Min-Max Output Integral Sliding Mode Control for Multiplant Linear

Uncertain Systems

F. J. Bejaranoa, A. Poznyaka and L. Fridmanb

Abstract— In this paper we consider the problem of using themin-max optimal control based on the LQ-index for a set ofsystems where only output information is available. We considerthat each system is affected by matched uncertainties, andwe propose to use an output integral sliding mode (OISM) tocompensate the matched uncertainties right after the beginningof the process. For the case when the number of inputs isless than the number of outputs, a hierarchical sliding modeobserver is proposed that converges to the original state withany small arbitrarily precision after any arbitrarily time.

I. INTRODUCTION

Here we presented the problem of controlling a set of sys-

tems using a unique control. The control used here is a robust

optimal control based in a min-max LQ-index proposed in [1]

and [2]. The optimal control based on the min-max LQ-index

is a control designed under two basic assumptions, namely,

the whole state is completely available and the system is free

of any uncertainty. Therefore, if we have output information

only, an observer for reconstructing the original states is

necessary to take advantage of the state feedback robust

optimal control. Furthermore, if each plant is influenced by a

matched uncertainty, before applying any control action, we

should ensure the compensation of such an uncertainty. In

[3] and [4] were proposed two different forms for resolving

the problem of matched uncertainty compensation for the

case of a control based on the min-max LQ-index in the

context of a multimodel system. In both cases were assumed

the availability of the complete vector state; this assumption

makes possible the application of the integral sliding mode

technique (see [5], [6], [7], [8]).

In the present paper we use the approach proposed in

[9]. There was suggested to use an output integral sliding

mode (OISM) to compensate the matched uncertainties right

after the beginning of the process. There was also suggested

to design a hierarchical observer using sliding modes. In

each step of the hierarchy one reconstruct a part of a vector

composed of multiplying an observability matrix by the

vector state. During the realization of such observer it is

needed to use some filters. However, there was shown that

the time of convergence and the observer error can be made

arbitrarily small just by modifying the sample time and the

filter constants used during the realization of the observer.

That is why we take advantage of the OISM to overcome

aCINVESTAV-IPN, Departamento de Control Automatico A.P. 14-740, CP 07000 Mexico D.F. javbejarano@yahoo.com.mx,apoznyak@ctrl.cinvestav.mx.

bFacultad de Ingenieria National Autonomous University ofMexico DEP-FI, UNAM, A. P. 70-256, CP 04510, Mexico, D.F.lfridman@verona.fi-p.unam.mx

the restrictions imposed in the design of the robust optimal

control.

A. Basic assumptions and restrictions

Since for each system the complete vector state is not

unavailable, in this work:

• we consider a finite set of plants whose trajectories are

suppose to be estimated;

• each plant of the system is described by a system of

linear time-invariant ODE (ordinary differential equa-

tions) with matched uncertainties which may be of a

nonlinear nature;

• the performance of each plant is characterized by a LQ-

index over a finite horizon;

• the optimal control action is assumed to be applied to

all plants simultaneously.

B. Main contribution

It is shown that it is possible to apply the robust optimal

control based on the min-max LQ-index even in the presence

of matched uncertainties and even if only the output (not

the complete state) of each system is measurable. To apply

the robust optimal control to the system, we use an output

integral sliding mode that allows to compensate the matched

uncertainties, and to estimate the state with any small preci-

sion.

II. PROBLEM STATEMENT

Let us consider a set of linear time invariant uncertain

systems

xα (t) = Aαxα (t) + Bαu (t) + Bαγ (t) + dα(t)yα (t) = Cαxα (t) , xα(0) = xα

0(1)

where α = 1, N , (N is a positive integer), xα (t) ∈ Rn

is the state vector at time t ∈ [0, T ], u (t) ∈ Rm is the

control and yα (t) ∈ Rp (1 ≤ p < n) is the output. The

vector dα(t) is assumed to be known for all t ∈ [0, T ]. The

current state xα (t) and the initial state xα0 are supposed to be

non available. Aα, Bα, Cα are known matrices of appropriate

dimension with rankBα = m and rankCα = p. Here all

the plants are running in parallel.

Throughout the paper we will assume that:

A1. (Aα, Bα) is controllable, (Aα, Cα) is observable.

A2. The vector γ (t) is upper bounded by a known

scalar function qa (t), that is,

‖γ (t)‖ ≤ qa (t) (2)

A3. It is known a bound for every vector xα0 , that is,

‖xα0 ‖ ≤ µ (3)

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A. The control design challenge

Before designing an optimal control, we must set free

the system from the effects of the matched uncertainties.

Therefore, the control design problem can be formulated as

follows: design the control u in the form

u = u0 + u1 (4)

where the control u1 will be designed to compensate the

uncertainty γ (t) from the initial time t = 0. And u0 (·) ≡u∗

0 (·), where u∗

0 (·) is a control function minimizing the LQ-

index:

minu0∈Rm

maxα∈1,N

hα (5)

hα :=1

2(xα (T ) , Gαxα (T ))

+1

2

T∫

t=0

[(xα (t) , Qαxα (t)) + (u0 (t) , Ru0 (t))] dt

Qα ≥ 0, Gα ≥ 0, R > 0

(6)

along the ‘nominal’ system trajectories

xα (t) = Aαxα (t) + Bαu0 + dα (7)

The exact solution of (5) requires the availability of all the

vector states xα (t) (see [2]) at any t ∈ [0, T ], and the system

must be free from any uncertainty. Therefore, to carried out

this optimal control we firstly should

1) ensure the compensation of the matched uncertainties

γ (t),2) design a state estimator that reconstruct every state

vector xα (t).

III. OUTPUT INTEGRAL SLIDING MODE (OISM)

Substitution of the control law (4) into (1) yields

xα (t) = Aαxα (t) + Bα (u0 + u1 + γ (t)) + dα(t) (8)

where α = 1, N . Let us define the following extended system

x(t) = Ax(t) + B (u0 + u1 + γ) + d

y (t) = Cx(t)(9)

where

x :=

x1

...

xN

, A :=

A1 0 · · · 0

0...

. . .

...

00 · · · 0 AN

, B :=

B1

...

BN

C =

C1 0 · · · 0

0...

. . .

...

00 · · · 0 CN

, d :=

d1

...

dN

(10)

Now, we will assume that

A4. rank (CB) = m

Thus, define the auxiliary affine sliding function s :RpN → RmN as follows

s (y (t)) := (CB)+

y (t) + σ (t) (11)

The term σ (t) includes an integral term which will be

defined below. Thus, for the time derivative s we have

s = (CB)+

CAx + u0 + u1 + γ + σ (12)

Define σ as

σ = − (CB)+

C [Ax + d] − u0

σ (0) = − (CB)+ y (0)(13)

The vector x represents an observer whose form will be

selected in the section IV. Substitution of σ into (12) gives

s = (CB)+

CA (x − x) + u1 + γ, s (0) = 0

The control u1 is designed in the following form

u1 = −β (t)s(t)

‖s(t)‖ (14)

with β (t) being a scalar gain that satisfies the condition

β (t) − qa (t) −∥∥∥(CB)

+CA

∥∥∥ ‖x − x‖ ≥ λ > 0

where λ is a constant. Notice that, by A3, an upper bound

of ‖x− x‖ always can be estimated. Selecting the Lyapunov

function as V =1

2‖s‖2

and in view of (14) and (2) one gets

V = (s, s)

=

(

s, (CB)+

CA (x − x) − βs

‖s‖ + γ

)

≤ −‖s‖[

β −∥∥∥(CB)

+CA

∥∥∥ ‖x − x‖ − qa

]

≤ −‖s‖λ ≤ 0

((s, s) := sT s). It means that V does not increase through

the time. Hence, since s(0) = 0, this implies

1

2‖s (t)‖2

= V (s (t)) ≤ V (s (0)) =1

2‖s (0)‖2

= 0

Thus, the identities

s (t) = s (t) = 0 (15)

hold for all t ≥ 0, i.e., there is no reaching phase.

From (12) and in view of the equality (15) the equivalent

control maintaining the trajectories on the surface is

u1eq = − (CB)+

CA (x − x) − γ (16)

Substitution of u1eq into (9) yields

x (t) = Ax (t) + B (CB)+

CAx (t) + Bu0 + d (t)y (t) = Cx (t)

(17)

where

A :=[

I − B (CB)+

C]

A (18)

Thus, our first objective was achieved, i.e., we have compen-

sated the uncertainty γ.

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IV. DESIGN OF THE OBSERVER

Now, with the system without uncertainties, we can recover

the state vector. In order to design the observer, (A,C)

must be observable. The following Lemma establishes the

conditions in terms of (A,B,C) to determine when (A,C)

is observable.

Lemma 1: The pair(

A,C)

is observable if and only if

the triple (A,B,C) has no invariant zeros, i.e.,

{s ∈ C : rank (P (s)) < n + m} = ∅ (19)

where P (s) is the Rosenbrock’s matrix system defined as

P (s) =

[sI − A −B

C 0

]

(20)

A proof of Lemma (1) has been given in [9].

Thus, henceforth we assumed that

A5. the triple (A,B,C) has no zeros.

It is known that when rank (CB) = m (assumption A4) and

p = m the triple (A,B,C) has invariant zeros; therefore, A4

and A5 imply that p > m.

The observer will be based on the recovering of the vectors

Cx (t), CAx (t) and so on until get CAl−1

x (t). After-

wards, the aim is to recover the vector Hx (t) where

HT =

[

CT(

CA)T

· · ·(

CAl−1

)T]

(21)

Here l is defined as the observability index that is the least

positive integer such that rank (H) = n (see, e.g., [10]).

Thus, after pre-multiplying Hx (t) by H+, the state vector

x (t) can be recover by x (t) = H+ Hx (t)︸ ︷︷ ︸

gotten online

, where H+ =

(HT H

)−1HT is the pseudo-inverse of H .

Before designing the observer we need to ensure a bound

required by the sliding mode algorithm. Design the following

dynamic system

�

x(t) = Ax(t) + Bu0(t) + B (CB)+

CAx (t)+L (y (t) − Cx (t))

(22)

where L must be designed such that the eigenvalues of A :=(A− LC) have negative real part. Let r (t) = x (t)− x (t),from (17) and (22), the dynamic equations governing r (t)are

r (t) =[

A − LC]

r (t) = Ar (t) (23)

Since the eigenvalues of A have negative real part, the

equation (23) is exponentially stable, i.e., there exist some

constants γ,η > 0 such that

‖r (t)‖ ≤ γ exp (−ηt)(√

Nµ + ‖x (0)‖)

(24)

Below it is shown that in the design of the observer we need

to know a bound of ‖r (t)‖. Thus, (24) ensures that we can

always satisfy such a requirement.

A. Auxiliary Dynamic Systems and Output Injections

The essential purpose in the design of the observer is to

recover the vectors

CAkx, k = 1, l − 1

Firstly, to recover CAx (t), let us introduce an auxiliary

vector state x1a (t) governed by the following dynamics

equations

x1a (t) = Ax (t) + B

[

u0(t) + (CB)+

CAx (t)]

+L(CL

)−1v1 (t) + d (t)

(25)

where x1a(0) satisfies Cx1

a(0) = y (0) and L is any matrix

such that det(CL

)6= 0. The vector x (t) represents the

observer we will design below. For the variable s1 ∈ RNp

defined by

s1(y (t) ,x1

a (t))

= Cx (t) − Cx1a (t) (26)

we have

s1(y (t) ,x1

a (t))

= CA (x (t) − x (t)) − v1 (t) (27)

with v1 (t) defined as v1 = M1s1

‖s1‖ . Here the scalar gain

M1 must satisfy the condition M1 >∥∥∥CA

∥∥∥ ‖r‖ to obtain the

sliding mode regime. A bound of ‖r‖ can be estimated using

(24). Then, repeating the procedure in III, we get s1 (t) = 0,

s1 (t) = 0 ∀t ≥ 0. Thus, from (26) we obtain that

Cx (t) = Cx (t) , ∀t ≥ 0 (28)

and from (27), the equivalent output injection is

v1eq (t) = CAx (t) − CAx (t) , ∀t > 0 (29)

Thus, CAx (t) is recovered from (29).

Now, the next step is to recover the vector CA2x (t). To

do that, let us design the second auxiliary state vector x2a(t)

generated by

x2a(t) = A2x(t) + ABu0(t) + L

(CL

)−1v2 (t)

+AB (CB)+

CAx (t) + d (t)

where x2a(0) satisfies CAx

1

a(0) + v1eq (0) − Cx2

a(0) = 0.

Again, for s2 ∈ RNp defined by

s2(v1eq,x

2a

)= CAx(t) + v1

eq (t) − Cx2a

and in view of (29), we have that s2 takes the form

s2(v1eq,x

2a

)= CAx (t) − Cx2

a (30)

Hence, the time derivative of s2 is

s2(v1eq,x

2a

)= CA

2x (t) − CAx(t) − v2 (t) (31)

Now, take the output injection v2 (t) as

v2 = M2s2

‖s2‖ , M2 >∥∥∥CA

2∥∥∥ ‖r‖ (32)

which implies that

s2 (t) = s2 (t) = 0 (33)

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In view of (33) and (31), v2eq (t) is

v2eq (t) = CA

2x (t) − CAx(t), t > 0 (34)

and the vector CA2x (t) can be recovered from (34).

Thus, iterating the same procedure, all the vectors CAix can

be recovered. In a summarizing form, the procedure above

goes as follows:

a) the dynamics of the auxiliary state xka(t) at the k-th

level is governed by

xka(t) = Akx(t) + Ak−1Bu0(t) + L

(CL

)−1vk

+Ak−1B (CB)+

CAx (t) + d (t)(35)

And the output injection vk at the k-th level is

vk = Mksk

‖sk‖ , Mk >∥∥∥CA

k∥∥∥ ‖r‖ (36)

where Mk is a scalar gain, and a bound of ‖r‖ can be found

using (24).

b) Define sk at the k-level of the hierarchy as:

sk (t) =

{

y − Cx1a, k = 1

vk−1eq + CA

k−1x− Cxk

a , k > 1(37)

where vk−1eq is the equivalent output injection whose general

expression will be obtained in the following Lemma, but

xka (0) should be chosen such that sk (0) satisfies

sk (0) = 0, k = 1, .., l − 1 (38)

Lemma 2: If the auxiliary state vector xka and the variable

sk are designed as in (35) and (37), respectively, then

vkeq (t) = CA

k[x (t) − x(t)] for all t ≥ 0 (39)

at each k = 1, l − 1.

Proof: It was shown that the following identity holds

v1eq (t) = CA [x (t) − x(t)] ∀t > 0

Now, suppose that the equivalent output injection vk−1eq is as

in (39). Then substitution of vk−1eq in (37) gives

sk(vk−1eq (t) ,xk

a (t))

= CAk−1

x (t) − Cxka (t) (40)

The derivative of (40) yields

sk (t) = CAk[x (t) − x (t)] − vk (t) (41)

Thus, selecting vk (t) as in (36) one gets

sk (t) ≡ 0, sk (t) ≡ 0 for all t ≥ 0 (42)

Therefore, (42) and (41) implies (39).

B. Observer in its Algebraic Form

Now, we can design an observer with the properties required

in the problem statement. From (28) and (39), we obtain the

following algebraic equations arrangement

Cx (t) = Cx(t) + y (t) − Cx(t)

CAx (t) = CAx(t) + v1eq (t)

...

CAl−1

x (t) = CAl−1

x(t) + vl−1eq

(43)

Thus, (43) yields the matrix equation

Hx (t) = Hx (t) + veq (t) , ∀t > 0 (44)

where H was defined in (21) and

vTeq =

[

(y (t) − Cx(t))T (v1eq

)T · · ·(vl−1eq

)T]

(45)

Since the pair(

A,C)

is observable, the matrix H has rank

n. Thus, the left multiplication of (44) by H+ yields

x (t) ≡ x (t) + H+veq (t) , ∀t > 0 (46)

That is why the observer can be designed as

x (t) := x (t) + H+veq (t) (47)

Therefore, we can formulate the following theorem.

Theorem 1: Under the assumptions A1-A5

x (t) ≡ x (t) ∀t > 0 (48)

Proof: It follows directly from (46) and (47).

C. Observer Realization

The achievement of the observer described by (47) requires

the availability of the equivalent output injection vkeq. How-

ever, the non idealities in the implementation of v(k) cause

the, so-called, chattering movements. Even though vkeq can

not be directly measured, it may be indirectly measured.

Namely, the first order-low pass-filter

τ vkav (t) + vk

av (t) = vk (t) ; vkav (0) = 0 (49)

gives an approach of vkeq (see [11]). That is, lim

τ→0∆/τ→0

vkav (t) =

vkeq

(t) , t > 0.

Where ∆ is proportional to the sampling time (the time that

vk lasts to pass from one state (M ) to other (−M )). So, we

can select τ = ∆η (0 < η < 1).

Hence, to realize the OISM observer we should:

1) use a sampling interval ∆ very small;

2) substitute vkeq

(t) into (37) and (45) by vkav (t);

3) chose xka (0) in such a way that

y (0) − Cx1a(0) = 0, for k = 1

CAk−1

x(0) − Cxka(0) = 0, for k > 1

so we ensure the identity sk (0) = 0, k = 1, ..., l − 1.

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V. OPTIMAL CONTROL DESIGN

We return in this section to the problem of the optimal control

u0 which resolves the problem (5). Substitution of (48) into

(17) yields the sliding motion equations for the state x that

takes the form

x(t) = Ax(t) + B (t)u0(x) + d

Now the solution for the min-max optimal problem (5) can

be given. Then, according [1], [2], the control solving (5)

for (7) is of the form:

u∗

0 (x) = −R−1B⊺ (Pλ∗x + pλ∗) (50)

where the matrix Pλ ∈ RnN×nN is the solution of the

parameterized differential matrix Riccati equation:

Pλ+PλA + ATPλ−PλBR−1BTPλ+ΛQ = 0Pλ (T ) = ΛG

(51)

and the shifting vector pλ satisfies

pλ+A⊺pλ−PλBR−1B⊺pλ + Pλd = 0; pλ (T ) = 0

where the weighting vector λ belongs to the simplex SN

SN = λ ∈ RN : λα ≥ 0,

N∑

α=1

λα = 1

and the matrices Q, G, and Λ denote the extended matrices

Q :=

Q1 0 · · · 0

0...

. . .

...

00 · · · 0 QN

, G :=

G1 0 · · · 0

0...

. . .

...

00 · · · 0 GN

Λ :=

λ1In×n 0 · · · 0

0...

. . .

...

00 · · · 0 λNIn×n

(52)

The matrix Λ = Λ (λ∗) is defined by (52) with the weight

vector λ = λ∗ solving the following finite dimensional

optimization problem

λ∗ = arg minλ∈SN

J (λ)

J (λ) := maxα=1,N

hα (53)

From (48), the estimated state x is used to realize the control

u0, i.e., the control u0 should be designed as

u0 (t) = u∗

0 (x) = −R−1B⊺ [Pλ∗ x + pλ∗ ] (54)

with x being designed according to (47).

VI. EXAMPLE

The following example shows the effectiveness of the sug-

gested control method. Consider a case of N = 3 where the

parameters are given by:

A1 =

−2 0.5 10.5 1.2 −21 2 −1.5

, A2 =

−0.3 1.5 −0.15−1 0.12 21 2 −3

A3 =

0.4 −1 0.30.5 −0.4 0.30.5 0.6 −1

, B1 =

0.511

, B2 =

1.5−21

B3 =

0.50.21

, C1,2,3 =

[1 0 00 1 0

]

γ (t) = sin (t) , d1,2,3 =

0.050.020.01

For this example the weights are λ∗

1 = 0.067, λ∗

2 = 0.301and λ∗

1 = 0.631, and the functional J (λ∗) = 502.800.

The trajectories for the three plants are shown in fig. 1,

fig. 2, and fig. 3; in these figures a comparison between the

trajectories of the original vector state and the trajectories of

the estimated state is made. The estimation error (e = x− x)

is graphed in fig. 4; since we know the first and the second

component of the state vector, then it is presented only the

third component of the error vector. The fig. 5 shows a

comparison between the control law u0 (x, t) when all the

state vector is available versus the control u (x, t) when only

output information is available.

0 2 4 6 8 10−10

−5

0

5

10

15

20

Time [s]

Sta

te x

1 a

nd

ob

serv

er

xe

1

0 0.02 0.04

−5

0

5x

1

xe1

Fig. 1. Trajectories of the original state and the estimated one for the firstplant.

CONCLUSIONS

Here we have shown that for the case when the number of

inputs is less than the number of outputs, the use of output in-

tegral sliding mode allows: first, to compensate the matched

uncertainties right after the initial time (independently of

the observation process), and second, as a consequence of

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0 2 4 6 8 10

−5

0

5

10

15

20

25

30

Time [s]

Sta

te x

2 a

nd

ob

serv

er

xe

2

0 0.01 0.02 0.03 0.04

−6

0

4x

2

xe2

Fig. 2. Trajectories of the original state and the estimated one for thesecond plant.

0 2 4 6 8 10−10

−5

0

5

10

15

20

Time [s]

Sta

te x

3 a

nd

ob

serv

er

xe

3

0 0.01 0.02 0.03 0.04

−5

0

5x

3

xe3

Fig. 3. Trajectories of the original state and the estimated one for the thirdplant.

0 2 4 6 8 10−1

0

1

Time [s]

err

or

e3 3

0 2 4 6 8 10−1

0

1

err

or

e1 3

0 2 4 6 8 10−1

0

1

err

or

e2 3

Fig. 4. Third component of the estimation error e = x − x.

0 2 4 6 8 10−4

−2

0

2

4

6

8

Time [s]

Co

ntr

ol

law

u0 0 0.01 0.02 0.03

3

4

5u

0 using xe

u0 using x

Fig. 5. Comparison between u (x, t) and u (x, t).

the first, to design a hierarchical observer that reconstructs

the system states. Using a low-pass filter for the observer

realization, we have shown that the estimation error depends

only on the sampling time and the filter time constant. It was

proven that the time of convergence for the observation error

can be made arbitrary small without any observer parameters

adjustment only by decreasing the sampling step and the filter

time constant. The use of an OISM might be a promising

technique not only for the compensation of the matched

uncertainties but also for making feasible the use of an

optimal control as we saw in this manuscript with the design

of an optimal control based on the min-max LQ-index using

only output information.

REFERENCES

[1] V. Boltyansky and A. Poznyak, “Robust maximum principle in mini-max control,” Int. J. of Control, vol. 72, pp. 305–314, 1999.

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