A novel approximate finite-time convergent guidance law with actuator fault

Abstract

In this paper, a novel approximate finite-time convergent guidance law for missiles is presented to accurately intercept maneuvering targets when the actuators cannot perform as well as usual. First, the governing equations are proposed according to the relationship between missile and target at engagement phase, and a typical nonlinear mathematic model of actuator efficiency is built. Then, an adaptive multiple-input-multiple-output guidance law based on super-twisting algorithm is proposed to drive the line-of-sight angular rate to converge in a small region around zero in finite time. According to the proposed adaptive law, the information about maneuverability of target is not necessary, and the parameter drift problem is also addressed by dead-time control technique. Moreover, the proposed formulation is validated by means of the detailed stability analysis with strict Lyapunov function and simulation results.

Appendix I : Proof of Theorem 1

Substituting Eq. (12) for Eq. (11) yields:

$$\begin{aligned} {\dot{{\varvec{x}}}}= & {} -k_1 \left( t \right) \frac{{\varvec{x}}}{\left\| {\varvec{x}} \right\| ^{1/2}}-k_2 \left( t \right) {\varvec{x}}+{{\varvec{\eta }}} \nonumber \\ {\dot{{\varvec{\eta }}}}= & {} {\dot{{\varvec{d}}}}-k_3 \left( t \right) \frac{{\varvec{x}}}{\left\| {\varvec{x}} \right\| ^{1/2}}-k_4 \left( t \right) {\varvec{x}} \end{aligned}$$

(19)

In order to facilitate the stability analysis, we introduced the following auxiliary vectors:

$$\begin{aligned} {\varvec{\xi }}_1 =\left( {\frac{L}{\left\| {\varvec{x}} \right\| }} \right) ^{1/2}{\varvec{x}}, {\varvec{\xi }}_2 ={\varvec{Lx}}, {\varvec{\xi }}_3 ={\varvec{\eta }} , (\forall x\ne 0) \end{aligned}$$

(20)

Computing the time derivative of Eq. (20) yields the following equation:

$$\begin{aligned} \left[ {{\begin{array}{l} {\dot{{\varvec{\xi }}}_1 } \\ {\dot{{\varvec{\xi }}}_2 } \\ {\dot{{\varvec{\xi }}}_3 } \\ \end{array} }} \right] =\left[ {{\begin{array}{c} {\frac{0.5\dot{{L}}{\varvec{x}}}{\left( {L\left\| {\varvec{x}} \right\| } \right) ^{1/2}}+\left( {\frac{L}{\left\| {\varvec{x}} \right\| }} \right) ^{1/2}\left( {{\varvec{I}}_{2\times 2} -\frac{{\varvec{xx}}^{T}}{2\left\| {\varvec{x}} \right\| ^{2}}} \right) \dot{{\varvec{x}}}} \\ {\dot{{L}}{\varvec{x}}+{L}\dot{{\varvec{x}}}} \\ {-c_3 \frac{L{\varvec{x}}}{\left\| {\varvec{x}} \right\| }-c_4 {L}^{2}{\varvec{x}}+\dot{{\varvec{d}}}} \\ \end{array} }} \right] \end{aligned}$$

(21)

It follows from transformation (20) that

$$\begin{aligned} \frac{{\varvec{x}}}{\left\| {\varvec{x}} \right\| }=\frac{{\varvec{\xi }}_1 }{\left\| {{\varvec{\xi }}_1 } \right\| }=\frac{{\varvec{\xi }}_2 }{\left\| {{\varvec{\xi }}_2 } \right\| },\left( {\frac{L}{\left\| {\varvec{x}} \right\| }} \right) ^{1/2}=\frac{L}{\left\| {{\varvec{\xi }}_1 } \right\| } \end{aligned}$$

(22)

Furthermore, according to Eqs. (19) and (22), it could be concluded that

$$\begin{aligned} \dot{{\varvec{x}}}=-c_1 {\varvec{\xi }}_1 -c_2 {\varvec{\xi }}_2 +{\varvec{\xi }}_3 \end{aligned}$$

(23)

From Eqs. (22) and (23), one can imply that

$$\begin{aligned} \left( {{\varvec{I}}_{2\times 2} -\frac{{\varvec{xx}}^{T}}{2\left\| {\varvec{x}} \right\| ^{2}}} \right) \dot{{\varvec{x}}}= & {} \left[ {{\varvec{\xi }}_3 -\frac{{\varvec{\xi }} _1 {\varvec{\xi }}_1^T {\varvec{\xi }}_3 }{2\left\| {{\varvec{\xi }}_1 } \right\| }} \right] \nonumber \\&-\sum _{i=1}^2 {c_i \left( {{\varvec{I}}_{2\times 2} -\frac{{\varvec{\xi }}_1 {\varvec{\xi }}_1^T }{2\left\| {{\varvec{\xi }}_1 } \right\| ^{2}}} \right) {\varvec{\xi }}_i } \nonumber \\= & {} \left[ {{\varvec{\xi }}_3 -\frac{{\varvec{\xi }}_1 {\varvec{\xi }}_1^T {\varvec{\xi }}_3 }{2\left\| {{\varvec{\xi }}_1 } \right\| }} \right] -\frac{c_1 {\varvec{\xi }}_1 +c_2 {\varvec{\xi }}_2 }{2}\nonumber \\ \end{aligned}$$

(24)

Substituting Eqs. (22)–(24) into (21) yields the following equation:

$$\begin{aligned} \dot{{\varvec{\xi }}}=-\frac{L}{2\left\| {{\varvec{\xi }}_1 } \right\| }{\varvec{A}}_1 {\varvec{\xi }} -{\varvec{LA}}_2 {\varvec{\xi }} +{\varvec{A}}_3 \end{aligned}$$

(25)

where ${\varvec{\xi }} =\left[ {{\varvec{\xi }}_1 \left| {{\varvec{\xi }}_2 \left| {{\varvec{\xi }}_3 } \right. } \right. } \right] \in {\mathbb {R}}^{6}$ is a six-dimensional column vector. The matrices ${\varvec{A}}_1$, ${\varvec{A}}_2$, ${\varvec{A}}_3$ are calculated below:

$$\begin{aligned} {\varvec{A}}_1= & {} \left[ {{\begin{array}{lll} {c_1 {\varvec{I}}_{2\times 2} }&{} {c_2 {\varvec{I}}_{2\times 2} }&{} {-2{\varvec{I}}_{2\times 2} } \\ {{\varvec{{0}}}_{2\times 2} }&{} {{\varvec{{0}}}_{2\times 2} }&{} {{\varvec{{0}}}_{2\times 2} } \\ {2c_3 {\varvec{I}}_{2\times 2} }&{} {{{\varvec{0}}}_{2\times 2} }&{} {{{\varvec{0}}}_{2\times 2} } \\ \end{array} }} \right] ,\nonumber \\ {\varvec{A}}_2= & {} \left[ {{\begin{array}{lll} {{{\varvec{0}}}_{2\times 2} }&{} {{{\varvec{0}}}_{2\times 2} }&{} {{{\varvec{0}}}_{2\times 2} } \\ {c_1 {\varvec{I}}_{2\times 2} }&{} {c_2 {\varvec{I}}_{2\times 2} }&{} {-{\varvec{I}}_{2\times 2} } \\ {{{\varvec{0}}}_{2\times 2} }&{} {c_4 {\varvec{I}}_{2\times 2} }&{} {{{\varvec{0}}}_{2\times 2} } \\ \end{array} }} \right] \nonumber \\ {\varvec{A}}_3= & {} \frac{\dot{L}}{2L}\left[ {{\begin{array}{l} {{\varvec{\xi }}_1 } \\ {2{\varvec{\xi }}_2 } \\ {{{\varvec{0}}}_2 } \\ \end{array} }} \right] +\left[ {{\begin{array}{l} {{{\varvec{0}}}_2 } \\ {{{\varvec{0}}}_2 } \\ {\dot{{\varvec{d}}}} \\ \end{array} }} \right] -\left[ {{\begin{array}{c} {L{\varvec{\xi }}_1 {\varvec{\xi }}_1^T {\varvec{\xi }} _3 /\left( {2\left\| {{\varvec{\xi }}_1 } \right\| ^{3}} \right) } \\ {{\varvec{0}}}_2 \\ {{\varvec{0}}}_2 \\ \end{array} }} \right] \nonumber \\ \end{aligned}$$

(26)

For system (21), the Lyapunov function is considered as follow,

$$\begin{aligned} V= & {} 2c_3 \left\| {{\varvec{\xi }}_1 } \right\| ^{2}+c_4 \left\| {{\varvec{\xi }}_2 } \right\| ^{2}\nonumber \\&+\frac{1}{2}\left\| {{\varvec{\xi }}_3 } \right\| ^{2}+\frac{1}{2}\left\| {c_1 {\varvec{\xi }}_1 +c_2 {\varvec{\xi }}_2 -{\varvec{\xi }}_3 } \right\| ^{2} \end{aligned}$$

(27)

which could be rewritten in the following matrix format,

$$\begin{aligned} V=\frac{1}{2}{\varvec{\xi }}^{T}{\varvec{P}}{\varvec{\xi }} \end{aligned}$$

(28)

where positive definite matrix ${\varvec{P}}\in {\mathbb {R}}^{6\times 6}$ is defined as

$$\begin{aligned} {\varvec{P}}=\left[ {{\begin{array}{ccc} {\left( {4c_3 +c_1^2 } \right) {\varvec{I}}_{2\times 2} }&{} {c_1 c_2 {\varvec{I}}_{2\times 2} }&{} {-c_1 {\varvec{I}}_{2\times 2} } \\ {c_1 c_2 {\varvec{I}}_{2\times 2} }&{} {\left( {2c_4 +c_2^2 } \right) {\varvec{I}}_{2\times 2} }&{} {-c_2 {\varvec{I}}_{2\times 2} } \\ {-c_1 {\varvec{I}}_{2\times 2} }&{} {-c_2 {\varvec{I}}_{2\times 2} }&{} {2{\varvec{I}}_{2\times 2} } \\ \end{array} }} \right] \end{aligned}$$

(29)

Evaluating the time derivative of V yields the following equation:

$$\begin{aligned} \dot{V}=-\frac{L}{2\left\| {{\varvec{\xi }}_1 } \right\| }{\varvec{\xi }}^{T}{\varvec{B}}_1 {\varvec{\xi }} -L{\varvec{\xi }} ^{T}{\varvec{B}}_2 {\varvec{\xi }} +\bar{{V}} \end{aligned}$$

(30)

where ${\varvec{B}}_1 =0.5\left( {{\varvec{A}}_1^T {\varvec{P}}+{\varvec{PA}}_1 } \right)$, ${\varvec{B}}_2 =0.5\left( {{\varvec{A}}_2^T {\varvec{P}}+{\varvec{PA}}_2 } \right)$, and $\bar{{V}}={\varvec{\xi }}^{T}{\varvec{PA}}_3$.

Let ${\varvec{A}}_{31} =\frac{\dot{L}}{2L}\left[ {{\varvec{\xi }}_1 \left| {2{\varvec{\xi }}_2 \left| {{{\varvec{0}}}_2 } \right. } \right. } \right]$, ${\varvec{A}}_{32} =\left[ {{{\varvec{0}}}_2 \left| {{{\varvec{0}}}_2 \left| {\dot{{\varvec{d}}}} \right. } \right. } \right]$, $A_{33} =-\left[ {\frac{L{\varvec{\xi }}_1 {\varvec{\xi }}_1^T {\varvec{\xi }}_3 }{2\left\| {{\varvec{\xi }}_1 } \right\| ^{3}}\left| {{{\varvec{0}}}_2 \left| {{{\varvec{0}}}_2 } \right. } \right. } \right]$, it is possible to verify that ${\varvec{A}}_3 = {\varvec{A}}_{31} +{\varvec{A}}_{32} +{\varvec{A}}_{33}$, then it is implied that $\bar{{V}}=\bar{{V}}_1 +\bar{{V}}_2 +\bar{{V}}_3$ with $\bar{{V}}_i ={\varvec{\xi }}^{T}{\varvec{PA}}_{3i}$, $i=1,2,3$, with

$$\begin{aligned} \bar{{V}}_1= & {} \frac{\dot{L}}{2L}\left[ \left( {4c_3 +c_1^2 } \right) \left\| {{\varvec{\xi }}_1 } \right\| ^{2}+3c_1 c_2 {\varvec{\xi }}_1^T {\varvec{\xi }}_2 -c_1 {\varvec{\xi }}_1^T {\varvec{\xi }}_3 \right. \nonumber \\&\left. +\,2\left( {2c_4 +c_2^2 } \right) \left\| {{\varvec{\xi }}_2 } \right\| ^{2}-2c_2 {\varvec{\xi }}_2^T {\varvec{\xi }}_3 \right] \end{aligned}$$

(31)

$$\begin{aligned} \bar{{V}}_2= & {} \dot{{\varvec{d}}}^{T}\left[ {2{\varvec{\xi }}_3 -c_1 {\varvec{\xi }}_1 -c_2 {\varvec{\xi }}_2 } \right] \end{aligned}$$

(32)

$$\begin{aligned} \bar{{V}}_3= & {} -\frac{L}{2\left\| {{\varvec{\xi }}_1 } \right\| }\left[ \left( {4c_3 +c_1^2 } \right) {\varvec{\xi }}_1^T {\varvec{\xi }}_3 \right. \nonumber \\&\left. +\frac{c_1 c_2 {\varvec{\xi }}_2^T {\varvec{\xi }}_1 {\varvec{\xi }}_1^T {\varvec{\xi }}_3 }{\left\| {{\varvec{\xi }}_1 } \right\| ^{2}}-\frac{c_1 {\varvec{\xi }}_3^T {\varvec{\xi }}_1 {\varvec{\xi }}_1^T {\varvec{\xi }}_3 }{\left\| {{\varvec{\xi }}_1 } \right\| ^{2}} \right] \end{aligned}$$

(33)

It follows from Eqs. (31) and (32) that

$$\begin{aligned} \bar{{V}}_1\le & {} \frac{\dot{L}}{2L}{\varvec{\xi }}^{T}{\varvec{P}}_1 {\varvec{\xi }} \end{aligned}$$

(34)

$$\begin{aligned} \bar{{V}}_2\le & {} \sqrt{4+c_1^2 +c_2^2 }\left\| {\dot{{\varvec{d}}}} \right\| \left\| {\varvec{\xi }} \right\| \le \sqrt{4+c_1^2 +c_2^2 }\left\| {\varvec{\xi }} \right\| \delta \end{aligned}$$

(35)

where ${\varvec{P}}_1 =\text {diag}\left( {\vartheta _1 ,\vartheta _2 ,\vartheta _3 } \right) \otimes {\varvec{I}}_{2\times 2}$, $\vartheta _1 =4c_3 +c_1^2 +1.5c_1 c_2 +0.5c_1$, $\vartheta _2 =4c_4 +2c_2^2 +1.5c_1 c_2 +c_2$, $\vartheta _3 =0.5c_1 +c_2$, and $\otimes$ stands for the Kronecker product. It also follows from Eq. (20) that ${\varvec{\xi }}_2 =\left\| {{\varvec{\xi }}_1 } \right\| {\varvec{\xi }}_1$. Moreover, considering the fact that

$$\begin{aligned} \frac{{\varvec{\xi }}_3^T {\varvec{\xi }}_1 {\varvec{\xi }}_1^T {\varvec{\xi }}_3 }{\left\| {{\varvec{\xi }}_1 } \right\| ^{2}}=\frac{\left( {{\varvec{\xi }}_1^T {\varvec{\xi }}_3 } \right) ^{2}}{\left\| {{\varvec{\xi }}_1 } \right\| ^{2}}\le \left\| {{\varvec{\xi }}_3 } \right\| ^{2} \end{aligned}$$

(36)

and it can then be concluded that

$$\begin{aligned} \bar{{V}}_3\le & {} -\frac{L}{2\left\| {{\varvec{\xi }}_1 } \right\| }\left[ {\left( {4c_3 +c_1^2 } \right) {\varvec{\xi }}_1^T {\varvec{\xi }}_3 -c_1 \left\| {{\varvec{\xi }}_3 } \right\| ^{2}} \right] \nonumber \\&-\frac{Lc_1 c_2 }{2}{\varvec{\xi }}_1^T {\varvec{\xi }}_3 \end{aligned}$$

(37)

Substituting Eqs. (34), (35) and (37) for (30) yields the following,

$$\begin{aligned} \dot{V}\le -\frac{L}{2\left\| {{\varvec{\xi }}_1 } \right\| }{\varvec{\xi }}^{T}{\varvec{P}}_2 {\varvec{\xi }} -L{\varvec{\xi }}^{T}{\varvec{P}}_3 {\varvec{\xi }} +\bar{{V}}_1 +\bar{{V}}_2 \end{aligned}$$

(38)

where

$$\begin{aligned} {\varvec{P}}_2= & {} \left[ {{\begin{array}{ccc} {c_1 \left( {c_1^2 +2c_3 } \right) }&{} 0&{} {-c_1^2 } \\ 0&{} {c_1 \left( {5c_2^2 +2c_4 } \right) }&{} {-3c_1 c_2 } \\ {-c_1^2 }&{} {-3c_1 c_2 }&{} {c_1 } \\ \end{array} }} \right] \otimes {\varvec{I}}_{2\times 2}\nonumber \\ \end{aligned}$$

(39)

$$\begin{aligned} {\varvec{P}}_3= & {} \left[ {{\begin{array}{ccc} {c_1 \left( {2c_1^2 +c_3 } \right) }&{} 0&{} 0 \\ 0&{} {c_2 \left( {c_2^2 +c_4 } \right) }&{} {-c_2^2 } \\ 0&{} {-c_2^2 }&{} {c_2 } \\ \end{array} }} \right] \otimes {\varvec{I}}_{2\times 2} \end{aligned}$$

(40)

It can be verified that matrices ${\varvec{P}}_1$, ${\varvec{P}}_2$ and ${\varvec{P}}_3$ are definitely positive. Then, one can imply that

$$\begin{aligned} \lambda _{\min } \left( {{\varvec{P}}_i } \right) \left\| {\varvec{\xi }} \right\| ^{2}\le {\varvec{\xi }}^{T}{\varvec{P}}_i {\varvec{\xi }} \le \lambda _{\max } \left( {{\varvec{P}}_i } \right) \left\| {\varvec{\xi }} \right\| ^{2} \end{aligned}$$

(41)

Similarly, for positive matrix ${\varvec{P}}$ in Eq. (28), it can be concluded that

$$\begin{aligned} \lambda _{\min } \left( {\varvec{P}} \right) \left\| {\varvec{\xi }} \right\| ^{2}\le 2V\le \lambda _{\max } \left( {\varvec{P}} \right) \left\| {\varvec{\xi }} \right\| ^{2} \end{aligned}$$

(42)

It follows from inequalities of Eqs. (36) and (37) that

$$\begin{aligned} \frac{2\lambda _{\min } \left( {{\varvec{P}}_i } \right) V}{\lambda _{\max } \left( {\varvec{P}} \right) }\le {\varvec{\xi }}^{T}{\varvec{P}}_i {\varvec{\xi }} \le \frac{2\lambda _{\max } \left( {{\varvec{P}}_i } \right) V}{\lambda _{\min } \left( {\varvec{P}} \right) } \end{aligned}$$

(43)

Substituting Eqs. (34), (35) and (43) for (38) yields

$$\begin{aligned} \dot{V}\le & {} \left[ {\frac{\dot{L}\lambda _{\max } \left( {{\varvec{P}}_1 } \right) }{L\lambda _{\min } \left( {\varvec{P}} \right) }-\frac{2L\lambda _{\min } \left( {{\varvec{P}}_3 } \right) }{\lambda _{\max } \left( {\varvec{P}} \right) }-\frac{L\lambda _{\min } \left( {{\varvec{P}}_2 } \right) }{\left\| {{\varvec{\xi }} _1 } \right\| \lambda _{\max } \left( {\varvec{P}} \right) }} \right] V\nonumber \\&+\sqrt{4+c_1^2 +c_2^2 }\left\| {\varvec{\xi }} \right\| \delta \end{aligned}$$

(44)

It follows from the fact $\left\| {{\varvec{\xi }}_1 } \right\| \le \left\| {\varvec{\xi }} \right\|$ and $\sqrt{\frac{2V}{\lambda _{\max } \left( {\varvec{P}} \right) }}\le \left\| {\varvec{\xi }} \right\| \le \sqrt{\frac{2V}{\lambda _{\min } \left( {\varvec{P}} \right) }}$ that

$$\begin{aligned} \dot{V}\le -\left( {L\delta _2 -\frac{\dot{L}}{L}\delta _1 } \right) V-\left( {L\delta _3 -\delta _4 \delta } \right) V^{1/2} \end{aligned}$$

(45)

where $\delta _1 =\frac{\lambda _{\max } \left( {{\varvec{P}}_1 } \right) }{\lambda _{\min } \left( {\varvec{P }}\right) }$, $\delta _2 =\frac{2\lambda _{\min } \left( {{\varvec{P}}_3 } \right) }{\lambda _{\max } \left( {\varvec{P}} \right) }$, $\delta _3 =\frac{\lambda _{\min } \left( {{\varvec{P}}_2 } \right) \sqrt{\lambda _{\min } \left( {\varvec{P}} \right) }}{\sqrt{2}\lambda _{\max } \left( {\varvec{P}} \right) }$, $\delta _4 =\sqrt{\frac{2\left( {4+c_1^2 +c_2^2 } \right) }{\lambda _{\min } \left( {\varvec{P}} \right) }}$.

Since $\dot{L}=m>0$ when $\left\| {\varvec{x}} \right\| >\varepsilon$, it can be determined that $\beta _1 =L\delta _2 -\frac{\dot{L}}{L}\delta _1 >0$, $\beta _2 =L\delta _3 -\delta _4 \delta >0$ can be calculated in finite time. After that, Eq. (45) can be reduced to

$$\begin{aligned} \dot{V}+\beta _1 V+\beta _2 V^{1/2}\le 0 \end{aligned}$$

(46)

Then, under the condition of $\left\| {\varvec{x}} \right\| >\varepsilon$, Lemma1 and state transformation (20), it is concluded that the system state ${\varvec{x}}$ would converge into a small area around zero in finite time. This completes the proof.