EE611 Deterministic Systems State Feedback Kevin D. Donohue Electrical and Computer Engineering University of Kentucky State Feedback Block diagram of a system with state feedback: ẋ x ut 1 rt b c + + s - A Reference input Given original system: ẋ=A xb ut y t =c x Show that state feedback y t k State feedback ẋ= A−bk xbr t system becomes: yt =c x Controllability and Feedback Can the application of state feedback to a system result in a loss of controllability? Theorem 8.1 Given pair (A-bk, b) and 1xn feedback vector k, then (A-bk, b) is controllable iff (A, b) is controllable. Controllable Canonical Form Given controllable system ẋ=A xb ut y t =c x with characteristic polynomial s=∣s I−A∣=sn1 s n−1 2 sn−2... n−1 s− n system can be transformed into controllable canonical form by x =P x [ ] 1 0 Q=P−1=[ b A b A 2 b A 3 b ... A n−1 b ] 0 ⋮ 0 1 1 0 ⋮ 0 2 1 1 ⋮ 0 ... n−1 ... n−2 ... n−3 ⋮ ⋮ ... 1 Controllable Canonical Form [ ][ ̇x 1 −1 − 2 ̇x 2 1 0 ⋮ = ⋮ ⋮ 0 0 ̇x n−1 0 0 ̇x n ][ ] [ ] ... − n−1 − n x 1 1 0 x 2 ... 0 0 ⋱ ⋮ ⋮ ⋮ ⋮ ut 0 ... 0 0 ̇x n−1 0 ... 1 0 x n y t =[ 1 2 ... n−1 n ] n−1 [] n−2 x 1 x 2 ⋮ x n−1 x n 1 s 2 s ⋯n−1 sn y s g s= = u s sn 1 sn−1 2 sn−2 ⋯ n−1 s n Eigenvalue Placement System behavior is modified by changing its eigenvalues through feedback. Theorem 8.3 Given n-dimensional controllable state equation: ẋ=A xb ut yt =c xd ut then state feedback u = r – kx with 1xn real-valued feedback vector k, the eigenvalues of A-bk can be arbitrarily assigned provided that complex conjugate eigenvalues are assigned in pairs. Example For unstable system below, find a feedback vector k to place eigenvalues at -4 and -2: [ ] [] ẋ= 2 24 1 x ut 1 0 0 y t =[ 1 2 ] x Show k = [ 8 32] and that observability is lost as a results of the feedback. Example Use controllable canonical form to find feedback vector k to place eigenvalues at -1±j1 and -(2)0.5 [ ] [] 2 1 0 0 ẋ= 0 2 0 x 1 ut 0 0 −1 1 y t =[ 2 −1 −1 ] x Show k = [ 6.41, 4.83, -1.77]P = [11.38, 6.37, .05] And for controllable canonical form transformation: [ 0 1 1 Q= 1 −1 −2 −1 −4 4 ] Regulation and Tracking Regulator Problem: Design feedback system to bring response to zero in response to a disturbance/initial condition. Tracking Problem: Given a step change in reference r to value a for t > 0, design a feedback system so that response y(t) approaches a as t approaches infinity (asymptotic tracking). Regulation As long as real parts of poles/eigenvalues are negative regulation is achieved. In cases where eigenvalues are too close to the jω axis, response can be improved by moving eigenvalues into desired range. y t=c exp A−bk t x 0 Tracking Poles/eigenvalues must be negative as in the case of regulation. In addition a gain (scaling) on reference signal must be applied to eliminate steady-state error in the tracking problem. ut = p rt −kx Gain p needs to be the reciprocal of the transfer function under the following limit: n 1 p=lim = n s 0 g f s Example Design a tracking system with state feedback such that system will settle in less than 2.5 seconds (within 2% of steady-state error) after a step change at time t=0, with 0 steady state error. [ ][] −0.75 1.25 −1.25 2 ẋ= −0.075 −1.375 0.5375 x 0.5 u t −0.25 −1.25 0.125 1 y t =[ 1 0 1 ] x Show eigenvalues −3.2 2 , −3.2± j 3.2 result in a k = [122.7, 273.2, -373.0] and p= 141.2 Stabilization If an unstable system is not controllable, it may still be possible to stabilize it with feedback if the unstable state variables are associated with the controllable part of the system [ ][ ][ ] [ ] c A 12 x x A bc ̇ c c = ut c x x c 0 A 0 ̇ c A partitioned state feedback vector can be introduced such that: x u=r−k x=r−k x=r−[ k c k c ] c xc [] and [ ][ c− 12 − A bc k c A bc k c ̇x c = c x 0 A ̇ c ][ ] [ ] x bc c ut x 0 c Example Determine if the unstable system can be stabilized. If so, find the feedback vector to stabilize the system. [ 0 0 0 ẋ= 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 −2 0 0 0 0 0 −1 1 0 0 0 −1 −1 0 0 0 0 0 1 0 0 0 0 −2 ] [] 0 1 0 1 0 0 0 x 0 ut 0 0 2 1 1 1 y t =[ 1 −1 2 0 1 0 1 ] x Show system is not controllable, but can be stabilized. If unstable modes have their real parts shifted left by 2, the feedback vector is k = [-2, -12, 0, 0, 0, 4, -6], this leaves the originally stables modes unaffected. Lecture Note Homework U11.1 Implement the system below in Simulink before and after feedback compensation (problem 8.7 in text) for a unit step response. Find feedback vector k such that resulting eigenvalues are at -2, -1±j and will track asymptotically any step reference input. 1 1 −2 1 y t =[ 2 0 0 ] x ẋ= 0 1 1 x 0 u t 0 0 1 1 [ ] [] Use the state space model for the original system, however add a state feedback loop using summers, gains, products, and constants. Hand in a copy of the simulation schematic with the corresponding plot of their outputs. Assume the system is relaxed at t = 0.