16.333 Lecture #10 State Space Control Basic state space control approaches
16.333 Lecture # 10 State Space Control • Basic state space control approaches
Fa2004 16.3339-1 State Space Basics State space models are of the forr i(t)= Ax(t)+ Bu(t) y(t)=Cx (t)+Du with associated transfer function G(s)=C(sI-A)B+D Note: must form symbolic inverse of matrix(sI-A), which is hard Time response: Homogeneous part =A., c(O) known Take Laplace transform (s)=(sI-A)-1x(0 (t)=C-1[(s-A)-]x0 But can show(sI- A)-1=l+4+43 C-1[(s/-A)=1+At+2(4t)2 Gives c(t)=eAtx(0) where eAt is Matrix Exponential ◇ Calculate in matlab using expm. m and not exp.m o Time response: Forced Solution -Matrix case i= Ar+ Bu Where r is an n-vector and u is a m-vector. Cam show )+/e4c-)Bu(T) 0 y(t)=Ceata(0)+/CeA(-T Bu(r)dr+Du(t) Ceat a(0) is the initial response Cea(t)B is the impulse response of the system Matlab is a trademark of the mathworks Inc
� � � � � � Fall 2004 16.333 9–1 State Space Basics • State space models are of the form x˙(t) = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) with associated transfer function G(s) = C(sI − A) −1 B + D Note: must form symbolic inverse of matrix (sI − A), which is hard. • Time response: Homogeneous part x˙ = Ax, x(0) known – Take Laplace transform X(s) = (sI − A) −1 x(0) ⇒ x(t) = L−1 (sI − A) −1 x(0) I A A2 – But can show (sI − A) −1 = + s2 + s3 + . . . s 1 so L−1 (sI − A) −1 = I + At + 2!(At) 2 + . . . = eAt – Gives x(t) = eAtx(0) where eAt is Matrix Exponential 3 1 Calculate in MATLAB�R using expm.m and not exp.m • Time response: Forced Solution – Matrix case x˙ = Ax + Bu where x is an nvector and u is a mvector. Cam show t x(t) = eAtx(0) + eA(t−τ ) Bu(τ )dτ 0 t y(t) = CeAtx(0) + CeA(t−τ ) Bu(τ )dτ + Du(t) 0 – CeAtx(0) is the initial response – CeA(t) B is the impulse response of the system. 1MATLAB�R is a trademark of the Mathworks Inc
Fa2004 16.3339-2 Dynamic Interpretation Since a= Tat-. then where we have written which is a column of rows Multiply this expression out and we get that At Assume A diagonalizable, then A a(0) given, has solution (t)=er 0)=Tet( ∑e:{nxrO)h ∑ State solution is a linear combination of the system modes vier eAit -Determines the nature of the time response Ui- Determines extent to which each state contributes to that mode Bi-Determines extent to which the initial condition excites the mode
� � � Fall 2004 16.333 9–2 Dynamic Interpretation • Since A = TΛT −1, then ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ T e At | | λ1t ... − w . . 1 − e = TeΛt T −1 = ⎣ v ⎦ 1 · · · vn ⎦ ⎣ ⎦ ⎣ . | | λnt T e − wn − where we have written ⎡ ⎤ − T w . 1 − T −1 = ⎣ . . ⎦ − T wn − which is a column of rows. • Multiply this expression out and we get that n At λit T e = e viwi i=1 • Assume A diagonalizable, then x˙ = Ax, x(0) given, has solution x(t) = eAtx(0) = TeΛt T −1 x(0) n = eλit vi{wi T x(0)} i=1 n λit = e viβi i=1 • State solution is a linear combination of the system modes vieλi eλit – Determines the nature of the time response vi – Determines extent to which each state contributes to that mode βi – Determines extent to which the initial condition excites the mode
Fa2004 16.3339-3 Note that the vi give the relative sizing of the response of each part of the state vector to the response mode 1 0 de 2 Clearly eid gives the time modulation A i real- growing/decaying exponential response Ai complex -growing/decaying exponential damped sinusoidal Bottom line: The locations of the eigenvalues determine the pole locations for the system, thus They determine the stability and or performance transient be- havior of the system It is their locations that we will want to modify with the controllers
� � � � Fall 2004 16.333 9–3 • Note that the vi give the relative sizing of the response of each part of the state vector to the response. 1 v1(t) = e−t mode 1 0 0.5 v2(t) = e−3t mode 2 0.5 • Clearly eλit gives the time modulation – λi real – growing/decaying exponential response – λi complex – growing/decaying exponential damped sinusoidal • Bottom line: The locations of the eigenvalues determine the pole locations for the system, thus: – They determine the stability and/or performance & transient behavior of the system. – It is their locations that we will want to modify with the controllers
Fa2004 16.3339-4 Full-state Feedback Controller Assume that the single-input system dynamics are given by Ar+ Bu y= Ca so that D=0 The multi-actuator case is quite a bit more complicated as we would have many extra degrees of freedom Recall that the system poles are given by the eigenvalues of A Want to use the input u(t) to modify the eigenvalues of A to change the system dynamics y A.B. C Assume a full-state feedback of the form u=r-Kx where r is some reference input and the gain K is rxn If r=0, we call this controller a regulator
Fall 2004 16.333 9–4 Fullstate Feedback Controller • Assume that the singleinput system dynamics are given by x˙ = Ax + Bu y = Cx so that D = 0. – The multiactuator case is quite a bit more complicated as we would have many extra degrees of freedom. • Recall that the system poles are given by the eigenvalues of A. – Want to use the input u(t) to modify the eigenvalues of A to change the system dynamics. r u y − OO / A, B, C / x K • Assume a fullstate feedback of the form: u = r − Kx where r is some reference input and the gain K is R1×n – If r = 0, we call this controller a regulator