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