Lecture abstract EE C128 / ME C134 – Feedback Control …ee128/fa14/Lectures/EEC128-chap11.pdf1....

EE C128 / ME C134 – Feedback Control Systems

Lecture – Chapter 11 – Design via Frequency Response

Alexandre Bayen

Department of Electrical Engineering & Computer Science

University of California Berkeley

September 10, 2013

Bayen (EECS, UCB) Feedback Control Systems September 10, 2013 1 / 38

Lecture abstract

Topics covered in this presentation

I FR compensator design advantages

I Lag, lead, & lag-lead compensators


Chapter outline

1 11 Design via frequency response11.1 Introduction11.2 Transient response via gain adjustment11.3 Lag compensation11.4 Lead compensation11.5 Lag-lead compensation


11 Design via FR 11.1 Intro




RL vs. FR design, [1, p. 626]

RL

I Stability & TR design via gainadjustment

I Repeated trials

I TR design via cascadecompensation

I Intuitive

I Steady-state error design viacascade compensation

I Repeated trials

FR

I Stability & TR design via gainadjustment

I Read gain from the plots

I TR design via cascadecompensation

I Repeated trials

I Steady-state error design viacascade compensation

I Design derivativecompensation & steady-stateerror jointly



FR design review, [1, p. 627]

Stability

I Nyquist criterion ! stabilityI CL stable if OL stable & OL magnitude FR has a gain less than 0 dB

at the frequency where the phase FR is 180�

TR

I # %OS / " �M

I " speed of response / " bandwidth

Steady-state error

I # steady-state error / " low-frequency magnitude responses


11 Design via FR 11.2 Transient response via gain adjustment




�M -TR-gain relation, [1, p. 627]

Concept

I ⇣ (& %OS) relate to �M

OL TF

G(s) =!2n

s(s+ 2⇣!n)

CL TF

T (s) =!2n

s2 + 2⇣!ns+ !2n

⇣-�M relation

�M = tan�1

✓2⇣q

�2⇣2+p

1+4⇣4

◆Figure: Bode plots showing gainadjustment for a desired �M




Procedure

1. Draw the Bode magnitude &phase plots for a convenientvalue of gain

2. Determine the required �M

from the %OS

Figure: Bode plots showing gainadjustment for a desired �M




Procedure

3. Find the frequency, !�M , onthe Bode phase diagram thatyields the desired �M

4. Change the gain by an amountto force the magnitude curve togo through 0 dB at !�M . Theamount of gain adjustment isthe additional gain needed toproduce the required �M

Figure: Bode plots showing gainadjustment for a desired �M



Example, [1, p. 628]

Example (TR design via gain adjustment)

IProblem: For the position control system, find the value of thepreamplifier gain, K, to yield %OS = 9.5% in the TR for a step input

ISolution: On the board

Figure: System


11 Design via FR 11.3 Lag compensation




Visualizing lag compensation, [1, p. 630]

Concept

I Improve the staticerror constant by "only thelow-frequency gainwithout anyresulting instability

I " �M of thesystem to yield thedesired TR

Figure: Visualizing lag compensation




Procedure

1. Set the gain, K, tothe value thatsatisfies thesteady-state errorspecification andplot the Bodemagnitude andphase diagrams forthis value of gain





Procedure

2. Find the frequencywhere �M is 5� to12� greater thanthe �M that yieldsthe desired TR.This compensatesfor the fact thatthe phase of thelag compensatormay still contributeanywhere from�5� to �12� ofphase at !�M





Procedure

3. Select a lagcompensatorwhose magnituderesponse yields acomposite Bodemagnitude diagramthat goes through0 dB at thefrequency found inStep 2





Procedure

3.1 Draw thecompensator’shigh-frequencyasymptote to yield0 dB for thecompensatedsystem at thefrequency found inStep 2





Procedure

3.2 Select the upperbreak frequency tobe 1 decade belowthe frequencyfound in Step 2





Procedure

3.3 Select thelow-frequencyasymptote to be at0 dB

3.4 Connect thecompensator’shigh- &low-frequencyasymptotes with a�20 dB/decadeline to locate thelower breakfrequency Figure: Visualizing lag compensation




Procedure

4. Reset the systemgain, K, tocompensate forany attenuation inthe lag network inorder to keep thestatic errorconstant the sameas that found inStep 1





Result

I Lag compensatorTF

GC(s) =s+ 1

T

s+ 1↵T

I ↵ > 1





Example (Lag compensation design)

IProblem: Use Bode diagrams to design a lag compensator to yield atenfold improvement in steady-state error over the gain-compensatedsystem while keeping %OS = 9.5%


Figure: System


11 Design via FR 11.4 Lead compensation




Visualizing lead compensation, [1, p. 635]

Concept

I Change the phase diagram

I " gain crossover / " bandwidth

I " �M / # %OS

I " �M / # Tp

I Implement a steady-state errorrequirement ! design a TR

Figure: Visualizing lead compensation




Concept

I Lead compensator TF

GC(s) =1

�

s+ 1T

s+ 1�T

I � > 1

I Frequency at maximum phaseshift angle

!max

=1

Tp� Figure: FR of a lead compensator for

various values of �




Concept

I Maximum phase shift angle

�max

= tan�1⇣1��2p�

⌘

= sin�1⇣1��1+�

⌘

I Magnitude at maximum phaseshift angle

|G(j!max

)| = 1p�

Figure: FR of a lead compensator forvarious values of �




Procedure

1. Find the CL bandwidth required to meet a Ts, Tp, or Tr requirement[1, p. 582]

!BW = !n

q(1� 2⇣2) +

p4⇣4 � 4⇣2 + 2

!n =4

Ts⇣and !n =

⇡

Tp

p1� ⇣2

2. Since the lead compensator has negligible e↵ect at low frequencies,set the gain, K, of the uncompensated system to the value thatsatisfies the steady-state error requirement

3. Plot the Bode diagrams for this value of gain and determine theuncompensated system’s �M




Procedure

4. Find the �M to meet ⇣ or %OS requirement. Then evaluate theadditional phase contribution required from the compensator.

5. Determine the value of � from the lead compensator’s required phasecontribution

GC(s) =1

�

s+ 1T

s+ 1�T

�max

= tan�1⇣1��2p�

⌘= sin�1

⇣1��1+�

⌘

6. Determine the compensator’s magnitude at the peak of the phasecurve

|G(j!max

)| = 1p�




Procedure

7. Determine the new !�M by finding where the uncompensatedsystem’s magnitude curve is the negative of the lead compensator’smagnitude at the peak of the compensator’s phase curve

8. Design the lead compensator’s break frequencies to find T and thebreak frequencies

GC(s) =1

�

s+ 1T

s+ 1�T

!max

=1

Tp�

9. Reset the system gain to compensate for the lead compensator’s gain

10. Check the bandwidth to be sure the speed requirement has been met

11. Simulate to be sure all requirements are met

12. Redesign if necessary to meet requirements




Example (Lead compensation design)

IProblem: Design a lead compensator to yield %OS = 20%,KV = 40, & Tp = 0.1 second


Figure: System


11 Design via FR 11.5 Lag-lead compensation




Intro, [1, p. 641]

Concept

I What we are not doing: separate lag & lead compensators

1. Design a lag compensator to lower the high-frequency gain, stabilizethe system, & improve the steady-state error

2. Design a lead compensator to meet the phase-margin requirements

I What we are doing: passive lag-lead networkI Eliminates the bu↵er amplifier that separates the lag network from the

lead network



Visualizing lag-lead compensation, [1, p. 641]

Concept

I Lag-lead passive compensator TF

GC(s) = GLead

(s)GLag

(s) =

✓s+ 1

TLead

s+ �TLead

◆✓s+ 1

TLag

s+ 1�T

Lag

◆

I 1st term in parentheses: lead compensatorI 2nd term in parentheses: lag compensatorI � replaces � & ↵ of lead & lag networks, respectively

GC(s) =1

�

s+ 1T

s+ 1�T

GC(s) =s+ 1

T

s+ 1↵T

I � & ↵ must be reciprocals of each otherI � > 1 & ↵ > 1 ! � > 1




Procedure

1. Using a 2nd-order approximation, find the CL bandwidth required tomeet Ts, Tp, or Tr [1, p. 582]

2. Set the gain, K, to the value required by the steady-state errorspecification

3. Plot the Bode diagrams for this value of gain

4. Using a 2nd-order approximation, calculate the �M to meet the ⇣ or%OS requirement [1, p. 590]

5. Select a new !�M near !BW

6. At the new !�M , determine the additional amount of phase leadrequired to meet the �M requirement. Add a small contribution thatwill be required after the addition of the lag compensator.




Procedure

7. Design the lag compensator by selecting the higher break frequencyone decade below the new !�M . The design of the lag compensatoris not critical, and any design for the proper �M will be relegated tothe lead compensator. The lag compensator simply providesstabilization of the system with the gain required for the steady-stateerror specification. Find the value of � from the lead compensator’srequirements. Using the phase required from the lead compensator,the phase response curve can be used to find the value of � = ��1.This value, along with the previously found lag’s upper breakfrequency, allows us to find the lag’s lower break frequency.




Procedure

8. Design the lead compensator. Using the value of � from the lagcompensator design and the value assumed for the new !�M , find thelower- and upper-break frequency for the lead compensator.

!max

=1

Tp�

9. Check the bandwidth to be sure the speed requirement has been met

10. Redesign if �M or TR specifications are not met, as shown byanalysis or simulation




Example (Lag-lead compensation design)

IProblem: Design a passive lag-lead compensator using Bode diagramsto yield %OS = 13.25%, Tp = 2 seconds, & Kv = 12

G(s) =K

s(s+ 1)(s+ 4)




Bibliography

Norman S. Nise. Control Systems Engineering, 2011.


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