Interleaved Arrays with Second-Order Difference Coarray Generation for Mutual Coupling Reduction

Abstract

Nonuniform linear arrays (NLAs) have attracted considerable attention owing to their capability to enhance the degrees of freedom (DOF) and increase the array aperture, which is essential for improving performance in terms of both the direction of arrival estimation and adaptive beamforming. This paper presents two new NLAs, namely interleaved coprime array (ICA) and augmented interleaved array (AIA). The ICA consists of two uniform linear subarrays, whose inter-element spacings are coprime integers. The ICA can obtain more uniform degrees of freedom (uDOF), while maintaining weak mutual coupling effects, because its configuration includes only minority sensors interleaved. The optimal configuration of the ICA is deduced using the principle of maximizing the uDOF. Moreover, the AIA geometry is obtained by removing a redundant element in the ICA and appropriately designing the location of the additional element. For a fixed total number of sensors, the AIA can gain more uDOF and DOF than the ICA, while exhibiting lower mutual coupling effects. The closed-form expressions for the properties of the proposed array structures involving an arbitrary number of sensors are also derived. Numerical simulations are performed to verify the superiority of the proposed arrays over other sparse arrays.

Appendices

Appendix A: Proofs of Property 1

Since the generated virtual sensors are symmetric at zero position, we only need to determine that the positive part of difference coarray contains consecutive lags in $[1,(N - P + 2)M - 2]$. The self-difference sets of subarrays 1 and 2 can be respectively expressed as follows:

$$ \begin{gathered} {\mathbb{D}}_{{{\mathbb{S}}_{1} ,{\mathbb{S}}_{1} }}^{\dag } = \left\{ {\left. {(n - 1)M} \right|n = 1,2, \ldots ,N} \right\}, \hfill \\ {\mathbb{D}}_{{{\mathbb{S}}_{2} ,{\mathbb{S}}_{2} }}^{\dag } = \left\{ {\left. {(m - 1)(M - 1)} \right|m = 1,2, \ldots ,M} \right\}. \hfill \\ \end{gathered} $$

(36)

Here, ${\mathbb{D}}_{{{\mathbb{S}}_{i} ,{\mathbb{S}}_{j} }}^{\dag }$ denote the non-negative elements of ${\text{diff}} \left( {{\mathbb{S}}_{i} ,{\mathbb{S}}_{j} } \right)$. The cross-difference set of the two subarrays can be described as

$$ {\mathbb{D}}_{{{\mathbb{S}}_{1} ,{\mathbb{S}}_{2} }}^{\dag } = \left\{ {\left. {\left| {(n - N + P - m)M + (m - 1)} \right|} \right|n = 1,2, \ldots ,N;m = 1,2, \ldots ,M} \right\}. $$

(37)

When $t_{1} = P$, $m = 2,3, \ldots ,t_{1}$, and $n = N - t_{1} + m$ are set, (37) becomes

$$ {\mathbb{D}}_{{{\mathbb{S}}_{1} ,{\mathbb{S}}_{2} }}^{\dag } = \left\{ {\left. {\left| {m - 1} \right|} \right|m = 2,3, \ldots ,P} \right\} = \left\{ {1,2, \ldots ,P - 1} \right\}. $$

(38)

If $t_{2} = 1$, $m = 1,2, \ldots ,P + t_{2}$, and $n = N - P + m - t_{2}$ are used, then

$$ {\mathbb{D}}_{{{\mathbb{S}}_{1} ,{\mathbb{S}}_{2} }}^{\dag } = \left\{ {\left. {\left| {M - m + 1} \right|} \right|m = 1,2, \ldots ,P + 1} \right\} = \left\{ {M,M - 1, \ldots ,M - P} \right\}. $$

(39)

The interval, $[1,M]$, should be continuous to maintain sufficient consecutive lags. According to (38) and (39), we can conclude that $M - P - (P - 1) = M - 2P + 1 \le 1$ must be satisfied. Then, $P \ge {M \mathord{\left/ {\vphantom {M 2}} \right. \kern-0pt} 2}$ can be got. Based on Sect. 3.3, $P$ is set to the minimum integer exceeding or equaling ${M \mathord{\left/ {\vphantom {M 2}} \right. \kern-0pt} 2}$, i.e., $P = \left\lfloor {{{(M + 1)} \mathord{\left/ {\vphantom {{(M + 1)} 2}} \right. \kern-0pt} 2}} \right\rfloor$.

When $t_{1} = P - 1$, $m = 2,3, \ldots ,t_{1}$, and $n = N - t_{1} + m$ are set, we can yield

$$ {\mathbb{D}}_{{{\mathbb{S}}_{1} ,{\mathbb{S}}_{2} }}^{\dag } = \left\{ {\left. {\left| {M + (m - 1)} \right|} \right|m = 2,3, \ldots ,P - 1} \right\} = \left\{ {M + 1,M + 2, \ldots ,M + P - 2} \right\}. $$

(40)

If $t_{2} = 2$, $m = 1,2, \ldots ,P + t_{2}$, and $n = N - P + m - t_{2}$ are utilized, the following formula can be gained:

$$ {\mathbb{D}}_{{{\mathbb{S}}_{1} ,{\mathbb{S}}_{2} }}^{\dag } = \left\{ {\left. {\left| {2M - (m - 1)} \right|} \right|m = 1,2, \ldots ,P + 2} \right\} = \left\{ {2M,2M - 1, \ldots ,2M - P - 1} \right\}. $$

(41)

As $P = \left\lfloor {{{(M + 1)} \mathord{\left/ {\vphantom {{(M + 1)} 2}} \right. \kern-0pt} 2}} \right\rfloor$, $2M - P - 1 - (M + P - 2) = M - 2P + 1 \le 1$ can be concluded, which implies that $[1,2M]$ is continuous. Similarly, when $t_{1} = 2$, $m = t_{1}$, and $n = N - t_{1} + m$ are set, ${\mathbb{D}}_{{{\mathbb{S}}_{1} ,{\mathbb{S}}_{2} }}^{\dag } = \left\{ {(P - 2)M + 1} \right\}$ can be obtained. If we employ $t_{2} = P - 1$, $m = 1,2, \ldots ,P + t_{2}$, and $n = N - P + m - t_{2}$, then

$$ \begin{gathered} {\mathbb{D}}_{{{\mathbb{S}}_{1} ,{\mathbb{S}}_{2} }}^{\dag } = \left\{ {\left. {\left| {(P - 1)M - (m - 1)} \right|} \right|m = 1,2, \ldots ,2P - 1} \right\} \hfill \\ = \left\{ {(P - 1)M,(P - 1)M - 1, \ldots ,(P - 1)M - (2P - 2)} \right\}. \hfill \\ \end{gathered} $$

(42)

Based on $P = \left\lfloor {{{(M + 1)} \mathord{\left/ {\vphantom {{(M + 1)} 2}} \right. \kern-0pt} 2}} \right\rfloor$, $(P - 1)M - (2P - 2) - [(P - 2)M + 1] = M - 2P + 1 \le 1$ can be derived. Therefore, as $t_{1}$ changes from $P$ to 2 while $t_{2}$ varies from 1 to $P - 1$, we can determine that $[1,(P - 1)M]$ is consecutive.

When $m = 2,3, \ldots ,M$, $n = N - M + m - t_{3}$, and $t_{3} = 0,1, \ldots ,N - M + 1$, it can be derived as

$$ \begin{gathered} {\mathbb{D}}_{{{\mathbb{S}}_{1} ,{\mathbb{S}}_{2} }}^{\dag } = \left\{ {\left. {\left| {(M - P + t_{3} )M - (m - 1)} \right|} \right|m = 2,3, \ldots ,M;t_{3} = 0,1, \ldots ,N - M + 1} \right\} \hfill \\ \left\{ {(M - P)M - 1,(M - P)M - 2, \ldots ,(M - P)M - (M - 1)} \right\} \cup \hfill \\ \left\{ {(M - P + 1)M - 1,(M - P + 1)M - 2, \ldots ,(M - P + 1)M - (M - 1)} \right\} \cup \hfill \\ \cdots \hfill \\ \left\{ {(N - P + 1)M - 1,(N - P + 1)M - 2, \ldots ,(N - P + 1)M - (M - 1)} \right\}, \hfill \\ \end{gathered} $$

(43)

where the elements of each set are generated as $m$ varies from 2 to $M$; meanwhile, different sets are obtained as $t_{3}$ changes from 0 to $N - M + 1$. Because $M - 2P \le 0$, $(M - P)M - (M - 1) - (P - 1)M = (M - 2P)M + 1 \le 1$ can be concluded. Consequently, $[1,(N - P + 1)M]$ is continuous. If we set $t_{3} = N - M + 2$, $n = N - M + m - t_{3} = m - 2$ can be yielded. Therefore, in this case, $m \ge 3$ (i.e., $m = 3,4, \ldots ,M$) needs to be satisfied. Then, the following formula is derived:

$$ {\mathbb{D}}_{{{\mathbb{S}}_{1} ,{\mathbb{S}}_{2} }}^{\dag } = \left\{ {(N - P + 2)M - 2,(N - P + 2)M - 3, \ldots ,(N - P + 2)M - (M - 1)} \right\}. $$

(44)

Thus, $[1,(N - P + 2)M - 2]$ is consecutive, and ${\text{uDOF}} = 2(N - P + 2)M - 3$ is obtained. This completes the proof.

Appendix B: Proofs of Property 2

The derivation of DOF is divided into two parts; one part involves the uDOF, and the other contains the discontinuous elements in the difference coarray, excluding uDOF. According to Appendix A, the number of positive consecutive lags, $(N - P + 2)M - 2$, is coincidentally the location of the third sensor in subarray 2. Thus, when $m = 4,5, \ldots ,M$ and $n = 1,2, \ldots ,M - 3$, the elements exceeding the uDOF in the cross-difference set are given by

$$ \left\{ {\begin{array}{*{20}l} {\left\{ {(N - P + 2)M - 3} \right\},\quad {\text{if}} \quad m = 4,n = 1} \hfill \\ {\left\{ {(N - P + 4)M - 4,(N - P + 3)M - 4} \right\},\quad {\text{if}} \quad m = 5,n = 1,2} \hfill \\ \ldots \hfill \\ {\left\{ \begin{gathered} (N - P + M - 2)M + 1,(N - P + M - 3)M + 1, \hfill \\ \ldots ,(N - P + 2)M + 1 \hfill \\ \end{gathered} \right\},\quad {\text{if}} \quad m = M,n = 1,2, \ldots ,M - 3.} \hfill \\ \end{array} } \right. $$

(45)

The number of elements in (45) is an equivariant series, and the sum of its terms is ${{(M - 2)(M - 3)} \mathord{\left/ {\vphantom {{(M - 2)(M - 3)} 2}} \right. \kern-0pt} 2}$. Moreover, there are $P - 2$ elements, which can be expressed as $\left\{ {(N - P + 2)M,(N - P + 3)M, \ldots ,(N - 1)M} \right\}$, in the self-difference sets of the two subarrays. In summary, the DOF of the proposed ICA configuration can be formulated as

$$ \begin{gathered} {\text{DOF}} = \underbrace {2(N - P + 2)M - 3}_{{\text{uDOF}}} + 2{{(M - 2)(M - 3)} \mathord{\left/ {\vphantom {{(M - 2)(M - 3)} 2}} \right. \kern-0pt} 2} + 2(P - 2) \hfill \\ \, = 2NM - 2PM - M + M^{2} + 2P - 1 \hfill \\ \, = 2NM + 2P(1 - M) + (M - 1)^{2} + 2(M - 1) - M \hfill \\ \, = 2NM + (M - 1)(M - 2P + 1) - M. \hfill \\ \end{gathered} $$

(46)

If $M$ is odd, then $M - 2P + 1 = 0$; otherwise, $M - 2P + 1 = 1$. This completes the proof.

Appendix C: Proofs of Property 4

For the ICA structure, the entries in the difference coarray, which are associated with the deleted element, can be expressed as

$$ \left\{ {\left. {\left| {nM} \right|} \right|n = 1,2, \ldots ,N - P} \right\} \cup \left\{ {\left. {\left| {m(M - 1)} \right|} \right|m = 1,2, \ldots ,M - 1} \right\}. $$

(47)

The above-mentioned sets can be represented by the self-difference sets of the proposed AIA. Therefore, the uDOF of the AIA will invariably be greater than or equal to the ICA. Further, we consider the cross-difference set between the additional element (i.e., the $M$-th sensor in subarray 2) and subarray 1, given as follows:

$$ \begin{array}{*{20}l} {\left\{ {\left. {\left| {(N - P + M - n + 1)M - 1} \right|} \right|n = 1,2, \ldots ,N - P} \right\} \cup } \hfill \\ {\left\{ {\left. {\left| {(N - P + M - n)M - 1} \right|} \right|n = N - P + 1,N - P + 2, \ldots ,N - 1} \right\},} \hfill \\ \end{array} $$

(48)

where $MM - 1$ is the interrupted term due to the deleted element. According to (19), we can get $\left| {{{3M} \mathord{\left/ {\vphantom {{3M} 2}} \right. \kern-0pt} 2} - (\tilde{N} + 2)} \right| \approx 0$, i.e., $(N - P + 2) \approx {{3M} \mathord{\left/ {\vphantom {{3M} 2}} \right. \kern-0pt} 2} \ne M$. Thus, $(N - P + 2)M - 1$ is not the interrupted term in (48).

Since $2 < M \le N$, $(M - P + 1)M - 1 < (N - P + 2)M - 1 < (N - P + M)M - 1$ can be derived. This implies that $(N - P + 2)M - 1$ is inclusive of (48). Hence, $[1,(N - P + 2)M - 1]$ is continuous. If $P = 2$, the term, $(N - P + 2)M$, does not exist in the difference coarray, and the uDOF is equal to $2(N - P + 2)M - 1$. When $P \ge 3$, $(N - P + 2)M$ is present. If $m = 3,4, \ldots ,M - 1$ and $n = m - 2$ are applied, the cross-difference set can be constructed as

$$ \begin{aligned} {\mathbb{D}}_{{{\mathbb{S}}_{3} ,{\mathbb{S}}_{4} }}^{\dag } = & \left\{ {\left. {\left| {(N - P + m - n + 1)M - m} \right|} \right|m = 3,4, \ldots ,M - 1} \right\} \\ = & \left\{ {(N - P + 3)M - 3,(N - P + 3)M - 4, \ldots ,(N - P + 2)M + 1} \right\}. \\ \end{aligned} $$

(49)

Therefore, $[1,(N - P + 3)M - 3]$ is consecutive, and ${\text{uDOF}} = 2(N - P + 3)M - 5$ can be gained. This completes the proof.

Appendix D: Proofs of Property 5

First, we consider the general case of $P \ge 3$. Similar to Appendix B, it is obvious that the number of positive consecutive lags, $(N - P + 3)M - 3$, is the position of the fourth element in subarray 2. Next, if $m = 5,6, \ldots ,M - 1$ and $n = 1,2, \ldots ,M - 4$, the elements exceeding the uDOF in the cross-difference set has the following general form:

$$ \left\{ {\begin{array}{*{20}l} {\begin{array}{*{20}l} {\left\{ {(N - P + 5)M - 5} \right\},} \hfill & {\quad {\text{if}} \quad m = 5,n = 1} \hfill \\ {\left\{ {(N - P + 6)M - 6,(N - P + 5)M - 6} \right\},} \hfill & {\quad {\text{if}} \quad m = 6,n = 1,2} \hfill \\ \end{array} } \hfill \\ \cdots \hfill \\ {\begin{array}{*{20}l} {\left\{ \begin{gathered} (N - P + M - 2)M + 1,(N - P + M - 3)M + 1, \hfill \\ \quad \ldots ,(N - P + 3)M + 1 \hfill \\ \end{gathered} \right\},} \hfill & {\quad {\text{if}} \, m = M - 1,n = 1,2, \ldots ,M - 4} \hfill \\ \end{array} .} \hfill \\ \end{array} } \right. $$

(50)

The total number of elements in (50) is ${{(M - 3)(M - 4)} \mathord{\left/ {\vphantom {{(M - 3)(M - 4)} 2}} \right. \kern-0pt} 2}$. Furthermore, the DOF generated by the M-th sensor of subarray 2 can be expressed in the following set:

$$ \left\{ \begin{gathered} (N - P + M)M - 1,(N - P + M - 1)M - 1, \hfill \\ \quad \ldots ,(N - P + 3)M - 1 \hfill \\ \end{gathered} \right\},\quad {\text{if}} \, m = M,n = 1,2, \ldots ,M - 2. $$

(51)

Similarly, for the self-difference sets of two subarrays, the number of elements exceeding the uDOF is $P - 3$, denoted as $\left\{ {(N - P + 3)M,(N - P + 4)M, \ldots ,(N - 1)M} \right\}$. Thus, the DOF of the AIA is summarized as

$$ \begin{aligned} {\text{DOF}} = & \underbrace {{2(N - P + 3)M - 5}}_{{{\text{uDOF}}}} + 2(M - 3)(M - 4)/2 + 2(M - 2) + 2(P - 3) \\ = & 2NM - 2PM + M + M^{2} + 2P - 3 = 2NM + 2P(1 - M) + (M - 1)^{2} + 2(M - 1) + M - 2 \\ = & 2NM + (M - 1)(M - 2P + 1) + M - 2. \\ \end{aligned} $$

(52)

If $M$ is odd, then ${\text{DOF}} = 2NM + M - 2$; otherwise, ${\text{DOF}} = 2NM + 2M - 3$.

Subsequently, the special case of $P = 2$ is considered. When $M = 3$ is used, the term exceeding the uDOF in the cross-difference set is given by $\left\{ {(N - P + M)M - 1} \right\},\quad {\text{if}} \, m = 3,n = 1$. Then, the DOF can be obtained as

$$ \begin{aligned} {\text{DOF}} = & \underbrace {{2(N - P + 2)M - 1}}_{{{\text{uDOF}}}} + 2 \\ = & 2NM + 1 = 2NM + M - 2. \\ \end{aligned} $$

(53)

In the case of $M = 4$, the elements beyond the uDOF are as follows:

$$ \left\{ {\begin{array}{*{20}l} {\left\{ {(N - P + 3)M - 3} \right\},\quad {\text{if}} \, m = 3,n = 1} \hfill \\ {\left\{ {(N - P + 4)M - 1,(N - P + 3)M - 1} \right\},\quad {\text{if}} \, m = 4,n = 1,2} \hfill \\ \end{array} .} \right. $$

(54)

Thus, the DOF can be calculated as

$$ \begin{aligned} {\text{DOF}} = & \underbrace {{2(N - P + 2)M - 1}}_{{{\text{uDOF}}}} + 6 \\ = & 2NM + 5 = 2NM + 2M - 3. \\ \end{aligned} $$

(55)

According to the above derivation, it can be observed that the approaches to calculating the DOF are consistent in all cases. This completes the proof.