1. 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 5.Obtain closed-loop TF 6.Obtain step response and compute specs 7.Decide if modification is needed nump=…; denp= …; sysp=tf(nump, denp); rlocus(sysp); use your program from several weeks ago to do all these syscl = feedback(sysc*sysp,1); Gpd=evalfr(sysp,pd); K=1/Gpd; sysc = K; [x,y]=ginput(1); pd=x+j*y;

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

3. 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 [x,y]=ginput(1); z=abs(x); phi1=angle(pd+z); phi2=phi1-phi; [x,y]=ginput(1); pd=x+j*y; Gpd=evalfr(sysp,pd) phi=pi - angle(Gpd) p=abs(real(pd))+abs(imag(pd)/tan(phi2)); K=abs((pd+p)/(pd+z)/Gpd); sysc=tf(K*[1 z],[1 p]); Hold on; rlocus(sysc*sysp);

4. 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 phipd=angle(pd); phi1=(phipd+phi)/2; phi2=phi1-phi;

5. Lag 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 Kdes = 1/ess; sysol = sysc*sysp; [nol, dol]=tfdata(sysol,'v'); dn0=dol(dol~=0); Kact=nol(end)/dn0(end); P

6. 7.If K act > K des, done else: pick 8.Re-compute 9.Closed-loop simulation & tuning as necessary z=-real(pd)/…; p=z*Kact/Kdes/(1+…); 0.05 or 0.1 K from 8 should be ~1, so 8 is normally skipped.

7. First PI design (a special Lag 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 PI Design steps

8. Alternative PI design Since PI = PD/s, Can first multiply system by 1/s Then design using PD The overall controller is the controller you designed divided by s

9. Overall controller design 1.Draw R.L. for G(s), hold graph 2.Draw desired region for closed-loop poles based on desired specs 3.If R.L. goes through region, pick p d on R.L. and in region. Go to step 7. C(s)G p (s) R(s)E(s)Y(s)U(s)

10. 4.Pick p d in region (near corner but inside region for safety margin) 5.Compute angle deficiency: 6.a. PD control, choose z pd such that then

11. 6.b. Lead control: choose z lead, p lead such that You can select z lead & compute p lead. Or you can use the “bisection” method to compute z and p. Then If  < 60~70 deg, a single stage of PD or lead will work. If  > 80~90 deg, use a two-lead or PD-lead controller.

12. 7.Compute overall gain: 8.If there is no steady-state error requirement, go to 14. 9.With K from 7, evaluate error constant that you already have:

13. The 0, 1, 2 should match p, v, a This is for lag control. For PI:

14. 10.Compute desired error const. from specs: 11.For PI : set K * a = K * d & solve for z pi For lag : pick z lag & let

15. 12.Re-compute K (this step may be unnecessary) 13. 14.Get closed-loop T.F. Do step response analysis. 15.If not satisfactory, go back to 3 and redesign.

16. If we have both PI and PD we have PID control:

17. If we have both Lead and Lag, we have lead-lag control:

18. Control System Implementation C(s)G p (s) R(s)E(s)Y(s)U(s) ControllerActuator Reference Command error output control Plant Sensor disturbance input noise + _ plant input

19. PC-in-the-loop Control Power Amp Actuator Reference Command output Plant Sensor disturbance input A/D D/A PC I/O All control algorithms implemented in PC (could be Matlab Real-Time) Needs data acquisition system, including A/D and D/A Needs power amplifier Signal conditioner and amplifier

20.  -Controller based control Power Amp Actuator Reference Command output Plant Sensor disturbance input  -Controller I/O Very similar architecture to PC-in-the-loop control All control algorithms implemented in  -controller  -controller has its own A/D and D/A, but make sure resolution is OK Still needs power amplifier, because  -controller outputs weak signal Signal conditioner and amplifier

21. Power electronic based control Op Amp circuit Actuator Reference Command output Plant Sensor disturbance input Difference amplifier Analog operation, fastest Inexpensive All algorithms in circuit hardware Op Amp circuits for various controllers are given in book No sampling and aliasing issues

22. Difference Amplifier Example circuit: e R1 R2 R3 R4 +−+− r y

23. Op-amp controller circuit: 1.Proportional: −+−+ −+−+ e R1 R2 R3 R4 u

24. 2.Integral: −+−+ −+−+ e R1 C1 R3 R4 u

25. 3.Derivative control: −+−+ −+−+ e C1 R3 R4 u R2

26. 4.PD controller: −+−+ −+−+ e C1 R3 R4 u R2 R1

27. 5.PI controller: −+−+ −+−+ e R1 C1 R3 R4 u R2

28. 6.PID controller: −+−+ −+−+ e R1 C2 R3 R4 u R2 C1

29. 7.Lead or lag controller: −+−+ −+−+ e R1 R3 R4 u R2 C1 C2

30. If R 1 C 1 > R 2 C 2 then z < p This is lead controller If R 1 C 1 p This is lag controller

31. 8.Lead-lag controller: −+−+ −+−+ e R1 R3 R6 u R2 C1 C2 R4 R5

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

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

36. 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

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

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

40. 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

42. 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 :

43. Also:

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

45. Design goal:

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