# Fault Diagnosis in Partially Observable Petri Nets with Quantum Bayesian Learning
## 4 Case Analysis and Verification
### Fault Diagnosis Results Based on QBPN  
The application of the quantum Bayesian inference in the QBPN model is essential to the fault probability of unobservable fault transitions calculation, which reflects the uncertainty of fault occurrence and the conditions of the quantum interference effect. Based on the POPN system model with an uncertain state, for the fault probability of the system estimation, the use of the QBPN fault diagnosis algorithm is vital. Table 4 shows the diagnosis results of the QBPN algorithm. In the test, we diagnose the fault transition \(t_{10}\) and fault transition \(t_{11}\). The observable transitions in the QBPN model are \(t_{0},t_{3},t_{4},t_{12},t_{14},t_{15}\). When the observable label probability transition sequence \(\sigma_{o}=(t_{0},1)(t_{4},0.883)(t_{3},0.926)\), the transition \(t_{10}\) is not in the probability transition set. The fault firing probability of \(t_{10}\) is 0. Its confidence is 0 and the fault state of the system is 0. So the system state is faultless. When \(\sigma_{o}=(t_{0},1)(t_{4},0.883)(t_{3},0.926)(t_{15},0.879)\), by observable transitions \(t_{0}t_{3}t_{4}\), the fault probability of \(t_{10}\) is \(Pr(t_{f}^{10}|t_{0},t_{4},t_{3})=0.1621\). According to the firing state of observ \begin{table}
\begin{tabular}{c c c c c c c} \hline \hline \(\sigma_{o}\) & \
\begin{tabular}{c} **Fault Transition Firing** \\ **Probability** \\ \end{tabular} & \(\theta\) & **QB Function** & \(F_{b}\) & \(Pr(\sigma_{o}\),\(\psi\),\(T_{f}^{*})\) & **W** &
\begin{tabular}{c} **Actual** \\ **System State** \\ \end{tabular} \\ \hline \((t_{0},1)(t_{4},0.883)\) & \(Pr(t_{f}^{10}|t_{0},t_{4}t_{3})\) & \(\pi/2\) & 0 & 0 & 0 & \(N\) & Normal \\ \hline \multirow{4}{*}{\((t_{0},1)(t_{4},0.883)\) \((t_{3},0.926)\)} & \(Pr(t_{f}^{10}|t_{0},t_{4}t_{3})\) & \(\pi/2\) & 0.1621 & & & & \\ \cline{2-8}  & \(Pr(t_{f}^{10}|t_{0},t_{4}t_{3})\) & 0 & 0.0471 & & & & \\ \cline{2-8}  & \(Pr(t_{f}^{11}|t_{0},t_{4}t_{3})\) & \(\pi/2\) & 0.1586 & & 0.0471 & \(N\) & Normal \\ \cline{2-8}  & \(Pr(t_{f}^{11}|t_{0},t_{4}t_{3})\) & 0 & 0.0076 & & & & \\ \hline \multirow{4}{*}{\((t_{0},1)(t_{4},0.883)\) \((t_{3},0.926)(t_{14},0.851)\) \((t_{15},0.879)\)} & \(Pr(t_{f}^{11}|t_{0},t_{4}t_{3},t_{14},t_{15})\) & 0 & 0.9263 & & & \\ \cline{2-8}  & \(Pr(t_{f}^{10}|t_{0},t_{4}t_{3},t_{14},t_{15},t_{12})\) & 0 & 0.9668 & & 0.9668 & \(F\) & Fault \\ \cline{2-8}  & \(Pr(t_{f}^{11}|t_{0},t_{4}t_{3})\) & \(\pi/2\) & 0.1586 & & & & \\ \cline{2-8}  & \(Pr(t_{f}^{11}|t_{0},t_{4},t_{3},t_{14},t_{15},t_{12})\) & 0 & 0.0006 & & & & \\ \hline \((t_{0},1)(t_{3},0.907)\) \((t_{13},0.736)\) & \(Pr(t_{f}^{11}|t_{0},t_{3},t_{13})\) & \(\pi/2\) & 1 & 1 & 1 & \(F\) & Fault \\ \hline \hline \end{tabular}
\end{table} 
Table 4: Fault diagnosis results of the quantum Bayesian fault algorithm based on POPN. Figure 11: Interference angle \(\{\theta_{3},\theta_{4}\}\) causes the change in the firing probability of \(t_{8}\) under the condition of \(t_{3}\). able transition \(t_{10}\) and Equation (7), we calculate the QB function of \(t_{10}\) and the modified fault probability is \(Pr(t_{f}^{10}|t_{0},t_{4},t_{3},t_{15})=0.0471\). Similarly, for the analysis of fault transition \(t_{11}\), the fault firing probability is \(Pr(t_{f}^{11}|t_{0},t_{4},t_{3},t_{15})=0.0076\). So the fault probability of the system is \(0.0076\) and the confidence is \(0.4\). So the fault state of the system is faultless. When \(\sigma_{o}=(t_{0},1)(t_{4},0.883)(t_{3},0.926)(t_{14},0.851)(t_{15},0.879)(t_ {12},0.463)\). Firstly, calculate the firing probability of fault transition \(t_{10}\). According to Equation (6), the conditional probability of fault transition \(t_{10}\) is \(Pr(t_{f}^{10}|t_{0},t_{4},t_{3})=0.1621\). According to the firing state of observable transition \(t_{14}\) and Equation (7), calculate the QB function. The modified firing probability of fault transition \(t_{10}\) is \(Pr(t_{f}^{10}|t_{0},t_{4},t_{3},t_{14})=0.7569\).