Plane geometry and convexity of polynomial stability regions - - PowerPoint PPT Presentation

Plane geometry and convexity

• f polynomial stability regions

Didier HENRION Michael ˇ SEBEK LAAS-CNRS

• Fac. Elec. Engr.
• Univ. Toulouse

Czech Tech. Univ. Prague France Czech Republic EIGOPT Workshop Leuven, Belgium June 2008

Plane SOF COMPLEIB: more than 100 problems of static output feedback (SOF) design compiled by F. Leibfritz www.compleib.de Given real matrices A, B, C, find real matrix K such that max real eig (A + BKC) < 0 Spectral abscissa, typically non-convex non-smooth 7 plane problems K ∈ R2 can be studied and solved graphically Motivation: provide geometric insight into SOF design problems

k1 k2 −1 −0.5 0.5 −0.5 −0.4 −0.3 −0.2 −0.1 0.1

k1 k2 −120 −100 −80 −60 −40 −20 20 10 20 30 40 50 60

k1 k2 −0.25 −0.2 −0.15 −0.1 −0.05 −1 −0.95 −0.9 −0.85 −0.8 −0.75 −0.7 −0.65 −0.6 −0.55 −0.5

k1 k2 1 2 3 4 5 6 7 8 9 40 45 50 55 60 65 70 75 80 85 90

k1 k2 20 40 60 80 100 120 140 160 180 200 10 20 30 40 50 60

k1 k2 −4 −3 −2 −1 1 2 −5 5 10 15 20 25 30 35 40

k1 k2 −1.5 −1 −0.5 0.5 −0.1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Intriguing observation 6 out of 7 plane SOF problems seem to have a convex feasibility set Can we explain why ? Can we detect convexity ? Can we (re)formulate the problem using convex programming ? Problem formally stated during an AIM worskhop in 2005

Algebraic problem statement Parametrized polynomial p(s, k) = p0(s) + k1p1(s) + k2p2(s) where pi(s) ∈ R[s] are given polynomials of s ∈ C and k ∈ R2 contains design parameters What are conditions on k1, k2 such that p(s, k) is (Hurwitz) stable i.e. all its roots lie in open left half-plane ?

Hermite stability condition Etienne B´ ezout Charles Hermite (1730-1783) (1822-1901) Symmetric version of Routh-Hurwitz criterion p(s) stable ⇐ ⇒ H(p) ≻ 0 Hermite matrix = Lyapunov matrix = B´ ezoutian matrix

B´ ezoutian resultant Let a(u), b(u) be polynomials of degree n B´ ezoutian matrix Bu(a, b) is the symmetric matrix of size n with entries bij satisfying linear equations a(u)b(v) − a(v)b(u) v − u =

n

• i=1

n

• j=1

bijui−1vj−1 Polynomial ru(a, b) = det Bu(a, b) is the resultant of a(u) and b(u) with respect to u Eliminates variable u from system of equations a(u) = b(u) = 0

Hermite matrix as a B´ ezoutian Hermite matrix of p(s) defined as B´ ezoutian matrix

• f real part and imaginary part of p(jω):

pR(ω2) = Re p(jω) ωpI(ω2) = Im p(jω) that is, H(p) = Bω(pR(ω2), ωpI(ω2)), the variable to be eliminated being frequency ω Polynomial p(s) is stable if and only if H(p) ≻ 0 Matrix H(p) is symmetric and quadratic in coefficients of p(s)

Example: SOF problem NN1 p0(s) = s(s2 − 13), p1(s) = s(s − 5), p2(s) = s + 1, pR(ω2) = −k1ω2 + k2, pI(ω2) = −ω2 − 13 − 5k1 + k2 Quadratic matrix inequality (QMI) H(k) =

  

k2(−13 − 5k1 + k2) −k2 k1(−13 − 5k1 + k2) − k2 −k2 k1

   ≻ 0

equivalent to

• k2(−13 − 5k1 + k2)

−k2 −k2 k1

• ≻ 0,

k1(−13 − 5k1 + k2) > 0 In general, such QMIs define non-convex regions.. However here it is convex

k1 k2 1 2 3 4 5 6 7 8 9 40 45 50 55 60 65 70 75 80 85 90

Geometric problem statement Define stability region S = {k ∈ R2 : p(s, k) stable} Is S convex ? If S is convex, give an LMI representation S = {k ∈ R2 : A0 + k1A1 + k2A2 ≻ 0} when possible, where the Ai are real symmetric matrices to be found

Along the boundary Define algebraic plane curve C = {k ∈ R2 : p(jω, k) = 0, w ∈ R} which is the set of parameters k for which stability may be lost since s = jω sweeps the imaginary axis Study of geometry of C = key idea of D-decomposition techniques (Neimark 1948) Used extensively in robust control (ˇ Siljak 1989, Ackermann et al. 1993, Barmish 1994, Bhattacharyya et al. 1995, Gryazina and Polyak, 2006)

Stability regions Curve C partitions C into connected components Si, i = 1, 2 . . . within which the number of stable roots of p remains constant Stability region S corresponds to the union of components containing exactly deg p stable roots, hence ∂S ⊂ C We study the geometry of C to understand the geometry of S

Elimination of frequency Note that p(jω, k) = 0 for some ω ∈ R if and only if pR(ω2, k) = p0R(ω2) + k1p1R(ω2) + k2p2R(ω2) = ωpI(ω2, k) = ωp0I(ω) + k1ωp1I(ω) + k2ωp2I(ω) = Eliminate variable ω via determinant of Hermite matrix h(k) = rω(pR(ω2, k), ωpI(ω2, k)) = det H(k) Implicit algebraic description C = {k : h(k) = 0}

Resultant factorisation Resultant can be factored as h(k) = l(k)g(k)2 with l(k) linear, hence C = L ∪ G = {k : l(k) = 0} ∪ {k : g(k) = 0} Equation of line L l(k) = rω(pR(ω), ω) = pR(0, k) = p0R(0) + k1p1R(0) + k2p2R(0) Defining polynomial of the other curve component G g(k) = rω(pR(ω, k), pI(ω, k))

Parametrising the other curve component From relations

• p1R(ω2)

p2R(ω2) ωp1I(ω2) ωp2I(ω2) k1 k2

• = −
• p0R(ω2)

ωp0I(ω2)

• we derive a rational parametrisation of G
• k1(ω2)

k2(ω2)

• =
• p2I(ω2)

−p2R(ω2) −p1I(ω2) p1R(ω2)

• p1I(ω2)p2R(ω2) − p1R(ω2)p2I(ω2)
• p0R(ω2)

p0I(ω2)

• =

  

q1(ω2) q0(ω2) q2(ω2) q0(ω2)

  

Using B´ ezoutians we can prove that symmetric pencil G(k) = Bω(q1, q2) + k1Bω(q2, q0) + k2Bω(q1, q0) is such that G = {k : det G(k) = 0}

Determinantal locus Defining C(k) = diag {l(k), G(k)}, algebraic curve C can be described as a determinantal locus C = {k : det C(k) = 0} with C(k) symmetric linear in k Recall that C partions C into connected components Si If C(k) ≻ 0 for some point k in the interior of Si for some i, then Si = {k : C(k) 0} is a convex LMI region Converse is false: Si may be convex, but not LMI..

Rigid convexity Convex sets which admit an LMI representation are called rigidly convex by Helton and Vinnikov (2002) Rigid convexity is stronger than convexity Algebraic characterisation: a generic line through the interior should intersect Zariski closure of the boundary at real points Equivalent to polynomial hyperbolicity (barrier functions, PDEs)

TV screen or Fermat quartic

−400 −300 −200 −100 100 200 300 400 −150 −100 −50 50 100 150 x y

Bivariate polynomial of degree 8

Trivariate polynomial of degree 3 (Cayley cubic)

LMI stability regions It may happen that

• Si is convex for some i, yet C(k) is not positive definite for

points k within Si

• Si is rigidly convex (= convex LMI) for some i, yet H(k) is not

positive definite for points k within Si If H(k) ≻ 0 and C(k) ≻ 0 for some point k in the interior of Si, then Si is an LMI region included in stability region S Quite often, on practical instances, we observe that S = Si for some i and moreover Si is convex LMI

SOF problem NN1 Hermite matrix H(k) =

  

k2(−13 − 5k1 + k2) −k2 k1(−13 − 5k1 + k2) − k2 −k2 k1

  

Hence h(k) = det H(k) = k2(−13k1 −k2 −5k12 +k1k2)2 and then l(k) = k2, g(k) = −13k1 − k2 − 5k12 + k1k2 Rational parametrization of curve G = {k : g(k) = 0} given by k1(ω2) = (ω2 + 13)/(ω2 − 5) k2(ω2) = ω2(ω2 + 13)/(ω2 − 5)

• btained with algcurves package of Maple

Symmetric affine determinantal representation of G = {k : det G(k) = 0} given by G(k) =

• 169 + 65k1 − 18k2

13 + 5k1 13 + 5k1 1 − k1

• btained via LinearAlgebra[BezoutMatrix] function of Maple

Symmetric affine determinantal representation of C = {k : det C(k) = 0} given by C(k) =

  

k2 169 + 65k1 − 18k2 13 + 5k1 13 + 5k1 1 − k1

  

LMI stability region S = {k : C(k) ≻ 0}

k1 k2 −5 5 10 −50 50 100

Example 14.4 from Ackermann (1993) p0(s) = s4 + 2s3 + 10s2 + 10s + 14 + 2a, p1(s) = 2s3 + 2s − 3/10, p2(s) = 2s + 1, with parameter a ∈ R We obtain C(k) = diag {l(k), G(k)} with l(k) = 140 + 20a − 3k1 + 10k2 G11(k) = 7920 + 4860a + 400a2 + (−1609 − 60a)k1 +(−270 + 200a)k2 G21(k) = −8350 − 2000a + 1430k1 + 130k2 G22(k) = 8370 − 1230k1 − 100k2 G31(k) = 900 + 200a − 130k1 G32(k) = −900 + 100k1 G33(k) = 100

k1 k2 −10 −5 5 10 15 20 −20 −15 −10 −5 5 10 15

When a = 1, S consists of two disconnected regions and the region including the origin is LMI

k1 k2 −10 −5 5 10 15 20 −20 −15 −10 −5 5 10 15

When a = 0, LMI region {k : C(k) ≻ 0} is not included in non-convex S

Conclusion Algebraic geometry explains why some plane stability regions can be convex Allows to detect convexity and find explicit LMI representations Relies on algebraic plane curve parametrizations and B´ ezoutians, hence extension to more than 2 parameters seems to be difficult Results not useful for optimisation purposes (H∞ etc) since bivariate LMIs can be solved algebraically

Determinantal representation Key issue: finding a symmetric affine determinantal representation of multivariate polynomials Very difficult in general, should exploit rational parametrisation

• f the corresponding hypersurface

Simplest 3rd degree case p(s, k) = s3 + k1s2 + k2s + k3 Find four symmetric real matrices A0, A1, A2, A3 such that det(A0 + A1k1 + A2k2 + A3k3) = k1k3 − k2