This presentation gives complete idea about time domain analysis of first and second order system, type number, time domain specifications, steady state error and error constants and numerical examples.

1. By
Dr.K.Hussain
Associate Professor & Head
Dept. of EE, SITCOE
Time Domain Analysis

2. Introduction
• In time-domain analysis the response of a dynamic system to
an input is expressed as a function of time.
• It is possible to compute the time response of a system if the
nature of input and the mathematical model of the system
are known.
• Usually, the input signals to control systems are not known
fully ahead of time.
• It is therefore difficult to express the actual input signals
mathematically by simple equation

3. Standard Test Signals
• The characteristics of actual input signals are a sudden shock,
a sudden change, a constant velocity, and constant
acceleration.
• The dynamic behavior of a system is therefore judged and
compared under application of standard test signals – an
impulse, a step, a constant velocity, and constant acceleration.
• The other standard signal of great importance is a sinusoidal
signal.

4. Standard Test Signals
• Impulse signal
– The impulse signal imitate the
sudden shock characteristic of
actual input signal.
– If A=1, the impulse signal is
called unit impulse signal.
0 t
δ(t)
A






00
0
t
tA
t)(

5. Standard Test Signals
• Step signal
– The step signal imitate
the sudden change
characteristic of actual
input signal.
– If A=1, the step signal is
called unit step signal






00
0
t
tA
tu )( 0 t
u(t)
A

6. Standard Test Signals
• Ramp signal
– The ramp signal imitate
the constant velocity
characteristic of actual
input signal.
– If A=1, the ramp signal
is called unit ramp
signal






00
0
t
tAt
tr )(
0 t
r(t)

7. Standard Test Signals
• Parabolic signal
– The parabolic signal
imitate the constant
acceleration characteristic
of actual input signal.
– If A=1, the parabolic
signal is called unit
parabolic signal.








00
0
2
2
t
t
At
tp )(
0 t
p(t)
parabolic signal with slope A
p(t)
Unit parabolic signal
p(t)

8. Relation between standard Test Signals
• Impulse
• Step
• Ramp
• Parabolic






00
0
t
tA
t)(






00
0
t
tA
tu )(






00
0
t
tAt
tr )(








00
0
2
2
t
t
At
tp )(


 dt
d
dt
d
dt
d

9. Laplace Transform of Test Signals
• Impulse
• Step






00
0
t
tA
t)(
AstL  )()}({ 






00
0
t
tA
tu )(
S
A
sUtuL  )()}({

10. Laplace Transform of Test Signals
• Ramp
• Parabolic
2
s
A
sRtrL  )()}({
3
)()}({
S
A
sPtpL 






00
0
t
tAt
tr )(








00
0
2
2
t
t
At
tp )(

11. Time Response of Control Systems
System
• The time response of any system has two components
• Transient response
• Steady-state response.
• Time response of a dynamic system response to an input
expressed as a function of time.

12. Time Response of Control Systems
• When the response of the system is changed from equilibrium it
takes some time to settle down. This is called transient response.
0 2 4 6 8 10 12 14 16 18 20
0
1
2
3
4
5
6
x 10
-3
Step Response
Time (sec)
Amplitude
Response
Step Input
Transient Response
SteadyStateResponse
• The response of the
system after the transient
response is called steady
state response.

13. Time Response of Control Systems
• Transient response depend upon the system poles only and not
on the type of input.
• It is therefore sufficient to analyze the transient response using a
step input.
• The steady-state response depends on system dynamics and the
input quantity.
• It is then examined using different test signals by final value
theorem.

14. First order system
• The first order system has only one pole.
• Where K is the D.C gain and T is the time constant
of the system.
• Time constant is a measure of how quickly a 1st
order system responds to a unit step input.
• D.C Gain of the system is ratio between the input
signal and the steady state value of output.
1

Ts
K
sR
sC
)(
)(

15. Example 1
• The first order system given below.
13
10


s
sG )(
5
3


s
sG )(
151
53


s/
/
• D.C gain is 10 and time constant is 3 seconds.
• For the following system
• D.C Gain of the system is 3/5 and time constant is 1/5
seconds.

16. Impulse Response of 1st Order System
• Consider the following 1st order system
1Ts
K
)(sC)(sR
0
t
δ(t)
1
1 )()( ssR 
1

Ts
K
sC )(

17. Impulse Response of 1st Order System
• Re-arrange following equation as
1

Ts
K
sC )(
Ts
TK
sC
/
/
)(
1

Tt
e
T
K
tc /
)( 

• In order to compute the response of the system in time domain
we need to compute inverse Laplace transform of the above
equation.
at
Ce
as
C
L 







1

18. Step Response of 1st Order System
• Consider the following 1st order system
1Ts
K
)(sC)(sR
s
sUsR
1
 )()(
 1

Tss
K
sC )(
1

Ts
KT
s
K
sC )(
• In order to find out the inverse Laplace of the above equation, we
need to break it into partial fraction expansion

19. Step Response of 1st Order System
• Taking Inverse Laplace of above equation








1
1
Ts
T
s
KsC )(
 Tt
etuKtc /
)()( 

• Where u(t)=1
 Tt
eKtc /
)( 
 1
  KeKtc 63201 1
.)(  
• When t=T (time constant)

20. Relation Between Step and impulse
response
• The step response of the first order system is
• Differentiating c(t) with respect to t yields
  TtTt
KeKeKtc //
)( 
 1
 Tt
KeK
dt
d
dt
tdc /)( 

Tt
e
T
K
dt
tdc /)( 


21. Example 2
• If initial conditions are not known then partial fraction
expansion is a better choice
12
6


SsR
sC
)(
)(
 12
6


Ss
sC )(
  1212
6


 s
B
s
A
Ss
s
sRsR
1
)(,)( inputstepaissince
  50
66
12
6
.

 ssSs
t
etc 50
66 .
)( 


22. Ramp Response of 1st Order System
• Consider the following 1st order system
1Ts
K
)(sC)(sR
2
1
s
sR )(
 12


Tss
K
sC )(
• The ramp response is given as
 Tt
TeTtKtc /
)( 


23. Parabolic Response of 1st Order System
• Consider the following 1st order system
1Ts
K
)(sC)(sR
3
1
s
sR )(
 13


Tss
K
sC )(Therefore,

24. Second Order System
• 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. 24

25. Second Order System
• A general second-order system is characterized by the
following transfer function.
22
2
2 nn
n
sssR
sC




)(
)(
25
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.


26. Example 3
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




)(
)(
Sol: Compare the numerator and denominator of the given transfer
function with the general 2nd order transfer function.
2 n ssn 22  
422 222
 ssss nn 
50.
1 n
26

27. 22
2
2 nn
n
sssR
sC




)(
)(
• Two poles of the system are
1
1
2
2




nn
nn
27
Second Order System

28. Second Order System Cont..
• 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
28

29. 1
1
2
2




nn
nn
2. Underdamped - when the system has two complex conjugate poles (0 < <1)
-a-b-c
δ
jω
29

30. 1
1
2
2




nn
nn
3. Undamped - when the system has two imaginary poles ( = 0).
-a-b-c
δ
jω
30

31. 1
1
2
2




nn
nn
4. Critically damped - when the system has two real but equal poles ( = 1).
-a-b-c
δ
jω
31

32. Underdamped System
32
For 0< <1 and ωn > 0, the 2nd order system’s response due to a
unit step input is as follows.
Important timing characteristics: delay time, rise time, peak
time, maximum overshoot, and settling time.


33. Delay Time
33
• The delay (td) time is the time required for the response to
reach half the final value the very first time.

34. Rise Time
• 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.

35. Peak Time
35
• The peak time is the time required for the response to reach
the first peak of the overshoot.
3535

36. Maximum Overshoot
36
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.

37. Settling Time
37
• 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%).

38. 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
38

39. 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








)(
39

40. 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
40

41. Step Response of underdamped System
tetetc d
t
d
t nn



 
sincos)( 


2
1
1









 
ttetc dd
tn




sincos)(
2
1
1
41
2
2
1
( ) 1 sin( )
1
1
tan
nt
d
e
c t t
where

 






  




42. Steady State Error
• If the output of a control system at steady state
does not exactly match with the input, the system
is said to have steady state error
• Any physical control system inherently suffers
steady-state error in response to certain types of
inputs.
• A system may have no steady-state error to a step
input, but the same system may exhibit nonzero
steady-state error to a ramp input.

43. Classification of Control Systems
• Control systems may be classified according to
their ability to follow step inputs, ramp inputs,
parabolic inputs, and so on.
• The magnitudes of the steady-state errors due
to these individual inputs are indicative of the
goodness of the system.

44. Classification of Control Systems
• Consider the unity-feedback control system
with the following open-loop transfer function
• It involves the term sN in the denominator,
representing N poles at the origin.
• A system is called type 0, type 1, type 2, ... , if
N=0, N=1, N=2, ... , respectively.

45. Classification of Control Systems
• As the type number is increased, accuracy is
improved.
• However, increasing the type number
aggravates the stability problem.
• A compromise between steady-state accuracy
and relative stability is always necessary.

46. Steady State Error of Unity Feedback Systems
• Consider the system shown in following figure.
• The closed-loop transfer function is

47. Steady State Error of Unity Feedback Systems
• The transfer function between the error signal E(s) and the
input signal R(s) is
)()(
)(
sGsR
sE


1
1
• The final-value theorem provides a convenient way to find
the steady-state performance of a stable system.
• Since E(s) is
• The steady state error is

48. Static Error Constants
• The static error constants are figures of merit of
control systems. The higher the constants, the
smaller the steady-state error.
• In a given system, the output may be the position,
velocity, pressure, temperature, or the like.
• Therefore, in what follows, we shall call the output
“position,” the rate of change of the output
“velocity,” and so on.
• This means that in a temperature control system
“position” represents the output temperature,
“velocity” represents the rate of change of the
output temperature, and so on.

49. Static Position Error Constant (Kp)
• The steady-state error of the system for a unit-step input is
• The static position error constant Kp is defined by
• Thus, the steady-state error in terms of the static position
error constant Kp is given by

50. Static Position Error Constant (Kp)
• For a Type 0 system
• For Type 1 or higher order systems
• For a unit step input the steady state error ess is

51. • The steady-state error of the system for a unit-ramp input is
• The static velocity error constant Kv is defined by
• Thus, the steady-state error in terms of the static velocity
error constant Kv is given by
Static Velocity Error Constant (Kv)

52. Static Velocity Error Constant (Kv)
• For a Type 0 system
• For Type 1 systems
• For type 2 or higher order systems

53. Static Velocity Error Constant (Kv)
• For a ramp input the steady state error ess is

54. • The steady-state error of the system for parabolic input is
• The static acceleration error constant Ka is defined by
• Thus, the steady-state error in terms of the static acceleration
error constant Ka is given by
Static Acceleration Error Constant (Ka)

55. Static Acceleration Error Constant (Ka)
• For a Type 0 system
• For Type 1 systems
• For type 2 systems
• For type 3 or higher order systems

56. Static Acceleration Error Constant (Ka)
• For a parabolic input the steady state error ess is

57. Summary of Steady State Errors

58. Example 4
• For the system shown in figure below evaluate the static
error constants and find the expected steady state errors
for the standard step, ramp and parabolic inputs.
C(S)R(S) -
))((
))((
128
52100
2


sss
ss

59. Example 4
))((
))((
)(
128
52100
2



sss
ss
sG
)(lim sGK
s
p
0










 ))((
))((
lim
128
52100
2
0 sss
ss
K
s
p
pK
)(lim ssGK
s
v
0










 ))((
))((
lim
128
52100
2
0 sss
sss
K
s
v
vK
)(lim sGsK
s
a
2
0












 ))((
))((
lim
128
52100
2
2
0 sss
sss
K
s
a
410
12080
5020100
.
))((
))((








aK

60. Example 4
pK vK 410.aK
0
0
090.

61. Ex 5: The open loop transfer function of a servo system
with unity feedback is given by
Determine the damping ratio, undamped natural
frequency of oscillation. What is the percentage
overshoot of the response to a unit step input.
SOLUTION: Given that
Characteristic equation
10
(s  2)(s  5)
G(s) 
10
H(s) 1
(s  2)(s 5)
G(s) 
1 G(s)H (s)  0

62. 10
 01
s2
 7s  20  0
(s  2)(s 5)
Compare with s2
 2 s 2
 0
n n
We get
*100 1.92%
16
1(0.7826)2

*0.7826
12

2* *4.472 7
  0.7826
 2
20
 e
20  4.472rad /sec
M p e
n 
n
2n  7  n 4.472rad /sec
  0.7826
M p 1.92%

63. H (s)  Ks
The damping factor of the system is 0.8. Determine the
overshoot of the system and value of ‘K’.
SOLUTION: We know that
12
G(s) 
s2
 4s 16
s2
 (4 16K)s 16  0
s2
 (4 16K)s 16R(s)
C(s)

1 G(s)H(s)
16
G(s)
R(s)
C(s)

is the characteristic eqn.
Ex 6: A feedback system is described by the following transfer
function

64. nn
2
n
2
Compare with
2
16
s  2 s   0
n  4rad /sec.
2n  4 16K
2*0.8*4  4 16K  K  0 . 1 5
*1001(0.8)2

M p 1.5%
*100 eM p  e
0.8
12


65. Ex 7: The open loop transfer function of unity feedback
system is given by
SOLUTION:
50
(1 0.1s)(s10)
G(s) 
Determine the static error coefficients Kp , Kv andKa
50
50
50
 0
 0
 5 lim
(1 0.1s)(s10)
(1 0.1s)(s10)
K  lims.G(s)H (s)
K  limG(s)H(s)
 lim s2
s0
2
Ka  s G(s)H (s)
 lims.
s0
s0
v
s0 (1 0.1s)(s 10)
s0
p