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Linear Control Systems Lecture # 12 State Feedback Control – p. 1/3

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discuss state variable feedback control, ackerman's formula, eigen values

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• Linear Control SystemsLecture # 12

State Feedback Control

p. 1/30

• The system

x = Ax+Bu

y = Cx+Du

is asymptotically stable if and only if

Re[i] < 0

for all eigenvalues of A

This condition also guarantees that the system is BIBOstable

The transient response is determined by exponentialmodes of the form tk1eit

p. 2/30

• The shape of the transient response is determined by thelocations of the eigenvalues, e.g.,

A small |Re[i]| produces slow convergence, while alarge |Re[i]| produces fast convergence

A real i produces a monotonic response, while acomplex i produces an oscillatory response

For a complex i, the exponential mode becomes moreoscillatory as the ratio |Im[i]/Re[i]| increases

If some eigenvalues of A do not have negative real parts orhave negative real parts but the transient response is notsatisfactory, we want to use feedback control to reassignthe eigenvalues

p. 3/30

• Feedback control is classified into

State Feedback: all state variables are measured andcan be used in feedback

Output Feedback: Only some output variables aremeasured and can be used in feedback

We start by studying state feedback

p. 4/30

• Consider the state feedback control

u = Fx+ v

F is an m n gain matrix and v(t) is an m-dimensionalexternal input

Closed-loop system

x = (A+BF )x+Bv

y = (C +DF )x+Dv

The stability and transient response of the closed-loopsystem are determined by the eigenvalues of (A+BF )

p. 5/30

• The eigenvalues of (A+BF ) are called the closed-loopeigenvalues, while those of A are called the open-loopeigenvalues

Can we choose F to arbitrarily assign the eigenvalues of(A+BF )?

Lemma: The uncontrollable eigenvalues of (A,B), if any,cannot be relocated by feedback

p. 6/30

• Proof: Suppose (A,B) is not controllable. Then there is anonsingular n n matrix P such that

A = P1AP =

[

A11 A12

0 A22

]

, B = P1B =

[

B1

0

]

where A11 is q q, B1 is q m, the pair (A11, B1) iscontrollable, and the eigenvalues of A22 are theuncontrollable eigenvalues

Partition FP asFP =

[

F1 F2

]

where F1 is m q and F2 is m (n q)

p. 7/30

• P1(A+BF )P = P1AP + P1BFP

=

[

A11 A12

0 A22

]

+

[

B1

0

]

[

F1 F2

]

=

[

A11 + B1F1 A12 + B1F2

0 A22

]

The closed-loop eigenvalues are the eigenvalues of(A11 + B1F1) and the eigenvalues of A22

Remark: There is a real F that assigns the eigenvalues of(A+BF ) at desired locations only if the desired complexeigenvalues are chosen in conjugate pairs

p. 8/30

• Theorem: Let (A,B) be a controllable pair and 1, . . . nbe an arbitrary set of complex numbers (subject only to theconstraint that complex numbers are in conjugate pairs).Then, there exists a real m n matrix F such that theeigenvalues of (A+BF ) are 1, . . . n

The proof is done by showing how to compute F

p. 9/30

• Single-input Systems

Since (A,B) is controllable, there is a nonsingular matrix Psuch that

Ac = P1AP, Bc = P

1B

Ac =

0 1 0...

... . . ....

0 0 1

0 1 n1

, Bc =

0...0

1

det(sI A) = sn + n1sn1 + + 1s+ 0

p. 10/30

• Let F = FcP1, where

Fc =[

f0 f1 fn1

]

Ac +BcFc =

0 1 0...

... . . ....

0 0 1

(0 f0) (1 f1) (n1 fn1)

det[sI (Ac +BcFc)] =

sn + (n1 fn1)sn1 + + (1 f1)s+ (0 f0)

p. 11/30

• If {1, . . . , n} are the desired eigenvalues, then thedesired characteristic polynomial is

d(s) = (s 1) (s n)

def= sn + dn1s

n1 + + d0

Choosefi = i di, for 0 i n 1

Hence, (Ac +BcFc) has the desired eigenvalues{1, . . . , n} and so does

(A+BF ) = P (Ac +BcFc)P1

because the eigenvalues are invariant under state(similarity) transformations

p. 12/30

• Example:

A =

0 1 0

0 0 1

1 0 1

, B =

0

0

1

det(sI A) = s3 s2 1

By Routh-Hurwitz criterion, the system is not asymptoticallystable. Design K to assign the eigenvalues of (A+BK)at 1, 1 j. Desired characteristic equation:

d(s) = (s+ 1)(s2 + 2s+ 2) = s3 + 3s2 + 4s+ 2

F =[

3 4 4]

p. 13/30

• Ackermans Formula: Let

d(s) = sn + dn1s

n1 + + d0

be the desired characteristic polynomial

d(Ac) = Anc + dn1A

n1c + + d0I = A

nc +

n1

i=0

diAic

By Cayley-Hamilton Theorem

Anc + n1An1c + + 0I = 0

Anc = n1An1c 0I =

n1

i=0

iAic

p. 14/30

• d(Ac) =

n1

i=1

(di i)Ai1c

Let ei be the unit vector with one in the ith element

eT1 =[

1 0 0]

, eTn =[

0 0 1]

Cc = [Bc, AcBc, . . . , An1c Bc] =

0 0 1

0 0 1 ...

......

0 1

1

eT1 Cc = eTn e

T1 = e

TnC1c

p. 15/30

• Ac =

0 1 0...

... . . ....

0 0 1

0 1 n1

eT1Ac = eT2 =

[

0 1 0 0]

eT1A2c = e

T2Ac = e

T3

eT1An1c = e

Tn

eT1d(Ac) = eT1

n1

i=0

(di i)Ai1c =

n1

i=0

(di i)eTi+1

p. 16/30

• eT1d(Ac) =

n1

i=0

(di i)eTi+1 = Fc

Fc = eTnC1c d(Ac)

RecallCc = P

1C C1c = C1P

Ac = P1AP Aic = P

1AiP, for i 0

d(Ac) = P1d(A)P

F = FcP1

p. 17/30

• F = FcP1

= eTnC1c d(Ac)P

1

= eTnC1PP1d(A)PP

1

= eTnC1d(A)

Ackermans Formula:

F = eTnC1d(A)

p. 18/30

• Example:

A =

1 0 1

2 1 1

0 1 1

, B =

1

0

1

Open-loop eigenvalues are 0.6348 j0.6916, 2.2695Desired closed-loop eigenvalues are 1, 1, 1

C =

1 0 1

0 3 4

1 1 4

, rank C = 3

p. 19/30

• d(s) = (s+ 1)3 = s3 + 3s2 + 3s+ 1

d(A) = A3 + 3A2 + 3A+ I =

6 2 5

10 2 3

4 1 2

eT3 =[

0 0 1]

F = eT3 C1d(A) =

[

0.8 1.4 1.2]

p. 20/30

• Multi-input systems

Suppose B has full rank. Since (A,B) is controllable, thereis a nonsingular matrix P such that

P1AP = Ac = Ac + BcAm, P1B = Bc = BcBm

Bm is nonsingular. Let F = FcP1

A+BF = P (Ac +BcFc)P1

Ac +BcFc = Ac + BcAm + BcBmFc

Take Fc = B1m (Adm Am)

where Adm is an m n matrix to be chosen

p. 21/30

• Ac +BcFc = Ac + BcAdm

Ac = Block diag

1. . .

1

0 0

ii

, i = 1, . . . ,m

Bc = Block diag

0...0

1

i1

, i = 1, . . . ,m

p. 22/30

• Letd(s) = s

n + dn1sn1 + + d0

be the desired characteristic polynomial

Choose the m n matrix Adm as follows:

the first (m 1) rows are chosen be zeros except thatthe jth row has one at column number 1 +

ji=1 i

the last row is[

d0 d1 dn1

]

p. 23/30

• Ac +BcFc =

0 1 0...

... . . ....

0 0 1

d0 d1 dn1

det[sI (A+BF )] = det[sI (Ac +BcFc)] = d(s)

Remark: For single-input systems, there is a unique matrixF that assigns the eigenvalues of (A+BF ) at 1, . . . , nFor multi-input systems, F is not unique

p. 24/30

• Example:

A =

1 0 1

2 2 2

1 0 3

, B =

1 0

0 2

1 1

Eigenvalues of A are 2, 0.7321, 2.7321Desired eigenvalues are 1, 1 j

d(s) = s3 + 3s2 + 4s+ 2

p. 25/30

• Matlab Calculations:W = ctrb(A,B); rank(W), rank(B)ans = 3ans = 2rank(W(:,1:3))ans = 3

1 = 2, 2 = 1

Cbar = [W(:,1) W(:,3) W(:,2)]; M = inv(Cbar);Q = [M(2,:); M(2,:)*A; M(3,:)]; P = inv(Q); PI = Q;Ac = PI*A*P; Bc = PI*B;Am = [Ac(2,:);Ac(3,:)]; Bm = [Bc(2,:);Bc(3,:)];Adm = [0 0 1;-2 -4 -3];Fc = inv(Bm)*(Adm - Am); F = Fc*PI;

p. 26/30

• Ac =

0 1 0

1 3 0.5

6 0 1

, Bc =

0 0

1 0.5

0 1

Am =

[

1 3 0.5

6 0 1

]

, Bm =

[

1 0.5

0 1

]

F =

[

0 1.5 5

0.6667 3.3333 4.6667

]

p. 27/30

• What if rank B = r < m?

There is a nonsingular matrix R such thatBR =

[

B1 0]

, where the n r matrix B1 has full rank

Find F1 to assign the eigenvalues of (A+B1F1). Take

F = R

[

F1

]

A+BF = A+BR

[

F1

]

= A+[

B1 0]

[

F1

]

= A+B1F1

p. 28/30

• Eigenvalue Assignment Using Matlab:Matlab has two commands

K = acker(A,B, p) assigns the eigenvalues of(ABK) at the elements of the vector p. It is usedonly with single-input systems

K = place(A,B, p) assigns the eigenvalues of(ABK) at the elements of the vector p. It works formulti-input systems, but does not allow multiple desiredeigenvalues

Using the place command in the last example gives

F =

[

0.1645 1.4989 4.7965

1.0000 1.3160 0.2639

]

p. 29/30

• Stabilizability: The system

x = Ax+Bu

is stabilizable if there exists a matrix F such that theclosed-loop system

x = (A+BF )x

is asymptotically stable

Theorem: (A,B) is stabilizable if and only if theuncontrollable eigenvalues of A, if any, have negative realparts

p. 30/30