# Two-Phase Non-Singular Terminal Sliding Mode Control of Nonlinear Systems
## 3 Problem Formulation  
In order to narrow down the settling time of the finite time systems, the traditional double power rate attracting laws need to be attenuated such that regions both close to and far away from the equilibrium point are equipped with a fast convergence rate. The key to the modification of the finite time systems lies in the requirement for the essential convergence process, where the idea of a finite-duration attractor is borrowed and proper improvement is made accordingly. In this paper, a new two-phase finite time system is constructed as follows:  
\[\dot{x} = \left\{ \begin{array}{ll}-k_{1}x^{\alpha}-k_{2}x^{2-\alpha},&|x| \geq x_{t}\\ -k_{1}x-k_{1}x^{\alpha},&|x|<x_{t}\end{array} \right. \tag{3}\]  
where \(k_{1}>0\) and \(0<\alpha<1\) are fractions, in which both numerator and denominator are positive odd integers, respectively. \(x_{t}\) is the transition state, satisfying \(|x_{t}|\leq|x_{0}|\), where \(x_{0}\triangleq x(0)\) is the initial state and different selections of \(x_{t}\) and \(x_{0}\) obtain a different upper bound of the settling time. **Lemma 1**. : _Consider nonlinear system (3); its origin is finite-duration stable and its settling time is bounded and varies with the transition state and the initial state._  
**Proof of Lemma 1**. : (i) \(|x|\geq x_{t}\)  
Let \(z=x^{1-\alpha}\), and the derivative of \(x\) can be rewritten as  
\[z+k_{1}(1-\alpha)(1+(\sqrt{\frac{k_{2}}{k_{1}}}z)^{2})=0 \tag{4}\]Integrating both sides of (4) yields  
\[\arctan(\sqrt{\frac{k_{2}}{k_{1}}}z(t))=\arctan(\sqrt{\frac{k_{2}}{k_{1}}}z(0))- \sqrt{k_{1}k_{2}}(1-\alpha)t \tag{5}\]  
Then, the settling time from \(x_{0}\) to \(x_{t}\) is obtained as  
\[t_{s,1}=\frac{\arctan(\sqrt{\frac{k_{2}}{k_{1}}}x_{0}^{1-\alpha})-\arctan( \sqrt{\frac{k_{2}}{k_{1}}}x_{t}^{1-\alpha})}{\sqrt{k_{1}k_{2}}(1-\alpha)} \tag{6}\]  
(ii) \(|x|\leq x_{t}\)  
Using the intermediate variable \(z=x^{1-\alpha}\), the system can be expressed as  
\[\dot{z}+k_{1}(1-\alpha)(1+z)=0 \tag{7}\]  
Solving the differential Equation (7),  
\[z(t)=e^{-k_{1}(1-\alpha)t}(1+z(0))-1 \tag{8}\]  
and setting \(z(t)=0\) gives the settling time from \(x_{t}\) to \(0\)  
\[t_{s,2} = \frac{1}{k_{1}(1-\alpha)}\ln(1+x_{t}^{1-\alpha}) \tag{9}\]  
Finally, the settling time of system (3) is written as  
\[t_{s} = \frac{\arctan(\sqrt{\frac{k_{2}}{k_{1}}}x_{0}^{1-\alpha})-\arctan (\sqrt{\frac{k_{2}}{k_{1}}}x_{t}^{1-\alpha})}{\sqrt{k_{1}k_{2}}(1-\alpha)}+ \tag{10}\] \[\frac{1}{k_{1}(1-\alpha)}\ln(1+x_{t}^{1-\alpha})\] \[\leq \frac{\pi}{2\sqrt{k_{1}k_{2}}(1-\alpha)}+\frac{x_{t}^{1-\alpha}}{ k_{1}(1-\alpha)}\]  
It is observed from Equation (10) that the settling time is dependent on both the transition state \(x_{t}\) and the initial state \(x_{0}\) visibly, where it covers a duration within \([0,t_{s}]\) and the width of the duration \([0,t_{s}]\) is finite for all \(x_{t}\) and \(x_{0}\). The proof is completed. **Remark 3**. : _Note that the duration \([0,t_{s}]\) can be determined by each given \(x_{t}\) and \(x_{0}\), and its width depends on \(x_{t}\) and \(x_{0}\). However, the duration \([0,t_{s}]\) is finite for all \(x_{0}\) since the upper bound of \(t_{s}\) is independent of \(x_{0}\), and the choice of the transition state has little impact on the duration width but has much benefit for the convergence process. Furthermore, the settling time is tailored by the designer, where the impact of the initial state is totally oriented by the requirements of the control objective, rather than ignored without consideration. Hence, the analysis of the finite duration of systems is of great value._  
In this paper, we apply the two-phase finite time system (3) to the sliding mode control design in order to bring benefits to the convergence rate and enhance the control performance. A class of nonlinear systems is considered and described as  
\[\left\{ \begin{array}{l}\dot{x}_{1}(t)=x_{2}(t)\\ \dot{x}_{2}(t)=f(x)+g(x)u+d(x,t)\end{array} \right. \tag{11}\]  
where \(V_{t}\triangleq V(s_{t})\) is the Lyapunov function with \(s_{t}\).