Chapter 7: Stability

We are familiar with the intuitive concept of stability in the context of system poles of the transfer function:

)()()()(

)(1

1

n

mpspszszsK

sH−−−−

=LL

If the s are:pi ' The system is:

all in the left half-plane stable

any in the right half-plane unstable

in the left half-plane except marginally stable for possibly non-repeated poles on the Im-axis

These rules do not always account for different types of stability, nor for state-variable descriptions. Except for the fact that we know the eigenvalues of a SISO system's A-matrix to be the poles of the transfer function through the relationship:

H s C sI A B D

C adj sI AsI A

B D

( ) ( )( )

det( )

= − +

= −−

+

−1

Same expression used to findeigenvalues.

Next we'll define several kinds of useful types of stability and discuss new tests for these in terms of state-variable descriptions.

especiallynonlinear systems

DEFINITION: For a system given by a general equation (not necessarily linear):

is an equilibrium point if, in the absence of any input,

)),(),(()( ttutxftx =&

xe

x t x t te( ) .= ≥ for all 0

Because is constant, this implies that xe

),0,(0)( txftx e==&

(Or, in discrete-time, )x k x k x ke( ) ( ) .+ = = ≥1 0 for all

"equilibrium point" = "critical point""equilibrium point" = "critical point"

( ) 0=tu

00

constant

===

ux

xe

&

Note that for LTI (Linear Time-Invariant) systems, this results in:

So either (when A is full rank), or , in which case A has at least one zero eigenvalue and there are an infinite number of actual equilibrium points lying in a subspace that must, of course, pass through the origin. We will consider only A's of full rank, so the only equilibrium points considered are

Ax e = 0

xe = 0 )(ANxe ∈

xe = 0

It is the equilibrium point that we will classify as stable or unstable; i.e., does a system stray from its equilibrium, and if so, how far?

xe

xexe

xe

xe

Definitions of different types of stability:

F

unstable

marginally stable stable

stable

unstable

Denote the solution of a zero-input system of equations as:

(which is just for our linear systems), and let the origin be the equilibrium point of interest.

Φ( ; , )t x t0 0

Φ( , )t t x0 0xe = 0

DEFINITION: The origin is a stable equilibrium if for any there exists a such that if , then for all

ε > 0δ ε( , )t0 0> x t( )0 < δ x t( ) < ε

t t> 0.

stable "in the sense of Lyapunov" (i.s.L.)

DEFINITION: The origin is asymptotically stable if it is stableand: there exists an such that whenever then

′ >δ ( )t0 0 x t t( ) ( )0 0< ′δlim ( )t

x t→∞

= 0

( ) 0=tu

x1

x2

Picture for a two variable system:

ε

δ

unstable

stable i.s.L.

asymptotically stable

Extra Conditions:If (or ) are independent of the initial time, then the stability is called "uniform."

If (or ) can be chosen arbitrarily large, then the equilibrium point is globally stable, or stable in the large.

δ

δ ′δ

′δ

DEFINITION: If there is a fixed constant M such that for every t (or k), then the input is bounded. If for every bounded input and every initial condition there exists a scalar such that , then the system is BIBS (bounded input, bounded state) stable.

Mu ≤

x t( )00))(,,( 00 >txtMN s sNtx ≤)(

DEFINITION: If input u is bounded above by constant and there exists a scalar such that for all time the system output satisfies , the system is BIBO (bounded input, bounded output) stable.

MN

Ny ≤

Now we consider linear systems: Recall that the solution of a linear system can be written:

[ ]

τττ+Φ=

ττττ−δ+ττΦ+Φ=

+ττττΦ+Φ=

ττττΦ+Φ=

∫

∫

∫

∫

dutHtxtttC

duDtBttCtxtttC

tutDduBttCtxtttCty

duBttxtttx

t

t

t

t

t

t

t

t

)(),()(),()(

)()()()(),()()(),()(

)()()()(),()()(),()()(

)()(),()(),()(

0

0

0

0

00

00

00

00

or

∆

"weighting matrix"This is also known as the impulse response matrixof the system.

Use properties ofdelta function

Consider the zero-input solution the the state variables:

By Cauchy's Theorem,

x t t t x t( ) ( , ) ( )= Φ 0 0

x t t t x t( ) ( , ) ( )≤ Φ 0 0

If there is a bound on the state-transition matrix such that , then by choosing we can prove stability i.s.L. by the definition.

This is a necessary and sufficient condition for stability i.s.L. of the zero equilibrium state. (Proof of necessity is by contradiction).

)( 0tκ Φ( , )t t0)(),( 00 ttt κ≤Φ

)(),(0

0 tt κε=εδ

If in addition, , then the equilibrium is asymptotically stable.

Φ( , )t t0 0→

( ) 0=tu

∞→t

Now re-consider the input-related types of stability: BIBS and BIBO:

All linear systems that are asymptotically stable are also exponentially stable. This means that the norm of the state transition matrix is bounded by an exponential function as it tends to zero.

All linear systems that are asymptotically stable are also exponentially stable. This means that the norm of the state transition matrix is bounded by an exponential function as it tends to zero.

Theorem: A system is BIBS stable iff there exists a finite fixed value such that

for all

)( 0tN

∞<≤τττΦ∫ )()(),( 00

tNdBtt

t

t t≥ 0 .

( ) )tu (assume M<

∞<τττΦ+Φ≤

ττττΦ+Φ≤

ττττΦ+Φ≤

∫

∫

∫

MdBttxtt

duBttxtt

duBttxtttx

t

t

t

t

t

t

0

0

0

)(),()(),(

)()(),()(),(

)()(),()(),()(

00

00

00

Stability i.s.L. is necessary for BIBS stability, which we can see by setting above, so we know that As long as and that

then the state will be bounded.

u t( ) = 0∞<Φ ),( 0tt

∞<≤τττΦ∫ )()(),( 00

tNdBtt

t

triangle & Cauchyinequalities

boundedness of the input

For BIBO stability, we consider any initial conditions as having resulted from some bounded past input, so that we may write:

τττ= ∫∞−

dutHtyt

)(),()(

Then because we are considering only boundedinputs, we can conclude that

is necessary and sufficient for BIBO stability.

∞<≤ττ∫∞−

H

tNdtH ),(

Recall that we can find the 2-norm of a matrix M by taking the square root of the largest eigenvalue of M MT .

From slide 243with −∞=0t

( ) Mtu <

In practice, this integral test is not often used. Instead, it can be shown that:

A time-invariant system is BIBO stable iff all of the poles of the reduced transfer function are in the open left-half complex plane, or, for discrete-time systems, inside the open unit circle.

Note that eigenvalues and poles are not necessarily the same thing! (i.e., for time-varying systems)

Time-Invariant systems:

Φ

Ψ

( , )

( , )

( )t t e

k A

A t t

k0

0

0

=

=

− (continuous time )

(discrete time )

The behavior of these state-transition matrices is obvious from the forms above. A summary of zero-input stability properties in terms of the eigenvalues of the A-matrix is shown in the table:

Unstable if ...

Stable i.s.L. if ...

Asymptotically stable if ...

Continuous time Discrete timeAxx =& x k Ax k( ) ( )+ =1

Any , or if is repeated.

Re( )λi >0 Re( )λi ≥0λi

Re( )λi ≤0All , or if is repeated.

Re( )λ i < 0λi

All Re( )λ i < 0

Any or if is repeated.

λi >1 λi ≥1 λi

All or if is repeated.

λi ≤ 1 λiλi < 1

All λi < 1

Modal Stability and Classification of Equilibria:

Recall that each eigenvector corresponds to an eigenvalue, and that the eigenvectors span invariant subspaces in the state-space. Then we can talk about the stability of these subspaces by considering the value of the eigenvaluecorresponding to each such subspace.

Recall the phase portraits in the last chapter that illustrated the "phase planes" of some 2-D systems. Consider how they are constructed from their modes:

{ }

=

−−=λ

−

−=

1001

41

4001

M

A

Stable Node

Large motion along vertical eigenvector; toward origin. “fast mode” corresp. to l = -4

Small motion along horizontal eigenvector; toward origin. “slow mode” corresp. to l = -1

Eigenvectors = invariant subspaces

e2

e1

x1

x2

{ }

−

=

−−=λ

−−−−

=

2111

41

3212

M

A

Stable Node

Large motion along second eigenvector; toward origin. “fast mode” corresp. to l = -4

Large motion along first eigenvector; toward origin. “slow mode” corresp. to l = -1

Eigenvectors = invariant subspaces

e2

e1

x1

x2

Similar pictures can be drawn for 3-D systems, where the invariant subspaces might then be planes also.

Higher dimensions work the same way, but are difficult to imagine.

Lyapunov Stability Methods: This can be a powerful stability testing technique. It applies to many nonlinear systems as well.

Consider the (unforced) mechanical system:

K

B M

0=++ KxxBxM &&&

The energy stored in this system is :2

212

21),( KxxMxxV += &&

Normally we expect a passive system to have positive energy, which would require (sufficiently) that :

M K, > 0

If the system is "stable", this energy should dissipate, so the rate of change of energy over time should be negative:

Evaluating this "along the trajectories of the system", i.e., according to the differential equation,

xxKxxMxxV &&&&&& +=),(

( )

2

2

),(

xB

xKxxKxxB

xxKxxMxxV xxx MK

MB

&

&&&

&&&&&&&&&

−=

+−−=

+= −−=

So if we expect this passive physical system to dissipate positive energy, then we must have

All three of the constants should therefore be positive in order for the energy to be positive and the change in energy to be negative. This is the idea behind Lyapunov's Direct Method.

B > 0.

What would we ask as conditions on the coefficients of

for a stable system? (Hint: Routh-Hurwitz

( )Ms Bs K2 + +

Usually, though, the state equations have little intuitive senseof "energy" to them, so we generalize the terminology.

QuadraticFormula

0,, >KBM

Notation: A function of a state vector, , is positive definite if, in a neighborhood of an equilibrium point, and ( for semidefinite). (reverse the signs to get negative (semi)definite).

V x( )V ( )0 0=

V x x( ) .> ≠0 0 for V x( ) ≥ 0

Lyapunov Theorem: For any system with equilibrium point , if a positive definite function can be found such that is negative semidefinite, then the system is stable in the sense of Lyapunov. The function is then called a "Lyapunov Function."

If the derivative is negative definite, then the origin is asymptotically stable.

)(xfx =&f ( )0 0= V x( )

)(xV&

V x( )

)( xV&

Notes: What these theorems do and don't say:

1. The neighborhood in which the Lyapunov function is defined is the only* neighborhood in which the initial condition can lie for which stability is guaranteed. A system might be unstable if we leave this neighborhood.

2. The theorems do not say how to find such Lyapunovfunctions. Doing so generally takes lots of experience and trial and error (especially for nonlinear systems).

3. If we cannot find a Lyapunov function (our candidates do not work), this does not imply the system is unstable!!! (It just means we haven't shown it is!)

Theorem: If a system is globally asymptotically stable, then aLyapunov function exists that is valid for all .x ≠ 0

*When linear systems are stable, they are globally stable.

Linear time-invariant systems give some interesting results: we choose a quadratic form as a "candidate"Lyapunov function:

with P being real, symmetric, and positive-definite. Now compute the derivative:

V x x PxT( ) ,=

[ ]xPAPAx

xPAxPxAx

xPxPxxxV

TT

TT

TT

+=

+=

+= &&& )(

Axx =&

We want this to be a negative definite function, so . . .

Lets’s examineLyapunov’s method

Obvious V(x) is PD

A P PA QT + = −

where Q is some positive definite matrix. This is a famous equation called a Lyapunov equation.

So we can guess a P matrix and compute Q to see if it is positive definite (or semidefinite for i.s.L. stability).

A more common (and reliable) technique is choose a positive definite Q and solve the Lyapunov equation for P (which is often not easy). Any Q will do, so is often used.Q I=

Theorem: A system with system matrix A is asymptoticallystable iff the solution P of the Lyapunov equation is positive definite whenever Q is positive definite.

Theorem: A system with system matrix A is asymptoticallystable iff the solution P of the Lyapunov equation is positive definite whenever Q is positive definite.

Note that any p.d. Q will show whether a system is stable in the reverse approach, whereas choosing a Pand solving for Q will show neither stability or instability if Q turns out not to be positive definite.

Note that any p.d. Q will show whether a system is stable in the reverse approach, whereas choosing a Pand solving for Q will show neither stability or instability if Q turns out not to be positive definite.

Interestingly, the Lyapunov equation has a closed-form solution, but it is of little help, because it requires knowledge of the state-transition matrix for the LTI system:

dtQeeP AttAT

∫∞

=0

Example: Prove that the nonlinear system:

is asymptotically stable. Consider the candidate Lyapunov function:

22

122112212

21

)( xxbxbxaxax

xx

+−−−=

=

&

&

V x a x x( ) = +2 12

22

a a1 2 0, >

So V(x) is obviously positive definite. First find equilibra:

0

0

2

1 2 2 1 1 2 2 12

2

=

= − − − +

x

a x a x b x b x x( )

The only equilibrium is 021 == ee xx

This is obviously:

==≠<

0000

2

2

xx

if if

[ ]

21221

22

221

21221

22212

221212

22

122112212212

22112

)(22

)(2222

)(22

22)(

xbxbxxa

xbxbxxxaxaxxa

xxbxbxaxaxxxa

xxxxaxV

+−−=

+−−−=

+−−−+=

+= &&&Now:

So at any time that , the system is asymptotically stable. If the system is stable i.s.L. This is the "region of stability".

x t2 0( ) =x t2 0( ) ≠

Stability of time-varying systems:

Interestingly, if we freeze time and compute theeigenvalues of A(t) at that instant, the results do not tell us whether the system is stable, except in special cases wherein the eigenvalues are changing "slowly."

Also, when a time-varying system has an unstableeigenvalue, that does not necessarily imply that the system is unstable overall!! (unless all of them are unstable).

Lyapunov techniques are better for time-varying systems, but are still very difficult.