CN103345275A - Single-shaft batch counteractive flywheel moment optimal distribution method based on angular momentum margin - Google Patents

Abstract

The invention discloses a single-shaft batch counteractive flywheel moment optimal distribution method based on an angular momentum margin, and belongs to the field of control distribution. The multi-objective optimization distribution method based on the angular momentum margin is designed for solving the problem that due to the fact that a traditional moment distribution method takes no account of self angular momentum of flywheels, the flywheels are saturated and cannot complete tasks. A weight coefficient optimization method is used, system energy consumption and output margins of the flywheels are integrally considered, a suitable weight coefficient is given to each item according to energy consumption and the output margin of each flywheel, the weight coefficients are adjusted in the process of moment distribution, therefore, energy consumption is reduced, and saturation of the flywheels is avoided. The single-shaft batch counteractive flywheel moment optimal distribution method based on the angular momentum margin is beneficial to improving flywheel system output capacity, can save fuel to a certain extent, and can be widely used for the field of rapid spacecraft attitude maneuver.

Description

Single shaft batch counteraction flyback moment based on angular momentum nargin is optimized distribution method

Technical field

The present invention relates to a kind of moment distribution method of flywheel, particularly a kind of moment optimization method of the single shaft counteraction flyback based on angular momentum nargin.Belong to control distribution field.

Background technology

Along with the develop rapidly of China's aerospace industry, the motor-driven and integrity problem of spacecraft rapid posture becomes main research contents at present.In this simultaneously, fault-tolerance plays crucial effects for military field and space application, and is extreme for these environment, the application scenario of inclement condition, and the reliability of spacecraft is a critical problem that can not be ignored.It is longer to consider that spacecraft is used the cycle, so often disposes the counteraction flyback of a consumed power on the spacecraft as the topworks of attitude control system.While is reliability and the fan-out capability thereof in order to improve spacecraft again, now often adopt the reaction wheel system of redundant configuration, how the steering order of expectation being assigned to that each limited counteraction flyback of redundant, moment and angular momentum gets on is one of key issue of required consideration when designing the Spacecraft Control allocation algorithm.

Now Chang Yong energy-optimised distribution method can satisfy the attitude control requirement of spacecraft, but it also exists certain shortcoming: the most important then is that the solution of utilizing energy-optimised distribution method to obtain might not be in the executable scope of counteraction flyback, may cause flywheel the saturated phenomenon of moment to occur, thereby cause motor-driven task to carry out.

Another method commonly used is quadratic programming, and QUADRATIC PROGRAMMING METHOD FOR can be handled the complexity control assignment problem under the confined condition well, and stronger physical significance is arranged.Consider the different qualities of topworks, more press close to engineering reality when topworks's turn rate limits under the inconsistent situation this method.But QUADRATIC PROGRAMMING METHOD FOR will be controlled assignment problem and be converted into nonlinear programming problem in the processing controls assignment problem, because traditional solution complexity and efficient are not high, therefore limit the use of quadratic programming.

Summary of the invention

Do not consider flywheel self angular momentum and may cause the problem that flywheel moment is saturated so that can't finish the work in order to overcome traditional moment distribution method, also in order to overcome the not high problem of QUADRATIC PROGRAMMING METHOD FOR solution complexity and efficient, the present invention proposes a kind of single shaft batch counteraction flyback moment based on angular momentum nargin and optimize distribution method.

Described single shaft batch counteraction flyback moment based on angular momentum nargin is optimized distribution method and be may further comprise the steps:

Step 1: set up the single shaft mathematical model of counteraction flyback moment distribution in batches.Suppose on certain axle of aircraft n counteraction flyback to be housed, then the installation matrix of this flywheel group can be written as System is expressed as u to this expectation moment _c, the instruction moment that each divided flywheel is fitted on is designated as

Then expect to satisfy u between the moment and instruction moment _c=I ^Tu _w, this formula also is the constraint condition of differentiate optimization problem simultaneously.

Step 2: the energy-optimised distribution method of deriving.Because the energy consumption of flywheel system and the output torque of each counteraction flyback are closely bound up, therefore, utilize the flywheel moment vector to be constructed as follows the energy indexes function

Then energy-optimised allocation strategy can be expressed as following optimization problem:

$\left\{\begin{array}{cc}& J_1=\frac{1}{2}{u}_w^T{u}_w=\min \\ s.t.& u_c=I^T{u}_w\end{array}\right.,$

Adopt method of Lagrange multipliers structure Lagrangian function to be

With Lagrangian function respectively to u _wAsk local derviation can draw following formula with λ,

$\left\{\begin{array}{c}\frac{\∂H}{\∂u_w}=u_w+\mathrm{I\λ}=0\\ \frac{\∂H}{\∂\mathrm{\λ}}=I^T{u}_w-u_c=0\end{array}\right.,$

Obtain after the arrangement: $\left\{\begin{array}{c}{u}_w=-\mathrm{I\λ}\\ I^T{u}_w=u_c\end{array}\right.,$

Simultaneous can obtain single shaft counteraction flyback moment optimization allocation strategy in batches, that is:

$u_w=\frac{I}{n}{u}_c$

Step 3: in conjunction with the concept of angular momentum nargin, derive and optimize distribution method in batches.The angular momentum initial value of supposing flywheel system is H _W0=[H _W01H _W02H _W0n], the angular momentum nargin vector that then defines the flywheel group is

H wherein _WmBe the maximum angular momentum that flywheel can reach, H _w=[H _W1H _W2H _Wn] be the output angle momentum of flywheel group, wherein:

For aerocraft system, because its attitude pace of change is slower, the characteristics that control cycle is short can be written as this formula: H _Wi=u _WiΔ t, Δ t are system's control cycle of aircraft, so H _w=u _wΔ t.

Then angular momentum nargin index is combined with energy indexes, utilizes the weight coefficient optimization method, propose to optimize in batches index: Wherein ρ ₁=diag (ρ ₁₁ρ ₁₂ρ _1n) be the energy weight coefficient of each flywheel, ρ ₂=diag (ρ ₂₁ρ ₂₂ρ _2n) be the angular momentum nargin weight coefficient of each flywheel.

Therefore, optimizing distribution method can be expressed as optimization problem in batches:

$\left\{\begin{array}{cc}& J=\frac{1}{2}{u}_w^T{\underset{\‾}{\mathrm{\ρ}}}_1{u}_w-\frac{1}{2}\mathrm{\Δ}{H}^T{\underset{\‾}{\mathrm{\ρ}}}_2\mathrm{\ΔH}=\min \\ s.t.& u_c=I^T{u}_w\end{array}\right.,$

Then Lagrangian function is H=J+ λ (I ^Tu _w-u _c), the formula that the embodies substitution Lagrangian function of each amount is obtained

$H=\frac{1}{2}{u}_w^T{\mathrm{\ρ}}_1{u}_w-\frac{1}{2}{\left(\frac{{\mathrm{IH}}_{\mathrm{wm}}-H_{w0}-H_w}{H_{\mathrm{wm}}}\right)}^T{\underset{\‾}{\mathrm{\ρ}}}_2\left(\frac{{\mathrm{IH}}_{\mathrm{wm}}-H_{w0}-H_w}{H_{\mathrm{wm}}}\right)+\mathrm{\λ}(I^T{u}_w-u_c)$

$=\frac{1}{2}{u}_w^T{\mathrm{\ρ}}_1{u}_w-\frac{1}{2}{\left(\frac{{\mathrm{IH}}_{\mathrm{wm}}-H_{w0}-u_w\mathrm{\Δt}}{H_{\mathrm{wm}}}\right)}^T{\underset{\‾}{\mathrm{\ρ}}}_2\left(\frac{{\mathrm{IH}}_{\mathrm{wm}}-H_{w0}-u_w\mathrm{\Δt}}{H_{\mathrm{wm}}}\right)+\mathrm{\λ}(I^T{u}_w-u_c)$

$=\frac{1}{2}{u}_w^T{\mathrm{\ρ}}_1{u}_w-\frac{1}{2}\left(\frac{H_{\mathrm{ws}}^T-u_w^T\mathrm{\Δt}}{H_{\mathrm{wm}}}\right){\underset{\‾}{\mathrm{\ρ}}}_2\left(\frac{H_{\mathrm{ws}}-u_w\mathrm{\Δt}}{H_{\mathrm{wm}}}\right)+\mathrm{\λ}(I^T{u}_w-u_c)$

$=\frac{1}{2}{u}_w^T{\mathrm{\ρ}}_1{u}_w-\frac{1}{2H_{\mathrm{wm}}^2}(H_{\mathrm{ws}}^T-u_w^T\mathrm{\Δt}){\underset{\‾}{\mathrm{\ρ}}}_2(H_{\mathrm{ws}}-u_w\mathrm{\Δt})+\mathrm{\λ}(I^T{u}_w-u_c)$

$=\frac{1}{2}{u}_w^T{\underset{\‾}{\mathrm{\ρ}}}_1{u}_w-\frac{1}{{2H}_{\mathrm{wm}}^2}(H_{\mathrm{ws}}^T{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}-H_{\mathrm{ws}}^T{\underset{\‾}{\mathrm{\ρ}}}_2{u}_w\mathrm{\Δt}-u_w^T{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}\mathrm{\Δt}+u_w^T{\underset{\‾}{\mathrm{\ρ}}}_2{u_w\mathrm{\Δt}}^2)+\mathrm{\λ}(I^T{u}_w-u_c)$

H wherein _Ws=IH _Wm-H _W0, with Lagrangian function respectively to u _wAsk local derviation to obtain with λ:

$\left\{\begin{array}{c}\frac{\∂H}{{\∂u}_w}={\underset{\‾}{\mathrm{\ρ}}}_1{u}_w+\frac{\mathrm{\Δ}t{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}}{H_{\mathrm{wm}}^2}-\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2{u}_w}{H_{\mathrm{wm}}^2}+\mathrm{I\λ}=0\\ \frac{\∂H}{\∂\mathrm{\λ}}=I^T{u}_w-u_c=0\end{array}\right.$

Obtain after the arrangement: $\left\{\begin{array}{c}{u}_w={(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}\·(\mathrm{I\λ}+\frac{\mathrm{\Δt}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}}{H_{\mathrm{wm}}^2})---\left(1\right)\\ I^T{u}_w=u_c---\left(2\right)\end{array}\right.$

(1) formula is launched and can be obtained:

$u_w={(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}\mathrm{I\λ}+{(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}\frac{\mathrm{\Δt}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}}{H_{\mathrm{wm}}^2}$

$=\mathrm{\λ}{(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}I+\frac{\mathrm{\Δt}}{H_{\mathrm{wm}}^2}{(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}---\left(3\right)$

To obtain in this formula substitution (2) formula

$u_c=I^T{u}_w={\mathrm{\λI}}^T{(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}I+\frac{\mathrm{\Δt}}{H_{\mathrm{wm}}^2}{I}^T{(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}$

Therefore $\mathrm{\λ}={\left[I^T{\left(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}\right)}^{-1}I\right]}^{-1}[u_c-\frac{\mathrm{\Δt}}{H_{\mathrm{wm}}^2}{I}^T{(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}]$

To can obtain optimizing in batches the analytical expression of distribution method in λ substitution (3) formula:

$u_w={\left[I^T{(\frac{{\mathrm{\Δt}}^2}{H_{\mathrm{wm}}^2}{\underset{\‾}{\mathrm{\ρ}}}_2-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}I\right]}^{-1}[u_c-\frac{\mathrm{\Δt}}{H_{\mathrm{wm}}^2}{I}^T{(\frac{{\mathrm{\Δt}}^2}{H_{\mathrm{wm}}^2}{\underset{\‾}{\mathrm{\ρ}}}_2-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}]{(\frac{{\mathrm{\Δt}}^2}{H_{\mathrm{wm}}^2}{\underset{\‾}{\mathrm{\ρ}}}_2-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}I$

$+\frac{\mathrm{\Δt}}{H_{\mathrm{wm}}^2}{(\frac{{\mathrm{\Δt}}^2}{H_{\mathrm{wm}}^2}{\underset{\‾}{\mathrm{\ρ}}}_2-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}$

In the following formula, Δ t is system's control cycle of aircraft.

Step 4: the choosing of weight coefficient; For the energy weight coefficient ρ ₁, for i flywheel, its energy weight coefficient is chosen for the ratio of output torque with the average of the output torque of last all flywheels of the moment of this flywheel of the moment, namely

And for initial time ρ ₁The unit's of being chosen for diagonal matrix, namely ρ ₁(t ₀)=diag (11 ... 1) _{N * n}For angular momentum nargin weight coefficient ρ ₂, for i flywheel, its angular momentum nargin weight coefficient was chosen for the ratio of average of angular momentum nargin of angular momentum nargin and all flywheels in last a period of time of this flywheel constantly, namely

And for initial time ρ ₂The unit's of being chosen for diagonal matrix, namely ρ ₂(t ₀)=diag (11 ... 1) _{N * n}

This method is utilized the weight coefficient optimization method, take all factors into consideration system capacity consumption and each flywheel output nargin, give suitable weight coefficient for each according to the output nargin of energy consumption and each flywheel, in the process that moment is distributed, regulate weight coefficient, when consuming to reach the minimizing energy, prevent that flywheel is saturated, not only be conducive to improve flywheel and be fan-out capability but also fuel saving to a certain extent.

Embodiment:

1. set up the single shaft mathematical model of counteraction flyback moment distribution in batches.Suppose on certain axle of aircraft n counteraction flyback to be housed, then the installation matrix of this flywheel group can be written as

System is expressed as u to this expectation moment _c, the instruction moment that each divided flywheel is fitted on is designated as Then expect to satisfy u between the moment and instruction moment _c=I ^Tu _w, this formula also is the constraint condition of differentiate optimization problem simultaneously.

2. the energy-optimised distribution method of deriving.Because the energy consumption of flywheel system and the output torque of each counteraction flyback are closely bound up, therefore, utilize the flywheel moment vector to be constructed as follows the energy indexes function

Then energy-optimised allocation strategy can be expressed as following optimization problem:

$\left\{\begin{array}{cc}& J_1=\frac{1}{2}{u}_w^T{u}_w=\min \\ s.t.& u_c=I^T{u}_w\end{array}\right.,$

Adopt method of Lagrange multipliers structure Lagrangian function to be

With Lagrangian function respectively to u _wAsk local derviation can draw following formula with λ,

$\left\{\begin{array}{c}\frac{\∂H}{\∂u_w}=u_w+\mathrm{I\λ}=0\\ \frac{\∂H}{\∂\mathrm{\λ}}=I^T{u}_w-u_c=0\end{array}\right.,$

Obtain after the arrangement: $\left\{\begin{array}{c}{u}_w=-\mathrm{I\λ}\\ I^T{u}_w=u_c\end{array}\right.,$

Simultaneous can obtain single shaft counteraction flyback moment optimization allocation strategy in batches, that is:

$u_w=\frac{I}{n}{u}_c$

3. in conjunction with the concept of angular momentum nargin, derive and optimize distribution method in batches.The angular momentum initial value of supposing flywheel system is H _W0=[H _W01H _W02H _W0n], the angular momentum nargin vector that then defines the flywheel group is H wherein _WmBe the maximum angular momentum that flywheel can reach, H _w=[H _W1H _W2H _Wn] be the output angle momentum of flywheel group, wherein:

For aerocraft system, because its attitude pace of change is slower, the characteristics that control cycle is short can be written as this formula: H _Wi=u _WiΔ t, Δ t are system's control cycle of aircraft.

Then angular momentum nargin index is combined with energy indexes, utilizes the weight coefficient optimization method, propose to optimize in batches index:

Wherein ρ ₁=diag (ρ ₁₁ρ ₁₂ρ _1n) be the energy weight coefficient of each flywheel, ρ ₂=diag (ρ ₂₁ρ ₂₂ρ _2n) be the angular momentum nargin weight coefficient of each flywheel.

Therefore, optimizing distribution method can be expressed as optimization problem in batches:

$\left\{\begin{array}{cc}& J=\frac{1}{2}{u}_w^T{\underset{\‾}{\mathrm{\ρ}}}_1{u}_w-\frac{1}{2}\mathrm{\Δ}{H}^T{\underset{\‾}{\mathrm{\ρ}}}_2\mathrm{\ΔH}=\min \\ s.t.& u_c=I^T{u}_w\end{array}\right.,$

Then Lagrangian function is H=J+ λ (I ^Tu _w-u _c), the formula that the embodies substitution Lagrangian function of each amount is obtained

$H=\frac{1}{2}{u}_w^T{\mathrm{\ρ}}_1{u}_w-\frac{1}{2}{\left(\frac{{\mathrm{IH}}_{\mathrm{wm}}-H_{w0}-H_w}{H_{\mathrm{wm}}}\right)}^T{\underset{\‾}{\mathrm{\ρ}}}_2\left(\frac{{\mathrm{IH}}_{\mathrm{wm}}-H_{w0}-H_w}{H_{\mathrm{wm}}}\right)+\mathrm{\λ}(I^T{u}_w-u_c)$

$=\frac{1}{2}{u}_w^T{\mathrm{\ρ}}_1{u}_w-\frac{1}{2}{\left(\frac{{\mathrm{IH}}_{\mathrm{wm}}-H_{w0}-u_w\mathrm{\Δt}}{H_{\mathrm{wm}}}\right)}^T{\underset{\‾}{\mathrm{\ρ}}}_2\left(\frac{{\mathrm{IH}}_{\mathrm{wm}}-H_{w0}-u_w\mathrm{\Δt}}{H_{\mathrm{wm}}}\right)+\mathrm{\λ}(I^T{u}_w-u_c)$

$=\frac{1}{2}{u}_w^T{\mathrm{\ρ}}_1{u}_w-\frac{1}{2}\left(\frac{H_{\mathrm{ws}}^T-u_w^T\mathrm{\Δt}}{H_{\mathrm{wm}}}\right){\underset{\‾}{\mathrm{\ρ}}}_2\left(\frac{H_{\mathrm{ws}}-u_w\mathrm{\Δt}}{H_{\mathrm{wm}}}\right)+\mathrm{\λ}(I^T{u}_w-u_c)$

$=\frac{1}{2}{u}_w^T{\mathrm{\ρ}}_1{u}_w-\frac{1}{2H_{\mathrm{wm}}^2}(H_{\mathrm{ws}}^T-u_w^T\mathrm{\Δt}){\underset{\‾}{\mathrm{\ρ}}}_2(H_{\mathrm{ws}}-u_w\mathrm{\Δt})+\mathrm{\λ}(I^T{u}_w-u_c)$

$=\frac{1}{2}{u}_w^T{\underset{\‾}{\mathrm{\ρ}}}_1{u}_w-\frac{1}{{2H}_{\mathrm{wm}}^2}(H_{\mathrm{ws}}^T{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}-H_{\mathrm{ws}}^T{\underset{\‾}{\mathrm{\ρ}}}_2{u}_w\mathrm{\Δt}-u_w^T{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}\mathrm{\Δt}+u_w^T{\underset{\‾}{\mathrm{\ρ}}}_2{u_w\mathrm{\Δt}}^2)+\mathrm{\λ}(I^T{u}_w-u_c)$

H wherein _Ws=IH _Wm-H _W0, with Lagrangian function respectively to u _wAsk local derviation to obtain with λ:

$\left\{\begin{array}{c}\frac{\∂H}{{\∂u}_w}={\underset{\‾}{\mathrm{\ρ}}}_1{u}_w+\frac{\mathrm{\Δ}t{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}}{H_{\mathrm{wm}}^2}-\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2{u}_w}{H_{\mathrm{wm}}^2}+\mathrm{I\λ}=0\\ \frac{\∂H}{\∂\mathrm{\λ}}=I^T{u}_w-u_c=0\end{array}\right.$

Obtain after the arrangement: $\left\{\begin{array}{c}{u}_w={(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}\·(\mathrm{I\λ}+\frac{\mathrm{\Δt}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}}{H_{\mathrm{wm}}^2})---\left(4\right)\\ I^T{u}_w=u_c---\left(5\right)\end{array}\right.$

(4) formula is launched and can be obtained:

$u_w={(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}\mathrm{I\λ}+{(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}\frac{\mathrm{\Δt}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}}{H_{\mathrm{wm}}^2}$

$=\mathrm{\λ}{(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}I+\frac{\mathrm{\Δt}}{H_{\mathrm{wm}}^2}{(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}---\left(6\right)$

To obtain in this formula substitution (5) formula

$u_c=I^T{u}_w={\mathrm{\λI}}^T{(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}I+\frac{\mathrm{\Δt}}{H_{\mathrm{wm}}^2}{I}^T{(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}$

Therefore $\mathrm{\λ}={\left[I^T{\left(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}\right)}^{-1}I\right]}^{-1}[u_c-\frac{\mathrm{\Δt}}{H_{\mathrm{wm}}^2}{I}^T{(\frac{{\mathrm{\Δt}}^2{\underset{\‾}{\mathrm{\ρ}}}_2}{H_{\mathrm{wm}}^2}-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}]$

To can obtain optimizing in batches the analytical expression of distribution method in λ substitution (6) formula:

$u_w={\left[I^T{(\frac{{\mathrm{\Δt}}^2}{H_{\mathrm{wm}}^2}{\underset{\‾}{\mathrm{\ρ}}}_2-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}I\right]}^{-1}[u_c-\frac{\mathrm{\Δt}}{H_{\mathrm{wm}}^2}{I}^T{(\frac{{\mathrm{\Δt}}^2}{H_{\mathrm{wm}}^2}{\underset{\‾}{\mathrm{\ρ}}}_2-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}]{(\frac{{\mathrm{\Δt}}^2}{H_{\mathrm{wm}}^2}{\underset{\‾}{\mathrm{\ρ}}}_2-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}I$ In the formula

$+\frac{\mathrm{\Δt}}{H_{\mathrm{wm}}^2}{(\frac{{\mathrm{\Δt}}^2}{H_{\mathrm{wm}}^2}{\underset{\‾}{\mathrm{\ρ}}}_2-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}$

Δ t gets system's control cycle.

4. weight coefficient chooses; For the energy weight coefficient ρ ₁, for i flywheel, its energy weight coefficient is chosen for the ratio of output torque with the average of the output torque of last all flywheels of the moment of this flywheel of the moment, namely

And for initial time ρ ₁The unit's of being chosen for diagonal matrix, namely ρ ₁(t ₀)=diag (11 ... 1) _{N * n}For angular momentum nargin weight coefficient ρ ₂, for i flywheel, its angular momentum nargin weight coefficient was chosen for the ratio of average of angular momentum nargin of angular momentum nargin and all flywheels in last a period of time of this flywheel constantly, namely

And for initial time ρ ₂The unit's of being chosen for diagonal matrix, namely ρ ₂(t ₀)=diag (11 ... 1) _{N * n}

So far, all variablees are known, can carry out moment to the batch flywheel and distribute.

Claims (1)

1. the single shaft batch counteraction flyback moment based on angular momentum nargin is optimized distribution method, it is characterized in that:

Described optimization distribution method comprises following step:

Step 1: set up the single shaft mathematical model of counteraction flyback moment distribution in batches; On certain axle of aircraft n counteraction flyback is housed, then the installation matrix of this flywheel group is written as $I={\left[\begin{array}{cccc}1& 1& ...& 1\end{array}\right]}_{n\×1}^T,$ The flywheel group is expressed as u to this expectation moment _c, the instruction moment that each divided flywheel is fitted on is designated as $u_w={\left[\begin{array}{cccc}{u}_{w1}& u_{w2}& ...& u_{\mathrm{wn}}\end{array}\right]}_{n\×1}^T,$ Should satisfy u between the expectation moment and instruction moment _c=I ^Tu _w

Step 2: utilize the flywheel moment vector to be constructed as follows the energy indexes function

Energy-optimised problem then:

$\left\{\begin{array}{c}{J}_1=\frac{1}{2}{u}_w^T{u}_w=\min \\ s.t.u_c=I^T{u}_w\end{array}\right.,$

Adopt method of Lagrange multipliers structure Lagrangian function to be

Utilize this function to obtain single shaft counteraction flyback instruction moment optimization allocation strategy in batches, that is:

$u_w=\frac{I}{n}{u}_c;$

Step 3: the angular momentum initial value of establishing flywheel system is $H_{w0}=\left[\begin{array}{cccc}{H}_{w01}& H_{w02}& \·\·\·& H_{w0n}\end{array}\right],$ The angular momentum nargin vector that then defines the flywheel group is

H wherein _WmThe maximum angular momentum that can reach for flywheel, $H_w=\left[\begin{array}{cccc}{H}_{w1}& H_{w2}& \·\·\·& H_{\mathrm{wn}}\end{array}\right]$ Be the output angle momentum of flywheel group, wherein: $H_{\mathrm{wi}}={\∫}_{t_1}^{t_2}{u}_{\mathrm{wi}}\mathrm{dt},$ This formula is abbreviated as: H _Wi=u _Wi△ t, △ t are system's control cycle of aircraft;

Angular momentum nargin index is combined with energy indexes, utilizes the weight coefficient optimization method, propose to optimize in batches index: $J=\frac{1}{2}{u}_w^T\underset{}{{\underset{\‾}{\mathrm{\ρ}}}_1}{u}_w-\frac{1}{2}{\mathrm{\ΔH}}^T{\underset{\‾}{\mathrm{\ρ}}}_2\mathrm{\ΔH},$ Wherein ${\underset{\‾}{\mathrm{\ρ}}}_1=\mathrm{diag}\left(\begin{array}{cccc}{\mathrm{\ρ}}_{11}& {\mathrm{\ρ}}_{12}& \·\·\·& {\mathrm{\ρ}}_{1n}\end{array}\right)$ Be the energy weight coefficient of each flywheel, ${\underset{\‾}{\mathrm{\ρ}}}_2=\mathrm{diag}\left(\begin{array}{cccc}{\mathrm{\ρ}}_{21}& {\mathrm{\ρ}}_{22}& \·\·\·& {\mathrm{\ρ}}_{2n}\end{array}\right)$ Angular momentum nargin weight coefficient for each flywheel;

Therefore, optimizing distribution method is expressed as optimization problem in batches:

$\left\{\begin{array}{c}{J}_1=\frac{1}{2}{u}_w^T{u}_w-\frac{1}{2}{\mathrm{\ΔH}}^T{\mathrm{\ρ}}_2\mathrm{\ΔH}=\min \\ s.t.u_c=I^T{u}_w\end{array}\right.,$ Use method of Lagrange multipliers, obtain Lagrangian function

$H=\frac{1}{2}{u}_w^T{\mathrm{\ρ}}_1{u}_w-\frac{1}{2}\mathrm{\Δ}{H}^T{\mathrm{\ρ}}_2\mathrm{\ΔH}+\mathrm{\λ}(I^T{u}_w-u_c),$ With the formula that the embodies substitution Lagrangian function of each amount, wherein H _Ws=IH _Wm-H _W0, obtain the analytical expression that batch is optimized distribution method:

$u_w={\left[I^T{(\frac{{\mathrm{\Δt}}^2}{H_{\mathrm{wm}}^2}{\underset{\‾}{\mathrm{\ρ}}}_2-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}I\right]}^{-1}[u_c-\frac{\mathrm{\Δt}}{H_{\mathrm{wm}}^2}{I}^T{(\frac{{\mathrm{\Δt}}^2}{H_{\mathrm{wm}}^2}{\underset{\‾}{\mathrm{\ρ}}}_2-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}]{(\frac{{\mathrm{\Δt}}^2}{H_{\mathrm{wm}}^2}{\underset{\‾}{\mathrm{\ρ}}}_2-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}I;$

$+\frac{\mathrm{\Δt}}{H_{\mathrm{wm}}^2}{(\frac{{\mathrm{\Δt}}^2}{H_{\mathrm{wm}}^2}{\underset{\‾}{\mathrm{\ρ}}}_2-{\underset{\‾}{\mathrm{\ρ}}}_1)}^{-1}{\underset{\‾}{\mathrm{\ρ}}}_2{H}_{\mathrm{ws}}$

Step 4: the choosing of weight coefficient; For the energy weight coefficient

For i flywheel, its energy weight coefficient is chosen for the ratio of output torque with the average of the output torque of last all flywheels of the moment of this flywheel of the moment, namely

And for initial time

The unit's of being chosen for diagonal matrix, namely ${\underset{\‾}{\mathrm{\ρ}}}_1\left(t_0\right)=\mathrm{diag}{\left(\begin{array}{cccc}1& 1& \·\·\·& 1\end{array}\right)}_{n\×n};$ For angular momentum nargin weight coefficient

For i flywheel, its angular momentum nargin weight coefficient was chosen for the ratio of average of angular momentum nargin of angular momentum nargin and all flywheels in last a period of time of this flywheel constantly, namely

And for initial time

The unit's of being chosen for diagonal matrix, namely

${\underset{\‾}{\mathrm{\ρ}}}_2\left(t_0\right)=\mathrm{diag}{\left(\begin{array}{cccc}1& 1& \·\·\·& 1\end{array}\right)}_{n\×n}.$