Finite-time attitude control for rigid spacecraft subject to actuator saturation

Abstract

This paper considers the problem of attitude stabilizing control for rigid spacecraft in the presence of input constraints. To address this problem, a smooth model is first presented such that its output is always in agreement with the constraints imposed by the actuators. Then, based on the “adding a power integrator” method and the backstepping technique, a novel finite-time attitude control scheme is proposed. The finite-time stability of the resultant closed-loop system is guaranteed by using the Lyapunov approach. Finally, the effectiveness of the control scheme derived here is illustrated by numerical simulations.

Appendix

1.1 Proof of Proposition 1

Define the dilation $\delta _\lambda ^r(\xi _1)=\lambda \xi _1$. By noting that $2-\alpha _1+\alpha _2=1+\alpha _1$, it follows that $U(\xi _1)$ is homogeneous of degree $1+\alpha _1$ with respect to the dilation $\delta _\lambda ^r(\xi _1)$. Consider the function $h(\xi _1, \alpha )=-\left[ \text{ sig }^{2-\alpha _1}(\xi _1)\right] ^TJ^{-1}\text{ sig }^{\alpha _2}(\xi _1)$, where $(\xi _1, \alpha ) \in S^3\times (0, \infty )$. Note that the function h is continuous and $\xi _1$ belongs to a compact set. Thus, the image of h is included in $(-\infty , -\lambda _{\min }(J^{-1}))$ for all $\xi _1\in S^3$ and $\alpha =1$, since $h(\xi _1, \alpha )=-\xi _1^TJ^{-1}\xi _1$ when $\alpha =1$. By using the tube lemma [42], it is concluded that there exists an $\epsilon >0$ such that for all $(\xi _1, \alpha )\in S^3\times (1-\epsilon , 1+\epsilon )$, $h(\xi _1, \alpha )\le -\lambda _{\min }(J^{-1})/2$. Then, we can obtain $U(\xi _1)\le -\lambda _{\min }(J^{-1})/2=-\lambda _{\min }(J^{-1})\left( \xi _1^T\xi _1\right) ^{\beta }/2$ for all $(\xi _1, \alpha )\in S^3\times (1-\epsilon , 1+\epsilon )$.

If $\xi _1=0$, it is clear that $U(\xi _1)\le -\lambda _{\min }(J^{-1})\left( \xi _1^T\xi _1\right) ^{\beta }/2$. For any $\xi _1\in R^3{\setminus }0$, by Lemma 6, we have that $\xi _1=\delta _\lambda ^r(\bar{\xi }_1)$, where $\bar{\xi }_1\in S^3$. Then, we can obtain that

$$\begin{aligned} U(\xi _1)&=U(\delta _\lambda ^r(\bar{\xi }_1))=\lambda ^{1+\alpha _1}U(\bar{\xi }_1)\nonumber \\&\le -\frac{1}{2}\lambda ^{1+\alpha _1}\lambda _{\min }(J^{-1})\left( \bar{\xi }_1^T\bar{\xi }_1\right) ^{\beta }\nonumber \\&=-\frac{\lambda _{\min }(J^{-1})}{2}\left( \xi _1^T\xi _1\right) ^{\beta }. \end{aligned}$$

(54)

This completes the proof.

1.2 Proof of Proposition 2

Note that

$$\begin{aligned} \frac{\partial \text{ sig }^{1/\alpha _2}(u_{di})}{\partial t}&=-\,k_2^{1/\alpha _2} \frac{\partial (\xi _{1i})}{\partial t}\nonumber \\&=-\,k_2^{1/\alpha _2}\left| \omega _i\right| ^{1/\alpha _1-1}\dot{\omega }_i +k_2^{1/\alpha _2}k_1^{1/\alpha _1}\dot{q}_i\nonumber \\&=-\,k_2^{1/\alpha _2}\left| \omega _i\right| ^{1/\alpha _1-1} \left( f_i(\omega )+(J^{-1}u)_i\right) \nonumber \\&\quad +k_2^{1/\alpha _2}k_1^{1/\alpha _1}\dot{q}_i, \end{aligned}$$

(55)

where $(\cdot )_i$ denotes the i-th element of a vector. If $\Vert q\Vert \le \Delta$ and $\Vert \omega \Vert \le \Delta$, it is easy to verify that

$$\begin{aligned}&\left| \omega _i\right| ^{1/\alpha _1-1}\left( f_i(\omega )+(J^{-1}u)_i\right) \le \bar{m}_{1}\sum _{j=1}^3|q_j|^{\alpha _1}\nonumber \\&\quad +\bar{m}_{2}\sum _{j=1}^3|\xi _{1j}|^{\alpha _1}+\bar{m}_3\sum _{j=1}^3|\xi _{2j}|^{\alpha _1} \end{aligned}$$

(56)

$$\begin{aligned} \dot{q}_i\le \bar{m}_{4}\sum _{j=1}^3|q_j|^{\alpha _1} +\bar{m}_{5}\sum _{j=1}^3|\xi _{1j}|^{\alpha _1}, \end{aligned}$$

(57)

where $\bar{m}_j (j=1, 2, \ldots , 5)$ are some positive constants, which implies that there exist some positive constants $m_1, m_2$ and $m_3$ such that

$$\begin{aligned}&\left| \frac{\partial \text{ sig }^{1/\alpha _2}(u_{di})}{\partial t}\right| \nonumber \\&\quad \le m_1\sum _{j=1}^3|q_j|^{\alpha _1} +m_2\sum _{j=1}^3|\xi _{1j}|^{\alpha _1}+m_3\sum _{j=1}^3|\xi _{2j}|^{\alpha _1}. \end{aligned}$$

(58)

This completes the proof.