Sliding-Mode-Based Fault Diagnosis and Fault-Tolerant Control for Quad-Rotors

Abstract

This chapter presents the design of a sliding-mode-based fault diagnosis and a fault-tolerant control for the trajectory tracking problem in quad-rotors. The problem considers external disturbances, and two different actuator faulty scenarios: multiple losses of rotor effectiveness or a complete rotor failure. The proposals are only based on measurable positions and angles. For the fault diagnosis strategy, a finite-time sliding-mode observer is proposed to estimate some state variables, and it provides a set of residuals. These residuals allow us the detection, isolation, and identification of multiple actuator faults even under the presence of a class of external disturbances. Using the proposed fault diagnosis, an actuator fault accommodation controller is developed to solve the trajectory tracking problem in quad-rotors under the effects of multiple losses of rotor effectiveness and external disturbances. The fault accommodation partially compensates the actuator faults allowing the usage of a baseline robust-nominal controller that deals with external disturbances. Additionally, in order to deal with the rotor failure scenario, an active fault-tolerant control is proposed. First, the rotor failure is isolated using the proposed fault diagnosis, and then, a combination of a finite-time sliding-mode observer, PID controllers, and continuous high-order sliding-modes controllers is proposed. Such a strategy allows the yaw angular velocity to remain bounded and the position tracking to be achieved even in the presence of some external disturbances. Numerical simulations and experimental results on Quanser’s QBall2 show the performance of the proposed strategies.

This chapter contains material reprinted from Romeo Falcón, Héctor Ríos, and Alejandro Dzul, A sliding-mode-based active fault-tolerant control for robust trajectory tracking in quad-rotors under a rotor failure, International Journal of Robust and Nonlinear Control. Copyright ©1999–2023 John Wiley & Sons, Inc. All rights reserved. Section 3 and Figs. 1, 2, 3, 4, 5, 6, 7, and 8 ©2022 IEEE. Reprinted with permission, from, R. Falcón, H. Ríos and A. Dzul, “A Robust Fault Diagnosis for Quad-Rotors: A Sliding-Mode Observer Approach,” in IEEE/ASME Transactions on Mechatronics, vol. 27, no. 6, pp. 4487–4496, Dec. 2022, DOI:10.1109/TMECH.2022.3156854. Section 4.1 ©2021 IEEE. Reprinted with permission, from, R. Falcón, H. Ríos and A. Dzul, “An Actuator Fault Accommodation Sliding-Mode Control Approach for Trajectory Tracking in Quad-Rotors,” 2021 60th IEEE Conference on Decision and Control (CDC), Austin, TX, USA, 2021, pp. 7100–7105, DOI:10.1109/CDC45484.2021.9682845.

Appendix

1.1 Proof of Theorem 2

The attitude-tracking error dynamics is given as

$$\begin{aligned} \dot{e}_\eta &=\varepsilon _{\eta },\\ \dot{\varepsilon }_{\eta }&=J\left( \tau -I_\eta Mf(t)\right) +\Xi w_{\eta }(\eta _{2})-\Lambda _\eta \eta _{2}+d_{\eta }(t)-\ddot{\eta }_d. \end{aligned}$$

Let us prove that the tracking error dynamics for the roll angle $\phi$ is UFTS when the controller (18b) is applied. Then, the closed-loop tracking error dynamics for $\phi$ is given as

$$\begin{aligned} \dot{e}_\phi &=\varepsilon _{\phi },\end{aligned}$$

(36a)

$$\begin{aligned} \dot{\varepsilon }_{\phi }&=\bar{\nu }_\phi -k_{\phi 1}\lceil e_{\phi }\rfloor ^{\frac{1}{3}}-k_{\phi 2}\lceil \varepsilon _{\phi }\rfloor ^{\frac{1}{2}},\end{aligned}$$

(36b)

$$\begin{aligned} \dot{\bar{\nu }}_\phi &=-k_{\phi 3}\lceil e_{\phi }\rfloor ^{0}-k_{\phi 4}\lceil \varepsilon _{\phi }\rfloor ^{0}+\bar{\Delta }_\phi (t), \end{aligned}$$

(36c)

where $\bar{\nu }_\phi =\nu _\phi -J_x^{-1}L(\tilde{f}_3(t)-\tilde{f}_4(t))+d_\phi (t)$ and $\bar{\Delta }_\phi (t)=-J_x^{-1}L(\dot{\tilde{f}}_3(t)-\dot{\tilde{f}}_4(t))+\dot{d}_\phi (t)$. It is given that $\dot{\tilde{f}}(t)=M^{-1}\zeta ^{-1}(\eta _{1})\dot{d}(t)+M^{-1}\dot{\zeta }^{-1}(\eta _{1})d(t)$, and thus

$$\begin{aligned} \dot{\tilde{f}}_1(t)&=\frac{m}{4c\phi c\theta }\dot{d}_z(t)+\frac{J_z}{4K_\tau }\dot{d}_\psi (t)+\frac{J_y}{2L}\dot{d}_\theta (t)+\rho (t),\\ \dot{\tilde{f}}_2(t)&=\frac{m}{4c\phi c\theta }\dot{d}_z(t)+\frac{J_z}{4K_\tau }\dot{d}_\psi (t)-\frac{J_y}{2L}\dot{d}_\theta (t)+\rho (t),\\ \dot{\tilde{f}}_3(t)&=\frac{m}{4c\phi c\theta }\dot{d}_z(t)-\frac{J_z}{4K_\tau }\dot{d}_\psi (t)+\frac{J_x}{2L}\dot{d}_\phi (t)+\rho (t),\\ \dot{\tilde{f}}_4(t)&=\frac{m}{4c\phi c\theta }\dot{d}_z(t)-\frac{J_z}{4K_\tau }\dot{d}_\psi (t)-\frac{J_x}{2L}\dot{d}_\phi (t)+\rho (t), \end{aligned}$$

where $\rho (t)=m(\dot{\theta }c\phi s\theta +\dot{\phi }s\phi c\theta )d_z(t)/4c^2\phi c^2\theta$. Then, based on the previous equalities, it follows that $\dot{\tilde{f}}_3(t)-\dot{\tilde{f}}_4(t)=J_xL^{-1}\dot{d}_\phi (t)$, and hence, (36c) can be rewritten as $\dot{\bar{\nu }}_\phi =-k_{\phi 3}\lceil e_{\phi }\rfloor ^{0}-k_{\phi 4}\lceil \varepsilon _{\phi }\rfloor ^{0}$. Hence, if the gains are selected as $k_{\phi 1}=25\varpi ^{\frac{2}{3}}$, $k_{\phi 2}=15\varpi ^{\frac{1}{2}}$, $k_{\phi 3}=2.3\varpi$ and $k_{\phi 4}=1.1\varpi$ with any $\varpi >0$, the finite-time convergence to zero is ensured for the tracking error dynamics (36) [42]. The same procedure can be followed to prove the finite-time convergence to zero for the tracking error dynamics of pitch and yaw angles, i.e., $(e_\eta ;\varepsilon _{\eta }) = 0$ is UFTS.

The position tracking error dynamics, taking into account the virtual control (18a), is given as

$$\begin{aligned} \dot{\epsilon }_\xi =A_k\epsilon _\xi +B\left[ d_\xi (t)-\frac{g_\xi (\eta _{1})}{m}\sum _{i=1}^{4}\tilde{f}_i(t)+w_\xi (\eta _{1},u_z,\nu ) \right] , \end{aligned}$$

(37)

$$\begin{aligned} A=\begin{pmatrix} 0_{6\times 3}&{}I_6\\ 0_3&{}0_{3\times 6} \end{pmatrix},\qquad B=\begin{pmatrix} 0_{6\times 3}\\ I_3 \end{pmatrix}, \end{aligned}$$

where $A_k:=A+BK_\xi \in \mathbb {R}^{9\times 9}$, $\epsilon _\xi :=(\bar{e}_\xi ^T,e_\xi ^T,\varepsilon _{\xi }^T)\in \mathbb {R}^{9}$ and $K_\xi =({\textbf {1}}K_{i\xi },{\textbf {1}}K_{p\xi },$ ${\textbf {1}}K_{d\xi })\in \mathbb {R}^{1\times 9}$, with ${\textbf {1}}:=(1,1,1)\in \mathbb {R}^{1\times 3}$. The nonlinear decoupling term $w_{\xi }=u_mg_{\xi }(\eta _{1})-G-\nu$ is Lipschitz in $\eta _{1}$ and continuous in $u_m$, then it follows that $||w_\xi ||_\infty \le L_\eta ||e_\eta ||_\infty$, for all $\eta _{1}$, $\nu \in \mathbb {R}^3$ and $u_m\in \mathbb {R}$, for some positive $L_\eta >0$. This implies that $w_\xi$ vanishes when $e_\eta =0$. Hence, the closed-loop tracking error dynamics (37), considering $w_\xi \equiv 0$, can be rewritten as

$$\begin{aligned} \dot{\epsilon }_\xi =A_k\epsilon _\xi +B\left[ d_\xi (t)-\frac{g_\xi (\eta _{1})}{m}\sum _{i=1}^{4}\tilde{f}_i(t) \right] . \end{aligned}$$

(38)

Let us propose a candidate Lyapunov function $V:\in \mathbb {R}^9\rightarrow \mathbb {R}$ as $V(\epsilon _\xi )=\epsilon _\xi ^TP\epsilon _\xi$, with $P = P^T > 0$. The time derivative of V, along the trajectories of the system (38), satisfies

$$\begin{aligned} \dot{V}(\epsilon _\xi )=\epsilon _\xi ^T(PA_k+A_k^TP)\epsilon _\xi +2\epsilon _\xi ^TPB\left[ d_\xi (t)-\frac{g_\xi (\eta _{1})}{m}\sum _{i=1}^{4}\tilde{f}_i(t) \right] . \end{aligned}$$

Since the pair (A, B) is controllable, there always exists $K_\xi$ such that $A_k$ is Hurwitz; and thus, it holds that $PA_k+A_k^TP=-Q$, with $Q = Q^T > 0$. Then, the time derivative of V is upper bounded as

$$\begin{aligned} \dot{V}(\epsilon _\xi )&\le -(1-\mu )\lambda _\text {min}\{Q\}||\epsilon _\xi ||^2,\\ \forall ||\epsilon _\xi ||&\ge \frac{2\lambda _\text {max}\{P\}}{\mu \lambda _\text {min}\{Q\}}\left( ||d_\xi ||_\infty +\frac{1}{m}\sum _{i=1}^{4}||\tilde{f}_i||_\infty \right) , \end{aligned}$$

for any $\mu \in (0,1)$. Then, it is proved that the position tracking error dynamics is ISS with respect to $d_\xi$ and $\tilde{f}$.$\Box$

1.2 Proof of Lemma 3

The convergence to zero of the attitude tracking error dynamics, when the Continuous Twisting Control is active (Algorithm 3—lines 2, 5, 9, and 12), is given in [2]. Then, only the convergence to zero of the attitude tracking error dynamics, when the positive (Algorithm 3—lines 6 and 11) and negative controllers (Algorithm 3—lines 3 and 8) are active, will be proven.

Let us assume that the first rotor has failed. Then, according to Algorithm 3—line 3, the angular moment $\tau _{\theta }$ must be negative, and thus, it is designed as

$$\begin{aligned} \tau _{\theta } & =\frac{J_{y}}{2}(\bar{\tau }_{\theta }+f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })-|\bar{\tau }_{\theta }+f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })|), \end{aligned}$$

(39)

where $f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })=-b_{\theta }\dot{\phi }\dot{\psi }+\frac{a_{\theta }}{J_{y}}\dot{\theta }-\hat{\eta }_{\theta }+\ddot{\theta }_{\star }$. Note that if $\bar{\tau }_{\theta }+f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })\le 0$, the angular moment $\tau _{\theta }$ given in (39) is rewritten as $\tau _{\theta }=J_{y}(\bar{\tau }_{\theta }+f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star }))$, where the Continuous Twisting Control is active, just as in Algorithm 3—lines 9 and 12. On the other hand, if $\bar{\tau }_{\theta }+f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })>0$, the control signal (39) is given by $\tau _{\theta }=0$, where the control effort is null, in order to avoid negative thrusts.

The closed-loop tracking error dynamics for $\theta$, taking into account (39), is written as

$$\begin{aligned} \dot{e}_{\theta } & =\varepsilon _{\theta },\\ \dot{\varepsilon }_{\theta } & =\frac{\bar{\tau }_{\theta }+f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })-|\bar{\tau }_{\theta }+f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })|}{2}-f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star }). \end{aligned}$$

Such a dynamics can be viewed as a state-dependent switched system where the switching surface is given by $\mathscr {S}:=\{(\bar{\tau }_{\theta },\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })\in \mathbb {R}\times \mathbb {R}^{2}\times \mathbb {R}:\bar{\tau }_{\theta }+f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })=0\}$, i.e.,

$$\begin{aligned} \dot{e}_{\theta } & =\varepsilon _{\theta },\end{aligned}$$

(40a)

$$\begin{aligned} \dot{\varepsilon }_{\theta } & =g_{\sigma (t)}(\bar{\tau }_{\theta },\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star }),\end{aligned}$$

(40b)

$$\begin{aligned} \sigma (t) & ={\left\{ \begin{array}{ll} 1, &{} \text {if}\ f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })\le -\bar{\tau }_{\theta },\\ 2, &{} \text {if}\ f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })>-\bar{\tau }_{\theta }, \end{array}\right. } \end{aligned}$$

(40c)

where $g_{1}(\bar{\tau }_{\theta },\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })=\bar{\tau }_{\theta }$ and $g_{2}(\bar{\tau }_{\theta },\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })=b_{\theta }\dot{\phi }\dot{\psi }-\frac{a_{\theta }}{J_{y}}\dot{\theta }+\hat{\eta }_{\theta }-\ddot{\theta }_{\star }$. In order to provide the convergence properties of the tracking error dynamics (40), the analysis is carried out for each operating mode, i.e., for each $\sigma =1,2$.

(1) Case $\mathbf {\sigma =1}$: In this case it holds that $f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })\le -\bar{\tau }_{\theta }$, implying that $g_{1}(\bar{\tau }_{\theta },\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })=\bar{\tau }_{\theta }$, and the tracking error dynamics (40) is rewritten as

$$\begin{aligned} \dot{e}_{\theta } & =\varepsilon _{\theta },\end{aligned}$$

(41a)

$$\begin{aligned} \dot{\varepsilon }_{\theta } & =v_{\theta }-k_{\theta 0}\left\lceil e_{\theta }\right\rfloor ^{\frac{1}{3}}-k_{\theta 1}\left\lceil \varepsilon _{\theta }\right\rfloor ^{\frac{1}{2}}, \end{aligned}$$

(41b)

$$\begin{aligned} \dot{v}_{\theta } & =-k_{\theta 2}\left\lceil e_{\theta }\right\rfloor ^{0}-k_{\theta 3}\left\lceil \varepsilon _{\theta }\right\rfloor ^{0}. \end{aligned}$$

(41c)

Then, at steady state ($e_{\theta },\varepsilon _{\theta },v_{\theta })=0$, according to [42], system (41) is UFTS. Thus, it follows that $\bar{\tau }_{\theta }(t)=v_{\theta }-k_{\theta 0}\left\lceil e_{\theta }\right\rfloor ^{\frac{1}{3}}-k_{\theta 1}\left\lceil \varepsilon _{\theta }\right\rfloor ^{\frac{1}{2}}=0$, for all $t\ge T_{\theta }$; and hence, the switching condition turns into $f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })\le 0$, implying that (39) is rewritten as

$$\begin{aligned} \tau _{\theta } & =J_{y}f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })=J_{y}(-b_{\theta }\dot{\phi }\dot{\psi }+\frac{a_{\theta }}{J_{y}}\dot{\theta }-\hat{\eta }_{\theta }+\ddot{\theta }_{\star }). \end{aligned}$$

(42)

Since $f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })\le 0$, then the control law (42) is negative. Recalling that $\hat{\eta }_{\theta }(t)=d_{\theta }(t)$, for all $t\ge T$ with $T<t_{f1}$, if $\hat{\eta }_{\theta }(t)=d_{\theta }(t)\ge \frac{a_{\theta }}{J_{y}}\dot{\theta }(t)-b_{\theta }\dot{\phi }(t)\dot{\psi }(t)+\ddot{\theta }_{\star }(t)$, holds for all $t\ge t_{f1}$, i.e., the constraint (30a), then $f_{\theta }(\eta _{2}(t),\hat{\eta }_{\theta }(t),\ddot{\theta }_{\star }(t))\le 0$, for all $t\ge t_{f1}$; and hence, system (41) never switches to the case $\sigma =2$ and ($e_{\theta },\varepsilon _{\theta },v_{\theta })=0$ is UFTS.

(2) Case $\mathbf {\sigma =2}$: In this case it holds that $f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })>-\bar{\tau }_{\theta }$, implying that $g_{2}(\bar{\tau }_{\theta },\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })=b_{\theta }\dot{\phi }\dot{\psi }-\frac{a_{\theta }}{J_{y}}\dot{\theta }+\hat{\eta }_{\theta }-\ddot{\theta }_{\star }$, and the tracking error dynamics (40) is rewritten as

$$\begin{aligned} \dot{e}_{\theta } & =\varepsilon _{\theta },\end{aligned}$$

(43a)

$$\begin{aligned} \dot{\varepsilon }_{\theta } & =b_{\theta }\dot{\phi }\dot{\psi }-\frac{a_{\theta }}{J_{y}}\dot{\theta }+d_{\theta }-\ddot{\theta }_{\star }. \end{aligned}$$

(43b)

If constraint (30a), i.e., $\hat{\eta }_{\theta }(t)=d_{\theta }(t)\ge \frac{a_{\theta }}{J_{y}}\dot{\theta }(t)-b_{\theta }\dot{\phi }(t)\dot{\psi }(t)+\ddot{\theta }_{\star }(t)$, holds for all $t\ge t_{f1}$, then it follows that $f_{\theta }(\eta _{2}(t),\hat{\eta }_{\theta }(t),\ddot{\theta }_{\star }(t))\le 0$, for all $t\ge t_{f1}$. Taking into account that $\varepsilon _{\theta }=\dot{\theta }-\dot{\theta }_{\star }$, (43b) can be written as follows

$$ \dot{\varepsilon }_{\theta }=b_{\theta }\dot{\phi }\dot{\psi }-\frac{a_{\theta }}{J_{y}}\dot{\theta }_{\star }-\frac{a_{\theta }}{J_{y}}\varepsilon _{\theta }+d_{\theta }. $$

The solution of the previous differential equation is given by

$$\begin{aligned} \varepsilon _{\theta }(t)=\varepsilon _{\theta }(t_{f1})e^{-\frac{a_{\theta }}{J_{y}}(t-t_{f1})} \\ +\int _{t_{f1}}^{t}e^{-\frac{a_{\theta }}{J_{y}}(t-\tau )}(b_{\theta }\dot{\phi }(\tau )\dot{\psi }(\tau )-\frac{a_{\theta }}{J_{y}}\dot{\theta }_{\star }(\tau )+d_{\theta }(\tau )-\ddot{\theta }_{\star }(\tau ))d\tau , \end{aligned}$$

and, since $f_{\theta }(\eta _{2}(t),\hat{\eta }_{\theta }(t),\ddot{\theta }_{\star }(t))\le 0$ implies that $b_{\theta }\dot{\phi }(t)\dot{\psi }(t)-\frac{a_{\theta }}{J_{y}}\dot{\theta }(t)+d_{\theta }(t)-\ddot{\theta }_{\star }(t)\ge 0$, for all $t\ge t_{f1}$, it is clear that

$$ \lim _{t\rightarrow \infty }\varepsilon _{\theta }(t)>0\Rightarrow \lim _{t\rightarrow \infty }e_{\theta }(t)>0. $$

Therefore, the previous statements imply that

$$ \lim _{t\rightarrow \infty }v_{\theta }(t)<0, $$

and thus

$$ \lim _{t\rightarrow \infty }\bar{\tau }_{\theta }(t)=\lim _{t\rightarrow \infty }(v_{\theta }-k_{\theta 0}\left\lceil e_{\theta }\right\rfloor ^{\frac{1}{3}}-k_{\theta 1}\left\lceil \varepsilon _{\theta }\right\rfloor ^{\frac{1}{2}})=-\infty . $$

On the other hand, note that, due to the convergence properties of the CTC, the roll angular velocity $\dot{\phi }$ converges to a bounded reference $\dot{\phi }_{\star }$; the yaw angular velocity $\dot{\psi }$ is bounded, as it is shown later by Lemma 4, and due to Assumption 1, the disturbance term $d_{\theta }$ is also bounded. Thus, it follows that $f_{\theta }(\eta _{2},\hat{\eta }_{\theta },\ddot{\theta }_{\star })=-b_{\theta }\dot{\phi }\dot{\psi }+\frac{a_{\theta }}{J_{y}}\dot{\theta }-d_{\theta }(t)+\ddot{\theta }_{\star }$ is bounded and $f_{\theta }(\eta _{2}(t),\hat{\eta }_{\theta }(t),\ddot{\theta }_{\star }(t))\le 0$, for all $t\ge t_{f1}$. Therefore, there always exists a finite time $t_{\sigma 1}$ such that $f_{\theta }(\eta _{2}(t),\hat{\eta }_{\theta }(t),\ddot{\theta }_{\star }(t))\le -\bar{\tau }_{\theta }(t)$ holds for all $t\ge t_{\sigma 1}$, and hence, system (41) always switches to the case $\sigma =1$, for which, ($e_{\theta },\varepsilon _{\theta },v_{\theta })=0$ is UFTS.

The previous analysis, together with the fact that ($e_{\phi },\varepsilon _{\phi })=0$ is UFTS, implies that, at steady state $(e_{\eta },\varepsilon _{\eta })=0$, the attitude tracking error dynamics is UFTS. The same procedure can be followed to analyze the convergence properties of Algorithm 1 when other rotors have failed. This concludes the proof.$\Box$

1.3 Proof of Lemma 4

In this proof, the yaw dynamics is analyzed in the occurrence of a rotor failure. With this aim, the results given by Lemmas 1–3 are considered; then, it is demonstrated that the yaw angular acceleration can be bounded at steady state. Next, the conditions to ensure the boundedness of the yaw angular velocity are obtained by means of the acceleration upper bound.

Consider the loss of the first rotor. Then, using (23), the yaw dynamics (31) can be rewritten as

$$\begin{aligned} \ddot{\psi }=-\frac{K_{\tau }}{J_{z}}(u_z+2\frac{\tau _{\theta }}{L})+b_{\psi }\dot{\phi }\dot{\theta }-\frac{a_{\psi }}{J_{z}}\dot{\psi }+d_{\psi }, \end{aligned}$$

(44)

and by substituting $u_z$, given in (17a), one obtains

$$\begin{aligned} \ddot{\psi }=-\dfrac{K_{\tau }}{J_{z}}(m\sqrt{\nu _{x}^{2}+\nu _{y}^{2}+(\nu _{z}+g)^{2}}+2\dfrac{\tau _{\theta }}{L})+b_{\psi }\dot{\phi }\dot{\theta }-\frac{a_{\psi }}{J_{z}}\dot{\psi }+d_{\psi }. \end{aligned}$$

(45)

As it was shown by Lemma 3, at steady state, $\tau _{\theta }=J_{y}(-b_{\theta }\dot{\phi }\dot{\psi }+\frac{a_{\theta }}{J_{y}}\dot{\theta }-\hat{\eta }_{\theta }+\ddot{\theta }_{\star })$ as in (42). Then, at steady state, (45) satisfies

$$\begin{aligned} \ddot{\psi }=-\dfrac{mK_{\tau }}{J_{z}}\sqrt{\nu _{x}^{2}+\nu _{y}^{2}+(\nu _{z}+g)^{2}}+b_{\psi }\dot{\phi }\dot{\theta } \nonumber \\ -\frac{a_{\psi }}{J_{z}}\dot{\psi }+d_{\psi }-\dfrac{2K_{\tau }J_{y}}{LJ_{z}}(\frac{a_{\theta }}{J_{y}}\dot{\theta }-b_{\theta }\dot{\phi }\dot{\psi }-\hat{\eta }_{\theta }+\ddot{\theta }_{\star }). \end{aligned}$$

(46)

Taking into account that the angular velocities and the disturbances are bounded, (46) satisfies

$$\begin{aligned} ||\ddot{\psi }||_{f_1}\le L_{\psi _1}+\left( \dfrac{2K_{\tau }J_{y}b_{\theta }\delta _{\phi }}{LJ_{z}}-\frac{a_{\psi }}{J_{z}}\right) \dot{\psi }, \end{aligned}$$

(47)

with $L_{\psi _1}=b_{\psi }\delta _{\phi }\delta _{\theta }+D_{\psi }-\dfrac{mK_{\tau }}{J_{z}}L_{\nu }+\dfrac{2K_{\tau }}{LJ_{z}}(a_{\theta }\delta _{\theta }+J_{y}(D_{\theta }+\delta _{\theta \star }))$. Subsequently, the solution of (47) satisfies

$$\begin{aligned} ||\dot{\psi }||_{f_1}\le ||\dot{\psi }(t_{f_{1}})||_{f_1}e^{\left( \frac{2K_{\tau }J_{y}}{LJ_{z}}b_{\theta }\delta _{\phi }-\frac{a_{\psi }}{J_{z}}\right) \left( t-t_{f_{1}}\right) } +L_{\psi 1}\left( 1-e^{\left( \frac{2K_{\tau }J_{y}}{LJ_{z}}b_{\theta }\delta _{\phi }-\frac{a_{\psi }}{J_{z}}\right) \left( t-t_{f_{1}}\right) }\right) , \end{aligned}$$

for all $t\ge t_{f_1}$. Therefore, if (32) holds, i.e., $\delta _{\phi }<\frac{LJ_{z}a_{\psi }}{2K_{\tau }J_{y}J_{z}b_{\theta }}$, for all $t\ge t_{f_1}$, then it is clear that

$$\begin{aligned} \lim _{t\rightarrow \infty }||\dot{\psi }||_{f_1}\le L_{\psi 1}. \end{aligned}$$

The same result is obtained if the second rotor fails and similar conclusions can be obtained considering a failure in the other rotors. This concludes the proof.$\Box$