Finite-time spacecraft attitude control under input magnitude and rate saturation

Abstract

This paper addresses attitude-stabilizing control for rigid spacecraft with actuator magnitude and rate saturation (MRS). Firstly, a continuous dynamical system is introduced to model the actuator MRS. Rigorous analysis is given to demonstrate that the model’s output can always meet the MRS constraints if the control scheme applied to the model derived here is continuous. Then, by using the proposed MRS model and the homogeneity property, two attitude-stabilizing control laws are presented. The measurement of angular velocity is required in the implementation of the first control law, but this requirement is unnecessary in the second one. The local finite-time stability of the resulting closed-loop system is ensured by employing the Lyapunov approach. Finally, several simulation examples are presented to validate the efficiency of the proposed method.

Appendices

Appendices

1.1 Proof of Theorem 1

By continuity of closed-loop solutions, $q(t), \omega (t), u(t), v(t)$, and consequently U(t) must be finite over some interval [0, T), where $T\in R_{>0} \cup \{+\infty \}$. Then, the proof is divided into two parts: (1) We show using Lyapunov analysis that while the states are finite, the states u and v must be uniformly bounded; (2) We show that the closed-loop system does not have finite escape time, such that from part (1), u and v are uniformly bounded for all $t \ge 0$.

Part 1: We prove that u(t) and v(t) are uniformly bounded on [0, T) and $|\tau _i(t)|\le \tau _M$, $|{\dot{\tau }}_i(t)|\le \tau _{dM}$$(i = 1, 2, 3)$, $\forall t \in [0, T)$. Note that

$$\begin{aligned} {\dot{v}}_i=\left[ 1-\left( \frac{|v_i|}{(1-c_1)\tau _{dM}}\right) ^{m_2}\right] U_i-c_2{\bar{v}}_i, \end{aligned}$$

where $i=1, 2, 3$. Then, we show that $|v_i(t)|\le (1-c_1)\tau _{dM}$ for all $t\in [0, T)$. To prove this, consider $V=v_i^2/2$. The time derivative of V is

$$\begin{aligned} {\dot{V}}=\left[ 1-\left( \frac{|v_i|}{(1-c_1)\tau _{dM}}\right) ^{m_2}\right] U_iv_i-c_2{\bar{v}}_iv_i. \end{aligned}$$

When $|v_i|=(1-c_1)\tau _{dM}$, we have ${\dot{v}}_i=-c_2{\bar{v}}_i$, and hence, ${\dot{V}}=-c_2{\bar{v}}_iv_i<0$. Since $|v_i(0)|<(1-c_1)\tau _{dM}$ (i.e., $V(0)\le (1-c_1)^2\tau _{dM}^2/2$) and ${\dot{V}}=-c_2{\bar{v}}_iv_i<0$ on $V=(1-c_1)^2\tau _{dM}^2/2$, we conclude that $V\le (1-c_1)^2\tau _{dM}^2/2$ is an invariant set [49], that is, $V(t)\le (1-c_1)^2\tau _{dM}^2/2$ for all $t \in [0, T)$, which in turn indicates that $|v_i(t)|\le (1-c_1)\tau _{dM}$ for all $t \in [0, T)$ if $|v_i(0)|<(1-c_1)\tau _{dM}$.

Similarly, when $|u_i|=\tau _M$, we have

$$\begin{aligned} {\dot{u}}_i=-c_1\frac{\tau _{dM}}{\tau _M^{m_1}}u_i^{m_1}, \end{aligned}$$

and we can verify that $|u_i(t)|\le \tau _M$ for all $t\in [0, T)$. Then, the upper bound of the rate ${\dot{\tau }}_i={\dot{u}}_i$ is given by

$$\begin{aligned} |{\dot{u}}_i|&=\left| v_i-g_{1i}v_i-\frac{c_1\tau _{dM}u_i^{m_1}}{\tau _M^{m_1}}\right| \\&\le |v_i|+\frac{c_1\tau _{dM}|u_i|^{m_1}}{\tau _M^{m_1}} \le \tau _{dM}. \end{aligned}$$

Therefore, we conclude that $|\tau _i(t)|\le \tau _{M}$ and $|{\dot{\tau }}_i(t)|\le \tau _{dM} (i=1, 2, 3), \forall t\in [0, T)$.

Part 2: It remains to show that $T=+\infty$. Suppose on the contrary, that $T>0$ is finite and that $\lim _{t\rightarrow T^{-}}|U(t)|=+\infty$. Consider the kinetic energy $V=\frac{1}{2}\omega ^\mathrm{T}J\omega$. For $t \in [0,T)$, ${\dot{V}}$ can be obtained as

$$\begin{aligned} {\dot{V}}&= \omega ^\mathrm{T}\tau \le \sum _{i=1}^3|\omega _i|\tau _M \le \frac{\Vert \omega \Vert ^2}{2}+\frac{3\tau _M^2}{2} \\&\quad \le \frac{V}{\lambda _{\min }(J)}+\frac{3\tau _M^2}{2}, \end{aligned}$$

where Young’s inequality has been used. By the comparison principle, we obtain

$$\begin{aligned} V(\omega (t))&\le \exp \left( \frac{t}{\lambda _{\min }(J)}\right) \left[ V(0)+\frac{3}{2}\tau _M^2\lambda _{\min }(J)\right] \\&\quad -\frac{3}{2}\tau _M^2\lambda _{\min }(J), \end{aligned}$$

and in particular, since the right-hand side is monotonically increasing,

$$\begin{aligned} V(\omega (t))&\le \exp \left( \frac{T}{\lambda _{\min }(J)}\right) \left[ V(0)+\frac{3}{2}\tau _M^2\lambda _{\min }(J)\right] \\&\quad -\frac{3}{2}\tau _M^2\lambda _{\min }(J) \end{aligned}$$

for all $t\in [0, T)$. Then, it follows that $\omega (t)$ is contained in a compact set over the interval [0, T). Since q(t) maintains unit length, and as already established, u(t) and v(t) are also contained in compact sets over [0, T), by continuity of the control law f, U(t) also remains in a compact set over the interval [0, T). This contradicts the assertion that $\lim _{t\rightarrow T^{-}}|U(t)|=+\infty$. Hence, $T=+\infty$.

1.2 Proof of Proposition 1

Let $\alpha \in (0, +\infty )$. Referring to [47], we can verify that the function ${\bar{V}}_{1}(x)$ is well defined, positive definite, radially unbounded, and of class $C^1(R^{12}, R)$.

(a) If $x=0$, we can obtain ${\bar{V}}_{1}(\delta _\lambda ^r(x))=\lambda ^2{\bar{V}}_{1}(x)=0$. For any $x\ne 0\in R^{12}$, since $V_{1}$ is continuous and $V_{1}(0)=0$, we can verify that $V_{1}(\delta _\rho ^{r}(x))\le 1$ for all $\rho \in [0, l)$, where $l>0$ is some constant. Thus, the integrand of ${\bar{V}}_{1}$ meets $\beta \left( V_{1}(\delta _\rho ^{r}(x))\right) /\rho ^3=0$, $\forall \rho \in (0, l)$. Then, because $V_{1}$ is radially unbounded, we can show that there exists an $L>0$ such that $V_{1}(\delta _\rho ^{r}(x))\ge 2$, $\forall \rho >L$. Hence, for this $x\in R^{12}$, we choose $0<l_m<l$ and $L_M>L$ and we have

$$\begin{aligned}&\int _{l_m}^{L_M}\frac{1}{\rho ^3}\beta \left( V_{1}\left( \delta _\rho ^{r}(x)\right) \right) \hbox {d}\rho =\int _{l}^{L}\frac{1}{\rho ^3}\beta \left( V_{1}\left( \delta _\rho ^{r}(x)\right) \right) \hbox {d}\rho \nonumber \\&\quad \quad +\int _{L}^{L_M}\frac{1}{\rho ^3}\hbox {d}\rho \nonumber \\&\quad =\int _{l}^{L}\frac{1}{\rho ^3}\beta \left( V_{1}\left( \delta _\rho ^{r}(x)\right) \right) \hbox {d}\rho + \frac{1}{2L^2}-\frac{1}{2L_M^2}. \end{aligned}$$

(25)

Because $0<l_m<l$ and $L_M>L$ were chosen arbitrarily, taking the limit as $l_m\rightarrow 0^+$ and $L_M\rightarrow \infty$ leads to

$$\begin{aligned} {\bar{V}}_{1}(x)=\int _{l}^{L}\frac{1}{\rho ^3}\beta \left( V_{1}\left( \delta _\rho ^{r}(x)\right) \right) \hbox {d}\rho + \frac{1}{2L^2}. \end{aligned}$$

(26)

It follows from the definition that

$$\begin{aligned}&{\bar{V}}_{1}\left( \delta _{\lambda }^r(x)\right) \nonumber \\&\quad = \lim _{l_m\rightarrow 0^+, L_M\rightarrow \infty }\int _{l_m}^{L_M}\frac{1}{\rho ^3}\beta \left( V_{1} \left( \delta _\rho ^{r}\left( \delta _{\lambda }^r(x)\right) \right) \right) \hbox {d}\rho \nonumber \\&\quad =\lim _{l_m\rightarrow 0^+, L_M\rightarrow \infty }\int _{l_m}^{L_M}\frac{1}{\rho ^3}\beta \left( V_{1} \left( \delta _{\rho \lambda }^{r}(x)\right) \right) \hbox {d}\rho . \end{aligned}$$

(27)

Considering the change of variables $s=\rho \lambda$ and noticing the fact that $\hbox {d}\rho /\hbox {d}s=1/\lambda$, we can obtain that

$$\begin{aligned}&{\bar{V}}_{1}\left( \delta _{\lambda }^r(x)\right) \nonumber \\&\quad =\lim _{l_m\rightarrow 0^+, L_M\rightarrow \infty }\int _{l_m}^{L_M}\frac{\lambda ^2}{s^3}\beta \left( V_{1} \left( \delta _{s}^{r}(x)\right) \right) \hbox {d}s\nonumber \\&\quad =\lambda ^2{\bar{V}}_{1}(x). \end{aligned}$$

(28)

(b) Note that

$$\begin{aligned} L_{g_{1}}{\bar{V}}_{1}(x)&=\int _{0^+}^\infty \frac{1}{\rho ^{3+(\alpha -1)/4}} \beta '\left( V_{1}(\delta _\rho ^r(x))\right) \nonumber \\&\quad \times L_{g_{1}}V_{1}\left( \delta _\rho ^r(x)\right) \hbox {d}\rho \end{aligned}$$

(29)

and

$$\begin{aligned}&L_{g_{1}}{\bar{V}}_{1}(\delta _{\lambda }^r(x))=\int _{0^+}^\infty \frac{1}{\rho ^{3+(\alpha -1)/4}} \beta '\left( V_{1}(\delta _\rho ^r(\delta _{\lambda }^r(x)))\right) \nonumber \\&\qquad \times L_{g_{1}}V_{1} \left( \delta _\rho ^r(\delta _{\lambda }^r(x))\right) \hbox {d}\rho \nonumber \\&\quad =\int _{0^+}^\infty \frac{1}{\rho ^{3+(\alpha -1)/4}} \beta '\left( V_{1}(\delta _{\rho \lambda }^r(x))\right) L_{g_{1}}V_{1} \left( \delta _{\rho \lambda }^r(x)\right) \hbox {d}\rho . \end{aligned}$$

(30)

Letting $s=\rho \lambda$ and noting that $\hbox {d}\rho /\hbox {d}s=1/\lambda$, we have

$$\begin{aligned}&L_{g_{1}}{\bar{V}}_{1}(\delta _{\lambda }^r(x))=\int _{0^+}^\infty \frac{1}{\rho ^{3+(\alpha -1)/4}} \beta '\left( V_{1}(\delta _{\rho \lambda }^r(x))\right) \nonumber \\&\qquad \times L_{g_{1}}V_{1} \left( \delta _{\rho \lambda }^r(x)\right) \hbox {d}\rho \nonumber \\&\quad = \int _{0^+}^\infty \frac{\lambda ^{(\alpha +7)/4}}{s^{3+(\alpha -1)/4}} \beta '\left( V_{1}(\delta _{s}^r(x))\right) L_{g_{1}}V_{1} \left( \delta _{s}^r(x)\right) \hbox {d}s\nonumber \\&\quad =\lambda ^{\frac{\alpha +7}{4}}L_{g_{1}}{\bar{V}}_{1}(x). \end{aligned}$$

(31)

By (a), we know that there exist $l>0$ and $L>0$ so that

$$\begin{aligned} L_{g_{1}}{\bar{V}}_{1}(x)&=\int _{l}^L\frac{1}{\rho ^{3+(\alpha -1)/4}} \beta '\left( V_{1}(\delta _\rho ^r(x))\right) \nonumber \\&\quad \times L_{g_{1}}V_{1}\left( \delta _\rho ^r(x)\right) \hbox {d}\rho \end{aligned}$$

(32)

for all $x\in S^{12}$. Define $h(x, \rho , \alpha )=L_{g_{1}}V_{1}\left( \delta _\rho ^r(x)\right)$, where $(x, \rho , \alpha )\in S^{12}\times \{\rho \in [l, L]\}\times (0, +\infty )$. Since h is a continuous function and $(x, \rho )$ belongs to a compact set, we can conclude that there exists $\varphi _1>0$ such that the image of h is included in $(-\infty , -\varphi _1)$ for $(x, \rho )\in S^{12}\times \{\rho \in [l, L]\}$ and $\alpha =1$. Using the tube lemma [50], it is concluded that there exists $\epsilon >0$ such that for all $(x, \rho , \alpha )\in S^{12}\times \{\rho \in [l, L]\}\times (1-\epsilon , 1+\epsilon )$, $h(x, \rho , \alpha )\le -\varphi _1$. Then, we can obtain

$$\begin{aligned} L_{g_{1}}{\bar{V}}_{1}(x)&\le -\varphi _1\int _{l}^L\frac{1}{\rho ^{3+(\alpha -1)/4}} \beta '\left( V_{1}(\delta _\rho ^r(x))\right) \hbox {d}\rho \nonumber \\&\le -\varphi \left( {\bar{V}}_1(x)\right) ^{\frac{\alpha +7}{8}} \end{aligned}$$

(33)

with $\varphi =\varphi _2/\varphi _3>0$, where $\varphi _2>0$ is the lower bound of $\varphi _1\int _{l}^L\frac{1}{\rho ^{3+(\alpha -1)/4}}$$\beta '\left( V_{1}(\delta _\rho ^r(x))\right) \hbox {d}\rho$ and $\varphi _3>0$ is the upper bound of $\left( {\bar{V}}_1(x)\right) ^{(\alpha +7)/8}$ for $(x, \alpha )\in S^{12}\times (1-\epsilon , 1+\epsilon )$.

When $x=0$, we have $L_{g_{1}}{\bar{V}}_{1}(x)\le -\varphi {\bar{V}}_{1}^{(\alpha +7)/8}(x)$. For any $x\in R^{12}\setminus \{0\}$, by Lemma 3, we can obtain that $x=\phi (\lambda , {\bar{x}})=\delta _\lambda ^r({\bar{x}})$, where ${\bar{x}}\in S^{12}$. Then, we have

$$\begin{aligned} L_{g_{1}}{\bar{V}}_{1}(x)&=L_{g_{1}}{\bar{V}}_{1}\left( \delta _\lambda ^r({\bar{x}})\right) =\lambda ^{\frac{\alpha +7}{4}}L_{g_{1}}{\bar{V}}_{1}({\bar{x}})\nonumber \\&\le -\varphi \lambda ^{\frac{\alpha +7}{4}}\left( {\bar{V}}_1({\bar{x}})\right) ^{\frac{\alpha +7}{8}}\nonumber \\&=-\varphi \left( {\bar{V}}_1(x)\right) ^{\frac{\alpha +7}{8}}. \end{aligned}$$

(34)

Therefore, $L_{g_{1}}{\bar{V}}_{1}(x)\le -\varphi {\bar{V}}_{1}^{(\alpha +7)/8}(x)$, $\forall x\in R^{12}$.

1.3 Proof of Theorem 3

Define $y=\hbox {col}(\bar{\tilde{q}}, {\tilde{\omega }})$, the dilation $\delta _\lambda ^r(y)=\hbox {col}(\lambda \bar{\tilde{q}}, \lambda ^{\alpha _1} {\tilde{\omega }})$, and $M_2=[-\gamma _1I_3/2, \; I_3/2; \; -\gamma _2I_3, \; 0]$. As $M_2$ is Hurwitz, there exists $N_2=N_2^\mathrm{T}>0$ so that $M_2^\mathrm{T}N_2+N_2M_2=-I_6$.

First of all, we show that the origin of the nominal system of (19), i.e., ${\dot{y}}=g_{2}(y)$, is finite-time stable, where $g_{2}(y)=\hbox {col}({\tilde{\omega }}/2-\gamma _1\hbox {sig}^{\alpha _1}(\bar{\tilde{q}})/2, -\gamma _2\hbox {sig}^{\alpha _2}(\bar{\tilde{q}}))$ is homogeneous of degree $(\alpha -1)/4<0$ in regard to the dilation $\delta _\lambda ^r(y)$.

Define the candidate Lyapunov function as:

$$\begin{aligned} {\bar{V}}_2(y)=\int _{0^+}^{+\infty }\frac{1}{\rho ^3}\beta (V_2(\delta _\rho ^r(y)))\hbox {d}\rho \end{aligned}$$

(35)

if $y\in R^{6}\setminus \{0\}$ and ${\bar{V}}_2(0)=0$, where $V_2(y)=y^\mathrm{T}N_2y$. Following the same steps of the proof of Proposition 1, we can obtain that ${\bar{V}}_2(y)$ is homogeneous of degree 2 in regard to the dilation $\delta _\lambda ^r(y)$, $L_{g_2}{\bar{V}}_2(y)$ is homogeneous of degree $(\alpha +7)/4$ in regard to the dilation $\delta _\lambda ^r(y)$, and

$$\begin{aligned} L_{g_2}{\bar{V}}_2(y)\le -\varphi {\bar{V}}_2^{\frac{\alpha +7}{8}}(y) \end{aligned}$$

(36)

for all $\alpha \in (1-\epsilon , 1)$ and for all $y\in R^{6}$, where $\varphi$ and $\epsilon$ are some positive constants. By noticing that $0<(7+\alpha )/8<1$, we conclude that the origin of the nominal system ${\dot{y}}=g_{2}(y)$ is finite-time stable. Furthermore, by noting that $\tilde{q}_4=\sqrt{1-\bar{\tilde{q}}^\mathrm{T}\bar{\tilde{q}}}$, it follows that $\lim _{\lambda \rightarrow 0}\frac{f_5(\lambda \bar{\tilde{q}}, \lambda ^{\alpha _1}{\tilde{\omega }})}{\lambda ^{\alpha _1}}=0$, $\lim _{\lambda \rightarrow 0}\frac{f_6(\lambda \bar{\tilde{q}}, \lambda ^{\alpha _1}{\tilde{\omega }})}{\lambda ^{\alpha _2}}=0$. By Lemma 2 in [27], we can obtain that the origin of system (19) is locally finite-time stable.

1.4 Proof of Theorem 4

Define $z=\hbox {col}({\bar{q}}, {\hat{\omega }}, u, v)$, $\eta =\hbox {col}(\bar{\tilde{q}}, {\tilde{\omega }}, {\bar{q}}, {\hat{\omega }}, u, v)$, and the dilations $\delta _\lambda ^r(z)=\hbox {col}(\lambda {\bar{q}}, \lambda ^{\alpha _1} {\hat{\omega }}, \lambda ^{\alpha _2} u, \lambda ^{\alpha _3} v)$ and $\delta _\lambda ^r(\eta )=\hbox {col}(\lambda \bar{\tilde{q}}, \lambda ^{\alpha _1} {\tilde{\omega }}, \lambda {\bar{q}}, \lambda ^{\alpha _1} {\hat{\omega }}, \lambda ^{\alpha _2} u, \lambda ^{\alpha _3} v)$.

Firstly, we prove that the nominal system of (21), i.e., ${\dot{\eta }}=g_{3}(\eta )$, is asymptotically stable, where

$$\begin{aligned}&g_{3}= \left( \begin{array}{l} \frac{{\tilde{\omega }}}{2}-\frac{1}{2}\gamma _1\hbox {sig}^{\alpha _1}(\bar{\tilde{q}})\\ -\gamma _2\hbox {sig}^{\alpha _2}(\bar{\tilde{q}}) \\ \frac{{\hat{\omega }}}{2}+\frac{1}{2}q_4{\tilde{\omega }}\\ J^{-1}u+\gamma _2\hbox {sig}^{\alpha _2}(\bar{\tilde{q}})\\ v\\ -k_1\hbox {sig}^{\alpha }({\bar{q}})-k_2\hbox {sig}^{\frac{\alpha }{\alpha _1}}({\hat{\omega }}) -k_3\hbox {sig}^{\frac{\alpha }{\alpha _2}}(u)-{\bar{k}}_4\hbox {sig}^{\frac{\alpha }{\alpha _3}}(v)\\ \end{array} \right) \end{aligned}$$

is homogeneous of degree $(\alpha -1)/4<0$ in regard to the dilation $\delta _\lambda ^r(\eta )$.

Consider the Lyapunov function candidate:

$$\begin{aligned} {\bar{V}}(\eta )={\bar{V}}_3(z)+K{\bar{V}}_2(y), \end{aligned}$$

(37)

where $K>0$ is a sufficiently large constant, ${\bar{V}}_2(y)$ is given by (35), and ${\bar{V}}_3(z)$ is defined by

$$\begin{aligned} {\bar{V}}_3(z)=\int _{0^+}^{+\infty }\frac{1}{\rho ^3}\beta (V_3(\delta _\rho ^r(z)))\hbox {d}\rho \end{aligned}$$

(38)

if $z\in R^{12}\setminus \{0\}$ and ${\bar{V}}_3(0)=0$, where $V_3(z)=z^\mathrm{T}N_1z$.

Following the same procedure of the proof of Proposition 1, it is concluded that there exists an $\epsilon _1>0$ such that the time derivative of ${\bar{V}}_3(z)$ along with the system ${\dot{z}}=g_{4}(z)=\hbox {col}({\hat{\omega }}/2, J^{-1}u, v, -k_1\hbox {sig}^{\alpha }({\bar{q}})-k_2\hbox {sig}^{\alpha /\alpha _1}({\hat{\omega }}) -k_3\hbox {sig}^{\alpha /\alpha _2}(u)-{\bar{k}}_4\hbox {sig}^{\alpha /\alpha _3}(v))$ is given by

$$\begin{aligned} L_{g_4}{\bar{V}}_3(z)\le -\varphi _1{\bar{V}}^{\frac{\alpha +7}{8}}_3(z) \end{aligned}$$

(39)

for all $\alpha \in (1-\epsilon _1, 1)$ and for all $z\in R^{12}$, where $\varphi _1>0$. Then, there exists an $\epsilon >0$ so that for all $\alpha \in (1-\epsilon , 1)$ and for all $\eta \in S^{18}$,

$$\begin{aligned}&L_{g_3}{\bar{V}}(\eta )\le -\varphi _1{\bar{V}}^{\frac{\alpha +7}{8}}_3(z)-K\varphi _2{\bar{V}}^{\frac{\alpha +7}{8}}_2(y)\nonumber \\&\qquad +\left( \frac{\partial {\bar{V}}_3(z)}{\partial z}\right) ^\mathrm{T}g_5(\eta )\nonumber \\&\quad \le -\frac{\varphi _1}{2}{\bar{V}}^{\frac{\alpha +7}{8}}_3(z) -\varphi _2\left( K-\frac{\varphi _3}{\varphi _2}\right) {\bar{V}}^{\frac{\alpha +7}{8}}_2(y), \end{aligned}$$

(40)

where $g_5(\eta )=\hbox {col}(q_4{\tilde{\omega }}/2, \gamma _2\hbox {sig}^{\alpha _2}(\bar{\tilde{q}}), 0, 0)$, and $\varphi _2$ and $\varphi _3$ are some positive constants. Since ${\bar{V}}(\eta )$ is homogeneous of degree of 2 in regard to the dilation $\delta _\lambda ^r(\eta )$, the above inequality holds for all $\eta \in R^{18}$. If we choose $K>\varphi _3/\varphi _2$, then we have $L_{g_3}{\bar{V}}(\eta )\le 0$. Thus, we conclude that the origin of the system ${\dot{\eta }}=g_{3}(\eta )$ is asymptotically stable. Because

$$\begin{aligned}&\lim _{\lambda \rightarrow 0}\frac{f_5(\lambda \bar{\tilde{q}}, \lambda ^{\alpha _1}{\tilde{\omega }})}{\lambda ^{\alpha _1}}=0,\\&\lim _{\lambda \rightarrow 0}\frac{f_6(\lambda \bar{\tilde{q}}, \lambda ^{\alpha _1}{\tilde{\omega }})}{\lambda ^{\alpha _2}}=0,\\&\lim _{\lambda \rightarrow 0}\frac{f_7(\lambda {\bar{q}}, \lambda ^{\alpha _1}{\hat{\omega }}, \lambda ^{\alpha _1}{\tilde{\omega }})}{\lambda ^{\alpha _1}}=0,\\&\lim _{\lambda \rightarrow 0}\frac{f_8(\lambda ^{\alpha _1}{\hat{\omega }})}{\lambda ^{\alpha _2}}=0,\\&\lim _{\lambda \rightarrow 0}\frac{f_9(\lambda ^{\alpha _2}u, \lambda ^{\alpha _3}v)}{\lambda ^{\alpha _3}}=0,\\&\lim _{\lambda \rightarrow 0}\frac{f_{10}(\lambda {\bar{q}}, \lambda ^{\alpha _1}{\hat{\omega }}, \lambda ^{\alpha _2}u, \lambda ^{\alpha _3}v)}{\lambda ^{\alpha }}=0, \end{aligned}$$

it follows from Lemma 2 in [27] that the local finite-time stability of the origin of system (21) is obtained. By noting that ${\tilde{\omega }}=\omega -{\hat{\omega }}$, we conclude that the angular velocity $\omega ={\tilde{\omega }}+{\hat{\omega }}$ converges to the origin in finite time.