# Simplified Two-Dimensional Generalized Partial Response Target of Holographic Data Storage Channel
## 2 2D GPR Target and Serial Detection  
### 2.1 2D GPR Target  
First, we estimate the coefficients of the 2D GPR target and 2D equalizer in the training process. The model for the training process is presented in Figure 1.  
To design and implement the 2D GPR target and equalizer, we use the GPR target \(\mathbf{G}\) with a size of \(3\times 3\), and the equalizer \(\mathbf{F}\) with a size of \(5\times 5\); these are presented as in the below matrices. \[\mathbf{G}=\left[ \begin{array}{cccc}g_{-1,-1}&g_{-1,0}&g_{-1,1}\\ g_{0,-1}&g_{0,0}&g_{0,1}\\ g_{1,-1}&g_{1,0}&g_{1,1}\end{array} \right], \tag{1}\]  
\[\mathbf{F}=\left[ \begin{array}{cccc}f_{-2,-2}&f_{-2,-1}&f_{-2,0}&f_{-2,1}&f_ {-2,2}\\ f_{-1,-2}&f_{-1,-1}&f_{-1,0}&f_{-1,1}&f_{-1,2}\\ f_{0,-2}&f_{0,-1}&f_{0,0}&f_{0,1}&f_{0,2}\\ f_{1,-2}&f_{1,-1}&f_{1,0}&f_{1,1}&f_{1,2}\\ f_{2,-2}&f_{2,-1}&f_{2,0}&f_{2,1}&f_{2,2}\end{array} \right]. \tag{2}\]  
With matrices \(\mathbf{G}\) and \(\mathbf{F}\), we can achieve the signals \(d[j,k]\) and \(z[j,k]\), as follows:  
\[d[j,k]=\sum_{m=-1}^{1}\sum_{n=-1}^{1}a[j-m,k-n]g_{m,n} \tag{3}\] \[=\left[ \begin{array}{cccc}a[j+1,k+1]&\dots&a[j,k]&\dots&a[j-1,k- 1]\end{array} \right]\left[ \begin{array}{c}g_{-1,-1}\\ g_{-1,0}\\ \vdots\\ g_{1,1}\end{array} \right]\]  
Figure 1: Training process model. \[z[j,k] =\sum\limits_{m=-2}^{2}\sum\limits_{n=-2}^{2}y[j-m,k-n]f_{m,n}\] \[=\left[ \begin{array}{ccccc}y[j+2,k+2]&\ldots&y[j,k]&\ldots&a[j-2,k -2]\end{array} \right]\left[ \begin{array}{c}f_{-2,-2}\\ f_{-2,-1}\\ \vdots\\ f_{2,2}\end{array} \right] \tag{4}\]  
We assign the vectors as follows:  
\[\mathbf{a}=\left[ \begin{array}{ccccc}a[j+1,k+1]&\ldots&a[j-1,k-1]\end{array}  \right]^{T}, \tag{5}\]  
\[\mathbf{g}=\left[ \begin{array}{ccccc}g_{-1,-1}&g_{-1,0}&\ldots&g_{1,1}\end{array}  \right]^{T}, \tag{6}\]  
\[\mathbf{y}=\left[ \begin{array}{ccccc}y[j+2,k+2]&\ldots&y[j-2,k-2]\end{array}  \right]^{T},\text{ and} \tag{7}\]  
\[\mathbf{f}=\left[ \begin{array}{ccccc}f_{-2,-2}&f_{-2,-1}&\ldots&f_{2,2}\end{array}  \right]^{T}. \tag{8}\]  
Using the definitions from (5) to (8), we can rewrite (3) and (4) as follows:  
\[d[j,k]=\mathbf{g}^{T}\mathbf{a}, \tag{9}\]  
\[z[j,k]=\mathbf{f}^{T}\mathbf{y}. \tag{10}\]  
We find \(\mathbf{g}\) and \(\mathbf{f}\) in order that \(z[j,k]\) is close to \(d[j,k]\). To implement this, we apply the minimum mean square error (MMSE) algorithm. This problem is presented as follows:  
\[\underset{\mathbf{f},\mathbf{g}}{\operatorname{argmin}}E\Big{\{}(z[j,k]-d[j, k])^{2}\Big{\}}=\underset{\mathbf{t},\mathbf{g}}{\operatorname{argmin}}E\bigg{\{} \left(\mathbf{g}^{T}\mathbf{a}-\mathbf{f}^{T}\mathbf{y}\right)^{2}\bigg{\}}, \tag{11}\]  
where \(E\) denotes the expectation. We can expand (11) as follows:  
\[\underset{\mathbf{f},\mathbf{g}}{\operatorname{argmin}}\Big{(}\mathbf{f}^{T}E \Big{\{}\mathbf{y}\mathbf{y}^{T}\Big{\}}\mathbf{f}-2\mathbf{f}^{T}E\Big{\{} \mathbf{ya}^{T}\Big{\}}\mathbf{g}+\mathbf{g}^{T}E\Big{\{}\mathbf{aa}^{T}\Big{\}} \mathbf{g}\Big{)} \tag{12}\]  
Since the trivial answer for minimizing (12) is \(\mathbf{f}=\mathbf{g}=0\), we must impose the constraint on \(\mathbf{f}\) or \(\mathbf{g}\). The constraint is represented by the below expression. \[\mathbf{E}^{T}\mathbf{g}=c, \tag{13}\]  
where \(c\) is an estimated multiplier of the center data and investigated in Section 4, and  
\[\mathbf{E}^{T}=\left[ \begin{array}{ccccccccc}0&0&0&0&1&0&0&0&0\end{array}  \right]. \tag{14}\]  
With the above constraints, we can rewrite the optimal problem (12) as follows:  
\[\underset{\mathbf{t},\mathbf{g}}{\operatorname{argmin}}\Big{(} \mathbf{f}^{T}E\big{\{}\mathbf{y}\mathbf{y}^{T}\big{\}}\mathbf{f}-2\mathbf{f}^ {T}E\big{\{}\mathbf{ya}^{T}\big{\}}\mathbf{g}+\mathbf{g}^{T}E\big{\{}\mathbf{ aa}^{T}\big{\}}\mathbf{g}\Big{)}\] (15) s.t.