Massive Access with Channel Statistical Information

Abstract

In this chapter, we study the problem of massive access in the 5G cellular IoT, where the channels are fast-varying. To address the challenging issue of channel state information (CSI) acquisition and beam design for a massive number of IoT devices over fast time-varying fading channels, we design a non-orthogonal beamspace multiple access framework. In particular, the user equipments (UEs) are non-orthogonal not only in the temporal-frequency domain, but also in the beam domain. We analyze the performance of the proposed non-orthogonal beamspace multiple access scheme, and derive an upper bound on the weighted sum rate in terms of channel conditions and system parameters. For further improving the performance, we propose three non-orthogonal beam construction methods with different beamspace resolutions. Finally, extensively simulation results show the performance gain of the proposed non-orthogonal beamspace multiple access scheme over the baseline ones.

Appendix The Proof of Theorem 1

Prior to proving Theorem 1, we first provide the following lemma [11]:

Lemma 3

If A , B , and X are symmetric positive semi-definite matrices, the matrix function f(X) in below is concave with respect to X.

$$\displaystyle \begin{aligned} f(\mathbf{X})=\log_2\det(\mathbf{I}+\mathbf{A}\mathbf{X})-\log_2\det(\mathbf{I}+\mathbf{B}\mathbf{X}).{} \end{aligned} $$

(5.43)

According to the definition, the achievable rate of the UE_m,n in (5.13) can be transformed as

$$\displaystyle \begin{aligned} \begin{array}{rcl} r_{m,n}&\displaystyle =&\displaystyle \log_2\left(\frac{\sum_{j=1,j\neq m}^{M}\sum_{i=1}^{N_j}|\bar{\mathbf{h}}_{m,n}^H{\mathbf{v}}_{m,n,j,i}|{}^2+\sum_{i=1}^{n}|\bar{\mathbf{h}}_{m,n}^H{\mathbf{v}}_{m,n,m,i}|{}^2+1} {\sum_{j=1,j\neq m}^{M}\sum_{i=1}^{N_j}|\bar{\mathbf{h}}_{m,n}^H{\mathbf{v}}_{m,n,j,i}|{}^2+\sum_{i=1}^{n-1}|\bar{\mathbf{h}}_{m,n}^H{\mathbf{v}}_{m,n,m,i}|{}^2+1}\right)\\ &\displaystyle =&\displaystyle \log_2\det\left(\bar{\mathbf{h}}_{m,n}\bar{\mathbf{h}}_{m,n}^H\left(\sum_{j=1,j\neq m}^{M}\sum_{i=1}^{N_m}{\mathbf{V}}_{m,n,j,i}+\sum_{i=1}^{n}{\mathbf{V}}_{m,n,m,i}\right)+\mathbf{I}\right)\\ &\displaystyle &\displaystyle -\log_2\det\left(\bar{\mathbf{h}}_{m,n}\bar{\mathbf{h}}_{m,n}^H\left(\sum_{j=1,j\neq m}^{M}\sum_{i=1}^{N_m}{\mathbf{V}}_{m,n,j,i}+\sum_{i=1}^{n-1}{\mathbf{V}}_{m,n,m,i}\right)+\mathbf{I}\right),\\{} \end{array} \end{aligned} $$

(5.44)

where ${\mathbf {v}}_{m,n,j,i}=\varLambda _{m,n}^{\frac {1}{2}}{\mathbf {P}}_{j,i}^{\frac {1}{2}}{\mathbf {s}}_{j,i}$ and ${\mathbf {V}}_{m,n,j,i}={\mathbf {v}}_{m,n,j,i}{\mathbf {v}}_{m,n,j,i}^H$. Equation (5.44) holds true due to the fact that $\det (\mathbf {I}+\mathbf {AB})=\det (\mathbf {I}+\mathbf {BA})$. Based on Lemma 3, r _m,n is a concave function of $\|\bar {\mathbf {h}}_{m,n}\|{ }^2$. Thus, applying the Jensen’s inequality yields

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mathrm{E}\{r_{m,n}\}&\displaystyle \leq&\displaystyle \log_2\det\left(\mathrm{E}[\bar{\mathbf{h}}_{m,n}\bar{\mathbf{h}}_{m,n}^H]\left(\sum_{j=1,j\neq m}^{M}\sum_{i=1}^{N_m}{\mathbf{V}}_{m,n,j,i}+\sum_{i=1}^{n}{\mathbf{V}}_{m,n,m,i}\right)+\mathbf{I}\right)\\ &\displaystyle &\displaystyle -\log_2\det\left(\mathrm{E}[\bar{\mathbf{h}}_{m,n}\bar{\mathbf{h}}_{m,n}^H]\left(\sum_{j=1,j\neq m}^{M}\sum_{i=1}^{N_m}{\mathbf{V}}_{m,n,j,i}+\sum_{i=1}^{n-1}{\mathbf{V}}_{m,n,m,i}\right)+\mathbf{I}\right)\\ &\displaystyle =&\displaystyle \log_2\det\left(\sum_{j=1,j\neq m}^{M}\sum_{i=1}^{N_m}{\mathbf{V}}_{m,n,j,i}+\sum_{i=1}^{n}{\mathbf{V}}_{m,n,m,i}+\mathbf{I}\right)\\ &\displaystyle &\displaystyle -\log_2\det\left(\sum_{j=1,j\neq m}^{M}\sum_{i=1}^{N_m}{\mathbf{V}}_{m,n,j,i}+\sum_{i=1}^{n-1}{\mathbf{V}}_{m,n,m,i}+\mathbf{I}\right).{} \end{array} \end{aligned} $$

(5.45)

Then, the upper bound of the weighted sum of the ergodic rates can be written as

$$\displaystyle \begin{aligned} \begin{array}{rcl} R_{\mathrm{ub}}&\displaystyle =&\displaystyle \sum_{m=1}^{M}\sum_{n=1}^{N_m}\alpha_{m,n}\left(\log_2\det\left(\sum_{j=1,j\neq m}^{M}\sum_{i=1}^{N_m}{\mathbf{V}}_{m,n,j,i}+\sum_{i=1}^{n}{\mathbf{V}}_{m,n,m,i}+\mathbf{I}\right)\right.\\ &\displaystyle &\displaystyle \left.-\log_2\det\left(\sum_{j=1,j\neq m}^{M}\sum_{i=1}^{N_m}{\mathbf{V}}_{m,n,j,i}+\sum_{i=1}^{n-1}{\mathbf{V}}_{m,n,m,i}+\mathbf{I}\right)\right)\\ &\displaystyle =&\displaystyle \sum_{m=1}^{M}\sum_{n=1}^{N_m}\sum_{c=1}^{N_t}\alpha_{m,n}\left(\log_2\left(\left(\sum_{j=1,j\neq m}^{M}\sum_{i=1}^{N_j}s_{j,i,c}p_{j,i,c}+\sum_{i=1}^{n}s_{m,i,c}p_{m,i,c}\right)\eta_{m,n,c}+1\right)\right.\\ &\displaystyle &\displaystyle \left.-\log_2\left(\left(\sum_{j=1,j\neq m}^{M}\sum_{i=1}^{N_j}s_{j,i,c}p_{j,i,c}+\sum_{i=1}^{n-1}s_{m,i,c}p_{m,i,c}\right)\eta_{m,n,c}+1\right)\right),{} \end{array} \end{aligned} $$

(5.46)

where Eq. (5.46) follows the fact that V _m,n,j,i is a diagonal matrix. The proof completes.