A polynomial is said to be stable if either:
The first condition provides stability for (or continuous-time) linear systems, and the second case relates to stability of discrete-time linear systems. A polynomial with the first property is called at times a Hurwitz polynomial and with the second property a Schur polynomial. Stable polynomials arise in control theory and in mathematical theory of differential and difference equations. A linear, time-invariant system (see LTI system theory) is said to be BIBO stable if every bounded input produces bounded output. A linear system is BIBO stable if its characteristic polynomial is stable. The denominator is required to be Hurwitz stable if the system is in continuous-time and Schur stable if it is in discrete-time. In practice, stability is determined by applying any one of several stability criteria.
- The Routh-Hurwitz theorem provides an algorithm for determining if a given polynomial is Hurwitz stable, which is implemented in the Routh–Hurwitz and Liénard–Chipart tests.
- To test if a given polynomial P (of degree d) is Schur stable, it suffices to apply this theorem to the transformed polynomial
obtained after the Möbius transformation which maps the left half-plane to the open unit disc: P is Schur stable if and only if Q is Hurwitz stable and . For higher degree polynomials the extra computation involved in this mapping can be avoided by testing the Schur stability by the Schur-Cohn test, the Jury test or the Bistritz test.
- Necessary condition: a Hurwitz stable polynomial (with real coefficients) has coefficients of the same sign (either all positive or all negative).
- Sufficient condition: a polynomial with (real) coefficients such that:
is Schur stable.
- Product rule: Two polynomials f and g are stable (of the same type) if and only if the product fg is stable.
- is Schur stable because it satisfies the sufficient condition;
- is Schur stable (because all its roots equal 0) but it does not satisfy the sufficient condition;
- is not Hurwitz stable (its roots are -1,2) because it violates the necessary condition;
- is Hurwitz stable (its roots are -1,-2).
- The polynomial (with positive coefficients) is neither Hurwitz stable nor Schur stable. Its roots are the four primitive fifth roots of unity
- Note here that
- It is a "boundary case" for Schur stability because its roots lie on the unit circle. The example also shows that the necessary (positivity) conditions stated above for Hurwitz stability are not sufficient.