Constrained Optimal Control. Constrained Optimal Control II

Transcription

1 Optimal Control, Guidance and Estimation Lecture 35 Constrained Optimal Control II Prof. Radhakant Padhi Dept. of Aerospace Engineering Indian Institute of Science - Bangalore opics: Constrained Optimal Control Motivation Pontryagin Minimum Principle ime Optimal Control of LI Systems ime Optimal Control of Double-Integral System Fuel Optimal Control Energy Optimal Control State Constrained Optimal Control

2 Summary of Pontryagin Minimum Principle Prof. Radhakant Padhi Dept. of Aerospace Engineering Indian Institute of Science - Bangalore Obective o find an "admissible" time history of control variable U t, t t, t f, + ( or, component wise, ) where U ( t) U U u ( t) U,which: ( ) 1) Causes the system governed by Xɺ = f t, X, U to follow an admissible traectory ) Optimizes (minimizes/maximizes) a "meaningful" performance index t f = ϕ ( f, f ) + (,, ) J t X L t X U dt t 3) Forces the system to satisfy "proper boundary conditions". ( ) 4

3 Solution Procedure of a given Problem Hamiltonian : Necessary Conditions : (,, λ) = (, ) + λ (, ) H X U L X U f X U H (i) State Equation: Xɺ = = f ( t, X, U ) λ H (ii) Costate Equation: ɺ λ = X (ii i) Optimal Control Equation: Minimize H with repect to U ( t) U ( ) ( λ) H ( X U λ ) i.e. H X, U,,, (iv) Boundary conditions: X ϕ = Specified, λ f = X f 5 ime Optimal Control of Control Constrained LI Systems Prof. Radhakant Padhi Dept. of Aerospace Engineering Indian Institute of Science - Bangalore

4 ime Optimal Control of LI Systems Problem : Minimize the time taken for an LI system to go from an arbitrary initial state to the desired final state. Note : By considering the desired final state as the origin of state space, it becomes time-optimal "regulator problem". System Dynamics (LI system) : Xɺ = AX + BU n m where, X R, U R are constant matrices 7 ime Optimal Control of LI Systems he control magnitude U satisfies U U U U U + or, component wise, u U, where = 1,,, m + where, U is the lower bound and U is the upper bound. Without loss of generality, it can be assumed that 1 U 1 or, component wise, u 1, where = 1,,, m Assumption : he system is state controllable, i.e. n 1 G B AB A B A B is of rank n. 8

5 ime Optimal Control of LI Systems Problem : Find optimal control which satisfies 1, takes the system from Initial state X to Final state (origin) in minimum time. Solution : t f ( ) = [,, ] t J U U u Step 1 : Selection of Performance Index ( f ) f L X U t dt = 1 dt = t t t (Note: t is fixed, but t is 'free') f t 9 ime Optimal Control of LI Systems Step : Hamiltonian (, λ, ) = 1 + λ [ + ] = 1+ [ AX ] λ + H X U AX BU U B λ Step 3 : State and Costate Equation ɺ H λ Boundary condition X = = AX + BU λ = = A f, H X X () = X ; X ( t ) = ; but t f is 'free' ɺ λ 1

6 ime Optimal Control of LI Systems Step 4 : Optimal Control 1 + (, λ, ) (, λ, ) H X U H X U [ AX ] λ + U 1 + [ AX ] B λ (1) If q >, then U = 1 () If q <, then U = + 1 min{ } λ + U B λ U q U q,, where, q B λ U q = U q U 1 Final Solution: { } sgn{ λ} u = sgn q = B th where B is colomn vector of B 11 ypes of ime-optimal Control ) ( ) 1 Normal ime - Optimal Control NOC System During the interval t, t f, a set of times t, t,, t t, t, 1 γ f where, γ = 1,,, = 1,,, m :, if and only if t t = γ q = B λ =, otherwise hen we have a normal time-optimal control problem. 1

7 Bang-Bang Control Reference: D. S. Naidu, Optimal Control Systems CRC Press, (Chap. 7) For NOC system, the Optimal control { } { } U = sgn q = sgn B λ t t, t f is a piecewise constant function of time (i.e. of Bang - Bang nature) 13 ypes of ime-optimal Control ) ( ) Singular ime - Optimal Control SOC [ ] 1 [ ] During the interval t, t f, one or more sub-intervals 1, : q = t, : Singularity Intervals hen we have a singular time-optimal control problem. Note : During singularity intervals, the time-optimal control is not defined Reference: D. S. Naidu, Optimal Control Systems CRC Press, (Chap. 7) 14

8 Condition for NOC System ( ) Assume λ = λ. hen A t { } { λ} { λ} A t u = { q } = { B e λ} U = sgn q = sgn B = sgn B e Component wise, sgn sgn [ ] Hence, if q = t 1,, hen all derivatives of q =, i.e. 15 Condition for NOC System q qɺ qɺɺ A t = B e λ = A t = B A e λ = A t B A e = λ = ( 1 ) ( 1) n n A t qɺ = B A e λ = Combining, one can write: G e A t λ = n 1 where, G B AB A B A B 16

9 Condition for NOC System Combining the results for = 1,, m, one can write: where A t G e λ = n 1 G B AB A B A B A t However, e and λ. Hence, for SOC problems, G must be singular. In other words, for NOC problems, G must be non-singular; i.e. the system should be completely state controllable. 17 Solution of OC System: Some Observations i Singular interval cannot exist for a completely controllable system i he necessary and sufficiency condition for NOC is that the system is completely controllable. Conversely, he necessary and sufficiency condition for SOC is that the system is uncontrollable. 18

10 Uniqueness and Number of Switches Uniqueness of Optiomal Control If the ime-optimal system is normal then the Optimal control solution is unique. Number of Switchings If original system is normal, and if all n eigenvalues are real hen U can switch (from 1 to + 1 or from + 1 to 1) at most ( n ) 1 times. 19 Structure of ime-optimal Control System Dr. Radhakant Padhi Associate Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore

11 Solution of OC System Problem : System Dynamics Xɺ = AX + BU subect to u 1, = 1,,, m Obective: X with minimum time under constraint Solution : Optimal control u { B λ} = sgn = sgn A t { B e λ} 1 Open-loop Structure of OC System Adopt an iterative procedure Assume Compute λ( t) Evaluate λ u Solve for system traectory X ( t) Monitor X ( t) and see if a t : X ( t ) =. hen the control is ime-optimal control. If not, then change λ and repeat the procedure. f f

12 Structure of OC System Reference: D. S. Naidu, Optimal Control Systems, CRC Press, (Chap. 7) 3 Structure of OC System Closed loop Structure : { h X } Optimal control law U = sgn ( ) ( ) However, an analytical/computational algorithm h( X ) = B λ X ( t) needs to be developed (demonstrated through a bench-mark example). 4

13 ime-optimal Control for Double Integral System Dr. Radhakant Padhi Associate Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore OC of a Double Integral system System Dynamics : mɺɺ y = f State variables : 1 x = y; x = yɺ 1 Double integral system described by ( ) xɺ = u where, u f / m u is constarined as u 1 t t, t f xɺ = x 6

14 OC of Double Integral System: Problem Statement Given Suect to Double integral system u ( ) x ( ) 1 1 t t, t f 1 Find an admissible control such that x [ ] x xɺ = x = u in minimum time. Assumption : Normal OC; i.e. No Singular control. 7 OC of Double Integral System Step 1 : Performance Index t f J = 1dt = t t t f Step : Hamiltonian ( ) H X, λ, u = 1+ λ x + λ u 1 Step 3 : Minimization of Hamiltonian ( λ ) H ( X u λ ) According to PMP, H X, u,,, λ u Optimal control λ u u = sgn { λ } 8

15 OC of a Double Integral System Step 4 : Costate Solution ɺ H λ1 = = x λ 1 ( t ) = λ ( ) 1 1 ɺ H λ = = λ = λ 1 1 x ( ) ( t) = ( ) t + ( ) λ λ λ 1 his results in four possibilities, depending on ( ) 1 ( ) λ ( ) λ ( ) the values of λ and λ. (assuming, ) 1 9 OC of a Double Integral System Step 5 : ime - Optimal Control Sequence Question : How to find h( X ), including the time of switching? 3

16 OC of a Double Integral System Step 6 : State raectories From state equation, xɺ = u and xɺ = x. Hence, ( ) 1 ( ) = + where, = = ± 1 x t x Ut U u 1 x1( t) = x1 ( ) + x ( ) t + Ut o Eliminate t, we observe that Hence, one can write: ( ( )) t = x x / U 1 1 x = x ( ) Ux ( ) + Ux 1 1 Note: U = ± 1 = 1/ U 31 OC of a Double Integral System Hence, finally ( ) ( ) t = x x if U = + 1, we have x = x ( ) x ( ) + x = C + x t = x x if U = 1, we have x1 = x1 ( ) + x ( ) x = C x where C, C are constants 1 ( x x ) Hence, it results in a family of parabolas in phase plane, 1 3

17 OC of a Double Integral System Reference: D. S. Naidu, Optimal Control Systems CRC Press, (Chap. 7) Arrow marks indicate evolution of traectories with increase of time. 33 OC of a Double Integral System ( ) ( ) At t = t, x t = x t =. f 1 f f Hence, from the phase-plane equation, 1 = x1 ( ) Ux ( ) + 1 x1 ( ) = Ux ( ) Collection of all such points define the "switching curve". 34

18 OC of a Double Integral System Step 7 : Switching Curve ( x x ) ( ) 1 ( ) wo curves γ and γ transfer any initial state to origin. Definitions: + he locus γ of all initial points,, by U = γ + = ( x1, x ) : x1 = x, x he locus γ of all initial points x. x, by U = 1 1 γ = ( x1, x ) : x1 = x, x ( ) ( ) 1 35 OC of a Double Integral System Step 7 : Switching Curve Complete switch curve γ as 1 γ = ( x1, x ) : x1 = x x = γ + γ Reference: D. S. Naidu, Optimal Control Systems CRC Press, (Chap. 7) 36

19 OC of a Double Integral System Step 8 : Phase Plane Regions R R + ( u = + 1) is the region to the left of γ, 1 R+ = ( x, x ) : x < x x 1 1 ( u = 1) is the region to the right of γ, 1 R = ( x, x ) : x > x x OC of a Double Integral System Control Strategy If the system is initially on i γ + then apply u = + 1 i γ then apply u = 1 { } { } i R+ then apply u = + 1, 1 i R then apply u = 1, + 1 Reference: D. S. Naidu, Optimal Control Systems, CRC Press, (Chap. 7) 38

20 OC of a Double Integral System Step 9 : Control Law if z, u 1 if z, u 1 ( x1 x ) ( x x ) + 1, γ + R+ Note: u = 1 1, γ R 1 o implement it, define z x + x x > = < = + 1. Next, if z =, then if x >, u = 1 and if x <, u = + 1. Note : he closed loop (feed back) control law is NONLINEAR even though the sytem is linear! 39 OC of a Double Integral System Step 1 : Minimum ime 1 x + 4x1 + x if ( x1, x ) R or x1 > x x 1 t = x + 4x + x if ( x, x ) R + or x < x x 1 x if ( x1, x ) γ or x1 = x x f

21 Close-loop Implementation of ime Optimal Control Law 41 hanks for the Attention.!! 4