Abstract

Given a real strictly Hurwitz polynomial H_{n}(s) = a_{0}\prod_{\upsilon = 1}^{n} (s - s_{\upsilon}), n = 3, 4, \cdots , the standard method of calculating the continued fraction expansion of \{[odd H_{n}(s)]/ [even H_{n}(s)]\}^{\pm 1} about its pole at infinity uses Routh's scheme or Hurwitz's determinants \Delta_{r}, r = 1, 2, \cdots , n , in the coefficients of H_{n}(s) (on the equivalence of the two, see [2]). In filter theory, cases are often encountered where knowledge of the zeros of H_{n}(s) precedes that of its coefficients, and one would then prefer to have formulas for the coefficients in the above continued fraction expansion directly in terms of the former rather than the latter. This is achieved by expressing \Delta_{r} as bialternants in the zeros of H_{n}(s) and reads \Delta_{r} = (-)^{r(r+1)/2} a_{0}^{r}A(0, 1, \cdots , n - r - 1, n - r + 1, \cdots , n + r - 1)/A(0, 1, \cdots , n - 1) , where the alternant in the denominator is the Vandermonde in s_{1}, s_{2}, \cdots , s_{n} , whereas the alternant in the numerator is obtained from it on replacing the exponents 0, 1, \cdots , n - 1 by 0, 1, \cdots, n - r - 1, n - r + 1, \cdots , n + r - 1 . Examples include H_{n}(s) = \prod_{\upsilon = 1}^{n} [s - j \exp (2 \upsilon - 1)j \pi /2n] and H_{n}(s) = (s + 1)^{n} .

Full Text
Paper version not known

Talk to us

Join us for a 30 min session where you can share your feedback and ask us any queries you have

Schedule a call

Disclaimer: All third-party content on this website/platform is and will remain the property of their respective owners and is provided on "as is" basis without any warranties, express or implied. Use of third-party content does not indicate any affiliation, sponsorship with or endorsement by them. Any references to third-party content is to identify the corresponding services and shall be considered fair use under The CopyrightLaw.