Download ECE 515, Fall 2007 - Problem Set #7 Solution: Pole Placement and Observer Design and more Assignments Electrical and Electronics Engineering in PDF only on Docsity!
ECE 515, Fall 2007 Problem Set #7 Solution
Solutions:
- Pole Placement
(a) Let K =
[
k 1 k 2
]
. Then A + BK =
[
2 + k 1 1 + k 2 −1 + 2k 1 1 + 2k 2
]
and so χA+BK (λ) = det(λI − (A + BK)) = λ^2 − (3 + k 1 + 2k 2 )λ + (3 − k 1 + 5k 2 ). Equate χA+BK (λ) = (λ + 1)(λ + 2) = λ^2 + 3λ + 2, then we have −(3 + k 1 + 2k 2 ) = 3, (3 − k 1 + 5k 2 ) = 2, which yields k 1 = − 4 , k 2 = −1.
(b) χA(λ) = λ^2 − 3 λ + 3 ⇒ A¯ =
[
]
, B¯ =
[
]
. P = CC¯ −^1 =
[
]
. In CCF,
want Acl =
[
]
⇒ K¯ =
[
]
[
]
. Transform back to original coordinate K = KP¯ =
[
]
(c) Observer dynamics: xˆ˙ = Axˆ + L(C ˆx − y) Error e = ˆx − x, error dynamics: e˙ = (A + LC)e. Want |e| → 0 as t → ∞ so chose L such that A + LC is Hurwitz.
i. Direct: Let L =
[
l 1 l 2
]
. Then A + LC =
[
2 + l 1 1 + l 1 −1 + l 2 1 + l 2
]
. So, χA+LC (λ) = det(λI − (A + LC)) = λ^2 − (3 + l 1 + l 2 )λ + (3 + 2l 1 + l 2 ). Equating χA+LC (λ) = (λ + 1 − i)(λ + 1 − i) = λ^2 + 2λ + 2 gives −(3 + l 1 + l 2 ) = 2, (3 + 2l 1 + l 2 ) = 2, which yields l 1 = 4, l 2 = −9. ii. Via OCF: Similar to the CCF form, if A¯ = P AP −^1 , C¯ = CP −^1 then C¯ A¯k^ = CAkP −^1 ∀k and so O¯ = OP −^1 ⇒ P = O¯−^1 O where O, O¯ are the observability matrices. In OCF, A¯ =
[
]
, C¯ =
[
]
. Want A¯ + L¯ C¯ =
[
]
, so
L¯ =
[
]
[
]
. Since A¯ + L¯ C¯ = P (A + LC)P −^1 = P AP −^1 + P LCP −^1 ,
we have L¯ = P L so in the original coordinate, L = P −^1 L¯ =
[
]
Comment: This problem demonstrates the essence of pole placement; you can do it either directly or via CCF form. The later is easier to work with when we have large dimension, and it is also easier to implement in computer as well. Regarding the actual calculation, just to make sure that the form you use (u = Kx or u = −Kx) is consistent throughout the calculation.
- (a) In this problem, you actually see how to put the controller and observer together and see the controller in action. Before you can build the simulation, you need to write down the equation first to see how the components connect together. The system:
P :
x˙ = Ax + Bu y = Cx
Note that we only have access to u and y only, not x. Design a feedback gain K such that A + BK is Hurwitz. Design an observer gain L such that A + LC is Hurwitz. The observer is
ˆx˙ = Aˆx + Bu + L(C xˆ − y)
which gives ˙e = (A + LC)e, e = ˆx − x. Putting together, the dynamic controller is
C :
xˆ = (A + LC)ˆx + Bu − Ly z = K xˆ
where z is the output of the dynamic controller. In the simulation, connect the B and C together. In the above scheme we use u = Kx and ˙e = (A + LC)e. If you use u = −Kx, ˙e = (A − LC)e, you need to write the dynamic controller accordingly:
C :
xˆ = (A − LC)ˆx + Bu + Ly z = −K ˆx
so that ˙e = (A − LC)e, e = ˆx − x. (b) With the feedback u = Kx + pr, the closed loop system is ˙x = (A + BK)x + Bpr so the transfer function from r to y is Hcl = C(sI − (A + BK))−^1 Bp. Want Hcl(0) = 1 ⇒ p = −(C(A + BK)−^1 B)−^1. Again, if you use u = −Kx + pr, then the closed-loop state matrix is A − BK instead of A + BK and p is then calculated accordingly. So far, we assume full state feedback. If the state is not known, we use an observer and the the control law is u = −K xˆ + pr where ˆx is the estimated state. See the design and simulation file hw7.mdl (in the simulation, we use u = −Kx, mainly because of the “place” function). Comment: If you don’t get your simulation running correctly, probably you have the wrong sign for the controller gain K (or L) in your SIMULINK. Thus, one needs to write down the dynamic equation first. The simulation will help to understand the overall control scheme (instead of simply calculating K and L as in other homework problems).
- State Feedback Properties
(a) The poles − 1 , 2 , −3 are controllable so if we place the closed-loop poles at − 2 , − 2 , −3, then the closed-loop TF is ˆg(s) = (s − 1)(s + 2)/((s + 2)(s + 2)(s + 3)) = (s − 1)/((s + 2)(s + 3)) due to pole/zero cancellation (feedback does not change zero, so we still have zero at 1, −2 and hence, we can cancel the zero at −2). The resulting system is BIBO stable because ˆg has poles − 2 , − 3 < 0. It is asymptotically stable because the eigenvalues are − 2 , − 2 , −3 (when we talk about asymptotic stability, we talk about state-space variable so we have dimension 3 system here. Feedback does not change dimension; the transfer function is ˆg but the underlying state space is still of dimension 3).
