2011
DOI: 10.1063/1.3663506
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Hadamard Factorization of Stable Polynomials

Abstract: The stable (Hurwitz) polynomials are important in the study of differential equations systems and control theory (see [7] and [19]). A property of these polynomials is related to Hadamard product. Consider two polynomials p, q ∈ R[x]:p(x) = a n x n + a n−1 x n−1 + · · · + a 1 x + a 0. Some results (see [16]) shows that if p, q ∈ R[x] are stable polynomials then (p * q) is stable, also, i.e. the Hadamard product is closed; however, the reciprocal is not always true, that is, not all stable polynomial has a fact… Show more

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Cited by 4 publications
(8 citation statements)
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“…Proof. The above inequalities follow from the stability of f and f g: conditions (8) are equivalent to (4)-( 5) and conditions ( 9)-(10) follow from (8) and from the identities:…”
Section: Main Results Letmentioning
confidence: 99%
See 1 more Smart Citation
“…Proof. The above inequalities follow from the stability of f and f g: conditions (8) are equivalent to (4)-( 5) and conditions ( 9)-(10) follow from (8) and from the identities:…”
Section: Main Results Letmentioning
confidence: 99%
“…Recall that a Hurwitz stable polynomial of degree n ≥ 1 has a Hadamard factorization if it is a Hadamard product of two Hurwitz stable polynomials of degree n. It is known that every stable polynomial of degree n ≤ 3 admits a Hadamard factorization (see Garloff and Shrinivasan [5]) and that for every n ≥ 4 there exists an n-th degree stable polynomial which is not Hadamard factorizable (see Bia las and Góra [2]). Some conditions for the existence of a Hadamard factorization can be found in Loredo-Villalobos and Aguirre-Hernández [8,9], but these conditions cannot be effectively applied in practice. In turn, some topological properties of the entire family of polynomials admitting a Hadamard factorization can be found in Aguirre-Hernández et al [1].…”
Section: Introductionmentioning
confidence: 99%
“…The necessary condition is immediate. To prove sufficient condition, we suppose that ( ) is Hadamardized, and then (see [14,15])…”
Section: The Kharitonov Theorem and The Set Of Hadamardized Hurwitz Pmentioning
confidence: 99%
“…It is shown in [12] that there are Hurwitz polynomials of degree equal to four that do not have a Hadamard factorization. In [13] necessary conditions for a Hurwitz polynomial to have a Hadamard factorization were obtained, and in [14,15] necessary and sufficient conditions for a Hurwitz polynomial of degree equal to four to admit a Hadamard factorization were obtained. Now in this paper we prove that H(Had) is an unbounded, nonconvex open set and arc-connected.…”
Section: Introductionmentioning
confidence: 99%
“…However in [18] it was shown that there are Hurwitz polynomials of degree 4 that do not have a Hadamard factorization in two Hurwitz polynomials. In [19,20], some conditions to Hadamard factorization existence were presented. The set of Hurwitz polynomials that admits a Hadamard factorization, we called Hadamardized Hurwitz polynomials, is denoted by H(Had) .…”
Section: Hadamardized Hurwitz Polynomialsmentioning
confidence: 99%