class wk14 16 frequency design - egr.msu.edu · – controller design on bode plot is simple. •...

2009 Spring ME451 - GGZ Week 14-15: Frequency Design

Control Design based upon

• RL (Root Locus)

• Frequency Response Design– Bode Diagram (BD)

– Nyquist Approach

Control Design (CD) Control Design (CD) –– Control Design ProcessControl Design Process

1. Modeling

Mathematical modelMathematical model

2. Analysis

ControllerController

3. Design/Synthesis

4. Implementation

PlantPlantOutputOutput

ActuatorActuatorControllerController

SensorSensor

InputInput

Control Method to be discussed:

• PID control

• Gain compensation

• Phase lead/lag compensation

• Lead/lag compensation


-

tt--domain:domain:

ss--domain:domain:

CD CD –– PID ControllerPID Controller

)(ty)(sP

)(tr

sKi/

pK

sKd

4342143421321

DerivativeIntegal

0alProportion

)()()()(

dt

tdeKdeKteKtu

d

t

ip++= ∫ ττ

++=++= sK

sKKsK

s

KKsC

D

I

pd

i

p

11)(

)(te)(tu


• Most popular in process and robotics industries

– Good performance

– Functional simplicity (Operators can easily tune.)

• To avoid high frequency noise amplification, derivative term

is implemented as

with τd much smaller than plant time constant.

• PI controller

• PD controller

CD CD –– PID Controller RemarksPID Controller Remarks

1+≈

s

sKsK

d

d

dτ

s

KKsC i

p+=)(

sKKsCdp

+=)(


• We plot y(t) for step reference r(t) with

– P controller

– PI controller

– PID controller

CD CD –– A Simple Example A Simple Example (1)(1)

-

)(ty)(sP

)(tr

sKi/

pK

sKd

)(te)(tu

3)1(

1)(

+=

ssP


• Simple

• Steady state error

– Higher gain gives

smaller error

• Stability


faster and more

oscillatory response

0 5 10 15 200

0.2

0.4

0.6

0.8

1

1.2

1.4

CD CD –– A Simple Example A Simple Example (P Controller (2))(P Controller (2))

pKsC =)(

1=p

K

2=p

K

5=p

K


0 5 10 15 200

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

• Zero steady state error

(provided that CL is

stable.)

• Stability


faster and more

oscillatory response

CD CD –– A Simple Example A Simple Example (PI Controller (3))(PI Controller (3))

s

KKsC i

p+=)(1=

pK

1=iK

5.0=iK

2.0=i

K

0=iK


0 5 10 15 200

0.5

1

1.5

2

• Zero steady state error

(due to integral control)

• Stability


more damped

response

• Too high gain worsen

performance.

CD CD –– A Simple Example A Simple Example (PID Controller (4))(PID Controller (4))

sKs

KKsC

d

i

p++=)(5.1,5.2 ==

ipKK

0

1

2

4

=

=

=

=

d

d

d

d

K

K

K

K


• Model-based

– Root locus

– Frequency response approach

– Useful only when a model is available

– Necessary if a system has to work at the first trial

• Empirical (without model)

– Ziegler-Nichols tuning rule (1942)

– Simple

– Useful even if a system is too complex to model

– Useful only when trial-and-error tuning is allowed

CD CD –– How to Turn PID ParametersHow to Turn PID Parameters


• Step response method (for only stable systems)

tt

OpenOpen--loop step responseloop step response

Steepest tangentSteepest tangent

PID parametersPID parameters

CD CD –– ZieglerZiegler--Nichols PID Tuning Rules Nichols PID Tuning Rules (1)(1)

)1

1()( sTsT

KsCd

I

p++=

)(ty

τ

α− ττα

τα

α

5.02/2.1

3/9.0

/1

Type

PID

PI

P

TTKDIP


• Ultimate sensitivity method

ClosedClosed--loop step responseloop step response

with a gain controller with a gain controller

Increase gain and find Increase gain and find KKcc

generating oscillation generating oscillation

(marginally stable case).(marginally stable case).

PID parametersPID parameters

CD CD –– ZieglerZiegler--Nichols PID Tuning Rules Nichols PID Tuning Rules (2)(2)

)1

1()( sTsT

KsCd

I

p++=

CCC

CC

C

DIP

TTKPID

TKPI

KP

TTK

125.05.06.0

8.04.0

5.0

Type

)(ty

CT

t


0 2 4 6 8 10 12 14 16 18 200

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

• Step response method • Ultimate sensitivity

0 2 4 6 8 10 12 14 16 18 200

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

PP

PIPI

PIDPID

PP

PIPI

PIDPID

CD CD –– A Simple Example A Simple Example (Revisited (5))(Revisited (5))


0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

0.790.79--0.280.28

CD CD –– OL Step Response for OL Step Response for “Step Response method”“Step Response method”


0 5 100

0.5

1

0 5 100

0.5

1

0 5 100

1

2

0 5 100

0.5

1

1.5

CD CD –– CL Step Responses forCL Step Responses for “Ultimate Sensitivity method”“Ultimate Sensitivity method”

1=P

K

4=P

K

2=P

K

8=P

K 8=C

K

7.3=C

T


• One example: Using OP amp

--

++--

++

Exercise: Derive this!Exercise: Derive this!

CD CD –– PID Controller Realization PID Controller Realization

++

+−

sCRsCR

C

C

R

R

21

12

2

1

1

2 1

)(tvi )(tv

o

1C 2

C

2R

1R

3R

4R


• PID control

– Most popular controller in industry

– Model-free methods for design are available.

– Simple controller structure

– Simple controller tuning

– Widely applicable

• Ziegler-Nichols tuning rules provide a starting point for

fine tuning, rather than final settings of controller

parameters in a single shot.

CD CD –– PID Control Summary and Exercise PID Control Summary and Exercise


• Z: # of CL poles in open RHP

• P: # of OL poles in open RHP (given)

• N: # of clockwise encirclement around -1

by Nyquist plot of OL transfer function L(s)

(counted by using Nyquist plot of L(s))

Remark: N = -1: a counter-clockwise encirclement

CD CD –– Nyquist Stability Criterion (Review)Nyquist Stability Criterion (Review)

0: stable is system CL =+=⇔ NPZ


• IF P=0 (i.e., if L(s) has no pole in open RHP or stable)

This fact is very important since openThis fact is very important since open--loop systems loop systems

in many practical problems have no pole in open RHP!in many practical problems have no pole in open RHP!

CD CD –– Nyquist Stability Criterion: A Special CaseNyquist Stability Criterion: A Special Case

0stable is system CL =+=⇔ NPZ:

0stable is system CL =⇔ N


ReRe

ImIm

ReRe

ImIm

CL stableCL stable CL unstableCL unstable

CD CD –– Examples with P = 0 (stable OL system) Examples with P = 0 (stable OL system)

)( ωjL )( ωjL


• Nyquist stability criterion gives not only absolute but also

relative stability.

– Absolute stability: Is the closed-loop system stable or

not? (Answer is yes or no.)

– Relative stability: How “much” is the closed-loop

system stable? (Margin of safety)

• Relative stability is important because a math model is

never accurate.

• How to measure relative stability?

– Use a “distance” from the critical point -1.

– Gain margin (GM) & Phase margin (PM)

CD CD –– Nyquist Stability Remarks Nyquist Stability Remarks


• Phase crossover

frequency ωp:

• Gain margin (in dB)

• Indicates how much OL

gain can be multiplied

without violating CL

stability.

Nyquist plot of L(s)Nyquist plot of L(s)

CD CD –– Nyquist Gain Margin (GM) Nyquist Gain Margin (GM)

180)( −=∠p

jL ω

)(

1log20 10

pjLGM

ω=


ReRe

ImIm

ReRe

ImIm

CD CD –– GM Examples GM Examples

dB6)(

1log20

2

10≈=

43421p

jLGM

ω

2

1)( −=

pjL ω

)( ωjL )( ωjL

3

1)( −=

pjL ω

dB5.9)(

1log20

3

10 ≈=

43421p

jLGM

ω


Same gain margin,Same gain margin,

but different relative stabilitybut different relative stability

Gain margin is often inadequateGain margin is often inadequate

to indicate relative stabilityto indicate relative stability

Phase margin!Phase margin!

CD CD –– Why GM Alone is Inadequate Why GM Alone is Inadequate


• Gain crossover

frequency ωg:

• Phase margin

• Indicates how much OL

phase can be added

without violating CL

stability.

Nyquist plot of L(s)Nyquist plot of L(s)

CD CD –– Nyquist Phase Margin (PM) Nyquist Phase Margin (PM)

1)( =∠g

jL ω

o

gjLPM 180)( +∠= ω


CD CD –– PM Example PM Example

)50)(5(

2500)(

++=

ssssL

dB80.14182.0

1log20

10==GM


• Advantages

– Nyquist plot can be used for study of closed-loop

stability, for open loop systems which is unstable and

includes time-delay.

• Disadvantage

– Controller design on Nyquist plot is difficult.

(Controller design on Bode plot is much simpler.)

We translate GM and PM on Nyquist plot We translate GM and PM on Nyquist plot

into those in Bode plot!into those in Bode plot!

CD CD –– Nyquist Plot Remarks Nyquist Plot Remarks


ωωgg

ωωpp

GMGM

PMPM

CD CD –– Bode Diagram Relative Stability Bode Diagram Relative Stability

)( ωjL∠

)(log2010

ωjL

ω

ω


• Advantages

– Without computer, Bode plot can be sketched easily.

– GM, PM, crossover frequencies are easily determined

on Bode plot.

– Controller design on Bode plot is simple.

• Disadvantage

– If OL system is unstable, we cannot use Bode plot for

stability analysis.

CD CD –– Bode Diagram Remarks Bode Diagram Remarks


100

101

102

103

-100

-50

0

100

101

102

103

-180

-270

ωωgg ωωpp

GMGM

PMPM

CD CD –– Bode Diagram Example Bode Diagram Example

)50)(5(

2500)(

++=

ssssL


10-1

100

-20

0

20

10-1

100

-180

-90

PMPM

GMGM

Time delay reducesTime delay reduces

relative stability!relative stability!

Delay timeDelay time

CD CD –– Bode Diagram Relative Stability Bode Diagram Relative Stability (time Delay)(time Delay)

)2)(1(

1)(

++=

ssssL

)2)(1()(

++=

−

sss

esL

s

lag phase deg 3.57 dTω


1 0-2

1 0-1

1 00

1 01

1 02

-1

-0 .5

0

0 .5

1

1 0-2

1 0-1

1 00

1 01

1 02

-6 0 0 0

-4 0 0 0

-2 0 0 0

0

• TF

Huge phase lag!Huge phase lag!

As can be explained with Nyquist stability criterion, As can be explained with Nyquist stability criterion,

this phase lag causes instability of the closedthis phase lag causes instability of the closed--loop system,loop system,

and hence, the difficulty in control.and hence, the difficulty in control.

(rad) )( , ,1)()( TjGjGesGTs ωωωω −=∠∀=⇒= −

CD CD –– Body Diagram of A Time DelayBody Diagram of A Time Delay


ωωggωωpp

GMGM

PMPM

CD CD –– Body Diagram Unstable CL Case Body Diagram Unstable CL Case

ω

ω

)( ωjL∠

)(log2010

ωjL


• Relative stability: Closeness of Nyquist plot to the critical

point -1

– Gain margin, phase crossover frequency

– Phase margin, gain crossover frequency

• Relative stability on Bode plot

• We normally emphasize PM in controller design.

CD CD –– Body Diagram Summary and Exercises Body Diagram Summary and Exercises


Design specifications in time domainDesign specifications in time domain

(Rise time, settling time, overshoot, steady state error, etc.)(Rise time, settling time, overshoot, steady state error, etc.)

Desired closedDesired closed--loop loop

pole location pole location

in sin s--domaindomain

Constraints on openConstraints on open--loop loop

frequency response frequency response

in sin s--domaindomain

Root locus shapingRoot locus shaping Frequency response shapingFrequency response shaping

(Loop shaping)(Loop shaping)

Approximate translationApproximate translation

CD CD –– Control Design Comparison Control Design Comparison


• Given G(s), design C(s) that satisfies time domain specs,

such as stability, transient, and steady-state responses.

• We learn typical qualitative relationships between open-

loop Bode plot and time-domain specifications.

PlantPlantControllerController

OL:OL:

CL:CL:

CD CD –– Feedback Control System Design Feedback Control System Design

)(ty)(tr)(sG)(sC

)()(:)( sCsGsL =

)(1

)(:)(

sL

sLsT

+=


SteadySteady--state accuracystate accuracy

SensitivitySensitivity

Disturbance rejectionDisturbance rejection

Noise Noise

reductionreduction

TransientTransient

Response speedResponse speed

Transient Transient

OvershootOvershootRelative stabilityRelative stability

Relative stabilityRelative stabilityNoise Noise

reductionreduction

CD CD –– Typical Desired OL Body DiagramTypical Desired OL Body Diagram

)( ωjL∠

)(log2010

ωjL


For steadyFor steady--state accuracy, state accuracy,

L should have high gain at low frequencies.L should have high gain at low frequencies.

yy((tt)) tracks tracks rr((tt)) composed of composed of

low frequencies very well.low frequencies very well.

CD CD –– Steady State Accuracy Steady State Accuracy (1)(1)


OL:OL:

CL:CL:

)(ty)(tr)(sG)(sC

)()(:)( sCsGsL =

)(1

)(:)(

sL

sLsT

+=

)( Large ωjL)(log2010

ωjL

1)(1

)()( ≈

+=

ω

ωω

jL

jLjY


• Step r(t)

Increase

• Ramp r(t)

Increase • Parabolic r(t)

Increase

For Kv to be nonzero,For Kv to be nonzero,

L must contain L must contain

at least one integrator.at least one integrator.

For Ka to be nonzero,For Ka to be nonzero,

L must contain L must contain

at least two integrators.at least two integrators.

<<--2020 <<--4040

CD CD –– Steady State Accuracy (2)Steady State Accuracy (2)

)0(: LKp

=

)(log2010

ωjL )(log2010

ωjL )(log2010

ωjL

ω ω ω

)(lim:0

ssLKs

v→

= )(lim: 2

0sLsK

sa

→=


Noise Noise

reductionreduction

TransientTransient


Transient Transient



reductionreduction




CD CD –– Typical Desired OL Body Diagram Typical Desired OL Body Diagram (Revisited)(Revisited)

)( ωjL∠

)(log2010

ωjL


• For illustration, we use the feedback system:

CD CD –– A 2A 2ndnd Order System ExampleOrder System Example


)(ty)(tr)(sG)(sC

)2()()(:)(

2

n

n

sssCsGsL

ςω

ω

+==

22

2

2)(1

)(:)(

nn

n

sssL

sLsT

ωςω

ω

++=

+=


0 5 10 150

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

For small percent overshoot, For small percent overshoot,

L should have larger phase margin.L should have larger phase margin.

10-1

100

101

-20

0

20

10-1

100

101

-180

-160

-140

-120

-100

CL step responseCL step responseOL Bode plotOL Bode plot

PMPM

CD CD –– Percent OvershootPercent Overshoot

1=n

ω5.0

3.0

1.0

=

=

=

ς

ς

ς


Noise Noise

reductionreduction

TransientTransient


Transient Transient



reductionreduction





)( ωjL∠

)(log2010

ωjL


0 5 10 150

0.2

0.4

0.6

0.8

1

1.2

1.4

100

-20

0

20

10-1

100

101

-180

-160

-140

-120

-100

For fast response, For fast response,

L should have larger gain crossover frequency.L should have larger gain crossover frequency.

CL step responseCL step responseOL Bode plotOL Bode plot

CD CD –– Response SpeedResponse Speed

3.0=ς3

2

1

=

=

=

n

n

n

ω

ω

ω


Noise Noise

reductionreduction

TransientTransient


Transient Transient



reductionreduction





)( ωjL∠

)(log2010

ωjL

ω

ω

ω


• We require adequate GM and PM for:

– safety against inaccuracies in modeling

– reasonable transient response

• It is difficult to give reasonable numbers of GM and PM for

general cases, but usually,

– GM should be at least 6dB

– PM should be at least 45deg

(These values are not absolute but approximate!)

• In controller design, we are especially interested in PM

(which typically gives good GM).

CD CD –– Relative StabilityRelative Stability


Noise Noise

reductionreduction

TransientTransient


Transient Transient



reductionreduction





)( ωjL∠

)(log2010

ωjL

ω

ω


yy((tt)) is not affected by is not affected by nn((tt))

composed of high frequencies.composed of high frequencies.

noisenoise

For noise rejection, For noise rejection,

L should have small gain at high frequencies.L should have small gain at high frequencies.

CD CD –– Noise RejectionNoise Rejection


)(ty)(tr)(sG)(sC

)(tn

)( Small ωjL)(log2010

ωjL

0)(1

)(

)(

)(≈

+−=

ω

ω

ω

ω

jL

jL

jN

jY

ω


• Reshape Bode plot of G(jω) into a

“desired” shape of

by a series connection of

appropriate C(s).

CD CD –– Frequency Shaping (Loop Shaping)Frequency Shaping (Loop Shaping)

noisenoisePlantPlantControllerController

)(ty)(tr)(sG)(sC

)(tn

)(tddisturbancedisturbance

)()(:)( ωωω jCjGjL =

ω

ω


• Bode plot of a series connection G1(s)G2(s) is the addition of

each Bode plot of G1 and G2.

– Gain

– Phase

• We use this property to design C(s) so that G(s)C(s) has a

“desired” shape of Bode plot.

CD CD –– Advantages of Body DiagramAdvantages of Body Diagram

)(log20)(log20)()(log202101102110

ωωωω jGjGjGjG +=

)()()()(1121

ωωωω jGjGjGjG ∠+∠=∠


• We use simple controllers for shaping.

– Gain

– Lead and lag compensators

CD CD –– Simple ControllersSimple Controllers

noisenoiseStable PlantStable PlantControllerController

)(ty)(tr)(sG)(sC

)(tn

)(tddisturbancedisturbance

KsC =)(

1

1

poly.)order -(1st

poly.)order -(1st)(

+

+==

ps

zs

sC


dBdB

degdeg

CD CD –– Bode Plot of a GainBode Plot of a Gain

KsC =)(

K10

log20


10-2

10-1

100

101

102

103

-100

-50

0

50

100

10-2

10-1

100

101

102

103

-180

-160

-140

-120

-100

In case of In case of K K > 1,> 1,

�� Gain increases Gain increases

uniformly, but phase does uniformly, but phase does

not change.not change.

�� Typically, Typically,

�� (Steady state) (Steady state) LL(0)(0)

�� (Speed) (Speed) ωωgg

�� (Stability & (Stability &

overshoot) PMovershoot) PM PMPM

CD CD –– Effect of a Gain C(s) of L(s) Effect of a Gain C(s) of L(s)

)0()( >= KsC

)(10)( sGsL =

)25(

2500)(

+=

sssG


10-2

10-1

100

101

102

103

-20

-15

-10

-5

0

10-2

10-1

100

101

102

103

-60

-40

-20

0

10-2

10-1

100

101

102

103

0

5

10

15

20

10-2

10-1

100

101

102

103

0

20

40

60

LeadLead compensatorcompensator LagLag compensatorcompensator

PHASE LEADPHASE LEAD PHASE LAGPHASE LAG

MEMORIZE THESE MEMORIZE THESE SHAPES!!!SHAPES!!!

CD CD –– Bode Diagrams of a Lead and Lag C(s)Bode Diagrams of a Lead and Lag C(s)

pz <

1

1)(

+

+=

ps

zs

sC

zp <


Noise Noise

reductionreduction

Relative stabilityRelative stability

Gain+lagGain+lag

LeadLead

LagLag

LeadLead

CD CD –– Guideline of a Lead and Lag DesignGuideline of a Lead and Lag Design

)( ωjL∠

)(log2010

ωjL

ω

ω


Destabilizing effectDestabilizing effect

•• Decreasing Decreasing ωωωωωωωωgg

Select z much (at least 1 Select z much (at least 1

decade) less than decade) less than ωωgg

10-2

10-1

100

101

102

103

-20

-15

-10

-5

0

10-2

10-1

100

101

102

103

-60

-40

-20

0

CD CD –– Effect of a Lag C(s) on L(s) Effect of a Lag C(s) on L(s)

p

z10

log20


30/1

3/1

1

110)(

s

s

LagsC

+

+=

10-2

10-1

100

101

102

103

-100

-50

0

50

100

10-2

10-1

100

101

102

103

-180

-160

-140

-120

-100

PM: 28 deg at PM: 28 deg at

ωωgg=47 rad/s=47 rad/s

PMPMPM: 27 deg at PM: 27 deg at


CD CD –– Lag + Gain C(s) DesignLag + Gain C(s) Design

)(sG)25(

2500)(

+=

sssG

)()( sCsGLag

)(sG

)()( sCsGLag


Noise Noise

reductionreduction


Gain+lagGain+lag

LeadLead

LagLag

LeadLead

CD CD –– Guideline of a Lead and Lag Design Guideline of a Lead and Lag Design (revisited)(revisited)

)( ωjL∠

)(log2010

ωjL

ω

ω


10-2

10-1

100

101

102

103

0

5

10

15

20

10-2

10-1

100

101

102

103

0

20

40

60

Stabilizing effectStabilizing effect

Increasing Increasing ωωωωωωωωgg

Select z&p around Select z&p around ωωgg

CD CD –– Effect of a Lead C(s) on L(s) Effect of a Lead C(s) on L(s)


101

102

103

-60

-40

-20

0

20

101

102

103

-180

-160

-140

-120

-100





CD CD –– Example of a Lead DesignExample of a Lead Design

)(sG

)()( sCsGLead

)()( sCsGLead

)(sG

1.94

21.38

1

1)(

s

s

LeadsC

+

+=

)25(

2500)(

+=

sssG


10-2

10-1

100

101

102

103

0

5

10

15

20

10-2

10-1

100

101

102

103

-60

-40

-20

0

20

40

CD CD –– LeadLead--Lag Compensator Lag Compensator

321321Lead

s

s

LagGain

s

s

sC1.94

21.38

30/1

3/1

1

1

1

110)(

+

+⋅

+

+⋅=

+


10-2

10-1

100

101

102

103

-100

-50

0

50

100

10-2

10-1

100

101

102

103

-180

-160

-140

-120

-100


ωωg g = 47 rad/s= 47 rad/s


ωωg g = 60 rad/s= 60 rad/s

CD CD –– Example of a LeadExample of a Lead--Lag DesignLag Design

)()( sCsG

)(sG

)(sG

)()( sCsG

)()()( sCsCsCLagLead

=

)25(

2500)(

+=

sssG


0 0.1 0.2 0.3 0.4 0.50

0.5

1

1.5

Uncompensated (Uncompensated (C(s) =C(s) =1)1)

LeadLead--lag compensatedlag compensated

Less overshoot is due to larger PM.Less overshoot is due to larger PM.

Faster response is due to larger wFaster response is due to larger wgg. .

CD CD –– Step ResponsesStep Responses


• Frequency shaping (Loop shaping) on Bode plot

• Effect of lead, lag, and lead-lag compensators

• Qualitative explanation

• In actual design, one needs to use Matlab.

• Next, more detail about

– Lag design

– Lead design

CD CD –– Loop Shaping SummaryLoop Shaping Summary





Noise Noise

reductionreduction

TransientTransient


Transient Transient


Relative stabilityRelative stabilityNoise reductionNoise reduction

• Next, frequency shaping (loop shaping) design

CD CD –– SummarySummary

)( ωjL∠

)(log2010

ωjL

ω

ω

ω


CD CD –– Body Diagram of a Lead/Lag C(s) (Review)Body Diagram of a Lead/Lag C(s) (Review)

10-2

10-1

100

101

102

103

-20

-15

-10

-5

0

10-2

10-1

100

101

102

103

-60

-40

-20

0

10-2

10-1

100

101

102

103

0

5

10

15

20

10-2

10-1

100

101

102

103

0

20

40

60

LeadLead compensatorcompensator LagLag compensatorcompensator

PHASE LEADPHASE LEAD PHASE LAGPHASE LAG

MEMORIZE THESE MEMORIZE THESE SHAPES!!!SHAPES!!!

pz <

1

1)(

+

+=

ps

zs

sC

zp <


dBdB +20+20

degdeg+45+45

dBdB

--2020

degdeg

--4545

Lead (z < p)Lead (z < p)

Lag (p < z)Lag (p < z)

dBdB+20+20

degdeg

+45+45

dBdB

--2020

degdeg

--4545

--4545 +45+45

CD CD –– StraightStraight--Line ApproximationsLine Approximations

1

1)(

+

+=

ps

zs

sG

1)( +=z

ssG

1

1)(

+=

ps

sG


• Design C(s) so that L(jω):=G(jω)C(jω) has a desired

shape.

• We study the design of simple compensators:

– Gain compensator (Today)

– Lag compensator (Today)

– Lead compensator (Next lecture)

CD CD –– Frequency Shaping (Loop Shaping)Frequency Shaping (Loop Shaping)

Stable PlantStable PlantControllerController

)(tr)(sG)(sC


Noise Noise

reductionreduction


Gain+lagGain+lag

LeadLead

LagLag

LeadLead

CD CD –– Guideline of LeadGuideline of Lead--Lag Design (Review)Lag Design (Review)

)( ωjL∠

)(log2010

ωjL

ω

ω


• Consider a system

• Analysis for C(s) = 1

– Stable

– PM at least 12 deg

– GM at least 3.5 dBThese values are too These values are too

small for good small for good

transient response!transient response!

10-2

10-1

100

101

102

-100

-50

0

50

10-2

10-1

100

101

102

-250

-200

-150

-100

CD CD –– An Example (LeadAn Example (Lead--Lag Design)Lag Design)


)(tr)(sG)(sC

)(ty

)2)(1(

4)(

++=

ssssG


• PM is specified to be 50 deg.

• In this example, to increase PM by gain compensation,

we need to lower the gain curve.

CD CD –– Gain Margin Compensation Gain Margin Compensation (Example (2))(Example (2))


10-2

10-1

100

101

102

-100

-50

0

50

10-2

10-1

100

101

102

-250

-200

-150

-100Uncompensated (C(s)=1)Uncompensated (C(s)=1)

Gain compensatedGain compensated

Low freq. gain Low freq. gain

decreases.decreases.

CD CD –– Bode Diagram for C(s) = 0.286 Bode Diagram for C(s) = 0.286 (Example (3))(Example (3))


0 5 10 150

0.2

0.4

0.6

0.8

1

1.2

1.4

K=0.455 (PM=35deg)K=0.455 (PM=35deg)

K=0.286 (PM=50deg)K=0.286 (PM=50deg)

K=0.158 (PM=65deg)K=0.158 (PM=65deg)

CD CD –– Step Responses Step Responses (Example (4))(Example (4))


dBdB

degdeg

--4545

--2020

10-2

10-1

100

101

102

103

-20

-15

-10

-5

0

10-2

10-1

100

101

102

103

-60

-40

-20

0

CD CD –– PhasePhase--Lag Compensator Lag Compensator (Review)(Review)

zpsGps

zs

<<+

+= 0,

1

1)(

z

p10log20


Step 1: To satisfy low frequency requirement, adjust DC gain of OL system by a constant gain K.

• Analysis for C(s) = 1

– Stable

– PM at least 12 deg

– GM at least 3.5 dB

We try to design phaseWe try to design phase--lag C(s) which giveslag C(s) which gives

•• PM 50degPM 50deg

•• Low frequency gain same as the original plant.Low frequency gain same as the original plant.

CD CD –– PhasePhase--Lag Compensator C(s) DesignLag Compensator C(s) Design


)(tr)(sG)(sC

)2)(1(

4)(

++=

ssssG


10-2

10-1

100

101

102

-100

-50

0

50

10-2

10-1

100

101

102

-250

-200

-150

-100

OKOK

CD CD –– PhasePhase--Lag Design Step 1 Lag Design Step 1 (C(s) = 1)(C(s) = 1)

1 with )( =KsKG


Step 2: Find the frequency ωg (which will become gain

crossover frequency after compensation) where

In this example,

Note: The reason of +5 deg is explained later.


4.0=g

ω

PM required : ,5180)(m

o

m

o

gjG φφω ++−=∠

{oooo

g

m

jG 125550180)( −=++−=∠φ

ω


10-2

10-1

100

101

102

-100

-50

0

50

10-2

10-1

100

101

102

-250

-200

-150

-100

PM=55PM=55


)(sKG

gω


Step 3:

dBdB

--2020

degdeg

--4545

For small phase lag at For small phase lag at ωωgg

For setting new gain crossover at For setting new gain crossover at ωωgg


)55.4

04.0(

)(

1.0==

g

g

jKGp

ω

ω

)04.0(1.0 ==g

z ω

)(

1log20log20

1010

gjKGz

p

ω=


10-2

10-1

100

101

102

-100

-50

0

50

10-2

10-1

100

101

102

-250

-200

-150

-100

PM=50PM=50

CD CD –– PhasePhase--Lag Design Step 3 Lag Design Step 3 (C(s) = C(C(s) = Claglag(s))(s))

)(sKG

1

1)(

0088.0

04.0

+

+=

s

s

LagsC

)()( sCsKGLag


0 5 10 150

0.5

1

1.5

2

C(s)=0.286 (PM=50deg, C(s)=0.286 (PM=50deg, ωωωωωωωωgg=0.5)=0.5)

C(s)=CC(s)=CLagLag(s) (PM=52.3deg, (s) (PM=52.3deg, ωωωωωωωωgg=0.4)=0.4)

Small overshoot is due to larger PM.Small overshoot is due to larger PM.

Slower response is due to smaller wSlower response is due to smaller wgg. .

C(s)=1C(s)=1 (PM=12deg, (PM=12deg, ωωωωωωωωgg=1.1)=1.1)

CD CD –– PhasePhase--Lag Design Step ResponsesLag Design Step Responses


• Gain controller design in Bode plot

– Gain changes uniformly over frequencies.

– Phase does not change.

• Lag compensator design in Bode plot

– Lag compensator can be used for

• Improving PM by maintaining low freq. gain, or

• Improving low freq. gain by maintaining PM

• Low freq. gain determines steady state error, disturbance

rejection, while PM does overshoot.

• Next, lead compensator design

CD CD –– PhasePhase--Lag Design SummaryLag Design Summary


• What is problematic?

– Nyquist stability criterion says that, for closed-loop

stability, Nyquist plot of open-loop system must

encircle -1 point.

– It is hard to translate this condition into Bode plot.

• To use FR technique…

UnstableUnstableControllerController

Stabilize first!Stabilize first!

CD CD –– If G(s) has OL RHP PolesIf G(s) has OL RHP Poles

)(sG)(sC

)(sH


1. Select z near uncompensated ωg.

In the example, ωg = 1.14. So, select, for example, z = 1.

2. Select p > z by trial-and-error.

3. Check PM and settling time. If not satisfactory, move the

pole p. If moving pole does not give the desired results,

try to move the zero z.

CD CD –– PhasePhase--Lead Design ProcedureLead Design Procedure


10-2

10-1

100

101

102

-100

-50

0

50

10-2

10-1

100

101

102

-250

-200

-150

-100

PM=50PM=50

CD CD –– PhasePhase--Lead Design ExampleLead Design Example

)(sG

1

1)(

50+

+=

sLead

ssC

)()( sCsGLead

57.1=g

ω


0 5 10 150

0.5

1

1.5

2C(s)=0.286 (PM=50deg, C(s)=0.286 (PM=50deg, ωωωωωωωωgg=0.5)=0.5)

C(s)=CC(s)=CLagLag(s) (PM=52.3deg, (s) (PM=52.3deg, ωωωωωωωωgg=0.4)=0.4)

C(s)=1C(s)=1 (PM=12deg, (PM=12deg, ωωωωωωωωgg=1.1)=1.1)

C(s)=CC(s)=CLeadLead(s) (PM=50deg, (s) (PM=50deg, ωωωωωωωωgg=1.6)=1.6)

Faster response is due to larger wFaster response is due to larger wgg..

(Compare crossover frequencies.) (Compare crossover frequencies.)

CD CD –– PhasePhase--Lead Design Step ResponsesLead Design Step Responses


69 69.2 69.4 69.6 69.8 7067

67.5

68

68.5

69

69.5

70

C(s)=0.286 (Kv=0.572)C(s)=0.286 (Kv=0.572)

Ramp referenceRamp reference

C(s)=CC(s)=CLagLag(s) (Kv=2)(s) (Kv=2)

C(s)=1C(s)=1 (Kv=2)(Kv=2)

C(s)=CC(s)=CLeadLead(s) (Kv=2)(s) (Kv=2)

CD CD –– PhasePhase--Lead Design Ramp ResponsesLead Design Ramp Responses


10-2

10-1

100

101

102

-100

-50

0

50

10-2

10-1

100

101

102

-250

-200

-150

-100

We recover the gain loss We recover the gain loss

at gain crossover freq.at gain crossover freq.

by adding a gain.by adding a gain.

CD CD –– Phase LeadPhase Lead--Lag Design Example (1)Lag Design Example (1)

)(sG

)()( sCsGLead

)()()( sCsCsGLagLead


10-2

10-1

100

101

102

-100

-50

0

50

10-2

10-1

100

101

102

-250

-200

-150

-100

PM=51PM=51

IncreasedIncreased

CD CD –– Phase LeadPhase Lead--Lag Design Example (2)Lag Design Example (2)

)(sG

)()(4)( sCsCsCLagLeadLL

=

)()( sCsGLL

43.1=g

ω


0 5 10 150

0.5

1

1.5

2

C(s)=0.286C(s)=0.286

C(s)=CC(s)=CLagLag(s)(s)

C(s)=1C(s)=1

C(s)=4CC(s)=4CLeadLead(s)C(s)CLagLag(s)(s)

CD CD –– Phase LeadPhase Lead--Lag Design Step ResponsesLag Design Step Responses


• Lead compensator can be used for improving

• Gain crossover frequency

• Phase margin

by maintaining low frequency gain,

• Lead-lag compensator can improve

– Transient (ωg for speed, PM for overshoot)

– Steady state (low frequency gain for error constant)

• Next, case studies

– Antenna azimuth position control

– Hard disk drive control

CD CD –– LeadLead--Lag Compensator SummaryLag Compensator Summary


• Characteristics

CD CD –– Design Example Design Example –– PD Controller (1)PD Controller (1)

KKKbKbsKsCDp

==+= , where),()(

When K = 1/b, we have

dB20

o90

b1.0 b10b

o0

dB0

φ

ω

ωφ

tan

,tan1

=

=−

b

b



)3)(1(

21)(

++=

ssssG

A Design Example (9.9)

71.23.5tan40

8

tan)(

81

===o

DsT φω

)(sG)(sC-

Design Target: %)2( sec)( 5.3 ,40 : ≤=sD

TPMoφ

Select crossover frequency (see text book page 373, Eq. 9-33):

0.3Select =cω

Bode Diagram (Blue Line next page):

)1)(1(

7)(

3++

=sss

sG

Step 1:

Step 2:



-100

-50

0

50

Ma

gnitu

de

(d

B)

10-1

100

101

102

-270

-225

-180

-135

-90

Pha

se

(d

eg)

Bode Diagram

Frequency (rad/sec)

Open Loop Bode Diagram (Blue Line):

3=c

ω

o8.206−o8.66

Phase Error

dB01.7Gain Error



ooo 8.661808.206 =+−=DMAX

Ph φ

Find Phase Margin at ωc:

284.1tan66.8

3

tan===

oφ

ωCb

8.206)( o−=COL

PM ω

PD Controller Compensate Phase to have desired PM:

444 3444 21Controller

)284.1()(

PD

sKsC +=

Step 3:

Step 4:



dBK 01.7log2010

−=

Plot Bode Diagram (Mag. only – Green) with K = 1.

And then find gain error:

2.242)()()( ==CCCGN

jLjCERR ωωω

Calculate the controller gain K:

)284.1(446.0)( += ssC

Step 5:

Step 6:

dB 7.01)( =CGN

ERR ω

)()( sLsC

446.010 20

01.7

==

−

K

Resulting PD controller:



Plot Bode Diagram (Red) with K = 0.446 results

o4.39=PM

3=Cω

)()( sLsC

)180 than lessdelay (Phase o∞=GM



)3)(1(

21)(

++=

ssssC

Design by Root Locus:

4.0100

40

100===

oo

Dφ

ς

)(sL)(sC-

Design Target: %)2( sec)( 5.3 ,40 : ≤=sD

TPMoφ

Select desired damp coefficient (text book page 328, Eq. 8-50):

Select desired frequency ωn:

5.34

=n

ςω

Step 1:

Step 2:

86.25.34.0

4=

⋅=

nω 86.23Select ≥=

nω



−−

+−=−±−=

)( 75.22.1

)( 75.22.11

2

12

C

C

nnCsj

sjjs ςωςω

Desired closed loop pole location:

ooo4434421 8.5629.38.9476.26.1133

21)75.22.1(

1

∠∠⋅∠=+−

Cs

jG

)3)(1(

21)()()()( Function Transfer Loop

G(s)

)( 44 344 2143421 +++==

sssbsKsGsCsL

sC

Calculate θPD:

To make closed loop pole (-1.2+2.75j) on the Root Locus

Step 3:

ooooo 2.858.568.946.113180 =+++−=PD

θ



43.12.12.85tan

75.2

tan

12

=+=+−

=on

PD

nb ςωθ

ςω

47.076.221

29.376.23=

⋅

⋅⋅=K

o

44344212.8576.275.22.1

1

∠=++− bj

Cs

Calculate b:

To make closed loop pole (-1.2+2.75j) on the Root Locus

Step 4:

129.376.23

2176.2)()()(

)(

)(

111

1

1

=⋅⋅

⋅==4434421

321

C

C

sG

sC

CCCKsGsCsL

Step 5:

Calculate K:

)43.1(47.0)( += ssC



Bode Diagram Design

Root Locus Design

)43.1(47.0)( += ssC

)284.1(446.0)( += ssC

-3 -2.5 -2 -1.5 -1 -0.5 0-5

-4

-3

-2

-1

0

1

2

3

4

5

0.84

0.050.110.170.240.340.46

0.62

0.84

1

2

3

4

5

1

2

3

4

5

0.050.110.170.240.340.46

0.62

Root Locus

Real Axis

Imagin

ary

Axi

s

-3 -2.5 -2 -1.5 -1 -0.5 0-5

-4

-3

-2

-1

0

1

2

3

4

5

0.84

0.050.110.170.240.340.46

0.62

0.84

1

2

3

4

5

1

2

3

4

5

0.050.110.170.240.340.46

0.62

Root Locus

Real Axis

Imagin

ary

Axi

s

2009 Spring ME451 - GGZ Page

100Week 14-15: Frequency Design

• Characteristics

CD CD –– Design Example Design Example –– Lead Controller (1)Lead Controller (1)

1 controller lead afor ,1

1)( >

+

+= α

τ

ατ

s

sKsC

When K = 1, we have the following Bode plot

α10

log20

o90

ατ

1

o0

dB0

1

1sin

+

−=

α

αφ

m

α10

log10

τ

1

τα

1

mφ

m

m

φ

φα

sin1

sin1

−

+=

Key Formula:




)2(

1)(

+=

sssG

A Design Example – Lead Controller

error SS %5

)(sG)(sC-

Design Target: %5error Ramp StateSteady ,45 : ≤= o

DPM φ

Determine control gain K by steady state error requirement:

2

1)()(lim20

05.0

1

0KsGssCK

sv

====→

Step 1:

Controller to be designed: s

sKsC

τ

ατ

+

+=

1

1)(

40=K




Draw Bode diagram of )1s(

20

)2(

40)(

2s +

=+

=ss

sG

Step 2:

dB40

OLC _ω

o90−

dB0

dB20

202

o18=OL

PM

dB20−

o180−

85.4 5.14

dB8.4−m

ω




Calculate α:

To get PM = 45° we target φm=30°, leading to PM =48°:

o30=m

φ

Step 3:

Step 4:

3-1

1

30sin1

30sin1

21

21

=+

=−

+=

o

o

α

)()( sLsC

44444 344444 21rCompensatoLeadtodueIncreaseGain

dB

10108.43log10log10 ==α

Find τ:

To find τ, we need to find ωm such that:

dBjGm

8.4log10)(log201010

−=−= αω




From Bode diagram, we have

0687.034.8

1==τ

85.40687.03

11=

⋅==

ατz

rad/s 4.8=m

ω

s

ssC

s

s

0687.01

2061.0140

)1(

)1(40)(

5.14

85.4

+

+=

+

+=

Therefore,

Then

8.41

==ατ

ωm

5.140687.0

11===

τp



CD CD –– Design Example Design Example –– Lead Controller Lead Controller (6)(6)

-100

-50

0

50M

agnitu

de (

dB

)

10-1

100

101

102

103

-180

-135

-90

Phase (

deg)

Bode Diagram

Frequency (rad/sec)

PM=39.4

ωωωωc=8.4

)1)(1(

)1(20)()(

5.142

85.4

ss

s

ssGsC

++

+=

class wk14 16 frequency design - egr.msu.edu · – controller design on bode plot is simple. •...

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