# Sensing And Actuation Strategies For Event Accounting Essay

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The problem of optimal control for .rst order stochastic systems with quadratic performance index over a .nite horizon is studied. The performance of three messaging policies for sensing combined with two hold circuits for actuation is compared based on optimization over the parameters of event detection and feedback control. The sampling rules include deterministic, level-crossing, optimal sampling, and the hold circuits include zero order hold (ZOH) and generalized hold (GH). The general results are established that level-crossing sampling performs more effectively than deterministic sampling and generalized hold outperforms zero order hold.

Index Terms

Event based systems, optimal control, stochastic control, sampled-data systems

I. INTRODUCTION In networked control systems, the control loops are closed through communication networks rather than wiring. The system elements are typically spatially isolated from one another and communicating over a wide area via both wired and wireless links [?], [?], [?]. The de.ning feature of such systems is re.ected in limited channel capacities, which are characterized by

constrained data rates through shared or wireless networks, thus limiting the average number of transmissions, and reducing the waiting time between transmissions. Simple analysis and design

This work was supported by NSERC and an iCORE PhD Recruitment Scholarship from the Province of Alberta.

X. Meng and T. Chen are with the Department of Electrical and Computer Engineering, University of Alberta, Edmonton, AB T6G 2V4, Canada (e-mail: [email protected]; [email protected]).

B. Wang is with the School of Electrical Engineering and Computer Science, University of Newcastle, Callaghan, NSW, 2308, Australia (e-mail: [email protected]).

make periodic sampling of the sensors and zero order hold of the actuators an attractive option for digital implementation [?]. Unfortunately, such methods are prone to too fast sampling. They often produce unnecessary data transmissions, and fail to work properly when high data rates are impossible in practice.

From practical considerations, however, it is nature that the sampling frequency must be high relative to the rate of change of the signals of interest, such as level-crossing sampling, send¿½on-delta [?], [?], adaptive sampling [?], and model based sampling [?]. The term, event based sampling, has been coined to characterize this type of sampling scheme [?]. The performances of event based sampling and periodic sampling are compared for .rst order [?], [?], and higher order [?] stochastic systems, which all turn out to favor the former. A number of authors have considered the problem of event-triggered control [?] and estimation [?], specially for networked control systems [?] and multi-agent systems [?], [?], [?], [?].

This paper discusses the problem of event-triggered sampling and control co-design for .rst order stochastic systems. A quadratic performance index is considered as the criterion for designing event detectors and controllers in a .nite time horizon. The controller co-locates with the actuator but resides on a separate node with the sensor in the network topology. The sensor is connected to the controller through a digital communication network medium. It is assumed that the sensor measures the state signal continuously and perfectly. Nevertheless a strict upper bound is imposed on the number of data transmissions from the sensor over a .nite time horizon, and the transmission is mediated by the event detector. There is a plethora of event detectors on the menu, but only three are presented here: deterministic sampling, level-crossing sampling and optimal sampling. Two types of hold circuits are described: zero order hold and generalized order hold. Their selection is based on their widespread use in practice as well as throughout the literature. For zero order hold, these measurements are directly used for feedback, whereas for generalized hold, these measurements permit us to perform a mean square estimation of the state, and the estimated states are subsequently used for feedback. The feedback gain and the parameter of the event detector are optimized based on a quadratic performance. The impulse hold where the control signal is an impulse, an extreme case of generalized hold, becomes optimal when the control weight of the quadratic performance index is zero. By comparing the different event-triggered sensing and actuation schemes, it is demonstrated that level-crossing sampling improves the performance signi.cantly, and generalized hold is desirable for event based control.

Notation: Rn denotes the n-dimensional Euclidean space. The symbols E and P mean the expectation and probability, respectively. For x . R, x+ denotes max (x, 0) and s . t means the minimum of s and t.

II. PROBLEM FORMULATION

Consider a .rst order linear diffusion process x (t) on a .nite time horizon [0,T ] described by the stochastic differential equation

dx (t)= ax (t) dt + u (t) dt + d. (t) ,x (0) = 0, (1)

where . (t) is a standard Wiener process or Brownian motion process, the drift coef.cient a is known, and the control process u (t) satis.es the standard assumptions to ensure that the solution x (t) exists and is unique in the sense of probability law.

The goal here is to minimize the quadratic performance criterion

T T

J = Ex 2 (t) dt+ .Eu 2 (t) dt(2)

00

with exactly one sample allowed in the interval [0,T ] by using different sampling strategies and hold circuits, respectively. Here . is the relative weight. Because of the important Markov property for diffusions, the control signals depend only on the received sample at the stopping time falling between the time interval [0,T ]. For zero order hold, the control signals are given by .

## .

0 if 0 = t < t,

u (t)= (3). Kx (t) if t = t = T.

For generalized hold, the control signals are given by

## . .

0 if 0 = t < t,

u (t)= (4). Kx¿½(t) if t = t = T,

where x¿½(t) is the mean square estimation of x (t) and obeys a linear ordinary differential equation dx¿½(t)

= ax¿½(t)+ u (t) (5)

dt for t = t = T , and x¿½(t)= x (t). Here the switching time t is a stopping time determined by sampling mechanisms including deterministic sampling, level-crossing sampling, and optimal sampling, which will be explained subsequently.

III. OPTIMAL DESIGN FOR ZERO ORDER HOLD

Assume a

=0. Now split the aggregate quadratic cost into three parts as follows:

tT T

J = E x 2 (t) dt + E x 2 (t) dt + .E K2 x 2 (t) dt . (6) 0 tt

The .rst term of the expression above is given by

t tt

e2at - 2at - 1

E x 2 (t) dt = E e 2a.d.dt = E 2 .

4a

0 00

The last two terms represent the part of the cost incurred on [t,T ], which can be expressed as

TT

E x 2 (t) dt + .E K2 x 2 (t) dt

tt

e2a(T -t) (a + K)2 - 4ea(T -t)K (a + K)+4K(a + K) - (a + K)2

= Ex 2 (t )

3

2a

x2 (t) K2 (T - t ) e2a(T -t) - 2a (T - t ) - 1

+E ++ .K2Ex 2 (t)(T - t )

22

a4a

e2a(T -t) - 2a (T - t) - 1 e2a(T -t) - 1

= E 2 + x 2 (t)+ x 2 (t) aK2 + x 2 (t) K¿½

4a2a

2

2a(T -t) - 1

e2a(T -t) - 2a (T - t) - 1 e¿½2 ¿½

= E+ x 2 (t) - x 2 (t )+ x 2 (t ) aK +

4a2 2a 4a 2a

where

2a(T -t) - 4e

ea(T -t) +3+2a (T - t)

a = . (T - t)+ ,

2a3 2a(T -t) - 2e

ea(T -t) +1

## ¿½ = .

a2

From the expression above and the fact a> 0, the optimal choice of the feedback gain K* is

2

a(T -t) - 1

ae

K * = - . (7)

2a(T -t)

2a3. (T - t )+2a (T - t)+3 - 4ea(T -t) + eRemark 1: For zero order hold, the optimal feedback gain K* cannot be predetermined. It is closely related to the sampling instant t but does not depend on the state x(t ) at the sampling instant directly. Whatever triggers the sampling, the controller has to wait for the sampling time to compute the optimal feedback gain. Let K* given by equation (7) be the optimal feedback gain. Then, the overall optimization problem is reduced to choose the pair (t,x (t)) such that the objective function e2at - 2at - 1 e2a(T -t) - 2a (T - t) - 1 e2a(T -t) - 1 ¿½2

J = E 2 + 2 + x 2 (t) - x 2 (t ) (8)

4a4a2a 4a

is minimized.

Next, two different classes of sampling strategies are going to be considered and the optimal design will be conducted within each class to minimize the performance measure in (8). The classes are deterministic sampling and level-crossing sampling.

A. Optimal Deterministic Sampling

Let us .rst minimize (8) over the class of deterministic sampling times. For deterministic sampling

2at - 1

e

E x 2 (t)= .

2a

The performance measure in equation (8) then can be written as follows

2at - 1 a(T -t) - 1 4

e2aT - 2aT - 1 ee

J = - .

4a2 4a2 2a3. (T - t)+ e2a(T -t) - 4ea(T -t) +3+2a (T - t)

The optimal deterministic sampling time t is the one that minimizes the above performance for a given weight .: t * = arg minJ

0=t=T

and the optimal feedback gain is given by

2

a(T -t*) - 1

ae

K *

## = - .

2a(T -t*)

2a3. (T - t*)+2a (T - t*)+3 - 4ea(T -t*) + e1) The Wiener Process Case: Let us now focus on the case a =0 which gives rise to a Wiener process. For a =0, the performance in (2) takes the form t2 32

t (T - t)2 (T - t)

J = E + t (T - t)+ K2 + Kt (T - t )+ + .K2t (T - t) .

23 2

The expression above allows us to see that the optimal choice of the feedback gain is

3(T - t)

K *

= - 2

2(T - t)+6.

and the objective function thus becomes

T 2 3t (T - t)3

J = - 2 .

2 4(T - t)+3.

It is a simple exercise to verify that the optimal sampling time satis.es

TT

= t * = .

42

The optimal design is to choose t to minimize the aggregate performance for a given weight

## .:

T 2 3t (T - t )3

t * = arg min - 2T 4 =t= T 2 2 4(T - t)+3.

and the optimal feedback gain is given by

3(T - t *)

K *

= - 2 .

2(T - t*)+6. Remark 2: For the special case of . =0, the solution to this optimization problem is given in [?] with

T 32.5T 2

t * K * J *

## = , = - , = .

2 T 8

Fig. 1 depicts the optimal performance and sampling time as a function of . for values of the parameter a = -1, 0, 1.

(a) Performance (b) Sampling time

Fig. 1: The optimal performance and sampling time of deterministic sampling and ZOH hold scheme.

B. Optimal Level-Crossing Sampling for the Brownian Motion Process

Level-crossing sampling is a simple event-triggered sampling scheme which generates a new sample whenever |x (t)| exceeds a speci.ed threshold d with the corresponding sampling instant de.ned as

td = inf {t ||x (t)| = d, x (0) = x0 . (-d, d)} .

t

Since a .nite horizon problem is considered, it may happen that td >T . Therefore, a time-out at the end of the time horizon is used and the sampling time is de.ned as t = td . T . Of course sampling at time t = T has absolutely no importance since such a sampling produces no contribution in the performance measure. It is important to .nd the probability of the threshold crossing before the time-out T . Start with deriving the moment generating function of the .rst hitting time t d.

Lemma 1: If td is the .rst hitting time of |x (t)| at the threshold d, then

v

cosh x0 2s

Ftd (s) _ E e -std = v .

cosh d 2s

Proof: From Lemma 7.4 in [?], it follows that

E[td] < 8.

Let

u (x0) _ E e -std .

Then by Proposition 7.2 in [?], the function u (x0) satis.es the elliptic equation 1 .2u (x0)

- su (x0) = 0; in x0 . (-d, d)

2 .x20

as well as the boundary condition

u (d)= u (-d)=1.

By solving the Dirichlet problem, the solution is given by

vv

x0 2s -x0 2s

e+ e

u (x0)= vv ,

d 2s + e-d 2s

e

from which the desired relation is obtained immediately.

Invoking the initial condition x0 =0, the probability density function of the random variable td is given by the following line integral, an integral formula for the inverse Laplace transform

.+jl st

11 e

f (t)= L-1 (Ftd (s)) = Ftd (s) e stds = lim v ds.

2pj2pj l.8 .-jl cosh d 2s

v

Since cosh (x) = cos(jx), the singularities of Ftd (s) which are also zeros of cosh d 2s are

2 p2

sk = - (2k + 1)k =0, 1, 2,...,

8d2 ,

then . can be set to zero. The probability of the threshold crossing before the time out T can be computed as

T T jl sT - 1

11 e

P[td <T ]= f (t) dt = Ftd (s) e stdt ds = lim v ds. 0 2pj 0 2pj l.8 -jl s cosh d 2s Note that the integrand does not in any way affect the status of the poles since s =0 is canceled by a single zero at zero on the numerator. In order to calculate this line integral over the complex plane, the standard methods of contour integration are employed. To wit, take a path that encloses the whole left half of the complex plane so that all the poles lie inside the contour. Then by the Cauchy residue theorem,

skT - 1

es - sk

P[td <T ] =lim v .

k=0 s.sk

sk cosh d 2s

In order to .nd the limit, apply the L¿½Hopital¿½s rule to the indeterminate form 0/0. Since sinh (x)= -j sin (jx) , then it can be shown that

s - sk p

lim v =(-1)k (2k + 1)

s.sk cosh d 2s 2d2 .

This expression permits the determination of the probability

-(2k+1)2

e. - 1 k p

P[td <T ]=2 p2 (-1)(2k + 1)

k=0 2d2

- (2k + 1)

8d2 -(2k+1)2.

4 14 k e

=(-1)k - (-1),

pk=0 2k +1 pk=0 2k +1

where

Tp2

. =8d2 .

The .rst series in the last equation converges although very slowly, and it produces an interesting value. In what follows, write 2k1+1 as an integral, and sum a geometric series effortlessly by adopting the technique described in [?]:

k 11 1

(-1)kk dx p

=(-1)x 2kdx =(-1)x 2kdx = 2 = . (9)

2k +1 1+ x4

000

k=0 k=0 k=0

Note that the computation is easy and straightforward. This leads the probability to

-(2k+1)2 Tp2

4 e

k 8d2

P[td <T ]=1 - (-1).

pk=0 2k +1 By letting T .8 or d =0, it can be obtained that P[td <T ]=1, which does make the common sense. The probability that a sample is generated before the time-out T as a function

of the threshold d is given in Fig. 2, and it monotonically decreases with the threshold d as seen from the .gure. Note that the probability depends on the ratio of the length of the time horizon T and the square of the threshold d. This suggests that, without loss of generality, the focus can be limited to T =1. The result for other T can be obtained from T =1 by scaling d2 to make the ratio invariant.

Fig. 2: Probability that a sample is generated as a function of the parameter d with T =1

Now the performance will be minimized by selecting the optimal threshold d and feedback gain K

td.TT T

J = E x 2 (t) dt + E x 2 (t) dt + .E K2 x 2 (t) dt .

0 td.Ttd.T

Applying the Ito differential rule gives d (T - t) x 2 (t)= -x 2 (t) dt +2(T - t) x (t) dx +(T - t) dt. (10) Taking integration in the sense of Ito and expectation on both sides leads to

td.T T 2 1 2

E x 2 (t) dt = E - (T - td . T )- (T - td . T ) x 2 (td . T )

22

0 T 2 1 2

= - E d2 (T - t d)+ +(T - td)+ ,

22

where the property of the Ito integral (see Theorem 3.2.1 in [?]) has been used so that the

expectation of the integration of the second term on the right hand in equation (10) vanishes.

Thus, the corresponding performance becomes

T 2 J = +1 K2d2E(T - td)+3+ Kd2E(T - td)+2+ .K2d2E (T - td)+ .

23

The next step is to write the above expression in terms of K and d alone. This requires

an evaluation of the .rst, second, and third moment: E (T - td)+ , E(T - td)+2, and

E(T - td)+3, which can be calculated as the way did in [?]:

k kk -.(2k+1)2

T 2p (-1). (-1)(-1)e

E (T - td)+ = - 3 + 3

k=0 k=0 k=0

2d2.2 2k +1 (2k + 1)(2k + 1)

-(2k+1)2.k

2d4 k+1 ed4 (-1)

E(T - td)+ = 512 p5 (-1)5 + 512 p55

(2k + 1)(2k + 1)

k=0 k=0 d2kk

T (-1)T 2 (-1)

- 64 +4

p3 (2k + 1)3 p 2k +1

k=0 k=0

-(2k+1)2.

33pT 4 k epT 4 k 1

E(T - td)+ =(-1)7 +(-1)

.4d2 (2k +1)2.d2 2k +1

k=0 k=0 3pT 4 13pT 4 1

- (-1)k +(-1)k

d23 .3d25

2.2(2k + 1)(2k + 1)

k=0 k=0 3pT 4 k 1

- (-1)7 .

.4d2

k=0 (2k + 1)

For positive integer values n, there is a formula

8 (-1)k 1 p 2n+1

= En , k=0 (2k + 1)2n+1 2 (2n)!2

where the Euler numbers En are the natural numbers de.ned according to:

246

E1xE2xE3x

sec x - 1= +++ ¿½¿½¿½

2! 4! 6!

The .rst three values of Euler numbers are

E1 =1,E2 =5,E3 = 61.

In particular,

8 k p3 8 k 5p5 8 k

(-1)(-1)(-1)61p7

## =; =; = .

35 7

k=0 (2k + 1)32(2k + 1)1536(2k + 1)184320

k=0 k=0

After substituting the series with their sums, the performance can be further reduced to the expression below:

-(2k+1)2.

T 2 1 12288d6 k e61

J =+ K2d2 (-1)+ T 3 - 3T 2d2 +5Td4 - d6

p77

23 k=0 (2k + 1)15

d4 -(2k+1)2.

k+1 e

+Kd2 512 (-1)5 +5 d4 - 2d2T + T 2 p5 k=0 (2k +1)3

k -.(2k+1)2

32d2 (-1)e

+.K2d2 T - d2 + 3 .

p3 k=0

(2k + 1)

The optimal d * will be computed by minimizing the performance for a given weight .:

d *

= arg min J.

d>0,K

Remark 3: Actually the resulting d optimized in this way is just a sub-optimal threshold for the zero order hold and level-crossing sampling strategy. Given .xed thresholds, the controller waits for the sampling time to compute the feedback gain, which performs better than any pre¿½determined feedback gains according to person-by-person optimality. The true optimal threshold and feedback gain should be optimized based on the performance index in (8), which are dif.cult to obtain. However, it is possible to obtain the analytic result for . =0 as shown in the following.

For . =0, the optimal feedback gain K* satis.es

3

K *

## = - .

2(T - td * )+

The corresponding performance measure then takes the form

T 2 3

J = - d2E (T - td)+

24

-.(2k+1)2

T 2 33 d4 (-1)k e

d2

= - T + d4 - 24 3 . 24 4 p3 k=0 (2k + 1)

The minimum performance incurred turns out to be

J * =0.2733T 2 ,

this being achieved by the choice

v

d *

=0.9389 T.

For this optimal threshold, the probability that the process reaches this threshold before the end time T is 68.59%.

Fig. 3(a) depicts the sub-optimal performance J as a function of the weight . and Fig. 3(b) the corresponding sub-optimal threshold.

(a) Performance (b) Sampling time

Fig. 3: The sub-optimal performance and threshold of level-crossing sampling and ZOH hold scheme.

IV. OPTIMAL DESIGN FOR GENERALIZED HOLD

The state of the controller in (5) is

(a+K)(t-t)

x¿½(t)= ex (t) , for t . [t,T ] .

Under the generalized hold, the state of (1) can be written as

tt

a(t-t)

x (t)= ex (t )+ e a(t-s)Ke(a+K)(s-t)x (t) ds + e a(t-s)d. (s)

tt t

(a+K)(t-t)

= ex (t )+ e a(t-s)d. (s) .

t

for t = t = T [?]. The aggregate quadratic performance can be decomposed as in the ZOH case:

tT T

J = E x 2 (t) dt + E x 2 (t) dt + .E K2 x¿½2 (t) dt .

0 tt

The second term of the expression above constitutes the part of state variance incurred from the switch time t up to the end time T , and it can be expressed as

T 2(a+K)(T -t) - 1

ee2a(T -t) - 2a (T - t) - 1

E x 2 (t) dt = E x 2 (t)+ .

t 2(a + K)4a2

As in the case of zero order hold, deterministic sampling and level-crossing sampling strategies are going to be considered and the optimal design will be conducted within each class to minimize the performance.

A. Optimal Deterministic Sampling

For deterministic sampling, the performance takes the form

2at + e2at - 1 2(a+K)(T -t) - 1 2at - 1 2(a+K)(T -t) - 1

e2a(T -t) - 2aT - 2 eee

J = ++ .K2 e

4a2 2a 2(a + K)2a 2(a + K) The optimal sampling time t* and feedback gain K* can be found by solving the following optimization problem:

minimize J subjec to 0 = t = T

For . =0, the optimal choice of K* and t* is

t * = T, and K * = -8.

2

The corresponding performance cost is

T 2 J = .

4

In this case, the optimal generalized hold becomes the impulse hold where the control signal is an impulse [?], [?], [?]. This is equivalent to reset the state to the origin at t =0.5T .

Fig. 4 depicts the optimal performance, sampling time, and feedback gain for values of the parameter a =1, 0, -1.

B. Optimal Level-Crossing Sampling for the Brownian Motion Process

Using the previous result yields

td.T

T 2 1 2

E x 2 (t) dt = - d2E (T - t d)+ - E (T - td)+

22

0

T 2K(T -td)+

e- 1(T - td)+2

E x 2 (t) dt = E x 2 (td . T )+ .

2K 2

td.T

Thus, the performance becomes

T 2 d2 1

2K(T -td)+

J = - d2E (T - td)+ ++ .K E e - 1 .

22 K

(a) Performance (b) Sampling time (c) Feedback Gain

Fig. 4: The optimal performance, threshold and feedback gain of deterministic sampling and generalized hold scheme as a function of . for systems with a = -1, a =0, and a =1 respectively.

2K(T -td)+

A numerical method of computing the term E e- 1 over a .xed .nite time interval is provided in [?]. Let W (x, t) be a continuous and bounded function de.ned on the region [-d, d] ¿½ [0,T ], and satisfy the partial differential equation

.W (x, t)1 .2W (x, t)

++ e -2Kt =0,

.t 2 .x2

along with the boundary and initial conditions

## .

. W (-d, t)= W (d, t)=0 for t . [0,T ) , . W (x, T )=0 for x . [-d, d] .

Then

td.T W (0, 0) = E e -2Ktdt .

0

A brief proof will be shown here. Applying standard It¿½o calculus on W (x, t) yields

td.T

E[W (x (td . T ) , td . T )] - W (0, 0) = E dW (x (t) , t)

0

= E td.T 0 .W (x, t) .t + 1 2 .2W (x, t) .x2 dt

td.T

= E -e -2Ktdt

0

If the process hits the boundary before time T , x (td . T )= ¿½d in which case W (x (td . T ) ,td . T )= W (¿½d, td)=0.

If the process does not reaches the threshold before T , then the sampling instant is td . T = T which yields W (x (td . T ) ,td . T )= W (x (T ) ,T )=0. Thus the above relationship can be concluded.

Note that W (0, 0) is still a function of d and K for .xed time horizon T . Therefore, de.ne

U (d, K) _ W (0, 0) .

Then the performance can be written as

T 2 k -.(2k+1)2

32 (-1)e

J = - d2T + d4 1 - 3

k=0

2 p3 (2k + 1)d2

+ 1+ .K e 2KT (1 - 2KU (d, K)) - 1 .

2 K The optimal d and K can be computed by minimizing the performance J.

Remark 4: The optimal d and K derived in this way are sub-optimal threshold, and feedback gain, respectively. Again, the true optimal result will be shown analytically below for the case . =0. For . =0, the optimal K* is

K *

= -8,

and then the performance takes the form as

T 2 k -.(2k+1)2

32d4 (-1)e

J =+ d4 - d2T - 3 .

k=0

2 p3 (2k + 1)

The minimum performance incurred turns out to be

J * =0.1977T 2 ,

this being achieved by the choice

v

d *

=0.9389 T.

V. PERFORMANCE COMPARISON FOR THE BROWNIAN MOTION PROCESS

To provide some insight, .rst consider the special case . =0; the results for this case are summarized in Table I. The table determines the ranking of the sampling and hold strategies. The optimal sampling is the best among all possible sampling rules, level-crossing sampling outperforms deterministic sampling, and generalized hold outperforms zero order hold, which are not surprising. Another important insight given by this table is that the optimal parameters for each sampling scheme are independent of hold circuits. This table also tells us that the hold contributes more to the performance improvement than the sampling.

TABLE I: Comparison of different sampling and control schemes

Performance J Triggering Instant t * Feedback Gain K *

Deterministic Sampling ZOH 0.3125T 2 t = 0.5T -3T -1

Level-Crossing Sampling ZOH 0.2733T 2 inf{t||x (t) | = 0.9389 v T } -1.5 (T - t )-1

Optimal Sampling ZOH [?] 0.2623T 2 inf{t|x 2 (t) = v 3 (T - t)} -1.5 (T - t )-1

Deterministic Sampling GH 0.2500T 2 t = 0.5T -8

Level-Crossing Sampling GH 0.1977T 2 inf{t||x (t) | = 0.9389 v T } -8

Optimal Sampling GH 0.1830T 2 inf{t|x 2 (t) = v 3 (T - t)} -8

The result on the optimal sampling and generalized hold scheme is given below. It is optimal in the sense that (K, t) minimizes the following performance measure among all possible sampling policy and feedback gain pairs:

2K(T -t) - 1

T 2 E x2 (t) e

J = - E x 2 (t)(T - t)+ .

22K

It makes intuitive sense to set K* to be -8. Because of this observation the performance

measure takes the form as below:

T 2

J = - E x 2 (t)(T - t) .

2

Next it is clear that in order to minimize the performance it is suf.cient to maximize the expected reward function:

E x 2 (t)(T - t) .

The optimal sampling rule is the symmetric quadratic envelope [?]:

v t * = inf{t|x 2 (t) = 3(T - t)}.

Now let us calculate the expected performance incurred by the optimal sampling scheme. At the sampling time, the inequality becomes an equality

vv

E x 2 (t)= 3T - 3E[t ]= E[t] .

Consequently,

v

3

E[t]= v T,

3+1

and the expect reward function becomes

v vv

E x 2 (t )(T - t)= 3E (T - t)2 = 3(3 - 2)T 2 + E t2

The next step is to evaluate the second moment of the sampling time. According to Theorem

8.5.8 in [?], the function x4 (t) - 6x2 (t) t +3t2 is a martingale. Since t is a bounded stopping time, then

E x 4 (t) - 6x 2 (t) t +3t2 =0.

This relation suggests

v vv E t 2 =2 3E[(T - t) t] - E (T - t )2 =23 - 1 T 2-(2 3+1)E t2

then from which it can be concluded

v

7 - 33

E t2 = T 2 .

4

Finally, the optimal performance becomes

vv

T 2 3 - 33 - 1

T 2 T 2

J = - = .

24 4 The realizations of a Wiener process with different event-triggered sensing and actuation strate¿½gies are shown in Fig. 5. The simulation was performed by choosing the same Wiener process.

Fig. 6 shows the optimal achievable cost J* for the four different event-triggered sampling and control strategies. It is notable that for the small weight ., the achievable performance of deterministic sampling and generalized hold is better than that of level-crossing sampling and zero-order hold, while the opposite is true when . becomes large. It is surprising that the sampling plays a more important role than the hold in performance improvement for large weights.

(a) Zero order hold (b) Generalized hold

Fig. 5: Control of a Wiener process with time horizon T =1.

Fig. 6: Optimal achievable performance J*

VI. CONCLUSION Three sampling schemes typically used for event detection combined with two hold circuits for control actuation have been considered for optimal design and performance comparison. The

three sampling schemes consist of deterministic sampling, level-crossing sampling and optimal sampling and the two hold circuits include zero order hold and generalized hold. The results

obtained in this paper now enable the performance ranking among the combination of different sampling and hold strategies.