基于連續(xù)干擾消除的毫米波MIMO系統(tǒng)混合預(yù)編碼算法
doi: 10.11999/JEIT180379
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南京郵電大學(xué)通信與信息工程學(xué)院 ??南京 ??210003
Hybrid Precoding Algorithm Based on Successive Interference Cancellation for Millimeter Wave MIMO Systems
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College of Telecommunications and Information Engineering, Nanjing University of Posts and Telecommunications, Nanjing 210003, China
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摘要:
該文研究多用戶毫米波MIMO系統(tǒng)的混合模數(shù)預(yù)編碼器和合并器設(shè)計(jì)。針對(duì)因信號(hào)傳播漫散射造成的多用戶間信號(hào)干擾問題，提出一種基于連續(xù)干擾消除(SIC)的魯棒混合預(yù)編碼算法。首先對(duì)信道矩陣進(jìn)行正交分解，以消除來自已知用戶信號(hào)的干擾，從而將含有非凸約束的多用戶鏈路優(yōu)化問題分解為多個(gè)單用戶鏈路優(yōu)化問題。然后采用相位提取算法逐個(gè)求解每個(gè)用戶的最優(yōu)傳輸鏈路，并結(jié)合最小均方誤差(MMSE)準(zhǔn)則求得多用戶混合預(yù)編碼矩陣。仿真結(jié)果表明，與現(xiàn)有的混合預(yù)編碼算法相比，所提算法在強(qiáng)干擾環(huán)境下具有顯著的性能優(yōu)勢。
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關(guān)鍵詞:
- MIMO系統(tǒng) /
- 毫米波 /
- 混合預(yù)編碼 /
- 連續(xù)干擾消除
Abstract:This paper investigates the design of hybrid analog and digital precoder and combiner for multi-user millimeter wave MIMO systems. Considering the problem of signal interference between multiple users due to diffuse scattering of signal propagation, a robust hybrid precoding algorithm based on Successive Interference Cancellation (SIC) is proposed. By deducing the orthogonal decomposition formula of the channel matrix to eliminate the interference from the known users’ signals, the multi-user links optimization problem with nonconvex constraints can be decompose into multiple single-user link optimization problems. The phase extraction algorithm is then used to search each user’s optimal transmission link one by one, and the multi-user hybrid precoding matrix is obtained in combination with Minimum Mean Square Error (MMSE) criterion. Simulation results show that the proposed algorithm has significant performance advantages compared with the existing hybrid precoding algorithms under severe interference conditions.
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表 1 基于連續(xù)干擾消除的混合預(yù)編碼算法
輸入：${{{H}}_{u}},{u} = 1 ,2, ·\!·\!· ,{U}\;$； 輸出：${{{F}}_{{\rm{BB}}}}$, ${{{F}}_{{\rm{RF}}}}$, ${{{w}}_{u}},{u} = 1 ,2, ·\!·\!· ,{U}\;$； 初始化：${{{F}}_{{\rm{RF}}}} = {\rm{Empty}}$；$\widehat {{H}} = {\rm{Empty}}$；${{{T}}_ 1 } = {{{H}}_ 1 }$； for ${u} = 1 :{U}\;$ (1) 對(duì)${{{T}}_{u}}$進(jìn)行SVD分解，得到${{{v}}_ 1 }$和${{{q}}_ 1 }$； (2) ${{{v}}_ 1 }$代入式(14)，得到${{{w}}_{u}}$； (3) ${{{q}}_ 1 }$代入式(15)，得到${{{f}}_{{\rm{RF}},{u}}}$； (4) $\widehat {{h}}_u^{\rm{H}} = {{w}}_u^{\rm{H}}{{{H}}_u}$, $\widehat {{H}} = {\left[{\widehat {{H}}^{\rm{H}}}{\rm{ }}{\widehat {{h}}_{u}}\right]^{\rm{H}}}$, ${{{F}}_{{\rm{RF}}}} = [{{{F}}_{{\rm{RF}}}}{\rm{ }}{{{f}}_{{\rm{RF}},u}}]$； (5) $\widehat {{H}}$和${{{F}}_{{\rm{RF}}}}$代入式(16)，得到${{{F}}_{{\rm{BB}}}}$； if ${u} < {U}\;$ (6) ${{{F}}_{u}} = {{{F}}_{{\rm{RF}}}}{{{F}}_{{\rm{BB}}}}$，代入式(17)，得到${{{T}}_{{u} + 1}}$； end if end for ${{{F}}_{{\rm{BB}}}}{\rm{ = }}\sqrt {\frac{P}{{\left\| {{{{F}}_{{\rm{RF}}}}{{{F}}_{{\rm{BB}}}}} \right\|_{\rm{F}}^{2}}}} {{{F}}_{{\rm{BB}}}}$ 
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