[1]-Tse, David, and Pramod Viswanath. Fundamentals of wireless communication. Cambridge university press, 2005.
文章目录
- 1-综述
- 2-系统模型
1-综述
misaligned OAC:预编码矩阵(含噪声)+ 没同步好
2-系统模型
θ ∈ R d \theta \in\mathbb{R}^d θ∈Rd全局模型.
M M M 个边缘设备.
大小为 B m B_m Bm的本地数据集 B m \mathcal{B}_m Bm,训练得到新模型 θ ~ m ∈ R d ; \tilde{\boldsymbol{\theta}}_m\in\mathbb{R}^d; θ~m∈Rd;
θ m ′ = B m ( θ ~ m − θ ) ∈ R d \theta_m^{\prime}=B_m(\tilde{\boldsymbol{\theta}}_m-\boldsymbol{\theta})\in\mathbb{R}^d θm′=Bm(θ~m−θ)∈Rd上传缩放的模型***.
θ + = ∑ m = 1 M θ m ′ \theta_+=\sum_{m=1}^M\theta_m^{\prime} θ+=∑m=1Mθm′服务器端估计算术和.
θ n e w = θ + 1 ∑ m B m θ + \theta_\mathrm{new}=\theta+\frac1{\sum_mB_m}\theta_+ θnew=θ+∑mBm1θ+是全局新权重.θ m ′ = [ ( s m t ) ⊤ , ( s m i ) ⊤ ] ⊤ \theta_m^{\prime}=[(s_m^\mathrm{t})^\mathrm{\top},(s_m^\mathrm{i})^\mathrm{\top}]^\mathrm{\top} θm′=[(smt)⊤,(smi)⊤]⊤将上行的权重转为复数.
s m ∈ C d / 2 : s m = s_m\in\mathbb{C}^{d/2}{:}s_m= sm∈Cd/2:sm= s m r + j s m i s_{m}^{\mathrm{r}}+js_{m}^{\mathrm{i}} smr+jsmi
s m = [ s m [ 1 ] s_m=[s_m[1] sm=[sm[1], s m [ 2 ] , … , s m [ L ~ ] ] ⊤ . s_{m}[2],\ldots,s_{m}[\tilde{L}]]^{\top}. sm[2],…,sm[L~]]⊤.每个时隙传 L L L个符号.
x m ( t ) = α m ∑ ℓ = 1 L s m [ ℓ ] p ( t − ℓ T ) , x_m(t)=\alpha_m\sum_{\ell=1}^Ls_m[\ell]p(t-\ell T), xm(t)=αm∑ℓ=1Lsm[ℓ]p(t−ℓT), 一个时隙,某台设备传输的信号.p ( t ) = 1 / 2 [ s g n ( t + T ) − s g n ( t ) ] p(t)~=~1/2\left[\mathrm{sgn}(t+T)-\mathrm{sgn}(t)\right] p(t) = 1/2[sgn(t+T)−sgn(t)].持续时间为 T T T的矩形脉冲.
α m \alpha_m αm信道预编码因子. α m = 1 / h ˉ m . \alpha_m=1/\bar{h}_m. αm=1/hˉm.r ( t ) = ∑ m = 1 M h ~ m x m ( t − τ m ) + z ( t ) , r(t)=\sum_{m=1}^M\tilde{h}_mx_m(t-\tau_m)+z(t), r(t)=∑m=1Mh~mxm(t−τm)+z(t), 预编码矩阵(含噪声)+ 没同步好.
h ~ m \tilde{h}_m h~m是平坦衰弱(相对于频率选择性衰弱而言,见【1,P44】)
τ m . \tau_m. τm.根据延时排序,e.g., τ 0 \tau_0 τ0没有延时第一个被接收.r ( t ) = ∑ m = 1 M h ~ m α m ∑ ℓ = 1 L s m [ ℓ ] p ( t − τ m − ℓ T ) + z ( t ) = ∑ ℓ = 1 L ∑ m = 1 M h m ′ s m [ ℓ ] p ( t − τ m − ℓ T ) + z ( t ) , r(t)=\sum_{m=1}^M\tilde{h}_m\alpha_m\sum_{\ell=1}^Ls_m[\ell]p(t-\tau_m-\ell T)+z(t)=\sum_{\ell=1}^L\sum_{m=1}^Mh_m^{\prime}s_m[\ell]p(t-\tau_m-\ell T)+z(t), r(t)=∑m=1Mh~mαm∑ℓ=1Lsm[ℓ]p(t−τm−ℓT)+z(t)=∑ℓ=1L∑m=1Mhm′sm[ℓ]p(t−τm−ℓT)+z(t),
h m ′ = h ~ m / h ˉ m h_m^{\prime}=\tilde{h}_m/\bar{h}_m hm′=h~m/hˉm不完美的CSI带来的.