CN102323754A - Robust gain scheduling control method based on regional pole assignment - Google Patents

Abstract

The invention provides a robust gain scheduling control method based on regional pole assignment. The realization principle is as follows: the controlled object is divided into a linear time invariant part and a parameter module by the method, wherein the parameter module is used for reflecting the nonlinear characteristic, time variant characteristic and certain uncertainty of the object; the nominal part is linear time invariant and the parametric variation part varies along with variation of the parameter module of the controlled object, thus ensuring the closed-loop system to satisfy the robust stability and robustness and have specified dynamic characteristics; the performance indexes are written into integral quadratic constraints under a unified framework according to the index requirements; in particular, the dynamic characteristics can be directly reflected by closed-loop poles on the position of the complex plane; therefore the index requirements of the dynamic characteristics are realized through regional pole constraints; and the constraints given by the performance index requirements are transformed into linear matrix inequality (LMI) systems and a solver in an LMI toolbox provided by a matrix laboratory (MATLAB) is utilized for solving, thus finally obtaining the controller satisfying the requirements.

Description

Robust gain scheduling control method based on regional pole allocation

Technical Field

The invention relates to a robust gain scheduling control method, in particular to a robust gain scheduling controller design method for realizing specified dynamic performance by using a pole allocation scheme, which can be widely applied to the field of controller design of a time-varying nonlinear multiple-input multiple-output system with higher requirements on dynamic performance.

Background

The design of a controller for a time-varying nonlinear multiple-input multiple-output system by adopting a classical control theory method has to carry out Jacobian linearization on each input and output channel at a series of working points. The resulting controller loses global characteristics and its stability cannot be guaranteed due to possible unstable zero-pole cancellation. The controllers designed based on modern control theory, such as linear quadratic regulators and the like, have poor robustness and extremely high requirements on model precision. The controller designed based on the robust control theory has good robust stability and robust performance, but has large conservative property, is difficult to be really used in engineering practice, and in addition, the performance mainly considered in the design process is H_∞/H₂And the global performance is equal, and the designed system is difficult to ensure local characteristics, particularly dynamic characteristics. In order to guarantee local characteristics, scholars propose a design method of robust gain scheduling. The method gives consideration to the global and local characteristics of the controller and meets the requirements of robust stability and robust performance. However, due to the complexity of the algorithm and the conservative property of the algorithm, the robust gain scheduling control method has difficulty in ensuring that the system has the specified dynamic characteristic.

Disclosure of Invention

The invention improves the traditional robust gain scheduling controller design method by using the regional pole allocation scheme, and ensures the local performance, especially the designated dynamic characteristic on the premise of ensuring the global robust stability and robust performance of the system.

The robust gain scheduling control method based on the pole allocation of the region comprises the following steps:

step 1: the controlled object is arranged into a standard form;

the controlled object is arranged into the following form

d_Δ＝Δe_Δ (2)

In the formula, the vector x represents the state of the controlled object, d_ΔRepresenting disturbance input, r representing reference input, u representing control input, e_ΔRepresenting the disturbance output, h representing the output of the performance channel, and y representing the measurement output; the matrix A is a system matrix, B₁、B₂、B₃Control matrices from disturbance input, reference input, control input, respectively, C₁、C₂、C₃Output matrices, D, leading to disturbance output channel, performance output channel, measurement output channel, respectively_ij(i, j ═ 1, 2, 3) a direct connection matrix between channels, where the direct connection term from control input u to measurement output y is 0; a parameter module where Δ represents nonlinear, time-varying characteristics and uncertainty;

step 2: establishing an expression of performance index constraint;

2.1 Global Performance: writing the global performance index requirement into an integral quadratic constraint form, wherein the specific index is obtained by a multiplication operator R according to the performance requirement_pGiven, having the form shown below:

where ε is any positive real number tending to 0; xi is the combination of the input vector and the output vector, and a reference input r and a performance output h are selected; zeta is input vector, choose reference input r; matrix R_pGeneral press

The form of the method is divided into four blocks, and numerical values of all parts are given according to specific index requirements;

2.2 local Properties

Limiting the poles to a certain area D on the complex plane, where D ═ z ∈ C | f_D(z) < 0}, and the region feature function is expressed as

<math> <mrow> <msub> <mi>f</mi> <mi>D</mi> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <mi>L</mi> <mo>+</mo> <mi>zM</mi> <mo>+</mo> <mover> <mi>z</mi> <mo>&OverBar;</mo> </mover> <msup> <mi>M</mi> <mi>T</mi> </msup> <mo>=</mo> <msub> <mrow> <mo>(</mo> <msub> <mi>L</mi> <mi>ij</mi> </msub> <mo>+</mo> <msub> <mi>M</mi> <mi>ij</mi> </msub> <mi>z</mi> <mo>+</mo> <msub> <mi>M</mi> <mi>ji</mi> </msub> <mover> <mi>z</mi> <mo>&OverBar;</mo> </mover> <mo>)</mo> </mrow> <mrow> <mn>1</mn> <mo>&le;</mo> <mi>i</mi> <mo>,</mo> <mi>j</mi> <mo>&le;</mo> <mi>m</mi> </mrow> </msub> <mo>,</mo> </mrow> </math> Wherein the coefficient matrix of the characteristic function

L＝L^T∈R^m×m，M∈R^m×m；

The pole region is a combination of the following simple regions:

1) semi-plane: re (z) < α:

then L is_-＝-2α，M_-1 is ═ 1; semi-plane:

Re(z)＞β：

then L is₊＝2β，M₊＝-1；

2) Circle with (-q, 0) as the center and r as the radius: <math> <mrow> <msub> <mi>f</mi> <msub> <mi>D</mi> <mi>r</mi> </msub> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mo>-</mo> <mi>r</mi> </mtd> <mtd> <mi>z</mi> <mo>+</mo> <mi>q</mi> </mtd> </mtr> <mtr> <mtd> <mover> <mi>z</mi> <mo>&OverBar;</mo> </mover> <mo>+</mo> <mi>q</mi> </mtd> <mtd> <mo>-</mo> <mi>r</mi> </mtd> </mtr> </mtable> </mfenced> </mrow> </math>

then $L_r=\left(\begin{array}{cc}-r& q\\ q& -r\end{array}\right),$ $M_r=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right);$

3) Cone with 2 θ as cone angle: <math> <mrow> <msub> <mi>f</mi> <msub> <mi>D</mi> <mi>c</mi> </msub> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mrow> <mo>(</mo> <mi>z</mi> <mo>+</mo> <mover> <mi>z</mi> <mo>&OverBar;</mo> </mover> <mo>)</mo> </mrow> <mi>sin</mi> <mi>&theta;</mi> </mtd> <mtd> <mrow> <mo>(</mo> <mi>z</mi> <mo>-</mo> <mover> <mi>z</mi> <mo>&OverBar;</mo> </mover> <mo>)</mo> </mrow> <mi>cos</mi> <mi>&theta;</mi> </mtd> </mtr> <mtr> <mtd> <mrow> <mo>(</mo> <mi>z</mi> <mo>-</mo> <mover> <mi>z</mi> <mo>&OverBar;</mo> </mover> <mo>)</mo> </mrow> <mi>cos</mi> <mi>&theta;</mi> </mtd> <mtd> <mrow> <mo>(</mo> <mi>z</mi> <mo>+</mo> <mover> <mi>z</mi> <mo>&OverBar;</mo> </mover> <mo>)</mo> </mrow> <mi>sin</mi> <mi>&theta;</mi> </mtd> </mtr> </mtable> </mfenced> </mrow> </math>

then $L_c=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right),$ <math> <mrow> <msub> <mi>M</mi> <mi>c</mi> </msub> <mo>=</mo> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mi>sin</mi> <mi>&theta;</mi> </mtd> <mtd> <mi>cos</mi> <mi>&theta;</mi> </mtd> </mtr> <mtr> <mtd> <mo>-</mo> <mi>cos</mi> <mi>&theta;</mi> </mtd> <mtd> <mi>sin</mi> <mi>&theta;</mi> </mtd> </mtr> </mtable> </mfenced> <mo>;</mo> </mrow> </math>

4) Horizontal strip-shaped area with the real axis as the symmetry axis and the width of 2 h:

<math> <mrow> <msub> <mi>f</mi> <msub> <mi>D</mi> <mi>h</mi> </msub> </msub> <mrow> <mo>(</mo> <mi>z</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mo>-</mo> <mi>h</mi> </mtd> <mtd> <mo>-</mo> <mi>z</mi> <mo>+</mo> <mover> <mi>z</mi> <mo>&OverBar;</mo> </mover> </mtd> </mtr> <mtr> <mtd> <mi>z</mi> <mo>-</mo> <mover> <mi>z</mi> <mo>&OverBar;</mo> </mover> </mtd> <mtd> <mo>-</mo> <mi>h</mi> </mtd> </mtr> </mtable> </mfenced> </mrow> </math>

then $L_h=\left(\begin{array}{cc}-h& 0\\ 0& -h\end{array}\right),$ $M_h=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right);$

Writing the pole region as the intersection of the above simple regions, i.e. D ═ I_σD_σThen L and M are written as L respectively_σAnd M_σHas a block diagonal form of L ═ blockdiag { Λ, L ═ L_σ，Λ}M＝blockdiag{Λ，M_σΛ, wherein the subscript mark σ ∈ { -, +, r, c, h }, which indicates that the pole region is composed of the simple region; and 3, step 3: judging whether a controller exists or not, so that the stability and the dynamic performance of the closed-loop system meet the requirements;

solving the following inequality group by using a feasp solver in an LMI toolbox provided by MATLAB:

in the set of inequalities, the unknown variables are: q, S, R,x, Y, where the symbol ker (Φ) represents the basis matrix of the null space of Φ, having Φ^Tker (Φ) ═ 0; mark

Represents the Kronecker product; if the inequality group (3) has a solution, the existence of the controller can enable the performance index of the closed-loop system to be met, and the subsequent steps are carried out to solve the controller; otherwise, the performance index cannot be met, and the original controller design problem is solved;

and 4, step 4: the multiplier P is constructed by first performing singular value decomposition as follows:

<math> <mrow> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mi>Q</mi> </mtd> <mtd> <mi>S</mi> </mtd> </mtr> <mtr> <mtd> <msup> <mi>S</mi> <mi>T</mi> </msup> </mtd> <mtd> <mi>R</mi> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <msup> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mover> <mi>Q</mi> <mo>~</mo> </mover> </mtd> <mtd> <mover> <mi>S</mi> <mo>~</mo> </mover> </mtd> </mtr> <mtr> <mtd> <msup> <mover> <mi>S</mi> <mo>~</mo> </mover> <mi>T</mi> </msup> </mtd> <mtd> <mover> <mi>R</mi> <mo>~</mo> </mover> </mtd> </mtr> </mtable> </mfenced> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <mo>=</mo> <mi>&Gamma;</mi> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mo>-</mo> <msub> <mi>&Sigma;</mi> <mo>-</mo> </msub> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <msub> <mi>&Sigma;</mi> <mo>+</mo> </msub> </mtd> </mtr> </mtable> </mfenced> <msup> <mi>&Gamma;</mi> <mi>T</mi> </msup> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>4</mn> <mo>)</mo> </mrow> <mo>,</mo> </mrow> </math> where Γ is the unitary matrix, - ∑_-Is a diagonal matrix formed by the singular values of the part less than zero, sigma₊Is a diagonal matrix formed by the singular values of the part which is larger than zero;

then, order $P_{}$ ${}_1=\left(\begin{array}{cc}Q& S\\ S^T& R\end{array}\right),$ <math> <mrow> <msub> <mi>P</mi> <mn>2</mn> </msub> <mo>=</mo> <mi>&Gamma;</mi> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <msubsup> <mi>&Sigma;</mi> <mo>-</mo> <mrow> <mn>1</mn> <mo>/</mo> <mn>2</mn> </mrow> </msubsup> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <msubsup> <mi>&Sigma;</mi> <mo>+</mo> <mrow> <mn>1</mn> <mo>/</mo> <mn>2</mn> </mrow> </msubsup> </mtd> </mtr> </mtable> </mfenced> <mo>,</mo> </mrow> </math> ${\mathrm{<figure-callout\; id="3"\; label="P"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">P</figure-callout>}}_3=\left(\begin{array}{cc}-\mathrm{<figure-callout\; id="0"\; label="I"\; filenames="HDA0000079904860000022.png"\; state="\{\{state\}\}">I</figure-callout>}& 0\\ 0& I\end{array}\right)---\left(5\right)$

Then there is $P=\left(\begin{array}{cc}{P}_1& P_2\\ P_2^T& P_3\end{array}\right)---\left(6\right);$

And 5, step 5: solving the linear matrix inequality set to enable the closed-loop system to realize performance indexes;

order toCan be combined with

Divided into four blocks, as shown by:

$\left(\begin{array}{cc}\hat{Q}& \hat{S}\\ {\hat{S}}^T& \hat{R}\end{array}\right):=I_e^T{\mathrm{PI}}_e---\left(7\right)$

order to <math> <mrow> <mover> <mi>Q</mi> <mo>&OverBar;</mo> </mover> <mo>:</mo> <mo>=</mo> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mover> <mi>Q</mi> <mo>^</mo> </mover> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <msub> <mi>Q</mi> <mi>p</mi> </msub> </mtd> </mtr> </mtable> </mfenced> <mo>,</mo> </mrow> </math> <math> <mrow> <mover> <mi>S</mi> <mo>&OverBar;</mo> </mover> <mo>:</mo> <mo>=</mo> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mover> <mi>S</mi> <mo>^</mo> </mover> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <msub> <mi>S</mi> <mi>p</mi> </msub> </mtd> </mtr> </mtable> </mfenced> <mo>,</mo> </mrow> </math> <math> <mrow> <mover> <mi>R</mi> <mo>&OverBar;</mo> </mover> <mo>:</mo> <mo>=</mo> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mover> <mi>R</mi> <mo>^</mo> </mover> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <msub> <mi>R</mi> <mi>p</mi> </msub> </mtd> </mtr> </mtable> </mfenced> <mo>,</mo> </mrow> </math> Solving the following inequality group by using a feasp solver in an LMI toolbox provided by MATLAB:

<math> <mrow> <mfenced open='' close='}'> <mtable> <mtr> <mtd> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <msub> <mi>H</mi> <mi>a</mi> </msub> <mo>+</mo> <msubsup> <mi>H</mi> <mi>a</mi> <mi>T</mi> </msubsup> </mtd> <mtd> <msub> <mi>H</mi> <mi>b</mi> </msub> </mtd> <mtd> <msubsup> <mi>H</mi> <mi>c</mi> <mi>T</mi> </msubsup> </mtd> </mtr> <mtr> <mtd> <msubsup> <mi>H</mi> <mi>b</mi> <mi>T</mi> </msubsup> </mtd> <mtd> <mover> <mi>Q</mi> <mo>&OverBar;</mo> </mover> </mtd> <mtd> <msubsup> <mi>H</mi> <mi>d</mi> <mi>T</mi> </msubsup> <mo>+</mo> <msup> <mrow> <mover> <mi>S</mi> <mo>&OverBar;</mo> </mover> <mover> <mi>R</mi> <mo>&OverBar;</mo> </mover> </mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mtd> </mtr> <mtr> <mtd> <msub> <mi>H</mi> <mi>c</mi> </msub> </mtd> <mtd> <msub> <mi>H</mi> <mi>d</mi> </msub> <mo>+</mo> <msup> <mover> <mi>R</mi> <mo>&OverBar;</mo> </mover> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> <msup> <mover> <mi>S</mi> <mo>&OverBar;</mo> </mover> <mi>T</mi> </msup> </mtd> <mtd> <mo>-</mo> <msup> <mover> <mi>R</mi> <mo>&OverBar;</mo> </mover> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mtd> </mtr> </mtable> </mfenced> <mo>&lt;</mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mi>L</mi> <mo>&CircleTimes;</mo> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <msub> <mi>X</mi> <mn>1</mn> </msub> </mtd> <mtd> <mi>I</mi> </mtd> </mtr> <mtr> <mtd> <mrow> <mtd> <mi>I</mi> </mtd> </mrow> </mtd> <mtd> <msub> <mi>Y</mi> <mn>1</mn> </msub> </mtd> </mtr> </mtable> </mfenced> <mo>+</mo> <mi>M</mi> <mo>&CircleTimes;</mo> <msub> <mi>H</mi> <mi>a</mi> </msub> <mo>+</mo> <msup> <mi>M</mi> <mi>T</mi> </msup> <mo>&CircleTimes;</mo> <msubsup> <mi>H</mi> <mi>a</mi> <mi>T</mi> </msubsup> <mo>&lt;</mo> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <msub> <mi>X</mi> <mn>1</mn> </msub> </mtd> <mtd> <mi>I</mi> </mtd> </mtr> <mtr> <mtd> <mi>I</mi> </mtd> <mtd> <msub> <mi>Y</mi> <mn>1</mn> </msub> </mtd> </mtr> </mtable> </mfenced> <mo>&GreaterEqual;</mo> <mn>0</mn> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>8</mn> <mo>)</mo> </mrow> </mrow> </math>

wherein, $H_a:=\left(\begin{array}{cc}{X}_{1}A+B_{k1}{C}_3& A_k\\ A+B_3{D}_{c11}{\mathrm{<figure-callout\; id="3"\; label="C"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_3& {\mathrm{AY}}_1^T+B_3{\mathrm{<figure-callout\; id="1"\; label="C\; k"\; filenames="HDA0000079904860000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_k\end{array}\right),$

$H_b:=\left(\begin{array}{ccc}{X}_1{\mathrm{<figure-callout\; id="1"\; label="B"\; filenames="HDA0000079904860000022.png"\; state="\{\{state\}\}">B</figure-callout>}}_1+B_{k1}{D}_{31}& B_{k2}& X_1{\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">B</figure-callout>}}_2+B_{k1}{D}_{32}\\ {\mathrm{<figure-callout\; id="1"\; label="B"\; filenames="HDA0000079904860000022.png"\; state="\{\{state\}\}">B</figure-callout>}}_1+B_3{D}_{c11}{D}_{31}& B_3{D}_{c12}& {\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">B</figure-callout>}}_2+B_3{D}_{c11}{D}_{32}\end{array}\right),$

$H_c:=\left(\begin{array}{cc}{\mathrm{<figure-callout\; id="1"\; label="C"\; filenames="HDA0000079904860000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_1+D_{13}{D}_{c11}{C}_3& C_1{Y}_1^T+D_{13}{C}_{k1}\\ D_{c21}{C}_3& C_{k2}\\ {\mathrm{<figure-callout\; id="2"\; label="C"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_2+D_{23}{D}_{c11}{C}_3& C_2{Y}_1^T+D_{23}{\mathrm{<figure-callout\; id="1"\; label="C\; k"\; filenames="HDA0000079904860000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_k\end{array}\right),$

$H_d:=\left(\begin{array}{ccc}{D}_{11}+D_{13}{D}_{c11}{D}_{31}& D_{13}{D}_{c12}& D_{12}+D_{13}{D}_{c11}{D}_{32}\\ D_{c21}{D}_{31}& D_{c22}& D_{c21}{D}_{32}\\ D_{21}+D_{23}{D}_{c11}{D}_{31}& D_{23}{D}_{c12}& D_{22}+D_{23}{D}_{c11}{D}_{32}\end{array}\right),$

the unknown variables are: a. the_k，B_k1，B_k2，C_k1，C_k2，D_c11，D_c12，D_c21，D_c22，X₁，Y₁；

If the optimal performance needs to be obtained, only the inequality group (8) is used as a constraint to optimize the unknown variable, so that the R is the sum of the R and the R_pThe expressed performance index is optimal;

and 6, step 6: solving a nominal part of the controller;

using already solved X₁，Y₁Performing singular value decomposition; the expression of the specific calculation method is as follows: I-X₁Y₁＝UΛV^TLet X₂＝UΛ^1/2 (9)

Y₂＝VΛ^1/2 (10)

U, V in the formula is unitary matrix, and A obtained by the solution is used_k，B_k1，B_k2，C_k1，C_k2，D_c11，D_c12，D_c21，D_c22，X₁，Y₁，X₂，Y₂Solving a linear equation set;

$\left.\begin{array}{c}{X}_1{B}_3{D}_{c11}+X_2{\mathrm{<figure-callout\; id="1"\; label="B\; c"\; filenames="HDA0000079904860000022.png"\; state="\{\{state\}\}">B</figure-callout>}}_c=B_{k1}\\ X_1{B}_3{D}_{c12}+X_2{\mathrm{<figure-callout\; id="2"\; label="B\; c"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">B</figure-callout>}}_c=B_{k2}\\ D_{c11}{C}_3{Y}_1^T+C_{c1}{Y}_2^T=C_{k1}\\ D_{c21}{C}_3{Y}_1^T+C_{c2}{Y}_2^T=C_{k2}\\ X_1(A+B_3{D}_{c11}{C}_3)Y_1^T+X_2{B}_{c1}{C}_3{Y}_1^T+X_1{B}_3{C}_{c1}{Y}_2^T+X_2{A}_c{Y}_2^T=A_k\end{array}\right\}---\left(11\right)$

wherein the unknown variable is A_c，B_c1，B_c2，C_c1，C_c2(ii) a A obtained by solving_c，B_c1，B_c2，C_c1，C_c2And D_c11，D_c12，D_c21，D_c22Together forming a nominal part of the controller;

and 7, step 7: solving a variable parameter part of the controller;

the following inequality was solved using the feasp solver in the LMI toolbox provided by MATLAB:

<math> <mrow> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mover> <mi>R</mi> <mo>^</mo> </mover> <mo>+</mo> <msup> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mi>&Delta;</mi> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <msub> <mi>&Delta;</mi> <mi>c</mi> </msub> </mtd> </mtr> </mtable> </mfenced> <mi>T</mi> </msup> </mtd> <mtd> <mover> <mi>S</mi> <mo>^</mo> </mover> <mo>+</mo> <msup> <mover> <mi>S</mi> <mo>^</mo> </mover> <mi>T</mi> </msup> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mi>&Delta;</mi> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <msub> <mi>&Delta;</mi> <mi>c</mi> </msub> </mtd> </mtr> </mtable> </mfenced> <msup> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mi>&Delta;</mi> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <msub> <mi>&Delta;</mi> <mi>c</mi> </msub> </mtd> </mtr> </mtable> </mfenced> <mi>T</mi> </msup> </mtd> </mtr> <mtr> <mtd> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mi>&Delta;</mi> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <msub> <mi>&Delta;</mi> <mi>c</mi> </msub> </mtd> </mtr> </mtable> </mfenced> </mtd> <mtd> <mo>-</mo> <msup> <mover> <mi>Q</mi> <mo>^</mo> </mover> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mtd> </mtr> </mtable> </mfenced> <mo>></mo> <mn>0</mn> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>12</mn> <mo>)</mo> </mrow> </mrow> </math>

in the formula, the unknown variable is Delta_cChanges along with the change of the parameter module delta;

and 8, step 8: a controller is constructed.

According to the previously determined A_c，B_c1，B_c2，C_c1，C_c2，D_c11，D_c12，D_c21，D_c22And Δ_cConstructing a controller, wherein a state equation is as follows:

d_cΔ＝Δ_ce_cΔ (14)

the realization principle of the invention is as follows: the method divides a controlled object into a linear time-invariant part and a parameter module, wherein the parameter module is used for reflecting the nonlinear characteristic, the time-variant characteristic and certain uncertainty of the object. The objective is to design a variable parameter controller, the nominal part of which is linear and time-invariant, and the variable parameter part changes along with the change of a controlled object parameter module, so that a closed-loop system meets the requirements of robust stability and robust performance and has specified dynamic characteristics. The performance indexes are written into a form of integral quadratic constraint according to the index requirements under a unified framework. It should be noted that, since the dynamic characteristics can be directly represented by the positions of the poles of the closed loop on the complex plane, these dynamic characteristics index requirements are achieved by the regional pole constraint. And converting constraints given by various performance index requirements into linear matrix inequality groups, and solving by using a solver in an LMI toolbox provided by MATLAB to finally obtain the controller meeting the requirements.

The invention has the beneficial effects that:

the robust gain scheduling controller design method based on the regional pole allocation can be widely applied to the design of a control system, and is particularly suitable for designing a controller for a time-varying nonlinear multiple-input multiple-output system. The controller designed by the method can meet the global performance index requirement and the local performance requirement, particularly the transition performance requirement.

Drawings

FIG. 1 is a block diagram of a model of a controlled object in the control method of the present invention;

FIG. 2 is a block diagram of a controller model in the control method of the present invention;

FIG. 3 is a diagram of an autopilot model matching architecture in the control method of the present invention;

FIG. 4 is a graph of a parameter change in an embodiment of the present invention;

FIG. 5 is a graph of square wave response in an embodiment of the present invention.

Detailed Description

Taking a certain missile as an example of longitudinal movement, an autopilot is designed to have the form of expressions (13) and (14), as shown in fig. 2.

To cope with time-varying characteristics and uncertainties, the controller is designed to be divided into two parts: linear time-invariant part K and time-variant parameter module Δ_cParameters can be automatically adjusted along with the change of missile dynamics. In order to ensure the stability of the guidance loop and achieve the required performance, an autopilot model matching design structure as shown in fig. 3 is constructed. In the figure, f_ycFor reference to the normal acceleration input command, u is the control input described in expression (1) above, δ_zFor rudder deflection angle, f_yNormal acceleration, q pitch angular velocity, f_yrNormal acceleration output for reference model, e_pIs the performance output. Will be controlledThe target missile is divided into two parts: a linear time-invariant portion G and a time-variant parameter block Δ; act represents steering engine dynamics; w_PA low pass filter representing the performance weighting function; and R represents a reference model for describing the required dynamic performance index. The design aim is to examine the low-frequency response characteristic of a closed-loop system formed by a controller, a steering engine and a missile. A closed-loop system formed by a controller, a steering engine and a missile is compared with a reference model, and if the low-frequency response of the closed-loop system is very close to that of the reference model, the closed-loop system can achieve the required dynamic performance index. As can be seen from the figure, the performance output e_pIs the low frequency part of the difference between the true value of normal acceleration and the reference model output value, we expect e_pAs close to 0 as possible. Thus, in the case of parameter variations, the global performance requirement is a robust secondary performance requirement with index γ, i.e.: ensure the robust stability of the system and have

<math> <mrow> <mfrac> <msub> <mrow> <mo>|</mo> <mo>|</mo> <msub> <mi>e</mi> <mi>p</mi> </msub> <mo>|</mo> <mo>|</mo> </mrow> <mn>2</mn> </msub> <msub> <mrow> <mo>|</mo> <mo>|</mo> <msub> <mi>f</mi> <mi>yc</mi> </msub> <mo>|</mo> <mo>|</mo> </mrow> <mn>2</mn> </msub> </mfrac> <mo>&lt;</mo> <mi>&gamma;</mi> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>15</mn> <mo>)</mo> </mrow> </mrow> </math>

According to H_∞Definition of Properties, expression (15) is practically equivalent to H with index γ_∞And (4) performance requirements. Note that γ should be as small as possible so that the problem is an optimal one; given an index value of γ, it is a sub-optimal feasibility problem, where γ is usually 1. In this example, global refers toMarking gamma as 1; the transition performance requirement overshoot is less than 5%, and the 2% stabilization time is less than 0.15 s; to limit the controller fast mode, the controller poles are required to be limited in the left half complex plane where the absolute values of the real and imaginary parts are less than 1000.

The controller design steps are detailed below.

1) The original missile longitudinal dynamics equation is subjected to quasi-linearization treatment, the design area of the example is selected as Ma belonging to [1.8, 3.6], H belonging to [500m, 6000m ], and a linear time invariant system and a variable parameter block which are connected in the mode of the figure 1 are obtained.

The state space expression of the linear time invariant system G is

Therein, there are

$A_g=\left(\begin{array}{cc}-1.4318& 1307.1288\\ 0.55& -3.8971\end{array}\right)$

${\mathrm{<figure-callout\; id="1"\; label="B\; g"\; filenames="HDA0000079904860000022.png"\; state="\{\{state\}\}">B</figure-callout>}}_{g_{}}=\left(\begin{array}{ccccc}-466.4594& 1.7891& 500.6380& 332.3441& -3.4649\\ 5.8017& 84.0866& 5.1033& -15.5201& -74.1971\end{array}\right),$ ${\mathrm{<figure-callout\; id="2"\; label="B\; g"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">B</figure-callout>}}_{g_{}}=\left(\begin{array}{c}-140.6400\\ 353.4696\end{array}\right),$

${\mathrm{<figure-callout\; id="1"\; label="C\; g"\; filenames="HDA0000079904860000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_{g_{}}=\left(\begin{array}{cc}0.0012& -1.3540\\ -0.0001& 0.0888\\ 0.0006& 0.0048\\ 0.0013& -1.3876\\ -0.0003& 0.2732\end{array}\right)$

$D_{g_{11}}=\left(\begin{array}{ccccc}-0.0003& -0.00003& -0.0003& -0.1111& -0.0065\\ -0.0040& 0.0005& -0.0037& -0.0247& -0.0693\\ -0.0002& 0.00003& -0.0003& 0.2509& -0.0285\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right),$ $D_{g_{12}}=\left(\begin{array}{c}0.1199\\ 0.9906\\ -0.0006\\ 0.2732\\ 1.3876\end{array}\right),$

${\mathrm{<figure-callout\; id="2"\; label="C\; g"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_{g_{}}=\left(\begin{array}{cc}1& 0\end{array}\right),$ $D_{g_{21}}=\left(\begin{array}{ccccc}0& 0& 0& 0& 0\end{array}\right),$ $D_{g_{22}}=0,$ ${\mathrm{<figure-callout\; id="3"\; label="C\; g"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_{g_{}}=\left(\begin{array}{cc}0& 1\end{array}\right),$ $D_{g_{31}}=\left(\begin{array}{ccccc}0& 0& 0& 0& 0\end{array}\right),$

$D_{g_{32}}=0;$

Order to

Andrepresenting normalized variables corresponding to Ma and H, i.e.

The time-varying parameter block delta is then of the form

<math> <mrow> <mi>&Delta;</mi> <mo>=</mo> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mover> <mi>Ma</mi> <mo>&OverBar;</mo> </mover> <msub> <mi>I</mi> <mrow> <mn>3</mn> <mo>&times;</mo> <mn>3</mn> </mrow> </msub> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <mover> <mi>H</mi> <mo>&OverBar;</mo> </mover> <msub> <mi>I</mi> <mrow> <mn>2</mn> <mo>&times;</mo> <mn>2</mn> </mrow> </msub> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>17</mn> <mo>)</mo> </mrow> </mrow> </math>

2) The reference model R is selected as

$R\left(s\right)=\frac{1600}{{\mathrm{<figure-callout\; id="2"\; label="s"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">s</figure-callout>}}^2+48s+1600}---\left(18\right)$

It is rewritten into a state space form

In the formula $A_r=\left(\begin{array}{cc}-48& -12.5\\ 128& 0\end{array}\right),$ $B_r=\left(\begin{array}{c}4\\ 0\end{array}\right),$ C_r＝(0 3.125)，D_r＝0。

3) Performance weighting function W_PIs selected as

$W_P\left(s\right)=\frac{0.5(s+840)}{s+3}---\left(20\right)$

It is rewritten into a state space expression of

In the formula A_w＝-3，B_w＝16，C_w＝26.1563，D_w＝0.5。

4) Steering engine dynamics Act is approximated as

$\mathrm{Act}\left(s\right)=\frac{170}{(s+170)}---\left(22\right)$

Rewriting it into a state space expression can be written as

In the formula A_a＝-170，B_a＝16，C_a＝10.625，D_a＝0。

5) G, R, W_PLinearly connected with Act to form a linear time-invariant part of the extended controlled object, and having the following form

Arranged in the form of (1), i.e.

$A=\left(\begin{array}{cccc}{A}_g& B_{g_2}{C}_a& 0& 0\\ 0& A_a& 0& 0\\ 0& 0& A_r& 0\\ B_w{C}_{g_2}& B_w{D}_{g_{22}}{C}_a& B_w{C}_r& A_w\end{array}\right)=\left(\begin{array}{cccccc}-1.4318& 1307.1288& -1494.3000& 0& 0& 0\\ 0.55& -3.8971& 3755.6145& 0& 0& 0\\ 0& 0& -170& 0& 0& 0\\ 0& 0& 0& -48& -12.5& 0\\ 0& 0& 0& 128& 0& 0\\ 16& 0& 0& 0& 50& -3\end{array}\right)$

$,{\mathrm{<figure-callout\; id="1"\; label="B"\; filenames="HDA0000079904860000022.png"\; state="\{\{state\}\}">B</figure-callout>}}_1=\left(\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="B\; g"\; filenames="HDA0000079904860000022.png"\; state="\{\{state\}\}">B</figure-callout>}}_{g_{}}\\ 0\\ 0\\ B_w{D}_{g_{21}}\end{array}\right)=\left(\begin{array}{ccccc}-466.4594& 1.7891& 500.6380& 332.3441& -3.4649\\ 5.8017& 84.0866& 5.1033& -15.5201& -7401971\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right),$

${\mathrm{<figure-callout\; id="2"\; label="B"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">B</figure-callout>}}_2=\left(\begin{array}{c}0\\ 0\\ B_r\\ -B_w{D}_r\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\\ 4\\ 0\\ 0\end{array}\right),$ ${\mathrm{<figure-callout\; id="3"\; label="B"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">B</figure-callout>}}_3=\left(\begin{array}{c}{B}_{g_2}{D}_a\\ B_a\\ 0\\ B_w{D}_{g_{22}}{D}_a\end{array}\right)=\left(\begin{array}{c}-140.6400\\ 353.4696\\ 16\\ 0\\ 0\\ 0\end{array}\right),$

${\mathrm{<figure-callout\; id="1"\; label="C"\; filenames="HDA0000079904860000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_1=\left(\begin{array}{cccc}{C}_{g_1}& D_{g_{12}}{C}_a& 0& 0\end{array}\right)=\left(\begin{array}{cccccc}0.0012& -1.3540& 1.2739& 0& 0& 0\\ -0.00012& 0.0888& 10.5247& 0& 0& 0\\ 0.00062& 0.0048& -0.0061& 0& 0& 0\\ 0.0013& -1.3876& 2.9029& 0& 0& 0\\ -0.00027& 0.2732& 14.7429& 0& 0& 0\end{array}\right),$

$D_{11}=D_{g_{11}}=\left(\begin{array}{ccccc}-0.0003& -0.00003& -0.0003& -0.1111& -0.0065\\ -0.0040& 0.0005& -0.0037& -0.247& -0.0693\\ -0.0002& -0.00003& -0.0003& -0.2509& -0.0285\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right),$ $D_{12}=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\end{array}\right),$

$D_{13}=D_{g_{12}}{D}_a=\left(\begin{array}{c}0\\ 0\\ 0\\ 0\\ 0\end{array}\right),$

${\mathrm{<figure-callout\; id="2"\; label="C"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_2=\left(\begin{array}{cccc}{D}_w{C}_{g_2}& D_w{C}_{g_{22}}{C}_a& -D_w{C}_r& C_w\end{array}\right)=\left(\begin{array}{cccccc}0.5& 0& 0& 0& -3.125& 26.1563\end{array}\right),$

$D_{21}=D_w{D}_{g_{21}}=\left(\begin{array}{ccccc}0& 0& 0& 0& 0\end{array}\right),$ D₂₂＝-D_wD_r＝0， $D_{23}=D_w{C}_{g_{22}}{D}_a=0,$

${\mathrm{<figure-callout\; id="3"\; label="C"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_3=\left(\begin{array}{cccc}0& 0& 0& 0\\ C_{g_2}& D_{g_{22}}{C}_a& 0& 0\\ C_{g_3}& D_{g_{32}}{C}_a& 0& 0\end{array}\right)=\left(\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 1& 0& 0& 0& 0& 0\\ 0& 1& 0& 0& 0& 0\end{array}\right),$

$D_{31}=\left(\begin{array}{c}0\\ D_{g_{21}}\\ D_{g_{31}}\end{array}\right)=\left(\begin{array}{ccccc}0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right),$ $D_{32}=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),$ $D_{33}=\left(\begin{array}{c}0\\ D_{g_{22}}{D}_a\\ D_{g_{32}}{D}_a\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right),$ $x=\left(\begin{array}{c}{x}_g\\ x_a\\ x_r\\ x_w\end{array}\right),$

r＝f_yc，h＝e_p， $y=\left(\begin{array}{c}{f}_{\mathrm{yc}}\\ f_y\\ q\end{array}\right).$

And delta is used as a variable parameter block of the controlled object after expansion as shown in expression (17).

Up to this point, the (generalized) controlled object that can be used for controller design, as represented by expressions (1) and (2), has been constructed. The global and local performance index requirements are then rewritten to a standard form to complete the controller design.

6) In this example, the design goal of the robust gain scheduling autopilot is to design an LPV controller so that the controlled closed-loop system meets the robust secondary performance index requirement of γ ═ 1, that is, the design goal is to design an LPV controller <math> <mrow> <msubsup> <mo>&Integral;</mo> <msub> <mi>t</mi> <mn>0</mn> </msub> <msub> <mi>t</mi> <mn>1</mn> </msub> </msubsup> <msup> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <msub> <mi>f</mi> <mi>yc</mi> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mi>e</mi> <mi>p</mi> </msub> </mtd> </mtr> </mtable> </mfenced> <mi>T</mi> </msup> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mo>-</mo> <mi>&gamma;</mi> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <mn>1</mn> <mo>/</mo> <mi>&gamma;</mi> </mtd> </mtr> </mtable> </mfenced> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <msub> <mi>f</mi> <mi>yc</mi> </msub> </mtd> </mtr> <mtr> <mtd> <msub> <mi>e</mi> <mi>p</mi> </msub> </mtd> </mtr> </mtable> </mfenced> <mi>dt</mi> <mo>&le;</mo> <mo>-</mo> <mi>&xi;</mi> <msubsup> <mo>&Integral;</mo> <msub> <mi>t</mi> <mn>0</mn> </msub> <msub> <mi>t</mi> <mn>1</mn> </msub> </msubsup> <msubsup> <mi>f</mi> <mi>yc</mi> <mn>2</mn> </msubsup> <mi>dt</mi> <mo>,</mo> </mrow> </math> <math> <mrow> <msub> <mi>R</mi> <mi>p</mi> </msub> <mo>=</mo> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mo>-</mo> <mi>&gamma;</mi> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <mn>1</mn> <mo>/</mo> <mi>&gamma;</mi> </mtd> </mtr> </mtable> </mfenced> <mo>=</mo> <mfenced open='(' close=')'> <mtable> <mtr> <mtd> <mo>-</mo> <mn>1</mn> </mtd> <mtd> <mn>0</mn> </mtd> </mtr> <mtr> <mtd> <mn>0</mn> </mtd> <mtd> <mn>1</mn> </mtd> </mtr> </mtable> </mfenced> <mo>.</mo> </mrow> </math>

Since the controller poles can be limited by limiting the closed-loop poles, the closed-loop pole constraints are written in the form of regional pole constraints, including

L_-＝0，M_-＝1；L₊＝-2000，M₊＝-1； $L_h=\left(\begin{array}{cc}-1000& 0\\ 0& -1000\end{array}\right),$ $M_h=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right).$

Therefore, it is

$L=\left(\begin{array}{ccc}{L}_{-}& 0& 0\\ 0& L_{+}& 0\\ 0& 0& L_h\end{array}\right)=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& -2000& 0& 0\\ 0& 0& -1000& 0\\ 0& 0& 0& -1000\end{array}\right),$ $M=\left(\begin{array}{ccc}{M}_{-}& 0& 0\\ 0& M_{+}& 0\\ 0& 0& M_h\end{array}\right)=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& -1& 0& 0\\ 0& 0& 0& -1\\ 0& 0& 1& 0\end{array}\right).$

7) A, B₁，B₂，B₃，C₁，D₁₁，D₁₂，D₁₃，C₂，D₂₁，D₂₂，D₂₃，C₃，D₃₁，D₃₂，R_pAnd substituting L and M into the inequality group (3), and solving the feasibility problem by using a feasp solver in an LMI toolbox provided by MATLAB.

Solving the feasibility problem and obtaining a feasible solution

$Q=\left(\begin{array}{cccccccccc}-2459.6& 0& 0& 0& 0& 0& 0& -1& 0& 0\\ 0& -2459.6& 0& 0& 0& 0& -1& 0& 0& 0\\ 0& 0& -2459.6& 0& 0& 0& 0& 0& -1& 0\\ 0& 0& 0& -2459.6& 0& -1& 0& 0& 0& 0\\ 0& 0& 0& 0& -2459.6& 0& 0& 0& 0& -1\\ 0& 0& 0& -1& 0& -0.00041& 0& 0& 0& 0\\ 0& -1& 0& 0& 0& 0& -0.00041& 0& 0& 0\\ -1& 0& 0& 0& 0& 0& 0& -0.00041& 0& 0\\ 0& 0& -1& 0& 0& 0& 0& 0& -0.00041& 0\\ 0& 0& 0& 0& -1& 0& 0& 0& 0& -0.00041\end{array}\right),$

$S=\left(\begin{array}{cccccccccc}-1.2979e-10& 0& 0& 0& 0& 0& 0& 0& 4.3042e-15& 0\\ 0& -1.2979e-10& 0& 0& 0& 4.3042e-15& 0& 0& 0& 0\\ 0& 0& -1.2979e-10& 0& 0& 0& 0& 4.3042e-15& 0& 0\\ 0& 0& 0& -6.0809e-11& 0& 0& 2.0165e-15& 0& 0& 0\\ 0& 0& 0& 0& -6.0809e-11& 0& 0& 0& 0& 2.0165e-15\\ 0& 0& 0& -2.4723e-14& 0& 0& 8.1984e-19& 0& 0& 0\\ 0& -5.2769e-14& 0& 0& 0& 1.7499e-18& 0& 0& 0& 0\\ -5.2769e-14& 0& 0& 0& 0& 0& 0& 0& 1.7499e-18& 0\\ 0& 0& -5.2769e-14& 0& 0& 0& 0& 1.7499e-18& 0& 0\\ 0& 0& 0& 0& -2.4723e-14& 0& 0& 0& 0& 8.1984e-19\end{array}\right),$

$R=\left(\begin{array}{cccccccccc}30155& 0& 0& 0& 0& 0& 0& 0& -1& 0\\ 0& 30155& 0& 0& 0& -1& 0& 0& 0& 0\\ 0& 0& 30155& 0& 0& 0& 0& -1& 0& 0\\ 0& 0& 0& 30155& 0& 0& -1& 0& 0& 0\\ 0& 0& 0& 0& 30155& 0& 0& 0& 0& -1\\ 0& -1& 0& 0& 0& 0.000033& 0& 0& 0& 0\\ 0& 0& 0& -1& 0& 0& 0.000033& 0& 0& 0\\ 0& 0& -1& 0& 0& 0& 0& 0.000033& 0& 0\\ -1& 0& 0& 0& 0& 0& 0& 0& 0.000033& 0\\ 0& 0& 0& 0& -1& 0& 0& 0& 0& 0.000033\end{array}\right),$

$X=\left(\begin{array}{cccccc}6.4985& 0.7574& 3.4& 9.1732& 1.7777& -0.2428\\ 0.7574& 206.6773& -674.3242& -4.6010& -0.9817& 0.1517\\ 3.4& -674.3242& 216240& 8.0901& 11.9220& -1.1219\\ 9.1732& -4.6010& 8.0901& 37051& 20812& -3900.8\\ 1.7777& -0.9817& 11.9220& 20812& 118890& -2466.9\\ -0.2428& 0.1517& -1.1219& -3900.8& -2466.9& 3252\end{array}\right),$

$Y=\left(\begin{array}{cccccc}696.4801& -11.0024& -0.6236& -30.0742& 106.62& 0.2413\\ -11.0024& 5.8708& -0.4571& 2.0474& 0.2423& -0.3173\\ -0.6236& -0.4571& 0.2475& 0.0914& 0.1068& 0.0775\\ -30.0742& 2.0474& 0.0914& 80.1077& -16.8275& -3.8082\\ 106.62& 0.2423& 0.1068& -16.8275& 21.406& 1.6258\\ 0.2413& -0.3173& 0.0775& -3.8082& 1.6258& 0.833\end{array}\right).$

8) Constructing P according to (4) - (6), and substituting into step 5

Solving an inequality group (8) by using a feasp solver in an LMI toolbox provided by MATLAB, and solving unknown variables

$A_k=\left(\begin{array}{cccccc}-1288.6& 18.7622& 60.8087& 45.9548& -21.9677& -27.0336\\ 1124.6& -1272.5& 281.9359& 38.436& 58.1303& -5.4829\\ 3586.6& 521.7326& -570.4227& -1368& 341.9625& 78.9544\\ -951.1645& -652.2056& 233.2493& 144.4435& -3.5302& -39.3907\\ -193.4034& -165.4955& 44.9922& -72.6537& 88.8478& 145.7109\\ -542.9429& 343.3224& -18.6637& 1552.6& -717.1845& -415.2463\end{array}\right),$

${\mathrm{<figure-callout\; id="1"\; label="B\; k"\; filenames="HDA0000079904860000022.png"\; state="\{\{state\}\}">B</figure-callout>}}_k=\left(\begin{array}{ccc}-73.1464& -5753.6& -8646.8\\ 36.6791& 5.5095& -146810\\ -58.9907& -294.518& -632.7225\\ -296370& 43756& -1269.9\\ -166500& 37004& -925.5923\\ 31204& -51489& 71.5198\end{array}\right),$

${\mathrm{<figure-callout\; id="2"\; label="B\; k"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">B</figure-callout>}}_k=\left(\begin{array}{ccccc}0.8754& 0.0312& -1.2337& 1.3276& -0.0326\\ -1.1894& 7.0742& 0.3250& 0.6043& -6.2428\\ 4.7209& -22.9192& -2.2694& -0.6529& 20.2209\\ 1.2658& -0.1535& -1.7461& 1.8525& 0.1285\\ 0.2462& -0.0343& -0.3388& 0.3588& 0.0289\\ -0.0332& 0.0011& 0.0463& -0.0496& -0.00079\end{array}\right),$

C_k1＝(39.5182 -3.7509 -11.4033 -8.7521 -5.0904 2.6514)，

${\mathrm{<figure-callout\; id="2"\; label="C\; k"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_k=\left(\begin{array}{cccccc}-2.2433e+5& -1.2982e+5& 7.7062e+4& 3.4794e+4& 3.4233e+4& 2.3731e+4\\ 4.1831e+5& -2.8208e+5& 4.0253e+4& -7.9968e+4& 3.0482e+3& 2.0239e+4\\ 8.5441e+3& 1.8336e+3& -268.5281& -605.249& 1.9332e+3& -9.1609\\ 4.3764e+5& -2.5419e+5& 2.7769e+4& -8.2202e+4& -200690e+3& 1.6088e+4\\ -3.6414e+5& -1.5583e+5& 1.0589e+5& 5.8106e+4& 4.8748e+4& 3.1804e+4\end{array}\right)$

，D_c11＝(0.0325-5.0983-22.6357)，

D_c12＝(0.00069 0.00087 -0.0011 0.0013 -0.00077)，

$D_{c21}=\left(\begin{array}{ccc}0.3813& -656.4937& 15156\\ -0.7336& 172.2624& -37326\\ 0.0046& 6652.1& -2492\\ -0.7664& -2750.6& -32309\\ 0.6476& -95.5463& 8590.4\end{array}\right),$

$D_{c22}=\left(\begin{array}{ccccc}0.339& 0.0054& -0.0033& 0.1030& 0.8391\\ -0.0583& 0.0172& 0.0785& -0.0819& -0.0149\\ -3.0926& 0.0045& 0.025& -0.0197& 0.3451\\ 1.3065& 0.0113& 0.1789& -0.0758& 0.0703\\ 0.0564& 0.0181& -0.0826& 0.091& -0.0163\end{array}\right),$

${\mathrm{<figure-callout\; id="1"\; label="X"\; filenames="HDA0000079904860000022.png"\; state="\{\{state\}\}">X</figure-callout>}}_1=\left(\begin{array}{cccccc}6.4985& 0.7574& 3.4& 9.1732& 1.7777& -0.2728\\ 0.7574& 206.6773& -674.3242& -4.6010& -0.9817& 0.1517\\ 3.4& -674.3242& 216240& 8.0901& 11.9220& -1.1219\\ 9.1732& -4.6010& 8.0901& 37051& 20812& -3900.8\\ 1.7777& -0.9817& 11.9220& 20812& 118890& -2466.9\\ -0.2428& 0.1517& -1.1219& -3900.8& -2466.9& 3252\end{array}\right),$

${\mathrm{<figure-callout\; id="1"\; label="Y"\; filenames="HDA0000079904860000022.png"\; state="\{\{state\}\}">Y</figure-callout>}}_1=\left(\begin{array}{cccccc}696.4801& -11.0024& -0.6236& -30.0742& 106.62& 0.2413\\ -11.0024& 5.8708& -0.4571& 2.0474& 0.2423& -0.3173\\ -0.6236& -0.4571& 0.2475& 0.0914& 0.1068& 0.0775\\ -30.0742& 2.0474& 0.0914& 80.1077& -16.8275& -3.8082\\ 106.62& 0.2423& 0.1068& -16.8275& 21.406& 1.6258\\ 0.2413& -0.3173& 0.0775& -3.8082& 1.6258& 0.833\end{array}\right).$

9) Calculated according to equations (9) and (10)

${\mathrm{<figure-callout\; id="2"\; label="X"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">X</figure-callout>}}_2=\left(\begin{array}{cccccc}-1.2706& 0.3484& -0.0480& 2.8605& -2.5636& 1.1823\\ 0.3544& -0.0155& 3.4521& -14.5029& 23.7837& 0.1351\\ 33.8124& 4.6740& -355.2026& 0.6478& 0.3330& 0.00075\\ -296.344& 1626.7& 0.1364& -5.6224& -0.7229& 0.0026\\ -3493.7& -139.9009& -3.5094& -0.1023& -0.0071& -0.00033\\ 39.1632& -175.5206& -5.3982& -52.0013& -6.6919& 0.0271\end{array}\right),$

${\mathrm{<figure-callout\; id="2"\; label="Y"\; filenames="HDA0000079904860000011.png,HDA0000079904860000022.png"\; state="\{\{state\}\}">Y</figure-callout>}}_2=\left(\begin{array}{cccccc}3451.4& -53.7335& -21.3585& -7.5083& 1.387& -0.0785\\ 22.7953& -46.1752& -287.9025& 19.3499& -9.958& -0.2797\\ 4.076& -2.5695& 151.8740& -8.9714& -16.5412& -0.7012\\ -27.7805& -1623.3& 28.7284& 6.5642& -0.2286& 0.0451\\ 617.9432& 225.4126& 127.7795& 40.2645& -7.9124& 0.4918\\ 29.0623& 73.3286& 51.5717& 27.9124& 13.4275& -0.7725\end{array}\right),$ The result is substituted into equation set (11) to solve, resulting in the nominal part of the controller, K in fig. 3:

10) solving according to expression (7) from the previously solved P

Solving for Delta by substituting inequality (12)_cThe variable parameter part of the controller is connected with the nominal part to form the controller. Note that Δ_cVaries with the change in delta. Since Delta is variable parameter and needs to be determined in real time according to working conditions, Delta_cIs also a real-time variable.

The simulation results are shown in fig. 4 and 5, wherein fig. 4 shows the actual parameter variation curve, and fig. 5 shows the square wave response output. From the normal acceleration curve in fig. 5, f can be seen_yAnd f_yrVery close, almost returnable, and overshoot less than 5%, 2% settling time less than 0.15s, both global and local performance requirements are met.

Claims (4)

1. The robust gain scheduling control method based on the pole allocation of the region is characterized by comprising the following steps of:

step 1: the controlled object is arranged into a standard form;

step 2: establishing an expression of performance index constraint;

2.1 Global Performance: writing the global performance index requirement into an integral quadratic constraint form, wherein the specific index is obtained by a multiplication operator R according to the performance requirement_pGiven, having the form shown below:

where ε is any positive real number tending to 0; xi is the combination of the input vector and the output vector, and a reference input r and a performance output h are selected; zeta is input vector, choose reference input r; matrix R_pGeneral press

The form of the method is divided into four blocks, and numerical values of all parts are given according to specific index requirements;

2.2 local Properties: limiting the poles to a certain area D on the complex plane, where D ═ z ∈ C | f_D(z) < 0}, and the region feature function is expressed as

Wherein the coefficient matrix L of the characteristic function is L^T∈R^m×m，M∈R^m×m；

And 3, step 3: judging whether a controller exists or not, so that the stability and the dynamic performance of the closed-loop system meet the requirements;

and 4, step 4: constructing a multiplier P; first, singular value decomposition is performed as follows:

where Γ is the unitary matrix, - ∑_-Is a diagonal matrix formed by the singular values of the part less than zero, sigma₊Is a diagonal matrix formed by the singular values of the part which is larger than zero; then, order

Then there is

And 5, step 5: solving the linear matrix inequality set to enable the closed-loop system to realize performance indexes;

order to

Can be combined with

Divided into four blocks, as shown by:

order to

Solving the following inequality group by using a feasp solver in an LMI toolbox provided by MATLAB:

wherein,

the unknown variables are: a. the_k，B_k1，B_k2，C_k1，C_k2，D_c11，D_c12，D_c21，D_c22，X₁，Y₁；

And 6, step 6: solving a nominal part of the controller;

using already solved X₁，Y₁Performing singular value decomposition; the expression of the specific calculation method is as follows: I-X₁Y₁＝UΛV^TLet X₂＝UΛ^1/2 (9) Y₂＝VΛ^1/2 (10)；

U, V in the formula is unitary matrix, and A obtained by the solution is used_k，B_k1，B_k2，C_k1，C_k2，D_c11，D_c12，D_c21，D_c22，X₁，Y₁，X₂，Y₂Solving a linear equation set;

wherein the unknown variable is A_c，B_c1，B_c2，C_c1，C_c2(ii) a A obtained by solving_c，B_c1，B_c2，C_c1，C_c2And D_c11，D_c12，D_c21，D_c22Together forming a nominal part of the controller;

and 7, step 7: solving a variable parameter part of the controller;

the following inequality was solved using the feasp solver in the LMI toolbox provided by MATLAB:

in the formula, the unknown variable is Delta_cChanges along with the change of the parameter module delta;

and 8, step 8: a controller is constructed.

According to the previously determined A_c，B_c1，B_c2，C_c1，C_c2，D_c11，D_c12，D_c21，D_c22And Δ_cConstructing a controller, wherein a state equation is as follows:

d_cΔ＝Δ_ce_cΔ (14) 。

2. the robust gain scheduling control method based on pole-in-area allocation as claimed in claim 1, characterized in that the standard form is sorted as follows:

d_Δ＝Δe_Δ (2)

in the formula, the vector x represents the state of the controlled object, d_ΔRepresenting disturbance input, r representing reference input, u representing control input, e_ΔRepresenting the disturbance output, h representing the output of the performance channel, and y representing the measurement output; the matrix A is a system matrix, B₁、B₂、B₃Control matrices from disturbance input, reference input, control input, respectively, C₁、C₂、C₃Output matrices, D, leading to disturbance output channel, performance output channel, measurement output channel, respectively_ij(i, j ═ 1, 2, 3) a direct connection matrix between channels, where the direct connection term from control input u to measurement output y is 0; Δ represents a parametric model of the non-linear, time-varying characteristics and uncertainty.

3. The robust gain scheduling control method based on the regional pole allocation as claimed in claim 1 or 2, wherein in step 3, the feasp solver in the LMI toolbox provided by MATLAB is used to solve the following inequality set to determine whether the controller exists:

in the set of inequalities, the unknown variables are: q, S, R,

x, Y, where the symbol ker (Φ) represents the basis matrix of the null space of Φ, having Φ^Tker (Φ) ═ 0; mark

Represents the Kronecker product; if the inequality group (3) has a solution, the existence of the controller can enable the performance index of the closed-loop system to be met, and the subsequent steps are carried out to solve the controller; otherwise, the performance index can not be met, and the design problem of the original controller is solved.

4. The robust gain scheduling control method based on pole-in-region allocation as claimed in claim 1 or 2, wherein the pole region in step 2 is a combination of the following simple regions:

1) semi-plane: re (z) < α:then L is_-＝-2α，M_-1 is ═ 1; semi-plane: re (z) > β:

then L is₊＝2β，M₊＝-1；

2) Circle with (-q, 0) as the center and r as the radius:

then

3) Cone with 2 θ as cone angle:

then

4) Horizontal strip-shaped area with the real axis as the symmetry axis and the width of 2 h:

then

Writing the pole region as the intersection of the above simple regions, i.e. D ═ I_σD_σThen L and M are written as L respectively_σAnd M_σHas a block diagonal form of L ═ blockdiag { Λ, L ═ L_σ，Λ}M＝blockdiag{Λ，M_σΛ, where the subscript σ ∈ { -, +, r, c, h }, indicates that the pole region is composed of the above simple regions.

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