## 3 A Modified HOSM Controller Using a PID-Sliding Surface for an ABS Laboratory Setup
### Design of the Modified HOSM Control Using a PID-Sliding Surface
\right)=-\frac{1}{2|s_{v}|^{1/2}}\Lambda_{1}\chi+\frac {1}{2}\Lambda_{2}\chi \tag{19}\]  
Then, the derivative of the Lyapunov function (17) using (19) is as follows:  
\[\dot{V}=-\frac{1}{2|s_{v}|^{1/2}}\chi^{T}P\Lambda_{1}\chi+\frac{1}{2}\chi^{T}P \Lambda_{2}\chi \tag{20}\]  
Solving the first term of (20), i.e., \(-\frac{1}{2|s_{v}|^{1/2}}\chi^{T}P\Lambda_{1}\chi\), we can rewrite it as follows:  
\[-\frac{1}{2|s_{v}|^{1/2}}\chi^{T}P\Lambda_{1}\chi =-\frac{1}{2|s_{v}|^{1/2}}\Big{(}\gamma_{1}(2\gamma_{3}+\gamma_{ 1}^{2})(\lfloor s_{v}\rceil^{1/2})^{2}-2\gamma_{1}^{2}\lfloor s_{v}\rceil^{1/ 2}\xi_{v} \tag{21}\] \[+\gamma_{2}(\gamma_{1}^{2}-2\gamma_{3})\lfloor s_{v}\rceil^{1/ 2}s_{v}-\gamma_{1}\gamma_{2}s_{v}\xi_{v}+\gamma_{1}\xi_{v}^{2}\Big{)}\]  
Since \(-\frac{1}{2|s_{v}|^{1/2}}\gamma_{2}(\gamma_{1}^{2}-2\gamma_{3})\lfloor s_{v} \rceil^{1/2}s_{v}=-\frac{1}{2}\gamma_{2}(\gamma_{1}^{2}-2\gamma_{3})(\lfloor s _{v}\rceil^{1/2})^{2}\), the first term (21) is as follows:  
\[-\frac{1}{2|s_{v}|^{1/2}}\chi^{T}P\Lambda_{1}\chi =-\frac{1}{2|s_{v}|^{1/2}}\chi^{T} \begin{pmatrix}\gamma_{1}(\gamma _{1}^{2}+2\gamma_{3})&0&-\gamma_{1}^{2}\\ 0&0&-\frac{1}{2}\gamma_{1}\gamma_{2}\\ -\gamma_{1}^{2}&-\frac{1}{2}\gamma_{1}\gamma_{2}&\gamma_{1}\end{pmatrix} \chi \tag{22}\] \[-\frac{1}{2}\chi^{T} \begin{pmatrix}\gamma_{2}(\gamma_{1}^{2}-2 \gamma_{3})&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix} \chi.\]  
Similarly, the second term \(\frac{1}{2}\chi^{T}P\Lambda_{2}\chi\) of (20) is as follows:  
\[\frac{1}{2}\chi^{T}P\Lambda_{2}\chi =\frac{1}{2}\Big{(}-\gamma_{2}(3\gamma_{1}+4\gamma_{3})(\lfloor s _{v}\rceil^{1/2})^{2}-2\gamma_{2}(\gamma_{2}^{2}+\gamma_{4})s_{v}^{2}-2\gamma_ {2}\xi_{v}^{2} \tag{23}\] \[-\gamma_{1}(5\gamma_{2}^{2}+2\gamma_{4})\lfloor s_{v}\rceil^{1/ 2}s_{v}+5\gamma_{1}\gamma_{2}\lfloor s_{v}\rceil^{1/2}\xi_{v}+4\gamma_{2}^{2 }s_{v}\xi_{v}\Big{)}\]  
If \(\lfloor s_{v}\rceil^{1/2}s_{v}=\frac{1}{|s_{v}|^{1/2}}s_{v}^{2}\) and \(\lfloor s_{v}\rceil^{1/2}\xi_{v}=\frac{1}{|s_{v}|^{1/2}}s_{v}\xi_{v}\), then the second term is rewritten in the matrix form:  
\[\frac{1}{2}\chi^{T}P\Lambda_{2}\chi=-\frac{1}{2}\gamma_{2}\chi^{T} \begin{pmatrix} 3\gamma_{1}^{2}+4\gamma_{3}&0&0\\ 0&2(\gamma_{2}^{2}+\gamma_{4})&-2\gamma_{2}\\ 0&-2\gamma_{2}&2\end{pmatrix} \chi-\frac{1}{2|s_{v}|^{1/2}}\gamma_{1}\chi^{T} \begin{pmatrix}0&0&0\\ 0&5\gamma_{2}^{2}+2\gamma_{4}&-\frac{5}{2}\gamma_{2}\\ 0&-\frac{5}{2}\gamma_{2}&0\end{pmatrix} \chi. \tag{24}\]  
Finally, using (22) and (24), one rewrites (20) as the following:  
\[\dot{V}=-\frac{1}{2|s_{v}|^{1/2}}\chi^{T}Q_{1}\chi-\chi^{T}Q_{2}\chi \tag{25}\]  
with  
\[Q_{1}=\frac{\gamma_{1}}{2} \begin{pmatrix}(\gamma_{1}^{2}+2\gamma_{3})&0&- \gamma_{1}\\ 0&2\gamma_{4}+5\gamma_{2}^{2}&-3\gamma_{2}\\ -\gamma_{1}&-3\gamma_{2}&1\end{pmatrix} ;\quad Q_{2}=\gamma_{2} \begin{pmatrix}2 \gamma_{1}^{2}+\gamma_{3}&0&0\\ 0&\gamma_{2}^{2}+\gamma_{4}&-\gamma_{2}\\ 0&-\gamma_{2}&1\end{pmatrix} .\]  
The derivative of the Lyapunov function (25) is negative definite if \(Q_{1},Q_{2}>0\). This condition is fulfilled if the gains are determined as in (15).