Time domain specifications of second order system

presentation slide on TIME DOMAIN SPECIFICATIONS OF SECOND ORDER

1. TIME DOMAIN SPECIFICATIONS OF
SECOND ORDER SYSTEM
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1SYED HASAN SAEED

2. SYED HASAN SAEED 2
BOOKS
1.AUTOMATICCONTROLSYSTEMKUO&
GOLNARAGHI
2.CONTROLSYSTEMANANDKUMAR
3.AUTOMATICCONTROLSYSTEMS.HASANSAEED

3. SYED HASAN SAEED 3
Considerasecondordersystemwithunitstepinputand
allinitialconditionsarezero.Theresponseisshowninfig.

4. 1.DELAYTIME(t
d):Thedelaytimeisthetimerequired
fortheresponsetoreach50%ofthefinalvaluein
firsttime.
2.RISETIME(t
r):Itistimerequiredfortheresponseto
risefrom10%to90%ofitsfinalvalueforover-
dampedsystemsand0to100%forunder-damped
systems.
Weknowthat:
SYED HASAN SAEED 4  






2
1
2
2
1
tan
1sin
1
1)(







t
e
tc
n
t
n
Where,

5. Letresponsereaches100%ofdesiredvalue.Putc(t)=1
SYED HASAN SAEED 5  
  01sin
1
1sin
1
11
2
2
2
2













t
e
t
e
n
t
n
t
n
n 0
t
n
e

Since,)sin())1sin((
0))1sin((
2
2


nt
t
n
n


Or,
Put n=1

6. SYED HASAN SAEED 62
2
1
)1(







n
r
rn
t
t 2
2
1
1
1
tan









n
rt
Or,
Or,

7. 3.PEAKTIME(t
p):Thepeaktimeisthetimerequired
fortheresponsetoreachthefirstpeakofthetime
responseorfirstpeakovershoot.
Formaximum
SYED HASAN SAEED 7  






t
e
tc
n
t
n
2
2
1sin
1
1)(
Since   
   )1(0
1
1sin
11cos
1
)(
0
)(
2
2
22
2










tn
n
nn
t
n
n
et
t
e
dt
tdc
dt
tdc








8. Since,
Equation can be written as
Equation (2) becomes
SYED HASAN SAEED 80
t
n
e
      


sin1
1sin11cos
2
222

 tt
nn
Put cos and      cos1sinsin1cos
22
 tt
nn 



cos
sin
))1cos((
))1sin((
2
2



t
t
n
n

9. SYED HASAN SAEED 9

nt
nt
pn
n


)1(
))1tan((
2
2
The time to various
peak
Where n=1,2,3,…….
Peak time to first overshoot, put n=12
1



n
pt
First minimum (undershoot) occurs at n=22
min
1
2



n
t

10. 4. MAXIMUM OVERSHOOT (M
P):
Maximum overshoot occur at peak time, t=t
p
in above equation
SYED HASAN SAEED 10  






t
e
tc
n
t
n
2
2
1sin
1
1)( 2
1



n
pt
Put, 






















2
2
2
1
1
.1sin
1
1)(
2
n
n
n
n
e
tc

11. SYED HASAN SAEED 112
2
1
2
2
1
2
1
1
1
1)(
1sin
sin
1
1)(
)sin(
1
1)(
2
2
2




























e
tc
e
tc
e
tc
Put,  sin)sin( 

12. SYED HASAN SAEED 122
2
2
1
1
1
11
1)(
1)(
















eM
eM
tcM
etc
p
p
p 100*%
2
1



eM
p

13. 5. SETTLING TIME (t
s):
Thesettlingtimeisdefinedasthetimerequiredforthe
transientresponsetoreachandstaywithinthe
prescribedpercentageerror.
SYED HASAN SAEED 13

14. SYED HASAN SAEED 14
Time consumed in exponential decay up to 98% of the
input. The settling time for a second order system is
approximately four times the time constant ‘T’.
6.STEADYSTATEERROR(e
ss):Itisdifferencebetween
actualoutputanddesiredoutputastime‘t’tendsto
infinity.n
sTt

4
4  )()( tctrLimite
t
ss 


15. EXAMPLE1:Theopenlooptransferfunctionofaservo
systemwithunityfeedbackisgivenby
Determinethedampingratio,undampednaturalfrequency
ofoscillation.Whatisthepercentageovershootofthe
responsetoaunitstepinput.
SOLUTION:Giventhat
Characteristicequation
SYED HASAN SAEED 15)5)(2(
10
)(


ss
sG 1)(
)5)(2(
10
)(



sH
ss
sG 0)()(1  sHsG

16. SYED HASAN SAEED 160207
0
)5)(2(
10
1
2




ss
ss
Compare with02
22

nn
ss  We get%92.1100*
7826.0
7472.4**2
sec/472.420
72
20
22
)7826.0(1
7826.0*
1
2


















eeM
rad
p
n
n
n %92.1
7826.0
sec/472.4


p
n
M
rad



17. EXAMPLE2:Afeedbacksystemisdescribedbythe
followingtransferfunction
Thedampingfactorofthesystemis0.8.determinethe
overshootofthesystemandvalueof‘K’.
SOLUTION:Weknowthat
SYED HASAN SAEED 17KssH
ss
sG



)(
164
12
)(
2 016)164(
16)164(
16
)(
)(
)()(1
)(
)(
)(
2
2





sKs
sKssR
sC
sHsG
sG
sR
sC
is the characteristic eqn.

18. Compare with
SYED HASAN SAEED 18K
ss
n
n
nn
1642
16
02
2
22





 .sec/4rad
n
 K1644*8.0*2  15.0K %5.1
100*100*
22
)8.0(1
8.0
1






p
p
M
eeM




19. EXAMPLE3:Theopenlooptransferfunctionofaunity
feedbackcontrolsystemisgivenby
Bywhatfactortheamplifiergain‘K’shouldbemultipliedso
thatthedampingratioisincreasedfrom0.3to0.9.
SOLUTION:
SYED HASAN SAEED 19)1(
)(
sTs
K
sG

 0
/
)(
)(
1.
)1(
1
)1(
)()(1
)(
)(
)(
2
2









T
K
T
s
s
T
K
T
s
s
TK
sR
sC
sTs
K
sTs
K
sHsG
sG
sR
sC
Characteristic Eq.

20. Compare the characteristic eq. with
Given that:
SYED HASAN SAEED 2002
22

nn
ss  T
K
T
n
n


2
1
2


We getTT
K1
2  T
K
n
 KT2
1

Or,9.0
3.0
2
1



 TK
TK
2
2
1
1
2
1
2
1





21. SYED HASAN SAEED 2121
2
1
2
1
2
2
1
9
9
1
9.0
3.0
KK
K
K
K
K











Hence, the gain K
1at which 3.0 Should be multiplied
By 1/9 to increase the damping ratio from 0.3 to 0.9