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7/29/2019 NonlinSys_pt2

1/18

Automatic Control III(Reglerteknik III)

fall 2012

4. Nonlinear systems, Part 2

(Chapter 12)

Hans Norlander

Systems and Control

Department of Information Technology

Uppsala University


2/18

REPETITION: EQUILIBRIA AND STABILITYNonlinear system:

= ( ), 0 = ( 0 0) ( 0 0) equilibrium point

Linearized model:

= + where =

( 0 0) =

( 0 0)

= 0 = 0

Let ( 0) be close to 0 and ( ) 0, then ( 0 0) is

stable if ( ) 0 is small for all 0

(globally) asymptotically stable if

(

)

0 as

(

(

0)) unstable if it is not stable (Definitions 12.12)

asymptotically stable/unstable if is asymptotically

stable/unstable (Theorems 12.12)

Fall 2012 Automatic Control III:4 part 2 Page 1


3/18

STABILITY, EXAMPLE

Case 1: Case 2:

=

2 31

1 32

=

2 + 31

1 + 32

For both cases = 0 is a stationary point with

=

0 1

1 0

eigenvalues

Theorems 12.12 does not apply what can be said about stability?



4/18

STABILITY, EXAMPLE, contdThe equilibrium point can be a stable or unstable focus, or a center.

The distance to the origin is

=

21 + 22

Examine the squared distance:

Set

(

) =

2

=

2

1 +

2

2

(

) = 2

1

1 + 2

2

2

Case 1:

( ) = 2 1( 2

31) + 2 2( 1

32) = 2(

41 +

42) 0

Case 2:

( ) = 2 1( 2 +

31) + 2 2( 1 +

32) = 2(

41 +

42) 0

The distance decreases for case 1, and increases for case 2.



5/18

STABILITY AND LYAPUNOV FUNCTIONS

Assume there is a function ( ), satisfying

(

) = 0

( ) 0 for =

( ) ( ) 0

( ) =

1

Interpret ( ) as a generalized distance from to

.

Along the solutions to the system = ( )

( ( )) =

( ( )) ( ) =

( ( )) ( ( )) 0

Such a function is called a Lyapunov function.



6/18

STABILITY AND LYAPUNOV FUNCTIONS, contd

Theorem 12.3: If a Lyapunov function satisfying

( ( )) ( ) = 0 for =

and ( ) as

can be found, then the equilibrium point

is globally

asymptotically stable.

Theorem 12.4: The first condition can be replaced by: No solution to

=

(

) lies completely in the area where

(

)

(

) = 0.

The tricky part is to find the Lyapunov function!



7/18

STABILITY AND LYAPUNOV FUNCTIONS, contdLet the properties of apply in a set .

Md

N

xo

Introduce the subset

:

= : ( )

Theorem 12.5: If (0)

, then 0 as long as ( )

. But

the solution cannot leave

. Hence, all solutions starting in

will converge to the stationary point.



8/18

STABILITY AND LYAPUNOV FUNCTIONS, contdFor the linear system

= ( 0 = 0 equilibrium)

try the Lyapunov function

( ) =

where is a positive definite matrix. Along the solutions

( ( )) = ( ) ( ) + ( ) ( )

= ( )( + ) ( )

If is a solution to the Lyapunov equation

+ = 0

then

(

(

)) =

(

)

(

)

0

0 = 0 asymptotically stable.



9/18


Lemma 12.1:

1. If has all eigenvalues strictly in the left half plane, then for

every matrix

=

0 (or

0), there is a matrix = 0 (or 0) that satisfies the Lyapunov equation.

2. If there are matrices = 0 and = 0 satisfying the

Lyapunov equation, and ( ) is detectable, then has all

eigenvalues strictly inside the left half plane.



10/18


Consider the nonlinear system

= + ( ) ( ) = ( )

Let have all eigenvalues in the left half plane, and let and be

positive definite matrices fulfilling

+

=

Then ( ) = is a Lyapunov function in an area around =

also for the nonlinear system. (Theorem 12.6)



11/18

REPETITION: SMALL GAIN THEOREM (Theorem 1.1)

1

2

-

-

1 1 1

2

2

2

Assume 1 and 2 stable. The closed loop system is input-output

stable if

1 2 1



12/18

THE CIRCLE CRITERIONConsider a linear system, with a nonlinear static feedback:

( )

(

)

1 1 1

2 2 2

Assume

(0) = 0 0 1 ( )

2 for = 0

Direct application of small gain theorem guarantees stability if

2 sup

(

) 1

which is too restrictive to be practical.



13/18

CIRCLE CRITERION, contd

Set

= 1 + 2

2

( ) = ( )

and

=

1 +

Then

( )

= 2 1

2



14/18

CIRCLE CRITERION, contdReformulation of the closed loop system:

(

)

(

)

( )

( )

( )

( )

P P

P

P

+ +

+

+

Cancel each other

Apply the small gain theorem for

( ) =

( )

1 + ( )and ( ) = ( )



15/18

CIRCLE CRITERION, contd

Applying the small gain theorem to the modified system description

gives a sufficient stability condition:

sup

(

) 1

that is

1

(

)

=

1 +

(

) (

)

=

1 (

)

+

Interpretation: The curve 1

(

) must not intersect the circle in the

complex plane with center in

and radius

.

More convenient and familiar if this can be reformulated to

conditions on the Nyquist curve!



16/18

THE NYQUIST CRITERIONLet - be a semicircle with radius

.

( ) maps - onto - .

ReRe

ImIm

-

-

( )

The Nyquist criterion: If-

encircles -1

times counter clockwise,the closed loop system has more poles in the RHP than

( ) has.

If - encircles -1 times clockwise, the closed loop system has fewer

poles in the RHP than

( ) has.



17/18

CIRCLE CRITERION, contdTheorem 12.7: Assume that ( ) has no poles in the right half plane,

and that (0) = 0, 1 ( ) 2 for = 0. Then the closed loop

system is input output stable if the Nyquist curve ( ), 0

does not enter, nor encircle the circle which intersects the negative

real axis (perpendicularly) in 1 1 and 1 2.

1/k1

1/k2

Re

Im



18/18

CIRCLE CRITERION, contdThe circle criterion also applies for unstable ( ) as long as ( ) is

stable.

Example: ( ) = 5 +2 5

2

2

+

, 1 = 0 5, 2 = stable closed loop.

3 2 1 0 1

2

1.5

1

0.5

0

0.5

1

1.5

2

1/k1

= 2

1/k2

= 0

Re

Im


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