Stable Polynomial - Examples

Examples

  • 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
\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.

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