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

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

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

Lead Compensator Design Summary

Lag Compensator Design Summary

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

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

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], [1 0.03 0.06]); >> sysOl = tf(num, den); >> step(sysOl)

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

Design Example – Root Locus Root Locus Plot

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 = -0.9755 + 0.4105i ans = 0.3260 Use K = 0.3260

Design Example – Root Locus Use proportional gain K = 0.326 to see if it gives satisfactory close- loop response: >> sysClP = feedback(0.326*sysOl, 1) Transfer function: 52.16 s^2 + 166.9 s + 91.28 -------------------------------------------- s^4 + 5.03 s^3 + 92.37 s^2 + 168.4 s + 93.68 >> step(sysClP) The overshoot is greater than 10 %, rise time is okay.

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.

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 = -8.7708 +14.3523i ans = 1.7599

Design Example – Root Locus Simulate the step response of the new closed-loop system: >> sysClLead = feedback(1.7599 * sysLead, 1) Transfer function: 281.6 s^3 + 1464 s^2 + 2295 s + 985.5 ------------------------------------------------------ s^5 + 25.03 s^4 + 422.4 s^3 + 2270 s^2 + 2327 s + 1034 >> step(sysClLead)

Design Example – Root Locus

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

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 ---------------------------- 0.06667 s^3 + 0.5667 s^2 + s >> [mag, phase, w] = bode(Gs); >> [GM, PM, Wcg, Wcp] = margin(mag, phase, w) GM = 0.8540 PM = -4.1158 Wcg = 3.8730 Wcp = 4.1910

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:

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 = 22.9367 Wcg = 11.5132 Wcp = 7.2994

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

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:

Design Example – Compensator Design from Freq. Response Let’s check the PM new compensated system: >> secondLead = tf([0.25 1], [0.025 1]); >> sysDoubleLead = secondLead * sysLead; >> [mag, phase, w] = bode(sysDoubleLead); >> [GM, PM, Wcg, Wcp] = margin(mag, phase, w) GM = 3.8669 PM = 45.6362 Wcg = 30.7887 Wcp = 13.7805

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')

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

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