Lecture 9:

Compensator Design in

Frequency Domain

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Objectives

• Recognize the relationship between: the overshoot (%OS), the damping ratio (ζ) , and the phase margin (ΦM ).

• Use two approaches employing Bode diagrams to design compensators (controllers) which achieve a desired phase margin:

Simple gain adjustment. Lag compensator. 2

Relationship between %OS, ζ , ΦM

• Consider a unity feedback system with the following open-loop transfer function

• The closed loop transfer function is the standard 2nd order system

3

)2()(

2

n

n

sssG

22

2

2)(

nn

n

sssT

The time response of the secondorder under-damped system

4

• The percentage overshoot, %OS, is given by

• Note that %OS is a function only of the damping ratio, ζ.

• The inverse is given by

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100% max

final

final

c

ccOS

1001

exp%2

OS

)100/(%ln

)100/ln(%22 OS

OS

• There is also a relationship between damping ratio and phase margin. The phase margin is obtained by solving |G(jω)| = 1 to obtain the frequency as

• The phase margin is

• There is a simpler formula

• Thus if we can vary the phase margin, we can vary the percent overshoot.

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421 412 n

42 412

2arctan

M

100M

The phase margin, can be varied via a simple gain adjustment as shown.

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Transient response via gain adjustment

Example

For the position control system shown below, find the value of preamplifier gain, K, to yield a 9.5% overshoot in the transient response for a step input. Use only frequency response methods.

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Solution

• The open-loop transfer function (with K = 1 for now) is

• We draw the Bode plot for this open-loop system.

9

)100)(36(

100)(

ssssG

Bode plots for the example

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• The value of ζ achieving %OS = 9.5% is

• In the Bode diagram we locate the point at which the phase is -120°. At this point, the frequency is 14.3 rad/sec and the gain is about -55dB.

• For this point to be the phase margin, the gain must be 0dB, so we need a gain of +55dB.

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60

6.0)095.0(ln

)095.0ln(

)100/(%ln

)100/ln(%2222

M

OS

OS

56310gain required

)20log(gain5dB5

20

55

K

Lag compensation

The function of the lag compensator is to increase the phase margin of the system to yield the desired transient response without affecting the low-frequency gain and hence it does not reduce system stability or steady error constant.

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Steady-state error constants

The steady error constants are:

position constant

velocity constant

acceleration constant

The value of the steady-state error decreases as the steady error constants increases.

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),(lim0

sGKs

p

),(lim0

ssGKs

v

),(lim 2

0sGsK

sa

Visualizing lag compensator

• The transfer function of the lag compensator is

• where z > p.

• The gain of this lag compensator at low frequency is unity and at high frequency is

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ps

zs

z

psGc

.)(

z

p

Visualizing lag compensator• For example, the frequency response of a lag compensator:

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01.0

1.0)(

s

ssGc

Visualizing lag compensator

In the figure below, the uncompensated system is unstable since the gain at -180° is greater than 0dB. The lag compensator, while not changing the low-frequency gain, does reduce the high frequency gain. The magnitude curve can be shaped to go through 0dB at the desired phase margin to obtain the desired transient response.

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Example

Given the following open-loop transfer function of a position control system,

use Bode diagram to design a lag compensator to yield a percent overshoot of 9.5%.

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,)10)(1(

100)(

ssssG

Solution • Before solving the problem, let us draw the step response of

the closed-loop system whose open loop transfer function is given.

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The response is too oscillatory!

What do you expect about the value of the damping ratio and phase margin of the system?

Solution

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Using the command margin, we draw the Bode diagram and find the phase margin to be only 1.58°.Too low value.

This system must be compensated!

Solution

• Now, let us start solving the problem.

• We want to design a lag compensator of the following transfer function

• First we need to plot the Bode diagram, see next.

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ps

zs

z

psGc

.)(

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Choose a point!

Solution

• From the previous example, we know that 9.5 %OS corresponds to a phase margin 60°. So, we should look for the point at which the phase passes -120°.

• However, as the lag compensator add some extra little phase lag, we will seek a phase margin of 70°. That is we look for the point at which the phase passes -110°.

• From the figure, we can locate this point at frequency 0.326 rad/sec at which the gain is 29.3dB.

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Solution

• At the frequency 0.326 rad/sec, the open loop has a gain 29.3dB.

• Therefore, it is required that the lag compensator, at frequency 0.326 rad/sec, to have a gain of

• Arbitrarily, locate the zero of the lag compensator at

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17.2910gain

)20log(gaindB3.29

20

29.3

0011.00326.010

326.0 pz

17.29

1

z

p

• Thus, the required lag compensator is

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0011.0

0326.0034.0

0011.0

0326.0.

0326.0

0011.0

.)(

s

ss

s

ps

zs

z

psGc

Solution

Checking the phase margin of the new system

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As a check, we use the command margin, to draw the Bode diagram again. Now the phase margin is 64.6° which is more than enough!!

Checking the step response of the new system

• It is interesting to plot the step response of the closed loop system after adding the lag compensator.

• As shown, the response is highly stable which is very good (compared to the oscillatory response of the original system) although it is sluggish too.

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