SCIENCE CHINA Information Sciences, Volume 63 , Issue 8 : 180302(2020) https://doi.org/10.1007/s11432-019-2921-x

Machine-learning-based high-resolution DOA measurement and robust directional modulation for hybrid analog-digital massive MIMO transceiver

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  • ReceivedDec 29, 2019
  • AcceptedMay 19, 2020
  • PublishedJul 15, 2020



This work was supported in part by National Natural Science Foundation of China (Nos. 61771244, 61871229).



Derivation of RAB ${\tilde{V}_{~RF}}$

Considering the DOA measurement error, the nonzero element of ${\boldsymbol~V}_{\rm~RF,RAB}$ can be represented as \begin{align}\hat{v}_{k,m}&=\int_{-\Delta\theta_\mathrm{{max}}}^{\Delta\theta_\mathrm{{max}}}{\rm e}^{-{\rm j}\frac{2\pi d}{\lambda}[(k-1)M+m-\frac{N+1}{2}]\cos(\hat{\theta}-\Delta\theta)}\cdot p(\Delta\theta){\rm d}(\Delta\theta) \\ &=\int_{-\Delta\theta_\mathrm{{max}}}^{\Delta\theta_\mathrm{{max}}}{\rm e}^{-{\rm j}\alpha_{k,m}\cos(\hat{\theta}-\Delta\theta)}\cdotp p(\Delta\theta){\rm d}(\Delta\theta) \\ &=\int_{-\Delta\theta_\mathrm{{max}}}^{\Delta\theta_\mathrm{{max}}}{\rm e}^{-{\rm j}\alpha_{k,m}[\cos(\hat{\theta})\cos(\Delta\theta)+\sin(\hat{\theta})\sin(\Delta\theta)]}\cdotp p(\Delta\theta){\rm d}(\Delta\theta), \tag{69} \end{align} where \begin{align}\alpha_{k,m}=\frac{2\pi d}{\lambda}\left[(k-1)M+m-\frac{N+1}{2}\right]. \tag{70} \end{align} By utilizing the second-order Taylor expansion, we can expand $\cos(\Delta\theta)$ and $\sin(\Delta\theta)$ at point $\Delta\theta=0$ as \begin{align} \cos(\Delta\theta)\approx1-\frac{1}{2}(\Delta\theta)^2, \sin(\Delta\theta)\approx\Delta\theta. \tag{71} \end{align} Substituting Eq. (71) in (69) yields \begin{align} \hat{v}_{k,m} &=\int_{-\Delta\theta_\mathrm{{max}}}^{\Delta\theta_\mathrm{{max}}}{\rm e}^{-{\rm j}\alpha_{k,m}[\cos(\hat{\theta})\cos(\Delta\theta)+\sin(\hat{\theta})\sin(\Delta\theta)]}\cdot p(\Delta\theta){\rm d}(\Delta\theta) \\ &=\int_{-\Delta\theta_\mathrm{{max}}}^{\Delta\theta_\mathrm{{max}}}{\rm e}^{-{\rm j}\alpha_{k,m}[\cos(\hat{\theta})-\cos(\hat{\theta})\cdot\frac{1}{2}(\Delta\theta)^2+\sin(\hat{\theta})(\Delta\theta)]} p(\Delta\theta){\rm d}(\Delta\theta) \\ &=\xi_{k,m}\int_{-\Delta\theta_\mathrm{{max}}}^{\Delta\theta_\mathrm{{max}}}{\rm e}^{{\rm j}\alpha_{k,m}[\cos(\hat{\theta})\cdot\frac{1}{2}(\Delta\theta)^2-\sin(\hat{\theta})(\Delta\theta)]}p(\Delta\theta){\rm d}(\Delta\theta) \\ &=\xi_{k,m}\int_{-\Delta\theta_\mathrm{{max}}}^{\Delta\theta_\mathrm{{max}}}\{\cos(\alpha_{k,m}\psi)+{\rm j}\sin(\alpha_{k,m}\psi)\}p(\Delta\theta){\rm d}(\Delta\theta) \\ &=\zeta_{k,m}+{\rm j}\eta_{k,m}, \tag{72} \end{align} where \begin{align}\xi_{k,m}={\rm e}^{-{\rm j}\alpha_{k,m}\cos(\hat{\theta})}, \tag{73} \end{align} and \begin{align}\psi=\left[\cos(\hat{\theta})\cdot\frac{1}{2}(\Delta\theta)^2-\sin(\hat{\theta})(\Delta\theta)\right]. \tag{74} \end{align} Similarly, by utilizing the second-order Taylor expansion, $\cos(\alpha_{k,m}\psi)$ can be represented as \begin{align} \cos(\alpha_{k,m}\psi)=&\,\cos\left(\alpha_{k,m}\left[\cos(\hat{\theta})\cdot\frac{1}{2}(\Delta\theta)^2-\sin(\hat{\theta})(\Delta\theta)\right]\right) \\ \approx&\,1-\frac{1}{2}\alpha_{k,m}^2\left[\cos(\hat{\theta})\cdot\frac{1}{2}(\Delta\theta)^2-\sin(\hat{\theta})(\Delta\theta)\right]^2 \\ =&\,1-\frac{1}{8}\alpha_{k,m}^2\cos(\hat{\theta})^2(\Delta\theta)^4-\frac{1}{2}\alpha_{k,m}^2\sin(\hat{\theta})^2(\Delta\theta)^2 \\ &+\frac{1}{2}\alpha_{k,m}^2\cos(\hat{\theta})\sin(\hat{\theta})(\Delta\theta)^3. \tag{75} \end{align} Since the last term of Eq. (75) is an odd function of $\Delta\theta$, then \begin{align}\int_{-\Delta\theta_\mathrm{{max}}}^{\Delta\theta_\mathrm{{max}}}\frac{1}{2}\alpha_{k,m}^2\cos(\hat{\theta})\sin(\hat{\theta})(\Delta\theta)^3\cdot p(\Delta\theta){\rm d}(\Delta\theta)=0. \tag{76} \end{align} Now, let us define \begin{align}\chi_1=&\,\int_{-\Delta\theta_\mathrm{{max}}}^{\Delta\theta_\mathrm{{max}}}(\Delta\theta)^4\cdot p(\Delta\theta){\rm d}(\Delta\theta) \\ =&\,\frac{2}{K_d\sqrt{2\pi \sigma^2}}\left\{-\sigma^2\cdot\Delta\theta_\mathrm{{max}}^3 {\rm e}^{-\frac{\Delta\theta_\mathrm{{max}}^2}{2\sigma^2}}-3\sigma^4\Delta\theta_\mathrm{{max}}{\rm e}^{-\frac{\Delta\theta_\mathrm{{max}}^2}{2\sigma^2}} +\frac{3\sqrt{2\pi}}{2}\sigma^5 \mathrm{erf}\left(\frac{\Delta\theta_\mathrm{{max}}}{\sqrt{2}\sigma}\right)\right\}, \tag{77} \end{align} and \begin{align}\chi_2&=\int_{-\Delta\theta_\mathrm{{max}}}^{\Delta\theta_\mathrm{{max}}}(\Delta\theta)^2\cdot p(\Delta\theta){\rm d}(\Delta\theta) \\ &=\frac{2}{K_d\sqrt{2\pi\sigma^2}}\left\{-\sigma^2\Delta\theta_\mathrm{{max}}\cdot {\rm e}^{-\frac{\Delta\theta_\mathrm{{max}}^2}{2\sigma^2}}+\frac{\sqrt{2\pi}}{2}\sigma^3{\rm erf}\left(\frac{\Delta\theta_\mathrm{{max}}}{\sqrt{2}\sigma}\right)\right\}. \tag{78} \end{align} Then the real part $\zeta_{k,m}$ of Eq. (72) can be expressed as \begin{align}\zeta_{k,m}=\xi_{k,m}\left(K_d-\frac{1}{8}\alpha_{k,m}^2\cos(\hat{\theta})^2\chi_1-\frac{1}{2}\alpha_{k,m}^2\sin(\hat{\theta})^2\chi_2\right). \tag{79} \end{align} In the same manner, we can have the $\eta_{k,m}$ as follows: \begin{align}\eta_{k,m}=\frac{1}{2}\xi_{k,m}\cdot\alpha_{k,m}\cos(\hat{\theta})\chi_2. \tag{80} \end{align} Until now, we complete the derivation of $\hat{v}_{k,m}$. That is, $\hat{v}_{k,m}=\zeta_{k,m}+j\eta_{k,m}$. Owing to the special structure of analog precoder, we only need the phase of $\hat{v}_{k,m}$. Therefore we can reformulated the analog $v_{k,m}$ as \begin{align}v_{{\rm RAB}, k,m}=\frac{1}{\sqrt{M}}\exp(j*\angle(\hat{v}_{k,m})), \tag{81} \end{align} which is what we need.


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