INC 341 PT & BPINC 341 PT & BP

INC341Design Using Graphical Tool

(continue)

Lecture 10

INC 341 PT & BP

Improving Both Steady-State Error and Transient Response

• PI, Lag improve steady-state error

• PD, Lead improve transient response

• PID, Lead-lag improve both

(PID = Proportional plus Intergal plus Derivative controller)

INC 341 PT & BP

PID Controller

s

K

Ks

K

KsK

s

sKKsKsK

s

KKsGc

)(

)( 3

2

3

1232

3213

21

INC 341 PT & BP

PID controller design

1. Evaluate the performance of the uncompensated system

2. Design PD controller to meet transient response specifications

3. Simulate and Test, redesign if necessary

4. Design PI controller to get required steady-state error

5. Find K constant of PID

6. Simulate and Test, redesign if necessary

INC 341 PT & BP

Example

Design PID controller so that the system can operatewith a peak time that is 2/3 of uncompensated system,at 20% OS, and steady-state error of 0 for a step input

INC 341 PT & BP

sec297.057.10

dpT

Step 1

• %OS = 20% damping ratio = 0.456

Ѳ = 62.87

• Search along ther line to find a point of 180 degree (-5.415±j10.57)

• Find a correspoding K=121.51

• Then find the peak time

INC 341 PT & BP

INC 341 PT & BP

• Decrease peak time by a factor of 2/3 get imaginary point of a compensator pole:

• To keep a damping ratio constant, real part of the pole will be at

• The compensator poles will be at -8.13±j15.867

867.15)297.0)(3/2(

p

d T

13.8)87.62tan(

d

Step 2

INC 341 PT & BP

92.55

)37.18tan(13.8

87.15

c

c

z

z

ดPD controller is (s+55.92)

• Sum of the angles from uncompensated poles and zeros to the test point (-8.13±j15.867) is

-198.37• The contribution angle for the compensator zero

is then 180-198.371 = 18.37

INC 341 PT & BP

INC 341 PT & BP

Step 3• Simulate the PD compensated system to

see if it reduces peak time and improves ss error

INC 341 PT & BP

s

ssGPI

5.0)(

Step 4• design PI compensator (one pole at origin

and a zero near origin; at -0.5 in this example)

• Find a new point along the 0.456 damping ratio line (-7.516±j14.67), with an associate gain of 4.6

INC 341 PT & BP

INC 341 PT & BP

s

ss

s

sss

ssKsGPID

)96.2742.56(6.4

)5.0)(92.55(6.4

)5.0)(92.55()(

2

K1 = 259.5, K2 = 128.6, K3 = 4.6

Step 5

• Evaluate K1, K2, K3 of PID controller

• Compare to s

sKKsKsGc

2321)(

INC 341 PT & BP

Step 6

INC 341 PT & BP

Lead-Lag Compensator Design

Same procedures as in designing PID:– Begin with designing lead compensator to get

the desired transient response– design lag compensator to improve steady-

state error

INC 341 PT & BP

Example

Design lead-lag compensator so that the system can operate with 20% OS, twofold reduction in settling time, and tenfold improvement in steady-state error for a ramp input

INC 341 PT & BP

Step 1

• %OS = 20% damping ratio = 0.456

Ѳ = 62.87

• Search along ther line to find a point of 180 degree (-1.794±j3.501)

• Find a correspoding K=192.1

• Then find the settling time

sec230.2794.1

44

sT

INC 341 PT & BP

INC 341 PT & BP

Step 2• Decrease settling time by a factor of 2 get a

real part of a compensator pole:

• To keep a damping ratio constant, imaginary part of the pole will be at

• The compensator poles will be at -3.588±j7.003

588.3)230.2)(2/1(

44

sT

003.7)87.62tan(588.3 d

INC 341 PT & BP

1.29

)35.15tan(588.3

003.7

c

c

p

p

Lead compensator is

• Select the compensator zero at -6 to coincide with the open-loop pole

• Sum of the angles from uncompensated poles and zeros to the test point (-3.588±j7.003) is

-164.65• The contribution angle for the compensator zero

is then 180-164.65 = 15.35

)1.29(

)6(

s

s

INC 341 PT & BP

Then find a new K at the design point (K=1977)

INC 341 PT & BP

Step 3Simulate the lead compensated system

INC 341 PT & BP

Step 4

• Originally the uncompensated system has the transfer function:

1.192,)10)(6(

)(

Ksss

KsG 201.3

106

1.192

vK

INC 341 PT & BP

• After adding the lead compensator, the system has changed to

• Static error constant, Kv, is then 6.794 (lead compensator has improved ss error by a factor of 6.794/3.201=2.122)

• So the lag compensator must be designed to improve ss error by a factor of 10/2.122=4.713

)1.29)(10(

1977)(

ssssGLC

713.4,)(

)()(

c

c

c

clag p

z

ps

zssG

INC 341 PT & BP

Step 5• Pick a pole at 0.01, then the associated zero will

be at 0.04713

• Lag-lead compensator

• Lag-lead compensatated open loop system

)01.0)(101.29(

)04713.0)(6()(

ss

sssG laglead

)01.0(

)04713.0()(

s

ssGlag

)01.0)(101.29)(10(

)04713.0()(

ssss

sKsGLLC

INC 341 PT & BP

INC 341 PT & BP

INC 341 PT & BP

Step 6

INC 341 PT & BP

Conclusions

INC 341 PT & BP

INC 341 PT & BP

Feedback Compensation

Put a compensator in the feedback path

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Tachometer

Tachometer generates a voltage output proportionalto input rational speed

Popular feedback compensator, rate sensor

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rate feedback

INC 341 PT & BP

ExampleDesign a feedback compensatorto decrease settling timeby a factor of 4 andkeep a constant %OS of 20

INC 341 PT & BP

%OS = 20% damping ratio = 0.456 Ѳ = 62.87

Search along the daping ratio line to get a summationof angle of 180 degrees at -1.809±j3.531

sec21.2809.1

44

sT

Find the corresponding K from the magnitude rule

settling time

841.257

531.3191.13531.3191.3531.3809.1 222222

K

Step 1

INC 341 PT & BP

Step 2

Reduce the Settling time by a factor of 4

236.7)21.2)(4/1(

44

sT

123.14)87.62tan(236.7 d

A new location of poles is at -7.236±j14.123

INC 341 PT & BP

At dominant pole -7.236±j14.123, KG(s)H(s) has a net angle of = -277.33 needs an additional angle from zero of 277.33-180 = 97.33

42.5

)33.97180tan(236.7

123.14

c

c

z

z

Find the corresponding K to the pole at -7.236+j14.123using the magnitude rule:

)15)(5(

)42.5(

sss

sK

K = 256.819

INC 341 PT & BP

211.1388

185.0

42.51

819.256

1

1

K

K

K

KK

f

f

f

Feedback block is 0.185(s+5.42)

INC 341 PT & BP

Physical System Realization

R1

R2C

Vi(s) Vo(s)

22

1

1

( )

sR CR

C sR s

PI Compensator

INC 341 PT & BP

R1

R2

C

Vi(s) Vo(s)

2 2

1 2

1 2

1

( )1

( )

c

sR R C

G sR R s

R R C

Lag Compensator

INC 341 PT & BP

R1

R2

C

Vi(s) Vo(s) 21

1( )cG s R C s

RC

PD Compensator

INC 341 PT & BP

R1

R2C

Vi(s) Vo(s)

1

1 2

1

( )1 1

,

c

cc c

c

sRC

G ssRC R C

s zz p

s p

Lead Compensator

INC 341 PT & BP

R1

R2C2

Vi(s) Vo(s)

C1

2 1 1 22 1

1 2

1

( )c

R C RCG s R C s

R C s

PID Compensator