State-space feedback 4 Ackermann’s approach J A Rossiter 1 Slides by Anthony Rossiter
State-space feedback 4 Ackermann’s approach
J A Rossiter
1
Slides by Anthony Rossiter
Introduction
• The previous videos showed how state feedback can place poles precisely as long as the system us fully controllable.
• Both relied on control canonical forms.
• This video shows how one can find a suitable state feedback without resorting to canonical forms within the design procedure.
Slides by Anthony Rossiter
2
xBKAxKxu
BuAxx
)(
Remark
Much of this resource derives and explains the result.
If you only want to know the result, go directly towards the end.
Slides by Anthony Rossiter
3
Pole polynomials
The approach given here relies on clear definitions of two polynomials:
• the open-loop pole polynomial.
• the desired closed-loop pole polynomial.
Slides by Anthony Rossiter
4
o
n
n
n
o asasasp
1
1
1
o
n
n
n
c sssp
1
1
1
Cayley-Hamilton Theorem
A matrix satisfies its own characteristic equation.
This implies that.
Slides by Anthony Rossiter
5
0
1
10 aaAI n
n
n
00
0
1
1
AaAaA n
n
n
The proof is omitted but relatively straightforward and available in most text
books.
Corollary
The following 2 identities are known to be true.
Slides by Anthony Rossiter
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001
1
1
IaAaAaA n
n
n
o
n
n
n
o asasasp
1
1
1
o
n
n
n
c sssp
1
1
1
0)()()( 01
1
1
IBKABKABKA n
n
n
Finding K such that this last
identify is satisfied implies achieving target
poles.
Ackermann’s approach
The intention is to exploit the latter of these identities and full rank nature of the controllability matrix.
First define:
The first concept is to show that pc(A) is a component of pc(A-BK) and exploit this in the algebra.
NOTE, by definition: Slides by Anthony Rossiter
7
IBKABKABKABKAp n
n
n
c 01
1
1 )()()()(
IAAAp n
c 01 )()()(
0)(
0)(
Ap
BKAp
c
c
Ackermann’s approach
The first concept is to expand pc(A-BK) and unpack the underlying structure. First consider each term.
Slides by Anthony Rossiter
8
BKABKA
BKBKAABKABKA
BKABfBKABKAABKA
BKABfBKABKAABKA
BKABfBKABKAABKA
n
nnnnnn
n
nnnnnn
n
nnnnnn
222
2
233322
1
122211
111
)()(
),()1()()1()(
),()1()()1()(
),()1()()1()(
Deliberate separation of terms with a leading ‘A’
and a leading ‘B’.
Ackermann’s approach The next step is to show that one can express each term using the controllability matrix.
Slides by Anthony Rossiter
9
n
c
v
nn
n
n
M
nnnn
n
nnnnnn
K
BKK
BKAf
BABAABBABKA
BKABfBKABKAABKA
21
12
111
)()1(
),()1(
],,,,[)(
),()1()()1()(
3
0
)(
),(
],,,,[)(
),()()(
3
1233
3
2233
v
M
nK
BKK
BKAf
BABAABBABKA
BKABfBKABKAABKA
c
The ZERO is key
Ackermann’s approach
Next expand pc(A-BK) and unpack the underlying structure. First consider each term.
Slides by Anthony Rossiter
10
1
2
22
1
11
)(
)(
)(
vMABKA
vMABKA
vMABKA
vMABKA
c
c
nc
nn
nc
nn
Only vn is non-zero in the bottom row!
IBKABKABKABKAp n
n
n
c 01
1
1 )()()()(
][
)(
1111
01
1
1
vvvM
IAAABKAp
nnnc
n
n
n
c
Ackermann’s approach
Show that pc(A) is a component of pc(A-BK) and exploit this in the algebra.
Slides by Anthony Rossiter
11
][
)(
1111
01
1
1
vvvM
IAAABKAp
nnnc
n
n
n
c
0)(
0)(
Ap
BKAp
c
c
][)(0 1111 vvvMAp nnncc
][)( 1111
1 vvvApM nnncc
Ackermann’s approach
Exploit the structure of vn by extracting only the last row of each side of the equation and thus extracting the feedback gain K.
Slides by Anthony Rossiter
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nv
nn
n
n
K
BKK
BKAf
21 )()1(
),()1(
][)( 1111
1 vvvApM nnncc
KApM cc )(100 1
,0100,0100 21 nn vv
As pc is by definition, one can find the required state feedback using this formulae.
Summary of Ackermann’s approach
Define the desired closed-loop pole polynomial.
The required feedback is given from the formula where Mc is the controllability matrix.
Slides by Anthony Rossiter
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KApM cc )(100 1
o
n
n
n
c sssp
1
1
1
EXAMPLES
Slides by Anthony Rossiter
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Example 1: Choose K to set the closed-loop poles at -1 and -2.
Slides by Anthony Rossiter
15
First find the required pole polynomial
43;2
1;
4.01
21
CBA
232 sspc
Find pc(A)
04.16.1
2.32
20
02
4.01
213
4.01
21)(
2
Apc
Example 1: Choose K to set the closed-loop poles at -1 and -2.
Slides by Anthony Rossiter
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Define the controllability matrix
;2
1;
4.01
21
BA
8.12
31],[ ABBM c
Use Ackermann’s formulae
95.031.0)(100 1 KApM cc
04.16.1
2.32)(Apc
Slides by Anthony Rossiter
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Example 2: Choose K to set the closed-loop poles at -0.5, -1 and -1.5.
Slides by Anthony Rossiter
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First find the required pole polynomial
431;
0
1
1
;
4.01.00
6.04.02.0
5.021
CBA
75.075.23 23 ssspc
Find pc(A) [Use MATLAB as tedious]
608.2257.004.0
742.1248.0046.0
715.156.023.0
75.075.23)( 23 IAAAApc
Example 2: Choose K to set the closed-loop poles at -1 and -2.
Slides by Anthony Rossiter
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Define the controllability matrix
02.01.00
1.06.01
25.211
],,[ 2BAABBM c
Use Ackermann’s formulae
57.2018.218.0)(100 1 KApM cc
;
0
1
1
;
4.01.00
6.04.02.0
5.021
BA
608.2257.004.0
742.1248.0046.0
715.156.023.0
)(Apc
Summary
Introduced concepts of pole placement state feedback without a control canonical form.
1. Show that, assuming full controllability, one can use Ackermann’s formulae to define the required feedback to achieve any desired pole polynomial.
2. Not paper/pen exercise in general as involves substantial matrix multiplication and inversion.
3. Numerically poorly conditioned for weakly controllable and large dimension systems.
Slides by Anthony Rossiter
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Kxu
KApM cc )(100 1
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Anthony Rossiter Department of Automatic Control and
Systems Engineering University of Sheffield www.shef.ac.uk/acse