## 4 Results and Discussion
### Flight Control System Assessment of a Single Flight  
Take the PDF calculation of the flap channel's main hydraulic pressure as an example. First, calculate the \(\mu\), \(\sigma^{2}\), \(C_{s}\) and \(C_{k}\) of the main hydraulic pressure, as shown in Table 3.  
Since the main hydraulic pressure distribution is single-peak and the skewness is not zero, the second extended \(\lambda-PDF\) is used to solve the unknown parameters. According to the equations listed in Section 3, we can obtain:  
\[\lambda=40;b_{0}=2.216\times 10^{3};b_{1}=-51.59;b_{2}=-0.1577, \tag{24}\]  
Then, the PDF of the main hydraulic pressure and its frequency histogram are obtained, as shown in Figure 4.  
Figure 4: PDF of the main hydraulic pressure. \begin{table}
\begin{tabular}{c c c c c c c c} \hline \hline
**Time** & **MHP/psi** & **AP/psi** & **AD/\({}^{\circ}\)** & **RD/\({}^{\circ}\)** & **ED/\({}^{\circ}\)** & **PP/mm** & **SCP/mm** \\ \hline
3:30:37 & 1887.207 & 1881.104 & \(-1.230\) & \(-0.469\) & \(-6.914\) & 3.795 & 30.537 \\
3:30:38 & 1899.414 & 1893.311 & \(-1.230\) & \(-0.469\) & \(-6.914\) & 4.241 & 30.537 \\
3:30:39 & 1905.518 & 1905.518 & \(-1.230\) & \(-0.469\) & \(-6.855\) & 4.018 & 30.537 \\
3:30:40 & 1923.828 & 1917.725 & \(-1.230\) & \(-0.469\) & \(-6.914\) & 4.241 & 30.537 \\
3:30:41 & 1936.035 & 1929.932 & \(-1.230\) & \(-0.469\) & \(-6.914\) & 4.018 & 30.537 \\
3:30:42 & 1948.242 & 1942.139 & \(-1.230\) & \(-0.469\) & \(-6.855\) & 4.018 & 30.537 \\ \(\cdots\) & \(\cdots\) & \(\cdots\) & \(\cdots\) & \(\cdots\) & \(\cdots\) & \(\cdots\) & \(\cdots\) \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Raw flight data (partly). \begin{table}
\begin{tabular}{c c} \hline \hline
**Statistics** & **Value** \\ \hline Mean, \(\mu\) & 2216.5 \\ Variance, \(\sigma^{2}\) & 32.0413 \\ Skewness, \(C_{s}\) & \(-0.177\) \\ Kurtosis, \(C_{k}\) & 2.8504 \\ \hline \hline \end{tabular}
\end{table} 
Table 3: Statistics of the main hydraulic pressure. With these PDF, the MCS method is used to conduct 10,000 simulations to generate 10,000 assessment input parameters with the same distribution. The next step is to calculate the health index of the flight control system. According to the theory in Section 2, the system will be divided into four channels: pitch, roll, yaw and flap, and their health status can be divided into four levels: healthy, slightly decline, severe decline and fault. Firstly, the relative degradation degree of the generated 10,000 input parameters is normalized. The health index of each channel is obtained by constructing the membership function, and then the weight of each channel relative to the flight control system is obtained through rough set reduction. Finally, the health index of all channels is weighted to obtain the whole system's health index. According to the maximum entropy, the first four-order statistical moments of this health index are counted firstly in Table 5:  
Then, the PDF of the health index is obtained according to the maximum entropy:  
\[ \begin{array}{rl}f(y)&=\exp\{1.4951+2.9369\times(y-0.7934)-58.9372\times(y-0.7934)^{2}\\ &\qquad-134.3976\times(y-0.7934)^{3}-250.9652\times(y-0.7934)^{4}\}\end{array}  \tag{25}\]  
The PDF and the original data histogram are drawn in Figure 8. It can be seen from the figure that the health index is mostly distributed between 0.75 and 1. Then, we can get the integration of PDF in (0.75, 1) as 0.7088, that is, the probability that the flight control system is in a healthy state is 70.88%, In the same way, the integration in (0, 0.25), (0.25, 0.5) and (0.5, 0.75) is 1.6338\(\times 10^{-10}\), 0.0016 and 0.2896, respectively, which indicates that the probability of fault state, severe decline state and slightly decline state is 0%, 0.16% and 28.96%. These results show that the flight control system in this flight is likely to be healthy, and does not require maintenance.