## 2 Problem Formulation and the Convergence of the Quantized ILC Law
A sector-bound expression can be describedas \(Q(v)=(1+\Delta(v))v\) by the given signal \(v\) and the quantizer \(Q(v)\), where \(\Delta(v)\) is a scalar function that satisfies \(|\Delta(v)|\leq\sigma\). We can calculate \(\sigma_{e}\) when the quantized density \(\eta_{e}\) is known, the sector-bound expression in the ILCQe law (4) can be expressed as  
\[Q(e_{k}(n+1))=(1+\Delta e_{k}(n+1))e_{k}(n+1), \tag{6}\]  
where \(|\Delta e_{k}(n)|\leq\sigma_{e}\) and \(\sigma_{e}=\frac{1-\eta_{e}}{1+\eta_{e}}\). Thus, the quantized ILC law (4) becomes  
\[u_{k+1}(n)=u_{k}(n)+\Gamma(1+\Delta e_{k}(n+1))e_{k}(n+1). \tag{7}\]  
For the sake of statement conciseness, a group of lifting vectors and matrix are denoted as  
\[\mathbf{e}_{k}=y_{d}-y_{k}=[e_{k}(1)|e_{k}(2)|\cdots|e_{k}(N)]^{\mathsf{T}},\]  
\[\Delta\mathbf{e}_{k}=[\Delta e_{k}(1)|\Delta e_{k}(2)|\cdots|\Delta e_{k}(N)]^{ \mathsf{T}},\]  
\[\mathbf{y}_{d}=[y_{d}(1)|y_{d}(2)|\cdots|y_{d}(N)]^{\mathsf{T}},\]  
\[\overline{\mathbf{\mathsf{E}}}_{k}=(\mathbf{\mathsf{I}}+\Delta\mathbf{e}_{k})\mathbf{e}_{k}.\]  
Then, the ILCQe law (7) with the data quantization is lifted as  
\[\mathbf{u}_{k+1}=\mathbf{u}_{k}+\Gamma\overline{\mathbf{\mathsf{E}}}_{k}. \tag{8}\]  
When the learning gain \(\Gamma\) is chosen properly, the output tracking error will be improved. The convergence of the output tracking error is discussed in the following sections. **Theorem 1**. : _Assume that the ILCQe law (8) is applied to the system (1), matrices \(\mathbf{\mathsf{B}}\) and \(\mathbf{\mathsf{C}}\), as well as the learning gain \(\Gamma\), satisfy the condition as \(\rho_{1}=|1-\Gamma\mathbf{\mathsf{C}}\mathbf{\mathsf{B}}|+\sigma_{e}|\Gamma\mathbf{ \mathsf{C}}\mathbf{\mathsf{B}}|<1\), where \(\sigma_{e}=\frac{1-\eta_{e}}{1+\eta_{e}}\), then the system's tracking error convergences to zero as \(k\to\infty\)._  
**Proof of Theorem 1. ** Taking Equations (7) and (8) into account, then  
\[\mathbf{e}_{k+1} =\mathbf{y}_{d}-\mathbf{y}_{k+1}=\mathbf{y}_{d}-\mathbf{y}_{k}-\left(\mathbf{y}_{k+1 }-\mathbf{y}_{k}\right) \tag{9}\] \[=\mathbf{e}_{k}-\Gamma\overline{\mathbf{\mathsf{H}}}_{k}\] \[=(\mathbf{\mathsf{I}}-\Gamma\mathbf{\mathsf{H}}-\Gamma\mathbf{\mathsf{H}}\Delta \mathbf{e}_{k})\mathbf{e}_{k}.\]  
Here,  
\[u_{k+1}(n)=u_{k}(n)+\Gamma(1+\Delta e_{k}(n+1))e_{k}(n+1). \tag{7}\]0 \[= \begin{bmatrix}1-\Gamma\mathbf{\mathsf{C}}\mathbf{\mathsf{B}}(1+\Delta \mathbf{e}_{k}(1))&0&\cdots&0\\ -\Gamma\mathbf{\mathsf{C}}\mathbf{\mathsf{A}}\mathbf{\mathsf{B}}(1+\Delta\mathbf{e}_{k}(2))&1- \Gamma\mathbf{\mathsf{C}}\mathbf{\mathsf{B}}(1+\Delta\mathbf{e}_{k}(1))&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ -\Gamma\mathbf{\mathsf{C}}\mathbf{\mathsf{A}}^{N-1}\mathbf{\mathsf{B}}(1+\Delta\mathbf{e}_{k}( N))&-\Gamma\mathbf{\mathsf{C}}\mathbf{\mathsf{A}}^{N-2}\mathbf{\mathsf{B}}(1+\Delta\mathbf{e}_{k}(N -1))&\cdots&1-\Gamma\mathbf{\mathsf{C}}\mathbf{\mathsf{B}}(1+\Delta\mathbf{e}_{k}(1))\\ \end{bmatrix} .\]  
Computing the norm on both sides of Equation (9), we find  
\[\big{|}\big{|}\mathbf{e}_{k+1}\big{|}\big{|}\leq\ \ \big{|}1-\Gamma\mathbf{\mathsf{C}}\mathbf{ \mathsf{B}}(1+\Delta\mathbf{e}_{k}(1))\big{|}\big{|}\big{|}\mathbf{e}_{k}\big{|}\big{|}\]  
Since \(|\Delta\mathbf{e}_{k}(1)|<\sigma_{e}\), then  
\[\big{|}\mathbf{e}_{k+1}\big{|}\big{|}\leq\big{(}\big{|}1-\Gamma\mathbf{\mathsf{C}}\mathbf{ \mathsf{B}}\big{|}+\sigma_{e}\big{|}\Gamma\mathbf{\mathsf{C}}\mathbf{\mathsf{B}}\big{|} \big{)}\big{|}\big{|}\mathbf{e}_{k}\big{|}\big{|}\]Let  
\[\rho_{1}=|1-\Gamma\mathbf{CB}|+\sigma_{e}|\Gamma\mathbf{CB}|,\]  
Then  
\[\big{|}\big{|}\epsilon_{k+1}\big{|}\big{|}=\rho_{1}||\epsilon_{k}||,\]  
where \(||\mathbf{H}||\leq b_{1}\). If \(\rho_{1}<1\) is satisfied, we can obtain  
\[\lim_{k\to\infty}\left|\left|\epsilon_{k+1}\right|\right|=0.\]  
This completes the proof. \(\square\)  
**Remark 1**.