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1

Design via Frequency Response

• How to use frequency response techniques to adjust the
gain to meet a transient response specification

• How to use frequency response techniques to design
cascade compensators to improve the steady-state error

cascade compensators to improve the transient response

cascade compensators to improve both the steady-state
error and the transient response

Dr Branislav Hredzak
Control Systems Engineering, Fourth Edition by Norman S. Nise
Copyright © 2004 by John Wiley & Sons. All rights reserved.

2

Transient Response via Gain Adjustment

Design Procedure:

1. Draw the Bode magnitude and phase plots for a
convenient value of gain.

2. Determine the required phase margin from the
percent overshoot.
− ln(%OS / 100) 2ζ
ζ = Φ M = tan −1
π 2 + ln 2 (%OS / 100) − 2ζ 2 + 1 + 4ζ 4

3. Find the frequency, ωΦM, on the Bode phase
diagram that yields the desired phase margin,
CD

4. Change the gain by an amount AB to force the
magnitude curve to go through 0 dB at ωΦM.


3

Lag Compensation
• improves the static error without any resulting instability

• increases the phase margin of the system to yield the desired transient
response

where α > 1.


4

Design Procedure

1. Set the gain, K, to the value that satisfies the steady-state error specification
and plot the Bode plots

2. Find the frequency where the phase margin is 50 to 120 greater than the phase
margin that yields the desired transient response

3. Select a lag compensator whose magnitude response yields a composite Bode
magnitude diagram that goes through 0 dB at the frequency found in step 2 as
follows:
• Draw the compensator's high-frequency asymptote to yield 0 dB at
the frequency found in step 2; select the upper break frequency to be
1 decade below the frequency found in step 2; select the low-
frequency asymptote to be at 0 dB; connect the compensator's high-
and low-frequency asymptotes with a -20 dB/decade line to locate
the lower break frequency.

4. Reset the system gain, K, to compensate for any attenuation in the lag network
in order to keep the static error constant the same as that found in step 1.


5

The transfer function of the lag compensator is

where α > 1.

Gc(s) = (s + 0.1)/(s + 0.01)


6

Problem Use Bode diagrams to design a lag compensator to yield a tenfold
improvement in steady-state error over the gain compensated system while
keeping the percent overshoot at 9.5%.

K=583.9

Solution
KV = 583.9 / 36 = 16.22, hence for a 10 fold improvement of steady state error
KV = 10 x 16.22 = 162.22

Therefore K = 583.9 x 10 = 5839 and


7
Bode plot for
K = 5839

2ζ
For 9.5% overshoot, ζ=0.6 and Φ M = tan −1 = 59.20
− 2ζ 2 + 1 + 4ζ 4
We increase phase margin by 100 to 69.20

Φ M = 69.2 0 occurs at a phase angle of - 1800 + 69.20 = −110.80
and the corresponding frequency is 9.8 rad/s


8

The magnitude at 9.8 rad/s is +24 dB
Compensator:
• high break frequency one decade below 9.8 rad/s, i.e., 0.98 rad/s

• low break frequency is found as -20 dB line intersection with 0 dB and
is 0.062 rad/s


9
s + 0.98
GC ( s ) =
s + 0.062
The compensator must have a dc gain of 1, hence, the gain of compensator
must be 0.062/0.98=0.063.

Then s + 0.98
GC ( s ) = 0.063
s + 0.062

obtained by
simulation


10

Lead Compensation
• increase the phase margin to reduce the percent overshoot
• increase the gain crossover to realize a faster transient response

Note: notice that the initial
slope, which determines the
steady-state error, is not
affected by the design


11

Lead Compensator Frequency Response

where β<1

• frequency,ωmax, at which the
maximum phase angle, φmax,
occurs can be found using

ωmax • the maximum phase angle φmax :

• compensator’s magnitude at
ωmax
ωmax is


12

Design Procedure

1. Find the closed-loop bandwidth required to meet the settling time, peak
time, or rise time requirement
2. Set the gain, K, of the uncompensated system to the value that satisfies the
steady-state error requirement.
3. Plot the Bode magnitude and phase diagrams for this value of gain and
determine the uncompensated system's phase margin.
4. Find the phase margin to meet the damping ratio or percent overshoot
requirement. Evaluate the additional phase contribution required from the
compensator.
5. Determine the value of β from the lead compensator's required phase
contribution.
6. Determine the compensator's magnitude at the peak of the phase curve
7. Determine the new phase-margin frequency by finding where the
uncompensated system's magnitude curve is the negative of the lead
compensator's magnitude at the peak of the compensator's phase curve.
8. Design the lead compensator's break frequencies
9. Reset the system gain to compensate for the lead compensator's gain.


13

Problem Design a lead compensator to yield a 20% overshoot and KV, = 40, with
a peak time of 0.1 second.

Solution
Tp = 0.1sec
ζ = 0.456 (20% overshoot)

ω BW = 46.6rad / s

In order to meet the specification Kv=40, K must be set 1440


14

• 20% overshoot implies PM=48.10
• original PM=340 at 29.6 rad/s
• Phase contribution from the
compensator = 48.1-34+10=24.10
• Using

for φmax= 24.10, β =0.42

• From

Gc ( jωmax ) dB = 3.76dB

The uncompensated system
passes through (-3.76) dB at
ωmax= 39 rad/s. Now, we select
39 rad/s as the new phase-
margin freq., which will result in
a 0dB crossover at 39 rad/s for the
compensatedDr Branislav Hredzak
system

15
• We find compensator break
frequencies from

β =0.42 ωmax= 39 rad/s

1/T = 25.3
1/(βT) = 60.2

- where 2.38 is the gain required to
keep the DC gain of the compensator
at unity


16

obtained by
simulation


17

Lag-Lead Compensation

• Design first the lag compensator to improve the steady-state error and
then design a lead compensator to meet the phase-margin


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