11 elec3114

Dr Branislav HredzakControl Systems Engineering, Fourth Edition by Norman S. Nise

Copyright © 2004 by John Wiley & Sons. All rights reserved.

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

• How to use frequency response techniques to design cascade compensators to improve the transient response

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



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.

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.

)100/(%ln)100/ln(%

22 OSOS

+

−=

πζ

42

1

412

2tanζζ

ζ

++−=Φ −

M



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 tenfoldimprovement 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

0

42

1 2.59412

2tan =++−

=Φ −

ζζ

ζMFor 9.5% overshoot, ζ=0.6 and

We increase phase margin by 100 to 69.20

rad/s 9.8 is frequency ingcorrespond theand8.1102.69180- of angle phase aat occurs 2.69 0000 −=+=ΦM

Bode plot for K = 5839



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

062.098.0)(

++

=sssGC

The compensator must have a dc gain of 1, hence, the gain of compensator must be 0.062/0.98=0.063.

Then 062.098.0063.0)(

++

=sssGC

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

• the maximum phase angle φmax :

• compensator’s magnitude at ωmax is

ωmax

ωmax



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 curve7. 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 frequencies9. 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

overshoot) (20% 456.0

sec1.0

=

=

ζpT

sradBW /6.46=ω

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

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 compensated system

dBjGdBc 76.3)( max =ω



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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11 elec3114