Sum and difference coarray based MIMO radar array optimization with its application for DOA estimation

Abstract

Based on the sum and difference coarrays, multiple-input multiple-output (MIMO) radar with minimum redundancy (MR) concept, referred to as MR MIMO, can considerably increase the spatial degrees of freedom (DOFs). However, traditional MR MIMO needs computational search to determine the position of each element. In this paper, a modified MR monostatic MIMO configuration is proposed, referred to as MMRM MIMO. In the proposed system, the MMRM MIMO radar is consisted of several levels of uniform linear array, which brings the advantage that the position of each element can be determined without computational search. Furthermore, it offers more than $N^{2}$ DOFs for an N-elemental array. In order to utilize the extended DOFs of MMRM MIMO radar for direction-of-arrival (DOA) estimation, an average Toeplitz approximation method (TAM) is employed, which achieves robust performance even under low signal-to-noise ratio, few snapshots and array error. Numerous simulation results are provided to demonstrate the effectiveness of the proposed method for DOA estimation.

Appendices

Appendix 1

Proof of Corollary 1

From (15), the last sensor location in MMRA is

$$\begin{aligned} x= & {} \left[ {N_2 \left( {N_3 +1} \right) \left( {N_1 +1} \right) +N_1 N_3 -1} \right] d \nonumber \\= & {} \left[ {\left( {N_3 +1} \right) N_1 N_2 +N_2 +N_A N_3 -1} \right] d \end{aligned}$$

(24)

where $N_{\mathrm{A}} =N_1 +N_2$ is fixed.

(i) $N_{\mathrm{A}} =N_1 +N_2$ is even:

If $N_1 =N_2$, the last sensor location in MMRA with $N_{\mathrm{A}}$ being even is

$$\begin{aligned} x_E \;=\;\left[ {\left( {N_3 +1} \right) N_1^2 +N_1 +N_{\mathrm{A}} N_3 -1} \right] d \end{aligned}$$

(25)

Hold $N_{\mathrm{A}}$ invariant, then change the number of sensors in first level and second level of Part I as ${N}'_{E,1} =N_1 -n$ and ${N}'_{E,2} =N_1 +n$ respectively, where $0<n<N_1$. Then (26) can be rewritten as

$$\begin{aligned} {x}'_E \;= & {} \left[ {{N}'_{E,2} \left( {N_3 +1} \right) \left( {{N}'_{E,1} +1} \right) +{N}'_{E,1} N_3 -1} \right] d \nonumber \\= & {} \left[ {\left( {N_3 +1} \right) \left( {N_1^2 -n^{2}} \right) +\left( {N_1 +n} \right) +N_{\mathrm{A}} N_3 -1} \right] d \nonumber \\= & {} \left[ {\left( {N_3 +1} \right) N_1^2 +\left( {n-n^{2}} \right) +\left( {N_1 -n^{2}N_3 } \right) +N_{\mathrm{A}} N_3 -1} \right] d\;\;\;<x_E \end{aligned}$$

(26)

Similarly hold $N_{\mathrm{A}}$ invariant, then change the number of sensors in first level and second level of Part I as ${N}''_{E,1} =N_1 +n$ and ${N}''_{E,2} =N_1 -n$ respectively, where $0<n<N_1$, then it can be obtained that

$$\begin{aligned} {x}''_E <{x}'_E <x_E \end{aligned}$$

(27)

In conclusion, if $N_{\mathrm{A}} =N_1 +N_2$ is even, the MMRA will extend the maximum degree of freedom when $N_1 =N_2$.

(ii) $N_{\mathrm{A}} =N_1 +N_2$ is odd:

If $N_2 =N_1 +1$, the last sensor location in MMRA with $N_{\mathrm{A}}$ being odd is

$$\begin{aligned} x_O= & {} \left[ {N_1 \left( {N_3 +1} \right) \left( {N_1 +1} \right) +\left( {N_1 +1} \right) +N_{\mathrm{A}} N_3 -1} \right] d \nonumber \\= & {} \left[ {\left( {N_3 +1} \right) N_1^2 +N_1 N_3 +N_{\mathrm{A}} N_3 +2N_1 } \right] d \end{aligned}$$

(28)

Hold $N_{\mathrm{A}}$ invariant, then change the number of sensors in first level and second level of Part I as ${N}'_{O,1} =N_1 -n$ and ${N}'_{O,2} =N_1 +1+n$ respectively, where $0<n<N_1$, then (24) can be rewritten as

$$\begin{aligned} {x}'_O= & {} \left[ {{N}'_{O,2} \left( {N_3 +1} \right) \left( {{N}'_{O,1} +1} \right) +{N}'_{O,1} N_3 -1} \right] d \nonumber \\= & {} \left[ {\left( {N_3 +1} \right) \left( {N_1 -n} \right) \left( {N_1 +1+n} \right) +\left( {N_1 +n+1} \right) +N_{\mathrm{A}} N_3 -1} \right] d\nonumber \\= & {} \left[ {\left( {N_3 +1} \right) N_1^2 +N_1 N_3 +N_{\mathrm{A}} N_3 +2N_1 -nN_3 -n^{2}\left( {N_3 +1} \right) } \right] d\;\;\;<x_O \end{aligned}$$

(29)

Similarly hold $N_{\mathrm{A}}$ invariant, then change the number of sensors in first level and second level of Part I as ${N}''_{O,1} =N_1 +n$ and ${N}''_{O,2} =N_1 +1-n$ respectively, where $0<n<N_1 +1$, then it is obvious that

$$\begin{aligned} {x}''_O <{x}'_O <x_O \end{aligned}$$

(30)

Therefore if $N_{\mathrm{A}} =N_1 +N_2$ is odd, the MMRA will enhance the maximum degree of freedom when $N_2 =N_1 +1$. $\square$

Appendix 2

Proof of Corollary 2

According to Corollary 1, if $N_{\mathrm{A}} =N_1 +N_2$ is even, the MMRA can extend the most DOFs when $N_1 =N_2$. Assume that$N_{\mathrm{A}}$ is even and the number of sensors in each level of Part I is $N_1$, then the number of sensors in Part II is

$$\begin{aligned} N_3 =N-N_{\mathrm{A}} =N-2N_1 \end{aligned}$$

(31)

Insert (31) into (25), then the last sensor location in MMRA is

$$\begin{aligned} x= & {} \left[ {\left( {N-2N_1 +1} \right) N_1^2 +N_1 +N_{\mathrm{A}} \left( {N-2N_1 } \right) -1} \right] d \nonumber \\= & {} \left[ {\left( {N_1^2 N-2N_1^3 } \right) +\left( {2N_{1} N-3N_1^2 } \right) +N_1 -1} \right] d \end{aligned}$$

(32)

Based on Corollary 1, if the number of sensors in Part I is $N_{\mathrm{A}} +1$ and $N_1$ is fixed, then ${N}'_2 =N_1 +1$. So keep N fixed and the number of sensors in Part II is

$$\begin{aligned} {N}'_3 =N-{N}'_{\mathrm{A}} =N-2N_1 -1 \end{aligned}$$

(33)

Insert (33) into (28), then the last sensor location in MMRA is

$$\begin{aligned} {x}'= & {} \left[ {{N}'_2 \left( {{N}'_3 +1} \right) \left( {N_1 +1} \right) +N_1 {N}'_3 -1} \right] d \nonumber \\= & {} \left[ {\left( {N_1^2 N-2N_1^3 } \right) +\left( {3N_1 N-6N_1^2 } \right) +\left( {N-3N_1 } \right) -1} \right] d \end{aligned}$$

(34)

Then the difference between (36) and (34) is

$$\begin{aligned} \Delta x_{O-E}= & {} {x}'-x \nonumber \\= & {} \left[ {\left( {N_1 N-3N_1^2 } \right) +\left( {N-4N_1 } \right) } \right] d \end{aligned}$$

(35)

Insert the known condition into (35), which leads to

$$\begin{aligned} \Delta x_{O-E} =\left\{ {{\begin{array}{ll} \left( {-2N_1 } \right) d<0,&{} \quad N=3N_1 -1; \\ \left( {-N_1 } \right) d<0,&{} \quad N=3N_1 ; \\ 1,&{}\quad N=3N_1 +1; \\ \end{array} }} \right. \end{aligned}$$

(36)

Next, if the number of sensors in Part I is $N_{\mathrm{A}} +2$ and ${N}'_2$ is fixed, then ${N}''_1 =N_1 +1$. Hence N is fixed and the number of sensors in Part II is

$$\begin{aligned} {N}''_3 =N-{N}''_{\mathrm{A}} =N-2N_1 -2 \end{aligned}$$

(37)

Insert (37) into (25), the last sensor location in MMRA is

$$\begin{aligned} {x}''= & {} \left[ {\left( {{N}''_3 +1} \right) {N}_1^{''2} +{N}''_1 +{N}''_{\mathrm{A}} {N}''_3 -1} \right] d \nonumber \\= & {} \left[ {\left( {N_1^2 N-2N_1^3 } \right) +\left( {4N_{1} N-9N_1^2 } \right) +\left( {3N-11N_{1} } \right) -5} \right] d \end{aligned}$$

(38)

Then the difference between (38) and (34) is

$$\begin{aligned} \Delta x_{E-O}= & {} {x}''-{x}' \nonumber \\= & {} \left[ {\left( {N_1 N-3N_1^2 } \right) +\left( {2N-8N_1 } \right) -4} \right] d \end{aligned}$$

(39)

Insert the known condition into (39), which can be rewritten as

$$\begin{aligned} \Delta x_{E-O} =\left\{ {{\begin{array}{ll} 0&{}\quad N=3\left( {N_1 +1} \right) -1; \\ N_1 +2>0,&{} \quad N=3\left( {N_1 +1} \right) ; \\ 2N_1 +4>0,&{}\quad N=3\left( {N_1 +1} \right) +1; \\ \end{array} }} \right. \end{aligned}$$

(40)

As observed in Sect. 3, the more $N_1$ is, the longer spacing between sensors in Part II is. Compared (36) to (40), we can obtain the conclusion: if $N=3N_1 -1$ or $3N_1$, then $N_{\mathrm{A}} =2N_1$ and $N_3 =N_1 -1$ or $N_1$; if $N=3N_1 +1$, then $N_{\mathrm{A}} =2N_1 +1$ and $N_3 =N_1$. $\square$