Relations to model followi The control law is Ru= Tuc-Sy № tice that Ro R ABm SBm Su mma ry Pole place me nt give s line ar e quations When do solutions e xist? T S ABm SBm S R Am Whe n do the y nat eXIst? ABmuc- R ()· What is the key difficulty? BAm E summation Proce ss model no distubances an(t-n bou(t-d (9) trod uang n bo Indirect ST r 1。 Estimation g(t-1)=(-y(t-1)…-y(t-n) (10) 2. An alo orithm The model be cores a re g re ssion y(t)="(t-1)9 Estimate S g iven by e(t)=6t-1)+K(t(t) c(t)=y(t)-g2(t-1)(t-1) P(t-1)p(t-1) K(t)=X+"(t-1)Pt-1)(t-(1) ()=÷(I-K(t)(t-1)Pt-1) C K.J. Astrom and BWittenmarkRelations to Model Following The control law is Ru = T uc ￾ Sy Notice that T R = AoB0m R = (AR0 + B￾ S)B0m AmR = ABm BAm + SBm RAm Hence u = T R uc ￾ S R y = ABm BAm uc + SBm RAm uc ￾ S R y = ABm BAm uc ￾ S R (y ￾ ym) (8) u y Σ Σ u c A B S R B A − 1 ym B m A m Summary  Pole placement gives linear equations  When do solutions exist?  When do they not exist?  What is the key diculty? Indirect STR 1. Estimation 2. An Algorithm 3. Examples Estimation Process model (No disturbances!) y(t) = ￾a1y(t ￾ 1) ￾ a2y(t ￾ 2) ￾ ::: ￾ any(t ￾ n) + b0u(t ￾ d0) + ::: + bmu(t ￾ d0 ￾ m) (9) Introducing  T = ( a1 a2 ::: an b0 ::: bm ) ' T (t ￾ 1) = (￾y(t ￾ 1)


￾ y(t ￾ n) u(t ￾ d0) : : :u(t ￾ d0 ￾ m)) (10) The model becomes a regression y(t) = 'T (t ￾ 1) Estimates given by ^ (t) = ^ (t ￾ 1) + K(t)"(t) "(t) = y(t) ￾ 'T (t ￾ 1)^ (t ￾ 1) K(t) = P (t ￾ 1)'(t ￾ 1)  + 'T (t ￾ 1)P (t ￾ 1)'(t ￾ 1) P (t) = 1  ￾ I ￾ K(t)'T (t ￾ 1)
P (t ￾ 1) (11) c K. J. Åström and B. Wittenmark 4