## 3 Control Protocol Design and Stability Analysis
We define the bipartite output error vectors as \(\zeta_{i}(t)=\sum_{j\in N_{i}}a_{ij}(y_{i}(t)-s_{ij}y_{j}(t))\) and \(\widehat{\zeta}_{i}(t)=\sum_{j\in N_{i}}a_{ij}(\widehat{y}_{i}(t)-s_{ij} \widehat{y}_{j}(t)),i=1,2,\ldots,B,B+1,\ldots,B+P\), i.e.,  
\[\zeta(t)=(S^{-1}LS\otimes I_{m})Y(t), \tag{15}\]  
and  
\[\widehat{\zeta}(t)=(S^{-1}LS\otimes I_{m})\widehat{Y}(t), \tag{16}\]  
where \(\zeta(t)=(\zeta_{1}^{\mathrm{T}}(t),\zeta_{2}^{\mathrm{T}}(t),\ldots,\zeta_{ B}^{\mathrm{T}}(t),\zeta_{B+1}^{\mathrm{T}}(t),\ldots,\zeta_{B+P}^{\mathrm{T}}(t))^{ \mathrm{T}}\) and \(\widehat{\zeta}(t)=(\widehat{\zeta}_{1}^{\mathrm{T}}(t),\,\widehat{\zeta}_{2 }^{\mathrm{T}}(t),\ldots,\)\(\widehat{\zeta}_{B}^{\mathrm{T}}(t),\widehat{\zeta}_{B+1}^{\mathrm{T}}(t),\ldots, \widehat{\zeta}_{B+P}^{\mathrm{T}}(t))^{\mathrm{T}}\). We denote the output observer error vectors as \(\overline{\zeta}_{i}(t)=\zeta_{i}(t)-\widehat{\zeta}_{i}(t),i=1,2,\ldots,B\), \(B+1,\ldots,B+P\). Then,  
\[\overline{\zeta}(t)=(S^{-1}LS\otimes I_{m})(Y(t)-\widehat{Y}(t)). \tag{17}\]  
We denote the exogenous disturbance observer error vectors as \(e_{i}(t)=\omega_{i}(t)-\widehat{\omega}_{i}(t)\), which indicates that  
\[e(t)=(I_{B+P}\otimes K)(d(t)-\widehat{d}(t)), \tag{18}\]  
and  
\[\widetilde{\widetilde{d}}(t)=(I_{B+P}\otimes H)\widehat{d}(t)-(I_{B+P}\otimes G _{2})e(t), \tag{19}\]  
where \(e(t)=(e_{1}^{\mathrm{T}}(t),e_{2}^{\mathrm{T}}(t),\ldots,e_{B+P}^{\mathrm{T}} (t))^{\mathrm{T}}\). Hence, we consider the Lyapunov function candidate as  
\[V(t)=\widehat{\xi}^{\mathrm{T}}(t)\widehat{\xi}(t)+\overline{\xi}^{\mathrm{T} }(t)\overline{\xi}(t)+e^{\mathrm{T}}(t)e(t), \tag{20}\]  
and we take the time derivative of \(V(t)\) with \(\widehat{\xi}(t)\), \(\overline{\xi}(t)\), and \(e(t)\) along the trajectories of (13), (14), and (19), respectively, as \[\dot{V}(t) =2\bar{\zeta}^{\mathsf{T}}(t)(S^{-1}LS\otimes C)[-\alpha(k_{1}+k_{2} \frac{\dot{\mu}(t)}{\mu(t)})(S^{-1}LS\otimes I_{n})\widehat{X}(t)+\widehat{F}\] \[\qquad-(I_{B+P}\otimes G_{\mathsf{I}}C)(X(t)-\widehat{X}(t))]\] \[-(I_{B+P}\otimes G_{1}C)(X(t)-\widehat{X}(t)). \tag{14}\]0 \[-(I_{B+P}\otimes G_{1}C)(X(t)-\widehat{X}(t)). \tag{14}\]1 \[-(I_{B+P}\otimes G_{1}C)(X(t)-\widehat{X}(t)). \tag{14}\]2 \[-(I_{B+P}\otimes G_{1}C)(X(t)-\widehat{X}(t)). \tag{14}\]3 \[-(I_{B+P}\otimes G_{1}C)(X(t)-\widehat{X}(t)). \tag{14}\]4 \[-(I_{B+P}\otimes G_{1}C)(X(t)-\widehat{X}(t)). \tag{14}\]5 \[-(I_{B+P}\otimes G_{1}C)(X(t)-\widehat{X}(t)). \tag{14}\]6 \[-(I_{B+P}\otimes G_{1}C)(X(t)-\widehat{X}(t)). \tag{14}\]7 \[-(I_{B+P}\otimes G_{1}C)(X(t)-\widehat{X}(t)). \tag{14}\]8 \[-(I_{B+P}\otimes G_{1}C)(X(t)-\widehat{X}(t)). \tag{14}\]9 \[\zeta(t)=(S^{-1}LS\otimes I_{m})Y(t), \tag{15}\]0  
Above, the following inequality is utilized:  
\[\zeta(t)=(S^{-1}LS\otimes I_{m})Y(t), \tag{15}\]1 \[\zeta(t)=(S^{-1}LS\otimes I_{m})Y(t), \tag{15}\]2 \[\zeta(t)=(S^{-1}LS\otimes I_{m})Y(t), \tag{15}\]3 \[\zeta(t)=(S^{-1}LS\otimes I_{m})Y(t), \tag{15}\]4  
According to (20)-(22), one has  
\[\zeta(t)=(S^{-1}LS\otimes I_{m})Y(t), \tag{15}\]5  
where \(\beta_{1}=\lambda_{min}(CC^{\mathsf{T}})k_{1}-2-\rho^{2}\) and \(\beta_{2}=\lambda_{min}(CC^{\mathsf{T}})k_{2}\). According to Lemma 1, we have \(V(t)\leqslant\mu^{-\beta_{2}}(t)\mathsf{e}^{-\beta_{1}(t-t_{0})}V(t_{0})\) for \(t\in[t_{0},t_{0}+T)\), which signifies that \(\lim\limits_{t\to t_{0}+T}V(t)=0\), and \(V(t)\equiv 0\) for \(t\in[t_{0}+T,+\infty)\). We can infer that for each \(i\), \(\lim\limits_{t\to t_{0}+T}\bar{\zeta}_{i}(t)=0\), \(\lim\limits_{t\to t_{0}+T}(\zeta_{i}(t)-\bar{\zeta}_{i}(t))=0\), and \(\widehat{\zeta}_{i}(t)\equiv 0\) under the definition of \(V(t)\) in (20), and \(\zeta_{i}(t)-\widehat{\zeta}_{i}(t)\equiv 0\) for \(t\in[t_{0}+T,+\infty)\), which means that \(\lim\limits_{t\to t_{0}+T}y_{i}(t)=s_{i\bar{\jmath}}y_{j}(t)\) and \(y_{i}(t)\equiv s_{i\bar{\jmath}}y_{j}(t)\) for \(t\in[t_{0}+T,+\infty)\). Therefore, system (1) is said to realize prescribed-time bipartite output consensus under distributed control protocol (10).