Alex Wang

43 Pages·2010·0.2 MB·English

by  Alex Wang

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