Design via Pole Placement
Determination of feedback gain by using controllable canonical form
1 1n n
Design via Pole Placement
Determination of feedback gain by Ackermann’s Formula
Determination of feedback gain by a causal approach
Design via Pole Placement
Example:
Determination of feedback gain by using controllable canonical form:
Determination of feedback gain by using Ackermann’s Formula:
Determination of feedback gain bycausal method:...........
Design via Pole Placement
Uncontrollable system:
Kalman Controllable Form
State Feedback
1 2(k 1) (k)
(k 1) (k)0
A B K A B Kx xc c cc cccx xAc cc
1 2(k) Kx(k) (k)u K K x 1K KV The modes of can
be arbitrarily assigned.
The modes of is not influenced by state feedback control
𝐴𝑐
Design via Pole Placement
Example:
Kalman Controllable Form
State Feedback
1 2(k 1) (k)
(k 1) (k)0
A B K A B Kx xc c cc cccx xAc cc
1 2(k) Kx(k) (k)u K K x 1K KV The modes of can
be arbitrarily assigned.
The modes of is not influenced by state feedback control
𝐴𝑐
Design via Pole Placement
Example:
Kalman Controllability Decomposition
Desired closed loop poles: -1, -2.5, -2.5
6.5 0.5 6.5 6.5 0 1 0
0.5 5.5 5.5 5.5 2 1 2(k 1) (k) (k)
0.5 0.5 0.5 6.5 3 4 3
0.5 0.5 5.5 0.5 3 2 3
x x u
6 1 0 0 1 1 1
0 6 0 0 1 1 1(k 1) (k) (k)
0 0 6 0 0 1 0
0 0 0 6 0 0 0
x x u
6 1 0 1 1 1
0 6 0 1 1 1
0 0 6 0 1 0c cA B
17.5 26 8.5
0 0 8.5
0 0 0
K
1 1
17.5 26 8.5 0
[ 0] 0 0 8.5 0
0 0 0 0
K K V V
Not: We do not feedback the uncontrollable mode
Design via Pole Placement
Example:
Kalman Controllability Decomposition
Desired closed loop poles: -1, -2.5, -2.5
6.5 0.5 6.5 6.5 0 1 0
0.5 5.5 5.5 5.5 2 1 2(k 1) (k) (k)
0.5 0.5 0.5 6.5 3 4 3
0.5 0.5 5.5 0.5 3 2 3
x x u
6 1 0 0 1 1 1
0 6 0 0 1 1 1(k 1) (k) (k)
0 0 6 0 0 1 0
0 0 0 6 0 0 0
x x u
6 1 0 1 1 1
0 6 0 1 1 1
0 0 6 0 1 0c cA B
17.5 26 8.5
0 0 8.5
0 0 0
K
1 1
17.5 26 8.5 0
[ 0] 0 0 8.5 0
0 0 0 0
K K V V
Not: We do not feedback the uncontrollable mode