Synthesis of the Optimal Control of the Spacecraft Orientation Using Combined Criteria of Quality

Abstract

The dynamic problem of a spacecraft’s (SC) rotation from an arbitrary initial angular position to a given final angular position is considered and solved. The case is investigated when the control is limited, and the minimized functional combines, in a given, proportion, the time of the maneuver and the integral of the energy of rotation. The construction of the optimal control is based on the quaternionic differential equation relating the vector of the angular momentum of the SC with the quaternion of the orientation of the body-related coordinate system. The control law is formulated in the form of an explicit dependence of the control variables on the phase coordinates. The analysis of the special control regime of the SC is carried out. Based on the conditions of transversality, as the necessary conditions for optimality, the optimal value of the kinetic energy of rotation when moving in a special control regime is determined. The created control algorithms allow turning of an SC with a kinetic rotation energy, which does exceed a predetermined level. For a dynamically symmetric SC, the problem of spatial reorientation is solved completely. The results of the mathematical modeling of the motion of an SC under the optimal control are presented, demonstrating the practical feasibility of the developed algorithm for controlling the spatial orientation of the SC.

APPENDIX

If an SC turns with the minimum value of integral (1.6), then the SC’s rotation can be stopped for a time not exceeding the known value (under constraint (1.3)), since minimization of functional (1.6) limits the kinetic energy of rotation.

We explain what has been said. The rate of change of the kinetic energy of rotation is ${{\dot {E}}_{{\text{k}}}}$ = L₁M₁/J₁ + L₂M₂/J₂ + L₃M₃/J₃. Under the constraints of (1.3), the optimal control in terms of the rapid quenching of the kinetic energy of rotation is [22]

$${{M}_{i}} = - \frac{{{{u}_{0}}{{L}_{i}}}}{{\sqrt {L_{1}^{2}{\text{/}}{{J}_{1}} + L_{2}^{2}{\text{/}}{{J}_{2}} + L_{3}^{2}{\text{/}}{{J}_{3}}} }},$$

((A.1))

in which ${{\dot {E}}_{{\text{k}}}} = - {{u}_{0}}\sqrt {2{{E}_{{\text{k}}}}}$ and ${{\ddot {E}}_{{\text{k}}}} = u_{0}^{2}$. To stop the rotation of the SC in the presence of constraint (1.3), the time is required to be not less than $\Delta t = {{\sqrt {2{{E}_{{\text{k}}}}(t)} } \mathord{\left/ {\vphantom {{\sqrt {2{{E}_{{\text{k}}}}(t)} } {{{u}_{0}}}}} \right. \kern-0em} {{{u}_{0}}}}$. Since during space flight one can stop the rotation of the SC at any time, it is necessary to focus on the value ${{t}_{{{\text{stop}}}}} = {{\sqrt {2{{E}_{{\max }}}} } \mathord{\left/ {\vphantom {{\sqrt {2{{E}_{{\max }}}} } {{{u}_{0}}}}} \right. \kern-0em} {{{u}_{0}}}}$. The solution of problem (1.1)–(1.6) allowed synthesizing the control, which provides the turn of the SC with the kinetic energy of rotation of not more than a₂/2a₁. If necessary (for example, in the event of a contingency urgently requiring us to stop the maneuver, or in other emergency cases) we will be able to extinguish the angular velocity within a time period of ${{\sqrt {{{a}_{2}}{\text{/}}{{a}_{1}}} } \mathord{\left/ {\vphantom {{\sqrt {{{a}_{2}}{\text{/}}{{a}_{1}}} } {{{u}_{0}}}}} \right. \kern-0em} {{{u}_{0}}}}$, since simultaneously with minimizing integral (1.6) the optimal control (2.12) limits the kinetic energy of rotation during the slew maneuver, and E_max ≤ a₂/2a₁.

In addition, optimization in accordance with criterion (1.6) minimizes the time the SC’s turns from the position Λ(0) = Λ_in to the position Λ(T) = Λ_f in the presence of restrictions E_k(t) ≤ a₂/2a₁ and (1.3). Let us prove the statement. Recall that

$$S = \int\limits_0^T {\sqrt {L_{1}^{2}{\text{/}}{{J}_{1}} + L_{2}^{2}{\text{/}}{{J}_{2}} + L_{3}^{2}{\text{/}}{{J}_{3}}} dt} $$

are functional “ways.” Consider the function ${{f}_{0}} = \sqrt {2{{E}_{{\text{k}}}}(t)}$. Accordingly, we have ${{f}_{{\max }}} = \sqrt {2{{E}_{{{\text{max}}}}}}$. Under constraint (1.3), the fastest possible spin-up of the SC occurs during control [22]

$${{M}_{i}} = \frac{{{{u}_{0}}{{L}_{i}}}}{{\sqrt {L_{1}^{2}{\text{/}}{{J}_{1}} + L_{2}^{2}{\text{/}}{{J}_{2}} + L_{3}^{2}{\text{/}}{{J}_{3}}} }},$$

((A.2))

for which ${{\dot {E}}_{{\text{k}}}} = {{u}_{0}}\sqrt {2{{E}_{{\text{k}}}}}$ and ${{\ddot {E}}_{{\text{k}}}} = u_{0}^{2}$, and the force moment M and the angular momentum L are equally directed (where |M| = const and ${\mathbf{\dot {M}}} = - {\boldsymbol{\omega }} \times {\mathbf{M}}$), which corresponds to control (2.12). For the fastest stop of rotation under constraint (1.3), it is necessary to control (A.1), at which the control torque M and the angular momentum L have opposite directions, |M| = const, and ${\mathbf{\dot {M}}} = - {\boldsymbol{\omega }} \times {\mathbf{M}}$, which corresponds to control (2.12). If u₀ → ∞, then the acceleration and braking time τ → 0 and during the whole turn E_k(t) = const = a₂/2a₁.

For any control satisfying constraint (1.3), two conditions are satisfied: f₀(t) ≤ u₀t if t ≤ T/2 and f₀(t) ≤ u₀(T – t) if t > T/2 (since L(0) = L(T) = 0). With the same value of the path functional S, the minimum value of T is reached if the time interval when E_k(t) = const = E_max is the longest and the time intervals when E_k(t) < E_max are as short as possible. Control (2.12) satisfies both requirements.

If u₀S ≤ a₂/a₁, then in laws (2.12) and (2.13) t₁ = t₂ = T/2 $= \sqrt {S{\text{/}}{{u}_{0}}}$ (relay control with one switching point is optimal); in this case, E_max = u₀S/2 and the restriction E_k(t) ≤ a₂/2a₁ is insignificant. The optimal turn time is T_opt$= 2\sqrt {S{\text{/}}{{u}_{0}}}$.

If u₀S > a₂/a₁, then t₂ > t₁ (t₁ = τ, t₂ = T – τ), and the turn with the special control regime is optimal (relay control with two switching points is implemented); τ is the acceleration (braking) time. In this case, with the optimal control (2.12), the pivot time is T_opt$= S\sqrt {{{a}_{1}}{\text{/}}{{a}_{2}}} + {{\sqrt {{{a}_{2}}{\text{/}}{{a}_{1}}} } \mathord{\left/ {\vphantom {{\sqrt {{{a}_{2}}{\text{/}}{{a}_{1}}} } {{{u}_{0}}}}} \right. \kern-0em} {{{u}_{0}}}}$. Of all the possible motions satisfying the conditions of turn (1.4) and (1.5) and constraint (1.3), we are interested in those that have the general property described by the inequality E_k(t) ≤ a₂/2a₁ with the optimal control (2.12). Whatever the control, the turning time T satisfies the inequality

$$T \geqslant {S \mathord{\left/ {\vphantom {S {\sqrt {2{{E}_{{\max }}}} }}} \right. \kern-0em} {\sqrt {2{{E}_{{\max }}}} }} + {{\sqrt {2{{E}_{{\max }}}} } \mathord{\left/ {\vphantom {{\sqrt {2{{E}_{{\max }}}} } {{{u}_{0}}}}} \right. \kern-0em} {{{u}_{0}}}}.$$

((A.3))

The right-hand side of this inequality is a monotonically decreasing function of the argument Е_max over the entire segment [0, u₀S/2] (E_max cannot be greater than u₀S/2 since the value E_max = u₀S/2 corresponds to the relay control with one switching point when phase of the SC’s rotation by inertia is absent). Other things being equal, the pivot time is minimal if the path functional S is minimal. Since for all controls of interest to us, E_max ≤ a₂/2a₁, the right-hand side of inequality (A.3) cannot be less than T_opt, since for control (2.12), the path functional S takes the minimum value. As a result, the pivot time is always T ≥ T_opt. Note that for a fixed turn time T the minimum value of f_max is ${{f}_{{\max }}} = ({{u}_{0}}T - \sqrt {u_{0}^{2}{{T}^{2}} - 4{{u}_{0}}S} ){\text{/}}2$. Accordingly, the minimum possible value of the maximum energy during turn E_max is

$${{E}_{{\max }}} = {{({{u}_{0}}T - \sqrt {u_{0}^{2}{{T}^{2}} - 4{{u}_{0}}S} )}^{2}}{\text{/}}8.$$

The value E_max is a monotonically decreasing function of the duration of the turn T. If T = T_opt, then E_max = a₂/2a₁. Therefore, if T < T_opt, then E_max > a₂/2a₁. This means that the kinetic energy of rotation E_k(t) does not exceed the level a₂/2a₁ and the time of the turn must be T ≥ T_opt. Therefore, T_opt is indeed the minimum possible time of a reorientation under the condition E_k(t) ≤ a₂/2a₁ and in the presence of constraint (1.3).

If u₀ → ∞ then the value of integral (1.6) is G = 2a₂T. Consequently, the time of the optimal turn is T = G/2a₂ and, minimizing integral (1.6), we obtain the turn of the SC for the minimum time.

Thus, it has been proved that minimization of functional (1.6) leads to the SC turning for the minimal time with the limited kinetic rotation energy, which is important in the practice of a space flight.