Lecture 3The State Transition MatrixFundamental Matrix ExampleThe State Transition Matrix FThe State Transition Matrix FSolution to the State EquationLTI Systems – Deriving FState Transition Matrix FExamplesAn Easy Way To Compute FExamine Transformation To A New Basis (LTI)DefinitionsSolution to the State EquationModal ExpansionModal ExpansionModal ExpansionModal Expansion (cont)Compute F Using Laplace TransformsCompute the Transfer Function Matrix G(s)Transfer Function MatrixReviewStructural Properties of the State Space SystemControllabilityControllability Tests for LTI SystemsControllability Tests for LTI SystemsControllability Tests for LTI SystemsThe Controllable SubspaceModal Approach to ControllabilityRows of B TestControllability is InvariantRows of B TestControllabilityControllabilityHautus TestExampleExampleExampleLecture 4Controllable SubspacesControllable SubspacesControllability DecompositionControllability DecompositionControllability DecompositionControllability DecompositionControllability DecompositionTime Varying CaseControllability for Time Varying SystemsControllability for Time Varying SystemsControllability for Time Varying SystemsControllability for Time Varying SystemsControllability for Time Varying SystemsObservabilityObservability: LTI CaseDualityDerive Tests Using DualityControllability Test SummaryObservability Test SummaryStabilizability/DetectabilityDegree of ControllabilityRepeated RootsKalman Decomposition TheoremMinimal RealizationsMinimal RealizationsTransmission ZerosTransmission ZerosTZ’s and Connection to Contr/ObsLecture 3K.A. Wise 2 The State Transition Matrix Chapter 9 ( ) ( )( ) ( )x Atx Btuy Ctx Dtu= += +nnnyxuxR uR yR∈∈∈Time Varying Systems ( )x Atx=Eq. (1) Theorem: There exists uniquely for every x(0) a x(t) that satisfies Eq. (1). Theorem: Solutions to Eq. (1) form a nx-dimensional vector space over . Def: - Fundamental matrix if its columns consist of nx linearly independent solutions to Eq. (1). nnxxU×1nn nxx xU UU×=RK.A. Wise 3 Fundamental Matrix Example 000xxt=( )( )( )1 1 0 10212 10 202xt xt Xx t tX X= == +( )( ) ( )1210xtx t tx t==Since the dimension of the vector space is 2, 2 linearly independent initial vectors can be found, and each one defines an unique solution. ( )1002001XxtX= =( )101Ut=()1002020XxtX= =( )222Utt=( )2021Utt=K.A. Wise 4 The State Transition Matrix Φ The solution to the state equation describes the transition of the initial state x(t0) state to the state x(t) at time t. ( )( )( )100,tt U tU t−Φ=Define as the state transition matrix. () ( )( )()( )( )( )( )( )( )( )1 2 121 11 10 1 10 2 20 2 20Let and be two fundamental matrices where .,U t U t U UPtt U tU t U tP P U t U tU t− −− −=Φ= = =( )0,ttΦexhibits the following properties: ( )( ) ( )1001) ,2) , ,tt Itt t t−Φ=Φ=Φ()0xt( )xt( )()( )( )( )20 21 10 0 1 203) , , ,4) , satisfies tt tt tt t t ttt x At xΦ =Φ Φ ≤≤Φ=( )( )( ) ( )( )( ) ( )( )( )( )110 0 00,,AtUttt U t U t AtU tU t At tt−−Φ= = =ΦK.A. Wise 5 The State Transition Matrix Φ From the previous example ( )() ( ) ( )( )2110200 022221102021002021,1011tUt Ut U t tttttt−−= = = Φ=−Now, consider the general solution to ( )( )x Atx Btu= +Theorem: The solution to is ( ) ( )x Atx Btu= +( )( ) ( )( ) ( ) ( )000,,ttxt tt xt t B u dτ τ ττ=Φ +Φ∫Zero Input Response Convolution IntegralK.A. Wise 6 Solution to the State Equation Proof: Differentiate the solution to show it satisfies the differential equation. ( )( ) ( ) ( )000,,ttx tt x t B u dtτ τ ττ∂=Φ +Φ∂∫()( )( ) ( )( ) ( )( ) ()() ( )000,,, ,ttAt tAt tt x tt Btut t B u dττ τ ττΦ= Φ +Φ + Φ∫( )( )( ) ( ) ( )( )( ) ( )000,,ttAt tt x t B u d Btutxtτ τ ττ= Φ +Φ +∫( ) ( )Atx Btu= +( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( )000,,ttyt Ct tt xt C t B D t u dτ τ τ τδ τ τ τ=Φ +Φ + −∫Impulse Response Matrix The output is given by:K.A. Wise 7 LTI Systems – Deriving Φ x Ax=Method 1 – Derive Φ using a Taylor Series. Expand about ()()00kkkttdxtxdt==( )xt0t( )( ) ( )2000 0 02xxt x xtt tt=+ −+ − +200 00 0dxx Ax x A xdt= = = ( )()( )00 000,!kkkAttxt x tt xk=−= = Φ∑( )( )( )0000,!kkAt tkAtttt ek−=−Φ= =∑( )( )00At txt e x−=K.A. Wise 8 State Transition Matrix Φ Check to see if it satisfies the properties of the state transition matrix ( )( ) ( )1001) ,2) , ,tt Itt t t−Φ=Φ=Φ( )()00At txt e x−=( )At teI−=( ) ( )00At t At tee I−−=( )( )( )( )( )20 21 1001203) , , , 4) , satisfies tt tt ttttttt x At xΦ =ΦΦ≤≤Φ=( )( ) ( )10 2021At t At tAt tee e−−−=( )( )( )( )100001!kkAt t At tkAttde Aedt k−−−=−= =−∑( )( )( )( )( )100100,At tAtAttt U tU tee e−−−Φ== =K.A. Wise 9 Examples 11120220000!00ktkkk t AtktketA A eekeλλλλλλλλ>= = = =∑1) Diagonal A-matrix. Assume real eigenvalues. 2) Jordan form. Assume real eigenvalues. ( )11 110110! 1!00kkkk kkktkkkttAttkttA A t tekke teeeλλλλλλλλλλλ−− −−≥≥= = = =−=∑∑K.A. Wise 10 An Easy Way To Compute Φ Assume A-matrix has distinct eigenvalues. Look at similarity transformation that transforms the A-matrix to diagonal form. ( )11110!kkAt T T tnxkTT tATT T v v e ek−−−Λ>Λ=Λ= ==∑Examine each term: ( )21 1 1 21TT TTTT T T− −− −Λ =Λ Λ=Λ( )1 11 1kkTT TT TT T T− −− −Λ =Λ Λ=Λ1Factor out and on left/rightTT−()1110!kkAt T T t tkte e T T Te Tk−Λ − Λ−>Λ= = =∑Diagonal matrix with on diagonal tieλK.A. Wise 11 Examine Transformation To A New Basis (LTI) ( )111 Substitute in for BAx Tx x T x x Tx xx Tx T Ax Bu TAx TBu TAT x TBu Ax Buy Cx Du CT x Du Cx Du−−−= = === +=+= + =+=+= +=+11 x Tx A TAT B TB C CT D D−−= = = = =( )( )11At TAT tCe B CT e TB−−=( ) ( )( )11At AtCT