Outline
•
•
•
•
Time Response
M. Sami Fadali
Professor of Electrical Engineering
University of Nevada
Pole/Zero locations and time response.
First order systems.
Second order systems.
Dominant poles.
1
Poles, Zeros, & System Response
K s  z1 s  z m 
K
s  p1 s  pn 
System Poles and Zeros
 (s  z j )
System Poles and Zeros: poles and zeros of its
transfer function
 (s  p )
Example
m
G(s) 
2
j 1
n
i
i 1
n G(s) Poles at pi
m G(s) Zeros at zi
zeros at: 1,2
poles at: 1,3,4
3
4
Time Response & Poles/Zeros
Example: System Response
ି௧
ିଶ௧
௡௔௧௨௥௔௟
௥௘௦௣௢௡௦௘
• The poles of the input give the forced response.
• The poles of the transfer function give the natural
response.
• Real-axis poles: Exponential decay (LHP) or
increase (RHP).
• Complex conjugate poles: oscillatory response.
• Zeros: Affect amplitudes of the different
components and therefore affect the overall
shape of the response.
ିଷ௧
௙௢௥௖௘ௗ
௥௘௦௣௢௡௦௘
5
4.3 First Order Systems
6
Step Response of 1st Order System
Transfer Function
Differential Equation
Unit Step Response
ି௔௧
7
8
Time Constant  s
Initial Slope
a) Time after which the decaying exponential e t/
reaches 37% of its initial value ( e 1 = 0.37)
• 3 time constants
e 3 0.05
• 4 time constants
e 4 0.02
• 5 time constants
e 5 0.01
b) Time after which the rising exponential 1e t/
reaches 63% of its final value ( 1e 1 = 0.63)
de  t / 
dt
 
e  t /
t0
d 1  e  t /  
dt


t0
 
t0
e  t /

1


t0
1

c) Time Constant
time at which a straight line at the
initial slope intersects the final value.
9
Specs. for 1st Order Systems
4.4 Second Order Systems
Rise Time: time to go from 10% to 90% of the final value.
 t1 /
 01
.
 t 2 /
 0.9
10%: 1 e
90%: 1 e
 e t 2  t1   
10
Transfer Function
G( s) 
0.9
9
01
.
 2n
 n2

s2  2 n s   n2  s   n 2   n2 1   2 
Differential Equation
Tr  t2  t1   ln 9  2.2
Settling Time: time to reach and stay within a specified
percentage of the final value.
Unit Step Response
Definitions:
c  2 n c   2n c   2n r
C (s)  G (s) R(s)
n = undamped natural frequency rad/s
 = damping ratio
d = damped natural frequency rad/s
௦
௦
௦
11

 n2
s 2  2 n s  n2 s
1
A s  A2
  2 1
s s  2 n s   n2
12
Time Responses of 2nd Order Systems
a) Oscillatory Response  = 0
ଶ
௡
ଶ
௡
ଶ
ଶ
௡
ଶ
௡
• The undamped natural frequency is the
frequency of oscillations when  = 0
13
14
b) Overdamped Response  > 1
Example: Oscillatory
G( s) 
n2
n2

s 2  2n s  n2 s  n 2  n2  2  1
C (s)  G (s) R(s) 

s  
 n2
n
1
K1
K2


s s  s1 s  s 2
2   n2  2  1 s

s1, 2   n   n  2  1   n    2  1
ଵ
ଶ
ଵ
ଶ
ଵ
15
ଶ
௦భ ௧

ଵ
ଵ
௦మ ௧
16
Ex. Overdamped System
Two real roots &
ଶ
௡
ଶ
௡
ଶ
௡
Example: Overdamped
>1
ଶ
௡
௡
Overdamped: two real poles
17
c) Underdamped Response  < 1
C ( s)  G ( s) R( s) 
n2
s    
2
n
18
Partial Fraction Expansion
1 s   n  d  n d 
C ( s)  
s
s   n 2  d2

 d2 s
1
A s  A2
  2 1
s s  2 n s  n2
Equate coefficients

1 s   n  d  1   2
 
s
s   n 2  d2
n2  s 2  2 n s  n2  A1s 2  A2 s




c(t )  1  e nt cos(d t ) 
sin(

t
)
d 
1  2


s 2 : 1  A1  0  A1  1
s : 2 n  A2  0  A2  2 n
19
20
Time Response



c(t )  1  e nt cos(d t ) 
sin(d t )
2
1 


Example: Underdamped
cos(AB)=cos(A)cos(B)+sin(A)sin(B)
sin(A+B)=sin(A)cos(B)+cos(A)sin(B)
c (t )  1 
e  nt
1  2
 1 
2
cos( d t )   sin( d t )
c(t )  1 
e nt
cos(d t   ),  sin 1 
2
1 

e  nt
 1
cos( d t   ),  sin 1 
2
1 
 1
e  nt
sin( d t   ),   cos 1    2  
2
1 
21
22
d) Critically Damped Response
 =1
Underdamped Poles
  cos 1 
 n2
C (s)  G (s)R(s) 
s   n 2 s
n
1
1
 

s s   n 2 s   n
c(t )  1  e nt nt  1, t  0
23
24
Specifications: 2nd Order Systems
Example: Critically Damped
1.Peak Time Tp : time to first peak (  < 1)
2. Percent Overshoot % OS
(  < 1)
௦௧
3.Settling time Ts : time to reach and stay within
2% of final value (5% and 1 % also used)
4. Rise Time Tr : time from 10% to 90% of final
value.
25
Specifications: 2nd Order
Step Response
26
Significance of 2nd Order Criteria
• Measures of Speed of Response Tr , Tp , n
Measures of Relative Stability % OS, 
• Mixed Measures Ts
• Tr and Tp increase together
• Use Tp since it has a simpler expression.
27
28
Time to First Peak Tp
Rise Time Tr
Tr  (1.76ζ 3  0.417ζ 2 + 1.039ζ + 1)/ n
Derivative of step response
1
_ c sC ( s )  s G ( s )  c  g (t )  impulse response
s


 n2 -1 
n2
d
c  _ -1  2
 _ 
2
2
2
 s  2 n s  n  d
s   n   d 

n
e  t sin d t   0 at max or min
2
1 
n
d t  l , l  0,1,2,  Tp 



d n 1   2
29
Evaluation of % OS
Effect of Changing 



c(t )  1  e nt cos(d t ) 
sin(d t )
2
1 


cmax  c(t ) t T    1  e
p
 n 


n 1 2 
d



sin




cos  
1  2


  

1 2 

c 1

%OS  max 100%  e 
1
 
 lnx 
 2  ln 2 x 
30
x
100 %
%OS
1
100
31
32
Ex. Mass-Spring-Damper
Evaluation of Ts
1
Y ( s)
 2
F ( s ) ms  bs  k
1m
 2
s  b m s  k m 
n2 k
G( s)  2
s  2 n s  n2
G( s) 
• Difficult since sinusoid not always at peak
• Obtain a rough estimate based on the exponential
decay
• Time constant of exponential decay = 1/n
 3

 n
 4
Ts  
 n
 5
  n
b
m
k
y
Equate Coefficients
5%
n2  k m  n  k m
2%
2 n  b m    b 2m 
  b 2 k m 
1%
k m
33
34
s-plane Contours
Underdamped Time Response
Constant damping ratio 
• Effect of moving pole on s-plane contour
on the time response.
• Estimate time response characteristics
from pole locations.
• Used later for design.
%OS  e
  


 1 2 
Tp 
100 %

d
Ts 
35
Constant imaginary part d
4
Constant real part n
 n
36
Example
Pole Location & Time Response
ଶ
௡
ଶ
Constant real part n :
ଶ
௡
௡
• For any gain
d changing, ܶ௦ fixed.
rad/s for any choice of gain
௡
Constant imaginary part d :
n changing, ܶ௣ fixed.
Constant damping ratio :
d is changing & n changing.
37
– Settling time:
• For
ସ
఍ఠ೙
௥௔ௗ
௦
௦
௡
s
ଶ
ఠ೙
• Increasing
(i)
decreases (increases OS%)
(ii)
increases ௡ and ௗ (decreases
faster response
௣)
:
38
Time Response: 3rd Order System
Effect of 3rd Pole on
Time Response
1.4
Slower response.
Decrease overshoot.
Reduced effect for pole farther in LHP.
Reduced effect if zero almost cancels
a pole.
Step Response
No Pole
1.2
Pole at -20
1
Amplitude
•
•
•
•
ଶ
0.8
0.6
G (s) 
0.4
Pole at -10
13a
s  4 s  13 s  a 


2
0.2
Pole at -5
0
0
0.5
1
1.5
2
2.5
Time (sec)
39
40
Time Response: 1st,2nd, 3rd Order
Time Response: 3rd Order System
Step Response
1
Amplitude
0.8 G(s)=2/(s2+2s+2)
ି఍ఠ೙ ௧
ௗ
0.6
0.4
G1(s)=G(s)/(10 s+1)
0.2
G2(s)=1/(10 s+1)
0
0
10
20
30
Time (seconds)
40
50
41
42
Time Response: 2nd Order System with Zero
Effect of Zero on Time Response
G ( s) 
Faster response.
Increased overshoot.
Reduced effect for zero farther in LHP.
Reduced effect if zero almost cancels a
pole.
s 2  4 s  13
Step Response
1
0.8
Amplitude
•
•
•
•
13 a s  a 
No zero
0.6
Zero at -20
0.4
Zero at -10
0.2
0
43
Zero at -5
0
0.5
1
1.5
Time (sec)
2
2.5
44
Nonminimum-phase Systems
Almost Canceling Pole and Zero
G( s) 
n2 s  a   
s 2  2n s  n2 s  a  ,   1
Undershoot in the time response.
n2 s  a   
1
G(s)  2
s s s  2 n s  n2 s  a 
1
 1
n2  a
1  a 
As  B

  2
 2
2 
2
s
 a  2 n a  n  s  a s  2 n s  n
a  2 n
2 n a  4 2n2  n2


,
B

2

A  1  2
n
a  2 n a  n2
a 2  2 n a  n2
Amplitude
0.8

Reduced effect if zero
almost cancels a pole.
Step Response
1.2
>> s=tf(‘s’)
>> g=2.6*(-s+5)/(s^2+4*s+13)
>> step(g)
0.6
0.4
0.2
0
-0.2
0
0.5
1
1.5
2
2.5
3
3.5
Time (seconds)
45
Dominant 2nd Order Pair
46
Dominant Pair Example
For systems with a 2nd order pair and with
zeros or additional poles that
(i) are all located far in the LHP, or
(ii) almost cancel,
then the time response is approximately the
same as that of the 2nd order pair.
• The 2nd order pair is said to be dominant.
RULE of THUMB: factor of 5 is enough.
•
1. Poles s1,2 = 5j5
s3 = 25
The underdamped pair is dominant since
exp(25 t) decays to zero quickly.
• RULE of THUMB factor of 5 is enough.
2. Poles s1,2 = 5j5, s3 = 5, Zero z = 5.01
The underdamped pair is dominant
Third pole almost cancels with zero.
47
48
Saturation
Nonlinearities & the Time Response
1
0.9
No saturation
0.8
Consider a feedback loop with one of the
following nonlinearities:
0.7
• Output cannot
exceed a fixed
value.
• Clipped output.
1. Saturation
2. Dead Zone
3. Backlash
0.6
0.5
With saturation
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
2.5
3
3.5
4
Time s
2
s+2
Step
Saturation
Transfer Fcn
Scope
2
s+2
Transfer Fcn1
49
50
Dead Zone
Backlash
No deadzone
0.8
• On reversing
directions, the output is
unchanged until the
input exceeds a
threshold value.
• Reversing directions
distorts the output.
0.7
• Output is zero
until the input
exceeds a
threshold value.
• Reduced output
amplitude.
0.6
0.5
0.4
0.3
0.2
With deadzone
0.1
0
0
1
2
3
4
5
6
7
8
9
2
Dead Zone
-0.5 to 0.5
s+2
1
s
Transfer Fcn2
Integrator
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
With backlash
0
1
2
2
1
s
Transfer Fcn1
Integrator1
4
5
6
7
8
9
10
Time s
2
Sine Wave
2 sin(2 t)
s+2
1
s
Transfer Fcn2
Integrator
Backlash
Scope
Scope
s+2
3
10
Time s
Sine Wave
2 sin(4 t)
No backlash
1.8
2
51
s+2
1
s
Transfer Fcn1
Integrator1
52