system transient/time responseguppy.mpe.nus.edu.sg/anpoo/fbcontrola/5system response.pdf ·...

ME2142/ME2142E Feedback Control Systems1

System Transient/Time ResponseSystem Transient/Time Response

ME2142/ME2142E Feedback Control SystemsME2142/ME2142E Feedback Control Systems


System responseSystem response

For stable systems, the magnitude of the transient response decreases with time and ultimately vanishes leaving only the steady-state response. It is always associated with the component

0with ae at


System Characteristic EquationSystem Characteristic Equation

Consider the system with the transfer function, Gc(s) as shown

Gc(s)System

Input OutputR(s) C(s)

The system’s characteristic equation is given by

0)( sDc

with

where Nc(s) and Dc(s) are polynomials of s.

)()(

)()()(

sDsN

sGsRsC

c

cc

Note that the characteristic equation is a property of the system and is not dependent on the input.



ExamplesExamples

Spring-mass-damper (Slide 9: Modelling of Physical Systems)

Transfer Function

Characteristic Eqn:

Spring-mass-damper (Slide 9: Modelling of Physical Systems)

Transfer Function

Characteristic Eqn:

KbsmsKbs

sXsX

sGi

o

2)()(

)(

02 Kbsms

R-C circuit (Slide 12: Modelling of Physical Systems)

Transfer Function

Characteristic Eqn:

R-C circuit (Slide 12: Modelling of Physical Systems)

Transfer Function

Characteristic Eqn:

11

RCsE

E

i

o

01 RCs

Closed-loop feedback system (Slide 8: Block Diagram Algebra)

Transfer Function

Characteristic Eqn:

Closed-loop feedback system (Slide 8: Block Diagram Algebra)

Transfer Function

Characteristic Eqn:

GHG

RC

1

01 GH



The roots of this equation are the closed-loop poles and they determine the transient response of the system.

Characteristic equation

0)( sDc

)()(

)()()(

sDsN

sGsRsC

c

cc

If all the roots, pr, are negative, then the transient response will eventually die away as t increases.

Each root, p, of this equation will contribute a term in the time response of the system. Or

ptAetc )(

pte

But if any of the roots is positive, then the transient response will grow without bounds as time increases. The system is then said to be unstable.


Given a dynamic system:Given a dynamic system:

System ResponseSystem Response

We use Standard test inputs to excite system and observe

response Classify systems with similar characteristics and

identify their performance characteristics with system parameters.

How do we specify the characteristics of the response required?

How do we compare it with another system?

How do we know whether it’s response will adequately meet our needs?

How will we know how it will respond to different inputs?


2) Ramp input2) Ramp input

000)(

ttAttr

2)(

sAsR

At

t = 0

r(t)

t

System Response –Test signalsSystem Response –Test signals

1) Step input1) Step input

000)(

ttAtr

sAsR )(

A

t = 0

r(t)

t

When A = 1, we have a unit step input.Used to study response to sudden changes in input.

When A = 1, we have a unit ramp input.Used to study response to gradual changes in input.


4) Sinusoidal input4) Sinusoidal input

000sin)(

tttAtr

t

r(t)

t = 0

3) Impulse input

is the unit-impulse function or Dirac delta function

3) Impulse input

is the unit-impulse function or Dirac delta function

)0()( Atr

AsR )(

A

t = 0

r(t)

t

System Response –Test signalsSystem Response –Test signals

When A = 1, we have a unit impulse input.Used to study response to sudden shocks or impacts.

Used for frequency response analysis.Important method. Will be discuss in the second half of course.

Using test signals (1) to (3) are often known as time response or transient response analysis while using test signal (4) is known as frequency response.


System Response – First-order systemsSystem Response – First-order systems

A first-order system can always be written in the standard form

T is known as the time constant and determines the speed of response.

A first-order system can always be written in the standard form

T is known as the time constant and determines the speed of response.

1)()(

TsK

sRsC

ExamplesExamples

Spring-damper system (Slide 7 of Modelling of Physical Systems)

.

Spring-damper system (Slide 7 of Modelling of Physical Systems)

. KbsK

sXsY

)()(

KbT

Ts

with

11

RC circuit (Slide 12 of Modelling of Physical Systems)

.

RC circuit (Slide 12 of Modelling of Physical Systems)

.RCT

TsRCsEE

i

o

with1

11

1

If the transfer function are the same, then the response y(t) and eo(t)will be the same for the same inputs in x(t) and ei(t)..



with

.

with

.

1)()(

TsK

sRsC

Response to a unit step inputResponse to a unit step input

ssR 1)(

Thus )(1

)( sRTs

KsC

sTs

TK 1)/1(

/

Ts

BsA

/1

Multiplying both sides by s and letting s=0 gives A = K

Multiplying both sides by (s+1/T) and letting s = -1/T gives B = -K

ThereforeTs

KsKsC

/1)(

Using tables

)1()( // TtTt eKKeKtc 0for t



Response to a unit step inputResponse to a unit step input )1()( / Ttetc For K = 1

Note: The smaller the time constant T, the faster the response.The shape is always the same.



with

.

with

.

1)()(

TsK

sRsC

Response to a unit ramp inputResponse to a unit ramp input

Thus )(1

)( sRTs

KsC

2

1)(s

sR

2

1)/1(

/sTs

TK

)/1(2 Ts

KTs

KTsK

For K = 1,

)1()( / TteTttc

with the error e(t) = r(t) – c(t)

Using tables

0for t)()( / TtTeTtKtc

)1( / TteT

r(t)

t = 0

r(t)

t

ess=Tc(t)

r(t)

t = 0

r(t)

t

ess=Tc(t)



with

.

with

.

1)()(

TsK

sRsC

Response to a unit impulse inputResponse to a unit impulse input

1)( sR

ThusTs

TKTs

KsC/1

/1

)(

Or TteTKtc /)(

For K = 1,

TteT

tc /1)(

r(t)

t = 0

1

t

r(t)

t = 0

1

t


The transient response all contains the term which is determined by the root of the characteristic equation and the parameter T.

The transient response all contains the term which is determined by the root of the characteristic equation and the parameter T.

Tte /

Response toResponse to

Unit Impulse

TteTKtc /

1 )(

System Response – Linear time-invariant systems

System Response – Linear time-invariant systems

1)()(

TsK

sRsCPropertiesProperties Characteristic EquationCharacteristic Equation

TsTs 101

Note that the unit step is the derivative of the unit ramp, and the unit impulse is the derivative of the unit step.

Note that similarly, c2(t) is the derivative of c3(t) and c1(t) is the derivative of c2(t) .

For linear time-invariant systems, the response to the derivative of an input can be obtained by taking the derivative of the response to the input.

Unit Step

)1()( /2

TteKtc

Unit Ramp

)()( /3

TtTeTtKtc


Block DiagramBlock Diagram

Permanent Magnet DC MotorPermanent Magnet DC Motor

e i

Ra La

eK J

bT

e i

Ra La

eK J

bT

The Permanent Magnet DC motor.The Permanent Magnet DC motor.

Governing equationsGoverning equations

eaa KdtdiLiRe

iKT t

bdtdJT

IsLRKE aae )(

IKT t

)( bJsT

+

-

E I

aa RsL 1

eK

tKT

bJs 1

eK

+

-

E

eK

eK

))(( bJsRsLK

aa

t


+

-

E

eK

eK

))(( bJsRsLK

aa

t

+

-

E

eK

eK

))(( bJsRsLK

aa

t

The Permanent Magnet DC motor.The Permanent Magnet DC motor.

Commonly

bJ

RL

a

a

La can then be neglected

Block diagram then becomesBlock diagram then becomes

E +

-

eK

bJsRK at

/E +

-

eK

bJsRK at

/

GHG

E

1

)(/

1

)(/

bJsRKK

bJsRK

aet

at

aet

at

RKKbJsRK

//

1

sK

aet

at

RKKbRK

K/

/with

aet RKKbJ

/

Permanent Magnet DC MotorPermanent Magnet DC Motor


aet RKKbJ

/

Speed Control of the DC MotorSpeed Control of the DC Motor

1sK

E

The response to a unit step input is first order with a time constant ofCan we make the response faster?

t = 0 t

K

With speed feedbackWith speed feedback

1sK

E Error+

-

VKc

Controller1

1

1

sKK

sKK

V c

c

KKs

KK

c

c

1 1''

sK

KKc

1'with

KKKK

Kc

c

1'

The resultant system is still first-order but the time constant is now much smaller, thus a much faster response.

Motor byitself

Motor byitself


System Response – Second-order systemsSystem Response – Second-order systems

A second-order system will be of the form

with a, b, c, d and e being constants.

A second-order system will be of the form

with a, b, c, d and e being constants.cbsasesd

sXsY

2)(

)(

Standard Form:

.

Standard Form:

. 22

2

21

2

221

2)(

12)()(

nn

n

nn

ssKsK

ssKsK

sRsC

We can re-writeWe can re-write

cbsasesd

sXsY

2)(

)(

acs

abs

sae

ad

2with and

ac

n 2ab

n 2



ExamplesExamples

RLC circuit (see Modelling of Physical Systems)

.

RLC circuit (see Modelling of Physical Systems)

.

11

)()(

2

RCsLCssEsE

i

o with andLCn12

LR

n 2

LCs

LRs

LC1

1

2

LCs

LRs

LC1

1

2

Spring-mass-damper.Spring-mass-damper.

KbsmsKbs

sXsX

i

o

2)(

)(

mKs

mbs

mKs

mb

2

with andmK

n 2

mb

n 2


Closed-Loop Position Feedback System(Servomechanism)

Closed-Loop Position Feedback System(Servomechanism)

V

1sK

s1

Gc

controller

E+

-

R

With Gc being a proportional gain Kp

E+

-

R)1( ss

KK p

with

KKpn 2

12 n

KK p

121

natural frequencydamping ratio

n

KKssKK

GHG

R p

p

21

In standard format

22

2

2 nn

n

ssR

ME2142/ME2142E Feedback Control Systems

Characteristic equation: 01 GH 0282

)2(51 2

ss

K

21

Examples: Determine the value of gain K for the closed-loop system to have anundamped natural frequency of 4. What will then be the damping factor?



2825

2 ss

010282 2 Kss

02514 222 nn ssKss

164)51( 22 Kn 3K

42 n 5.0


Transfer function: GHG

RC

1

22




2825

2 ss

164)51( 22 Kn 3K

42 n 5.0

KssK

102825

2

)51(45.2

2 KssK


Consider Consider 22

2

2 nn

n

ssR

Time Response – Second-order systemsTime Response – Second-order systems

For , the roots are equal and the system is said to be critically damped.For , the roots are equal and the system is said to be critically damped.

1 np 2,1

The roots of the characteristic equation are

122,1 nnp

For , the roots are both real and unequal and the system is said to be overdamped.For , the roots are both real and unequal and the system is said to be overdamped.

1 122,1 nnp

For , the roots are a pair of complex conjugates10

where is called the damped natural frequency and the response is underdamped.

dn jp 2,1

21 nd


Step Response – Second-order systemsStep Response – Second-order systems

Therefore

withwith ssR 1)( 22

2

2 nn

n

ssR

sss nn

n 1)2( 22

2

This represents a decaying oscillatory response depending upon with a frequency of oscillation of This represents a decaying oscillatory response depending upon with a frequency of oscillation of

Underdamped ResponseUnderdamped Response 10

d

From tables (Entry 24 in Tables), we haveFrom tables (Entry 24 in Tables), we have

)sin(1

1)(2

tetc d

tn

20 0t

2

1 1tan

21 nd


From tables (Entry 18 in Transform tables with ), we haveFrom tables (Entry 18 in Transform tables with ), we havena

)1(1)()(

122

atat ateea

tcass

25


We have We have

sss nn

n 1)2( 22

2

Critically damped ResponseCritically damped Response 1

ss n

1)( 2

2

This represents a non-oscillatory response with an exponentially decaying transient component and a zero steady-state error. The speed of decay of the transient component depends upon the parameter .

This represents a non-oscillatory response with an exponentially decaying transient component and a zero steady-state error. The speed of decay of the transient component depends upon the parameter .n

tn

t nn teetc 1)( 0tgiving for


We use Entry 17 in Transform tables. We use Entry 17 in Transform tables.


sss nn

n 1)2( 22

2

Overdamped ResponseOverdamped Response 1

sss nnnn

n

)1)(1( 22

2

0t

btat eCeC 211

for , C1 and C2 being constants.

The response is non-oscillatory, starts initially with and exponentially rises to . The response is non-oscillatory, starts initially with and exponentially rises to .

0)0( c1)( c

If , then and the first exponential term will decay much faster than the second. The pole can then be neglected and the system behaves like a first-order system.

If , then and the first exponential term will decay much faster than the second. The pole can then be neglected and the system behaves like a first-order system.

ba 1)( as

))((

2

bsassn

12 nna 12 nnbwith and

2nab We have

)

2

(11)( btat aebebaab

tcn

so that



Normalized response curves

For fast response,is usually

desirable.7.0

If no overshoot is required, is usually used.

1


Transient Response SpecificationsTransient Response Specifications

Maximum (percent) overshoot:

%100)(

)()(

cctc

M pp

Delay time

Rise time:10% - 90%, or5% - 95%, or0% - 100%

Peak time

Settling time: time to reach and stay within specified limits, usually 2% or 5%.

Five measures of transient performance – based on 2nd-order underdamped responseFive measures of transient performance – based on 2nd-order underdamped response


Measures of transient performanceMeasures of transient performance

We haveWe have

)sin(1

1)(2

tetc d

tn

2

1 1tan

1)( rtc giving 0)sin( rd t or 0 rd t

Thus rd t

2

1 1tan

2

1 1tan

Rise TimeRise Time rt

drt

giving

21 nd


We haveWe have

)sin(1

1)(2

tetc d

tn

2

1 1tan

Peak TimePeak Time pt

21 nd

01

)(sin)(2

pn

p

tnpd

tt

etdt

tcd

giving 0sin pd t or ,3,2,,0 pd t

Therefore for the first peak. d

pt



We haveWe have

)sin(1

1)(2

tetc d

tn

2

1 1tan

Maximum OvershootMaximum Overshoot

21 nd

pM

1)( pp tcM

])/(sin[1 2

)/(

dd

dne

)sin(1 2

)1/( 2

e

As 21)sin(

Therefore )1/( 2 eM p



We haveWe have

)sin(1

1)(2

tetc d

tn

2

1 1tan

Settling TimeSettling Time

21 nd

st

The curves gives the

envelope curves of the transient response. )1/(1 2 tne

is found to be approximately

where “time constant”

st

Tts 4

Tts 3

(2% criterion)

(5% criterion)

n

T

1



End

system transient/time responseguppy.mpe.nus.edu.sg/anpoo/fbcontrola/5system response.pdf ·...

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