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-AFall 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)