In this section, we first introduce the structure of MTNA, which can be used directly for VCAM method to obtain the SDCA with maximum continuous DOFs. Then, under the condition of maximizing the continuous DOFs, we give the positions of redundant sensors existed in MTNA. After eliminating the redundancy, NNAs are finally proposed. Compared with other sparse arrays, NNAs possess larger physical apertures and continuous DOFs.

#### 3.1. Introduction of Modified Translational Nested Array (MTNA)

From Reference [

25], we know that the sensor positions of translational NA can be expressed as

where

${\mathbb{S}}_{1,{a}_{1}}$ and

${\mathbb{S}}_{2,{a}_{2}}$ represent subarray 1 and subarray 2 of translational NA, respectively.

${N}_{1}\in {\mathbb{Z}}^{+}$ and

${N}_{2}\in {\mathbb{Z}}^{+}$ respectively denote the sensor number of

${\mathbb{S}}_{1,{a}_{1}}$ and

${\mathbb{S}}_{2,{a}_{2}}$, while

${a}_{1}$ and

${a}_{2}$ are the corresponding translation distance. Note that

${\mathbb{Z}}^{+}$ denotes the positive integer set.

It is obvious that the prototype NA is a special case of translational NA with

${a}_{1}={a}_{2}=0$. According to Theorem 1 in [

25], we know that

${a}_{1}$ and

${a}_{2}$ should satisfy the following relationships for making the SDCA of translational NA possess the maximum continuous DOFs.

R1: ${a}_{1}={a}_{2}=\lfloor \left(\left({N}_{1}+1\right){N}_{2}-2\right)/2\rfloor $.

R2: ${a}_{1}={a}_{2}=-\lfloor \left({N}_{1}+1\right)\left(2{N}_{2}+1\right)/2\rfloor $.

R3: ${a}_{1}=\left({N}_{1}+1\right){N}_{2}+{a}_{2}$, ${a}_{2}=-\lceil \left({N}_{1}+1\right)/2\rceil $.

R4: ${a}_{1}=\left({N}_{1}+1\right){N}_{2}+{a}_{2}$, ${a}_{2}=-\lfloor \left(\left({N}_{1}+1\right)\left(3{N}_{2}+2\right)-2\right)/2\rfloor $.

For the above four cases,

$\lfloor \ast \rfloor $ and

$\lceil \ast \rceil $ denote the rounding to integer operations, where

$\lfloor \ast \rfloor \le \ast $ and

$\lceil \ast \rceil \ge \ast $. However, to make the SDCA keep the full continuous characteristic, only the values of

${a}_{1}$ and

${a}_{2}$ provided in R2 and R3 can be utilized. In addition, corollary 1 in [

25] indicates that translational NA structures in R2 and R3 are mirror symmetric about zero point. Therefore, we just need to consider R3 in this paper for the convenience of analysis. Then, Equation (14) can be denoted as:

where

$a=-\lceil \left({N}_{1}+1\right)/2\rceil $. When the equivalent received array

$\tilde{\mathbb{S}}$ is the union of

${\mathbb{S}}_{1,a}$ and

${\mathbb{S}}_{2,a}$, it is clear from Definition 1 that SDCA can be expressed directly as:

where

$M={N}_{1}{N}_{2}+{N}_{1}+{N}_{2}$. Obviously, SDCA expressed in Equation (16) is completely continuous. From Definition 2 we know that the number of continuous DOFs of SDCA is:

Although the translational NA denoted by (15) can generate a continuous SDCA, it cannot be used directly as the received array. The reason has been mentioned in

Section 2 that only the equivalent received array

$\tilde{\mathbb{S}}$ has a direct connection with SDCA. Specifically, since the elements of

$\tilde{\mathbb{S}}$ are obtained by performing the difference operation on those of

$\mathbb{S}$, it is obvious that at least one element of

$\mathbb{S}$ should be selected as the subtrahend. Nevertheless, observing Equation (15), we can find that all of elements in

${\mathbb{S}}_{1,a}$ and

${\mathbb{S}}_{2,a}$ are always greater than or equal to one. Thus, when

$\mathbb{S}$ is the union of

${\mathbb{S}}_{1,a}$ and

${\mathbb{S}}_{2,a}$,

$\tilde{\mathbb{S}}$ cannot have the same form as

$\mathbb{S}$ regardless of what the index set

$\mathbb{N}$ is selected as. In this way, the resulting SDCA can no longer possess the maximum continuous DOFs. To solve the above problem, we modify the translational NA in (15) as follows.

**Definition** **3.** (

MTNA).

Let the physical sensor number of subarrays in translational NA are ${N}_{1}$ and ${N}_{2}$, respectively, then the MTNA can be defined aswhere the elements of ${\mathbb{S}}_{MTNA}$ are sorted in ascending order. ${\mathbb{S}}_{1,a}$ and ${\mathbb{S}}_{2,a}$ are defined in Equation (15), and $a=-\lceil \left({N}_{1}+1\right)/2\rceil $.It is obvious that MTNA is the union of

$\left\{0\right\}$,

${\mathbb{S}}_{1,a}$, and

${\mathbb{S}}_{2,a}$. So, the total number of sensors in MTNA is

$N={N}_{1}+{N}_{2}+1$. In addition, based on the description in

Section 2, it is easy to define the index sets as

$\mathbb{M}=\left\{2,3,\cdots ,\mathrm{N}\right\}$ and

$\mathbb{N}=\left\{1,1,\cdots ,1\right\}$, where

$\mathbb{N}$ contains

${N}_{1}+{N}_{2}$ identical elements, i.e., 1. Combining (18) with its index sets, we have

$\tilde{\mathbb{S}}={\mathbb{S}}_{1,a}{\displaystyle \cup}{\mathbb{S}}_{2,a}$, where the corresponding SDCA possesses the maximum continuous DOFs.

Next, we consider an example with ${N}_{1}={N}_{2}=3$ to illustrate the above analysis. In this example, sensor positions of subarrays in translational NA can be expressed as ${\mathbb{S}}_{1,-2}=\left\{11,12,13\right\}$ and ${\mathbb{S}}_{2,-2}=\left\{2,6,10\right\}$, respectively. From Definition 3, we can get ${\mathbb{S}}_{\mathrm{MTMA}}=\left\{0,2,6,10,11,12,13\right\}$. Then, letting $\mathbb{M}=\left\{2,3,4,5,6,7\right\}$ and $\mathbb{N}=\left\{1,1,1,1,1,1\right\}$, we have $\tilde{\mathbb{S}}=\left\{2,6,10,11,12,13\right\}$ and ${\mathbb{V}}_{\mathrm{SDCA}}=\{-26,-25,\cdots ,25,26\}$. It is obvious that the resulting SDCA is fully continuous, and the corresponding number of continuous DOFs is equal to 53.

#### 3.2. Redundancy Analysis of MTNA

From the previous subsection, we know that SDCA of MTNA has the maximum continuous DOFs. However, to achieve this goal, the optimal selections of

${N}_{1}$ and

${N}_{2}$ need to be determined first when we know the total number of sensors

$N$. Accordingly, we build the following optimization problem:

Since the specific value of $a$ is related with the parity of ${N}_{1}$, we can obtain multiple different solutions of Equation (19) provided below.

S1: If $N=4k$, we have ${N}_{1}=2k-1$ and ${N}_{2}=2k$. Then, ${L}_{\mathrm{c}}=16{k}^{2}+4k-3$.

S2: If $N=4k+1$, we have ${N}_{1}=2k$ and ${N}_{2}=2k$, or ${N}_{1}=2k-1$ and ${N}_{2}=2k+1$. Then, ${L}_{\mathrm{c}}=16{k}^{2}+12k-3$.

S3: If $N=4k+2$, we have ${N}_{1}=2k$ and ${N}_{2}=2k+1$, or ${N}_{1}=2k+1$ and ${N}_{2}=2k$. Then, ${L}_{\mathrm{c}}=16{k}^{2}+20k+1$.

S4: If $N=4k+3$, we have ${N}_{1}=2k+1$ and ${N}_{2}=2k+1$. Then, ${L}_{c}=16{k}^{2}+28k+9$.

Note that $k$ is a positive integer for the above solutions. Since both S2 and S3 can be divided into two different solutions, it is clear that there exist six different selections about ${N}_{1}$ and ${N}_{2}$ to maximize ${L}_{\mathrm{c}}$. Observing ${N}_{1}$ and ${N}_{2}$ in S1–S4 again, we find that they can also be divided into four different cases from the view of parity property. Accordingly, we derive the following property of MTNA involving redundant sensors.

**Property** **1.** For MTNA, its redundant sensors can be analyzed under four different combinations of${N}_{1}$and${N}_{2}$, i.e.,

C1: If${N}_{1}\ge 4$and${N}_{2}\ge 4$are even,$\left\{{\mathbb{S}}_{1,a}\left(i\right)\right|i=1,2,\cdots ,{N}_{1}/2-1\}$or$\left\{{\mathbb{S}}_{1,a}\left(i\right)\right|i={N}_{1}/2+1,{N}_{1}/2+2,\cdots ,{N}_{1}-1\}$, and$\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i=2,3,\cdots ,{N}_{2}/2,{N}_{2}\}$or$\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i={N}_{2}/2+1,{N}_{2}/2+2,\cdots ,{N}_{2}\}$in MTNA are redundant sensors. Then, the total number of redundant sensors is$\left({N}_{1}+{N}_{2}-2\right)/2$.

C2: If${N}_{1}\ge 4$is even and${N}_{2}\ge 5$is odd,$\left\{{\mathbb{S}}_{1,a}\left(i\right)\right|i=1,2,\cdots ,{N}_{1}/2-1\}$or$\left\{{\mathbb{S}}_{1,a}\left(i\right)\right|i={N}_{1}/2+1,{N}_{1}/2+2,\cdots ,{N}_{1}-1\}$, and$\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i=2,3,\cdots ,\left({N}_{2}-1\right)/2,{N}_{2}\}$or$\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i=\left({N}_{2}+3\right)/2,\left({N}_{2}+5\right)/2,\cdots ,{N}_{2}\}$in MTNA are redundant sensors. Then, the total number of redundant sensors is$\left({N}_{1}+{N}_{2}-3\right)/2$.

C3: If${N}_{1}\ge 3$is odd and${N}_{2}\ge 4$is even,$\left\{{\mathbb{S}}_{1,a}\left(i\right)\right|i=1,2,\cdots ,\left({N}_{1}-1\right)/2\}$or$\left\{{\mathbb{S}}_{1,a}\left(i\right)\right|i=\left({N}_{1}+3\right)/2,\left({N}_{1}+5\right)/2,\cdots ,{N}_{1}\}$, and$\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i=2,3,\cdots ,{N}_{2}/2,{N}_{2}\}$or$\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i={N}_{2}/2+1,{N}_{2}/2+2,\cdots ,{N}_{2}\}$in MTNA are redundant sensors. Then, the total number of redundant sensors is$\left({N}_{1}+{N}_{2}-1\right)/2$.

C4: If${N}_{1}\ge 3$and${N}_{2}\ge 5$are odd,$\left\{{\mathbb{S}}_{1,a}\left(i\right)\right|i=1,2,\cdots ,\left({N}_{1}-1\right)/2\}$or$\left\{{\mathbb{S}}_{1,a}\left(i\right)\right|i=\left({N}_{1}+3\right)/2,\left({N}_{1}+5\right)/2,\cdots ,{N}_{1}\}$, and$\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i=2,3,\cdots ,\left({N}_{2}-1\right)/2,{N}_{2}\}$or$\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i=\left({N}_{2}+3\right)/2,\left({N}_{2}+5\right)/2,\cdots ,{N}_{2}\}$in MTNA are redundant sensors. Then, the total number of redundant sensors is$\left({N}_{1}+{N}_{2}-2\right)/2$.

From Property 1, we can find that although there exist redundant sensors in C1–C4, the number of redundant sensors in C3 is the largest compared to the other three cases, which implies that we can remove more redundant sensors as long as

${N}_{1}$ is odd and

${N}_{2}$ is even. In order to visually illustrate this interesting phenomenon,

Figure 1 depicts two examples, where the total number of sensors is fixed to be 10. According to S3, we can confirm that there exist two different solutions, i.e.,

${N}_{1}=4$ and

${N}_{2}=5$, or

${N}_{1}=5$ and

${N}_{2}=4$. Then, based on Property 1, if

${N}_{1}=4$ and

${N}_{2}=5$, the redundant sensors of MTNA as shown in

Figure 1a can be expressed as

$\left\{23\right\}$ or

$\left\{25\right\}$, and

$\left\{7,22\right\}$ or

$\left\{17,22\right\}$. It is clear that the total number of redundant sensors is 3. Conversely, if

${N}_{1}=5$ and

${N}_{2}=4$, the redundant sensors of MTNA as shown in

Figure 1b can be denoted as

$\left\{22,23\right\}$ or

$\left\{25,26\right\}$, and

$\left\{9,21\right\}$ or

$\left\{15,21\right\}$, where the total number of redundant sensors is 4. Note that, the number of rest of sensors for MTNAs shown in

Figure 1a,b is 7 and 6, respectively. And, by constructing the time average vectors as described in

Section 2, the remaining sensors of MTNAs shown in

Figure 1a,b can generate a same SDCA with the maximum continuous DOFs. Hence, for a known total sensor number

$N$, when

${N}_{1}$ and

${N}_{2}$ are respectively odd and even, sensors except for redundant ones in MTNA can generate the SDCA with the largest continuous DOFs.

#### 3.3. The Proposed Novel Nested Arrays (NNAs)

As aforementioned, if ${N}_{1}$ is odd and ${N}_{2}$ is even, the number of removable sensors in MTNA becomes the largest. Observing C3 mentioned in Property 1, we know that both ${\mathbb{S}}_{1,a}$ and ${\mathbb{S}}_{2,a}$ contain two-part alternative redundant sensors. So, there exist four different combinations for the rest of sensors in ${\mathbb{S}}_{1,a}$ and ${\mathbb{S}}_{2,a}$. Based on this, four kinds of novel nested arrays (NNAs) are defined below.

**Definition** **4.** (NNAs). Given parameters ${N}_{1}\ge 3$ and ${N}_{2}\ge 4$, where ${N}_{1}$ is odd and ${N}_{2}$ is even, then four kinds of NNAs are defined as follows:

(1): If$\left\{{\mathbb{S}}_{1,a}\left(i\right)\right|i=1,2,\cdots ,{N}_{1}/2-1\}$and$\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i=2,3,\cdots ,{N}_{2}/2,{N}_{2}\}$in${\mathbb{S}}_{\mathrm{MTNA}}$are removed, then the first kind of NNA (i.e., NNA-I) can be expressed as:where${\overline{\mathbb{S}}}_{1,a}=\left\{{\mathbb{S}}_{1,a}\left(i\right)\right|i=\left({N}_{1}+1\right)/2,\left({N}_{1}+3\right)/2,\cdots ,{N}_{1}\}$,

${\overline{\mathbb{S}}}_{2,a}=\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i=1,{N}_{2}/2+1,{N}_{2}/2+2,\cdots ,{N}_{2}-1\}$.

(2): If$\left\{{\mathbb{S}}_{1,a}\left(i\right)\right|i=1,2,\cdots ,{N}_{1}/2-1\}$and$\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i={N}_{2}/2+1,{N}_{2}/2+2,\cdots ,{N}_{2}\}$in${\mathbb{S}}_{\mathrm{MTNA}}$are removed, then the second kind of NNA (i.e., NNA-II) can be expressed as:where${\overline{\mathbb{S}}}_{1,a}=\left\{{\mathbb{S}}_{1,a}\left(i\right)\right|i=\left({N}_{1}+1\right)/2,\left({N}_{1}+3\right)/2,\cdots ,{N}_{1}\}$,${\overline{\mathbb{S}}}_{2,a}=\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i=1,2,\cdots ,{N}_{2}/2\}$. (3): If$\left\{{\mathbb{S}}_{1,a}\left(i\right)\right|i=\left({N}_{1}+3\right)/2,\left({N}_{1}+5\right)/2,\cdots ,{N}_{1}\}$and$\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i=2,3,\cdots ,{N}_{2}/2,{N}_{2}\}$in${\mathbb{S}}_{\mathrm{MTNA}}$are removed, then the third kind of NNA (i.e., NNA-III) can be expressed as:where${\overline{\mathbb{S}}}_{1,a}=\{-{\mathbb{S}}_{1,a}\left(i\right)|i=1,2,\cdots ,\left({N}_{1}-1\right)/2\}{\displaystyle \cup}\left\{{\mathbb{S}}_{1,a}\left(\left({N}_{1}+1\right)/2\right)\right\}$,${\overline{\mathbb{S}}}_{2,a}=\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i=1,{N}_{2}/2+1,{N}_{2}/2+2,\cdots ,{N}_{2}-1\}$. (4): If$\left\{{\mathbb{S}}_{1,a}\left(i\right)\right|i=\left({N}_{1}+3\right)/2,\left({N}_{1}+5\right)/2,\cdots ,{N}_{1}\}$and$\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i={N}_{2}/2+1,{N}_{2}/2+2,\cdots ,{N}_{2}\}$in${\mathbb{S}}_{\mathrm{MTNA}}$are removed, then fourth kind of NNA (i.e., NNA-IV) can be expressed as:where${\overline{\mathbb{S}}}_{1,a}=\{-{\mathbb{S}}_{1,a}\left(i\right)|i=1,2,\cdots ,\left({N}_{1}-1\right)/2\}{\displaystyle \cup}\left\{{\mathbb{S}}_{1,a}\left(\left({N}_{1}+1\right)/2\right)\right\}$,${\overline{\mathbb{S}}}_{2,a}=\left\{{\mathbb{S}}_{2,a}\left(i\right)\right|i=1,2,\cdots ,{N}_{2}/2\}$. It should be noted that, elements of the above NNAs are sorted in ascending order of their respective absolute values. Meanwhile, different index sets $\mathbb{M}$ and $\mathbb{N}$ corresponding to the above four kinds of NNAs are defined as follows.

**Definition** **5.** (

Index Sets).

For NNA-I and NNA-II, the index sets are collectively defined as:whereFor NNA-III and NNA-IV, the index sets are collectively defined as:where Apparently, combining NNAs with their respective index sets, according to the construction principle in Definition 1, we can construct the specific time average vectors so as to obtain the equivalent received array

$\tilde{\mathbb{S}}$, and then the satisfying SDCA with maximum continuous DOFs can be obtained. Nevertheless, although NNAs and index sets are already given in Definitions 4 and 5, the relationship among

${N}_{1}$,

${N}_{2}$, and total number of sensors of NNAs is still indistinct. Hence, before using NNAs for DOA estimation, we need to address this problem first. Note that, it is apparent from Definition 4 that the sensor number of NNAs is

$\overline{N}=\left({N}_{1}+{N}_{2}+3\right)/2$. From Equation (17), we know that the number of continuous DOFs of SDCA is

${L}_{\mathrm{c}}=4\left(M+a\right)+1$, where

$M={N}_{1}{N}_{2}+{N}_{1}+{N}_{2}$ and

$a=-\lceil \left({N}_{1}+1\right)/2\rceil $. Hence, the optimization problem can be constructed as follows:

Combining C3 in Property 1 with Equation (26), it is easy to get the relationship among

$\overline{N}$,

${N}_{1}$,

${N}_{2}$, as well as

${L}_{\mathrm{c}}$, which is as shown in

Table 1.

Then, according to

Table 1 and Definition 4, physical apertures of the proposed four kinds of NNAs can be summarized as follows.

**Property** **2.** For NNA-I and NNA-II with$\overline{N}$sensors, their physical apertures are identical and can be expressed as: While for NNA-III and NNA-IV with$\overline{N}$sensors, their physical apertures are also identical and can be expressed as: It is obvious from Property 2 that NNA-III and NNA-IV possess larger physical aperture than NNA-I and NNA-II for the same sensor number, which means that the former can realize better DOA estimation performance than the latter.

Next, to illustrate the exploitation of the proposed NNAs and index sets more clearly, two examples of NNA-I and NNA-IV are provided as shown in

Figure 2. Let

$\overline{N}$ be 10, then the optimal

${N}_{1}$ and

${N}_{2}$ are 9 and 8, respectively. According to Definition 4, the sensor position sets of NNA-I and NNA-IV can be given as

${\mathbb{S}}_{\mathrm{NNA}-\mathrm{I}}=\left\{0,5,45,55,65,80,81,82,83,84\right\}$ and

${\mathbb{S}}_{\mathrm{NNA}-\mathrm{IV}}=\{0,5,15,25,35,-76,-77,-78,-79,80\}$, respectively. It is evident that physical apertures of NNA-I and NNA-IV are respectively equal to 84 and 159. Shown in

Figure 2a is the physical structure of NNA-I, while

Figure 2b shows the physical structure of NNA-IV. Besides, from Definition 5, we can determine their respective index sets as

${\mathbb{M}}_{1}=\left\{1,1,1,1,1,1,1,1,1,2,2,2,2,2,3,4,5\right\}$,

${\mathbb{N}}_{1}=\{2,3,4,5,6,7,8,9,10,6,7,8,9,10,6,6,6\}$,

${\mathbb{M}}_{2}=\left\{1,1,1,1,1,2,3,4,5,6,7,8,9,6,7,8,9\right\}$, and

${\mathbb{N}}_{2}=\left\{2,3,4,5,10,10,10,10,10,1,1,1,1,2,2,2,2\right\}$. As shown in

Figure 2c, we then obtain the equivalent received arrays of NNA-I and NNA-IV, which are identical and can be expressed as

$\tilde{\mathbb{S}}=\left\{5,15,25,35,45,55,65,75,76,\cdots ,84\right\}$. Finally, according to Equation (12), we can easily obtain this fully continuous SDCA. It is obvious from

Figure 2d that the number of continuous DOFs of SDCA is equal to 337. However, in both of these examples, physical aperture of NNA-IV is larger than that of NNA-I. Thus, we can infer that NNA-IV has better DOA estimation performance than NNA-I.