control systems - chibum lee · pdf file2014-08-05 · chibum lee -seoultech control...

Control Systems

Transient and Steady State Response

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Control SystemsChibum Lee -Seoultech

Outline

Time Domain Analysis

1st order systems

• Unit step response

• Unit ramp response

• Unit impulse response



After the mathematical model of the system is obtained,

analysis of system performance is needed.

• Input signal to a control system is unknown

but what if to use known test input signal

• A correlation b/w the response characteristics of a system to a

typical test input signal and the capability of the system to cope

with actual input signals

SystemInput r(t) Output y(t)



Typical input test signal

• Step function

• Ramp function

• Acceleration function

• Impulse function

• Sinusoidal function

• White noise



Time response

• Transient response: goes from IC to final state

• Steady state response: the system output behaves as t

approaches infinity

)()()( tytyty sstr

SystemInput(t) Output(t)


1st Order Systems

Standard form ryyT

1

1

)(

)(

)()()1(

TssRsY

sRsYTs

TF of 1st order system

Y(s) Y(s)


1st Order Systems –Step response

Unit step response w/ zero initial conditions

Tss

TsT

ssTssY

1

11

1

1

1

1

1)(

ssR

tUtr s

1)(

)()(

0t,1)( / Ttety

1


1st Order Systems –Step response

Characteristic feature of a 1st order system:

• At t=T,

ie. the response y(t) is 63.2% of its total change

• At t=0, the slope of the response

632.0 368.00.1 1)()( 1 eTyty

Te

Tdtdy

t

Tt

t

11

0

/

0

T : time constant

y(t)


1st Order Systems –Ramp response

Unit ramp response w/ zero initial conditions

1

1

1

1

1)(

2

2

2

TsT

sT

s

sTssY

0 t,)( / TtTeTtty

2

1)(

)()(

ssR

ttUtr s


1st Order Systems –Ramp response

The error between the reference & the output.

• Smaller T smaller ess

ess= T

Te(t)e)e()eT( TeTtt

y(t)r(t)e(t)

tss

t/T

t/T

lim

1

y(t)

y(t)


1st Order Systems –Impulse response

Unit impulse response

1

1)(

TssY

0t,1

)( / TteT

ty

1)()()( sRttr

y(t)

y(t)


1st Order Systems –Impulse response

Compare

• Time response to an impulse reference signal is identical to

an initial condition response with zero reference.

TyyyT 1

)0( w/ 0

1

1)(

1)()1(

0)()0()(

1

sTsY

sYsTsYYTssTY

T

0)0( w/ (0) yyyT

1

1)(

1)()1(

1)()(

sTsY

sYsTsYssTY


Outline

2nd order systems

• Unit step response

• Unit ramp response


2nd Order System

Example: Servo system

y (=)

)(

1

)()()(

2

BJssT(s)θ(s)

sTsBssJs

TBJ


2nd Order System

Closed-loop

22

2

22

2

)(

)(

nn

n

ss

JKs

JBsJK

KBsJsK

sRsY

Y(s) (=(s))

Standard form of 2nd order system

JKB

JK

n2

,


2nd Order System

Solve the following differential equation

• Case 1.

• Case 2.

0)0(,0)0(,2 22 xxxxx nnn

0)0(,2

1)0(,02 2 xxxxx

nnn


2nd Order System

Dynamic behavior of 2nd order system can be described

in terms of damping ratio and natural frequency n

• Characteristic equation (denominator of TF=0)

• Poles (root of characteristic equation)

22

2

2)(

)(

nn

n

sssRsY

02 22 nnss

2

21 1, nn jss

10


2nd Order System

Pole location

1, ,1 If

, ,1 If

1, ,10 If

, ,0 If

2

21

21

2

21

21

nn

n

nn

n

ss

ssjss

jss


2nd Order System-Step response

For underdamped case,

22222

22

2

)(1)(

1

1

2)(

dn

d

dn

n

nn

n

sss

s

ssssY

10

)1

(tan re whe)sin(1

1

sin1

cos1)(

21

2

2

te

tetety

d

t

dt

dt

n

nn

21 nd

1)(lim

tyt



For no damped case,

2222

2 11)(

nn

n

ss

ssssY

0

tty ncos1)(

1)(lim

tyt

y(t)



For critically damped case,

ssssssY

n

n

nn

n 1

)(

1

2)(

2

2

22

2

1

)1(1)( tety ntn

1)(lim

tyt



For overdamped case,

)1()1(

1

1

)1)(1()(

22

22

2

nn

nnnn

n

sss

ssssY

212

)1(

22

)1(

22

21

22

121

)1(12

1

)1(12

11)(

se

se

eety

tstsn

tt nn

1)(lim

tyt

,1

1, 2

21 nnss


2nd Order System-Impulse response

Impulse response or Initial condition response

For

For

22

2

2)(

nn

n

sssY

tety dtn n

sin

1)(

2

0for 0)(lim

tyt

10

1

tn

ntety 2)(


Outline

Higher order systems

• Dominant poles


Higher Order Systems

Unit step response

01

1

1

01

1

1

)(

)(

asasasabsbsbsb

sRsY

nn

nn

mm

mm

)2(2

1)(

1)(

1

response ord 2nd

22

2

1

response ord1st

01

1

1

01

1

1

nrqss

csbps

asa

sasasasabsbsbsbsY

r

k kkk

kkkkjkq

j j

j

nn

nn

mm

mm

r

kkk

tkkk

tk

q

j

tpj tectebeaaty kjkjj

1

22

1

1sin1cos)(


Higher Order Systems

• Response of higher order system

=Responses of 1st order systems

+ Responses of 2nd order systems

• Dominant closed-loop poles

Closed-loop poles that will dominant effects on the transient

response behavior

Closed-loop poles nearest the 𝑗𝜔 axis


Example – Dominant pole

Ex.

)25.6)(256(

)5.2(5.62

)(

)(2

ssss

sRsY

)256(

)5.2(10

)(

)(2

sss

sRsY


Outline

Transient-response specification

Parameter selection

Root location and transient response


Transient-Response specification (2nd ord. sys.)

Transient Response Characteristics to a unit step input

• Delay time, td : time required for reach 50%

• Rise time, tr : time required for rise from 10% to 90% or

from 0% to 100%

• Peak time, tp : time required for reach peak value

• Maximum overshoot, Mp :

• Settling time, ts: time required for reach

and stay 2% or 5% of

final value

y(t)

)()( yty p



Rise time:

tr

y(t)

2

2

2

1tansin

1cos

)sin1

(cos11)(

rdrdrd

rdrdt

r

ttt

ttety rn

tetety dt

dt nn

sin1

cos1)(2



01

tan1

01

tan01

10For

1tan

1

21

21

2

21

dr

dr

t

t

???



• From system pole

22

2

2)(

)(

nn

n

sssRsY

2

21 1, nn jss

)(11

tan1

2

1

ddrt

n

nsangle2

1

1

1tan)(


y(t)


Peak Time

tp

,3,2,,00)sin(0)sin(

0)sin()cos()cos()sin(

0

)sin()cos()cos()sin()(

pdpddd

nnpd

t

pddpdnpdpdd

nn

t

pddpdd

ndtpdpd

d

ntnp

ttte

tttte

ttettety

pn

pn

pnpn

tetety dt

dt nn

sin1

cos1)(2


y(t)

tetety dt

dt nn

sin1

cos1)(2


Maximum

Overshoot

Mp

%100)(

)()(PO)Overshoot(%

)sin(1

)cos(

1)(1)(

2211

2

yyty

eee

ytyM

p

dpp

nn

dn


y(t)


Settling

Time

ts

• Approximation (comes from envelop function )

criterion 5%for 5

criterion 2%for 4

ns

ns

t

t

tetety dt

dt nn

sin1

cos1)(2

211

tne



y(t)

)1

(tan re whe)sin(1

1

sin1

cos1)(

21

2

2

te

tetety

d

t

dt

dt

n

nn


Parameter Selection –example1

Ex. Servo system with

velocity feedback

Determine 𝐾 and 𝐾ℎ so

that the unit step response

𝑀𝑝=0.2, 𝑡𝑝=1 sec.

Find 𝑡𝑟, 𝑡𝑠.

(J=1 kg m2, B=1 Nm/rad/s)

Y(s)

Y(s)

JK

KJKKb

ssKsKKbJsK

sRsY

nh

nn

n

h

,2

2)(

)(22

2

2



criterion) (5% sec 86.13

, criterion) (2% sec 48.24

sec 65.01

1tan

sec 178.02

m,N 5.12,2

53.31

11

456.061.11

2.0

2

21

2

22

2

1 2

ns

ns

ndr

hnnh

n

ndp

p

tt

t

KBKJKJK

JK

KJKKb

t

eM



Ex. Select the gain 𝐾 and the parameter 𝑝

the transient response to a step input

criterion) (2%4

%3.4

stPO

s

trequiremen thesatisifies 22, exmaple,For

144

707.0%3.4)(

)(

2,2)(

)(

2

2

11

2

22

2

2

Kp

st

ey

yePO

pKssKpss

KsRsY

nn

s

nnnn

n


Root Location and Transient Response

Impulse/ IC response for root locations

01

1

1

01

1

1

)(

)(

asasasabsbsbsb

sRsY

nn

nn

mm

mm


Outline

Steady state errors in feedback systems


Steady-State Errors in Feedback Systems

Unity feedback system

)()()(1

lim)(lim)(lim

)()()(1

1)()()(

)()()(1

)()()(

00sR

sGsGsssEtee

sRsGsG

sYsRsE

sRsGsG

sGsGsY

pcsstss

pc

pc

pc

number) type:())(())((

))(())(()()(

121

121 Npspspspss

zszszszsKsGsGll

Nmm

pc


Unit step input


0or ,1For

1

1

)()(

)()(lim)()(lim ,0For

constant)error position :(1

1

)()(lim1

1

1

)()(1lim)(lim)(lim

1

0

1

00

0

00

ssp

pss

l

m

spcsp

pppcs

pcsstss

eKNK

e

pspsszszsKsGsGKN

KKsGsG

ssGsGsssEtee


Unit ramp input


0or)()(

)()(lim ,2For

1

)()(

)()(lim ,1For

or

0)()(


constant)error velocity :(1

)()(lim

1

)()(

1lim

1

)()(1lim)(lim)(lim

1

1

0

1

1

1

0

1

0

1

00

0

0

200

ssl

Nm

sv

vss

l

m

sv

ss

l

m

spcsv

vvpcspc

s

pcsstss

epspss

zszsKsKN

Ke

pspsszszsKsKN

epspsszszsKssGssGKN

KKsGssGsGssGs

ssGsGsssEtee



Type 1 system response to a ramp input

y(t)

y(t)


Unit parabolic(acceleration) input


0or)()(

)()(lim ,3For

1

)()(

)()(lim ,2For

or

0)()(


constant)error on accelerati:(1

)()(lim

1

)()(

1lim

1

)()(1lim)(lim)(lim

1

12

0

1

2

12

0

1

12

0

2

0

2

0

220

300

ssl

Nm

sa

ass

l

m

sa

ss

lN

m

spcsa

aapcspc

s

pcsstss

epspss

zszsKsKN

Ke

pspsszszsKsKN

epspss

zszsKssGsGsKN

KKsGsGssGsGss

ssGsGsssEtee


Type 2 system response to a parabolic input


y(t)

y(t)

control systems - chibum lee · pdf file2014-08-05 · chibum lee -seoultech control...

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