Example
- Let (for the sake of simplicity we take real coefficients) where (to avoid a root in zero so that we can use the Routh–Hurwitz theorem). First, we have to calculate the real polynomials and :
- Next, we divide those polynomials to obtain the generalized Sturm chain:
- yields
- yields and the Euclidean division stops.
Notice that we had to suppose b different from zero in the first division. The generalized Sturm chain is in this case . Putting , the sign of is the opposite sign of a and the sign of by is the sign of b. When we put , the sign of the first element of the chain is again the opposite sign of a and the sign of by is the opposite sign of b. Finally, -c has always the opposite sign of c.
Suppose now that f is Hurwitz-stable. This means that (the degree of f). By the properties of the function w, this is the same as and . Thus, a, b and c must have the same sign. We have thus found the necessary condition of stability for polynomials of degree 2.
Higher-order example
A tabular method can be used to determine the stability when the roots of a higher order characteristic polynomial are difficult to obtain. For an nth-degree polynomial
the table has n + 1 rows and the following structure:
| | |
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| | |
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| | |
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| | |
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| | |
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where the elements and can be computed as follows:
When completed, the number of sign changes in the first column will be the number of non-negative roots.
0.75
| 1.5
| 0
| 0
|
-3 | 6 | 0 | 0
|
3 | 0 | 0 | 0
|
6 | 0 | 0 | 0 |
In the first column, there are two sign changes (0.75 → −3, and −3 → 3), thus there are two non-negative roots where the system is unstable.
The characteristic equation of a servo system is given by:[6]
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0 |
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0 |
0 |
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= |
0 |
0 |
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0 |
0 |
0 |
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0 |
0 |
0 |
for stability, all the elements in the first column of the Routh array must be positive. So the conditions that must be satisfied for stability of the given system as follows:[6]
[6]
We see that if
then
Is satisfied.
- [7]
We have the following table :
1 |
11 |
200 |
0 |
1 |
1 |
0 |
0 |
1 |
20 |
0 |
0 |
-19 |
0 |
0 |
0 |
20 |
0 |
0 |
0 |
there are two sign changes. The system is unstable, since it has two right-half-plane poles and two left-half-plane poles. The system cannot have jω poles since a row of zeros did not appear in the Routh table.[7]
Sometimes the presence of poles on the imaginary axis creates a situation of marginal stability. In that case the coefficients of the "Routh array" in a whole row become zero and thus further solution of the polynomial for finding changes in sign is not possible. Then another approach comes into play. The row of polynomial which is just above the row containing the zeroes is called the "auxiliary polynomial".
We have the following table:
1 | 8 | 20 | 16
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2 | 12 | 16 | 0
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2 | 12 | 16 | 0
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0 | 0 | 0 | 0
|
In such a case the auxiliary polynomial is which is again equal to zero. The next step is to differentiate the above equation which yields the polynomial . The coefficients of the row containing zero now become
"8" and "24". The process of Routh array is proceeded using these values which yield two points on the imaginary axis. These two points on the imaginary axis are the prime cause of marginal stability.[8]