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Feedback Control Systems (FCS) Lecture-26-27-28-29 State Space Canonical forms Dr. Imtiaz Hussain email: [email protected] URL :http://imtiazhussainkalwar.weebly.com/ Lecture Outline – Canonical forms of State Space Models • Phase Variable Canonical Form • Controllable Canonical form • Observable Canonical form – Similarity Transformations • Transformation of coordinates – Transformation to CCF – Transformation OCF Canonical Forms • Canonical forms are the standard forms of state space models. • Each of these canonical form has specific advantages which makes it convenient for use in particular design technique. • There are four canonical forms of state space models – – – – – Phase variable canonical form Controllable Canonical form Companion forms Observable Canonical form Diagonal Canonical form Modal forms Jordan Canonical Form • It is interesting to note that the dynamics properties of system remain unchanged whichever the type of representation is used. Phase Variable Canonical form • The method of phase variables possess mathematical advantage over other representations. • This type of representation can be obtained directly from differential equations. • Decomposition of transfer function also yields Phase variable form. Phase Variable Canonical form • Consider an nth order linear plant model described by the differential equation −1 + 1 −1 + ⋯ + −1 + = () • Where y(t) is the plant output and u(t) is the plant input. • A state model for this system is not unique but depends on the choice of a set of state variables. • A useful set of state variables, referred to as phase variables, is defined as: 1 = , 2 = , 3 = , −1 ⋯ , = −1 Phase Variable Canonical form 1 = , 2 = , 3 = , −1 ⋯ , = −1 • Taking derivatives of the first n-1 state variables, we have 1 = 2 , 2 = 3 , 3 = 4 ⋯ , −1 = = − 1 − −1 2 − ⋯ − 1 + () x1 0 x 2 0 x n 1 0 x a n n 1 0 0 1 0 0 a n 1 a n3 0 x1 0 0 x2 0 u 1 x n 1 0 a 1 x n 1 Phase Variable Canonical form 1 = , 2 = , 3 = , −1 ⋯ , = −1 • Output equation is simply y 1 0 0 x1 x2 0 x n 1 x n Phase Variable Canonical form y u (t ) y ( n 1 ) (n) ∫ ＋ xn ∫ … ∫ y x2 y x1 ∫ ＋ y a1 (n2) x n 1 a2 an 8 Phase Variable Canonical form u 1 x n 1 x 1 n 1 x s s n2 1 x1 x 2 s 1 1 1 s y x1 2 3 n 1 n 9 Phase Variable Canonical form (Example-1) • Obtain the state equation in phase variable form for the following differential equation, where u(t) is input and y(t) is output. 3 2 2 3 +4 2 +6 + 8 = 10() • The differential equation is third order, thus there are three state variables: 1 = 2 = 3 = • And their derivatives are (i.e state equations) 1 = 2 2 = 3 3 = −41 − 32 − 23 + 5() Phase Variable Canonical form (Example-1) 1 = 1 = 2 2 = 3 = 2 = 3 3 = −41 − 32 − 23 + 5() • In vector matrix form x1 0 x 0 2 x 3 4 y ( t ) 1 0 1 0 3 0 x1 0 1 x 0 u (t ) 2 2 x 3 5 x1 0 x2 x 3 Home Work: Draw Sate diagram Phase Variable Canonical form (Example-2) • Consider the transfer function of a third-order system where the numerator degree is lower than that of the denominator. () 2 + 1 + 2 = 3 () + 1 2 + 2 + 3 • Transfer function can be decomposed into cascade form () 1 3 + 1 2 + 2 + 3 () 2 + 1 + 2 () • Denoting the output of the first block as W(s), we have the following input/output relationships: () 1 = 3 () + 1 2 + 2 + 3 () = 2 + 1 + 2 () Phase Variable Canonical form (Example-2) () 1 = 3 () + 1 2 + 2 + 3 () = 2 + 1 + 2 () • Re-arranging above equation yields 3 = −1 2 − 2 − 3 ()+ () () = 2 () + 1 () + 2 () • Taking inverse Laplace transform of above equations. = −1 − 2 − 3 ()+ u() () = + 1 + 2 () • Choosing the state variables in phase variable form 1 = 2 = 3 = Phase Variable Canonical form (Example-1) • State Equations are given as 1 = 2 2 = 3 3 = −3 1 − 2 2 − 1 3 + () • And the output equation is () = 2 1 + 1 2 + 3 1 2 1 2 3 Phase Variable Canonical form (Example-1) • State Equations are given as 1 = 2 2 = 3 3 = −3 1 − 2 2 − 1 3 + () • And the output equation is () = 2 1 + 1 2 + 3 1 2 −1 −2 −3 Phase Variable Canonical form (Example-1) • State Equations are given as 1 = 2 2 = 3 3 = −3 1 − 2 2 − 1 3 + () • And the output equation is () = 2 1 + 1 2 + 3 • In vector matrix form x1 0 x 0 2 x 3 a 3 y ( t ) b 2 b1 1 0 a2 0 x1 0 1 x 0 u (t ) 2 a 1 x 3 1 x1 bo x 2 x 3 Companion Forms • Consider a system defined by n n 1 n n 1 y a 1 y a n 1 y a n y b o u b1 u b n 1u b n u • where u is the input and y is the output. • This equation can also be written as () + 1 −1 + ⋯ + −1 + = () + 1 −1 + ⋯ + −1 + • We will present state-space representations of the system defined by above equations in controllable canonical form and observable canonical form. Controllable Canonical Form () + 1 −1 + ⋯ + −1 + = () + 1 −1 + ⋯ + −1 + • The following state-space controllable canonical form: x1 0 x 2 0 x n 1 0 x a n n representation 1 0 0 1 0 0 a n 1 an2 is called 0 x1 0 0 x2 0 u 1 x n 1 0 a 1 x n 1 a Controllable Canonical Form () + 1 −1 + ⋯ + −1 + = () + 1 −1 + ⋯ + −1 + y b n a n b o b n 1 a n 1b o b2 a 2 bo x1 x2 b1 a 1b o b o u x n 1 x n Controllable Canonical Form d d dt dt … ＋ d dt u (t ) ＋ v (n) ∫ ∫ … ∫ ＋ 1 2 n v x2 ∫ v x1 1 n 1 n Controllable Canonical Form (Example) () +3 = 2 () + 3 + 2 • Let us Rewrite the given transfer function in following form () 0 2 + + 3 = 2 () + 3 + 2 a2 2 a1 3 b2 3 b1 1 x1 0 x 2 a 2 1 x1 0 u a1 x 2 1 x1 0 x 2 2 1 x1 0 u 3 x2 1 bo 0 Controllable Canonical Form (Example) () 0 2 + + 3 = 2 () + 3 + 2 a2 2 a1 3 b2 3 y b 2 a 2 b o y 3 b1 1 bo 0 x1 b1 a 1b o b o u x2 x1 1 x2 Controllable Canonical Form (Example) () +3 = 2 () + 3 + 2 • By direct decomposition of transfer function Y (s) s3 s 3s 2 2 U (s) Y (s) U (s) s 1 s s 2 2 P (s) P (s) P (s) 3s P (s) 3s 1 2 P (s) P (s) 2s 2 P (s) • Equating Y(s) with numerator on the right hand side and U(s) with denominator on right hand side. Y (s) s 1 P (s) 3s U (s) P (s) 3s 1 2 P ( s )......... .......( 1) P (s) 2s 2 P ( s )......... .......( 2 ) Controllable Canonical Form (Example) • Rearranging equation-2 yields P (s) U (s) 3s • 1 P (s) 2s 2 P ( s )......... .......( 3 ) Draw a simulation diagram using equations (1) and (3) P (s) U (s) 3s 1 P (s) 2s 2 Y (s) s P (s) 1 P (s) 3s -2 -3 x 2 U(s) P(s) 1/s x 2 x1 1/s 1 3 x1 Y(s) 2 P (s) Controllable Canonical Form (Example) -2 -3 x 2 U(s) P(s) 1/s x 2 x1 1/s 3 x1 Y(s) 1 • State equations and output equation are obtained from simulation diagram. x1 x 2 x 2 U ( s ) 3 x 2 2 x1 Y ( s ) 3 x1 x 2 Controllable Canonical Form (Example) x1 x 2 • x 2 U ( s ) 3 x 2 2 x1 Y ( s ) 3 x1 x 2 In vector Matrix form x1 0 x 2 2 y 3 1 x1 0 f (t ) 3 x2 1 x1 1 x 2 Observable Canonical Form () + 1 −1 + ⋯ + −1 + = () + 1 −1 + ⋯ + −1 + • The following state-space representation is called an observable canonical form: x1 0 x 2 1 x n 1 0 x 0 n 0 0 0 0 0 0 0 1 a n x1 b n a n b o x2 b n 1 a n 1b o a n 1 u a 2 x n 1 b 2 a 2 b o a 1 x n b1 a 1b o Observable Canonical Form () + 1 −1 + ⋯ + −1 + = () + 1 −1 + ⋯ + −1 + y 0 0 0 x1 x2 1 b o u x n 1 x n Observable Canonical Form (Example) () +3 = 2 () + 3 + 2 • Let us Rewrite the given transfer function in following form () 0 2 + + 3 = 2 () + 3 + 2 a2 2 a1 3 x1 0 x 2 1 b2 3 b1 1 bo 0 a 2 x1 b 2 a 2 b o u a 1 x 2 b1 a 1b o x1 0 x 2 1 2 x1 3 u 3 x 2 1 Observable Canonical Form (Example) () 0 2 + + 3 = 2 () + 3 + 2 a2 2 a1 3 b2 3 y 0 y 0 b1 1 x1 1 b o u x2 x1 1 x2 bo 0 Similarity Transformations • It is desirable to have a means of transforming one state-space representation into another. • This is achieved using so-called similarity transformations. • Consider state space model x ( t ) Ax ( t ) Bu ( t ) y ( t ) Cx ( t ) Du ( t ) • Along with this, consider another state space model of the same plant x ( t ) A x ( t ) B u ( t ) y (t ) C x (t ) D u (t ) • Here the state vector , say, represents the physical state relative to some other reference, or even a mathematical coordinate vector. Similarity Transformations • When one set of coordinates are transformed into another set of coordinates of the same dimension using an algebraic coordinate transformation, such transformation is known as similarity transformation. • In mathematical form the change of variables is written as, x (t ) T x (t ) • Where T is a nonsingular nxn transformation matrix. • The transformed state () is written as x (t ) T 1 x (t ) Similarity Transformations • The transformed state () is written as x (t ) T 1 x (t ) • Taking time derivative of above equation x ( t ) T x ( t ) T x ( t ) T x ( t ) T A T 1 1 1 1 AT 1 x (t) Ax ( t ) Bu ( t ) x ( t ) Ax ( t ) Bu ( t ) AT x ( t ) Bu ( t ) x (t ) T x (t ) AT x ( t ) T 1 Bu ( t ) B T 1 B x ( t ) A x ( t ) B u ( t ) Similarity Transformations • Consider transformed output equation y (t ) C x (t ) D u (t ) • Substituting = −1 () in above equation y (t ) C T 1 x (t ) D u (t ) • Since output of the system remain unchanged [i.e. = ()] therefore above equation is compared with = + () that yields C CT D D Similarity Transformations • Following relations are used to preform transformation of coordinates algebraically A T 1 C CT AT B T 1 D D B Similarity Transformations • Invariance of Eigen Values sI A sI T sT T 1 1 1 AT T T 1 AT sI A T sI A sI A sI A 1 T T I Transformation to CCF • Transformation to CCf is done by means of transformation matrix P. P CM W • Where CM is controllability Matrix and is given as = −1 ⋯ and W is coefficient matrix a n 1 a n2 W a1 1 an2 a1 a n3 1 1 0 0 0 1 0 0 0 Where the ai’s are coefficients of the characteristic polynomial − = + 1 −1 + 2 −2 + ⋯ + −1 s+ Transformation to CCF • Once the transformation matrix P is computed following relations are used to calculate transformed matrices. A P 1 AP 1 B P B C CP D D Transformation to CCF (Example) • Consider the state space system given below. 1 1 2 1 1 1 2 = 0 1 3 2 + 0 () 3 1 1 1 3 1 • Transform the given system in CCF. Transformation to CCF (Example) 1 1 2 1 1 1 2 = 0 1 3 2 + 0 () 3 1 1 1 3 1 • The characteristic equation of the system is − 1 −2 −1 − = 0 − 1 −3 = 3 − 3 2 − − 3 −1 −1 − 1 1 = −3, a2 W a1 1 2 = −1, a1 1 0 1 1 0 3 0 1 3 = −1 3 1 0 1 0 0 Transformation to CCF (Example) 1 1 2 1 1 1 2 = 0 1 3 2 + 0 () 3 1 1 1 3 1 • Now the controllability matrix CM is calculated as = 1 = 0 1 2 2 10 3 9 2 7 • Transformation matrix P is now obtained as 1 = × = 0 1 2 10 −1 3 9 −3 1 2 7 3 −1 = 0 3 0 −1 1 0 1 −3 1 1 0 0 0 Transformation to CCF (Example) • Using the following relationships given representation is transformed into CCf as A P 1 state space 1 B P B AP 0 1 A P AP 0 3 1 0 1 0 1 3 0 1 B P B 0 1 − = 3 − 3 2 − − 3 Transformation to OCF • Transformation to CCf is done by means of transformation matrix Q. 1 Q (W OM ) • Where OM is observability Matrix and is given as = −1 ⋯ and W is coefficient matrix a n 1 a n2 W a1 1 an2 a1 a n3 1 1 0 0 0 1 0 0 0 Where the ai’s are coefficients of the characteristic polynomial − = + 1 −1 + 2 −2 + ⋯ + −1 s+ Transformation to OCF • Once the transformation matrix Q is computed following relations are used to calculate transformed matrices. A Q 1 AQ B Q 1 B C CQ D D Transformation to OCF (Example) • Consider the state space system given below. 1 1 2 1 1 1 2 = 0 1 3 2 + 0 () 3 1 1 1 3 1 1 () = 1 1 0 2 3 • Transform the given system in OCF. Transformation to OCF (Example) 1 1 2 1 1 1 2 = 0 1 3 2 + 0 () 3 1 1 1 3 1 • The characteristic equation of the system is − 1 −2 −1 − = 0 − 1 −3 = 3 − 3 2 − − 3 −1 −1 − 1 1 = −3, a2 W a1 1 2 = −1, a1 1 0 1 1 0 3 0 1 3 = −1 3 1 0 1 0 0 Transformation to OCF (Example) 1 () = 1 1 0 2 3 • Now the observability matrix OM is calculated as 1 1 2 1 1 1 2 = 0 1 3 2 + 0 () 3 1 1 1 3 1 = 1 1 = 1 3 5 6 2 0 4 10 • Transformation matrix Q is now obtained as = × −1 0.333 = −0.333 0.166 −0.166 0.166 0.166 0.333 0.666 0.166 Transformation to CCF (Example) • Using the following relationships given representation is transformed into CCf as A Q 1 AQ 0 1 A Q AQ 1 0 3 1 B Q B 2 1 B Q 0 0 1 1 B state C CQ D D 3 1 3 C CQ 0 space 0 1 Home Work • Obtain state space representation of following transfer function in Phase variable canonical form, OCF and CCF by – Direct Decomposition of Transfer Function – Similarity Transformation – Direct Approach () 2 + 2 + 3 = 3 () + 5 2 + 3 + 2 To download this lecture visit http://imtiazhussainkalwar.weebly.com/ END OF LECTURES-26-27-28-29