Feedback Control: Stabilizing Unstable Systems with Pole Placement, Exams of Signals and Systems

Download Feedback Control: Stabilizing Unstable Systems with Pole Placement and more Exams Signals and Systems in PDF only on Docsity! EECS 216 FEEDBACK CONTROL Given: Fixed LTI system with impulse response h(t) = e2tu(t) (unstable!) Goal: Somehow connect a controller to the system to make system stable. Open Use cascade controller (compensator) C(s) in series with given system Loop: H(s)=L{e2tu(t)}= 1s−2=→ C(s)= 1H(s)=s–2→ c(t) ∗ x(t)=dxdt − 2x(t). But: This is a BAD idea: Roundoff error→won’t cancel pole {2} exactly! Closed Use a feedback controller G(s) = K (constant gain) in feedback loop: Loop: x(t) → ⊕ → |h(t)| → • → y(t) equivalent to Y (s)X(s) = H(s) 1−G(s)H(s) . տ |g(t)| ւ (feedback connection) Then: HCLOSED LOOP (s)=HCL(s)= Y (s) X(s)= H(s) 1−KH(s)=[ 1 s−2 ]/[1 − K 1s−2 ] = 1s−2−K . Pole: K+2<0 if K<–2. In fact, K=–102→ hCLOSED LOOP (t)=hCL(t)=e −100tu(t)! Note: We need K<0→negative feedback to stabilize an unstable system. Given: Mechanical actuator has a very slow step response s(t)=(1–e−t)u(t). Goal: Use feedback to speed up step response so output tracks input rapidly. Have: S(s)=1s– 1 s+1= 1 s(s+1) → H(s)= 1s+1 → HCL(s)=[ 1s+1 ]/[1–K 1s+1 ]= 1s+1−K Then: Using K=–99→ HCL(s)= 1s+100 → SCL(s)= 1s(s+100)=0.01s – 0.01s+100 And: sCL(t)=.01u(t)–.01e −100tu(t). Much faster! Also need amplify ×100. Given: h(t)=cos(t)u(t)→H(s)= s s2+1 → HCL(s)=[ ss2+1 ]/[1−K ss2+1 ]= ss2−Ks+1 . Poles: K2 ± √ (K2 ) 2 − 1. Need K < 0 (negative feedback again) to stabilize. |K| < 2 : Poles: K2 ± j √ 1 − (K2 )2 both lie on the unit circle at e±j cos −1(K/2). |K| > 2 : Poles: K2 ± √ (K2 ) 2 − 1 both lie on the real axis (their midpoint is K2 ). K=–2: Double pole at –1→ HCL(s)= s(s+1)2 → hCL(t)=(1–t)e−tu(t) (try s−1s2 ) Note: H(s)=N(s)D(s) → HCL(s)= N(s)/D(s) 1−KN(s)/D(s)= N(s) D(s)−KN(s)=polynomial ratio. So: Zeros unchanged; Poles move from roots of D(s)=0 to D(s)–KN(s)=0 Problem: What choices of poles? Need:plot of roots of D(s)–KN(s)=0 as K varies Solution: This is called a root locus. Learn about it in EECS 460. Above example: EX: K decreases from 0 to −∞: Poles move along unit circle from ±j to -1; coalesce at –1; split apart; move in opposite directions along real axis. One pole becomes “fast,” other becomes “slow.” Fastest for both=–1.