State Variable Analysis of Linear Dynamical Systems

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1 Chapter 6 State Variable Analysis of Linear Dynamical Systems 6 Preliminaries In state variable approach, a system is represented completely by a set of differential equations that govern the evolution of the system states The state of a dynamical system is a set of physical quantities, that specifies the evolution of the system completely However, these physical quantities need not be unique What is unique is the number of states (called system order) 62 State Equations In general there are n state equations for an nth order dynamical system dx i (t) state input {}}{{}}{ f i [ x (t),x 2 (t),,x n (t), r (t),r 2 (t),,r p (t), i, 2,,n Let the variable C (t),c 2 (t),,c q (t) betheq output variables of the system C j (t) g j [x (t),,x n (t),r (t),,r p (t), j, 2,, q 2

2 Lecture Notes on Control Systems/D Ghose/22 2 The above two equations together form the set of dynamical equations representing the system In the vector form, x(t) x (t) x 2 (t) x n (t) ; r(t) f[x(t), r(t) c(t) g[x(t), r(t) f f 2 f ; g f n g g 2 g q r (t) r 2 (t) r p (t) ; c(t) C (t) C 2 (t) C q (t) If the system is linear and time-invariant, the dynamic equations can be written as, ; State Eqn: Output Eqn: Ax(t)+Br(t) c(t) Dx(t)+Er(t) where, A {a ij } n n D {d ij } q n B {b ij } n p E {e ij } q p are constant matrices Check that this system of equations is linear 63 State Transition Matrix Consider the homogenous part of the state equation (dropping the forcing function) So, the homogeneous state equation is, Ax(t) The state transition matrix is a matrix that satisfies the above linear homogeneous state equation Let Φ(t) ben n state transition matrix Then, dφ(t) AΦ(t)

3 Lecture Notes on Control Systems/D Ghose/22 22 should be satisfied Let x() denote the initial state at t (say) Then Φ(t) is also defined by the matrix equation, x(t) Φ(t)x() which is the solution of the homogeneous state equation for t How can we determine Φ(t)? Take Laplace transform on both sides, Ax(t) where, I is an n n identity matrix, and six(s) x() AX(s) X(s) (si A) x() X(s) X (s) X 2 (s) X n (s) with X i (s) L[x i (t) Further, we assume that matrix (SI-A) is nonsingular Taking inverse Laplace transform on both side, and so, x(t) L [(si A) x() Φ(t) L [(si A), (A) Another solution is x(t) e At x() Φ(t) e At I + At + 2! A2 t 2 + 3! A3 t 3 +, (B) So, (A) and (B) are the two expressions for the state transition matrix Note: State transition matrix represents the free response of the system It shows the response obtained due to the initial conditions only 2 Φ(t) depends only on matrix A

4 Lecture Notes on Control Systems/D Ghose/22 23 Properties of the State Transition Matrix Φ() I 2 Φ (t) Φ( t) 3 Φ(t 2 t )Φ(t t )Φ(t 2 t ), for any t,t,t 2 So, a state transition process can be divided into several sequential transitions 4 [Φ(t) k Φ(kt), (k is an integer) 64 State Transition Equation The state transition equation is defined as the solution of the linear non-homogeneous state equation Consider the LTI state equation, Ax(t)+Br(t) We can obtain the expression for the state transition equation using the Laplace transform method sx(s) x(o) AX(s)+BR(s) X(s) (si A) x() + (si A) BR(s) x(t) L [(si A) x(o)+l [(si A) BR(s) This is when t In general, t x(t) Φ(t)x() + Φ(t τ)br(τ)dτ, t }{{} using the convolution integral x(t) Φ(t t o )x(t o )+ t t Φ(t τ)br(τ)dτ

5 Lecture Notes on Control Systems/D Ghose/22 24 The output is then c(t) Dx(t)+Er(t) Example DΦ(t t )x(t o )+D t t Φ(t τ)br(τ)dτ + Er(t) [ [ x (t) 2 (t) 2 3 [ x (t) x 2 (t) [ r(t) Now, what is x(t) [x (t) x 2 (t) T when the input is a unit step input? ẋ Ax + Bu [ [ A ; B 2 3 Φ(t) L [(si A) [ [ si A s 2 3 [ (si A) s +3 s(s +3)+2 2 s [ s +3 s 2 +3s +2 2 s [ s +3 (s +)(s +2) 2 s L [(si A) The state transition equation is s+3 (s+)(s+2) 2 (s+)(s+2) (s+)(s+2) s (s+)(s+2) ( 2 s+ s+2 ) ( s+ s+2 ) 2( ) ( + 2 s+ s+2 s+ 2e t e 2t e t e 2t 2e t +2e 2t t [ s 2 s +3 ) s+2 e t +2e 2t x(t) Φ(t t )x(t )+ Φ(t τ)br(τ)dτ t Assuming t and knowing that r(τ), [ [ x (t) 2e x(t) t e 2t e t e 2t [ x () x 2 (t) 2e t +2e 2t e t +2e 2t x 2 ()

6 Lecture Notes on Control Systems/D Ghose/ [ t 2e (t τ) e 2(t τ) e (t τ) e 2(t τ) [ 2e (t τ) +2e 2(t τ) e (t τ) +2e 2(t τ) dτ [ [ x () t e + (t τ) e 2(t τ) x 2 () e (t τ) +2e 2(t τ) dτ [ [ x () e + (t τ) t 2 e 2(t τ) t x 2 () e (t τ) t + e 2(t τ) t [ [ x () e t e 2t x 2 () +e t + e 2t [ x () x 2 () + [ 2 e t + 2 e 2t e t e 2t, t Hence, x (t) ( 2e t e 2t) x () + ( e t e 2t) x 2 () + 2 e t 2 e 2t x 2 (t) ( 2e t +2e 2t) x () ( e t +2e 2t) x 2 () + e t e 2t How do we express the output of a n-th order system as a state equation/ output equation? Consider a linear time-invariant system described by the n-th order differential equation: d n c(t) n + a n d n c(t) n + + a 2 dc(t) where, c(t) is the output and r(t) is the input Let + a c(t) r(t) So, x (t) c(t) x 2 (t) dc(t) x 3 (t) d2 c(t) 2 x n dn c(t) x dx (t) dx 2(t) dx n (t) dx n (t) dn c(t) a x (t) a 2 x 2 (t) a n x n + r(t)

7 Lecture Notes on Control Systems/D Ghose/22 26 The output equation is c(t) x (t) These can be written in the matrix form as, The output equation is, Ax(t)+Br(t) A a a 2 a 3 a n n n c(t) Dx(t) D [ n ; B Note: The above form of A and B matrices is called the phase variable canonical form Example: Phase variable canonical form Consider a system given by a differential equation Rewrite as, d 3 c(t) 3 +5 d2 c(t) 2 + dc(t) +2c(t) r(t) Let, d 3 c(t) 3 5 d2 c(t) 2 dc(t) 2c(t)+r(t) be the state variables Then we have, (t) x 2 (t) 2 (t) x 3 (t) x (t) c(t) x 2 (t) dc(t) x 3 (t) d2 c(t) 2 3 (t) 5x 3 (t) x 2 (t) 2x (t)+r(t)

8 Lecture Notes on Control Systems/D Ghose/22 27 which can be written in the matrix form as (t) 2 (t) 3 (t) 2 5 The output equation being c(t) x (t) [ x (t) x 2 (t) x 3 (t) x (t) x 2 (t) x 3 (t) + r(t) The phase variable canonical form has certain advantages with regards to controllability and pole placement design through state feedback There is no guarantee that a given system in the state space form will be the phasevariable canonical form Question: Is it possible to transform a system to the phase variable canonical form? Answer: If S [ B AB A 2 B A n B is non-singular then these exists a non singular matric Q so that y(t) Qx(t) so that x(t) Q y(t) We can transform the original equation as follows: ẋ(t) Ax(t)+Br(t) Q ẏ(t) AQ y(t)+br(t) ẏ(t) QAQ y(t)+qbr(t) A y(t)+b r(t) where, A [ Phase variable canonical form B and A QAQ B QB

9 Lecture Notes on Control Systems/D Ghose/22 28 The transforming matrix Q is given by Q Q Q A Q A n where Q [ [ B AB A 2 B A n B Example: Transformation to phase variable canonical form Let ẋ(t) Ax(t)+Br(t) [ A ; B [ First check the matrix S [ B AB [ which has determinate S andsos is a non-singular matrix the original system can be transformed to the phase variable canonical form We obtain Q as Q [ [ S [ [ [ [ Q Q Q A [ [ A QAQ [ [ [ [ [ [ B QB [ [

10 Lecture Notes on Control Systems/D Ghose/22 29 Another Example Note: This example shows that when the RHS of the differential equation contains derivatives of r(t) (the input) it is not easy to define the state variables just by inspection Consider the system given by the differential equation, d 3 c(t) +5 d2 c(t) + dc(t) +2c(t) dr(t) +2r(t) In order to define the state variables we rewrite the equation as, d 3 c(t) dr(t) 5 d2 c(t) dc(t) 2c(t)+2r(t) 3 2 x (t) c(t) x 2 (t) dc(t) x 3 (t) d2 c(t) r(t) 2 The state equation are (t) x 2 (t) 2 (t) x 3 (t)+r(t) 3 (t) 5[x 3 (t)+r(t) x 2 (t) 2x (t)+2r(t) 5x 3 (t) x 2 (t) 2x (t) 3r(t) (t) (t) ẋ(t) 2 (t) 2 (t) + r(t) 3 (t) (t) 3 Note that his is not in the phase variable canonical form and needs to be transformed S [ B AB A 2 B AB A 2 B S S ( ) ( ) + ( 3)(4 9)

11 Lecture Notes on Control Systems/D Ghose/22 22 Hence it is possible to convert it to the phase variable canonical form ( work it out yourself!)

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