1. Controller design by R.L.
Typical setup:
C(s)
Gp(s)
Root Locus Based Controller Design Goal:
Select poles and zero of C(s) so that R.L. pass through desired region
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
Draw R.L. for given plant
Draw desired region for poles from specs
Pick a point on R.L. and in desired region
Use ginput to get point and convert to complex #
Compute K using abs
and polyval
Obtain closed-loop TF
Obtain step response and compute specs
Decide if modification is needed
nump=…; denp= …; sys_p=tf(nump,denp); rlocus(sys_p);
sys_cl = feedback(sys_c*sys_p,1);
use your program from several weeks ago to do all these

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

5. From specs draw region for desired c.l. poles Select pd from region
Approximation to PD
Same usefulness as PD
Lead Control:
Draw R.L. for G
From specs draw region for desired c.l. poles
Select pd from region
Let Pick –z somewhere below pd on –Re axis Let Select
[x,y]=ginput(1); pd=x+j*y;
Gpd=evalfr(sys_p,pd)
phi=pi - angle(Gpd)
[x,y]=ginput(1); z=abs(x);
phi1=angle(pd+z); phi2=phi1-phi;
p=abs(real(pd))+abs(imag(pd)/tan(phi2));
K=abs((pd+p)/(pd+z)/Gpd);
sys_c=tf(K*[1 z],[1 p]);
Hold on;
rlocus(sys_c*sys_p);

6. Alternative Lead Control
Draw R.L. for G
From specs draw region for desired c.l. poles
Select pd from region
Let
Select
phipd=angle(pd);
phi1=(phipd+phi)/2; phi2=phi1-phi;

7. Lag Design steps Draw R.L. for G(s).
From specs, draw desired pole region
Select pd on R.L. & in region
Get
With that K, compute error constant (Kpa, Kva, Kaa) from KG(s)
From specs, compute Kpd, Kvd, Kad
sys_ol = sys_c*sys_p;
[nol,dol]=tfdata(sys_ol,'v');
dn0=dol(dol~=0); Kact=nol(end)/dn0(end);
Kdes = 1/ess;

8. If K#a > K#d , done else: pick
Re-compute
Closed-loop simulation & tuning as necessary
z=-real(pd)/…;
p=z*Kact/Kdes/(1+…);
0.05 or 0.1

9. PI Design steps First design: design PD for G(s)/s Second design:
Draw R.L. for G(s)
From specs, draw desired region
Pick pd on R.L. & in region
i. Choose ii. Choose
Simulate & tune

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

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

13. Draw R.L., it passes through the desired region.
Pick pd on R.L. & in Region
pick pd = – j160
Now choose z to meet Ka:

14. Also:

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

16. Design goal:

17. If tr = 0.0105 not satisfactory we need to reduce tr by ≈ 5%

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

19. Pick pd in region (near corner but inside region for safety margin)
Compute angle deficiency:
a. PD control, choose zpd such that then

20. b. Lead control: choose zlead, plead such that You can select zlead & compute plead. Or you can use the “bisection” method to compute z and p. Then

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

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

23. Compute desired error const. from specs:
For PI : set K*a = K*d & solve for zpi For lag : pick zlag & let

24. Re-compute K
Get closed-loop T.F. Do step response analysis.
If not satisfactory, go back to 3 and redesign.

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

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

27. Lead-lag design example
Too much overshoot, too slow & ess to ramp is too large.
R(s)
E(s)
C(s)
U(s)
Gp(s)
Y(s)

29. Draw R.L. for G(s) & the desired region

30. Clearly R.L. does not pass through desired region. need PD or lead.
Let’s do lead. Pick pd in region

31. Now choose zlead & plead.
Could use bisection.
Let’s pick zlead to cancel plant pole s + 0.5

32. Use our formula to get plead
Now compute K :
Now evaluate error constant Kva

34. Could re-compute K, but let’s skip:
do step response.

39. Control System Implementation
R(s)
E(s)
C(s)
U(s)
Gp(s)
Y(s)
disturbance
input
Reference
Command
output
error
Controller
control
Actuator
Plant + _
plant
input
Sensor
noise

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

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

42. Power electronic based control
disturbance
input
Reference
Command
Difference
amplifier
output
Op Amp
circuit
Actuator
Plant
Sensor
Analog operation, fast
Inexpensive
All algorithms in circuit hardware
No sampling and aliasing issues

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

44. Op-amp controller circuit: Proportional:− + e − + u

45. Integral: C1 R4 R1 R3 − + e − + u

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

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

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

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

50. Lead or lag controller:− + e − + u C1

51. If R1C1 > R2C2 then z < p This is lead controller
If R1C1 < R2C2 then z > p This is lag controller

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