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Lecture 5 Inputoutput stability

or

How to make a circle out of the point 1 + 0i, and differentways to stay away from it ...

r yG(s)

f()

y

k1y

k2y f(y)

1k1

1k2

G(i)

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Course Outline

Lecture 1-3 Modelling and basic phenomena

(linearization, phase plane, limit cycles)

Lecture 4-6 Analysis methods

(Lyapunov, circle criterion, describing functions)

Lecture 7-8 Common nonlinearities

(Saturation, friction, backlash, quantization)

Lecture 9-13 Design methods

(Lyapunov methods, Backstepping, Optimal control)

Lecture 14 Summary

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Todays Goal

To understand

signal norms

system gain

bounded input bounded output (BIBO) stability

To be able to analyze stability using

the Small Gain Theorem,

the Circle Criterion,

Passivity

Material

[Glad & Ljung]:Ch 1.5-1.6, 12.3 [Khalil]:Ch 57.1

lecture slides

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History

r yG(s)

f()

y

f(y)

For what G(s) and f() is the closed-loop system stable?

Lure and Postnikovs problem (1944)

Aizermans conjecture (1949) (False!)

Kalmans conjecture (1957) (False!)

Solution by Popov (1960) (Led to the Circle Criterion)

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Gain

Idea: Generalize static gain to nonlinear dynamical systems

u yS

The gain of Smeasures the largest amplification from uto y

Here Scan be a constant, a matrix, a linear time-invariant

system, a nonlinear system, etc

Question: How should we measure the size of u and y?

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Norms

A norm measures size.

Anormis a function from a space to R+, such that for all

x,y

x 0 and x = 0 x = 0

x + y x + y

x = x, for all R

Examples

Euclidean norm: x =x21 + + x

2n

Max norm: x = max{x1, . . . , xn}

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Signal Norms

A signal x(t) is a function from R+ to Rd.A signal norm is a way to measure the size of x(t).

Examples2-norm (energy norm):x2=

0

x(t)2dt

sup-norm: x= suptR+ x(t)

The space of signals with x2< is denotedL

2.

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Parsevals Theorem

TheoremIf x,y L2 have the Fourier transforms

X(i) =

0

eitx(t)dt, Y(i) =

0

eity(t)dt,

then

0

yT(t)x(t)dt = 12

Y(i)X(i)d.

In particular

x22

=

0

x(t)2dt = 1

2

X(i)2d.

x2< corresponds to bounded energy.

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System Gain

A system Sis a map between two signal spaces: y = S(u).

u yS

The gain of Sis defined as (S) = supuL2

y2u2

= supuL2

S(u)2u2

ExampleThe gain of a static relation y(t) = u(t) is

() = supuL2

u2u2

= supuL2

u2u2

=

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ExampleGain of a Stable Linear System

G

= supuL2

Gu2u2

= sup(0,)

G(i)

101

100

101

102

101

100

101

(G) G(i)

Proof:Assume G(i) K for (0,). Parsevals theoremgives

y22 = 1

2

Y(i)2d

= 12

G(i)2U(i)2d K2u22

This proves that (G) K. See [Khalil, Appendix C.10] for aproof of the equality.

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2 minute exercise: Show that(S1S2) (S1)(S2).

u y

S2 S1

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ExampleGain of a Static Nonlinearity

f(x) Kx, f(x) = K x

u(t) y(t)f() x

x

K xf(x)

y22 =

0

f2

u(t)

dt

0

K2u2(t)dt = K2u22

foru(t) =

x 0 t 10 t > 1

one has y2= Ku2 = Ku2

= (f) = sup

uL2

y2

u2

= K.

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BIBO Stability

u yS (S) = sup

uL2

y2u2

DefinitionSis bounded-input bounded-output (BIBO) stable if (S) < .

Example: If x = Axis asymptotically stable thenG(s) = C(sI A)1B + D is BIBO stable.

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The Small Gain Theorem

r1

r2

e1

e2

S1

S2

Theorem

AssumeS1 and S2 are BIBO stable. If

(S1)(S2) < 1

then the closed-loop map from (r1, r2) to (e1,e2) is BIBO stable.

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Proof of the Small Gain Theorem

Existence of solution (e1,e2) for every (r1, r2) has to be verified

separately. Then

e12 r12 + (S2)[r22 + (S1)e12]

gives

e12r12 + (S2)r221 (S2)(S1)

(S2)(S1) < 1, r12< , r22< give e12< .Similarly we get

e22r22 + (S1)r12

1 (S1)(S2)

so also e2 is bounded.

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Linear System with Static Nonlinear Feedback (1)

r yG(s)

f()

y

K y

f(y)

G(s) = 2

(s + 1)2 and 0

f(y)

y K

(G) = 2and (f) K.

The small gain theorem gives that K [0,1/2) implies BIBOstability.

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The Nyquist Theorem

G(s)

2 1.5 1 0.5 0 0.5 1

2

1.5

1

0.5

0

0.5

1

1.5

2

G()

Theorem

The closed loop system is stable iff the number of

counter-clockwise encirclements of 1by G(

) (note:increasing) equals the number of open loop unstable poles.

Th S ll G i Th b C i

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The Small Gain Theorem can be Conservative

Let f(y) = K yfor the previous system.

1 0.5 0 0.5 1 1.5 2

1

0.5

0

0.5

1

G(i)

The Nyquist Theorem proves stability when K [0,).The Small Gain Theorem proves stability when K [0,1/2).

Th Ci l C it i

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The Circle Criterion

Case 1: 0 < k1k2<

r y

G(s)

f()

y

k1y

k2yf(y)

1k1

1k2

G(i)

TheoremConsider a feedback loop with y = Guandu = f(y) + r. Assume G(s) is stable and that

0 < k1 f(y)

y k2.

If the Nyquist curve of G(s) does not intersect or encircle thecircle defined by the points 1/k1 and 1/k2, then theclosed-loop system is BIBO stable from rto y.

Oth

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Other cases

G: stable system

0 < k1

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Linear System with Static Nonlinear Feedback (2)

y

K yf(y)

1 0.5 0 0.5 1 1.5 2

1

0.5

0

0.5

1

1

KG(i)

The circle is defined by 1/k1= and 1/k2= 1/K.

min Re G(i) = 1/4so the Circle Criterion gives that if K [0,4) the system isBIBO stable.

Proof of the Circle Criterion

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Proof of the Circle Criterion

Let k = (k1 + k2)/2and

f(y) = f(y) ky. Then

f(y)y k2 k12 =: R

y1

y2

r1

r2

e1

e2

G(s)

f()

r1 G

G

k

y1

r2 f()

r1= r1 kr2

Proof of the Circle Criterion (contd)

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Proof of the Circle Criterion (cont d)

r1r2

G(s)

f()

k

R

1G(i)

SGT gives stability for

G(i)R < 1with

G=

G

1 + kG.

R < 1

G(i) = 1G(i)+ k

Transform this expression through z 1/z.

Lyapunov revisited

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Lyapunov revisited

Original idea: Energy is decreasing

x = f(x), x(0) = x0

V(x(T)) V(x(0)) 0

(+some other conditions on V)

New idea: Increase in stored energy added energy

x = f(x,u), x(0) = x0

y = h(x)

V(x(T)) V(x(0)) T0

(y,u) external power

dt (1)

Motivation

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Motivation

Will assume the external power has the form(y,u) = yTu.

Only interested in BIBO behavior. Note that

V 0with V(x(0)) = 0and (1)

T0

yTu dt 0

Motivated by this we make the following definition

Passive System

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Passive System

u yS

DefinitionThe system Sispassivefromuto yif

T0

yTu dt 0, for alluand allT> 0

andstrictly passivefromuto yif there > 0such that T0

yTu dt (y2T+ u2T), for alluand all T> 0

A Useful Notation

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A Useful Notation

Define thescalar product

y,uT =

T0

yT(t)u(t) dt

u yS

Cauchy-Schwarz inequality:

y,uT yTuT

where yT = y,yT. Note that y= y2.

2 minute exercise

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2 minute exercise

AssumeS1 and S2 are passive. Are then parallel connectionand series connection passive? How about inversion; S11 ?

u yS1

S2

u y

S1 S2

u yS11

2 minute exercise

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2 minute exercise

AssumeS1 and S2 are passive. Are then parallel connection

and series connection passive? How about inversion; S11

?

u yS1

S2

u yS1 S2

Passive Not passive

u,y = u,S1(u) + u,S2(u) 0 E.g., S1= S2= 1s

u yS11

Passive

u,y = S1(y),y 0

Feedback of Passive Systems is Passive

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Feedback of Passive Systems is Passive

r1

r2

y1

y2

e1

e2

S1

S2

If S1 and S2 are passive, then the closed-loop system from

(r1, r2) to (y1,y2) is also passive.

Proof: y, rT= y1, r1T+ y2, r2T

= y1, r1 y2T+ y2, r2 + y1T

= y1,e1T+ y2,e2T 0

Hence, y, rT 0if y1,e1T 0and y2,e2T 0

Passivity of Linear Systems

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Passivity of Linear Systems

TheoremAn asymptotically stable linear system G(s) is

passiveif and only if

Re G(i) 0, > 0

It isstrictly passiveif and only if there exists > 0such that

Re G(i) (1 + G(i)2), > 0

Example

G(s) = s + 1

s + 2 is passive and

strictly passive,

G(s) = 1

s is passive but not

strictly passive. 0 0.2 0.4 0.6 0.8 10.4

0.2

0

0.2

0.4

0.6

G(i)

A Strictly Passive System Has Finite Gain

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A Strictly Passive System Has Finite Gain

u y

S

If Sis strictly passive, then (S) < .

Proof:Note that y2 = limT yT.

(y2T+ u2T) y,uT yT uT y2 u2

Hence, y2T y2 u2, so lettingT gives

y2 1

u2

The Passivity Theorem

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y

r1

r2

y1

y2

e1

e2

S1

S2

TheoremIf S1

is strictly passive and S2

is passive, then the

closed-loop system is BIBO stable from rto y.

Proof of the Passivity Theorem

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y

S1 strictly passive and S2 passive give

y12T+ e12T y1,e1T+ y2,e2T= y, rTTherefore

y12T+ r1 y2, r1 y2T

1

y, rT

or

y12T+ y2

2T 2y2, r2T+ r1

2T

1

y, rT

Finally

y2T 2y2, r2T+1

y, rT

2 +

1

yTrT

LettingT gives y2 Cr2 and the result follows

Passivity Theorem is a Small Phase Theorem

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y

r1

r2

y1

y2

e1

e2

S1

S2

21

ExampleGain Adaptation

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p p

Applications in channel estimation in telecommunication, noise

cancelling etc.

Model

Processu

(t)

G(s)

G(s)

y

ym

Adaptation law:

d

dt = u(t)[ym(t) y(t)], > 0.

Gain AdaptationClosed-Loop System

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p p y

u

s

(t)

G(s)

G(s)

y

ym

Gain Adaptation is BIBO Stable

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u S

( )u ym y

s

G(s)

Sis passive (Exercise 4.12), so the closed-loop system is BIBO

stable if G(s) is strictly passive.

Simulation of Gain Adaptation

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Let G(s) = 1

s + 1

+ , = 1,u = sin t, (0) = 0and = 1

0 5 10 15 202

0

2

0 5 10 15 200

0.5

1

1.5

y, ym

Storage Function

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Consider the nonlinear control system

x = f(x,u), y = h(x)

Astorage functionis a C1 functionV : Rn R such that

V(0) = 0 and V(x) 0, x = 0

V(x) uTy, x,u

Remark:

V(T) represents the stored energy in the system

V(x(T)) stored energy att = T

T0

y(t)u(t)dt absorbed energy

+ V(x(0)) stored energy att = 0

,

T> 0

Storage Function and Passivity

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Lemma: If there exists a storage function Vfor a system

x = f(x,u), y = h(x)

with x(0) = 0, then the system is passive.

Proof:For all T> 0,

y,uT = T

0

y(t)u(t)dt V(x(T)) V(x(0)) =V(x(T)) 0

Lyapunov vs. Passivity

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Storage function is a generalization of Lyapunov function

Lyapunov idea: Energy is decreasing

V 0

Passivity idea: Increase in stored energy Added energy

V uTy

Example KYP Lemma

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Consider an asymptotically stable linear system

x = Ax + Bu, y = Cx

Assume there exists positive definite symmetric matrices P, Q

such that

ATP + PA = Q, and BTP =C

Consider V = 0.5xTPx. Then

V = 0.5( xTPx + xTP x) = 0.5xT(ATP + PA)x + uTBTPx

= 0.5xTQx + uTy < uTy, x = 0 (2)

and hence the system is strictly passive. This fact is part of the

Kalman-Yakubovich-Popov lemma.

Next Lecture

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Describing functions (analysis of oscillations)