Fall 2001 16.319-1 Topic #9 16.31 Feedback Control State-Space Systems What are the basic properties of a state-space model and how do we analyze these? ● ss to te Copyright 2001 by Jonathan How
Fall 2001 16.31 9–1 Topic #9 16.31 Feedback Control State-Space Systems • What are the basic properties of a state-space model, and how do we analyze these? • SS to TF Copyright 2001 by Jonathan How
Fall 2001 16.319-1 sS→TF In going from the state space model i(t)=A.(t)+ Bu(t y(t)= Ca(t)+ Du(t) to the transfer function G(s)=C(sI -A)-B+D need to form the inverse of the matrix(sI- A)-a symbolic inverse- not easy at all For simple cases, we can use the following a1 a2 a4 a3 a a1 a4 For larger problems, we can also use Cramer's rule Turns out that an equivalent method is to form I-4-B C D G(s)=C(SI-A)B+D= det(sI-A Reason for this will become more apparent later when we talk about how to compute the "zeros"of a state-space model(which are the roots of the numerator Example from before 1-a2-C A=100|,B=0,C=[bb2by then sta1 a2 a3 1 00 G(s) b3+b28+b1s det(sl-A 01s0 det(sl-a b b2 b3 and det(sI-A)=s+a182+a28+s3 Key point: Characteristic equation of this system given by det(sl-A
Fall 2001 16.31 9–1 SS ⇒ TF • In going from the state space model x˙(t) = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) to the transfer function G(s) = C(sI − A)−1B + D need to form the inverse of the matrix (sI − A) – a symbolic inverse – not easy at all. • For simple cases, we can use the following: a1 a2 a3 a4 −1 = 1 a1a4 − a2a3 a4 −a2 −a3 a1 For larger problems, we can also use Cramer’s Rule • Turns out that an equivalent method is to form: G(s) = C(sI − A) −1 B + D = det sI − A −B C D det(sI − A) – Reason for this will become more apparent later when we talk about how to compute the “zeros” of a state-space model (which are the roots of the numerator) • Example from before: A = −a1 −a2 −a3 100 010 , B = 1 0 0 , C = b1 b2 b3 T then G(s) = 1 det(sI − A) s + a1 a2 a3 −1 −1 s 0 0 0 −1 s 0 b1 b2 b3 0 = b3 + b2s + b1s2 det(sI − A) and det(sI − A) = s3 + a1s2 + a2s + s3 • Key point: Characteristic equation of this system given by det(sI − A)
Fall 2001 16319-2 Time Response Can develop a lot of insight into the system response and how it is modeled by computing the time response a(t) Ho omogeneous part Forced solution ● Homogeneous part O known Take Laplace transform X(s)=(s-A)-x(0) so that (t)=C-[ But can show (s-A) [(sI-A)-=I+At+o(At)2+ x(t)=ex(0) eAt is a special matrix that we will use many times in this course Transition matri Matri exponential Calculate in MATLAB( using expm. m and not expm I ote that e 证AB=B4 We will say more about eat when we have said more about A(eigenvalues and eigenvectors) Computation of eAt=C-i(sI-A)-) straightforward for a 2-state system I mATLAB is a trademark of the mathworks Inc
Fall 2001 16.31 9–2 Time Response • Can develop a lot of insight into the system response and how it is modeled by computing the time response x(t) – Homogeneous part – Forced solution • Homogeneous Part x˙ = Ax, x(0) known – Take Laplace transform X(s)=(sI − A) −1 x(0) so that x(t) = L−1 (sI − A) −1 x(0) – But can show (sI − A) −1 = I s + A s2 + A2 s3 + ... so L−1 (sI − A) −1 = I + At + 1 2!(At) 2 + ... = eAt – So x(t) = eAtx(0) • eAt is a special matrix that we will use many times in this course – Transition matrix – Matrix Exponential – Calculate in MATLABr using expm.m and not exp.m 1 – Note that e(A+B)t = eAteBt iff AB = BA • We will say more about eAt when we have said more about A (eigenvalues and eigenvectors) • Computation of eAt = L−1{(sI −A)−1} straightforward for a 2-state system 1MATLABr is a trademark of the Mathworks Inc.
Fall 2001 16.319-3 Example: i= Ar, with 01 (s-A) 2s+3 s+31 ) (s+2)(s+ 2 s+1s+2s+1s+2 +2s+1 26 2e+ 2e
Fall 2001 16.31 9–3 • Example: ˙x = Ax, with A = 0 1 −2 −3 (sI − A) −1 = s −1 2 s + 3 −1 = s +3 1 −2 s 1 (s + 2)(s + 1) = 2 s + 1 − 1 s + 2 1 s + 1 − 1 s + 2 −2 s + 1 + 2 s + 2 −1 s + 1 + 2 s + 2 eAt = 2e−t − e−2t e−t − e−2t −2e−t + 2e−2t −e−t + 2e−2t
