Control Systems - Chibum Lee · PDF file13-03-2014 · Chibum Lee -Seoultech Control...

Control Systems

Control Systems Design

Lead-Lag Compensator

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Control SystemsChibum Lee -Seoultech

Outline

Lead compensator design in frequency domain

Lead compensator design steps.

Example on lead compensator design.


Frequency Domain Design

Frequency response approach can impose the transient

response performance indirectly

Bode diagrams: useful for the frequency domain

specification

Nyquist plot: useful for system (relative) stability analysis.


Lead Compensator

Appropriate improvement in transient response and

a small change in steady-state accuracy

Polar plot (KC = 1)

• 𝛼 determines the maximum phase

angle 𝜙𝑚 which occurs at 𝜔𝑚

)10(1

1

1

1)(

Ts

Ts

KTs

TsKsG ccc Zero:-

1

𝑇Pole: -

1

𝑇

1

1)(

Tj

TjKjG cc

112sin

1 1

2

m


Lead Compensator

Bode diagram (KC = 1, 𝛼=0.1)

• Lead compensator: High-pass filter

• Goal of the lead compensator: to provide sufficient phase-lead

angle to offset the excessive phase lag in the uncompensated

system

11.0

11.0)(

Tj

TjjGc

Tm

1

1

1)(

Tj

TjKjGc


Lead Compensator

The maximum phase lead angle 𝜙𝑚 occurs at 𝜔𝑚

• To increase phase margin, magnitude of the compensated system to

cross 0 dB at this frequency 𝜔𝑚

• but need to care for magnitude distortion added by lead compensator

-100

-50

0

50

Magnitude (

dB

)

0.01 0.1 1 10 100-270

-225

-180

-135

-90

-45

Phase (

deg)

Bode Diagram

Gm = 9.93 dB (at 4.01 rad/s) , Pm = 104 deg (at 0.443 rad/s)

Frequency (rad/s)𝜔𝑐: Gain crossover frequency

Phase Margin )()(180 cc jHjGPM


Lead Compensator

The magnitude of the lead controller at:

• This is the amount that the lead compensator will shift the

magnitude plot. To have the gain crossover point (0 dB) at the

right point, we have to make sure that the geometric frequency

mean falls at the point where the open-loop uncompensated

system is 20 log10( 𝛼) below 0 dB.

• If is absorbed into the control gain, then the previous maximum

magnitude would be

1

1

1

1)(

1 j

j

Tj

TjjG

T

c

m

Tm

1

1


Lead Compensator Design

Step 0: Assume following lead compensator

• Open-loop Plant:

• Compensator:

• Loop Gain of Compensated System:

with

( )G s

1

1 1( ) ( ) ( ) ( )

1 1C

Ts TsG s G s K G s G s

Ts Ts

1( ) ( )G s KG s

)(1

1

1

1)( KK

Ts

TsK

Ts

TsKsG ccc


Lead Compensator Design Steps

Step 1: Determine K to satisfy static error constants

(Kp or Kv)

Step 2: Using this K, draw a Bode diagram of G1(s), and

evaluate the phase margin

Step 3: Determine the necessary phase angle needed to

meet design specs.



Step 4: Determine by using

( 𝜙𝑚=required phase lead angle + 5~12deg)

Find the frequency where:

Select this frequency as the new gain crossover frequency, where the maximum phase shift occurs

Step 5: Determine the lead compensator

1sin

1m

1 10

1( ) 20logG s

Tcm

1

Zero @ -1

𝑇Pole @ -

1

𝑇



Step 6: Solve for

Step 7: Check the gain margin. If it is not satisfactory,

one may have to iterate.

KKc


Example

Ex.

Design a lead compensator to give Kv = 20 s-1,

PM 50 deg, and GM 10 dB

• Step 0: Find low-frequency gain:

Using

4

( )2

G ss s

1( )

1C C

TsG s K

Ts

1

4( ) ( )

2C

KG s KG s K K

s s


Example

• Step 1:

Gain requirements:

• Step 2: draw a Bode diagram of G1(s),

and evaluate the phase margin

0 0

1 4lim ( ) lim 2

1 2C

s s

Ts KKv sG s G s s K

Ts s s

2 20 10K K

)2(

40)(1

jjjG


Example

• Step 3: Determine the necessary phase angle needed to meet design specs.

𝜙𝑚= 33 (necessary)+5 (additional) deg

(extra 5 deg for compensation of the shift in gain crossover frequency.)

• Step 4: Determine

1 sin 381 sin1sin 0.24

1 1 sin 1 sin 38

mm

m


Example

• Step 5: Determine the corner frequencies

First we find 𝜔𝑐 by noting that

The graph gives

at

The zero at 1/T

The pole at 1/T

12.04 6.2 dB

1 10

1( ) 20logG s

rad/sec9 cm

dBjG 2.6)(1

1 1 9 .24 4.41 /C Cw w rad secTT

1 9 / .24 18.4 /Cwrad sec

T


Example

• Step 6: Solve for KC = K/

• Step 7: Verify the design

PM 50 deg and GM infinite…

10/.24 41.7CK K

4.41( ) 41.7

18.4C

sG s

s


Example

• Nyquist


Lecture Outline

Lag compensator design in frequency domain

Lag compensator design steps.

Example on lag compensator design.


Lag Compensator

An appreciable improvement in steady-state accuracy

at the expense of increasing the transient-response time

Polar plot (KC = 1)

)1(1

1

1

1)(

Ts

Ts

KTs

TsKsG ccc

Zero:-1

𝑇Pole: -

1

𝑇

1

1)(

Tj

TjKjG cc


Lag Compensator

Bode diagram (KC = 1, 𝛽=10)

• Lead compensator: Low-pass filter

• Goal of the lead compensator: to provide attenuation in high

frequency range to give a system sufficient phase margin

110

110)(

Tj

TjjGc

1

1)(

Tj

TjKjG cc


Lag Compensator Design Steps

Step 0: Assume the following lag compensator

• Compensator:

• Loop Gain of Compensated System:

with

Step 1: Determine K to satisfy static error constants.

1( ) ( )G s KG s

)(1

1

1

1)( KK

Ts

TsK

Ts

TsKsG ccc

)(1

1)(

1

1)()( 1 sG

Ts

TssG

Ts

TsKsGsGc



Step 2: Find C for G1(s)

• Check PM and GM to see if they meet specs

• If not, find the frequency

where = -180 deg + required PM

• Required PM = specified PM + 5~12 deg. for safety margin

• This frequency is the new gain crossover frequency C

Step 3: Choose the corner frequency =1/T of the zero

• We want to change the magnitude plot without changing the

phase plot at the new crossover frequency

• Therefore, choose the zero at 1/T to be around 1 decade below

the new corner frequency C

1( )G jw



Step 4: Determine and the pole location...

• We now examine to find out how much it is greater

than 0 dB.

• Choose and then the pole is at 1/T

Step 5: Form the lag compensator...

• The actual compensator gain

)(1 cjG

101 log20)(dB0 cjG

KKc


Example

Ex.

Design a lag compensator Kv = 5 sec-1,

PM > 40 degrees and GM > 10 dB

• Step 0:

)1(,1

1

)(

Ts

Ts

KsG cc

)15.0)(1(

1)(

ssssG

)15.0)(1()()(1

sss

KsKGsG


Example

• Step 1:

Gain

• Step 2: draw a Bode diagram

of G1(s)

5

)15.0)(1(lim)(lim

)(1

1lim)()(lim

01

0

100

K

Ksss

sKssG

sGTs

TsssGssGK

ss

sc

sv

)15.0)(1(

5)(1

ssssG


Example

• Step 3:

The phase margin is –20 degrees which means the unity feedback of the gain adjusted uncompensated system G1(s) is unstable.

If we want to have a phase margin of 40 deg, we should add another 5~12 deg to be safe.

Choose corner frequency =1/T=0.1 rad/sec (zero of the lag compensator)

The phase of G1(s) is –128 degaround 0.5 rad/sTherefore, the new crossover frequency c should be about 0.5 rad/s.


Example

• Step 4:

Based on the new cross over

frequency, the zero should be placed

at = 1/T = 0.05 rad/s about a

decade below c.

The gain of G1(s) at c = 0.5 rad/s is

about 20 dB so that’s how much the

lag compensator must attenuate.

The pole of the lag compensator

10201

log20 10

rad/sec 01.01

T


Example

• Step 5:

100

110

1

)(

s

s

KsG cc

5.010

5

KKc

)15.0)(1)(1100(

)110(5)()(

ssss

ssGsGc


Example

Lag controller causes slower transients


Outline

Lead-Lag compensator

Qualitative system responses.


Lead-Lag Compensators

Whether to use the Lead or Lag controller depends on

the nature of your plant

• Lead for improved transient performance

• Lag for improved steady-state performance

when you need improved transient performance and

steady-state tracking lead-lag compensator

)1,1(1

11

)(

2

2

1

1

Ts

Ts

Ts

Ts

KsG cc


Lead Lag Compensators

Phase lead adds phase at the uncompensated gain

crossover frequency thereby increasing the phase margin

Phase lag provides attenuation allowing an increase in

gain at low frequency to improve steady-state

performanceKc = 1, γ = β = 10, T2 = 10T1.

Kc = 1, γ = β.


Lead Lag Compensators

Qualitative system responses

uncompensated lead lag lead-lag

)(ty )(ty )(ty )(ty

)(ty )(ty )(ty )(ty

Control Systems - Chibum Lee · PDF file13-03-2014 · Chibum Lee -Seoultech Control...

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Transcript of Control Systems - Chibum Lee · PDF file13-03-2014 · Chibum Lee -Seoultech Control...