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Lur'e Problem and Absolute Stability MCE693/793: Analysis and Control of Nonlinear Systems Classical Lur’e Problem and Absolute Stability Hanz Richter Mechanical Engineering Department Cleveland State University Introduction Many nonlinear control problems involve a feedback loop involving a linear system and a static nonlinearity: 0 e G(s) y − φ(y) A widely-studied case is to consider G(s) without direct transmission term: x˙ = Ax + Be y = CT x (D =0 in the state-space representation). −1 G(s) = C(sI A) B − e = Φ(y) − 2 / 25 Introduction Suppose Φ(y) = k, some constant gain. Recalling the Nyquist criterion, if G(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 as an approximation and the “critical point” 1/N(A, w). − This method is approximate and aimed at limit cycle analysis, but does not give 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’e and Postnikov, Stability theory of regulating systems, (Russian) Prikladnaya Matematika i Mekhanika (1944). 3 / 25 Sector-Bounded Nonlinearities A sector (also sectoric, sector-bounded) nonlinearity is a continuous function Φ: R R satisfying 7→ Φ(y) k1 k2 ≤ y ≤ for some k1 0, k2 0 and for all y =0. The definition implies Φ(0) = 0 ≥ ≥ 6 and yΦ(y) 0 (first and third quadrants). ≥ Φ(y) k2y k1y y 4 / 25 Absolute Stability and Aizerman Conjecture The Nyquist criterion gives a conclusive answer about stability whenever Φ(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 the sector bounds) so that the system is U.A.S.I.L.? In late-40’s Soviet research they thought that for a sector nonlinearity, stability could be ensured if the Nyquist criterion held for all k [k1,k2]. This was ∈ known as the Aizerman Conjecture. 5 / 25 Falsehood of the Aizerman Conjecture - Taylor Example The Aizerman conjecture is false, because limit cycles can exist. Take 10s2 G(s) = s4 +0.04s3 +2.021s2 +0.0404s +0.9803 3 and k1 =0, k2 = , with Φ(y) = y . ∞ However, it has been shown that the Aizerman conjecture is true for all first and second-order systems and higher-order systems of special forms. 6 / 25 The Kalman Conjecture Kalman’s conjecture (1957) uses the maximum and minimum slopes of Φ(y) instead of the sector bounds. Suppose the slope of Φ satisfies dΦ M M ≤ dy ≤ Kalman conjecture: The closed-loop system is asymptotically stable if the Nyquist plot of G(s) does not enclose the critical point 1/k for all values − M k M. ≤ ≤ The Kalman conjecture is also false, as a counterexample by Taylor shows. Taylor also showed that replacing the nonlinearity by its describing function and using the Nyquist criterion is not a valid method in general. 7 / 25 The Popov Criterion Popov’s criterion is only a sufficient condition. Assumptions: ■ G(s) is asymptotically stable. In a state-space realization (A,B,C), we require A to be Hurwitz and (A, B) controllable (a minimal realization, no pole-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 ≥ 1 Re [(1 + jwq)G(jw)] + > 0 k then the closed-loop system is asymptotically stable. If we separate G(jw) = G1(w) + jG2(w), we obtain the equivalent form: 1 G1(w) wqG2(w) + > 0 − k 8 / 25 Popov Criterion- Graphical Interpretation 1 The criterion requires G1(w) > + qwG2(w). Call x = G1(w) =Re(G(jw)) − k 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 1 x qy + =0 − k This is called a Popov plot (note that w is taken in [0, )). Im(W (jw)) = wIm(G(jw)) ∞ 1 q Re(W (jw)) = Re(G(jw)) − 1 k W (jw) Popov line 9 / 25 Ranges for q and k Asymptotic stability is obtained when the inequality holds for the given k and some q. In some cases (next example), q =0 makes the inequality hold for the sector [0, ]. This means that the system is asymptotically stable for any ∞ nonlinearity 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, Hsu and 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 no ∞ information about Φ is available, the inequality must hold with q =0. 10 / 25 Examples Take s +3 G(s) = s2 +7s + 10 We check that the system will be asymptotically stable for any nonlinearity Φ (0, ] ∈ ∞ Now take 1 G(s) = (s + 1)(s + 4)(s + 5) We see that for Φ = k, the Routh-Hurwitz criterion gives the range (0, 270) for k. For other nonlinearities, we formulate the problem of finding the maximum sector bound k for which there is a feasible q and solve it graphically. 11 / 25 Pole-Shifting Method The Popov criterion holds only for G(s) asymptotically stable. When this fails, we can use a loop transformation: ′ α¯ G (s) − 0 + + y Φ G(s) − + − ′ Φ α¯ Constant α¯ must be chosen so that G′(s) = G(s)/(1+αG ¯ (s)) becomes asymptotically 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. ≥ 12 / 25 Example Take 1 G(s) = s(s + 1) Use pole shifting and the Popov criterion to analyze the absolute stability of the corresponding Lur’e system. 13 / 25 The Circle Criterion The Popov criterion does not take advantage of information about the low sector bound when available. The circle criterion includes information about both 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 simple ≥ statement: The Lur’e system is absolutely stable (asymptotically stable for all nonlinearities Φ [kl,kh]) if Φ(u, t) satisfies: ∈ Φ kl <cl cr <kr ≤ u ≤ and the Nyquist plot of G(jw) does not touch or enclose D(cl,cr). Recall that the Nyquist plot is traced with −∞ < w < ∞. 14 / 25 Circle Criterion - Graphical Interpretation Im (G(jw)) − 1 − 1 cl cr Re (G(jw)) The circle is not unique in general, and circles close to the imaginary axis yield wider sectors. Example: Saturation nonlinearity. The slope of the function within the 1 saturation band determines , and we can take 1/cl . − cr − → −∞ 15 / 25 Circle Criterion - More General Systems The pole shifting method can be used for unstable systems, but a more general version 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 16 / 25 Circle Criterion - More General Systems The system is asymptotically stable if the condition matching the previous cases holds: ■ Case 1: The Nyquist plot of G does not enter the circle C(k1,k2) and encircles it ρ times counterclockwise. ■ Case 2: The Nyquist plot of G is contained in the half-space Re(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. 17 / 25 Example Again, take 1 G(s) = s(s + 1) Use pole shifting and the circle criterion to analyze the absolute stability of the corresponding Lur’e system. 18 / 25 Positive-Real Transfer Functions and the Lur’e Problem Suppose the linear system in the Lur’e problem is asymptotically stable. What additional conditions can be imposed on the system so that the closed-loop system remains stable for any nonlinearities in the sector [0, 1]? Note can use pole shifting and/or scaling to find a more general sector. Let G(s) have state-space realization (A,B,C, 0) and use the quadratic Lyapunov function: V (x) = xT Px with P = P T > 0. We show in class that 2 V˙ xT (AT P + P A)x +2xT (CT PB)Φ 2Φ ≤ − − For this, note that the sector condition is equivalent to φ(y)(φ(y) ky) 0 − ≤ 19 / 25 Lur’e Problem and Kalman-Yakubovich-Popov Lemma We found 2 V˙ xT (AT P + P A)x +2xT (CT PB)Φ 2Φ ≤ − − If we can find L, ǫ > 0 so that AT P + P A = LT L ǫP − − BT P = C √2L − we can show that 2 V˙ ǫxT Px xT LT Lx +2√2xT LT Φ 2Φ ≤ − − − so 2 V˙ ǫxT Px Lx √2Φ ǫxT Px ≤ − − || − || ≤ − which shows global asymptotic stability.
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