Routh–Hurwitz stability criterion


In control system theory, the Routh–Hurwitz stability criterion is a mathematical test that is a necessary and sufficient condition for the stability of a linear time invariant control system. The Routh test is an efficient recursive algorithm that English mathematician Edward John Routh proposed in 1876 to determine whether all the roots of the characteristic polynomial of a linear system have negative real parts. German mathematician Adolf Hurwitz independently proposed in 1895 to arrange the coefficients of the polynomial into a square matrix, called the Hurwitz matrix, and showed that the polynomial is stable if and only if the sequence of determinants of its principal submatrices are all positive. The two procedures are equivalent, with the Routh test providing a more efficient way to compute the Hurwitz determinants than computing them directly. A polynomial satisfying the Routh–Hurwitz criterion is called a Hurwitz polynomial.
The importance of the criterion is that the roots p of the characteristic equation of a linear system with negative real parts represent solutions ept of the system that are stable. Thus the criterion provides a way to determine if the equations of motion of a linear system have only stable solutions, without solving the system directly. For discrete systems, the corresponding stability test can be handled by the Schur–Cohn criterion, the Jury test and the Bistritz test. With the advent of computers, the criterion has become less widely used, as an alternative is to solve the polynomial numerically, obtaining approximations to the roots directly.
The Routh test can be derived through the use of the Euclidean algorithm and Sturm's theorem in evaluating Cauchy indices. Hurwitz derived his conditions differently.

Using Euclid's algorithm

The criterion is related to Routh–Hurwitz theorem. From the statement of that theorem, we have where:
By the fundamental theorem of algebra, each polynomial of degree n must have n roots in the complex plane. Thus, we have the condition that ƒ is a stable polynomial if and only if pq = n. Using the Routh–Hurwitz theorem, we can replace the condition on p and q by a condition on the generalized Sturm chain, which will give in turn a condition on the coefficients of ƒ.

Using matrices

Let f be a complex polynomial. The process is as follows:
  1. Compute the polynomials and such that where y is a real number.
  2. Compute the Sylvester matrix associated to and .
  3. Rearrange each row in such a way that an odd row and the following one have the same number of leading zeros.
  4. Compute each principal minor of that matrix.
  5. If at least one of the minors is negative, then the polynomial f is not stable.

    Example

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 . 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.

Routh–Hurwitz criterion for second and third order polynomials

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:
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.
In the first column, there are two sign changes, thus there are two non-negative roots where the system is unstable.
The characteristic equation of a servo system is given by :
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 :
We see that if
then
Is satisfied.
We have the following table :
1112000
6 16 100
10 1200 2000
-19000
20000

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.
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:
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 following 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.