## 5 Improved Tracking Differentiator (SYSTD)
### Design and Stability Judgment of SYSTD  
The key to constructing a new type of tracking differentiator lies in designing an acceleration function with both linear and nonlinear characteristics. Since sign functions and piece-wise functions can increase switching noise, they are not suitable for application. Therefore, this paper employs elementary functions, which are defined as follows:  
\[sys(x,a_{4},a_{5})=\frac{x}{\sqrt{a_{4}+a_{5}x^{2}}} \tag{19}\]  
where \(a_{4}>0\), \(a_{5}>0\), \(a_{4}\in R\), \(a_{5}\in R\). The following can be deduced:  
\[\left\{ \begin{array}{l}\lim\limits_{x\rightarrow+\infty}sys=\frac{1}{\sqrt{ a_{5}}},\lim\limits_{x\rightarrow-\infty}sys=\frac{-1}{\sqrt{a_{5}}}\\ sys(x)+sys(-x)=0\end{array} \right. \tag{20}\]  
It can be seen that \(sys(x)\) is a smooth and continuous odd function passing through the first and third quadrants, and it is bounded. Its basic characteristics are shown in Figure 5. As can be seen from Figure 5a, the parameter \(a_{4}\) can adjust the curve change rate. The smaller the value of \(a_{4}\), the greater the curve change rate. When \(x>1\) or \(x<-1\), the function curve gradually approaches saturation; when \(-1<x<1\), the function is approximately linearly changing, and, the smaller the value of \(a_{4}\), the larger the slope. It can also be seen from Figure 5a that the parameter \(a_{5}\) can adjust the amplitude of the curve, that is, the saturation boundary. This indicates that the TD based on the \(sys(x)\) function has good tracking and differential characteristics, and the \(sys(x)\) function requires fewer parameters to be adjusted. Therefore, the following second-order system S\({}_{0}\) is constructed:  
Figure 4: Structure diagram of the HCS. \[\left\{ \begin{array}{l}\dot{z}_{1}(t)=z_{2}(t)\\ \dot{z}_{2}(t)=-sys(z_{1}(t),a_{41},a_{51})-sys(z_{2}(t),a_{42},a_{52})\end{array} \right. \tag{21}\]  
To design SYSTD, Lemma 1 needs to be introduced first. **Lemma 1** (2015). : _For a time-invariant system \(S\):_  
\[\left\{ \begin{array}{l}\dot{x}_{1}=x_{2}\\ \dot{x}_{2}=f(x_{1},x_{2})\end{array} \right.\;\;x_{1},x_{2}\in D\subset R^{2} \tag{22}\]  
_If in the real domain \(R^{2}\):_  
_(1) There exists a positive-definite function (or negative-definite function) \(V(x)\) that increases infinitely:_  
_(2) In the entire phase space, the total derivative of \(V(x)\) along the solution of system \(S\) with respect to time \(t\) is always negative (or always positive);_  
_(3) The set of points that satisfy only contains the origin._  
Then the zero-solution of the system is globally asymptotically stable. **Theorem 1**. : _When the parameters \(a_{41}\), \(a_{51}\), \(a_{42}\), and \(a_{52}\) are all greater than zero, the system \(S_{0}\) satisfies the Lyapunov global asymptotic stability condition:_  
\[ \begin{array}{l}\underset{t\rightarrow\infty}{\lim}z_{1}(t)=0\\ \underset{t\rightarrow\infty}{\lim}z_{2}(t)=0\end{array}  \tag{23}\]  
Proof. : First, construct the Lyapunov function as follows:  
\[V(z_{1},z_{2})=\int_{0}^{z_{1}}sys(\xi,a_{4},a_{5})d\xi+\frac{1}{2}z_{2}^{2} \tag{24}\]  
when \(a_{4}>0\) and \(a_{5}>0\), because the function \(sys(\xi,\,a_{4},\,a_{5})\) is an odd function and is monotonically increasing in the first and third quadrants, the variable \(\xi\) and the function \(sys(\xi,a_{4},a_{5})\) always have the same sign. Therefore, the following is true:  
\[\int_{0}^{z_{1}}sys(\xi,a_{4},a_{5})d\xi\geq 0 \tag{25}\]  
Equation (25) holds with the equal sign only when \(z_{1}=0\). When \(z_{1}\
eq 0\), according to Equations (24) and (25), \(V(z_{1},z_{2})>0\) always holds. Therefore, there exists a positive-definite function \(V(z_{1},z_{2})\) that increases infinitely. Secondly, take the derivative of the function \(V(z_{1},z_{2})\) with respect to time \(t\):  
Figure 5: (**a**) The curve of the function \(sys(x)\) when the parameter \(a_{4}\) changes and the parameter \(a_{5}\) remains unchanged. (**b**) The curve of the function \(sys(x)\) when the parameter \(a_{5}\) changes and the parameter \(a_{4}\) remains unchanged.