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

Plane geometry and convexity of polynomial stability regions Michael ˇ Didier HENRION 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 ∈ R 2 can be studied and solved graphically Motivation: provide geometric insight into SOF design problems

0.1 0 −0.1 k 2 −0.2 −0.3 −0.4 −0.5 −1 −0.5 0 0.5 k 1

60 50 40 30 k 2 20 10 0 −120 −100 −80 −60 −40 −20 0 20 k 1

−0.5 −0.55 −0.6 −0.65 −0.7 k 2 −0.75 −0.8 −0.85 −0.9 −0.95 −1 −0.25 −0.2 −0.15 −0.1 −0.05 0 k 1

90 85 80 75 70 k 2 65 60 55 50 45 40 0 1 2 3 4 5 6 7 8 9 k 1

60 50 40 k 2 30 20 10 0 0 20 40 60 80 100 120 140 160 180 200 k 1

40 35 30 25 20 k 2 15 10 5 0 −5 −4 −3 −2 −1 0 1 2 k 1

0.9 0.8 0.7 0.6 0.5 k 2 0.4 0.3 0.2 0.1 0 −0.1 −1.5 −1 −0.5 0 0.5 k 1

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 ) = p 0 ( s ) + k 1 p 1 ( s ) + k 2 p 2 ( s ) where p i ( s ) ∈ R [ s ] are given polynomials of s ∈ C and k ∈ R 2 contains design parameters What are conditions on k 1 , k 2 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 B u ( a, b ) is the symmetric matrix of size n with entries b ij satisfying linear equations n n a ( u ) b ( v ) − a ( v ) b ( u ) b ij u i − 1 v j − 1 � � = v − u i =1 j =1 Polynomial r u ( a, b ) = det B u ( 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 of real part and imaginary part of p ( jω ): p R ( ω 2 ) = Re p ( jω ) ωp I ( ω 2 ) = Im p ( jω ) that is, H ( p ) = B ω ( p R ( ω 2 ) , ωp I ( ω 2 )), 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 p 0 ( s ) = s ( s 2 − 13), p 1 ( s ) = s ( s − 5), p 2 ( s ) = s + 1, p R ( ω 2 ) = − k 1 ω 2 + k 2 , p I ( ω 2 ) = − ω 2 − 13 − 5 k 1 + k 2 Quadratic matrix inequality (QMI)   k 2 ( − 13 − 5 k 1 + k 2 ) 0 − k 2 H ( k ) = 0 k 1 ( − 13 − 5 k 1 + k 2 ) − k 2 0  ≻ 0    − k 2 0 k 1 equivalent to � � k 2 ( − 13 − 5 k 1 + k 2 ) − k 2 ≻ 0 , k 1 ( − 13 − 5 k 1 + k 2 ) > 0 − k 2 k 1 In general, such QMIs define non-convex regions.. However here it is convex

90 85 80 75 70 k 2 65 60 55 50 45 40 0 1 2 3 4 5 6 7 8 9 k 1

Geometric problem statement Define stability region S = { k ∈ R 2 : p ( s, k ) stable } Is S convex ? If S is convex, give an LMI representation S = { k ∈ R 2 : A 0 + k 1 A 1 + k 2 A 2 ≻ 0 } when possible, where the A i are real symmetric matrices to be found

Along the boundary Define algebraic plane curve C = { k ∈ R 2 : 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 S i , 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 p R ( ω 2 , k ) p 0 R ( ω 2 ) + k 1 p 1 R ( ω 2 ) + k 2 p 2 R ( ω 2 ) = = 0 ωp I ( ω 2 , k ) = ωp 0 I ( ω ) + k 1 ωp 1 I ( ω ) + k 2 ωp 2 I ( ω ) = 0 Eliminate variable ω via determinant of Hermite matrix h ( k ) = r ω ( p R ( ω 2 , k ) , ωp I ( ω 2 , 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 ) = r ω ( p R ( ω ) , ω ) = p R (0 , k ) = p 0 R (0) + k 1 p 1 R (0) + k 2 p 2 R (0) Defining polynomial of the other curve component G g ( k ) = r ω ( p R ( ω, k ) , p I ( ω, k ))

Parametrising the other curve component From relations p 1 R ( ω 2 ) p 2 R ( ω 2 ) p 0 R ( ω 2 ) � � � � � � k 1 = − ωp 1 I ( ω 2 ) ωp 2 I ( ω 2 ) ωp 0 I ( ω 2 ) k 2 we derive a rational parametrisation of G p 2 I ( ω 2 ) − p 2 R ( ω 2 ) � � q 1 ( ω 2 )   − p 1 I ( ω 2 ) p 1 R ( ω 2 ) k 1 ( ω 2 ) p 0 R ( ω 2 ) � � � � q 0 ( ω 2 ) = =   k 2 ( ω 2 ) p 0 I ( ω 2 ) q 2 ( ω 2 ) p 1 I ( ω 2 ) p 2 R ( ω 2 ) − p 1 R ( ω 2 ) p 2 I ( ω 2 )   q 0 ( ω 2 ) Using B´ ezoutians we can prove that symmetric pencil G ( k ) = B ω ( q 1 , q 2 ) + k 1 B ω ( q 2 , q 0 ) + k 2 B ω ( q 1 , q 0 ) 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 S i If C ( k ) ≻ 0 for some point k in the interior of S i for some i , then S i = { k : C ( k ) � 0 } is a convex LMI region Converse is false: S i 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

150 100 50 0 y −50 −100 −150 −400 −300 −200 −100 0 100 200 300 400 x Bivariate polynomial of degree 8

Trivariate polynomial of degree 3 (Cayley cubic)

LMI stability regions It may happen that • S i is convex for some i , yet C ( k ) is not positive definite for points k within S i • S i is rigidly convex (= convex LMI) for some i , yet H ( k ) is not positive definite for points k within S i If H ( k ) ≻ 0 and C ( k ) ≻ 0 for some point k in the interior of S i , then S i is an LMI region included in stability region S Quite often, on practical instances, we observe that S = S i for some i and moreover S i is convex LMI

SOF problem NN1 Hermite matrix   k 2 ( − 13 − 5 k 1 + k 2 ) 0 − k 2 H ( k ) = 0 k 1 ( − 13 − 5 k 1 + k 2 ) − k 2 0     − k 2 0 k 1 Hence h ( k ) = det H ( k ) = k 2 ( − 13 k 1 − k 2 − 5 k 12 + k 1 k 2 ) 2 and then l ( k ) = k 2 , g ( k ) = − 13 k 1 − k 2 − 5 k 12 + k 1 k 2 Rational parametrization of curve G = { k : g ( k ) = 0 } given by ( ω 2 + 13) / ( ω 2 − 5) k 1 ( ω 2 ) = ω 2 ( ω 2 + 13) / ( ω 2 − 5) k 2 ( ω 2 ) = obtained with algcurves package of Maple

Symmetric affine determinantal representation of G = { k : det G ( k ) = 0 } given by � � 169 + 65 k 1 − 18 k 2 13 + 5 k 1 G ( k ) = 13 + 5 k 1 1 − k 1 obtained via LinearAlgebra[BezoutMatrix] function of Maple Symmetric affine determinantal representation of C = { k : det C ( k ) = 0 } given by   0 0 k 2 C ( k ) = 0 169 + 65 k 1 − 18 k 2 13 + 5 k 1     0 13 + 5 k 1 1 − k 1 LMI stability region S = { k : C ( k ) ≻ 0 }

100 50 k 2 0 −50 −5 0 5 10 k 1

Example 14.4 from Ackermann (1993) p 0 ( s ) = s 4 + 2 s 3 + 10 s 2 + 10 s + 14 + 2 a , p 1 ( s ) = 2 s 3 + 2 s − 3 / 10, p 2 ( s ) = 2 s + 1, with parameter a ∈ R We obtain C ( k ) = diag { l ( k ) , G ( k ) } with l ( k ) = 140 + 20 a − 3 k 1 + 10 k 2 7920 + 4860 a + 400 a 2 + ( − 1609 − 60 a ) k 1 G 11 ( k ) = +( − 270 + 200 a ) k 2 G 21 ( k ) = − 8350 − 2000 a + 1430 k 1 + 130 k 2 G 22 ( k ) = 8370 − 1230 k 1 − 100 k 2 G 31 ( k ) = 900 + 200 a − 130 k 1 G 32 ( k ) = − 900 + 100 k 1 G 33 ( k ) = 100

15 10 5 0 k 2 −5 −10 −15 −20 −10 −5 0 5 10 15 20 k 1 When a = 1, S consists of two disconnected regions and the region including the origin is LMI