# High-Performance Flux Tracking Controller for Reluctance Actuator
## 5 Conclusions and Discussion
### Conflicts of Interest:  
The authors declare no conflict of interest. **Appendix A. A Convergence Analysis of Disturbance Observer Theorem A1 [37]. _For the disturbance observer (10), there is a positive constant m, if \(|d(k+1)-d(k)|<m\), then there is \(k_{0}\), when \(k>k_{0}\), we can draw a conclusion that \(\overset{\sim}{d}(k)<m/g\), where, \(0<g<1\)._  
Proving: First, decompose \(\overset{\sim}{d}(k)\) into:  
\[\overset{\sim}{d}(k)=\overset{\sim}{d}_{1}(k)+\overset{\sim}{d}_{2}(k)\] (A1)  
Setting \(\overset{\sim}{d}_{1}(0)=0\), and \(\overset{\sim}{d}_{2}(0)=\overset{\sim}{d}_{1}(0)\), from the expression (A1), we can see that  
\[\overset{\sim}{d}(k+1)=\overset{\sim}{d}_{1}(k+1)+\overset{\sim}{d}_{2}(k+1)\] (A2)  
According to Equation (12), it can be obtained  
\[\overset{\sim}{d}(k+1)=(1-g)\overset{\sim}{d}_{2}(k)\] (A3)  
Next, to prove \(\overset{\sim}{d}_{1}(k)<m/g\), induction is applied and the details are as follows,  
1. When \(k=0\), \(\overset{\sim}{d}_{1}(0)=0<m/g\)
2. Assuming \(\left|\overset{\sim}{d}_{1}(k)\right|<m/g\), and combining the Equation (A2) and \(0<g<1\), when \(k+1\), we can see that, \[ \begin{array}{rl}\left|\overset{\sim}{d}_{1}(k+1)\right|&\leq(1-g)\left| \overset{\sim}{d}_{1}(k)\right|+|d(k+1)-d(k)|\\ &\hskip 142.26378 pt<(1-g)\frac{m}{g}+m=\frac{m}{g}\end{array} \] (A4) From what has been mentioned, we can draw a conclusion that \[\left|\overset{\sim}{d}_{1}(k)\right|<\frac{m}{g},\;k\geq 0\] (A5) By considering Equation (A3) and \(0<1-g<1\), we obtain \[\overset{\sim}{d}_{2}(k+1)=(1-g)\overset{\sim}{d}_{2}(k)\leq(1-g)\left| \overset{\sim}{d}_{2}(k)\right|<\overset{\sim}{d}_{2}(k)\] (A6) Thus, it is obvious that \(\overset{\sim}{d}_{2}(k)\) is a decreasing function, that is, there is \(k_{0}^{\prime}\), when \(k>k_{0}^{\prime}\), \(\overset{\sim}{d}_{2}(k)\) can be any small. Eventually, we can verify a result as shown in Equation (A7). Specifically, there is \(k_{0}\) and when \(k>k_{0}\), we obtain  
\[\left|\overset{\sim}{d}(k)\right|=\left|\overset{\sim}{d}_{1}(k)+\overset{ \sim}{d}_{2}(k)\right|\leq\left|\overset{\sim}{d}_{1}(k)\right|+\left| \overset{\sim}{d}_{2}(k)\right|<\frac{m}{g}\] (A7)  
Appendix B Stability Analysis Theorem A2 [37]. _For the disturbance observer (10), the stability conditions of the closed loop system are as follows:_  
1. \(0<q<1,0<g<1\)There is a positive constant \(m\), \(|d(k+1)-d(k)|<m\)
3. \(0<C^{T}B<\eta\) The detailed proof is as follows: Set \(v(k)=C^{T}B\frac{m}{g}d(k)\), and obtain \[|v(k)|<C^{T}B\frac{m}{g}<\eta\] (100) That is, \[-C^{T}B\frac{m}{g}<v(k)<C^{T}B\frac{m}{g}\] (101) Equation (12) can be written as \[s(k+1)=qs(k)-\eta\text{sgn}(s(k))+v(k)\] (102) Then, we will analyze the stability of the system from the following four aspects: Case 1: \(s(k)\geq C^{T}B\frac{m}{g}+\eta>0\) \[s(k+1)-s(k)=(q-1)s(k)-\eta+v(k)<0\] (103) \[ \begin{array}{rl}s(k+1)+s(k)&=(q+1)s(k)-\eta+v(k)\\ &\geq(q+1)(C^{T}B\frac{m}{g}+\eta)-\eta+v(k)\\ &=q(C^{T}B\frac{m}{g}+\eta)+C^{T}B\frac{m}{g}+v(k)>0\end{array} \] (104) Obviously, \[s(k+1)^{2}<s(k)^{2}\] (105) Case 2: \(s(k)\leq-C^{T}B\frac{m}{g}-\eta<0\) \[s(k+1)-s(k)=(q-1)s(k)+\eta+v(k)>0\] (106) \[ \begin{array}{rl}s(k+1)+s(k)&=(q+1)s(k)+\eta+v(k)\\ &<s(k)+\eta+v(k)\\ &\leq-C^{T}B\frac{m}{g}-\eta+\eta+v(k)\\ &=C^{T}B\frac{m}{g}+v(k)<0\end{array} \] (107) Then, \[s(k+1)^{2}<s(k)^{2}\] (108) Case 3: \(0<s(k)<C^{T}B\frac{m}{g}+\eta\) \[s(k+1) =qs(k)-\eta+v(k)<q(C^{T}B\frac{m}{g}+\eta)-\eta+v(k)\] (109) \[ \begin{array}{rl}s(k+1)&=(C^{T}B\frac{m}{g}+\eta)<C^{T}B\frac{m}{g}+ \eta\\ \end{array} \] (110) Then, \[|s(k+1)|<C^{T}B\frac{m}{g}+\eta\] (111) Case 4: \(-C^{T}B\frac{m}{g}-\eta<s(k)<0\) \[ \begin{array}{rl}s(k+1)&=qs(k)+\eta+v(k)>s(k)+\eta+v(k)\\ &>-C^{T}B\frac{m}{g}-\eta+\eta+v(k)>-C^{T}B\frac{m}{g}-\eta\\ \end{array} \] (111) \[ \begin{array}{rl}s(k+1)&=qs(k)+\eta+v(k)<\eta+v(k)<C^{T}B\frac{m}{g}+ \eta\\ \end{array} \] (112)Then,  
\[|s(k+1)|<C^{T}B\frac{m}{g}+\eta \tag{10}\]  
From what has been mentioned, we can draw a conclusion as follows,  
1.