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Energy Efficiency of Massive Random Access in MIMO Quasi-Static Rayleigh Fading Channels with Finite Blocklength (2210.11970v1)

Published 21 Oct 2022 in cs.IT and math.IT

Abstract: This paper considers the massive random access problem in MIMO quasi-static Rayleigh fading channels. Specifically, we derive achievability and converse bounds on the minimum energy-per-bit required for each active user to transmit $J$ bits with blocklength $n$ and power $P$ under a per-user probability of error (PUPE) constraint, in the cases with and without \emph{a priori} channel state information at the receiver (CSIR and no-CSI). In the case of no-CSI, we consider both the settings with and without knowing the number $K_a$ of active users. The achievability bounds rely on the design of the ``good region''. Numerical evaluation shows the gap between achievability and converse bounds is less than $2.5$ dB in the CSIR case and less than $4$ dB in the no-CSI case in most considered regimes. When the distribution of $K_a$ is known, the performance gap between the cases with and without knowing the value of $K_a$ is small. For example, in the setup with blocklength $n=1000$, payload $J=100$, error requirement $\epsilon=0.001$, and $L=128$ receive antennas, compared to the case with known $K_a$, the extra required energy-per-bit when $K_a$ is unknown and distributed as $K_a\sim\text{Binom}(K,0.4)$ is less than $0.3$ dB on the converse side and $1.1$ dB on the achievability side. The spectral efficiency grows approximately linearly with $L$ in the CSIR case, whereas the growth rate decreases with no-CSI. Moreover, we study the performance of a pilot-assisted scheme, which is suboptimal especially when $K_a$ is large. Building on non-asymptotic results, when all users are active and $J=\Theta(1)$, we obtain scaling laws as follows: when $L=\Theta \left(n2\right)$ and $P=\Theta\left(\frac{1}{n2}\right)$, one can reliably serve $K=\mathcal{O}(n2)$ users with no-CSI; under mild conditions with CSIR, the PUPE requirement is satisfied if and only if $\frac{nL\ln KP}{K}=\Omega\left(1\right)$.

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