CHAPTER III - Stability and Robustness
NONLINEAR SYSTEMS
JOSE C. GEROMEL
DSCE / School of Electrical and Computer EngineeringUNICAMP, CP 6101, 13081 - 970, Campinas, SP, Brazil,
Campinas, Brazil, August 2008
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CHAPTER III - Stability and Robustness
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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CHAPTER III - Stability and Robustness
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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CHAPTER III - Stability and Robustness
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
Describing functions
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
Describing functions
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
Describing functions
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
Describing functions
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
Describing functions
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
Describing functions
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
Describing functions
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
The Nyquist criterion
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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CHAPTER III - Stability and Robustness
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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CHAPTER III - Stability and Robustness
The Lyapunov criterion
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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CHAPTER III - Stability and Robustness
The Lyapunov criterion
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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CHAPTER III - Stability and Robustness
The Lyapunov criterion
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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CHAPTER III - Stability and Robustness
The Lyapunov criterion
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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CHAPTER III - Stability and Robustness
The Lyapunov criterion
Time-varying systems
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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CHAPTER III - Stability and Robustness
The Lyapunov criterion
Time-varying systems
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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CHAPTER III - Stability and Robustness
The Lyapunov criterion
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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CHAPTER III - Stability and Robustness
The Lyapunov criterion
Switched linear systems
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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CHAPTER III - Stability and Robustness
The Lyapunov criterion
Switched linear systems
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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CHAPTER III - Stability and Robustness
The Lyapunov criterion
Switched linear systems
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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CHAPTER III - Stability and Robustness
The Lyapunov criterion
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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CHAPTER III - Stability and Robustness
The Lyapunov criterion
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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CHAPTER III - Stability and Robustness
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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CHAPTER III - Stability and Robustness
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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CHAPTER III - Stability and Robustness
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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CHAPTER III - Stability and Robustness
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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CHAPTER III - Stability and Robustness
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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CHAPTER III - Stability and Robustness
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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