ProblemFormulation KnownResults LinearPartialPolePlacement NewConditions Output Feedback Pole Placement: A Best Linear Condition Alex Wang DepartmentofMathematics&Statistics TexasTechUniversity Workshop in Celebration of the Life, Mathematics and Memories of Christopher I. Byrnes ProblemFormulation KnownResults LinearPartialPolePlacement NewConditions Outline 1 Problem Formulation 2 Known Results 3 Linear Partial Pole Placement 4 New Conditions ProblemFormulation KnownResults LinearPartialPolePlacement NewConditions Outline 1 Problem Formulation 2 Known Results 3 Linear Partial Pole Placement 4 New Conditions ProblemFormulation KnownResults LinearPartialPolePlacement NewConditions Let x˙ = Ax +Bu y = Cx be an m-input, p-output linear system of McMillan degree n, where A, B, C are n×n, n×m, and p×n matrices, respectively. If we apply an output feedback u = Ky then the closed loop system becomes x˙ = (A+BKC)x y = Cx ProblemFormulation KnownResults LinearPartialPolePlacement NewConditions The map K 7→ det(sI −(A+BKC)) is called the pole placement map. Let G(s) = C(sI −A)−1B be the transfer function of the original system, and G(s) = D−1(s)N (s) = N (s)D−1(s) l l r r be the left and right coprime (polynomial) decompositions of the transfer function, respectively. ProblemFormulation KnownResults LinearPartialPolePlacement NewConditions We can write the pole placement map as det(sI −A−BKC) = det(sI −A)det(I −(sI −A)−1BKC) = det(sI −A)det(I −C(sI −A)−1BK) = det(sI −A)det(I −G(s)K) = detD (s)det(I −D−1(s)N (s)K) l l l = det(D (s)−N (s)K). l l Also det(sI −A−BKC) = det(I −BKC(sI −A)−1)det(sI −A) = det(I −KC(sI −A)−1B)det(sI −A) = det(I −KG(s))det(sI −A) = det(I −KN (s)D−1(s))detD (s) r r r = det(D (s)−KN (s)) r r ProblemFormulation KnownResults LinearPartialPolePlacement NewConditions Pole Placement Map So we have 4 forms of the pole placement map (cid:20) (cid:21) (cid:2) (cid:3) I det(sI −A−BKC) = det D (s) −N (s) (1) l l K (cid:20) (cid:21) D (s) −N (s) = det l l (2) −K I (cid:20) (cid:21) = det(cid:2) −K I (cid:3) Nr(s) (3) D (s) r (cid:20) (cid:21) I N (s) = det r (4) K D (s) r ProblemFormulation KnownResults LinearPartialPolePlacement NewConditions Let Grass(m,m+p) be the set of equivalence classes of all (m+p)×m full rank matrices under the equivalence relation that two matrices are equivalent if their column spans are the same. Similarly, let Grˆass(p,m+p) be the set of equivalence classes of all p×(m+p) full rank matrices under the equivalence relation that two matrices are equivalent if their row spans are the same. For any M ∈ Grass(m,m+p), we define its dual Mˆ ∈ Grˆass(p,m+p) to be the matrix such that MˆM = 0. ProblemFormulation KnownResults LinearPartialPolePlacement NewConditions Note that (cid:20) (cid:21) (cid:2) −K I (cid:3) ∈ Grˆass(m,m+p), I ∈ Grass(p,m+p), K they are dual to each other, and they define the kernel and image representations of the compensator (cid:20) (cid:21) (cid:20) (cid:21) (cid:20) (cid:21) (cid:2) (cid:3) y y I −K I = 0, = y. u u K ProblemFormulation KnownResults LinearPartialPolePlacement NewConditions and (cid:20) (cid:21) (cid:2) D (s) −N (s) (cid:3) ∈ Grˆass(p,m+p), Nr(s) ∈ Grass(m,m+p) l l D (s) r for all s, they are dual to each other, and they define the kernel and image representations of the system (cid:20) (cid:21) (cid:20) (cid:21) (cid:20) (cid:21) (cid:2) D (s) −N (s) (cid:3) y = 0, y = Nr(s) v, l l u u D (s) r where v = D−1u. r