## 2 System Model
### ZF Beamforming for MU-MIMO Wireless Networks  
Consider the downlink of an MU-MIMO system with a single Access Point (AP) operating over a single band using \(N\) transmit antennas and a single antenna receiver for each of the \(M\) Mobile Stations (MSs). The combined vector channel can be described as:  
\[\mathbf{y}=\mathbf{H}\mathbf{W}\mathbf{x}+\mathbf{z}. \tag{3}\]where \(\mathbf{y},\mathbf{x},\mathbf{z}\in\mathbb{C}^{M\times 1}\), a channel gain matrix \(\mathbf{H}\in\mathbb{C}^{M\times N}\), and a beamforming matrix \(\mathbf{W}\in\mathbb{C}^{N\times M}\).  
Let:  
\[\mathbf{H}^{*}=\mathbf{Q}\mathbf{R} \tag{4}\]  
be a \(\mathbf{Q}\mathbf{R}\) decomposition of \(\mathbf{H}^{*}\)[47]. Let \(\mathbf{L}=\mathbf{R}^{*}\), and let \(\mathbf{H}=\mathbf{L}\mathbf{Q}^{*}\). Assume that \(\mathbf{W}=\mathbf{Q}\). The equivalent channel is given by:  
\[\mathbf{y}=\mathbf{L}\mathbf{x}. \tag{5}\]  
Note that at this stage, each user is not subject to interference by users with a higher index [48]. To complete the ZF beamforming, we need to invert \(\mathbf{L}\) and normalize the total transmit power. Practically, since \(\mathbf{L}\) is a lower triangular matrix, we can use forward substitution to compute the transmitted vector. Furthermore, we can normalize the total power of the transmitted vectors prior to beamforming with \(\mathbf{Q}\), since \(\mathbf{Q}\) is unitary. Let \(\mathbf{s}\in\mathbb{C}^{M\times 1}\) be the vector of symbols that are required to be transmitted to the different mobile stations.  
We also use the following assumptions:  
* The AP has perfect Channel State Information (CSI).
* Transmitted data symbols (\(s_{k}\)) are uncorrelated
* The power allocated to each user is constant. To simplify notation, we assume that the same power is allocated to all users, i.e., \(E\{s_{k}^{2}\}=1\) and \(E\{x_{k}^{2}\}=\mathcal{P}\).
* Let \(\mathbf{x}=\mathbf{L}^{-1}\mathbf{G}\ \mathbf{s}\) where \(\mathbf{G}\) is a diagonal \(M\times M\) real gain matrix. Since \(L\mathbf{x}=\mathbf{G}\mathbf{s}\), then \(\mathbf{x}\) and \(\mathbf{G}\) can be calculated using forward substitution as described below: \[\sum_{j=1}^{k}l_{kj}x_{j}=g_{kk}s_{k}\implies x_{k}=\frac{1}{l_{kk}}\left(g_{ kk}s_{k}-\sum_{j=1}^{k-1}l_{kj}x_{j}\right).\] (6) Assuming \(\mathbf{A}\), then: \[E\{x_{k}^{2}\}=\frac{1}{l_{kk}^{2}}E\left\{(g_{kk}s_{k}-\sum_{j=1}^{k-1}l_{kj}x_ {j})^{2}\right\}=\frac{1}{l_{kk}^{2}}\left(g_{kk}^{2}+\mathcal{P}\sum_{j=1}^{k -1}l_{kj}^{2}\right)=\mathcal{P}.\] (7) From Equation (7), it can be derived that: \[\begin{split}& g_{kk}=\sqrt{\mathcal{P}\left((l_{kk}^{2}-\beta_{k} \sum_{j=1}^{k-1}l_{kj}^{2})\right)},\\ &\text{where }\beta_{k}\in[0,1]\text{ and }g_{k,k}\text{ is always real.}\end{split}\] (8) Define \(\mathbf{x}\) by: \[x_{k}=\frac{1}{l_{kk}}\left(\sqrt{\mathcal{P}(l_{kk}^{2}-\beta_{k}\sum_{j=1}^{ k-1}l_{kj}^{2})}s_{k}-\sum_{j=1}^{k-1}l_{kj}x_{j}\right).\] (9) The energy that is allocated to transmit \(s_{k}\) is: \[\mathcal{P}\left(1-\frac{\beta_{k}}{l_{kk}^{2}}\sum_{j=1}^{k-1}l_{kj}^{2}\right).\] (10)Therefore, the SINR as a function of \(\beta_{k}\) is given by:  
\[SINR(\beta_{k})=\frac{\mathcal{P}\left(1-\frac{\beta_{k}}{l_{kk}}\sum\limits_{j=1 }^{k-1}l_{kj}^{2}\right)}{l_{kk}^{2}(1-\beta_{k})\sum\limits_{j=1}^{k-1}l_{kj}^ {2}\mathcal{P}+\sigma_{k}^{2}}. \tag{11}\]  
The optimal \(\beta_{k}\) solves the following problem:  
\[\beta_{k}=arg\max\limits_{\beta_{k}\in[0,1]}(SINR(\beta_{k})). \tag{12}\]  
The achievable bit rate for MS \(k\) is given by:  
\[r_{k}=\log_{2}(1+SINR(\beta_{k})). \tag{13}\]