# Data-Driven Model-Free Adaptive Sliding Mode Control Based on FFDL for Electric Multiple Units
## 3 Model-Free Adaptive Sliding Mode Controller for EMUs
### Stability Analysis
 \tag{25}\]  
Since the step size factor \(\beta\in(0,2]\), it is not difficult to obtain:  
\[\frac{\beta\parallel\Delta\mathbf{\hat{H}}(t-1)\parallel^{2}}{\mu+\parallel \Delta\mathbf{\hat{H}}(t-1)\parallel^{2}}<\frac{\beta\parallel\Delta\mathbf{ \hat{H}}(t-1)\parallel^{2}}{\parallel\Delta\mathbf{\hat{H}}(t-1)\parallel^{2}}=2. \tag{26}\]  
Then, it can be obtained from (25) and (26):  
\[\left\|\delta(t-1)\left(I-\frac{\beta\Delta\mathbf{\hat{H}}(t-1)\Delta\mathbf{ \hat{H}}^{T}(t-1)}{\mu+\parallel\Delta\mathbf{\hat{H}}(k-1)\parallel^{2}} \right)\right\|^{2}<\|\delta(t-1)\|^{2}. \tag{27}\]  
From (27), there must be \(0<M<1\) such that the following inequality holds:  
\[\left\|\delta(t-1)\left(I-\frac{\beta\Delta\mathbf{\hat{H}}(t-1)\Delta \mathbf{\hat{H}}^{T}(t-1)}{\mu+\parallel\Delta\mathbf{\hat{H}}(t-1)\parallel ^{2}}\right)\right\|\leq M\|\delta(t-1)\|. \tag{28}\]  
Substituting (28) into (24), one obtains:  
\[ \begin{split}\|\delta(t)\|\leq& M\|\delta(t-1)\|+2b \leq M(M\|\delta(t-2)\|+2b)+2b\\ \leq&\cdots\leq M^{t-1}\|\delta(1)\|+\frac{1-M^{t-1 }}{1-M}2b.\end{split}  \tag{29}\]  
It can be obtained from (29) that the parameter estimation error \(\delta(t)=[\delta_{1}(t),\delta_{2}(t)]\) is bounded, and since \(\mathbf{\Phi}(t)\) is bounded, the estimation value \(\mathbf{\hat{\Phi}}(t)\) is bounded too. The proof of conclusion (1) is completed. \(\qed\)  
Since \(\mathbf{\Phi}_{1}(t),\mathbf{\Phi}_{2}(t)\) is unknown, \(\delta_{1}(t),\delta_{2}(t)\) cannot be accurately obtained. The introduction of \(\mathbf{\Phi}_{1}(t),\mathbf{\Phi}_{2}(t)\) prevents the control input from being too large, and for the convenience of the controller design, let the element in \(\delta_{1}(t),\delta_{2}(t)\) be adjustable constants. **Proof 2**. : \(\underset{t\rightarrow\infty}{\lim}\|\boldsymbol{e}(t)\|=0\)_._  
Substituting (7) and (14) into (15) yields:  
\[\boldsymbol{e}(t+1) =[I-\frac{\rho(\hat{\boldsymbol{\Phi}}_{2}(t)-\delta_{2}(t))( \hat{\boldsymbol{\Phi}}_{2}(t)-\delta_{2}(t))^{T}}{\lambda+\left\|\hat{ \boldsymbol{\Phi}}_{2}(t)-\delta_{2}(t)\right\|^{2}}][\boldsymbol{e}(t)-(\hat {\boldsymbol{\Phi}}_{1}(t)-\delta_{1}(t))\Delta v(t)]. \tag{30}\]  
Since \(\rho\in(0,1]\) and \(\lambda>0\), we have:  
\[\left\|I-\frac{\rho(\hat{\boldsymbol{\Phi}}_{2}(t)-\delta_{2}(t))(\hat{ \boldsymbol{\Phi}}_{2}(t)-\delta_{2}(t))^{T}}{\lambda+\left\|\hat{\boldsymbol {\Phi}}_{2}(t)-\delta_{2}(t)\right\|^{2}}\right\|= \epsilon\in(0,1). \tag{31}\]  
Since \(\boldsymbol{\Phi}_{1}(t)-\delta_{1}(t)\) is bounded, then combining (30), (31), and Assumption 3, we have:  
\[\underset{t\rightarrow\infty}{\lim}\|\boldsymbol{e}(t+1)\|= \underset{t\rightarrow\infty}{\lim}\|\boldsymbol{e}(t)\|_{\mathcal{E}}\in(0,1). \tag{32}\]  
This means that \(\underset{t\rightarrow\infty}{\lim}\|\boldsymbol{e}(t)\|=0\). The proof of conclusion (2) is completed. Proof 3**. : The output sequence and the input sequence are bounded. Since \(\boldsymbol{e}(t)\) and \(\boldsymbol{v}_{d}\) are bounded, it is easy to find that the output sequence is bounded. According to (15) and (32), using \(\sqrt{ab}\leq a+b\) and \(\lambda_{\min}<\lambda\), we have:  
\[ \begin{split}\|\Delta\boldsymbol{u}_{MFA}(t)\|\leq& \left\|\rho\frac{\left(\hat{\boldsymbol{\Phi}}_{2}(t)-\delta_{2}(t) \right)^{T}}{\lambda+\left\|\hat{\boldsymbol{\Phi}}_{2}(t)-\delta_{2}(t) \right\|^{2}}\boldsymbol{e}(t)\right\|\leq\left|\rho\frac{\left\|\hat{ \boldsymbol{\Phi}}_{2}(t)-\delta_{2}(t)\right\|}{\lambda+\left\|\hat{ \boldsymbol{\Phi}}_{2}(t)-\delta_{2}(t)\right\|^{2}}\right|\|\boldsymbol{e}( t)\|\\ \leq&\left|\frac{\rho\|\hat{\boldsymbol{\Phi}}_{2}(t )-\delta_{2}(t)\|}{2\sqrt{\lambda\left\|\hat{\boldsymbol{\Phi}}_{2}(t)-\delta _{2}(t)\right\|^{2}}}\right|\|\boldsymbol{e}(t)\|\leq\left|\frac{\rho}{2\sqrt {\lambda_{\min}}}\right|\|\boldsymbol{e}(t)\|\end{split}  \tag{33}\]  
Further, from (32) and (33), we have:  
\[