Topic #15 16.31 Feedback Control State-Space Systems . Full-state Feedback Control How do we change the poles of the state-space system? Or, even if we can change the pole locations Where do we put the poles? Linear Quadratic Regulator Symmetric Root Locus How well does this approach work? Copyright [2001 by JOnathan dHow. O
Topic #15 16.31 Feedback Control State-Space Systems • Full-state Feedback Control • How do we change the poles of the state-space system? • Or, even if we can change the pole locations. • Where do we put the poles? — Linear Quadratic Regulator — Symmetric Root Locus • How well does this approach work? Copyright 2001 by Jonathan How. 1
Fall 2001 16.3115-1 Pole placement e So far we have looked at how to pick k to get the dynamics to have some nice properties(i. e stabilize A 入(A)~A1(A-BK) Classic Question: where should we put these closed-loop poles? Of course we can use the time-domain specifications to locate the dominant poles- roots of +2cnS2=0 Then place rest of the poles so they are "much faster"than the dominant behavior. For example Could keep the same damped frequency wd and then move the real part to be 2-3 times faster than real part of dominant pole Just be careful moving the poles too far to the left because it takes a lot of control effort
Fall 2001 16.31 15—1 Pole Placement • So far we have looked at how to pick K to get the dynamics to have some nice properties (i.e. stabilize A) λi(A) ; λi(A − BK) • Classic Question: where should we put these closed-loop poles? • Of course we can use the time-domain specifications to locate the dominant poles — roots of: s2 + 2ζωns + ω2 n = 0 • Then place rest of the poles so they are “much faster” than the dominant behavior. For example: — Could keep the same damped frequency wd and then move the real part to be 2—3 times faster than real part of dominant poles ζωn • Just be careful moving the poles too far to the left because it takes a lot of control effort
Fall 2001 16.3115-2 Could also choose the closed-loop poles to mimic a system that has similar performance to what you would like to achieve Just set pole locations equal to those of the prototype system Various options exist Bessel Polynomial Systems of order k -Gp(s)=Bs Step response of Gp for various k k=1 0.6 04 0.2 1.0 3.0 Time(seconds) Roots of normalized Bessel polynomials corresponding to a settling time of one second of Bk (s) 4.6200 24.0530±/23400 3-50093,39668±37845 4-4.0156±50723.55281±j6553 546448041104±63142.59268/30813 642169±尸530062613±44018,7.1205±14540 -80271,4.361±A8.7519,6.5714±.6786,-7.6824±28081 844554±9715,6855+1692788.1682±A41057,8.7693±3616 99658545696±/11871145815.85962±/5365594013±2665 104683512403373609±377,898:6057,49657±9342.104278±13071 All scaled to give settling times of 1 second, which ou can change to ts by dividing the poles by ts
Fall 2001 16.31 15—2 • Could also choose the closed-loop poles to mimic a system that has similar performance to what you would like to achieve: — Just set pole locations equal to those of the prototype system. — Various options exist • Bessel Polynomial Systems of order k → Gp(s) = 1 Bk(s) • All scaled to give settling times of 1 second, which you can change to ts by dividing the poles by ts
Fall 2001 6.3115-3 e Procedure for an nn order system Determine the desired settling time t Find the k=n polynomial from the table Divide pole locations by ts Form desired characteristic polynomial a(s)and use acker/place to determine the feedback gains Simulate to check performance and control effort Example G(s s(s+4)(s+1) with 5-40 100B 100 so that nm=k=3 Want t=2 sec. So there are 3 50093/2=-25047and (-3.96837845)/2=-1.9834±1.8922 Use these to form a(s and find the gains using acker e The Bessel approach is fine, but the step response is a bit slow
Fall 2001 16.31 15—3 • Procedure for an nth order system: — Determine the desired settling time ts — Find the k = n polynomial from the table. — Divide pole locations by ts — Form desired characteristic polynomial Φd(s) and use acker/place to determine the feedback gains. — Simulate to check performance and control effort. • Example: G(s) = 1 s(s + 4)(s + 1) with A = −5 −4 0 1 0 0 0 1 0 B = 1 0 0 so that n = k = 3. — Want ts = 2 sec. So there are 3 poles at: −5.0093/2 = −2.5047 and (−3.9668 ± 3.7845i)/2 = −1.9834 ± 1.8922i — Use these to form Φd(s) and find the gains using acker • The Bessel approach is fine, but the step response is a bit slow
Fall 2001 163115-4 Another approach is to select the poles to match the nn polyno- mial that was designed to minimize the itae "integral of the time multiplied by the absolute value of the error JItae t e(tl dt in response to a step function Both bessel and itae are tabulated in FPE-508 Comparison for k=3(Given for wo 1 rad/sec, so slightly different than numbers given on previous page d=(s+0.920)(s+0.7465±07112) o4B=(s+0.7081)(s+0.5210±1.068 So the itae poles are not as heavily damped Some overshoot Faster rise-times Problem with both of these approaches is that they completely ig. nore the control effort required the designer must iterate
Fall 2001 16.31 15—4 • Another approach is to select the poles to match the nth polynomial that was designed to minimize the ITAE “integral of the time multiplied by the absolute value of the error” JIT AE = Z ∞ 0 t |e(t)| dt in response to a step function. • Both Bessel and ITAE are tabulated in FPE-508. — Comparison for k = 3 (Given for ω0 = 1 rad/sec, so slightly different than numbers given on previous page) φB d = (s + 0.9420)(s + 0.7465 ± 0.7112i) φIT AE d = (s + 0.7081)(s + 0.5210 ± 1.068i) • So the ITAE poles are not as heavily damped. — Some overshoot — Faster rise-times. • Problem with both of these approaches is that they completely ignore the control effort required — the designer must iterate