TIME RESPONSE
OF
SECOND ORDER SYSTEM
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1SYED HASAN SAEED
REFERENCE BOOKS:
1.AUTOMATICCONTROLSYSTEMKUO&GOLNARAGHI
2.CONTROLSYSTEMANANDKUMAR
3.AUTOMATICCONTROLSYSTEMS.HASANSAEED
SYED HASAN SAEED 2
SYED HASAN SAEED 3
Block diagram of second order system is shown in fig.
R(s) C(s)
_
+)2(
2
n
n
ss
s
sR
sR
sssR
sC
A
sssR
sC
nn
n
nn
n
1
)(
)(
2)(
)(
)(
2)(
)(
22
2
22
2
For unit step input
SYED HASAN SAEED 422
22
2
2
)1(
2
.
1
)(
nn
nn
n
ss
sss
sC
Replace by)1()(
222
nn
s
Break the equation by partial fraction and put)1(
222
nd 1
)3(
)()(
.
1
2222
2
A
s
B
s
A
ss
dndn
n
)2(
)1()(
.
1
)(
222
2
nn
n
ss
sC
SYED HASAN SAEED 5
22
)(
dn
s
Multiply equation (3) by and put)2()(
))((
)(
2
2
2
nnn
nd
dndn
dnn
dn
n
n
dn
ssB
sj
jj
j
B
j
B
s
B
js
Equation (1) can be written as
SYED HASAN SAEED 6)4(
)(
.
)(
1
)(
)(
1
)(
2222
22
dn
d
d
n
dn
n
dn
nn
ss
s
s
sC
s
s
s
sC
Laplace Inverse of equation (4))5(sin.cos.1)(
tetetc
d
t
d
n
d
t
nn
2
1
nd
Put
SYED HASAN SAEED 7 tt
e
tc
ttetc
dd
t
dd
t
n
n
sincos.1
1
1)(
sin.
1
cos1)(
2
2
2
)sin(
1
1)(
1
tan
cos
sin1
2
2
2
t
e
tc
d
t
n
Put
SYED HASAN SAEED 8)6(
1
tan)1(sin
1
1)(
2
12
2
t
e
tc
n
t
n
Put the values of d
&)7(
1
tan)1(sin
1
)(
)()()(
2
12
2
t
e
te
tctrte
n
t
n
Error signal for the system
The steady state value of c(t)1)(
tcLimite
t
ss
Thereforeatsteadystatethereisnoerrorbetween
inputandoutput.
=naturalfrequencyofoscillationorundamped
naturalfrequency.
=dampedfrequencyofoscillation.
=dampingfactororactualdampingor
dampingcoefficient.
Forequation(A)twopoles(for )are
SYED HASAN SAEED 9n
d
n
2
2
1
1
nn
nn
j
j 10
Depending upon the value of , there are four cases
UNDERDAMPED ( ): When the system has two
complex conjugate poles.
SYED HASAN SAEED 10 10
Fromequation(6):
Timeconstantis
Responsehavingdampedoscillationwithovershootand
undershoot.Thisresponseisknownasunder-damped
response.
SYED HASAN SAEED 11n
/1
UNDAMPED ( ): when the system has two
imaginary poles.
SYED HASAN SAEED 120
From equation (6)
Thus at the system will oscillate.
Thedampedfrequencyalwayslessthantheundamped
frequency()becauseof.Theresponseisshownin
fig.
SYED HASAN SAEED 13ttc
ttc
n
n
cos1)(
)2/sin(1)(
0
For n
n
SYED HASAN SAEED 14
CRITICALLY DAMPED ( ): When the system has
two real and equal poles. Location of poles for
critically damped is shown in fig. 1
SYED HASAN SAEED 15)(
11
)(
)(
)(
2
.
1
)(
1
2
2
2
2
22
2
n
n
nn
n
n
n
nn
n
ssssss
ss
sC
sss
sC
For
After partial
fraction
Take the inverse Laplace )8()1(1)(
1)(
tetc
etetc
n
t
t
n
t
n
nn
SYED HASAN SAEED 16
From equation (6) it is clear that is the actual
damping. For , actual damping = . This actual
damping is known as CRITICAL DAMPING.
The ratio of actual damping to the critical damping is
known as damping ratio . From equation (8) time
constant = . Response is shown in fig.n
1 n
n
/1
OVERDAMPED( ):whenthesystemhastworeal
anddistinctpoles.
SYED HASAN SAEED 171
Response of the
system
From equation (2)
SYED HASAN SAEED 18)9(
)1()(
.
1
)(
222
2
nn
n
ss
sC )1(
222
nd
Put )10(
)(
.
1
)(
22
2
dn
n
ss
sC
We get
Equation (10) can be written as)11(
))((
)(
2
dndn
n
sss
sC
After partial fraction of equation (11) we get
SYED HASAN SAEED 19
Put the value of d
)12(
112
1
112
11
)(
22
22
dn
dn
s
ss
sC
)13(
)1(112
1
)1(112
11
)(
222
222
nn
nn
s
ss
sC
Inverse Laplace of equation (13)
From equation (14) we get two time constants
SYED HASAN SAEED 20)14(
)1(12)1(12
1)(
22
)1(
22
)1(
22
tt
nn
ee
tc n
n
T
T
)1(
1
)1(
1
2
2
2
1
SYED HASAN SAEED 21)15(
)1(12
1)(
22
)1(
2
t
n
e
tc
Fromequation(14)itisclearthatwhenisgreaterthan
onetherearetwoexponentialterms,firsttermhastime
constantT
1andsecondtermhasatimeconstantT
2.T
1<
T
2.Inotherwordswecansaythatfirstexponentialterm
decayingmuchfasterthantheotherexponentialterm.
Sofortimeresponseweneglectit,then)16(
)1(
1
2
2
n
T