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%.
17
,)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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