(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 ...
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.
Pole Placement M. S. Fadali Professor EE University of Nevada, Reno
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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 detI 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 detI 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 n1 0 1 0 0 0 1 0 0 0 t 2n b Tc1 Cc C 1 = 0 0 1 t n 2 ,n 0 1 t 2 n t n1,n 1 t 2 n t3n t nn
an1 1 1 0 0 0 0 0
Ab ... An1 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 nd1 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 an1 z n1 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
aid1 ai 1 kc ,i ,
01n 1 I n 1 01n 1 Ac b c k Tc k c ,1 k c , 2 k c ,n a0 a1 an 1 1 01n 1 I n 1 a0 k c ,1 a1 k c , 2 an 1 k c , n I 0 d 1n d1 n 1 d a0 a1 an 1
i 1,2,, n
kc ,i aid1 ai 1 k c ad a a a0
a d a0d
a1 an 1
T
a1d
and1
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 c1 T 1 C c C 1
் ିଵ
் ்
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் ିଵ 19
1 1 z z adj zI n Ab P0b P1b Pn1b T z n1 z n1 T P0b P1b b , Pn1 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 Tc1
• The desired characteristic polynomial is 3 0.92 + 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 0.9 110 0.0577 0.625 0.1154 0.0173 0.3125 0.1654 3
10 85 40 25
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