4 - more time response (read-only)

06/03/2017

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

UWE Bristol

Industrial ControlUFMF6W-20-2

Control Systems EngineeringUFMEUY-20-3

Lecture 4: Second OrderTime Response

Today’s Lecture

• Last Week:– Inputs: Step, Ramp and Pulse– Response in the Time Domain (First order)– Final Value Theorem

• This Week:– Second order time response

Second Order Systems

• Many physical systems (even with higher order) can be approximated by a second order model

• Tools and specs are well developed and standardised

• Characterised by ζ (damping ratio) and ωn(natural frequency)

22

2

2

2 221

1)(nn

n

nn

sssssG

wzww

wwz ++

=++

=


• Unit step response (with unity gain):

22

2

2

2 221

1)(nn

n

nn

sssssG

wzww

wwz ++

=++

=

21

where

cossin1)(

zww

wwwwzzw

-=

÷÷ø

öççè

æ+-= -

nd

ddd

nt ttety n

Second Order System

y(t)

Time0

1- e-zw nt

1+ e-zw nt

1- e-zw nt zwn

wd

sinwd t + coswd tæ

è ç

ö

ø ÷

Time response(underdamped)

Second Order System

y(t)

Time0

Periodic Time(frequency is ωd rad/s)

2pwd

06/03/2017


Second Order System

y(t)

Time0

Effect of ζ

Increasing ζ

Under-damped ζ<1

Over-damped ζ>1

Critically damped ζ=1

ζ<1: under-damped(oscillatory response)(ζ=0: sustained oscillations)

ζ=1: critical damping(fastest response with no overshoot)

ζ>1: over-damped(Cannot overshoot)

Second Order System

y(t)

Time0 tr tp ts

y(tp) –yss

95%

5% settling time

yss

Performance Criteria• Rise time, tr

– Time to reach steady state value (for underdamped systems)

– Time to go from 10% to 90% aplitude for overdampedsystems

• Peak time, tp– Time to initial overshoot

• Peak overshoot, y(tp) –yss– Initial overshoot, above

steady state value

• Settling time, ts– Time for response to reach and

remain in specified ratio (±5% in this case)

• Number of oscillations to ts


• Rise time– By setting y(tr) = 1 in response (and much

manipulation):

– Low damping ratio gives faster response (small tr), but also produce oscillatory responses.

– Compromise necessary (0.4<ζ<0.8 a good start)

÷÷

ø

ö

çç

è

æ --= -

zz

pw

21 1

tan1

drt

Second Order System

• Time to peak:– Time to peak: from inspection, for oscillatory

system, time to peak is one half periodic response:

– Confirmed by setting first derivative of y to 0 (and much manipulation)

ddpt w

pwp==

221

Second Order System

• Peak overshoot– Use tp and evaluate using time response

– Percentage maximum peak value:

( )%100

valuefinal valuefinal - peak value

´÷÷ø

öççè

æ -=

ss

ssp

yyty

( ) ÷÷ø

öççè

æ+-= -

pdpdd

ntp ttety pn ww

wwzzw cossin1

Second Order System

• Settling Time– Decaying exponential term describes

envelope of the oscillation– Equate to 0.05 gives the 5% settling time

( )pnte zw-

( )

ns

snt

t

te sn

zw

zwzw

3%)5(

so305.0ln05.0

=

-==-®=-

06/03/2017


Second Order System

• Number of oscillations before settling time

timePeriodic timeSettling nsoscillatio ofNumber =

Example

• Mass-Spring-Damper system

m

k

c

f(t)

x(t)

( ) ( )( ) 42

11

22 ++

=++

==ss

mks

mcsm

sFsXsG

Mass, m = 1 kgSpring coefficient, k = 4 N/mDamping coefficient, c = 2 Ns/m

34321

21

rad/s 24

2 ==-=

=

==

zww

z

w

nd

n

Example

• Mass-Spring-Damper

m

k

c

f(t)

x(t)

( )42

12 ++

=ss

sG

Mass, m = 1 kgSpring coefficient, k = 4 N/mDamping coefficient, c = 2 Ns/mInput Force = unit step

To put in standard form, numerator must equal ωn

2 so we multiply by 1/4

( )42

441

2 ++=

sssG

( ) úû

ùêë

é÷ø

öçè

æ +-= - ttetc t 3cos3sin311

41

34321

21

rad/s 24

2 ==-=

=

==

zww

z

w

nd

n

Example

• Mass-Spring-Damper– Periodic Time:

– Rise time:

– Time to peak:

34321

21

rad/s 24

2 ==-=

=

==

zww

z

w

nd

n

( ) sec 209.13tan3

11tan1 1

21 =-=

÷÷

ø

ö

çç

è

æ --= -- p

zz

pwd

rt

sec 628.33

22Time Periodic ===p

wp

d

sec 804.13

221

====p

wp

wp

ddpt

Example

• Mass-Spring-Damper– Peak overshoot

• Apply tp to equation

– Percentage increase

34321

21

rad/s 24

2 ==-=

=

==

zww

z

w

nd

n

( )

%6100%125.0

25.0291.0

00%1 valuefinal

valuefinal - peak value

=´÷øö

çèæ -

=

´÷÷ø

öççè

æ -=

ss

ssp

cctc

( ) 291.0)804.1(3cos)804.1(3sin311

41 804.1 =ú

û

ùêë

é÷ø

öçè

æ +-= -etc p

Example

• Mass-Spring-Damper– 5% Settling Time

– Number of oscillations

34321

21

rad/s 24

2 ==-=

=

==

zww

z

w

nd

n

( ) sec 3)2(5.0

33%)5( ===n

st zw

827.0628.33

timePeriodic timeSettling nsoscillatio ofNumber ===

06/03/2017


Example

• Mass-Spring-Damper system

s 628.3Time Periodic =

s 3=st

s 209.1=rts 804.1=pt

%)16( 291.0overshootPeak =

Today’s lecture• Step response of Second Order System• Can be used to represent many (higher order)

systems• Defined by natural frequency and damping ratio• Damping ratio relates to oscillation in response• Performance criteria

– Rise time– Peak time and amplitude– Settling time

• Tutorial Sheet 4: Identifying performance criteria

4 - more time response (read-only)

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