Stability, Pole Placement, Observers and StabilizationStability of autonomous systems The pole...

Stability of autonomous systems The pole placement problem Stabilization by state feedback State observers Pole placement and

Stability, Pole Placement, Observers and

Stabilization

H.L. Trentelman1

1University of Groningen, The Netherlands

DISC Course Mathematical Models of Systems

H.L. Trentelman University of Groningen

Stability, Pole Placement, Observers and Stabilization


Outline

1 Stability of autonomous systems

2 The pole placement problem

3 Stabilization by state feedback

4 State observers

5 Pole placement and stabilization by dynamic output feedback




Stability of autonomous Systems

Autonomous systems

Let P(ξ) ∈ Rq×q[ξ], with det P(ξ) 6= 0, i.e., detP(ξ) is not the zero

polynomial.Consider the system of differential equations P( d

dt)w = 0.

This represents the dynamical system Σ = (R, Rq,B) withB = {w : R → R

q | w satisfies P(

d

dt

)

w = 0 weakly}.Since det P(ξ) 6= 0, the resulting system is autonomous. Hence B

is finite-dimensional and each weak solution of is a strong one.




Stability of autonomous systems

All solutions

All solutions are infinitely differentiable. In fact, w ∈ B if and onlyif w = w1 + w2 + · · · + wN , with wks associated with one of thedistinct roots λ1, λ2, . . . , λN of det P(ξ). This wk is given by

wk(t) =

(

nk−1∑

`=0

Bk`t`

)

eλk t ,

where nk is the multiplicity of the root λk of det P(ξ), Bk`s aresuitable constant complex vectors.




Stability of autnomous systems

Stability definitions

The linear dynamical system Σ = (R, Rq,B) is said to be

1 stable if all elements of its behavior B are bounded on thehalf-line [0,∞), i.e, if (w ∈ B) ⇒ (there exists M ∈ R suchthat ‖w(t)‖ ≤ M for t ≥ 0). Of course, this bound M

depends on the particular solution w ∈ B,

2 unstable if it is not stable,

3 asymptotically stable if all elements of B approach zero fort → ∞ (i.e, if (w ∈ B) ⇒ (w(t) → 0 as t → ∞)).

Important note: If Σ is stable or asymptotically stable then it mustbe autonomous. Hence without loss of generality B = ker P( d

dt)

with P(ξ) nonsingular.





Definition of simple and semisimple roots

Let P(ξ) ∈ Rq×q[ξ] be nonsingular. Then

1 The roots of P(ξ) are defined to be those of the scalarpolynomial detP(ξ). Hence λ ∈ C is a root of P(ξ) if andonly if rank P(λ) < q,

2 The root λ is called simple if it is a root of det P(ξ) ofmultiplicity one,

3 semisimple if the rank deficiency of P(λ) equals themultiplicity of λ as a root of P(ξ) (equivalently, if thedimension of ker P(λ) is equal to the multiplicity of λ as aroot of det P(ξ)).

Clearly, for q = 1 roots are semisimple if and only if they are simple,but for q > 1 the situation is more complicated.





Example

λ = 0 is a root of multiplicity 2 for both the polynomial matrices

(

ξ 00 ξ

)

and

(

ξ 10 ξ

)

. (1)

This root is semisimple in the first case, but not in the second.





Theorem

Let P ∈ Rq×q[ξ] be nonsingular. The system represented by

P( d

dt)w = 0 is:

1 asymptotically stable if and only if all the roots of detP(ξ)have negative real part;

2 stable if and only if for each λ ∈ C that is a root of P(ξ),either (i) Re λ < 0, or (ii) Re λ = 0 and λ is a semisimpleroot of P(ξ).

3 unstable if P(ξ) has a root with positive real part and/or anonsemisimple root with zero real part.





Examples

1 Scalar first-order system aw + ddt

w = 0. Associatedpolynomial P(ξ) = a + ξ. Root −a. Hence this system isasymptotically stable if a > 0, stable if a = 0, and unstable ifa < 0. Note: behavior B = {Ae−at | A ∈ R}.

2 Scalar second-order system aw + d2

dt2 w = 0. Associatedpolynomial P(ξ) = a + ξ2. Roots λ1,2 = ±

√−a for a < 0,

λ1,2 = ±i√

a for a > 0, and λ = 0 is a double, not semisimpleroot when a = 0. Thus we have (a < 0 ⇒ instability),(a > 0 ⇒ stability), and (a = 0 ⇒ instability). Indeed:B = {Ae

√−at + Be−

√−at | A,B ∈ R}, if a < 0,

B = {A cos√

at + B sin√

atA | A,B ∈ R}, if a > 0,B = {A + Bt | A,B ∈ R}, if a = 0.





Special case: stability of state equations

Autonomous state system d

dtx = Ax , with A ∈ R

n×n.Polynomial matrix P(ξ) = Iξ − A. Roots: the eigenvalues of A.An eigenvalue λ of A is called semisimple if λ is a semisimple rootof det(Iξ − A), i.e:

the multiplicity of λ is equal to dim ker(λI − A).

Note: dim ker(λI−A) is always equal to the number of independenteigenvectors associated with the eigenvalue λ.





Corollary

The system defined by d

dtx = Ax is:

1 asymptotically stable if and only if the eigenvalues of A havenegative real part,

2 stable if and only if for each λ ∈ C that is an eigenvalue of A,either (i) Reλ < 0, or (ii) Re λ = 0 and λ is a semisimpleeigenvalue of A,

3 unstable if and only if A has either an eigenvalue with positivereal part or a nonsemisimple one with zero real part.




The pole placement problem

Consider the linear time-invariant dynamical system in state formdescribed by

d

dtx = Ax + Bu,

with x : R → Rn and u : R → R

m the state and the input trajectory,and with A ∈ R

n×n and B ∈ Rn×m.

Allow linear state feedback controllers of the form u = Nx , withN ∈ R

m×n called the feedback gain matrix.Closed loop equation:

d

dtx = (A + BN)x .

This represents an autonomous dynamical system.




Stability of Autonomous Systems

u = Nx

Plantddt

x = Ax + Bu

Feedback

processor





We call the eigenvalues of A + BN the closed loop poles.Question: what closed loop pole locations are achievable by choosingthe feedback gain matrix N?Closed loop characteristic polynomial:

χA+BN(ξ) := det(ξI − (A + BN))

Pole placement problem

Let A ∈ Rn×n and B ∈ R

n×m be given. What is the set of polyno-mials χA+BN(ξ) obtainable by choosing the matrix N ∈ R

m×n?





The pole placement theorem (W.M. Wonham, 1969)

Consider the system d

dtx = Ax +Bu. For any real monic polynomial

r(ξ) of degree n there exists N ∈ Rm×n such that χA+BN = r(ξ) if

and only if the system d

dtx = Ax + Bu is controllable.

Monic polynomial of degree n: p(ξ) = ξn+pn−1ξn−1+. . .+p1ξ+p0.

W.M. Wonham, ”On pole assignment in multi-input controllablelinear systems”, IEEE Transactions on Automatic Control, 1967.




Proof of the pole placement theorem





System similarity

Σn,m := the family of all systems d

dtx = Ax +Bu with n state and

m input variables.We say: (A,B) ∈ Σn,m.Let (A1,B1), (A2,B2) ∈ Σn,m. We call (A1,B1) and (A2,B2)similar if there exists a nonsingular matrix S ∈ R

n×n such thatA2 = SA1S

−1 and B2 = SB1.

If in the state space of d

dtx = Ax + Bu we change the coordinate

basis by defining z(t) = Sx(t), then the dynamics of z are governedby

d

dtz = SAS−1z + SBu.

Hence similarity corresponds to changing the basis in the state space.





Lemma

Assume (A1,B1), (A2,B2) ∈ Σn,m are similar. Let r(ξ) ∈ R[ξ] be amonic polynomial. Then there exists a matrix N1 ∈ Rm×n such thatχA1+B1N1

(ξ) = r(ξ) if and only if there exists a matrix N2 ∈ Rm×n

such that χA2+B2N2(ξ) = r(ξ).

Proof

Let A2 = SA1S−1 and B2 = SB1. Assume χA1+B1N1

(ξ) = r(ξ).Define N2 = N1S

−1. Then A2 + B2N2 = S(A1 + B1N1)S−1, so

A1 +B1N1 and A2 +B2N2 are similar, and therefore χA2+B2N2(ξ) =

r(ξ) as well. Also the converse.





Lemma

The system (A,B) ∈ Σn,m is similar to a system (A′,B ′) ∈ Σn,m

with A′,B ′ of the form

A′ =

(

A′11 A′

12

0 A′22

)

, B ′ =

(

B ′1

0

)

and with (A′11,B

′1) controllable.





Proof

Two cases:

1 if (A,B) is controllable then A′11 = A′ = A and B ′

1 = B ′ = B ,and A′

12, A′22 and the 0-matrices are ’void’.

2 If (A,B) is not controllable, thenR := im(B AB A2B . . . An−1B) 6= R

n, so is a proper subspaceof R

n. Choose a basis of Rn adapted to R. The matrices with

respect to this new basis are of the form (A′,B ′) becauseAR ⊂ R and im B ⊂ R.





Pole placement ⇒ Controllability

Assume (A,B) is not controllable. Then it is similar to (A′,B ′) ofthe form

A′ =

(

A′11 A′

12

0 A′22

)

, B ′ =

(

B ′1

0

)

with n1, the dimension of A′11, less than n. Now apply N ′ = (N ′

1 N ′2),

with N ′1 ∈ R

m×n1 and N ′2 ∈ R

m×(n−n1), on this system. Then

A′ + B ′N ′ =

(

A′11 + B ′

1N′1 A′

12 + B ′1N

′2

0 A′22

)

.

Characteristic polynomial is given by χA′

11+B′

1N′

1(ξ)χA′

22(ξ). There-

fore, the characteristic polynomial χA′+B′N′(ξ), and henceχA+BN(ξ), has, regardless of what N is chosen, χA′

22(ξ) as a factor.





Theorem

Let r(ξ) be a monic polynomial of degree n. Assume (A,B) con-trollable, and m = 1. Let F ∈ R

1×n be the solution of the systemof linear equations

F (B AB · · ·An−2B An−1B) = (0 0 · · · 0 1).

Then N = −F r(A) yields χA+BN(ξ) = r(ξ). Here r(A) := r0I +r1A + · · · + rn−1A

n−1 + An.

Note: this yields a proof of the implication Controllability ⇒ Poleplacement for the special case m = 1 !





Notation

Σcontn,m ⊂ Σn,m are all controllable systems with state space R

n andinput space R

m.

Heymann’s lemma (1968)

Let (A,B) ∈ Σcontn,m . Assume K ∈ R

m×1 such that BK 6= 0.Then there exists a N ′ ∈ R

m×n such that (A + BN ′,BK ) ∈ Σcont1,n .

M. Heymann, ”Comments to: Pole assignment in multi-input con-trollable linear systems”, IEEE Transactions on Automatic Control,1968.





Controllability ⇒ Pole placement, m > 1: coup de grace

Choose K ∈ Rm×1 such that BK 6= 0, by controllability, B 6= 0,

hence such a K exists.Choose N ′ ∈ R

m×1 such that (A + BN ′,BK ) controllable.We are now back in the case m = 1.Choose N ′′ ∈ R

1×n such that A + BN ′ + BKN ′′ has the desiredcharacteristic polynomial r(ξ).Finally, take N = N ′ + KN ′′.Then A + BN = A + BN ′ + BKN ′′ so χA+BN(ξ) = r(ξ).





Algorithm

Data: A ∈ Rn×n, B ∈ R

n×m, with (A,B) controllable; r(ξ) ∈ R[ξ]with r(ξ) monic and of degree n.Required: N ∈ R

m×n such that χA+BN(ξ) = r(ξ).Algorithm:

1 Find K ∈ Rm×1 and N ′ ∈ R

m×n such that (A + BN ′,BK ) iscontrollable.

2 Put A′ = A + BN ′, B ′ = BK , and compute F from

F [B ′,A′B ′, . . . , (A′)n−1B ′] = [0 0 . . . 0 1].

3 Compute N ′′ = −F r(A′).

4 Compute N = N ′ + KN ′′.

Result: N is the desired feedback matrix.




Stabilization

So: if (A,B) is controllable then all polynomials can be ”placed”.Question: what polynomials can be placed if (A,B) is not control-lable?.Recall (A,B) ∈ Σn,m is similar to

A′ =

(

A′11 A′

12

0 A′22

)

, B ′ =

(

B ′1

0

)

.

The matrix A′22 characterizes the noncontrollable behavior of the

system. Its characteristic polynomial χA22(ξ) is called the uncontrol-

lable polynomial of the system. Its roots are called the uncontrollablepoles (”modes”).




Stabilization

Theorem

Consider the system (A,B). Let χu(ξ) be its uncontrollable polyno-mial. There exists a feedback matrix N ∈ R

n×m such that

χA+BN(ξ) = r(ξ)

if and only if r(ξ) is a real monic polynomial of degree n that hasχu(ξ) as a factor.




Stabilization

Proof

Take (A′,B ′) similar to (A,B). Partition N ′ conformably as N ′ =(N ′

1 N ′2). Then

A′ + B ′N ′ =

(

A′11 + B ′

1N′1 A′

12 + B ′1N

′2

0 A′22

)

.

Hence χA′+B′N′(ξ) = χA′

11+B′

1N′

1(ξ)χA22

(ξ) = χA′

11+B′

1N′

1(ξ)χu(ξ).

Let r(ξ) have χu(ξ) as a factor. Then r(ξ) = r1(ξ)χu(ξ) for somemonic r1(ξ). Since (A′

11,B′1) is controllable, χA′

11+B′

1N′

1(ξ) can be

made equal to r1(ξ). Then with N ′ = (N ′1 0) we have χA′+B′N′(ξ) =

r1(ξ)χu(ξ) = r(ξ). Also the converse.




Stabilization


dtx = Ax + Bu with the control law u = Nx .

The closed loop system d

dtx = (A + BN)x is asymptotically stable

if and only if A + BN is Hurwitz.Question: when does there exist N ∈ R

m×n such that A + BN isHurwitz?

Corollary

There exists N such A + BN is Hurwitz if and only if the uncontrol-lable polynomial χu(ξ) of (A,B) is Hurwitz.




Stabilization

Definition

We call the system d

dtx = Ax +Bu, or, equivalently, the pair (A,B),

stabilizable if its uncontrollable polynomial χu(ξ) is Hurwitz.

Theorem (Hautus test)

The system d

dtx = Ax + Bu is stabilizable if and only if

rank(

λI − A B)

= n

for all λ ∈ C+ := {λ ∈ C | Re(λ) ≥ 0}.




Stabilization




State observers

Consider the following plant

d

dtx = Ax + Bu , y = Cx

x is the state, u the input, and y the output.System parameters: A ∈ R

n×n, B ∈ Rn×m, and C ∈ R

p×n.Denote the class of such systems by Σn,m,p.We also write (A,B ,C ) ∈ Σn,m,p.The external (manifest) signals u and y are measured.Aim: deduce the internal (latent) signal x from these measurements.




State observers

D. Luenberger, 1963




State observers

An algorithm that estimates x from u and y is called a state observer.Denote the estimate of x by x , and define the estimation error as

e := x − x .

A state observer is a dynamical system with u and y as input, x asoutput, and that makes e = x − x small in some sense.Here we focus on the asymptotic behavior of e(t) for t → ∞.




State observers

Plant

Observer

y

x

u

+

−

e

u

x




State observers

Proposed state observer:

d

dtx = (A − LC )x + Bu + Ly ,

Combining this equation with

d

dtx = Ax + Bu , y = Cx

yields the following equation for e = x − x :

d

dte = (A − LC )e.

d

dte = d

dtx − d

dtx = (Ax + Bu) − ((A − LC )x + Bu + Ly) =

(Ax+Bu)−((A−LC )x+Bu+LCx) = (A−LC )(x−x) = (A−LC )e.




State observers

We want that e(t) → 0 as t → ∞ for all e(0) = x(0) − x(0), i.e.,we want A − LC to be Hurwitz.Often we need a certain rate of convergence for e(t). This leads tothe following question:What eigenvalues (”observer poles”) can we achieve for A − LC bychoosing the observer gain matrix L?In linear algebra terms:Let A ∈ R

n×n and C ∈ Rp×n be given. What is the set of polyno-

mials χA−LC (ξ) obtainable by choosing the matrix L ∈ Rn×p?




State observers

Theorem


dtx = Ax + Bu , y = Cx . There exists for

every real monic polynomial r(ξ) of degree n a matrix L such thatχA−LC (ξ) equals r(ξ) if and only if the system is observable.

Proof

(A,C ) observable pair if and only if (AT ,CT ) controllable pair.Note: for any real square matrix M, χM(ξ) = χMT (ξ). Assume(A,C ) observable. By the pole placement theorem, there exists forall r(ξ) a matrix N ∈ R

p×n such that χAT +CTN(ξ) = r(ξ). ThusχA−LC (ξ) = r(ξ) with L = −NT .Conversely: Assume there exists for all r(ξ) a matrix L ∈ R

n×p suchthat χA−LC (ξ) = r(ξ). Then χAT +CT (−L)T (ξ) = r(ξ) so (AT ,CT )controllable, whence (A,C ) observable.




State observers

What if (A,C ) is not observable?The dynamical systems (A1,B1,C1) ∈ Σn,m,p and (A2,B2,C2) ∈Σn,m,p are called similar if there exist a nonsingular S such thatA1 = SA2S

−1, B1 = SB2, C1 = C2S−1.

Theorem

The system (A,B ,C ) ∈ Σn,m,p is similar to a system of the form(A′,B ′,C ′) in which A′ and C ′ have the following structure:

A′ =

(

A11 0A12 A22

)

, C ′ = (C1 0),

with (A11,C1) observable.All such decompositions lead to matrices A22 that have the samecharacteristic polynomial.




State observers

The polynomial χA22(ξ) is called the unobservable polynomial of

(A,C ). Notation: χ0(ξ)Its roots are called the unobservable eigenvalues (modes).

Theorem

Consider the system (A,B ,C ) ∈ Σn,m,p. Let χ0(ξ) be its unobserv-able polynomial. There exists L ∈ R

n×p such that χA−LC (ξ) = r(ξ)if and only if r(ξ) is a real monic polynomial of degree n that hasχ0(ξ) as a factor.

Proof

(

A11 0A12 A22

)

+

(

L1

L2

)

(C1 0) =

(

A11 + L1C1 0A12 + L2C1 A22

)




State observers

Corollary

There exists an observer d

dtx = (A− LC )x + Bu + Ly such that for

all initial states x(0) and x(0)

limt→∞

x(t) − x(t) = 0,

i.e. such that A − LC is Hurwitz, if and only if the unobservablepolynomial χ0(ξ) of (A,C ) is Hurwitz.

Definition

The system (A,B ,C ) ∈ Σn,m,p is called detectable if the the unob-servable polynomial χ0(ξ) of (A,C ) is Hurwitz.




State observers

Theorem (Hautus test)

The system (A,B ,C ) ∈ Σn,m,p is detectable if and only if

rank

(

λI − A

C

)

= n

for all λ ∈ C+ := {λ ∈ C | Re(λ) ≥ 0}.




Dynamic output feedback

Plant:d

dtx = Ax + Bu , y = Cx .

Linear time-invariant feedback controller:

d

dtz = Kz + Ly , u = Mz + Ny ,

with z : R → Rd the state of the controller, K ∈ R

d×d , L ∈ Rd×p,

M ∈ Rm×d , and N ∈ R

m×p the parameter matrices specifying thecontroller.State dimension d ∈ N is called the order of the controller. Is adesign parameter: typically, we want d to be small.





By combining the equations of the plant and controller we obtainthe closed loop system

d

dt

(

x

z

)

=

(

A + BNC BM

LC K

)(

x

z

)

, y = Cx , u = Mz + Ny

Compact form: with xe := col(x , z) (the extended state), Ae, Ce,and He defined in the obvious way, this yields the closed loop systemequations

d

dtxe = Aexe , y = Cexe , u = Hexe.

This is an autonomous dynamical system. We call the eigenvaluesof Ae the closed loop poles and χAe

(ξ) the closed loop characteristicpolynomial.





Question

What closed loop pole locations are achievable by choosing(K ,L,M,N)?More precisely:Let A ∈ R

n×n, B ∈ Rn×m, and C ∈ R

p×n be given. Determinethe set of polynomials χAe

(ξ) obtainable by choosing d ∈ N andK ∈ R

d×d , L ∈ Rd×p, M ∈ R

m×d , N ∈ Rm×p, and where Ae is

given by

Ae =

(

A + BNC BM

LC K

)

.

Full solution to this problem is unknown. We will describe a veryuseful partial result.





We have already seen how to proceed if C = I , i.e., if the full statevector is measured. Let

u = N ′x

be a memoryless state feedback control law obtained this way.We have also seen how we can estimate the state x of from (u, y).Let

d

dtx = (A − L′C )x + Bu + L′y

be a suitable observer.Separation principle: combine an observer with a state controllerand use the same controller gains as in the case in which the stateis measured.





This yields the feedback controller:

d

dtx = (A − L′C )x + BN ′x + L′y , u = N ′x .

This is, of course, a feedback processor with order d = n,K = A − L′C + BN ′, L = L′, M = N ′, and N = 0.

Closed loop system obtained by using this feedback controller:

d

dt

(

x

x

)

=

(

A BN ′

L′C A − L′C + BN ′

)(

x

x

)

,

u = N ′x , y = Cx .





N

Plant

Controller

u

xObserver





We are interested in the characteristic polynomial of Ae .Define a similarity transformation S ∈ R

2n×2n by

S :=

(

I 0I −I

)

Note: S−1 = S .Also

(

A + BN ′ −BN ′

0 A − L′C

)

= S

(

A BN ′

L′C A − L′C + BN ′

)

S−1

Hence the closed loop characteristic polynomial equals the productof χA+BN′(ξ) and χA−L′C (ξ).





Theorem (Pole placement by dynamic output feedback)

Consider the system (A,B ,C ) and assume that (A,B) is control-lable and that (A,C ) is observable.Then for every real monic polynomial r(ξ) of degree 2n, factoriz-able into two real polynomials of degree n, there exists a feedbackcontroller (K ,L,M,N) of order n such that the closed loop systemmatrix Ae has characteristic polynomial r(ξ).

Proof

Take d = n, K = A − L′C + BN ′, L = L′, M = N ′, andN = 0. Choose N ′ such that χA+BN′(ξ) = r1(ξ) and L′ such thatχA−L′C (ξ) = r2(ξ), where r1(ξ) and r2(ξ) are real factors of r(ξ)such that r(ξ) = r1(ξ)r2(ξ).





Theorem (Stabilization by dynamic output feedback)

Consider the system (A,B ,C ) and let χu(ξ) be its uncontrollablepolynomial, χ0(ξ) its unobservable polynomial. Then

1 For any real monic polynomials r1(ξ) and r2(ξ) of degree n

such that r1(ξ) has χu(ξ) as a factor and r2(ξ) has χ0(ξ) as afactor, there exists a feedback controller (K ,L,M,N) of ordern such that the closed loop system matrix Ae hascharacteristic polynomial r(ξ) = r1(ξ)r2(ξ).

2 There exists a feedback controller (K ,L,M,N) as such thatthe closed loop system is asymptotically stable if and only ifboth χu(ξ) and χ0(ξ) are Hurwitz, i.e., if and only if (A,B) isstabilizable and (A,C ) is detectable.



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