leadlagbode handout

7/23/2019 LeadLagBode Handout

1/27

EE 3CL4, 91 / 3 0

Tim Davidson

FrequencyDomain

Approach toCompensatorDesign

LeadCompensators

LagCompensators

EE3CL4:Introduction to Linear Control Systems

Section 9: Design of Lead and Lag Compensators using

Frequency Domain Techniques

Tim Davidson

McMaster University

Winter 2014


2/27

EE 3CL4, 92 / 3 0

Tim Davidson

FrequencyDomain


LeadCompensators

LagCompensators

Outline

1 Frequency Domain Approach to Compensator Design

2 LeadCompensators

3 Lag

Compensators


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4/27

EE 3CL4, 95 / 3 0

Tim Davidson

FrequencyDomain


LeadCompensators

LagCompensators

Bode diagram

L(j) = 1

j(1 +j)(1 +j/5)

Gain margin

15 dB

Phase margin 43


5/27

EE 3CL4, 96 / 3 0

Tim Davidson

FrequencyDomain


LeadCompensators

LagCompensators

Compensators and Bode

diagram

We have seen the importance of phase margin IfG(s)does not have the desired margin,

how should we chooseGc(s)so thatL(s) =Gc(s)G(s)does?

To begin, how doesGc(s)affect the Bode diagram Magnitude:

20 log10|Gc(j)G(j)

|=20 log10(|Gc(j)|+ 20 log10|G(j)|

Phase:

Gc(j)G(j) = Gc(j) + G(j)


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EE 3CL4, 98 / 3 0

Tim Davidson

FrequencyDomain


LeadCompensators

LagCompensators

Lead Compensators

Gc(s) = Kc(s+z)s+p , with |z| < |p|, alternatively, Gc(s) = Kc

1+s1+s , wherep=1/and =p/z>1

Bode diagram (in the figure, K1=Kc/):


7/27

EE 3CL4, 99 / 3 0

Tim Davidson

FrequencyDomain


LeadCompensators

LagCompensators

Lead Compensation

What will lead compensation, do? Phase is positive: might be able to increase phasemarginpm

Slope is positive: might be able to increase thecross-over frequency,c, (and the bandwidth)


8/27

EE 3CL4, 910/30

Tim Davidson

FrequencyDomain


LeadCompensators

LagCompensators

Lead Compensation

Gc(s) = Kc

1+s1+s

By making the denom. real, can show thatGc(j) =atan

(1)1+()

2 Max. occurs when =m= 1

=zp

Max. phase angle satisfies tan(m) = 12 Equivalently, sin(m) = 1

+1

At =m, we have |Gc(jm)| =Kc/


9/27

EE 3CL4, 911/30

Tim Davidson

FrequencyDomain


LeadCompensators

LagCompensators

Bode Design Principles

Set the loop gain so that desired steady-state errorconstants are obtained

Insert the compensator to modify the transientproperties:

Damping: through phase margin Response time: through bandwidth

Compensate for the attenuation of the lead network, ifappropriate

To maximize impact of phase lead, want peak of phase near

cof thecompensatedopen loop


10/27

EE 3CL4, 912/30

Tim Davidson

FrequencyDomain


LeadCompensators

LagCompensators

Design Guidelines1 For uncompensated (i.e., proportionally controlled)

closed loop, set gainKpso that steady-state error

constants of the closed loop meet specifications

2 Evaluate the phase margin, and the amount of phase

lead required.

3 Add a little safety margin to the amount of phase lead

4 From this, determine using sin(m) =

1

+15 To maintain steady-state error consts, setKc=Kp6 Determine (or approximate) the frequency at which

KpG(j)has magnitude 10 log10().7 If we setmof the compensator to be this frequency,

thenGc(jm)G(jm) =1 (or 1). Hence, thecompensator will provide its maximum phase

contribution at the appropriate frequency

8 Choose=1/(m). Hence,p=m

.

9 Setz=p/.

10 Compensator: Gc(s) = Kc(s+z)s+p .


11/27

EE 3CL4, 913/30

Tim Davidson

FrequencyDomainApproach toCompensatorDesign

LeadCompensators

LagCompensators

Example

Type 1 plant of order 2: G(s) = 5

s(s+2) Design goals: Steady-state error due to a ramp input less than 5% of

velocity of ramp Phase margin at least 45 (implies a damping ratio)

Steady state error requirement implies Kv=20. For prop. controlled Type 1 plant: Kv =lims0 sKpG(s).

HenceKp=8.

To find phase margin of prop. controlled loop we need

to findc, where |KpG(jc)| = 40

jc(jc+2)= 1

c 6.2rad/s Evaluate KpG(j) = 90 atan(/2)at =c Hencepm, prop =18


12/27

EE 3CL4, 914/30

Tim Davidson


LeadCompensators

LagCompensators

Example

pm, prop=18

. Hence, need 27 of phase lead Lets go for a little more, say 30 So, want peak phase of lead comp. to be 30 Solving 1

+1 =sin(30)yields=3. SetKc=3 8

Since 10 log

10(3

) =4.8 dB we should choose

mto be

where 20 log10 40jm(jm+2)

= 4.8 dB Solving this equations yieldsm=8.4rad/s Thereforez=m/

=4.8,p=z=14.4

Gc(s) = 24(s+4.8)

s+14.4

Gc(s)G(s) = 120(s+4.8)s(s+2)(s+14.4) , actualpm=43.6

Goal can be achieved by using a larger target foradditional phase, e.g., = 3.5


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EE 3CL4, 915/30

Tim Davidson


LeadCompensators

LagCompensators

Bode Diagram


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EE 3CL4, 916/30

Tim Davidson


LeadCompensators

LagCompensators

Step Response

C


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EE 3CL4, 917/30

Tim Davidson


LeadCompensators

LagCompensators

Ramp Response

EE 3CL4 9


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EE 3CL4, 918/30

Tim Davidson


LeadCompensators

LagCompensators

Ramp Response, detail

EE 3CL4 9


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EE 3CL4, 920/30

Tim Davidson


LeadCompensators

LagCompensators

Lag Compensators

Gc(s) = Kc(s+z)s+p , with |p| < |z|, alternatively,

Gc(s) = Kc(1+s)

1+s , wherez=1/ and = z/p>1 Bode diagrams of lag compensators for two differents, in the case whereKc=1/

EE 3CL4 9


18/27

EE 3CL4, 921/30

Tim Davidson


LeadCompensators

LagCompensators

What will lag compensation do?

Since zero and pole are typically close to the origin,phase lag aspect is not really used.

What is useful is the attenuation above =1/:gain is 20 log10(), with little phase lag

Can reduce cross-over frequency,c, without addingmuch phase lag

Tends to reduce bandwidth

EE 3CL4 9


19/27

EE 3CL4, 922/30

Tim Davidson


LeadCompensators

LagCompensators

Qualitative example

Uncompensated system has small phase margin

Phase lag of compensator does not play a large role Attenuation of compensator does:creduced by about a factor of a bit more than 3

Increased phase margin is due to the natural phasecharacteristic of the plant

EE 3CL4 9


20/27

EE 3CL4, 923/30

Tim Davidson

FrequencyDomainApproach to

CompensatorDesign

LeadCompensators

LagCompensators

Bode Design Principles

For lag compensators:

Set the loop gain so that desired steady-state errorconstants are obtained

Insert the compensator to modify the phase margin: Do this by reducing the cross-over frequency Observe the impact on response time

Basic principle: Set attenuation to reduce cfar enough so

that uncompensated open loop has desired phase margin

EE 3CL4, 9


21/27

EE 3CL4, 924/30

Tim Davidson


CompensatorDesign

LeadCompensators

LagCompensators

Design Guidelines1 For uncompensated (i.e., proportionally controlled)

closed loop, set gainKpso that steady-state error

constants of the closed loop meet specifications2 Evaluate the phase margin, analytically, or using a

Bode diagram. If that is insufficient. . .

3 Determinec, the frequency at which theuncompensated open loop,K

pG(j), has a phase

margin equal to the desired phase margin plus 5.4 Design a lag comp. so that the gain of the compensated

open loop,Gc(j)G(j), at =cis 0 dB

ChooseKc=Kp/so that steady-state error constsare maintained

Place zero of the comp. around c/10 so that at

cweget almost all the attenuation available from the comp.

Choose so that 20 log10() =20 log10(|KpG(jc)|).With that choice andKc=Kp/, |Gc(jc)G(jc)| 1

Place the pole atp=z/

Compensator: Gc(s) = Kc(s+z)s+p

EE 3CL4, 9


22/27

, 25/30

Tim Davidson


CompensatorDesign

LeadCompensators

LagCompensators

Example, same set up as lead

design

Type 1 plant of order 2: G(s) = 5s(s+2) Design goals:

Steady-state error due to a ramp input less than 5% ofvelocity of ramp

Phase margin at least 45 (implies a damping ratio)

Steady state error requirement implies Kv=20. For prop. controlled Type 1 plant: Kv =lims0 sKpG(s).

HenceKp=8.

To find phase margin of prop. controlled loop we need

to findc, where |KpG(jc)| = 40jc(jc+2)

= 1 c 6.2rad/s Evaluate KpG(j) = 90 atan(/2)at =c

Hencepm, prop =18

EE 3CL4, 9


23/27

26/30

Tim Davidson


CompensatorDesign

LeadCompensators

LagCompensators

Example

Since want phase margin to be 45, we set csuch thatG(jc) = 180 + 45 + 5 = 130. = c 1.5

To make the open loop gain at this frequency equal to 0 dB,the required attenuation is 20 dB. Actual curves are around2 dB lower than the straight line approximation shown

Hence = 10. SetKc=Kp/= 0.8 Zero set to be one decade below c; z=0.15 Pole isz/= 0.015.

HenceGc(s) = 0.8(s+0.15)

s+0.

015

EE 3CL4, 9


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Tim Davidson


CompensatorDesign

LeadCompensators

LagCompensators

Example: Compd open loop

Compensated open loop: Gc(s)G(s) = 4(s+0.15)s(s+2)(s+0.015)

Numerical evaluation: newc=1.58 new phase margin= 46.8

By design,Kvremains 20

EE 3CL4, 9


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Tim Davidson


CompensatorDesign

LeadCompensators

LagCompensators

Step Response

EE 3CL4, 929/30


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Tim Davidson


CompensatorDesign

LeadCompensators

LagCompensators

Ramp Response

EE 3CL4, 930/30


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Tim Davidson


CompensatorDesign

LeadCompensators

LagCompensators

Ramp Response, detail

leadlagbode handout

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