INC341 Design Using Graphical Tool (continue)

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INC 341 PT & BPINC 341 PT & BP

INC341Design Using Graphical Tool

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Lecture 10

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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)

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PID Controller

sKKsKsK

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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

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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

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sec297.057.10

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

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• 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(

13.8)87.62tan(

Step 2

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)37.18tan(13.8

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

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Step 3• Simulate the PD compensated system to

see if it reduces peak time and improves ss error

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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

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ssKsGPID

)96.2742.56(6.4

)5.0)(92.55(6.4

)5.0)(92.55()(

K1 = 259.5, K2 = 128.6, K3 = 4.6

Step 5

• Evaluate K1, K2, K3 of PID controller

• Compare to s

sKKsKsGc

2321)(

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Step 6

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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

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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

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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

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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(

003.7)87.62tan(588.3 d

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)35.15tan(588.3

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(

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

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Step 3Simulate the lead compensated system

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Step 4

• Originally the uncompensated system has the transfer function:

1.192,)10)(6(

KsG 201.3

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• 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,)(

clag p

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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()(

sssG laglead

)01.0(

)04713.0()(

ssGlag

)01.0)(101.29)(10(

)04713.0()(

sKsGLLC

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Step 6

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Conclusions

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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

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

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%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

Find the corresponding K from the magnitude rule

settling time

841.257

531.3191.13531.3191.3531.3809.1 222222

Step 1

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Step 2

Reduce the Settling time by a factor of 4

236.7)21.2)(4/1(

123.14)87.62tan(236.7 d

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

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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

)33.97180tan(236.7

123.14

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

)15)(5(

)42.5(

K = 256.819

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211.1388

819.256

Feedback block is 0.185(s+5.42)

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Physical System Realization

Vi(s) Vo(s)

C sR s

PI Compensator

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Vi(s) Vo(s)

sR R C

G sR R s

Lag Compensator

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Vi(s) Vo(s) 21

1( )cG s R C s

PD Compensator

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Vi(s) Vo(s)

( )1 1

G ssRC R C

s zz p

Lead Compensator

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Vi(s) Vo(s)

2 1 1 22 1

R C RCG s R C s

PID Compensator

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