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*To*: AUDITORY@xxxxxxxxxxxxxxx*Subject*: Re: Hilbert envelope bandwidth*From*: Yadong Wang <ydwang@xxxxxxxxxxx>*Date*: Mon, 27 Sep 2004 11:49:11 -0400*Comments*: cc: Christof Faller <cfaller@AGERE.COM>*Delivery-date*: Mon Sep 27 12:29:34 2004*Importance*: Normal*In-reply-to*: <1AF46B69-107D-11D9-A6A9-000A95D9E156@agere.com>*Reply-to*: Yadong Wang <ydwang@xxxxxxxxxxx>*Sender*: AUDITORY Research in Auditory Perception <AUDITORY@xxxxxxxxxxxxxxx>

If bandwidth of bandpass filtered signal is B Then: The envelope (as defiend by Hilbert transform), log-envelope, instantaneous frequency (time derivative of phasee) are not band-limited. But it can be shown that: envelope squre and intensity weighted instantaneous frequency (IWIF) are bandlimited with bandwidth = B. Best. Yadong ------------------------------------------------------------------------ - Yadong Wang, Postdoctoral Fellow Cognitive Neuroscience of Language Lab Dept. of Linguistics 1401 Marie Mount Hall University of Maryland College Park MD 20742 (o) (301) 405-2587 -----Original Message----- From: AUDITORY Research in Auditory Perception [mailto:AUDITORY@LISTS.MCGILL.CA] On Behalf Of Christof Faller Sent: Monday, September 27, 2004 7:02 AM To: AUDITORY@LISTS.MCGILL.CA Subject: Hilbert envelope bandwidth Dear list, I am struggling with the following question: Given a signal x(n) with X(f) = 0 for |f| < f1 or |f| > f2 (bandpass filtered signal with bandwidth B = f2-f1) e(n) is the Hilbert envelope of x(n) which can then be written as: x(n) = e(n)y(n), where y(n) is the "temporally flattened" version of x(n). The spectrum of e(n) satisfies: E(f) = 0 for |f| > f3 (Due to its DC offset, the evelope e(n) contains frequencies down to zero). ==> Can f3 be expressed as a function of B (the bandwidth of signal x)? Any comments/suggestions are appreciated. Thanks, Christof Faller

**References**:**Hilbert envelope bandwidth***From:*Christof Faller

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