Time domain analysis

This presentation gives complete idea about time domain analysis of first and second order system, type number, time domain specifications, steady state error and error constants and numerical examples.

1. By
Dr.K.Hussain
Associate Professor & Head
Dept. of EE, SITCOE
Time Domain Analysis

2. Introduction
•Intime-domainanalysistheresponseofadynamicsystemto
aninputisexpressedasafunctionoftime.
•Itispossibletocomputethetimeresponseofasystemifthe
natureofinputandthemathematicalmodelofthesystem
areknown.
•Usually,theinputsignalstocontrolsystemsarenotknown
fullyaheadoftime.
•Itisthereforedifficulttoexpresstheactualinputsignals
mathematicallybysimpleequation

3. Standard Test Signals
•Thecharacteristicsofactualinputsignalsareasuddenshock,
asuddenchange,aconstantvelocity,andconstant
acceleration.
•Thedynamicbehaviorofasystemisthereforejudgedand
comparedunderapplicationofstandardtestsignals–an
impulse,astep,aconstantvelocity,andconstantacceleration.
•Theotherstandardsignalofgreatimportanceisasinusoidal
signal.

4. Standard Test Signals
•Impulse signal
–Theimpulsesignalimitatethe
suddenshockcharacteristicof
actualinputsignal.
–IfA=1,theimpulsesignalis
calledunitimpulsesignal.
0
t
δ(t)
A





00
0
t
tA
t


)(

5. Standard Test Signals
•Step signal
–Thestepsignalimitate
thesuddenchange
characteristicofactual
inputsignal.
–IfA=1,thestepsignalis
calledunitstepsignal





00
0
t
tA
tu


)(
0
t
u(t)
A

6. Standard Test Signals
•Ramp signal
–Therampsignalimitate
theconstantvelocity
characteristicofactual
inputsignal.
–IfA=1,therampsignal
iscalledunitramp
signal





00
0
t
tAt
tr


)(
0
t
r(t)

7. Standard Test Signals
•Parabolic signal
–Theparabolicsignal
imitatetheconstant
accelerationcharacteristic
ofactualinputsignal.
–IfA=1,theparabolic
signaliscalledunit
parabolicsignal.







00
0
2
2
t
t
At
tp


)(
0
t
p(t)
parabolic signal with slope A
p(t)
Unit parabolic signal
p(t)

8. Relation between standard Test Signals
•Impulse
•Step
•Ramp
•Parabolic





00
0
t
tA
t


)( 





00
0
t
tA
tu


)( 





00
0
t
tAt
tr


)( 







00
0
2
2
t
t
At
tp


)(    dt
d dt
d dt
d

9. Laplace Transform of Test Signals
•Impulse
•Step





00
0
t
tA
t


)( AstL )()}({ 





00
0
t
tA
tu


)( S
A
sUtuL )()}({

10. Laplace Transform of Test Signals
•Ramp
•Parabolic2
s
A
sRtrL )()}({ 3
)()}({
S
A
sPtpL  





00
0
t
tAt
tr


)( 







00
0
2
2
t
t
At
tp


)(

11. Time Response of Control Systems
System
•Thetimeresponseofanysystemhastwocomponents
•Transientresponse
•Steady-stateresponse.
•Timeresponseofadynamicsystemresponsetoaninput
expressedasafunctionoftime.

12. Time Response of Control Systems
•Whentheresponseofthesystemischangedfromequilibriumit
takessometimetosettledown.Thisiscalledtransientresponse.0 2 4 6 8 10 12 14 16 18 20
0
1
2
3
4
5
6
x 10
-3


Step Response
Time (sec)
Amplitude
Response
Step Input
Transient Response
Steady State Response
•Theresponseofthe
systemafterthetransient
responseiscalledsteady
stateresponse.

13. Time Response of Control Systems
•Transientresponsedependuponthesystempolesonlyandnot
onthetypeofinput.
•Itisthereforesufficienttoanalyzethetransientresponseusinga
stepinput.
•Thesteady-stateresponsedependsonsystemdynamicsandthe
inputquantity.
•Itisthenexaminedusingdifferenttestsignalsbyfinalvalue
theorem.

14. First order system
•Thefirstordersystemhasonlyonepole.
•WhereKistheD.CgainandTisthetimeconstant
ofthesystem.
•Timeconstantisameasureofhowquicklya1
st
ordersystemrespondstoaunitstepinput.
•D.CGainofthesystemisratiobetweentheinput
signalandthesteadystatevalueofoutput.1

Ts
K
sR
sC
)(
)(

15. Example 1
•Thefirstordersystemgivenbelow.13
10


s
sG)( 5
3


s
sG)( 151
53


s/
/
•D.Cgainis10andtimeconstantis3seconds.
•Forthefollowingsystem
•D.CGainofthesystemis3/5andtimeconstantis1/5
seconds.

16. Impulse Response of 1
st
Order System
•Consider the following 1
st
order system1Ts
K )(sC )(sR
0
t
δ(t)
11)()( ssR 1

Ts
K
sC)(

17. Impulse Response of 1
st
Order System
•Re-arrange following equation as1

Ts
K
sC)( Ts
TK
sC
/
/
)(
1
 Tt
e
T
K
tc
/
)(


•Inordertocomputetheresponseofthesystemintimedomain
weneedtocomputeinverseLaplacetransformoftheabove
equation.at
Ce
as
C
L








1

18. Step Response of 1
st
Order System
•Consider the following 1
st
order system1Ts
K )(sC )(sR s
sUsR
1
)()( 1

Tss
K
sC)( 1

Ts
KT
s
K
sC)(
•InordertofindouttheinverseLaplaceoftheaboveequation,we
needtobreakitintopartialfractionexpansion

19. Step Response of 1
st
Order System
•Taking Inverse Laplace of above equation







1
1
Ts
T
s
KsC)(  
Tt
etuKtc
/
)()(


•Where u(t)=1 
Tt
eKtc
/
)(

1   KeKtc 63201
1
.)( 

•When t=T (time constant)

20. Relation Between Step and impulse
response
•The step response of the first order system is
•Differentiating c(t) with respect to t yields 
TtTt
KeKeKtc
//
)(

1  
Tt
KeK
dt
d
dt
tdc
/)(

 Tt
e
T
K
dt
tdc
/)(



21. Example 2
•Ifinitialconditionsarenotknownthenpartialfraction
expansionisabetterchoice12
6


SsR
sC
)(
)(  12
6


Ss
sC)(   1212
6


 s
B
s
A
Ss s
sRsR
1
)(,)( input step a is since   50
66
12
6
.

 ssSs t
etc
50
66
.
)(



22. Ramp Response of 1
st
Order System
•Consider the following 1
st
order system1Ts
K )(sC )(sR 2
1
s
sR)(  1
2


Tss
K
sC)(
•Therampresponseisgivenas 
Tt
TeTtKtc
/
)(



23. Parabolic Response of 1
st
Order System
•Consider the following 1
st
order system1Ts
K )(sC )(sR 3
1
s
sR)(  1
3


Tss
K
sC)(
Therefore,

24. Second Order System
•Wehavealreadydiscussedtheaffectoflocationofpolesandzeroson
thetransientresponseof1
st
ordersystems.
•Comparedtothesimplicityofafirst-ordersystem,asecond-ordersystem
exhibitsawiderangeofresponsesthatmustbeanalyzedanddescribed.
•Varyingafirst-ordersystem'sparameter(T,K)simplychangesthespeed
andoffsetoftheresponse
•Whereas,changesintheparametersofasecond-ordersystemcan
changetheformoftheresponse.
•Asecond-ordersystemcandisplaycharacteristicsmuchlikeafirst-order
systemor,dependingoncomponentvalues,displaydampedorpure
oscillationsforitstransientresponse. 24

25. Second Order System
•Ageneralsecond-ordersystemischaracterizedbythe
followingtransferfunction.22
2
2
nn
n
sssR
sC




)(
)(
25
un-dampednaturalfrequencyofthesecondordersystem,
whichisthefrequencyofoscillationofthesystemwithout
damping.n
dampingratioofthesecondordersystem,whichisameasure
ofthedegreeofresistancetochangeinthesystemoutput.

26. Example 342
4
2


sssR
sC
)(
)(
•Determinetheun-dampednaturalfrequencyanddampingratio
ofthefollowingsecondordersystem.4
2

n 22
2
2
nn
n
sssR
sC




)(
)(
Sol:Comparethenumeratoranddenominatorofthegiventransfer
functionwiththegeneral2
nd
ordertransferfunction.2
n
 ss
n22  422
222
 ssss
nn 50. 1
n
26

27. 22
2
2
nn
n
sssR
sC




)(
)( •Twopolesofthesystemare1
1
2
2




nn
nn
27
Second Order System

28. Second Order System Cont..
•Accordingthevalueof ,asecond-ordersystemcanbesetinto
oneofthefourcategories:1
1
2
2




nn
nn 
1.Overdamped-whenthesystemhastworealdistinctpoles(>1).
-a-b-c
δ
jω
28

29. 1
1
2
2




nn
nn 2.Underdamped-whenthesystemhastwocomplexconjugatepoles(0<<1)
-a-b-c
δ
jω
29

30. 1
1
2
2




nn
nn 3.Undamped-whenthesystemhastwoimaginarypoles(=0).
-a-b-c
δ
jω
30

31. 1
1
2
2




nn
nn 4.Criticallydamped-whenthesystemhastworealbutequalpoles(=1).
-a-b-c
δ
jω
31

32. Underdamped System
32
For 0<<1 and ω
n> 0, the 2
nd
order system’s response due to a
unit step input is as follows.
Important timing characteristics: delay time, rise time, peak
time, maximum overshoot, and settling time.

33. Delay Time
33
•Thedelay(t
d)timeisthetimerequiredfortheresponseto
reachhalfthefinalvaluetheveryfirsttime.

34. Rise Time
34
•Therisetimeisthetimerequiredfortheresponsetorisefrom10%
to90%,5%to95%,or0%to100%ofitsfinalvalue.
•Forunderdampedsecondordersystems,the0%to100%risetimeis
normallyused.Foroverdampedsystems,the10%to90%risetimeis
commonlyused.
34

35. Peak Time
35
•The peak time is the time required for the response to reach
the first peak of the overshoot.
3535

36. Maximum Overshoot
36
Themaximumovershootisthemaximumpeakvalueofthe
responsecurvemeasuredfromunity.Ifthefinalsteady-state
valueoftheresponsediffersfromunity,thenitiscommonto
usethemaximumpercentovershoot.Itisdefinedby
Theamountofthemaximum(percent)overshootdirectly
indicatestherelativestabilityofthesystem.

37. Settling Time
37
•Thesettlingtimeisthetimerequiredfortheresponsecurve
toreachandstaywithinarangeaboutthefinalvalueofsize
specifiedbyabsolutepercentageofthefinalvalue(usually2%
or5%).

38. Step Response of underdamped System222222
2
21
nnnn
n
ss
s
s
sC




)(
•Thepartialfractionexpansionofaboveequationisgivenas22
2
21
nn
n
ss
s
s
sC




)(  
2
2
n
s  
22
1
n    
222
1
21





nn
n
s
s
s
sC)( 22
2
2
nn
n
sssR
sC




)(
)(  
22
2
2
nn
n
sss
sC



)(
Step Response
38

39. Step Response of underdamped System
•Aboveequationcanbewrittenas   
222
1
21





nn
n
s
s
s
sC)(  
22
21
dn
n
s
s
s
sC




)( 2
1 
nd
•Where ,isthefrequencyoftransientoscillations
andiscalleddampednaturalfrequency.
•TheinverseLaplacetransformofaboveequationcanbeobtained
easilyifC(s)iswritteninthefollowingform:   
2222
1
dn
n
dn
n
ss
s
s
sC








)(
39

40. Step Response of underdamped System   
2222
1
dn
n
dn
n
ss
s
s
sC








)(    
22
2
2
22
1
11
dn
n
dn
n
ss
s
s
sC












)(    
22
2
22
1
1
dn
d
dn
n
ss
s
s
sC










)( tetetc
d
t
d
t
nn





sincos)(



2
1
1
40

41. Step Response of underdamped Systemtetetc
d
t
d
t
nn





sincos)(



2
1
1 










ttetc
dd
t
n





sincos)(
2
1
1
412
2
1
( ) 1 sin( )
1
1
tan
n
t
d
e
c t t
where








  




42. Steady State Error
•Iftheoutputofacontrolsystematsteadystate
doesnotexactlymatchwiththeinput,thesystem
issaidtohavesteadystateerror
•Anyphysicalcontrolsysteminherentlysuffers
steady-stateerrorinresponsetocertaintypesof
inputs.
•Asystemmayhavenosteady-stateerrortoastep
input,butthesamesystemmayexhibitnonzero
steady-stateerrortoarampinput.

43. Classification of Control Systems
•Controlsystemsmaybeclassifiedaccordingto
theirabilitytofollowstepinputs,rampinputs,
parabolicinputs,andsoon.
•Themagnitudesofthesteady-stateerrorsdue
totheseindividualinputsareindicativeofthe
goodnessofthesystem.

44. Classification of Control Systems
•Considertheunity-feedbackcontrolsystem
withthefollowingopen-looptransferfunction
•Itinvolvestheterms
N
inthedenominator,
representingNpolesattheorigin.
•Asystemiscalledtype0,type1,type2,...,if
N=0,N=1,N=2,...,respectively.

45. Classification of Control Systems
•Asthetypenumberisincreased,accuracyis
improved.
•However,increasingthetypenumber
aggravatesthestabilityproblem.
•Acompromisebetweensteady-stateaccuracy
andrelativestabilityisalwaysnecessary.

46. Steady State Error of Unity Feedback Systems
•Consider the system shown in following figure.
•The closed-loop transfer function is

47. Steady State Error of Unity Feedback Systems
•ThetransferfunctionbetweentheerrorsignalE(s)andthe
inputsignalR(s)is)()(
)(
sGsR
sE


1
1
•Thefinal-valuetheoremprovidesaconvenientwaytofind
thesteady-stateperformanceofastablesystem.
•SinceE(s)is
•Thesteadystateerroris

48. Static Error Constants
•Thestaticerrorconstantsarefiguresofmeritof
controlsystems.Thehighertheconstants,the
smallerthesteady-stateerror.
•Inagivensystem,theoutputmaybetheposition,
velocity,pressure,temperature,orthelike.
•Therefore,inwhatfollows,weshallcalltheoutput
“position,”therateofchangeoftheoutput
“velocity,”andsoon.
•Thismeansthatinatemperaturecontrolsystem
“position”representstheoutputtemperature,
“velocity”representstherateofchangeofthe
outputtemperature,andsoon.

49. Static Position Error Constant (K
p)
•The steady-state error of the system for a unit-step input is
•The static position error constant K
pis defined by
•Thus, the steady-state error in terms of the static position
error constant K
pis given by

50. Static Position Error Constant (K
p)
•For a Type 0system
•For Type 1or higher order systems
•For a unit step input the steady state error e
ssis

51. •The steady-state error of the system for a unit-ramp input is
•The static velocity error constant K
vis defined by
•Thus, the steady-state error in terms of the static velocity
error constant K
vis given by
Static Velocity Error Constant (K
v)

52. Static Velocity Error Constant (K
v)
•For a Type 0system
•For Type 1systems
•For type 2 or higher order systems

53. Static Velocity Error Constant (K
v)
•For a ramp input the steady state error e
ssis

54. •The steady-state error of the system for parabolic input is
•The static acceleration error constant K
ais defined by
•Thus,thesteady-stateerrorintermsofthestaticacceleration
errorconstantK
aisgivenby
Static Acceleration Error Constant (K
a)

55. Static Acceleration Error Constant (K
a)
•For a Type 0system
•For Type 1systems
•For type 2systems
•For type 3or higher order systems

56. Static Acceleration Error Constant (K
a)
•For a parabolic input the steady state error e
ssis

57. Summary of Steady State Errors

58. Example4
•Forthesystemshowninfigurebelowevaluatethestatic
errorconstantsandfindtheexpectedsteadystateerrors
forthestandardstep,rampandparabolicinputs.
C(S)R(S)
-))((
))((
128
52100
2


sss
ss

59. Example 4))((
))((
)(
128
52100
2



sss
ss
sG )(limsGK
s
p
0
 










 ))((
))((
lim
128
52100
2
0 sss
ss
K
s
p 
p
K )(limssGK
s
v
0
 










 ))((
))((
lim
128
52100
2
0 sss
sss
K
s
v 
vK )(lim sGsK
s
a
2
0
 










 ))((
))((
lim
128
52100
2
2
0 sss
sss
K
s
a 410
12080
5020100
.
))((
))((












aK

60. Example 4
p
K 
vK 410.
aK 0 0 090.

61. Ex 5: The open loop transfer function of a servo system
with unity feedback is given by
Determinethedampingratio,undampednatural
frequencyofoscillation.Whatisthepercentage
overshootoftheresponsetoaunitstepinput.
SOLUTION: Given that
Characteristicequation
10
(s2)(s5)
G(s)
10
H(s)1
(s2)(s5)
G(s)
1G(s)H(s)0

62. 10
01
s
2
7s200
(s2)(s5)
Comparewiths
2
2s
2
0
n n
Weget
*1001.92%
16
1(0.7826)
2

*0.7826
1
2

2**4.4727
0.7826

2
20
e
204.472rad/sec
M
p e

n 
n
2
n 7 
n
4.472rad /sec
0.7826
M
p 1.92%

63. H (s) Ks
The damping factor of the system is 0.8. Determine the
overshoot of the system and value of ‘K’.
SOLUTION: We know that
12
G(s)
s
2
4s16
s
2
(416K)s160
s
2
(416K)s16R(s)
C(s)

1G(s)H(s)
16
G(s)
R(s)
C(s)

is the characteristiceqn.
Ex 6: A feedback system is described by the following transfer
function

64. nn
2
n
2
Compare with
2
16
s2s0

n 4rad/sec.
2
n4 16K
2*0.8*4 4 16K K0.15
*100
1(0.8)
2

M
p1.5%
*100 eM
pe
0.8
1
2


65. Ex 7: The open loop transfer function of unity feedback
system is given by
SOLUTION:
50
(10.1s)(s10)
G(s)
DeterminethestaticerrorcoefficientsK
p,K
vandK
a
50
50
50
0
0
5lim
(10.1s)(s10)
(10.1s)(s10)
Klims.G(s)H(s)
KlimG(s)H(s)
lim s
2
s0
2
K
a s G(s)H (s)
lims.
s0
s0
v
s0 (10.1s)(s10)
s0
p