State_space_modeling__analysis.pdf

State Space Modelling & Analysis Dr. Mohammed Obaid Mustafa Department of Electrical Eng. College of Engineering University of Mosul 2022-2023  Lecture Outline – Basic Definitions – State Equations – State Diagram – Transfer Matrix...

State Space Modelling & Analysis Dr. Mohammed Obaid Mustafa Department of Electrical Eng. College of Engineering University of Mosul 2022-2023  Lecture Outline – Basic Definitions – State Equations – State Diagram – Transfer Matrix – Solution of State Equation – State Controllability – State Observability – Output Controllability  Definitions State of a system: We define the state of a system at time t0 as the amount of information that must be provided at time t0, which, together with the input signal u(t) for t  t0, uniquely determine the output of the system for all t  t0. State Variable: The state variables of a dynamic system are the smallest set of variables that determine the state of the dynamic system. State Vector: If n variables are needed to completely describe the behaviour of the dynamic system then n variables can be considered as n components of a vector x, such a vector is called state vector. State Space: The state space is defined as the n-dimensional space in which the components of the state vector represents its coordinate axes.  Definitions Let x1 and x2 are two states variables that define the state of the system completely. dx x2 dt Velocity State (t=t1) State (t=t1) State Vector x1 Position x Two Dimensional State space State space of a Vehicle 4  State Space Equations In state-space analysis we are concerned with three types of variables that are involved in the modeling of dynamic systems: input variables, output variables, and state variables. The dynamic system must involve elements that memorize the values of the input for t> t1. Since integrators in a continuous-time control system serve as memory devices, the outputs of such integrators can be considered as the variables that define the internal state of the dynamic system. Thus the outputs of integrators serve as state variables. The number of state variables to completely define the dynamics of the system is equal to the number of integrators involved in the system.  State Space Equations Assume that a multiple-input, multiple-output system involves 𝑛 integrators. Assume also that there are 𝑟 inputs 𝑢1 𝑡 , 𝑢2 𝑡 , ⋯ , 𝑢𝑟 𝑡 and 𝑚 outputs 𝑦1 𝑡 , 𝑦2 𝑡 , ⋯ , 𝑦𝑚 𝑡. Define 𝑛 outputs of the integrators as state variables: 𝑥1 𝑡 , 𝑥2 𝑡 , ⋯ , 𝑥𝑛 𝑡. Then the system may be described by 𝑥1 𝑡 = 𝑓1 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡) 𝑥2 𝑡 = 𝑓2 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡) 𝑥𝑛 𝑡 = 𝑓𝑛 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡)  State Space Equations The outputs 𝑦1 𝑡 , 𝑦2 𝑡 , ⋯ , 𝑦𝑚 𝑡 of the system may be given as. 𝑦1 𝑡 = 𝑔1 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡) 𝑦2 𝑡 = 𝑔2 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡) 𝑦𝑚 𝑡 = 𝑔𝑚 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡) If we define 𝑥1 𝑓1 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡) 𝑥2 𝑓 (𝑥 , 𝑥 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡) 𝒙 𝑡 = ⋮ 𝒇 𝒙, 𝒖, 𝑡 = 2 1 2 ⋮ 𝑢1 𝑥𝑛 𝑓𝑛 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡) 𝑢2 𝒖 𝑡 = ⋮ 𝑦1 𝑢𝑟 𝑔1 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡) 𝑦2 𝑔2 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡) 𝒚 𝑡 = ⋮ 𝒈 𝒙, 𝒖, 𝑡 = ⋮ 𝑦𝑚 𝑔𝑚 (𝑥1 , 𝑥2 , ⋯ , 𝑥𝑛 ; 𝑢1 , 𝑢2 , ⋯ , 𝑢𝑟 ; 𝑡)  State Space Modeling State space equations can then be written as 𝒙 𝑡 = 𝒇(𝒙, 𝒖, 𝑡) State Equation 𝒚 𝑡 = 𝒈(𝒙, 𝒖, 𝑡) Output Equation If vector functions f and/or g involve time t explicitly, then the system is called a time varying system.  State Space Modeling If above equations are linearised about the operating state, then we have the following linearised state equation and output equation: x (t )  A(t ) x(t )  B (t )u (t ) y (t )  C (t ) x(t )  D(t )u (t )  State Space Modeling If vector functions f and g do not involve time t explicitly then the system is called a time-invariant system. In this case, state and output equations can be simplified to x (t )  Ax (t )  Bu (t ) y (t )  Cx (t )  Du (t ) D B C A  Example-1 Consider the mechanical system shown in figure. We assume that the system is linear. The external force u(t) is the input to the system, and the displacement y(t) of the mass is the output. The displacement y(t) is measured from the equilibrium position in the absence of the external force. This system is a single-input, single- output system. From the diagram, the system equation is 𝑚𝑦(𝑡) + 𝑏𝑦(𝑡) + 𝑘𝑦(𝑡) = 𝑢(𝑡) This system is of second order. This means that the system involves two integrators. Let us define state variables 𝑥1 (𝑡) and 𝑥2 (𝑡) as 𝑥1 𝑡 = 𝑦(𝑡) 𝑥2 𝑡 = 𝑦(𝑡)  Example-1 𝑥1 𝑡 = 𝑦(𝑡) 𝑥2 𝑡 = 𝑦(𝑡) 𝑚𝑦(𝑡) + 𝑏𝑦(𝑡) + 𝑘𝑦(𝑡) = 𝑢(𝑡) Then we obtain 𝑥1 𝑡 = 𝑥2 (𝑡) 𝑏 𝑘 1 𝑥2 𝑡 = − 𝑦 𝑡 − 𝑦 𝑡 + 𝑢 (𝑡) 𝑚 𝑚 𝑚 Or 𝑥1 𝑡 = 𝑥2 (𝑡) 𝑏 𝑘 1 𝑥2 𝑡 = − 𝑥2 𝑡 − 𝑥1 𝑡 + 𝑢 (𝑡) 𝑚 𝑚 𝑚 The output equation is 𝑦 𝑡 = 𝑥1 𝑡  Example-1 𝑏 𝑘 1 𝑥1 𝑡 = 𝑥2 (𝑡) 𝑥2 𝑡 = − 𝑥2 𝑡 − 𝑥1 𝑡 + 𝑢 (𝑡) 𝑦 𝑡 = 𝑥1 𝑡 𝑚 𝑚 𝑚 In a vector-matrix form,  x1 (t )   0 1   x (t )   0   x (t )   k b   1    1 u (t )    x2 (t )    2   m m m  x1 (t )  y (t )  1 0   2 x (t )   Example-1 State diagram of the system is 𝑥1 𝑡 = 𝑥2 (𝑡) 𝑏 𝑘 1 𝑥2 𝑡 = − 𝑥2 𝑡 − 𝑥1 𝑡 + 𝑢 (𝑡) 𝑚 𝑚 𝑚 𝑦 𝑡 = 𝑥1 𝑡 -k/m -b/m 𝑥2 𝑢(𝑡) 1/m 1/s 1/s 𝑥1 𝑦(𝑡) 𝑥2 = 𝑥1  Example-1 State diagram in signal flow and block diagram format -k/m -b/m 1/m 𝑥2 𝑥1 𝑢(𝑡) 1/s 1/s 𝑦(𝑡) 𝑥2 = 𝑥1  Transfer Matrix (State Space to T.F) Now Let us convert a space model to a transfer function model. x (t )  Ax(t )  Bu (t ) (1) y(t )  Cx(t )  Du(t ) (2) Taking Laplace transform of equation (1) and (2) considering initial conditions to zero. sX ( s )  AX ( s )  BU ( s ) (3) Y ( s )  CX ( s )  DU ( s ) (4) From equation (3) ( sI  A) X ( s )  BU ( s ) X ( s )  ( sI  A) 1 BU ( s ) (5)  Transfer Matrix (State Space to T.F) Substituting equation (5) into equation (4) yields 1 Y ( s )  C ( sI  A) BU ( s )  DU ( s ) Y ( s )  C ( sI  A) 1  B  D U (s) Y (s) 1  C ( sI  A) B  D U ( s)  Example#3 Convert the following State Space Model to Transfer Function Model if K=3, B=1 and M=10;  x   0 1   x  0       K B      1  f (t )   v     v   M M M   x y (t )  0 1  v   Example#3 Substitute the given values and obtain A, B, C and D matrices.   x  0 1   x  0    v   3 1      1  f (t )    10   v    10  10   x y (t )  0 1  v   Example#3  0 1  0 A 3 1 B1   10   10  10  D0 C  0 1 Y (s)  C ( sI  A) 1 B  D U ( s)  Example#3  0 1  0 A 3 1 B1   10   10  10  D0 C  0 1 1   s 0  0 1  0  0 1  1  Y (s)   3 1 U (s)  0 s      10    10 10    Example#3 1   s 0  0 1  0  0 1  1  Y (s)   3 1 U (s)  0 s      10    10 10   1  s 1   0  0 1  3 1  Y (s) 1 U (s)  s   10   10 10    1   s  10 1  0   0 1 Y (s) 1   1 3 s  10  U (s) 1 3 s(s  )    10 10  10   Example#3  1   s  10 1  0   0 1 Y (s) 1   1 3 s  10  U (s) 1 3 s(s  )    10 10  10  Y (s) 1  3  1  0   10 s   U (s) s(s  1 )  3   10  10 10 Y (s) 1 s  U ( s ) s ( s  1 )  3 10 10 10  Example#3 Y (s) 1 s  U ( s ) s ( s  1 )  3 10 10 10 Y (s) s  U ( s ) s (10 s  1)  3  Home Work Obtain the transfer function T(s) from following state space representation. Answer  Solution of State Equations Consider the state equation given below x (t )  Ax(t ) (1) Taking Laplace transform of the equation (1) sX ( s )  x(0)  AX ( s ) sX ( s )  AX ( s )  x(0) sI  AX ( s)  x(0) X ( s )  sI  A x(0) 1 1 X (s)  x ( 0) sI  A  Solution of State Equations 1 X (s)  x ( 0) sI  A Taking inverse Laplace x(t )  e At x(0)  (t )  e At State Transition Matrix  Example-4 Consider RLC Circuit obtain the state transition matrix ɸ(t).  1 iL Vo vc   0   v   1  + +  i    1 C   c    C u(t ) Vc  L  R  iL      0 - - L L R  3, L  1 and C  0.5 vc  0  2 vc  2  i      i    u(t )  L  1  3  L  0   Example-4 (cont...) vc  0  2 vc  2  i      i    u(t )  L  1  3  L  0  State transition matrix can be obtained as   S 0  0  2   1  (t )   1[( SI  A)1 ]   1          0 S  1  3    Which is further simplified as  S  3 2     1 ( S  1)( S  2 ) ( S  1)( S  2 )  (t )     1 S    ( S  1)( S  2) ( S  1)( S  2)   Example-4 (cont...)  S  3 2     1 ( S  1)( S  2 ) ( S  1)( S  2 )  (t )     1 S    ( S  1)( S  2) ( S  1)( S  2)  Taking the inverse Laplace transform of each element ( 2e  e ) ( 2e  2e ) t 2 t t 2 t  ( t )   t  2 t t  2t   ( e  e ) ( e  2e )   Home Work Compute the state transition matrix if  1 0 0  A  0 2 0    0 0  3 Solution  (t )   [( SI  A) ] 1 1  State Space Trajectories The unforced response of a system released from any initial point x(to) traces a curve or trajectory in state space, with time t as an implicit function along the trajectory. Unforced system’s response depend upon initial conditions. x(t )  Ax(t ) Response due to initial conditions can be obtained as x(t )   (t )x(0)  State Transition Any point P in state space represents the state of the system at a specific time t. x2 P(x1, x2) x1 State transitions provide complete picture of the system x2 t0 t1 t6 t2 t3 x1 t5 t4  Example-5 For the RLC circuit of example-4 draw the state space trajectory with following initial conditions. vc ( 0) 1    Solution  iL ( 0 )   2  x(t )  (t )x(0) vc  (2e t  e 2t ) (2e t  2e 2t ) 1  i    t  2 t t  2t     L   ( e  e ) (  e  2e )   2  vc   3e  3e  t 2 t vc  3e t  3e 2t  i    t  2t  iL  e t  3e 2t  L    e  3e   Example-5 (cont...) Following trajectory is obtained State Space Trajectory of RLC Circuit 2 1.5 t-------->inf 1 0.5 iL 0 -0.5 -1 -1 -0.5 0 0.5 1 1.5 2 Vc  Example-5 (cont...) State Space Trajectories of RLC Circuit 2 1.5 0  1  1   0.5 1  0   1 0  iL 0     -0.5 0 -1   1   -1.5 -2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Vc  Equilibrium Point The equilibrium or stationary state of the system is when x(t )  0 State Space Trajectories of RLC Circuit 2 1.5 1 0.5 0 iL -0.5 -1 -1.5 -2 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Vc  Solution of State Equations Consider the state equation with u(t) as forcing function x (t )  Ax(t )  Bu (t ) (1) Taking Laplace transform of the equation (1) sX ( s )  x(0)  AX ( s )  BU ( s ) sX ( s )  AX ( s )  x(0)  BU ( s ) sI  AX ( s)  x(0)  BU ( s) x(0)  BU ( s ) X (s)  sI  A  Solution of State Equations x(0)  BU ( s ) X (s)  sI  A x(0) BU ( s ) X (s)   sI  A sI  A Taking the inverse Laplace transform of above equation. t x(t )   (t ) x(0)    (t   )u ( )dt 0 Natural Response Forced Response  Example#6 Obtain the time response of the following system:  x1   0 1   x1  0  x    2      u (t )  3  x2  1  2  Where u(t) is unit step function occurring at t=0. consider x(0)=0. Solution Calculate the state transition matrix  (t )   [( SI  A) ] 1 1  Example#6 Obtain the state transition equation of the system t x(t )   (t ) x(0)    (t   )u ( )dt 0 STATE DIAGRAM From Differential Equations to State Diagrams Eigenvalues & Eigenvectors  Example-2 State space representation of armature Controlled D.C Motor. Ra La B  ia ea eb T J  ea is armature voltage (i.e. input) and  is output. dia ea  Ra ia  La  eb dt T  J  B  Example-2 T  K t ia eb  K b J  B-K t ia  0 dia La  Ra ia  K b  ea dt Choosing 𝜃, 𝜃 𝑎𝑛𝑑 𝑖𝑎 as state variables 0 1 0 𝐵 𝐾𝑡 0 𝑑 𝜃 0 − 𝜃 0 𝜃 = 𝐽 𝐽 𝜃 + 1 𝑒𝑎 𝑑𝑡 𝑖 𝐾𝑏 𝑅𝑎 𝑖𝑎 𝑎 0 − − 𝐿𝑎 𝐿𝑎 𝐿𝑎 Since 𝜃 is output of the system therefore output equation is given as 𝜃 𝑦 𝑡 = 1 0 0 𝜃 𝑖𝑎  State Controllability A system is completely controllable if there exists an unconstrained control u(t) that can transfer any initial state x(to) to any other desired location x(t) in a finite time, to ≤ t ≤ T. uncontrollable controllable  State Controllability Controllability Matrix CM CM  B AB A2 B  An 1 B  System is said to be state controllable if rank(CM )  n  State Controllability (Example) Consider the system given below  1 0  1  x    x   u  0  3 0 y  1 2x State diagram of the system is U (s ) 1 s -1 x1 1 Y (s ) 1 2  3 x2 s -1  State Controllability (Example) Controllability matrix CM is obtained as CM  B AB 1  1 B  AB    0  0 Thus 1  1 CM    0 0  Since 𝑟𝑎𝑛𝑘(𝐶𝑀) ≠ 𝑛 therefore system is not completely state controllable.  State Observability A system is completely observable if and only if there exists a finite time T such that the initial state x(0) can be determined from the observation history y(t) given the control u(t), 0≤ t ≤ T. observable unobservable  State Observability Observable Matrix (OM)  C   CA    Observability Matrix OM   CA 2      CA n 1  The system is said to be completely state observable if rank(OM )  n  State Observability (Example) Consider the system given below 0 1  0  x    x   u  0  2  1  y  0 4x OM is obtained as C  OM    CA Where C  0 4 0 1  CA  0 4   0  12 0  2   State Observability (Example) Therefore OM is given as 0 4  OM     0  12 Since 𝑟𝑎𝑛𝑘(𝑂𝑀) ≠ 𝑛 therefore system is not completely state observable. 4 U (s ) 1 s -1 x2 s -1 x1 Y (s ) 2  Output Controllability Output controllability describes the ability of an external input to move the output from any initial condition to any final condition in a finite time interval. Output controllability matrix (OCM) is given as  OCM  CB CAB CA 2 B  CA n 1 B   Home Work Check the state controllability, state observability and output controllability of the following system 0 1  0  A  , B   , C  0 1 1 0 1