《现代控制理论》课程教学资源课程教学资源——教学课件_Chap1_1.6 Obtaining a Jordan Canonical Form by State Transformation

CHAPTER1 STATE SPACE MODELCONTENT> 1.1 Definition of State Space> 1.2 Obtaining State Space Model from I/O Model> 1.3ObtainingTransferFunctionMatrixfromStateSpace Model>1.4 ModelofCompositeSystems> 1.5 State Transformation of the LTI system>1.6Obtaininga Jordan CanonicalFormby StateTransformation

CHAPTER1 STATE SPACE MODEL • CONTENT  1.1 Definition of State Space  1.2 Obtaining State Space Model from I/O Model  1.3 Obtaining Transfer Function Matrix from State Space Model  1.4 Model of Composite Systems  1.5 State Transformation of the LTI system  1.6 Obtaining a Jordan Canonical Form by State Transformation

1.6 Obtaining a Jordan Canonical Form byState TransformationConsider the LTI system such asy(t) = CX(t) + Du(t)X(t) = AX(t) + Bu(t)In this section, we will find the nonsingular transformationX(t) = Px(t), by which the state space description can betransform to Jordan canonical form from some general formCase 1 The eigenvaluesof A are all distinct

1.6 Obtaining a Jordan Canonical Form by State Transformation

1.6 Obtaining a Jordan Canonical Form byState TransformationCase 1 The eigenvalues of A are all distinct.Let ,2,..-2, be the distinct eigenvalues of A, and letV, be the eigenvector of A associated with the eigenvalue2, (i=1,2,...,n) ..Then, the matrix P =, V, ... V, I is anonsingular matrixSince AV, = a,VAP =[AV[ava.y.1...AV]=00[00M0000=P=[VV.2002n00

1.6 Obtaining a Jordan Canonical Form by State Transformation

1.6 Obtaining a Jordan Canonical Form byState TransformationCase 1 The eigenvalues of A are all distinct.AP =[AVa.VAV,]=[[avD00<M[000000=P=[VV.F00^n00MT0000P-1AP =Hence.002

1.6 Obtaining a Jordan Canonical Form by State Transformation

1.6 Obtaining a Jordan Canonical Form byState TransformationCase1 The eigenvalues of A are all distinct.y(t) = CX(t) + Du(t)X(t) = AX(t) + Bu(t)P-[iv, ... V,]LetorandX(t) = PX(0)X(t) = P-1 X(t)It can transform the general state space model into thediagonalcanonicalformX(t) = P-1APX()+ P-1Bu(t) = AX(t)+ Bu(t)y(t) = CPX(t) + Du(t) = CX(t) + Du(t)

1.6 Obtaining a Jordan Canonical Form by State Transformation

1.6 Obtaining a Jordan Canonical Form byState TransformationCase 1 The eigenvalues of A are all distinct.y(t) = CX(t)+ Du(t)X(t) = AX(t) + Bu(t)P=[ViV, ...V]LetandX(t) = PX(t)orX(t) = P-1 X(t)X(t) = P-1 APX(t) + P- Bu(t) = AX(t)+ Bu(t)y(t) = CPX(t) + Du(t) = cX(t) + Du(t)[07diagonal canonical formwhere A=P-AP-元,0D=D=CPB = P-IB

1.6 Obtaining a Jordan Canonical Form by State Transformation

Example 1.18 The state spacemodel of a system is[7][2-1-1X=00y=[1 0 1]x-1X+2u[021[3Determine the transformation matrix P and transform thestate space model into the diagonal canonical form by thestate transformation X(t) = Px(t) .Solution(a) The characteristic equation1[2-2100[2I-A=1+1= (a-2)(-1)( +1)= 00-2元-1yields the eigenvalues 2 = 2, 2, =l, 2, =-1

Example 1.18 The state space model of a system is[7][2-1-1X=00-1y=[1 0 1]xX+2u0231Solutiontheeigenvalues2=2,,=l,2,=-1(b) Since the eigenvalues are distinct, the state equation canbe converted into a diagonal canonical form by the means ofa state transformation X(t) = Px(t)

Example 1.18 The state space model of a system is[7][2-1-1X=00y=[1 o 1]x-1X+2u[02[31Solutiontheeigenvalues=2,2,=1,2,=-1when =2， the equationthe1 yieldsAV=AV[c]0VofAcorresponding tocharacteristicvector=0[1]10A,and Vi =is selectedhere.[0

Example 1.18 The state space model of a system is[7][2-1-1X=00y=[1 o 1]x-1X+2u0231Solutiontheeigenvalues2=2,元,=l,2,=-1When , =l, the equationthe2,V2 =AV2 yields[1]characteristicvector V,=oof A correspondingto 2.1