一種基于有限內(nèi)存擬牛頓法的混合波束成形算法
doi: 10.11999/JEIT230656 cstr: 32379.14.JEIT230656
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杭州電子科技大學(xué)通信工程學(xué)院 杭州 310000
基金項(xiàng)目: 國(guó)家自然科學(xué)基金(U21A20450, 62301204)
A Hybrid Beamforming Algorithm Based on Limited-Broyden-Fletcher-Goldfarb-Shanno
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School of Communication Engineering, Hangzhou Dianzi University, Hangzhou 310000, China
Funds: The National Natural Science Foundation of China (U21A20450, 62301204)
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摘要: 針對(duì)現(xiàn)有混合波束成形算法運(yùn)行時(shí)間長(zhǎng)、頻譜效率低、誤碼率高的問(wèn)題,該文提出一種基于有限內(nèi)存擬牛頓法的混合波束成形算法(LBFGS)。該算法首先通過(guò)數(shù)字預(yù)編碼器的最小二乘解構(gòu)建單變量目標(biāo)函數(shù);然后采用目標(biāo)函數(shù)的梯度近似黑塞矩陣的逆得到搜索方向并沿搜索方向更新模擬預(yù)編碼器,直到滿(mǎn)足停止條件;最后固定模擬預(yù)編碼器得到數(shù)字預(yù)編碼器。MATLAB仿真結(jié)果表明,LBFGS算法較現(xiàn)有MO算法減少了28%的運(yùn)行時(shí)間,頻譜效率提高了1.05%,誤碼率降低了1.06%。
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關(guān)鍵詞:
- 毫米波 /
- 大規(guī)模MIMO /
- 混合波束成形 /
- 重疊天線陣列
Abstract: To solve the problems of long runtime, low spectral rate and high bit error rate, which exist in conventional hybrid beamforming schemes, a hybrid beamforming algorithm based on Limited-Broyden-Fletcher-Goldfarb-Shanno (LBFGS) is proposed. Firstly, a single variable objective function is constructed through the least squares solution of the digital precoder. Then, the gradient of the objective function is used to approximate the inverse of the Hessian matrix for obtaining the search direction and the analog precoder is updated along the search direction until the stop condition is satisfied. Finally, the analog precoder is fixed to obtain the digital precoder. The MATLAB simulation analysis indicate that LBFGS algorithm reduces the running time by 28%, increases spectral rate by 1.05%, and reduces bit error rate by 1.06%, compared to MO algorithm.-
Key words:
- Millimeter wave /
- Massive MIMO /
- Hybrid beamforming /
- Overlapped antenna arrays
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算法1 計(jì)算搜索方向 輸入:${\gamma _0}{\text{ = }}1$,${\mathrm{grad}}{{\boldsymbol{\mathcal{J}}}_{{{\boldsymbol{v}}_k}}}$,已存儲(chǔ)的梯度向量
$ \{ {\boldsymbol{s}}_i^{(k)},{\boldsymbol{y}}_i^{(k)}\} _{i = k - m}^{k - 1} $輸出: 搜索方向${{\boldsymbol{\eta}} _k}$ ${\boldsymbol{\mathcal{H}}}_k^0 = {\gamma _k}$; $ {\boldsymbolq7j3ldu95} \leftarrow {\mathrm{grad}}{{\boldsymbol{\mathcal{J}}}_{{v_k}}} $; If $m \ne 0$ for $i = k - 1,k - 2, \cdots ,k - m$do ${\boldsymbol{\alpha}} \leftarrow {\rho _i} < {\boldsymbol{s}}_i^{(k)},{\boldsymbolq7j3ldu95} > $ ${\boldsymbolq7j3ldu95} \leftarrow {\boldsymbolq7j3ldu95} - {\alpha _i}{\boldsymbol{y}}_i^{(k)}$ end for ${\boldsymbol{e}} \leftarrow {\boldsymbol{\mathcal{H}}}_k^0{\boldsymbolq7j3ldu95}$ for $i = k - m,k - m + 1, \cdots ,k - 1$ $\beta \leftarrow {\rho _i} < {\boldsymbol{y}}_i^{(k)},{\boldsymbol{e}} > $ $ {\boldsymbol{e}} \leftarrow {\boldsymbol{e}} + {\boldsymbol{s}}_i^{(k)} < {\varepsilon _i} - \beta > $ end for ${{\boldsymbol{\eta}} _k} = - {\boldsymbol{e}}$ else ${{\boldsymbol{\eta}} _k} = - {\boldsymbol{\mathcal{H}}}_k^0{\boldsymbolq7j3ldu95}$ end if 下載: 導(dǎo)出CSV
算法2 基于有限內(nèi)存擬牛頓法的模擬預(yù)編碼器算法 輸入:最優(yōu)全數(shù)字矩陣${{\boldsymbol{V}}_{\rm{opt}}}$,初始模擬波束成形矩陣${\boldsymbol{V}}_{\rm{RF}}^0$,內(nèi)存
容量$\forall M \in \mathbb{Z}$且$M > 0$。輸出:${{\boldsymbol{V}}_{\rm{RF}}}$ 初始化:內(nèi)部循環(huán)次數(shù)$k = 0$,內(nèi)存占用量$m = 0$ 根據(jù)式(10)計(jì)算黎曼梯度${\mathrm{grad}}{{\boldsymbol{\mathcal{J}}}_{{{\boldsymbol{v}}_0}}}$ while梯度的范數(shù)達(dá)到閾值${\mu _2}$ 根據(jù)算法1計(jì)算搜索方向${{\boldsymbol{\eta}} _k}$ 線搜索并回縮得到${{\boldsymbol{v}}_{k + 1}}{\text{ = }}{{{R}}_{{{\boldsymbol{v}}_k}}}({\alpha _k}{{\boldsymbol{\eta}} _k})$ 計(jì)算黎曼梯度${\mathrm{grad}}{{\boldsymbol{\mathcal{J}}}_{{{\boldsymbol{v}}_{k + 1}}}}$ 計(jì)算$ {\boldsymbol{s}}_k^{(k + 1)} $,$ {\boldsymbol{y}}_k^{(k + 1)} $,${\rho _{k + 1}} = 1/ < {\boldsymbol{s}}_k^{(k + 1)},{\boldsymbol{y}}_k^{(k + 1)} > $ if 滿(mǎn)足存儲(chǔ)條件 計(jì)算${\gamma _{k + 1}} = < {\boldsymbol{s}}_k^{(k + 1)},{\boldsymbol{y}}_k^{(k + 1)} > /||{\boldsymbol{y}}_k^{(k + 1)}|{|^2}$ if溢出 丟棄$ {\boldsymbol{s}}_{k - M}^{(k)} $,$ {\boldsymbol{y}}_{k - M}^{(k)} $; end if 歷史梯度向量傳輸 存儲(chǔ)$ {\text{\{ }}{\boldsymbol{s}}_k^{(k + 1)},{\boldsymbol{y}}_k^{(k + 1)}{\text{\} }} $ 若$m < M$,那么$m = m + 1$,否則$m = M$ else ${\gamma _{k + 1}} = {\gamma _k}$; end if $k \leftarrow k + 1$ end 下載: 導(dǎo)出CSV
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