CN102519473A - Mixed sine maneuvering path guiding method for high-paddle fundamental frequency satellite - Google Patents

Abstract

The invention relates to a mixed sine maneuvering path guiding method for a high-paddle fundamental frequency satellite. The mixed sine maneuvering path guiding method for a high-paddle fundamental frequency satellite is characterized in that based on reasonable distribution of maneuvering time and stabilization time, according to a moment and an angular momentum capacity of an execution mechanism, an angular acceleration curve is designed, wherein the angular acceleration curve shows an angular acceleration change rule that an angular acceleration is changed into zero from a positive value, then is changed into a negative value from zero and then is changed into zero from the negative value; and one half of a whole period of a sine curve mixed step function is a positive value curve part and the other half is a negative value curve part. The mixed sine maneuvering path guiding method for a high-paddle fundamental frequency satellite retains the advantage of small excitation of a sine maneuvering path on a flexible paddle, combines the advantage of time optimality of a step function, harmonizes the contradiction between maneuvering rapidity and paddle excitation smoothness, is especially suitable for providing path program for a large-angle, rapid and high-paddle fundamental frequency satellite in imaging breadth increasing, instant observation on a emergency area or stereo imaging rapid maneuvering, and satisfies requirements of rapid and stable maneuvering performances of a high-paddle fundamental frequency satellite.

Description

A kind of sinusoidal motor-driven path guide method of mixing that is suitable for the higher satellite of windsurfing fundamental frequency

Technical field

The invention belongs to spacecraft attitude control field, the path guide method when relating to a kind of flexible satellite fast reserve.

Background technology

Aspect the satellite fast reserve, in order to increase the fabric width that forms images, the accident area to be realized instant observation, when perhaps carrying out three-dimensional imaging, required regular side-sway automotive and pitching motor-driven.Big when the motor-driven angle of satellite, stabilization time short, when the degree of stability index is high, the windsurfing flexible vibration that mobile process evokes will become the principal element of restriction mobility.For the lower satellite of windsurfing fundamental frequency, sinusoidal motor-driven path can effectively reduce the excitation of large angle maneuver to flexible appendage, obtains higher stability.Although sinusoidal motor-driven path is compared with time optimal Bang-Bang path; Required motor-driven arrival time will be grown; But for the lower satellite of windsurfing fundamental frequency, delaying on the time kept in reserve can be contained in the outstanding benefit that obtains aspect the vibration of reduction windsurfing in sinusoidal motor-driven path.But, for the higher situation faster of motor-driven arrival time that requires simultaneously of fundamental frequency, adopt sinusoidal motor-driven path to vibrate the effect that obtains aspect less fully and compare with sacrificing motor-driven arrival time at windsurfing, not obvious to the contribution of fast reserve performance.Therefore need to the higher new solution of situation searching faster of motor-driven arrival time that requires simultaneously of fundamental frequency.

Summary of the invention

Technology of the present invention is dealt with problems and is: overcome the deficiency of prior art, provide that a kind of response speed is fast, good stability mix the motor-driven path guide method of satellite of stepped curve based on sine.

Technical solution of the present invention is: a kind of sinusoidal motor-driven path guide method of mixing that is suitable for the higher satellite of windsurfing fundamental frequency, and step is following:

(1) confirms time kept in reserve t according to the motor-driven and stable time index of the attitude of satellite _Max

(2) confirm maximum motor-driven angular acceleration a according to the topworks's ability that is disposed on the satellite _Max

(3) confirm maximum motor-driven limited angular velocity omega according to topworks's angular momentum capacity that is disposed on the satellite and sensor range _Max

(4) attitude of satellite is the motor-driven path time that is divided into is [t ₁t ₂] the rate of acceleration section of increasing progressively, time be [t ₂t ₃] even accelerating sections, time be [t ₃t ₄] rate of acceleration decline fraction, time be [t ₄t ₅] at the uniform velocity coasting-flight phase, time be [t ₅t ₆] the rate of deceleration section of increasing progressively, time be [t ₆t ₇] even braking section, time be [t ₇t ₈] rate of deceleration decline fraction, t wherein ₁Be the motor-driven start time of satellite,

t ₃=t ₂+ Δ T, t ₅=t ₄+ Δ t,

t ₇=t ₆+ Δ T,

T is the cycle of sine function, and the initial value of T does

Δ T is the even acceleration time, and the initial value of Δ T does

Δ t is a coasting time, and the initial value of Δ t does $\mathrm{\Δt}=\frac{{2\mathrm{\π\ω}}_{\max }}{(\mathrm{\π}-2)a_{\max }}+\frac{4\left|{\mathrm{\θ}}_d\right|}{(\mathrm{\π}-2){\mathrm{\ω}}_{\max }}-\frac{6-\mathrm{\π}}{\mathrm{\π}-2}{t}_{\max };$

(5) be a according to the cycle T of sine function, maximum motor-driven angular acceleration _MaxAnd even acceleration time Δ T, the coasting time Δ t of satellite calculate in each section satellite respectively based on the attitude angle acceleration a of sine mixing step function, and corresponding attitude angular velocity ω and attitude angle θ, wherein:

(51) the rate of acceleration section of increasing progressively, $a=a_{\max }\sin \left(\frac{2\mathrm{\&pi;}}{T}(t-t_1)\right),$ $\mathrm{\&omega;}=\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}T}{2\mathrm{\&pi;}}(1-\cos \left(\frac{2\mathrm{\&pi;}}{T}(t-t_1)\right)),$

θ wherein _dBe motor-driven angle; If, then upgrade according to Δ T＜0 that initial value calculates

Δ T=0,

If the T+2 Δ T＜T that calculates according to initial value _Min, then upgrade T=T _Min, the renewal maximum angular acceleration does

Δ T=0, Δ t=0, T _MinBe the shortest patient time kept in reserve;

(52) even accelerating sections, a=a _Max, $\mathrm{\&omega;}=\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}T}{2\mathrm{\&pi;}}+a_{\max }(t-t_2),$

$\mathrm{\&theta;}=\frac{a_{\max }{T}^2(\mathrm{\&pi;}-2)}{8{\mathrm{\&pi;}}^2}+\frac{a_{\max }T(t-t_2)}{2\mathrm{\&pi;}}+\frac{a_{\max }{(t-t_2)}^2}{2},$ Wherein the value of T, Δ T, Δ t is with the rate of acceleration section of increasing progressively;

(53) rate of acceleration decline fraction, $a=a_{\max }\sin \left(\frac{2\mathrm{\&pi;}}{T}(t-{\mathrm{<figure-callout\; id="3"\; label="t"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">t</figure-callout>}}_3+\frac{T}{4})\right),$

$\mathrm{\&omega;}=a_{\max }\mathrm{\&Delta;T}+\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}T}{2\mathrm{\&pi;}}(1-\cos \left(\frac{2\mathrm{\&pi;}}{T}(t-{\mathrm{<figure-callout\; id="3"\; label="t"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">t</figure-callout>}}_3+\frac{T}{4})\right)),$ $\mathrm{\&theta;}\left(t_4\right)=\frac{a_{\max }{\mathrm{\&Delta;<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{2}+\frac{a_{\max }\mathrm{\&Delta;TT}}{4}+\frac{a_{\max }\mathrm{T\&Delta;<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}{2\mathrm{\&pi;}}+\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}{T}^{}{}^2}{4\mathrm{\&pi;}},$

Wherein the value of T, Δ T, Δ t is with the rate of acceleration section of increasing progressively;

(54) coasting-flight phase at the uniform velocity, a=0, $\mathrm{\&omega;}=\frac{a_{\max }T}{\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T},$

$\mathrm{\&theta;}=\frac{a_{\max }{\mathrm{\&Delta;<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{2}+a_{\max }\mathrm{T\&Delta;T}(\frac{1}{4}+\frac{1}{2\mathrm{\&pi;}})+\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}{T}^{}{}^2}{4\mathrm{\&pi;}}+(\frac{a_{\max }T}{\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T})(t-t_4),$ Wherein the value of T, Δ T, Δ t is with the rate of acceleration section of increasing progressively;

(55) rate of deceleration section of increasing progressively, $a=a_{\max }\sin \left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;t}-\mathrm{\&Delta;T})\right),$

$\mathrm{\&omega;}=a_{\max }\mathrm{\&Delta;T}+\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}T}{2\mathrm{\&pi;}}(1-\cos \left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;t}-\mathrm{\&Delta;T})\right)),$

$\mathrm{\&theta;}=\frac{a_{\max }{\mathrm{\&Delta;<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{2}+a_{\max }\mathrm{T\&Delta;T}(\frac{1}{4}+\frac{1}{2\mathrm{\&pi;}})+\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}{T}^{}{}^2}{4\mathrm{\&pi;}}+\frac{a_{\max }T}{\mathrm{\&pi;}}\mathrm{\&Delta;t}+a_{\max }\mathrm{\&Delta;T}(t-t_5+\mathrm{\&Delta;t})$

$+\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}T}{2\mathrm{\&pi;}}(t-t_5)-\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}{T}^{}{}^2}{4{\mathrm{\&pi;}}^2}\sin \left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;t}-\mathrm{\&Delta;T})\right),$

If, then upgrade

Δ t=0 according to Δ t＜0 that initial calculation goes out; If Δ T＜0 that calculates according to initial value; Then upgrade

Δ T=0 $\mathrm{\&Delta;\; t}=\frac{\mathrm{\&pi;}}{a_{\mathrm{Max}}T}\left(\right|{\mathrm{\&theta;}}_d|-\frac{\left|a_{\mathrm{Max}}\right|{\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{2\mathrm{\&pi;}});$ If $\mathrm{\&Delta;\; t}=\frac{\mathrm{\&pi;}}{a_{\mathrm{Max}}T}\left(\right|{\mathrm{\&theta;}}_d|-\frac{\left|a_{\mathrm{Max}}\right|{\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{2\mathrm{\&pi;}})<0,$ Then upgrade $T=\sqrt{2\mathrm{\&pi;}\left|\frac{{\mathrm{\&theta;}}_d}{a_{\mathrm{Max}}}\right|},$ Δ T=0, Δ t=0;

(56) even braking section, a=-a _Max, $\mathrm{\&omega;}=\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}T}{2\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T}-a_{\max }(t-t_6),$

$\begin{array}{c}\mathrm{\&theta;}=\frac{a_{\max }{\mathrm{\&Delta;<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{2}+\frac{a_{\max }\mathrm{\&Delta;TT}(\mathrm{\&pi;}+1)}{2\mathrm{\&pi;}}+(\frac{a_{\max }T}{\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T})\mathrm{\&Delta;t}+\frac{a_{\max }{T}^2(3\mathrm{\&pi;}+2)}{{8\mathrm{\&pi;}}^2}\\ +(\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}T}{2\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T})(t-t_6)-\frac{a_{\max }{(t-t_6)}^2}{2}\end{array},$ Wherein the value of T, Δ T, Δ t is with the rate of deceleration section of increasing progressively;

(57) rate of deceleration decline fraction, $a=a_{\max }\sin \left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;t}-2\mathrm{\&Delta;T})\right),$

$\mathrm{\&omega;}=\frac{a_{\max }\&CenterDot;\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}{2\mathrm{\&pi;}}(1-\cos \left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;t}-2\mathrm{\&Delta;T})\right)),$

$\mathrm{\&theta;}=\frac{a_{\max }\mathrm{\&Delta;TT}(\mathrm{\&pi;}+2)}{2\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T\&Delta;t}+\frac{{3a}_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}{T}^{}{}^2}{8\mathrm{\&pi;}}+a_{\max }{\mathrm{\&Delta;<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2$

$+\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}T}{2\mathrm{\&pi;}}(t-t_7+2\mathrm{\&Delta;t})-\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}{T}^{}{}^2}{{4\mathrm{\&pi;}}^2}\sin \left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;t}-2\mathrm{\&Delta;T})\right),$

Wherein the value of T, Δ T, Δ t is with the rate of deceleration section of increasing progressively;

(6) the described path of tracking step (5) during the motor-driven control of satellite.

The present invention's advantage compared with prior art is: in the fast reserve process that has the complicated satellite of the higher flexible appendage of fundamental frequency; Adopt the motor-driven path of the sine mixing step of the inventive method; On the basis of reasonable distribution time kept in reserve and stabilization time; According to the moment and the angular momentum capacity of topworks, design earlier after just be zero again for negative be zero angular acceleration curve again, value is the sinusoidal curve mixing step function curve that the part of positive and negative is respectively half period.The inventive method is compared with traditional bang-bang path and standard sine path, has both weakened that flexible appendage has shortened the time kept in reserve again to the influence of satellite degree of stability under the large angle maneuver, thereby has ensured the fast and stable time after the fast reserve.

Description of drawings

Fig. 1 is the FB(flow block) of the inventive method;

The motor-driven path of Fig. 2 for adopting the inventive method to obtain;

Fig. 3 is the motor-driven path of a bang-bang satellite axis of rolling attitude angular velocity curve;

Fig. 4 is the motor-driven path of a standard sine satellite axis of rolling attitude angular velocity curve;

The satellite axis of rolling attitude angular velocity curve of Fig. 5 for adopting the inventive method to obtain;

Fig. 6 is a satellite sailboard flexible mode coordinate displacement under the motor-driven path of bang-bang;

Fig. 7 is a satellite sailboard flexible mode coordinate displacement under the sinusoidal motor-driven path;

The satellite sailboard flexible mode coordinate displacement of Fig. 8 for adopting the inventive method to obtain.

Embodiment

Present surveillance satellite has wide-angle fast reserve and stable demand, and motor-driven angle is big, and stabilization time is short, and stability index is high, and satellite improves the windsurfing height usually in order to satisfy the fast reserve ability, and fundamental frequency is higher.Therefore, need to adopt to guarantee the time kept in reserve rapidity, can guarantee the motor-driven path of less excitation windsurfing flexible vibration again.The inventive method has kept sinusoidal motor-driven path to the less advantage of flexible windsurfing excitation; Combine simultaneously the time optimal characteristics of step function again; When slowing down the motor-driven jigger panel vibration that puts in place; Optimize the time kept in reserve as far as possible, coordinated motor-driven rapidity and windsurfing and encouraged the contradiction between the mild property, be particularly suitable for wide-angle, quick, the motor-driven path design of the higher satellite of windsurfing fundamental frequency.

As shown in Figure 1, be the process flow diagram of the inventive method.

Among the present invention, angular acceleration adopts the sinusoidal step function form of mixing, and promptly becomes accelerating sections

Even accelerating sections a=a _Max, wherein a is an angular acceleration, a _MaxBe the motor-driven angular acceleration of maximum, be subject to topworks's ability and sensor range; T is the planning acceleration cycle; The even acceleration time is Δ T.Therefore, are acceleration time and deceleration time

Consider the time kept in reserve constraint, suppose that the maximum time kept in reserve is t _Max, the situation that large angle maneuver and maximum angular rate are limited supposes that limited maximum angular velocity is ω _Max, the time of at the uniform velocity sliding between accelerating sections and the braking section is Δ t.

According to the attitude angular velocity after the accelerating sections end

Can know Confirm the funtcional relationship between trigonometric function cycle T and the even acceleration time Δ T according to this inequality, promptly

The maximum time kept in reserve is t _Max, can know time kept in reserve T+2 Δ T+ Δ t≤t _MaxConfirm the funtcional relationship between trigonometric function cycle T, even acceleration time Δ T, the coasting time Δ t, i.e. T+2 Δ T+ Δ t=t according to this inequality _Max

If T+2 Δ T>=T _Min, can further motor-driven as required angle θ _dConfirm coasting time Δ t, $\mathrm{\&Delta;\; t}=\frac{{2\mathrm{\&pi;\; \&omega;}}_{\mathrm{Max}}}{(\mathrm{\&pi;}-2)a_{\mathrm{Max}}}+\frac{4\left|{\mathrm{\&theta;}}_d\right|}{(\mathrm{\&pi;}-2){\mathrm{\&omega;}}_{\mathrm{Max}}}-\frac{6-\mathrm{\&pi;}}{\mathrm{\&pi;}-2}{t}_{\mathrm{Max}},$ T in the formula _MinBe the minimum period of setting according to the time kept in reserve, θ _dBe motor-driven angle, and then calculate $T=\frac{2\mathrm{\&pi;}(t_{\mathrm{Max}}-\frac{{\mathrm{\&omega;}}_{\mathrm{Max}}}{a_{\mathrm{Max}}}-\frac{\left|{\mathrm{\&theta;}}_d\right|}{{\mathrm{\&omega;}}_{\mathrm{Max}}})}{\mathrm{\&pi;}-2},$ $\mathrm{\&Delta;\; T}=\frac{{\mathrm{\&omega;}}_{\mathrm{Max}}}{a_{\mathrm{Max}}}-\frac{T}{\mathrm{\&pi;}};$ If coasting time Δ t＜0 that calculates is obviously inadvisable, then make Δ t=0, according to time function relation

And the motor-driven angle that puts in place $\mathrm{\&theta;}\left(t_8\right)=\frac{a_{\mathrm{Max}}\mathrm{\&Delta;\; <figure-callout\; id="2"\; label="TT"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">TT</figure-callout>}}{2}+(\frac{a_{\mathrm{Max}}T}{\mathrm{\&pi;}}+a_{\mathrm{Max}}\mathrm{\&Delta;\; T})\mathrm{\&Delta;\; t}+\frac{a_{\mathrm{Max}}{\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{2\mathrm{\&pi;}}+\frac{a_{\mathrm{Max}}\mathrm{T\&Delta;\; T}}{\mathrm{\&pi;}}+a_{\mathrm{Max}}{\mathrm{\&Delta;\; <figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2,$ Get with Δ t=0 homographic solution

If T+2 Δ T＜T _Min, show maximum motor-driven angular acceleration a _MaxShould be littler than the angular acceleration under topworks's ability and the sensor limit of range, therefore upgrade a _Max, make T=T _Min, Δ T=0, Δ t=0 calculates

The derivation in motor-driven path is following:

(1) the rate of acceleration section of increasing progressively: [t ₁t ₂], t ₁Be the motor-driven start time,

With angular acceleration a at time period [t ₁T] go up time integral, can get through calculating

Same angle speed ω is at time period [t ₁T] go up time integral, can get

At time t ₁And t ₂Point has a (t respectively ₁)=0, a (t ₂)=a _Max, ω (t ₁)=0, $\mathrm{\&omega;}\left(t_2\right)=\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}T}{2\mathrm{\&pi;}},$ θ(t ₁)＝0， $\mathrm{\&theta;}\left(t_2\right)=\frac{a_{\max }{T}^2(\mathrm{\&pi;}-2)}{{8\mathrm{\&pi;}}^2}.$

(2) even accelerating sections: [t ₂t ₃], t ₃=t ₂+ Δ T; A=a _MaxWarp is at time period [t ₂T] go up angular acceleration a and carry out integration and quadratic integral respectively, can get attitude angular velocity and attitude angle is respectively $\mathrm{\&omega;}=\frac{a_{\mathrm{Max}}\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}{2\mathrm{\&pi;}}+a_{\mathrm{Max}}(t-t_2);$ $\mathrm{\&theta;}=\frac{a_{\mathrm{Max}}{T}^2(\mathrm{\&pi;}-2)}{{8\mathrm{\&pi;}}^2}+\frac{a_{\mathrm{Max}}T(t-t_2)}{2\mathrm{\&pi;}}+\frac{a_{\mathrm{Max}}{(t-t_2)}^2}{2},$ At time t ₃The point, $\mathrm{\&omega;}\left(t_3\right)=\frac{a_{\mathrm{Max}}\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}{2\mathrm{\&pi;}}+a_{\mathrm{Max}}\mathrm{\&Delta;\; T},$ $\mathrm{\&theta;}\left(t_3\right)=\frac{a_{\mathrm{Max}}\&CenterDot;T^2(\mathrm{\&pi;}-2)}{{8\mathrm{\&pi;}}^2}+\frac{a_{\mathrm{Max}}\mathrm{T\&Delta;\; <figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}{2\mathrm{\&pi;}}+\frac{a_{\mathrm{Max}}\&CenterDot;{\mathrm{\&Delta;\; <figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{2}.$

(3) rate of acceleration decline fraction: [t ₄t ₄], $t_4={\mathrm{<figure-callout\; id="3"\; label="t"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">t</figure-callout>}}_3+\frac{T}{4};$ $a=a_{\mathrm{Max}}\mathrm{Sin}\left(\frac{2\mathrm{\&pi;}}{T}(t-{\mathrm{<figure-callout\; id="3"\; label="t"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">t</figure-callout>}}_3+\frac{T}{4})\right);$ With angular acceleration a at time period [t ₃T] go up time integral, can get through calculating $\mathrm{\&omega;}=a_{\mathrm{Max}}\mathrm{\&Delta;\; T}+\frac{a_{\mathrm{Max}}\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}{2\mathrm{\&pi;}}(1-\mathrm{Cos}\left(\frac{2\mathrm{\&pi;}}{T}(t-{\mathrm{<figure-callout\; id="3"\; label="t"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">t</figure-callout>}}_3+\frac{T}{4})\right));$ Angle speed ω is at time period [t ₃T] go up time integral, $\mathrm{\&theta;}=\frac{a_{\mathrm{Max}}{\mathrm{\&Delta;\; <figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{2}+a_{\mathrm{Max}}\mathrm{\&Delta;\; T}(t-t_3)+\frac{a_{\mathrm{Max}}\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}{2\mathrm{\&pi;}}(t-{\mathrm{<figure-callout\; id="3"\; label="t"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">t</figure-callout>}}_3+\frac{T}{4}+\mathrm{\&Delta;\; T})-\frac{a_{\mathrm{Max}}{\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{{4\mathrm{\&pi;}}^2}\mathrm{Sin}\left(\frac{2\mathrm{\&pi;}}{T}(t-{\mathrm{<figure-callout\; id="3"\; label="t"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">t</figure-callout>}}_3+\frac{T}{4})\right),$ At time t ₄Point, a (t ₄)=0, $\mathrm{\&omega;}\left(t_4\right)=\frac{a_{\mathrm{Max}}T}{\mathrm{\&pi;}}+a_{\mathrm{Max}}\mathrm{\&Delta;\; T},$ $\mathrm{\&theta;}\left(t_4\right)=\frac{a_{\mathrm{Max}}{\mathrm{\&Delta;\; <figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{2}+\frac{a_{\mathrm{Max}}\mathrm{\&Delta;\; TT}}{4}+\frac{a_{\mathrm{Max}}\mathrm{T\&Delta;\; <figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}{2\mathrm{\&pi;}}+\frac{a_{\mathrm{Max}}{\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{4\mathrm{\&pi;}}.$

(4) coasting-flight phase at the uniform velocity: [t ₄t ₅], t ₅=t ₄+ Δ t; A=0; Through calculating,

$\mathrm{\&theta;}=\frac{a_{\mathrm{Max}}{\mathrm{\&Delta;\; <figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{2}+a_{\mathrm{Max}}\mathrm{\&Delta;\; TT}(\frac{1}{4}+\frac{1}{2\mathrm{\&pi;}})+\frac{a_{\mathrm{Max}}{\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{4\mathrm{\&pi;}}+(\frac{a_{\mathrm{Max}}T}{\mathrm{\&pi;}}+a_{\mathrm{Max}}\mathrm{\&Delta;\; T})(t-t_4),$ At time t ₅The point $\mathrm{\&omega;}\left(t_5\right)=\frac{a_{\mathrm{Max}}T}{\mathrm{\&pi;}}+a_{\mathrm{Max}}\mathrm{\&Delta;\; T},$ $\mathrm{\&theta;}\left(t_5\right)=\frac{a_{\mathrm{Max}}{\mathrm{\&Delta;\; <figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{2}+\frac{a_{\mathrm{Max}}\mathrm{T\&Delta;\; T}}{4}+\frac{a_{\mathrm{Max}}{\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{4\mathrm{\&pi;}}+(\frac{a_{\mathrm{Max}}T}{\mathrm{\&pi;}}+a_{\mathrm{Max}}\mathrm{\&Delta;\; T})\mathrm{\&Delta;\; t}.$

(5) rate of deceleration section of increasing progressively: [t ₅t ₆], $t_6=t_5+\frac{T}{4};$ $a=a_{\mathrm{Max}}\&CenterDot;\mathrm{Sin}\left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;\; t}-\mathrm{\&Delta;\; T})\right);$ With angular acceleration a at time period [t ₅T] go up time integral, can get through calculating $\mathrm{\&omega;}=a_{\mathrm{Max}}\&CenterDot;\mathrm{\&Delta;\; T}+\frac{a_{\mathrm{Max}}\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}{2\mathrm{\&pi;}}(1-\mathrm{Cos}\left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;\; t}-\mathrm{\&Delta;\; T})\right));$ With angular velocity omega at time period [t ₅T] upper integral, $\begin{array}{c}\mathrm{\&theta;}=\frac{a_{\mathrm{Max}}{\mathrm{\&Delta;\; <figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{2}+a_{\mathrm{Max}}\mathrm{T\&Delta;\; T}(\frac{1}{4}+\frac{1}{2\mathrm{\&pi;}})+\frac{a_{\mathrm{Max}}{\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{4\mathrm{\&pi;}}+\frac{a_{\mathrm{Max}}T}{\mathrm{\&pi;}}\mathrm{\&Delta;\; t}+a_{\mathrm{Max}}\mathrm{\&Delta;\; T}(t-t_5+\mathrm{\&Delta;\; t})\\ +\frac{a_{\mathrm{Max}}\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}{2\mathrm{\&pi;}}(t-t_5)-\frac{a_{\mathrm{Max}}{\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{4{\mathrm{\&pi;}}^2}\mathrm{Sin}\left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;\; t}-\mathrm{\&Delta;\; T})\right)\end{array};$ At time point t ₆On, a (t ₆The a of)=- _Max, $\mathrm{\&omega;}\left(t_6\right)=\frac{a_{\mathrm{Max}}\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}{2\mathrm{\&pi;}}+a_{\mathrm{Max}}\mathrm{\&Delta;\; T},$

$\mathrm{\&theta;}\left(t_6\right)=\frac{a_{\max }{\mathrm{\&Delta;<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{2}+\frac{a_{\max }\mathrm{\&Delta;<figure-callout\; id="2"\; label="TT"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">TT</figure-callout>}}{2}+\frac{a_{\max }\mathrm{\&Delta;<figure-callout\; id="2"\; label="TT"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">TT</figure-callout>}}{2\mathrm{\&pi;}}+(\frac{a_{\max }T}{\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T})\mathrm{\&Delta;t}+\frac{{3a}_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}{T}^{}{}^2}{8\mathrm{\&pi;}}+\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}{T}^{}{}^2}{4{\mathrm{\&pi;}}^2}.$

(6) even braking section: [t ₆t ₇], t ₇=t ₆+ Δ T; A=-a _MaxWith angular acceleration a at time period [t ₆T] upward can get time integral

With angular velocity omega at time period [t ₆T] upper integral, can get through calculating

$\mathrm{\&theta;}=\frac{a_{\max }{\mathrm{\&Delta;<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2}{2}+\frac{a_{\max }\mathrm{\&Delta;TT}(\mathrm{\&pi;}+1)}{2\mathrm{\&pi;}}+(\frac{a_{\max }T}{\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T})\mathrm{\&Delta;t}+\frac{a_{\max }{T}^2(3\mathrm{\&pi;}+2)}{{8\mathrm{\&pi;}}^2}$ ；

$+(\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}T}{2\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T})(t-t_6)-\frac{a_{\max }{(t-t_6)}^2}{2}$

At time point t ₇On, calculating can get,

$\mathrm{\&theta;}\left(t_7\right)=\frac{a_{\max }\mathrm{\&Delta;<figure-callout\; id="2"\; label="TT"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">TT</figure-callout>}}{2}+(\frac{a_{\max }T}{\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T})\mathrm{\&Delta;t}+\frac{{3a}_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}{T}^{}{}^2}{8\mathrm{\&pi;}}+\frac{a_{\max }\mathrm{T\&Delta;T}}{\mathrm{\&pi;}}+a_{\max }{\mathrm{\&Delta;<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2+\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}{T}^{}{}^2}{{4\mathrm{\&pi;}}^2}.$

(7) rate of deceleration decline fraction: [t ₇t ₈], $t_8=t_7+\frac{T}{4};$ $a=a_{\mathrm{Max}}\mathrm{Sin}\left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;\; t}-2\mathrm{\&Delta;\; T})\right);$ With angular acceleration a at time period [t ₇T] go up time integral, can get through calculating, $\mathrm{\&omega;}=\frac{a_{\mathrm{Max}}\&CenterDot;\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}{2\mathrm{\&pi;}}(1-\mathrm{Cos}\left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;\; t}-2\mathrm{\&Delta;\; T})\right));$ With angular velocity omega at time period [t ₇T] upper integral, can get through calculating

$\mathrm{\&theta;}=\frac{a_{\max }\mathrm{\&Delta;TT}(\mathrm{\&pi;}+2)}{2\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T\&Delta;t}+\frac{{3a}_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}{T}^{}{}^2}{8\mathrm{\&pi;}}+a_{\max }{\mathrm{\&Delta;<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2$

$+\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}T}{2\mathrm{\&pi;}}(t-t_7+2\mathrm{\&Delta;t})-\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}{T}^{}{}^2}{{4\mathrm{\&pi;}}^2}\sin \left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;t}-2\mathrm{\&Delta;T})\right);$

ω (t wherein ₈)=0, $\mathrm{\&theta;}\left(t_8\right)=\frac{a_{\max }\mathrm{\&Delta;<figure-callout\; id="2"\; label="TT"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">TT</figure-callout>}}{2}+(\frac{a_{\max }T}{\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T})\mathrm{\&Delta;t}+\frac{a_{\mathrm{<figure-callout\; id="2"\; label="max\; T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">max</figure-callout>}}{T}^{}{}^2}{2\mathrm{\&pi;}}+\frac{a_{\max }\mathrm{T\&Delta;T}}{\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;}{\mathrm{<figure-callout\; id="2"\; label="T"\; filenames="HSA00000633196800021.png"\; state="\{\{state\}\}">T</figure-callout>}}^2.$

The acceleration calculation in above-mentioned each stage is only considered the situation of forward angular movement, need get negative like the motor-driven angle of reality for negative then acceleration.

Embodiment 1

Path design with the motor-driven 70 ° of processes of certain typical complicated satellite axis of rolling is an example, supposes satellite moment of inertia 2962kg.m ², topworks is a control-moment gyro, and the maximum moment that can provide is 20Nm, and the angular momentum capacity is Nms, and sensor is a gyro, maximum angle measurement speed is 2.5 °/s, requires motor-driven and be 80s stabilization time.The maximum time kept in reserve of then obtaining satellite according to step (1) is 40s, and the maximum motor-driven angular acceleration that obtains satellite according to step (2) is 0.3869 °/s ², obtaining maximum angular rate according to step (3) is 2.5 °/s.Consider in the Project Realization and need leave certain surplus, so to choose maximum motor-driven angular acceleration during design path be a for gyro to measure and topworks _Max=0.3 °/s ², maximum angular rate ω _Max=2.4 °/s.According to step (4), calculate the sine trigonometric function cycle $T=\frac{2\mathrm{\&pi;}(t_{\mathrm{Max}}-\frac{{\mathrm{\&omega;}}_{\mathrm{Max}}}{a_{\mathrm{Max}}}-\frac{{\mathrm{\&theta;}}_d}{{\mathrm{\&omega;}}_{\mathrm{Max}}})}{\mathrm{\&pi;}-2}=15.5943s.$ Choose minimum period T _Min=8s, the even acceleration time

According to step (5), calculate coasting time $\mathrm{\&Delta;\; t}=\frac{{2\mathrm{\&pi;\; \&omega;}}_{\mathrm{Max}}}{(\mathrm{\&pi;}-2)a_{\mathrm{Max}}}+\frac{4\left|{\mathrm{\&theta;}}_d\right|}{(\mathrm{\&pi;}-2){\mathrm{\&omega;}}_{\mathrm{Max}}}-\frac{6-\mathrm{\&pi;}}{\mathrm{\&pi;}-2}{t}_{\mathrm{Max}}=18.3333s.$ Suppose t ₁=0s, it is following to calculate motor-driven path:

The rate of acceleration section of increasing progressively, time [0 3.8986] s, a=0.3sin (0.1283 π t) °/s ²,

ω＝0.7446(1-cos(0.1283πt))，

θ＝0.7446(t-2.4819sin(0.1283πt))；

Even accelerating sections, time [3.89866.9348] s, a=0.3 °/s ²,

ω＝0.7446+0.3(t-3.8986)°/s，

θ＝1.0548+7.446(t-3.8986)+0.15(t-3.8986) ²°；

The rate of acceleration decline fraction, time [6.934810.8334] s, a=0.3sin (0.1283 (t-3.0362)) °/s ²,

ω＝0.9109+0.7446(1-cos(0.1283(t-6.9348)))°/s，

θ＝3.6434+1.6555(t-6.9348)-1.8480sin(0.4029(t-6.9348)))°；

Coasting-flight phase, time [10.833429.1667] s, a=0 °/s ²,

ω＝2.4001°/s，

θ＝13+2.4(t-10.8334)°；

The rate of deceleration section of increasing progressively, time [29.166733.0653] s, a=0.3sin (0.1283 π (t-21.3695)) °/s ²,

ω＝1.6555-0.7446cos(0.1283π(t-21.3695))°/s，

θ＝56.9999+1.6554(t-29.1667)-1.848sin(0.1283π(t-21.3695))；

Even braking section, time [33.065336.1015] s, a=-0.3 °/s ²,

ω＝1.6555-0.3(t-33.0653)°/s，

θ＝65.3019+1.6554(t-33.0653)-0.15(t-33.0653) ²；

Rate of deceleration decline fraction, time [36.101540.0001] s, a=0.3sin (0.1283 π (t-24.4057)) °/s ²,

ω＝0.7446-0.7446cos(0.1283π(t-24.4057))°/s，

θ＝67.0974+0.7446(t-36.1015)-1.848sin(0.1283π(t-24.4057))。

The direction of attitude angle acceleration a is identical with the actual motor-driven direction of satellite.

Shown in Figure 2 is attitude angle acceleration, the attitude angular velocity according to the design of above result of calculation, the reserve road diametal curve of attitude angle.

Embodiment 2

With 70 ° of processes of certain typical complicated satellite axis of rolling 80s fast reserve is example, supposes satellite moment of inertia 2962kg.m ², flexible windsurfing fundamental frequency 0.5Hz, control cycle is 0.1s, through calculating, chooses sine trigonometric function cycle T=20s, maximum angular acceleration a _Max=0.3 °/s ², maximum angular rate ω _Max2.4 °/s.Find that through mathematical simulation as shown in Figure 3, under the motor-driven path of bang-bang, the maximum tracking error of attitude angular velocity is 1.2 °/s in the mobile process; As shown in Figure 4, under sinusoidal motor-driven path, the maximum tracking error of attitude angular velocity is 0.7 °/s in the mobile process; As shown in Figure 5, mix based on sine under the motor-driven path of stepped curve, the maximum tracking error of attitude angular velocity is 0.9 °/s in the mobile process; Significantly reduce than the motor-driven path of bang-bang.As shown in Figure 6, under the motor-driven path of bang-bang, the modal coordinate displacement of flexible windsurfing is to the maximum about 0.14; As shown in Figure 7, be to the maximum about 0.08 based on the modal coordinate displacement of sinusoidal motor-driven path downwarp property windsurfing.As shown in Figure 8, the modal coordinate displacement that mixes the motor-driven path downwarp property windsurfing of stepped curve based on sine is to the maximum about 0.08.According to computational analysis, the motor-driven required time that puts in place is 41.7332 under the sinusoidal motor-driven path, and is as shown in Figure 2, and the motor-driven required time that puts in place is 40.0001 under this method.It is thus clear that; This method has bigger improvement aspect the vibration of the motor-driven path of standard bang-bang flexible windsurfing in reducing mobile process; Reducing aspect the time kept in reserve bigger improvement is arranged than the motor-driven path of standard sine, and stability is more or less the same, thereby improved the quick performance of satellite.

The content of not doing to describe in detail in the instructions of the present invention belongs to those skilled in the art's known technology.

Claims (1)

1. sinusoidal motor-driven path guide method of the mixing that is suitable for the higher satellite of windsurfing fundamental frequency is characterized in that step is following:

(1) confirms time kept in reserve t according to the motor-driven and stable time index of the attitude of satellite _Max

(2) confirm maximum motor-driven angular acceleration a according to the topworks's ability that is disposed on the satellite _Max

(3) confirm maximum motor-driven limited angular velocity omega according to topworks's angular momentum capacity that is disposed on the satellite and sensor range _Max

(4) attitude of satellite is the motor-driven path time that is divided into is [t ₁t ₂] the rate of acceleration section of increasing progressively, time be [t ₂t ₃] even accelerating sections, time be [t ₃t ₄] rate of acceleration decline fraction, time be [t ₄t ₅] at the uniform velocity coasting-flight phase, time be [t ₅t ₆] the rate of deceleration section of increasing progressively, time be [t ₆t ₇] even braking section, time be [t ₇t ₈] rate of deceleration decline fraction, t wherein ₁Be the motor-driven start time of satellite,

t ₃=t ₂+ Δ T, t ₅=t ₄+ Δ t, t ₇=t ₆+ Δ T,

T is the cycle of sine function, and the initial value of T does

Δ T is the even acceleration time, and the initial value of Δ T does Δ t is a coasting time, and the initial value of Δ t does $\mathrm{\&Delta;t}=\frac{{2\mathrm{\&pi;\&omega;}}_{\max }}{(\mathrm{\&pi;}-2)a_{\max }}+\frac{4\left|{\mathrm{\&theta;}}_d\right|}{(\mathrm{\&pi;}-2){\mathrm{\&omega;}}_{\max }}-\frac{6-\mathrm{\&pi;}}{\mathrm{\&pi;}-2}{t}_{\max };$

(5) be a according to the cycle T of sine function, maximum motor-driven angular acceleration _MaxAnd even acceleration time Δ T, the coasting time Δ t of satellite calculate in each section satellite respectively based on the attitude angle acceleration a of sine mixing step function, and corresponding attitude angular velocity ω and attitude angle θ, wherein:

(51) the rate of acceleration section of increasing progressively, $a=a_{\mathrm{Max}}\mathrm{Sin}\left(\frac{2\mathrm{\&pi;}}{T}(t-t_1)\right),$ $\mathrm{\&omega;}=\frac{a_{\mathrm{Max}}T}{2\mathrm{\&pi;}}(1-\mathrm{Cos}\left(\frac{2\mathrm{\&pi;}}{T}(t-t_1)\right)),$ $\mathrm{\&theta;}=\frac{a_{\mathrm{Max}}\&CenterDot;T}{2\mathrm{\&pi;}}((t-t_1)-\frac{T}{2\mathrm{\&pi;}}\mathrm{Sin}\left(\frac{2\mathrm{\&pi;}}{T}(t-t_1)\right)),$ θ wherein _dBe motor-driven angle; If, then upgrade according to Δ T＜0 that initial value calculates $T=\left|\frac{{\mathrm{\&pi;\; \&omega;}}_{\mathrm{Max}}}{a_{\mathrm{Max}}}\right|,$ Δ T=0, $\mathrm{\&Delta;\; t}=\frac{\mathrm{\&pi;}}{a_{\mathrm{Max}}T}\left(\right|{\mathrm{\&theta;}}_d|-\frac{\left|a_{\mathrm{Max}}\right|T^2}{2\mathrm{\&pi;}});$ If the T+2 Δ T＜T that calculates according to initial value _Min, then upgrade T=T _Min, the renewal maximum angular acceleration does

Δ T=0, Δ t=0, T _MinBe the shortest patient time kept in reserve;

(52) even accelerating sections, a=a _Max, $\mathrm{\&omega;}=\frac{a_{\max }T}{2\mathrm{\&pi;}}+a_{\max }(t-t_2),$

$\mathrm{\&theta;}=\frac{a_{\max }{T}^2(\mathrm{\&pi;}-2)}{8{\mathrm{\&pi;}}^2}+\frac{a_{\max }T(t-t_2)}{2\mathrm{\&pi;}}+\frac{a_{\max }{(t-t_2)}^2}{2},$ Wherein the value of T, Δ T, Δ t is with the rate of acceleration section of increasing progressively;

(53) rate of acceleration decline fraction, $a=a_{\max }\sin \left(\frac{2\mathrm{\&pi;}}{T}(t-t_3+\frac{T}{4})\right),$

$\mathrm{\&omega;}=a_{\max }\mathrm{\&Delta;T}+\frac{a_{\max }T}{2\mathrm{\&pi;}}(1-\cos \left(\frac{2\mathrm{\&pi;}}{T}(t-t_3+\frac{T}{4})\right)),$ $\mathrm{\&theta;}\left(t_4\right)=\frac{a_{\max }{\mathrm{\&Delta;T}}^2}{2}+\frac{a_{\max }\mathrm{\&Delta;TT}}{4}+\frac{a_{\max }\mathrm{T\&Delta;T}}{2\mathrm{\&pi;}}+\frac{a_{\max }{T}^2}{4\mathrm{\&pi;}},$

Wherein the value of T, Δ T, Δ t is with the rate of acceleration section of increasing progressively;

(54) coasting-flight phase at the uniform velocity, a=0, $\mathrm{\&omega;}=\frac{a_{\max }T}{\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T},$

$\mathrm{\&theta;}=\frac{a_{\max }{\mathrm{\&Delta;T}}^2}{2}+a_{\max }\mathrm{T\&Delta;T}(\frac{1}{4}+\frac{1}{2\mathrm{\&pi;}})+\frac{a_{\max }{T}^2}{4\mathrm{\&pi;}}+(\frac{a_{\max }T}{\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T})(t-t_4),$ Wherein the value of T, Δ T, Δ t is with the rate of acceleration section of increasing progressively;

(55) rate of deceleration section of increasing progressively, $a=a_{\max }\sin \left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;t}-\mathrm{\&Delta;T})\right),$

$\mathrm{\&omega;}=a_{\max }\mathrm{\&Delta;T}+\frac{a_{\max }T}{2\mathrm{\&pi;}}(1-\cos \left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;t}-\mathrm{\&Delta;T})\right)),$

$\mathrm{\&theta;}=\frac{a_{\max }{\mathrm{\&Delta;T}}^2}{2}+a_{\max }\mathrm{T\&Delta;T}(\frac{1}{4}+\frac{1}{2\mathrm{\&pi;}})+\frac{a_{\max }{T}^2}{4\mathrm{\&pi;}}+\frac{a_{\max }T}{\mathrm{\&pi;}}\mathrm{\&Delta;t}+a_{\max }\mathrm{\&Delta;T}(t-t_5+\mathrm{\&Delta;t})$

$+\frac{a_{\max }T}{2\mathrm{\&pi;}}(t-t_5)-\frac{a_{\max }{T}^2}{4{\mathrm{\&pi;}}^2}\sin \left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;t}-\mathrm{\&Delta;T})\right),$

If, then upgrade

Δ t=0 according to Δ t＜0 that initial calculation goes out; If Δ T＜0 that calculates according to initial value; Then upgrade

Δ T=0 $\mathrm{\&Delta;\; t}=\frac{\mathrm{\&pi;}}{a_{\mathrm{Max}}T}\left(\right|{\mathrm{\&theta;}}_d|-\frac{\left|a_{\mathrm{Max}}\right|T^2}{2\mathrm{\&pi;}});$ If $\mathrm{\&Delta;\; t}=\frac{\mathrm{\&pi;}}{a_{\mathrm{Max}}T}\left(\right|{\mathrm{\&theta;}}_d|-\frac{\left|a_{\mathrm{Max}}\right|T^2}{2\mathrm{\&pi;}})<0,$ Then upgrade $T=\sqrt{2\mathrm{\&pi;}\left|\frac{{\mathrm{\&theta;}}_d}{a_{\mathrm{Max}}}\right|},$ Δ T=0, Δ t=0;

(56) even braking section, a=-a _Max, $\mathrm{\&omega;}=\frac{a_{\max }T}{2\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T}-a_{\max }(t-t_6),$

$\begin{array}{c}\mathrm{\&theta;}=\frac{a_{\max }{\mathrm{\&Delta;T}}^2}{2}+\frac{a_{\max }\mathrm{\&Delta;TT}(\mathrm{\&pi;}+1)}{2\mathrm{\&pi;}}+(\frac{a_{\max }T}{\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T})\mathrm{\&Delta;t}+\frac{a_{\max }{T}^2(3\mathrm{\&pi;}+2)}{{8\mathrm{\&pi;}}^2}\\ +(\frac{a_{\max }T}{2\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T})(t-t_6)-\frac{a_{\max }{(t-t_6)}^2}{2}\end{array},$ Wherein the value of T, Δ T, Δ t is with the rate of deceleration section of increasing progressively;

(57) rate of deceleration decline fraction, $a=a_{\max }\sin \left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;t}-2\mathrm{\&Delta;T})\right),$

$\mathrm{\&omega;}=\frac{a_{\max }\&CenterDot;T}{2\mathrm{\&pi;}}(1-\cos \left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;t}-2\mathrm{\&Delta;T})\right)),$

$\mathrm{\&theta;}=\frac{a_{\max }\mathrm{\&Delta;TT}(\mathrm{\&pi;}+2)}{2\mathrm{\&pi;}}+a_{\max }\mathrm{\&Delta;T\&Delta;t}+\frac{{3a}_{\max }{T}^2}{8\mathrm{\&pi;}}+a_{\max }{\mathrm{\&Delta;T}}^2$

$+\frac{a_{\max }T}{2\mathrm{\&pi;}}(t-t_7+2\mathrm{\&Delta;t})-\frac{a_{\max }{T}^2}{{4\mathrm{\&pi;}}^2}\sin \left(\frac{2\mathrm{\&pi;}}{T}(t-t_1-\mathrm{\&Delta;t}-2\mathrm{\&Delta;T})\right),$

Wherein the value of T, Δ T, Δ t is with the rate of deceleration section of increasing progressively;

(6) the described path of tracking step (5) during the motor-driven control of satellite.

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