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ISAC-RIS系统下基于条件生成对抗网络的信道估计研究

刘钰 郑泽林 刘罡

刘钰, 郑泽林, 刘罡. ISAC-RIS系统下基于条件生成对抗网络的信道估计研究[J]. 电子与信息学报. doi: 10.11999/JEIT251168
引用本文: 刘钰, 郑泽林, 刘罡. ISAC-RIS系统下基于条件生成对抗网络的信道估计研究[J]. 电子与信息学报. doi: 10.11999/JEIT251168
LIU Yu, ZHENG Zelin, LIU Gang. Conditional Generative Adversarial Network-Based Channel Estimation for RIS-Assisted ISAC System[J]. Journal of Electronics & Information Technology. doi: 10.11999/JEIT251168
Citation: LIU Yu, ZHENG Zelin, LIU Gang. Conditional Generative Adversarial Network-Based Channel Estimation for RIS-Assisted ISAC System[J]. Journal of Electronics & Information Technology. doi: 10.11999/JEIT251168

ISAC-RIS系统下基于条件生成对抗网络的信道估计研究

doi: 10.11999/JEIT251168 cstr: 32379.14.JEIT251168
详细信息
    作者简介:

    刘钰:女,博士,讲师,研究方向为智能信号处理和干扰消除

    郑泽林:女,硕士生,研究方向为无线通信和信号处理

    刘罡:男,工程师,硕士生导师,研究方向为深度学习和移动通信

    通讯作者:

    刘罡 liugang@cwxu.edu.cn

  • 中图分类号: TN929.5

Conditional Generative Adversarial Network-Based Channel Estimation for RIS-Assisted ISAC System

  • 摘要: 通感一体化(ISAC)技术作为未来无线通信发展的关键趋势,旨在通过频谱资源的高效利用,实现通信与感知功能的融合与协同。当智能反射表面(RIS)被引入ISAC系统后,可重构无线传播环境,从而显著提升通信质量及感知精度。然而,准确的信道估计对于保障可靠运行是至关重要的。尽管传统的深度学习方法在一定程度上能够应对信道估计问题,但在面对多用户复杂信道环境时,其泛化能力和估计精度仍存在不足。针对上述问题,该文对于RIS辅助多用户ISAC系统提出一种基于条件生成对抗网络(CGAN)的两阶段信道估计方法。该方法通过调整RIS的开关状态,分阶段完成对直射信道与反射信道的估计,以提高信道估计的准确性和稳定性。通过生成网络与判别网络的对抗训练,不仅能够学习观测信号与真实信道之间的映射关系,还能根据判别网络的反馈来不断优化输出,从而有效提升训练效率与估计精度。仿真结果表明,与传统深度学习方法相比,所提基于CGAN的方案在信道估计性能上均表现出显著优势。该结果验证了CGAN方法在RIS辅助ISAC系统下信道估计的应用潜力,并为实现更精准和可靠的系统部署奠定了基础。
  • 图  1  RIS辅助的多用户ISAC系统模型

    图  2  导频传输协议

    图  3  数据增强参数敏感性实验

    图  4  加权因子对NMSE的影响

    图  5  所提出的基于CGAN的估计框架

    图  6  M = 6且L = 30时SAC信道估计的NMSE与SNR的关系

    图  7  在SNR = 0 dB和10 dB下且L = 30时,SAC信道估计的NMSE与M的关系

    图  8  在SNR = 0 dB和10 dB下且M = 6时,通信信道估计的NMSE与L的关系

    表  1  CGAN网络的具体参数

    模型 网络 大小 激活函数
    CGAN 生成器 输入层 $ 2M{P}^{{{\mathrm{S}}_{l}}} $ -
    FFL 100 LeakyReLU
    FFL 200 LeakyReLU
    输出层 $ 2{M}^{2} $ -
    判别器 输入层 $ 2{M}^{2} $ -
    FFL 100 LeakyReLU
    FFL 200 LeakyReLU
    输出层 1 -
    下载: 导出CSV

    表  2  两个阶段的总子帧持续时间

    时隙持续时间($ \mu \text{s} $) 时隙个数 子帧持续时间($ \mu \text{s} $) 子帧个数 两个阶段的总子帧持续时间($ \mu \text{s} $)
    $ \mathit{\mathrm{\mathit{S}}}_1 $ $ {T}_{\mathrm{P}}=0.52 $ $ {P}^{{{\mathrm{S}}_{1}}}=M+K=12 $ $ \begin{aligned}T_{\mathrm{F}}^{{\mathrm{S}}_{1}}&={T}_{\mathrm{P}}{P}^{{{\mathrm{S}}_{1}}}\\&=6.24\end{aligned} $ $ {C}^{{{\mathrm{S}}_{1}}} $ $ {T}_{\mathrm{E}}={C}^{{{\mathrm{S}}_{1}}}T_{\mathrm{F}}^{{\mathrm{S}}_{1}}+\left({C}^{{{\mathrm{S}}_{2}}}-{C}^{{{\mathrm{S}}_{1}}}\right)T_{\mathrm{F}}^{{\mathrm{S}}_{2}}=99.84 $
    $ \mathrm{\mathit{S}}_2 $ $ {T}_{\mathrm{P}}=0.52 $ $ {P}^{{{\mathrm{S}}_{2}}}=\max \left\{M,K\right\}=6 $ $ \begin{aligned}T_{\mathrm{F}}^{{\mathrm{S}}_{2}}&={T}_{\mathrm{P}}{P}^{{{\mathrm{S}}_{2}}}\\&=3.12\end{aligned} $ $ {C}^{{{\mathrm{S}}_{2}}}-{C}^{{{\mathrm{S}}_{1}}} $
    下载: 导出CSV

    表  3  SAC信道的路径损耗

    距离(m) 路径损耗指数 路径损耗
    BS-目标-
    BS链路
    $ {d}_{\mathrm{S}} = 150 $ $ {\beta }_{\mathrm{S}} = 3 $ $ {\xi }_{\mathrm{S}} = {\xi }_{0}{\left({d}_{\mathrm{S}}/{d}_{0}\right)}^{-{{\beta }_{\mathrm{S}}}} $
    RIS-BS
    链路
    $ {d}_{\mathrm{IB}} = 50 $ $ {\beta }_{\mathrm{IB}} = 2.3 $ $ {\xi }_{\mathrm{IB}} = {\xi }_{0}{\left({d}_{\mathrm{IB}}/{d}_{0}\right)}^{-{{\beta }_{\mathrm{IB}}}} $
    $ \mathrm{\mathit{U}}_k $-BS
    链路
    $ {d}_{{{\mathrm{U}}_{k}}\mathrm{B}} = 50 $ $ {\beta }_{{{\mathrm{U}}_{k}}\mathrm{B}} = 3.5 $ $ {\xi }_{{{\mathrm{U}}_{k}}\mathrm{B}} = {\xi }_{0}{\left({d}_{{{\mathrm{U}}_{k}}\mathrm{B}}/{d}_{0}\right)}^{-{{\beta }_{{{\mathrm{U}}_{k}}\mathrm{B}}}} $
    BS-$ {D}_{j} $
    链路
    $ {d}_{\mathrm{B}{{\mathrm{D}}_{j}}} = 50 $ $ {\beta }_{\mathrm{B}{{\mathrm{D}}_{j}}} = 3.5 $ $ {\xi }_{\mathrm{B}{{\mathrm{D}}_{j}}} = {\xi }_{0}{\left({d}_{\mathrm{B}{{\mathrm{D}}_{j}}}/{d}_{0}\right)}^{-{{\beta }_{\mathrm{B}{{\mathrm{D}}_{j}}}}} $
    $ \mathit{\mathit{\mathrm{\mathit{U}}}}_k $-RIS
    链路
    $ {d}_{{{\mathrm{U}}_{k}}\mathrm{I}} = 2 $ $ {\beta }_{{{\mathrm{U}}_{k}}\mathrm{I}} = 2 $ $ {\xi }_{{{\mathrm{U}}_{k}}\mathrm{I}} = {\xi }_{0}{\left({d}_{{{\mathrm{U}}_{k}}\mathrm{I}}/{d}_{0}\right)}^{-{{\beta }_{{{\mathrm{U}}_{k}}\mathrm{I}}}} $
    RIS-$ {D}_{j} $
    链路
    $ {d}_{\mathrm{I}{{\mathrm{D}}_{j}}}=2 $ $ {\beta }_{\mathrm{I}{{\mathrm{D}}_{j}}}=2 $ $ {\xi }_{\mathrm{I}{{\mathrm{D}}_{j}}}={\xi }_{0}{\left({d}_{\mathrm{I}{{\mathrm{D}}_{j}}}/{d}_{0}\right)}^{-{{\beta }_{\mathrm{I}{{\mathrm{D}}_{j}}}}} $
    下载: 导出CSV

    表  4  训练时间(s)

    ELM FNN CGAN
    ISAC BS $ \mathit{\mathrm{\mathit{S}}}_1 $:$ \boldsymbol{A} $,$ {\boldsymbol{b}}_{k} $ 2.54 10.31 436.09
    $ \mathrm{\mathit{\mathit{\mathit{\mathit{\mathit{\mathit{\mathit{\mathit{\mathrm{\mathit{S}}}}}}}}_{\mathrm{2}}}}} $:$ {\boldsymbol{B}}_{k} $ 14.76 485.26 618.24
    下行$ {D}_{j} $ $ \mathrm{\mathit{S}}_1 $:$ {\boldsymbol{d}}_{j} $ 2.38 3.94 493.52
    $ \mathrm{\mathit{S}}_2 $:$ {\boldsymbol{D}}_{j} $ 9.41 88.82 522.06
    下载: 导出CSV
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出版历程
  • 收稿日期:  2025-11-05
  • 修回日期:  2026-01-22
  • 网络出版日期:  2026-03-04

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