To simplify description, the output SNR at each RS for the MS-to-RS link is denoted as, and the output SNR at BS for the RS-to-BS link can be similarly denoted as.
The instantaneous transmission rate in the BS-RS link at slot t is R ( t ) = W log 2 1 + P ( t ) N ( t ) bits/s.
In the nontransparent mode, each TDD frame is divided into downlink and uplink intervals, each of which is further divided into an access link interval for the MS (i.e., used for the BS-to-MS and RS-to-MS links) and a relay link interval for the RS (i.e., used for the BS-to-RS link) [8].
Even if the buffer at the RS is large enough to store all the data for the MS, the resources on the BS-RS link have been wasted when the MS performs a handover to another RS or BS.
The coverage probability for the MBS-RS link p c _RU is given by p c _RU = 2 R 2 ∫ 0 R exp − πλ r 2 p c _MR μ Γ RU, α × exp − Γ RU r α σ 2 1 + Γ RU P R r d r (10).
The amount of frames within a superframe where the RS is serving MSs depends on the capacity of the BS-RS link and the RS-MS links.
As the capacity of the BS-RS link is very high, the best result was achieved when the RS serves its MSs in five out of eight frames.
Moreover, due to the bottleneck of the BS-RS link and the availability of the non-conflict unlicensed frequency band, the RS may only serve a limited number of the MSs concurrently, i.e., the RS has limited channel resource for RS-MS links.
aHere we only consider the RS-to-train link, because the RS is assumed to be in the fixed place, such that the BS-to-RS link will not change much.
where γsr(t) and γrd(t), respectively, are the SNRs of the BS-to-RS link and the RS-to-train link at time slot t, and r and r′, respectively, are the current RS and the RS that the train may handoff to.
The coverage probability for the MBS-RS link p c _MR is given by p c _ MR = 2 π λ M ∫ r > 0 exp − π r 2 λ M + λρ Γ MR, α × exp − Γ MR r α σ 2 / P MR r d r (8).
