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Stable polynomial
Characteristic polynomial whose associated linear system is stable

In the context of the characteristic polynomial of a differential equation or difference equation, a polynomial is stable if all its roots lie either in the open left half-plane or the open unit disk. The first condition ensures stability for continuous-time systems, known as Hurwitz-stable polynomials, while the second applies to discrete-time systems and is called Schur stability. These concepts are fundamental in control theory and the theory of differential equations. A linear, time-invariant system is BIBO stable if its characteristic polynomial is stable. Stability is commonly verified using various stability criteria.

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Properties

Q ( z ) = ( z − 1 ) d P ( z + 1 z − 1 ) {\displaystyle Q(z)=(z-1)^{d}P\left({{z+1} \over {z-1}}\right)} obtained after the Möbius transformation z ↦ z + 1 z − 1 {\displaystyle z\mapsto {{z+1} \over {z-1}}} which maps the left half-plane to the open unit disc: P is Schur stable if and only if Q is Hurwitz stable and P ( 1 ) ≠ 0 {\displaystyle P(1)\neq 0} . 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 f ( z ) = a 0 + a 1 z + ⋯ + a n z n {\displaystyle f(z)=a_{0}+a_{1}z+\cdots +a_{n}z^{n}} with (real) coefficients such that
a n > a n − 1 > ⋯ > a 0 > 0 , {\displaystyle a_{n}>a_{n-1}>\cdots >a_{0}>0,} 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.
  • Hadamard product: The Hadamard (coefficient-wise) product of two Hurwitz stable polynomials is again Hurwitz stable.1

Examples

  • 4 z 3 + 3 z 2 + 2 z + 1 {\displaystyle 4z^{3}+3z^{2}+2z+1} is Schur stable because it satisfies the sufficient condition;
  • z 10 {\displaystyle z^{10}} is Schur stable (because all its roots equal 0) but it does not satisfy the sufficient condition;
  • z 2 − z − 2 {\displaystyle z^{2}-z-2} is not Hurwitz stable (its roots are −1 and 2) because it violates the necessary condition;
  • z 2 + 3 z + 2 {\displaystyle z^{2}+3z+2} is Hurwitz stable (its roots are −1 and −2).
  • The polynomial z 4 + z 3 + z 2 + z + 1 {\displaystyle z^{4}+z^{3}+z^{2}+z+1} (with positive coefficients) is neither Hurwitz stable nor Schur stable. Its roots are the four primitive fifth roots of unity
z k = cos ⁡ ( 2 π k 5 ) + i sin ⁡ ( 2 π k 5 ) , k = 1 , … , 4 . {\displaystyle z_{k}=\cos \left({{2\pi k} \over 5}\right)+i\sin \left({{2\pi k} \over 5}\right),\,k=1,\ldots ,4\,.} Note here that cos ⁡ ( 2 π / 5 ) = 5 − 1 4 > 0. {\displaystyle \cos({{2\pi }/5})={{{\sqrt {5}}-1} \over 4}>0.} 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.

Stable matrices

Just as stable polynomials are crucial for assessing the stability of systems described by polynomials, stability matrices play a vital role in evaluating the stability of systems represented by matrices.

Hurwitz matrix

Main article: Hurwitz-stable matrix

A square matrix A is called a Hurwitz matrix if every eigenvalue of A has strictly negative real part.

Schur matrix

Schur matrices is an analogue of the Hurwitz matrices for discrete-time systems. A matrix A is a Schur (stable) matrix if its eigenvalues are located in the open unit disk in the complex plane.

See also

References

  1. Garloff, Jürgen; Wagner, David G. (1996). "Hadamard Products of Stable Polynomials Are Stable". Journal of Mathematical Analysis and Applications. 202 (3): 797–809. doi:10.1006/jmaa.1996.0348. https://doi.org/10.1006%2Fjmaa.1996.0348