Active IRS versus Passive IRS

Wireless Communication Aided by Intelligent Reflecting Surface: Active or Passive?. Changsheng You et.al. IEEE Wireless Communications Letters, Dec. 2021 (pdf) (Citations 9)

Quick Overview

a single antenna access point communicates with a single-antenna user

rate-maximization with passive/active IRS:

active IRS should be deployed closer to the RECEIVER with the decreasing amplification power of the active IRS （known in Lemma1）
passive IRS is known to be near the transmitter or receiver to minimize the path-loss (known in my-note section 3.2)

optimal the placement:

passive IRS outperform active IRS when the number of element is sufficiently large/ the amplification power of the active IRS is too small.(known in my-note section 3.3)
optimal placement to maximize the weighted sum-rate of uplink and downlink communications. Passive is more likely to achieve superior rate performance.(known in my-note section 3.4)

Model

Main Work

Active IRS

\[ y_{\mathrm{act}}=\boldsymbol{h}_{\mathrm{IU}}^{H} \eta \boldsymbol{\Theta}\left(\boldsymbol{h}_{\mathrm{AI}} s+\boldsymbol{n}_{\mathrm{F}}\right)+n \]

achievable rate = $\log_2(1+\text{SNR})=\log _{2}\left(1+\frac{P_{\mathrm{A}}\left|\boldsymbol{h}_{\mathrm{IU}}^{H} \eta \boldsymbol{\Theta} \boldsymbol{h}_{\mathrm{AI}}\right|^{2}}{\left\|\boldsymbol{h}_{\mathrm{IU}}^{H} \eta \boldsymbol{\Theta}\right\|^{2} \sigma_{\mathrm{F}}^{2}+\sigma^{2}}\right)$

得到优化目标（P1）：

然后由于（4a）恒成立（是e的指数次项），且（4e）等价形式$P_{\mathrm{F}} \geq N P_{\mathrm{A}} \beta /\left(D^{2}/4+H^{2}\right)+N \sigma_{\mathrm{F}}^{2}$

（原文这个等式有错误，$D^2$没有乘系数1/4）

得到等效的（P2）:

因为信道建模为： \[ \boldsymbol{h}_{\mathrm{AI}}=h_{\mathrm{AI}} \boldsymbol{a}_{\mathrm{r}}\left(\theta_{\mathrm{AI}}^{\mathrm{r}}, \vartheta_{\mathrm{AI}}^{\mathrm{r}}, N\right), \text { where } h_{\mathrm{AI}} \triangleq \sqrt{\beta / d_{\mathrm{AI}}^{2}} e^{-j \frac{2 \pi}{\lambda} d_{\mathrm{AI}}} \] 优化目标就是max(SINR)

整理得到优化目标（P3）：

可以得到Lemma1:AP-IRS的距离$x_{AI}$随$P_F$的增大而单调递减；随$N$非减。(随$P_F$的单调递减证明没看懂)

然后有给定的lemma2：

给定$H\ll D$: \[ \max \left\{C_{1} d_{\mathrm{AI}}^{2}, C_{2} d_{\mathrm{IU}}^{2}\right\} \gg C_{3} d_{\mathrm{AI}}^{2} d_{\mathrm{IU}}^{2}, \forall x_{\mathrm{AI}} \in[0, D] \] 如果有： \[ \sqrt{P_{A} \beta} / \sigma_{\mathrm{F}}+\sqrt{P_{F} \beta} / \sigma \gg D \]

所以由Lemma2可以得到优化目标（P4）：

这就是个高中数学题：

Passive IRS

与active IRS 类似，得到$\mathrm{SNR}_{\text {pas }}^{(\mathrm{DL})}=\frac{P_{\mathrm{A}} \beta^{2} N^{2}}{d_{\mathrm{AI}}^{2} d_{\mathrm{IU}}^{2} \sigma^{2}}$，然后由参考文献给出，对于Passive IRS，$x_{AI}=0\quad or\quad D$，得到： \[ \mathrm{SNR}_{\text {pas }}^{(\mathrm{DL}) *} \approx \widetilde{\mathrm{SNR}}_{\text {pas }}^{(\mathrm{DL})} \triangleq \frac{P_{\mathrm{A}} \beta^{2} N^{2}}{H^{2}\left(D^{2}+H^{2}\right) \sigma^{2}} \]

Active IRS versus Passive IRS

即：

给定较小的$H$，当Active IRS的功率$P_F$太小或者IRS装配了大量的elements（$N$很大），Passive IRS 往往优于 Active IRS。

Joint Downlink and Uplink communication

maximize the weighted sum-rate of uplink and downlink communications.

对于一个较小的$P_F$，active IRS 在上行时应该接近AP，在下行接近User（see Lemma1）。（需要权衡）

而Passive IRS，IRS放置于AP或者User，对于上行/下行都是最佳（因为上行的$x_{AI}=0 orD$和下行的最佳$x_{AI}=Dor0$是相同的）。（直接最佳）

Simulation

参数设置:

N=400; % number of IRS elements
H=1.5; % IRS altitude
fc =0.75e9; % carrier wave frequency
lambda = 0.4; % wave length 
D = 50; % horizontal distance between AP and User
% Other parameters
PA_dBm = 20; % the power of transmitter (DownLink)
PU_dBm = 15; % the power of receiver (UpLink)
sigma_2_dBm = -80; % noise power from IRS to User
sigma_F_2_dBm = -70; % noise power from AP to IRS 
beta_dB = -30; % reference path loss
% 转换为功率比
PA = 10^((PA_dBm-30)/10);
PU = 10^((PU_dBm-30)/10);
sigma_2 = 10^((sigma_2_dBm-30)/10);
sigma_F_2 = 10^((sigma_F_2_dBm-30)/10);
% beta = (lambda/(4*pi))^2;
beta = 10^(beta_dB/10);

Fig. 2

对比Optimal位置优化和Suboptimal位置优化，证明作者提出的Suboptimal是有效的：

首先是Optimal way，对应于（P3）

function out = optimal_way(PA,PU,sigma_2,sigma_F_2,beta, N, H, D)
out = [];
    for PF_dBm = 0:1:10
        PF = 10^((PF_dBm-30)/10);
        x0 = sqrt(max(0, N*PA*beta/(PF-N*sigma_F_2)-H^2));
        max_obj=0;
        max_indx = 0;
        for xa = x0:1:D % one-dimensional search to find the best placement
            C1 = beta*sigma_F_2;
            C2 = PA*beta*sigma_2/PF;
            C3 = sigma_2*sigma_F_2/PF;
            d_AI_2 = xa^2+H^2;
            d_IU_2 = (D-xa)^2+H^2;
            obj = PA*beta^2*N/(C1*d_AI_2+C2*d_IU_2+C3*d_AI_2*d_IU_2);
            if obj>max_obj
                max_obj = obj;
                max_indx = xa;
            end
        end
        out = [out max_indx];
    end
end

然后是Suboptimal way，对应于（P4）:

function out = suboptimal_way(PA,PU,sigma_2,sigma_F_2,beta, N, H, D)
    out = [];
    for PF_dBm = 0:1:10
        PF = 10^((PF_dBm-30)/10);
        x0 = sqrt(max(0, N*PA*beta/(PF-N*sigma_F_2)-H^2));
        out = [out max(sigma_2*PA/(sigma_2*PA+sigma_F_2*PF)*D, x0)];
    end
end

activet_IRS_1_2

Fig. 3

Active IRS的achievable rate需要先算SNR，再由$R=\log_2(1+\text{SNR})$得到。为了方便计算，这里我用Suboptimal优化位置。对应于（eqn.11）

eqn.11

function R_act = ActiveIRS_achievable_rate(PA,PU,sigma_2,sigma_F_2,beta, H, D, PF)
    R_act = [];
    for N=100:10:500
        % 用Suboptimal的位置优化结果求achievable rate
        x0 = sqrt(max(0, N*PA*beta/(PF-N*sigma_F_2)-H^2));
        threshold = sigma_2*PA*D/(sigma_2*PA+sigma_F_2*PF);
        if x0<=threshold
            SNR_act = N*beta/(D^2)*(PA/sigma_F_2+PF/sigma_2);
        else
            disp('2')%发现一个问题，所有的x0都小于threshold
            C1 = beta*sigma_F_2;
            C2 = PA*beta*sigma_2/PF;
            SNR_act = PA*beta^2*N/(C1*x0^2+C2*(D-x0)^2);
        end
        R_act = [R_act log2(1+SNR_act)];
    end
end

Passive IRS同理：

eqn.15

function R_pas = PassiveIRS_achievable_rate(PA,PU,sigma_2,sigma_F_2,beta, H, D)
    R_pas = [];
    for N=100:10:500
        SNR_pas = PA*beta^2*N^2/(H^2*(H^2+D^2)*sigma_2);
        R_pas = [R_pas log2(1+SNR_pas)];
    end
end

activet_IRS_1_3

Fig. 4

计算weighted sum-rate，意味存在两个分配问题：

IRS位置
上行、下行分配的时间（我是这么理解$w^{(UL)}$和$w^{(DL)}$）

对于Active IRS，对应（P6）：

function out = ActiveIRS_Weighted_Sum_rate(PA,PU,sigma_2,sigma_F_2,beta, N, H, D, PF)
    out = [];
    for w_dl = 0:0.02:1
        w_ul = 1-w_dl;
        max_obj = 0;
        for xa=0:1:D
            d_AI_2 = xa^2+H^2;
            d_IU_2 = (D-xa)^2+H^2;
            R_act_UL = log2(1+PU*beta^2*N^2/(beta*N*sigma_F_2*d_IU_2+PU*beta*N*sigma_2/PF*d_AI_2+N*sigma_2*sigma_F_2/PF*d_AI_2*d_IU_2));
            R_act_DL = log2(1+PA*beta^2*N^2/(beta*N*sigma_F_2*d_AI_2+PA*beta*N*sigma_2/PF*d_IU_2+N*sigma_2*sigma_F_2/PF*d_AI_2*d_IU_2));
            obj = w_ul*R_act_UL+w_dl*R_act_DL;
            max_obj = max(max_obj, obj);
        end
        out = [out , max_obj];
    end
end

对于Passive IRS，对应（P7）：

function out = PassiveIRS_Weighted_Sum_rate(PA,PU,sigma_2,sigma_F_2,beta, N, H, D, PF)
    out = [];
    for w_dl = 0:0.02:1
        w_ul = 1-w_dl;
        R_obj = w_ul*log2(1+PU*beta^2*N^2/(H^2*(D^2+H^2)*sigma_2))+w_dl*log2(1+PA*beta^2*N^2/(H^2*(D^2+H^2)*sigma_2));
        out = [out R_obj];
    end
end

activet_IRS_1_4

Fig. 5（结果和论文中有一定差异）

考虑$H$对weighted sum-rate 的影响，Active IRS：

w_dl = 0.5;
w_ul = 1-w_dl;
out = [];
for H = 1.5:0.5:10
    max_obj = 0;
    for xa=0:1:D
        d_AI_2 = xa^2+H^2;
        d_IU_2 = (D-xa)^2+H^2;
        R_act_UL = log2(1+PU*beta^2*N^2/(beta*N*sigma_F_2*d_IU_2+PU*beta*N*sigma_2/PF*d_AI_2+N*sigma_2*sigma_F_2/PF*d_AI_2*d_IU_2));
        R_act_DL = log2(1+PA*beta^2*N^2/(beta*N*sigma_F_2*d_AI_2+PA*beta*N*sigma_2/PF*d_IU_2+N*sigma_2*sigma_F_2/PF*d_AI_2*d_IU_2));
        obj = w_ul*R_act_UL+w_dl*R_act_DL;
        max_obj = max(max_obj, obj);
    end
    out = [out , max_obj];
end

Passive IRS：

out = [];
for H = 1.5:0.5:10
    R_obj = w_ul*log2(1+PU*beta^2*N^2/(H^2*(D^2+H^2)*sigma_2))+w_dl*log2(1+PA*beta^2*N^2/(H^2*(D^2+H^2)*sigma_2));
    out = [out, R_obj];
end

activet_IRS_1_5

Question

complex channel gain到底指什么?

Passive IRS 为什么x=D/x=0?

Paper Reading > Active IRS

#IRS #Active IRS

Active IRS versus Passive IRS

https://lcjoffrey.top/2022/06/13/activeIRS-1/

作者

Joffrey

发布于

2022年6月13日

更新于

2022年6月14日

许可协议

合作IRS-一个IRS拆分为两个上一篇

博士阶段的研究路线-20220524 下一篇