ECE 680 Fall Test #2 Solutions. 1. Use Dynamic Programming to find u(0) and u(1) that minimize. J = (x(2) 1) u 2 (k) x(k + 1) = bu(k),

Size: px

Start display at page:

Download "ECE 680 Fall Test #2 Solutions. 1. Use Dynamic Programming to find u(0) and u(1) that minimize. J = (x(2) 1) u 2 (k) x(k + 1) = bu(k),"

Gwenda Bell
5 years ago
Views:

1 ECE 68 Fall 211 Test #2 Solutions 1. Use Dynamic Programming to find u() and u(1) that minimize subject to 1 J (x(2) 1) u 2 (k) k x(k + 1) bu(k), where b. Let J (x(k)) be the minimum cost of transfer a state x(k) to some final state x(2). Then, and J (x(1)) min u(1) J (x(2)) (x(2) 1) 2 (bu(1) 1) 2, ( 2u 2 (1) + J (x(2)) ) ( min 2u 2 (1) + (bu(1) 1) 2). u(1) There are no constraints on u(1). Therefore, we compute u (1) by solving J(x(1)) u(1). We have Hence, J(x(1)) u(1) 4u(1) + 2b(bu(1) 1). u (1) Substituting the above into J (x(1)) yields Continuing, we obtain J (x()) min u() b 2 + b 2 J (x(1)) b 2. ( 2u 2 () + J (x(1)) ) ( min 2u 2 () + 2 ). u() 2 + b 2 1

2 Hence, u () 2. Use the HJB equation to find u that minimizes J ( 3x 2 + u 2) dt subject to Assume ẋ x + u. J 1 2 p(t)x2, and p(t) ẇ(t) w(t). Then, solve for w(t), find a general solution for p(t). Eliminate the integration constants from p(t), and write down the expression for u as a time-varying state-feedback controller. The Hamiltonian function is H 1 ( 3x 2 + u 2) + J ( x + u). 2 x There are no constraints on u, so we compute u by solving H u. We have Hence H u u + J x. u J x. 2

3 Since 2 H 1 >, u is a strict minimizer of H. We substitute u into the HJB u 2 equation to get J t x2 + 1 ( ) J 2 J ( ) J 2 2 x x x. x Simplifying yields We assume J t x2 1 2 ( ) J 2 J x J 1 2 p(t)x2. x x. Hence, J t 1 2ṗx2 and J x px. Substituting the above into the HJB equation, we obtain 1 2ṗx x2 1 2 p2 x 2 px 2 Thus, we have to solve the equation We assume that Hence, ( 1 2ṗ p2 p) x 2. ṗ + 3 p 2 2p. (1) ṗ Substituting (2) and (3) into (1) yields ẅw + ẇ 2 + 3w 2 ẇ 2 + 2ẇw w 2 Thus, we have to solve the differential equation The characteristic equation of (4) is p ẇ w. (2) ẅw + ẇ2 w 2. (3) ( ẅ + 3w + 2ẇ)w w 2. ẅ 2ẇ 3w. (4) λ 2 2λ 3 (λ + 1)(λ 3). 3

4 Hence, and w C 1 e t + C 2 e 3t, ẇ C 1 e t + 3C 2 e 3t. Substituting the above two expressions into (2) yields The boundary condition is p(1). Hence, p(t) C 1e t + 3C 2 e 3t C 1 e t + C 2 e 3t. (5) C 1 e 1 + 3C 2 e 3, that is, C 1 3C 2 e 4. (6) We substitute (6) into (5) to get p(t) 3C 2e 4 e t 3C 2 e 3t 3C 2 e 4 e t + C 2 e 3t 3 (e4 t e 3t ) 3e 4 t + e 3t. Hence, the optimal state-feedback control law has the form u J x px 3 (e4 t e 3t ) x. 3e 4 t + e 3t 3. Determine the optimal state-feedback controller, u k(t)x, that minimizes J 1 2 x2 (1) + 1 u 2 dt subject to ẋ u. 4

5 We form the Hamiltonian matrix H A BR 1 B Q A 1/2. We next find the inverse of the characteristic matrix of the Hamiltonian matrix, e Ht L ( 1 [si H] 1) L 1 s 1/2 1 s L t s 2s 2 1 s. We next compute e H(1 t) 1 1 (1 t) 2 1 Φ 11 Φ 12 Φ 21 Φ 22. We have, p(1) Fx(1) x(1). Hence, p P(t)x Therefore, (Φ 22 FΦ 12 ) 1 (FΦ 11 Φ 21 )x ( ) 1 2 (1 t) x 2 3 t x u R 1 B p t x 1 3 t x 5

6 4. For the continuous model, construct ẋ 1 x + 1 u, (i) (5 pts) the Euler discrete model for the sampling period h 1/2; (ii) (1 pts) the exact discrete model for the sampling period h 1/2. (i) The Euler discrete model is obtained by approximating ẋ as ẋ x((k + 1)h) x(kh) h Taking the above into account, we obtain x[k + 1] x[k]. h (ii) The exact discrete model has the form x[k + 1] (I 2 + ha)x[k] + hbu[k] 1 h x[h] + u[h] 1 h 1.5 x[h] + u[h]. 1.5 x[k + 1] Φx[k] + Γu[k] ( ) h e Ah x[k] + e Aη dη bu[k]. Note that In our case Hence, e Ah I 2 + Ah A2 h 2 + Φ e Ah I 2 + Ah 6 A 2 O. 1 h

7 Next, Γ ( ) h e Aη dη b h 1 η dη 1 1 h h2 /2 h2 /2 h 1 h For the discrete-time model, x[k + 1] 1 x[k] + 1 u[k] y[k] [ 1 ] x[k]. construct the augmented discrete-time system that you would use in the design of a model predictive controller. The augmented discrete-time system has the form x a [k + 1] Φ a x a [k] + Γ a u[k] y[k] C a x a [k], where the augmented state vector is defined as x a [k] x[k] y[k], and Φ a Φ O CΦ I p 1 1 1, 7

8 and Γ a Γ 1 CΓ, C a [ ] [ O I p 1 ]. 6. Consider the following model of a discrete-time system, x(k + 1) 2x(k) + u(k), x(), k 2 Use the Lagrange multiplier approach to calculate the optimal control sequence {u(),u(1),u(2)} that transfers the initial state x() to x(3) 7 while minimizing the performance index J 1 2 u(k) 2 2 k We begin by defining the composite input vector u [ u() u(1) u(2) ]. Then the performance index J can be represented as J 1 2 u u. Next, write the plant model in the form mu f, where m R 1 3 and f is a scalar. We proceed as follows. First we write x(2) 2x(1) + u(1) 4x() + 2u() + u(1). 8

9 Using the above, we obtain x(3) 7 2x(2) + u(2) 8x() + 4u() + 2u(1) + u(2). We represent the above in the format mu f as follows [ ] u() u(1) 7 u(2) Thus we formulated the problem of finding the optimal control sequence as a constrained optimization problem min 1 2 u u subject to mu f To solve the above problem, we form the Lagrangian l(u,λ) 1 2 u u + λ (mu f), where λ is the Lagrange multiplier. Applying the Lagrange s first-order condition we get u + m λ and mu f. From the first of the above conditions, we calculate, u m λ. Substituting the above into the second of the Lagrange conditions gives λ ( mm ) 1 f. Combining the last two equations, we obtain a closed-form formula for the optimal input sequence u m ( mm ) 1 f In our problem, u u() u(1) u(2) f (mm ) m

10 7. (15 pts) Find all the points that satisfy the KKT conditions for the following optimization problem: minimize x x 2 2 subject to x x We form the Lagrangian function, l(x,µ) x x µ(4 x 2 1 2x 2 2). The KKT conditions take the form, Dxl(x,µ) [ 2x 1 2µx 1 µ(4 x 2 1 2x 2 2) 8x 2 4µx 2 ] µ 4 x 2 1 2x 2 2. From the first of the above equality, we obtain (1 µ)x 1 (2 µ)x 2. We first consider the case when µ. Then, we obtain the point which does not satisfy the constraints. x (1), The next case is when µ 1. Then we have to have x 2 and using µ(4 x 2 1 2x 2 2) gives x (2) 2 and x (3) 2. For the case when µ 2, we have to have x 1 and we get x (4) and x (5)

An Introduction to Model-based Predictive Control (MPC) by

ECE 680 Fall 2017 An Introduction to Model-based Predictive Control (MPC) by Stanislaw H Żak 1 Introduction The model-based predictive control (MPC) methodology is also referred to as the moving horizon