# Bounded Attitude Control with Active Disturbance Rejection Capabilities for Multirotor UAVs
## Appendix B Proof of Proposition 2  
Proof. : Consider the candidate Lyapunov function \(V:\mathbb{S}^{3}\times\mathbb{R}^{3}\to\mathbb{R}\), which is positive definite for \(\mathbb{S}^{3}\times\mathbb{R}^{3}\setminus(1\,0^{T}\,0^{T})^{T}\)  
\[ \begin{split} V&=\frac{1}{2}\omega_{e}^{T}\mathbf{ J}\omega_{e}+\kappa((1-\bar{q}_{0})^{2}+\tilde{\mathbf{q}}_{v}^{T}\tilde{ \mathbf{q}}_{v})\\ &=\frac{1}{2}\omega_{e}^{T}\mathbf{J}\omega_{e}+2\kappa(1-\bar{q }_{0})\end{split} \] (A7)  
where \(\mathbf{J}\) is defined as before, and \(\kappa>0\) must be determined. The derivative of (A7) after using (10) is given by  
\[ \begin{split}\dot{V}&=\mathbf{\omega}_{e}^{T}\mathbf{ J}\omega_{e}-2\kappa\dot{\bar{q}}_{0}\\ &=\mathbf{\omega}_{e}^{T}(\mathbf{I}+\xi)+\kappa\tilde{\mathbf{q}}_{ v}^{T}\mathbf{\omega}_{e}\\ &=\underbrace{\omega_{e_{1}}(\Gamma_{1}+\xi_{1})+\kappa\tilde{q}_{ 1}\omega_{e_{1}}}_{\dot{V}_{1}}+\underbrace{\omega_{e_{2}}(\Gamma_{2}+\xi_{2} )+\kappa\tilde{q}_{2}\omega_{e_{2}}}_{\dot{V}_{2}}+\underbrace{\omega_{e_{3}}( \Gamma_{3}+\xi_{\bar{q}_{3}})+\kappa\tilde{q}_{3}\omega_{e_{3}}}_{\dot{V}_{3}} \end{split} \] (A8)  
\(\dot{V}\) is the sum of the three terms \((\dot{V}_{1},\dot{V}_{2},\dot{V}_{3})\). First \(\dot{V}_{1}\) is analyzed. From \(\Gamma_{1}\) in (16) and Equation (A8), the following is obtained:  
\[\dot{V}_{1}=\omega_{e_{1}}\big{(}-\sigma_{M_{2}}\big{(}\hat{\xi}_{1}+\sigma_{ M_{12}}(\tilde{\lambda}_{1}[\omega_{e_{1}}+\rho_{1}\tilde{q}_{1}])\big{)}+\xi_{1} \big{)}+\tilde{q}_{1}\omega_{e_{1}}\] (A9)  
If we choose \(K_{0}<M_{12}-M_{11}\), \(\sigma_{M_{12}}\) is always operating in its linear region so \(\dot{V}_{1}\) becomes  
\[\dot{V}_{1}=-\omega_{e_{1}}\sigma_{M_{12}}(\tilde{\lambda}_{1}[\omega_{e_{1}} +\rho_{1}\tilde{q}_{1}])+\kappa\tilde{q}_{1}\omega_{e_{1}}+\omega_{e_{1}}( \underbrace{\xi_{1}-\hat{\xi}_{1}}_{\alpha_{1}})\] (A10)  
Assuming that \(|\omega_{e_{1}}|>2\rho_{1}\), that is, \(\omega_{e_{1}}\in]2\rho_{1},+\infty[\). Since \(|\bar{q}_{1}|\leq 1\), it follows that \(|\omega_{e_{1}}+\rho_{1}\tilde{q}_{1}|\geq\rho_{1}+\epsilon\) for any \(\epsilon>0\) that is sufficiently small. Therefore, \(\omega_{e_{1}}+\rho_{1}\tilde{q}_{1}\) has the same sign as \(\omega_{e_{1}}\). From Proposition 1, the error \(\Delta_{1}\) is bounded, and this bound will be denoted by \(\bar{r}\). From Equation (A10) and the norm condition on the quaternion, \(\dot{V}_{1}\) takes the following form:  
\[ \begin{split}\dot{V}_{1}&=-\omega_{e_{1}}\sigma_{M_ {12}}(\tilde{\lambda}_{1}[\omega_{e_{1}}+\rho_{1}\tilde{q}_{1}])+\kappa\tilde{ q}_{1}\omega_{e_{1}}+\omega_{e_{1}}\Delta_{1}\\ &\leq-|\omega_{e_{1}}|\sigma_{M_{11}}(\tilde{\lambda}_{1}(\rho_{1 }+\epsilon))+\kappa|\omega_{e_{1}}|+\bar{r}|\omega_{e_{1}}|\end{split} \] (A11)  
Taking  
\[\kappa+\bar{r}<\min(M_{11},\tilde{\lambda}_{1}\rho_{1}+\epsilon)\] (A12)one can assure the decrease of \(V_{1}\), i.e., \(\dot{V}_{1}<0\). Consequently, \(\omega_{e_{1}}\) enters \(\Phi_{1}=\{\omega_{e_{1}}:|\omega_{e_{1}}|\leq 2\rho_{1}\}\) in finite time \(t_{1}\) and remains in it thereafter. In this case, \((\omega_{e_{1}}+\rho_{1}\tilde{q}_{1})\in[-3\rho_{1},3\rho_{1}]\). Let \(M_{11}\) verify the next inequality \(M_{11}\geq 3\tilde{\lambda}_{1}\rho_{1}\). Equation (A12) then becomes:  
\[\kappa+\bar{r}<\tilde{\lambda}_{1}\rho_{1}+\epsilon\] (A13)  
For \(t_{2}>t_{1}\), the argument of \(\sigma_{M_{11}}\) will be bounded as follows  
\[|\tilde{\lambda}_{1}(\omega_{e_{1}}+\rho_{1}\tilde{q}_{1})|\leq 3\tilde{ \lambda}_{1}\rho_{1}\leq M_{11}\] (A14)  
Consequently, \(\sigma_{M1}\) operates in a linear region  
\[\Gamma_{1}=-\tilde{\lambda}_{1}[\omega_{e_{1}}+\rho_{1}\tilde{q}_{1}]\] (A15)  
As a result, (A11) becomes  
\[ \begin{split}\dot{V}_{1}&=-\tilde{\lambda}_{1} \omega_{e_{1}}^{2}-\tilde{\lambda}_{1}\rho_{1}\omega_{e_{1}}\tilde{q}_{1}+ \kappa\omega_{e_{1}}\tilde{q}_{1}+\Delta_{1}\omega_{e_{1}}\\ &=-\tilde{\lambda}_{1}\omega_{e_{1}}^{2}-(\underbrace{\tilde{ \lambda}_{1}\rho_{1}-\kappa}_{k_{1}})\omega_{e_{1}}\tilde{q}_{1}+\Delta_{1} \omega_{e_{1}}\end{split}