A Sparse Recovery Method for DOA Estimation Based on the Sample Covariance Vectors

Abstract

In this paper, a computation-efficient method utilizing sparse recovery technique is proposed to address the problem of direction of arrival (DOA) estimation based on sample covariance matrix vectors. In the development of the new method, the DOA estimation problem is reformulated in a way that each column of the sample covariance matrix is reintroduced as pseudo-measurements. With this reformulation, multiple candidates of the DOA estimation are obtained by utilizing sparse recovery concept in which an explicit formula of the threshold parameter is provided. The optimal DOA estimation is then selected by employing the maximum likelihood estimation criterion from these multiple candidates. The proposed approach not only has higher resolution and ability of processing coherent sources without the need of decorrelation preprocessing, but also exhibits robust performance, especially in the case of low signal-to-noise ratio and/or small number of snapshots. Numerical studies confirm the effectiveness of the proposed method.

Appendices

Appendix 1: Derivation of $E({{\mathbf{w}}_i})=\sigma _n^2{{\mathbf{I}}_M}(:,i)$ and $E({{\mathbf{w}}_i}{\mathbf{w}}_i^{\mathrm{H}}) = \frac{1}{L}\sigma _n^2{\mathbf{R}}(i,i){{\mathbf{I}}_M} + \sigma _n^4{\mathbf{E}}_M^{(i,i)}$

With the assumption that the entries in the noise vector ${\mathbf{n}}(t)$ are spatially uncorrelated, circularly symmetric complex Gaussian random process with zero mean and equal variance $\sigma _n^2$, ${\mathbf{n}}(t)$ can be represented by [6, 18],

$$\begin{aligned} {\mathbf{n}}(t) = \frac{1}{{\sqrt{2} }}{{\varepsilon }}(t) + i \cdot \frac{1}{{\sqrt{2} }}\varsigma (t), \end{aligned}$$

(25)

where ${{\varepsilon }}(t)$ and ${{\varsigma }}(t)$ are real Gaussian distributed random vectors with zero mean and variance $\sigma _n^2\mathbf I$. The expectation of ${\mathbf{w}}_i$ is calculated as

$$\begin{aligned} \begin{array}{l} E({{\mathbf{w}}_i})\\ \quad = E\left\{ {\frac{1}{L}\sum \limits _{t = 1}^L {{\mathbf{n}}(t){{[{{\tilde{x}}_i}(t) + {n_i}(t)]}^{\mathrm{*}}}} } \right\} \\ \quad = \frac{1}{L}\sum \limits _{t = 1}^L {E\left\{ {{\mathbf{n}}(t){{[{{\tilde{x}}_i}(t) + {n_i}(t)]}^{\mathrm{*}}}} \right\} } \end{array}. \end{aligned}$$

(26)

Since $\frac{1}{L}\sum _{t = 1}^L {E\left\{ {{n_j}(t){{[{{\tilde{x}}_i}(t) + {n_i}(t)]}^{\mathrm{*}}}} \right\} } = \left\{ {\begin{array}{*{20}{c}} {0 \quad j \ne i}\\ {\sigma _n^2 \;j = i} \end{array}} \right.$, we have $E({{\mathbf{w}}_i})=\sigma _n^2{{\mathbf{I}}_M}(:,i)$.

The correlation of ${\mathbf{w}}_i$ and ${{\mathbf{w}}_i}^{\mathrm{H}}$ is computed as

$$\begin{aligned} E({{\mathbf{w}}_i}{\mathbf{w}}_i^{\mathrm{H}})= & {} \frac{1}{{{L^2}}}\sum \limits _{t = 1}^L {\sum \limits _{l = 1}^L {E\left\{ {{\mathbf{n}}(t){{[{{\tilde{x}}_i}(t) + {n_i}(t)]}^*}{{\mathbf{n}}^{\mathrm{H}}}(l)[{{\tilde{x}}_i}(l) + {n_i}(l)]} \right\} } } \nonumber \\= & {} \frac{1}{{{L^2}}}\sum \limits _{t = 1}^L {\sum \limits _{l = 1}^{t - 1} {E\left\{ {{\mathbf{n}}(t){{[{{\tilde{x}}_i}(t) + {n_i}(t)]}^*}{{\mathbf{n}}^{\mathrm{H}}}(l)[{{\tilde{x}}_i}(l) + {n_i}(l)]} \right\} } } \nonumber \\&+ \frac{1}{{{L^2}}}\sum \limits _{t = 1}^L {\sum \limits _{l = t + 1}^L {E\left\{ {{\mathbf{n}}(t){{[{{\tilde{x}}_i}(t) + {n_i}(t)]}^*}{{\mathbf{n}}^{\mathrm{H}}}(l)[{{\tilde{x}}_i}(l) + {n_i}(l)]} \right\} } } \nonumber \\&+ \frac{1}{{{L^2}}}\sum \limits _{t = 1}^L {E\left\{ {{\mathbf{n}}(t){{[{{\tilde{x}}_i}(t) + {n_i}(t)]}^*}{{\mathbf{n}}^{\mathrm{H}}}(t)[{{\tilde{x}}_i}(t) + {n_i}(t)]} \right\} }. \end{aligned}$$

(27)

Utilizing the mutual independency of the snapshots, the sum of the first and the second terms in Eq. (27) can be expressed as

$$\begin{aligned}&\frac{1}{L}\sum \limits _{t = 1}^L {E[{\mathbf{n}}(t)\tilde{x}_i^*(t) + {\mathbf{n}}(t)n_i^*(t)]} \cdot \left\{ \frac{1}{L}\sum \limits _{l = 1}^{t - 1} {E[{{\mathbf{n}}^{\mathrm{H}}}(l)\tilde{x}_i^*(l) + {{\mathbf{n}}^{\mathrm{H}}}(l)n_i^*(l)]}\right. \nonumber \\&\quad \left. + \frac{1}{L}\sum \limits _{l = t + 1}^L {E[{{\mathbf{n}}^{\mathrm{H}}}(l)\tilde{x}_i^*(l) + {{\mathbf{n}}^{\mathrm{H}}}(l)n_i^*(l)]} \right\} \nonumber \\&\qquad = \sigma _n^2{{\mathbf{I}}_M}(:,i) \cdot \frac{{L - 1}}{L}\sigma _n^2{({{\mathbf{I}}_M}(:,i))^{\mathrm{H}}}\nonumber \\&\qquad = \frac{{L - 1}}{L}\sigma _n^4{\mathbf{E}}_M^{(i,i)}. \end{aligned}$$

(28)

And the third term in Eq. (27) can be further written as

$$\begin{aligned}&\frac{1}{{{L^2}}}\sum \limits _{t = 1}^L {E\left\{ {{\mathbf{n}}(t){{[{{\tilde{x}}_i}(t) + {n_i}(t)]}^*}{{\mathbf{n}}^{\mathrm{H}}}(t)[{{\tilde{x}}_i}(t) + {n_i}(t)]} \right\} } \nonumber \\&\quad = \frac{1}{{{L^2}}}\sum \limits _{t = 1}^L {E[{\mathbf{n}}(t)\tilde{x}_i^*(t){{\mathbf{n}}^{\mathrm{H}}}(t){{\tilde{x}}_i}(t) + {\mathbf{n}}(t)n_i^*(t){{\mathbf{n}}^{\mathrm{H}}}(t){n_i}(t)]} \nonumber \\&\quad = \frac{1}{L}{{{\tilde{\mathbf{R}}}}_s}(i,i) \cdot \sigma _n^2{{\mathbf{I}}_M} + \frac{1}{{{L^2}}}\sum \limits _{t = 1}^L {E[{\mathbf{n}}(t)n_i^*(t){{\mathbf{n}}^{\mathrm{H}}}(t){n_i}(t)]}, \end{aligned}$$

(29)

where ${{\tilde{\mathbf{R}}}_{\mathbf{s}}} = \frac{1}{L}E[{\tilde{\mathbf{x}}}(t){{\tilde{\mathbf{x}}}^{\mathrm{H}}}(t)]$ and ${{\tilde{\mathbf{R}}}_s}(i,i)$ denotes the ${(i,i)^{th}}$ entry of ${{\tilde{\mathbf{R}}}_{\mathbf{s}}}$.

Since

$$\begin{aligned}&\frac{1}{{{L^2}}}\sum \limits _{t = 1}^L {E\left[ {n_i}(t)n_i^*(t)n_i^*(t){n_i}(t)\right] } \\&\quad = \frac{1}{{{L^2}}}\sum \limits _{t = 1}^L {E\left[ \frac{1}{2}{{{\varepsilon }}^2}(t) + \frac{1}{2}{{{\varsigma }}^2}(t)\right] } \cdot \left[ \frac{1}{2}{{{\varepsilon }}^2}(t) + \frac{1}{2}{{{\varsigma }}^2}(t)\right] \\&\quad = \frac{1}{{{L^2}}}\sum \limits _{t = 1}^L {E\left[ \frac{1}{4}{{{\varepsilon }}^4}(t) + \frac{1}{4}{{{\varsigma }}^4}(t) + \frac{1}{2}{{{\varepsilon }}^2}(t){{{\varsigma }}^2}(t)\right] } \\&\quad = \frac{1}{L}\left( \frac{3}{4}\sigma _n^4 + \frac{3}{4}\sigma _n^4 + \frac{1}{2}\sigma _n^4\right) = \frac{2}{L}\sigma _n^4 \end{aligned}$$

and

$$\begin{aligned}&\frac{1}{{{L^2}}}\sum \limits _{t = 1}^L {E[{n_j}(t)n_i^*(t){\mathbf{n}}(t){n_i}(t)]} = \left\{ {\begin{array}{*{20}{c}} {\frac{2}{L}\sigma _n^4{{\mathbf{I}}_M}(:,i)\quad j = i}\\ {\frac{1}{L}\sigma _n^4{{\mathbf{I}}_M}(:,j)\quad j \ne i} \end{array}} \right. , \end{aligned}$$

the second term in Eq. (29) can be simplified as

$$\begin{aligned} \frac{1}{{{L^2}}}\sum \limits _{t = 1}^L {E[{\mathbf{n}}(t)n_i^*(t){{\mathbf{n}}^{\mathrm{H}}}(t){n_i}(t)]} = \frac{1}{L}\sigma _n^4{{\mathbf{I}}_M} + \frac{1}{L}\sigma _n^4E_M^{(i,i)}. \end{aligned}$$

(30)

Substituting Eq. (30) into Eq. (29), we have

$$\begin{aligned}&\frac{1}{{{L^2}}}\sum \limits _{t = 1}^L {E\left\{ {{\mathbf{n}}(t){{[{{\tilde{x}}_i}(t) + {n_i}(t)]}^*}{{\mathbf{n}}^{\mathrm{H}}}(t)[{{\tilde{x}}_i}(t) + {n_i}(t)]} \right\} } \nonumber \\&\quad = \frac{1}{L}{{{\tilde{\mathbf{R}}}}_s}(i,i) \cdot \sigma _n^2{{\mathbf{I}}_M} + \frac{1}{L}\sigma _n^4{{\mathbf{I}}_M} + \frac{1}{L}\sigma _n^4{\mathbf{E}}_M^{(i,i)}\nonumber \\&\quad =\frac{1}{L}{\mathbf{R}}(i,i) \cdot \sigma _n^2{{\mathbf{I}}_M} + \frac{1}{L}\sigma _n^4{\mathbf{E}}_M^{(i,i)}. \end{aligned}$$

(31)

Now combining Eqs. (28) and (31), we can calculate $E({{\mathbf{w}}_i}{\mathbf{w}}_i^{\mathrm{H}})$ as

$$\begin{aligned}&E({{\mathbf{w}}_i}{\mathbf{w}}_i^{\mathrm{H}}) = \frac{{L - 1}}{L}\sigma _n^4{\mathbf{E}}_M^{(i,i)} + \frac{1}{L}{\mathbf{R}}(i,i) \cdot \sigma _n^2{{\mathbf{I}}_{M \times M}} + \frac{1}{L}\sigma _n^4{\mathbf{E}}_M^{(i,i)}\nonumber \\&\quad = \frac{1}{L}{\mathbf{R}}(i,i) \cdot \sigma _n^2{{\mathbf{I}}_M} + \sigma _n^4{\mathbf{E}}_M^{(i,i)}. \end{aligned}$$

(32)

Appendix 2: Proof of ${{\mathbf{w}}_i} \sim {\mathrm{AsN(}}\sigma _n^2{{\mathbf{I}}_M}(:,i),\frac{1}{L}\sigma _n^2{\mathbf{R}}(i,i){{\mathbf{I}}_M}{\mathrm{)}}$

According to Eq. (11), let ${\varvec{\Gamma }}(t) = {\mathbf{n}}(t){[{\tilde{x}_i}(t) + {n_i}(t)]^{\mathrm{*}}}$, and with the assumptions made in Sect. 2, the time correlation matrix of ${\varvec{\Gamma }}(t)$ can be expressed as follows

$$\begin{aligned} {{\mathbf{C}}_{\varvec{\Gamma }}} = E\left\{ {\varvec{\Gamma }}(t){\varvec{\Gamma }}{(l)^{\mathrm{H}}}\right\} = {{\mathbf{0}}_{M \times M}}\quad for\;t \ne l, \end{aligned}$$

(33)

where ${{\mathbf{0}}_{M \times M}}$ denotes $M \times M$ zero matrix. The Eq. (33) implies that the random vectors ${\varvec{\Gamma }}(t)$ for different t are mutually independent. Meanwhile, it can be readily deduced that the entries of ${\varvec{\Gamma }}(t)$ are independent with each other for any t, and the random vectors ${\varvec{\Gamma }}(t)$ for different t have same probability distribution.

Based on the analysis above, and according to the Lindeberg–Levy central limit theorem in [11], ${{\mathbf{w}}_i}=\frac{1}{L}\sum \limits _{t = 1}^L {{\varvec{\Gamma }}(t)}$ satisfies an independent asymptotic Gaussian distribution, i.e., the probability distribution of ${{\mathbf{w}}_i}$ asymptotically approaches the Gaussian distribution as L increases. With the previous results in “Appendix 1,” it can be easily concluded that ${{\mathbf{w}}_i} \sim {\mathrm{AsN(}}\sigma _n^2{{\mathbf{I}}_M}(:,i),\frac{1}{L}\sigma _n^2{\mathbf{R}}(i,i){{\mathbf{I}}_M}{\mathrm{)}}$.