Optimal placement of barrier coverage in heterogeneous bistatic radar sensor networks

Abstract

Barrier Coverage is an important sensor deployment issue in many industrial, consumer and military applications.The barrier coverage in bistatic radar sensor networks has attracted many researchers recently. The Bistatic Radars (BR) consist of radar signal transmitters and radar signal receivers. The effective detection area of bistatic radar is a Cassini oval area that determined by the distance between transmitter and receiver and the predefined detecting SNR threshold. Many existing works on bistatic radar barrier coverage mainly focus on homogeneous radar sensor networks. However, cooperation among different types or different physical parameters of sensors is necessary in many practical application scenarios. In this paper, we study the optimal deployment problem in heterogeneous bistatic radar networks.The object is how to maximize the detection ability of bistatic radar barrier with given numbers of radar sensors and barrier’s length. Firstly, we investigate the optimal placement strategy of single transmitter and multiple receivers, and propose the patterns of aggregate deployment. Then we study the optimal deployment of heterogeneous transmitters and receivers and introduce the optimal placement sequences of heterogeneous transmitters and receivers. Finally, we design an efficient greedy algorithm, which realize optimal barrier deployment of M heterogeneous transmitters and N receivers on a L length boundary, and maximizing the detection ability of the barrier. We theoretically proved that the placement sequence of the algorithm construction is optimal deployment solution in heterogeneous bistatic radar sensors barrier. And we validate the algorithm effectiveness through comprehensive simulation experiments.

Appendix

Proof

(Proof 1) When a placement sequence consist of one transmitter and one receiver, the coverage length reaches the maximum when the distance between them is $2\sqrt {{c_{i}}}$.

Let d₁ = ∥T_iR_j∥, we consider three possible cases as follows.

• Case 1: when ${d_{1}} < 2\sqrt {{c_{i}}}$, the coverage range is presented in Figure 19.

  We can divide the range into 3 parts: part $\overline {A{T_{i}}}$, part $\overline {{T_{i}}{R_{j}}}$, part $\overline {{R_{j}}B}$. Let ∥AT_i∥ = ∥R_jB∥ = x, because of ∥AT_i∥ ×∥AR_j∥ = c_i, then x (x + d₁) = c_i, so the length of coverage range is ${L_{1}} = \sqrt {{d_{1}^{2}} + 4{c_{i}}}$.

• Case 2: when ${d_{1}} = 2\sqrt {{c_{i}}}$, the coverage range is presented in Figure 20.

  Similar to Case 1, the length of coverage range is ${L_{2}} = \sqrt {{d_{1}^{2}} + 4{c_{i}}}$, then L₂ > L₁, because of $d_{1}^{(2)} > d_{1}^{(1)}$.

• Case 3: when ${d_{1}} > 2\sqrt {{c_{i}}}$, the coverage range is presented in Figure 21.

Figure 19

Detection area when ${d_{1}} < 2\sqrt {{c_{i}}}$

Figure 20

Detection area when ${d_{1}} = 2\sqrt {{c_{i}}}$

Figure 21

Detection area when ${d_{1}} > 2\sqrt {{c_{i}}}$

Because $d_{1}^{(3)} > d_{1}^{(2)}$, then ∥AT_i∥⁽³⁾ = ∥R_jD∥⁽³⁾ < ∥AT_i∥⁽²⁾ = ∥R_jB∥⁽²⁾, and ∥T_iB∥⁽³⁾ = ∥R_jC∥⁽³⁾ < ∥T_iO∥⁽²⁾ = ∥R_jO∥⁽²⁾, so we get that L₃ = 2 (∥AT_i∥⁽³⁾ + ∥T_iB∥⁽³⁾) < L₂ = 2 (∥AT_i∥⁽²⁾ + ∥T_iO∥⁽²⁾). □

Proof

(Proof 2) Optimal deployment interval when the number of deployed receivers is n = 2, as shown in Figure 22.

Figure 22

Optimal deployment interval when n = 2

As shown in Figure 21, let d₁ = ∥T₁R₁∥, d₂ = ∥R₁R₂∥. When ${d_{1}} = 2\sqrt {{c_{i}}}$ and d₂ is the solution of $\frac {1}{2}{d_{2}} \times \left ({\frac {1}{2}{d_{2}} + {d_{1}}} \right ) = {c_{i}}$, the coverage length of the placement sequence is the longest. It is because the following discussion.

When ∥R₁R₂∥ < d₂, it is shown in Figure 23.

Figure 23

Deployment situation when $\left \| {{R_{1}}{R_{2}}} \right \| < {d_{2}}$

Here, compared with Figure 23, ∥T₁D^∗∥ < ∥T₁D∥¹ (the superscript 1 denotes Figure 23), and ∥AT₁∥ remains unchanged, therefore the coverage length of the placement sequence becomes shorter.

When ∥R₁R₂∥ > d₂, it is shown in Figure 24.

Figure 24

Deployment situation when $\left \| {{R_{1}}{R_{2}}} \right \| > {d_{2}}$

∥DE∥ < ∥CD∥¹ (the superscript 1 denote the Figure 22), so the coverage length turns shorter compared with Figure 22. Therefore, if we deploy two receivers on one side of a transmitter, the coverage length is longest when ∥T₁R₁∥ = d₁ and ∥R₁R₂∥ = d₂. □

Proof

(Proof 3) When a placement sequence consists of one transmitter and n receivers, the coverage length is longest when we deploy the n receivers on both sides of the transmitter equally and the deploy intervals satisfied the optimal deployment intervals.

As shown in Figure 25, we assume that we have two deployment strategies: strategy1 and strategy2.

Figure 25

Two deployment strategies

strategy1: |left₁ − right₁| ≤ 1, left₁ + right₁ = n

strategy2: |left₂ − right₂| > 1, left₂ + right₂ = n

left denotes the number of receivers that are deployed on the left of the transmitter, and the right denotes the number of receivers that are deployed on the right of the transmitter.

In Proof 2, if we deploy receivers on one side of the transmitter, the coverage length is the longest if the deployment intervals are the optimal deployment intervals. Therefore, we assume that the deployment intervals in strategy1 or strategy2 all satisfy the requirement of the optimal deployment intervals.

For convenience, we assume left₁ ≤ right₁, left₂ ≤ right₂, we get left₁ > left₂, right₁ < right₂ and left₁ − left₂ = right₂ − right₁. The optimal interval array d[] is a decreasing array, we assume.

sum₁ = d[left₂] + d[left₂ + 1] + ... + d[left₁]

sum₂ = d[right₁] + d[right₁ + 1] + ... + d[right₂]

Therefore, sum₁ > sum₂, and the range from left₂ to right₁ is a shared part of both strategy1 and strategy2. So the coverage length obtained from strategy1 is longer than that from strategy2. □

Proof

(Theorem 1) There are two placement sequences, S₁ and S₂. S₁ satisfies the converge deployment and S₂ don’t satisfies the converge deployment.

Our proof is reduction ad absurdum. We assume S₂ is the optimal placement sequence. Its coverage range covered is the longest. Because S₂ doesn’t satisfy the converge deployment, there is at least one transmitter T_i, the deployment sequence with this transmitter (and its receivers set) doesn’t satisfy the optimal interval deployment. However, S₁ satisfies the converge deployment, the deployment sequence with this transmitter (and its receivers set) satisfies the optimal interval deployment. We obtain the following expression from section 4. 2, ${\text {cov}} er({~}^{1}S_{{T_{i}}}^{{K_{i}}}) > {\text {cov}} er({~}^{2}S_{{T_{i}}}^{{K_{i}}})$ and ${\text {cov}} er({~}^{1}S_{{T_{j}}}^{{K_{j}}}) > {\text {cov}} er({~}^{2}S_{{T_{j}}}^{{K_{j}}}),j = 1,2,...M{\mathrm { and j}} \ne {\mathrm {i}}$. K_i denotes the number of receivers in T_i’s receivers set. ${~}^{1}S_{i}^{{K_{i}}}$ denotes the placement sequence with T_i and its K_i receivers. Therefore, ${\text {cov}} er({S_{1}}) > {\text {cov}} er({S_{2}})$. S₂ is not set up for the optimal placement sequence. □

Proof

(Theorem 2) If there are two placement sequences with the same transmitter parameters of the, and the number of receivers owned by each transmitter is the same. If both placement sequences satisfy the converge deployment, that is to say they have the same basic coverage patterns, therefore their coverage ranges are equal. We give an example to illustrate this conclusion, It is shown in Figure 26.

Figure 26

Comparison of different deployment sequences satisfying converge deployment structure

Both placement sequences are composed of the same three basic coverage patterns. But the arrangement sequences of these three basic coverage patterns are different, Therefore they are two different placement sequences. But the coverage range covered by each basic coverage patterns of the two placement sequences correspond are equal. The coverage range covered by these two placement sequences are equal. □