ContSys1 L6 SDOF Control

8/12/2019 ContSys1 L6 SDOF Control

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19/03/20141

Mechanical Engineering Science 8

Dr. Daniil Yurchenko

Performance enhancement


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Introduction

In this section we will examine ways ofimproving the performance of the

proportional action remote positioncontroller (RPC)

NOTE: The use of the RPC is onlyintended to be an example. The techniquesdeveloped here can be applied to any formof control system.


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Example

Determine the value of gain K so that themaximum overshoot in the unit step

response is 0.2. Assume that J=1kg m2

,F=1Nm/rad/sec

By definition, the maximum overshoot is

given as

%1002

1

eMp


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Example

2.021

e61.1

2.0

1ln2.0ln

1 2

)1(61.1 222 456.0

456.0

2

KJ

F 456.0

2

1

K

202.1912.0

1 2

K 096.1

J

Kn

22.31 2

n

pt


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Example

456.02

KJ

F 202.1

912.0

1 2

K

096.1JK

n 22.31 2

n

pt

But what if we need peak time to be equal to 1sec?

Can we fulfil this requirement?


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Optimisation to unit step

21

n

rt

21

eMp


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8

Negative velocity feedback

This technique feeds back the output velocity

as well as the output itself to modify

transients.

dt

dTu

umdt

dF

dt

dJ

O

OO

2

2

dt

dTm

dt

dF

dt

dJ OOO

2

2

m

dt

dTF

dt

dJ OO

)(

2

2

)()()()( 002 sMssTFsJs

sTFJsM )(

12

0

Applying L-transform


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9


We can see that the velocity signal is calculatedfrom the derivative of the output signal.

We can produce the velocity signal in a numberof ways, e.g. a tachometer attached to the outputshaft, electronically through an operationalamplifier

However we do it there will be a gain termassociated with the differentiation which we callthe derivative time constant T


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Negative velocity feedbackThe system block diagram without NVF was:

K+

-

o

Proportional

Controller

FsJs 21Mi

K+-

o

Proportional

Controller

FJs1M

i

s1


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Negative velocity feedbackThe system block diagram with NVF was:

K+

-

o

Proportional

Controller

FJs1Mi

s

1

T

+-

KTFJs

K

FJs

KT

FJs

K

1GH1

GFunctionTransfer


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12


+

-

o

KTFJs

K

i

s

1

KsKTFJs

K

sKTFJs

K

sKTFJs

K

i

2

0

1GH1

G


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13


+

-

o

KTFJs

K

i

s

1

iii

i

KsKTFJs

sKTFJs

KsKTFJs

KE

KsKTFJs

K

2

2

20

20

1


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We see that the coefficient of the firstderivative has changed from F to (F+KT)

proportional

Negative velocity FB

KFsJs

K

i

2

0

KFsJs

FsJs

E

i

2

2

KsKTFJs

K

i

2

0

KsKTFJs

sKTFJs

E

i

2

2


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Comparing the two characteristic equationswith the standard form we see:

KJF

2 proportional action controller

KJ

KTF

2

proportional and negative velocity feedback

controller

Thus we can artificially alter the damping ratio through

the time constant T without changing the friction F


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This means that the transients can beoptimised independently of other systemparameters.


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Again from the two characteristicequations we see that the steady state

errors for a ramp input are:

K

F

n

eSteadtStat

2proportional action only

K

KTF

neSteadtStat

2 proportional and negative velocity

feedback controller

Hence the steady state error (velocity lag) remains and

may in fact be increased.Class to show no effect on

droop


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Proportional + derivative (PD) controlThis is another attempt to improve the performance

K+

-

o

FsJs 21i

KDs

++

+

-

o

FsJs 21i

)1( DsK


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PD controlThis is another attempt to improve the performance

+

-

o

FsJs 21i

)1( DsK

KsKDFJs

DsK

FsJsDsK

FsJs

DsK

i

2

2

20 )1(

)1(1

)1(

GH1

G

iii

KsKDFJs

FsJs

KsKDFJs

DsKE

2

2

20

)1(1


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PD control

Thus the expression for damping ratio isidentical to that for the proportional +negative velocity feedback system.

Hence it possible to artificially alter the

damping ratio by adjusting time constant Twithout changing load friction.

Therefore transients may be optimised

independently of other system parameters.

KJ

KDF

2


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PD control

Since:

Thus we can see that the steady state errorfor a ramp input is given by:

This is the same as for the proportional

only system

iKsKDFJs

FsJsE

2

2

KF

sKsKDFJsFsJssE

sstst

22

2

0

1lim


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PD control

As a result we can optimise the transientsindependently of the steady state error by usingproportional plus derivative action control.

However we still have no way of eliminating thesteady state error following a ramp input, wecan only minimiseit using this technique.

Similarly the steady state error following a step

load disturbance (droop) is unaffected by theintroduction of proportional + derivative action.

Class to show this.


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Summary Both proportional + negative velocity feedback and

proportional + derivative action systems deliver ameans of optimising transient response independentlyof the mechanical system (i.e. through adjustment of

time constant T) However this also increases the steady state error

resulting from a ramp input in the case of negativevelocity feedback.

Both techniques produce a steady state error from aramp input and both suffer from droop following a stepload torque disturbance.


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Proportional + Integral (PI)Control

Steady state errors can generally be eliminatedby incorporation of integral action within thecontroller. You can think of an integrator giving

a gradual increase in output with time. Therefore,for a small input, as this magnitude is integratedover time its effect increases and the system isforced to take account of it and correct the error.

We introduce integral action mathematically as a1/s term, i.e. the reciprocal of differentiation.


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PI Control

Practically this means that when weemploy proportional + integral action weincorporate a term:

Into the feed forward path

The constantRis the integral timeconstantand it arises from the same sourceas the derivative time constants in theprevious cases

sKRK

sRK

1

Proportional term

Integral term


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PI ControlThis is another attempt to improve the performance

K+

-

o

FsJs 21i

s

RK

++

+

-

o

FsJs 21i

s

RK 1


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PI ControlThis is another attempt to improve the performance

KRKsFsJs

RsK

sRKFsJs

sRK

FsJssRK

FsJs

sRK

i

232

2

20 )(

/1

)/1(

)/1(1

)/1(

iii KRKsFsJs

FsJs

KRKsFsJs

RsKE

23

23

230

)(1

+

-

o

FsJs 21i

s

RK 1


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PI Control

From the above we see that we haveincreased the order of the characteristicequation by 1 from 2 to 3. Thus, anadditional root is introduced. Increasing

the order has a tendency topush thesystem towards instability---Not Good!But what effects does this have on thesteady state errors?

023 KRKsFsJs


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PI Control

Since:

Thus we can see that the steady state errorfor a ramp input is given by:

Therefore steady state error becomes =0!

i

KRKsFsJs

FsJsE

23

23

01lim223

23

0

sKRKsFsJsFsJssE

sstst


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Proportional + Integral Controllers We have seen that the introduction of integral action

can remove steady state errors such as droop andvelocity lag. However, we also observe in the twoexamples given that adding integral action to the

controller raises the order of the characteristicequation by one. Hence there is an additional root tothe complimentary function part of the solution. Wethen need to ensure that the combination of systemconstants and controller constants does not produce

an unstable systemone in which the responseincreases monotonically or increases in an oscillatoryfashioneither of which is exemplified by a positivereal part to a root of the characteristic equation.


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Type Transfer Function Error

P

PNVF

PD

PI

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Comparison

KFsJs

K

i

2

0

KFsJs

FsJs

E

i

2

2

KsKTFJsK

i

2

0 KsKTFJs

sKTFJs

E

i

2

2

KsKDFJsDsK

i

2

0

)1(

KsKDFJs FsJsEi

2

2

KRKsFsJs

RsK

i

23

0 )(

KRKsFsJs

FsJs

E

i

23

23


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Type Advantage Disadvantage

P Influence the systems natural

frequency and damping.

Steady state error to a ramp input is

not zero

PNVF Improve dampingReduce maximum overshoot

Reduce rise time

Reduce settling time

Steady state error to a ramp input isnot zero

PD Improve damping

Reduce maximum overshoot

Reduce rise time

Reduce settling time

Steady state error to a ramp input is

not zero

PI Improve damping

Reduce maximum overshoot

Steady state error to a ramp input is 0

Increase rise time

Push the system towards instability

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Comparison


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Comparison of PNVF and PD

KsKTFJsK

i

2

0

KsKDFJsDsK

i

2

0 )1(

PNVF PD

21

eMp 221

212

nn

PD

p DDeM

211

exp221 211

2

nn

PD

p DDex

PNVF

p

PD

p MM


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Comparison of PNVF and PD

KsKTFJsK

i

2

0

KsKDFJsDsK

i

2

0 )1(

PNVF PD

n

nnPD

s

DDt

/

02.0

21ln

22

n

st

4%)2(

n

st3%)5(

n

nnPD

s

DDt

/

05.0

21ln

22

PNVF

s

PD

s tt


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PID control

CLASS, DERIVE EXPRESSIONS FOR THE

TRANSFER FUNCTION AND ERROR.

K+

-

o FsJs 2 1i

KDs

++

s

RK

+


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Example

2.021

e61.1

2.0

1ln2.0ln

1 2

)1(61.1 222 456.0

456.0

2

KJ

F 456.0

2

1

K

202.1912.0

1 2

K 096.1

J

Kn

22.31 2

n

pt


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Uncontrolled Response

Zoom in


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Example

Determine the value of gain K so that themaximum overshoot in the unit step

response is 0.2 and peak time is 1 sec.Assume that J=1kg m2, F=1Nm/rad/sec

We COULD NOT satisfy both

criteria simply adjusting the

gain value!


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Introduce PNVF control

KsKTFJsK

i

2

0

22

2

0

2 nn

n

i ss

KTJKTF

n 12 KJ

Kn

2.0

21

e

456.0 0.11

2

n

pt

53.3n 46.122 nK Tn 46.12122.32

178.046.12

22.2

T


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Controlled Response


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PI Controller

KRKsFsJs

RsK

i

23

0 )(023 KRKsFsJs

Two cases possible: 1) All roots are real

2) 1 real and 2 complex-conjugate roots

The roots of the characteristic equation that cause the dominant

transient response of the system are called the dominant roots

In other words, the dominant roots are the roots with the

smallest real part


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PI ControllerFor example

If 5,21 32,1 sis

are the dominant roots2,1s

If 1.0,21 32,1 sis

is the dominant root3s

S


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Consider a simply positioning system

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Example: Positioning System

E l P i i i S


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The block diagram of open loop system

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)1(11

bsass

as1

1

yu

bs1

1

E l P iti i S t


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The block diagram of closed loop system

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)1(11

bsass

+

-

yuK

as11+

-

yuK

)1(

1

bs s

1+

-

H1

H2

E l Wi d T bi C t l


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Stall Control System

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Example: Wind Turbine Control

E l Wi d T bi C t l


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Pitch Control System

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E l Wi d T bi C t l


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E l Wi d T bi C t l


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E l Wi d T bi C t l


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ContSys1 L6 SDOF Control

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