## 3 Research Methodology
### System Modeling
#### 3.2.5 Controller Design  
To formulate the architecture of the observer, [PERSON] et al. have provided the mathematical model in [32]. We are using the same model to explain the observer's design. \[\dot{x}=Ax+Bu \tag{21}\]  
\[y=Cx+Du \tag{22}\]  
\[\dot{\overline{x}}=A\overline{x}+Bu \tag{23}\]  
\[\overline{y}=C\overline{x} \tag{24}\]  
where \"\(x\)\" represents the state vector, \"\(u\)\" represents the input vector, and \"\(y\)\" represents the output vector. The system matrices are \(A\), \(B\), \(C\), and \(D\). \[(\dot{\overline{x}}-\dot{x})=A(\overline{x}-x) \tag{25}\]  
\[(\overline{y}-y)=C(\overline{x}-x) \tag{26}\]  
\[\dot{\overline{x}}=A\overline{x}+Bu+L(\overline{y}-y) \tag{27}\]  
In (27), \(L\) is the feedback gain. \[\dot{\overline{x}}-\dot{x}=A(\overline{x}-x)+L(\overline{y}-y) \tag{28}\]  
\[(\overline{y}-y)=C(\overline{x}-x) \tag{29}\]  
\[\dot{\overline{x}}-\dot{x}=(A+LC)(\overline{x}-x) \tag{30}\]  
\[y=Cx+Du \tag{22}\]0  
\[y=Cx+Du \tag{22}\]1  
FDI does not declare any fault when \"\(e_{x}\)\" goes to zero. If \"\(e_{x}\)\" is out of bound, an error is identified, and the faulty value will be replaced. The complex structure of AFTCS, and its slow response time due to excessive computations, are two of its main drawbacks [33]. The nonlinear observer design equation we obtain from (27) follows:  
\[y=Cx+Du \tag{22}\]2  
where \"\(g(x,\,u,t)\)\" is a nonlinear function and assumed to be globally Lipschitz. \(\overline{L}\) is the gain for the nonlinear observer. Let \(e_{x}(t)\) be the error vector:  
\[y=Cx+Du \tag{22}\]3  
From Equations (33) and (34), we obtain:  
\[y=Cx+Du \tag{22}\]4  
**Lemma 1. **_The error \(e_{x}(t)\) asymptotically approaches 0 if we can obtain a matrix \(R\), \(X\), and scalar \(\mu\) such that \(R=R^{T}>0\) and \(\mu>0\) in order to meet the linear matrix inequality (LMI):_  
\[ \begin{bmatrix}RA+A^{T}R+XC+C^{T}X^{T}+\mu\lambda^{2}I&R\\ R&-\mu I\end{bmatrix} <0 \tag{36}\]  
_The reliability of each sensor is denoted by \(R\)._  
The following equation can be used to select the observer gain matrix:  
\[\overline{L}=R^{-1}X \tag{37}\]  
We can validate the choice of observer gain matrix by evaluating the following Lyapunov function and proving its derivative to be zero:  
\[V(t)=e_{x}^{T}(t)Re_{x}(t) \tag{38}\]  
Next, we need to check \(\dot{V}(x)<0\forall xeD-\{0\}\), as listed below:  
\[ \begin{array}{l}\dot{V}(t)=e_{X}^{T}\big{(}RA+R\overline{L}C+A^{T}R+C^{T}L^{ -T}R\big{)}e_{x}+2e_{x}^{T}R(g(\overline{x}\mu,k)-g(x,u,k)\big{)}\\ \leq e_{X}^{T}\big{(}RA+R\overline{L}C+A^{T}R+C^{T}L^{-T}R\big{)}e_{x}+1/\mu e _{x}^{T}R^{2}e_{x}+\mu\|g(\overline{x}\mu,k)-g(x,u,k)\|^{2}\\ \leq e_{X}^{T}\big{(}RA+R\overline{L}C+A^{T}R+C^{T}L^{-T}R\big{)}e_{x}+1/\mu e _{x}^{T}R^{2}e_{x}+\mu\lambda^{2}\|e_{x}\|^{2}\\ =e_{X}^{T}\big{(}RA+R\overline{L}C+A^{T}R+C^{T}L^{-T}R\big{)}+\mu\lambda^{2}I+ 1/\mu R^{2}\big{)}e_{x}\end{array}  \tag{39}\]  
Substituting (38) in (39), we obtain:  
\[\dot{V}(t)\leq e_{X}^{T}\big{(}\big{(}RA+XC+A^{T}R+C^{T}X^{T}\big{)}+\mu \lambda^{2}I+1/\mu R^{2}\big{)}e_{x} \tag{40}\]  
If the inequality is given in (40) exits, \(e_{x}\) asymptotically approaches 0. \[(RA+XC+A^{T}R+C^{T}X^{T}+\mu\lambda^{2}I+1/\mu R^{2})<0 \tag{41}\]  
This inequality becomes equal to (41) by applying the Schur complement, which completes the proof. **Theorem 1. **_The error \(e_{x}(t)\) approaches to zero exponentially with rate \(k/2\) if there exist matrices \(R\), \(X\), and scalars \(\mu\), \(k\) such that \(R=R^{T}>0\) and \(\mu,k>0\) to satisfy the following:_  
\[ \begin{bmatrix}RA+A^{T}R+XC+C^{T}X^{T}+\mu\lambda^{2}I+kR&R\\ R&-\mu I\end{bmatrix} <0 \tag{42}\]  
To prove this, consider (40) and (42), which give  
\[\dot{V}(t)\leq-k{e_{x}^{T}}Re_{x}=-kV(t) \tag{43}\]  
Hence, we can write  
\[V(t)\leq e_{x}^{T}V(0) \tag{44}\]From (38) we obtain  
\[\lambda_{min}\ (R)\ \|e_{x}(t)\|^{2}\ \leq e^{-kt}\lambda_{max}\ (R)\ \|e_{x}(0)\|^{2} \tag{45}\]  
where \(\lambda_{min}\) and \(\lambda_{max}\) are the minimum and maximum eigenvalues of \(R\), respectively.