MCE693/793: Analysis and

Control of Nonlinear Systems

Classical Lur’e Problem and Absolute Stability

Hanz Richter

Mechanical Engineering Department

Cleveland State University

Introduction

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Many nonlinear control problems involve a feedback loop involving a linearsystem and a static nonlinearity:

0−

y

φ(y)

G(s)e

A widely-studied case is to consider G(s) without direct transmission term:

x = Ax+Be

y = CTx

(D = 0 in the state-space representation).

G(s) = C(sI −A)−1B

e = −Φ(y)

Introduction

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Suppose Φ(y) = k, some constant gain. Recalling the Nyquist criterion, ifG(s) is stable and the Nyquist graph does not enclose the critical point −1/k,we conclude asymptotic stability.

When Φ(y) is a nonlinearity, we have used the describing function method asan approximation and the “critical point” −1/N(A,w).

This method is approximate and aimed at limit cycle analysis, but does notgive definitive results when applied to stability analysis.

The general (exact) stability problem for the above systems with stable G(s)and static nonlinearity is known as the Lur’e Problem after the paper by Lur’eand Postnikov, Stability theory of regulating systems, (Russian) PrikladnayaMatematika i Mekhanika (1944).

Sector-Bounded Nonlinearities

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A sector (also sectoric, sector-bounded) nonlinearity is a continuous functionΦ : R 7→ R satisfying

k1 ≤Φ(y)

y≤ k2

for some k1 ≥ 0, k2 ≥ 0 and for all y 6= 0. The definition implies Φ(0) = 0and yΦ(y) ≥ 0 (first and third quadrants).

y

Φ(y)

k1y

k2y

Absolute Stability and Aizerman Conjecture

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The Nyquist criterion gives a conclusive answer about stability wheneverΦ(y) = k (squashed sector). The Absolute Stability question (Lur’e problem)is:

If Φ is a sector nonlinearity, what conditions must G(s) satisfy (in terms of thesector bounds) so that the system is U.A.S.I.L.?

In late-40’s Soviet research they thought that for a sector nonlinearity, stabilitycould be ensured if the Nyquist criterion held for all k ∈ [k1, k2]. This wasknown as the Aizerman Conjecture.

Falsehood of the Aizerman Conjecture - Taylor

Example

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The Aizerman conjecture is false, because limit cycles can exist. Take

G(s) =10s2

s4 + 0.04s3 + 2.021s2 + 0.0404s+ 0.9803

and k1 = 0, k2 = ∞, with Φ(y) = y3.

However, it has been shown that the Aizerman conjecture is true for all firstand second-order systems and higher-order systems of special forms.

The Kalman Conjecture

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Kalman’s conjecture (1957) uses the maximum and minimum slopes of Φ(y)instead of the sector bounds. Suppose the slope of Φ satisfies

M ≤ dΦ

dy≤ M

Kalman conjecture: The closed-loop system is asymptotically stable if theNyquist plot of G(s) does not enclose the critical point −1/k for all valuesM ≤ k ≤ M .

The Kalman conjecture is also false, as a counterexample by Taylor shows.Taylor also showed that replacing the nonlinearity by its describing functionand using the Nyquist criterion is not a valid method in general.

The Popov Criterion

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Popov’s criterion is only a sufficient condition. Assumptions:

■ G(s) is asymptotically stable. In a state-space realization (A,B,C), werequire A to be Hurwitz and (A,B) controllable (a minimal realization, nopole-zero cancellations).

■ The nonlinearity is in the sector [0, k]

Popov’s criterion: If there exists a constant q such that for all w ≥ 0

Re [(1 + jwq)G(jw)] +1

k> 0

then the closed-loop system is asymptotically stable. If we separateG(jw) = G1(w) + jG2(w), we obtain the equivalent form:

G1(w)− wqG2(w) +1

k> 0

Popov Criterion- Graphical Interpretation

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The criterion requires G1(w) > − 1

k+ qwG2(w). Call x = G1(w) =Re(G(jw))

and y = wG2(w) = wIm(G(jw)). Also define W (jw) = x+ jy.

Then the Re-Im plot of W (jw) needs to be on one side of the Popov line

x− qy +1

k= 0

This is called a Popov plot (note that w is taken in [0,∞)).Im(W (jw)) = wIm(G(jw))

Re(W (jw)) = Re(G(jw))

−

1

k

1q

Popov lineW (jw)

Ranges for q and k

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Asymptotic stability is obtained when the inequality holds for the given k andsome q. In some cases (next example), q = 0 makes the inequality hold for thesector [0,∞]. This means that the system is asymptotically stable for anynonlinearity in that sector, including time-varying nonlinearities.

More precise information about the feasible intervals for q has been obtained(Narendra and Taylor, Frequency Domain Criteria for Absolute Stability, Hsuand Meyer, Modern Control Principles and Applications). For instance:

■ Single-valued, time-invariant nonlinearity in (0,∞): q ∈ R. But if k = ∞(relay): q ∈ [0,∞)

■ Time-independent hysteresis in (0,∞): q ∈ (−∞, 0]

■ General, time-dependent nonlinearity in (0,∞]: q = 0. That is, if noinformation about Φ is available, the inequality must hold with q = 0.

Examples

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Take

G(s) =s+ 3

s2 + 7s+ 10

We check that the system will be asymptotically stable for any nonlinearityΦ ∈ (0,∞]

Now take

G(s) =1

(s+ 1)(s+ 4)(s+ 5)

We see that for Φ = k, the Routh-Hurwitz criterion gives the range (0, 270) fork. For other nonlinearities, we formulate the problem of finding the maximumsector bound k for which there is a feasible q and solve it graphically.

Pole-Shifting Method

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The Popov criterion holds only for G(s) asymptotically stable. When this fails,we can use a loop transformation:

α

α

Φ G(s)+

−+

−

+

−

0 y

Φ′

G′(s)

Constant α must be chosen so that G′(s) = G(s)/(1 + αG(s)) becomesasymptotically stable and Φ′ = Φ− α is still in the first and third quadrants.

Note that this requires α ≥ α, where α is the low sector bound of Φ.

Example

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Take

G(s) =1

s(s+ 1)

Use pole shifting and the Popov criterion to analyze the absolute stability ofthe corresponding Lur’e system.

The Circle Criterion

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The Popov criterion does not take advantage of information about the lowsector bound when available. The circle criterion includes information aboutboth sector bounds [kl, kh] for Φ.

Let D(cl, cr) denote a circle centered on the real axis such that −1/cl and−1/cr are the real axis intersections.

For G(s) asymptotically stable and kl ≥ 0, the criterion has a simplestatement:

The Lur’e system is absolutely stable (asymptotically stable for allnonlinearities Φ ∈ [kl, kh]) if Φ(u, t) satisfies:

kl < cl ≤Φ

u≤ cr < kr

and the Nyquist plot of G(jw) does not touch or enclose D(cl, cr).

Recall that the Nyquist plot is traced with −∞ < w < ∞.

Circle Criterion - Graphical Interpretation

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Im (G(jw))

Re (G(jw))−

1

cl−

1

cr

The circle is not unique in general, and circles close to the imaginary axis yieldwider sectors.

Example: Saturation nonlinearity. The slope of the function within thesaturation band determines − 1

cr, and we can take −1/cl → −∞.

Circle Criterion - More General Systems

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The pole shifting method can be used for unstable systems, but a more generalversion of the criterion is available (see S&L). Assumptions:

■ G(s) has no purely imaginary poles and ρ poles with positive real parts.

■ The nonlinearity is in [k1, k2]

Then there are four cases, according to the location of [k1, k2] relative to zero:

1. 0 < k1 ≤ k2

2. 0 = k1 < k2

3. k1 < 0 < k2

4. k1 < k2 < 0

Circle Criterion - More General Systems

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The system is asymptotically stable if the condition matching the previouscases holds:

■ Case 1: The Nyquist plot of G does not enter the circle C(k1, k2) andencircles it ρ times counterclockwise.

■ Case 2: The Nyquist plot of G is contained in the half-spaceRe(z) > −1/k2.

■ Case 3: The Nyquist plot of G is in the interior of C(k1, k2)

■ Case 4: The Nyquist plot of −G does not enter the circle C(−k1,−k2)and encircles it ρ times counterclockwise.

Example

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Again, take

G(s) =1

s(s+ 1)

Use pole shifting and the circle criterion to analyze the absolute stability of thecorresponding Lur’e system.

Positive-Real Transfer Functions and the Lur’e Problem

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Suppose the linear system in the Lur’e problem is asymptotically stable. Whatadditional conditions can be imposed on the system so that the closed-loopsystem remains stable for any nonlinearities in the sector [0, 1]? Note can usepole shifting and/or scaling to find a more general sector.

Let G(s) have state-space realization (A,B,C, 0) and use the quadraticLyapunov function:

V (x) = xTPx

with P = P T > 0. We show in class that

V ≤ xT (ATP + PA)x+ 2xT (CT − PB)Φ− 2Φ2

For this, note that the sector condition is equivalent to

φ(y)(φ(y)− ky) ≤ 0

Lur’e Problem and Kalman-Yakubovich-Popov Lemma

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We found

V ≤ xT (ATP + PA)x+ 2xT (CT − PB)Φ− 2Φ2

If we can find L, ǫ > 0 so that

ATP + PA = −LTL− ǫP

BTP = C −√2L

we can show that

V ≤ −ǫxTPx− xTLTLx+ 2√2xTLTΦ− 2Φ2

soV ≤ −ǫxTPx− ||Lx−

√2Φ||2 ≤ −ǫxTPx

which shows global asymptotic stability.

SPR Transfer Functions and the KYP Lemma

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KYP Lemma: The above LMIs are always feasible wheneverG(s) = C(sI −A)−1B is strictly positive real, or SPR.

(See S&L, Sect. 4.6). A transfer function G(s) is positive real if

Re (G(s)) ≥ 0 whenever Re(s) ≥ 0

Also, G(s) is strictly positive real (SPR) if G(s− δ) is PR for some δ > 0.

That is, G(s) maps the closed half-space Re(z) ≥ 0 into itself.

Example: 1/(s+ a) is PR, in fact SPR.

SPR Transfer Functions : Frequency Domain

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Theorem: A transfer function G(s) is SPR if and only if:

1. G(s) is asymptotically stable

2. Re(G(jw) > 0 for all w ≥ 0.

The second condition indicates that SPR functions have Nyquist plots entirelycontained in the open right half of the complex plane. This also means thatthe phase shift must be within ±90◦.

Further, the relative degree (difference between number of poles and zeros)can be only 0 or 1, and G(s) must be minimum-phase (real parts of all zeroesmust be negative).

Example

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Note that the KYP lemma is equivalent to

ATP + PA = −Q

BTP = C

with Q = QT > 0. A direct way to work with the KYP lemma is to select Cfor any feasible P (always exists, since A is Hurwitz). We use a numericalexample to show stability against an arbitrary sector-bounded, time-varyingnonlinearity.

Lur’e Problem - General Setting

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Consider the MIMO system

x = Ax+Bp

q = Cx

pi = φi(qi)

where i = 1, 2..np. Suppose the nonlinearities are in the [0, 1] sector:

0 ≤ yφi(y) ≤ y2

or equivalentlyφi(y)(φi(y)− y) ≤ 0

Note the nonlinearities are still single-input, single output, and that positivefeedback is assumed.

MIMO Lur’e Problem and LMI Feasibility

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It can be shown (see Boyd, et. al. Linear Matrix Inequalities in System and

Control Theory) that closed-loop stability is equivalent to feasibility of theLMI:

ATP + PA PB +ATCTΛ+ CTT

BTP + ΛCA+ TC ΛCB +BTCTΛ− 2T

< 0

with P = P T > 0, Λ ≥ 0 and T ≥ 0, where Λ and T are diagonal matrices.