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EE 4314

Examples on Compensator Design

Spring 2011



Stability Margins

Gain margin: (GM): Factorby which the gain can beraised 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 measuresof how close the Nyquistplot comes to encircling

the -1 point



Lead Compensator Design Summary



Lag Compensator Design Summary



Design Example – Root Locus (p. 261)

G s s e s

160 s 2.5 s 0.7

s2 5 s 40 s2 0.03 s 0.06

The transfer function between theelevator input and the pitch attitude

is

where

is the pitch angle* (degrees)

is the elevator angle (degrees)

e

Design the controller so that

1. Rise time of 1 sec or less2. Overshoot less than 10%

* see sec 10.3 for details



Design Example – Root Locus

According to the specifications:

From (3.60) p.116:

t r

1.8

n

From (3.64) p.118:

M p e / 1 2

n 1.8 rad /sec

0.6

As a result, desired closed-loop pole should be at:

p j d

where

n

d n 1 2 (3.56) p.112




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)




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




Root Locus Plot




Using “rlocfind” command to find the gain K that yield the closed-looppoles 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




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 than10 %, rise time is okay.




Since proportional control is not enough, we can use leadcompensator – as a lead compensator would shift the root locusfurther to the left (lowering rise time and decreasing the transientovershoot (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 poleof the lead compensator should be 5 to 20 times of the value of thezero. Let’s pick z = 2 (as Wn = 1.8) and p = 2 * 10 = 20.

D s K s 2

s 20




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




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)




D s 1.7599 s 2

s 20



Design Example – Compensator Design from Freq. Response

Example 6.17 (p. 358) – a Type 1 Servomechanism System

KG s K 10 s s /2.5 1 s /6 1

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

K v lim s0

sK G s K 100 1 0 1

K 1

D i E l C t D i f F R




Step 2: Evaluate the PM of the uncompensated system using Kobtained 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




Step 3: Add extra margin (about 5-10°) and determine the phase lead

max 45o

5o

(4o

)

54o

Step 4: Determine from (6.40) p. 349:

1 sin max

1 sin max

0.1

Step 5: Pick at the crossover frequency: ma x

1/T m ax

T 0.5

D s Ts 1

Ts 1

0.5 s 1

0.05 s 1




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




We can use “sisotool” command to find GM, PM, Wcp, Wcg:

>> sisotool(sysLead)




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 thePM of 23°.

So, we need to repeat Step 1 – Step 5 again to add additional leadcompensator.

Assuming the second iteration gives:

D s Ts 1

Ts 1

0.25 s 1

0.025 s 1




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




The overall compensator becomes:

D s 0.5 s 1(0.05 s 1)

0.25 s 1(0.025 s 1)

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')




Step responses of the single lead compensator system vs doublelead 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

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