## Pole Placement Definition of Pole Placement State

(Pole Assignment, Pole Allocation). Placing the poles or eigenvalues of the closed-loop system at specified locations. â¢ Poles can be arbitrarily placed if and only ...

Definition of Pole Placement • (Pole Assignment, Pole Allocation) Placing the poles or eigenvalues of the closed-loop system at specified locations. • Poles can be arbitrarily placed if and only if the system is controllable. • Pole placement is easier if the system is given in controllable form.

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State Feedback

Closed-Loop Dynamics

• Use the state vector to compute the control action for specified system dynamics.

• Substitute for the control

x(k 1)  Ax(k )  B Kx(k )  v( k )

x(k  1)  Ax(k )  Bu( k )

  A  BK x(k )  Bv( k )

y( k )  Cx(k )

u( k )   Kx(k )  v( k ) v(k)

x(k+1)

u(k) +

B

2

+

T

• Closed-loop state matrix Acl  A  B K x(k)

y(k)

x(k 1)  Acl x(k )  Bv( k )

C

y( k )  Cx(k )

A

K 3

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Output Feedback

Pole Placement

• Output measurements y must then be used to obtain the control u u( k )   K y y( k )  v( k ) x(k 1)  Ax(k )  B K y Cx(k )  v( k )  A  BK y C x(k )  Bv( k )

v(k)

x(k+1)

u(k) +

B

+

Ay  A  BK y C

x(k) T

• Choose the gain matrix K or Ky to assign the system eigenvalues to an arbitrary set {i , i = 1, ... , n}. • Easier with state feedback but output feedback is more realistic.

=  K y Cx(k )  v( k )

y(k) C

A

K 5

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Procedure 9.1: Pole Placement by Equating Coefficients

Theorem 9.1 State Feedback If the pair (A, B) is controllable, then there exists a feedback gain matrix K that arbitrarily assigns the system poles to any set {i , i = 1, ... , n}. Furthermore, if the pair (A, B) is stabilizable, then the controllable modes can all be arbitrarily assigned.

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1. Evaluate the desired characteristic polynomial from the specified eigenvalues using the ௡ expression ௗ௖ ௜ ௜ୀଵ 2. Evaluate the closed-loop characteristic polynomial using the expression ௖௟ 3. Equate the coefficients of the two polynomials to obtain equations to be solved for the entries of the matrix . 8

Solution

Example 9.1 • Assign the eigenvalues {0.3  j 0.2} to the pair

0 1  A , 3 4  

• For the given eigenvalues the desired characteristic polynomial is dc ( )    0.3  j 0.2  0.3  j 0.2  2  0.6  0.13

0  b  1 

• The closed-loop state matrix 0 1  0  A  bk T      k1 3 4 1

 0 k2    3  k1

1  4  k 2 

• The closed-loop characteristic polynomial is 1    detI n   A  bk T   det    3  k1    4  k 2   2  4  k 2   3  k1  9

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Lemma

Equating Coefficients

1    detI n  A  bk T   det    3  k1    4  k 2   2  4  k 2   3  k1 

• Gives two equations  (i) 4  k2 = 0.6 (ii) 3 + k1 = 0.13 

For any controllable pair matrix an invertible

k2 = 3.4 k1 = 3.13

where

ିଵ

, there exists such that

ିଵ

is in controllable form.

i.e. kT = [3.13, 3.4] 11

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Transformation

Transformation C௖

௡ିଵ ௖ ௖

௖ ௖

ିଵ

ିଵ

ିଵ

ିଵ ିଵ

ିଵ ௡ିଵ

ିଵ

௡ିଵ

C ିଵ

C௖ Cିଵ

CCିଵ ௖

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Theorem For any controllable single-input pair (A, b), and any set of real or complex conjugate values {i, i = 1, 2, …, n}, there exists a unique 1 by n vector kT such that the eigenvalues of the closed-loop matrix (A b kT) are placed at  Remark

 a1 a2 a a3  2 1 n 1 Tc  C Cc = b Ab ... A b     a 1  n1 0  1 0 0 0 1    0 0 0  t  2n          b Tc1  Cc C 1 =  0 0 1  t n  2 ,n    0 1 t 2 n  t n1,n    1 t 2 n t3n  t nn 

 an1 1  1 0      0 0   0 0

Ab ... An1 b 

1

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Proof The set can be used to construct the desired characteristic polynomial p d ( z )   z  id   z n  a nd1 z n 1    a1d z  a0d n

i 1

By the Lemma, any controllable system with characteristic polynomial n

1- For controllable multiple-input systems, a r by n matrix K exists but is not unique. 2- For stabilizable systems, the controllable modes can be arbitrarily selected but the uncontrollable modes are invariant. 15

p ( z )    z  i   z n  an1 z n1    a1 z  a0 i 1

can be written in controllable form 0 1 n 1 I n 1   Ac      a 0  a 1   a n 1 

0 1 n 1  bc     1  16

Pole Placement

Feedback Gains

• For the desired pole placement, change the characteristic polynomial of the system to pd(z) using the feedback gain vector k Tc  k c ,1

The desired characteristic polynomials is implemented if the following equalities hold

kc , 2  kc ,n 

aid1  ai 1  kc ,i ,

01n 1 I n 1   01n 1  Ac  b c k Tc    k c ,1 k c , 2  k c ,n    a0  a1   an 1   1  01n 1 I n 1       a0  k c ,1   a1  k c , 2    an 1  k c , n  I 0     d 1n d1 n 1 d    a0  a1   an 1 

i  1,2,, n

kc ,i  aid1  ai 1 k c  ad  a a  a0

a d  a0d

a1  an 1 

T

a1d

 and1 

T

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State Feedback For Any Form

Transformation Matrix T Obtain the transformation matrix T from the controllability matrices or from the adj[zI  A]

• By the Lemma, there exists a similarity transformation to controllable canonical form. is unique. ܿ is unique and therefore ௖

் ௖ ௖

ିଵ

ିଵ

T  CC c1 T 1  C c C 1

் ିଵ ௖

் ்

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் ିଵ ௖ 19

 1   1   z   z     adj  zI n  Ab  P0b P1b  Pn1b  T        z n1   z n1      T  P0b P1b  b , Pn1  I n 20

MATLAB

Design Procedure

>> A = [0, 1; 3, 4]; >> B = [0; 1]; >>poles=[-0.3+j*0.3,-0.3-j*0.3]; % Desired poles >> K=place(A, B, poles) % Gain matrix place: ndigits= 16 K= 3.18 4.6 ndigits is a measure of the accuracy of pole placement.

1. Calculate the characteristic polynomial of the plant and obtain the vector . 2. Calculate the desired characteristic polynomial ௗ ଶ

ௗ ଵ

ௗ ௡

and the vector

݀.

4. Calculate the transformation matrix using either (i) , or (ii) T  1 = Cc C  1 5. Calculate the feedback gain matrix ܶ

݀

ିଵ

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MATLAB: Basic Principles

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Example 9.2

• Generate the characteristic polynomial of a matrix: >> poly(A) • Obtain the coefficients of the characteristic polynomial from a set of desired eigenvalues given as the entries of a vector poles >> desired= poly(poles)

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• Design a state feedback for the pair to obtain the eigenvalues {0.1, 0.4  j 0.4}  0.1 0 0.1 A   0 0.5 0.2   0.2 0 0.4

 0.01  b 0    0.005

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Solution

Solution (cont.)

• Characteristic polynomial of the state matrix 3 2 + 0.27   0.01 i.e. a2 = 1, a1 = 0.27, a0 = 0.01 • The transformation matrix Tc1

• The desired characteristic polynomial is 3 0.92 + 0.4   0.032 i.e. ad2 = 0.9, ad1 = 0.4, ad0 = 0.032 • Feedback gain vector

1

1  0 0 10 1.5 0.6 1 3 Tc  0 1 1   10  0 1 1.3      1 1 0.73  5 4 1.9  1.25  0.3846  0.1923 3  10  0.0577 0.625 0.1154     0.0173 0.3125 0.1654 

k T  a0d  a0

a1d  a1

a2d  a2 Tc

1

1.25  0.3846  0.1923    0.032  0.01 0.4  0.27  0.9  110  0.0577 0.625 0.1154     0.0173 0.3125 0.1654  3

  10 85 40 25

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