## 4 Control Method Design of the Pump
### Backstepping Sliding Control Method  
The steps of designing the backstepping sliding mode control method for pressure tracking of the pump are as follows:  
Step1. According to the pump outlet pressure \(x_{1}\) and the command signal \(x_{1d}\), the tracking error of the system is defined as:  
\[e_{1}=x_{1}-x_{1d}. \tag{24}\]  
After derivation the above pressure error, then,  
\[\dot{e}_{1}=\dot{x}_{1}-\dot{x}_{1d}=a_{1}x_{2}-a_{2}x_{1}-\dot{x}_{1d}. \tag{25}\]  
Define the Lyapunov function of the first state variable of the system as,  
\[V_{1}=0.5{e_{1}}^{2}. \tag{26}\]  
Derivation the Lyapunov function of the first state variable of the system,  
\[\dot{v}_{1}=e_{1}\dot{e}_{1}=e_{1}\big{(}\dot{x}_{1}-\dot{x}_{1d}\big{)}=e_{1} \big{(}a_{1}x_{2}-a_{2}x_{1}-\dot{x}_{1d}\big{)}. \tag{27}\]  
Let \(x_{2}=-\frac{1}{a_{1}}\big{(}c_{1}e_{1}-a_{2}x_{1}-\dot{x}_{1d}\big{)}+e_{2}\), if \(c_{1}>0\). \(e_{2}\) is virtual controller, then  
\[e_{2}=x_{2}+\frac{1}{a_{1}}\big{(}c_{1}e_{1}-a_{2}x_{1}-\dot{x}_{1d}\big{)}. \tag{28}\]  
Substituting it into the above formula and then:  
\[\dot{V}_{1}=-{c_{1}}{e_{1}}^{2}+a_{1}e_{1}e_{2}. \tag{29}\]  
At this condition, if \(e_{2}=0\) and then \(\dot{V}_{1}\leq 0\), the stability of Lyapunov is satisfied under these conditions. In order to make \(e_{2}=0\), the next design is needed. Step 2. Define the Lyapunov function of the second state variable of the system as:  
\[V_{2}=V_{1}+0.5{e_{2}}^{2}. \tag{30}\]  
Substitute \(e_{2}=x_{2}+\frac{1}{a_{1}}\big{(}c_{1}e_{1}-a_{2}x_{1}-\dot{x}_{1d}\big{)}\) and the state space equation of \(x_{2}\) into the above equation, and then:  
\[\dot{V}_{2}=-{c_{1}}{e_{1}}^{2}+a_{1}e_{1}e_{2}+e_{2}\bigg{(}x_{3}+\frac{{a_{ 2}}^{2}}{a_{1}}x_{1}-a_{2}x_{2}-\frac{1}{a_{1}}\ddot{x}_{1d}+\frac{1}{a_{1}}c _{1}\dot{e}_{1}\bigg{)}. \tag{31}\]  
Let  
\[x_{3}=-\frac{{a_{2}}^{2}}{a_{1}}x_{1}+a_{2}x_{2}+\frac{1}{a_{1}}\ddot{x}_{1d}- \frac{1}{a_{1}}c_{1}\dot{e}_{1}-a_{1}e_{1}-c_{2}e_{2}+e_{3}, \tag{32}\]  
and  
\[e_{3}=\frac{{a_{2}}^{2}}{a_{1}}x_{1}-a_{2}x_{2}-\frac{1}{a_{1}}\ddot{x}_{1d}+ \frac{1}{a_{1}}c_{1}\dot{e}_{1}+a_{1}e_{1}+c_{2}e_{2}+x_{3}, \tag{33}\]  
\[\dot{e}_{1}=\dot{x}_{1}-\dot{x}_{1d}=a_{1}x_{2}-a_{2}x_{1}-\dot{x}_{1d}. \tag{25}\]0  
At this condition, if \(e_{3}=0\) and then \(\dot{V}_{2}\leq 0\), the stability of Lyapunov is satisfied under these conditions. In order to make \(e_{3}=0\), the next design is needed. Step 3. Define the Lyapunov function of the third state variable of the system as:  
\[\dot{e}_{1}=\dot{x}_{1}-\dot{x}_{1d}=a_{1}x_{2}-a_{2}x_{1}-\dot{x}_{1d}. \tag{25}\]1Derivation the Lyapunov function of the third state variable of the system, then  
\[\dot{e}_{1}=\dot{x}_{1}-\dot{x}_{1d}=a_{1}x_{2}-a_{2}x_{1}-\dot{x}_{1d}. \tag{25}\]2  
Substitute \(e_{3}\) and the state space equation of \(x_{3}\) into the above equation, and then  
\[\dot{V}_{3}=-c_{1}{e_{1}}^{2}-c_{2}{e_{2}}^{2}-c_{3}{e_{3}}^{2}+e_{3}\Bigg{(} \begin{array}{c}\big{(}b_{1}+{a_{2}}^{2}\big{)}x_{2}+(b_{2}-a_{2})x_{3}+\Big{(} b_{3}-\frac{{a_{2}}^{2}}{a_{1}}\Big{)}x_{1}+b_{4}x_{4}+b_{5}\\ -\frac{1}{a_{1}}\dddot{x}_{1d}+\frac{c_{1}}{a_{1}}\ddot{e}_{1}+a_{1}\dot{e}_{1 }+c_{2}\dot{e}_{2}\end{array} \Bigg{)} \tag{37}\]  
Letbe  
\[x_{4}=-\frac{1}{b_{4}}\Bigg{(} \begin{array}{c}\big{(}b_{1}+{a_{2}}^{2}\big{)} x_{2}+(b_{2}-a_{2})x_{3}+\Big{(}b_{3}-\frac{{a_{2}}^{2}}{a_{1}}\Big{)}x_{1}\\ +b_{5}-\frac{1}{a_{1}}\dddot{x}_{1d}+\frac{c_{1}}{a_{1}}\ddot{e}_{1}+a_{1} \dot{e}_{1}+c_{2}\dot{e}_{2}+e_{2}+c_{3}\dot{e}_{3}\end{array} \Bigg{)}+e_{4}, \tag{38}\]  
Then  
\[\dot{V}_{3}=-c_{1}{e_{1}}^{2}-c_{2}{e_{2}}^{2}-c_{3}{e_{3}}^{2}+b_{4}e_{3}e_{4}, \tag{39}\]  
At this condition, if \(e_{4}=0\) and then \(\dot{V}_{3}\leq 0\), the stability of Lyapunov is satisfied under these conditions. In order to make \(e_{4}=0\), the next design is needed. Step 4. Define the Lyapunov function of the last state variable of the system as:  
\[V_{4}=V_{3}+0.5\sigma^{2}, \tag{40}\]  
where  
\[\sigma=k_{1}e_{1}+k_{2}e_{2}+k_{3}e_{3}+e_{4}. \tag{41}\]  
Let  
\[S=\frac{1}{b_{4}}\Bigg{[}