Fall 2001 6.3110-1 Topic #10 16.31 Feedback Control State-Space Systems What are the basic properties of a state-space model, and how do we analyze these? e Time Domain Interpretations System Modes Copyright 2001 by Jonathan How
Fall 2001 16.31 10–1 Topic #10 16.31 Feedback Control State-Space Systems • What are the basic properties of a state-space model, and how do we analyze these? • Time Domain Interpretations • System Modes Copyright 2001 by Jonathan How
Fall 2001 6.3110-1 ● Forced solution Consider a scalar case . I+bu, a (0) gi ()=eta()+/ ea(-Tbu(r)dr where did this come from? Ciaa]=d(e-atar(t))=e-atbu(t de-at.(r)d (1)-x(0)=0 u(TaT forced solution- Matrix case A r+ Bu where r is an n-vector and u is a m-vector Just follow the same steps as above to get (t)=e4x(0)+ Bu()d and if y= Ca+ Du, then y(t)=Ce x(0)+Ceae-T Bu(r)dr+ Du(t) Ceatc(0)is the initial response CeA()B is the impulse response of the system
Fall 2001 16.31 10–1 • Forced Solution – Consider a scalar case: x˙ = ax + bu, x(0) given ⇒ x(t) = eatx(0) + t 0 ea(t−τ) bu(τ )dτ where did this come from? 1. ˙x − ax = bu 2. e−at [ ˙x − ax] = d dt(e−atx(t)) = e−atbu(t) 3. t 0 d dτ e−aτx(τ )dτ = e−atx(t) − x(0) = t 0 e−aτ bu(τ )dτ • Forced Solution – Matrix case: x˙ = Ax + Bu where x is an n-vector and u is a m-vector • Just follow the same steps as above to get x(t) = eAtx(0) + t 0 eA(t−τ) Bu(τ )dτ and if y = Cx + Du, then y(t) = CeAtx(0) + t 0 CeA(t−τ) Bu(τ )dτ + Du(t) – CeAtx(0) is the initial response – CeA(t) B is the impulse response of the system.
Fall 2001 16.3110-2 Have seen the key role of eat in the solution for (t) Determines the system time response But would like to get more insight Consider what happens if the matrix A is diagonalizable, i. e. there exists a T such that 入1 TAT=A which is diagonal m= The where Follows since eAt=I+ At+ 2 (At) and that a=TAT-1. so we can I+At+(4)2+ I+TAT-t+e(TAT- t)+ Ten This is a simpler way to get the matrix exponential, but how find T and A? Eigenvalues and Eigenvectors
Fall 2001 16.31 10–2 • Have seen the key role of eAt in the solution for x(t) – Determines the system time response – But would like to get more insight! • Consider what happens if the matrix A is diagonalizable, i.e. there exists a T such that T −1 AT = Λ which is diagonal Λ = λ1 ... λn Then eAt = T eΛt T −1 where eΛt = eλ1t ... eλnt • Follows since eAt = I + At + 1 2!(At)2 + ... and that A = TΛT −1, so we can show that eAt = I + At + 1 2!(At) 2 + ... = I + TΛT −1 t + 1 2!(TΛT −1 t) 2 + ... = T eΛt T −1 • This is a simpler way to get the matrix exponential, but how find T and λ? – Eigenvalues and Eigenvectors
Fall 2001 16.3110-3 Eigenvalues and eigenvectors Recall that the eigenvalues of A are the same as the roots of the characteristic equation (page A is an eigenvalue of A if det(AI-A)=0 which is true iff there exists a nonzero v(eigenvector) for which Au=入 Repeat the process to find all of the eigenvectors. Assuming that the n eigenvectors are linearly independent A AT=TA TTAT=A . One word of caution: Not all matrices are diagonalizable 01 det(sI-A)= only one eigenvalue s=0 (repeated twice). The eigenvectors solve 「01 0 00 eigenvectors are 71 of the form0/,n≠0→ would only be one Need the Jordan Normal Form to handle this case(section 3.7.3)
Fall 2001 16.31 10–3 Eigenvalues and Eigenvectors • Recall that the eigenvalues of A are the same as the roots of the characteristic equation (page 9–1) • λ is an eigenvalue of A if det(λI − A)=0 which is true iff there exists a nonzero v (eigenvector) for which (λI − A)v = 0 ⇒ Av = λv • Repeat the process to find all of the eigenvectors. Assuming that the n eigenvectors are linearly independent Avi = λivi i = 1,...,n A v1 ··· vn = v1 ··· vn λ1 ... λn AT = TΛ ⇒ T −1 AT = Λ • One word of caution: Not all matrices are diagonalizable A = 0 1 0 0 det(sI − A) = s2 only one eigenvalue s = 0 (repeated twice). The eigenvectors solve 0 1 0 0 r1 r2 = 0 eigenvectors are of the form r1 0 , r1 = 0 → would only be one. • Need the Jordan Normal Form to handle this case (section 3.7.3)
Fall 2001 6.3110-4 Mechanics ● Consider A s+1-1 det(sI-A)=(s+1)(s-5)+8=s2-4s+3=0 so the eigenvalues are s1=l and S2= 3 Eigenvectors(sI-Au=0 (s1I-A)1= 8s-5 U21 2-1 U11 02011-021=0,→21=2011 Ul is then arbitrary (0), so set v1=1 4-1 (s21-A 4 Confirm that Av;= XiI
Fall 2001 16.31 10–4 Mechanics • Consider A = −1 1 −8 5 (sI − A) = s + 1 −1 8 s − 5 det(sI − A)=(s + 1)(s − 5) + 8 = s2 − 4s +3=0 so the eigenvalues are s1 = 1 and s2 = 3 • Eigenvectors (sI − A)v = 0 (s1I − A)v1 = s + 1 −1 8 s − 5 s=1 v11 v21 = 0 2 −1 8 −4 v11 v21 =0 2v11 − v21 = 0, ⇒ v21 = 2v11 v11 is then arbitrary (= 0), so set v11 = 1 v1 = 1 2 (s2I − A)v2 = 4 −1 8 −2 v12 v22 =0 4v12 − v22 = 0, ⇒ v22 = 4v12 v2 = 1 4 • Confirm that Avi = λivi