Time Responce Analysis
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Transcript of Time Responce Analysis
Closed Loop Control System
C(s)= G(s).E(s)
B(s)= C(s).H(s)
E(s) = R(s)-B(s)
B(s)=G(s).E(s).H(s)
B(s)/E(s) = G(s).H(s) Open loop T.F.
C(s)= G(s).E(s)
=G(s)[R(s)-B(s)]
=G(s)R(s)-G(s)B(s)
= G(s)R(s)-G(s) .C(s).H(s)
C(s)[1+G(s)H(s)]=G(s)R(s)
C(s)/R(s)=G(s)/[1+G(s)H(s)] Closed Loop T.F.(with –ve feedback)
C(s)/R(s)=G(s)/[1-G(s)H(s)] Closed Loop T.F.(with +ve feedback)
Step Response of underdamped System
222222 2
21
nnnn
n
ss
s
ssC
)(
• The partial fraction expansion of above equation is given as
22 2
21
nn
n
ss
s
ssC
)(
2ns
22 1 n
2221
21
nn
n
s
s
ssC )(
22
2
2 nn
n
sssR
sC
)(
)(
22
2
2 nn
n
ssssC
)(
Step Response
Step Response of underdamped System
• Above equation can be written as
2221
21
nn
n
s
s
ssC )(
22
21
dn
n
s
s
ssC
)(
21 nd• Where , is the frequency of transient oscillations and is called damped natural frequency.
• The inverse Laplace transform of above equation can be obtained easily if C(s) is written in the following form:
2222
1
dn
n
dn
n
ss
s
ssC
)(
Step Response of underdamped System
2222
1
dn
n
dn
n
ss
s
ssC
)(
22
2
2
22
111
dn
n
dn
n
ss
s
ssC
)(
222221
1
dn
d
dn
n
ss
s
ssC
)(
tetetc dt
dt nn
sincos)(
21
1
Step Response of underdamped System
tetetc dt
dt nn
sincos)(
21
1
ttetc ddtn
sincos)(21
1
n
nd
21
• When 0
ttc ncos)( 1
Step Response of underdamped System
ttetc ddtn
sincos)(21
1
sec/. radn and if 310
0 2 4 6 8 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
Step Response of underdamped System
ttetc ddtn
sincos)(21
1
sec/. radn and if 350
0 2 4 6 8 100
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response of underdamped System
ttetc ddtn
sincos)(21
1
sec/. radn and if 390
0 2 4 6 8 100
0.2
0.4
0.6
0.8
1
1.2
1.4
Step Response of underdamped System
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
b=0
b=0.2
b=0.4
b=0.6
b=0.9
Step Response of underdamped System
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
wn=0.5
wn=1
wn=1.5
wn=2
wn=2.5
Time-Domain Specification
30
For 0< <1 and ωn > 0, the 2nd order system’s response due to a unit step input looks like
Time-Domain Specification
31
• The delay (td) time is the time required for the response to reach half the final value the very first time.
Time-Domain Specification
32
• The rise time is the time required for the response to rise from 10% to 90%, 5% to 95%, or 0% to 100% of its final value.
• For underdamped second order systems, the 0% to 100% rise time is normally used. For overdamped systems, the 10% to 90% rise time is commonly used.
32
Time-Domain Specification
33
• The peak time is the time required for the response to reach the first peak of the overshoot.
33 33
Time-Domain Specification
34
The maximum overshoot is the maximum peak value of the response curve measured from unity. If the final steady-state value of the response differs from unity, then it is common to use the maximum percent overshoot. It is defined by The amount of the maximum (percent) overshoot directly indicates the relative stability of the system.
Time-Domain Specification
35
• The settling time is the time required for the response curve to reach and stay within a range about the final value of size specified by absolute percentage of the final value (usually 2% or 5%).
Time Domain Specifications of Underdamped system
Time Domain Specifications (Rise Time)
ttetc ddtn
sincos)(21
1
equation above in Put rtt
rdrdt
r ttetc rn
sincos)(21
1
1)c(tr Where
rdrdt
tte rn
sincos21
0
0 rnte
rdrd tt
sincos
21
0
Time Domain Specifications (Rise Time)
as writen-re be can equation above
0
1 2
rdrd tt
sincos
rdrd tt
cossin
21
21rd ttan
21 1
tanrd t
Time Domain Specifications (Rise Time)
21 1
tanrd t
n
n
drt
21 11
tan
drt
b
a1 tan
Time Domain Specifications (Peak Time)
ttetc ddtn
sincos)(21
1
• In order to find peak time let us differentiate above equation w.r.t t.
ttettedt
tdcd
ddd
tdd
tn
nn
cossinsincos)(
22 11
tttte dd
dddn
dntn
cossinsincos22
2
11
0
tttte dn
dddn
dntn
cossinsincos2
2
2
2
1
1
1
0
Time Domain Specifications (Peak Time)
tttte dn
dddn
dntn
cossinsincos2
2
2
2
1
1
1
0
0
1 2
2
tte ddd
ntn
sinsin
0 tne 0
1 2
2
tt dddn
sinsin
0
1 2
2
d
nd t
sin
Time Domain Specifications (Peak Time)
0
1 2
2
d
nd t
sin
0
1 2
2
d
n
0tdsin
01 sintd
d
t
,,, 20
• Since for underdamped stable systems first peak is maximum peak therefore,
dpt
Time Domain Specifications (Maximum Overshoot)
pdpd
t
p ttetc pn
sincos)(
21
1
1)(c
1001
1
12
pdpd
t
p tteM pn
sincos
equation above in Putd
pt
100
1 2
dd
ddp
dn
eM
sincos
Time Domain Specifications (Maximum Overshoot)
100
1 2
dd
ddp
dn
eM
sincos
100
1 2
1 2
sincosn
n
eM p
1000121
eM p
10021
eM p
equation above in Put 21-ζωω nd
Time Domain Specifications (Settling Time)
ttetc ddtn
sincos)(21
1
12 nn
n
T
1
Real Part Imaginary Part
Time Domain Specifications (Settling Time)
n
T
1
• Settling time (2%) criterion • Time consumed in exponential decay up to 98% of the input.
ns Tt
44
• Settling time (5%) criterion • Time consumed in exponential decay up to 95% of the input.
ns Tt
33
Summary of Time Domain Specifications
ns Tt
44
ns Tt
33 100
21
eM p
21
nd
pt21
nd
rt
Rise Time Peak Time
Settling Time (2%)
Settling Time (4%)
Maximum Overshoot
Example#5 • Consider the system shown in following figure, where
damping ratio is 0.6 and natural undamped frequency is 5 rad/sec. Obtain the rise time tr, peak time tp, maximum overshoot Mp, and settling time 2% and 5% criterion ts when the system is subjected to a unit-step input.
Example#5
ns Tt
44
10021
eM p
dpt
drt
Rise Time Peak Time
Settling Time (2%) Maximum Overshoot
ns Tt
33
Settling Time (4%)
Example#5
drt
Rise Time
21
1413
n
rt.
rad 9301 2
1 .)(tan
n
n
str 550
6015
9301413
2.
.
..
Example#5
nst
4
dpt
Peak Time Settling Time (2%)
nst
3
Settling Time (4%)
st p 78504
1413.
.
sts 331560
4.
.
sts 1560
3
.
Example#5
10021
eM p
Maximum Overshoot
1002601
601413
.
..
eM p
1000950 .pM
%.59pM
Example#5 Step Response
Time (sec)
Am
plit
ude
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60
0.2
0.4
0.6
0.8
1
1.2
1.4
Mp
Rise Time