The Routh Stability Criterion
When consider the design and analysis of feedback
control systems, stability is of the utmost importance.
From a practical point of view, a closed-loop feedback
system that is unstable is of little value.
Using feedback, we can stabilize unstable plants and
then with a judicious selection of controller parameters,
we can adjust the transient performance.
The Concept of Stability
A system that works at its original equilibrium states is effected by a disturbance (noise) . When the disturbance go off , the system can come back to the original equilibrium states the system is stable. Or else the system is unstable.
Definition of stability
The Concept of Stability
A stable system is a dynamic system with a
bounded response to a bounded input.
or
The unity-step response analysis of the high-order control systems
Assume : nm
the zeros of the transfer function
the poles of the transfer function
We can get the unity-step response :
)()(1
10
1
10 sRasasa
bsbsbsC
n
nn
m
mm
sssssss
zszsk
n
m 1
)())((
)()(
21
1
ssR
ttr
1)(
)(1)(
;,,
;,,
21
21
n
m
sss
zzz
n
i
ts
i
n
m ieAa
btc
1
)(
(1) If all poles are in the left of the s-plane,the system is stable .
(2) The transient portions of the response are exponentially decayed.corresponding to the poles are at the real of the left s-plane .
The transient portions of the response are the decayed oscillatory motions. corresponding to the complex poles in the left of the s-plane.
(3) The more far the poles in the left s-plane go from the imaginary axis.
The more quick the transient portions of the response are decayed.
The unity-step response analysis of the high-order control systems
(4) The more far the more poles in the left s-plane go from
the real axis. the more large the overshoot of the
response is.
(5)The zeros of the system only affect the coefficient of the
transient portions.
(6) The poles,which are the most close to the imaginary axis
of the s-plane,are of the most affecting to the transient
portions of the response.
dominant poles.
The unity-step response analysis of the high-order control systems
Suppose a transfer function
The Concept of Stability
n
nn
m
mm
asasa
bsbsb
sR
sC
1
10
1
10
)(
)(
We can always factor as
n
i
i
m
j
j
ps
zsK
sR
sC
1
1
)(
)(
)(
)(
The closed-loop system is stable if i , 0)( ipR
Note: It might turn out that there are pole-zero
cancellations, that is, z j = p i for some i. j.
If this happens, system is still unstable if 0)( ipR
The sufficient and necessary conditions of
the stability for a linear system
Im
Re
S-plane
The relationship between the systems stability and the position of poles in S-plane.
The Concept of Stability
Graphic representation
The Routh criterion represents a method of determining
the location of roots of a polynomial with constant real
coefficients with respect to the left half and right half of
the s-plane, without solving for the roots.
The Routh Criterion
The characteristic equation is:
0)( 11
10
nn
nn asasasasD
The roots are, of course nppp , , 21
Important question:
Can we tell if the system is stable, without actually
solving for the roots?
To ensure the equation does not have roots with positive real
parts, it is necessary (but not sufficient) that the following
conditions hold:
(1). All the coefficients of the equation have the same sign.
(2). None of the coefficients vanish.
These conditions are based on the laws of algebra, which
relate the coefficients of the equation .
0)( 11
10
nn
nn asasasasD all the coefficients are real.
The Routh Criterion
063.3sin101137.2237.245 ssess
The characteristic equation is short of one or more than one items The system must be unstable .
Example :
unstable, be short of the item
0663.3sin101137.2237.245 sssess
The coefficient of the characteristic equation are
different in sign The system must be unstable.
Example :
unstable, The coefficient are different in sign.
4a s
The Routh Criterion
The two necessary conditions for Equation to have no roots in the right-half s-plane
can easily be checked by inspection of the equation. However, these conditions are
not sufficient, for it is quite possible that an equation with all its coefficients nonzero
and of the same sign still may not have all the roots in the left half of the s-plane.
Routh Array
0)( 11
10
nn
nn asasasasD
nn
n
n
n
n
acs
ccccs
ccccs
aaaas
aaaas
1,1
0
44342414
3
43332313
2
7531
1
6420
The characteristic equation ( ) 00 a
The first two rows come directly from the polynomial D(s). Each
subsequent row is formed by operations on the two rows above:
The Routh Criterion
Routh Array
0)( 11
10
nn
nn asasasasD
nn
n
n
n
n
acs
ccccs
ccccs
aaaas
aaaas
1,1
0
44342414
3
43332313
2
7531
1
6420
1
302113
a
aaaac
The Routh Criterion
00 a
Routh Array
0)( 11
10
nn
nn asasasasD
nn
n
n
n
n
acs
ccccs
ccccs
aaaas
aaaas
1,1
0
44342414
3
43332313
2
7531
1
6420
1
504123
a
aaaac
The Routh Criterion
00 a
Routh Array
nn
n
n
n
n
acs
ccccs
ccccs
aaaas
aaaas
1,1
0
44342414
3
43332313
2
7531
1
6420
0)( 11
10
nn
nn asasasasD
nn
n
n
n
n
acs
ccccs
ccccs
aaaas
aaaas
1,1
0
44342414
3
43332313
2
7531
1
6420
1
706133
a
aaaac
The Routh Criterion
00 a
Routh Array
nn
n
n
n
n
acs
ccccs
ccccs
aaaas
aaaas
1,1
0
44342414
3
43332313
2
7531
1
6420
0)( 11
10
nn
nn asasasasD
nn
n
n
n
n
acs
ccccs
ccccs
aaaas
aaaas
1,1
0
44342414
3
43332313
2
7531
1
6420
13
23131314
c
caacc
The Routh Criterion
00 a
Routh Array
0)( 11
10
nn
nn asasasasD
nn
n
n
n
n
acs
ccccs
ccccs
aaaas
aaaas
1,1
0
44342414
3
43332313
2
7531
1
6420
13
33151324
c
caacc
The Routh Criterion
00 a
Routh Array
0)( 11
10
nn
nn asasasasD
nn
n
n
n
n
acs
ccccs
ccccs
aaaas
aaaas
1,1
0
44342414
3
43332313
2
7531
1
6420
13
43171334
c
caacc
The Routh Criterion
00 a
Routh Array
nn
n
n
n
n
acs
ccccs
ccccs
aaaas
aaaas
1,1
0
44342414
3
43332313
2
7531
1
6420
the number of unstable roots of D(s) is the number of
sign changes in the first column of the Routh array.
0)( 11
10
nn
nn asasasasD
nn
n
n
n
n
acs
ccccs
ccccs
aaaas
aaaas
1,1
0
44342414
3
43332313
2
7531
1
6420
0)( 11
10
nn
nn asasasasD
The Routh Criterion
00 a
Conclusion(Routh Criterion):
Necessary, Sufficient condition
All elements of the first column of the Routh-
table(array) are positive .
The system is and must is stable.
The Routh Criterion
the number of unstable roots of D(s) is the number of
sign changes in the first column of the Routh array.
The characteristic equation of a second-order system is
The Routh array is written as
Therefore, the requirement for a stable second-order
system is simply that all the coefficients be positive.
0)( 212
0 asasasD
0
0
2
0
1
1
20
2
as
as
aas
Examples
0105323 sss
0205323 sss
Stable
10
03
5
103
51
0
1
2
3
s
s
s
s
20
03
5
203
51
0
1
2
3
s
s
s
s
Example 1
Examples
Example 2
First column has two sign changes!
Unstable . 2 roots with positive real parts
0832 23 sss
8
01
082
031
0
1
2
3
s
s
s
s
Example 3
Examples
First column has two sign changes!
There are two unstable poles. In fact, the roots are:
-2.2483
0.1241 + 1.8822i
0.1241 - 1.8822i
Examples Stability vs. Parameter Range
Its much easier to use a calculator or Matlab to find roots.
So why use Routh? Routh allows us to determine
symbolically what values of a parameter will lead to
stability/instability.
Examples Stability vs. Parameter Range
For what values of k is the following system stable?
Solution: The Closed Loop transfer function is:
Ks
K
s
K
s
K
sKG
sKGsT
3
3
3
)1(
)1(1
)1(
)(1
)()(
0133)( 23 KssssD
Example 4
For stability, need first column to be positive,
so that K -1.
If K < -1, first column is + + + - , so there is 1 unstable pole.
If K > 8, first column is + + - + , so there are 2 unstable poles.
Routh Array 013323 ksss
ks
ks
ks
s
1
03
8013
031
0
1
2
3
Examples Stability vs. Parameter Range
Example 4
Possible problems:
If the first element of a row is zero, process fails.
Solution: Replace 0 by , a small positive number. (If a whole row is zero, must replace row as explained in the book. This happens whenever there are complex conjugate pairs of roots on the imaginary axis. )
The Routh Criterion
0122234 ssss
02
222
s4 1 1 1
s3 2 2 0
s2 0 1
s1 0
s0 1
Unstable. 2 roots with positive real parts.
Note:
Can use a infinitesimal >0 substituting the zero element in
the first column.
>0
Examples
Example 5
If a whole row is zero, must replace the row as explained in the book.
See 2-13 Routh_Rurwitz Criterion example 2.13.3
This happens whenever there are complex conjugate pairs of roots on
the imaginary axis.
)1025.0)(11.0()(
sss
KsG
G(s)
Determine the range of K for a stable system.
The characteristic equation
035.0025.0 23 Ksss
The range of K for a stable system is
( )(0.1 1)(0.025 1)
KG s
s s s
035.0025.0 23 Ksss
0K
14 0025.035.0 KK
140 K
Ks
Ks
Ks
s
0
1
2
3
035.0
025.035.0035.0
01025.0
Examples Axis shift
Example 6
The characteristic equation
The range of K for a stable system is
Now, it is desired that the real part of all roots should
be less than -1, what is the range of K ?
Let , then 11 ss
035.0025.0 23 Ksss
11 ss
0)2740(1511 12
1
3
1 Ksss
0.675K
8.4 , 0)2740(1511 KK
8.4675.0 K
2740
011
)2740(1511
0274011
0151
0
1
1
1
2
1
3
1
Ks
Ks
Ks
s
Example 6
Examples Axis shift
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