LECTURE (7) STATE-SPACE REPRESENTATION OF LTI SYSTEMS Assist. Prof. Amr E. Mohamed لما نصدق اننا ِن ْْقدَ ر ،،، نِـ ْقـدَ ر Agenda State Variables of a Dynamical System State Variable Equation Why State space approach Derive Transfer Function from State Space Equation Time Response and State Transition Matrix 2 Introduction The classical control theory and methods (such as root locus) that we have been using in class to date are based on a simple input-output description of the plant, usually expressed as a transfer function. These methods do not use any knowledge of the interior structure of the plant, and limit us to single-input single-output (SISO) systems, and as we have seen allows only limited control of the closed-loop behavior when feedback control is used. Modern control theory solves many of the limitations by using a much “richer” description of the plant dynamics. The so-called state-space description provide the dynamics as a set of coupled first-order differential equations in a set of internal variables known as state variables, together with a set of algebraic equations that combine the state variables into physical output variables. 3 Definition of System State u1 (t ) u2 (t ) ur (t ) y1 (t ) Inner state variables x1 , x2 , xn y2 (t ) y p (t ) State: The state of a dynamic system is the smallest set of variables (𝒙𝟏 , 𝒙𝟐 , … … , 𝒙𝒏 ) (called State Variables or State Vector) such that knowledge of these variables at 𝑡 = 𝑡0, together with knowledge of the input for 𝑡 ≥ 𝑡0 , completely determines the behavior of the system for any time t to t0 . The number of state variables to completely define the dynamics of the system is equal to the number of integrators involved in the system (System Order). Assume that a multiple-input, multiple-output system involves n integrators (State Variables). Assume also that there are r inputs u1(t), u2(t),……. ur(t) and p outputs y1(t), y2(t), …….. yp(t). 4 General State Representation State equation: Output equation: x (t ) A x(t ) B u (t ) y (t ) C x(t ) D u (t ) 𝑥 = 𝑆𝑡𝑎𝑡𝑒 𝑉𝑒𝑐𝑡𝑜𝑟 𝑥= 𝑑 𝑥(𝑡) 𝑡 = 𝐷𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑆𝑡𝑎𝑡𝑒 𝑉𝑒𝑐𝑡𝑜𝑟 𝑢 = 𝐼𝑛𝑝𝑢𝑡 𝑉𝑒𝑐𝑡𝑜𝑟 𝑦 = 𝑂𝑢𝑡𝑝𝑢𝑡 𝑉𝑒𝑐𝑡𝑜𝑟 𝐴 = 𝑆𝑡𝑎𝑡𝑒 𝑀𝑎𝑡𝑟𝑖𝑥 = 𝑆𝑦𝑠𝑡𝑒𝑚 𝑀𝑎𝑡𝑟𝑖𝑥 𝐵 = 𝐼𝑛𝑝𝑢𝑡 𝑀𝑎𝑡𝑟𝑖𝑥 𝐶 = 𝑂𝑢𝑡𝑝𝑢𝑡 𝑀𝑎𝑡𝑟𝑖𝑥 𝐷 = 𝐹𝑒𝑒𝑑𝑏𝑎𝑐𝑘 𝑚𝑎𝑡𝑟𝑖𝑥 Dynamic equations State-Space Equations (Model) State equation: Output equation: State variable x1 (t ) x (t ) x(t ) 2 x ( t ) n n1 State Vector A nn x (t ) A x(t ) B u (t ) Dynamic equations y (t ) C x(t ) D u (t ) y1 (t ) y (t ) 2 y (t ) y ( t ) p p1 u1 (t ) u (t ) u (t ) 2 u ( t ) r r1 Input Vector B nr C x1 (0) x (0) x(0) 2 x ( 0 ) n n1 Output Vector pn D pr 6 Block Diagram Representation Of State Space Model D u (t ) B x (t ) + - 1 s x(t ) + C + y (t ) A 7 Input/Output Models vs State-Space Models State Space Models: consider the internal behavior of a system can easily incorporate complicated output variables have significant computation advantage for computer simulation can represent multi-input multi-output (MIMO) systems and nonlinear systems Input/Output Models: are conceptually simple are easily converted to frequency domain transfer functions that are more intuitive to practicing engineers are difficult to solve in the time domain (solution: Laplace transformation) 8 Some definitions System variable: any variable that responds to an input or initial conditions in a system State variables: the smallest set of linearly independent system variables such that the values of the members of the set at time t0 along with known forcing functions completely determine the value of all system variables for all t ≥ t0 State vector: a vector whose elements are the state variables State space: the n-dimensional space whose axes are the state variables State equations: a set of first-order differential equations with b variables, where the n variables to be solved are the state variables Output equation: the algebraic equation that expresses the output variables of a system as linear combination of the state variables and the inputs. General State Representation 1. Select a particular subset of all possible system variables, and call state variables. 2. For nth-order, write n simultaneous, first-order differential equations in terms of the state variables (state equations). 3. If we know the initial condition of all of the state variables at 𝑡0 as well as the system input for 𝑡 ≥ 𝑡0 , we can solve the equations State-Space Representation of nth-Order Systems of Linear Differential Equations Consider the following nth-order system: (𝒏) (𝒏−𝟏) 𝒚 + 𝒂𝟏 𝒚 + … + 𝒂𝒏−𝟏 𝒚 + 𝒂𝒏 𝒚 = 𝒖 where y is the system output and u is the input of the System. The system is nth-order, then it has n-integrators (State Variables) Let us define n-State variables 11 State-Space Representation of nth-Order Systems of Linear Differential Equations (Cont.) Then the last Equation can be written as 12 State-Space Representation of nth-Order Systems of Linear Differential Equations (Cont.) Then, the stat-space state equation is where 13 State-Space Representation of nth-Order Systems of Linear Differential Equations (Cont.) Since, the output equation is Then, the stat-space output Equation is where 14 Example #1 From the diagram, the system Equation is 𝑀𝑦 + 𝐵𝑦 + 𝐾𝑦 = 𝑓(𝑡) y K M f(t) B This system is of second order. This means that the system involves two integrators (State Variables). Let us define the state variables 𝑥1 = 𝑦 𝑥2 = 𝑦 Then, we obtain 𝑥1 = 𝑦 = 𝑥2 1 1 −𝐵 𝐾 1 𝑥2 = 𝑦 = −𝐵𝑦 + 𝐾𝑦 − 𝑓 𝑡 = 𝑥2 − 𝑥1 − 𝑓 𝑡 𝑀 𝑀 𝑀 𝑀 𝑀 15 Example #1 (Cont.) Then, the State Space equation is 0 1 𝑥 0 𝑥1 1 = −𝐾 −𝐵 𝑥 + 1 𝑓(𝑡) 𝑥2 2 𝑀 𝑀 𝑀 The output Equation is 𝑥1 𝑦= 1 0 𝑥 2 The System Block diagram is 16 Example #2 + Remark : the choice of states is not unique. ei (t ) di(t ) 1 t Ri (t ) L i(t )dt ei (t ) dt c 0 let x1 (t ) i (t ) x2 (t ) i(t )dt y(t ) i(t ) R x1 L x 2 1 1 LC 0 x1 y (t ) 1 0 x2 x1 1 x L ei (t ) 2 0 c i (t ) - let + L R ec (t ) - xˆ1 (t ) i (t ) xˆ2 (t ) ec (t ) y(t ) i(t ) R R xˆ1 L L 1 0 L xˆ2 xˆ1 y (t ) 1 0 xˆ2 xˆ1 1 xˆ L ei (t ) 2 0 17 y2 Example #3 K M2 B3 y1 M1 f (t ) B1 B2 M 1 y1 B1 y1 B2 ( y1 y 2 ) K ( y1 y 2 ) f (t ) M 2 y2 B3 y 2 B2 ( y 2 y1 ) K ( y 2 y1 ) 0 x1 y1 let x2 y 2 x3 y1 x4 y 2 x1 x2 x3 x4 x1 x2 f (t ) x3 x4 18 Example #4 Find the state space model for a system that described by the following differential equation c 9c 26c 24c 24r Solution: The system is 3rd order, then it has three states as follows x1 c differentiation x1 x2 x2 c x2 x3 x3 c x3 24 x1 26 x2 9 x3 24r The output equation is y c x1 Example #4 1 0 x1 0 x 0 0 1 2 x3 24 26 9 x1 y 1 0 0 x2 x3 x1 0 x 0 r 2 x3 24 State-Space Representations of Transfer Function Systems 21 State-Space Representation in Canonical Forms We here consider a system defined by where u is the control input and y is the output. We can write this equation as we shall present state-space representation of the system defined by (1) and (2) in controllable canonical form, observable canonical form, and diagonal canonical form. 22 Controllable Canonical Form We consider the following state-space representation, being called a controllable canonical form, as Note that the controllable canonical form is important in discussing the pole-placement approach to the control system design. 23 Observable Canonical Form We consider the following state-space representation, being called an observable canonical form, as 24 Diagonal Canonical Form Diagonal Canonical Form greatly simplifies the task of computing the analytical solution to the response to initial conditions. We here consider the transfer function system given by (2). We have the case where the dominator polynomial involves only distinct roots. For the distinct root case, we can write (2) in the form of 25 Diagonal Canonical Form (Cont.) The diagonal canonical form of the state-space representation of this system is given by 26 Example #5 Obtain the state-space representation of the transfer function system (16) in the controllable canonical form. Solution: From the transfer function (16), we obtain the following parameters: b0 = 1, b1 = 3, b2 = 3, a1 = 2, and a2 = 1. The resulting state-space model in controllable canonical form is obtained as 27 Example #6 Find the state-space representation of the following transfer function system (13) in the diagonal canonical form. Solution: Partial fraction expansion of (13) is Hence, we get A = −1 and B = 3. We now have two distinct poles. For this, we can write the transfer function (13) in the following form: 28 State Space model to Transfer Function 29 State Space model to Transfer Function The state space model x Ax Bu y Cx Du by Laplace transform sX s AX s BU s X s sI A BU s Y s CX s DU s Y s C sI A B D U s Then, the transfer function is Y s 1 T s C sI A B D U s 1 1 Example (2) Find the transfer function from the following transfer function Solution: 1 0 0 10 x 0 0 1 x 0 u 1 2 3 0 y 1 0 0 x s 1 0 sI A 0 s 1 1 2 s 3 ( s 2 3 s 2) s3 1 1 s ( s 3 ) s 2 s ( 2 s 1 ) s adj ( sI A) 1 sI A det(sI A) s 3 3s 2 2 s 1 Example (2) T s C sI A B D 1 ( s 2 3s 2) s3 1 1 0 0 1 s ( s 3) s 2 s ( 2 s 1 ) s T (s) s 3 3s 2 2 s 1 10( s 2 3s 2) T s 3 s 3s 2 2s 1 10 0 0 System Poles from State Space model poles and check the stability of the following state space Example find the System model 2 0 5 x x u 1 3 0 Solution: Since y 1 0 x s 2 sI A 1 s 3 sI A s 2 s ( s 3) 2 0 To find the poles Then the poles are {-1, -2 }, the system is stable 1 s3 State-Space Modeling with MATLAB 34 State-Space Modeling with MATLAB MATLAB uses the controllable canonical form by default when converting from a state space model to a transfer function. Referring to the first example problem, we use MATLAB to create a transfer function model and then convert it to find the state space model matrices: 35 State-Space Modeling with MATLAB Note that this does not match the result we obtained in the first example. See below for further explanation. No we create an LTI state space model of the system using the matrices found above: 36 State-Space Modeling with MATLAB we can generate the observable and controllable models as follows: 37 State transition matrix 38 Introduction d x(t ) Ax (t ) Bu (t ) dt y (t ) Cx(t ) Du(t ) The behavior of x(t) and y(t): 1) Homogeneous solution of x(t). 2) Non-homogeneous solution of x(t). 39 Homogeneous solution x (t ) Ax (t ) sX ( s) x(0) AX ( s) 1 x(t ) L [( sI A) ]x(0) e At x(0) X ( s) ( sI A) 1 x(0) 1 1 1 (t ) e L [( sI A) ] State transition matrix At x(t0 ) e At 0 x(0) x(0) e At 0 x(t0 ) At At 0 x(t ) e e x(t0 ) e A ( t t 0 ) x(t0 ) (t t0 ) x(t0 ) 40 State Transition Matrix Properties 1 1 (t ) e L [(sI A) ] At 1. ( 0) I 2. 1 (t ) (t ) 3. x(0) (t ) x(t ) 4. (t 2 t1 ) (t1 t0 ) (t 2 t0 ) 5. (t ) (kt ) k 41 Non-homogeneous solution d x(t ) Ax (t ) Bu (t ) dt y (t ) Cx(t ) Du(t ) sX ( s) x(0) AX ( s) BU ( s ) ( sI A) X ( s) x(0) BU ( s ) 1 1 X ( s) ( sI A) x(0) ( sI A) BU ( s ) x(t ) L1[( sI A) 1 ]x(0) L1[( sI A) 1 BU ( s)] t x(t ) (t ) x(0) (t ) Bu ( )d Convolution 0 Homogeneous 42 Non-homogeneous solution (Cont.) t x(t ) (t ) x(0) (t ) Bu ( )d 0 t x(t ) (t t0 ) x(t0 ) (t ) Bu ( )d t0 t y (t ) C(t t0 ) x(t0 ) C(t ) Bu ( )d Du(t ) t0 Zero-input response Zero-state response 43 Example 1 1 x1 0 x1 0 x 2 3 x 1u (t ) 2 2 let x(0) 0 0 u (t ) unit step T t 2t 2 e e (t ) L1[(sI A) 1 ] e At t 2t 2 e 2 e x(t ) (t ) x(0) Ans: e 1 e 2t t 2t e 2e t (t ) Bu( )d 0 1 1 x1 e t e 2t 2 x 2 2 e t e 2t L1[(sI A) 1 BU (s)] 44 How to find State transition matrix (t ) e At L1[(sI A) 1 ] Methode 1: (t ) L1[(sI A) 1 ] Methode 2: (t ) e At Methode 3: Cayley-Hamilton Theorem 45 Method 1: 1 1 (t ) L [(sI A) ] 1 0 x1 0 0 x1 0 x 0 4 3 x 1 0 u1 2 2 u x3 1 1 2 x3 0 1 2 x1 y1 (t ) 1 0 0 y (t ) 0 0 1 x2 x 2 3 s 2 6 s 11 s 2 3 adj ( sI A) 1 1 2 ( sI A) 3 s 2 3s sI A s ( s 4)( s 2) 3 3s 2 s4 s 1 s 4 s 46 Method 2: (t ) e At 0 x1 1 x1 1 0 x 0 2 0 x 1 u1 2 2 u x3 0 0 3 x3 1 2 x1 y1 (t ) x 6 6 1 y (t ) 2 2 x3 e t At (t ) e 0 0 0 e 2t 0 diagonal matrix 0 0 3t e 47 Diagonization linear system by Meiling CHEN 48 Diagonization linear system by Meiling CHEN 49 Case 1: i distinct 1 0 A 3 4 1 3 4 1 3 2 1 ( 3)( 1) depend 1 v1 3 (1 I A)V1 0 3 3 4 v2 1 1 v1 (2 I A)V2 0 3 3 v2 P V1 v1 1 v2 3 v1 1 v2 1 1 1 3 0 1 V2 P AP 3 1 0 1 50 In the case of A matrix is phase-variable form and 1 2 3 n P v1 v2 1 1 1 2 n vn 1 n 1 n 1 n 1 2 n 1 1 2 1 P AP t e Pe P At 3 Vandermonde matrix for phase-variable form 4 1 51 Case 2: i distinct 1 0 1 A 0 1 0 0 0 2 1 I A 0 0 1 1 0 0 0 2 ( 1)( 1)( 2) 1 2 0 0 1 v1 (1 I A)V1 0 0 0 v2 0 0 0 1 v3 v1 1 0v1 0v2 v3 0 v2 0 v3 0 depend v1 0 0v1 0v2 v3 0 v2 1 v3 0 V1 V2 52 3 2 1 0 1 v1 (3 I A)V3 0 1 0 v2 0 0 0 0 v3 v1 1 v1 0v2 v3 0 v2 0 v3 1 P V1 V2 1 0 1 1 0 0 V3 0 1 0 P 1 AP 0 1 0 0 0 1 0 0 2 53 Case 3: i distinct Jordan form 1 2 3 P v1 v2 v3 P AP Jordan 1 form Generalized eigenvectors (1 I A)v1 0 (1 I A)v2 v1 (1 I A)v3 v2 e Aˆ t e 1t 1 1 1 ˆ P AP A 1 1 1 te 1t e 1t e 1t 1t te e 1t t2 2 54 Example: 3 1 A 1 1 3 1 1 1 ( 2) 2 1 1 v11 (1 I A)V1 0 1 1 v12 v11 1 v12 1 1 1 v21 1 (1 I A)V2 1 1 v22 1 P V1 v21 1 v22 0 1 1 2 1 1 ˆ V2 P AP A 1 0 0 2 e 2t e Aˆ t te 2t At Aˆ t 1 e Pe P 2t e 55 Method 3: 56 An an 1 An 1 a1 A a0 I 0 An an 1 An 1 a1 A a0 I An 1 an 1 An a1 A2 a0 A an 1 (an 1 An 1 a1 A a0 I ) a1 A2 a0 A any f ( A) k0 I k1 A k2 A2 kn An f ( A) 0 I 1 A 2 A n 1 A 2 n 1 n 1 k 0 k A k 57 Example: 100 A ? 1 2 A 0 1 f ( A) A100 0 I 1 A let 1 2 0 2 ( 1)( 2) 0 , 1 1, 2 2 f (1 ) 100 1 f (2 ) 100 2 0 11 1 100 0 12 2 100 0 2 2100 1 2100 1 101 1 0 1 2 1 2 2 100 100 100 f ( A) A (2 2 ) (2 1) 0 1 0 1 1 0 58 Example: 3 1 A 2 0 e ? At 3 1 0 , 1 1, 2 2 2 f (1) e t 0 11 0 1 0 2e t e 2t f (2) e 2t 0 12 0 1 2 1 e 2 t e t t e 2e e At 2t 1 1 0 2t t 3 0 1 ( e e ) 2 0 2e 2t e t 2t t 2 e 2 e e 2 t e t 2t t e 2e 59 60 61 62 linear system by Meiling CHEN 63 linear system by Meiling CHEN 64 linear system by Meiling CHEN 65 Controllability and Observability 66 Introduction The main objective of using state-space equations to model systems is the design of suitable compensation schemes to control these systems. Typically, the control signal u(t) is a function of several measurable state variables. Thus, a state variable controller, that operates on the measurable information is developed. State variable controller design is typically comprised of three steps: Assume that all the state variables are measurable and use them to design a full-state feedback control law. In practice, only certain states or combination of them can be measured and provided as system outputs. An observer is constructed to estimate the states that are not directly sensed and available as outputs. Reduced-order observers take advantage of the fact that certain states are already available as outputs and they don’t need to be estimated. Appropriately connecting the observer to the full-state feedback control law yields a state-variable controller, or compensator. 67 Introduction a given transfer function G(s) can be realized using infinitely many state-space models certain properties make some realizations preferable to others one such property is controllability 68 Motivation1: Controllability x1 2 1 x1 1 x 0 1 x 0 u (t ) 2 2 x1 y 1 0 x2 uncontrollable u x2 (0) s s 1 x2 x 2 1 x1 (0) s 1 x1 s 1 x1 1 y 2 1 controllable 69 Controllability and Observability Plant: x Ax Bu , x R n y Cx Du Definition of Controllability A system is said to be (state) controllable at time t 0 , if there exists a finite t1 t0 such for any x(t0 ) and any x1 , there exist an input u[t0 ,t1 ] that will transfer the state x(t0 ) to the state x1 at time t1 , otherwise the system is said to be uncontrollable at time t 0 . 70 Controllability Matrix Consider a single-input system (u ∈ R): The Controllability Matrix is defined as A, B Controllable rank (C ) n, C ( A, B) B AB A2 B An 1 B We say that the above system is controllable if its controllability matrix 𝐶(𝐴, 𝐵) is invertible. As we will see later, if the system is controllable, then we may assign arbitrary closed-loop poles by state feedback of the form 𝑢 = −𝐾𝑥. Whether or not the system is controllable depends on its state-space realization. 71 Example: Computing 𝐶(𝐴, 𝐵) Let’s get back to our old friend: Here, Is this system controllable? 72 Controllability Matrix det(U ) 0 if u R Controllability Matrix U B AB A2 B An1B Example: An Uncontrollable System 1 0 1 x x u 0 3 0 y 1 2x ※ State x2 is uncontrollable. U (s) 1 s -1 x1 1 1 s -1 Y (s) x2 2 3 73 Proof of controllability matrix xk 1 Axk Buk xk 2 Axk 1 Buk 1 xk 2 A( Axk Buk ) Buk 1 A2 xk ABuk Buk 1 xk n An xk An 1 Buk An 2 Buk 1 ABuk ( n 2 ) Buk ( n 1) xk n An xk An 1 Buk An 2 Buk 1 ABuk ( n 2 ) Buk ( n 1) uk xk n An xk An 1 B AB B uk ( n 2 ) uk ( n 1) Initial condition 74 Motivation2: Observability x1 2 0 x1 3 x 0 1 x 1 u (t ) 2 2 x1 y 1 0 x2 u 1 x2 (0) s s 1 x 2 x2 x1 (0) s x1 1 3 s 1 x1 1 y 2 observable unobservable 75 Controllability and Observability Plant: x Ax Bu , x R n y Cx Du Definition of Observability A system is said to be (completely state) observable at time t 0 , if there exists a finite t1 t0 such that for any x(t0 ) at time t 0 , the knowledge of the input u[t t ] and the output y[t t ] over the time interval [t0 , t1 ] suffices to determine the state x0 , otherwise the system is said to be unobservable at t 0 . 0, 1 0, 1 76 Observability Matrix A, C Observable rank (V ) n if y R det(V ) 0 C CA 2 Observability M atrix V CA CAn 1 Example: An Unobservable System 0 1 0 x x u 0 2 1 y 0 4x ※ State x1 is unobservable. U (s) 1 4 s -1 x2 s -1 x1 Y (s) 2 77 Proof of observability matrix xk 1 Ax k Bu k yk Cxk Duk (1) yk 1 Cxk 1 Duk 1 yk 1 C ( Ax k Bu k ) Duk 1 CAx k CBu k Duk 1 (2) yk n 1 CAn 1 xk CAn 2 Bu k CAn 3 Bu k 1 CBu k ( n 2) Duk ( n 1) (n) C CA xk (1), (2), (n) n 1 CA yk Duk yk 1 CBu k Duk 1 CABu k ( n 3) CBu k ( n 2 ) Duk ( n 1) Inputs & outputs 78 Example Plant: x Ax Bu , x R n y Cx Du 0 1 0 A , B , C 0 1 1 0 1 0 1 Controllability Matrix V B AB 1 0 C 0 1 Obervability Matrix N CA 1 0 rank (V ) rank ( N ) 2 Hence the system is both controllable and observable. 79 Controllability and Observability Theorem I xc (t ) Ac xc (t ) Bc u(t ) Controllable canonical form Controllable A system in Controller Canonical Form (CCF) is always controllable!! Theorem II xo (t ) Ao xo (t ) B o u (t ) y (t ) Co xo (t ) Observable canonical form Observable A system in Observable Canonical Form (OCF) is always controllable!! 80 Example s2 T ( s) ( s 1)(s 2) Controllable canonical form 0 1 1 3 1 C 2 V CA 2 1 U B AB Observable canonical form 2 2 U B AB 1 1 C 0 1 V CA 1 3 1 0 0 xc xc u 2 3 1 y 2 1xc rank[U ] 2 n 0 2 2 xo xo u 1 3 1 y 0 1xo rank[U ] 1 n rank[V ] 1 n rank[V ] 2 n 81 Linear system (Analysis) Theorem III x (t ) Jx (t ) Bu (t ) Jordan form y (t ) Cx(t ) Du(t ) Jordan block J1 J J2 C C1 C2 J 3 C3 B1 B B2 B3 First column has no zero column Least row has no zero row 82 Example 2 1 0 b11 x 0 2 0 x b12 u 0 0 1 1 y c11 c12 3x If b12 0 uncontrollable If c11 0 unobservable 83 1 1 1 1 x 1 2 1 2 0 1 0 0 x 1 0 1 0 2 0 0 b11 0 0 b12 1 0 b13 0 1 1 2 u 1 0 0 1 b21 1 1 2 0 0 2 0 y 1 0 1 2 0 1 1 x 1 0 2 3 0 2 2 C11 C12 C13 C21 84 b11 C11 b12 b13L.I . b21L.I . C13L.I . C21L.I . controllable C12 observable In the previous example b11 C11 b12 b13L.I . b21L.I . C13L.I . C21L.D. controllable C12 unobservable 85 Example 2 x 2 y 1 1 1 2 2 2 2 1 3 1 1 2 1 1 1 L.I. 2 2 1 x 3 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 2 1u 0 0 0 1 0 0 L.I. L.I. L.D. 86 Kalman Canonical Decomposition x Ax Bu Diagonalization: All the Eigenvalues of A are distinct, i.e. 1 2 3 n & y Cx Du There exists a coordinate transform z Tx such that 0 1 . Am T 1 AT where Am n 0 bm1 System in z-coordinate becomes Bm T 1B z Am z Bmu bmn y Cm z Cm CT cm1 cmn Homogeneous solution of the above state equation is z(t ) v1e t z1 (0) vn e t z n (0) 1 If bmi 0 and cmi 0, mode i is controllable and observable n 87 How to construct coordinate transformation matrix for diagonalization All the Eigenvalues of A are distinct, i.e. 1 2 3 n Eigenvecto rs, v1, v2 , ,vn , are independent. The coordinate matrix for diagonalization T [v1, v2 , ,vn ] Consider diagonalized system z1 1 z1 bm1u z2 2 z2 bm 2u zn λn zn bmn u y cm1z1 cm2 z2 cmn zn 88 Transfer function is n H ( s) i 1 cmi bmi cm1bm1 c b mn mn s i s 1 s n If bmi 0 or cmi 0, mode i is uncontrollable or unobservable, H(s) has pole-zero cancellation. bm1 cm1 1 u (t ) bm 2 cm 2 2 ∑ y (t ) bmn n cmn 89 Kalman Canonical Decomposition SC O u(t ) SCO SC O SCO y (t ) SCO : Controllable, Observable Subsystem SCO : Controllable, Unobservable Subsystem SCO : Uncontrollable, Observable Subsystem SCO : Uncontrollable, Unobservable Subsystem 90 Kalman Canonical Decomposition: State Space Equation xCO ACO x 0 CO xC O 0 xC O 0 y CCO 0 0 ACO 0 0 AC O 0 0 0 CC O 0x 0 xCO BCO 0 xCO BCO u 0 xC O 0 AC O xC O 0 (5.X) 91 Example Plant: 1 x 0 0 y c11 0 2 0 c12 0 b11 0 x b12 u 3 b13 c13 x If b13 0, mode 3 is uncontrollable. If c13 0, mode 3 is unobservable. The same reasoning may be applied to mode 1 and 2. 92 Pole-zero Cancellation in Transfer Function From Sec. 5.2, state equation x Ax Bu y Cx Du may be transformed to n H ( s) i 1 Cmi bmi Cm1bm1 C b mn mn s i s 1 s n If bmi 0 , mode i is uncontrollable and vanishes in T.F.. If cmi 0, mode i is unobservable and vanishes in T.F.. Hence, the T.F. represents the controllable and observable parts of the state variable equation. 93 Example 4 0 2 x x u 1 2 1 Plant: 1 2, 2 4 y 0 1x Transfer Function Y ( s) 1 H ( s) C sI A B U ( s) 0 2 s 2 1 0 1 s 4 1 s 2s 4 1 2s 2 0 1 s 2s 4 s 2 1 s4 1 Mode"-2" vanishes in T.F.. 94 Example 5.6 0 6 2 x x u 1 1 1 Plant: 1 2, 2 -3 y 0 1x Transfer Function Y ( s) 1 1 T ( s) C sI A B U ( s) s3 Mode"2" vanishes in T.F.. 95 Minimum Realization Realization: Realize a transfer function via a state space equation. Example Realization of the T.F. T ( s ) 1 s3 Method 1: U (s) 1 Y ( s) 1 T ( s) U ( s) s3 Method 2: s -1 1 Y (s) 3 U (s) 1 Y ( s) 1 s2 T ( s) U ( s) s3 s2 s -1 3 s There is infinity number of realizations for a given T.F. . 1 Y (s) 1 -1 2 96 Minimum Realization Minimum realization: Realize a transfer function via a state space equation with elimination of its uncontrollable and unobservable parts. Example 5.8 Realization of the T.F. T ( s ) 5 s3 Y ( s) 5 T ( s) U ( s) s3 U (s) 1 s -1 5 Y (s) 3 97 98