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Unformatted text preview: Closedloop pole locations Closedloop pole locations Conﬁnuoustime intuition: risetime, damping ratio, settling time. Linear Quadratic Regulators (LQR) Optimal estimation (Kalman ﬁltering) zplane ”.5 I H :1
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1—s/a 33(3) 2 (pole = a, time constant 2 1/a ﬂ = 4.67" Settling time (to within 1%): at its: zplane Closedloop pole locations Imaginary 2T) 1 Real Discrete equivalent: z!y = ‘3'” Roy Smith, ECE 147b, 16: 2 Closedloop pole locations Closedloop pole locations Imaginary z—plane Imaginary
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e m =n/T splane 0 Dominant complex poles: .§, §* Natural frequency: mm = abs(§)
Damping ratio: C = sin(0) Rise time (10% to 90%): 73,. x w— Overshoot: M = e‘TFC/V "<2 Roy Smith, ECE 147b, 16: 3 Closedloop pole locations Closedloop pole locations zplane “
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I! Roy Smith, ECE 147b, 16: 4 Closedloop pole locations Pole placement State feedback: u(k:) = —Kz(k)
Closedloop dynamics: a:(k+ 1) = (A — BK) :1:(k) Closedloop poles given by: eig(A — BK)
Ackermann’s formula: K = [0 0 1]C_1'yC(A) ‘7. Example by hand
Z desired closedloop poles are p1 and p2 I2 = eye(2,2); Ctrl = ctrb(A,B); "A controllability matrix
gamma_c = (A  pl*I2) * (A  p2*12); % polynomial in A K = [O , 1] * inv(Ctrl) * gamma_c; Z Example with place K = place(A,B, [p1; p2]); Roy Smith, ECE 147b, 16: 5 Closedloop pole locations Linear Quadratic Regulator (LQR)
State feedback: u(k) = —K:c(k)
Closed100p dynamics: xclpUc + 1) = (A — BK) xdp(k:) Closedloop poles given by: eig(A — BK)
00 Deﬁne a cost function: JLQR = ZachUcV Qx(.1p(k) + u(k)T Ru(k)
k=1 Weighting matrices: Q (state) and R (input) Find the K that minimizes JLQR Note that this guarantees that A — BK is stable. Why?
K = dlqr(A,B,Q,R); Roy Smith, ECE 147b, 16: (7 Closedloop pole locations Linear Quadratic Regulator (LQR) JLQR = Zmr:lp(k)TQTclp(k) + ”(k)TR’LL(k)
k3=l Weighting matrices: Q (state) and R (input)
Q is symmetric and positive deﬁnite: Q = QT and xTQsc > 0 for all as R is symmetric and positive deﬁnite: R 2 RT and uTRu > 0 for all u Q E Rdi111(:1:) )< <lim(:l;) and R E Rdi1u(’u) )< (lim('u,) Initial choice (Bryson’s rule): Q is diagonal with Q” = W
R is dia onal with R — 4
g ‘ mum/cm Roy Smith, ECE 147b, 16: 7 Closedloop pole locations Linear Quadratic Estimation (Kalman ﬁltering)
How to choose pole locations for the estimation errors: A — LC Plant model: $(k + 1) = Aa:(k) + Bu(k) + 010(k)
3106) = C 173(k) + C(19)
Process noise: 111(k) (zero—mean White noise)
Measurement noise: v(k) (zeromean white noise) Estimator: :i'(k+l) = A5606) + Bu(k) + L(y(k) — 622(k)) Estimator error: riﬂe) = m(k:) — at(k) :E(k+1) = (A — LC)5c(k) + Gw(k) — Lv(k) Roy Smith, ECE 147b, 16: 8 Closedloop pole locations Linear Quadratic Estimation (Kalman ﬁltering) Plant model: :I:(k+ 1) = Ach) + Bu(k) + Gw(k)
.7109) = 096(k) + ”(79) What is the optimal choice of L to balance between the two noise sources: 111(k) and 1106)?
Estimation error: i:(k+1) = (A — LC) £09) + Gw(k:) — Lv(k:)
Objective: min £{i‘(k)TaE(k)} (minimize estimation error variance) To do this we must know (or at least approximate): 8{w(k) w(k)T} : QN (QN E Rdim(ur)xdim(w)) £{v(k)v(k)T} = RN (RN 6 Rdim<y>xdim<y>) L = kalman(Pz,QN,RN) Roy Smith, ECE 147b, 16: 9 ...
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 Spring '07
 RODWELL

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