Modern Control System
EKT 308
Steady-State and Stability
Quick Review
1. Laplace transform
2. Poles and zeros , transfer function
3. Simplification complex block diagram, signal flow
diagram
4. State space modeling
5. Modeling physical systems
6. Time response of first and second order systems
Topic to cover
1. Steady State Error
2. Routh Hurwitz Stability Criterion
Review : Transient and Steady-State Response
Analysis
Step response of a control system
Review: Performance Measures for step
response
 Delay Time Td : Time to reach half of the final value for the
first time.
Rise Time Tr : Time to rise to the final value.
Underdamped (0% 100%)
Overdamped (10%  90%)
 Peak Time T p : Time to reach the first peak of the overshoot.
 Percentage overshoot,
P.O. 

M pt  f v
fv
100%
y (T p )  y ()
y ( )
100%
Steady State Error
R(s)
E(s)
+
B(s) -
From the diagram
Consider
G (s )
Y(s)
H (s )
1
E ( s) 
R( s )
1  G( s) H ( s)
Ta1 s  1Ta 2 s  1...Tai s  1
G (s)  K n
s Tb1 s  1Tb 2 s  1...Tbj s  1
H ( s) 
Tx1 s  1Tx 2 s  1...Txk s  1
T
y1
s  1T y 2 s  1...T yl s  1
Use the FINAL VALUE THEOREM and define steady state error,
that is given by
and
ess  lim e(t )  lim sE (s)
t 
s0
ess
Unit step
1
R( s) 
s
Unit step input,
From
E ( s) 
Steady state error,
1
R( s)
1  G ( s) H ( s)
s
1
s0 1  G ( s ) H ( s ) s
ess  lim
We define step
error coefficient,
K s  lim G(s) H (s)
s0
1
e ss 
Thus, the steady state error is
1 KS
G ( s) H ( s)
By knowing the type of open-loop transfer function,
We could determine step error coefficient and thus the steady state error
K s  lim G(s)H (s)
s0
 lim K
s0
Ta1s  1Ta 2 s  1...Tai s  1 T
s n Tb1s  1Tb 2 s  1...Tbj s  1 T
s  1Tx 2 s  1...Txk s  1
y1s  1Ty 2 s  1...Tyl s  1
x1
1
1 K
For open-loop transfer function of type 0:
Ks  K
For open-loop transfer function of type 1:
K s  , ess  1  0
1 
For open-loop transfer function of type 2:
K s  , ess  1  0
ess 
1 
STEADY STATE ERROR 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 step error coefficient,
Knowing the open-loop transfer function, then
K s  Lim G ( s) H ( s)  Lim
s 0
And steady state error of
Its final value is
s 0
5
5
9s  1
1
1
1
ess 


1 KS 1 5 6
y ss  1  1 6  5 6
Ks
Unit Ramp
As in the previous slide, we know that
r (t )  t
R( s ) 
, while its Laplace form is
E ( s) 
1
R( s )
1  G( s) H ( s)
Thus, its steady state error is
Define ramp error coefficient,
1
s2
s
1
2
s0 1  G ( s ) H ( s ) s
ess  lim
K r  lim sG(s) H (s)
s0
Which the steady state error is e ss 
1
Kr
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  
Just like 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.
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
(iii) Determine the steady state error, if the input is a unit ramp.
Compensatot
i
+
-
K
DC motor
0.01
s(0.4s  1)
m
K  10 3
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
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
Hence steady state error of
s 0
10
 10
s(0.4 s  1)
ess 
1
 0.1
Kr
K  10 3
Steady State Error of Feedback Control System
Stability
Stability
Stability
Routh-Hurwitz Stability Criteria
If a polynomial is given by
T (s)  an s n  an 1s n 1  .....  a1s  a0  0
Where,
an , an 1....., a1, a0
are constants and
n  1...
Necessary condition for stability are:
(i)
.
All the coefficients of the polynomial
are of the same sign. If not, there are poles
on the right hand side of the s-plane
(ii) All the coefficient should exist accept for the
a0
For the sufficient condition, we can now formed a Routh-array,
Routh’s Array
sn
s n 1
an
an  2 an  4
an  6
an 1
an  3
an  5
an  7
b1
b2
b3
b4
c1
c2
c3
c4






s3
h1
h2
s2
i1
i2
s1
j1
s0
k1
s n2
s n 3









  
  
  
  
Routh Array elements
an an  2
a
a
a a a a
b1   n 1 n 3  n 1 n  2 n n 3
an 1
an 1
an an  4
a
a
a a a a
b2   n 1 n 5  n 1 n  4 n n 5
an 1
an 1
an 1 an 3
b
b2
ba a b
c1   1
 1 n 3 n 1 2
b1
b1
an 1 an 5
b
b3
ba a b
c2   1
 1 n 5 n 1 3
b1
b1
j1  
h1
i1
h2
i2
i1
k1  i2

i1h2  h1i2
i1
Routh-Hurwitz Criteria states
that the number of roots of
charateristic equation is the
same as the numberof sign
changed of the first column.
Case 1: No zero on the first column
After the Array has been tabled, all the elements on the first column are
not equal to zero.
If there is no sign changed, all the poles are in the LHP. While the number
of poles on the RHP is equal to number of sign change on the first column
of the Routh’s array.
Example:
Consider a fourth order characteristic equation
D( s)  2s  s  12s  8s  2  0
4
3
2
Example
Solution:
Form the Routh’s array
s4
2
12
s3
1
8
12  16
 4
1
2
s
2
s1
 32  2
 8.5
4
s0
2
2
There are two sign change on rows 2 and 3. Hence, there two
poles on the RHP (Right-half f s-palne).
Scilab solution
CE=poly([2 1 12 8 2],'s','c');
roots(CE)
ans =
0.0885283 + 2.4380372i
0.0885283 - 2.4380372i
- 0.3385283 + 0.2311130i
- 0.3385283 - 0.2311130i
Case 2: Coeffiecient of the first column is zero but not the others.
Change the zero element by a small positive number, . The number
of pole on the RHP will depend on the number of sign change.
Example:
Consider a fifth order characteristic equation D( s)  s 5  2s 4  3s 3  6s 2  5s  3  0
Solution:
Form the
Routh’s array
5
1
3
5
s4
2
6
3
s
s3
s
2
s1
s
0
66
 0,
2
=1
6  7

 1
10  3
 3 .5
2
3
42  49  6 2
 6.5
12  14
3
, is a small positive number there are two sign change at row 3 and 4,   
and also at row 4 and 5     . Hence, there two poles on the RHP.
If   1
Scilab Solution
-->CE=poly([3 5 6 3 2 1],'s','c')
-->roots(CE)
ans =
0.3428776 + 1.5082902i
0.3428776 - 1.5082902i
- 1.6680888
- 0.5088331 + 0.7019951i
- 0.5088331 - 0.7019951i
Case 3: All the coefficients on a row are zeros.
Form an auxiliary equation from the row above it and replace the
coefficient of the row with the differentiated coefficient of the auxiliary
equation. For this case, if there is no sign change, the characteristic
equation has a pair of poles with opposite sign of real component or/and a
pair of conjugate poles on the imaginary axis.
Example:
Consider this fifth order
characteristic equation
D( s)  s 5  7 s 4  6s 3  42s 2  8s  56  0
Formed Routh’s array
s5
s4
s3
1
6
8
7
42
56
42  42
0
7
56  56
0
7
84
28
s2
1
21
9.3
s
s0
56
56
Form the auxillary equation on the second row:
Differentiate the equation:
P( s )  7 s 4  42s 2  56
dP ( s )
 28s 3  84 s
ds
As there is no sign change,
there is a pair of conjugate poles on the axis and/or a pair of poles with opposite
sign of real component. To be sure we can use Scilab
-->CE=poly([56 8 42 6 7 1],'s','c')
-->roots(CE)
ans =
- 7.
- 8.049D-16 + 2.i
- 8.049D-16 - 2.i
1.4142136i
- 1.4142136i
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.
j
STABLE
UNSTABLE

Example
A closed-loop transfer function is given by
C (s)
K
 2
R ( s ) s s  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.
Solution:
Charactristic equation is
Expand the equation
Form the Routh’s array
ss 2  s  1s  2  K  0
s 4  3s 3  3s 2  2s  K  0
s4
1
3
3
3
2
92 7

3 3
K
s
s
1
s
s0
2
14 3  3K 14  9 K

73
7
K
K
To ensure that there is no poles on the RHP of the s-plane, there must
be no sign change on the first column of the Routh’s Array, therefore for no sign
change:
14  9 K
Refering to row 4:
0
7
which gives
K  14 9
and row 5
Hence its range
K 0
0  K  14 9
 7 3 2  14 9  0
Oscillating frequency
23
rad.s-1.