5System Response

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ME2142 Feedback Control Systems1

System Transient/Time Response

ME2142 Feedback Control Systems


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System Response

For a stable system, the system response consists of two parts. One is transientresponse, and the other is steady-state response. The magnitude of the transientresponse decreases with time and ultimately vanishes leaving only the steady-stateresponse.


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For an unstable system, the magnitude of the system response increases with time.

t

Output

System Response


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System Characteristic Equation

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

Gc(s)

System

Input OutputR(s) C(s)

The systems characteristic equationis given by

0)( sDc

with

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

)()()(

)()(

sDsNsG

sRsC

c

cc

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


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Examples

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

Transfer Function

Characteristic Eqn:

Kbsms

Kbs

sX

sXsG

i

o

2)(

)()(

02 Kbsms

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

Transfer Function

Characteristic Eqn:

1

1

RCsE

E

i

o

01RCs

Closed-loop feedback system (Slide 9: Block Diagram Algebra)

Transfer Function

Characteristic Eqn:

GH

G

R

C

1

01 GH


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The rootsof this equation are the closed-loop poles.

Characteristic equation

0)( sDc

)(

)()(

)(

)(

sD

sNsG

sR

sC

c

c

c

If all the roots, pi, are negative, then the transient responsewill eventually die away as tincreases.

Consider a typical case, in which each root, xi=piof thisequation is a real number and is distinct from each other. Thuseach root will contribute a term in the time response ofthe system

But if anyof the roots is positive real value, then thetransient response will grow without boundsas timeincreases. The system is unstable.

tpie

.. ... .)( tpiAetc


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For a given dynamic system, we attempt to answer thefollowing questions:

System Response

We will Use standard test inputsto excite system and

analyze the response.

Identify the relationship between the performancecharacteristics and the system parameters.

How to determine the characteristics of the systemresponse?

How to compare it with that of another system?

How to identify whether the system response willadequately meet our needs (stability, error, etc.)?

How to know the system response subject todifferent inputs?


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4) Sinusoidal input

00

0sin)(

t

ttAtr

t

r(t)

t = 0

3) Impulse input

is an impulsefunctionor a Dirac deltafunction

)0()( Atr

AsR )(

A

t = 0

r(t)

t

System Response Test Signals

WhenA= 1, we have a unit impulseinput.

We can study the system response to suddenshocks or impacts.

We can analyze the system response in frequency domain.(the main topic in the second half of course)

Using test signals (1) to (3) are often known as time responseortransient responseanalyses, while using test signal (4) is known as

frequency response analyses.



System Response First-order Systems

A first-ordersystem can always be written in the standard form

T is known as the time constant, which may determine the speed ofresponse.

1)(

)(

Ts

K

sR

sC

Examples

Spring-damper system (Slides 7-8 of Modelling of Physical Systems)

.Kbs

K

sX

sY

)(

)(

KbT

Ts

with

1

1

RC circuit (Slide 12 of Modelling of Physical Systems)

.

RCTTsRCsE

E

i

o

with1

1

1

1




with

.

1)(

)(

Ts

K

sR

sC

Response to a unit step input

ssR 1)(

Thus )(1

)( sRTs

KsC

sTs

TK 1

)/1(

/

Ts

B

s

A

/1

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

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

Therefore

Ts

K

s

KsC

/1

)(

Using tables

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




Response 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.




with1)(

)(

Ts

K

sR

sC

Response to a unit ramp input

Thus )(1

)( sRTs

KsC

2

1)(

ssR

2

1

)/1(

/

sTs

TK

)/1(2 Ts

KT

s

KT

s

K

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)




with1)(

)(

Ts

K

sR

sC

Response to a unit impulse input

1)( sR

ThusTs

TK

Ts

KsC

/1

/

1)(

Or TteT

Ktc /)(

For K = 1,

TteT

tc /1)(

r(t)

t = 0

1/T

t


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The transient response all contains the term which is determined by

the root of the characteristic equation and the parameter T.

Tte /

Response to

Unit Impulse

Tte

T

Ktc /

1 )(

1)(

)(

Ts

K

sR

sCProperties Characteristic Equation

TsTs 1

01

Note that the unit stepis the derivative of the unit ramp, and the unit

impulseis 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 a linear time-invariant system, the response to the derivative of an

input can be obtained by taking the derivative of the response to the input.

Unit Step

)1()( /2TteKtc

Unit Ramp

)()( /3TtTeTtKtc



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Block diagram

Permanent Magnet DC Motor

e i

Ra

La

eK J

bT

The Permanent Magnet DC motor.

Governing equations

eaa Kdt

diLiRe

iKT t

bdt

dJT

IsLRKE aae )(

IKT t

)( bJsT

+

-

E I

aa RsL

1

eK

tKT

bJs

1

eK

+

-

E

eK

eK

))(( bJsRsL

K

aa

t



17

+

-

E

eK

eK

))(( bJsRsL

K

aa

t

The Permanent Magnet DC motor.

Commonly

La is small, and can then be neglected.

Block diagram then becomes

E +

-

eK

bJs

RK at

/

GH

G

E

1

)(

/1

)(

/

bJs

RKK

bJs

RK

aet

at

aet

at

RKKbJs

RK

/

/

1

s

K

aet

at

RKKb

RKK

/

/with

aet RKKb

J

/

Permanent Magnet DC Motor



18

aet RKKb

J

/

Speed Control of the DC Motor

1s

K

E

The respo nse to a unit step inpu t is

shows here.

Quest ions : Can we make the respo nse

faster? t = 0 t

K

With speed feedback

1sK

E

Error

+

-

V Kc

Controller

11

1

s

KKs

KK

V c

c

KKs

KK

c

c

1

1'

'

s

K

KKc1

'with

KK

KKK

c

c

1

'

The characteristic equation is still first-order, but the time constant is now

smaller, thus the system has faster response.

Motor by

itself



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System Response Second-order Systems

A second-ordersystem will be of the form

where a, b, c, d and e are constants.cbsas

esd

sX

sY

2)(

)(

Standard Form:

We can re-write

cbsas

esd

sX

sY

2)(

)(

a

cs

a

bs

sa

e

a

d

2

with anda

cn 2

a

bn 2

22

2

212

)()(

)(

nn

n

ssKsK

sR

sC



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Examples

RLC circuit (see Modelling of Physical Systems)

.

1

1

)(

)(2

RCsLCssE

sE

i

o with andLC

n

12 L

Rn 2

LCs

L

Rs

LC1

1

2

Spring-mass-damper.

Kbsms

Kbs

sX

sX

i

o

2)(

)(

m

Ks

m

bs

m

Ks

m

b

2 with andm

K

n2

m

b

n

2


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Closed-Loop Position Feedback System

V

1s

K

s

1 G

c

controller

E+

-

R

where Gcis a proportional gainKp

E+

-

R

)1( ss

KKp

with

KK

p

n

2

12

n

KKp

1

2

1

natural frequency

damping ratio

n

KKss

KK

GH

G

R p

p

2

1 In standard format

22

2

2 nn

n

ssR


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Characteristic equation: 01 GH 0282

)2(512

ss

K

22

Examples: Determine the value of gain Kfor the closed-loop system, which has anundamped natural frequency of 4. What is the damping factor?


R C

-

+

2

K282

52

ss

010282 2 Kss

02514

222 nnssKss

164)51( 22 Kn 3K

42 n 5.0


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Consider22

2

2 nn

n

ssR

Time Response Second-order Systems

For , the roots are equal and the system is so calledcritically 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 so called overdamped.

1 122,1 nnp

For , the roots are a pair of complex conjugates10

where is called the damped natural frequencyand

the response is underdamped.

dn jp 2,1

21 nd


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Step Response Second-order Systems

Therefore

with an inputs

sR 1)( 22

2

2 nn

n

ssR

sss nn

n 1

)2( 22

2

This represents a decaying oscillatoryresponse depending upon with

a frequency of oscillation of .

Underdamped Response 10

d

From tables (Entry 24 in Tables), we have

)sin(1

1)(2

te

tc d

tn

20 0t

2

1 1

tan

21 nd


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

)1(1

)()(

122

atatatee

atc

ass

25


We have

sss nn

n 1

)2( 22

2

Critically Damped Response 1

This represents a non-oscillatoryresponse with an exponentially

decaying transient component and a zero steady-state error. The speed

of decay of the transient response depends upon the parameter .n

t

n

t nn teetc

1)( 0tgiving for

ss n

n 12

2


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We use Entry 17 in Transform tables.


sss nn

n 1

)2( 22

2

Overdamped Response 1

sss nnnn

n

)1)(1( 22

2

0t

btat eCeC 211

for , C1andC2are constants.

The response is non-oscillatory, starts initially with andexponentially 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.

ba1)( as

))((

2

bsass

n

12 nna 1

2 nnb with and

2

nab We have

)

2

(1

1)( btat aebebaab

tc n

so that


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Step Response Second-order systems

Normalized response curves

For fast response,

is usually

desirable.

7.0

If no overshoot is

required,

is usually used.

1


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Transient Response Specifications

Maximum overshoot:

%100)(

)()(

c

ctcM

p

p

Delay time

Rise time:

10% - 90%, or

5% - 95%, or

0% - 100%

Peak time

Settling time: time to

reach and stay within

specified limits, usually

2% or 5%.

Five measures of transient performancebased on 2nd-order underdamped response


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Measures of Transient Performance

We have

)sin(11)(2

t

e

tc d

tn

2

1 1

tan

1)( rtc giving 0)sin( rdt

0 rdtThus we have

Rise Time rt

which gives

2

1 nd

or rdt

,0d .02

, which can not be satisfied, since

,0rt

d

rt


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We have

)sin(11)(2

t

e

tc d

tn

2

1 1

tan

Peak Time pt

2

1 nd

01

)(sin)(2

pn

p

tnpd

tt

etdt

tcd

giving 0sin pdt or ,3,2,,0 pdt

Therefore for the first peak.d

pt



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We have

)sin(11)(2

t

e

tc d

tn

2

1 1

tan

Maximum Overshoot

2

1 nd

pM

As

Therefore


1 pp tcM

dd

dne/sin

1 2

/

sin1 2

1/ 2

e

21/

eMp

21sinsin


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We have

)sin(11)(2

t

e

tc d

tn

2

1 1

tan

Settling Time

2

1 nd

st

The curves gives the

envelope curves of the transient response.

)1/(1 2 tne

is found to be approximately

where

st

Tts 4Tts 3

(2% criterion)

(5% criterion)

n

T

1



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Summarization First-order Systems

Transient response is exponential.

One dynamic parameter: time constant T.The smaller this is, thefaster the response.

At t=Tsec, response will reach 63.2%of final step.

At t=4Tsec, response will settle to within 2%of final value.


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Summarization Second-order systems

Normalized response curves

Two dynamicparameters:

Undamped naturalfrequency,

Damping ratio.

S i i i S ifi i


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Summarization - Transient Response Specifications

Maximum overshoot:

%100)(

)()(

c

ctcM

p

p

Settling time: 4T to

within 2% of the final

value.

Five measures of transient performancebased on 2nd-order underdamped response

n

T

1


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