NONLINEAR SYSTEMS - Unicampgeromel/non_stab.pdf · NONLINEAR SYSTEMS JOS´E C. GEROMEL DSCE / School of Electrical and Computer Engineering UNICAMP, CP 6101, 13081 - 970, Campinas,

CHAPTER III - Stability and Robustness

NONLINEAR SYSTEMS

JOSE C. GEROMEL

DSCE / School of Electrical and Computer EngineeringUNICAMP, CP 6101, 13081 - 970, Campinas, SP, Brazil,

[email protected]

Campinas, Brazil, August 2008

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Contents

1 CHAPTER III - Stability and RobustnessNote to the readerPreliminariesThe Nyquist criterion

Describing functions

The Lyapunov criterionRobustnessTime-varying systemsSwitched linear systems

The Popov criterionProblems

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Note to the reader

Note to the reader

This text is based on the following main references :

J. C. Geromel e R. H. Korogui, “Controle Linear de SistemasDinamicos : Teoria, Ensaios Praticos e Exercıcios (inPortuguese), Edgard Blucher Ltda, 2011.

J. C. Geromel and P. Colaneri, “Robust Stability of TimeVarying Polytopic Systems”, Systems & Control Letters, Vol.55, No 1, pp. 81-85, 2006.

J. C. Geromel and P. Colaneri, “Stability and Stabilization ofContinuous-time Switched Linear Systems”,SIAM Journal on

Control and Optimization, Vol. 45, No 5, pp. 1915-1930,2006.

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Preliminaries

Preliminaries

Stability and robustness analysis of dynamic systems (linear ornot) can be done with the following celebrated criteria:

Nyquist stability criterion - A frequency domain criteriondeveloped to deal with linear systems. However, it can begeneralized to cope with nonlinearities by adopting the socalled first harmonic approximation.Lyapunov criterion - A time domain criterion, sufficientlygeneral to cope with any dynamic system. The Popov criterionis a possible generalization to deal with a particular class ofnonlinear systems named Lur’e systems. Time-varying dynamicsystems can also be treated.

This chapter is devoted to study these criteria in theframework of nonlinear systems.

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The Nyquist criterion


The Nyquist criterion is based on the “Cauchy’s ResidueTheorem” applied to some function of complex variablef (z) : C → C defined in a domain D ⊂ C. It states that

1

2πj

∮

C

f (z)dz =

r∑

k=1

R(f , zk)

where :

z1, · · · , zr are isolated singular points of f (z).the closed contour C ⊂ C contains all points z1, · · · , zr in itsinterior.

⇓

The residue R(f , zk) is calculated by partial decomposition off (z)

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The Cauchy’s Residue Theorem is applied to prove that thefollowing equality holds

1

2πj

∮

C

g ′(z)

g(z)dz = Nz − Np

where Nz is the number of zeros of g(z) inside the closedcontour C ∈ C and Np is the number of poles of g(z) insidethe same contour.

Important fact : The isolated singular points of the function

f (z) :=g ′(z)

g(z)

are the poles and the zeros of g(z). Hence f (z) fails to beanalytic at the poles and zeros of g(z).

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Assume that z0 is a zero of multiplicity m0 of g(z), locatedinside the closed contour C . Hence,

g(z) = (z − z0)m0p(z)

where p(z) is analytic at z0 and p(z0) 6= 0 which provides

f (z) =m0

z − z0+

p′(z)

p(z)

However, since p′(z)/p(z) is analytic at z0 it can bedeveloped in Taylor series yielding the conclusion thatR(f , z0) = m0. Doing the same for all poles and zeros insideC the previous equality follows.

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The line integral can also be calculated from

∮

C

g ′(z)

g(z)dz =

∮

C

d ln(g(z))

= jarg (g(z))|C

which provides the final formula

Fact (Main formula)

1

2π∆C arg(g(z)) = Nz − Np

where Nz and Np are, respectively, the number of zeros and poles

of g(z) inside the closed contour C ∈ C.

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Example

Consider the function g(z) = 1(z+0.5)(z−2) and the closed

contours A, B and C as indicated below. Notice the poles ofg(z) indicated by × and the zeros of h(z) = 0.6 + g(z)indicated by ◦.

−4 −3 −2 −1 0 1 2 3−4

−3

−2

−1

0

1

2

3

4

Im(z)

Re(z)

AB

C

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Example

The figure below shows the closed contours obtained from A,B and C through the mapping of g(z). Notice the indicatedpoints (0, 0) and (−0.6, 0).

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1Im(g

(z))

Re(g(z))

A

B

C

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Example

The function g(z) has two poles {−0.5, 2} and no zeros.Hence, from the contours A, B and C we haveNz = 0,Np = 1, Nz = 0,Np = 1 and Nz = 0,Np = 2respectively.

Looking at the point (0, 0) we have (1/2π)∆A = −1,(1/2π)∆B = −1 and (1/2π)∆C = −2 respectively.

The function h(z) has two poles {−0.5, 2} and two zeros{0.75 ± j0.3227}. Hence, from the contours A, B and C wehave Nz = 2,Np = 1, Nz = 0,Np = 1 and Nz = 2,Np = 2respectively.

Looking at the point (−0.6, 0) we have (1/2π)∆A = 1,(1/2π)∆B = −1 and (1/2π)∆C = 0 respectively.

⇓

The main formula holds

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y

H(s)ur

κ+

−

e

Consider the linear system given above with κ > 0. Theclosed loop system characteristic equation is

1 + κH(s) = 0

From the mapping of C through H(s), looking at the criticalpoint −1/κ+ j0, we determine Ncrit = (1/2π)∆C arg(H(s)).

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Defining the closed contour C

Im

Re

∞

C

Denote Np the number of poles of H(s) inside C .From the mapping of C through H(s), looking at the criticalpoint −1/κ+ j0, the number of encirclements (with sign) Ncrit

is determined.Using the main formula we determine Nz = Ncrit + Np, thenumber of zeros of 1 + κH(s) inside C .

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The following conclusions can be drawn:

Nz = 0 indicates that the system is asymptotically stable.There are no roots of 1 + κH(s) = 0 inside C .Nz > 0 indicates that the system is unstable. There are Nz

roots of 1 + κH(s) = 0 inside C .

It may occur that

H(jωc ) = −1

κc

for some (κc , ωc). In this case, the mapping of H(s) passesover the critical point and Nz becomes undetermined. This isonly possible if s = ±jωc are roots of the characteristicequation 1 + κcH(s) = 0. The system oscillates!

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The following points are important as far as the Nyquistcriterion is concerned.

It is a necessary and sufficient condition for stability of lineartime invariant systems Nz = 0.For H(s) of interest we have

lim|s|→∞

H(s) = const

the mapping of C through H(s) is determined exclusively bythe frequency response

H(jω) ∀ω ∈ R

H(s) may be nonrational. For instance, it may contain termsof the form e−τ s when dealing with time-delay systems.

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y

H(s)ur

f (·)+

−

e

The main goal is to study the stability of the above closedloop system where f (·) : R → R is a memoryless nonlinearfunction.

It is also important to detect the existence of limit cycles, thatis periodic solutions.

The Nyquist criterion can be used, provided the nonlinearity isreplaced by its describing function.

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As for the frequency response of linear systems, consider thesinusoidal input e(t) = asin(ωt) with a > 0 and recall thatthe output of the nonlinear block u(t) = f (asin(ωt)) being aperiodic function with period T = 2π/ω such that

∫ π/ω

−π/ωu(t)dt = 0

can be expanded in Fourier series, whose truncation up to thefirst term is

u(t) ≈ αsin(ωt) + βcos(ωt) ∈ R

where

α =ω

π

∫ π/ω

−π/ωu(t)sin(ωt)dt , β =

ω

π

∫ π/ω

−π/ωu(t)cos(ωt)dt

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Defining the describing function associated to f (·) as

L(a) =α+ jβ

a∈ C

the immediate conclusion is that corresponding to the inputasin(ωt) we have the output

u(t) ≈ a|L(a)|sin(ωt +∠L(a))

That is, the describing function L(a) plays the same role fornonlinear systems than the transfer function does for LTIsystems.

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Some important remarks are as follows:

L(a) does not depend on ω ∈ R since adopting the change ofvariable θ = ωt we obtain

α =1

π

∫

π

−π

f (asin(θ))sin(θ)dθ , β =1

π

∫

π

−π

f (asin(θ))cos(θ)dθ

Whenever f (·) is an odd function then β = 0 and consequently

L(a) =1

aπ

∫

π

−π

f (asin(θ))sin(θ)dθ

is always a real function.

The describing function is merely a function L(a) : a > 0 → C

that can be numerically calculated for any given nonlinearity.

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Example

Example 1 : For the linear function f (e) = κe we obtain

L(a) =κ

π

∫ π

−πsin(θ)2dθ

= κ

As expected, in the linear case, the describing function isconstant and coincides with the function.

Example 2 : For the sign function (relay) defined as

f (e) =

{

M , e > 0−M , e < 0

assuming any value in the interval −M ≤ f (0) ≤ M.

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Example

Since this is an odd function, we have

L(a) =2

πa

∫ π

0Msin(θ)dθ

=4M

πa

Example 3 : For the (symmetric) hysteresis of the formf (e)

e

M

h

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Example

The describing function is determined from

f (asin θ)

θ

M

π

τ

−π

τ − π

where a sin(τ) = h. Hence, it is defined only for a ≥ h.

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Example

Consequently, simple integration provides

α =4M

πcos(τ) =

4M

πa

√

a2 − h2

and

β = −4M

πsin(τ) = −

4M

πah

yielding

L(a) =4M

πa2

√

a2 − h2 − j4M

πa2h

It is important to observe that, in this case, the describingfunction associated to the hysteresis is complex.

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y

H(s)ur

L(a)+

−

e

This figure shows the approximated block diagram where thenonlinear function is replaced by its describing function. TheNyquist criterion may be applied to the characteristic equation

1 + L(a)H(s) = 0

As before, the mapping of the closed contour C through H(s)is obtained and the critical point to be considered is −L(a)−1.

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In the complex plane ReH(jω) × ImH(jω) if we draw themapping H(jω) and the set of critical points −L(a)−1 for alla > 0, a point of interest is the one defined by the pair(ωc , ac) such that

H(jωc ) = −L(ac)−1

The consequence is interesting. Assuming thaty(t) = acsin(ωct), from the block diagram we obtain

y(t) ≈ ac |L(ac )||H(jωc )|sin(ωct + ∠L(ac) + ∠H(jωc ) + π)

≈ acsin(ωct)

showing that it is a solution of the system underconsideration. With the Nyquist criterion, the stability of thisperiodic solution can be established.

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The validity of this method does not depend on the describingfunction approximation alone. Indeed, from the block diagramit is seen that L(a)H(jω) must be a good approximation forthe frequency response of the operator

y(t) = h(t) ∗ f (e(t))

Even though L(a) is not a good approximation for thefrequency response of f (e(t)) the former condition still mayhold whenever

|H(jω)| << 1 , ∀ω > ωc

That is, errors introduced by the describing functionapproximation can be attenuated by the linear transferfunction H(jω).

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Example

Example 4 : Consider the previous block diagram where f (·)represents the hysteresis already treated and

H(s) =1

s(s + 1)

Simple calculations show that

−L(a)−1 = −π

4M

√

a2 − h2 − jπh

4M

which indicates that only its real part depends on a > 0. Foran hysteresis with M = h = 1, the next figure shows thecurves H(jω) obtained from the mapping of C which containthe origin s = 0 and −L(a)−1 for all a > 1.

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Example

−2 −1 0 1 2−2

−1

0

1

2

P

A

B

Point P is characterized by H(jω) = −L(a)−1 which provides

ωc = 0.7866 , ac = 1.2723

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Example

Noticing that Np = 0 due to the fact that both poles of H(s)are outside C , applying the Nyquist criterion we obtain:

Point A : Ncrit = 1 yields Nz = 1, the system is unstable. Inthis case, a increases and A → P .Point B : Ncrit = 0 yields Nz = 0, the system is stable. In thiscase, a decreases and B → P .

⇓

The conclusion is that P defines a stable limit cycle. Its firstharmonic approximation is

y(t) = 1.2723 sin(0.7866t + φ)

with φ arbitrary.

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Example

0 5 10 15 20 25−10

−8

−6

−4

−2

0

2

4

6

8

10

−10 −5 0 5 10−3

−2

−1

0

1

2

3

y(t) × t y × y

The above figures show the trajectories of the system and thefirst harmonic approximation of the limit cycle with φ = 0. Itis confirmed that the limit cycle is stable.

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Example

Example 5 : To apply the describing function approach tothe Van der Pol’s equation

y − µ(1− y2)y + y = 0

where µ > 0, we first rewrite it as

y − µy + y = µu , u = (−y)3/3

to put in evidence that it can be represented by the previousblock diagram with r = 0, e = −y and

H(s) =µs

s2 − µs + 1, f (e) = e3/3

Notice that H(s) is unstable and f (e) is an odd function.

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Example

The describing function associated to f (·)

L(a) =a2

3π

∫ π

−πsin4(θ)dθ

= a2/4

is real and positive. On the other hand, the equalityH(jωc ) = −L(ac)

−1 holds for

ωc = 1 , ac = 2

yielding a first harmonic approximation of the limit cycle asbeing y(t) ≈ 2sin(t). Notice that it does not depend on theparameter µ > 0. Consequently, this approximation may bevery crude for certain values of this parameter.

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Example

−2 −1.5 −1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

P

A B

Np = 2 since H(s) has two poles inside C . We have:Point A : Ncrit = 0 yields Nz = 2, the system is unstable. Inthis case, a increases and A → P .Point B : Ncrit = −2 yields Nz = 0, the system is stable. Inthis case, a decreases and B → P .

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Example

−2 0 2−10

−8

−6

−4

−2

0

2

4

6

8

10

0 20 40 60−3

−2

−1

0

1

2

3

4

5

y(t) × t y × y

The limit cycle defined by point P is stable. The above figuresshow the trajectories of the system and the first harmonicapproximation of the limit cycle for µ = 5.

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The Lyapunov criterion

Robustness

An important class of linear systems subject to parameteruncertainty is defined as

x = Aλx

where x ∈ Rn and Aλ =

∑Ni=1 λiAi . The vector

λ ∈ Λ ={

λ ∈ RN : λ1 + · · ·+ λN = 1, λi ≥ 0

}

which defines a particular convex combination is unknown.The stability conditions follow from the quadratic in the stateLyapunov function

v(x) = x ′Px

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Robustness

Indeed, assuming that P ∈ Rn×n is parameter independent,

the time derivative written in the form

v(x) =

N∑

i=1

λix′(A′

iP + PAi )x

provides the celebrated quadratic stability condition:

Lemma

If there exists a symmetric matrix P > 0 satisfying the LMIs

A′iP + PAi < 0, i = 1, · · · ,N

then the system is asymptotically stable.

It is important to keep in mind that this is a sufficient stabilitycondition valid even though λ ∈ Λ is time varying.

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Robustness

A more accurate result, for λ ∈ Λ unknown but time invariantis obtained with the Lyapunov function

v(x) = x ′Pλx

where, as before, Pλ =∑N

i=1 λiPi .

Lemma

If there exist symmetric matrices Pi > 0 and Qi for i = 1, · · · ,Nsatisfying the LMIs

A′iPj + PjAi + Qi − Qj < 0 i , j = 1, · · · ,N

the system is asymptotically stable.

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Robustness

Actually, multiplying those N2 LMIs by λiλj , and summing allterms it is verified that

A′λPλ + PλAλ < 0

holds, implying that v(x) < 0, for all λ ∈ Λ.

Imposing Pi = P > 0 and Qi = Q for all i = 1, · · · ,N , thequadratic stability condition is recovered as a special case.For i = j = 1, · · · ,N we have

A′iPi + PiAi < 0

requiring, once again, that matrices Ai for all i = 1, · · · ,Nmust be asymptotically stable.

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Time-varying systems

Assume that the unknown parameter is time-varying, that isλ(t) ∈ Λ for all t ≥ 0 and satisfies λ = Πλ for some λ(0) ∈ Λ.To be sure that λ(t) ∈ Λ for all t ≥ 0 we need

N∑

i=1

πij = 0 , j = 1, · · · ,N

Moreover, adopting the parameter dependent Lyapunovfunction v(x) = x ′Pλx , asymptotical stability is assuredwhenever

Pλ > 0 , A′λPλ + PλAλ + PΠλ < 0

for all λ ∈ Λ.

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A possible way to impose that inequality is given in the nextlemma:

Lemma

If there exist symmetric matrices Pi > 0 and Qi for i = 1, · · · ,Nsatisfying the LMIs

A′iPj + PjAi + Qi − Qj +

N∑

k=1

πkiPk < 0 , i , j = 1, · · · ,N

then the system is asymptotically stable.

This is true becauseN∑

i=1

N∑

k=1

λiπkiPk =

N∑

k=1

λkPk

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Considering i = j = 1, · · · ,N in the previous stabilityconditions we obtain

A′iPi + PiAi +

N∑

j=1

πjiPj < 0

which does not require matrices Ai , i = 1, · · · ,N beasymptotically stable, since for each i there always exits anindex j = 1, · · · ,N such that πji ≤ 0.

Generally but not always, matrix Π belongs to the well knownclass of Metzler matrices M with all nonnegative off diagonalelements, that is

πij ≥ 0, ∀i 6= j

In this case, πii = −∑N

j 6=i=1 πji ≤ 0, for all i = 1, · · · ,N.

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Switched linear systems

A switched linear system is characterized by the state spacerealization

x(t) = Aσ(t)x(t), x(0) = x0

where x(t) ∈ Rn is the state and σ(t) : R → {1, · · · ,N} is

the switching rule. Notice that σ(·) is a control variable to bedetermined in order to preserve asymptotic stability and solve

minσ

∫ ∞

0x(t)′Qσ(t)x(t)dt

where matrices Qi ≥ 0 for all i = 1, · · · ,N. It is recognizedthat this is a very hard to solve problem and so onlyguaranteed cost solutions (minimum upper bound) aresearched.

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The main difficulty is that σ(t) is a discontinuous function.The next result follows from the continuous butnon-differentiable Lyapunov function v(x) = mini x

′Pix .

Theorem

If there exist matrices Pi > 0 and Π ∈ M such that

A′iPi + PiAi +

N∑

j=1

πjiPj + Qi < 0 , i = 1, · · · ,N

then with σ(t) = argmini x(t)′Pix(t) the closed loop system is

globally asymptotically stable and

∫ ∞

0x(t)′Qσ(t)x(t)dt < min

ix ′0Pix0

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Some remarks are important:

The so called Lyapunov-Metzler inequalities are very hard tosolve due to the product of variables πjiPj .Since πii ≤ 0 for all i = 1, · · · ,N the existence of a feasiblesolution does not require Ai be asymptotically stable.

A more conservative but easier to solve condition is as follows:

Corollary

The previous theorem holds whenever the Lyapunov-Metzler

inequalities are replaced by

A′iPi + PiAi + γ(Pj − Pi ) + Qi < 0 , i 6= j = 1, · · · ,N

where Pi > 0 for all i = 1, · · · ,N and γ > 0.

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It is clear that the minimum guaranteed cost is obtained fromthe solution of the problem

mini=1,··· ,N

infγ,P1,··· ,PN∈Φ

x ′0Pix0

where Φ is the set of all solutions of the modifiedLyapubnov-Metzler inequalities. The inner minimization iscalculated using any LMI solver coupled to a line search. Theouter minimization is determined exhaustively.

In this framework, several generalizations are possible,including those related to concepts similar to H2 and H∞

norms.

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Example

Let us consider a switched linear system composed by twosubsystems given by

A1 =

[

0 12 −9

]

, A2 =

[

0 1−2 2

]

The previous optimization problem has been solved forQ1 = Q2 = 0 and γ = 11.8, yielding the positive definitematrices

P1 =

[

6.7196 1.62931.6293 1.0222

]

, P2 =

[

6.0826 2.12932.1293 2.2206

]

which is used to implement the stabilizing state feedbackswitching control.

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Example

−5 −4 −3 −2 −1 0 1 2 3 4 5−5

−4

−3

−2

−1

0

1

2

3

4

5

x1

x2

The above figure shows the phase plane of the closed loopsystem with the switching control

σ(t) = arg mini=1,2

x(t)′Pix(t)

which is very effective for global stabilization.

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The Popov criterion

The Popov criterion

The celebrated Popov stability criterion applies to a class ofnonlinear systems denominated Lur’e systems, with statespace realization

x = Ax + Bw

y = Cx

w = −f (y)

where x ∈ Rn, w ∈ R and y ∈ R. The linear part (A,B ,C ) is

a SISO LTI system. Generalizations to cope with proper andMIMO systems exist. The nonlinearity is such that:

f (0) = 0.It belongs to the sector (0, κ > 0) that is (f (y)− κy)f (y) ≤ 0.

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The Popov criterion

The Popov criterion

The key idea to get the stability condition is to consider thePopov-Lyapunov function

v(x) = x ′Px + 2θ

∫ y

0f (ξ)dξ

where P = P ′ > 0 and θ ≥ 0. Notice that v(x) > 0 for allx 6= 0 ∈ R

n and v(0) = 0. Hence, imposing

v(x) < (w + κy)′w + w ′(w + κy)

for all [x ′ w ]′ 6= 0 ∈ Rn+1 we obtain the classical Popov’s

stability criterion expressed in terms of an LMI with respect tothe matrix P > 0 and the scalar θ ≥ 0.

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Problems

Problems

1. Apply the Nyquist criterion for the stability study of thealgebraic equation

1 + κH(s) = 0

where κ > 0, considering the following cases:

H(s) = s+2s+10

H(s) = 1(s+10)(s+2)2

H(s) = (s+10)(s+1)(s+100)(s+2)3

H(s) = (s+1)s2(s+10)

H(s) = s+1s(s+2)

H(s) = 1(s+2)(s2+9)

Moreover, in each case, determine the critical value κc .

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Problems

Problems

2. For the following nonlinear functions

f (e) =

M , e > ∆/20 , |e| < ∆/2

−M , e < −∆/2

f (e) =

e + 1 , e > 1/23e , |e| < 1/2

e − 1 , e < −1/2

f (e) =

κ(e −∆/2) , e > ∆/20 , |e| < ∆/2

κ(e +∆/2) , e < −∆/2

determine the describing functions and draw in the complexplane the critical path −L(a)−1.

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Problems

Problems

3. Consider the nonlinear system with the structure used in thestudy of describing functions where

H(s) =1− s/4

s2 + s/2 + 1

and f (·) is the nonlinearity given in the item c) of Problem 1.Verify the existence of a limit cycle. In the affirmative, caseverify if it is stable and determine its amplitude and frequencyby the describing function approximation.

4. Consider the nonlinear system with the same structure wherethe nonlinearity is an hysteresis.

For H(s) = 1/(s + 1), a limit cycle exists ?For H(s) = 1/(s2 + 2ξs + 1), determine ξ > 0 for theexistence of a stable limit cycle. Determine its amplitude andfrequency by the describing function approximation.

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Problems

Problems

5. Consider the nonlinear system with the same structure wherethe nonlinearity is the saturation function

f (e) =

κ , e > 1κe , |e| < 1−κ , e < −1

and H(s) has the state space realization

x =

0 1 00 0 10 −1 −2

x +

001

u

y =[

1 0 0]

x

Determine κ > 0 to assure the existence of a stable limitcycle. Determine its amplitude and frequency by thedescribing function approximation.

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NONLINEAR SYSTEMS - Unicampgeromel/non_stab.pdf · NONLINEAR SYSTEMS JOS´E C. GEROMEL DSCE / School of Electrical and Computer Engineering UNICAMP, CP 6101, 13081 - 970, Campinas,