Distributed group cooperative guidance for multiple missiles with fixed and switching directed communication topologies

Abstract

Distributed group cooperative guidance (DGCG) problems for multiple missiles with fixed and switching directed communication topologies are dealt with. In contrast to traditional cooperative guidance utilized to intercept one single target, the group cooperative guidance is applicable to multiple targets. In the group cooperative attack, multi-missile system is classified into multiple subgroups, and each subgroup attacks its respective target. Missiles in the same subgroup are required to reach the target simultaneously, while different subgroups may cooperate successively in terms of the impact time. To realize the group cooperative attack, a two-step guidance strategy is developed. The first step adopts a DGCG law based on local neighboring information for multiple missiles to realize group consensus on range-to-go and leading angle. The second step begins when the group consensus of missiles is achieved. During this step, each missile attacks the target under the proportional navigation guidance law. Derived from feedback linearization and convergence analysis, sufficient conditions for multiple missiles to realize the group cooperative attack with fixed and switching directed communication topologies are proposed, respectively. Finally, numerical examples with two subgroups are given to show the effectiveness of the two-step guidance strategy.

Appendix

Fig. 11

Variables used in the proof of Lemma 2

Proof of Lemma 2

Inspired by [40], geometric arguments are used in the proof of Lemma 2. Note that $Re (\rho ) < 0$, and any $\rho$ located in the third quadrant can be regarded as a complex conjugate of $\rho$ located in the second quadrant. Consequently, we only need to consider $\rho$ located in the second quadrant of the complex plane. The variables in Fig. 11 are given as follows

$$\begin{aligned}&\left\{ {\begin{array}{*{20}{c}} \begin{aligned} &{}{{\xi _1} = {k_c}\beta \rho }\\ &{}{{\xi _2} = \sqrt{{{\left( {{k_c}\beta \rho } \right) }^2} + 4{k_c}\rho } } \end{aligned} \end{array}} \right. \end{aligned}$$

(A1)

$$\begin{aligned}&\left\{ {\begin{array}{*{20}{c}} \begin{aligned} &{}{{\zeta _1} = {k_c}\beta \rho - {k_c}\alpha }\\ &{}{{\zeta _2} = \sqrt{{{\left( {{k_c}\beta \rho - {k_c}\alpha } \right) }^2} + 4{k_c}\rho } } \end{aligned} \end{array}} \right. \end{aligned}$$

(A2)

$$\begin{aligned}&\left\{ {\begin{array}{*{20}{c}} \begin{aligned} &{}{{\chi _1} = {{\left( {{k_c}\beta \rho } \right) }^2}}\\ &{}{{\chi _2} = {{\left( {{k_c}\beta \rho } \right) }^2} + 4{k_c}\rho }\\ &{}{{\chi _3} = 4{k_c}\rho } \end{aligned} \end{array}} \right. \end{aligned}$$

(A3)

$$\begin{aligned}&\left\{ {\begin{array}{*{20}{c}} \begin{aligned} &{}{{\delta _1} = {{\left( {{k_c}\beta \rho - {k_c}\alpha } \right) }^2}}\\ &{}{{\delta _2} = {{\left( {{k_c}\beta \rho - {k_c}\alpha } \right) }^2} + 4{k_c}\rho }\\ &{}{{\delta _3} = 4{k_c}\rho } \end{aligned} \end{array}} \right. \end{aligned}$$

(A4)

In the following, it will be verified that if ${k_c} > 0$, $Re \left( \rho \right) < 0$, $\alpha \ge 0$ and

$$\begin{aligned} \beta > \sqrt{\frac{2}{{{k_c}\left| \rho \right| \cos \left( {{{\tan }^{ - 1}}\frac{{\hbox { Im } (\rho )}}{{ - Re(\rho )}}} \right) }}} \end{aligned}$$

then

$$\begin{aligned}&Re ({\lambda _ \pm }) \\&\quad = Re \left( {\frac{{{k_c}\beta \rho - {k_c}\alpha \pm \sqrt{{{\left( {{k_c}\beta \rho - {k_c}\alpha } \right) }^2} + 4{k_c}\rho } }}{2}} \right) < 0 \end{aligned}$$

The following two cases will be studied.

• (i) $\alpha = 0$

Considering $\rho$ locates in the second quadrant of the complex plane, $Re({\lambda _ \pm }) = Re \left( {{\xi _1} \pm {\xi _2}} \right) < 0$ can be verified by showing that $Re ({\xi _2}) < Re ({\xi _1})$. To cope with this, the triangle composed of ${\chi _1}$, ${\chi _2}$ and ${\chi _3}$ is first considered. From Fig. 11, ${\xi _1}$ and ${\chi _1}$ can be expressed in polar coordinates as $\left( {\left| {{\xi _1}} \right| ,\arg ({\xi _1})} \right)$ and $\left( {\left| {{\chi _1}} \right| ,\arg ({\chi _1})} \right)$, respectively. Note that ${\gamma _1} = \arg ({\xi _1})$ and ${\gamma _1} + {\varphi _2} = \arg ({\xi _1}) + {\gamma _2} = \pi$, one obtains ${\varphi _2} = {\gamma _2}$, meaning that

$$\begin{aligned} {\varphi _2} = {\tan ^{ - 1}}\frac{{\hbox { Im } (\rho )}}{{ - Re(\rho )}} \end{aligned}$$

(A5)

Based on the law of cosines, one knows

$$\begin{aligned} {\left| {{\chi _2}} \right| ^2} = {\left| {{\chi _1}} \right| ^2} + {\left| {{\chi _3}} \right| ^2} - 2\left| {{\chi _1}} \right| \left| {{\chi _3}} \right| \cos {\varphi _2} \end{aligned}$$

(A6)

Meanwhile, it can be verified from Eqs. (A1), (A3) and (A5) that if ${k_c} > 0$, $Re \left( \rho \right) < 0$, $\alpha \ge 0$ and

$$\begin{aligned} \beta > \sqrt{\frac{2}{{{k_c}\left| \rho \right| \cos \left( {{{\tan }^{ - 1}}\frac{{\hbox { Im } (\rho )}}{{ - Re(\rho )}}} \right) }}} \end{aligned}$$

then

$$\begin{aligned} {\left| {{\chi _3}} \right| ^2} - 2\left| {{\chi _1}} \right| \left| {{\chi _3}} \right| \cos {\varphi _2} < 0 \end{aligned}$$

(A7)

Substitute Eqs. (A7) into (A6), one obtains $\left| {{\chi _2}} \right| < \left| {{\chi _1}} \right|$, that is

$$\begin{aligned} \left| {{\xi _2}} \right| < \left| {{\xi _1}} \right| \end{aligned}$$

(A8)

Furthermore, note that ${\chi _2} = {\chi _1} + {\chi _3}$, and ${\chi _3}$ locates in the second quadrant, one has $\arg ({\chi _2}) < \arg ({\chi _1})$, meaning that

$$\begin{aligned} \arg ({\xi _2}) < \arg ({\xi _1}) \end{aligned}$$

(A9)

Thus, based on Eqs. (A8) and (A9), one knows

$$\begin{aligned} Re \left( {{\lambda _ \pm }} \right)= & {} Re \left( {\frac{{{\xi _1} \pm {\xi _2}}}{2}} \right) \\= & {} Re \left( {\frac{{{k_c}\beta \rho \pm \sqrt{{{\left( {{k_c}\beta \rho } \right) }^2} + 4{k_c}\rho } }}{2}} \right) < 0 \end{aligned}$$

• (ii) $\alpha > 0$

This case is similar to the case of $\alpha = 0$. $Re \left( {{\zeta _1} \pm {\zeta _2}} \right) < 0$ can be verified by showing that $Re ({\zeta _2}) < Re ({\zeta _1})$. From Fig. 11, ${\xi _1}$, ${\zeta _1}$, ${\chi _1}$ and ${\delta _1}$ can be expressed in polar coordinates as $\left( {\left| {{\xi _1}} \right| ,\arg ({\xi _1})} \right)$, $\left( {\left| {{\zeta _1}} \right| ,\arg ({\zeta _1})} \right)$, $\left( {\left| {{\chi _1}} \right| ,\arg ({\chi _1})} \right)$ and $\left( {\left| {{\delta _1}} \right| ,\arg ({\delta _1})} \right)$. Based on Eqs. (A1)–(A4), one can obtain that $\left| {{\chi _1}} \right| = {\left| {{\xi _1}} \right| ^2}$, $\left| {{\delta _1}} \right| = {\left| {{\zeta _1}} \right| ^2}$, $\arg ({\chi _1}) = 2\arg ({\xi _1})$ and $\arg ({\delta _1}) = 2\arg ({\zeta _1})$. Since $\alpha> 0, {k_c} > 0$ and ${\zeta _1} = {\xi _1} - {k_c}\alpha$, one gets $\arg ({\zeta _1}) > \arg ({\xi _1})$, meaning that $\arg ({\delta _1}) > \arg ({\chi _1})$. Furthermore, note that ${\chi _3} = {\delta _3} = 4{k_c}\rho$, one can obtain ${\psi _2} < {\varphi _2}$. From Fig. 11, by comparing the triangles composed of ${\chi _1},{\chi _2},{\chi _3}$ and ${\delta _1},{\delta _2},{\delta _3}$, it can be verified that $\left| {{\delta _2}} \right| < \left| {{\delta _1}} \right|$. Then, via an analysis similar to that described in the case of $\alpha = 0$, one knows

$$\begin{aligned} Re \left( {{\lambda _ \pm }} \right)= & {} Re \left( {\frac{{{\zeta _1} \pm {\zeta _2}}}{2}} \right) \\= & {} Re \left( {\frac{{{k_c}\beta \rho - \alpha {k_c} \pm \sqrt{{{\left( {{k_c}\beta \rho - \alpha {k_c}} \right) }^2} + 4{k_c}\rho } }}{2}} \right) \\< & {} 0 \end{aligned}$$

The proof of Lemma 2 is completed. $\square$