Lecture 6 Stability

ENIN 811System Analysis and SynthesisSystem Analysis and Synthesis

Dr. Wei Peng

Lecture 6:

Stability

ENIN 811 2014 Winter 1 Dr. Wei Peng

1 Introduction1 Introduction

System are designed to perform some System are designed to perform some task or to process signals. If a system is not stable, the system may burn out, disintegrate or saturate when a signal disintegrate, or saturate when a signal , no matter how small, is applied. Therefore an unstable system is useless in practice and stability is a basic in practice and stability is a basic requirement for all system.

ENIN 811 2014 Winter Dr. Wei Peng2


The response of linear systems can The response of linear systems can always be decomposed as the zero-state response and the zero-input response. We will introduce the BIBO (boundedWe will introduce the BIBO (bounded-input bounded-output) stability for the zero-state response, and marginal and asymptotic stability for the zero input asymptotic stability for the zero input response.


2 Input-Output Stability of LTI Systems

Consider a SISO linear time-invariant (LTI) system described byy y

( ) ( ) ( ) ( ) ( ) == tt dgtudtguty (causal)( ) ( ) ( ) ( ) ( ) == dgtudtguty 00 (causal) where g(t) is the impulse response and the

t i li ti i i t l d system is linear time-invariant, causal, and relaxed at t=0.



Definition: A system is said to be BIBO stable (bounded-input bounded-output stable) if every ( p p ) ybounded input excites a bounded output.

Bounded input ( )


Theorem 5.1 A SISO system described by (5.1) is BIBO stable if and only if g(t) is absolutely y g( ) yintegrable in [0,) , i.e. there exists a constant M such thatM such that

( )


Proof: (i) let u(t) be a input with utu )( Proof: (i) let u(t) be a input with for all

tt

mutu )(0t

( ) ( ) ( ) ( ) ( )= tt dgtudgtuty 00 ( ) ( )


(ii) BIBO stable ?(ii) BIBO stable ?We know Suppose( )


Theorem If a input-output system with impulse response, has:p p ,

(1) Step input (or a constant input), the output converges to a constant functionconverges to a constant function.

(2) Sinusoidal input, the output converges to a sinusoidal function.

(3) Any input not converging to 0 the output (3) Any input not converging to 0, the output will converge to the unstable input mode.



Th 5 2 If t ith i l Theorem 5.2 If a system with impulse response g(t) is BIBO stable, then, as t :1. The output excited by u(t)=a, for t 0,

approaches g(0)*aapproaches g(0) a.2. The output excited by u(t) = sin0t, for t 0,

approaches

with is the Laplace transform of g(t), i.e.



Proof: If u(t)=a for all t0 then (5 1) becomesProof: If u(t)=a for all t0, then (5.1) becomes

Which implies:

If u(t) = sin0t, then (5.1) becomes:( ) 0 , ( )



Thus we have as t Thus we have, as t ,

Replacing s by jw in (5.2) yields

So we have

d and



Substituting these into (5 3) givesSubstituting these into (5.3) gives

Theorem 5.3A SISO t ith ti l t f A SISO system with proper rational transfer function is BIBO stable if and only if every

l f h i l pole of has a negative real part or, equivalently, lies inside the left-half s-plane


Pole and Reduce FormPole and Reduce Form A rational function is a ratio of polynomials A rational function is a ratio of polynomials

N(s)/D(s) If the numerator N(s) and the denominator D(s)

have no roots in common, then the rational function N(s)/D(s) is in Minimal FormExample (s 2)/(s2 4) is not in reduced form Example. (s-2)/(s2 - 4) is not in reduced form, because s = 2 is a root of both numerator and denominator. We can rewrite this in reduced form as


Pole and Reduce FormPole and Reduce Form Poles: For a rational function in Minimal Form Poles: For a rational function in Minimal Form

the poles are the values of s where the d i t i l t denominator is equal to zero


ExampleExample

Consider the positive feedback system shown Consider the positive feedback system shown in above figure, Its impulse response was computed as

where the gain a can be positive or negative. W hWe have

and


SolutionSolutionThus we conclude that the positive feedback Thus we conclude that the positive feedback

system is BIBO stable if and only if the gain a has a magnitude less than 1 has a magnitude less than 1.

The transfer function of the system was d computed as

It is an irrational function of s and Theorem It is an irrational function of s and Theorem 5.3 is not applicable. In this case, it is simpler to use Theorem 5 1 to check its simpler to use Theorem 5.1 to check its stabilityENIN 811 2014 Winter Dr. Wei Peng18

BIBO Stability of MIMO SystemsBIBO Stability of MIMO Systems

Theorem 5 M1Theorem 5.M1 A multivariable system with impulse

response matrix is BIBO stable if and only if every is absolutely integrable in [0, ).

Theorem 5.M3Theorem 5.M3 A multivariable system with proper rational

transfer matrix is BIBO stable if and transfer matrix is BIBO stable if and only if has a negative real part.


BIBO Stability of a State Space EquationBIBO Stability of a State Space Equation


Poles of G(s) and Eigenvalues of APoles of G(s) and Eigenvalues of A


3 Internal Stability 3 Internal Stability


Marginal stability(stability in the sense of Lyapunov)

(1) Definition: Every finite excites a bounded state response

0x( )txp(2) A d ffi i bl di i

( ) 0,00

Asymptotic stabilityAsymptotic stability

(1) Definition: marginal stable + ( )( ) 0lim =txmarginal stable + ( )

(2) A necessary stable condition: ( ) 0lim txt

( ) .0Re

Minimal Polynomial Minimal Polynomial


Example Example

Consider

BecauseBecause


Solution Solution


4 Lyapunov Theory4 Lyapunov TheoryMotivations:Motivations: In the previous section, we determine stability of

LTI systems by analytically solving an ODE LTI systems by analytically solving an ODE (ordinary differential equation). Then we determine stability according to the obtained solution For stability according to the obtained solution. For general nonlinear systems, it is difficult, or impossible to get the desired analytic solution impossible, to get the desired analytic solution. Then how to determine stability?Lyapunov (direct) method direct means that we Lyapunov (direct) method --direct means that we doesnt need to solve the ode first. We just directly determine stability from the ODEdirectly determine stability from the ODE.


Homogeneous systemHomogeneous system

Homogeneous system and equilibrium stateThe system with no control input is called The system with no control input is called

homogeneous system. Usually, the time varying homogeneous system can be varying homogeneous system can be described by


4 Lyapunov Theory4 Lyapunov Theory


Equilibrium StateEquilibrium State


Stability in the Sense of LyapunovStability in the Sense of Lyapunov

Definition 5.3 (Stability in the sense of Lyapunov) Suppose that xe is a isolated equilibrium state. pp e qWe say that xe is stable in the sense of Lyapunovif for any given there exists a real number if, for any given , there exists a real number r such that the motion excited by initial state x which satisfiesby initial state x0 which satisfies

(5.15)meets with ( ); tttt

( )

(5 16)ENIN 811 2014 Winter Dr. Wei Peng36

( ) 000 ,,; ttxtxt e (5.16)

Stability in the Sense of Lyapunovy y p


Asymptotical StabilityAsymptotical Stability Definition 5 4 (Asymptotical stability) A isolated Definition 5.4 (Asymptotical stability) A isolated

equilibrium state xe of dynamic system (5.13) is ll d t ti ll t bl if called asymptotically stable if

(i) xe is stable in the sense of Lyapunov;( ) e y p ;(ii) for given in definition 5.3 and any , there exists a real number such that there exists a real number , such that the motion excited by initial state x0which satisfies (5.15) also meets with

( ) )(; tTttxtxt + (5 17)ENIN 811 2014 Winter Dr. Wei Peng38

( ) ),,(,,; 0000 tTttxtxt e + (5.17)

Asymptotical StabilityAsymptotical Stability(5 17) is equivalent to(5.17) is equivalent to

The concept of Definition 5.4 can be explained by the Fig.5.2

ENIN 811 2014 Winter Dr. Wei Peng39Fig.5.2 Asymptotical stable equilibrium state

Global Asymptotical StabilityGlobal Asymptotical Stability

Definition 5.5 (Global asymptotical stability) if a motion excited by stability) if a motion excited by any un-zero state vector x0 in state

fspace is bounded and satisfies

then we say that original equilibrium state xe =0 is whole asymptotically stablestable.


Global Asymptotical StabilityGlobal Asymptotical Stability


Example -- Cannot Use Eigenvalues in LTV System

Eigenvalues can not be used to determine the Eigenvalues can not be used to determine the stability of LTV systems

Consider: xe

xtAxt

== 1)(2

&Consider: xxtAx ==

10)(

Then:

( ) ( ) ( ) ( ) ttt eeettt 5.001 ( ) ( ) ( ) ( )=== tettt 00,,121 ENIN 811 2014 Winter Dr. Wei Peng42

Example p-- Cannot Use Eigenvalues in LTV System

( )( ) ( ) ( ) ( )

===

t

ttt

eeee

ttt0

5.00,,121

B l t f ( t t) g ith t b d

e0Because element of (et e-t) grows without bound,

the equation is neither asymptotically stable nor marginally stable. This example shows that even though the eigenvalues can be defined for A(t) at g g ( )every t , the concept of eigenvalues is not useful in the time-varying casein the time varying case.


Global Asymptotical StabilityFor Dynamic System

Theorem 5 5 (global asymptotical stability) For Theorem 5.5 (global asymptotical stability) For dynamic system (5.13), if there is a scalar f ti V( t) ith fi t d ti function V(x, t) with first-order continuous partial derivative and V(0,t)=0 satisfy:

(i) V(x,t) is positive definite and bounded, that is there exit two continuous non-decrease is, there exit two continuous non decrease scalar functions and with and such that for any t t and any and such that for any t t0 and any x0, following inequalities holding



(ii) the derivative of V(x,t) with respect to tis negative definite and bounded that is negative definite and bounded, that is, there is a continuous non-decrease

fscalar function with such that for any t t0 and x0, we havet at o a y t t0 a d 0, e a e


Global Asymptotical Stability forTime-Invariant System



Theorem 5.6 For time-invariant system (5.18), if there exists a continuous scalar function V(x) with first-order derivative and V(0)=0,satisfies(i) V(x) is positive definite;(ii) is negative definite;(ii) is negative definite;(iii) as , we have then the original equilibrium state 0 is globally asymptotically stable.


ExampleExample Consider time-invariant system Consider time-invariant system

Obviously the originalObviously the original

is its solo equilibrium state Consider a scalar is its solo equilibrium state. Consider a scalar function of


Solution Solution we know that (i) V(x) is positive definite;we know that (i) V(x) is positive definite;(ii) By the computation of the derivative, we have


Solution Solution

(iii) as ,

we have

by Theorem 5 6 we know that the original by Theorem 5.6, we know that the original equilibrium state is globally asymptotically t blstable.



Theorem 5.7 For time-invariant system (5.18), if there exists a continuous scalar function V(x) with first-order derivative and V(0)=0, satisfiessatisfies(i) V(x) is positive definite;(ii) is negative semi-definite;(iii) the set contains no

trajectories other than trivial trajectory(iv) as( )



then the original equilibrium state 0 is globallyasymptotically stableasymptotically stable. An equivalent expression of (iii)

The solution of with respective to x(t) is not any trajectory (solution) of (5.18) (t) s ot a y t ajecto y (so ut o ) o (5 8)excited by any initial state except to 0.


Semi DefinitionSemi-Definition

A continuously differentiable function f(x) is A continuously differentiable function f(x) is negative semidefinite ( or positive semidefinite ) on a neighborhood of the origin, if f(0) = 0, and f(x) 0 ( 0) for every non-zero x.( ) ( ) y

if f(0) = 0, and f(x) < 0 ( > 0) for every non-zero x. f(x) is negative definite (positive definite).zero x. f(x) is negative definite (positive definite).


Example Example

Consider time-invariant nonlinear system

Obviously, 0 is the solo equilibrium state.Select we haveSelect , we have(i) V(x) is positive definite;(ii)


Solution Solution

(ii)(ii)

is negative semi-definite (why?). So, at t l l d th t 0 ipresent, we can only conclude that 0 is

stable. Is it still asymptotically stable? We need to verify condition (iii) in Theorem 5.7.


Solution Solution

(iii) Suppose that we have(iii) Suppose that , we have(a) x1 is arbitrary but x2 = 0(b) x1 is arbitrary but x2 = -1

For situation (a), suppose that isa solution of original system excited by someinitial state this impliesinitial state, this implies


SolutionSolutionFor situation (b) we suppose that is a For situation (b), we suppose that is a

solution of original system excited by some initial state and we have state and we have

which is a contradictive result.Now we can conclude that 0 is asymptotically stable

since condition (iii) of Theorem 5.7 is satisfied.since condition (iii) of Theorem 5.7 is satisfied.(iv) Obviously, condition (iv) is also met with, so we say

that 0 is global asymptotically stablethat 0 is global asymptotically stable.ENIN 811 2014 Winter Dr. Wei Peng58

Lyapunov Theorem for Linear SystemLyapunov Theorem for Linear System

This part we apply the Lyapunov Theorem to a This part we apply the Lyapunov Theorem to a special case: linear system of

Definition 5 6 Definition 5.6 We call square matrix A stable if all the eigenvalues of A have negative real parts



Theorem 5 8Theorem 5.8All eigenvalues of A have negative real parts if and only if for any given positive definite symmetric matrix N, the Lyapunov equationsymmetric matrix N, the Lyapunov equationATM+MA= -N (5.19)

has a unique symmetric solution M and M is positive definite.positive definite.


Lyapunov Theorem for Linear System



For 3) Suppose there are two solutions MFor 3), Suppose there are two solutions M1and M2. Then we haveAT(M1-M2)+(M1-M2)A=0Multiplying from left side and from Multiplying from left side and from right side of above equation leads to

Integrating above equation from 0 to yields


Complex Conjugate TransposeComplex Conjugate Transpose

where A =

E l Example:


Lecture 6 Stability

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