## 3 Iterative Learning Observer-Based Thruster Fault Reconstruction Approach
### Stability Analysis of the Iterative Learning Observer  
In this subsection, the stability and convergence of the proposed ILO was proved. The following is the theorem that was used. **Theorem 1**. : _Suppose that Assumption 1 holds if there exist positive definite symmetric matrices \(\mathbf{P}\in\mathbb{R}^{6\times 6}\), \(\mathbf{Q}_{1}\in\mathbb{R}^{3\times 3}\) and matrices \(\mathbf{L}_{2}\in\mathbb{R}^{3\times 6}\), \(\mathbf{Y}\in\mathbb{R}^{6\times 6}\) and a positive scalar \(\gamma\) such that the following relations hold_  
\[\mathbf{P}\mathbf{A}+\mathbf{A}^{T}\mathbf{P}-\mathbf{Y}\mathbf{C}-\mathbf{C}^{T}\mathbf{Y}^{T}<0 \tag{8}\]  
\[(\mathbf{L}_{2}\mathbf{C})^{T}=\mathbf{P}\mathbf{H} \tag{9}\]  
\[(\eta+\gamma_{1}){\mathbf{Q}_{1}}^{2}-\mathbf{I}_{3}\leq 0 \tag{10}\]  
\[\gamma\geq d_{\max} \tag{11}\]  
_where \(\gamma_{1}>0\), and \(\eta=1+\mu\) with \(\mu\geq 0\). In this way, the state estimation error \(\mathbf{e_{q}}(t)\) and fault-reconstructing error \(\mathbf{e_{F}}(t)\) are uniformly ultimately bounded, and matrix \(\mathbf{L}_{1}\) can be determined by \(\mathbf{L}_{1}=\mathbf{P}^{-1}\mathbf{Y}\)._  
**Proof of Theorem 1. ** Define  
\[\left\{ \begin{array}{l}\Delta\mathbf{f}(t)=\mathbf{F}(t)-\mathbf{Q}_{1}\mathbf{F}(t-\tau)\\ \Delta\mathbf{f}(t)=\mathbf{F}(t)-\mathbf{Q}_{1}\mathbf{F}(t-\tau)\end{array} \right. \tag{12}\]  
then, we have  
\[\Delta\mathbf{f}(t)-\Delta\mathbf{\hat{y}}(t)=\mathbf{e_{F}}(t)-\mathbf{Q}_{1}\mathbf{e_{F}}(t-\tau) \tag{13}\]Using (13), the fault-reconstructing error can be described by  
\[\mathbf{e_{F}}(t)=\mathbf{Q}_{1}\mathbf{e_{F}}(t-\tau)+\Delta\mathbf{f}(t)-\Delta\mathbf{\hat{f}}(t) \tag{14}\]  
Further, (6) can be converted into the following form:  
\[\hat{\mathbf{e}}_{q}(t)=(\mathbf{A}-\mathbf{L}_{1}\mathbf{C})\mathbf{e}_{q}+\mathbf{H}(\mathbf{Q}_{1}\mathbf{e_{ F}}(t-\tau)+\Delta\mathbf{f}(t)-\Delta\mathbf{\hat{f}}(t))+\mathbf{E}\mathbf{d}(t)-\mathbf{\gamma} \text{Csgn}(\mathbf{e}_{q}) \tag{15}\]  
According to (5) and (12), one has  
\[\mathbf{L}_{2}\mathbf{C}\mathbf{e}_{q}=\frac{1}{Q_{2}}[\hat{\mathbf{F}}(t)-\mathbf{Q}_{1}\hat{\mathbf{F }}(t-\tau)]=\frac{1}{Q_{2}}\Delta\hat{f}(t) \tag{16}\]  
Consider the following Lyapunov-Krasovski function candidate  
\[V(t)=\mathbf{e}_{q}^{T}\mathbf{P}\mathbf{e}_{q}+\beta\int_{t-\tau}^{t}\mathbf{e}_{F}^{T}\mathbf{e} _{F}ds \tag{17}\]  
where \(\mathbf{P}\) is a positive definite symmetric matrix and \(\beta>0\). The derivative of the Lyapunov functional with respect to time can be given by  
\[ \begin{array}{l}\dot{V}(t)=\mathbf{e}_{q}^{T}[(\mathbf{A}-\mathbf{L}_{1}\mathbf{C})^{T}\mathbf{P }+\mathbf{P}(\mathbf{A}-\mathbf{L}_{1}\mathbf{C})]\mathbf{e}_{q}+2\mathbf{e}_{q}^{T}\mathbf{P}\mathbf{H}[\mathbf{Q }_{1}\mathbf{e}_{F}(t-\tau)+\Delta\mathbf{f}(t)-\Delta\mathbf{\hat{f}}(t)]\\ -2\gamma\mathbf{e}_{q}^{T}\mathbf{P}\text{C}\text{sgn}(\mathbf{e}_{q})+2\mathbf{e}_{q}^{T}\bm {P}\mathbf{E}\mathbf{d}(t)-\beta\mu\mathbf{e}_{F}^{T}(t)\mathbf{e}_{F}(t)+\beta(1+\mu)\mathbf{e}_{ F}^{T}(t)\mathbf{e}_{F}(t)\\ -\beta\mathbf{e}_{F}^{T}(t-\tau)\mathbf{e}_{F}(t-\tau)\end{array}  \tag{18}\]  
where \(\eta=1+\mu\) with \(\mu>0\). According to (14), it follows that  
\[ \begin{array}{l}\beta\eta\mathbf{e}_{F}^{T}(t)\mathbf{e}_{F}(t)=\beta\eta[\mathbf{e}_{F }^{T}(t-\tau)\mathbf{Q}_{1}{}^{T}\mathbf{Q}_{1}\mathbf{e}_{F}(t-\tau)+\Delta\mathbf{\hat{f}}^{ T}(t)\Delta\mathbf{f}(t)+\Delta\mathbf{\hat{f}}^{T}(t)\Delta\mathbf{\hat{f}}(t)\\ +2\mathbf{e}_{F}^{T}(t-\tau)\mathbf{Q}_{1}{}^{T}\Delta\mathbf{f}(t)-2\Delta\mathbf{\hat{f}}^{T}( t)\Delta\mathbf{f}(t)-2\Delta\mathbf{\hat{f}}^{T}(t)\mathbf{Q}_{1}\mathbf{e}_{F}(t-\tau)] \end{array}  \tag{19}\]  
Using [PERSON]'s inequality [27], one can obtain  
\[2\eta\mathbf{e}_{F}{}^{T}(t-\tau)\mathbf{Q}_{1}{}^{T}\Delta\mathbf{f}(t)\leq\gamma_{1} \mathbf{e}_{F}^{T}(t-\tau)\mathbf{Q}_{1}{}^{T}\mathbf{Q}_{1}\mathbf{e}_{F}(t-\tau)+\frac{\eta^ {2}}{\gamma_{1}}\Delta\mathbf{\hat{f}}^{T}(t)\Delta\mathbf{f}(t) \tag{20}\]  
where \(\gamma_{1}>0\). Substituting (20) into Equation (19) leads to  
\[