1. Examples on Compensator Design Spring 2011
EE 4314
Examples on Compensator Design
Spring 2011

2. Stability Margins
Gain margin: (GM): Factor by which the gain can be raised before instability results
Phase margin (PM): Amount by which the phase of G(jω) exceeds -180° when |KG(jω)|=1

3. Stability Margins
GM and PM are measures of how close the Nyquist plot comes to encircling the -1 point

4. Lead Compensator Design Summary

5. Lag Compensator Design Summary

6. Design Example – Root Locus (p. 261)
The transfer function between the elevator input and the pitch attitude is
where
is the pitch angle* (degrees)
is the elevator angle (degrees)
Design the controller so that
Rise time of 1 sec or less
Overshoot less than 10%
* see sec 10.3 for details

7. Design Example – Root Locus
According to the specifications:
From (3.60) p.116:
From (3.64) p.118:
As a result, desired closed-loop pole should be at:
where
(3.56) p.112

8. Design Example – Root Locus
Let’s look at open loop response of the system, by using Matlab:
>> num = 160 .* conv([1 2.5], [1 0.7]);
>> den = conv([1 5 40], [ ]);
>> sysOl = tf(num, den);
>> step(sysOl)

9. Design Example – Root Locus
Now, look at the root locus of the system:
>> Wn = 1.8;
>> DR = 0.6;
>> sigma = DR * Wn;
>> Wd = Wn * sqrt(1 - DR^2);
>> p = -sigma + j*Wd;
>> q = -sigma - j*Wd;
>> rlocus(sysOl)
>> hold on
>> plot(p, '*')
>> plot(q, '*')
>> grid on

10. Design Example – Root Locus
Root Locus Plot

11. Design Example – Root Locus
Using “rlocfind” command to find the gain K that yield the closed-loop
poles that locate as close to the desired poles location as possible:
>> rlocfind(sysOl)
Select a point in the graphics window
selected_point = i
ans =
0.3260
Use K =

12. Design Example – Root Locus
Use proportional gain K = to see if it gives satisfactory close-
loop response:
>> sysClP = feedback(0.326*sysOl, 1)
Transfer function:
52.16 s^ s
s^ s^ s^ s
>> step(sysClP)
The overshoot is greater than
10 %, rise time is okay.

13. Design Example – Root Locus
Since proportional control is not enough, we can use lead
compensator – as a lead compensator would shift the root locus
further to the left (lowering rise time and decreasing the transient
overshoot (p. 249).
From p. 250, the zero of the lead compensator should be placed in
the neighborhood of the desired closed-loop Wn, and the pole
of the lead compensator should be 5 to 20 times of the value of the
zero. Let’s pick z = 2 (as Wn = 1.8) and p = 2 * 10 = 20.

14. Design Example – Root Locus
Now pick gain K of the lead compensator:
>> lead = tf([1 2], [1 20]);
>> sysLead = lead * sysOl;
>> rlocus(sysLead)
>> hold on
>> plot(p, '*')
>> plot(q, '*')
>> grid on
>> hold off
>> rlocfind(sysLead)
Select a point in the graphics window
selected_point = i
ans =
1.7599

15. Design Example – Root Locus
Simulate the step response of the new closed-loop system:
>> sysClLead = feedback( * sysLead, 1)
Transfer function:
281.6 s^ s^ s
s^ s^ s^ s^ s
>> step(sysClLead)

16. Design Example – Root Locus

17. Design Example – Compensator Design from Freq. Response
Example 6.17 (p. 358) – a Type 1 Servomechanism System
Design a lead compensator so that the PM = 45° and Kv = 10.
From p. 356:
Step 1: Pick K to satisfy error constant
from p. 181

18. Design Example – Compensator Design from Freq. Response
Step 2: Evaluate the PM of the uncompensated system using K
obtained from step 1
>> Gs = tf(10, conv([1 0], conv([1/2.5 1], [1/6 1])))
Transfer function: 10
s^ s^2 + s
>> [mag, phase, w] = bode(Gs);
>> [GM, PM, Wcg, Wcp] = margin(mag, phase, w)
GM =
0.8540
PM =
Wcg =
3.8730
Wcp =
4.1910

19. Design Example – Compensator Design from Freq. Response
Step 3: Add extra margin (about 5-10°) and determine the phase lead
Step 4: Determine from (6.40) p. 349:
Step 5: Pick at the crossover frequency:

20. Design Example – Compensator Design from Freq. Response
Step 6: Draw the compensated frequency response and check the PM:
>> lead = tf([0.5 1], [0.05 1]);
>> sysLead = lead * Gs;
>> [mag, phase, w] = bode(sysLead);
>> [GM, PM, Wcg, Wcp] = margin(mag, phase, w)
GM =
2.3204
PM =
Wcg =
Wcp =
7.2994

21. Design Example – Compensator Design from Freq. Response
We can use “sisotool” command to find GM, PM, Wcp, Wcg:
>> sisotool(sysLead)

22. Design Example – Compensator Design from Freq. Response
Step 6: Iterate on the design until all specifications are met.
Add an additional lead compensator if necessary.
We can see that with only on lead compensator, we can only get the
PM of 23°.
So, we need to repeat Step 1 – Step 5 again to add additional lead
compensator.
Assuming the second iteration gives:

23. Design Example – Compensator Design from Freq. Response
Let’s check the PM new compensated system:
>> secondLead = tf([0.25 1], [ ]);
>> sysDoubleLead = secondLead * sysLead;
>> [mag, phase, w] = bode(sysDoubleLead);
>> [GM, PM, Wcg, Wcp] = margin(mag, phase, w)
GM =
3.8669
PM =
Wcg =
Wcp =

24. Design Example – Compensator Design from Freq. Response
The overall compensator becomes:
Let’s take a look at the step responses of the compensated system:
>> sysLeadCl = feedback(sysLead, 1);
>> sysDoubleLeadCl = feedback(sysDoubleLead, 1);
>> step(sysLeadCl)
>> hold on
>> step(sysDoubleLeadCl)
>> hold off
>> grid on
>> legend('Single Lead Compensator', 'Double Lead Compensator')

25. Design Example – Compensator Design from Freq. Response
Step responses of the single lead compensator system vs double
lead compensator system

26. Compensator Design Conclusion
Lead Compensator => PD Controller
Speed up system response
Lowering the rise time
Decreasing the overshoot
Lag Compensator => PI Controller
Improve steady-state accuracy