eddins@uicbert.eecs.uic.edu (Steve Eddins) (05/24/91)
wilf@sce.carleton.ca (Wilf Leblanc) writes: >jefft@phred.UUCP (Jeff Taylor) writes: >> [mock conversation about zeros of linear phase filter] >> "WHERE ARE THE ZEROS? >> "On the unit circle." >"... but the zeros are on the unit circle" ?????? > [counterexample deleted] >The zeros are not on the unit circle. To clarify: an FIR filter with coefficient symmetry (and therefore linear phase) may only have zeros with the following forms, if the coefficients are also assumed to be real: 1. zero at 1 2. zero at -1 3. zeros at e^{j\omega} and e^{-j\omega} (conjugate pair on the unit circle) 4. zeros at \alpha e^{j\omega}, \alpha e^{-j\omega}, \alpha^{-1} e^{j\omega}, and \alpha^{-1} e^{-j\omega} (conjugate reciprocal quad) 5. zeros at \alpha and \alpha^{-1} These forms may be derived from the following considerations: a) To get real coefficients, any complex roots must occur in conjugate pairs. b) The Z transform in this case is a symmetric polynomial in z^{-1}. It can be shown that if z_0 is a root of a symmetric polynomial, then so is 1/z_0. So zeros of an FIR filter *can* appear on the unit circle, but not always. -- Steve Eddins eddins@brazil.eecs.uic.edu (312) 996-5771 FAX: (312) 413-0024 University of Illinois at Chicago, EECS Dept., M/C 154, 1120 SEO Bldg, Box 4348, Chicago, IL 60680