time response second order
TRANSCRIPT
TIME RESPONSE
OF
SECOND ORDER SYSTEM
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1SYED HASAN SAEED
REFERENCE BOOKS:
1. AUTOMATIC CONTROL SYSTEM KUO & GOLNARAGHI
2. CONTROL SYSTEM ANAND KUMAR
3. AUTOMATIC CONTROL SYSTEM S.HASAN SAEED
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Block diagram of second order system is shown in fig.
R(s) C(s)_
+)2(
2
n
n
ss
ssR
sRsssR
sC
AsssR
sC
nn
n
nn
n
1)(
)(2)(
)(
)(2)(
)(
22
2
22
2
For unit step input
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22
22
2
2
)1(2
.1
)(
nn
nn
n
ss
ssssC
Replace by )1()( 222 nns
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
sssC
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22)( dns Multiply equation (3) by and put
)2()(
))((
)(2
2
2
nnn
nd
dndn
dnn
dn
n
n
dn
ssB
sj
jj
jB
jB
sB
js
Equation (1) can be written as
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)4()(
.)(
1)(
)(
1)(
2222
22
dn
d
d
n
dn
n
dn
nn
ss
s
ssC
s
s
ssC
Laplace Inverse of equation (4)
)5(sin.cos.1)(
tetetc d
t
d
nd
t nn
21 ndPut
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tte
tc
ttetc
dd
t
dd
t
n
n
sincos.11
1)(
sin.1
cos1)(
2
2
2
)sin(1
1)(
1tan
cos
sin1
2
2
2
te
tc d
tn
Put
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)6(1
tan)1(sin1
1)(2
12
2
te
tc n
tn
Put the values of d &
)7(1
tan)1(sin1
)(
)()()(
212
2
te
te
tctrte
n
tn
Error signal for the system
The steady state value of c(t)
1)(
tcLimitet
ss
Therefore at steady state there is no error betweeninput and output.
= natural frequency of oscillation or undampednatural frequency.
= damped frequency of oscillation.
= damping factor or actual damping ordamping coefficient.
For equation (A) two poles (for ) are
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n
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.
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10
From equation (6):
Time constant is
Response having damped oscillation with overshoot andunder shoot. This response is known as under-dampedresponse.
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n/1
UNDAMPED ( ): when the system has two imaginary poles.
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0
From equation (6)
Thus at the system will oscillate.
The damped frequency always less than the undampedfrequency ( ) because of . The response is shown infig.
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ttc
ttc
n
n
cos1)(
)2/sin(1)(
0
For
n
n
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CRITICALLY DAMPED ( ): When the system has two real and equal poles. Location of poles for critically damped is shown in fig.
1
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)(
11
)(
)()(
2.
1)(
1
2
2
2
2
22
2
n
n
nn
n
n
n
nn
n
ssssss
sssC
ssssC
For
After partial fraction
Take the inverse Laplace
)8()1(1)(
1)(
tetc
etetc
n
t
t
n
t
n
nn
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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 ( ): when the system has two realand distinct poles.
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1
Response of the system
From equation (2)
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)9()1()(
.1
)(222
2
nn
n
sssC
)1( 222 ndPut
)10()(
.1
)(22
2
dn
n
sssC
We get
Equation (10) can be written as
)11())((
)(2
dndn
n
ssssC
After partial fraction of equation (11) we get
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Put the value of d
)12(
112
1
112
11)(
22
22
dn
dn
s
sssC
)13(
)1(112
1
)1(112
11)(
222
222
nn
nn
s
sssC
Inverse Laplace of equation (13)
From equation (14) we get two time constants
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)14()1(12)1(12
1)(22
)1(
22
)1( 22
tt nn eetc
n
n
T
T
)1(
1
)1(
1
22
21
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)15()1(12
1)(22
)1( 2
tnetc
From equation (14) it is clear that when is greater thanone there are two exponential terms, first term has timeconstant T1 and second term has a time constant T2 . T1 <T2 . In other words we can say that first exponential termdecaying much faster than the other exponential term.So for time response we neglect it, then
)16()1(
1
22
n
T
THANK YOU
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