Post on 27-Mar-2021
Automatic Control Systems (FCS)
Lecture-6Time Domain Analysis of 2nd Order Systems
Introduction• We have already discussed the affect of location of pole and zero on the
transient response of 1st order systems.
• Compared to the simplicity of a first-order system, a second-order systemexhibits a wide range of responses that must be analyzed and described.
• Varying a first-order system's parameters (T, K) simply changes the speedand offset of the response
• Whereas, changes in the parameters of a second-order system canchange the form of the response.
• A second-order system can display characteristics much like a first-ordersystem or, depending on component values, display damped or pureoscillations for its transient response.
Introduction• A general second-order system (without zeros) is
characterized by the following transfer function.
22
2
2 nn
n
sssR
sC
)(
)(
)2()(
2
n
n
sssG
Open-Loop Transfer Function
Closed-Loop Transfer Function
Introduction
un-damped natural frequency of the second order system,which is the frequency of oscillation of the system withoutdamping.
22
2
2 nn
n
sssR
sC
)(
)(
n
damping ratio of the second order system, which is a measureof the degree of resistance to change in the system output.
Example#1
42
42
sssR
sC
)(
)(
• Determine the un-damped natural frequency and damping ratioof the following second order system.
42 n
22
2
2 nn
n
sssR
sC
)(
)(
• Compare the numerator and denominator of the given transferfunction with the general 2nd order transfer function.
sec/radn 2ssn 22
422 222 ssss nn 50.
1 n
Introduction
22
2
2 nn
n
sssR
sC
)(
)(
• The closed-loop poles of the system are
1
1
2
2
nn
nn
Introduction
• Depending upon the value of , a second-order system can be setinto one of the four categories:
1
1
2
2
nn
nn
1. Overdamped - when the system has two real distinct poles ( >1).
-a-b-cδ
jω
Introduction
• According the value of , a second-order system can be set intoone of the four categories:
1
1
2
2
nn
nn
2. Underdamped - when the system has two complex conjugate poles (0 < <1)
-a-b-cδ
jω
Introduction
• According the value of , a second-order system can be set intoone of the four categories:
1
1
2
2
nn
nn
3. Undamped - when the system has two imaginary poles ( = 0).
-a-b-cδ
jω
Introduction
• According the value of , a second-order system can be set intoone of the four categories:
1
1
2
2
nn
nn
4. Critically damped - when the system has two real but equal poles ( = 1).
-a-b-cδ
jω
Time-Domain Specification
11
For 0< <1 and ωn > 0, the 2nd order system’s response due to a unit step input looks like
Time-Domain Specification
12
• The delay (td) time is the time required for the response toreach half the final value the very first time.
Time-Domain Specification
13
• 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 isnormally used. For overdamped systems, the 10% to 90% rise time iscommonly used.
13
Time-Domain Specification
14
• The peak time is the time required for the response to reach the first peak of the overshoot.
1414
Time-Domain Specification
15
The maximum overshoot is the maximum peak value of theresponse curve measured from unity. If the final steady-statevalue of the response differs from unity, then it is common touse the maximum percent overshoot. It is defined by
The amount of the maximum (percent) overshoot directlyindicates the relative stability of the system.
Time-Domain Specification
16
• The settling time is the time required for the response curveto reach and stay within a range about the final value of sizespecified by absolute percentage of the final value (usually 2%or 5%).
S-Plane
δ
jω
• Natural Undamped Frequency.
n
• Distance from the origin of s-plane to pole is naturalundamped frequency inrad/sec.
S-Plane
δ
jω
• Let us draw a circle of radius 3 in s-plane.
3
-3
-3
3
• If a pole is located anywhere on the circumference of the circle thenatural undamped frequency would be 3 rad/sec.
S-Plane
δ
jω
• Therefore the s-plane is divided into Constant NaturalUndamped Frequency (ωn) Circles.
S-Plane
δ
jω
• Damping ratio.
• Cosine of the angle betweenvector connecting origin andpole and –ve real axis yieldsdamping ratio.
cos
S-Plane
δ
jω
• For Underdamped system therefore, 900 10
S-Plane
δ
jω
• For Undamped system therefore,90 0
S-Plane
δ
jω
• For overdamped and critically damped systems therefore,
0
1
S-Plane
δ
jω
• Draw a vector connecting origin of s-plane and some point P.
P
45
707045 .cos
S-Plane
δ
jω
• Therefore, s-plane is divided into sections of constant damping ratio lines.
Example-2• Determine the natural frequency and damping ratio of the poles from the
following pz-map.
-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0
-1.5
-1
-0.5
0
0.5
1
1.50.220.420.60.740.840.91
0.96
0.99
0.220.420.60.740.840.91
0.96
0.99
0.511.522.533.54
Pole-Zero Map
Real Axis (seconds-1)
Imagin
ary
Axis
(seconds-1
)
Example-3
• Determine the naturalfrequency and damping ratio ofthe poles from the given pz-map.
• Also determine the transferfunction of the system and statewhether system isunderdamped, overdamped,undamped or critically damped.
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2-3
-2
-1
0
1
2
3
0.420.560.7
0.82
0.91
0.975
0.5
1
1.5
2
2.5
3
0.5
1
1.5
2
2.5
3
0.140.280.420.560.7
0.82
0.91
0.975
0.140.28
Pole-Zero Map
Real Axis (seconds-1)
Imagin
ary
Axis
(seconds-1
)
Example-4
• The natural frequency of closedloop poles of 2nd order system is 2rad/sec and damping ratio is 0.5.
• Determine the location of closedloop poles so that the dampingratio remains same but the naturalundamped frequency is doubled.
42
4
2 222
2
sssssR
sC
nn
n
)(
)(-2 -1.5 -1 -0.5 0 0.5 1
-3
-2
-1
0
1
2
3
0.280.380.5
0.64
0.8
0.94
0.5
1
1.5
2
2.5
3
0.5
1
1.5
2
2.5
3
0.080.170.280.380.5
0.64
0.8
0.94
0.080.17
Pole-Zero Map
Real Axis
Imagin
ary
Axis
Example-4• Determine the location of closed loop poles so that the damping ratio remains same
but the natural undamped frequency is doubled.
-8 -6 -4 -2 0 2 4-5
-4
-3
-2
-1
0
1
2
3
4
5
0.5
0.5
24
Pole-Zero Map
Real Axis
Imagin
ary
Axis
S-Plane
1
1
2
2
nn
nn
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
)(
22 ns
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 oscillationsand is called damped natural frequency.
• The inverse Laplace transform of above equation can be obtainedeasily 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 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 peaktherefore,
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 abovein Put d
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 5rad/sec. Obtain the rise time tr, peak time tp, maximumovershoot Mp, and settling time 2% and 5% criterion ts whenthe 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 TimeSettling 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
Example#5Step 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
Example#6• For the system shown in Figure-(a), determine the values of gain K
and velocity-feedback constant Kh so that the maximum overshootin the unit-step response is 0.2 and the peak time is 1 sec. Withthese values of K and Kh, obtain the rise time and settling time.Assume that J=1 kg-m2 and B=1 N-m/rad/sec.
Example#6
Example#6
Nm/rad/sec and Since 11 2 BkgmJ
KsKKs
K
sR
sC
h
)()(
)(
12
22
2
2 nn
n
sssR
sC
)(
)(
• Comparing above T.F with general 2nd order T.F
Kn K
KK h
2
1 )(
Example#6
• Maximum overshoot is 0.2.
Kn K
KK h
2
1 )(
2021
.ln)ln(
e
• The peak time is 1 sec
dpt
245601
1413
.
.
n
21
14131
n
.
533.n
Example#6
Kn K
KK h
2
1 )(
963.n
K533.
512
533 2
.
.
K
K
).(.. hK512151224560
1780.hK
Example#6963.n
nst
4
nst
3
21
n
rt
str 650. sts 482.
sts 861.
Example#7When the system shown in Figure(a) is subjected to a unit-step input,the system output responds as shown in Figure(b). Determine thevalues of a and c from the response curve.
)( 1css
a
Example#8Figure (a) shows a mechanical vibratory system. When 2 lb of force(step input) is applied to the system, the mass oscillates, as shown inFigure (b). Determine m, b, and k of the system from this responsecurve.
Example#9Given the system shown in following figure, find J and D to yield 20%overshoot and a settling time of 2 seconds for a step input of torqueT(t).
Example#9
Example#9
Step Response of critically damped System ( )
• The partial fraction expansion of above equation is given as
22
n
n
ssR
sC
)(
)(
22
n
n
sssC
)(
Step Response
22
2
nnn
n
s
C
s
B
s
A
ss
211
n
n
n ssssC
)(
teetct
nt nn 1)(
tetc ntn
11)(
1
Step Response of overdamped and undamped Systems
• Home Work
71
72
Example 10: Describe the nature of the second-order systemresponse via the value of the damping ratio for the systems withtransfer function
Second – Order System
128
12)(.1
2
sssG
168
16)(.2
2
sssG
208
20)(.3
2
sssG
Do them as your own revision