ContSys1 L5 SDOF RespAll

8/12/2019 ContSys1 L5 SDOF RespAll

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Mechanical Engineering Science 8

Dr. Daniil Yurchenko

Response of a second

order system


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Remote position controller

A device which is used to position aninertia load according to a desired angular

position from a remote point.

Input, i

Angular

position,

Output, o

Position

transducer

Error unit

Electric

Motor Inertia, J

Viscous friction, F

Torque, mo

motor

Inertial load


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Desired angular position input to error unit

Error unit subtracts actual angular position togive error signal

Error signal passed to electric motor whichgenerates a torque directly proportional to errorvoltage applied to it

Such a system is known as a proportional action

controllersince the controlling action (motortorque in this case) is directly proportional tosystem error


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Remote position controller Open loop descriptions of system elements

Motor

Where K is a constant of proportionality & m isthe motor torque

Thus motor torque is proportional to error.

Leads to

Km

Km


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Applying Newtons second law to inertialload

Where F is viscous friction constant and Jis mass moment of inertia of load

Re-arranging

2

2

dt

dJ

dt

dFm OO

mdt

dF

dt

dJ OO

2

2


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Applying L-transform and assuming zero ICs

This may be represented by the system blockdiagram in the next slide

FsJsM

2

0 1

][2

2

mLdt

dF

dt

dJL OO

)()()( 002 sMsFssJs


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The system closed loop equation

GH1

GFunctionTransfer

is

K

+

-

o

Proportional

Controller

FsJs 21Mi

1;2

HFsJs

KG

Oi

KFsJs

K

FsJs

KFsJs

K

i

2

2

20

1


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8


The relationship between system outputand the error is

and

Re-arrange to give system characteristicequation:

)()()( sEss iO

)()( 2 sFsJs KsO

)()( 22 sFsJssEKFsJs i


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9

Standard forms

Similar to the standard form concept westudied when we worked with vibrating

systems. In this case system naturalfrequencies and damping ratios will bemade up of a mixture of mechanical andelectronic components rather than springs,

masses & dampers


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10

Characteristic equation

Or

Leads to standard form

iFsJssEKFsJs 22 )(

)()(

22

ssJ

F

ssEJ

K

sJ

F

s i

)(2)(2 222

ssssEss innn

where

J

Kn

and

KJ

F

J

Fn

22


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11

Similarly the governing equation becomes

Or

Leading to the standard form

)()(2

sKsKFsJs iO

)()(2 222 ssss inOnn

)()(2

sJ

K

sJ

K

sJ

F

s iO


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12

Control system response

This can be defined using 2 parameters,the error and the output O.

The error response can be obtained fromsolving the characteristic equation for aspecified input and the output responsecomes from solving the governingequation for a specific input.


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Both the error and the output can each beconsidered to be the sum of two components:

a transient response and a steady stateresponse.

The transient response is the complementaryfunction solution to the describing equationand the steady state response is the particularintegral solution (c.f. first order systems)


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Thus for the error:

For a satisfactory control system weshould have:

And

statesteadytransienttotal _

0statesteady _

transientShould be of minimum duration and

may need to have limited overshoot


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Control system response 1)unit step

As before the input is defined as:

We will use the remote position controlleras an example.

First we will investigate the error responseof this system.

00 tti , 01 tti ,


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Repeating the characteristic equation:

L-transform of the unit step response is:

Rearranging:

s

tULsi1

)()(

)()( 22 sFsJssEKFsJs i

sKFsJs

FsJssE

1)(

2

2


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Thus:

The steady-state error can be easily obtainedby using Theorem 3 (Lecture 3):

Thus,

sKFsJsFsJssE 1)(

2

2

)(lim)()(lim0

ssQqtqst

0lim)(lim)(2

2

00

KFsJs

FsJsssEe

ssstst

0statesteady


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The system response will depend on thevalue of

To show that lets get the characteristicequation, which is obtained by assumingzero input:

Thus:02 22 nnss

0)(2 22

sss Onn


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This is a quadratic with solution:

Consider 3 cases:1

2 nns

10

1

Underdamped

Critical

1 Overdamped


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Case

Both roots are real:

Then

1,1 222

1 nnnn ss

1

11

2)(

)(

22

2

22

2

nnnn

n

nn

n

i

O

ss

sss

s


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For unit step response

Which can be written as

Using the table of inverse L-transform

1

ssss

nnnn

nO

1

11

)(22

2

bsasss nO

2

)(

btatnOO aebe

baab

sLt 1

1)()(2

1


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Which after simplification gives:

Then the error is:

And the

1

212

21

121)(

se

set

tsts

nO

212

0

21

12)()()(

s

e

s

ettttsts

ni

0)(lim

tt

statesteady

No Vibrations


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Then

Which leads to

And for unit step input

1

nnns 12

2,1

22

22

2

2)(

)(

n

n

nn

n

i

O

ssss

s

ss

s

n

nO

1)(

2

2


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Then

Then the error is:

And

1

tettt nt

in 1)()()( 0

0)(lim

tt

statesteady

tesLt nt

OOn 11)()( 1

No Vibrations


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Then we have two complex conjugate roots

Which leads to

And for unit step input

10

22

2

2)(

)(

nn

n

i

O

sss

s

ssss

nn

nO

1

2

)(22

2

222,1 111 nnnn is


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Then

te

sLt n

t

OO

n

2

2

1 1sin1

1)()(

10

21 1tan

21

nd

The system response is oscillatory with

damped natural frequency and decays

exponentially

d

ttet dd

t

On

sin

1cos1)(

2


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The error is

And

t

e

ttt n

t

i

n

2

20 1sin1)()()(

0)(lim

tt

statesteady

10

0

2sin1)(

tt nO

2

sin)( tt n


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21 1tan

n

ncos


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samethestaysst

samethestayspM

samethestaysn


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Mp - maximum percent overshoot

where tp - peak time

td - delay time. The time required for theresponse to reach half of the final value.

tr - raise time. The time required for theresponse to rise from 10% to 90% or from0 to 100% of its final value

%100)(

)()(

0

00

pp

t

M


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tssettling time. The time required for theresponse curve to reach and stay within arange about the final value

(usually 2% or 5%)

In control system design we are largely

interested in obtaining a rapid responsewhilst minimising overshoot.


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The peak time may be obtained bydifferentiation

tetetete

ttette

ttedt

d

dt

td

dnt

dd

t

d

tnd

t

n

d

d

dd

t

dd

t

n

dd

tO

nnnn

nn

n

cos1

1sinsin

1cos

cos1sinsin1cos

sin1

cos1)(

2

2

2

2

22

2


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Equating to zero yields

The first peak corresponds to

p

nn

ttdd

t

d

tnO tete

dt

td

sinsin

1

0)(

2

2

0sin1 2

2

pdd

n t

0sin pdt ,...2,1,0, nntpd

d

pt


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The first peakd

pt


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The maximum overshoot occurs at tp=/d

sin1

sin1

1sin1

1)()(

22

2

d

n

d

n

dn

ee

et

d

dOO


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2

22

221 1tan

1tan

1tan

22

2

2

2

2

2

2

2 1sin11

1

1

tan1tansin

dn

dn

eet OO

sin1

)()(2

%100

21

eMp


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By increasing K we increase the overshoot

%10021

eMp KJF

2 K

K


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The rise time tr ,

rd

t

rO t

e

t

rn

sin111)( 2

1)( rO t

0sin1 2

rd

t

te rn

0sin rdt

,...2,1,0, nntrd 21

nd

rt


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By increasing K we decrease the rise time

K

21

nd

rt


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In MATLAB

dcsbsa

s

s

i

O

2

)(

)(

][

]00[

dcbden

anum

),( dennumstep


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Unit ramp response of RPC

In this case we present the system with aunit ramp input:

The L-transformtti )(

2

1)()(

s

tLs ii

22

2 1)(

sKFsJs

FsJssE


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222

2 1

2

2)(

sss

sssE

nn

n

J

Kn

KJ

F

2

02

2

2lim

)(lim)(

222

2

0

0

nnn

n

s

sstst

s

s

ss

ss

ssEe

This is aconstant

steady

state error

)(lim)()(lim0

ssQqtqst


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Transient error response.

To find this we need to examine thecomplementary solution to the characteristic

equation. i.e. the solution to:

This is identical to the equation we used to find

the complementary function solution for the stepinput. Therefore the complementary solution isidentical to that case

02 22 nnss


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For two real roots

No oscillatory response.

1

2222 1

11)(

ssss

nnnn

nO

22

2

2)(

)(

nn

n

i

O

sss

s


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For two real roots

It results in

No oscillatory response

1

nnn

s 122,1

22

2 1

)( sss n

n

O


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Then we have two complex conjugate roots

Which leads to

Partial fraction expansion

10

222

2 1

2

)(

sss

s

nn

nO

222,1 111 nnnn is

222222

2

2

1

2 nnnn

n

ss

DCs

s

BAs

sss


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10

222

2322

222

2

)2(

)2)((

2 sss

DsCsssBAs

sss nn

nn

nn

n

220

21

2

3

:

02:

02:

0:

nn

nn

n

Bs

BAs

DBAs

CAs


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10

222

2 2

242)(

nnn

nn

n

nO

ss

s

s

ss

212

2

2

2

2

2222

2

2

1

1tan,1sin

1

2

1sin1

142

2

2

2

1421)(

te

tet

ss

s

ssssLt

n

t

n

n

t

nn

nnnnnn

O

n

n


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When t goes to infinity the second term

vanishes

10

12cos

,1sin1

2)(

21

2

2

te

tt n

n

t

n

O

n

te

ttt nn

t

n

i

n

2

20 1sin

1

2)()()(


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The error is

And

n

tstatesteady t

2)(lim

10

te

ttt nn

t

n

i

n

2

2

0 1sin

1

2)()()(

J

Kn

KJ

F

2

K

F

K

J

KJ

F

n

stst

2


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At t=0

If then

0sin1

120 2

nn

0 0t

te

ttt nn

t

n

i

n

2

20 1sin

1

2)()()(

)1(4sin,12cos 222222


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Either the output response or the errorresponse may be plotted against time

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ContSys1 L5 SDOF RespAll