## 3 Aggressive Control of Quadrotors with Extended Finite-Time Observer
### Extended Finite-Time Observer
The Equation (21) can be rewritten as \(\hat{x}_{i}^{d}=A_{o}\hat{x}_{i}^{d}-B_{o}(X_{i}^{d})^{(5)}\), where \(|(X_{i}^{d})^{(5)}|<\rho\) is bounded, since \(X_{i}^{d}(t)\in C^{5}(\mathbb{R}^{+},\mathbb{R}^{2})\). Choose an invertible matrix \(M\), which transforms \(A_{o}\) to Jordan standard form \(J_{o}=MA_{o}M^{-1}\). We process the variable transformation \(\epsilon=M\hat{x}_{i}^{d}\), then we have \(\dot{\epsilon}=J_{o}\epsilon-MB_{o}(X_{i}^{d})^{(5)}\). The Lyapunov function candidate is chosen by \(V(\epsilon)=\frac{1}{2}\epsilon^{T}\epsilon\). Its derivative yields, \(\dot{V}(\epsilon)=\epsilon^{T}J_{o}\epsilon-\epsilon^{T}MB_{o}(X_{i}^{d})^{(5)}\). We suppose that the Jordan matrix \(J_{o}\) has \(l\) Jordan blocks, the size of the blocks is \(m_{i}\), \(i=\{1,\ldots,l\}\). The eigenvalue of each block is represented by \(\lambda_{i}(I_{o})\). The Lyapunov function candidate can be rewritten by  
\[V(\epsilon)=\sum_{i=1}^{m_{1}}\frac{1}{2}\epsilon_{i}^{T}\epsilon_{i}+\cdots+ \sum_{j=m_{1}+\cdots+m_{l-1}+1}^{m_{1}+\cdots+m_{l}}\frac{1}{2}\epsilon_{i}^{T} \epsilon_{i}=\sum_{i=1}^{l}V_{i}(\epsilon)\]  
The quadratic term \(\epsilon^{T}J_{o}\epsilon\) in \(\hat{V}(\epsilon)\) yields,  
\[\epsilon^{T}J_{o}\epsilon=\lambda_{1}(J_{o})\sum_{j=1}^{m_{1}}\epsilon_{j}^{2 }+\sum_{j=1}^{m_{1}-1}\epsilon_{j}\epsilon_{j+1}+\ldots+\lambda_{l}(J_{o}) \sum_{j=m_{1}+\cdots+m_{l-1}+1}^{m_{1}+\cdots+m_{l}}\epsilon_{j}^{2}+\sum_{j=m _{1}+\cdots+m_{l-1}+1}^{m_{1}+\cdots+m_{l-1}-1}\epsilon_{j}\epsilon_{j+1} \tag{23}\]  
where the intersection term \(\sum_{j=m_{1}+\cdots+m_{l}-1}^{m_{1}+\cdots+m_{l}-1}\epsilon_{j}\epsilon_{j+1}\) is null, when \(m_{i}=1\). We note that \(\sum_{j=m_{1}+\cdots+m_{l-1}+1}^{m_{1}+\cdots+m_{l-1}+1}\epsilon_{j}^{2}-\sum_ {j=m_{1}+\cdots+m_{l-1}+1}^{m_{1}+\cdots+m_{l-1}+1}\epsilon_{j}\epsilon_{j+1}= (1/2)\epsilon_{m_{1}+\cdots+m_{l-1}+1}^{2}+(1/2)\sum_{j=m_{1}+\cdots+m_{l-1}+1 }^{m_{1}+\cdots+m_{l-1}-1}(\epsilon_{j}-\epsilon_{j+1})^{2}+(1/2)\epsilon_{m_ {1}+\cdots+m_{l}}^{2}\geq 0\). Thus, \(\sum_{j=m_{1}+\cdots+m_{l-1}+1}^{m_{1}+\cdots+m_{l-1}+1}\epsilon_{j}^{2}\geq \sum_{j=m_{1}+\cdots+m_{l-1}+1}^{m_{1}+\cdots+m_{l-1}+1}\epsilon_{j+1}\), i.e., \(\sum_{j=m_{1}+\cdots+m_{l-1}+1}^{m_{1}+\cdots+m_{l-1}-1}\epsilon_{j}\epsilon_{j+ 1}\leq 2V_{i}(\epsilon)\). Add both side from \(i=1\) to \(l\), then we have  
\[\sum_{j=1}^{m_{1}-1}\epsilon_{j}\epsilon_{j+1}+\ldots+\sum_{j=m_{1}+\cdots+m_ {l-1}+1}^{m_{1}+\cdots+m_{l-1}-1}\epsilon_{j}\epsilon_{j+1}\leq 2V(\epsilon) \tag{24}\]  
According to (23) and (24), we have  
\[\epsilon^{T}J_{o}\epsilon\leq\sum_{i=1}^{l}2\lambda_{i}(J_{o})V_{i}(\epsilon) +2V(\epsilon)\leq 2(1-\gamma)V(\epsilon) \tag{25}\]  
Then, the derivative of the Lyapunov function candidate yields,  
\[\hat{V}(\epsilon)=\epsilon^{T}J_{o}\epsilon-\epsilon^{T}MB_{o}(X_{i}^{d})^{(5) }\leq 2(1-\gamma)V(\epsilon)-\epsilon^{T}MB_{o}(X_{i}^{d})^{(5)}\]  
Therefore,  
\[\hat{V}(\epsilon) \leq 2(1-\gamma)V(\epsilon)+\bar{\lambda}|(X_{i}^{d})^{(5)}|\| \epsilon\|=(1-\gamma)\|\epsilon\|^{2}+\bar{\lambda}\rho\|\epsilon\|=[(1- \gamma)\|\epsilon\|+\bar{\lambda}\rho]\|\epsilon\| \tag{26}\] \[=\sqrt{2}[(1-\gamma)\|\epsilon\|+\bar{\lambda}\rho]\,V^{1}( \epsilon)=-\sqrt{2}[(\gamma-1)\|\epsilon\|-\bar{\lambda}\rho]\,V^{1}(\epsilon)\]  
When \(\|\vec{x}_{i}^{d}\|\geq\beta\), then, \(\|\epsilon\|=\|M\vec{x}_{i}^{d}\|\geq\underline{\lambda}\|\vec{x}_{i}^{d}\| \geq\underline{\lambda}\beta\). Note that \(\gamma>1\), thus, \((\gamma-1)\|\epsilon\|-\bar{\lambda}\rho\geq(\gamma-1)\underline{\lambda}\beta -\bar{\lambda}\rho\). Thus, \(\hat{V}(\epsilon)\leq-\sqrt{2}[(\gamma-1)\underline{\lambda}\beta-\bar{ \lambda}\rho]V^{1}(\epsilon)\). Hence, \(\hat{V}(\epsilon)\) is negative definite, since \(\beta>\frac{3\rho}{\underline{\lambda}(\gamma-1)}\). In addition, \(\hat{V}(\epsilon)=0\), if and only if \(V(\epsilon)=0\). Note that \(V(\epsilon)=\frac{1}{2}\epsilon^{T}\epsilon=0\), only if \(\epsilon=0\) which is equivalent to that the observer error \(\vec{x}_{i}^{d}=0\). Therefore, the observer error is asymptotic stable.