1. Controller design by R.L. Typical setup: C(s)G p (s) Root Locus Based Controller Design Goal: 1.Select poles and zero of C(s) so that R.L. pass through desired region 2.Select K corresponding to a good choice of dominant pole pair

2. Types of classical controllers Proportional control –Needed to make a specific point on RL to be closed-loop system dominant pole Proportional plus derivative control (PD control) –Needed to “bend” R.L. into the desired region Lead control –Similar to PD, but without the high frequency noise problem; max angle contribution limited to < 75 deg Proportional plus Integral Control (PI control) –Needed to “eliminate” a non-zero steady state tracking error Lag control –Needed to reduce a non-zero steady state error, no type increase PID control –When both PD and PI are needed, PID = PD * PI Lead-Lag control –When both lead and lag are needed, lead-lag = lead * lag

3. Proportional control design 1.Draw R.L. for given plant 2.Draw desired region for poles from specs 3.Pick a point on R.L. and in desired region Use ginput to get point and convert to complex # 4.Compute K using abs and polyval 5.Obtain closed-loop TF 6.Obtain step response and compute specs 7.Decide if modification is needed

4. Design steps: 1.From specs, draw desired region for pole. Pick from region, not on RL 2.Compute 3.Select 4.Select: Gpd=evalfr(sys_p,pd) phi=pi - angle(Gpd) z=abs(real(pd))+abs(imag(pd)/tan(pi-phi)) Kd=1/abs(pd+z)/abs(Gpd) PD controller design

5. Drawbacks of PD Not proper : deg of num > deg of den High frequency gain → ∞: High gain for noise Saturates circuits Cannot be implemented physically

6. Lead Control: 1.Draw R.L. for G 2.From specs draw region for desired c.l. poles 3.Select p d from region 4.Let Pick –z somewhere below p d on –Re axis Let Select Approximation to PD Same usefulness as PD

7. Alternative Lead Control 1.Draw R.L. for G 2.From specs draw region for desired c.l. poles 3.Select p d from region 4.Let Select

8. Lag control design It has “destabilizing” effect (lag) Not used for improving M p, t r, t p, t s, … Use it to reduce a non-zero e ss Use it when R.L. of G(s) go through the desired region but e ss is too large.

9. Design steps 1.Draw R.L. for G(s). 2.From specs, draw desired pole region 3.Select p d on R.L. & in region 4.Get 5.With that K, compute error constant (K pa, K va, K aa ) from KG(s) 6.From specs, compute K pd, K vd, K ad

10. 7.If K #a > K #d, done else: pick 8.Re-compute 9.Closed-loop simulation & tuning as necessary

11. Example: Want: Solution: C(s)G p (s)

12. Draw region

13. Draw R.L. Pick p d on R.L. & in Region pick p d = – 0.35 + j0.5 Since there is one in G(s)

17. A better tuning may be to go back and re-pick p d

18. Lag control can improve e ss, but cannot eliminate e ss Use PI control to eliminate e ss PI :

19. Only advantage of PI: remove e ss It has destabilizing effect May ↑ M P, ↑ t s, etc. “Sluggish” settling, just like Lag Needs trial and error tuning of K p and K I

20. This way, just use your PD design program.

21. Second design: 1.Draw R.L. for G(s) 2.From specs, draw desired region 3.Pick p d on R.L. & in region 4.i. Choose ii. Choose 5. 6.Simulate & tune

22. Example: Want: Solution: Draw R.L. C(s)G p (s)

24. Clearly, R.L. pass through desired region. Pick (right on boundary) Choose

25. Step response: e ss = 0 No M P (no overshoot) fast rise to 0.85, then very sluggish to 1 Tune 1: K P ↑ to 2.5 K P

27. None unique solution Design is a creative process guided by science

28. Example: Want: C(s)G (s)

29. Sol:G(s) is type 1 Since we want finite e ss to unit acc, we need the compensated system to be type 2 C(s) needs to have in it

31. Draw R.L., it passes through the desired region. Pick p d on R.L. & in Region pick p d = – 180 + j160 Now choose z to meet K a :

32. Also:

33. Pick z = 0.03 Do step resp. of closed-loop: Is it good enough?

34. Design goal:

35. If t r = 0.0105 not satisfactory we need to reduce t r by ≈ 5%

36. Label each Root Locus with one TF choice. F E J D A BC GH I