Analytical Solution of Constrained LQ Optimal Control for Horizon 2 Jos´ e De Don´ a September 2004 Centre of Complex Dynamic Systems and Control
Outline A Motivating Example 1 Cautious Design Serendipitous Design Tactical Design Explicit Solutions 2 Explicit vs Implicit A Simple Problem with N = 2 3 Dynamic Programming 4 Motivating Example Revisited Conclusions 5 Centre of Complex Dynamic Systems and Control
A Motivating Example Consider the double integrator plant: d 2 y ( t ) = u ( t ) . dt 2 The zero-order hold discretisation with sampling period 1 is: x k + 1 = Ax k + Bu k , y k = Cx k , with � � � � 1 1 0 . 5 � � A = , B = , C = 1 0 . 0 1 1 Assume the actuator has a maximum and minimum allowed value (saturation) of ± 1. Thus, the controller has to satisfy the constraint: | u k | ≤ 1 for all k . Centre of Complex Dynamic Systems and Control
A Motivating Example The schematic of the feedback control loop is shown in the Figure. u k x k linear controller sat system Figure: Feedback control loop. “sat” represents the actuator modelled by the saturation function ⎧ 1 if u > 1 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ sat( u ) = u if | u | ≤ 1 , ⎪ ⎪ ⎪ ⎪ ⎪ − 1 if u < − 1 . ⎩ Centre of Complex Dynamic Systems and Control
(i) Cautious Design � � . Suppose the initial condition is x 0 = − 6 0 We start with a “low gain” linear state feedback controller (LQR), computed so as to minimise the objective function: ∞ V ∞ ( { x k } , { u k } ) = 1 � � � x k Qx k + u k Ru k , 2 k = 0 � � 1 0 with weighting matrices Q = C C = and R = 20. 0 0 The solution is then computed by solving the algebraic Riccati equation P = A PA + Q − K ( R + B PB ) K , where K = ( R + B PB ) − 1 B PA . We obtain the linear state feedback law: � � u k = − Kx k = − 0 . 1603 0 . 5662 x k . Centre of Complex Dynamic Systems and Control
(i) Cautious Design � � 1 0 “Slow gain” control: Q = C C = , R = 20, u k = − Kx k . 0 0 The resulting input and output sequences are shown in the Figure. 1 0.8 We can see from the figure that 0.6 u k 0.4 the input u k satisfies the given 0.2 0 constraint | u k | ≤ 1, for all k ; for −0.2 −0.4 k 0 5 10 15 20 25 this initial condition . 1 0 −1 The response is rather slow. The y k −2 −3 “settling time” is of the order of 8 −4 −5 samples. −6 0 5 10 k 15 20 25 Centre of Complex Dynamic Systems and Control
(ii) Serendipitous Design � � 1 0 “High gain” control: Q = C C = , R = 2, u k = sat( − Kx k ) . 0 0 The resulting input and output sequences are shown in the Figure. We can see from the figure that 3 the input u k satisfies the given 2 u k 1 constraint | u k | ≤ 1, for all k . The 0 controller makes better use of the −1 0 5 10 k 15 20 25 available control authority. 1 0 −1 The amount of overshoot is y k −2 −3 essentially the same. The −4 −5 “settling time” is of the order of 5 −6 0 5 10 k 15 20 25 samples. Centre of Complex Dynamic Systems and Control
(ii) Serendipitous Design � � 1 0 “High gain” control: Q = C C = , R = 0 . 1, u k = sat( − Kx k ) . 0 0 The resulting input and output sequences are shown in the Figure. 6 We can see from the figure that 4 2 the input u k satisfies the given u k 0 constraint | u k | ≤ 1, for all k . The −2 −4 control sequence stays saturated −6 0 5 10 k 15 20 25 much longer. 4 2 0 y k Significant overshoot. The −2 “settling time” blows out to 12 −4 −6 samples. 0 5 10 k 15 20 25 Centre of Complex Dynamic Systems and Control
Recapitulation Going from R = 20 → 2: same overshoot, faster response. Going from R = 2 → 0 . 1: large overshoot, long settling time. Let us examine the state space trajectory corresponding to the serendipitous strategy R = 0 . 1, u k = sat( − Kx k ) : The control law u = − sat( Kx ) 4 partitions the state space into 3 three regions. Hence, the 2 controller can be characterised 1 x 2 k as a switched control strategy: 0 −1 R 2 ⎧ − Kx if x ∈ R 0 , ⎪ ⎪ ⎪ −2 ⎪ R 0 ⎪ ⎨ u = K ( x ) = 1 if x ∈ R 1 , R 1 −3 ⎪ ⎪ ⎪ ⎪ ⎪ − 1 if x ∈ R 2 . ⎩ −4 x 1 −6 −4 −2 0 2 4 6 k Centre of Complex Dynamic Systems and Control
Recapitulation 4 Heuristically, we can think,in this example, of x 2 as “velocity” and 3 x 1 as “position.” Now, in our 2 attempt to change the position 1 x 2 k rapidly (from − 6 to 0), the velocity 0 has been allowed to grow to a −1 R 2 relatively high level ( + 3). This −2 R 0 would be fine if the braking action −3 R 1 were unconstrained. However, −4 x 1 −6 −4 −2 0 2 4 6 k our input (including braking) is ⎧ limited to the range [ − 1 , 1 ] . − Kx if x ∈ R 0 , ⎪ ⎪ ⎪ ⎪ ⎪ Hence, the available braking is ⎨ u = K ( x ) = 1 if x ∈ R 1 , ⎪ ⎪ inadequate to “pull the system ⎪ ⎪ ⎪ − 1 if x ∈ R 2 . ⎩ up,” and overshoot occurs. Centre of Complex Dynamic Systems and Control
(iii) Tactical Design We will now start “afresh” with a formulation that incorporates constraints from the beginning in the design process. A sensible idea would seem to be to try to “look ahead” and take account of future input constraints (that is, the limited braking authority available). We now consider a Model Predictive Control law with prediction horizon N = 2. At each sampling instant i and for the current state x i , the two-step objective function: i + 1 V 2 ( { x k } , { u k } ) = 1 i + 2 Px i + 2 + 1 � � � 2 x x k Qx k + u k Ru k , (1) 2 k = i is minimised subject to the equality and inequality constraints: x k + 1 = Ax k + Bu k , k = i , i + 1 , (2) | u k | ≤ 1 , k = i , i + 1 . Centre of Complex Dynamic Systems and Control
(iii) Tactical Design In the objective function (1), we set, as before, Q = C C , R = 0 . 1. The terminal state weighting matrix P is taken to be the solution of the algebraic Riccati equation P = A PA + Q − K ( R + B PB ) K , where K = ( R + B PB ) − 1 B PA . As a result of minimising (1) subject the constraints (2), we obtain an optimal fixed-horizon control sequence { u , u i + 1 } i We then apply the resulting value of u to the system in a i receding horizon control form. We can see that this strategy has the ability to “look ahead” by considering the constraints not only at the current time i , but also one step ahead i + 1. Centre of Complex Dynamic Systems and Control
(iii) Tactical Design � � 1 0 MPC: Q = C C = , R = 0 . 1, Horizon N = 2 0 0 The resulting input and output sequences are shown in the Figure. 6 Dashed line: control u k = − Kx k . 4 2 Solid line: MPC u k 0 −2 −4 We can see from the figure that −6 0 5 10 k 15 20 25 the output trajectory with 4 constrained input now has 2 0 y k minimal overshoot. Thus, the −2 idea of “looking ahead” has paid −4 −6 dividends. 0 5 10 k 15 20 25 Centre of Complex Dynamic Systems and Control
Recapitulation As we will see in this part of the course, the receding horizon control strategy (MPC) we have used can be described as a partition of the state space into different regions in which affine control laws hold. Serendipitous strategy Receding horizon tactical R = 0 . 1, u k = sat( − Kx k ) . design R = 0 . 1, N = 2. 4 4 R 3 3 3 R 2 2 2 R 0 1 1 x 2 k x 2 k 0 0 −1 R 2 −1 R 1 −2 −2 R 0 R 4 −3 R 1 −3 −4 −4 x 1 x 1 −6 −4 −2 0 2 4 6 −6 −4 −2 0 2 4 6 k k Centre of Complex Dynamic Systems and Control
Explicit vs Implicit Before we start studying how to find an explicit characterisation of the MPC solution, let us first define what procedures we would call explicit and which ones we would call implicit. Implicit (numerical) Explicit solution solution ( p = parameter, ( p = parameter, a , b = constants). a , b = constants). f p ( z ) = z 2 + 2 apz + bp f p ( z ) = z 2 + 2 apz + bp z ∂ f p . . . z z k z 0 z k + 1 ∂ z = 0 ⇒ z ( p ) = − ap p p p Centre of Complex Dynamic Systems and Control
Problem Setup We consider the discrete time system x k + 1 = Ax k + Bu k , where x k ∈ R n and u k ∈ R . The pair ( A , B ) is assumed to be stabilisable. We consider the following fixed horizon optimal control problem: V P N ( x ) : N ( x ) = min V N ( { x k } , { u k } ) , (3) subject to: x 0 = x , x k + 1 = Ax k + Bu k for k = 0 , 1 , . . . , N − 1 , u k ∈ U = [ − ∆ , ∆ ] for k = 0 , 1 , . . . , N − 1 , where ∆ > 0 is the input constraint level. The objective function is: N − 1 V N ( { x k } , { u k } ) = 1 N Px N + 1 � � � 2 x x k Qx k + u k Ru k . (4) 2 k = 0 Centre of Complex Dynamic Systems and Control
Problem Setup Let the control sequence that achieves the minimum in (3) be: { u 0 , u 1 , . . . , u N − 1 } The Receding horizon control law, which depends on the current state x = x 0 , is K N ( x ) = u 0 . Can we obtain an explicit expression for K N ( · ) defined above, as a function of the parameter x ? Centre of Complex Dynamic Systems and Control
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