Transcript Time Domain Analysis of 2nd Order Systems

Feedback Control Systems (FCS)
Lecture-22-23-24
Time Domain Analysis of 2nd Order Systems
Dr. Imtiaz Hussain
email: [email protected]
URL :http://imtiazhussainkalwar.weebly.com/
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 firstorder system or, depending on component values, display damped
or pure oscillations for its transient response.
Introduction
• A general second-order system is characterized by
the following transfer function.
C(s)
R( s )
2

n
2
2
s  2 n s   n
Introduction
C(s)
R( s )
n

2

n
2
2
s  2 n s   n
un-damped natural frequency of the second order system,
which is the frequency of oscillation of the system without
damping.
damping ratio of the second order system, which is a measure
of the degree of resistance to change in the system output.
Example#1
• Determine the un-damped natural frequency and damping ratio
of the following second order system.
C(s)
R(s)
4

s
2
 2s  4
• Compare the numerator and denominator of the given transfer
function with the general 2nd order transfer function.
C(s)
R( s )
2
n  4
s
2
2

n
2
  n  2 rad / sec
 2  n s 
2
n
 s
2
2
s  2 n s   n
 2s  4
 2 n s  2 s
 
n
1
   0 .5
Introduction
C(s)
R( s )
2

n
2
2
s  2 n s   n
• Two poles of the system are

2
1
  n   n 
2
1
  n   n
Introduction
  n   n

2
1
  n   n

2
1
• According the value of 
one of the four categories:
, a second-order system can be set into
1. Overdamped - when the system has two real distinct poles (  >1).
jω
-c
-b
-a
δ
Introduction
  n   n

2
1
  n   n

2
1
• According the value of 
one of the four categories:
, a second-order system can be set into
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 
one of the four categories:
, a second-order system can be set into
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 
one of the four categories:
, a second-order system can be set into
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
11
Time-Domain Specification
• 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.
Time-Domain Specification
• The peak time is the time required for the response to reach
the first peak of the overshoot.
14
14
Time-Domain Specification
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
• 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
• 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
• For Underdamped system 0     90  therefore, 0    1
jω
δ
S-Plane
• For Undamped system   90  therefore,   0
jω
δ
S-Plane
• For overdamped and critically damped systems   0 
therefore,   0
jω
δ
S-Plane
• Draw a vector connecting origin of s-plane and some point P.
jω
P
45

δ
  cos 45

 0 . 707
S-Plane
• Therefore, s-plane is divided into sections of constant damping
ratio lines.
jω
δ
Example-2
• Determine the natural frequency and damping ratio of the poles from the
following pz-map.
Pole-Zero Map
1.5
0.91
0.84
0.74
0.6
0.42
0.22
0.96
Imaginary Axis (seconds-1)
1
0.5
0.99
4
0
-0.5
3.5
3
2.5
2
1.5
1
0.5
0.99
-1
0.96
-1.5
-4
0.91
-3.5
0.84
-3
-2.5
0.74
-2
-1.5
0.6
-1
0.42
0.22
-0.5
0
Example-3
Pole-Zero Map
3
0.7
• Determine
the
natural
frequency and damping ratio of
the poles from the given pzmap.
0.42
0.28
0.14
2.5
2
2 0.82
1.5
0.91
Imaginary Axis (seconds-1)
• Also determine the transfer
function of the system and state
whether
system
is
underdamped,
overdamped,
undamped or critically damped.
0.56
3
1
1
0.975
0.5
0.975
0.5
0
-1
1
0.91
1.5
-2 0.82
2
0.7
-3
-3
0.56
-2.5
-2
0.42
-1.5
0.28
-1
0.14
-0.5
2.5
30
Example-4
• 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
C(s)
R( s )
2

2
s  2 n s 
2
0.64
n
2
n

4
2
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
0.5
-3
-2
0.38
-1.5
0.28
-1
0.17
0.08 2.5
-0.5
30
Real Axis
Example-4
• 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

Step Response of underdamped System
C(s)
R( s )
2

n
2
s  2 n s 
2
Step Response
C(s) 
2
n

n
2
2
s s  2 n s   n

• The partial fraction expansion of above equation is given as
C(s) 
1

s
s  2 
n
2
2
s  2  n s   n
2

n 1 
s  2 n 2
C(s) 
1
s

s  2 
2
2
n
2
2
2

2
2
s  2  n s    n   n    n
C(s) 
1
s

s  2 
 s   n 2
n
2
 n 1  

2

Step Response of underdamped System
C(s) 
1

s
s  2 
 s   n 2
n
2

 n 1  
2

• Above equation can be written as
C(s) 
1
s

s  2 
 s   n 2
n
2
 d
• 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:
C(s) 
1
s

s  
 s   n 2
n
2
 d


n
2
 s   n 
2
 d
Step Response of underdamped System
1
C(s) 

s
s  
 s   n 2
n
2
 d


 s   n 

1
C(s) 

s
C(s) 
1
s

s  
 s   n 
s  
s   n 
c(t )  1  e
  n t
2

n
2

2
d
n

2
d

cos  d t 
n
2
1
2
n 1  
 s   n 2
2
2
 d
d

1

1
2
 d
2
 s   n 2
2
e
  n t
2
 d
sin  d t
Step Response of underdamped System
c(t )  1  e
  n t
cos  d t 

1

  n t
 cos  t 
c(t )  1  e
d


• When
  n t

1
  0
d  n 1  
2
e
2
 n
c ( t )  1  cos  n t
2
sin  d t

sin  d t 


Step Response of underdamped System

  n t
 cos  t 
c(t )  1  e
d


if   0 . 1
and

1
2

sin  d t 


 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

  n t
 cos  t 
c(t )  1  e
d


if   0 . 5
and

1
2

sin  d t 


 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

  n t
 cos  t 
c(t )  1  e
d


if   0 . 9
and

1
2

sin  d t 


 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
Time Domain Specifications (Rise Time)
c(t )  1  e
Put
  n t

 cos  t 
d



1
2

sin  d t 


t  t r in above equation
c(t r )  1  e
Where
c(t r )  1
0  e
e
  n t r
  n t r
 0
  n t r

 cos  t 
d r



 cos  t 
d r



0   cos  d t r 



1

1
2
2

sin  d t r 



1

sin  d t r 


2

sin  d t r 


Time Domain Specifications (Rise Time)

 cos  t 
d r


above equation

sin  d t r   0



1
2
can be re - writen as
sin  d t r  
1

tan  d t r  

1 
 d t r  tan



2
cos  d t r
1
2

1

2




Time Domain Specifications (Rise Time)

1 
 d t r  tan



tr 
1
d
1
2


n 1  
1 
tan


 n

tr 




2





 
d
  tan
1
a
b
Time Domain Specifications (Peak Time)
c(t )  1  e
  n t

 cos  t 
d



1
2

sin  d t 


• In order to find peak time let us differentiate above equation w.r.t t.
dc ( t )
dt
0  e
  n e
  n t
  n t





  n t

0  e



 cos  t 
d



1
2


  n t
   sin  t 
sin  d t   e
d
d




2
n
cos  d t 
 n
1
2
n
cos  d t 
1
2
sin  d t   d sin  d t 

d
1

sin  d t   d sin  d t 
2
 n

d
1
n
1
1
2
2
2
2

cos  d t 



cos  d t 



cos  d t 


Time Domain Specifications (Peak Time)

  n t

0  e


2
n
cos  d t 
 n
1
2
sin  d t   d sin  d t 



2


  n t
n

e
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



n
1
1
2
2

cos  d t 


Time Domain Specifications (Peak Time)


2
 n
sin  d t 
 d   0
 1 2





2
 n

 d   0
 1 2



sin  d t  0
1
 d t  sin
t 
0
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  t 
d p



1
2

sin  d t p 


c(  )  1
M
Put
p
tp 
M


  n t p 
 1  e
cos  d t p 






d
p

1
2



sin  d t p  1  100




in above equation


  n
d
  e





cos  d


d


1
2
sin  d

 
 100

 d 

Time Domain Specifications (Maximum Overshoot)
M
Put
p


  n
d
  e


ω d  ω n 1-ζ
M
p
2



cos  d


d


1
sin  d
2

 
 100

 d 

in above equation

  n

n
 e



M
p

1 
2


cos  





 e




1 

M
p
 e
2

1
2


sin 
 100
 
 

 1  0   100



1 
2
 100
Time Domain Specifications (Settling Time)

  n t
 cos  t 
c(t )  1  e
d


T 
1
2

sin  d t 


  n   n 
1


n
Real Part
2
1
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

T 
n
1

n
• Settling time (5%) criterion
• Time consumed in exponential decay up to 95% of the input.
t s  3T 
3

n
Summary of Time Domain Specifications
Rise Time
tr 
 
d
Peak Time
 

n 1  
tp 
2

d


n 1  
Settling Time (2%)
t s  4T 
t s  3T 
4

Maximum Overshoot
n

3

M
n
Settling Time (4%)
p
 e

1 
2
 100
2
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
Rise Time
tr 
Peak Time
 
tp 
d
Settling Time (2%)
t s  4T 
t s  3T 

n
3

n
Settling Time (4%)
d
Maximum Overshoot
4


M
p
 e

1 
2
 100
Example#5
Rise Time
 
tr 
tr 
d
3 . 141  
n 1  
  tan
tr 
1
(

2
n 1  

2
)  0 . 93 rad
n
3 . 141  0 . 93
5 1  0 .6
2
 0 . 55 s
Example#5
Peak Time
tp 
tp 
Settling Time (2%)

ts 
d
3 . 141
4
ts 
 0 . 785 s
Settling Time (4%)
ts 
ts 
3

3
0 .6  5
n
 1s
4

4
0 .6  5
n
 1 . 33 s
Example#5
Maximum Overshoot

M
p
 e

M
p
p
M
1 
2
 100
3 . 141  0 . 6
1 0 .6
 e
M

2
 100
 0 . 095  100
p
 9 .5 %
Example#5
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#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.
Example#6
Example#6
Since J  1 kgm
2
and
C(s)
R( s )
B  1 Nm/rad/sec

K
2
s  (1  KK h ) s  K
• Comparing above T.F with general 2nd order T.F
C(s)
R( s )
n 
2

K
n
2
2
s  2 n s   n
 
(1  KK h )
2 K
Example#6
n 
K
• Maximum overshoot is 0.2.
 
(1  KK h )
2 K
• The peak time is 1 sec
tp 
ln( e
n 1  

1 
d
3 . 141
1


2
)  ln 0 . 2 
n 
2
3 . 141
1  0 . 456
 n  3. 53
2
Example#6
 n  3. 96
n 
K
3. 53 
3 . 53
2
K
 K
K  12 . 5
 
(1  KK h )
2 K
0 . 456  2 12 . 5  (1  12 . 5 K h )
K h  0. 178
Example#6
 n  3. 96
tr 
 
n 1  
ts 
2
4

n
t s  2. 48 s
t r  0. 65 s
ts 
3

n
t s  1. 86 s
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.
a
s ( cs  1 )
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.
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).
Example#9
Example#9
Step Response of critically damped System ( 
C(s)
R( s )
2

n
s   n 
2
Step Response
C(s) 
2
n
s s   n 
• The partial fraction expansion of above equation is given as
2
n
s s   n 

2
A
s
C(s) 
1
s


B
s  n
1
s  n
c( t )  1  e
 n t
c(t )  1  e
1


C
 s   n 2
n
 s   n 2
 ne
 nt
 nt
1   n t 
t
2
)
Step Response of overdamped and
undamped Systems
• Home Work
71
Second – Order System
Example 10: Describe the nature of the second-order system
response via the value of the damping ratio for the systems with
transfer function
1. G (s) 
2.
G (s) 
3. G (s) 
12
s  8 s  12
2
16
s  8 s  16
2
Do them as your own
revision
20
s  8 s  20
2
72
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END OF LECTURES-22-23-24