06/03/2017

© 2017 University of the West of England, Bristol. 1

UWE Bristol

Industrial ControlUFMF6W-20-2

Control Systems EngineeringUFMEUY-20-3

Lecture 3: Modelling SystemsTime Response

© 2017 University of the West of England, Bristol.

Today’s Lecture

• Inputs: Step, Ramp and Pulse• Response in the Time Domain (First order)

– Step– Ramp

• Final Value Theorem

© 2017 University of the West of England, Bristol.

Step Input• Consider:

00

for 0

)(<³

îíì

=ttA

tf

( )sA

sAe

sAe

sAedtAesF

stst =÷÷

ø

öççè

æ---=-==

-¥-¥-¥-ò

0

00

If A = 1, it is called a unit step function

© 2017 University of the West of England, Bristol.

Ramp Input

00

for 0

)(<³

îíì

=ttAt

tf

( )

2

stepunit 000

00

0)(sAdte

sAdt

se

steAsF

dtetAdtAtesF

ststst

se

vdtedv

dtdutu

dv

st

u

st

stst

=+=úúû

ù

êêë

é+-=

==

òò

òò

¥-

¥ -¥-

-=®=

=®=

¥-

¥-

--

ò ò ×-×=× duvvudvu :remember

© 2017 University of the West of England, Bristol.

Pulse Input

0

00

and 00

for 0

)(ttt

tttAtf

><££

îíì

=

t0

• Area under function, A

• If t0 approaches zero à impulse• If A=1:

• Unit pulse or

• Unit impulse δ(t)

• Example:

• Physiology: shot of drug

• Hydraulics: dumping of fluid

Area = A

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Pulse Input

0

00

and 00

for 0

)(ttt

tttAtf

><££

îíì

=

( )000

1)(0000 0

stt

t

ststt

estAe

stAdte

tAsF -

=

-- -=-

== ò

( )A

sseA

stdtd

edtd

AsteAsF

st

t

st

t

st

t=ú

û

ùêë

é=

úúúú

û

ù

êêêê

ë

é -=ú

û

ùêë

é -=

-

®

-

®

-

®

0

0

0

0

0

0 00

0

0

00

0lim

1lim1lim)(

For an impulse, set t0 à 0

For a unit impulse, L[δ(t)] = 1L’Hôpital’s Rule© 2017 University of the West of

England, Bristol.

06/03/2017

© 2017 University of the West of England, Bristol. 2

Inputs

• Step Input:

• Ramp Input:

• Impulse:

• For unit response, replace A with 1

( ) 2sAsF =

( )sAsF =

( ) AsF =

© 2017 University of the West of England, Bristol.

Applying Inputs

• Transfer function:

• R(s) is input• Step input: • Ramp input:• Impulse:

G(s)R(s) C(s)

)()( sGsAsC =

)()( 2 sGsAsC =

)()( sAGsC =

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Time Response

• Determine Time Response– 1. Apply Input to Transfer Function– 2. Simplify (if possible)– 3. Use Inverse Laplace Transform Tables to

convert from the s-domain back to the time-domain

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First Order: Unit Step

• Unit Step Input: R(s) = 1/s• Transfer Function:

• Output:• Not listed on Inverse LT tables!

asA

sssG

+=

+=´

+=

ttg

tt

tg

111

1)(

ttg 1 and where == aA

)(11)(ass

AasA

ssC

+=

+´=

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Partial Fraction Expansion

• A0 = 0 (numerator and denominator have different orders)

asA

sAA

assAsC

+++=

+= 21

0)()(

( )atat eaAe

aA

aAtc

asaA

saA

assAsC -- -=-=®

+-=

+= 1)(11

)()(

( ) sAasAAas

AsA

assAsC 21

21

)()( ++=®

++=

+=

( ) aAAAsAsAaAsAA

121

211

++=++=

aAAaAA =®= 11

aAAAA -=®+= 2210

groupcommon terms

© 2017 University of the West of England, Bristol.

First Order: Unit Step

• We can use equation to estimate response

time Output

0 0

τ 0.632%

3τ 0.95%

5τ 0.99%

10τ 0.999%

( )ateaAtc --= 1)(

ttg 1 and where == aA ÷÷

ø

öççè

æ-=

-tgt

etc 1)(

© 2017 University of the West of England, Bristol.

06/03/2017

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Example: Unit Step

• Unit Step

1 1 == tg

÷÷ø

öççè

æ-=

-t

tg t

etc 1)(

63.2%

95%

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First Order: Unit Step

• Any first order system represented by:

reaches 95% of final value after three time constants

• The time constant, τ, must be small for a fast response

• First order lag

÷÷ø

öççè

æ-=

-tt

eKtc 1)(

© 2017 University of the West of England, Bristol.

First Order: Unit Ramp

• Unit Ramp Input: R(s) = 1/s2

• Transfer Function:

• Output:• Not listed on Inverse LT tables!

asA

sssG

+=

+=´

+=

ttg

tt

tg

111

1)(

ttg 1 and where == aA

)(11)( 22 ass

AasA

ssC

+=

+´=

© 2017 University of the West of England, Bristol.

Partial Fraction Expansion

• A0 = 0 (numerator and denominator have different orders)

• Solution with:

asA

sA

sAA

assAsC

++++=

+= 32

21

02 )()(

÷÷ø

öççè

æ--=

-ttt

ettc 1)(

1 1 == tg

© 2017 University of the West of England, Bristol.

Example: Unit Ramp

• Unit Ramp

÷÷ø

öççè

æ--=

-ttt

ettc 1)(

4τ

τ

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Example: First Order

• Unit Impulse Input: R(s) = 1• Transfer Function:

• Output:• Inverse LT:

asA

sssG

+=

+=´

+=

ttg

tt

tg

111

1)(

ttg 1 and where == aA

asAsC+

´=1)(

t

tg t

at eAetcasAsC

-- ==®+

= )()(

© 2017 University of the West of England, Bristol.

06/03/2017

© 2017 University of the West of England, Bristol. 4

Example: First Order

• Unit Impulse with 1 1 == tg

t

tg t

etc-

=)(

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Final Value Theorem

• To find the final steady state value

• Example:

)(lim)(lim0

ssFtfst ®¥®

=

5)(lim)(

33

5)(

s3F(s)input ;

35)(

)()(

==¥+

=

=+

==

ssXxss

sX

ssG

sFsX

© 2017 University of the West of England, Bristol.

Today’s lecture• Inputs: Pulse (impulse), Step, Ramp• To determine time response: get transfer

function in s-domain, apply input, then go back to time domain using:– Inverse Laplace transforms– Partial Fraction technique then ILT

• Step response• Ramp response• Final Value Theorem

© 2017 University of the West of England, Bristol.