matrix $\mathbf{R}$ are perfectly known at the BS. According to and , the downlink channel matrix defined in can be rewritten as[^5]$$\begin{split}
\mathbf{G}&=\hat{\mathbf{G}}+\tilde{\mathbf{G}}\\
&=\mathbf{D}^{1/2}\hat{\mathbf{H}}\mathbf{R}^{1/2}+\mathbf{D}^{1/2}\tilde{\mathbf{H}}\mathbf{R}^{1/2}.
\end{split}$$Upon invoking the estimated channel matrix $\hat{\mathbf{G}}$ and the ZF precoding, the transmitted vector $\mathbf{x}\in\mathbb{C}^{M\times 1}$ is written as $$\label{eq_zf_0}\begin{split}
\mathbf{x}&=\gamma\hat{\mathbf{G}}^{\dagger}\mathbf{s}\\
&=\gamma\hat{\mathbf{G}}^H(\hat{\mathbf{G}}\hat{\mathbf{G}}^H)^{-1}\mathbf{s},
\end{split}$$where $\gamma$ is the power controlling factor, $\mathbf{s}=[s_1,\cdots,s_K]^T$ is the $K\times 1$ information-bearing symbol vector and $s_k$ represents the symbol intended to the $k$-th active-UE. Denoting the available total transmit power at the BS as $P$, the long-term power constraint is given by $$\text{E}[\text{tr}(\mathbf{x}\mathbf{x}^H)]\leq P.$$ Thus, $\gamma$ can be calculated as$$\label{eq_pc_0}\gamma=\sqrt{\frac{P}{\text{tr}(\hat{\mathbf{G}}\hat{\mathbf{G}}^H)^{-1}}}.$$It should be noted that we employ the equal power allocation among active-UEs for the sake of low computational complexity. Relying on , the $K\times 1$ dimensional received signal vector at the $K$ active-UEs is expressed as $$\begin{split}\mathbf{y}&=\mathbf{G}\mathbf{x}+\mathbf{n}\\&=\gamma\mathbf{s}+\gamma\tilde{\mathbf{G}}\hat{\mathbf{G}}^{\dagger}\mathbf{s}+\mathbf{n},\end{split}$$where $\mathbf{n}=[n_1,\cdots,n_k]^T\in\mathbb{C}^{K\times 1}$ is the additive white Gaussian noise (AWGN) vector at the UEs, and $n_k\sim\mathcal{CN}(0,\sigma_n^2)$ represents the noise at the $k$-th active-UE. Additionally, the received signal received at the $k$-th active-UE is given by $$y_k=\gamma s_k+\gamma\sqrt{cd_k^{-\alpha}}\tilde{\mathbf{h}}_k\mathbf{R}^{1/2}\hat{\mathbf{G}}^{\dagger}\mathbf{s}+n_k.$$Then, we can write the SINR recorded at the $k$-th active-UE as $$\label{eq_sinr_0}\text{SINR}_k=\frac{\gamma^2}{\sigma_n^2+cd_k^{-\alpha}\gamma^2\tilde{\mathbf{h}}_k\mathbf{R}^{1/2}\hat{\mathbf{G}}^{\dagger}\mathbf{s}\mathbf{s}^H(\hat{\mathbf{G}}^{\dagger})^H(\mathbf{R}^{1/2})^H\tilde{\mathbf{h}}_k^H}.$$Then, the sum rate $\mathcal{R}$ is given by $$\label{eq_r_0}\begin{split}
\mathcal{R}&=\left(1-\frac{K}{T}\right)\sum_{k=1}^K\mathcal{R}_k\\
&=\left(1-\frac{K}{T}\right)\sum_{k=1}^K\log_2(1+\text{SINR}_k),
\end{split}$$where $\mathcal{R}_k$ is the rate of the $k$-th 