Automatic Control Theory
CSE 322
Lec. 7
Time Domain Analysis
( 2nd Order Systems )
Dr. Tamer Samy Gaafar
Course Web Site
 www.tsgaafar.faculty.zu.edu.eg
 Email:
tsgaafar@yahoo.com
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 system
exhibits a wide range of responses that must be analyzed and described.
 Varying a first-order system's parameters (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.
Introduction
 A general second-order system (without zeros) is characterized by the
following transfer function.
G(s) 

2
n
s ( s  2 n )
 n2
C( s )
 2
R( s ) s  2 n s   n2
Open-Loop Transfer Function
Closed-Loop Transfer Function
Introduction
 n2
C( s )
 2
R( s ) s  2 n s   n2

n
damping ratio of the second order system, which is a measure of
the degree of resistance to change in the system output.
un-damped natural frequency of the second order system, which
is the frequency of oscillation of the system without damping.
Example -1
 Determine the un-damped natural frequency and damping ratio of the
following second order system.
C( s )
4
 2
R( s ) s  2s  4
• Compare the numerator and denominator of the given transfer function
with the general 2nd order transfer function.
 n2
C( s )
 2
R( s ) s  2 n s   n2
n2  4
 n  2 rad / sec
s 2  2 n s  n2  s 2  2s  4
 2 n s  2s
  n  1
   0.5
Introduction
 n2
C( s )
 2
R( s ) s  2 n s   n2
• The closed-loop poles of the system are
  n   n  2  1
  n   n  2  1
Introduction
  n   n  2  1
  n   n  2  1
• Depending upon the value of
of the four categories:

, a second-order system can be set into one
1. Overdamped - when the system has two real distinct poles (
jω
-c
-b
-a
δ

>1).
Introduction
  n   n  2  1
  n   n  2  1
• According the value of
four categories:

, a second-order system can be set into one of the
2. Underdamped - when the system has two complex conjugate poles (0 <  <1)
jω
-c
-b
-a
δ
Introduction
  n   n  2  1
  n   n  2  1
• According the value of
four categories:

, a second-order system can be set into one of the
3. Undamped - when the system has two imaginary poles (  = 0).
jω
-c
-b
-a
δ
Introduction
  n   n  2  1
  n   n  2  1
• According the value of
four categories:

, a second-order system can be set into one of the
4. Critically damped - when the system has two real but equal poles (  = 1).
jω
-c
-b
-a
δ
Time-Domain Specification
For 0< <1 and ωn > 0, the 2nd order system’s response due to a unit
step input looks like ( underdamped )
12
Time-Domain Specification
• The delay (td) time is the time required for the response to reach
half the final value the very first time.
13
Time-Domain Specification
• The rise time (tr) 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.
Time-Domain Specification
• The peak time (tp) is the time required for the response to reach
the first peak of the overshoot.
15
15
Time-Domain Specification
The maximum overshoot ( Mp ) 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.
16
Time-Domain Specification
• The settling time (ts) 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%).
17
Time Domain Specifications
Rise Time
 
 
tr 

2
d
n 1  
Peak Time


tp 

d  1   2
n
Settling Time (2%)
t s  4T 
t s  3T 
4
 n
3
 n
Settling Time (5%)
Maximum Overshoot

Mp e

1 2
 100
Time-Domain Specification
19
S-Plane
• Natural Undamped Frequency.
jω
• Distance from the origin of splane to pole is natural
undamped frequency in rad/sec.
n
δ
S-Plane
• Let us draw a circle of radius 3 in s-plane.
• If a pole is located anywhere on the circumference of the circle the natural
undamped frequency would be 3 rad/sec.
jω
3
-3
3
-3
δ
S-Plane
• Therefore the s-plane is divided into Constant Natural Undamped
Frequency (ωn) Circles.
jω
δ
S-Plane
• Damping ratio.
• Cosine of the angle between
vector connecting origin and
pole and –ve real axis yields
damping ratio.
jω

  cos
δ
S-Plane
0   1
• For Underdamped system 0    90  therefore,
jω
δ
S-Plane
• For Undamped system   90  therefore,
 0
jω
δ
S-Plane
• For overdamped and critically damped systems   0  therefore,
 1
jω
δ
S-Plane
• Draw a vector connecting origin of s-plane and some point P.
jω
P
45 
  cos 45  0.707
δ
S-Plane
• Therefore, s-plane is divided into sections of constant damping ratio lines.
jω
δ
Example-2
• The natural frequency of closed loop
poles of 2nd order system is 2 rad/sec
and damping ratio is 0.5.
Pole-Zero Map
3
0.5
0.38
0.28
0.17
0.64
1.5
1
Imaginary Axis
1
0.94
0.5
0.94
0.5
0
-1
1
0.8
1.5
-2
n2
C( s )
4
 2
 2
2
R( s ) s  2 n s   n
s  2s  4
0.08 2.5
2
2
0.8
• Determine the location of closed loop
poles so that the damping ratio remains
same but the natural undamped
frequency is doubled.
3
2
0.64
0.5
-3
-2
0.38
-1.5
0.28
-1
0.17
0.08 2.5
-0.5
30
Real Axis
Example-2
• Determine the location of closed loop poles so that the damping ratio remains same but the
natural undamped frequency is doubled.
Pole-Zero Map
5
4
0.5
3
Imaginary Axis
2
1
4
0
2
-1
-2
-3
0.5
-4
-5
-8
-6
-4
-2
0
2
4
S-Plane
 n  n  2  1
 n  n  2  1
n  2  1

n
Step Response of underdamped
System
 n2
C( s )
 2
R( s ) s  2 n s   n2
Step Response
n2
C ( s)  2
ss  2n s  n2 
• The partial fraction expansion of above equation is given as
s  2 n
1
C( s )   2
s s  2 n s   n2

n2 1   2
s  2 n 2
s  2 n
1
C( s )   2
s s  2 n s   2 n2   n2   2 n2
s  2 n
1
C( s )  
s s   n 2  n2 1   2



Step Response of underdamped
System
s  2 n
1
C( s )  
s s   n 2  n2 1   2
• Above equation can be written as


s  2 n
1
C( s )  
s s   n 2  d2
• Where d  n 1   2 , 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:
s   n
 n
1
C( s )  

2
2
s s   n   d s   n 2  d2
Step Response of underdamped
System
s   n
 n
1
C( s )  

2
2
s s   n   d s   n 2  d2

n 1   2
1 2
s   n
1
C( s )  

2
2
s s   n   d
s   n 2  d2
s   n
d
1

C( s )  

2
2
2 s   2   2
s s   n    d
1
n
d
c(t )  1  e
 n t
cos  d t 

1 2
e  nt sin  d t
Step Response of underdamped
System
c(t )  1  e
 n t
cos  d t 

1 2
e  nt sin  d t



c(t )  1  e  nt cos d t 
sin d t 
2


1




• When   0
d  n 1   2
 n
c(t )  1  cos  n t
Step Response of underdamped
System



c(t )  1  e  nt cos d t 
sin d t 
2


1




if   0.1 and n  3 rad / sec
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
2
4
6
8
10
Step Response of underdamped
System



c(t )  1  e  nt cos d t 
sin d t 
2


1




if   0.5 and n  3 rad / sec
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
2
4
6
8
10
Step Response of underdamped
System



c(t )  1  e  nt cos d t 
sin d t 
2


1




if   0.9 and n  3 rad / sec
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
2
4
6
8
10
Step Response of underdamped
System
2
b=0
b=0.2
b=0.4
b=0.6
b=0.9
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
1
2
3
4
5
6
7
8
9
10
Step Response of underdamped
System
1.4
1.2
1
0.8
wn=0.5
wn=1
wn=1.5
wn=2
wn=2.5
0.6
0.4
0.2
0
0
1
2
3
4
5
6
7
8
9
10
Time Domain Specifications of
Underdamped system
Summary of Time Domain Specifications
Rise Time
 
 
tr 

2
d
n 1  
Peak Time


tp 

d  1   2
n
Settling Time (2%)
t s  4T 
t s  3T 
4
 n
3
 n
Settling Time (5%)
Maximum Overshoot

Mp e

1 2
 100
Example -3
 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-3
Rise Time
Peak Time

tp 
d
 
tr 
d
Settling Time (2%)
t s  4T 
t s  3T 
Maximum Overshoot
4
 n
3
 n
Settling Time (5%)

Mp e

1 2
 100
Example -3
Rise Time
 
tr 
d
tr 
3.141  
n 1  

2
2

1


  tan 1( n
)  0.93 rad
 n
tr 
3.141  0.93
5 1  0.6 2
 0.55 s
Example -3
Peak Time
Settling Time (2%)

tp 
d
3.141
tp 
 0.785 s
4
ts 
4
 n
4
ts 
 1.33 s
0.6  5
Settling Time (5%)
ts 
3
 n
3
ts 
 1s
0. 6  5
Example -3
Maximum Overshoot

Mp e

Mp e

1 2
 100
3.1410.6
1 0.6 2
 100
Example -3
Step Response
1.4
1.2
Mp
Amplitude
1
0.8
0.6
0.4
Rise Time
0.2
0
0
0.2
0.4
0.6
0.8
Time (sec)
1
1.2
1.4
1.6
Example -4
 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 unitstep 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.
Example -4
Example -4
Since J  1 kgm2 and
B  1 Nm/rad/sec
C( s )
K
 2
R( s ) s  (1  KK h )s  K
• Comparing above T.F with general 2nd order T.F
 n2
C( s )
 2
R( s ) s  2 n s   n2
n  K
 
(1  KK h )
2 K
Example -4
n  K
• Maximum overshoot is 0.2.
 
(1  KK h )
2 K
• The peak time is 1 sec

tp 
d
1

ln( e

1 2
)  ln 0.2
3.141
n 1   2
n 
3.141
1  0.456 2
n  3.53
Example -4
n  3.53
n  K
3.53 
K
3.53 2  K
K  12 .5
 
(1  KK h )
2 K
0.456  2 12 .5  (1  12 .5K h )
K h  0.178
Example -4
tr 
 
n  3.53
ts 
n 1   2
4
 n
t s  2.48 s
t r  0.65 s
ts 
3
 n
t s  1.86 s
Example -4
•Repeat part (a) without the velocity feedback.
•What is your observations ?
Further Reading
Time Domain Specifications (Rise Time)



c(t )  1  e  nt cos d t 
sin d t 
2


1




Put t  t r in above equation



c(t r )  1  e  ntr cos d t r 
sin d t r 
2


1




Where c(tr )  1



0  e  ntr cos d t r 
sin d t r 
2


1




 e  nt r  0



0  cos d t r 
sin d t r 
2


1




Time Domain Specifications (Rise Time)



cos d t r 
sin d t r   0
2


1




above equation can be re - writen as
sin  d t r  
1 2

tan  d t r  
cos  d t r
1 2

2

1


d t r  tan 1  







Time Domain Specifications (Rise Time)
2

1


d t r  tan 1  







  1 2
tr 
tan 1   n
d
 n


1
 
tr 
d





  tan
1
a
b
Time Domain Specifications (Peak Time)



c(t )  1  e  nt cos d t 
sin d t 
2


1




• In order to find peak time let us differentiate above equation w.r.t t.





dc(t )

 n t 


t
d
  n e
cos d t 
sin d t   e n  d sin d t 
cos d t 




dt
1 2
1 2






2



n
d
0  e  nt  n cos d t 
sin d t  d sin d t 
cos d t 
2
2


1


1




2


2

1




n
0  e  nt  n cos d t 
sin d t  d sin d t  n
cos d t 
2
2


1


1




Time Domain Specifications (Peak Time)
2


2

1




 nt 
n
n
0e
 n cos d t 
sin d t  d sin d t 
cos d t 
2
2


1


1




  2

n
e  nt 
sin d t  d sin d t   0
 1 2



e  n t  0
  2

n

sin d t  d sin d t   0
 1 2



  2

n
sin d t 
 d   0
 1 2



Time Domain Specifications (Peak Time)
  2

n
sin d t 
 d   0
 1 2



  2

n

 d   0
 1 2



sin d t  0
d t  sin 1 0
t
0,  , 2 , 
d
• Since for underdamped stable systems first peak is maximum peak therefore,

tp 
d
Time Domain Specifications (Maximum
Overshoot)
c(t p )  1  e
 n t p



cos d t p 
sin d t p 
2


1




c()  1


 

 t
M p  1  e n p  cos  d t p 
sin  d t p   1  100

2

 
1



 

Put

tp 
in above equation
d
Mp
  n 
d
  e






 
 cos   
sin

 100
d
d


2
d
 d 
1




Time Domain Specifications (Maximum
Overshoot)
Mp
Put
  n 
d
  e






 
 cos   
sin  d
 100
d


2
d
 d 
1




ωd  ωn 1-ζ 2 in above equation
  n

n
M p   e






1 2 
  100
cos


sin



2
1


  


1 2
 1  0  100
M p   e



Mp e

1 2
 100
Time Domain Specifications (Settling Time)



c(t )  1  e  nt cos d t 
sin d t 
2


1




T
1
 n
  n   n  2  1
Real Part
Imaginary Part
Time Domain Specifications (Settling Time)
• Settling time (2%) criterion
• Time consumed in exponential decay up to 98% of the input.
t s  4T 
4
 n
T
1
 n
• Settling time (5%) criterion
• Time consumed in exponential decay up to 95% of the input.
t s  3T 
3
 n
Step Response of critically damped System
(  1 )
 n2
C( s )

R( s ) s   n 2
Step Response
C( s ) 
s s   n 
• The partial fraction expansion of above equation is given as
 n2
s s   n 
2
A
B
C
 

s s   n s   n 2
n
1
1
C( s )  

s s  n s  n 2
c(t )  1  e nt  n e nt t
c(t )  1  e nt 1  n t 
 n2
2
68