CHAPTER 3
Time
Response.
1
3.2 Poles and Zeros and System Response.
K
s 
C (s)  G (s) R(s)
where
s2
G (s) 
s5
1
R( s) 
s
G (s) 
K
s(s   )
( s  2)
A
B
C (s) 
 
s ( s  5) s s  5
2/ 5 3/ 5


s
s5
C (s) 
2 3 5t
c(t )   e
5 5
c(t )  K1  K 2e t
Figure 3.1: (a) System showing input and output;
(b) Pole-zero plot of the system;
2
(c) Evolution of a system response.
3.3 First Order System.
 The transfer function,
G(s) 
C ( s)
K

R ( s ) ts  1
C (s) 
1 K
A
B
 
s ts  1 s ts  1
 For a unit step of; 1/s,
 Its response,
c(t )  A  B' e  t t
A, B and B’ are constant. For K=1 and t=1/a then,
c(t )  1  e  at
 Time constant (1/a), is defined as the time for e-at to decay to 37%of its
initial value or the time it takes for step response to reach 63% of its
final value.
3
Cont’d…
 Step response,
K
C ( s) 
s( s  a)
c(t )  K1  K 2e  at
Figure 3.2: (a) First Order Response to
a Unit Step.
4
Cont’d…



The pole of the transfer function is at –a, the farther the pole
from the imaginary axis, the faster the transient response.
Rise time (Tr), the time the response to go from the 0.1 to 0.9
of its final value. Tr=2.2/a.
Settling time (Ts), time range when the response to reach and
stay within 2% of its final value. Let c(t) = 0.98 then the
Ts.=4/a.
5
3.4 Second Order System.
 The transfer function,
 For Impulse response,
 Where,
 Standard Form,
a0
C ( s)

R ( s ) s 2  b1 s  b0
C ( s) 
1
2

s  1 s   2
K n2
C ( s)

R( s ) s 2  2 n s   n2
Where K is the dc gain,
 is the damping ratio, n is the undammped natural frequency.
s 2  2 n s   n2  0
Where
s1, 2  n  n  2  1
6
Cont’d…
nd
 Example of 2 order system responses.
Figure 3.3: Second Order System, pole plots and Step Response.
7
Cont’d…
General 2nd Order System.
C ( s)  R( s)G ( s)
C ( s) 
b
s( s 2  as  b)
Natural Frequency (n),
 n2
G (s)  2
s  2 n s   n2
b   n2
a  2 n
n  b
 
a
2 n
Damping Ratio (),
Example 3.1: Find the Natural Frequency (n) and Damping Ratio
(),
G( s) 
36
( s 2  4.2s  36)
Solution:
n = 6 and =0.35.
8
Cont’d…
 From previous page,
  kos1
 n2
b
G( s)  2

s  as  b s 2  2 n s   n2
C ( s)  R( s)G ( s)
C ( s) 
b
s( s 2  as  b)
c(t )  K1  K 2e t cos(t   )
Figure 3.4: Left; Plot for an underdamped 2nd Order System.
Right; Step Response for 2nd Order System Damping Cases.
9
Cont’d…
 From previous page,
Figure 3.5: Second Order Response as a Function of Damping Ratio.
10
3.4.1 Over Damped Response.
 The transfer function,
C (s) 
9
9

s ( s 2  9s  9) s ( s  7.854)( s  1.146))
 Poles at the origin from the unit step and two real poles
from the system.
 Constant force response and force response.
c(t )  K1  K 2 e 7.854t  K 3 e 1.146t
11
Example 3.1: Over Damped Response.
Find the step response of the system.
Solution:
Expand the partial fraction.
Take the inverse Laplace Transform.
12
3.4.2 Under Damped Response.
 Under Damped transfer function,
C (s) 
 When 0 < < 1 The transfer function is,
C ( s)
 The Pole position is,

G(s) 

s 2  2 n s   n2
2
n
R( s )
9
s ( s 2  2 s  9)
K n2
s   n  j d s   n  j d 
d  n 1  2
K
K*


s   d  j d s   d  j d
13
Cont’d…
 From previous page,
Figure 3.6: Second Order Response as a Function of Damping Ratio.
14
Cont’d…
 Performance Measures.
Tp 
Ts 

n 1  
4
 n
2
Peak Time
Settling Time
Overshoot
%M p 
c t p   c 
c 
 100%
   1 2 




1

e

1
  100%



1


Figure 3.7: (Top) The 2nd Order
Underdamped Response Specification.
(bottom) Percent overshoot versus
15
damping factor
Cont’d…
 Performance Measures.
Tp 

n 1   2
Ts 

4
 n
%OS  100e

d


4
d

1 2
 Poles position
 n2
G(s)  2
s  2 n s   n2
K
K*


s   d  j d s   d  j d
Figure 3.8: Lines of constant peak time,
Tp , settling time, Ts , and percent
overshoot, %OS Note: Ts2 < Ts1 ; Tp2 <
Tp1; %OS1 < %OS2
16
 Pole Placement.
Cont’d…
 d  PO 
 d  PO  Ts 
 d  Ts 
Figure 3.9: Step responses of second-order
underdamped systems as poles move: (a)
with constant real part; (b) with constant
imaginary part;
17
(c) with constant damping ratio
3.4.3 Critically Damped.
 The transfer function,
 n2
C ( s)

R( s) s   n 2
18
Example 3.3: Critically Damped Response.
Find the step response of the system.
Solution:
Expand the partial fraction.
19
Dominant Pole.

The formula that describing %OS, ts, tp were
derived only for system with two complex poles and
no zeros.

A system with more than two poles or zeros can be
approximated as a second order system that has just
two complex dominant poles.
20
Dominant Pole.
Cont’d…
Will approach second
order system
Cannot be represented
as second order
system
Figure 3.11: Component responses of a three-pole system: (a) pole plot; (b)
component responses: non-dominant pole is near dominant second-order pair
(Case I), far from the pair (Case II), and at infinity (Case III).
21
Cont’d…
Effect of adding a zero to a two-pole system


The closer is the zero to dominant poles, the greater its
effect on transient response.
As the zero move away from dominant poles, the
response approaches that of the two pole system.
Starting at poles 1±j2.828,
then consecutively add
zeros at -3, -5, -10.
22
3.5 Stability.
(i) Stable system.
 Natural response approaches
zero.
 Poles in LHP.
(ii) Unstable system.
 Natural response grows.
 Poles in RHP.
(iii) Marginally stable system.
 Natural response neither
grows/approaches zero.
 Poles on j axis.
Figure 3.12: Closed-loop poles and
response: a. stable system; b. unstable
system
23
3.6 Routh-Hurwitz Stability Criteria.
What is Routh-Hurwitz Criterion (RHC)?
 Through the RHC method we can tell how many closeloop system poles are in the left half plane, in the right halfplane and on the j-axis. We can find the number of poles
in each section of the s-plane, but cannot find their
coordinate.
 The number of roots of the polynomial that are in the
right half-plane is equal to the number of changes in the
first column.
 The RHC method requires two steps;
(1) Generate the data table called Routh table.
(2) Interpret the Routh table to tell number of close loop
system poles in the left half plane, in the right half-plane
and on the j-axis.
24
Cont’d…
 The Close-Loop Transfer function.
 Initial layout for the Routh-Hurwitz Table.
 Completed Routh Table.
25
Example 3.4: Routh-Hurwitz.
Make a Routh table from the system shown below.
Solution:
Find the equivalent close loop system.
Figure (b) above.
Interpretation:
There are two sign changes in the first column.
1  -72  103
The system is unstable, two poles exist in the right half plane.
26
Example 3.5: Routh-Hurwitz.
P(s) = s3 + 10s2 + 31s + 1030
s3
s2
s1
1
10

b1 =
s0
1
31
1
103
1

c1 =
31
1030

=
- 72
1
103
 72
0
 72
b2 =

=
c2 =
0
0
1
0
1
0
1
1
0
 72
0
 72

=0
b3 =

=0
c3 =
1
0
1
0
1
1
0
 72
0
 72
=0
=0
103
 The number of RHP poles = The number of SIGN
CHANGES of COL 1
 TWO sign changes:
 RHP Poles =2
P( s)  s 3  10s 2  31s  1030
P( s)  ( s  13.4136)( s  1.7068  j8.595)( s  1.7068  j8.595)
27
Example 3.6: Routh-Hurwitz.
T (s) 
Solution:
200
s  6 s  11s 2  6 s  200
4
3
s4
1
11
200
s3
6 1
6 1
0
s2
10 1
200 20
0
s1
-19
0
0
s0
20
0
0
 Two sign changes: 2 RHP (UNSTABLE)
 Poles: 2 LHP and 2 RHP
P( s)  s 4  6s 3  11s 2  6s  200
P( s)  ( s  4.27  j 2.54)( s  4.27  j 2.54)( s  1.27  j 2.54)( s  1.27  j 2.54)
28
Example 3.7: Routh-Hurwitz.
T (s) 
Solution:
10
s  2 s  3s 3  6 s 2  5 s  3
5
4
s5
1
3
5
s4
2
6
3
s3
0 
7/2
0
s2
6  7
3
0

s1
42  49  6 2
12  14
0
0
s0
3
0
0
 Assume  is small POSITIVE : TWO sign changes
 Poles: 2 RHP, 3 LHP
P( s)  s 5  2s 4  3s 3  6s 2  5s  3
P( s)  ( s  1.66)( s  .34  j1.5)( s  .34  j1.5)( s  .51  j 0.7)( s  .51  j 0.7)
29
Example 3.8: Routh-Hurwitz.
T ( s) 
Solution:
1
2s 5  3s 4  2s 3  3s 2  2s  1
T (s) 
1
2 s  3 s  2 s 3  3s 2  2 s  1
s5
2
2
2
s4
3
3
1
s3
0 
4/3
0
s2
3  4
1
0
5
4

s1
12  16  3 2
9  12
0
0
s0
1
0
0
 Assume  is small positive: Two sign changes
 Poles: 2 RHP, 3 LHP
P( s)  2s 5  3s 4  2s 3  3s 2  2s  1
P( s)  ( s  1.33)( s  .33  j.89)( s  .33  j.89)( s  .41  j 0.5)( s  .41  j 0.5)
30
Example 3.9: Routh-Hurwitz.
T (s) 
s5
Solution:
s4
P( s )  s  6s  8
4
2
dP( s)
 4s 3  12s  0
ds
1
7
10
s  7 s  6 s  42 s 2  8s  56
5
4
3
6
1
42
8
6
56
8
s3
0 4 1
0 12 3
0 0 0
s2
3
8
0
s1
1/3
0
0
s0
8
0
0
 NO sign changes: No RHP (STABLE)
 Row of ZEROS indicate existence of complex poles &
Symmetric Equations
 Poles: 1 LHP and 4 on jw axis
P( s)  s 5  7 s 4  6s 3  42s 2  8s  56
P( s)  ( s  7)( s  j 2)( s  j 2)( s  j1.414)( s  j1.414)
31
Entire row is zero



When a purely even or odd polynomial is a factor of the
original polynomial.
Even polynomial only have roots that are symmetrical
about the origin.
The symmetry can occur under 3 conditions:
1.
2.
3.
The roots are symmetrical and real.
The roots are symmetrical and imaginary.
The roots are quadrantal.
j
3
2
1
1

3
2
32
Example 3.10: Routh-Hurwitz.
T (s) 
128
Solution:
s  3s  10 s  24 s  48 s  96 s  128 s  192 s  128
8
7
6
5
4
3
s8
1
10
48
128
s7
3 1
24 8
96 32
192 64
s6
2 1
16 8
64 32
128 64
s5
0 6 3
s4
8/3 1
64/3 8
s3
-8 -1
-40 -5
s2
3 1
24 8
s1
3
s0
8
0 32 16 0 64 32
2
128
P( s)  s 6  8s 4  32s 2  64
dP( s)
 6s 5  32s 3  64s  0
ds
0 0 0
64 24
 2 sign changes: 2 RHP (symmetric)
 Poles: 2 RHP, 4 LHP and 2 on j axis
P( s)  s 8  3s 7  10s 6  24s 5  48s 4  96s 3  128s 2  192s  128
P( s)  ( s  2)( s  1)( s  j 2)( s  j 2)( s  1  j1.73)( s  1  j1.73)( s  1  j1.73)( s  1  j1.73)
33
Use of Routh Hurwitz Criteria
 Main use is to determine the position of the poles, which
in turns can determine the stability of the response.
Example
A closed-loop transfer function is given by
C ( s)
K

R( s) s s 2  s  1 s  2  K


Determine the range for K for the system to be always
stable and its oscillating frequency before it becomes
unstable.
34
Solution:
 Characteristic equation is:
ss 2  s  1s  2  K  0
 Expand the equation
s 4  3s 3  3s 2  2s  K  0
 Form the Routh’s array
s4
1
3
s3
s2
3
2
K
s1
s0
92 7

3
3
14 3  3K 14  9 K

73
7
K
K
35
Solution:
 For no sign change
 Referring to row 4
14  9 K
0
7
 which gives, K  14 9
 and row 5, K  0
 Hence its range,
0  K  14 9
 7 3 2  14 9  0
rad
 Oscillating frequency, 2 3
s
36
Steady state
R(s)
E(s)
+
B(s)
-
Y(s)
G (s )
H (s)
 From the diagram
1
E ( s) 
R( s )
1  G( s) H ( s)
 Consider
G( s)  K
Ta1s  1Ta 2 s  1...Tai s  1
s n Tb1s  1Tb 2 s  1...Tbj s  1
 And
Tx1 s  1Tx 2 s  1...Txk s  1
H ( s) 
Ty1s  1Ty 2 s  1...Tyl s  1
 Use the final value theorem and define steady state error, ess
that is given by
ess  lim e(t )  lim sE (s)
t 
s0
37
Unit step
 Unit step input,
 From
E ( s) 
1
R( s) 
s
1
R( s)
1  G ( s) H ( s)
 Steady state error, e 
ss lim
s
1
s0 1  G ( s ) H ( s ) s
 We define step error coefficient,
K s  lim G(s) H (s)
s0
 Thus, the steady state error is e 
ss
1
1 KS
 By knowing the type of open-loop transfer function,
G ( s) H ( s)
 we can know step error coefficient and thus the steady state
error
Ta1s  1Ta2 s  1...Tai s  1 Tx1s  1Tx 2 s  1...Txk s  1

lim
K
K s  lim G(s)H (s) s0 n
s Tb1s  1Tb 2 s  1...Tbj s  1 Ty1s  1Ty 2 s  1...Tyl s  1
s0
38
Unit step
 For open-loop transfer function of type 0:
1
e

K s  K , ss 1  K
 For open-loop transfer function of type 1:
1
0
K s  , ess 
1 
 For open-loop transfer function of type 2:
Ks
 , ess 
1
0
1 
39
Unit step
Example:
A first order plant with time constant of 9 sec and dc gain of 5 is negatively
feedback with unity gain,
determine the steady state error for a unit step input and the final value of the
output.
Solution:
The block diagram of the system is
R(s)
+
-
5
9s  1
Y(s)
As we are looking for a steady state error for a step input, we need to know, K s
5
Knowing the open-loop transfer function, then K s  lim G s H s   lim
s 0
s 0 9 s  1
1
1
1


And steady state error of, ess 
1 KS 1 5 6
Its final value is, y ss  1  1 6  5 6
5
40
Unit Ramp
As in the above section, we know that r ( t )  t , while its Laplace form is R( s ) 
E ( s) 
1
R( s )
1  G( s) H ( s)
1
s2
Thus, its steady state error is
ess  lim s
s 0
1
1
1  G( s) H ( s) s 2
Define ramp error coefficient,
K r ; Kr  lim sG(s) H (s)
Which the steady state error as
1
ess 
Kr
s0
Just like for the unit step input we can conclude the steady state error for a unit ramp through the type
of the open-loop transfer function of the system.
For open-loop transfer function of type 0:
Kr  0
For open-loop transfer function of type 1:
Kr  K
For open-loop transfer function of type 2:
Kr  
41
Unit Ramp
Example:
A missile positioning system is shown.
(i) Find its closed-loop transfer function
 m (s)
 i ( s)
(ii) Determine its undamped natural frequency and its damping ratio if
K  10 3
(iii) Determine the steady state error, if the input is a unit ramp.
(iv) Cadangkan satu kaedah bagi menghapuskan ralat keadaan mantap untuk (iii).
Compensator
i
+
-
K
DC motor
0.01
s(0.4s  1)
m
42
Unit Ramp
Solution:
(a) By Mason rule, the closed-loop transfer function is
0.01
 m ( s)
0.01K
0.025K
s(0.4 s  1)


 2
2
 i ( s ) 1  K . 0.01
0.4 s  s  0.01K s  2.5s  0.025K
s(0.4 s  1)
K.
0.01K
0.4 s 2  s  0.01K
0.025K
 2
s  2.5s  0.025K

,
3
(b) If K  10
 m ( s)
25
 2
 i ( s ) s  2.5s  25
Comparing with a standard second order transfer function
 m ( s)
K n2
 2
 i ( s) s  2 n s   n2
43
Unit Ramp
Comparing
 n2  25
Thus undamped natural frequency
n  5
rad.s-1
and
2 n  2.5
damping ratio of
  0.25
(c) To determine the ramp error coefficient, we must obtain its open-loop transfer function
Go ( s )  K .
0.01
s(0.4s  1)
As it is a type 1, the system will have a finite ramp error coefficient, putting
Go ( s ) 
10
s(0.4 s  1)
K r  lim sGo ( s)  lim s.
s0
K  10 3
s0
10
 10
s(0.4s  1)
Hence steady state error of
ess 
1
 0.1
Kr
44
Unit Parabola
Its time function r (t )  t 2,while its LaplaceR( s ) 
s
2
s 0 1  G( s )H ( s ) s 3
2
,thus its steady state error is
s3
ess  lim
Define parabolic error coefficient, K pa
K pa  lim s 2G( s )H ( s )
s 0
Similarly we can determine its steady state error by knowing the type of the open-loop transfer function
For open-loop transfer function of type 0:
K pa  0
For open-loop transfer function of type 1:
K pa  0
For open-loop transfer function of type 2:
K pa  K
45
In summary we can make a table of the steady state error for the above input
Unit step
Type 0
1
1 Ks
Type 1
0
Type 2
0
Unit ramp
Unit parabolic


1
Kr
0

2
K pa
46