Transmit antenna selection for millimeter-wave communications using multi-RIS with imperfect transceiver hardware

Abstract

This article presents a comprehensive exploration of the synergy between transmit antenna selection (TAS) and reconfigurable intelligent surfaces (RISs) in millimeter-wave (MW) communication systems, considering the impact of practical conditions. Notably, it accounts for imperfect transceiver hardware (ITH) at both the transmitter and receiver. Additionally, real-world channel models and receiver noise statistics are integrated into the analysis, providing a realistic representation of wireless systems in future networks. Mathematical formulas of outage probability (OP) and system throughput (ST) of the multi-RIS-assisted MW communications with ITH and TAS (shortened as the considered communications) are derived for analyzing the system behaviors. These formulas facilitate a comprehensive examination of system behavior. Through a series of comparative scenarios, including evaluations of OP and ST with and without TAS, with and without RISs, and with and without ITH (where the absence of ITH is denoted as perfect transceiver hardware, or PTH), the study substantiates the substantial advantages of TAS and RISs while shedding light on the significant influence of ITH. It is demonstrated that even in the presence of ITH, MW communication performance can be dramatically enhanced by optimizing the number of transmit antennas, selecting suitable carrier frequencies and RIS placements, and utilizing appropriate bandwidth. Ultimately, the derived formulas are rigorously validated through Monte-Carlo simulations, reinforcing the credibility of the findings.

Appendix

In this appendix, all mathematical derivations are detailedly presented to derive the OP formula of the considered communications with TAS and ITH in (24).

Since $\bar{c}_{su}$ is the single channel from S to U, its hth moment can be computed as [64]

$$\begin{aligned} \Delta _{\bar{c}_{su}}(h) \triangleq \mathbb {E}\{(\bar{c}_{su})^h\} = \int _{0}^{\infty } y^h f_{\bar{c}_{su}}(y) dy. \end{aligned}$$

(27)

After using (13), (27) becomes

$$\begin{aligned} \Delta _{\bar{c}_{su}}(h) = \frac{\Gamma (m_{su}+h/2)}{\Gamma (m_{su})} \Big ( \frac{m_{su}}{\Omega _{su}}\Big )^{-h/2}. \end{aligned}$$

(28)

From (28), the 1st and 2nd moments of $\bar{c}_{su}$ are, respectively, given as

$$\begin{aligned} \Delta _{\bar{c}_{su}}(1)&=\frac{\Gamma (m_{su}+1/2)}{\Gamma (m_{su})} \sqrt{\frac{\Omega _{su}}{m_{su}}}, \end{aligned}$$

(29)

$$\begin{aligned} \Delta _{\bar{c}_{su}}(2)&= \Omega _{su}. \end{aligned}$$

(30)

Since $\mathcal {X}_{lk} = \bar{a}_{lk} \bar{b}_{lk}$, its PDF is [65]

$$\begin{aligned} f_{\mathcal {X}_{lk}}(\rho ) = \int _{0}^{\infty } \frac{1}{y} f_{\bar{a}_{lk}} \Big (\frac{\rho }{y}\Big ) f_{\bar{b}_{lk}} (y) dy. \end{aligned}$$

(31)

Utilizing (13), (31) is now expressed as

$$\begin{aligned} \nonumber f_{\mathcal {X}_{lk}}(\rho ) =&\, \frac{4}{\Gamma (m_{b_l}) \Gamma (m_{a_l}) } \big (\frac{m_{b_l}}{ \Omega _{b_l}}\big )^{m_{b_l}} \big (\frac{m_{a_l}}{ \Omega _{a_l}}\big )^{m_{a_l}} \alpha ^{2 m_{a_l}-1} \\&\times \int _{0}^{\infty } y^{2 m_{b_l} - 2 m_{a_l} -1} \exp \big ( -\frac{\rho ^2 m_{a_l}}{\Omega _{a_l} y^2} - \frac{m_{b_l} y^2}{\Omega _{b_l}}\big ) dy. \end{aligned}$$

(32)

Applying [66, Eq. (3.478.4)], (32) becomes

$$\begin{aligned} \nonumber f_{\mathcal {X}_{lk}}(\rho ) =&\frac{4 }{\Gamma (m_{a_l}) \Gamma (m_{b_l}) } \Big (\frac{m_{a_l} m_{b_l} }{\Omega _{a_l} \Omega _{b_l} }\Big )^\frac{{m_{a_l} + m_{b_l}}}{2} \rho ^{ m_{a_l} + m_{b_l} -1} \\&\times \mathcal {K}_{m_{b_l} - m_{a_l}} \Bigg (2 \rho \sqrt{\frac{m_{a_l} m_{b_l} }{\Omega _{a_l} \Omega _{b_l} }} \Bigg ). \end{aligned}$$

(33)

Based on (33), the hth moment of $\mathcal {X}_{lk}$ is calculated as

$$\begin{aligned} \Delta _{\mathcal {X}_{lk}}(h) \triangleq \mathbb {E}\{\mathcal {X}_{lk}^h\} = \int _{0}^{\infty } y^h f_{\mathcal {X}_{lk}}(y) dy. \end{aligned}$$

(34)

After utilizing [66, Eq. (6.561.16)], (34) becomes

$$\begin{aligned} \Delta _{\mathcal {X}_{lk}}(h) = \Big (\frac{m_{a_l} m_{b_l} }{\Omega _{a_l} \Omega _{b_l} }\Big )^\frac{-h}{2} \times \frac{\Gamma (m_{a_l}+h/2) \Gamma (m_{b_l}+h/2)}{\Gamma (m_{a_l}) \Gamma (m_{b_l})}. \end{aligned}$$

(35)

Based on (35), the specific moments of $\Delta _{\mathcal {X}_{lk}}$ can be straightforwardly derived, i.e.,

$$\begin{aligned} \Delta _{\mathcal {X}_{lk}}(1)&= \sqrt{\frac{\Omega _{a_l} \Omega _{b_l} }{m_{a_l} m_{b_l} }} \times \frac{\Gamma (m_{a_l}+1/2) \Gamma (m_{b_l}+1/2)}{\Gamma (m_{a_l}) \Gamma (m_{b_l})}, \end{aligned}$$

(36)

$$\begin{aligned} \Delta _{\mathcal {X}_{lk}}(2)&= \frac{\Omega _{a_l} \Omega _{b_l} }{m_{a_l} m_{b_l} } \times \frac{\Gamma (m_{a_l}+1) \Gamma (m_{b_l}+1)}{\Gamma (m_{a_l}) \Gamma (m_{b_l})} = \Omega _{a_l} \Omega _{b_l}. \end{aligned}$$

(37)

From the above moments, the CDF of $\mathcal {X}_{lk}$ is obtained as [51]

$$\begin{aligned} F_{\mathcal {X}_{lk}}(\rho ) \approx \frac{1}{\Gamma \Big ( \frac{[\Delta _{\mathcal {X}_{lk}}(1)]^2 }{\Delta _{\mathcal {X}_{lk}}(2) - [\Delta _{\mathcal {X}_{lk}}(1)]^2} \Big )} \gamma \Bigg ( \frac{[\Delta _{\mathcal {X}_{lk}}(1)]^2 }{\Delta _{\mathcal {X}_{lk}}(2) - [\Delta _{\mathcal {X}_{lk}}(1)]^2}, \frac{\Delta _{\mathcal {X}_{lk}}(1) \rho }{\Delta _{\mathcal {X}_{lk}}(2) - [\Delta _{\mathcal {X}_{lk}}(1)]^2} \Bigg ) \cdot \end{aligned}$$

(38)

It is because $\mathcal {Y}_{l} = \sum _{k=1}^{G_l} \mathcal {X}_{lk}$, its CDF is formulated as [67]

$$\begin{aligned} F_{\mathcal {Y}_{l}}(\rho ) \approx \frac{1}{\Gamma \Big ( \frac{ [\Delta _{\mathcal {X}_{lk}}(1)]^2 G_l }{\Delta _{\mathcal {X}_{lk}}(2) - [\Delta _{\mathcal {X}_{lk}}(1)]^2} \Big )} \gamma \Bigg ( \frac{ [\Delta _{\mathcal {X}_{lk}}(1)]^2 G_l }{\Delta _{\mathcal {X}_{lk}}(2) - [\Delta _{\mathcal {X}_{lk}}(1)]^2}, \frac{\Delta _{\mathcal {X}_{lk}}(1) \rho }{\Delta _{\mathcal {X}_{lk}}(2) - [\Delta _{\mathcal {X}_{lk}}(1)]^2} \Bigg ) \cdot \end{aligned}$$

(39)

Now, the hth moment of $\mathcal {Y}_{l}$ is formulated [68]

$$\begin{aligned} \nonumber \Delta _{\mathcal {Y}_{l}}(h) \triangleq \mathbb {E}\{\mathcal {Y}_{l}^h\} =&\sum _{h_1=0}^{h} \sum _{h_2=0}^{h_1} \cdots \sum _{h_{G_l -1}=0}^{h_{G_l -2}} {h\atopwithdelims ()h_1} {h_1 \atopwithdelims ()h_2} \cdots {h_{G_l -2} \atopwithdelims ()h_{G_l -1}} \\&\times \Delta _{\mathcal {X}_{l1}}(t -h_1) \Delta _{\mathcal {X}_{l2}}(h_1 -h_2) \cdots \Delta _{\mathcal {X}_{lG_l}}(h_{G_l -1}). \end{aligned}$$

(40)

On the other hand, since ${\mathcal {Z}}= \sum _{l=1}^{M} \mathcal {Y}_{l}$, its hth moment is also formulated as

$$\begin{aligned} \nonumber \Delta _{{\mathcal {Z}}}(h) \triangleq \mathbb {E}\{{\mathcal {Z}}^h\} =&\sum _{h_1=0}^{h} \sum _{h_2=0}^{h_1} \cdots \sum _{h_{M -1}=0}^{h_{M -2}} {t \atopwithdelims ()h_1} {h_1 \atopwithdelims ()h_2} \cdots {h_{M -2} \atopwithdelims ()h_{M -1}} \\&\times \Delta _{\mathcal {Y}_{1}}(t -h_1) \Delta _{\mathcal {Y}_{2}}(h_1 -h_2) \cdots \Delta _{\mathcal {Y}_{M}}(h_{M -1}). \end{aligned}$$

(41)

Now, based on (35), (40), and (41), the specific moments of ${\mathcal {Z}}$ can be derived, i.e.,

$$\begin{aligned} \Delta _{\mathcal {Z}}(1)&= \sum _{l=1}^{M} \sum _{k=1}^{G_l} \Delta _{\mathcal {X}_{lk}}(1) , \end{aligned}$$

(42)

$$\begin{aligned} \nonumber \Delta _{\mathcal {Z}}(2) =&\sum _{l=1}^{M} \Big [ \sum _{k=1}^{G_l} \Delta _{\mathcal {X}_{lk}}(2) +2 \sum _{k=1}^{G_l} \Delta _{\mathcal {X}_{lk}}(1) \sum _{k'=k+1}^{G_l} \Delta _{\mathcal {X}_{lk'}}(1) \Big ] \\&+ 2 \sum _{l=1}^{M} \Big [ \sum _{k=1}^{G_l} \Delta _{\mathcal {X}_{lk}}(1) \Big ] \sum _{l'=l+1}^{M} \Big [ \sum _{k=1}^{L_{l'}} \Delta _{\mathcal {X}_{l'k}}(1) \Big ] , \end{aligned}$$

(43)

where $\Delta _{\mathcal {X}_{lk}}(1)$ and $\Delta _{\mathcal {X}_{lk}}(2)$ are, respectively, given in (36) and (37).

Then, the hth moment of ${\mathcal {T}} = {\mathcal {Z}} + \bar{c}_{su}$ can be formulated as

$$\begin{aligned} \nonumber \Delta _{{\mathcal {T}}}(h)&\triangleq \mathbb {E}\{( {\mathcal {Z}} +\bar{c}_{su})^h\} = \mathbb {E}\left\{ \sum _{i=0}^{h} {h \atopwithdelims ()i} \bar{c}_{su}^i {\mathcal {Z}}^{h-i} \right\} \\&= \sum _{i=0}^{h} {h \atopwithdelims ()i} \Delta _{\bar{c}_{su}}(i) \Delta _{\mathcal {Z}}(h-i). \end{aligned}$$

(44)

Based on (44), the specific moments of $\mathcal {T}$ can be derived, i.e.,

$$\begin{aligned} \Delta _\mathcal {T}(1)&= \Delta _{\mathcal {Z}}(1) + \Delta _{\bar{c}_{su}}(1) , \end{aligned}$$

(45)

$$\begin{aligned} \Delta _\mathcal {T}(2)&= \Delta _{\mathcal {Z}}(2) + \Delta _{\bar{c}_{su}}(2) + 2 \Delta _{\mathcal {Z}}(1) \Delta _{\bar{c}_{su}}(1) . \end{aligned}$$

(46)

Now, the CDF of ${\mathcal {T}}$ can be obtained from its moments [51], thus, we obtain

$$\begin{aligned} \nonumber F_{\mathcal {T}}(\rho )&= \frac{1}{\Gamma \Big ( \frac{[\Delta _\mathcal {T}(1)]^2 }{\Delta _\mathcal {T}(2) - [\Delta _\mathcal {T}(1)]^2} \Big )} \gamma \Bigg ( \frac{[\Delta _\mathcal {T}(1)]^2 }{\Delta _\mathcal {T}(2) - [\Delta _\mathcal {T}(1)]^2}, \frac{\Delta _\mathcal {T}(1) \rho }{\Delta _\mathcal {T}(2) - [\Delta _\mathcal {T}(1)]^2} \Bigg ) \\&= 1 - \frac{1}{\Gamma \Big ( \frac{[\Delta _\mathcal {T}(1)]^2 }{\Delta _\mathcal {T}(2) - [\Delta _\mathcal {T}(1)]^2} \Big )} \Gamma \Bigg ( \frac{[\Delta _\mathcal {T}(1)]^2 }{\Delta _\mathcal {T}(2) - [\Delta _\mathcal {T}(1)]^2}, \frac{\Delta _\mathcal {T}(1) \rho }{\Delta _\mathcal {T}(2) - [\Delta _\mathcal {T}(1)]^2} \Bigg ). \end{aligned}$$

(47)

Then, the OP from (23) is expressed as

$$\begin{aligned} \nonumber \mathcal {P}_\text {out}&= \Bigg [ \text {Pr} \Bigg \{\mathcal {T}^2< \frac{\sigma ^2_\text {U} \rho _\text {th}}{\Big ( 1 - \big [(\alpha _\text {S}^\text {t})^2 + (\alpha _\text {U}^\text {r})^2 \big ] \rho _\text {th} \Big ) P_\text {S}} \Bigg \} \Bigg ]^N \\ \nonumber&= \Bigg [ \text {Pr} \Bigg \{\mathcal {T} <\sqrt{\frac{\sigma ^2_\text {U} \rho _\text {th}}{\Big ( 1 - \big [(\alpha _\text {S}^\text {t})^2 + (\alpha _\text {U}^\text {r})^2 \big ] \rho _\text {th} \Big ) P_\text {S}}} \Bigg \} \Bigg ]^N \\&= \Bigg [F_{\mathcal {T}} \Bigg (\sqrt{\frac{\sigma ^2_\text {U} \rho _\text {th}}{\Big ( 1 - \big [(\alpha _\text {S}^\text {t})^2 + (\alpha _\text {U}^\text {r})^2 \big ] \rho _\text {th} \Big ) P_\text {S}}} \Bigg ) \Bigg ]^N. \end{aligned}$$

(48)

Applying (47), (48) is solved as

$$\begin{aligned} \mathcal {P}_\text {out} =&\Bigg [F_{\mathcal {T}} \Bigg (\sqrt{\frac{\sigma ^2_\text {U} \rho _\text {th}}{\Big ( 1 - \big [(\alpha _\text {S}^\text {t})^2 + (\alpha _\text {U}^\text {r})^2 \big ] \rho _\text {th} \Big ) P_\text {S}}} \Bigg ) \Bigg ]^N \nonumber \\ =&\Bigg [1 - \frac{1}{\Gamma \Big ( \frac{[\Delta _\mathcal {T}(1)]^2 }{\Delta _\mathcal {T}(2) - [\Delta _\mathcal {T}(1)]^2} \Big )} \end{aligned}$$

(49)

$$\begin{aligned}&\times \Gamma \Bigg ( \frac{[\Delta _\mathcal {T}(1)]^2 }{\Delta _\mathcal {T}(2) - [\Delta _\mathcal {T}(1)]^2}, \frac{\Delta _\mathcal {T}(1)}{\Delta _\mathcal {T}(2) - [\Delta _\mathcal {T}(1)]^2} \sqrt{\frac{\sigma ^2_\text {U} \rho _\text {th}}{\Big ( 1 - \big [(\alpha _\text {S}^\text {t})^2 + (\alpha _\text {U}^\text {r})^2 \big ] \rho _\text {th} \Big ) P_\text {S}}} \Bigg ) \Bigg ]^N. \end{aligned}$$

(50)

Utilizing the binomial theorem, the OP formula of the considered communications with TAS is derived as (24) in Theorem. The proof is now complete. $\square$