Root Locus Design. MEM 355 Performance Enhancement of Dynamical Systems

Transcription

1 Root Locus Design MEM 355 Performance Enhancement of Dynamical Systems Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University

2 Outline The root locus design method is an iterative, graphical procedure for selecting and tuning compensators. Basic Compensator Types The Root Locus Design process Example: BWR Pressure Control Hard Control Problems a brief introduction

3 Basic Compensators G s c ( ) Name Effect on Ultimate State Error Effect on Stability K K s + α s P (uncompensated) PI Improves Degrades K s 2 + α s 1 + α 0 s s + α K, α > β s + β s + α K, α < β s + β PID Improves Improves somewhat lag Improves somewhat Degrades somewhat lead Degrades somewhat Improves somewhat ( α ) K s + Rate feedback (PD) Degrades Improves s K s + 2ρωs+ ω ρω 2 2 s + ω2 Notch Neutralizes plant resonance

4 Procedure 1) Uncompensated system (proportional) Root locus Ultimate state error/step response 2) Compensated system Choose/modify compensator Root locus Ultimate state error/step 3) Repeat step 2)

5 Example: BWR Pressure Controller

6 BWR Pressure Control Model G R ( s) ( ) ( ) Reactor Pipe Dipole ( ) ( ) s zr s zp s + 2ξν 1 1s+ ν 1 s + 2ξν N Ns+ ν N = γ s pr s pp s + 2ρω 1 1s+ ω1 s + 2ρNωNs+ ωn Pipe Dipole Acoustics (first N harmonics)

7 BWR Transfer Functions ( ) ( + 8) ( ) ( ) 2 25 s s s s+ 144 Gp ( s) = ( s 5) s 0.1 s 4 s s s + 2 PI Gc ( s) = K s 2 s + 2 s s+ 64 PI plus Notch Gc ( s) = K s 2 s

8 Uncompensated (P) >> s=tf('s'); >> Gp=15*(25/(s+5)^2)*((s+0.5)/(s+0.1))*((s+0.25)/(s+4))*((s^2+2*.05*12*s+144)/(s^2+2*.25*8*s+64)); >> rlocus(gp) >> sgrid >> [K,Poles]=rlocfind(G) Select a point in the graphics window sgrid selected_point = i K = Poles = i i i i >> Gce=1/(1+K*Gp)

9 Uncompensated (P) 1 lim sge ( s) = s 0 s

10 Uncompensated (P), error response to step

11 PI Stability limit >> [K,Poles]=rlocfind(G) Select a point in the graphics window selected_point = i K = Poles = i i i i i i

12 PI >> [K,Poles]=rlocfind(G) Select a point in the graphics window selected_point = i K = Poles = i i i i i i

13 PI + Notch >> [K,Poles]=rlocfind(G) Select a point in the graphics window selected_point = i K = Poles = i i i i i i i i

14 PI + Notch

15 PI + Notch Gain backed off a bit to achieve ~0.7 damping ratio >> [K,Poles]=rlocfind(G) Select a point in the graphics window selected_point = i K = Poles = i i i i i i i i

16 Hard Control Problems Control design problems can be difficult for many reasons: Demanding specifications Nonlinearity Actuator/sensor constraints Coupling Implementation constraints to name a few. But here are some deceptively simple issues involving linear SISO systems that can give a designer headaches. Non-collocated Nuclear plant pressure control Large spacecraft attitude Flexible rocket attitude Right half plane zero Most tail lift aircraft (space shuttle) Steam plant level control Bicycle with front wheel steering Right half plane pole Rotorcraft High performance aircraft (F-16) Some high performance ground vehicles Most missiles Both RHF pole and zero Some aircraft (X-29) Bicycle with rear wheel steering Inverted pendulum

17 Example: Collocated vs Non-collocated The goal is to position the load. A position sensor can be placed on motor or load. 2 s + k 1 J2 Θ 1 = T J1 2 2 ( J1+ J2) k s s + JJ 1 2 k 1 Θ 2 = T JJ ( J1+ J2) k s s + JJ 1 2 J = J = 1, k = T θ1 θ2 motor load flexible shaft

18 Example Cont d G θ 1 = s s 2 ( s + 2) Im Im G 1 θ = s ( s + 2) Re Re Collocated Noncollocated

19 Lead Compensation Collocated Non-collocated

20 Lead Response (collocated case) θ2 e = θ θ 1

21 Lead + Notch Collocated Non-collocated

22 Right Half Plane Zero First, let s look at the effect of a zero on the step response: ( ) = G( s) = G( s) = G s s y 1 1 s + 1 s 1 + 2s+ 1 s + 2s+ 1 s + 2s The RHP zero causes the output to initially move in the opposite direction t

23 Example: Simple Nonminimum Phase System Uncompensated Consider the process G p ( s) 10s + 1 s s ( + 1) Try a PD compensator c = ( ) = ( + 1.1) G s K s

24 Example with PD Poles look decent, but look at the initial error.

25 Example: Check Initial Error G G e e K = 1 1 = 1+ G 1+ K s+ 1.1 ( ) ( ) ( + 1) 2 ( s + s) + K( s+ 1.1)( 10s+ 1) ( + 1) 10s + 1 s s s( s ) 2 ( ) ( ) s s = = 1 10K s K s+ 1.1K (from MATLAB) s s Ge = s s e( 0 ) = lim sge ( s) = s s Initial Value Theorem

26 Comments on Positive Feedback Root Locus Interpreting the root locus presents a problem because the system involves positive feedback (equivalently, negative K when using a negative feedback loop). This is because of the negative sign in the transfer function. Note that computing the root locus with MATLAB does not present a problem. Positive feedback (or K<0) root locus rules are slightly different from the negative feedback rules. Here are the two changes. 4. Real-axis segments: For K < 0, real axis segments to the left of an even number of finite real axis poles and/or zeros are part of the root locus.

27 Positive Feedback Root Locus, Cont d 5. Behavior at infinity: The root locus approaches infinity along asymptotes with angles: 2kπ θ =, k = 0, ± 1, ± 2, ± 3, # finite poles # finite zeros Furthermore, these asymptotes intersect the real axis at a common point given by σ = finite poles finite zeros # finite poles # finite zeros

28 F-16, X-29 G p ( s) = ( )( s ) s s ( s )( s )( s s ) Unstable Pole p ( ) = ( ) G s G s s 26 s 6 RHP pole-zero pair Minimum phase part