Guidance law with circular no-fly zone constraint

Abstract

The proposed guidance law is used for guiding a missile against a maneuvering target while satisfying a circular no-fly zone (NFZ) constraint. It consists of two parts: virtual-target guidance (VTG) and boundary-constraint handling scheme (BCHS). In order that the missile avoids the NFZ, VTG first maps the actual target to a virtual one, then obtains the relative motion between the virtual target and missile, and finally uses proportional navigation to steer the missile to the virtual target. The missile also hits the actual target when it hits the virtual target because the virtual and actual targets are coincident at this moment. In some cases, especially when the initial velocity vector of the missile points toward the center of the NFZ, if the evasive action taken by VTG is found to be insufficient, then BCHS will be enabled to keep the missile from entering the NFZ unless the target enters the NFZ.

Appendix

Proposition 1

If $H_0 >0,\sigma _0 \in [-\pi /2,\;\pi /2]$, and the system S1 described by Eqs. (57–58) is under the control of BCHS described by Eqs. (59–61), then the trajectory ($H$, $\sigma$) of the system S1 converges to the stable node $(0, 0)$ while satisfying that $H\ge 0.$

Proof

From Eqs. (57–59), we get a closed-loop system S2 described by

$$\begin{aligned} \dot{H}&= V_M \sin (\sigma ) \end{aligned}$$

(87)

$$\begin{aligned} \dot{\sigma }&= -2\xi \omega _n \sin (\sigma )-\frac{\omega _n^2 H\cos (\sigma )}{V_M } \end{aligned}$$

(88)

The polar coordinates of the system S2 are

$$\begin{aligned} r&= \sqrt{H^{2}+\sigma ^{2}} \end{aligned}$$

(89)

$$\begin{aligned} \theta&= \left\{ {\begin{array}{l} \arctan \left( {\frac{\sigma }{H}} \right) ;\;\quad \quad \quad \hbox {if}\,\, H\ge 0 \\ \arctan \left( {\frac{\sigma }{H}} \right) +\pi ;\;\quad \hbox {if} H<0 \,\,\hbox {and}\,\, \sigma \ge 0 \\ \arctan \left( {\frac{\sigma }{H}} \right) -\pi ;\;\quad \hbox {if} H<0 \,\,\hbox {and} \,\,\sigma <0 \\ \end{array}} \right. \end{aligned}$$

(90)

The time derivatives of $r$ and $\theta$ are

$$\begin{aligned} \dot{r}&= \frac{1}{r}\left[ {HV_M \sin (\sigma )-2\xi \omega _n \sin (\sigma )\sigma -\frac{\omega _n^2 H\cos (\sigma )\sigma }{V_M }} \right] \end{aligned}$$

(91)

$$\begin{aligned} \dot{\theta }&= \frac{1}{r^{2}}\left[ {-2\xi \omega _n H\sin (\sigma )-\frac{\omega _n^2 H^{2}\cos (\sigma )}{V_M }-V_M \sin (\sigma )\sigma } \right] \nonumber \\ \end{aligned}$$

(92)

Now Proposition 1 is proved in two steps:

Step 1: Prove that $H(t)\ge 0$for every $t \ge t_{0}$.

We need to prove that the following sub-proposition is true first.

Sub-proposition 1

The Area $\Omega _1$ described by the following inequalities is an invariant set of the autonomous system S2.

$$\begin{aligned} \left\{ {\begin{array}{l} H\ge 0 \\ -\pi /2\le \sigma \le \pi /2 \\ \sigma \ge -\frac{(\sqrt{\xi ^{2}-1}+\xi )\omega _n }{V_M }H \\ \end{array}} \right. \end{aligned}$$

(93)

Note that the line

$$\begin{aligned} \sigma =-\frac{\left( \sqrt{\xi ^{2}-1}+\xi \right) \omega _n }{V_M }H \end{aligned}$$

(94)

is parallel to one of the eigenvectors of the Jacobian matrix of the system S2 at the origin. Let’s observe the movements of the trajectories starting from the four sides of $\Omega _1$:

When $H\ge 0$ and $\sigma =\pi /2$, from Eq. (88), we get

$$\begin{aligned} \dot{\sigma }=-2\xi \omega _n <0 \end{aligned}$$

(95)

When $H=0$ and $0\le \sigma \le \pi /2$, from Eq. (87), we get

$$\begin{aligned} \dot{H}=V_M \sin (\sigma )\ge 0 \end{aligned}$$

(96)

When $0<H\le {\pi V_M }/{[2(\sqrt{\xi ^{2}-1}+\xi )\omega _n ]}$ and $\sigma ={-(\sqrt{\xi ^{2}-1}+\xi )\omega _n H}/{V_M }$, by substituting $H={-V_M \sigma }/{[(\sqrt{\xi ^{2}-1}+\xi )\omega _n ]}$ into Eq. (92), we can obtain

$$\begin{aligned} \dot{\theta }=\frac{V_M \sigma }{r^{2}(\sqrt{\xi ^{2}-1}+\xi )^{2}}\left[ {\sin (\sigma )-\sigma \cos (\sigma )} \right] \end{aligned}$$

(97)

Observe the function $f(\sigma )=\sin (\sigma )-\sigma \cos (\sigma )$ defined for $-\pi /2\le \sigma <0$. The derivative of $f(\sigma )$ is

$$\begin{aligned} \hbox {d}f(\sigma )\hbox {/d}\sigma =\sigma \sin (\sigma )>0 \end{aligned}$$

(98)

So $f(\sigma )$ is an increasing function of $\sigma$, and then we have that $f(\sigma )<f(0)=0$. Therefore, we have that $\dot{\theta }>0$.

When $H>{\pi V_M }/{[2(\sqrt{\xi ^{2}-1}+\xi )\omega _n ]}$ and $\sigma =-\pi /2$, from Eq. (88), we get

$$\begin{aligned} \dot{\sigma }=2\xi \omega _n >0 \end{aligned}$$

(99)

From the above analysis, it can be concluded that all trajectories starting from the boundary of $\Omega _1$ move into the area $\Omega _1$. So, if the initial point $(H_0 ,\;\sigma _0 )\in \Omega _1$, then the corresponding phase trajectory will remain in the area $\Omega _1$, i.e., $(H(t),\;\sigma (t))\in \Omega _1$for every $t \ge t_{0}$. Thus, Sub-proposition 1 has already been proved.

Now we need to prove that $(H_0 ,\;\sigma _0 )\in \Omega _1$. Since $H_0 >0$ and $\sigma _0 \in [-\pi /2,\;\pi /2]$ are the conditions of Proposition 1, we only need to prove that $(H_0 ,\;\sigma _0 )$ satisfies the third inequality of Eq. (93). Equation (61) can be converted into

$$\begin{aligned} -\sqrt{\xi ^{2}-1}\le \frac{V_M \sigma _0 }{2\omega _n H_0 }\le \sqrt{\xi ^{2}-1} \end{aligned}$$

(100)

Then

$$\begin{aligned} -\frac{(\sqrt{\xi ^{2}-1}+\xi )\omega _n }{V_M }H_0 \le -\frac{2\omega _n H_0 }{V_M }\sqrt{\xi ^{2}-1}\le \sigma _0 \end{aligned}$$

(101)

Thus, we have that $(H_0 ,\;\sigma _0 )\in \Omega _1$. From the above analysis, it can be deduced that $H(t)\ge 0$ for every $t \ge t_{0}$.

Step 2: Prove that if $(H_0 ,\;\sigma _0 )\in \Omega _1$, then $(H,\;\sigma )$ converges to (0, 0).

$\Omega _2$ represents an area on the phase plane, and is determined by

$$\begin{aligned} \left\{ {\begin{array}{l} H\ge 0 \\ -\pi /2\le \sigma \le 0 \\ \sigma \ge -\frac{(\sqrt{\xi ^{2}-1}+\xi )\omega _n }{V_M }H \\ \end{array}} \right. \end{aligned}$$

(102)

where it is obvious that $\Omega _2 \subset \Omega _1$. Now we need to prove that the following sub-proposition is true.

Sub-proposition 2

If $(H_0 ,\;\sigma _0 )\in \Omega _2$, then $(H,\;\sigma )$ converges to (0, 0).

Since two sides of $\Omega _2$ are the same as that of $\Omega _1$, we only need to observe the movements of the trajectories starting from the points satisfying that$H\ge 0$ and $\sigma =0$. When $H\ge 0$ and $\sigma =0$, from Eq. (88), we get

$$\begin{aligned} \dot{\sigma }=-\frac{\omega _n^2 H}{V_M }\le 0 \end{aligned}$$

(103)

Thus, $\Omega _2$ is also an invariant set of the autonomous system S2. Define a differentiable function $V(H,\;\sigma )$ as

$$\begin{aligned} V(H,\;\sigma )=\frac{1}{2}[V_M \sin (\sigma )]^{2}+\frac{1}{2}(\omega _n H)^{2} \end{aligned}$$

(104)

When $(H,\;\sigma )\in \Omega _2$, $V(H,\;\sigma )\ge 0$ with equality if and only if $H=\sigma =0$. From Eq. (89), the equivalent propositions will result: when $(H,\;\sigma )\in \Omega _2$,  $r\rightarrow \infty \Leftrightarrow H\rightarrow \infty \Leftrightarrow V(H,\sigma )\rightarrow \infty$. The time derivative of $V(H,\;\sigma )$ is

$$\begin{aligned} \begin{aligned} \dot{V}&=\frac{\partial V}{\partial H}\dot{H}+\frac{\partial V}{\partial \sigma }\dot{\sigma } \\&=\omega _n V_M H\sin ^{2}(\sigma )[-2\xi V_M \cos (\sigma )+\omega _n \sin (\sigma )] \\ \end{aligned} \end{aligned}$$

(105)

When $(H,\;\sigma )\in \Omega _2$, $\dot{V}\le 0$ with equality if and only if $\sigma =0$. If $(H_0 ,\;\sigma _0 )\in \Omega _2$, then $(H,\;\sigma )\in \Omega _2$ for $t\ge t_0$ since $\Omega _2$ is an invariant set. Then $V\ge 0$ and $\dot{V}\le 0$ for every $t\ge t_0$, and then from [32], it can be concluded that $V$ has a finite limit as $t\rightarrow \infty$. Taking time derivative of Eq. (105) yields

$$\begin{aligned} \begin{aligned} \ddot{V}&=\frac{\partial \dot{V}}{\partial H}\dot{H}+\frac{\partial \dot{V}}{\partial \sigma }\dot{\sigma } \\&=-2\xi \omega _n V_M^3 \sin ^{3}(\sigma )\cos (\sigma )+\omega _n^2 V_M^2 \sin ^{4}(\sigma ) \\&\quad +8\xi ^{2}\omega _n^2 V_M^2 H\sin ^{2}(\sigma )\cos ^{2}(\sigma )-4\xi ^{2}\omega _n^2 V_M^2 H\sin ^{4}(\sigma ) \\&\quad -6\xi \omega _n^3 V_M H\sin ^{3}(\sigma )\cos (\sigma )+4\xi \omega _n^3 V_M H^{2}\sin (\sigma )\cos ^{3}(\sigma ) \\&\quad -2\xi \omega _n^3 V_M H^{2}\sin ^{3}(\sigma )\cos (\sigma )-3\omega _n^4 H^{2}\sin ^{2}(\sigma )\cos ^{2}(\sigma ) \\ \end{aligned} \end{aligned}$$

(106)

Then

$$\begin{aligned} \begin{aligned} |\ddot{V}|&\le |2\xi \omega _n V_M^3 \sin ^{3}(\sigma )\cos (\sigma )|+|\omega _n^2 V_M^2 \sin ^{4}(\sigma )| \\&\quad +|8\xi ^{2}\omega _n^2 V_M^2 H\sin ^{2}(\sigma )\cos ^{2}(\sigma )| \\&\quad +|4\xi ^{2}\omega _n^2 V_M^2 H\sin ^{4}(\sigma )| \\&\quad +|6\xi \omega _n^3 V_M H\sin ^{3}(\sigma )\cos (\sigma )| \\&\quad +|4\xi \omega _n^3 V_M H^{2}\sin (\sigma )\cos ^{3}(\sigma )| \\&\quad +|2\xi \omega _n^3 V_M H^{2}\sin ^{3}(\sigma )\cos (\sigma )| \\&\quad +|3\omega _n^4 H^{2}\sin ^{2}(\sigma )\cos ^{2}(\sigma )| \\&\le (\omega _n V_M^2 +6\xi \omega _n^2 V_M H+3\omega _n^3 H^{2})(2\xi V_M +\omega _n ) \\ \end{aligned} \end{aligned}$$

(107)

When $(H,\;\sigma )\in \Omega _2$, from Eq. (87), it is clear that $\dot{H}\le 0$. Thus, if $t\ge t_0$, then $0\le H\le H_0$. Then we get

$$\begin{aligned} |\ddot{V}| \le (\omega _n V_M^2 +6\xi \omega _n^2 V_M H_0 +3\omega _n^3 H_0^2 )(2\xi V_M +\omega _n ) \end{aligned}$$

(108)

It can be seen that for $t\ge t_0$,$\ddot{V}(t)$ is bounded, and thereby $\dot{V}(t)$ is uniformly continuous. Now we need to use Barbalat Lemma [33] as follows

Barbalat Lemma

If the differentiable function $f(t)$ has a finite limit as $t\rightarrow \infty$, and if $\dot{f}$ is uniformly continuous, then $\dot{f}(t)\rightarrow 0$ as $t\rightarrow \infty .$

According to Barbalat Lemma, since $V(t)$ has a finite limit as $t\rightarrow \infty$ and $\dot{V}(t)$ is uniformly continuous, it can be concluded that $\dot{V}(t)\rightarrow 0$ as $t\rightarrow \infty$. The set of all points in $\Omega _2$ such that $\dot{V}=0$ is {$(H, \sigma ){\vert }\sigma =0$ and $H\ge 0$}. However no trajectory can converge to the point satisfying that $H>0$ and $\sigma =0$ because $\dot{\sigma }<0$. Therefore, as $t\rightarrow \infty$, all trajectories starting from the points in $\Omega _2$ converge to the origin where $\dot{H}=0$ and $\dot{\sigma }=0$. Hence, Sub-proposition 2 has been proved.

Define $\Omega _3$ as the relative complement of $\Omega _2$ in $\Omega _1$. $\Omega _3$ is determined by

$$\begin{aligned} \left\{ {\begin{array}{l} H\ge 0 \\ 0<\sigma \le \pi /2 \\ \end{array}} \right. \end{aligned}$$

(109)

Now we need to prove that the Sub-proposition 3 is true.

Sub-proposition 3

If $(H_0 ,\;\sigma _0 )\in \Omega _3$ , the corresponding phase trajectory will enter $\Omega _2$ in finite time.

From Eqs. (87–88), when $(H,\;\sigma )\in \Omega _3$, we get that $\dot{H}>0$ and $\dot{\sigma }<0$. So there must exist a time $t_1 >t_0$ such that $H_1 >H_0 \ge 0$ and $0<\sigma _1 <\sigma _0 \le \pi /2$. Here $H_1$ and $\sigma _1$ are the states at the time $t_{1}$. Before the phase trajectory enters $\Omega _2$, we have that if $t>t_1$, then $H>H_1 >0$ and $0<\sigma <\sigma _1 <\pi /2$. Hence, from Eq. (88), we get

$$\begin{aligned} \begin{aligned} \dot{\sigma }&=-2\xi \omega _n \sin (\sigma )-\frac{\omega _n^2 H\cos (\sigma )}{V_M } \\&\quad <-\frac{\omega _n^2 H\cos (\sigma )}{V_M }<-\frac{\omega _n^2 H_1 \cos (\sigma _1 )}{V_M } \\ \end{aligned} \end{aligned}$$

(110)

So there must exists a time $t_{2}$ such that $\sigma _2 =0$. From Eq. (110), we get

$$\begin{aligned} t_2 <t_1 +\frac{V_M \sigma _1 }{\omega _n^2 H_1 \cos (\sigma _1 )} \end{aligned}$$

(111)

Thus, it is proved that Sub-proposition 3 is true. By combining Sub-propositions 2 and 3 together, it is proved that if $(H_0 ,\;\sigma _0 )\in \Omega _1$, then $(H,\;\sigma )$ converges to (0, 0) as time goes to infinity. So Proposition 1 has been proved.