time domain specifications of second order system
TRANSCRIPT
TIME DOMAIN SPECIFICATIONS OF SECOND ORDER SYSTEM
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1SYED HASAN SAEED
SYED HASAN SAEED 2
BOOKS
1. AUTOMATIC CONTROL SYSTEM KUO &GOLNARAGHI
2. CONTROL SYSTEM ANAND KUMAR
3. AUTOMATIC CONTROL SYSTEM S.HASAN SAEED
SYED HASAN SAEED 3
Consider a second order system with unit step input andall initial conditions are zero. The response is shown in fig.
1. DELAY TIME (td): The delay time is the time requiredfor the response to reach 50% of the final value infirst time.
2. RISE TIME (tr): It is time required for the response torise from 10% to 90% of its final value for over-damped systems and 0 to 100% for under-dampedsystems.
We know that:
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21
2
2
1tan
1sin1
1)(
te
tc n
tn
Where,
Let response reaches 100% of desired value. Put c(t)=1
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01sin1
1sin1
11
2
2
2
2
te
te
n
t
n
t
n
n
0 tneSince,
)sin())1sin((
0))1sin((
2
2
nt
t
n
n
Or,
Put n=1
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2
2
1
)1(
n
r
rn
t
t
2
21
1
1tan
n
rt
Or,
Or,
3. PEAK TIME (tp): The peak time is the time requiredfor the response to reach the first peak of the timeresponse or first peak overshoot.
For maximum
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te
tc n
tn
2
21sin
11)(Since
)1(01
1sin
11cos1
)(
0)(
2
2
22
2
tnn
nn
t
n
n
et
te
dt
tdc
dt
tdc
Since,
Equation can be written as
Equation (2) becomes
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0 tne
sin1
1sin11cos
2
222
tt nn
Put cosand
cos1sinsin1cos 22 tt nn
cos
sin
))1cos((
))1sin((
2
2
t
t
n
n
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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=1
21
n
pt
First minimum (undershoot) occurs at n=2
2min
1
2
n
t
4. MAXIMUM OVERSHOOT (MP):
Maximum overshoot occur at peak time, t=tp
in above equation
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te
tc n
tn
2
21sin
11)(
21
n
ptPut,
2
2
2
1
1.1sin
11)(
2
n
n
n
n
etc
SYED HASAN SAEED 11
2
2
1
2
2
1
2
1
11
1)(
1sin
sin1
1)(
)sin(1
1)(
2
2
2
etc
etc
etc
Put,
sin)sin(
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2
2
2
1
1
1
11
1)(
1)(
eM
eM
tcM
etc
p
p
p
100*%21
eM p
5. SETTLING TIME (ts):
The settling time is defined as the time required for thetransient response to reach and stay within theprescribed percentage error.
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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. STEADY STATE ERROR (ess): It is difference betweenactual output and desired output as time ‘t’ tends toinfinity.
n
s Tt
44
)()( tctrLimitet
ss
EXAMPLE 1: The open loop transfer function of a servosystem with unity feedback is given by
Determine the damping ratio, undamped natural frequencyof oscillation. What is the percentage overshoot of theresponse to a unit step input.
SOLUTION: Given that
Characteristic equation
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)5)(2(
10)(
sssG
1)(
)5)(2(
10)(
sH
sssG
0)()(1 sHsG
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0207
0)5)(2(
101
2
ss
ss
Compare with 02 22 nnss 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
EXAMPLE 2: A feedback system is described by thefollowing transfer function
The damping factor of the system is 0.8. determine theovershoot of the system and value of ‘K’.
SOLUTION: We know that
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KssH
sssG
)(
164
12)(
2
016)164(
16)164(
16
)(
)(
)()(1
)(
)(
)(
2
2
sKs
sKssR
sC
sHsG
sG
sR
sC
is the characteristic eqn.
Compare with
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K
ss
n
n
nn
1642
16
02
2
22
.sec/4radn
K1644*8.0*2 15.0K
%5.1
100*100*22 )8.0(1
8.0
1
p
p
M
eeM
EXAMPLE 3: The open loop transfer function of a unityfeedback control system is given by
By what factor the amplifier gain ‘K’ should be multiplied sothat the damping ratio is increased from 0.3 to 0.9.
SOLUTION:
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)1()(
sTs
KsG
0
/
)(
)(
1.)1(
1
)1(
)()(1
)(
)(
)(
2
2
T
K
T
ss
T
K
T
ss
TK
sR
sC
sTs
K
sTs
K
sHsG
sG
sR
sC
Characteristic Eq.
Compare the characteristic eq. with
Given that:
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02 22 nnss
T
K
T
n
n
2
12
We get
TT
K 12
T
Kn
KT2
1Or,
9.0
3.0
2
1
TK
TK
2
2
1
1
2
1
2
1
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21
2
1
2
1
2
2
1
9
9
1
9.0
3.0
KK
K
K
K
K
Hence, the gain K1 at which 3.0 Should be multiplied
By 1/9 to increase the damping ratio from 0.3 to 0.9