1. No category

2. ROUTH STABILITY CRITERION AND GERSCHGORIN
CIRCLES
2.1 Introduction
In the Routh stability criterion to decide the stability of the system,
characteristic polynomial with all the co efficient positive is considered. To compute
the characteristic polynomial using the determinant (
from a system matrix,
takes lot of computations. Here it has been explained to identify the sign of the
characteristic polynomial using the Gerschgorin circles.
In our method we consider the system matrix, and for this matrix we draw
the Gerschgorin circles and we observe the bound, based on the bound we decide the
sign of the characteristic polynomial. This is a graphical and heuristic method and
this will be explained through examples which are given below.
2.2 Heuristic approach in determination of the sign of the
characteristic polynomial of a system matrix using Gerschgorin
circles
Example 2.2.1: Given a system matrix of order
.
5
4
1
1
4
5
1
1
1
1
4
2
1
1
2
4
32
(2.2.1)
Gerschgorin circle of the above matrix is drawn and is shown in the figure 2.1
Fig 2.1: Gerschgorin bound [ -1 , 11 ]
Since the bound on the right hand side of the s-plane is very much greater
than the bound on the left hand side of the s-plane, it guarantees that there exist an
eigenvalue on the right hand side of the s-plane which implies that there exist
change in sign of the characteristic polynomial.i.e., all the coefficient of the
characteristic polynomial is not positive.
Probability 0.98 there exist change in sign in the coefficient of the
characteristic polynomial.
To verify the above conclusion the characteristic polynomial is calculated
and given here.
Actual characteristic polynomial is
33
Example 2.2.2: Given a system matrix A of order
and the trace of the matrix
is zero.
0
0
0
0
1
0
0
1
1
1
0
0
1
1
1
1
0
0
-1
1
0
0
0
-1
1
(2.2.2)
Gerschgorin circle of the above matrix is drawn and is shown in figure 2.2
Fig 2.2: Gerschgorin bound [-4, 4]
Here the Gerschgorin bound on the left hand is equal to Gerschgorin bound
on the right hand. Since trace of the matrix is zero, it guarantees the existence real
eigenvalue on the right hand side of the s-plane. Hence there exist changes in sign in
the coefficient of the characteristic polynomial.
Probability 0.999 there exist change in sign of the characteristic polynomial.
To verify the above conclusion the characteristic polynomial is calculated
and given here.
34
Actual characteristic polynomial is
Example 2.2.3: Given a system matrix
and the trace of the
matrix is not zero.
-3
1
1
3
-1
-1
1
1
-1
(2.2.3)
Gerschgorin circle of the above matrix is drawn and is shown in figure 2.3
and the trace of the matrix is negative.
Fig 2.3: Gerschgorin bound [-5, 1]
The Gerschgorin bound on the left hand side is greater than the bound on the
right hand side of the s-plane. Hence probability 0.3 the system is unstable. Hence
there is no change in sign in the characteristic polynomial.
Here trace of the matrix is equal to the left Gerschgorin bound implies with
all certainty the eigenvalues lie on real axis of the left hand side of the s-plane
barring the possibility of existence of eigenvalues at the origin. But here in the
example, there an eigenvalue at the origin making the determinant zero. Hence all
the coefficient of the characteristic polynomial is positive.
35
Example 2.2.4: Given the system matrix of order
The Gerschgorin circle is
drawn and is shown in figure 2.4.the trace of the matrix is negative.
-2
-3
0
0
-2
-3
-3
0
-2
2.2.4
Fig 2.4: Gerschgorin bound [-5, 1]
The Gerschgorin bound on the left hand side is very much greater than the
Gerschgorin bound on the right hand side of the s-plane and it is very likely that the
coefficient of the characteristic polynomial is positive.
To verify the above conclusion the characteristic polynomial is calculated
and given here
Actual characteristic polynomial is
36
Example 2.2.5: Given the system matrix of order
The Gerschgorin circles are
drawn and shown in figure 2.5 .the trace of the matrix is negative.
-2
-2
0
0
-2
-2
-2
0
-2
(2.2.5)
Fig 2.5: Gerschgorin bound
Gerschgorin bound lies entirely on the left hand side, hence all the
coefficient of characteristic polynomial are positive
To verify the above conclusion the characteristic polynomial is calculated
and given here
is
2.3 Alternative Method to Routh Criterion for testing stability of a
system matrix using Gerschgorin circles
In the Routh criterion, characteristic polynomial with all the coefficient
positive is considered. The Routh table is applied for this characteristic polynomial.
If there exist two changes in sign in the first column of the Routh array, then there
exist complex conjugate eigenvalues with the positive real part, which in turn
implies that the system is unstable.
37
An attempt has been made to find the stability of the system using
Gerschgorin circles which is an alternative method for the Routh stability criterion.
Consider a characteristic polynomial where all the co efficient are positive
and then apply the Routh test. If there exist two changes in sign of the first column
of the Routh array then there exist complex conjugate eigenvalues with the positive
real part. For this characteristic polynomial obtain the companion form of the matrix
and draw the Gerschgorin circles. Compute the determinant
at the left
Gerschgorin bound and origin. If there exist changes in sign of the determinant
, then there exist real eigenvalues on the left half of the s-plane.
Compute the real eigenvalues by applying the Bisection method/Secant at the
Gerschgorin bound. The difference between the trace and real eigenvalues gives the
presence of complex conjugate eigenvalues with positive real part. Also the real
eigenvalue on the left half of the s-plane will be greater than the real part of the
complex conjugate eigenvalue with the positive real part.
Since we have computed the real eigenvalues and also the real part of the
complex conjugate, considering the polynomial
(
)
(
)
where λ =- the real eigenvalue.
-2aλ+
)) = 0
–
By equating the coefficient of
and the constant term, we can obtain the
value of the imaginary part. Hence the complex conjugate eigenvalue with the
positive real part can be computed.
Routh stability criterion: [53]
Consider the characteristic polynomial
3
+10
+5
(2.3.1)
+ 5 λ +2 =0
38
Table 2.1: Routh test applied to the given polynomial
Applying the Routh test
3
5
10
5
2
2
1
2
Examining the first column of the Routh array, it is found that there are two
changes in sign (from 3.5 to -0.5/3.5 and from – 0.5/3.5 to 2) .Therefore the system
under consideration is unstable having complex conjugate eigenvalues with positive
real part.
Gerschgorin circle Method:
Consider the companion form of the above characteristic polynomial (2.3.1)
--
0
1
0
0
0
0
1
0
0
0
0
1
-
-
-
The Gerschgorin circle for the matrix is drawn and is shown fig 2.6
39
Fig 2.6: Gerschgorin bound [-4.33, 1]
jEigenvalues of the matrix obtained by applying Bisection method at the
Gerschgorin bound [-4.33, 0] are
λ = -2.927754
λ = - 0.444232
Since the given polynomial satisfies the Routh test, there exist no real
eigenvalues on the right hand side of the s-plane.
We know that
-3.33 = -2. 927754 - 0.44232 +
-3.33 +3.71986 =
= 0.03865 >0
The remaining two eigenvalues are complex conjugate pairs with positive real part.
40
Real part of the complex conjugate pairs are
>0
j
Real
σ
• Complex conjugate eigenvalues real part
•
•
σ
•
The real part of the complex conjugate eigenvalues is very much less
than the absolute value of one of the real eigenvalue which belongs to left hand
side of s-plane.
0.019325<< |0.4452|
Conclusion: The presence triangle in the Gerschgorin circle represents the
changes in sign in the first column of the Routh array.
Examples 2.3.1:
Consider the characteristic polynomial whose all coefficient are positive.
 (2.3.2)
Applying the Routh Stability Criterion for the above polynomial (2.2)
Routh Stability Criterion method:
Table 2.2: Routh test table
1
4
0
2
26
0
-9
0
26
41
In the first column of the Routh array there exists change. Hence there exist
complex conjugate eigenvalues with the positive real part.
Gerschgorin circle method:
The companion form of the above polynomial (2.3.2)
0
1
0
0
0
1
-26
-4
-2
The Gerschgorin circle is drawn for the above matrix and is shown in
figure 2.7
j
s-plane
•
•
•
Fig 2.7: Gerschgorin bound
42
Eigenvalues of the above matrix obtained by applying SECANT METHOD
at the Gerschgorin bound
i.e., the left hand side of the s-plane we get
We know that the
-2 = -3.241019 +
There exist complex conjugate pairs with positive real part.
The real part of the complex conjugate being 0.6205
The imaginary part of the complex conjugate eigenvalue can be computed.
Also consider the general polynomial
Where c is the negative real eigenvalue and
are the real and
imaginary part of the complex conjugate eigenvalues with positive real part.
Equating the coefficient of
and constant term we obtain the value
of the imaginary part as 2.7639
Remark: In the Gerschgorin circle approach both real and imaginary part of
the complex number can be computed.
Example 2.3.2:
Consider the characteristic polynomial
(2.3.3)
43
Routh Stability Criterion method:
Table 2.3: Routh test table
1
20
144
8
71
0
11
144
0
-33
0
144
In the first column of the Routh array there exists change in sign of the first
column of the Routh array. Hence there exist complex conjugate eigenvalues with
the positive real part.
Gerschgorin circle method:
The companion form of the above characteristic polynomial (2.3.3)
0
1
0
0
0
0
1
0
0
0
0
1
-71
-20
-8
-144
The Gerschgorin circle is drawn for the above matrix and is shown in figure 2.8
Fig 2.8: Gerschgorin bound
44
Eigenvalues of the above matrix obtained by applying Secant method at the
Gerschgorin bound are
Since there are two real eigenvalues on the left hand side of the s-plane, we obtain
jω
the remaining eigenvalues as follows
•
•
•
σ
•
The other two eigenvalues are complex conjugate pairs with the positive real part
with the real part being 0.27976 > 0
Example 2.3.3:
Consider the characteristic polynomial
(2.3.4)
Routh Stability Criterion method:
Table 2.4 : Routh test table
1
52
51
2
82
388
11
-143
-108
388
103
0
0
388
45
In the first column of the Routh array there exists changes in sign of the first
column of the Routh array. Hence there exist complex conjugate eigenvalues with
the positive real part.
Gerschgorin circle method:
Consider the companion form of the above characteristic polynomial (2.3.4)
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
-51
-82
-52
-2
-388
The Gerschgorin circles is drawn for the above matrix and is shown in
figure 2.9
Fig 2.9: Gerschgorin bound
46
Since there exist one change in sign of the determinant
real eigenvalue in the Gerschgorin bound
by
applying
the
SECANT
METHOD
there exist one
The eigenvalues are computed
in
the
Gerschgorin
bound.
jω
•
We know that the trace of the matrix is given by
•
α•
σ
•
•
There exists one pair of complex conjugate eigenvalues with positive real part.
Example 2.3.4: Consider the characteristic polynomial
(2.3.5)
jω
•
Table 2.5: Routh test table
•
Routh Stability Criterion method:
1
2
8
2
4
0
0
8
0
σ
•
47
•
Table 2.6: Routh table replacing zero by
Replace the zeros in the last row by small number
1
2
8
2
4
0
and continue the Routh test
8
→ -16
0
8
0
0
0
There exists changes in sign in the second, third and fourth row since
very small the number
is
is negative. Hence there exists complex conjugate
eigenvalues with positive real part. Hence the system is unstable.
Remark: The presence of rectangular structure in the Gerschgorin circles
represents the changes in sign in the first column of the Routh array.
Gerschgorin circle method:
Consider the companion form of the above polynomial (2.3.5)
0
1
0
0
0
0
1
0
0
0
0
1
-8
-4
-2
-3
The Gerschgorin circles is drawn for the above matrix and is shown in
figure 2.10
48
Fig 2.10: Gerschgorin bound [-8, 8]
There exist no changes in sign of the determinant (
), hence there
exist no real eigenvalues. The system is unstable since
1. The Gerschgorin circles are concentric in the Gerschgorin bound hence the
eigenvalues are present on both side of the s-plane.
2. Compute determinant (
at the Gerschgorin bound
3. Since the Gerschgorin bound are equal, there exists complex conjugate pairs
with positive.
Example 2.3.5: Consider the characteristic polynomial where one of the coefficient
are not present
(2.3.6)
49
Table 2.7: Routh table for the polynomial where all the coefficient are not
present Routh Stability Criterion method:
1
0
0
2
8
0
-4
0
0
8
0
Gerschgorin circle method:
Consider the companion form of the above polynomial (2.3.6)
0
1
0
0
0
1
-8
0
-2
The Gerschgorin circle is drawn for the above matrix and is shown in figure 2.11
Fig 2.11: Gerschgorin bound
Applying the Bisection method in the Gerschgorin bound
50
we get
We know that
- 2= -2.931641 +
Hence the other two eigenvalues are complex conjugate pairs with positive real part
the real part being 0.46582
Consider the polynomial
Equating the coefficients of the given polynomial and above polynomial we get
The above result is verified using MATLAB
(2.3.7)
Example 2.3.6: Consider the polynomial
Routh stability criterion:
Table 2.8: Routh test
1
2
3
1
2
5
-
-2
5
5
51
Since in Routh array, the first element in the third is zero and hence it is
replaced by a small number
which is a positive number. Hence there exist a
change in sign in the first column of the Routh array, which in turn implies that there
exist a complex conjugate eigenvalues with the positive real part.
Gerschgorin circle method:
Consider the companion form of the above polynomial (2.2.7)
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
-5
-3
-2
-2
-1
The Gerschgorin circle of the above matrix is drawn and is as shown in fig 2.12
Fig 2.12: Gerschgorin bound
52
Applying the Bisection method at the Gerschgorin bound
we get
The system is unstable since
1. The Gerschgorin circles are concentric in the Gerschgorin bound and hence
the eigenvalues are spread on both side of the imaginary axis.
2. Since
3. Since trace –real eigenvalue is positive; there exists complex conjugate
eigenvalues with positive real part.
The above conclusion are verified using MATLAB
Example2.3.7: Consider the characteristic polynomial
(2.3.8)
Routh stability criterion [53]
Table 2.9: Routh test
1
8
20
2
12
16
1
6
8
2
12
16
1
6
8
0
0
16
53
Consider the auxiliary equation
(2.3.9)
A=
=4
Table 2.10: Routh test using auxiliary equation
1
8
20
2
12
16
1
6
8
2
12
16
1
6
8
1
3
0
3
8
0
16
0
8
There exist no changes in sign in the first column of the Routh array. Hence
there exist no complex conjugate eigenvalues with the positive real part. Since all
the coefficient of the characteristic polynomial are positive, there exit no real
eigenvalues on the right hand side of the s-plane.
There exist eigenvalues with complex conjugate pairs on the imaginary axis,
which in turn implies that system is critical stable.
Companion form of the above polynomial (2.2.8)
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
-16
-16
-20
-12
-8
-2
54
The Gerschgorin circle is drawn for the above matrix and is shown in figure 2.13
Fig 2.13 :Gerschgorin bound
Compute the determinant
at the Gerschgorin bound
.
= 95387539 >0
16 >0
Hence there exist no changes in sign of the determinant
at
There exist no real eigenvalues on the left hand side of the s-plane.
Since all the coefficient of the characteristic polynomial is positive there
exist no real eigenvalue on the right hand side of the s-plane.
Hence all the eigenvalues are complex conjugate pairs with positive negative
real part.
Since there is no real eigenvalue we cannot find the real part of the complex
conjugate using the trace of the matrix.
55
We know that
Since trace is negative, there exist one pair of complex conjugate eigenvalue
with negative real part, and the value of the negative real part must be greater than
the value of the real part of the complex conjugate on the positive side.
Remark: In the above we cannot find the exact location of eigenvalues using
Gerschgorin circles and also hence we cannot decide about the absolute stability of
the system.
2.4 Condition for all the coefficients of the characteristic polynomial
of a system matrix to be positive.
Consider a system matrix
Let λ =
of order
be the eigenvalues of the matrix
jω
•
•
σ
•
Consider
56
Condition for the coefficients of all the characteristic polynomial to be positive
If the characteristic polynomial is given equating these three conditions with the
coefficient of the given characteristic polynomial, obtain the value of
value of
knowing the
and by the above algorithm.
Consider the following example
Example 2.4.1:
Given the system matrix is of order
1
-3
0
0
-1
-3
-3
0
-1
The Gerschgorin circle of the above matrix is drawn and is shown in figure 2.14
Fig 2.14: Gerschgorin bound
57
By applying the Secant method in the Gerschgorin bound [-4, 0], we get
is one of the real eigenvalue.
Since
this implies that there exist complex conjugate eigenvalues with
positive real part.
Let
and
i.e.,
=0.5 the real of the complex conjugate eigenvalues.
Consider the characteristic polynomial of the above matrix is
And the general characteristic polynomial
Equating the coefficient of
and λ we get
Solving the above equations we get the value of
Actual value of
=
58
=
Hence given the system matrix and its characteristic polynomial the
imaginary part can be computed.
If the characteristic polynomial is not given we can determine
the condition for
not exactly but
can be given by the following relation.
Hence the valve of the imaginary part will be greater than 1.936.
2.5 Routh’s stability criterion for the characteristic polynomial of
the form
The Routh stability criteria cannot be applied when all the elements in the
first row becomes zero. But an attempt has been made to apply the Routh test , by
replacing the zero elements in the first row by a small number
and proceeding
further it was found that there exists two changes in sign in the first column of the
Routh array which implies that there exist complex conjugate eigenvalues with the
positive real part.
Routh test
Consider a polynomial
(2.5.1)
Apply the Routh‟s test for the polynomial as follows
59
Table 2.11: Routh test
1
0
1
0
0
0
In the above table 2.8 the first row does not contain all non zero elements
and the second row elements are all zeros.
2.7 Authors test
We continue the table 2.11, by replacing all the zero elements of the second
row by
Table 2.12: Routh table for Authors test
1
-1
0
1
0
0
0
1
0
0
In the above table there exist two changes in sign of the first column. Hence
the systems characteristic polynomial possesses two roots on the right half of the splane. The roots are complex conjugate pairs with positive real part.
The eigenvalues are
;
;
;
We get two sets of complex conjugate eigenvalues , one set on the right hand
side of the s-plane and the another set on the left hand side of the s-plane.
60
Remark: If we adopt the Routh‟s test using the existing special cases for the above
example, the solution is as follows
Consider the polynomial (2.5.1)
Table 2.13: Routh table for Authors test
1
0
1
1
0
1
Now consider the auxiliary equation
Replace second row by 4, 0, 0 and continue .Then in the third row, the first
element is zero and hence replace by
and continue. Observe the first column of
the table 2.10. There exists two changed in sign .Hence two roots are present on the
right half of the s-plane.
Conclusions: Based on the above five examples discussed we can almost
heuristically decide the sign of the coefficient of the characteristic polynomial.
Since it is a graphical approach it requires no computations in deciding the sign
of the coefficient of the characteristic polynomial.
The Routh stability criterion is tested for the characteristic polynomial
which is obtained from the system matrix and it takes lots of computation.
61
The Gerschgorin circles approach is a graphical technique, which starts with
a given system matrix and the presence of triangular structure in the
Gerschgorin circle represents the change in sign of the first column of the
Routh array. In this method we also compute the real eigenvalues and for the
matrix of order
, the complex conjugate eigenvalues can also be
computed. This method of computing the complex conjugate eigenvalues can be
extended to the higher order matrix also. The characteristic polynomial where
all the coefficients are not present are considered and we observe that the
stability cannot be decided by the Routh test where as in Gerschgorin circle
technique the stability is decided and the eigenvalues are also computed.
Also we observe that if all the eigenvalues are complex conjugate pairs
for the system matrix, the change in sign of the first column of the Routh array
is represented by the rectangular structure.
62