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SCIENCE CHINA Information Sciences, Volume 64 , Issue 10 : 200302(2021) https://doi.org/10.1007/s11432-021-3281-7

Low-cost intelligent reflecting surface aided Terahertz multiuser massive MIMO: design and analysis

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  • ReceivedMar 4, 2021
  • AcceptedJun 4, 2021
  • PublishedSep 7, 2021

Abstract


Acknowledgment

This work was supported by National Key RD Program of China (Grant No. 2020YFB1805703), National Natural Science Foundation of China (Grant No. 61871344), and Zhejiang Provincial Natural Science Foundation of China (Grant No. LR20F010002).


Supplement

Appendix

Lemmas 1–3

łemma Consider the vector ${\boldsymbol~x}$, which can be expressed as \begin{equation}{\boldsymbol x} = \boldsymbol{A}_x {\boldsymbol v} \tag{67}\end{equation} with $\boldsymbol{A}_x~\in~\mathcal{C}^{M~\times~N}~$ and ${\boldsymbol~v}~\sim~\mathcal{N}_c(\mathbf{0},~\sigma_x^2~{\boldsymbol~I}_N~)~$. It holds true that \begin{equation}\mathrm{E} [ {\boldsymbol x} {\boldsymbol x}^{\rm H} ] = \sigma_x^2 \boldsymbol{A}_x \boldsymbol{A}_x^{\rm H}, \tag{68}\end{equation} and \begin{equation}\mathrm{E} [ {\boldsymbol x}^{\rm H} {\boldsymbol B} {\boldsymbol x} ] = \sigma_x^2 \mathrm{tr} ( {\boldsymbol B} \boldsymbol{A}_x \boldsymbol{A}_x^{\rm H} ) \tag{69}\end{equation} with ${\boldsymbol~B}~\in~\mathcal{C}^{M~\times~M}$.

Proof. Please refer to Appendix B.

łemma Consider the matrix ${\boldsymbol~Y}$, which can be expressed as \begin{equation}{\boldsymbol Y} = \boldsymbol{A}_1 {\boldsymbol U} \boldsymbol{A}_2, \tag{70}\end{equation} where $\boldsymbol{A}_1~\in~\mathcal{C}^{M~\times~L}$, $\boldsymbol{A}_2~\in~\mathcal{C}^{L~\times~N}$, and ${\boldsymbol~U}~=~\mathrm{diag}~(~\alpha_1,\ldots,\alpha_L~)$, $\alpha_i~\sim~\mathcal{N}_c(0,\sigma_y^2)$. It holds true that \begin{equation}\mathrm{E} [ {\boldsymbol Y} {\boldsymbol B}_1 {\boldsymbol Y}^{\rm H} ] = \sigma_y^2 \boldsymbol{A}_1 \mathrm{diag}( \boldsymbol{A}_2 {\boldsymbol B}_1 \boldsymbol{A}_2^{\rm H} ) \boldsymbol{A}_1^{\rm H}, \tag{71}\end{equation} and \begin{equation}\mathrm{E} [ {\boldsymbol Y}^{\rm H} {\boldsymbol B}_2 {\boldsymbol Y} ] = \sigma_y^2 \boldsymbol{A}_2^{\rm H} \mathrm{diag} (\boldsymbol{A}_1^{\rm H} {\boldsymbol B}_2 \boldsymbol{A}_1 ) \boldsymbol{A}_2 \tag{72}\end{equation} with ${\boldsymbol~B}_1~\in~\mathcal{C}^{N~\times~N}$ and ${\boldsymbol~B}_2~\in~\mathcal{C}^{M~\times~M}$.

Proof. Please refer to Appendix B.

łemma Consider the $N$-dimensional random vector ${\boldsymbol~h}$ with $\mathrm{E}[~{\boldsymbol~h}~{\boldsymbol~h}^{\rm~H}~]~=~{\boldsymbol~R}$ and ${\boldsymbol~R}~\in~\mathcal{C}^{N~\times~N}$. It holds true that \begin{equation}\mathrm{E} [ {\boldsymbol h}^{\rm H} {\boldsymbol B} {\boldsymbol h} ] = \mathrm{tr}({\boldsymbol B} {\boldsymbol R}) \tag{73}\end{equation} with ${\boldsymbol~B}~\in~\mathcal{C}^{N~\times~N}$.

Proof. Please refer to Appendix B.

The proof of Lemmas 1–3

The proof of Lemma 1

For the term $\mathrm{E}[{\boldsymbol~x}~{\boldsymbol~x}^{\rm~H}]$, we have \begin{equation}\mathrm{E} [ {\boldsymbol x} {\boldsymbol x}^{\rm H} ] = \mathrm{E} [ \boldsymbol{A}_x {\boldsymbol v} {\boldsymbol v}^{\rm H} \boldsymbol{A}_x^{\rm H} ] = \boldsymbol{A}_x \mathrm{E} [ {\boldsymbol v} {\boldsymbol v}^{\rm H} ] \boldsymbol{A}_x^{\rm H} = \sigma_x^2 \boldsymbol{A}_x \boldsymbol{A}_x^{\rm H}. \tag{74}\end{equation} As for the term $\mathrm{E}[~{\boldsymbol~x}^{\rm~H}~{\boldsymbol~B}~{\boldsymbol~x}~]$, we have \begin{equation}\mathrm{E} [ {\boldsymbol x}^{\rm H} {\boldsymbol B} {\boldsymbol x} ] = \mathrm{E} [ {\boldsymbol v}^{\rm H} \boldsymbol{A}_x^{\rm H} {\boldsymbol B} \boldsymbol{A}_x {\boldsymbol v} ]. \tag{75}\end{equation} Without loss of generality, we assume that \begin{equation}{\boldsymbol v} = [v_1,\ldots,v_n,\ldots,v_N ], \tag{76}\end{equation} \begin{equation}\boldsymbol{A}_x^{\rm H} {\boldsymbol B} \boldsymbol{A}_x = \left[ \begin{array}{ccccc} w_{1,1} & \cdots & w_{1,n} & \cdots & w_{1,N} \\ \vdots & & \vdots & & \vdots \\ w_{n,1} & \cdots & w_{n,n} & \cdots & w_{n,N} \\ \vdots & & \vdots & & \vdots \\ w_{N,1} & \cdots & w_{N,n} & \cdots & w_{N,N} \end{array} \right], \tag{77}\end{equation} Then, we can obtain \begin{equation}\mathrm{E}[ {\boldsymbol v}^{\rm H} \boldsymbol{A}_x^{\rm H} {\boldsymbol B} \boldsymbol{A}_x {\boldsymbol v} ] = \mathrm{E} [ w_{1,1}'v_1+\cdots+w_{1,n}'v_n+\cdots+w_{1,N}'v_N ] \tag{78}\end{equation} with \begin{equation}w_{1,n}'=v_1^{\rm H} w_{1,n} + \cdots + v_n^{\rm H} w_{n,n} + \cdots + v_N^{\rm H} w_{N,n}. \tag{79}\end{equation} It holds true that $\mathrm{E}[v_i^{\rm~H}~w~v_j]~=~0,\forall~i~\neq~j$ because the elements in ${\boldsymbol~v}$ are independent of each other. Thus, we have \begin{equation}\mathrm{E} [ {\boldsymbol x}^{\rm H} {\boldsymbol B} {\boldsymbol x} ] = \mathrm{E}[ {\boldsymbol v}^{\rm H} \boldsymbol{A}_x^{\rm H} {\boldsymbol B} \boldsymbol{A}_x {\boldsymbol v} ] = \sigma_x^2 ( w_{1,1} + \cdots + w_{n,n} + \cdots + w_{N,N} ) = \sigma_x^2 \mathrm{tr} (\boldsymbol{A}_x^{\rm H} {\boldsymbol B} \boldsymbol{A}_x ). \tag{80}\end{equation} The proof of Lemma 1 finished.

The proof of Lemma 2

For the first term $\mathrm{E}~[~{\boldsymbol~Y}~{\boldsymbol~B}_1~{\boldsymbol~Y}^{\rm~H}~]$, we have \begin{equation}\mathrm{E} [ {\boldsymbol Y} {\boldsymbol B}_1 {\boldsymbol Y}^{\rm H} ] = \mathrm{E} [ \boldsymbol{A}_1 {\boldsymbol U} \boldsymbol{A}_2 {\boldsymbol B}_1 \boldsymbol{A}_2^{\rm H} {\boldsymbol U}^{\rm H} \boldsymbol{A}_1^{\rm H} ]. \tag{81}\end{equation} Without loss of generality, we assume that \begin{equation}\boldsymbol{A}_2 {\boldsymbol B}_1 \boldsymbol{A}_2^{\rm H} = \left[ \begin{array}{ccccc} p_{1,1} & \cdots & p_{1,n} & \cdots & p_{1,L} \\ \vdots & & \vdots & & \vdots \\ p_{n,1} & \cdots & p_{n,n} & \cdots & p_{n,L} \\ \vdots & & \vdots & & \vdots \\ p_{L,1} & \cdots & p_{L,n} & \cdots & p_{L,L} \end{array} \right]. \tag{82}\end{equation} Then, ${\boldsymbol~U}~\boldsymbol{A}_2~{\boldsymbol~B}_1~\boldsymbol{A}_2^{\rm~H}~{\boldsymbol~U}^{\rm~H}$ can be computed as \begin{equation}{\boldsymbol U} \boldsymbol{A}_2 {\boldsymbol B}_1 \boldsymbol{A}_2^{\rm H} {\boldsymbol U}^{\rm H} = \left[ \begin{array}{ccccc} \alpha_1 p_{1,1} \alpha_1^{\rm H} & \cdots & \alpha_1 p_{1,n} \alpha_n^{\rm H} & \cdots & \alpha_1 p_{1,L} \alpha_L^{\rm H} \\ \vdots & & \vdots & & \vdots \\ \alpha_n p_{n,1} \alpha_1^{\rm H} & \cdots & \alpha_n p_{n,n} \alpha_n^{\rm H} & \cdots & \alpha_n p_{n,L} \alpha_L^{\rm H} \\ \vdots & & \vdots & & \vdots \\ \alpha_L p_{L,1} \alpha_1^{\rm H} & \cdots & \alpha_L p_{L,n} \alpha_n^{\rm H} & \cdots & \alpha_L p_{L,L} \alpha_L^{\rm H} \end{array} \right]. \tag{83}\end{equation} $\mathrm{E}~[~p~\alpha_i~\alpha_j~]~=~0$ $(\forall~i\neq~j)$ holds true due to the fact that the elements $\alpha_1,~\ldots,~\alpha_L$ are independent of each other. Thus, we have \begin{eqnarray}\mathrm{E}[ {\boldsymbol Y} {\boldsymbol B}_1 {\boldsymbol Y}^{\rm H} ] &=& \boldsymbol{A}_1 \mathrm{E} [{\boldsymbol U} \boldsymbol{A}_2 {\boldsymbol B}_1 \boldsymbol{A}_2^{\rm H} {\boldsymbol U}^{\rm H} ] \boldsymbol{A}_1^{\rm H} \tag{84} \\ &=& \sigma_y^2 \boldsymbol{A}_1 \left[ \begin{array}{ccccc} p_{1,1} & 0 & \cdots & \cdots &0 \\ 0 & \ddots & & & \vdots \\ \vdots & & p_{n,n} & & \vdots \\ \vdots & & & \ddots & 0 \\ 0 & \cdots & \cdots & 0 & p_{L,L} \end{array} \right] \boldsymbol{A}_1^{\rm H} \\ &=& \sigma_y^2 \boldsymbol{A}_1 \mathrm{diag}[\boldsymbol{A}_2 {\boldsymbol B}_1 \boldsymbol{A}_2^{\rm H}] \boldsymbol{A}_1^{\rm H}. \tag{85} \end{eqnarray} Similarly, we can obtain \begin{equation}\mathrm{E} [ {\boldsymbol Y}^{\rm H} {\boldsymbol B}_2 {\boldsymbol Y} ] = \sigma_y^2 \boldsymbol{A}_2^{\rm H} \mathrm{diag} (\boldsymbol{A}_1^{\rm H} {\boldsymbol B}_2 \boldsymbol{A}_1 ) \boldsymbol{A}_2. \tag{86}\end{equation} The proof of Lemma 2 finished.

The proof of Lemma 3

Without loss of generality, we assume that \begin{equation}{\boldsymbol h}=[h_1,\ldots,h_n,\ldots,h_N], \tag{87}\end{equation} \begin{equation}{\boldsymbol R} = \mathrm{E}[ {\boldsymbol h} {\boldsymbol h}^{\rm H} ] = \left[ \begin{array}{ccccc} r_{1,1} & \cdots & r_{1,n} & \cdots & r_{1,N} \\ \vdots & & \vdots & & \vdots \\ r_{n,1} & \cdots & r_{n,n} & \cdots & r_{n,N} \\ \vdots & & \vdots & & \vdots \\ r_{N,1} & \cdots & r_{N,n} & \cdots & r_{N,N} \end{array} \right], \tag{88}\end{equation} and \begin{equation}{\boldsymbol B} = \left[ \begin{array}{ccccc} b_{1,1} & \cdots & b_{1,n} & \cdots & b_{1,N} \\ \vdots & & \vdots & & \vdots \\ b_{n,1} & \cdots & b_{n,n} & \cdots & b_{n,N} \\ \vdots & & \vdots & & \vdots \\ b_{N,1} & \cdots & b_{N,n} & \cdots & b_{N,N} \end{array} \right]. \tag{89}\end{equation} Thus, the term $\mathrm{E}[{\boldsymbol~h}^{\rm~H}~{\boldsymbol~B}~{\boldsymbol~h}]$ can be computed as \begin{eqnarray}\mathrm{E}[{\boldsymbol h}^{\rm H} {\boldsymbol B} {\boldsymbol h}] &=& \mathrm{E} [(h_1^{\rm H} b_{1,1} + \cdots + h_N^{\rm H} b_{N,1})h_1 + \cdots + ( h_1^{\rm H} b_{1,N} + \cdots + h_N^ H b_{N,N} )h_N ] \\ &=& \sum_{i=1}^N \sum_{j=1}^N r_{i,j} b_{i,j}. \tag{90} \end{eqnarray} Then, the term $\mathrm{tr}({\boldsymbol~B}~{\boldsymbol~R})$ can be expressed as \begin{eqnarray}\mathrm{tr} ( {\boldsymbol B} {\boldsymbol R} ) &=& (b_{1,1}r_{1,1}+\cdots+b_{1,N}r_{N,1})+\cdots+( b_{N,1}r_{1,N} + \cdots + b_{N,N} r_{N,N}) \\ &=& \sum_{i=1}^N \sum_{j=1}^N r_{i,j}b_{i,j}. \tag{91} \end{eqnarray} Hence, we can obtain \begin{equation}\mathrm{E} [ {\boldsymbol h}^{\rm H} {\boldsymbol B} {\boldsymbol h} ] = \mathrm{tr} ( {\boldsymbol B} {\boldsymbol R} ) \tag{92}\end{equation} for arbitrarily distributed random vector ${\boldsymbol~h}$. The proof of Lemma 3 finished.

The computation of autocorrelation matrix $R_i^e$

First of all, we begin with calculating the autocorrelation matrix of ${\boldsymbol~h}_k$: \begin{align}{\boldsymbol R}_k =&\, \mathrm{E} [ {\boldsymbol h}_k {\boldsymbol h}_k^{\rm H} ] \\ =&\, \mathrm{E} [ {\boldsymbol H} \mathbf{\Theta} {\boldsymbol h}_{r,k} ({\boldsymbol H} \mathbf{\Theta} {\boldsymbol h}_{r,k})^{\rm H} ] \\ =&\, \mathrm{E} [ {\boldsymbol H} (\bar{\mathbf{\Theta}} + \tilde{\mathbf{\Theta}} ) ( \bar{{\boldsymbol h}}_{r,k} + \tilde{{\boldsymbol h}}_{r,k} ) ( \bar{{\boldsymbol h}}_{r,k} + \tilde{{\boldsymbol h}}_{r,k} )^{\rm H} (\bar{\mathbf{\Theta}}^{\rm H} + \tilde{\mathbf{\Theta}}^{\rm H}) {\boldsymbol H}^{\rm H} ] \\ =&\, \mathrm{E} [ {\boldsymbol H} (\bar{\mathbf{\Theta}} \bar{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H} \bar{\mathbf{\Theta}}^{\rm H} \!+\! \bar{\mathbf{\Theta}} \tilde{{\boldsymbol h}}_{r,k} \tilde{{\boldsymbol h}}_{r,k}^{\rm H} \bar{\mathbf{\Theta}}^{\rm H} ) {\boldsymbol H}^{\rm H} ] \!+\! \mathrm{E} [ {\boldsymbol H} (\bar{\mathbf{\Theta}} \bar{{\boldsymbol h}}_{r,k} \tilde{{\boldsymbol h}}_{r,k}^{\rm H} \tilde{\mathbf{\Theta}}^{\rm H} \!+\! \bar{\mathbf{\Theta}} \tilde{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H} \tilde{\mathbf{\Theta}}^{\rm H} ) {\boldsymbol H}^{\rm H} ] {} \\ &+ \mathrm{E} [ {\boldsymbol H} (\tilde{\mathbf{\Theta}} \bar{{\boldsymbol h}}_{r,k} \tilde{{\boldsymbol h}}_{r,k}^{\rm H} \bar{\mathbf{\Theta}}^{\rm H} \!+\! \tilde{\mathbf{\Theta}} \tilde{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H} \bar{\mathbf{\Theta}}^{\rm H} ) {\boldsymbol H}^{\rm H} ] \!+\! \mathrm{E} [ {\boldsymbol H} (\tilde{\mathbf{\Theta}} \bar{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H} \tilde{\mathbf{\Theta}}^{\rm H} \!+\! \tilde{\mathbf{\Theta}} \tilde{{\boldsymbol h}}_{r,k} \tilde{{\boldsymbol h}}_{r,k}^{\rm H} \tilde{\mathbf{\Theta}}^{\rm H} ) {\boldsymbol H}^{\rm H} ] \tag{93} \\ =&\, \mathrm{E} [ {\boldsymbol H} \bar{\mathbf{\Theta}} {\boldsymbol R}_{r,k} \bar{\mathbf{\Theta}}^{\rm H} {\boldsymbol H}^{\rm H} ] + 2\mathrm{Re}( \mathrm{E} [ {\boldsymbol H} \bar{\mathbf{\Theta}} \bar{{\boldsymbol h}}_{r,k} \tilde{{\boldsymbol h}}_{r,k}^{\rm H} \tilde{\mathbf{\Theta}}^{\rm H} {\boldsymbol H}^{\rm H} ] {} \\ &+ \mathrm{E} [ {\boldsymbol H} \bar{\mathbf{\Theta}} \tilde{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H} \tilde{\mathbf{\Theta}}^{\rm H} {\boldsymbol H}^{\rm H} ] ) + \mathrm{E} [ {\boldsymbol H} \tilde{\mathbf{\Theta}} {\boldsymbol R}_{r,k} \tilde{\mathbf{\Theta}}^{\rm H} {\boldsymbol H}^{\rm H} ]. \tag{94} \end{align} Eq. (93) holds true because that ${\boldsymbol~H}$ is independent of $\mathbf{\Theta}$ and ${\boldsymbol~h}_{r,k}$ and $\mathrm{E}[\tilde{\mathbf{\Theta}}]~=~\mathrm{E}[\tilde{{\boldsymbol~h}_{r,k}}]~=~0~$. According to Lemmas 1–3 and by applying the property of channel hardening for massive MIMO systems, we can obtain (94) from (93). Next, we can compute the first term of (94) as follows: \begin{eqnarray}\mathrm{E}[ {\boldsymbol H} \bar{\mathbf{\Theta}} {\boldsymbol R}_{r,k} \bar{\mathbf{\Theta}}^{\rm H} {\boldsymbol H}^{\rm H} ] &=& \mathrm{E} [ \bar{{\boldsymbol H}} \bar{\mathbf{\Theta}} {\boldsymbol R}_{r,k} \bar{\mathbf{\Theta}}^{\rm H} \bar{{\boldsymbol H}}^{\rm H} ] + \mathrm{E} [ \tilde{{\boldsymbol H}} \bar{\mathbf{\Theta}} {\boldsymbol R}_{r,k} \bar{\mathbf{\Theta}}^{\rm H} \tilde{{\boldsymbol H}}^{\rm H} ] \tag{95} \\ &=& \bar{{\boldsymbol H}} \bar{\mathbf{\Theta}} {\boldsymbol R}_{r,k} \bar{\mathbf{\Theta}}^{\rm H} \bar{{\boldsymbol H}}^{\rm H} + \sigma_{\alpha}^2 \boldsymbol{A}_r \mathrm{diag}( \boldsymbol{A}_t^{\rm H} \bar{\mathbf{\Theta}} {\boldsymbol R}_{r,k} \bar{\mathbf{\Theta}}^{\rm H} \boldsymbol{A}_t ) \boldsymbol{A}_r^{\rm H}. \tag{96} \end{eqnarray} Eq. (95) holds true due to that $\mathrm{E}~[~\tilde{{\boldsymbol~H}}~]~=~0~$ and Eq. (96) holds true based on Lemma 2. Similarly, the second term of (94) can be computed as follows: \begin{eqnarray}& &2\mathrm{Re}( \mathrm{E} [ {\boldsymbol H} \bar{\mathbf{\Theta}} \bar{{\boldsymbol h}}_{r,k} \tilde{{\boldsymbol h}}_{r,k}^{\rm H} \tilde{\mathbf{\Theta}}^{\rm H} {\boldsymbol H}^{\rm H} ] + \mathrm{E} [ {\boldsymbol H} \bar{\mathbf{\Theta}} \tilde{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H} \tilde{\mathbf{\Theta}}^{\rm H} {\boldsymbol H}^{\rm H} ] ) \\ & &= 2\mathrm{Re} \Bigg\{ \frac{\eta \sqrt{\rho_k}}{ \sqrt{\sum_{i=1}^K \rho_i \mathrm{E} [ \| {\boldsymbol h}_{r,i} \|^2 ]} } ( \mathrm{E} [ {\boldsymbol H} \bar{\mathbf{\Theta}} \bar{{\boldsymbol h}}_{r,k} \mathrm{vec} ( \mathrm{diag} ( {\boldsymbol R}_{r,k} - \bar{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H} ) )^{\rm T} {\boldsymbol H}^{\rm H} ] {} \\ & & + \mathrm{E} [ {\boldsymbol H} \bar{\mathbf{\Theta}} ({\boldsymbol R}_{r,k} - \bar{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H}) \mathrm{diag} ( \bar{{\boldsymbol h}}_{r,k}^{\rm H} ) {\boldsymbol H}^{\rm H} ] ) \Bigg\} \\ & &= 2\mathrm{Re} \Bigg\{ \frac{\eta \sqrt{\rho_k}}{ \sqrt{\sum_{i=1}^K \rho_i \mathrm{tr}( {\boldsymbol R}_{r,i} )} } \big( \bar{{\boldsymbol H}} \bar{\mathbf{\Theta}} \bar{{\boldsymbol h}}_{r,k} \mathrm{vec} ( \mathrm{diag} ( {\boldsymbol R}_{r,k} - \bar{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H} ) )^{\rm T} \bar{{\boldsymbol H}}^{\rm H} {} \\ & & + \sigma_{\alpha}^2 \boldsymbol{A}_r \mathrm{diag} ( \boldsymbol{A}_t^{\rm H} \bar{\mathbf{\Theta}} \bar{{\boldsymbol h}}_{r,k} \mathrm{vec} ( \mathrm{diag} ( {\boldsymbol R}_{r,k} - \bar{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H} ) )^{\rm T} \boldsymbol{A}_t ) \boldsymbol{A}_r^{\rm H} {} \\ & & + \bar{{\boldsymbol H}} \bar{\mathbf{\Theta}} ( {\boldsymbol R}_{r,k} - \bar{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H} ) \mathrm{diag}(\bar{{\boldsymbol h}}_{r,k}^{\rm H}) \bar{{\boldsymbol H}}^{\rm H} {} \\ & & + \sigma_{\alpha}^2 \boldsymbol{A}_r \mathrm{diag}( \boldsymbol{A}_t^{\rm H} \bar{\mathbf{\Theta}} ( {\boldsymbol R}_{r,k} - \bar{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H} ) \mathrm{diag} ( \bar{{\boldsymbol h}}_{r,k}^{\rm H} ) \boldsymbol{A}_t )\boldsymbol{A}_r^{\rm H} \big) \Bigg\}. \tag{97} \end{eqnarray} Then, we compute the third term of (94) as follows: \begin{eqnarray}\mathrm{E} [ {\boldsymbol H} \tilde{\mathbf{\Theta}} {\boldsymbol R}_{r,k} \tilde{\mathbf{\Theta}}^{\rm H} {\boldsymbol H}^{\rm H} ] &=& \frac{ \eta^2 }{ \sum_{i=1}^K \rho_i \mathrm{E}( \| {\boldsymbol h}_{r,i} \|^2 ) } \Bigg( \sum_{i=1}^K \rho_i \mathrm{E} [ \bar{{\boldsymbol H}} ( ( {\boldsymbol R}_{r,i} - \bar{{\boldsymbol h}}_{r,i} \bar{{\boldsymbol h}}_{r,i}^{\rm H} ) \odot {\boldsymbol R}_{r,k} ) \bar{{\boldsymbol H}}^{\rm H} ] {} \\ & &+ \sum_{i=1}^K \rho_i \mathrm{E} [ \tilde{{\boldsymbol H}} ( ( {\boldsymbol R}_{r,i} - \bar{{\boldsymbol h}}_{r,i} \bar{{\boldsymbol h}}_{r,i}^{\rm H} ) \odot {\boldsymbol R}_{r,k} ) \tilde{{\boldsymbol H}}^{\rm H} ] \Bigg) \\ &=& \frac{ \eta^2 }{ \sum_{i=1}^K \rho_i \mathrm{tr}( {\boldsymbol R}_{r,i} ) } \Bigg( \bar{{\boldsymbol H}} ( ( {\boldsymbol R}_{r,i} - \bar{{\boldsymbol h}}_{r,i} \bar{{\boldsymbol h}}_{r,i}^{\rm H} ) \odot {\boldsymbol R}_{r,k} ) \bar{{\boldsymbol H}}^{\rm H} {} \\ & &+ \sum_{i=1}^K \rho_i \sigma_{\alpha}^2 \boldsymbol{A}_r \mathrm{diag}( \boldsymbol{A}_t^{\rm H} ( ( {\boldsymbol R}_{r,i} - \bar{{\boldsymbol h}}_{r,i} \bar{{\boldsymbol h}}_{r,i}^{\rm H} ) \odot {\boldsymbol R}_{r,k} )\boldsymbol{A}_t) \boldsymbol{A}_r^{\rm H} \bigg). \tag{98} \end{eqnarray} Combining (94)–(98), we can obtain the autocorrelation matrix ${\boldsymbol~R}_{r,k}$ as \begin{eqnarray}{\boldsymbol R}_{r,k} &=& \bar{{\boldsymbol H}} \bar{\mathbf{\Theta}} {\boldsymbol R}_{r,k} \bar{\mathbf{\Theta}}^{\rm H} \bar{{\boldsymbol H}}^{\rm H} + \sigma_{\alpha}^2 \boldsymbol{A}_r \mathrm{diag}( \boldsymbol{A}_t^{\rm H} \bar{\mathbf{\Theta}} {\boldsymbol R}_{r,k} \bar{\mathbf{\Theta}}^{\rm H} \boldsymbol{A}_t ) \boldsymbol{A}_r^{\rm H} {} \\ & &+ 2\mathrm{Re} \Bigg\{ \frac{\eta \sqrt{\rho_k}}{ \sqrt{\sum_{i=1}^K \rho_i \mathrm{tr}( {\boldsymbol R}_{r,i} )} } \big( \bar{{\boldsymbol H}} \bar{\mathbf{\Theta}} \bar{{\boldsymbol h}}_{r,k} \mathrm{vec} ( \mathrm{diag} ( {\boldsymbol R}_{r,k} - \bar{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H} ) )^{\rm T} \bar{{\boldsymbol H}}^{\rm H} {} \\ & &+ \sigma_{\alpha}^2 \boldsymbol{A}_r \mathrm{diag} ( \boldsymbol{A}_t^{\rm H} \bar{\mathbf{\Theta}} \bar{{\boldsymbol h}}_{r,k} \mathrm{vec} ( \mathrm{diag} ( {\boldsymbol R}_{r,k} - \bar{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H} ) )^{\rm T} \boldsymbol{A}_t ) \boldsymbol{A}_r^{\rm H} {} \\ & &+ \bar{{\boldsymbol H}} \bar{\mathbf{\Theta}} ( {\boldsymbol R}_{r,k} - \bar{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H} ) \mathrm{diag}(\bar{{\boldsymbol h}}_{r,k}^{\rm H}) \bar{{\boldsymbol H}}^{\rm H} {} \\ & &+ \sigma_{\alpha}^2 \boldsymbol{A}_r \mathrm{diag}( \boldsymbol{A}_t^{\rm H} \bar{\mathbf{\Theta}} ( {\boldsymbol R}_{r,k} - \bar{{\boldsymbol h}}_{r,k} \bar{{\boldsymbol h}}_{r,k}^{\rm H} ) \mathrm{diag} ( \bar{{\boldsymbol h}}_{r,k}^{\rm H} ) \boldsymbol{A}_t )\boldsymbol{A}_r^{\rm H} \big) \Bigg\} {} \\ & &+ \frac{ \eta^2 }{ \sum_{i=1}^K \rho_i \mathrm{tr}( {\boldsymbol R}_{r,i} ) } \Bigg( \bar{{\boldsymbol H}} ( ( {\boldsymbol R}_{r,i} - \bar{{\boldsymbol h}}_{r,i} \bar{{\boldsymbol h}}_{r,i}^{\rm H} ) \odot {\boldsymbol R}_{r,k} ) \bar{{\boldsymbol H}}^{\rm H} {} \\ & &+ \sum_{i=1}^K \rho_i \sigma_{\alpha}^2 \boldsymbol{A}_r \mathrm{diag}( \boldsymbol{A}_t^{\rm H} ( ( {\boldsymbol R}_{r,i} - \bar{{\boldsymbol h}}_{r,i} \bar{{\boldsymbol h}}_{r,i}^{\rm H} ) \odot {\boldsymbol R}_{r,k} )\boldsymbol{A}_t) \boldsymbol{A}_r^{\rm H} \Bigg). \tag{99} \end{eqnarray}

Then, we start to compute the expectation term $\mathrm{E}[\|~\mathbf{\Gamma}_k^{\rm~H}~{\boldsymbol~h}_k~\|^2]$: \begin{equation}\mathrm{E}[\| \mathbf{\Gamma}_k^{\rm H} {\boldsymbol h}_k \|^2] = \mathrm{E}[ \| {\boldsymbol h}_k^{\rm H} \mathbf{\Gamma}_k \mathbf{\Gamma}_k^{\rm H} {\boldsymbol h}_k \| ] = \mathrm{tr} ( {\boldsymbol R}_k {\boldsymbol R}_k^{\rm sub} ) \tag{100}\end{equation} with \begin{equation}{\boldsymbol R}_k^{\rm sub}=\left[ \begin{array}{ccc} \displaystyle \frac{{\boldsymbol R}_{1,k}}{\mathrm{tr}({\boldsymbol R}_{1,k})} & \cdots & \mathbf{0} \\ \vdots & \ddots & \vdots \\ \mathbf{0} & \cdots & \displaystyle\frac{{\boldsymbol R}_{N_s,k}}{\mathrm{tr}({\boldsymbol R}_{N_s,k})} \end{array} \right]. \tag{101}\end{equation} As for the term $\mathrm{E}[~(\mathbf{\Gamma}_k^{\rm~H}~{\boldsymbol~h}_k~)^{\rm~H}~\mathbf{\Gamma}_{k,e}^{\rm~H}~{\boldsymbol~h}_k~]$, we have \begin{equation} \mathrm{E}[ (\mathbf{\Gamma}_k^{\rm H} {\boldsymbol h}_k )^{\rm H} \mathbf{\Gamma}_{k,e}^{\rm H} {\boldsymbol h}_k ] = \mathrm{E} [ {\boldsymbol h}_k^{\rm H} \mathbf{\Gamma}_k \mathbf{\Gamma}_{k,e}^{\rm H} {\boldsymbol h}_k ] = {\rm e}^{-\frac{\sigma_{\nu}^2}{2}} \mathrm{E} [ \mathrm{tr} ( {\boldsymbol R}_k {\boldsymbol R}_k^{\rm sub} ) ] . \tag{102}\end{equation} Eq. (102) holds true because the analog precoding error $\mathbf{\Delta}$ is statistically independent of CSI. According to the previous work 2), we have \begin{equation}\mathrm{E} [ \mu_{m,n} {\rm e}^{{\rm j} \nu_{m,n}} ] = {\rm e}^{-\sigma_{\nu}^2/2}, \tag{103}\end{equation} and \begin{equation}\mathrm{E} [ \mu_{m,n} {\rm e}^{{\rm j} \nu_{m,n}} \mu_{i,j}^{\rm H} {\rm e}^{-{\rm j} \nu_{i,j}} ] = \left\{ \begin{aligned} &\sigma_{\mu}^2 + 1, & & \forall (m,n)=(i,j), \\ &{\rm e}^{-\sigma_{\nu}^2}, & & \forall (m,n)\neq (i,j). \end{aligned} \right. \tag{104}\end{equation} Then, we focus on the expectation term $\mathrm{E}~[~\|~(~\mathbf{\Gamma}_k^{\rm~H}~{\boldsymbol~h}_k~)^{\rm~H}~\mathbf{\Gamma}_{k,e}^{\rm~H}~{\boldsymbol~h}_k~\|^2~]$. Note that the number of antennas at the BS will be sufficiently large in THz multiuser massive MIMO systems, we apply channel hardening to $\mathrm{E}~[~\|~(~\mathbf{\Gamma}_k^{\rm~H}~{\boldsymbol~h}_k~)^{\rm~H}~\mathbf{\Gamma}_{k,e}^{\rm~H}~{\boldsymbol~h}_k~\|^2~]$ and we can obtain \begin{equation}\mathrm{E} [ \| ( \mathbf{\Gamma}_k^{\rm H} {\boldsymbol h}_k )^{\rm H} \mathbf{\Gamma}_{k,e}^{\rm H} {\boldsymbol h}_k \|^2 ] = \mathrm{E} | \mathrm{tr}( {\boldsymbol R}_k {\boldsymbol R}_k^{\rm sub} \mathbf{\Delta}^{\rm H} ) |^2 = {\rm e}^{-\sigma_{\nu}^2} | \mathrm{tr}( {\boldsymbol R}_k {\boldsymbol R}_k^{\rm sub} ) |^2. \tag{105}\end{equation} Similarly, we can compute the term $\mathrm{E}~[~\|~(~\mathbf{\Gamma}_k^{\rm~H}~{\boldsymbol~h}_k~)^{\rm~H}~\mathbf{\Gamma}_{k,e}^{\rm~H}~{\boldsymbol~h}_i~\|^2~]$ as follows: \begin{equation}\mathrm{E} [ \| ( \mathbf{\Gamma}_k^{\rm H} {\boldsymbol h}_k )^{\rm H} \mathbf{\Gamma}_{k,e}^{\rm H} {\boldsymbol h}_i \|^2 ] = \mathrm{tr} ( {\boldsymbol R}_k {\boldsymbol R}_k^{\rm sub} {\boldsymbol R}_i^e {\boldsymbol R}_k^{\rm sub} ) \tag{106}\end{equation} with \begin{equation}({\boldsymbol R}_i^e)_{m,n} = \left\{ \begin{aligned} &(1+\sigma_{\mu}^2)({\boldsymbol R}_i)_{m,n}, & & \forall m=n, \\ &{\rm e}^{-\sigma_{\nu}^2}({\boldsymbol R}_i)_{m,n}, & & \forall m \neq n. \end{aligned} \right. \tag{107}\end{equation} Finally, we compute the expectation term $\mathrm{E}[~\|~(\mathbf{\Gamma}_k^{\rm~H}~{\boldsymbol~h}_k)^{\rm~H}~\mathbf{\Gamma}_{k,e}^{\rm~H}~{\boldsymbol~n}~\|^2~]$ as follows: \begin{equation}\mathrm{E}[ \| (\mathbf{\Gamma}_k^{\rm H} {\boldsymbol h}_k)^{\rm H} \mathbf{\Gamma}_{k,e}^{\rm H} {\boldsymbol n} \|^2 ] = \sigma_u^2 \mathrm{E} [ {\boldsymbol h}_k^{\rm H} \mathbf{\Gamma}_k \mathbf{\Gamma}_{k,e}^{\rm H} \mathbf{\Gamma}_{k,e} \mathbf{\Gamma}_k^{\rm H} {\boldsymbol h}_k ] = \sigma_u^2 (\sigma_{\mu}^2 + 1) \mathrm{tr} ( {\boldsymbol R}_k {\boldsymbol R}_k^{\rm sub} {\boldsymbol R}_k^{\rm sub} ). \tag{108}\end{equation}

Wang W J, Liu A, Zhang Q, et al. Robust multigroup multicast transmission for frame-based multi-beam satellite systems. IEEE Access, 2018, 6: 46074–46083


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