CN102353301B - Guidance method with terminal restraint based on virtual target point - Google Patents

Abstract

The invention discloses a guidance method with terminal restraint based on a virtual target point and belongs to the technical field of aircraft guidance and control systems. By using a target virtualization method, an aircraft can fly in the optimum height to attack a target; a virtualized target is positioned, above a real target, in the height, namely a detonation height which is required for a damage element; by designing a guidance law, when the aircraft reaches an imaginary target, a warhead detonates; and the conventional mode of attacking the real target by flying is changed into a guidance mode of flying to the virtualized target and detonating according to a required attitude angle. By adoption of the guidance law of the method, the aim that the aircraft with the explosion-formed damage element flies and attacks in an expected height can be fulfilled, the requirement of the optimum detonation height of the damage element is met, the problem of the attitude angle of the aircraft with the explosion-formed damage element during end attacking can be solved, and the requirement of the detonation moment of the damage element on the attitude angle of the aircraft is met.

Description

Based on the guidance method with end conswtraint of virtual target point

Technical field

The invention belongs to aircraft guidance and Control System Design field, relate to the Design of Guidance Law problem of being furnished with explosive forming and injuring first aircraft, specially refer to and how in flight guided procedure, adopt the mode of virtual target point to realize aircraft to plunder and fly to hit top, and moment of detonating make aircraft keep particular pose angle injuring unit.

Background technology

Explosive forming is injured unit and is referred to after warhead activation under the effect of flexible linear-shaped charge, metal liner is forged into a penetration body that is similar to bullet, this penetration body has stable flight characteristics, enters in target with kinetic energy penetration at a high speed, realizes precisely strike mission efficiently.The Design of Guidance Law that is equipped with explosive forming to injure first aircraft is faced with new problem: because the unidirectional narrow beam of first outgoing is injured in explosive forming, in addition plunder the attack pattern that flies to hit top, it is closely related with attitude of flight vehicle and position of aircraft that unit is injured in the moment explosive forming that makes to detonate; Particularly for different targets of attack, in order to obtain the best effect of injuring, requirement have different detonate pattern with form expect injure n-ary form n (as stock jet or explosive forming bullet), and the different patterns of detonating has strict demand to the height in warhead activation moment.Therefore,, for bonding Shu Juneng injures the critical position that unit can pinpointing, require aircraft to there is the attitude angle and the flying height that are conducive to injure first hit in the warhead activation moment.

To injure first aircraft with explosive forming and realize maximum and injure effect in order to realize, need the end guiding stage of aircraft to meet following 2 points:

(1) warhead activation moment aircraft transient posture is controlled, and the moment attitude of detonating is by being required attitude;

(2) height of warhead activation moment aircraft is controlled, reaches the requirement for height in the moment of detonating.

The proportional guidance law of conventional aircraft guidance method and improved form thereof, optimal guidance law, differential Game Guidance Law, what technology was comparatively ripe at present is proportional guidance law or its improved form.Traditional guidance law does not add end conswtraint angle and the limitation in height in the moment of detonating.For research herein injure first aircraft with explosive forming, can realize satisfied unit the detonate attitude of moment to aircraft and the part of flying height requirement of injuring in order to realize the best effect of injuring, just must to consider adding in Guidance Law Design.

Summary of the invention

For meeting the requirement of injuring first aircraft and realize precision strike with explosive forming in guidance process, the object of this invention is to provide a kind of guidance method with end conswtraint based on virtual target point.

The present invention utilizes the method for virtual target to realize aircraft to plunder and fly target of attack at optimum height, in real target, have an imaginary target in vain, in order to injure, unit is desired detonates highly the height of imagination target in spatial domain, guidance law is established on top makes aircraft warhead activation in the time flying to imaginary target, now original employing is plunderred the mode that flies to attack real goal become fly to virtual target and by the guide mode that detonates of requirement attitude angle, the moment is found when target endways to realize aircraft, the autonomous tracking target of aircraft also reduces flying height, in the time injuring unit's desired height of formation and attitude of flight vehicle according to expectation, ignite warhead, thereby complete the object precisely striking target.

In the aircraft flight process the present invention relates to, Aircraft Angle of Attack is ignored, and being approximately trajectory inclination angle is attitude of flight vehicle angle.

End trajectory is divided into two stages by the present invention, adopts the method design guidance law of control with changed scale coefficient.

The present invention adopts the method for optimum control to be write the conversion of trajectory equation as state equation, and realizes the constraint to the trajectory angle of fall by method for optimally controlling.

Described method for optimally controlling is realized as follows to the optimization detailed process of the Guidance Law terminal angle of fall:

The first step: determine the feasible moment point position of detonating;

Second step: the mathematical modeling of setting up aircraft and target relative motion relation;

Aircraft initial position is M point, and target is positioned at T point, and the distance of aircraft and target is r, and θ is trajectory inclination angle:

rsinθ＝y (1)

(1) formula is carried out to secondary differentiate,

$\stackrel{\·\·}{r}\sin \mathrm{\θ}+2\stackrel{\·}{r}\stackrel{\·}{\mathrm{\θ}}\cos \mathrm{\θ}-r{\stackrel{\·}{\mathrm{\θ}}}^2\sin \mathrm{\θ}+r\stackrel{\·\·}{\mathrm{\θ}}\cos \mathrm{\θ}=a_{\mathrm{my}}---\left(2\right)$

Adopt Taylor series that (2) formula is launched, carry out linearization process, by formula (2)

sin θ, cos θ exist respectively

θ ₀place carries out single order Taylor series expansion,

sinθ＝sinθ ₀+cosθ ₀×(θ-θ ₀) (3)

cosθ＝cosθ ₀-sinθ ₀×(θ-θ ₀) (4)

${\stackrel{\·}{\mathrm{\θ}}}^2={\stackrel{\·}{\mathrm{\θ}}}_0^2+2{\stackrel{\·}{\mathrm{\θ}}}_0\×(\stackrel{\·}{\mathrm{\θ}}-{\stackrel{\·}{\mathrm{\θ}}}_0)---\left(5\right)$

Full scale equation (2) becomes

$\stackrel{\·\·}{r}\×(\sin {\mathrm{\θ}}_0+\cos {\mathrm{\θ}}_0\×(\mathrm{\θ}-{\mathrm{\θ}}_0))+2\stackrel{\·}{r}\stackrel{\·}{\mathrm{\θ}}\×(\cos {\mathrm{\θ}}_0-\sin {\mathrm{\θ}}_0\×(\mathrm{\θ}-{\mathrm{\θ}}_0))-$

$r\×({\stackrel{\·}{\mathrm{\θ}}}_0^2+2{\stackrel{\·}{\mathrm{\θ}}}_0\×(\stackrel{\·}{\mathrm{\θ}}-{\stackrel{\·}{\mathrm{\θ}}}_0))\×(\sin {\mathrm{\θ}}_0+\cos {\mathrm{\θ}}_0\×(\mathrm{\θ}-{\mathrm{\θ}}_0))---\left(6\right)$

$+r\stackrel{\·\·}{\mathrm{\θ}}\×(\cos {\mathrm{\θ}}_0-\sin {\mathrm{\θ}}_0\×(\mathrm{\θ}-{\mathrm{\θ}}_0))=a_{\mathrm{my}}$

Formula (6) becomes through abbreviation

$\mathrm{a\θ}+b\stackrel{\·}{\mathrm{\θ}}+m\stackrel{\·\·}{\mathrm{\θ}}+q+k=a_{\mathrm{my}}---\left(7\right)$

$a=\stackrel{\·\·}{r}\cos {\mathrm{\θ}}_0+r{\stackrel{\·}{\mathrm{\θ}}}_0^2\cos {\mathrm{\θ}}_0$

$b=2\stackrel{\·}{r}(\cos {\mathrm{\θ}}_0+{\mathrm{\θ}}_0\sin {\mathrm{\θ}}_0)-2r{\stackrel{\·}{\mathrm{\θ}}}_0(\sin {\mathrm{\θ}}_0-{\mathrm{\θ}}_0\cos {\mathrm{\θ}}_0)$

m＝r(cosθ ₀+θ ₀sinθ ₀)

$q=\stackrel{\·}{\mathrm{\θ}}\mathrm{\θ}(-2\stackrel{\·}{r}\sin {\mathrm{\θ}}_0-2\stackrel{\·}{r}\cos {\mathrm{\θ}}_0)-\stackrel{\·\·}{\mathrm{\θ}}\mathrm{\θ}r\sin {\mathrm{\θ}}_0$

$k=\stackrel{\·\·}{r}(\sin {\mathrm{\θ}}_0+{\mathrm{\θ}}_0\cos {\mathrm{\θ}}_0)+r{{\stackrel{\·}{\mathrm{\θ}}}_0}^2(\sin {\mathrm{\θ}}_0-{\mathrm{\θ}}_0\cos {\mathrm{\θ}}_0);$

The 3rd step: get endways two on trajectory with reference to the stage: one is initial time

the point in moment; Another elects end moment θ as ₀,

in the first stage, due to θ,

very little,, cause q, k and other amount in formula (7) to compare in a small amount, therefore formula (7) is reduced to

$a_{\mathrm{my}1}=2\stackrel{\·}{r}\stackrel{\·}{\mathrm{\θ}}+r\stackrel{\·\·}{\mathrm{\θ}}+\stackrel{\·\·}{r}\mathrm{\θ}---\left(8\right)$

And for second stage, in the q in formula (7), contain non-linear relation, and be microvariations methods due to what adopt, the θ in q can think the θ that second stage adopts ₀therefore formula (7) is reduced to

$a_{\mathrm{my}2}=\mathrm{a\θ}+c\stackrel{\·}{\mathrm{\θ}}+d\stackrel{\·\·}{\mathrm{\θ}}+k---\left(9\right)$

$c=2\stackrel{\·}{r}(\cos {\mathrm{\θ}}_0+{\mathrm{\θ}}_0\sin {\mathrm{\θ}}_0)-2r{\stackrel{\·}{\mathrm{\θ}}}_0(\sin {\mathrm{\θ}}_0-{\mathrm{\θ}}_0\cos {\mathrm{\θ}}_0)+{\mathrm{\θ}}_0(-2\stackrel{\·}{r}\sin {\mathrm{\θ}}_0-2\stackrel{\·}{r}\cos {\mathrm{\θ}}_0)$

d＝r(cosθ ₀+θ ₀sinθ ₀)-θ ₀rsinθ ₀；

Adopt this kind of variable element guidance law, select Proportional coefficient K ₁, K ₂carry out weights distribution, make K ₁+ K ₂=1, draw,

a _my＝K ₁a _my1+K ₂a _my2

$a_{\mathrm{my}}=\mathrm{l\θ}+g\stackrel{\·}{\mathrm{\θ}}+w\stackrel{\·\·}{\mathrm{\θ}};$

$l=k_{2}a+\stackrel{\&CenterDot;\&CenterDot;}{r}{\mathrm{<figure-callout\; id="1"\; label="k"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">k</figure-callout>}}_1;$ $g=2\stackrel{\&CenterDot;}{r}{\mathrm{<figure-callout\; id="1"\; label="k"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">k</figure-callout>}}_1+\mathrm{<figure-callout\; id="2"\; label="c\; k"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">c</figure-callout>}{k}_{}{}_2;$ w＝rk ₁+dk ₂；

The 4th step: set up the optimal guidance law with terminal angle restriction;

θ is trajectory tilt angle, order $\begin{array}{c}\mathrm{\&theta;}={\mathrm{\&theta;}}_x+{\mathrm{\&theta;}}_f\\ \stackrel{\&CenterDot;}{\mathrm{\&theta;}}={\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_x\end{array},$ θ _xfor variable, θ _fterminal trajectory tilt angle for expecting:

Order $\begin{array}{c}{x}_1={\mathrm{\&theta;}}_x\\ x_2={\stackrel{\&CenterDot;}{x}}_1\end{array},$ Row are write state equation and are obtained $\begin{array}{c}{\stackrel{\&CenterDot;}{x}}_1=x_2\\ {\stackrel{\&CenterDot;}{x}}_2=\frac{-x_{1}l-x_{2}g+a_{\mathrm{my}}}{w}\end{array},$

Abbreviation is $\stackrel{\&CenterDot;}{X}=\mathrm{AX}+\mathrm{BU},$

$A=\left[\begin{array}{cc}0& 1\\ p& n\end{array}\right]$

$B=\left[\begin{array}{c}0\\ \frac{1}{w}\end{array}\right]$

p＝-l/w

n＝-g/w

u＝a _my

Initial conditions t=t ₀time, x ₁(t ₀)=θ (t ₀)+θ _f;

End conswtraint condition t=t _ftime, x ₁(t _f)=0; x ₂(t _f)=0; Now θ=θ _f;

Select quadratic performance index $J=X^T\mathrm{FX}+1/2{\&Integral;}_{t_0}^{t_f}{u}^2\mathrm{dt},$

$x=\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right];$ $F=\left[\begin{array}{cc}{f}_{11}& f_{12}\\ f_{21}& f_{22}\end{array}\right];$ $A=\left[\begin{array}{cc}0& 1\\ p& n\end{array}\right];$ $B=\left[\begin{array}{c}0\\ \frac{1}{w}\end{array}\right],$ R＝1，Q＝0。

Obtaining optimal control law by the theory of optimal control is

$u^{*}=-R^{-1}{B}^{T}P{X}^{*}=-\left[\begin{array}{cc}0& \frac{1}{r}\end{array}\right]\left[\begin{array}{cc}{P}_{11}& P_{12}\\ P_{21}& P_{22}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=-\frac{1}{r}(P_{11}{x}_1+P_{22}{x}_2)$

In formula, P meets the matrix of Riccati equation, because of || F|| → ∝, separates Riccati equation and obtains

$\left\{\begin{array}{c}{\stackrel{\&CenterDot;}{P}}^{-1}-{\mathrm{AP}}^{-1}-P^{-1}{A}^T+{\mathrm{BR}}^{-1}{B}^T-P^{-1}{\mathrm{QP}}^{-1}=0\\ P^{-1}\left(t_f\right)=F^{-1}=0\end{array}\right.$

Order $P^{-1}=\left[\begin{array}{cc}{q}_{11}& q_{12}\\ q_{21}& q_{22}\end{array}\right],$ Wherein q ₁₂=q ₂₁

Solve $P=\left[\begin{array}{cc}2\mathrm{bt}& b\\ b& -\mathrm{nb}-\mathrm{mbt}\end{array}\right],$ Wherein $t_{\mathrm{go}}=\frac{r}{\stackrel{\&CenterDot;}{r}},$ $b=\frac{1}{(3p-2n^2-2\mathrm{pnt}){\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">r</figure-callout>}}^2}$

$P=\left[\begin{array}{cc}\frac{n+\mathrm{pt}}{b(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}& \frac{1}{b(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}\\ \frac{1}{b(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}& \frac{-2t}{b(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}\end{array}\right]$

Obtaining optimal control law is

$u^{*}=-\frac{1}{r}(P_{11}{x}_1+P_{22}{x}_2)=-\frac{(\mathrm{pt}+n)(\mathrm{\&theta;}+{\mathrm{\&theta;}}_f)}{\mathrm{br}(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}+\frac{2t\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}{\mathrm{br}(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}$

In formula $t_{\mathrm{go}}=\frac{r}{\stackrel{\&CenterDot;}{r}};$ $b=\frac{1}{(3p-2n^2-2\mathrm{pnt}){\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">r</figure-callout>}}^2};$ p＝-l/w；n＝-g/w； $l=k_{2}a+\stackrel{\&CenterDot;\&CenterDot;}{r}{\mathrm{<figure-callout\; id="1"\; label="k"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">k</figure-callout>}}_1;$ $g=2\stackrel{\&CenterDot;}{r}{\mathrm{<figure-callout\; id="1"\; label="k"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">k</figure-callout>}}_1+\mathrm{<figure-callout\; id="2"\; label="c\; k"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">c</figure-callout>}{k}_{}{}_2;$ w＝rk ₁+dk ₂； $a=\stackrel{\&CenterDot;\&CenterDot;}{r}\cos {\mathrm{\&theta;}}_0+r{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0^2\cos {\mathrm{\&theta;}}_0;$

d＝r(cosθ ₀+θ ₀sinθ ₀)-θ ₀rsinθ ₀。

The optimal control law finally obtaining is

$u^{*}=-\frac{1}{r}(P_{11}{x}_1+P_{22}{x}_2)=-\frac{(\mathrm{pt}+n)(\mathrm{\&theta;}+{\mathrm{\&theta;}}_f)}{\mathrm{br}(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}+\frac{2t\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}{\mathrm{br}(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}.$

Good effect of the present invention is:

(1) solved with explosive forming injure first aircraft Desired Height realize plunder fly attack problem, met and injured first Optimal Burst requirement for height;

(2) solved the problem of injuring first aircraft the end game moment attitude angle with explosive forming, met and injured the unit's requirement of moment to attitude of flight vehicle angle of detonating.

Brief description of the drawings

Fig. 1 is aircraft flight trajectory schematic diagram.

Fig. 2 is the aircraft flight schematic diagram based on virtual target point.

Fig. 3 is the relation of moment attitude of flight vehicle and flight position of detonating.

Fig. 4 is aircraft and target relative motion geometrical relationship figure.

Fig. 5 a be when proportionality coefficient be K ₁=0.7, K ₂=0.3 makes, flying height temporal evolution curve.

Fig. 5 b be when proportionality coefficient be K ₁=0.6; K ₂=0.4 o'clock, flying height temporal evolution curve.

Form 1 is different proportion COEFFICIENT K ₁, K ₂trajectory inclination effect is contrasted.

Detailed description of the invention

Below in conjunction with accompanying drawing, the present invention is described in detail, detailed description of the invention is as follows:

As seen from Figure 1, this aircraft is after presumptive area is thrown in, and wing launches, engine start, and the flight of highly cruising in 200m left and right with the speed of about 100m/s, searches for target.Aircraft is found after target, to adopt different attack patterns according to target characteristic, as: if treat that target of attack is heavy armored target, adopt the extension rod-type pattern of detonating; As treated, target of attack is lightweight armor target, adopts the pattern of detonating of explosive forming penetration body.The different patterns of injuring requires the warhead activation moment to have different requirement for height, therefore for injure first aircraft with explosive forming, in order to hit the mark accurately, the aircraft elevation information in the moment of detonating must be added in the design of Guidance Law.Therefore the present invention proposes the concept of virtual target: false section, in the overhead of realistic objective, has a virtual target, need aircraft to hit virtual target with the angle of fall of design.

As seen from Figure 2, aircraft, from A point, treats that target of attack original position is B point.The mobility of considering target, hypothetical target moves to B _ftime, now require the aircraft A that should fly _fpoint, A _fpoint is the virtual target point of design.Aircraft is at A _fwhen point detonates warhead, choosing of required attitude angle is to be determined by the relation between flying height h and position of aircraft information.In order to discuss conveniently, below in the Design of Guidance Law with the constraint of the terminal angle of fall, all suppose that injuring first aircraft with explosive forming detonates at virtual impact point, be converted into the Guidance Law Design problem with the constraint of the terminal angle of fall therefore meet the Guidance Law Design of aircraft altitude and Gesture.

The present invention adopts method for optimally controlling to realize the optimization to the Guidance Law terminal angle of fall.Specific implementation process is as follows:

1. the feasible moment point location positioning that detonates

Position, attitude, velocity magnitude and the direction of warhead activation moment aircraft, and the static first flying speed etc. of injuring all can affect final strike effect.In other parameter one timings, injure first point of impact ordinate and be monotone variation with the aircraft angle of pitch in the moment of detonating, in the time that the moment of detonating, aircraft had the positive angle of pitch, injure first point of impact and will be ahead of at the projecting direction on ground the moment body projected position on the ground that detonates along speed.The ordinate of the point of impact causing due to the moment aircraft angle of pitch difference of detonating changes fairly obvious, therefore in the time selecting Detonating Time, need to consider that the aircraft angle of pitch is on injuring the longitudinally impact of accuracy at target of unit.Fig. 3 expresses, and the radius of the feasible zone that detonates is R, the A1 point if aircraft now flies, if expect hit, the angle of pitch that aircraft must have is θ.

In the present invention, ignore angle of attack impact, realize by the constraint of moment trajectory tilt angle that end is detonated the task that target is precisely attacked.

2. the mathematical modeling of aircraft and target relative motion relation

In Fig. 4, aircraft initial position is M point, and target is positioned at T point, and the distance of aircraft and target is r, and θ is trajectory inclination angle.

Known by geometrical relationship

rsinθ＝y (1)

(1) formula is carried out to secondary differentiate,

$\stackrel{\&CenterDot;\&CenterDot;}{r}\sin \mathrm{\&theta;}+2\stackrel{\&CenterDot;}{r}\stackrel{\&CenterDot;}{\mathrm{\&theta;}}\cos \mathrm{\&theta;}-r{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}^2\sin \mathrm{\&theta;}+r\stackrel{\&CenterDot;\&CenterDot;}{\mathrm{\&theta;}}\cos \mathrm{\&theta;}=a_{\mathrm{my}}---\left(2\right)$

Owing to having occurred in model (2) the nonlinear elements such as sin θ, cos θ, adopt conventional method to analyze system, and the present invention adopts Taylor series that (2) formula is launched, and carries out linearization process.

By in formula (2)

sin θ, cos θ exist respectively

θ ₀place carries out single order Taylor series expansion,

sinθ＝sinθ ₀+cosθ ₀×(θ-θ ₀) (3)

cosθ＝cosθ ₀-sinθ ₀×(θ-θ ₀) (4)

${\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}^2={\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0^2+2{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0\&times;(\stackrel{\&CenterDot;}{\mathrm{\&theta;}}-{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0)---\left(5\right)$

Full scale equation (2) becomes

$\stackrel{\&CenterDot;\&CenterDot;}{r}\&times;(\sin {\mathrm{\&theta;}}_0+\cos {\mathrm{\&theta;}}_0\&times;(\mathrm{\&theta;}-{\mathrm{\&theta;}}_0))+2\stackrel{\&CenterDot;}{r}\stackrel{\&CenterDot;}{\mathrm{\&theta;}}\&times;(\cos {\mathrm{\&theta;}}_0-\sin {\mathrm{\&theta;}}_0\&times;(\mathrm{\&theta;}-{\mathrm{\&theta;}}_0))-$

$r\&times;({\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0^2+2{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0\&times;(\stackrel{\&CenterDot;}{\mathrm{\&theta;}}-{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0))\&times;(\sin {\mathrm{\&theta;}}_0+\cos {\mathrm{\&theta;}}_0\&times;(\mathrm{\&theta;}-{\mathrm{\&theta;}}_0))---\left(6\right)$

$+r\stackrel{\&CenterDot;\&CenterDot;}{\mathrm{\&theta;}}\&times;(\cos {\mathrm{\&theta;}}_0-\sin {\mathrm{\&theta;}}_0\&times;(\mathrm{\&theta;}-{\mathrm{\&theta;}}_0))=a_{\mathrm{my}}$

Formula (6) becomes through abbreviation

$\mathrm{a\&theta;}+b\stackrel{\&CenterDot;}{\mathrm{\&theta;}}+m\stackrel{\&CenterDot;\&CenterDot;}{\mathrm{\&theta;}}+q+k=a_{\mathrm{my}}---\left(7\right)$

Here

$a=\stackrel{\&CenterDot;\&CenterDot;}{r}\cos {\mathrm{\&theta;}}_0+r{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0^2\cos {\mathrm{\&theta;}}_0$

$b=2\stackrel{\&CenterDot;}{r}(\cos {\mathrm{\&theta;}}_0+{\mathrm{\&theta;}}_0\sin {\mathrm{\&theta;}}_0)-2r{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0(\sin {\mathrm{\&theta;}}_0-{\mathrm{\&theta;}}_0\cos {\mathrm{\&theta;}}_0)$

m＝r(cosθ ₀+θ ₀sinθ ₀)

$q=\stackrel{\&CenterDot;}{\mathrm{\&theta;}}\mathrm{\&theta;}(-2\stackrel{\&CenterDot;}{r}\sin {\mathrm{\&theta;}}_0-2\stackrel{\&CenterDot;}{r}\cos {\mathrm{\&theta;}}_0)-\stackrel{\&CenterDot;\&CenterDot;}{\mathrm{\&theta;}}\mathrm{\&theta;}r\sin {\mathrm{\&theta;}}_0$

$k=\stackrel{\&CenterDot;\&CenterDot;}{r}(\sin {\mathrm{\&theta;}}_0+{\mathrm{\&theta;}}_0\cos {\mathrm{\&theta;}}_0)+r{{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0}^2(\sin {\mathrm{\&theta;}}_0-{\mathrm{\&theta;}}_0\cos {\mathrm{\&theta;}}_0)$

θ ₀,

the selection of two parameters has a great impact for guidance precision.

The trajectory that aircraft follows the trail of the objective can be divided into two stages: in the starting stage, and trajectory tilt angle θ ≈ 0,

variation little, can be similar to and be seen as

in the time that aircraft approaches target, trajectory change of pitch angle is larger, θ with

all can not think 0 °.But consider, the target to be attacked of discussion of the present invention is ground maneuver target, not one is made large maneuvering target, therefore can suppose that trajectory inclination angle is to change within the specific limits.Therefore get endways two on trajectory with reference to the stage: one is initial time

the point in moment; Another elects end moment θ as ₀,

should be noted that the reference point of end

θ ₀choose for the precision of whole guidance and there is significant impact, during with application, should choose setting in actual Design of Guidance Law depending on concrete condition.

In the first stage, due to θ,

very little, cause q, k and other amount in formula (7) to compare in a small amount, therefore formula (7) is reduced to

$a_{\mathrm{my}1}=2\stackrel{\&CenterDot;}{r}\stackrel{\&CenterDot;}{\mathrm{\&theta;}}+r\stackrel{\&CenterDot;\&CenterDot;}{\mathrm{\&theta;}}+\stackrel{\&CenterDot;\&CenterDot;}{r}\mathrm{\&theta;}---\left(8\right)$

And for second stage, in the q in formula (7), contain non-linear relation, and be microvariations methods due to what adopt, the θ in q can think the θ that second stage adopts ₀therefore formula (7) is reduced to

$a_{\mathrm{my}2}=\mathrm{a\&theta;}+c\stackrel{\&CenterDot;}{\mathrm{\&theta;}}+d\stackrel{\&CenterDot;\&CenterDot;}{\mathrm{\&theta;}}+k---\left(9\right)$

Here,

$c=2\stackrel{\&CenterDot;}{r}(\cos {\mathrm{\&theta;}}_0+{\mathrm{\&theta;}}_0\sin {\mathrm{\&theta;}}_0)-2r{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0(\sin {\mathrm{\&theta;}}_0-{\mathrm{\&theta;}}_0\cos {\mathrm{\&theta;}}_0)+{\mathrm{\&theta;}}_0(-2\stackrel{\&CenterDot;}{r}\sin {\mathrm{\&theta;}}_0-2\stackrel{\&CenterDot;}{r}\cos {\mathrm{\&theta;}}_0)$

d＝r(cosθ ₀+θ ₀sinθ ₀)-θ ₀rsinθ ₀

In the derivation of said process, because the k value of formula (7) is very little, therefore ignored.

Due to choosing of stage one and stage two, play control action for the trajectory of whole system, so adopt this kind of variable element guidance law, select Proportional coefficient K ₁, K ₂carry out weights distribution, make K ₁+ K ₂=1, draw,

a _my＝K ₁a _my1+K ₂a _my2

$a_{\mathrm{my}}=\mathrm{l\&theta;}+g\stackrel{\&CenterDot;}{\mathrm{\&theta;}}+w\stackrel{\&CenterDot;\&CenterDot;}{\mathrm{\&theta;}};$

$l=k_{2}a+\stackrel{\&CenterDot;\&CenterDot;}{r}{\mathrm{<figure-callout\; id="1"\; label="k"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">k</figure-callout>}}_1;$ $g=2\stackrel{\&CenterDot;}{r}{\mathrm{<figure-callout\; id="1"\; label="k"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">k</figure-callout>}}_1+\mathrm{<figure-callout\; id="2"\; label="c\; k"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">c</figure-callout>}{k}_{}{}_2;$ w＝rk ₁+dk ₂；

4. there is the optimal guidance law design of terminal angle restriction

θ is trajectory tilt angle, in order to introduce the concept of optimum control, can make

θ＝θ _x+θ _f

$\stackrel{\&CenterDot;}{\mathrm{\&theta;}}={\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_x$

θ _xfor variable, θ _ffor the terminal trajectory tilt angle of expecting.

Order

x ₁＝θ _x

x ₂＝x ₁

Row are write state equation and are obtained

${\stackrel{\&CenterDot;}{x}}_1=x_2$

${\stackrel{\&CenterDot;}{x}}_2=\frac{-x_{1}l-x_{2}g+a_{\mathrm{my}}}{w}$

Abbreviation is

$\stackrel{\&CenterDot;}{X}=\mathrm{AX}+\mathrm{BU}$

$A=\left[\begin{array}{cc}0& 1\\ p& n\end{array}\right]$

$B=\left[\begin{array}{c}0\\ \frac{1}{w}\end{array}\right]$

p＝-l/w

n＝-g/w

u＝a _my

Initial conditions t=t ₀time,

x ₁(t ₀)＝θ(t ₀)+θ _f；

$x_2\left(t_0\right)=\stackrel{\&CenterDot;}{\mathrm{\&theta;}}\left(t_0\right);$

End conswtraint condition t=t _ftime,

x ₁(t _f)＝0；

x ₂(t _f)＝0；

Now θ=θ _f

Select quadratic performance index

$J=X^T\mathrm{FX}+1/2{\&Integral;}_{t_0}^{{\mathrm{<figure-callout\; id="2"\; label="t\; f\; u"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_f}{u}^{}{}^2\mathrm{dt},$

$x=\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right];$

$F=\left[\begin{array}{cc}{f}_{11}& f_{12}\\ f_{21}& f_{22}\end{array}\right]$

$A=\left[\begin{array}{cc}0& 1\\ p& n\end{array}\right];$

$B=\left[\begin{array}{c}0\\ \frac{1}{w}\end{array}\right],$ R＝1，Q＝0。

Obtaining optimal control law by the theory of optimal control is

$u^{*}=-R^{-1}{B}^{T}P{X}^{*}=-\left[\begin{array}{cc}0& \frac{1}{r}\end{array}\right]\left[\begin{array}{cc}{P}_{11}& P_{12}\\ P_{21}& P_{22}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=-\frac{1}{r}(P_{11}{x}_1+P_{22}{x}_2)$

In formula, P meets the matrix of Riccati equation.

Cause || F|| → ∝, separates Riccati equation and obtains

$\left\{\begin{array}{c}{\stackrel{\&CenterDot;}{P}}^{-1}-{\mathrm{AP}}^{-1}-P^{-1}{A}^T+{\mathrm{BR}}^{-1}{B}^T-P^{-1}{\mathrm{QP}}^{-1}=0\\ P^{-1}\left(t_f\right)=F^{-1}=0\end{array}\right.$

Order

$P^{-1}=\left[\begin{array}{cc}{q}_{11}& q_{12}\\ q_{21}& q_{22}\end{array}\right]$

Wherein q ₁₂=q ₂₁

Solve

$P=\left[\begin{array}{cc}2\mathrm{bt}& b\\ b& -\mathrm{nb}-\mathrm{mbt}\end{array}\right]$

Wherein

$t_{\mathrm{go}}=\frac{r}{\stackrel{\&CenterDot;}{r}},$ $b=\frac{1}{(3p-2n^2-2\mathrm{pnt}){\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">r</figure-callout>}}^2}$

$P=\left[\begin{array}{cc}\frac{n+\mathrm{pt}}{b(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}& \frac{1}{b(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}\\ \frac{1}{b(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}& \frac{-2t}{b(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}\end{array}\right]$

Obtaining optimal control law is

$u^{*}=-\frac{1}{r}(P_{11}{x}_1+P_{22}{x}_2)=-\frac{(\mathrm{pt}+n)(\mathrm{\&theta;}+{\mathrm{\&theta;}}_f)}{\mathrm{br}(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}+\frac{2t\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}{\mathrm{br}(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}$

In formula

$t_{\mathrm{go}}=\frac{r}{\stackrel{\&CenterDot;}{r}};$

$b=\frac{1}{(3p-2n^2-2\mathrm{pnt}){\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">r</figure-callout>}}^2};$

p＝-l/w；

n＝-g/w；

$l=k_{2}a+\stackrel{\&CenterDot;\&CenterDot;}{r}{\mathrm{<figure-callout\; id="1"\; label="k"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">k</figure-callout>}}_1;$

$g=2\stackrel{\&CenterDot;}{r}{\mathrm{<figure-callout\; id="1"\; label="k"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">k</figure-callout>}}_1+\mathrm{<figure-callout\; id="2"\; label="c\; k"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">c</figure-callout>}{k}_{}{}_2;$

w＝rk ₁+dk ₂；

$a=\stackrel{\&CenterDot;\&CenterDot;}{r}\cos {\mathrm{\&theta;}}_0+r{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0^2\cos {\mathrm{\&theta;}}_0;$

d＝r(cosθ ₀+θ ₀sinθ ₀)-θ ₀rsinθ ₀。

The optimal control law finally obtaining is

$u^{*}=-\frac{1}{r}(P_{11}{x}_1+P_{22}{x}_2)=-\frac{(\mathrm{pt}+n)(\mathrm{\&theta;}+{\mathrm{\&theta;}}_f)}{\mathrm{br}(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}+\frac{2t\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}{\mathrm{br}(2\mathrm{nt}+2{\mathrm{<figure-callout\; id="2"\; label="pt"\; filenames="HDA0000091628040000022.png,HDA0000091628040000023.png"\; state="\{\{state\}\}">pt</figure-callout>}}^2+1)}$

Realize angle from engineering, what the aircraft the present invention relates to adopted is laser radar target seeker, so r,

all can record; Remaining time t _gobe there is to considerable influence in guidance precision, due to r,

can record, so according to

can directly obtain t _goit is remaining time.Guidance precision is determined by laser radar target seeker.For

although laser radar target seeker can not record, can apply Robust Kalman Filter device and estimate

5. simulation example

In pitch plane, the original position of aircraft, speed parameter are:

x _m0＝0m，y _m0＝100m；

v _m0＝120m/s；

Target initial position, speed parameter are:

x _t0＝2000m，y _t0＝0m；

v _t0＝30m/s；

θ _f＝0.3rad；

(1)K ₁＝0.7；K ₂＝0.3

Adopt the Guidance Law described above simulation result that follows the trail of the objective as follows, trajectory as shown in Figure 5 a.

(2)K ₁＝0.6；K ₂＝0.4

Adopt the Guidance Law described above simulation result that follows the trail of the objective as follows, trajectory as shown in Figure 5 b.

Table 1 has provided different proportion COEFFICIENT K ₁, K ₂time, gained trajectory inclination angle transformation relation.

Table 1

Simulation result shows, the Guidance Law of the present invention's design has been realized the control to flight terminal trajectory tilt angle; K as seen from Figure 5 ₁, K ₂determine the shape of trajectory.

The final design requirement that the method for designing with end conswtraint angle of visible this kind based on virtual target point met.

Claims (5)

1. the guidance method with end conswtraint based on virtual target point, it is characterized in that: utilize the method for virtual target to realize aircraft and plunder and fly target of attack at optimum height, in real target, have an imaginary target in vain, in order to injure, unit is desired detonates highly the height of imagination target in spatial domain, default guidance law makes aircraft warhead activation in the time flying to imaginary target, now original employing is plunderred the mode that flies to attack real goal become fly to virtual target and by the guide mode that detonates of requirement attitude angle, the moment is found when target endways to realize aircraft, the autonomous tracking target of aircraft also reduces flying height, in the time injuring unit's desired height of formation and attitude of flight vehicle according to expectation, ignite warhead, thereby complete the object precisely striking target.

2. the guidance method with end conswtraint based on virtual target point as claimed in claim 1, is characterized in that: in above-mentioned aircraft flight process, Aircraft Angle of Attack is ignored, being approximately trajectory inclination angle is attitude of flight vehicle angle.

3. the guidance method with end conswtraint based on virtual target point as claimed in claim 1, is characterized in that: end trajectory is divided into two stages, adopts the method design guidance law of control with changed scale coefficient.

4. the guidance method with end conswtraint based on virtual target point as described in claim 1 or 2 or 3, it is characterized in that: adopt the method for optimum control to be write the conversion of trajectory equation as state equation, and realize the constraint to the trajectory angle of fall by method for optimally controlling.

5. the guidance method with end conswtraint based on virtual target point as claimed in claim 4, is characterized in that: described method for optimally controlling is realized as follows to the optimization detailed process of the Guidance Law terminal angle of fall:

The first step: determine the feasible moment point position of detonating;

Second step: the mathematical modeling of setting up aircraft and target relative motion relation;

Aircraft initial position is M point, and target is positioned at T point, and the distance of aircraft and target is r, and θ is trajectory inclination angle, and aircraft is y at the height of vertical plane;

rsinθ＝y （1）

(1) formula is carried out to secondary differentiate,

$\stackrel{\&CenterDot;\&CenterDot;}{r}\sin \mathrm{\&theta;}+2\stackrel{\&CenterDot;}{r}\stackrel{\&CenterDot;}{\mathrm{\&theta;}}\cos \mathrm{\&theta;}-r{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}^2\sin \mathrm{\&theta;}+r\stackrel{\&CenterDot;\&CenterDot;}{\mathrm{\&theta;}}\cos \mathrm{\&theta;}=a_{\mathrm{my}}$

Adopt Taylor series that (2) formula is launched, carry out linearization process, by formula (2)

sin θ, cos θ are respectively at trajectory tilt angle initial time

θ ₀place carries out single order Taylor series expansion,

sinθ＝sinθ ₀+cosθ ₀×(θ-θ ₀) （3）

cosθ＝cosθ ₀-sinθ ₀×(θ-θ ₀) （4）

${\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}^2={\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0^2+2{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0\&times;(\stackrel{\&CenterDot;}{\mathrm{\&theta;}}-{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0)---\left(5\right)$

Full scale equation (2) becomes

$\begin{array}{c}\stackrel{\&CenterDot;\&CenterDot;}{r}\&times;({\sin \mathrm{\&theta;}}_0+{\cos \mathrm{\&theta;}}_0\&times;(\mathrm{\&theta;}-{\mathrm{\&theta;}}_0))+2\stackrel{\&CenterDot;}{r}\stackrel{\&CenterDot;}{\mathrm{\&theta;}}\&times;({\cos \mathrm{\&theta;}}_0-{\sin \mathrm{\&theta;}}_0\&times;(\mathrm{\&theta;}-{\mathrm{\&theta;}}_0))-\\ r\&times;({\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0^2+2{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0\&times;(\stackrel{\&CenterDot;}{\mathrm{\&theta;}}-{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0))\&times;({\sin \mathrm{\&theta;}}_0+{\cos \mathrm{\&theta;}}_0\&times;(\mathrm{\&theta;}-{\mathrm{\&theta;}}_0))\\ +r\stackrel{\&CenterDot;\&CenterDot;}{\mathrm{\&theta;}}\&times;({\cos \mathrm{\&theta;}}_0-{\sin \mathrm{\&theta;}}_0\&times;(\mathrm{\&theta;}-{\mathrm{\&theta;}}_0))=a_{\mathrm{my}}\end{array}---\left(6\right)$

Formula (6) becomes through abbreviation

$\mathrm{a\&theta;}+b\stackrel{\&CenterDot;}{\mathrm{\&theta;}}+m\stackrel{\&CenterDot;\&CenterDot;}{\mathrm{\&theta;}}+q+k=a_{\mathrm{my}}---\left(7\right)$

$a=\stackrel{\&CenterDot;\&CenterDot;}{r}{\cos \mathrm{\&theta;}}_0+r{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0^2{\cos \mathrm{\&theta;}}_0$

$b=2\stackrel{\&CenterDot;}{r}({\cos \mathrm{\&theta;}}_0+{\mathrm{\&theta;}}_0{\sin \mathrm{\&theta;}}_0)-2r{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0({\sin \mathrm{\&theta;}}_0-{\mathrm{\&theta;}}_0{\cos \mathrm{\&theta;}}_0)$

m＝r(cosθ ₀+θ ₀sinθ ₀)

$q=\stackrel{\&CenterDot;}{\mathrm{\&theta;}}\mathrm{\&theta;}(-2\stackrel{\&CenterDot;}{r}{\sin \mathrm{\&theta;}}_0-2\stackrel{\&CenterDot;}{r}{\cos \mathrm{\&theta;}}_0)-\stackrel{\&CenterDot;\&CenterDot;}{\mathrm{\&theta;}}\mathrm{\&theta;r}{\sin \mathrm{\&theta;}}_0$

$k=\stackrel{\&CenterDot;\&CenterDot;}{r}({\sin \mathrm{\&theta;}}_0+{\mathrm{\&theta;}}_0{\cos \mathrm{\&theta;}}_0)+r{{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0}^2({\sin \mathrm{\&theta;}}_0-{\mathrm{\&theta;}}_0{\cos \mathrm{\&theta;}}_0);$

The 3rd step: get endways two on trajectory with reference to the stage: one is initial time; Another elects the end moment as; In the first stage, due to θ,

very little, cause q, k and other amount in formula (7) to compare in a small amount, therefore formula (7) is reduced to

$a_{\mathrm{my}1}=2\stackrel{\&CenterDot;}{r}\stackrel{\&CenterDot;}{\mathrm{\&theta;}}+r\stackrel{\&CenterDot;\&CenterDot;}{\mathrm{\&theta;}}+\stackrel{\&CenterDot;\&CenterDot;}{r}\mathrm{\&theta;}---\left(8\right)$

And for second stage, in the q in formula (7), contain non-linear relation, and be microvariations methods due to what adopt, the θ in q can think the θ that second stage adopts ₀therefore formula (7) is reduced to

$a_{\mathrm{my}2}=\mathrm{a\&theta;}+c\stackrel{\&CenterDot;}{\mathrm{\&theta;}}+d\stackrel{\&CenterDot;\&CenterDot;}{\mathrm{\&theta;}}+k---\left(9\right)$

$c=2\stackrel{\&CenterDot;}{r}({\cos \mathrm{\&theta;}}_0+{\mathrm{\&theta;}}_0{\sin \mathrm{\&theta;}}_0)-2r{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0({\sin \mathrm{\&theta;}}_0-{\mathrm{\&theta;}}_0{\cos \mathrm{\&theta;}}_0)+{\mathrm{\&theta;}}_0(-2\stackrel{\&CenterDot;}{r}{\sin \mathrm{\&theta;}}_0-2\stackrel{\&CenterDot;}{r}{\cos \mathrm{\&theta;}}_0)$

d＝r(cosθ ₀+θ ₀sinθ ₀)-θ ₀rsinθ ₀；

Adopt this kind of variable element guidance law, select Proportional coefficient K ₁, K ₂carry out weights distribution, make K ₁+ K ₂=1, draw,

a _my＝K ₁a _my1+K ₂a _my2

$a_{\mathrm{my}}=\mathrm{l\&theta;}+g\stackrel{\&CenterDot;}{\mathrm{\&theta;}}+w\stackrel{\&CenterDot;\&CenterDot;}{\mathrm{\&theta;}};$

$l=k_{2}a+\stackrel{\&CenterDot;\&CenterDot;}{r}{k}_1;$ $g=2\stackrel{\&CenterDot;}{r}{k}_1+c{k}_2;$ w＝rk ₁+dk ₂；

The 4th step: set up the optimal guidance law with terminal angle restriction;

θ is trajectory tilt angle, order $\begin{array}{c}\mathrm{\&theta;}={\mathrm{\&theta;}}_x+{\mathrm{\&theta;}}_f\\ \stackrel{\&CenterDot;}{\mathrm{\&theta;}}={\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_x\end{array},$ θ _xfor variable, θ _ffor the terminal trajectory tilt angle of expecting;

Order $\begin{array}{c}{x}_1={\mathrm{\&theta;}}_x\\ x_2={\stackrel{\&CenterDot;}{x}}_1\end{array},$ Row are write state equation and are obtained $\begin{array}{c}{\stackrel{\&CenterDot;}{x}}_1=x_2\\ {\stackrel{\&CenterDot;}{x}}_2=\frac{-x_{1}l-x_{2}g+a_{\mathrm{my}}}{w}\end{array},$

Abbreviation is $\stackrel{\&CenterDot;}{X}=\mathrm{AX}+\mathrm{BU},$

$A=\left[\begin{array}{cc}0& 1\\ p& n\end{array}\right]$

$B=\left[\begin{array}{c}0\\ \frac{1}{w}\end{array}\right]$

p＝-l/w

n=-g/w

u＝a _my

Initial conditions t=t ₀time, x ₁(t ₀)=θ (t ₀)+θ _f;

End conswtraint condition t=t _ftime, x ₁(t _f)=0; x ₂(t _f)=0; Now θ=θ _f;

Select quadratic performance index $J=X^T\mathrm{FX}+1/2{\&Integral;}_{t_0}^{t_f}{u}^2\mathrm{dt},$

$x=\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right];$ $F=\left[\begin{array}{cc}{f}_{11}& f_{12}\\ f_{21}& f_{22}\end{array}\right];$ $A=\left[\begin{array}{cc}0& 1\\ p& n\end{array}\right];$ $B=\left[\begin{array}{c}0\\ \frac{1}{w}\end{array}\right],$ R=1,Q=0

Obtaining optimal control law by the theory of optimal control is

$u^{*}=-R^{-1}{B}^T{\mathrm{PX}}^{*}=-\left[\begin{array}{cc}o& \frac{1}{r}\end{array}\right]\left[\begin{array}{cc}{P}_{11}& P_{12}\\ P_{21}& P_{22}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=-\frac{1}{r}(P_{11}{x}_1+P_{22}{x}_2)$

In formula, P meets the matrix of Riccati equation, R=1; For ensureing guidance precision, because of || F|| → ∝, separates Riccati equation and obtains

$\left\{\begin{array}{c}{\stackrel{\&CenterDot;}{P}}^{-1}-{\mathrm{AP}}^{-1}-P^{-1}{A}^T+{\mathrm{BR}}^{-1}{B}^T-P^{-1}{\mathrm{QP}}^{-1}=0\\ P^{-1}\left(t_f\right)=F^{-1}=0\end{array}\right.$

Order $P^{-1}=\left[\begin{array}{cc}{q}_{11}& q_{12}\\ q_{21}& q_{22}\end{array}\right],$ Wherein q ₁₂=q ₂₁

Solve $P=\left[\begin{array}{cc}2\mathrm{bt}& b\\ b& -\mathrm{nb}-\mathrm{mbt}\end{array}\right],$ Wherein $t_{\mathrm{go}}=\frac{r}{\stackrel{\&CenterDot;}{r}},$ $b=\frac{1}{(3p-{2n}^2-2\mathrm{pnt})r^2}$

$P=\left[\begin{array}{cc}\frac{n+\mathrm{pt}}{b(2\mathrm{nt}+2p{t}^2+1)}& \frac{1}{b(2\mathrm{nt}+2p{t}^2+1)}\\ \frac{1}{b(2\mathrm{nt}+2p{t}^2+1)}& \frac{-2t}{b(2\mathrm{nt}+2p{t}^2+1)}\end{array}\right]$

Obtaining optimal control law is

$u^{*}=-\frac{1}{r}(P_{11}{x}_1+P_{22}{x}_2)=-\frac{(\mathrm{pt}+n)(\mathrm{\&theta;}+{\mathrm{\&theta;}}_f)}{\mathrm{br}(2\mathrm{nt}+2p{t}^2+1)}+\frac{2t\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}{\mathrm{br}(2\mathrm{nt}+2p{t}^2+1)}$

In formula $t_{\mathrm{go}}=\frac{r}{\stackrel{\&CenterDot;}{r}};$ $b=\frac{1}{(3p-2n^2-2\mathrm{pnt})r^2};$ p＝-l/w；n=-g/w； $l=k_{2}a+\stackrel{\&CenterDot;\&CenterDot;}{r}{k}_1;$ $g=2\stackrel{\&CenterDot;}{r}{k}_1+c{k}_2;$ w＝rk ₁+dk ₂； $a=\stackrel{\&CenterDot;\&CenterDot;}{r}{\cos \mathrm{\&theta;}}_0+r{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_0^2{\cos \mathrm{\&theta;}}_0;$

d＝r(cosθ ₀+θ ₀sinθ ₀)-θ ₀rsinθ ₀

The optimal control law finally obtaining is

$u^{*}=-\frac{1}{r}(P_{11}{x}_1+P_{22}{x}_2)=-\frac{(\mathrm{pt}+n)(\mathrm{\&theta;}+{\mathrm{\&theta;}}_f)}{\mathrm{br}(2\mathrm{nt}+2p{t}^2+1)}+\frac{2t\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}{\mathrm{br}(2\mathrm{nt}+2p{t}^2+1)}.$

Images