## 2 System Model
### Problem Formulation  
In the DD domain, the two-dimensional convolution of the input signal with the DD domain channel yields the output signal. To facilitate channel estimation, Equation (18) is converted to the following form:  
\[\mathbf{y}=\boldsymbol{\mathcal{X}}\mathbf{h}+\widetilde{\mathbf{w}} \tag{19}\]  
where \(\mathbf{y}=\text{vec}(\mathbf{Y}^{\text{T}})\in\mathbb{C}^{MN\times 1}\) is serial receiving signal in the DD domain, and \(\mathbf{h}\in\mathbb{C}^{MN\times 1}\) is the vector form of the DD channel response. In addition, \(\boldsymbol{\mathcal{X}}=[\boldsymbol{\mathcal{X}}_{0},\ldots,\boldsymbol{ \mathcal{X}}_{I},\ldots,\boldsymbol{\mathcal{X}}_{I-1}]\in\mathbb{C}^{MN\times MNL}\) is composed of the transmission vector of the DD domain, in which \(\boldsymbol{\mathcal{X}}_{I}\in\mathbb{C}^{MN\times MN}\) is a block diagonal matrix, including \(M\) small matrix \(\boldsymbol{\Delta}_{i}\) of dimension \(N\times N\). Define a new index \(r_{l}=\text{cirshift}(1:M,l)\in\mathbb{C}^{1\times M}\), and let \(X(r_{l}(i),:)=[h_{i}^{l}(0),\ldots,h_{i}^{l}(N-1)]\) to obtain the following:  
\[\begin{array}{ll}\Delta_{i}&=\text{circ}(h_{i}^{l}(0),\ldots,h_{i}^{l}(N-1)) \\ &=\left[\begin{array}{cccc}h_{i}^{l}(0)&h_{i}^{l}(N-1)&\cdots&h_{i}^{l}(1)\\ h_{i}^{l}(1)&h_{i}^{l}(0)&\cdots&h_{i}^{l}(2)\\ \vdots&\vdots&\ddots&\vdots\\ h_{i}^{l}(N-1)&h_{i}^{l}(N-2)&\cdots&h_{i}^{l}(0)\end{array}\right]\end{array} \tag{20}\]  
From Equation (19), it is found that the dimension of the parameter to be solved \(\mathbf{h}\) is larger than the dimension of the observed value \(\mathbf{y}\), which suggests that the equation is an underdetermined equation and not amenable to direct solution. To estimate \(\mathbf{h}\), we employ the BEM to reduce the number of parameters to be solved, which can use a fixed set of basis functions and a linear combination of its basis coefficients to fit the time-domain channel response. In this paper, DPS-BEM [23] is adopted to model the time-domain channel tap as follows:  
\[g_{l}(n)=\sum_{q=0}^{Q-1}b_{q}(n)c_{q}(l)+\varepsilon_{l}(n) \tag{21}\]  
where \(\mathbf{b}_{q}\in\mathbb{C}^{MN\times 1}\), \(\mathbf{c}_{l}\in\mathbb{C}^{Q}\), and \(\varepsilon_{l}\in\mathbb{C}^{N}\) denote the basis vector, basis coefficients and error vector, respectively. Based on the relationship between the DD domain channel and the time domain, we can obtain the following matrix expression:  
\[\begin{array}{ll}\mathbf{h}_{l}&=\frac{1}{\sqrt{N}}(\mathbf{F}_{N}\otimes \mathbf{I}_{M})\mathbf{B}\mathbf{c}_{l}+\frac{1}{\sqrt{N}}(\mathbf{F}_{N} \otimes\mathbf{I}_{M})\mathbf{\varepsilon}_{l}\\ &=\mathbf{A}\mathbf{c}_{l}+\widetilde{\mathbf{\varepsilon}}_{l}\end{array} \tag{22}\]  
By substituting Equation (22) into Equation (19), the relationship between the observed signal and the basis coefficients can be obtained as follows:  
\[\begin{array}{ll}\mathbf{y}&=\boldsymbol{\mathcal{X}}(\mathbf{I}_{\text{L} }\otimes\mathbf{A})\mathbf{c}+\widetilde{\mathbf{w}}\\ &=\boldsymbol{\Phi}\mathbf{c}+\widetilde{\mathbf{w}}\end{array} \tag{23}\]  
where \(\mathbf{c}=[\mathbf{c}_{0},\ldots,\mathbf{c}_{l},\ldots,\mathbf{c}_{L-1}]^{ \text{T}}\in\mathbb{C}^{QL}\) denotes the basis coefficients matrix, \(\boldsymbol{\Phi}\triangleq\boldsymbol{\mathcal{X}}(\mathbf{I}_{\text{L}} \otimes\mathbf{A})\in\mathbb{C}^{MN\times QL}\).