Uplink SINR and rate analysis in massive MIMO systems with two-layer linear receiver

Abstract

In a previous work, a two-layer linear receiver was proposed as a low-complexity scheme that can achieve a good trade-off between performance and complexity in a massive multiple-input multiple-output (MIMO) context. For this scheme, the antenna array is split up into a number of subsets and multi-cell minimum mean-square-error (M-MMSE) combining is applied in each subset at the first layer. Then, the resulting outputs are combined using maximal-ratio combining (MRC) at the second layer. This paper presents a more complete performance analysis. First, we provide an entirely analytical derivation of the distribution of the output signal-to-noise-and-interference ratio (SINR) of first-layer processing based on multivariate statistical theory. The analysis includes both cases where the subset size is smaller or greater than the total number of signals. Furthermore, upper bounds and approximations on the average uplink SINR and achievable rate are derived. Numerical simulations confirm our analysis results.

Appendices

Appendix A

In this appendix, we provide the derivation of (27). From (26), the expression for the PDF of $\gamma _{mk}$ can be expanded by applying the binomial theorem to the polynomial $\left( x-t\right) ^{N-d_{s}-1}$ in the integrand, i.e.

$$\begin{aligned} f_{\gamma _{mk}}(x)= & {} \dfrac{\Gamma (N\!+\!1)\,P_{0}\, \sigma _{n}^{2(N-d_{s})}\,e^{- \frac{\sigma _{n}^{2}}{P_{0}}x}}{\Gamma (d_{s}) \Gamma (N\!\!-\! d_{s}\!+\!1) \Gamma (N\!-\!d_{s}) \alpha _{s}^{N\!-d_{s}+1}} \nonumber \\{} & {} \; \times \, \sum \limits _{i=0}^{N\!-\!d_{s}\!-\!1} \! (-1)^{i} \left( {\begin{array}{c}N\!\!-\!d_{s}\!-\!1\\ i\end{array}}\right) x^{N-d_{s}-i-1} \nonumber \\[7pt]{} & {} \; \times \, \int _{0}^{x} \frac{t^{d_{s}+i-1}}{\left( \frac{P_{0}}{\alpha _{s}}+t \right) ^{N+1}} \, e^{\frac{\sigma _{n}^{2}}{P_{0}}t} \; dt. \ \end{aligned}$$

(A1)

Let

$$\begin{aligned} I= \int _{0}^{x} \frac{t^{d_{s}+i-1}}{\left( \frac{P_{0}}{\alpha _{s}}+t\right) ^{N+1}} \, e^{\frac{\sigma _{n}^{2}}{P_{0}}t} \; dt. \end{aligned}$$

(A2)

Then, the integral can be further expanded using the power series representation of the exponential function, i.e.

$$\begin{aligned} {} I= & {} \sum \limits _{i=0}^{\infty } \frac{1}{j!} \left( \frac{\sigma _{n}^{2}}{P_{0}}\right) ^{j} \int _{0}^{x} \frac{t^{d_{s}+i+j-1}}{\left( \frac{P_{0}}{\alpha _{s}}+t\right) ^{N+1}} \; dt \nonumber \\[3pt]\overset{(a)}{=} & {} \sum \limits _{i=0}^{\infty } \frac{1}{j!} \left( \frac{\sigma _{n}^{2}}{P_{0}}\right) ^{j} \left( \frac{\alpha _{s}}{P_{0}}\right) ^{N+1} \frac{x^{d_{s}+i+j}}{d_{s}+i+j} \times \nonumber \\ {}{} & {} {}_2F_1\left( N{+}1,d_{s}\!{+}\!i\!+\!j;d_{s}\!{+}\!i\!+\!j+1;-\frac{\alpha _{s}}{P_{0}}x\right) , \end{aligned}$$

(A3)

where the integral is solved in step (a) by virtue of [28, eq. 3.194-1]. We note that ${}_2F_1\left( a,b;c;z\right)$ is the Gaussian hypergeometric function. Finally, substituting (A3) in (A1), we obtain the desired result (27).

Appendix B

From (40), we apply the change of variables $s=t-\sigma _{n}^{2}x$ to obtain

$$\begin{aligned} {} \! \! \! \! \! f_{\gamma ^{u}_{mk}}(x)= & {} \dfrac{ e^{ -\frac{ \sigma _{n}^{2}x}{P_{0}} } }{\Gamma (N)\,\Gamma (d_{s}\!-\!N\!+\!1)\;\alpha _{s}^{d_{s}\!-\!N\!+\!1}P_{0}^{N}\, x^{d_{s}\!-\!N\!+2}} \times \nonumber \\{} & {} \; \int _{0}^{\infty } \, s^{d_{s}-N} \Big (s+\sigma _{n}^{2}x\Big )^{N} \! e^{-s \left( \frac{1}{P_{0}}+\frac{1}{\alpha _{s}x}\right) } \,ds. \end{aligned}$$

(B4)

Performing a binomial expansion of the polynomial in the integrand yields

$$\begin{aligned} {} \! \! \! \! \! \! f_{\gamma ^{u}_{mk}}(x)= & {} \dfrac{ e^{ -\frac{ \sigma _{n}^{2}x}{P_{0}} } }{\Gamma (N)\,\Gamma (d_{s}\!-\!N\!+\!1)\;\alpha _{s}^{d_{s}\!-\!N\!+\!1}P_{0}^{N}\, x^{d_{s}\!-\!N\!+2}} \times \nonumber \\ {}{} & {} \sum \limits _{i=0}^{N} \left( {\begin{array}{c}N\\ i\end{array}}\right) \left( \sigma _{n}^{2}x\right) ^{i} \int _{0}^{\infty } \! s^{d_{s}-i} e^{-s \left( \frac{1}{P_{0}}+\frac{1}{\alpha _{s}x}\right) } ds, \end{aligned}$$

(B5)

where the resulting integral corresponds to the definition of the Gamma function. Therefore, with some manipulations, we obtain the desired result (41).