State space modelState transition matrix: eAtExampleSlide 4I/O model to state spaceSlide 6Slide 7Characteristic valuesSlide 9Slide 10Solution of state space modelS.S to T.F.Slide 13But don’t use those for hand calculationSlide 15Eigenvalues, eigenvectorsSlide 17Slide 18Slide 19Slide 20Slide 21Slide 22State space modelState space model:linear:where: u: input y: output x: state vectorA,B,C,D, or F,G,H,J are const matrices ),( :eqOutput ),( :eq StateuxhyuxfxDuCxyBuAxxIf dim(x) = n, we say system order = n A is nxn matrix, called system matrixSystem property largely determined by properies of A.State transition matrix: eAt•eAt is an nxn matrix •eAt =ℒ-1((sI-A)-1), or ℒ (eAt)=(sI-A)-1• eAt= AeAt= eAtA•eAt = Inxn+At+ A2t2+ A3t3 + •eAt is invertible: (eAt)-1= e(-A)t•eA0=I•eAt1 eAt2= eA(t1+t2)dtddef!21!31Example,0000AssssAsI100100)(11)(00)()1()0()0()1())((111111tutussAsIeAtLLLLL,0021aaA211211100100)(asasasasAsI)(00)(100121211tuetueasasetataAtLExample,1011A110)1(1111011)1(11011)(2211ssssssssAsI101100ssAsI)(0)()(110)1(11121tuetutetuesssetttAtLI/O model to state space•Infinite many solutions, all equivalent.•Controller canonical form: uxbbbyuxaaaxubudtdbudtdbyaydtdaydtdaydtdnnnnnnnnnn]0[1000001000001000010 1101100111101111Examplettrydydtdydtyddtyd02233)(235d tdrydtd ydtyddtydd tyddtd 235:223344n=4 a3 a2 a1 a0 b1 b0=b2=b3=0    xyuxx001010005312100001000010>> n=[1 2 3];d=[1 4 5 6];>> [A,B,C,D]=tf2ss(n,d)A = -4 -5 -6 1 0 0 0 1 0B = 1 0 0C = 1 2 3D = 0>> tf(n,d) Transfer function: s^2 + 2 s + 3---------------------s^3 + 4 s^2 + 5 s + 6DuCxyBuAxxudtdudtudydtdydtyddtyd32654222233Characteristic values•Char. eq of a system isdet(sI-A)=0the polynomial det(sI-A) is called char. pol.the roots of char. eq. are char. valuesthey are also the eigen-values of Ae.g. ∴ (s+1)(s+2)2 is the char. pol. (s+1)(s+2)2=0 is the char. eq. s1=-1,s2=-2,s3=-2 are char. values or eigenvaluesuxx0102001200012)2)(1(200120001det)det(  sssssAsI2100)2(12100011 )2)(1(001)2)(1(000)2()2)(1(1)(2221ssssssssssssAsI)(00)()(000)( ) (2221tuetutetuetueettttAtLcanAttteee 100set t=022I0001∴NocanAtttteetee 0at t=0:1001 1101 yes,01101 0Aeteeeteeedtdttttttt???√√Solution of state space modelRecall: sX(s)-x(0)=AX(s)+BU(s) (sI-A)X(s)=BU(s)+x(0) X(s)=(sI-A)-1BU(s)+(sI-A)-1x(0) x(t)=(ℒ-1(sI-A)-1))*Bu(t)+ ℒ-1(sI-A)-1) x(0) x(t)= eA(t-τ)Bu(τ)d τ+eAtx(0) y(t)= CeA(t-τ)Bu(τ)d τ+CeAtx(0)+Du(t)DuCxyBuAxxt0t0S.S to T.F.X(s)=(sI-A)-1BU(s)Y(s)=C(sI-A)-1BU(s)+DU(s) =(D+ C(sI-A)-1B)U(s) ∴ T.F. H(s)= D+ C(sI-A)-1BIn matlab: ss2tfeigrootspolyuse help to find out how to use these•In Matlab:>> A=[0 1;-2 -3];>> B=[0;1];>> C=[1 3];>> D=[0];>> [n,d]=ss2tf(A,B,C,D)n = 0 3.0000 1.0000d = 1 3 2>>tf(n,d)23130)(22sssssHBut don’t use those for hand calculationuse:X(s)=(sI-A)-1BU(s)+(sI-A)-1x(0) x(t)=ℒ-1{(sI-A)-1BU(s)}+{ℒ-1 (sI-A)-1} x(0)& Y(s)=C(sI-A)-1BU(s)+DU(s)+C(sI-A)-1x(0) y(t)= ℒ-1{C(sI-A)-1BU(s)+DU(s)}+C{ℒ-1 (sI-A)-1} x(0)e.g. xyuxx01,012001u= unit step10)0(x2111121111 10210011101210011)(ssssssssssssX)()()()(2tuetuetutxtt 111 102111101 )()()(sssssssDUsCXsY)()()( tuetutytNote: T.F.=D+ C(sI-A)-1B 1101210011010sssEigenvalues, eigenvectorsGiven a nxn square matrix A, p is an eigenvector of A if Ap p∝i.e. λ s.t. Ap= λpλis an eigenvalue of AExample: ,Let ,∴p1 is an e-vector, & the e-value=1 Let ,∴p2 is also an e-vector, assoc. with the λ =-2 2001A011p1101012001pAp 102p102201020012ApIn Matlab>> A=[2 0 1; 0 2 1; 1 1 4];>> [P,D]=eig(A)P = 0.6280 0.7071 0.3251 0.6280 -0.7071 0.3251 -0.4597 -0.0000 0.8881 p1 p2 p3D =1.2679 0 0