R16 Type II量化反馈码本的产生

文章目录

    • 前言
    • 1. R15 TypeII Codebook:简短回顾
      • 1.1 W 1 \boldsymbol{W}_1 W1:Beam Selection
      • 1.2 W 2 \boldsymbol{W}_2 W2:Linear Combination Coefficients Quantization
    • 2 R16 eTypeII Codebook
      • 2.1 空/频域基底
      • 2.2 系数矩阵 W ~ 2 \tilde{\boldsymbol{W}}_2 W~2的量化
      • 2.3 生成R16 Codebook的步骤
      • 2.4 rank=2
    • 3 仿真设置建议
      • 3.1 设置
      • 3.2 Metric
    • 主要参考文献

前言

R16 Type II codebook是R15的增强版,R15仅考虑在空域上压缩,R16同时在空域和频域上压缩,所以一般把R16 Codebook 称为 R16 enhanced Type II codebook。下面,我们将简短回顾R15 Type II codebook,再深入介绍R16 eType II codebook。如果没有特殊说明,下面的介绍只关注单流(rank/layer=1)

1. R15 TypeII Codebook:简短回顾

NR R15 codebook is a two-stage precoder in the form
W = W 1 W 2 \boldsymbol{W} = \boldsymbol{W}_1 \boldsymbol{W}_2 W=W1W2

(关于Wideband Matrix W 1 \boldsymbol W_1 W1是不是layer-common这一点存疑)

1.1 W 1 \boldsymbol{W}_1 W1:Beam Selection

W 1 ∈ C P × 2 L \boldsymbol{W}_1 \in \mathbb{C}^{P \times 2L} W1CP×2L是一个块对角矩阵 ( P = 2 N 1 N 2 ) ({P}=2 N_1 N_2) (P=2N1N2)表示双极化的CSI-RS port数, N 1 N_1 N1 N 2 N_2 N2分别表示水平方向和垂直方向的天线数。
W 1 = [ B 0 0 B ] \boldsymbol{W}_1 = \left[ \begin{matrix} \boldsymbol{B}& \boldsymbol{0}\\ \boldsymbol{0}& \boldsymbol{B}\\ \end{matrix} \right] W1=[B00B]

其中 B = [ b 0 , ⋯   , b L − 1 ] ∈ C N 1 N 2 × L \boldsymbol{B}=[\boldsymbol{b}_0, \cdots, \boldsymbol{b}_{L-1}]\in \mathbb{C}^{N_1 N_2 \times L} B=[b0,,bL1]CN1N2×L的列向量 { b i } i = 0 L − 1 {\{\boldsymbol{b}_i\}}_{i=0}^{L-1} {bi}i=0L1是正交的,从两维的过采样DFT矩阵中选出 L L L个正交的基底。水平方向和垂直方向的过采样DFT向量可以表示为( O 1 O_1 O1 O 2 O_2 O2分别表示水平方向和垂直方向的过采样因子):
Horizontal:  u m = [ 1 , e j 2 π m O 1 N 1 , ⋯   , e j 2 π m ( N 1 − 1 ) O 1 N 1 ] ∈ C N 1 × 1 Vertical:  v l = [ 1 , e j 2 π l O 2 N 2 , ⋯   , e j 2 π l ( N 2 − 1 ) O 2 N 2 ] ∈ C N 2 × 1 \begin{aligned} \text{Horizontal: } \boldsymbol{u}_m &= [1, e^{\frac{j 2 \pi m}{O_1 N_1}}, \cdots, e^{\frac{j 2 \pi m(N_1 - 1)}{O_1 N_1}}] \in \mathbb{C}^{N_1 \times 1} \\ \text{Vertical: }\boldsymbol{v}_l &= [1, e^{\frac{j 2 \pi l}{O_2 N_2}}, \cdots, e^{\frac{j 2 \pi l(N_2 - 1)}{O_2 N_2}}] \in \mathbb{C}^{N_2 \times 1} \end{aligned} Horizontal: umVertical: vl=[1,eO1N1j2πm,,eO1N1j2πm(N11)]CN1×1=[1,eO2N2j2πl,,eO2N2j2πl(N21)]CN2×1

矩阵 B \boldsymbol{B} B每个列向量 b l , m \boldsymbol{b}_{l,m} bl,m可以表示为:
b m , l = u m ⊗ v l ∈ C N 1 N 2 × 1 \boldsymbol{b}_{m,l} = \boldsymbol{u}_{m} \otimes \boldsymbol{v}_l \in \mathbb{C}^{N_1N_2 \times 1} bm,l=umvlCN1N2×1


m = O 1 n 1 ( i ) + q 1 ,    0 ≤ n 1 ( i ) < N 1 , 0 ≤ q 1 < O 1    ( 0 ≤ m ≤ N 1 O 1 − 1 ) m=O_1 n^{(i)}_1 + q_1, \ \ 0 \leq n^{(i)}_1 < N_1, 0 \leq q_1m=O1n1(i)+q1,  0n1(i)<N1,0q1<O1  (0mN1O11)

l = O 2 n 2 ( i ) + q 2 ,    0 ≤ n 2 ( i ) < N 2 , 0 ≤ q 2 < O 2    ( 0 ≤ m ≤ N 2 O 2 − 1 ) l=O_2 n^{(i)}_2 + q_2, \ \ 0 \leq n^{(i)}_2 < N_2, 0 \leq q_2l=O2n2(i)+q2,  0n2(i)<N2,0q2<O2  (0mN2O21)

1.2 W 2 \boldsymbol{W}_2 W2:Linear Combination Coefficients Quantization

W 1 \boldsymbol{W}_1 W1可以看作是空域上的压缩基底( 2 L < 2 N 1 N 2 2L < 2N_1N_2 2L<2N1N2个基底), W 2 \boldsymbol{W}_2 W2构成了这 2 L 2L 2L个基底上的系数(坐标),我们要做的就是对这 2 L 2L 2L个系数做量化。 W 2 ∈ C 2 L × 1 \boldsymbol{W}_2 \in \mathbb{C}^{2L \times 1} W2C2L×1的每个元素 w ℓ w_{\ell} w 0 ≤ ℓ < L 0 \leq {\ell}0<L)可以被表示(量化)为
w ℓ = λ ˉ ℓ λ ℓ e j 2 π ϕ ℓ w_{\ell} = \bar \lambda_{\ell} \lambda_{\ell} e^{j 2 \pi \phi_{\ell}} w=λˉλej2πϕ

其中
λ ˉ ℓ \bar \lambda_{\ell} λˉ is a wideband reference amplitude
λ ℓ \lambda_{\ell} λ is a subband differentia amplitude
ϕ ℓ \phi_{\ell} ϕ is a subband phase term

2 R16 eTypeII Codebook

2.1 空/频域基底

R16考虑到了子带(时延)之间的相关性,因此反馈的系数可以进一步减少,我们令 N 3 N_3 N3为子带的数量(# frequency units),R16 codebook可以表示为
W = W 1 W ~ 2 W f H ∈ C P × N 3 \boldsymbol{W} = \boldsymbol{W}_1 \tilde{\boldsymbol{W}}_2 \boldsymbol{W}^H_f \in \mathbb C^{P \times N_3} W=W1W~2WfHCP×N3

UE需要反馈:

  • W 1 = { b i } i = 0 L − 1 \boldsymbol{W}_1={\{\boldsymbol{b}_i \}}_{i=0}^{L-1} W1={bi}i=0L1 空域选择的基底
  • W f = { f k } k = 0 M − 1 \boldsymbol{W}_f = {\{\boldsymbol{f}_k\}}_{k=0}^{M-1} Wf={fk}k=0M1 频域选择的基底
  • 线性组合系数 W ~ 2 = { c i , f } ,    i = 0 , 1 , ⋯   , L − 1 ,    k = 0 , 1 , ⋯   , M − 1 \tilde{\boldsymbol{W}}_2={\{c_{i,f}\}}, \ \ i = 0,1,\cdots,L-1, \ \ k=0,1,\cdots,M-1 W~2={ci,f},  i=0,1,,L1,  k=0,1,,M1

注意, W f ∈ C N 3 × M \boldsymbol{W}_f \in \mathbb{C}^{N_3 \times M} WfCN3×M是由 M M M个DFT基底构成的矩阵: W f = [ f 1 . ⋯   , f M ] \boldsymbol{W}_f=[\boldsymbol{f}_1. \cdots, \boldsymbol{f}_M] Wf=[f1.,fM],其中
f k = [ 1 , e − 2 π k N 3 , ⋯   , e − 2 π k ( N 3 − 1 ) N 3 ] ∈ C N 3 × 1 ,    0 ≤ k < N 3 \boldsymbol{f}_k = [1,e^{- \frac{2 \pi k}{N_3}},\cdots,e^{- \frac{2 \pi k(N_3-1)}{N_3}}] \in \mathbb{C}^{N_3 \times 1}, \ \ 0 \leq k < N_3 fk=[1,eN32πk,,eN32πk(N31)]CN3×1,  0k<N3

注意,频域上的压缩矩阵 W f \boldsymbol{W}_f Wf的选取对于不同流/layer是独立的的。

具体来看, N 3 N_3 N3 M M M的取值如何选取:
M M M: Number of DFT bais vectors
N 3 N_3 N3: Number of Frequency domain (FD) units

我们有
N 3 = N s b × R N_3 = N_{sb} \times R N3=Nsb×R

M = p v × N 3 R M = p_v \times \frac{N_3}{R} M=pv×RN3

其中

  • N s b N_{s_b} Nsb is the number of subband.
  • R R R indicates the number of FD units contained in each subband.
  • p v p_v pv is configured by high-level signaling to determin the number of DFT basis vectors.

2.2 系数矩阵 W ~ 2 \tilde{\boldsymbol{W}}_2 W~2的量化

W ~ 2 ∈ C 2 L × M \tilde{\boldsymbol{W}}_2 \in \mathbb{C}^{2L \times M} W~2C2L×M是两维的归一化线性组合系数, W ~ 2 \tilde{\boldsymbol{W}}_2 W~2可以被拆分为:
W ~ 2 = [ λ p 0 I L 0 0 λ p 1 I L ] W ˉ 2 \tilde{\boldsymbol{W}}_2 = \left[ \begin{matrix} \lambda _{p_0}\boldsymbol{I}_L& \boldsymbol{0}\\ \boldsymbol{0}& \lambda _{p_1}\boldsymbol{I}_L\\ \end{matrix} \right] \bar{ \boldsymbol{W}}_2 W~2=[λp0IL00λp1IL]Wˉ2

其中

  • λ p 0 , λ p 1 ∈ C R e f \lambda _{p_0}, \lambda _{p_1} \in \mathcal{C}_{Ref} λp0,λp1CRef correspond to the reference amplitude quantization values for each of the two polarizations ( p 0 p_0 p0 and p 1 p_1 p1)
  • W ~ 2 \tilde{\boldsymbol{W}}_2 W~2 is normalized such that max ⁡ { λ p 0 , λ p 1 } = 1 \max \{ \lambda _{p_0}, \lambda _{p_1}\}=1 max{λp0,λp1}=1

另一部分, W ˉ 2 \bar{ \boldsymbol{W}}_2 Wˉ2可以表示为:
W ˉ 2 = Λ ⊙ e j 2 π Φ \bar{ \boldsymbol{W}}_2 = \boldsymbol{\Lambda} \odot e^{j 2 \pi \boldsymbol{\Phi}} Wˉ2=Λej2πΦ

其中 ⊙ \odot 表示Hadamard积,and each of the elements of Λ \boldsymbol \Lambda Λ and Φ \boldsymbol{\Phi} Φ represent the differential amplitude ( ∈ C A \in \mathcal{C}_A CA) and phase values ( ∈ C P \in \mathcal{C}_P CP) of 2 L × M 2L \times M 2L×M linear combination coefficients.

另外,为了进一步降低CSI反馈的开销, 2 L M 2LM 2LM个线性组合系数只有 2 β L M ( 0 < β < 1 ) 2 \beta LM(0< \beta<1) 2βLM(0<β<1)个系数被反馈。

  • C R e f = { 1 , ( 1 2 ) 1 4 , ( 1 2 2 ) 1 4 , ( 1 2 3 ) 1 4 , ⋯   , ( 1 2 14 ) 1 4 , 0 } \mathcal{C}_{Ref}=\{1, (\frac{1}{2})^{\frac{1}{4}}, (\frac{1}{2^2})^{\frac{1}{4}}, (\frac{1}{2^{3}})^{\frac{1}{4}}, \cdots, (\frac{1}{2^{14}})^{\frac{1}{4}}, 0\} CRef={1,(21)41,(221)41,(231)41,,(2141)41,0}
  • C A = { 1 , 1 2 , 1 2 , 1 2 2 , 1 4 , 1 4 2 , 1 8 , 1 8 2 } \mathcal{C}_{A} =\{1,\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2 \sqrt{2}}, \frac{1}{4}, \frac{1}{4 \sqrt{2}}, \frac{1}{8}, \frac{1}{8 \sqrt{2}}\} CA={1,2 1,21,22 1,41,42 1,81,82 1}
  • C P \mathcal{C}_P CP: 16-PSK/ 8-PSK

3GPP给出了 ( L , p v , β ) (L,p_v, \beta) (L,pv,β)三个参数的配置。

R16 Type II量化反馈码本的产生_第1张图片

2.3 生成R16 Codebook的步骤

假设一共有$N_3个子带,我们考虑信号模型为

y U E ( f n ) = H ( f n ) W C S I − R S s C S I − R S + z \boldsymbol{y}^{UE}(f_{n})=\boldsymbol{H}(f_{n}) \boldsymbol{W}_{CSI-RS} \boldsymbol{s}_{CSI-RS} + \boldsymbol{z} yUE(fn)=H(fn)WCSIRSsCSIRS+z

有效信道
H e ( f n ) = H ( f n ) W C S I − R S ∈ C N r × P \boldsymbol{H}_e(f_{n}) = \boldsymbol{H}(f_{n}) \boldsymbol{W}_{CSI-RS} \in \mathbb{C}^{Nr \times P} He(fn)=H(fn)WCSIRSCNr×P

基于上述模型和等效信道,我们可以通过以下步骤构造生成R16 Codebook precoder

  • Step-1: UE估计下行有效信道 H e ( f n ) ∈ C N r × P \boldsymbol{H}_e(f_{n})\in \mathbb{C}^{Nr \times P} He(fn)CNr×P

  • Step-2: 在全频带是上计算协方差矩阵(求均值)
    R D L = E f n [ H e H ( f n ) H e ( f n ) ] ∈ C P × P \boldsymbol{R}_{DL}= \mathbb{E}_{f_n} \left [ \boldsymbol{H}^H_e(f_{n}) \boldsymbol{H}_e(f_{n}) \right] \in \mathbb{C}^{P \times P} RDL=Efn[HeH(fn)He(fn)]CP×P

  • Step-3:将下行协方差矩阵映射到wideband beam matrix W 1 ∈ C P × 2 L \boldsymbol{W}_1 \in \mathbb{C}^{P \times 2L} W1CP×2L(通过遍历的方式寻找最优解)
    W 1 = arg ⁡ max ⁡ W ~ 1 ∈ C W 1 ∥ W ~ 1 H R D L W ~ 1 ∥ F 2 \boldsymbol{W}_1 = \arg\max_{\tilde{\boldsymbol{W}}_1 \in \mathcal{C}_{\boldsymbol{W}_1}} {\Vert \tilde{ \boldsymbol{W}}}_1^H \boldsymbol{R}_{DL} \tilde{\boldsymbol{W}}_1 \Vert_F^2 W1=argW~1CW1maxW~1HRDLW~1F2

  • Step-4: 将整个频带上的所有特征向量记为(我们实际要去量化的)
    V f = [ v f 1 , v f 2 , ⋯   , v f N 3 ] ∈ C P × N 3 \boldsymbol{V}^f = [\boldsymbol{v}^{f_1}, \boldsymbol{v}^{f_2}, \cdots, \boldsymbol{v}^{f_{N_3}}] \in \mathbb{C}^{P \times N_3} Vf=[vf1,vf2,,vfN3]CP×N3

    其中 v f n \boldsymbol{v}^{f_n} vfn表示第 n n n个子带对应的协方差矩阵 R D L ( f n ) = H e H ( f n ) H e ( f n ) \boldsymbol{R}_{DL}(f_n)=\boldsymbol{H}^H_e(f_{n}) \boldsymbol{H}_e(f_{n}) RDL(fn)=HeH(fn)He(fn)的主特征向量。空域上线性组合系数构成的矩阵 W 2 \boldsymbol{W}_2 W2可以表示为
    W 2 = W 1 H V f ∈ C 2 L × N 3 \boldsymbol{W}_2 = \boldsymbol{W}^H_1 \boldsymbol{V}^f \in \mathbb{C}^{2L \times N_3} W2=W1HVfC2L×N3

  • Step-5: 从 N 3 N_3 N3个DFT基向量( N 3 N_3 N3维空间)中选取 M M M个基底作为频域上的压缩基底。
    基于最大化能量准则,我们计算
    k 1 , ⋯   , k M = arg max ⁡ k 1 , ⋯   , k M ∑ k m ∥ W 2 f k m ∥ 2 k_1,\cdots,k_M = \argmax_{k_1,\cdots,k_M } \sum_{k_m}{ \Vert \boldsymbol{W}_2 \boldsymbol{f}_{k_m} \Vert}_2 k1,,kM=k1,,kMargmaxkmW2fkm2

    N 3 N_3 N3个基底中找到能量最大的 M M M个基底,或者表述为:
    ∥ W 2 f k 1 ∥ 2 ≥ ∥ W 2 f k 2 ∥ 2 ≥ ∥ W 2 f k 3 ∥ 2 ≥ ⋯ ≥ ∥ W 2 f k M ∥ 2 ≥ ⋯ { \Vert \boldsymbol{W}_2 \boldsymbol{f}_{k_1} \Vert}_2 \geq { \Vert \boldsymbol{W}_2 \boldsymbol{f}_{k_2} \Vert}_2 \geq { \Vert \boldsymbol{W}_2 \boldsymbol{f}_{k_3} \Vert}_2 \geq \cdots \geq { \Vert \boldsymbol{W}_2 \boldsymbol{f}_{k_M} \Vert}_2 \geq \cdots W2fk12W2fk22W2fk32W2fkM2

    由此得到频域压缩的基底
    W f = [ f k 1 , f k 2 , ⋯   , f k M ] ∈ C N 3 × M \boldsymbol{W}_f=[\boldsymbol{f}_{k_1},\boldsymbol{f}_{k_2},\cdots,\boldsymbol{f}_{k_M}] \in \mathbb{C}^{N_3\times M} Wf=[fk1,fk2,,fkM]CN3×M

  • Step-6:计算空域-频域的系数矩阵 W ~ 2 ∈ C 2 L × M \tilde{\boldsymbol{W}}_2 \in \mathbb{C}^{2L \times M} W~2C2L×M:
    W ~ 2 = W 2 W f ∈ C 2 L × M \tilde{\boldsymbol{W}}_2=\boldsymbol{W}_2 \boldsymbol{W}_f \in \mathbb{C}^{2L \times M} W~2=W2WfC2L×M

  • Step-7: 对 W ~ 2 ∈ C 2 L × M \tilde{\boldsymbol{W}}_2 \in \mathbb{C}^{2L \times M} W~2C2L×M进行量化:首先,在 2 L M 2LM 2LM个线性组合系数找出模最大的 2 β L M ( 0 < β < 1 ) 2 \beta LM(0< \beta<1) 2βLM(0<β<1)个系数,其它置为0。
    7.1 分别找到 W ~ 2 \tilde{\boldsymbol{W}}_2 W~2两个极化方向上模最大的元素和索引:

    + 4 5 ° +45^{\degree} +45°方向的索引记为 ( l 0 , m 0 ) ,    1 ≤ l 0 ≤ L , 1 ≤ m 0 ≤ M (l_0,m_0), \ \ 1 \leq l_0 \leq L, 1 \leq m_0 \leq M (l0,m0),  1l0L,1m0M
    − 4 5 ° -45^{\degree} 45°方向的索引记为 ( l 1 , m 1 ) ,    L + 1 ≤ l 1 ≤ 2 L , 1 ≤ m 1 ≤ M (l_1,m_1), \ \ L+1 \leq l_1 \leq 2L, 1 \leq m_1 \leq M (l1,m1),  L+1l12L,1m1M

    W ~ 2 \tilde{\boldsymbol{W}}_2 W~2归一化为:
    W ~ 2 n o r m = 1 max ⁡ { ∣ W ~ 2 ( l 0 , m 0 ) ∣ , ∣ W ~ 2 ( l 1 , m 1 ) ∣ } W ~ 2 \tilde{\boldsymbol{W}}^{norm}_2=\frac{1}{\max\{|\tilde{W}_2(l_0,m_0)|,|\tilde{W}_2(l_1,m_1)|\}} \tilde{\boldsymbol{W}}^{}_2 W~2norm=max{W~2(l0,m0),W~2(l1,m1)}1W~2

    7.2 对参考幅度值做量化:
    + 4 5 ° +45^{\degree} +45°方向: λ p 0 = arg min ⁡ λ p 0 ∈ C r e f ∣ W ~ 2 ( l 0 , m 0 ) − λ p 0 ∣ \lambda_{p_0}=\argmin_{\lambda_{p_0} \in \mathcal{C}_{ref}}|\tilde{W}_2(l_0,m_0) - \lambda_{p_0}| λp0=argminλp0CrefW~2(l0,m0)λp0
    − 4 5 ° -45^{\degree} 45°方向: λ p 1 = arg min ⁡ λ p 1 ∈ C r e f ∣ W ~ 2 ( l 1 , m 1 ) − λ p 1 ∣ \lambda_{p_1}=\argmin_{\lambda_{p_1} \in \mathcal{C}_{ref}}|\tilde{W}_2(l_1,m_1) - \lambda_{p_1}| λp1=argminλp1CrefW~2(l1,m1)λp1

    其中 C R e f = { 1 , ( 1 2 ) 1 4 , ( 1 2 2 ) 1 4 , ( 1 2 3 ) 1 4 , ⋯   , ( 1 2 14 ) 1 4 , 0 } \mathcal{C}_{Ref}=\{1, (\frac{1}{2})^{\frac{1}{4}}, (\frac{1}{2^2})^{\frac{1}{4}}, (\frac{1}{2^{3}})^{\frac{1}{4}}, \cdots, (\frac{1}{2^{14}})^{\frac{1}{4}}, 0\} CRef={1,(21)41,(221)41,(231)41,,(2141)41,0}

    7.3 将 W ~ 2 n o r m \tilde{\boldsymbol{W}}^{norm}_2 W~2norm拆分为
    W ~ 2 n o r m = [ λ p 0 I L 0 0 λ p 1 I L ] W ˉ 2 n o r m ∈ C 2 L × M \tilde{\boldsymbol{W}}^{norm}_2 = \left[ \begin{matrix} \lambda _{p_0}\boldsymbol{I}_L& \boldsymbol{0}\\ \boldsymbol{0}& \lambda _{p_1}\boldsymbol{I}_L\\ \end{matrix} \right] \bar{ \boldsymbol{W}}^{norm}_2 \in \mathbb{C}^{2L \times M} W~2norm=[λp0IL00λp1IL]Wˉ2normC2L×M

    7.4 对 W ˉ 2 n o r m \bar{ \boldsymbol{W}}^{norm}_2 Wˉ2norm中的非零项做量化:
    W ˉ 2 q u a n ( l , m ) = λ e j 2 π ϕ = arg max ⁡ λ ∈ C A , ϕ ∈ C P ∣ W ˉ 2 n o r m ( l , m ) − λ e j 2 π ϕ ∣ \bar{ {W}}^{quan}_2(l,m) = \lambda e^{j 2 \pi \phi}=\argmax_{\lambda \in \mathcal{C}_A, \phi \in \mathcal{C}_P} |\bar{ {W}}^{norm}_2(l,m) - \lambda e^{j 2 \pi \phi}| Wˉ2quan(l,m)=λej2πϕ=λCA,ϕCPargmaxWˉ2norm(l,m)λej2πϕ

    最终,量化后的码本为:
    W q u a n = W 1 [ λ p 0 I L 0 0 λ p 1 I L ] W ˉ 2 q u a n W f H = W 1 W ~ 2 W f H \begin{aligned} \boldsymbol{W}^{quan} &= \boldsymbol{W}_1 \left[ \begin{matrix} \lambda _{p_0}\boldsymbol{I}_L& \boldsymbol{0}\\ \boldsymbol{0}& \lambda _{p_1}\boldsymbol{I}_L\\ \end{matrix} \right] \bar{ \boldsymbol{W}}^{quan}_2 \boldsymbol{W}^H_f \\ &= \boldsymbol{W}_1 \tilde{ \boldsymbol{W}}_2 \boldsymbol{W}^H_f \end{aligned} Wquan=W1[λp0IL00λp1IL]Wˉ2quanWfH=W1W~2WfH

2.4 rank=2

R16 Type II量化反馈码本的产生_第2张图片

3 仿真设置建议

3.1 设置

(1) R15 Type I
rank=1:遍历所有码本,根据最大化RSRP准则找到量化最优的码本
rank=2:遍历所有码本,根据最大化RSRP准则找到量化最优的码本

(2) R16 Type II
基本配置

  • 每个subband包含4 / 8个RB
  • 直接配置 N 3 N_3 N3为子带数
  • paramcombination-r16取5,即 L = 4 ,   p v = 1 / 4 ,   β = 3 / 4 L=4,\ p_v=1/4, \ \beta=3/4 L=4, pv=1/4, β=3/4,可以计算出,频域压缩的基底数 M = p v ⋅ N 3 = N 3 / 4 M=p_v \cdot N_3=N_3 / 4 M=pvN3=N3/4,如果没有除尽,我们考虑上取整 M = ⌈ N 3 / 4 ⌉ M=\lceil N_3 / 4 \rceil M=N3/4
  • 空域压缩基底:layer-common
  • 对于rank=2,频域压缩基底:layer-independent

额外说明

  • 假设UE已知完美信道,不考虑估计误差,直接对完美信道进行量化,因此只考虑量化误差
  • 假设已知外层权矩阵 W C S I − R S ∈ C N B S × P \boldsymbol{W}_{CSI-RS} \in \mathbb{C}^{N_{BS} \times P} WCSIRSCNBS×P,这可以通过遍历8个外层权来选择最大化夹角余弦所对应的矩阵。

3.2 Metric

(1) R15 Type I: 空域压缩,各个子带单独反馈
(1.1) rank=1
R D L s u b = λ 1 v 1 v 1 H + ∑ k > 1 λ k v k v k H R^{sub}_{DL} = \lambda_1 \boldsymbol{v}_1 \boldsymbol{v}^H_1 + \sum_{k>1} \lambda_k \boldsymbol{v}_k \boldsymbol{v}^H_k RDLsub=λ1v1v1H+k>1λkvkvkH

其中 λ 1 ≥ λ 2 ≥ ⋯ ≥ λ P \lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_P λ1λ2λP。若令R15 Type I量化后的码本为 w ∈ C P × 1 \boldsymbol{w} \in \mathbb{C}^{P \times 1} wCP×1,考虑以下两种metric:
square of cosine : ∣ < v 1 ∥ v 1 ∥ 2 , w ∥ w ∥ 2 > ∣ 2 MSE : ∥ v 1 ∥ v 1 ∥ 2 − w ∥ w ∥ 2 ∥ 2 2 \begin{aligned} \text{square of cosine} &: {\left \vert <\frac{\boldsymbol{v}_1}{{\Vert \boldsymbol{v}_1 \Vert}}_2, \frac{\boldsymbol{w}}{{\Vert \boldsymbol{w} \Vert}}_2 > \right \vert}^2 \\ \text{MSE} &: \left \Vert \frac{\boldsymbol{v}_1}{{\Vert \boldsymbol{v}_1 \Vert}}_2- \frac{\boldsymbol{w}}{{\Vert \boldsymbol{w} \Vert}}_2 \right \Vert^2_2 \end{aligned} square of cosineMSE: <v1v12,ww2> 2: v1v12ww2 22

(1.1) rank=2
R D L s u b = λ 1 v 1 v 1 H + λ 2 v 2 v 2 H + ∑ k > 2 λ k v k v k H R^{sub}_{DL} = \lambda_1 \boldsymbol{v}_1 \boldsymbol{v}^H_1 + \lambda_2 \boldsymbol{v}_2 \boldsymbol{v}^H_2 + \sum_{k>2} \lambda_k \boldsymbol{v}_k \boldsymbol{v}^H_k RDLsub=λ1v1v1H+λ2v2v2H+k>2λkvkvkH

其中 λ 1 ≥ λ 2 ≥ ⋯ ≥ λ P \lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_P λ1λ2λP。若令R15 Type I量化后的码本为 W = [ w 1 , w 2 ] ∈ C P × 2 \boldsymbol{W} = [\boldsymbol{w}_1, \boldsymbol{w}_2]\in \mathbb{C}^{P \times 2} W=[w1,w2]CP×2,量化的metric与rank=1等价, w 1 \boldsymbol{w}_1 w1 w 2 \boldsymbol{w}_2 w2分别评估。

(2) R16 Type II: 空/频域压缩
回顾整个频带的特征向量(我们实际要去量化的):
layer-1:
V 1 f = [ v 1 f 1 , v 1 f 2 , ⋯   , v 1 f N 3 ] ∈ C P × N 3 \boldsymbol{V}^f_1 = [\boldsymbol{v}^{f_1}_1, \boldsymbol{v}^{f_2}_1, \cdots, \boldsymbol{v}^{f_{N_3}}_1] \in \mathbb{C}^{P \times N_3} V1f=[v1f1,v1f2,,v1fN3]CP×N3

layer-2:
V 2 f = [ v 2 f 1 , v 2 f 2 , ⋯   , v 2 f N 3 ] ∈ C P × N 3 \boldsymbol{V}^f_2 = [\boldsymbol{v}^{f_1}_2, \boldsymbol{v}^{f_2}_2, \cdots, \boldsymbol{v}^{f_{N_3}}_2] \in \mathbb{C}^{P \times N_3} V2f=[v2f1,v2f2,,v2fN3]CP×N3

量化的metric与R15 rank=1类似,量化后各个子带、各个layer单独评估,再综合。

主要参考文献

[1] Zhihua Shi, et,al., Chapter 8 - Multiple-input multiple-output enhancement and beam management. 5G NR and Enhancements, Elsevier, 2022,

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