## 3 Parameter-Optimal Iterative Learning Control with Quantized Information
\tag{26}\]  
Therefore,  
\[\left(w_{k}\mathbf{I}_{r_{k}}+\widetilde{\mathbf{M}}_{k}\left(\widetilde{\mathbf{ H}}_{k}^{\mathsf{T}}\widetilde{\mathbf{H}}_{k}+\widetilde{\mathbf{H}}_{k}^{ \mathsf{A}\mathsf{T}\mathsf{A}}\widetilde{\mathbf{H}}_{k}\right)\widetilde{ \mathbf{M}}_{k}\right)\widetilde{\mathbf{Y}}_{k}=\widetilde{\mathbf{M}}_{k} \widetilde{\mathbf{H}}_{k}^{\mathsf{T}}\widetilde{\mathbf{e}}_{k}+\widetilde{ \mathbf{M}}_{k}\widetilde{\mathbf{H}}_{k}^{\mathsf{T}}\widetilde{\mathbf{e}}_{k}. \tag{38}\]  
\[w_{k}\mathbf{I}_{N-r_{k}}\hat{\mathbf{Y}}_{k}=\mathbf{0}. \tag{39}\]  
As \(w_{k}\
eq 0\), Equation (39) results in  
\[\hat{\mathbf{Y}}_{k}=0. \tag{40}\]  
Because the matrix \(\left(w_{k}\mathbf{I}_{r_{k}}+\widetilde{\mathbf{M}}_{k}\left(\widetilde{ \mathbf{H}}_{k}^{\mathsf{T}}\widetilde{\mathbf{H}}_{k}+\widetilde{\mathbf{H}} _{k}^{\mathsf{A}\mathsf{T}\mathsf{A}}\widetilde{\mathbf{H}}_{k}\right) \widetilde{\mathbf{M}}_{k}\right)\) is positive definite and thus invertible, Equation (38) have a unique solution as  
\[\widetilde{\mathbf{Y}}_{k}=\left(w_{k}\mathbf{I}_{r_{k}}+\widetilde{\mathbf{M} }_{k}\left(\widetilde{\mathbf{H}}_{k}^{\mathsf{A}\mathsf{T}\mathsf{A}} \widetilde{\mathbf{H}}_{k}+\widetilde{\mathbf{H}}_{k}^{\mathsf{A}\mathsf{T} \mathsf{A}}\widetilde{\mathbf{M}}_{k}\right)^{-1}(\widetilde{\mathbf{M}}_{k} \widetilde{\mathbf{H}}_{k}^{\mathsf{A}\mathsf{T}}\widetilde{\mathbf{e}}_{k}+ \widetilde{\mathbf{M}}_{k}\widetilde{\mathbf{H}}_{k}^{\mathsf{A}\mathsf{T}} \widetilde{\mathbf{e}}_{k})=\widetilde{\mathbf{L}}_{k}(\widetilde{\mathbf{M}}_ {k}\widetilde{\mathbf{H}}_{k}^{\mathsf{T}\mathsf{A}}\widetilde{\mathbf{e}}_{k }+\widetilde{\mathbf{M}}_{k}\widetilde{\mathbf{H}}_{k}^{\mathsf{A}\mathsf{T}} \widetilde{\mathbf{e}}_{k}), \tag{41}\]  
where  
\[\widetilde{\mathbf{L}}_{k}=\left(w_{k}\mathbf{I}_{r_{k}}+\widetilde{\mathbf{M} }_{k}\left(\widetilde{\mathbf{H}}_{k}^{\mathsf{A}\mathsf{T}\mathsf{A}} \widetilde{\mathbf{H}}_{k}+\widetilde{\mathbf{H}}_{k}^{\mathsf{A}\mathsf{T} \mathsf{A}}\widetilde{\mathbf{H}}_{k}\right)\widetilde{\mathbf{M}}_{k}\right)^ {-1}. \tag{42}\]  
For simplicity, denote  
\[\widetilde{\mathbf{\Pi}}_{k}=\widetilde{\mathbf{M}}_{k}\left(\widetilde{ \mathbf{H}}_{k}^{\mathsf{T}}\widetilde{\mathbf{H}}_{k}+\widetilde{\mathbf{H}} _{k}^{\mathsf{A}\mathsf{T}\mathsf{A}}\widetilde{\mathbf{H}}_{k}\right) \widetilde{\mathbf{M}}_{k}.\]  
Then,  
\[\widetilde{\mathbf{L}}_{k}=\left(w_{k}\mathbf{I}_{r_{k}}+\widetilde{\mathbf{ \Pi}}_{k}\right)^{-1}. \tag{43}\]  
Therefore, Equations (29), (40), and (41) lead to  
\[\mathbf{Y}_{k}=\mathbf{R}_{k}^{-\mathsf{T}}\left[ \begin{array}{c}\widetilde{ \mathbf{Y}}_{k}\\ \widetilde{\mathbf{Y}}_{k}\end{array} \right]=\mathbf{R}_{k}\left[ \begin{array} []{c}\widetilde{\mathbf{Y}}_{k}\\ \mathbf{0}\end{array} \right]=\mathbf{R}_{k}\left[ \begin{array}{cc}\widetilde{ \mathbf{L}}_{k}\widetilde{\mathbf{M}}_{k}\widetilde{\mathbf{H}}_{k}^{ \mathsf{A}\mathsf{T}}&\widetilde{\mathbf{L}}_{k}\widetilde{\mathbf{M}}_{k} \widetilde{\mathbf{H}}_{k}^{\mathsf{A}\mathsf{T}}\\ \mathbf{0}&\mathbf{0}\end{array} \right] \begin{bmatrix}\widetilde{\mathbf{e}}_{k }\\ \widetilde{\mathbf{e}}_{k}\end{bmatrix} . \tag{44}\]  
Pre-multiplying Equation (15) by \(\mathbf{R}_{k}^{\mathsf{T}}\) results in  
\[\mathbf{R}_{k}^{\mathsf{T}}\mathbf{e}_{k+1} =\left[ \begin{array}{c}\widetilde{\mathbf{e}}_{k+1}\\ \widetilde{\mathbf{e}}_{k+1}\end{array}