On the attitude stabilization of a rigid spacecraft using two skew control moment gyros

Abstract

The attitude control of a rigid spacecraft with two skew single-gimbal control moment gyros (CMGs), which is subject to an underactuated nonholonomic constraint, is investigated. Nonlinear control theory is used to show that the combined dynamics of the spacecraft-CMG system are small-time locally controllable (STLC) from and feedback stabilizable to any equilibrium where two CMGs never encounter certain special configurations. Specially, the attitude stabilization issue is approached under the restriction that the total angular momentum of the spacecraft-CMG system is zero, which not only guarantees that the feasible equilibrium attitude can be any orientation but also renders STLC for these attitudes. In order to overcome the troublesome singular problem of two skew CMGs, a nonlinear approximation of the full attitude equations is derived for control law design by assuming that the spacecraft angular velocity is small. A novel singular quaternion stabilization law is then proposed to stabilize the spacecraft attitude with bounded angular velocities, which in turn ensures the satisfaction of the small angular velocity assumption during the entire control process. Numerical examples and experimental results validate the effectiveness of the proposed control method in stabilizing the full spacecraft-CMG system.

Appendix: Proof of Claim 1

Proof

We first show that $\bar{{\varvec{q}}}(t)$ will enter the region $\left| \Delta \right| \le m_\Delta$ in finite time. Assuming $\left| \Delta \right| >m_\Delta$ for all $t\ge 0$, Eqs. (55–57) indicate that $\lim _{t\rightarrow \infty } y(t)=\infty$ and hence $\lim _{t\rightarrow \infty } \rho (t)=\infty$ since $m>2/3$ and $mg-k>0$. This leads to a contradiction, since $\rho (t)\le 1$ for all $t\ge 0$. Therefore, $\bar{{\varvec{q}}}(t)$ leaves the region $\left| \Delta \right| >m_\Delta$ and enters the region $\left| \Delta \right| \le m_\Delta$ in finite time.

According to the preceding analysis, let us assume $\left| \Delta \right| \le m_\Delta$. The time derivative of $\Delta ^{2}$ along the closed-loop trajectories is

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}\Delta ^{2}=-\frac{1}{\rho }\left( {\rho \left| \Delta \right| G\left( {\left| \Delta \right| } \right) +F_1 +F_2 } \right) \end{aligned}$$

(64)

where the function $G(\cdot )$ is given in (40) and

$$\begin{aligned} F_1&= (q_2 q_3 +(q_0 q_1 +q_2 q_3 )\Delta ^{2})\omega _{e1} (t)\end{aligned}$$

(65)

$$\begin{aligned} F_2&= (-q_1 q_3 +(q_0 q_2 -q_1 q_3 )\Delta ^{2})\omega _{e2} (t) \end{aligned}$$

(66)

Noting that $m_\Delta >2/3$ according to Proposition 2, it follows from the analysis in the proof of Proposition 1 that $G(m_\Delta )>0$. Hence, one can further deduce that

$$\begin{aligned} \mathop {\lim }\limits _{\left| \Delta \right| \rightarrow m_\Delta } \rho \left| \Delta \right| G\left( {\left| \Delta \right| } \right) =\rho m_\Delta G\left( {m_\Delta } \right) >0 \end{aligned}$$

(67)

In addition, the time derivative of $q_3^2$ satisfies

$$\begin{aligned} \frac{\hbox {d}}{\hbox {d}t}q_3^2&= -g\rho \Delta \tanh (\Delta )-q_2 q_3 \omega _{e1} (t)+q_1 q_3 \omega _{e2} (t) \nonumber \\&\ge -gq_3^2 -q_2 q_3 \omega _{e1} (t)+q_1 q_3 \omega _{e2} (t) \end{aligned}$$

(68)

It follows from (53), (54a), and (68) that $\omega _{e1}$ and $\omega _{e2}$ are decaying faster than $q_3^2$. Noting that in the region $\left| \Delta \right| \le m_\Delta$ we have $\rho >m_\Delta ^{-2} q_3^2$, then $\omega _{e1}$ and $\omega _{e2}$ are also decaying faster than $\rho$ and, therefore, it follows from (64–67) that there exist some $t_2 >0$ such that $\lim _{\left| \Delta \right| \rightarrow m_\Delta } {\hbox {d}\Delta ^{2}}/{\hbox {d}t}<0$ for all $t\ge t_2$. This means that the vector field of the closed-loop system on the boundary $\left| \Delta \right| =m_\Delta$ points into the interior of the region $\left| \Delta \right| \le m_\Delta$. Therefore, $\bar{{\varvec{q}}}(t)$ will stay in the region $\left| \Delta \right| \le m_\Delta$ for all $t\ge t_2$ and thus $\lim _{t\rightarrow \infty } {\omega _{ei} (t)}/\rho (t)=0, i=1,2$.

On the other hand, it follows from (43) and (64) and $\lim _{t\rightarrow \infty } {\omega _{ei} (t)}/\rho (t)=0, i=1,2$, that

$$\begin{aligned} \mathop {\lim }\limits _{t\rightarrow \infty } \left( {\frac{\hbox {d}}{\hbox {d}t}\Delta ^{2}} \right) =-\left| \Delta \right| G\left( {\left| \Delta \right| } \right) \le -(g-kq_0 )\Delta ^{2}\nonumber \\ \end{aligned}$$

(69)

which implies that $\lim _{t\rightarrow \infty } \Delta (t)=0$. Letting $m_\Delta =1$ and using (50), we can further deduce that

$$\begin{aligned} \mathop {\lim }\limits _{t\rightarrow \infty } \left( {\frac{\hbox {d}}{\hbox {d}t}\frac{\Delta ^{2}}{\sqrt{\rho }}} \right)&= -\frac{1}{2\sqrt{\rho }}\left( 2\left| \Delta \right| G\right. \nonumber \\&\quad \left. +\Delta ^{2}(-kq_0 +g\Delta \tanh (\Delta )) \right) \nonumber \\&\le -\frac{(2g-3kq_0 )}{2}\frac{\Delta ^{2}}{\sqrt{\rho }} \end{aligned}$$

(70)

which means $\lim _{t\rightarrow \infty } {\Delta ^{2}(t)}/{\sqrt{\rho (t)}}=0$ since $g\ge {3k}/2>0$. This completes the proof. $\square$