## 4 Fault Diagnosis Based on Traditional and Interval Analytical Redundancy Relations  
### Traditional Analytical Redundancy Relations Method  
Considering there is no such thing as an absolutely perfect system without degradation, fault-free modeling of the system accounts for leaky fluid resistances, stuck fluid resistances, etc. To simulate and inject faults, only the pertinent resistive element parameters need to be altered. Minor, general, serious, and deadly faults can be classified based on the degree of degradation of faults, and abrupt and progressive faults based on the degree of progression of faults. Table 3 shows that faults of types A, B, and C are studied in this study. To describe the system's operational state, define the Boolean logic variable \(\beta\):  
\[\beta=[\text{A},\text{B},\text{C}]\quad(\text{A}/\text{B}/\text{C}\,=0\;or\;1) \tag{11}\]  
For example, \(\beta=[0,0,1]\) means that only type C fault exist and no type A and B.  
ARR at the co-potential node (0-node) can be summarized as \"Same potential variable divides the flow variable\", and at the co-flow node (1-node) can be summarized as \"Same flow variable divides the potential variable\":  
\[0-\text{node}:\left\{ \begin{array}{ll}p_{1}=p_{2}=p_{3}=\cdots p_{n}\\ \text{ARR}=\sum\limits_{i=1}^{n}\alpha_{i}Q_{i}\end{array} \right.\qquad i=1,2,3,\cdots n \tag{12}\]  
Figure 13: System diagnostic bond graph considering uncertain interference factors. \[1-\text{node}:\left\{ \begin{array}{ll}Q_{1}=Q_{2}=Q_{3}=\cdots Q_{n}\\ \text{ARR}=\sum\limits_{i=1}^{n}\alpha_{i}p_{i}\end{array} \right.\quad i=1,2,3, \cdots n \tag{13}\]  
where \(n\) is the node's bond excluding the virtual sensor; \(\alpha_{i}\) is the power flow coefficient, for the bond with the half-arrow pointing to the node \(\alpha_{i}=1\) and for the bond with the half-arrow departing from the node \(\alpha_{i}=-1\). According to Figure 5, analytical redundancy relations (ARRs) and residuals \(r_{i}\) are obtained. **ARR1 for node \(0_{(4)}\)**:  
\[0_{(4)}-\text{node}:\left\{ \begin{array}{ll}p_{4}=p_{8}=p_{9}\\ \text{ARR}_{1}=Q_{4}+Q_{8}-Q_{9}=0\end{array} \right. \tag{14}\]  
\[\left\{ \begin{array}{ll}Q_{4}=Q_{2}=Q_{5}=\frac{p_{8}}{k_{PA}}=\frac{p_{2}-p _{4}}{k_{PA}}=\frac{Sp-SSp_{1}}{k_{PA}}\\ Q_{8}=Q_{17}=Q_{21}+Q_{20}-Q_{19}-Q_{18}=SSQ_{1}+\frac{SSQ_{2}}{A_{Rad}}-C_{ Rod}\frac{d(SSp_{3})}{dt}-\frac{SSp_{2}}{k_{Ex-l}}\\ Q_{9}=Q_{10}=Q_{11}=\frac{p_{10}}{k_{AT}}=\frac{p_{9}-p_{11}}{k_{AT}}=\frac{ SSP_{1}-p_{T}}{k_{AT}}\end{array} \right. \tag{15}\]  
\[r_{1}=\frac{Sp-SSp_{1}}{R_{PA}}+SSQ_{1}+\frac{SSQ_{2}}{A_{Rad}}-C_{Rad}\frac{d (SSp_{3})}{dt}-\frac{SSp_{3}}{R_{Ex-l}}-\frac{SSp_{1}-p_{T}}{R_{AT}} \tag{16}\]  
**ARR2 for node \(0_{(5)}\)**:  
\[0_{(5)}-\text{node}:\left\{ \begin{array}{ll}p_{7}=p_{12}=p_{16}\\ \text{ARR}_{2}=Q_{7}-Q_{12}-Q_{16}=0\end{array} \right. \tag{17}\]  
\[\left\{ \begin{array}{ll}Q_{7}=Q_{3}=Q_{6}=\frac{p_{6}}{k_{PB}}=\frac{p_{3}-p _{7}}{k_{PB}}=\frac{Sp-SSp_{2}}{k_{PB}}\\ Q_{12}=Q_{13}=Q_{14}=\frac{p_{13}}{k_{ST}}=\frac{p_{12}-p_{14}}{k_{ST}}=\frac{ SSp_{2}-p_{T}}{k_{ST}}\\ Q_{16}=Q_{22}=Q_{23}+Q_{24}+Q_{25}=SSQ_{1}+A_{Roldess}SSQ_{2}+C_{Roldess}\frac{d(SSp_{4} )}{dt}\end{array} \right. \tag{18}\]  
\[r_{2}=\frac{Sp-SSp_{2}}{R_{PB}}-\frac{SSp_{2}-p_{T}}{R_{BT}}-SSQ_{1}-A_{[PERSON] }SSQ_{2}-C_{[PERSON]}\frac{d(SSp_{4})}{dt} \tag{19}\]  
**ARR3 for node \(1_{(13)}\)**:  
\[1_{(13)}-\text{node}:\left\{ \begin{array}{ll}Q_{23}=Q_{21}=Q_{26}\\ \text{ARR}_{3}=p_{23}-p_{21}-p_{26}=0\end{array} \right. \tag{20}\]  
\[\left\{ \begin{array}{ll}p_{21}=SSp_{3}\\ p_{23}=SSp_{4}\\ p_{26}=R_{ln-l}SSQ_{1}\end{array} \right. \tag{21}\]  
\[r_{3}=SSp_{4}-SSp_{3}-R_{ln-l}SSQ_{1} \tag{22}\]  
**ARR4 for node \(1_{(14)}\)**:  
\[1_{(14)}-\text{node}:\left\{ \begin{array}{ll}Q_{27}=Q_{28}=Q_{29}\\ \text{ARR}_{4}=p_{28}-p_{27}-p_{29}=0\end{array} \right. \tag{23}\]