DMA covert communication

DMA隐蔽通信的新思路

拆分DMA的RF链

思路的主要来源:
@article{Ci2021HybridBD,
title={Hybrid Beamforming Design for Covert Multicast mmWave Massive MIMO Communications},
author={Wei Ci and Chenhao Qi and Geoffrey Ye Li and Shiwen Mao},
journal={2021 IEEE Global Communications Conference (GLOBECOM)},
year={2021},
volume={},
pages={}
}

文章的仿真图,注意这篇文章是多用户的,纵轴是最小隐蔽速率, 横轴是最大干扰功率和PcMAX是?
Multiuser Covert Communication 多合法用户隐蔽通信?
思考: DMA可以用于远场吗?DMA用于远场的模型

下一步是首先看看多用户的隐蔽,然后看看能不能把这个文章复现出来.

$$\begin{array}{r}r\left({\mathbf{p}}_i\right)={\mathbf{a}}_i^H\mathbf{s}+n_i,i=b_j\text{ or }w,\forall j=1,\mathrm{2...}K\end{array}$$

The digital input to the DMA is given by $\mathbf{w}={\left[\sqrt{P_c}{\mathbf{w}}_b^T{x}_b,\sqrt{P_J}{\mathbf{w}}_w^T{x}_w\right]}^T\in {\mathbb{C}}^{N_d\times 1}$ , and thus by (1) the channel input is

$$\mathbf{s}=\mathbf{HQ}\mathbf{w}$$

$\mathbf{s}\in N_d{N}_e\times 1$

$$\begin{array}{r}r\left({\mathbf{p}}_i\right)={\mathbf{a}}_i^H\mathbf{HQ}\mathbf{w}+n_i,i=b\text{ or }w\end{array}$$

$$\begin{array}{r}\mathbf{H}=\left[\begin{array}{cc}{\mathbf{H}}_b& 0\\ 0& {\mathbf{H}}_w\end{array}\right]\end{array}$$

$\mathbf{H}\in {\mathbb{C}}^{N_d{N}_e\times N_d{N}_e}$
${\mathbf{H}}_b\in {\mathbb{C}}^{N_{dc}{N}_e\times N_{dc}{N}_e}$
${\mathbf{H}}_w\in {\mathbb{C}}^{N_{dj}{N}_e\times N_{dj}{N}_e}$

$$\begin{array}{r}\mathbf{Q}=\left[\begin{array}{cc}{\mathbf{Q}}_b& 0\\ 0& {\mathbf{Q}}_w\end{array}\right]\end{array}$$

$\mathbf{Q}\in {\mathbb{C}}^{N_d{N}_e\times N_d}$ 不涉及发射功率分量
${\mathbf{Q}}_b\in {\mathbb{C}}^{N_{dc}{N}_e\times N_{dc}}$
${\mathbf{Q}}_w\in {\mathbb{C}}^{N_{dj}{N}_e\times N_{dj}}$

$\mathbf{w}={\left[\sqrt{P_c}{\mathbf{w}}_b^T{x}_b,\sqrt{P_J}{\mathbf{w}}_w^T{x}_w\right]}^T\in {\mathbb{C}}^{N_d\times 1}$

${\mathbf{w}}_b\in {\mathbb{C}}^{N_{dc}\times 1}$
${\mathbf{f}}_b={\mathbf{H}}_b{\mathbf{Q}}_b{\mathbf{w}}_b$
$||{\mathbf{f}}_b||=||{\mathbf{H}}_b{\mathbf{Q}}_b{\mathbf{w}}_b|{|}_2^2=1$

${\mathbf{w}}_w\in {\mathbb{C}}^{N_{dj}\times 1}$
${\mathbf{f}}_w={\mathbf{H}}_w{\mathbf{Q}}_w{\mathbf{w}}_w$
$||{\mathbf{f}}_w||=||{\mathbf{H}}_w{\mathbf{Q}}_w{\mathbf{w}}_w|{|}_2^2=1$

$P_w$ is the power of the jamming signal $e_w$ , which obeys the uniform distribution in
$P_w\sim \left[0,P_{wmax}\right]$

$P_b\le P_{bmax}$
where $b$ and $w$ denote the legitimate user and Willie, respectively, $n_i\sim \mathcal{N}\left(0,{\sigma }_i^2\right)$ denote the noise at the legitimate user or Willie.

有人工噪声

这里不对，需要对信道矩阵也要分割

${\mathcal{H}}_0:r\left({\mathbf{p}}_w\right)=\sqrt{P_w}{\mathbf{a}}_w^H{\mathbf{H}}_w{\mathbf{Q}}_w{\mathbf{w}}_w+n_w$

${\mathcal{H}}_1:r\left({\mathbf{p}}_w\right)=\sqrt{P_b}{\mathbf{a}}_w^H\mathbf{HQ}{\mathbf{w}}_b{e}_b+\sqrt{P_w}{\mathbf{a}}_w^H\mathbf{HQ}{\mathbf{w}}_w{e}_w+n_w$

Willie有足够的时间来积累能量

H 0 : T 0 = ∣ r ( p w ) ∣ 2 = P w ∣ a w H H Q w w e w ∣ 2 + σ w 2 \mathcal{H}_{0}:T_0 = | r\left(\mathbf{p}_{w}\right) |^2 = P_w \left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right|^2 + \sigma_{w}^2 H0​:T0​=∣r(pw​)∣2=Pw​ ​awH​HQww​ew​ ​2+σw2​

H 1 : T 1 = ∣ r ( p w ) ∣ 2 = P b ∣ a w H H Q w b e b ∣ 2 + P w ∣ a w H H Q w w e w ∣ 2 + σ w 2 \mathcal{H}_{1}: T_1 = |r\left(\mathbf{p}_{w}\right)|^2 =P_b\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2 + P_w\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 + \sigma_{w}^2 H1​:T1​=∣r(pw​)∣2=Pb​ ​awH​HQwb​eb​ ​2+Pw​ ​awH​HQww​ew​ ​2+σw2​

Optimal detection threshold: $\tau$

Willie makes a binary decision on whether the received signal is noise or signal plus noise.

The test statistic is given by : $T\left({\mathbf{y}}_w\right)=|r\left({\mathbf{p}}_w\right){|}^2$

Denote $D_0$ and $D_1$ as the decisions that the received signal is noise and that the received signal is signal plus noise, respectively.

$P_{\mathrm{FA}}=\mathbb{P}\left(D_1\mid {\mathcal{H}}_0\right)=Pr(T_0\ge \tau )$
$P_{\mathrm{MD}}=\mathbb{P}\left(D_0\mid {\mathcal{H}}_1\right)=Pr(T_1<\tau )$

P F A = Pr ⁡ ( T 0 ≥ τ ) = { 1 , τ < σ w 2 1 − τ − σ w 2 P w max ⁡ ∣ a w H H Q w w e w ∣ 2 , σ w 2 ≤ τ < σ w 2 + P w max ⁡ ∣ a w H H Q w w e w ∣ 2 0 , τ ≥ σ w 2 + P w max ⁡ ∣ a w H H Q w w e w ∣ 2 \mathcal{P}_{\mathrm{FA}}=\operatorname{Pr}\left(T_{0} \geq \tau\right)=\left\{\begin{array}{ll} 1, & \tau<\sigma_{\mathrm{w}}^{2} \\ 1-\frac{\tau-\sigma_{\mathrm{w}}^{2}}{P_{\mathrm{w} \max } \left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 }, & \sigma_{\mathrm{w}}^{2} \leq \tau<\sigma_{\mathrm{w}}^{2}+P_{\mathrm{w} \max } \left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 \\ 0, & \tau \geq \sigma_{\mathrm{w}}^{2}+P_{\mathrm{w} \max } \left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 \end{array}\right. PFA​=Pr(T0​≥τ)=⎩ ⎨ ⎧​1,1−Pwmax​∣awH​HQww​ew​∣2τ−σw2​​,0,​τ<σw2​σw2​≤τ<σw2​+Pwmax​ ​awH​HQww​ew​ ​2τ≥σw2​+Pwmax​ ​awH​HQww​ew​ ​2​

P M D = Pr ⁡ ( T 1 < τ ) = { 0 , τ < σ w 2 + P b ∣ a w H H Q w b e b ∣ 2 τ − σ w 2 − P b ∣ a w H H Q w b e b ∣ 2 P w max ⁡ ∣ a w H H Q w w e w ∣ 2 , σ w 2 + P b ∣ a w H H Q w b e b ∣ 2 ≤ τ < σ w 2 + P b ∣ a w H H Q w b e b ∣ 2 + P w max ⁡ ∣ a w H H Q w w e w ∣ 2 1 , τ ≥ σ w 2 + P b ∣ a w H H Q w b e b ∣ 2 + P w max ⁡ ∣ a w H H Q w w e w ∣ 2 \mathcal{P}_{\mathrm{MD}}=\operatorname{Pr}\left(T_{1}<\tau\right)=\left\{\begin{array}{ll} 0, & \tau<\sigma_{\mathrm{w}}^{2}+P_{\mathrm{b}} \left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2 \\ \frac{\tau-\sigma_{\mathrm{w}}^{2}-P_{\mathrm{b}} \left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2 }{P_{\mathrm{w} \max } \left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right|^2 }, & \sigma_{\mathrm{w}}^{2}+ P_{\mathrm{b}} \left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right|^2 \leq \tau<\sigma_{\mathrm{w}}^{2} + P_{\mathrm{b}} \left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right|^2 +P_{\mathrm{w} \max } \left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right|^2 \\ 1, & \tau \geq \sigma_{\mathrm{w}}^{2} + P_{\mathrm{b}} \left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right|^2 + P_{\mathrm{w} \max } \left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right|^2 \end{array}\right. PMD​=Pr(T1​<τ)=⎩ ⎨ ⎧​0,Pwmax​∣awH​HQww​ew​∣2τ−σw2​−Pb​∣awH​HQwb​eb​∣2​,1,​τ<σw2​+Pb​ ​awH​HQwb​eb​ ​2σw2​+Pb​ ​awH​HQwb​eb​ ​2≤τ<σw2​+Pb​ ​awH​HQwb​eb​ ​2+Pwmax​ ​awH​HQww​ew​ ​2τ≥σw2​+Pb​ ​awH​HQwb​eb​ ​2+Pwmax​ ​awH​HQww​ew​ ​2​

In the worst case, Willie has a perfect knowledge of the channel from Alice to Willie:

$${\mathcal{P}}_{\mathrm{e}}\triangleq {\mathcal{P}}_{\mathrm{FA}}+{\mathcal{P}}_{\mathrm{MD}}$$

if ${\mathcal{P}}_{\mathrm{e}}=0$, due to the expression of ${\mathcal{P}}_{\mathrm{FA}}=0$ and ${\mathcal{P}}_{\mathrm{MD}}=0$ :

P w max ⁡ ∣ a w H H Q w w e w ∣ 2 ≤ P b ∣ a w H H Q w b e b ∣ 2 P_{w\max}\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 \le P_b\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2 Pwmax​ ​awH​HQww​ew​ ​2≤Pb​ ​awH​HQwb​eb​ ​2

So we can’t make P w max ⁡ ∣ a w H H Q w w e w ∣ 2 ≤ P b ∣ a w H H Q w b e b ∣ 2 P_{w\max}\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 \le P_b\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2 Pwmax​ ​awH​HQww​ew​ ​2≤Pb​ ​awH​HQwb​eb​ ​2 hold

To ensure ${\mathcal{P}}_{\mathrm{e}}\ne 0$: We should make:

P w max ⁡ ∣ a w H H Q w w e w ∣ 2 > P b ∣ a w H H Q w b e b ∣ 2 P_{w\max}\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 > P_b\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2 Pwmax​ ​awH​HQww​ew​ ​2>Pb​ ​awH​HQwb​eb​ ​2

P e = { 1 − τ − σ w 2 P w max ⁡ ∣ a w H H Q w w e w ∣ 2 , σ w 2 ≤ τ < σ w 2 + P b ∣ a w H H Q w b e b ∣ 2 1 − P b ∣ a w H H Q w b e b ∣ 2 P w max ⁡ ∣ a w H H Q w w e w ∣ 2 , σ w 2 + P b ∣ a w H H Q w b e b ∣ 2 ≤ τ < σ w 2 + P w max ⁡ ∣ a w H H Q w w e w ∣ 2 τ − σ w 2 − P b ∣ a w H H Q w b e b ∣ 2 P w max ⁡ ∣ a w H H Q w w e w ∣ 2 , σ w 2 + P w max ⁡ ∣ a w H H Q w w e w ∣ 2 ≤ τ < σ w 2 + P b ∣ a w H H Q w b e b ∣ 2 + P w max ⁡ ∣ a w H H Q w w e w ∣ 2 1 ,  else  (21) \mathcal{P}_{\mathrm{e}}=\left\{\begin{array}{ll} 1-\frac{\tau-\sigma_{\mathrm{w}}^{2}} { P_{w\max}\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 }, & \sigma_{\mathrm{w}}^{2} \leq \tau<\sigma_{\mathrm{w}}^{2}+P_b\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2 \\ 1-\frac{ P_b\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2 }{ P_{w\max}\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 }, & \sigma_{\mathrm{w}}^{2}+ P_b\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2 \leq \tau<\sigma_{\mathrm{w}}^{2}+P_{w\max}\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2\\ \frac{\tau-\sigma_{\mathrm{w}}^{2}-P_b\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2}{ P_{w\max}\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 }, & \sigma_{\mathrm{w}}^{2}+ P_{w\max}\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 \leq \tau<\sigma_{\mathrm{w}}^{2}+P_b\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2+ P_{w\max}\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 \\ 1, & \text { else } \end{array}\right. \tag{21} Pe​=⎩ ⎨ ⎧​1−Pwmax​∣awH​HQww​ew​∣2τ−σw2​​,1−Pwmax​∣awH​HQww​ew​∣2Pb​∣awH​HQwb​eb​∣2​,Pwmax​∣awH​HQww​ew​∣2τ−σw2​−Pb​∣awH​HQwb​eb​∣2​,1,​σw2​≤τ<σw2​+Pb​ ​awH​HQwb​eb​ ​2σw2​+Pb​ ​awH​HQwb​eb​ ​2≤τ<σw2​+Pwmax​ ​awH​HQww​ew​ ​2σw2​+Pwmax​ ​awH​HQww​ew​ ​2≤τ<σw2​+Pb​ ​awH​HQwb​eb​ ​2+Pwmax​ ​awH​HQww​ew​ ​2 else ​(21)

对${\mathcal{P}}_{\mathrm{e}}$求导，然后可以发现：

Optimal $\tau$ achieved by (21b)

P ^ e = 1 − P b ∣ a w H H Q w b e b ∣ 2 P w max ⁡ ∣ a w H H Q w w e w ∣ 2 , \widehat{\mathcal{P}}_{\mathrm{e}}=1 - \frac{ P_b\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2 }{ P_{w\max}\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 }, P e​=1−Pwmax​∣awH​HQww​ew​∣2Pb​ ​awH​HQwb​eb​ ​2​,

SINR at Bob
γ b ≜ P b ∣ a b H H Q w b e b ∣ 2 P w ∣ a b H H Q w w e w ∣ 2 + σ b 2 (23) \gamma_{b} \triangleq \frac{ P_b\left| \mathbf{a}_{b}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2 }{ P_w\left| \mathbf{a}_{b}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 + \sigma_{b}^{2} } \tag{23} γb​≜Pw​ ​abH​HQww​ew​ ​2+σb2​Pb​ ​abH​HQwb​eb​ ​2​(23)

We need to maximize ${\gamma }_b$

The covert rate :

$$R_b={\log }_2\left(1+{\gamma }_b\right)$$

maximize the covert rate:
Optimization problem about ${\mathbf{w}}_w$,${\mathbf{w}}_b$,$\mathbf{Q}$, $P_b$

$$\begin{array}{rl}\underset{{\mathbf{w}}_w,{\mathbf{w}}_b,\mathbf{Q},P_b}{\max }& R_b,\\ \text{ s.t. }& {\mathcal{P}}_{\mathrm{e}}\ge 1-ϵ,\\ & P_b\le P_{bmax},\\ & ||{\mathbf{w}}_b|{|}_2^2=1,\\ & ||{\mathbf{w}}_w|{|}_2^2=1,\\ & \mathbf{Q}的约束\end{array}\text{\hspace{1em}(24)}$$

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(Constraint 1 is equal to)

$$P_{\mathrm{b}}\le g\left({\mathbf{w}}_{\mathrm{b}},{\mathbf{w}}_{\mathrm{w}},\mathbf{Q}\right)\text{\hspace{1em}(25)}$$

$g\left({\mathbf{w}}_{\mathrm{b}},{\mathbf{w}}_{\mathrm{w}},\mathbf{Q}\right)$ is expressed as:

g ( w b , w w , Q ) ≜ ϵ P w max ⁡ ∣ a w H H Q w w e w ∣ 2 ∣ a w H H Q w b e b ∣ 2 (26) g\left( \mathbf{w}_{\mathrm{b}}, \mathbf{w}_{\mathrm{w}}, \mathbf{Q} \right) \triangleq \frac{\epsilon P_{w\max}\left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 }{ \left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2 } \tag{26} g(wb​,ww​,Q)≜∣awH​HQwb​eb​∣2ϵPwmax​ ​awH​HQww​ew​ ​2​(26)

Note that (23) monotonically increases with $P_b$.

The optimal $P_b$:

$$P_{\mathrm{b}}=\min \left\{P_{\mathrm{bmax}},g\left({\mathbf{w}}_{\mathrm{b}},{\mathbf{w}}_{\mathrm{w}},\mathbf{Q}\right)\right\}.\text{\hspace{1em}(27)}$$

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(25) (26) (27) 联合表述了第一个约束和第二个约束

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针对$P_b$问题的优化：

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$$\underset{{\mathbf{w}}_{\mathrm{b}},{\mathbf{w}}_{\mathrm{w}},\mathbf{Q}}{\max }g\left({\mathbf{w}}_{\mathrm{b}},{\mathbf{w}}_{\mathrm{w}},\mathbf{Q}\right).\text{\hspace{1em}(28)}$$

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主目标函数，最大化Bob处的信噪比

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Due to (23), we need to maximize ∣ a b H H Q w b e b ∣ 2 \left| \mathbf{a}_{b}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2 ​abH​HQwb​eb​ ​2 and minimize ∣ a b H H Q w w e w ∣ 2 \left| \mathbf{a}_{b}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 ​abH​HQww​ew​ ​2

Hopefully there are feasible solutions for

$${\mathbf{a}}_b^H\mathbf{HQ}{\mathbf{w}}_w{e}_w=0\text{\hspace{1em}(29)}$$

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Then the optimization of ${\mathbf{w}}_w$ and ${\mathbf{w}}_b$ is decoupled. (24) is converted into the following two independent optimization problems as

max ⁡ w w ∣ a w H H Q w w e w ∣ 2  s.t.  ∥ w w ∥ 2 2 = 1 , a b H H Q w w e w = 0 , (30) \begin{aligned} \max _{\mathbf{w}_{\mathrm{w}}} & \left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} \right| ^2 \\ \text { s.t. } & \left\|\mathbf{w}_{\mathrm{w}}\right\|_{2}^{2}=1,\\ & \mathbf{a}_{b}^{H} \mathbf{H Q} \mathbf{w}_{w} e_{w} = 0, \end{aligned} \tag{30} ww​max​ s.t. ​ ​awH​HQww​ew​ ​2∥ww​∥22​=1,abH​HQww​ew​=0,​(30)

max ⁡ w b ∣ a b H H Q w b e b ∣ 2 σ b 2  s.t.  ∣ a w H H Q w b e b ∣ 2 = 0 , ∥ w b ∥ 2 2 = 1 (31) \begin{aligned} \max _{\boldsymbol{w}_{\mathrm{b}}} & \frac{ \left| \mathbf{a}_{b}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2 }{\sigma_{b}^{2}} \\ \text { s.t. } & \left| \mathbf{a}_{w}^{H} \mathbf{H Q} \mathbf{w}_{b} e_{b} \right| ^2 = 0, \\ & \left\|\mathbf{w}_{\mathrm{b}}\right\|_{2}^{2}=1 \end{aligned} \tag{31} wb​max​ s.t. ​σb2​ ​abH​HQwb​eb​ ​2​ ​awH​HQwb​eb​ ​2=0,∥wb​∥22​=1​(31)

where (30a) and (31b) are based on (28) and the transformation from (24a) to (31a) is based on (29).

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