## Appendix A  
_Stability Analysis_  
We consider the Lyapunov function \(V_{t}\) as below:  
\[V_{t}=\frac{1}{2}s_{t}^{2}+\frac{\omega_{t}{}^{2}}{2\bar{a}_{B}}. \tag{1}\]  
The time derivative of Equation (1) can be computed as  
\[\dot{V}_{t}=s_{t}\dot{s}_{t}+\frac{\omega_{t}\dot{\omega}_{t}}{\bar{a}_{B}}. \tag{2}\]  
Substituting Equation (3) and the time derivative of Equation (8) into Equation (2) yields  
\[\dot{V}_{t}=s_{t}\big{(}\bar{q}_{d,t}-N_{t}-\bar{M}^{-1}\tau_{t}^{B}+K_{s}\dot{ \epsilon}_{t}\big{)}+\frac{\omega_{t}\dot{\omega}_{t}}{\bar{a}_{B}}. \tag{3}\]  
Substituting Equation (9) into Equation (3) yields  
\[\dot{V}_{t}=s_{t}\big{[}\Lambda_{t}\delta_{t}-E_{t}-\beta s_{t}-\bar{K}\text{ sgn}(s_{t})\big{]}+\frac{\omega_{t}\dot{\omega}_{t}}{\bar{a}_{B}} \tag{4}\]  
where the TDE error \(E_{t}=N_{t}-N_{t-L}\) is bounded as \(E^{*}\) according to Lemma 1. \(\delta_{t}=-\bar{q}_{t-L}+\bar{q}_{d,t}+K_{s}\dot{\epsilon}_{t}+\beta s_{t}+\bar {K}\text{sgn}(s_{t})\) and \(\Lambda_{t}=1-\bar{M}\dot{M}_{t}\leq 0\). Then, \(\delta_{t}\) can be summarized by adding \(\pm\bar{q}_{t}\) as follows:  
\[\delta_{t} =\bar{q}_{t}-\bar{q}_{t-L}+\ddot{\epsilon}_{t}+K_{s}\dot{\epsilon }_{t}+\beta s_{t}+\bar{K}\text{sgn}(s_{t})\] \[=\ddot{q}_{t}-\ddot{q}_{t-L}+\dot{s}_{t}+\beta s_{t}+\bar{K}\text {sgn}(s_{t}) \tag{5}\]where \(\dot{s}_{t}=\ddot{e}_{t}+K_{s}\dot{e}_{t}\). Substituting Equation (A5) into Equation (A4), we have  
\[\dot{V}_{t}\leq |s_{t}||\Lambda_{t}||\Delta_{t}|+s_{t}\Lambda_{t}\dot{s}_{t}+\beta \Lambda_{t}s_{t}^{2}+\Lambda_{t}\tilde{R}|s_{t}|+(E^{*}-\tilde{R})|s_{t}|-\beta s _{t}^{2}+\frac{\omega_{t}\dot{\omega}_{t}}{\bar{a}_{B}}\] \[\leq |s_{t}||\Lambda_{t}|(\lambda^{*}-\tilde{R})+s_{t}\Lambda_{t}\dot{ s}_{t}+(E^{*}-\tilde{R})|s_{t}|-\beta(1-\Lambda_{t})s_{t}^{2}+\frac{\omega_{t} \dot{\omega}_{t}}{\bar{a}_{B}}\] (A6)  
where \(\dot{s}_{t}=\ddot{e}_{t}+K_{s}\dot{e}_{t}\) and \(|E_{t}|\leq E^{*}\). \(\Delta_{t}=\ddot{q}_{t}-\ddot{q}_{t-L}\) is bounded from the reasonable assumption, i.e., \(|\ddot{q}_{t}-\ddot{q}_{t-L}|\leq\lambda^{*}\) for a certain positive value \(\lambda^{*}\). If the switching gain \(\tilde{R}\) can be chosen to be  
\[\tilde{R}>\max\{E^{*},\lambda^{*}\}\] (A7)  
resulting in  
\[\dot{V}_{t}\leq-\beta(1-\Lambda_{t})s_{t}^{2}+s_{t}\Lambda_{t}\dot{s}_{t}+ \frac{\omega_{t}\dot{\omega}_{t}}{\bar{a}_{B}}.\] (A8)  
Multiplying both sides of Equation (A2) by \(\Lambda_{t}\), we have  
\[\dot{V}_{t}=s_{t}\dot{s}_{t}+\frac{\omega_{t}\dot{\omega}_{t}}{\bar{a}_{B}}. \tag{2}\]0 (A9)  
Solving for \(s_{t}\Lambda_{t}\dot{s}_{t}\) yields  
\[\dot{V}_{t}=s_{t}\dot{s}_{t}+\frac{\omega_{t}\dot{\omega}_{t}}{\bar{a}_{B}}. \tag{2}\]1 (A10)  
Substituting Equation (A10) into Equation (A8), it follows that  
\[\dot{V}_{t}=s_{t}\dot{s}_{t}+\frac{\omega_{t}\dot{\omega}_{t}}{\bar{a}_{B}}. \tag{2}\]2 (A11)  
From Equation (A11), it can be represented as below:  
\[\dot{V}_{t}=s_{t}\dot{s}_{t}+\frac{\omega_{t}\dot{\omega}_{t}}{\bar{a}_{B}}. \tag{2}\]3 \[\dot{V}_{t}=s_{t}\dot{s}_{t}+\frac{\omega_{t}\dot{\omega}_{t}}{\bar{a}_{B}}. \tag{2}\]4 (A12)  
For \(\theta_{t}\geq 0\), it follows that  
\[\dot{V}_{t}=s_{t}\dot{s}_{t}+\frac{\omega_{t}\dot{\omega}_{t}}{\bar{a}_{B}}. \tag{2}\]5 (A13)  
For any positive \(\bar{e}\), the Lyapunov function \(V_{t}\) decreases with a non-zero speed if \(|s_{t}|>\delta+\bar{e}\) where \(\delta=\frac{\max\{\omega_{t}\}}{\beta}\). It implies that the sliding variable \(s_{t}\) enters the set \(\mathbf{S}^{U}=\{s_{t}||s_{t}|\leq\delta+\bar{e}\}\) within a finite time \(t_{f}>0\). The sliding variable \(s_{t}\) exists then within the set \(\mathbf{S}^{U}\), and the Lyapunov function of Equation (A1) has an upper bound \(V^{*}\):  
\[\dot{V}_{t}=s_{t}\dot{s}_{t}+\frac{\omega_{t}\dot{\omega}_{t}}{\bar{a}_{B}}. \tag{2}\]6 (A14)  
where \(|\omega_{t}|\leq\omega^{*}\).