首页> 外文会议>INFOCOM '94. Networking for Global Communications., 13th Proceedings IEEE >Throughput performance of multiuser detection in unslottedcontention channels
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Throughput performance of multiuser detection in unslottedcontention channels

机译:未开槽的多用户检测的吞吐量性能竞争渠道

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Analyzes the throughput of a centralized packet receiver whichutilizes multiuser detection in an unslotted contention channel. Thereceiver is capable of demodulating all overlapping, unit-lengthpackets, provided that the T-length packet headers do not overlap, andat most K packets overlap at any one time. If one of these conditions isviolated, the remainder of the busy period is assumed to beunresolvable. Assuming that the aggregate packet arrival process isPoisson, the authors analyze the throughput as a function of the totaloffered load (λ), the maximum number of resolvable overlappingpackets (K), and the fractional header duration (T). They derive aniterative expression for the throughput for K=2 or 3, generalizable tolarger K, which permits solution of the throughput to arbitraryaccuracy. A closed form solution is given for the case K=2 and T=0.5.The authors provide bounds on the throughput for arbitrary K, λ,and T, which are more convenient than the exact approach for T>0 orlarge K. Numerical results illustrate the improvement in throughput dueto multiuser detection, and suggest a technique for finding the minimumcomplexity multiuser receiver (minimum K) as a function of other systemparameters
机译:分析集中式数据包接收器的吞吐量 在未分配时隙的竞争信道中利用多用户检测。这 接收器能够解调所有重叠的单位长度 前提是T长度数据包头不重叠,并且 任何时候最多K个数据包重叠。如果这些条件之一是 违反,则假定繁忙时段的剩余时间为 无法解决。假设总的数据包到达过程是 泊松(Poisson),作者分析了吞吐量作为总量的函数 提供的载荷(λ),可解决的最大重叠数 封包(K)和小数标头持续时间(T)。他们得出一个 K = 2或3的吞吐量的迭代表达式,可推广为 较大的K,可以将吞吐量解为任意 准确性。对于K = 2和T = 0.5的情况,给出了一个封闭形式的解决方案。 作者为任意K,λ, 和T,这比T> 0或T> 0时的精确方法更方便 大K。数值结果说明了吞吐量的提高,这是由于 进行多用户检测,并提出一种寻找最小值的技术 复杂度多用户接收器(最小K)作为其他系统的函数 参数

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