Discrete Time Systems by Mario A. Jordan and Jorge L. Bustamante - HTML preview

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i= q j= si

+ 2 εT

k

A − LC

P Ad − LdC εk−d + 2 εTk A − LC P ∑ ∑ BHijζij

i=1 j=1

T

i= q j= ri

T

i= q j= si

+ 2 εT

ζd ⎠ +

k

A − LC

P

∑ ∑ BHdij ij

2 εTk−d Ad − LdC P ∑ ∑ BHijζij

i=1 j=1

i=1 j=1

⎞ ⎛

T

T

i= q j= ri

i= q j= si

i= q j= si

+ 2 εT

ζd ⎠ + ⎝

k−d Ad − LdC

P

∑ ∑ BHdij ij

∑ ∑ BHijζij P ∑ ∑ BHijζij

i=1 j=1

i=1 j=1

i=1 j=1

T

i= q j= ri

i= q j= ri

+ ⎝∑ ∑ BHdζd⎠ ⎝

ζd

ij ij

P

∑ ∑ BHdij ij

i=1 j=1

i=1 j=1

(58)

where

ζij = hij( k) χi, ζd =

(

ij

hdij k) χdi.

(59)

index-44_1.png

index-44_2.png

index-44_3.png

index-44_4.png

index-44_5.png

index-44_6.png

32

Discrete Time Systems

From (36) and (37), we have

i= q j= si

∑ ∑ ζT 1 1 ζ

ij

h

b

ij ≥ 0

(60)

i=1 j=1

ij

ij

i= q j= ri

∑ ∑ ( ζd) T 1 1 ζd ≥

ij

ij

0

(61)

i=1 j=1

hd

bd

ij

ij

Using (43) and (59), the inequalities (60) and (61) become, respectively,

i= q j= si

i= q j= si

∑ ∑

T

εT

1

Hi − KiC

ζij − ∑ ∑

ζTζ

b

ij ij ≥ 0

(62)

i=1 j=1

i=1 j=1

ij

i= q j= ri

i= q j= ri

∑ ∑

T

εT

ζd −

1 ( ζd) Tζd ≥

k−d Hd

i

KdiC

ij

∑ ∑

ij

ij

0

(63)

i=1 j=1

i=1 j=1 bdij

Consequently,

⎤ ⎡

ε

T

k

Γ11 Γ12 Γ13 Γ14

εk

ε

⎥ ⎢

⎥ ⎢ ε

Δ

( ) Γ

V ≤ k−d

22 Γ23 Γ24

k−d

⎥ ⎢

⎥ ⎢

ζ ⎦ ⎣

⎦ ⎣

(64)

k

( ) ( ) Γ33 Γ34

ζk

ζd

( ) ( ) ( ) Γ

ζd

k

44

k

where

T

Γ11 = A − LC P A − LC − P + Q

(65)

T

Γ12 = A − LC P Ad − LdC

(66)

T

Γ13 = M T( K 1, ..., Kq) + A − LC PΣ

(67)

T

Γ14 = A − LC PΣ d

(68)

T

Γ22 = Ad − LdC P Ad − LdC − Q

(69)

T

Γ23 = Ad − LdC PΣ

(70)

T

Γ24 = N T( Kd 1, ..., Kdq) + Ad − LdC PΣ d

(71)

Γ33 = Σ TPΣ Υ

(72)

Γ34 = Σ TPΣ d

(73)

Γ44 = (Σ d) TPΣ d − Υ d

(74)

ζk = [ ζT

] T

11, ..., ζ T

1 s , ζ T

(75)

1

21, ..., ζ T

qsq

ζd = [( ζd ) T

) T

) T

) T] T

k

11

, ..., ( ζd 1 r

, ( ζd

, ..., ( ζd

(76)

1

21

qrq

index-45_1.png

index-45_2.png

index-45_3.png

index-45_4.png

Observers Design for a Class of Lipschitz Discrete-Time Systems with Time-Delay

33

and M( K 1, ..., Kq), Σ, Υ are defined in (49), (53) and (55) respectively.

Using the Schur Lemma and the notation R = LTP, the inequality (48) is equivalent to

Γ11 Γ12 Γ13 Γ14

⎢( ) Γ

22 Γ23 Γ24

( ) ( ) Γ

< 0.

(77)

33 Γ34

( ) ( ) ( ) Γ44

Consequently, we deduce that under the condition (48), the estimation error converges

asymptotically towards zero. This ends the proof of Theorem 4.2.

Remark 4.3. Note that we can consider a more general observer with more degree of freedoms as

follows :

i= q

l= d

ˆ xk+1 = A ˆ xk + Adxk−d + ∑ Beq( i) fi( vi

) +

k, wik

Ll yk−l − C ˆ xk−l

(78a)

i=1

l=0

l= d

vi =

k

Hi ˆ xk + ∑ Ki, l yk−l − C ˆ xk−l

(78b)

l=0

l= d

wi =

k

Hdi ˆ xk−d + ∑ Kdi, l yk−d − C ˆ xk−d

(78c)

l=0

This leads to a more general LMI using the general Lyapunov-Krasovskii functional :

j= d i= j

Vk = εT

εT

k Pεk + ∑ ∑ k−iQjεk−i

j=1 i=1

4.3 Numerical example

Now, we present a numerical example to show the performances of the proposed method. We

consider the modified chaotic system introduced in Cherrier et al. (2006), and described by :

˙ x = Gx + F( x( t), x( t − τ))

(79)

where

−α α 0

−αδ tanh( x 1( t))

G = ⎣ 1 1 1 ⎦ , F( x( t), x( t − τ)) = ⎣

0

0 −β −γ

sin( σx 1( t − τ))

Since the proposed method concerns discrete-time systems, then we consider the discrete-time

version of (79) obtained from the Euler discretization with sampling period T = 0.01. Hence,

we obtain a system under the form (1a) with the following parameters :

−αδT 0

A = I

3 + TG, Ad = 0R3 × 3, B =

0

0

0

T

and

f ( xk, xk−d) =

tanh( x 1( k))

sin( σx 1( k − d)

34

Discrete Time Systems

that we can write under the form (35) with

H 1 = 1 0 0 , Hd =

1

0 0 0

H 2 = 0 0 0 , Hd = σ

2

0 0

Assume that the first component of the state x is measured, i.e. : C = 1 0 0 .

The system exhibits a chaotic behavior for the following numerical values :

α = 9, β = 14, γ = 5, d = 2

δ = 5, = 1000, σ = 100

as can be shown in the figure 1.

The bounds of the partial derivatives of f are

100

50

0

x 3

−50

−100

10

5

0

0

−10

x

−5

x

2

1

Fig. 1. Phase plot of the system

a 11 = 1, b 11 = 1, ad =

=

21

1, bd 21

1

According to the remark 4.1, we must solve the LMI (48) with

0 0

0

˜ bd =

=

21

bd 21

ad 21

2,

˜

Ad = 0 0

0

0 0 −T σ

Hence, we obtain the following solutions :

1.3394

0

L = ⎣ 4.9503 ⎦ , Ld = ⎣

0

⎦ , K 1 = 0.9999, K 2 = 0.0425, Kd =

=

1

1.792 × 10 13, Kd 2

100

40.8525

1000

The simulation results are shown in figure 2.

Observers Design for a Class of Lipschitz Discrete-Time Systems with Time-Delay

35

1

3

x

x

0.8

1

2

the estimate of x

2

the estimate of x2

0.6

1

0.4

1

0.2

0

0

Magnitude

−1

Magnitude

−0.2

−0.4

−2

−0.6

−3

−0.8

−1

−4

0

50

100

150

0

50

100

150

Tim (k)

Time (k)

(a) The first component x 1 and its estimate ˆ x 1

(b) The second component x 2 and its estimate ˆ x 2

30

20

10

0

−10

Magnitude

−20

x

−30

3

the estimate of x3

−40

0

50

100

150

Tim (k)

(c) The third component x 3 and its estimate ˆ x 3

Fig. 2. Estimation error behavior

5. Conclusion

This chapter investigates the problem of observer design for a class of Lipschitz nonlinear

time-delay systems in the discrete-time case. A new observer synthesis method is proposed,

which leads to a less restrictive synthesis condition. Indeed, the obtained synthesis condition,

expressed in term of LMI, contains more degree of freedom because of the general structure

of the proposed observer. In order to take into account the noise (if it exists) which affects

the considered system, a section is devoted to the study of H∞ robustness. A dilated LMI

condition is established particularly for systems with differentiable nonlinearities. Numerical

examples are given in order to show the effectiveness of the proposed results.

A. Schur Lemma

In this section, we recall the Schur lemma and how it is used in the proof of Theorem 2.1.

index-48_1.png

36

Discrete Time Systems

Lemma A.1. Boyd et al. (1994) Let Q 1, Q 2 and Q 3 be three matrices of appropriate dimensions such

that Q 1 = QT 1 and Q 3 = QT 3 . Then, the two following inequalities are equivalent :

Q 1 Q 2 < 0,

(80)

QT 2 Q 3

Q 3 < 0 and Q 1 − Q 2 Q− 1

<

3 QT

2

0.

(81)

Now, we use the Lemma A.1 to demonstrate the equivalence between M1 + M2 < 0 and

M4 < 0.

We have

⎤ ⎡

−P + Q 0

˜

AT PB

˜

AT P ˜

A ˜

AT P ˜

Ad 0

M

⎦ ⎣

1 + M2 =

( )

−Q

˜

AT PB

+

( )

˜

d

AT P ˜

A

d

d 0

( )

( ) BTPB − αIq

( )

( ) 0

(82)

M T M

M

M

M

15

15 + M T

16

16 M T

15

25 + M T

16

26 0

+ 1 ⎣

( )

M T M

M

⎦ .

αγ 2

26

26 + M T

25

25 0

f

( )

( )

0

By isolating the matrix

P

0

0

Λ = ⎢

⎣0 αγ 2 I

0

f s 1

0

0

αγ 2 I

f s 2

we obtain

˜

AT M T M T

˜

15

16

A

˜

Ad 0

−P + Q 0

˜

ATPB