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

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In this section, LMIs-based conditions of delay-dependent stability analysis will be

considered for discrete-time systems with multiple time-varying delays. The following

result gives sufficient conditions to guarantee that the system (1) for u = 0, k ≥ 0

k

is stable.

Theorem 1: For a given set of upper and lower bounds d , i id for corresponding time-varying

delays d

n× n

n× n

∈ℜ

∈ℜ

ki , if there exist symmetric and positive-definite matrices 1

P

, i

Q

and

n n

=

i

R

×

∈ℜ

, i 1,…, N and general matrices 2

P and 3

P such that the following LMIs hold:

N

N

⎢∑ Q + ∑( d d )

T

T

+

i

i

i

i

R

1

P

A 2

P

2

P A

*

*

*

* ⎥

i=1

i=1

T

T

+

+

2

P

3

P A

1

P

3

P

3

P

*

*

*

= ⎢

T

T

M

<

d

A 1 2

P

d

A 1 3

P

1

Q

*

*

0 (3)

T

T

d

A 2 2

P

d

A 2 3

P

0

2

Q

*

0

* ⎥

T

T

dN

A

2

P

dN

A

3

P

0

0

N

Q

<

i

Q

i

R

Stability Criterion and Stabilization of

Linear Discrete-time System with Multiple Time Varying Delay

285

Terms denoted by * are deduced by symmetry. Then the system (1) is stable.

Proof: Consider the following change of variables:

N

x

=

= − +

+

k+1

y , 0

k

yk Axk

(4)

d

A ixkdki

i=1

Define x = [ T

T

T

T

T

k

xk yk xkd 1

x − ]

k dN

, and consider the following Lyapunov-Krasovskii

candidate functional:

V( x ) =

+

+

1

V ( x )

2

V ( x )

3

V ( x )

k

k

k

k (5)

with

T T

=

1

V ( x ) x E Px ,

k

k

k

N

k−1

T

=

2

V ( x )

k

∑ ∑ xl i

Q xl

i=1 l= kdki

and

N d +1

i

k−1

T

=

3

V ( x ) ∑ ∑

x R x ,

k

m i m

i=1 l=− d +2 m= k+ l−1

i

where

0

i

Q > and

0

i

R > , and E and P are, respectively, singular and nonsingular matrices

with the following forms:

I 0 0

0⎤

P

0 0

0⎤

1

0 0 0

0⎥

⎢ 2

P

3

P

0

0⎥

E = ⎢0 0 0 0 0⎥ , P = ⎢ 0

0

I 0 0⎥

0

0

⎢0 0 0

0⎥

⎢ 0 0 0

I

where 1

P is a symmetric and positive-definite matrix.

The difference Δ V( x )

k is given by

V

Δ ( x ) = Δ

+ Δ

+ Δ

1

V ( x )

2

V ( x )

3

V ( x )

k

k

k

k (6)

Let us now compute Δ 1

V ( x )

k :

T

T

T T

Δ

=

=

1

V ( x )

k

1

V ( xk+1)

1

V ( x )

k

x E Px +

x E Px

k+1

k 1

k

k

⎡ 1 ⎤

x

⎢2 k

⎢ 0

T

T

T

T

= y

=

− ⎡

k 1

P yk xk 1

P xk yk 1

P y

2 x

0 0

0

k

k

⎦ 1

P ⎢ 0 ⎥

⎢ 0 ⎥

which has the following equivalent formulation using the fact that

N

0 = − y +

+

k

Axk

as

d

A ixkdki

i=1

286

Discrete Time Systems

Δ

=

1

V ( x )

k

⎡ 1

⎤ ⎡ 1

⎤ ⎤

⎢⎡0 0 0

0⎤

I

0

0

0

T

I

A

0

0

⎥ ⎢

⎥ ⎥

⎢⎢

2

2

⎥ ⎢

⎥ ⎥

0

⎢⎢

1

P

0

0⎥

⎥ ⎢ 0

I

0

0

A

I A

A

⎥ ⎥

(7)

T

T

d 1

x ⎢⎢0 0 0 0 0

dN

⎥ −

⎥ − ⎢

⎥ ⎥

k

P

P x

⎢⎢

0

0

0

0

0

T

k

⎥ ⎢

d

A 1 0

0⎥ ⎥

⎢⎢

0

0⎥

⎥ ⎢

⎥ ⎥

⎢⎢⎣0 0 0 0 0⎥

⎥ ⎢

⎥ ⎥

⎣ 0

0

0

0 ⎥⎦ ⎢⎣ 0

T

A

0

0⎥ ⎥

dN

⎦ ⎦

The difference Δ 2

V ( x )

k is given by

N

k

N

k−1

T

T

Δ

=

= ∑ ∑

2

V ( x )

k

2

V ( xk+1)

2

V ( x )

k

xl i

Q xl ∑ ∑ xl i

Q xl

i=1 l= k+1− d

i=1

ki

l= kdki

Note that

N

k

N

kid

N

k−1

N

T

T

T

T

∑ ∑ x

= ∑ ∑

+ ∑ ∑

+

l

i

Q xl

xl i

Q xl

xl i

Q xl xk i

Q xk

i=1 l= k+1− d

i=1 l= k+1− d

i=1 l= k+1− d

i=1

ki

ki

i

N

k−1

N

k−1

N

T

T

T

∑ ∑ x

= ∑ ∑

+

l

i

Q xl

xl i

Q xl xkd Q x

ki

i kdki

i=1 l= kd

i=1 l= k+1− d

i=1

ki

ki

Using this, Δ 2

V ( x )

k can be rewritten as

N

N

N

kid

T

T

T

Δ

= ∑

− ∑

+

2

V ( x )

k

xk i

Q xk

xkd Q x

∑ ∑ x Q x

ki

i k dki

l

i l

i=1

i=1

i=1 l= k+1− dki

(8)

N

k−1

N

k−1

T

T

+∑ ∑ x Q x −∑ ∑ x Q x .

l

i l

l

i l

i=1 l= k+1− d

i=1 l= k+1

i

dki

For Δ 3

V ( x ),

k

we have

N d +1

k

N d +1

i

i

k−1

T

T

Δ

= ∑ ∑ ∑

3

V ( x )

k

xm i

R xm ∑ ∑

xm iRxm

i=1 l=− d +2 m= k+ l

i=1 l=− d +2 m= k+ l−1

i

i

N d +1

i

k−1

k−1

= ∑ ∑ [

T

T

T

T

x

+

− ∑

m i

R xm xk i

R xk

xm i

R xm xk+ l−1 i

R xk+ l−1]

(9)

i=1 l=− d +2

= +

= +

i

m k l

m k l

N d +1

i

N

kid

= ∑ ∑ [ T

T

x

T

T

= ∑

k

i

R xk xk+ l−1 i

R xk+ l−1]

[( d

d ) x R x

x R x ].

i

i

k

i k

l

i l

i=1 l=− d +2

=

i

i 1

l= k+1− i