Xref: utzoo comp.dsp:1700 sci.electronics:20408 Path: utzoo!utgpu!news-server.csri.toronto.edu!rpi!think.com!spool.mu.edu!uwm.edu!ux1.cso.uiuc.edu!uicbert.eecs.uic.edu!eddins From: eddins@uicbert.eecs.uic.edu (Steve Eddins) Newsgroups: comp.dsp,sci.electronics Subject: Re: Linear phase zeros (was: 180 deg phase shift) Message-ID: <1991May24.150419.18850@uicbert.eecs.uic.edu> Date: 24 May 91 15:04:19 GMT Article-I.D.: uicbert.1991May24.150419.18850 References: <1991May8.222501.19572@syd.dms.CSIRO.AU> <1991May10.003817.5593@milton.u.washington.edu> <625@fudd.dataco.UUCP> <1991May15.055011.5823@milton.u.washington.edu> <3404@phred.UUCP> Organization: EECS Dept., University of Illinois at Chicago Lines: 49 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