## 3 Structured Sparse Bayesian Approach for Channel Estimation
### Structured Sparse Bayesian Channel Estimation Problem Formulation
\begin{split}&\hat{\alpha}_{l,k}^{(i+1)}=\arg\max_{\alpha_{l,k}} \mathcal{G}(\alpha_{l,k},\hat{\gamma}^{(i)},\mathbf{y}|\hat{\alpha}_{l,k}^{(i)}, \hat{\gamma}^{(i)})\\ &\hat{\gamma}^{(i+1)}=\arg\max_{\alpha_{l,k}}\mathcal{G}(\hat{ \alpha}_{l,k}^{(i+1)},\gamma,\mathbf{y}|\hat{\alpha}_{l,k}^{(i)},\hat{\gamma}^{(i)} )\end{split}  \tag{22}\]  
where \(\left\{\hat{\alpha}_{l,k}^{(i)},\hat{\gamma}^{(i)}\right\}\) represents \(\alpha_{l,k}\), \(\mathit{the}\gamma\) estimated value in the \(i\)-th iteration. After a series of mathematical calculations, the specific algorithm is as follows. (1) Step 1: A substitute for the objective function \(-\log p(\alpha_{l,k},\gamma,\mathbf{y})\) is required at the corresponding step of the \((i+1)\)th iteration. In this study, we make use of the surrogate function proposed by [21], which meets the requirements stated in (21), that is, a simple proof is provided in Appendix B. \[ \begin{split}&\mathcal{G}(\alpha_{l,k},\gamma,\mathbf{y}|\hat{ \alpha}_{l,k}^{(i)},\hat{\gamma}^{(i)})=\\ &-\int p(\mathbf{h}|\mathbf{y},\hat{\alpha}_{l,k}^{(i)},\hat{\gamma} ^{(i)})\log\frac{p(\alpha_{l,k},\gamma,\mathbf{h},\mathbf{y})}{p(\mathbf{h}|\mathbf{y}, \hat{\alpha}_{l,k}^{(i)},\hat{\gamma}^{(i)})}d\mathbf{h}\\ &\geq-\log p(\alpha_{l,k},\gamma,\mathbf{y})\end{split}  \tag{23}\]  
where the final inequation obeys [PERSON]'s inequality, and when the inequation satisfies \((\alpha_{l,k},\gamma)=(\alpha_{l,k}^{(i)},\hat{\gamma}^{(i)})\), the equal sign is true. By substituting (15) into (23), we can obtain  
\[ \begin{split}\mathcal{G}(\alpha_{l,k},\gamma,\mathbf{y}|\hat{\alpha} _{l,k}^{(i)},\hat{\gamma}^{(i)})&=-\log p(\alpha_{l,k})-\log p( \gamma)\\ &-\mathbb{E}_{[\log p(\mathbf{h}|\alpha_{l,k},\gamma)]}+const\end{split}  \tag{24}\]  
where the expectation is in relation to \(p(\mathbf{h}|y,\hat{\alpha}_{l,k}^{(i)},\hat{\gamma}^{(i)})\). \(const\) denotes that it is not related to \(\alpha_{l,k}\) or \(\gamma\). (2) Step 2: To minimize the surrogate function, the parameters \(\alpha_{l,k}\) are changed sequentially in the corresponding step of the \((i+1)\)th iteration. To update \(\alpha_{l,k}\); \(\gamma\) is regarded as a constant when \(\alpha_{l,k}\) is estimated. By substituting (13) and (18) into (24), we obtain  
\[ \begin{split}&\mathcal{G}(\alpha_{l,k},\hat{\gamma}^{(i)},\mathbf{y}| \hat{\alpha}_{l,k}^{(i)},\hat{\gamma}^{(i)})\\ &=-\sum_{k=1}^{2k_{\max}+2Q+1}(a\log\alpha_{l,k}-b\alpha_{l,k}) \\ &-\log\lvert diag(\mathbf{\gamma})\times diag(\mathbf{D}_{l})\rvert\\ &+tr((diag(\mathbf{\gamma})\times diag(\mathbf{D}_{l}))\times(\sum+\hat{ \mathbf{\mu}}\hat{\mathbf{\mu}}^{H}))\\ &+const\end{split}  \tag{25}\]  
where Equation (17) is used to compute \(\mathbf{\hat{\sum}}\) and \(\hat{\mathbf{\mu}}\), and \((\alpha_{l,k},\gamma)\) is the result obtained from the \(i\)-th iteration. Next, we consider the derivative of (25) with regard to \(\alpha_{l,k}\) to analyse the optimality condition. \[ \begin{split}\frac{\partial\mathcal{G}(\alpha_{l,k},\hat{\gamma}^ {(i)},\mathbf{y}|\hat{\alpha}_{l,k}^{(i)},\hat{\gamma}^{(i)})}{\partial\alpha_{l, k}}&=\frac{a}{\alpha_{l,k}}-b-\hat{\gamma}^{(i)}\mathbf{g}\\ &+(\xi_{k-1}+\xi_{k}+\xi_{k+1})\end{split}  \tag{26}\]  
where  
\[\mathbf{g}=\sum_{j=1}^{L}\] \[\left\{ \begin{array}{l}\left[\mathbf{\hat{\Sigma}}\right]_{[j,l]}+ \left\lvert\left[\mathbf{\hat{\mu}}\right]_{j}\right\rvert^{2}\\ +\left[\mathbf{\hat{\Sigma}}\right]_{[[j+MAX]_{L},[j+MAX]_{L}]}+\left\lvert\left[ \mathbf{\hat{\mu}}\right]_{[j+MAX]_{L}}\right\rvert^{2}\\ +\left\lvert\left[\mathbf{\hat{\mu}}\right]_{[\max,MAX+j]_{L},[j_{\max}-MAX+j]_{L }]}\right\rvert^{2}\\ +\left\lvert\left[\mathbf{\hat{\mu}}\right]_{[\max,MAX+j]_{L}}\right\rvert^{2} \end{array} \right\} \tag{27}\]  
\[\xi_{k}=\frac{1}{\alpha_{l,k-1}+\alpha_{l,k}+\alpha_{l,k+1}} \tag{28}\]  
Specifically, when \(k=1\)  
\[\xi_{k}=\frac{1}{\alpha_{1}+\alpha_{2}}\] (28a) when \[k=\max \tag{28b}\]  
where \(MAX=2k_{max}+2Q+1\).