received observations y0, . . . , yk, the controllerneeds to construct the input u(t), t ≥ 0. We consider thefollowing assumption.

Assumption 3: The control input is constant during [kT,kT +∆] for every k:

u(t) = uk, ∀t ∈ [kT, kT +∆], ∀k ∈ Z+,

and u(t) does not depend on the future measurements {yk :kT +∆ ≥ t}.This assumption guarantees causality, but it also implies thatduring the intervals in which the sensor may measure the plantstate, the controller holds the input. During the period t ∈(kT + ∆, (k + 1)T ) for which the state has been measured,we allow the controller to take any value.

Our objective is to clarify how large ∆ can be to ensure thestabilizability of the system.

Remark 1: Assumption 3 is required to prove the necessityresults in the paper. We note that the sufficient conditions forstabilizability that we present in Sections IV and VI hold evenif we limit the class of controllers to piecewise constant ones,for which Assumption 3 trivially holds.

III. COMPUTATION OF ESTIMATION SETS

In Sections III and IV, we consider a feedback systemwith no quantization and study the effect of clock mismatchesbetween the sensor and the controller on the stabilizability ofthe system.

Here, we introduce the notion of a tight state estimation set,which will be instrumental to establish the necessity results inthe paper. Let I−k denote the set of plant states x(kT ) that are

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compatible with all the measurements y0, y1, . . . , yk−1 takenbefore time kT . We shall call this the estimation set of x(kT ).Given I−k , we now study how to update the estimation set ofx(kT ) using the new observation yk. From (3), yk is given by

yk = eAδkx(kT ) +

∫ kT+δk

kT

eA(kT+δk−τ)Bu(τ)dτ, (4)

which depends on the unknown parameter δk ∈ [0,∆] and thetrue state x(kT ). Thus, once yk is known, x(kT ) must lie inthe following set Jk:

Jk =

{e−Aδkyk −

∫ kT+δk

kT

eA(kT−τ)Bu(τ)dτ : δk ∈ [0,∆]

}(5)

={

e−Aδk(yk +A−1Buk

)−A−1Buk : δk ∈ [0,∆]

}.

(6)

Here, the second equality follows by Assumptions 1 and 3.With this set Jk, after receiving yk, we obtain an updatedestimation set Ik that now also considers the measurement ykand can be defined by

Ik := I−k ∩ Jk. (7)

The estimation set I−k+1 at the next time x((k + 1)T ) isconstructed from Ik using the model (1) and is given by

I−k+1 =

{eATxk +

∫ (k+1)T

kT

eA((k+1)T−τ)Bu(τ)dτ : xk ∈ Ik

}.

(8)

Since I−0 = Rn, (7) leads to I0 = J0 and, because of (8),the initial estimation set of x(T ) is

I−1 ={

e−Aδ0a0 + b0 : δ0 ∈ [0,∆]}, (9)

where

a0 := eAT (y0 +A−1Bu0),

b0 := −eATA−1Bu0 +

∫ T

0

eA(T−τ)Bu(τ)dτ.

After receiving y1, we obtain the new estimation set

J1 ={

e−Aδ1a1 + b1 : δ1 ∈ [0,∆]}, (10)

where a1 := y1 +A−1Bu1 and b1 := −A−1Bu1.The set Ik defined in (7) is a tight estimation set of x(kT ) in

the following sense: For every xk ∈ Ik, there exists a trajectoryfor the state x(t), t ∈ [0, kT +∆], that produces y0, y1, . . . , ykwith possible delays {δi ∈ [0,∆]}ki=0 and satisfies x(kT ) =xk. We formally define tightness of an estimation set below.

Definition 1: An estimation set Ik ⊂ Rn of x(kT ), which isconstructed from I−0 , u(t), t ∈ [0, kT+∆], and y0, y1, . . . , yk,is said to be tight if for every xk ∈ Ik, there exist x0 ∈ I−0and {δi ∈ [0,∆]}ki=0, such that

eAkTx0 +

∫ kT

0

eA(kT−τ)Bu(τ)dτ = xk, (11)

eA(iT+δi)x0 +

∫ iT+δi

0

eA(iT+δi−τ)Bu(τ)dτ = yi,

∀i ∈ {0, 1, . . . , k}. (12)

The following result is a direct consequence of the construc-tion outlined above for estimation sets.

Proposition 1: The estimation set Ik defined in (7) is tightin the sense of Definition 1.

IV. STABILIZABILITY WITH INFINITE BIT-RATE

We start with the case of a scalar plant, for which we derivea tight bound on the maximum offset ∆ for stabilizability. Wethen show that for the vector plants case, clock offsets do notnecessarily limit stabilizability.

We employ the following stabilizability definition.Definition 2: The plant (1) is stabilizable if for any sequence

of offsets {δk}∞k=0, δk ∈ [0,∆], and any initial state x(0) ∈I−0 , there exists a feedback controller such that the closed-loopstate converges to the origin, i.e., limt→∞ x(t) = 0.

A. Single eigenvalue case

Consider the following controllable, unstable scalar system:

x(t) = λx(t) + bu(t). (13)

From controllability in Assumption 1, we assume without lossof generality that b = 1 and, from instability, that λ > 0.

Theorem 1: Let Assumptions 1–3 hold. Then the the scalarplant (13) is stabilizable in the sense of Definition 2 if andonly if the bound on the offset satisfies 0 ≤ ∆ < ∆, where

∆ :=

{T if 0 < T ≤ 1

λ ln 3,

T − 1λ ln(eλT − 2) if T > 1

λ ln 3.(14)

The bound T − {ln(eλT − 2)}/λ in the lower branch of(14) is monotonically decreasing with respect to λ and T . Thisobservation is consistent with the intuition that a larger growthrate eλT during one sampling period will result in a tighterrequirement on the clock accuracy. The limitation ∆ = T forthe case 0 < T < (ln 3)/λ arises directly from Assumption2. Fig. 3 shows the value of ∆ versus the sampling period Tfor the plant (13) with λ = 1.0.

Our approach to prove Theorem 1 is based on the analysis ofthe estimation sets introduced in Section III. For scalar plants,by (6), the estimation set Jk computed from yk can be writtenas

Jk ={

e−λδk(yk + λ−1uk

)− λ−1uk : δk ∈ [0,∆]

}. (15)

Thus, Jk becomes a bounded interval in R. Therefore, with (7)and (8), Ik and I−k+1, k ∈ Z+, also result in bounded intervalsin R. Notice that we have assumed I−0 = R.

Let lk be the length of the interval Ik, which is the updatedestimation set of x(kT ) defined in (7). In the following, westudy whether or not lk → 0 from which we will be able toconclude stabilizability. Suppose that Ik−1 is obtained basedon y0, y1, . . . , yk−1 at t = (k − 1)T + ∆. Then, one cancompute I−k from (8). The length lk is determined by I−kand Jk, thus lk depends on the true state xk ∈ I−k , the jitterδk ∈ [0,∆], and the input u(t), t ∈ ((k − 1)T + ∆, kT ). To

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0 1 2 30

1

2

3

Nominal sampling interval T

Upper

boundon∆

(ln 3)/λ(ln 2)/λ

Fig. 3: Upper bound on ∆ versus sampling period T (λ = 1.0):The solid line is the active bound ∆ and the vertical dottedlines indicate T = (ln 2)/λ (left) and T = (ln 3)/λ (right),respectively.

prove that we are able to stabilize the closed loop for all pos-sible xk and δk, we have to consider the worst case trajectory.Hence, we need to determine under which conditions we have

minu(t),t∈((k−1)T+∆,kT )

maxxk∈I−

k ,δk∈[0,∆]lk → 0 as k → ∞,

from which we would conclude that no controller could resultin a decrease of the interval Ik to a single point and thereforethat there are trajectories for which xk does not converge. Thefollowing lemma is the key technical result needed to proveTheorem 1 as it gives the expansion rate of lk.

Lemma 1: Given an estimation set Ik−1 with width lk−1,for every k ≥ 1, it follows that

minu(t),t∈((k−1)T+∆,kT )

maxxk∈I−

k ,δk∈[0,∆]lk =

(1− e−λ∆)eλT

2lk−1,

(16)

where the minimum is achieved with any input that placesthe interval

{xk + λ−1uk : xk ∈ I−k

}symmetrically about the

origin.Proof: For simplicity of notation, we first introduce biased

sets as follows: Define yk := yk + λ−1uk and let Jk := {x+λ−1uk : x ∈ Jk}. Then, from (15), this biased estimation setJk can be simply represented as

Jk =

[e−λ∆yk, yk

]if yk > 0,

{0} if yk = 0,[yk, e−λ∆yk

]if yk < 0.

(17)

Similarly, we define the biased sets

I−k :={xk + λ−1uk : xk ∈ I−k

},

Ik := I−k ∩ Jk ={xk + λ−1uk : xk ∈ Ik

}.

Note that the length lk of Ik is the same as that of Ik and theanalysis in this proof holds for any bias λ−1uk. In Fig. 4, weillustrate the update procedure for the estimation set.

State

Time

(a) (b) (c)

Fig. 4: Update procedure of the estimation set: (a) Theinitial estimation set I−k expands during the sensing interval[kT, kT +∆]; (b) After receiving yk, the controller computesthe set Jk of the candidates of the state at kT which may resultin yk somewhere in [kT, kT +∆]; (c) The updated estimationset Ik is obtained as the intersection of I−k and Jk.

We now show that the input u(t), t ∈ ((k− 1)T +∆, kT ),achieving the minimum in (16) is the one that makes I−ksymmetric about the origin. In what follows, we evaluate theworst case length maxxk,δk lk for a given I−k . We start bydescribing the range of yk under the knowledge that xk ∈ I−kand δk ∈ [0,∆]. With (4), the set of all possible yk is asfollows:

yk ∈{

eλδkxk + λ−1(eλδk − 1)uk : xk ∈ I−k , δk ∈ [0,∆]},

which results in

yk ∈{

eλδkz : δk ∈ [0,∆], z ∈ I−k

}. (18)

Thus, we can rewrite the maximization term in (16) as

maxxk∈I−

k ,δk∈[0,∆]lk = max

yk∈{eλδkz:δk∈[0,∆],z∈I−k }

lk.

Next, we evaluate maxyklk for each variation of I−k , and

then compute its minimum over u(t), t ∈ ((k−1)T +∆, kT ).In the rest of the proof, we assume yk = 0 since it followsfrom (17) that yk = 0 does not maximize lk. Denote the lowerbound and the upper bound of I−k by p and q, respectively:I−k = [p, q]. Suppose that

|p| ≤ |q|, (19)

i.e., the midpoint of I−k is nonnegative. For the case |p| > |q|,one can apply the same discussion below by flipping signs.

Consider the following two cases.(i) p ≤ 0: In this case, it follows that q ≥ 0 from (19) and

thus, I−k contains the origin. By (17), we have

Ik =

{[e−λ∆yk,min(q, yk)

]if yk > 0,[

max(p, yk), e−λ∆yk]

if yk < 0.

Taking maximums in both cases yk > 0 and yk < 0, and using(19), one can see that the length lk is maximum when yk = q

maxyk∈{eλδkz:δk∈[0,∆],z∈I−

k }lk = (1− e−λ∆)q.

The right-hand side takes its minimum under (19) and (i) forthe case in which q = −p = eλT lk−1/2, and the minimum is

minu(t),t∈((k−1)T+∆,kT )

maxxk∈I−

k ,δk∈[0,∆]lk=(1− e−λ∆)

eλT

2lk−1.

(20)

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(ii) p > 0: In this case, noticing that all possible yk arepositive from (18), we obtain

Ik =[max(p, e−λ∆yk),min(q, yk)

].

Therefore,

maxyk∈{eλδkz:δk∈[0,∆],z∈I−

k }lk =

{(1− e−λ∆)q if e−λ∆q ≥ p,

q − p if e−λ∆q < p.

(21)

From the condition (ii) and (8), we have q > q−p = eλT lk−1.Thus, both branches in (21) are greater than the right-hand sideof (20). Hence, I−k satisfying (ii) cannot minimize maxyk

lk,which concludes the proof of Lemma 1.

Proof of Theorem 1: (Sufficiency) If ∆ < ∆, then wehave that γ := (1− e−λ∆)eλT /2 < 1. Thus, from Lemma 1,for a given estimation set Ik−1 of x((k − 1)T ), there existsa control input resulting in lk < γlk−1 for all δk and xk−1 ∈Ik−1. Such control input can be constructed as follows: Dividethe interval ((k−1)T +∆, kT ) into two parts of equal length.In the first time period, the controller applies a constant inputwhich makes I−k symmetric about λ−1uk. As we see in theproof of Lemma 1, this control input minimizes maxxk,δk lk,which is the length of the estimation set Ik in the worst case.In the second time period, the control input is simply set to0. This control input satisfies Assumption 3 and simplifies thecalculation of the estimation set Jk given by (6). Repeatingthis procedure for each sampling period, we can make thesequence lk converge to 0, which implies the stability of thefeedback system.

(Necessity) When γ := (1−e−λ∆)eλT /2 ≥ 1, the sequencelk does not converge to zero, no matter how we choose u(t).This means that there always exist a positive constant ϵ and asequence of integers k1, k2, . . . such that lki ≥ ϵ, ∀i. Since Ikis a tight estimation set, this means that there exists a possibletrajectory of x(t) such that |x(tki)| ≥ ϵ/2, ∀i and thereforex(t) does not converge to zero.

We now present a numerical example to illustrate thebehavior of the system.

Example 1: Consider the system (13) with λ = 1.0 and thenominal sampling period T = 2.0. The initial state is takento be x(0) = 1 and the δk are chosen uniformly randomlyfrom [0,∆]. We employ the control law mentioned in theproof of Theorem 1, which minimizes the length of the worstcase estimation set. We first consider the case of a smalloffset bound that satisfies the stability condition in Theorem1: suppose ∆ = 0.30, which is less than ∆ ≈ 0.32. In Fig. 5a,we illustrate the state x(t) and the input u(t) by the solid lineand the dashed line, respectively. Moreover, the observationsyk are depicted by the cross marks and the upper and lowerbounds of the estimation sets Ik are represented by △ and▽, respectively. One can observe that, as time progresses, theupper bounds and the lower bounds of Ik become closer, andthe state successfully converges to 0. Next, we consider thecase ∆ = 0.45, which violates the stability limit ∆ ≈ 0.32.Fig. 5b shows the behavior of the system in this case. We seethat the state (solid line) oscillates and the estimation set Ik(triangle marks) does not shrink to a single point.

0 4 8 12 16 20−3

−2

−1

0

1

2

3

t

x(t),u(t)

x(t)u(t)

(a) ∆ = 0.30 < ∆ ≈ 0.32

0 4 8 12 16 20−3

−2

−1

0

1

2

3

tx(t),u(t)

x(t)u(t)

(b) ∆ = 0.45 > ∆ ≈ 0.32

Fig. 5: Behavior of the system (λ = 1.0, T = 2.0, ∆ ≈0.32): The plant state x(t) (solid), the input u(t) (dashed), theobservations yk (cross), and the estimation sets Ik (△: upperbounds, ▽: lower bounds).

B. Multiple eigenvalues case

In this section, we consider plants with multiple eigenvaluesand study the maximum offset tolerable to achieve stability.

Theorem 2: Let Assumptions 1–3 hold. If A has at leasttwo distinct real eigenvalues, then for any ∆ satisfying (2),the plant (1) is stabilizable in the sense of Definition 2 andthe state can be taken to the origin in a finite time.

Theorem 2 follows from the lemma below and the control-lability of the system.

Lemma 2: If A has at least two distinct real eigenvalues, thenfor any initial state and for any ∆ satisfying (2), there existsa control input such that the state of (1) can be reconstructedprecisely with a finite number of measurements.

In view of Theorem 2, and unlike in the scalar case,there is no practical limitation on the clock offset if theplant dynamics has two distinct real eigenvalues. Intuitively,we can explain this result as follows: At the time afterreceiving the measurement yk = x(kT+δk), the controller hasn+k+1 unknown variables, i.e., x(0) and δ0, . . . , δk. On theother hand, the measurements y0, y1, . . . , yk provide 2(k+1)equations that constrain these unknowns. If these equationswere linear and independent, then one could determine theexact state in k steps as soon as 2(k+1) ≥ n+k+1. In realitythey are not, but when A has two distinct real eigenvalues,using an appropriate input signal u(t) they provide enough“independence” to recover the initial state. Moreover, since

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the plant is controllable, once its state is precisely known,there exists an input that drives the state to the origin in afinite time interval.

The proof of Lemma 2 consists of two steps. We firstconsider a subsystem of (1) associated with the two distinctreal eigenvalues. Since there exist eigenvalues λ1, λ2 ∈ Rof A such that λ1 = λ2, via an appropriate coordinatetransformation, we have

V −1x(t) =

λ1λ2

∗

V −1x(t) + V −1Bu(t),

for some nonsingular matrix V ∈ Rn×n. Extracting the firsttwo rows of the above system, we obtain the subsystem

ξ(t) = Λξ(t) + ν(t), ζk = ξ(kT + δk), (22)

where Λ := diag(λ1, λ2), ξ and ν are the corresponding statesand the inputs, respectively, and ζk is the output, i.e., the firsttwo elements of V −1yk.

For the first step of the proof, we compute estimation setsfor the state ξ(kT ) based on the measurements y0, . . . , yk.Instead of the estimation sets of x represented by Jk, Ik, andI−k+1, which are defined in (6)–(8), we denote the correspond-ing estimation sets of ξ by Jk, Ik, and I−k+1. In light of (9),the estimation set I−1 of ξ(T ) is expressed as

I−1 ={

e−Λδ0 a0 + b0 : δ0 ∈ [0,∆]}.

Here, a0 and b0 are defined as follows: Let νk be the first twoelements of V −1Buk. Then, a0 and b0 are defined as

a0 := eΛT (ζ0 + Λ−1ν0),

b0 := −eΛTΛ−1ν0 +

∫ T

0

eΛ(T−τ)ν(τ)dτ. (23)

Note that Λ is nonsingular since A is invertible. Moreover, by(10), it follows that

J1 ={

e−Λδ1 a1 + b1 : δ1 ∈ [0,∆]},

where

a1 := ζ1 + Λ−1ν1, b1 := −Λ−1ν1. (24)

We claim that the intersection of I−1 and J1, which is theupdated estimation set I1, is a set containing only a finitenumber of points. This claim is formally stated in the followinglemma.

Lemma 3: There exists a control input ν(t), t ∈ [0, T +∆]that makes the updated estimation set I1 = I−1 ∩ J1 contain,at most, a finite number of points.

To prove this lemma, we need to consider the followingproblem: Let a, b, and c be nonzero vectors in R2; alsofix [T1, T2] ⊂ R and [T3, T4] ⊂ R. Define functions f :[T1, T2] → R2 and g : [T3, T4] → R2 as

f(t) := eΛta, g(t) := eΛtb+ c (25)

and find the intersection of {f(t) : t ∈ [T1, T2]} and{g(t) : t ∈ [T3, T4]}. The following lemma states that this

intersection can be a set of finite points if a, b, and c arechosen appropriately.

Lemma 4: If at least one of the vectors a, b, and c is notan eigenvector of Λ, then the intersection of the sets F :={f(t) : t ∈ [T1, T2]} and G := {g(t) : t ∈ [T3, T4]} contains,at most, a finite number of points.

Proof: See Section IV-C.We now prove Lemma 3 using Lemma 4.

Proof of Lemma 3: Consider the statement of Lemma 4with a = a0, b = a1, and c = b0 − b1. Then, I1 correspondsto the intersection of F and G. Note that a0 and a1 can beassumed to be nonzero since if either a0 = 0 or a1 = 0, thenI−1 or J1 becomes a single point, and the result is triviallytrue.

In what follows, we show that there exists an input suchthat b0− b1 is not equal to 0 nor is an eigenvector of Λ. FromAssumption 3, ν(t) takes a constant value ν0 for t ∈ [0,∆],thus it follows from (23) and (24) that

b0 − b1 = −Λ−1eΛ(T−∆)ν0 +

∫ T

∆

eΛ(T−τ)ν(τ)dτ + Λ−1ν1.

Note that the first and the third terms in the right-hand side areknown to the controller. Furthermore, the integral term is thezero-state-response of (22). Because of the controllability ofthe original system (1), x(kT ) or ξ(kT ) can be set arbitrarily,and hence b0− b1 can be made an arbitrary vector by selectingan appropriate input ν(t), t ∈ (∆, T ). Thus, with an input ν(t)such that b0 − b1 is not 0 nor an eigenvector of Λ, it followsfrom Lemma 4 that I1 becomes a set of finite points.

The second step, which will complete the proof of Lemma2, consists of providing a procedure to determine the value ofthe state once I1 contains only a finite number of points.

Proof of Lemma 2: Consider the subsystem (22) of (1).From Lemma 3, we can pick the control signal so that I1 is aset with a finite number of points. For each element of I1, wecan determine corresponding δ1 and x(2T ) as follows. Pick apoint ξ1 ∈ I1. From (22), we have that

ζ1 = eΛδ1ξ1 +

∫ δ1

0

eΛ(δ1−τ)ν(τ + T )dτ

= eΛδ1(ξ1 + Λ−1ν1

)− Λ−1ν1.

Thus, if ξ1 = −Λ−1ν1, δ1 is uniquely determined fromvariables known to the controller. When ξ1 = −Λ−1ν1, weselect the input so that ξ2 := eΛT ξ1 +

∫ T

0eΛ(T−τ)ν(τ +T )dτ

satisfies ξ2 = −Λ−1ν2, which is always possible in view ofcontrollability. In this case, the set I2 would continue to befinite, and we would be able to determine the correspondingδ2 using a similar procedure. For simplicity, in the remainderof the proof we shall assume that we have determined δ1, butthe reasoning would be similar if we had δ2 instead. For thevalue of δ1 so obtained, we can compute x(T ) by solving

y1 = eATx(T ) +

∫ δ1

0

eA(δ1−τ)Bu(τ + T )dτ

and can compute x(2T ) using the variation of constantsformula. From controllability, the controller can bring thispoint x(2T ) to the origin by an appropriate open-loop input

7

−70 −60 −50 −40 −30 −20 −10 00.5

1

1.5

2

2.5

3

3.5

4

4.5

5

x(T )

x(0)

y1

J1I−1

I0

y0

Fig. 6: Estimation sets I0, I−1 , and J1 in the state spaceR2 (horizontal axis: first element of x, vertical axis: secondelement of x).

u(t) for t ∈ (2T +∆, 3T ). The next output y3 takes a nonzerovalue only if the selected point ξ1 that we used to estimatex(2T ) does not correspond to the true state trajectory. Hence,we reduce the number of elements of Ik at least by one elementin one sampling interval (or two periods, in case we had toconsider δ2 ). Repeating this procedure, we eventually obtainyk = 0, which means that the origin has been reached.

We now show an example where I1 becomes a single point.In this example, we can make the state converge to zero usingonly two observations y0 and y1.

Example 2: Consider the following system:

A =

[2 00 1

], B =

[11

], T = 1.5, ∆ = 1.0.

Notice that ∆ is greater than the bound (14) in the scalarcase for each eigenvalue, 2 and 1, of A since T −{ln(e2T − 2)

}/2 ≈ 0.052 and T − ln(eT − 2) ≈ 0.59. In

Fig. 6, we plot the estimation sets I0, I−1 , and J1 computedbased on y0 and y1 in the state space R2. Since we assumeI−0 = R2, I0 is equal to J0. Note that the unknown statesx(0) and x(T ) are not used to compute the estimation sets.The figure shows that I−1 ∩ J1 = I1, which contains a singlepossible value for x(T ).

C. Proof of Lemma 4

Let us define h : [T1, T2]× [T3, T4] → R2 as

h(t, s) := f(t)− g(s) = aeΛt − beΛs − c, (26)

where f(·) and g(·) are defined in (25). If F ∩G contains in-finitely many points, this function h has infinitely many zeros,i.e., there exists a sequence {(tk, sk) ∈ [T1, T2]× [T3, T4]}∞k=1

such that (tk, sk) = (tk′ , sk′) if k = k′ and h(tk, sk) = 0, ∀k.Since [T1, T2] × [T3, T4] is closed and bounded, there must

exist a subsequence {(tk(i), sk(i))}∞i=1 of {(tk, sk)}∞k=1 and apoint (t0, s0) ∈ [T1, T2]× [T3, T4] such that h(tk(i), sk(i)) = 0, ∀i ∈ {0, 1, . . . },

(tk(i), sk(i)) = (tk(j), sk(j)) if i = j,(tk(i), sk(i)) → (t0, s0) as i→ ∞.

(27)

Without loss of generality, we assume that {(tk(i), sk(i))}∞i=1

equals {(tk, sk)}∞k=1. In what follows, we show that if at leastone of the vectors a, b, and c is not an eigenvector of Λ, thenthere is no such sequence and point, and thus F ∩G is a finiteset.

The rest of the proof is separated into two lemmas. First, inthe following lemma we consider a zero (t0, s0) of h at whichthe direction vectors df/dt and dg/ds are not parallel, andprove that there is no sequence {(tk, sk)}∞k=1 satisfying (27)that converges to such a zero. Then, Lemma 6, we considera zero (t0, s0) at which the directions are parallel and showthat, also in this case, there is no sequence satisfying (27).

Lemma 5: Consider h defined in (26) with nonzero vectorsa, b, and a zero (t0, s0) ∈ [T1, T2]× [T3, T4] of h: h(t0, s0) =0. Suppose that

df(t0)dt

= αdg(s0)

dt. ∀α ∈ R. (28)

Then, there exists a neighbor Bϵ of (t0, s0) such that h(t, s) =0, ∀(t, s) ∈ Bϵ\ {(t0, s0)}.

Proof: We start by rewriting h(t, s) as follows:

h(t, s) = (t− t0)h1(t)− (s− s0)h2(s), (29)

where h1 and h2 are given by

h1(t) :=

{eΛ(t−t0)−I

t−t0eΛt0a if t = t0,

ΛeΛt0a if t = t0,

h2(s) :=

{eΛ(s−s0)−I

s−s0eΛs0b if s = s0,

ΛeΛs0b if s = s0.

Note that h1 and h2 are continuous. Moreover, from (28), weconclude that γ := infα∈R ∥h1(t0)− αh2(s0)∥ > 0.

We show that the vectors h1(t) and h2(s) are linearlyindependent in a neighborhood of (t0, s0). Pick some ϵ > 0and consider the ball Bϵ := {(t, s) : ∥(t, s) − (t0, s0)∥ < ϵ}.Suppose for contradiction that there exist (t, s) ∈ Bϵ andα ∈ R such that

h1(t) = αh2(s). (30)

Let l1(t) := h1(t0) − h1(t) and l2(s) := h2(s0) − h2(s).Since h1 and h2 are continuous, we have that limt→t0 l1(t) =lims→s0 l2(s) = 0. For these vectors, it follows that

∥l1(t)− αl2(s)∥ ≤ ∥l1(t)∥+ |α| ∥l2(s)∥.

Here, since ∥h1(t)∥ and ∥h2(s)∥ are bounded by positiveconstants respectively from above and below on Bϵ, by (30),|α| is bounded from above. Thus, for a sufficiently small ϵ,the right-hand side of the above inequality can be arbitrarilysmall. However, with (30), it follows that

∥l1(t)− αl2(s)∥ = ∥h1(t0)− αh2(s0)∥ (31)

8

and the right-hand side of (31) is bounded by γ as

∥h1(t0)− αh2(s0)∥ ≥ γ.

This results in a contradiction for a sufficiently small ϵ,from which we conclude that h1(t) and h2(s) are linearlyindependent in a sufficiently small ball Bϵ. Recalling (29), wehave that h(t, s) = 0 on Bϵ only at (t, s) = (t0, s0).

Now we consider zeros at which the direction vectors areparallel, i.e., F is tangent to G. The following lemma showsthat if we choose a, b, or c not to be eigenvectors of Λ, thenthe zeros must be isolated.

Lemma 6: Consider h defined in (26) with nonzero vectorsa, b, and c. Suppose that there exist a sequence {(tk, sk)}∞k=1

satisfying (27) and a scalar α ∈ R for which

df(t0)dt

= αdg(s0)

dt. (32)

Then, the vectors a, b, and c must all be eigenvectors of Λ.Proof: Let uk := tk− t0 and vk := sk− s0. By (27), and

since a and b are nonzero vectors, we have that limk→∞ uk =limk→∞ vk = 0, uk = 0, vk = 0, and

f(t0) = g(s0), f(t0 + uk) = g(s0 + vk) (33)

for every k ≥ 1. From (25) and (32), both sides of the aboveequality can be respectively written as

f(t0 + uk) = f(t0) + (eΛuk − I)eΛt0a

= f(t0) + α(eΛuk − I)eΛs0b,

g(s0 + vk) = g(s0) + (eΛvk − I)eΛs0b.

Thus, with (33), we obtain

α(eΛuk − I)b = (eΛvk − I)b. (34)

Now, let us assume by contradiction that b is not an eigen-vector of Λ = diag(λ1, λ2), i.e., b ∈

{[b1 0]⊤ : b1 = 0

}∪{

[0 b2]⊤ : b2 = 0

}. Since λ1 = λ2, all elements of b are

nonzero. Thus, (34) implies that α(eλiuk − 1) = eλivk − 1holds for i ∈ {1, 2}, and hence it follows that

vk = vλ1(uk) = vλ2(uk), ∀k ≥ 1, (35)

where vλ(u) is give by

vλ(u) :=1

λln{α(eλu − 1) + 1

}.

Below, we show that there is no sequence {uk}∞k=1 satisfyinglimk→∞ uk = 0,

uk = 0,

(35).(36)

Taking derivative of vλ(u), we have

dvλdu

=αeλu

α(eλu − 1) + 1,

d2vλdu2

=αλeλu(1− α)

{α(eλu − 1) + 1}2.

Thus, by Taylor’s theorem, vλ(u) can be written as

vλ(u) = αu+1

2α(1− α)λu2 + o(u2),

Plant

Controller

Sensor

u(t)

x(t)

yk=x(kT+�k)

Quantizer Yk

Fig. 7: Feedback system with quantizer.

where ψ(x) = o(ϕ(x)) implies that ψ(x)/ϕ(x) → 0 as x→ 0.Let us define d(u) as

d(u) := vλ1(u)− vλ2(u)

=1

2α(1− α)(λ1 − λ2)u

2 + o(u2).

We show that d(u) has an isolated zero at u = 0. To this end,define d(u) as

d(u) :=

{u−2d(u) if u = 0,12α(1− α)(λ1 − λ2) if u = 0.

Then, d(u) is continuous at u = 0. Recall that λ1 = λ2.Moreover, we have α = 1 since it follows from (25), (32),and (33) that

c = (α− 1)eΛs0b, (37)

and c is nonzero. Thus, we have that d(0) = 0. Therefore, fromcontinuity of d(u), there exists ϵ > 0 such that d(u) = 0,∀u ∈ {u : |u| < ϵ}. This means that d(u) = 0, that is,vλ1(u) = vλ2(u), ∀u ∈ {u : 0 < |u| < ϵ}, which contradictsthe existence of {uk}∞k=1 satisfying (36). Thus, we have thatb is an eigenvector of Λ, and from (32) and (37), a and c areeigenvectors as well.

V. PRELIMINARIES FOR THE CASE WITH QUANTIZATION

So far, we have assumed that real-valued vectors yk canbe transmitted from the sensor to the controller. However,signals in networked control systems are often quantized todiscrete values because of the sensor’s limited capabilitiesand/or finite capacity in the communication channels. In therest of the paper, we extend the setup in Section II to dealwith quantization.

Instead of the original feedback system in Fig. 1, considerthe system depicted in Fig. 7. The plant and controller arethe same as those in Section II, but the observed state yk ∈Rn is quantized before being sent to the controller. We shallconsider static quantization laws that partition the state-spaceRn into a discrete family of sets and, at the kth sampling timetransmit a symbol r(yk) that represents the set to which ykbelongs. For simplicity of notation, we denote by Yk both thesymbol and the set corresponding to the measurement yk. Inthe next section we present results for logarithmic and uniformquantization sets.

In the quantization case, we impose the following assump-tion instead of Assumption 3.

9

Assumption 4: The control input is zero during [kT, kT+∆]for every k, i.e.,

uk = 0, ∀k ∈ Z+,

and u(t) does not depend on the future measurements {Yk :kT +∆ ≥ t}.

Remark 2: Assumption 4 simplifies computing the effect ofthe input uk during the sensing period [kT, kT+∆], while theexact value of yk may not yet be known to the controller. Thisassumption can be lifted if the quantizer knows the control law.

We now redefine the estimation set Jk in (6) to fit thequantization setup:

Jk ={

e−Aδk yk : δk ∈ [0,∆], yk ∈ Yk}. (38)

The estimation sets Ik and I−k+1 are defined as in (7) and (8)respectively, but with Jk defined above instead of (6). Notethat the estimation set Ik is also tight in the following sense,which is a natural extension of Definition 1.

Definition 3: An estimation set Ik ⊂ Rn of x(kT ), which isconstructed from I−0 , u(t), t ∈ [0, kT+∆], and Y0, Y1, . . . , Yk,is said to be tight if for every xk ∈ Ik, there exist x0 ∈ I−0 ,{δi ∈ [0,∆]}ki=0, and {yi ∈ Yi}ki=0, such that

eAkTx0 +

∫ kT

0

eA(kT−τ)Bu(τ)dτ = xk,

eA(iT+δi)x0 +

∫ iT+δi

0

eA(iT+δi−τ)Bu(τ)dτ = yi,

∀i ∈ {0, 1, . . . , k}.

VI. STABILIZABILITY UNDER CLOCK OFFSETS ANDQUANTIZATION

A. Single eigenvalue case: Logarithmic quantizer

Consider the scalar system in (13). In this subsection, weemploy the logarithmic quantizer proposed in [15]: Givenα0 > 0 and ρ ∈ (0, 1), the quantizer r is given by Yk = r(yk),∀k ∈ Z+, with

r(y) =

(ρi+1α0, ρ

iα0

]if y > 0,

{0} if y = 0,[−ρiα0,−ρi+1α0

)if y < 0,

(39)

where i is the integer satisfying ρi+1α0 < |y| ≤ ρiα0.Here, the ratio ρ expresses the coarseness of the logarithmicquantizer; the quantization levels become fine as ρ → 1 andbecome coarse as ρ→ 0.

For the system given in Fig. 7 with this logarithmic quan-tizer, the following theorem gives a necessary condition anda sufficient condition for stability, expressed in terms of theparameter ρ that defines the coarseness of the quantizer.

Theorem 3: Let Assumptions 1, 2, and 4 hold. Consider thefeedback system in Fig. 7 with the scalar plant (13) and thelogarithmic quantizer (39). If the system is stabilizable in thesense of Definition 2, then

ρ ∈

{(0, 1) if eλT ≤ 2,(max

{0, eλ∆ − 1

eλT /2−1

}, 1)

if eλT > 2.(40)

0 0.1 0.2 0.3 0.4 0.5 0.60

0.2

0.4

0.6

0.8

1

Stabilizable

Not stabilizable

Fig. 8: Bounds on coarseness ρ of quantization for stabilizabil-ity versus the maximum clock offsets ∆: the necessary condi-tion in (40) (dashed), the sufficient condition in (41) (solid),and the upper bound on the clock offset for stabilizability in(14) (dash-dot).

On the other hand, the feedback system is stabilizable if

ρ ∈(max

{0, eλ∆

(1− 1

eλT /2

)}, 1

). (41)

We illustrate the bounds on the coarseness ρ given in (40)and (41) through an example.

Example 3: Consider a scalar plant (13) with λ = 1.0 andfix the nominal sampling interval as T = 1.5. In Fig. 8, weplot the necessary lower bound on ρ for stabilizability givenby (40) and the sufficient bound given by (41). The dash-dotline represents the maximum clock offset tolerable for stability∆ ≈ 0.59 without quantization given by Theorem 1. We seethat, as the clock offset ∆ approaches ∆, the ratio ρ of theendpoints of a quantization cell goes to 1 and hence, very finequantization is needed for stabilizability.

The approach used to prove Theorem 3 also relies ondetermining the length lk of the estimation set Ik and constructupper and lower worst-case bounds for this length.

Lemma 7: Consider the feedback system in Fig. 7 with thescalar plant (13) and a logarithmic quantizer (39). Given anestimation set Ik−1 of width lk−1, k ≥ 1, the worst-case widthlk of Ik satisfies the following upper and lower bounds:

1− ρe−λ∆

1 + (1− ρ)e−λ∆

eλT

2lk−1 ≤ min

u(t),t∈((k−1)T+∆,kT )

maxxk∈I−

k ,δk∈[0,∆]lk

≤(1− ρe−λ∆

) eλT

2lk−1, (42)

where the second inequality holds when the control inputplaces I−k symmetrically about the origin.

Proof: Let p and q denote the lower bound and theupper bound for I−k , respectively, i.e., p := infξ∈I−

kξ and

q := supξ∈I−kξ. Assume that

|q| ≥ |p|, (43)

that is, the center of I−k is nonnegative. If this is not thecase, the analysis below can be adapted by flipping signs andreplacing q with p.

10

As we discussed in the proof of Lemma 1, the min-max oflk in (42) can be computed by finding the largest lk, as weranging over all possible quantized outputs Yk. To obtain anexplicit formula of maxYk

lk, we consider the following twocases.

(i) p ≤ 0 < q: From (4), the set of possible Yk compatiblewith xk ∈ I−k and δk ∈ [0,∆] is given by Yk ∈ Y−∪{0}∪Y+,where Y− and Y+ are defined as

Y− :=

[−ρiα0,−ρi+1α0

): i ≥

ln eλ∆|p|α0

ln ρ

, i ∈ Z

,

Y+ :=

(ρi+1α0, ρ

iα0

]: i ≥

ln eλ∆qα0

ln ρ

, i ∈ Z

.

We note that the interval (ρi+1α0, ρiα0], i = ⌊ln y

α0/ ln ρ⌋, is

the quantization cell that contains y > 0, i.e., ρi+1α0 < y ≤ρiα0.

In what follows, we evaluate maxYklk on each subset {0},

Y+, and Y− to describe maxYk∈Y−∪{0}∪Y+ lk. If Yk = {0},we have Ik = {0} and hence lk = 0. Next, suppose thatYk ∈ Y+. Then, from (38), the estimation set Jk is given by

Jk =(e−λ∆ρi+1α0, ρ

iα0

], i ≥

ln eλ∆qα0

ln ρ

, i ∈ Z,

where, i represents the index of a quantization cell. Taking theintersection of Jk and I−k , which is bounded by q from above,we have

lk =

{q − e−λ∆ρi+1α0 if ρiα0 ≥ q,(1− e−λ∆ρ

)ρiα0 if ρiα0 < q.

Let us consider the maximum of lk over i . If ρiα0 ≥ q,i.e., i ≤ ln q

α0/ ln ρ, then lk is increasing with i, and hence

the maximum in this case occurs when i = ⌊ln qα0/ ln ρ⌋.

If ρiα0 < q, or i > ln qα0/ ln ρ, since lk is decreasing for

this case, the maximum occurs when i = ⌊ln qα0/ ln ρ⌋ + 1.

Therefore, we have that

maxYk∈Y+

lk = max{q − e−λ∆

¯q,

(1− e−λ∆ρ

)¯q}, (44)

where¯q := ρ⌊ln

qα0

/ ln ρ⌋+1α0, which is the infimum of thequantization cell containing the upper bound q of I−k . For thecase Yk ∈ Y−, following the same discussion for Yk ∈ Y+,we obtain

maxYk∈Y−

lk = max{|p| − e−λ∆

¯p,

(1− e−λ∆ρ

)¯p},

where¯p := ρ⌊ln

|p|α0

/ ln ρ⌋+1α0. In view of (43), we have thatthe maximum of lk over Yk ∈ Y− ∪ {0} ∪ Y+ is equal to theright-hand side of (44).

(ii) 0 < p < q: The estimation sets Jk constructed from Ykwhich are compatible with (ii) are given by

Jk=(e−λ∆ρi+1α0, ρ

iα0

],

ln eλ∆qα0

ln ρ

≤ i ≤

⌊ln p

α0

ln ρ

⌋, i ∈ Z.

The maximum of lk over Yk can be obtained by followingthe discussion above for the case Yk ∈ Y+ in (i). The onlydifference is that, in the current case, the lower bound p ofI−k may be greater than that of Jk. This reasoning eventuallyleads to

maxYk

lk=

{max

{q − e−λ∆

¯q,(1−e−λ∆ρ

)¯q}

if p ≤ e−λ∆

¯q,

q − p if p > e−λ∆

¯q,

(45)

where¯q := ρ⌊ln

qα0

/ ln ρ⌋+1α0, which is the same in (44). WhenI−k lies far from the origin, and thus p > e−λ∆

¯q, then there

exists a case that the estimation set Jk becomes a superset ofI−k for some Yk.

As the second step of the proof, we evaluate the minimumof maxYk

lk obtained in (44) and (45) over the input u(t).Note that u(t) changes the position of I−k with respect tothe origin. We first establish that minu(t) s.t. (ii) maxYk

lk ≥minu(t) s.t. (i) maxYk

lk as follows. Let us compare (45) with(44). If p ≤ e−λ∆

¯q, then maxYk

lk takes the same form forboth cases (i) and (ii), and that is increasing with q. Since q inthe case (ii) cannot be smaller than that in (i), the claim is true.Otherwise, if p < e−λ∆

¯q, the right-hand side of (45) equals

q − p = eλT lk−1. This is greater than minu(t) s.t. (i) maxYklk

since it is possible to make minu(t) s.t. (i) maxYklk less than

or equal to eλT lk−1/2 with the input such that p = −q =eλT lk−1/2.

From the above discussion, we have that it suffices toconsider the case (i) for the evaluation of minu(t) maxxk,δk lkin (42). A simple calculation shows that the worst case lengthmaxYk

lk in (44) is bounded from below and above by thefollowing functions, both linear to q:

1− ρe−λ∆

1 + (1− ρ)e−λ∆q ≤ max

{q − e−λ∆

¯q,

(1− e−λ∆ρ

)¯q}

≤(1− ρe−λ∆

)q. (46)

Since both the lower bound and the upper bound in the aboveare increasing with q, the bounds are minimized when q =−p = eλT lk−1/2, which means that I−k is symmetric aboutthe origin. Substituting this into (46) leads to (42).

Proof of Theorem 3: (Sufficiency) If ρ satisfies (41),then there exists a logarithmic quantizer resulting in (1 −ρe−λ∆)eλT /2 < 1. Therefore, from Lemma 7, with the controlinput which places I−k symmetrically about the origin, we canmake Ik narrower than the previous estimation set for anyx(0) ∈ I−0 and δk ∈ [0,∆]. Thus, repeating this procedurefor each sampling period, we can achieve lk → 0 as k → ∞,which implies the stability of the feedback system.

(Necessity) When eλT ≤ 2, every ρ ∈ (0, 1) satisfies thesufficient condition (41) and hence the system is stabilizable.Consider the case of eλT > 2. Since Ik is a tight estimation set,when the system is stabilizable, the length lk must converge to0 for every x(0) and {δk}∞k=0. From Lemma 7, it is requiredfor the convergence of lk that

1− ρe−λ∆

1 + (1− ρ)e−λ∆

eλT

2< 1.

The above inequality is equivalent to (40) when eλT > 2.

11

B. Single eigenvalue case: Uniform quantizer

In this subsection, we consider the infinite- and finite-rangeuniform quantizers that divide the state space uniformly. Giventhe width w ∈ (0,∞) of the quantization cells, the infinite-rage uniform quantizer is defined by

r(y) = [iw, (i+ 1)w ) , (47)

where i is the integer such that iw ≤ y < (i+ 1)w.Theorem 4: Let Assumptions 1, 2, and 4 hold. Consider the

feedback system in Fig. 7 with the scalar plant (13) and theinfinite-range uniform quantizer (47). When 0 ≤ ∆ < ∆, with∆ given by (14), there exists a feedback controller such that,along solutions to the closed-loop, we have

lim supt→∞

|x(t)| ≤

{0 if eλT < 2,12

eλTw1−(1−e−λ∆)eλT /2

if eλT ≥ 2,(48)

for every initial condition x(0) ∈ R and sampling offsets{δk ∈ [0,∆]}∞k=0.

Furthermore, if

eλT ≥ 2

1− e−λ∆ (1− e−λ∆), (49)

then for any causal control law satisfying the setup in SectionV, there exist an initial condition x(0) ∈ R and a sequence{δk ∈ [0,∆]}∞k=0 such that

lim supt→∞

|x(t)| ≥ 1

2

eλ(T−∆)w

1− (1− e−λ∆) eλT /2. (50)

Remark 3: We can see from (48) that for any ∆ ∈ [0,∆),the uniform quantizer leads to a bounded solution, regardlessof the width w of the quantizations, but with the caveat that theupper bound in the lower branch of (48) increases as both wand ∆ increase. We also see from (48) that asymptotic stabilityis possible provided that eλT < 2, but for larger values of eλT

the state x(t) may not converge to zero. In particular, wheneλT exceeds the bound in (49), we can conclude from (50) thatx(t) will surely not converge to zero for some initial conditionsand δk sequences.

Before we turn to the proof of Theorem 4, we study a morerealistic scenario in which the quantizer’s input range is a finiteinterval that does not cover the entire state space R. Let usconsider a finite-range quantizer that partitions (−σ, σ), σ > 0,into an even number N cells of the same width w = 2σ/Nand is defined by Yk = r(yk), ∀k ∈ Z+, with

r(y) =

[⌊

yw

⌋w,

⌊yw + 1

⌋w)

if − σ + w ≤ y < σ,

(−σ,−σ + w) if − σ < y < −σ + w,

∅ else.(51)

With the use of the finite-range quantizer (51), globalstabilizability (as defined in Section IV) is not possible, sothe control objective must be relaxed to containability, whichhas been studied in [32].

Definition 4: The feedback system in Fig. 7 with the N -levelquantizer (51) is containable if for any sphere S centered at theorigin, there exist an open neighborhood M of the origin and

0 0.1 0.2 0.3 0.4 0.5 0.60

2

4

6

8

Containable

Not containable

Fig. 9: Base-2 logarithms of the bounds of the number N ofquantization cells versus the maximum clock offsets ∆: thesufficient condition in (52) (solid), the necessary condition in(53) (dashed), and the upper bound on the clock offset forstabilizability in (14) (dash-dot).

a feedback control law such that if x(0) ∈ M then x(t) ∈ Sfor all t ≥ 0.

Theorem 5: Let Assumptions 1, 2, and 4 hold. Consider thefeedback system in Fig. 7 with the scalar plant (13) and theuniform quantizer (51). The feedback system is containable ifeλT < 2 or

0 ≤ ∆ < ∆, N >eλT

1− (1− e−λ∆)eλT /2, (52)

where ∆ is given in (14).Conversely, if the systems is containable and (49) holds,

then it follows that

0 ≤ ∆ < ∆, N >eλ(T−∆)

1− (1− e−λ∆)eλT /2. (53)

The following example illustrates this result.Example 4: Consider the scalar plant (13) with λ = 1.0 and

fix the nominal sampling interval as T = 1.5. Fig. 9 showsthe sufficient bound on N in (52) and the necessary boundin (53) for containability. The vertical axis is log2N , whichcorresponds to the number of bits needed to send one of Nsymbols. From the figure, we see that for a clock offset largerthan ∆ (i.e., on the right side of the dash-dot line), the boundson N goes to infinity and the system is not containable.

Theorems 4 and 5 can be derived using the followinglemma, which is the analogous of Lemma 7 in the logarithmiccase. See the Appendix for the proofs of Lemma 8 andTheorem 4.

Lemma 8: Consider the feedback system in Fig. 7 with thescalar plant (13) and a uniform quantizer (51). For k ∈ Z+,suppose that the estimation set I−k satisfies

{eλtxk : xk ∈ I−k , t ∈ [0,∆]

}⊂ (−σ, σ), (54)

12

i.e., I−k is contained in the quantizer’s input during the kthsampling, and let l−k be the length of I−k . Then, the length lkof Ik satisfies

minu(t),

t∈((k−1)T+∆,kT )

maxxk∈I−

k ,δk∈[0,∆]lk ≤ κe−λT l−k + β, (55)

where

κ :=(1− e−λ∆

) eλT

2, β := e−λ∆w. (56)

Furthermore, it holds that

minu(t),

t∈((k−1)T+∆,kT )

maxxk∈I−

k ,δk∈[0,∆]lk

≥

{l−k2 if l−k < 2w,

κe−λT l−k + e−λ∆β else,(57)

where the equality holds if l−k < 2w. In (55) and (57), theminimum occurs when I−k is placed symmetrically about theorigin.

Proof of Theorem 5: (Sufficiency) We fix a specificvalue for σ > 0 and show that there exists an intervalI−0 = [−l−0 /2, l

−0 /2], l

−0 > 0, such that x(t) can be contained

in (−σ, σ) for all t ≥ 0. First, suppose that (52) holds. Then,the constant κ defined by (56) satisfies κ < 1 and there existsa constant l−0 such that w = 2σ/N and

σ >eλ(T+∆)β

2(1− κ)>

eλ∆

2l−0 . (58)

With such a quantizer and I−0 , we have that |x(t)| <eλ(T+∆)β/{2(1 − κ)} for t ∈ [0,∆]. Furthermore, we showbelow that the state can be bounded by this upper bound forall t > ∆.

Notice that I−0 = [−l−0 /2, l−0 /2] satisfies (54) from (58)

and is symmetric about the origin. Thus, from (55) and (58),we have that l0 < β/(1 − κ). Therefore, in light of (55),κ < 1, and l−k = eλT lk−1, by a control input which placesI−k symmetrically about the origin, we can make Ik so thatlk < β/(1− κ) and

|x(kT +∆)| ≤ eλ(T+∆)lk−1

2<

eλ(T+∆)β

2(1− κ), ∀k ∈ Z+, (59)

where l−1 := e−λT l−0 .Moreover, we can bound |x(t)| by the far right-hand side

of (59) also for t ∈ (kT +∆, (k + 1)T +∆) as follows:From (59), there exists t1 ∈ (kT +∆, (k + 1)T ) such that|x(t1)| < eλ(T+∆)β/{2(1−κ)}. Pick any t2 ∈ (t1, (k + 1)T )and consider the control input which takes the constant during[t1, t2] to bring the center of the set {eλ(t1−kT )xk : xk ∈ Ik}to the origin and zero for the rest. Then, since |x((k+1)T+∆)|satisfies the bound in (59), |x(t)| is also bounded by the sameupper bound for t > ∆.

On the other hand, when eλT < 2 is satisfied, there existsI−0 such that l−0 < 2w. Then, from (57), l0 = l−0 /2, whichresults in l−1 = eλT l0 < l−0 by the above control input, andthis concludes containability for this case as well.

(Necessity) When the system is containable, there existsa control input so that the quantizer does not saturate, i.e.,

all xk ∈ Ik are bounded as |xk| < σ for all time k.Otherwise, we lose track of some states compatible with thepast measurements. Thus, when (49) holds, the right-handside of (50) in Theorem 4 must be smaller than 2σ fromcontainability. This fact and N = 2σ/w lead us to (53).

C. Multiple eigenvalue case

For a specific class of general order systems, we canapply some of the results presented above for scalar systems.Consider the feedback system in Fig. 7 with a plant (1) forwhich the unstable part of A is real diagonalizable, i.e., A issimilar to the matrix [

Au 00 As

],

where Au := diag (λ1, . . . , λnu), λi > 0, i = 1, . . . , nu,and As ∈ R(n−nu)×(n−nu) has no unstable eigenvalue. Thestabilization of this system boils down to stabilizing the scalarsystems ξi(t) = λiξi(t) + νi(t), i = 1, . . . , nu, based on thequantized values of ξi(kT + δk). If all the scalar systems arestabilizable with logarithmic quantizers (or containable withuniform quantizers), then the feedback system is stabilizablein the sense of Definition 2 (containable in the sense ofDefinition 4).

However, the results in Theorem 2 indicate that such condi-tions induced by the scalar case results may be conservative.The main difficulty in extending Theorem 2 to the quantizationcase lies in the evaluation of the volume of the estimationset Ik. Unlike in the no-quantization case, with quantizationthe observed state comes as a set in the state space and theset Jk becomes a band of exponential functions. Then, Ik isthe intersection of exponential bands and obtaining a loweror upper bound of its volume analytically remains an openproblem.

VII. CONCLUSION

We studied the stability of a linear system using an asyn-chronous sensor and controller connected with a communi-cation channel. We first considered the quantization-free caseand derived conditions on the clock offset for stabilizability.The condition for scalar plants gives a tight limitation on theoffset, which depends on the level of instability of the plantand the sampling period. For vector plants, we showed that ifthe plant has at least two different real poles, then the systemis always stabilizable in finite time.

We then study for the quantization case. For scalar plantswith a logarithmic or uniform quantizer, we derived a nec-essary condition and a sufficient condition for stabilizabilityor containability, in terms of the coarseness of the quantizers.Although a stabilizing algorithm has been given in the proofs,this algorithm is generally not suitable for practical controllers.It remains an important topic for future work the constructionof practical control algorithms.

Acknowledgment: The authors would like to thank Prof.Kenji Kashima at Kyoto University for the proof of Lemma 6.

13

APPENDIX

We first present the proof of Lemma 8. Then, Theorem 4is proved using Lemma 8.

Proof of Lemma 8: We follow the approach in the proofof Lemma 7 and evaluate the maximum of lk over all possibleYk. Let p and q denote the lower bound and the upper boundfor I−k , respectively, i.e., p := infξ∈I−

kξ and q := supξ∈I−

kξ,

and assume that |q| ≥ |p|.The proof consists of three steps. First, we obtain an expres-

sion for maxYklk in terms of p and q. To do so, we consider

the two cases (i) p < 0 < q and (ii) 0 ≤ p < q. We then es-tablish that minu(t) s.t. (i) maxYk

lk ≤ minu(t) s.t. (ii) maxYklk.

Finally, the inequalities (55) and (57) are proved using thebounds on minu(t) maxYk

lk obtained in the previous step.Step 1: Consider the following two cases.(i) p < 0 < q: If in addition q < w, then lk = q or lk = |p|.

Thus, from the assumption |q| ≥ |p|, we have

maxYk

lk = q. (60)

Otherwise, i.e., if q ≥ w, we show that

maxYk

lk = max{q − e−λ∆

¯q,

(1− e−λ∆

)¯q + e−λ∆w

}(61)

where¯q is the lower bound of the cell containing q, which

is given by¯q := ⌊q/w⌋w. In the case of (i), the set of all

possible Yk is given by Y+ ∪ Y−, where

Y+ :=

{[iw, (i+ 1)w ) : 0 ≤ i ≤

⌊eλ∆qw

⌋, i ∈ Z

},

Y− :=

{[iw, (i+ 1)w ) :

⌊eλ∆pw

⌋≤ i ≤ −1, i ∈ Z

}.

When Yk ∈ Y+, it follows that

lk = min {q, (i+ 1)w} − e−λ∆iw, 0 ≤ i ≤⌊

eλ∆qw

⌋, i ∈ Z.

Regarding the above lk, for Yk such that q < (i + 1)w, orequivalently i ≥ ⌊q/w⌋, we have lk = q − e−λ∆iw. Since lkis decreasing with i, it takes the maximum q−e−λ∆

¯q when i =

⌊q/w⌋; otherwise, i.e., if i ≤ ⌊q/w⌋−1, then lk = (i+1)w−e−λ∆iw, and hence its maximum is (1 − e−λ∆)

¯q + e−λ∆w.

Therefore, (61) holds for Yk ∈ Y+.Following the same discussion, for Yk ∈ Y−, we have that

maxYk

lk = max{|p| − e−λ∆

⌊|p|w

⌋w,

(1− e−λ∆

) ⌊ |p|w

⌋w + e−λ∆w

}.

Noticing that |q| ≥ |p|, this is smaller than or equal to theright-hand side of (61). Thus, (61) expresses the maximumover Y+ ∪ Y−.

From (60) and (61), maxYklk with I−k satisfying (i) is given

as

maxYk

lk =

{q − e−λ∆

¯q if

¯q = 0 or q −

¯q ≥ e−λ∆w,(

1− e−λ∆)¯q + e−λ∆w else.

(62)

(ii) 0 ≤ p < q: In this case, the range of possible Yk isgiven by

Yk ∈{[iw, (i+ 1)w ) :

⌊ pw

⌋≤ i ≤

⌊eλ∆qw

⌋, i ∈ Z

}.

Hence, it follows that

lk = min {q, (i+ 1)w} −max{p, e−λ∆iw

},⌊ p

w

⌋≤ i ≤

⌊eλ∆qw

⌋, i ∈ Z.

Following a similar analysis to (i), we have that maxYklk with

I−k satisfying (ii) is given by

maxYk

lk = max{q −max

{p, e−λ∆

¯q},

¯q −max

{p, e−λ∆

(q − w

)} }. (63)

Step 2: In this step, we show that minu(t) s.t. (i) maxYklk ≤

minu(t) s.t. (ii) maxYklk. From (62), the minimum of

minu(t) s.t. (i) maxYklk can be computed as

minu(t) s.t. (i)

maxYk

lk=

L if L < w,

max{L− e−λ∆

⌊Lw

⌋w,(

1− e−λ∆) ⌊

Lw

⌋w + e−λ∆w

}else,

(64)

where L := l−k /2. Meanwhile, it is difficult to obtainminu(t) s.t. (ii) maxYk

lk directly from (63). Thus, in what fol-lows, we prove that for any I−k satisfying (ii), maxYk

lkbecomes greater than (64).

Suppose that I−k corresponds to the case (ii) above andconsider two cases (a) L < w and (b) L ≥ w.

(a) L < w: We prove that, for this case, it holds thatmaxYk

lk ≥ L, while minu(t) s.t. (i) maxYklk = L from (64).

To do so, three cases (a-1)–(a-3) are examined depending onthe size of I−k :

(a-1) p ≥¯q: In this case, I−k is contained in the quantization

cell[q,¯q + w

). Thus, it is true that maxYk

lk = 2L ≥ L.(a-2)

¯q − w ≤ p <

¯q: From (63), we have that

maxYk

lk = max{q −max

{p, e−λ∆

¯q},¯q − p

}≥ max

{q −

¯q,

¯q − p

}, (65)

where the inequality follows since¯q > p by (a-2) and

¯q ≥

e−λ∆

¯q. The far right-hand side of (65) is greater than or equal

to L since q −¯q < L implies that

¯q − p = 2L− (q −

¯q) > L.

(a-3) p <¯q − w: From the condition, the quantization cell[

q − w,¯q)

becomes a subset of I−k . Such Yk results in Iksuch that lk ≥ w, which is greater than L by the condition(a).

(b) L ≥ w: To obtain a simpler expression to (63), considertwo cases (b-1) and (b-2):

(b-1) p ≤ e−λ∆(¯q − w): From (63), it follows that

maxYk

lk = max{q − e−λ∆

¯q, (1− e−λ∆)

¯q + e−λ∆w

}.

Noticing that¯q > 0, we have that the above expression is the

same as the right-hand side of (62). Since q and¯q cannot be

14

smaller than or equal to those in the case (i), it holds thatmaxYk

lk > minu(t) s.t. (i) maxYklk.

(b-2) p > e−λ∆(¯q − w): In this case, we have

maxYk

lk = max{q − e−λ∆

¯q,

¯q − p

}≥¯q − p = 2L−

(q −

¯q)

> 2L− w ≥ L.

Here, the second inequality holds by the fact that q −¯q < w

and the third inequality is obtained by the condition (b). On theother hand, from (64), it follows that minu(t) s.t. (i) maxYk

lk <L and thus maxYk

lk > minu(t) s.t. (i) maxYklk for this case as

well.Step 3: From Step 2, we have that minu(t) maxYk

lk equalsminu(t) s.t. (i) maxYk

lk in (64) and the minimum occurs whenq = −p = L. With κ and β in (56), the right-hand side of(64) is bounded from above as

minu(t),

t∈((k−1)T+∆,kT )

maxYk

lk ≤ κe−λT l−k + β,

which concludes (55). Moreover, since (64) is bounded frombelow as

minu(t),

t∈((k−1)T+∆,kT )

maxYk

lk ≥

{l−k2 if l−k < 2w,

κe−λT l−k + e−λ∆β else,

we have (57).

Now we are ready to prove Theorem 4.Proof of Theorem 4: (Sufficiency) In the setup of

Theorem 4, we employ the infinite-range uniform quantizer(47). Thus, (54) in Lemma 8 follows for all k. Since it followsthat κ < 1 by ∆ < ∆ and from (55), we can make Ik so thatfor any x(0) ∈ I−0 and {δk ∈ [0,∆]}∞k=0,

lim supk→∞

lk ≤ β

1− κ

by placing I−k symmetrically about the origin. Thus, using thecontrol input stated in the proof of Theorem 5, we have

lim supk→∞

|x(t)| ≤ eλ(T+∆)

2

β

1− κ,

which establishes the lower branch of (48). Furthermore, ifeλT < 2, then it follows that β/(1 − κ) < 2e−λTw. Hence,there exists k′ ∈ Z+ such that lk′ < 2e−λTw. By applying theabove stated input, we obtain from (57) that lk′+1 = eλT lk′/2,which implies limk→∞ lk = 0 and thus we have (48).

(Necessity) If (49) holds, then e−λ∆β/(1− κ) ≥ 2e−λTw.Thus, by (57), there exists a possible path of the state resultingin

lim supk→∞

lk ≥ e−λ∆β

1− κ.

The inequality (50) follows from this.

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