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Chapter 2 A Simple Optimal Control Problem In this Chapter, we consider a simple optimal control problem for linear discrete-time systems. 2.1 System Description. We begin with a continuous-time linear time-invariant system described by dx = Ax + Bu, x(0) = xo , (2.1) dt where A and B are n×n and n×m constant matrices respectively. The admissible controls u(·) are assumed to be piecewise continuous functions of t taking their values in Rm . For any admissible control, the solution to (2.1) is given by Z t xu (t; xo , 0) = exp(At)xo + exp(A(t − τ ))Bu(τ )dτ. (2.2) 0

Consider the corresponding discrete-time system with uniform sampling period ∆: Z (k+1)∆ xu ((k + 1)∆; x(k ∆), k∆) = exp(A∆)x(k∆) + exp(A((k + 1)∆ − τ ))Bu(τ )dτ. k∆

(2.3) We assume that the control u is held constant over the time interval [k ∆, (k + 1)∆], i.e. u(τ ) ≡ u(k) for all τ ∈ [k ∆, (k + 1)∆]. Let x(k∆) be denoted by x(k ). Then (2.3) can be rewritten in the form: x(k + 1) = Hx(k) + Gu(k ), where H = exp(A∆),

G=

Z

x(0) = xo ,

(2.4)

∆

exp(Aτ )Bdτ.

(2.5)

0

For the foregoing discrete-time system, the set of all admissible controls Uad is taken to be the set of all control sequences {u(0), u(1), . . . , u(K − 1)} such that u(k) ∈ Rm for any k = 0, 1, . . .. 2.2 Problem Statement and Basic Considerations. The optimal control problem can be stated as follows: Given xo ∈ Rn at k = 0, and a target set G = {xf } with xf being a given point in Rn , find the smallest integer K ≥ 0 and an admissible control sequence {u(0), u(1), . . . , u(K− 1)} which steers xo to G at time step K . The foregoing problem is often called the minimum-time regulator problem. Before we proceed to solve this problem, the following basic question should be considered: Given xo , xf ∈ Rn , does there exists an admissible control sequence {u(0), . . . , u(K − 1)} which steers xo to xf for some finite K ? This question pertains to controllability. 8

Without loss of generality, we let xf = 0. We shall answer the foregoing question by constructing the sets of initial states which can be steered to xf = 0 in a finite number of time steps. For one-step steering, we require: 0 = x(1) = Hxo + Gu(0). Since H is nonsingular, hence xo = −H−1 Gu(0). Let us define the set of all initial states which can be steered to 0 in at most one time step by: ∆

Γ1 ={x ∈ Rn : x = −H−1 Gu(0); u(0) ∈ Rm } = span{[H−1 G]i , i = 1, . . . , m}, where [F]i denotes the i-th column of the matrix F. Similarly, for two-step steering, we require: 0 = x(2) = H2 xo + HGu(0) + Gu(1), or xo = −H−1 Gu(0) − H−2 Gu(1). We define ∆

Γ2 ={x ∈ Rn : x = −H−1 Gu(0) − H−2 Gu(1); u(0), u(1) ∈ Rm } = span{[H−1 G]i1 , [H−2 G]i2 , i1 , i2 = 1, . . . , m}. Continuing the foregoing process, we can define the set of all initial states which can be steered to 0 in at most K time steps: K−1 n

ΓK = {x ∈ R : x = −

X

H−K+i Gu(K − i − 1); u(k) ∈ Rm , k = 0, 1, . . . , K − 1}

i=0

= span{[H−1 G]i1 , . . . , [H−K G]iK ; i1 , . . . , iK = 1, . . . , m}.

(2.6)

Evidently, the Γk ’s have the following telescopic inclusion property: Γ1 ⊂ Γ2 ⊂ · · · ⊂ ΓK .

(2.7)

For complete controllability, we require Γn = Rn , which implies that the columns of the matrix [H−1 G, . . . , H−n G] or the controllability matrix Q = [Hn−1 G, . . . , HG, G]

(2.8)

must span Rn , or equivalently, Rank Q = n. 2.3 Solution for Scalar-control Case. Consider the case with scalar control u. Here, (2.4) has the form: x(k + 1) = Hx(k) + gu(k), 9

x(0) = xo ,

(2.9)

where u(k) ∈ R and g ∈ Rn defined by g=

Z

∆

exp(Aτ )bdτ,

0

where B = b is a constant vector in Rn . Proposition 2.1. Let vi = H−i g. If Rank[v1 , . . . , vn ] = n,

(2.10)

then given any initial state xo ∈ Rn , the following admissible control sequence will steer xo ∈ Rn to the zero state 0 in at most n time steps: u(0) = −η1 (0), u(1) = −η2 (0) = −η1 (1), .. . u(n − 1) = −ηn (0) = −η1 (n − 1), where x(k) =

n X

ηi (k)vi .

(2.11) (2.12)

i=1

We verify the foregoing result for the case where n = 2. Since Rank[v1 , v2 ] = 2 by assumption, {v1 , v2 } forms a basis for R2 . For any xo ∈ R2 , we can write xo = η1 (0)v1 + η2 (0)v2 . Thus, x(1) = H(η1 (0)v1 + η2 (0)v2 ) + gu(0) = (Hv1 )η1 (0) + (Hv2 )η2 (0) + gu(0). Since Hv1 = g,

Hv2 = H−1 g = v1 ,

we have x(1) = η1 (0)g + η2 (0)v1 + gu(0). If we set u(0) = −η1 (0), then x(1) = η2 (0)v1 = η1 (1)v1 . Now, x(2) = Hx(1) + gu(1) = Hv1 η1 (1) + gu(1) = (η1 (1) + u(1))g.

10

(2.13)

Evidently, if we set u(1) = −η1 (1),

(2.14)

then x(2) = 0, or xo is steered to the zero state in two time steps by controls (2.13) and (2.14). The foregoing result can be interpreted geometrically (see Fig.2.1). 1. xo is resolved into its components with respect to basis vectors v1 and v2 . 2. The effect of the control u(0) is to (i) knockout the component of xo with respect to v1 that “rotated” into the subspace S = span{g}; (ii) rotate the component of xo with respect to v2 into the subspace Γ1 = span{v1 }. 3. The effect of the control u(1) is to knockout the component of x(1) with respect to v1 “rotated” into S so that the state is at the origin of R2 . span{v } 2

xo 2

u(1)g = -

(0)v

2

(1)g

1

v u(0)g = -

(0)g

(0)v

1

2

span{v }

1

1

1

2

(0)v

1

v1

(0)g

1

(1)g

2

S = span{g}

Figure 1: Geometric interpretation of control actions.

In the general case, xo ∈ Rn has components η1 (0), . . . , ηn (0) with respect to basis {v1 . . . , vn }. With the control sequence given by (2.11), after the first time step, the state can be expressed as a linear combination of v1 , . . . , vn−1 , i.e. the state lies in Γn−1 . After the second time step, the state lies in Γn−2 and so on. After n time steps, the state lies at the origin of Rn . 2.4 Optimal Feedback Controls (scalar control case). The optimal control sequences given in (2.11) are open-loop controls. However, we note from (2.11) that u(k) = −η1 (k),

11

(2.15)

i.e. u(k) is −1 times the first component of x(k) with respect to the basis vector v1 . Now,     x1 (k) η1 (k) n X    ..  = (2.16) x(k) =  ηi (k)vi = P  ...  , .  i=1 xn (k) ηn (k) where P = [v1 | . . . |vn ]. Thus, we obtain the required optimal feedback control (or control law) given by n X (2.17) u(k ) = −η1 (k ) = − (P−1 )1i xi (k ), i=1

where (P )1i denotes the i-th element in the first row of P−1 . The difference equation of the corresponding feedback control system is −1

x(k + 1) = (H − g(P−1 )1 )x(k ),

(2.18)

where (F)1 denotes the first row of the matrix F. This control law is a linear function of the state x(k ). A block diagram of the optimal feedback control system is shown in Fig.2.2.

Figure 2: Block diagram of optimal feedback control system.

Exercise 2.1 Show that the matrix (H − g(P−1 )1 ) is nilpotent, i.e. (H − g(P−1 )1 )k = zero matrix for some finite positive integer k . 2.5 Minimum-time Tracking Problem. The minimum-time regulator problem discussed earlier can be extended to the case where the target set G consists of a single point which varies with time. To fix ideas, we consider again the case with scalar control. We begin with the continuous-time system: dx = Ax + bu(t), dt

x(0) = xo .

(2.19)

Let the target set be specified by G(t) = {R(t)}, where R(t) = (r(t), 0, . . . , 0)T , and r = r(t) is a given function of t specifying the target position at any time t. The 12

continuous-time version of the minimum-time tracking problem can be stated as: given any initial state xo ∈ Rn , find an admissible control u∗ = u∗ (t) which steers xo to the target point (r (t∗ ), 0, . . . , 0)T in minimum time t∗ . Let ˆx(t) = R(t) − x(t). Then, the equation for x, ˆ in view of (2.19), has the form: dˆ x dR = Aˆ x − AR(t) − bu + . dt dt

(2.20)

First, consider the special case where r(t) = ro U (t) with U = U (t) being the unit-step function starting at t = 0, and ro an arbitrary real number. We need to consider two distinct cases. Case 1: The system has a pure integrator. Let x1 be the output of the pure integrator (i.e. no feedback from x1 to other parts of the system), then the matrix A has the form:   0 1 0 ... 0  0    A =  .. .  .  nonzero matrix 0 In this case, AR(t) = 0, and dR/dt = 0 for all t > 0. Thus, (2.20) reduces to dˆx ˆ = Aˆ x + bu, dt

(2.21)

where bˆ = −b. Evidently, we may discretize (2.21) in time and arrive at an equation having the same form as that of (2.9). Thus, the optimal feedback control law for the corresponding discrete-time system is given by u(k) = −

n n X X (P−1 )1i xˆi (k) = −(P−1 )11 (r(k) − x1 (k)) + (P−1 )1i xi (k ),

where P = [ˆ v1 , . . . , v ˆ n ],

(2.22)

i=2

i=1

ˆi = H g v ˆ, −i

g ˆ=

Z

∆

ˆ exp(Aτ )bdτ.

(2.23)

0

Case 2: The system does not have a pure integrator. In this case, a non-zero constant u is needed to hold x1 at ro . The following required steady-state control uss can be determined from the transfer function between x1 and u, and applying the Final-value Theorem: uss = ro /G(0),

G(0) = [1, 0, . . . , 0](sI − A)−1 b|s=0 .

(2.24)

The equilibrium state z for the corresponding discrete-time system with u(k) = uss can be found by z = Hz + guss , 13

or z = (I − H)−1 guss .

(2.25)

Note that since the system has no pure integrator, A is nonsingular. Consequently, H = exp(A∆) has no eigenvalue equal to one. Thus, (I − H) is invertible. Now, we define ˜x(k) = x(k) − z,

u˜(k ) = u(k ) − uss .

(2.26)

The difference equation for x(k) ˜ is given by ˜x(k + 1) = H˜ x(k) + g˜ u(k).

(2.27)

The above equation has the same form as that of (2.9). Thus, the optimal control law is given by n X (P−1 )1i x˜i (k), u˜(k) = − i=1

or u(k) = −

n X

(P−1 )1i (xi (k) − zi ) + uss ,

(2.28)

i=1

where zi is the i-th component of z. Remark: We observe that u(k) given by (2.28) depends on uss which in turn depends on H. Thus, the accuracy of z depends on H, an undesirable feature from the practical standpoint. This type of system is sometimes called a “calibrated system”. Finally, similar result can be obtained for the case where r(t) can be written in the form of a K-th degree polynomial in t: K X α i ti , r(t) = i!

(2.29)

i=0

where K < n. Assume that the system has K pure integrators whose outputs correspond to x1 , . . . , xK . We define   x1 − r   x2 − dr/dt   ..     .   x ˆ =  xK − dK−1 r/dtK−1  . (2.30)   xK+1     ..   . xn

Then the differential equation for xˆ has the form (2.21). Thus, the minimum-time tracking control law for the corresponding discrete-time system has the form: ¶ µ n K X X di−1 r (k) + βi xi (k), (2.31) u(k) = βi xi (k) − i−1 dt i=K+1 i=1 where the coefficients βi depend on A and b. 14...