Topic并16 16.31 Feedback Control State-Space Systems Open-loop Estimators Closed-loop Estimators Observer Theory (no noise)-Luenberger IEEE TAC Vol 16, No. 6, pp. 596-602, December 1971 Estimation Theory(with noise)-Kalman Copyright [2001 by JOnathan dHow. O
Topic #16 16.31 Feedback Control State-Space Systems • Open-loop Estimators • Closed-loop Estimators • Observer Theory (no noise) — Luenberger IEEE TAC Vol 16, No. 6, pp. 596—602, December 1971. • Estimation Theory (with noise) — Kalman Copyright 2001 by Jonathan How. 1
Fall 2001 16.3116-1 Estimators/observers . Problem: So far we have assumed that we have full access to the state a(t)when we designed our controllers Most often all of this information is not available Usually can only feedback information that is developed from the sensors measurements Could try“ output feedback” K KI ame as the proportional feedback we looked atat the beginning of the root locus work This type of control is very difficult to design in general Alternative approach: Develop a replica of the dynamic sys- tem that provides an estimate of the system states based on the measured output of the system ● New plan 1. Develop estimate of a(t) that will be called a(t) 2. Then switch from u=-Ka()to u=-kit ●1 wo key questions How do we find i(t)? Will this new plan work?
Fall 2001 16.31 16—1 Estimators/Observers • Problem: So far we have assumed that we have full access to the state x(t) when we designed our controllers. — Most often all of this information is not available. • Usually can only feedback information that is developed from the sensors measurements. — Could try “output feedback” u = Kx ⇒ u = Ky ˆ — Same as the proportional feedback we looked at at the beginning of the root locus work. — This type of control is very difficult to design in general. • Alternative approach: Develop a replica of the dynamic system that provides an “estimate” of the system states based on the measured output of the system. • New plan: 1. Develop estimate of x(t) that will be called ˆx(t). 2. Then switch from u = −Kx(t) to u = −Kxˆ(t). • Two key questions: — How do we find ˆx(t)? — Will this new plan work?
Fall 2001 6.3116-2 Estimation schemes . assume that the system model is of the form Ac+ Bu, a(0) unknown Where 1. A.B. and C are known 2. u(t) is known 3. Measurable outputs are y(t) from C+I e Goal: Develop a dynamic system whose state for all time t20. Two primary approaches Open-loop Closed-loop Open-loop Estimator Given that we know the plant matrices and the inputs, we can just perform a simulation that runs in parallel with the system (t)=A+B(t) Then i(t=a(t)v t provided that i(0=a(o Major Problem: We do not know a(0)
Fall 2001 16.31 16—2 Estimation Schemes • Assume that the system model is of the form: x˙ = Ax + Bu , x(0) unknown y = Cx where 1. A, B, and C are known. 2. u(t) is known 3. Measurable outputs are y(t) from C 6= I • Goal: Develop a dynamic system whose state xˆ(t) = x(t) for all time t ≥ 0. Two primary approaches: — Open-loop. — Closed-loop. Open-loop Estimator • Given that we know the plant matrices and the inputs, we can just perform a simulation that runs in parallel with the system ˙ xˆ(t) = Axˆ + Bu(t) • Then ˆx(t) ≡ x(t) ∀ t provided that ˆx(0) = x(0) • Major Problem: We do not know x(0)
Fall 2001 16.3116-3 X() SYSTEM A B.c y() u(七 DBSERVER A,8.C 文( Analysis of this case i(t)= Ac+Bu( i(t)=A.c+Bu(t) Define the estimation error (t)=a(t-i(t) Now want c(t)=0Vt. But is this realistic?) Subtract to get dt x-)=Ax-)→(t)=A which has the solution (t) At Gives the estimation error in terms of the initial error
Fall 2001 16.31 16—3 • Analysis of this case: x˙(t) = Ax + Bu(t) ˙ xˆ(t) = Axˆ + Bu(t) • Define the estimation error x˜(t) = x(t) − xˆ(t). Now want ˜x(t)=0 ∀ t. (But is this realistic?) • Subtract to get: d dt(x − xˆ) = A(x − xˆ) ⇒ x˜˙(t) = Ax˜ which has the solution x˜(t) = eAtx˜(0) — Gives the estimation error in terms of the initial error
Fall 2001 16.3116-4 Does this guarantee that a=0v t? Or even that→0ast→∞?( which is a more realistic goal - Response is fine if (0)=0. But what if (0)+0? ● If a stable,then→0ast→oo, but the dynamics of the estima- tion error are completely determined by the open-loop dynamics of the system(eigenvalues of A) Could be very slow no obvious way to modify the estimation error dynamics Open-loop estimation does not seem to be a very good idea Closed-loop Estimator An obvious way to fix this problem is to use the additional informa- tion available How well does the estimated output match the measured output? Compare:y= C with y=Ci Then form y=y-y≡Cx
Fall 2001 16.31 16—4 • Does this guarantee that ˜x = 0 ∀ t? Or even that ˜x → 0 as t → ∞? (which is a more realistic goal). — Response is fine if x˜(0) = 0. But what if x˜(0) 6= 0? • If A stable, then ˜x → 0 as t → ∞, but the dynamics of the estimation error are completely determined by the open-loop dynamics of the system (eigenvalues of A). — Could be very slow. — No obvious way to modify the estimation error dynamics. • Open-loop estimation does not seem to be a very good idea. Closed-loop Estimator • An obvious way to fix this problem is to use the additional information available: — How well does the estimated output match the measured output? Compare: y = Cx with ˆy = Cxˆ — Then form ˜y = y − yˆ ≡ Cx˜