Recursive blind equalization for the bounded noise case under different modulations

Moussa, A.; Pouliquen, Mathieu; Frikel, Miloud; Bedoui, S.; Abderrahim, Kamel; M’Saad, M.

doi:10.1109/med.2017.7984304

Cited by 3 publications

(1 citation statement)

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“…Thirdly, the implementation of the algorithm requires only one design parameter, namely an upper bound on noise magnitude. It is worth noticing that the proposed blind equalisation algorithm is a suitable extension of the blind equalisation algorithm proposed in [16] that has been progressively elaborated in [17][18][19]. This extension allowed to increase the convergence speed while reducing the variance of the equaliser output error.…”

Section: Contributions Of This Papermentioning

confidence: 99%

Blind equalisation in the presence of bounded noise

Moussa

Pouliquen

Frikel

et al. 2018

IET signal process.

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This study addresses the blind equalisation problem in the presence of bounded noise using an optimal bounding ellipsoid algorithm. This provides an adequate blind equalisation algorithm with an accurate parameter estimation. A fundamental analysis of the involved equaliser is performed to emphasise its underlying properties. This fundamental result is corroborated by promising simulation results. where n h , h i , {e k }, and {s k } are, respectively, the order, the complex channel tap weight, the unknown noise, and the input symbols. The channel can be a single-input multi-output channel with p ≥ 1 outputs. This means that x k ∈ ℂ p × 1 , e k ∈ ℂ p × 1 , and h i ∈ ℂ p × 1. The following assumptions complete the description of the problem: A.1 The input symbols {s k } are independent, identically distributed. They are drawn from a known constellation C [quadrature amplitude modulation (QAM), phase-shift keying (PSK), or amplitude-shift keying (ASK)]. A.2 The noise sequence is bounded, i.e. |e k (j) | ≤ δ e (j) for all j ∈ {1, …, p}, where e k (j) is the jth component of e k and |. | is the complex modulus. A.3 The sequence of variables {s k } and {e k (j) } are independent for all j ∈ {1, …, p}.

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Section: Contributions Of This Papermentioning

confidence: 99%