Time domain control system

1. Control Systems (CS)
Ali Raza
Assistant Professor
Indus University Karachi, Pakistan
email: ali_raza@faculty.muet.edu.pk
Lecture-14-15
Time Domain Analysis of 2nd Order Systems
1

2. Introduction
• We have already discussed the affect of location of poles and zeros on
the transient response of 1st order systems.
• Compared to the simplicity of a first-order system, a second-order system
exhibits a wide range of responses that must be analyzed and described.
• Varying a first-order system's parameter (T, K) simply changes the speed
and offset of the response
• Whereas, changes in the parameters of a second-order system can
change the form of the response.
• A second-order system can display characteristics much like a first-order
system or, depending on component values, display damped or pure
oscillations for its transient response. 2

3. Introduction
• A general second-order system is characterized by the
following transfer function.
22
2
2 nn
n
sssR
sC




)(
)(
3
un-damped natural frequency of the second order system,
which is the frequency of oscillation of the system without
damping.
n
damping ratio of the second order system, which is a measure
of the degree of resistance to change in the system output.


4. Example#1
42
4
2


sssR
sC
)(
)(
• Determine the un-damped natural frequency and damping ratio
of the following second order system.
42
n
22
2
2 nn
n
sssR
sC




)(
)(
• Compare the numerator and denominator of the given transfer
function with the general 2nd order transfer function.
sec/radn 2 
ssn 22  
422 222
 ssss nn 
50. 
1 n
4

5. Introduction
22
2
2 nn
n
sssR
sC




)(
)(
• Two poles of the system are
1
1
2
2




nn
nn
5

6. Introduction
• According the value of , a second-order system can be set into
one of the four categories:
1
1
2
2




nn
nn

1. Overdamped - when the system has two real distinct poles ( >1).
-a-b-c
δ
jω
6

7. Introduction
• According the value of , a second-order system can be set into
one of the four categories:
1
1
2
2




nn
nn

2. Underdamped - when the system has two complex conjugate poles (0 < <1)
-a-b-c
δ
jω
7

8. Introduction
• According the value of , a second-order system can be set into
one of the four categories:
1
1
2
2




nn
nn

3. Undamped - when the system has two imaginary poles ( = 0).
-a-b-c
δ
jω
8

9. Introduction
• According the value of , a second-order system can be set into
one 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-c
δ
jω
9

10. Time-Domain Specification
10
For 0< <1 and ωn > 0, the 2nd order system’s response due to a
unit step input looks like


11. Time-Domain Specification
11
• The delay (td) time is the time required for the response to
reach half the final value the very first time.

12. Time-Domain Specification
• 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.

13. Time-Domain Specification
13
• The peak time is the time required for the response to reach
the first peak of the overshoot.
1313

14. Time-Domain Specification
14
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.

15. Time-Domain Specification
15
• 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%).

16. S-Plane
δ
jω
• Natural Undamped Frequency.
n
• Distance from the origin of s-
plane to pole is natural
undamped frequency in
rad/sec.
16

17. 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 the
natural undamped frequency would be 3 rad/sec.
17

18. S-Plane
δ
jω
• Therefore the s-plane is divided into Constant Natural
Undamped Frequency (ωn) Circles.
18

19. S-Plane
δ
jω
• Damping ratio.
• Cosine of the angle between
vector connecting origin and
pole and –ve real axis yields
damping ratio.

 cos
19

20. S-Plane
δ
jω
• For Underdamped system therefore,
900   10  
20

21. S-Plane
δ
jω
• For Undamped system therefore,
90 0
21

22. S-Plane
δ
jω
• For overdamped and critically damped systems
therefore,

0
0
22

23. S-Plane
δ
jω
• Draw a vector connecting origin of s-plane and some point P.
P

45
707045 .cos  

23

24. S-Plane
δ
jω
• Therefore, s-plane is divided into sections of constant damping
ratio lines.
24

25. 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
1.5
0.220.420.60.740.840.91
0.96
0.99
0.220.420.60.740.840.91
0.96
0.99
0.511.522.533.54
Pole-Zero Map
ImaginaryAxis(seconds-1
)
25

26. Example-3
• Determine the natural
frequency and damping ratio of
the poles from the given pz-
map.
• Also determine the transfer
function of the system and state
whether system is
underdamped, overdamped,
undamped or critically damped.
-3 -2.5 -2 -1.5 -1 -0.5 0
-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
ImaginaryAxis(seconds-1
)
26

27. Example-4
• The natural frequency of closed
loop poles of 2nd order system is 2
rad/sec and damping ratio is 0.5.
• Determine the location of closed
loop poles so that the damping
ratio remains same but the natural
undamped frequency is doubled.
42
4
2 222
2




sssssR
sC
nn
n


)(
)(
-2 -1.5 -1 -0.5 0
-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
ImaginaryAxis
27

28. 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
ImaginaryAxis
28

29. S-Plane

1
1
2
2




nn
nn
29

30. Step Response of underdamped System
222222
2
21
nnnn
n
ss
s
s
sC




)(
• The partial fraction expansion of above equation is given as
22
2
21
nn
n
ss
s
s
sC




)(
 2
2 ns 
 22
1  n
   222
1
21





nn
n
s
s
s
sC )(
22
2
2 nn
n
sssR
sC




)(
)(
 22
2
2 nn
n
sss
sC



)(
Step Response
30

31. Step Response of underdamped System
• Above equation can be written as
   222
1
21





nn
n
s
s
s
sC )(
  22
21
dn
n
s
s
s
sC




)(
2
1   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
s
sC








)(
31

32. Step Response of underdamped System
    2222
1
dn
n
dn
n
ss
s
s
sC








)(
    22
2
2
22
1
11
dn
n
dn
n
ss
s
s
sC












)(
    22222
1
1
dn
d
dn
n
ss
s
s
sC










)(
tetetc d
t
d
t nn



 
sincos)( 


2
1
1
32

33. Step Response of underdamped System
tetetc d
t
d
t nn



 
sincos)( 


2
1
1









 
ttetc dd
tn




sincos)(
2
1
1
n
nd



 2
1
• When 0
ttc ncos)(  1
33

34. Step Response of underdamped System









 
ttetc dd
tn




sincos)(
2
1
1
sec/. radnandif 310  
0 2 4 6 8 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
34

35. Step Response of underdamped System









 
ttetc dd
tn




sincos)(
2
1
1
sec/. radnandif 350  
0 2 4 6 8 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
35

36. Step Response of underdamped System









 
ttetc dd
tn




sincos)(
2
1
1
sec/. radnandif 390  
0 2 4 6 8 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
36

37. Step Response of underdamped System









 
ttetc dd
tn




sincos)(
2
1
1
37

38. Step Response of underdamped System
0 1 2 3 4 5 6 7 8 9 10
0
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
38

39. Step Response of underdamped System
0 1 2 3 4 5 6 7 8 9 10
0
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
39

40. Time Domain Specifications
n
s Tt

4
4 
n
s Tt

3
3  100
2
1





eM p
2
1 





n
d
pt
2
1 








n
d
rt
Rise Time Peak Time
Settling Time (2%)
Settling Time (4%)
Maximum Overshoot
40

41. 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.
41

42. Example#5
n
s Tt

4
4 
100
2
1





eM p
d
pt



d
rt

 

Rise Time Peak Time
Settling Time (2%) Maximum Overshoot
n
s Tt

3
3 
Settling Time (4%) 42

43. Example#5
d
rt

 

Rise Time
2
1
1413





n
rt
.
rad930
1 2
1
.)(tan 

 
n
n



str 550
6015
9301413
2
.
.
..




43

44. Example#5
n
st

4

d
pt



Peak Time
Settling Time (2%)
n
st

3

Settling Time (4%)
stp 7850
4
1413
.
.

sts 331
560
4
.
.



sts 1
560
3



. 44

45. Example#5
100
2
1





eM p
Maximum Overshoot
100
2
601
601413
 


.
..
eM p
1000950  .pM
%.59pM
45

46. Example#5
Step Response
Time (sec)
Amplitude
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Mp
Rise Time
46

47. Example#6
• For the system shown in Figure-(a), determine the values of gain K
and velocity-feedback constant Kh so that the maximum overshoot
in the unit-step response is 0.2 and the peak time is 1 sec. With
these 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.
47

48. Example#6
48

49. Example#6
Nm/rad/secandSince 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
KKh
2
1 )( 

49

50. Example#6
• Maximum overshoot is 0.2.
Kn 
K
KKh
2
1 )( 

 20
2
1
.ln)ln( 




e
• The peak time is 1 sec
d
pt



2
45601
1413
.
.

n
2
1
1413
1
 

n
.
533.n
50

51. Example#6
Kn 
K
KKh
2
1 )( 

963.n
K533.
512
533 2
.
.


K
K
).(.. hK512151224560 
1780.hK
51

52. Example#6
963.n
n
st

4

n
st

3

2
1 




n
rt
str 650. sts 482.
sts 861.
52

53. Example#7
When the system shown in Figure(a) is subjected to a unit-step input,
the system output responds as shown in Figure(b). Determine the
values of a and c from the response curve.
)( 1css
a
53

54. Example#8
Figure (a) shows a mechanical vibratory system. When 2 lb of force
(step input) is applied to the system, the mass oscillates, as shown in
Figure (b). Determine m, b, and k of the system from this response
curve.
54

55. Example#9
Given 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 torque
T(t).
55

56. Example#9
56

57. Example#9
57

58. Example # 10
• Ships at sea undergo
motion about their roll axis,
as shown in Figure. Fins
called stabilizers are used to
reduce this rolling motion.
The stabilizers can be
positioned by a closed-loop
roll control system that
consists of components,
such as fin actuators and
sensors, as well as the ship’s
roll dynamics.
58

59. Example # 10
• Assume the roll dynamics, which relates the roll-angle output, 𝜃(𝑠),
to a disturbance-torque input, TD(s), is
• Do the following:
a) Find the natural frequency, damping ratio, peak time, settling
time, rise time, and percent overshoot.
b) Find the analytical expression for the output response to a unit
step input.
59

60. Step Response of critically damped System ( )
• The partial fraction expansion of above equation is given as
 2
2
n
n
ssR
sC




)(
)(
 2
2
n
n
ss
sC



)(
Step Response
   22
2
nnn
n
s
C
s
B
s
A
ss 






 2
11
n
n
n sss
sC


 


)(
teetc t
n
t nn 
 
 1)(
 tetc n
tn

 
11)(
1
60

61. 61

62. 62
Example 11: Describe the nature of the second-order system
response via the value of the damping ratio for the systems with
transfer function
Second – Order System
128
12
)(.1 2


ss
sG
168
16
)(.2 2


ss
sG
208
20
)(.3 2


ss
sG
Do them as your own
revision

63. Example-12
• For each of the transfer
functions find the locations of
the poles and zeros, plot them
on the s-plane, and then write
an expression for the general
form of the step response
without solving for the inverse
Laplace transform. State the
nature of each response
(overdamped, underdamped,
and so on).
63

64. Example-13
• Solve for x(t) in the system shown in Figure if f(t) is a unit step.
64

65. END OF LECTURES-14-15
65