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Re-Transmission Diversity Multiple Access based on SIC and HARQ-IR

Jinho Choi The author is with School of Electrical Engineering and Computer Science, Gwangju Institute of Science and Technology (GIST), Gwangju, 61005, Korea (Email: [email protected]). This work was supported by the GIST Research Institute (GRI) in 2016.
Abstract

We consider multiple access with re-transmission diversity (RTxD) based on a hybrid automatic request (HARQ) protocol with incremental redundancy (IR) in this paper. In order to mitigate multiple access interference (MAI), we employ successive interference cancellation (SIC) at a receiver and derive conditions that all the signals from active users can be successfully decoded within a certain number of re-transmissions of IR blocks. We consider two special cases, and based on them we derive two multichannel random access schemes that have sub-channels in the power and rate domains. Through the analysis and simulations, we can show that the proposed RTxD random access schemes can outperform other existing similar RTxD random access schemes.

Index Terms:
random access; re-transmission diversity; HARQ; successive interference cancellation

I Introduction

There have been various coordinated multiple access schemes to share a common radio resource block (in the frequency or time domain) in wireless communications, especially for uplink transmissions [1]. As opposed to coordinated multiple access, it is often desirable to consider uncoordinated multiple access to accommodate a number of users of low activity with limited radio resource for packet communications using random access schemes [2]. For example, slotted ALOHA [3] and carrier sense multiple access (CSMA) [4] are widely studied for random access. While random backoff plays a key role in mitigating multiple access interference (MAI), which results in packet collision, for most random access schemes (such as slotted ALOHA and CSMA), multiuser detection and other signal processing approaches can be used to mitigate MAI when packets from multiple users are collided in random access. Similar to IDMA, network-assisted diversity multiple access (NDMA) [5] employs multiuser detection with re-transmissions for diversity combining. In particular, if there are MM collided packets at a time, users re-transmit the collided packets MM times in total so that a receiver is able to separate them.

While re-transmission diversity111RTxD is used as the abbreviation of re-transmission diversity to differentiate it from repetition time diversity (RTD). (RTxD) is exploited to separate collided signals in NDMA, it has also been employed in link-layer protocols for single-user transmissions. For example, in hybrid automatic repeat request (HARQ) protocols, RTxD together with channel coding plays a crucial role in providing reliable transmissions over fading channels [6, 7]. HARQ can also be applied to random access [8], where RTxD with incremental redundancy (IR) blocks allows to achieve the channel capacity. It is shown in [9] that the rate optimization can improve the performance of ARQ protocols in terms of outage probability and throughput. In [10], a performance analysis of HARQ protocols is carried out under block-fading, where closed-form expressions are derived for the throughput and outage probability. In [11], closed-form expressions are obtained for the throughputs of various HARQ protocols including those in [9]. HARQ is applied to a multiuser system with multiple orthogonal channels in [12], where resource allocation is considered with HARQ to improve the network outage and fairness.

The notion of RTxD is also used to improve the probability of transmission success in slotted ALOHA [13]. In conjunction with RTxD, signal processing approaches at a receiver such as successive interference cancellation (SIC) can be used to suppress the MAI from collided packets [14, 15].

In this paper, we consider multiple access with both HARQ-IR and SIC that can exploit RTxD and mitigate MAI, respectively. The resulting RTxD multiple access scheme is similar to that in [8] except for SIC. It can also be seen as a generalization of that in [14] with re-transmissions of IR blocks instead of identical coded blocks for better performances. The proposed scheme can not only take advantage of IR to have a smaller number of re-transmissions, but also exploit SIC to effectively mitigate MAI. With the proposed scheme, we find the conditions to guarantee successful decoding of all active signals within a certain number of re-transmissions of IR blocks from active users. While we mainly focus on the application of the proposed scheme to random access in this paper, the proposed scheme can also be employed for coordinated multiple access where conditions for successful decoding of all active signals within a certain number of re-transmissions can be imposed by a scheduler. In this case, we can show that the spectral efficiency can be higher than that of orthogonal multiple access due to SIC.

The main contributions of the paper are as follows: i) We apply HARQ-IR to multiple access with SIC at a receiver to derive an RTxD multiple access scheme, which outperforms other existing random access schemes, and find conditions for successful decoding within a certain number of re-transmissions; ii) From those conditions with some special cases, we further derive two multichannel random access schemes that have sub-channels in the power domain and rate domain.

The rest of the paper is organized as follows. In Section II, we present the system model and the proposed scheme for RTxD multiple access with HARQ-IR and SIC. We analyze the length of frame or required number of re-transmissions for successful decoding and consider two special cases of the proposed scheme in Section III. With multiple power levels, we propose a multichannel random access scheme in Section IV. We also derive another multichannel random access scheme with multiple rates in Section V. Both random access schemes are based on two special cases studied in Section III. We present simulation results in Section VI and conclude the paper with some remarks in Section VII.

Notation: Matrices and vectors are denoted by upper- and lower-case boldface letters, respectively. 𝐈{\bf I} stands for the identity matrix. 𝔼[]{\mathbb{E}}[\cdot] and Var(){\rm Var}(\cdot) denote the statistical expectation and variance, respectively. 𝒞𝒩(𝐚,𝐑){\cal C}{\cal N}({\bf a},{\bf R}) (𝒩(𝐚,𝐑){\cal N}({\bf a},{\bf R})) represents the distribution of circularly symmetric complex Gaussian (CSCG) (resp., real-valued Gaussian) random vectors with mean vector 𝐚{\bf a} and covariance matrix 𝐑{\bf R}.

II System Model

In this section, we consider a system consisting of a base station (BS) and multiple users for uplink transmissions by sharing a common channel. Denote by KK the number of users. We assume that a frame consists of multiple slots. At each frame, a subgroup of users can be active and send signals to the BS. Each active user employs HARQ-IR. For RTxD, an active user keeps sending IR blocks in every slots until the BS sends a positive acknowledgment (ACK). Note that in the conventional HARQ-IR scheme, each user may have an orthogonal radio resource block so that there is no interference from other users. On the other hand, in the proposed scheme, multiple active users share a common radio resource block for multiple access. Thus, the received signal during time slot tt is given by

𝐱t=khk𝐬k,t+𝐧t,{\bf x}_{t}=\sum_{k\in{\cal I}}h_{k}{\bf s}_{k,t}+{\bf n}_{t}, (1)

where hkh_{k} and 𝐬k,t{\bf s}_{k,t} represent the channel coefficient and IR block, respectively, from user kk\in{\cal I} over time slot tt to the BS, and 𝐧t𝒞𝒩(0,𝐈){\bf n}_{t}\sim{\cal C}{\cal N}(0,{\bf I}) is the background noise (note that the noise variance is normalized for convenience). In addition, {\cal I} denotes the index set of active users in the current frame. We assume that {\cal I} is invariant within a frame and denote by MM the number of active users, i.e., M=||M=|{\cal I}|. For convenience, throughout the paper, we assume that 𝐬k,t𝒞𝒩(𝟎,Pk𝐈){\bf s}_{k,t}\sim{\cal C}{\cal N}({\bf 0},P_{k}{\bf I}), i.e., Gaussian codebooks are used for IR blocks, where PkP_{k} is the transmission power of user kk and each IR block, 𝐬k,t{\bf s}_{k,t}, is a codeword. In addition, for convenience, we let αk=|hk|2\alpha_{k}=|h_{k}|^{2} and assume that {αk}\{\alpha_{k}\} is known at the BS.

It is clear that conventional HARQ-IR becomes a special case with ||=1|{\cal I}|=1. That is, the above multiple access scheme reduces to conventional HARQ-IR if there is only one user at a time, say user kk. In this case (i.e., conventional HARQ-IR), the minimum number of IR blocks required for successful decoding under the assumption of Gaussian codebooks can be given by [8, 16]

TCH=min{T|t=1Tlog2(1+αkPk)Rk},T_{\rm CH}=\min\left\{T\,|\,\sum_{t=1}^{T}\log_{2}(1+\alpha_{k}P_{k})\geq R_{k}\right\}, (2)

where RkR_{k} is the (initial) rate of user kk and TT represents the number of re-transmissions. Note that TCHT_{\rm CH} is a random variable (stopping time). As mentioned earlier, multiple users can transmit signals using HARQ-IR without MAI as in (2) if they are assigned to orthogonal radio resource blocks.

At each time slot, the BS is to decode signals from users. As usual in HARQ protocols, if the BS fails to decode signals, it sends a negative acknowledgment (NACK). If the BS can decode certain users’ signals, it broadcasts ACK with the indices of those users. The users whose signals are decoded by the BS do not send signals in the rest of the current frame, while the other active users keep sending IR blocks. The decoding process at the BS is divided into multiple stages, and in each stage, we assume that the BS succeeds to decode signals from an active user with SIC. As a result, we can have

(n)=(n1)k(n),{\cal I}(n)={\cal I}(n-1)\setminus k(n), (3)

where \setminus represents the set difference (i.e., BA={xB|xA}B\setminus A=\{x\in B\,|\,x\notin A\}), (n){\cal I}(n) denotes the index set of active users whose signals are not decoded at stage nn, and k(n)k(n) represents the index of the user whose signals are successfully decoded at stage nn. Clearly, we have (0)={\cal I}{(0)}={\cal I} and the number of stages is MM (in this case, at each stage, we assume that only one user’s signal is decoded).

III The Length of Frame

In this section, we study the length of frame, denoted by TFT_{\rm F}, or the required number of re-transmissions of IR blocks from the last active users for successful decoding in the proposed RTxD multiple access scheme. Clearly, TFT_{\rm F} is a random variable that depends on the number of active users, their channel coefficients, and signal powers.

III-A The Length of Frame

The number of IR blocks required for stage 1 or the first successful decoding is given by

τ(1)=mink(0)τk(1),\tau{(1)}=\min_{k\in{\cal I}(0)}\tau_{k}(1),

where

τk(1)=min{T|t=1Tlog2(1+γk(0))Rk},k(0).\tau_{k}(1)=\min\left\{T\,|\,\sum_{t=1}^{T}\log_{2}(1+\gamma_{k}(0))\geq R_{k}\right\},\ k\in{\cal I}(0).

Here, γk(0)\gamma_{k}(0) is the signal-to-interference-plus-noise ratio (SINR) before SIC or at stage 0, which is given by

γk(0)=αkPkm(0)kαmPm+1,k(0),\gamma_{k}(0)=\frac{\alpha_{k}P_{k}}{\sum_{m\in{\cal I}(0)\setminus k}\alpha_{m}P_{m}+1},\ k\in{\cal I}(0), (4)

where m(0)kαmPm\sum_{m\in{\cal I}(0)\setminus k}\alpha_{m}P_{m} is the power of MAI when the signal from user kk is to be decoded. As mentioned earlier, since k(1)k(1) represents the index of the first user whose signals can be successfully decoded at slot T(1)T{(1)} in stage 1, we have k(1)=argmink(0)Tk(1)k(1)=\operatorname*{\arg\!\min}_{k\in{\cal I}(0)}T_{k}(1).

Denote by 𝐱t(n){\bf x}_{t}(n) the received signal after SIC at stage nn during slot tt. Let 𝐱t(0)=𝐱t{\bf x}_{t}(0)={\bf x}_{t}. At stage nn, up to time slot t=τ(n)t=\tau(n), after SIC, 𝐱t(n){\bf x}_{t}(n) can be updated from 𝐱t(n1){\bf x}_{t}(n-1) as follows:

𝐱t(n)\displaystyle{\bf x}_{t}(n) =𝐱t(n1)hk(n)𝐬k(n),t\displaystyle={\bf x}_{t}(n-1)-h_{k(n)}{\bf s}_{k(n),t} (5)
=m(m1)k(n)hm𝐬m,t+𝐧t\displaystyle=\sum_{m\in{\cal I}(m-1)\setminus k(n)}h_{m}{\bf s}_{m,t}+{\bf n}_{t} (6)
=m(m)hm𝐬m,t+𝐧t,t=1,,T(n).\displaystyle=\sum_{m\in{\cal I}(m)}h_{m}{\bf s}_{m,t}+{\bf n}_{t},\ t=1,\ldots,T{(n)}. (7)

Note that since the BS sends ACK to user k(n)k(n), this user does not send any IR blocks during the rest of the current frame. That is, there is no signal from user k(n)k(n) for t>T(n)t>T{(n)}. Thus, we can claim that the resulting received signal at stage nn (after SIC) is given by

𝐱t(n)=m(n)hm𝐬m,t+𝐧t{\bf x}_{t}{(n)}=\sum_{m\in{\cal I}{(n)}}h_{m}{\bf s}_{m,t}+{\bf n}_{t} (8)

for all tt within the current frame. From this, we can define the SINR at stage nn as

γk(n)=αkPkm(n)kαmPm+1,k(n).\gamma_{k}{(n)}=\frac{\alpha_{k}P_{k}}{\sum_{m\in{\cal I}{(n)}\setminus k}\alpha_{m}P_{m}+1},\ k\in{\cal I}{(n)}. (9)

The number of re-transmissions of IR blocks required for stage n+1n+1 is now defined as

τ(n+1)=mink(n){τk(n+1)},\tau{(n+1)}=\min_{k\in{\cal I}(n)}\{\tau_{k}(n+1)\}, (10)

where

τk(n+1)=min{T|t=1Tlog2(1+γk(n))Rk},k(n).\tau_{k}(n+1)=\min\left\{T\,|\,\sum_{t=1}^{T}\log_{2}(1+\gamma_{k}{(n)})\geq R_{k}\right\},\ k\in{\cal I}{(n)}. (11)

In addition, we have k(n+1)=argmink(n)τk(n+1)k(n+1)=\operatorname*{\arg\!\min}_{k\in{\cal I}(n)}\tau_{k}(n+1).

For example, assume that M=2M=2. Then, we have τ(1)=min{τ1(1),τ2(1)}\tau(1)=\min\{\tau_{1}(1),\tau_{2}(1)\}, where

τ1(1)\displaystyle\tau_{1}(1) =min{T|t=1Tlog2(1+α1P1α2P2+1)R1}\displaystyle=\min\left\{T\,|\,\sum_{t=1}^{T}\log_{2}\left(1+\frac{\alpha_{1}P_{1}}{\alpha_{2}P_{2}+1}\right)\geq R_{1}\right\}
τ2(1)\displaystyle\tau_{2}(1) =min{T|t=1Tlog2(1+α2P2α1P1+1)R2}.\displaystyle=\min\left\{T\,|\,\sum_{t=1}^{T}\log_{2}\left(1+\frac{\alpha_{2}P_{2}}{\alpha_{1}P_{1}+1}\right)\geq R_{2}\right\}.

If τ1(1)<τ2(2)\tau_{1}(1)<\tau_{2}(2), user 1’s signal can be decoded first and canceled at time τ(1)=τ1(1)\tau(1)=\tau_{1}(1). In this case, at stage 1, the BS sends ACK to user 1. After SIC, the number of IR blocks for stage 2 becomes

τ(2)=min{T|t=1Tlog2(1+α2P2)R2}.\tau(2)=\min\left\{T\,|\,\sum_{t=1}^{T}\log_{2}\left(1+\alpha_{2}P_{2}\right)\geq R_{2}\right\}.

Note that if τ(2)τ(1)=τ1(1)\tau(2)\leq\tau(1)=\tau_{1}(1), the BS sends ACK to user 2 too. Thus, at the end of slot τ(1)\tau(1), the BS can send ACK to users 1 and 2 simultaneously in this case. Otherwise, user 2 keeps sending IR blocks until the BS sends ACK at the end of slot τ(2)\tau(2) (>τ1(1))(>\tau_{1}(1)).

From (10), the length of frame can be found as

TF=maxn{τ(n)}.T_{\rm F}=\max_{n}\{\tau(n)\}. (12)

If M=1M=1 (i.e., conventional HARQ-IR), we have TF=τ(0)=TCHT_{\rm F}=\tau(0)=T_{\rm CH}. We can also see that if τ(n+1)τ(n)\tau(n+1)\leq\tau(n), we have TF=τ(0)T_{\rm F}=\tau(0).

For comparison purposes, we can consider the length of frame when SIC is not used, which is similar to NDMA [5] with HARQ-IR or the approach in [8] where each active user runs his/her HARQ-IR protocol independently, but no SIC is used at the BS. Without SIC, the SINRs remain unchanged for all stages. Thus, the length of frame becomes

TF,nosic=maxk{τk(1)},T_{\rm F,no-sic}=\max_{k\in{\cal I}}\left\{\tau_{k}(1)\right\}, (13)

which is the maximum of the required numbers of IR blocks for all active users. It is easy to see that TF=TF,nosicT_{\rm F}=T_{\rm F,no-sic} if M=1M=1. Furthermore, due to SIC, in general, we can show that TFTF,nosicT_{\rm F}\leq T_{\rm F,no-sic}.

Note that if identical coded blocks (rather than IR blocks) are repeatedly transmitted as in [14], (11) is modified as

τ^k(n+1)=min{T|log2(1+t=1Tγk(n))Rk}.\hat{\tau}_{k}(n+1)=\min\left\{T\,|\,\log_{2}\left(1+\sum_{t=1}^{T}\gamma_{k}(n)\right)\geq R_{k}\right\}. (14)

Using Jensen’s inequality, we can show that τ^k(n+1)τk(n+1)\hat{\tau}_{k}(n+1)\geq\tau_{k}(n+1), which results in a longer length of frame than that in (12). We may regard the approach222Note that in [14], the length of frame is fixed and an active user does not transmit packets for all slots in a frame, which are different from the proposed scheme in this section. Furthermore, no feedback within a frame is considered. Despite such differences, there are some common key features such as the use of SIC and multiple transmissions of packets from an active user within a frame. in [14] as a special case of the proposed scheme in this paper, since it can be obtained by replacing IR blocks with identical coded blocks.

III-B Special Cases

In this subsection, we consider two special cases of the proposed scheme. Based on those, we can further derive two variations of the proposed RTxD multiple access scheme in Sections IV and V.

III-B1 Equalized Overall Channel Gains

For convenience, let βk=αkPk\beta_{k}=\alpha_{k}P_{k}, which is referred to as the overall channel (power) gain of user kk. Suppose that all the overall channel gains are equalized as

βk=β,k,\beta_{k}=\beta,\ k\in{\cal I}, (15)

i.e., the power of active user kk is given by Pk=βαk,kP_{k}=\frac{\beta}{\alpha_{k}},\ k\in{\cal I}.

Lemma 1

With equalized overall channel gains, the length of frame becomes the minimum positive integer of TT that satisfies

R(n)Tlog2(1+β(Mn)β+1),R_{(n)}\leq T\log_{2}\left(1+\frac{\beta}{(M-n)\beta+1}\right), (16)

where the R(n)R_{(n)}’s are the ordered rates as follows:

R(1)R(M).R_{(1)}\leq\ldots\leq R_{(M)}.
Proof:

See Appendix A. ∎

This result demonstrates that it is possible to guarantee a certain length of frame (or access delay) by assigning rates properly with equalized overall channel gains as in (16). We will discuss a random access scheme based on this in Section V in detail.

III-B2 Equal Rate

We assume that the rates are all the same, i.e., Rk=R,kR_{k}=R,\ k\in{\cal I}. Denote by β(m)\beta_{(m)} the mmth largest βk\beta_{k} in {\cal I}, i.e.,

β(1)β(M).\beta_{(1)}\geq\ldots\geq\beta_{(M)}. (17)
Lemma 2

With equal rate, the length of frame becomes the minimum positive integer of TT that satisfies

RTlog2(1+β(n)B(n)+1),n=1,,M,R\leq T\log_{2}\left(1+\frac{\beta_{(n)}}{B_{(n)}+1}\right),\ n=1,\ldots,M, (18)

where B(n)B_{(n)} is the MAI at stage nn, which is given by

B(n)=m=n+1Mβ(m).B_{(n)}=\sum_{m=n+1}^{M}\beta_{(m)}. (19)
Proof:

See Appendix B. ∎

Clearly, it is possible to guarantee a certain length of frame (or access delay) by assigning powers properly for a fixed RR. We will discuss a random access scheme based on the power allocation in the next section, i.e., Section IV.

We can also have the following result from Lemma 2.

Corollary 1

If there exists a positive integer TT that satisfies

2RT1ζ(n)<2RT+11,n=1,,M,2^{\frac{R}{T}}-1\leq\zeta_{(n)}<2^{\frac{R}{T+1}}-1,\ n=1,\ldots,M, (20)

where

ζ(n)=β(n)B(n)+1,\zeta_{(n)}=\frac{\beta_{(n)}}{B_{(n)}+1}, (21)

we have τ(n)=T\tau(n)=T for all nn and TF=TT_{\rm F}=T.

According to Corollary 1, we can see that if the SINRs lies in the region [2RT1,2RT+11)[2^{\frac{R}{T}}-1,2^{\frac{R}{T+1}}-1), all the signals can be decoded within TT re-transmissions.

IV Power-Domain Multiple Access

In this section, based on Corollary 1, we derive a multichannel random access scheme with distributed power allocation which suits to random access under the following assumption.

  • A1)

    Each user knows his/her channel state information (CSI), i.e., channel power gain, αk\alpha_{k}.

Note that A1) is valid when the coherence time is sufficiently long (the channel gains remain unchanged within a frame) with time division duplexing (TDD) mode. In this case, each user is able to estimate his/her channel coefficient through downlink training (i.e., a pilot signal transmission from the BS) prior to the beginning of a frame based on the channel reciprocity.

IV-A Power Allocation for Random Access

We define multiple power levels, denoted by {vl}\{v_{l}\}, as

vl=Γ(Vl+1),v_{l}=\Gamma(V_{l}+1), (22)

where Γ\Gamma is the target SINR and Vl=m=l+1LvmV_{l}=\sum_{m=l+1}^{L}v_{m}. Here, LL is the number of power levels, which is a design parameter, and VL=0V_{L}=0. Clearly, we have v1>v2>>vLv_{1}>v_{2}>\ldots>v_{L} for any Γ>0\Gamma>0. In addition, after some manipulations, we can find that

vl=Γ(Γ+1)Ll.v_{l}=\Gamma(\Gamma+1)^{L-l}. (23)

For convenience, let 𝒱={v1,,vL}{\cal V}=\{v_{1},\ldots,v_{L}\}. Define the regions for the channel (power) gain, αk\alpha_{k}, as {𝒜l}\{{\cal A}_{l}\} that satisfy l=1L𝒜l=[0,)\bigcup_{l=1}^{L}{\cal A}_{l}=[0,\infty) and 𝒜l𝒜l=,ll{\cal A}_{l}\cap{\cal A}_{l^{\prime}}=\emptyset,\ l\neq l^{\prime}. For convenience, 𝒜l{\cal A}_{l} is called the llth channel power region. Suppose that the transmission power of user kk can be decided as follows:

Pk=P(αk)=vlαk,αk𝒜l.P_{k}=P(\alpha_{k})=\frac{v_{l}}{\alpha_{k}},\ \alpha_{k}\in{\cal A}_{l}. (24)

In order to lower the transmission power by taking advantage of known CSI, we can assume 𝒜1>𝒜2>>𝒜L{\cal A}_{1}>{\cal A}_{2}>\ldots>{\cal A}_{L}, where the inequality of 𝒜l>𝒜l{\cal A}_{l}>{\cal A}_{l^{\prime}} implies x>yfor all x𝒜l and y𝒜lx>y\ \mbox{for all $x\in{\cal A}_{l}$ and $y\in{\cal A}_{l^{\prime}}$}.

The power allocation in (24) is performed at each user without any coordination by the BS. With this distributed power allocation, a random access scheme can be considered. In this random access, a user decides the transmission power as in (24). If all active users decide different powers, all the signals can be decoded within a certain number of re-transmissions as follows.

Lemma 3

Suppose that the power allocation in (24) is employed for all active users and Rk=RR_{k}=R for all kk. If MLM\leq L and each active user’s αk\alpha_{k} lies in a different 𝒜l{\cal A}_{l}, the length of frame is bounded as

TFΛ(R,Γ)=Rlog2(1+Γ).T_{\rm F}\leq\Lambda(R,\Gamma)\buildrel\triangle\over{=}\frac{R}{\log_{2}(1+\Gamma)}. (25)
Proof:

See Appendix C. ∎

The resulting random access is referred to as power-domain (or power division) multiple access (PDMA) in this paper. PDMA can be seen as a multichannel random access scheme with multiple sub-channels in the power-domain. The number of sub-channels in the power-domain is LL.

Note that as in (25), the number of re-transmissions is bounded by Λ(R,Γ)\Lambda(R,\Gamma) and can be shorter than this (when the number of active users is less than LL) by taking advantage of HARQ-IR (i.e., if the BS succeeds to decode all the signals before Λ(R,Γ)\Lambda(R,\Gamma) re-transmissions, it sends ACK signals to all the active users and can proceed to the next frame).

To illustrate a specific power allocation scheme, we consider the following assumption.

  • A2)

    The channel power gains are iid and they have the following distribution:

    αkf(x)=ex/γγ,x0,\alpha_{k}\sim f(x)=\frac{e^{-x/\gamma}}{\gamma},\ x\geq 0, (26)

    where γ\gamma is the average channel power gain, i.e., 𝔼[|hk|2]=𝔼[αk]=γ{\mathbb{E}}[|h_{k}|^{2}]={\mathbb{E}}[\alpha_{k}]=\gamma. The resulting channels are independent Rayleigh fading channels.

Suppose that the probability that αk\alpha_{k} lies in 𝒜l{\cal A}_{l} is the same for all ll, i.e.,

Pr(αk𝒜l)=1L.\Pr(\alpha_{k}\in{\cal A}_{l})=\frac{1}{L}. (27)

In this case, under A2), it follows

𝒜l={[A1,),l=1;[Al,Al1),l=2,,L.\displaystyle{\cal A}_{l}=\left\{\begin{array}[]{ll}[A_{1},\infty),&l=1;\cr[A_{l},A_{l-1}),&l=2,\ldots,L.\cr\end{array}\right. (30)

where Al1>AlA_{l-1}>A_{l}. If the AlA_{l}’s are decided to hold (27) with AL=0A_{L}=0, this power region assignment can result in a high transmission power for αk𝒜L\alpha_{k}\in{\cal A}_{L} under A2). To see this clearly, we can consider the average transmission power conditioned on αk𝒜L\alpha_{k}\in{\cal A}_{L}. From (24), we have

𝔼[Pk|αk𝒜L]\displaystyle{\mathbb{E}}[P_{k}\,|\,\alpha_{k}\in{\cal A}_{L}] =vL𝔼[1αk|αk𝒜L]\displaystyle=v_{L}{\mathbb{E}}\left[\frac{1}{\alpha_{k}}\,\bigl{|}\,\alpha_{k}\in{\cal A}_{L}\right]
0AL11xex/γ𝑑x=.\displaystyle\propto\int_{0}^{A_{L-1}}\frac{1}{x}e^{-x/\gamma}dx=\infty.

In order to avoid this problem, the user of the channel power gain less than a threshold may not transmit signals. For example, for a given AL>0A_{L}>0, if αk<AL\alpha_{k}<A_{L}, user kk may not transmit signals. In this case, under A2), the conditional pdf of αk\alpha_{k} becomes

f(αk|αkAL)=CLγeαkγ,αkAL,f(\alpha_{k}\,|\,\alpha_{k}\geq A_{L})=\frac{C_{L}}{\gamma}e^{-\frac{\alpha_{k}}{\gamma}},\ \alpha_{k}\geq A_{L}, (31)

where CL=eALγC_{L}=e^{\frac{A_{L}}{\gamma}}. In addition, (27) is modified as

Pr(αk𝒜l|αkAL)=1L.\Pr(\alpha_{k}\in{\cal A}_{l}\,|\,\alpha_{k}\geq A_{L})=\frac{1}{L}. (32)

Then, after some manipulations, we can decide the AlA_{l}’s as follows:

Al1=Al+γlnCLLCLLeAlγ.\displaystyle A_{l-1}=A_{l}+\gamma\ln\frac{C_{L}L}{C_{L}L-e^{\frac{A_{l}}{\gamma}}}. (33)

In Fig. 1, we illustrate the division of the power domain with the AlA_{l}’s for a given distribution of αk\alpha_{k}, f(x)f(x).

Refer to caption
Figure 1: An illustration of the division of the power domain with the AlA_{l}’s, where the probability that αk\alpha_{k} lies within [Al,Al1)[A_{l},A_{l}-1) is equal for all ll.

In PDMA, we can have the following result for the average transmission power per user.

Lemma 4

The average transmission power per user is upper-bounded as

𝔼[Pk|αkAL](Γ+1)LLAL.\displaystyle{\mathbb{E}}[P_{k}\,|\,\alpha_{k}\geq A_{L}]\leq\frac{(\Gamma+1)^{L}}{LA_{L}}. (34)
Proof:

See Appendix D. ∎

In (34), we can see that the average transmission power can be high for a large LL as the upper-bound grows exponentially with LL. While it is not desirable to have a large LL in terms of the transmission power, more signals can be successfully transmitted for a larger LL in PDMA.

IV-B Outage Events

In this subsection, we consider outage events in PDMA. There are two different cases of outage events that result in a longer length of frame than Λ(R,Γ)\Lambda(R,\Gamma) as follows:

  • The first outage event is the event that M>LM>L. Then, there have to be multiple active users choosing the same power level. This event is referred to as overflow event.

  • The second outage event is that there are multiple users of the same power levels although MLM\leq L. This event is referred to as power collision.

In order to see the performance in terms of the probability of outage events, we consider the following assumptions.

  • A3)

    A user becomes active independently with access probability pap_{a}.

According to A3), the probability of overflow event becomes

Pout,1=Pr(M>L)=m=L+1K(Km)pam(1pa)Km.\displaystyle P_{{\rm out},1}=\Pr(M>L)=\sum_{m=L+1}^{K}\binom{K}{m}p_{a}^{m}(1-p_{a})^{K-m}. (35)

Thus, in order to have a low probability of overflow event, it is desirable to have L>KpaL>Kp_{a}.

For the case of (32), the probability of power collision events conditioned on MM active users is given by [17]

Pout,2(M)=1i=1M(1mL)P~out,2(M)=1eM22L,\displaystyle P_{{\rm out},2}(M)=1-\prod_{i=1}^{M}\left(1-\frac{m}{L}\right)\approx\tilde{P}_{{\rm out},2}(M)=1-e^{-\frac{M^{2}}{2L}}, (36)

where the approximation is reasonable when MLM\ll L.

Lemma 5

Under A3), an upper bound on the expectation of P~out,2(M)\tilde{P}_{{\rm out},2}(M), (i.e., the average probability of power collision) is given by

𝔼[P~out,2(M)]1e𝔼[M22L]=1eKpa2L((K1)pa+1).\displaystyle{\mathbb{E}}[\tilde{P}_{{\rm out},2}(M)]\leq 1-e^{{\mathbb{E}}\left[-\frac{M^{2}}{2L}\right]}=1-e^{-\frac{Kp_{a}}{2L}\left((K-1)p_{a}+1\right)}. (37)
Proof:

See Appendix E. ∎

From (37) and (34), we can observe a trade-off between the transmission power and outage probability in terms of LL. That is, a larger LL results in a lower probability of power collision, while a small LL is desirable to have a low transmission power.

We have a few remarks as follows.

  • According to Lemma 3, it is expected that all the signals from active users can be decoded within Λ(R,Γ)\Lambda(R,\Gamma) re-transmissions if there is no outage event. However, there could be outage events and the BS may need to choose one of the following two options:

    1. 1.

      the BS can allow the users to transmit IR blocks until all the signals from the users can be decoded;

    2. 2.

      the BS can send the final ACK/NACK messages after Λ(R,Γ)\Lambda(R,\Gamma) re-transmissions and proceed to the next frame.

    In the second option, the active users experiencing outages become backlogged users together with the users of low channel gains (i.e., αk<AL\alpha_{k}<A_{L}). Those users (of low channel gains and experiences outage events) can try to transmit IR blocks in the next frame immediately. This approach is called fast re-trial [18, 19], which can only be employed in multichannel random access schemes. Note that fast re-trial is possible as there are multiple sub-channels in the power domain in PDMA.

  • While we mainly focus on random access in this section, it is possible to exploit multiple sub-channels in the power domain for coordinated multiple access with a scheduler to avoid outage events. For example, a scheduler at the BS can assign unique power levels to up to LL users at the cost of signaling overhead, which can guarantee a maximum length of frame, Λ(R,Γ)\Lambda(R,\Gamma).

V Rate-Domain Multiple Access over Block-Fading Channels

In this section, we consider a variation of the RTxD random access based on HARQ-IR in Section II over block-fading channels that vary from one slot to another. In this case, (1) becomes

𝐱t=khk,t𝐬k,t+𝐧t,{\bf x}_{t}=\sum_{k\in{\cal I}}h_{k,t}{\bf s}_{k,t}+{\bf n}_{t}, (38)

where hk,th_{k,t} is the channel coefficient from user kk to the BS at time slot tt. We assume that hk,th_{k,t} is independent. Note that we do not consider Assumption A1) in this section. Thus, instantaneous CSI is not exploited by users.

As in (4), for fast block-fading channels, we can have the initial SINR at stage 0 as follows:

γk,t(0)=αk,tPkm(0)kαm,kPm+1,\gamma_{k,t}(0)=\frac{\alpha_{k,t}P_{k}}{\sum_{m\in{\cal I}(0)\setminus k}\alpha_{m,k}P_{m}+1}, (39)

where αk,t=|hk,t|2\alpha_{k,t}=|h_{k,t}|^{2}. For convenience, we assume that PkP_{k} is decided to have the same average receive power at the BS for all kk as follows:

𝔼[αk,tPk]=γkPk=U,for all k,{\mathbb{E}}[\alpha_{k,t}P_{k}]=\gamma_{k}P_{k}=U,\ \mbox{for all $k$},

where γk=𝔼[αk,t]\gamma_{k}={\mathbb{E}}[\alpha_{k,t}]. Furthermore, we consider the following assumption:

  • A2a)

    We assume that αk,tPk\alpha_{k,t}P_{k} has the same distribution for all kk, i.e., the αk,tPk\alpha_{k,t}P_{k}’s are assumed to be iid.

Then, for a given TT, the average sum rate becomes

𝔼[t=1Tlog2(1+γk,t(0))]\displaystyle{\mathbb{E}}\left[\sum_{t=1}^{T}\log_{2}\left(1+\gamma_{k,t}(0)\right)\right] =T(ψM(U)ψM1(U)),\displaystyle=T\left(\psi_{M}(U)-\psi_{M-1}(U)\right), (40)

where ψM(x)=𝔼[log2(1+k=1Mαk,tx)]\psi_{M}(x)={\mathbb{E}}\left[\log_{2}\left(1+\sum_{k=1}^{M}\alpha_{k,t}x\right)\right]. If there is an active user of a rate R(1)T(ψM(U)ψM1(U))R_{(1)}\leq T\left(\psi_{M}(U)-\psi_{M-1}(U)\right), the average number of re-transmissions for successful decoding is equal to or less than TT. The signals from this user can be removed using SIC after successful decoding. If there is another user of a rate R(2)T(ψM1(U)ψM2(U))R_{(2)}\leq T\left(\psi_{M-1}(U)-\psi_{M-2}(U)\right), the signals from this users can be decoded and this user’ average number of re-transmissions for successful decoding is equal to or less than TT. Based on this, we can consider a random access scheme based on HARQ-IR in the rate domain.

Suppose that there are LL rates that are given by

ηl=T(ψLl+1(U)ψLl(U)δ)l=1,,L,\eta_{l}=T\left(\psi_{L-l+1}(U)-\psi_{L-l}(U)-\delta\right)\ l=1,\ldots,L, (41)

where δ>0\delta>0 is a design parameter and ψ0(U)=0\psi_{0}(U)=0. The average sum rate (after TT re-transmissions) is given by

η¯=1Ll=1Lηl=TL(ψL(U)δ).\displaystyle\bar{\eta}=\frac{1}{L}\sum_{l=1}^{L}\eta_{l}=\frac{T}{L}(\psi_{L}(U)-\delta). (42)

An active user can randomly choose one of the rates in ={η1,,ηL}{\cal R}=\{\eta_{1},\ldots,\eta_{L}\} and transmits IR blocks encoded at the selected initial rate. The resulting multichannel random access scheme is referred to as rate-domain multiple access (RDMA).

Lemma 6

In RDMA, suppose that the following two conditions hold:

η1<η2<<ηL\displaystyle\eta_{1}<\eta_{2}<\ldots<\eta_{L} (43)
Pr(ZlTηl)>1ϵ,\displaystyle\Pr\left(\frac{Z_{l}}{T}\geq\eta_{l}\right)>1-\epsilon, (44)

where Zl=log2(1+k=1Ll+1αk,tPk)log2(1+k=1Llαk,tPk)Z_{l}=\log_{2}\left(1+\sum_{k=1}^{L-l+1}\alpha_{k,t}P_{k}\right)-\log_{2}\left(1+\sum_{k=1}^{L-l}\alpha_{k,t}P_{k}\right). If MLM\leq L and each active user’s rate is different, the number of re-transmissions is equal to or less than TT with a probability higher than or equal to (1ϵ)M(1-\epsilon)^{M}.

Proof:

See Appendix F. ∎

For example, we may consider Rayleigh fading channels for hk,th_{k,t} as in A2). In this case, from [20, 21], we can show that ψM(x)=1ln2m=1Me1xEm(1x)\psi_{M}(x)=\frac{1}{\ln 2}\sum_{m=1}^{M}e^{\frac{1}{x}}E_{m}\left(\frac{1}{x}\right), l=1,,Ll=1,\ldots,L, where Em(x)=1exttm𝑑tE_{m}(x)=\int_{1}^{\infty}e^{-xt}t^{-m}dt is the exponential integral function of order mm. Then, the rates in (41) become

ηl=T(1ln2e1UELl+1(1U)δ).\eta_{l}=T\left(\frac{1}{\ln 2}e^{\frac{1}{U}}E_{L-l+1}\left(\frac{1}{U}\right)-\delta\right). (45)

Since Em(x)>Em+1(x)E_{m}(x)>E_{m+1}(x), we can claim that ηl<ηl+1\eta_{l}<\eta_{l+1}, which implies that the condition in (43) can hold for Rayleigh fading channels.

The condition in (44) is valid under Assumption A2a) based on the weak law of large numbers [22]. In particular, for a sufficiently large TT and a small δ\delta, there might be a small ϵ\epsilon satisfying (44). Consequently, for independent Rayleigh fading channels, the signals from MM active users in RDMA can be decoded within TT re-transmissions with a high probability if MLM\leq L and all the active users choose different initial rates.

Note that as in PDMA, there might be two outage events. Their probabilities (and approximation) are the same as those in (35), (36), and (37). An advantage of RDMA over PDMA is that the transmission power may not be high provided that all the γk\gamma_{k}’s are sufficiently high.

We can consider an asymptotic performance of RDMA for a large TT to compare its achievable rate to the achievable rate of orthogonal multiple access with HARQ-IR as follows.

Lemma 7

Suppose that TT is sufficiently large so that ϵ\epsilon and δ\delta can be sufficiently small. Under A2a), the gain of RDMA over orthogonal multiple access in terms of rate can be bounded as

R¯rdraR¯omaψL(U)ψ1(LU),\displaystyle\frac{\bar{R}_{\rm rdra}}{\bar{R}_{\rm oma}}\leq\frac{\psi_{L}(U)}{\psi_{1}(LU)}, (46)

where R¯rdra\bar{R}_{\rm rdra} and R¯oma\bar{R}_{\rm oma} represent the average transmission rates of RDMA and orthogonal multiple access (that has the number of IR blocks in (2)), respectively, that can be achieved with the same average transmission power. Note that the upper-bound is achievable if M=LM=L and there are no outage events (i.e., if RDMA is used with a scheduler to keep M=LM=L and to avoid choosing the same rate by multiple users).

Proof:

See Appendix G. ∎

In (46), we see that RDMA can have a higher spectral efficiency than orthogonal multiple access (scheduled one without any interference) if ψL(U)ψ1(LU)\psi_{L}(U)\geq\psi_{1}(LU). The following result shows that this is true.

Lemma 8

Under A2a), we have

ψL(U)ψ1(LU).\psi_{L}(U)\geq\psi_{1}(LU). (47)
Proof:

See Appendix H. ∎

In Fig. 2, we show ψL(U)\psi_{L}(U) and ψ1(LU)\psi_{1}(LU) for different values of LL when U=3U=3 dB for independent Rayleigh fading channels. We can see that RDMA can provide a higher spectral efficiency than orthogonal multiple access.

Refer to caption
Figure 2: ψL(U)\psi_{L}(U) and ψ1(LU)\psi_{1}(LU) for different values of LL when U=3U=3 dB for independent Rayleigh fading channels.

We have the following remarks.

  • If channels are invariant within a frame, RDMA can exploit the channel reciprocity under A1) to take advantage of known CSI. In this case, as in PDMA, each active user can choose one of the available rates in {\cal R} according to the channel gain.

  • PDMA can also be used for block-fading channels where channel gains are varying for each slot. In this case, the transmission power is randomly chosen from a set of pre-determined powers.

VI Simulation Results

In this section, we present simulation results under A2) (i.e., for independent Rayleigh fading channels). For performance comparisons, we consider an RTxD random access scheme without SIC (but HARQ-IR is employed) [8], which has the length of frame in (13), and another random access scheme of re-transmissions of identical coded blocks based on [14], which has the length of frame in (14). For convenience, the former scheme is referred to as multiple access without SIC (MA without SIC) and the latter scheme is referred to as repetition with SIC.

Throughout this section, for PDMA, we assume that ALA_{L} is decided to have Pr(αk<AL)=0.1\Pr(\alpha_{k}<A_{L})=0.1. In addition, for RDMA, δ\delta is set to δ=η15\delta=\frac{\eta_{1}}{5}.

Fig. 3 shows the average lengths of frame of PDMA, MA without SIC, and repetition with SIC for different values of LL when γ=0\gamma=0 dB, K=50K=50, pa=0.1p_{a}=0.1, R=4R=4, and Λ(R,Γ)=10\Lambda(R,\Gamma)=10 (note that the value of Γ\Gamma is decided for given values of RR and Λ(R,Γ)\Lambda(R,\Gamma) from (25)). It is shown that PDMA can provide a smaller length of frame than the other schemes and the lengths of frame of all the schemes decrease with LL at the cost of the transmission power (as will be shown in Fig. 4 (a)). Since Λ(R,Γ)\Lambda(R,\Gamma) is set to 10, it is expected that the average length of frame of PDMA is smaller than 10. If LL is large (i.e., L10L\geq 10), we can see that the average length of frame of PDMA can be smaller than 10. However, if LL is not large, there might be outage events that result in a longer length of frame than the desired one, 10.

Refer to caption
Figure 3: Average lengths of frame of PDMA and MA without SIC for different values of LL when γ=0\gamma=0 dB, K=50K=50, pa=0.1p_{a}=0.1, R=4R=4, and Λ(R,Γ)=10\Lambda(R,\Gamma)=10.

In Fig. 4 (a), the average transmission power is shown for different values of LL. The transmission power increases with LL, which can be predicted by (34). As mentioned earlier, the lengths of frame of MA without SIC and repetition with SIC also decrease with LL as their transmission powers333For simulations of MA without SIC, we assume the same transmission power as that in PDMA. increase with LL. The probability of power collision decreases with LL in Fig. 4 (b) as expected by (37) (we note that (37) is a reasonable upper-bound on 𝔼[Pout,2(M)]{\mathbb{E}}[P_{{\rm out},2}(M)]).

Refer to caption

(a)                               (b)

Figure 4: Performance of PDMA for different values of LL when γ=0\gamma=0 dB, K=50K=50, pa=0.1p_{a}=0.1, R=4R=4, and Λ(R,Γ)=10\Lambda(R,\Gamma)=10: (a) the average transmission power; (b) the probability of power collision.

Fig. 5 shows the performances of PDMA for different values of Λ(R,Γ)\Lambda(R,\Gamma) when γ=0\gamma=0 dB, K=50K=50, pa=0.1p_{a}=0.1, R=4R=4, and L=20L=20. We can see that a large Λ(R,Γ)\Lambda(R,\Gamma) is desirable for a smaller length of frame of PDMA. In addition, the transmission power can be low at the cost of a longer transmission time.

Refer to caption

(a)                               (b)

Figure 5: Performance of PDMA for different values of Λ(R,Γ)\Lambda(R,\Gamma) when γ=0\gamma=0 dB, K=50K=50, pa=0.1p_{a}=0.1, R=4R=4, and L=20L=20: (a) the average lengths of frame of PDMA and MA without SIC; (b) the average transmission power.

To see the impact of KK on the performances of PDMA, MA without SIC, and repetition with SIC, different values of KK are considered and simulation results are shown in Fig. 6 when γ=0\gamma=0 dB, L=20L=20, pa=0.1p_{a}=0.1, R=4R=4, and Λ(R,Γ)=10\Lambda(R,\Gamma)=10. As KK increases, the average lengths of frame of all the schemes increase as there are more active users, while PDMA could provide a lower length of frame than Λ(R,Γ)\Lambda(R,\Gamma) up to K=100K=100 users. Note that since Λ(R,Γ)\Lambda(R,\Gamma) is fixed, as shown in Fig. 6 (b), the average transmission power is invariant with respect to KK.

Refer to caption

(a)                               (b)

Figure 6: Performance of PDMA for different values of KK when γ=0\gamma=0 dB, L=20L=20, pa=0.1p_{a}=0.1, R=4R=4, and Λ(R,Γ)=10\Lambda(R,\Gamma)=10: (a) the average lengths of frame of PDMA and MA without SIC; (b) the average transmission power.

We now consider the performance of RDMA. Fig. 7 shows the performances of RDMA and MA without SIC for different values of LL when U=3U=3 dB, K=50K=50, pa=0.1p_{a}=0.1, and T=10T=10. RDMA performs better than MA without SIC in terms of the average length of frame as shown in Fig. 7 (a). In Fig. 7 (b), it is shown that the average initial rate decreases with LL, where the theoretical one is obtained from (42).

Refer to caption

(a)                               (b)

Figure 7: Performance of RDMA for different values of LL when U=3U=3 dB, K=50K=50, pa=0.1p_{a}=0.1, and T=10T=10: (a) the average lengths of frame of PDMA and MA without SIC; (b) the average transmission power.

To see the impact of TT on the performances of RDMA and MA without SIC, different values of TT are considered and simulation results are shown in Fig. 8 when U=3U=3 dB, K=50K=50, pa=0.1p_{a}=0.1, and L=10L=10. The gap between the average lengths of frame of RDMA and MA without SIC increases with TT as shown in Fig. 8 (a), which shows that RDMA performs much better than MA without SIC, especially for a large TT. We also see that the average length of frame of RDMA can be smaller than TT for large TT’s, which demonstrates that RDMA can more likely guarantee that the required number of re-transmissions for successful decoding is less than or equal to TT when TT is sufficiently large.

Refer to caption

(a)                               (b)

Figure 8: Performance of RDMA for different values of TT when U=3U=3 dB, K=50K=50, pa=0.1p_{a}=0.1, and L=10L=10: (a) the average lengths of frame of PDMA and MA with SIC; (b) the average transmission power.

Finally, in Fig. 9, we show the normalized spectral efficiencies (i.e., the (average) initial rate per user over average length of frame) of PDMA, RDMA, and MA without SIC for different transmission powers with L=20L=20, γ=0\gamma=0 dB, K=50K=50, pa=0.1p_{a}=0.1, and Λ(R,Γ)=T=10\Lambda(R,\Gamma)=T=10. We can see that PDMA performs better than RDMA. This performance gain results from the power allocation with known CSI (in (24)) in PDMA. It is interesting to note that this gain vanishes when the transmission power is high. In both PDMA and RDMA, the spectral efficiency increases with the average transmission power. This implies that SIC can effectively mitigate interference from other active users. On the other hand, in MA without SIC, the increase of spectral efficiency becomes saturated when the transmission power increases. This saturation indicates that the spectral efficiency mainly depends on the MAI in MA without SIC and, as a result, the increase of the transmission power does not help improve the spectral efficiency unless some interference cancellation techniques (e.g., SIC) are employed.

Refer to caption
Figure 9: Normalized spectral efficiency versus (average) transmission power (with L=20L=20, γ=0\gamma=0 dB, K=50K=50, pa=0.1p_{a}=0.1, and Λ(R,Γ)=T=10\Lambda(R,\Gamma)=T=10).

VII Concluding Remarks

In this paper, we studied a multiple access scheme that are based on HARQ-IR and SIC to exploit RTxD and mitigate MAI, respectively. We found conditions that can guarantee a certain number of re-transmissions in the proposed multiple access scheme. Based on the derived conditions and two special cases, we considered two multichannel random access schemes, PDMA and RDMA, that have sub-channels in the power and rate domains, respectively. The analysis results showed that the proposed multiple access scheme can have a shorter length of frame than other existing similar schemes, which implies that the proposed multiple access scheme outperforms other existing ones in terms of the spectral efficiency. We also demonstrated that PDMA and RDMA can be used for coordinated multiple access. In this case, RDMA can have a higher spectral efficiency than orthogonal multiple access.

While we focused on some fundamental properties of the proposed multiple access schemes, we did not consider some possible extensions in this paper. For example, we may consider the case that the BS is equipped with multiple antennas (which may increase the spectral efficiency) and a hybrid approach of PDMA and RDMA in the future as further research topics.

Appendix A Proof of Lemma 1

The initial SINR is β(M1)β+1\frac{\beta}{(M-1)\beta+1}. The number of re-transmissions for stage 1 is

τ(1)\displaystyle\tau(1) =minkminT{T|Tlog2(1+β(M1)β+1)Rk}\displaystyle=\min_{k}\min_{T}\left\{T\,\bigl{|}\,T\log_{2}\left(1+\frac{\beta}{(M-1)\beta+1}\right)\leq R_{k}\right\}
=minT{T|Tlog2(1+β(M1)β+1)R(1)}.\displaystyle=\min_{T}\left\{T\,\bigl{|}\,T\log_{2}\left(1+\frac{\beta}{(M-1)\beta+1}\right)\leq R_{(1)}\right\}.

Thus, in stage 1, the signals of rate R(1)R_{(1)} can be removed. Subsequently, we can show that

τ(n)=minT{T|Tlog2(1+β(Mn)β+1)R(n)}.\tau(n)=\min_{T}\left\{T\,\bigl{|}\,T\log_{2}\left(1+\frac{\beta}{(M-n)\beta+1}\right)\leq R_{(n)}\right\}.

From this, we can see that the length of frame is the minimum positive integer of TT that satisfies (16).

Appendix B Proof of Lemma 2

From (17), it can be shown that the SINRs at stage 0 are given by

β(1)Bβ(1)+1β(2)Bβ(2)+1β(M)Bβ(M)+1,\displaystyle\frac{\beta_{(1)}}{B-\beta_{(1)}+1}\geq\frac{\beta_{(2)}}{B-\beta_{(2)}+1}\geq\ldots\geq\frac{\beta_{(M)}}{B-\beta_{(M)}+1},

where B=mβ(m)B=\sum_{m}\beta_{(m)}, which demonstrates that the active user who will be decoded first is the user that has the highest channel gain, β(1)\beta_{(1)}, since the rates are the same. Removing the signals from this user, we can also show that

β(2)(Bβ(1))β(2)+1β(M)(Bβ(1))β(M)+1.\frac{\beta_{(2)}}{(B-\beta_{(1)})-\beta_{{(2)}+1}}\geq\ldots\geq\frac{\beta_{(M)}}{(B-\beta_{(1)})-\beta_{{(M)}+1}}.

From this, it is clear that the decoding order is the same as that in (17) (i.e., the order of the overall channel gains). Thus, at stage nn, from (19), the SINR of the signal to be decoded is given in (21). From this, we can see that the length of frame is the minimum positive integer of TT that satisfies (18).

Appendix C Proof of Lemma 3

Due to the power determination in (24), βk\beta_{k} has to be one of 𝒱{\cal V}, i.e., βk𝒱\beta_{k}\in{\cal V}. In addition, since the αk\alpha_{k}’s lie in different channel power regions and MLM\leq L, βk\beta_{k}, kk\in{\cal I}, should have different values. In addition, for the ordered overall channel gains, β(1)>>β(M)\beta_{(1)}>\ \ldots\ >\beta_{(M)}, since β(m)𝒱\beta_{(m)}\in{\cal V}, we have β(n)B(n)+1Γ\frac{\beta_{(n)}}{B_{(n)}+1}\geq\Gamma. Note that the equality holds if β(1)=v1,,β(n)=vn\beta_{(1)}=v_{1},\ldots,\beta_{(n)}=v_{n}. Thus, according to Lemma 2, the length of frame, TFT_{\rm F}, is less than or equal to Λ(R,Γ)\Lambda(R,\Gamma), which completes the proof.

Appendix D Proof of Lemma 4

Since if αk𝒜l\alpha_{k}\in{\cal A}_{l}, from (24), we have the following bounds: vlAl1<PkvlAl\frac{v_{l}}{A_{l-1}}<P_{k}\leq\frac{v_{l}}{A_{l}}. Using the upper-bound on PkP_{k} and from (32), we can obtain the following upper-bound on the average transmission power:

𝔼[Pk|αkAL]1Ll=1LvlAl.\displaystyle{\mathbb{E}}[P_{k}\,|\,\alpha_{k}\geq A_{L}]\leq\frac{1}{L}\sum_{l=1}^{L}\frac{v_{l}}{A_{l}}. (48)

Noting that AL<AL1<<A1A_{L}<A_{L-1}<\ldots<A_{1} and using (23), it follows l=1LvlAl1ALl=1Lvl=1ALl=1LΓ(1+Γ)Ll\sum_{l=1}^{L}\frac{v_{l}}{A_{l}}\leq\frac{1}{A_{L}}\sum_{l=1}^{L}v_{l}=\frac{1}{A_{L}}\sum_{l=1}^{L}\Gamma(1+\Gamma)^{L-l}. Since

l=1L(1+Γ)Ll\displaystyle\sum_{l=1}^{L}(1+\Gamma)^{L-l} =(1+Γ)L1l=0L1(1+Γ)l\displaystyle=(1+\Gamma)^{L-1}\sum_{l=0}^{L-1}(1+\Gamma)^{-l}
(1+Γ)L11111+Γ=(Γ+1)LΓ,\displaystyle\leq(1+\Gamma)^{L-1}\frac{1}{1-\frac{1}{1+\Gamma}}=\frac{(\Gamma+1)^{L}}{\Gamma},

we have l=1LvlAl(Γ+1)LAL\sum_{l=1}^{L}\frac{v_{l}}{A_{l}}\leq\frac{(\Gamma+1)^{L}}{A_{L}}. Substituting this into (48), we can have (34).

Appendix E Proof of Lemma 5

Using Jensen’s inequality, we can show that

𝔼[eM22L]=eln𝔼[eM22L]e𝔼[M22L]=e12L𝔼[M2].\displaystyle{\mathbb{E}}[e^{-\frac{M^{2}}{2L}}]=e^{\ln{\mathbb{E}}[e^{-\frac{M^{2}}{2L}}]}\geq e^{{\mathbb{E}}\left[-\frac{M^{2}}{2L}\right]}=e^{-\frac{1}{2L}{\mathbb{E}}\left[M^{2}\right]}.

Under A3), since MM is a binomial random variable, we have 𝔼[M2]=Kpa(1pa)+(Kpa)2{\mathbb{E}}[M^{2}]=Kp_{a}(1-p_{a})+(Kp_{a})^{2}. From this, we can have (37).

Appendix F Proof of Lemma 6

Suppose that the active user who has the lowest initial rate is user kk. It can be shown that the sum rate of this user after TT re-transmissions is given by

t=1Tlog2(1+(αk,t+mkαm,t)Pk)\displaystyle\sum_{t=1}^{T}\log_{2}\left(1+\left(\alpha_{k,t}+\sum_{m\in{\cal I}\setminus k}\alpha_{m,t}\right)P_{k}\right)
log2(1+(mkαm,t)Pk),\displaystyle-\log_{2}\left(1+\left(\sum_{m\in{\cal I}\setminus k}\alpha_{m,t}\right)P_{k}\right),

which has the same statistical properties as those of ZLM+1Z_{L-M+1} as ||=M|{\cal I}|=M. Furthermore, user kk has the lowest rate among MM active users and all the rates are different, the rate of user kk has to be equal to or less than ηLM+1\eta_{L-M+1}, i.e., RkηLM+1R_{k}\leq\eta_{L-M+1}. From (44), we have

Pr(ZLM+1TRk)Pr(ZLM+1TηLM+1)1ϵ.\displaystyle\Pr\left(\frac{Z_{L-M+1}}{T}\geq R_{k}\right)\geq\Pr\left(\frac{Z_{L-M+1}}{T}\geq\eta_{L-M+1}\right)\geq 1-\epsilon. (49)

Thus, at time tt (or after TT re-transmissions), the signals from user kk can be decoded and removed by SIC with a probability equal to or higher than 1ϵ1-\epsilon.

Once user kk is removed, the user of the second lowest initial rate, which is less than or equal to ηLM+2\eta_{L-M+2}, may have M1M-1 interfering signals. The sum rate of this user might be ZLM+2Z_{L-M+2}. Thus, as in (49), we can show that the signals from this user can be decoded and removed by SIC with a probability equal to or higher than 1ϵ1-\epsilon. Consequently, all the signals from MM active users can be decoded within TT re-transmissions with a probability equal to or higher than (1ϵ)M(1-\epsilon)^{M}.

Appendix G Proof of Lemma 7

Suppose that the number of active users in RDMA, MM, becomes LL and all users have different rates. Then, from (41), under A2a), the average rate of RDMA per user becomes η¯=𝔼[ηl]=1Ll=1Lηl=TL(ψL(U)δ)\bar{\eta}={\mathbb{E}}[\eta_{l}]=\frac{1}{L}\sum_{l=1}^{L}\eta_{l}=\frac{T}{L}(\psi_{L}(U)-\delta). This is the maximum rate of RDMA as there are no outage events, which implies that

R¯rdraη¯.\bar{R}_{\rm rdra}\leq\bar{\eta}. (50)

For orthogonal multiple access, we assume that the time slot is divided into LL sub-slots and in each sub-slot, one user can transmit signals using HARQ-IR without any interference. In this case, according to [8], for a large TT, the average (achievable) rate per user (with TT re-transmissions) becomes

R¯oma=TL(𝔼[log2(1+αk,tPk)]δ)=TL(ψ1(LU)δ)\displaystyle\bar{R}_{\rm oma}=\frac{T}{L}({\mathbb{E}}[\log_{2}(1+\alpha_{k,t}P_{k})]-\delta)=\frac{T}{L}(\psi_{1}(LU)-\delta) (51)

with a high probability (1ϵ)(\geq 1-\epsilon), where PkP_{k} is decided to be LL times higher than that in RDMA (i.e., Pk=LUγkP_{k}=L\frac{U}{\gamma_{k}}) to have the same total power. From (50) and (51), we can have (46), which completes the proof.

Appendix H Proof of Lemma 8

Consider an optimization problem as follows:

min{xk}𝔼[log2(1+k=1Lαkxk)]\displaystyle\min_{\{x_{k}\}}{\mathbb{E}}[\log_{2}(1+\sum_{k=1}^{L}\alpha_{k}x_{k})] (52)
subject tok=1Lxk=L,xk0,\displaystyle\mbox{subject to}\ \sum_{k=1}^{L}x_{k}=L,\ x_{k}\geq 0, (53)

where the αk\alpha_{k}’s are iid and αk0\alpha_{k}\geq 0 as in A2a). For convenience, assume that 𝔼[αk]=1{\mathbb{E}}[\alpha_{k}]=1 for all kk. If xk=Ux_{k}=U for all kk, 𝔼[log2(1+k=1Lαkxk)]{\mathbb{E}}[\log_{2}(1+\sum_{k=1}^{L}\alpha_{k}x_{k})] becomes ψL(U)\psi_{L}(U). On the other hand, if x1=Lx_{1}=L and x2==xL=0x_{2}=\ldots=x_{L}=0, it becomes ψ1(LU)\psi_{1}(LU).

Since 𝔼[log2(1+k=1Lαkxk)]{\mathbb{E}}[\log_{2}(1+\sum_{k=1}^{L}\alpha_{k}x_{k})] is concave, the maximum exists. To find the optimal solution, with a Lagrange multiplier, λ\lambda, we can show that the following equality has to be satisfied for all kk:

ddxk(𝔼[log2(1+k=1Lαkxk)]λk=1Lxk)\displaystyle\frac{d}{dx_{k}}\left({\mathbb{E}}\left[\log_{2}\left(1+\sum_{k=1}^{L}\alpha_{k}x_{k}\right)\right]-\lambda\sum_{k=1}^{L}x_{k}\right)
=1ln2𝔼[αk1+k=1Lαkxk]λ=0.\displaystyle=\frac{1}{\ln 2}{\mathbb{E}}\left[\frac{\alpha_{k}}{1+\sum_{k=1}^{L}\alpha_{k}x_{k}}\right]-\lambda=0.

From this, since 𝔼[log2(1+k=1Lαkxk)]{\mathbb{E}}[\log_{2}(1+\sum_{k=1}^{L}\alpha_{k}x_{k})] is symmetric with respect to {xk}\{x_{k}\}, we can see that the optimal solution becomes xk=1x_{k}=1. Thus, ψL(U)\psi_{L}(U) is the maximum, which implies (47).

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